geometry
rectangle
ratio
3D geometry
calculus
integration
inequalities

1. A while ago in a laboratory in an african country, a dangerous bacteria named "Ishangololo" was made by an accident. After maturity, this bacteria splits into two bacterias like itself. The age of maturity might not be the same in all bacterias. We say that the bacteria A is a child of the bacteria B if A has came into existence by the division of the bacteria B of by the division of one of B's children. Right now, we have 3k bacterias in the lab. Prove that there is a bacteria whose number of children among the current bacterias is a number greater or equal than k and smaller or equal than 2k.

2. Around a circle, we have written n numbers each either 0 or 1 and one of these n numbers is underlined. In each round, we can do one of the below :
i) Remove the line below the underlined number and underline one of the numbers next to it.
ii) Change the underlined number from 0 to 1 and vice versa.
Find the least number of rounds needed so that we can change any first condition to another.

3. A 8x8 square on the plane divided into 64 squares 1x1must be covered by 64 black and 64 white isosceles triangles (each square must be covered by two triangles). A covering is "fine" if any two neighboring (having a common side) triangles are of different colors. How many different "fine" colorings are there?

4. A journalist is looking for a person named Z at a meeting of N persons. He has been told that Z knows all the other people at the meeting but none of them knows Z. The journalist may ask any person X : 'Do you know that man?' about any Y.
a) Can he be certain to find Z by asking less than N questions?
b) What minimal number of questions are needed to find Z? (Prove that a lesser number won't work.)

5. A rectangle axb (a>b) is cut into right triangles so that any common side of two triangles is a leg (cathetus) for one of them and a hypotenuse for the other. (Every two triangles have a common side or a common vertex or no common points.) Prove that the ratio a/b is not less than 2.

6. 8 students were asked to solve 8 problems. (the same ones)
a) Each problem was solved by 5 students. Prove that one can find two students so that each problem was solved by at least one of them.
b) If each problem was solved by 4 students, then it is possible that no such a pair of students exists. Prove this.

7. The integers 1,2, ... , n each appear n times in an nxn matrix. Show that there is a row or column containing at least n1/2 distinct integers.

8. There is a race car on a circular racetrack of n miles. There are fuel stations around the track, such that the total amount of fuel at all the stations is just enough to travel n miles. Show that there is a point on the track at which the race car may start, such that it can complete a circuit by picking up fuel along the way.

9. Define a sequence of strings as follows: Let R0 = 1 , and let Rn be the string obtained by replacing each instance of 'i' in Rn-1 with '123..i' and finally adding 'n+1' at the end. So for instance, R1 = 12 , R2 = 1123 , R3 = 11121234 . Show that for n>0 , if we write Rn down and Rn backwards under that, then each column will contain exactly one '1' .

10. Let m and n be positive integers. Let x1 , x2, ... , xn be non negative integers, whose average is less than or equal to m+1, and let y1 , y2, ... , yn be non negative integers, whose average is less than or equal to n+1. Show that some non empty subset of the xi's has sum equal to some non empty subset of the yj's.

11. 8 singers take part in an art festival. The organizer wants to plan m concerts. 4 singers go on stage in every concert. Restrict that the times of which every two singers go on stage in a concert are all the same. What is the minimal of m? (Make a design.)

12. Between 2n-1 positive integers, prove that we can choose n of them in which the sum of these n integers is divisible by n.

13. On the island of Camelot, live 13 gray, 15 brown and 17 crimson chameleons. If two chameleons of different colors meet, they both simultaneously change color to the third color. Is it possible that they will eventually all be the same color?

14. The numbers x1 , x2 , ... , xn are such that x1 + x2 + ... + xn = 0 and x12 + x22 + ... + xn2 = 1 . Prove that there are two numbers among them whose product is no greater than -1/n.

15. We say that some positive integer m "covers" the number 1998, if 1, 9, 9, 8 appear in this order as digits of m. (For instance 1998 is covered by 215993698 but not by 213326798.) Let k(n) be the number of positive integers that cover 1998 and have exactly n digits (n>4), all different from 0. What is the remainder of k(n) in division by 8?

16. The cube S with edge 2 consists of 8 unit cubes. We will call the cube S with one unit cube removed, a "piece". The cube T with edge length 2n consists of (2n)3 unit cubes. Prove that if one unit cube is removed from T, then the remaining solid can be built with pieces.

17. Peter has three accounts in a bank, each with an integral number of dollars. He is only allowed to transfer money from one account to another so that the amount of money in the latter is doubled.
a) Prove that Peter can always transfer his money into two accounts.
b) Can Peter always transfer his money into one account?

18. On a straight line, n red points and n blue points (not necessarily distinct) are marked arbitrarily. Prove that the sum of the distances between every two points of different colors is not less than the sum of the distances between every two points of the same color.

19. A 7x7 square is made up of 16 '1x3' tiles and 1 '1x1' tile. Prove that the 1x1 tile lies either at the center of the square or adjoins one of its boundaries.

20. At a round table are 1994 girls playing a deck of n cards. Initially, one girl holds all of the cards. In each turn, if at least one girl holds at least two cards, one of these girls must pass a card to each of her two neighbors. The game ends when and only when each girl is holding at most one card.
a) Prove that if n > 1993, then the game cannot end.
b) Prove that if n < 1994, then the game must end.

[hide="1"]
The problem statement is equivalent to the following: Given a full binary tree with $ 3n$ leaves, show there exists a full binary subtree with at most $ 2n$ leaves and at least $ n$ leaves.

The proof of this is simple.  Take the largest full proper subtree of our tree.  Clearly it has at least half of the leaves.  If the number of leaves is too large, take the largest full proper subtree of that.  Keep doing this until we are in the desired interval.  Because each subsequent subtree has at least half of the leaves of the tree before it, we cannot skip through the interval.  By the Well Ordering Principle, we must be in the interval at some point.[/hide]

[hide="2"] Uh...$ 2n\minus{}1$?  Assuming the places of the numbers matters.[/hide]

[hide="Idea on 3"]
Depending on the orientation of the diagonal separating the two isoceles triangles in a 1 x 1 square, assign that square either 0 or 1.  The sum of the numbers in a 2 x 2 square must be even.  So we can calculate the number of ways to number the squares of a chessboard with 0s and 1s so that the sum of the numbers in any 2 x 2 square is even, and then multiply by two (because black and white are interchangeable).[/hide]

[hide="19"]
Number the square in two ways as follows:

1 2 3 1 2 3 1   -1-2-3-1-2-3-1
2 3 1 2 3 1 2   -2-3-1-2-3-1-2
3 1 2 3 1 2 3   -3-1-2-3-1-2-3
1 2 3 1 2 3 1   -1-2-3-1-2-3-1
2 3 1 2 3 1 2   -2-3-1-2-3-1-2
3 1 2 3 1 2 3   -3-1-2-3-1-2-3
1 2 3 1 2 3 1   -1-2-3-1-2-3-1

Every 1 x 3 tile covers one of each kind of number.  But there are 16 2's, 16 3's, and 17 1's.  Therefore, the 1 x 1 tile is on a 1.  Similarly, the 1 x 1 tile is on a -1.  This yields the desired result.

This double numbering trick is very useful when you have a symmetric shape.[/hide]

[b]11.[/b] Every singer sings with 3k other singers in total. But that number is divisible by 7 because he sings with each of the remaining 7 singers equal number of times. Thus there are minimal 7 concerts in which one singer sings. That gives 56 concerts in total, but every concert is sung by 4 of them so the real number of concerts is minimal 14. example for 14: take singers a,b,c,d,e,f,g,h and concerts as follows: abcd, abef, abgh, aceg, acfh, adeh, adfg, bcef, bcgh, bdeg, bdfh, cdeh, cdfg, efgh.

[b]13.[/b] Let a, b, c be numbers of gray, brown and crimson chameleons. a, b, c will always make the complete system of residues modulo 3 (easy to check) so the statement is impossible.

What do you mean in Problem 7?

I claim that you mean:

7. The integers $ 1,2,\dots, n$ each appear $ n$ times in an $ n\times{n}$ matrix. Show that there is a row or column containing at least $ \sqrt{n}$ distinct integers.

You can give counterexamples for $ \frac{n}{2}$, $ n\plus{}\frac{1}{2}$, and $ \frac{n\plus{}1}{2}$.  So I believe you mistyped it and it should be $ n^{1/2}$.

[hide="Solution"]
Let $ r_{i}$ and $ c_{i}$ be the $ i$-th row and the $ i$-th column for each $ i \equal{} 1,2,\dots,n$.  Denoting by $ R$ and $ C$ are the sets $ R: \equal{}\{r_{1},r_{2},\dots,r_{n}\}$ and $ C: \equal{}\{c_{1},c_{2},\dots,c_{n}\}$.  Let $ G$ be a bipartite graph with two sets of vertices $ V_{1}\equal{} R\cup{C}$ and $ V_{2}\equal{}\{1,2,\dots,n\}$. A vertex $ v\in V_{1}$ is connected to a vertex $ k$ in $ V_{2}$ iff $ k$ belongs to the row or the column for which $ v$ is the representative.  (You actually don't need to use Graph Theory.  It is essentially the Double Counting Technique.)

Suppose for each $ k$, the number of the neighbors of $ k$ in $ R$ is $ x_{k}$ and the number of the neighbors of $ k$ in $ C$ is $ y_{k}$.  Therefore, $ x_{k}y_{k}\geq n$ (because each integer occurs $ n$ times).  Using the AM-GM Inequality, one obtains $ \deg(k) \equal{} x_{k}\plus{}y_{k}\geq 2\sqrt{x_{k}y_{k}}\geq 2\sqrt{n}$.  Thus,
\[ \sum_{v\in{V_{1}}}\deg{v}\equal{}\sum_{k\in{V_{2}}}\deg(k)\geq 2n\sqrt{n}\,.\]
Hence, $ \frac{1}{2n}\sum_{v\in{V_{1}}}\deg{v}\geq\sqrt{n}$ and according to the Pigeonhole Principle, there must exist a row or a column with more than $ \sqrt{n}$ distinct numbers.

[/hide]

19. A $ 7\times7$ square is made up of $ 16$ '$ 1\times3$' tiles and $ 1$ '$ 1\times1$' tile. Prove that the $ 1\times{1}$ tile lies either at the center of the square or adjoins one of its boundaries.

[hide="Solution Using Generating Function"]
Let $ P(x,y) \equal{} (1\plus{}x\plus{}x^{2}\plus{}\cdots\plus{}x^{6})(1\plus{}y\plus{}y^{2}\plus{}\cdots\plus{}y^{6})$.  The term $ x^{i}y^{j}$ represents the square at the intersection of the $ (i\plus{}1)$-st row and $ (j\plus{}1)$-st column.  Suppose that we can cover the $ 7\times7$ board with sixteen $ 1\times3$ tiles and one $ 1\times1$ tile.  If the $ 1\times1$ tile is situated at the $ (a\plus{}1)$-st row and the $ (b\plus{}1)$-st column, then,
\[ P(x,y) \equal{} x^{a}y^{b}\plus{}(1\plus{}x\plus{}x^{2})Q(x,y)\plus{}(1\plus{}y\plus{}y^{2})R(x,y)\,,\]
for some polynomial $ Q$ and $ R$.

Let $ \omega \equal{}\frac{\minus{}1\plus{}\sqrt{\minus{}3}}{2}$ be the primitive cube root of unity.  Thus,
\[ P(\omega,\omega) \equal{}\omega^{a\plus{}b}\,.\]
On the other hand,
\[ P(\omega,\omega) \equal{}\left(1\plus{}(\omega\plus{}\omega^{4})(1\plus{}\omega\plus{}\omega^{2})\right)\left(1\plus{}(\omega\plus{}\omega^{4})(1\plus{}\omega\plus{}\omega^{2})\right)\equal{}1\,.\]
Hence, $ \omega^{a\plus{}b}\equal{} 1$ or $ a\plus{}b$ is a multiple of $ 3$.

Similarly, $ \omega^{a\minus{}b}\equal{} P\left(\omega,\frac{1}{\omega}\right) \equal{} 1$.  Thus, $ 3$ divides $ a\minus{}b$.  Consequently, $ a$ and $ b$ are multiples of $ 3$ and hence, the $ 1\times1$ tile is at the square $ (1,1)$, $ (1,4)$, $ (1,7)$, $ (4,1)$, $ (4,4)$, $ (4,7)$, $ (7,1)$, $ (7,4)$, or $ (7,7)$.  Hence, it is either on the boundary or at the center of the board.
[/hide]

[quote="Batominovski"]Are you sure about the 16th problem?

16. The cube $ S$ with edge $ 2$ consists of $ 8$ unit cubes. We will call the cube $ S$ with one unit cube removed, a "piece". The cube $ T$ with edge length $ 2n$ consists of $ (2n)^{3}$ unit cubes. Prove that if one unit cube is removed from $ T$, then the remaining solid can be built with pieces.

