I got... 43. I'm going to states anyway, so I don't care.

I got n, where n is a point not on either axis in the complex plane.

my school doesn't base teams on the school round

either 38 or 36. i don't remember i just remember i.....FAILED. i almost got a lower score in this year's school than last year's. i feel stupid.

I got 43 - Somehow, I made a mistake on the last target and i got 29 on the sprint.

fufufufufufufufufufufufu fourty five

I got #12 wrong on the sprint. Was one off from the correct answer.

[quote="cognos599"][quote="#H34N1"]perfect[/quote]

r u in 8th grade or below this year????!?!?!?!?!?!?!?!?!?!?[/quote]'

8th grade

46

My school doesn't care about them either, its just a practice test.

[quote="#H34N1"]perfect[/quote]

this is the same person who claimed to have gotten 45 on last year's chapter test, while 34 was the highest in his state i may add

I got 24/14.  I got second in my school (though we have one amazing math person at our school).

28/16... I made a stupid mistake and another stupid mistake.

Yay, I got $ x$ where $ x \equal{} \lim_{x\rightarrow\infty}{\frac{46x^2\minus{}127x\plus{}255}{x^2\minus{}2x\plus{}1}}$

[hide="For those who do not know limits"]46[/hide]
 :first:
The second place person in my school got a 35.

LOL 35?! :rotfl:
o well i got a 46 and destroyed second place by at least 15 points
but things were different at chapter
and things are WAY different in tx :(

I got 45 and destroyed second place (37)

At chapter was similar

I got 44 and destroyed second place (35).

Can't talk about state.

I got 46 and tied with 3 other people in bay area California.  I wish i was in your chapter area.  It would be so much less nerve-wracking.

[quote="Yongyi781"]Yay, I got $ x$ where $ x \equal{} \lim_{x\rightarrow\infty}{\frac {46x^2 \minus{} 127x \plus{} 255}{x^2 \minus{} 2x \plus{} 1}}$

[hide="For those who do not know limits"]46[/hide]
 :first:
The second place person in my school got a 35.[/quote]
Oh yeah, well I got $ \lim_{x\rightarrow \infty}{|(\frac {xi \plus{} 29969}{666\cdot xi}i)^{2}|}$
Oh....
Pwnt.

peijin, you don't even know how to take rational limits, let alone those involving complex numbers

[quote="ZhangPeijin"][quote="Yongyi781"]Yay, I got $ x$ where $ x = \lim_{x\rightarrow\infty}{\frac {46x^2 - 127x + 255}{x^2 - 2x + 1}}$

[hide="For those who do not know limits"]46[/hide]
 :first:
The second place person in my school got a 35.[/quote]
Oh yeah, well I got $ \lim_{x\rightarrow \infty}{|(\frac {xi + 29969}{666\cdot xi}i)^{2}|}$
Oh....
Pwnt.[/quote]

So you're telling me you got $ \frac {1}{443556}$. That's a terrible score, seeing that it is between 0 and 1, and the maximum being 46. :P

Pwnt... whatever that means...

Edit: Oh yeah, and for those who want to know...

$ \lim_{x\rightarrow \infty}{|(\frac {xi + 29969}{666\cdot xi}i)^{2}|} = \lim_{x\rightarrow \infty}{\frac{x^2 + 898140961}{443556 \cdot x^2} = \frac{1}{443556}}$

Meh, I just made some random junk in a limits problem but whatever, glad you could solve it.

[quote="Lawrence Wu"]i predict ra5249 and stevenmeow will get perfect state scores...they got close last year[/quote]

ra? But weren't the top 4 46-45-40-40? That would make stevenmeow far above anyone else eligible this year.

I would guess that ra was one of the 40's

Me too...but then I would be closer to ra then he was to stevenmeow...I had a 38.


Solve system equations
\[ \large x = \sin{y}^{\cos{\sqrt {y}}
}\]

\[ \large y = \sin{x}^{\cos{\sqrt {x}}
}\]
For $ \large x,y\in (0,\pi)$

Solve equations
\[{ \large (2\sqrt {2} + 3)^{\cos{x}} - 2^{\sin(\frac {x}{2}})} = 1
\]

One of solutions $ x\equal{}y\equal{}0$
Let $ f(x)\equal{}sin x^{cos\sqrt x }$, then $ x,y$ are roots $ f(f(z))\equal{}z.$ Obviosly $ 0\le x,y\le 1$ (because sinz<=1) and is easy to chek, that $ f'(x)>0$ for $ 0\le x \le 1$. From $ y\equal{}f(x)\le f(1)\equal{}sin 1,x\equal{}f(y)\le 1$ we get $ x,y<0.78365$ and in these interval $ f''<0$. It give only one solution $ x\equal{}y\equal{}0$, because $ f'(0)\equal{}1$.

but I don't think you completed your solution  :(
What do you think about the following roots?
\[ x\equal{}y\equal{}0.830116974449655\]

[quote="zaizai-hoang"]but I don't think you completed your solution  :(
What do you think about the following roots?
\[ x \equal{} y \equal{} 0.830116974449655
\]
[/quote]
If $ x\equal{}0.830116974449655$, then $ y\equal{}f(x)\equal{}0.778429559673872...$  :lol:

but my computer give a correct result  :maybe:

May be you calculate $ f(x)\equal{}(sin x)^{cos\sqrt x },$ but I consider $ f(x)\equal{}sin(x^{cos\sqrt x }).$

[quote="Rust"]May be you calculate $ f(x) \equal{} (sin x)^{cos\sqrt x },$ but I consider $ f(x) \equal{} sin(x^{cos\sqrt x }).$[/quote]
yeah, I calculate $ f(x) \equal{} (sin x)^{cos\sqrt x }$ and It is still true  :blush:

Because f(x) monotonen (increase or decrease) system $ y\equal{}f(x),x\equal{}f(y)$ had only solution $ x\equal{}y\equal{}z$, were $ z\equal{}f(z)$.
Because f'>0 all solution $ f(z_*)\equal{}z_*$ may be find by procedure $ z_0<z_*,z_{n\plus{}1}\equal{}f(z_n)$, then $ z_*\equal{}lim_{n\to \infty }z_n$.


superior algebra
superior algebra unsolved

Supposed that we have a field $ F$, and we know that the order of the field is $ 16$. Is $ p(x)\equal{}x^3\plus{}x\plus{}1$ is irreducible over $ F$ ? Explain your reason . Thanks in advance.

Is it irreducible over $ \mathbb{F}_2$?  What field would we get by adjoining a root?


[size=150]Find the numerical value for this infinite multiplication :

$ S \equal{}\prod_{k\equal{}2}\frac{k^{3}\minus{}1}{k^{3}\plus{}1}\equal{}\frac{7}{9}\times\frac{26}{28}\times\frac{63}{65}\times ...$

please, don't just write down the answer, mention your methods. Thanks in advance,,,[/size]

[hide=" hint"]$ k^{3}\minus{}1 \equal{} (k\minus{}1)(k^{2}\plus{}k\plus{}1)$
$ (k\plus{}1)^{3}\plus{}1 \equal{} (k\plus{}2)(k^{2}\plus{}k\plus{}1)$ [/hide]
It is quite easy .

I didnt check everything but i know its the right answer from memory


geometry
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if its a target question, program crammers along with shoelace into your calc so its easy to find the solutions. or just use rref but i like crammers better. m8tr1c13s r pwnage.

Well, if it was target, it wouldn't be much of a problem concerning time, but I've seen this question as sprint and team round questions (as for the latter, if you're doing a team round by yourself, it helps to do it much more quickly :wink: ).  Of course, I wouldn't need a program or matrices to solve the problem, but I still would like to know a formula, if anyone knows one. :)

rref is faster.

And I found this type of question in AoPS FTW. I only had 30 seconds to solve it though so I didn't bother  :( . Is there a non-calculator method?

Um, could someone help, concerning the formula?  rref is basically a matrix "solver," the best word that I can use in this case.

If the equations are $ a_ix\plus{}b_iy\equal{}c_i$ ($ i\equal{}1,2,3$), then the formula is
\[ S\equal{}\frac 12\left|\frac{D^2}{D_1D_2D_3}\right|\]
where $ D$ is the determinant of the full matrix of the system and $ D_i$ is the $ i$-th $ 2\times 2$ minor of the coefficient matrix. If you are familiar with 2 by 2 and 3 by 3 determinants, this may be a bit faster than solving 3 linear systems and finding one 2 by 2 determinant (also known as shoelace formula).

Example: $ x\plus{}2y\equal{}5$, $ x\minus{}y\equal{}3$, $ 2x\plus{}y\equal{}7$.
The full matrix is $ \begin{pmatrix}1&2&5\\1&\minus{}1&3\\2&1&7\end{pmatrix}$
$ D\equal{}\minus{}7\plus{}12\plus{}5\minus{}(\minus{}10\plus{}3\plus{}14)\equal{}3$
$ D_1\equal{}1\plus{}2\equal{}3$
$ D_2\equal{}1\minus{}4\equal{}\minus{}3$
$ D_3\equal{}\minus{}1\minus{}2\equal{}\minus{}3$
Answer: $ \frac 1{6}$.

If you are really quick in mental arithmetic, you have a chance to do this example in under 30 seconds but I agree that as a question for FTW, it is, erm, "hardly appropriate".

[quote="unimpossible"]whats rrf?[/quote]
You spelled it incorrectly, unimpossible; rref is what you mean.  Rref (say it out loud; it's fun) means "reduced row-echelon form."  You have to have a somewhat solid understanding of matrices if you want to understand rref.  But here's a definition...

A matrix is in reduced row-echelon form if it satisfies the following conditions:
1.  The first nonzero number in each row (reading from left to right) is 1.  This is called the leading entry.
2.  The leading entry in each row is to the right of the leading entry in the row immediately above it.
3.  All rows consisting entirely of zeros are at the bottom of the matrix.
4.  Every number above and below each leading entry is a 0.

This is basically the upper-triangular form of a matrix that follows special conditions, allowing for swift back-substitution when solving the system of equations.

Matrices are not very fun... I HATED the chapter in our math textbook about it.  You can rigorously solve a matrix to be in rref, but using a calculator is stupendously less time-consuming.

Back to the original question, I don't know of a terribly fast way of solving such problems as the one you mention, Math Geek.  However, you could try inscribing the triangle formed by the bounded region in a rectangle; this way you can take the area of the rectangle minus the sum of the areas of the newly formed triangles (if I had Asymptote drawing skills, I'd show you).
By "inscribe," I mean to make sure that at least one vertex of the triangle is a vertex of the rectangle.  Meanwhile, the other two vertices of the triangle lie somewhere on the rectangle.  This will ALWAYS result in 2 or 3 additional triangles (depending on how you inscribe).  I prefer having 3 resulting additional triangles; this means that only one vertex of the triangle is a vertex of the rectangle.  However, this also means that you can situate the rectangle such that the sides of the rectangle are parallel to or are part of the x and y axes.  This allows for FAST computation of side lengths of the rectangle and triangles, allowing the areas to be calculated swiftly as well.

Come to think of it, I know of a link that describes the process that I am mentioning (it also describes the shoelace method as well).

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

Just read the first page of the article; the rest is about the shoelace method.

[quote="Ihatepie"][quote="Math Geek"]Is there a formula for the area of a triangle in a cartesian plane given the equations of the three lines that bound it?  Yes, I've searched this, but no one has properly answered this question.  This type of problem keeps coming up in MC, and I always have to find their intersection points and use shoelace.  Is there an alternative, much more elegant/easier method, cuz it wastes valuable time on a single question.  Thanx![/quote]

I'm sorry guys, but it's kind of ridiculous how all of you are asking formulas for everything. Just do it by doing work. There are some essential formulas, like n(n+1)/2 or shoelace, but no one needs to know the equation of how to find the area of a trapezoid from its sides or how to find the amount of partitions a number has.[/quote]

Um, it's not ridiculous to ask for formulas if there indeed exists an easier method.  I do know how to do it with "essential formulas," but they take too long to use for an ordinary sprint round question.  Unless if, of course, you have a formula to solve this, but seeing as you don't, plz don't complain.

