Euler
vector
trigonometry
geometry
radical axis
geometry proposed

Erect similar isosceles triangles $BA_1C$, $CB_1A$, $AC_1B$ on the sides of triangle ABC (with bases BC, CA, AB; erect either all of them inwards or all of them outwards). Denote $a_1=B_1C_1$, $b_1=C_1A_1$, $c_1=A_1B_1$. Draw the circles with centers A, B, C and radii $a_1$, $b_1$, $c_1$. Prove that the radical center of these circles lies on the Euler line of triangle ABC.

  Darij

I think the only approach is to compute
    Let a be the radical aixs of (B) and (C) and it intersects BC at U. Call A' and M be the projections of H and O on BC. We will prove $\frac{A'U}{UM}$ are symmetric with a, b, c.
   Do you have any nicer solution, Darij ?  :P

The proposed solution uses vectors and is rather short; however, I haven't it at hand now, hence I post my solution (which is not synthetic either):

It is well-known that a point P lies on the Euler line of triangle ABC if and only if $\left( b^{2}-c^{2}\right) \cdot AP^{2}+\left( c^{2}-a^{2}\right) \cdot BP^{2}+\left( a^{2}-b^{2}\right) \cdot CP^{2}=0$. (This can be proved using vectors.) Now, if we denote by R the radical center of our circles with centers A, B, C and radii $a_1$, $b_1$, $c_1$, then we have to prove that this radical center R lies on the Euler line of triangle ABC; in other words, we have to prove

$\left( b^{2}-c^{2}\right) \cdot AR^{2}+\left( c^{2}-a^{2}\right) \cdot BR^{2}+\left( a^{2}-b^{2}\right) \cdot CR^{2}=0$.

But this is equivalent to

$\left( BR^{2}-CR^{2}\right) \cdot a^{2}+\left( CR^{2}-AR^{2}\right) \cdot b^{2}+\left( AR^{2}-BR^{2}\right) \cdot c^{2}=0$.

Since the point R is the radical center of the circles with centers A, B, C and radii $a_1$, $b_1$, $c_1$, its powers with respect to these circles are equal, i. e. we have $AR^{2}-a_{1}^{2}=BR^{2}-b_{1}^{2}=CR^{2}-c_{1}^{2}$. In other words, $BR^{2}-CR^{2}=b_{1}^{2}-c_{1}^{2}$, $CR^{2}-AR^{2}=c_{1}^{2}-a_{1}^{2}$ and $AR^{2}-BR^{2}=a_{1}^{2}-b_{1}^{2}$. Hence, instead of proving

$\left( BR^{2}-CR^{2}\right) \cdot a^{2}+\left( CR^{2}-AR^{2}\right) \cdot b^{2}+\left( AR^{2}-BR^{2}\right) \cdot c^{2}=0$,

it will be enough to show

$\left( b_{1}^{2}-c_{1}^{2}\right) \cdot a^{2}+\left( c_{1}^{2}-a_{1}^{2}\right) \cdot b^{2}+\left( a_{1}^{2}-b_{1}^{2}\right) \cdot c^{2}=0$.

In order to prove this, we call $t = \measuredangle A_1BC = \measuredangle A_1CB = \measuredangle B_1CA = \measuredangle B_1AC = \measuredangle C_1AB = \measuredangle C_1BA$ the common base angle of the similar isosceles triangles $BA_1C$, $CB_1A$, $AC_1B$ (this angle t shall be positive if these isosceles triangles are erected outwardly, and negative if they are erected inwardly). Then, $\measuredangle CB_{1}A=180^{\circ }-\measuredangle B_{1}CA-\measuredangle B_{1}AC=180^{\circ }-t-t=180^{\circ }-2t$, so that $\sin \measuredangle CB_{1}A=\sin \left( 2t\right) $. Now, by the Sine Law in triangle $CB_1A$, we have

$AB_{1}=CA\cdot \frac{\sin \measuredangle B_{1}CA}{\sin \measuredangle CB_{1}A}=b\cdot \frac{\sin t}{\sin \left( 2t\right) }=b\cdot \frac{\sin t}{2\sin t\cos t}=\frac{b}{2\cos t}=bu$

with $u=\frac{1}{2\cos t}$. Similarly, $AC_{1}=cu$.

Finally, $\measuredangle B_{1}AC_{1}=\measuredangle B_{1}AC+\measuredangle CAB+\measuredangle C_{1}AB=t+A+t=A+2t$. Hence, we can apply the Cosine Law to triangle $B_1AC_1$ and get

$a_{1}^{2}=B_{1}C_{1}\;^{2}=AB_{1}\;^{2}+AC_{1}\;^{2}-2\cdot AB_{1}\cdot AC_{1}\cdot \cos \measuredangle B_{1}AC_{1}$
$=\left( bu\right) ^{2}+\left( cu\right) ^{2}-2\cdot bu\cdot cu\cdot \cos \left( A+2t\right) $
$=u^{2}\cdot \left( b^{2}+c^{2}-2bc\cos \left( A+2t\right) \right) $
$=u^{2}\cdot \left( b^{2}+c^{2}-2bc\left( \cos A\cos \left( 2t\right) -\sin A\sin \left( 2t\right) \right) \right) $
$=u^{2}\cdot \left( b^{2}+c^{2}-2bc\cos A\cos \left( 2t\right) +2bc\sin A\sin \left( 2t\right) \right) $.

If we call S the area of triangle ABC, then $S=\frac{1}{2}bc\sin A$, so that 2 bc sin A = 4S, and we get

$a_{1}^{2}=u^{2}\cdot \left( b^{2}+c^{2}-2bc\cos A\cos \left( 2t\right) +4S\sin \left( 2t\right) \right) $.

Similarly,

$b_{1}^{2}=u^{2}\cdot \left( c^{2}+a^{2}-2ca\cos B\cos \left( 2t\right) +4S\sin \left( 2t\right) \right) $

and

$c_{1}^{2}=u^{2}\cdot \left( a^{2}+b^{2}-2ab\cos C\cos \left( 2t\right) +4S\sin \left( 2t\right) \right) $.

Hence,

$b_{1}^{2}-c_{1}^{2}$
$=u^{2}\cdot \left( c^{2}+a^{2}-2ca\cos B\cos \left( 2t\right) +4S\sin \left( 2t\right) \right) -u^{2}\cdot \left( a^{2}+b^{2}-2ab\cos C\cos \left( 2t\right) +4S\sin \left( 2t\right) \right) $
$=u^{2}\cdot \left( c^{2}-2ca\cos B\cos \left( 2t\right) -b^{2}+2ab\cos C\cos \left( 2t\right) \right) $
$=u^{2}\cdot \left( c^{2}-b^{2}+\cos \left( 2t\right) \cdot \left( 2ab\cos C-2ca\cos B\right) \right) $.

But after the Cosine Law, $2ca\cos B=c^{2}+a^{2}-b^{2}$ and $2ab\cos C=a^{2}+b^{2}-c^{2}$, so that

$2ab\cos C-2ca\cos B=\left( a^{2}+b^{2}-c^{2}\right) -\left( c^{2}+a^{2}-b^{2}\right) =2\left( b^{2}-c^{2}\right) $.

Therefore,

$b_{1}^{2}-c_{1}^{2}=u^{2}\cdot \left( c^{2}-b^{2}+\cos \left( 2t\right) \cdot 2\left( b^{2}-c^{2}\right) \right) $
$=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot \left( c^{2}-b^{2}\right) $.

Similarly,

$c_{1}^{2}-a_{1}^{2}=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot \left( a^{2}-c^{2}\right) $

and

$a_{1}^{2}-b_{1}^{2}=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot \left( b^{2}-a^{2}\right) $,

so that

$\left( b_{1}^{2}-c_{1}^{2}\right) \cdot a^{2}+\left( c_{1}^{2}-a_{1}^{2}\right) \cdot b^{2}+\left( a_{1}^{2}-b_{1}^{2}\right) \cdot c^{2}$
$=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot \left( \left( c^{2}-b^{2}\right) \cdot a^{2}+\left( a^{2}-c^{2}\right) \cdot b^{2}+\left( b^{2}-a^{2}\right) \cdot c^{2}\right) $
$=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot \left( c^{2}a^{2}-a^{2}b^{2}+a^{2}b^{2}-b^{2}c^{2}+b^{2}c^{2}-c^{2}a^{2}\right) $
$=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot 0=0$,

and we are done.

  Darij

Don't worry about it. I like your solution, Darij. :cool:


τι χρώμα πρέπει να έχει το όνομα του stedes στη λίστα? Στείλτε και τις δικές σας προτάσεις...
[size=200][b]s[color=white]t[/color]e[color=white]d[/color]e[color=white]s[/color][/b][/size]
[color=darkred][size=200][b]stedes[/b][/size][/color]
[color=red][size=200][b]stedes[/b][/size][/color]
[color=orange][size=200][b]stedes[/b][/size][/color]
[color=green][size=200][b]stedes[/b][/size][/color]
[color=cyan][size=200][b]stedes[/b][/size][/color]
[color=blue][size=200][b]stedes[/b][/size][/color]
[color=indigo][size=200][b]stedes[/b][/size][/color]
[color=#f8f8f8][size=200][b]stedes[/b][/size][/color]

ΚΑΤΙ ΣΕ ΦΟΥΞΙΑ ΔΕΝ ΥΠΑΡΧΕΙ??
ο STEDES θα το εχει δει σιγουρα (σαν guest),χωρις να κανει log in.....
Ρε συ...,φτωχο χρωματολογιο εχεις.....Περιμενα να το γραψεις σε 64000000 χρωματα (s-p-s)A,pew moy ti shmainei to p-s-p,giati o stelios den mu lei.... Lei oti βγαινει απο το κωδικας da vinci....μπουρδες.....Μηπως εχει καμια σχεση με ατομικα τροχιακα????.............

[color=#f114ee][b][size=200]stedes[/size][/b][/color]
εννοεις το S-ps, τα τέλεια τετράγωνα? αν ναι καμία σχέση και οι 2 προτάσεις


[quote="ursula"]I'm intrigued about the regional distribution of the people who believe in something. Actually I kniw the continent where people believe the most is Africa  but we don't see much of them here. So there it goes, people in this forum are not a good statistical sample but nevertheless...

Ursula[/quote]

Nice idea! We'll discuss the results after a week...

[quote="moogra"]what if you're asian but live in North America, does that count as N.A. or asia[/quote]
I would assume that counts as N.A., since the poll wasn't about ethnicity

[quote="moogra"]what if you're asian but live in North America, does that count as N.A. or asia[/quote]
I would think asia because it asks where you are from.

Normally, when the content of posts on a poll thread consists of people stating what they voted, I just move it to G&FF.  In this instance, since this poll was created with a tight relationship to another thread in this forum, I'm just going to delete posts of that sort and let hsiljak's suggestion control: any real discussion of some accumulated results can stay, but posts that merely list how you voted will be disappeared.

I wounder if a mathematician could be an atheist (if it is defined as the denial of theism, taken from wikipedia). Wouldn't that be like saying "no that theorem is false" without prooving it? I think a mathematician could as best/worse be an agnostics. This also goes the other way around.
I come to think of this joke.

[quote]A Mathematician, an engineer and a physicist were traveling through Scotland when they saw a black sheep through the window of the train.
"Aha", says the engineer, "I see that Scottish sheep are black."
"Hmm", says the physician, "You mean that some Scottish sheep are black".
"No", says the mathematician, "All we know is that there is at least one sheep in Scotland, and that at least one side of that one sheep is black!"[/quote]

LoL, I find many studious people are atheist.
For example, I'm in gifted classes in highschool and I can assure that more than half of the people are atheist, while the rest of the school is mostly religious.

[quote="Andreas"][quote="kostas"]I do belive in GOD.every person that I know and does not believe is just a looser and wants a reason for this[/quote]
Wow, this sentence is a good example of how careless is the respect for other positions into the christian moral (or better into the christian church)  :([/quote]
well sorry ror being abit rude. I would like to hear the other side

I seem to be the only Asian non-believer till now, as against five believers.

Then would you please do me the favour of voting so?

Africa includes Egypt and South Africa, and there are plenty of smart people there.
Of the 5 freshmen in advanced math courses at my school, 4 are religious and believe in some God from a major religion, and 1 atheist (not me)

[quote="Ignite168"]How come some countries in South America aren't included in this poll? And why is Africa included? There are no smart people on that continent.[/quote]
I think its fairly obvious that Brazil goes under Latin America, no?

What about Belize, Trinidad and Tobago, Guyana, Jamaica, Barbados and Suriname?

If you did your geography well, you would have known that ALL of them come under Latin America. In fact, with only the exception of USA and Canada, the entire America is Latin.

And Africans are way smarter than other people, and too smart to be bothered to be interacting with them much. Only when the others come to them to learn or for help, they oblige out of pity, though with utter contempt, before going back to their tasks at hand.

I often think that they should be less contemptuous of us and more friendly, but that is life.

[quote="Ignite168"]How come some countries in South America aren't included in this poll? And why is Africa included? There are no smart people on that continent.[/quote]

Since I happen to know many africans, not from egypt or south africa and never heard them say anything of the like, I must conclude that at least they are smarter than you. Which clearly means nothing nevertheless

I made the poll, and decided not to include a category for every country in the world for practical reasons. There isn't either a category for Australia or Borneo. I hope they will not be offended

Ursula

It's interesting that as I post this the non-believers outnumber the believers. In the entire world i am pretty sure more people believe in some sort of god than do not, but in a sample of math-oriented (supposedly) people it's the other way around. Why would that be?

People who like math are more logical and use what they know to make conclusions.

Somebody like me does not like to be presented with a theorem without seeing a proof.  (And then, I figure out the proof is too much for my brain to handle).

