AMC
AIME
geometry
USA(J)MO
USAMO
modular arithmetic
function

This is the thread where we will discuss the Mock AIME (http://www.tjhsst.edu/~tmildorf/math/PracticeAIME.pdf), solutions to the problems, results, etc. Obviously, the time for submitting answers has ended.

The Solutions may be found at http://www.tjhsst.edu/~tmildorf/math/PracticeAIMESolutions.pdf.

Answers in spoiler:[hide]

1. 680

2. 161

3. 110

4. 801

5. 118

6. 504

7. 320

8. 177

9. 576

10. 113

11. 006

12. 338

13. 443

14. 352

15. 141[/hide]

This was my opinion of the Mock AIME problem distribution:

3 Easy/6 Medium/6 Hard

5 Algebra/3.5 Combinatorics/2.5 Number Theory/4 Geometry

1. Easy - Combinatorics

2. Easy - Algebra/Geometry

3. Easy - Algebra

4. Medium - Number Theory

5. Medium - Algebra

6. Medium - Algebra

7. Medium - Combinatorics

8. Medium - Geometry

9. Medium - Number Theory

10. Hard - Algebra/Geometry

11. Hard - Combinatorics/Number Theory

12. Hard - Geometry

13. Hard - Algebra

14. Hard - Combinatorics

15. Hard - Geometry

Leaderboard (Those with scores above average):

MysticTerminator 11

Pork_Chop8, GeoMar 7

super008 6

Teddy 5

jeffrey, Rep123max, white_horse_king88, pi^2, mandy_pal, JGeneson, dts, JSRosen3, interesting_move 4

Miscellaneous Statistics:

Average Score - 3.5769 [hide]= 93/26[/hide]

Mode Score - 4

Median Score - 4

Easiest Question - 3

Hardest Question - Tie: 10, 12, 14, 15

Score Distribution:

15

14

13

12

11 *

10

9

8

7 **

6 *

5 *

4 *********

3 ***

2 *****

1 **

0 **

# of Solvers Problem by Problem

1 ******

2 *******

3 **********************

4 ***********

5 ***************

6 ***

7 *****

8 ********

9 **************

10

11 *

12

13 *

14

15

Mock USAMO Index Max(AMC A, AMC B) + 10 * AIME

1. Mystic Terminator 235

2. Super008 200.5

3. Teddy 183.5

4. JGeneson 170

5. dts 154.5

6. Rep123max 153.5

7. pi^2 145.5

8. White_Horse_King88 144

9. JSRosen3 143

10. Fermatprime 141

Beta 140

Nosepuppy 132

T3hg0suazn 123

Yif Man 12 122

Mlee06 118.5

Genius1601 116

joml88 113

Dnas 108 *

Zzzz3660 96 *

RC-7th 94.5 *

Confuted 91

Lucky707 90

takuanitromars36 89 *

[1] Pork_Chop8 210.5 !

[1] GeoMar 70 ()

[3] jeffrey 40 ()

[3] mandy_pal 40 ()

[3] interesting_move 40 ()

* - No AIME entry

() - No AMC score / AMC score not in top scorer lists merged together.

! - Based on AMC C score. This is the third AMC and will not change this group USAMO qualifiers.

The top 10 indices qualify for the Mock USAMO, as well as the top 5 AIME scores independent of grade (excluding top indices), indicated by #. and [#] respectively.

For a bonus point: [hide]What was the typo in the .pdf version of the Mock AIME?[/hide]

The Participants (I did not receive answers from everyone):

crayshot97 (Editor)

jeffrey (i)

albertfermat (i)

Rep123max (i)

white_horse_king88 (i)

Pork_Chop8 (i)

Lucky707 (i)

era (i)

confuted (i)

Teddy (i)

GeoMar (i)

nosepuppy (i)

BHorseMath (i)

mlee06 (i)

T3hg0suazn

genius1601

joml88

Fermatprime

pi^2

mandy_pal

yif man12

super008

JGeneson

dts

ginasboys

Ultimatemathgeek

JSRosen3

beta

MysticTerminator

interesting_move

(i) - Independent; took contest early.

Thanks to all who participated and especially to crayshot97 for reviewing the contest; I was glad to have a second opinion, particularly on #14.

Hmm.........tough test.

I bombed AMC A and B, so It logically follows that I bombed this one too.
One the AIME Poll, I clicked "Easy" accidently. LOL

hm. I thought it was very, very difficult...the first problem took me at least 15 minutes (compared to like 2 minutes average) and I still doubt I got it right...

I did really bad on this :(, it was a lot harder than 2004 AIMEs

Very hard test.  Will be happy if I get 5+.

-interesting_move

This feels like one of my tests in english this year.  :( The one I got a 59 on.

It was hard... I didn't think that it was much harder than the average AIME though.  I answered 5 of them legitimately but problem messed up a bunch of those.  So my score will be $\leq 5$.  I thought the questions were good and I really want to see how to solve some of them.

I also thought it was very hard.  I made much less progress on this test than on the 2004 AIME A.  I don't know, maybe I've just been spoiled by taking an easy AIME.  I liked the solutions for numbers 5 and 6, however.

This was very very very hard.  I'll be happy if I even get a question right, considering I gave up after half an hour, and didn't get any solutions.

I got 3-5 without guessing, and narrowed a few more down to where I could make an educated guess.  :) I guess that's pretty good (no pun intended. :P)

Oh yeah, when are we gonna get solutions?

We can start discussing our solutions/answers now, right? Or do we have to wait until results are posted?

[quote="interesting_move"]I finished #1 and #2 in my first hour...Pretty sad.  Then I answered 7 more, 2 of which I know definitely that I missed.  So it's looking like a score of 5-6 for me....

-interesting_move[/quote]

Hey, that's a [heck] of a lot better than most of us got. I got a 5, max. (Unless my vague intuitions were right on #3 and #7.)

[quote="Mildorf"]
* There are at least two typos in the solutions and one in the problems. Beta found the typo in the .pdf of the problems so he is officially unofficially qualified to take the Mock USAMO if we have it.

[/quote]

Can I assume that you did not correct them yet?  (like you said on previous posts)

That wouldn't make much sense, would it? After all, these first few mock tests were designed to mimic the actual ones, and the actual rules state that one can take two AMC contests and one AIME contest.

[quote="Mildorf"]

* There are at least two typos in the solutions and one in the problems. Beta found the typo in the .pdf of the problems so he is [b]officially unofficially [/b]qualified to take the Mock USAMO if we have it.

[/quote]
Wow. Wouldn't it be a worse thing to be officially unofficial than to be regularly unofficial because officially unofficial refers to some judicator formally deciding that some event is unofficial. I'm just joking of course. However, if the print was [b] officially, unofficially qualified [/b] it might have meant what I said. I just found this interesting though.

Bah, I just deleted all of the AIME answer submissions to empty the inbox but then I realized that at the end of their answers one person had sent a request for books to study etc.  Could that person resend a PM? I have forgotten who it was.

Could someone tell me where I can find more info about using phi to solve mod problems?  Is there anything in aops vol 2?

[quote="Ultimatemathgeek"][quote="Mildorf"]* Ok, I have fixed the known typos in the problem statements and the solutions. No the internal SSH was not fixed, but even better - I installed TeX on my computer. :D[/quote]

So all the problems are fixed?  (I don't know what "the internal SSH" is, explain to me?)

So where can I get this fixed pdf's?[/quote]

* LaTeX stopped working on the remote service machine in the computer lab, so I would SSH to another computer in the lab on which TeX worked properly.  For whatever reason, the latter part of this process stopped working.  This is all irrelevant now since I will be working strictly on my computer and putting only the final product online.

* The old .pdf files have been overwritten by the new, fixed files.

[quote="Mildorf"]
* LaTeX stopped working on the remote service machine in the computer lab, so I would SSH to another computer in the lab on which TeX worked properly.  For whatever reason, the latter part of this process stopped working.  This is all irrelevant now since I will be working strictly on my computer and putting only the final product online.

* The old .pdf files have been overwritten by the new, fixed files.[/quote]

Thanks.

One more question.

What is the difference between LaTex you are referring to and Tex?      (I know it is a weird question, but give me an answer please)
I know both of them have to do with writing math stuff

[quote="Mildorf"]I'm no authority on TeX (pronounced "Tech", not what it looks like) and the like, so if someone who is knowledgeable reads this please check me, but I don't think there's really any difference. TeX refers to the general language in which these "mathy" papers are written, LaTeX is just a flavor of TeX.  For example, the command line command for compiling TeX files is usually "latex <filename>.tex" which creates a .dvi file.  This is converted into a .pdf by using "dvipdfm <filename>.dvi" which creates "<filename>.pdf"[/quote]

so, it is like QBasic and Visual Basic is a flavor of Basic, and LaTex is a flavor of Tex?

are there some variations in the lanaguae of Tex and LaTex?

"For example, the command line command for compiling TeX files is usually "latex <filename>.tex" which creates a .dvi file.  This is converted into a .pdf by using "dvipdfm <filename>.dvi" which creates "<filename>.pdf""

I don't get what you are trying to say.  That part seems totally unrelevent to my question.

Thanks

Is it just me or is problem 14 just like #3 on the first Yellow/Red MOP test in 2002?

Very nice problems, by the way.

I got basic background of circle..
Can any tell me how to find the greatest y/x of the circle...
Will appreciate if someone could draw that out...

Sorry for revive, but PDF is not working...


this is my code to solve a problem on project euler[code]public class ProjectEuler1 {
	public int[] threeMultiple() {
		int[] a = new int[(1000/3) + 1];
		a [0] = 0;
		int i = 1;
		while(i <= 1000/3) {
			try {
				a[i] = a[i-1]+3*i;
				i++;
			}

			catch(Exception e) {
				System.out.println("ja;lkdj");
			}

		}
		return a;
	}
	public int[] fiveMultiple() {
		int[] b = new int[(1000/5) + 1];
		b [0] = 0;
		int i = 1;
		while(i <= 1000/3) {
			try {
				b[i] = b[i-1]+3*i;
				i++;
			}

			catch(Exception f) {
				System.out.println("asdlkfja;lsdkf");
			}

		}
		return b;
	}

	public static void main(String[] args) {
		ProjectEuler1 something = new ProjectEuler1();
		int[] a = something.threeMultiple();
		int[] b = something.fiveMultiple();
		System.out.println(a[1000/3]);
		System.out.println(b[1000/5]);
	}
}[/code]

but, it always comes up with an exception, please help :D
[/url]

Well, I could just tell you where the error is, but the cool thing about java exceptions is that you can work it out yourself..

Firstly, get rid of the try + catch bits. You don't have to catch them, and if you leave java to deal with it, it will print out information on where the error is (you can do that manually as well but its easier not doing so).

If you do that, it will say:

java.lang.ArrayIndexOutOfBoundsException: 201
        at ProjectEuler1.fiveMultiple(ProjectEuler1.java:xxx)

where xxx is the line number.

You look at that line, and see thats the line which says b[i] = b[i-1]+3*i;.
The exception tells you it is trying to access b[201] which is out of bounds, because the b array was declared as int[201].

Why is it trying to access index 201? Answer that, and you've fixed the exception.

If you also read that line I mentioned before closely, you'll spot another obvious mistake too ;)


The United States uses zip codes for all mail delivery addresses. These are of the form abcde-fghi , where  abcde and fghi represent digits. How many zip codes are possible in which a is not 0? Express your answer in scientific notation, using  for multiplication. (For example, write "", not "" or "".)
The thing between the quotation marks was latex. I didn't know how to express it. Do you do something like $ 91.0^8$ because it didn't seem to work.

problems are better off in the Alcumus problem section
I would think this to be 9^9...

It's not $ 91\cdot 0^8$, but $ 9\cdot 10^8$ (maybe a typo). This is easily seen: $ a$ may only take the $ 9$ values from $ 1$ to $ 9$, while all other $ 8$ digits may take full values from $ 0$ to $ 9$. As they take independently these values, the correct answer ensues.

oh hmmph i read wrong, i thougt all teh digits couldn't use 0. :(

You express it with the little star symbol, "*" so it should be 9*10^8 (what you would type in).

Lol I made a typo... do you need to use latex with the dollars around it? EDIT: nvm.. thanks jjfun you express it like 9*10^8, I thought you do 9.10^8 because the question said to use . for multiplication.

[quote="mavropnevma"]Don't use the "dot for multiplication" in the same position as the normal period. In latex, there is the simple period, giving $ 9.10$, and the command [code]\cdot[/code], giving $ 9\cdot 10$. As you see, it's a "centered" dot. The more that the normal period is used as decimal separator, so $ 9.10$ reads as $ 9$ and one tenth, while $ 9\cdot 10$ reads as $ 9$ time $ 10$.[/quote]

Just use the star symbol, the problem is worded badly.


inequalities
trigonometry
function
geometry
search
trapezoid
congruent triangles

Given a triangle $ ABC$. Let $ X$, $ Y$, $ Z$ be points on the segments $ \left[BC\right]$, $ \left[CA\right]$, $ \left[AB\right]$, respectively, such that $ AZ\equal{}BX\equal{}CY$ and $ \widehat{BAX}\equal{}\widehat{CBY}\equal{}\widehat{ACZ}$.
Prove that triangle $ ABC$ is equilateral.

Your problem is similar in this section may be you can find it here
http://www.mathlinks.ro/Forum/viewtopic.php?t=88459&sid=054d15afb04dfb937fc30177f5c70423

[quote="borislav_mirchev"]You can see different approach to this problem in the following link:
http://www.math10.com/forumbg/viewtopic.php?p=17148#17148.
it is the most beautiful thing I've seen recently!

