induction
function
algebra unsolved
algebra

[b]Problem 1 solution:[/b]

Put $ x\equal{}0$ we have $ f(0)\equal{}0$ so the statement is true for $ x\equal{}0,1$

$ f(x) \le f(x)\plus{}f(y) \le f(x\plus{}y) \Rightarrow f(x)$ is increasing.

Put $ x\equal{}y\equal{}t$ we have $ f(t)\equal{}\frac{f(2t)}{2}$, for any $ t \in [0,\frac{1}{2}]$
So by induction we have $ f(\frac{1}{2^n}) \le \frac{f(1)}{2^n}\equal{}\frac{1}{2^n}$ for any $ n \in \mathbb{N}$

Now for any $ x \in (0,1)$ , $ \exists n \in \mathbb{N}$ such that $ \frac{1}{2^n} < x \le \frac{1}{2^{n\minus{}1}}$

then $ f(x) \le f(\frac{1}{2^{n\minus{}1}}) \le 2(\frac{1}{2^n})<2x$ , done.

[quote="stephencheng"][b]Problem 1 solution:[/b]

Put $ x \equal{} 0$ we have $ f(0) \equal{} 0$ so the statement is true for $ x \equal{} 0,1$
[hide]
$ f(x) \le f(x) \plus{} f(y) \le f(x \plus{} y) \Rightarrow f(x)$ is increasing.

Put $ x \equal{} y \equal{} t$ we have $ f(t) \equal{} \frac {f(2t)}{2}$, for any $ t \in [0,\frac {1}{2}]$
So by induction we have $ f(\frac {1}{2^n}) \le \frac {f(1)}{2^n} \equal{} \frac {1}{2^n}$ for any $ n \in \mathbb{N}$

Now for any $ x \in (0,1)$ , $ \exists n \in \mathbb{N}$ such that $ \frac {1}{2^n} < x \le \frac {1}{2^{n \minus{} 1}}$

then $ f(x) \le f(\frac {1}{2^{n \minus{} 1}}) \le 2(\frac {1}{2^n}) < 2x$ , done.[/hide][/quote]

Many thanks! :lol: Good solution.Seem that I must learn from your solution!

[b][size=150]About prob.3[/size]

since I got  $ \left|a\plus{}2b\right|\leq8$
$ \left|a\plus{}b\right|\leq2$
Then I got  $ b\leq10$
But I can't give a example of $ (a,c)$
I considered $ f(1),f(0),f(\frac{1}{2})$ during my working.[/b]

[size=150][b]Problem4[/b][/size]

I can't explain the reasons of this problem clearly.So,I turn to you.
Thanks for helping. :)

[quote="admire9898"][b][size=150]About prob.3[/size]

since I got  $ \left|a \plus{} 2b\right|\leq8$
$ \left|a \plus{} b\right|\leq2$
Then I got  $ b\leq10$
But I can't give a example of $ (a,c)$
I considered $ f(1),f(0),f(\frac {1}{2})$ during my working.[/b][/quote]

What do you mean by Card in problem 2?

In problem 3, you can check that $ 4f(\frac{1}{2})\minus{}f(1)\minus{}3f(0)\equal{}b$.  You can do this by expanding out the function and computing coefficients of $ a, b, c$.  Anyway, we plug in the maximum value for $ f(\frac{1}{2})$ and minimums for $ f(1), f(0)$ and we get that $ b\leq 8$.  The values for $ a$ and $ c$ that work for $ b\equal{}8$ are $ a\equal{}\minus{}8$ and $ c\equal{}\minus{}1$.  This GeoGebra app shows this.  [geogebra]75311dbf8981235d910793b0545cba1d52919b50[/geogebra]

I'm sorry, but I can't get it to show the bottom of the box.  Just take my word for it, or graph it yourself.

Thanks to [b]tenniskidperson3[/b].I have realized what I have done wrong.Thanks for your solutions.

[quote]What do you mean by Card in problem 2?[/quote]

[color=darkblue][size=150] [b]Card[/b] means the amount of elements which in the set {E}[/size][/color]
 :)

Any thoughts toward Any Pro2&4?
Need help.

Hint to problem 4:
If $ a_1,a_2,...,a_n$ are the numbers in $ S$ an $ a_1 \geq a_2 \geq...\geq a_n$, then $ \prod a_k$ = $ \sum a_k$ $ \leq$ $ n.a_n$.


conics
hyperbola
geometry proposed
geometry

Let $ABC$ be a triangle. construct the equilateral triangles $BCX$ , $CAY$ and $ABZ$ in the exterior of $ABC$.
Prove that:
(i) $AX$ , $BY$ and $CZ$ meet at a point.
(ii)$AX=BY=CZ$

[quote="mecrazywong"]Indeed, it is well known.
A more general assertion for (i) would be the following:
Construct similar isosceles triangles BCX,CAY,ABZ with the sides of triangle ABC as the base sides. Then AX,BY,CZ are concurrent.[/quote]

The point where they concur describes the [url=http://mathworld.wolfram.com/KiepertHyperbola.html]Kiepert hyperbola[/url] when you modify the shape of the isosceles triangles.


The number 115 can be written as $ 12q \plus{} r$ where $ q$ and $ r$ are
integers and $ 0\leq r <12$. What is the value of $ q \minus{} r$ ?

The greatest possible integral value for $ q$ is $ 9$.  Then, $ 115 \minus{} 12 \cdot 9$, is $ 7$.  So the answer is $ 9\minus{}7$, or $ \boxed{2}$.


Let $ ABCD$ be a trapezoid with $ AB\parallel CD,AB>CD$ and $ \angle{A} \plus{} \angle{B} \equal{} 90^\circ$. Prove that the distance between the midpoints of the bases is equal to the semidifference of the bases.

[quote="limes123"]M - midpoint of AB, N - of CD. K,L intersection points of line parell to AD through point N and line parell to BC through point N with AB. KLN is right triangle => MK=MN=ML=1/2(AB-CD) ;)[/quote]

[color=blue][b]Nice and simple solution :)  ![/b][/color]

[color=red] [b]HINT[/b][/color]
With the above notation (last post), if $ O$ is the common point of $ AD,BC$ , then $ O,M,N$ are collinear and triangle $ OAB$ is right. But

  $ OM \equal{} \frac {AB}{2} , ON \equal{} \frac {CD}{2}$  and  $ NM \equal{} OM \minus{} ON$ .

 Babis
XXXXXXXXXXXXXXXXXXXXX

[quote="YaMohammad"]Hi
Here's my solution.
If $ AB $  and $ CD $ intersect at $ E $ then $ AEB=90 $
Let $ M $ and $ N $ be the midpoint of $ DC $ and $ AB $.
Now we have that $ M $ and $ N $ and $ E $ are collinear.
So we have $ EM = DC/2 $ and $ EN= AB/2 $ so the problem is solved.[/quote]
$AB$ is parallel to $CD$

you mean if $AD$ and $BC$ intersect at $E$  ...


[quote="dduclam"]Prove that for all $ a,b,c$ be nonnegative real numbers, we have
\[ \frac {b \plus{} c \minus{} a}{5a^2 \plus{} 4bc} \plus{} \frac {c \plus{} a \minus{} b}{5b^2 \plus{} 4ca} \plus{} \frac {a \plus{} b \minus{} c}{5c^2 \plus{} 4ab}\ge\frac1{a \plus{} b \plus{} c}
\]
[/quote]
After expanding we obtain:
$ \sum_{sym}(20a^5b\plus{}25a^4b^2\minus{}45a^3b^3\minus{}8a^4bc\plus{}36a^3b^2c\minus{}28a^2b^2c^2)\geq0,$ which is equivalent to
$ \sum_{cyc}ab(a\minus{}b)^2(28c^2\minus{}8(a\plus{}b)c\plus{}20a^2\plus{}65ab\plus{}20c^2)\geq0,$ which is obvious.

[quote="dduclam"]Prove that for all $ a,b,c$ be nonnegative real numbers, we have
\[ \frac {b \plus{} c \minus{} a}{5a^2 \plus{} 4bc} \plus{} \frac {c \plus{} a \minus{} b}{5b^2 \plus{} 4ca} \plus{} \frac {a \plus{} b \minus{} c}{5c^2 \plus{} 4ab}\ge\frac1{a \plus{} b \plus{} c}\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-82-d-d-lam.html


Show that

$ \begin{vmatrix}
1\plus{}a_1^2 & a_1a_2 & \ldots & a_1a_n\\
a_2a_1 & 1\plus{}a_2^2 & \ldots & a_2a_n\\
\vdots & \vdots & \ddots & \vdots\\
a_na_1 & a_na_2 & \ldots & 1\plus{}a_n^2
\end{vmatrix}\equal{}1\plus{}a_1^2\plus{}a_2^2\plus{}\cdots \plus{}a_n^2.$

Actually I get the feeling I've seen this problem before, but I didn't find it in my books. Can somebody tell me if it's a well-known one or something? Thanks.

It's probably well-known, but I don't have an immediate reference or name.

I'm assuming that we're working over $ \mathbb{R}$ here - although that may not be important.

Let's phrase this this way: let $ a$ be a column vector. The claim is that
\[ \det(aa^T\plus{}I)\equal{}1\plus{}a^Ta\equal{}1\plus{}\|a\|^2\]
Note that the matrix $ aa^T$ is a rank 1 matrix. Hence $ n\minus{}1$ of its $ n$ eigenvalues are $ 0.$ The sum of the eigenvalues equals the trace and $ \text{tr}\,(aa^T)\equal{}\text{tr}\,(a^Ta)\equal{}a^Ta\equal{}\|a\|^2.$

So the eigenvalues of $ aa^T$ are $ \|a\|^2,0,0,\dots,0.$ Add $ I$ and we get $ aa^T\plus{}I$ with eigenvalues $ \|a\|^2\plus{}1,1,1,\dots,1.$ The product of those eigenvalues is the determinant, and that product is $ \|a\|^2\plus{}1.$ We're done.

If we were working over a field other than $ \mathbb{R},$ I wouldn't have used the notation $ \|a\|^2$ for $ a^Ta,$ but the proof works anyway. (Note that $ a$ is an eigenvector of $ aa^T$ with eigenvalue $ a^Ta$ over any field, at long as $ a\ne 0$, and the $ a\equal{}0$ case is trivial anyway.)

The only squirrely thing that can happen outside $ \mathbb{R}$ is that $ a^Ta$ might be zero- in which case, the matrix isn't diagonalizable, but the answer still holds.

i have a different solution... let $ A$ be the given matrix... if some $ a_i \equal{} 0$, then expanding the determinant through the i-th column we would obtain a $ (n \minus{} 1)\times (n \minus{} 1)$ matrix which "looks" (don't know how to explain) like $ A$... so an inductive argument will prove the result...

now suppose, that $ a_i\neq 0$ for all $ i$... note that any eigenvector $ v$ of $ A \minus{} I$ satisfies the equation $ a_iBv \equal{} 0$, where $ B$ is the row matrix $ (a_1 \ a_2 \ ... \ a_n)$... this implies that $ 1$ has geometric multiplicity greater or equal than $ n \minus{} 1$... since the algebraic multiplicity is greater than or equal to the geometric multiplicity this would leave at most one more distinct eigenvalue $ \lambda$...

note that the sum of the eigenvalues equals $ tr (A) \equal{} n \plus{} a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2$... since we assumed that $ a_i\neq 0$ we can't have all $ n$ eigenvalues to be equal to $ 1$... therefore, we should have $ n \minus{} 1$ of them being equal to $ 1$ and the other one equal to $ 1 \plus{} a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2$...

since the determinant equals to the product of the eigenvalues we conclude that $ \det A \equal{} 1 \plus{} a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2$  :D

(i think this goes something similar to the solution of Prof. Kent)


integration calculus trigonometry quadratics algebra polynomial logarithms

\[ \ I\equal{}\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{xsinx dx}{1\plus{}sin^{2}x}.
I\equal{}?\]
 :rotfl:  :rotfl:  :rotfl:

hello, this integral is not elementary.
Sonnhard.

No, you're wrong. It's really easy peasy to calculate, as you can see [url=http://integrals.wolfram.com/index.jsp?expr=%28x%28sin%28x%29%29%29%2F%281%2B%28sin%28x%29%29%5E2%29]here[/url] :D  :what?:

More seriously, i think it may be just a typing mistake, the real integral being $ I \equal{} \int\limits_{\frac {\pi}{6}}^{\frac {\pi}{4}}\frac {sinx dx}{1 \plus{} sin^{2}x}$ which is... well, still too difficult for me, but more reasonable than the first one.
Otherwise it seems quite a hard one :surrender:

hello, set $ z\equal{}\tan\left(\frac{x}{2}\right)$ with $ dx\equal{}\frac{2dz}{1\plus{}z^2}$ and
$ \sin(x)\equal{}\frac{2z}{1\plus{}z^2}$.
Sonnhard.

