Find the 6-digit number beginning and ending in the digit 2 that is the product of three consecutive even integers.

Is the answer... [hide]64*66*68=287232?  I got this first by checking the units digit.  Of the three consecutive even numbers, 4*6*8 is the only combination that gives 2 as the units digit.  Next, I checked  the hundred thousand digit.  The only cube that would result in 200000 was 60^3.  Therefore, the three numbers are 64, 66, and 68[/hide]

Im interested in knowing how you would reach that the tens digit is 6 without a calculator. Surely, you wouldn't sit there and start multiplying, or would u?

Hmm, only 6 digits..

Anyway, is there a way?

You can approximate it. 60^3=216000, and 70^3=343000, so the numbers are going to be in the 60s.


Solve the system of the equations:
$ 3(x^{2}\plus{}y^{2}\plus{}z^{2})\equal{}1$
$ x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2}\equal{}xyz(x\plus{}y\plus{}z)^{3}$

No one  :maybe:  is this problem difficult or has no interest to solve  :huh:

[quote="arqady"][quote="galois01"]Solve the system of the equations:
$ 3(x^{2} \plus{} y^{2} \plus{} z^{2}) \equal{} 1$
$ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \equal{} xyz(x \plus{} y \plus{} z)^{3}$[/quote]
$ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \equal{} xyz(x \plus{} y \plus{} z)^{3}\Leftrightarrow$
$ \Leftrightarrow3(x^2 \plus{} y^2 \plus{} z^2)(x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2}) \equal{} xyz(x \plus{} y \plus{} z)^{3}.$
But $ xyz(x \plus{} y \plus{} z)\geq0,$  $ 3(x^2 \plus{} y^2 \plus{} z^2)\geq(x \plus{} y \plus{} z)^2$ and $ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \geq xyz(x \plus{} y \plus{} z).$
If $ 3(x^2 \plus{} y^2 \plus{} z^2) > (x \plus{} y \plus{} z)^2$ then $ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \equal{} xyz(x \plus{} y \plus{} z) \equal{} 0.$
Hence, $ (x,y,z)\in\{(\pm\frac {1}{\sqrt3},0,0),(0,\pm\frac {1}{\sqrt3},0),(0,0,\pm\frac {1}{\sqrt3})\}.$
If $ 3(x^2 \plus{} y^2 \plus{} z^2) \equal{} (x \plus{} y \plus{} z)^2$ then $ x \equal{} y \equal{} z \equal{} \pm\frac {1}{3}.$[/quote]
Thank's for your interesting arqady very nice solution  :)


$ a,b,c \geq 0, ab\plus{}bc\plus{}ca\equal{}1$ prove that:
$ \frac{1}{(a\minus{}b)^2\plus{}1}\plus{}\frac{1}{(b\minus{}c)^2\plus{}1}\plus{}\frac{1}{(c\minus{}a)^2\plus{}1} \geq 2$

[size=150]does holds.
try:
$ a\equal{}\frac{17}{9},b\equal{}\frac{1}{6},c\equal{}\frac{1}{3}$
BQ[/size]

[quote="xzlbq"][size=150]does holds.
try:
$ a \equal{} \frac {17}{9},b \equal{} \frac {1}{6},c \equal{} \frac {1}{3}$
BQ[/size][/quote]
Thank you very much! How about this ineq:
Let $ a,b,c \geq 0$ and $ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \equal{} 2$
Prove that:$ \frac {a}{(a \plus{} b)(a \plus{} c)} \plus{} \frac {b}{(b \plus{} c)(b \plus{} a)} \plus{} \frac {c}{(c \plus{} a)(c \plus{} b)} \leq 1$

[quote="mathstarofvn"][quote="xzlbq"][size=150]does holds.
try:
$ a \equal{} \frac {17}{9},b \equal{} \frac {1}{6},c \equal{} \frac {1}{3}$
BQ[/size][/quote]
Thank you very much! How about this ineq:
Let $ a,b,c \geq 0$ and $ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \equal{} 2$
Prove that:$ \frac {a}{(a \plus{} b)(a \plus{} c)} \plus{} \frac {b}{(b \plus{} c)(c \plus{} a)} \plus{} \frac {c}{(c \plus{} a)(c \plus{} b)} \leq 1$[/quote]

There is a typo in your problem. The second term in LHS will be $ \frac{b}{(b\plus{}c)(b\plus{}a)}$ instead of $ \frac{b}{(b\plus{}c)(c\plus{}a)}$

Hey! What about the proofs ?

[quote="xzlbq"][quote="mathstarofvn"]$ a,b,c \geq 0, ab \plus{} bc \plus{} ca \equal{} 1$ prove that:
$ \frac {1}{(a \minus{} b)^2 \plus{} 1} \plus{} \frac {1}{(b \minus{} c)^2 \plus{} 1} \plus{} \frac {1}{(c \minus{} a)^2 \plus{} 1} \geq 2$[/quote]

[size=150]$ 3\geq \frac {1}{(a \minus{} b)^2 \plus{} 1} \plus{} \frac {1}{(b \minus{} c)^2 \plus{} 1} \plus{} \frac {1}{(c \minus{} a)^2 \plus{} 1} \geq 1$
holds.
But I have not a proof.
BQ[/size][/quote]

$ a,b,c \ge 0,ab \plus{} bc \plus{} ca \equal{} 1.$

$ \sum{\frac {1}{(a \minus{} b)^2 \plus{} 1}} \ge 1$

$ \iff \sum{\frac {1}{(a \minus{} b)^2 \plus{} ab \plus{} bc \plus{} ca}} \ge \frac {1}{ab \plus{} bc \plus{} ca}$

$ \iff \frac {\sum_{sym}{(a^4b^2 \plus{} a^4bc \plus{} a^3b^3 \plus{} 2a^3b^2c \plus{} 2a^2b^2c^2)}}{(ab \plus{} bc \plus{} ca)\prod{((a \minus{} b)^2 \plus{} ab \plus{} bc \plus{} ca)}} \ge 0$,

clearly;

$ \sum{\frac {1}{(a \minus{} b)^2 \plus{} 1}} \le 3$

$ \iff \sum{\frac {1}{(a \minus{} b)^2 \plus{} ab \plus{} bc \plus{} ca}} \le \frac {3}{ab \plus{} bc \plus{} ca}$

$ \iff \frac {\sum_{sym}{(2a^5b \plus{} a^4b^2 \minus{} 4a^4bc \minus{} a^3b^3 \plus{} 6a^3b^2c \minus{} 4a^2b^2c^2)}}{(ab \plus{} bc \plus{} ca)\prod{((a \minus{} b)^2 \plus{} ab \plus{} bc \plus{} ca)}} \ge 0$,

by Muirhead Inequality, we have $ \sum_{sym}{(a^4b^2 \minus{} a^3b^3)} \ge 0$ and $ \sum_{sym}{(4a^3b^2c \minus{} 4a^2b^2c^2)} \ge 0$, so we just need to prove that

$ \sum_{sym}{(2a^5b \minus{} 4a^4bc \plus{} 2a^3b^2c)} \ge 0,$

by AM-GM, we have

$ \sum_{sym}{(2a^5b \plus{} 2a^3b^2c)} \ge \sum_{sym}{4a^4b^{\frac {3}{2}}c^{\frac {1}{2}}},$

so just need to prove

$ \sum_{sym}{a^4b^{\frac {3}{2}}c^{\frac {1}{2}}} \ge \sum_{sym}{4a^4bc},$

which is true by Muirhead Inequality.

done.

$ \sum_{cyc}\frac 1{1\plus{}(a\minus{}b)^2}\leq 3$ can be easily proved, since
$ \sum_{cyc}\frac 1{1\plus{}(a\minus{}b)^2} \equal{} \sum_{cyc}\left[1\minus{}\frac{(a\minus{}b)^2}{1\plus{}(a\minus{}b)^2}\right]$
$ \equal{}3\minus{}\sum_{cyc}\frac {(a\minus{}b)^2}{1\plus{}(a\minus{}b)^2}\leq 3$...
In order to prove $ \sum_{cyc}\frac1{1\plus{}(a\minus{}b)^2}\geq 1$, I cannot find an easy way... :oops:

[quote="xzlbq"]
[size=150]$ \frac {1}{(a \minus{} b)^2 \plus{} 1} \plus{} \frac {1}{(b \minus{} c)^2 \plus{} 1} \plus{} \frac {1}{(c \minus{} a)^2 \plus{} 1} \le 3$
BQ[/size][/quote]
$ LHS \le 1\plus{}1\plus{}1 \equal{}3$  :|


AMC
AIME
USA(J)MO
USAMO
geometry
AoPS Books

I took the AMC for the first time this year, and qualified for the AIME. I took the AIME and got utterly pwnt (I think i got a two).
I went into the AIME utterly blind. I barely even glanced at previous year's problems.

So, for next year, what would you suggest that I study?
As in which AoPS book(s)? Anything else worth my time?
 I have a copy of the Art and Craft of Problem Solving (by Paul Zeitz) which i inherited from my brother, although it seems more oriented towards  the USAMO and IMO.

I don't remember perfectly what I did. But I will at least outline a strategy which *I* think should be successful.

If you haven't already, AoPS Volumes I & II (available in this site's Bookstore) are very good, and have enormous popularity. Work through both of these meticulously, i.e. actually read and understand slowly; then work the problems at the end of the chapter (all, or at least most of them - but as we have a whole year to prepare now, you might as well do them all). Work the problems throroughly, be sure you are right, and then look at the proposed solutions. (That's one thing that many people mess up - you never nibble a problem, think "oh, I think I [i]would[/i] just do something like this," and then look at the solution - that teaches you nothing; work all problems (that you can) fully, and if you can't, do as much as you can before you become overly frustrated)

After you have worked through these, if you still have brain cells left, you can start focusing on subjects. Where do you feel weakest? Many people feel that they suck at geometry - so they pick up an Introduction to Geometry book, etc. After reading both the AMC books, you should probably have adequate knowledge to be working past year's AIME"s. Again, take these tests just as you took the tests in the AoPS books - work the problems thoroughly, and then check the solutions.

And you have correctly identified ACoPS by Zeitz to be more proof oriented - it is. I used it in preparation for the USAMO last year, and I must say that it is a great book - for USAMO.

So that's my two cents. Good luck! And if anybody disagrees with my approach (and would like to add to that hazy step at the end) feel free!


i dunno how to prove that the following limit exists:

$ \lim_{x\to0}\frac {e^x \minus{} x \minus{} 1}{x^2}.$

We have: $ \frac{e^{x}\minus{}x\minus{}1}{x^{2}} \equal{} \frac{1}{x^2}\left(\sum_{k \equal{}0}^{\infty} \frac{x^k}{k!} \minus{} x \minus{} 1\right) \equal{} \sum_{k \equal{}2}^{\infty} \frac{x^{k}}{k!} \equal{} 0$ as $ x \to 0$.

the limit is $ \frac{1}{2}$. It is found by L' Hospital rule

akech, you forgot to divide by $ x^2,$ and the limit is not zero.

I encourage my own students not to rely so much on formal sigma notation but just to write things out with ... . It's easier to see the pattern. That is, write,

$ \frac{e^x\minus{}x\minus{}1}{x^2}\equal{}\frac{1\plus{}x\plus{}\frac{x^2}2\plus{}\cdots \minus{}x\minus{}1}{x^2}.$

You should be able to look at this and see what the limit is.

yes, i've seen those solutions before, and there's another one which doesn't use power series.

since i know how to compute the limit, i wanted to know how would you guys prove that it does exist by not computing it.

$ L\equal{}\lim_{x\to0}\frac{e^x\minus{}x\minus{}1}{x^2} \equal{} \left(\frac{0}{0}\right) \\ \\
\text{Use l'Hopital's Rule for the Indeterminate form: } \\ \\
L \equal{}\lim_{x\to0} \frac{e^x\minus{}1}{2x} \equal{} \left(\frac{0}{0}\right) \\ \\
\text{Again use l'Hopital's Rule for the Indeterminate form: } \\ \\
\equal{}\lim_{x\to0}\frac{e^x}{2} \equal{} \frac{1}{2}$

[quote]i wanted to know how would you guys prove that it does exist by [b]not computing it.[/b][/quote]

If the limit starts off as an indeterminate form, then you cannot be certain if it exists or not until it is computed.

yes, that's true, i think we can make a conjecture by first assuming that the limit equals some number then proving it by definition.

[quote="Stephen"][quote="Rajiv"]$ L \equal{} \lim_{x\to0}\frac {e^x \minus{} x \minus{} 1}{x^2} \equal{} \left(\frac {0}{0}\right) \\
 \\
\text{Use l'Hopital's Rule for the Indeterminate form: } \\
 \\
L \equal{} \lim_{x\to0} \frac {e^x \minus{} 1}{2x} \equal{} \left(\frac {0}{0}\right) \\
 \\
\text{Again use l'Hopital's Rule for the Indeterminate form: } \\
 \\
\equal{} \lim_{x\to0}\frac {e^x}{2} \equal{} \frac {1}{2}$[/quote]

You are right, but first, I think that you must prove that the lim exists.(No?)[/quote]
No you dont .

you may want to look at L'Hospital again http://en.wikipedia.org/wiki/L%27Hospital_rule


The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

$ \textbf{(A)} \ 50 \qquad \textbf{(B)} \ 52 \qquad \textbf{(C)} \ 54 \qquad \textbf{(D)} \ 56 \qquad \textbf{(E)} \ 58 \qquad$

[asy]unitsize(3mm);
defaultpen(linewidth(0.8pt));

path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0);
path p2=(0,1)--(1,1)--(1,0);
path p3=(2,0)--(2,1)--(3,1);
path p4=(3,2)--(2,2)--(2,3);
path p5=(1,3)--(1,2)--(0,2);
path p6=(1,1)--(2,2);
path p7=(2,1)--(1,2);
path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7;
for(int i=0; i<3; ++i)
{
for(int j=0; j<3; ++j)
{
draw(shift(3*i,3*j)*p);
}
}[/asy]

[hide]looking at the big square, we see that it can be composed of $9$ smaller, unit squares. In the picture, we see that there are only $4$ unit squares. So, the area that the squares take up is equal to $\frac{4}{9}$ of the big square. So, the area that the pentagons take up is $\frac{5}{9}$ of the big square. Which is about $55.55555...$%, which is closest to $56$%, or $\boxed{D}$[/hide]

[hide="Answer"]The larger square is $3\times 3$, and each smaller one is $1\times 1$, so the area of the combined pentagons is $9-4=5$, and the percentage is $\frac{5}{9}=0.\overline{5}\Rightarrow \boxed{D}$.[/hide]

Yes, there are, but read the question again. It says, "The [i]percent of the plane[/i] that is enclosed by the pentagons is closest to..."
The pentagons do not have the same area as the squares :)

Ignoring the pentagons, and assuming the squares are $1 \times 1$, we can see that the square is a $9 \times 9$ square with area $9^2 = 81.$ Since the squares are unit and there are $36$ of them, the total area occupied by the squares is $\frac{36}{81} = \frac{4}{9}.$ So the remaining area is, of course, tiled by pentagons so the total area occupied by pentagons is $1-\frac{4}{9} = \frac{5}{9} \approx 56\% \Longrightarrow D.$

This question appears to be underspecified. Everyone else is assuming that the larger squares are $3\times3$, but they could be larger or smaller and the pentagons would still be congruent to each other. For example, if the larger squares are $3.1\times3.1$, then the pentagons (with side lengths $1, 1.1, 1, 1.1/\sqrt{2}, 1.1/\sqrt{2}$) would enclose $1-\frac{4}{3.1^2} = 0.5838$ of the plane and E would be correct. How are you sure that the bases of the pentagons are $1$ and not $1.1$?

