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The polynomial $ W(x)\equal{}x^2\plus{}ax\plus{}b$ with integer coefficients has the property that for every prime $ p$ there is an integer $ k$ such that both $ W(k),W(k\plus{}1)$ are divisible by $ p$. Show that there is an integer $ m$ such that $ W(m)\equal{}W(m\plus{}1)\equal{}0$.

we take $ p$ prime
$ \exists k: \ p|w(k \plus{} 1),p|w(k)$
so $ 2k \equal{} \minus{} 1 \minus{} a \plus{} w(k \plus{} 1) \minus{} w(k)\equiv \minus{} 1 \minus{} a[p]$ and $ (2k)^2 \plus{} 2a(2k) \plus{} 4b \equal{} 4w(k)\equiv 0[p]$
gives $ (a \plus{} 1)^2 \plus{} 2a( \minus{} 1 \minus{} a) \plus{} 4b\equiv (2k)^2 \plus{} 2a(2k) \plus{} 4b[p]\equiv 0[p]$
then $ p|(a \plus{} 1)^2 \plus{} 2a( \minus{} 1 \minus{} a) \plus{} 4b,\forall p\ prime$ gives $ (a \plus{} 1)^2 \plus{} 2a( \minus{} 1 \minus{} a) \plus{} 4b \equal{} 0$
so we can take $ m\equal{}\frac{\minus{}a\minus{}1}{2}$ (because $ 2|a \plus{} 1$)
we get $ w(m) \equal{} \frac {(a \plus{} 1)^2 \plus{} 2a( \minus{} 1 \minus{} a) \plus{} 4b}{4} \equal{} 0$ and $ w(m \plus{} 1) \equal{} w(m) \plus{} 2m \plus{} a \plus{} 1 \equal{} 0 \plus{} 0 \equal{} 0$

An approach  solution with using following lemma .
 [u]Lemma [/u] Let $ a$ is a natural number . Suppose exist a prime $ p_o$ such that for all $ p$ is a prime and $ p\geq p_o$ then a is a quadric reside mod $ p$ then
$ a$ is a perfect square.
It is an well know lemma . You can search it on forum  :)
[b]Problem [/b] $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$
Suppose that for each prime $ p$ exist an integer $ n$ such that $ p|P(n)$
Prove that $ P(x)$ have rational  roots .
[u]Proof[/u]
Apply the lemma we have $ \Delta$ is a perfect square .
So $ f(x)$ have rational roots .
Now apply this problem to your problem  .
Because $ a \equal{} 1$ so all rational roots of $ f(x)$ must be integer .
Call roots are $ x_1,x_2$ then
$ f(x) \equal{} (x \minus{} x_1)(x \minus{} x_2)$
Because for all prime p then exist an positive integer k such that p$ f(k),f(k \plus{} 1)$
$ p|k \minus{} x_1 ,p|k \plus{} 1 \minus{} x_2$ (1)
or $ p|k \minus{} x_2,p|k \plus{} 1 \minus{} x_1$ (2)
One of two condition $ (1),(2)$ must occur for infinite prime p .
So must have  $ |x_1 \minus{} x_2| \equal{} 1$
This problem claim .
[u]Note [/u] The general problem:
$ f(x)\equal{}ax^2\plus{}bx\plus{}c$
Prove that if for any $ m\in Z$ then exist an integer $ k$ such that $ m|f(k)$ then $ f(x)$ has an integer root .

[quote="TTsphn"]
It is an well know lemma . You can search it on forum  :)
[/quote]

Can you show?

Also, can you show your general problem?

Have some ways to solve this lemma.
We only need to prove for $ a$ is a square free .
From conditional we have for any $ n$ such that all prime divisor of $ n$ are great than $ p_0$ then
$ (\frac{a}{p})\equal{}1$
Where $ (\frac{a}{n})$ is Jacobi's symbol .
[u]Case 1 :[/u] $ a\equiv 1 (\mod 2 )$
$ a\equal{}\prod_{i\equal{}1}^r p_i$ ,$ a>1$
Chose $ m\equiv 1 (\mod 4)$ then
$ (\frac{a}{n})\equal{}(\frac{n}{a})\equal{}\prod_{i\equal{}1}^r (\frac{n}{p_i})$
Chose $ n\equiv 1 (\mod p_i),i\equal{}1,..,r\minus{}1$
$ n\equiv n_0 (\mod p_r)$ where $ n_0$ is a non quadric reside mod $ p_r$
So $ (\frac{n}{a})\equal{}\minus{}1$
It follows that there exist an $ p$ such that $ (\frac{a}{p})\equal{}\minus{}1$ and $ p>p_0$
[u]Case 2 [/u] $ a\equal{}2a'$
Chose $ n\equiv 1 (\mod 8)$
$ (\frac{a}{n})\equal{}(\frac{a'}{n})$
It is similar the case 1 .
So problem claim .
An other solution uses Diriclet theorem about prime of the form $ an\plus{}b$

[quote="Carlez Tevos"][quote="TTsphn"]
[u]Note [/u] The general problem:
$ f(x) \equal{} ax^2 \plus{} bx \plus{} c$
Prove that if for any $ m\in Z$ then exist an integer $ k$ such that $ m|f(k)$ then $ f(x)$ has an integer root .[/quote]

Thanks. Can you show this as well?[/quote]
Assume the hypothesis: given any $ m \in \mathbb{Z}$, there exists an integer $ k$ such that $ m | f(k)$; that is, there exists an integer $ l$ such that $ lm \equal{} f(k)$. So, in the case that $ m \equal{} 0$, there exists an integer $ k_0$ such that $ f(k_0) \equal{} 0$. Hence $ k_0$ is an integer root of $ f(x)$. QED.


AMC
AIME
USA(J)MO
USAMO
USAMTS
AMC 12
Putnam

*sniffles*...we feel ya Fiery...

but there's next year right?

hmm...methinks you are overreacting.

Life does not revolve around one math test, I'm sorry that you did not make the usamo, but it is pretty insignificant in the grand scheme of things.  Personally, I never made the usamo, but that does not deter me from doing problems in my free time as a hobby.  I know that you could not have put so much effort into the aime if you didn't enjoy what you were doing, so just go with it.

[quote="Fierytycoon"]The journey has ended. It's all over. I pulled a 5 on the AIME II a couple days ago. A mere 5. 5. I won't be making USAMO this year, with my 126.5 + 10*5 = 176.5 index. I don't know what I did wrong. I just broke, cracked under the pressure of the event. I couldn't really think clearly. I always seemed to feel so desperate during those 3 hours, guessing around, hoping to stumble upon a correct answer. Desperate.[/quote]

I felt exactly the same way after coming out of the AIME I, having missed problems that everyone else on AoPS seemed to have gotten right.  During the test, I only did one problem in a somewhat nice way, #9.  The rest, I did brute-force arithmetic, and made a mistake in virtually every one.  At the end, I came out with a measly 2 (your 5 is excellent compared to this), after several months of obsessing about the AIME anyway.

[quote="Fierytycoon"]I feel so empty inside now. So plain. So dull. As if there is nothing to do anymore, no purpose in my life. I guess I could go do something leisurely and fun, maybe something that I had always wanted to do. Maybe I will learn the art of pencil spinning, or perhaps learn to moonwalk, or perhaps flirt with girls. I don't know. My life seems meaningless right now.[/quote]

I recommend spending a week or so taking a break from math.  Read a good book, watch some movies, listen to music, maybe spend some time just talking to people.  Keep your mind off of math, and you'll feel better.

Hugs are good.  I'm sure there are plenty of girls who would give you a hug if you tell them you're having a bad week, or whatnot.  Get hugs.

And when you've recovered, remember that it [i]doesn't[/i] end here.  There's USAMTS, Mandelbrot, Cody Bowl, and so on.  They're not as prestigious or well-recognized as AMC/AIME/USAMO, but they're still fun.  And you still have next year.  And college.  And the rest of your life.  Even when the last competition is over, you can still have fun with math.  Start a competition of your own.  Write a book.  Do research.  Solve the Riemann Hypothesis.

[quote="Fierytycoon"]People like Simon, Joel, Ravi B, gauss202, to name only a few. They don't really need any help with math from any of the other users here. They just come here during their leisure time and give some younger kids a little boost..just because they feel like it.[/quote]

I agree that they deserve our thanks.  But you're forgetting someone: yourself.  You worked hard.  You solved problems.  You stayed home and studied when you could have gone partying, gotten drunk, gotten arrested, and thrown in prison.  (note the sarcasm).  You got goodly at maths, and that's all that matters.  (Except for the prison thing.)

So go have some fun.  Collect those hugs.  Flirt your little heart out.  But keep the maths inside your heart.

[quote="mathfanatic"]Solve the Riemann Hypothesis.[/quote]

:-) I keep thinking they should put the RH, Goldbach's conjecture, or some similiar problem as a question on some math test, like the USAMO or something, and just see what happens.

Anyway, I sorta felt the same way after my AIME.  I was very upset after a friend of mine, Mark, scored higher than I did.  He did the same on the AMC 12.  I do much more math than he does, and love it so much more than he does.  But he still did better than I.  It just makes me so aggravated, knowing that there are a zillion kids out there just like that.  I'm pretty competitive btw, if you can't tell :-).

But I just resolved myself to work harder.  I bought AoPS (which arrived yesterday, yay!) and decided I'll make USAMO next year.  It'll take some doing, with my 100.5 and 4.  But thinking about how much I've grown since even last year, it's doable.

So, I hope you're  :le: a junior so you can try again next year, and if not, life goes on.  You can love math, and even do something great with it, without making USAMO.

for some reason i know if i saw those counting problems outside of the AIME, i'd get them. but come test day, i miss number 4, number 5, and number 8. so instead of a possible 132.5+80, 212.5, i'm at a 132.5+50, a 182.5. i know how you feel fiery.

am I the only one that's feeling really apathetic about actual school right now?
it just sucks so much right now.

[quote="Magnara"]I keep thinking they should put the RH, Goldbach's conjecture, or some similiar problem as a question on some math test, like the USAMO or something, and just see what happens.[/quote]

I have two stories like that. A while ago there was a thread asking for people's favorite problem. I said my favorite problem at the advanced level was a certain problem from a famous textbook, Concrete Mathematics by Graham, Knuth, and Patashnik. The problem is a restatement of the Riemann hypothesis. At least it is in the section of textbook problems headed "Research Problems."

The other story is of the founder of linear programming, who as a graduate student came late to class one day and saw two problems on the blackboard, but didn't hear his professor's introduction to the problems. He took the problems home to do as homework, and thought, "Wow! These problems are hard." The problems, of course, were famous unsolved problems the professor had mentioned in the lecture that this student missed, NOT that week's homework. But he solved the problems, and his solutions turned into his Ph.D. thesis. I'll add details to both of these stories with edits tomorrow.

Fiery, I know how you feel and I grief with you. I grieved with my son recently too. He has broken down only two times in recent memory - once when he was a 6th grader getting beat up by bully 8th graders, and the second time was after this year's AIME I) Nothing any of us can say at this time will make you feel too much better. But you know what, time does heal. It still hurts, but my son no longer grieves.

Meanwhile many students have gone on to the college of their choice wondering what the obsession with this USAMO thingy (or MOP, or IMO) was all about. They have never heard of any of those terms and apparently life goes on for them too (kind of  :lol:  ). By the way, 126.5 on AMC12 was an excellent score for this year, and 5 on AIME is commendable any year. So, maybe you'll get even big time with AMC next year, or maybe you won't. Either way, you can forever be a respectable leader in the AoPS community helping to guide the younger group to enjoy math as much as you do, but without the misguidance that they must do well in the AMC contests to matter. (This is not to say that AMC should be under-appreciated for, in my opinion, its great contribution to society by helping to increase public interest in math.)

mathfanatic wrote:Fierytycoon wrote:The journey has ended. It's all over. I pulled a 5 on the AIME II a couple days ago. A mere 5. 5. I won't be making USAMO this year, with my 126.5 + 10*5 = 176.5 index. I don't know what I did wrong. I just broke, cracked under the pressure of the event. I couldn't really think clearly. I always seemed to feel so desperate during those 3 hours, guessing around, hoping to stumble upon a correct answer. Desperate.

I felt exactly the same way after coming out of the AIME I, having missed problems that everyone else on AoPS seemed to have gotten right.  During the test, I only did one problem in a somewhat nice way, #9.  The rest, I did brute-force arithmetic, and made a mistake in virtually every one.  At the end, I came out with a measly 2 (your 5 is excellent compared to this), after several months of obsessing about the AIME anyway.

Fierytycoon wrote:I feel so empty inside now. So plain. So dull. As if there is nothing to do anymore, no purpose in my life. I guess I could go do something leisurely and fun, maybe something that I had always wanted to do. Maybe I will learn the art of pencil spinning, or perhaps learn to moonwalk, or perhaps flirt with girls. I don't know. My life seems meaningless right now.

I recommend spending a week or so taking a break from math.  Read a good book, watch some movies, listen to music, maybe spend some time just talking to people.  Keep your mind off of math, and you'll feel better.