And so I get $ \int \frac {sinx dx}{1 + sin^{2}x} = \int \frac {\frac {2z}{1 + z^2}}{1 + \left(\frac {2z}{1 + z^2}\right)^2}\cdot\frac {2}{1 + z^2} dz$, that with simple algebra is reducible to $ 2\int\frac {2z}{z^4 + 6z^2 + 1}dz$. Now I set $ z^2 = m$: this implies $ \dm = 2zdz$ and I get $ 2\int\frac {dm}{m^2 + 6m + 1}$. The quadratic polinomial below the line has two real roots, $ - 3 + 2\sqrt2$ and $ - 3 - 2\sqrt2$ (anybody knows how can I get the plus-minus symbol?), so the fraction is equal to the sum of two fractions with a linear denominator: now I have $ \frac {1}{2\sqrt2}\left(\int\frac {dm}{m - 2\sqrt2 + 3} - \int\frac {dm}{m + 2\sqrt2 + 3}\right)$ and finally $ \frac {1}{2\sqrt2}\ln\left|\frac {z^2 + 3 - 2\sqrt2}{z^2 + 3 + 2\sqrt2}\right|$, where $ z = \tan\left(\frac {x}{2}\right)$. So, if my supposition about the possible mai quoc thang's mistake is correct $ I = \frac {1}{2\sqrt2}\ln\left(\frac {22 - 14\sqrt2 - 12\sqrt3 + 8\sqrt6}{30 - 12\sqrt3 - 6\sqrt2}\right)$ and this was not that difficult (but it was way too long  :D )

Bye
GioMott


geometry
circumcircle
parallelogram
geometry unsolved

what are $ O_1$ and $ O_2$? I think you mean that $ O_1$ is the circumcircle of ABC and $ O_2$ the circumcircle of CJI.

The altitude fro C in ABC meet $ (O_2)$ on M. Clearly 'cause ABC and JIC are similar and cause the circumcenter and the ortocenter are isogonal cojugate we have that CM pass through $ O_2$ and that $ CO_1 \perp IJ$. But 'cause $ IJ$ is the radical exes between $ (O)$ and $ (O_2)$ we have $ OO_2 \perp IJ$ and so $ OO_2 \parallel CO_1$. But $ OO_1 \parallel CO_2$, so $ O_1OO_2C$ is a parallelogram and then $ \overline{OO_1} \equal{} \overline{CO_2}$. Then also $ O_2O_1OM$ is a parallelogram and so $ MO \parallel O_1O_2$. But $ O_1O_2 \perp CL$ and so $ OM$ pass through $ L$. And then we conclude that $ O_1O_2 \parallel OL$

[quote="encyclopedia"]Let $ (O)$ be acircle pass through $ A,B$ of triangle $ ABC$. $ I,J$ are interesections of $ (O)$ with $ AC,BC$, resp. The circumcircles of $ ABC$ and  $ IJC$ intersect at $ B$ and $ L$ prove that $ O_1O_2//OL$.[/quote]
[hide="Solution"]
By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366594#366594, we have that $ OL\perp LC$.  Also, $ O_1$ and $ O_2$ both are centers of circles containing chord $ LC$, which means that they both lie on the perpindicular bisector of $ LC$.  Hence, $ O_1O_2$ is the perpindicular bisector of $ LC$ and it follows that $ O_1O_2\perp LC$.  Thus, $ O_1O_2\parallel OL$. [/hide]

Yes you are right, sorry I miss the name of $ O_1,O_2$, I checked your solution they are true thanks!


Does anyone know a program that allows u to draw nice geometric shapes?

Look here if you want a free one [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14734[/url]

Geometer's sketchpad is really good if you are willing to pay money,  at least that is what i have heard-I don't acually have the program.

Btw, is there any way to get that one program into English, Darij?


Given the set {1,2,3,5,8,13,21,34,55}, how many integers between 3 and 89 cannot be written as the sum of exactly two elements of the set?

Can anyone help me out with this problem?

Thanks in advance. [/quote]

Are you allowed to use a specific member more than once?

you mean two distinct elements, right? if so...

(below, the $ F_i$ refer to elements of the *abbreviated* fibonacci sequence, 1, 2, 3, 5, ..., since this is what the problem's dealing with anyway)

the key is to notice that if $ m\neq n$, $ p\neq q$, then $ F_m \plus{} F_n \equal{} F_p \plus{} F_q\Rightarrow \{m,n\} \equal{} \{p,q\}$... assume WLOG that $ m < n$, $ p < q$. if $ q > n$, then $ q \minus{} 1\geq n$ and $ q \minus{} 2\geq n \minus{} 1\geq m$, so, since the fibonacci sequence is strictly increasing, $ F_p \plus{} F_q \equal{} F_p \plus{} F_{q \minus{} 1} \plus{} F_{q \minus{} 2}\geq F_p\plus{}F_n \plus{} F_m > F_m \plus{} F_n$, a contradiction. similarly, we can't have $ n>q$. so $ n\equal{}q$, from which it's also clear that $ m\equal{}p$.

Er, thanks for the help, but that just confused me even more. I'm looking for a number answer.

If it helps, here are the multiple choice answers:

A) 51
B) 57
C) 55
D) 34
E) 43

The answer should be (A)

He has proven that all of the sums of two distinct numbers in the set are different, so there $ \binom{9}{2} \equal{} 36$ integers are in this form in the range of $ [3,89]$. The result then follows.


geometry
parallelogram
geometry proposed

It's easy. Let $ \vec{AB} \equal{} a,\vec{AD} \equal{} b,BP\cap DN \equal{} {L}$ and $ \frac {BO}{OD} \equal{} k$.
 Then $ \vec{AO} \equal{} \frac {a \plus{} kb}{k \plus{} 1}$ and $ \vec{AL} \equal{} \frac {(k \plus{} 1)(a \plus{} kb)}{1 \plus{} k(k \plus{} 1)}$.
Witch means that $ A,O,K$ collinear. And the problem is finished.

[quote="Jutaro"]Through a point $ O$ on the diagonal $ BD$ of a parallelogram $ ABCD$, segments $ MN \parallel AB$ and $ PQ \parallel AD$ are drawn, with $ M$ on $ AD$ and $ Q$ on $ AB$. Prove that the lines $ AO$, $ BP$ and $ DN$ are concurrent.[/quote]
[hide="Solution"]
Let $ AM \equal{} BN \equal{} w$, $ MD \equal{} MC \equal{} z$, $ AQ \equal{} DP \equal{} x$, and $ QB \equal{} PC \equal{} y$.  Thus, we can find that $ ON \equal{} y$, $ MO \equal{} x$, $ QO \equal{} w$, and $ OP \equal{} z$.  Now, let $ AO$ intersect $ ND$ at $ R$, let $ PB$ meet $ ND$ at $ R'$, let $ BP$ meet $ MN$ at $ T$, and $ DO$ meet $ AN$ at $ S$.  Since $ AB\parallel MN$, we have that $ \triangle ASB\sim \triangle NSO\implies \frac {AS}{SN} \equal{} \frac {AB}{NO} \equal{} \frac {x \plus{} y}{y}$.  Hence, Ceva's Theorem on $ \triangle AND$ gives that $ \frac {AS}{SN}\cdot \frac {NR}{RD}\cdot \frac {MD}{AM} \equal{} 1$, so $ \frac {z}{w}\cdot \frac {NR}{DR}\cdot \frac {x \plus{} y}{y} \equal{} 1$ and therefore,
\[ \frac {DR}{NR} \equal{} \frac {z}{w}\cdot \frac {x \plus{} y}{y}
\]
Since $ TN\parallel DC$, we have that $ \triangle BTN\sim \triangle BPC$, so $ \frac {PC}{TN} \equal{} \frac {w \plus{} z}{w}$.  Notice that $ \triangle DR'P\sim \triangle NR'T$, $ PC \equal{} y$ and $ DP \equal{} x$, so
\[ \frac {DR'}{R'N} \equal{} \frac {DP}{TN} \equal{} \frac {x}{TN} \equal{} \frac {y}{TN}\cdot \frac {x}{y} \equal{} \frac {w \plus{} z}{w}\cdot \frac {x}{y}
\]
Since $ O$ lies on $ BD$, we have that $ \triangle BON\sim \triangle BDC\implies \frac {y}{x \plus{} y} \equal{} \frac {ON}{DC} \equal{} \frac {BN}{BC} \equal{} \frac {w}{w \plus{} z}\implies \frac {x}{x \plus{} y} \equal{} \frac {z}{z \plus{} w}$.  Therefore, $ \frac {z}{w}\cdot \frac {x \plus{} y}{y} \equal{} \frac {w \plus{} z}{w}\cdot \frac {x}{y}$ after canceling denominators.  Hence, $ \frac {DR}{NR} \equal{} \frac {DR'}{NR'}\implies R \equal{} R'$. So, $ BP$ and $ AO$ meet $ ND$ at the same place, which implies that $ AO$, $ BP$, and $ ND$ are concurrent at $ R$. [/hide]

[quote="Jutaro"] [color=darkred]Through a point $ O$ on the diagonal $ BD$ of a parallelogram $ ABCD$, segments $ MN \parallel AB$ and $ PQ \parallel AD$
are drawn, with $ M$ on $ AD$ and $ Q$ on $ AB$. Prove that the lines $ AO$, $ BP$ and $ DN$ are concurrent.[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ \{\begin{array}{ccc}
QA=x & , & QB=y\\\\
MA=u & , & MD=v\end{array}$ and $ \{\begin{array}{c}
X\in AO\cap DN\\\\
Y\in BP\cap DN\end{array}$ . Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \|\begin{array}{cc}
\overline {AOX}\mathrm {\ in\ }\triangle DMN\ : \ & \frac {AM}{AD}\cdot \frac {XD}{XN}\cdot\frac {ON}{OM}=1\implies\frac {XD}{XN}=\frac {x(u+v)}{yu}\\\\
\overline {BYP}\mathrm {\ in\ }\triangle CDN\ : \ & \frac {BN}{BC}\cdot\frac {PC}{PD}\cdot\frac {YD}{YN}=1\implies\frac {YD}{YN}=\frac {x(u+v)}{yu}\end{array}\|$ $ \implies$ $ \frac {XD}{XN}=\frac {YD}{YN}$ $ \implies$ $ AO\cap BP\cap DN\ne\emptyset$ . [/color]


quadratics
geometry
3D geometry
modular arithmetic
algebra
factorization
difference of cubes

Let $m\in Z^{+}$ and $k\in Z^{+}$ such that $0\leq k\leq 3^{m}-1$. Show that $3^{m+1}$ doesn't divide $(-2)^{k}-1$.

[quote="barasawala"]Let $m\in Z^{+}$ and $k\in Z^{+}$ such that $0\leq k\leq 3^{m}-1$. Show that $3^{m+1}$ doesn't divide $(-2)^{k}$.[/quote]
Are you sure? It is trivial  :huh:

...but -2 is a cubic nonresidue mod 9, done?

To elaborate: because -2 is a cubic nonresidue mod $3^{m+1}$, its order must exceed $3^{m}$ mod $3^{m+1}$, because it is at most a quadratic residue (we note 2, 3 are the only primes dividing $\varphi(3^{m+1)}$, and $\mathbb{U}_{3^{m+1}}$ has generators).

I don't understand much.

Can it be done easier than this?

[hide]
Suppose $k=a\cdot 3^{b}$ where $a$ is not divisible by 3. Then difference of cubes creates a massive factorization
\[(-2)^{k}-1=((-2)^{a}-1)\prod_{j=0}^{b-1}((-2)^{2a\cdot 3^{j}}+(-2)^{a\cdot 3^{j}}+1)\]
There are $b+1$ factors, and none of them are divisible by 9. To see this, the order of $-2$ modulo 9 is 3, that takes care of the first factor, while the other factors all have the form $x^{2}+x+1$. If $x^{2}+x+1\equiv 0\pmod{9}$, then $(2x+1)^{2}\equiv-3\pmod{9}$, which is clearly not possible because it implies that $(2x+1)^{2}$ has exactly one factor of 3 dividing it.

The above stuff shows that $(-2)^{k}-1$ is divisible by at most $b+1$ factors of 3. So if $(-2)^{k}-1$ were divisible by $3^{m+1}$, then $b\geq m$ and $k=a\cdot 3^{b}\geq 3^{m}$, a contradiction.
[/hide]


inequalities
trigonometry
inequalities proposed

In triangle $ ABC$  the following is true :
$ \tan{\frac{A}{4}}$ and $ \tan{\frac{B}{4}}$ are roots of the equation $ x^2 \plus{}a_1 x \plus{}b_1 \equal{}0$ .
$ \tan{\frac{B}{4}}$ and $ \tan{\frac{C}{4}}$ are roots of the equation $ x^2 \plus{}a_2 x \plus{}b_2 \equal{}0$
$ \tan{\frac{C}{4}}$ and $ \tan{\frac{A}{4}}$ are roots of the equation $ x^2 \plus{}a_3 x \plus{}b_3 \equal{}0$

Find the maximum value of: $ M \equal{}(1\minus{}a_1 \plus{}b_1)(1\minus{}a_2 \plus{}b_2)(1\minus{}a_3 \plus{}b_3)$

[quote="Sophors"]In triangle $ ABC$  the following is true :
$ \tan{\frac {A}{4}}$ and $ \tan{\frac {B}{4}}$ are roots of the equation $ x^2 \plus{} a_1 x \plus{} b_1 \equal{} 0$ .
$ \tan{\frac {B}{4}}$ and $ \tan{\frac {C}{4}}$ are roots of the equation $ x^2 \plus{} a_2 x \plus{} b_2 \equal{} 0$
$ \tan{\frac {C}{4}}$ and $ \tan{\frac {A}{4}}$ are roots of the equation $ x^2 \plus{} a_3 x \plus{} b_3 \equal{} 0$

Find the maximum value of: $ M \equal{} (1 \minus{} a_1 \plus{} b_1)(1 \minus{} a_2 \plus{} b_2)(1 \minus{} a_3 \plus{} b_3)$[/quote]

$ M\equal{}\prod{(1\plus{}\tan{\frac {A}{4}}\plus{}\tan{\frac {B}{4}}\plus{}\tan{\frac {A}{4}}\tan{\frac {B}{4}})}\equal{}\prod{(1\plus{}\tan{\frac {A}{4}})^2}$

note that
\[ \left( {1 \plus{} \tan \frac{B}{4}} \right)\left( {1 \plus{} \tan \frac{C}{4}} \right) \le \left( {1 \plus{} \tan \frac{{B \plus{} C}}{8}} \right)^2\]
\[ \Leftrightarrow \frac{{\sin ^2 \frac{{B \minus{} C}}{8}\sin \frac{{\pi  \minus{} B \minus{} C}}{4}}}{{2\sqrt 2 \sin \left( {\frac{\pi }{4} \minus{} \frac{B}{8}} \right)\sin \left( {\frac{\pi }{4} \minus{} \frac{C}{8}} \right)\sin \left( {\frac{\pi }{4} \plus{} \frac{B}{8}} \right)\sin \left( {\frac{\pi }{4} \plus{} \frac{C}{8}} \right)\cos ^2 \frac{{B \plus{} C}}{8}}} \ge 0,\]
is true, thus, $ M$ get maximum value must be $ A\equal{}B\equal{}C$, so, ...


calculus
integration
function
calculus computations

$ \int \limits_{0}^{\frac{\pi}{2}} cos^n{x}dx$

Integration by parts or Beta Function may work.

