induction algebra polynomial logarithms function calculus derivative

Given $x_1 + .. + x_n = a_1 + .. + a_n$ with $x, a > 0$

Prove that the product $P = {x_1}^{a_1}..{x_n}^{a_n}$ attain max value when $x_i = a_i$.

Note: $a$ is fixed, $x$ can be changed.

just a thought, could you use strong induction on $n$? i am not sure that it would be rigirous through, assume for n and 1, multiply them for n+1

That's what I tried, but it looks ugly since I had to used the polynomial expansion.  I wonder if there's another way that look better.

[quote="t0rajir0u"]I tried Jensen's, but it only allows you to conclude $x_n, a_n$ similarly sorted, but they aren't necessarily equal...[/quote]

Hmm, can you show me a case when they are not equal?

Note that $\ln{P} = \ln{(x_1^{a_1} \cdots x_n^{a_n})} = \sum a_i\ln{x_i}$.

Take the concave function $f(x) = \ln{x}$. Let $\sum x_i = \sum a_i = A$ for convenience.

By Jensen's with weights $\frac{a_i}{A}$, we have

$\frac{\ln{P}}{A} = \sum \frac{a_i}{A}\ln{x_i} \le \ln{\left(\frac{1}{A}\sum a_ix_i\right)}$.

So it is only necessary to maximize $\sum a_ix_i$. But by Cauchy,

$\sum a_ix_i \le \sqrt{(a_1^2+\cdots+a_n^2)(x_1^2+\cdots+x_n^2)}$

which has equality at $a_i = x_i$ (because $a_i$'s are constant).

EDIT: On second thought, since $\sum x_i^2$ isn't constant, that Cauchy step might not work...

EDIT2: Maybe this approach will work better... Assume it has a maximum at $x_i = a_i$. Then show that by changing any two terms, we have $(a_j-\delta)^{a_j}(a_k+\delta)^{a_k} \le a_j^{a_j}a_k^{a_k}$. I think we can show this by analyzing the derivative of the function $f(\delta) = (a_j-\delta)^{a_j}(a_k+\delta)^{a_k}$ and showing that it's always negative...

$f'(\delta) = -\ln{a_j} \cdot (a_j-\delta)^{a_j} - \ln{a_k} \cdot (a_k+\delta)^{a_k}$ if I did that correctly.

Thank you, the problem is finally solved.  Taking the derivative is a good approach :)

I already considered the 'changing two terms' possibility, but I don't think it works because the maximum might be attainable by changing more than two terms simultaneously... I'm also very suspicious of the Cauchy solution because I don't think you can show that the Jensen's and Cauchy parts have the same maximum case.  :(

Laputa:  For example $x_1 = 1, x_2 = 3, x_3 = 6$ and $a_1 = 1, a_2 = 2, a_3 = 7$ have the same sum and are similarly sorted, but $x_i \neq a_i$.

Edit:  [b]WAIT.[/b]  The problem as stated is incomplete!  In the above example, $\sum x_i = \sum a_i = 10$ and we are supposed to show that the maximum comes for $x_i = a_i$, however, $1^1 \times 3^2 \times 6^7 > 1^1 \times 3^3 \times 6^6$ !  There has to be some extra condition.  (In fact, the maximum comes when $x_1 = x_2 = 1, x_3 = 8$ and $a_1 = a_2 = 1, a_3 = 8$, I think.  But even so, now the wording of the problem confuses me.)

I don't think you're understanding the problem quite correctly... The given $a_i$ are FIXED. As in numbers. Then we want to show to maximize $P$ we have to set all the $x_i = a_i$.

For instance, if $a_1 = 1, a_2 = 2, a_3 = 7$ as in your example, then we want to show that $x_1 = 1, x_2 = 2, x_3 = 7$. I think it's true.

I'm pretty sure the Cauchy/Jensen solution doesn't work; I'm still working out the specifics of the derivative solution...

Maybe if we take $f(\delta) = \prod (a_i-m_i\delta)^{a_i}$ for constants $m_i$ such that $\sum m_i = 0$. That should cover all possibilities. The derivative of that is a beast, but I think it's always negative (of course given the condition that $a_i > m_i\delta$).

Well, if you're going to use derivatives, it's easiest to use partial derivatives having in mind the given conditions. Then you directly get $x_i=a_i.$

Couldn't we use the rearrangemente inequality between base and exponent, saying that P is maximum when $x_i>x_j\rightarrow a_i>a_j$?