Hugs are good.  I'm sure there are plenty of girls who would give you a hug if you tell them you're having a bad week, or whatnot.  Get hugs.

And when you've recovered, remember that it doesn't end here.  There's USAMTS, Mandelbrot, Cody Bowl, and so on.  They're not as prestigious or well-recognized as AMC/AIME/USAMO, but they're still fun.  And you still have next year.  And college.  And the rest of your life.  Even when the last competition is over, you can still have fun with math.  Start a competition of your own.  Write a book.  Do research.  Solve the Riemann Hypothesis.

Fierytycoon wrote:People like Simon, Joel, Ravi B, gauss202, to name only a few. They don't really need any help with math from any of the other users here. They just come here during their leisure time and give some younger kids a little boost..just because they feel like it.

I agree that they deserve our thanks.  But you're forgetting someone: yourself.  You worked hard.  You solved problems.  You stayed home and studied when you could have gone partying, gotten drunk, gotten arrested, and thrown in prison.  (note the sarcasm).  You got goodly at maths, and that's all that matters.  (Except for the prison thing.)

So go have some fun.  Collect those hugs.  Flirt your little heart out.  But keep the maths inside your heart.

Sorry for the massive quote.

He's right though. Life doesn't revolve around a math test. I'm sure you've worked very hard on this test, much harder than perhaps some of the rest of us.

It's what I felt like walking out of MC State last yr. Been to state three years, edging closer to Nats each year. Final year, I'm in top 10 and miss nats qualification by 1 point (and 1st place by 3 points). However, in countdown I make 3rd (qualifying onto the team). Minor problem: our state doesn't think countdown was important enough to be official so I didn't get to go to Nats. anyways. Like for the trip home, I looked back on three years, three teams, hundreds of after school practices, pizza parties, inside jokes, math club stuff. It was as if all of a sudden you felt like you've wasted the last 3 years of your life. Nothing mattered for a bit: friends, family, life...nothing. But of course after a while, I learned to get past it. You will too. It's not the end of everything. It's just the beginning of another year of hard work, fun, trials, and lessons. And if you don't get into USAMO...so what? About 1 in every 10,000 kids that take AMC do. It's natural. You're already at the top. Be proud of where you are. Keep going with life. Finals are coming up: do well, make everyone proud.

[hide]Anyways, to continue the MC story. This year, I find out that due to some national policy, all states must now take countdown as official. See...now THAT just REALLY bites. I felt like tearing up to the state office and yelling at someone for a few hours. I had "wasted" 3 years of my life only to find that they changed the rules a few months too late.[/hide]

This year I had really decided not to spend too much effort on math and spend more time on other fun like music and debate. I decided to put off Calc BC til next year and just not do math this year. Then I got on AoPS and that changed. I still put off Calc BC and didn't take curricular math, but at least I didn't just drop out of competing completely.

I think these last two weeks I've done almost everything I could (besides studying overmuch) for this test. Lots of buddies wished me good luck and whatever the results are for USAMO, I'll be happy.

If I make USAMO, great! (There'll be a chance for MOP too *sigh*) If not, that's kool also. There's still next year. And the year after. And the year after that.

Have fun. Enjoy life while you're still young.

Well, look here, in my opinion, I'm proud enough of even making AIME freshman year, top that off, it's only my first year ever even doing AMCs, I didn't even think I'd make AIME.  I mean, you shouldn't feel bad, not at all.  I really don't understand why you would feel bad if you liked math anyways, one test doesn't make or break you.

Fiery, I think you did a great job.  Just qualifying for AIME is an accomplishment.  Solving 5 such questions is way higher than average.  I also have been impressed by the work ethic that you have exhibited on this site.  Such work will pay off if you decide to pursue the mathematical sciences in college.  Even if you don't, I find the logical and problem-solving skills I gained from math to be useful in non-math endeavors too.

By the way, in college, there are still math competitions!  The Putnam exam is my favorite of all the competitions, whether high school or college.  First, it deemphasizes geometry, which for some of us is a good thing. :)   Second, it introduces calculus, which allows for a refreshing variety of new problems.  Third, you only need to solve 3 or 4 questions out of 12 to receive national recognition.

Oh, and thanks for the compliment.  I appreciate it.

It's okay.. you'll get over it.. I was crestfallen after the AIME 1, but I've realized now that there isn't much point in being sad.. the only thing I can do, and must tell myself to do, is start working hard to prepare for next year..

Fiery - when you're a sophomore in college, and have way more free time than most people because you're breezing through most of your classes, you'll have this year spent learning how to think about tough problems to thank for it.  Knocking off problems under time (and other) pressure is not the only way problem solving ability shows, nor is it the most important application of it.

Doing well on the AIME, while the most immediate goal of working now, is not nearly the most valuable possible result of your work now.  Hardly consoling or obvious, but true.

The pursuit of mathematics is far more rewarding than the pursuit of the AIME.  As you grow older, this will become more and more clear and the rewards will become more and more present (and more tangible than a USAMO invitation).

Same thing happened to me this year on AMC. I felt very empty too after knowing that I didn't pass. I felt stupid. I felt that my life was a waste. It felt really sad after spending so much time but didn't make it. Fiery, at least you made AIME, I didn't even make it.

You will get over it. I am feeling really sad too at now since everybody is talking about AIME, which I didn't even qualify to take.

[quote="rippledance"]*nod* There [i]is[/i] life after AIME, Fiery.[/quote]

Although that fact may not be apparent within 24 hours after the test.

[quote="mathfanatic"]Are you allowed to bring, say, a piece of paper numbered from 1 to 25 into the AMC?  I mean, it's not an aid at all (unless you don't know how to count to 25...), but the rules say "nothing other than a calculator, a ruler, ..."[/quote]

It's probably safer not to bring it; just number down one of your scrap papers.

And no, the fact that there is life after is DEFINITELY not apparant within the first 24 hours after.. maybe not even the next 24 days.. or weeks...

[quote]And no, the fact that there is life after is DEFINITELY not apparant within the first 24 hours after.. maybe not even the next 24 days.. or weeks...[/quote]

...especially when USAMO would have gotten you a guaranteed ticket to MOP.

[quote="mathfanatic"]...especially when USAMO would have gotten you a guaranteed ticket to MOP.[/quote]

Mop wasn't that much fun :)

As for the numbered paper -- are we so computer-dependant that we can't write ourselves the list of numbers from 1 to 25?  (Circling on the question paper works, too, but I usually write all over it and make it messy.)

[quote="ComplexZeta"]I don't understand the objection to simply filling in the bubbles as you go. I never even considered doing anything else. And it's easy enough to erase if you mess up.[/quote]

I think I'm with Simon on this one - I've heard all sorts of horror stories from 'bubble at the end' types.  Never heard one from 'bubble as you go'ers.

...which is exactly why we bubble-at-the-end. We like to live on the edge.

I felt sorry for you
[quote="Fierytycoon"]The journey has ended. It's all over. I pulled a 5 on the AIME II a couple days ago. A mere 5. 5. I won't be making USAMO this year, with my 126.5 + 10*5 = 176.5 index. I don't know what I did wrong. I just broke, cracked under the pressure of the event. I couldn't really think clearly. I always seemed to feel so desperate during those 3 hours, guessing around, hoping to stumble upon a correct answer. Desperate.

I feel so empty inside now. So plain. So dull. As if there is nothing to do anymore, no purpose in my life. I guess I could go do something leisurely and fun, maybe something that I had always wanted to do. Maybe I will learn the art of pencil spinning, or perhaps learn to moonwalk, or perhaps flirt with girls. I don't know. My life seems meaningless right now.

I feel like a gray wisp, sitting here at my desk, contemplating, facing the harshness of reality and the heaviness of my responsibilities in society. And what heavy burdens those are, those burdens that society places on individuals. College admissions, social obligations, romantic difficulties. Sigh. But wisps should not feel weight, should they?

I don't know how frequently I will be visiting this forum in the near future. I don't know what I'm going to do. I just want to thank everyone on this board who has helped me out at one point or another in my journey. I particularly want to thank Mr. Rusczyk and Mr. Crawford for all of the help they have given me, and for all of the wonderful resources they have provided with this site and with their books. I also want to thank all of the users who, out of their own good will, come here to help younger math students like me with their problems. People like Simon, Joel, Ravi B, gauss202, to name only a few. They don't really need any help with math from any of the other users here. They just come here during their leisure time and give some younger kids a little boost..just because they feel like it.

And this is not one of those artificial "thanks" that people often say out of mere courtesy when someone holds the door for them, or when someone gives them driving directions. I really mean it. I wouldn't have the same versatility in thinking as I have now, the same agility in solving problems as I have now, and the same love and passion of mathematics that I have now without everyone's help.

Thank you.

*emotional tears*[/quote]

Don't worry about it.  AIME is important, but there is still other math competitions there such as ARML.  If you can do good on those it is very good, too.  Besides,  I am sure next year you will get into USAMO if you continue work hard.

[quote="Fierytycoon"]The journey has ended. It's all over. I pulled a 5 on the AIME II a couple days ago. A mere 5. 5. I won't be making USAMO this year, with my 126.5 + 10*5 = 176.5 index. I don't know what I did wrong. I just broke, cracked under the pressure of the event. I couldn't really think clearly. I always seemed to feel so desperate during those 3 hours, guessing around, hoping to stumble upon a correct answer. Desperate.

I feel so empty inside now. So plain. So dull. As if there is nothing to do anymore, no purpose in my life. I guess I could go do something leisurely and fun, maybe something that I had always wanted to do. Maybe I will learn the art of pencil spinning, or perhaps learn to moonwalk, or perhaps flirt with girls. I don't know. My life seems meaningless right now.

I feel like a gray wisp, sitting here at my desk, contemplating, facing the harshness of reality and the heaviness of my responsibilities in society. And what heavy burdens those are, those burdens that society places on individuals. College admissions, social obligations, romantic difficulties. Sigh. But wisps should not feel weight, should they?

I don't know how frequently I will be visiting this forum in the near future. I don't know what I'm going to do. I just want to thank everyone on this board who has helped me out at one point or another in my journey. I particularly want to thank Mr. Rusczyk and Mr. Crawford for all of the help they have given me, and for all of the wonderful resources they have provided with this site and with their books. I also want to thank all of the users who, out of their own good will, come here to help younger math students like me with their problems. People like Simon, Joel, Ravi B, gauss202, to name only a few. They don't really need any help with math from any of the other users here. They just come here during their leisure time and give some younger kids a little boost..just because they feel like it.

And this is not one of those artificial "thanks" that people often say out of mere courtesy when someone holds the door for them, or when someone gives them driving directions. I really mean it. I wouldn't have the same versatility in thinking as I have now, the same agility in solving problems as I have now, and the same love and passion of mathematics that I have now without everyone's help.

Thank you.

*emotional tears*[/quote]

I've tried bubbling at the end, but I always either forget to or somehow bubble wrong.  I see an A and bubble a B.  So I've given up on that.  Erasings fine with me.

I bubble in on the AIME about an hour from the end, because I change my answer on my paper so frequently that I don't want to take the chance of a misread from the scanner.

I've never bubbled incorrectly either.

On the AMC, I always check my answers (including bubbling) at least once briefly to reduce the chance of careless mistakes.  Since I rarely have time to look back, I figure that I might as well check while I still know how to do the problem.

On the AIME, I still think it's a good idea to check while you're on that problem, because it's still fresh in the mind, and if you're as messy as I am, you won't be able to even see what your numbers were afterward.  Since the AIME is a whole 3 hours, you might as well double-, triple-, and quadruple-check your bubbling; usually, 5 or so minutes lost doesn't affect your score at all, except perhaps save you a few mistakes.

Thanx.  That's useful advice for me.  I've always checked right after I've done the problem and that hasn't worked so well.

[quote="Nukular"]I don't think you should check immediately after doing the problem, since it IS still fresh in your mind and chances are you are going to make the same mistake over again. I never catch a mistake immediately after doing the problem.[/quote]

I disagree, but mostly because on the AMC it is quite possible to do every question 2 different ways.


MATHCOUNTS AMC AIME FTW LaTeX inequalities geometry

Many of you participated in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208578]10 Round Tournament[/url] last year, hosted by isabella2296.  Since I quite enjoy stealing ideas, and summer has started (or is just around the corner), I am hosting one.

[size=200]THE RULES[/size]

10 rounds, 10 problems each.  Simple enough.  Each round consists of 8 short-answer problems and 2 problems that require a solution or proof.  Short-answer problems are worth 2 points (all-or-nothing) and proof problems are worth 5 points (partial credit may be granted).  Each round, a certain number of people will be eliminated.  Your round score is out of 26, and exactly half of it carries on to the next round.

For example, suppose Danny Oh scored 24 points in the first round.  Exactly half of that (12) carries to the second round.  In the second round, Danny scored 20 points.  His total round two score is 12+20=32, and exactly half of that (16) carries on to the next round.