The definition of "Latin America" is indeed extremely fuzzy.  The idea is that Latin refers to Romance languages. However, the most common definition I hear is that it's all American nations without Canada and the USA.
Which doesn't make sense because in Quebec French is spoken (and extremely well protected). And at the same time English is spoken in Guyana and Belize, and Dutch in Surinam, and both of them are Germanic languages.

Ignite's comment about the intelligence of African people is just a provocation.

By the way, why was ursula's topic moved to Games and Fun, it used to be in Round Table? Ursula is one of the people who make Round Table more interestig, I think it's kind of a weird signal if topics about people drinking too much water aren't moved but discussions about religion are.

[quote="kstan013"]I always thought of Latin America as just Spanish and Portugese :maybe:[/quote]
Well yeah that was the point, that it would refer to South America, which happens to be mostly former Spanish or Portuguese territory.  But as I said, it's not really a good term.

I would like to point out that there are several degrees to religion.  Here in Flanders, more than half of the kids go to catholic schools, and about everyone is baptized but.... most people I know don't go to church every week (and those who do are extremely bored and so was I when I still had to when I was in elementary school)  Do most people believe here? Most people I know refuse to answer that question, including my parents.  Kind of weird when you consider they chose to baptize me.

[quote="ProtestanT"]it is quite interesting how there are more nonbelievers than believers. it shows that these days people are skeptical of anything foreign to them.[/quote]
Religion is far from foreign to me. I was forced to come, I was forced to watch and I left. :|

cuz u were skeptical! (not that its bad or anything)


suppose A and B are two angles such that
sin A + sin B = 1
cos A + cos B = 0
find the value of 12 cos 2A + 4 cos 2B.[/code]

If $\{A,B\}\subset (0,\pi )$ then $\cos A+\cos B=0\Longleftrightarrow A+B=\pi$.

You can remark the mid point of $P(\cos A,\ \sin A),\ Q(\cos B,\ \sin B)$ on $x^{2}+y^{2}=1$.

[hide="Solution"]
$\cos A+\cos B = 0 \iff \cos A =-\cos B \iff B=180^\circ-A \ \text{or}\ 180^\circ+A$
$\sin A+\sin B = \sin A+\sin (180^\circ \pm A) = \sin A \pm \sin A = 2\sin A \ \text{or}\ 0$
But, $\sin A+\sin B = 2\sin A \ \ (\because \sin A+\sin B = 1 \neq 0)$
Thus, $\sin A=\frac{1}{2}\iff A=30^\circ \ \text{or}\ 150^\circ$
$\therefore (A, B)=(30^\circ, 150^\circ) \ \text{or}\ (150^\circ, 30^\circ)$

Thus,
$12 \cos (2 \times 30^\circ)+4 \cos (2 \times 150^\circ) = 6+2 = 8$
$12 \cos (2 \times 150^\circ)+4 \cos (2 \times 30^\circ) = 6+2 = 8$
[/hide]

I thougtht that here $A$ and $B$ are values of the angles of a triangle $ABC$ !


Hi
Im more of a math forum type, but for our chemistry homework we were asked to evaluate the precison of a few instruments by looking at a table of experimental measurements, and then the correct measurement.
What is a good mathematical way to evaluate precision when given a table of experimental measurements?
Thanks,
Mathemonster


Where's the best place to buy a rubik's cube (durability is key)?

Try Toys R us. Look for the 3*3*4's, 5*5*5's, 2*2*2's, etc.

The absolute best place is [url=http://www.rubiks.com]www.rubiks.com[/url]

That link is not false.

So I guess this: https://secure.rubiks.com/lvl3/index_lvl3.cfm?lan=eng&lvl1=produc&lvl2=rubbrn&lvl3=clasic&lvl4=cubcls#
will suffice?

I've been hearing that the 25th Anniversary cubes have problems with the stickers falling off, that's why I'm concerned.

i have a 25th anniverary cube, and yeah the green stickers aren't too terribly durable.  but i'm not sure if they're worse than your average rubik's cube.

i have a 25th anniverary cube also [i]it is not[/i] durable i [i]do not[/i] reccomend them i'm to scared to play with mine now...

I had a 25th anniversary cube that I bought at Mathpath. It only lasted a few days [size=59]because I lost it[/size] :rotfl:.

Then I got a little one from Mathcounts, and it's really annoying.

The 25th anniversary cubes do have a big problem with the sticker's falling off.  I have one, and the stickers are wearing off a lot (peeling, etc) already, even though I haven't used it much.

[quote="perfectnumber628"]the trick is to buy replacement stickers (4 sets of them, one for each size  :lol: ) and then who cares if all the stickers fall off?

and then once you start carrying around a ton of cube stickers people really think you're odd.  :roll:[/quote]

I'll second that- I was restickering my 25th anniversary cube during friends' graduation parties, and people started laughing when I took the cube stickers out of my wallet (yes, I keep cube stickers in my wallet.  Right by the key cards from the past 2 ARMLs.

[quote="perfectnumber628"]is "restickering" a word?  :D[/quote]
Since when have I ever cared about that?  English is much more fun when treated as a wikilanguage (anyone can edit it.)  FYI, "wikilanguage" isn't a word, either.

I got one (25th Aniversary). Why is it so difficult to twist (hard to "speedcube")? Does it become slick over time or do I have to wax it or something?

Argg... I forgot the last layer method(s) I usually use.

New Question:
PVC stickers are generally more durable (for color) than plastic stickers, right?

My friend uses WD-40 to grease his Rubik's Cube.  :rotfl:

And the little one I got at MathCounts broke after about 10 minutes.  :(


calculus
integration
logarithms
calculus computations

After substitution $ u\equal{}xy\qquad I\equal{}\int_{0}^{1}\int_{0}^{y}\frac{1}{\ln u\;\; (1\plus{}u^{2})}\frac{du}{y}dy$ .

Then use Fubini: $ I\equal{}\int_{0}^{1}\underbrace{\int_{u}^{1}\frac{1}{y}dy}_{\minus{}\ln u}\frac{1}{\ln u\;\; (1\plus{}u^{2})}du\equal{}\int_{0}^{1}\frac{\minus{}1}{1\plus{}u^{2}}\, du\equal{}\minus{}\frac{\pi}{4}$


geometry
trapezoid
ceiling function
USAMTS
AMC
USA(J)MO
USAMO

Eh, I feel sort of the same way, but I'm taking it differently from you. I'm annoyed at it, though I see where they're coming from, and so next time I'm going to assume the grader is a 2nd grader who happens to know advanced math, but still needs every triviality explained to them. Keep on doing it; don't give up because your grader happened to be grumpy and nitpicking when he graded your paper.

EDIT: I'm annoyed at losing 2 points on the first problem because I didn't explicitly show the fourth digit had to be 2 and the number couldn't be more than 4 digits long.

Agreed. Regardless of what you scored, you still gained experience in problem solving which no grader could possibly take away, and you hopefully enjoyed the problem solving process as much as I did. Besides, nature has a way of balancing everything out. Maybe a grader will overlook your next few mistakes and you shall gain back those 3 points.

Oh, I'll still work the problems, don't worry. And type up my solutions. But I don't see why I should turn them in any more. I'd have to get practically a perfect on round 3 and 4 (only 1 point grace now) in order to get Gold. And I don't know if I could pull that off.

Whatever. I might end up trying anyway. Don't know yet.

Just wanted to share my emotions a little bit here  :lol:\

EDIT: Well, come to think of it, I may end up appealing #4 after all. Because I asked my friend, and he also said that there was nothing wrong with my proof (that I had really shown that the result holds; that the grader may have overlooked that)...

Time to think. Should I really make a fuss about this? Yes. Later though. Warcraft now  :blush:

[quote="Mathlete9366"]I don't really want to dispute any grading, just rather post my comments about the grading of Round 2.

I feel that I was slighted 3 points for not a very clear reason. I apparently lost 2 points on #2 for not proving that (the point formed from the extensions of the sides of the trapezoid), (M), and (N) are collinear; this is just a trivial result from similar triangles. My understanding is that the grading scale is 4 for minor holes, 3 for major holes. I don't think that not stating the fact that these triangles are similar warrants losing 40% credit on the solution.

Also on #4, I showed a "counterexample" for $n=k^{2}+k-2$ (that is, the conditions in the problem were not satisfied). Yet in my grader comments, the grader said that I had not proven that $n<k^{2}+k-1$ was not possible or whatever; but shouldn't showing a construction of $n=k^{2}+k-2$ be sufficient, as if we choose any $n<k^{2}+k-2$ we just don't include those elements in the construction? (it would still work...)

Anyway, I think that I lost points for minor details that truly did not invalidate the proof. I feel that I deserve the points that I lost, yet there is no way for me to get them back as the grader really didn't overlook anything.

So, goodbye USAMTS.

See you next year.[/quote]
Decreasing the value by one isn't the only way to decrease the number, a formula less than that which might work would be um, $k^{2}+\lceil .99k \rceil-1$
I got 2 points docked for not showing that the points were collinear, and although I thought that was obvious I guess I was wrong. The thing is, the USAMTS people can grade this how they want and if they think that what you are missing is a big deal then it doesn't matter if we agree...

[quote="Mathlete9366"]Oh, I'll still work the problems, don't worry. And type up my solutions. But I don't see why I should turn them in any more. I'd have to get practically a perfect on round 3 and 4 (only 1 point grace now) in order to get Gold. And I don't know if I could pull that off.[/quote]

#1: You could get 2 perfect rounds.
#2: Silver still gets prizes.  :lol:

Yeah I loss 3 points in the first wrong from retarded algebraic mistakes on typing up....

But I figure might as well continue and jack some free prizes for revenge... :rotfl: (even bronze gets prizes, com'on)

And yeah, it can also train you to explaining every little detail, so if you tutor others it'll be less frustrating.  Plus who knows if the USAMO grader isn't as grumpy, if not worse...

It's free, so might as well take advantage of it :rotfl:

[quote="Mathlete9366"]Oh, I'll still work the problems, don't worry. And type up my solutions. But I don't see why I should turn them in any more. I'd have to get practically a perfect on round 3 and 4 (only 1 point grace now) in order to get Gold. And I don't know if I could pull that off.[/quote]

I'm in the same senario except I messed up on Round 1. Annoying  pyramid.

When I did mail-in contests (USAMTS and Pomona) in high school, what did I gain? I learned mathematical writing skills, [i]especially[/i] when my solutions didn't receive full credit. The problems weren't hard for me, but there was nowhere else to go for feedback on proofs.
My biggest problem with the USAMTS was that it was too easy to get 5 on a problem; the Pomona grading was harder on my solutions, and pushed me more.

Losing points isn't a reason to stop participating. It's a reason to learn from your mistakes and write better next time.

[quote]
I feel that I was slighted 3 points for not a very clear reason. I apparently lost 2 points on #2 for not proving that (the point formed from the extensions of the sides of the trapezoid), (M), and (N) are collinear; this is just a trivial result from similar triangles. My understanding is that the grading scale is 4 for minor holes, 3 for major holes. I don't think that not stating the fact that these triangles are similar warrants losing 40% credit on the solution.
[/quote]

Indeed, while this is a simple result using Homothecy (something of a generalization of similar triangles), it still needs to be proven.  Most math students (assuming they hadn't seen the question before) would not think it obvious that "the midpoints of the parallel sides of a trapezoid are collinear with the intersection of the extensions of the non-parallel sides."  It's obvious after a little thought, sure, but it doesn't just jump out at you in the same way that the triangle inequality does.

As for the second comment, yes, you're right that it does still work if you exclude the final element, and you're also right that this one is pretty obvious, but you still have to prove that it's true.  Part of the USAMTS involves getting used to rigorous proof.  When on a timed contest, we often use hand-wavy methods and make half-baked conjectures since all that we need is the answer.  However, when doing math, it's little things like that invalidate grand proofs.  Consider Unique Prime Factorization.  Although it seems a trivial fact, the fatal flaw behind early proofs of Fermat's Last Theorem was an assumption that a generalization of prime numbers, known as "prime elements," were also unique up to negativity.

As for quitting USAMTS just because your chances of being a gold winner are now low... um... so?  So what if you don't make gold?  You're still set to get above a 90, which is definitely a respectable score.  Stay with it.  The most helpful part of the USAMTS is the push to provide completely rigorous proofs.  Why give that up?

[quote="ehehheehee"]What is the Pomona?  I googled it and it returned a college... Is the competition still around?

Is the USAMTS grading like the USAMO grading?[/quote]
:rotfl:
USAMO, from what I've heard, is so strict in grading that a minor error in which one might even be able to get a five on the USAMTS, would constitute a minus 4 or something on the USAMO. The graders are so strict, partial credit is almost impossible from what I've heard.

[quote="eyefragment"][quote]
I feel that I was slighted 3 points for not a very clear reason. I apparently lost 2 points on #2 for not proving that (the point formed from the extensions of the sides of the trapezoid), (M), and (N) are collinear; this is just a trivial result from similar triangles. My understanding is that the grading scale is 4 for minor holes, 3 for major holes. I don't think that not stating the fact that these triangles are similar warrants losing 40% credit on the solution.
[/quote]

Indeed, while this is a simple result using Homothecy (something of a generalization of similar triangles), it still needs to be proven.  Most math students (assuming they hadn't seen the question before) would not think it obvious that "the midpoints of the parallel sides of a trapezoid are collinear with the intersection of the extensions of the non-parallel sides."  It's obvious after a little thought, sure, but it doesn't just jump out at you in the same way that the triangle inequality does.[/quote]

I pretty much stated some statements about similar angles, then said:
[quote]This is sufficient to prove that $\Delta AED$ and $\Delta BEC$ are similar right triangles and are homothetic with point $E$ as the center of homothecy.
So, $E$, $M$, and $N$ are collinear.[/quote]
This was enough to prove that E, M, and N are collinear. While it was fairly obvious to me after I had drawn the diagram, it is better to prove too much than not enough.