There's actually an easy proof of this.  You're given a sum, and you're trying to maximize a product.  Does anything come to mind?

I have a solution like this, waiting for any comments:

[b]1. [/b] Case ${a^a}{b^b}$

Consider the the function $f(x) = {a*ln(a-x) + b*ln(b+x)}$ with $0 \leq x \leq a$

we want to maximize the product so $a*ln(a-x) + b*ln(b+x)$ should be maximized as well.

$f'(x) = \frac{b}{b+x} - \frac{a}{a-x}$

Since $\frac{b}{b+x} \leq 1$ & $\frac{a}{a-x}\geq 1, f'(x) \leq 0$, so $x = 0$ to attain the maximum value.

[b]2.[/b] Consider the case of of ${a^a}{b^b}{c^c}$:

let $f(x_1, x_2) = {a^a}{b^b}{c^c} - {(a-x_1-x_2)^a}{(b+x_1)^b}{(c+x_2)^c}$

So $f(x_1, x_2) = [{a^a}{b^b}{c^c} - {(a-x_1)^a}{(b+x_1)^b}{c^c}] + [{(a-x_1)^a}{(b+x_1)^b}{c^c} - {(a-x_1-x_2)^a}{(b+x_1)^b}{(c+x_2)^c}]$

From 1. above, we know that the derivative of the first square bracket is $\leq 0$, so consider the second bracket:

Let $a\" = a-x_1$ then

$(2nd Bracket) = {(b+x_1)^b}[{(a\")^a}{c^c} - {(a\"-x_2)^a}{(c+x_2)^c}]$

Again, consider the function $g(x_2) = a*ln(a\"-x_2) + c*ln(c+x_2)$

$g'(x) = \frac{c}{c+x} - \frac{a}{a\"-x}$ with $0 \leq x \leq a\"$

Since $a = a\" + x \geq a\"-x$, $\frac{a}{a\"-x} \geq 1$ while $\frac{c}{c+x} \leq 1$ so $g'(x) \leq 0$.

--> The derivative of $f(x_1, x_2) \leq 0$

So we can conclude that if we decrease from one variable and increase in other variables then the product will be less than before.  So this problem is proven.

Note: Please put any comment on my proff on this page:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=419269#p419269[/url]

[quote="zeus_three"]Apply the weighted AM-GM inequality to this?[/quote]

Yes.


algebra
difference of squares
special factorizations
AMC

How many right triangles have integer leg lengths $ a$ and $ b$ and a hypotenuse of length $ b\plus{}1$, where $ b<100$?

$ \textbf{(A)}\ 6 \qquad
\textbf{(B)}\ 7 \qquad
\textbf{(C)}\ 8 \qquad
\textbf{(D)}\ 9 \qquad
\textbf{(E)}\ 10$

[hide]the difference of squares:
between b^2 and (b+1)^2
(a+b)(a-b)
a-b is always 1
a+b=2b+1

a^2+b^2=(b+1)^2

so basically... a^2=2b+1
you just have to find all the b's possible where 2b+1=a square number... 2b+1 = odd number so all the odd squares from 3-199 work.

3^2,5^2,7^2,9^2,11^2,13^2

that's 6
Ans:A[/hide]

[hide="Solution"]
(Inspired by one of the Proofs Without Words things on the AoPS sidebar)

Say $ b$ is the side length of a square, and the hypotenuse, $ b \plus{} 1$, is the side length of another square.

Since $ (b \plus{} 1)^2 \minus{} b^2 \equal{} a^2$,

$ \frac{a^2 \minus{} 1}{2} \equal{} b$.

Now all we have to do is determine for which positive values of $ a$ that $ 0 < b < 100$

Note that $ a$ needs to be an odd number for $ b$ to be an integer

$ 3 \leq a \leq 13$

$ \frac{13 \minus{} 3}{2} \plus{} 1 \equal{} \boxed{6}$, so $ A$.
[/hide]

[quote="madshock"][hide="Lucky Way"]
If you know some of your pythagorean triplets, then you know:
3, 4, 5
5, 12, 13
7, 24, 25
if you watch, the numbers seem increasing by a significant amount so I assumed that there would be no more than 6, which is the smallest answer choice.
the next ones would somewhere around 50-55 for b and then 70-75, and then 90-95
(as a guess)
too bad I didn't realize this simple and lucky way DURING the test  :wallbash:
[/hide][/quote]

Wow, great solution!

[hide]$ a^2\plus{}b^2 \equal{} b^2\plus{}2b\plus{}1 \implies a^2 \equal{} 1\plus{}2b$, where $ 0<b<100$.