The amount of people eliminated each round will be decided when the sign-ups end.  It would also be helpful if I had a helper to help run this tournament.

[b]Sign-ups start now, and will end on June 14, 11:59:59 PM PST.  Please do not sign-up unless you are committed to doing the problems.[/b]

[size=50]
good luck, kids[/size]

I'll join!

[size=59]gee, kids...[/size]

[quote="yaofan"]I join i guess. What level are the problems?[/quote]

The first round will look somewhat like this:

1. MOEMS level problem
2. Easy AMC 8 level problem
3. Medium AMC 8 level problem
4. Medium AMC 8 level problem
5. Medium State MATHCOUNTS level problem
6. Medium State MATHCOUNTS level problem
7. Medium State MATHCOUNTS level problem
8. Hard State MATHCOUNTS level problem
9. Easy - Medium AIME level problem
10. Medium AIME level problem

Each round will get a little harder than the previous one.

[quote="PowerOfPi"]	I think I will join. When will it start? Does it start on June 15? I am currently on vacation and will get back home on June 11. If it starts before June 11, then I can't do that part.[/quote]

Yes, the tournament starts on June 15.

[quote="rd5493"]yeah, i'll join, though I will be on vacation during the week of june 15th, but i should have a laptop handy. will u PM us the problems or what?[/quote]

Problems will be posted in-thread, and a reminder will be PMed to everyone.

[b][hide="Signups (13)"]
$ \begin{tabular}{|c|c|c|} \hline Place & Username & Score \\
\hline 1 & ytrewq & 0 \\
\hline 1 & batteredbutnotdefeated & 0 \\
\hline 1 & yaofan & 0 \\
\hline 1 & PowerOfPi & 0 \\
\hline 1 & Fermatsmymom & 0 \\
\hline 1 & rd5493 & 0 \\
\hline 1 & akhil0422 & 0 \\
\hline 1 & isabella2296 & 0 \\
\hline 1 & mathfreak123 & 0 \\
\hline 1 & nikeballa96 & 0 \\
\hline 1 & varunrocks & 0 \\
\hline 1 & jerry125 & 0 \\
\hline 1 & pytheagle & 0 \\
\hline \end{tabular}$
[/hide][/b]

[b]Helper[/b]: Twin Prime Conjecture

[size=75]By the way, does anyone else think it's amusing that the sub-forum "Middle School Tournaments" is under "High School Contests & Programs"?[/size]

I think credit to me should be bigger. Like, size-24 font. :D Oh, and I join.

[size=9]good luck, kids[/size]

[quote="akhil0422"]
I wanna join.

And can I be a helper?[/quote]

You can either join [b]xor[/b] be a helper.

... You'll probably be better off [b]without[/b] me as a helper, and though I going to fail at this, I'll join.

i might actually do this.
[size=9]good luck, kids [/size]
JOIN.

I will join.
I AM GOING TO TOTALLY FAIL.

No, they're never allowed.  I doubt they'd give you much of an advantage on this round anyway...

Still waiting on the first submission...

I'm sorry... I have really limited access to internet at AMSP... I can't print anything or use a computer for longer than 15 minutes at a time... so... i think i'm gonna have to drop out b/c i'm never going to be able to take the test. Worrying about the class tests is hard enough!!

@AIME15: I might post the tests once I get home... or one by one on my blog... i'm too lazy to do it right now. Here is a quick overview.

Comp Geo: easy for the first week. your teacher will be Zuming's mom, who is a really nice teacher. you will be doing mc problems though...

Mod Arithmetic: a little challenging... not really hard... takes paying attention to in class, though.

I have to drop out. I don't really have time for this anymore.

Blargh, you guys don't need to post to drop out, just don't send in solutions.

[hide="Problems and Solutions"]
1. In triangle $ ABC$, $ D$ is the midpoint of $ \overline{BC}$.  Given that $ \angle BAD=15^\circ$ and $ \angle CAD=30^\circ$, evaluate $ \tan C$.

[hide="Solution"]
Let $ BD=CD=a$ and $ AC=b$.  Applying the Law of Sines to triangles $ ACD$ and $ ABCD$ leads to the equations $ 2a=\frac{b}{\sin (C+30)}$ and $ 2a\sqrt{2}=\frac{b}{\sin (C+45)}$.  Dividing these gives $ \sqrt{2}=\frac{\sin (C+30)}{\sin (C+45)}$.  Expanding and simplifying gives $ \cos C=(\sqrt{3}-2)\sin c \implies \tan C=\frac{1}{\sqrt{3}-2}=\boxed{-2-\sqrt{3}}$.
[/hide]

2. Let set $ S=\{a_1,a_2,a_3,\dots,a_{12} \}$ where all $ 12$ elements are distinct.  We wish to form sets each of which contain one or more of the elements of set $ S$ (including the possibility of using all the elements of $ S$).  The only restriction is that the subscript of each element in a specific set must be an integral multiple of the smallest subscript in the set.  For example, $ \{a_2,a_6,a_8\}$ is an acceptable set, as is $ \{a_6\}$.  How many such sets can be formed?

[hide="Solution"]
If the element of lowest subscript in a set is $ a_1$, we can accept or reject each of the other $ 11$ elements, which produces $ 2^{11}$ possible sets where $ a_1$ has the lowest subscript.  If the element of the lowest subscript is $ a_2$, we can accept or reject $ a_4,a_6,a_8,a_{10},a_{12}$ yielding $ 2^5$ possible sets.  If $ a_3$ has the lowest subscript, there are $ 2^3$ sets - with $ a_4$, we have $ 2^2$ sets, with $ a_5$ or $ a_6$ we have $ 2^1$ sets each, and for $ a_7$ through $ a_{12}$ we have $ 2^0$ sets each.

The answer is thus $ 2^{11}+2^5+2^3+2^2+2(2^1)+6(2^0)=\boxed{2102}$.
[/hide]

3. For each nonnegative integer $ n$, define the nonnegative integer $ \overline{n}$ as follows:

(a) $ \overline{0}=0,\overline{1}=2,\overline{2}=1$.
(b) When $ n$ is represented in base three as $ n=a_0+a_1 \cdot 3^1 + a_2 \cdot 3^2 + \dots + a_r \cdot 3^r$, where $ a_i \in \{0,1,2\}$, then $ \overline{n}=\overline{a_0}+\overline{a_1} \cdot 3^1 + \overline{a_2} \cdot 3^2 + \dots + \overline{a_r} \cdot 3^r$.

Prove that for all nonnegative integers $ n$, $ \frac{1}{2}n \le \overline{n} \le 2n$.

[hide="Solution"]
Let $ I$ be the set of $ i$ such that $ a_i \neq 0$.  Then:

\[ n+\overline{n}=\sum_{i \in I} a_i3^i + \sum_{i \in I}\overline{a_i}3^i=3\sum_{I}3^i \le 3n\]

with equality iff $ n$ has all $ 0$'s and $ 1$'s as digits.  Subtracting $ n$ from both sides of the inequality $ n+\overline{n} \le 3n$ yields $ \overline{n} \le 2n$.

Similarly,

\[ n+\overline{n}=\frac{3}{2}\sum_{I}2 \cdot 3^i \ge \frac{3}{2}n \implies \overline{n} \ge \frac{1}{2}n.\]
[/hide]

4. In triangle $ ABC$, choose any points $ K \in BC, L \in AC, M \in AB, N \in LM, R \in MK, F \in KL$.  If $ E_1,E_2,E_3,E_4,E_5,E_6,E$ denote the area of triangles $ AMR,CKR,BKF,ALF,BNM,CLN,ABC$ respectively, show that $ E \ge 8\sqrt[6]{E_1E_2E_3E_4E_5E_6}$.

[hide="Solution"]
Let $ E(XYZ)$ stand for the area of a triangle $ XYZ$.  We have:

\[ \frac{E_1}{E}=\frac{E(AMR)}{E(AMK)} \cdot \frac{E(AMK)}{E(ABK)} \cdot \frac{E(ABK)}{E(ABC)}=\frac{MR}{MK} \cdot \frac{AM}{AB} \cdot \frac{BK}{BC}\]

so

\[ \left (\frac{E_1}{E} \right )^{1/3} \le \frac{1}{3} \left (\frac{MR}{MK}+\frac{AM}{AB}+\frac{BK}{BC} \right )\]

Similarly, we obtain $ \left (\frac{E_2}{E} \right )^{1/3} \le \frac{1}{3} \left (\frac{KR}{MK}+\frac{BM}{AB}+\frac{CK}{BC} \right )$.

Therefore, $ \left (\frac{E_1}{E} \right )^{1/3} + \left (\frac{E_2}{E} \right )^{1/3} \le 1 \implies \sqrt[3]{E_1}+\sqrt[3]{E_2} \le \sqrt[3]{E}$.  Analogously, $ \sqrt[3]{E_3}+\sqrt[3]{E_4} \le \sqrt[3]{E}$ and $ \sqrt[3]{E_5}+\sqrt[3]{E_6}\le \sqrt[3]{E}$, so:

\begin{align*}
8\sqrt[6]{E_1E_2E_3E_4E_5E_6}&=2\left (\sqrt[3]{E_1}\sqrt[3]{E_2} \right )^{1/2} \cdot 2\left (\sqrt[3]{E_3}\sqrt[3]{E_4} \right )^{1/2} \cdot 2\left (\sqrt[3]{E_5}\sqrt[3]{E_6} \right )^{1/2}
\\ & \le \left (\sqrt[3]{E_1}+\sqrt[3]{E_2} \right ) \cdot \left (\sqrt[3]{E_3}+\sqrt[3]{E_4} \right ) \cdot \left (\sqrt[3]{E_5}+\sqrt[3]{E_6} \right )
\\ & \le E.
\end{align*}
[/hide]

5. The matrix

\[ \left ( \begin{tabular}{ccc}a_{11} & \cdots & a_{1n} \\ \vdots & \cdots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{tabular} \right )\]

satisfies the inequality $ \sum_{j=1}^{n}|a_{j1}x_1+\dots+a_{jn}x_n | \le M$ for each choice of numbers $ x_i$ equal to $ \pm 1$.  Show that

\[ |a_{11}+a_{22}+\dots+a_{nn}| \le M.\]

[hide="Solution"]
We use induction.  The statement is trivial for $ n=1$.  Suppose it holds for $ n=k$.  We consider $ n=k+1$.  From the given condition, we have

\[ \sum_{j=1}^k |a_{j,1}x_1+\dots+a_{j,k}x_k+a_{j,k+1} |+ |a_{k+1,1}x_1+\dots+a_{k+1,k}x_k+a_{k+1,k+1}| \le M\]

and

\[ \sum_{j=1}^k |a_{j,1}x_1+\dots+a_{j,k}x_k-a_{j,k+1} |+ |a_{k+1,1}x_1+\dots+a_{k+1,k}x_k-a_{k+1,k+1}| \le M\]

for each choice of $ x_i = \pm 1$.  Since $ |a+b|+|a-b| \ge 2|a|$ for all $ a,b$, we have

\[ 2\sum_{j=1}^k |a_{k1}x_1+\dots+a_{jk}x_k|+2|a_{k+1,k+1}| \le 2M\]

so

\[ \sum_{j=1}^k | a_{j1}x_1+\dots+a_{jk}x_k| \le M-|a_{k+1,k+1} |.\]

By the inductive assumption, $ \sum_{j=1}^k|a_{jj}| \le M-|a_{k+1,k+1}|$, which is equivalent to the desired inequality.
[/hide]
[/hide]

[hide="Scoreboard"]
\begin{tabular}{|c|c|c|c|}\hline Place & Username & Score & Round 4 Score \\
\hline 1 & fortenforge & 25.125 & 13 \\
\hline 2 & Thunder365 & 20.25 & 4 \\
\hline 3 & Obscuron & 19.75 & 0 \\
\hline 4 & PerpetualMotion & 16 & 7 \\
\hline 5 & connaissance & 13.75& 0 \\
\hline 6 & vijaymenon & 11.875 &  0\\
\hline 7 & lotsofmath & 11.75 & 0 \\
\hline 8 & srfan & 10.375 & 0 \\
\hline\end{tabular}
[/hide]

[hide="Analysis"]
[hide="Problem 1"]
No submissions received at all for this problem.  Personally, I didn't think it was too difficult - mostly plug and chug formulas.
[/hide]
[hide="Problem 2"]
All 3 submitters answered this problem - only one of them correctly.  Once you realize you can count these based on the smallest subscript, the problem becomes arithmetic.
[/hide]
[hide="Problem 3"]
Two submissions.  They used different methods (a little similar, though) but none were completely rigorous.
[/hide]
[hide="Problem 4"]
No submissions.  The notation obviously scared off everyone.
[/hide]
[hide="Problem 5"]
Thunder365 submitted a solution to this - very unrigorous, but the idea worked.
[/hide]
[/hide]

Given the lack of submissions, I'm going to call this a finished tournament.

Congratulations to fortenforge, Thunder365, and Obscuron for placing in the top 3!  Unfortunately, I am too poor to give you guys smiley trophies, so you guys will have to do with :P.

dam, i thought it was due the 15th, could you post round 5 if u already finished writing it as a practice set of problems?