You can see the solution here.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=80515[/url]

[quote="valeriummaximum"]i think there is a reduction formula for this ....[/quote]Check kunny's link, it's all in there.


AMC AIME MIT college search USA(J)MO USAMO

Ok...so I dont know if this is a great time to ask...but I am going to ask before May 11, where they will announce who did well on the USAMO.

Anyways, I just remembered this, but there was that option on the AIME that said something about "colleges and other institutions sometimes request names and address of high AIME scorers" and you had to check "yes" to give the AMC permission to release your name.

Well, I think I forgot to check yes.  However, I want to change it now, because I ended up getting a really high AIME score (so that will look good), but more importantly, I havent received anything from top universities/summer math programs through the PSATs, and I think I never will because I never checked yes on the form (and my PSATs were <220).  However, I am a junior, so it would be nice to get info. from colleges that I will be applying to next year (especially the ones that care like Caltech and MIT) or from a summer math program this year.

Anyways, I was just wondering if anyone else receives summer math related activites/letters from top colleges because of AIME scores (or do you guys not know how you receive them?).  Do you guys remember from last year??? (I forgot to check yes last year too I think--even if I did, I was a sophomore so I dont think it would have mattered).  Or is it not that necessary to check "yes", as many colleges/math programs dont really request your name and stuff (I would think they do since AIME >>> PSAT for top colleges)??

I contacted the AMC director, but obviously hes really busy, so I am just going to ask other students who received high AIME scores/remember from last year?

Thanks

For college applications (including MIT/Caltech), you self-report your AMC/AIME score. It won't matter at all if you didn't check the box on the AMC/AIME answer sheets.

[quote="worthawholebean"]You don't get info about colleges from the AIME box as far as I know. It only goes to summer programs and such. The PSAT box, on the other hand, causes you to receive trees worth of college spam.[/quote]
Agreed!!

Ya I know you self report your score (im not worried that they wont believe my score because they can just search my name online on the USAMO qualfiers page).  However, I havent actually gotten anything from any programs or colleges that I would like to go to (only from second level colleges--bad PSAT score or maybe I just wrote my address/email wrong?).  But if you dont get things from colleges for the AIME, then I guess it isnt that big of a deal for that.  Also, I was also hoping that, if I dont get to do research this summer, I might apply to some math program in the summer (wont make MOP).  I really do not have the time to up every program because I am very busy right now.

I will just resign "yes" up anyways, as it doesnt hurt to.  The more information, the better.

thanks guys.

[quote="the future"]However, I havent actually gotten anything from any programs or colleges that I would like to go to (only from second level colleges--bad PSAT score or maybe I just wrote my address/email wrong?).[/quote]

Don't worry about this.  The colleges you would like to go to probably won't send you information based on your PSAT score no matter how good it is.  Also, trust me, once you start getting random stuff from colleges you'll stop reading it entirely pretty soon.

I checked those boxes on the AMC and AIME this year, and all I've received is stuff from like Mathcamp, Awesome Math, Mathzoom, etc.  those summer programs

I have checked that box in the past, and the stuff I have received is similar to what Quickster94 said, especially Awesome Math.

As for the PSAT, I would say this is really more of a College Confidential topic, but I think lower tier schools opt to buy the contact info of most students (i.e. for X cents a name, you can buy the names of the top BLAH scorers), so even if you get stuff from places in which you are interested, you will still get the others.  I did get mail from Caltech and MIT, but this may have been due to SAT or ACT, not PSAT, since I ended up with a consistent stream of college mail for about 2 years.

I think amc/aime/usamo probably will not cause you to receive these college e-mails. Of all these college e-mails I've been getting since the past psat scores were released, there are a few from cal-tech; no mit, though. (My psat score was 220+)

I haven't really read most of them except the first a few days and a few colleges I have in mind.

If you want to go to a summer math program just go to the website.

Mathcamp, Awesome Math, PROMYS, HCCSiM, Ross are the main ones to look into (hopefully I didn't miss any).

Note that, at least for this summer, the application deadlines have passed for many of these. I know PROMYS at least is still accepting applications.


inequalities
function
inequalities theorems

[quote="Jakey"]Hi, I need help on one point of this solution. The problem goes like this:
and oh yeah, y1 means y sub 1 and everything like this follows.

Show that if the values y1,y2,...,yk are all nonnegative, and if m1,m2,...mk are positive integers, then:

(m1y1 +m2y2+...+mkyk)/(m1+m2+...+mk) >= (y1y2...yk)^1/(m1+m2+...+mk).

i know how to do this, I could just use AM-GM or its modified version which is the weighted means which makes it easier to do this. but there's a connected problem that i dont seem to figure yet:

Show consequently that is r1,r2,...,rn are proper fractions satisfying r1+ r2+ ... +rk=1, then
r1y1 + r2y2 +...+rkyk >= y1^(r1+r2+...+rk). I know how to do the first few steps of this, just using the weighted means which arives me to this point: r1y1 + r2y2 +...+rkyk >= y1y2....yk. but how do you prove that it is >= to y1?

i need help thanks[/quote]
Have many way to solve . You can consider it as a function of $ y_1$ and use derviverty to find end point of this function + induction.
Solution 2 : You prove it in $ Q$ and it true for in $ R$
let $ n_i\equal{}\frac{p_1}{N}$ where $ p_1,N\in N$ and $ N\equal{}\sum_{i\equal{}1}^n p_1\i$


minh la newmember, hy vong co bac nao chi ho cac danh Latex

Bạn có thể vào trang web ddtoanhoc hoặc onthi.hcmup.edu.vn để download luôn tài liệu có hướng dẫn bằng tiếng Việt.

Cac ban tai file nay ve de cai dat tieng viet Cho PCTEXv5

chao toi la thanh vien moi .Toi cung ko biet latex cac ban co the giup toi ko :D Toi rat cam on  :lol:


24. In how many ways can a,b,c,and d be chosen from {0,1,2..9} so that a<b<c<d and a+b+c+d is a multiple of three?

A) 54
B) 64
C) 63
D) 90
E) 72

25. $\angle BAC$ is said to be "laceable" if distinct points $X_1,X_2,...X_{2n}$ can be found that:
$\Rightarrow X_{2k-1}$ is on the AC for each values of k,
$\Rightarrow X_{2k}$ is on AB for each value of k, and
$\Rightarrow AX_{1} = X_1*X_2 = X_2*X_3 = .....=X_{2n-1}*X_{2n} = X_{2n}*A$

For example, angle 20 degrees is laceable, as shown. The number of laceable acute angles, whose sizes in degrees are integers, is:

A) 3
B) 4
C) 5
D) 6
E) 7

Image is attached.

For 24, make the set into a collection mod 3: {0, 1, -1, 0, 1, -1, 0, 1, -1, 0}
Then do it case-by-case:
Four 0's = 1 way
Three 0's, 1 other = 0 ways
Two 0's, 1 +1, 1 -1: Number of ways = 4C2*3C1*3C1= 6*3*3=54
One 0, either 3 +1's or 3 -1's.  Therefore number of ways= 4C1*2 = 8
Zero 0s, then 2 +1's and 2 -1's: Number of ways = 3C2*3C2 = 3*3 = 9
Total number of ways=sum of all those=72

25.  This one is hard to get.  Do you have the solutions?  It'd be easier to follow the solution, with the pictures, drawing it out as you go, to understand this one.

[quote="Silverfalcon"]$AX_{1} = X_1*X_2 = X_2*X_3 = .....=X_{2n-1}*X_{2n} = X_{2n}*A$[/quote]This seems like a very interesting problem but I'm not sure what the notation $P*Q$ means for points $P$ and $Q$. You left out the $*$ between $A$ and $X_1$. Was that a typo or are the equalities refering to lengths of some kind?

Wumbate provides a correct counting solution to #24.  If time is short, an approximation might work:
[hide]I expect about one-third of the sums will be divisible by 3.

${10\choose 4}=210$ total combinations.  So, about 70 will be divisible by 3.

Any answer far from 70 fails my sanity-check.

Here is where experience and "number sense" come into play.  Are you comfortable eliminating all multiple-choice answers, except "72"?  If so, then answer "72" and move on.[/hide]

By $X_1*X_2$ he means the length between X_1 and X_2.  The Cayley can be found here [url=http://cemc.uwaterloo.ca/english/contests/cayley.shtml]http://cemc.uwaterloo.ca/english/contests/cayley.shtml[/url].  Download the 2003 one and look at the last problem for the pictures.

Your approximation is good.  It's originally what I did but 70 wasn't a choice.  If it went from 1 to 9 then that method would work, I think.  The answer choices were 54, 64, 63, 90, 72.  If you feel comfortable guessing 72 from your approximation, then do it.  It would've worked on this one, but often 70 would be a choice to trick people who do that.


vector
abstract algebra
superior algebra
superior algebra unsolved

I'm working on a problem, but i would need some ideas & insights from you guys.

I've been thinking about under what situations I have the result "If I can embed X in Y and Y in X then X and Y are *morphic." I know if my embeddings are just surjective maps, then the Schroder-Bernstein theorem gives me this result for set-isomorphism.

So, I thought that [0,1] embeds in R and R embeds in [0,1]. I notice that R is not ???morphic to [0,1].

is this correct ?

An analogous result is easily true in finite linear algebras, where if I have 1-1 linear maps from X into Y and Y into X, then X and Y are isomorphic. I think a proof could easily be done for finite linear algebras too (using the rank+nullity
theorem, right?) I have no idea if the result is true for all algebraic embeddings.

For some cathegory $ K$ we can define the following property:

For any objects $ A,B$ and any Monomorphisms $ \alpha: A \to B$, $ \beta: B \to A$ we can find an isomorphism $ \gamma: A \to B$.
We call this property $ CSB_i(K)$.

Similar:
For any objects $ A,B$ and any Epimorphisms $ \alpha: A \to B$, $ \beta: B \to A$ we can find an isomorphism $ \gamma: A \to B$.
We call this property $ CSB_s(K)$.

$ CSB_i(K)$ and $ CSB_s(K)$ are true for $ K \equal{}$:
- sets.
- vector spaces over some field $ k$.
- some types of modules over some types of rings.

They are false for $ K \equal{}$:
- groups.
- rings.
- fields.
- general modules.
- topological spaces (as you mentioned above).

Thanks Zetax,

I started with a theorem. Could u please check my demonstration?

Theorem: Let X and Y be finite linear spaces. Suppose that I have a map f which embeds X into Y and a map h which embeds Y into X, then X and Y are isomorphic.

Let f and h be 1-1 linear maps (as they are embeddings).
We know that hof is 1-1, and a linear map from X to X (an endomorphism).
Because hof is 1-1, we know it has rank equal to |X|.
But then we know that it's nullity is 0 by the rank+nullity theorem,
and hence hof(0) = 0 and for no other value is hof(v) = 0.

Let {b_1, b_2,..., b_n} be a basis for X.
We know that the expression A = (a_1)*(b_1) + (a_2)*(b_2) + ... + (a_n)*(b_n) is a nonzero for all a_i,
and we know then that hof(A) is a nonzero for all a_i  from the rank+nullity result above.
This means that (a_1)*hof(b_1) + (a_2)*hof(b_2) + ... + (a_n)*hof(b_n) is a nonzero for all a_i,
because hof is a linear transformation.

But this means that the image of the basis is linearly independent.

To show that hof is surjective, it suffices to suppose that alpha doesn't belong to hof(X)
and get a contradiction.
Given our supposition, we may conclude a_0*(alpha) doesn't belong to hof(X) for all a_0.
Likewise, it can't be the case that (a_1)*hof(b_1) + (a_2)*hof(b_2) + ... + (a_n)*hof(b_n) = (a_0)*(alpha)
for any group of a_i.

Thus a_1*hof(b_1) + a_2*hof(b_2) + ... + (a_n)*hof(b_n) + (a_0)*(alpha),which is nonzero for all a_i.
That means that {b_1,b_2,...,b_n,alpha} are linearly independent.
So, we know |X| >= n+1, but we were given |X| = n
by the above, which is a contradiction.

Hence, we know that hof is an isomorphism. Moreover, we know that h is surjective,
which means that h is an isomorphism. QED

Hopefully my proof is correct

Looks correct (you sometimes forget to exclude some numbers to be $ 0$, but thats no big deal), but you can make it shorter (you did something similar):

If $ f: X \to Y$ is an injective linear map, then a basis of $ X$ will be lineary independent in $ f(X)$ again. Thus $ |X| \equal{} |f(X)| \leq |Y|$.
Doing the same with $ g: Y \to X$, we get $ |Y| \leq |X|$. In total we get $ |X|\equal{}|Y|$, and spaces of same dimension are isomorphic.
Or alternatively, you can see that $ |X| \equal{} |f(X)| \leq |Y| \leq |f(X)|$, and the only subspace of $ Y$ having dimension $ |Y|$ is $ Y$ itself. Thus $ f(X)\equal{}Y$ is surjective.

