geometry
3D geometry
pyramid
ratio
probability
factorial
placeholder

[quote="james4l"]Basically, you gain as much time that separates the last post and your post.

[color=red]So e.g.

A: 4:00 (first)
B: 4:23
C: 4:26
B: 4:41
A: 5:50

A has 1:09, B has 23+15=0:38, and C has a mere 3 min.

Additional rules:
1) You win if you satisfy both of the following: a) you have OVER NINE THOUSAND minutes = [b]175[/b] hours = 6.[b]5[/b] days, and b) your accumulated time is at least twice as much [b]and 5 hours more than[/b] the runner-up.
2) No doubleposting ever. (you can however [b]edit[/b] your previous post  but only if you were the last to post)
3) No moderator abusements.
4) [b]YES[/b] editing.
5) Keep a running total scoreboard in hide tags. [/quote]

Have fun![/color]

+1:08
[hide]1. andrewjjiang97: 1:12
2. blucarneal: 1:08[/hide]

[b]You do realize that editing doesn't work, right? Because if you were the last to post, it doesn't show when you edited, so you can't confirm when you edited, basically meaning you can make up any time you want. So in order for it to run on this forum, editing needs to be turned...off. [/b]

[hide="+0:04"]1. blucarneal ~ 2:58
2. PowerOfPi ~ 1:13
3. andrewjjiang97 ~ 1:12
4. 15!!! ~ 0:29
5. eqjj168 ~ 0:12
6. PhireKaLk6781 ~ 0:04[/hide]

[quote="PhireKaLk6781"][b]You do realize that editing doesn't work, right? Because if you were the last to post, it doesn't show when you edited, so you can't confirm when you edited, basically meaning you can make up any time you want. So in order for it to run on this forum, editing needs to be turned...off. [/b][/quote]
You can just delete your last post and post later if you were last to post.
[b][color=red]
[i]Ex.[/i]: Bob posts at 10:00 a.m. and records a time of +1:00.
He comes back at 2:00 p.m. to find no one has posted after him.
He can then delete his last post and post at that time and record a new time of +5:00[/color][/b]

[hide="+0:26"]1. blucarneal ~ 2:58
2. PowerOfPi ~ 1:13
3. andrewjjiang97 ~ 1:12
4. eqjj168 ~ 0:38
5. 15!!! ~ 0:29
6. PhireKaLk6781 ~ 0:04[/hide]

[quote="eqjj168"]
[b][color=red]
[i]Ex.[/i]: Bob posts at 10:00 a.m. and records a time of +1:00.
He comes back at 2:00 p.m. to find no one has posted after him.
He can then delete his last post and post at that time and record a new time of +5:00[/color][/b][/quote]

That has always been allowed. But bluecarneal is saying you can edit too, but that isn't an option.

[hide="+0:26"]1. blucarneal ~ 2:58
2. PowerOfPi ~ 1:13
3. andrewjjiang97 ~ 1:12
4. eqjj168 ~ 0:38
5. PhireKaLk6781 ~ 0:34
6. 15!!! ~ 0:29
[/hide]

:P Until someone stops me...
[hide="+1:49"]
1. everyone else - 220:03:04
2. chenhsi - 144:17:37
[/hide]

Hey, hey... YOU WIN! Game over, mods please lock, or make new thread with final standings and delete this one.  363 Days.

I take credit for longest Necropost type game ever made, longest game with a definite end ever finished, and second place.

Congratulations, although I wish you had waited for it to be a full year.

I shouldn't be bumping this, but wow. This was the second game of reaper... back then called necropost. (Pun?) Wow. You can see the link to the thread where I found the links to these long forgotten games [url=here]https://artofproblemsolving.com/community/c19451h1826745_reaper_1[/url] You can even find the link to the  first game of reaper. Wow.  (Sadly that link has been locked.)

[quote=ObjectZ][quote=babyzombievillager]I shouldn't be bumping this, but wow. This was the second game of reaper... back then called necropost. (Pun?) Wow. You can see the link to the thread where I found the links to these long forgotten games [url=https://artofproblemsolving.com/community/c19451h1826745_reaper_1]here[/url] You can even find the link to the  first game of reaper. Wow.  (Sadly that link has been locked.)[/quote]
FTFY[/quote]

Thank you so much! And @2above, haha. However, why did this get moved to FF? I think Games forum is a better place. FF is too spammy nowadays...

And wow. Nowadays reaps are counted by the minutes. Back then it was counted by the hours... wow.

what the heck, you won by [i]nearly[/i] 10 years.


What is the external normal of an ellipsoid at a certain point (x,y,z)? How do I compute it?

Thanks,
Yuval

Suppose you have a surface given by an equation $ g(x,y,z)\equal{}c$ for some constant $ c.$ (It should be pretty easy to come up with that for an ellipsoid.) Then at any point on this surface at which the gradient of $ g$ is nonzero, the normal vector is given by the normalized gradient: $ \vec{\mathbf{n}}\equal{}\pm\frac{\nabla g}{\|\nabla g\|}.$

That $ \pm$ sign gives you two choices; you can be the judge of which one counts as "external."


topology
linear algebra
matrix
quadratics
conics
hyperbola
vector

system: $\dot {x}=-y$  $\dot {y}=-x$, the system can be decoupled and solved by introducing $u=x+y$ and $v=x-y$ Rewrite the system in terms of $u$ and $v$ and solve for $u(t)$ and $v(t)$ starting from arbitrary initial condition $(u_0,v_0)$ . I rewrote the system as $\dot {x}=(v-u)/2$ and $\dot {y}=-(v+u)/2$ but i dont know how to solve for $u(t)$ and $v(t)$ also im not sure how to get the equations for the stable and unstable manifolds in terms of $u$ and $v$. thanks for the help

I'm assuming that this is just a two dimensional problem. If it's not, the what I'm about to say could look fairly silly.

We can write it as $\frac{d}{dt}\begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}0&-1\\ -1&0\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}$

What you've done with your substitution is to begin diagonalizing that matrix; you didn't complete the diagonalization to render it entirely in $u$ and $v.$

The eigenvalues of that matrix are $1$ and $-1.$ The system is unstable, and the stationary point at the origin is a saddle point.

A quadratic form suggested by the diagonalized form: Let $w=x^2-y^2.$ Then $\dot{w}=2x\dot{x}-2y\dot{y}=-xy+yx=0.$ So $w$ is constant, and the solutions travel along the hyperbolas that are the level curves of $w.$

Another way to look at this is with the matrix exponential. Let $y$ be a vector and $A$ a square matrix. The solution to the linear homogeneous constant coefficient system $\dot{y}=Ay$ is $y(t)=e^{tA}y(0).$

In this case, $A=\begin{bmatrix}0&-1\\ -1&0\end{bmatrix}$ and that has a particularly nice exponential:

$e^{tA}=\begin{bmatrix}\cosh t & -\sinh t \\ -\sinh t&\cosh t\end{bmatrix}.$


Solve: $ \begin{cases} 2y(x^2\minus{}y^2)\equal{}3x\\x(x^2\plus{}y^2)\equal{}10y \end{cases}$


Given 2 sets of 4-sided dice all of which are fair.  The numbers on die A are 6,6,6,1 and on die B are 4,4,4,9.  When two (or more) dice are thrown, their numbers are summed for their result.

1st, one of die A is thrown against one of die B 100 times, and it is found that die A beats die B more often.

2nd, three of die A are thrown against three of die B 100 times, and it is found that the A dice beat the B dice more often also.

3rd, is there a type of "Squeeze Theorem" that guarantees that if two of die A are thrown against two of die B (and in this case for 100 times again), that the A dice will beat the B dice more often too, or not?  If not, then why not?  Or how not?


Umm....if you don't want proof questions, the Olympiad section is really, not the place for you. Try getting started.

[quote="Computer Nuke"]Hi people, I'm new   :D . I live in Florida and we do Mu Alpha Theta, don't know if you ever heard of it before, but I seriously need some geometry tests. I don't mean hard ones and not the proof ones, I just want a regular, moderately difficult geometry test. Well.......Maybe not one but it would be great if anybody could give me one.

Thanks.[/quote]

hi dear computer.

i dont think that u can find any test, here. :wink:

ask admins about it.


Define a sequence of reals by $ x_1 \equal{} 1$ and $ x_{n\plus{}1} \equal{} x_n \plus{}\sqrt[3]{x_n}$ for $ n\geq 1$. Prove that there exist $ a, b\in R$ so that $ \lim_{n\rightarrow\infty}\frac{x_n}{an^b}\equal{}1$

Nobody like this problem? My result is $ b \equal{} \frac32$


Find all ordered pairs $(a,b)$ for real $a,b$

$a+b-10\sqrt{a+b}+25=9$
$a-b-10\sqrt{a-b}+49=25$

vish, check case 1

adunk, nice approach


Αφου η προθεσμια αποστολης καποιας λυσης της ασκησης εχει ληξει, μηπος μπορει καποιος να γραψει τη λυση της τελευταιας ασκησης του μηνα; Γιατι η εκφωνηση της ασκησης μου φαινεται πολυ μυστηρια και δεν μπορω να καταλαβω τι ακριβως ζηταει... Βεβαια εστειλα πριν 1 μηνα περιπου μια λυση αλλα δε νομιζω να ειναι σωστη

Ριξτε μια ματια οποιος θελει: http://www.hms.gr/eme/modules/wfsection/article.php?articleid=642&lang=greek

Ούτε εγώ είμαι σίγουρος. Anyway, here it is:

Είναι $ V_{tot} \equal{} 2,4\cdot2,2\cdot3 \equal{} 15,84 \ \ m^3$ ή $ 15.840\ \ l$
και $ V_{tsim} \equal{} 0,2\cdot0,3\cdot0,06 \equal{} 0,0036 \ \ m^3$ ή $ 3,6\ \ l$

Ο όγκος που απορροφά ο τσιμεντόλιθος είναι το $ \frac {1}{17}$ του όγκου του.
Δηλαδή, ουσιαστικά, με κάθε τσιμεντόλιθο που προσθέτουμε $ \frac {16}{17}\cdot V_{tsim} \equal{} \frac {16}{17}\cdot3,6 \equal{} \frac {57,6}{17} \ \ l$

Ο όγκος νερού είναι $ 11.250\ \ l$, οπότε μένουν $ 15.840 \minus{} 11.250 \equal{} 4.590\ \ l$

Άρα απαιτούνται $ \frac {4.590}{\frac {57,6}{17}} \equal{} 1.354,6875$

Οπότε απαιτούνται $ 1.355$ τσιμεντόλιθοι ώστε το νερό να την καλύψει πλήρως.

Παρ’όλα αυτά, θα χυθεί απ’έξω και κάποια ποσότητα νερού.

[quote="Orfikoss"]Ούτε εγώ είμαι σίγουρος. Anyway, here it is:

Είναι $ V_{tot} \equal{} 2,4\cdot2,2\cdot3 \equal{} 15,84 \ \ m^3$ ή $ 15.840\ \ l$
και $ V_{tsim} \equal{} 0,2\cdot0,3\cdot0,06 \equal{} 0,0036 \ \ m^3$ ή $ 3,6\ \ l$

Ο όγκος που απορροφά ο τσιμεντόλιθος είναι το $ \frac {1}{17}$ του όγκου του.
Δηλαδή, ουσιαστικά, με κάθε τσιμεντόλιθο που προσθέτουμε $ \frac {16}{17}\cdot V_{tsim} \equal{} \frac {16}{17}\cdot3,6 \equal{} \frac {57,6}{17} l$

Ο όγκος νερού είναι $ 11.250 l$, οπότε μένουν $ 15.840 \minus{} 11.250 \equal{} 4.590 l$

Άρα απαιτούνται $ \frac {4.590}{\frac {57,6}{17}} \equal{} 1.354,6875$

Οπότε απαιτούνται $ 1.355$ τσιμεντόλιθοι ώστε το νερό να την καλύψει πλήρως.

Παρ’όλα αυτά, θα χυθεί απ’έξω και κάποια ποσότητα νερού.[/quote]

εγω απο την αλλη την εστειλα γραφοντας οτι δεν γνωριζουμε το σχημα της δεξαμενης (αφου δεν διευκρινιζει οτι ειναι σχηματος ορθ παραλληλεπιπεδου) και οτι προφανως το μεγιστο ογκο θα τον ειχε αν ηταν ορθογωνιου παραλληλεπιπεδου σχηματος η δεξαμενη, αρα η μεγιστη ποσοτητα τσιμεντολιθων που μπορουν να τοποθετηθουν για να πετυχουμε το ζητουμενο ειναι 1.354... Τελος παντων θα το μαθουμε οταν βγουν τα αποτελεσματα


summer program
MathPath
MATHCOUNTS
AMC
AIME
USAMTS
USA(J)MO

One of my jobs as a co-moderator of the Getting Started forum is to post easy problems that welcome beginners. People who are active in problem-solving soon get better, and forget how hard it is to write an easy problem. I will take most of the problems in the Getting Started Series from published sources with full solutions. The problems are meant to be easy for beginners in problem-solving.

I invite all participants on AoPS who are still at the getting started level to try to solve the Getting Started Series challenge problems. Adults (all people age eighteen or older), [b]please[/b] let the beginners try these. I suggest that everyone try to show steps to their solution of the problem, as best they can, because that is an important skill to practice. Please put your answers and solutions in hidden text, that is, highlight the words in your solution and then click on the Hide button in your message composition window, or put codes like this [ hide ] and this [ / hide ] (except without the extra spaces) around the text you want to hide. People can read hidden text by clicking on the link that reveals the hidden text, [b]after[/b] they have tried to solve the problem themselves. Here is an example of hidden text, which you can practice reading:

[hide]This is in hidden text to keep from spoiling the surprise.[/hide]

As I post each Getting Started Series challenge problem, I will link back to this set of instructions, so I ask you all for your kind suggestions on how I can make these instructions more clear and helpful to new problem-solvers. Enjoy, and best wishes in your study of math.