@above then the smaller congruent squares wouldn't be squares, or congruent.

@thedoge: They would be squares of side $1$, hence congruent.

[quote=TheMaskedMagician]Why isnt the answer $50$? There are $36$ pentagons and $36$ squares.[/quote]
The question asks about percent area, not number of each.

but I agree that the problem is underspecified
unless it is explicitelty [u]written[/u] or coded that AB = BC the tiling could be as well, or any variant :

[quote=CantCount]Is this really a high school math problem? It looks like from Singapore's 4th grade fractions lesson.[/quote] I was thinking the same.

It's normal to have a few easy problems to start the AMC exams so that participating is not too discouraging for the majority of students who are just trying the exam. Some are easy because they only involve prealgebra while some are easy because they are not tricky at all, and some are easy both ways. For students who are likely to qualify for the AIME (as most of my students are), part of the challenge is to save time on these easy problems so that they can spend more time on problems 16-25.

By contrast, the easy Putnam problems are often out of reach of the newer competitors, and the median score is sometimes 0/120 which people find discouraging.


trigonometry
logarithms
function
inequalities theorems
inequalities

use
$ \prod_{n \in \mathbb{N}} (1 \minus{} \frac {\theta^{2}}{n^{2}\pi^{2}}) \equal{} \frac {\sin \theta}{\theta}$
take $ \ln$ on both the sides expand using the formula of $ \ln(1 \plus{} x),|x| < 1$ and $ \sin x$
Hence compare coefficients

[quote="sepehr ahmadzadeh"]now if you factor $ p(x)$ you get :
$ p(x) \equal{} (1 \minus{} \frac {x}{\pi})(1 \minus{} \frac {x}{ \minus{} \pi})(1 \minus{} \frac {x}{2\pi})(1 \minus{} \frac {x}{ \minus{} 2\pi})...$ [/quote]
Why not for example $ p(x) \equal{} e^x(1 \minus{} \frac {x}{\pi})(1 \minus{} \frac {x}{ \minus{} \pi})(1 \minus{} \frac {x}{2\pi})(1 \minus{} \frac {x}{ \minus{} 2\pi})...$ $ ?$ :wink:


limit
Functional Analysis
real analysis
real analysis unsolved

Suppose that $ X$ is a Banach space, $ x_1 , x_2 , x_3, ... \in X$ and $ x_1 \plus{} x_2 \plus{} ... \in X$. $ T$ is a bounded linear functional on $ X$.
Does the following equality holds?

$ T(x_1 \plus{} x_2 \plus{} ... ) \equal{} T(x_1) \plus{} T(x_2) \plus{} ...$

Thanks :)

Yes. Define $ x\equal{}\sum_{k\equal{}1}^{\infty}x_k$ and $ y_n\equal{}\sum_{k\equal{}1}^nx_k$ We are given that $ \lim_{n\to\infty}\|y_n\minus{}x\|\equal{}0.$ (We know this because there's no other thing that the notation $ x_1\plus{}x_2\plus{}\cdots\in X$ could mean.)

$ T(y_n)\equal{}\sum_{k\equal{}1}^nT(x_k)$ simply because $ T$ is linear.

But then $ |T(y_n)\minus{}T(x)|\le \|T\|\cdot\|y_n\minus{}x\|$ by the definition of "bounded." That implies that $ \lim_{n\to\infty}|T(y_n)\minus{}T(x)|\equal{}0.$

We interpret this as $ \sum_{k\equal{}1}^{\infty}T(x_k)\equal{}T(x)\equal{}T\left(\sum_{k\equal{}1}^{\infty}x_k\right).$


function
calculus
derivative
algebra proposed
algebra

Determine function derivative  $ f\colon \mathbb{I}\to (0;\plus{}\infty )$,$ I\in  R$,such that $ f(0)\equal{}1$,$ f^{3}(x)\plus{}f^{`}(x)\equal{}0$ .

_______________________________________
Azerbaijan Land of the Fire  :P

[quote="Pirkuliyev Rovsen"]Determine function derivative  $ f\colon \mathbb{I}\to (0; \plus{} \infty )$,$ I\in R$,such that $ f(0) \equal{} 1$,$ f^{3}(x) \plus{} f^{`}(x) \equal{} 0$ .[/quote]

$ f(x) \equal{} \frac 1{\sqrt {2x \plus{} 1}}$ but only if $ 0\in\mathbb I$ and $ (\minus{}\infty,\minus{} \frac 12]\cap \mathbb I\equal{}\emptyset$


group theory
abstract algebra
superior algebra
superior algebra unsolved

For n>= 5 prove that every subgroup of A_n of index n is isomorphic to A_n.

Are you sure? A subgroup U of index > 1 in a finite group G has less elements than G and hence can't U be isomorphic to G.

Sorry it should be isomorphic to  A_(n-1) .


Hexagon ABCDEF is inscribed in a circle such that $ AB\equal{}CD\equal{}EF$ and $ AD, BE, CF$ are concurrent.  Prove that $ AB^3\equal{}BC\cdot DE\cdot FA$.

Eh... even you said it yourself on the source... 'usamo'... it has 'mo' in it, this is an olympiad level problem. It doesn't belong here in Intermediate T, should be at least at the level of Pre-Olympiad or higher.

its like the last part of the problem so its like $ \frac{usamo}{2}$

[quote="jli"]Hexagon ABCDEF is inscribed in a circle such that $ AB \equal{} CD \equal{} EF$ and $ AD, BE, CF$ are concurrent.  Prove that $ AB^3 \equal{} BC\cdot DE\cdot FA$.[/quote]
[hide="Solution"]
Let the diagonals intersect at $ P$.  It follows that because $ ABCDEF$ is cyclic, $ \frac {AB}{DE} \equal{} \frac {BP}{PD}$, $ \frac {FE}{BC} \equal{} \frac {PE}{PC}$, and $ \frac {CD}{AF} \equal{} \frac {PD}{PF}$.  Hence,\[ \frac {AB\cdot FE\cdot DC}{DE\cdot BC\cdot AF} \equal{} \frac {BP}{PD}\cdot \frac {PF}{BP}\cdot \frac {PD}{PF} \equal{} 1\]so\[ AB^3 \equal{} AB\cdot FE\cdot DC \equal{} DE\cdot BC\cdot AF\]because $ AB \equal{} CD \equal{} EF$. [/hide]


Hi, is there a formula for sinA-cosB? If so, how's it derived? thanks!

[quote="ocha"]this should be HSB or something...
It's easy if you know double angle formula

$ \sin(a \plus{} b) \minus{} \sin(a \minus{} b) \equal{} (\sin(a)\cos(b) \plus{} \sin(b)\cos(a)) \minus{} (\sin(a)\cos(b) \minus{} \sin(b)\cos(a))$

$ \equal{} 2\sin(b)\cos(a)$

let $ a \equal{} \frac {x \plus{} y}{2}, b \equal{} \frac {x \minus{} y}{2}$

$ \sin(x) \minus{} \sin (y) \equal{} 2\sin(\frac {x \minus{} y}{2})\cos(\frac {x \plus{} y}{2})$

let $ y \equal{} \frac {\pi}{2} \minus{} z$

$ \sin (x) \minus{} \cos(z) \equal{} 2\sin(\frac {x \plus{} z}{2} \minus{} \frac {\pi}{4})\cos(\frac {x \minus{} y}{2} \plus{} \frac {\pi}{4})$[/quote]

right, right, i only knew sinA-sinB,sinA+sinB, cosA+cosB, cosA-cosB. thanks!:)


Well, in fact moroccan team had a problem in the plane reservation, so we were obliged to take the train to Algaziras and then the boat to Tangiers, Cool :D .

"nice place" it is too weak said,I would say one of the most beautiful places in the world...well...maybe just very beautiful...


probability...
1) A AND B PLAY A SERIES OF GAMES WHICH CANNOT BE DRAWN AND "p" AND "q" ARE THEIR RESPECTIVE CHANCES OF WINNING A SINGLE GAME.WHAT IS THE CHANCE THAT A WINS "m" GAMES BEFORE B WINS "n" GAMES?

2) A REAL ESTATE MAN HAS EIGHT MASTER KEYS TO OPEN SEVERAL NEW HOMES. ONLY ONE MASTER KEY WILL OPEN ANY GIVEN HOUSE. IF 40% OF THESE HOMES ARE USUALLY LEFT UNLOCKED, WHAT IS THE PROBABILITY THAT THE REAL ESTATE MAN CAN GET INTO A SPECIFIC HOUSE IF HE SELECTS THREE MASTER KEYS AT RANDOM BEFORE LEAVING THE OFFICE?

please help...............

Turn caps lock off.

1 is not well explained. Do one of $ A$ and $ B$ always have to win? Is $ p\plus{}q\equal{}1$? Defining two separate variables for their chance of winning kinda confuses things.

2 is the complement of the chance the home is locked times the chance he doesn't pick the correct key. The chance the home is locked is $ 3/5$. The chance he doesn't get the right key is $ 5/8$. So the answer is $ 1 \minus{} (3/5)(5/8) \equal{} \boxed{5/8}$.

[hide]
Why is the chance of not getting right key is $ \frac{5}{8}$ ?[/hide]


LaTeX ARML search AMC USA(J)MO USAMO induction

We're trying to compile a top 10 list for AoPS.

You know you're an AoPSer if . . .

(we'll be doing the same for ML soon).

.. you stay awake all night-long trying to improve the site's software :P ?

You learn to use  Latex with your paws...

... you wonder why LaTeX is sold in paint stores.

... you think the AMC is a math contest, not a movie channel.

... you ask your school whether the "art of problem solving" counts as the required art class.

[quote="Ragingg"]You spend more time on the AoPS forum than you do on your homework.
Your homepage is http://www.artofproblemsolving.com.
You dream about what other AoPSers look like.
You consider your primary language LaTeX.
Any time you hear people talk about "the art of" you think problem solving.
You classify every math problem you see into the basic, intermediate, or advanced forum.
You get more AoPS notifications than you do spam.
You'll skip your prom to go to the Olympiad Problem Solving Math Jam.[/quote]

These are all quality.

[quote="peter"][quote="Ragingg"]You get more AoPS notifications than you do spam.[/quote]
Is this possible? :?[/quote]Yup! :D

You're logged into AoPS on every computer in your school's labs.

You dread clearing your cookies because you will have to retype in your AoPS login.
You address AoPSers by their usernames... in real life.
You click reload while reading the forums hoping to find a new proof without words flash.
You know other AoPSers favorite factoring tricks.  :lol:

[quote="peter"][quote="Ragingg"]You get more AoPS notifications than you do spam.[/quote]
Is this possible? :?[/quote]

Used to happen to me back in the days I was a lot more active in posting

[quote="Ravi B"]... you wonder why LaTeX is sold in paint stores.

... you think the AMC is a math contest, not a movie channel.
[/quote]

First one, nice.

Second one, AMC has channels now? I thought they were still in the theatre-only days.

You pronounce sin like sign.
You expect "iff" to be in the dictionary.
It takes more time for your teacher to find a formula than it does for you to derive it.
Your idea of a perfect score is 42 points.
Success is measured by one's USAMO index.
The average PSAT score of your friends is 234.6.
You wonder why inductions have to be done through a ceremony.
The majority of your instant messenger contacts think of math as a hobby.
Chatrooms can be used for math discussions too.
Those 1 gigabyte inboxes just don't seem big enough since you added the "AoPS notifications" folder.
If someone says Cauchy, you say Schwartz.
Your favorite subject is Combinatorics.

...you tell your friends that your girlfriend's name is AIME.
...you've written a mock contest and posted it on the forum
...you resort to spamming to have a competitive post count
...you introduce yourself in public using your username
...you wear an AoPS sign to ARML
...you back up and raise your eyebrows whenever you see a gopher hole

[quote="MithsApprentice"][quote="Ravi B"]... you wonder why LaTeX is sold in paint stores.

... you think the AMC is a math contest, not a movie channel.
[/quote]

First one, nice.

Second one, AMC has channels now? I thought they were still in the theatre-only days.
[/quote]
I was referring to the [url=http://www.amctv.com]American Movie Channel[/url].  I guess that one wasn't very funny.

Ah...I guess the American Movie Channel is just not popular down south. But if we change "channel" to "theatre" above, it works splendidly.

...You know how to use more than 4 functions on your graphing calculator.

[quote="perfectnumber628"]...whenever you see a decimal representation of $\pi$, you read each digit to make sure nobody's trying to pass off an incorrect approximation[/quote]

Yeah, I have a friend who claims to know like... 200? 300? 800? digits of pi, but I could've sworn he said something like 3.1415926535897932sday3846...

[quote="B-tone_Party"][quote="shadysaysurspammed"]If your teacher says 'art of solving problems',you start sweating because you want the site to be a secret from others  and you randomly change the topic:D[/quote]

My personal philosophy is "Spread the AoPS Love"- I've already introduced three people to this site.[/quote]

Not here brother,with the greed and competition,naha.My friend and i kept this site a secret since last year,he just told one guy who never uses uses it and now all my friends have gone to a place for coaching where they have no computers at all :lol: So the secret is safe!!Only I and my friend used it and now only me.

...you have artofproblemsolving.com set as your home page
...you hate it when people think pi is "3.141592654" because of the rounding on their calculator
...your screen name contains a math reference
...you celebrate at least 3 math-related holidays

Edit:  Wow.   I just realized that half of those are on the t-shirt.  That doesn't make them anything less true, though.

