[quote="TALENTO"]Hi

Find all the four-digit positive integers $ N \equal{} abcd$ such that $ abcd$ is equal to
product  $ a^b \cdot{} c^d$.

Greetings  :D[/quote]

Brute force : $ 2592$

Yes, but how? pco as I come to this result and shows that it is the only one.

Greetings  :D

[quote="TALENTO"]Yes, but how? [/quote]

As I said : brute force. Just 9000 tests to do. And there is a unique solution.

How would a formal approach?

Greetings  :D

Hello please help with this problem as I see no solutions.

Greetings  :D

Hello please if I can lend a hand to give a formal solution.

Greetings  :D


[hide][b]162[/b]
There are 27 different ways for 3 dices to be rolled and the sum to be 10. There are 6 arrangements of colors for the dices, so
27*6 = 162[/hide]

The answer key says 27, I don't think the different colors thing matters (although it should).

[hide]
Counting duplicates - 27
136
145
154
163

244
235
253
226
262

316
361
325
352
334
343

415
451
424
442
433

514
541
523
532

613
631
622

I would split this into two groups, the first die, and the other two dice. If the first die is a 1, then the other two must add up to 9, what combinations of 2 dice makes 9? If the first die is a 2, then the other two must add up to 8, what combinations of 2 dice makes 8? Just repeat this simplification for 1-6 for the first die and you should be all set.
[/hide]

I wonder why it mentions that the dice are of different color though? I thought that was to indicate that different orders count as separate combinations, i.e. 613 is not the same as 316.

[quote="easyas3.14159..."]Three fair, standard six faced dice of different colors are rolled. In how many ways can the dice be rolled such that the sum of the numbers is 10?[/quote]

[hide]
1,3,6
1,4,5
2,2,6
2,3,5
2,4,4
3,3,4

6 ways[/hide]

Biff: She only counted combinations.

I haven't seen the problem but the color part of it was probably to make it so 424, 244, and 442 would count as three different ways, not one.

Actually i just realized that the answer really is 27 because if youre looking for permutations and not combinations, then it already takes care of the colors.
EX:
4,3,3

You have
4,3,3
3,4,3
3,3,4

The first one denotes the first color, second is second color, third is third color.
So you can't multiply by 6 because you have already arranged out the colors[/b]

[quote="magixter"][hide][size=75]Actually, the Answer is can be found in a simple table, there is also a pattern![/size]

1 3 6
1 4 5
1 5 4
1 6 3

2 2 6
2 3 5
2 4 4
2 5 3
2 6 2

3 1 6
3 2 5
3 3 4
3 4 3
3 5 2
3 6 1

so on.....

As you can see, it goes 4+5+6, then it will reverse itself and go 6+5+4. So the answer is (4+5+6)+(4+5+6) or (4+5+6)*2
or 15*2=30.

The Answer is 30 WAYS![/hide][/quote]

Incorrect, here's why:
[hide]
1. Your pattern is incorrect, it goes 4, 5, 6, 5, 5, 4, 3.
2. You didn't account for the fact that 622, 433, 244 will have less combinations because two of the dice will be the same.[/hide]

Sorry, that is correct,

The pattern is

4+5+6+5+4+3=27

My mistake!!

 ;)

Is there any way to find the answer without listing the combinations out?!

Well, kind of.
Think of it this way.

If the first is 1, we have to make 9 with the other 2. 4 ways to do this
First is 2 - we have to make 8. 5 ways.
first is 3- we have to make 7. 6 ways.
first is 4- we have to make 6. 5 ways.
First is 5- we have to make 5. 4 ways.
First is 6- we have to make 4. 3 ways.

And finding out how many ways each could be made is fairly simple, it's a huge part of dice probability.

[hide]first is 6, 3 ways
first is 5, 4 ways
first is 4, 5 ways
first is 3, 6 ways
first is 2, 5 ways
first is 1, 4 ways

27 total[/hide]


number theory
least common multiple
LaTeX

Every 3 guests used a dish for rice between them, every 4 a dish for broth, every 2 a dish for meat there are 65 dishes.Can you figure how many guests there were?

Enjoy  :coolspeak:  :w00tb:
Please no  :spam:

[quote="cgyao15"]actually there are many solutions.
If you wanted the least amount, we have
[hide] the LCM{2,3,4}=12 so twelve guests[/hide][/quote]
You are actually wrong, [hide] 60 guests, remember 65 dishes[/hide]

A little $ \text{\LaTeX}$ touch up:

\begin{align*}
\dfrac{x}{3} + \dfrac{x}{4} + \dfrac{x}{2} &= 65\\
\dfrac{13}{12}x &= 65\\
x &= 60
\end{align*}

Thanks, what did I do wrong to mess up the format?

You didn't use the align* environment or the \frac command.

I just tried it in a preview in the Test froum. Thanks for the help on the $ \text{\LaTeX}$.

EDIT- Hello 250 posts and next level!


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Given $ A \in M_{n}(K)$.Prove that:if A is not invertible then we can find $ B \in M_{n}(K),B \not \equal{} 0$ such that $ AB \equal{} BA \equal{} 0$

$ A\cdot\text{adj }A \equal{} \text{adj }A\cdot A \equal{} \det A\cdot I_n$, what is $ 0$ if $ A$ is non-invertible (because then, $ \det A \equal{} 0$)...

[EDIT: wrong]

  darij

That's not really enough, since $ \text{rank} \, A \leq n \minus{} 2 \, \implies \, \text{adj} \, A \equal{} 0$.

OK,because for any invertible matrix $ T$,
we have:$ TAT^{\minus{}1}TBT^{\minus{}1}\equal{}0 \iff AB\equal{}0$
$ TBT^{\minus{}1}TAT^{\minus{}1} \iff BA\equal{}0$
So we may consider the Jordan Form:
Consider the $ 0$'s block is at the end.
Let $ B\equal{}E_{n1}$ :D

Something easier:

The minimal polynomial of $ A$ is of the form $ X \cdot P(X)$. Now just choose $ B \equal{} P(A)$. :)

[EDIT] Darij saved the day :showoff:

Or take the smallest integer $ k$ such that the left upper $ k\times k$ minor of the matrix $ A$ has zero determinant, and define $ B$ to be the adjugate of this $ k\times k$ minor, filled up with $ 0$'s in the remaining $ n\minus{}k$ rows and $ n\minus{}k$ columns.

  Darij

In a finite-dimensional commutative algebra, every element is either a unit or a zero-divisor. Apply to $ K[A]$. owk

Ok,Can we prove more:
Given $ A \in M_n(K)$ with $ rank(A)\equal{}k$,Then there exist a matrix $ B$ with $ rank(B)\equal{}n\minus{}k$,such that $ AB\equal{}BA\equal{}0$

My proof is not right in fact,because we need that $ K$ is algebrically closed.And  perfect_radio,darij grinberg's proof seems not adapt to the new problem.

[quote="zhaobin"]
Given $ A \in M_n(K)$ with $ rank(A) \equal{} k$,Then there exist a matrix $ B$ with $ rank(B) \equal{} n \minus{} k$,such that $ AB \equal{} BA \equal{} 0$
.[/quote]

I  think this proposition is right,Because you can consider the Smith's canonical form instead of Jordan's canonical form. :)

You do not need any canonical forms, just choose an isomorphism $ K^n/\mathrm{im}\,A\cong\ker A$ and let $ B$ describe the composition $ K^n\to K^n/\mathrm{im}\,A\to\ker A\subseteq K^n$. owk


let a,b,c>0 prove that lower part inequality for 0<x<min{a,b,c} is ture                                               $ \frac{x(a-x)}{x^2+ab-bx}\ + \frac{x(b-x)}{x^2+bc-cx}\ + \frac{x(c-x)}{x^2+ca-ax}\ \leq\ 1$

Very nice one!

[color=blue][b]Problem.[/b] If a, b, c and x are positive reals, and $x\leq\min\left(a;\;b;\;c\right)$, then

$\frac{x\left(a-x\right)}{x^2+ab-bx}+\frac{x\left(b-x\right)}{x^2+bc-cx}+\frac{x\left(c-x\right)}{x^2+ca-ax}\leq 1$.[/color]

[i]Solution.[/i] Since $x\leq\min\left(a;\;b;\;c\right)$, we have $x\leq a$, and thus $x^2+ab-bx\geq x^2+ab-ba\geq x^2\geq 0$, and similarly $x^2+bc-cx\geq 0$ and $x^2+ca-ax\geq 0$. Thus the denominators of the fractions in the above inequality are all $\geq 0$. Now,

$1-\left(\frac{x\left(a-x\right)}{x^2+ab-bx}+\frac{x\left(b-x\right)}{x^2+bc-cx}+\frac{x\left(c-x\right)}{x^2+ca-ax}\right)$
$=\frac{\left(x^3-\left(a-x\right)\left(b-x\right)\left(c-x\right)\right)^2}{\left(x^2+ab-bx\right)\left(x^2+bc-cx\right)\left(x^2+ca-ax\right)}\geq 0$,

and thus $\frac{x\left(a-x\right)}{x^2+ab-bx}+\frac{x\left(b-x\right)}{x^2+bc-cx}+\frac{x\left(c-x\right)}{x^2+ca-ax}\leq 1$, so the problem is solved.

  darij


A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only $ 1/3$ of the marbles in the bag are blue. Then yellow marbles are added to the bag until only $ 1/5$ of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

$ \textbf{(A)}\ \frac {1}{5}\qquad \textbf{(B)}\ \frac {1}{4}\qquad \textbf{(C)}\ \frac {1}{3}\qquad \textbf{(D)}\ \frac {2}{5}\qquad \textbf{(E)}\ \frac {1}{2}$

Say after the yellow marbles are added, there are $ 20$ marbles that are blue.  So if we double this, we get $ \frac{40}{120}$, or $ \boxed{\frac13}$.

This is stupid for problem number 14. YOu can just plug in values 1R1B.2R1B.2R1B2Y.2R2B2Y. which is 1/3

[size=75]I disagree with ernie and varunrocks.

After all the extra marbles are added except the blue, $ 1/3$ of the marbles are red and $ 1/5$ are yellow. So $ 1\minus{}1/3\minus{}1/5$, or $ 7/15$ are blue before you add the extra blues. So out of $ 15x$, $ 7x$ are blue, $ 5x$ are red, and $ 3x$ are yellow. Now when you double the amount of blues, $ 14$ are blue out of $ 22x$ total marbles. So $ 14$ out of $ 22$ are blue after the amount of blues is doubled. So $ 7/11$ are blue.

Unfortunately this isn't one of the options, so... I don't know what I did wrong. Will someone please tell me why? :oops: [/size]

Ok nevermind...I misread the problem but now I understand it. I agree with ernie and varunrocks.  :blush:

it's not A...

Just use small cases start with 1 blue, add 2 red so now 1/3 of the bag is blue.

Now to make 1/5 of the back blue you have to add 2 yellow.

If you double to blues, it will be 2,2,2 or 1/3 of the bag is blue

I think it's $ \frac{1}{3}$

This is cause, after you add red marbles, the ratio would be 1:2

Then after you add yellow marbles, the ratio is Blue:Red:Yellow---> 1:2:2 or 1:5 for Blue:Total

So...after you double, it's 2:2:2 like Ernie said. $ \frac{2}{6}\equal{}\boxed{\frac{1}{3}}$

$ \frac 15$ of the marbles are blue. We add another $ \frac 15$ blue marbles to the bag and the blue marbles are now $ \frac{\frac 25}{\frac 65} \equal{} \frac 13$

The answer is $ \textbf{(C)}\ \frac{1}{3}$


Prove that there isn't $ n \in \mathbb{N}$ such that: $ 1\plus{}2\plus{}...\plus{}n\equal{}a \cdot 1111$, where $ a \in \{ 1,2,...,9 \}$.

[hide="Bashing..."]
$ n(n\plus{}1)\equal{}a\cdot 2222$.
$ n^2\plus{}n\minus{}a\cdot 2222\equal{}0$.

Examining the discriminant, $ \sqrt{1\plus{}8888a}$ must be a perfect square for $ n$ to be an integer.

Checking, none of $ a\in\{1,2,...,9\}$ make this a perfect square, so $ n$ is not a natural number.[/hide]

There must be a better way.

[hide="logic"]alternatively, we could use logic and examine the prime factorization of 2222a, which is 2*11*101. For this to satisfy n(n+1)=2222a, there would have to be a factor which is either 1 above 101*11 (1112) or 1 below 101*11 (1110). this is not possible with numbers 1 through 9, so our (not by any means rigorous) proof is complete[/hide]

sorta like the guy b4 me said n(n+1)=a*2*11*101
2*a*11 cant become 102 or 100 if a is from 1-9 so yea....


What is the smallest of three positive consecutive odd integers if
the product of the second and third integers is 63?

I was told that x + x+ 1 + x+2 = three odd consecutive integers.

Can I equate x + x+ 1 + x+2 to 63 and still get 5?

It's impossible for x, x + 1, and x + 2 to be three consecutive odd integers (assuming x is an integer). At least one of these terms must be even. Perhaps you mean x, x + 2, and x + 4?

Let the numbers be x-2,x,and x+2. We are given that x(x+2)=63. The prime factorization of 63 is $3^27$. By playing around with the numbers a bit(as in glancing at them...), we see that x could be 7 and x+2=9. The smallest number would be 5 in this case.


Let $ a$, $ b$, $ c$, and $ d$ be real numbers such that $ a\leq b$ and $ c\leq d$. What is wrong with $ a\minus{}c\leq b\minus{}d$? Make a similar inequality.

Heh I made that mistake in my Intermediate Algebra Challenge Set when proving classical inequalities.

The easiest counter example is $ a \equal{} \minus{}1$, $ b \equal{} 1$, $ c \equal{} \minus{}2$, and $ d \equal{} 2$.

We have:

$ \minus{}1 \le 1\\
\minus{}2 \le 2\\
\minus{}1 \plus{} 2 \le \minus{}1\\
1 \le \minus{}1$

Obviously that is not true, so the subtraction is invalid. I wonder if the subtraction works if you flip the sign.

Anyways this is why it doesn't work. First, multiply the second inequality by $ \minus{}1$.

$ \minus{}c \ge \minus{}d$.