Here in the United States, the television talk show host and humorist David Letterman has used a top-ten list format for years to make jokes about various subjects. With his example in mind, I'd like to share a top-ten list about how you can know you're not a Getting Started level mathlete anymore.

Top Ten Signs You're Not Just Getting Started in Math Problem-Solving Anymore

10. You've been to [url=http://www.mathpath.org]MathPath[/url].

9. You've gone on to nationals in [url=http://www.mathcounts.org]Mathcounts[/url].

8. You've aced the [url=http://www.unl.edu/amc/e-exams/e4-amc08/amc8.html]AMC 8[/url].

7. You've qualified for [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] twice.

6. You've had two commended problems in [url=http://www.usamts.org/]USAMTS[/url].

5. You've qualified for [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]USAMO[/url] (or the equivalent, in your country).

4. You've been to [url=http://www.unl.edu/amc/a-activities/a6-mosp/mosp.html]MOP[/url] (or the equivalent, in your country).

3. You think the problems on [url=http://www.artofproblemsolving.com/Forum/index.php?f=150]AoPS Intermediate Topics forum[/url] are awfully easy.

2. You've been a member of your country's [url=http://olympiads.win.tue.nl/imo/]IMO[/url] team.

And the number-one sign you're not just getting started in math problem-solving anymore is

(in hidden text, to build suspense)

[hide="This really is the top sign."]
1. You're a co-author of the book series [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]The Art of Problem Solving[/url]. [/hide]

Now that we have [url=http://www.artofproblemsolving.com/Forum/index.php?c=1][b]five[/b] levels of problem-solving forums[/url], there should be no reason for putting challenge problems that are too difficult here on this Getting Started forum. After discussing the matter with the site managers, my agenda is make this forum more inviting for real beginners in problem solving. Fortunately, few people who actively participate here stay at the Getting Started level for very long, but new members will continue to register for this site and I'd like the easiest problem-solving forum on the site to be an inviting place for them to develop their skills.

Here are some friendly suggestions:

1) If you would like to post a challenge problem for other participants to solve, please post it with the term "Challenge Problem" in the subject title. It would be especially great if you put an indication of the math topic of the problem in the title, for example, "Challenge Problem: Rate and Distance" or "Challenge Problem: Probability" or whatever. All participants are welcome to post challenge problems here as long as they are at the appropriate (easy) level for this board. Any problem that one might see on an AMC 8 test, or on a school, chapter, or state test in Mathcounts, is not too hard. (I think national Mathcounts test questions properly belong on the Intermediate Topics forum or a higher forum, as do nearly all problems that are in the form of "Prove . . . ")

2) If you are a getting started problem-solver yourself, you are very welcome to post solutions to challenge problems as public replies. Note that the key word here is "solutions"--you should always be able to show some step(s) that got you to your answer. If you are an experienced problem-solver you should leave answering the challenge problems to the beginners. Everyone is welcome to post informational questions (for example, "'What is the formula for . . . ?" "What does the term . . . mean?") and everyone is welcome to answer informational questions that they are able to answer correctly.

3) Please post solutions in hidden text, which prevents giving away the answer to other people who want to try to solve the problem. You can put your solution in hidden text by highlighting the text and clicking the Hide control at the top of every message composition screen. Below is an example of hidden text, which you can read by clicking on the link text:

[hide]Using hidden text to keep solutions invisible is a good example of polite, mature behavior and the mark of a skilled problem-solver.[/hide]

4. [b]Do not[/b], under any circumstances, post "solutions" that consist only of the output of computer programs, especially if that posting is in non-spoiler text. What we are practicing here is the use of the human brain, not pushing buttons on computers. Of course you are welcome to discuss, in detail as part of your reply, how trying numerical examples with a computer program might give you ideas on how to solve a problem, but this forum is for thoughtful people, not for computers, to post on, and I will ruthlessly delete any such reply I discover on this forum (remembering a notorious past example that has not been repeated for a while).

5. [b]Please[/b] consider the level of your challenge problem postings carefully. The Intermediate Topics forum and the Pre-Olympiad forum are both available for fans of more challenging problems. The same netiquette of labeling your problem in a user-friendly way and posting solutions in hidden text is welcome on those forums too, I'm sure.

6. Try your best in spelling, grammar, and choice of problem presentation when you post. I am aware that several people who post great posts here are second-language users of English, and I will be tolerant of language-learner mistakes. I will attempt to correct, with my moderation powers, spelling errors and other errors of English expression in other people's posts, because the learners here can learn best from correctly written mathematical English. Please don't be offended by any editing I do to your posts--I appreciate my occasional typing mistakes being pointed out, so that I can correct them by editing my posts. Good-quality writing comes from good-quality editing.

Please let me know by private message about any improvements you desire in this forum, and good luck to all of you Getting Started forum participants in your future mathematics studies.


Prove that if  $ab\geqq1$ ,then  $a^2+b^2\geqq a+b$

I'll hide this one, because it's pretty straightforward

[hide]Say $a,b\geq 0$

$ab\geq 1\Rightarrow a+b\geq 2$

But then $a^2+b^2\geq \frac {1}{2}(a+b)^2\geq a+b$ as desired.

Say that $a,b<0$, then the RHS stays the same and the LHS decreases.  Thus, we're done.
[/hide]

[quote="Gyan"][quote="blahblahblah"]

$ab\geq 1\Rightarrow a+b\geq 2$

 [/quote]

Not to nitpick but  be just a little careful as no one said a and b are non-negative.  say a=b= -2  and  we do   have a+b LESS than 2 (while ab=4) .

(The original ineq   needs, a and b to be real but  does not require both to be non-negative..)[/quote]
Gyan, please note that blahblahblah has assumed that $a,b\geq0$.

Indeed, by smoothing, if it's true for $a,b\geq 0$, it's true for $a,b<0$.

Since $a^2+b^2+1\geqq ab+a+b$,we are done.


Are there any particularly good books around dealing with olympiad number theory? I searched for a topic but didn't find anything that good.

The kind of difficulty level that I am interested in is around the easy IMO questions, perhaps heading towards q2/5 difficulty as well. I used to be able to do most of the easiest IMO/ISL number theory, but I am getting worse and am looking for a book to reverse this decline.  :)

Basically, what I mean is a number theory equivalent of Geometry Revisited.

I'm looking for something like this, but more problem solving-algebra oriented. I like the big, messy algebra problems where you must cleverly factor or substitute or telescope to get the answer. While a book really wouldn't be appropriate for this, does anyone have a collection of problems that use these techniques?

Also, what a good Geometry Revisted-esque reference for algebra? Such as functional equations and systems of odd equations? I don't think such a think is possible, because every one I have seen uses some fairly problem-specific observations, but still, I would be glad to see a problem collection.

(In my mind, when I say problem collection, I mean something like "102 combinatorial problems". That is a great book that uses many instructive examples, and really expanded my ideas of combinatorics.)

In response to the question: I don't know if there is one big one that many AoPSers use, but I know if you PM chess64 he can get you some ebooks for number theory at the college level.

Well there is a book called "101 Algebra Problems..."

[quote="PenguinIntegral"]Could you offer a reputable bookseller that offers this book?[/quote]
"101 Problems in Algebra"
Also was written by the authors (Titu Andreescu, Zuming Feng) of "102 combinatorial problems".

And for number theory
"104 Number Theory Problems" by Titu Andreescu, Dorin Andrica, Zuming Feng will be released in December this year, we can pre-order it   :wink:

Check both of them here:
http://www.amazon.com
Somehow the Algebra ones are not available now... :maybe:


Putnam
calculus
function
integration
topology
real analysis
conics

For someone who is already familiar with BC Calculus, would seriously studying real analysis help substantially with doing Putnam problems? I understand that the Putnam has "analysis" problems, but students who have not actually taken analysis could still do these problems I suppose. For example, there may be a problem that requires the mean value theorem, which is part of real analysis, but students to whom the theorem was introduced in their single variable calculus classes could still do the problem with having taken an analysis course.

Thanks for any input you may have.

You may occasionally need to know other things, like the fact that continuous functions achieve a max and min on compact sets, and that the image of connected sets under continuous mappings are also connected.  But generally it's not overly technical.

[quote="Fierytycoon"] would seriously studying real analysis help substantially with doing Putnam problems?[/quote]

Yes.  The general habit of estimation and approximation is very useful, as is the exposure to some standard examples and counterexamples with series, integrals, point set topology and so on.    If by "seriously studying" you mean mastering Lebesgue measure theory, functional analysis, Fourier series and the other material that appears later in the textbooks, then this mostly won't appear on Putnam-level competitions, but the basic ideas from some proof or body of theory might show up.

I'll expand a little on what blahblahblah said. What is the difference between calculus and real analysis? Calculus [i]is[/i] analysis, after all. But in a course entitled "Calculus," it may be too easy to fall into the trap of thinking that this is about formulas and closed forms and symbolic manipulations. There's a mindset that comes with taking the analysis course that brushes these superficialities aside: it's all about estimation and inequalities.

For example, this was a problem on a recent Putnam (I think 2001, but someone could correct me on that.)

A parabola intersects a circle of radius 1. Can the arclength of the portion of the parabola inside the circle exceed 4?

You see right away - nothing too theoretical or technical there. This is a [i]calculus[/i] problem. But if you're hung up on closed forms and exact calculations, you're probably not going to get there. You must estimate, and the cleverer your estimate, the quicker and cleaner you'll make this.

One warning- under Putnam time pressure, that parabola problem was brutally hard. I know I didn't get it.


[quote="JasonSullivan"]Find the radius and centre of the circle described by the equation
x2 + y2 − 2x − 4y + 1 = 0
by writing it in the form (x − a)2 + (y − b)2 = c2 for suitable a, b and c.[/quote]

This is a middle school problem.

$ (x^2\minus{}2x\plus{}1)\minus{}1\plus{}(y^2\minus{}4y\plus{}4)\minus{}4\plus{}1\equal{}0\\
(x\minus{}1)^2\plus{}(y\minus{}2)^2\equal{}4.$
Hence $ r\equal{}2$ and $ C\equal{}[1,2]$.

[quote="JasonSullivan"]Right, well it was from an oxford uni (England) worksheet so don't blame me for putting it in the wrong one :)

How do you justify c=2, given what is left is 1?[/quote]

The circle with the centre $ C\equal{}[m,n]$ and with the radius $ r>0$ is given by the equation
$ (x\minus{}m)^2\plus{}(y\minus{}n)^2\equal{}r^2$.

Hence $ r^2\equal{}4$ which implies $ r\equal{}2$ (denoted as $ c$ by you).

yes, I did not phrase that very well.

$ (x\minus{}1)^2 \plus{} (y\minus{}2)^2\equal{} x^2 \plus{} y^2 \minus{}2x \minus{}4y \plus{}1 \plus{}4$

In your solution you took away this 1 and 4 and added 1: all fine. But then we only have 1 left if we compare it to the original question, therefore $ r^2\equal{}\minus{}1$ ?

[quote="JasonSullivan"]yes, I did not phrase that very well.

$ (x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 \equal{} x^2 \plus{} y^2 \minus{} 2x \minus{} 4y \plus{} 1 \plus{} 4$

In your solution you took away this 1 and 4 and added 1: all fine. But then we only have 1 left if we compare it to the original question, therefore $ r^2 \equal{} \minus{} 1$ ?[/quote]

$ \begin{array}{rcl} (x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 & \equal{} & 4\quad\Leftrightarrow \\
(x^2 \minus{} 2x \plus{} 1) \plus{} (y^2 \minus{} 4y \plus{} 4) & \equal{} & 4\quad\Leftrightarrow \\
x^2 \minus{} 2x \plus{} y^2 \minus{} 4y \plus{} 1 \plus{} 4 \minus{} 4 & \equal{} & 0\quad\Leftrightarrow \\
x^2 \minus{} 2x \plus{} y^2 \minus{} 4y \plus{} 1 & \equal{} & 0. \end{array}$

Hence indeed $ r\equal{}2$.

:blush: Clearly. *shakes head at self*

thanks!


geometry
circumcircle
IMO Shortlist
reflection

In an acute triangle $ ABC$ segments $ BE$ and $ CF$ are altitudes. Two circles passing through the point $ A$ and $ F$ and tangent to the line $ BC$ at the points $ P$ and $ Q$ so that $ B$ lies between $ C$ and $ Q$. Prove that lines $ PE$ and $ QF$ intersect on the circumcircle of triangle $ AEF$.

[i]Proposed by Davood Vakili, Iran[/i]

[quote="April"]In an acute triangle $ ABC$ segments $ BE$ and $ CF$ are altitudes. Two circles passing through the point $ A$ anf $ F$ and tangent to the line $ BC$ at the points $ P$ and $ Q$ so that $ B$ lies between $ C$ and $ Q$. Prove that lines $ PE$ and $ QF$ intersect on the circumcircle of triangle $ AEF$.[/quote]
[hide="Solution"]We need to prove that $ \angle QFB \equal{} \angle PEC$
Since the line $ BC$ tangent to the circle $ (AFQ)$ and $ (AFP)$at $ P$ and $ Q$, $ BP^2 \equal{} BQ^2 \equal{} BF\cdot BA \Rightarrow BP \equal{} BQ$
$ \angle BQF \equal{} \angle BAQ \Rightarrow \angle QFB \equal{} \angle AQB$
Denote $ H$ be the orthocenter of triangle $ ABC$, $ AH$ is perpendicular to $ BC$ at $ D$.Then,
$ CE\cdot CA \equal{} CD\cdot CB \equal{} CB^2 \minus{} BD\cdot BC \equal{} CB^2 \minus{} BF\cdot BA \equal{} CB^2 \minus{} BQ^2 \equal{} (CB \plus{} BQ)(CB \minus{} BP) \equal{} CQ\cdot CP$
This impiles that the quadrilateral $ AEPQ$ is concylic, hence $ \angle PEC \equal{} \angle AQB \equal{} \angle QFB$.
Since this equality of angles, we have concurrence of $ PE,QF$ and circle $ (AEF)$[/hide]

[hide]
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I have same soloutions as Mr.danh's.
I have a question everybody : Isnt G4 a little easier than G3?