[quote="AIME15"]Given the lack of submissions, I'm going to call this a finished tournament.[/quote]
4 round tournament

YES!!!! I knew this was going to happen :) I made a bet with my friend that it would end before 5th round :) I knew it!

Haha, give me half the money :P

Who was the bet with? pingala? MChow? bowser? ilovepink? miller4math? EcstaticPotter? Some other Miller kid?

heh, maybe u didn't realize that perpetual motion scored a 23?

[quote="AIME15"]

Who was the bet with? pingala? MChow? bowser? ilovepink? miller4math? EcstaticPotter? Some other Miller kid?
[/quote]

hmm, did i mention that i don't go to miller???  :P


Anyone kno the results for the Indiana ISSMA State Marching Band Competition???

that's last year...o well...i guess i can wait like everyoen else and find out in the paper 2moro morning. lol


Find determinat of matrix $D_n=[a_{ij}]$ which $a{}_i{}_j=\binom{n+i}{2j-1}$. Find $\det D_n$

If $k>m,$ should we interpret ${m \choose k}$ as zero?

Preliminary computations - no claim of general result yet.

$n=1: \ |2|=2$

$n=2: \ \left|\begin{array}{cc}3&1\\4&4\end{array}\right|=8$

$n=3: \ \left|\begin{array}{ccc}4&4&0\\5&10&1\\6&20&6\end{array}\right|=64$

$n=4: \ \left|\begin{array}{cccc}5&10&1&0\\6&20&6&0\\ 7&35&21&1\\ 8&56&56&8\end{array}\right|=1024$

That suggests a conjecture: that the determinant is $2^{T_n}$ where $T_n=\frac{n(n+1)}2.$

I'm trying to figure out how to apply [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=57722]this[/url], but I haven't figured out how to count the nonintersecting paths. The model I'm using:
Consider a lattice in the plane. Put $A_i$ at $(0,n-i)$. Put $B_j$ at $(2j-1,2n-2j+1)$. The allowed paths have $x$ and $y$ both increasing.

Let $L_n$ be an alternating-sign lower-triangular chunk of Pascal's triangle. For instance:

$L_4=\begin{bmatrix}1&0&0&0\\-1&1&0&0\\1&-2&1&0\\-1&3&-3&1\end{bmatrix}.$

Then $L_nD_n$ has some nice properties.

For instance, $L_4D_4=B=\begin{bmatrix}1&10&5&0\\5&10&1&0\\0&1&10&5\\ 0&5&10&1\end{bmatrix}$

In this case the eigenvalues of $B$ are $2,4,8,16.$

I conjecture that this pattern continues. I don't have the proof of that.


geometry
geometric transformation
reflection
induction
symmetry
3D geometry
sphere

Tetris is awesome. So simple, yet so fun.

Ive always wondered something about it. See, there's been several variations on this theme, pieces with more blocks. In the original, there were only 7 pieces: straight, Z, Z-reverse, L, L-reverse, square, and T.

With 3-block pieces there are 2 ways,
With 4-block pieces there are 7 ways,
With 5-block pieces there are 12 ways,
With 6-blocks how many?
7-blocks? 100 blocks?

Is there a formula for n ways to arrange x blocks all touching?

I don't think i would want to play 100-block tetris.

[quote="Hexaditidom"]With 6-blocks how many?
7-blocks? 100 blocks? [/quote]

[url=http://www.mathpuzzle.com/polyom.htm]This[/url] Web site should answer some of your questions.  Way down the page, they claim "A polyomino is a figure made of solidly connected squares.  There is 1 omino (a square), 1 domino, 2 triominoes, 5 tetrominoes, 12 pentominoes, 35 hexominoes, 108 heptominoes, 369 octominoes."

This is sequence [url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000105]A000105[/url] in the Encyclopedia of Integer Sequences. I couldn't find any sequences beginning with 1, 1, 2, 7, 12.

Hm.  Hexa's question is interesting, because unlike the standard polyomino problem, mirror reflections are counted as different.  I wonder if it's possible to get an asymptotic formula : probably one would come from one for the polyominoes, because almost all polyominoes are asymmetric.

[quote="ComplexZeta"]Ok this sequence actually starts 1, 1, 2, 7, 19 I think. Can anyone figure out the sixth term?[/quote]

Actually it's 1, 1, 2, 7, 18. Reference: http://mathworld.wolfram.com/Polyomino.html

In general this question is very hard. I don't think there's a formula for the exact number of distict n-polyominoes (either with or without reflections).

When I was in high school I tried finding asyptotic bounds on the 3-dimensional version of this problem.. with very little success.

One possible idea is to use induction.  I don't know how to do it, but it's a possibility...

Edit (reply to JBL): I don't mean strictly induction, but the general idea, but anyways, it's true that it's hard to add squares, because how many ways are there to add a square to different polyominoes and end up with the same polyomino?  There could be a lot...

Induction only works if you know what kind of answer you're looking for.  Also, adding squares seems like a not-too-nice inductive step: there are many, and not always the same number, of ways to get each piece from smaller pieces.

Here's a question, though.  After you get to n = 8, a new problem arises:  can you have holes in pieces?  At least from a Tetris point of view, that's a rather significant question.  I wonder how much it gets limited if we require that our pieces be topologically equivalent to a disk, or whatever.

Ok so the sequence is [url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000988]A000988[/url].

We can even sort of have holes with n=7. It would be annoying to play tetris with the pieces with holes, but I think pieces that look like Z or S are annoying anyway.

"I play tetris with only the square blocks" :)

I shall go look at this sequence and see.

I had worked on this problem quite a bit once. It always surprises me that nobody has been able to find a formula for this question (or symmetry not allowed or holes allowed or etc etc). It seems so "simple" in a way, but theres no formula. Theres a real good book about polyominoes: its called "Polyominoes". :D. But really, its a good book, by Solomon Golomb.

[quote="Hexaditidom"]100 blocks?[/quote]

I don't think that they make numbers this big....

Ah, they make numbers much bigger. There's a sequence that begins with 1, 6, 21, 107, but then the fifth term is known to be at least 47176870, and the sixth term is known to be at least 8690333381690951. More fun stuff can be found [url=http://www.cs.berkeley.edu/~aaronson/bignumbers.html]here[/url].

I'm pretty sure Busy Beaver numbers are bigger than Ramsey numbers. I need convincing that Busy Beaver numbers are larger than Ackermann numbers though. That's probably because I don't know anything about the halting problem.

Ah.  I was wondering whether the numbers you posted were the Ramsey numbers, actually.  Busy Beaver numbers ... remember the name, not what they do -- something with Turing machines, right?  I can't imagine anything that goes up much like Ackermann numbers, though.

Hm. I wonder if Van der Waerden's bounds for his proof of the Van der Waerden theorem increase anything like that fast.

I just thought of something related. Anyone play Tetrisphere, the N64 game?

Its not played like normal tetris. You have a sphere covered in 7 different blocks, with 1-unit squares to fill in the gaps. You get an aerial view from the piece being dropped, and you have to drop it on an identical piece that touches one other identical piece. If you cant find any, you can slide pieces and break the 1-unit squares to bring them together. Theyre in layers and your goal is to expose enough of the core to blow the rest of it apart.

Anyway. Since they're square pieces, and they sit on a sphere, i assume the sphere is a cube with the corners squashed in. Then, wouldn't there be a spot somewhere where the corners were, with 3 sides border each other? I never found that spot. Unless it isnt a sphere at all...

Not if they're spherical squares, necessarily -- the sphere doesn't necessarily have to come from a cube like that.  You might be able to completely cover a sphere with four-sided shapes with 4 meeting at each corner.  It just wouldn't have to come from a cube.

It seems I was right-- it isn't a sphere at all.
It's really a 2 dimensional map, which wraps around at the sides. Since it's shaped like a ball, you can only see a small part of this map, which makes a clever illusion.


geometry
circumcircle
geometric transformation
homothety
ratio
angle bisector
perpendicular bisector

A line connecting the feet of perpendiculars from orthocenter to the angle bisectors
   of angles A passes thru the midpoint of BC in triangle ABC.

   Col. Pestich.

A quick angle chase reveals that this line is parallel to $AO$, where $O$ is the circumcenter, and since it passes through the midpoint of $AH$, it must be the image of the diameter of $(ABC)$ passing through $A$ through th4e homothety of center $H$ and ratio $\frac 12$, meaning that it's the diameter of the nine-point circle which passes through the midpoint of $AH$. From here we get the conclusion that it passes through the midpoint of $BC$.

[quote="pestich"]A line connecting the feet of perpendiculars from orthocenter to the angle bisectors
   of angles A passes thru the midpoint of BC in triangle ABC.[/quote]

I remember having posted a solution of this problem, but either I can't find it again or it got lost during the hacker attack. Anyway, the problem is really eeeeaaasy:

Let H be the orthocenter and A', B', C' the feet of the three altitudes of a triangle ABC. Then, the problem will be trivial as soon as we will succeed to show that the following three facts hold:

[b]1.[/b] The foot of the perpendicular from the orthocenter H of triangle ABC to the internal angle bisector of the angle CAB lies on the perpendicular bisector of the segment B'C'.
[b]2.[/b] The same holds for the foot of the perpendicular from the orthocenter H of triangle ABC to the external angle bisector of the angle CAB.
[b]3.[/b] The perpendicular bisector of the segment B'C' passes through the midpoint of the side BC of triangle ABC.

Indeed, proving [b]1.[/b] is very simple: If we call W the foot of the perpendicular from the orthocenter H of triangle ABC to the internal angle bisector of the angle CAB, then we have < AWH = 90°; togehter with < AB'H = 90° and < AC'H = 90°, this yields that the points W, B' and C' lie on the circle with diameter AH. Since the point W lies on the angle bisector of the angle CAB, we have < C'AW = < WAB'; in other words, the chordal angles of the chords C'W and B'W in the circle with diameter AH are equal. Hence, these chords themselves are also equal; in other words, we have C'W = B'W, and we conclude that the point W lies on the perpendicular bisector of the segment B'C'.

The proof of [b]2.[/b] is analogous to the proof of [b]1.[/b].

Finally, establishing [b]3.[/b] is more than trivial: Since < BB'C = 90° and < BC'C = 90°, the points B' and C' lie on the circle with diameter BC; hence, the perpendicular bisector of the chord B'C' of this circle must pass through the center of this circle, i. e. the midpoint of the segment BC.

And the problem is solved.

  Darij


geometry
3D geometry
number theory
greatest common divisor

How would you solve problems like these:

1) For $ 5a^3 \equal{} 6b^4$ find the minimun value of (a,b) with a and b positive integers. (Yes; it is on the classes section)

2)Solve $ 3n^2\equal{}m^3$  for m and n positive integers.

I tried to solve them using modular algebra. Can you solve it step by step and clearly?

Thanks :lol:  :lol:  :lol:

[hide="hint"]make a series of substitutions until you can draw a conclusion[/hide]

For the first

[hide]

$ a$ will have to be equal to $ (6^{3n})(5^{3m})$ and $ b$ will have to be equal to $ (6^{4x})(5^{4y})$ because $ a$ is a perfect cube and $ b$ is to the fourth power.

When you multiply two numbers with the same base, the exponents add.  So you need to examine the equations $ 3n \equal{} 4x \plus{} 1$ and $ 3m \plus{} 1 \equal{} 4y$ and find the smallest values you can for $ (n,x)$ and $ (m,y)$

[/hide]

winternights...I don't understand your point. For example substituding "a" for 2 we get 40 which is not a multiple of a power of 6. Substituding by 3 we obtain 135 which isn't a multiple of any power of 6.

What do you mean it's in the classes section?
Just to clarify, $ (a,b)$ is the gcd of $ a$ and $ b$?

For both questions, you need to use the theorem: If $ p$ is a prime such that $ p|ab$, then $ p|a$ or $ p|b$. Then remember what it means for one number to divide another, write more equations and substitutions, and look for stuff to cancel.

@winternights: I think you reversed your thinking or made a typo. $ a$ itself is not necessarily a third power; we are interested in the third power of $ a$. Similar comment for $ b$.

Bictor717, I mean this: http://www.artofproblemsolving.com/Classes/IntroNumbers/PostTest.pdf

I can't still get to the solution :wink:  :wink:  do someone have the solution?

What about the second problem?

[hide="1)"]
$ a\equal{}6m$
$ b\equal{}5n$

$ 5\cdot 6^3m^3\equal{}6\cdot5^4n^4$

$ 6^2m^3\equal{}5^3n^4$

$ m\equal{}5k$
$ n\equal{}6l$

$ 6^2\cdot 5^3k^3\equal{}5^3\cdot 6^4 l^4$

$ k^3\equal{}6^2l^4$

now the left side is a perfect cube, so the right side must be as well and least integer satisfying the condition is $ l\equal{}6$ implying $ k\equal{}36$. Going back through substitutions gives us $ a\equal{}1080, \ b\equal{}180$[/hide]

[hide="2)"]
$ m\equal{}3k$

$ 3n^2\equal{}3^3k^3$

$ n^2\equal{}3^2k^3$

$ n\equal{}3l$

$ 9l^2\equal{}9k^3$

$ l^2\equal{}k^3$

We have equality only if $ l\equal{}p^3, \ k\equal{}p^2$ which gives us all ordered pairs $ (m,n)\equal{}(3p^3,3p^2), p \in \mathbb{N}$[/hide]

Yea... I can see that being confusing.  When I wrote $ a$ and $ b$ I meant the third power of $ a$ and the fourth power of $ b$.

