MATHCOUNTS
probability
analytic geometry
articles

wrong..

From a facsimile of an answer in a MATHCOUNTS book...

"A point is selected at random from the interval -10 is less than or equal to x is less than equal to 10. What is the probability that the coordinate of the point is a solution of x is equal to or greater than 7? Express your answer as a common fraction."

ANswer: 3/20.

However, I disagree.. From -10 to 10, there are 21 units, not 20 units. And from 7 to 10, there are 4 units, not 3 units. Thus, the answer should be 4/21.

Does anyone else agree with me?

ALso, another question... (copied from a reminder to myself)

AoPS: Ask question about whats best math to do in head when idling around when I dont have scratch paper. Or whats best math to do when I have scratch paper. For a beginner at mental math. And not just multiplying two-digit numbers in my head (sadly, I am not the Gaussian prodigy).

You are thinking that the point you must pick is an integer, when it can be any number in that interval. Look at it terms of line lengths, not counting the integer values of x. You should find the solutions manual to be correct.

I have no idea about the mental math...

Yup, they're not just integers. There was a problem similar to this one in this year's Mathcounts Nationals, i think 5 or 7 on the target, except it asked for probability that a product was negative. I thought it was integers then and got it wrong.

Oh!! Now I see! I misinterpreted the question again and am ruined because of it! (Yes, I have read the article in AoPS about that..)

Ok, one more question... How do the hardest problems in textbooks compare to competition-style problems? I'm especially interested in Houghton-Mifflin textbooks.

Hmm.. I m not really sure about how they compare.  I feel that textbooks give you the raw tools to do well on competition problems(usually) but to truly do well you must become creative with these tools.  Becoming creative takes practice and experience.  You can usually find a few problems out of each chapter that build your creativity.  Most textbooks are composed mostly of formulas and exercises.  There might be certain textbooks that focus on competition type problems... but I am not the person who would know.

Thanks for the response, joml! :) Yeah; it really varies on textbook. I think that Houghton Mifflin's textbooks have many problems that require creativity to solve. (Many type C problems; there are even problems labeled as specifically hard so they must be designed for some people seeking greater challenge).


geometry
parallelogram
analytic geometry
AMC
USA(J)MO
USAMO
summer program

[quote="mr. math"]how would you do this problem?

The coordinates of the vertices of a parallelogram are (!0,1), (7,-2), (4,1) and (x,y). what is the sum of the distinct possible values for x?

 :spider:  :sleeping:[/quote]
[hide]If you're given three points, there are three possible (x,y) points as the other one. If the line connecting (x,y) and (0,1) are parallel to the other line, then x=-3 or x=3. The other point is (11,-2) connecting with (7,-2) and (4,1). The sum of the x values is $\boxed{11}$?[/hide]

[hide]
I graphed quickly and found the three x coordinates to be -3, 3, and 11. Add to get $\boxed{11}$.[/hide]

This is actually an easy one to do mentally, probably not CD though :pilot:

[hide]I mentally did this under 45 seconds and I believe the answer should be 11.[/hide]

Yeah but [b]most[/b] CD questions are done under 10 seconds. :heli:

Well, I took 9 seconds to do the question and that is under 10 seconds so hence I did this question under 10 seconds
Q.E.D.

Thanks

That is how I am going to do well on the USAMO in future years...hopefully. :lol:

On usamos, when they say, prove so and so, respond with the following.

This statement was typed on a mathematics contest, so it must be true. Hence, so and so holds.
Q.E.D.

7 points right there.

[quote="samath"]Well.....my USAMO stuff was quite a bit...um....more rigorous than that, and I got only a 5.....but have fun trying :P![/quote]

That's beastly man, I only got a 2 or a 4, not so sure though.

[quote="SplashD"]You guys are so lucky, my school doens't even offer AMC contests,  :( but I'm trying to change that[/quote]
You're not alone. The teacher who was supposed to be running the contest said it put too much 'strain' on the students since they had to do other contests too.  :mad:

[quote="mathgeniuse^ln(x)"]Okay, I did the question under 10 seconds.  What can you do about that?[/quote]

Liar :P

OK if you really did it in under 10 sec. can you show me a method or does your mind just work 5 times as fast as everyone elses?

I calculated the vertices possible were 11, 3, and -3, I think, quickly and got 11.  If my vertices are wrong, I am lucky that I got it right.  It took me 9 seconds

robinhe, do you know the people from Michigan who are going to MathPath?

[quote="robinhe"][quote="mathgeniuse^ln(x)"]Okay, I did the question under 10 seconds.  What can you do about that?[/quote]OK if you really did it in under 10 sec. can you show me a method or does your mind just work 5 times as fast as everyone elses?[/quote]Add up the three given x-coordinates.  That works as long as the three given points have distinct x-coordinates and are not collinear.

Yeah, that's the way I did it as well...jk.

But, yeah, I never noticed that.

Thanks

Wow, that should have been obvious to solve it.  Then, it would have only taken about 3 seconds (the time it takes to read the problem) to solve it.

It took me around 30 seconds, so this probably could be a countdown round question.

What exactly does the function $!n$ mean?  I just assumed the (!0,1) was a typo and it meant (0,1).

It probably was a typo, because I have never seen a factorial sign before a number.

Unless, you know, it was supposed to go after it...

Quick, [b]solve the the problem now with the point (0!,1) instead of (!0,1)[/b]

[quote="b-flat"]It is a function on my calculator (added with Omnicalc), but I don't know what it does.  Some inputs and outputs:
!n(0)=1
!n(1)=0
!n(2)=1
!n(3)=2
!n(4)=9
!n(5)=44
!n(6)=265[/quote]It looks like the "derangements" function: the number of permutations of $1..n$ that have no fixed points.

Fixed points are related with circular permutations :huuh:

A fixed point is a number that is not moved by the permutation.  For example, the permutation 1432 has two fixed points: 1 and 3.

Thanks, I think I get it now.

So it is the number of combinations in which all of the numbers change position.

So for 4 numbers (1,2,3,4), they are:
(4,3,2,1)
(2,1,4,3)
(3,4,1,2)
(3,4,2,1)
(4,3,1,2)
(4,1,2,3)
(2,3,4,1)
(2,4,1,3)
(3,1,4,2)

You guys sure he didn't mean 10 and not 1? As ! = 1 with shift and he probably held it too long after making the (

Think fast, [b]solve the the problem now with the point (10,1) instead of (!0,1) [/b] :ninja:

sweet result.
Non-collinear is required of course, but "distinct x-coordinates" doesn't seem to be required.  Note that given points have repeated y-coordinates, but method does work for y-coord as well. (sum=-2+-2+4=1+1+-2=0)
Any idea why?
I noticed that the 3 new points form a triangle with sides parallel to original triangle and 4x as large.  Also, this result shows that the two triangles have the same centroid or center of mass (take the mean of the coordinates).

The only reason that "distinct $x$-coordinates" is required is that the problem asked for the sum of the [i]distinct[/i] possible values of $x$.

To see why the method works, let $A$, $B$, and $C$ be the three given vertices.  Then the three possible values for the fourth vertex are (using vector notation)
\begin{eqnarray*} D_1 & = & A + B - C \\ D_2 & = & A - B + C \\ D_3 & = & -A + B + C \end{eqnarray*}
When you add up those three points, their sum is $A + B + C$.  That's why adding up the coordinates works.

Let $G$ be the centroid of triangle $ABC$; in other words
\[ G = \frac{A + B+ C}{3} \, .  \]
From the equations above, we confirm your observation that the centroid of triangle $D_1 D_2 D_3$ is also $G$.

Also, note that $A$ is the midpoint of $\overline{D_1D_2}$; $B$ is the midpoint of $\overline{D_1D_3}$; and $C$ is the midpoint of $\overline{D_2D_3}$.  That observation confirms your other results about triangle $D_1D_2D_3$.

Oh, yeah, distinct, I need to keep that in mind.  But it is pretty easy to check if things are the same or different, right?


function
AMC
AIME
geometry
complex numbers

A function f  is defined on the complex numbers by f(z)=(a+bi)z where a  and b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin.

[b]What is the image of each point (is that just f(z)??)?  What is the point it is equidistant to?[/b]

Btw that wasnt the whole question, if you want to do it,it also has that a^2+b^2=8^2 and then find b^2

[quote="the future"][b]What is the image of each point (is that just f(z)??)?[/b][/quote]

Yes.

[quote="the future"][b]What is the point it is equidistant to?[/b][/quote]

The origin, just like it says.  By the way, I'm pretty sure there is already a thread about this question.

t0rajir0u, should we also look for the old threads before ask questions? Not all of us are as experienced as you.
Is it rude if I post a problem from an older AMC or AIME without looking for a thread about it?

Sometimes the title won't be obvious

Geometry Problem
Aime # blah
Chords
Help with
...  could all be the subject lines for the same problem.

guys sorry i posted this twice (the other one wouldnt delete)

Anyways, I also posted this just in case anyone wanted to solve (the other post about this didnt have the other numbers like a^2+b^2=81)

sorry again and thanks guys


I say [hide]Tommorow = 27-3 = 24

Today = 23

My Birthday = 23-5 = 18[/hide]

[quote="math92"]My birthday was 5 days ago.  If three days afrer tommorow is the 27th what is the date of my birthday.

i keep getting diffrent answers each time i try this :?[/quote]
[hide]
Say today is $x$

$x+4=27$

$x=23$

Today is the 23rd.

$23-5=18th$

Your birthday is on the 18th.[/hide]


limit
real analysis
real analysis unsolved

Let $x_{n+1}=\sqrt[n]{1+nx_n}-1$. Find $\lim x_n$ and $\lim nx_n$at infinity.

[quote="Kent Merryfield"]Numerical work suggests this conjecture: $\lim_{n\to\infty}n^2x_n=4.$[/quote]

I think your conjecture is wrong.

First of all let`s prove that $x_n\to 0$. Suppose it doesn`t converge to $0$ ( it`s easy tosee that it connverges from Bernoulli) then $\ell+1=\lim_{n\to\infty}\sqrt[n]{1+nx_n}=\lim_{n\to\infty}\frac{1+(n+1)x_{n+1}}{1+nx_n}=1$ so $\ell+1=1$ from where $\ell=0$ contradiction!. So $x_n\to 0$.
For the second part I used Cesaro-Stolz and improved Bernoulli : $\lim_{n\to\infty}nx_n=\lim_{n\to\infty}\frac{x_n\cdot x_{n+1}}{x_n-x_{n+1}}$ then $\frac{x_n\cdot x_{n+1}}{x_n-x_{n+1}}\leq \frac{x_n}{x_{n+1}}\frac{2}{n-1}$ a few more computations and eventually $\frac{x_n\cdot x_{n+1}}{x_n-x_{n+1}}\leq \frac{2n}{n-1} (\frac{1}{n}+x_n) \to 0$.

[quote="Kent Merryfield"]Numerical work suggests this conjecture: $\lim_{n\to\infty}n^2x_n=4.$[/quote]I can also see this. Would be interesting to have a proof.

but I thought that was just proved incorrect? :huh:

It wasn't. spix didn't address that question; he just solved the original problem, with results consistent with the conjecture.


How do I make cylinders and cones in 3D? And I how do I shade something so it looks 3D?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215899]this topic[/url], as well as Mr. Ivaldi's site for lots of examples.


Find all functions $ f$, defined on the real numbers and taking real values, which satisfy the equation $ f(x)f(y)\equal{}f(x\plus{}y) \plus{}xy$ for all real numbers $ x$ and $ y$.

Putting $ x\equal{}y\equal{}0$ we get $ f(0)^2 \equal{} f(0)$, so $ f(0)\equal{}0$ or $ f(0)\equal{}1$.

Now, putting $ x\equal{}0$ we get $ f(0)f(y)\equal{}f(y)$. If there is some $ y$ such that $ f(y) \neq 0$, then $ f(0)\equal{}1$. But if there isn't any, then $ f(x)\equal{}0$ for all $ x$, which is not a solution since $ xy$ is not always equal to $ 0$. Hence $ f(0)\equal{}1$.

Putting $ y\equal{}\minus{}x$ we get $ f(x)f(\minus{}x) \equal{} 1 \minus{} x^2$. So, one of $ f(1)$ and $ f(\minus{}1)$ must equal $ 0$.

*Suppose $ f(1)\equal{}0$. Then, putting $ y\equal{}1$ we get $ f(x\plus{}1) \plus{} x \equal{} 0$ for all $ x$, or equivalently, $ \boxed{f(x) \equal{} \minus{}x\plus{}1}$.
*Suppose $ f(\minus{}1)\equal{}0$. Then, putting $ y\equal{}\minus{}1$ we get $ f(x\minus{}1) \minus{} x \equal{} 0$ for all $ x$, or equivalently, $ \boxed{f(x) \equal{} x\plus{}1}$.