Thanks to Davood Vakili for a good problem.

[quote]In an acute triangle $ ABC$ segments $ BE$ and $ CF$ are altitudes. Two circles passing through the point $ A$ anf $ F$ and tangent to the line $ BC$ at the points $ P$ and $ Q$ so that $ B$ lies between $ C$ and $ Q$. Prove that lines $ PE$ and $ QF$ intersect on the circumcircle of triangle $ AEF$.

[i]Proposed by Davood Vakili, Iran[/i][/quote]
Let $ H$ be the orthocenter of $ \triangle ABC$ and $ D$ is the projection of $ A$ onto $ BC$. Then consider the inversion through pole $ H$, power $ k \equal{} \overline {HA}.\overline {HD}$ $ \equal{} \overline {HE}.\overline {HB}$ $ \equal{} \overline {HC}.\overline {HF}$. We have $ \mathcal {I}(H,k)$ maps $ P$ into $ P'$ and $ Q$ is taken into $ Q'$, combine with the result that $ \mathcal {I}(H,k)$ maps $ BC$ into $ AEF$. Thus, $ P'$ and $ Q'$ both lie on $ (AEF)$. Therefore, obviously, $ H$ will be also the orthocenter of $ \triangle AQP$, and $ Q',$ $ P'$ respectively are the projections of $ P,$ $ Q$ onto $ AQ$ and $ AP$.

Since $ AF$ is the radical axes of the two circles that pass through $ A,$ $ F$ and tangent to $ BC$ at $ P,$ $ Q$ respectively. Therefore, $ AF$ will goes to midpoint $ B$ of $ PQ$. In the other hand, we also have $ BP^2$ $ \equal{} BQ^2$ $ \equal{} \overline {BF}.\overline {BA}$ $ \equal{} \overline {BD}.\overline {BC}$. Then if $ C'$ is the intersection of $ P'Q'$ with $ PQ$, we have $ (QPDC')$ is harmonic. And by [i]MacLaurin identity[/i], $ BQ^2$ $ \equal{} BP^2$ $ \equal{} \overline {BD}.\overline {BC'}$, which implies $ C$ must coincide with $ C'$. Or we can say $ P',$ $ Q'$ and $ C$ are collinear. From $ (CDPQ)$ is harmonic, then $ \overline {CA}.\overline {CE}$ $ \equal{} \overline {CP}.\overline {CQ}$ $ \equal{} \overline {CD}.\overline {CB}$ $ \Longrightarrow$ $ E$ $ \in$ $ (AQP)$. The same argument holds for $ F$ $ \in$ $ (HPQ)$.

Finally, we have $ \angle (FQ,EP)$ $ \equal{} \angle (FQ,QP) \plus{} \angle (QP,EP)$ $ \equal{} \angle (HF,HP') \plus{} \angle (AQ,AE)$ $ \equal{} \angle (AF,AP') \plus{} \angle (AP',AE)$ $ \equal{} \angle (AF,AE)$. Thus, if $ W$ is the intersection of $ EP$ with $ FQ$, then $ W$ $ \in$ $ (AEF)$. Our proof is completed then. $ \square$

Let $H$ be the orthocenter of  $\triangle ABC$ and let $D$ the foot of the A-altitude. $B$ lying on the radical axis $AF$ of both circles has equal power with respect to them. Thus $BQ^2 \equal{} BP^2 \equal{} BF \cdot BA \equal{} BH \cdot BE \equal{} BD \cdot BC$ $\Longrightarrow$ circle $ \odot(PHE)$ is tangent to $ BC$ at $P,$ yielding $\angle EPQ \equal{} \angle BHP$ and also due to Newton's theorem, the cross ratio $(D,C,P,Q)$ is harmonic $\Longrightarrow$ $CP \cdot CQ \equal{} CD \cdot CB \equal{} CH \cdot CF,$ i.e. $FHPQ$ is cyclic $ \Longrightarrow$ $\angle FQP \equal{} \angle CHP.$ Therefore if $R\equiv QF \cap PE,$ we have $\angle QRP \equal{} \angle BAC$ $\Longrightarrow$ $ R \in \odot(AFE).$

Here is my solution:
First inversion with center A and radius AF. Now it is enough to prove that circumcircles of  $AP_{1}E_1$ and $AQ_{1}F$ intersects at point R which belongs to the line $ E_{1}F$. This is equivalent to
$\angle{AP_{1}E_{1}}=\angle{AQ_{1}F}$. We see $\angle{AP_{1}E_{1}}=\angle{P_{1}AF}+\angle{AFP_{1}}+\angle{P_{1}E_{1}F}+\angle{P_{1}FE_{1}}$
$=\angle{B_{1}P_{1}F}+\angle{B_{1}FP_{1}+\angle{LP_{1}F}+\angle{P_{1}FL}}$

(here we used that $FP$ is tangent to circumcircle of $AQB$ and then it is also tangent to  circumcircle of $ELP_1$ ; $L$ belongs to $E_{1}F_{1}$  and $AL$ is perpendicular to ${E_{1}F_1}$)
Because $\angle{AQ_{1}P_1}=\angle{B_{1}P_{1}F}+\angle{B_{1}FP_{1}}$. So we see that we need to show
that ${P_{1}Q_{1}FL}$ is cyclic. That is true and we'll prove this if we take $S$ as midpoint of $AH$ (H is orthocenter of AEF). It holds that $\angle{SP_{1}F}=\angle{SQ_{1}F}=90=\angle{SLF}$. Done! :)

We have to prove that $ \widehat{BFQ}=\widehat{PEC} $.We have $ BQ^2=BA \cdot BF $,triangles $ BFQ,BQA $ are similar therefore $ \widehat{BFQ}=\widehat{BQA} $.It remains to prove that $ \widehat{PEC}=\widehat{BQA} $. Since $ BP=BQ $ and $ BP^2=BD \cdot BA $,$ D $ is inverse of $ C $ wrt circle $ (B,BP) $ so we have $ (C,D;P,Q)=-1 $.By a known theorem we have $ CP \cdot CQ=CD \cdot CB $,but $ EABD $ is cyclic so $ CE \cdot CA=CD \cdot CB=CP \cdot CQ $.Thus, $ PEAQ $ is cyclic,and $  \widehat{PEC}=\widehat{BQA} $.

Dear Mathlinkers,
in order to have a more upper view for each problem, we can image that each proposed problem are built on a squeletton i.e. based on a well known situation.
In this case, I have though to the Boutin's theorem which originated a chain of well known results and conduct to the result in question.
This is only a personnal... to be developped...
Sincerely
Jean-Louis

My Solution :
Consider $X$ such that $X\equiv(FQ)\cap(PE)$. If $X$ lies in $(AEF)$ then $\angle{XEA}=\angle{XFA}\Leftrightarrow \angle{CEP}=\angle{BFQ}$.
Since $(AFQ)$ is tangent to $BC$ and $(AFP)$ si tangent to $BC$ then we have : $BQ^{2}=BF.BA=BP^{2}$. Otherwise, we have : $\angle{BFQ}=\angle{ABC}-\angle{FQC}=\angle{ABC}-\angle{BAQ}$ and $\angle{CEP}=\pi-\angle{ACB}-\angle{CPE}$. Hence we are left to prove that : $\angle{CAQ}=\angle{CPE}$ which means it suffices to prove that $\triangle{ACQ} \sim \triangle{CPE}$.
From the fact that $\triangle{CFB} \sim \triangle{CDH}$ and $\triangle{BHD} \sim \triangle{BEC}$, we have : $\frac{BH}{BC}=\frac{BD}{BE}$ and $\frac{CH}{CB}=\frac{CD}{CF}$ which give us : $BH.BE+CH.CF=BC(CD+BD)=BC^{2}$. But we have : $BP^{2}=BF.BA=BH.BE$ and $CH.CF=CE.CA$, then we get : $CE.CA=BC^{2}-BP^{2}=(CB-BP)(CB+BQ)=CP.CQ$ and hence : $\frac{CA}{CQ}=\frac{CP}{CE}$ which means that $\triangle{CAQ} \sim \triangle{CPE}$ which give us the desired result.

It's obvious that if we prove that AEPQ is cyclic then we will be done.
Now take inversion in A which maps circle BCEF to itself and our thesis is equivalent to that $Q', P', C$ are collinear, but this is well-known result about polars.

Let us invert the figure wrt $A$ with power $AF.AC=AE.AB$.So $BC$ goes to $\odot AEF$.Now $\odot APF,\odot AQF$ goes to the tangents drawn from $C$ to $\odot AEF$.So the inverses of $P,Q$ (suppose $P',Q'$ ).So we just need to show that the circles $\odot AP'B, \odot AQ'C$ concur on $BC$.For this note that $P'Q'$ is the polar of $C$ wrt $\odot AEF$.So $B,P',Q'$ are collinear. Now the rest is just angle chasing.

Let M be the intersection point of QF and PE. We need to prove that ∠QMP =
∠BAC. Since ∠MQP = ∠QAB (QB is a tangent to the circle around △QFA),
it is enough to prove that ∠QAB+∠BAC = ∠QMP+∠MQP, or equivalently,
∠QAE =∠EPC. Thereforewe need to prove that AQPE is a cyclic quadrilateral.
From $BQ^2 = BF .BA = BP^2$ we get BP = BQ. Adding $BF .BA = BP^2$ to AF · AB = AE ·AC (which holds since BCEF is cyclic) we get $AB^2 = AE .AC+BP^2.$
From the Pythagorean theorem we have $AB^2 = AE^2+BE^2 = AE^2+BC^2-CE^2$,
from which we get $BC^2-CE^2 = AE .EC+BP^2$ This implies that $BC^2-BP^2 =
CE2+AE .EC$, or equivalently
CE . (CE +AE) = (BC+BP)(BC−BP) =CQ·CP.
Thus CE ·CA =CP·CQ and QPEA is cyclic.  :D

I used trigonometry to prove that $QD:DH=AD:DP\Leftrightarrow QH\perp AP$, so $H$ is the orthocenter of $\triangle{APQ}$, thus $PH\perp AQ$. $BP^2=BF\cdot BA=BH\cdot BE$ implies $\angle{BPH}=\angle{BEP}$, and saecula saeculorum says $\angle{QFB}=\angle{AQB}$. Let the intersection be $X$, and get $\angle {XFH}=\frac{\pi}{2}-\angle {AQB}=\angle{HPB}=\angle{BEP}$. Thus $F,H,E,X$ cyclic.

[hide]
[img]http://s16.postimg.org/golltq96d/Untitled.png[/img]

Let $D$ be the foot of perpendicular from $A$ on $BC,$ and let $PE$ and $QF$ intersect at $X$. Note that by Power of a Point,
\[BP = BQ = \sqrt{BF \times AB }.\] Also, since $AD \perp BC$ and $BE \perp AC,$ $ABDE$ is cyclic with diameter $AB.$ By Power of a Point,
\[AC \times CE = BC \times CD.\]
Analogously we obtain the relations
\[AF \times AB = AE \times AC \\
BD \times CD = BF \times AB. \]
Now, note that
\[\begin{array}{lcl}
CP \times CQ & = & (BC+QP)(BC- BP) \\
& = & (BC+\sqrt{BF \times AB})(BC-\sqrt{BF \times AB}) \\
& = & BC^2 - BF \times AB \\
& = & BC^2 - BD \times BC \\
& = & BC(BC-BD) \\
& = & BC \times CD \\ & = & AC \times CE, \end{array}\]
so $AEPQ$ is cyclic by converse of Power of a Point.

Let $QE$ and $PF$ intersect at $X.$ Note that
\[\angle QAE = 180^{\circ} - \angle EPB= \angle QXP + \angle XQP,\]
so
\[\angle QAB + \angle BAC = \angle QXP + \angle XQP.\]
Since $BC$ is tangent to the circumcircle of $\triangle AFQ,$ $\angle XQP = \angle QAB.$ Thus, $\angle BAC = \angle QXP,$ which implies $AEFX$ is cyclic, as desired. $\blacksquare$ [/hide]

My solution with just a bit harmonic quadruples, that gives a few additional properties of the given construction:
Let $H$ be the orthocenter of $ABC$ . Consider the circumcircle $\omega$ of $AEHF$, let  $AP\cap\omega=U$, $AQ\cap\omega=V$.
We will prove $BU$ and $BV$ are tangent to $\omega$ at $U$ and $V$ respectively.
By power of point $B$ we get that $BQ^2=BF.BA=BP^2$, so $BQ=BP$.
$$\angle FBC=\angle FEA=\angle QVF$$ so $VFBQ$ is cyclic, similarly so is $UFBP$.
Now we have that $$\angle FQB=\angle QAF=\angle BVF$$ so $BV$ is tangent to $\omega$. Similarly, $BU$ is also tangent to $\omega$.
From the above cyclic quadrilaterals by power of point $A$ we have $$AV.AQ=AF.AB=AU.AP=AE.AC$$ so $VUPQ$ is cyclic. As $BP=BQ$ and $BU=BV$, $B$ is the circumcenter of $(VUPQ)$, so $\angle QUP = 90 ^{\circ}$. But also $\angle HUP = 90 ^{\circ}$, so $Q, H$ and $U$ are colinear. Similarly, $P, H$ and $V$ are colinear.
Because $BU$ and $BV$ are tangents to $\omega$, $UVHE$ is harmonic, so$$(U,V,H,E)=-1$$. Projecting through $A$ onto $BC$ gives us that $$(Q,P,D,C)=-1$$, so $C, U$ and $V$ are colinear by the complete quadrilateral $QPUV$.
Finally, a quick angle-chase finishes our solution:
Let $FQ\cap PE=W$. Then $\angle AEW=\angle PEC= \angle PUC =\angle VUA = \angle QVB = \angle QFB = \angle AFW$, so $AFEW$ is concyclic. Q.E.D.