Well, the one I posted is the only one available :(. You can shorten the computation a bit if you recall how the 3 by 3 determinant is expanded with respect to the last column (you need to compute the corresponding 2 by 2 determinants anyway) but still it is a computation requiring some time (about 40 seconds for small integer coefficients, the nice thing being that you do not need to work with fractions up to the very last moment).


You are given a standard kilogram mass and a tuning fork that is calibrated in Hz. You are also provided with a complete collection of laboratory equipment, but none of it is calibrated in SI units. You do not know the values of any fundamental constants. Which of the following quantities could you measure in SI units?
A. The acceleration due to gravity.
B. The speed of light in a vacuum.
C. The density of room temperature water.
D. The spring constant of a given spring.
E. The air pressure in the room.

a) has units m/s^2
b) has units m/s
c) has units kg/m^3
d) has units (kg*m/s^2)/m=kg/s^2
e) has units (kg*m/s^2)/m^2=kg/(s^2m)

Therefore d has to be the right one. Just measure it how you would typically, then you should have units A^2/B, where A is a measure of mass, and B is a measure of time. Then you figure out the ratios kg/A by weighing the mass, and s/B by measuring the frequency of the tuning fork, and you can then convert A^2/B into kg/s^2.

Therefore, the answer is d)

Cheers,

Rofler

i understand why d is correct, but could someone offer a possible experiment that actually does this measurement?


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Here is the prob:
Maria has a jar of pennies. When she arranges the pennies in rows of 15 she finds that there are 7 left over. When she arranges them in rows of 19 she finds that there are 8 left over. What is the smallest number of pennies that could be in the jar?

I thought that i could do this with the Division Algorithim like:
\documentclass{article}
\begin{document}
$ x \equal{} 19q \plus{} 8$
$ x \equal{} 15q \plus{} 8$
\end{document}
But when i tried to solve for x i got a negative number, or i just might be solving wrong........ :?:

There is a theorem called the Chinese Remainder Thm if you know modular arithmetic.... Look it up on the AOPSWiki...   :lol:
Otherwise, here:

[hide="Solution"]

$ x\equal{}15p\plus{}7$
$ x\equal{}19q\plus{}8$
$ 15p\plus{}7\equal{}19q\plus{}8$
$ p\equal{}\frac{19q\plus{}1}{15}$
We look for the smallest integral solutions for x, so minimize p.
Testing q 0 through 10 gives nonintegral p, but q=11 works. So q=11, p=14, and x=[b]217[/b].

[/hide]

[hide="answer"] I got $ 217$ but as I have little to no experience with modular arithmetic, I'm sure my solution was fairly inefficient.

I had $ x \equal{} 15n \plus{} 7$ and $ x \equal{} 19k \plus{} 8$ which gave me $ 15n \equal{} 19k \plus{} 1$

So basically I just looked for multiples of 19 with units digit 4 or 9 and checked if 1 more than that number was divisible by 15. Luckily it only took me two tries since $ 19\cdot11 \plus{} 1 \equal{} 210 \equal{} 15\cdot14$

Therefore we have $ x \equal{} 210 \plus{} 7 \equal{} 217$

[/hide]

Beaten as well.

Meh. Beaten. Mine is a bad method, anyways.  :blush:

[hide="Solution"]
Hmm...erm. First, the equations.
$ x \equal{} 19q \plus{} 8$
$ x \equal{} 15q \plus{} 7$ (7, not 8)

Well, I'm not entirely sure what the division algorithim is, but the number of rows doesn't necessarily have to be the same.

Hmm. I guess that you could look at it logically. Or brute force it....

[u]Multiples of fifteen:[/u]
15, 30, 45, 60, 75....

[u]Adding 7:[/u]
22, 37, 52, 67, 82.

[b]So the number has to end in a 2 or 7.[/b]

[u]Multiples of 19:[/u]
19, 38, 57, 76, 95, 114....

[u]Adding 8[/u]
27, 46, 65, 84, 103, 122...

[b]This is not as helpful. It tells us that it could end in a 7, 6, 5, 4...:([/b]

Continuing to examine the 19's pattern (since it's far pickier...) gives us:
$ 19 * 1 \equal{} 19, 19 \plus{} 8 \equal{} 27$
$ 19 * 6 \equal{} 19, 114 \plus{} 8 \equal{} 122$
$ 19 * 11 \equal{} 19, 209 \plus{} 8 \equal{} 217$
$ 19 * 16 \equal{} 19, 304 \plus{} 8 \equal{} 312$

You can see that you have to add 5 to whatever you're multiplying by, or just as a better general idea, multiply 19 by something that ends in a 1 or a 6, so that when you add 8, you get something that ends in a 2 or a 7. This will help satisfy the previous condition set down by the rows of 15 withaa remainder of 7.

Then you could brute force it from there, I suppose; looking for an overlap with the multiples of 15 and 19 (plus the 7 and 8, respectively.) However, now that you've figured out the guidelines to go by, it's a lot easier. Since you know some numbers for 19, find multiples for 15 that are close by. This is where your mad estimation skills come in. :wink:

$ 19 * 1 \equal{} 19, 19 \plus{} 8 \equal{} 27$
$ 19 * 6 \equal{} 19, 114 \plus{} 8 \equal{} 122$
$ 19 * 11 \equal{} 19, 209 \plus{} 8 \equal{} 217$
$ 19 * 16 \equal{} 19, 304 \plus{} 8 \equal{} 312$

$ 15 * 1 \equal{} 15, 15 \plus{} 7 \equal{} 22$
$ 15 * 2 \equal{} 30, 30 \plus{} 7 \equal{} 37$
$ 15 * 8 \equal{} 120, 120 \plus{} 8 \equal{} 127$
$ 19 * 14 \equal{} 19, 210 \plus{} 7 \equal{} 217$

[b]The answer is:[/b] $ 217$

...Couldn't you also use mods to look at this problem?

$ x \equal{} 19 mod 8$
$ x \equal{} 15 mod 7$

Yeah, that's about as far as I can go with mods. :blush:
Perhaps someone else could elaborate on that.
[/hide]

And about your LaTeX, on the AoPS board, just enclose your commands in dollar signs. You don't need to begin a new document class or whatever. :lol:

Eww... I don't like bashing with CRT.

[hide]
\[ x\equiv8\pmod{19}\equiv7\pmod{15}\implies x\minus{}7\equiv1\pmod{19}\equiv0\pmod{15}\]

It is trivial to prove that $ 1\pmod{19}$ must end with 0 or 5. The smallest integer that satisfies the above expression is $ x\minus{}7\equal{}210\rightarrow\boxed{x\equal{}217}$
[/hide]

but with CRT there's no need to guess anything. :)
[hide]
Using the extended Euclid's algorithm (or guess and check) we see that $ 4(19)\minus{}5(15)\equal{}1.$
then $ x\equiv7\cdot(4)(19)\minus{}8\cdot(5)(15)\mod(15\cdot19)\equiv 217\mod(15\cdot19).$[/hide]

[quote="hunter34"]It is trivial to prove that $ 1\pmod{19}$ must end with 0 or 5[/quote]

Hmm, $ 39\equiv1\pmod{19}$, but $ 39$ does not end in $ 0$ or $ 5$...

I'm not really sure that I follow your logic...

Anyway, just to avoid any possible confusion, the Chinese Remainder Theorem does not solve this problem, it simply guarantees that there is a solution.  It says that the guessing and checking part of this problem will work out eventually.

[quote="Poincare"]Can someone explain to me what mathhunter just did? Plz explain without modular arithmetic in order to explain modular arithmetic, lol.....

thank you,
my real name is to weird to put in here but oh well,
Dhaivat Pandya :D[/quote]

Hmm, that might be difficult, considering that this is a modular arithmetic problem and the only non-modular arithmetic solutions are basically guess & check (mods make that part a bit easier & more rigorous).

Mods are fairly simple, and if you want to go very far in Number Theory, you will have to learn them.

[b]Definition of mods:[/b]
$ a\equiv b\pmod{c}\Leftrightarrow c|(a\minus{}b)$
$ c$ is a factor of $ (a\minus{}b)$.

Let $ N\equal{}19a\plus{}8\equal{}15b\plus{}7$

Then, we have
$ 19a\plus{}1\equal{}15b$
$ 19a\equiv\minus{}1\pmod{15}$
$ 19a\equiv209\pmod{15}$
$ a\equiv11\pmod{15}$

Let $ a\equal{}15k\plus{}4$, so $ N\equal{}19(15k\plus{}11)\plus{}8\equal{}285k\plus{}217$, so the smallest positive value of $ N$ is 217.

As to what hunter34 did, I have no idea, because the part about it ending in a 0 or 5 is not correct...

[quote="b-flat"][quote="Poincare"]Can someone explain to me what mathhunter just did? Plz explain without modular arithmetic in order to explain modular arithmetic, lol.....

thank you,
my real name is to weird to put in here but oh well,
Dhaivat Pandya :D[/quote]

Hmm, that might be difficult, considering that this is a modular arithmetic problem and the only non-modular arithmetic solutions are basically guess & check (mods make that part a bit easier & more rigorous).

Mods are fairly simple, and if you want to go very far in Number Theory, you will have to learn them.

[b]Definition of mods:[/b]
$ a\equiv b\pmod{c}\Leftrightarrow c|(a \minus{} b)$
$ c$ is a factor of $ (a \minus{} b)$.

Let $ N \equal{} 19a \plus{} 8 \equal{} 15b \plus{} 7$

Then, we have
$ 19a \plus{} 1 \equal{} 15b$
$ 19a\equiv \minus{} 1\pmod{15}$
$ 19a\equiv209\pmod{15}$
$ a\equiv11\pmod{15}$

Let $ a \equal{} 15k \plus{} 4$, so $ N \equal{} 19(15k \plus{} 11) \plus{} 8 \equal{} 285k \plus{} 217$, so the smallest positive value of $ N$ is 217.

As to what hunter34 did, I have no idea, because the part about it ending in a 0 or 5 is not correct...[/quote]

I don't understand how you went from $ 19a \plus{} 1 \equal{} 15b$ to $ 19a\equiv \minus{} 1\pmod{15}$
please help

thanks,
Dhaivat Pandya

^Since $ b$ is an integer, $ 19\plus{}1$ is a multiple of $ 15$; $ 15$ divides the difference of $ 19$ and $ \minus{}1$

So $ 19a\equiv{\minus{}1}\pmod{15}$

[quote="hunter34"]
$ x \minus{} 7\equiv1\pmod{19}\equiv0\pmod{15}$
It is trivial to prove that $ 1\pmod{19}$ must end with 0 or 5.
[/quote]
I think Hunter34 meant that if x-7 is 1 mod 19 [b]and[/b] 0 mod 15, then x-7 must end in a 0 or 5.

Also, there is no need for guess and check with CRT, because the extended Euclidean algorithm basically gives the answer...sometimes it's easier to just guess and check, but if the problem involves really big numbers it might take a while to guess the solutions.

the advantage of not guessing and checking is that it doesn't matter what numbers you have, they can all be solved just as easily.
for example since we know that 4(19)-5(15)=1, a general solution to
$ x\equiv a\mod15$
$ x\equiv b\mod19$
is $ x\equiv a\cdot (4)(19)\plus{}b\cdot(\minus{}5)(15)\mod(15\cdot 19).$

[quote="pi^2/6"]the advantage of not guessing and checking is that it doesn't matter what numbers you have, they can all be solved just as easily.
for example since we know that 4(19)-5(15)=1, a general solution to
$ x\equiv a\mod15$
$ x\equiv b\mod19$
is $ x\equiv a\cdot (4)(19) \plus{} b\cdot( \minus{} 5)(15)\mod(15\cdot 19).$[/quote]

Most of the guessing and checking comes from finding the $ 4(19)\minus{}5(15)\equal{}1$ part...