Now you can't add the two inequalities because they make different comparisons, so you can't subtract.


If anyone could show me how this type of problem is done, it would be greatly appreciated!

For the given plane curve, find the unit vector T and the unit normal vector N.

x = t-sint, y = 1-cost, where t = pi/2

So your curve is $ \vec{r}(t)=<t-\sin(t), 1-\cos(t)>$ so the tangent is $ r'(t)=<1-\cos(t), \sin(t)>$ and thus at $ t=\pi/2$ we have $ r'(\pi/2)=<1, 1>$ to make this into the unit tangent divide by the length $ \vec{T}(t)=\frac{r'(t)}{|r'(t)|}$ but since we already evaluated we know $ \vec{T}(\pi/2)=<1/\sqrt{2}, 1/\sqrt{2}>$.

Also,
$ \vec{N}(t)=T'(t)=\frac{d}{dt}\frac{<1-\cos(t), \sin(t)>}{\sqrt{2-2\cos(t)}}$


All of you plz write down your IITJE registration nos..

Mine is 5071016...

mine is 9747833 :rotfl: :D

I dont think neone would wanna do this except soumya :D

i agree with madness :D :D :rotfl:

(but i hope against it )

rohit's roll no was a troll.. :rotfl:  :rotfl:  :mad:

u expected it to be correct?  :huh:   :P  :rotfl:

i forget that it was madness :oops:  :rotfl:

Dint you read this?  :lol:  :D

[quote="madness"]
[b]I dont think neone would wanna do this except soumya [/b][/quote]

mysteriously my username has transformed into Da Vinci  :spider:

[hide] :clap: [/hide]

we can perhaps do it ourselves here.. :rotfl:  :roll:

i hacked aops and now i know all ur passwords  :P  :P  :D

it's great posting here after so long!!! :D
i still rule see my posts!!! wow :roll:

can we do sth about flags in JOI?????? :)  :D  :)

I thought everyone is supposed to be indian there :huh: Do you want state flags? :huh: There aren't any as far as i know :maybe:

karnataka has a state flag  :huh:  :huh:  :huh:

yes i can have thema t JOI but wud it really matter?


calculus
integration
calculus computations

[hide="Answer"]$ \left(x e^{x^k}\right)' \equal{} e^{x^k} \plus{} x \cdot kx^{k\minus{}1} e^{x^k} \equal{} (1 \plus{} kx^k)e^{x^k}$

Therefore, $ \int_0^1 (1 \plus{} kx^k)e^{x^k}\,dx \equal{} \int_0^1 \left(x e^{x^k}\right)'\,dx \equal{} \left[x e^{x^k}\right]_0^1 \equal{} e$.[/hide]

That's correct answer. I made the problem as follows.

$ \int_0^1 e^{x^{2n}}dx\equal{}\left[xe^{x^{2n}}\right]_0^1\minus{}\int_0^1 x\cdot 2nx^{2n\minus{}1}e^{x^{2n}}dx$

$ \equal{}e\minus{}2n\int_0^1 x^{2n}e^{x^{2n}}dx$, yielding $ \int_0^1 (1\plus{}2nx^{2n})e^{x^{2n}}dx\equal{}e.$

$ \therefore \int_0^1 (1\plus{}2008x^{2008})e^{x^{2008}}dx\equal{}e.$


Let $ (a_n)_{n\ge 1}$ a geometric progression and let $ S_k\equal{}\sum_{i\equal{}1}^ka_i$ . Prove that $ S_n,S_{2n}\minus{}S_n$ and $ S_{3n}\minus{}S_{2n}$ form a geometric progression .

[quote="Pain rinnegan"]Let $ (a_n)_{n\ge 1}$ a geometric progression and let $ S_k \equal{} \sum_{i \equal{} 1}^ka_i$ . Prove that $ S_n,S_{2n} \minus{} S_n$ and $ S_{3n} \minus{} S_{2n}$ form a geometric progression .[/quote]

$ a_i\equal{}a_0r^i$

1) If $ r\equal{}1$, $ S_n\equal{}na_0$, $ S_{2n}\equal{}2na_0$ and $ S_{3n}\equal{}3na_0$. Then $ S_n\equal{}na_0$, $ S_{2n}\minus{}S_n\equal{}na_0$ and $ S_{3n}\minus{}S_{2n}\equal{}na_0$ form a geometric progression (ratio $ 1$).

2) If $ r\neq 1$, $ S_n\equal{}a_0r\frac{r^n\minus{}1}{r\minus{}1}$, $ S_{2n}\equal{}a_0r\frac{r^{2n}\minus{}1}{r\minus{}1}$ and $ S_{3n}\equal{}a_0r\frac{r^{3n}\minus{}1}{r\minus{}1}$. Then $ S_n\equal{}a_0r\frac{r^n\minus{}1}{r\minus{}1}$, $ S_{2n}\minus{}S_n\equal{}a_0r\frac{r^{2n}\minus{}r^n}{r\minus{}1}$ and $ S_{3n}\minus{}S_{2n}\equal{}a_0r\frac{r^{3n}\minus{}r^{2n}}{r\minus{}1}$ form a geometric progression (ratio $ r^n$).

And I dont think this might be in an olympiad forum.

pco, you've just beaten me to it. (I forgot to consider r=1.)


Solve the following equations simultaneously: \[x^y = z,\;y^z = x,\;z^x = y.\]

If you proposed it in the number theory section I assume that $x,y,z$ are integers, am I wrong?
Clearly, they have to be positive.
Now, $z=x^y \geq x$.
Similarly $x \geq y$ and $y \geq z$, thus $x=y=z$.
The system reduces to $x^x = x$ which gives $x=y=z=1.$

Pierre.

Starting with $x^y=z$, raising both sides to the $x$ power we get
\[\displaystyle x^{xy}=z^x=y\]
Raising both sides to the $z$ power we get
\[\displaystyle x^{xyz}=y^z=x\]
Similarly
\[\displaystyle y^{xyz}=y\]
\[\displaystyle z^{xyz}=z\]
Multiplying these together
\[\displaystyle (xyz)^{xyz}=xyz\]
For $x,y,$ and $z$ integers this implies $xyz=1$ or $xyz=-1$ which when considered with the other equations implies $x=y=z=\pm 1$


Suppose that $ \sum_{n\equal{}0}^\infty a_n$ diverges and $ a_n>0$. Show that there exist $ b_n$ with $ \frac{b_n}{a_n}$ tending to zero as n tends to infinity and $ \sum_{n\equal{}0}^\infty b_n$ divergent.

[quote="JohnPentax"]Suppose that $ \sum_{n \equal{} 0}^\infty a_n$ diverges and $ a_n > 0$. Show that there exist $ b_n$ with $ \frac {b_n}{a_n}$ tending to zero as n tends to infinity and $ \sum_{n \equal{} 0}^\infty b_n$ divergent.[/quote]

You may find it helpful to think about $ N_1<N_2< ...$ such that $ \sum_{n\equal{}N_k}^{N_{k\plus{}1}\minus{}1}a_n > 1$ for all $ k$.


geometry
geometric transformation
reflection
analytic geometry
algebra
polynomial
function

I couldn't believe this year's Problems would be this easy. Got all the solutions in 3 hrs. The Competition ended $ 1\frac{3}{4}$ hours ago.

1. Let $ ABC$ be triangle, $ I$ its in-center; $ A_1,B_1,C_1$ be the reflections of $ I$ in $ BC, CA, AB$ respectively. Suppose the circum-circle of triangle $ A_1B_1C_1$ passes through $ A$. Prove that $ B_1,C_1,I,I_1$ are concylic, where $ I_1$ is the in-center of triangle $ A_1,B_1,C_1$.

2. Find all triples$ (p,x,y)$ such that $ p^x\equal{}y^4\plus{}4$, where $ p$ is a prime and $ x,y$ are natural numbers.

3. Let $ A$ be a set of real numbers such that $ A$ has at least four elements. Suppose $ A$ has the property that $ a^2\plus{}bc$ is a rational number for all distinct a.b.c in $ A$. Prove that there exists a positive integer $ M$ such that $ a\sqrt{M}$ is a rational number for every $ a$ in $ A$

4. All the points with integer coordinates in the $ xy$-Plane are coloured using three colours, red, blue and green, each colour being used at least once. It is known that the point $ (0,0)$ is red and the point $ (0,1)$ is blue. Prove that there exist three points with integer coordinates of distinct colours which form the vertices of a right-angled triangle.

5. Let $ ABC$ be a triangle; $ \lambda_A,\lambda_B,\Lambda_c$ be three equal disjoint circles inside $ ABC$ such that $ \lambda_A$ touches $ AB$ and $ AC$ ,$ \lambda_B$ touches $ BC$ and $ BAC$,$ \lambda_C$ touches $ CA$ and $ CB$. Let $ \lambda$ be a circle circles $ \lambda_A$,$ \lambda_B$,$ \lambda_C$. Prove that the line joining the circum-center $ O$ and the in-center $ I$ of triangle $ ABC$ passes through the center of $ \lambda$.
[Note: I Have changed the symbol tao to lambda($ \lambda$) because I am not very familiar with latex.]

6.Let $ P(x)$ be a Polynomial with integer coefficients. Prove that there exists two polynomials $ Q(x)$ and $ R(x)$, again with integer coefficients, such that $ (i)$ $ P(x)Q(x)$ is a polynomial in $ x^2$; and $ (ii)$ $ P(x)R(x)$ is a polinomial in $ x^3$

What is India Section? How to go there? (I am new here)

[quote="pardesi"]can someone move his to india section[/quote]

Why move to India section? Why not share with others?  :wink:

By the way, guess where problem 3 is from: http://www.mathlinks.ro/viewtopic.php?t=111253

Solutions could be discussed here. As regards results and other discussions see the India sub forum.

Case1: If x is even, x=2k
(p^k-y^2)*(p^k+y^2)=4=1*4=2*2
If p^k -y^2 is odd, then p^k+y^2 is odd and vice versa
So (p^k-y^2,p^k+y^2)cant take (1,4) and as y is natural number, they cant even take (2,2)
Case2:If x is odd
p^x = 4 mod y
=> As 4 is square,p^x should be square modulo y
=> x mod T(y)= odd as T(y) is even for all y>2(T(y):totient function of y)
If y=2, p^x = 20, So, y^4+4 can't have only one distinct prime factor
If y=1,p^x=1+4=5=>p=5,x=1.
The only solution is (5,1,1)
Thanks for reminding y=1 case

But $ (p,x,y)\equal{}(5,1,1)$ is a solution.

How is x(mod T(y)) odd???

In that case x is odd and T(y) is even.
So, (odd) mod (even) is always odd

Case 1: $ p$ is odd

$ p^x \equal{} (y^2\plus{}2y\plus{}2)(y^2\minus{}2y\plus{}2)$

If $ p$ divides $ y^2\plus{}2y\plus{}2$ and $ y^2\minus{}2y\plus{}2$, $ p$ divides $ 4y$, so $ y \equiv 0 (\mod p)$ and $ p^x \equiv y^4\plus{}4 \equiv 4 (\mod p)$.

It's contradiction. So $ p$ doesn't divide $ y^2\minus{}2y\plus{}2$.

$ y^2\minus{}2y\plus{}2$ is a factor of $ p^x$, so it must be 1. $ y \equal{} 1$. $ (5,1,1)$ is a solution.

Case 2: $ p$ is odd

By modulo 8, $ x \equal{} 2$. But it's not a solution.

[quote="Akashnil"]5. Let $ ABC$ be a triangle; $ \lambda_A,\lambda_B,\Lambda_c$ be three equal disjoint circles inside $ ABC$ such that $ \lambda_A$ touches $ AB$ and $ AC$ ,$ \lambda_B$ touches $ BC$ and $ BAC$,$ \lambda_C$ touches $ CA$ and $ CB$. Let $ \lambda$ be a circle circles $ \lambda_A$,$ \lambda_B$,$ \lambda_C$. Prove that the line joining the circum-center $ O$ and the in-center $ I$ of triangle $ ABC$ passes through the center of $ \lambda$.
[Note: I Have changed the symbol tao to lambda($ \lambda$) because I am not very familiar with latex.]
[/quote]
[quote="limsk1"]I think there are some errata during writing problem No.5^^;;;
Please explain me what is exact problem...[/quote]
Yes. It will be

5. Let $ ABC$ be a triangle; $ \Gamma_A,\Gamma_B,$$ \Gamma_C$ be three equal disjoint circles inside $ ABC$ such that $ \Gamma_A$ touches $ AB$ and $ AC$ ,$ \Gamma_B$ touches $ BC$ and $ BA$,$ \Gamma_C$ touches $ CA$ and $ CB$. Let $ \Gamma$ be a circle [b]touching[/b] $ \Gamma_A$,$ \Gamma_B$,$ \Gamma_C$ externally. Prove that the line joining the circum-center $ O$ and the in-center $ I$ of triangle $ ABC$ passes through the center of $ \Gamma$.

Thank you Pardesi. :wink:

the code for $ \Gamma$ is
[code]\Gamma[/code]

u can find latex codes [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url] at the top of any page u vist on Math Links/AoPs

[quote="Akashnil"]
5. Let $ ABC$ be a triangle; $ \lambda_A,\lambda_B,\Lambda_c$ be three equal disjoint circles inside $ ABC$ such that $ \lambda_A$ touches $ AB$ and $ AC$ ,$ \lambda_B$ touches $ BC$ and $ BAC$,$ \lambda_C$ touches $ CA$ and $ CB$. Let $ \lambda$ be a circle circles $ \lambda_A$,$ \lambda_B$,$ \lambda_C$. Prove that the line joining the circum-center $ O$ and the in-center $ I$ of triangle $ ABC$ passes through the center of $ \lambda$.
[/quote]
This problem is almost same with imo 1981/5


Prove that for any $a,b,c>0$,

$\displaystyle (a+b)(b+c)(c+a) \ge 8abc,$

and determine when equality holds.