Thanks, what's a good intro text on category theory?

mac lane, categories for the working mathematician

besides, the finiteness of the dimension is superfluous. as zetax pointed out, CSB holds in the category of $ k$-vector spaces.

an interesting discussion on that: http://sbseminar.wordpress.com/2007/10/30/theme-and-variations-schroeder-bernstein/


calculus
integration
real analysis
real analysis unsolved

Let f be integrable with int(f(x))dx from 0->1 = int(xf(x))dx from 0->1 = 1.
Show that int((f(x))^2)dx from 0->1 >= 4

Is your problem like this?

[quote="KelvinNg"]Let f be integrable with $ \int_0^1 f(x)\ dx \equal{} \int_0^1 xf(x)\ dx \equal{} 1.$
Show that $ \int_0^1 (f(x))^2\ dx \geq 4.$[/quote]

That minimum value is achieved for $ f(x)\equal{}1\plus{}3\cdot(2x\minus{}1)\equal{}6x\minus{}2$, and it's minimal by an orthogonality argument. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146373]here[/url] for a complete argument, generalized to arbitrary degrees.

I read kunny's hint but seems a little labour some. Here is another one.(concerning Holder ineq)

$ \int_{0}^{1} [a(x - 1) + 1]f(x)dx = 1\leq (\int_{0}^{1}[ax - (a - 1)]^2dx)(\int_{0}^{1}f^2(t)dt)$ (this is holder ineq or Cauchy ineq)
but $ \int_{0}^{1}[ax - (a - 1)]^2dx = \frac {1}{3}a^2 - a + 1\geq \frac {1}{4}$ and set $ a = \frac {3}{2}$ the "$ \eq$" stands.
Substitution, we get  $ \int_{0}^{1}f^2(t)dt \geq \frac {1}{\frac {1}{3}a^2 - a + 1} =4$


MATHCOUNTS
geometry
trapezoid
Pythagorean Theorem

I am not geniuses like most of you, although I am in the top math class. This makes it very hard for me to understand/complete some questions. This week, we are on Warm Up 9 (a mathcounts paper [specifially mathcounts 2006-2007]), and I am having trouble with number 7. The question is as follows:
[b]The scale drawing shown represents a field in the shape of an isosceles trapezoid. What is the area of the actual field? Express your answer in simplest radical form.[/b]
 somehow i cannot figure out how to post the picture onto this but you can go to this link:
http://74.125.113.132/search?q=cache:H1WrlU44Y-AJ:sch.ci.lexington.ma.us/~jfrost/MC/2006%2520-%25202007%2520Warmups.pdf+warm+up+9+page+41+mathcounts+2006+2007&hl=en&ct=clnk&cd=1&gl=us&client=firefox-a
scroll down until you see "warm up 9" at the top of the page and then go to number seven, it should show you the measurements of the trapezoid.
I CANNOT figure this question out!!!!!
PLEASE HELP!!!!

Draw the heights of the trapezoid.  Since this is isosceles, we have cut base into 3 parts, in this case, equal.  They are $ 2$, so we can use the Pythagorean Theorem on the right triangle we have created with $ c\equal{}12$, and $ a\equal{}2$.  We have:
\[ 2^2 \plus{} b^2 \equal{} 12^2
\]
\[ 4 \plus{} b^2 \equal{} 144
\]
\[ b^2 \equal{} 140
\]
\[ b \equal{} \sqrt{140}
\]
\[ b \equal{} 2\sqrt{35}
\]

The height is thus $ 2\sqrt{35}$.  Use the formula to find the area of a trapezoid and you get:
\[ \frac{2\sqrt{35}(8)}{2}
\]
\[ 8\sqrt{35}
\]

We now need to convert it.  We have found $ \text{cm}^2$, so we convert into $ \text{m}^2$, we need to multiply by $ 100$.  Our final answer is:
\[ \boxed{\boxed{\boxed{800\sqrt{35}}}}
\]

Thanks so much!!! That helped me a lot with the homework- I knew it had something to do with the "a squared plus b squared" theory but I couldn't figure out what!!! Again, thanks for  your help!
  -not so genius


Anyone going to SIMUW 2008, either as a camper or as a TAC?

I'll be back for a second year as a TAC. I was also a participant back in the day, and let me assure you, you're going to have an awesome summer.

That's what I hear!
I guess we'll be working together then; I'm going to be a TAC. I'm stoked.


[i]You are given $ 50$ black tokens and $ 50$ white tokens, which you are free to place in two identical
containers in any way you wish, with each container containing as many tokens of each color as you
decide. After you distribute the tokens, you will be blindfolded, then asked to choose one of the two
containers and draw a single token from the chosen container. If you draw a white token, you will win
$ \$10,000$, but if you draw a black token, you will win nothing. How should you distribute the tokens in
order to maximize the probability that you draw a white token? Give a full explanation of your answer.[/i]
[hide]
If there are more whites than blacks in container 1, there are more blacks than whites in container 2. Thus it is impossible to have a probability over $ \frac12$ for both containers.
If the probability for both is $ \frac12$, then we have an overall probability of $ \frac12$.
Otherwise, consider the container with smaller probability of drawing white. Call the probablity $ P_<\equal{}\frac{a}{a\plus{}b}$, where $ a$ is the number of white tokens and $ b$ is the number of black ones. Then \[ \frac12\minus{}\frac{a}{a\plus{}b}\equal{}\frac{b\minus{}a}{2(a\plus{}b)}\geq\frac{1}{2(a\plus{}b)}\geq\frac{1}{198}\]as $ a<b\leq50$.
Thus $ \frac{a}{a\plus{}b}\leq\frac{49}{99}$. Since the probability in the other container is $ \leq1$, the upper bound on overall probability is \[ \frac{1\plus{}\frac{49}{99}}{2}\equal{}\frac{74}{99}\]

We can achieve this probability by placing one white token in container 1, and the rest of the tokens in container 2 $ \square$
[/hide]

[hide="a start..."]Label the containers A and B. There is 1/2 probability that you choose A, and 1/2 probability that you choose B. Now you just need to put the black and white tokens in. Let $ A_b$ and $ B_b$ be the number of black balls in A and B, respectively. Let $ A_w$ and $ B_w$ be the number of white balls in A and B, respectively. We want to maximize

$ \frac{1}{2}(\frac{A_w}{A_w\plus{}A_b}\plus{}\frac{B_w}{B_w\plus{}B_b})$, where $ A_b\plus{}B_b\equal{}A_w\plus{}B_w\equal{}50$.[/hide]

Now we just need to maximize the sum of two fractions given two sums.

If no more limitants are given I would do something like this:

Put $ 25$ black tokens in each container fisrt, then covering all the black tokens you put $ 25$ white tokens in each container, so basically you can always win. :rotfl:

 :P

OK, then assume the containers a shaken well :P


Does anyone  here once hear about Makoto Kobayashi,the Japanese physicist who got Nobel prize for the discovery of the origin of the broken symmetry ?
From what I've heard from my friends and some referential websites,there are two points  make me suprising about him
First,he is not good at English (but may be better than me  :D ) so he must express the Nobel thesis in Japanese .
and secondly,he have never travelled out of Japan until when he go out to get the Nobel prize.
Are all infomation above completely correct ?
I am very curious why he can get the Nobel prize without the ability of speaking English and exchanging knowledge with someone outside his country.
I am hope that my bad English won't make all nice friends here uncomfortable,
I had difficluties of writing in English and had to spend about one hour writing this so  I 'll appreciate all replies :)

A lot of people can read and write in English but have a hard time speaking it, especially in Japan.  But I have no idea whether that's the case here.

Not traveling outside Japan definitely does not imply not exchanging knowledge with anyone outside Japan. Especially with the internet, you don't have to be a globe trotter to be a good physicist. From my impression of Japan, there are enough smart people to collaborate with there. And physics and math are kind of a universal language. Not speaking English might make it harder to understand work that comes from American universities, but you could still use a translator or just look at the equations.

What makes you think that his inability to speak english might hurt his chances of getting Nobel prize?

That might turn out to be a drawback, if the testers don't know japanese, not vice versa, clearly. Plus, even if he doesn't know english at all, he could ask someone to translate his works or...well, you get the point, right?

Dang! Why isn't there a Nobel prize for Mathematics? All these fields especially  physics and to an extent both Chemistry, Computer Science, and Economics apply mathematical principles in order to discover some new result. I sometimes find that the mere notion of not having a Nobel prize for Math as being ludicrous, it utterly lacks logic (Whatever Nobel's will might say); I think it is high-time they include a Nobel prize for Math just like they have included for Economics.

Some potential reasons listed [url=http://en.wikipedia.org/wiki/Nobel_prize#Lack_of_a_Nobel_Prize_in_Mathematics]here[/url].

there is the Fields medal, which is only awarded to mathematicians under 40.  there are also the [url=http://en.wikipedia.org/wiki/Abel_Prize]Abel prize[/url] and the [url=http://en.wikipedia.org/wiki/Wolf_Prize_in_Mathematics]Wolf prize[/url].  Though young, I think that it is fairly safe to consider the Abel prize the mathematical equivalent of the Nobel prize.

Certainly, the people who award the Abel prize are hoping that it will be so-considered ;)

[quote="nkht-tk14"]Does anyone  here once hear about Makoto Kobayashi,the Japanese physicist who got Nobel prize for the discovery of the origin of the broken symmetry ?
From what I've heard from my friends and some referential websites,there are two points  make me suprising about him
First,he is not good at English (but may be better than me  :D ) so he must express the Nobel thesis in Japanese .
and secondly,he have never travelled out of Japan until when he go out to get the Nobel prize.
Are all infomation above completely correct ?
[/quote]

Wrong!

It would probably be helpful if, instead of just asserting that someone was wrong, you mentioned what part of what they said was wrong, what you believe the truth is, and perhaps some explanation of how you know that  :wink:

Sorry, the Physicist is Masukawa, he dislike learning English.

Any reason in particular?  Seems kind of stubborn.


Didn't there used to be a thread stickied which had a bunch of sites with AMC resources listed?  What happened to that thread?

It's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=amc+resources&t=8715]here[/url]; I removed the sticky because it was cluttering up the forum and nothing had been posted in it for a long time.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="michael.jackson_1958"]if $ %Error. "display" is a bad command.
\ G$ be a finite group and $ %Error. "display" is a bad command.
\ N\lhd\ G$ and $ %Error. "display" is a bad command.
\ P$ be a sylow p_subgroup of $ %Error. "display" is a bad command.
\ G$.then $ %Error. "display" is a bad command.
\ N\cap\ P$ is a sylow p_subgroup of $ %Error. "display" is a bad command.
\ N$and $ %Error. "display" is a bad command.
\dfrac{NP}{N}$ is a sylow p_subgroup of $ %Error. "display" is a bad command.
\dfrac{G}{N}$.[/quote]

Hello !

Proof of the first statement.
Let us choose a Sylow $ p$-subgroup $ T_{H}$ of $ N$. This is a $ p$-group, thus it is contained in a Sylow $ p$-subgroup of $ G$, in other words, it is contained in a conjugate of $ P$. Thus, there exists an element $ g$ such that $ gT_{H}g^{-1}$ is contained in $ P$. Since $ N$ is normal, $ gT_{H}g^{-1}$ is contained in $ N$ and thus in $ P \cap N$.
Now, $ gT_{H}g_{-1}$ has the same order as $ T_{H}$, thus is a $ p$-subgroup of Sylow of $ N$. Since $ P \cap N$ contains this $ p$-subgroup of Sylow of $ N$ and is a $ p$-subgroup of $ N$, it must be equal to this $ p$-subgroup of Sylow of $ N$, thus $ P \cap N$ is a $ p$-subgroup of Sylow of $ N$.
So, your first statement is true.
(I take this proof in Bourbaki, Algebra, I, § 6, coroll. 3, b). I wonder if it is the simplest proof.)

Let us prove the second statement.
$ %Error. "display" is a bad command.
\dfrac{NP}{N}$ is isomorphic to $ %Error. "display" is a bad command.
\dfrac{P}{N \cap P}$, thus its order is a power of $ p$. Thus, it is sufficient to prove that the power of $ p$ in the order of $ %Error. "display" is a bad command.
\dfrac{NP}{N}$ is the same as in the order of $ %Error. "display" is a bad command.
\dfrac{G}{N}$.
The power of $ p$ in the order of $ %Error. "display" is a bad command.
\dfrac{NP}{N}$ is the quotient of the power of $ p$ in $ NP$ by the power of $ p$ in $ N$ and the power of $ p$ in the order of $ %Error. "display" is a bad command.
\dfrac{G}{N}$ is the quotient of the power of $ p$ in $ G$ by the power of $ p$ in $ N$.
Thus, all amounts to proving that the power of $ p$ in the order of $ NP$ is the same as in the order of $ G$. It is true, since $ NP$ contains $ P$ and the power of $ p$ in the order of $ P$ is the same as in the order of $ G$.

Friendly,
P.


function
induction
algebra unsolved
algebra

Let function $\large f: N^{*} N^{*}$ satisfying:
$\large f(f(n))+f(n)=2n$
Prove that:
$\large f(n)=n$


floor function
modular arithmetic
number theory solved
number theory

The characteristic equation is $x^2-2x-1 = 0$, which after solving yields $a_n = \frac{1}{2 \sqrt{2}}{(1+\sqrt{2})}^n - \frac{1}{2 \sqrt{2}}{(1-\sqrt{2})}^n$, and which after simplifying gives $a_n = {n \choose 1} + 2{n \choose 3} + 2^2 {n \choose 5} + \ldots + 2^{\lfloor \frac{n-1}{2} \rfloor} {n \choose 2\lfloor \frac{n-1}{2} \rfloor + 1}$.