That said, it doesn't seem like too big of a leap in imagination to piece together what I meant.

It gives the same answer as flame's approach.

Well our approaches are essentially the same. The difference is I take it step by step, and you find it immediately using modular arithmetic.


All terms are positive integers

1.Prove that for primitive pythagorean triples (where a^2+b^2=c^2 and a b and c share no common factor), a=m^2-n^2, b=2mn, and c=m^2+n^2 for some integers m and n.

2. Prove fermat's last theorem for case 4(a^4+b^4  <> c^4).

3. Prove that a circle inscribed in a pythagorean right triangle has a integer radius

4. Prove that sqrt(a)+sqrt(b) <> c when a isnt a perfect square

5. Prove that sqrt(a)+sqrt(b)+sqrt(c) <> d when a isnt a perfect square(
In case any of these are by chance contest problems, the first 3 are from Excursions in Number Theory (written in 1991), and the last two are made up by me. The last two arent really number theory but o well
-edit- last two problems edited

yif man12 wrote:I don't think number 3 is right...

[hide]We know that the triangle has a 90 degree angle. Thus, the arc that the two legs make have an angle of 180 degrees. This means the line connecting the endpoints of the legs is the diameter. That means the hypotenuse is equal to the diameter. But the radius is half the diameter, which means it could be a fraction, like in a 3-4-5 triangle.[/hide]

[hide]Actually, I think #3 is right, assuming that the original poster meant a pythagorean triple (i.e. a right triangle with integer side lengths).  Recall that all such triples can be generated with suitable choices for integers m > n, where a = m^2 - n^2, b = 2mn, and c = m^2 + n^2.  (Actually, this is #1.)  From the familiar formula rs = area, where r is the radius and s is the semiperimeter, we have r = (ab/2)/[(a+b+c)/2] = ab/(a+b+c) = [(m^2-n^2)*2mn]/(m^2 - n^2 + 2mn + m^2 + n^2) = [2mn(m+n)(m-n)]/(2m^2 + 2mn) = n(m-n), which is an integer if m and n are integers.

[/hide]

[quote="yif man12"]Oh! The circle's inscribed in the triangle, not the other way around! My bad, my bad.[/quote]

Hm, now we're even.

tetrahedr0n wrote:Quote:4. Prove that sqrt(a)+sqrt(b)  sqrt(c) when a isnt a perfect square

Is this for arbitrary integers a,b,c? Because

[hide] sqrt(2) + sqrt(8) = sqrt(18) [/hide]

Am I missing something?

problems edited

Few comments: Well known:

Prob 3: Actaully it is even stronger ;  a^4+b^4  can not be a perfect square ( let alone 4th power)

Prob 3:   [hide] r= (a+b-c)/2  (and of course (a+b-c)  is even)  [/hide]


I know this is a worn-out topic, and you'd think anything to say has already been said. however, today I got my pins, (i'm assuming there's 57, but I haven't counted) and I was wondering if anyone else has gotten thiers yet. The illinois ones have a picture of the sears tower at night on them. Have you gotten your pins (if you're going to nats)? If so describe them!

Haven't gotten them yet. However, last year there was a skier pin, Rocky Mountains pin, and a Colorado flag pin. Can you guess which state I'm from, then?

We haven't even talked about them yet, or anything else we're actually going to do at Nationals for that matter, I think it's because we don't practice enough.

[quote="Phelpedo"]Good luck trying to trade your pins, Ignite. If I know CT, you'll be getting the same "We're full of surprises!" pins from 10 years ago. Kinda ironic, eh?[/quote]

They give the same pins every year to CTers? -_- I might bring my Odyssey of the Mind pins just to shake things up a bit :P

Does anybody that went to nats last year have a Utah pin?  Well, if you do, then you know just how lame they were.  I myself had about 15-20 pins left over after the competition.  When I helped at the chapter competition this year, the state coordinator gave me two of the new pins.  This year they might actually be worth something.  First off, they're metal.  They also look much better than a white plastic pin shaped like Utah with [color=red][b][i]Utah[/i][/b][/color] written on it in italics...

If you really want to know how they look, just pm me.


Find all $n$ such that the number $\underbrace{77 \ldots 7}_{n \text{ times}}1$ is prime.

$7n+1 = n+1 (mod 3)$.  Therefore, $n \neq 2 (mod 3)$.

Not sure where to go from there.

I don't think a general guarantee of primality can be made.

[quote="Zellex"]so, cincodemayo, can you tell us your solution?[/quote]

I do not have a solution myself... I just noticed that a few smaller numbers lead to a prime number ($n=1, \ 12$), so that led me to wonder whether or not there is a finite set of numbers $n$ (and what they are).

So... I probably have less of a clue as to how to do this problem than most of you do.  :D

Let the given number be $n$.

By a well known lemma, all primes greater than 3 are congruent to $\pm{1}\bmod{6}$.

We have:
[b]
Case 1: $n \equiv{1}\bmod{6}$[/b]

$7(10^{n}+10^{n-1}+\ldots+1) \equiv{1}\bmod{6}\Rightarrow 7\cdot \frac{10^{n+1}-1}{9}\equiv{7}\equiv{1}\bmod{6}$

$10^{n+1}-1 \equiv{9}\bmod{6}\Rightarrow 10^{n}\equiv{10}\bmod{6}\Rightarrow 10^{n}\equiv{1^{n}}\equiv{1}\bmod{3}$

[b]Case 2: $n \equiv{5}\bmod{6}$[/b]

$7\cdot \frac{10^{n+1}-1}{9}\equiv \frac{10^{n+1}}{9}\equiv{5}\bmod{6}\Rightarrow 10^{n}\equiv 1 \bmod{3}$, which is always true.

Is there a problem in my proof?? Because I get that all possible solutions exist, and this is clearly not the case, because $771$ is not prime...

EDIT: Heheh, I figured out what went wrong... not all numbers in the form $6n\pm 1$ are prime... if that were the case, I'd be rich since I would've solved the Riemann Hypothesis  :D But the above proof shows that all such proposed numbers, $n$, are congruent to $\pm{1}\bmod{6}$

$\frac{70}{9}(10^{n+1}-1)+1$ is relatively prime to any given integer $d$ with $gcd(30, d) = 1$ infinitely often.  It doesn't constitute a guarantee of primality.


geometry
3D geometry
trigonometry
email
algorithm
Mafia

:D
Now let's hope I get one too.  I didn't have school today...so maybe? lol.  Hopefully?  *sigh*

Maybe we should bug Ian.

hehe I'm going too...
not that anyone knows who i am... :ninja:

Still waiting for the invitation :(
Hopefully I'm invited :)

[quote="otaku"]of course you will be invited Mr. IMO[/quote]
When did my last name changed to IMO? :D

If I remember correctly, Jenn started [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30483]this thread[/url] a year ago too. Must be obsessed with the Waterloo Seminar. :D

[quote="lightrhee"]
 Must be obsessed with the Waterloo Seminar. :D[/quote]

How can you not be?!?!?! Its a blast! Hehehehe. It just might be a little different this year, but always fun.

:lol: Im so excited! Last year was awesome.
I still remember walking into the lounge on the first day and seeing a huge crowd of people.  And in the middle . . . David solving a Rubiks cube blindfolded. :o
 :rotfl: And that was my introduction to the "math world"

Just hoping they don't devote the week to trig again. They couldn't have picked a topic I'm worse at, I was so lost  :?

[quote="billzhao"]Just so you know, I've decided not to attend Waterloo Seminar this year. So I won't be able see some of you guys for a long time ...

From a practical perspective, that means that you need to find someone to be the new photographer  :P[/quote]

 :o

Do you realize the amount of people who rely on your stash of pictures to remember some of those moments?!  It just won't be the same!!! Who else isn't lazy enough to take the amount of pictures you do!!!!

Jenn posts on here once per year... to start the seminar thread ;)

Of course, I'll still be in class, so I'll see you all.  I'll definitely make a few visits.  Are you staying in Renison again?

Somebody send me the plan of what's going on when; that way I can make SURPRISE APPEARANCES!

They didn't send me an invitation yet either. I think the evil organizers are just creating suspense and pulling at the strings of our souls. But yay, Seminar! Will be my third, and last. I wonder what they'll replace Sportsworld with.

[quote="billzhao"]Just so you know, I've decided not to attend Waterloo Seminar this year. So I won't be able see some of you guys for a long time ...

From a practical perspective, that means that you need to find someone to be the new photographer  :P[/quote]
Why? :?

[quote]I wonder what they'll replace Sportsworld with.[/quote]

On my invitation, it said that Sportsworld would be on the Friday.  So has it been closed down?

And do Grade 12s get invited to this (I'm in Grade 11)?

Thanks :D

lol... my second post this year!! *pointed glare at Elyot*
To answer some of the questions,

- Elyot: The seminar is between June 11th and 17th, and I think we're staying at Renison again.. Surprise appearances? I thought you would basically stay with us all week!! AND what happened to being the houseparent?
- Sarah: I hope they send you one soon.. If not, we need to start bugging Ian again.. email me sometime!!
- Yufei: ARGH!!!! We need you!!!! The photographs!!
- David: I'm not more obssessed than you guys.. I just happen to live in Waterloo and get the invitations earlier than you since it's through mail..
- Others: Sportsworld is replaced by Bingemans, which is basically a waterpark. I've never been there, but it's probably just as exciting as Sportsworld.. *sarcastic laugh*

Anyways, looking forward to seeing you guys again and meeting new people as well!

think back to seminar... :rotfl:
anyways, if you see Will, ask him about a Minesweeper Flags game...

Me and otaku will keep the forum alive by posting random things in random topics. :rotfl:

you're starting to sound like peng shi!  :rotfl:

uh oh, three consecutive rofl's...

[quote="Elyot"]
Actually, yes they do.  My invitation was mass-emailed to a bunch of former IMO people... I recognized Ralph, Olena, and maybe a few other email addresses from the list.  So perhaps Jacob and the others will be there too.  Of course, some of them might be doing other things, or might be on co-op terms.[/quote]

What email address was it?  If I had ever received a message I would have come, but I never even knew which week it was (last year I stopped by a week early).

Dunno... that email is long gone.  If you've changed your email in the last couple of years, they wouldn't know.

I vaguely remember waiting for you to show up last year, then you never did, and we were EXTREMELY disappointed..  :D

I guess that counts as an excuse.. sorta..

Nice to see you join this forum, Ralph!!

[quote="rem"][quote="andy_tok"]Seminar was quite fun.[/quote]
quite fun?!
Are you kidding me? IT WAS SO MUCH FUN!!!! :)

By the way, who are you? You seem to be new to this forum.[/quote]

I presume his name is Andy  :D

yes. Using the contacts list, I am 99% sure that he is Andy Zhao.
Hi Andy! I am glad you joined Aops :)

Sorry, but I'm not Andy zhao. I'm Yifan Wang.

LOL. :rotfl:  You were wrong, Jenn!
Anyways, welcome to the forum!

it's a misleading username then  :D  :blush:

[quote="andy_tok"]sorry :blush:[/quote]
Welcome to Aops/mathlinks Yifan! :lol:


Hi, I'm BOGTRO. I'm running a tournament, as you can see. See Tourney 5 for the setup. First 8 people get to play. If you sign up, post something along the lines of "Hi, I'm <name>. My highest rating is <number>, and my current is <number>"

Go.

Hi, I don't know what my name is. :( Maybe you can help me figure it out?

My highest rating was 1779, current 1705.

No, if i can count i see izzy joining and maybach. I suggest that joining has a good chance to be disappointed because boggy is kinda inactive.

[quote="Dillinger"]Can i join? highest rating is 1415. Currently is 1167.[/quote]

Okay dude, can you read the posts above yours?

Not enough joiners. I'm about to cancel.

i join.highest 1451 current 1362.im not in america so i hav a different clock with america.check northern marianas island.saipan.

I will join. My highest rating is 1733. Current is 1496.

[quote="cgyao15"][quote="boggy"]
Not enough joiners. I'm about to cancel
[/quote]
READ THE ABOVE POSTS.[/quote]

Wow. If the tourney is being cancelled, mods, please lock this thread. Thanks.


Prove that if $ a, b$ are two positive integers and $ \sqrt{a}$ is irrational then $ \sqrt{a}+\sqrt{b}$ is irrational.