Yes the USAMO grades alot stricter than the USAMTS. That is because the USAMO is a competition to determine the smartest math students in the U.S. The USAMTS is a learning experience rather than a competition. The USAMTS is willing to give points if you merely show that you understand the problem whereas, the USAMO requires that you solved at least part of the problem before you get any points.

I believe that if you solve the problem completely, they start with a score of 7 and deduct points for things like clarity and elegance. If you didn't solve the problem completely or have a hole in the solution, they start with a score of 0 and add points for significant progress.

[quote="Xantos C. Guin"]I believe that if you solve the problem completely, they start with a score of 7 and deduct points for things like clarity and elegance. If you didn't solve the problem completely or have a hole in the solution, they start with a score of 0 and add points for significant progress.[/quote] I believe USAMO doesn't deduct for "non-elegance." Since there was an arguement over the Mock USAMO grading last year, when they awarded 7 for only correct, elegant solutions.

[quote="ehehheehee"]What is the Pomona?  I googled it and it returned a college... Is the competition still around?

Is the USAMTS grading like the USAMO grading?[/quote]

Yes it is a college.  It hosts a proof competition somewhat like USAMTS, but one can register as a school team.

If you are in Southern California you can definitely check it out.

[url=http://www.math.pomona.edu/problemsets.html]Pomona[/url], which uses the same questions as [url=http://www.math.wisc.edu/~talent/problems.html]wisconsin mts[/url]

How much work is it?  Is it the same as the USAMTS?  (I probably need to wait till next year if I want to join.)

Alex, I got docked 2 points for boxing my $\theta$ on number 2 (66 degrees) and I think the grader missed my $\angle MNA$ which was right under it... unfortunately I subtracted incorrectly and put $\angle MNA = 113$!

Also, on number 5 I again boxed the wrong thing... $AF$ instead of $EF$.  I had found x to be 6.5 (so 1.5 follows trivially) but I neglected to put in down, so another 2 points down the drain.

You're not alone buddy, I'm sure we all have made some pretty stupid mistakes that cost us more points than we probably should have lost, but remember that the points are really just for fun if you think about it... it's not too serious.  What matters is the fact that you solved the problems.

Yeah, I lost points on number 1 for not showing that the number couldn't be more than 4 digits... oops haha?  Although I have an idea.  Let's all mope together.  We'll all be happier that way!

[quote="bpms"][quote="Phelpedo"][quote="bpms"]USAMO, from what I've heard, is so strict in grading that a minor error in which one might even be able to get a five on the USAMTS, would constitute a minus 4 or something on the USAMO. The graders are so strict, partial credit is almost impossible from what I've heard.[/quote]
Uh...where are you hearing these things from?[/quote]
I thought someone said this around the time of the "what do you think your USAMO score is" poll. Perhaps they were exaggerating, or my memory exaggerated. Probably the second :blush:[/quote]

no, it was the first. i remember when someone (other than you) said that.

If a minor error would get a 3, how could you get a 6?

Does that mean if I have 30 minutes left, I should look at my proofs for holes instead of trying to hack one more question down?  If I ever make it, that is.

Definitly make sure that your proofs work if you haven't already done so. But if you have one question left, you might want to attempt that one if you think that you can solve it.

Hmm.... I am scared. Well, I am one of the stupidest eighth graders who stumbled upon USAMTS with my friend's suggestion. My best score so far is [b]17[/b] on the second round (I thought that one was relatively easy) with 5 5 5 1 1. (I totally misunderstood #4 and had a brain infarction with #5 :oops: )

Well, as a simple-minded middle school student, I try to specify EVERY SINGLE POINT that even my little brother would understand. The fact that some people argue about losing 1 or 2 points for thinking something as trivial is amazing. I know, I know. You all think I'm an idiot. But just a word of the wise...

USAMTS is by far very subjective competition... Just like the new SAT's with stupid essay sections. What I believe is that the people who score 25's in USAMTS has their say in all these problems. SO let's just stop fussing about little problem in the grading. ANYWAY, you're not going to die because you didn't win USAMTS, which is not such a big competition compared to the Olympiads. USAMTS is free, and no other competition let you solve problems at your own pace. This is a great start for writing proofs.

Edit: My brain is going nuts and it is going through a serious ache with round 3  :(

I was nowhere near that good in 8th grade.  My brain was trained to do Mathcounts and speed.

You are not supposed to comment on the difficulty of the round...

I think USAMTS is less subjective than you think.  Itis easier to see if a student proves something than judge the quality of writing.  Your score probably could differ by a point at most depending on the grader, but that should even out after 4 rounds...

I don't think people are idiots unless it is either me, or the person does not try.

[quote="Andrew Kim"]Hmm.... I am scared. Well, I am one of the stupidest eighth graders who stumbled upon USAMTS with my friend's suggestion. My best score so far is 17 on the second round (I thought that one was relatively easy) with 5 5 5 1 1. (I totally misunderstood #4 and had a brain infarction with #5 Embarassed )

USAMTS is by far very subjective competition... Just like the new SAT's with stupid essay sections. What I believe is that the people who score 25's in USAMTS has their say in all these problems. SO let's just stop fussing about little problem in the grading. ANYWAY, you're not going to die because you didn't win USAMTS, which is not such a big competition compared to the Olympiads. USAMTS is free, and no other competition let you solve problems at your own pace. This is a great start for writing proofs.

Edit: My brain is going nuts and it is going through a serious ache with round 3  :([/quote]
Don't worry about your score, just keep trying to solve problems. You will get better after doing a few rounds. At least you were aware of the USAMTS in 8th grade.

Be happy that you are getting free proof writing experience and feedback through this competition. Not many other competitions exist that are free and you don't have to qualify for.

My brain is also aching from this round, or at least a certain problem on this round.


I just want to make sure if this is correct..

1: 3log(b) X - log(b) Y = log(b) (x^3/Y) ??

(b) means that there is a small b underneath the log.

yes, because blog(a)=log(a^b), and logb-loga=log(b/a)


vector
geometry
3D geometry
advanced fields
advanced fields unsolved

Let $K$ be a convex subset of a normed vector space, $\mathrm{int}\,K \neq \varnothing$. Prove that $\mathrm{int}\,K$ is dense in $K$.

// $\mathrm{int}\,K$ := interior of $K$.

take a point $x \in K$. One wants to show that there exists points $y \in int(K)$ arbitrary close to $x$. Because $int(K) \neq \varnothing$, there exists an open ball $B=B(c,r) \subset int K$. Then the open cone $C=\{\lambda x+(1-\lambda) y : 0<\lambda<1, y \in B\}$ is included in $int(K)$, so the conclusion follows.

I solved the problem in a more complicated way: I was proving that $\mathrm{cl}\,\mathrm{int}\,K = \mathrm{cl}\,K$. Thanks!

By the way, it is enough to consider $\bigcup_{0 < \lambda < 1}(\lambda x+(1-\lambda) \mathrm{int}\,K)$ instead of $C$, so the introduction of $B$ is redundant.

So I offer the following problems. Let $X$ be a topological vector space, $K$ a convex subset of $X$ with $\mathrm{int}\,K \neq \varnothing$. Prove that

1) $\mathrm{cl}\,\mathrm{int}\,K = \mathrm{cl}\,K$.
2) $\mathrm{int}\,\mathrm{cl}\,K = \mathrm{int}\,K$.


If you roll a dice six times, what is the probability that you obtain EXACTLY 5 fives?

I don't think it's (1/6)^5 or (1/6)^5 * (5/6).

Thanks!

Well, there are C(6,5)=6 ways to choose 5 of the 6 rolls to be fives. So using the binomial distribution probability formula, the probability that exactly 5 rolls out of 6 would produce fives is $ 6(\frac{1}{6})^5(\frac{5}{6})\equal{}\frac{5}{6^5}\equal{}\boxed{\frac{5}{7776}}$.


symmetry
trigonometry
Pythagorean Theorem
geometry

You are making one of those baby toys where you have to put pieces of certain shapes through the same shaped hole.

1) There is a square hole with side length 20. Find the minimum side length the equilateral triangular piece can have so that it cannot fit through the square hole in any way.

2) There is a square piece with side length 5. Find the minimum side length the equilateral triangular hole can have so that the the square piece cannot fit through it in any way.

For both questions, assume you cannot turn the piece sideways.

[quote="SplashD"]1) There is a square hole with side length 20. Find the minimum side length the equilateral triangular piece can have so that it cannot fit through the square hole in any way.[/quote]
Is this the same question as "Find the maximum side length the equilateral triangular piece can have so that it can fit through the square hole."
[hide="1"] We have the square $ABCD$. Then we draw the equilateral triangle so that one vertex coincides with a vertex of the square. So we have triangle $AEF$ with $E$ on $BC$ and $F$ on $CD$. We call the side length of the triangle $s$. $EC=FC=\frac{s\sqrt2}{2}$. So, $BE=20-\frac{s\sqrt2}{2}$, $AB=20$ and $AE=s$.
Using the Pythagorean Theorem:

\begin{eqnarray*}20^{2}+(20-\frac{s\sqrt2}{2})^{2}&=&s^{2}\\400+400-20\sqrt2s+\frac{s^{2}}{2}&=&s^{2}\\s^{2}+40\sqrt2s-1600&=&0\end{eqnarray*}

$s=\frac{-40\sqrt2\pm\sqrt{3200+6400}}{2}=\frac{-40\sqrt2\pm40\sqrt6}{2}=\boxed{20\sqrt6-20\sqrt2}$ [/hide]

[hide="Generalization, I guess..."]
Put the side length of the square as $s$, and the side length of the triangle as $t$. Then, to maximize $t$ while the triangle still can fit into the square, one vertex of both shapes must coincide and there must be a line of symmetry down the diagonal containing these vertices.

Thus, ${t = \frac{s}{\sin 75^\circ}= \frac{4s}{\sqrt 6+\sqrt 2}= \boxed{s(\sqrt 6-\sqrt2}})$.[/hide]

Oh, I forgot to mention that their centers of the triangle and the square have to coincide. :oops:


There was definitely a required cover sheet, but only an administrator/principal had to sign it.  But it was required to have the cover sheet.

[quote="andersonw"]Heh, I'm on the opposite side of the spectrum. On day 2, I didn't submit anything for 5 or 6 but there's a Y next to my number for those. (day 1 is correct, though, since I didn't submit anything for 2 or 3 and it said YNN).

As for your question #4, I really doubt it.
And only the principal/proctor has to sign the cover sheet (T-III).[/quote]
For your 5 and 6, your protector would have to submit a empty paper for answer sheet.
They ask you to submit all problem

[quote="Scrambled"]Edit: Nevermind, sorry.
[hide]Hey everyone,
I checked online for USAMO solution confirmation, and it seems for some reason that nobody taking the USAMO at our school got their solutions confirmed for day 2 (and some students did not get any confirmed for day 1).  Just a few questions regarding this...
1) Was there a required cover sheet? I distinctly recall there being one that we had to sign last year, but our proctor didn't provide one this year for any of us.
2) When was the last time the confirmation list was updated?
3) what else (besides for mailing delays or some difficulty on our end) could be the cause of this problem?
and finally
4) If we do find out that the solutions were misplaced or forgotten-to-be-mailed or something like that, can we still submit them?

Thanks,
Scrambled.[/hide][/quote]

Seems like you guys has lots of people taken it, so I guess your exam manager is overnighting the solution instant of faxing them. Ask him


Does someone understand what all that means ? :(

Pierre.

[quote="ehsan2004"]what was your meant ? find min(m(S))...? :huh:[/quote]
We consider all the such sets S .And find min m(S) :)

This 's a beautiful sum . Why aren't you interested it   :mad


Find all functions $f: \mathbb{R}\to\mathbb{R}$ satisfy the condition:
$\frac{f(x)+f(y)}{2}\ge f(\frac{x+y}{2})+|x-y|$

i remember solving this on USAMO 2000. the answer is there is no such function. we'll prove by induction that

$\frac{f(x)+f(y)}{2} - f(\frac{x+y}{2}) \geq 2^n |x-y|$ for any $n \in \mathbb N$ (*)

from which it is obvious there can be no such function because if $x\neq y$ are fixed then the RHS can be as large as we want while the LHS is fixed.

the case $n = 0$ is the assumption of the problem. assume (*) holds for $n \geq 0$. then

$\frac{f(x)+f(\frac{x+y}{2})}{2} - f(\frac{3x+y}{4}) \geq 2^{n} |\frac{x-y}{2}|$

$2[\frac{f(\frac{x+3y}{4}) + f(\frac{3x+y}{4})}{2} - f(\frac{x+y}{2})] \geq 2 \cdot 2^{n} |\frac{x-y}{2}|$

$\frac{f(\frac{x+y}{2}) + f(y)}{2} - f(\frac{x+3y}{4}) \geq 2^{n} |\frac{x-y}{2}$

adding gives precisely

$\frac{f(x)+f(y)}{2} - f(\frac{x+y}{2}) \geq 2^{n+1} |x-y|$


calculus
integration
calculus computations

Evaluate:
                    $I=\int_{0}^{\pi}\frac{dx}{\sqrt{2+cosx}}$

Let  u= 2 + cosx       then  we  have  du  =  -sinx dx     dx =  - du/sinx
   And   sin2x + cos2x = 1    sinx=    ,   sinx 0  because  x
   So   we  have  -sinx = -        for   x= π   u = 1     for  x= 0   u = 3     and  the  integral  I is  been  transformed  to  the

            =


percent
geometry
3D geometry
calculus
integration
modular arithmetic
arithmetic sequence

1. Find the values of x + y if x2 + y2 = 36 and xy = -10.

2. When six gallons of gasoline are put into a car's tank, the indicator goes from 1/4 of a tank to 5/8. What is the total capacity of the gasoline tank?