Now it's easy to verify that both functions fulfill the conditions:

$ (\minus{}x\plus{}1)(\minus{}y\plus{}1) \equal{} xy \minus{} x \minus{} y \plus{} 1 \equal{} (\minus{}x\minus{}y\plus{}1)\plus{}xy$,
$ (x\plus{}1)(y\plus{}1) \equal{} xy \plus{} x \plus{} y \plus{} 1 \equal{} (x\plus{}y\plus{}1)\plus{}xy$.


prove that

$a^2 = b^2 + b + 1$

has no integral solution.

There are integral solutions (a,b) in ( :pm: 1,0).  Do you mean for positive integers?

oops. sorry. you are right. PROVE THAT THE EQUATION HAS NO POSITIVE INTEGRAL SOLUTIon.

a,b > 0

b^2 < b^2 + b + 1 < b^2 + 2b + 1
b^2 < b^2 + b + 1 < (b+1)^2

Since b^2 + b + 1 is between two consecutive perfect squares, it's not a perfect square.  Therefore there are no integral solutions in a^2 = b^2 + b + 1.


A sequence is defined as follows:

$a_1=1$, $a_{n+1}=3a_n+2^n$

Find the explicit formula of this sequence.

Yes! Found it!  :D

[hide]The explicit formula is: \[3^{n}-2^{n}\]

Woow...  :P

Here's how I found it:
Terms are: $1,5,19,65$
$t_{3}=3(3(1)+2^{1}) + 2^{2}=19$
So after some searching we get the explicit formula:
$3^{n-1}+3^{n-2}.2+3^{n-3}.2^{2}+... +3.2^{n-2}+2^{n-1}$
This is the correct solution, but to make it look better, we can write it as:
$\frac{3^{n}-2^{n}}{3-2}$
And that of course equals:
$3^{n}-2^{n}$
Done![/hide]

You are right, Jan.

Here is my solution.

$a_{n+1}=3a_n+2^n\Longleftrightarrow a_{n+1}+2^{n+1}=3(a_n+2^n) $

Therefore we obtain $a_n+2^n=3^{n-1}(a_1+2)=3^n$,yielding Jan's result.

kunny

Nice solutions, guys :)

My solution was similar to kunny's, althuogh harder.

Dividing both sides by $2^{n+1}$, we get $\frac{a_{n+1}}{2^{n+1}}=\frac{3}{2}\left(\frac{a_n}{2^n}\right)+\frac{1}{2}$

Let $b_n=\frac{a_n}{2^n}$

$b_{n+1}=\frac{3}{2}b_n+\frac{1}{2}  \Longleftrightarrow b_{n+1}+1=\frac{3}{2}(b_n+1)$
So, $b_n+1=\left(\frac{3}{2}\right)^n$
$b_n=\left(\frac{3}{2}\right)^n-1=\frac{3^n-2^n}{2^n}$

Therefore, $a_n=2^n \cdot b_n=3^n-2^n$


we did this on a practice test we took to study for district, me and this other person both got 29, but it's not 29, so I don't get it...

How many positive, even three-digit numbers exist such that the sum of hte hundreds digit and hte tens digit equals the units digit?

Try going through it in cases. What happens when the ones digit is 2? 4?

I got 20 numbers, but dont know if thats right

[quote="LuCy4EvA"]um... this is what I did:

if it ends in 2, there's 2 possibilities,
if it ends in 4, there's 4 possiblities,
if it ends in 6, there's 6 possibilities,
if it ends in 8, there's 8 possibilities,
if it ends in 0, there's 9 possibilities
so 2+4+6+8+9 = 29
so I dont' get what we did wrong[/quote]

How did you get 9 posssibilites for 0??? I got 0 possibilities for 0.

yep, that's the answer... that's strange, doesn't the numbers ending in 0 count?

um... ooooooo, never mind, I only counted the unit digit, I counted the ones that adds up to 10... thank you!

[quote="thematrix"]maybe I'm wrong,but if it ends in 0,there are no possibilities[/quote]
I agree

sry!!! :blush: I did this in a hurry and forgot that 0 is not the same as 10 in this case... I'm sorry!

I got 20 also; the numbers that worked were:
$202, 112 \Rightarrow 2 $
$404, 314, 224, 134 \Rightarrow 4 $
$606, 516, 426, 336, 246, 156 \Rightarrow 6 $
Continue the pattern (ending in 8 should have 8) If you don't believe me:
$808, 718, 628, 538, 448, 358, 268, 178 \Rightarrow 8$

$2+4+6+8 = 20$


Let $ 0<a,b,c<1$. Prove that $ \frac{1\minus{}a}{1\plus{}b\plus{}c}\plus{}\frac{1\minus{}b}{1\plus{}c\plus{}a}\plus{}\frac{1\minus{}c}{1\plus{}a\plus{}b} \ge 3(1\minus{}a)(1\minus{}b)(1\minus{}c)$

[quote="thanhnam2902"]Let $ 0 < a,b,c < 1$. Prove that $ \frac {1 \minus{} a}{1 \plus{} b \plus{} c} \plus{} \frac {1 \minus{} b}{1 \plus{} c \plus{} a} \plus{} \frac {1 \minus{} c}{1 \plus{} a \plus{} b} \ge 3(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$[/quote]
we have
$ (1\plus{}b\plus{}c)(1\minus{}b)(1\minus{}c)\leq1$(cauchy)
so $ \frac{1\minus{}a}{1\plus{}b\plus{}c}\geq (1\minus{}a)(1\minus{}b)(1\minus{}c)$
Q>E>D

[quote="tuandokim"][quote="thanhnam2902"]Let $ 0 < a,b,c < 1$. Prove that $ \frac {1 \minus{} a}{1 \plus{} b \plus{} c} \plus{} \frac {1 \minus{} b}{1 \plus{} c \plus{} a} \plus{} \frac {1 \minus{} c}{1 \plus{} a \plus{} b} \ge 3(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$[/quote]
we have
$ (1 \plus{} b \plus{} c)(1 \minus{} b)(1 \minus{} c)\leq1$([size=150]cauchy[/size])
so $ \frac {1 \minus{} a}{1 \plus{} b \plus{} c}\geq (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$
Q>E>D[/quote]
I think you mean AM-GM,


inequalities
geometry
logarithms
probability
conics
analytic geometry
function

I read the section on the trivial inequality in AOPS, but I didn't understand it.  Could you guys help me out a bit, and explain what the importance of it is?

And the following problems I need help on.

417. A ball is thrown upward from the top of a tower.  If its height is described by -t^2+60t+700, what is the greatest height the ball attains?

426. If the smallest value of y satisfying the equation y=3x^2+6x+k is 4, find the value of k.

Thanks a lot!  I'm studying for the AMC 10 as of now, and looking at some of the past years' test, I can consistenly get more than half right!  :)   Sometimes even 20/25 when I'm lucky..

What areas should I focus on?  Will you have to know the laws of logarithms (change of base..)?  In addition, are there lots of probablity and counting questions where you have to identify everything that's going on/multiple cases?  Those really kill me.  Oh yeah, last question, do they ask questions on number bases?

Are you familiar with conics and their graphs?  That first problem's equation graphs a downward-opening parabola.  to find the maximum value of y (height), you need to find the parabola's vertex.

AMC 10 rarely has logs on it.  It can have number bases, but they aren't all that common.

For 417, the equation is just a parabola. Find the vertex x coordinate ($\frac{-b}{2a}$) and plug it into the function.
For 426, plug in 4 for y and subtract it over. Then take the descriminant ($b^2-4ac$) and set it equal to 0. Note that c is in terms of k.

As for your AMC questions, I wouldn't know. I've never taken it before, this year will be my first time. Check out the previous tests and see for yourself.

for 426, the tangent line thing is unnecessary, all you need to do is complete the square, and notice that the (x+a)^2 term must be greater than or equal to 0, so the other part will be equal to 4, and you can solve for k

417. A ball is thrown upward from the top of a tower. If its height is described by -t^2+60t+700, what is the greatest height the ball attains?

Yo have to find the point where the function have the maximal value (in spanish call "vértice").

-b/2a=-60/2*-1=-60/-2=30.

Replacing t for 30 in the function whe have:

-(30^2)+60*30+700=-900+180+700=1600.

So, the greatest height the ball attains is 1600.

Sorry, I made a mistake: in the last part, when I replaced t by 30, I put:

-(30^2)+60*30+700=-900+180+700=1600, but it was

-(30^2)+60*30+700=-900+1800+700=1600.

The result was ok, but I wrote 180 instead of 1800.

The trivial inequality states that $x^2 \ge 0$ for any real $x$.  The inequality can be used to prove a number of useful results:

$x = a-b$.  We have $(a - b)^2 \ge 0 \implies a^2 - 2ab + b^2 \ge 0 \implies a^2 + b^2 \ge 2ab$, or, written in more common form, $\boxed{ \sqrt{ \frac{a^2 + b^2}{2} } \ge \sqrt{ab} }$.  This is known as the Quadratic Mean-Geometric Mean inequality, or QM-GM (not very common).

$x = a-b$.  We have $a^2 - 2ab + b^2 \ge 0 \implies a^2 + 2ab + b^2 \ge 4ab \implies (a+b)^2 \ge 4ab \implies a+b \ge 2 \sqrt{ab}$, or $\boxed{ \frac{a+b}{2} \ge \sqrt{ab} }$.  This is known as the Arithmetic Mean-Geometric Mean inequality, or AM-GM.

$x = a-b$.  We have $a^2 + b^2 \ge 2ab$ as before.  Now, consider $2(a^2 + b^2)$.  Applying QM-GM we have $a^2 + b^2 + 2ab \le 2(a^2 + b^2) \implies \frac{ (a+b)^2}{4} \le \frac{a^2 + b^2}{2} \implies \boxed{ \frac{a+b}{2} \le \sqrt{ \frac{a^2 + b^2}{2} } }$, known as the Arithmetic Mean-Quadratic Mean inequality, or AM-QM.

Now, if only I could remember how to prove AM-HM... :(

For t0rajir0u

[hide]$(a-b)^2\geq 0\implies a^2-2ab+b^2\geq 0\implies a^2+2ab+b^2\geq 4ab\implies$
$\implies (a+b)^2\geq 2\cdot 2ab\implies \frac{a+b}{2}\geq\frac{2ab}{a+b}$[/hide]

Of course, AM-HM and RMS-AM (also called AM-QM) are just special cases of Cauchy, which states

$(u_1v_1+u_2v_2+\cdots+u_nv_n)^2\leq (u_1^2+u_2^2+\cdots+u_n^2)(v_1^2+v_2^2+\cdots+v_n^2)$.

AM-HM: $v_n=\frac{1}{u_n}$ and RMS-AM: $v_n=1$.

The trivial inequality proof for this works like this: $(v_1x-u_1)^2+(v_2x-u_2)^2+\cdots+(v_nx-u_n)^2\geq 0$ for all $x$, and if there is equality, it is when $x=\frac{u_1}{v_1}=\frac{u_2}{v_2}=\cdots =\frac{u_n}{v_n}$.

Expanding this out gives a quadratic with at most one real root, so it must have a nonpositive discriminant.

This becomes $ax^2-2bx+c$, where $a=v_1^2+v_2^2+\cdots+v_n^2$, $b=u_1v_1+u_2v_2+\cdots+u_nv_n$, and $c=u_1^2+u_2^2+\cdots+u_n^2$. Set the discriminant to less than or equal to 0 to arrive at the inequality.


[quote="gotztahbeazn"]For the problem, how many positive integers satisfy both of the following:

1) All of the number's digits are either 1,2, or 3
2) The sum of the digits is 10

The solution states that if [hide] $F(n)$ is the number of positive integers consisting of 1,2 or 3, then $F(n+3) = F(n+2) + F(n+1) + F(n)$. Why is this true? can anyone explain this better? [/hide][/quote]
BTW, what is $n$?

oh sorry, $n$ is the sum of the digits

so in this case, we wish to find $f(10)$, so the number of positive integers whose digits are 1,2,3 and with digits sum to $10$

if $n$ is the sum of the digits, and only $1,2,3$ are allowed, then...
let's say you wanted to find $f(n)$.  That recursively means you count all of the ones that begin with 1, then 2, then 3.
So if it begins with 1 the number of ways is $f(n-1)$.
If it begins with 2 the number of ways is $f(n-2)$.
If it begins with 3 the number of ways is $f(n-3)$.
Thus, $f(n) = f(n-1)+f(n-2)+f(n-3) \rightarrow f(n+3) = f(n+2)+f(n+1)+f(n)$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

If $ P^{\minus{}1}AP\equal{}B$, where $ P , A ,B \in M_{n}(\mathbb{R})$

can we have:

$ \sum_{1\leq i<j\leq n}\begin{vmatrix}
a_{ii} &a_{ij} \\
a_{ji} &a_{jj}
\end{vmatrix}\equal{}\sum_{1\leq i<j\leq n}\begin{vmatrix}
b_{ii} &b_{ij} \\
b_{ji} &b_{jj}
\end{vmatrix}$

If I am understanding correctly, you are saying that the sum of the principal minors of order 2 is the same for both A and B. Then this is correct not only for the order 2 minors but also for the principal minors of orders 1, ..., n. Since A and B are similar, they have the same characteristic polynomial $ \phi(\lambda) \equal{} \lambda^n \minus{} s_1 \lambda^{n\minus{}1} \plus{} s_2 \lambda^{n\minus{}2} \minus{} ... \plus{} (\minus{}1)^n s_n$. But $ s_i$ is the sum of all principal minors of order $ i$.