Thus, $ a^2$ can be any odd square where b is in that range. $ 1<a^2<199$, so squares $ 3^2,...,13^2$ work, or 6 squares, A.[/hide]


calculus
integration
function
real analysis
real analysis theorems

hi everyone

I have this question :

Can anyone explain me why when we have for example the function $ f(x)\equal{}\int^{x\plus{}1}_{x}e^{t^2}dt$ then it holds

$ x\leq t \leq x\plus{}1$ ?

Because we are taking the integral on the interval $ [x,x\plus{}1]$  :lol:


superior algebra
superior algebra solved

let I,J be 2 ideals of ring R.show residue of any element of IJ in R/I inter. J
 is nilpotent.

In fact every such element is 0 since IJ is a subset of $I\cap J$


inequalities
search
inequalities unsolved

1.$ (x,y,z > 0)$            prove that     $ x^2 \plus{} y^2 \plus{} z^2\ge\sqrt {2}(xy \plus{} yz)$
2.$ (c\ge b\ge a\ge 0)$  prove that     $ (a \plus{} 3b)(b \plus{} 4c)(c \plus{} 2a)\ge 60abc$

to 1. It's easy!
$ L\equal{}(x^2\plus{}\frac{y^2}{2})\plus{}(\frac{y^2}{2}\plus{}z^2)\ge \sqrt{2}(xy\plus{}yz)\equal{}R$ (by AM-GM)
It's good!

but 2.?

[quote="sxgfly"]to 1. It's easy!
$ L \equal{} (x^2 \plus{} \frac {y^2}{2}) \plus{} (\frac {y^2}{2} \plus{} z^2)\ge \sqrt {2}(xy \plus{} yz) \equal{} R$ (by AM-GM)
It's good!

but 2.?[/quote]

it is done by AM-GM and the fact that $ c\ge b\ge a\ge 0$, $ LHS \ge 60 a^{\frac{11}{12}}b^{\frac{19}{20}}c^{\frac{17}{15}} \ge 60 abc$


geometry
3D geometry
tetrahedron
circumcircle
geometry unsolved

Let $ ABCD$ is a tetrahedron and let $ AB\equal{}a, CD\equal{}b, AC\equal{}AD\equal{}BC\equal{}BD\equal{}1$. Find radius of the circumsphere of $ ABCD$.

There are "ready made" expressions for that question but, once the tetrahedron is not "very irregular' since all the faces are isosceles, we'll try to make it from basics.

The center of the circumsphere, $ O$ , will be the intersection of the perpendiculars $ JO$ and $ KO$ to the faces $ BCD$ and $ ACD$ , with $ J$ and $ K$ as their circumcenters, in the medians $ BM$ and $ AM$ of length $ h \equal{} \sqrt {1 \minus{} c^2}$ where $ c \equal{} {b\over2}$.
The distance from $ J$ and $ K$ to $ M$ , is $ y \equal{} {{h^2 \minus{} c^2}\over{2h}}$.
Prolonging $ MO$ it'l meet $ BA$ , perpendicularly , at it's center $ N$ , so that the rectangular triangles, $ JMO$ and $ MNB$ are similar, and the dinstance between $ M$ and $ O$ is $ x \equal{} {{hy}\over{\sqrt {h^2 \minus{} d^2}}} \equal{} {{h^2 \minus{} c^2}\over{2\sqrt {h^2 \minus{} d^2}}}$, here $ d \equal{} {a\over2}$.

Now the distance from $ O$ to any vertex will be the circumradius; for instance $ R \equal{} \sqrt {c^2 \plus{} x^2} \equal{} {{4(h^2 \minus{} d^2)c^2 \plus{} {(h^2 \minus{} c^2)}^2}\over{4(h^2 \minus{} d^2)}}$.


Below are a number of statements:
1.Precisely one of these statements is untrue.
2. Precisely two of these statements are untrue.
3. Precisely three of these statements are untrue.
4. Precisely four of these statements are untrue.
5 .Precisely five of these statements are untrue.
6. Precisely six of these statements are untrue.
7. Precisely seven of these statements are untrue.
8. Precisely eight of these statements are untrue.
9. Precisely nine of these statements are untrue.
10. Precisely ten of these statements are untrue.