It is obviously not true when $ n\equiv 0,3,5,6 (\mathrm{mod}\,7)$.  (For $ (2n)^{3}\equiv\minus{}1 (\mathrm{mod}\,7)$ for these numbers.)

I think the problem is:
16. The cube $ S$ with edge $ 2$ consists of $ 8$ unit cubes. We will call the cube $ S$ with one unit cube removed, a "piece". The cube $ T$ with edge length $ 2^{n}$ consists of $ (2^{n})^{3}$ unit cubes. Prove that if one unit cube is removed from $ T$, then the remaining solid can be built with pieces.

If this is the case, then the solution is quite simple.  Just induct on $ n$.[/quote]

I assumed the question was "Find all $ n$ for which this is possible"

12. is the Erdos-Grinburg-Ziv theorem  :wink:

18 is leningrad some years ago,I'll send the exact year and the solution later...


Let $ f$ and $ g$ be functions that are differentiable on the interval $ (a,b)$.  Let $ h(x) \equal{} \begin{cases} f(x) ;& a < x < c \\  g(x);& c \leq x< b \end{cases}$. Show that if $ f(c) \equal{} g(c)$ and $ f'(c) \equal{} g'(c)$ then $ h$ is differentiable at $ c$.


function
calculus
trigonometry
algebra
domain
absolute value
inequalities unsolved

$a^{2}+b^{2}= 1, |a+\frac{a}{b}+b+\frac{b}{a}| >/= 2-\sqrt{2}$

I tried letting a=sin x and b= cos x, and tried to find the global minimum of the function, but the absolute value signs make things harder...Anyway, there is meant to be a purely algebraic method with no calculus involved. Help please. The >/= is meant to be more or equal to btw.

A hint : Put $y=\sin x+\cos x$, we have $\frac{\cos{x}}{\sin{x}}+\frac{\sin{x}}{\cos{x}}=\frac{1}{\sin{x}\cos{x}}=\frac{2}{y^{2}-1}$.

$| y+\frac{2}{y^{2}-1}|$ So what should I do now? The Absolute value signs are still annoying when finding the global minimum >.<

Now you can use real analysis with $f(y)=y+\frac{2}{y^{2}-1}$ on $[-\sqrt{2},\sqrt{2}]\setminus \{-1,1\}$.  :wink:

I have no idea what you mean by real analysis, or why you choose that domain or range. Sorry im new at this

[quote="Gib Z"]I have no idea what you mean by real analysis[/quote]
$f'(y)=\frac{y^{4}-2y^{2}-4y+1}{(y^{2}-1)^{2}},...$,

[quote] or why you choose that domain or range.[/quote]
Because $y=\sqrt{2}\cdot\sin (x+\frac{\pi}{4})$ and $y^{2}-1\not = 0$.
Note: I see that the equation $f'(y)=0$ isn't easy! Sorry  :blush:

Ahh perhaps I should start over, this is meant to be a precalc question anyway, it an elementary solution..


Another math question for whoever wants to answer it for me. Thank you.

 The speed of light is 299 793 km/s. If it takes 8 mins. for light to reach the earth from the sun, which of the following is the best estimate of the distance (in kilometres) between the earth and the sun?

A. 300 000 x 8 x 60

B. 300 000 x 8

C. [u]300 000[/u]
                      x 60
             8

D. [u]300 000[/u]
             8

E. [u]300 000[/u]
                     x 8
            60

so approximately light goes 300000 km in a second.

8 minutes = 8*60 seconds

so the answer would be A (300000*8*60)


Find $n$ such that the sum of the first $n$ positive integers is a three digit number having equal digits.

No solution required! The problem was proposed by Professor Tiberiu Roman, who had the idea of an international math competition. He was the one who invented the IMO. The first two IMO's, as everybody knows, were organized by Romania in 1959 and 1960.
I felt that prof. Roman's name has to be mentioned somewhere in this site.

[quote="enescu"]
I felt that prof. Roman's name has to be mentioned somewhere in this site.[/quote]

 :)

Pierre.

Is Professor Tiberiu Roman still alive ? Tell us more about his life, connections to mathematical olympiads and how the idea of an first international mathematical olympiad evolved in Romania.  :)

Well, unfortunately I don't have detailed information about this. As far as I know, prof. Roman was involved in math olympiads until late 60's. After that he retired. Nevertheless, in 1999, when the IMO was held in Romania, prof. Roman was invited to the opening ceremony and got a big hand from all the participants. I was there and I reckon I was deeply touched. I will look for further information and keep you posted.


linear algebra
matrix
linear algebra unsolved

Let $ A\in M_{n\times m}(F),B\in M_{m\times n}(F)$,where $ F$ is a field.If $ AB$ is Idempotent.(Which means $ (AB)^{2}\equal{} AB$),and $ rank(AB) \equal{} rank(BA)$.
Prove:$ BA$ is Idempotent.

$ spectrum(AB)\equal{}\{0_{1},\cdots,0_{p},1_{1},\cdots,1_{q}\}$ with $ dim(ker(AB))\equal{}p,dim(ker(AB\minus{}I))\equal{}q.$
$ det(BA\minus{}\lambda{I})\equal{}\pm{det(AB\minus{}\lambda{I})\lambda^{m\minus{}n}}$, thus $ spectrum(BA)\equal{}\{0_{1},\cdots,0_{p\plus{}m\minus{}n},1_{1},\cdots,1_{q}\}.$
$ q\equal{}rank(AB)\equal{}rank(BA)$ thus $ dim(ker(BA))\equal{}m\minus{}q\equal{}m\minus{}n\plus{}p$.
Moreover: $ dim(ker(BA\minus{}I))\equal{}dim(ker(AB\minus{}I))\equal{}q$ (cf:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=894873#894873 ).
Conclusion: $ BA$ is diagonalizable; thus it's a projector.  QED.

Thank you,loup blanc.
My proof is completely different from you.
By $ (AB)^{2}\equal{}AB$,we can easily get $ rank(BAB)\equal{}rank(AB)\equal{}rank(BA)$
thus there exist a matrix $ C$ such that $ BABC\equal{}BA$
Multiply $ A$ both side,we have: $ ABABC\equal{}ABA\equal{}ABC$
Thus $ BA\equal{}B(ABC)\equal{}B(ABA)$


not allowing me to change my blog background. =(

Sometimes, it doesn't come right away, so try hitting CTRL+F5 when viewing your blog. I had the same problem and ctrl+F5 usually fixes it.


geometry
rectangle
analytic geometry
trapezoid

Let $ABCD$ be a rectangle with center $O.$ Let $N\in (AO),$ let $M$ be the midpoint of $[AD]$ and let $P= MN \cap CD,\, E= OP\cap BC.$ Prove that $NE\perp BC.$

[hide]Easy with coordinate geometry... though I'm sure there has to be a better solution. If nobody comes up with something I'll post my solution.[/hide]

[hide]Let $PM\cap AB=Q$, $PO\cap AB=F$.
Since $AB\parallel PD$, $\triangle AQM \sim\triangle DPM$, and $AM = MD\Longrightarrow AQ = PD$. $\triangle POD\sim\triangle FOB$, and $DO = OB\Longrightarrow PD=BF$. Then $AQ = BF$.
$\triangle AQN \sim\triangle CPN$ and $\triangle BFE \sim\triangle CPE$. $PC = PC$ and $AQ=BF$, so $\frac{AN}{NC}=\frac{BE}{EC}$, which means that $NE\parallel AB\Longrightarrow NE\perp BC$.[/hide]

[quote][color=darkred][b][u]Generalization.[/u][/b] Let $ABCD$ be a trapezoid ($AD\parallel BC$). Denote the middlepoints $O$, $M$ of the segments $[BD]$, $[AD]$ respectively.
For a point $N\in AC$ define the intersections $P\in NM\cap CD$ and $R\in PO\cap BC$.  Prove that $NR\parallel AB.$[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the intersections $S\in NP\cap BC$ and $T\in OP\cap AD$. Observe that $TD=BR$ and

$\left\{\begin{array}{c}\frac{PD}{PC}=\frac{TD}{RC}=\frac{RB}{RC}\\\\ \frac{PD}{PC}=\frac{MD}{SC}=\frac{MA}{SC}=\frac{NA}{NC}\end{array}\right\|$ $\Longrightarrow$ $\frac{RB}{RC}=\frac{NA}{NC}$ $\Longrightarrow$ $NR\parallel AB$.

[b][u]Commentary.[/u][/b] From the [u]hypothesis of the April's problem[/u] I used in the my proof only the following essential deeds :

$1\blacktriangleright AD\parallel BC$, i.e. the quadrilateral $ABCD$ is a trapezoid ;
$2\blacktriangleright$ the point $O$ is the middlepoint of the diagonal $[BD]$ ;
$3\blacktriangleright$ the point $M$ is the middlepoint of the side $[AD]$ ;
$4\blacktriangleright$ the point $N$ belongs to the line $AC$.

Thus, you can extend this problem to a trapezoid ! You must learn to "clean" the hypothesis, to remove the useless things. [/color]

[quote][color=darkred][b][u]Generalization II.[/u][/b] Let $ABCD$ be a trapezoid ($AD\parallel BC$). Denote two points $O\in BD$, $M\in AD$ for which $OM\parallel AB$.
For a point $N\in AC$ define the intersections $P\in NM\cap CD$ and $R\in PO\cap BC$.  Prove that $NR\parallel AB.$[/color][/quote][hide="Proof."]$S\in NP\cap BC$ and $T\in OP\cap AD\Longrightarrow\left\{\begin{array}{c}OM\parallel AB\Longleftrightarrow \frac{MD}{MA}=\frac{OD}{OB}\\\\\frac{PD}{PC}=\frac{TD}{RC}=\frac{TD}{RB}\cdot \frac{RB}{RC}=\frac{OD}{OB}\cdot\frac{RB}{RC}\\\\ \frac{PD}{PC}=\frac{MD}{SC}=\frac{MD}{MA}\cdot\frac{MA}{SC}=\frac{MD}{MA}\cdot\frac{NA}{NC}\end{array}\right\|$ $\Longrightarrow$ $\frac{RB}{RC}=\frac{NA}{NC}$ $\Longrightarrow$ $NR\parallel AB$.[/hide]

Thank you for two interesting problems and useful advice, [color=indigo][b]Mr. Virgil Nicula[/b][/color]  :lol:

[hide="A small present (bonus) for April."][quote="Virgil Nicula"][color=darkred][b][u]An equivalent enunciation of the second generalization.[/u][/b]

Given are in space two triangles $ABC$, $A'BC$ and the points $\{\begin{array}{c}M\in AB\ ,\ N\in AC\\\ M'\in A'B\ ,\ N'\in A'C\end{array}$ so that $MN\parallel M'N'$.

Define the intersection $I\in MM'\cap NN'$. Prove that $I\in AA'\Longleftrightarrow MN\parallel BC$.[/color][/quote]

[color=darkblue][b]The latest commentary.[/b] This problem is easily, but it is a very good example for a development of a lesson in which the students will understand
that one problem isn't abandoned when they found already at least a proof. Only from here the creation (innovation) is starting and has the last word.[/color][/hide]

[quote="Virgil Nicula"][quote][color=darkred][b][u]Generalization II.[/u][/b] Let $ABCD$ be a trapezoid ($AD\parallel BC$). Denote two points $O\in BD$, $M\in AD$ for which $OM\parallel AB$.
For a point $N\in AC$ define the intersections $P\in NM\cap CD$ and $R\in PO\cap BC$.  Prove that $NR\parallel AB.$[/color][/quote][/quote] Generalization:
Let $ABCD$ be [color=darkred]any quadrilateral[/color]. Denote two points $O\in BD$, $M\in AD$ for which $OM\parallel AB$.
For a point $N\in AC$ define the intersections $P\in NM\cap CD$ and $R\in PO\cap BC$.  Prove that $NR\parallel AB$.[hide="Proof"]
Apply Menelaus for $\triangle ADC$ and transversal $NMP$, and for $\triangle BDC$ and transversal $ROP$. Hope i am right :) [/hide]

Yes, that right. [color=red]Very nice problem and nice solution too![/color]

Thanks.  :P

For the initial problem: Denote $Q=MN\cap AB, R=PE\cap AB$; as $OM$ is midline in triangle $\Delta PQR \implies QR=2\cdot OM=AB\iff AQ=RB$. Next, $\frac{QN}{NP}=\frac{AQ}{CP}=\frac{RB}{CP}=\frac{RE}{PE}$, so $\frac{RE}{PE}=\frac{QN}{NP}\iff NE\parallel AB\iff NE\bot BC$.

