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I'm a junior starting the college search and was just wondering what you all thought...  :)

As I have heard,it is hardest to get into Harvard, probably, because Harvard concidered to be the most prestigious university in the world,so I voted for Harvard.

  By the way,I have applied to Harvard, Princeton and MIT :blush: ,and I hope I will get into, at least,one of them,anyhow :blush: .

P.S. My vote was the first one :P  :D .

This poll might be slightly skewed, because the people that use this forum are primarily math/science people, and that could make getting into MIT/Caltech seem easier to get into than Harvard/Princeton etc.

Also, while it's interesting to talk about which schools are the hardest to get into, I would advise that you focus more on which school you feel will be right for you.

Granted, numbers don't tell the whole story, but they're still a good starting point:

\[\begin{tabular}{|c|c|c|}\hline \multicolumn{3}{|c|}{Class of '10 Admission Statistics}\\ \hline University & \# Applicants & \% Admitted\\ \hline Harvard & 22,753 & 9.3\\ \hline MIT & 11,373 & 13.0\\ \hline Princeton & 17,563 & 10.9\\ \hline Yale & 21,099 & 8.6\\ \hline Brown & 18,313 & 14.6\\ \hline Columbia & 17,148 & 9.6\\ \hline Dartmouth & 13,937 & 15.4\\ \hline Stanford & 22,000 & 10.9\\ \hline Cornell & 28,097 & 24.7\\ \hline UPenn & 20,479 & 17.7\\ \hline \end{tabular}\]

Source: http://ivysuccess.com/admission_stats_2010.html

Unfortunately, that data doesn't tell you much.  You'd rather need to know about the quality of the applicant pool at each school.

Thanks for your input.  I'm trying to decide which schools would be considered my "reach" schools, and thought I'd seek a little advice.

By the way, good luck, Tiks!  :)

I see there is quite a consensus here about what college (of those listed in the poll) is hardest to get into. What do you think is the hardest aspect of admission to Harvard?

Also, some of those depend on which program you are applying to.

For example, Wharton drags down UPenn's acceptance rate a considerable amount.

I think that Harvard is the most prestigious university in that most people in the country think of it as [i]the[/i] top university.  College rankings aren't necessarily in accordance with prestige not to mention that rankings vary from source to source.

As for tokenadult's question, I think Harvard's hard to get into because of its prestige.  Going to an Ivy League school is to many scholars the ultimate dream.

[quote="joml88"]As for tokenadult's question, I think Harvard's hard to get into because of its prestige.[/quote]

So in your hypothesis, the mechanism is

a) Harvard is the most prestigious college (for whatever reason)

therefore

b) the most strong applicants apply to Harvard

therefore

c) the applicant pool for Harvard is more competitive than at any other college,

right?

That sounds plausible to me. I am still learning about how great the magnitude of difference might be between Harvard and "peer" colleges as to strength of applicant pool. It is plain enough from real-world observation that each year there are some applicants who are admitted to one college but rejected from another, in almost all the possible pairings of colleges.

IIT, Peking U, Ecole Polytechnique, Moscow State U, and others are harder to get into, but for the list given above...

[quote="tokenadult"]What do you think is the hardest aspect of admission to Harvard?[/quote]

Among general-purpose universities (so excluding MIT and Caltech), for any given specialization that many applicants have, Harvard almost always has a stronger pool of such specialists compared to the other schools.  For many, perhaps most, applicants, they are not competing for one of 1600 seats, they are competing for one of 50 to 200 seats against similarly categorized candidates.  Those similar candidates are a stronger bunch for admission to Harvard.

The lower admissions rate at Yale reflects the smaller size of the school.  Statistically the class that enrolls at Harvard tends to be stronger than the one at Yale, but weaker than Caltech.  I don't remember the comparison to MIT.

[quote="juicybooty911"]Some of my mom's friends (parents) tells me Havard's undergraduate schools aren't as strong as Princeton's or MIT's.[/quote]

Are they saying that about the most math-oriented programs only, or about the undergraduate degree programs in general?

[quote="pieterminate"]Harvard's the most prestigious school?  Weird....then why was Harvard behind Princeton in US Top Colleges rankings for the last 4 years.....? Other is a good choice since it includes stanford, cal tech, ucberkeley, duke, etc...[/quote]

Well, if US News says so, then it must be true.

Seriously, though, it would be difficult to dispute that Harvard is most prestigious *university* in the country.  It would be impossible to dispute that it's at least as prestigious as Princeton and Yale.

But the question is about "hardest to get into," which is not the same question as "most prestigious," although they're obviously related.  "How hard it is to get into" is mostly a function of the strength of the applicant pool (vaguely related to size of applicant pool) and the number of students who get admitted (which itself is a function of the size of desired freshman class and the percentage of admitted students who get accepted).

Mathematically speaking, one ought to define what it means for A to be harder to get into than B, which is not as obvious as it sounds.  One reasonable definition:  Out of the students who apply to both A and B, fewer get into A than B.  (Exercise: This does not necessarily define an ordering, although in practice it probably does.)

I think that the answer is probably Harvard, and I do have some basis for believing this.  There was a scholarly study done on where students choose to go when they have a choice, and they overwhelmingly choose Harvard over other schools.  (Sorry I don't have the link.)  This very likely makes Harvard more difficult to get into.

On the other hand, there are wildcards like the fact that Caltech has a very small class.  For example, Cooper Union is very difficult to get into despite the fact that most people have never heard of it.

[quote="statistical"]IIT, Peking U, Ecole Polytechnique, Moscow State U, and others are harder to get into, but for the list given above...
[/quote]
The Peking University is hard to get into only because there are so many people in China, and it is just better than almost all of the universities in Mainland China.

I'm not at all against the idea of volunteering, I just find the notion that people put on the guise of "helping people and volunteering to support a cause" when in reality they're mind is set to the fact of "this will help me for college so I'll do this."

I've done some things that are volunteer, too, but always as an indirect effect of some other action (Scouts, mainly..).

Sadly, I do have an enormous passion for staring at a TV for hours a day. oh well.

Well, your avatar does sort of imply that :P  :roll:

[quote="tokenadult"]How do we know that AoPSers are lacking in "leadership, community service, and other non-academic extracurricular activities"? [/quote]

Because it's obvious.  Mathematical competition is mostly an individual activity.  Surfing AoPS is an individual activity.  Reading math books and competition preparation are mostly individual activities.  A few math competitors might undertake large-scale organizational work for a mathematical event, or teach mathematic to others in a significant and documented way.  But for most, the math indeed contributes nothing to the "leadership, community service, and other nonacademic extra activities" and subtracts from the available time for other pursuits that do fit in those categories.

It is also true that mathematicians tend to be more introverted than the average.  AoPSers are, more than likely, nerdier than average.  IMO medalists, though an interesting bunch, are clearly nerdier than average, and there are few from any country who did any form of "leadership and community service".   The same is true for math majors, math PhDs, physics students, computer programmers, engineers, and university professors.   There is no point trying to deny these completely obvious facts; even if not a logical consequence of interest in mathematics, there is definitely some statistical correlation going beyond mere stereotypes.

[quote]And, by the way, what is an example of a "non-academic" extracurricular activity? You can lead and serve the community with academic extracurricular activities, can't you?[/quote]

Sports, the school newspaper, drama.  Most of these are activities whose output can be enjoyed by the whole school community, much more so than an individual seeking a competition medal by writing an exam.  In that sense they are "non-academic extracurricular activities"  that contribute to the community rather than an individual's resume.

I don't agree with or in any way justify the ideology of US university admissions in these matters.  But it is at least clear what the terms such as "leadership" refer to, as used in that context.

[quote="Statistical"]
Because it's obvious.
[/quote]

Great way to start your argument.  I don't think I even needed to read any more beyond that.

Sure, since one is spending time on math (s)he will have less time to spend on other things, but there are 24 hours in a day and 365 days a year.   No one spends all that time doing one thing only.  Believe me, I know a lot of people who are really good at competition math.  In fact, I feel that most of them are more inclined to be leaders, do commserv, and other EC's than your average student.

If you can actually show me a reputable statistical study that shows some sort of correlation that you speak of, then I'll probably be more hospitable to what you've said.

what about valentin...good at maths ,has leadership abilities (adminstartor ML),social service(ML , AoPs),extracurricular activities .... :D

just may be an exception.... :maybe:

Richard Rusczyk volunteers a lot these days.

Recent Time piece: talks about   ""A Reunion at the "MIT of India"  and gives its perspective:

[quote]IIT has often been called the MIT or Harvard of India, but there's a big difference — IIT is a lot more selective than the top Ivy League schools. About 250,000 Indian students take the first screening exam for a spot at an IIT; 100,000 make it to the next round; but only 4,000 are eventually selected. Even if they could make the cut at IIT, however, the brightest young American students are less likely now than they were a generation ago to choose engineering.[/quote]
 From:
  http://www.time.com/time/business/article/0,8599,1641232,00.html
[quote]Google, Microsoft and General Electric came to Santa Clara, Calif., last weekend, and all but begged graduates of one of the world's top engineering schools to work for them. Google spent $ \$$200,000 to be the lead sponsor of the four-day-long reunion of 3,500 alumni. Microsoft's research center in Hyderabad came calling. The CEO of GE, Jeff Immelt, already employs 1,500 graduates and says he needs more. Stanford? MIT? Harvard? Nope. This was a gathering of graduates of the Indian Institutes of Technology.[/quote]

BTW ...Hillary Clinton  was one of the speakers ... (last time the key note speaker was Bill Gates...)

[quote="shbmsc90"]4,000 out of 250,000 is indeed a small percentage, but do all of the 250,000 students have to pay an $ \$60-85$ fee to apply (i.e. do they have a chance)? I'm sure 100,000 people would apply to Harvard but they don't wish to waste money when they know they don't have a chance...[/quote]
True,    application fees ( and tuition) for  IIT is much less than Harvard  (indeed when I went to IIT  many decades ago it was virtually free) but  most of the applicants invest much (much – much more) more than typical application fee…  thousands invest a hefty application fee just to take test  … to qualify for  the training schools which train them to take the admission  test  to IIT.. .. almost a madness but there are coaching schools which coach you to get into a coaching school  which coach you for IIT/JEE tests ..:) !!! … Hard to believe for those who are  not familiar,  but many parents invest much more than  a typical yearly tuition for   a  regular school  just to ‘prepare’ for that entrance test.

As 60 minutes  (American TV –CBS show) piece  once said, “you put MIT, Harvard and Princeton all together and then you start getting some idea how much prestige   IIT’ have in India” …  in that  piece there was also an interesting piece…there are people who keep  ivy league schools as  “safety” in case they do not get into  IIT…(Not that uncommon to see  people who get into top schools here in US not being able to get into IIT’s)

Unless you are familiar…IIT craze is really  something unbelievable

[quote="Gyan"]... there are coaching schools which coach you to get into a coaching school  which coach you for IIT/JEE tests)[/quote]
That certainly makes the SAT prep courses in US look minor league  :), but the  idea that willing and able (or sometimes not so able) parents doing a lot for their children’s education is universal.

[quote="Gyan"]... there are people who keep ivy league schools as  “safety” in case they do not get into  IIT…(Not that uncommon to see  people who get into top schools here in US not being able to get into IIT’s)[/quote]
Supply and demand?

Let say the top 20-30 (by whatever meaningless ranking method one chooses) colleges in US get combined/compressed into one single college with only 3000 spaces, while the rest of the colleges remain helplessly downgraded by those meaningless ranking methods. Can we possibly imagine the craziness then?

Hey, a (Texas?) mother tried to murder a cheerleader just so that her daughter can make the squad. I shudder to think ... :D

It's not that Americans are less crazy about these kinds of things than Indian people or Chinese people. Americans just have more different things that they can focus on. We don't have one entrance exam, like for IIT or the Chinese universities. We have several different "entrance exams" and tons of other things that people do to make themselves look qualified. And we have a large number of prestigious universities instead of just one or a few.

[quote] . We don't have one entrance exam, like for IIT or the Chinese universities. We have several different "entrance exams" and tons of other things that people do to make themselves look qualified. And we have a large number of prestigious universities instead of just one or a few.[/quote]

To be honest, neither  does India ("have one entrance exam" etc..). just like US, there are many Indian universities having "different " admission criteria...and just like in US,  if someone does not get into one top school, there are other choices.. (after all majority of those who do not get into IIT do go to some other schools :) ..as far as total number goes,  IIT, obviously  produces only a small fraction of engineers in India..

I am sure  many think  that  there are "only one or a  few"  (or none - if you go by, say US News and World report - Many ratings I have seen   do not rate IIT anywhere in the top or indeed even in first tier schools) " prestigious universities" out side US of A  but at least IMO, there are plenty first rate universities in India good enough to produce  students who do fairly well  as a graduate student in, say a  top university  in USA  .. look at graduate students in  your own Princeton, there you will find gradutes from non-IIT Indian universities too..

 One thing which may be a little different  about   IIT is "one dimensional"  "merit based" (in general but with some exceptions - there are or going to be  reserved seats aka  affirmative actions etc) admission criteria --.just the results of JEE (except for minimum required GPA in high school) ... no essay .. no legacy etc. .. and since number of seats are only few thousands.. competition for getting in is rather tough..  ..

Anyway this might be OT ..except  for the title of the topic ("hardest to get into") ..:)

If you go on College Confidential, it seems like Stanford is the  hardest to get into. However, here: [url]http://encarta.msn.com/encnet/departments/college/?article=10schoolsToughesttoGetInto07[/url]

it says the toughest college to get into is MIT.

Also, this other site I went to- MIT vs. Harvard (which I cannot find now) said that the applicants for MIT are already brilliant... and some more. I forgot.  :oops:

ooh it was somewhere on everything2.com: [url]http://www.everything2.com/index.pl?node_id=50515[/url]


A car drives in a circular path with a diameter of 1200 feet. If the car completes exactly two laps each minute, what is the car's speed, in feet per second? Express your answer to the nearest whole number.