 Which of these statements is true?

since untrue=false,[hide]only the ninth one is true[/hide]

[hide="Explanation"]

  Firstly, its impossible for there to be 0 untrue statements, it would mean everyone is telling the truth, a contradiction.  Now, we know that the number of untrue statements varies from:

1 - 10

  We also know one and only one number of untrue statements is correct.  Thus, only one statement is true, and the other nine are false.  Which means the 9th statement is true. [/hide]

[quote="mathnerd314"]But [u][i][b]noooooooooooooooooooooooo[/b][/i][/u]. The statement that "below are a number of statements" is true. Therefore there are eight untrue statements.[/quote]

Nice try, but there's still 9 untrue statements ;)

Oh. So... uh... what's the correct answer?!

[quote="mathnerd314"]So... uh... what's the correct answer?![/quote]

#9.


What is the value of $ 100\minus{}81\plus{}64\minus{}49\plus{}36\minus{}25\plus{}16\minus{}9\plus{}4\minus{}1$?

hello, we have $ \sum_{i\equal{}1}^{10}(\minus{}1)^i\cdot i^2\equal{}55$.
Sonnhard.

Which doesn't really explain the solution.

Consider the terms in pairs.

19+15+11+7+3, this is an arithmetic sequence with 5 terms and a middle term of 11. Therefore the sum is 5*11=55.

hello, this will explain my solution it comes from the sum $ \sum_{i\equal{}1}^n(\minus{}1)^i\cdot i^2\equal{}\frac{1}{2}(\minus{}1)^nn(n\plus{}1)$.Sonnhard.


This is from ASHME 1954 (also problem #80 chapt. 4 in aops book 1)

If x varies as the cube of y, and y varies as the fifth root of z, then x varies as the nth power of z. what is n?

does varies mean that x is inversly proportional to y?  what does varies mean.

Thanks

hmmm... I think it means x is proportional to y^3, etc.
[hide="so here is my solution"]$x=ky^3; y=nz^{1/5} \implies x=k\cdot n^3 \cdot z^{3/5} = mz^{3/5}$ So x varies as the 3/5 power of z, I belive[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that for any group $ G, \, \, |G/Z(G)| \neq 91.$

[hide="The hint in the book says ..."] that assuming $ |G/Z(G)| = 91$, prove that $ G/Z(G)$ is cyclic.[/hide]

if $ H$ is a group of order $ 7*13$, sylow theorems yield normal subgroups of order $ 7$ and $ 13$. they have to intersect trivially, thus their product is $ H$. then $ H \cong \mathbb{Z}/7 \times \mathbb{Z}/13 \cong \mathbb{Z}/7*13$, i.e. $ H$ is cyclic.

if $ G/Z(G)$ is cyclic, then $ G$ is abelian (easy exercise), i.e. $ G/Z(G)$ is trivial.

How does one get  $ H \cong \mathbb{Z}/7 \times \mathbb{Z}/13 \cong \mathbb{Z}/91$?
Thanks.

well think over it, then the only non-obviousl fact is that the elements of the two normal subgroups commute with each other. this comes from a general fact:

let $ N,M$ be normal subgroups of a group $ G$ with trivial intersection. then $ mn=nm$ for all $ m \in M, n \in N$ (and thus $ NM \cong N \times M$).

this follows from $ mnm^{-1}n^{-1}\in M \cap N$.


Given $ \pi_{v} \in L(V \oplus W, V)$, $ S \in L(U,V)$, and $ T \in L(U,W)$, show that there is a natural linear transformation $ R \equal{} S \times T \in L(U, V \oplus W)$ such that $ \pi_{v} \circ R \equal{} S$. Also, $ \pi_{v}(x,y) \equal{} x$, where $ x$ and $ y$ are vectors.

I proceeded with this problem by  toying around with $ S$ and $ T$. Since we know that $ S: U \longrightarrow V$ and $ T: U\longrightarrow W$ then i just wrote

$ S(u) \equal{} ( v_{1},v_{2})$ and $ T(u) \equal{} (w_{1}, w_{2}).$ where $ u \in U$ and $ (v_{1}, v_{2}) \equal{} v \in V$ and similarly with $ w\in W$

Since i m trying to construct $ R: U \longrightarrow V \oplus W$ I added $ S(u)$ and $ T(u)$ to get

$ R \equal{} S(u) \plus{} T(u) \equal{} (v_{1} \plus{} w_{1}, v_{2} \plus{} w_{2})$.

then to show $ \pi_{v} \circ R \equal{} S$, I simply applied $ \pi_{v}$  on $ R$.Thus,

$ \pi_{v}(R) \equal{} \pi_{v}(v_{1} \plus{} w_{1}, v_{2} \plus{} w_{2}) \equal{} v_{1} \plus{} w_{1} \in S(u)$???