Bonnie and Clyde are sliding 380 KG bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 420 N of force while Bonnie pulls forward on a rope with 295 N of force.

What is the safe's coefficient of kinetic friction on the bank floor?

A 2.0 Kg wood block is launched up a wooden ramp that is inclined at a 35 degrees angle. The block’s initial speed is 10m/s .
 What vertical height does the block reach above its starting point?

What speed does it have when it slides back down to its starting point?

A 50 Kg  box hangs from a rope.
What is the tension in the rope if the box moves up at a steady 5.0 m/s ?
What is the tension in the rope if the box has  v= 5.0  m/s and is slowing down at 5.0 m/s^2?

I'll answer the first one for you.  First draw a free-body diagram.  We see that there is a total force of 715 N directed to the right (I'm choosing this to be the positive direction of motion).  Since the floor is horizontal (it is, yes?), the normal force supplied by the floor is 380 kg.  We know that $ F_{f}\equal{}\mu_{k}N\equal{}\mu_{k}(380g\text{N})\approx3724\mu_{k}\text{N}\equal{}715\text{N}$ (why?).  Hence $ \mu_{k}\equal{}.192$.


geometry
geometric transformation
reflection
vector
rotation
absolute value
angle bisector

Let ACE be a triangle, B on AC and D on AE. Let F be the intersection of CD and BE. Prove if AB+BF=AD+DF , then AC+CF=AE+EF.

Indeed, AB + BF = AD + DF holds if and only if the quadrilateral ABFD has an excircle, and AC + CF = AE + EF holds if and only if the quadrilateral ACFE has an excircle. And both conditions are just equivalent to the existence of a circle touching the lines AC, AE, CF and EF.

The result is called [b]Urquhart Theorem[/b], for M. L. Urquhart (1902-1966). See

Dan Pedoe, [i]The Most "Elementary" Theorem of Euclidean Geometry[/i], Mathematics Magazine, 49 (1976) p. 40-42.

  Darij

[quote="darij grinberg"]Indeed, AB + BF = AD + DF holds if and only if the quadrilateral ABFD has an excircle, and AC + CF = AE + EF holds if and only if the quadrilateral ACFE has an excircle. And both conditions are just equivalent to the existence of a circle touching the lines AC, AE, CF and EF.
  Darij[/quote]

are you sure about AB+BF=AD+DF <==> ABDF has excircle? I dont think this is true when ABDF is convex, e.g. take such an ABDF with an excircle. Reflect D in BF, call this point D'. then the quadrilatrel ABD'F also satisfies the condition, but i dont think it can have an excircle. Also i think a kite is a counter-example to that statement; since a kite has no excircle.  :?  :? (or does it? whats the formal definition of excircle?) so i dont think that proof works, since ABFD is always convex :(.

darij was not quite exact saying "excircle", but we can always make it correct saying a circle that touches every side, probably with imaginary radius for your cases where two sides cross :D (which is not important for the task anyway).

Peter

I knew my posting would entail a hell of confusion. Indeed, different forms of quadrilaterals and different kinds of in/excircles have different conditions of existence. When I was studying Mannheim's theorem, I had been trying to classify all types of quadrilaterals - convex, non-convex, self-intersecting - with all forms of tangent circles, but my patience was at the end very quickly. What I came up with is an application of the theory of directed circles providing a way of working with such quadrilaterals without having to distinguish dozens of cases.

We fix the orientation of a plane, i. e. we introduce a "clockwise" and a "counterclockwise" sense of rotation.

At first, if k is any circle in the plane, and d is a number from the set {-1; 1}, then we will call the pair (k; d) a [b]directed circle[/b]. The [b]radius[/b] of this directed circle is defined as the radius of the circle k, multiplied with d (so, the absolute value of the radius of the directed circle (k; d) equals the radius of k, and the sign is that of d).

Moreover, we consider directed lines. If g is a line in the plane, then there exist two unit vectors $\overrightarrow{e_1}$, $\overrightarrow{e_2}$ parallel to this line. We have $\overrightarrow{e_1}=-\overrightarrow{e_2}$. Now, if we pick out one of these two unit vectors - let's say, $\overrightarrow{e_1}$ -, then we can introduce directed segments on the line g by defining the [b]directed distance[/b] PQ between two points P and Q on g to be the real number k such that $\overrightarrow{PQ} = k \cdot \overrightarrow{e_1}$. It's clear that such a number exists and is unique. So we can call the pair (g; $\overrightarrow{e_1}$) a [b]directed line[/b]; on any directed line, we can measure directed distances.

Now we are going to define the [b]directed distance from a point to a directed line[/b]:

If P is a point, and (g; $\overrightarrow{e_1}$) is a directed line, then we consider the perpendicular p from the point P to the line g. This perpendicular p meets the line g at a point Q. If we now rotate the unit vector $\overrightarrow{e_1}$ about 90? in [i]clockwise[/i] direction (i. e. about -90?, mathematically speaking), then the resulting vector $\overrightarrow{e_3}$ is parallel to p, and hence it defines, together with p, a directed line (p; $\overrightarrow{e_3}$). Now, on this directed line, we measure the directed distance PQ. This directed distance will be called the [b]directed distance from the point P to the directed line g[/b].

Finally, we will say that a directed circle is [b]tangent[/b] to a directed line if and only if the directed distance of the center of the circle to the directed line equals the radius of the directed circle.

Actually, you will see that if a directed circle (k; d) is tangent to a directed line (g; $\overrightarrow{e_1}$), then the non-directed circle k must obviously touch the non-directed line g, but the converse is not true: If the non-directed circle k touches the non-directed line g, then the directed distance from the center of k to the directed line (g; $\overrightarrow{e_1}$) is either equal to the radius of the directed circle (k; d), or it has the same absolute value but different sign. In the first case, the directed circle is tangent to the directed line; in the second case, it is not.

In order not to lose the contact to earth in Bourbakist excesses, I will tell you how you can intuitively imagine the tangency of a directed circle with a directed line: You have a directed circle (k; d) and a directed line (g; $\overrightarrow{e_1}$). Consider a stone rotating uniformly and continuously along the circle. Namely, it should rotate counterclockwise if d = 1 and clockwise if d = -1. Suddendly, at the point where the circle k meets the line g, the stone flies away tangentially. Of course, it then flies along the line g. Now, if the direction in which the stone gets catapulted is equal to the direction of $\overrightarrow{e_1}$, then the directed circle (k; d) is tangent to the directed line (g; $\overrightarrow{e_1}$).

Now we are ready to formulate the quadrilateral theorem:

If ($g_1$, $\overrightarrow{e_1}$), ($g_2$, $\overrightarrow{e_2}$), ($g_3$, $\overrightarrow{e_3}$), ($g_4$, $\overrightarrow{e_4}$) are four directed lines in the plane, then there exists a directed circle tangent to all of these four directed lines if and only if the equation $A_{12}A_{23}+A_{34}A_{41} = A_{23}A_{34}+A_{41}A_{12}$ holds, where $A_{ij}$ is the point of intersection of $g_i$ and $g_j$ (cyclic indices), and the distances $A_{41}A_{12}$, etc. are directed distances measured along the directed lines ($g_1$, $\overrightarrow{e_1}$), ($g_2$, $\overrightarrow{e_2}$), ($g_3$, $\overrightarrow{e_3}$), ($g_4$, $\overrightarrow{e_4}$).

This is an arrangement-independent version of the well-known fact that a quadrilateral ABCD has an incircle if and only if AB + CD = BC + DA, and of all pendants of this fact related to excircles and concave quadrilaterals. You can also prove it in an arrangement-independent way (using the angle bisector between two directed lines - the set of all points having equal directed distances to these two lines).

Now you have the arsenal to prove Urquhart's theorem for all possible and impossible  :D  cases at once. It is, BTW, a nice exercise on directed lines.

Concerning your comment, Peter, one doesn't need imaginary radii, but it's indeed impossible to distinguish between "incircles" and "excircles" in general.

  Darij


function
algebra
polynomial
rational function
absolute value
advanced fields
advanced fields unsolved

Two questions:

What is the general form of a rational function with absolute value $1$ on the circle $\vert z\vert=1$.  How are the zeroes and poles related to each other?

If a rational function is real on $\vert z\vert=1$, how are the zeroes and poles situated?

You can map lines and circles to each other through inversion, so both problems can be made equivalent to the question of which rational functions are real on the real axis. The inversion $z\to \frac{z-i}{1-iz}$ maps the real line to the unit circle; applying this transformation to the argument and the value of a real rational function gives a rational function with unit magnitude on the unit circle.

How would we find the general form though?

[quote="blahblahblah"]Two questions:

What is the general form of a rational function with absolute value $1$ on the circle $\vert z\vert=1$.  How are the zeroes and poles related to each other?[/quote]

Moebius transformations $f(z) = \frac{z-w}{1-{\bar{w}}z}$, with $|w| < 1$,  and their reciprocals.

You can take any rational function that is $\mathbb{R}\to\mathbb{R}$ and compose it with a Mobius transformation sending $\mathbb{R}$ to $\partial\mathbb{D}$.

You get all of the finite Blaschke products, multiplied by the reciprocals of other Blaschke products. Assume $w_1, w_2, \dots,w_m,v_1,v_2,\dots,v_n\in\{z: |z|<1\},$ with not all of the $w_j$ necessarily distinct. Then we can have

$f(z)=\alpha z^p\prod_{j=1}^m\frac{z-w_j}{1-\overline{w_j}z}\prod_{k=1}^n \frac{1-\overline{v_k}z}{z-v_k}$ where $|\alpha|=1$ and $p\in\mathbb{Z}.$

The relationship of zeros and poles should be this: if there's a zero of order $k$ at $w$, then there's a pole of order $k$ at $\frac1{\bar{w}}.$ (This includes zeros and poles at zero and infinity.)

You know, you don't need all those cases: $z=\frac{z-0}{1-0z}$, $\frac{1-\overline{v}z}{z-v}=\frac{\overline{v}}{v} \cdot\frac{z-w}{1-\overline{w}z}$, where $w=\frac1{\overline{v}}$

[quote="blahblahblah"]We derived that transformation in class.  But the polynomials $f(z)=z^n$ and their reciprocals satisfy the above conditions and are also rational functions.  So is there a similar result for rational functions of higher order?[/quote]

If you allow several (or multiple) roots and poles you can divide them out one at a time by the relevant Moebius transformations, ending up with a constant function.  This is how one proves the statement about Blaschke products exhausting the possibilities.


trigonometry
inequalities unsolved
inequalities

\[\begin{gathered}\Pr ove\,\,\,that \hfill \\ \frac{{9\sin A}}{{\sin B.\sin C}}+\frac{{16\sin B}}{{\sin C.\sin A}}+\frac{{25\sin A}}{{\sin A.\sin B}}\geqslant 2\sqrt{769}\,\,\,\,\,\,\,\,\,\forall \vartriangle ABC \hfill \\ \end{gathered}\]
HELP ME, BY  cauchy_schwarz PLZ !!!  :(

[quote="andy_tok"]Suppose A>=B, then ${sinA\geq sinB\Rightarrow{\frac{9sinA}{sinBsinC}}\geq 9}$ The equality never holds.
I guess there must be more than one typo in your question. :)[/quote]
Sorry, but your solution is false.
I sure that the qualtity is holds.
$A\ge B$ then we can't conclude $sinA\ge sin B$
I had a solution but i want the solution by cauchyschwarz.  :(  :|  :mad:  :)  :D

:Lol: Sorry for my careless. I edited my question. $2\sqrt{769}$ not 5. Lol, so far  :o  :blush:

[quote="Svejk"][quote="andy_tok"][quote="kcbu"]
$A\ge B$ then we can't conclude $sinA\ge sin B$
[/quote]
 :) yes you can. sine law[/quote]

this is true only if $A,B \in [0,\frac{\pi}{2}]$ because on this interval sin is increasing.[/quote]
 :) A little bit different here. ABC is a trangle. So even if$A\geq \frac{\pi}{2}$,
$\sin{B}<\sin{(\pi-A)}<\sin{A}$ cuz $A+B<\pi$


$2^{15}= 32768$and the sum of its digits is$3+2+7+6+8 = 26$.