[hide]if the diameter is 1200, the circumference is $ 1200\pi$ Every minute it goes $ 1200\pi*2 \equal{} 2400\pi$ feet per second

every second it goes $ \frac {2400\pi}{60} \equal{} 40\pi$ feet per second. Rounded to the nearest whole number, this is equivalent to 126ftps[/hide]

thx ernie

however, i believe the answer is 126.  :rotfl:

how?

however, i believe you are right and i suk :rotfl:


Find all positive integer solutions x and n of the equation:
$x^2+615=2^n$

Note that $615$ is divisible by $3$, so since $2^n$ is not, $x$ is not divisible by $3$; It follows that $x^2 = 1 \mod [3]$, so that $2^n = 1 \mod [3]$. Therefore $n = 2k$ is even.
The equation becomes $615 = (2^k - x)(2^k + x)$. with $2^k + x > 2^k - x$. Since $2^k + x > 0$, we then have $2^k - x$.
Then, $2^k + x$ and $2^k - x$ are two positive divisors of $615$ whose product is $615$.
The positive divisors of $615 = 3 \cdot 5 \cdot 41$ are $1,3,5,5, 41, 123, 205, 615$.
Thus $(2^k + x,2^k - x)$ is one of the couples $(615,1),(205,3),(123,5),41,15)$.
Direct checking shows that the only solution is $k=6$ and $x=59$, which leads to the solution $x=59$ and $n=12$.

Pierre.


Two rectangles have integer dimensions, and both have a perimeter of 144 cm. What is the greatest possible difference between the areas of two such rectangles?


I like this problem. :)

Write $z + 1 = \cos \theta + i \sin \theta + 1 = 2 \cos^2 \frac{\theta}{2} + 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \cos \frac{\theta}{2} (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})$.  Hence it suffices to show that $\left| \cos \theta \right| + \left| \cos 2\theta \right| + \left| \cos 3\theta \right| \geq 1$ for $0 \leq \theta \leq \frac{\pi}{2}$.

For $0 \leq \theta \leq \frac{\pi}{6}$, $\cos \theta + \cos 2\theta + \cos 3\theta \geq \frac{\sqrt{3}}{2} + \frac{1}{2} > 1$.

For $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{4}$, $\cos \theta + \cos 2\theta - \cos 3\theta = 2 \cos \frac{3\theta}{2} \cos \frac{\theta}{2} - 2\cos^2 \frac{3\theta}{2} + 1 > 1$.

For $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$, $\cos \theta - \cos 2\theta - \cos 3\theta \geq 1 \Leftrightarrow 2 \cos^2 \theta + \cos \theta - 2 \geq 0$ which is easily shown to be true. :)

Here's another way to prove it:
$|\cos \theta |+|\cos 2\theta |+|\cos 3\theta |\ge$
$|\cos \theta \sin 3\theta|+|\cos 2\theta |+|\cos 3\theta \sin \theta |  \ge$
$|\cos \theta \sin 3\theta -\cos 3\theta \sin \theta |+|\cos 2\theta | =$
$|\sin 2\theta |+|\cos 2\theta |\ge |\sin 2\theta |^2+|\cos 2\theta |^2=1.$

Or, even better:
$|z+1|+|z^2+1|+|z^3+1|\ge |z^3+1-(z+1)|+|z^2+1|=|z||z^2-1|+|z^2+1|=$
$|z^2-1|+|z^2+1|\ge |z^2+1-(z^2-1)|=2.$


integration
real analysis
real analysis unsolved

How to prove that the Sobolev space $H^{1}(\Omega)$ is complete.

Proof by definition: Sobolev spaces are commonly defined as the completion of $C^{\infty}(\Omega)$ under a certain norm.

Hint:
The Sobolev space is defined as:
\[H^1(\Omega)=\{u: \Omega \to \mathbb{R} | D^\alpha u \in L^2(\Omega), \forall | \alpha| \le 1\}\]
with norm
\[||u||_{H^1(\Omega)}^{2}=\int_{\Omega} \sum_{|\alpha| =0}^{1} |D^\alpha u|^2 \mathrm{d}x\]

+Step 1: Let $\{u_n \}$ be a Cauchy sequence in $H^1 ( \Omega)$, we prove that
$u_n \to u$ in $L^2(\Omega)$ and $D^\alpha u_n \to v_{\alpha}$ in $L^2(\Omega)$.

+Step 2: Prove that: $v_\alpha=D^{\alpha}u $. Ok :)


Arrange the ff from greatest to least:

2^110, 3^75, 5^49

[b]^[/b] means raised to (an exponent)

I really need it. Hope you could help me.

How could I use modular method to solve that?

IS that a proper method for this? Or just comparison?

I wouldn't say there is a "proper" method. Generally for these types of problems you just re-write them like he did, and compare.


In math class our teacher wants us to have our own calculator, preferably a TI-30 or TI-34. She said the TI-30 is $10 and the TI-34 is $20. What is the difference between the two calculators? And is it worth the extra $10 for the TI-34?

ummmmmm

i think you should get a ti30 because its the same as the ti34 and for half the price.

my math teacher says no calcs on tests but calcs for class so i tried the 34 and 30

the ti30 is the best.

You could just use a regular scientific calculator, instead of buying another one.

[quote="Math Champion"]You could just use a regular scientific calculator[/quote]
They are regular scientific calculators.

What class is this for and what will you need it for? I would go with the TI-30 because I would think they do have the same functions.

i think you should get the ti 30 because its half the price and i dont think thers much difference in between.

[quote="athunder"]i say get ti 89 or ti 84. o.o
once you get to harder stuff you will need them, and they can easily be used for the easier stuff.[/quote]

I don't need them right now and I'm doing fine.

I have a TI-34 but a TI-30 is almost the same, the read quality is a little worse but nothing big. I say TI-30, but only if you don't have any yet. If you have a TI-34 don't waste time getting another.

And if your in 6th grade + you should really consider a scientific calculator.  You can get those things for like 7 dollars at Wal Mart.

Actually I recall getting a TI-34 for $ \frac{1}{2}$ the price of the battery pack we bought, only \$5.18

[quote="tennis123"]@ PowerofPi: I'm taking algebra II.
another question- for classes after algebra II, what type of calculator will I need? Is there a calculator similar to the TI-30 or TI-34 that I can use now and later? Thanks for any help
EDIT: I'd probably also use this calculator for MathCounts[/quote]

You'll probably need a graphing calculator for Pre-Calculus and/or Calculus (in my school district, graphers are [b]required[/b] for Algebra II and above).  A TI-83 or TI-84 should definitely be fine for school math.

Yes, a graphing calculator is required later on (i'm not sure what math class you are in) but if your teacher recommends a smaller, more basic calculator, it is probably better to have one.

I also like ti-36 and it is also 20 dollars here http://www.amazon.com/Texas-Instruments-TI36X-Scientific-Calculator/dp/B00000JBNM/ref=sr_1_1?ie=UTF8&s=electronics&qid=1253134508&sr=8-1

There is  free shipping if you buy something more than 25 dollars including that calculator. Then you have an excuse for buying something you want lol.

I just got a ti-84 silver plus a few days ago and its a pain to walk to classes w/ because i don't want to scratch it.

i agree with athuner and you should get a ti-84 o.o
it's been very helpful to me so far, i do everything in class WAY faster then the TI-15s there.


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that $ x^2+y^2-1$ is irreducible in $ %Error. "texbb" is a bad command.
{Q}[x, y].$

[quote="rghthndsd"]Interpret this as being a polynomial in $ \mathbb{Q}[y][x]$, remembering that $ \mathbb{Q}[x,y] \cong \mathbb{Q}[y][x]$.[/quote]

[b]rghthndsd[/b], thank you, I solved this problem a while ago.  However, I looked back at this problem tonight and was unsure how we know $ \mathbb{Q}[x,y] \cong \mathbb{Q}[y][x]$.  I know their definitions and what they are, it is just I don't know how to show that.

Anyway, back to the problem at hand.  I showed it was irreducible by taking $ x \mapsto x\plus{}2$ because automorphisms preserve irreducibility.  Then I got $ (x\plus{}2)^2\plus{}y^2\minus{}1 \equiv y^2\plus{}3 \text{ mod } (x)$.  So now we know $ y^2\plus{}3$ is irreducible in $ \mathbb{Q}[x,y]$ by Eisenstein Criterion with $ p\equal{}3$.  I just don't know why $ \mathbb{Q}[x,y] \cong \mathbb{Q}[y][x]$.  Thanks!

why is $ \mathbb{Q}[x,y] \cong \mathbb{Q}[y][x]?$

Check definitions: Both sides are polynomials in two variables. There is also a canonical homomorphism $ A[y][x] \to A[x,y]$ that is injective and surjective.
Or just use the universal property of polynomial rings: both fulfill them.


trigonometry
calculus
derivative
function
geometry proposed
geometry

Please don't say approximate value.
[hide][img]http://img237.imageshack.us/img237/3408/eay305nesdfredjhj5.gif[/img][/hide]

I calculate this and it is $2^{1/2}$

is this right? if yes, say me, to me post my solution, because in the actual form is a little bit confuse

this picture shows how to solve the problem
I omit some trivial things.

I set the length $AB= 1$ just to simplify. since we are interested in the ratios. and the lengths of this triangle in proportional to the side $AB$

It's easy to calculate the point $A$, since we have an angle, a point ($B$) and a length.

then. we calculate the Equation of $BD$, it's easy, since we have a point and an angle.

and after it we calculate the equation of $AD$. It's easy too. since we have an angle and a point.

we make the intersection of the $2$ previosly equations and find $D$

for the point $C$ it's very simple, since we have an angle and a point ($A$)

Since we have the points is easy to find the lengths.

i generalize this problem because this problem is very beautiful and when i put an angle of six degrees on my problem the angle is too much small :)

c ya

Thank you very much. :) But i don't understand how did you calculate all this equations. :)

These are the other solutions.

Solution 1
Owned by Murat CEYHAN.

Create AKC equilateral triangle.Than BKA will be isosceles.Than we use special triangle near picture.Complete the diagram to pentagon.EBDK is cyclic.Than we use lentgh of 36-36-108 and 30-60-90 and find sqrt2.

[hide][img]http://img82.imageshack.us/img82/5955/45t45t5xm0.jpg[/img][/hide]

Solution 2
Owned by Murat Yalçın
Create pentagon.Draw AE.We find BED=12(Diagram 2).Than let the perpendicular from D to EC.We find sqrt2.

[hide][img]http://img526.imageshack.us/img526/8141/168pf6.gif[/img][/hide]

Shahzada, I prefer use analytical Geometry.
So let's some basical things.

I have a Point $P$ and an angle, of a line that pass throught $P$, formed with the $x$ axis.
So the equation of the line will be

${P=\{\alpha ,\beta \}}$

${y-\beta =\tan (\theta ) (x-\alpha )}$

We can deduce it from the fact that the derivative of the line is constant, and the derivative gives the tangent of a curve

${y'(x)=\frac{\Delta y}{\Delta x}=\frac{y-\beta }{y-\alpha }}$

The derivative it's equal to the average of the function

At first i deduce the most simple equation, of $BD$ and After $BA$, and after $AD$ and finally $AC$.

With this, we can solve the generalization of the problem with just one image, that's why i think this is a beautiful problem.

c ya


Set A consists of the numbers in the arithmetic sequence $12, 18, 24,\dots$ and Set B consists of numbers in the arithmetic sequence $15, 24, 33\dots$. What is the sum of the five smallest numbers common to A and B?

$12+6x = 15+9y$
$6x = 3+9y$
$2x = 1+3y$
$0 = 1+y \pmod 2$
$y = 1 \pmod 2$
$y = 1, 3, 5, 7, 9$

So substitute these possible values of y into $15+9y$ and add them up.

24
42
60
78
96

Total = 300


calculus
integration
trigonometry
symmetry
logarithms
trig identities
real analysis

Integrate tan x tan 2x tan 3x with respect to x.

I wonder we could use the equality: $ \tan (a\plus{}b\plus{}c)\equal{}....$.

Substitute $ x\equal{}\arctan t$. Apply various trig identities. It becomes a rational function.

$ \tan(2x)\equal{}\frac{2t}{1\minus{}t^2}$
$ \tan(3x)\equal{}\frac{3t\minus{}t^3}{1\minus{}3t^2}$

The new integral is $ \int\frac{6t^3\minus{}2t^5}{(1\minus{}t^2)(1\minus{}3t^2)(1\plus{}t^2)}\,dt$.
Apply partial fractions. By even-odd symmetry, we only get $ \frac{t}{1\plus{}at^2}$ terms: $ \int\frac{\minus{}t}{1\minus{}t^2}\plus{}\frac{2t}{1\minus{}3t^2}\plus{}\frac{\minus{}t}{1\plus{}t^2}\,dt$
These are easy to integrate: $ \frac12\ln(1\minus{}t^2)\minus{}\frac13\ln(1\minus{}3t^2)\minus{}\frac12\ln(1\plus{}t^2)\plus{}C$
Translating back into the original, a form of the general antiderivative (valid near zero*) is $ \frac12\ln(1\minus{}\tan^2 x)\minus{}\frac13\ln(1\minus{}3\tan^2 x)\minus{}\frac12\ln(1\plus{}\tan^2 x)\plus{}C$.

It's too complicated to be a good exercise in a standard calculus class, but the techniques are at that level.

*I don't like the $ \ln|\cdot|$ formulation.

or, $ \tan x\tan 2x \tan 3x \equal{} \frac {\sin x}{\cos x}\cdot \frac{\sin 2x}{\cos 2x}\cdot \frac {\sin 3x}{\cos 3x}$.