I m new to linear transformations, so can anyone tell me how to actually do this correctly? better yet, if it's all wrong, please tell me the best way to start. Thanks.

it's all wrong (remember the definitions), you have to define R(u)=(S(u),T(u)).


Asymptote algebra function domain vector LaTeX

(Maybe this belongs in a lower level of mathematics but still found it in a high school textbook)

When we are required to draw graphs (e.g. using excel), do we have to show all the
-asymptotes
-breaking points in the graph if it is not continuous at at least one point?

If so how much should we show of each?  If the graph is starting to curve down at one side and up the other side at a certain value, can we just show the start and say that it is an asymptote?

Also if the graph stops at a point, (i.e. Domain of the function being drawn is x<10)  Do we have to draw the graph upto x=10 or upto x=10.1 to show that the graph in discontinuous at x=10?

Usually it depends on the requirement. What are the graphs for? A class assignment? Personal clarification?

If you wish to leave out the asymptote, you may, but you'll have to note somewhere that there is an asymptote and whatever value of $ x$ or $ y$. Generally, you should always display holes as a simple open circle in your graph.

but can we actually show those using excel and nothing else?

I'm not sure; I'm terrible at Excel other than graphing simple data graphs for science labs.

You can however, learn [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]Asymptote[/url], which is a programming language similar to LaTeX, in which you can easily express asymptotes and such on graphs.


Let $ a,b,c$ be positive real numbers. Prove that:
$ \sum_{cyclic}\frac{a^3}{a^3\plus{}b^3\plus{}abc} \ge 1$

-Nguyen Van Thach

I can't find any better way than expanding  :(
Expanding, we get
$ a^6b^3 \plus{} b^6c^3 \plus{} c^6a^3 \ge a^5b^2c^2 \plus{} a^2b^5c^2 \plus{} a^2b^2c^5$
Let $ a^2b \equal{} x$, $ c^2a \equal{} y$, $ b^2c \equal{} z$, then this is equivalent to
$ x^3 \plus{} y^3 \plus{} z^3 \ge x^2y \plus{} y^2z \plus{} z^2x$
But $ x^3 \plus{} x^3 \plus{} y^3 \ge 3x^2y$, so adding this cyclically gives the desired result.

Normalize to $ abc\equal{}1,a^3\equal{}x,b^3\equal{}y,c^3\equal{}z,xyz\equal{}1$

$ \sum \frac{x}{x\plus{}y\plus{}1}\ge 1$

This is better for expanding, get to similar as dgreen, and finish it off.

[quote="dgreenb801"]I can't find any better way than expanding ....[/quote]
this what i did substitute $ x\equal{}\frac{a}{b}$ and similarly define $ x,y,z$ the inequality becomes direct cauchy also i think it can be solved by writing the one in the RHS as $ \frac{a^3\plus{}b^3\plus{}c^3}{a^3\plus{}b^3\plus{}c^3}$ and taking it to the LHS and using sos.

[hide=" solution"]

the $ LHS$ is homogeneous so let $ a^3b^3c^3 \equal{} 1$

let $ a^3 \equal{} \frac {x}{y}, b^3 \equal{} \frac {y}{z}, c^3 \equal{} \frac {z}{x}$

therefore
$ \sum_{cyclic}\frac {a^{3}}{a^{3} \plus{} b^{3} \plus{} abc} \equal{} \sum_{cyclic}\frac {xz}{xz \plus{} y^2 \plus{} yz} \equal{} \sum_{cyclic}\frac {(xz)^2}{(xz)^2 \plus{} xy^2z \plus{} xyz^2}$

No by cauchy

$ (xy \plus{} yz \plus{} zx)^2 \left(\sum_{cyclic}\frac {(xz)^2}{(xz)^2 \plus{} xy^2z \plus{} xyz^2} \right) \ge (xy \plus{} yz \plus{} zx)^2$

and we're done

[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $H$ be a subgroup of $G$ with $h$ elements. Suppose $\exists a\in G,\ \forall x\in H\ (xa)^{3}=1$. Let $P=\{x_{1}ax_{2}a\dots x_{n}a|x_{i}\in H, n\in \mathbb N\}$. Find an upper bound for $|P|$.
PS This is a simple problem. It might be helpful if you consider the case $G$ being abelian first.

Ok, here is the final [hide="answer"]$3h^{2}$[/hide].Now, any idea?