Best regards,
sunken rock


Enrollment is open for kids who will be in 6th or 7th grade in September 2005.
The full time school is free for the families of students.
Teachers are also welcome to apply.

http://www.amsacs.org

I would suggest that you also post this in the Massachusetts forum.


calculus
integration
abstract algebra
function
real analysis
real analysis unsolved

can anyone find it please

i've got a result from this page:

http://wims.unice.fr/wims/wims.cgi?module=tool/analysis/function.en

but i doubt if it's right

It's not an elementary function. There might be expressions for it if we add enough additional functions to the system.

There are some ways to estimate definite integrals of $x^{x}$.  See for example
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34003[/url].

Buffalo's method was the nicest and most accurate.


a,b,c sides of a triangle with a^2 + b^2 + c^2 = s

prove

(root(s-2ab)+root(s-2bc)+root(s-2ca))/root(s) < j

for j = 2root(2), or improve on it [smaller j]

yes,

although, from your proof, j= root(6) is a better bound since,

6(x^2 + y^2 + z^2) <= 3((x+y)^2 + (y+z)^2 + (z+x)^2)

I think we could get better though?


let $a,b,c$ be positive real   such that :
$(a+b-c)(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})=4$
 find minimal of expression P
                    $P=(a^4+b^4+c^4)(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4})$
 I want to have a solution complete


algebra
polynomial
Vieta
geometry
3D geometry
AMC
AIME

If a cubic polynomial like $ 8x^3\plus{}1001x\plus{}2008\equal{}0$ has no x-squared term, then the sum of the roots is 0.  Why is this the case?  Can this rule be generalized to other polynomials?

Thanks!

It can be applied to all $ d$-degree polynomials $ P(x)$ that do not have a $ d \minus{} 1$-degree term.

Therefore, for example, in a fourth-degree polynomial of the form
$ P(x) \equal{} ax^{4} \plus{} bx^{2} \plus{} cx \plus{} d$ that doesn't have a cubic term (of the form $ kx^{3}$)
the sum of its roots is $ 0$.
This is simply an application of [url=http://mathworld.wolfram.com/VietasFormulas.html]Vieta's Formulas[/url]

In direct response to your question, we can consider a polynomial with roots $ r,s,t$

$ P(x) \equal{} a(x \minus{} r)(x \minus{} s)(x \minus{} t) \equal{} a(x^3 \minus{} (r \plus{} s \plus{} t)x^2 \plus{} (rs \plus{} st \plus{} tr)x \minus{} rst)$

So if there is no $ x^2$ term then we must have $ r \plus{} s \plus{} t \equal{} 0$

And this then generalises to Vieta's Formula

Learning a lot about Vieta's is really useful (and pretty simple too). You can do a lot with it like find the sum of the squares of the roots, cubes of the roots, etc. and the sum of the reciprocals of the roots, sum of the squares of the reciprocals, etc.

I think that's the polynomial from an AIME II 2008 question.


$ a,b,c$natural $ a!b!\equal{}a!\plus{}b!\plus{}c!$  find $ a,b,c$

See here : http://www.mathlinks.ro/viewtopic.php?t=169922


I'm stuck with this problem and i can't solve it right now, while i did in the past. :maybe:  :mad:
Simplify the: $ \sqrt[3]{{26 \plus{} 15\sqrt 3 }} \plus{} \sqrt[3]{{26 \minus{} 15\sqrt 3 }}$

The answer as i have in my old papers is the number 4, but now i can't prove it.

[quote="ffao"][hide]
Let $ a \equal{} \sqrt [3]{{26 \plus{} 15\sqrt 3 }} \plus{} \sqrt [3]{{26 \minus{} 15\sqrt 3 }}$, $ b \equal{} \minus{} \sqrt [3]{{26 \plus{} 15\sqrt 3 }}$ and $ c \equal{} \minus{} \sqrt [3]{{26 \minus{} 15\sqrt 3 }}$.

Clearly, $ a \plus{} b \plus{} c \equal{} 0$, and so, $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \equal{} 0$.

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} 0$
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$
$ a^3 \minus{} (26 \plus{} 15\sqrt 3) \minus{} (26 \minus{} 15\sqrt 3) \equal{} 3a\sqrt [3]{(26 \plus{} 15\sqrt 3)(26 \minus{} 15\sqrt 3)}$
$ a^3 \minus{} 52 \equal{} 3a\sqrt [3]{676 \minus{} 675}$
$ a^3 \minus{} 52 \equal{} 3a$
$ a^3 \minus{} 3a \minus{} 52 \equal{} 0$

We're looking to factor $ a^3 \minus{} 3a \minus{} 52$. From the form of that expression, it's probably $ (a \minus{} r_1)(a^2 \plus{} ar_1 \plus{} r_2)$. We have that

$ r_1r_2 \equal{} 52$
$ r_1^2 \minus{} r_2 \equal{} 3$

From which we conclude that $ r_1 \equal{} 4$, $ r_2 \equal{} 13$.

$ (a \minus{} 4)(a^2 \plus{} 4a \plus{} 13) \equal{} 0$

Clearly, $ a \in \mathbb{R}$, so $ a \equal{} 4$, the only real solution to that equation.
[/hide][/quote]
Ingenious solution! Many thanks.

 I don't think that i had solved it that way back then, so if you or anyone else have different solutions please post them too. :oops:

If you don't see the previous solution you can always expand the cube:

[hide="Solution #2"]
Let $ a \equal{} \sqrt[3]{{26 \plus{} 15\sqrt 3 }} \plus{} \sqrt[3]{{26 \minus{} 15\sqrt 3 }}$, as previously.

$ a^3 \equal{} 26 \plus{} 15\sqrt 3 \plus{} 3\sqrt[3]{(26 \plus{} 15\sqrt 3)^2(26 \minus{} 15\sqrt 3)} \plus{} 3\sqrt[3]{(26 \plus{} 15\sqrt 3)(26 \minus{} 15\sqrt 3)^2} \plus{} 26 \minus{} 15\sqrt 3$

$ a^3 \equal{} 52 \plus{} 3\sqrt[3]{{26 \plus{} 15\sqrt 3 }} \plus{} 3\sqrt[3]{{26 \minus{} 15\sqrt 3 }}$
$ a^3 \equal{} 52 \plus{} 3(\sqrt[3]{{26 \plus{} 15\sqrt 3 }} \plus{} \sqrt[3]{{26 \minus{} 15\sqrt 3 }})$
$ a^3 \equal{} 52 \plus{} 3a$
[/hide]


geometry
perimeter
real analysis
real analysis unsolved

How do you divide a circular cake into n pieces of equal area while minimizing the total length of all the cuts? I can't figure out how you go about doing it.

How about a slightly simpler beginning question: assuming a radius of $ 1,$ for what values of $ n$ is it possible to cut the cake into equal area pieces with total length of cut strictly less than $ n$? (You can get $ n$ from equally spaced radial cuts.)

[asy]size(100);
int n=10;
D(CR((0,0),10));
D(CR((0,0),3),red);
for (int k=0;k<n;++k) D(3*dir(k*360/n)--10*dir(k*360/n),red);[/asy]
This is better than using the radial cuts for $ n\ge 5$ ;).

Is your inner ring there intended to be a circle or a regular polygon?

It was originally intended to be a circle but you are right: a regular polygon may be even better, so, perhaps, you can go down to $ n\equal{}4$ as well :)

If we use Fedja's idea (with the circle, because it makes for easy computation) we can see further improvements: with 10 slices, it's better to cut the central disk in half and leave 8 slices in the outer annulus than to do what Fedja diagrammed; for 11, 12 and 13 slices it's better to cut the central disk into 3 slices (in particular, if every slice has area 1 then the picture Fedja drew requires cuts of length $ 10(\sqrt {11} \minus{} 1) \plus{} 2\pi \approx 29.45$ while dividing the central disk into three pieces requires cuts of length $ 3\sqrt {3} \plus{} 8(\sqrt {11} \minus{} \sqrt {3}) \plus{} 2\pi\sqrt {3} \approx 28.76$, while the "obvious" cutting requires $ 11\sqrt {11} \approx 36.48$); for 14 and 15 slices it's better to quarter the central disk.  Then for 16 or more slices, you get lower perimeter by using three rings instead of only two (best for three rings with 16 slices is 1 central disk, 5 slices in the middle ring and 10 slices in the outer ring).  With 26 or more slices, you can get improvement in a three-ring situation by beginning to cut the central disk (3-9-14 is best), then for 33 or more slices you do best with four rings, and so on.  Maybe if we linearize these arrangements per Kent's suggestion we can get even better values, although with more than 2 rings that gets fairly complicated.

For very large $ n,$ we could tile the cake with regular hexagons, doing whatever we need to to fit it in around the edges. This would result in a total length asymptotic to $ \sqrt [4]{12}\sqrt {n},$ if my arithmetic is right.

I got
$ \sqrt {\frac {3\pi n}{2 \minus{} \sqrt {3}}}$
for the regular hexagons, but I may have made a mistake.

I should have said $ \sqrt[4]{3}\sqrt{2\pi n}.$

A hexagon of side length $ 1$ has area $ \frac{3\sqrt{3}}2.$ We want hexagons of area $ \frac{\pi}{n},$ which gives them side length $ \frac{\sqrt{2\pi}}{\sqrt[4]{27}\sqrt{n}}.$

The total length of sides should be $ 3n$ times the length of one side.

Again, this is asymptotic for large $ n,$ as we haven't dealt with the edges of the pattern.

I redid it and got that. I'm not sure how I got that other number the first time


This is my present for you, VasC.
Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Determine the greatest value of the following expression
\[ P\equal{}\frac{a\sqrt{k^2b^2\plus{}c^2}\plus{}b\sqrt{k^2c^2\plus{}a^2}\plus{}c\sqrt{k^2a^2\plus{}b^2}}{(a\plus{}b\plus{}c)^2}\]
where $ k$ is a real numbers.

P/s: We have a nice solution with the Cauchy Schwarz. :)

[quote="can_hang2007"]This is my present for you, VasC.
Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Determine the greatest value of the following expression
\[ P \equal{} \frac {a\sqrt {k^2b^2 \plus{} c^2} \plus{} b\sqrt {k^2c^2 \plus{} a^2} \plus{} c\sqrt {k^2a^2 \plus{} b^2}}{(a \plus{} b \plus{} c)^2}
\]
where $ k$ is a real numbers.

P/s: We have a nice solution with the Cauchy Schwarz. :)[/quote]

Do you use the technique like this?

http://www.mathlinks.ro/viewtopic.php?t=194955

Not yet, Loc. I meant Cauchy Schwarz Inequality, we can use it. :)


[quote="mdk"]Prove that (sin(x)+acos(x))(sin(x)+bcos(x)) ge 1+((a+b)/2)^2[/quote]

It would be a lot nicer if you learned to use LaTex.
\[(\sin(x)+a\cos(x))(\sin(x)+b\cos(x))\ge 1+\left(\frac{a+b}{2}\right)^{2}.\]


A three-digit number "abc" has the property that the product of a and b is equal to c. What is the greatest possible value of the three-digit number?

[hide]abc... a*b =c

so we want to maximize a, meaning that we would have to use 1 for b. so we get a = c =9.

so 919.[/hide]


2. In the game of primes, players roll a fair cube whose faces are labeled 2, 3, 5, 7, 11, and 13. What is the smallest integer sum greater than 2 that cannot be made by rolling the die once, twice or three times?

Is it 30?The message is too small. Please make the message longer before submitting.

Assume that n is the sum of 1,2, or 3 primes, so n must be equal to or less than 13.3=39.
If n is odd, n must be sum of at most 3 odd primes (it is a theorem, but I can't remember whose :blush: ).
If n is even, n must be greater than 2+13+13=28.
And we need to find the smallest value of n, then n must be 30.
(My English is too poor, so I can't express the solution well. Sorry  :blush: )


Twenty percent of Poe M.S. students walks to school. If the number of ``walkers'' doubles, and there is no change in the number of non-walkers, what fraction of the students is now walkers? Express your answer as a common fraction.

Let's say there are $ 100$ students at Poe M.S. That means that there would be $ 20$ "walkers", since 20% of 100 is $ 20$.

The number of walkers doubles, but there is no change in the number of non-walkers. So currently, there are $ 80$ non-walkers, and $ 20$ walkers. When the number of walkers doubles, that gives us $ 40$ walkers, but we still have $ 80$ non-walkers.

We have to be careful not to do $ \frac{40}{100}$, since the number of non-walkers stays the same. Instead, it's $ \frac{40}{120}$, which simplifies to $ \frac{1}{3}$.

If you have any questions, please do not hesitate to ask.

You could pretend there a two walkers, alternatively.


How many different sets of positive integers {a,b,c} are there such that a*b*c = 2310?

We can try prime factorization, so

$2310=231\times 10=3\times 77\times 10=2\times 3\times 5\times 7\times 11$.

You can go on from there, is it ordered sets?  If there are ordered sets, then just do combinations.

I can find the recent posting of this, but...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22517 is an older solution.