$(a+b)(b+c)(a+c) \geq 8abc$
$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc \geq 8abc$
$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 6abc$

which ( I think) follows by Muirheads inequality on a,b,c since (2,1,0) majorizes (1,1,1)

The equality obviously holds when $a=b=c$

$\displaystyle (a+b)(b+c)(c+a) \ge 8abc,$
$\displaystyle a^2b+a^c+b^a+b^c+c^a+c^b+2abc \ge 8abc$
$\displaystyle a(c^2-2bc+b^2)+b(a^2-2ac+c^2)+c(a^2-2ab+b^2) \ge 0$
$\displaystyle a(b-c)^2+b(a-c)^2+c(a-b)^2 \ge 0$

[quote="edriv"]$\displaystyle a^2b+a^c+b^a+b^c+c^a+c^b+2abc \ge 8abc$[/quote]

This step seems strange.  Did you forget some exponents?   :huh:

[quote="shobber"]$a+b \geq 2\sqrt{ab}$. Multiply them togather and we are done.[/quote]

Probably the fastest solution :D  The $8$ on the RHS motivates it.

[hide]
WLOG let $a+b+c=1$ because the expressions are homogenous. Then we have
\[ \begin{eqnarray*} (a+b)(b+c)(c+a)&=&(1-c)(1-b)(1-a)\\ &=&ab+bc+ca-abc\\ &=&\left( \frac 1a +\frac 1b +\frac 1c -1\right) abc \]

By AM-HM, $\frac 1a +\frac 1b +\frac 1c\geq 9$ which proves the inequality.
[/hide]

Substitute $x=a+b; y=b+c; z=a+c$:
\[ xyz \geq (x-y+z)(x+y-z)(y+z-x) \]
Which is true by rearrangement for products.  :)

[quote="Jan"]
Which is true by rearrangement for products.  :)[/quote]

Please, could you explain this to me?

Well, $\left\{x+y,x+z,y+z\right\}$ and $\left\{-z,-y,-x\right\}$ are sorted equal. So it's literally rearrangement for products.  :)

\[ (x+z-z)(x+y-x)(y+z-y) \geq (x-y+z)(x+y-z)(y+z-x) \]

[quote="shobber"]$a+b \geq 2\sqrt{ab}$. Multiply them togather and we are done.[/quote]

Simple the best!  :)


inequalities
invariant
geometry
geometric transformation
inequalities solved

If a<b<c<d prove that

(a+b+c+d)^2 > 8(ac+bd)

Source:

[i]Niels Hendrik Abel Contest 1993, Norway[/i] :)

The official solution is really boring.

Can you please post the official solution ?

P.S. I have found a very neat and simple solution.

The official solution:

Let F = (a + b + c + d)2 - 8(ac + bd).
When a varies, and b, c, d are fixed, F attains its minimum for a = 3c - b - d.
This minimum equals 8(d - c)(c - b) > 0.

My solution.

This inequality is invariant by translation A=a +x, B=b+x, C=c+x, D=d+x.

If we get x=-b we have A= a-b<0, B=0 , C=c-b>0 ,D=d-b>0 and the inequality become

 (A+B+C)^2>8AC   true as AC<0.

I hope it is correct. Do you think it is more simple or more difficult than official solution?

well I just thought about this problem and I think I came up with a nicer way: apply AM-GM to get
(a+b+c+d)2 \geq 4(a+d)(b+c) > 8(ac+bd) <=>
ab+ac+bd+cd > 2ac +2bd <=> ab-ac-bd+cd >0 <=> (a-d)(b-c)>0 which is true from the condition a<b<c<d and we are done.  nice problem.

Your solution, Valentin, is very nice and ,above all, more natural and simple than mine.

A quite natural solution is the following:
  Consider f(x)=(x-a)(x-c)+(x-b)(x-d). Then f(b)<0, thus the equation f(x)=0 has real roots, from where the conclusion follows.

don't get me wrong harazi, but that solution, in my opinion, is far from being natural. it is not something anyone can come up to anytime ...
i repeat, it is just my opinion :)

Well, in fact it is quite natural after you prove 3 or 4 inequalities (for example Aczel) using this method.

The  official  condition  was  b<c<d  and  wanted  to  prove  the  inequality
(a+b+c+d) 2 > 8(ac+bd)  for  all  reals  a.


If $a,b,c$ are positive numbers such that
$\frac a{b}+\frac b{c}+\frac c{a}=5$,
then
$\frac b{a}+\frac c{b}+\frac a{c}\ge \frac{17}{4}$.

Let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
Then $xyz=1, x+y+z=5$. ETS : $xy+yz+zx\ge \frac{17}{4}$.
By Lagrange multiplier,
If $xy+yz+zx$ is minumum then $y+z=\lambda_{1}+\lambda_{2}yz , ...$.
Case 1) $(x-y)(y-z)(z-x)\not=0$
$(x-y)(z-\lambda_{1})=0$ -> $x=y=z=\lambda_{1}$. Contradiction!
Case 2) WLOG y=z.
$xy^{2}=1, x+2y=5$ -> $\frac{1}{y^{2}}+2y=5$ -> $(2y-1)(y^{2}-2y-1)=0$.
-> $y=\frac{1}{2}$ or $1+\sqrt{2}$ or $1-\sqrt{2}$.
now, it is easy to show $xy+yz+zx \ge \frac{17}{4}$.

[hide]
So substituting $x=\frac{a}{b}$ etc. as in above solution, let
\[P(t)=4(t-x)(t-y)(t-z)=4t^{3}-20t^{2}+kt-4\]
be the cubic with leading coefficient 4 and roots $x,y,z$. Then $k=4(xy+yz+zx)$. Suppose $k<17$. Then for all positive $t$,
\[P(t)=(t-4)(2t-1)^{2}+(k-17)t<(t-4)(2t-1)^{2}\]
In particular,
\[(x-4)(2x-1)^{2},\ (y-4)(2y-1)^{2},\ (z-4)(2z-1)^{2}\]
are all positive, so $x,y,z>4$. But then $x+y+z>12>5$, a contradiction! Therefore $k\geq 17$ and $xy+yz+zx\geq \frac{17}{4}$.
[/hide]

The idea from this inequality is similar to that of Vietnam TST 2004. We put $a/b=x, b/c=y, c/a=z$, then $xyz=1$ and $x+y+z=5$. From this and AM-GM inequality we get
\[(5-z)^{2}\geq\frac4z.\]
Note that $z=4$ is one factor of $z(5-z)^{2}=4$. Then we have from the restriction $3-\sqrt8\leq z\leq 4$ or $z\geq3+\sqrt8$. The desired inequality reads
\[xy+yz+zx\geq\frac{17}4.\]
The rest is, I think, not too hard.

This is the same as (entirely the same in solution) an old unsolved problem of mine posted before by my friend Gackiem in Mathlinks

Let $a,b,c$ be positive real numbers such that $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10.$ Prove that
\[\frac{13+\sqrt{5}}{2}\ge \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{13-\sqrt{5}}{2}.\]

[i]Hungkhtn , can you post your solution?[/i] :(

[quote="hungkhtn"]This is the same as (entirely the same in solution) an old unsolved problem of mine posted before by my friend Gackiem in Mathlinks

Let $ a,b,c$ be positive real numbers such that $ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} 10.$ Prove that
\[ \frac {13 \plus{} \sqrt {5}}{2}\ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \frac {13 \minus{} \sqrt {5}}{2}.
\]
[/quote]
It is unsolved since it is wrong (for the right hand). Try $ a\equal{}\frac{10\minus{}\sqrt{10}\plus{}5\sqrt{2}\minus{}2\sqrt{5}}{20},b\equal{}\frac{\sqrt{10}}{10},c\equal{}\frac{10\minus{}\sqrt{10}\plus{}2\sqrt{5}\minus{}5\sqrt{2}}{20}$ ;)
Actually, the following holds :)
Let $ a,b,c$ be positive real numbers such that $ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} 10.$ Prove that
\[ \frac {7\plus{}8\sqrt{2}\minus{}5\sqrt{5}}{2}\ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \frac {7\plus{}5\sqrt{5}\minus{}8\sqrt{2}}{2}.
\]
We can prove it using AM-Gm Inequality. :)

[quote="can_hang2007"]It is unsolved since it is wrong (for the right hand). Try $ a \equal{} \frac {10 \minus{} \sqrt {10} \plus{} 5\sqrt {2} \minus{} 2\sqrt {5}}{20},b \equal{} \frac {\sqrt {10}}{10},c \equal{} \frac {10 \minus{} \sqrt {10} \plus{} 2\sqrt {5} \minus{} 5\sqrt {2}}{20}$ ;)
Actually, the following holds :)
Let $ a,b,c$ be positive real numbers such that $ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} 10.$ Prove that
\[ \frac {7 \plus{} 8\sqrt {2} \minus{} 5\sqrt {5}}{2}\ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \frac {7 \plus{} 5\sqrt {5} \minus{} 8\sqrt {2}}{2}.
\]
We can prove it using AM-Gm Inequality. :)[/quote]
The counter example of mine is urgly. I write it since it is the equality case of my inequality (for the right hand)


Λοιπόν βάζω μια ανισότητα  πολύ ωραία του φίλου μου Νίκου .
Λοιπόν $a,b,c>0$  να δείξετε ότι
  $(a^2+b^2+c^2)^2\geq\frac{3}{2}(a(b^3+c^3)+b(c^3+a^3)+c(a^3+b^3))$

Μου φάνηκε πολύ ωραία και για να τη λύσω φώναξα σήμα κυνδύνου  ;)

Σιλουανέ

 μακάρι να το χειριζόμουνα το Latex όπως εσύ .Στέλνω τη λύση μου ως συνήθως στο συνημμένο .

 Με τον ίδιο τρόπο την έκανες και εσύ η αλλιώς ?

 Ν.Ρ

Silouan exeis grapsei lanthsmeni ekfonisi.Stin parenthesi toy 2ou melous eina a^3(b+c)+....kyklika.
Alla mallon kai se autin tin periptosi vgainei me ton  tropo tou Romanidi!
Geia sou Niko!

[quote="Nick Rapanos"]Silouan exeis grapsei lanthsmeni ekfonisi.Stin parenthesi toy 2ou melous eina a^3(b+c)+....kyklika.
[/quote]

Ρε Νίκο το ίδιο πράγμα έγραψα απλά άλλαξα τις παρενθέσεις

Τώρα αφού αποδείχθηκε εύκολα αυτή για δοκιμάστε αυτή .
$(a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)$     :D

I teleutai isodynama metatrepetai:

1/2*symmetriko athroisma:(a^2-2ab+bc-c^2+ca)^2   >=0 profanes!

Σωστά  :D  . Λοιπόν η λύση μου στην ανισότητά σου ήταν η εξής
$(a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)$  kai
$(a^2+b^2+c^2)^2\geq 3(ab^3+bc^3+ca^3)$

Προσθέτουμε και το ζητούμενο .
****Σημείωση η ισότητα στην ανισότητα $(a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)$  ισχύει μεν όταν a=b=c αλλά και όταν  $a: b: c=\sin^2\frac{4\pi}{7}: \sin^2\frac{2\pi}{7}: \sin^2\frac{\pi}{7}$ και κυκλικά .

Και μπορούμε επίσης να δείξουμε ότι
$a^4+b^4+c^4+(3r^2-1)(a^2b^2+b^2c^2+c^2a^2)+3r(1-r)abc(a+b+c)\geq 3r( a^3b+b^3c+c^3a)$

όπου $r\in R$

Kai ligo pio pano leei oti tha itan mia poly kali askisi na apodeikseis pote isxyei to ison!!!!!!!!!!!!!!!!!

Nai πρέπει να λύσεις το σύστημα
$a^2-2ab+bc-c^2+ca=0$ και τα άλλα συμμετρικά τους ίσα με μηδέν


I am stuck when trying to prove this theorem. Would you please help ?

Suppose a(n)z^n and b(n)z^n are 2 power series with radii of convergence R and S respectively. Provided that R>S, show that (a(n)+b(n))z^n has radius of convergence S.

What I have done so far is :

If |z|<S then i showed that (a(n)+b(n))z^n converges absolutely.
When it comes to the case |z|>S i don't know how to proceed. I don't know how to make use of the condition R>S although i could find counter example if R=S.

Thank you very much.

Suppose that $\sum (a_{n}+b_{n})z^{n}$ has radius of convergence $T>S$. Subtracting $\sum a_{n}z^{n}$, get a power series with the radius of convergence (at least) $\min(T,R)>S$. A contradiction.


geometry
geometric transformation
reflection
circumcircle
parallelogram
geometry proposed

[color=darkred]Let $ABC$ be a triangle inscribed in the circle $C(O)$. Denote : the points $E\in AB$ and $F\in AC$ for which $O\in EF$ and $OE=OF$: the reflection $A'$ of the vertex $A$ w.r.t. the circumcenter $O$; the projection $P$ of the point $A'$ to the line $EF$. Prove that :
$1.\blacktriangleright\widehat{BPA'}\equiv\widehat{CPA'}$.
$2.\blacktriangleright$ the lines $EF$, $BC$ and the tangent-line $A'A'$ in the point $A'\in C$ to the circle $C$ are concurrently.[/color]

[quote="hunter_cvp_ic"]It is easy!!
we have AEA'F be paralelogram then $\widehat{A'FC}=\widehat{BEA'}$ other A'BPF and A'CEP are cyclic then we have done[/quote]
[color=darkblue]Nice ! Sorry, I forgot to add the second point of the conclusion.
Now I added it in the initial enunciation.[/color]

denote $X$ the intersection of $FA'$ and $BC$ and $Y$ of $A'A'$ and $BC$.

we have $(BXCY)$ h. div. and $A'A'$ tangent therefore $A'F$ is the polar of $Y$, so $A'F \perp OY$, but $A'F \perp EF$ , so $EF,BC,A'A'$ are concurent in $Y$.

Other sol you can use buterfly lemma you can prove that point $A_{2}$ symetric of $A_{1}$ w.r.t $O$ is on the tanget from $A$of circle $C$ ( $A_{1}$ be intersection of $A'A'$ with $BC$ then $A'A_{1}$ be tanget of circle $(O)$


vector
real analysis
LaTeX
linear algebra
linear algebra unsolved

An immediate result:
If Sp(A) is included in [a,b] and Sp(B) in [c,d] then Sp(A+B) is included in [a+c,b+d].
proof:  for x a vector, Sup(x*(A+B)x)<=Sup(x*Ax)+Sup(x*Bx) <=     (b+d)x*x.
      Friendly.