Where have you this problem from?[/quote]

I have also tried this approach. It does not lead to a correct solution. You can't WLOG assume that $ a\geq b\geq c$, since the situation is not symmetric. So we don't have a solution to the above problem yet.

  darij

Darij, you are wrong this time. Solution in math10.com is correct and is very beautiful. I took whole day to analyze it and this approach really works. It is solved by some great matematician. I don't know him. Please spent more time to see it is correct. I'm interested about this problem because I invented it independently. There is a solution and is a great solution. But takes a lot time to be discovered. At first time I also didn't believed. I think if you and I cannot solve this problem by using some approach it is not enough to say no one cannot solve it.

P.S. Some details aren't described very good and sometimes there is not enough details but the solution is very good and beutiful. It is for all triangles. If you don't understand something, please ask me :) I suppose bulgarian is not your native language.

It is not a matter of belief. There are non-equilateral triangles $ ABC$ for which $ a^2b(a \minus{} b)(a \plus{} b) \plus{} b^2c(b \minus{} c)(b \plus{} c) \plus{} c^2a(c \minus{} a)(c \plus{} a) \equal{} 0$. I tend to think (I have never checked, though) that these are obtuse-angled; if this is the case, at least for acute-angled triangles the solution is correct. However great mathematician he is, his solution is flawed. Feel free to tell him, I think he would be more open towards criticism than you are (for whatever reason).

  Darij

No, darij,
the following equality is true for every equaliteral triangle. It is not true for every not equaliteral triangle. But from given conditions it follows.
Then he makes the following:
From given conditions he get: $ G(a,b,c)\equal{}0$ For value of G(a,b,c) - see below.
(I'll sketch the following case $ a \geq b \geq c$)
$ G(a,b,c)\equal{}a^{2}b(a\minus{}b)(a\plus{}b)\plus{}b^{2}c(b\minus{}c)(b\plus{}c)\plus{}c^{2}a(c\minus{}a)(c\plus{}a)$
Then he must divide G(a,b,c) by sum of the smallest sides: (b+c).
$ H(a,b,c) \equal{} a^{2}b(a\minus{}b)\plus{}b^{2}c(b\minus{}c)\plus{}c^{2}a(c\minus{}a)$
After dividing G(a,b,c) by (b+c) it follows that $ * H(a,b,c) \leq 0$
Then he represent: $ *** H(a,b,c)\equal{}(a\plus{}b\minus{}c)(a\minus{}b)(a\minus{}c)b\plus{}a(b\plus{}c\minus{}a)(b\minus{}c)^{2}$
$ ****$ But a,b,c are sides of a triangle and $ a \geq b \geq c$. So
$ ** H(a,b,c) \geq 0$ From $ * ** follows H(a,b,c)\equal{}0$ But from this $ *** and ****$ follows a=b=c.
This is the technique used and it works for all six cases checkedat least 3 times manually and with software. Any questions?

Well, consider
$ G(a,b,c) \equal{} a^{2}b(a \minus{} b)(a \plus{} b) \plus{} b^{2}c(b \minus{} c)(b \plus{} c) \plus{} c^{2}a(c \minus{} a)(c \plus{} a)$.
Take $ a\equal{}20\plus{}x$, $ b\equal{}1\plus{}x$ and $ c\equal{}21$. These $ a$, $ b$, $ c$ are the sidelengths of a triangle for every $ x>0$. And $ G\left(a,b,c\right)$ has zeros for $ x$ being approximately $ 3. 413239828$ or approximately $ 7. 414128648$. As you see the corresponding triangles are far from equilateral.

I have already said one time that the assumption $ a\geq b\geq c$ can't be made "WLOG" and I don't like repeating myself. It would be better if people would check their proofs, particularly of inequalities, for mistakes.

  Darij

Questions:
1. Why do you think sides of the triangle should be in the form: 20+x, 1+x, 21?
2. May this to be the reason for your opinion?
3. Is the situation same if you find sides in the form: 9-x, 6, 3+x?
4. Are you sure you didn't make a mistake?

[quote="borislav_mirchev"]1. Why do you think sides of the triangle should be in the form: 20+x, 1+x, 21?[/quote]

Why not?

[quote="borislav_mirchev"]2. May this to be the reason for your opinion?[/quote]

I already said that I had tried the same approach a long time ago, and found a counterexample. I had forgotten it since, and in the last post I found a new counterexample. Besides I mentioned a flaw in your proof which you have still not fixed. Reason enough?

[quote="borislav_mirchev"]3. Is the situation same if you find sides in the form: 9-x, 6, 3+x?[/quote]

Who cares? Throughout mathematics, one counterexample is enough to confute assertions.

[quote="borislav_mirchev"]4. Are you sure you didn't make a mistake?[/quote]

Probably not as sure as you are. But excessive surety is not necessarily a virtue.

  darij

I think by adding your substitution for the sides you added additional constraints. It is the reason for you to find a contra-example and maybe you are wrong. When I've more time I'll translate whole solution but you should know you should spend few hours in deep thinking. Until that I think the problem is open. And I believe my solution is ok. I think it seems to be an easy problem but it isn't at all. Also I think it is human to err.

[quote="borislav_mirchev"]I think by adding your substitution for the sides you aded added additional constraints.[/quote]

Yes, I did. It was not my aim to find all counterexamples. One is enough, as I said.

[quote="borislav_mirchev"]It is the reason for you to find a contra-example and maybe you are wrong. When I've more time I'll translate whole solution but you should know you should spend few hours in deep thinking. Until that I think the problem is open.[/quote]

Sorry but I have no time for thinking about this problem, which is, indeed, open - at least for me, because I have not seen any solution to it yet. But for me the case of your "solution" is closed - it is flawed, so please do some modifications/corrections to it before you ask me again about it, or tell me why my counterexample is wrong.

[quote="borislav_mirchev"]I think it seems to be an easy problem but it isn't at all. Also I think it is human to err.[/quote]

Such problems can turn out arbitrarily hard. And yes, making mistakes is very human, but one can reduce the damage done by them if one listens to criticism.

  Darij

[quote="borislav_mirchev"]Darij, forgot for the triangle for a moment.
x+7 and x+21 cannot be equal never no matter what value have x. Consider their difference diff = x+21 - (7+x) = 14 a constant. Your substitution is wrong. Say am I a genious?[/quote]

What do you want to say by this?

  darij

I want to say that  with this substitution no mater what equality is given for the sides you will every time wil have a difference between sides 14. It is impossible for all substitutions  to get equaliteral triangle. Try with Pythagor Theorem for example or with cosine law for a triangle with angle 60 degrees.

[quote="borislav_mirchev"]I want to say that  with this substitution no mater what equality is given for the sides you will every time wil have a difference between sides 14. It is impossible for all substitutions  to get equaliteral triangle.[/quote]

Yes, and what is the problem? I wanted to show that there exist [b]non-equilateral[/b] triangles $ ABC$ satisfying $ G\left(a,b,c\right)\equal{}0$.

  darij

[quote="borislav_mirchev"]If you are right where the error then comes from?! I think if you are right you've found an example where triangle ABC is not equaliteral and given conditions are satisfied and it is indeed impossible to prove ABC is equaliteral by the given conditions -  is it true?[/quote]

It is impossible to prove that $ ABC$ is equilateral knowing only $ G\left(a,b,c\right)\equal{}0$. But it is not necessarily impossible to prove that $ ABC$ is equilateral given that $ X$, $ Y$, $ Z$ are points on the segments $ \left[BC\right]$, $ \left[CA\right]$, $ \left[AB\right]$, respectively, such that $ AZ\equal{}BX\equal{}CY$ and $ \widehat{BAX}\equal{}\widehat{CBY}\equal{}\widehat{ACZ}$. Id est, the problem may be correct (but may also be not - as I said, a solution remains to be found).

  Darij

In my solution cotg(ACM) is determined by the sides of the triangle. So if your example is true it will be very interesting. How did you solve your equation and what is your equation? I tried by using http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic
...
if we both are right there may be a software problem. It is written that all mine 4 roots are real but there is an imaginery part ... it looks very strange.

P.S. My equation is $ 40x^{4}\plus{}821x^{3}\minus{}10876x^{2}\minus{}169997x\plus{}511980\equal{}0$

Darij, my equation is not correct. Correct is $ 59x^4 \plus{} 1600x^3 \minus{} 2516x^2 \minus{} 162397x \plus{} 511980 \equal{} f(x) \equal{} 0$ I tried with a correct one but I think there is a software problem. It is with big numbers only - your example and for example: a=19+x, b=1+x, c=20.
Try:
I. a=7+x, 1+x, 8
II. a=14+x, b=1+x, c=15
In the following site:
http://www.quickmath.com/www02/pages/modules/equations/index.shtml
They gave two complex and two reals but negative solutions. They are not good examples. Please, give me one example that is common to your software and to:
http://www.quickmath.com/www02/pages/modules/equations/index.shtml.
If I'm right - we are both right. And the conclusion from my programm experience is true: "No code - no problems" and more general "No people, no problems". Please give me an example with smaller numbers. If it is correct it most probably that the triangle with these condition may not be equaliteral. Thank you for the time spent.

[b]I think - you are may not be wrong with 85 % of probability.[/b]
But from this solution we have relation between sides of the triangle that probably may be used.

P.S. I entered your approximations in f(x) - there is a small difference from 0 - after 5-th sign - but there is such a difference - it may be from the precision. If you can, please, give me example with smaller numbers for the sides.

Sorry, I don't have time to use QuickMath now (it seems to require a lot of transforming equations by hand), but here is how you can see that there are non-equilateral triangles $ ABC$ with $ G\left(a,b,c\right) \equal{} 0$ without solving algebraic equations:

Consider the function $ x\mapsto G\left(20 \plus{} x,1 \plus{} x,21\right)$. This is a continuous function. It takes a positive value for $ x \equal{} 8$ and a negative value for $ x \equal{} 7$ (this is just integer computation). Thus, it takes the value $ 0$ somewhere inbetween, i. e. there exists an $ x\in\left[7,8\right]$ such that $ G\left(20 \plus{} x,1 \plus{} x,21\right) \equal{} 0$. Of course, the triangles with sides $ 20 \plus{} x$, $ 1 \plus{} x$ and $ 21$ is non-equilateral.

  Darij

Thank you , Darij! I think you are right in 99.99999 %. But our time is not lost. This time I think I've a solution - my, own, using only pure geometry and without inequalities. But I'll need your help. [b]Please answer to the following question[/b] before post my solution:
[b]Question:[/b] It is given triangle ABC on side BC is taken point P such that $ AB^{2} \equal{} BP.BC$. Is it true that exist circle k such that AB is tangent to k and A, P , C lie on k?

[quote="borislav_mirchev"]But I'll need your help. [b]Please answer to the following question[/b] before post my solution:
[b]Question:[/b] It is given triangle ABC on side BC is taken point P such that $ AB^{2} \equal{} BP.BC$. Is it true that exist circle k such that AB is tangent to k and A, P , C lie on k?[/quote]

This is correct, if the points $ P$ and $ C$ lie on one side of the point $ B$.

Looking forward to see your solution!

  Darij

My solution is in the file attached. Please read it carefully and ask me all questions you have. If it is correct. May you write a solution for the cases U,V,W coiscides and A or B or C > 90 degrees.

Why no one want to verify my solution?!  :( I think it is correct and took me a long time to discover it. Please check it.

[quote="BaBaK Ghalebi"]it would be better if you post the whole solution here,instead of attaching it,because it would be easier to read and in long time,the file you've attached will be lost...[/quote]

Why will it be lost? It's attached as a file on MathLinks, not posted on Imageshack or something similar...

As for the solution, I am sorry to disappoint you again, but $ 2t \equal{} \frac {a}{\sin\alpha}$ should be $ 2t \equal{} \frac {b}{\sin\alpha}$ (because $ AC$ is $ b$, not $ a$), what is not symmetric anymore.

I am moving the topic into Open Questions.

  Darij

I discovered this morning my error  :( . I'm sorry. This problem is not easy at all but I'll continue to explore it... who knows maybe some day I'll have correct arguments. If someone have any ideas let helps. I've one more question. May you using some drawing program to check if the triangle with these properties is equaliteral? I'm wondering also how such triangle may be constructed?

Note [b]It is true that if A+Fi, B+Fi, C+Fi < 90 degrees - triangle is equaliteral. But I cannot succeed in other cases.[/b] (Fi common equal angle)

If A,B,C > 2Fi or A,B,C < 2Fi Using b=xsin(A+Fi)/sin(Fi), c=xsin(B+Fi)/sinFi, a=xsin(C+Fi)/sinFi. By checking different cases for a, b, c (a>c>b for example) also may be proved - triangle is equaliteral.

Dear all my friends.

I would like to present a simpler approach, with a ''constructive'' way, probably not so strict as a mathematical solution, but I think may be valid.

The basic idea is to search how to we can construct a configuration as the problem states.

It is easy to show that in every equilateral triangle $ \bigtriangleup ABC,$ we can define three points $ X,$ $ Y,$ $ Z,$ on its side segments $ BC,$ $ AC,$ $ AB$ respectively, such that $ AZ \equal{} BX \equal{} CY$ and simultaneously such that $ \angle BAX \equal{} \angle CBY \equal{} \angle ACZ$ $ ($ it is enoufg the one of two conditions, in order to be true the other also $ ).$

We consider now, the case when $ \bigtriangleup ABC,$ is an arbitrary acute triangle, with $ BC \equal{} a > AC \equal{} b > AB \equal{} c.$ As well known we have also $ \angle A > \angle B > \angle C.$

$ \bullet$ At first, we can say that it is easy to define the points $ X,$ $ Y,$ $ Z,$ on $ BC,$ $ AC,$ $ AB$ respectively,

such that $ AZ \equal{} BX \equal{} CY \equal{} d \leq c$ and it is enough to only prove that $ \angle BAX > \angle CBY.$

Then the conclusion of impossibility of the simultaneously condition of $ \angle BAX \equal{} \angle CBY \equal{} \angle ACZ,$ it follows.