Don't forget that you can use the number 1 also


Find the solutions in $M_n$ for the equation $Com(A)=B$,where $B$ is a matrice with $rang=1$ and $com(A)$ is the comatrice of $A$.

Every matrice $B$ of rang 1 can be written in the form $B=P*E_{11}*Q$,where $P,Q$ are inversible.
Now let the equation:$Com(A)=E_{ij}$.We have $A*E_{ij}=E_{ij}*A=0$.So the $i^{th}$ column and $j^{th}$ line of $A$ are 0.For A to verify the equation we should have that the minor obtained by elyminating this column and line has determinant 1.The comatrice of such a matrix is equal with $E_{ij}$.Hence any matrix with the $i^{th}$ column and $j^{th}$ line equal to 0 and the with the minor $(i,j)$ having determinant 1 verifies the equation.For $Com(A)=P$ where P is inversible we can get easily the solution,so we've got the general solution.

Killer, I am not sure it's fair what you do!!! ;)


function
algebra
functional equation
algebra proposed

Find all continuous functions f:R->R s.t.

(i) f(xy)=f(x)f(y)+f(-x)f(-y) for all x,y in R
(ii) f(x+y)<=f(x)+f(y) for all x,y in R
(iii) f(1999)=1000

ok time for a partial solution :D
note that f can not be a constant function.putting x=-1,y=1 we get
f(-1)=2f(1)f(-1)
case 1) f(-1)=0 and f(0)=1/2 or 0 (i'll consider f(0)=1/2 )
note that putting x=x and y=0 we get f(x)+f(-x)=1==>f(1)=1
now consider the function g(x) =2f(x)-x.after plugging it into the functional equation we get g(xy)+xy=[g(x)+x-1][g(y)+y-1] + 1
now putting x=x,y=1 we get g(x)=1-x (not possible by initial conditions) or g(y)=1 ==> f(x) =(x+1)/2 and this can be verified
 f(0) = 0 about this i'm still thinking .mayb i'm overlooking some obvious fact.anyways something is better than nothing
 :)


function
analytic geometry
real analysis
real analysis unsolved

[quote="Duncreggan"]Let X be a finite set and Y a countable set. Let F(X, Y) the set of functions defined in X with values in Y. Prove that F(X, Y) is countable.

I need some help, please![/quote]

There's a very simple identification of $ F(X,Y)$ with $ N^k$.

Do you mean F(X,Y) to be the set of functions from X to Y? (the word 'in' confused me, "defined in X"). Such a function is a special kind of relation from A to B, i.e. a subset of the cartesian product AxB (namely such that every first coordinate has a unique second coordinate). Write $ Y\equal{}\{y_1,y_2,...\}$.

To get the idea, first assume |X|=1, say X={x}. Then all functions from X to Y are of the form $ \{(x,y_i)\}$, i.e. $ F(X,Y)\equal{}\bigcup_{i\in\mathbb{N}}\{(x,y_i)\}$. We know that the countable union of countable (in this case even finite) sets is again countable, so the conclusion follows.

Now assume $ |X|\equal{}n\in\mathbb{N}$, write $ X\equal{}\{x_1,...,x_n\}$. Then all functions from X to Y are of the form $ \{(x_1,y_{i_1}),...,(x_1,y_{i_n})\}$ with $ i_1,..,i_n\in\mathbb{N}$. So $ F(X,Y)\equal{}\bigcup_{(i_1,...,i_n)\in\mathbb{N}^n}\{(x_1,y_{i_1}),...,(x_1,y_{i_n})\}$. Since $ \mathbb{N}^n\equal{}\mathbb{N}\times...\times\mathbb{N}$ ($ n$ times) is again countable (assuming you know this), this again is the countable union of countable (finite) sets, so the conclusion follows.

I would like to know if my solution is correct.

Since X is finite we can write X = $ \{x_1,...,
x_n\}$. Then, define the map $ \phi : \mathcal{F}(X, Y) \rightarrow Y^n$ by
\[ \phi(f) \equal{} (f(x_1),..., f(x_n)).\]
Note that $ \phi$ is a one-to-one function. Since $ Y^n \equal{} Y\times ... \times Y$ is
countable because Y is countable, we have $ \mathcal{F}(X, Y)$ is
countable. (The result follows by the injection of $ \phi$).


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

How can i show that the polynomial x^5 + 20x +16 in Q[X]
>is not solvable by radicals by showing it is the Galois group A5
>
>I have tried but I can only show that the splitting field is of degree 5
>so it contains a 5-cycle and since it has 1 real root and 4 complex ones
>it also contains a 2-cycle but this leads me to S5 and that is apparently wrong I should be getting A5
Can anyone help me see how there are no 2-cycle but a product of two 2-cycle

Thanks

The conjugation switches two pairs of roots. Thus it is of type $ (23)(45)$ where $ 2,3,4,5$ stand for complex roots (and $ 1$ being the real root).

For a full proof that the group is contained in $ A_5$, you would need to show that the discriminant $ D$ is a perfect square. Even permutations of the roots fix $ \sqrt{D}$, while odd permutations send it to $ \minus{}\sqrt{D}$.

Of course, $ S_5$ isn't solvable either; that would have been an easier exercise.


Point E lies on the base AD of the trapezoid ABCD. The triangles ABE, BCE and CDE have equal perimeters.
Prove that $|BC| = |\frac{AD}{2}|$.

my work seems kind of wrong

[hide]Let altitude from E of triangle BEC meet BC at F

Let altitude from B of triangle ABE meet AE at G

Let altitude from C of triangle CDE meet ED at H

by AAS triangle BGE is congruent to triangle BFE

by AAS triangle EFC is congruent to triangle ECH

let the length of EF by a.

let the length of AG be b.

let the length of FC by c.

because the perimeters of the three triangles are equal, b+sqrt(b^2+a^2)=c+sqrt(c^2+a^2) and c and b must be equal.  same goes for the lengths of HE and BF.  Since all the triangles are congruent and AD is two bases while BC is only one base of the same triangle, AD/2 is equal to BC[/hide]

I think you are on the right track, and there seems to be nothing wrong, just a little more explanation I think. Here it is: since AB=FC by what you have done with the right triangles, and when you drew the altitudes of the triangles, you have BF=GE and FC=EH, which means (since AB=FC), AB=EH, and when you proved once again that EH=BF, you also have EH=BF=GE, so therefore all six length are equal and there are two on top and four on bottom so therefore BC=AD/2.


Indeed I can do it for generic k  ;) .

Again, see my post #8 at http://mathlinks.ro/Forum/viewtopic.php?t=16360 .

  Darij


AMC USA(J)MO USAMO Harvard college MIT Princeton

Why is it that the people who are always winning math competitions always come from one of these three schools?

Does someone who makes it to USAMO and goes to one of these schools really deserve it? After all, they have MUCH more resources than most other students. The disparity between what students at these schools and regular run-of-the-mill public schools is enormous. Then, not only are these people rewarded for their "genius" etc, etc, they go onto Harvard, MIT, Princeton, etc, etc.

I am sure that most people here will disagree with this, but this is the impression that I have been getting ever since I started participating in all this math competition and stuff. It is really frustrating to know that you have no next to no chance because you go to a lousy school and started late because there was no math team at your school.

On this note, I would like to say how greatly appreciative I am of Art of Problem Solving. While many people have stated how much they owe to AoPS, this praise is definately not undeserved.

I curious as to what people think about what I have said because this has been running through my head a lot for the past year or so.

there was discussion on this before... maybe called 'usamo and college applications' or something like that.  check the college forum.

But i think most people actually share your beliefs...

First, I am pleased that you appreciate the Art of Problem Solving.  I personally think that there is enough material on this site to make up for a lot of the extra resources that students at the top school have access to.  You may find other sources for mathematics education that help you in your pursuit as well.

Second, we are currently running a Math Jam (right now) on the AMC B problems from this year.  Going over the solutions might be a good way to supplement your education.  We run these free Math Jams 2-3 times per month usually.

Third, do not despair.  You can find resources outside of your school.  You can also work to help your own school understand how best to build a program.  Many of us had this experience as students.  The desire to move forward is much of the inspiration for the Art of Problem Solving as it is today.

Mathematics is first and foremost about curiosity.  "How do I figure this out?  How do I solve this problem?  Why does this work the way it works?"  If you stay on the inquisitive course and keep working then in the end you will find yourself amongst the serious problem solvers.

Hrmph ...

I'm upset that Stuy isn't on that list ...

anyhow, with respect to what you actually said:

I hope this doesn't sound mean, but that's life -- there is no place on earth and never has been where everyone has equal opportunities.  I don't think that's a good thing (and I find the disparity in opportunity in mathematics among different groups of people to be quite upsetting), but on the large scale there isn't really much to be done about it, at least on a short-term basis.  The closest thing to a solution for the short term is to do your best to make up for it.  I'm all for people starting math teams in their own schools -- if you find a teacher willing to meet with 5-6 students once or twice a week, you have a math team.  There are lots of resources on line, including AoPS and many others that I won't try to name.

One final point about the schools you named -- all the people who go to TJ, AAST, Exeter, or Stuyvesant are chosen to go there.  Now, there are distinct limitations on that choosing, whether it be geographical or financial or whatever.  But to say that they (we) are "rewarded for our 'genius'" is silly -- the schools pick groups of students who are highly likely to succeed already.  The fact that Stuy is a good school has much more to do with the student body than the faculty.*  So again, I think the best solution is to try and find the people who are (or might be) interested and band together with them in doing things like starting a math team or whatever else.

*Note that this is coming from someone who thinks very highly of the Stuyvesant faculty (particularly in math).

Yeah, your personal motivation is much more important that the resources that your school can happen to provide. Even if a student's school does not have nearly as many resources as Exeter or TJ or, in Joel's case, Stuy, if that student is truly motivated and eager to succeed in mathematics and math competions during their high school years, they will not be deterred at all. I am sure that any of the students on past US IMO teams who went to very good high schools would still have accomplished what they did had they had the same motivation, but had gone to a less encouraging school.

I don't know. I went to a normal public high school that didn't offer all the resources that a math magnet school would although we were much better off than most other public schools, but I had the good fortune to be the second-best mathematician at my school when I was a freshman. I think that was my primary motivation. I have never been a big fan of being the second-best, and so I worked hard to try to change it around. Had I been the best, I doubt I would be nearly as good at math as I am now.

[quote] I had the good fortune to be the second-best mathematician at my school when I was a freshman. I think that was my primary motivation. I have never been a big fan of being the second-best, and so I worked hard to try to change it around. Had I been the best, I doubt I would be nearly as good at math as I am now.[/quote]

Really? But even if you had been the best at your school, you would certainly not be the best in the US, or the world. There are always higher goals to reach for, right?

I go to AAST, and I have worked at least an hour a day on math since about 8th grade. We have all the past AMC's, AIMEs, and tons of national and international olympiads in storage so we can practice. I think our successes come from the fact that we draw the best math students from the county, and that we have a lot of study and practice resources available to us. On top of that, we devote a <i>lot</i> of time to it.

i forgot i think mr. president bush went to either Philips andover or Philips Exeter. i received a 100 page book from them the SCHOOL COSTS A LOT. but that doesn't make him smart. his SAT is about 200 lower than Clinton's

[quote]but that doesn't make him smart. his SAT is about 200 lower than Clinton's[/quote]

true..but having a higher SAT doesn't make Clinton smarter either  :wink:

Cliton provided millions of jobs and he reduced the debt. Then look at Bush reversing everything good Clinton did.

churchhill, that avatar fooled me into thinking you were mr. crawford until i noticed the poor grammar

[quote="churchilljrhigh"]We have different signatures so there isn't really any confusion.[/quote]

Are you trying to pose as one of the AoPS instructors? Why not make an avatar based on Jiang Zemin's photograph?

Well, fubu on AOPS told me to change my old one.

what are your top 3 scores in AMC 12A and 12B?

[quote]Why is it that the people who are always winning math competitions always come from one of these three schools [AAST, Exeter, TJ]? . . . I curious as to what people think about what I have said because this has been running through my head a lot for the past year or so.[/quote]

It's unfair to say that its solely the schools that are responsible for the success.  There is a lot of talent in the selection pool for the schools (as an AAST junior, i would know).

Students at AAST spend their weekends in school, practicing math.  Mr Holbrook, the math team coach for the school, effectively has more than one copy of AoPS 1 and 2 per student.  He has a massive treasure trove of mathematical problems and books.

While the availability of problems is an advantage, it would not do justice to my fellow students to say that this is the sole reason for success.  There is a lot of dedication on the part of the students.  As mentioned above, many kids spend an extra [b] 8 - 16 hours a week [/b] in school doing math [b] [i] (on the weekends). [/i] [/b]  It is this type of dedication that brings victories.  Mathematics is not something that is learned haphazardly.  It takes hours upon hours of studying and math work in order to reach the level we are currently at.