[color=red]Three equal circles with a comon point H, are intersecting two by two in the points A,B,C.Prove that the circumcircle of the triangle ABC is equal with the three circles mentioned.(This problem is known as the Titeica's theorem[/color])
[i]Proof 1[/i] $ BH$ is the segment of intersection $ C(O_3,R)$ and $ C(O_1,R)$. The triangle $ O_3BH$ is isoscel, and alsow  $ O_1BH$ is isoscel,and results that $ O_3BO_1H$ romb$ \Longrightarrow$ and therefore $ O_3B_1\equal{}O_1B_1$ where $ B_1$ is the midle of the segment $ BH$.In the same way $ O_2HCO_1$ romb $ \Longrightarrow$ $ O_2C_1\equal{}O_1C_1$ where $ C_1$ is the midle of the segment $ CH$,and $ O_2A_1\equal{}O_3A_1$.  $ C(O_1)\equal{}C(O_2)$ and $ C(O_1) \cap C(O_2)\equal{}CH$ $ \Longrightarrow$ $ AB$ is the comon tangent to this circles $ \Longrightarrow$ $ O_1B \bot AB$ , $ O_2A \bot AB$ \Longrightarrow AO_2O_1B$ is a paralelogram with$<B=90$ .In the same way$O_3O_2CB is a paralelogram with <C=90, the same as $ AO_3O_1C$ As a consequence H becomes the ortocenter of ABC, and it was the center of $ C(O_1O_2O_3)$.But $ \Delta ABC \equiv \Delta O_1O_2O_3 \Longrightarrow C(ABC)\equal{}C(O_1O_2O_3)$. But $ C(O_1O_2O_3)\equal{}C(O_1)\equal{}C(O_2)\equal{}C(O_3)$ and results that $ C(ABC)\equal{}C(O_1)\equal{}C(O_2)\equal{}C(O_3)$
.This was a clasic demonstration.
[i]Proof 2[/i] Here's an elegant one:D. $ A_1 , B_1 , C_1$ are on the Euler's circle in $ \Delta O_1O_2O_3$. We had proved that  H is the ortocenter of the triangle $ ABC \Longrightarrow A_1, B_1, C_1$ are on the Euler's circle in $ \Delta ABC$.But $ A_1,B_1, C_1$ are on the Euler's circle of $ \Delta O_1O_2O_3$ and in the same time on the Euler's circle of $ \Delta ABC$ $ \Longrightarrow C(ABC)\equal{}C(O_1O_2O_3)\equal{}C(O_1)\equal{}C(O_2)\equal{}C(O_3)$
[i]Proof 3[/i]
Here's a complex solution:D. Consider a cartezian system, and $ O_1(z_1), O_2(z_2), O_3(z_3)$

Continuing: results that $ A,B,C$ will have the afixez $ z_2\plus{}z_3,z_1\plus{}z_3,z_1\plus{}z_2$. \[ AB\equal{} | (z_1\plus{}z_3)\minus{}(z_2\plus{}z_3)|\equal{}|z_1\minus{}z_2|\equal{}O_1O_2\] In the same way we obtain $ BC\equal{}O_2O_3$, $ AC\equal{}O_1O_3$ .Therefore $ \Delta ABC \equiv \Delta O_1O_2O_3$ so their circumcircles will be equal.
If we note $ \omega\equal{}z_1\plus{}z_2\plus{}z_3$ and let Q be a point with the afix $ \omega$.We have $ |z_1|\equal{}|z_2|\equal{}|z_3|\equal{}r$ we can prove that Q is the center of the circumscribe circle of the triangle ABC.

I'm expecting more solutions to this beautifull problem!

The second problem is a problem of Mihail Popescu.
[color=darkblue]If three equal equilateral triangles $ OA_1B_1, OA_2B_2,OA_3B_3$ have point O in comon then:
a). $ C(OA_1B_1),C(OA_2B_2),C(OA_3B_3)$ are intersectating two by two in three points who form a triangle and the circumcircle of that triangle is equal to the circles mentioned.
b). $ C(OA_1B_2), C(OA_2B_3), C(OA_3B_1)$ are intersectating two by two In three points who form a equilateral triangle equal whith those that are mentioned.The same property  takes place  for the circles $ (OA_1B_3),(OA_2B_1),(OA_3B_2)$.[/color]

I will post the solution this night.I'm interested in a complex solution wich I haven't discovered yet:D