What is the sum of the digits of the number $2^{1000}$?

$1366$ using a calculator program. I'm not sure if there is even a clean solution to this...


complex numbers
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

1. $ C$ is the algebra of complex numbers, how to show $ C \otimes _R C$ is isomorphic to $ C \oplus C$?
2. $ A$ and $ B$ are real algebra, how to show: $ A \otimes _R B \otimes _R C\equal{}(A \otimes _R C) \otimes _C (B \otimes _R C)$?

$ \mathbb{C}\cong \mathbb{R}[X]/(X^2\plus{}1)$, so $ \mathbb{C}\otimes_\mathbb{R}\mathbb{C}\cong \mathbb{C}[X]/(X^2\plus{}1)\cong \mathbb{C}[X]/(X\minus{}i)\times \mathbb{C}[X]/(X\plus{}i)\cong \mathbb{C}\times\mathbb{C}$.
Explicitly we get $ \mathbb{C}\otimes_\mathbb{R}\mathbb{C}\to \mathbb{C}\times\mathbb{C}, z\otimes w\mapsto (zw,z\bar{w})$ a well-defined $ \mathbb{R}$-iso.

2. use the universal property of the tensor product (and nothing else). and replace $ R \to C$ by an arbitrary ring homomorphism ;).


$ 0<x,y,z<\frac{\pi}{2}$ and $ x\plus{}y\plus{}z\equal{}\pi$  , $ n,m,k$ integer number
Find min
$ T\equal{}n*tanx\plus{}m*tany\plus{}k*tanz$

Unless $ n, m, k$ are also positive we don't have a minimum.

sorry $ n,m,k > 0$

if                  $ n\equal{}1 , m\equal{}2, k\equal{}3$ min(T)=?


Yeah.

RESULTS:

1st Place. AIME15  :first:
2nd Place. isabella2296 :winner_second:
3rd Place. vallon22 :winner_third:
4th Place. yaofan
5th Place. jxl28
6th Place. arb95
7th Place. Tied - Dillinger and jjx1

Were the matches decided randomly??

Yaofan wins 2-0(4-3, 4-2)

Total fail on first game, last question.

So vallon22 and Dillinger have...3 weeks to complete their matches, while (nearly) everyone else waits on them...can't we shorten the deadline to a week, maximum?  3 week deadlines kinda ruin the whole objective of smaller tournaments.

i've pmed Dillinger
waiting for a reply with a time

Eh, didn't expect everyone to be so active. Deadline is in 1 week.

First round is best of 3, 2nd round is best of 5, Finals is best of 7.
PLAYERS:

AIME15 (2023)
jjx1(1725)
isabella2296(1650ish?)
vallon22 (1650ish?)
jxl28(1617)
yaofan(1607)
arb95(1462)
Dillinger(1233)

ROUND 1:

AIME15(2023) vs. jxl28(1617)
jjx1(1725) vs. yaofan(1607)
isabella2296(1650ish) vs. arb95(1462)
vallon22(1650ish) vs. Dillinger(1233)

(1.1) AIME15(2023) vs. jxl28(1617) [AIME15 wins]
(1.2) jjx1(1725) vs. yaofan(1607) [yaofan wins]
(1.3) isabella2296(1650ish) vs. arb95(1462) [isabella2296 wins]
(1.4) vallon22(1650ish) vs. Dillinger(1233)

(2.1) AIME15(2023) vs. Winner(1.4)
(2.2) isabella2296(1650ish) vs. yaofan(1607)
(2.3) jxl28(1617) vs. Loser(1.4)
(2.4) jjx1(1725) vs. arb95(1462)

(3.1) Winner(2.1) vs. Winner(2.2)
(3.2) Loser(2.1) vs. Loser(2.2)
(3.3) Winner(2.3) vs. Winner(2.4)
(3.4) Loser(2.3) vs. Loser(2.4)

Notes: Deadlines shorted to 1 week for contact and 1.5 weeks to play games. In deciding an unfinished match:

(1) Player making contact wins
(2) If both players make contact, but one person makes almost no effort, other person wins.
(3) If both players make an effort, more effort wins.
(4) If both players make an equal amount of effort, but cannot agree on a time, random.org will be used.

Waiting on Dillinger.

arb95 and I are tied 1-1.

Is this 2 out of 3 or 3 out of 5.

@jjx1: best of 5 for round 2, my computer's not working right now so I'll come back a little later to play our games

got a reply

gonna play our game on monday

jxl28 defeats Dillinger 3-0 (4-0, 4-0, 4-0)

yaofan and I are tied 1-1, I believe, so far.

vallon gets pwnd by AIME 3-0 (4-1,4-3,4-2) <-something like that

First round is best of 3, 2nd round is best of 5, Finals is best of 7.
PLAYERS:

AIME15 (2023)
jjx1(1725)
isabella2296(1650ish?)
vallon22 (1650ish?)
jxl28(1617)
yaofan(1607)
arb95(1462)
Dillinger(1233)

ROUND 1:

AIME15(2023) vs. jxl28(1617)
jjx1(1725) vs. yaofan(1607)
isabella2296(1650ish) vs. arb95(1462)
vallon22(1650ish) vs. Dillinger(1233)

(1.1) AIME15(2023) vs. jxl28(1617) [AIME15 wins]
(1.2) jjx1(1725) vs. yaofan(1607) [yaofan wins]
(1.3) isabella2296(1650ish) vs. arb95(1462) [isabella2296 wins]
(1.4) vallon22(1650ish) vs. Dillinger(1233) [vallon22 wins]

ROUND 2:

(2.1) AIME15(2023) vs. vallon22(1650ish) [AIME15 wins]
(2.2) isabella2296(1775) vs. yaofan(1607) [1-1 in progress]
(2.3) jxl28(1617) vs. Dillinger(1233) [jxl28 wins]
(2.4) jjx1(1725) vs. arb95(1462) [arb95 wins]

ROUND 3:

(3.1) AIME15(2023) vs. Winner(2.2)
(3.2) vallon22(1650ish) vs. Loser(2.2)
(3.3) jxl28(1617) vs. arb95(1462) [jxl28 wins]
(3.4) Dillinger(1233) vs. jjx1(1725)

Waiting on isabella2296-yaofan.

FINAL RESULTS:

1. Winner(3.1)
2. Loser(3.1)
3. Winner(3.2)
4. Loser(3.2)
5. jxl28!
6. arb95!
7. Winner(3.4)
8. Loser(3.4)

Edited to update results.

[quote="jxl28"]Sorry if this was too early, but arb95 and I already finished our 3rd round game....
[hide="Results"]
jxl28 defeats arb95 4-2 (1-4, 3-4, 4-0, 4-0, 4-0, 4-2)
[/hide][/quote]

jxl28 defeated arb95  :)

actually right now I lead isabella2296 2-1, i think

First round is best of 3, 2nd round is best of 5, Finals is best of 7.
PLAYERS:

AIME15 (2023)
jjx1(1725)
isabella2296(1775)
vallon22 (1650ish?)
jxl28(1617)
yaofan(1607)
arb95(1462)
Dillinger(1233)

ROUND 1:

AIME15(2023) vs. jxl28(1617)
jjx1(1725) vs. yaofan(1607)
isabella2296(1650ish) vs. arb95(1462)
vallon22(1650ish) vs. Dillinger(1233)

(1.1) AIME15(2023) vs. jxl28(1617) [AIME15 wins]
(1.2) jjx1(1725) vs. yaofan(1607) [yaofan wins]
(1.3) isabella2296(1650ish) vs. arb95(1462) [isabella2296 wins]
(1.4) vallon22(1650ish) vs. Dillinger(1233) [vallon22 wins]

ROUND 2:

(2.1) AIME15(2023) vs. vallon22(1650ish) [AIME15 wins]
(2.2) isabella2296(1775) vs. yaofan(1607) [isabella2296 wins]
(2.3) jxl28(1617) vs. Dillinger(1233) [jxl28 wins]
(2.4) jjx1(1725) vs. arb95(1462) [arb95 wins]

ROUND 3:

(3.1) AIME15(2023) vs. isabella2296(1775) [FIRST PLACE MATCH]
(3.2) vallon22(1650ish) vs. yaofan(1607) [THIRD PLACE MATCH]
(3.3) jxl28(1617) vs. arb95(1462) [jxl28 wins]
(3.4) Dillinger(1233) vs. jjx1(1725)

REMEMBER: FOR THE LAST ROUND, THE MATCHES ARE BEST OF 7.

For the sake of completing this tourney, jjx1 and Dillinger are being given only 3 days to complete their match.

I can't seem to be able to play the last game with Dillinger. He's always afk.

srry to post again, but considering the first 3 places are determined, i'm gonna mark this tourney as finished

hope nobody minds =)


inequalities
function
inequalities proposed

Let $x,y,z$ be postive real numbers such that $x+y+z=1$, prove that
\[ \frac x{\sqrt{1+x}}+\frac y{\sqrt{1+y}}+\frac z{\sqrt{1+z}}\leq\frac{\sqrt3}2.  \]
I independently created it today. But I think that it was proposed by someone else long before.

[quote="Lovasz"]Very easy.
By Bunhiakovsky's inequality:
    $LHS\leq\sqrt{(x+y+z)(\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z})} =\sqrt{3-(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z})}$
And:
        $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\geq\frac{9}{3+x+y+z}=\frac{9}{4}$
The proof ...[/quote]
I apply Jensen's inequality for concave function $f(t)=\frac t{\sqrt{1+t}}$.


[color=red]deleted my hint because the problem is from a contest.[/color]

[quote]Let $ a$, $ b$, $ c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} \frac {9}{4}$.
Find maximum value of the sum
$ S \equal{} \left(a \plus{} \sqrt {a^2 \plus{} 1}\right)^b\left(b \plus{} \sqrt {b^2 \plus{} 1}\right)^c\left(c \plus{} \sqrt {c^2 \plus{} 1}\right)^a$.[/quote]

This is a contest problem, please delete your post.

[Moderator edit: Exact source = Maths&Youth T7/374, deadline 15 oct 2008]

Svejk, you're free to post your solution again. (Sorry for keeping this topic hidden for so long; it seems that nobody seems to care for the temp forum.)


How come in TeXnicCenter some commands that work on AoPS don't work? For example, when I type in \binom{9}{6}, it shows up as an error and in the PDF as 96.

I don't have LaTeX installed in my PC now, but I think that you should use the amsmath package. :|

[quote="Centy"]Have use include the the \usepackage cmmand at the start of your TeX file?[/quote]

Here's what I typed:
[code]\documentclass[11pt]{article}
\begin{document}
$\usepackage{amsmath}$
$\binom{9}{6}$
\end{document}[/code]
But that still doesn't work.  :|

Look at [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#packages]Additional Packages[/url] and you'll see that \usepackage{amsmath} must go [i]before[/i] \begin{document}. What induced you to put it between dollar signs which is only for [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#mathmode]mathematics mode[/url]?

You will find that reading the AoPS sections linked above on using LaTeX will help you to understand the structure of a tex document.

:oops: Sorry...I guess I'll try to stop being lazy then :dry:


AMC
USA(J)MO
USAMO
geometry
geometric transformation
homothety
algorithm

Due to the fact that I've been doing mostly Olympiad problems this summer I've decided to start a USAMO marathon where AOPSers will have a chance to solve proof-type problems and receive feedback on their solutions from their peers so they have a feel to what they might have scored had it been the actual USAMO.

[b]  [size=150]Problem 1[/size] [/b]

[b] There are 9 delegates at a conference, each speaking at most three languages. Given any three delegates, at least 2 speak a common language. Show that there are three delegates with a common language. [/b]

Hmm, at first sight, I'd think some set theory could be applied to that.

This is one is pretty straight forward although set theory might be applicable.

I think simple case working will do. There's probably a better way though.

Lemma: If a certain delegate can understand (and speak to) at least four other delegates, then there will be three delegates with a common language.

Proof: Suppose the originally named delegate one can understand delegates two, three, four, and five. Because delegate one can speak at most 3 languages, at least one language most be spoken by 3 delegates, falling from Pigeonhole.

Now we can move on to tackle the actual problem.

Assume, for the sake of contradiction, that there does not exist three delegates with a common language. For a set delegate A, he or she can understand at most three other delegates, from the lemma. Without loss of generality, let delegate B be one of the other five delegates who cannot understand delegate A. Again, from the lemma, delegate B can only understand at most three other delegates. Thus, there will be at least one person who cannot understand either delegate A or delegate B. But then, this person, delegate A, and delegate B form a group of three that does not share a common language, which is a contradiction. Hence, it is concluded that at least one of delegate A and delegate B can understand at least four other delegates, so there will be three delegates with a common language, from the lemma.

This is exactly correct! I'd say full points for this proof. :lol:

tell me if i have interpreted this correctly:

let $n=2^k$ for some positive integer k. Prove that given a set of $2n-1$ distinct arbitrary integers, it is always possible to find $n$ integers within this set such that their sum is divisible by n.