3. A merchant paid $30 for an article. He wishes to place a price tag on it so that he can offer a 10 percent discount on the price marked on the tag and still make a profit of 20 percent on the cost. What price should he mark on the tag?

4. For a given arithmetic sequence, the sum of the first fifty terms is 200, and the sum of the next fifty terms is 2700. What is the first term of the sequence?

5. Let r be the result of doubling both the base and the exponent of ab, with b not equal to zero. If r equals the product of ab and xb, what is the value of x?

6. How many perfect squares are divisors of the product 1! x 2! x 3! x ... x 9! ?

2. When six gallons of gasoline are put into a car's tank, the indicator goes from 1/4 of a tank to 5/8. What is the total capacity of the gasoline tank?

[hide]The 6 gallons took up 3/8 of the tank. 1/8 of the tank holds 2 gallons. 2*8=16. The tank holds 16 gallons of gasoline.[/hide]

gauss202 wrote:1. Find the values of x + y if x2 + y2 = 36 and xy = -10.

[hide](x+y)(x+y)=x :^2:+2xy+y :^2:
                =36+2(-10)
                =36-20
                =16                    the squareroot of 16 is 4[/hide]gauss202 wrote:3. A merchant paid $30 for an article. He wishes to place a price tag on it so that he can offer a 10 percent discount on the price marked on the tag and still make a profit of 20 percent on the cost. What price should he mark on the tag?

[hide]30(1.2)=36 this is the amount he must sell it for in the end to make 20% profit

.9x=36     .9 of the price tag price (x) must be 36 dollars

9x=360

x=40[/hide][/hide]

Dandelion wrote:gauss202 wrote:1. Find the values of x + y if x2 + y2 = 36 and xy = -10.

[hide](x+y)(x+y)=x :^2:+2xy+y :^2:
                =36+2(-10)
                =36-20
                =16                    the squareroot of 16 is 4[/hide]

are there any other answers?

dandelion wrote:[hide]the squareroot of 16 is 4[/hide]

[hide] :sqrt: 16= :pm: 4[/hide]

That is not entirely correct.  :rt2: only denotes the POSITIVE square root of 2; if you wish to express the NEGATIVE square root of 2, it has to be written as -:rt2:. However, when you're solving x2 = 2, the solutions are :pm::rt2:.

Ya ya this is always confusing.  When you study the nth root theorem you will have fun with this.

x^2 = 16 has 2 solutions based on the fundamental theorem of algebra.  x= :pm: 4

 x =:sqrt: 16 denotes the positive square root of 16, so x=4

"the square root of 16" denotes both  :pm: 4

So if someone asks you find "the cube root of 8" there are three solutions.

But if they ask 3 :sqrt: 8 there is only one solution, 2

Ah, but what about imaginary solutions?

MODERATOR NOTE:  I know I am going beyond intermediate level.

[quote="JS1527"]
"the square root of 16" denotes both  :pm: 4
[/quote]

I'm pretty sure the square root of 16 only denotes +4.

If you say "the square roots of 16" or "the cube roots of 8," then you are encompassing the various negative and complex roots.  I would venture that "the square root" when applied to a positive real number does in fact imply that we're talking about only the positive root.  It doesn't really matter very much, though -- the basic point has been made that numbers have multiple roots and it is always important to check how many of them your problem allows for.  In this case, it allowed for both.

[quote="JMI"]What do you mean three solutions? I only see one. The cube root of 8 is 2and only two because -2 cubed equals -8.[/quote]

De Moivre.  Google it.

[quote]De Moivre. Google it.[/quote]

Actually you mean nth root theorem.  De Moivre's theorem is for finding powers of complex numbers, not roots of complex numbers.

Wouldn't they be the same thing since taking the 3rd root of something is the same as raising it to the 1/3(if I understand this conversation correctly)?

Well I have only used it for positive integral powers.  Yes and no.  The formulas are different.

z^n = r^n * cis(n :theta: )     <---- De Moivre's Theorem

z^(1/n) = r^(1/n) * cis ( ( :theta: +2k :pi: )/n ) Where k=0,1,2,3...(n-1)   <---- nth root theorem.

Fine.  But in my defense, if he looked for De Moivre, he'd almost certainly find n-th root.  </derail>

If you had stated what you meant, instead of telling him to google it, it would have been a lot more helpful.

lol, this thread has totally gone away from just answering the given problems

Can somebody post the solutions to these problems or if i get the answers i will post the solutions

[quote="gauss202"]

4. For a given arithmetic sequence, the sum of the first fifty terms is 200, and the sum of the next fifty terms is 2700. What is the first term of the sequence?

[/quote]

Lets say the common difference is $ a$. Th first term is $ x$, the second $ x \plus{} a$, third $ x \plus{} 2a$, and the $ n^{th}$ term is $ x \plus{} (n \minus{} 1)a$. When we subtract $ 2700 \minus{} 200 \equal{} 2500$, we get
($ 99a \plus{} 98a . . . \plus{} 51a \plus{} 50a$) - ($ 49a \plus{} 48a . . . \plus{} 2a \plus{} a$) = $ 50a * 50$ = $ 2500a$. Therefore, $ a \equal{} 1$. Now that we know what $ a$ is, then, $ 50x$ + ($ 49a \plus{} 48a . . . \plus{} 2a \plus{} a$) = 200. Plug into $ a \equal{} 1$,
$ 50x$ + ($ 50a$ * $ \frac {49}{2}$) = $ 50x$ + $ 1225a$ = $ 50x$ + $ 1225$ = $ 200$.
Then,
$ 50x \equal{} \minus{} 1025$ so $ x \equal{} \frac { \minus{} 1025}{50}$ = $ \frac { \minus{} 41}{2}$
Your first term, although a rather odd answer is $ \boxed{\frac { \minus{} 41}{2}}$.

I know that this is wrong because it must be $ \boxed{\minus{}20}$, after plugging in, but I don't know what went wrong. Probably some simply arithmetic. Can you please try to show my error??
True explanation provided would be great.
Thanks.

5. Let r be the result of doubling both the base and the exponent of a^b, with b not equal to zero. If r equals the product of a^b and x^b, what is the value of x?

We have $ r \equal{} (2a)^{2b} \equal{} 2^{2b} \cdot a^{2b} \equal{} a^b \cdot x^b$.  Cancelling $ a^b$, we have $ a^b \cdot 2^{2b} \equal{} x^b$.  Taking the $ b$ root, we get $ a \cdot 4 \equal{} x$.  So $ \boxed{4a}$?

[quote="10000th User"][quote="fermat007"][quote="gauss202"]

4. For a given arithmetic sequence, the sum of the first fifty terms is 200, and the sum of the next fifty terms is 2700. What is the first term of the sequence?

[/quote]

Lets say the common difference is $ a$. Th first term is $ x$, the second $ x \plus{} a$, third $ x \plus{} 2a$, and the $ n^{th}$ term is $ x \plus{} (n \minus{} 1)a$. When we subtract $ 2700 \minus{} 200 \equal{} 2500$, we get
($ 99a \plus{} 98a . . . \plus{} 51a \plus{} 50a$) - ($ 49a \plus{} 48a . . . \plus{} 2a \plus{} a$) = $ 50a * 50$ = $ 2500a$. Therefore, $ a \equal{} 1$. Now that we know what $ a$ is, then, $ 50x$ + ($ 49a \plus{} 48a . . . \plus{} 2a \plus{} a$) = 200. Plug into $ a \equal{} 1$,
$ 50x$ + ($ 50a$ * $ \frac {49}{2}$) = $ 50x$ + $ 1225a$ = $ 50x$ + $ 1225$ = $ 200$.
Then,
$ 50x \equal{} \minus{} 1025$ so $ x \equal{} \frac { \minus{} 1025}{50}$ = $ \frac { \minus{} 41}{2}$
Your first term, although a rather odd answer is $ \boxed{\frac { \minus{} 41}{2}}$.

I know that this is wrong because it must be $ \boxed{ \minus{} 20}$, after plugging in, but I don't know what went wrong. Probably some simply arithmetic. Can you please try to show my error??
True explanation provided would be great.
Thanks.[/quote]That's indeed the answer. $ \boxed{ \minus{} 20}$ is [b]incorrect[/b]; the sum adds up to 225 the first 50 terms with $ x \equal{} \minus{} 20,a \equal{} 1$.[/quote]

Oh, well thanks for the reply. I don't know why I assumed it was $ \minus{}20$ when I plugged in. That was probably the error. So, it turns out that $ \boxed{\frac{\minus{}41}{2}}$  is the correct answer.

[quote="yif man12"][quote="Dandelion"][quote="gauss202"]1. Find the values of x + y if x2 + y2 = 36 and xy = -10. [/quote]

(x+y)(x+y)=x :^2:+2xy+y :^2:
                =36+2(-10)
                =36-20
                =16                    the squareroot of 16 is 4[/quote]
are there any other answers?[/quote]

Correction for someone else's prior post.

It should be the equivalent of this:

$ x^2 \plus{} y^2 \equal{} 36$

$ xy \equal{} \minus{} 10$

$ x^2 \plus{} (2xy) \plus{} y^2 \equal{} 36 \plus{} (2xy)$

Note:  $ 2xy \equal{} 2( \minus{} 10) \equal{} \minus{} 20$

$ x^2 \plus{} 2xy \plus{} y^2 \equal{} 36 \plus{} ( \minus{} 20)$

$ (x \plus{} y)^2 \equal{} 16$

$ x \plus{} y \equal{} \pm\sqrt {16}$

$ x \plus{} y \equal{} \boxed{\pm4}$

I think that it's up to problem [hide="#8."]Let $ x \plus{} y \equal{} 4$ and $ x^2 \plus{} y^2 \equal{} 10$. Let $ n \equal{} 2010!$. Find the $ n$th digit to the right of the decimal place in $ \frac{x^{2010} \plus{} y^{2010}}{7}$.[/hide]

LOL this fails (I better have got this right)

$ x\equal{}1, y\equal{}3$
$ \frac{1\plus{}3^{2010}}{7}$
Now we want to find the remainder when $ 3^{2010}$ is divided by 7.
There is a pattern, 3, 2, 6, 4, 5, 1, so for 2010 its 1.
Now its the 2010!th digit after 2/7=.285714...
2010! is divisible by 6 (no duh) so its [b]4[/b]

Correct. One could also note that $ 3^3\equiv\minus{}1\pmod{7}$, and so $ 3^{2010}\equal{}(3^3)^{670}\equiv(\minus{}1)^{670}\equiv1\pmod{2010}$.


1. What is the general formula for determining two unknown vectors, a and b, that are perpendicular to a know vector in 3 space?

Move this? Seems a little difficult for intermediate.

isn't the cross product performed with two known vectors to find a vector that is perpendicular to them?

All right, let's tweak it. Given $v$, choose $v'$ not parallel to $v$. Then $v\times v'=w$ and $v\times w=u$ are perpendicular to $v$.

There is no smooth globally applicable formula for vectors with this property- see the "Hairy Ball Theorem".

I misread the question initially- I hope this helps.

The cross product is a very specific and peculiar geometric object. There's another way to do this that fits more with elementary linear algebra.

Suppose your vector is $v=[a,b,c]^T.$ What you're actually trying to do is to find a basis for the solution space of the homogeneous equation $\langle v,x\rangle=0,$ Written out in detail, this is simply

$ax+by+cz=0.$

Continuing the linear algebra formalities: we have a $1\times 3$ coefficient matrix which is of rank $1.$ (It had better have rank $1$ - rank $0$ would make it the zero vector, which is presumably ruled out by the structure of the problem.)

Let's try this (leaving the transposes out of the notation):

Let $v=[2,-1,4]$. We're trying to solve $2x-y+4z=0.$

This equation is already in row-echelon form. There is one pivot in the first column, so the second and third columns are associated with free variables. We now make two intependent choices of those free variables - a canonical way to make that choice would be to make them alternately zero and one.

Let $y=1$ and $z=0.$ The $2x=1$ or $x=\frac12.$ That gives us $w=\left[\frac12,1,0\right],$ or if you prefer, we could clear the fractions and use $w=[1,2,0].$

Let $y=0$ and $z=1.$ Then $2x+4=0$ or $x=-2.$ That gives us $u=[-2,0,1].$

This $u$ and $w$ are both orthogonal to $v,$ and they span the (2-dimensional) space of all vectors orthogonal to $v.$

We don't have $u$ orthogonal to $w$ - that was one advantage of using the cross product. But there is a standard linear algebra answer: if you'd prefer an orthogonal set, start with this independent set and use the Gram-Schmidt process.

One thing to make very clear: there is nothing remotely close to uniqueness here. This problem has infinitely many different correct answers.


MATHCOUNTS ARML email AMC AIME calculus trigonometry

Sign up for the new COUNTDOWN COMPETITION. This'll be using either AIM or Yahoo, whichever works better. Sign up now. Post any questions now. Further rules will be decided later. Please post your grade. 9th graders may participate. I may divide up the competition into sections.

I would do it, but I can t use any Instant Messengers on my mom s computer because "it will ruin the computer."

I'll do it.

when exactly will this take place?

I may or may not be in, but if you guys want to use g-mail chat, I can give invitations.  However, if I'm in, AIM works for me.

If it starts after the school year begins, I probably won't be able to participate.

Well, first of all, I understand that school is about to start, and therefore, everyone will have a lot less time. Thus, this competition will probably take more time to determine a victor, maybe. On the other hand, we might also take a couple hours someday for the tournament/competition, so please tell me when your school starts if you might have conflicts with school. suggestions are welcome.

If i can do it, it has to be this weekend, otherwise i can t join.