Easy if you know humpty configurations.

Rephrase the problem wrt $APQ$. We know $F$ is the humpty point, and $E$ is the queue point; so the problem is equivalent to:
Let $ABC$ be a triangle and $X$ be it's humpty point. $Y$ is a point on $BC$ such that $XY \perp AX$. If $M$ is the midpoint of $BC$, then let $Z$ be a point on $AY$ such that $\angle AZM=90$. Show that $BZ$ and $CX$ concur on $(AZX)$.

Notice that $\angle BCX=\angle XAC$ as $X$ is the humpty. If $T$ is the desired intersection, notice that $Z$ lies on $(ABC)$, as it is the $A-$Queue point (the intersection of $HM$ and $(ABC)$, where $H$ is the ortocenter of $ABC$, and this point lies on $(AHX)$). Now, as $Z \in (ABC)$, then $\angle ZBC=\angle TBC=180-\angle ZAC=180-\angle ZAX-\angle XAC=180-\angle ZAX-\angle XCB=180-\angle ZAX-TCB$, but looking at triangle $TBC$: $=\angle ZTX = \angle TBC+\angle TCB=180-\angle ZAX$ by the angles above, and thus $T \in (AZX)$.

Let $D$ be the foot from $A$ and $H$ the orthocenter. Note that $(F,A)$, $(H,E)$, $(D,C)$ are inverse pairs about $(PQ)$. We have

$$CP\cdot CQ = CB^2 - BP^2 = CB^2 - BD\cdot BC = CD\cdot CB$$

so $PQAE$ is cyclic. To finish we just angle chase:

$$\measuredangle QPE = \measuredangle QAE = \measuredangle QAB + \measuredangle BAC = \measuredangle PQE + \measuredangle BAC$$

which proves the desired concurrency.

Solved with some ideas from [b] fuzimiao2013[/b]

Invert at $A$ to swap $(AEF)$ and $BC$. The problem then becomes:
[quote=G4 inverted]Let $P, Q$ be points on $(AEF)$ such that $PB, QB$ tangent to $(AEF)$. Let $X$ be the second intersection of $(ABQ)$ and $(APC)$. Prove $X \in BC$.[/quote]

Let $\omega$ be the circle centered at $B$ passing through $P, Q$. Then:
$$pow_{\omega}(C) = a^2 - BP^2 = a^2 - (BF*c) = a^2 - (a \cos(B))c = a^2 - \frac{a^2+c^2-b^2}{2} = \frac{b^2 + a^2 - c^2}{2} = ab\cos(C)$$

Meanwhile:
$$pow_{AEF}(C) = CE*CA = (a\cos(C))b$$

These are equal, so $C \in PQ$. As a result:
$$\angle AQB = \pi - \angle APQ = \angle CPA$$

So,
$$\angle CXA + \angle AXB = \angle CPA + (\pi - \angle AQB) = \pi.$$

Thus, $X \in BC$ as desired.

First, we note that $BF\cdot BA = BP^2  = BQ^2$, so $B$ is the midpoint of $PQ$, so $\odot(PQ)$ and $\odot(AEF)$ are orthogonal. Thus, $P$ lies on the polar of $Q$ w.r.t. $\odot(AEF)$. Now, let $PE$ and $PF$ intersects $\odot(AEF)$ again at $E_1$ and $F_1$, and let $Q' = E_1F\cap F_1E$. By Pascal's theorem on $E_1EEF_1FF$, we have $P$, $Q'$, $EE\cap FF$ (which is midpoint of $BC$) are collinear, so $Q'\in BC$. Moreover, by Brokard, $Q'$ also lies on the polar of $P$, so $Q=Q'$, and we are done.

[hide = solution]
Invert about $A$ with radius $\sqrt{AE \cdot AC}$. In the inverted diagram, let us denote the image of a point $X$ with $X^*$. Observe that $P^*$ and $Q^*$ are simply the points where the two tangents from $B$ to $\odot(AEF)$ meet it. And we need to only show that circles $\odot(AP^*C)$ and $\odot(AQ^*B)$ meet on line $BC$. After obvious angle chasing, this is equivalent to $\angle AP^*C = \angle AQ^*B$. Observe that $$\angle AQB^* = 180^{\circ}-\angle AFQ^* = \angle AP^*Q^*$$ and thus we need only show that $C$ lies on $P^*Q^*$. But if we let $H$ be the orthocentre of $ABC$ this is trivial by Brokard on $AFHE$.
[/hide]

UwU This problem was soo gooooddd!!! Loved it!! But why do I not see Inversion anymore these days. %T__T

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Let $H$ denote the orthocenter of $\triangle ABC$. Now let $R$ and $S$ be the second intersections of $\odot(B,BP)$ with $AB$.

Firstly, from PoP we get that $BF\cdot BA=BP^2=BQ^2$ which gives us that $\odot(B,BP)$ is orthogonal to $\odot(AEF)$. Now perform an Inversion $\mathbf{I}(A,\sqrt{AF\cdot AB})$. Note that the condition $BF\cdot BA=BP^2$ gives us that $A$ is the inverse of $F$ w.r.t. $\odot(B,BP)$.  This by using EGMO lemma 9.17 gives us that $AF\cdot AB=AR\cdot AS$. So we get that $\mathbf{I}(A,\sqrt{AF\cdot AB})$ fixes $\odot(PQR)$.

Now note that the Inversion swaps $\odot(AEF)\leftrightarrow BC$. This thus gives that the intersections of $\odot(AEF)\cap\odot(PQR)$ and that of $\odot(PQR)\cap BC$ get swapped under this Inversion. Let the second intersections of $AP$ and $AQ$ with $\odot(PQR)$ be $P^*$ and $Q^*$ respectively. Now as $PQ$ is the diameter of $\odot(PQR)$, we get that $\measuredangle PP^*Q=90^\circ$. Similarly as $AH$ is the diameter of $\odot(AEF)$, we get that $\measuredangle AP'H=90^\circ$. This thus finally gives that $\overline{P^*-H-Q}$ are collinear. Similarly we also get that $\overline{Q^*-H-P}$ are also collinear.

To finally finish, we have,\[-1=(P,Q;B,\infty_{BC})\overset{A}{=}(P^*,Q^*;F,A)\overset{Q}{=}(H,A;QF\cap\odot(AH),Q').\]We also have,\[-1=(P,Q;B,\infty_{BC})\overset{H}{=}(Q',P';E,H)\overset{P}{=}(H,A;PE\cap\odot(AH),Q').\]

This finishes as $PE\cap\odot(AH)\equiv QF\cap\odot(AH)$ and we are done. :stretcher:

-----

[b]Remark:[/b] [quote=Wizard_32]Cute result! I wonder if there is a way to accurately construct the point $P.$[/quote]

yes there indeed is! :yup: you can take the circle that is orthogonal to $\odot(AEF)$ and consider its intersections with $BC$ to get $P$ and $Q$ as I did in my diagram. :)

By well-known properties, it is clear that $F$ is the A-Humpty Point of $\triangle APQ$, from which it follows that $H$ is also the orthocenter of $\triangle APQ$. As such, $(QP; DC) = -1$, so $CE \cdot CA = CD \cdot CB = CP \cdot CP$, so $AEPQ$ is cyclic. Angle chasing using the fact that $\angle EPQ = 180^{\circ} - \angle EAQ$ and $\angle FQB = \angle BAQ$ gives that $\angle FAE = \angle FGE$, as desired.

Let $D$ be the foot of the $A$-altitude and $H$ be the orthocenter in $\triangle ABC$. Because $BQ^2 = BP^2 = BF \cdot BA$, we find that $F$ is the $A$-humpty point of $\triangle APQ$, and since $\overline{HF} \perp \overline{AF}$, we also see that $H$ is the orthocenter in $\triangle APQ$. Therefore, if we let $P_1$ be the foot from $P$ onto $\overline{AQ}$, then $AEHFP_1$ is cyclic. Thus, letting $X$ be the second intersection of $QF$ with $(AEHFP_1)$, we see that
$$QF \cdot QX = QP_1 \cdot QA = QD \cdot QP,$$
as $AP_1DP$ is cyclic, so $FXPD$ is cyclic too. Thus,
$$\measuredangle FXP = \measuredangle FDP = \measuredangle FDC = \measuredangle FAC = \measuredangle FAE = \measuredangle FXE,$$
which implies that $X$ lies on line $EP$, as needed.

Let $D$ be the foot from $A$. Note that by Power of a Point, $$BP^2=BQ^2=BF\cdot BA.$$ Notably, $BP=BQ.$

Claim 1: $(QP;DC)=-1$. We have $$BD\cdot BC=BF\cdot BA=BP^2,$$ since $AFDC$ is cyclic, hence shown since $B$ is the midpoint of $PQ$.\\

Claim 2: $AEPQ$ is cyclic. Note that we have $$CE\cdot CA=CD\cdot CB=CB(CB-BD)=CB^2-CB\cdot BD$$ and $$CP\cdot CQ=CB^2-BP^2.$$ Thus, it suffices to show that $$CB^2-CB\cdot BD=CB^2-BP^2,$$ which is just $$BP^2=CB\cdot BD$$ which is clearly true since $(QP;DC)=-1$.\\

Now, since $AEPQ$ is cyclic, we have $$\angle CET=\angle AQP=\angle QFB=\angle AFT,$$ so $AETF$ is cyclic and we are done. Note that $\angle AQP=\angle QFB$ is due to $(AFQ)$ being tangent to $BC$.

Solved with [b]OronSH[/b] and [b]GrantStar[/b]:

Let $X$ be the intersection of lines $PE$ and $QF$.

Claim: $QFHP$ is cyclic.
[hide = Proof 1]
Let $D$ be the foot from $A$ to $BC.$ Since $CDFA,BDEA$ are cyclic, by Power of a Point we have $BF \cdot BA+CE \cdot CA=BD \cdot BC+CD \cdot BC=BC^2.$ From here, we have $BP^2=BF \cdot BA,$ so $CH \cdot CF=CE \cdot CA=BC^2-BF \cdot BA=BC^2-BP^2=CP \cdot CQ,$ so the claim is proven.
[/hide]
[hide = Proof 2]
Let $M$ be the midpoint of $BC.$ By Linearity of Power of a Point on points $B,C,M$ with respect to $(AFHE)$ and the circle with radius zero at $M,$ we get $BF \cdot BA + CE \cdot CA=BC^2,$ by the Three Tangents lemma. From here, proceed as in the first proof.
[/hide]

Since $BF \cdot BA = BP^2$ and $BF \cdot BA = BH \cdot BE$, we have $BH \cdot BE = BP^2$, hence $(BHE)$ is tangent to $\overline{BC}$.

Now, we have that
\begin{align*}
= ~ & \angle XFH + \angle HEX \\
= ~ & (180^{\circ} - \angle QFH) + (180^{\circ} - \angle HEP) \\
= ~ & (180^{\circ} - \angle QFH) + (180^{\circ} - \angle HPQ) \\
= ~ & 180^{\circ},
\end{align*}
hence $X$ lies on $(EHF)$ as desired.

First we obviously have
\[BP^2=BF \cdot BA = BQ^2\]
and thus, $BP=BQ$. Now, since $B$ is the midpoint of $PQ$ and
\[BP^2=BF \cdot BA = BD \cdot BC\]
we must have $(CD;PQ)=-1$. But this means,
\[CP \cdot CQ = CD \cdot CB = CE \cdot CA\]
and thus, $AEPQ$ must be cyclic. Now, let $R = \overline{QF} \cap \overline{PE}$. This gives us that,
\begin{align*}
    \measuredangle AFR &= \measuredangle AQF + \measuredangle FAQ \\
    &= \measuredangle AQF + \measuredangle FQP\\
    &= \measuredangle AQP\\
    &= \measuredangle AEP \\
    &= \measuredangle AER
\end{align*}
and thus, $R$ must lie on the circumcircle of $\triangle AEF$ as required.

Checking all 90+ solutions above, I think this method is new (albeit rather assymmetrical). The motivation is trying to draw $P$ and $Q$ with just ruler and compass.

Note that $F$ is the $A$-humpty point of $APQ$, so to construct $P$ and $Q$ we can proceed as follows;

Let $G$ is reflection of $F$ across $B$, and $J$ is the reflection of $F$ across $BC$, then $P$ and $Q$ are merely the intersections of $(AGJ)$ with $BC$.

Claim: $AEPQGJ$ cyclic.

Proof: It suffice to prove $AEGJ$ cyclic. The main idea is that $DF$ and $DE$ are reflections across $BC$, so $J, D, E$ are colinear. Hence $\angle EJF=\angle FDA=90-\angle A$, so $\angle EJG=180-\angle A=\angle GAE$, and this proves the claim.