[quote="alekk"]suppose that $A,B \in M_n(\mathbb{C})$ are two hermitian matrix. If one knows the eigenvalues of the two matrices $A$ and $B$, what can be said about the eigenvalues of $A+B$ ? Can it take arbitrary values ? Wich ones ? etc ..[/quote]That's not an easy question. Obviously the spectrum $\sigma_{A+B}$ can be anywhere in the interval $[\min \sigma_A + \min \sigma_B, \max \sigma_A + \max \sigma_B]$. The details depend on the eigenvectors of $A$ and $B$. A recent discussion can be found in

W. Fulton, Eigenvalues of sums of Hermitian matrices (after A. Klyachko), Seminaire  Bourbaki, expose 845, Juin 1998.

However, as $n$ becomes large, $\sigma_{A+B}$ becomes "almost" independent of the eigenvectors and depends only on $\sigma_A$ and $\sigma_B$, $\sigma_{A+B} \sim \sigma_A \boxplus \sigma_B$, the "free convolution" of the two spectra (in the sense of measures on the real line). [i]Hey, I just guessed the LaTeX code for $\boxplus$, and it worked :)[/i]
Details about free convolutions may be found in a [url=http://www.msri.org/communications/preprints/online/1998-040.html]MSRI preprint[/url]by P Biane.


[quote="ayush_2008"]Out of four numbers a,b,c,d first three are in AP and last three are in GP.
and a-d=30.
find the sum a+b+c+d[/quote]

So the system :
$ a\minus{}d\equal{}30$
$ a\plus{}c\equal{}2b$
$ bd\equal{}c^2$

easy to solve :

$ (a,b,c,d)\equal{}(a,\frac{5a\minus{}30\plus{}\epsilon\sqrt{(3a\minus{}50)^2\minus{}1600}}8,$ $ \frac{a\minus{}30\plus{}\epsilon\sqrt{(3a\minus{}50)^2\minus{}1600}}4,$ $ a\minus{}30)$ where $ \epsilon\equal{}\pm 1$ and $ a\ge 30$ or $ a\le\frac{10}3$

If you add a supplementary constraint (not in the current problem) $ a,b,c,d\in\mathbb Z$, you need to solve $ (3a\minus{}50)^2\minus{}1600\equal{}e^2$ and so the set of solutions :
$ (a,b,c,d)\in\{(\minus{}117,\minus{}27,63,\minus{}147),$ $ (\minus{}18,\minus{}3,12,\minus{}48),(0,0,0,\minus{}30),$ $ (30,15,0,0),(36,24,12,6),$ $ (\minus{}18,\minus{}27,\minus{}36,\minus{}48),(3,\minus{}3,\minus{}9,\minus{}27),$ $ (45,15,\minus{}15,15),(84,24,\minus{}36,54)\}$

If you add a supplementary constraint (not in the current problem) $ a,b,c,d\in\mathbb N$, you get :
$ (a,b,c,d)\in\{(36,24,12,6)\}$

[quote="ayush_2008"]sir any simple solution.[/quote]

I dont understand what you mean with "any simple solution", sorry.

My solution :

If the problem is to be solved in $ \mathbb R$ : $ a\plus{}b\plus{}c\plus{}d\equal{}\frac {23a \minus{} 330 \plus{} 3\epsilon\sqrt {(3a \minus{} 50)^2 \minus{} 1600}}8$ where $ \epsilon \equal{} \pm 1$ and $ a\ge 30$ or $ a\le\frac {10}3$

If the problem is to be solved in $ \mathbb Z$ : $ a\plus{}b\plus{}c\plus{}d\in\{\minus{}129,\minus{}111,\minus{}57,\minus{}36,\minus{}30,\plus{}45,\plus{}60,\plus{}78,\plus{}126\}$

If the problem is to be solved in $ \mathbb N$ : $ a\plus{}b\plus{}c\plus{}d\equal{}78$


geometry
parallelogram
Charles Leytem
homothety
circumcircle
IMO
IMO 2007

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

[i]Author: Charles Leytem, Luxembourg[/i]

Interesting to have the converse of [url=http://www.mathlinks.ro/viewtopic.php?t=54065]South Africa 1999[/url] as an IMO problem...

  Darij

Let $ P$, $ M$ and $ N$ be the midpoints of $ BD$, $ CF$ and $ CG$ respectively. Since $ ABCD$ is a parallelogram, $ AC$ also passes through $ P$. Since $ EC \equal{} EF$, $ EM\perp CF$; similarly $ EN\perp CG$.

Consider the homothety centred at $ C$ of ratio $ \frac{1}2$. Clearly $ A$, $ F$ and $ G$ are mapped to $ P$, $ M$ and $ N$ respectively. Hence $ P$, $ M$ and $ N$ are collinear.

Let $ P'$ be the foot of perpendicular from $ E$ to $ BD$. Since $ E$ lies on the circumcircle of $ \triangle BCD$, $ P'$, $ M$ and $ N$ are collinear (Simson line). Therefore, $ P$ and $ P'$ coincide, or $ E$ lies on the perpendiculr bisector of $ BD$.

Since $ \angle BPE \equal{}\angle CME \equal{} 90^{\circ}$ and $ \angle PBE \equal{}\angle MCE$ (angles in the same segment), $ \triangle PEB$ is similar to $ \triangle MEC$. Now $ \angle MNC \equal{}\angle MEC \equal{}\angle PEB \equal{}\frac{1}2\angle DEB \equal{}\frac{1}2\angle DCB \equal{}\frac{1}2\ angle DAB$.

We have proved that the line $ l$ and the Simson line $ PMN$ are related by a homothety, so they are parallel. Therefore the angle between $ l$ and $ AD$ is equal to the angle between $ MN$ and $ BC$, implying that $ l$ is the bisector of $ \angle DAB$.

Here is a ‘philosophical’ solution based on converse mentioned by Darij.
Let us suppose the converse is proved. Lets move line l and look for moving of point
E, the circumcenter of FGC. Its easy to show (or obvious for more experienced people:))
that E passes some quadratic curve which must
be hyperbola since it has an asympthote and its unbounded.
In our interval (while l intersect CD in ints interior) it is some connected part of hyperbola.
Also it is all at the same side of line CD so there are at most
two situations when E is on circle BCD. The first is our desired and the second is trivial
E=F=G=C.

I only want to rewrite the problem and make solution of leepakhin clearly

[b]Problem:[/b][color=blue]Given a parallelogramen $ ABCD$ and a line $ l$ is through $ A$ intersects segment $ CD$ at $ F$ and $ BC$ at $ G$.If exist $ E$ is a point on circumcircle of triangle $ BCD$ such that $ E$ is circumcenter of $ CFG$. Prove that $ l$ is bisector of $ \angle DAB$.[/color]

[b]Solution:[/b]Let $ P,M,N$ be the midpoints of $ BD, CF, CG$ resp.$ ABCD$ is a parallelogram, $ AC$ also passes through $ P$. Easily seen $ P,M,N$ are colinear (by homothety or they are in the same line parallel with $ l$) Since $ EC=EF\Rightarrow EM\perp CF$ similarly $ EN\perp CG$. Thus $ MN$ is simson line of $ E$ w.r.t $ BCD$ but $ MN\cap BD=\{P\}\Rightarrow EP\perp BD$ but $ P$ is midpoint of $ BD\Rightarrow EB=ED\Rightarrow \angle DBE=\angle BDE\ (1)$ But $ \angle DBE=\angle DCE,\angle BDE=\angle ACG\ (2)$($ ACED$ is cyclic). Combine $ (1),(2)$ we get $ \angle DCE=\angle ACG\Rightarrow \angle CEF=\angle CEG\Rightarrow EC$ is midline of $ FG\Rightarrow$ triangle $ CFG$ is isoceles $ \Rightarrow\angle DAG=\angle AGC=\frac{180^{0}-\angle FCG}{2}=\frac{\angle BCD}{2}=\frac{\angle BAD}{2}$

Note: Invense problem: Given a parallelogramen $ ABCD$ and a line $ l$ is bisector of $ BAD$, $ l$ intersect lines $ CD$ and $ BC$ at $ F,G$.Prove that circumcenter of triangle $ CFG$ lie on circumcircle of triangle $ BCD$.  :)

i have solved this problem with some easy trigonometry(at home :( )

I solved it by showing that BEG and DED are similar.
What do you think was this problem on the shortlist? (I think some G3 )

This problem can be done using complex numbers, too.

let (ABD)=a (DBE)=b (EBC)=c and EC=sinb.k
from sin laws DC=sin(b+c).k BC=sina.k FC=2.sinc.cosb.k GC=2.sinc.cos(a+c).k
on the other hand we have FC/AB=GC/GB from similarity
so writing the above results yields
(sina+2.sinc.cos(a+c)).cosb=cos(a+c).sin(b+c)
(sina+sin(2c+a)-sina).cosb=cos(a+c).sin(b+c)
sin(2c+a+b)+sin(2c+a-b)=sin(2c+a+b)+sin(b-a)
sin(b-a)=sin(2c+a-b)
1) b-a+2c+a-b=180 c=90 b=m(DEF)+c so m(ABC)>180 contradiction
2) b-a=2c+a-b so b=a+c hence FC=CG and AB=BG which proves our problem.

[quote="Sepp"][quote="Albanian Eagle"]I solved it by showing that BEG and DED are similar.
[/quote]
You mean DEC  :wink:

Since $ AD \parallel BG$, then $ \angle DAG = \angle BGA$. So we have to prove $ ABG$ is isosceles.
$ F, C, G$ all lie on the circle with center $ E$ and radius $ EF$. Let $ K$ be the intersection between this circle and the circumcircle $ \mathcal O$ of $ BCD$.
As proved early $ E$ is the midpoint of arc $ BD$ on $ \mathcal O$. Then $ BCKD$ is a trapezoid and $ \angle ECK = \angle EBC$.
Since $ BD \parallel KC$, then $ \angle EBD = \angle ECG = \angle EGC$ and by LAL $ \Delta EDC \equiv \Delta EBG$.
Finally $ ABG$ is isosceles and the thesis follows.[/quote]
yeah!
that's exactly my solution. Quite simple and straight foreward. You get the idea by seeing the proof of the other direction :)

[quote="Sepp"][quote="Albanian Eagle"]I solved it by showing that BEG and DED are similar.
[/quote]
You mean DEC  :wink:

Since $ AD \parallel BG$, then $ \angle DAG = \angle BGA$. So we have to prove $ ABG$ is isosceles.
$ F, C, G$ all lie on the circle with center $ E$ and radius $ EF$. Let $ K$ be the intersection between this circle and the circumcircle $ \mathcal O$ of $ BCD$.
As proved early $ E$ is the midpoint of arc $ BD$ on $ \mathcal O$. Then $ BCKD$ is a trapezoid and $ \angle ECK = \angle EBC$.
Since $ BD \parallel KC$, then $ \angle EBD = \angle ECG = \angle EGC$ and by LAL $ \Delta EDC \equiv \Delta EBG$.
Finally $ ABG$ is isosceles and the thesis follows.[/quote]
I want to ask you how do you get $ E$ is the midpoint of arc $ BD$ on $ \mathcal O$

Here is an overcomplicated solution I found which involves some phantom points.

Let $(BCD)$ intersect $AB$ and $AD$ at $P$ and $Q$ where $P \ne B, Q \ne D$.

[color=#3D85C6][b]Claim: [/b][/color]$PF$ and $QG$ intersect on $(BCD)$ at some point $H$.

[color=#3D85C6][b]Proof: [/b][/color]It suffices to prove for some point $H'$, if we let $F' = PH \cap CD$ and $G' = QH \cap CB$, then $A, F', G'$ is collinear which follows from Pascal's Theorem on $BCDQH'P$. $\square$

[color=#3D85C6][b]Claim: [/b][/color]$F, H, E, G$ are concyclic.

[color=#3D85C6][b]Proof: [/b][/color]Notice that $$\measuredangle{FEG} = 2\measuredangle{FCG} = 2\measuredangle{FCB} = 2\measuredangle{DCB} = 2\measuredangle{DAB}$$ and $$\measuredangle{FHG} = \measuredangle{FHC}+\measuredangle{CHG} = \measuredangle{PHC}+\measuredangle{CHQ} = \measuredangle{PBC}+\measuredangle{CDQ} = \measuredangle{ABC}+\measuredangle{CDA} = 2\measuredangle{DAB}$$ so $\measuredangle{FEG} = \measuredangle{FHG}$ as desired. $\square$

Therefore, we see that $(FEG)$ intersects $(BCD)$ at two points $H$ and $H'$, with $H'$ being on the same side of $CD$ as $B$ while $H$ is on the opposite side. However, because $G$ lies on the opposite side of $CD$ as $B$ and $F$ lies on $CD$, this means the center of $(FEG)$ must be on the same side of $CD$ as $G$, which means $E = H$. It is easy to check that this only works when $CF = CG$ which also means that $l$ is the angle bisector of $\angle{DAB}$. $\blacksquare$

[color=#f00][b]Remarks: [/b][/color]This problem was decent, but I spent like over 2 hours on this problem for some reason and did not think about homothety at all. :oops_sign:

Quite easy for a G4

Define $N$ as the midpoint of $GC$, $M$ as the midpoint of $CF$, and $P$ as the midpoint of $BD$. Observe that $P$ is also the midpoint of $AC$; by taking a homothety centered at $C$ with ratio $\frac12$, we have $P, M, N$ collinear. Furthermore, $N$ and $M$ are feet from $E$ to $BC$ and $CD$, so by simson line, $P$ is the foot from $E$ to $BD$ and thus $EB = ED$. We now have
\[\angle ECF = \angle EBD = \angle EDB = \angle ECG\]
Now this means $CG = CF$, so
\[\angle BAF = \angle AFD = \angle CGF = \angle GAD\]
We conclude that $\ell$ is the angle bisector.