[quote="perfectnumber628"][quote="B-tone_Party"]...you hate it when people think pi is "3.141592654" because of the rounding on their calculator[/quote]
semi-related side comment: every week or so, my math teacher gets some math related cartoon to hang on the door; this week's said something corny about pi and pie, which apparently equals 3.1415926535898  :huh:

come on, you DON'T round pi like that.[/quote]
I agree.  That's just horrible.

My sixth grade math teacher used to insist that we use pi=3.14 :roll:

...if you've worked on a single math problem for more than a year

...you've gotten a math-related injury at a math competition (true story)

When you spend longer trawling through the search function to make sure your question hasn't already been answered than actually posting your question.

[quote="perfectnumber628"]you've used latex to write something for school[/quote]

Oohoohooh! I did that on a math assignment... but then I ended up spending half an hour translating it back into ugly notation because I didn't think my math teacher had a LaTeX compiler... :mad: :D

-if, in the set of signatures at this forum that contained maths, you've a personnel collection of more than 30% of them, or you memorized more than 10 of those.
-if when you use your calculator, you solely use the special function keys to do [b]all[/b] calculation(i.e. not using key0-9 and/or +-*/ at all).
-or even further, if you write propositions on representation of intergers using special function and 0(which is the default number when you turn a calculator on) alone.
-...or the default number that shows up when you turn on your calculator isn't 0 ?!
-if you routinely do problems on AoPS at the class, and after solving a (difficult) problem your maths teacher ask you to do a simple integral on the blackboard... and you failed to do it!
-or that doesn't matter now, because the problem you've done interested your teacher so much that he abadoned the schedule and discuss that problem with the class instead.

...when aops gets a new server you notice about 407 things different

and then when it says the website is having technical difficulties and you can't view anything or post anything, you feel helpless and lost  ;)


Given $ x$ varies directly as $ y^2$ and $ x \equal{} 3$ when $ y\equal{}5$, find $ x$ when $ y \equal{} 6$. Express your answer as a common fraction.

[quote="GameBot"]Given $ x$ varies directly as $ y^2$ and $ x \equal{} 3$ when $ y \equal{} 5$, find $ x$ when $ y \equal{} 6$. Express your answer as a common fraction.[/quote]
Set 'x' equal to '$ y^2$'.
$ ax \equal{} y^2$
Input y/x values to find the constant of variation.
$ 3a \equal{} 5^2 \equal{} 25$
$ a \equal{} \frac {25}{3}$
Therefore, the constant of variation is $ \frac {25}{3}$. Input 6 for 'y', and solve for 'x'.
$ \frac {25}{3}x \equal{} 6^2 \equal{} 36$
$ x \equal{} \frac {3}{25} \times 36 \equal{} \frac {108}{25}$.
Therefore, $ \boxed{\frac {108}{25}}$.


number theory
greatest common divisor
number theory unsolved

Denote by d the gcd of the coefficients of (a+b)^n except the first and the last . prove that d=p if n is the power of a prime p , and d=1 otherwise

[quote="lajanugen"]Denote by d the gcd of the coefficients of (a+b)^n . prove that d=p if n is the power of a prime p , and d=1 otherwise[/quote]
$ gcd(C_n^0,C_n^1,...,C_n^n)\equal{}1$, because $ C_n^0\equal{}C_n^n\equal{}1$
$ gcd(C_n^1,...,C_n^{n\minus{}1})\equal{}\binom{1,n\not\equal{}p^k}{p,n\equal{}p^k}$ posted before.


inequalities
calculus
integration
function
complex analysis
complex numbers
absolute value

Show that for all integers $n \geq 0$ the inequality:

${| e^{z}-\sum_{i=0}^{n}\frac{z^{i}}{i!}}| \leq \frac{|z|^{n+1}}{(n+1)!}$

holds for all complex numbers $z$ such that $\Re z \leq 0$.  Conversely, show that if for a given $z$ the above inequality holds for infinitely many positive integers $n$, then $\Re z \leq 0$.

P.S.  I do not enjoy these types of problems...  :(

Well, this has almost nothing to do with complex analysis. Using the Taylor formula with the integral form for the remainder, we conclude that the difference on the left equals $\frac{z^{n+1}}{n!}\int_{0}^{1}e^{tz}(1-t)^{n}$, so the first part of the problem just asks to show that $\text{Re\,}z\le 0$ implies that the integral is not greater than $\frac 1{n+1}$ in absolute value, which is obvious since $|e^{tz}|\le 1$. The second part is more interesting, so I leave it to somebody else ;)

Fedja, thanks for the quick reply.  I do not disagree with you that the first part of the problem is easy.  I just included the whole statement of the problem for completeness.  For what its worth, the problem does come from an old complex analysis exam...

[quote="edoo"]Show that for all integers $n \geq 0$ the inequality:

${| e^{z}-\sum_{i=0}^{n}\frac{z^{i}}{i!}}| \leq \frac{|z|^{n+1}}{(n+1)!}$

holds for all complex numbers $z$ such that $\Re z \leq 0$.
[/quote]
Fedja gave you one answer. Here is how I would do it since I dont like the Integral form of the Remainder.

[b]Taylor Theorem (Lagrange Form):[/b] Given an infinitely differentiable function on $(a,b)$ with $a<0<b$. Given $T_{n-1}(x)$ as its $n-1$ degree Taylor polynomial. Given the remainder $R_{n}(x) = f(x)-T_{n-1}(x)$. Then if $x\in (a,b)$ and non-zero we have that $R_{n}(x) = \frac{f^{(n)}(y)}{n!}x^{n}$ for some $y$ strictly between $0$ and $x$ (between the annuli). $\blacksquare$

We note that $\sum_{j=0}^{n}\frac{z^{j}}{j!}= T_{n}(z)$. So the remainder depends on $R_{n+1}(z)$. By Lagrange's form of the remainder we see that:
$R_{n+1}(z) = \frac{f^{(n+1)}(y)}{(n+1)!}z^{n+1}= \frac{e^{y}}{(n+1)!}z^{n+1}$ for $y$ between the annuli of $z$ and $0$.
Now just find an upper bound on this error term.
$|R_{n+1}(z)| \leq | \frac{z^{n+1}}{(n+1)!}| = \frac{|z|^{n+1}}{(n+1)!}$ since $|e^{y}| \leq 1$ because $\mathfrak{Re}(z)\leq 0$

[quote="fedja"]Are you sure that there is such a thing as "remainder in the Lagrange form" for complex-valued functions? ;)[/quote]
No  :rotfl:
I never even seen a remainder term for the Taylor series in my Complex Analysis book, because its pretty easy to show the holomorphic functions always have a Taylor series so it doesnt use it.

But I did think about whether I can do what I did. I wasnt afraid to take a gamble.

Actually for the first part of the problem, I argued like Fedja using an integral form of the remainder.  But Sylow's error made me question that approach.  Looking at Wikipedia, I see that the Complex Integral remainder is of the following form:

$R_{n}(x)=\frac{1}{2 \pi i}\oint_{C}\frac{f(z)}{(z-a)^{n+1}(z-x)}dz$ where $C$ is a circle surrounding $a$.

This may mean that I got more than I bargained for with this question.  Sometimes the easier problems are more difficult than they appear.   :)


Find the area of the region between y=x^2-2x-3 and the x-axis.

Isn't this simply a definite integral? Since $ x^2\minus{}2x\minus{}3\equal{}(x\minus{}3)(x\plus{}1)$, we have

$ \int_{\minus{}1}^{3} x^2\minus{}2x\minus{}3 dx$.

Evaluate those results, and subtract the second $ (x\equal{}\minus{}1)$ from the first $ (x\equal{}3)$. (Somebody correct me if I'm wrong).


Solve for all real x: (2x^2  x +4)^2  6(2x^2  x) = 19.

MY bad...

Isn't it just straightforward algebra?

the answer I'm getting...

[hide]unless there's a trick the answer I'm getting is -3[/hide]

To solve this one:

[hide]let 2x^2-x=y.

Then, you get a quadratic for y.

All downhill from there[/hide]


mods, maybe you should move this to the intermediate topics place.

[quote="mgao"]$ \frac{4z}{2x-1}+\frac{1}{z}=\frac{4x}{2y-1}+\frac{1}{x}=\frac{4y}{2z-1}+\frac{1}{y}$
$ x+y+z=0\Longrightarrow$ $ x+y=-z$,$ y+z=-x$,$ x+z=-y$.
Find $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
(I'm not even sure if this really has one solution)[/quote]
Actually here is the problem:
$ \frac{1}{z}-\frac{4z}{2x-1}=\frac{1}{x}-\frac{4x}{2y-1}=\frac{1}{y}-\frac{4y}{2z-1}$
$ x+y+z=0\Longrightarrow$ $ x+y=-z$,$ y+z=-x$,$ x+z=-y$.
Solve for $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.


AMC
AIME
USA(J)MO
USAMO
ARML
probability
AMC 12

I just found out about the Olympiad tests (amc, aime, etc.). I was wondering if the online classes offered at AoPS about the amc and aime are good. Did anyone take them? Was it helpful(as in improvements on your scores)? Which class(es) would you recommend to someone who wants to qualify for USAMO and currently can only do about a few of the AMC 12 problems and almost all till the last amc 10 problems?

...AMC 12 Problem Series, and AIME Problem Series after that, I guess...

I took AMC 12 Problem Series, and I thought it was good.  If you can only solve a few of the problems on the AMC 12, then you should consider taking the AMC 10 Problem Series, even if you're over 10th grade.  Last year, they had a Math Jam about the two AMC problem series, so you should go to that to find out more.

As for improvements, well... I did improve by a lot on the AMC this year, but there were many, many reasons that I improved besides just the problem series - AoPS books, local math contests, ARML practices, this site, etc.  Don't expect the problem series alone to raise your score by more than about 10 points.  Going from a few problems on AMC to USAMO qualification is a huge jump.

They don't have AMC-12 problem solving series.

My question:
If you did several things to prepare(AoPS books, local math contests, ARML practices, this site, etc), do you feel that if you didn't take the classes you'd have still made it (with no extra things added)?

Also, do you feel the classes a good value? Information wise and finantial wise? Do they pack a lot of helpful information into a small, cheap but qualitied, package?

I can attest that AoPS's classes' "financial value" are among the highest. Not only are the prices for the classes extremely small compared to other online offerings, the material is also unique, as nowhere else can you find so many courses dedicated to solely problem solving.

[quote="woot45"]My question:
If you did several things to prepare(AoPS books, local math contests, ARML practices, this site, etc), do you feel that if you didn't take the classes you'd have still made it (with no extra things added)?

Also, do you feel the classes a good value? Information wise and finantial wise? Do they pack a lot of helpful information into a small, cheap but qualitied, package?[/quote]

I would still have made what?  I definitely would have qualified for AIME even if I hadn't taken the AMC 12 Problem Series.  But the AMC 12 Problem Series helped me learn how to solve some of the more difficult AMC problems and some easy AIME problems, so it did help me qualify for USAMO.

I think the courses are pretty good and not very expensive.
I plan to take an AoPS class to prepare and study really hard in this summer in order to make the USAMO next year. So do you have any suggestion about what online class should I take that properly would improve my score? Im really bad at counting and probability (i couldnt not do any problem of this type in AMC this year), but I also want to take AIME series since my other fields are not very strong in order to make USAMO though (I didnt pass the AMC, but I can solve like 5 problems in the AIME when I take it at home.) I failed the AMC 10 this year with stupid mistakes and counting/probability problems, so I was advised to take the AMC 10 series. So Im totally confused now. What do you think ? I can afford money for 1 class only. :-(
Thanks in advance.

In my opinion, a subject class will help you more than a problem series.  I would take the Intro or Intermediate Counting and Probability course.

I'm pretty sure I can attest to the effectiveness of the classes :).  I'm taking the AIME problem series class, and I've already learned a number of methods I'd never thought of.  Then again, we've really only had one class so far (imagine what it'll be like 10 weeks from now!).

If you don't want to spend the money on classes, you can always study on your own. If you can only get a few AMC12s then print them out from the contests section and spend some hours doing them. When you can get 120 or better yet 130+ then try doing AIME's.

Also have the AoPS vol 1 and 2 (but itll cost ya but VERY well worth it).

Join your regional ARML, do problems from the problem solving board here. You can learn from other peoples' solutions too.

My opinion is that if you don't have the money for both, get the AoPS books before you enroll in classes.


algebra
polynomial
calculus
derivative
function
calculus computations

How can I use the "Mean value Thereom " in Calculus to prove that

[color=red]A polynomial with real coefficients has at most n real roots.[/color]

Thanks in advance.

When a polynomial with degree $ n$ has $ m > n$ roots, how many roots does the derivative of $ f$ have? Use the intermediate value theorem.

Maybe it's useful to prove, via Mean Value Theorem, that between two consequential roots of the derivative $ f^{\prime}$ of a differentiable function $ f$, there is, at most, one root of $ f$.
If $ p(x)$ is a polynomial of degree $ n$, then the $ (n \minus{} 2)$-th derivative of $ p(x)$ is $ p^{(n \minus{} 2)}(x) \equal{} a_2{x^2} \plus{} a_1{x} \plus{} a_0$, which has, at most, 2 real roots, say $ \rho_1 < \rho_2$. So, according to the proposition above, the $ (n \minus{} 3)$-th derivative has at most one root  in the interval $ (\rho_1,\rho_2)$.
It remains, but i think that's easy, to prove that in the intervals $ ( \minus{} \infty, \rho_1)$ and $ (\rho_2, \plus{} \infty)$ there are 2, at most, roots of $ p^{(n \minus{} 3)}(x)$.
And so on..

EDIT [ P.S. The solution that suggests PeterP it is more direct. We just post our solutions at the same time. ]


If $a$ is in the interval $[m,n]$, prove that there exist two positive reals $s$ and $t$ such that $s + t = 1$ and $a = sm + tn$.

this problem looks obvious at first but looking closer, it is not only saying, one can see that it is true. nevertheless heres a solution:
[hide]
$sm+tn=sm+tm+tn-tm=m+t(n-m)$
now we know that $t\in[0,1]$ and $n-m$ is the length of the interval. so we can add every fraction of the difference by choosing $t$. hence $sm+tn$ can represent every $a\in[m,n]$
[/hide]
peeta


<----

I play on KGS too. This should probably be moved to High School -> Other Problem Solving Topics -> Math & Strategy Games

I started in late october (my friend got me into it) and I now play on KGS from time to time (though I rarely get time to do so).  I'm right now rated at an 18kyu there, but I don't know what my actual rating would be.