By the claim, we get $\angle FQP+\angle EPQ=\angle QAF+180-\angle QAE=180-\angle A$, so $QF$ meet $PE$ at $(AEF)$.
[img]https://cdn.discordapp.com/attachments/895734686230401046/1186650730640191519/image.png?ex=6594059d&is=6581909d&hm=66df0a6a6c5e8676e2c20f136fe395d29aefa9b377787177b71b2e1ea607219e&[/img]

Observe HFQP cyclic
Then by power of point observe AEPQ cyclic
Then angle chase
We are done

From $\angle QAF = \angle QFB$ and $\angle PAF = \angle PFB$ we get $F$ is Humpty point of $\triangle APQ$

As $FH \perp AB$ and $AH \perp PQ$ we get $H$ is orthocenter of $\triangle APQ \implies Q,F,H,P$ is cyclic.

by power of point
$$CP.CQ=CH.CF=CE.CA$$

we get $A,E,P,Q$ cyclic.
$$\angle FQP + \angle EPQ = \angle QAF + 180 - \angle QAE = \angle BAC$$ $\blacksquare$

Edit :[hide] I forget to mention some IMP point like $B$ is midpoint of $PQ$ which is obvious from power of point.[/hide]


email
AwesomeMath
summer program
geometry

Did you get in?

I just got my acceptance email!

I got 8 answered by the way.

[quote="Aegor"]I got mine as well.  However, I did not receive a score on my test.  Was that in your email?[/quote]
No.
I'm just saying, so we can see what a "floor" would be like.
But I think they also base it on grade :|

Oh.  My score was either similar or the same, so yeah...  I was really worried for a moment about AMC scores... Dr. T probably has easy access to them, and my 2007A score was less than formidable. :P  Bad day.

[quote="13375P34K43V312"]I got in, I only answered 6 and somehow looked over a few solutions on the diophantine. So with really good reccomendation letters, I'm assuming 5 will be the floor.[/quote]
Well you're in 8th grade.
Someone in 10th or 11th will probably need a score of like 8/9 minimum.

I made it also; the people at AwesomeMath get these decisions done pretty quickly!  (I think they base acceptance on more than just the admissions test).

[quote="bpms"]
Well you're in 8th grade.
Someone in 10th or 11th will probably need a score of like 8/9 minimum.[/quote]

I'm in 10th grade, and I didn't really try 10 and I also missed a few solutions for 6, so I guess that that's enough.  Depending on which ones you do, I could see someone getting in with six or seven correct who's my age...  It just depends on which ones you did, why you got the others wrong, and the rest of your application.  They want to see HOW you solve problems.  Solving them obviously helps, but it's not everything.

I made it as well.

I answered seven (skipped the two geometry questions and the diophantine).  I missed two, but both were really stupid mistakes.

[quote="Aegor"][quote="mathgeniuse^ln(x)"]Actually I made it.

I thought I didn't make it cause I didn't get my result as quickly as you all did.[/quote]

Good for you! I was also worried when I read the first post; I thought that they would send out acceptance letters all out once, then rejection letters later.   :P[/quote]
Well my email came to me yesterday at 11:53 PM

i got in, i answered 9 of them (skipped 10)

i got in.

my email came really late so i had assumed i didnt...

I made it!!!

I think I got most of the problems right, although I made a very careless error on #5, causing me to get 18 as my answer instead of 19...

I received the brochure late, so wasn't aware of the program until a few days into the test period.  I still managed to answer five of the problems, but haven't sent in the rest of my application yet.  Of course, this also means that I have a chance to improve my score on the next test (and do it in ten days instead of about two) :wink: .

One of my solutions was displayed on the web site as an official solution, so that's probably a good sign.

[quote="kyyuanmathcount"]The people at AwesomeMath get these decisions done pretty quickly! (I think they base acceptance on more than just the admissions test).[/quote]
They get the decisions done quickly because they are small enough that they only have to sort through a few hundred applications and don't need to be extremely rigorous about scoring (It's not just about how many problems you answered correctly).

[quote]
One of my solutions was displayed on the web site as an official solution, so that's probably a good sign.
[/quote]

That's really cool!  My solutions were too long, almost every single one of them took up the whole page and I wrote very tiny.  I wasn't sure how much detail they wanted, so I just wrote everything down.  So it took me about four hours total to write down all of the solutions, even though my solutions were the same as those on the website.  :P

yay AMP was great last year, have fun guys (my parents aren't gonna spend $\$$3000 again for a math camp, lol)

Note: Gabriel is a bit incomprehensible, but very smart nevertheless

Gabriel is not going back this year...I think those evaluation papers we filled out at the end of camp had something to do with that.

However, a good majority of the teaching staff and the mentors are returning (which is a very good thing) so all the people planning on going this year should have one hell of a time.

[quote="b-flat"]I got in, skipping the geometry problems (my huge weakness).[/quote]
It's my main weakness as well, and probably not a coincidence: schools in the U.S. tend to practically ignore geometry (my school district has a single, one-year geometry course, which I took in $8^{th}$ grade).

[quote="Max300"]Gabriel is not going back this year...I think those evaluation papers we filled out at the end of camp had something to do with that.

However, a good majority of the teaching staff and the mentors are returning (which is a very good thing) so all the people planning on going this year should have one - of a time.[/quote]

WHAT!?

lucky them

but they have no one to get mad at for doing bad

BTW THIS IS MY 400TH POST w00t!!!!!!!

if you got in last year, do you have to take the test again?

not_trig... too bad that ur not going back  :(

good times

lol i remember gabriel

I got in on test A with a 9/10 (didn't do number 10)


For which $ n\in \{3900,3901,...,3909\}$ can the set $ \{1,2,...,n\}$ be partitioned into disjoint triples in such a way that in each triple one of the number equals the sum of the other two?

$ n$ must be a multiple of 3, and $ 1\plus{}2\plus{}...\plus{}n$ must be even, so there is only one $ n$, which is $ 3900$?!

Our work is to draw such an partition.

[quote="greentreeroad"]I think 3903 is also a number with this property, right?[/quote]

it is  :D

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=462158&p2591974#p2591974, this is something more general, i think i have a solution, but only for $n\equiv0,3mod(48)$.


search
superior algebra
superior algebra unsolved

Prove that, if for every element $ a$ of a ring there exists a natural number $ n>\equal{}2$ such that $ a^{n}\equal{}a$, then the ring is commutative.

Some hints appear [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=912581235&t=77695]here[/url] (in the last post, which talks about Herstein's [i]Noncommutative Rings[/i]).


[quote="hanschinaman"]x,y,z>=0 satisfying x+y+z=1

find the maxinum of $\sum{\sqrt{\frac{1-x}{1+x}}}$[/quote]
The maximum is more than $\frac{3}{\sqrt2}.$ :D

The maximum is 1+2/3*3^(1/2).
See:
 http://www.irgoc.org/bbs/dispbbs.asp?boardid=4&replyid=90&id=36&page=1&skin=0&Star=3


An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?

$ \textbf{(A)}\ 0.5 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 1.5 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 2.5$

The same exact problem and solution has been posted already:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132475

Due to the base of the aquariam stays the same, we shall depend on that to find the solution.
(100 x 40) x 40 = the volume of the water =  160000
after the brick is added the amout of volume is increase- 40 x 20 x 10 = 8000
now add the both of the numbers - 8000 + 160000 = 168000
Now use the base - 168000/(40 x 100)= 42
Subract 42 - 40= 2
Answer is  [b](D.) 2[/b][/b]


[quote="arqady"][quote="pvthuan"]Prove that if $a,b,c>0$, then
\[ \frac{ab+c^2}{a^2+b^2}+\frac{ca+b^2}{c^2+a^2}+\frac{bc+a^2}{b^2+c^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}\leq4.  \][/quote]
This is not true: $a\to\infty.$ ;)[/quote]
Yes, it is not true. It should be

\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq4.  \]

[quote="pvthuan"] It should be
\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq4.  \][/quote]
I think this is true! :lol: But my proof is very ugly. :ninja:

[quote="pvthuan"]

\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq4.  \][/quote]

I have a nice solution I think. Using Cauchy in Engel form it suffices to prove that

$\frac{(a+b+c)^2}{ab+bc+ca}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq 4$

which is equivalent to show that
$(a^2+b^2+c^2)^2+(ab+bc+ca)^2\geq 2(a^2+b^2+c^2)(ab+bc+ca)$

which is true by AM-GM

Silouan, that is a simple and good solution. Congratulations!

Thank you very much .I think that all inequalities are beautiful because at the 99%
of the cases we can solve them in a beautiful and easy way :D

[quote="arqady"]I think $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+2\sqrt2\cdot\frac{ab+ac+bc}{a^2+b^2+c^2}\geq3+2\sqrt2$ where $a>0,b>0,c>0,$
but I am not sure. ;)[/quote]

Is that the best constant, argady?

[quote="pvthuan"]
Is that the best constant, argady?[/quote]
Yes! ;)
Id est, if $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k \cdot\frac{ab+ac+bc}{a^2+b^2+c^2}\geq3+k$ is true for $a>0,b>0,c>0$ then $k\leq2\sqrt2.$

[quote="arqady"][quote="pvthuan"]
Is that the best constant, argady?[/quote]
Yes! ;)
Id est, if $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} k \cdot\frac {ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2}\geq3 \plus{} k$ is true for $ a > 0,b > 0,c > 0$ then $ k\leq2\sqrt2.$[/quote]
$ k_{best}\equal{}3\sqrt[3]{4}\minus{}2.$
[quote="pvthuan"] It should be
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\geq4.\]
[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \left(\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\right)^2\geq4.\]


Prove that for any natural $ n$ the number
$ \dfrac{11^n\minus{}1}{7^n\minus{}1}$ is not natural.


hallo,
Find all Funktions $f: R\to R$ such that
$f(x-y)-xf(x) \leq  1-x$
for all $x, y \in R$.

Let $x=0 \Rightarrow f(-y)\leq 1 \Rightarrow f(z)\leq1$ for each $z$
Now let $y=0 \Rightarrow f(x)-xf(x)\leq 1-x \Rightarrow (f(x)-1)(1-x)\leq 0 \Rightarrow$ if $x>1$ then $f(x)\geq 1 \Rightarrow f(x)=1$ for each $x>1$.
Now let $y=x-2$ then $1-xf(x)\leq 1-x \Rightarrow 0\leq x(f(x)-1) \Rightarrow$ if $x>0$ then $f(x)=1$ and if $x\leq 0$ then $f(x)\leq 1$
Lets prove that all that functions satisfy the condition:
$f(x-y)\Leftrightarrow f(z)$
$f(z)-xf(x)\leq 1-xf(x)\leq 1-x$


[quote="Vasc"]Let $x,y,z$ be positive reals such that $xyz=1$. Prove that

$\frac{x}{x^{-8}+y+z}+\frac{y}{x+y^{-8}+z}+\frac{z}{x+y+z^{-8}} \ge 1$.[/quote]
$\sum_{cyc}\frac{x}{x^{-8}+y+z}=\sum_{cyc}\frac{x^6}{y^3z^3+x^5y+x^5z}\geq\frac{(x^3+y^3+z^3)^2}{\sum\limits_{cyc}(x^5y+x^5z+x^3y^3)}\geq1$,
where the last inequality is equivalent to $\sum_{cyc}(x-y)^2(x^4-x^2y^2+y^4)\geq0$, which is obvious.
It seems that $\sum_{cyc}\frac{x}{x^{-10}+y+z}\geq1$ is also true and very strong,
but $\sum_{cyc}\frac{x}{x^{-11}+y+z}\geq1$ is wrong already.


Find the area of the triangle formed by the points

$(5,2)$, $(7,8)$, and $(2,1)$...

WITHOUT using Heron's Formula.

[hide="hint"]Try it first, don't be so lazy. [url=http://wangsblog.com/jeffrey/]See most recent post[/url][/hide]

[hide]Shoelace Theorem; also pick's theorem

Take cross multiply each
5 2
7 8
2 1
5 2

(40+7+4)-(14+16+5)=16
16/2=8

8[/hide]

[hide]similiar to the shoelace theorem,
given verticies of a triangle $(x_n,y_n)$ for $n=0,1,2$

the area is given by:

$A_{\triangle}=\frac{1}{2} \left| \begin{tabular}{|ccc|} x_0&y_0&1\\ x_1&y_1&1\\ x_2&y_2&1\\ \end{tabular} \right|$

this is how the shoelace theorem is derived...

the rest is easy from here...[/hide]

[quote="bookaholic"]I don't think any of that is necessary...
[hide]Consider the right triangle with vertices (2,1), (7,1), and (7,8). Its area is 35/2. Then you subtract the area of the right triangle with vertices (5,2), (7,2), and (7,8), which is 6, and the area of the trapezoid with vertices (2,1), (7,1), (7,2), and (5,2), which is 7/2. Then you get 8.[/hide][/quote]

that is exactly what the formula does...


Find all vectors of $n$ real numbers $(x_1,x_2,\ldots,x_n)$ such that
\[ \left\{ \begin{array}{ccc} x_1 & = & \dfrac 1{x_2} + \dfrac 1{x_3} + \cdots + \dfrac 1{x_n } \\ x_2 & = & \dfrac 1{x_1} + \dfrac 1{x_3} + \cdots + \dfrac 1{x_n} \\ \ & \cdots & \ \\ x_n & = & \dfrac 1{x_1} + \dfrac 1{x_2} + \cdots + \dfrac 1{x_{n-1}} \end{array} \right.  \]
[i]Mircea Becheanu[/i]

Let's have a go for this one !

Let $S = \sum x_i$

Summing all equations and using $S$, we obtain the equivalent  system of equation :

$x_i = S \times (n-1) - 1/x_i$ or ${x_i}^2 - S/(n-1) \times x_i + 1 = 0$

From this we can deduct that all $x_i$ are $a$ or $1/a$ for some $a$ to be found.