CRT only tells you that a solution exists.  The algorithm that actually finds that solution is called the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]extended Euclidean algorithm[/url].  In particularly, we want to solve

$ 15x \plus{} 7 \equal{} 19y \plus{} 8 \implies$
$ 15x \minus{} 19y \equal{} 1$

Similar problems are presented and discussed in this (fairly lengthy) thread:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80124&sid=df9439496046e0ff9f97cfc644408395]Taken from the Wiki[/url]


five 'bots tangling with pasta
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and seeing Montana's green heights
--------------------------------------------------------
Are any of you in FIRST? :)

FLL?yuuuup.
[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=200[/url] [size=59](my blog!!! :D )[/size]


Prove that  $ \quad a,b\in\mathbb{R}$

$ \frac{\lvert a\plus{}b\rvert}{1\plus{}\lvert a\plus{}b\rvert}\le\frac{\lvert a\rvert}{1\plus{}\lvert a\rvert}\plus{}\frac{\lvert b\rvert}{1\plus{}\lvert b\rvert}$

[quote="ocha"][hide]
First note that if $ \frac {|a|}{1 \plus{} |a|} \le \frac {|b|}{1 \plus{} |b|} \Longrightarrow |a| \plus{} |ab|\le |b| \plus{} |ab| \Longrightarrow |a|\le |b|$

Since $ |a \plus{} b| \le |a| \plus{} |b|$ we have

$ \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|} \le \frac {|a| \plus{} |b|}{1 \plus{} |a| \plus{} |b|} \equal{} \frac {|a|}{1 \plus{} |a| \plus{} |b|} \plus{} \frac {|b|}{1 \plus{} |a| \plus{} |b|} \le \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}$

 :) [/hide][/quote]

$ \boxed{\frac {|a|}{1 \plus{} |a|} \le \frac {|b|}{1 \plus{} |b|} \Longrightarrow |a| \plus{} |ab|\le |b| \plus{} |ab| \Longrightarrow |a|\le |b|}$

Should it change into

$ \frac {|a|}{1 \plus{} |a|} \le \frac {|b|}{1 \plus{} |b|} \iff |a| \plus{} |ab|\le |b| \plus{} |ab| \iff |a|\le |b|$?


geometry
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[color=cyan]Joel is sitting around doodling when he draws a rectangle and an equilateral triangle.  He notices two things: the side of the equilateral triangle is exactly equal to one side of the rectangle, and both the rectangle and triangle have the property that their area and perimeter are equal.  (That is, the rectangle's perimeter is numerically equal to its area, and the triangle's perimeter is equal to its area.)  What is the other side of the rectangle?

What if the rectangle's perimeter equals the triangle's area and vice versa?  (You have to find both sides of the rectangle for this one.)[/color]

[hide] Let a side of the triangle be s, and the other side of the rectangle be t. Then s^2 V3 / 4 = 3s, so s=4V3

Then 8V3 + 2t = 4t V 3, and we can solve for x easily from here (note that I'm too lazy to actually find x lol)

For the second one, we have 3s = sb

so b = 3, and since

2 (s + b) = s^2 V3 / 4

we can solve for s.[/hide]

JBL wrote:Joel is sitting around doodling when he draws a rectangle and an equilateral triangle.  He notices two things: the side of the equilateral triangle is exactly equal to one side of the rectangle, and both the rectangle and triangle have the property that their area and perimeter are equal.  (That is, the rectangle's perimeter is numerically equal to its area, and the triangle's perimeter is equal to its area.)  What is the other side of the rectangle?

What if the rectangle's perimeter equals the triangle's area and vice versa?  (You have to find both sides of the rectangle for this one.)

[hide]for the second, we let a=one side of the triangle and b=the other side of the rectangle. Then 2a+2b=a:^2::rt3:/4, and 3a=ab => b=3, as a is not 0 (if it is, we get the equally valid solution a=0, b=0). Plugging b=3 into the first equation gives 8a+24=a:^2::rt3: => :rt3:a:^2:-8a-24=0 =>quadratic formula gives a=(8:pm::sqrt:(64+96:rt3:))/2:rt3: which probably simplifies to something.[/hide]


linear algebra
matrix
algebra
polynomial

Let $A,B$ two matrices of order $n$ such that $ det(A+X)=det(B+X)$ for any matrix $X$ of order $n$.
Prove that $A=B$. ;)

By letting $X$ be the zero matrix, we can see that $\det A=\det B$.
By letting $X$ have one element equal to 1 and all others 0, we have that the sum of the determinant and one $(n-1)\times(n-1)$ minor are equal for the two matrices.
Thus all $(n-1)\times(n-1)$ minors are equal. By letting $X$ be the identity on some $k\times k$ submatrix and zero elsewhere, we have that the complement minor plus the sum of various larger minors is the same for $A$ and for $B$.

Working downward, the $k\times k$ minors are equal for each $k$. When we reach the $1\times 1$ minors, we are done, as these are simply the elements of $A$ and $B$.

Let $X=(x_{ij})$

$det(A+X)$ is a polynomial of total degree $n$ in $n^2$ variables $x_{ij}$
in the expansion of $det(A+X) = det(B+X)$
The monome of degree total $n-1$ with the same variables in $det(A+X)$
and $det(B+X)$ are equal
 $a_{i_{l}j_{k}}x_{i_1,j_1}...x_{i_{n-1},j_{n-1}}=b_{i_{l}j_{k}}x_{i_1,j_1}...x_{i_{n-1},j_{n-1}}$

$a_{i_{l}j_{k}}=b_{i_{l}j_{k}}$

$A=B$

Take $ C = A-B $. Then we have for every $ Y $ that $ det(C+Y) =
det(Y) $. Let us prove that $ C = 0 $. If not, then there exists $
v \neq 0 $, such that $ w:= Cv \neq 0 $. Take invertible $ Y $,
such that $ Yv = -w $ ( we can do it, completing $ v $ to a basis
and $ -w $ to a basis, and sending elements form one basis to
another ). Then $ (C+Y)v = 0 $, so $ det(C+Y) = 0 $ , but $ Y $ is
invertible, so $ det(Y) \neq 0 $ and we come to contradiction.

[quote="Leva1980"] Then we have for every $ Y $ that $ det(C+Y) =
det(Y) $[/quote] here another proof that $C=0$

Suppose that $C\neq 0$
$r=rank(C)$ $
C$ is equivalent to $J_r$
there exist $P,Q$ invertible matrices

$C=P\left(\begin{array}{cc}J_r&0\\0&0\end{array}\right)Q$

Choose $Y=P\left(\begin{array}{cc}0&0\\0&I_{n-r}\end{array}\right)Q$

then $det(C+Y)=det(PQ)\neq 0$
$det(Y)=0$ contradiction so $rank(C)=0$
$C=0$


induction
analytic geometry
combinatorics unsolved
combinatorics

A triomino is an L-shaped pattern made from three square tiles. A $ 2^k*2^k$ chessboard, whose squares are the same size as the tiles, has one of its squares painted puce. Show that the chessboard can be covered with triominoes so that only the puce square is exposed.

By induction on $ k$

In the case $ k \equal{} 1$, the shape of the remaining part when one square is coloured is the shape of a triominoe, so the problem is solved.

For $ k > 1$ divide the board into four quarters of size $ 2^{k\minus{}1} * 2^{k\minus{}1}$ the coloured square lies in one of the four quarters.
Place a single trionimoe in the central four squares so that it lies in the three other quarters. Now each of the four quarters have all
but one square reamaining  that need to be covered with triominoes. By induction on $ k$ this must be possible.

Another inductive approach comes from two facts (where we take the board to have integer coordinates from $ 0$ to $ 2^k\minus{}1$ in both directions):
- one chosen tile and a L-piece can always be arranged to be a $ 2 \times 2$ square in four different ways, one of them such that both coordinates of the lower left square are even.
- four L-pieces can be arranged to give a L-piece of twice the original size.

Combining those, one can induct by rescaling the board by a factor of two.


analytic geometry
graphing lines
slope
logarithms
function
algebra
domain

For what values of $ a$ are $ y\equal{}a^x$ and $ y\equal{}1\plus{}x$ tangent at $ x\equal{}0$?

Lines are tangent at some point if the tangents are perpendicular in this point. Therefore $ k_{1}\equal{}\minus{}\frac{1}{k_{2}}$, where $ k_{1}$ and $ k_{2}$ are the slopes of the lines.

$ a\ln{a}\equal{}\minus{}1 \Leftrightarrow a^{a}\equal{}\frac{1}{e}$

I disagree!! They are tangent if the slopes are equal. Just think about the explicit formula of a tangent. It's coefficient is exactly the same as the derivation of the function in a point. Therefore, the slope of $ a^x$ in $ x \equal{} 0$ needs to be the same as the slope of $ x \plus{} 1$. This is:

$ \ln a \equal{} 1 \Rightarrow a \equal{} e$

Picture:

[url=http://img171.imageshack.us/my.php?image=slopemr3.png][img]http://img171.imageshack.us/img171/8105/slopemr3.th.png[/img][/url]

And to make things a bit more clearer, the solution you gave is complex, so the tangent would not occur in the domain of real numbers.

But, even further, if your equation was used it would give the equation:

$ \ln a \equal{} \minus{}1$

and not the one you gave. :)

[quote="milin"]I disagree!! They are tangent if the slopes are equal.[/quote]

Sorry for giving a wrong answer. I confused "being tangent" with "intersecting at right angle".

[quote="milin"]But, even further, if your equation was used it would give the equation:

$ \ln{a}\equal{}\minus{}1$[/quote]

I also made computational mistake using $ x\equal{}1$ instead of $ x\equal{}0$.


function
algebra
polynomial
geometry
number theory
prime numbers
national olympiad

[b]First day[/b]

1) Find all $ f: Q^ \plus{} \to\ Z$ functions that satisfy $ f(\frac {1}{x}) \equal{} f(x)$ and $ (x \plus{} 1)f(x \minus{} 1) \equal{} xf(x)$ for all rational numbers that are bigger than 1.

2) Quadrilateral $ ABCD$ has an inscribed circle which centered at $ O$ with radius $ r$. $ AB$ intersects $ CD$ at $ P$; $ AD$ intersects $ BC$ at $ Q$ and the diagonals $ AC$ and $ BD$ intersects each other at $ K$. If the distance from $ O$ to the line $ PQ$ is $ k$, prove that $ OK\cdot\ k \equal{} r^2$.

3) Within a group of $ 2009$ people, every two people has exactly one common friend. Find the least value of the difference between the person with maximum number of friends and the person with minimum number of friends.

[b]Second day[/b]

4) For which $ p$ prime numbers, there is an integer root of the polynominal $ 1 \plus{} p \plus{} Q(x^1)\cdot\ Q(x^2)\ldots\ Q(x^{2p \minus{} 2})$ such that $ Q(x)$ is a polynominal with integer coefficients?

5) In a triangle $ ABC$ incircle touches the sides $ AB$, $ AC$ and $ BC$ at $ C_1$, $ B_1$ and $ A_1$ respectively. Prove that $ \sqrt {\frac {AB_1}{AB}} \plus{} \sqrt {\frac {BC_1}{BC}} \plus{} \sqrt {\frac {CA_1}{CA}}\leq\frac {3}{\sqrt {2}}$ is true.

6) In a class of $ n\geq 4$ some students are friends. In this class any $ n \minus{} 1$ students can be seated in a round table such that every student is sitting next to a friend of him in both sides, but $ n$ students can not be seated in that way. Prove that the minimum value of $ n$ is $ 10$.

1) [b]Claim: (1) Let $ f$ be a function with the given properties, then for $ p,q\in\mathbb{N}$ with $ (p,q)\equal{}1$ we have $ f\left(\frac{p}{q}\right)\equal{}\frac{p\plus{}q}{2}f(1)$ and $ f(1)$ must be even.
(2) Conversely, for arbitrary $ m\in\mathbb{Z}$ the function $ f_m$ defined by $ f_m\left(\frac{p}{q}\right)\equal{}m(p\plus{}q)$ for $ p,q\in\mathbb{N}$ with $ (p,q)\equal{}1$ has the required properties.[/b]

[i]Proof of claim[/i]: Assume assertion (1) is not true and let $ (p,q)$ be a minimal counterexample with regard to $ p\plus{}q$. Because of $ f(x)\equal{}f(x^{\minus{}1})$ we can assume $ p>q$. Application of the second identity with $ x\equal{}\frac{p}{q}>1$ yields $ f\left(\frac{p}{q}\right)\equal{}f\left(\frac{p\minus{}q}{q}\right)\frac{p\plus{}q}{p}$. Since $ (p\minus{}q,q)\equal{}(p,q)\equal{}1$ and $ p\minus{}q,q\in\mathbb{N}$ with $ p\minus{}q\plus{}q<p\plus{}q$ hold, we have by minimality $ f\left(\frac{p\minus{}q}{q}\right)\equal{}\frac{p}{2}f(1)$ and we get $ f\left(\frac{p}{q}\right)\equal{}\frac{p\plus{}q}{2}f(1)$ contradicting the existence of a minimal counterexample. Choosing $ p\equal{}2,q\equal{}1$ yields $ f(1)$ even.