It is unfair for those who go to local schools and don't have the resources that a public magnet school could provide.  However, there is a plethora of resources available for those whose schools do not provide these resources.  For instance, an amazing web site, [url]http://www.kalva.demon.co.uk[/url] has a whole slew of olympiad problems, including all the IMO's, USAMO's, AIME's, and other math competitions.  MAA writes great books that you can order.  ART OF PROBLEM SOLVING is the bible of mathematics :)

As for eighth graders competing, half the time they are better than freshmen.  For motivated eighth-graders, there are always programs available.  One must search and find the programs.

[quote]But I agree with the fact that the magnet schools do have more of an advantage over a regular public school. Our school selects all of the best students from northern NJ, and the ones dedicated to math spend four to eight hours every weekend at math team practice, not to mention the many hours during the day and at home.[/quote]

Ignoring variables like sickness, performance is directly related to time spent working on mathematics.  Students like Lun Mu devote a large part of their free time to mathematics.  Thus, it is unfair to whine about the performance of Exeter or TJ or AAST without appreciating the amount of time devoted to mathematics.  I do not have personal experience in TJ or Exeter, but as I am told the TJ math team is mostly a student-run club, spearheaded by incredibly devoted and talented students with little assistance from teachers.

(If you got bored reading the long paragraphs, just jump down here)  No one can carp about someone performing better than another unless the niggler recognizes and appreciates the dedication and effort put into mathematics (or in any other field).

Does AAST have 8th graders?  I was under the impression that it goes from 9th to 12th grade.

AAST does [b] not [/b] have eighth-graders.  AAST is a high school.  There is a program called "Academy After Hours" where the math team coach teachers night classes to motivated youngsters.  Our coach allows these youngsters, if they are good enough, to compete in math olympiads.

As for ARML, there are 15 people to a team.  Just like how some home-schooled kids form a team, just like how entire states send teams, we allow interested eighth-graders an opportunity to compete by bringing them together.  This can be compared to the NYC teams, a combination of high-school students from different public schools that are together for the purpose of ARML.

[quote]the AAST 8th graders do not really compare with teams that have seniors on them.[/quote]

It is unfair to compare our 8th graders to division A or division B teams at ARML.  You are trying to compare kids with a gap of 4 years, so it is expected that they do not compare.  However, their relative performance is impressive, noting that this is the first time for these kids, whereas seniors have competed for four years.  Furthermore, you are trying to compare first-string teams from regions like Eastern Mass to eighth-graders from the small Bergen county.  If you wish to rank the AAST 8th graders, do not compare them to first-string seniors with years and awards under their belts.  You are trying to compare fledglings and aces.

True, although if you'd continued reading what I wrote, you would note that I went on to say

[quote="JBL"]They tend to do pretty terribly overall, although not so badly considering that most people there are a couple or three years older than them.[/quote]

Also, it's worth pointing out that "little Bergen County" has a population of roughly 900,000 people and that it is one of the richest counties in one of the richest states in the country.

I don't know if they still do, but in the early 90s Maryland brought a team of 8th graders that did very well at ARML. I know they came in ahead of several states A Teams. And in 2000 an 8th grader came in first place at ARML - Tiankai Liu.

In all honesty, the only advantage we have over other schools is that we pull good students out of a large pool of people. But we only get people with potential; we still put in a lot of hours (just as much, if not more, than other schools).

[quote]
Being able to put in many hours is a serious advantage -- many schools can't find (or pay for) teachers who will do that for their kids. [/quote]

The ability to put in extra hours is dependent on the student, not the teacher.  One can have a teacher willing to devote every waking moment to mathematics, but it means little if the student is not willing to devote time to mathematics.  It might be nice to have a teacher, but many of the greatest IMO team members (e.g. Reid Barton) are homeschooled, so it is unfair to attack the school.

The presence of a math coach (or lack thereof) does not determine success.  It is all about how much time you are willing to put into mathematics outside of school (and no teacher can change how much effort you devote ouside of the school)

[quote="God"]The presence of a math coach (or lack thereof) does not determine success.  It is all about how much time you are willing to put into mathematics outside of school (and no teacher can change how much effort you devote ouside of the school)[/quote]

I respectfully disagree, particularly with the second point.  Thousands of students every year have the love of mathematics (and plenty of other subjects) squashed out of them by teachers who don't care about the subject and don't allow it to demonstrate its beauty.  The quality of a teacher has very much to do with invigorating students -- particularly in the non-magnet school situation where the vast majority of the student body isn't already interested on their own.  (Remember, you're coming from an atypical school in many respects.)

I do agree with you, of course, that the will of the student is important in all of this, but it's not the only important thing.

And, just to be clear, I'm not trying in any way to disparage AAST -- I think your performance is quite impressive, and I commend you for it.  But it's not replicable everywhere by any old random group of kids who like math.  There are a lot of factors involved that other places simply cannot replicate.

[quote="JBL"]Being able to put in many hours is a serious advantage -- many schools can't find (or pay for) teachers who will do that for their kids.  The NYC team can't do it because we aren't chosen soon enough and because it's just hard to organize when there's no good structure in place to contact everyone, etc., and no one being paid to do it.[/quote]

They can't do it.. yet.. Wait till you see the totally automated system I"m about to put up....


algebra
polynomial
inequalities proposed
inequalities

Given $ a,b,c \ge 0$.Prove that:
(a) $ (a \plus{} b \plus{} kc)(b \plus{} c \plus{} ka)(c \plus{} a \plus{} kb) \ge \frac {(k \plus{} 2)^3}{9\sqrt {3}}(a \plus{} b \plus{} c)^2\sqrt {ab \plus{} bc \plus{} ca}$
for k such that:$ 54k \ge (k \plus{} 2)^3$
(b) $ (a \plus{} b \plus{} kc)(a \plus{} c \plus{} kb)(b \plus{} c \plus{} ka) \ge 2(k \minus{} 1)\sqrt {k}(a \plus{} b \plus{} c)^2\sqrt {ab \plus{} bc \plus{} ca}$
for all k>1.Equality occur if and only $ k \ge 3\plus{}\sqrt{8}$.For $ k \equal{} 3 \plus{} \sqrt {8}$,equality occur if and only a=b,c=0 or any cyclic permution.
(c)Prove that:,we don't have k>0 such that:
$ (a \plus{} b \plus{} kc)(b \plus{} c \plus{} ka)(c \plus{} a \plus{} kb) \le \frac {(k \plus{} 2)^3}{9\sqrt {3}}(a \plus{} b \plus{} c)^2\sqrt {ab \plus{} bc \plus{} ca}$
(d)Prove that for a,b,c is side length in triangle,
$ (a \plus{} b \plus{} kc)(b \plus{} c \plus{} ka)(c \plus{} a \plus{} kb) \le \frac {(k \plus{} 2)^3}{9\sqrt {3}}(a \plus{} b \plus{} c)^2\sqrt {ab \plus{} bc \plus{} ca}$
for k such that:$ 9k(\sqrt {12} \plus{} 3) \le (k \plus{} 2)^3$


Find the coefficient of $ x^2$ in the expansion of \[ \left(x\plus{}\frac{1}{x}\right)^6\].

hello, the coefficient is $ \binom{6}{2}x^4\left(\frac{1}{x}\right)^2\equal{}15x^2.$
Sonnhard.


Let $ (u_{n})$ be a sequence given by $ u_{1}$ and the formula $ u_{n\plus{}1}\equal{}\frac{k\plus{}u_{n}}{1\minus{}u_{n}}$ ,where $ k\geq 0;n\equal{}1;2;...$
Given that $ u_{1}\equal{}u_{13}$.Find value of $ k$

[quote="Rust"]Posted 3-4 days before.[/quote]

And killed because it was a live topic in some Vietnam math review.
Don't answer this problem.


Find $ (\log_2{x})^2$ if $ \log_2{(\log_8{x})} \equal{} \log_8{(\log_2{x})}$.  Please clearly show all algebraic manipulation.

Thanks in advance!

hello, we have $ \log_8 x \equal{} \frac {\ln(x)}{\ln(8)} \equal{} \frac {\ln(x)}{3\ln(2)}$, so we have $ \log_2\left(\frac {\ln(x)}{3\ln(2)}\right) \equal{} \log_8\left(\frac {\ln(x)}{\ln(2)}\right)$, setting $ \frac {\ln(x)}{\ln(2)} \equal{} u$ so we get
$ \log_2 \left({\frac {u}{3}}\right) \equal{} \log_8 u$, this is equivalent with $ \frac {\ln(\frac {u}{3})}{\ln(2)} \equal{} \frac {\ln(u)}{3\ln(2)}$, simplifying we get $ \ln(\frac {u}{3}) \equal{} \frac {\ln(u)}{3} \Leftrightarrow 3\ln(u) \minus{} 3\ln(3) \equal{} \ln(u)$ or $ \ln(u) \equal{} \ln(3^{\frac {3}{2}})$ and so we have $ u \equal{} 3^{\frac {3}{2}}$, our $ u \equal{} \log_2 x$ so we have $ \log_2 x \equal{} 3^{\frac {3}{2}}$ and from here follows $ \left(\log_2 x\right)^2 \equal{} 3^3 \equal{} 27$.
Sonnhard.


Prove by combinatorial reasoning that

$ \sum_{k\equal{}1}^{n}k\binom{n\plus{}1}{k\plus{}1}\equal{}(n\minus{}1)2^{n}\plus{}1$.

You've already posted this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168173[/url]

Do not double post.


[quote="colts18"]Yeah, that's what I got.[/quote]
how did you get that?  YOu don't even know what a hypotenuse is.  Or even how to say it. :P

They probably can't do the sqrt function on latex.

Basically that's $\sqrt{16-9}$ or $\sqrt{7}$

yup if u make 4 the hypotenuse, c squared is 16, and 3 squared is 9, so its the root of (c squared minus 9 (sqrt 16-9) which is 7... making the other side 2.6457513110645905905016157536393 (calculator..)

To clarify the definition of hypotenuse, it is the longest side of a right triangle, or the side opposite the right angle. The other two sides are known as legs.

[quote="gr8sk8r"][quote="jimli"]make four the hypotenuse
so it's 7^.5[/quote]
Why do you raise 7 by .5??[/quote]

the radical of 7 (sqrt(7)) is the same thing as 7^1/2=7^0.5 because, squareroot is the same thing as the 2nd root of x-variable. Some example will clarify this:

7^1/2=squareroot of 7
7^1/3=cubedroot of 7
7^2/2=squareroot of 7^2
7^3/2=squareroot of 7^3

Therefore, sqrt(4^2-3^2)=sqrt(16-9)=sqrt(7)=7^1/2 or 7^0.5

-Masoud Zargar

If it is writen in the form $b^{\frac{x}{y}}$ it is the same as
$\sqrt[y]{b^x}$

[quote="math92"]If the lengths of two sides of a right triangle are 3 and 4, what is the least possible length of the third side?[/quote]

[hide]If $4$ was the hypotenuse, then the other leg would be

$\sqrt{4^2-3^2}$

$\sqrt{(4-3)(4+3)}$

$\sqrt{1*7}$

$\sqrt{7}$[/hide]


geometry
3D geometry
probability
circumcircle
similar triangles
angle bisector

20. In acute triangle ABC, points D and E lie on BC and AC, respectively. If AD and BE intersect at T so that AT/DT=3 and BT/ET=4, what is CD/BD?

23. Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

24. In triangle ABC, we have AB=7, AC=8, and BC=9. Point D is on the circumcircle of the triangle so that AD pisects angle BAC. What is the value of AD/CD?

[hide="20"] Draw TX parrallel to AC so that X is on BC.  Thus, we obtain two pairs of similar triangles: BTX and BEC and DTX and DAC.  Now, call DX=a.  So, by the second pair of similar triangles, we have that
\[ \frac {DX}{XC} \equal{} \frac {TD}{AT}
\]

\[ XC \equal{} 3a
\]
So, by the first pair,
\[ \frac {BX}{XC} \equal{} \frac {BT}{TE}
\]

\[ \frac {BX}{3a} \equal{} 4
\]

\[ BX \equal{} 12a
\]

\[ BD \equal{} 11a
\]

\[ DC \equal{} 4a
\]

\[ \frac {CD}{BD} \equal{} \boxed{\frac {4}{11}}
\]
[/hide]

[hide="24"] Say that AD and BC intersect at P.  So, by the Angle Bisector Theorem,
\[ \frac {BP}{CP} \equal{} \frac {AB}{AC}
\]

\[ \frac {BP}{CP} \equal{} \frac {7}{8}
\]

\[ BP \plus{} CP \equal{} BC \equal{} 9
\]

\[ BP \equal{} \frac {21}{5}
\]

\[ CP \equal{} \frac {24}{5}
\]
Now, ACDB is cyclic, so $ \angle DAC \equal{} \angle BCD$ and $ \angle ADC \equal{} \angle PDC$, so triangle ADC is similar to triangle CDP.  Thus,
\[ \frac {AC}{PC} \equal{} \frac {AD}{CD}
\]

\[ \frac {AD}{CD} \equal{} \frac {8}{\frac {24}{5}} \equal{} \frac {40}{24} \equal{} \boxed{\frac {5}{3}}
\]
[/hide]

23.