Changing the product of the denominator and numerator into the form of sum might be solved,

:(

$ \tan\,3x\equal{}\frac{\tan\,x\plus{}\tan\,2x}{1\minus{}\tan\,x\;\tan\,2x}$

$ \Rightarrow \tan\,3x \cdot \left( 1\minus{}\tan\,x\;\tan\,2x\right) \equal{} \tan\,x\plus{}\tan\,2x$

$ \Rightarrow \tan\,x\;\tan\,2x\;\tan\,3x \equal{} \tan\,3x\minus{}\tan\,2x\minus{}\tan\,x$

so,

$ \int \tan\,x\;\tan\,2x\;\tan\,3x \;dx\equal{}\int \left(\tan\,3x\minus{}\tan\,2x\minus{}\tan\,x\right) \;dx$

$ \equal{} \minus{}\frac{1}{3} \ln\,\cos \, 3x \plus{} \frac{1}{2} \ln\,\cos \, 2x \plus{} \ln\,\cos \, x \plus{} C$

misan, you should add the absolute value to anti-logarithm.

... or at least acknowledge it, as I did.

Exercise in trig identities: show that misan's form is equivalent to mine. It's even effectively the same three terms.


Find all polynomial $ P(x) = a_nx^n+a_{n-1}x^{n-1}+..+a_1x+a_0$ with $ n \geq 0$ and P(x) satisfy these conditions:
        P(x) has n roots $ x_i \leq -1$ with  i=1..n;
         \[ a_0^2+a_1a_n=a_n^2+a_0a_{n-1}\]

Perhaps because it is false....The equality does not holds in general (look at $P(x)=(x+1)(x+2)$.
I think the LHS is at least the RHS. Please check your statement and correct it.

Pierre.

It is not wrong. u can find the Poland Tst 1990 and check the statement, I think it 's allright.
    And, the question is " to find" , so if the equality does not holds then the answer is "there 're no polynomials" OK ?
    I've tried Viete theorem, but working with the sum/product of n roots is not easy, can u help me?

We can WLOG assume $a_{n}=0$. Denote $-x_{i}$ as $x_{i}$. We have $a_{0}=x_{1}...x_{n}$, $a_{1}=x_{1}..x_{n}\left(\frac 1{x_{1}}+...+\frac{1}{x_{n}}\right)$, $a_{n-1}=x_{1}+...+x_{n}$.

Case $n=1$ is evident.
Case $n=2$ leads to equality $x_{1}^{2}x_{2}^{2}+x_{1}+x_{2}=1+x_{1}x_{2}(x_{1}+x_{2})$ which is equivalent to $(x_{1}x_{2}-1)(x_{1}-1)(x_{2}-1)=0$. Hence the only equality case assuming $x_{i}\geq 1$ is $x_{1}=1$ or $x_{2}=1$. Note that actually we proved $(x_{1}x_{2}-1)(x_{1}-1)(x_{2}-1)\geq 0$.
Case $n\geq 3$ is the following: \[x_{1}^{2}...x_{n}^{2}+x_{1}..x_{n}\left(\frac 1{x_{1}}+...+\frac{1}{x_{n}}\right)=1+x_{1}...x_{n}(x_{1}+...+x_{n}).\] Note that if we change $x_{1}$ to $x_{1}x_{2}$ and $x_{2}$ to $1$ then $LHS-RHS$ non-increses, and it remains the same iff $x_{1}=1$ or $x_{2}=1$.
Suppose not all of $x_{i}$ except one are 1 . Perfoming described changes of variables we will come to situation $x_{2}=...=x_{n}=1$ and at least 1 time $LHS-RHS$ decreased. But $x_{1}^{2}+x_{1}(n-1+1/x_{1})=1+x_{1}(n-1+x_{1})$. It follows that the only possible way to satisfy the initial equality is \[P(x)=a(x-b)(x-1)^{n-1}.\]


I never really understood this theorem. Con someone help me figure this problem out:

The collection of intervals {(2^-n, 1/n]: n=1,2,...} covers the interval (0,1]. Is this a Vitali cover?

I've read what a "Vitali cover" is only five minutes ago so I may well say huge mistake but $1$ is only covered by )1/2,1] which has diameter $1/2$ : this is not a Vitali cover.


Would it be possible to get a mod? Ever since I made the first post in there (other than MCrawford) all of a sudden, tens of CTers started posting most of which, we didn't even know were from CT.

[hide="A list of AoPSers from CT"]

akalra1
algebrainic
brianjg1971
cherrypi
coach sue
codykaz
dbkhoa
edursho
Fedge Matthews
Hcb163
Ignite168
imsocool15
jan4iceskate
Jhonny Ruiz
jrshoch
JSRosen3
JZF
kidnappedturtle
kine
lotrfan
maniacman384
mintedcoin
mtms5467
Naismith
ngade
pillango
rrodgers
seltang
sfrick
snufflingqwert
Timecat8
trig_is_fun88
worthawholebean
Xtrememath88
Yasha B K
yif man12
zabZeno

Ex-CTers:
Phelpedo (In MA)
Yasha B K (In NC)[/hide]

[hide="Posters within the past year"]
akalra1
algebrainic
boo
cherrypi
gkodumudi
hockeyadi23
Ignite168
jan4iceskate
Jhonny Ruiz
kidnappedturtle
led14pa2
lotrfan
maniacman384
mtms5467
nicegina123
ngade
rrodgers
snufflingqwert
Timecat8
worthawholebean
xtrememath8
yif man12

Phelpedo[/hide]

[hide="CTers that post in CT Forum"]
akalra1
algebrainic
Ignite168
jan4iceskate
mtms5467
ngade
nicegina123
rrodgers
Timecat8
yif man12

Phelpedo[/hide]

As for actual mod canindates, I'd be happy with me, mtms5467, or yif man12 if he starts posting again.

I'm glad to see that I have been discrimated against in your list  :D  :wink: .

no, u are better than everyone, so u have ur own section
- :lol: jorian

[quote="redcomet46"][quote="Ignite168"]

Ex-CTers:
Phelpedo (In MA)
Yasha B K (In NC)[/hide]
[/quote]

Isn't his name Yakov?[/quote]

No?


[quote="akscola"]Hi,
consider f(x)=ax+exp(x) .How many stationary poits it has and why?[/quote]

So a stationary, or fixed point, is where f(b)=b
That would give us:

b=ab+exp(b)
Solving for a we get

a=(b-exp(b))(b)

So any fixed point must meet the conditions of this equation.. (we should first note that
b=0 is clearly not a fixed point as f(0) = 1 ).

Perhaps you can do something with the function
f(x)=(x-exp(x))/x
to show something about the # of fixed points.

[quote="akscola"]Hi,
Stationary point is a point where derevative is zero (f'(x)=0).[/quote]

Okay, well thats alot easier. Have you found the derivative?
I have the answer if you want to double check.

If you set the derivative equal to zero,
$a+e^x=0$ gives $x=ln(-a)$
To answer your question, there is 1 such point if a is negative otherwise there are 0 such points.


How can I prove that every function monotone on closed interval $ [a,b]$ is integrable. If $ f$ is monotone on $ [a,b]$ is $ \inf f(x)(x\in[a,b])\equal{}f(a)$ and $ \sup f(x)(x\in[a,b])\equal{}f(b)$?

We're talking about proving Riemann integrability.

Yes, you can say that about the sup and inf. You can also say the same thing about any subinterval $ [x_{i\minus{}1},x_{i}]$ of a partition.

The classic proof, by visually-oriented hint: Use an equal-width partition, use upper and lower sums, and "stack the boxes."


algebra
function
domain
complex analysis
complex analysis unsolved

Suppose f(z) is analytic on domain G, except the for poles in G. prove that the only singularities of g(z)=$ \frac{f'(z)}{f(z)\minus{}A}$ are simple poles at all the poles of f(z) and all the points z in G such that f(z)=A

You've already asked this. Please don't double post.


This is a question I would like an opinion by MattMcGann or someone from MIT, as well as fellow AoPSers.

How much emphasis does MIT put on distinguishing regular and honor classes (ex: Alg 2/ Alg 2 honors) or how fast I move through the high school math curriculum?

I am in the 8th grade now. When I was in 7th grade, I skipped PreAlgebra and took Algebra 1. It turns out that I had a schedule conflict for Alg 1 honors, so I took regular.

Next year (now) I tried to skip Geometry Honors. I gave up on the idea after I was told I was not allowed to get credit for the Honors class and would have to take Alg 2 regular. So I find Geometry Honors very easy.

I wanted to learn Calculus at latest next school year because I am doing Precalc outside of school. However, I am told that I am not allowed to skip anything Alg 2 or harder.

How much will this situation at high school hold me back from getting into MIT?

There's no set path for what you have to take in high school.  The general philosophy is that you should be taking the hardest classes you can reasonably handle and that your school will let you take, but admissions officers take into account your resources or lack thereof.

Don't feel like you have to take the highest-level math courses you possibly can.  If you rush through the curriculum you might miss more than you realize.  And there are lots of other ways to enrich your mathematical experience, such as going to summer programs.

My opinion is that you should do more math rather than skip classes. The math at school does not include some important topics like problem solving and discrete math. For example, see [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]this[/url] and [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_DiscreteMath.php]this[/url].


geometry
incenter
geometric transformation
homothety
circumcircle
analytic geometry
graphing lines

The exam was not difficult this year. How did u guys fare????
I got 3 problems perfectly right and 4th one i don't know if examiners will give me credit for...

The general opinion in my region is that everybody got atleast 2 (number theory was just useless)

I think cut-off will be high this year considering the level of difficulty of problems. The highest in my region is 5 perfectly right and a scratchy 6th (by a 12th std guy)

Post your responses guys..

I did 1,2,3,4,6 in three hours...

And what about the rest one hour... :mad:

I read the 5th question wrong (didn't notice 'equal circles') and went round and round with variables and variables (variable radii r1,r2,r3... :P )

And later after exam I realised that the problem was actually quite easy...

@pardesi (about 5th prob)
You don't need inversion..
the problem is actually quite easy without inversion... (using trigo)
So i am pretty sure i would have got that question in half-an-hour had i not made an ass out of myself...

Although i did good but i wanted to have a perfect paper... :(  :(

[quote="SOUMYASHANT NAYAK"]I did 1,2,3,4,6 in three hours...

And what about the rest one hour... :mad:

I read the 5th question wrong (didn't notice 'equal circles') and went round and round with variables and variables (variable radii r1,r2,r3... :P )

And later after exam I realised that the problem was actually quite easy...

@pardesi (about 5th prob)
You don't need inversion..
the problem is actually quite easy without inversion... (using trigo)
So i am pretty sure i would have got that question in half-an-hour had i not made an - out of myself...

Although i did good but i wanted to have a perfect paper... :(  :([/quote]

But 5th question is hardly 5 lines. the triangle formed by the centres of the 3 circles and the given triangle are homothetic and incentre is centre of homothety

my paper was a ****
did 1,5,6 :) soumya how good is the trigo solution

[quote="hell_ever"]I

But 5th question is hardly 5 lines. the triangle formed by the centres of the 3 circles and the given triangle are homothetic and incentre is centre of homothety[/quote]
can u post ur solution hell_ever

Hmmm...Lot of words have been banned even ones like d.a.m.n.

Anyways, how much marks is each question?

Call the centres of the 3 circles which touch the sides P, Q, R
P near B, Q near A and R near C
Easily prove that $ PQ\parallel{}AB, QR\parallel{}CA, RP\parallel{}BC$
I is the incentre of both PQR and ABC--->homothetic
CAll the centre of circle which touches these 3 externally M.
$ MP\equal{}r(\tau_{A})\plus{}r(\tau)\equal{}MQ\equal{}MR$,
So M is the circumcentre of PQR

i don't have any idea the marks/question but normally 5 and 6 are high scoring

@soumyashant

Will ur paper be evaluated too with the other 12Ths.

[quote="SOUMYASHANT NAYAK"]I read the 5th question wrong (didn't notice 'equal circles')[/quote]
So sorry to hear. :o I too read it wrong at first but later realized the mistake.Finally I got all 6.What is your region? I'm from WB.

how did u do the 5th on Akashnil and which class are u from

[quote="Cyber-Shreyas"]Can someone post the questions for the benefit of the general public.. And can our beloved mod edit the first post and put them there.[/quote]
Just Copy it from the "Recent topics" area.

@Alchemist: Sanjith, how much are the people from our school who wrote getting???

:D I got 3 and expect a lil' bit more than that in terms of partial credit.im in 11.
 here's a proof of the lemma Ai^2 is rational.
Its natural and quite eezy(justsix) steps to prove that A+B/C is rat.(rational)
hence A+B/C + B+C/A is also rat.
on simplifying and using given condition we get AC is rat. :|
hence BC is rat and hence A/C is rat.multiply by AC we get A^2 is rat for all A in the set.From here the soln. is 3 steps away.if 1 of the elememt is rat then all are rat since AC is rat for all A and C.then choose M to be  a perfect square.
suppose A is irrat.then A=root k where k is rat.
since AC is rat then all C in the set r of the form t(root k) where t is rat. then it is eezy to find M such that M*k is a perfect square which completes the proof.  :!:

Sulfur oxide you in chennai...class 11...do i know you? I am Sanjith...P.S.Senior..

i am so dead....sh**, sh**,.... (million times)
I misread the combinatorics problem. Instead of (0,1) i read (1,0) and the entire problem is now wrong....(puts head in toilet)...my chances have dimmed by about 1000 candela..... :(  :(  :(  :(

and now what if the one checking assumes the problem u assumed to be the correct problem...is the chance now anyways atleast 550 Cd increased

i don't think i ought to be too optimistic... all hope now rests on the 2 geometry (both of wich matched with official solution :) ) and no. theory....

beleive me they check worse than NSEP.....i have seen people solving 1 or even half getting thru :)

thanks anyways and best of luck to everyone who wrote.... for the results :wink:

our regional co-ordinator told us to expect it between mid-april and april-end

why so late...they declare it normally within 25 days of the exam...last time the exam was on 5th Feb...we got results by 27th Feb...the previous year it was on Feb. 6th and we got it by 28th ...