(2) is verified by straightforward calculation.

In the 6th question n's minimum value is 10 not maximum, I told orl to fix. Sorry for this mistake.


modular arithmetic
logarithms
number theory

What is the remainder when the 1492nd Fibonacci number is divided by 25? We are defining the first to be 1 and the second to be 1 as well.

[hide]$x^{2}=x+1$

$x^{3}=x^{2}+x=2x+1$

and the pattern continues so

$x^{n}=F_{n}x+F_{n-1}$

The two roots are $\phi_{1}$ and $\phi_{2}$ so

$F_{n}=\frac{{\phi_{1}}^{n}-{\phi_{2}}^{n}}{\phi_{1}-\phi_{2}}$

Using this you can use a computer to find the desired number and see that \[F_{1492}\equiv 0\bmod 25,\] but there must be a nicer way to do it.[/hide]

[quote="Elemennop"]I think unless you actually stumble immediately across the elegant way to do this, it actually is faster just to straight up list the remainders mod 25 until they start repeating...they can't have a period more than like, 25 anyways.[/quote]

Actually, the period seems to be $100$ from towersfreak's answer.

Okay, 100 still isn't bad anyways...it'd take maybe 2 minutes max to list it all out

These sequences are  fast to calculate  with the method very analogous to how we calculate $x^{m}$   by repeated squaring.

(This is why  Fermat's  little theorem's test to prove a number is not prime is WAY faster than trying all the divisions)

So   even if you don't know the cycle ( If you are curious the cycle is called Pisano Period (google it to know more) , and for 25 it indeed is 100) to calculate $F_{1492}$ would only require about $\log_{2}1500$ or about 12 steps  (Not 1400 steps).

The point is it's very fast (specially using mod math) to calculate $F_{2n}$ from values around  $F_{n}$ (without going through each F's in middle) .. You  write 1492 in binary and combine values of ($F_{1024}, F_{256},F_{12}$ etc.
One  standard way is to use Lucas number and formulas like:
$F_{2n}= F_{n}L_{n}$
and $L_{2}n = (L_{n})^{2}-2(-1)^{n}$
Hope this helps.

Of course,  for 25 [b](or any power of 5)[/b] you don't have to go through all this,
the result is VERY Easy to find ... Hint Hint...! :)[hide]  (Look at Binnet's formula..: - 5 is really special!)[/hide]

[hide]Noticing the pattern we see that, it goes like this (mod 25)

1,1,2,3,5,8,13,21,9,5,14,19,8,2,10,12,22,9,6,15,21,11,7,18,0

Then the next one will just each be 18 times the first sequence as in

18, 18, 36, 54, ...

So basically $1492=25*59+17$.

Note that $F_{17+25k}(mod 25)=18^{k}*22$

So we need to find the remainder when $18^{59}*22$ is divided by 25.

Using Fermat's Little Theorem we see that $18^{20}$ is congruent to $1(mod 25)$, so $18^{-1}$ is congruent to $18^{59}mod 25$.

So $18^{-1}$ is also congruent to $7 mod 25$.  And $22$ is congruent to $22 mod 25$ so the answer is congruent to $154 mod 25$, which is $4$.

$4$

There is another way, that I think Jclarke knows, but I won't spoil it.

Sorry if it is unclear[/hide]

If you figure out the trick or shortcut for powers of five,

Here are two problems: Really cute,

1.  Prove if  $n=2 (5^{k})$  (k is positive  integer) the Fibonacci numbers would cycle in $6n$ numbers.
2.   If $n=24 (5^{k})$  (k is natural number), the cycle is $n$
3. for $n= 5^{k}$  the cycle is $4n$

(For example the cycle for mod 24 would be 24 and for  mod 10, it will be 60,
for 50 it will be 300 etc... and for 25 as the case given above, it will be 100)

[img]http://mathworld.wolfram.com/images/gifs/FoxTrot20051011.gif[/img]


[size=134]Find all solutions to $ x^2 \plus{} 1 \equal{} 0$ in the ring $ \mathbb{H}$ of quarternions.

Why is the answer not at odds with Proposition 1.7?

Proposition 1.7 says

If $ K$ is an extension of $ F$ and $ f(x) \in F[x],$ with $ \deg f(x) \equal{} n,$
then $ f(x)$ has at most $ n$ roots in $ K.$[/size]

Calculate $ (a\plus{}bi\plus{}cj\plus{}dj)^2$ to find that this equals $ 1$ iff $ a\equal{}0, b^2\plus{}c^2\plus{}d^2\equal{}1$.

This does not contradict proposition 1.7 as this isn't a field.


trigonometry
function
real analysis
real analysis theorems

Hello, I have a question.
If $ u(t)$ is the unit step function. Can I found the Laplace transform of $ u(\sin \omega t \minus{} x(t))$ in function of the laplace transform of $ x(t)$ ?

Not in any usable way. Sure, the Laplace transform is essentially invertible- but you would have to go fully through the inversion process to find where that inner expression crosses zero.


I'll just sketch it, and I hope it works :).

The product on the right is equal to $\prod_{n\ge 1}\frac{(1-x^n)^2}{1-x^{2n}}$, and the identity is true iff what we get by multiplying both sides with $\prod_{n\ge 1}(1-x^{2n})$ holds. This means that we have to prove $\left(\sum_{n\in\mathbb Z}(-1)^nx^{n^2}\right)\cdot\prod_{m\ge 1}(1-x^{2m})=\prod_{p\ge 1}(1-x^p)^2\ (*)$.

We have Euler's identity, which has a nice combinatorial proof that I won't post here :): $\prod_{m\ge 1}(1-x^m)=\sum_{n\in\mathbb Z}(-1)^nx^{\frac{3n^2+n}2}$. We can apply this to both $x$ and $x^2$ and replace the two products in $(*)$ with the sums we get. All we need to do now is check that $\left(\sum_{n\in\mathbb Z}(-1)^nx^{n^2}\right)\cdot\left(\sum_{m\in\mathbb Z}(-1)^mx^{3m^2+m}\right)=\left(\sum_{p\in\mathbb Z}(-1)^px^{\frac{3p^2+p}2}\right)^2$, and I'm sure this poses no major difficulties.


geometry
inradius
inequalities
trigonometry
geometry unsolved

A triangle $ABC$ is inscribed in a circle with center $O$ and radius $R$. If the inradii of the triangles $OBC,OCA,OAB$ are $r_{1}, r_{2}, r_{3}$, respectively, prove that $\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\ge\frac{4\sqrt{3}+6}{R}$

I do not think so. You certainly do not say that $\triangle ABC$ is acute. Take isosceles $\triangle ABC$ with the base angles $\angle A = \angle B = 30^\circ.$ $\triangle OBC,\ \triangle OCA$ are equilateral, $r_{1}= r_{2}= \frac{R \sqrt{3}}{6}.$ Area of $\triangle OAB$ is the same as areas of the other 2 equilateral triangles, $\triangle_{3}= \triangle_{2}= \triangle_{1}= \frac{R^{2}\sqrt{3}}{4},$ semiperimeter $s_{3}= R+\frac{R\sqrt{3}}{2},$ inradius $r_{3}= \frac{\triangle_{3}}{s_{3}}= \frac{R \sqrt{3}}{2(2+\sqrt{3})},$

$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}= \frac{6}{R\sqrt{3}}+\frac{6}{R\sqrt{3}}+\frac{2(2+\sqrt{3})}{R\sqrt{3}}= \frac{16+2 \sqrt{3}}{R \sqrt{3}}=$

$= \frac{\frac{16}{3}\cdot \sqrt{3}+2}{R}= \frac{4 \sqrt{3}+4 \cdot \frac{\sqrt{3}}{3}+2}{R}< \frac{4 \sqrt{3}+6}{R}$

[color=green][b][u]Yetti[/u][/b] proved that the triangle $ABC$ [u]must be acute ![/u] Therefore, here is the new enunciation.[/color]

[quote][color=darkred]An [u]acute[/u] triangle $ABC$ is inscribed in a circle with center $O$ and radius $R$.
If the inradii of the triangles $OBC,OCA,OAB$ are $r_{1}, r_{2}, r_{3}$, respectively,
prove that $\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\ge\frac{4\sqrt{3}+6}{R}\ .$[/color][/quote]

[color=darkred][b]Lemma (well-known).[/b] In any [u]acute[/u] triangle $ABC$ exist the following inequalities :[/color]

$\{\begin{array}{ccc}\sum \cos A\le \frac{3}{2}& \mathrm{\ and\ }& \sum\frac{1}{\cos A}\ge 6\ .\\\\ \sum \sin 2A\le \frac{3\sqrt 3}{2}& \mathrm{\ and\ }& \sum\frac{1}{\sin 2A}\ge 2\sqrt 3\ .\end{array}$

[color=darkblue][b]Proof.[/b] $\frac{1}{2}\cdot (2R+a)\cdot r_{1}=[BOC]=\frac{1}{2}\cdot a\cdot R\cos A$ $\implies$ $\frac{1}{r_{1}}=\frac{2}{R}\cdot (\frac{1}{\sin 2A}+\frac{1}{2}\cdot\frac{1}{\cos A})$ $\implies$

$\sum\frac{1}{r_{1}}=\frac{2}{R}\cdot(\sum\frac{1}{\sin 2A}+\frac{1}{2}\cdot\sum\frac{1}{\cos A})\ge \frac{2}{R}\cdot(2\sqrt 3+\frac{1}{2}\cdot 6)$ $\implies$ $\sum\frac{1}{r_{1}}\ge \frac{2(3+2\sqrt 3)}{R}\ .$[/color]

[quote="yetti"]Beat did not write "acute", you added it to the quote.[/quote][b][u]Yetti[/u][/b], you are right ! Now is it O.K. ?

Except I did not claim that the triangle must be acute, I only gave a counter-example for an obtuse one. For a given $\angle C,$ $\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}$ is minimum when $\triangle ABC$ is isosceles. Assuming $\angle C > 90^\circ$ is obtuse, $\triangle ABC$ isosceles, and putting

$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}= \frac{s_{1}}{\triangle_{1}}+\frac{s_{2}}{\triangle_{2}}+\frac{s_{3}}{\triangle_{3}}= \frac{4(1+\cos \frac{C}{2})}{R \sin C}-\frac{2(1+\sin C)}{R \sin 2C}= \frac{4 \sqrt{3}+6}{R}$

this ugly trigonometric equation has 2 roots in $\angle C \in (90^\circ,\ 180^\circ),$ namely $\angle C_{1}\approx 109.3^\circ$ and $\angle C_{2}\approx 147.5^\circ.$ If $\angle C$ is obtuse and out of this interval, the inequality holds.


geometry
parallelogram
geometry unsolved

Consider five points $ A, B, C, D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a
cyclic quadrilateral. Let $ l$ be a line passing through $ A$. Suppose that $ l$ intersects the interior
of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF \equal{} EG \equal{} EC.$
Prove that $ l$ is the bisector of angle $ DAB$.

I tried drawing a diagram but I'm not sure how could $ l$ intersects the interior of the segment DC and can also interset segment BC.  This seems impossible from my drawing, if I understand "intersects the interior" correctly.  By "interior", does it mean at some point on the segment?
A diagram would help a lot.

I think you can find solutions from the IMO #2 2007 topic in the IMO 2007 Forum.