I guess I'm attracted by the sheer simplicity of the rules, and the idea of balance, and having to make open-ended decisions.

I have never actualy played go, owever I am hoping to begin playing at some point.(getting a set for christmas!)

[quote="curl"]Are there any other Go players here? I started this past september, and since then I've been playing fairly often. It's a seemingly simple but very intense strategic game. Some say a Go game is like a conversation.

Anyways, if anyone is interested, we can play on KGS!  :)[/quote]

go baduk!

I play Go a little bit from time to time. I first started playing at Chinese school, where I took a class in it, though I didn't become very good at it  :lol: Mainly played on the $9\times9$ boards, since there wasn't enough time for games on the regular $19\times 19$.

Anyone play on Yahoo?

I've played Go once, and I sucked, but my cousins are all good. I usually play chess because I'm more familliar with the game.

jka did you revive a dead topic? :huh:

I had been playing it very actively once, about four years ago, in a local go club. I even participated in a couple of tournaments in my country. I stopped because the club closed and I didn't have much time for active practice as well  :(  The last time I played, I was ranked 17-18 kyu (now I'm probably back to 24...)

Go is a great game. I play on tom go. :) I used to go to a local club, but got bored when i totally pwn their "2d"s.

Yip! Go is fun but challenging. I went to a class until we moved away. TT They had comics, I tell you! Comics! And right now this boy younger than me in my neighberhood could probably own me. Oh well. That's the way it is. *sigh*

... i play go, inspired by Hikaru no GO XD lol... ne ways, i play on kgs, rated 7k, and username: hikaru999.  Message me on there if you want to play

I play go as well.... though I haven't played in quite a while... I find go a very challenging and unique game...  maybe I'll get back into it again...  :lol:

i used to play go a lot.  my highest ranking was 4k.  how about u guys?

It is a board game with black and white pieces, with the objective to surround and "capture" your opponent's color.

I play go. Although not much at all nowadays.

I am the 2001 Canadian Go champ, believe it or not :P
Could've represented Canada at the World Ing Cup, but I wasn't a Canadian citizen so I couldn't =(


Extensive results, not just #1 individual and #1 state.

I'm pretty sure they have it on the website, because I've seen it before.  Browse around and you might get lucky  ;) .

Do you want the list of Nats-goers? Or the top 50 at nats? http://www.mathcounts.org/webarticles/articlefiles/536-2005_Natl_Comp_Rankings.pdf is the 05 Nat rankings for the top 50
http://www.mathcounts.org/webarticles/articlefiles/536-State%20Winners%20for%20Web%20PDF.pdf is the list of State Winners, in somewhat random order.

If you search for other such pages, I know that at least 2003,4,5 are on the internet, and I think 02 is too.

On the MATHCOUNTS website, you can find fall newsletters from years back really far. The results (top 10) are on them

Or,
     You can google "finalists for XXXX MATHCOUNTS competition" and it should come up with something.

i got no results for http://www.google.com/search?hl=en&q=%22finalists+for+XXXX+MATHCOUNTS+competition%22+&btnG=Google+Search

What I really want to know are the scores... i just got a 31 on the 2004 competition (17/7) and I think that would put me in the top 50, but I'm not sure. Trouble is, I'm probably not even going to make nationals this year with Indiana's difficulty, ugh.

Yeah, that would put you somewhere around 42nd. I should know, that's what I got on it when I was actually in the competition. :P

there should be I saw one last year :P it's on their site:lol:

If it was on the Mathcounts site, then check http://www.archive.org .

It has archives of lots of sites listed by date.


integration
logarithms
trigonometry
calculus
function
calculus computations

Integrate $ x^2/(1\plus{}x^n).$
Try the particular case with $ n\equal{}5.$

Are you asking for a closed form or what?

$ I_n \equal{} \int \frac{x^2}{1\plus{}x^n} dx$

$ I_1 \equal{} \int \frac{x^2 \plus{}x \minus{}x \minus{}1 \plus{}1}{1\plus{}x} dx \equal{} \int x \minus{} 1 \plus{} \frac{1}{1\plus{}x} dx \equal{} \frac{x^2}{2} \minus{} x \plus{} \ln (1\plus{}x) \plus{}C$

$ I_2 \equal{} \int \frac{x^2 \plus{}1 \minus{}1}{1\plus{}x^2} dx \equal{} \int 1 \minus{} \frac{1}{1\plus{}x^2} dx \equal{} x \minus{} \tan^{\minus{}1} x \plus{}C$

$ I_3 \equal{} \int \frac{x^2}{1\plus{}x^3} dx \equal{} \frac{1}{3} \ln (1\plus{}x^3) \plus{}C$

$ I_4 \equal{} \int \frac{x^2}{1\plus{}x^4} dx\equal{} \int \frac{ (u\minus{}1)^2}{1\plus{}(u\minus{}1)^4}\equal{}\int \frac{u^2 \minus{}2u \plus{}1}{u^4\minus{}4u^3\plus{}6u^2\minus{}4u} du$

$ \equal{} \int \frac{u^2\minus{}2u\plus{}1}{u(u^3\minus{}4u^2\plus{}6u\minus{}4)}dx \equal{} \int \frac{u^2\minus{}2u\plus{}1}{u(u\minus{}2)(u^2\minus{}2u\plus{}2)}$

$ \equal{}\int \frac{1}{20(u\minus{}2)} \minus{} \frac{1}{4u} \plus{} \frac{2(u\minus{}1)\minus{}1}{10((u\minus{}1)^2\plus{}1)} du$

$ \equal{}\frac{\ln (u\minus{}2)}{20}\minus{}\frac{\ln u}{4} \plus{} \ln((u\minus{}1)^2\plus{}1) \minus{}\frac{\tan^{\minus{}1}(u\minus{}1)}{10}$

$ \equal{}\frac{\ln (x\minus{}1)}{20}\minus{}\frac{\ln (x\plus{}1)}{4} \plus{} \ln(x^2\plus{}1) \minus{}\frac{\tan^{\minus{}1}x}{10}\plus{}C$

$ I_5 \equal{} \int \frac{x^2}{1\plus{}x^5} dx \equal{} \int \frac{ (u\minus{}1)^2}{1\plus{}(u\minus{}1)^5} dx \equal{} \int \frac{u^2 \minus{}2u \plus{}1}{u^5\minus{}5u^4\plus{}10u^3\minus{}10u^2\plus{}5u} du$

$ \equal{} \int \frac{u^2 \minus{} 2u \plus{}1 }{u (u^4\minus{}5u^3\plus{}10u^2\minus{}10u\plus{}5)}du$ if you can factor it you can do the same thing?

hello, Maple 12 gives the following answer
$ \int\frac {x^2}{1 \plus{} x^n}dx \equal{} \frac {1}{3}x^3hypergeom\left(\left[1,\frac {3}{n}\right],\left[1 \plus{} \frac {3}{n}\right], \minus{} x^n\right)$.
Sonnhard.

i bet i partial fractioned wrong or something.. but what is hypergeom =(

for n=6 i think the integral is a substitution in the antiderivative of arctan. Expressing such things in elementary functions is in general no easy task, or not possible. An interval might have been nicer.

[quote="kenn4000"].. but what is hypergeom =([/quote]

[url=http://mathworld.wolfram.com/HypergeometricFunction.html]hypergeometric function in Mathworld[/url]

[url=http://mathworld.wolfram.com/EulersHypergeometricTransformations.html]EulersHypergeometricTransformations in Mathworld[/url]

and how we calculate the simple case [int] x/(1+x²) dx ?

hello, do you mean $ \int \frac{x}{1\plus{}x^2} dx$?
Sonnhard.

We can always integrate $ \int \frac {x^m}{1 + x^n} \, dx$ in terms of elementary functions.

We can even find an explicit form. Let $ m, n$ be integers such that $ 0 \leq m < n$. Put $ \theta_k = \tfrac{2k - 1}{n} \pi$, and

$ A_k (x) = \begin{cases} \tan^{ - 1}\left(\frac {x - \cos \theta_{k}}{\sin \theta_{k}} \right) & \sin \theta_k \neq 0 \\
0 & \sin \theta_k = 0 \end{cases}$

$ B_k (x) = \frac {1}{2} \log \left(x^{2} - 2 \cos \theta_{k}x + 1\right)$.

Then we have

$ \int \frac {x^{m}}{1 + x^{n}} \, dx = \frac {1}{n}\sum_{k = 1}^{n} \sin (m + 1) \theta_{k} \, A_k (x) - \cos (m + 1) \theta_{k} \, B_k (x)$

This comes from the partial fraction expansion: $ \frac {x^m}{1 + x^n} = - \frac {1}{n} \sum_{k = 1}^{n} \frac {e^{i(m + 1)\theta_k}}{x - e^{i\theta_k}}$.


A rectangular box $ P$ is inscribed in a sphere of radius $ r$. The surface area of $ P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $ r$?

$ \textbf{(A)}\  8 \qquad \textbf{(B)}\  10 \qquad \textbf{(C)}\  12 \qquad \textbf{(D)}\  14 \qquad \textbf{(E)}\  16$

[quote="JesusFreak197"][hide="Answer"]$4(x+y+z)=112\Rightarrow x+y+z=28$
$2(xy+xz+yz)=384$
$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=784\Rightarrow x^2+y^2+z^2=400$...[/hide][/quote]

AGH! I got that far, but didn't realize that $r=\sqrt{x^2+y^2+z^2}/2$ :mad:

btw, can someone explain why that [i]is[/i] true? :?

The box is inscribed in the sphere, so all vertices touch the sphere. Any space diagonal of the box is a diameter of the sphere, so you find the space diagonal then divide by two.

Oh thanks. I should think sometimes. :wallbash_red:


Prove that, if you double the units digit of a square, you don't get another units digit of a square unless the units digit of the original square is a multiple of 5.

[hide]The only possible units digits of any square are 1, 4, 9, 6, 5, or 0 and the only two numbers that can be doubled to get another number on the list are 5 and 0. [/hide]

[hide]Quadratic residues mod 10 are $0,1,4,9,5,6$. Doubling each gives $0,2,8,8,0,2$, and only if the units digit is $0$ or $5$ (thus a multiple of 5) is it possible to get a square.[/hide]


Solve this equation in integer set:        \[ \frac {x^7 \minus{} 1}{x \minus{} 1} \equal{} y^5 \minus{} 1\]

It's ISL 2006, see for example [url=http://www.mathlinks.ro/viewtopic.php?p=780855#780855]here[/url]. :wink:


What is the area of the region bounded by x+y=32, 2x+3y=42, and 3x-y=21?
Please explain your answer.

graph, separate into right triangles with lattice vertices, add/subtract
i'm too lazy to solve right now though

hello, solving the system $ x+y=32,2x+3y=42$ so we get say $ A(54,-22)$ analogously we get $ B(53/4,75/4)$ and $ C(105/11,84/11)$, now we compute the sidelength of the triangle $ \Delta ABC$, namely
$ \overline{AB}=\frac{163}{4}\sqrt{2}=c$,$ \overline{BC}=\frac{163}{44}\sqrt{10}=a$,$ \overline{CA}=\frac{163}{11}\sqrt{13}=b$ so we get finaly after an ugly computation the area of our triangle as
$ {A={\frac {26569}{7744}}\,\sqrt { \left( 11\,\sqrt {2}+\sqrt {2}\sqrt {5}
+4\,\sqrt {13} \right)  \left( -11\,\sqrt {2}+\sqrt {2}\sqrt {5}+4\,
\sqrt {13} \right)  \left( 11\,\sqrt {2}-\sqrt {2}\sqrt {5}+4\,\sqrt {
13} \right)  \left( 11\,\sqrt {2}+\sqrt {2}\sqrt {5}-4\,\sqrt {13}
 \right) }}$
Sonnhard.

I guess I would find the pointsof intersection and then use the shoelace algorithm to find the area

:O
Um, thanks for the explanation, but if this was a MC problem, what would be a fast and easy way to do it?

Well you could find the intersection points of each of the 2 lines the use shoelace to find the area but there really isn't an easy way

Could you explain the "shoelace"?
Thanks.

The bottom of [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=287748]this page[/url] details how to use the Shoelace Theorem.


hi, i'm stuck in this question:
15. how many different planes are determined by pairs of four different lines AP, BP, CP,and DP no three of which are coplanar?

my answer was 4 but the answer in the book is 6 and i dunno why..

my answer was i put point A, point B, point C, point D in 4 planes...the 4 planes intersect and the intersection is a line... the line has the point P... through two points there is a line.

am i wrong??

hello
sorry but u r wrong.
no need to think much
it is C(4,2)=6.
every plane should have two lines .So u just have to choose the two lines out of four.
Thank u.

so you're saying that because there is an intersection of 2 lines, then there is exactly one plane that contains the lines...and there is an intersection of 4 lines, so one plane contains 2 and another contains 2 and we have 2 more planes to make it into a total of 6 planes?

i don't understand what you mean by C(4,2)

thank you for helping

hello
every plane should contain any two lines
1 plane :APB which have line AP and BP.
2 plane :APC
3.plane APD
4.plane BPD
5.plane BPC
6.plane CPD

C(4,2) :: it is $ \frac{4!}{2!* 2!}$ = 6.
I hope it would be clear now.
Thank u.

[quote="kabi"]hello
every plane should contain any two lines
1 plane :APB which have line AP and BP.
2 plane :APC
3.plane APD
4.plane BPD
5.plane BPC
6.plane CPD

C(4,2) :: it is $ \frac {4!}{2!* 2!}$ = 6.
I hope it would be clear now.
Thank u.[/quote]
ur wonderful...thanks a lot  :)


There is a 7x7 board of square units. Each square is colored either white or black. A square may be white only if it is at the edge of the board, or if it is adjacent to a black square. Find the minimum number of black squares.

Simple casework and common sense gives...

[hide]
adjacent = share a side
4? (at the squares diagonally adjacent to the center square)

adjacent = share a point
7??

[/hide]

 :maybe: Feels like MATHCOUNTS.

Ok sorry for this bad post.


This thread will be discussing fibonacci sequence identities and all that. But here is the catch: you must quote the previous post.

Edit: Ignore the quoting rule, or it will be spam.

Wait, spam disguised as math? Remove the spam part and this thread is fine. Otherwise, it will most certainly be locked within the next few days.

EDIT: On second thought, it won't really be fine, since it will still either

1. Die

or

2. Be closed

The thread doesn't really have much point and does not really have a topic. People don't exactly go, "Hey, did you know this identity? Haha, yeah, it's cool." If they want help proving an identity, it should probably not go in FF, and the answer can probably be found online.