[hide="Solution"] Suppose $ \sqrt{a}+\sqrt{b}= r \Leftrightarrow \sqrt{a}= r-\sqrt{b}$.  Then $ \sqrt{a}$ has the minimal polynomial over the rationals

$ x^{2}-a$

From the LHS, but it also has the minimal polynomial over the rationals

$ (x-r)^{2}-b$

Which, since it has a linear term, is clearly different (unless $ r = 0$, but since $ \sqrt{a}+\sqrt{b}> 0$ this is absurd).  But the minimal polynomial is unique; contradiction. [/hide]


[hide="I hate solving quadratics, so I won't"]The height of the triangle at $x$ is $1-x/9$ for $1\leq x\leq 9$. The first unit has a height of only $1/2-1/18=4/9$, so $A(x)=x-x^{2}/18-1/2$ $A(9)=4$ so we need $A(x)=2$. Multiplying by $18$ we get $x^{2}-18x+27+0$.[/hide]

[quote="Ignite168"]A vertical line divides the triangle with vertices (0,0) (1,1) and (9,1) in the xy plane into two regions of equal area. Find the equation of the vertical line.[/quote]

[hide="solution"]The area of the triangle is $4$ so half the area is $2$.

The equation of the line passing through $(0,0)$ and $(9,1)$ is $y=\frac{1}{9}x$

Let the equation of the vertical line be $x=a$ so the height of the triangle is $\frac{9-a}{9}$ and the base is $9-a$.

$\frac{(9-a)^{2}}{9}=4$

$a=3,15$ but it can be 15 so $\boxed{x=3}$

All these posts are starting to remind me of "Interval" except the problems are actually cooler.[/hide]


An organic compound A on treatment with ethyl alcohol gives a carboxylic acid B and compound C. Hydrolysis of C under acidic condition gives B and D. Oxidation of D with $ KMnO_4$ also gives B. B on heating with $ Ca(OH)_2$ gives E ($ C_3H_6O$) E does not gives Tollen's test and doesnot reduce Fehling's solution but forms 2,4-dinitrophenyl hydrazone. Identify A to E.

A=acetic anhydride
B=acetic acid
C=ethyl ethanoate
D=ethyl alcohol
E=acetone


what does center of a graph mean?i mean what is it? :?:

This is a good modelling exercise, I give you what I think constitues most of the answer .
Suppose you have a finite simple (no orientation) connected graph.
Informally it means you can move from any vertex to any other using only edges).
Now between any pair of vertices {x,y} you can define a distance
as the "smallest way" from x to y (smallest is measured in terms of edges or steps).
You may check yourself that this defines a distance (obeys the distance's axioms) .

Now thinking topologically observe that on any space E finite (compact...) having a distance D, you can define :
 diameter(E) = MAX D(x,y) where x and y run on E
 radius(x)   = MAX D(x,y) where y runs on E
 It can be checked that the same definitions work for a connected graph.
 Now try to find the center definition ( bear in mind that it may be more than one point.

Note1 :  In a tree there are two concepts a bit different : center and centroid look in google.
Note2 : You may also define a distance between edges.
Hope it helps


Well, cool is $clo_{2}$, which is chlorine and 2 oxygen, so why are you looking people who like bleach??? :P

Define "cool". It's not in the Websters Dictionary, so I'm confused.

P.S. If everyone in Mathcounts is cool, then I'm in if this isn't  a cult or something...

[quote="kstan013"]hey everyone i'm starting a cool club and you can only be in it if you're cool so post here anyone who's cool and if you're not cool then BACK OFF[/quote]

Lol people can actually be offended by this club. If you want to start a club then ask the admins to make a "Group" and only send invites to the people you think are "cool".

good idea demon hey are you still a mod cuz if you are you can make the cool group!!!

[quote="junggi"]i think kstan is sugar high[/quote]

ditto. Let's give him sleeping drugs. :P

I'll take them as long as they're not anal

Ok. They're not anal.

*kstan eats them*

Did I say NOT anal? I meant to say that they ARE anal. Oops. :oops:

Well, could someone at least lock this? It's getting spammy.

BACK TO NOSPAMALOT

who wants to be in cool club

oh yes and you arnt cool unles you read a calc book or visit my blog

ok. I visited your blog AND read a calc book. Could you post this in your blog and lock this thing???

DUDE

I'LL MAKE A LOSER CLUB!!!

Or I'll join this cool club.

kstan, I think you meant "cule" club.

uhh... if he did, I'm gonna quit...

P.S. kstan, please put this in your BLOG, and Miths Apprentice, please lock this...

do you want to get banned from the cool club?

Well, what are the benefits of this club??? :roll:

fame...cuz everyone goes oooooooooooooooooo...that 1=2 character is mighty cool

uuhh.... I'm sort of known over AoPS... but... ok.

But will everyone go "OMG kstan is cool!!!! I gotta get into the club!!!!! Please Please Please!!!!!"?

I'm just messing with your mind. Let's have a private message discussion about this....

Dude
i can picture a certain sibling who'd join this club saying "kstan is so cool let me join this club"

Ugh. If you want, make two groups sort of like Red vs Blue. Mass PM all members of people who want to be in the group (group invites). And see if they'll set up a "blue" or "red" forum that is pass word protected or something. This topic is getting spammy and crazy. Locked.


The warm winter weather meant that the average temperature at noon for the school week was 11 degrees warmer than the average temperature of 34 degrees on record for the same 5-day period. The temperature rose throughout the 5 days.

    * Friday's temperature was 10 degrees warmer than twice Monday's temperature.
    * Wednesday's temperature was 3 degrees warmer than the average for the week, but 8 degrees colder than Thursday.
    * Tuesday's temperature was three-fourths of Wednesday's temperature.

What was the noon temperature for each of the 5 days?

This really belongs in classroom math, not brainteasers but...
[hide]
M:25, T:36, W:48, R:56, F:60
[/hide]

In my opinion the fundamental tenets inclusive of the puzzle under reference are fully in conformity with the various provisions governing this sub-forum.

The solution to the problem is provided hereunder as follows.

[hide]NOON TEMPERATURES

Monday.........25 degrees;
Tuesday........36 degrees;
Wednesday....48 degrees;
Thursday:.. ...56 degrees;
Friday............60 degrees [/hide]


number theory
least common multiple
prime factorization

A group of 25 friends were discussing a large positive integer. "It can be divided by 1," the first friend said. "It can be divided by 2," said the second friend. "And by 3," said the third friend. "And by 4," added the fourth friend. This continued until everyone had made such a comment. If exactly two friends were incorrect, and those two friends said consecutive numbers, what was the least possible integer they were discussing?

[hide="Solution"]
It is likely that one first notes that $ 23$ is the largest single number that can be eliminated. However, even if one of its neighbors were removed, their factors would still be contained somewhere else, and thus it would not decrease the number further. We then note that we could eliminate a power of $ 2$ by taking away the $ 16$ and $ 17$, which would divide the number by $ 34$. A $ 3$ can't be eliminated, since both $ 9$ and $ 18$ have two powers of $ 3$, so taking one away would not work, and $ 25 < 3^3 \equal{} 27$, so we can't consider taking away $ 27$. A $ 5$ can only be eliminated if we take away a $ 24$ and $ 25$, but that would only divide by $ 5$. A $ 7$ can't be eliminated by similar logic to that of the $ 3$'s not being able to be eliminated, and similarly, $ 11$ can't be. $ 13$ can be eliminated if we take $ 13$ and then $ 12$ or $ 14$, but that would be useless, since we would still only be dividing by $ 13$. Finally, we get to $ 17$, which we have seen to give a division of $ 34$, and $ 19$, which won't help (only divides by $ 19$). The least common multiple is $ \text{lcm}(1,2,\ldots,25) \equal{} 2^4\cdot 3^2\cdot5^2\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23$, so taking away a $ 2$ and $ 17$ gives a product of $ 2^3\cdot3^2\cdot5^2\cdot7\cdot11\cdot13\cdot19\cdot23 \equal{} \boxed{787386600}$. Darn, I think making and running a program would have been faster.
[/hide]

ugh.. that looks kinda fishy... with that large of an integer...

But i kind of understand your logic! (that's something that doesn't happen very often...  :D )

Thanks!

Umm does anyone else have a solution (less complex, please) to this problem?

I had this today in my mc practice.. It's kind of hard.. I think it should be in the ALL TIME GREATEST MC PROBLEMS...

[quote="O M G"]um it says 2 people are wrong, right...
Oh and math if you take away 2 and 17 that makes 13 people wrong since 12 numbers in between prime factorization has a 2.
But, I could be wrong  :oops_sign:  :wink:[/quote]
[quote="ThinkFlow"]He means 16 and 17.[/quote]

No, I meant "$ 2$ and $ 17$" as in ultimately, you will take away a factor of $ 2$ and a factor of $ 17$. Sorry for not being clear, though.


Let $ P$ be an interior point of $ \triangle ABC$ such that the lines $ AA_{1},BB_{1},CC_{1}$ are concurrent at $ P$ and the points $ A_{1},B_{1},C_{1}$ lie on $ BC,CA,AB$ respectively.Find using vectors or otherwise value of
$ \frac{PA_{1}}{AA_{1}}\plus{}\frac{PB_{1}}{BB_{1}}\plus{}\frac{PC_{1}}{CC_{1}}$

let [..] denote area of triangle

obviously, $ \frac {PA_1}{AA_1} \equal{} \frac {[PBC]}{[ABC]}$, $ \frac {PB_1}{BB_1} \equal{} \frac {[PAC]}{[ABC]}$, $ \frac {PC_1}{CC_1} \equal{} \frac {[PAB]}{[ABC]}$ summing up three expressions gives a total of 1.

Thank you.Silly of me to miss the point.


re paidia sorry pou mplekome sta podia sas kai stin spoudaia douleia pou kanete alla thelo an mporeite na mou lusete kati. loipon;
exo 18 mpales idiou sxhmatos opou  oi protes 9 exoun to idio varos kai oi ypoloipes to idio alla pio ligo.
theloume me 3 zugiseis na apodeiksoume ston apenanti mas, o opoios gnvrizei oti oi mpales exoun an 9 exoun diaforetiko varos, oti oi mpales me to perissotero varos einai oi protes.
euxaristo ek ton proteron

mporis ligo na eksigisis kalitera...?
[quote]o opoios gnvrizei oti oi mpales exoun an 9 exoun diaforetiko varos[/quote]
+episis na kses oti mporis na stilis parapano apo 1 post sto idio topic...
dn xriazete na anigis kenurgio topic...

 :D

[quote="tasevago"]o οποίος γνωρίζει ότι οι μπάλες έχουν ανα 9 διαφορετικό βάρος.
δηλαδή ο άλλος ξέρει ότι οι πρώτες 9 έχουν ίσο βάρος και λίγο περισσότερο απο τις υπόλοιπες 9 που έχουν επίσης το ίδιο βάρος[/quote]

Θα μετρήσω τις πρώτες εννιά και μετά τις άλλες εννιά και οι πρώτες εννιά θα είναι πιο βαριές.
(Προφανώς και δεν είναι η λύση αλλά καλό θα ήταν να εξηγήσεις την εκφώνηση καλύτερα!!!!!!!!!!!!!!!!!!!!!!)


geometry
geometric transformation
dilation
ratio
parallelogram
geometry solved

[quote="barasawala"]Let $ AA_{1},BB_{1},CC_{1}$ be the altitudes in acute triangle $ ABC$, and let $ X$ be an arbitrary point. Let $ M,N,P,Q,R,S$ be the feet of the perpendiculars from $ X$ to the lines $ AA_{1},BC,BB_{1},CA,CC_{1},AB$. Prove that $ MN,PQ,RS$ are concurrent.[/quote]

This is Theorem 2 from the note "On the paracevian perspector" on my website (which currently lives at http://www.cip.ifi.lmu.de/~grinberg/ or, archived, at http://sites.google.com/site/darijgrinberg/website ). In that note, I also prove a generalization of your problem (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=14692 for a discussion of that generalization).

I will leave to you the fun of finding the short solution of the above problem which does [i]not[/i] apply to the generalization. ;)

  Darij

[quote="barasawala"]Let $ AA_{1},BB_{1},CC_{1}$ be the altitudes in acute triangle $ ABC$, and let $ X$ be an arbitrary point. Let $ M,N,P,Q,R,S$ be the feet of the perpendiculars from $ X$ to the lines $ AA_{1},BC,BB_{1},CA,CC_{1},AB$. Prove that $ MN,PQ,RS$ are concurrent.[/quote]

This problem also has been presented last time in Hyacinthos Forum on Aug 8, 2006, by Quang Tuan Bui and there was an extensive discussion, about also the generalization of the problem, as it has been presented first time by Eric Daneels (''[b]paracevian perspector[/b] ?''), on Jul 23,2004.

You can see at: http://groups.yahoo.com/group/Hyacinthos/message/10135 and http://groups.yahoo.com/group/Hyacinthos/message/13907

An alternative (than Darij’s in his website) proof, it has already been posted also in Hyacinthos Forum at
http://groups.yahoo.com/group/Hyacinthos/message/14034 and some notes at   http://groups.yahoo.com/group/Hyacinthos/message/14088  for a simpler yet proof.