[quote="M4RI0"]Se define la sucesión: $x_{1}=x_{2}=1$ y $x_{n+2}=2x_{n+1}+8x_{n}-1$, para todo $n \geq 1$. Demostrar que $x_{n}$ es un cuadrado perfecto para todo $n$.[/quote]
vamos a comprobar que: $x_{n}=a_{n}^{2}$, donde $a_{1}=a_{2}=1$.
$a_{n+2}=a_{n+1}+2a_{n}...(1)$, para todo $n\geq 1$.
entonces, esta otra recurrencia es mucho mas facil. Donde obtemos que:
$a_{n}=(2^{n}-(-1)^{n})/3....(2)$, reemplazando (1) en el problema.
$< = >x_{n+2}=a_{n+1}^{2}+4a_{n}^{2}+4a_{n}a_{n+1}=8a_{n}^{2}+2a_{n+1}^{2}-1$
operando esto y factorizando tenemos que vastara con que se cumpla lo siguiente:
$< = > 1=(a_{n+1}-2a_{n})^{2}$ahora en esta expresion reemplazamos (2), y operando con facilidad tenemos que la igualdad se cumple. por lo tanto
$x_{n}=a_{n}^{2}$, para todo $n\geq 1$

Viendo los primeros casos se hace obvio, pero el completo lo q a mi no se me habia ocurrido. :blush:

BUENO ACA LES PONGO UN PROBLEMA QUE TIENE UNA SOLUCION QUE SI ME AGRADO.
PROBLEMA.-
sean $n\geq 3$ numeros reales positivos, donde la suma de todos ellos, es menor o igual que 3,y la suma de los cuadrados de cada uno es mayor o igual que 1. Demostrar que existen 3 numeros talque su suma es mayor que igual que 1.

[quote="aev5peru"][quote="R.G.A.M."][color=blue]La idea de considerar la función de recurrencia $a_{n+2}=a_{n+1}+2a_{n}$, ¿cómo se te ocurrió?
¿Existe algún proceso que lleva a utilizar esa función o fue, simplemente, intuición?[/color][/quote]
Viendo los primeros, casos como indica MARIO, vi el camino intuitivo, ya que se formaba $1^{2}, 1^{2}, 3^{2}, 5^{2}, 11^{2}$.  Y generalise.[/quote]
[color=blue]Hallé una solución utilizando el razonamiento inductivo, a partir de la misma peculiaridad que habías notado, $aev5peru$. ¿Les gustaría observar el método?[/color]

[color=blue]Como mencioné, utilicé el razonamiento inductivo, tal vez de una forma similar a la solución ya enviada.

Analizando del modo siguiente:
$\begin{equation*}\begin{aligned}x_{1}&= 1^{2}= 2 \times 0+1 \\ x_{2}&= 1^{2}= 2 \times 1-1 \\ x_{3}&= 3^{2}= 2 \times 2+1 \\ x_{4}&= 5^{2}= 2 \times 5-1 \\ \end{aligned}\end{equation*}$
Puedo asumir, como hipótesis de inducción, que para todo $n \in \mathbb{N}$, $x_{n}= a_{n}^{2}$, y consideramos además $x_{n+1}= a_{n+1}^{2}$, agregando a la hipótesis de inducción.
Nótese que la sucesión $a_{n}$ cumple con las siguientes características:
$\begin{equation*}\begin{aligned}a_{1}&= 1 \\ a_{2}&= 1 \\ a_{n+1}&= 2a_{n}+\left({-1}\right)^{n}\\ \end{aligned}\end{equation*}$
Esta serie se cumplirá siempre. Para el paso siguiente, utilicé la deducción siguiente:
$\begin{equation*}\begin{aligned}a_{n}&= \frac{{a_{n+1}-\left({-1}\right)^{n}}}{2}\\ &= \frac{{a_{n+1}+\left({-1}\right)^{n+1}}}{2}\\ \end{aligned}\end{equation*}$
Prosigamos con la solución, teniendo en cuenta todo lo anteriormente mencionado.
$\begin{equation*}\begin{aligned}x_{n+2}&= 2x_{n+1}+8x_{n}-1 \\ &= 2a_{n+1}^{2}+8a_{n}^{2}-1 \\ &= 2a_{n+1}^{2}+2a_{n+1}^{2}+\left({-1}\right)^{n+1}4a_{n+1}+1 \\ &= 4a_{n+1}^{2}+4\left({-1}\right)^{n+1}a_{n+1}+1 \\ &= \left[{2a_{n+1}+\left({-1}\right)^{n+1}}\right]^{2}\\ \end{aligned}\end{equation*}$
Nótese que $a_{n+2}^{2}= \left[{2a_{n1}+\left({-1}\right)^{n+1}}\right]^{2}$. Entonces, $x_{n+2}= a_{n+2}^{2}$ y, por tanto, quedan demostrados la hipótesis inductiva y, como consecuencia, el enunciado.[/color]

[color=blue]He aquí otra solución al primer enunciado de este tópico.

Dada la igualdad $x_{n+2}= 2x_{n+1}+8x_{n}-1$, inmediatamente deducimos $9x_{n+2}= 18x_{n+1}+72x_{n}-9$ $\Rightarrow \left({9x_{n+2}-1}\right) = \left({18x_{n+1}-2}\right)+\left({72x_{n}-8}\right)$.
Consideremos, entonces, $u_{n}= 9x_{n}-1$. Inmediatamente $u_{1}=u_{2}=8$ y $u_{n+2}= 2u_{n+1}+8u_{n}$ y, aplicando el método de la ecuación característica (a falta de un nombre más apropiado) obtenemos que $u_{n}= 4^{n}-2 \times \left({-2}\right)^{n}$.
Luego $9x_{n}-1 = 4^{n}-2 \times \left({-2}\right)^{n}$ $\Rightarrow x_{n}= \frac{{4^{n}-2 \times \left({-2}\right)^{n}+1}}{9}$ $= \frac{{\left({-2}\right)^{2n}-2 \times \left({-2}\right)^{n}+1}}{9}$ $= \left[{\frac{{1-\left({-2}\right)^{n}}}{3}}\right]^{2}$, y queda demostrado el enunciado.

[b]P.D.:[/b] Agradezco a $aev5peru$ por ayudarme a trabajar con las series de recurrencias al presentarse un coeficiente independiente en la fórmula de recurrencia.[/color]

[quote="R.G.A.M."][quote="aev5peru"]BUENO ACÁ LES PONGO UN PROBLEMA QUE TIENE UNA SOLUCIÓN QUE SI ME AGRADO.
PROBLEMA.-
sean $n\geq 3$ números reales positivos, donde la suma de todos ellos, es menor o igual que 3,y la suma de los cuadrados de cada uno es mayor o igual que 1. Demostrar que existen 3 números talque su suma es mayor que igual que 1.[/quote]
[color=blue]Creo que no está del todo bien el enunciado. Recomiendo que se revise. Explicaré el porqué.

Sea $a_{i}$ la sucesión, con $1 \leqslant i \leqslant n$. Si algún valor de la sucesión es mayor o igual que $1$, satisface, por lo tanto, consideraremos que cada término es menor que $1$. Nótese entonces que $a_{i}> a_{i}^{2}\forall i$.
Supongamos que todos los términos de la sucesión son iguales a $a$. Entonces $3 \geqslant na$ $\Rightarrow \tfrac{3}{n}\geqslant a$ $\Rightarrow \tfrac{9}{n}\geqslant 3a$.
Si $n > 9$, entonces es imposible que la suma tres términos de esta sucesión sea mayor o igual que $1$ y, por lo tanto, no verifica.[/color][/quote]
Veo que tienes un error muy grave.
tu consideras que todos son iguales a el termino $a<1$.,$\Rightarrow \tfrac{3}{n}\geqslant a$ , entonces elevemos al cuadrado.
$\Rightarrow \tfrac{9}{n^{2}}\geqslant a^{2}$->entonces multiplicamos por $n$,$\Rightarrow \tfrac{9}{n}\geqslant na^{2}\geq 1$, entonces $9\geq n$ $(\to <-)$ ya que $n>9$, entonces para que se cumpla las condiciones del problema no es posible que todos sean iguales.


calculus
integration
inequalities proposed
inequalities

For a>0, and 0 < x < a, what is a tight upper and lower bound for
$ \int_0^a \frac{tan^{\minus{}1}(a\minus{}x)}{1\plus{}x^2} dx$

upper bound is attained: $ \int_0^a \frac{tan^{\minus{}1}(a\minus{}x)}{1\plus{}x^2} dx \le \frac{2}{\pi}tan^{\minus{}1}(\frac{2}{\pi}a)$.

What about a lower bound? A general expression of the integral is:
$ \int_0^a \frac{tan^{\minus{}1}(b(a\minus{}x))}{1\plus{}x^2} dx$ for $ 0<b<1$


function
integration
calculus
real analysis
real analysis theorems

Suppose m(X)<oo and define p(f)=int_X |f|/(1+|f|)dm
Then the map (f,g)->p(f-g) is a metric on the space of complex-valued measurable functions on X (if we identify 2 functions which are equal a.e.) and p(f_n-f)->0 iff fn->f in measure.

The verification of metric is easy. I prove the second part here.

[b]If:[/b] Suppose $f_n - f$ converges to 0 in measure.
Let $A_n = X[|f_n-f| > \sigma]$ and $B_n = X[|f_n-f| < \sigma]$
Then $mA_n\to 0$ and $mB_n\to mX$ as $x\to\infty$.

Given $\epsilon > 0$. We can find $\sigma$ such that $\frac{\sigma}{1+\sigma}mX < \frac \epsilon2$ and $N$ large enough, such that $mA_n < \frac \epsilon2$ for $n>N$. Then we have
\begin{eqnarray*} p(f_n - f) &=& \int_{A_n} \frac{|f_n-f|}{1+|f_n-f|}dm + \int_{B_n} \frac{|f_n-f|}{1+|f_n-f|}dm \\ &\leq& \int_{A_n} 1 dm + \int_{B_n} \frac{\sigma}{1+\sigma}dm\\ &=& m(A_n) + \frac{\sigma}{1+\sigma}m(B_n)\\ &<& \frac \epsilon2 + \frac{\sigma}{1+\sigma}mX < \epsilon \end{eqnarray*}
for $n>N$.
Thus $p(f_n - f) \to 0$.

[b]Only If: [/b] For any $\sigma$,
$p(f_n - f) = \int_{A_n} \frac{|f_n-f|}{1+|f_n-f|}dm + \int_{B_n} \frac{|f_n-f|}{1+|f_n-f|}dm$
tends to 0. Since both of the integrals on the right-hand side are non-negative, so both of them converge to 0.
The first integral $\int_{A_n} \frac{|f_n-f|}{1+|f_n-f|}dm > \frac{\sigma}{1+\sigma}mA_n$, hence $\frac{\sigma}{1+\sigma}mA_n \to 0$, and $mA_n\to 0$. It is exactly that $f_n$ converges to $f$ in measure.


$ (a^2 \plus{} b^2 \plus{} c^2 \equal{} 1,a,b,c>0)$
prove that   $ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\ge4\sqrt {3}$

Weighted AmGm. $ a \plus{} b \plus{} c \plus{} 9\frac {1}{9abc} \ge 12\sqrt [12]{\frac {abc}{(9abc)^9}} \equal{} 12\sqrt [12]{\frac {1}{3^{18}(abc)^8}} \ge 4\sqrt {3} \Leftrightarrow \sqrt [3]{(abc)^2} \le \frac {1}{3}$, which follows from $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.

[b]Generalization:[/b]

Let $ a_1,\ a_2,\ \dots ,\ a_n > 0$ such that $ a_1^{2}\plus{}a_2^{2}\plus{}\dots\plus{}a_n^{2}\equal{}1$.

Prove that: $ a_1\plus{}a_2\plus{}\dots\plus{}a_n\plus{}\frac{1}{a_1a_2... a_n}\ge \sqrt{n}\plus{}\sqrt{n^n}$ .