Yes, you have interpretted this correctly.

Shame on me for posting twice in a row but here as some other problems.

[size=150] [b] Problem 2 [/b] [/size]

[b] Show that if $n=2^k$ then we can choose $n$ integers among $2n-1$ integers such that their sum is divisible by $n$.[/b]

[b][size=150]Problem 3[/size][/b]

[b] Prove that among 15 positive coprime integers greater than 1 and less than 1992 there is at least one prime. [/b]

[b][size=150]Problem 4[/size][/b]

[b] Let $d(n)$ be the sum of the digits of $n \in N$. Solve $n+d(n)+d(d(n))=1997$ [/b]

Problem 3 should specify "pairwise relatively prime" or "pairwise coprime". The statement is obviously false if the other interpretation is chosen.

problem 4
Let n = 1000w + 100x + 10y + z
I proved that w = 1, x = 9.
I got it down to 86 = 11y +2z + d (x+ y)

Now I can't figure out what y and z are. :mad:

[quote="sirorange"][b][size=150]Problem 3[/size][/b]

[b] Prove that among 15 positive coprime integers greater than 1 and less than 1992 there is at least one prime. [/b]

Meh, this is because the square root of 1991 is 44, and the 15th prime is 47. But of course I'm too lazy to do the rest  :P[/quote]

This is nice observation but "the rest" is what really gets you the points. :P  :P

Problem 3

[b]Points of interest[/b]
Since all the numbers are pairwise relatively prime, every number must have its own prime divisors which no other number has. Thus, there exists a set of fifteen prime integers, denoted S, such that for every prime in the set, one and only one of the fifteen numbers is divisible by. Note that the first 15 primes are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47. Also note that 47*47=2209>1992. Finally, label the original set of 15 numbers N.

Now we can assume all of the fifteen numbers in N are composite, since they are all greater than 1, for the sake of contradiction.

Suppose there exists n (n<15, n nonnegative) integers in S smaller than 47. Then there exists 15-n primes in S greater than 47 and 14-n primes not in S that are smaller than 47. For each of the 15-n primes in S greater than 47, to satisfy the following conditions: [list]the element in N divisible by this largest element is composite
all the integers in N are pairwise relatively prime and less than 1992[/list]...the number in N divided by the prime in S must also be divisible by another prime, not a divisor of any other element in N, smaller than 47. Since there are only 14-n primes not in S smaller than 47 and 15-n primes in S greater than 47, S must have 15 integers smaller than 47, which clearly cannot happen, since there are only 14 primes smaller than 47. Consequently, at least one number in N is prime.

I've thought about number 2 for a while. I don't seem to have gotten anywhere...

Any particular direction I could go?

Thanks for any help.

For number two try a simple induction and then see if you can pigeonhole it.

As for towersfreak's solution, overall it is correct and nicely done. The only comments that I have are that there are a few wording issues and perhaps using <=47, >=47 instead of >,<47 would be better (though it makes little difference since 47*47>1992) but I think it would still merit a 7.

Here is a solution VERY similar to towersfreak's but it applies a simple ordering of elements:

Call the integers in $N$ $a_1, a_2, a_3, ..., a_{15}$. Assume they are all composite and $<1992$. Let $p_i$ be the smallest prime factor of $a_i$. Since each pair of $a_i$ is coprime then all of the $p_i$ must be distinct. Let $m$ be the largest of the $p_i$. Since the 15th prime is 47 then we can conclude $m \geq 47 \Rightarrow m^2>1992$. Contradiction!

Is it just me or are there no solutions for #4...

Obviously, $n < 1997$. So we know $d(n) < 27$ and $d(d(n)) < 10$. These give us very rough bounds.

So $n > 1960$. We also note that $n < 1980$ because just $n+d(n)$ would exceed $1997$.

That doesn't give us very many numbers to check, so we do it by hand. Some are obviously too small or too big so we can skip those. I ended up checking only four numbers, all of which turned out to be $1998$: $1963, 1968, 1971, 1974$.

You are correct in there being no solution to #4. :lol: If no one posts a solution I will show a really nice solution I found. Think about an invariant...

[quote="DPopov"]You are correct in there being no solution to #4. :lol:[/quote]

Ah. It's quite simple. Just notice that $n \equiv d(n) \mod{3}$. So $n+d(n)+d(d(n)) \equiv 0 \mod{3}$, but $1997 \equiv 2 \mod{3}$.

This is exactly right!!! :lol:  :lol:  :lol: Excellent solution, the same as mine! I would say full points.

[quote="tetrahedr0n"]I used the stronger condition

$n \equiv d(n) \pmod{9}$

(the proof, of course, is the exact same as the one for $3$.) But this condition is more useful, such as for the find $d(d(d(4444^{4444})))$ problem.

Anyone going to post some algebra or inequalities?[/quote]

Ah, yes, of course, that's much more useful indeed. I was hoping to see some inequalities as well (there better be one on USAMO 2006).

[b][size=150]Problem 5[/size]

Prove that if $x_i>0$ and $x_iy_i-z_i^2>0$ for $i \leq n$ then

\[\frac{n^3}{\left(\sum_{i=1}^n x_i\right) \left(\sum_{i=1}^n y_i \right)-\left(\sum_{i=1}^n z_i \right)^2} \leq \sum_{i=1}^n \frac{1}{x_iy_i-z_i^2}\][/b]

Keep in mind that Problem 2 still remains unsolved.

For completeness, problem 2 is a special case of the Erdos-Ginsberg-Ziv theorem ($n$ can be any natural number.) I have proved that result elsewhere on the forum.

Also, should that inequality be homogenous?

[quote="blahblahblah"]Hah - I didn't know that the $n+d(n)+d(d(n))$ problem was a USAMO problem!  I ran a problem seminar at one of the Canadian regional math camps for grade 9s and 10s and I assigned it as a bonus question. :blush:[/quote]

I don't think these are all USAMO problems. Some of them, like the one you mentioned, I would think far too easy to be on a USAMO.

[quote="tetrahedr0n"]For the inequality, taking $n=1$ gives $z \leq z^2$ which isn't true on $(0,1)$. Maybe the $z_i$ should be squared on the LHS (which homogenizes, like Mildorf says.)[/quote]

No, the sum of the $z_i$'s should be squared.

Yeah, sorry about that! Yes, some of these questions are not USAMO level blahblahblah. So here is one that you might find a bit more challenging! ;)

Take a sequence of numbers where $a_j \geq 0$ and $\sum a_{j}^2 < \infty$.
Construct another sequence $b_j=\frac{1}{2} \left(a_j+a_{j+1}\right)-\frac{1}{8} \left(y_{j+1}-2y_j+y_{j-1}\right)$ where $y_j=min(a_j,a_{j+1})$.

Prove that $\sum b_{j}^2 \leq \sum a_{j}^2$

[b][size=150]Problem 6[/size]
An $n$ x $n$ matrix whose entries come from the set $S=[1, 2, ..., 2n-1]$ is called a silver matrix if for each $i=1, 2, ..., n$, the ith row and the ith column together contain all the members of $S$. Show that silver matrices exist for infinitely many values of $n$.

[size=150]Problem 7[/size]

The positive real number $x_0, ..., x_{1995}$ and

\[x_{i-1}+\frac{2}{x_{i-1}}=2x_i+\frac{1}{x_i}\]

for $i=1, ..., 1995$. Find the maximum value that $x_0$ can have.

[size=150]Problem 8[/size]

Let acute $\triangle ABC$ have orthocenter and circumcenter $H$ and $O$ respectively. For $X$, $Y$, $Z$ on $BC$, $AX$ passes through $H$, $AY$ bisects $\angle A$, and $Z$ bisects $BC$. The perpendicular from $T$ on $AY$ meets $AX$ at $H$. Prove $AO||TZ$

[size=150]Problem 9[/size]

Find a set with $n$ integers such that for every two numbers $x$ and $y$ in the set, $(x-y)|(x+y)$. [/b]

yay a geometry problem! (I refer to #8, of course)

As O and H are isogonal conjugates, T is the nine-point center; the homothety about the centroid with factor -2 takes TZ to AO; thus, they are parallel.

9

bleah
induct
I dunno whether I'm supposed to FIND a set or prove one exists. I'll do the latter, since I'm lazy. I'll impose an additional condition on my set: it has to contain nonnegative integers. (This'll make it easier for me later.) So...basecase!!!! $n = 1$. Take the set {1}. Now, assume we have a set S for $n = k-1$ and we're trying to get a set for $n = k$. Consider all the pairwise differences in S (thingies of the form x - y, where x and y are distinct elements in S), and multiply them all together; call this P. Add P to every element in S and then throw a 0 in there; call this new set S'. As P is obviously positive and S had nonnegative elements, S' (before throwing in the 0) has strictly positive elements, so we can throw in the 0 without fear of repetition and whatnot. Now, S' has n = k elements, and for any pair of elements $(x', y')$, if the pair includes 0, it obviously works, so assume it doesn't. Then $(x', y')$ came from a pair $(x, y)$ in S by the relation $x' = x + P, y' = y + P$. Now, $x' - y' = x - y, x - y | P$ (by the construction of P), and $x - y | x + y$ (by the fact that S was a valid set for n = k-1), so we have $x - y | (x + y) + 2P$, or in other words, $x' - y' | x' + y'$. YESSS WE WIN!!!! The sets that this algorithm generates for the first few n follow:

{1}, {0, 1}, {0, 1, 2}, {0, 2, 3, 4}, {0, 48, 50, 51, 52} or something, etc. If you actually wanted to find a set for all n, you could try to figure out a closed form for these sets here.

[b][size=150]Problem 10[/size]

In $\triangle ABC$, $\angle C=90$, $\angle A=30$, and $BC=1$. Find the minimum length of the length of the longest side of a triangle inscribed in $\triangle ABC$ (that is, one such that each side of $\triangle ABC$ contains a different vertex of the triangle). [/b]


Find limes for $ w\geqslant 0$

$ \lim_{n\to\infty}\sqrt[n]{w^{n}\plus{}(w\plus{}\frac{1}{2})^{\minus{}n}\plus{}1}$

For $ w>1$, $ \left( w \plus{} \frac {1}{2}\right)^{ \minus{} n} \plus{} 1 \equal{} \mathcal{O}(1)$, so we can ignore that part, and the limit is just $ \lim_{n\to \infty}(w^n)^\frac {1}{n} \equal{} w$.  For $ \frac {1}{2} < w < 1$ the constant $ 1$ dominates the other two terms, so the limit is $ 1$.  Similarly, for $ 0\leq w < \frac {1}{2}$ the $ \left(w \plus{} \frac {1}{2}\right)^{ \minus{} n}$ term is dominant, so the limit is $ \left(w \plus{} \frac {1}{2}\right)^{ \minus{} 1}$.  In summary,
\[ \lim_{n\to\infty}\sqrt [n]{w^{n} \plus{} (w \plus{} \frac {1}{2})^{ \minus{} n} \plus{} 1} \equal{} \begin{cases} w & w\geq 1 \\
1 & \frac {1}{2} < w < 1 \\
\frac {2}{2w \plus{} 1} & 0\leq w\leq \frac {1}{2} \end{cases}
\]


Prove that every metric space with countable number of elements is non-connected

Consider an element $x$ and an open ball around $x$ of radius not equal to any of the distances from the other elements to $x$. This set is both open and closed.

We can even go a step beyond and say that any such space is totally disconnected; for any given element $y$ other than $x$, we can choose the distance less than $d(x,y)$, separating the two points.


function
induction
real analysis
real analysis unsolved

Find all continous functions $f: \Re\to(0,\infty)$ such that $\frac{f(4x)}{f(3x)} =3^x$.

$g(x)=ln(f(x)$ is continuous and satisfies $g(x)-g(\frac{3}{4}x)=\frac{ln(3)}{4}x$.
Hence, an easy induction shows that, $g(x)-g((\frac{3}{4})^nx)=ln(3)(1-(\frac{3}{4})^n)x$.
Continuity implies that: $g(x)=g(0)+ln(3)x$. Hence $f(x)=K3^x$ for a constant $K>0$


calculus
derivative
real analysis
real analysis unsolved

I heard this weird contradictory thing somewhere and was hoping someone could explain why it is so:

if you have $ x\plus{}x\plus{}x\plus{}...\plus{}x$, where $ x$ is added $ x\minus{}$times, then it is the same as $ x^2$.
But once you take the derivative, you get simply $ x$ on one end and $ 2x$ on the other. Why does this work this way and where is the problem?

[quote="emant777"]I heard this weird contradictory thing somewhere and was hoping someone could explain why it is so:

if you have $ x \plus{} x \plus{} x \plus{} ... \plus{} x$, where $ x$ is added $ x \minus{}$times, then it is the same as $ x^2$.
But once you take the derivative, you get simply $ x$ on one end and $ 2x$ on the other. Why does this work this way and where is the problem?[/quote]

It is falcacious because we have that the number of x's on the left hand side is obviously a variable.