$ \bullet$ At second, we can say that it is easy to define the points $ X,$ $ Y,$ $ Z,$ on $ BC,$ $ AC,$ $ AB$ respectively,

such that $ \angle BAX \equal{} \angle CBY \equal{} \angle ACZ \equal{} \angle \omega \leq \angle C$ and it is enough to only prove that $ CY > BX.$

Then the conclusion of impossibility of the simultaneously condition of $ AZ \equal{} BX \equal{} CY,$ it follows.

So, we conclude that when the two conditions as the problem states, exist simultaneously in the configuration of a triangle, then this triangle may be only equilateral.

$ \bullet$ I am looking forward in your remarks and comments.

Best regards, kostas vittas.

It is a very interesting approach. It is true as a logic but I've two additional questions:
1. If $ AZ \equal{} BX \equal{} CY \equal{} d \leq c$ how did you prove that: $ \angle BAX > \angle CBY$?
2. If $ \angle BAX \equal{} \angle CBY \equal{} \angle ACZ$ how did you prove that: $ CY > BX$?

P.S. If your approach is true - it probably may be used for all triangles maybe, using the fact: every triangle with an angle not less than 90 degrees have two acute angles.

vittasko's approach doesn't work :(. It seem right for some situations but may be found other - that it isn't correct. But it was a good attempt - till this moment no one cannot find something better.

[quote="borislav_mirchev"]
1. If $ AZ \equal{} BX \equal{} CY \equal{} d \leq c$ how did you prove that: $ \angle BAX > \angle CBY$?
2. If $ \angle BAX \equal{} \angle CBY \equal{} \angle ACZ$ how did you prove that: $ CY > BX$?[/quote]
Sorry dear [b][size=100]borislav_mirchev[/size][/b], because I can’t post quickly replies.

I will try to describe the ''constructive'' way I mentioned and I hope that it is not wrong.

[b][size=100](a)[/size][/b]. – See the [b][size=100]schema t=99674(a)[/size][/b].

We define the points $ X,$ $ Y,$ $ Z,$ on $ BC,$ $ AC,$ $ AB$ respectively such that $ AZ \equal{} BX \equal{} CY \equal{} d \leq AB$ and we will prove that $ \angle BAX > \angle CBY.$

Let $ B'$ be, the point on $ BC,$ such that $ BB' \equal{} AB,$ between $ B,$ $ C,$ because of $ BC > AB.$

Through $ B',$ $ C,$ we draw two parallel lines such that $ \angle BB'X' \equal{} \angle BCY' \equal{} \angle B,$ where $ X',$ $ Y',$ their intersection points from the line segment $ BY.$

We define the point $ X'',$ on the line segment $ B'X',$ such that $ B'X'' \equal{} CY \equal{} d$ $ ,(1)$

From $ BC > BB'$ $ \Longrightarrow$ $ CY' > B'X'$ $ ,(2)$

From $ \angle BCY < \angle BCY'$ $ \Longrightarrow$ $ CY > CY'$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ B'X'' > BX'$ $ ,(4)$

From $ (4)$ we conclude that $ \angle B'BX'' \equal{} \angle BAX > \angle B'BX' \equiv CBY$ $ ($ because of the congruent triangles $ \bigtriangleup ABX$ and $ \bigtriangleup BB'X''$ $ )$ and the proof of the part [b][size=100](a)[/size][/b] is completed.

[b][size=100](b)[/size][/b]. – See the [b][size=100]schema t=99674(b)[/size][/b].

We define the points $ X,$ $ Y,$ $ Z,$ on $ BC,$ $ AC,$ $ AB$ respectively, such that $ \angle BAX \equal{} \angle CBY \equal{} \angle ACZ \equal{} \angle \omega \leq \angle C$ and we will prove that $ CY > BX.$

Let $ B'$ be, the point on $ BC$ as before, such that $ BB' \equal{} AB,$ between $ B,$ $ C,$ because of $ BC > AB.$

Through $ B',$ $ C,$ we draw two parallel lines such that $ \angle BB'X' \equal{} \angle BCY' \equal{} \angle B,$ where $ X',$ $ Y',$ their intersection points from the line segment $ BY.$

From the congruent triangles $ \bigtriangleup ABX$ and $ \bigtriangleup BB'X',$ we have that $ B'X' \equal{} BX$ $ ,(1)$

From $ BC > BB'$ $ \Longrightarrow$ $ CY' > B'X'$ $ ,(2)$

From $ \angle BCY < \angle BCY'$ $ \Longrightarrow$ $ CY > CY'$ $ ,(3)$

From $ (1),$ $ (2)$ $ (3)$ $ \Longrightarrow$ $ CY > B'X' \equal{} BX$ and the proof of the [b][size=100]part (b)[/size][/b], is completed.

Kostas Vittas.

It is serious work and very good ideas. It seems to be OK, but I'm not sure about (a) (3) and (b) (3). (b) (3) is more suspicious (If B+w>90 it may not be true that CY>CY' - they for example may be equal). What do you think?

[quote="borislav_mirchev"]It is serious work and very good ideas. It seems to be OK, but I'm not sure about (a) (3) and (b) (3). (b) (3) is more suspicious (If B+w<90 it may not be true that CY>CY' - they for example may be equal). What do you think?[/quote]
Thank you dear [b][size=100]borislav_mirchev[/size][/b], for your interesting remark.

I think that when the point $ X',$ is between $ B,$ $ Y,$ or $ X'\equiv Y,$ then we can easy to say that $ CY > B'X' \equal{} BX.$

$ \bullet$ Because of $ \angle A > \angle B > \angle C,$ we conclude that $ \angle C \leq 60^{o}$ and then we  can easy to show that we have the maximum rate of the segment $ B'X',$ when $ CA \equal{} CB$ and $ \angle \omega \equal{} \angle C.$

In this case, indeed we have $ \angle B \plus{} \angle \omega > 90^{o}$ and also the point $ X',$ lies on the extension of $ BY,$ towards the point $ Y$ $ ($ $ Y,$ between $ B$ and $ X'$ $ ).$ It is also true the result that $ CY > B'X' \equal{} BX$ and, I think, may be proved based on the bellow [b][size=100]Lemma[/size][/b].

[b][size=100][color=DarkBlue]LEMMA. – An isosceles triangle $ \bigtriangleup ABC$ is given $ ($ $ AB \equal{} AC$ $ )$ and let $ D$ be, an arbitrary point between $ A,$ $ C.$ The line through vertex $ C$ and parallel to $ BD,$ intersects the sideline $ AB,$ at one point so be it $ E.$ Prove that $ DE > BC$[/color][/size][/b].

So, I can’t think the case with $ CY \equal{} B'X' \equal{} BX,$ you mentioned.

Best regards, Kostas Vittas.

PS. Give me please little time for more details, because I am a little confused and very tired.

To Kostas Vittas: I want your solution to be correct. It is a difficult problem I think for solution of from long time ago without results. My arguments to write that when B+w>90 we may have CY=CY' are the following. When CY' is perpendicular to BY it is easy to see CY>CY'. But when BY' is "long enough" CY' increases and may be far longer than CY. So when CY>CY' and CY<CY' may be true - why not CY=CY'? If you have any questions, please ask me. I don't understand your solutions to the last detail because you use very interesting constructive approach. Please write it with as many details as it is possible.

To Darij Grinberg: Darij, what do you think about that solution? Is it correct? Please, check all details - especially the case (b) (3). I think vittasko is very good with geometry but the scepticism is infectious and I want to be sure with all details.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. – An isosceles triangle $ \bigtriangleup ABC$ is given $ ($ $ AB \equal{} AC$ $ )$ and let $ D$ be, an arbitrary point between $ A,$ $ C.$ The line through vertex $ C$ and parallel to $ BD,$ intersects the sideline $ AB,$ at one point so be it $ E.$ Prove that $ DE > BC$[/color][/size][/b][/quote]
[b][size=100]PROOF[/size][/b]. – See the [b][size=100]schema t=99674(c)[/size][/b].

Through vertex $ B,$ we draw the line $ BS,$ such that $ \angle SBD \equal{} \angle ADB \equal{} \angle ACE,$ which intersects the line segment $ CE$ at a point, so be it, $ E'.$

Because of $ \angle ADB \equal{} \angle ACB \plus{} \angle CBD,$ we conclude that $ \angle ADB \equal{} \angle SBD > \angle ABD$ and so, the point $ E',$ lies between $ C,$ $ E.$

Hence, we conclude that $ DE > DE' \equal{} BC$ $ ($ from the isosceles trapezium $ CDBE$ $ )$ and the proof is completed.

$ \bullet$ Return now to our problem, we will try to show how it is changed the segment $ B'X'.$

The point $ B',$ is always between $ B,$ $ C$ and since the point $ X'$ lies on a line passes through $ B',$ The segment $ B'X'$ becomes maximum, when the angles $ \angle B$ and $ \angle \omega,$ take their maximum rate $ ($ we remind that we have $ \angle A > \angle B > \angle C$ and then, $ \angle C \leq 60^{o}$ and $ \angle B \leq \angle A.$ $ ).$

That is, when $ \angle B \equal{} \angle A$ and $ \angle \omega \equal{} \angle C$ then, the segment $ B'X',$ becomes maximum.

In this case (see the [b][size=100]schema t= 99674(d)[/size][/b]), we can also prove that $ CY > B'X' \equal{} BX,$ based on the above [b][size=100]Lemma[/size][/b] as follows:

Through the point $ X',$ considered between $ Y,$ $ Y',$ that is on the extension of $ BY,$ towards $ Y,$ we draw the line parallel to $ B'Y,$ which intersects $ BC,$ at a point so be it $ X'',$ between $ B',$ $ C$ ( because of the point $ X',$ is between $ Y,$ $ Y'$ ).*

So, based on the above [b][size=100]Lemma[/size][/b], we have that $ YX'' > B'X'$ $ ,(1)$

It is easy to show that $ CY > YX''$ $ ,(2)$ because of the acute angles $ \angle BB'Y$ and $ \angle BCY \equiv \angle C.$

Hence, from $ (1),$ $ (2),$ we conclude that $ CY > B'X' \equal{} BX$ and the proof is completed.

Kostas Vittas.

(*) This point is always true if the point $ X',$ is between $ Y$ and $ Y'',$ where $ Y'',$ is the intersection point of the line segment $ BY,$ from the line through $ C$ and parallel to $ B'Y.$ So, it must be searched if the point $ X',$ lies always between $ Y,$ $ Y'',$ when it lies on the extension of $ BY,$ towards $ Y.$

It is very hard work. I'll read carefully all materials you sent and say my opinion. I'm interested about opinions from all skilled (as Darij Grinberg, pontios, Virgil Nicula /I'm sorry if I skip someone - I don't know all people in mathlinks.ro/) people that read our discussion and if this approach work for triangles with one of its angles not less than 90 degrees.

I think it is better to use the below [b][size=100]Lemma 1[/size][/b], instead the one I mentioned, when the segment $ B'X',$ becomes maximum.

[b][size=100][color=DarkBlue]LEMMA 1. – An isosceles triangle $ \bigtriangleup ABC$ is given, with $ AB \equal{} AC > BC$ and let $ D,$ $ E,$ two point on its side segments $ AC,$ $ AB$ respectively, such that $ \angle CBD \equal{} \angle ACE \equal{} \angle BAC \equal{} \angle \omega.$ Prove that $ AE > CD$[/color][/size][/b].

[b][size=100]PROOF[/size][/b]. – See the [b][size=100]schema t=99674(e)[/size][/b].

We denote as $ F,$ the intersection point of $ DB,$ $ CE.$

It is easy to show that the triangles $ \bigtriangleup BCD,$ $ \bigtriangleup CDF$ are isosceles ones, with $ BC \equal{} BD$ and $ CD \equal{} CF$ $ ,(1)$

From the isosceles triangle $ \bigtriangleup EAC,$ we have $ AE \equal{} CE$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ AE \equal{} CE > CF \equal{} CD$ and the proof is completed.

$ \bullet$ Return again to our problem, see the [b][size=100]schema t=99674(d)[/size][/b], it is better now to show that $ CY > BX \equal{} B'X',$ from the isosceles triangle $ \bigtriangleup CAB$ $ ($ $ CA \equal{} BB$ $ ),$ as a direct application of the above [b][size=100]Lemma 1[/size][/b].

kostas Vittas.

1. Is the first lemma always true? See c.pdf. What if $ \angle ADB \equal{} 165^{o} \angle ABC \equal{} \angle BCA \equal{} 85^{o} \angle BAC \equal{} 10^{o}$ ? (In this case E' is not on CE - between C and E.)
2. I'm not sure but I think with your lemmas you prove only that the statement is true when B'X' achieves its maximum.

Strongest proof I've is when a>=b>=c the triangle is equaliteral but in the case a>=c>=b I think it may not be equaliteral or if it is equaliteral - no one can prove it is equaliteral.

[quote="borislav_mirchev"]1. Is the first lemma always true? See c.pdf. What if $ \angle ADB \equal{} 165^{o} \angle ABC \equal{} \angle BCA \equal{} 85^{o} \angle BAC \equal{} 10^{o}$ ? (In this case $ E'$ is not on $ CE$ - between $ C$ and $ E$).[/quote]

I think that the first [b][size=100]Lemma[/size][/b], is always true. That is, always is true the result that $ DE > BC.$

Indeed, in the case you mentioned, with $ \angle ADB \equal{} 165^{o}$ and $ \angle ABC \equal{} \angle BCA \equal{} 85^{o},$ $ \angle BAC \equal{} 10^{o}$ then, the point $ E',$ lies on the extension of $ BC$ towards $ C,$ $ ($ not between $ C,$ $ E$ $ ).$

But in this case, it is clear that we don’t need the [b][size=100]Lemma[/size][/b], because of we have $ DE > DB > BC,$ from $ \angle BCD \equal{} 85^{o} > \angle BDC \equal{} 15^{o}.$

[quote="borislav_mirchev"]2. I'm not sure but I think with your lemmas you prove only that the statement is true when B'X' achieves its maximum.[/quote]

I think this is not correct. When the point $ X',$ lies between $ B,$ and $ Y,$ or $ X'\equiv Y,$ then it has already been proved that

$ CY > CY' > B'X' \equal{} BX,$ $ ,(1)$ without using any of the [b][size=100]Lemmas[/size][/b].