It's exactly the same method that leads to the solution: weighted AmGm: we have
$ \sum_{i \equal{} 1}^{n} a_i \plus{} n^{\frac {n \plus{} 1}{2}}\cdot \frac {1}{n^{\frac {n \plus{} 1}{2}}\prod_{i \equal{} 1}^{n}a_i} \ge (n \plus{} n^{\frac {n \plus{} 1}{2}}) \sqrt [n \plus{} n^{\frac {n \plus{} 1}{2}}]{\prod_{i \equal{} 1}^{n}a_i \cdot \frac {1}{(n^{\frac {n \plus{} 1}{2}}\prod_{i \equal{} 1}^{n}a_i)^{n^{\frac {n \plus{} 1}{2}}}}} \equal{} (n \plus{} n^{\frac {n \plus{} 1}{2}})\sqrt [n \plus{} n^{\frac {n \plus{} 1}{2}}]{\frac {1}{(\prod_{i \equal{} 1}^{n}a_i)^{n^{\frac {n \plus{} 1}{2} \minus{} 1}}}} \ge n \plus{} n^{\frac {n}{2}},$
which is equivalent to $ \frac {1}{n} \ge (\prod_{i \equal{} 1}^{n}a_i)^{\frac {2}{n}}$, which follows from $ \sum_{i \equal{} 1}^{n} a_i^2 \equal{} 1$.

[quote="sxgfly"]
a,b,c>0   so  $ (a \plus{} b \plus{} c)((a \plus{} b)^{2} \plus{} 3c^{2})$[color=red]>[/color]$ c(a \plus{} b)^{2}$ !!!!!!    yes or no?[/quote]

Well , yeah , but it doesn't matter , the solution is correct . In the main inequality equality happens when $ a\equal{}b\equal{}c$ .


geometry
trapezoid
geometric transformation
reflection
homothety
perpendicular bisector
power of a point

Let ABCD be an isosceles trapezoid with bases AD and BC. Let a circle which is tangent to both AB and AC intersect segment BC at points M and N. Now, consider $\omega$, the incircle of triangle BCD. Let X and Y be the intersections (closer to D) of DM and DN with $\omega$. Prove that XY is parallel to BC.

It's clear that nothing changes if, instead of taking $\omega$ to be the incircle, we take it to be the reflection of the circle tangent to $AB,AC$ in the perpendicular bisector of $BC$ (all we need is for $\omega$ to be a circle tangent to $DB,DC$), so let's do that.

$\omega$ cuts $BC$ in the points $M',N'$, the reflections of $M,N$ (respectively) in the midpoint of $BC$. Assume it touches $AB,AC$ in $B',C'$ respectively, and consider the circle $\omega_1$ tangent to $DB,DC$ in $B'',C''$ s.t. $B$ lies between $B',B''$, the same for $C$, and $BB''=CC',CC''=BB'$. After making the computations which say that the power of $B$ wrt $\omega$ is equal to the power of $C$ wrt $\omega_1$ and vice-versa, we see that $\omega_1$ cuts $BC$ precisely in $M,N$, so $X,Y$ are the homothetic images of $M,N$ through the homothety centered at $D$ which maps $\omega_1$ to $\omega$, and this shows that $XY\|MN=BC$ (the last equality means equality of lines).

[quote="grobber"]After making the computations which say that the power of $B$ wrt $\omega$ is equal to the power of $C$ wrt $\omega_1$ and vice-versa, we see that $\omega_1$ cuts $BC$ precisely in $M,N$...[/quote]

Grobber, how do you use the equal powers to deduce that the circle intersects $BC$ at $M, N$? Perhaps I am missing something obvious....

Could anyone post a more detailed proof to this ?

Let $\mathcal C$ be the circle tangent to AB, AC at S, T and intersecting BC at M, N. $BS^{2}= BM \cdot BN,\ CT^{2}= CN \cdot CM.$ Let $S' \in DB,\ T' \in DC$ be points such that BS' = BS and CT' = CT. We also require that S, S' are on the opposite sides of the line BC and the same for T, T'. WLOG, assume $AS < AB,$ $AT < AC,$ hence $DS' > DB,\ DT' > DC.$ A circle $\mathcal K_{1}$ tangent to DB at S' and passing through M passes also through N. Likewise, a circle $\mathcal K_{2}$ tangent to DC at T' and passing through N also passes through M.  Assuming the circles $\mathcal K_{1},\ \mathcal K_{2}$ intersecting at M, N are not identical, MN is their radical axis. The tangent length to $\mathcal K_{1}$ from D is equal to $DS' = DB+BS' = DB+BS = DB+BA-AS,$ while the tangent length from D to $\mathcal K_{2}$ is equal to $DT' = DC+CT' = DC+CT = DC+CA-AT.$ Since the sides $BA = DC$ and the diagonals $DB = CA$ of the isosceles trapezoid ABCD are equal, and since tangent lengths $AS = AT$ to the circle $\mathcal C$ from A are also equal, we get $DS' = DT'.$ This means that D is on the radical axis $MN \equiv BC$ of $\mathcal K_{1},\ \mathcal K_{2},$ which is a contradiction. The only way to avoid this contradiction is to have the intersecting circles $\mathcal K_{1}\equiv \mathcal K_{2}$ identical, so that their radical axis does not exist. Call this unique circle $\mathcal K.$ D is the external similarity center of the circle $\mathcal K$ and the incircle $\omega$ of $\triangle BCD,$ the triangles $\triangle DXY \sim \triangle DMN$ are centrally similar with the same similarity center, therefore their corresponding sides are parallel, $XY \parallel MN \equiv BC.$

I reduced this problem to the one I mentioned in this post [url]https://artofproblemsolving.com/community/u408889h2496197p21040775[/url] so if someone can take a look at it and try to finnish my progress I would be very thankful.


algebra
polynomial
calculus
integration
induction
number theory unsolved
number theory

This is a strange (at least for me) problem that I made up yesterday, and have no idea how to go about solving it. I have no real idea about its difficulty. It may in fact be rather trivial and/or already known to you, but anyhow, here goes:

Determine the maximum $n$ such that there exist $n$ polynomials $P_1, P_2, ..., P_n$ with integer coefficients such that

(i) $P_i(k)-P_{i-1}(k)=a$, for some constant $a \neq 0$, any integer $k$ and any $i, 2 \leq i \leq n$, and

(ii) for each $i, 1 \leq i \leq n$ , $P_i$ can be written as the product of two or more polynomials with integer coefficients and each with degree at least 1.

[quote="Severius"]This is a strange (at least for me) problem that I made up yesterday, and have no idea how to go about solving it. I have no real idea about its difficulty. It may in fact be rather trivial and/or already known to you, but anyhow, here goes:

Determine the maximum $n$ such that there exist $n$ polynomials $P_1, P_2, ..., P_n$ with integer coefficients such that

(i) $P_i(k)-P_{i-1}(k)=a$, for some constant $a$, any integer $k$ and any $i, 2 \leq i \leq n$, and

(ii) for each $i, 1 \leq i \leq n$ , $P_i$ can be written as the product of two or more polynomials with integer coefficients and each with degree at least 1.[/quote]
Is NOT trivial!?  take $a=0$, $n$ an integer and $P_i(x)=P_1(x)$ for every $i$ satisfy the second condition, so the bound is INFINITY-.

Well, I thought it was obvious in that context, but of course I'm looking for answers with $a \neq 0$. I'll edit the first post.

If there exist such polynomials then $a$ must be integer.

Now let $n>1$ be abitrary,we wish to construct those  polynomials which have the given properties.

I think you would agree with me if i take:

 $P_i(x)=(x-a)(x-2a)...(x-na)+x-ia , \forall i=1,2,...,n$

That was so simple. Thanks.

Now I know there are arbitrarily long sequences of such polynomials, I wonder if that holds in case we ask for $k$ polynomial factors, with $k>2$...

Maybe these arguments can help you :Assume that $n\in\mathbb N$,$n>1$ then:
 For abitrary integer $k>1$,there exists an integral polynomial $P_k(x)$ s.t:
    $P_k(x)+a,P_k(x)+2a,...,P_k(x)+na$ each has representation as product of $k$ integral polynomials.
 We construct $P_k(x)$ by induction on $k$.
 The existence of $P_2(x)$ has been proved.
  Now assume that we have $P_k(x)$,put
   $P_{k+1}(x)=P_k(x)+{\left(P_k(x)+a\right)}...{\left(P_k(x)+na\right)}$


John has 5 quarters and David has 5 dimes. John owes David
35 cents. John should give David $ m$ quarters and John should get
$ n$ dimes from David so that John has paid his debt exactly. What
is the value of $ m \plus{} n$?

John should give David 3 quarters (75 cents) and should receive 4 dimes (40 cents) in return.
$ 75\minus{}40\equal{}35$, so the debt is paid. $ (m,n)\equal{}(3,4)$
$ 3\plus{}4\equal{}\boxed{7}$


[quote="Makaveli"]Hey people, I want from you to share some experience with me about what to do a day before the contest, what to do immediately before the contest, what to do and think during it, how to maintain maximal concentration during all 4 hours and those things[/quote]

Try to be calm, cool, happy and of course optimistic.
you should also believe that you can succeed.
The day of the contest Have much [i]self-confidence[/i]

By the way, a math competition must be fun, so dont forget to enjoy it....

Good luck

 :)  :)

[quote="socrates"]By the way, a math competition must be fun, so dont forget to enjoy it....[/quote]

That's the most important thing!!!!!!!

[quote="José"][quote="dule_00"]I heard that eating chocolate would be helpful?[/quote]

Yes, and I was told that sugar too... I don't know how to explain it in english

[hide="in spanish"] El combustible del sistema nervioso es la glucosa, por lo que, mientras más azúcar consumas (nada de excesos, obvio) más "atento" vas a estar[/hide][/quote]

carbohydrates, $C_nH_{2n}O_n$, enter your body and give you energy, which apparently highly activates your brain cells and so you mentally function better. :P

Masoud Zargar

[quote="José"][quote="dule_00"]I heard that eating chocolate would be helpful?[/quote]

Yes, and I was told that sugar too... I don't know how to explain it in english

[hide="in spanish"] El combustible del sistema nervioso es la glucosa, por lo que, mientras más azúcar consumas (nada de excesos, obvio) más "atento" vas a estar[/hide][/quote]

[hide="In English:"] The fuel for the nervous system is glucose, so the more sugar you consume (not in excess, obviously), the more "attentive" you will be.[/hide]


Suppose that $a_{1},a_{2},\dots,a_{k}\in\mathbb C$ that for each $1\leq i\leq k$ we know that $|a_{k}|=1$. Suppose that \[\lim_{n\to\infty}\sum_{i=1}^{k}a_{i}^{n}=c.\] Prove that $c=k$ and $a_{i}=1$ for each $i$.

[quote="Omid Hatami"]Suppose that $a_{1},a_{2},\dots,a_{k}\in\mathbb C$ that for each $1\leq i\leq k$ we know that $|a_{k}|=1$. Suppose that \[\lim_{\n\rightarrow\infty}\sum_{i=1}^{k}a_{i}^{n}=c.\] Prove that $c=k$ and $a_{i}=1$ for each $i$.[/quote]

By Kronecker's theorem for any natural $N >1$ and $\epsilon >0$ there exist $n > N$ such that $|n \arg(a_{i})-2\pi k_{i}| < \arccos (\epsilon)$ for some integers $k_{i}$, $i=1,2,...,k$ so that $\cos(n \arg{ a_{i}}) > 1-\epsilon$ for all $i=1,2,...,k$, so the $k \geq |\sum_{i=1}^{k}a_{i}^{n}| \geq k-\epsilon$, so we get that $c=k$ and the result follows.

Now solve the following beautiful problem given at oral ens ulm: suppose that the series of general term $a_{n}$ is absolutely convergent and that the series of general term $a_{n}^{p}$ has zero sum for all natural p. Then all terms are 0. What happens if we drop the first hypothesis?

For the first part of the question, just notice that all sums of the form $S_{k}=\sum_{n_{1}<\ldots<n_{k}}a_{n_{1}}\cdot\ldots\cdot a_{n_{k}}$ are zero, and hence the function $z\mapsto\prod_{n}(1-a_{n}z)$, which is entire and whose only zeros are $\frac 1{a_{n}}$ (we may assume WLOG that $a_{n}\ne 0,\ \forall n$), is actually identically equal to zero. We have a contradiction.