This reminded me of the logic students use when they first encounter the problem of deriving $ x^x$ (that was in high school, if I remember correctly). Some use the derivative of $ x^a$, some use the derivative of $ a^x$, they get different answers-and both are incorrect (until someone gets the idea of applying logarithms...)


geometry
circumcircle
incenter
angle bisector
geometry unsolved

Let $ABC$ be a triangle, $(O)$ be the circumcircle of $ABC$. Let $(O_1)$ be the circle tangent to $AB,AC$ at $A_1,A_2$, and tangent to $(O)$. Suppose $BA_2$ intersect $CA_1$ at $A_0$, similarly we have $B_0,C_0$. Prove that $AA_0,BB_0,CC_0$ are concurrent.
Help me! (If it is not clearly ,please tell me,thanx) ;)

Ok!
I did it. Just use inversion at center $A$ to obtain via Ceva theorem that
$BX:XC=\frac{(a-c+b)b}{(a-b+c)c}$,
where $X$ is a point of intersection of $AA_0$ and $BC$.
Now I think you see...
 :D  :D  :D  :D

Since I (the incenter of ABC) is the midpoint of A_1A_2 and, obviously, this line is perpendicular to the line AI (angle bisector of <BAC), the problem can be re-worded accordingly.
Further, use Ceva 3x and, finally, its converse.

[b]NOTE: [i]It seems the problem is valid when the three circles are all externally tangent to (ABC) as well, this time I_a being the midpoint of A_1A_2, a.s.o.[/i][/b]

Best regards,
sunken rock

$ I$ is incentre of $ ABC$. Then $ AI$ is perpendicular to $ A_1A_2$ and $ I \in A_1A_2$
$ K_a \equal{} AA_O \cap BC ; L_a \equal{} A_1A_2 \cap BC$ ... sililar : $ K_b;K_c$ and $ L_b;L_c$
Because:  $ (L_a,K_a,B,C) \equal{} (L_b,K_b,C,A) \equal{} (L_c,K_c,A,B) \equal{} \minus{} 1$ , then :
$ AA_0,BB_0,CC_0$ are concurrent
$ < \equal{} > \frac {K_aB}{K_aC}.\frac {K_bC}{K_bA}.\frac {K_cA}{K_aB} \equal{} 1$
$ < \equal{} > \frac {L_aB}{L_aC}.\frac {L_bC}{L_bA}.\frac {L_cA}{L_aB} \equal{} 1$
$ < \equal{} > L_a , L_b , L_c$  are in a straight line

Which is  obvious because $ L_aL_bL_c$ is orthopolar of $ I$ WRT $ \Delta ABC$

you can see it here : http://www.mathlinks.ro/viewtopic.php?t=273212
[img]http://img16.imageshack.us/img16/9743/98443748.jpg[/img]

Dear Mathlinkers,
for a synthetic proof, see for example
http://perso.orange.fr/jl.ayme   vol. 4  A newmixtilinear incircle adventure II p. 18
Sincerely
Jean-Louis


Mr. R's high school had a math tournament today, did anybody here go? How did you do? Supposedly there are some people in my area that do classes here, but they never respond. (MarkSmith888, did you go?) Speak now!

Me: I got 5th in alg2, blah (arithmetic mistakes, as always)

Yea, I went.  Wow, we kind of met each other I suppose, I was the Grissom guy who got first.

No way, you were in alg2? Good job. I'll say hi next time; I'm the guy in the blue hat, lol.]

Oh BTW-Didn't you get a 95 or so? Which problem did you miss, the function?

Actually I think I missed number 7 (can't tell for sure since I didn't write down my answers), but I remember I got three Cs in a row.  I put C instead of D (didn't read the denominator, lol).  Are you going to Auburn?  If you are, good luck and maybe we'll finally meet each other.

Ah, that stinks. Oh well, you still won, lol.

Yep, I am going to Auburn, but don't count on me placing. That test is so strange, the geometry test last year...I placed, but only because I was good at guessing.

Down with Chris Harris! =D

haha, at least you placed, I didn't.  Yea, the Geometry test last year was kind of weird, I got a lower score on that one than on the Vestavia Geometry test.  I'm guessing Chris Harris is the guy who beat everyone by like 8 points on the Vestavia Alg II Test this school year.

Yes, and he got a 95 on the Grissom test. I remember the first Auburn test I took, some of the questions were REALLY dumb, like adding up 10 random large numbers.

What happened to the Austin High tournament and Auburn High tournament?

I am in my 1st year and have never heard of it :(

[quote="now a ranger"]What happened to the Austin High tournament and Auburn High tournament?
[/quote]

Austin High School is my old high school. . .  They still have a tournament?


vector
linear algebra
linear algebra unsolved

:maybe:

Let u, v be distinct vectors of a vector space V.
Show that if {u,v} is a basis for V and a and b are nonzero scalars, then both {u+v, au} and {au, bv} are also bases for V.

I don't necessarily want the solution... but I would like a few guiding steps to get me in the right direction.

so far all I have is this:

for a w in V, w = xu + yv for scalars x, y.
and for       0 = xu + yv, x=y=0

...

yeah, so basically I have nothing at all... someone please help![/code][/hide]

woops, ignore this thread.. i got the solution (i think :) )


geometry
MATHCOUNTS
analytic geometry
ratio
area of a triangle
Heron\u0027s formula
similar triangles

I can't figure out these problems.....

1. What is the radius of the circle inscribed in triangle ABC if AB = 12, AC = 14, BC = 16? Express your answer in simplest form.

2. In triangle ABC points D and E lie on sides AB and BC, respectively, so that $ \frac{AD}{DB} \equal{} \frac{3}{5}$ and $ \frac{CE}{EB}  \equal{} {1}{2}$.  Segments CD and AE intersect at Point F.  What is the value of $ \frac{CD}{FD}$?  Express your answer as a common fraction.

For the first problem, you can use the formula A=rs, where A is the area, r is the radius, and s is the semiperimeter.  Use Heron's Formula to find A.

[hide="1)"]
[hide="Hint"]
Use AIME15's hint. $ [ABC]=rs$ where $ s$ is the semiperimeter and $ r$ is the inradius. Use Heron's Formula to find $ [ABC].$[/hide]

[hide="Solution"]
First find the semiperimeter. This is $ \frac{12+14+16}{2}=21.$ \begin{align*}[ABC]&=21r\\\sqrt{21(21-12)(21-14)(21-16)}&=21r\\\sqrt{3^3\cdot5\cdot7^2}&=21r\\21\sqrt{15}&=21r\\r&=\boxed{\sqrt{15}}.\end{align*}[/hide][/hide]

Hmmmmm does anyone know how to do number 2? Maybe you can solve this using similar triangles or something.

EDIT: wait misread the question, nevermind
EDIT: Wait, I still dont get it

You can use Menelaos theorem.

I think $ \frac{CE}{EB}\equal{}\frac{1}{2}$.

no, it says that $ \frac {CE}{EB} \equal{} 12$.

Wait, do you think there is a typo in the question?

EDIT: hmm I googled Menelaus' theorem and found that it only dealt with proving collinear points on a triangle. Can that help us with the question?

I remember a very similar MATHCOUNTS problem that only require similar triangles. I can't seem to be able to draw the right lines though...

[hide="Cheap Solution Outline"]
Assume $ \triangle ABC$ is right with hypotenuse $ AC$, $ AB\equal{}8, BC\equal{}13$.  Set this on the coordinate grid.  Then $ A\equal{}(0,8), B\equal{}(0,0), C\equal{}(13,0), D\equal{}(0,5), E\equal{}(1,0)$.  Find the equation of the line from $ A$ to $ E$, and similarly for $ C$ to $ D$.  The intersection is $ F$.  Use the corresponding $ x$-coordinates for $ D,F,C$ to find $ \frac{CD}{FD}$.
[/hide]


geometry
incenter
combinatorics unsolved
combinatorics

10 points are marked inside a circle of diameter 5. Prove that it is always possible to find 2 points such that distance between them isn't greater than 2.

try to find an example to disprove that statement:
first construct a circle with diameter 2 and incenter M (M is the incenter of this circle we actually are talkin about). Either there is no or one point P inside this circle with r=1.
let $C=A \setminus B$ the "remainder". with A the area inside the circle with d=5 and B the area the area inside the circle with d=2.
cut this area C into 8 kongruent parts. If such an example is possible than in at least one of these 8 kongruent parts there have to be 2 points.
Its easy now to show that this could never be possible, because the maximum distance between two of such points is always smaller than 2!

could you solve this problem using expected value? if so, how?


calculus
integration
derivative
Putnam
function
real analysis
real analysis unsolved

Show that

\[ \int_{0}^{\ell}\int_{0}^{t}f\left(x\right)\,dx\,dt=\int_{0}^{\ell}\left(\ell -x\right)f\left(x\right)\,dx. \]

Write that $\ell-x$ as $\int_x^\ell 1\,dy$: the second integral is $\int_0^\ell \int_x^\ell f(x)\,dy\,dx$.

Do you see what's going on now?

Let $g(\ell) =\int_{0}^{\ell}\int_{0}^{t}f(x)\,dx\,dt-\int_{0}^{\ell}(\ell -x)f(x)\,dx$.
Differentiating with respect to $\ell$ we get:

\[ g'(\ell)= \int_{0}^{\ell}f(x)\,dx+ (\ell \int_{0}^{\ell} f(x)\,dx)'-(\int_{0}^{\ell} xf(x)\,dx)'= \]
\[ \int_{0}^{\ell}f(x)\,dx+\ell f(\ell )-\int_{0}^{\ell} f(x)\,dx-\ell f(\ell)=0 \]
And since $g(0)=0$ implies that $g \equiv 0$. q.e.d.

I wsed the fact that $(\int_0^x f(t)dt)'=f(x)$ if the differentiation iswith respect to $x$. jmerry, could you finish your solution? I don't see how to continue it and seems to be a usefull method. Thanks.

jmerry is just using Fubini's theorem.  It's probably a bit clearer if you draw the square $[0,\ell]^2$ and look at what changes you will get in the limits if you interchange the order of integration.

Exactly. Whenever you see an iterated integral, the first thing you should think is "What happens when I switch the order?" When you have something that isn't obviously a multiple integral, consider introducing another variable to smooth it out. (see Putnam 2004 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20503]A6[/url] for a nice use of this)

Can somebody show me how exactly works on this example. I'm a little confused.  If the function under the integral is bounded and continuous can I swich the order of integrals? Do I need to change $dx$ with $dy$? It cannot figure it out.. :?  :?

Extend the region of integration to the whole square $[0,\ell]$, and multiply $f$ by the characteristic function of the triangle. The function isn't continuous now, but it's still bounded (on a compact region), so Fubini's theorem applies. Switching the order of integration, we have the desired result.

Here's what I understood:

$\int_0^\ell \int_0^tf(x)dxdt=\int_0^\ell (\int_0^tf(x)\cdot 1 dx+\int_t^1 f(x)\cdot 0 dx)dt\int_0^\ell \int_0^\ell f(x)\chi(x,t)dxdt$
Where $\chi(x,t)=1 \Leftrightarrow 1\ge t\ge x\ge 0$.

The last integral is equal from "Fubini" theorem with:

\[ \int_0^\ell \int_0^\ell f(x)\chi (x,t)dt dx=\int_0^\ell f(x)\int_0^\ell \chi(x,t) dt dx= \]
\[ \int_0^\ell f(x)(\int_0^x\chi(x,t)dt+\int_x^\ell \chi(x,t)dt)dx=\int_0\ell (\int_0^x 0 dt+\int_x^\ell 1 dt)dx= \]
\[ \int_0^\ell f(x)\int_x^\ell dydx=\int_0^\ell f(x)(\ell-x)dx \]

Tell me if i'm wrong this is my first use of Fubini theorem.


Βρείτε όλες τις λύσεις $ (x, y, z, t)$ με $ x, y, z, t$ $ \in{\mathbf{N}}$ του συστήματος

 $ x \plus{} y \equal{} zt$
 $ z \plus{} t \equal{} xy$
 $ x\leq{y}$
 $ x\leq{z}\leq{t}$

και δείξτε  ότι , για $ x, y, z, t$$ \in{\mathbf{Z}}$ , το ίδιο σύστημα έχει άπειρο πλήθος λύσεων .