We use one of the [b][size=100]Lemmas[/size][/b] and better the [b][size=100]Lemma 1[/size][/b], in order to prove that this result is also true, when the segment $ B'X',$ becomes maximum.

So, we conclude that the result $ (1)$ is also true, when the point $ X'$ lies on the extension of $ BY$ towards $ Y,$ till the segment $ B'X'$ to be maximum.

$ \bullet$ I think that the follows configuration, help us to show better that the result $ (1),$ is true in all cases of an acute (or not acute) triangle.

Let $ \bigtriangleup ABC$ be, an equilateral triangle.

We define the point $ A',$ as variable between $ A,$ $ B$ and then, we have the configuration of an acute triangle $ \bigtriangleup A'BC$ $ ($ not acute when $ A',$ lies between $ B$ and midpoint $ M$ of the side segment $ AB$ $ ),$ where $ BC > A'C > A'B$ and never mind ( for a moment ) because $ \angle A'BC \equal{} 60^{o}.$

We define two points $ X,$ $ Y,$ on side segments $ BC,$ $ AC$ of $ \bigtriangleup ABC$ respectively, such that

$ \angle BAX \equal{} \angle CBY \equal{} \angle \omega \leq \angle BCA'$ $ ($ not $ \angle BCA$ $ ).$

Through the point $ A',$ we draw the parallel line to the line segment $ AX,$ which intersects $ BC,$ at a point so be it $ X'$ and we denote the point $ Y'\equiv BY\cap A'C.$

It is easy to show that for every place of $ A',$ between $ A,$ $ B$ and for $ 0^{o} < \angle \omega \leq \angle BCA',$ we have $ CY' > CY \equal{} BX > BX'$

$ \Longrightarrow$ $ CY' > BX'$ $ ($ because of the acute angles $ \angle BCY > \angle BCY'$ $ ).$

$ \bullet$ Now, how may be variable the point $ B$ on the line segment $ BC,$ when the point $ A',$ has been defined?

The point $ B,$ may be variable only between the points $ B',$ and $ B'',$ where $ \angle CB'A' \equal{} \angle CA'B'$ and $ A'B'' \equal{} A'C,$

$ ($ we remind that we have supposed that in the triangle $ \bigtriangleup A'BC,$ we have always $ BC > A'C > A'B$ and so, $ \angle A' > \angle B > \angle C$ and then must be $ A'B \leq A'C$ and $ \angle B \leq \angle A'$ and $ \angle \omega \leq C \leq 60^{o}$ $ ).$

In the case of the isosceles triangle $ \bigtriangleup CA'B'$ $ ($ $ \angle CB'A' \equal{} \angle CA'B'$ $ ),$ it has already been proved that it is true the result $ (1).$

In the case of the isosceles triangle $ \bigtriangleup A'B''C$ $ ($ $ A'B'' \equal{} A'C$ $ ),$ it is also true the result $ (1),$ and it is easier to prove it.

Hence, we conclude that the result $ CY > BX,$ is true, in all cases of an acute, or not acute triangle.

Best regards, Kostas vittas.

PS. I enjoyed the problem dear [b][size=100]borislav_mirchev[/size][/b] and thank you for it. But I think that I have not to say something more.

Thank you, vittasko. I don't understand your solution but it is not a reason to be wrong or correct. I think I don't understand your solution because it is in many posts and it is "not standard" and uses very interestning approaches. If your ideas are correct you are genious and it is very beautiful solution. I'm happy also to see an opinion that the statement is true for all triangles.
Darij, what do you think about the solution? Is it correct? If it is may you (or someone else) write it in systematized way (and with descriptions for people like me).

From this time to now this problem may be called "The theorem of Bobi".

Take a look at this: http://www.mathlinks.ro/viewtopic.php?p=1131812#1131812


There are n piles of stones in a line. If a pile contains at least two stones more than the pile on its right, then a stone can be moved to the pile on its right. Initially, all the piles are empty except the leftmost pile which has n stones. If more than one move is possible, then a possible move is chosen arbitrarily. Show that after a finite number of moves, no more moves are possible and that the final configuration is independent of the moves made. Describe the final configuration. For example, we n = 6, one might get successively: 5 1, 4 2, 4 1 1, 3 2 1.

[hide="Proof that it eventually stops"]
Let pile P_i be the pile that is ith from the left. Then, consider the number n*(number of stones in pile P_1) + (n-1)*(number of stones in pile P_2) + ... + 1*(number of stones in pile P_n.) Every time we perform the operation, this number decreases by 2. It starts out as a finite number, and it always must be positive, so finitely many operations can be performed.
[/hide]


Putnam
induction
algorithm
inequalities
college contests

=============================
Let a1, a2, ..., a_n and b1, b2, ..., b_n be nonnegative real numbers.
Show that
(a1 a2 ... a_n)^{1/n} + (b1 b2 ... b_n)^{1/n} <=
( (a1+b1) (a2+b2) ... (a_n + b_n) )^{1/n}
=============================

well I think problem two is trivial since amfulger (i'm not sure) already posted somewhere yesterday. the solution follows easily by induction after 2^k, k being a positive integer. After that it's easy to see that n+1 imples n, (just take an+1 = n\sqrt a1a2...an and bn+1 = n\sqrt b1b2...bn). :D

Is this problem trivial because I solved it?

no. I also solved it with the same algorithm some time ago. it took me a while to remember where did I found it, but eventually http://www.kalva.demon.co.uk/irish/irish92.html :D

it certainly isn't a putnam problem. is it just me, or this year's putnam isn't what it's supposed to be? :)

Indeed, the problems of this year are easier than those from other years. But this is because two problems were already well-known: A2 and A-6. I don't know if you solved that problem without knowing it. I find it very difficult. And we shouldn't forget problems A5 and B6, which are quite difficult. Especially problem B6 is extremely beautiful and difficult.

[quote="harazi"]I don't know if you solved that problem without knowing it. I find it very difficult. [/quote]
I did solve it without knowing the problems' solution when I first saw it (a few years ago). It seems extremely easy to me. :D

It's easy only because you knew a similar proof for the AM-GM inequality.

yes of course. :) but who goes to putnam should know at least how to prove AM-GM for n numbers ... ! :D

If you read carefully my post, then you will find that I didn't say I find difficult problem A2, but problem A6. Indeed, A2 is not difficult-god knows I have solved lots of much more difficult inequalities- but it's not trivial.

AM-GM: what is it, and when should we have learned that?

This year's Putnam was not kind to me - I earned a 4 after getting a 28 my Freshmen year

[quote="tom11784"]AM-GM: what is it, and when should we have learned that?
[/quote]
Arithematic-mean is not smaller than Geometric-mean, i.e.,
\[
\frac{a_1+a_2+\cdots+a_n}{n} \geq \sqrt[n]{a_1a_2\cdots a_n}
\]
for $a_i\geq 0$.


Find $ \lambda$ (element of R) for which the equation $ (x\plus{}1)^4 \minus{} \lambda(x\plus{}1)^2.(x^2 \plus{}2x \plus{}2) \plus{} (\lambda \minus{}1).(x^2 \plus{}2x \plus{}2)^2 \equal{}0$ has at least one real root.

Let (x+1)^2=y. That "element of R" be a  :D
So we can rewrite the problem:
y^2 -ay(y+1)  + (a-1)(y+1)^2 = 0
Or: y^2  - ay^2 - ay + (a-1)y^2 + 2(a-1)y + a-1 =0
 => y(2a-2-a) + a-1 = 0
 => y(a-2) + a-1 = 0
 => a can't be 2 and y=(1-a)/(a-2)
Because y>=0 so a must be in [1;2]

I did the same but the book says $ \lambda$ lies in $ (\minus{}\inf , 2)$

[quote="Conjurer"]I did the same but the book says $ \lambda$ lies in $ ( \minus{} \inf , 2)$[/quote]
How can $ \lambda$ lie in $ (\minus{}\infty,2)$?
Can you post the book's solution?
I think Nguyen Ngoc Linh is right.

Who is the author of that book?  :D  :D  :D

Subtract $ 1$ from each side and factor: $ (\lambda \minus{} 2) ((x \plus{} 1)^2 \plus{} 1) \equal{} \minus{} 1$

$ \minus{} (\text{min} (x^2 \plus{} 1))^{ \minus{} 1} \leq \lambda \minus{} 2 \leq \minus{} (\text{max} (x^2 \plus{} 1))^{ \minus{} 1}$

$ \minus{} 1 \leq \lambda \minus{} 2 < 0$      [i]this became an open interval because the max in the previous line is unattainable[/i]

$ \boxed{1 \leq \lambda < 2}$

The book must have thought that $ (x \plus{} 1)^2 \plus{} 1$ could be any positive number?


calculus
integration
trigonometry
logarithms
LaTeX
calculus computations

integral of (dx/sqrt(9+x^2))

i tried letting x = 9tan(theta) but it didn't work.  can anyone help me?

how do i get this math font that you guys type with?

Look for the parts of this site that refer to "Latex" (or, $\LaTeX$). There are tutorials, and there's a forum for discussing it.

The short version. Put what you want between dollar signs. Code words start with backslashes, and many things are fairly easily mnemonic: \int for integral, \sqrt for root, \frac for fraction, \theta for $\theta.$ Enclose things that need to be enclosed with brackets: {stuff}.

If I type (dollar)\int \frac{dx}{\sqrt{9+x^2}} (dollar) only put in real dollar signs, I get:

$\int \frac{dx}{\sqrt{9+x^2}}$

To see the code in someone else's post, just hold your mouse over the expression or click on it. If you're trying it for yourself, be sure to use the preview button: everyone, even the experienced people, make mistakes with this.

It's called LaTeX. I know what it's like not to know it...but I read a few of the LaTeX posts by Valentin and he made it as easy as $\{x \in \mathbb{N} \mid 1\le x \le 3 \}$, if you know what I mean.  :)


logarithms
real analysis
real analysis unsolved

How would you solve the equation
$x^{x}= ax$
for x using the Lambert W function.

I have spent more than half-an-hour on this problem and I just can't get it to the form $C = f(x)e^{f(x)}$.

$(x-1) \ln x = \ln a$.

Also, if $a = C$, then $\ln f(x)+f(x) = (x-1) \ln x$.

Thus, $\ln (f(x) / x) = x \ln x-f(x)$.

Therefore, $\frac{f(x)}{x}= e^{x \ln x-f(x)}$.

Idk where to go from there...

That's pretty much what I was doing. But that doesn't seem to lead us anywhere. Anyone else got some smart ideas/tricks?


algebra
polynomial
Vieta
geometry
3D geometry
algebra unsolved

Complete the cube and then use the method of vieta to find all roots of x^3 + 6x^2 +9x + 3 = 0 in the form a cos(w) + b for suitable a,b and w

This is a double post and has been answered elsewhere.

Here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=187053

Pierre.


Has anyone here tried the books, the Art of Problem Solving VI, VII?

Any comments?

Are they useful for competitions?  :wink:

Oh, never mind, V means volume right?? Okay then, sorry about that.

They are extremely useful... Im not doing this for brownie points either... They are, in my opinion, the best starting books for a competitive math student.

EPIC NECROPOST BATMAN!!!!!!!! :trampoline:  :ddr:   :stretcher:  :wow:

[quote="Chinaboy"]And make sure you completely understand the 1st book before you move on to the 2nd.[/quote]

That's not necessary.
The first time I read VII a few years ago, there were definitely some problems in V1 I couldn't solve.

I understood about half of V2 the first time through; after a approximately one month aime training session I read V2 again, this time much more slowly; I underestood approximately 90% of the material.

Only after I went through ACoPS and started olympiad training did I actually understand some of the topics in there like inversion the last geo chapter in V2.

Very useful books, i went from a 100ish AMC 10 to a 140 in from just the 1st book, and with the other AOPS curriculum and V2, i get around 9 on the AIME, and these are problems which 5-6 months ago i looked at and wanted to go cry in a corner.

Granted, don't for a minute think these books teach "shortcuts" and "tricks".  These aren't like ACT prep books.  No, these books teach you how to THINK and that's what's most important for math competitions.


Which rhymes more, Missouri and curry or Texas and breakfast? Lol, random.

Of course Texas and Breakfast!

Just use Soulja boy!

honestly, texas and breakfast DO NOT rhme even with any accent. this is because the kf sound in beakfast does not ever match up with the x sound in texas even with an accent. THIS DOES NOT MEAN IT DOES NOT WORK IN A SONG.

the "ouri" in MIssouri actually does have the same sound as the "urry" in curry, so these work MORE. and sont use the argument that texas and breakfast works in a song, b/c missouri/curry also works, and it works even more

oops i didnt see i double posted until now... could a mod delete this?

Misouri and curry actually do make a good rhyme, even though it would be really hard to use them in a song. In my oppinion Texas and breakfast don't rhyme at all.

[quote="stevenmeow"]oops i didnt see i double posted until now... could a mod delete this?[/quote]

If you were the last poster, you can always delete your own post.  :wink:

Why are there a lot of people that are saying texas and breakfast rhyme?

if u pronounce them with a japanese/asian accent, they're like this:

texas: teh-kuu-SAA-su
breakfast: buu-lek-fast

missouri: mee-su-o-li
curry: kuu-li

so i think missouri and curry ryme mor if u hav bad pronounciation. also i like curry  :lol:

[quote="tinytim"]Everyone from texas except stevenmeow probably said texas and breakfast rhyme[/quote]

I'm from TX and I know Texas doesn't even come close to rhyming with breakfast. That being said, if purple, orange, and silver are the only reasonable words that have no rhymes, what does rhyme with breakfast?