Okay, sure. Read the edit: deleting a rule.

Shouldn't this be moved now that it's just math and not spam?

Okay, you are right. I will ask the mod to move it to the appropriate forum.


$p > 3$  is a prime. Find all integers $a$, $b$, such that $a^2 + 3ab + 2p(a+b) + p^2 = 0$.

Treating the equation as a quadratic in $a$,
\[ a=\frac{-3b-2p\pm\sqrt{9b^2+4pb}}{2}, \] so $9b^2+4bp=m^2$ for some integer $m$.  Then
$81b^2+36bp+4p^2=9m^2+4p^2$, so
$(9b+2p)^2=(3m)^2+(2p)^2$.  Hence $\{3m,2p,9b+2p\}$ is a solution to the Pythagorean equation, so $\{3m,2p\}=\{u^2-v^2,2uv\}$ for some integers $u$, $v$, with $u^2+v^2=\pm(9b+2p)$.

In fact, $2p\equiv 2 \text{ (mod 4) }\not\equiv u^2-v^2$, so $2p=2uv$ and $u=\pm p$,$v=\pm 1$.  Hence
\begin{eqnarray*} m&=&\frac{u^2-v^2}{3}=\frac{p^2-1}{3}\\ b&=&\frac{\pm(u^2+v^2)-2p}{9}=\frac{\pm(p^2+1)-2p}{9}=\left(\frac{p-1}{3}\right)^2 \text{ or } -\left(\frac{p+1}{3}\right)^2 \end{eqnarray*}
If the former,
\begin{eqnarray*} a&=&\frac{-3\frac{p^2+1-2p}{9}-2p\pm\frac{p^2-1}{3}}{2}=\frac{-1-2p}{3} \text{ or } \frac{-p^2-2p}{3}\\ \end{eqnarray*}
If the latter,
\begin{eqnarray*} a&=&\frac{3\frac{p^2+1+2p}{9}-2p\pm\frac{p^2-1}{3}}{2}=\frac{p^2-2p}{3} \text{ or } -\frac{1-2p}{3}\\ \end{eqnarray*}
To make these integers we pick the first set of solutions if $p\equiv 1 \text{ (mod 3)}$, the second set otherwise.

[hide]there is another way too
$b^{2}-ab=(a+b+p)^{2}$
or,   (a+2b+p)(-a-p)=ab
or,    by comparing and taking cases we see that there are infinite solutions to this equation
of the form (a , b , p)  =  (-p , 0 , p)
where p is any prime >3
[/hide]

No need for quadratics, nor Pythagorean triples. Write the equation as $b(3a+2p) + (a+p)^2 = 0$. Since $p>3$, we have $3a+2p\neq 0$ (but even if we had $p=3$, then $3a+2p = 0$ will imply $a=-2$, which does not satisfy). So we need $3a+2p \mid (a+p)^2$.

1. One possibility is $3a+2p = \pm 1$, so $a = \frac {-2p \pm 1} {3}$, according to the residue of $p$ modulo $3$ (no solution for $p=3$).

2. If $|3a+2p|>1$, let a prime $q$ be such that $q\mid \gcd(3a+2p, (a+p)^2)$; then $q\mid a$ and so $q\mid p$, therefore $q=p$. So $a = kp$, and we need $3k+2 \mid (k+1)^2p$. Since $\gcd(3k+2,k+1) = 1$, it follows $3k+2 \mid p$.
One possibility now is $3k+2=-1$, whence $k=-1$, and so $a=-p$ (leading to $b=0$), the infinite family of solution found by [b]rohitx007[/b] (who missed however all the other), but which family was missed by [b]Agrippina[/b]). Otherwise, we need $3k+2 =\pm p$, so $k = \frac {-2 \pm p} {3}$, according to the residue of $p$ modulo $3$ (no solution for $p=3$), thus $a = \frac {-2p \pm p^2} {3}$, which together with the solutions at point 1. were duly found by [b]Agrippina[/b].

[b]A nice approach[/b]

\[a^2+3ab+2p(a+b)+p^2=0\]
\[\Rightarrow\ b=-(p+a)^2/(3a+2p)\]
\[\Rightarrow\ (3a+2p)\mid\ (a+p)^2\]
\[\Rightarrow\ (3a+2p)\mid\ (3a+3p)^2\]
\[\Rightarrow\ (3a+2p)\mid\ (3a+2p)^2+2p(3a+2p)+p^2\]
\[\Rightarrow\ (3a+2p)\mid\ p^2\]
Hence the possible values of 3a+2p are $1,-1,p,-p.p^2,-p^2$.
Considering the various cases [b]modulo 3[/b] we get the solutions.

it's dibyo_99's approach , he showed it at the ISI camps.  :) ...nice....isn't it??
can you tell us your motivation  dibyo_99??


function
Euler
linear algebra
matrix
Olimpiada de matematicas

Aca abro el topic para geo

[color=blue]$Problema$ $1$

Sea $ABCD$ un cuadrilátero cíclico, y sea $O$ la intersección de sus diagonales. Suponer que:

$(AB)(OC)=(AC)(OB)=(BC)(OA)$

Demostrar que $BD \geq 4 BO$  (MOP, 1998)[/color]

SOLUCION.
Como $(AB)(OC)=(AO)(BC)=(BO)(AC)...(1)$,
ENTONCES $<ABO=<OBC$, por el inscriptible $<CAD=<ACD\Rightarrow AD=CD=n$
Sea $AB=a\to AO=ak$, $BC=b\to OC=bk$, $BO=c$, despejando $c$ en $(1)$, en funcion de $a$ y $b$, $c=ab/(a+b)...(2)$
por la semejansa entre los triangulos $AOD$ y $BOC$: $(c)(d)=(ak)(bk)$, reemplazando $(2)$
tenemos que $d=(a+b)k^{2}...(3)$
Aplicando el Teorema de Ptolomeo.
$(c+d)(a+b)k=n(a+b)\to (c+d)k=n...(4)$
Por la semejansa en los triangulos $AOB$ y $BAD$, tenemos que
$n^{2}=d(c+d)$, reemplazando $(4)$.
$k^{2}(c+d)=d$,reemplazando $(3)$
$c+d=a+b\to d=a+b-c$ reemplazando $(3)$ y $(4)$.
$(a+b)k^{2}=(a+b)-ab/(a+b)\Rightarrow k^{2}=(a^{2}+b^{2}+ab)/(a+b)^{2}$
reemplazando en lo que nos pide.
$BD\geq 4BO$
$< = >d\geq 3c$
$< = >k^{2}(a+b)\geq 3ab/(a+b)$
$< = >(a^{2}+b^{2}+ab)/(a+b)^{2}\geq 3ab/(a+b)^{2}$
$< = >(a-b)^{2}\geq 0$, lo cual es verdadero.
$Q.E.D.$

[color=blue]p_1)( nivel 2 , rioplatense 2006)

sea $ABC$ un triangulo rectangulo en $A$ . considere todos los triangulos $XYZ$ rectangulos isosceles en $X$, donde esta sobre el segmento $BC$ , $Y$ sobre el segmento $AB$, y $Z$ sobre el segmentos $AC$.
determinar el lugar geometrico de los puntos de las hipotenusas $YZ$ de tales triangulos $XYZ$[/color]

Aqui mi solucion para el problema $1$:

Denotemos: $AB=m,BC=n,OA=a,OB=b,OC=c$

Luego:$(m)(c)=(a+c)(b)=(n)(a) \Rightarrow \frac{m}{a}=\frac{n}{c}\Rightarrow \measuredangle ABO = \measuredangle OBC \Rightarrow \boxed{b^{2}=mn-ac}$

Además:$\frac{c}{a+c}=\frac{b}{m}$ y $\frac{a}{a+c}=\frac{b}{n}$ $\Rightarrow$ $\frac{b}{m}+\frac{b}{n}= 1$ $\Rightarrow$ $\boxed{b(m+n)=mn}$

ABCD es cíclico, de lo cual tenemos: $\boxed{ac=bd}$

Por $MA$ y $MH$:$\frac{m+n}{2}\geq \frac{2}{ \frac{1}{m}+\frac{1}{n}}= 2b$

Usaremos las tres relaciones encontradas para llegar a la desigualdad pedida:

$\Rightarrow m+n \geq 4b \Rightarrow b(m+n) \geq 4b^{2}\Rightarrow mn\geq 4b^{2}$
$\Rightarrow b^{2}+ac\geq 4b^{2}\Rightarrow ac\geq 3b^{2}\Rightarrow bd\geq 3b^{2}\Rightarrow d\geq3b \Rightarrow b+d\geq 4b$
$\Rightarrow BD\geq 4BO$

demostremos que $<ABE=<QBD$, vemos que si esto se cumple , esto se hara analogo para los puntos $A$ y $C$ , y como $AD, BE, FC$ concurren se cumple el ceva trigonometrico , con ello tambein se cumplira para $BQ, AP, CR$, vemos que $<ABE+<BAE = 90$, y por inscriptible $<BAE = <BDQ$ ademas, $<BDQ+<QBD = 90$ porlo tanto $<ABE=<QBD$,
 QED

[quote="carlosbr"][color=blue]Problema 03:

Sea $ABC$ un triangulo acutangulo y sean $D,E,F$ los pies de las alturas trazadas desde $A,B,C$.
Sean $P,Q,R$ los pies de las perpendicuares desde $A,B,C$ hacia $EF, FD, DE$, respectivamente.

Probar que las lineas $AP, BQ, CR$ son concurrentes.[/color]

Carlos Bravo[/quote]
Bueno, como dice marco, como son isogonales, concurren en el circunetro de $ABC$.

bueno tuve un pequeño error al tipear el problema  :blush: :

P_1) sea $ABC$ un triangulo rectangulo en $A$. considere todos los triangulos $XYZ$ rectangulos isosceles en $X$, donde $X$ esta sobre el segmento $BC$, $Y$ sobre el segmento $AB$, y $Z$ sobre el segmento $AC$.
determinar el lugar geometrico de los puntos medios de las hipotenusas $YZ$ de tales triangulos $XYZ$.

Ahora si tiene sentido. Para el anterior el lugar creo q era todo el triangulo :rotfl:

[quote="M4RI0"][color=blue]$Problema$ $4$

Una circunferencia es tangente al circuncírculo de un triangulo $ABC$ y ademas tangentes a los lados $AB$, $AC$ en $P$, $Q$ respectivamente. Demostrar que el punto medio de $PQ$ es el incentro del triangulo $ABC$. [/color]

Pueden encontrar este problema en el Matescopio con el numero $200.4$[/quote]
Este problema sale rapido, usando el teorema de Pascal.

Postea tu solucion por favor. Yo tengo una ubicando primero un punto $X$ sobre $AB$ tal que $XM$ es perpendicular a $AB$, luego con trigonometria hallamos $AX$ que resulta ser $p-a$($p:$ semiperimetro) y como ademas $M$ se encuentra sobre la bisectriz del angulo $A$, concluimos que $M$ es el incentro de $ABC$.

Mas vale tarde que nunca.....Sea $X$ el punto de tangencia entre el circuncirculo y la otra circunferencia. $XP$ interseca a la circunferencia circunscrita C en $P_{1}$, $XQ$ interseca a C en $Q_{1}$, es sencillo demostrar que $P_{1}$ es el punto que divide al arco AB en partes iguales(por homotecia entre las circunferencias), luego $CP_{1}$, $BQ_{1}$ son bisectrices y por T. PASCAL en $P_{1},A,Q_{1},C,X,B$, las bisectrices se cortan en un punto sobre $PQ$, luego como $AQ=AP$, al trazar la bisectriz de A, cortara a PQ en su punto medio.

algunos problemas q pueden salir por potencia de un punto :

1) Sean $S_{1}$ y $S_{2}$ dos circulos y sea $O$ el punto de corte de las dos tangentes externas comunes a los dos circulos . demostrar que la potencia de $O$ con respecto a cualquier circulo tangente externamente a $S_{1}$ y $S_{2}$ es constante.

2) Sea $S$ un circulo ; sean $A$ y $B$ dos puntos . una recta variable a traves de $A$ corta a $S$ en $M$ y $N$ . demostar que los circuncirculos de los triangulos $MNB$ pasan por otro punto fijo , distinto de $B$

3) En un triangulo $ABC$ , sea $I$ el incentro y sean $X$ , $Y$ y $Z$ los puntos de tangencia del incirculo con los lados $BC$ ,  $AC$ y $AB$ respectivamente . la recta $AX$ corta al incirculo nuevamente en $P$ , y la recta $AI$ corta a $YZ$ en $Q$ . demostrar que $X$ , $I$ , $Q$ y $P$ estan sobre una misma circunferencia

[quote="mhuarancca"]
3) En un triangulo $ABC$ , sea $I$ el incentro y sean $X$ , $Y$ y $Z$ los puntos de tangencia del incirculo con los lados $BC$ ,  $AC$ y $AB$ respectivamente . la recta $AX$ corta al incirculo nuevamente en $P$ , y la recta $AI$ corta a $YZ$ en $Q$ . demostrar que $X$ , $I$ , $Q$ y $P$ estan sobre una misma circunferencia[/quote]
A esto le falta algo, ya que si $AB=AC$, entonces, $P,Q,I,X$ seran colineales, considerando $AB$ diferente de $AC$.
Tomando a la circunferencia inscrita al triangulo $ABC$ como circunferencia de Magnus (inversion), invirtiendo la recta $AX$.
$A,P,X$, se invierten en $Q,P,X$ respectivamente, como $I$ es el centro de esta  circunferencia. $P,Q,I,X$, son puntos conciclicos.

en un triangulo $ABC$ , un circulo corta a $BC$ en $L$ y $L"$ , a $AC$ en $M$ y $M"$ y a $AB$ en $N$ y  $N"$ . demostrar que si $AL$ , $BM$ y $CN$ son concurrentes, entonces $AL"$ , $BM"$ y $CN"$ tambien son concurrentes

Voy a mostrar un lema más general del cual sigue el problema inmediatamente:

[u][b]Lema: [/b][/u] [i]Si $ (AL, BM, CN)$ y $ (AL', BM', CN')$ son dos trios de cevianas concurrentes con $ L,L',M,M',N,N'$ sobre los lados de un triangulo. Los puntos $ L,L',M,M',N,N'$ están sobre una cónica.[/i]

[u][i]Demostración:[/i][/u] Sean $ X\in MN\cap BC$, $ Y\in LN'\cap AC$ y $ Z\in L'M'\cap AB$
Por Ceva:
$ (\frac{BN}{NA}) (\frac{AM}{MC}) (\frac{CL}{LB})\equal{}1$
$ (\frac{BN'}{N'A}) (\frac{CL'}{L'B})(\frac{AM'}{M'C})\equal{}1$

Por Menelao:
$ \frac{BX}{XC}\equal{}(\frac{BN}{NA}) (\frac{AM}{MC})$
$ \frac{CY}{YA}\equal{}(\frac{BN'}{N'A}) (\frac{CL}{LB})$
$ \frac{AZ}{ZB}\equal{}(\frac{CL'}{L'B}) (\frac{AM'}{M'C})$

Entonces
$ (\frac{BX}{XC})(\frac{CY}{YA})(\frac{AZ}{ZB}) \equal{}(\frac{BN}{NA}) (\frac{AM}{MC})(\frac{BN'}{N'A}) (\frac{CL}{LB}) (\frac{CL'}{L'B}) (\frac{AM'}{M'C})\equal{}1$
Que por Menelao $ X,Y,Z$ son colineales.
Por el teorema de Pascal, como los lados opuestos del hexagono $ LL'MM'NN'$ se cortan en puntos colineales los puntos $ L,L',M,M',N,N'$ están sobre una cónica.