Ομορφη ασκηση. Λοιπον εχουμε

$ z\plus{}t\equal{}xy>\equal{}y$  αφου οι αριθμοι ειναι φυσικοι
$ zt\equal{}x\plus{}y<\equal{}2z\plus{}t<\equal{}3t$
Οποτε $ z<\equal{}3$

Για $ z\equal{}1,3$ δεν εχουμε λυσεις ενω για $ z\equal{}2$ παιρνουμε τις τετραδες
$ (x,y,z,t)\equal{}(2,2,2,2)$ και $ (x,y,z,t)\equal{}(1,5,2,3)$ οι οποιες ικανοποιουν το συστημα

Στους ακεραιους τωρα θετουμε $ y\equal{}0$ και παιρνουμε την απειρια λυσεων
$ (x,y,z,t)\equal{}(\minus{}c^2,0\minus{}c,c)$ οπου c φυσικος αριθμος


abstract algebra
group theory
superior algebra
superior algebra unsolved

Is there a $ \mathbb{Z}$-module homomorphism $ f: \mathbb{Z}/20\mathbb{Z}\to \mathbb{Z}/14\mathbb{Z}$ which maps $ [1]$ to $ [2]$?

no.

where would [7] go? what about [21]=[1]?

If $ \varphi$ is a group homomorphism from $ \mathbb{Z}_{20}$ to $ \mathbb{Z}_{14}$, then the image of $ \varphi$ is both a subgroup of $ \mathbb{Z}_{14}$ and isomorphic to a factor group of $ \mathbb{Z}_{20}$, namely $ \mathbb{Z}_{20}/\text{ker}(\varphi)$.

In particular, the order of the image divides both 20 and 14, therefore 2. So the image is a subgroup of $ \mathbb{Z}_{14}$ of order 1 or 2, therefore either the trivial subgroup or $ 7\mathbb{Z}_{14}\equal{}\{ [0],[7]\}$. So nothing can be mapped to 2, period.


Do you think we will ever be able to travel between the stars?

YES, YES,YES,YES!!![size=59]and I'll be the first one out there :D [/size]

I believe it is possible, but I am also sure that I won't be able to expirence it in my life...

I believe it might be possible, but I don't see it being of any use, so it certainly won't be prioritized. I put "no."

Until we find a more efficient method, no.  Because at the moment we don't even travel at close to light speed, it would take much too long to travel between the stars.  But someday, I think we will be able to.

Heh.  I put no.

It's really pointless for humans to travel between the stars.  By the time it's really necessary to do so, humans would be called something else.  ;)

[quote="b-flat"]I think we will eventually be able to travel among the stars, but I don't know how soon.

Advantages of traveling amoung the stars:

1.  When Earth fills up with trash, gets destroyed by war or becomes over-populated, we will have another place to live.
2.  Because there are so many billions of planets, there are bound to be many more intellegent spieces such as in Star Wars.  Then, we could meet with them.
3.  On Earth, there are limited resources, if we travel among the stars, we might be able to find more resources to use such as fossil fuels.
4.  Why not?[/quote]

Okay, here's why I don't think those will help:

1) Okay, this might be a possibility, but it's certainly not any time soon. Plus, we'd have to spend way too much money practically building a habitable ecosystem, assuming we have a planet in a relatively circular orbit around a medium-sized star with no supernovae or other damaging stellar events around for many years...it'd be much easier to fix the earth by then. :P

2) I disagree. There are SOOOOO many variables that must be perfect for anything at all to develop. Plus, the probability for life is amazingly remote once everything's as good as it can get. When you get to intelligent life, well, even according to evolutionary theory/ages, you need billions of years of undisturbed time to get to intelligent life. Assuming the (doubtful) theory of evolution, you basically need another earth somewhere else in the universe. It doesn't get much better than here. ;) And Star Wars isn't really evidence...

3) AFAIK, fossil fuels come from dead organisms, so you'd need organisms to start with. See above.

4) Because it would cost likely quadrillions of dollars just to spend a manned space ship out there, without any intent of colonizing a planet or anything. You'd need fuel, food, etc. for several decades, and people already go crazy from a year in the space station.

Space isn't all it's made out to be.

[quote="b-flat"]I think we will eventually be able to travel among the stars, but I don't know how soon.

Advantages of traveling amoung the stars:

1.  When Earth fills up with trash, gets destroyed by war or becomes over-populated, we will have another place to live.
2.  Because there are so many billions of planets, there are bound to be many more intellegent spieces such as in Star Wars.  Then, we could meet with them.
3.  On Earth, there are limited resources, if we travel among the stars, we might be able to find more resources to use such as fossil fuels.
4.  Why not?[/quote]

1. By the time we have the technology, we would probably have a way to get rid of the trash. However, overpopulation is a major problem.
2. I totally agree, the way life was created on Earth was probably one of the many possible ways of creating life.
3. Maybe we could use the Sun  :D
4. It'll need the cooperation of many countries for this to succeed.

Happy New Year!!! ( I wish it would snow, but it is raining!!! In case if your wondering, I live in Utah where it should be snowing now...)

Really, how close has the world come to destroying itself in the cold war?  Is that not going to happen in... say... the next 1000 years?

i was reading something about it in a book by S. hawkings... it seems that to be able to do it so that the people inside the spaceship feel the trip to be short enough so that they reach the closest star in their lifetimes, you need speeds that aproach a lot to lightspeed.  the amount of energy to reach such speeds... well, it's just stupid (also considering you want to accelerate in space)... i doubt very much that it'll happen.

And when we get there, we'll be able to copulate with gorgeus alien chicks to put all earthling females to shame.
And no one will ever have to work for a living anymore. And with the alien chicks fantastically advanced medical science we'd be able to live forever.
And the alien chick species will have no males because that would make the situation unpleasent.
And the alien chicks shall enlighten us as to the formula for the overlap area between 2 triangles.
Also Energy will be very cheap because of the fusion reactors we will have had up and running by then...

There will be no wars, no malice between people, no hate, no fear, no cold, no hunger...

And no one will ever die alone...

Sure. I can see it happening in my lifetime... (NOT!!!)

seriously that won't happen
whenever i say that there is always someone around saying , you don't what they will invent you

yeah, uhm, the year 2000 looks a lot more exciting when you watch terminator (1984), space odyssey (1968),etc...
when apollo 11 was flying to the moon, some senator said he hoped to get usa on mars by 2000, now the best bet is about 2018

technology does not advance as fast as we would wish, that is just wishful thinking because of a brief period during the space race when everything seemed possible

[quote="Peter VDD"]Tell me what's the point of traveling between stars?

Im glad they postpone the star programs and focus on the more useful technology...[/quote]

I think you're right. Other than the problem of overpopulation, theres no other reason...

hmm if i get my history correctly one saturn V to launch three  astronauts to the moon (and back) costed 400 million dollar
with their current budget i think nasa has something like 15 billion dollars a year
if you were to do a massive evacuation to mars, not needing to get them back, it still is not financially possible

[quote="Peter VDD"][quote="jimhu"]Other than the problem of overpopulation, theres no other reason...[/quote]And how would it solve that? There's no food generation on the moon, neither there's air or something. Unless you did think of that, but then there are much ethical ways of killing the extras. :P

We could dump our waste there, but that's about it I guess.

@Fredbel... well, I do not think the cost of such a launch is lineair in the number of passengers ;) but what if they land on the moon? Same problem, even if we can run, there's nowhere to hide or land.[/quote]
Yeah. It's sort of scary that we depend so much on nature, the sun, and human nature to keep us alive.

Then comes another question...

This might be unrelated to the topic but kinda related:

How long can new technologies come out until one day technology can't get better, and what might be the cause?

why are you waiting for that , we have reached that point, now evil people are just making it annoying for us
in belgium we are taunted with endless ringtune commercials, commercials that simply say sms us ! (they used to say what you could win, now it just means, so we can endlessly spam your cellphone at your expense) , telephone games with ridiculously simple word permutations at night on commercial stations that make living hells of handicapped, elderly, workless, lonely (they promise that one user who gets in 15000 euros, but they mean that is the maximum, after a phone bill of 1400 euros you win a hundred euros) , digital television which is an excuse for our distribution channel to throw of decent channels to force you into getting digital television, machines that get virusses instead of obeying et cetera

my digital television example is a good example of advancements not making a better alternative for something, but being sold after cutting of supply to the original decent old product

I'm sorry i can't complain live is good in belgium, but i just cannot stand how much worthless commercial stuff is around nowadays

Absent a disconunity, such as the return of Christ, the end of time, the destruction of man on earth by a space object, I believe that we will have effective travel between the stars at some point.

Addressing the point that there is no new developing technology: about 1900 the head of the US Patent office formally asked the US Congress to eliminate his office as everything useful that would ever be invented had already been invented.  It is easy to be overwhelmed by the mundane majority and overlook the insightful minority of developments.

Addressing the persistent idea that the US space program was a waste (I don't have the figures for other countries' programs):  this is simply untrue.  The space program is the only large program that paid for itself from an economic prespective.  This means that the new technologies spun off (integrated circuits, velcro, silly putty (inside the launch couches), space pens, better understanding of metalurgy, cell phone, satalites, etc.) have generated not only sufficient income to replace the value in the economy, they have created sufficient income so that the increased taxes paid to the US government from the increased economic activity is greater than the money spent on the space program.

Note I am not even adding anything in this evaluation for the great value of being a way for the Soviet Union and the United States to compete, rather than fighting a war, or two, and killing folks.

Let me now turn to the silly idea that we are running out of space to store/dispose of garbage, or for people to live.  These ideas were the offspring of the ideas in the mid 1960s that the world's food supply would run out and we needed to limit the number of children to avoid this upcomming disaster.  It didn't happen because we developed much more productive farming techniques.  We already have even more productive methods, but since we have a worldwide glut of food, the price of food is so low that these new techniques are not being currently implemented. (It's that downward sloping demand curve again.)

Not that we would do either, but simply to add perspective let me relate two facts:  If we placed all of the disposed of garbage the United States produces into the Grand Canyon, it would take something like 20 years to fill it in.  Except in specific geographies, there simply is no shortage of space.  Places like Japan are incinerating their garbage, receiving something like a 99% reduction in volume.  This talk about shortages is generated by the professional environmentalists who want governments and people to give them control over economic activity in answer to this unfounded fear.

An often quoted number in this effort to dominate folks is that we only have x years of permitted landfill space left.  Generally, this is true.  The professional environmentalists have convinced the governments that landfills should use the best technology, that technology is evolving and permits should not be granted for more than about 10 years, as the government can require the newest technology (liner seals and the like) when approving the new permit.  Thus, we by definition have less than 10 years of permitted landfill space left.  Isn't a shortage, its just the results of the laws.

(Also, one must consider the view of the reviewing governmental agency.  Do they want more or less to do?  More means a larger staff, more stability in your job, perhaps a larger job, more prestige as you are managing the future of the city/industry more often.  (If I don't approve your permit in the next year, you don't have an approved landfill and your city/industry is in trouble.))

Let me end with what for me was a chilling fact.  When the US space shuttle first orbited, the cost reported to put a pound in orbit was the same, inflated adjusted, as it cost to place a pound in Virigina when Jamestown was founded by the British.  (I know that there was air to breath, water to drink and food to eat in Virgina; where none of them are in orbit, let alone on the moon.)  However, just consider how long it took for a sustainable presence to develop and don't feel we are doing too bad in the space development area.  Also note the prize won for the two space entries in a short time last summer.  I am told that the computers abord the Ford Torus have greater computation power than those abord the space shuttle.  Technology is advancing, rapidly.  Consider where we were ten years ago, 100 years ago. . . .

"That's all I have to say about that."

We'll definitely travel between the stars. See:[url]http://home.comcast.net/~mbmcneill7/[/url]

Anybody tried to read the Heim theory papers?
(Reported in New Scientist a few weeks back, look for Heim Theory in yahoo!)
The maths is too confusing for physisists?
It a G.U.T. thing that has some (little) credibility.
Apparently faster-than-light travel is possible?
Once you've had a look, consider this:

What if rather than going into a 'parallel' space that allows you
to travel above light speed, you just cause the fabric of space-time
to expand again, i.e. you end up a long way from where you started,
but no nearer to where you want to be?!
I wouldn't want to be the John Kryten (Farscape) for this one!...

Hmm... i'd say interstellar travel is possible and likely but not in the near future.

Why travel above lightspeed if you can travel at lightspeed? Time does not pass at lightspeed..

And i stick to my theory that all of that will not happen

Travelling to the moon is possible
it has been proven, the people who can do it are still around, do we do it?
no only one nation has done it succesfully , six times,

If we don't even do that, do we really expect to be on Mars in 2020?
I am sure we won't be back on the moon before the last of the now remaining nine moonwalkers have passed away

Sorry for the pessimism

One factor that might limit scientific developments would be the length of someone's life. Even if you studied for a hundred years, there is only so much you can learn, so eventually, no one will even be able to understand any of the inventions that we will someday have. Unless each invention is made up of smaller subinventions, and every inventor in the world picks a random subset from these to combine to make a new invention....

Edit: However, I think that we will reach the stars before we reach that point.