[quote="tinytim"]Everyone from texas except stevenmeow probably said texas and breakfast rhyme[/quote]

hey! you can't just assume that i think tx and breakfast rhymes

Ahh, this argument again.

Ok. I would like to point out that Missouri and Curry, in fact, do not rhyme.

I'm Indian. I know what curry is, and how it's pronounced. the u in curry is pronounced like the uh in "Uh, where the heck did I leave my Chuck Norris action figure?"

Curry then is like cuhry.

Missouri's ou is pronounced like errr... so like yeah.

They dun rhyme.

Which leaves Texas and Breakfast.

FRIGGIN PUT IT IN SOULJA BOY IT WORKS!!!

it all depends on the way u pronaunce it. like in chinese curry is cuu-lai.

Uh, yeah, but...

CURRY IS INDIAN.

Don't we get first dibs on pronunciation?

ur flaggy says ur singaporean! (or whatever it is.) plus, japan has Japanese Curry that isnt the same as indian curry!

True, I was just born in Singapore.

Dude. Brown ppl invented curry.

When you think Indian, u think curry. :P

When you think Japanese, u think, uh, sushi. Yeah.

Racial Stereotypes 101 :D

[quote="tinytim"]everyone except dingzhou and stevenmeow. :lol:[/quote]

hey! you can't just assume i don't think tx and breakfast rhymes.

When I think Japanese I think Sky City. It's freakin' awesome!

Texas and breakfast sound so similar, and I don't care if they don't rhyme. "Watch all of Texas eat their breakfast!" sounds good.

@arvind_sn:
when i think Japan, i think samurai and curry. actually i never heard of indian curry until 4th grade. i likes japanese curry more. although it's true that indians were first to invent curry.

@1=2:
what is sky city???

http://dsc.discovery.com/convergence/engineering/skycity/interactive/interactive.html

Image: [img]http://www.takenaka.co.jp/takenaka_e/engi_e/c02/img/c02_1_03.jpg[/img]

Size: 2/3 mile high, 14 subsections. http://www.youtube.com/watch?v=3-bBuBT1oYU is the first part of the episode.

Back on topic now.

1. Not yet.
2. [size=200]PLEASE DON'T POST IN TOPICS THAT ARE MORE THAN THREE MONTHS OLD LIKE THIS ONE![/size]


use a coordinate proof to show that the three altitudes of a triangle always intersect at one point. in other words proof the existance of orthocenter.

this shouldn't be too hard for you guys since ur all so good.  :rotfl:

Why do you want to use analytic?

Let $B$ and $C$ be on the x-axis and $A$ on the y-axis. The show that the altitude of $AB$ and $AC$ cut the y-axis at one point.


When a positive number is multiplied by the sum of twice the
number and half the number, the result is the original number.
What is the number? Express your answer as a common fraction.

We translate the English into:

$ x(2.5x) \equal{} x$.  Solving gives
$ x \equal{} \boxed{\frac{5}{2}}$


search quadratics algebra polynomial calculus integration derivative

suppose that $ P,Q\in \mathbb{Z}[x]$.do you know anything about solving equation $ P(x)\equal{}Q(y)$ in integers?is it open problem?in which special case there is a solution?
for example $ x^k\minus{}dy^k\equal{}D$
I am very interested to everything about this equation and it's special cases...

Even the case that $ \text{deg } P, Q \le 3$ or $ \text{deg } P \equal{} 2, \text{deg } Q \equal{} 4$ is the problem of finding integer points on [url=http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=Hjo&q=integer+points+on+elliptic+curves&btnG=Search]elliptic curves[/url], a deep problem about which much is known, but progress is far from complete.

Indeed, already $ x^3\minus{}dy^3\equal{}c$ is hard. Note that up to "obvious" exceptions (singular curves), all equations of degree $ 3$ or higher have only finitely many integer solutions (unlike linear ones and by Pell quadratic ones).

thanks a lot friends
which equations are obvious?can you say in mare detail?what do you mean by 'equations of degree $ 3$ or higher'?

The "obvious" exceptions include examples like $ x^3 \equal{} 27y^3$ which obviously have infinitely many solutions because the polynomial $ x^3 \minus{} 27y^3$ is reducible.  ZetaX means any polynomial in two variables containing a term of degree $ 3$ or higher, and the result to which he refers is [url=http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points]Siegel's theorem[/url].

There are certain easy cases where it is possible to prove that there are no or few integer points by elementary arguments ranging from the appropriate modulus to some work over a field extension.  For example, you can show that $ x^3 \plus{} y^4 \equal{} 19^{19}$ has no integer solutions by working $ \bmod 13$.  This does not help at all in elucidating the general case, though.

what about $ x^4 \equal{} y^5$?it has infinitely many solutions,but $ x^4 \minus{} y^5$ is irreducible.
is the following true?
$ P,Q\in \mathbb{Z}[X]$ and $ max\{deg(P),deg(Q)\}\ge3$ and  the equation $ P(x) \equal{} Q(y)$ have infinitely many solutions,then there exist $ F,H\in \mathbb{Q}[X]$
such that $ P(F(x)) \equal{} Q(H(x))$ for all $ x$.

The $ x^2\minus{}y^3$ or the $ x^4\minus{}y^5$ example shows that we really need to use nonsingular curves. That means: there are no multiple points or cusps.
Effectively, it means we can find a unique tangent for every point on the curve (see http://en.wikipedia.org/wiki/Non-singular ).

is it true that every nonsingular curve with $ deg\ge3$ have only finitely many integer point?
suppose that $ P\in \mathbb{Z}[X,Y]$ then $ P$ is singular if there exist $ (a,b)$ such that  partial derivatives of $ P$ with respect to both x and y vanish at $ (a,b)$.am i right?

by the way $ y^3\plus{}y\minus{}x^3\minus{}x\equal{}0$ is nonsingular curve and therefore the equation $ y^3\plus{}y\equal{}x^3\plus{}x$ have only finitely many solution,which is obviously wrong,since for $ x\equal{}y$ equation hold.where am i wrong?

There is a singularity $ x \equal{} y \equal{} \frac {i}{\sqrt 3}$. Note that smooth is defined over an algebraic closure of the field in question, so the singular points are allowed to have algebraic coordinates (not only rational ones).

sorry,I don't understand something:(
for every curve of the form $ P(x) \minus{} Q(y) \equal{} 0$,with $ deg(P),deg(Q)\ge2$ there exist $ (a,b)\in\mathbb{C}^2$ such that partial derivatives of $ P\minus{}Q$ with respect to both x and y vanish.and so every curve of the form $ P(x) \minus{} Q(y) \equal{} 0$ is singular...?!
is there an elementary way to find when one curve is singular?

That's not true.  (Have you actually tried working out an example?).  One well-known class of nonsingular curves are those of the form $ y^2 \equal{} x^3 \plus{} ax \plus{} b$ where the polynomial on the RHS doesn't have repeated roots; these are elliptic curves in Weierstrass form.  Another example is the Fermat curve $ x^n \plus{} y^n \equal{} 1$ (although here we're interested in rational rather than integer points).

One way to tell if a cubic curve is singular is to write it in Weierstrass form (there is a general way to do this) and then check if the cubic on the RHS has repeated roots or not.

ali666: you need to have $ (a,b)$ to be on (complex extension of) the curve (not arbitrary in $ \mathbb C$).

yes,now I get it,sorry:blush:
what about $ x\equal{}y^3$?it is nonsingular and with $ deg\ge3$ and have infinitely many integer point...

Correct me if I'm wrong, those who know better, but I believe that curve has a singularity at infinity.  In other words, embedded in the projective complex plane (where it is given by $ XZ^2 \equal{} Y^3$) it has a singularity at $ (1 : 0 : 0)$.  This is a somewhat subtle issue.


cool i'll be seeing you guys there.

So I accidentally deleted the email with all the forms we needed to fill out. Could any of you guys forward me a copy? I'll pm you guys about it.

Just wondering, what did you guys get for the kangaroo problem?

The problems are reused, so I don't think you should discuss them.

Alright.

The message is too small. Please make the message longer before submitting.


Let n>4 be an integer. Prove that among any n distinct integers chosen from 1,2,...,2n there exist two whose lowest common mulitple is at most 3n+6.

Bomb

See problem 204, pp3, [url=http://www.math.ust.hk/excalibur/v9_n3.pdf]volume 9.3[/url], Mathematical Excalibur.


linear algebra
matrix
linear algebra unsolved

Let $A^2 = A$ and $B^2 = B$.
Show that the eigenvalues of $AB$ are real and belong to the segment $[0, 1]$.

As stated, it's not correct (or not entirely, at least). Take $A=\begin{pmatrix}1&0\\0&0\end{pmatrix},\ B=\begin{pmatrix}2&-1\\2&-1\end{pmatrix}$, and $x=\begin{pmatrix}1\\0\end{pmatrix}$. You have $ABx=2x$, so the eigenvalues certainly aren't in $[0,1]$.

Edit:

It does, however, work if you have the extra condition that $A,B$ are [i]orthogonal[/i] projections, i.e. $A^*=A,B^*=B$, in addition to $A^2=A,B^2=B$.

Let $\lambda\ne 0$ be an eigenvalue of $AB$, and let $x\ne 0$ be an eigenvector corresponding to $\lambda$. Also let $(\cdot,\cdot)$ denote the usual scalar product in $\mathbb C^n$. We have $ABx=\lambda x$. Multiply this with $A$ to the left to get $ABx=\lambda Ax$. Since $\lambda\ne 0$, we get $Ax=x$. Now $\lambda\|x\|^2=(ABx,x)=(Bx,Ax)=(Bx,x)=(B^2x,x)=(Bx,Bx)=\|Bx\|^2$, and all the assertions follow: $\lambda$ is real, obviously $\ge 0$, and it must also be $\le 1$ because the norm of any orthogonal projection (such as $B$) is $1$, so $\|Bx\|\le\|x\|$.


Are there any states not currently represented in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=249]Local Communities[/url] forums that an AoPSer would be interested in moderating?

[quote="EUCLA"]italy  :D
but it doesn't matter[/quote]

You'll have to request that one from Val, but the nation communities will grow more slowly here than the state communities simply because there is a much greater amount of work behind establishing each.

Well I wouldn't like to create a community that is not used too much. But since I'm in a good mood  :P and the Hungarian community is already created, I'll just changed that to Italian :D

[quote="EUCLA"]thanks...now i'll wait for someone to talk  :)[/quote]You would have to start talking yourself, otherwise all people will wait for someone else to talk first, and no one will talk :D

taht's a good idea-- i'm really stupid :lol:

[quote="Valentin Vornicu"]But since I'm in a good mood  :P and the Hungarian community is already created, I'll just changed that to Italian :D[/quote]

:lol: I wonder how you got this wonderful idea, Valentin! (I'm wondering whether to put inverted commas around "wonderful" or not. ;) )

[quote="Valiowk"]:lol: I wonder how you got this wonderful idea, Valentin! (I'm wondering whether to put inverted commas around "wonderful" or not. ;) )[/quote] :?  :?  :huh:


induction
combinatorics solved
combinatorics
duplicate

Lagrangia posted this problem a day or two ago and asked for some ideas :

A (2k+1) x (2k+1) chessboard in coloured in white and black in the usual way such that the four corners are black. For which k is it possible to cover some squares on the table with trominoes (L-shaped figures made up from 3 squares) such that all the black squares are covered ? For these values of k , what is the minimal number of trominoes ?

I have a solution but I will post it tonight because I don't have time now.Anybody else solved it?

[color=#cc0000][moderator says: see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=17337 ][/color]

It goes like this :

I claim that it's true for k>=3 (so the board must be at least 7x7) and that the minimum number is (k+1)^2. Let f(k)=the minimum number of trominoes needed for a (2k+1) x (2k+1) board.

First , colour the board like this , starting from the upper left corner (this example is for k=3) :

http://br.crashed.net/~slash/zzz/fig1.jpg

Notice that there are exactly (k+1)^2 letters A on the board and each of them is on a black square . Also a tromino covers AT MOST one A so we need at least (k+1)^2 trominoes.Thus , f(k)>=(k+1)^2 .

But each tromino occupies three squares and since they don't overlap we need : 3*f(k)<=(2k+1)^2 <=> 3k^2 + 6k + 3 <= 4k^2 + 4k + 1 , or (k-1)^2 >= 3 , which means (k-1)^2 >= 4 , or k>=3 .

Next , we give this arrangement for a 7 x 7 board :

http://br.crashed.net/~slash/zzz/fig2.jpg

Now we prove by induction that f(k)=(k+1)^2 . Consider a (2k+3)x(2k+3) board and cover the lower left (2k+1)x(2k+1) "corner" by induction with (k+1)^2 trominoes. We are left something on the side , which we cover like in the next picture :

http://br.crashed.net/~slash/zzz/fig3.jpg

The number of trominoes we have added is exactly the number of A's we
would have added , so it's (k+2)^2 - (k+1)^2 .
Thus , f(k+1) <= (k+2)^2 - (k+1)^2 + f(k) and since  we also have
f(k)=(k+1)^2 and f(k+1)>=(k+2)^2 , we get that f(k+1)=(k+2)^2.

Nice problem!

(p.s. : sorry for the lousy hand-drawing :D )


quadratics
abstract algebra
number theory unsolved
number theory

There are n students seated in a circle during the break. The teacher starts distributing chocolates in a clockwise fashion starting from say student1. He then skips 1 student and then gives a chocolate to student 3 . He then skips 2 students and gives a chocolate to student 6 and so on in such a pattern .
Find all values of n for which all students get atleast a chocolate after a finite number of rounds of distribution.