Let $X,Y,$ and $Z$ be the midpoints of $\overline{BD},\overline{DC},$ and $\overline{BC}$ respectively, and note that $X,Y,$ and $Z$ are collinear by a homothety with scale factor $2$ about $C.$ Since $\overline{EY}\perp\overline{DC}$ and $\overline{EZ}\perp\overline{BC},$ $\overline{EX}\perp\overline{BD}$ by Simson Line. Hence, $\triangle EBD$ is isosceles with $EB=ED.$ Since $XEZB$ and $YEZC$ are cyclic, $$\angle GAD=\angle FGC=\angle YZC=\angle XEB=\angle XED=90-\angle XDE$$ and $$90-\angle ECZ=\angle CEZ=\angle CYZ=\angle CFG=\angle BAF.$$ Since $\angle XDE=180-\angle ECB=\angle ECZ$ we are done. $\square$

c o m p l e x  b a s h

Set $e=0$ with $(CFG)$ as the unit circle. Then since there is a homothety with scale factor $k>1$ centered at $G$ mapping $\triangle FCG$ to $\triangle ABG$, we can set
\[a=g+k(f-g)\]
\[b=g+k(c-g)\]
\[d=a+c-b=c+k(f-c).\]
Then since $\measuredangle DBE=\measuredangle DCE=\measuredangle FCE$ it follows that
\[\frac{c-f}{c}\div \frac{c+k(f-c)-g-k(c-g)}{g+k(c-g)}=\frac{c-f}{c}\cdot \frac{kc-(k-1)g}{(-2k+1)c+kf+(k-1)g}\]
is a real number, which implies that it is equal to its conjugate of
\[\frac{\tfrac{1}{c}-\tfrac{1}{f}}{\tfrac{1}{c}}\div \frac{\tfrac{1}{c}+k\left(\tfrac{1}{f}-\tfrac{1}{c}\right)-\tfrac{1}{g}-k\left(\tfrac{1}{c}-\tfrac{1}{g}\right)}{\tfrac{1}{g}+k\left(\tfrac{1}{c}-\tfrac{1}{g}\right)}=\frac{f-c}{f}\cdot \frac{kfg-(k-1)cf}{(-2k+1)fg+kcg+(k-1)cf}.\]
We can cancel out a factor of $c-f$ to obtain
\[\frac{kc-(k-1)g}{(-2k+1)c^2+kcf+(k-1)cg}=\frac{(k-1)cf-kfg}{(-2k+1)f^2g+kcfg+(k-1)cf^2}\]
and cross-multiplying gives that
\[(-2k^2+3k-1)c^3f+(2k^2-k)c^2fg+(k^2-k)c^2f^2-k^2cf^2g+(k^2-2k+1)c^2fg-(k^2-k)cfg^2\]
is equal to
\[(-2k^2+k)cf^2g+(2k^2-3k+1)f^2g^2+k^2c^2fg-(k^2-k)cfg^2+(k^2-k)c^2f^2-(k^2-2k+1)cf^2g.\]
Now we subtract the second expression from the first and factor to get
\[(2k^2-3k+1)(cf^2g+c^2fg-c^3f-f^2g^2)=f(2k-1)(k-1)(g-c)(c^2-fg)=0.\]
Now since $k>1,g\neq c,f\neq 0$ we obtain $c^2=fg$, which implies that $CF=CG$. Then $\measuredangle FAB=\measuredangle GFC=\measuredangle CGF=\measuredangle DAF$, so $\ell$ is the bisector of $\angle DAB$. $\blacksquare$

It is sufficient to prove that $\triangle CGF$ is isosceles. Let $H,I,J$ be the midpoints of $CA,CF,CG$. Clearly $H,I,J$ are collinear. Since $EJ\perp CG, EI\perp CF$, by the simson line theorem we can say $EH\perp BD$. Meaning $E$ is the midpoint of the arc $\widehat{BCD}$ because $H$ is also the midpoint of $BD$. So $\angle GEC=\angle BDE=\angle DBE=\angle DCE$.

Let $X$ be the second intersection of $AB$ and $BCED$, and $S$ be the second intersection of $(FCG)$ with $(BCED)$. It is easy to see that $DX=DA$.
\\
$\angle XSF=\angle XSC-\angle FSC=\angle BAD-\angle FGC=\angle XAD- \angle FAD=\angle XAF\Rightarrow XASF$ is cyclic.
\\
$\angle DES=\angle DCS=\frac{\angle FES}{2}\Rightarrow DE$ bisects $\angle FES$, and since $FE=ES$ and $DE=DE$, we have that triangles $DEF$ and $DES$ are congruent, hence $DS=DF$. By the fact that $AXFS$ is cyclic and $DA=DX$ and $DF=DS$ implies that $D$ is the center of $AXFS$. By the incenter lemma we have that $F$ is the incenter of triangle $CSX$. Now, $\angle XAF=\angle XSF=\angle FSC=\angle FGC=\angle FAD$ and we are done.

We first prove the converse, i.e. if $\ell$ is the bisector, then the center of $(CFG)$ lies on $(BCD):=\Gamma$.
In this case, I claim that our $E$ is the arc midpoint of $BD$. Let $H$ be the point on $\Gamma$ such that $BDHC$ is an isosceles trapezoid, which by symmetry gives $EC=EH$. Equivalently, $H$ is the reflection of $A$ over $\overline{BD}$. Angle chasing gives
$$\angle CGF=\angle FAD=\angle BAF=\angle DFA=\angle CFG,$$
hence $CF=CG$. Further,
$$\angle GCE=180^\circ-\angle BCE=\angle BDE=\angle DBE=\angle FCE,$$
hence $\overline{CE}$ bisects $\angle FCG$. Combining these two facts, it follows that $E$ lies on the perpendicular bisector of $\overline{FG}$, i.e. $EF=EG$.
Now, note that $DH=DA=DF$ (where the second equality comes from $\angle FAD=\angle DFA$). Further, $\overline{DE}$ is the bisector of $\angle FDH=\angle CDH$, as $E$ is the midpoint of minor arc $CH$. Hence $F$ and $H$ are reflections of each other over $\overline{ED}$, so $EF=EH$. Since $EC=EH$ and $EF=EG$, it follows that $EF=EG=EC$, so $E$ is indeed the circumcenter.

We now use the converse (!!) to prove that there cannot exist two choices of $\ell$ such that the center of $(CFG)$ lies on $\Gamma$. This can probably be done with hyperbolas but we don't know anything about hyperbolas so we use a synthetic geometry proof instead. Suppose we had some $\ell'$ intersecting $\overline{CD}$ at $F'$ and $\overline{BC}$ at $G'$, such that the center $E'$ of $(CF'G')$ was on $\Gamma$ as well. Note that $E'$ lies the perpendicular bisector of $F'G'$, which intersects $\overline{CE}$ on $(CF'G')$—precisely, on the midpoint of arc $F'G'$ not containing $C$. Further, as $F'$ moves from $C$ to $D$, the length of $CE'$ strictly increases.
If $F'$ lies on segment $\overline{CF}$, then $E'$ should lie on minor arc $CE$ of $\Gamma$, as it is closer to $C$ than $E$ is. On the other hand, said minor arc lies on the opposite side of $\overline{CE}$ as the perpendicular bisector of $F'G'$ does. This is because $CG'<CF' \iff DA=DF<DF'$ (from $\triangle F'CG' \sim \triangle F'DA$) which is true, so the angle bisector theorem implies that the midpoint of $\overline{F'G'}$ lies on the opposite side of $\overline{CE}$ as the minor arc does, and then our previous intersection finding implies that the perpendicular bisector will not "switch sides" within $(CF'G')$.
Likewise, if $F'$ lies on segment $\overline{DF}$, then $E'$ should lie on major arc $CE$ of $\Gamma$, as it is farther from $c$ than $E$ is. On the other hand, this time we have $CG'>CF'$, so the midpoint of $\overline{F'G'}$ again lies on the opposite side of $\overline{CE}$ as the major arc does. These two cases capture all possibilities, so $\ell'$ cannot exist, as desired. $\blacksquare$

Let $\Gamma = (BCED)$, $P = DA \cap \Gamma$, $Q = BA \cap \Gamma$. We have $F = \lambda \cap CD$, and define $X = QF \cap \Gamma$.

By Pascal on $PXQBCD$, we have $A = BQ \cap DP$, $F = XQ \cap CD$, and $BC \cap XP$ collinear. Hence they are collinear on $\lambda$, and since $G = \lambda \cap BC$ then $G$ lies on $XP$. \\

[b]Claim 1:[/b] $X$ is the $C$-dumpty point of $\triangle CGF$.
[i]Proof:[/i] $\angle XGC = \angle PGC = \angle GPD = \angle XPD = \angle XCD = \angle XCF$. Similarly, $\angle GCX = \angle BPX = \angle BQX = \angle CFX$. Hence, by AA we have $\triangle XCG \sim \triangle XFC$, and $X$ is the $C$-dumpty point. $\square$

It's well-known that $CX \perp EX$. Hence, since $C, X, E$ lie on $\Gamma$, either $X = E$ or $E$ is the $C$-antipode in $\Gamma$. \\

[u]Case 1:[/u] $X = E \Rightarrow \angle ECG = \angle EFC = \angle FCE \Rightarrow CE$ is the angle bisector of $\angle FCG$. Hence $\triangle CGF$ is isosceles. Since $\triangle FDA \sim \triangle FCG$ by AA we have $\angle FAB = \angle AFD = \angle DAF$, and $\lambda$ is the bisector of $\angle DAB$.

[u]Case 2:[/u] $E$ is the $C$-antipode. Then $(CGF)$ is the image of $\Gamma$ under homothety scale factor 2 at $C$, and so no other point on $(CGF)$ lies inside $\Gamma$. But $F \in (CGF)$ lies on segment $CD$ within $\Gamma$, contradiction.

Note that the Simson line of $E$ wrt $\triangle BCD$ is homothetic to $\ell$ wrt $C$ with factor $1/2,$ and therefore passes through the center of $ABCD$. This yields $|EB|=|ED|\implies \angle FCE=\angle DBE=\angle BDE=\angle GCE,$ so it's trivial that $|CF|=|CG|$ and hence $\angle BAF=\angle CFG=\angle CGF=\angle DAF,$ as desired.

The key is to construct the Simson Line of $E$ with respect to triangle $BCD$. By taking a homothety at $C$ with ratio $2$, it follows that this line is parallel to $\ell$.

On the other hand, it is well-known that the the Simson Line is parallel to $\overline{CJ}$, where $J$ is the point at which the altitude at $E$ to $\overline{BD}$ meets $(BCD)$. Thus, it suffices to show $J$ is the arc midpoint.

But the aforementioned homothety implies that the Simson Line bisects $\overline{AC}$, so it bisects $\overline{BD}$ too. This common midpoint $O$ is thus both the midpoint and foot from $E$ in triangle $EAB$, implying that it is isosceles. Thus, $\overline{EJ}$ bisects $\widehat{BD}$, which is enough.

Let $M,N,O$ be midpoints of $CG$, $CF$, and $BD$, respectively. Note that $EN\perp CD$ and $EM\perp BC$. Note that the perpendicular from $E$ onto $BD$ must lie on $MN$. Since $O,M,N$ are midpoints of $AC$, $FC$, and $GC$ respectively, $O$ lies on $MN$. Therefore, $EO\perp BD$.

Now, $EB=ED$, $\angle EMB=\angle END$ and $\angle NDE=\angle EBM$ so $\triangle BME\cong \triangle DNE$. Therefore, $EM=EN$ which implies $CG=CF$. Thus, \[\angle FAD=\angle CGF=\angle CFG=\angle BAF\] as desired.

Let $H_{b}$ and $H_{d}$ be the midpoints of $CG$ and $CF$, respectively. Notice that $\triangle ADF \sim \triangle GCF$ and $\triangle EH_{d}D \sim \triangle EH_{b}B$. The first follows from $ABCD$ being a parallelogram and the second follows from $\angle EDC = \angle EBC$ combined with $\overline{EH_{d}} \perp \overline{FC}$ and $\overline{EH_{b}} \perp \overline{CG}$. These give:
\[\frac{\frac{1}{2}CF\cdot\tan\angle EFC}{\frac{1}{2}CG\cdot\tan\angle ECG} = \frac{EH_{d}}{EH_{b}} = \frac{DH_{d}}{BH_{b}} = \frac{DF + FH_{d}}{BC + CH_{b}} = \frac{DF + \frac{1}{2}CF}{BC + \frac{1}{2}CG} = \frac{CF}{CG}\cdot \frac{\frac{DF}{CF}+\frac{1}{2}}{\frac{AD}{CG}+\frac{1}{2}} = \frac{CF}{CG}\]
where the last equality follows from $\triangle ADF \sim \triangle GCF$. Now $\tan\angle EFC = \tan\angle ECG$ and the angles are acute, hence $\angle EFC = \angle ECG$. This implies that $\triangle EFC \cong \triangle EGC$, so $CF = CG$ and therefore $AD = DF$, implying that $\ell$ is the angle bisector of $\angle BAC$ as desired.

    Got a hint to construct midpoints, but I think the finish is pretty unique!
    Let J,K,L be the midpoints of GC,FC,BD. Notice that by a homothety of 1/2 ratio at C we get AFG is a line implies LKJ is a line; Since EJ perp. BC, EK perp. CD, by converse of Simson line we get that EL perp. BD. Now note by pop we get $$\frac{DF}{FC}=\frac{AD}{CG}=\frac{BC}{CG}\implies\frac{DF}{BC}=\frac{FC}{CG}=\frac{AB}{BG}=\frac{CD}{BG}\implies BC\cdot CD=DF\cdot BG\implies^{BC\cdot BG=DF\cdot CD}CD=BG,BC=DF\implies BAG=BGA=GAD,$$ as desired. $\blacksquare$

Let $M$ be the midpoint of $\overline{BD}$, so by a homothety at $C$, $M$ is collinear with the feet of the altitudes from $E$ to $\overline{CB}$ and $\overline{CD}$. On the other hand, by Simson lines, these two feet are also collinear with the foot from $E$ to $\overline{BD}$, hence this foot is just $M$, so $E$ is the midpoint of arc $BCD$.