[hide]The total permutation of painting is $ 2^6\equal{}64$

Case 1: all 6 faces are the same color, which always satisfies the condition; permutation: $ 1*2\equal{}2$

Case 2: 5 faces are the same color, which also always satisfies the condition; permutation: $ 6*2\equal{}12$

Case 3: 4 faces are the same color, which satisfies the condition if and only if the other 2 faces are opposite; so it has a permuation $ 3*2\equal{}6$

Hence, the probability is: $ (2\plus{}12\plus{}6)/64\equal{}5/16$[/hide]

P.S.: At last I know how to use LaTeX...thanks @Icy and the creators...it looks much clearer now :)

[hide="23"]Count the complement:
If 5 of one color, then, 0 ways such that there are no 4 square "ring" of some same color.
If 4, then one for every edge, 12.
If 3, then either one for every vertex, or one for every pair of edges that are on opposite sides of a face square, 12.

Then, $ \frac{12\plus{}20\plus{}12}{2^6}\equal{}5/16$.[/hide]

[hide="24"]Let AD intersect BC at E. Since they bisect the same arc, angle ADC = angle ABC. By AA, $ \triangle ABE ~ \triangle ADC$. From angle-bisector, BE=7*9/15. AD/DC=AB/BE=5/3.[/hide]


Lol can someone offer a explanation please.

hello, do you mean $ x\plus{}\tan(x)\equal{}0$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean $ x \plus{} \tan(x) \equal{} 0$?
Sonnhard.[/quote]
Ha that makes more sense.

yes))

$ x\plus{}tan(x)\equal{}0$

sorry)

hello, what do you want to know about it, the roots of the equation?
Sonnhard.

yes. if it has roots.. and it's given that $ 0 < \equal{} x < \equal{} \pi/2$

or just find the maximum of

$ \sin (2 \alpha) \left(800 \minus{} \frac{500}{\pi}\alpha \right)$

if $ 0<\equal{}\alpha<\equal{} \pi/2$

First of all $ \tan(\frac {\pi}{2})$ is undefined, so we have $ 0 \le x < \frac {\pi}{2}$.

Over this interval, we have $ \tan(x) \ge 0$.

So, the only possible zero occurs when $ x \equal{} \tan(x) \equal{} 0$, which is true for $ x \equal{} 0$. Thus, $ x \equal{} 0$ is the only solution.

[b]Xantos C. Guin[/b]

I got it  :)

can you help me to find the max $ f(\alpha)$ if $ 0\le \alpha \le \frac {\pi}{2}$

$ f(\alpha) = \sin (2 \alpha) \left(800 - \frac {500}{\pi}\alpha \right)$

I tried by this way but got another problem...
$ f(\alpha)'=0$

$ 1600cos2\alpha-frac{500}{\pi}sin2\alpha+2\alpha cos2\alpha=0$

$ (1600-\frac{1000\alpha}{\pi})cos2\alpha=\frac{500}{\pi}sin2\alpha$

$ tan2\alpha=\frac{500}{1600\pi-1000\alpha}$

or $ tan2\alpha=\frac{5}{16\pi-10\alpha}$

hello, differentiate your $ f(\alpha)$ with respect to $ \alpha$ we get
$ f'(\alpha)\equal{}2\cos(2\alpha)\left(800\minus{}\frac{500}{\pi}x\right)\plus{}\sin(2\alpha)\left(\minus{}\frac{500}{\pi}\right)$.
Now you must solve the equation $ f'(\alpha)\equal{}0$.
Sonnhard.

hello, i have got $ 2\cos(2\alpha)\left(800\minus{}\frac{500\alpha}{\pi}\right)\equal{}\frac{500}{\pi}\sin(2\alpha)$
$ \frac{2\pi}{500}\left(800\minus{}\frac{500\alpha}{\pi}\right)\equal{}\tan(2\alpha)$
$ \frac{16\pi}{5}\minus{}2\alpha\equal{}\tan(2\alpha)$
Sonnhard.

[b]Dr Sonnhard Graubner[/b]

your one is true) i made a mistake there))) there must be 1/tan(2$ \alpha)$

and how to find the roots now?

hello, i think it can only be solved by a numerical way.
Sonnhard.


So who's seen Napoleon Dynamite?  I haven't, but when I do I probably won't laugh because everyone in my school quotes something from it every day, and it kind of gets annoying after a while.  But what I don't understand is why the movie is getting so much attention after it came out on DVD, and not while it was in theaters (At least I think that's what happenned).

Eh, I saw it.

It was fun, but the stupidist movie I ever watched EVER.

I dont' see the point in making such a pointless movie...

that movie has no right to exist

it's like the ultimate feel-bad movie (hey! look at all these dweebs we found! what do you think we can make them do??????)

HAHA.... That was such a funny movie but really pointless and stupid.

The people at my school are like the people at hs_potter's.  There's alwyas a bunch of people saying stuff like "Tina you fat lard" or something like that, I've never seen that movie.

we had to watch it in our school.
it was really messed up (funny, but messed up)

i found it to be the dumbest most pointless movie i have ever seen. i just sat there the whole time and waited for them to snap out of it! and it didn't happen! i was like...oh my word

dont be hatin'. this movie was awesome. especially since:
a). i was reminded of several people i knew.
b). i cracked up laughing at all the understated funnies.
c). it was an independent film that made it big.

i would have to agree with everyone else: it was really dumb but still highly amusing.  it's kinda nice to see something original even if it is that pointless.  Also, I think it'll die off pretty soon.

[quote="mcalderbank"]Yes it was stupid, and was very one-dimensional, but it was still funny and I agree with bleumoose.

Also it will die off once another stupid funny movie takes its place.[/quote]

i honestly don't think that any movie can be dumber than napoleon dynamite.  :)

there is no plot so theres really nothing much to say

funny/absurd/gross scene:

Guy shoots a cow while a schoolbus of children passes by.

Sound effects: Kids screaming, 'meeuh"

[quote="jli"]It was so dumb it was funny[/quote]

Lol... You got that right! People are always asking why it was so funny. I'm like because it was stupid.

[quote="Danbert"]Someone brought it along to a XC-ski meet and we tried to watch it on the way back, but the bus's DVD player was screwed up, so we saw it in 3-second segments separated by 10 seconds of black.  It was still funny.

Last Wednesday, I met a kid they could have based the movie on.[/quote]

Oh you know thats not actually the DVD players fault. It was a promotion video and to prevent illegal coping they make it that way. I saw in that version also. Same way I saw Anchorman.

[quote="Jhyon8149"]On our way back home from HMMT, we watched Napoleon Dynamite. I heard from people that this is a great movie. But you know what? Napoelon Dynamite is the WORST MOVIE EVER! There is no plot, it's just retarded....... I don't know why people came up to me saying it's the funniest movie in the world.... Maybe I missed something?[/quote]

boy did u hit the nail squarely on the head! i couldn't agree more. i kno i kno i already posted my opinions here but this was so right on i had to say something. :)

[quote="Tokyo"]can some1 actually tell me exactly is this movie about?
Like some scenes or something[/quote]

It's um about Napoleon. And really about nothing and everything all at once.

[quote="Tokyo"]can some1 actually tell me exactly is this movie about?
Like some scenes or something[/quote]
It consists of a fair amount of scenes of someone just randomly staring at a wall.  And then some Mexican dude gets elected school president.  Thats basically the whole plot.

haha :D come on, you guys, theres a bit more to the movie than that.

I thought it was funny because I could relate to a lot of the people in the movie, and the scene where Napoleon dances is pretty cool


geometry
circumcircle
incenter
geometry open

Triangles ABC and A'B'C' lie on the same circle and have the same orthocenter. AA', BB',CC' intersect at points X,Y,Z. Prove that the circumcenter of XYZ lie on OH. ( H is a common orthocenter and O is circumcenter)

I have nice solution for this problem
but it so long and my picture not same to your  picture
In my figure A,A',B,B',C,C' are on the circle wiht its order
I use some lemma :
 1) triangle ABC ,I,O be incenter and circumcenter of ABC ,I touch side of ABC at D,E,F then orthocenter of DEF ,O,I are conlinear
 2) Let M,N,P be intersection of tangent of(O) at A,B,C
           M',N',P'  be intersection of tangent of(O) at A',B',C'
 then M.N,P,M',N',P' are concyclic
 3) AA',BB',CC' met at X,Y,Z on MM',NN',PP' and  O'X perpendicular to MM'...etc...
 4) two circle with ccenter O,O' .line $\delta$ met (O) and (O') at A,B,C,D( a,b on (O) and C,D on (O')) let M be intersection of tangent at A,C of (O) and (O') ...similar ly we have N,P,Q then M,N,P,Q are concyclic wiht center O'' such O,O'and O'' are conlinear .
So you have proof

[quote="Mamat"]Someone replied :). I'll check your solution, I hope it's correct. Because mine is awful and too long. Thank you very much.[/quote]
I and two my friend was proved this problem and it correct -I think ( we take 6 day to prove it
are you have solution for your problem? :)

Nice :). Now I have a good solution for this problem.

if you have time you may post so lution for this pro MAMAT :)

Ok, thank you hunter_cvp_ic for nice solution

[quote="Mamat"]Finally I have some time to check your solution. What is Q? Which order do you choose when you find M,N,P?[/quote]
 well ! You want to ask point Q on lemma 4 ,is't it?
If i'm correct then
 M be intersecttion of tangents from A and C of circle (O) and (O') N be intersection of tangents from A and D of (O) and (O') ,P be be intersection of tangents from B and C of (O) and (O') ,Q be be intersection of tangents from B and D of (O) and (O') .  :lol:
and you can choose M,N,P on any order for which they be intersection of one tangent of (O) and one of (O')

How did you prove that circumcircle of XYZ is tangent to MM', NN', PP'?

Sorry ! it so long
then may be i can't post it now ..I will post it latter( some days)

It's ok, because I think that I have the proof, but it's really very long (if it's true).


Phillip flips an unfair coin eight times. This coin is twice as likely to come up heads as tails. How many times as likely is Phillip to get exactly three heads than exactly two heads?

The probability that he gets heads on any flip is $ \frac23$ and $ \frac13$ for tails. Hence, the probability that he gets exactly three heads is $ \binom83\left(\frac23\right)^3\left(\frac13\right)^5$ and the probability that he gets exactly two heads is $ \binom82\left(\frac23\right)^2\left(\frac13\right)^6$, so the ratio is $ \frac {\binom83\left(\frac23\right)^3\left(\frac13\right)^5}{\binom82\left(\frac23\right)^2\left(\frac13\right)^6} \equal{} \frac {\binom83\cdot\frac23}{\binom82\cdot\frac13} \equal{} \frac {56\cdot2}{28} \equal{} 2\cdot2 \equal{} \boxed4$.

[hide=Alternative solution]
So it's twice as likely to be heads than tails, as given in the problem.
C(8,3)=8*7*6/3*2*1
C(8,2)=8*7/2*1

C(8,3)/C(8,2)=6/3=2

Thus there are twice as many ways and each of them is twice as likely as tails.
2*2=4

Wait, why is this in Intermediate Algebra?

[hide=Solution]The probability he gets heads is $\frac{2}{3}$ and the probability he gets tails is $\frac{1}{3}$. The answer is hence $\frac{\binom{8}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^5}{\binom{8}{2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^6} = \boxed{4}$.[/hide]

[hide=Solution.]We use generating functions. Denote by $1$ flipping tails and by $x$ flipping heads. Since the coin is twice as likely to come up heads, we need $2x,$ not $x.$ Thus, the generating function for one flip is $(1+2x).$ For eight flips, it is $(1+2x)^8.$ When expanded, the coefficient of $x^3$ is $\binom83(2)^3(1)^5$ and the coefficient of $x^2$ is $\binom82(2)^2(1)^6.$ This means our ratio is $\frac{\binom83(2)^3(1)^5}{\binom82(2)^2(1)^6}=\boxed4.$

[quote=Blue_banana4][hide=long solution]There are $\dbinom{8}{3}=56$ ways to get exactly $3$ heads. There is a $\dfrac{2}{3}$ chance of Phillip getting a heads and a $\dfrac{1}{3}$ chance of tails. For each case there is a $\left(\dfrac{2}{3}\right)^3\cdot\left(\dfrac{1}{3}\right)^5=\dfrac{8}{6561}$ chance of it occuring. So, there is a $56\cdot\dfrac{8}{6561}=\dfrac{448}{6561}$ of getting exactly $3$ heads. We can do the same for $2$ heads: $\dbinom{8}{2}=28$ ways, $\left(\dfrac{2}{3}\right)^2\cdot\left(\dfrac{1}{3}\right)^6=\dfrac{4}{6561}$. $28\cdot\dfrac{4}{6561}=\dfrac{112}{6561}$. $\dfrac{448}{6561}\div\dfrac{112}{6561}=\boxed{4}$.[/hide][/quote]

Me: gets same problem again, goes to 'Page Feed, ' looks at my own solution and inputs correct answer.