The problem is equivalent to find all $ n$  such that if $ 0\leq r<n$, there is $ m$ with $ \frac{m(m\plus{}1)}{2}\cong (r) mod(n)$.
$ \frac{m(m\plus{}1)}{2}\cong (r) mod(n)$ iff $ (2m\plus{}1)^{2}\cong (8r\plus{}1) mod(8n)$
If $ n\equal{}2^{k}$:
Let $ x,y$ be odd numbers such that $ x^{2}\minus{}y^{2}\equal{}(x\plus{}y)(x\minus{}y)$ is multiple of $ 8n\equal{}2^{k\plus{}3}$, then one of $ x\plus{}y$ and $ x\minus{}y$ should be multiple of $ 2$ but not of $ 4$, an the other one should be multiple of $ 2^{k\plus{}2}$.
So there is $ 2^{k}$ odd quadratic residues module $ 2^{k\plus{}3}$ and they have the form $ 8r\plus{}1$.

If $ n\equal{}2^{k}m$, $ m>1$ odd:
Let $ t$ be a nonquadratic residue module $ m$, and $ r$ such that $ 8r\plus{}1\cong (t)mod(m)$ (we can do this by the chinese remainder theorem), then we have that there is not $ m$ such that $ (2m\plus{}1)^{2}\cong (8r\plus{}1) mod(2^{k\plus{}3}m)$.


Good job Diogene, I moved it to solved problem


For $n \in \mathbb{N}$ and $\alpha \in \mathbb{R}$, let $S_n(\alpha) = \sum_{k=0}^n (-1)^{[k \alpha]}$. Prove that if $\alpha$ is irrational then $S_n(\alpha)=0$ for infinitely many $n \in \mathbb{N}$.


number theory
least common multiple
number theory unsolved

Let $ m$,$ n$ are positive integer and $ m > n$  $ \implies$ Prove that;
$ [m,n] \plus{} [m \plus{} 1,n \plus{} 1] > \frac {2mn}{\sqrt{mn}}$

first note that $ [a,b] \equal{} \frac {ab}{(a,b)}$. Let $ (m,n) \equal{} d$, $ m \equal{} da,n \equal{} db$.
Since $ x \equal{} (m \plus{} 1,n \plus{} 1) \equal{} (da \plus{} 1,db \plus{} 1)|a \minus{} b \equal{} \frac {m \minus{} n}{d}$ we get $ [m \plus{} 1,n \plus{} 1]\geq \frac {d(m \plus{} 1)(n \plus{} 1)}{m \minus{} n}\geq d$.
Now $ [m,n] \equal{} \frac {mn}{d}$ so the claim follows from AM-GM
In fact the stronger bound
\[ [m,n] \plus{} [m \plus{} 1,n \plus{} 1]\geq \frac {2mn}{\sqrt {m \minus{} n}}\]
Still holds


limit
real analysis
real analysis unsolved

I guess the answer is $+ \infty$
but can we prove $2^{a_1}3^{b_1}-2^{a_2}3^{b_2}=k$
(where k is  a positive integer) has finite pairs of $(a_1,b_1,a_2,b_2)$??


geometry
perimeter
email
geometry unsolved

Are you sure it was not a your friend's joke?

Joke or not.I don't mind it.Any solution?

[quote="mathmaniack"]Let ABC be a triangle and P is an interior point of ABC.If PA=5,PB=6,PC=7
cm find the maximum and minumum value of perimeter of ABC.Can you give a general formula?[/quote]

I have just got two emails asking me about this problem. Is it possible that there is no closed formula for the solution?

Desperately,
  Darij


If $ (x\plus{}1)^{\frac{1}{2}} \equal{} 2^{\frac{1}{3}}$, find the value of $ (x\plus{}1)^2$ in simplest radical form.

hello, a small hint, $ \sqrt[3]{16}\equal{}\sqrt[3]{8\cdot2}\equal{}2\sqrt[3]{2}$.
Sonnhard.

To express this on FTW, one must type:

[code]2sqrt[3]{2}[/code]

Is there any simpler way?

[quote="dragon96"]In FTW, is \sqrt[3]{16} acceptable?[/quote]

I don't think so because it is not simplified.


logarithms conics ellipse calculus derivative search function

[hide="Part 1"]$4x-2x^{2}= y^{2}\ge 0$
$4x \ge 2x^{2}$
$2x \ge x^{2}$

If x is positive: $2 \ge x$
If x is negative: $2 \le x$ which is impossible

I'm missing something?[/hide]

[hide="Beginning of a solution for part 2"]We complete the square: $2x^{2}-4x+2+y^{2}=2$
$2(x-1)^{2}+y^{2}=2$
$(x-1)^{2}+\frac{y^{2}}{2}=1$
This is an ellipse centered at $(1,0)$ with $a=\sqrt{2}$, $b=1$, $c=1$
Uhhhh...[/hide]

[quote="ifai"]Let $x,y$ be positive numbers satisfying $2x^{2}+y^{2}-4x=0$.
(1) Find the range of $x$;
(2) Find the maximum of $\log x+\log y$.[/quote]

[hide="Solution"]

(1) $y^{2}= 4x-2x^{2}\ge 0 \Leftrightarrow 0 < x \le 2$
(2) Let us maximize $xy$ to maximize $\log xy.$ We have that $y = \sqrt{4x-2x^{2}},$ since $y$ is positive. So, we must maximize $f(x) = x \sqrt{4x-2x^{2}}$ when $x \in (0, 2].$ If we take the derivative and set it equal to zero, we get
\[\frac{x(2-2x)}{\sqrt{4x-2x^{2}}}+\sqrt{4x-2x^{2}}=0\\ \iff x(2-2x)+4x-2x^{2}= 0\\ \iff 2x(3-2x) = 0\\ \iff x = 0, 3/2 \]
Now, note that $x \neq 0, 2,$ because this causes $f(x) = 0.$ Now, note that since $f'(x) = \frac{6x-4x^{2}}{\sqrt{4x-2x^{2}}},$ we have that $f'(1) > 0, f'(5/3) < 0 \iff f''(3/2) < 0,$ so a maximum is attained when $x = 3/2.$ Thus, the maximum of $f(x)$ when $x \in (0, 2]$ is $f(3/2) = \frac{3\sqrt{6}}{4}.$ Thus, the maximum of $\log xy$ is $\frac{3}{2}\log \frac{3}{2}.$

I'll try to search for a non-calculus based solution...

[/hide]

[quote="Karth"]So, we must maximize $f(x) = x \sqrt{4x-2x^{2}}$ when $x \in (0, 2].$[/quote]
Note that the maximum is achieved at the same point as the function $4x^{3}-2x^{4}$ achieves the maximum because sqrt is increasing function. This simplifies the calculation of the maximum point a lot.

Consulting the idea of mathnerd314...
(2) $2x^{2}+y^{2}-4x=0\,\Rightarrow\,(x-1)^{2}+\frac{y^{2}}2=1$
Then $x=1+\cos\theta$, $y=\sqrt2\sin\theta$, $0<\theta<\pi/2$.
Thus
$\log x+\log y$
$=\log xy$
$=\log[(1+\cos\theta)(\sqrt2\sin\theta)]$
$\leq...$

[quote="ifai"][quote="Karth"]
(1) $y^{2}= 4x-2x^{2}\ge 0 \Leftrightarrow 0 < x \le 2$
[/quote]
Well, $y$ is also a positive number...[/quote]

Oh, right... forgot about the rhs limit. So $0 < x < 2.$

[quote="ifai"]Let $x,y$ be positive numbers satisfying $2x^{2}+y^{2}-4x=0$.
(1) Find the range of $x$;
[/quote]

Range means formally that is is the set of all $y$ so that the equation $2x^{2}+y^{2}-4x = 0$ has a solution in at least one real $x$.

Now rewrite this equation as,
$2x^{2}-4x+y^{2}=0$
In order to has at least on real solution in $x$ for a given $y$ we need the discrimanat to be non-negative.
$16-8y^{2}\geq 0$
So,
$y^{2}\leq 2$
Solve,
$-\sqrt{2}\leq y \leq \sqrt{2}$

[quote="Karth"]It's not asking for the range, but the range of $x$  :wink:[/quote]

...in other words, the domain.


algebra
polynomial
number theory unsolved
number theory

$P(x) = x^3-x^2-5x+2$
Find all Integer $n$ such that $(P(n))^2$ are square of prime number

Observe that
\begin{eqnarray*} P ( x ) & = & ( x + 2 ) \cdot ( x^2 - 3 x + 1 ) \end{eqnarray*}
Therefore $P ( n ) \in \mathbb{P} \Longrightarrow \left[ n^2 - 3 n + 1 = \pm 1 \mbox{ or } n + 2 = \pm 1^{^{}} \right]$
Equivalently we have $n = - 3$ or $n = - 1$ or $n = 3$ or $n = 0$ or $n = 1$ or $n = 2$.
Checking each of these possibilities we conclude that
\begin{eqnarray*} P ( n )^2 = p^2 \mbox{ for } \mbox{ some } p \in \mathbb{P} & \Longleftrightarrow & n \in \{ - 1, 0, 1, 3 \} \end{eqnarray*}

Actually in any problem asking you about prime values of a polynomial, the polynomial must be reducible.
Proving that there is an infinite number of prime values taken by irreducible polynomials is a hard open problem. You guess that finding them all should be even harder. It is safe to assume it won't be asked during a contest :). That is the polynomial will always be reducible.

I don't understand why write :$P(n)^2$ is square of prime

To avoid to handle negatives of primes (but yes, thats a bit funny ),


geometry
geometric transformation
reflection
geometry solved

[quote="Rushil"]$ ABCDEF$ is a convex hexagon. Consider the following statements.

(1) $ AB$ is parallel to $ DE$.

(2) $ BC$ is parallel to $ EF$.

(3) $ CD$ is parallel to $ FA$.

(4) $ AE \equal{} BD$.

(5) $ BF \equal{} CE$.

(6) $ CA \equal{} DF$.

Show that if all six of these statements are true then the hexagon is cyclic. Prove that, in fact, five of the six statements suffice.[/quote]

The result seems to be false, indeed.
Take an equilateral triangle $ \triangle ABC$, and consider its reflection $ \triangle A'B'C'$ in a line paralel
to $ BC$, close to $ BC$ and between $ A$ and $ BC$. Then the hexagon $ AB'BA'CC'$ satisfies all the
statements in the problem, but is not cyclic.

Ociretsih is right: Not even all six statements (let alone five of them) yield that the hexagon is cyclic.

What is actually true is: If a convex hexagon ABCDEF satisfies AB || DE, BC || EF, CD || FA and AD = BE = CF, then it is cyclic. This was proven on http://www.cip.ifi.lmu.de/~grinberg/Lamoen/Support.html .

  Darij

Hey I just saw this question. Its a bit wrong. The correct question ( which I hope the mods will kindly replace the [url=http://www.mathlinks.ro/resources.php?c=78&cid=46&year=2002]original one posted in the resources section.....[/url])

For a convex hexagon $ ABCDEF$ in which each pair of opposite sides is unequal, consider the following statements.

($ a_1$) $ AB$ is parallel to $ DE$.             ($ a_2$)$ AE \equal{} BD$.

($ b_1$) $ BC$ is parallel to $ EF$.              ($ b_2$)$ BF \equal{} CE$.

($ c_1$) $ CD$ is parallel to $ FA$.              ($ c_2$) $ CA \equal{} DF$.

$ (a)$ Show that if all six of these statements are true then the hexagon is cyclic.

$ (b)$ Prove that, in fact, five of the six statements suffice.

Hey can anyone answer this question....???? :maybe:  :huh:  :|

[quote="Rijul saini"]
For a convex hexagon $ ABCDEF$ in which each pair of opposite sides is unequal, consider the following statements.

($ a_1$) $ AB$ is parallel to $ DE$.             ($ a_2$)$ AE \equal{} BD$.

($ b_1$) $ BC$ is parallel to $ EF$.              ($ b_2$)$ BF \equal{} CE$.

($ c_1$) $ CD$ is parallel to $ FA$.              ($ c_2$) $ CA \equal{} DF$.

$ (a)$ Show that if all six of these statements are true then the hexagon is cyclic.

$ (b)$ Prove that, in fact, five of the six statements suffice.[/quote]

here is the [url=http://www.isid.ac.in/~rbb/inmo2002_sol.pdf]official solution....[/url]

We can prove that each of the parallel sides form an isosceles trapezium. Further all the diagonals are concurrent at a point this proves that it can be a cyclic.


It is clear that x_n tends to 0 ( it is bounded and decresing and its limit verifies l=l-l^2). Also, from Cesaro- sTolz we find that nx_n=n/(1/x_n) tends to 1/(1/x_n+1-1/x_n), which tends to 1. So, nx_n tends to 1. Then n(1-nx_n)/lnn=nx_n* (1/x_n-n)/lnn tends to [(1/x_n)-n]/lnn which tends by Cesaro-Stolz to [1/(x_n+1)-1/x_n-1]/(ln(1+1/n)), which tends to n[ 1/(x_n-(x_n)^2)-1/x_n-1], which is nx_n/(1-x_n), which is 1.

Just one small typo this time, Harazi :)
[quote][1/(x_n+1)-1/x_n-1]/(ln(1+1/n))[/quote]
It should be $[1/x_{n+1}-1/x_{n}-1]/\ln(1+1/n)$.

I think not many people keep Cesaro-Stolz in their mind. So I will refresh our memory.

[b]Theorem[/b] (Cesaro-Stolz Lemma)
Let $\{x_{n}\}$ and $\{y_{n}\}$ be two sequences of real numbers satisfying two properties:
1) $0<y_{1}<y_{2}<y_{3}<...$ and $\lim y_{n}=+\infty$;
2) $L=\lim \frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}$ exists, finite or infinite.
Then the limit $\lim \frac{x_{n}}{y_{n}}$ exists and is $L$.

Guess what, Namdung? This was published lon, long time ago in AMM. :wink:


No one at the contest got this question correct. Can you  :) ?

How many ways are there to fill a four-by-four grid with four $1$'s, $2$'s, $3$'s, and $4$'s so that no two adjacent squares have a composite sum?