Nice!!! My solution used actually the result posted by Omid.


ABCD is a convex quadrilateral. We draw its diagnals to divide the quadrilateral to four triabgles. P is the intersection of diagnals.  I1,I2,I3,I4 are excenters of  PAD,PAB,PBC,PCD(excenters corresponding vertex P)
PROVE: I1,I2,I3,I4 are concyclic if and only if ABCD is a tangential quadrilateral

lemma http://www.mathlinks.ro/Forum/viewtopic.php?t=21758

Denote  ∠PBC=2∠7  ∠PCB=2∠7’            ∠PCB=2∠8 ∠PBC=2∠8’
∠PDA=2∠5  ∠PAD=2∠5’            ∠PAB=2∠6  ∠PBA=2∠6’
PI7/PI3=r7/R7=1/(tan∠7 * tan∠7’)  =>  PI3=PI7 * tan∠7 * tan∠7’
Analogously  PI4=PI8 * tan∠8 * tan∠8’
             PI2=PI6 * tan∠6 * tan∠6’
                PI1=PI5 * tan∠5 * tan∠5’
PI1*PI3= PI5 * tan∠5 * tan∠5’ *  PI7 * tan∠7 * tan∠7’
PI2*PI4= PI6 * tan∠6 * tan∠6’ *  PI8 * tan∠8 * tan∠8’

 From lemma we have   AB+CD=AD+BC 〈=〉 I5I6I7I8 are concyclic  〈=〉 tan∠5 * tan∠5’ *  tan∠7 * tan∠7’= tan∠6 * tan∠6’ * tan∠8 * tan∠8’ 〈=〉 PI5*PI7=PI6*PI8

In generic case  we have that
                    PI5=AP*PD tan∠5 * tan∠5’ PI7=PB*PC tan∠7 * tan∠7’
             PI6=AP*BP tan∠6 * tan∠6’ PI8= PD*CP tan∠8 * tan∠8’
PI5*PI7 * tan∠5 * tan∠5’ *  tan∠7 * tan∠7’= PI6*PI8 * tan∠6 * tan∠6’ * tan∠8 * tan∠8’ 〈=〉 PI5*PI7 * tan∠5 * tan∠5’ *  tan∠7 * tan∠7’ * AP*PD * PB*PC= PI6*PI8 * tan∠6 * tan∠6’ * tan∠8 * tan∠8’* AP*BP* PD*CP 〈=〉 （PI5*PI7）^2=( PI6*PI8)^2

So AB+CD=AD+BC 〈=〉 PI5*PI7 * tan∠5 * tan∠5’ *  tan∠7 * tan∠7’= PI6*PI8 * tan∠6 * tan∠6’ * tan∠8 * tan∠8’

so we have  I1I2I3I4 are concyclic〈=〉 PI1*PI3 = PI2*PI4 〈=〉
(PI5 * tan∠5 * tan∠5’ *  PI7 * tan∠7 * tan∠7’)( PI6 * tan∠6 * tan∠6’ *  PI8 * tan∠8 * tan∠8’) 〈=〉AB+CD=AD+BC
thus we are done!

In the fact, with the help of lemma (as [b]plane geometry[/b] mentioned), we can solve this problem easier than his solution ( his proof has many trigonometric transformations  :) ). However, the lemma is a hard problem. I am not pleased with the solution in Crux. And I am finding a synthetic solution to the lemma - a wonderful problem.
I'm happy if someone has a synthetic proof ?  :lol:
Good luck!

Wow! I saw two nice solutions of [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=5895]yetti[/url] and [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=3758]juancarlos[/url]. Thanks, guy.
By the same way, [b]plane geometry[/b]'s problem can be solved. This problem is the first problem of Iran TST 2002.
Look at [url=http://www.mathlinks.ro/viewtopic.php?p=1249276#1249276]a new problem[/url]

wow,dear mr.danh,canyou post the link of  solutions of yetti and juancarlos?


function
calculus
derivative
calculus computations

Given this information:
$ f(1) \equal{} 3$

$ f'(1) \equal{} \minus{}2$

$ g(1) \equal{} \minus{}3$

$ g'(1) \equal{} 4$

and
$ h(x) \equal{} (2f(x)\plus{}3)(1\plus{}g(x))$, then what is the value of $ h'(1)$?

Hint: Expand the product and then use the product rule of differentiation to write the general form of h'(x). Then replace x with 1 and you should get the answer.

[quote="math92"]Given this information:
$ f(1) \equal{} 3$

$ f'(1) \equal{} \minus{} 2$

$ g(1) \equal{} \minus{} 3$

$ g'(1) \equal{} 4$

and
$ h(x) \equal{} (2f(x) \plus{} 3)(1 \plus{} g(x))$, then what is the value of $ h'(1)$?[/quote]
$ h(x)\equal{}2f(x)\plus{}2f(x)g(x)\plus{}3\plus{}3g(x)$
$ h'(x)\equal{}2f'(x)\plus{}2f'(x)g(x)\plus{}2f(x)g'(x)\plus{}0\plus{}3g'(x)$
$ h'(x)\equal{}2*\minus{}2\plus{}2*\minus{}2*\minus{}3\plus{}2*3*4\plus{}0\plus{}3*4$
$ h'(x)\equal{}\minus{}4\plus{}12\plus{}24\plus{}12$
$ h'(x)\equal{}44$
Hm?

You also could do it without expanding, just using the product rule and sum/constant multiple derivative rules.

$ h(x) \equal{} (2f(x) \plus{} 3)(1 \plus{} g(x))$
$ h'(x) \equal{} 2f'(x)(1 \plus{} g(x)) \plus{} (2f(x) \plus{} 3)g'(x)$
So $ h'(1) \equal{} 2f'(1)(1 \plus{} g(1)) \plus{} (2f(1) \plus{} 3)g'(1) \equal{} 2(\minus{}2)(\minus{}2) \plus{} (9)(4) \equal{} 44$


How many six digit numbers are there whose digits alternate between even and off numbers (even-odd-even-odd-even-odd or odd-even-odd-even-odd-even)?

[hide]There are 5 possible even digits and 5 possible odd digits, so the answer is $2*5^{5}=\boxed{6250}$[/hide]

wrong, the answer is
[hide]
$28125$
[/hide]

I actually just figured this out myself heh, heres my answer
[hide]
since the first number will be even or odd there are 9 possibilties since there are 5 odd numbers and 4 even numbers available, we count the rest as just 5 because the first number determines what set the next number will be from and it will be from only one and not both.
so we have:

$9 \cdot 5^{5}= \boxed{28125}$
[/hide]


abstract algebra
number theory unsolved
number theory

[quote="RaduSorici"]Let $x,y,p,n,k > 0$ be integer numbers such that $x^n+y^n=p^k$. Prove that if $n$ is odd and greater than $1$, and $p$ is an odd prime number we have that $n=p^s$ where $s$ is an integer $> 0$[/quote]
We can assume $p\not |x,y$ and  $n$ an odd prime, then $x+y=p^\mu$ becauise $p$ is prime, and $\mu\in \mathbb{Z}_+$, then (because $n>1$ then $\mu<k$) then taking module $p^{\mu+1}$ we get
\[
p^{\mu}ny^{n-1}=0 \ mod \ p^{\mu+1},
\]
and as $p\net |y$ we get that $n=p$. Now, if $n$ is composite, for every prime $Q$ we can write the same ecuation considering $x^n=\left(x^{n/Q}\right)^Q$ and hence,  using the last argument,  we conclude that $Q=p$ then $n$ is a power of $p$, i.e. $n=p^s$ with $s>0$ because $n>1$.

RobertuX


geometry
geometric transformation
reflection
homothety
angle bisector
geometry unsolved

Let the lines PA and PB are tangent to a circle C at A and B ,where P is outside the circle.A line through P intersects the circle at two distinct points,say C and D, and AC and AD intersects the line through B parallel to PA at E and F.prove that EB=BF.

Let Q be the intersection of the tangents at C and D. Since P lies on the polar of Q, Q is on the polar of P: A,B,Q are on a line.
Then, AQ is the A-symmedian of ADC.
Now note that $\angle ADC = \angle CAP = \angle AEF$ and $\angle ACD = \angle AFE$. Then, the reflection on the angle bisector of CAD and a homotety will being the triangle ADC in AEF. Then, the symmedian of CAD is the median of AEF: AB is the median of AEF!

I hope it is clear, bye bye

Let $M\equiv AB\cap PD$ be and, as well known, we have that the points $P,$ $M,$ are harmonic conjugates of $C,$ $D.$

Hence, the pencil $A.PMCD,$ is in harmonic conjugation and because of $EF\parallel PA,$ we conclude that $BE = BF$ and the proof is completed.

$($ [b][size=109]THEOREM.[/size][/b] - When a line is parallel to one of the rays of a pencil in harmonic conjugation, then it is intersected from the other three it's rays, at three points formed two congruent segments $).$

Kostas Vittas.

is there any link that i could learn about harmonic conjugation.i am very curious about that.


Let $P$ be a point inside a triangle $ABC$ such that $\angle PBC = \angle PCA < \angle PAB$. The line $PB$ meets the circumcircle of triangle $ABC$ at a point $E$ (apart from $B$). The line $CE$ meets the circumcircle of triangle $APE$ at a point $F$ (apart from $E$). Show that the ratio $\frac{\left|APEF\right|}{\left|ABP\right|}$ does not depend on the point $P$, where the notation $\left|P_1P_2...P_n\right|$ stands for the area of an arbitrary polygon $P_1P_2...P_n$.

Since the point E lies on the circumcircle of triangle ABC, we have < EAC = < EBC. But < EBC = < PBC = < PCA; thus, < EAC = < PCA. Hence, the lines AE and PC are parallel, so that |APE| = |ACE|. Consequently, |APEF| = |APE| + |AEF| = |ACE| + |AEF| = |ACF|.

Now, since the points A, P, E, F lie on one circle, < AFE = 180° - < APE; in other words, < AFC = 180° - < APE = < APB. Also, since the points A, E, C, B lie on one circle, < ACE = < ABE, what rewrites as < ACF = < ABP. From < AFC = < APB and < ACF = < ABP, it follows that the triangles ACF and ABP are similar, and thus the ratio of their areas equals to the ratio of the squares of their corresponding sidelengths: $\frac{\left|ACF\right|}{\left|ABP\right|}=\frac{AC^2}{AB^2}$. Since |APEF| = |ACF|, this becomes $\frac{\left|APEF\right|}{\left|ABP\right|}=\frac{AC^2}{AB^2}$, and since this ratio obviously does not depend on P, the problem is thus solved.

  Darij

Maybe renounce at the condition $m(\widehat {PCA})<m(\widehat {PAB})$. If $\widehat {PCA}\equiv \widehat {PAB}$ then $E\equiv F$, i.e. the line $CE$ is tangent to the circumcircle of the triangle $APE$ and if $m(\widehat {PCA})>m(\widehat {PAB})$ then $F\in (CE)$. In the both cases, the conclusion is truly.
[b]Remark.[/b] [i]Dear Darij[/i], please (if is possibly) [u]renounce[/u] at the much too details in your solutions. Otherwise, you lose from the brightness of the demonstration !

[hide="Here I offer briefly (three rows) the nice Darij-s proof because it has much too details."] $\boxed {\mathrm {Start\ of\ proof}}$
$\blacktriangleright\widehat {EAC}\equiv \widehat {EBC}\equiv \widehat {PBC}\equiv \widehat {PCA}\Longrightarrow$$AE\parallel PC\Longrightarrow [APC]=[EPC]\Longrightarrow [APEF]=[ACF]\ .$
$\blacktriangleright\widehat {BPA}\equiv \widehat {CFA}\ ,\ \widehat {ABP}\equiv \widehat {ACF}\Longrightarrow \triangle ACF\sim\triangle ABP\ .$
Therefore, $\frac{[APEF]}{[ABP]}=\frac{[ACF]}{[ABP]}=\left(\frac{AC}{AB}\right)^2=\left(\frac bc\right)^2$, what is constant.
$\boxed {\mathrm{End\ of\ proof}}$[/hide]

[quote="Virgil Nicula"]
[b]Remark.[/b] [i]Dear Darij[/i], please (if is possibly) [u]renounce[/u] at the much too details in your solutions. Otherwise, you lose from the brightness of the demonstration !