 :(

I'll do it. AIM. School starts next week, XC is already practicing, so I'll have limited time too.

we can set up a chat room in AIM i guess

thats the best plan

just wondering, how are people gonna get on during thanksgiving?
-jorian

[quote="fireemblem13"]due to popular demand... not really... we'll continue. I've been busy like  :cursing: recenty, that and i thought people had lost interest, apparently i was wrong. Week of Thanksgiving, Thursday, 2pm EST.

about the problems, yea. :| i guess i could try making up problems.[/quote]

just wanted to make sure this is still happening.

Is it okay if some problems come from Number Sense?

@now a ranger: i think this already started :)

have a happy thanksgiving,
jorian

[quote="fireemblem13"]Number Sense is like mental math. a set of 80 questions that you do in your head for which you have 10 minutes. of course, i'll present the questions one at a time like in mathcounts. here's a site: [url]http://www.texasmath.org[/url] go to Downloads on the left tab and click Number Sense Tests. then click on one of them.

@now a ranger: yea, in fact it's been a couple months since it started. :roll:[/quote]

Those are way too trivial, even for Mathcounts.

[quote="fireemblem13"]what's trivial mean? if it means easy, some problems near the end are somewhat similar to those on countdown mathcounts.[/quote]

sorry, i only looked at the first few problems

some have calculus or trig though at the end

if you wanted to use those problems, you really shouldn't have told us so that that we won't look at them beforehand.

it's your decision

I think number sense shows up the most often among all other categories, probably arond 40%.

could someone give me a link to a national countdown. i'll change the numbers around. i'm just interested in the forms of the problems.

you have to buy them, they're not on the internet. well at least i don't think so.

yay i won.

it only took less than 15 minutes?

-jorian

Pretty much. Actually it took like 5 minutes, we spent 10 minutes diddlydaddling.

Notice that a mod deleted all of our posts concerning food. We should stop making posts like that.


function
real analysis
real analysis unsolved

Let $ (X,\bold{X},\mu)$ a finite measure space. The sequence $ (f_n)$  converges in measure to $ f$ and $ (g_n)$ converges in measure to $ g$

Prove that $ (f_n g_n)$ converges in measure to $ fg$

[quote="Tyl"]Note that $ f g \equal{} \frac14 \left( (f \plus{} g)^2 \minus{} (f \minus{} g)^2 \right)$.[/quote]

ok ok, but how i prove the measure convergence? i don't see this "very easy"

Do it under the assumption that all functions are bounded by some big constant $ M$ first. Then note that the set $ \{|f|\plus{}|g|>M\}$ has an arbitrarily small measure when $ M$ is large ;).


Here's a pretty sweet problem that I mentioned as a physics problem but is essentially a math problem... all the physics you need to know is Distance = Rate <i.e. velocity> * time

[b]A motorboat going downstream overcame a raft at Point A; t=60 min later it turned back and after some time passed the raft at a distance l=6.0 km from the point A. Find the flow velocity assuming the duty of the engine to be constant.[/b]

Enjoy!
IW.OAV

PS And yes, I do realize that velocity is really displacement v. time but there's no need to be a sticky picky on this problem though.

the raft does not move relative to the flow right?

I think I've solved the problem; my answer is [hide]3 km/hr[/hide]

Here's how I arrived at this answer:

[hide]

Let the boat's rate equal x km/hr

and the river's rate equal y km/hr

When the boat goes downstream, its rate is x+y and when it goes up, its rate is x-y.

Since it takes the boat 1 hour to go downstream, it must have traveled x+y km.

Since it goes upstream the same distance less 6 km, it goes up x+y-6 km.

Thus, it takes the boat 1 hr down and (x+y-6)/(x-y) hr up (since t=d/r).

Therefore, the total amount of time elapsed between the two meetings with the raft is 1 + (x+y-6)/(x-y) = (x+y-6)/(x-y) + (x-y)/(x-y) = (2x-6)/(x-y)

The raft has traveled 6 km.  So its rate is 6/[(2x-6)/(x-y)] = 6(x-y)/(2x-6)

Since the raft's rate is the same as the river rate, y = 6(x-y)/(2x-6)

Therefore y(2x-6) = 6x - 6y

2xy - 6y = 6x - 6y

2xy = 6x

y = 3

So the river rate is 3 km/hr

[/hide]

Is this right?

-mathfanatic

Yeah, I think thats right. An even easier way of solving it:

[hide]

Since everything in the river is affected the same way by the flow, the flow has no effect at all on the time the boat takes to get back to the raft. So assuming the flow was 0, the boat takes 2 hours to return to the raft. So the raft has travelled 6 km in 2 hours, and so the river is flowing at 3 km/hr.

[/hide]

Awww...shoulda seen that.  Very nice, TripleM.  I like the way you temporarily assume one thing in order to establish that no matter what, it'll take 2 hours for the boat to get there and back.


Given that $ f(x)$ is a polynomial with degree $ 5$ and the coefficient of highest degree term is $ 3$.If $ f(1)\equal{}1$,$ f(2)\equal{}2$,$ f(3)\equal{}3$,$ f(4)\equal{}4$,$ f(5)\equal{}5$.Compute the value of $ f(6)$.

hello, from the conditions we get
$ f(x)\equal{}3x^5\minus{}45x^4\plus{}225x^3\minus{}675x^2\plus{}823x\minus{}360$ and from here we get $ f(6)\equal{}\minus{}6114$
Sonnhard.

The polynomial $ f(x)\minus{}x$ vanishes at $ x\equal{}1, 2, 3, 4, 5$ so it is divisible by $ (x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)$.  Hence with the given conditions we have:
$ f(x)\minus{}x\equal{}3(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)$
$ f(6)\equal{}3*5!\plus{}6\equal{}366$


Solve it(make it normal):

$ \frac {1}{\sqrt{a_1} \plus{} \sqrt{a_2} \plus{} ... \plus{} \sqrt{a_n}} \equal{} ?$

For example,
The normal type of $ \frac {1}{\sqrt {3} \plus{} \sqrt {2}}$ is $ \sqrt {3} \minus{} \sqrt {2}$

hello, in the case of $ n=3$ i have found
$ {\frac{1}{\sqrt{a_1}+\sqrt{a_2}+\sqrt{a_3}}=-\frac{(\sqrt{a_1}+\sqrt{a_2}-\sqrt{a_3})(-a_1-a_2+a_3+2\sqrt{a_1a_2})}{a_1^2+a_2^2+a_3^2-2a_1a_2-2a_1a_3-2a_2a_3}}$.
Is there a general formula?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, in the case of $ n = 3$ i have found
$ {\frac {1}{\sqrt {a_1} + \sqrt {a_2} + \sqrt {a_3}} = - \frac {(\sqrt {a_1} + \sqrt {a_2} - \sqrt {a_3})( - a_1 - a_2 + a_3 + 2\sqrt {a_1a_2})}{a_1^2 + a_2^2 + a_3^2 - 2a_1a_2 - 2a_1a_3 - 2a_2a_3}}$.
Is there a general formula?
Sonnhard.[/quote]
Yes,when $ n=2,3,4$,it's a bit easier.
but when $ n>=5$,it's quite difficult!

hello, do you have a solution for $ n\equal{}4$?
Sonnhard.


geometry perimeter conics hyperbola Asymptote

a convex quadrilateral is cut by its diagonals into four triangles of equal perimeter .

What can u infer for the shape of this quadrilateral.?

Let X be any point in the plane of the quadrilateral ABCD (convex or not), not necessarily inside of the quadrilateral, such that the perimeters of the triangles $\triangle ABX, \triangle BCX, \triangle CDX, \triangle DAX$ are equal. Denote a = AB, b = BC, c = CD, d = DA the quadrilateral sides. Then

AB + BX + AX = BC + BX + CX = CD + CX + DX = DA + DX + AX

Adding the terms with AB, CD and the terms with BC, DA

AB + CD + AX + BX + CX + DX = BC + DA + BX + CX + DX + AX

AB + CD = BC + DA

which means that the quadrilateral ABCD is tangential with an incircle. The vertices B, D of a tangential quadrilateral lie on the same branch $g_1$ of a hyperbola g with the focal points A, C, because DC - DA = BC - BA. Similarly, the vertices A, C lie on the same branch $h_1$ of a hyperbola h with the focal points B, D, because CD - CB = AD - AB. The point X is the intersection of the other branches $g_2, h_2$ of these 2 hyperbolas, the ones not passing through the vertices B, D resp A, C, because

$XC - XA = p(\triangle BCX) - (BC + BX) - p(\triangle ABX) + (AB + BX) =$

$= BA - BC = -(BC - BA)$

$XD - XB = p(\triangle DAX) - (DA + AX) - p(\triangle ABX) + (AB + AX) =$

$= AB - AD = -(AD - AB)$

The diagonals AC, BD intersecting at a point E are chords of the hyperbola branches $g_1, h_1$, hence, they cannot intersect the hyperbola branches $g_2, h_2$, if g, h are proper hyperbolas (just like a line cannot intersect a circle in more than 2 points). The only way the points X, E can coincide is when both hyperbolas, together with their asymptotes, degenerate into the perpendicular bisectors of the diagonals AC, BD. But then the quadrilateral ABCD has perpendicular diagonals cutting each other in half, i.e., it is a rhombus.


Update on the prize mailing: we are still waiting for the T-shirts to arrive.  Once they do (in the next couple of weeks I hope), we'll start mailing out the prizes.

So then prize mailing might be delayed until late June or early July?

So when are we going to recieve mathematica?

If you selected Mathematica as a prize, you will be contacted separately by Wolfram Research (the makers of Mathematica), and they are responsible for delivering it.  You will received your other prize(s) (if applicable) and your T-shirt direct from us when we do our mailing in the next 2-4 weeks.


[b]Prove that a quadrilateral ABCD has maximum area when it is cyclic.[/b]

(maybe there are some bits missing.  I'm guessing that the other condition is that [b]the perimeter of the quadrilateral is constant[/b])

I think there are at least two types of solutions.  Simple geometry and calculus.
As many solutions would be appreciated.

Also if it is possible could I have some advice on how to prove these kinds of questions and how to get started? i.e. what to consider first to start off.  I'm completely lost on this

$ \mbox{If } |AB| \equal{} a, \ |BC| \equal{} b, \ |CD| \equal{} c, \ |DA| \equal{} d \mbox{ and } p \equal{} \frac {a \plus{} b \plus{} c \plus{} d}{2}, \mbox{ then: } \\
S_{ABCD} \equal{} \sqrt {(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d) \minus{} abcd.cos^2\frac {1}{2}.(B \plus{} D)}\Rightarrow \\
\Rightarrow \max{S_{ABCD}} \equal{} \sqrt {(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d)}\Leftrightarrow cos\frac {1}{2}.(B \plus{} D) \equal{} 0\Leftrightarrow \\
\Leftrightarrow \frac {1}{2}.(B \plus{} D) \equal{} 90^0\Leftrightarrow B \plus{} D \equal{} 180^0 \Leftrightarrow \mbox{ABCD\minus{}cyclic.}$

I guess the missing part is that all the sides' lengths of the quadrilateral are fixed, not just the perimeter.


geometry
3D geometry
function
combinatorics proposed
combinatorics

Let $n\in\mathbb N$.

For every permutation $p$ of $\left\{ 1,2,...,n \right\}$, let $I(p)$ be the number of inversions of $p$. (An [i]inversion of $p$[/i] is a pair $\left(i,j\right)$ of elements of $\left\{ 1,2,...,n \right\}$ such that $ i<j$ and $ p(i)>p(j)$.) And let $ a_k$ be the number of permutations $p$ of $\left\{ 1,2,...,n \right\}$ such that $ I(p) = k$. Prove that $ \sum _{i = 0}^{\infty} a_{3i + 1} = \sum _{i = 0}^{\infty} a_{3i + 2}$.

[color=red][[b]Moderator edit:[/b] As the post by kevinatcausa below shows, this holds only if $n\neq 2$.][/color]

So if I understand the setup correctly $ n$ is fixed?

If so, and $ n\equal{}2$, wouldn't $ a_{0}\equal{}a_{1}\equal{}1$, all other $ a_i\equal{}0$ (so the sums are different)?

Well, here's some overkill: for any $ n$, we have
\[ F(q) : \equal{} \sum_{p \in S_n} q^{\text{inv}(p)} \equal{} \prod_{i \equal{} 0}^{n \minus{} 1} \sum_{j \equal{} 0}^i q^j \equal{} 1 \cdot (1 \plus{} q) \cdot (1 \plus{} q \plus{} q^2) \cdots.
\]
Now apply a roots-of-unity filter to this expression: with $ w, z$ the complex cube roots of $ 1$, we have that $ \sum_{i \geq 0} a_{3i} \equal{} \frac {F(1) \plus{} F(w) \plus{} F(z)}{3} \equal{} \frac {n!}{3}$, $ \sum_{i \geq 0} a_{3i \plus{} 1} \equal{} \frac {\frac {F(1) \minus{} 1}{1} \plus{} \frac {F(w) \minus{} 1}{w} \plus{} \frac {F(z) \minus{} 1}{z}}{3} \equal{} \frac {n! \minus{} 1 \minus{} \frac {1}{w} \minus{} \frac {1}{z}}{3} \equal{} \frac {n!}{3}$, and so also $ \sum_{i \geq 0} a_{3i \plus{} 2} \equal{} \frac {n!}{3}$.  Note in simplifying these expressions we're using that $ F(q)$ is divisible by $ 1 \plus{} q \plus{} q^2$, and this is only true for $ n \geq 3$.

It's not immediately obvious from this method how far this stretches, though there are some obvious conjectures to make.  (Specifically, that $ n \geq m$ if and only if the inversion number of permutations in $ S_n$ is uniformly distributed modulo $ m$.)  I feel like there should be a clean probabilistic proof, or perhaps something nice and bijective.