-JHMT

[hide="My attempt"]
For the purpose of the question, composite$\equiv$even, since the possible sums are from 3 to 7.

No two adjacent squares can have the same parity: but any two squares with opposing parity are ok.

Therefore we must assign the even numbers to the leading diagonal and the two parallel diagonals each with two squares in them. (or the odds in these, and the evens in the other).
There are $\binom{8}{4}$ ways we can distribute the evens around, and $\binom{8}{4}$ ways the odds are distributed. Double this, to account for the fact we can swap odds and evens around.

$\binom{8}{4}\cdot\binom{8}{4}\cdot2=\boxed{9800}$.

(Ok, this seems like a wild overcount, but whatever, I probably made a mistake somewhere)[/hide]

Yeah, at the contest I was playing around with filling in the grid and suddenly came to what I thought was a contradiction, so I put down 0.  :rotfl:

Nice solution tim1234133.


Prove that any three consecutive numbers multiplied by themselves their product will be multiple of 6.

I assume that number means integer. Among three consecutive integers, there is one divisible by three and at least one even. As (2,3)=1, the product is divisible by six.


Let $ a,b,c$ be side of triangle prove that

$ (3a\minus{}b\minus{}c)(b\minus{}c)^2\plus{}(3b\minus{}c\minus{}a)(c\minus{}a)^2\plus{}(3c\minus{}a\minus{}b)(a\minus{}b)^2\ge 0$

Is it trivial  :blush: ?

Hi, that nice but actually you have solve inequality $ p^2\ge 16Rr - 5r^2$ it is equivalent to my inequality but the hard problem is show they are equivalent  :P , thank to Virgil Nicula to help me :)

$ 9IG^2 = p^2 - 16Rr + 5r^2$
\[{ aGA^2 + bGB^2 + cGC^2 = (a + b + c)GI^2 + abc\ge abc\Leftrightarrow \sum\frac {GA^2}{bc}\ge 1\Leftrightarrow \sum\frac {2(b^2 + c^2) - a^2}{bc}\ge 9\Leftrightarrow \sum 2(\frac {b}{c} + \frac {c}{b})\ge\frac {a^3 + b^3 + c^3}{abc} + 9\Leftrightarrow \sum 2(\frac {b}{c} + \frac {c}{b} - 2)\ge\frac {a^3 + b^3 + c^3}{abc} - 3\Leftrightarrow \sum 2\frac {(b - c)^2}{bc}\ge \frac {a + b + c}{2abc}(\sum (b - c)^2)}
\]
and

$ \Leftrightarrow$ the above inequality and $ \Leftrightarrow$ trivial  :D !

[quote="Mathias_DK"][quote="encyclopedia"]Let $ a,b,c$ be side of triangle prove that

$ (3a \minus{} b \minus{} c)(b \minus{} c)^2 \plus{} (3b \minus{} c \minus{} a)(c \minus{} a)^2 \plus{} (3c \minus{} a \minus{} b)(a \minus{} b)^2\ge 0$

Is it trivial  :blush: ?[/quote]
Assume wlog $ a \ge b \ge c$. Then $ (c \minus{} a)^2 \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2$.
Thus:
$ (3a \minus{} b \minus{} c)(b \minus{} c)^2 \plus{} (3b \minus{} c \minus{} a)(c \minus{} a)^2 \plus{} (3c \minus{} a \minus{} b)(a \minus{} b)^2$ $ \ge (3a \minus{} b \minus{} c \plus{} 3b \minus{} c \minus{} a)(b \minus{} c)^2 \plus{} (3c \minus{} a \minus{} b \plus{} 3b \minus{} c \minus{} a)(a \minus{} b)^2 \ge 0$ $ \iff$
$ 2(a \plus{} b \minus{} c)(b \minus{} c)^2 \plus{} 2(b \plus{} c \minus{} a)(a \minus{} b)^2 \ge 0$.

To the question if it is trivial. It took me two minutes to solve it and write the solution. (I timed it, because I saw you question :) )[/quote]

Sorry for me a question, the above equivalent means $ (c \minus{} a)^2 \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2\Leftrightarrow (3b \minus{} c \minus{} a)(c \minus{} a)^2 \ge (3b \minus{} c \minus{} a)((a \minus{} b)^2 \plus{} (b \minus{} c)^2)$  :maybe:  is it true when we don't have $ 3b>c\plus{}a$?

$ a\geq b \geq c$

$ b\geq c$
$ 2b\geq b \plus{} c > a$

Sum them and you have $ 3b > c \plus{} a$. As for the $ p^2\geq 16Rr \minus{} 5r^2$ one way to prove it is to use Blandan's theorem (I'm not sure if this is how one writes his name in English) which states that $ p, r$ and $ R$ satisfy the following inequalities

$ 2R^2 \plus{} 10Rr \minus{} r^2 \minus{} 2(R \minus{} 2r)\sqrt {R^2 \minus{} 2Rr}\leq p^2$

$ p^2\leq 2R^2 \plus{} 10Rr \minus{} r^2 \plus{} 2(R \minus{} 2r)\sqrt {R^2 \minus{} 2Rr}$ (crazy theorem ain't it  :roll: )

To my knowledge this theorem gives the strongest inequalities between $ p, r$ and $ R$. By the way how does one prove that $ 9IG^2 \equal{} p^2 \minus{} 16Rr \plus{} 5r^2$ ?

As to why your inequality is equivalent to $ p^2\geq 16Rr \minus{} 5r^2$ one way to show this is to have $ a \equal{} x \plus{} y, b \equal{} y \plus{} z$ and $ c \equal{} z \plus{} x$ where $ x \equal{} p \minus{} b, y \equal{} p \minus{} c$ and $ z \equal{} p \minus{} a$. Then $ \sigma_1 \equal{} x \plus{} y \plus{} z \equal{} p, \sigma_2 \equal{} xy \plus{} yz \plus{} zx \equal{} 4Rr \plus{} r^2$ and $ \sigma_3 \equal{} pr^2$.

Edit: The theorem is called Blundon's theorem and here is it's proof [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164095[/url]

[quote="encyclopedia"] [color=darkred]Let $ a,b,c$ be the lengths of the sides of $ \triangle ABC$ . Prove that $ \sum (3a \minus{} b \minus{} c)(b \minus{} c)^2\ge 0$ .[/color] [/quote]
[color=darkblue][size=117][u][b]Nice inequality[/b][/u][/size] and its proof by [b]Encyclopedia[/b] ! In my opinion, it is not trivial.

It is equivalently with the strong [u][b]Blundon[/b]'s inequality[/u] $ \boxed {\ p^2\plus{}5r^2\ge 16Rr\ }\ .$ Indeed,

$ \sum [3a \minus{} (b \plus{} c)]\left[\left(b^2\plus{}c^2\right)\minus{}2bc\right]\equal{}$ $ 3\cdot \sum a\left(b^2\plus{}c^2\right)\minus{}18abc\minus{}\sum (b\plus{}c)\left(b^2\plus{}c^2\right)\plus{}2\cdot\sum bc(b\plus{}c)\equal{}$

$ 5\cdot\sum bc(b\plus{}c)\minus{}18abc\minus{}2\cdot\sum a^3\minus{}\sum bc(b\plus{}c)\equal{}$ $ 4\cdot\sum bc(b\plus{}c)\minus{}18abc\minus{}2\left(s_1^3\minus{}3s_1s_2\plus{}3s_3\right)\equal{}$

$ 4s_1s_2\minus{}12s_3\minus{}18s_3\minus{}2s_1^3\plus{}6s_1s_2\minus{}6s_3\equal{}$ $ 10s_1s_2\minus{}36s_3\minus{}2s_1^3\equal{}$ $ 20p(p^2\plus{}r^2\plus{}4Rr)\minus{}144Rrp\minus{}16p^3\equal{}$

$ 4p^3\plus{}20pr^2\minus{}64Rpr\equal{}4p(p^2\plus{}5r^2\minus{}16Rr)\ge 0\ .$

[b][u]Remark[/u].[/b] I used the notations $ \left\|\begin{array}{c}
s_1\equal{}a\plus{}b\plus{}c\equal{}2p\\\\
s_2\equal{}ab\plus{}bc\plus{}ca\equal{}p^2\plus{}r^2\plus{}4Rr\\\\
s_3\equal{}abc\equal{}4Rpr\end{array}\right\|$ and identity  $ a^3\plus{}b^3\plus{}c^3\equal{}s_1^3\minus{}3s_1s_2\plus{}3s_3\ .$[/color]

[quote="encyclopedia"][quote="Mathias_DK"][quote="encyclopedia"]Let $ a,b,c$ be side of triangle prove that

$ (3a \minus{} b \minus{} c)(b \minus{} c)^2 \plus{} (3b \minus{} c \minus{} a)(c \minus{} a)^2 \plus{} (3c \minus{} a \minus{} b)(a \minus{} b)^2\ge 0$

Is it trivial  :blush: ?[/quote]
Assume wlog $ a \ge b \ge c$. Then $ (c \minus{} a)^2 \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2$.
Thus:
$ (3a \minus{} b \minus{} c)(b \minus{} c)^2 \plus{} (3b \minus{} c \minus{} a)(c \minus{} a)^2 \plus{} (3c \minus{} a \minus{} b)(a \minus{} b)^2$ $ \ge (3a \minus{} b \minus{} c \plus{} 3b \minus{} c \minus{} a)(b \minus{} c)^2 \plus{} (3c \minus{} a \minus{} b \plus{} 3b \minus{} c \minus{} a)(a \minus{} b)^2 \ge 0$ $ \iff$
$ 2(a \plus{} b \minus{} c)(b \minus{} c)^2 \plus{} 2(b \plus{} c \minus{} a)(a \minus{} b)^2 \ge 0$.

To the question if it is trivial. It took me two minutes to solve it and write the solution. (I timed it, because I saw you question :) )[/quote]

Sorry for me a question, the above equivalent means $ (c \minus{} a)^2 \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2\Leftrightarrow (3b \minus{} c \minus{} a)(c \minus{} a)^2 \ge (3b \minus{} c \minus{} a)((a \minus{} b)^2 \plus{} (b \minus{} c)^2)$  :maybe:  is it true when we don't have $ 3b > c \plus{} a$?[/quote]
$ 3b \minus{} c \minus{} a \ge b \plus{} 2c \minus{} c \minus{} a \equal{} b \plus{} c \minus{} a > 0$ since by the assumption $ b \ge c$ :) Don't be sorry to ask :)

@Virgil: I would have great difficulties proving $ p^2 \plus{} 5r^2 \ge 16Rr$ if it wasn't rewritten in the form encyclopedia did. (If I were to solve it, I would probably expand it myself) Are there any simple geometric intepretation of $ p^2 \plus{} 5r^2 \minus{} 16Rr$?


real analysis
function
calculus
integration
limit
geometry
rectangle

is it true that if $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is separately continuous, then it's riemann-integrable?
and is it lebesgue-integrable?

no one? :)
fedja? moubinool? kent? myth? blahblahblah?

Quite clearly not Riemann Integrable. You can create a function that is not bounded on a compact interval but is still seperately continuous.

A concrete example:
$f(x,y) = \frac{(xy)^{\frac13}}{x^{2}+y^{2}}\quad (x,y) \ne 0 \quad f(0,0) = 0$

Unless I am thinking incorrectly this is a separately continuous function, but it is unbounded on for example the line $(x,y) = (t,t)$ when $t$ is small.

For Lebesgue I was under the impression that every function was lebesgue integrable if you chose the right set of axioms.

uhm...
actually, i think i meant riemann-measurable, and lebesgue-measurable..
anyway, you can add the hypothesis of boundedness and positiveness, if you want :)
and i don't think you need so many axioms :) good luck, and good work :)

THe question stopped making sense now. What is it that you want exactly?

Lebesgue measurable: yes. (Assuming "separately continuous" means it is continuous on every line...) I had this for homework in my analysis class :P.

So the idea we had was to pick some connected open set $O \in \mathbb{R}$, and we want the inverse image to be measurable. So we divide $\mathbb{R}^{2}$ into a grid of squares of unit length. All the squares that touch the inverse image of $O$ are marked. Those for which all four sides are entirely contained in the inverse image are "kept." We know that the entire square must be in the inverse image because the function is continuous along the boundary, and $O$ is connected, and the inverse image of a connected set under a continuous function is still connected. The marked squares that aren't good are then divided into squares of half the side length as before, and the process repeats. Eventually, we have countable unions of squares in $R^{2}$ that approximate the inverse image of $O$ to within $\epsilon$ from the inside and from the outside, so $F^{-1}(O)$ is measurable.

I hope that's clear....  :maybe: A picture might help.

As for the other part: I don't know what "Riemann measurable" means.

ok, sorry, i've been unclear (and not at all precise).
let me reformulate the problem:
$f: [0,1]^{2}\to [0,1]$ is separately continuous (meaning that for every $x,y \in [0,1]$ the functions $f(x,\cdot)$ and $f(\cdot, y)$ are continuous).
does this imply that $f$ is riemann integrable?

Xevarion, i have to take a deeper look into your solution, and see if it can be adapted to the "new" problem...
sorry again for "incomplete" and "incorrect" statement...

In that case- Kalle already gave a counterexample. Every Riemann integrable function is bounded, so it suffices to construct an unbounded function which is separately continuous.