En el problema original como cinco puntos definen una cónica los otro tres puntos de corte del ciruclo deben determinar cevianas concurrentes.

[quote="skuge"]Voy a mostrar un lema más general del cual sigue el problema inmediatamente:

[u][b]Lema: [/b][/u] [i]Si $ (AL, BM, CN)$ y $ (AL', BM', CN')$ son dos trios de cevianas concurrentes con $ L,L',M,M',N,N'$ sobre los lados de un triangulo. Los puntos $ L,L',M,M',N,N'$ están sobre una cónica.[/i]

[u][i]Demostración:[/i][/u] Sean $ X\in MN\cap BC$, $ Y\in LN'\cap AC$ y $ Z\in L'M'\cap AB$ .[/quote]
creo q podemos reducir la demostracion.
Vemos que $ {A, N', B, Z}y tambien,{X, B, L, C}$, son cuaterna armonica. Ahora desde $ Y$ tenemos: -$ YC, YL, YB, YX$ y -$ YA, YN', YB, YZ$ son rayos armonicos...entonces $ X,Z,Y$ COLINEALES!!. :D

[quote="skuge"]En qué número de problema vamos?? Bueno acá va el siguiente, uno fácil de Estonia:

Sean $ M,N$ y $ K$ los puntos de tangencia del incirculo de $ \triangle ABC$. Si $ Q$ es el centro del circulo que pasa por los puntos medios de $ MN, NK$ y $ KM$,
Muestre que $ Q$ está sobre la linea que une el incentro y el circuncentro de $ \triangle ABC$.[/quote]
SOL.
Consideremos a el incirculo de $ ABC$ como la circunferencia de inversion.
vemos que los puntos medios de $ MN,NK,KM$, son los puntos inversos de $ A,B,C$, entonces la circunferencia q pasa por $ ABC$ se invierte en la circunferencia q pasa por los puntos medios de $ MN,NK,KM$, ademas sabemos que cuando se invierten una circunferencia $ T$ ,sea $ T*$ la circunferencia invertida respecto de la de inversion, entonces los centros de la circunferencia de inversion, de $ T$ y $ T*$ son colineales. Entonces $ Q$, incirculo y circuncirculo son colineales.

Bien, esa esta linda... cuando tenga tiempo la proxima semana pongo mi solución...
Otro problemita de calentamiento.

En un triangulo $ ABC$ con $ \angle A\equal{}60$ muestre que si la linea de euler corta los lados $ BA$ y $ CA$ en $ P$ y $ Q$ respec. $ HP\equal{}OQ$

(con H y O la notacion de siempre)

buen prob. ahi esta mi sol.
[hide]Solucion:  $ < BOC \equal{} 2 < A \equal{} 120 \equal{} < B\plus{}< C \equal{} < BHA_{1}\plus{}< CHA_{1}$
$ \equal{} < BHC \equal{} 120 \equal{} 90\plus{}\frac{ < A}{2}\equal{} < BIC$, donde $ A_{1}$ es el pie de la altura desde $ A$ a $ BC$, luego, $ CHIOB$:ciclico, el centro de la circunferencia que pasa por estos 5 puntos($ X$) esta en $ AI$(la interseccion de AI y la circunferencia circunscrita a ABC), ademas $ H, O$ son isogonales conjugadas, esto es $ < BAH \equal{} CAO$ y $ < HAI \equal{} < IAO$, por otro lado: $ AH \equal{} 2RcosA \equal{} R$(calculo simple), luego $ AH \equal{} AO$, de esto se tiene que H y O son simetricos con respecto a AI, como AI: bisectriz de A; P y Q: simetricos respecto de AI, luego $ XP \equal{} XQ$, por potencia de P respecto a la circ. por CHOB, $ (PH)(PO) \equal{} (PO)^{2}\minus{}r^{2}$ donde r es $ XI$, tambien $ (QH)(QO) \equal{} (QO)^{2}\minus{}r^{2}\equal{} (PO)^{2}\minus{}r^{2}$, luego $ (PO)(PH) \equal{} (QO)(QH)$, como $ OH$ es fijo, al remplazar en ambos lados se obtiene $ HP \equal{} OQ$[/hide]

[quote="skuge"]Bien, esa esta linda... cuando tenga tiempo la proxima semana pongo mi solución...
Otro problemita de calentamiento.

En un triangulo $ ABC$ con $ \angle A \equal{} 60$ muestre que si la linea de euler corta los lados $ BA$ y $ CA$ en $ P$ y $ Q$ respec. $ HP \equal{} OQ$

(con H y O la notacion de siempre)[/quote]
Consideraremos sin perder generalidad $ AB<AC$ ( ya que si $ AB\equal{}AC$, entonces $ ABC$ equilatero y su recta de euler no esta definida).
Ahora $ <BHC\equal{}<BOC\equal{}120$, entonces $ BHOC$ inscriptible. Sabemos que la recta de euler pasa por $ H$ y $ O$ ( $ P,O,H y Q$ colineales). Ahora $ <HBA\equal{}30$, Como $ BHOC$ inscriptible, $ <PHB\equal{}<OCB\equal{}HCA$ (lo ultimo puesto que $ O y H$ son isogonales), pero $ <HCA\equal{}30$,  luego $ <PHB\equal{}30$, entonces, $ <APH\equal{}60$, ENTONCES $ AQP$ equilatero. Luego $ <PAH\equal{}<OAC$, entonces $ APH$ y $ AQO$ son congruentes $ (A.L.A.)$,  Con lo cual $ PH\equal{}OQ$ que es lo que queriamos demostrar.

Mi solución es parecida a la de cuenca:
Es conocido que la reflexión de la altura sobre la bicectriz pasa por el circuncírculo y es fácil mostrar que $ AH \equal{} AO$ entonces $ \triangle AHO$ es isóseles y $ AI\bot HO$, mirnado angulitos es obvio que $ PQA$ es equilatero y esto termina el problema porque la bicectriz pasa por el punto medio.

[quote="skuge"]Muestre que la reflexión sobre los lados opuestos de las cevianas que pasan por el punto de fermat concurren.[/quote]
No entiendo....reflexion respecto de que?....Bueno si fuera de la bisectriz del angulo respectivo de donde sale la ceviana.Entonces creo que con Ceva Trigonometrico ( la relacion de lados que nos indica el teorema de ceva lo expresamos en relacion de senos de el angulo formado por las cevianas.. :wink: ). tendriamos lo pedido.

[quote="aev5peru"]
No entiendo....reflexion respecto de que?....Bueno si fuera de la bisectriz del angulo respectivo de donde sale la ceviana.Entonces creo que con Ceva Trigonometrico ( la relacion de lados que nos indica el teorema de ceva lo expresamos en relacion de senos de el angulo formado por las cevianas.. :wink: ). tendriamos lo pedido.[/quote]

Ok. El problema es el siguiente:
Si $ F$ es le punto de fermat del $ \triangle ABC$ sean $ x, y$ y $ z$ las reflexiones de las rectas $ AF$, $ BF$ y $ CF$ sobre los lados $ BC$, $ CA$ y $ AB$ respectivamente.
~Mostrar que $ x,y,z$ concurren.

Lo primero que se me ocurrio fue coord. baricentricas, pero me parecio muy operativo; demostrare por coordenadas trilineales que x, y, z concurren en el conjugado isogonal del punto de fermat.
$ A_{1}$ es el reflejo de A en BC, por un calculo simple $ A_{1}=(-1: 2cosC: 2cosB)$, ademas es conocido que el punto de fernat es $ (\frac{1}{sen(A+60)}:\frac{1}{sen(B+60)}:\frac{1}{sen(C+60)})$, luego si $ A_{2}$ es la interseccion de la rescta AF(F:punto de fermat) con BC, entonces $ A_{2}=(0:\frac{1}{sen_{(}B+60)}:\frac{1}{sen(C+60)})$, luego para hallar la ecuacion de la recta A1A2, se necesita que el determinante de:
$ \begin{pmatrix}x&y&z\\0&y_{1}&z_{1}\\-1&2cosC&2cosB\end{pmatrix}$ es 0(y1=csc(b+60),z1=csc(c+60)); de esto: $ xsen(C-B)+ysen(B+60)-zsen(C+60)=0$, por la simetria en las demas ecuaciones, lo que se puede hacer es hallar el determinante del sistema y demostrar que es 0(demora mucho), yo intui que era el conj. isogonal de F, y lo grafique en computadora y en efecto lo era :) ; para finalizar basta probar que el conj. isogonal de F=$ F_{1}(sen(A+60): sen(B+60): sen(C+60))$, verifica la ecuacion, y en efecto; como:
sen(A+60)=sen(B+C-60), la ecuacion queda: $ -cos(B+C).sen(B-C)+senBcosB-senCcosC=0$ lo cual es evidente por identidades de arcos compuestos. Luego cada recta reflejada pasa por un punto fijo, luego todas concurren. :wink:
Nota: Esto es muy operativo, ¿has encontrado algo sintetico, skuge?
P.D. Si hay algo que no se entiende o alguna correcion, por favor decirme para corregirlo o explicarlo mejor :blush:

Mi turno de proponer un problema:
En un triangulo $ ABC$, sea $ M$ un punto arbitrario, demostrar la siguiente desigualdad:
$ AMsenBMC\plus{}BMsenAMC\plus{}CMsenAMB\leq s$, donde $ s$ es el semiperimetro del triangulo $ ABC$ y averiguar para que punto $ M$ se da la igualdad.

Mi solucion:

Primero tracemos las perpendicular a MA, MB , MC por los pntos A, B,C respectivamente. Dichas perpendiculares forman con sus intersecciones un triangulo TUV (T es el vertice opuesto a A, U opuesto a B, y V opuesto a C).
Ahora, el problema se transforma en demostrar que:
$ AMsenT\plus{}BMsenU\plus{}CMsenV leq s$ lo cual es equivalente a
AM.VU/2 + BM.VT/2  CM.UT/2 $ /leq$ R(AB+BC+CA) (R=circunradio de TUV) y esto es
(ABC) $ /leq$ R(AB+BC+CA).
Ahora, como sabemos que el triangulo pedal de menor perimetro es el ortico, entonces si demostramos que
(ABC) $ /leq$ R(DE+EF+FD) (donde D, E, F son los pies de las alturas del triangulo TUV; D es opuesto a T, E es opuesto a U, y F es opuesto a V) ya habremos acabado.
Pero esta ultima desigualdad, de hecho es una igualdad. Por lo que el problema ha sido probado.

Un problema muy interesante.

Ah, se me olvidaba. La igualdad se da cuanto M es ortocentro de TUV, es decir, cuando es incentro de ABC. :lol:

[quote="mhuarancca"]Sea $ ABC$ un triángulo rectángulo en $ A.$ Considere todos los triángulos $ XYZ$ rectángulos isosceles en $ X,$ donde $ X$ esta sobre el segmento $ BC,$ $ Y$ sobre el segmento $ AB$ y $ Z$ sobre el segmento $ AC.$ Determinar el lugar geométrico de los puntos medios de las hipotenusas $ YZ$ de tales triángulos $ XYZ.$[/quote]
Como $ XY \equal{} XZ$ y $ \angle YXZ \equal{} \angle YAZ \equal{} 90^{\circ },$ se sigue que el cuadrilátero $ AYXZ$ es cíclico y que los arcos $ XY$ y $ XZ$ de su circuncírculo son conguentes $ \Longrightarrow$ $ AX$ es bisectriz de $ \angle BAC.$ Así  $ X$ es fijo en $ BC$ siendo el pie de la A-bisectriz interior.

Si $ P$ es la proyección ortogonal de $ X$ en el cateto $ AB$ y $ M$ el punto medio del segmento $ YZ,$ el cuadrilátero $ MYPX$ es cíclico en vista que los ángulos $ \angle YPX$ y $ \angle YMX$ son ambos rectos. Luego se tendrá $ \angle XPM \equal{} \angle XYZ \equal{} 45^{\circ}.$

El lugar geométrico de $ M$ es pues el segmento de recta que conecta los pies de las perpendiculares de $ X$ a $ AB,AC.$

[quote="mhuarancca"] Sean $ \mathcal{S}_{1}$ y $ \mathcal{S}_{2}$ dos círculos y sea $ O$ el punto de corte de las dos tangentes externas comunes a los dos círculos. Demostrar que la potencia de $ O$ con respecto a cualquier circulo tangente externamente a $ \mathcal{S}_{1}$ y $ \mathcal{S}_{2}$ es constante.[/quote]
Sea $ \mathcal{S}$ una circunferencia arbitraria tangente externamente a $ \mathcal{S}_{1}$ y $ \mathcal{S}_{2}$ en $ A,B,$ respectivamente. Los puntos $ A$ y $ B$ son los centros de homotecia negativa que transforman $ \mathcal{S}_{1} \mapsto \mathcal{S}$ y $ \mathcal{S}_{2} \mapsto \mathcal{S},$ respectivamente. $ O$ es por supuesto el centro de la homotecia positiva que transforma $ \mathcal{S}_{1} \mapsto \mathcal{S}_{2},$ así por el teorema de Monge-d'Alembert los puntos $ A,B,O$ están alineados. $ A,B$ son homólogos en la inversión que trasforma $ \mathcal{S}_{1}$ en $ \mathcal{S}_{2}$ con mismo centro $ O,$ por consiguiente la potencia de $ O$ con respecto a la circunferencia  $ \mathcal{S}$ es constante e igual a la potencia de inversión que transforma $ \mathcal{S}_{1}$ en $ \mathcal{S}_{2}.$

[quote="mhuarancca"]En un triángulo $ ABC,$ sea $ I$ el incentro y sean $ X,$ $ Y$ y $ Z$ los puntos de tangencia del incírculo con los lados $ BC,$  $ AC$ y $ AB,$ respectivamente. La recta $ AX$ corta al incírculo nuevamente en $ P$ y la recta $ AI$ corta a $ YZ$ en $ Q.$ Mostrar que $ X,$ $ I,$ $ Q$ y $ P$ están sobre una misma circunferencia.[/quote]
Por teorema del cateto en el triángulo $ \triangle AYI$ se tiene $ AY^2 \equal{} AQ \cdot AI,$  pero la potencia de $ A$ con respecto al incírculo $ (I)$ es $ AY^2 \equal{} AP \cdot AX$ $ \Longrightarrow$ $ AQ \cdot AI \equal{} AP \cdot AX$ $ \Longrightarrow$ $ I,Q,P,X$ son concíclicos.