If $2^k \mid n$, then since ${n \choose i} = \frac{n}{i} {n-1 \choose i-1}$, for odd $i$, we have ${n \choose i} \equiv 0 \pmod{2^k}$ and we are done.

As for the case of $2^k \mid a_n$, by considering $\mod 2^{i+1}$ for the equation $a_n = {n \choose 1} + 2{n \choose 3} + 2^2 {n \choose 5} + \ldots + 2^{\lfloor \frac{n-1}{2} \rfloor} {n \choose 2\lfloor \frac{n-1}{2} \rfloor + 1}$, if $2^i \mid n$, and if $2^{i+1} \mid a_n$, then $2^{i+1} \mid n$. Now, $2^0 = 1$ obviously divides $a_n$, so inductively, we have $2^k \mid a_n \Rightarrow 2^k \mid n$, and we are done.


symmetry
geometry
geometric transformation
reflection
trapezoid
combinatorics unsolved
combinatorics

Find the smallest $n \in \mathbb{N}$ such that if any 5 vertices of a regular $n$-gon are colored red, there exists a line of symmetry $l$ of the $n$-gon such that every red point is reflected across $l$ to a non-red point.

Call "sequence" the way the colored points appear, reverse the sequence, if we put the reversed sequence over the points of the n_agon starting anywhere, and it doesnt overlap the colored points, then we may reflect the n-agon by the line  perpendicular to the midpoint of the line joining the starting points of the original sequence and the reversed one.

So what we have is that there is a starting position for the reversed sequence that doesnt overlap our original sequence. Every point of our original sequence overlaps with $5$ reversed sequences, so we have at most $25$ overlaps, more over, every pair of points of the original sequence have a common overlap, so we have at most $25-10=15$ overlaped  reversed sequences, so we get for $n>15$ a non overlaping reversed sequence!

Now, number the vertices from $0$ to $14$ clockwise, and put colored points in vertex: $0,1,3,6,10$ it is easy to see the overlaping arguments gives equality here.

your example for $ n\equal{}15$ is wrong. take the symmetry line that passes from the vertex $ 7$.
i can't find an example for $ n\equal{}15$. i think there isn't any. if somebody with a full solution please write it.

Two different red points "eliminate" one line from the candidates to be the line of symmetry (the one that reflects one of them to another) and every point eliminates one line (the one that the point lies on). So the number of eliminated lines is $ \binom{5}{2}\plus{}5\equal{}15$. So if we had the solution for $ n\equal{}15$ we must have chosen each line once.
  Now consider the "sequence" of points and the sequence of the distances between them (the distance is the minimum of the number of points we have to go though going clockwise or counterclockwise).  If any two pairs of different (in a pair) points have the same distance we have a trapezoid (or a triangle) and we choose a line twice, so no two distances can be equal. We have $ \binom{5}{2}\equal{}15$ distances and the maximal distance is 7, the minimal is 0, so we have 2 equal distances. Therefore we choose a line twice, thus we do not choose some line and this line is the line of the symmetry.

my answer is $ 17$ but i can't prove it, i tried it a few times and i saw it is not valid for 15, nor 16. And could you please give a full solution to this problem, folks?

[quote="Pascual2005"]Call "sequence" the way the colored points appear, reverse the sequence, if we put the reversed sequence over the points of the n_agon starting anywhere, and it doesnt overlap the colored points, then we may reflect the n-agon by the line  perpendicular to the midpoint of the line joining the starting points of the original sequence and the reversed one.

So what we have is that there is a starting position for the reversed sequence that doesnt overlap our original sequence. Every point of our original sequence overlaps with $ 5$ reversed sequences, so we have at most $ 25$ overlaps, more over, every pair of points of the original sequence have a common overlap, so we have at most $ 25 \minus{} 10 \equal{} 15$ overlaped  reversed sequences, so we get for $ n > 15$ a non overlaping reversed sequence!

Now, number the vertices from $ 0$ to $ 14$ clockwise, and put colored points in vertex: $ 0,1,3,6,10$ it is easy to see the overlaping arguments gives equality here.[/quote]
I don't think so. Please examine.

Please read my previous post before reading this.
Number the points in the clockwise direction. Now I will write the 5 points for with the thesis is false.
For n=10, numbers are 0,1,4,7,8.
For n=11, numbers are 0,1,3,4,9.
For n=12, numbers are 0,1,3,5,6.
For n=13, numbers are 0,1,5,8,12.
For n=14 we have 14 symmetry lines. We take the distances between adjoining points and the points having a common adjoining point. There are 10 distances such and the maximal distance is 7 and minimal is 1. So there are 3 pairs of the distances of the same length, so there at least 2 lines are eliminated twice. As we eliminate 15 lines and we eliminate 2 lines twice, so we really eliminate $ 15\minus{}2\equal{}13$ lines, so we don't eliminate some line and this line is the line of the symmetry which we were looking for.

If you have any questions please write.

[quote="polarb1"]Two different red points "eliminate" one line from the candidates to be the line of symmetry (the one that reflects one of them to another) and every point eliminates one line (the one that the point lies on). So the number of eliminated lines is $ \binom{5}{2} \plus{} 5 \equal{} 15$. So if we had the solution for $ n \equal{} 15$ we must have chosen each line once.
  Now consider the "sequence" of points and the sequence of the distances between them (the distance is the minimum of the number of points we have to go though going clockwise or counterclockwise).  If any two pairs of different (in a pair) points have the same distance we have a trapezoid (or a triangle) and we choose a line twice, so no two distances can be equal. We have $ \binom{5}{2} \equal{} 15$ distances and the maximal distance is 7, the minimal is 0, so we have 2 equal distances. Therefore we choose a line twice, thus we do not choose some line and this line is the line of the symmetry.[/quote]

Sorry. Actually we have only 10 distances (we consider only distances between adjoining points and points having a common adjoining point), but it doesn't matter.

This problem seems much easier if you think about it in terms of numbers rather than reflections.

If you identify the vertices of an $ n\minus{}$gon with the residues mod $ n$, then the reflections of the $ n\minus{}$gon take the form $ f_a(x)\equal{}a\minus{}x$ for some $ a$.  A reflection maps the red point $ x$ to the red point $ y$ iff $ x\plus{}y\equal{}a$.  So we can recast the question as follows:

"What is the smallest $ n$ such that there is not a subset $ S$ of 5 residues modulo $ n$ such that every $ a$ can be written as the sum of two residues in $ S$?

The example above shows that $ S$ exists for $ n\equal{}13$.  If $ n\equal{}14$, then there must be 7 odd residues, each of which must be the sum of an odd residue in $ S$ and an even residue in $ S$.  But there are at most 6 choices of an odd and an even residue, so this is impossible and we are done.

For all $n$ up to $13$ examples are possible and they are trivial to find.For $n=14$ we number points with $1, 2 ... 14$ in clockwise order.We call line of  symmetry l good if it sends one of red points to one of red points and bad otherwise.For every pair of red points there can exist exactly one good line.In case $n=14$ we have $7$ lines of symmetry and if we have at most $6$ good lines so there exist one bad line  and we are done.Now if two red points have numbers with same parity then there does not exist good line for these two points(trivial).Number of good lines(that can exist) is $10- (aC2)-((a-5)C2)$ where a is number of red points on with even number and $aC2$ is "from $a$ we chose $2$ "(idk how these goes in latex).Now $10- (aC2)-((a-5)C2)<=6$ (min when $a=2$ or $a=3$) so there is a bad line and we are done .


I have an unfair coin that lands as head with probability of $ \frac{2}{3}$. If I flip the coin 5 times, what is the probability that I get exactly two heads?

Not sure:

The permutation method is 2/3 * 2/3 * 1/3 * 1/3 * 1/3 = 4/243. Then I think you mutliply by 5C2, so the answer is 40/243 (maybe).

you have 5 flips and 2 to be heads so that is 5C2=10. The probability for getting 2 heads, 3 tails is 2/3x2/3x1/3x1/3x1/3 so this equals 40/243

Thanks.

Ben rolls four fair 20-sided dice. What is the probability that exactly two of the dice show an even number?

As above, we have $ \dbinom42\frac1{2^4}\equal{}\boxed{\frac38}$.

I flip 6 coins.  What's the probability that I get exactly 3 heads?

Please stop posting these problems, AIME15. We all understand the solution, and there is no need for us to plug and chug even more. Thanks.


symmetry
combinatorics proposed
combinatorics

Let d(n) be the number of partitions of a set of n elements (put d(0)=1).
Prove that $ d(n)\equal{}\sum_{k\equal{}0}^{n\minus{}1} {{n\minus{}1}\choose k} d(k)$

It is straightforward.
Let $ a$ be one of the elements. Just choose the subset which contains $ a$ (it has $ k$ elements, where $ 0\leq k \leq n\minus{}1$). Then, you only have to consider a partition of the remaining set, which contains $ n\minus{}1\minus{}k$ elements.
Then, you only have to change the variable in the sum ($ k'\equal{}n\minus{}1\minus{}k$) and use the symmetry of the binomial coefficients.

Pierre.

Good, that's my own solution  :wink:

The second step is to find a closed form expression for d(n), but I'm afraid it is not an "Olympiad Section" problem  :lol:


Find all polynomial $ P(z)$ such that
$ P(z)P(z\minus{}1)\equal{}P(z^2)$
for all $ z\in C$

[hide="solution"]
$ P$ has finite degree, but if $ P(z)\equal{}0$ then $ P(z^2\equal{}0)$. So $ P(z)$ has only $ 0$ as a root or is a constant. If it only has $ 0$ as a root then $ P(z)\equal{}z$ but that fails to satisfy the equation. If it is a constant then we have $ c^2\equal{}c$ implying $ c\equal{}0$ or $ c\equal{}1$.
[/hide]

Actually, Watson, the "If $ P(z) \equal{} 0$ then $ P(z^2) \equal{} 0$" just reduces the possible zeros to the roots of unity and zero.

Indeed.  Posted before.

[hide="Solution"] Nonconstant $ P(x)$ are clearly monic.  Write $ P(z) \equal{} \prod_{i\equal{}1}^{n} (x \minus{} r_i)$.  The given equality tells us $ \prod_{i\equal{}1}^{n} (x \minus{} r_i)(x \minus{} (r_i \plus{} 1)) \equal{} \prod_{i\equal{}1}^{n} (x^2 \minus{} r_i)$; in other words, the square roots of the roots are the roots themselves plus the roots plus one.  As Mathnerdmo observed, the first condition is only satisfied by the roots of unity.  The only roots of unity such that $ r_i \plus{} 1$ are also roots of unity are the primitive third roots of unity, so the only nonconstant solution is $ P(z) \equal{} z^2 \plus{} z \plus{} 1$. [/hide]


$ A$ is a subset of$ X$, topological space

 let$ f: A \to Y$is continuous where $ Y$ is Hausdorff Space

then denote continuous function $ g$ by $ g: \bar{A} \to Y$ determined by expanding $ f$

Show that $ g$ is uniquely determined by $ f$

Let $ f,g: X\rightarrow Y$ be continuous maps between topological spaces where $ Y$ is Hausdorff.  The set $ \text{Eq}(f,g)\equal{}\{x\in X\mid f(x)\equal{}g(x)\}\equal{}(f,g)^{\minus{}1}(\Delta)$ is a closed subset of $ X$ since it is the preimage under the continuous map $ (f,g): X\rightarrow Y\times Y$ of the diagonal $ \Delta\equal{}\{(y,y)\in Y\times Y\mid y\in Y\}$, which is closed since $ Y$ is Hausdorff.  Then if $ f$ and $ g$ agree on a subset $ A\subseteq X$, then they also agree on $ \bar{A}$ since $ A\subseteq\text{Eq}(f,g)\Rightarrow\bar{A}\subseteq\text{Eq}(f,g)$.  In particular, if $ f$ and $ g$ agree on a dense subset of $ X$, they are equal as any continuous map into a Hausdorff space is determined by its values on a dense subset.


geometry
incenter
ratio
angle bisector
geometry proposed

[size=110][color=darkred]Let $\triangle ABC$ and the point $D\in (BC)$ . Denote the incircles $w_1=C(I_1,r_1)$ and $w_2=C(I_2,r_2)$ of the triangles

 $ABD$ and $ACD$ respectively. Prove that the quadrilateral $BI_1I_2C$ is cyclic $\Longleftrightarrow \frac{DA+DB}{DA+DC}=\frac{AB}{AC}$.[/color][/size]

Here is my solution ( !? ):

  Incenter I1 divides angle bisector of ADB in ratio

    ( DA + DB ) / AB  = DI1 / I1X       where X is the trace of DI1 on AB

    similarly

    ( DA + DC ) / AC = DI2 / I2Y        Y is the trace of DI2 on AC

    Angle bisector from B in triangle BDX  divides DX proportionate to sides

    DI1 / I1X =  DB / BX  ,  similarly  DI2 / I2Y =  DC / CY

    For points B, I1, I2, C to be concyclic, the  line from X into  trace of Y onto BC

    has to be parallel to I1J2,   where J2 is trace of  I2  onto  BC.

   The same for line from Y into trace of X has to be parallel to I2J1.

   This makes DC / CY = DB / BX    and now

   ( DA + DB ) / AB  =  ( DA + DC ) / AC

    and  the problem is solved.

   Thank you.

    M.T.