I guess if $ n \geq m$, we can divide the permutations into classes as follows:

Think of a permutation as a reordering of the numbers from $ \{1, \dots n\}$, and remove the smallest number appearing in the first $ m$ elements of the list.  What's left is a list of $ n\minus{}1$ of the numbers with one missing number.

This function is an $ m$ to $ 1$ map, as we can take a list of $ n\minus{}1$ numbers and add in the missing number in any of the first $ m$ spots.  Furthermore, each preimage of this function has a different number of inversions modulo $ m$ (moving the smallest number one spot right increases the inversions by one), so we get a uniform distribution modulo $ m$.

Okay, very nice.  It happens that the generating-functional proof carries through as well:  let $ w_0 \equal{} 1, w_1, \ldots, w_{m \minus{} 1}$ be the $ m$th roots of unity.  Then we have $ \sum_{0 \leq j \leq m \minus{} 1} w_j^n$ is equal to $ 0$ unless $ m | n$ in which case it's equal to $ m$.  Set $ F_n(q) : \equal{} \sum_{p \in S_n} q^{\text{inv}(p)} \equal{} (1 \plus{} q \plus{} \ldots \plus{} q^{m \minus{} 1})\cdot (\text{other stuff})$.  For $ 0 \leq k \leq m \minus{} 1$, we have
\[ \frac {F_n(q)}{q^k} \equal{} (\text{stuff with small negative powers of } q) \plus{} \sum_{i \geq 0} (\#\{p \in S_n \mid \text{inv}(p) \equal{} k \plus{} i\}) q^i
\]
and so $ \sum_{0 \leq j \leq m \minus{} 1} \frac {F_n(w_j)}{w_j^k} \equal{} 0 \plus{} m \cdot \sum_{i \geq 0} (\#\{p \in S_n \mid \text{inv}(p) \equal{} k \plus{} i\})$.  But we have $ F_n(w_j) \equal{} 0$ for $ j > 0$ and $ F_n(w_0) \equal{} F_n(1) \equal{} n!$, so we're done.

Question: can we show that for $ n < m$, these numbers are [i]not[/i] evenly distributed?


algebra
polynomial
number theory proposed
number theory

Suppose $a_1,a_2,a_3,b_1,b_2,b_3$ are distinct positive integers such that
\[(n \plus{} 1)a_1^n \plus{} na_2^n \plus{} (n \minus{} 1)a_3^n|(n \plus{} 1)b_1^n \plus{} nb_2^n \plus{} (n \minus{} 1)b_3^n\]
holds for all positive integers $n$. Prove that there exists $k\in N$ such that $ b_i \equal{} ka_i$ for $ i \equal{} 1,2,3$.

Hi,hxy09
Have you tried the method in  http://www.mathlinks.ro/viewtopic.php?t=26797 ?
Or http://www.mathlinks.ro/viewtopic.php?t=4556 ?

My idea was same to you,and I am familiar to these two links :)
But I didn't solve it even I have worked on it for more than two and a half hours!!!
Unfortunately,it was said that the solution was not "analysis" as the previous two links.

And then maybe you can check if my solution is right(or 1 point?).I am really confused about it.
Sorry for not using Latex.
Let c1>c2>c3, d1>d2>d3,and {c1,c2,c3}={a1,a2,a3},{d1,d2,d3}={b1,b2,b3}
Let {f1(n),f2(n),f3(n)}={f4(n),f5(n),f6(n)}={n+1,n,n-1}
And f1(n)c1^n+f2(n)c2^n+f3(n)c3^n=(n+1)a1^n+na2^n+(n-1)a3^n
f4(n)d1^n+f5(n)d2^n+f6(n)d3^n=(n+1)b1^n+nb2^n+(n-1)b3^n
Then do something as metioned above...
We can assume that (f4(n)d1^n+f5(n)d2^n+f6(n)d3^n)/(f1(n)c1^n+f2(n)c2^n+f3(n)c3^n)=sum(Pi(n)/Qi(n)*ri^n)      i=1,2,...,k
where Pi(x),Qi(x) (i=1,2,...,k) are rational coefficient polynomial ,and Qi(x) (i=1,2,...,k) are monic.
And r1>r2>...>rk>=1 ,which are rational.
Then do something like Gauss's Lemma([url=http://en.wikipedia.org/wiki/Gauss's_lemma_(polynomial)]here[/url])
We know Qi(x)=1 (i=1,2,...,k)
Then f4(n)d1^n+f5(n)d2^n+f6(n)d3^n=(f1(n)c1^n+f2(n)c2^n+f3(n)c3^n)sum(Pi(n)*ri^n)  i=1,2,...,k
Then we know f1(x)=f4(x),f3(x)=f6(x),f2(x)=f5(x)
We can get the result by simple calculation.

I will give you more details if you want.
I just wonder if my solution is correct. :maybe:

I was thinking about analysis all the time during the contest...
Indeed,we can play the same trick as http://www.mathlinks.ro/viewtopic.php?p=466215#p466215
Not that difficult,right? :D

Wow,thank you for your excellent first step!!! :lol:
And could you please post the remaining steps?

P.S Very sorry for my late reply,for I was very busy during the last two weeks

Here is a number-theoretic solution.

[b]Claim: [/b] We have $(a_1^k+a_2^k+a_3^k)(b_1^k-b_3^k) = (b_1^k+b_2^k+b_3^k)(a_1^k-a_3^k)$ for any $k\in\mathbb{Z}^+$.

[b]Proof: [/b] Let $T$ denote that expression. Take any prime $p > 2010(a_1^k+a_2^k+a_3^k)$ and by CRT, pick $n$ such that
\begin{align*}
n&\equiv k \pmod{p-1} \\
n&\equiv -\frac{a_1^k-a_3^k}{a_1^k+a_2^k+a_3^k} \pmod p
\end{align*}
Then by FLT we have $(n+1)a_1^n+na_2^n+(n-1)a_3^n \equiv n(a_1^k+a_2^k+a_3^k) + (a_1^k-a_3^k)\equiv 0\pmod p$. Thus
\begin{align*}
n(a_1^k+a_2^k+a_3^k) + (a_1^k-a_3^k) &\equiv 0\pmod p \\
n(b_1^k+b_2^k+b_3^k) + (b_1^k-b_3^k) &\equiv 0\pmod p
\end{align*}
Multiply the first equation by $(b_1^k+b_2^k+b_3^k)$ and subtract to $(a_1^k+a_2^k+a_3^k)$ times the second equation. We deduce that $p\mid T$ for any large prime $p$. Hence $T=0$.
-----
Back to the main problem. From the claim, we get
$$(a_2b_1)^t+2(a_3b_1)^t+(a_3b_2)^t = (a_1b_2)^t+2(a_1b_3)^t+(a_2b_3)^t$$
For any $t\in\mathbb{Z}^+$. This is enough to imply that $a_3b_1=a_1b_3$ and $\{a_2b_1, a_3b_2\} = \{a_1b_2, a_2b_3\}$. The latter imply $a_1b_2=a_2b_1$ and $a_3b_2=a_2b_3$. Thus let $a_i=ac_i$ and $b_i=bc_i$ where $\gcd(a,b)=1$. Plugging back, we deduce that $a\mid b$ so we are done.

Here is the analysis solution along the lines of user [b]wrongthings[/b]. Let $$x_n = \frac{(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n}{(n + 1)a_1^n + na_2^n + (n - 1)a_3^n}$$

The main claim is that there exists rational functions $t_1(n), \cdots, t_N(n)$ (fraction of the form $t_j(n)=\frac{P(n)}{Q(n)}$ ) and reals $q_1, \cdots, q_k$ at least 1 such that $x_n = \sum_{k=1}^N t_k(n) q_k^n$ for all sufficiently large $n$.

Let $a= \max\{a_1, a_2, a_3\}$. Then for some rational functions $s_j(n)$, I can write this as

$$x_n = \frac{s_1(n) (\frac{b_1}{a})^n + s_2(n) (\frac{b_2}{a})^n + s_3(n) (\frac{b_3}{a})^n}{1 + s_4(n) u^n + s_5(n) v^n}$$

Where $1,u,v$ is a permutation of $\frac{a_1}{a}, \frac{a_2}{a}, \frac{a_3}{a}$

I can rewrite $$x_n = \left( s_1(n) \left(\frac{b_1}{a}\right)^n + s_2(n) \left(\frac{b_2}{a}\right)^n + s_3(n) \left(\frac{b_3}{a}\right)^n \right) \left( \sum_{e\ge 0} \left( -s_4(n)u^n - s_5(n)v^n\right)^e\right) = \sum_{k\ge 1} r_k(n) q_k^n$$

Here $|q_j|$ is monotonically decreasing (The reordering of terms is due to absolute convergence.)

Let's say $|q_{N+1}| < 1$. Let $f\colon \mathbb{R} \to [-1/2, 1/2)$ such that $f(x)\equiv x(\bmod\; 1)$. Then $$f\left( \sum_{k=1}^N r_k(n)q_k^n \right) = poly(n) q_{M+1}^n (1+o(1))$$

Let $P(n)$ such that $P(n)r_k(n)=Q_k(n) \in \mathbb{Z}[x]$ then $$f\left( \sum_{k=1}^N Q_k(n)q_k^n \right) = f\left( \sum_{k=1}^N P(n)r_k(n)q_k^n \right) = poly(n) q_{M+1}^n (1+o(1))\to 0$$

If $y_n = \sum_{k=1}^N Q_k(n)q_k^n$ then $y_n$ obeys some linear recursion of the form $\sum_{j=1}^T c_jy_{n+j}=0 \forall n$. It follows that $\sum_{j=1}^T c_jf(y_{n+j})=0 \forall n$ sufficiently large, since $|\sum_{j=1}^T c_jf(y_{n+j})|$ is an integer and is less than 1. It follows that $y_{n+j}$ also obeys a recursion with characteristic roots $r_1, \cdots, r_N$. In this case, $r_1, \cdots, r_N$ are all distinct positive rationals at least 1, so we can see that $f(y_{n+j})=k_1r_1^n +\cdots+k_Nr_N^n = 0$ for all sufficiently large $n$ (induct on $j$ to see $k_j=0$). This actually implies that $$x_n = \sum_{k=1}^N r_k(n)q_k^n$$

Now if I expand LHS (as functions without plugging $n$ in) in $$\sum_{k=1}^M s_k(n)p_k^n = \left(\sum_{k=1}^N r_k(n)q_k^n\right) \left((n + 1)a_1^n + na_2^n + (n - 1)a_3^n\right) = (n + 1)b_1^n + nb_2^n + (n - 1)b_3^n$$

(Where each $s_k(n)p_k^n$ is of the form $r_k(n) (n+2-j) (q_ka_j)^n$) We can see that since $$\sum_{k=1}^M s_k(n)p_k^n=(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n$$ holds for all sufficiently large $n$, they are actually the same function!

Suppose $p_1 \ge \cdots \ge p_M$, then $p_1 = \max\{a_1,a_2,a_3\}q_1 = \max\{b_1,b_2,b_3\}p_M$, $p_3= \min\{a_1,a_2,a_3\}q_M=\min\{b_1,b_2,b_3\}$ because these are the only way to get $q_ka_j$ to be equal to $p_1, p_M$.

We know from $$x_n = \left( \frac{n+1}{n+2-l} \left(\frac{b_1}{a}\right)^n + \frac{n}{n+2-l}  \left(\frac{b_2}{a}\right)^n + \frac{n-1}{n+2-l}  \left(\frac{b_3}{a}\right)^n \right) \left( \sum_{e\ge 0} \left( -s_4(n)u^n - s_5(n)v^n\right)^e\right) = \sum_{k\ge 1} r_k(n) q_k^n
 (\ast)$$ that $r_N(n)$ is of the form $(n+2-l_1)^{c_1} (n+2-l_2)^{c_2} (n+2-l)^{-c_1-c_2}$ where $\{l_1,l_2\} = \{-1,0,1\} \backslash \{l\}$.

Claim: If $l=\pi(1), a_{\pi(1)} > a_{\pi(2)} > a_{\pi(3)}$ then $b_{\pi(1)} > b_{\pi(2)} > b_{\pi(3)}$

Proof: From that expression it is easy to show that $r_1(n)=1$. The key claim is that $r_N(n)=1$ which implies the $(n+2-\pi(3))(a_{\pi(3)})^n$ gets multiplied by $1\times q_N^n$ to get $(n+2-\pi(3))(b_{\pi(3)})^n$ which implies $b_{\pi(3)}$ is minimum. Suppose $b_{\pi(3)}$ is not minimum, then $(n+2-\pi(3))(a_{\pi(3)})^n r_N(n) q_N^n = (n+2-\pi(2)) b_{\pi(2)}^n$. We know that all $r_N(n)$ is of the form $\frac{R(n)}{(n+2-\pi(1))^t}$. This implies that $t=0$ and $r_N(n)=1$

Claim: $a_{\pi(1)}b_{\pi(2)} = a_{\pi(2)}b_{\pi(1)}$

Proof: We know that $q_2 = \max \{ \frac{b_{\pi(2)}}{a_{\pi(1)}}, \frac{b_{\pi(1)}a_{\pi(2)}}{a_{\pi(1)}^2}\}$. If $a_{\pi(1)}b_{\pi(2)} \ne a_{\pi(2)}b_{\pi(1)}$ then it's clear that $\frac{b_{\pi(2)}}{a_{\pi(1)}} > \frac{b_{\pi(1)}a_{\pi(2)}}{a_{\pi(1)}^2}$ because otherwise there is one unique way to generate $R_j(n) (\frac{b_{\pi(1)}a_{\pi(2)}}{a_{\pi(1)}^2})^2 \in (b_{\pi(2)}, b_{\pi(1)})$ so this term doesn't get cancelled out. Therefore,

$$b_{\pi(1)} = a_{\pi(1)} q_1, b_{\pi(2)} = a_{\pi(1)} q_2, b_{\pi(3)} = a_{\pi(3)} q_N, \frac{b_{\pi(2)}}{a_{\pi(1)}} > \frac{b_{\pi(1)}a_{\pi(2)}}{a_{\pi(1)}^2}$$

Note this implies that $a_{\pi(1)}q_2 \ne a_{\pi(2)}q_1$. Consider the third largest number among $a_kq_j$. There are two possibilities: $a_{\pi(2)}q_1$ and $a_{\pi(1)}q_3$. Their coefficients must cancel, but $l=\pi(1), a_{\pi(1)} > a_{\pi(2)} > a_{\pi(3)}$ then $b_{\pi(1)} > b_{\pi(2)} > b_{\pi(3)}$ gives contradiction.