There are $n$ $'a'$ and $(n-k)$ $'1/a'$, for some k and $S = k \times a + (n-k)/a$
Replacing and solving equation gives : $a = \pm {\dfrac{\sqrt{\left (k-1\right )\left (k+1 -n\right )}}{k-1}}$.
Roots are real (and $\neq 0$) iff $k = n$, i.e. $a = \pm 1/\sqrt{n-1}$

Solutions are then :
$x_1 = \pm \sqrt{n-1}$, $x_j = \pm 1/\sqrt{n-1}$ and all permutations (with same $\pm$).

subtract any 2 of the equations and get either x(i)=1/x(j)or x(i)=x(j)
let any x(1)=x(2)=...=x(k)=1/x(k+1)=....=1/x(n)=a (this is the only possibility)
then a=(k-1)/a +(n-k)a
so a^2=(k-1)/(1+k-n)
this is positive so k>=n
but obviously k<=n
so k=n
all the numbers are equal
so, a=(n-1)/a
 we are through


Let $ \angle xOy < 180^{o}$ and M is inside it. Construct a secant through M intersects Ox, Oy at A and B respectively such that:

$ a) \frac {1}{MA} \plus{} \frac {1}{MB}$ max

$ b) S_{OAB}$ min.

Denote $ X'O'Y'$ is an angle which is symmetric to $ XOY$ by point M. The points of intersection of this angles is B and C.
The points of intersection line $ l$ ($ M$ lies on $ l$) with this angles is $ B_1,B_1',C_1,C_1'$. Now if $ l$ and $ BC$ are different then
\[ 2S_{OB_1C_1}\equal{}S_{OB_1C_1}\plus{}S_{O'B_1'C_1'}\equal{}S_{OBO'C}\plus{}S_{BB_1B_1'}\plus{}S_{CC_1C_1'}>S_{OBO'C}\] which is double area of $ OB_1C_1O'$ in case $ l\equal{}BC$.

I don't think that there exist a beautiful solution.
Counting solution is:
Denote $ \angle OAM\equal{}\alpha, \angle AOM\equal{}\beta, \angle MOB\equal{}\gamma$. Then by the sinus theorem $ \frac {OM} {MA}\plus{}\frac{OM}{MB}\equal{}\frac {sin \alpha} {sin \beta}\plus{}\frac{sin {\alpha\plus{}\beta}}{sin \gamma}\equal{}sin \alpha (\frac{1}{sin \beta}\plus{} \frac{cos \beta}{sin \gamma})\plus{}cos \alpha \frac{sin \beta}{sin \gamma}\equal{}$ $ Asin \alpha \plus{} Bcos \alpha$. You can construct segments with length A and B, so you can construct $ sin \alpha$ and $ cos \alpha$ for which $ Asin \alpha \plus{} Bcos \alpha$ takes minimum value. So you can constract $ AB$.
 :ninja:


If f(x) = 2 cos2 x + sin x + 1, find the value of f( PI/2).

Great stuff on this site thanks to caring tutors like yourself.

[quote="fourierseries"]I think stuff like this should go in the Getting Started Forum instead![/quote]

Calm down.  Due to the level of the problems being posted in Getting Started (some of them exceed the intended difficulty level) moderators tend not to demote problems to lower forums unless it is something like a mathcounts problem in Pre Olympiad or something.

Thanks


There are 16 characters (not all distinct) in the following expression, which is not mathematically correct.  Move exactly
two characters to another part of the expression to form a different expression which is also mathematically legitimate, and
its value will have the largest real value possible of all possible new expressions.

What is this new 16 character expression?

Note:  i = square root of -1

Expression:
---------------

(i - i^2)! + 4^(1 - i)

Is the value obtained restricted to real?

Since we cannot compare complex numbers.

modeler,

YES, the value obtained is restricted to reals.

However, I HAD the word "real" already in the headline as
in the phrase "largest real value," and you see that before you click into my problem.

That said, I edited my problem with what you stated in mind
and inserted the word "real" in the body of my problem for clarity (at the sake of possible redundancy).

Goodnight.

==  The problem is still open.  ==

***  Spoilers ahead  ***
.......(Hiding it wasn't working for me this time.)
*
*
*
*
*
*
*
*
*
*
*
*
*
*

(i-i^2)!+ 4^(1-i)

My offering is...
Move the "^2" group, and in this case they just happen to
be side by side.  Split them up.  Put the "2" immediately to the
right of the only "4" to make a two-digit number.  Put the "^"
immediately to the right of the only "1" present.  The new
expression would be (i-i)! + 42^(1^-i)  [Answer]
--> 0! + 42^1 = 1 + 42 = 43

You moved 3 characters, instead of 2 of them.
[hide]There was a space between the last "-" and the last
"i" characters in the given expression, so it cannot be removed.
(In general, the retyping of the expression with fewer spaces is wrong.)

Your solution is not valid.

*** The problem is still open. ***
[/hide]

[quote="trisect"]There are 16 characters (not all distinct) in the following expression[/quote]

Please count the number of characters in your expression, if spaces are counted.

No, spaces are not being used as characters here.  The 16 characters you see TYPED in the expression are all the total characters I'm referring to.

[hide="solution"]original expression:

$(i-i^{2})!+4^{1-i}$

We move the ^2 and get the expression

$(i-i)!+4^{1-i^{2}}=1+16=17$[/hide]

No, by moving the "^2" group immediately to the right of the final "i," you moved the final close parenthesis
(of the expression), which is a total of one too many moves.

Also, showing the expression in Latex (or anything other than the symbols I chose) by you or anyone else is misleading because the number of characters in those systems is
different, or likely to be different than the 16 constant number of characters I used.

When you move a character (s) and reposition it (them), you have to take into account whether or not it (they) moves/move (pushes) another already existing character out of its place, WHICH IS ITSELF "a move."

Also, I counted six spaces in the original expression, but certain length spaces might be created with a two-character move, providing it creates a valid expression.

The problem is still open to be solved.

Here are three expressions that tie for the same value with the spacing requirements, as best I can see:
[hide]i(-i^2)! + 4^(1)- i
        i(-i^2)! +(4^1 - i)
       (i(-i^2)! + 4^1 - i),
       All = 4    [/hide]


Let $ k$ be an integer, $ k \geq 2$, and let $ p_{1},\ p_{2},\ \ldots,\ p_{k}$ be positive reals with $ p_{1} \plus{} p_{2} \plus{} \ldots \plus{} p_{k} \equal{} 1$. Suppose we have a collection $ \left(A_{1,1},\ A_{1,2},\ \ldots,\ A_{1,k}\right)$, $ \left(A_{2,1},\ A_{2,2},\ \ldots,\ A_{2,k}\right)$, $ \ldots$, $ \left(A_{m,1},\ A_{1,2},\ \ldots,\ A_{m,k}\right)$ of $ k$-tuples of finite sets satisfying the following two properties:
(i) for every $ i$ and every $ j \neq j^{\prime}$, $ A_{i,j}\cap A_{i,j^{\prime}} \equal{} \emptyset$, and
(ii) for every $ i\neq i^{\prime}$ there exist $ j\neq j^{\prime}$ for which $ A_{i,j} \cap A_{i^{\prime},j^{\prime}}\neq\emptyset$. Prove that
\[ \sum_{b \equal{} 1}^{m}{\prod_{a \equal{} 1}^{k}{p_{a}^{|A_{b,a}|}}} \leq 1.
\]

I proved it by induction about $ m$.

Let $ S = \sum_{b = 1}^{m}\prod_{a = 1}^{k} p_{a}^{|A_{b,a}|}$.

The case $ m = 1$ is trivial.

Suppose that the proposition is true for all $ m$ less than $ M$. If $ m = M$,
If there exist a element $ x$ which appears in only $ A_{a,n}$ for fixed $ n$,
we can remove all $ x$s.    ( this removing doesn't affect the satisfaction of the condition )
But then the value of $ S$ is increased. So we can assume that there doesn't exist such $ x$.  $ (*)$

Choose a element in some sets. Denotes it $ x$.
Let ${ B_{i} = \{j | x \in A_{j,i}}\}$ and $ f(X) = \sum_{b\in X}\prod_{a = 1}^{k} p_{a}^{|A_{b,a}|}$.
Because of the condition (i), $ B_{i}$s are disjoint sets. Let $ C = \{1,\cdots,m\} - B_{1} - \cdots - B_{k}$.

For each $ B_{i}$, sets "with respect to" $ B_{i}$ , that is $ A_{a,b}$s such that $ a\in B_{i}$, also satisfy the condition. Removing $ x$ from those sets also doesn't affect this. ( see the argument for $ (*)$ ) .If we did this, $ f(B_{i})$ turns to $ \frac {1}{p_{i}}f(B_{i}) = b_{i}$.
Furthermore, If we add some sets "with respect to" $ C$, this collection of sets remain to satisfy the condition. Because there exist least two nonempty $ B_{i}$ (see $ (*)$), we can apply induction hypothesis to this. If $ f(C) = X$, We get $ b_{i} + X \le 1 \Rightarrow b_{i} \le 1 - X$.

We're goal is to show that $ S = p_{1}b_{1} + \cdots + p_{k}b_{k} + X \le 1$.
But this can be proved now. $ p_{1}b_{1} + \cdots + p_{k}b_{k} + X \le (1 - X)(p_{1} + \cdots + p_{k}) + X = 1 - X + X = 1$.
Therefore induction step is ended.

$ Q.E.D.$

N.B. My work on this problem was based around drawing tables, and I couldn't find a way to upload images, so I strongly advise the marker draw a table as instructed to help them visualise the ideas of this proof, though I hope the explanation is sound enough anyway.

Firstly, let $ S$ be the union of all sets $ A_{i,j}$. Write it in the form $ S \equal{} \{a_1,a_2,....,a_n\}$ (a union of finitely many finite sets is finite).

Now draw a table with $ a_1,a_2,....,a_n$ along the top and the numbers $ 1,2,....,m$ down the left hand side. By condition (i), for some $ i$ there exists at most one $ j$ such that any single $ a_s \in A_{i,j}$. If such a $ j$ exists, we signify this in our table by writing $ j$ in the cell in column $ a_s$ and row $ i$. If no such $ j$ exists, leave the cell blank. We call the [i]sum[/i] of a table the quantity given in the question to estimate.

We now use the following algorithm to 'fill in' all the blank cells without affecting the value we are trying to estimate.

Suppose that in row $ i$ the cells in columns $ a_{s_t}$ are blank for $ t \equal{} 1,2,....,u$. Then replace row $ i$ with $ k^u$ rows with the following properties:
- For $ s \not\in \{s_t\}$ let the value in column $ a_s$ be constant for all our new rows and equal to the original value from 'row $ i$'.
- Let our $ k^u$ rows all be different and let the ordered $ u$-tuple formed by the values of the cells in columns $ a_{s_1}, a_{s_2},... ,a_{s_u}$ take all values in $ \{1,2,...,k\}^u$.

Doing this replacement procedure to a single row $ i$, noting that this leaves every term of the sum except that where $ b \equal{} i$ invariant, replaces $ \prod_{a \equal{} 1}^k p_a^{|A_{i,a}|}$ with the value (where the sum is taken over all $ k$-tuples $ (t_v)$ of nonnegative integers such that $ t_1 \plus{} t_2 \plus{} ... \plus{} t_k \equal{} u$, corresponding to adding up how many times each of $ \{1,2,...,k\}$ appears in the previously empty cells of a given row) given by
\[ \sum_{(t_v)}\binom{u}{t_1}\binom{u \minus{} t_1}{t_2}...\binom{u \minus{} t_1 \minus{} t_2 \minus{} ... \minus{} t_{k \minus{} 1}}{t_k}\prod_{a \equal{} 1}^k p_a^{|A_{i,a}| \plus{} t_a}
\]
This can be re-written as
\[ \prod_{a \equal{} 1}^k p_a^{|A_{i,a}|}\sum_{(t_v)}\binom{u}{t_1}\binom{u \minus{} t_1}{t_2}...\binom{u \minus{} t_1 \minus{} t_2 \minus{} ... \minus{} t_{k \minus{} 1}}{t_k}\prod_{a \equal{} 1}^k p_a^{t_a}
\]

\[ \equal{} \prod_{a \equal{} 1}^k p_a^{|A_{i,a}|}(p_1 \plus{} p_2 \plus{} ... \plus{} p_k)^u \equal{} \prod_{a \equal{} 1}^k p_a^{|A_{i,a}|}
\]
So this operation does indeed leave the sum invariant, and does not create any more rows with empty cells. We can therefore perform this operation on all rows containing empty cells. Furthermore, if condition (ii) does not hold for the new table, since we have *added* elements to our sets, it quite clearly cannot have held for the original table.
Better Explanation: If for $ i'$ there were originally $ j\not \equal{} j'$ for which $ A_{i,j} \cap A_{i',j'} \not \equal{} \emptyset$ then in the new situation for any $ i$ value corresponding to a newly created row, if $ i'$ corresponds to an old row, this old relationship still holds (because all the elements of $ S$ previously in 'row $ i$' are now in all our rows. If $ i'$ also corresponds to a newly-created row, the permutations among the previously empty cells being different imply that there is a column $ a_s$ in which content $ j, j'$ of cells $ i$ and $ i'$ differ, corresponding to a nonempty intersection between sets $ A_{i,j}$ and $ A_{i',j'}$ with $ j\not \equal{} j'$ (nonempty because they contain $ a_s \in S$.

We therefore perform this operation to every row in the original table, to yield a new table with the same sum, but with no empty cells.