Using his $f(x,y)=\frac{(xy)^{\frac13}}{x^{2}+y^{2}}$, we look at $f_{x}(y)=f(x,y)$ as a function of $y$. When $x=0$, $f_{x}(y)=0$. When $x\neq0$, $f_{x}(y)=x^{\frac13}\cdot y^{\frac13}\cdot \frac1{x^{2}+y^{2}}$, and each piece is continuous in $y$.
As $t\to0^{+}$, $f(t,t)=\frac{t^{\frac23}}{2t^{2}}=\frac12t^{-\frac43}\to\infty$, and $f$ is unbounded.

why do you say that every riemann integrable function is bounded?
isn't $x^{-1/2}$ riemann integrable on $(0,1)$? and, anyway, in my last post i've asked for $f$ to take values in $[0,1]$...
moreover, i think the function you mentioned actually _is_ riemann integrable, just its integral is $+\infty$... that's why i've said "measurable" and not integrable...
anyway, again, i'm sorry for misundertandings, it's really my fault this time  :o

The improper integral $\lim_{t\to0}\int_{t}^{1}\frac1{\sqrt{x}}\,dx$ exists, but that doesn't mean $\frac1{\sqrt{x}}$ is Riemann integrable on the unit interval. The Riemann integral is defined as a limit over partitions, and those values can be arbitrarily large for any partition when the function is unbounded. $\infty$ is not a value for Riemann purposes, so any such function is not Riemann integrable.

As for the function taking values in $[0,1]$; yeah, that kills this example.

right, right... i've forgot the finiteness of the partition of the intervall, and made confusion..

anyway, doesn't it exist a definition for riemann measurability similar to the one for lebesgue? i mean, shouldn't just ask for $\chi_{\{f>t\}}$ to be integrable be enough to say that $f$ is integrable (at least for bounded functions)?

Even if $f$ is bounded, it may fail to be Riemann integrable. Indeed, let $K\subset [0,1]$ be a Cantor set of positive measure and let $(x_{j},y_{j})$ be a dense sequence in $K\times K$ such that no $x_{j}$ or $y_{j}$ is an endpoint of a complementary interval to $K$ and each point that occurs in this sequence occurs in it infinitely many times. Now, for each $(x_{j},y_{j})$ choose the constituting intervals $I_{j}$ and $J_{j}$ from the complement of $K$ that are different from all previously chosen intervals and such that $\mathrm{dist\,}(I_{j},x_{j}), \mathrm{dist\,}(J_{j},y_{j})<\frac{1}{j}$. At last, for each $j$, construct a continuous function $f_{j}$ that is supported by $I_{j}\times J_{j}$ and such that $\max f_{j}=1$. Put $f=\sum_{j}f_{j}$. Then $f$ is discontinuous on the set $K\times K$ of positive measure and, thereby, not Riemann integrable. But each vertical or horizontal line intersects at most one rectangle $I_{j}\times J_{j}$ and, therefore, $f$ is continuous in each variable. I wonder whether it is possible to construct a similar example of a function whose restriction to [b]every[/b] line is continuous.

actually, i seem to remember the counterexample i was shown was continuous on every line... and i think it's possible to ask for it to be continuous on every polynomial line: just take the the sets $A = \{(x, \exp(-1/x^{2}) | x\in (0,1]\}$, $B= \{(x, k\exp(-1/x^{2}) | x\in (0,1], k\in (1/2, 3/2)\}$; then consider a function which takes value $1$ on $A$, $0$ outiside $B$, and continuous in $(0,1)^{2}$. this is continuous on every polynomial line everywhere in the square $[0,1]^{2}$, but it's obviously discontinuous in $(0,0)$.
iterating the construction of this function, and rescaling it in the right way, you should get what you ask for...


Oscar buys 13 pencils and 3 erasers for $ \$1.00$. A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?

$ \textbf{(A) } 10\qquad \textbf{(B) } 12\qquad \textbf{(C) } 15\qquad \textbf{(D) } 18\qquad \textbf{(E) } 20$

[hide]

$\$.07$ for pencils and $\$.03$ for erasers would cost $\$1.00$.
($91+9$)

So, together, one pencil and one eraser would be $\$.10$

A
[/hide]


can anyone prove or disprove that -

In a cyclic polygon, for any triangulation, the sum of inradii of each of the triangles in a specific triangulation is constant.

It is Japanese Theorem! You can see it at [url]http://mathworld.wolfram.com/JapaneseTheorem.html[/url]

Oh! Yes, Sangaku problem, Edo era, 1600 some year.

thanks you! hadnt heard of it before.. the idea itself came across while imagining lines to be fixed at one vertex and rotated to pass through the other verices each time forming different triangles whose inradii seemed to be increasing and decreasing. so i thought the sum could be constant.

See also here: [url]http://www.mathlinks.ro/viewtopic.php?p=863233#863233[/url].


geometry
incenter
circumcircle
trigonometry
geometry proposed

Let $\triangle{ABC}$ have the property that $|BC|=\frac{|AC|+|AB|}{2}$. $\mathcal{C}$ is the circle passing from $A$ and the midpoints of $[AC],[AB]$. Let us draw the tangents from the centroid $G$ to $\mathcal{C}$ show that one of the tangency points is the incenter of $\triangle{ABC}$

Lets change the condition :
$|BC|=\frac{|AC|+|AB|}{2}\leftrightarrow IG \parallel BC$

[color=darkred][b][u]Rather,[/u][/b] $\boxed{\ IA\perp IO\Longleftrightarrow b+c=2a\Longleftrightarrow GI\parallel BC\Longrightarrow GI=\frac{1}{6}\cdot |b-c|\ }\ \ (*)\ !$[/color] [color=darkblue] [b]Proof.[/b]

$1\blacktriangleright IA\perp IO\Longleftrightarrow$ $IA^{2}+IO^{2}=OA^{2}$ $\Longleftrightarrow$ $\frac{bc(p-a)}{p}+R^{2}-2Rr=R^{2}$ $\Longleftrightarrow$ $bc(p-a)=2Rpr$ $\Longleftrightarrow$ $2bc(p-a)=abc$ $\Longleftrightarrow$ $b+c=2a\ .$

$2\blacktriangleright GI\parallel BC\Longleftrightarrow$ $h_{a}=3r\Longleftrightarrow$ $3ar=ah_{a}\Longleftrightarrow$ $3ar=2pr$ $\Longleftrightarrow$ $3a=2p$ $\Longleftrightarrow b+c=2a\ .$

$3\blacktriangleright$ Suppose w.l.o.g. $b\ne c\ .$ Denote $M\in AG\cap BC$ and $D\in AI\cap BC\ .$ Then $DB=\frac{ac}{b+c}$ , $MB=\frac{a}{2}$ and $DM=|MB-DB|\Longrightarrow$ $DM=\frac{a|b-c|}{2(b+c)}\ .$

Therefore, $b+c=2a\Longleftrightarrow$ $GI\parallel BC$ $\Longleftrightarrow$ $\frac{GI}{DM}=\frac{AG}{AM}$ $\Longleftrightarrow$ $GI=\frac{a|b-c|}{3(b+c)}$ and $b+c=2a$ $\implies$ $GI=\frac{|b-c|}{6}\ .$

[b]Proof of the proposed problem.[/b] Denote the middlepoints $E$ , $F$ of the sides $[CA]$ , $[AB]$ respectively and the circumcircles $w=C(O,R)$ ,

$w_{1}=C(O_{1},R_{1})$ of the triangles $ABC$ , $AEF$ respectively. Prove easily that the segment $[OA]$ is a diameter in the circle $w_{1}$ and $R_{1}=\frac{R}{2}\ .$

Must prove [u]the first[/u] $GO_{1}>R_{1}$ and [u]the second[/u] $I\in w_{1}\ !$ Indeed,

$\blacktriangleright GO_{1}>R_{1}\Longleftrightarrow$ $4\cdot GO^{2}_{1}>R^{2}$ $\Longleftrightarrow$ $2\cdot\left(GA^{2}+GO^{2}\right)-R^{2}>R^{2}$ $\Longleftrightarrow$ $GA^{2}+GO^{2}>R^{2}$ $\Longleftrightarrow$ $\frac{4m^{2}_{a}}{9}+R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}>R^{2}$ $\Longleftrightarrow$

$4m^{2}_{a}>a^{2}+b^{2}+c^{2}$ $\Longleftrightarrow$ $2(b^{2}+c^{2})-a^{2}>a^{2}+b^{2}+c^{2}$ $\Longleftrightarrow$ $2\left(b^{2}+c^{2}\right)>(2a)^{2}$ $\Longleftrightarrow$ $2\left(b^{2}+c^{2}\right)>(b+c)^{2}\ ,$ what is truly iff $b\ne c\ .$

$\blacktriangleright$ From the chain $(*)$ obtain $b+c=2a$ $\Longleftrightarrow$ $IA\perp IO$ $\Longleftrightarrow$ $I\in w_{1}\ ,$ i.e. $2\cdot IO_{1}=R\ .$

[u]Come back ![/u] $GI\perp IO_{1}$ $\Longleftrightarrow$ $GI^{2}+IO^{2}_{1}=GO^{2}_{1}$ and $2\cdot IO_{1}=R$ , $GI=\frac{|b-c|}{6}$  from the chain $(*)$ $\Longleftrightarrow$ $(b-c)^{2}+9R^{2}=9\left[2\left(GA^{2}+GO^{2}\right)-AO^{2}\right]$

$\Longleftrightarrow$ $(b-c)^{2}+9R^{2}=8m^{2}_{a}+2\left[9R^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right]-9R^{2}$ $\Longleftrightarrow$ $(b-c)^{2}=2\left[2\left(b^{2}+c^{2}\right)-a^{2}\right]-2\left(a^{2}+b^{2}+c^{2}\right)$ $\Longleftrightarrow$

$(b-c)^{2}=2\left(b^{2}+c^{2}\right)-4a^{2}$ $\Longleftrightarrow$ $(b-c)^{2}=2(b^{2}+c^{2})-(b+c)^{2}\ ,$ what is truly. In conclusion, $IG\perp IO_{1}\ ,$ i.e. the line $GI$ is tangent to the circle $w_{1}\ .$

[b]Remark 1.[/b] $\left\{\begin{array}{c}\boxed{\ b+c=2a\Longleftrightarrow\cos\frac{B-C}{2}=\sqrt{\frac{2r}{R}}\Longleftrightarrow\sin \frac{A}{2}=\sqrt{\frac{r}{2R}}\Longleftrightarrow\cos A=1-\frac{r}{R}\Longleftrightarrow\cos B+\cos C=\frac{2r}{R}\ }\\\\ \boxed{\ b+c=2a\Longleftrightarrow (a+b)(a+c)s_{a}=bcS\Longleftrightarrow p^{2}+9r^{2}=18Rr\ }\end{array}\right\}$

[u]Generally[/u], $\cos\frac{B-C}{2}\ge \sqrt{\frac{2r}{R}}\ ,$ but $\sin \frac{A}{2}\ge\sqrt{\frac{r}{2R}}\Longleftrightarrow\cos A\le 1-\frac{r}{R}\Longleftrightarrow b+c\le 2a\ !$

[b]Remark 2.[/b] I used the well-known relations : $2S=ah_{a}=2pr$ ; $4RS=abc$ ; $4m^{2}_{a}=2\left(b^{2}+c^{2}\right)-a^{2}$ ;

$OI^{2}=R^{2}-2Rr$ ; $IA^{2}=\frac{bc(p-a)}{p}$ ; $GO^{2}=R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}\ .$[/color]

Erhe I wanted to post the complete solution , you were faster.

[color=darkblue][b]A nice and "snapshot" proof.[/b] Denote the diameter $A_{1}A_{2}$ of the circumcircle $w$ for which $A_{1}\in AI\ .$ Therefore,

$b+c=2a$ $\Longrightarrow$ $\{\begin{array}{c}OI\perp AA_{1}\mathrm{\ ,\ i.e.\ I\ is\ the\ middlepoint\ of\ [AA_{1}]}\\\ IG\parallel BC\mathrm{\ ,\ i.e.\ }IG\perp A_{1}A_{2}\end{array}$ $\Longrightarrow$ $\widehat{OIG}\equiv\widehat{AA_{1}A_{2}}$ with the common value $\frac{|B-C|}{2}$ $\Longrightarrow$

$\widehat{OIG}\equiv\widehat{IAO}$ $\Longrightarrow$ the line $GI$ is tangent to the circumcircle of the triangle $AIO\ ,$ i.e. the circle $w_{1}\ !$

[b]Remark.[/b] In this case the value of the acute angle between the lines $OI$ and $BC$ is $\frac{|B-C|}{2}\ .$ If $\{D,M\}\subset (BC)$ , where $ID\perp BC$ and $MB=MC\ ,$

then $DM=\frac{|b-c|}{2}$ and $DM=IO\cdot\cos\frac{B-C}{2}\ ,$ i.e. $|b-c|=2\cdot \sqrt{R^{2}-2Rr}\cdot \cos\frac{B-C}{2}\ .$ But $\cos\frac{B-C}{2}=\sqrt{\frac{2r}{R}\ .}$ Thus, $b+c=2a$ $\Longrightarrow$

$|b-c|=2\sqrt{2r(R-2r)}\ ,$ i.e. $\boxed{\ b+c=2a\Longrightarrow (b-c)^{2}+(R-4r)^{2}=R^{2}\ \ (*)}\ .$ Indeed, if you have a bit of pacience, $|b-c|=2\cdot\sqrt{2r(R-2r)}$ $\Longleftrightarrow$

$p(b-c)^{2}=8Rpr-16pr^{2}$ $\Longleftrightarrow$ $p(b-c)^{2}=2abc-16(p-a)(p-b)(p-c)$ $\Longleftrightarrow$ $\ldots\ldots$ $\Longleftrightarrow$ $[2a^{2}-(b+c)a-3(b-c)^{2}][2a-(b+c)]=0\ .$ [b][u]

The implication $(*)$ isn't truly and reciprocally ![/u][/b] [u]What is the geometrical interpretation of the relation[/u] $2a=b+c+\frac{3(b-c)^{2}}{a}\ ?$

[b]Commentary.[/b] Here is an example of problem which is difficulty at the first sight, but "reading" with the more attention,

it becomes a source of the another interesting questions. In the my opinion, [b][u]you never have to abandon quickly a problem, even if you solved it.  [/u][/b][/color]