I read somewhere that interstellar transport is so energy consuming that a smart race would never attempt it. Going to the nearest star and back would take 8 years. Our telescopes have already scanned the local stars so much that there is no hope of ever discovering a system that could sustain life that is easily reached.

Personally, I believe that science fiction stories such as Halo and Star Wars are impossible.

Sorry for the topic revival.


Prove that the number of primes is infinite by use of the formula determining the toitent of n

Well, first thing that comes to mind is consider m=the product of all primes.  phi(m)=m*zeta(1) by the product representation of the euler function, and that value is undefined.


Suppose $ x_1, x_2, . . . , x_{2n}$ are real numbers such that $ |x_{i\plus{}1}\minus{}x_i| \leq 1$ for any $ i$, $ 1 \leq i \leq 2n \minus{} 1$.

Prove that
$ |x_1| \plus{} |x_2| \plus{} ... \plus{} |x_{2n}| \plus{} |x_1 \plus{} x_2 \plus{} ... \plus{} x_{2n}| \geq n(n \plus{} 1)$.

There is something wrong... choose $ x_i \equal{} 0$ for each $ i$ and the inequality does not hold.

Pierre.


An art collector went into an art dealer. He asked to have 2 items valued. One was an old violin and the other, an oil painting. The experts studied them for days before confirming their remarkable findings. The collector was told that the items were an original Stradivarius and a previously unknown work by Vincent van Gogh. At first, the collector was thrilled, but later he became very dejected. Why?

There are frequently many answers that one can give to this type of problem.  Which is the one you have in mind?

Stradivarius painted the picture, and van Gogh made the violin? Just kiddin'.  :rotfl:

hsiljak, you were right... were you really kidding?

I knew the trick before. But wasn't 100% sure, so I went with a dose of precaution  :lol:

But wouldn't a painting by Stadivarius and a violin by Van Gogh probably fetch a lot anyway? :wink:

[quote]Search Van Gogh on wikipedia, then ctrl-F violin. Not found. Therefore, it is possible that Van Gogh never made violins. (Violins are hard and tedious to make in the first place...)

In the same vein it is somewhat unlikely that Stradivarius painted an oil painting.[/quote]
This is a riddle/puzzle and is not meant to be real.  Everyone knows Van Gogh did not produce violins and everyone knows Stradivarius did not paint.


AMC USA(J)MO USAMO function geometry \/closed

Tonight (Friday, June 10th) we will be having a special [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url] Math Jam!

The Jam is at 11 PM Eastern / 8 PM Pacific.

Please join us and learn all about [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url]! :w00t:

[quote="bubala"]If I were planning to do AoPS independent study next year, should I do WOOT instead?  How hard is WOOT?  I understand it is aimed at MOP qualifiers, but what about USAMO qualifiers (people who barely qualified like me).  Is it too hard?  HOw long does it take?  Ooh I have sooooo many questions?  And what's the MATHJAM about tonight?  How hard is it?[/quote]The Math Jam is scheduled to answer all of your questions.

The Math Jam will consist of one Olympiad problem that's slightly on the easy side, followed by a discussion of WOOT.  Please post any WOOT questions you will likely ask tonight here so we can prepare our answers ahead of time.

[quote="The Original Pi Guy"]hm... what happened to the generating functions math jam.

I was looking forward to that...[/quote]

It will be rescheduled - we will still hold it!

Hopefully, it's not at 11 like this one... Those of us on the east coast (especially those of us who go to bed at 9:30) are not very well accommadated. I guess I'll just have to look at them afterwards.

Hi folks,

Why is the mathjam scheduled so late? I wanted to attend, but I've got the ACT tommorow, so I'm going to bed early. Not that it's such a big deal, since I'll read the transcript.

[quote="tetrahedr0n"]Why is the mathjam scheduled so late? [/quote]Tonight's Math Jam is being simulcast at the meeting of the Las Vegas Futurists, where MCrawford is speaking about online education.  That's why it's unusually late.

[quote="Peter VDD"]Is this open for non-US students as well? (that includes, will the times be fit for worldwide public?)[/quote]If you are asking about the program, yes. That is why it's called [b]Worldwide[/b] Online Olympiad Training :)

[quote="MysticTerminator"]how do MOPpers register?

edit

oh there's a mathjam. heh. forgot about those thingies.[/quote]

Y'all will receive a registration form at MOP.

Oh so the generating functions math jam [i]will[/i] indeed be held at a later time

And here I was thinking it was just a cover up for the WOOT math jam...

That was a good math jam, now I might have to consider getting a temporary summer job or have my dad hire me for something lol, but right now I just gotta study :)

[quote="hello"]Even though I dont really care, I wanted to go to the mathjam but the green button didnt appear to me for some reason, I was wondering why? I dont think my laptop is horrible[/quote]

You can get to the Math Jam in general by just clicking directly on the Classroom link on the sidebar.

[quote="orl"]I would like to read the WOOT transcript but there isn't any online.[/quote]

There is now.

[quote="Valentin Vornicu"][quote="The Original Pi Guy"]hm... what happened to the generating functions math jam.

I was looking forward to that...[/quote]It was postponed due to WOOT Math Jam. Sorry  :blush:[/quote]

Darn!  :mad: When is it rescheduled to?  :)

[quote="fleeting_guest"]Was the geometry problem in the June 10 WOOT transcript taken from some thread on AoPS?
There is more that one can say about it.[/quote]

It was taken from an old National Olympaid, and there is indeed plenty more one can say about it, and other approaches.


Suppose a, b, and c are distinct digits for which the number 708,a6b,8c9 is a multiple of 99. find a+b+c

give solutions in spoiler please

[hide]16.

38 + 16 = 54 (multiple of 3 so divisible by 3)

38 - 16 = 22 (i'ts kinda hard - the adding the alternate digits and subtracting them... well it's like 454: 4-5+4 = 3 (not divisible by 11, so 454 isn't divisible...)[/hide]

9's divisbility rule: sum of digits have to be divisbible by 9

11's divisibility rule: difference in sums of alternating digit sums is a multiple of 11.

both of these combien to give u 16 is the only possibility

[quote="Syntax Error"]Suppose a, b, and c are distinct digits for which the number 708,a6b,8c9 is a multiple of 99. find a+b+c
[/quote]

[hide]well for 708a6b8c9 to be divisible by 99, then it has to be divisible by 9 and 11 because 9*11=99
div by 9=
7+0+8+a+6+b+8+c+9=38+a+b+c might equal =54 (i actually guess and checked but that takes a while to write)
54-38=16=a+b+c
div by 11=
7-0+8-a+6-b+8-c+9=38-a-b-c=38-16=22
a+b+c=16

16 YAY!!!! :pilot: [/hide]


The polynomials $P$, $Q$, $R$ with real coefficients, one of which is degree $2$ and two of degree $3$, satisfy the equality $P^2+Q^2=R^2$.  Prove that one of the polynomials of degree $3$ has three real roots.

now that I have thought a little bit more about it ... i figured that the problem may have asked to prove that q and r cannot both have all the roots real numbers ...

[quote=Valentin Vornicu]Are you sure that this is the full statement of the problem? Because if it is so ... Then the polynomials:
$p(x)=x^2+2x$
$q(x)=x^3+2x^2+3/2x$
$r(x)=x^3+2x^2+5/2x$
Satisfy the relationship but neither $q(x)$ nor $r(x)$ have all real roots.

And this is how this all began:
It seemed to me that if $x_0$ is a root of $r(x)$ then it is also a root for $q(x)$ and $p(x)$.  Then the equality becomes something like (dividing by the polynomial $(x-x_0)^2$:
$(ax+b)^2+(q_1(x))^2=(r_1(x))^2$ and further more:
$(ax+b)^2=(r_1(x))^2-(q_1(x))^2$ where $a,b$ are reals, and $q_1(x)$ and $r_1(x)$ are real polynomials of degree $2$. Thus, if we suppose that $r(x)$ and $q(x)$ neither have all the roots reals, then $q_1(x)$ and $r_1(x)$ have no real roots and thus they have constant signs (1).

We write again:
$(ax+b)^2=(r_1(x)+q_1(x) )( r_1(x)-q_1(x) )$ but because of (1) one of these parenthesis have constant sign, and thus cannot be have real roots, such as $x=-b/a$. Then it must be the other parenthesis that has this double root $x=-b/a$. Let's say WLOG that $r_1(x)+q_1(x) >0$ for all $x$, and $r_1(x)-q_1(x) =k(x+b/a)^2$, where $k$ is a non-zero real number.
Then, looking at the degrees of the polynomials we find that $r_1(x)+q_1(x)  = a^2/k > 0$
This gives us the following:
$2r_1(x)=k(x+b/a)^2+a^2/k$  and
$2q_1(x)=k(x+b/a)^2-a^2/k$
Computing the discriminant of these two second degree polynomials we get:
$\Delta_r=a^2-k^2(b/a)^2$ and $\Delta_q=-a^2-k^2(b/a)^2$
At this point I realized that for some cleverly chosen a,b and k we can make it so that $\Delta_r$ and $\Delta_q$ are both negative. For instance take $a=1, b=2$ and $k=1$. Then our polynomials are:
$(x+2)^2=(x^2+2x+5/2)^2-(x^2+2x+3/2)^2$ and by multyplying this with let's say $x^2$ we get the polynomials I was talking about when I first started out.

Any questions?
(last time I said this there was a mistake in my proof ... :) )[/quote]

Why $x_0$ is root $p$

[quote=haha0201][quote=Valentin Vornicu]Are you sure that this is the full statement of the problem? Because if it is so ... Then the polynomials:
$p(x)=x^2+2x$
$q(x)=x^3+2x^2+3/2x$
$r(x)=x^3+2x^2+5/2x$
Satisfy the relationship but neither $q(x)$ nor $r(x)$ have all real roots.

And this is how this all began:
It seemed to me that if $x_0$ is a root of $r(x)$ then it is also a root for $q(x)$ and $p(x)$.  Then the equality becomes something like (dividing by the polynomial $(x-x_0)^2$:
$(ax+b)^2+(q_1(x))^2=(r_1(x))^2$ and further more:
$(ax+b)^2=(r_1(x))^2-(q_1(x))^2$ where $a,b$ are reals, and $q_1(x)$ and $r_1(x)$ are real polynomials of degree $2$. Thus, if we suppose that $r(x)$ and $q(x)$ neither have all the roots reals, then $q_1(x)$ and $r_1(x)$ have no real roots and thus they have constant signs (1).

We write again:
$(ax+b)^2=(r_1(x)+q_1(x) )( r_1(x)-q_1(x) )$ but because of (1) one of these parenthesis have constant sign, and thus cannot be have real roots, such as $x=-b/a$. Then it must be the other parenthesis that has this double root $x=-b/a$. Let's say WLOG that $r_1(x)+q_1(x) >0$ for all $x$, and $r_1(x)-q_1(x) =k(x+b/a)^2$, where $k$ is a non-zero real number.
Then, looking at the degrees of the polynomials we find that $r_1(x)+q_1(x)  = a^2/k > 0$
This gives us the following:
$2r_1(x)=k(x+b/a)^2+a^2/k$  and
$2q_1(x)=k(x+b/a)^2-a^2/k$
Computing the discriminant of these two second degree polynomials we get:
$\Delta_r=a^2-k^2(b/a)^2$ and $\Delta_q=-a^2-k^2(b/a)^2$
At this point I realized that for some cleverly chosen a,b and k we can make it so that $\Delta_r$ and $\Delta_q$ are both negative. For instance take $a=1, b=2$ and $k=1$. Then our polynomials are:
$(x+2)^2=(x^2+2x+5/2)^2-(x^2+2x+3/2)^2$ and by multyplying this with let's say $x^2$ we get the polynomials I was talking about when I first started out.

Any questions?
(last time I said this there was a mistake in my proof ... :) )[/quote]

There is no mistake (which is good since a 2003 person wouldn't come back and fix that).[/quote]

Try not to bump old threads!!

Solution of Valentin Vornicu is wrong, because $p^2+q^2-r^2=x^4$ not zero.

Solution of problem
Let $p$ has degree 2, and $q,r$ has degree $3$ and coeficients of $x^3$ are positive
Then $p^2=r^2-q^2=(r-q)(r+q)$
$r+q$ has 3 degree, so $r-q$ has first degree, so $p(x)-q(x)=a(x-x_1) \to (x-x_1)|p(x) \to (x-x_1)^2|p(x)^2 \to x-x_1 |r(x)+q(x)$
So $r(x)=r_1(x)(x-x_1),q(x)=q_1(x)(x-x_1),p(x)=b(x-x_1)(x-x_2)$ and $r_1(x)-q_1(x)=a$
$b^2(x-x_2)^2=a(r_1(x)+q_1(x))=a(2q_1(x)+a) \to q_1(x)=\frac{b^2(x-x_2)^2-a^2}{2a}$ has two roots.
So $q(x)$ has total $3$ roots