Now, it is clear that the rows of the table form some subset $ U$ of the set of vectors $ V \equal{} \{1,2,...,k\}^n$. We cannot have two identical rows, else condition (ii) would be violated, because in such identical rows $ i,i'$ an element is in $ A_{i,j}$ if and only if it is in $ A_{i',j}$, so for $ j\not \equal{} j', A_{i,j}\cap A_{i',j'}$ would be empty.
For a vector $ v \in V$ let $ f_d(v)$ denote the number of entries in $ v$ with value $ d$. The original definition of a table's sum is clearly equivalent to the following expression:
\[ \sum_{u \in U} \prod_{a \equal{} 1}^k p_a^{f_a(u)}
\]
The $ p_i$ are positive reals, so each term in this sum is positive. The sum is therefore maximal when the set $ U$ is as large as it can possibly be. But $ U$ is a subset of $ V$, so $ U \equal{} V$ is this maximal case, and here it is clear that (with each vector corresponding to 'choosing a number from each bracket')
\[ \sum_{v \in V} \prod_{a \equal{} 1}^k p_a^{f_a(v)} \equal{} (p_1 \plus{} p_2 \plus{} ..... \plus{} p_k)^n \equal{} 1
\]
Thus
\[ \sum_{u \in U} \prod_{a \equal{} 1}^k p_a^{f_a(u)} \leq 1
\]
And this was the result required.

Equality occurs if and only if (once all the blank spaces have been filled) every possible partition of $ S$ into $ k$ subsets corresponds to a $ k$-tuple $ (A_{i,1},A_{i,2},....,A_{i,k})$ in our collection.


Question a was:

Find a unit vector from the point P = (1, 2) and toward the point Q = (3, 6).

my answer is:
2/sqrt(20)i + 4/sqrt(20)j

the second question asks

Find a vector of length 15 pointing in the same direction, how can I find this?

You've already got a unit vector. If you multiply that unit vector by the scalar $ c,$ what does its length become?

well if I multiply the unit vector with a scalar c then I can either get a larger/smaller magnitude vector with the same direction, I know that.. problem is how do I make sure that the length is 15

[quote="Roosta"]Point here is that a unit vector has length 1, so multiplying it by $ c$ gives a vector with what length?

[hide="Hint"]When multiplied by a scalar, a vector's length gets multiplied by that scalar and the direction doesn't change.[/hide][/quote]

[quote="Roosta"]Point here is that a unit vector has length 1, so multiplying it by $ c$ gives a vector with what length?

I understand this point, so if I multiply 2/sqrt(20)i + 4/sqrt(20)j  by 2 then I will get something of 4/sqrt(20)i + 8/sqrt(20)j, clearly the length of this vector is not 15, so are you saying that I should find a c that will result in the length of the vector to be 15?

When you multiply by 2 you get a vector of length 2: new length = old length $ \times$ c. To see this, we have for $ \vec{v} \equal{} a\hat{\imath} \plus{} b\hat{\jmath}$:

$ \mbox{Length} \equal{} \sqrt {a^2 \plus{} b^2} \equal{} \sqrt {\left(\frac {4}{\sqrt {20}}\right)^2 \plus{} \left(\frac {8}{\sqrt {20}}\right)^2} \equal{} 2$

Try to do this with the same unit vector (multiplying be $ c$ that is) and you should find that the length is:

[hide="Answer"]$ \mbox{Length} \equal{} c$[/hide]

[quote="Roosta"]When you multiply by 2 you get a vector of length 2: new length = old length $ \times$ c. To see this, we have for $ \vec{v} \equal{} a\hat{\imath} \plus{} b\hat{\jmath}$:

$ \mbox{Length} \equal{} \sqrt {a^2 \plus{} b^2} \equal{} \sqrt {\left(\frac {4}{\sqrt {20}}\right)^2 \plus{} \left(\frac {8}{\sqrt {20}}\right)^2} \equal{} 2$

Try to do this with the same unit vector and you should find that the length is:

[hide="Answer"]$ \mbox{Length} \equal{} c$[/hide][/quote]

sorry, but I don't get what you mean here.. what do you mean by the same unit vector?

$ \frac{2}{\sqrt {20}}\imath \plus{} \frac{4}{\sqrt {20}}\jmath$


Nobody?

It is easy to show for all $n$ that $a=b=c$ is a local minimum, i.e. the ineq is true in a neighborhood of (a,a,a). Maybe that helps  :roll:


Let $ x>0$. Find maximum of $ f(x)\equal{}\frac{x}{(x\plus{}2008)^2}$

$ f(x)\equal{}\frac{x}{(x\plus{}2008)^2}\leq \frac{x}{4\cdot x\cdot 2008}\equal{}\frac{1}{8032}$. The equality holds when $ x\equal{}2008$.

To clarify, Kunny used $ (a\minus{}b)^2 \implies a^2\plus{}b^2 \geq 2ab$ on the denominator. In particular, $ (x\plus{}2008)^2 \equal{} x^2\plus{}2008^2 \plus{} 2 \cdot x \cdot 2008 \geq 2 \cdot x \cdot 2008 \plus{} 2 \cdot x \cdot 2008 \equal{} 4 \cdot x \cdot 2008$


Find all functions $ f: \mathbb{R}^\plus{}\rightarrow \mathbb{R}^\plus{}$ such that

(i) $ f(x\plus{}yf(x))\equal{}f(x)f(y)$ for all $ x,y>0$
(ii) there are at most finitely many $ x$ such that $ f(x)\equal{}1$

If there exists an $ a < b$ s.t. $ f(a)\equal{}f(b)\equal{}m$

Then plugging in x=a, then x=b gives $ f(a\plus{}ym)\equal{}f(b\plus{}ym)$ (*)

Now, if there are only finitely many values such that $ f(x)\equal{}1$ then $ f(a\plus{}ym)\equal{}f(a)$ can only have finitely may solutions (since $ f(a\plus{}ym)\equal{}f(a)f(y)$).

But, $ f(a)\equal{}f(b)\equal{}f(2b\minus{}a)\equal{}f(3b\minus{}2a)\equal{}...$ by applying $ f(a\plus{}(b\minus{}a)k)\equal{}f(b\plus{}(b\minus{}a)k)$ which is gotten by letting
$ y\equal{}(b\minus{}a)k/m$ in (*).

Hence, we derive a contradiction. Thus f is injective, so since $ f(x\plus{}yf(x))\equal{}f(y\plus{}xf(y))$ we must have for all $ x,y>0$ $ x\plus{}yf(x)\equal{}y\plus{}xf(y)$ Plugging in $ x\equal{}1$ gives $ 1\plus{}yf(1)\equal{}y\plus{}f(y)$ so $ f(y)\equal{}(f(1)\minus{}1)y\plus{}1$ Therefore, we must have $ f(y)$ being linear.

Letting $ f(x)\equal{}ax\plus{}b$ We get $ a(x\plus{}y(ax\plus{}b))\plus{}b\equal{}(ax\plus{}b)(ay\plus{}b)$

$ ax\plus{}a^2yx\plus{}aby\plus{}b\equal{}a^2xy\plus{}abx\plus{}aby\plus{}b^2$

$ ax\plus{}b\equal{}abx\plus{}b^2$

$ ax\plus{}b\equal{}b(ax\plus{}b)$

$ (b\minus{}1)(ax\plus{}b)\equal{}0$

$ b\minus{}1\equal{}0$

$ b\equal{}1$

Hence, it is easy to check, that all functions are therefore of the form $ f(x)\equal{}ax\plus{}1$ with $ a>0$


parameterization
trigonometry
calculus
calculus computations

How i can make a sketch of theese curves?  There are some standard way to aproach theese problems? Thanks in advance

$1) \, \alpha(t)=(-1+t^{2},2-t^{2}), \ t \in \mathbf{R}$

$\alpha$ is the "half" parametrization of the line parametrized by $\beta(t)=(-1+t,2-t)=(-1,2)+t(1,-1), \ t \in \mathbf{R}$ because $t^{2}\geq 0$ (this is right?)

have no idea on the above problems

$2) \, \alpha(t)=(\cos^{2}t, \, \sin^{2}t), \ t \in \mathbf{R}$

$3) \, \alpha(t)=(\sin t, \, \cos{2t}), \ t \in [0, 2\pi]$

$4) \, \alpha(t)=(\cos t, \,-3+\sin 2t), \ t \in [0,2\pi ]$

$5) \, \alpha(t)=(2, \, 1, \, t), \ t \in \mathbf{R}$

$6) \, \alpha(t)=(2\cos t, \, 3\sin t, \, 5), \ t \in [0,2\pi]$

$7) \, \alpha(t)=(t^{2}-1, \, 2, \, t), \ t \in \mathbf{R_{+}}$

In one method, you may let $\alpha(t) = (x, y) = (x(t), y(y))$ and find the relationship between $x$ and $y$. Not always, but often it reduces a parametrized curve into more familiar form. For example

(2) $x, y \in [0, 1]$ and $x+y = \cos^{2}t+\sin^{2}t = 1$, so the image of $\alpha$ is the line segment joining $(0, 1)$ and $(1, 0)$.

(3) $y = \cos 2t = 1-2 \sin^{2}t = 1-2x^{2}$.

If you want to apply this method to (4), maybe you have to split this curve and consider each parts.

From (5) to (7), at least one component of $\alpha$ is constant (for example, third component of $\alpha$ in (6) is constant), so they are just plane curves and you may discard a constant component to regard them as a curve in $\mathbb{R}^{2}$.


trigonometry
geometry
trig identities
Law of Sines
exterior angle

When you have $A = \theta_1$ and $a = s_1$ You are given $c = s_2$

Now you use law of sines to find angle $C$, $\theta_2$

$\theta_2 = \frac{(sin \theta_1)(s_2)}{sin\theta_1}$

If $\theta_2< 180^\circ - \theta_1$, there are $2$ possible triangles. However I am perplexed as how this can be. Wouldn't the different angles of $\theta_2$ produce $2$ different side lengths for $c$? This is only logical because one of the two possible angles must be larger and thus produce a longer opposite side $s_2$. Both longer and shorter sides can't be the same side.

Am I simply overlooking something?

Help is much appreciated.

well, do you know the identity $\sin \theta=\sin 180-\theta$?

$\theta_2 = \frac{(sin \theta_1)(s_2)}{sin\theta_1}$. Are you sure that's correct?

I don't understand how you're getting your result.

From the Law of Sines, we have

$\frac{\sin A}{a} = \frac{\sin C}{c} \implies \frac{\sin \theta_1}{s_1} = \frac{\sin \theta_2}{s_2}$
$\sin \theta_2 = \frac{s_2}{s_1} \sin \theta_1$
$\theta_2 = \arcsin \left( \frac{s_2}{s_1} \sin \theta_1 \right)$

$\theta_2 < 180^{\circ} - \theta_1$ must always be true because $\theta_1 + \theta_2 + \theta_3 = 180^{\circ}$.

You are given two sides and an angle opposite one of the sides.  There are $2$ triangles that this could be because there are two possible values of $\theta_2$ owing to the identity $\sin \theta = \sin (180^{\circ} - \theta)$.  One is an acute angle, and the other is an obtuse angle (or they are both right angles).

Edit:  In other words, $\arcsin \left( \frac{s_2}{s_1} \sin \theta_1 \right)$ gives two different values of $\theta_2$ for $0 < \theta_2 < 180^{\circ}$.

It's the reason while the A-S-S triangle congruency identity doesn't work.

Geometric construction so you can see why there are 2 such triangles: Given acute triangle $ABC$ (draw it in such a way side $BC$ is base of triangle). You said given angle $A$, sides $BC=a,AB=c$. Now, with compass, draw a circle $\Gamma$ with radius $AC$ and center at $A$. This circle will intersect at another point $D$ along the [b]extended[/b] line of $BC$ and draw the line $AD$ (in other words, if circle $\Gamma$ intersects $BC$ then $AD$ is inside triangle, if not, $\Gamma$ will intersect $BC$ at the extended line of $BC$ in which of course $AD$ lies outside of the triangle. The 2 triangles are $ABC$ and $ABD$ with $\sin C=\sin ABD$; both triangles will have the same Law of Sine ratio. The idea follows from the theorem that the exterior angle is the sum of the 2 remote interior angles. Also, it is always the case that one angle is acute and the other one is obtuse, so you can actually find the type of triangle you want easily. Like Treething said too, this shows that SSA is not a congruency condition.

I understand this now. I was visioning the triangle in my mind wrong. I was looking at the wrong side. But yes, now everything is very clear and it works out very nice.


The equation $z^6 + z^3 + 1$ has one complex root with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane.  Determine the degree measure of $\theta$.

[hide="Solution"]Letting $x=z^3$, we have $x^2+x+1=0$

The roots of these are $x=\cis{120^\circ}$ and $x=\cis{240^\circ}$. We now apply Euler's Theorem on $z=x^{\frac{1}{3}}$.
From $z^3= \cis{120}$, we see
$z= \cis{40^\circ},\cis{(40+120)^\circ},\cis{(40+240)^\circ}$ $=\cis{40^\circ},\cis{160^\circ},\cis{280^\circ}$

From $z^3= \cis{240}$, we see
$z= \cis{80^\circ},\cis{(80+120)^\circ},\cis{(80+240)^\circ}$ $=\cis{80^\circ},\cis{200^\circ},\cis{320^\circ}$.