I will post here next time the two alternative synthetic proofs I have in mind of the ''paracevian perspector '' theorem.

Kostas Vittas.

[quote="probability1.01"]I did not check the above links, but here is a way to prove it:

Dilate with a factor of 2 about X. The image of MN becomes the line $A_{1}X$ reflected about $AA_{1}$, and analogous results hold for the other two sides. Since the altitudes of ABC are the angle bisectors of its orthic triangle, the images of MN, PQ, and RS concur at the isogonal conjugate of X in the orthic triangle. Thus MN, PQ, and RS themselves must also concur.[/quote]

This is the simple proof I spoke of :) .

The problem also appears as problem 5.81 in Prasolov's Planimetry problems book (see http://www.math.su.se/~mleites/ and http://michaj.home.staszic.waw.pl/prasolov.html for English and http://www.mccme.ru/free-books/ for Russian versions), with the same proof.

  Darij

[b][size=117]THE PARACEVIAN PERSPECTOR THEOREM.[/size][/b]

 – A triangle $\bigtriangleup ABC$ is given and  let $H$ be, an arbitrary fixed point, and let $\bigtriangleup HaHbHc$ be, it’s cevian triangle.

- Let $P$ be, an arbitrary also point and $Pa,$ $Pb,$ $Pc,$ the traces on sidelines $BC,$ $CA,$ $AB,$  from the lines through $P$ and parallel to $AHa,$ $BHb,$ $CHc,$ respectively.

- Let $Ap,$ $Bp,$ $Cp$ be, the traces on segment lines $AHa,$ $BHb,$ $CHc,$ from the lines through $P$ and parallel to the sidelines $BC,$ $CA,$ $AB,$ respectively.

- Prove that the lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent at one point so be it, $Q,$ as the midpoint of the segment between the point $P$ and the $H-Ceva$ conjugate of $P.$

[b][size=117]PROOF  1.[/size][/b] - $($ In my drawing $AB = 15.3,$$AC = 14.2,$ $BC = 11.2,$  $AH = 12.8,$ $BH = 4.2,$ $AP = 8.2,$ $CP = 6.8$ $).$

$1)$ – We denote the intersection points of sidelines of $\bigtriangleup ABC,$ from the segment lines $PAp,$ $PBp,$ $PCp,$ as follows:

- As $D,$ $D',$ the intersection points of $AC,$ $AB$ respectively, from the segment line $PAp.$

- As $E,$ $E',$ the intersection points of $AB,$ $BC$ respectively, from the segment line $PBp.$

- As $F,$ $F',$  the intersection points of $BC,$ $AC$ respectively, from the segment line $PCp.$

Also we denote:

- As $A',$ the intersection point of $DE',$ $D'F.$

- As $B',$ the intersection point of $DE',$ $EF'.$

- As $C',$ the intersection point of $EF',$ $D'F.$

     Based on the below proposition, we will prove that the lines $AP,$ $BP,$ $CP,$ pass through the points $A',$ $B',$ $C',$ respectively.

$2)$ – [b][size=117]PROPOSITION 1.[/size][/b] - A triangle $\bigtriangleup ABC$ is given and let $P$ be, an arbitrary point inwardly to it. We denote as $D,$ $D',$ the intersection points of $AC,$ $AB$ respectively, from the line through $P$ and parallel to $BC.$ Also we denote as $E',$ $F,$ the intersection points of $BC,$ from the lines through $P$ and parallel to $AC,$ $AB$ respectively. Prove that the lines $DE',$ $D'F,$ $AP,$ are concurrent.

$2a)$ - [b][size=117]PROOF.[/size][/b] -  We denote as $A',$ the intersection point of $DE',$ $D'F$ and as $P',$ the one of $BC,$ from the segment line $AP.$

From $DD'//BC,$ $\Longrightarrow$ $\frac{PD}{P'C}= \frac{AP}{AP'}= \frac{PD'}{P'B}$

$\Longrightarrow$ $\frac{PD}{(P'C)-(PD)}= \frac{PD'}{(P'B)-(PD')}$  $\Longrightarrow$ $\frac{PD}{P'E'}= \frac{PD'}{P'F}$ $,(1)$ $($ because of $PD = E'C$ and $PD' = FB$ $).$

From $(1)$ $\Longrightarrow$ $\frac{PD}{PD'}= \frac{P'E'}{P'F}$ and so, the points $P,$ $P',$ $A',$ are collinear. Hence the lines $DE',$ $D'F,$ $AP,$ are concurrent and the proof of [b][size=109]proposition 1,[/size][/b] is completed.

$3)$ – In our configuration now, we conclude that the line $AP,$ passes through the point $A'$ $($ intersection point of $DE',$ $D'F$ $)$ and similarly the lines $BP,$ $CP,$ pass through the points $B',$ $C',$ respectively.

     By the same way, from the triangle $\bigtriangleup AHaC$ and the point $P$ inwardly to it, we conclude that the lines $ApPa,$ $DE',$ $AP,$ are concurrent. Hence, the line $ApPa$ passes through the point $A'$ and similarly the lines $BpPb,$ $CpPc,$  pass through the points $B',$ $C',$ respectively.

$4)$ – We denote as $K,$ $L,$ $M,$ the intersection points of the line $CC',$ trom the segment lines $AB,$ $A'Ap,$ $AHa$ respectively and as $N,$ the intersection point of $EE',$ from the line $B'L.$ We will prove that the points $B,$ $M,$ $N,$ are collinear.

It is enough to prove that $(E, N, P, E') = (K, M, P, C).$

We consider the pencils $B'.C'LPT$ and $A'.C'LPT,$ where $T,$ is the intersection point of $CC',$ $A'B'$ and then we have that

$(B'.C'LPT) = (A'.C'LPT)$ $,(2)$  $($ congruent double ratios of these bundles of lines $).$

 $(B'.C'LPT) = (E, N, P, E')$ $,(3)$ $($ intersection of $B'.C'LPT,$ from the line $EE'$ $).$

$(A'.C'LPT) = (D', Ap, P, D)$ $,(4)$ $($ intersection of $A'.C'LPT,$ from the line $DD'$ $).$

from $(2),$ $(3),$ $(4),$ $\Longrightarrow$ $(E, N, P, E') = (D', Ap, P, D)$ $,(5).$

The lines $CC',$ $DD',$ intersect the pencil $A.BHaPC$ and so, we have that $(K, M, P, C) = (D', Ap, P, D)$ $,(6).$

From $(5),$ $(6),$ $\Longrightarrow$ $(E, N, P, E') = (K, M, P, C)$ $,(7).$

From $(7),$ we conclude that the lines $EK,$ $MN,$ $CE',$ are concurrent. So, the line $MN$ passes through the vertex $B$ of $\bigtriangleup ABC$ $($ as the intersection point of $EK,$ $CE'$ $).$ Hence, the points $B,$ $M,$ $N,$ are collinear.

$5)$ – We will prove now, that the lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent. Let $Q$ be, the intersection point of $BpPb,$ $CpPc$ and it is enough to prove that this point lies on $A'L$ $($ because of it has already been proved that the line $ApPa,$ passes through the point $A'$ $).$ We will prove that the pencils $B'.C'LBpA',$ $C'.B'LCpA',$ have congruent double ratios. That is, we will prove that

$(B'.C'LBpA') = (C'.B'LCpA').$

$(B'.C'LBpA') = (E, N, Bp, E')$ $,(8).$

$(C'.B'LCpA') = (F', P, Cp, F)$ $,(9).$

We consider the bundles of lines ( = pencils ) $B.ENBpE'$ and $C.F'PCpF.$ They have the sideline $BC$ of $\bigtriangleup ABC,$ as their common ray and the points $A$ $($ as the intersection point of $BE,$ $CF'$ $),$ $M$ $($ as the intersection point of $BN,$ $CP$ $)$ and $H$ $($ as the intersection point of $BBp,$ $CCp$ $),$ lie on the same line $AHa.$ So, we have that

$(B.ENBpE') = (C.F'PCpF)$ $,(10).$

From $(10)$ $\Longrightarrow$ $(E, N, Bp, E') = (F', P, Cp, F)$ $,(11).$

From $(8),$ $(9),$ $(11),$ $\Longrightarrow$  $(B'.C'LBpA') = (C'.B'LCpA')$ $,(12).$

From $(12)$ and because of $B'C'$ is the common ray of pencils $B'.C'LBpA'$ and $C'.B'LCpA',$ we conclude that the points $L$ $($ as the intersection point of $B'L,$ $C'L$ $),$ $Q$ $($ as the intersection point of $B'Bp,$ $C'Cp$ $)$ and $A'$ $($ as the intersection point of $B'A',$ $C'A'$ $),$ are collinear. Hence, the lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent at one point, so be it $Q.$

$6)$ – We will prove now, that the concurrency point $Q,$ of the lines $ApPa,$ $BpPb,$ $CpPc,$ is the midpoint of the segment between the point $P$ and the $H-Ceva$ conjugate of $P.$

This result has been mentioned by Quang Tuan Bui, at last discussion in Hyacinthos Forum  http://groups.yahoo.com/group/Hyacinthos/message/13938 and first time by Bernard Gibert http://groups.yahoo.com/group/Hyacinthos/message/10136

Let $S$ be, the intersection point of $AA',$ $B'C'.$ So, from the parallelogram $AEPF',$ we have that $PS = SA.$ Hence, the line through vertex $A$ of $\bigtriangleup ABC$ and parallel to $B'C',$ intersects the line $PB'$ at a point so be it $B'',$ such that $PB' = B'B''.$

Similarly the line through vertex $C$ of $\bigtriangleup ABC$ and parallel to $A'B',$ intersects the line $PB',$ at the same point $B'',$ because of $PT = TC,$ from the parallelogram $CDPE'.$

Hence, the lines through vertices $A,$ $B,$ $C,$ of $\bigtriangleup ABC$ and parallel to $B'C',$ $A'C,$ $A'B'$ respectively, intersect each other at points $A'',$ $B'',$ $C'',$ lie on the lines $AA',$ $BB',$ $CC'$ respectively. The triangle $\bigtriangleup A''B''C'',$ is the anticevian triangle of $\bigtriangleup ABC,$ with respect to the point $P.$

$6a)$ – We will prove now, that $B''Hb\parallel BpPb.$  In triangle $\bigtriangleup B'BBp$ and from $PPb\parallel BBp$ we have that

$\frac{PPb}{BBp}= \frac{B'P}{B'B}$ $\Longrightarrow$ $\frac{B'B''}{B'B}= \frac{BpHp}{BBp}$ $,(13)$  $($ because of $B'P = B'B''$  and  $PPb = BpHb$ $).$

From $(13),$ $\Longrightarrow$ $B''Hb\parallel B'Bp$ and so, we conclude that the line $B''Hb$ connecting the vertices $B'',$ $Hb,$ of the triangles $\bigtriangleup A''B''C'',$ $\bigtriangleup HaHbHc$ respectively, intersects the segment line $PQ,$ at a point so be it $R,$ such that $PQ = QR$ $($ because of $PB' = B'B''$ $).$

By the same way we can prove that the lines $A''Ha,$ $C''Hc,$ are parallel to $A'Ap,$ $C'Cp$ respectively and intersect the segment line $PQ,$ at the sane point $R.$ So, the triangles $\bigtriangleup A''B''C'',$ $\bigtriangleup HaHbHc,$ are perspective with perspector the point $R,$ lies on the segment line $PQ$ and such that $PQ = QR.$

Because of the triangle $\bigtriangleup A''B''C'',$ is the anticevian triangle of $\bigtriangleup ABC,$ with respect to the point $P$ $($ since $\bigtriangleup ABC$ is the cevian triangle of $\bigtriangleup A''B''C''$ wrt $P$ $),$ we conclude that the point $Q$ $($ the concurrency point of $ApPa,$ $BpPb,$ $CpPc$ $),$ is the midpoint of the segment between the point $P$ and the point $R,$ as the $H-Ceva$ conjugate of $P$ and the proof is completed.

$\bullet$ This proof is dedicated to [b]Babis Stergiu.[/b]

Kostas Vittas.

PS. I will post here next time, the simpler proof of the ''paracevian perspector'' theorem, which has been arisen from the discussion in Hyacinthos Forum, last summer.

[b][size=117]PROOF  2.[/size][/b] - $($ see the figure t=126138(c) $).$

It has already been proved that the segment lines $PAp,$ $PBp,$ $PCp,$ as parallel to the sidelines $BC,$ $AC,$ $AB$ of $\bigtriangleup ABC$ respectively, intersect them at pairs of points $D,$ $D'$ and $E,$ $E'$ and $F,$ $F',$ such that the intersection points $A'\equiv DE'\cap D'F,$ $B'\equiv DE'\cap EF',$ $C'\equiv EF'\cap D'F,$ lie on the segment lines $AP,$ $BP,$ $CP,$ respectively.