Find the values for each function below after plugging f(x + 1).

(1) f(x) = sqrt{x^2 + 2}

(2) f(x) = 1 - (1/(x + 2)^2

Yes, I needed to learn HOW to do the questions and your reply shows me how.

Thanks!

well basically if it says find f(x+1), you just put x+1 in all the places where you see x

example:
$f(x)=x^{2}$
find $f(x+1)$
then you plug $(x+1)$ into $X$
so you get
$(x+1)^{2}$

do you have a book or something? all these problems are trivial and are covered in a algebra book

To altheman:

Yes, I have many books but none of the books are easy to follow, which is very common in terms of math, physics and astronomy.


Show that there do not exist positive integers $ x$,$ y$,$ u$, and $ v$ such that $ (x, y) \equal{} 1$,$ (u, v) \equal{} 1$, $ xy \equal{} uv$, and $ x^2 \plus{} y^2 \equal{} u^2 \minus{} v^2$.

Just set $ v\equal{}xy/u$ and get $ (xy)^2$+$ (yu)^2$+$ (xu)^2$=$ u^4$.

    In page 3 there is an exercise I' ve posted, according to which the previous equation doesn't have solutions in positive integers.

[hide="panos_lo"]
ΣΥΓΧΑΡΗΤΗΡΙΑ ΓΙΑ ΤΗΝ ΕΠΙΤΥΧΙΑ ΣΟΥ ΣΤΗΝ JΒΜΟ :!:  :!:
[/hide]

Page 3?

And Dimitris X, what are you saying?

[quote="Zhero"]Page 3?[/quote]

Yep, page 3: http://www.mathlinks.ro/viewtopic.php?t=284204

[quote="Zhero"]And Dimitris X, what are you saying?[/quote]

From what I gather, he's wishing panos success in the JBMO (should things like this really be said in a problem thread?)


There is no pass but please tell highest rating. If you did not start playing, go to http://www.artofproblemsolving.com/Edutainment/g1/index.php
I will probably play from 9:00-9:30 Eastern Time.

Oops. I forgot to ask for your current rating.

Maybe you wanna post this in the FTWOS forum?

[quote="Dillinger"]But I'm asking people in Algebra 1 class.[/quote]

But you're talking about FTW, right? ;)

Why don't you just pm them? Agreed that it's in the wrong forum though.


geometry
3D geometry
number theory solved
number theory

Find all natural number $n$ such that $3^n-71$ is perfect cube.

[hide]$3^{n} - 71 = x^3$.

We try $\mod 8$.

$3^{n} \equiv 1, 3$.

$x^3 \equiv 0, 1, 3, 5, 7$.

$71 \equiv 7$.

We have only two cases.
1. $x^3 \equiv 1 - 7 \equiv 2$.

2. $x^3 \equiv 3 - 7 \equiv 4$.

But $2$ and $4$ are not cubic residues $\mod 8$. No solutions.[/hide]


Let n be a natural number such that n= a^2 + b^2 + c^2, for some natural numbers a,b,c. Prove that 9n = (p1a + q1b + r1c)^2 + (p2a + q2b + r2c)^2 + (p3a + q3b + r3c)^2, where pi, qi, ri are all non zero integers. Further,if 3 does not divide at least one of a,b,c, prove that 9n can be expressed in the form x^2 + y^2 +z^2, where x,y,z are natural numbers none of which is divisble by 3.

[quote="N.T.TUAN"]Let $n$ be a natural number such that $n= a^{2}+b^{2}+c^{2}$, for some natural numbers $a,b,c$ . Prove that $9n = (p_{1}a+q_{1}b+r_{1}c)^{2}+(p_{2}a+q_{2}b+r_{2}c)^{2}+(p_{3}a+q_{3}b+r_{3}c)^{2}$, where $p_{i}, q_{i}, r_{i}$ are all non zero integers. Further,if $3$ does not divide at least one of $a,b,c,$ prove that $9n$ can be expressed in the form $x^{2}+y^{2}+z^{2}$, where $x,y,z$ are natural numbers none of which is divisble by $3$.

Now, I only see $9n=(a+2b+2c)^{2}+(2a-2b+c)^{2}+(2a+b-2c)^{2}$ :)[/quote]
How can you say that each term is not divisible by 3???????

Well The correct solution must be
if $3$ does not divide $(a+b+c)$ then $9n$ can be expressed as
$9n=(-a+2b+2c)^{2}+(2a-b+2c)^{2}+(2a+2b-c)^{2}=x^{2}+y^{2}+z^{2}$where $x,y,z$ are natural numbers none of which is divisble by $3$.
but if $3|(a+b+c)$ then this method fails as each of $x,y,z$ are divisible by $3$
WLOG assume 3 does not divide $c$ (it's given).
let $a+b+c=3k$ for some k being a natural number , substituting
$c=3k-a-b$ we get that, $9n$ can be expressed as
$9n=(a+b+3k)^{2}+(a+4b-6k)^{2}+(4a+b-6k)^{2}=x^{2}+y^{2}+z^{2}$where $x,y,z$ are natural numbers none of which is divisble by $3$.
because each term is congruent to $(a+b)$i.e $-c$ i.e non zero modulo $3$.


cos^-1(x(sqrt3)) + cosx = pi/2

solve for x

hello, do you mean $ \cos^{\minus{}1}(x\sqrt{3})\plus{}\cos(x)\equal{}\frac{\pi}{2}$?
Sonnhard.


algebra
polynomial
calculus
integration
function
domain
linear algebra

Let $ A$ and $ B$ be two commutative rings with unity such that $ A\subseteq B$. Let $ u\in B$ and $ v\in B^{\times}$. Assume that $ u\in A\left[v\right]\cap A\left[v^{ \minus{} 1}\right]$.

[b](a)[/b] Prove that $ u$ is integral over $ A$.

(This is [url=http://jmilne.org/math/CourseNotes/math676.html]J. S. Milne, [i]Algebraic Number Theory[/i], version 3.02[/url], exercise 2-5, without the useless "integral domains" condition.)

[b](b)[/b] If $ P\in A\left[X\right]$ and $ Q\in A\left[X\right]$ are polynomials satisfying $ u \equal{} P\left(v\right) \equal{} Q\left(v^{ \minus{} 1}\right)$, then show that $ u$ is the root of a monic polynomial of degree $ \deg P \plus{} \deg Q$ over $ A$. (Of course, this is a stronger version of part [b](a)[/b].)

[b](c)[/b] Modify the above statements to replace the condition $ v\in B^{\times}$ by $ v\in B$.

([i]Remark.[/i] While [b](c)[/b] is easy, [b](b)[/b] seems to be harder than [b](a)[/b]. I know two solutions for [b](a)[/b], but only one of them still shows [b](b)[/b].)

Let the degrees of the polynomials be $ n$ and $ m$ respectively. I think (b) can be proven reasonably easily by simply writing the two equations for $ u$ (clearing denominators in the second equation) then multiplying the first equation by $ v^{i}$ for all $ i<m$, multiplying the second equation by $ v^{j}$ for all $ j<n$ and then writing the resulting $ n\plus{}m$ equations in matrix form. You get an equation of the form $ Mx\equal{}0$ where $ M$ is an $ (n\plus{}m)$ by $ (n\plus{}m)$ matrix and $ x\in B^{n\plus{}m}$ is the vector with i-th entry $ v^{i\minus{}1}$. The equation $ \det(M)\equal{}0$ is the desired equation satisfied by $ u$.

Nice - I know this solution (it is the same as in Milne, and the same as I gave in [url=http://www.cip.ifi.lmu.de/~grinberg/Integrality.pdf][i]A few facts on integrality[/i][/url], as a proof to Corollary 3 - the real argument is in the proof of Theorem 2), but I didn't look at it from the same angle as you. I considered the main idea to be to show that there is a faithful $ A$-submodule of $ B$ generated by $ n \plus{} m$ elements such that multiplication by $ u$ maps this module to itself. The way you wrote up the solution, it now looks different: The matrix $ M$ is the Sylvester matrix of the polynomials $ P\left(X\right) \minus{} u$ and $ Q\left(X^{ \minus{} 1}\right)X^m \minus{} uX^m$, and its determinant is the resultant of these two polynomials.

For [b](c)[/b], all I did was just replacing the condition $ Q\left(v^{ \minus{} 1}\right) \equal{} u$ by $ Q\left(v^{ \minus{} 1}\right)v^m \equal{} uv^m$, where $ m \equal{} \deg Q$ (note that $ Q\left(v^{ \minus{} 1}\right)v^m$ means the polynomial $ Q\left(X^{ \minus{} 1}\right)X^m$ evaluated at $ X \equal{} v$, which is defined for every $ v$ and not just for invertible $ v$). You seem to be aware of this generalization, too (as you say you clear denominators in the second equation), and all proofs I know for the $ v\in B^{\times}$ case also work in this one.

I mentioned a different proof of [b](a)[/b] that does not help showing [b](b)[/b]. Here is an outline:

Let $ u \equal{} \sum_{i \equal{} 0}^n s_iv^i \equal{} \sum_{i \equal{} 0}^m t_iv^{ \minus{} i}$ for some elements $ s_0$, $ s_1$, ..., $ s_n$, $ t_0$, $ t_1$, ..., $ t_n$ of $ A$. Define $ n \plus{} m \plus{} 1$ elements $ a_0$, $ a_1$, ..., $ a_{n \plus{} m \plus{} 1}$ of $ A$ by

$ a_i \equal{} \left\{\begin{array}{c} t_{m \minus{} i},\text{ if }i < m; \\
t_0 \minus{} s_0,\text{ if }i \equal{} m; \\
\minus{} s_{i \minus{} m},\text{ if }i > m\end{array}\right.$ for all $ i\in\left\{0,1,...,n \plus{} m\right\}$.

Then, $ \sum_{i \equal{} 0}^{n \plus{} m}a_iv^i \equal{} 0$ and $ \sum_{i \equal{} 0}^n a_{i \plus{} m}v^i \equal{} u$. So it is enough to prove the following fact:

[color=blue][b]Theorem 2.[/b] Let $ A$ and $ B$ be two rings (commutative, with unity) such that $ A\subseteq B$. Let $ N\in\mathbb{N}$. Let $ a_0$, $ a_1$, ..., $ a_N$ be $ N \plus{} 1$ elements of $ A$, and let $ v\in B$ be such that $ \sum_{i \equal{} 0}^N a_iv^i \equal{} 0$. Then, $ \sum_{i \equal{} 0}^{N \minus{} k} a_{i \plus{} k}v^i$ is integral over $ A$ for every $ k\in\left\{0,1,...,N\right\}$.[/color]

And this can be proven by backward induction over $ k$:

For $ k \equal{} N$, we have $ \sum_{i \equal{} 0}^{N \minus{} k} a_{i \plus{} k}v^i \equal{} a_N$, which is obviously integral over $ A$.

Set $ \rho_k \equal{} \sum_{i \equal{} 0}^{N \minus{} k} a_{i \plus{} k}v^i$ for every $ k\in\left\{0,1,...,N\right\}$. Then, $ \rho_{k \minus{} 1} \equal{} \sum_{i \equal{} 0}^{N \minus{} k} a_{i \plus{} k \minus{} 1}v^i$ satisfies $ \rho_{k \minus{} 1} \equal{} \rho_k v \plus{} a_{k \minus{} 1}$. Now, $ \sum_{i \equal{} 0}^N a_iv^i \equal{} 0$ yields

$ \sum_{i \equal{} 0}^{k \minus{} 1}a_i\rho_k^{k \minus{} 1 \minus{} i}\left(\rho_k v\right)^i \plus{} \left(\rho_k v\right)^k \equal{} \rho_k^{k \minus{} 1}\left(\sum_{i \equal{} 0}^{k \minus{} 1}a_iv^i \plus{} \rho_k v^k\right)$
$ \equal{} \rho_k^{k \minus{} 1}\left(\sum_{i \equal{} 0}^{k \minus{} 1}a_iv^i \plus{} \sum_{i \equal{} 0}^{N \minus{} k} a_{i \plus{} k}v^i v^k\right) \equal{} \rho_k^{k \minus{} 1}\cdot\underbrace{\sum_{i \equal{} 0}^N a_iv^i}_{ \equal{} 0} \equal{} 0$,

so that $ \rho_k v$ is integral over $ A\left[\rho_k\right]$. Thus, $ \rho_{k \minus{} 1} \equal{} \rho_k v \plus{} a_{k \minus{} 1}$ is integral over $ A\left[\rho_k\right]$ as well. Hence, if $ \rho_k$ is integral over $ A$, then so is $ \rho_{k \minus{} 1}$, and we can proceed by induction.