Our desired root has measure of $\boxed{160^\circ}$

Whoa, the cis's didn't show up. Everything in front of the degrees except the last line should have cis's. [/hide]

[hide="Answer"]$y=z^3$, $y^2+y+1=0\Rightarrow y=\frac{-1\pm i\sqrt{3}}{2}=\cos \theta +i\sin \theta\Rightarrow \theta =\frac{2\pi}{3}+2k\pi$
$z=e^{i\theta}\Rightarrow y=z^3=e^{3i\theta}$
$z=\cos \frac{\pi (2+6k)}{9}+i\sin \frac{\pi (2+6k)}{9}$
$90\le \theta \le 180$, $\theta =\frac{8\pi}{9}=\boxed{160}$[/hide]

[quote="4everwise"]The equation $z^6+z^3+1$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in thet complex plane. Determine the degree measure of $\theta$.[/quote][hide="Solution"]We desire the roots of $P(z)=z^6+z^3+1$. Notice that \[ (z^3-1)P(z)=z^9-1, \] which just has the ninth roots of unity. However, in multiplying by $(z^3-1)$, the extraneous third roots of unity have also been included. Thus, the roots of $P(z)$ are just the ninth roots of unity which are not third roots of unity.

The ninth roots of unity are: \[ \cos(\theta)+i\cdot\sin(\theta), \] for $\theta=\left(\frac{360^\circ}{9}\right)(n)$ for $n=0,1,2,3,4,5,6,7,8$ and $i=\sqrt{-1}$.
The third roots of unity are \[ \cos(\phi)+i\cdot\sin(\phi), \] for $\phi=\left(\frac{360^\circ}{3}\right)(m)$ for $m=0,1,2$.

The only ninth root of unity which has argument $90^\circ\le\theta\le180^\circ$ which is not a third root of unity is $\cos(160^\circ)+i\cdot\sin(160^\circ)$. Hence, the desired answer is $\boxed{160}$.[/hide]

\cis is [b]not[/b] a $\LaTeX$ command.  cis is shorthand notation.  The only way around it is to use \text{cis}, e^{i\theta}, or \cos +i\sin.

It's worth noting that $\exp$ is a latex command.

I doubt everyone viewing this forum knows what $\exp$ means, so just for the record $\exp(x)=e^x.$  Thus, $\text{cis}\theta=\exp(i\theta).$

It's a typesetting savior - there is no way to make something like $e^{e^x}$ look nice - especially if, say, it's in a fraction.

[hide]
$0 = z^6 + z^3 + 1$
Multiply by $z^3$.
$0 = z^9 + z^6 + z^3$
Subtract the equations.
$0 = z^9 - 1$, or $z^9 = 1$, so $z=e^{2\pi i k/9}$. The argument is $2\pi k/9$. In degrees, the one in the range 90 to 180 is is $\boxed {160}$. (it can't be a third root of unity because then $z^6+z^3 + 1 \ne 0$.)[/hide]

Let $z^3 = x$. The two roots of $x^2+x+1=0$ are $\cos 120 + i \sin 120$ and $\cos 240 + i \sin 240$. Adding $360$ to the arguments and using DeMoivre, the first gives a result of $\boxed{160}$ and the second gives a result of $200$ - but out of these only the first is in the desired range.

The equation can be written as $\frac{z^9-1}{z^3-1}$, so the solutions are the roots of unity of $9$ minus the roots of unity of $3$. The two roots of unity of 9 between $90^{\circ}$ and $180^{\circ}$ are $120^{\circ}$ and $160^{\circ}$, but $120^{\circ}$ is a root of unity of $3$. Thus, the answer is $\boxed{160}$.

If this were a problem today, it would probably be #4 at most, imo. The AIME tries to keep the first few questions so that they don't use hard topics, but this is one of the easiest problems you could make in a somewhat hard topic.

$$z^6+z^3+1=\frac{z^9-1}{z^3-1}$$

For which the roots need to be written as $\frac{2\pi k}{9}$ and $k$ cannot be divisible by $3$. Furthermore, by the range provided, $k$ must be between $5$ and $2$ and cannot be $3$. This gives $160^{\circ}$ as our answer.

Treating it as a quadratic equations in $z^3$ gives
$$z^3 = \frac{-1 \pm \sqrt3 i}{2} = \cos \theta + i\sin \theta.$$
Hence $\theta = 120^\circ + n360^\circ$ or $240 ^\circ + n 360^\circ$ for some integer $n$.
By DeMoivre's
$$z = \cos \frac{\theta}{3} + i \sin \frac{\theta}{3}$$
so the only value of $\frac{\theta}{3}$ in the range is $\boxed{160^\circ}$ when $\theta = 120^\circ + 360^\circ = 480^\circ$.

Surprised this hasn't been posted yet. (Exponential form ftw)
-----
Set $y=z^3$. Now, we just have $$y^2+y+1=0 \rightarrow \frac{y^3-1}{y-1}=0.$$ However, these are just the cube roots of unity except for $1$. They are $e^{\frac{2i\pi}{3}}$ and $e^{\frac{4i\pi}{3}}$. Now, setting $z=re^{i\theta}$ and equating, we find $e^{\frac{2i\pi}{9}}\cdot e^{\frac{2i\pi}{3}}$ (which is the Root of Unity) gives us of $e^{\frac{8i\pi}{9}}$. Finally, converting the argument to degrees gives $\frac{8i\pi}{9}=\boxed{160^{\circ}}$.

$z^6+z^3+1=\frac{z^9-1}{z^3-1}$, so the solution needs to be a ninth root of unity, but not a third root of unity, so this is just $\boxed{160}.$ (120 doesn't work)


inequalities
geometry
inequalities proposed

For an arbitrary triangle of area S with medians M_a,M_b,M_c which are denoted by p.q,r respectively (for easier typing) prove

S<=3abcpqr/[a(b+c)qr +b(c+a)rp +c(a+b)pq],with equality if and only if the triangle is equilateral.

This inequality is proved in the Mathematical Gazette Volume 84,2000, page
423.It is part of an article on inequalities connecting medians, angle bisectors and altitudes of a triangle.

This one is very interesting. Here is my proof:

Use the formula: abc=4RS the inequality is equivalent to
         sum(a(b+c)M_b.M_c) <=12RM_aM_bM_c
         sum (a(b+c)/M_a) <= 12R  (*)

[b]Lemma[/b]:  2M_a >= (b+c)cos(A/2)   (**)
        2.(2b^2+2c^2-a^2) >= (1+cosA).(b+c)^2
        (b-c)^2 >= (b-c)^2.cosA  (obviously true)
(Because a^2=b^2+c^2-2bccosA)

Apply this lemma, now we have to prove this one:
             sin(A/2) + sin(B/2) +sin(C/2)  <= 3/2
But sin(A/2) + sin(B/2) +sin(C/2) <= 3sin (pi/6) = 3/2
The inequality is proved.


Don't title "bet no one can get this."  This is AoPS.  Somebody can get it unless it's an open question, and even then possibly somebody can get it anyway.  :D  It's a pointless challenge.  Describe the problem in your title instead.

They probably do it so that people click on it.

[quote="bthings_2000"]You are wrong. There are problems that cannot be solved, and there are problems on these forums that have gone unsolved.[/quote]

Those tend to be problems placed in the wrong forum.  Some of them may be very hard, yes, but they don't tend to actually be unsolvable.

torajirou's right bthings.If people don't reply to a question,it doesn't mean that no one's got it.Some of the more experienced solvers leave it for others to try before solving it themselves,like toirajirou :D

Funny, I was honestly going to make this exact thread...


I have some tables and chairs. If I place two chairs at each table, I have one extra chair. If I place three chairs at each table, I  have one table with no chairs. What is the sum of the number of tables and the number of chairs?

Let $ t$ be the number of tables and $ c$ be the number of chairs, so that $ 2t \plus{} 1 \equal{} c$ and

$ 3(t \minus{} 1) \equal{} 3t \minus{} 3 \equal{} c\implies2t \plus{} 1 \equal{} 3t \minus{} 3\implies t \equal{} 4\implies c \equal{} 9\implies c \plus{} t \equal{} \boxed{13}$.


function
algebra
rational function
real analysis
real analysis solved

Does there exists any sufficiently exact lower estimation of the norm for rational function using it's coefficients
$\displaystyle max_{[0,1]}\frac{P_n(x)}{Q_n(x)}\geq...$something using coefficients of P,Q and n

How about $\frac {S(P)}{S(Q)}e^{-Cn}$ where $S(P)$ is the sum of absolute values of coefficients of $P$ and $C>0$ is some number? One can get more precise, but first see whether the estimate of this kind really helps in your problem or it gives nothing at all...

Thanks,fedja. I think that it's not enough for my problem, because of the very fast growing factor $e^{-Cn}$. I would be very good if it was something not so fast growing. Maybe if you know any better estimation, you would show me it.
I'm just wondering whether you given estimation is exact-i meant: Does the equality holds?

[quote="eugene"]
I'm just wondering whether you given estimation is exact-i meant: Does the equality holds?[/quote]  Unfortunately, it is almost exact (up to the value of $C$): just take $Q(x)=1$ and $P(x)=\left(x-\frac12\right)^n$


Find the least value of the expression
P=(3+ 1/a +1/b)(3+ 1/b +1/c)(3+ 1/c +1/a)
where the positive numbers a,b,c satisfy the condition a+b+c<=3/2
(this problem from 'Mathematics and youth magazine')

[hide]
okay, because you want to minimize the value of the expression you want a, b, and c to be as large as possible.  Also, because $1/x$ appears twice for each variable, you want $a=b=c$ which means $a+b+c=\frac{3}{2}$
$a = b = c = \frac{1}{2}$
After filling his in you have
$(3+\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2}})*(...)*(...)$
$7*7*7$
Therefore my answer is
$\boxed{343}$
[/hide]

[hide]For minimum, a=b=c. so, $3a=\frac{3}{2},$ $a=\frac{1}{2}$

Plugging in, we get $(3+2+2)^3=343$[/hide]


calculus
derivative
topology
Functional Analysis
real analysis
real analysis unsolved

The following problem is nice and useful. Let $ G$ be an open set in Banach space $ X$. Assume that $ f : G \to X$ is a compact operator and $ x_0 \in G$. If $ f'(x_0)$ exists, then $ f'(x_0)$ is also a compact operator.  :)

It is pleasure if someone can post some applications.

[quote="alekk"]what is a compact operator ? (I mean, is it linear ?)[/quote]

...then $ f'(x_0)$ is also a [b]linear [/b] compact operator...

The reason is that $ f'(x_0) \in \mathcal{L}(G,X)$.

I am asking if $ f$ is linear (this is the only definition I know about compact operator). In this case $ f\equal{}f'$, isn't it ?

[quote="alekk"]I am asking if $ f$ is linear (this is the only definition I know about compact operator). In this case $ f \equal{} f'$, isn't it ?[/quote]

Oops, if $ A \in \mathcal L(X, Y)$ then $ (x \mapsto Ax)'(x_0) \equal{} A$ for every $ x_0\in X$. In fact we cannot write down $ f' \equiv f$ since $ f' : X \to \mathcal L(X, Y)$ and $ f : X \to Y$.

$ \boxed{ f : G \to X \text{ is compact}} \xrightarrow{{f'\left( {x_0 } \right)\text{ exists}}} \boxed{f'(x_0) : X \to X \text{ is compact}}$

of course $ f'\equal{}f$ means here $ f'(x_0)\equal{}f$ for every $ x_0 \in X$. I still do not get the point (if you are talking about Fréchet derivative, then $ f'(x_0): X \to Y$ if $ f: X \to Y$) What is your definition of derivative ?

Probably, what was meant is the following: $ f: X\to Y$ is a non-linear map that maps, say, the unit ball of $ X$ to a compact set in $ Y$ and has Frechet derivative at $ 0$. Prove that this derivative is a compact linear operator.

I guess we are speaking about nonlinear compact maps, i.e. maps that take bounded sets onto relatively compact ones.

WLOG $ x_0 = 0$ and $ G$ contains a (closed) unit ball. Suppose, contrary to our claim, that $ Df(0)$ is not compact. Then there exists a sequence $ (x_n)$, $ \|x_n\|\le1$, and a number $ \epsilon > 0$ such that
\[ \epsilon \le \|Df(0)(x_m - x_n)\| \quad \text{ for any } m,n \in \mathbf N.
\]
Set
\[ R(x) : = f(x) - f(0) - Df(0)x
\]
and find $ \delta > 0$ so that for any $ \|x\|\le\delta$,
\[ \|R(x)\| \le a\delta
\]
with some $ a < \epsilon/2$ (this is possible since $ f$ is differentiable at 0).
Then we must have that
\begin{align*} \|f(\delta x_m) - f(\delta x_n)\| & = \|\delta Df(0)(x_m - x_n) + R(\delta x_m) + R(\delta x_n)\| \\
& \ge \epsilon\delta - \left( \|R(\delta x_m)\| + R(\delta x_n)\| \right) \\
& \ge (\epsilon - 2a)\delta > 0 \end{align*}
for any $ m,n \in \mathbf N$. This contradicts the compactness of $ f$.


induction
linear algebra
matrix
combinatorics proposed
combinatorics

hi evrone,

Let S be the set of permutation of [[1,n]] whit no fix point. Is there more such permutation with odd signature or even ?  :P  (hint:easy)

Quite simple induction shows that $d(n)=(-1)^{n-1}(n-1)$, where $d(n)$ is the difference between the number of even permutations and odd permutations with no fixed points. This means that there are more even permutations if $n$ is odd and vice-versa.

nice sol. Here is another quite simple sol:consider the determinant of the following matrix:
 0111..11
 1011..11
 1101..11
 ..       ..
 111...10  and conclude ..  :D


If $ p(n)\equal{}\overline{abc}$ is an integer such that $ 17a\plus{}8b\plus{}945c\equal{}n$. Find S if:

$ S\equal{}p(2005)\plus{}p(2006)\plus{}p(2007)\plus{}p(2008)$

Notice that $ 17 a\plus{}8 b\le 17 . 9\plus{}8 . 9$ then $ c\equal{}2$ for those $ n$s from $ 2005$ to $ 2008$
getting: $ S\equal{}382\plus{}462\plus{}542\plus{}622\equal{}2008$