[size=120][color=darkred]Let $\triangle ABC$ , the point $D\in (BC)$ and the incircles $w_{1}=C(I_{1},r_{1})\ ,\ w_{2}=C(I_{2},r_{2})$ of $\triangle ABD$ ,

$\triangle ACD$ respectively. Prove that the quadrilateral $BI_{1}I_{2}C$ is cyclic $\iff \frac{DA+DB}{DA+DC}=\frac{AB}{AC}$.[/color][/size]

[size=110][color=darkblue][b][u]Proof.[/u][/b] Denote $M\in AB\cap I_{1}I_{2}$, $N\in AC\cap I_{1}I_{2}$, $P\in MN\cap AD$. The quadrilateral $BI_{1}I_{2}C$ is cyclically $\iff$ $\widehat{MI_{1}B}\equiv\widehat{DCI_{2}}$

($\widehat{CI_{2}N}\equiv\widehat{I_{1}BD}$).  Therefore $\widehat{AMN}\equiv\widehat{ANM}$, $AM=AN$, $\frac{I_{1}P}{I_{1}M}=\frac{AP}{AM}=\frac{AP}{AN}=\frac{I_{2}P}{I_{2}N}$,  i.e. $\frac{I_{1}P}{I_{1}M}=\frac{I_{2}P}{I_{2}N}$ what means

$\frac{PM}{I_{1}M}=\frac{PN}{I_{2}N}$. Denote $h_{1}=\delta_{AB}(D)$, $h_{2}=\delta_{AC}(D)$ and $d_{1}=\delta_{AB}(P)$, $d_{2}=\delta_{AC}(D)$ where $\delta_{d}(X)$- the distance of $X$ to $d$.

Thus, $\frac{h_{1}}{h_{2}}=\frac{d_{1}}{d_{2}}$ and $\frac{d_{1}}{r_{1}}=\frac{MP}{MI_{1}}=\frac{NP}{NI_{2}}=\frac{d_{2}}{r_{2}}$ $\Longleftrightarrow$ $\frac{d_{1}}{d_{2}}=\frac{r_{1}}{r_{2}}$ $\Longleftrightarrow$ $\frac{h_{1}}{r_{1}}=\frac{h_{2}}{r_{2}}$ $\Longleftrightarrow$ $\frac{ch_{1}}{cr_{1}}=\frac{bh_{2}}{br_{2}}$ $\Longleftrightarrow$ $\frac{[ABD]}{cr_{1}}=\frac{[ADC]}{br_{2}}$ $\Longleftrightarrow$

$\frac{c+AD+BD}{c}=\frac{b+AD+DC}{b}$ $\Longleftrightarrow$ $\boxed{\ \frac{DA+DB}{c}=\frac{DA+DC}{b}\ }\ .$[/color][/size]

I believe we can use the fact that a quadrilateral $ABCD$ is cyclic if and only if $\frac{a-b}{c-b}: \frac{c-d}{a-d}\in\mathbb{R},\ \frac{a-b}{c-b}, \frac{c-d}{a-d}\in\mathbb{C}/\mathbb{R}$, i.e. cross-ratio.

This is mine .
The reverse results easely from my proof .
And I denoted v=(1/2)*<ABC and t=(1/2)*<ACB .
I also denoted I_1=E and I_2=F .

Dear Virgil and Mathlinkers,
can you please precise your reference about Crux...
Thank in advance.
Jean-Louis

Dear Mathlinkers,
the problem being solved, the question is:

construct the point D. Who want to begin...

Sincerely
Jean-Louis


Is there somebody in this forum selected for INAO Junior 2008? Can somebody throw some light w.r.t questions asked in this exam  :?:

Regards

A 100 marks exam
= 20 objective questions(2 marks each) + 6 problems (10 marks each)

The MCQs are from physics and mathematics till XIth(easier parts)
The long answers are mostly visualisation type questions and are quite easy...

Ex,
How much teapowder is consumed in Mumbai every day??
(you get credit for your ability to make appropriate assumptions validating your answer...)

How fast should an aircraft travel such that it will always see sunset??
What is the minimum number of mirrors required in a box to make it look as if the light has not undergone any change in path in the box i.e. it comes out in the same direction as it entered in..?

Two-three physics questions which are in elementary IX,X,XI CBSE textbook...
How long should a 60 W bulb of 10% efficiency glow to melt the ice on the Himalayas...??

1 mathematic question:
Ex,
find all x,y,z such that x+y+z=xyz
find all integer coefficient polynomials which pass through three given points...

best of luck.

P.S. [b]You don't need to have knowledge of astronomy for NSEA. [/b]That is exactly why you learn lots and lots of things in the camp. this is the only olympiadwhere you learn far more from the camp than you ever learnt by yourself in your life...because astro is not a part of school curriculumIt is an easy exam but the competition is also stiff and cut-off is generally about 65-70/100...

and yes there will be a question to find the units place of $ (XYZSWEWSWF)^{kjsfgajf}$
where all the alphabets are for  some necessary variables .the answer will be 1 :D

[quote="SOUMYASHANT NAYAK"]

Ex,
How much teapowder is consumed in Mumbai every day??
(you get credit for your ability to make appropriate assumptions validating your answer...)

[/quote]

Errr....I'm sorry but what is the relevance of this question wrt to Astronomy?? You guys pulling that poor kid's leg or what?

[quote="pardesi"]no shreya snot at all ...
i hope i had such questions in my 10th class senior paper...where we had terrbley difficult questions(they seemed difficult at 10th)
last year we had this question in our senior paper  :)[/quote]

Okay but how is this related to astronomy??

[quote="SOUMYASHANT NAYAK"]

P.S. [b]You don't need to have knowledge of astronomy for NSEA. That is exactly why you learn lots and lots of things in the camp. this is the only olympiad where you learn far more from the camp than you ever learnt by yourself in your life...because astro is not a part of school curriculum[/b] It is an easy exam but the competition is also stiff and cut-off is generally about 65-70/100...[/quote]

i hope you read this.. :P

actually it tests ur creativity and how well and logically u can approximate(the best answer for an unexplained thing goes to pardesi  :D )

i wrote the inao last time, what answer did ya get guys for the problem
How much teapowder is consumed in Mumbai every day??
if i remember correctly my answer was approx 2 lac.

My answer was 18 million g

As I said, the answer is the least important thing...
(the justification of your answer is more important..)

I got 8/10 in that question..

In reply to rnmanu's message,

Hi !
I am [b]Karan Singh[/b] from [b]Bhubaneswar[/b].
And surely enough i am scribling here because i am selected thru NSEA and now into INAO.

And sadly because, there's no materail ( book / previous QPs ) for preparation.

Furthermore, the kind of questions i heard here are even more depressing because my answers don't match with answers here.

MY ANSWER:

Population of mumbai : 12 million = 12 x ( 10^6 )

Say out of this, say , 1 in every 4 persons do not drink it for ,
         say health problem.
         or drink coffee ( cold / hot )
         or cold drinks.
So 3 out of every 4 people drink tea.

Mathematically, that is 12million x 3 / 4 = 9million

Say 3 million people drink tea 1 time a day.
and 6 million drink it 2 times a day.

So no of tea cups a day = 3million + ( 2 x 6million )
                                   = 15 million

No of teaspoons required for 1 cup = 1
Mass of 1 teaspoon of tea = 4 g

So mass of tea consumed daily in mumbai = 15million x 4 g = 60 million g = 60 thousand kg = 60 tonnes

ANSWER that i get is [b]60 TONNES[/b] a day.

Is that correct ???
( i believe no. )

soumya has already clearly stated taht the answers don't matter just the procedure


function
trigonometry
algebra unsolved
algebra

[quote="marsupilami"][b]Problem 1:[/b]
Determine all functions [b]f[/b] satisfying:
1. $ x_1 > x_2 \Rightarrow f(x_1) < f(x_2)$    $ \forall x\in[0,2\pi]$
2. $ f(x) \geq sinx$    $ \forall x\in[0,2\pi]$

[b]Problem 2:[/b]
Determine all functions [b]f[/b] satisfying:
1. $ x_1 > x_2 \Rightarrow f(x_1) < f(x_2)$    $ \forall x\in[0,4]$
2. $ f(x) \geq 4x^3 \minus{} 3x \minus{} 1$    $ \forall x\in[0,4]$[/quote]
Huh ????
Are you sure the problem is correct ?

There are obviously infinitely many solutions and I dont think there is a general form.
For example, for the first problem : any decreasing function greater than $ 1$ over $ [0,2\pi]$ is solution :

$ 1 \plus{} \frac 1{x^2\plus{}1}$

$ 2 \minus{} \sin(\frac x4)$

$ 1 \plus{} \sqrt {2\pi} \minus{} \sqrt {|x|}$

... and a lot of others ...


AMC AIME USA(J)MO USAMO calculus integration

What do you think about cheating in a math olympiad? I've heard people that save formulas in their calculators, and things like that, or they buy some calculators in order to cheat.............. what's your opinion?

My opinion: Good mathematicians don't need calculators or formulas or tricks to cheat.... they just need their mind.

I never knew calculators were allowed into math olympiads.

Umm calculators aren't allowed in math olympiads... :huh:

[quote="Treething"][quote="math92"]One of my classmates puts programs on his calculator that deals with the test.  He also puts formulas and stuff on it to... :([/quote]

sounds like me[/quote]
lol, but my classmate is a real calculator geek and takes his calculator everywhere he goes.  He took it to Egypt on vacation, he takes it to lunch at school, and he keeps it in his jaket pocket.

I don't understand this thread at all. Usually, math olympiad problems can't be solved with a calculator since most of the problems are generalizations anyway.  :?

Besides repeating the fact that you can't use calculators on most olympiads... I don't think cheating is a problem. At least on the levels at which cheating can take place (like AMC, AIME, USAMO) you're not going to get far beyond that if you don't really know the stuff. And they can obviously tell if you know how to solve problems. So at least in the US cheating at the USAMO level won't get you far.

As Iversonfan2005 and paladin8 mentioned, using calculator(s) in math olympiad does not make a big difference. The only problem is, though, that you can put tons of formulas into your calculator, which [i]might[/i] help you.

I think that this calculator issue is more important in smaller math contests. In some math competitions, having a powerful calculator has a large influence on your score. Some problems become very trivial if you have a powerful calculator like TI-89. I don't think this is very fair.

[quote="José"]Please, suppose that you CAN use calculators, like in Argentina...[/quote]

A good math olympiad problem should not be solvable with a calculator so cheating shouldn't be a big deal, but frt makes a good point about smaller math contests, like the AMC. A TI-89 (though I have one) trivializes a decent number of problems, especially on calculus-based tests (basically any elementary derivative/integral). Personally I think calculators should just not be used on contests - the only real thing you need a calculator for is like graphing and brute arithmetic, stuff that is not necessary on contests.

All right, nobody's answer me, so I'll change the question: What about cheating at ANY exam???? maybe a school one

I thought I [i]did[/i] answer your question... Anyway, as I mentioned in my previous post, some problems are completely trivialized if you use your calculator. I basically believe that teachers shouldn't allow students to use any calculator during quizzes, tests, etc. The ability to think mathematically has nothing to do with the ability to use calculators.

In Japan, we aren't allowed to use any calculator at all. In fact, no one (including teachers) has any calculator because we don't need it. When I came to the U.S., I was surprised by the fact that everyone had their own calculator. I had never seen a calculator that could plot graphs, operate matrices, find the numerical values of derivatives/integrals, and even store programs (and games!).

Here in our Math Olympiads we only use common calcs, nothing about integrals or graphs, just +, -, *, /, a^b and sqrt [a] {b}, nothing else

I think that it depends on the contest.  If they say that you aren't allowed to put programs into your calculator, then obviously its wrong.  However, if they don't say anything about it, it isn't cheating.   Many SAT prep books even encourage students to program formulas and such into their calculators.  But it doesn't really matter, because nothing that you could put on a calculator could ever overcome the elegan solution on an olympiad style test.


How many groups( up to isomorphism) are there of order $429=3.11.13$ , and describe them.

Using Sylow theorem and the fact that the number of groups of order $3$ for instance must divide $11.13$, you can show that $G$ has a unique $p$-Sylow for every prime number $p=3,11,13$. From there you know that those $p$-Sylows are in direct sum, and then $G$ is cyclic with order $429$.

That's not everything- the subgroup of order 39 can be nonabelian.

Why is there only one subgroup of order three, is thirteen not an option too? :o

Could u exhibit more? By Sylow theorem... How?
Thx in advance.


Chord $ \left[AD\right]$ is perpendicular to the diameter $ \left[BC\right]$ of a circle. Let $ E$ and $ F$ be the midpoints of the arcs $ AC$ and $ CD$, respectively. If $ AD\bigcap BE\equal{}\left\{G\right\}$, $ AF\bigcap BC\equal{}\left\{H\right\}$ and $ m(AC)\equal{}\alpha$, find the measure of angle $ BHC$ in terms of $ \alpha$.

$\textbf{(A)}\  90{}^\circ \minus{}\frac{\alpha }{2} \qquad\textbf{(B)}\  60{}^\circ \minus{}\frac{\alpha }{3} \qquad\textbf{(C)}\  \alpha \minus{}30{}^\circ \\ \qquad\textbf{(D)}\  15{}^\circ \plus{}\frac{\alpha }{2} \qquad\textbf{(E)}\  \frac{180{}^\circ \minus{}2\alpha }{3}$

Review the statement. As it is, $ H\in BC$, so the angle is $ 180^\circ$.

You're right. It asks for $ BHG$ not $ BHC$.