Well, correction is not that bad actually. They check papers closely for rigour and that is partly the reason why people with one and one and a half get through. Its rigour, proofs and presentation that matter, not merely getting the solution. And yes, results are normally out in February.
Sathej

Whenevr th results cum out...
jus post th link to th results here as soon as ny1 of u all get it...

please close this topic
mr moderator
as inmo results r already declared


calculus
integration
floor function
calculus computations

using integral calculus alone, show  :D

$ \sec \frac{2\pi}{17} \plus{} \sec \frac{4\pi}{17} \plus{} \sec \frac{6\pi}{17}  \plus{} \ldots \plus{} \sec \frac{14\pi}{17}  \plus{} \sec \frac{16\pi}{17} \equal{} 8$

Using the same logic we can make a generalization of this.

\[ \sum _{i\equal{}1}^n \sec \left(\frac{2 i \pi }{2 n\plus{}1}\right)\equal{}2 (\minus{}1)^n \left\lfloor \frac{n\plus{}1}{2}\right\rfloor\]

Your case is just $ n\equal{}8$


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

[quote="Lagrangia"]Let $A = |a_{ij}| \in M_{n}(C),$ $B = |a_{ij}| \in M_{n,n+1}(C)$ [/quote]

Hi Lagrangia

For the entries of $A$ is it the modulus of $a_{ij}$ ?

$B$ is a matrix of n-rows and n+1-columns, but with the above definition
the entries are the same as $A$. Seems not clear !!

I don't think it's modulus... it would have no sense that $A\in M_n(\mathbb C)$...
and the entries of $B$ are the same with $A$, except that the $n+1$ column has the elements $a_{1\ n+1},a_{2\ n+1},\ldots,a_{n\ n+1}$ with the property that $a_{k\ n+1}=-\sum_{i=1}^na_{ki}$ (because the sum of the elements of $B$ on each row is 0).

[quote="Lagrangia"]Let $A = |a_{ij}| \in M_{n}(C),$ $B = |a_{ij}| \in M_{n,n+1}(C)$ with the property that the sum on each of the lines of $B$ is equal to zero. Prove that: $det(BB^{T}) = (n+1) [det(A)]^{2}.$[/quote]

I was sure I wrote it like:

Let $A = ||a_{ij}|| \in M_{n}(C),$ $B = ||a_{ij}|| \in M_{n,n+1}(C)$ with the property that the sum on each of the lines of $B$ is equal to zero. Prove that: $det(BB^{T}) = (n+1) [det(A)]^{2}.$

but then again  :blush:

this is the problem like in the book.

cheers! :D :D

and...

[quote="Moubinool"] $B$ is a matrix of n-rows and n+1-columns, but with the above definition
the entries are the same as $A$. Seems not clear !![/quote]

is what Kuba suggested.

cheers! :D :D

$B=[A|U]$ where $U\in M_{n\times 1}(\mathbb{C})$.
Then $BB^t=AA^t+UU^t$.

But the sum of the elements of each of the lines of $B$ is $0$, so $U=AE$ where $E\in M_{n\times 1}(\mathbb{C})$, $E=(e_{i,1})_{i=\overline{1,n}}$ and $e_{i,1}=-1 \forall \ i=\overline{1,n}$.

So $BB^t=AA^t+(AE)(AE)^t=AA^t+AEE^tA^t=A(I_n+EE^t)A^t$.
So $\det(BB^t)=\det(I_n+EE^t)\cdot \det A^2$.

By a result you can find in the Theorems and Formulas,
$P_{EE^t}(X)=X^{n-1}\cdot P_{E^tE}$, where $P_D(X)=\det(XI_n-D)$ for some square $n\times n$ matrix D is the characteristic polynomial of D.
$E^tE=n$, so $P_{EE^t}(X)=X^{n-1}(X-n)$. So the eigenvalues of $I_n+EE^t$ are $1$ with multiplicity $n-1$ and $n+1$ with multiplicity $1$, so $\det(I_n+EE^t)=n+1$ and we are done.


[quote="RoFlLoLcOpT"]Whatever answer you get, you can express it in a decimal first. If that doesn't work, then express it in a fraction (since you get two tries). Then report the problem :P[/quote]

That does not take into account the possibility that the answer is to be expressed in percent form, which in this case seems like the form implied for, as percents were used in the writing of the problem.

Hi nick42,

From http://www.artofproblemsolving.com/Alcumus/Instructions.php#answering

[quote="Instructions"]Probabilities, decimal values, and ratios should always be expressed as simplified common fractions unless otherwise specified.[/quote]
Looking at this, the wording probably needs to be tweaked to say percentage values or something like that. However, in the particular problem you presented, there would be no way to represent the value as a percent or decimal as it would be repeating.

According to the directions, to enter two times pi, use: 2\pi.

I needed to enter 1/16 times pi as an answer, so I entered: 1/16\pi which was wrong. I'm guessing it parsed it as 1/(16 * pi). How do you enter (1/16) * pi?
Thanks.

pi/16

Do you have to do \pi or just pi?

[quote="r15s11z55y89w21"]On FTW, the \ can be omitted. On Alcumus, I believe you need to use \pi/16[/quote]

Alcumus accepted \pi/16 as a correct answer. I knew the correct answer mathematically - but I wasn't sure how to enter it (nor did I know how to try different representations of my answer without getting the question wrong). Thanks for your help!


The Asian elephant has an average gestation period of 609 days. How many weeks is this gestation period?

Since there are $ 7$ days in a week, a gestation period of $ 609$ days is $ \frac{609}{7}\equal{}100\minus{}13\equal{}\boxed{87}$ weeks.


Fie a,b numere reale pozitive astfel incat a+b=1.Sa se arate ca :
(a+1/a)^2 + (b+1/b)^2>=25/2

Avem din Cauchy : $(a+\frac1a)^{2}+(b+\frac1b)^{2}\geq\frac{(a+b+\frac1a+\frac1b)^{2}}{2}=\frac{(1+\frac1{ab})^{2}}{2}$ si cum $ab\leq\frac14$[Din Medii] rezulta ca \[(a+\frac1a)^{2}+(b+\frac1b)^{2}\geq\frac{(1+4)^{2}}{2}=\frac{25}{2}\] End of story

mdea frumoasa solutie Andrei!!!Eu nu am facut-o asa.... :blush: Am facut-o putin mai muncitoreste

Se dau numerele reale pozitive$a_{1},a_{2}\dots a_{n}$ cu suma 1.Sa se demonstreze inegalitatea: \[(a_{1}+\frac1{a_{1}})^{2}+(a_{2}+\frac1{a_{2}})^{2}+\dots+(a_{n}+\frac1{a_{n}})^{2}\geq n^{3}+2n+\frac1n\]

[quote="andyciup"]Se dau numerele reale pozitive$a_{1},a_{2}\dots a_{n}$ cu suma 1.Sa se demonstreze inegalitatea: \[\left(a_{1}+\frac1{a_{1}}\right)^{2}+\left(a_{2}+\frac1{a_{2}}\right)^{2}+\dots+\left(a_{n}+\frac1{a_{n}}\right)^{2}\geq n^{3}+2n+\frac1n\] [/quote]
Sa nu uitam de parantezele mari!Parca pare mai frumos asa. :)

Avem $\left( \left( x+\frac1{x}\right)^{2}\right)^{\prime \prime}= 2+\frac6{x^{4}}$, deci $x \mapsto \left( x+\frac1{x}\right)^{2}$ este o functie convexa pe $(0,1)$.

Din definitie/Jensen ies ambele inegalitati.

[hide="Remarca"]
Problema originala este echivalenta cu faptul ca $x \mapsto \left( x+\frac1{x}\right)^{2}$ este o functie semi-convexa pe $(0,1)$. Asta inseamna ca dupa ce am demonstrat problema originala (e interesanta rezolvarea lui andy), generalizarea iese tot cu Jensen (dar fara derivate).
Ori asta, ori Cauchy.
[/hide]


Find all positive integers x,y such that  $ y(y\plus{}1)^{2}\plus{}x(x\plus{}1)^{2}\equal{}8xy$

_______________________________________--
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$ :rotfl:

$ (x\plus{}1)^2 \ge 4x \Longleftrightarrow x(x\plus{}1)^2 \ge 4x^2$
$ y(y\plus{}1)^2 \ge 4y^2$.

So $ 8xy \ge 4x^2\plus{}4y^2 \Longleftrightarrow 4(x\minus{}y)^2 \le 0$.

So $ x\equal{}y$.

So $ 2x(x\plus{}1)^2\equal{}8x^2 \Longleftrightarrow (x\plus{}1)^2\equal{}4x \Longleftrightarrow (x\minus{}1)^2\equal{}0$.

So $ x\equal{}y\equal{}1$.


number theory
relatively prime
greatest common divisor
number theory unsolved

Prove that if p,q are positive integers primes such that r = $ \frac{p^{2}\plus{}q^{2}}{p\plus{}q}$  is integer, then r is prime.

Since $ p\plus{}q$ divides $ (p\plus{}q)^2 \equal{} p^2 \plus{} 2pq \plus{} q^2$, we have that $ r$ is an integer $ \iff p\plus{}q$ divides $ 2pq$.

If $ p\equal{}q$, this clearly happens (because $ 2p$ divides $ 2p^2$), and $ r\equal{}\frac{2p^2}{2p}\equal{}p$ is a prime.

If $ p \neq q$, $ p\plus{}q$ is relatively prime with both $ p$ and $ q$, so we have $ p\plus{}q|2$, which is not possible.

Hence $ p\equal{}q$ and the result follows.

I'm sorry, I'm begginer, and I don't understand very well your logic. Please, tell me more detalhed.

Thanks  :lol:

OK.

We know $ (p\plus{}q)^2 \equal{} p^2 \plus{} 2pq \plus{} q^2$. So, we can say $ p^2 \plus{} q^2 \equal{} (p\plus{}q)^2 \minus{} 2pq$.

Then, $ r \equal{} \frac{(p\plus{}q)^2 \minus{} 2pq}{p\plus{}q} \equal{} \frac{(p\plus{}q)^2}{p\plus{}q} \minus{} \frac{2pq}{p\plus{}q} \equal{} p\plus{}q \minus{} \frac{2pq}{p\plus{}q}$.

If $ r$ is an integer, since $ p\plus{}q$ is an integer, we have that $ \frac{2pq}{p\plus{}q}$ is also an integer. Hence, $ p\plus{}q$ [b]divides [/b]$ 2pq$. Moreover, if $ p\plus{}q$ does NOT divide $ 2pq$, $ r$ can't be an integer. So $ r$ is an integer [b]if and only if[/b] $ p\plus{}q$ divides $ 2pq$.

If $ p\equal{}q$, we have $ p\plus{}q\equal{}2p$ and $ 2pq\equal{}2p^2$. Clearly, $ 2p$ divides $ 2p^2$, since $ 2p^2 \equal{} p(2p)$. In this case, $ r \equal{} \frac{p^2\plus{}p^2}{p\plus{}p} \equal{} \frac{2p^2}{2p} \equal{} p$, which is a prime, as the problem says.

Now we will suppose that there are some [b]distinct [/b]primes $ p$ and $ q$ such that $ r$ is an integer. This will lead to a contradiction, proving that $ r$ is an integer if and only if $ p\equal{}q$. And we have already shown that in this case $ r\equal{}p$ which is a prime, so when we prove this, the solution will be complete.

We said $ r$ is an integer if and only if $ p\plus{}q$ divides $ 2pq$. If $ p \neq q$, $ p\plus{}q$ is not a multiple of $ p$ (because $ q$ can't be a multiple of $ p$, since it is a prime and it's $ \neq p$). The same way, we know $ p\plus{}q$ is not a multiple of $ q$ either. Since $ p$ and $ q$ are primes, this means $ p\plus{}q$ is [b]relatively prime[/b] with $ p$ and $ q$, that means, their greatest common divisor is $ 1$.

Now, $ p\plus{}q$ divides $ 2pq$ implies $ p\plus{}q$ divides $ 2$. But this is not possible, because $ p\plus{}q > 2$. Then, there aren't two distinct primes $ p$ and $ q$ such that $ r$ is an integer, as we wanted to prove.

Is it better now?  :D

Thanks  :lol:
Yes, much better. You're a genius. Thanks for your help (And thanks for patience).
I'm sorry if I bored you.
This question is from Brazilian Olympiad in mathematics. (OBM/2008).


Set $x = a-b, y = b-c, z = c-a$ and assume WLOG $a \le b \le c$.  It is clear we can calculate $a, b, c$ given $x, y, z$.

Next, multiply by 3 to get $3xyz = \frac{10^{6012} - 1}{9} \times 10$, which has, er, a lot of solutions, so do we find them all or count how many there are?

t0rajir0u, we must also have x+y+z=0, so I don't know if there are so many solutions. :)


What do you mean? The solution seems perfectly valid (although yes, perhaps not suitable for this forum... but thats ok) to me.

[quote="sheepwarrior"]...one egg left in the carton? [/quote]

i think he means the "carton" part of the question, sicne the carton is taken :)

[quote="amirhtlusa"][quote="sheepwarrior"]dunno if this goes here are games and fun factory...

there are six eggs in a carton. 6 people come and take one each.
no two people take the same egg and no egg is added.
how is it possible for there to still be one egg left in the carton?

[hide="the answer"]
5 people each take one egg and last person takes the carton  :lol: [/hide][/quote]
your answer is not possible, because your question asks how is it possible that there would be still one egg left [b]in the carton!!! [/b]
your question should be something like this: how is it possible for there  to still be one egg left?[/quote]

huh??