[quote="skuge"]Muestre que la reflexión sobre los lados opuestos de las cevianas que pasan por el punto de fermat concurren.[/quote]
Ya que tal concurrencia es también cierta para el segundo punto de Fermat, se puede enunciar el problema así:

[b]Proposición:[/b] $ F$ es un punto de Fermant de $ \triangle ABC$ y $ F'$ el correspondiente punto Isodinámico. $ F_a,F_b,F_c$ son los simétricos de $ F$ en $ BC,CA,AB$ y $ \triangle A''B''C''$ es el triángulo ceviano de $ F.$ Entonces $ \triangle A''B''C''$ y $ \triangle F_aF_bF_c$ son perspectivos con centro de perspectividad $ F'.$

Como el triángulo pedal de $ F'$ respecto a $ \triangle ABC$ es equilátero, entonces los puntos simetricos $ A',B',C'$ de $ F'$ respecto a $ BC,CA,AB$ forman otro triángulo equilátero homotético al triángulo pedal de $ F'.$ Las perpendiculares desde $ A,B,C$ a los lados del triángulo pedal de $ F'$ concurren en su conjugado isogonal $ F.$ Entonces $ AX \perp B'C' ,$ $BY \perp A'C'$ y $ CZ \perp A'B'$ (estando X,Y,Z en B'C',C'A', A'B') concurren en $ F.$ Note que $ BC,CA,AB$ pasan a ser mediatrices de $ F'A,F'B,F'C$ $ \Longrightarrow$ $ AX,$ $BY,$ $CZ$ son mediatrices de $ B'C',C'A',A'B'.$ Consiguientemente resulta $ A' \in AF,$ $ B' \in BF$ y $ C' \in CF.$ Las simétricas $ F_aF',$ $F_bF',$ $ F_cF'$ de las rectas $ AF,BF,CF$ con respecto a $ BC,CA,AB$ pasan entonces por $ A'',B'',C''$ $ \Longrightarrow$ $ \triangle A''B''C''$ y $ \triangle F_aF_bF_c$ son perspectivos a través de $ F'.$


Find the convergence radius of the serie
\[ \sum_1^\infty \frac{x^{a_n}}{n^3} \]
Where $a_n$ is the number of $0's$ when $n$ is expanded in the third base. (coeff when writen as sum of power of 3)

It is a good one. It's a pity nobody answered :(. Here is a small [hide="Hint"]The answer is $25=3^3-2$ [/hide]

It is reeadily that the radius of convergence won't change if we replace the denominator with $3^{[log_3n]}^3$.
Ok, now our series $f(x)=\sum \frac{x^{a_m}}{m^3}=\sum_{n\geq 0} (coefficient of x^n)x^n=\sum_{n\geq 0} \sum_{m: a_m=n}\frac{1}{m^3} x^n=(*)$.
Now the coefficient $s_n$ of $x^n$ can be computed in the following way: since $a_m=n$, $k=[log_3 m] \geq n$ and the number of such $m$ with fixed number of digits $k+1 \geq n$ is ${2^{k+1-n}C_k^n}$.
Thus $s_n=\sum_{k \geq n} \frac{{2^{k+1-n}C_k^n}}{27^k}$.
It is easy to prove that $\sum_{k \geq n}\frac{C_k^n}{a^k}=\frac{1}{(a-1)^{n+1}}$ for $|a|>1$-in our case we need $a$ to be $a=\frac{27}{2}>1$. So useing the above assertions $s_n=\frac{(\frac{1}{27/2-1})^{n+1}}{2^n-1}=\frac{4}{25^{n+1}}$, which means that the radius of convergence is equal to $R=(\frac{1}{s_n})^{\frac{1}{n}}=25$.


In how many ways can we divide 5 people into teams so that each team has at least one person? (Seriously, how do you do this?)  Remember that there can be any number of teams from 1-5.

[quote="Walk Around The River"]hmm, my method would be to use balls and urns for 1, 2, 3, 4, and 5, and then add, but im sure theres a better way...[/quote]
Use $\sum^{k}_{n=0}\binom{k}{n}=2^{k}$
[hide="generalization"]
Dividing into k people where k is 1 mod 4:
$\binom{k-1}{0}+\binom{k-1}{1}+\binom{k-1}{2}+...+\binom{k-1}{\frac{k-1}{2}}$
$2^{k-2}+\frac{\binom{k-1}{\frac{k-1}{2}}}{2}$
So we get $8+3=11$[/hide]
I'm probably wrong though :(

ABCDE
A BCDE
B ACDE
C ABDE
D ABCE
E ABCD
AB CDE
AC BDE
AD BCE
AE BCD
BC ADE
BD ACE
BE ACD
CD ABE
CE ABD
DE ABC
Unless i over counted there should be 15

I meant 16, i didnt see the first one

didn't he say there could be any number of teams?  So

A, B, C, D, E would be a way with 5 teams

I'm thinking :

[hide]
If 5 teams then only $1 way$
If 4 teams then one team must have 2 people, so $5 \choose 2$ then spread the rest out = $10 ways$
If 3 teams we have a choice.  Either 2-2-1 or 3-1-1.
If the teams are 2 to 2 to 1, then ${5 \choose 2 }\cdot{3\choose 2}\cdot 1 = 10 \cdot 3 \cdot 1 = 30 ways$
If the teams are 3 to 1 to 1, then ${5 \choose 3 }\cdot 1 \cdot 1 = 10 \cdot 3 \cdot 1 = 10 ways$
If 2 teams we have a choice.  Either 4 to 1 or 3 to 2.
If the teams are 4 to 1, then ${5 \choose 4}\cdot 1 = 5 \cdot 1 = 5 ways$
If the teams are 3 to 2, then ${5 \choose 3}\cdot 1 = 10 \cdot 1 = 10 ways$
If 1 team, then just $1 way$

That's a total of 67 ways to make teams.  I don't like the sound of that answer, what do you all think?

[/hide]

This is alot like ARML 2006 Individual problem 8, which I may mention 3 (I think?) people in the country got right. Unfortunately the boring casework is the way to go here.

[quote="alanchou"]Actually, is it possible that someone could list the possibilities with people A,B,C,D,E?[/quote]
If this was a Team question, it might be possible to list all cases.  Pretty tedious, though.

[quote="frost13"][hide]That's a total of 67 ways to make teams.[/hide]what do you all think?[/quote]
I think you overcounted one of your cases.

[quote="13375P34K43V312"]This is alot like ARML 2006 Individual problem 8.[/quote]
Yes, there's some similarity in casework.  But, I think the ARML problem is much harder than this question.  [BTW, please don't discuss the ARML problem, here.]

[quote="rcv"]
[quote="frost13"][hide]That's a total of 67 ways to make teams.[/hide]what do you all think?[/quote]
I think you overcounted one of your cases.
[/quote]

Yup, double counted the 3 teams with the 2 - 2 - 1 distribution.  Just need to divide by 2 for that case.  So actual answer.....

[hide]divide the 30 ways for that case by 2 to get 15.  Subtracting those extra 15 from the 67 gives us 52.

A little birdy pointed out my flaw.  Thanks birdy :lol: [/hide]

This is why my signature says I'm still learning and always will be!!!


There's this problem I have while doing the AIME. The last 15 minutes or so are wasted because i'm too lazy to follow through some ideas by the end. I think I just wear out. I've done lots of practice, and after a couple of sittings, it's a little better, but does anyone have any other tips?

Well, think about it this way - you get ONE chance. You might as well just force yourself to focus, I mean - ONE chance a year. Other things that might be on your mind - they can wait because you are using your ONE chance wisely.  :D

I just did the 2003 AIME1 and I managed to solve a problem in the last few minutes! So I got a nine instead of my usual 7 or 8.Have to tell ye, my brain was almost dead in the end. Need sleep...


calculus
integration
trigonometry
calculus computations

The solution below uses elementary techniques. I will use $a$ instead of $\alpha$ to save myself the trouble of typing extra characters.

$I =-\int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}}{\cos}^{-1}\left( \frac{\cos a}{\sin\theta}\right) d(\cos\theta)$
$=-\left[{\cos}^{-1}\left( \frac{\cos a}{\sin\theta}\right) \cos\theta \Big|_{\frac{\pi}{2}-a}^{\frac{\pi}{2}}-\int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}}\frac{{\cos}^{2}\theta \sin\theta{d\theta}}{\sqrt{{\sin}^{2}a-{\cos}^{2}\theta}(1-{\cos}^{2}\theta)}\right]$ ... (Using integration by parts)

Make the substitution $\cos\theta = \sin a \sin t$. We get

${I = \cos a \int_{0}^{\frac{\pi}{2}}\frac{{(\sin a \sin t)}^{2}dt}{1-{(\sin a \sin t)}^{2}}=-\cos a\int_{0}^{\frac{\pi}{2}}dt+\cos a \int_{0}^{\frac{\pi}{2}}\frac{dt}{(1-\sin a \sin t)(1+\sin a \sin t)}}$

$=-\frac{\pi}{2}\cos a+\frac{\cos a}{2}\int_{0}^{\frac{\pi}{2}}\left( \frac{1}{1+\sin a \sin t}+\frac{1}{1-\sin a \sin t}\right) dt$

$=-\frac{\pi}{2}\cos a+I_{1}+I_{2}$ (say)

Now, use $\sin t = \frac{2\tan\frac{t}{2}}{1+{\tan}^{2}\frac{t}{2}}$ and make the substitution $y = (\tan \frac{t}{2}+\sin a)$ in $I_{1}$; so we have

$I_{1}= \cos a\int_{\sin a}^{1+\sin a}\frac{dy}{y^{2}+{\cos}^{2}a}={\tan}^{-1}\left( \frac{y}{\cos a}\right)\Big|_{\sin a}^{1+\sin a}$

${= \tan}^{-1}\left( \frac{1+\sin a}{\cos a}\right)-a$

In a similar way, we obtain ${I_{2}= \tan}^{-1}\left( \frac{1-\sin a}{\cos a}\right)+a$

${{\therefore I_{1}+I_{2}= \tan}^{-1}\left( \frac{1+\sin a}{\cos a}\right)+\tan}^{-1}\left( \frac{1-\sin a}{\cos a}\right) = \frac{\pi}{2}$
... [ Use ${\tan}^{-1}x+{\tan}^{-1}y ={\tan}^{-1}\left( \frac{x+y}{1-xy}\right)$]

$\Rightarrow I =-\frac{\pi}{2}\cos a+I_{1}+I_{2}= \boxed{\frac{\pi}{2}(1-\cos a)}$.  :lol:


Euler
geometry
circumcircle
geometry proposed

Let $ P$ be a point in the plane of triangle $ ABC$. Let $ Q$ be the isogonal conjugate of $ P$ with respect to triangle $ ABC.$ Prove that line $ PQ$ is parallel to the Euler line of triangle $ ABC$ if and only if the triangle formed by the circumcenters of triangles $ PBC,PCA,PAB$ and triangle $ ABC$ are perspective.

I think this problem is wrong. See below

Dear Han-sol Shin
I think it is true! also see the following picture:

The problem should state that:

the triangle formed by the circumcenters of triangles $ PBC$, $ PCA$, $ PAB$ and triangle $ ABC$ are perspective if and only if the line $ PQ$ is parallel to the Euler line of triangle $ ABC$ or one of the points $ P$ and $ Q$ lies on the circumcircle of triangle $ ABC$.

(The locus of all points $ P$ such that the line $ PQ$ is parallel to the Euler line of triangle $ ABC$ is called the [i]Neuberg cubic[/i] of triangle $ ABC$. The above is one of the properties of this cubic. I don't remember a synthetic proof, but many problems have been solved after I left geometry, so this doesn't mean no proof is known...)

  darij

According to the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=373375]Find locus[/url], $\triangle ABC$ and the triangle formed by the circumcenters of $\triangle PBC,$ $\triangle PCA$ and $\triangle PAB$ are perspective $\Longleftrightarrow$ Either $P$ lies on the Neuberg cubic $\mathcal{N}$ of $\triangle ABC$ or it lies on the circumcircle $\mathcal{O}.$ Thus, it remains to show that $PQ$ is parallel to the Euler line of $\triangle ABC$ $\Longleftrightarrow$ $P \in \mathcal{N}.$

Equation of line $\tau$ passing through $P\equiv (u:v:w)$ and $Q \equiv \left ( \frac{a^2}{u}:\frac{b^2}{v}:\frac{c^2}{w} \right)$ is:

$\tau \equiv u(c^2v^2-b^2w^2)x+v(a^2w^2-c^2u^2)y+w(b^2u^2-a^2v^2)z=0$

$\tau$ is parallel to the Euler line of $\triangle ABC$ $\Longleftrightarrow$ $\tau$ goes through its infinity point $X_{30}$

$(S_AS_B+S_AS_C-2S_BS_C: S_BS_C+S_BS_A-2S_CS_A: S_CS_A+S_CS_B-2S_AS_B)$

Hence, setting $(u,v,w) \longrightarrow (x,y,z),$ locus $f(x,y,z)=0$ of $P$ is

$\sum_{\text{cyclic}} (S_AS_B+S_AS_C-2S_AS_B)(c^2y^2-b^2z^2)x=0,$ i.e. the Neuberg cubic $\mathcal{N}.$

About this problem there are many equivalent propositions，and it have been totally solved in useing geometric method by chinese expert zhonghao Ye，I hope him or his student can come here to explan this Problems.