If you try to get an estimate on the degree of $ \rho_k$ from this proof, you get something like $ \left(n \minus{} k\right)!$...

  darij

Nice notes, I learned a lot from them actually - thanks. (I like $ a\in A$ implies $ a$ is 1-integral in particular  :P). Seriously though, thm 2 is rather nice.

Well, I wrote them up just to have the proofs written out with every detail, so I can think about how they generalize to the $ \left(A,I\right)$-integral case. This is the case when $ A\subseteq B$ is a ring extension, and $ I$ is an ideal of $ A$, and we call some $ u\in B$ integral over $ \left(A,I\right)$ if and only if there exists some $ n\in\mathbb{N}$ and elements $ i_k\in I^k$ for every $ k\in\left\{1,2,...,n\right\}$ such that $ u^n \plus{} \sum_{k \equal{} 1}^ni_ku^{n \minus{} k} \equal{} 0$. Note that this generalizes two well-known things: integrality over a ring (when $ I \equal{} A$) and integrality over an ideal within the same ring (when $ B \equal{} A$). If I have not made mistakes (what is quite a strong assumption), we have the following facts:

[b]Analogue of Theorem 4.[/b] Let $ A$ and $ B$ be two rings such that $ A\subseteq B$. Let $ v\in B$ and $ u\in B$. Let $ m\in\mathbb{N}$ and $ n\in\mathbb{N}$. Assume that $ v$ is $ m$-integral over $ A$ (not necessarily over $ \left(A,I\right)$ !), and that $ u$ is $ n$-integral over $ \left(A\left[v\right],IA\left[v\right]\right)$. Then, $ u$ is $ nm$-integral over $ \left(A,I\right)$.

[b]Analogue of Theorem 5.[/b] Let $ A$ and $ B$ be two rings such that $ A\subseteq B$.
[b](a)[/b] Let $ a\in I$. Then, $ a$ is $ 1$-integral over $ \left(A,I\right)$.
[b](b)[/b] Let $ x\in B$ and $ y\in B$. Let $ m\in \mathbb{N}$ and $ n\in \mathbb{N}$. Assume that $ x$ is $ m$-integral over $ \left(A,I\right)$, and that $ y$ is $ n$-integral over $ \left(A,I\right)$. Then, $ x \plus{} y$ is integral (and the degree is $ nm$) over $ \left(A,I\right)$.
[b](c)[/b] Let $ x\in B$ and $ y\in B$. Let $ m\in \mathbb{N}$ and $ n\in \mathbb{N}$. Assume that $ x$ is $ m$-integral over $ \left(A,I\right)$, and that $ y$ is $ n$-integral over $ A$ (not necessarily over $ \left(A,I\right)$ !). Then, $ xy$ is integral (and the degree is $ nm$) over $ \left(A,I\right)$.

Note that Theorem 4 can be proven using resultants, but the analogue of Theorem 5 doesn't follow from it anymore as nicely as Theorem 5 followed from Theorem 4. But at least, the analogue of Theorem 4 allows you to reduce the general case to the case $ A \equal{} I$ and the case $ B \equal{} A$. The first case is known, and the second one is thoroughly studied in [url=http://people.reed.edu/~iswanson/book/index.html]Irena Swanson and Craig Huneke, [i] Integral Closure of Ideals, Rings, and Modules[/i], 2006-2009[/url]. Now what somewhat frightens me is that I don't see any literature about the general case. Anyway, if you are still reading this, I could as well tell you the most general case: Let $ A\subseteq B$ be a ring extension, and let $ \left(I_k\right)_{k\in\mathbb{N}}$ be a filtration of ideals in $ A$ (that is, $ I_1\subseteq I_2\subseteq I_3\subseteq ...$ are ideals of $ A$ satisfying $ I_aI_b\subseteq I_{a \plus{} b}$ for all $ a$ and $ b$). An element $ u\in B$ will be called integral over $ \left(A,\left(I_k\right)_{k\in\mathbb{N}}\right)$ if and only if there exists some $ n\in\mathbb{N}$ and elements $ i_k\in I_k$ for every $ k\in\left\{1,2,...,n\right\}$ such that $ u^n \plus{} \sum_{k \equal{} 1}^ni_ku^{n \minus{} k} \equal{} 0$. How do elements of integral over $ \left(A,\left(I_k\right)_{k\in\mathbb{N}}\right)$ behave with respect to addition, multiplication and integrality over integrality? See the new version of [url=http://www.cip.ifi.lmu.de/~grinberg/Integrality.pdf]my integrality notes[/url] for some answers.

  darij


The ratio of losses to wins for Kyle's team is 3 to 2. If the team had played the same number of games, but had won twice as many of its games, what would the ratio of losses to wins have been? Express your answer as a common fraction.

This may be represented as 3 losses for every 2 wins. If the number of games is to stay the same but the number of wins is to double, then it would be 1 loss for every 4 wins. This is written in ratio as 1:4 or 1/4


algebra
polynomial
quadratics
number theory unsolved
number theory

Let [tex]p>4[/tex] be a prime number. Show that [tex]p^2 | \sum_{j = 1}^{p - 1} \frac{( p - 1 ) !}{j}[/tex].

The RHS is equal to $(2)(3)(4)...(p-1)+(1)(3)(4)...(p-1)+(1)(2)(4)...(p-1)+...+(1)(2)(3)...(p-2)$

The first term plus the last term is divisible by p.  The second term plus the second to last term is divisible by p, etc.  Rewriting and dividing by p, we have that the RHS divided by p is equal to:

$(2)(3)(4)...(p-2)+(1)(2)(3)...(p-4)(p-3)(p-1)+...+(1)(2)...(.5(p-3))(.5(p+3))(.5(p+5))...(p-1)$

We want this too be congruent to 0 mod p.  Dividing by (p-1)! mod p, we get:

$1^{-1}(p-1)^{-1}+2^{-1}(p-2)^{-1}+...+(.5(p-1))^{-1}(.5(p+1))^{-1}$

Equivalent to (mod p again):
$(-1)(1^{2}+2^{2}+...(.5(p-1))^{2})$
Which is, by the well known formula for the sum of the first n squares, divisible by p... anyone got an easier way?

I guess I could have made the first couple steps more clear... the point is that:

$(p-1)!/j + (p-1)!/(p-j) = (p-j + j) (p-1)!/(j(p-j)) = (p) (p-1)!j^{-1}(p-j)^{-1}$
And then too, the sum for j=1...(p-1)/2 of $(p-1)!j^{-1}(p-j)^{-1}$ is divisible by p, so the original bit was divisible by pp.

Anyhow, the reason it only works for p>4 is that the sum of the first (3-1)/2 squares is not divisible by three, while for all other odd primes, the sum of the first (p-1)/2 squares is divisible by p.

The RHS is the coefficient of $x$ in the polynomial $Q(x)=(x-1)(x-2)\ldots(x-(p-1))$ (well, give or take a sign change, which doesn't matter anyway). $Q(p)=(p-1)!$, so $p^3|Q(p)-Q(0)$.

Now, all the coefficients of $Q$ (except for the first one and the last one, of course) are divisible by $p$ (this is easy to prove by reducing $Q$ modulo $p$), so all the monomials of $Q$ of degree at least $2$ are divisible by $p^3$ (when the argument is $p$). This means that the monomial of $Q$ of degree $1$ is also divisible by $p^3$, so the coefficient of $x$ must be divisible by $p^2$.

P.S.:

It's also easy to see why the above doesn't work for $p=3$: in that case, the degree of $Q$ is $2$, so it's not true that all the monomials of $Q$ of degree $\ge 2$ are divisible by $p^3$.


what did your team place at county? we got 1st :)

We got 7th, and yeah, we really stink. cuz everyone made ME do ALL of the problems on the team round alone, and so we probably failed it

We came 2nd;stupid mistake on team round.

well, at states, we got 6 and placed 9th cuz we changed 3 right answers to wrong answer at the last sec. literaly

He asked for county :wink:
But we came 3rd at states.

I'm pretty sure we have the hardest chapter in MI...

your location says ohio. did you just move?

so you live in michigan. not ohio....or vise versa?

No, I live in Ohio, and moved from Michigan.
We're getting off topic.

oh you got me confused when you said "have". ok

This discussion is getting pretty off-topic, and we have plenty of threads about results / scores already, so I'm going to lock this topic.


induction
real analysis
real analysis unsolved

Can one find continuum many disjoint subsets of $\mathbb{R}$ all of positive measure ?

If by "measure" you mean "outer measure", then I believe so. I think I even started a topic on the subject, in which fedja answered.

The proof I have in mind involves well-ordering the $2^{\aleph_{0}}$ open sets of measure $1$, say, and building our $2^{\aleph_{0}}$ sets by transfinite induction, making sure that they are disjoint and not contained in any of the open sets of measure $1$. That way you get sets of outer measure $>1$. I don't have the time to go through this in detail though.


find the area contained by the points of intersection of the graphs of $ 12x^2\minus{}20xy\plus{}7y2\equal{}0$ and $ 2x\minus{}3y\plus{}4\equal{}0$. all answers are in square units

*btw, this is a comp. test, so no calculus should be needed*

[quote="DaReaper"]find the area contained by the points of intersection of the graphs of $ 12x^2 \minus{} 20xy \plus{} 7y2 \equal{} 0$ and $ 2x \minus{} 3y \plus{} 4 \equal{} 0$. all answers are in square units

*btw, this is a comp. test, so no calculus should be needed*[/quote]
Actually DaReaper...I wouldn't be surprised if the solution contained calculus...this year the problems had some calculus in it...especially ciphering...a possible suggestion would be to find the intersection points and take the integral of one subtracted from the other from the intersection points.
I couldn't find a non-calculus method when I was taking the test :|

so there is no way to do this w/o calc??

Using calculus is by far the simplest method.  The region is very irregular, so it would be difficult to calculate.


The $ 2n$ move chess game has the same rules as the regular one,with only one exception each player has to make $ 2n$ consecutive moves at a time. Prove that white (who plays first) always has a non losing strategy.

Here $ n$ is a natural number!

Let's suppose second player has a winning strategy.

White moves knight $ 2n$ times, back and forth, back and forth, to give black the same starting position.

Then white must have at this point a non-losing strategy because the option for a consecutive move draw has been left open, and second player (now white) from absolute start has a winning strategy.

For $ n\ge 2$ prove that white has a winning strategy :D  :P

[quote="manjil"][quote="ChessTal"][quote="manjil"]Ok lets say I make a condition that we are not to repeat two moves at a time!Then?[/quote]
Then what? I can't find a solution then. Is there any?[/quote]

Well that's my question right![/quote]
OK then:

So you are asking to prove that:
[i] The $ 2n$ move chess game has the same rules as the regular one,with only one exception each player has to make $ 2n$ consecutive moves at a time. Prove that white (who plays first) always has a non losing strategy but with not repeating two moves at a time*.[/i]

*I guess by "at a time" you mean "per turn".
 There is also one ambiguity about what we would do with checks. I mean if in a 6(n=3) Chess game, white for example has in a move a 6-moves move, then what happens if in the 4th move of the series for e.g, he gives a check? Does his turn ends(in the 4th move stopping the 6 move series) and it's the opponent's turn to play his 6 moves or he would capture the King in his 5th move and win? The last case obviously is not a valid Chess game since in Chess there is no King capture.

[u] Having the last thing in mind and by adopting the second non-Chess case of a [b]"captured King or checkmated King=win condition"[/b] we have that:[/u]

•For $ n\ge2$ both sides have $ 2n\ge4$ moves per turn to play. So white just plays at the start as it was mentioned before 1.e4, Bc4, Qf3, Qxf7 mate!
So for $ n\ge2$ this game guarantees always white a win.

•For $ n \equal{} 1$ both sides have 2 moves per turn to play. So white starts with 1.Nc3, e3
In his second move and always depending on black's answer, he plays 2.Qf3, Bc4
Now threats for black to be mated or captured(his King) are too strong and i haven't found saving moves for him, but there might be. And obviously this in not a complete solution to the $ n \equal{} 1$ case.

And i'm not even sure if you mean this kind of game. So you need to clarify the ambiguity....

I doubt there's a winning strategy for White in the $ n \equal{} 1$ variant (Alessandro Castelli wouldn't agree: he'd say 1. e4/Nf3 or 1. d4/Nf3 are theoretically 1:0 situations  :)  ).

The reason is: both [url=http://www.chessvariants.org/multimove.dir/marseill.html]Marseillais[/url]and [url=http://www.chessvariants.org/multimove.dir/doublemove.html]Doublemove[/url]chess are still played for fun, even by GMs. If there was such a strategy, the games would be abandoned (although-White has a considerable advantage in the game).

Maybe I'll return to this some day,when I'm not too busy and analyze it further. ;)