[quote="JBL"]amirhtlusa is having the same initial reaction I did, which is that one person took the carton, *instead of an egg.*  The "real" answer is that one person took the carton, *along with his egg,* thus clearly leaving one egg still in the carton.[/quote]

sry for being unclear i just meant first 5 were taken out so the last person took the las egg by taking the carton which held the last egg  :blush:

[quote="amirhtlusa"][quote="sheepwarrior"]dunno if this goes here are games and fun factory...

there are six eggs in a carton. 6 people come and take one each.
no two people take the same egg and no egg is added.
how is it possible for there to still be one egg left in the carton?

[hide="the answer"]
5 people each take one egg and last person takes the carton  :lol: [/hide][/quote]
your answer is not possible, because your question asks how is it possible that there would be still one egg left [b]in the carton!!! [/b]
your question should be something like this: how is it possible for there  to still be one egg left?[/quote]

no.... although the last guy takes the carton the egg is still left in the carton, right? The question is how can one egg still [b]be left in the carton[/b]?


inequalities search function AMC USA(J)MO USAMO induction

For a,b,c positive real numbers prove that

$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\geq(a+b+c)^3$

We have \[ x^5 - x^2 + 3 \geq x^3 + 2\] for real $x$ so it suffices to prove that \[(a^3 + 2)(b^3 + 2)(c^3 + 2) \geq (a + b + c)^3,\] but the latter inequality is an immediate consequence of the generalized Hölder inequality.

please use the Search function before posting a problem, especially those with known sources, such as olympiads.

The same problem is located here:
http://www.mathlinks.ro/viewtopic.php?t=5383&highlight=usamo+2004

what is the generalized Holder inequality?
sorry for my ignorance

A more flexible variant of Cauchy-Schwarz is the following.

[b]Hölder's inequality[/b]

Let $w_{1}, \dots, w_{n}$ be positive real numbers whose sum is 1. For any positive real numbers $a_{ij}$,
\[
\prod_{i=1}^{n} \left( \sum_{j=1}^{m} a_{ij} \right)^{w_{i}}
\geq \sum_{j=1}^{m} \prod_{i=1}^{n} a_{ij}^{w_{i}}.
\]

[b]Proof[/b]

By induction, it suffices to do the case $n=2$, in which case we'll write $w_{1} = 1/p$ and $w_{2} = 1/q$. Also without loss of generality, we may rescale the $a_{ij}$ so that $\sum_{j=1}^{m} a_{ij} = 1$ for $i=1,2$. In this case, we need to prove
\[
1 \geq \sum_{j=1}^{m} a_{1j}^{1/p} a_{2j}^{1/q}.
\]
On the other hand, by weighted AM-GM,
\[
\frac{a_{1j}}{p} + \frac{a_{2j}}{q} \geq a_{1j}^{1/p} a_{2j}^{1/q}
\]
and the sum of the left side over $j$ is $1/p + 1/q = 1$, so the proof is complete. (The special case of weighted AM-GM we just used is sometimes called Young's inequality.)

Here's another example of an application of Hölder's inequality.

Prove that for $w,x,y,z \geq 0$,
\[ w^{6}x^{3} + x^{6} y^{3} + y^{6}z^{3} + z^{6}x^{3} \geq
w^{2}x^{5}y^{2} + x^{2}y^{5}z^{2} + y^{2}z^{5}w^{2} + z^{2}x^{5}y^{2}.\]
(This can be done either with Hölder or with weighted AM-GM; try it both ways.)

i cant see how to solve it ...
could you please post a solution


please help me with this problem.

p&q are prime numbers. find all(p,q,n)wich

are the answers of :

p^2+P+1=q^n+q^(n-1)+....+q+1

(n>2)


Given rectangle $ MNPQ$ inscribed triangle $ ABC$.$ MP\cap NQ\equal{}H$. Find locus of $ H$

Maybe it's here: http://www.mathlinks.ro/Forum/viewtopic.php?t=152019


Let S = {1, 2, 3, ..., 3000}. Let A be a subset of S with |A| = 2000.
Prove that A has a subset B which satisfies :
(1) |B| = 1000 ;
(2) when the elements of B are put in increasing order, then they alternate in parity, i.e. they are odd, even, odd, even, odd ... or even, odd, even, odd, even ...

Think about how many odd and even elements the A and B sets might have in them in best-case and worst-case scenarios.

Yeah, I know to solve it myself you know ...

I just posted it because it's interesting.

A hint maybe :

[hide]Use the pigeonholeprinciple. (No joke this time)[/hide]

[hide] observe the pairs of consecutive numbers (1,2)(3,4)...(2999,3000). There are 1500 pairs: since there are 2000 numbers in set A, there must be at least 500 of the pairs in the set. Let set B be the 1000 numbers from the consecutive number pairs in set A. [/hide]


floor function
search
algebra unsolved
algebra

Let I be an open interval of length 1/n, where n ∈ N. Find the maximum
possible number of rational numbers of the form a/b, where 1 ≤ b ≤ n,
that lie in I.

I must've made mistake somewhere or the question is incorrect (or trivial as the 3rd possibility).

We can suppose that our interval is $(0, \frac{1}{n})$ as if $\frac{a}{b} \in (m,m+\frac{1}{n})$ then $\frac{a-mb}{b} \in (0,\frac{1}{n})$. But now we see that $\frac{1}{n} \leq \frac{1}{b} \leq \frac{a}{b}$ so there are no rational numbers fulfilling the conditions

[quote="Rafal"]Maybe I don't understand something but it's obvious that there are infinitely many open intrvals of length $\frac{1}{n}$ which have exactly one factor of the form $\frac{a}{b}$ where $1 \leq b \leq n$ so the answer is $n$.[/quote]
Could you prove it Rafal?

[quote="Diarmuid"]Megus, I think you made a mistake in assuming that $I$ could be expressed in the form $\left(m,m+\frac{1}{n}\right)$ for some integer $m$.[/quote]

Thanks for pointing out - I was just silly :blush:

[quote="Diarmuid"]Megus, I think you made a mistake in assuming that $I$ could be expressed in the form $\left(m,m+\frac{1}{n}\right)$ for some integer $m$.

I think the answer to the question is actually $\left\lfloor\frac{n+1}{2}\right\rfloor$, assuming that the given rationals have to be distinct, but I'm not sure...[/quote]
A prove for this?

I search for this problem one month and now I find a solution :lol:
We divide the total of the fractions $\frac{a}{b}$ at $I$ in to totals.
So,one total is $M_1={\frac{k_i}{l_i}}$ with $i=1,2,,,r $ and ${1\leq l_i\leq\frac{n}{2}}$ and $M_2={\frac{x_i}{y_i}}$ and ${\frac{n}{2}<y_i\leq n}$
For each $l_i$ there exist $c_i\in Z$ such that $\frac{n}{2}\leq c_1 l_i\leq n$
If $y_{s+1}=c_i l_i$ and $x_{i+1}=c_i k_i$ then $y_1=y_k$ where $1\leq k\leq r+s$ and $1\leq j\leq r+s$ so $|\frac{x_i}{y_i}-\frac{x_k}{y_k}|\leq\frac{1}{y_j}\leq\frac{1}{n}$ which is a contradiction. So the max of the fractions is $r+s\leq n-[\frac{n}{2}]\leq [\frac{n+1}{2}]$


AMC
USA(J)MO
USAMO
search
Diophantine equation

What is the sum of the first 4 positive integers such that 1+2+3..+n is a perfect square

$ n(n+1)$ must be twice a square number, and since $ n$ and $ n+1$ are consecutive numbers, one must be an odd square while the other must be twice an square number.

So basically, we need to find the smallest cases of $ k$ where $ k^{2}$ and $ 2k^{2}\pm1$ are square. Casework bash shows that the smallest four values of $ n$ are 1, 8, 49, and 288. The sum of those four numbers is 346.

...and there's probably a ten times easier solution than this.

This is a classic example of a pell equation. Search.

That's pretty interesting, I guess.

So this means if I can reduce a USAMO problem to something like that, I can state that it's a case of Pell's equation and stop there?

Huh? No. The rest of the problem is trivial if it reduces to a pell equation, but you still have to finish the problem; Pell equations have a 'general' solution that you can cite.

Also you didn't make it into a pell equation. I was asking you to search for a solution because this problem has been posted a million times.

Yeah, that's what I was trying to say (guess I misphrased it, lol). Sounds like it might come in handy.

Okay here are some links that can help:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=419880129&t=86928[/url]

[url]http://mathworld.wolfram.com/SquareTriangularNumber.html[/url]


hi :)

Prove that if $2^{n}-1$ is prime then $n$ is prime.

[hide]Assume for the sake of contradiction that $n$ is not prime.  Let $d$ be the smallest prime factor of $n$, so that $1<d<n$.  Then $2^{d}-1|(2^{d})^{n/d}-1$, so $2^{n}-1$ is not prime, a contradiction.[/hide]

[quote="Phelpedo"]To clarify, it is a well-known fact that $a|b \rightarrow 2^{a}-1 | 2^{b}-1$ (is this an iff by the way?). Also, $|b \rightarrow 2^{a}+1 |2^{b}-1$ (again, is this an iff?)[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112287[/url]. :)


geometry
circumcircle
trapezoid
trigonometry
geometric transformation
homothety
ratio

Nice Problem!

We know that the Area of one quadrilateral ABCD is $\frac{AC.BD.sen(k)}{2}$, where k is the angle between the diagonals!

So, we have to prove that $\frac{AD}{sen(BMC)}= \frac{BE}{sen(CMD)}= \frac{CF}{sen(AMB)}$.

Let $A_{1}, B_{1}, ... , F_{1}$ the circuncenters of $MAB, MBC, ..., MFA$.
We have that $A_{1}B_{1}\parallel D_{1}E_{1}$, $B_{1}C_{1}\parallel E_{1}F_{1}$ and $F_{1}A_{1}\parallel C_{1}D_{1}$, and $A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}$ is cyclic!
So, $A_{1}B_{1}C_{1}$ is congruent to $D_{1}E_{1}F_{1}$,  $A_{1}C_{1}= D_{1}F_{1}$ and $A_{1}F_{1}= D_{1}C_{1}$.
This implies that $A_{1}C_{1}D_{1}F_{1}$ is one rectangle. Let A' and D' the midpoints of MA and MD. We have $AD = 2A'D'$
So, $\frac{AD}{2sen(BMC)}= \frac{A'D'}{sen(BMC)}= \frac{A_{1}D_{1}}{sen(180-BMC)}= \frac{A_{1}D_{1}}{sen(A_{1}B_{1}C_{1}}$

$= 2R$, where $R$ is the ray of $A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}$ circuncircle!

Do the same for the other fractions, and we will get the same result (2R)

We denote $ B = A_2,C = A_3,D = A_4,E = A_5,F = A_6$ for brevity. Let $ P_i$ denote the midpoint of $ MA_i$ and $ O_i$ the circumcenter of $ \triangle MA_iA_{i + 1}$, where the index is taken modulo $ 6$. $ O_iO_{i - 1}$ is the perpendicular bisector of $ MA_i$. Therefore $ O_iO_{i + 1}\parallel O_{i + 3}O_{i + 4}$. Hence $ O_iO_{i + 1}O_{i + 3}O_{i + 4}$ is an isosceles trapezoid, implying $ O_iO_{i + 3} = k$ (constant).

Now consider the forces $ \overrightarrow{P_1P_4},\overrightarrow{P_5P_2},\overrightarrow{P_3P_6}$ acting at point $ M$. Let $ R$ denote the circumradius of circle $ O_1O_2O_3O_4O_5O_6$. We have
\begin{align*} & \frac {P_{i + 2}P_{i + 5}}{\sin\angle P_iMP_{i + 1}} = \frac {P_{i + 2}P_{i + 5}}{\sin\angle P_iO_iP_{i + 1}} = \frac {P_{i + 2}P_{i + 5}\cdot 2R}{O_{i + 1}O_{i + 5}} = 2R\sin\angle O_{i + 1}O_{i+5}O_{i + 4} = O_{i + 1}O_{i + 4} = k.\end{align*}
Hence the forces are in equilibrium, and so
\begin{align*}\overrightarrow{P_1P_4} + \overrightarrow{P_5P_2} + \overrightarrow{P_3P_6} = 0 \\
\Rightarrow\overrightarrow{P_2P_5}\times (\overrightarrow{P_1P_4} + \overrightarrow{P_5P_2} + \overrightarrow{P_3P_6}) = 0 \\
\Rightarrow\overrightarrow{P_2P_5}\times (\overrightarrow{P_1P_4} + \overrightarrow{P_3P_6}) = 0.\end{align*}
Therefore $ |\overrightarrow{P_2P_5}\times\overrightarrow{P_1P_4}| = |\overrightarrow{P_2P_5}\times\overrightarrow{P_6P_3}|$ implying $ [P_1P_2P_4P_5] = [P_2P_3P_5P_6]$. Similarly we can show that $ [P_1P_2P_4P_5] = [P_3P_4P_6P_1]$. Now the homothety with center $ M$ and ratio $ 2$ maps $ P_i$ to $ A_i$. Hence the conclusion.

[quote="e.lopes"]This implies that $ A_{1}C_{1}D_{1}F_{1}$ is one rectangle...[/quote]

I think this is not true...

it's similar to an OC's problem.
let $O_1,O_2,...,O_6$ be six centers,$O_1O_2$ intersects $O_5O_6$ at $O_7$
by OC's theorem,$\frac{O_7O_2}{FC}=\frac{1}{2sinBMC}=\frac{O_7O_5}{BE}$
since $O_1,...,O_6$ are cyclic,hence
$\frac{O_7O_6}{O_7O_1}=\frac{sinAMB}{sinAMF}=\frac{FC}{BE}=\frac{O_7O_2}{O_7O_5}$
so $BEsinAMB=FCsinAMF$ yielding $(ABDE)=(CDFA)$.