I'm back with the solution.For the point a, from the equality of the triangles results the equality of the circumcircles and the problem reduces to Titeica's theorem.
For point b,let $ C(O,R)$ the circle who passes the rest of the top points of the equilateral triangles mentioned.Let A",B",C" the centers of $ C(OA_3B_2)$ , $ C(OA_1B_3)$,$ C(OA_2B_1)$ and $ A$,$ B$,$ C$ the points of intersection of these circles. $ D$,$ E$,$ F$ the midle of the segments $ B_2A_3$, $ B_3A_1$, $ B_1A_2$ with $ M$,$ N$,$ P$,$ Q$ the midles of segments $ OC,OB,OA_3,OB_3$ and $ <A_3OB_2\equal{}\alpha$,$ <A_1OB_3\equal{}\beta$,$ <A_2OB_1\equal{}\gamma$, $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi/2$.
$ <B'FD\equal{}<B'DF\equal{}<B_3B_1A_3\equal{}\pi/6$ and results that B' is the center of the circumscribe circle of the equilateral triangle build on DF as a side.Same A' and C'. We all know that the centers of the circumscribe circles of the equilateral triangles build in the exterior by the sides of a triangle,are forming a equilateral triangle, such as A'B'C'.
The two quadrilaterals $ A'A_3PA''$ and $ B'B_3QB''$ are inscriptibles, therefore $ OA' \cdot OA''\equal{}OB' \cdot OB''\equal{}R^2/2$ from where A'B' and A"B" are antiparalels in the $ \Delta OA'B'$, from where $ <OB''A''\equal{}<OA'B'$
.In the same way $ <OC''A''\equal{}<OA'C'$ and $ <OB''A''\plus{}<OC''A''\equal{}\pi/3$.In the triangle A"B"C" , $ <A''\equal{}\beta\plus{}\gamma$.Consider the circle of diameter OA" and M,N,P points on this circle.$ <MA''N\equal{}<PA''O$ therefore $ MN\equal{}OP\equal{}R/2$ and $ BC\equal{}2MN\equal{}R$.In the same way $ AB\equal{}AC\equal{}R$ and it results that the triangle ABC is equilateral, equal with the three triangles given.


conics
parabola
geometry
geometric transformation
reflection
analytic geometry
graphing lines

how do you find the focus of a parabola?  I know what it means, but how do you find it in a parabola?  it would really help if you told me how to do this.  Thanks!!!

It depends on how your parabola is constructed.

$ \blacktriangleright$ If your parabola is in the Cartesian plane and has an central axis parallel to either the x or y axis of the plane, then you can write the equation in [url=http://en.wikipedia.org/wiki/Parabola#Analytic_geometry_equations]Focus-Directrix[/url] form. This is likely what you are talking about.

$ \blacktriangleright$ In general, the parabola has an interesting property that any ray parallel to the axis of the parabola that is coming from the inside will reflect off the parabola and travel in a new ray that contains the focus. Thus, if two different rays parallel to the parabola reflect off the inside, then the intersection of the two reflected rays will be the focus.

[hide="Answer"]If a parabola is in the form:

$ y\equal{}a(x\minus{}h)^2\plus{}k$

Then the focus is the point:

$ (h,k\plus{}\frac{1}{4a})$

If a parabola is in the form:

$ x\equal{}a(y\minus{}k)^2\plus{}h$

Then the focus is the point:

$ (h\plus{}\frac{1}{4a},k)$[/hide]
[hide="Note"] Also note that a parabola will not have a focus if it is degenerate (it will either be a line or two parallel lines). An example of a degenerate parabola would be $ x^2\minus{}x\equal{}0$, it is degenerate because there isn't a $ y$ value therefore the graph would be the line $ x\equal{}0$.

[/hide]

In conics, the equation of a parabola could also be $ (y\minus{}k)^2\equal{}4a(x\minus{}h)$ with the vertex at $ (h\plus{}a, k)$. (Interchange x and h with y and k for the 2nd equation, with vertex at $ (h, k\plus{}a)$.)

In the form $ f(x)\equal{}ax^2\plus{}bx\plus{}c$, the vertex is at $ (\minus{}\frac{b}{2a},f(\minus{}\frac{b}{2a}))$.

I'm going to throw out three claims without proof -- you're invited to explore/text/prove them yourself.

[b]Claim 1:[/b] It is possible to arrange the Cartesian equation of any parabola into the form $ (ax \plus{} by)^2 \equal{} cx \plus{} dy \plus{} e$
[i][u]Hint:[/u] Use the [url=http://en.wikipedia.org/wiki/Conic#Cartesian_coordinates]discriminant[/url] of a conic section[/i]

$ \blacktriangleright$ I define the [i]family[/i] of a line to be the set of all lines on the Cartesian plane that have the same slope.