We can similarly show that $\frac{b_j}{a_j}$ is constant so we are done.

Trying to analyze MarkBcc168's solution for its motivation.

[b][size=150]Step 1: Applying a hard skill / well-known theorem.[/size][/b]
Well, clearly we need to simplify the expression given. Let's set
\[(n+1)a_1^n+na_2^n+(n-1)a_3^n=n(a_1^n+a_2^n+a_3^n)+(a_1^n-a_3^n).\]
Now we try to apply some more well-known theorems. If we [b]commit to trying modulos/divisibility[/b], then one of the most obvious theorems to use should be Fermat's Little Theorem (aka [b]the theorem that reduces exponents for free[/b]). I think [b]this is the most important step[/b], trying to organize all the thoughts (since there's a lot to try) cohesively by applying specific steps and [b]committing to something[/b]. We need to find a prime $p$ where
\[p\mid n(a_1^n+a_2^n+a_3^n)+(a_1^n-a_3^n)\]
and we can say $n\equiv k\pmod {p-1}$ which lets us write
\[n(a_1^n+a_2^n+a_3^n)+(a_1^n-a_3^n)\equiv n(a_1^k+a_2^k+a_3^k)+(a_1^k-a_3^k)\equiv 0\pmod p\]
hence we get the equivalence in MarkBcc168's solution.
[b][size=150]Step 2: Proving a result by bounding.[/size][/b]
I think this is probably a well-known hard skill: [b]To prove $T=0$, it suffices to show $p\mid T$ for sufficiently large $p$.[/b]
If you make heuristic estimates, you realize this has to be what you need to use (and you [b]must coagulate your ideas into a solution path[/b], or else you'll be confused and give up, rip).

So we have that
\[n(a_1^k+a_2^k+a_3^k)+(a_1^k-a_3^k)\equiv 0\pmod p\]
\[n(b_1^k+b_2^k+b_3^k)+(b_1^k-b_3^k)\equiv 0\pmod p\]
and the point now is that
\[T=(a_1^k+a_2^k+a_3^k)(b_1^k-b_3^k)-(b_1^k+b_2^k+b_3^k)(a_1^k-a_3^k)\equiv 0\pmod p\implies T=0.\]
[b][size=150]Step 3: Finishing off.[/size][/b]
After this, there was an approach I tried where you realize that now
\[a_1^n+a_2^n+a_3^n\mid b_1^n+b_2^n+b_3^n\]
but this doesn't really generate traction. Let's [b]commit to something else.[/b] Obviously, many terms in the expansion of $T$ should cancel, so expanding gives
\[(a_2b_1)^k+2(a_3b_1)^k+(a_3b_2)^k=(a_1b_2)^k+2(a_1b_3)^k+(a_2b_3)^k.\]
Looking at this equation, we just need to show that all the corresponding parts are actually equal, which seems obviously true.
Oh wait, it is. We just need some asymptotics. Consider the maximum in $\{a_2b_1,a_3b_1,a_3b_2\}$ and $\{a_1b_2,a_1b_3,a_2b_3\}$. Evidently these are both equal. Notice that each of the six products has a coefficient $c\in \{1,2\}$ associated with it; we have two cases.
[list]
[*]If the coefficients assigned to the maximums are different, then we'll get, say $m=a_3b_1=a_1b_2=a_2b_3$. Then we trivially find $n=a_2b_1=a_3b_2=a_1b_3$, hence
\[\frac{m}{n}=\frac{a_3}{a_2}=\frac{a_1}{a_3}=\frac{a_2}{a_1}\implies a_1=a_2=a_3,\]
a contradiction.
[*]If the coefficients assigned to the maximums are the same, we can split into two subcases. When $c=2$, we get $a_3b_1=a_1b_3$ and by the condition on distinct integers it follows that $a_3b_2=a_2b_3,a_2b_1=a_1b_2$. This implies $b_i=ka_i$. When $c=1$, we can WLOG $a_2b_1=a_1b_2$, hence
\[2(a_3b_1)^k+(a_3b_2)^k=2(a_1b_3)^k+(a_2b_3)^k\]
now by the same maximum argument two numbers, one on each side, should be equal. We've already covered $a_3b_1=a_1b_3$, if $a_3b_1=a_2b_3$ or $a_1b_3=a_3b_2$ we'll obtain all four products equal, contradiction.
[/list]
All cases are exhausted and we may conclude.


geometry
inequalities
circumcircle
algebra
polynomial
3D geometry
pyramid

[b]Please block this topic and do not reply to it. This contest is currently running. This topic is created so everyone can do the problems silently at home and recognize cheating attempts.[/b]

The limits are:
[list]
Problems 1-4: till 3rd October
Problems 5-8: till 3rd November
Problems 9-12: till 3rd December
[/list]

Problem 1.
On some fields of a $ m\times n$ ($ m,n\geq 2$) chessboard rooks are being placed. It is known that every rook is attacked by at most two other rooks.
Determine the largest possible number of placed rooks for which such a situation is possible.

Problem 2.
Let $ n \geq 2$. Let $ r_1,r_2,\ldots ,r_{n\minus{}1}$ be remainders of numbers $ 1,1\plus{}2,1\plus{}2\plus{}3,\ldots ,1\plus{}2\plus{}\ldots \plus{}(n\minus{}1)$ modulo $ n$ respectively. Determine all such values of $ n$ for which the sequence $ (r_1,r_2,\ldots ,r_{n\minus{}1})$ is a permutation of sequence $ (1,2,\ldots ,n\minus{}1)$.

Problem 3.
The incircle of triangle $ ABC$ is tangent to the sides $ BC,CA,AB$ at points $ D,E,F$ respectively. Points $ M,N,J$ are centers of incircles of triangles $ AEF,BDF,DEF$. Show that points $ F$ and $ J$ are symmetrical with respect to line $ MN$.

Problem 4.
Show that for all nonnegative real values $ a,b,c$ an inequality occurs:
$ 4(\sqrt{a^3b^3}\plus{}\sqrt{b^3c^3}\plus{}\sqrt{c^3a^3})\leq 4c^3\plus{}(a\plus{}b)^3$.

Problem 5.
For every integer $ n\geq 1$ determine the largest possible number of subsets of set $ \{1,2,\ldots n\}$ with following condition satisfied:
every two of those subsets are disjoint or one is contained in the second.

Problem 6.
A triangle $ ABC$ is given, where $ AB\equal{}AC$. On rays $ AB$ and $ AC$ points $ K$ and $ L$ are chosen not on the sides of the triangle, so $ 4\cdot BK\cdot CL\equal{}BC^2$. Point $ M$ is the center of segment $ BC$. Lines $ KM$ and $ LM$ intersect the circumcircle of triangle $ AKL$ second time at $ P$ and $ Q$ respectively. Show that lines $ PQ$ and $ BC$ are parallel.

Problem 7.
A sequence of integer numbers $ f_i$ is defined as follows: $ f_0\equal{}0,f_1\equal{}1,f_n\equal{}f_{n\minus{}1}\plus{}f_{n\minus{}2}$ for $ n\geq 2$.
Find all polynomials $ W$ with integer coefficients satisfying following condition:
for every $ n\equal{}0,1,2,\ldots$ there exists such a integer number $ k$ that $ W(k)\equal{}f_n$.

Problem 8.
Diagonals of a basis $ ABCD$ of a pyramid $ ABCDS$ intersect at point $ H$ and are perpendicular. Furthermore, segment $ SH$ is the altitude of the pyramid from point $ S$. Let $ K,L,M,N$ be the projections of point $ H$ onto planes $ ABS,BCS,CDS,DAS$ respectively. Show that lines $ KL,MN,AC$ are parallel or concurrent.

Problem 9.
A table $ 2008\times 2008$ is given. Two players alternately make moves. A move consist of choosing a white or black pawn and placing it on a chosen field. A player wins if after his move there is a sequence of five pawn in the same color in a line (horizontal, vertical or diagonal). Determine if the first player has a winning strategy.

Problem 10.
Point $ P$ is the center of the smaller arc $ BC$ of the circumcircle of triangle $ ABC$, where $ \angle BAC \equal{}60^\circ$. Point $ M$ is the center of segment connecting the centers of outcircles of this triangles, tangent to sides $ AB$ and $ AC$. Show that $ PM\equal{}2\cdot BP$.

Problem 11.
Show that for all integer numbers $ k>m\geq 1$ an inequality occurs:
$ \frac{\sqrt[k]{k!}}{\sqrt[m]{m!}}<\frac{k}{m}$.

Problem 12.
A prime number $ p$ is given. On the left side of the blackboard numbers $ 1,2,\ldots ,p\minus{}1$ are written, on the right just one $ 0$. We consider a sequence of $ p\minus{}1$ moves, each of them as follows: we choose a number from the left side, add it to all he numbers on the blackboard and erase it. Determine for which values of $ p$ one can choose such a sequence of moves so the number remaining on the blackboard after all of them will be divisible by $ p$.

PM me please or any other moderator of the Olympiad forums if any of the problems still gets posted.

Yea guys please don't spoil this preliminary competition, it makes no sense in the end. Try to do the problems on your own. Anyone caught discussing them on the Forum will be banned.


geometry
trapezoid
trigonometry
rectangle
Pythagorean Theorem

Consider the two circles with centers at O1 and O2, which are externally tangent at P.
Line L is a common tangent to both these circles at T1 and T2 respectively. The bigger circle has a diameter of a units and the smaller one has a diameter of b units. Find the length of the line segment connecting T1 and T2.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=808[/img]
[hide="Answer"]Answer is $\sqrt{ab}$, but I don't know why.[/hide]

[quote="frt"][hide="Hint"]
$O_1O_2T_2T_1$ is a trapezoid, where $O_1T_1\perp L$ and $O_2T_2 \perp L$.

Break it into a right triangle and a rectangle.
[/hide][/quote]

[hide]yes, and then u can use trig to find the base of the triangle, which equals the other side of the rectangle.[/hide]

[hide]Draw perpendicular from line O1T1 to O2.
$x=\sqrt{(a+b)^2-(a-b)^2}$
$x=\sqrt{(2a)(2b)}$ by difference of squares.
$x=2\sqrt{ab}$[/hide]

[quote="frt"][quote="anirudh"]The bigger circle has a [b]diameter[/b] of a units and the smaller one has a [b]diameter[/b] of b units.
[/quote]
 ;)[/quote]

Then, using the same logic, the answer is:
[hide]$\sqrt{ab}$[/hide]

yeah, just do it with $\frac{a}{2}$ and $\frac{b}{2}$...

haha, i made the same mistake. I keep incorrectly thinking radius instead of diameter ;;p

[hide]
$(a+b)^{2}=(a-b)^{2}+(2x)^{2} \Rightarrow x=(ab)^{\frac{1}{2}}$

[/hide]
a less trivial problem would be with regular N-gons.

[hide]
Since the line is a tangent, it is perpendicular to the radii.
Thus, O1O2T2T1 is a trapezoid.
We break up this trapezoid into a rectangle and a right triangle.
The height of the triangle is $\frac a2-\frac b2$.
The hypotenuse of the triangle is $\frac a2+\frac b2$.
So by the Pythagorean Theorem, the other leg is:
$(\frac a2+\frac b2)^2-(\frac a2-\frac b2)^2$
$=(\frac a2+\frac b2+\frac a2-\frac b2)(\frac a2+\frac b2-\frac a2+\frac b2)$
$=ab$
Edit: Oops, I forgot to take the square root.
[/hide]


ok whats the difference between cation and an anion....??

"Physics is life"


ratio
limit
inequalities
geometric series
real analysis
real analysis unsolved

Show that if $ Lim|\frac{a_{n\plus{}1}}{a_{n}}| \equal{} L > 1, as\; n\rightarrow \infty,$ then $ |a_{n}|\rightarrow \infty$ as $ n\rightarrow\infty$
Also, from that, deduce that $ a_{n}$ does not approach 0 as $ n \rightarrow \infty$.
The book suggests showing there is some number r>1 such that for some number N, $ |a_{n\plus{}1}|> r|a_{n}|$ for all n [u]>[/u]N.

I was able to show that the series $ |a_{N\plus{}1}|$ diverges toward infinity, but I don't think that's very helpful.

Any help would be great!

The proofs of a these convergence tests often boil down to making the series look like a geometric series, and using the fact that $ \sum r^n$ converges for $ \minus{}1 < r < 1$ and diverges otherwise. The root test also uses this method. I can get more specific if you like.

I am able to prove the Ratio Test itself, its this side sort of proof that I can't do.  I can show that $ |a_{n}|$ does not go to 0, but I cannot show that it goes towards $ \infty$

Suppose that $ \lim |a_{n\plus{}1}/a_n| \equal{} L > 1$. For large enough $ n$, $ |a_{n \plus{} 1}/a_n|$ is "bounded away" from $ 1$, so that we can say $ |a_{n \plus{} 1}/a_n| \ge 1 \plus{} \varepsilon$ for some number $ \varepsilon > 0$. Then $ |a_{n \plus{} 1}| \ge (1 \plus{} \varepsilon)|a_n| \ge (1 \plus{} \varepsilon)^2|a_{n \minus{} 1}| \ge \ldots$. Maybe you'll want to be a little more rigorous than this but it's the idea.