Then $\angle FCE=\angle GCE$, so $CF=CG$ and thus $\angle GFC=\angle FGC \implies \angle GAB=\angle FAD$, which finishes the problem. $\blacksquare$

[b]EDIT:[/b] bro ive done this problem before too whatttttttttttt

had to get hint ;-; my brain

Just draw Simson line of $E$ to $BCD$, which passes through midpoint of $BD$. So $E$ is arc midpoint, finish since $\triangle ECF\sim\triangle EBD\sim \triangle EGC$


[hide]

Make the substitution . Then the RHS is , while the right side is

Then dividing through and reducing the RHS, we have , so . Plugging this back into  yields .

[/hide]


LaTeX
analytic geometry
graphing lines
slope
geometry

I'm not sure if this has been posted before, but how do you make a footer or header when making a latex document?  I've looked on the web, but found nothing pertaining to a header.  I found how to make a footer

But when I type in [code]\footnote{text}[/code] all I get is a footer but
1) it's not at the very bottom of the page and
2) the footer is only half a page long.  Isn't a footer supposed to extent the entire width of the page?  It only reaches the center....

Also, I'm not sure if this has been posted as well, but how do you make a graph in latex?  I've (kind of) found how to make a line (with a slope and stuff) but how do I get the graph part?  The x and y axis?

You seem to be confusing a footnote with a footer. A footer is always at the bottom of the page and might have, for example, a page number or just be blank. A footnote is a numbered text which explains something in the text. It is left-aligned after the last paragraph and goes [i]above[/i] the footer, and can contain lots of numbered text and can even spill over to the next page.

The layout of the page depends on which document class you are using but if you wish to alter the headers and footers ([i]not[/i] the footnotes) then use the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]fancyhdr[/url] package.

The x and y axes are just lines so you draw them in the same way. See the Producing Mathematical Graphics section of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url]

[quote="stevem"]You seem to be confusing a footnote with a footer. A footer is always at the bottom of the page and might have, for example, a page number or just be blank. A footnote is a numbered text which explains something in the text. It is left-aligned after the last paragraph and goes [i]above[/i] the footer, and can contain lots of numbered text and can even spill over to the next page.

The layout of the page depends on which document class you are using but if you wish to alter the headers and footers ([i]not[/i] the footnotes) then use the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]fancyhdr[/url] package.
[/quote]Hi Steve,

I've got \ usepackage{fancyhdr} written in the preamble, however is it necessary to write all the information which shows up with the link you gave?

Also is it ok to use more than one package? I've got this written down:
\ usepackage{fancyhdr}
\ usepackage{amsfonts}
\ usepackage{amsmath}
\ usepackage[pdftex]{graphicx}.

You see I'm having the same problem. I have \ cfoot{\thepage}, but unlike the header this is not near the bottom of the page. (about 4.5cm above the end of the page!) :o Is there any way I can get this to move lower down?

Cheers.

[quote="beans"]I've got \ usepackage{fancyhdr} written in the preamble, however is it necessary to write all the information which shows up with the link you gave?[/quote]No, you only need to use the commands that are relevant to you

[quote="beans"]Also is it ok to use more than one package? I've got this written down:
\ usepackage{fancyhdr}
\ usepackage{amsfonts}
\ usepackage{amsmath}
\ usepackage[pdftex]{graphicx}[/quote]LaTeX documents often use dozens of packages which are fine as most don't conflict with one another as they do different things. But I would replace
\ usepackage{amsfonts}
with
\ usepackage{amssymb}
as amssymb loads amsfonts automatically and allows commands like \therefore (not that you will be using this one much given our previous conversations :)) and others listed in [url=ftp://tug.ctan.org/pub/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]The Comprehensive LaTeX Symbol List[/url].

[quote="beans"]You see I'm having the same problem. I have \ cfoot{\thepage}, but unlike the header this is not near the bottom of the page. (about 4.5cm above the end of the page!) :o Is there any way I can get this to move lower down?
[/quote]You need to adjust the margins of the page which can be done by [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#layout]Layout of the Page[/url] but the simplest way is to use yet another very useful package called the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]geometry package[/url]. Nothing to do with the mathematical subject but makes page layout easy. Thus
\usepackage[margin=1.5in]{geometry}
puts a margin of 1.5in all the way round. This the easiest of all the options but there's a lot more you can do eg writing
\usepackage[top=4cm, bottom=3cm]{geometry}
which makes the top margin 4cm and the bottom margin 3cm.

[quote="stevem"]No, you only need to use the commands that are relevant to you[/quote] Cool thanks.

[quote="stevem"]LaTeX documents often use dozens of packages which are fine as most don't conflict with one another as they do different things. But I would replace
\ usepackage{amsfonts}
with
\ usepackage{amssymb}
as amssymb loads amsfonts automatically and allows commands like \therefore (not that you will be using this one much given our previous conversations :)) and others listed in [url=ftp://tug.ctan.org/pub/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]The Comprehensive LaTeX Symbol List[/url]. [/quote] Haha, well one can live in hope ;) It's good to have though, just in case! And wow- that document has a lot of symbols!

[quote="stevem"]You need to adjust the margins of the page which can be done by [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#layout]Layout of the Page[/url] but the simplest way is to use yet another very useful package called the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]geometry package[/url]. Nothing to do with the mathematical subject but makes page layout easy. Thus
\usepackage[margin=1.5in]{geometry}
puts a margin of 1.5in all the way round. This the easiest of all the options but there's a lot more you can do eg writing
\usepackage[top=4cm, bottom=3cm]{geometry}
which makes the top margin 4cm and the bottom margin 3cm.[/quote] Once again, thanks! It worked a treat (tweaked it a little so my file didnt change too drastically) but it's looking good now. :D


[quote="erik-the-red"]In terms of difficulty, how does Part I of the Michigan Mathematics Prize Competition compare to the AMC 12?[/quote]

IMO (that's "in my opinion," not Int'l Math Olympiad ;)), MMPCI is *far* easier.  I scored 34/35 (back in '99) on MMPCI, but didn't fare as well on AMC12.  The point, I think, is that MMPCI is all trivial problems that you can learn to solve very efficiently, while some AMC12 may contain some slightly trickier problems.

Edit: My memory is apparently not so good.  I missed two problems and it's out of 40, not 35.


induction
geometry
rectangle
inequalities open
inequalities

There is a question that really bothers me:
  Prove that for all x_1,...,x_n>0 with product 1 and for all n>3 we have
 1/sqrt(1+x_1)+1/sqrt(1+x_2)+....+1/sqrt(1+x_n)<n-1, but that n-1 cannot be replaced by a smaller number. The fact that n-1 cannot be replaced by a smaller one is trivial, but the other problem I find extremely difficult. Of course, I may be wrong, but I would really like to see a solution. Thank you very much.

I will bring this one to your attention since it is a very interesting and hard inequality. I hope that someone will be so smart to prove it. I find it one of the hardest inequalities.

I think this problem can be solved on two steps:
1) Prove it for n = 4, i.e if x1.x2.x3.x4 = 1 then
             Sum 1/sqrt(1+xi) < 3
2) Use induction to show for all n: Suppose that if x1...xn = 1 then
             Sum{1..n} 1/sqrt(1+xi) < n-1
For x1...xn.x_(n+1) = 1, we can suppose that x_(n+1) <= 1 => x1...xn >= 1 => Sum{1..n} 1/sqrt(1+xi) < n-1 (obviously)
        => Sum{1..n+1} 1/sqrt(1+xi) < n.
So all we have to do is 1). I think someone in our site can help me in finding sln for this one.

Namdung

I have a geometric idea for this one.
    Now we replace x1 = tga1, x2 = tga2, x3 = tgx3, x4 = tga4.(a1,a2,a3,a4 <90).
    So we have tga1*tga2*tga3*tga4 = 1.
    Let ABCD is a rectangle and M is a point in it such that  the angle MAD = a1, MBA = a2, MCD= a3. Therefore MDC = a4.
   All we have to do is to provethat cosa1 + cosa2+cosa3+cosa4 <3.
   I think that there can't be any number m <3 such that  \sum cosa1 < m. It 's easy to see that. But I have trouble to prove this inequality. Please continue,thanks.

I have found a solution to this one, but I will not post it since a nice consequence could be a nice problem. So, have patience to discuss a little bit with Valentin, to whom I have sent a related problem, for Mathlinks Contest.


So, I did a quick search of the forum. So there is that type of question where you are asked about certain properties of tiling a rectangle of $ m\times n$ or something using for example a domino or some other shape. Pretty much all the questions are solved using recursion. So my question is how is recursion actually used because I have no idea. Plus is there a way to determine the number of ways to tile a $ m \times n$ rectangle using dominos?


A sequence a_n is a nondecreasing sequence of integers from 0 to k. Prove that the number of such sequences for given n is C(n+k,n)=C(n+k,k).

[hide]We picture this geometrically, graphing an on the y-axis and n on the x-axis. We are given the two points (0,0) and (n,k). Furthermore, we can only move right or up. We must make n right-ward moves total and k up-ward moves total. The number of ways to choose in which order we do this is the given binomial coefficient.[/hide]

Interesting approach! Here's the more conventional way to do it, at least in my experience:

Consider bm=m+am. Then we think about the terms we can have for the bms. They must all be distinct, but other than that, we're free to select any numbers from 1 to n+k. Thus there are (n+k choose k) possible sequences.


[quote="qwerty430"]:bye:   the degree of difficulty  of  IMO-sl 2008 - G8

... sorry about my english.kk..[/quote]

[quote="test20"]I fully agree.
Problem G8 on the IMO'2008 ShortList must be one of the most difficult problems
that has ever been proposed for a mathematical highschool competition.
The jury made a good decision by not selecting it.[/quote]

[quote="orl"]You should not be discussing ISL 2008 related information yet.[/quote]

[quote="qwerty430"]sorry...2007 short _G8..

:P ...hihihi[/quote]

saved for posterity.

sorry, i'm just curious. Could someone post the problem statement of supposedly geometry problem 8 for IMO-07shortlist?


Evaluate:

$ \lim_{x \to \infty} ~\left[~\left(kx^n \plus{} \sum_{i\equal{}0}^{n\minus{}1} a_{i}x^{i}\right)^{\frac{1}{n}} \minus{} \left(kx^n \plus{} \sum_{i\equal{}0}^{n\minus{}1} b_{i}x^{i}\right)^{\frac{1}{n}}~\right]$

where $ k > 0$ is a real number and $ a_i$ and $ b_i$ are sequences of real numbers.

This is somewhat interesting to work out.

[quote="Zakary"]Evaluate:

$ \lim_{x \to \infty} ~\left[~\left(kx^n \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} a_{i}x^{i}\right)^{\frac {1}{n}} \minus{} \left(kx^n \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} b_{i}x^{i}\right)^{\frac {1}{n}}~\right]$

where $ k > 0$ is a real number and $ a_i$ and $ b_i$ are sequences of real numbers.

This is somewhat interesting to work out.[/quote]
there is no nothing interesting it's standart problem and can be solved by Taylor
$ \lim_{x \to \infty} ~\left[~\left(kx^n \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} a_{i}x^{i}\right)^{\frac {1}{n}} \minus{} \left(kx^n \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} b_{i}x^{i}\right)^{\frac {1}{n}}~\right] \equal{} \lim _{x \minus{} > \plus{} oo} (xk^{1/n}(1 \plus{} \frac {a_{n \minus{} 1}}{x} \plus{} o(1/x)) \minus{} xk^{1/n}(1 \plus{} \frac {b_{n \minus{} 1}}{x} \plus{} o(1/x))) \equal{} k^{1/n} \frac {(a_{n \minus{} 1} \minus{} b_{n \minus{} 1})}{n}$
if you understand right then this willl be good example for you:
find the limit :
$ \lim_{x \to \infty} ~ x \left[~\left(kx^n \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} a_{i}x^{i}\right)^{\frac {1}{n}} \minus{} \left(kx^n \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} b_{i}x^{i}\right)^{\frac {1}{n}}     \minus{}   k^{1/n} \frac {(a_{n \minus{} 1} \minus{} b_{n \minus{} 1})}{n}  ~\right]$

another way by the identity $ a\minus{}b\equal{}\frac{a^n\minus{}b^n}{a^{n\minus{}1}\plus{}....\plus{}ab^{n\minus{}1}}$

[quote="drapt@"]another way by the identity $ a \minus{} b \equal{} \frac {a^n \minus{} b^n}{a^{n \minus{} 1} \plus{} .... \plus{} ab^{n \minus{} 1}}$[/quote]
how to use this identity?  :what?:

yes, but after is not obviusly to me, it's hard way


Find all rational solutions of $ y^2 \equal{} x(x\plus{}9)(x\minus{}16)$.

The obvious rational points by inspection are x=0,-9,16, however using GP-Pari and mwrank shows that this is a rank 0 Z2xZ4 curve, so all the rational points are only the 8 torsion points:

[36,-180] [16,0] [36,180] [-9,0] [-4,20] [0,0] [-4,-20] and the point at infinity [0/0,1/0]

Proving that the curve is rank 0 is a harder task, I do trust that mwrank got it right in this case because of 2nd descents over the elliptic curve and its phi image.

OK, but this has also a moduli space interpretation. Find it.


calculus
integration
trigonometry
calculus computations

Find

         $\int \frac{\sqrt{\sin x}}{{\sqrt{\sin x}} + {\sqrt{\sin x}}}dx$

[quote="Treething"]I think you made a mistake :roll: .

x/2 + C.[/quote]

That's the beauty of the problem. I have not made any mistake. If you want then I can provide you a hint.

Bye :)

[quote="Beginner"]Maybe you mean $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ ?[/quote]
Yes you are right.
 :ninja:

[quote="Kalle"]This primitive took Maple 53 seconds to evaluate and it is several pages long. Are you sure this can be done?[/quote]

Who is "Maple". I did in half a page.

[quote="Virgil Nicula"]$I=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx=\int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} dx,\ x\in \left(0,\frac{\pi}{2}\right)$.

$t=a(x)=\sqrt{\tan x}\Longrightarrow I=2F(\sqrt{\tan x})$, where $F=\int \frac{t^2}{(1+t)(1+t^4)} dt$

[b]a.s.o.[/b]

[b]I think is a defined integral ![/b]: $I=\int^{\frac{\pi}{2}}_0 f(x) dx$, where $f(x)=\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.$

$I=\int^{\frac{\pi}{2}}_0 f(x) dx,\ J=\int^{\frac{\pi}{2}}_0 f\left(\frac{\pi}{2}-x\right) dx\Longrightarrow I=J,\ I+J=\frac{\pi}{2}\Longrightarrow I=\frac{\pi}{4}.$[/quote]

I too did the same way. The question was of indefinite integral.