[quote="littletush"]by the way,I'd like to know what kind of contest Zhautykov is?
why is its 2006 identical to Kazakhstan 2006?[/quote]
it's international MO,held in Kazakhstan :)

can someone please post another solution

Solved with Elliott Liu, Isaac Zhu, Jeffrey Chen, Kevin Wu, Luke Robitaille, and Raymond Feng.

Let \(U\), \(V\), \(W\), \(X\), \(Y\), \(Z\) be the circumcenters of \(\triangle MFA\), \(\triangle MAB\), $\ldots$, \(\triangle MEF\). Then \(\overline{UV}\) and \(\overline{XY}\) are both perpendicular to \(\overline{AD}\), so cyclic hexagon \(UVWXYZ\) has opposite sides parallel. Since \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) are the reflections of \(M\) over the sides of \(UVWXYZ\), take a homothety at \(M\) with scale factor \(1/2\) sending \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) to the projections of \(M\) onto sides \(UV\), \(VW\), \(WX\), \(XY\), \(YZ\).

[asy]         size(6cm); defaultpen(fontsize(10pt));         pair U,V,Y,X,Z,WW,M,A,B,C,D,EE,F;         U=dir(120);         V=reflect( (0,0),(0,1))*U;         Y=dir(250);         X=reflect( (U+V)/2,origin)*Y;         Z=dir(190);         WW=reflect( (U+Z)/2,origin)*X;         M=0.1*dir(10);         A=foot(M,U,V);         B=foot(M,V,WW);         C=foot(M,WW,X);         D=foot(M,X,Y);         EE=foot(M,Y,Z);         F=foot(M,Z,U);

draw(unitcircle);         draw(A--D,linewidth(.4)+dashed);         draw(B--EE,linewidth(.4)+dashed);         draw(C--F,linewidth(.4)+dashed);         draw(U--V--WW--X--Y--Z--cycle);

dot("\(U\)",U,U);         dot("\(V\)",V,V);         dot("\(W\)",WW,WW);         dot("\(X\)",X,X);         dot("\(Y\)",Y,Y);         dot("\(Z\)",Z,Z);         dot("\(A\)",A,NW);         dot("\(B\)",B,NE);         dot("\(C\)",C,dir(C-F));         dot("\(D\)",D,S);         dot("\(E\)",EE,W);         dot("\(F\)",F,unit(F-C));         dot("\(M\)",M,dir(120));     [/asy]

Observe that \begin{align*}     [ABDE]&=\tfrac12AD\cdot BE\cdot\sin\angle(\overline{AD},\overline{BE})\\     &=\tfrac12AD\cdot BE\cdot\sin\angle UZW\\     &=\frac{AD\cdot BE\cdot CF}{2\cdot WZ}, \end{align*} and we are done upon noting that \(UX=VY=WZ\) by isosceles trapezoids \(UVXY\), \(VWYZ\), \(WXZU\).

[quote=pohoatza]Let $ABCDEF$ be a convex hexagon and it`s diagonals have one common point $M$. It is known that the circumcenters of triangles $MAB,MBC,MCD,MDE,MEF,MFA$ lie on a circle.
Show that the quadrilaterals $ABDE,BCEF,CDFA$ have equal areas.[/quote]

I know this problem, the author is Nairi Sedrakyan. His geometry problems are usually very hard, but most of them are extremely interesting. He is also the author of 1996 IMO geometry problem 5, none of the strongest IMO teams (China, USA, Russia) was able to solve that problem correctly, moreover Chinese IMO team received 0 points on that problem (the funny thing is: that problem is considered as an easy problem for Nairi Sedrakyan’s geometry standards).


trigonometry
geometry
incenter
function
circumcircle
cyclic quadrilateral
angle bisector

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

A very nice problem!

Construct the point  $T$ in the outside such that $\measuredangle{TAD}=72^{\circ}$ and $\measuredangle{TDA}=\measuredangle{ABC}$.

This means that $T, A, B$ are collinear and $\triangle{TAD}\sim\triangle{CAB}$.

It results that $\frac{TA}{AC}=\frac{DA}{AB}\Longrightarrow \frac{TA}{DA}=\frac{CA}{BA}$. And because $\measuredangle{TAC}=\measuredangle{DAB}$, from the last relation we obtain $\triangle{TAC}\sim\triangle{DAB}.$

Continuing, let the lines $TC$ and $BD$ intersect in point $O$.

Because triangles $TAC$ and $DAB$ are similar, we obtain $\measuredangle{ABO}=\measuredangle{ACO}$, thus the quadrilateral $AOCB$ is cyclic.

This means that $\measuredangle{OAC}=\measuredangle{OBC}=18^{\circ}\Longrightarrow \measuredangle{DAO}=18^{\circ}$.

Also, let $\measuredangle{ADB}=x\Longrightarrow \measuredangle{ABD}=72^{\circ}-x\Longrightarrow\measuredangle{OCA}=72^{\circ}-x$.

Also, in cyclic quadrilateral $OABC$, we have $\measuredangle{OAB}=90^{\circ}\Longrightarrow \measuredangle{OCB}=90^{\circ}$. But $\measuredangle{DCB}=126^{\circ}\Longrightarrow \measuredangle{OCD}=36^{\circ}.$

Now we have calculated all the angles around point $O$ and the triangle $DAC$, thus from the generalised Ceva trig, we obtain
\[\frac{\sin\measuredangle{DAO}}{\sin\measuredangle{OAC}}\cdot\frac{\sin\measuredangle{ACO}}{\sin\measuredangle{OCD}}\cdot\frac{\sin\measuredangle{ODC}}{\sin\measuredangle{ODA}}=1\Longrightarrow \frac{\sin18^{\circ}}{\sin18^{\circ}}\cdot\frac{\sin(72^{\circ}-x)}{\sin36^{\circ}}{\cdot\frac{\sin36^{\circ}}{\sin x}}=1 \]

\[\Longrightarrow \sin (72^{\circ}-x)=\sin x\Longrightarrow x=72^{\circ}-x\Longrightarrow x=36^{\circ}\]
This means that $\measuredangle{ADP}=36^{\circ}\Longrightarrow \measuredangle{APD}=108^{\circ}$!!

very nice solution andy but you've got a little typo at the very end : APD is 108. very nice solution ...

this is a little similar to andyciup solution. let $\angle ADB=\alpha$. then, $\angle ABD=72-\alpha, \angle ACB=18+\alpha, \angle ACD=108-\alpha$. apply trig ceva to the quad ABCD, then,
$\frac{\sin 36 \sin 36 \sin (72-\alpha) \sin (18+\alpha)}{\sin \alpha\sin 72\sin 18\sin (108-\alpha)}=1$.

now, $\frac{\sin 36\sin 36}{\sin 72\sin 18}=2\sin 36$ and $\sin (72-\alpha) \sin (18+\alpha)=\frac{1}{2}\sin (36+2\alpha)$. this implies that

$\frac{\sin 36\sin (36+2\alpha)}{\sin \alpha \sin (72+\alpha)}=1$.

clearing the denominator and applying prod-to-sum formulas it follows that

$\cos 2\alpha-\cos(72+2\alpha)=\cos 72-\cos (72+2\alpha)$ or $\cos 2\alpha=\cos 72$. since $\alpha<72<90$ it follows that $2\alpha=72$ or $\angle APD=180-36-\alpha=108$.

btw, nice problem, and it seems a little similar in its form to bmo 07 p1. is the author of this problem the same as the one from the bmo?

Let $E$ be the point on $AC$, different from $P$, such that $\angle EBC = 18$. Then $\angle DAE = \angle DBE = 36$, then ABED is cyclic and $\angle EDB = \angle EAB = 72$. Now $\angle EDC = \angle BDC = 36$, therefore C is the incenter of BED, and $\angle BEC = \angle DEC = 36$. So, $\angle ABD = \angle AED = 36$ and $\angle APD = \angle PAB+\angle PBA = 72+36 = 108$.  :)

[quote="pohoatza"][color=darkred]Let $ ABCD$ be a convex quadrilateral with $ \angle{DAC}= \angle{BDC}= 36^\circ$ , $ \angle{CBD}= 18^\circ$ and $ \angle{BAC}= 72^\circ$. Determine the measure of $ \angle{APD}$, where $ P\in AC\cap BD$.[/color][/quote]

[quote][color=darkred][b]Lemma (extension).[/b] Let $ ABCD$ be a convex quadrilateral. Prove that $ \{\begin{array}{c}m(\widehat{CBD})=x\ ,\ m(\widehat{BDC})=2x\\\ m(\widehat{CAD})=y\ ,\ m(\widehat{CAB})=2y\\\ P\in AC\cap BD\ ,\ m(\widehat{APD})=z\end{array}$ $ \implies$ $ \boxed{\cos (z+x-y)+\cos (z+y-x)+\cos (x+y-z)=0}\ \ (*)$.[/color][/quote][hide="Proof of the lemma."][color=darkblue]Observe that $ \{\begin{array}{c}m(\widehat{ABD})=z-2y\ ,\ m(\widehat{ACD})=z-2x\\\ m(\widehat{ACB})=180^{\circ}-x-z\ ,\ m(\widehat{ADB})=180^{\circ}-y-z\end{array}$. Therefore,

$ \sin\widehat{ABD}\cdot\sin\widehat{DAC}\sin\widehat{CDB}\cdot\sin\widehat{BCA}=$ $ \sin\widehat{CBD}\cdot\sin\widehat{BAC}\cdot\sin\widehat{ADB}\cdot\sin\widehat{DCA}$ $ \Longleftrightarrow$

$ \sin (z-2y)\cdot\sin y\cdot\sin 2x\cdot\sin (z+x)=\sin x\cdot\sin 2y\cdot\sin (z+y)\cdot\sin (z-2x)$ $ \Longleftrightarrow$

$ \cos x\cdot [\cos (x+2y)-\cos (2z+x-2y)]=\cos y\cdot [\cos (2x+y)-\cos (y+2z-2x)]$ $ \Longleftrightarrow$

$ \cos (2x+2y)+\cos 2y-\cos (2x+2z-2y)-\cos (2z-2y)=$

$ \cos (2x+2y)+\cos 2x-\cos (2y+2z-2x)-\cos (2z-2x)$ $ \Longleftrightarrow$

$ \sin (x+y)\cdot\sin (x-y)+\sin 2z\cdot\sin (2x-2y)+\sin (2z-x-y)\cdot\sin (x-y)=0$ $ \Longleftrightarrow$

$ \sin (x+y)+2\sin 2z\cdot\cos (x-y)+\sin (x+y-2z)=0$ $ \Longleftrightarrow$

$ \sin z\cdot\cos (x+y-z)+\sin 2z\cdot\cos (x-y)=0$ $ \Longleftrightarrow$

$ \cos (x+y-z)+2\cos z\cdot\cos (x-y)=0$ $ \Longleftrightarrow$

$ \cos (x+y-z)+\cos (z+x-y)+\cos (z+y-x)=0$.[/color][/hide]
[color=darkblue][b]Remark.[/b] The relation $ (*)$ is symmetrically in the variables $ x$, $ y$. Thus, I can propose the following problem :[/color]
[quote][color=darkred]Let $ ABCD$ be a convex quadrilateral with $ \angle{DAC}=18^{\circ}$, $ \angle{BDC}= 72^\circ$ and $ \angle{CBD}=\angle{BAC}= 36^\circ$. Determine the measure of $ \angle{APD}$, where $ P\in AC\cap BD$.[/color][/quote]
[color=darkblue][b] Proof of the proposed problem.[/b] For $ \{\begin{array}{c}x=18^{\circ}\\\ y=36^{\circ}\end{array}$ the relation from the above lemma becomes $ \cos (54-z)+\cos (z-18)+\cos (z+18)=0$, i.e. $ f(z)\equiv\sin (36+z)+2\cos z\cdot\cos 18=0$. Observe that $ z>90^{\circ}$ and on the interval $ (\frac{\pi}{2},\pi )$ the function $ f$ is strict decreasing and $ f(108^{\circ})=0$. In conclusion, $ m(\widehat{APD})=108^{\circ}$.[/color]

We prove the circumcenter O of triangle BCD superimposes with A, ffrom where will find angle APD=108. Suppose the contrary.
DOC=DAC=36 so AOCD is cyclic. COB=CAB=72 so AOBC is cyclic as well. From cyclic AOCD we find CDO=CAO=72=CAB so O lies on AB and lies on the circumscribed circle of triangle ABC so O=A or O=B. But O is circumcenter of BCD so O cannot superimpose with B.

[b]Proof[/b] extend $ CA$ to point $ E$ such that $ AE\equal{}AB$ then $ \angle{AEB}\equal{}\angle{ABE}\equal{} 36^{\circ}\equal{}\angle{CDB}$
then $ CBED$ is concyclic. so $ \angle{CED}\equal{}\angle{CBD}\equal{} 18^{\circ}$ and $ \angle{ADE}\equal{}\angle{DAC}\minus{}\angle{CED}\equal{} 36^{\circ}\minus{}18^{\circ}\equal{}18^{\circ}$
which mean $ AD\equal{}AE\equal{}AB$. hence $ A$ is circumcenter of quadrilateral $ BCDE$  then $ AD\equal{}AC$ and $ \angle{ACD}\equal{}72^{\circ}$
$ \angle{APD}\equal{}\angle{ACD}\plus{}\angle{BDC}\equal{}36^{\circ}\plus{}72^{\circ}\equal{}\boxed{108^{\circ}}$

Tummy  :lol:

Same solution of Nayel. One question though, just because the angles are twice as much can we conclude that A is the center of the circle?