[hide]Denote $ K \equal{} BE\cap AF$. By the initial conditions, $ \angle ABE \equal{} \angle EBC \equal{} \angle CBF \equal{} \angle FBD \equal{} {\alpha\over 4}$, and $ \angle AFB \equal{} {{180^\circ \minus{} \alpha}\over 2} \equal{} 90^\circ \minus{} {\alpha\over 2}$. Since $ \angle EBF \equal{} {\alpha\over 2}$, from the $ \triangle KBF$ we get $ {BK\perp AH}$. But $ AD\perp BH$ by the construction, hence $ G$ is the orthocenter of $ \triangle AHB$, so $ HG\perp AB$. If $ L\equal{}HG\cap AB$, from $ \triangle LHB$ we get $ \angle BHG \equal{} 90^\circ \minus{} {\alpha\over 2}$, so the answer is $ A$.[/hide]


Prove that for arbitrary positive integers $x,y,z$,the form $4xyz-x-y$is not a perfect square.

[quote="spider_boy"]Actually the equation $4mn-m-n\not=k^2$ is EULER's equation.
I solved it last year.[/quote]
the case $z=1$ is easy,but here $z$ is an arbitrary integer.

[quote="pavel kozlov"]This problem is solved with help of Jacobi-symbols.[/quote]

How to use it?Show your solution,please.

I mean this
[quote="suweijie"]Prove that for arbitrary positive integers $ x,y,z$,the form $ 4xyz \minus{} x \minus{} y$is not a perfect square.[/quote]


im using texnicenter as latex editor but when i build the program, nothing appears in the pdf output.

[quote="stevem"]You need to give much more information than that. Are you using LaTeX=>PDF to compile? Is this true for any document or just one? If the latter, what is the content? Are there any error messages? What is in the log file? What does the log say when you compile
[code]\documentclass{article}
\begin{document}
Hello world
\end{document}[/code][/quote]

Yes, I am using Latex => PDF and it happens to all documents. Well, i found a way to remedy it. I used DVI and the Print it to Adobe PDF. Thanks anyway.


So what did you get on Round 2?
Vote 19-20 if you got an 18  :blush:

I got:

25/25

#2 Commended

[quote="calc rulz"][quote="mustafa"]15354

 :mad:

And then to add insult to injury, there isn't even an option for my 18 in your poll[/quote]

Oops, sorry  :blush:[/quote]

You should be sorry :mad: . You crushed my spirit :noo:


[quote="churchilljrhigh"]What is maml?[/quote]

MAML = Massachusetts Association of Mathematics Leagues; website is http://www.maml.org; it seems to be down right now.

That was the old website. It hadn't been updated at all this year (or even midway through last year I think).

yeah, that was the old website, but does anyone know of a new website?? (the last time that was updated was after the maml2 results from last year... june 2003 that is)

i thought the test was pretty cool... here's one of the problems...

2.) a. Prove that among any 10 consecutive positive integers there is at least 1 that is not divisibe by any of {2,3,5,7}.

b. Prove that among any 14 consecutive positive integers there is at least 1 that is not divisible by any of {2,3,5,7,11}.

c. Find 13 consecutive positive integers, each of which is divisible by at least one of {2,3,5,7,11}.

I impressed myself and actually got those....

here's another..
3.) a. let a and b be positive real numbers. Prove that...

(a^2 + b^2) / 2 >= ab

b. let a,b,c,d be positive real numbers with ab<cd.
Prove:
2ab < bc + ad.

i found part a to be really easy but couldn't get part b.

for anyone who took it, i think my score was roughly 75, what about u guys?

For 3b, here's a hint that may point you in the right direction.  Spoiler: [hide]First try to prove it given the equality condition ab = cd (relaxing the final inequality from < to :le:).[/hide]


Can anyone put me in touch with Alabama's ARML coach?

or does anyone have the following information for Alabama's team:

1) How many people fly to ARML
2) Where are they flying from?
3) Where are they flying to?
4) Current rates obtained
5) Current airlines used
6) Whether any special negotiated rate is already being used. (and if so, any
contact(s) that you can share would be greatly appreciated).

Thanks,
Jack

sup

I think the ARML coach is either Kay Tipton or Todd Taylor, both at Vestavia....it's probably your best bet to send an email to Tipton asking her...

as for the other Q's;;

1 - 15-20ish...depends on how many people pay their way as alternates. Minimum of 16, I think.

2 - BHM airport

3 - Probably las vegas airport, since we're goign to the vegas site.

4-5) dunno

6 - people who qualify for the team get a reduced rate, and everyone else pays their own airfare..

so you payin your way?? I dont' remember a Jack being on the qualifying lists that they made up, but I haven't checked them in a while, so eh

I'm curious why you're interested in Alabama's team as it seems you're involved with the NY team. Anyway... Mrs. Tipton's email address is tiptonkw at vestavia . k12 . al . us, and she will know about airline stuff.

Hi, are we leaving from Birmingham? I was told that we are all traveling to the Atlanta airport and leaving from there. Change of plans?

actually, you might be right, I'm just remembering last year we went through BHM...but it does make sense to go through ATL, I'm not even sure if BHM has a line to vegas


What is the coefficient of the term x^9 when (1 + x)(1 + x2)(1 + x3)(1 + x4)(1 + x5) is expanded.
Explain your steps.

$ (1 \plus{} x^4)(1 \plus{} x^5) \equal{} 1 \plus{} x^4 \plus{} x^5 \plus{} x^9$

We can see that there is no way for the $ 1$ to be made into an $ x^9$ after multiplication, so we can discard that.

Now we have $ x^4 \plus{} x^5 \plus{} x^9 \equal{} x^4(1 \plus{} x \plus{} x^5)$, so we can change our problem into finding the coefficient of the $ x^5$ term.

$ (1 \plus{} x^3)(1 \plus{} x \plus{} x^5) \equal{} 1 \plus{} x \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} \cdots$ (the rest is not important)

There is no way for the $ 1$ or $ x$ to be made into an $ x^5$, so we can ignore them.

$ x^3 \plus{} x^4 \plus{} x^5 \equal{} x^3(1 \plus{} x \plus{} x^2)$, so we can just find the $ x^2$ term.

$ (1 \plus{} x^2)(1 \plus{} x \plus{} x^2) \equal{} 1 \plus{} x \plus{} 2x^2 \plus{} \cdots$

$ (1 \plus{} x)(1 \plus{} x \plus{} 2x^2) \equal{} 1 \plus{} 2x \plus{} \boxed3x^2 \plus{} \cdots$

Edit:

Alternatively, note that $ (1 \plus{} x)(1 \plus{} x^2)(1 \plus{} x^4) \equal{} 1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^7$, and

$ (1 \plus{} x^3)(1 \plus{} x^5) \equal{} 1 \plus{} x^3 \plus{} x^5 \plus{} x^8$.

Now, multiplying these, we can see that if we use the $ 1$ term on from the second expansion,

there are no ways, and if we use any other term in the second expansion,

we can get $ x^9$ in one way, so the coefficient of $ x^9$ is $ \boxed3$.

Using a combinatorial method, we can see that this problem is the same as counting the number of ways for a combination of numbers from the set 1, 2, 3, 4, 5 (each used at most once) to add to 9..

We have:
1. 4 + 5 = 9
2. 2 + 3 + 4 = 9
3. 1 + 3 + 5 = 9

There are three ways, thus the coefficient is 3.


http://www.ioi2005.pl/news/IOI-Newsletter-Issue3.pdf
Page 2, Can-Moo in Poland. [color=red][edited by Mod][/color]

(August 24, 2005) Each of the four Canadians won a medal at the 2005 IOI in Nowy Sacz, Poland. Richard Peng received a gold medal for his 17th-place finish. A silver medal went to Simon Parent (62nd) while bronze medals went to Aidin Kashigar (85th) and to Qiyu Zhu (109th). Congratulations! The next IOI will be held from August 17th to 24th in Merida, Mexico.

AMAZING JOB GUYS!!! :D

[quote="TheDreamer"]
AMAZING JOB GUYS!!! :D[/quote]
Thanks...we were quite lucky (especially me) :blush:
So that's 2 golds, 3 silvers, 4 bronzes for Can-Moo this year....that MUST be a record. :moose: [color=red][edited by Mod][/color]

Richard!  Since when was it EVER Can MUU??!?!?!?!

Well I agree that Can-moo is more reminding of a cow than a moose  :ninja:


The text i follow mention that if $ A$ is a closed and bounded set then any infinite subset of it $ A$ must have a limit point in S.
It also gives a proof of the statement using heine-Borel theorem.
i cant understand the flaw in this argument of mine regarding this claim.My argument is since A is a subset of S any limit point of A must also be a limit point of S since S is closed S must contain all its limit points so infact what one ends up proving is all limit points of A are in S.
What is the flaw in the arguement?

Please state again what is your problem. What are $ A$ and $ S$? Which is closed and bounded? Do we have $ A \subseteq S$? The sequence has its terms in $ A$, but we look for an accumulation point in $ S$?


[color=purple] [b] 1[/b] Let triangle ABC and M,N,P move in [BC],[CA],[AB].Prove that
  $abMN^{2}+bcNP^{2}+caMP^{2}\ge 4S^{2}$
  [b]2 [/b] Prove that
 $\frac{a^{2}{m_{a}}^{2}+b^{2}{m_{b}}^{2}+c^{2}{m_{c}}^{2}}{\sqrt{3}(a^{2}+b^{2}+c^{2})}\ge S$
  Kill it please :D [/color]


analytic geometry
LaTeX
rotation
trigonometry

Another question: Anyone seen a package that lets u create polar-co-ord graph paper.

[url=http://tug.org/PSTricks/main.cgi?file=Examples/polar]PSTricks[/url] allows the use of polar coordinates in LaTeX, but if you wish to just print plain polar graph paper then [url=http://pharm.kuleuven.be/pharbio/gpaper.htm]Graph paper printer[/url] is a free Windows program that will do so.

You can do this in [url=http://www.tug.org/metapost.html]MetaPost[/url].  For example:
[code]u := 0.5cm;
pickup pencircle scaled 0.5pt;
n := 6;
pair A;
A := (n*u,0);
for i=1 upto n:
 draw fullcircle scaled (2*u*i);
endfor;
for i=1 upto 6:
  A := A rotated 30;
  draw -A--A;
endfor;

%% a cardiod
path p;
p := (2*u,0);
for i=1 upto 360:
  r := 2 + 2*(sind(i));
  x := r * (cosd(i));
  y := r * (sind(i));
  p := p..(x,y)*u;
endfor;
pickup pencircle scaled 1pt;
draw p;
[/code]
produces the cardiod $r = 2 + 2 \sin\theta$ on a polar-graph-paper background:
[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/9ca707e4775aceb75f79d09cfc43c8ec.png[/img]

That is a nice graph.  Thxs for the help.

Nice graph Metapost looks really nice, I need to do some RTFM-ing on PSTricks and MetaPost. :)


Given that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{a\plus{}b\plus{}c}$

Prove that $ \forall$ odd $ n$

$ \left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)^n\equal{}\frac{1}{a^n\plus{}b^n\plus{}c^n}\equal{}\frac{1}{(a\plus{}b\plus{}c)^n}$

[hide]We have that
\begin{align*} \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{a + b + c} & = 0 \\
\frac {a + b}{ab} + \frac {a + b}{c(a + b + c)} & = 0 \\
\frac {(a + b)(b + c)(c + a)}{abc(a + b + c)} & = 0 \\
\end{align*}

Hence, one of $ a + b,b + c,c + a$ is equal to zero, and the result follows.[/hide]

timwu showed that one of $ a \plus{} b,b \plus{} c,c \plus{} a$ is zero.

for all $ n$ we have that
\[ \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^n \equal{} \left(\frac {1}{a \plus{} b \plus{} c}\right)^n \equal{} \frac {1}{(a \plus{} b \plus{} c)^n}
\]
without loss of generality assume $ a \plus{} b \equal{} 0$ with $ a,b\not \equal{} 0$.  it (should be) clear that $ (a \plus{} b \plus{} c)^n \equal{} a^n \plus{} b^n \plus{} c^n$ holds if $ n$ is odd, for then both sides of the equation are equal to $ c^n$, but this is not necessarily true if $ n$ is even (i.e., $ (a\plus{}b)^n \equal{} a^n \plus{} b^n \equal{} 0$ does not hold)

Oh, never mind.  I just realized that from
$ a^n \minus{} b^n \equal{} (a\minus{}b)(a^{n\minus{}1} \plus{} a^{n\minus{}2}b \plus{} a^{n\minus{}3}b^2 \plus{} ... \plus{} a b^{n\minus{}2} \plus{} b^{n\minus{}1})$, we get

$ a^n \plus{} b^n$ with $ a\plus{}b$ as a factor by replacing $ b$ by $ \minus{}b$ only when $ n$ is odd; that is what I was missing.

factorizations aren't even really necessary, if $ n$ is odd then $ a\plus{}b\equal{}0\implies a\equal{}\minus{}b\implies a^n\equal{}(\minus{}b^n)\implies a^n \equal{} \minus{}b^n \implies a^n\plus{}b^n\equal{}0$.


how do you report inapporite behavior in the classroom

Please report it to your teacher or assistant. This can be done via chat once the classroom is moderated or via Private Message.

who do we contact for inapporrite behavior in introchat

Please report it to your teacher via Private Message.

I often see people loitering in the classroom during free classroom time posting things like
"lalalalalalalalaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
And large blobs of LaTeX for no reason.
I understand that in some cases, they may be testing their LaTeX, because I do that sometimes, but I usually say "Testing LaTeX" before I post it. So is this normal? It is kind of a bother when we try to meet in the classroom to discuss something and we have to scroll down through 30 posts of thick LaTeX.
Am I supposed to be doing something about this?  :maybe:


Are the $ 3*3$matrices over $ \mathbb{C},\ \left(\begin{array}{ccc}1 &1 &0\\ 0 &1 &0\\ 0 &0 &1\end{array}\right)$ and $ \left(\begin{array}{ccc}1 &0 &0\\ 0 &\minus{}1 &0\\ 0 &0 &1\end{array}\right)$ similar?