This "bounded away" concept is a useful tidbit to remember, you can state it more formally if you like and it follows readily from the definition of convergence.

So with that, you can show that $ |a_{n\plus{}1}|$ is always growing larger, does that mean a) that it goes to infinity and b) that $ |a_{n}|$ also goes to infinity?

Can anyone confirm my last post, more specifically, the part b) of my thoughts please?

Thanks.

$ |a_{n\plus{}1}|$ and $ |a_n|$ are the same sequence - the only difference is in how you labeled them. Of course they have the same behavior in the limit, since they are the same sequence.

But to get back to a): saying that a sequence is "always growing larger" does not mean it tends to $ \infty.$ "Always growing larger" sounds to me like "increasing.' Note that $ \arctan n$ or $ 1\minus{}\frac1n$ are increasing sequences that do not tend to $ \infty.$

The reason that $ |a_{n\plus{}1}|$ tends to infinity is that it is larger than another sequence - in hadjumrad's notation, $ (1\plus{}\epsilon)^n|a_1|$ - that is known to tend to infinity.

The ratio test and the root test are both merely comparison tests. Both are semi-automated comparisons to [b]geometric[/b] series. Any attempt to prove the tests must make that comparison explicit. There must be an inequality with something times $ r^n$ for some $ r\ne 1$ on one side of the inequality.


function
calculus
integration
logarithms
real analysis
calculus computations

1. Is there any increasing function defined on an interval which is not differentiable at any point in that interval?

2.How do you do integrals like $ \int_{0}^{ \infty}ln(x)e^{\minus{}x^2}dx$ or $ \int_{0}^{ \infty}sin(x)e^{\minus{}x^2}dx$?

[quote="0714446459923"]
2.How do you do integrals like $ \int_{0}^{\infty}ln(x)e^{ \minus{} x^2}dx$ or $ \int_{0}^{\infty}sin(x)e^{ \minus{} x^2}dx$?[/quote]

$ \int_{0}^{\infty}ln(x)e^{ \minus{} x^2}dx\equal{}\{x^2\equal{}t,dx\equal{}\frac{1}{2\sqrt{t}}dt\}\equal{}\frac{1}{4}\int_{0}^{\plus{}\infty}\ln{t}~e^{\minus{}t}~t^{\frac{1}{2}\minus{}1}dt\equal{}\frac{1}{4}\Gamma^{(1)}(\frac{1}{2})$.

Any ideas for the second?

[quote="0714446459923"]1. Is there any increasing function defined on an interval which is not differentiable at any point in that interval?
[/quote]
$ f(x)\equal{}x[x],x \in \mathbb{R}^{\plus{}}$
clearly is strictly increasing and not differentiable at any point

what is [x]? do you mean the floor function? if so, wouldn't x[x] be differentiable at any point other than a whole number?

yes it is the greatest integer function
no i don't think

how is it not differentiable at say, x=3/2?

every individual section of the function is linear and hence differentiable except at the endpoints of the segment

The answer to (1) is "No". Moreover, any monotone function on an interval is differentiable at almost every (in the sense of Lebesgue measure) point of that interval. The proof is too long to outline it here but you can find it in every decent measure theory textbook.


Prove that $sin\frac{\pi}{14}$ is a root of $8x^3-4x^2-4x+1=0$. Find the other roots.

thoughts :

it's equivalent to showing cos(3pi/7) is a root.  If cos(x) is a root, we have 8cos(x)^3-4cos(x)^2-4cosx+1=0, which can be written as 2cos(3x)-2cos(2x)+2cosx-1=0.  then maybe we can use complex numbers from here...

edit: i think sum-to-product formulas are more helpful, but im stuck... maybe this isnt the right route.  after using sum-to products it comes down to (1-2sin(x)^2)(-2sin(x)+1)=1/2...

[hide]For $\theta=\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}$ or $\pi$,
we have $\cos{4\theta}+cos{3\theta}=0$
letting $x=\cos{\theta}$,
and since $\cos{4\theta}=8x^4-8x^2+1$, $cos{3\theta}=4x^3-3x$,
we have $8x^4+4x^3-8x^2-3x+1=0$
since $x=-1$ is a root, dividing by $x+1$ gives $8x^3-4x^2-4x+1$.
So the other roots are $\cos{\frac{\pi}{7}}$ and $\cos{\frac{5\pi}{7}}$.[/hide]

how do you know cos(3x)+cos(4x)=0 for those values you gave?

[quote="cats..."]
$2\cos(3x)-2\cos(2x)+2\cos x-1=0$.  then maybe we can use complex numbers from here...
[/quote]
if i recall correctly, this identity was an imo problem... quite an aged one, maybe 1967...
it's quite straightforward:
you just have to remember that $ z = \cos(\frac{\pi}{7}) + i\sin(\frac{\pi}{7})$ is a root of $ z^7 - 1 = 0$ different from $1$. so, dividing by $z-1$, and taking real parts of other complex roots, you get that identity.
(i think this has been one of the first imo problems i've solved, that's why i remember it).

since for $\theta=\frac{\pi}{7}$,
$4\theta=\pi-3\theta$
so $\cos{4\theta}=-\cos{3\theta}$
and similarly for the others...


Support
induction
algorithm
combinatorics unsolved
combinatorics

A horizontal board is a support for four vertical pegs. One of the pegs passes through holes in $n$ disks of increasing size, the largest disk being at the bottom, the smallest one on the top. We may take disks off a peg, always one at a time, and put each disk on another peg, but never in such a way that a disk is put on top of a smaller disk. Prove that the minimum number of transfers one should make to transfer all the disks to another peg is at least $2^{\sqrt n-1}$ but less than $2^{2 \sqrt n+2}$.

[i]Komal, A. 250, November 2000[/i]

Greetings to all the Vietnam mathlinks members for the problem...  :D

[i]Edited by Myth[/i]

Maybe this helps (maybe it doesn't :D ). I think that for 3 pegs the minimum number of transfers is 2 n -1. I never thought  about a proof, but I had a computer game and every time I could change the peg in 2 n -1 moves I would get the message "Congratulations! You used the minimum number of moves!". I couldn't choose every n, of course. The options were n=1, 2, 3, 4, 5, 6, 7.

I'll think about this problem, which is, I think, much more challenging. The strategy for the 3 peg version is, I think, obvious after you play the game for a while.

Actually, there's lots of info about this game on the Net. Here's just a link:

[url]http://www.lucas-nussbaum.net/hanoi.php?lang=en[/url]

However, I'm beginning to think that the classic game (3 pegs) doesn't help too much...

Sorry I'm making so many virtually useless posts on this topic, but here's something that we could really use:

[url]http://pages.prodigy.net/mathguy/tower.htm[/url]

It's a Java applet for the general game (you can use any number of pegs, from 3 to 50 and any number of disks from 3 to 200!).

Just another nice problem to move from Algebra to Combinatorics :)
(and a reason to remind someone to finally solve the problem in order to justify 3 flood messages :D )

I will prove upper bound.

Consider for simplicity the problem for perfect squares.
By induction, consider $(n+1)^{2}$ discs. Our algorithm will be the following:
1. Move $n^{2}$ upper discs on another peg using $2^{2n+2}$ operations.
2. Move the remaining $2n+1$ disc to one of two remaining pegs using classical Towers of Ha Noi. It will take $2^{2n+1}$ operations.
3. Move $n^{2}$ to the same peg. It will take $2^{2n+2}$ operations again.
Summary, we used less than $2^{2n+4}$ opearions.

that's not very difficult really
solution canbe dound after you enter here
[url]http://www.cut-the-knot.org/recurrence/hanoi.shtml[/url]

hello every vietnamese. I am vietnamese too

This forum has another purpose. I like to solve problems and I like then other people solve them.
Don't want to say rude words.

Oh I'm very sorry really
I means "can be found", it's my typing mistake
I don't know the word mean but I don't have a purpose to say rude word
i'm very sorry


1. ABCD is a parallelogram.  E is a point on BC and F is a point on CD.  AE cuts BF at G, AF cuts DE at H and BF cuts DE at K.
Prove that [AGKH] = [BEG] + [CEKF] + [DFH]

2. Let ABCD be a parallelogram. A line from C cuts the diagonal BD in E and AB in F.  If F is the midpoint of AB, and the area of [BEC] is 100 find [AFED].

New to the site, later I'll add symbols and diagrams.

I figured out #1.
[hide="Solution"]
Extend AD to Z such that CZ is perpendicular to AD.
-----
Parallelogram, therefore BC = AD.
[BEG] + [CKEF] = [BFC] - [KEG], [BFC] = (CX)(BC) or (CX)(AD)
So [BEG] + [CKEF] = (CX)(AD) - [KEG] <-- (1)
-----
[DFH] = [ADF] - [ADH], [ADF] = (XZ)(AD), [ADH] = (YZ)(AD)
So [DFH] = (XZ)(AD) - (YZ)(AD) <-- (2)
-----
(1) + (2) = [BEG] + [CEKF] + [DFH]
= (XZ)(AD) - (YZ)(AD) + (CX)(AD) - [KEG]
= AD(XZ - YZ + CX) - [KEG]
= AD(CY) - [KEG] <-- (3)
-----
[AGKH] = [AED] - [ADH] - [KEG], [AED] = (CZ)(AD), [ADH] = (YZ)(AD)
So [AGKH] = (CZ)(AD) - (YZ)(AD) - [KEG]
= AD(CZ - YZ) - [KEG]
= AD(CY) - [KEG] <-- (4)
-----
By (3) and (4) (or since (3) = (4)),
[AGKH] = [BEG] + [CEKF] + [DFH]
[/hide]
Yah, crude stuff.  In the future I'll try to get pics etc in there.  And maybe have it structured a little better.

FEB is similar to DEC(AAA similarity)
Since FB/BC=1/2,
EB/ED=1/2----------------------(1)
AreaFEB/AreaDEC=(FB/BC)^2=1/4------------------(2)
Ar(BEC)/ Ar(EDC) =BE/ED=1/2 [using 1]
So,ar(EDC)=200
Ar(DCB)=200+100=300=Ar(ABD)------------------(3)
Using (2),
Ar(FEB)=1/4*200=50
Ar(AFED)= Ar(ABD)-Ar(FEB)=300-50=250 :huh:
Being NEW as you I too dont know how to attach pictures. SORRY.


Is there anyway to prove that a whole number (whose digit's sum is a multiple of 3) then the whole number itself is a multiple of 3.

example: 12  its digits are 1 and 2 therefore the sum of those digits are 3 and it is divisble by three. Is it true for all numbers and if it is then prove it.

Any positive integer can be represented thus:

$N=\sum_{k=0}^{n}{a_k10^k}$, where $a_k$ are its digits. Write the sum this way

$N=\sum_{k=0}^{n}\left(a_k+a_k(10^k-1)\right)=\sum_{k=0}^{n}a_k+\sum_{k=0}^{n}a_k(10^k-1)$

The second term is divisible by $3$ in any case, since the numbers $10^k-1$ consist of repeated nines (or are equal to zero for $k=0$). Therefore, the number $N$ is divisible by $3$ IF AND ONLY IF the number $\sum_{k=1}^{n}a_k$ is divisible by $3$, and that number is the sum of its digits.

Sorry, i have know idea what your answer means. I'm only in grade 11 math.

:)
make it easier to explain, let's just explain what Farenhajt said with a little example.

Let's say your have a four-digit number abcd, where a,b,c and d are the digits respectively. so the number abcd = a*1000 + b*100+c*10+d. manipulate it slightly, abcd = a*999 + b*99 + c*9 + a + b +c + d now.. Clearly if a+b+c+d is divisible by 3, or even 9, the four digit number abcd is also divisible by 3 or 9.

So yeah, relate this to what Farenhajt said (and that sigma sign means summation of the terms as k increases by 1 from k= 0 till k=n), you should see it...

Still learning LaTeX... sry for the poor presentation... :blush:


linear algebra
matrix
geometry
analytic geometry

can u tell me what is a compatible determinate system?...please give me an example..

This isnt the right place to post that. (I dont know what the right place is because I have no idea what that is)

I think he's talking about a square matrix and its determinant which I truly forgot how to do because I haven't used it for the past half year except for a $2 \times 2$ in the form:
$|a b|$
$|c d|$
which is ad-bc

moderator, please move to getting started?

the determinant of matrix

a b
c d

= ad-bc

I think... It's more complicated with the 3 by 3 matrix though, if you need more, pm me.

just use a TI z80 calculator, it's so easy.

$3\times3$ matrix:
$|abc|$
$|def|$
$|ghi|$
$=$
$aei+bfg+cdh-afh-bdi-ceg$

[quote="SplashD"]He wants the general form n by n[/quote]
OK then... http://astronomy.swin.edu.au/~pbourke/other/determinant/index.html  :maybe:

lol were doing matrixes and determinants in math now

|a b|
|c d|

the determinate would be ad-cb

the determinate is used in Cramer's Rule and useful for solving 3 variable equations and areas of triangles on the coordinate plane

LOL get Prof. Strang's book on linear algebra for an easier way...
if you wanted to with numbers(i don't know how to express it with variables) you could have used forward substitution and then mult. the pivots...

2 3
4 3

we eliminate by mult. by -2 in the 1st row and subtract that from the 2nd
2 0
0 -3

2*-3=-6 so that's our det.

notice we could have done
6-12=-6

This was from 2 years ago. Please do not revive old threads. Thank you.