You can reference here：[b]Neuberg曲线[/b][url]http://bbs.cnool.net/topic_show.jsp?id=6265276&thesisid=494&flag=topic1[/url]


linear algebra
matrix
linear algebra unsolved

[quote="mareleG"][this one may be known, but it's nice for sure]

Let $A \in M_n(R), a_{ij} = gcd(i,j)$. Prove or disprove $det(A) > 0$.[/quote]Not only is the matrix invertible, it's actually true that $\det A = \prod_{k=0}^n \phi(n)$ where $\phi(n)$ is the totient function.

[quote="dryga"][quote="mareleG"][this one may be known, but it's nice for sure]

Let $ A\in M_{n}(R), a_{ij}\equal{} gcd(i,j)$. Prove or disprove $ det(A) > 0$.[/quote]Not only is the matrix invertible, it's actually true that $ \det A \equal{}\prod_{k \equal{} 0}^{n}\phi(n)$ where $ \phi(n)$ is the totient function.[/quote]
Could you provide a proof of this fact

[color=red][Moderator edit: Proofs have been merged into http://www.mathlinks.ro/Forum/viewtopic.php?t=37587 .][/color]


Find all $ x\in R\plus{}$ such that
$ 10\minus{}x\equal{}xlnx$

If $ y \equal{}  ln (ex)$, then the equation can be written as $ y e^y \equal{} 10$. Then we use lambert W function.


Euler
function
number theory
relatively prime
number theory proposed

let [b][i]n [/i][/b] be an odd integer and [b][i]m[/i][/b] be n's euler function (the number of positive integers less than n and relatively prime to it). prove that there is a prime [b][i]p[/i][/b] such that it divides 2^m-1, but does not divide m.

A few times I thought I solved it, but in the end I found a mistake... i hope someone can tell me where can I find a solution... i try all web, but could'n find it. Thank you
Matija[/list][/tex]

nice but why you post in proposed and... btw  here is the proof:
$n=\prod_{i=1}^{k}p_i^\alpha_i$ it's trivial that $2^k|m$so we have $2^m-1=t{}^{2{}^k}-1$s.t $t>3$ and is even,so:
$2^m-1=t{}^{2{}^k}-1=(t-1)\prod_{i=1}^{k-1}(t{}^{2{}^i}+1)$it means that $2^m-1$ has k+1 factor in this form ,t-1 and $t{}^{2{}^i}+1$s.t they are coprime but n has just k prime factor so we are done.

I wasn't careful enough while reading some topics and didn't pay attention where I am sending it... sorry...
but i don't get why are you sure that this prime isn't p-1, where p is some prime divisor of n... m can have more than k prime divisors...
thanks... M

:? I can't understand what's wrong ,could you be more explicite?[quote]i don't get why are you sure that this prime isn't p-1, where p is some prime divisor of n[/quote]you mean p-1 is prime number or...

it seems to me that you proved that p doesn't divide n, not m...

in my previous post p isn't the prime we're looking for, but some prime divisor of n- you marked those with indexes i=1..k

thank you
Matija


geometry
geometric transformation
reflection
symmetry
perpendicular bisector

In $ \triangle ABC$, $ D$ is on $ AC$ s.t. $ AB\equal{}CD$, $ \angle A\equal{}80$, $ \angle C\equal{}20$. Find $ \angle DBC$.

It trivially follows from [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=271741]this problem[/url] that the answer is $ 180^\circ \minus{} 150^\circ \minus{} 20^\circ \equal{} \boxed{10^\circ}$ (let $ C\equal{}A$, $ B\equal{}C$, $ A\equal{}B$, and $ D\equal{}M$ to get from your problem to the one in the link).

Dang, I tried about a million reflections and symmetries, never worked until I saw that.

Okay, here is my solution: EDIT:made it shorter
Let F be the reflection of A over the perpendicular bisector of BC. Then $ DC\equal{}CF$ and $ \angle DCF\equal{}60$, so $ \triangle DCF$ is equilateral. Since $ \triangle BCF$ is isoceles, by symmetry D is on the bisector of $ \angle CBF$, so $ \angle CBD\equal{}10$.


inequalities
trigonometry
induction
inequalities unsolved

Let be given $2n$ [b]positive real[/b] numbers $a_1,a_2,...,a_n$ and $b_1,b_2,...b_n$ satisfying: $a_1+a_2+...+a_n=b_1+b_2+...+b_n=\pi$.
Prove that:$\sum_{i=1}^{n}\frac{\cos a_i}{\sin b_i}\le\sum_{i=1}^{n}\cot b_i$

that's quite complex.you may try induction

I have already but I can't.I have done anything that I can do before posting this topic

Help me please!!!!!!!!!!!!!!!!!!


Ring Theory
superior algebra
superior algebra solved

In a commutative ring of characteristic 2, prove that the idempotents (a = a^2) form a subring.
where a subring is where a-b and ab are in S when a and b are in S.

$(a\pm b)^2=a^2+b^2+ab+ba=a^2+b^2+2ab=a^2+b^2=a+b$ (where we used characteristic $2$ and commutativity)

$(ab)^2=abab=a^2b^2=ab$ (where we used commutativity again)

So $a\pm b\in S$, $ab\in S$.

As blahblahblah is well aware, the use of the symbol "$\pm$" is a ring of characteristic 2 is a redundancy.


A convex polygonal has $ n$ edge lie in a square 1X1.
Prove:  Exist 3 summit $ A,B,C$ of the polygonal satisfy $ S(ABC) < \frac{16pi}{n^3}$

This might be clear to other people, but could you explain your notation?

What is the S function? $ S(ABC)$ represents what?
And what is a "summit" of a polygon?

Just some notation discrepancies (at least to me).

My interpretation of this question is as follows:

A convex n-gon is fully contained within a unit square. Prove that there exists at least one group of three vertexes such that the area of the triangle formed by these vertexes is less than $ \frac{16}{n^3} \pi$


and this was number 7:

7.	Joe and Suzy are standing at diametrically opposite points on a             7. __________
      circular track.  They start running towards each other(each is moving
      at a constant speed).  They first meet once Joe has run $ 3\pi$ miles.  Joe
      runs 10 mph slower than Suzy.  What is the area of the circular track?
      Express your answer in terms of $ \pi$ .

and this is number 17

17.  $ \frac{\log_{6} {8}} {\log_{6} {144}}$ can be expressed as $ \log_{a} {b}$. What is the value of  a*b?                           17. __________

and this is number 23

what is the value of $ (\sin{1337})^2\plus{}(\cos{257})^2$                                      23. __________

and then look at number 28(compared to everything else)

28	Joey and David run track and field.  Joey runs at an                  28. _________
      average speed of 10 mph while David runs at an average
      speed of 5 miles per hour.  Joey give David a 4 hour head
      start on a 200 mile race, how many more minutes will it
      take David to complete the race than Joey assuming no
      brakes are necessary?

is it just me or are these insanely hard?

[color=white]Haven't tried them yet... will edit this post when I do.[/color]
but could you post the whole sprint?
I would appreciate it a lot.
I think.
Barely looked at it, but... yeah they're insanely hard.  :lol: But, this is MC. No trig. And no logs. So at least those two aren't MC level... Except the answer to 23 is obviously 1.  :lol:

17 is ambiguous. It could be 144*8, or 144*144*8*8, or stuff.


We have $ f(s^2\plus{}f(t))\equal{}f((\minus{}s)^2\plus{}f(t)) \Rightarrow (2t\minus{}s)^2 f(2t\plus{}s)\equal{}(2t\plus{}s)^2 f(2t\minus{}s)$. Now for every $ p,q$ we can find $ s,t$ such that $ p\equal{}2t\minus{}s,q\equal{}2t\plus{}s$.So $ p^2 f(q)\equal{}q^2 f(p)$ for all $ p,q$.
Case 1: If there is a $ p \neq 0$ such that $ f(p)\equal{}0$, we will have all $ f(q)\equal{}0$. And it's easy to check that $ f(x)\equal{}0$ for all $ x$ is a solution.
Case 2: If not , we have for $ x \neq 0$, $ \frac{f(x)}{x^2}\equal{}k$, where $ k$ is a non-zero constant , $ \Rightarrow f(x)\equal{}kx^2$ for all $ x$
Substitute $ f(x)\equal{}kx^2$ into the equation we have $ k(s^2\plus{}kt^2)^2\equal{}k(s^2\minus{}4t^2)^2 \Rightarrow s^2\plus{}kt^2\equal{}s^2\minus{}4t^2$ or $ \minus{}(s^2\minus{}4t^2)$ (rejected) for all $ x$ $ \Rightarrow k\equal{}\minus{}4$, so $ f(x)\equal{}\minus{}4x^2$ is the solution to this case.


inequalities
LaTeX
trigonometry
inequalities proposed

[quote="mateivld"]Prove that in any acute triangle(p semiperimeter),with standard notations we have $\displaystyle \frac{9}{2(2+\sqrt 3)}\leq \sum \frac{Rp}{2aR+bc}<\frac{3}{2}$[/quote]

  Using Schwaz inequality only,combine with: $a^2+b^2+c^2 \leq 9R^2$  I'll post my solution tomorrow ..Ok

[quote="bodom"]it doesn't look so nice :? i guess i should stop looking for an elegant solution and try something more like brute force huh?[/quote]
what an idiot i am. :blush:  :blush:  :blush: of course it is almost trivial.from am-hm we have that sigma(1/(2aR+bc))>=9/(sigma(2aR+bc)).now we must prove that 2(2+sqrt3)Rp>=2R(sigma(a))+sigma ab<=>2(2+sqrt3)Rp>=4Rp+sigma ab <=>2sqrt3 *Rp>=sigma ab
but(sigma a)^2>=1/3 * sigma ab
so it is enough to prove that 6sqrt3 *Rp>=4*p^2<=>
3sqrt3/2 * R>=p and that is well known :P

[quote="socrates"]I think this problem was given in a (not recent ) mathlinks contest.
Check it[/quote]
i don't think so.It looks far to easy for a  ML contest.

The left part is easy,but what about the right part? :)

for the moment i do not have the time to try it but i am almost sure that something like brute force works in here.do u have a more elegant solution mateivld?

[quote="bodom"][quote="bodom"].....(sigma a)^2>=1/3 * sigma ab.......:P[/quote][/quote]I am sorry, Bodom, it is wrong. It is right such: $\left(\sum a\right)^2\ge 3\sum ab$. No problem, although you wrote wrong, however you thought right !. I recommend you to learn LaTeX as well as the English language. I will transcribe the your solution ($2s=a+b+c$).

$2s\le 3R\sqrt 3\Longrightarrow 3(ab+bc+ca)\le (a+b+c)^2\le 6Rs\sqrt 3$$\Longrightarrow \sum bc\le 2Rs\sqrt 3\Longrightarrow$

$2R\sum a+\sum bc\le 4Rs+2Rs\sqrt 3\Longrightarrow \sum (2aR+bc)\le 2(2+\sqrt 3)Rs\Longrightarrow$

$9\le \sum (2aR+bc)\cdot \sum \frac{1}{2aR+bc}\le 2(2+\sqrt 3)Rs\cdot \sum \frac{1}{2aR+bc}\Longrightarrow$

$\boxed {\ \frac{9}{2(2+\sqrt 3)}\le Rs\cdot \sum \frac{1}{2aR+bc}\ }.$

Like you how I wrote the your text ? The conclusion is clearly: my dear Bodom, [b]learn LaTeX ![/b]

$\ \boxed {\mathrm {\ We\ should\ do\ right\ what\ we\ do\ at\ all\ }}$

[quote="Virgil Nicula"][ I recommend you to learn LaTeX as well as the English language.quote]
yes i know i must learn latex but i can not download it and i do not have the time to solve this problem.About my english i do not know what do you mean :? .And about my solution it is only a typo.it should be indeed (sigma a)^2>=3sigma ab.but i only wrote wrong because you see that further in my solution i used the correct form

^[hide="Here is the solution of the right inequality."] [u]Solution.[/u] I suppose w.l.o.g. $a\le b\le c$. Prove easily that $a+h_a\le b+h_b\le c+h_c$, i.e.

$\frac{1}{c+h_c}\le \frac{1}{b+h_b}\le \frac{1}{a+h_a}$. Thus, $\boxed {\ \sum \frac{1}{a+h_a}\le \frac{3}{a+h_a}\ }\ \ (1)$.

Therefore, $C<90^{\circ}\Longrightarrow \cos C>0\Longrightarrow a^2+b^2>c^2\Longrightarrow \boxed {\ a^2>c^2-b^2\ }\ \ (2)$;

$\ a>c-b\Longrightarrow \boxed {\ a^2>a(c-b)\ }\ \ (3)$.

From the sum of the relations $(2)$ and $(3)$ results the relation $2a^2>(c-b)(a+b+c)\ \ (4)$.

From the product between the relation $a>c-b$ and the relation $(4)$ results the relation

$\boxed {\ 2a^3>(c-b)^2(a+b+c)\ }$. Prove easily that this relation is equivalently with the relation

$4s(s-b)(s-c)>a^2(s-a)\Longleftrightarrow 4S^2>a^2(s-a)^2\Longleftrightarrow$

$2S>a(s-a)\Longleftrightarrow h_a>s-a\Longleftrightarrow \boxed {\ a+h_a>s\ }\ \ (5)$. At last,

$Rs\sum \frac{1}{2aR+bc}=Rs\sum \frac{1}{2aR+2Rh_a}=\frac s2\sum \frac{1}{a+h_a}<\frac s2\cdot \frac{3}{a+h_a}<\frac 32$

(from the relations $(1)$ and $(5)$) $\Longrightarrow$$\boxed {\ Rs\sum \frac{1}{2aR+bc}<\frac 32\ }$

Try you finding a non-strict right inequality ? It is a provocation !

[u]Remark.[/u] $a+h_a\le b+h_b\Longleftrightarrow a^2 b+b\cdot ah_a\le ab^2+a\cdot bh_b\Longleftrightarrow$

$ab(a-b)\le 2S(a-b)\Longleftrightarrow (a-b)(ab-2S)\le 0\Longleftrightarrow a-b\le 0\Longleftrightarrow a\le b$

because $ab-2S=ab-ab\sin C=ab(1-\sin C)\ge 0$. Thus,

$a\le b\le c\Longleftrightarrow a+h_a\le b+h_b\le c+h_c$.[/hide]

My solution is based on the fact that $\displaystyle\frac{pR}{2aR+bc}\leq \frac{1}{2}$in any acute triangle . :)