The lines through vertices $A,$ $B,$ $C,$ of $\bigtriangleup ABC$ and parallel to $B'C',$ $A'C',$ $A'B'$ respectively, intersect each other at the points $A'',$ $B'',$ $C'',$ also lie on the segment lines $AP,$ $BP,$ $CP,$ respectively.

So, we have the configuration of the triangle $\bigtriangleup ABC,$ as the Cevian triangle of $\bigtriangleup A''B''C'',$ with respect to the point $P$ $($ Hence the triangle $\bigtriangleup A''B''C'',$ is the Anticevian triangle of $\bigtriangleup ABC,$ wrt $P$ $),$

Because of the segment lines $AHa,$ $BHb,$ $CHc$ are concurrent at one point $($ here the point $H$ $)$ and also the segment lines $A''A,$ $B''B,$ $C''C$ are concurrent at one point $($ here the point $P$ $)$, based on the well known [b][size=109]proposition 2,[/size][/b] we conclude that the segment lines $A''Ha,$ $B''Hb,$ $C''Hc,$ are concurrent at one point, so be it $R,$ which in a such our configuration, is called the $H-Ceva$ conjugate of $P,$ with respect to the triangle $\bigtriangleup ABC.$

It has already been proved $($ see the proof  1 $),$ that the segment lines $ApPa,$ $BpPb,$ $CpPc,$ pass through the points $A',$ $B',$ $C',$ as the midpoints of the segments $PA'',$ $PB'',$ $PC''$ respectively.

Because of $ApPa\equiv A'Ap\parallel A''Ha$ and $BpPb\equiv B'Bp\parallel B''Hb,$ and $CpPc\equiv C'Cp\parallel C''Hc,$ we conclude that the segment lines $ApPa,$ $BpPb,$ $CpPc,$ intersect the segment $PR,$ at the same point, so be it $Q,$ as it’s midpoint. That is the segment lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent at the point $Q,$ as the midpoint of the segment between $P$ and $R$ as the $H-Ceva$ conjugate of $P$ with respect to the triangle $\bigtriangleup ABC$ and the proof is completed.

$\bullet$ This proof is dedicated to [b]Vladimir Zajic.[/b]

Kostas Vittas.

PS. [b][size=117]PROPOSITION  2.[/size][/b] - ( well known ) - A triangle $\bigtriangleup ABC$ is given and let $\bigtriangleup DEF$ be, the Cevian triangle of an arbitrary point $P,$ in the plane. If $D',$ $E',$ $F',$ are three points on the sidelines $EF,$ $DF,$ $DE$ of $\bigtriangleup DEF$ respectively, such that the segment lines $DD',$ $EE',$ $FF',$ to be concurrent at one point, prove that the segment lines $AD',$ $BE',$ $CF',$ are also concurrent at one point.

Dear Mathlinkers,
a new proof of "The paracevian perspector" has been put on my website
http://perso.orange.fr/jl.ayme vol. 4
Sincerely
Jean-Louis

I can't understand why some of the last lines of the text of the [b][color=#000]proof 1[/color][/b] which I have been presented before ( post #6) doesn't viewed correctly.

So, I post here the text of these lost lines of the text, beginning from the point of $(13)$.

From $(13),$ $\Longrightarrow$ $B''Hb\parallel B'Bp$ and so, we conclude that the line $B''Hb$ connecting the vertices $B'',$ $Hb,$ of the triangles $\bigtriangleup A''B''C'',$ $\bigtriangleup HaHbHc$ respectively, intersects the segment line $PQ,$ at a point so be it $R,$ such that $PQ = QR$ $($ because of $PB' = B'B''$ $).$

By the same way we can prove that the lines $A''Ha,$ $C''Hc,$ are parallel to $A'Ap,$ $C'Cp$ respectively and intersect the segment line $PQ,$ at the sane point $R.$ So, the triangles $\bigtriangleup A''B''C'',$ $\bigtriangleup HaHbHc,$ are perspective with perspector the point $R,$ lies on the segment line $PQ$ and such that $PQ = QR.$

Because of the triangle $\bigtriangleup A''B''C'',$ is the anticevian triangle of $\bigtriangleup ABC,$ with respect to the point $P$ $($ since $\bigtriangleup ABC$ is the cevian triangle of $\bigtriangleup A''B''C''$ wrt $P$ $),$ we conclude that the point $Q$ $($ the concurrency point of $ApPa,$ $BpPb,$ $CpPc$ $),$ is the midpoint of the segment between the point $P$ and the point $R,$ as the $H-Ceva$ conjugate of $P$ and the proof is completed.

$\bullet$ This proof is dedicated to [b]Babis Stergiu.[/b]

Kostas Vittas.

PS. I will post here next time, the simpler proof of the ''paracevian perspector'' theorem, which has been arisen from the discussion in Hyacinthos Forum, last summer.


I play piano.  And I'm trying to teach myself flute.  But it's not going so well...  I have like zero points for flute technique.   :)

I think this belongs in the music and arts forum.  Does someone want to move it there?

Thanks.

On the offhand, I play the piano. Stereotypically speaking...

For reasons beyond my control, I play piano and flute.

i play sax, guitar, and I'm learning piano. I used to play viola and I sometimes play around on my friend's drum set, so maybe i'm all of the above?

Hey guys, look at this place!
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171023]CLICK ME[/url] :wow:

actually, no one look at it.......especially if youve posted here

I play the violin, "the pen-the pencil-and the book" , the recorder, and somewhat the harmonica.

I've played the violin for 3 years, and quit this year.
I've played the the PPB, for about 12 years, and I'm still playing it.
I've played the recorder for 2 years, then quit.
I sporadically play the harmonica.

I can play the recorder a little bit. But not much. (It was from 4th grade, and I'm in 7th grade now).

I play the flute. And I guess the recorder, too.  :huh:

I can kinda play recorder too.  But I haven't played since like 4th grade...  4 years....   And I used to play harmonica...  But I don't practice those anymore.  Oh, and I used to play guitar, too.  Now I just have time for piano and a little flute.

What is with all the flutes?

I also play piano.

what is with all the flute players???

oh and i play trumpet!!!!!

[quote="monkeygirl13"]I play Clarient the awesomest instrument on the planet.[/quote]

you spelled that wrong :rotfl:

[quote="monkeygirl13"]I play Clarient the awesomest instrument on the planet.[/quote]

Eh, not quite the awsomest in my opinion. I think piano is awesome. Franz Liszt believed that one piano and 10 fingers could produce the same sound that a full orchestra could. And he actually trascribed the whole 4th Mvt. of Beethoven's 9th Symphony. But it's very hard, so many octaves and big chords...

Some strings, like violin and cello, are also nice. Great music, nice sound, etc.

Well, I like clarinet too, but not that fun.

Anythin blow into? -.-'

Well, I'll say...
Piano 9 years
Recorder 7 years in intervals
Alto Sax 6 yeas
S-Sax 5 years
T-Sax 4 years
Flute 3 years (failed out)
Air guitar 2 years
Air drums 1 year :D

[quote="remy1140"]I have been playing the piano for 9 years and I used to play the cello (3 years) and clarinet (1/2 years) but quit when I was in middle school. All classical music..

My dad is a cellist/music professor, and my grandfather was one of the first people in Korea who started playing the cello, so yea I was heavily influenced by music  :)[/quote]
Yah cellists!

I've played cello for 6 years and piano for 7. Too bad this poll only allows one vote.


geometry
circumcircle
trigonometry
geometric transformation
geometry unsolved

In a convex quadrilateral $ ABCD$ with $ \angle ABC\equal{}\angle ADC\equal{}135^{\circ}$, points $ M,N$ are taken on rays $ AB,AD$ respectively such that $ \angle MCD\equal{}\angle NCB\equal{}90^{\circ}$. The circumcircles of triangles $ AMN,ABD$ cut again at $ K$. Prove that $ AK$ is perpendicular to $ KC$.

We will use the easy and well-known fact:

The center of spiral similarity which maps the segment $ AB$ to segment $ CD$ (order of points is important here) is the second intersection of the circumcircles of $ ACS$ and $ BDS$, where $ S$ is the intersection of lines $ AB$ and $ CD$.

Ok, so note that $ \angle BAD \plus{} \angle BCD \equal{} 90^{\circ}$, it means that $ \angle BAC \equal{} \angle BCM \equal{} \angle DCN$. Since $ \angle CBM \equal{} \angle CDN \equal{} 45^{\circ}$ triangles $ CBM$ and $ CDN$ are similar. Let $ E, F$ be the projections of $ C$ on $ DN$ and $ BM$ respectively. Because of the similarity points $ E, F$ divide segments $ DN, BM$ in the same ratio. Now, consider the spiral similarity which takes $ DN$ to $ BM$, clearly $ K$ is the center of this similarity. But $ K$ is also the center of the similarity which takes $ ED$ to $ FB$, so $ K$ lies on the circumcircle of $ AEF$, so it follows that $ \angle AKC \equal{} 90^{\circ}$.

Very nice solution. But I am not sure that this fact is well-known.
I noticed that given angles of $ 135^\circle$ and perpendicularity are not very important, since you only need similarity of 2 triangles. Right?

Yes. But since we must have $ \angle ABC \equal{} \angle ADC$ the angle $ 135^{\circ}$ fits very well here.

it was a really hard problem,a few contestants managed to solve it the year it was proposed,anyway as far as I remember it has also a nice solution using inversion.

Firstly, I will give a lemma:

[b][color=#f00]Lemma:[/color][/b] The $ O_1 $, $ O_2 $, $ O_3 $ centers of the three fixed circles passing through the fixed $ D $ and $ E $ points ensure that the $ DE $ line is the same. Cut an arbitrary line passing through the $ E $ point at $ A $, $ B $, $ C $, containing circles centered at $ O_1 $, $ O_2 $, $ O_3 $. In this case
$$ \dfrac {| AB |}{| BC |} $$
rate is fixed.

[b][color=#f00]Proof: [/color][/b] The radii of the circles with the center of $ O_1 $, $ O_2 $, $ O_3 $ are fixed $ r_1 $, $ r_2 $, $ r_3 $ respectively. The centers $ O_1 $, $ O_2 $, $ O_3 $ are located on the perpendicular bisector line of $ [DE] $. Let $ F $ intersect with the line passing through the $ E $ point and the line $ O_1O_3 $. From the force theorem in the circle
[center][img]https://cdn.artofproblemsolving.com/attachments/3/d/8918776634fbeb5d67b3eeb81cd96602b3c8ef.png[/img][/center]
$$ | FE | \cdot | FA | = | FO_1 | ^ 2 - r_1 ^ 2 = c_1 $$
$$ | FE | \cdot | FB | = | FO_2 | ^ 2 - r_2 ^ 2 = c_2 $$
$$ | FE | \cdot | FC | = | FO_3 | ^ 2 - r_3 ^ 2 = c_3 $$

($ c_i $'s constant) equations are written. From here, $ \dfrac {| FE | \cdot | FB |} {| FE | \cdot | FA |} = \dfrac {c_2} {c_1} $ and $ \dfrac {| FB |} {| FA |} $ it will be constant.

Now we looking main problem.

[b][color=#f00]Solution:[/color][/b]
[center][img]https://cdn.artofproblemsolving.com/attachments/d/e/dbebd8917bbe4ba4da8be301182c08f595088b.png[/img][/center]
Let's draw circumcircle of triangle $ AKC$  (red circle). $ \angle ATC \equal{} x, \angle AUC \equal{} 180 ^{\circ} \minus{} x$. We have to prove $ \angle ATC \equal{} \angle AUC \equal{} 90 ^{\circ}$. Clearly, triangles $ DCN\sim BCM$ are similar and so,  $$ \dfrac {|CM|}{|CN|} \equal{} \dfrac {|CB|}{|CD|} \dots (1) $$ According to the sinus formula $ \dfrac {|TC|}{\sin TAC} \equal{} \dfrac{|AC|}{\sin ATC}$, $ \dfrac {|NC|}{\sin NAC} \equal{} \dfrac {|AC|}{\sin ANC}$, $ \dfrac {|UC|}{\sin UAC} \equal{} \dfrac {|AC|}{\sin AUC}$ and $ \dfrac {|MC|}{\sin MAC} \equal{} \dfrac {|AC|}{\sin AMC}$. Hence, $$ \dfrac {|TC|}{|CD|} \equal{} \dfrac {|UC|}{|CB|} \dots (2) $$

From relations $(1)$ and $(2)$, $ \dfrac {|TC|}{|CN|} \equal{} \dfrac {|UC|}{|CM|}$. $ \angle TNC \equal{} \angle UMC$ and by the [b][color=#f00]Lemma[/color][/b], $\dfrac{|BM|}{|MU|}=\dfrac{|DN|}{|NT|}$. Hence triangles $ NCT\sim MCU$ (side-angle-side).

By the smilarity, $ 180 ^{\circ} \minus{} x \equal{} x$ and $ x \equal{} 90 ^{\circ}$. $ \angle AKC \equal{} \angle AUC \equal{} 90 ^{\circ}$

Lokman GÖKÇE