[/quote]

  M.T.

[i]Armpist[/i], you are [u]badly[/u] ! I militate in favour of a universal mathematical language which can be used at least only in the writing. Can you help me with a some links where I can find the about all math symbols which are used in the ... your western country ?  In the my eastern country the mathematics are at its home, my dear !

Virgil, I know that my way of writing up proofs is rather suboptimal, but it's a matter of habit and I don't pretty like using logical symbols in proofs. Also, noting down every detail is my personal way of preventing mistakes (of course, this doesn't completely eliminate all mistakes, but significantly decreases their number). And please don't take Armpist too seriously, you have probably noted that he uses to have good and surprising ideas on geometry but likes to post some pretty abstruse polemics.

  Darij

I am sorry Darij, I prize highly the nice Armpist's proposed problems and solutions, but now I don't prize the human being Armpist.
Presumption of innocence: he is still a young man.

[i]Dear Armpist[/i], I answer you punctually:
1.  Evidently, I think you are badly ... at least with mine (a little, there).
2.  [b]Le[/b]lia is my wife and I am [b]Vi[/b]rgil (Nicula)$\Longleftrightarrow$ [b]Levi[/b].
It is O.K. ? Where is the contradiction ?
[b]P.S.[/b] (postscript). Please, read the signature at the bottom of my posts. In each year, I am six months in Yakima, WA, USA. Can I meet you in the ... wild Nord-West (Green State Washington). I wish you become a my [u]good[/u] friend.
Now I wait an answer only from you (but through a private message) !

[quote="armpist"][quote="Virgil Nicula"]
[b]Remark.[/b] [i]Dear Darij[/i], please (if is possibly) [u]renounce[/u] at the much too details in your solutions. Otherwise, you lose from the brightness of the demonstration ![/quote]
Do not do it under any circumstances  or circumcircles!
The common language here is English, not a collection of [color=red]some Eastern European
symbols pretending to have something in common with international geometry.[/color]Thank you. M.T.[/quote]Very, very nice, the western man ! Know you what is the international geometry ? If you know, please say and me. You don't understand that many Mathlinks-users don't know the English language like you don't know the another some language. It is O.K. (this word is used and in a eastern country). Thank you. V.N.

As our initial assumptions let $\angle{PBC}=\angle{PCA}=\theta,\angle{BAP}=x$.Note that $\angle{PBA}=\angle{EBA}=\angle{ECA}=\angle{FCA}=B-\theta$.$\angle{PAC}=A-x\angle{PAF}=\angle{PEC}=\angle{BEC}=\angle{BAC}=A$.Thus $\angle{CAF}=x$.This implies that $\triangle{ACF} \sim \triangle{ABP} \Rightarrow \frac{BP}{FC}=\frac{AB}{AC}$.$\angle{APF}=\angle{AEF}=\angle{ABC}=B \Rightarrow \triangle{APF} \sim \triangle{ABC} \Rightarrow \frac{AF}{AP}=\frac{AC}{AB}$.
Combining these two equalities we get $\frac{AP \times BP}{AF \times FC}=\frac{AB^2}{AC^2}$.
Also we have $\angle{APB}=\angle{AFC}=180-B-x+\theta$.Thus $\frac{[APB]}{[ACF]}=\frac{AB^2}{AC^2}$
.As a final observation we note that
$\angle{AEF}=B=B-\theta+\theta=\angle{EBA}+\angle{EBC}=\angle{ECA}+\angle{PCA}=\angle{PCE} \Rightarrow PC \parallel AE$.
Thus the result follows:
$\frac{AC^2}{AB^2}=\frac{[ACF]}{[APB]}=\frac{[AEF]+[AEC]}{[APB]}=\frac{[AEF]+[APE]}{[APB]}=\frac{[APEF]}{[APB]}$
which is clearly independent of $P$.


The hour hand of a watch moves at the correct speed, but the minute hand moves one and a half times as fast as it should. Yesterday, it showed the correct time at 3 p.m. When
did it next show the correct time?

Prove your answer.

[hide]
Since the hour hand will always be correct, it remains only to find when the minute hand will be at the right time.
The minute hand will always be ahead of the real time until it is exactly one hour ahead of the real time, at which point it shows the right time again.
This takes 3 hours, so 6pm yesterday.[/hide]

[hide]Yes, the minute hand TRAVELS 3 hours, but only 2 hours pass, so 5 o clock.[/hide]

Yeah I got the same answer as samath,
I found that the time when the clock shows the correct time would be:
$t=2(k-l)$ hours where $k$ and $l$ are the number of revolutions of the correct and the wrong clocks.
As $k\geq0$ and $l\geq0$ and they have to be integral values
So, minimum is $t=2$ hours

Yup, 5:00 is the answer.

1.5x2 = 3 is the first positive integer value formed by 1.5 x n.

Hence, n=2 and 3+2 = 5.

5:00 pm

Yes, Samath is right. The number in my signature increases by 1  :oops:


superior algebra
superior algebra unsolved

Hi, i need to show that $ f$ is an isomorphism.

$ f: C \minus{} > R$ (complex to real numbers)

$ f(x \plus{} y*i) \equal{} x \plus{} y$

need to find $ ker(f)$ and $ im(f)$ to show that $ f$ is isomorphic.

$ ker(f) \equal{}$ identity (under addition - i.e. zero)

so far i've got: $ x \plus{} y \equal{} 0, x \equal{} \minus{} y$

not sure how to use this to show that $ f$ is isomorphic.

any help would be appreciated. thanx.

[quote="mavropnevma"]How could $ f$ be injective, when $ f(x(1 \minus{} i)) \equal{} 0$ for all real $ x$ ?[/quote]

sorry, my mistake. i was supposed to find out if $ f$ is isomorphic, not to show that it's isomorphic.

i know that $ im(f) \equal{} f / ker(f) \equal{} (x\plus{}(\minus{}x)*i) / (x \plus{} (\minus{}x))$

it does not seem to work...


[quote="mathwizarddude"]If $ 2n^2 \plus{} 1 \equal{} m^2$, prove that $ (nm)^2$ is a triangular number.[/quote]

$ n^2m^2\equal{}\frac{(2n^2)(m^2)}{2}\equal{}\frac{(m^2\minus{}1)(m^2)}{2}$, done.


and after he told them why, everyone believed him. Why?

stupid solution:

[hide="in here"]You multiply both sides by 0. Since the results are equal, then 10=50.[/hide]

Because the "10" is in base 50, while the "50" is in base 10.

Because the people he told were extremely gullible.

Does it have anything to do with the guy being named Bob?

Silly question :rotfl:
[hide]
Do you believe in him, kstan013?
[/hide]

Is this actually mathematical, or do the 10 and 50 symbolize something else?

Is Bob/Steve held in any particularly high regard among his community?

If it is mathematical, if Bob/Steve told this to you, would you believe him?

[b]Is this actually mathematical, or do the 10 and 50 symbolize something else?[/b]mathematical

[b]Is Bob/Steve held in any particularly high regard among his community?[/b]nope. just your average joe

[b]If it is mathematical, if Bob/Steve told this to you, would you believe him?[/b]yes i would...and not because I'm gullible

[b]is it like 0=1? divide by zero error?[/b]no connection whatsoever

Does 10 mean ten?
Does = mean equals?
Does 50 mean fifty?

In the normal base-ten sense, that is.

The answer to all of mathnerd's questions are YES
Bpms, nope, but you've probably got the best IDEA, even though your answer is nowhere near right

I was thinking along the lines of "One Zero = Five Zeros", but if the answers to mathnerd's questions were yes, then this can't be right. Is it anywhere close, or is it completely off-track?

does this have anything to with modular numbers
or division?

nothing to do with mods , but it kind of has something to do with division (but not completely)

Joke Answer

[hide]Everyone he told had no basic knowledge of numbers and didn't know what 10 or 50 was! Bob, being a respected member of society, was believed by everyone else.[/hide]

Did he give a long proof to this "identity" :rotfl: or did he just say one thing?

[quote="kstan013"]The answer to all of mathnerd's questions are YES[/quote]

This is certainly stretching it.  If = means "equals" then it certainly doesn't mean "equal" in the normal arithmetic sense.

Does this have anything to do with Roman numerals?

[quote="box of rocks"]If this puzzle does not rely on what has been becoming a common misleading difference in the use of units (dimensions), or what looks here to the ABSENCE of REQUIRED units between the two quantities in question, then I would be very surprised indeed.

This problem is trying to have it BOTH ways (apparently), instead of it being nonmisleading.[/quote]You are very close box of rocks.

No roman numerals
He could have said more to clarify
He didn't need a long proof; if I would tell you, you would go: Oh. Duh.
Even the geniuses believe him.

[quote="mathnerd314"][quote="bigboy82892"][hide="Guess."]$10^{o}C=50^{o}F$?[/hide][/quote]

D'oh. That's probably right.

:roll:[/quote]and it is! I said you would say...[quote="kstan013"]Oh. Duh.[/quote] :D

take that t0rajir0u and all of your skepticism!


algebra
polynomial
real analysis
real analysis unsolved

All coefficients of polinomial $P (z) \in C [X]$ are $0$ or $1$ and $P (0) = 1$. Prove that for every root $x \in C$ of polynomial is $x \geq \frac{-1 + \sqrt{5}}{2}$.

alexilic

what about $P(x)=x+1$. And what's with $x\geq...$ in complex numbers..?

[quote="alexilic"]All coefficients of polinomial $P (z) \in C [X]$ are $0$ or $1$ [/quote]
I even can't understand this :(

Let $z^n+a_{n-1}z^{n-1}+...+a_1z+a_0=0, a_i\in[0,1],0\leq i \leq{n-1}$ and $z\in\mathcal l C$. Prove that $Re(z) \leq \frac{1+\sqrt 5}{2}$

I actually think that he wanted $|x|>\ge \frac{-1+\sqrt 5}{2}$ and the problem is for sufficiently big degree of polinomials (2,3 or greater). I think that you should try Algebra subforum in Advanced Section for more answers. The proof have not much in common with calculus.


Let $p>5$ be a prime number. Show that, in decimal notation, that either $p^{2}+1$, $p^{2}-1$ ends in $0$.

[hide]Clearly both $p^{2}+1$ and $p^{2}-1$ are even. Hence either $p^{2}+1$ or $p^{2}-1$ is divisible by $5$.
The quadratic residues $\bmod 5$ are $-1, 0, 1$, and $p > 5$ so it's proved.[/hide]

One thing is for sure that a prime number $p$ such that $p > 5$
will not have $5$ as its unit digit because then it would become divisible by $5$. Also $p$ would be always odd.

So the other possibilities would be $1,3,7,9$ for the unit's digit of the prime number $p$.

Now $p^{2}$ would inevitably have unit's digit as $1$ or $9$
as $1^{2}= 1, 3^{2}= 9, 7^{2}= 9,9^{2}= 1$.
Now if the last digit of $p^{2}$ is one of $1,9$ then out of $p^{2}+1$ or $p^{2}-1$ one would have its unit digit as $0$.
Hence proved.


For any three positive reals a, b, c, we have

$\frac{a\sqrt{a^2+bc}}{(b+c)^2}+\frac{b\sqrt{b^2+ca}}{(c+a)^2}+\frac{c\sqrt{c^2+ab}}{(a+b)^2}\ge\frac{3\sqrt{2}}{4}$

I look it first time and I have a solution,but maybe your solution is easier:

I prove a stronger:

$\sum\frac{a\sqrt{a^2+3bc}}{(b+c)^2} \ge \frac{3}{2}$

First,by Trebusep:

$3\sum\frac{a\sqrt{a^2+3bc}}{(b+c)^2} \ge (\sum a\sqrt{a^2+3bc})(\sum\frac{1}{(a+b)^2})$

Use two inequality:

$\sum a\sqrt{a^2+3bc}\ge 2(ab+bc+ca)$

$(ab+bc+ca)(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}\frac{1}{(c+a)^2}) \ge \frac{9}{4}$

And we have done!