@ [b]youarebad[/b], I am not sure your claim is obvious (you need either $AC=AD$, or $AB=AC$ as well), but:
If the interior angle bisector of $\angle BAC$ intersects $BD$ at $E$, then from $\angle EAC=\angle EDC=36^\circ, AECD$ is cyclic and $CE=CD$ (from $\angle CAE=\angle DAC$), so $\angle CED=\angle CAD=36^\circ$, consequently $\angle BCE=\angle CBE=18^\circ$ and $BE=CE$; $E$ being inside the triangle $\triangle ABC$, lying on the interior angle bisector of $\angle BAC$ and on the perpendicular bisector of $BC$, the triangle $ABC$ is isosceles, so $\angle BEC=180^\circ-36^\circ=144^\circ=2\cdot m(\angle BAC)$ implies $E$ is circumcenter of $\triangle ABC$, hence $AE=CE$; from $AECD$ cyclic, we get $BD$ angle bisector of $\angle ADC\implies \angle ADB=36^\circ$; with $\angle CAD=36^\circ$, we get the desired result, $\angle APD=108^\circ$.

Best regards,
sunken rock

[quote="youarebad"]Yeah, simple solution. We know that, the circumcircle of $\triangle BCD$ have center at point $A$. So, $\triangle ABD$ is isosceles triangle. So, $\angle ABD = \frac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}$.

So, $APD = 180^{\circ} - 36^{\circ} - 36^{\circ} = \boxed{108^{\circ}}$[/quote]

Imho the bulk of the problem is proving your assumption.

[hide="Proof that the circumcenter of triangle BCD is A"]
Let the circumcenter of $\triangle BCD$ be $A'$.  Then we know that $\angle BA'C=2\times angle BDC=72^{\circ}$, and since $\angle BAC=72^{\circ}$, $A'$ is on the circumcircle of $\triangle BAC$.  Similarly, we deduce that $A'$ is on the circumcircle of $\triangle DAC$.  Their intersection points are $A$ and $C$; clearly $A'$ isn't $C$ so therefore $A'=A$. [/hide]

The rest is trivial.

Let  $C_{1}$  be the circumcircle of triangle BCD . Let E be the point  where AC intersects with  $C_{1}$ .

The quadrilateral BCDE is cyclic , so we have $<DEC=<DBC=18^{\circ}$ and $<CEB=<BDC=36^{\circ}$

Aso we have $<DAE=<ADE+<DEA $<=> $<ADE=18^{\circ}$ therefore  triangle ADE is isosceles .

Similar we get  triangle ABE isosceles and we conclude that A is te circumcenter of  triangle BDE and also

of triangle BDC (because BCDE is cyclic) . Hence we have that  triangle ADC is isosceles and

$<ADC=<ACD$ <=> $<ADB=36^{\circ}$  and we have that $<APD=108^{\circ}$

[quote="nayel"]note that $\angle BCD=180-18-36=126$. now $\angle BCD+\frac{1}{2}\angle BAD=180$. also since $\angle DAC=2\angle DBC$ and $\angle BAC=2\angle BDC$, we conclude that the circle with center at $A$ of radius $AC$ passes through $B$ and $D$. so $AB=AC=AD$. hence $\angle ABD=\angle ADB=36$ and thus $\angle APD=72+36=108$.[/quote]

NICE ONE!!!!

[b]Solution:[/b]

[b]Construction:[/b] Let $X$ be on $BD$ such that $ADCX$ is cyclic. Let $Y$ be on $BD$ such that $BCYA$ is cyclic.

[b]Claim:[/b] $\angle BCY = 90^{\circ}$.

[b]Proof:[/b]
$\angle XCB = \angle DXC - 18^{\circ} = \angle DAC - 18^{\circ} = 18^{\circ}$.
$\angle XCY = 180^{\circ} - 36^{\circ} - \angle XYC = 144^{\circ} - 72^{\circ} = 72^{\circ}$.
So, $\angle BCY=90^{\circ}$.

[b]Claim:[/b] $YX=XB=AX$

[b]Proof:[/b] $YX=XB$ is trivial by the above angle chase.
As $\angle YAB = 90^{\circ}$, $AX$ is the median. So, $YX=XB=AX$.

So, $\angle APD = 180^{\circ} - 18^{\circ} - (18+36)^{\circ} = 108^{\circ}$. $\blacksquare$

Let the circle $\odot (ABD)$ intersect $AC$ second time at $E$. It follows that $\angle BDE=72^\circ$ and $\angle DBE=36^\circ$, i.e. $C$ is the incenter of $\triangle BED$ and $AC$ the bisector of $\angle BED$, hence $AD=AB$, making the required angle $\angle APD=108^\circ$.

Best regards,
sunken rock

I think this is a different solution :maybe:
[hide=Solution]I am a big fan of cyclic quadrilaterals so why not make some  :-D !
Let the angle bisector of $\angle BAC$ intersect $BD$ at $E$
We have that $\angle EAC = \angle EDC = 36^{\circ} \implies AECD$ is cyclic.
So $$\angle EAC = \angle DEC = \angle EDC = \angle DAC = 36^{\circ} \implies CE=DC$$ We also have that $\angle BCD = 126^{\circ}$ and $$\angle ECD = 180^{\circ} - \angle EAD = 108^{\circ} \implies \angle ECB = 18^{\circ} = \angle EBC \implies CE=BE$$
Now focusing on $\triangle AEB$ and $\triangle AEC$ we get that $\angle BAE = \angle CAE = 36^{\circ}$ and $BE=CE$ and $AE=AE$
So by sine law we know that $\sin(\angle ABE)=sin(\angle ACE) \implies \angle ABE = \angle ACE$ or $\angle ABE = 180^{\circ} - \angle ACE$
But if we follow the second case we have $$\angle ABC+ \angle  ACB+ \angle  BAC = 180^{\circ} = 72^{\circ}+18^{\circ}+18^{\circ}+ \angle  ACE + \angle  ABE = 288^{\circ} > 180^{\circ}$$ which is a clear contradiction
So we must have $\angle  ACE = \angle  ABE = 36^{\circ}$
Now $\angle  APD = \angle  EPC = 180 - \angle CEP - \angle ACE = 108^{\circ}$ and so we are done! :D $\blacksquare$ [/hide]

btw @(someone_i_know_but_u_dont) that IMO 2017 P4 was too good oof, i reflected almost everything across $S$ except $A$ i was so high and also ggb is too bad for angle chase, too hard to write angles on screen, ig have to switch to pen and paper sadge
[hide=storage]
$\angle BAC = 2 \cdot \angle BDC$ and $\angle DAC = 2 \cdot \angle DBC \implies$ the center of $\odot DBC$ lies on both $\odot BAC$ and $\odot CAD \implies$ center of $\odot BDC \in \{ A, C \}$(now $C$ obviously can't be the center since the center must be distinct from the vertices of the triangle) $\implies \overline{AD} = \overline{AC} \implies \angle ADC = \angle ACD = 72^{\circ} \implies \angle ADB = 72^{\circ} - 36^{\circ} \implies \angle APD = 180^{\circ} - 36^{\circ} - 36^{\circ} = 108^{\circ}$ and we are done :D :blush:.
[/hide]

[quote=proshi]Someone please check this as my solution does not incorporate the fact that $\angle CBD=18^\circ.$

[hide=Solution]

Construct the angle bisector of $\angle PAB$ and let it intersect $DB$ at $Q.$ So $\angle PAQ=\angle QAB=36^\circ.$ Since $\angle PAQ=\angle CDB=36^\circ,$ $AQCD$ is cyclic. Thus, $\angle DAP=\angle DQC=36^\circ.$ Because $\angle DPA=\angle CPQ,$ $\triangle DPA\sim \triangle CPQ.$ This implies that $DA||CQ.$

So, $AQCD$ is an isosceles trapezoid which means that $\angle PDA=36^\circ$ which implies $\angle DPA=108^\circ.$ [/hide][/quote]

If I am not mistaken, $\triangle DPA \sim \triangle CPQ$ does not imply $DA \parallel CQ$.

Let $O$ be the circumcenter of triangle $BCD$; clearly $O\not\equiv D$. We compute $\angle BCD = 180^{\circ} - \angle DBC - \angle BDC = 126^{\circ}$ and $\angle BOD = 360^{\circ} - 2\angle BCD = 108^{\circ}$, as well as $\angle DOC = 2\angle DBC = 36^{\circ}$. Hence $\angle DOC = \angle DAC$ and $\angle BAD = \angle BOD$ and if we suppose that $O\not\equiv A$, then it would follow that the circumcircles of triangles $ACD$ and $ABD$ have $O$ as a third common intersection point, contradiction. Therefore $O\equiv A$, hence $AC = AD$, thus $\angle ACD = 90^{\circ} - \frac{\angle DAC}{2} = 72^{\circ}$ and $\angle APD = \angle PCD + \angle PDC = 108^{\circ}$.


I have lurked at AoPS for awhile and finally created an account after meeting Richard at an awards dinner last week.  I have collected some of my favorite puzzles on my web site [url]http://www.agmath.com[/url].

A direct link to the puzzle page:  [url]http://www.agmath.com/7801.html[/url]

Enjoy.

Here is a sample, and one of my favorites... a lot of the puzzles I have collected have likely been discussed on this forum already, but I rarely see this one:

What is the fewest number of strictly acute triangles into which an obtuse triangle can be dissected, or... prove that it is impossible to dissect an obtuse triangle into any number of acute triangles.

A better visual explanation is on my site's puzzle page:  http://www.agmath.com/7801.html

It is puzzle number 9.


$ n$ points on the circumference of the circle are such that no three chords formed by the points are concurrent. Prove that chords partition the interior of the triangle into $ k$ parts.
\[ k\equal{}{n\choose 4}\plus{}{n\choose 2}\plus{}1\]

cf.
[url=http://www.amazon.com/dp/0471801550/]Problems in Combinatorics and Graph Theory[/url] - amazon.com
4.32


Please give me the C++ code for solving equation of degree 3rd and 4th. Thanks


geometry
Stanford
college
ARML
Berkeley Math Circle

Nobody ever posts in the bay area or stanford communities!!!  Could we start a california community?  (either that or get more posters in the bay area!!!)

Just for the record, there is also a "Southern California ARML" forum, which we keep hidden. Anyone who wants to join that should contact me.

Besides the items mentioned above, the SF Bay area has  the SFBA ARML team and the Berkeley Math Circle (including BAMO).

There are four ARML teams in California:

SFBA (contact Ted Alper)

Northern California (contact the UC Davis Math. Dept.)

Southern California (contact me)

San Diego (contact Richard Rusczyk or one of the other AoPS administrators)


Find the smallest positive integer $n$ such that $n\sin1>5\cos1+1$.

[quote="M4RI0"]$1rad$ or $1^\circ$ :?:[/quote]

The original problem doesn't mention that. I am not sure either. :(

[quote="Iversonfan2007"][quote="M4RI0"]$1rad$ or $1^\circ$ :?:[/quote]

he means degree most likely[/quote]

I thought it's $1^\circ$. By that assumption, now how can I go any further??

Well, I would do something like this:

For shortness, put $a=\sin 1^\circ$. Then we have to find the smallest $n$ such that $n>\frac{5\sqrt{1-a^{2}}+1}{a}$.

For small $x$ we can take $\sin x\approx x$, hence $a\approx\frac{\pi}{180}$ ($x$ is understood to be in radians, of course). Moreover, since $a^{2}\ll 1$, we can use the approximation $\sqrt{1-a^{2}}\approx 1-\frac{a^{2}}{2}\approx 1-\frac{\pi^{2}}{2\cdot 180^{2}}$.

Combining all this, we get $\frac{5\sqrt{1-a^{2}}+1}{a}\approx \frac{5\left(1-\frac{\pi^{2}}{2\cdot 180^{2}}\right)+1}{\frac{\pi}{180}}=\frac{1080}{\pi}-\frac{\pi}{72}$. Since $\pi>3.14$, the first term is not exceeding $\frac{1080}{3.14}=343.9...$, and the second term lowers that estimate further, but not for more than $0.05$ since ${\pi\over 72}<{3.6\over 72}=0.05$. Therefore $n=344$. Electronic calculation of the original expression confirms this as correct.


Putnam
algebra
polynomial
college contests

Let $ n$ be a positive integer. Find the number of pairs $ P,Q$ of polynomials with real coefficients such that
\[ (P(X))^2\plus{}(Q(X))^2\equal{}X^{2n}\plus{}1\]
and $ \text{deg}P<\text{deg}{Q}.$

This was my favorite.

There are $ 2^{n \plus{} 1}$ solutions.

We can rewrite the equation $ P(x)^2 \plus{} Q(x)^2 \equal{} x^{2n} \plus{} 1$ as $ (Q(x) \plus{} iP(x))(Q(x) \minus{} iP(x)) \equal{} x^{2n} \plus{} 1$. This makes $ Q \plus{} iP$ and $ Q \minus{} iP$ complex factors of $ x^{2n} \plus{} 1$. By the degree condition, we must have $ Q(x) \equal{} \pm x^n \plus{} (\text{lower degree})$, so $ Q \plus{} iP$ is degree $ n$ with leading coefficient $ \pm 1$.
This makes $ Q(x) \plus{} iP(x) \equal{} \pm(x \minus{} r_1)(x \minus{} r_2)\cdots (x \minus{} r_n)$ where $ r_1,r_2,\dots,r_n$ are some $ n$ of the zeros of $ x^{2n} \plus{} 1$, and $ Q(x) \minus{} iP(x) \equal{} \pm(x \minus{} \overline{r_1})(x \minus{} \overline{r_2})\cdots (x \minus{} \overline{r_n})$. The $ r_i$ and their conjugates must cover the zeros evenly; this means that each conjugate pair of roots $ \exp\left(\pm \frac {k \plus{} \frac12}{n}\pi i\right)$ contains exactly one of the $ r_i$. There are $ 2^n$ ways to split up these $ n$ pairs. Along with the choice of overall sign, this gives us $ 2^{n \plus{} 1}$ possible choices for $ Q \plus{} iP$. Splitting these into real and imaginary parts, we have $ 2^{n \plus{} 1}$ solutions of the original equation.