[quote="mffmartinelli"]No this is an indefinite integral really. First we ca\m make this:
   $\int \frac{\sqrt{\sin x}}{{\sqrt{\sin x}} + {\sqrt{\cos x}}}dx=\int \frac{\sqrt{\sin x}+\sqrt{\cos x}}{{\sqrt{\sin x}} + {\sqrt{\cos x}}}dx-\int \frac{\sqrt{\cos x}}{{\sqrt{\sin x}} + {\sqrt{\cos x}}}dx=x-\int \frac{\sqrt{\cos x}}{{\sqrt{\sin x}} + {\sqrt{\cos x}}}dx=x-\int \frac{1}{ 1+ {\sqrt{\tan x}}}dx$
   And now make a change of variables. Let $u=\sqrt{\tan x}$. And now we resolve our problem.[/quote]

I also solved the problem in same manner.


A sit of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?

$ \text{(A)}\ 10 \qquad
\text{(B)}\ 11 \qquad
\text{(C)}\ 18 \qquad
\text{(D)}\ 19 \qquad
\text{(E)}\ 20$

[hide="Brute force"]
The pattern becomes clear immediately.
\[ \begin{tabular}{|r|r|r|} \hline \# & \text{Removed} & \text{Left} \\
\hline 1 & 10 & 90 \\
2 & 9 & 81 \\
3 & 9 & 72 \\
4 & 8 & 64 \\
5 & 8 & 56 \\
6 & 7 & 49 \\
7 & 7 & 42 \\
8 & 6 & 36 \\
9 & 6 & 30 \\
10 & 5 & 25 \\
11 & 5 & 20 \\
12 & 4 & 16 \\
13 & 4 & 12 \\
14 & 3 & 9 \\
15 & 3 & 6 \\
16 & 2 & 4 \\
17 & 2 & 2 \\
\boxed{18} & 1 & 1 \\
\hline \end{tabular}
\]
[/hide]

[hide="Solution"]
Given $ n^2$ tiles, a step removes $ n$ tiles, leaving $ n^2 \minus{} n$ tiles behind. Now,
\[ (n \minus{} 1)^2 \equal{} n^2 \minus{} n \plus{} (1 \minus{} n) < n^2 \minus{} n < n^2
\]
so in the next step $ n \minus{} 1$ tiles are removed. This gives $ (n^2 \minus{} n) \minus{} (n \minus{} 1) \equal{} n^2 \minus{} 2n \plus{} 1 \equal{} (n \minus{} 1)^2$, another perfect square, and the process repeats.

Thus each two steps we cycle down a perfect square, and in $ (10 \minus{} 1)\times 2 \equal{} 18$ steps, we are left with $ 1$ tile.
[/hide]

I did it by brute force, doesn't take that long :lol:


You were 8 minutes late. It was posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110695[/url].  :wink:


MATHCOUNTS
AMC
AIME
geometry
AMC 12
probability
articles

AOPS 1 and 2 are good to study from. I think just volume one can get you to AIME if you know it fairly well.

I have the AoPS 1&2, Intro Geometry, Into Counting and Probab., and Intro Number Theory.
Volume 1 and the Introduction books are pretty good for getting you throught AMCs, but for AIMEs and harder stuff, refer to Intermediate Classes and Volume 2.

Also, make sure that you have printed out all the old tests. Since you have plenty of time before the next AMC try to get through all of them twice.

I have personally found that just doing a lot of contests problems helps me more than reading books.

what is a good amount of time to study per day?

it depends on how far you want to go. you can never study too much.

[quote="soltixnj"]Don't set apart an amount of time each day.  Rather, do math only as much as you see fit, with a little bit of motivation.   If you tell yourself "Oh, I'll do 1 hour of math each day", math really turns into a bore and you probably won't want to learn anything.

On another note, for the AMC 10, a strategy that many use is to answer exactly 21 questions, which leaves room for exactly 2 questions wrong to get a 120.
The max score using this strategy is a 132, which sets you off well for the AIME.
19 (6) + 4(1.5) = 20(6) = 120.

For the AMC 12, it's common to answer exactly 17 questions, which leaves room for exactly 2 questions wrong.
15(6) + 8(1.5) = 90 + 12 = 102, a passing score.
The max score here is only 114, but remember that the amc12 takers do determine the floor score.[/quote]

but the cutoff this year wont be 120 and 100 (they werent for the A and im guessing that wont be the case for the B)...
So you couldve done 22 questions on the 10 and missed 3 to still get a 118.5 and pass...but then again you cnat predict the score so just do as many as you can...if you really want to pass, always assume the cutoff is above 114.

Don't look at the test and decide that the cutoff score will be lowered from 120/100.
If it actually doesn't get lowered, then you would have just made a terrible mistake.

Interesting sidenote:

Let there be two unanswered questions. P: stands for probability of.
P: guess both right is 1/25, which yields a 12 point reward.
P: guess one right is 8/25, which yields a 6 point reward.
P: guess both wrong is 16/25, which yields a 0 point reward.

[12 + 6(8) + 16(0)]/25 = 4 point expected return.  This is better than just skipping and receiving 3 points.

If you had ONE unanswered question, it is favorable for you to leave it blank.  If you had TWO unanswered questions, it would be advantageous for you to guess ON BOTH.

only if 60/25=4, which is true in base -10, I guess

[quote="solafidefarms"][quote="AMC37.5"][quote="soltixnj"]
Interesting sidenote:

Let there be two unanswered questions. P: stands for probability of.
P: guess both right is 1/25, which yields a 12 point reward.
P: guess one right is 8/25, which yields a 6 point reward.
P: guess both wrong is 16/25, which yields a 0 point reward.

[12 + 6(8) + 16(0)]/25 = 4 point expected return.  This is better than just skipping and receiving 3 points.

If you had ONE unanswered question, it is favorable for you to leave it blank.  If you had TWO unanswered questions, it would be advantageous for you to guess ON BOTH.[/quote]

That can't be right. [12 + 6(8) + 16(0)]/25 = 2.56 < 3.[/quote]

Can nobody do math anymore? 12+6(8)=60, divided by 25 is 2.4, which is neither 4 nor 2.56.[/quote]

The reason why none of these statements are correct is because a correct answer doesn't give you a 6 point reward, but actually only a [b]4.5[/b] point reward, since not answering the question would have given you 1.5 points.  Similarly, getting a question wrong would give you a 1.5 point deduction, and the AMC is designed to make guessing unprofitable.  The expected value of points earned on a guessed question is $ \frac{\minus{}1.5(4)\plus{}6(1)}{5}\equal{}0$, which is less than the 1.5 if left unanswered.

As for advice, learning from my traumatic score this year, I'd say if you ever have to do the test at a bad time of day (in my case in the afternoon right after school) do the problems from last to first, or #25-1, so you don't get tired near the end and end up not being able to do the last few problems.

[i]Instead of trying to analyze exactly how many questions you will answer, get wrong, and leave blank[/i], just study hard every day (and especially in the summer) and try to answer as many as possible on the AMC.

Besides, if you choose to aim higher than just scraping by the cutoff to the AIME, you'll need a high index to get into USAMO.  Getting 3 more problems right on the AIME is harder than getting 5 more problems right on an AMC.  Make sure your AMC score is high so that you won't have more pressure on the AIME.

After all, you can learn a lot in a year.  If you get your level ramped up to 138-150 and you're just 1 problem away from a perfect on the real contest, it's best to make an educated guess on the last problem.  If you're lucky, a 150 is definitely better than a 145.5.

I would actually say that doing past AIMEs is better than doing past AMCs. They have a much better range of difficulty, all of the problems require thought (as opposed to the first 5-10 AMC problems which are basically plugging in numbers), and you'll then have experience for the AIME as well. I also believe that anyone with a reasonable mathematics background can solve the first 1 or 2 AIME problems if they put in enough thought, so difficulty should not be a problem.

[quote="davidyko"]Just do lots of old contest problems, hang around the forums, and look at AoPS vol. I and II, and Zeitz later on.
And eh, I don't really do that whole "answer exactly 17 problems" thing. I reckon just doing my best is good enough.[/quote]
i completely agree. do as many as you can. make is so that you don't have to rely on strategic approaches over knowledgeable approaches.

[quote="mathemonster"]I am fairly new to problem solving. i did mathcounts and was like 8 at states and got about a 110 on the AMC 10 without putting in any study time, or knowing how much some people actually studied. Being on Aops has motivated me to really try to improve and i am commited to study. Wat do u think the recommended books and plan of study is for someone at my level. My goal is to comfortably get to the aime next year when I am a sophmore. I have Art of Problm solving v.1 is that a good book? Have all of you guys read all the aops intro books? I would really appreciate some feedback on good books, and a plan of study. Thanks alot :)[/quote]

seems like you've got a lot of potential.
i don't really think you need books, but rather work on problems to build your skills and learn new things. take aimes and amc's or whatever and try to destroy the problems.

thanks... I think im going to do aops v1 first steps for math olympians and do some practice tests, as well as the aops amc 10 c0urse......

Don't do FSMO. Nearly everything there is covered in AoPS Volume 1, and AoPS definitely teaches it better, if you're willing to spend the time.

haha im in pretty much the same situation as you. i didnt even know about mathcounts and the amc until around the start of february this year when i was searching the internet and came across this site. its inspired me also to really do well on the amc's. i got a 109.5 on the 10b with about 2 weeks of studying aops vol. 1. though i only read the most important chapters i think that if i didnt bother to read it at all i would be scoring much lower. so yeah aops vol. 1 really helps. good luck next year. we'll qualify for aime together.

Before the talk gets transferred about AIME (which has often happened and probably will)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=70754#70754

 :D

[quote="miller4math"]ive seen so many of these "HOW DO I IMPROVE MY ___ SCORES?" threads
that im beginning to feel that they're spam - like

someone should create some "how to get better at ____" articles in the resources section

this is just my opinion but i feel that many feel the same way[/quote]

That is an excellent opinion and I'll eventually do that for next AMC. But really, the whole point is this: math competition is like a marathon. You cannot just wake up one day and run several miles. It takes days and days of work and giving up other stuffs to practice and practice. So, do the problems on and on.

Now, that is easier said than done. What I've noticed from people who had huge jumps on their contests often involved participating in other math competitions.  Also, increasing your level of comprehension of English helps too. Sometimes, I've seen kids getting problems wrong because they either read wrong or did not understand the problem correctly. Those are easy points to fix so do them!


Sorry, if this was already posted here before, (It is  a well know particular form of Beal's conjecture)

Prove that there are no solutions for :

3^x+4^y = 5^z

where  (x,y,z) are  integers  and greater than 2.

Shouldn't this be at least in the Pre-Olympiad level, considering the fact that the equation is a buffling one for intermediates?

Moderators - Can you move this to the other form. Thanks.

hmm...  I wonder how you solve this.

I didnt get anywhere but heres where I got up to which I presume every1 got further than me.

$5^z$ must have a unit digit of 5.

Unit digit of 3 is repeated every 4 times (3, 9, 7, 1)
Unit digit of 4 is repeated every 2 times (4, 6)

and in order to make a 5, we need to combine either 9 and 6 or 1 and 4.

So we get,

$3^{2a}+2^{2b}$ and $3^{4a}+2^{4y-2}$

where a,b equal any natural number.

I just complicated the problem.  Any ideas?

[quote="nodrog2002"]I just complicated the problem.  Any ideas?[/quote]
Try to find solution instead of proving its non-existence.  ;)

In my opinion, it is quite hard problem if you see it for first time :)

Very first thoughts that came to me:

Mod 3:
1 = (-1)^z mod 3 -> z is even; z=2c

Mod 4:
(-1)^x = 1 mod 4 -> x is even; x=2b

4^y + 9^b = 25^c

(2^y)^2 + (3^b)^2 = (5^c)^2

which relembles the Pythagorean Theorem.  It is well known that for two integers r and s, (r^2-s^2,2rs,r^2+s^2) is the set of all Pythagorean triples.  Now 2rs could either be 3^b or 2^y.  2rs=3^b is impossible since 3 is odd.  2rs = 2^y -> r=2^a and s=2^b -> 4^a + 4^b = 25^c, also impossible since 25 is odd.  QED.

Is this right?  My first-ever attempt at doing a pre-olympiad problem :)

@nr1337. What if s=1?

And I must say that I though you are solving $x^3+y^4=z^5$, that is why I posted my previous message  :blush:

Anyway, you can try it:
a) Find at least one solution in positive integer numbers: $x^3+y^4=z^5$
b) Prove that there are infinitely many solutions

[quote="Myth"]@nr1337. What if s=1?[/quote]

s=1 -> r=2^(y-1), and

4^(y-1) + 1 = 25^c
4^(y-1) = 25^c - 1 = (24)(biggeometricseriesstuff)

24 = 2^3 * 3.  But 4^(y-1) doesn't have a factor of 3!  Contradiction.

I was solving 3^x + 4^y = 5^z because that was what the first post said and I didn't read ahead :D

[quote="nr1337"]2rs = 2^y -> r=2^a and s=2^b -> 4^a + 4^b = 25^c, also impossible since 25 is odd.[/quote]
Oooppsss... It should be $4^a + 4^b = 5^c$ !!! See carefully!

nr1337 - very good!
[quote]Very first thoughts that came to me:

Mod 3:
1 = (-1)^z mod 3 -> z is even; z=2c

Mod 4:
(-1)^x = 1 mod 4 -> x is even; x=2b [/quote]

(BTW  You have to be a little careful,  z can be  1 when x=0, as 3^0+4^1=5^1 .. but the conclusion is ok as long as x  is >0)

Once you reach the conclusion x, and z are even, rest follows:

[hide]2^(2a) = (5^c+3^b)(5^c-3^b)
Since 2 is prime, each factor in RHS  is a power of 2
so let us say  2^m = 5^c+3^b
and               2^n = 5^c-3^b
(m>n)
so  2^m+2^n = 2*5^c
and 2^m - 2^n = 2*3^b

or 2^p(2^q-1) = 3^b
where p=n-1, q=m-n

since  3 is prime,  p  has to be zero.
and we get  2^q - 1 = 3^b
now   you do  have a solution when q=2 but if q>2;  we take modulo 8, LHS is   -1 (or 7) mod 8; but RHS can never be -1 (it is always 3 or 1 mod 8);

QED.[/hide]

[quote="Myth"]@

Anyway, you can try it:
a) Find at least one solution in positive integer numbers: $x^3+y^4=z^5$
b) Prove that there are infinitely many solutions[/quote]

Not related to  the original problem, but ....