I think theyt are not similar beacuse their det. are not equal...

Let's make the question more interesting.

Are $ \left(\begin{array}{ccc}1 &1 &0\\ 0 &1 &0\\ 0 &0 &\minus{}1\end{array}\right)$ and $ \left(\begin{array}{ccc}1 &0 &0\\ 0 &\minus{}1 &0\\ 0 &0 &1\end{array}\right)$ similar?

Invariant multiples of the first matrix are $ 1,\, (\lambda\minus{}1)^{2},\,\lambda\plus{}1$ and the second matrix's are $ \lambda\minus{}1,\,\lambda\minus{}1\,\lambda\plus{}1$. So they are not equal and hence these amtrices can not be similar.

Is there another way finding out the similarity of matrices?
 :maybe:

These matrix should have equal Jordan form, but actually they are already in Jordan form and they are different.So they aren't similar .Am I right?


When Roslyn writes her name repeatedly in a 5 by 5 table as shown, the bottom right corner contains the letter R. If she repeats the process in a 20 by 20 table, what letter would occupy the bottom right corner?

ROSLY
NROSL
YNROS
LYNRO
SLYNR

[hide=" :nhl: "]
Her name is 6 letters long.
Thus for the first row, she will write ROSLYNROSLYNROSLYNRO
Then the first letter on the next row is S.
Thus she will keep writing 2 letters up on the start of the next row.
In terms of the first letter in the row, the rows are
R
S
Y
continuing for 6 times.
That takes care of 18 rows.
then theres
R
S
with S being row 20.
Note that from the first row, the last letter in the row is 1 after the first.
Thus the answer is $L$[/hide]

[hide]Roslyn has 6 letters.
There are 400 squares in a 20 by 20 table.
$400\equiv 4 \pmod{6}$
The forth letter of Roslyn is $L$[/hide]


modular arithmetic
quadratics
number theory unsolved
number theory

Let $n$ be a positive integer. Prove that the number $2^n + 1$ has no prime divisor of the form $8 \cdot k - 1$, where $k$ is a positive integer.

It is $2^n+1$ and the problem has already been solved. It is direct consequence of formulae for Legendre symbol for 2.

I see: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3599.

[color=#FF0000][moderator edit: link edited.][/color]

The link that dieuhuynh provided does not work...
Here is a proof:

Let us suppose on the other hand that there exists $p$ such that $p|2^n+1$ and $p=7\pmod{8}$. Note that $\left(\frac{-1}{p}\right)=-1$ since $2\nmid\frac{p-1}{2}$ so $n$ cannot be even.
Hence $n+1$ is even so $2^{n+1}=-2\pmod{p}$ implies $-2=-1\cdot 2$ is a quadratic residue mod p i.e. $1=\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p^2-1}{8}}$ so $\frac{p-1}{2}+\frac{p^2-1}{8}$ is even.
Since $2|p-1$ and $8|p+1$, $\frac{p^2-1}{8}=\frac{(p-1)(p+1)}{8}$ must be even. Then $2|\frac{p-1}{2}$, a contradiction.

Assume the contrary, then $\left(\frac{-1}{p}\right)=-1$ and $\left(\frac{2}{p}\right)=1$, so
$$1=\left(\frac{2}{p}\right)^n=\left(\frac{2^n}{p}\right)=\left(\frac{-1}{p}\right)=-1,$$
that's a contradiction.


Last year, it said on the site that 7th to 10th graders that scored at least 9 on the AIME got to take the USAMO.  Do you think they'll do that this year too?

This year it should be 7 or 8. So anyone who isn't an upperclassman and scored a {Insert floor here}+ makes the USAMO. They've done that since 2000 I think.

i don't know....7 or 8 is tricky, because this year's AIME was pretty easy again. Then again, they ARE expanding...


Not that we know of.

Adobe Reader is [b]free[/b] at [url]http://www.adobe.com/products/acrobat/readstep2.html[/url]

There are perhaps other options...go to Google and search for "pdf reader"

I believe [url=http://www.cs.wisc.edu/~ghost/]Ghostscript/GSView[/url] (you'll need both - click "[url=http://www.cs.wisc.edu/~ghost/doc/AFPL/get814.htm]Obtaining AFPL Ghostscript 8.14[/url]" and "[url=http://www.cs.wisc.edu/~ghost/gsview/get46.htm]GSView 4.6[/url]") can be used to open PDF files.  I can't imagine why you can't use Adobe Reader, but if you can't, Ghostscript and GSView are a good bet.

...but if that doesn't work, maybe the site administrators could also post a JPG version of the file by taking a screenshot of the PDF and editing it with Paint.

[quote="JP"]if you do not have adobe reader, is there any other way to open up a USAMTS problems[/quote]

The LaTeX source file is now available on the same page as the PDF file, as a courtesy to those who want to submit their solutions in LaTeX.  You could copy the LaTeX file and process it into a dvi file on your own computer. Though I don't see why you would be able download a LaTeX processor onto a computer yet not be able to download Acrobat Reader. You would also need to download the relevant style files and erase the command to print the USAMTS logo, which is a PDF image.

If you are for some reason restricted to text formats, you could read the LaTeX file as plain text by skipping anything that starts with a backslash and ignoring all the dollar signs around numbers. But in that case the diagrams are very difficult to visualize.

Erin Schram
USAMTS grader

You can submit the entry form with your first set of solutions, which are due on October 4.

[quote="theone853"]Also, when are the entry forms due by becuase I just discovered them an hour ago. :blush:  :blush:[/quote]

[quote="theone853"]Would it be okay if I made a pdf with adobe acrobat 6.0 instead, so I don't have to Tex everything up?[/quote]
Sure.

You are [b]not[/b] required to submit your solutions in TeX.  You can handwrite your solutions if you wish.  If you do so, we ask that you write your solutions on copies of the following page (starting on a new page for each problem):

[url]http://www.artofproblemsolving.com/USAMTS/Downloads/USAMTS_solutions.pdf[/url]

Make sure that your name, username, and USAMTS ID # appear on [b]each[/b] page, along with the round and problem number (we're currently in Round 1).


This Math Jam has been postponed to next Wednesday due to power failures in our region resulting from Southern California wildfires...

Power failures? How come this site's OK?

hmm...I doubt he meant that the server was in trouble...just that his house sometimes has power failures...

Yes, the site is hosted elsewhere and yes, Richard is having power failure due to the fires.


Do you know how to make a fire extinguisher?
which can be made at home ?

Why would you want to make one at home?  It's pretty dangerous.

A common fire extinguisher has dry ice (solid carbon dioxide) in it - which is not very simple to get at home. However, a good fire extinguisher is sand.

if you can think of reactions where co2 is released and the reaction is fast, maybe we can make it

Well carbon dioxide is a product of standard combustion.  The matter is speeding it up, though.

:o i cant burn fuel for making fire extinguisher, its quite dangerous.
suggest  any other equation?

The problem is, burning almost anything is dangerous, as is making a fire extinguisher.  You'd want something you have the most control over.

hey i got a idea.should have got before
bicarbonat+acid    arrow      co2+........
so, now help in finding the fastes of these reactions. :roll:

The reaction of bicarbonates with acids is fast. Now you just have to collect the liberated CO2 and put it under pressure.

ok i 'll think of that, :|
suggest the bicarbonate and acid easily and cheaply available.
carbonate can be baking soda  and some acid thats not that strong
  hence, we can make a fire extinguisher at home .....amazing :D  :D

the above reaction would also get water. so, it would not be good in case of electrical fires.

But carbon dioxide is liberated as a gas, while water as a liquid.

if we wil squeeze co2 out with pressure, water may come along

And who talked about "squeeze co2 out with pressure"? If you want to take out any water that may go along CO2 then just pass the gas through a tube filled with magnesium sulphate, or some similar arrangement. Then, when you have collected the gas, you have to put under pressure. Did I made myself clear?

yes , you made yourself clear.
goud idea.
the main thing is pressure. we can put the setup in a bottle in case of water case. then pressure can be made and gas wuold be released when there is adequate pressure.
  however, in case of mgso4 how do we build pressure?


hello, i think there is no closed form for the solution of your equation.
Sonnhard.

Express a solution as $ x(y) \equal{} \ldots$. Maybe it's not easy but at least doable.


Can anyone prove that "if the sum of the angles subtended by two chords at the circumference is 180 degree.then the two chords are equal. :blush:  :blush: "For source see the solution of I.F.shyarigin problem 252.  :P


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$ . Prove that :

$ \frac{ab}{c\plus{}1}\plus{}\frac{bc}{a\plus{}1}\plus{}\frac{ca}{b\plus{}1}\le \frac{1}{4}$

The inequality is equivalent with:

$ \frac{ab}{a\plus{}b\plus{}2c}\plus{}\frac{bc}{b\plus{}c\plus{}2a}\plus{}\frac{ca}{c\plus{}a\plus{}2b}\le \frac{a\plus{}b\plus{}c}{4}$

Now using that $ \frac{1}{x\plus{}y}\le\frac 14\left(\frac 1x\plus{}\frac 1y\right),\ \forall\ x,y>0$ we have:

$ \frac{ab}{a\plus{}b\plus{}2c}\plus{}\frac{bc}{b\plus{}c\plus{}2a}\plus{}\frac{ca}{c\plus{}a\plus{}2b}\le$ $ \frac 14\left(\frac{ab}{b\plus{}c}\plus{}\frac{ab}{c\plus{}a}\plus{}\frac{bc}{c\plus{}a}\plus{}\frac{bc}{a\plus{}b}\plus{}\frac{ca}{a\plus{}b}\plus{}\frac{ca}{b\plus{}c}\right)\equal{}\frac 14(a\plus{}b\plus{}c)$.

[hide="Constantin"]Aceeiasi solutie am dat si eu in concurs  :lol: [/hide]


So...

Does anyone know how well you need to do on the USAMO to get into MOSP this year? How many people are they accepting?

I know seniors still have to place top twelve, but fortunately, I'm a junior.

[quote]How many people are they accepting?
[/quote]

Top 30 would be the cutoff for you.

Last year, you would have had to get about 3 solutions completely correct to make top 30, maybe a point or two less.

I know a student who was a junior last year and got a 20 and was not invited to MOsP.  I think he was probably just after the cutoff.

[quote="masamune2387"]I know the problems are proof-type, but do they give partial credit? (e.g. right method but arithmetic error, etc.)[/quote]

Just as a quick answer, yes, they do. However, I do not know much about the details of the scoring and how rigorous the solutions have to be either.

I don't know 100% how they score the problems, but this is my take on how my own problems seemed to be scored at MOP: if you make some headway on the solution you get 1, maybe 2 points.  If you make very good progress -- all of the important ideas are there, but you fail to connect the dots, you get 3 or 4.  If you solve the problem completely except for a minor error or two that could be very simply corrected, you get 5 or 6.  If your proof is complete and without errors, you get a 7.

I think entry into MOP is calculated by the following index

(AMC Score) + 10* ( AIME Score) + 9*(USAMO Score)

Last year I had 144, 9 and 17.. so that could be a possible minimum

There are probably exceptions for high USAMO scores...

According to AMC site "MOSP Factoids" (last updated 9/03) 24-30 are selected by:

1. top 12 USAMO scores
2. non-seniors from next top 17 USAMO scores
3. if less than 24 selected from 1st 2 criteria, bring it up to 24 from  non-seniors based on USAMO index (AMC score + 10xAIME) + 9xUSAMO score

According to the post on 2/09/04 by AMC director Steve Dunbar in the topic "Mathimatical Olympiad Program, 2004" the rules are instead:

1. top 12 USAMO scores
2. 12-18 of the highest remaining non senior "USAMO index" (presumably AMC score + 10xAIME + 9xUSAMO score)


[quote="gotztahbeazn"]Find all solutions to: $\displaystyle a^{2003}  \equiv  1 (mod 17)$[/quote]

by fermat's little theorem, if $(a, 17) = 1$ (which it has to be for $a^{2003} \equiv 1 (\mod 17)$), $a^{16} \equiv 1 (\mod 17)$; so $a^{2000} = (a^{16})^{125} \equiv 1 (\mod 17)$; so you're left with solving $a^3 \equiv 1 (\mod 17)$...

$17|a^3 - 1 = (a-1)(a^2+a+1)$, so either $ a \equiv 1$ or $a^2+a+1 \equiv 0$, and the solutions to this can be found by inspection...


Let $ \triangle{ABC}$ and $ (I)$ is the incenter.Denote:
$ (BIC)\cap AB,AC \equal{} C_1,B_2$
$ (CIA)\cap BC,BA \equal{} A_1,C_2$
$ (AIB)\cap CA,CB \equal{} B_1,A_2$
Let $ P \equal{} AA_1\cap{BB_2},Q \equal{} CC_2\cap{BB_1},R \equal{} AA_2\cap{CC_1}$.Construction$ : PP_0\perp{AB},QQ_0\perp{BC},RR_0\perp{CA}$.
Prove that $ : PP_0,QQ_0,RR_0$ concurrent
Sorry,I poor English  :lol:

[quote="sunken rock"]The triangle ABP and the similar ones have I as orthocenter!

Best regards,
sunken rock[/quote]
I agree with you.You can post full solution ?

Call D the midpoint of arc BC opposite to A in (ABC); obviously this is the circumcenter of (BIC), and it belongs to the bisector of <BAC, hence AB_2 = AB and AC_1 = AC, consequently BB_2 and AI are perpendicular... Analogously AA_1 and BI are perpendicular, so I is the orthocenter of triangle ABP, a.s.o.

Best regards,
sunken rock