As I'm uncertain about how the papers are marked, I want to ask:
My solution is basically the same as jmerry's one, but for mysterious reason I took $ Q$ to be monic, that said I got the answer to be $ 2^n$. How many marks will be deducted usually?

Yeah! I'm curious about that as I also took Q to be monic for misterious reasons and got $ 2^n$... :(

My experience with minor mistakes tends to be that they are vicious and want us to cry.

hi, jmerry, just for the record, would you mind correcting the typo in your proof please.  It's very minor, but it can trip inexperienced students such as myself.

the typo is in the third paragraph and fifth line.  the parethetheiss are messed up.

[quote="jmerry"] ... and
 $ Q(x \minus{} iP(x)) \equal{} \pm(x \minus{} \overline{r_1})(x \minus{} \overline{r_2})\cdots (x \minus{} \overline{r_n})$.  ...[/quote]

Since I'm not a moderator, I no longer have the ability to edit that post.

[color=green][But I am. Is that OK? - K.M.][/color]

Can you please explain the part about the ri's covering the zeros evenly and how you find the expression for them?

it just means that $ r_i$ is a root of $ iP\plus{}Q$ implies that its conjugate will be a root of $ \minus{}iP\plus{}Q$

The answer is $2^{n+1}$.

Let the roots of $x^{2n}+1$ be $r_1, \overline{r_1}, \cdots, r_n, \overline{r_n}$. Then observe the way to factor it into irreducibles in $\mathbb{R}[x]$ is $(x-r_j)(x-\overline{r_j})$.

Since $\deg P<\deg Q=n$ it follows that $\deg(P+iQ)=n$.

Claim: $P+iQ$ is divisible by exactly one of $r_i, \overline{r_i}$ for each $1\le i\le n$.

Proof: If it is divisible by both then $x^{2n}+1$ has double root. If it is divisible by neither, $x^{2n}+1$ is not divisible by $r_i$.

Thus, pick a set from $\{1,\cdots,n\} = [n]$ then $P(x)+iQ(x) = \pm \prod\limits_{j\in S} (x-r_j) \prod\limits_{j\in [n]\backslash S} (x-\overline{r_j})$ work and only these work. This leaves $2^{n+1}$ possibilities for $P+iQ$ so $2^{n+1}$ possibilities for a pair $(P,Q)$.

[b]Extension.[/b] What if $P,Q\in \mathbb{Z}[x]$?

About the @canbankan 's extension. What if $P,Q\in \mathbb{Z}[x]$?
We write the decomposition in $\mathbb{Z}[i][x]$ of $x^{2n}+1$.
For example, using Maple, consider the case $n=9$.
In the sequel, we assume that $Q$ is monic and that the leading coefficient of $P$ is $>0$; such a solution $(P,Q)\in (\mathbb{Z}[x])^2$,  called basic solution, gives birth to $4$ distinct couples solutions.
$x^{18}+1=(x+i)(x^2+ix-1)(x^6+ix^3-1)\overline{(x+i)(x^2+ix-1)(x^6+ix^3-1)}=RST\overline{RST}$.
Here the number of factors -over $\mathbb{Z}[i]$- is $2p$ (here $p=3)$; then the number of basic solutions is $2^{p-1}$ (here $4$).
Indeed, to obtain all the values of $Q+iP$, it suffices to consider the products $RST ,R\overline{S}T,RS\overline{T},R\overline{ST}$.
We obtain $(2x^8-2x^4-2x^2+1)i+x^9-2x^7-2x^5+2x,(2x^6-1)i+x^9-2x^3,$
$(-2x^8+2x^6-2x^4+2x^2-1)i+x^9-2x^7+2x^5-2x^3+2x,x^9+i$.
If $n$ is a power of $2$, then $p(n)=1$.
Other values: $p(15)=4,p(3\times 15)=6$, $\max_{n\leq 100}p(n)=6$, $p(7\times 15)=8$,  $\max_{n\leq 200}p(n)=8$, $p(15\times 15)=9$, $p(21\times 15)=12$.
I don't know if the value of $p(n)$ can be written in a reduced way.


ratio
geometry
summer program
MathPath
AoPS Books

what is the ratio of the area of the black region to the area of the whole triangle if the red lines are trisectors.

[quote="shobber"][hide]
$\frac{1}{7}$[/hide][/quote]

why!?

[quote="WindSlicer"]Well there are 7 spaces on the diagram.. only one is shaded. So that's why I think he put $\frac{1}{7}$[/quote]

whoaa...not all the spaces are the same area though

This was on the Mathpath thing.
[hide]Place one unit if pressure on $A$ and 2 on $B$ thus forcing $C$ to have 4 giving a total weight of 6 on $F$. Hence

$FI/IA = 1/6$

$EI/IC = 4/3$

So

$FI/IG/GA = 1/3/3$

$EH/HI/IC=1/3/3$

$[AGD] = [EBH] = [FCI] = \frac 1{21} [ABC]$

Since $[EBC] = \frac 13 [ABC]$ then the area of quadrilateral $HIFB= DGIC = \frac 13 - \frac 2{21} = \frac 5{21} [ABC]$

Thus $[GIH] = \frac {2}{3} - \frac 1{21} - \frac {10}{21} = \frac 17 [ABC]$[/hide]

He's using mass point geometry.  And, I should have realized without even thinking, that they are side trisectors.

Can this be done without using mass point geometry?  I don't think I have ever heard of it, hehe.

Another way to do it is to put it on a cartesian plane then use shoestring to take the small triangle's area over the largest.

What in the world?  mass point geometry?  units of pressure?  shoestrings?!

Geez... maybe I should look in my AoPS books more often.  My geometry is more limited than I had thought.

[quote="SnowStorm"]Hmm, is it $\frac{1}{9}$ by chance?[/quote]
no

[quote="SnowStorm"]Unit of pressure?  Are these angle trisectors or side trisectors?[/quote]

and they are side trisectors

[quote="JBL"]Neither of those is really a "getting started" technique -- anyone who uses them with that facility is really intermediate level.  Anyhow, there are other techniques that can solve it, they just take longer -- keep working at it![/quote]
Do you consider area ratio a getting started technique?

In tarquin's graph, we can see that:
$\frac{\triangle{ABC}}{\triangle{AGB}}=\frac{\triangle{BCG}+\triangle{ACG}+\triangle{ABG}}{\triangle{ABG}}= 2+\frac{1}{2}+1=\frac{7}{2}$

Hence: $\frac{\triangle{ABG}}{\triangle{ABC}}=\frac{2}{7}$
Similarly, we have:
$\frac{\triangle{AIC}}{\triangle{ABC}}=\frac{2}{7}$

$\frac{\triangle{BHC}}{\triangle{ABC}}=\frac{2}{7}$

So $\triangle{IGH}=\triangle{ABC}-\triangle{ABG}-\triangle{BHC}-\triangle{AIC}=\frac{1}{7}$

[quote="tarquin"]This was on the Mathpath thing.
$[AGD] = [EBH] = [FCI] = \frac 1{21} [ABC]$
[/quote]

I get the rest of your solution, but how did you get that part?

$\frac 17 \cdot \frac 13$ and all the lines are in the same ratios.

I don't think anyone should have posted a problem on the MathPath qualifying quiz...then again it's probably too late anyway.

This is a classical problem. Should have appeared centuries ago...

i like your proof but I don't get the first step

I agree.

I saw this similar problem in Cayley once and exact problem on my geometry book twice. :)


calculus
derivative
function
calculus computations

How do you go about finding $ f(5)$ and $ {f}'(5)$ of $ f(t)\equal{}100e^t$?

The first one is trivial, just plug in $ t\equal{}5$: $ f(5)\equal{}100e^{5}$.  The second one is just as trivial, since the derivative of $ 100e^{t}$ is $ 100e^{t}$.  So $ f'(5)\equal{}100e^{5}$.

Ok well I thought it was too easy to be true...


I have a worksheet about polygonal numbers I need to answer.

The question I need help on is: What patterns can you find in pyramidal numbers?

If there was a pyramid with three objects at the bottom, with two on top of these three, with one on top of all of these, then there would be six of the objects. The formula for finding the number of objects in a pyramid with $ n$ objects at the bottom is;

$ \frac{n(n\plus{}1)}{2}$


Let $ A_1A_2...A_{10}$ be a regular decagon. Find the number of obtuse triangles whose vertices are among $ A_1$, $ A_2$, ..., $ A_{10}$.

[hide="Outline"]Divide diagonals (chords of circumcircle) into 4 types, then count. Answer: 60. [/hide]


Evaluate:
$ \sum_{i\equal{}1}^{50000} \begin{cases} \log i & \text{if } 0 \equiv 100000 \pmod{i} \\ 0 & \text{if } 0 \not\equiv 1000000 \pmod{i} \end{cases}$

[hide="Hint"]The messy problem statement is equivalent to
$ (\sum_{d \mid 100000} \log d) \minus{} \log 100000$
Logarithms and prime factorizations should now take over.[/hide]

[hide="Hint 2"] $ \prod_{d|n}d\equal{}n^{\frac{\tau(n)}{2}}$[/hide]

Er, I think I may have phrased the question incorrectly or too confusingly, as that's an incorrect solution.

[hide="Longer solution"]
Since $ 1000000\equal{}2^65^5$, the divisors are the following:

\[ \noindent 2^05^0,2^05^1,2^05^2,...,2^05^6\\
\noindent 2^15^0,2^15^1,2^15^2,...,2^15^6\\
\noindent...\\
\noindent 2^65^0,2^65^1,2^65^2,...,2^65^6\\ \]

The sum of the logarithms of all the divisors equals the logarithm of the product of all the divisors. The product of all the divisors can be found to be and simplified to $ (2^{21})^7(5^{21})^7\equal{}10^{147}$.

$ \log 10^{147}\equal{}\boxed{147}$
[/hide]

Your phrasing was confusing because it contained a typo: you have one instance of 100000 and another of 1000000 in the original problem.  (You also have a typo in your proposed solution.)  AviateurPilot gave most of a (correct) solution using 100000.


If I measure the sand lines around a bunker and come up with a measurement in feet, what is the formula to find the square footage of the area. The bunker refers to a golf coures sand bunker.

do u want a special case like an eclipse etc. or even want to take the dent of the bunker into the calculation ?

if u only have the perimeter u are not able to get the area u can only say $A\le \frac{p^2}{2\pi}$ for an area in the plane, but otherwise i dont know what u mean ...
peeta

Thanks for your effort.  I found the answer.


[quote="dinoman"]Dr. Worm flew 80 km with his jetback. He flew the first 4 minutes at half speed and the second 4 minutes at full speed. What was the full speed of Dr. Worm?[/quote]

[hide]80 = 4(1/2x) + 4x = 6x   x = 13 and 2/3 miles per minute[/hide]

are you sure you did your arithmetic right?
cause 80 divided by 6 isn't 13 and 2/3 in my book
your answer is dreadfully close....good try

also I forgot to say this but put answers in km per hour

[hide="aaaaaaaaaah"]
It's 13 1/3 in my book of doom.
[/hide]

[hide="do not look"]13 and 1/3[/hide]

[color=darkred]You can't put apostrophes in a hide command :P .[/color]


function
limit
floor function
logarithms
real analysis
real analysis unsolved

Let $ f$ and $ g$ be two real-valued functions such that $ f(x) \minus{} \frac {1}{9}f\left(\frac {x}{3}\right) \equal{} g(x) \qquad$ for every $ x \in ( \minus{} \infty,0) \cup (0,\infty).$

Suppose that $ g(x) \to 1$ as $ x \to 0.$ Does it necessarily follow that $ \lim_{x \to 0} f(x)$ exists?

Hope you guys can shed some light on this inquiry of mine.

I thank you in advance for your support.  :)

No: let $ g(x) \equal{} 1$ for all $ x$, let $ f(x) \equal{} \frac {9}{8}$ for $ x \leq 0$, and for $ x > 0$ let $ f(x) \equal{} \frac {9}{8} \plus{} \sum_{i \equal{} \minus{} \infty}^{ \minus{} \lfloor \log_3 x \rfloor} 9^i$.  I believe this satisfies everything you asked, but $ \lim_{x \to 0^ \plus{} } f(x) \equal{} \infty$.

1) [b]Wow![/b] How did you think of those functions, man?  :o

2) Which conditions should we impose on the pair $ \{f, g\}$ in order for $ \lim_{x \to 0} f(x)$ to exist? Any ideas?

To define $ f$ for $ x > 0$, we can define it arbitrarily on the interval $ (1, 3]$, and the given equation extends it uniquely to the rest of $ {\bf R}_{ > 0}$.  So, we really have a ridiculous amount of freedom in constructing examples.  I just chose the simplest example that this idea suggests.

As for the second question, I have no idea :)  Perhaps requiring that $ f$ be bounded near 0 together with $ g$ having a limit is sufficient (at least, it destroys examples like mine), but I'm not very confident of this.  But that's probably where I would start.  (Of course, it may be helpful to define $ f$ based on some interval $ (x, 3x]$ for $ x$ sufficiently tiny that $ g$ is near to its limit value rather than on $ (1, 3]$.)


For $n=4$, this is my inequality published in Gazeta Matematica, 10, 2002.
The inequality is also valid for $n=5$, but it is not true for $n \geq 6$ (see the case $a_1=...=a_6$).

Could you give the solution of the case $n=5$ Vasc?

Only a hint:
Multiply the inequalities
$2(1+x^3)^5 \geq (1+x^5)(1+x^2)^4$   for $x=a_1, ... , x=a_5$,
then use Holder's (Huygens's) inequality.