One trivial solution would be when k=0,1,2 ....

[hide]x =  2^(20k+8)
y  = 2^(15k+6)
z  = 2^(40k+5)[/hide]

(256^3+64^4=32^5 etc...)

[quote="Gyan"][quote]s=1 -> r=2^(y-1), and

4^(y-1) + 1 = 25^c
4^(y-1) = 25^c - 1 = (24)(biggeometricseriesstuff)

24 = 2^3 * 3. But 4^(y-1) doesn't have a factor of 3! Contradiction. [/quote]
Smaill point:  here  RHS (if you  follow the logic  given before  where the sides of  a primitive right angle triangle is takes as  2rs, r^2-s^2 and r^2+s^2)   ought to be 5^c and not 25^c .. you can still prove what you want but it is just a little harder than when you take it as  25^c[/quote]
I have already written about it.


Prove that any n-gon can be cut into triangles by non-intersecting diagonals

Prove that the number of triangles obtained by cutting an n-gon by non-intersecting diagonals is equal to n-2

[quote="anncar"]Prove that any n-gon can be cut into triangles by non-intersecting diagonals [/quote]

[hide="Sketch"] The set of cuts we want forms a "zigzag" pattern.  That is, number the vertices $ A_0, A_1, A_2, ...$.  Then draw the lines from $ A_0$ to $ A_2$ to $ A_{n\minus{}1}$ to $ A_3$ to $ A_{n\minus{}2}$ until every vertex (except one) has been reached.  This is most rigorous by induction, but both results should be provable directly. [/hide]

Are you sure that algorithm should necessarily work on concave polygons?

(An attribute "simple" should be added to the statement, since it doesn't hold for non-simple polygons.)

i dont understand the soution. Could you go into more detail? and how about the second problem?

[quote="Farenhajt"]Are you sure that algorithm should necessarily work on concave polygons?[/quote]

I'd considered that, but wasn't sure if the result even held for concave polygons (although it does).  The proof by induction is much easier, then.

[hide="Solution"] It's easy to verify that an arbitrary quadrilateral can be divided into $ 2$ triangles; at least one of the two diagonals will be contained inside the quadrilateral, so use that one.

Suppose that we have decomposed an arbitrary $ k$-gon into $ k \minus{} 2$ triangles.  An arbitrary $ k \plus{} 1$-gon can be constructed by replacing one of the sides with two sides.  The triangle that was there is now a quadrilateral; apply the above argument again, and now we have divided a $ k\plus{}1$-gon into $ k\minus{}1$ triangles.  QED. [/hide]


chess64 - 6
The Original Pi Guy - 6*
Shogia - 6*
dangvh - 4
BornForMath - 4
JesusFreak197 - 3
Xantos C. Guin - 3
eryaman - 3
tarquin - 2
Secret Asian - 2
Snowstorm - 1

Honorable Mention for Solution: BornForMath
Reason: He gave the possible answers for #6.

* denotes the extra credit.

The solution is made below for you to use.

Here are two last points :D (This was something you didn't expect, did you?)

Because of the fact that #5 was badly worded and #6-7 were very hard (#6 was actually weird), here is what I decided to do:

If you can do two things below (only the fastest person gets the credit so not all people are going to get them):

You may do both.

Extra Credit 1: If you look at the solution packet, I used formula $\frac{ab(a-1)(b-1)}{4}$ for the number of rectangles. Show me a proof on how this works.

Extra Credit 2: With fully written solution, show me the work for #7. The correct answer is 6044 but since nobody got this one, I'll give credit for anyone who gives me FULL solution that is LaTeXed (It doesn't have to be document. You can just send as private message but it must be LaTeXed.)

#6 from Honorable Mention, BornForMath, (he doesn't go to normal ones since he didn't send me the full solution.)

[quote="BornForMath"]Alright here are the answers i got for number 6:
$(a, b, c, d)$ as $(7, 63, 180, 420), (7, 60, 252, 315), (9, 21, 140, 1260)$, so the answers are $\boxed{670, 634, 1430}$[/quote]

[quote="Silverfalcon"]Extra Credit #1 is taken by The Original Pi Guy.

His proof was basically this:

First, we find number of ways to choose 2 lines from $a$ parallel lines and 2 lines from $b$ perpendicular lines. The number of rectangles is product of this.

$\binom{a}{2} \cdot \binom{b}{2} = \frac{a!}{(a-2)!2!} \cdot \frac{b!}{(b-2)!2!} = \frac{a(a-1)}{2} \cdot \frac{b(b-1)}{2} = \frac{ab(a-1)(b-1)}{4}$

QED[/quote]

What?  :mad: I was just about to send you that! I just got home from school! No fair  :mad:

[quote="SnowStorm"]Argh! Why didn't you change the pdf?  I had no clue you had made changes to the problems.[/quote]

Eh.. I don't know. ;)

And um, I know it sounds not fair but I can't give credit to everybody because it's "extra-credit."

Still, extra credit 2's not taken.. :roll:

heh, yea ive been trying the 2nd one, but i havent bbeen getting very far

[quote="The Original Pi Guy"]Heh. I sent him that solution in between presentations during Chemistry.
Heh.  :lol:  :lol:  :lol:  :lol:[/quote]

 :cursing:  :furious:  :wallbash_red:  :wallbash_red: nooooooooooo...  :lol: Oh well. As long as the original pi guy doesn't get extra credit #2 also, I'm fine.  :first:  :D

If I send you the solution for #5, will I get extra credit? Please?  :lol:

I scored 3. Why am I not on the list?
EDIT: Thanks.

I found out that two people did get #7 right :blush: and I marked them wrong.

They were The Original Pi Guy and Shogia.

But this doesn't mean they are the ONLY ones who are eligible for the extra credit. It's still on and if you send me the correct answer with solutions (fully written, must be Latex), I'll give that person extra credit.

 :)

P.S. Thanks to Shogia for pointing out this.

I thought something was wrong...

WAIT, I GOT #7 RIGHT?????!!!!!

Yes. :)

So I gave you original 7-question credit. But extra credit is for whoever gives me fully-written solution for that problem.

What? Noooooooooooooo...  :mad:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red: darn, darn, darn....

Don't worry chess64. I solved the first sequence, but I guessed on the second sequence, and I have absolutly no idea how to do the second sequence... :blush:

Sorry, extra credit 2 is now taken by Shogia.

[quote="Shogia"]First we need to find the value of the first part: $\prod^{2004}_{r=1} \sin (\frac{2r\pi}{4009})$
$\sin(\frac{2r\pi}{4009})=\frac{e^\frac{2ri\pi}{4009}-e^\frac{-2ri\pi}{4009}}{2i}$
To simplify things we can let $e^\frac{2ri\pi}{4009}=\zeta$ (chosen since we will be working with a cyclotomic equation.)
This means that we have $\sin(\frac{2r\pi}{4009})=\frac{\zeta^r-\zeta^{-r}}{2i}$ also we will say that $x^{4009}-1=1+x+...+x^{4008}=\prod^{4008}_{r=1} x-\zeta^r$
Looking at the coefficients we find that $\prod^{4008}_1 \zeta^r=1$
By setting $x$ equal to $1$ and $-1$ respectively we find that $-\sin(\frac{2r\pi}{4009})=\sin(\frac{2\pi(4009-k)}{4009})$
Finally we may begin to simplify this *pants*
$(\prod^{2004}_{r=1} \sin (\frac{2r\pi}{4009}))^2=\prod^{4008}_{r=1}=\frac{1}{2^{4008}}\prod^{4008}_{r=1}\zeta^{2r}-1=\frac{1}{2^{4008}}\prod^{4008}_{r=1} (\zeta^r-1)(\zeta^r+1)=\frac{4009}{2^{4008}}$
Because we need the square root of the original statement in the above string of simplifications our answer is$ \sqrt \frac{4009}{2^{4008}}=\frac{\sqrt{4009}}{2^{2004}}$

Now to solve the summation part of our equation: (Yeah I'm long winded, I'm sure there is an easier way but I dont think like that)
$\sum^\infty_{s=0} \sum^\infty_{t=0} \frac {((-1)^s + (-1)^t)\cos(s)\sin(t)}{s!t!}$  For initial simplicity I will set this equal to everyones favorite variable, x.
Because the bounds of the summation are the same with respect to both s and t we can switch thier places in the equation and using the identity for the $sin(x+y)$ we find that $2x=\sum^\infty_{s=0} \sum^\infty_{t=0} \frac {((-1)^s + (-1)^t)\sin(s+t)}{s!t!}=2\sum^\infty_{s=0} \sum^\infty_{t=0} \frac {( (-1)^s)\sin(s+t)}{s!t!}$ again because of the equality of s and t.  Now for one final substitution $s+t=y$ and splitting up the summation we find that this is equivalent to $2\sum^\infty_{y=0} \frac{sin(y)}{s!t!}\sum^y_{s=0}(-1)^s{y\choose s}$
The summation on the left is a little bit weird but it doesnt matter since the summation on the right is equivalent to the expansion of $(1-1)^y$ which is $0$ and makes the whole sum $0$.

Therefore our final answer is $\frac{\sqrt{4009}}{2^{2004}}+0+\frac{1573}{2^{2004}}=\frac{1573+\sqrt{4009}}{2^{2004}}$, so  $a=1573$, $b=1$, $c=4009$, $d=2$, and $e=2004$.  Which means that $a+b+c+d+e=7589$ giving us our final answer.  I'm sorry if the explanations leave something to be desired.[/quote]

Wow, nice job.  I don't think I ever would have gotten that heh.

I was lucky that I had seen a similar equation to the second part before and was able to do that.  The first part I liked because of my fascination with demoivre numbers and roots of unity.

[quote="pkothari13"]yikes, that problem was kinda a far stretch for a "getting started" forum though, eh?[/quote]

From difficult ones in Round 1 and Round 2, their levels can go up to far as Intermediate. Some Round 3 may reach easy Pre-Olympiad. I'm not sure for exact how hard it can go but my contest isn't mean to create easy anyway.

[quote="Shogia"]The first problem of round 3 reminds me a lot of the olympiad inequality problems.[/quote]

that doesn't sound good  :(  :read:  :help: uh-oh...  :P

Number two isn't a lot better.

I'm still working on drawing and proof. :D I mean, I have it on the paper but making it in $\LaTeX$ is another story. :roll:

and you expect me with my little knowledge of latex to be able to do it, lol.  I will see what I can manage.

Gotta learn it some time or another, eh?

[quote="Shogia"]The first problem of round 3 reminds me a lot of the olympiad inequality problems.[/quote]

I think those type of problems are created "backwards", or taken from another source.


In how many ways may we write the number 10 as the sum of three positive integer summands? Here order counts, so, for example, 1 + 8 + 1 is to be regarded different from 8 + 1 + 1

hi. there are two cases, that is if you are talking about non-negative or posititive numbers.

Case 1: Includes zeros
Consider x + y + z = 10. We can imagine x, y and z as containers where * can be linearly aligned. For example,

***+**+***** = 3 + 2 + 5.
+**+******** = 0 + 2 + 8.

Notice that you can have a one-one correspondence between the integral solutions and the linear arrangement (permutation) of 10 * and 2 +'s sings which equals 12 objects.  But * and + are repeated. Hence you have a permutation in which 10 objects are alike and 2 objects are alike. Therefore  $ \frac{12!}{10!2!}$.

This equals to $ \binom{10+2}{10}$

In general, in the equation $ x_{1}+x_{2}+...+x_{n}= r$
The number of non negative integer solutions is $ \binom{n+r-1}{n}$

Now, maybe you can try case 2.   :)

is this as simple as it looks? first we start with the first number 1. if we have one, the other 2 digits must have a sum of 9. theres 1,8 (note we cant use zero) 2,9..... and we get 8 of them. with 2, there are 7 different ways to make 8(plus the 2), 6 different ways to make 7...and so on. so now we have $ 1+2+.......+7+8$. using the sum formula, we get 36.

@guile: since zero is not a positive integer, your case one doesn't work.

@abacadaea: I got the same answer, but I just listed all the options :oops:

[hide]Since each integer must be positive, we have to split 7 between the three integers. $ \binom{9}{2}=36$[/hide]

Sorry, my mistake. I was in a hurry.  :blush:

[quote="NeverOddOrEven"][quote="guile"]
In general, in the equation $ x_{1}+x_{2}+...+x_{n}= r$
The number of non negative integer solutions is $ \binom{n+r-1}{n}$
[/quote]

I think you got the number of solutions wrong. Shouldn't it be $ \binom{r+n-1}{n-1}$?[/quote]

yeah, you're right


some of the writing that i wrote after a certain part doesnt show up anymore and one of my pictures wont come up

Make sure your image document is called satelitedesign.png and that it is in the same directory as your tex file.

i did that and they are in the same foldr and i cant post my picture the document size is too big aparently

Try the document with another picture first. Maybe the png is damaged?

Are you using pdflatex, i.e. do you compile directly to pdf? Perhaps attach your logfile, the information of it could tell us more.
Your code works for me with a sample image that I've given the appropriate file name.

Stefan

it compilles and the first half of the document shows up perfectly but the second half with pictures and words doesnt show up

The fact that you cannot post your picture because of too big size makes me think that it just doesn't fit the page and $ \text{\LaTeX}$ goes nuts trying to place it. Try to scale your picture down a bit and see if it helps.

Check and/or post the logfile, if something's wrong there should be errors or warnings listed.

Stefan

i put the latex on a usb to take to  my moms house and then it got completely screwed up and at that  time i figured the algebra for the latex wpould take a while so i have completely erased the project and will start again later so no more help needed on this topic