[b]Claim 2:[/b] The family of lines that contains the axis of a particular parabola is the only family in which every member line intersects that parabola only once.

[b]Claim 3:[/b] If a parabola's equation is written in the form $ (ax \plus{} by)^2 \equal{} cx \plus{} dy \plus{} e$, then the line $ ax \plus{} by \equal{} 0$ is in the same family as the axis.

From here, we know the slope of the axis, which means that we can emulate the light-ray approach suggested in the second triangular bullet of my first post.

[b]Example:[/b] The equation of a parabola is $ (x \minus{} y)^2 \equal{} x \plus{} y \plus{} 2$. Find the focus.

The family that contains the axis is $ x \minus{} y \equal{} C_1$, where $ C_1$ is an arbitrary constant. Thus, the family that contains the tangent to the vertex is $ x \plus{} y \equal{} C_2$. Substituting this form into our original equation and looking for a value of $ C_2$ that gives only one point of contact, we find that the vertex is $ ( \minus{} 1, \minus{} 1)$. Since the axis contains the vertex, the axis must be $ x \minus{} y \equal{} 0$, or $ x \equal{} y$. We know the focus is on this line.

Now, consider the line $ x \minus{} y \equal{} \minus{} 1$. This intersects the parabola at the point $ ( \minus{} 1,0)$. We can use calculus to find that the slope of the normal line at this point is $ \minus{} \frac {1}{3}$, so the normal line at this point is $ y \equal{} \minus{} \frac {1}{3}(x \plus{} 1)$. The reflected ray is a portion of the line formed by reflecting $ x \minus{} y \equal{} \minus{} 1 \iff y \equal{} (x \plus{} 1)$ across the line $ y \equal{} \minus{} \frac {1}{3}(x \plus{} 1)$, which results in $ y \equal{} \minus{} 7 (x \plus{} 1)$.

Finally, we seek the intersection between $ y \equal{} \minus{} 7(x \plus{} 1)$ (reflected ray) and $ x \equal{} y$ (axis) to find that the focus is $ \left( \minus{} \frac {7}{8} , \minus{} \frac {7}{8} \right)$

[b]Note:[/b] In this case, it would have also been straightforward to rotate the parabola 45º, since the rotation of the parabola is clear from the equation.

Here's how I would "prove" claim 2. Because the slope of any parabola (when rotated so the axis of symmetry is parallel with the y-axis) is infinitely growing, any line other than a vertical one would eventually intersect the parabola.

I'm sure that there are "better", more algebraic ways to prove it, but this is a simple intuitive proof.


logarithms
trigonometry
calculus
calculus computations

Answer the following Question.

[1] $\frac{d}{dx} \ln \sqrt{1+\cos ^ 2 \frac{x}{2}}$

[2] ${\frac{d}{dx} (\sqrt{x})^x}$

[3] $\frac{d}{dx} \ln _x (\ln x)$

[4] $e^x\frac{d}{dx}\left(x^2\frac{d}{dx}(x^2e^{-x})\right)$

[5] Find the values of $a,b$ such that $\frac{d}{dx}\{x\sqrt{x^2+3}+a\ln (x+\sqrt{x^2+3})\}=b\sqrt{x^2+3}$.

1
$\frac{-\sin{x}}{4(1+\cos^2{(\frac{x}{2})}}$
2
$(\sqrt{x})^x(\ln{\sqrt{x}}+\frac{1}{2})$
3
i don't get what that x is behind the second ln, is just x times the ln?
4
$6x^2+2x^3-x^4$

[quote="amirhtlusa"]1
$\frac{-\sin{x}}{4(1+\cos^2{(\frac{x}{2})}}$ :)$ =-\frac{\sin x}{2(\cos x +3)}$
2
$(\sqrt{x})^x(\ln{\sqrt{x}}+\frac{1}{2})$ :)$=\frac{1}{2}(\sqrt{x})^x (\ln x+1)$
3
i don't get what that x is behind the second ln, is just x times the ln? x [color=blue]expresses the base of logarithm.
[/color]
4
$6x^2+2x^3-x^4$ :( [/quote]

How about for 5?

[quote="Jimmy"][quote="amirhtlusa"]
i don't get what that x is behind the second ln, is just x times the ln?
[/quote]

He means $\frac{d}{dx} \left(\frac{\ln(\ln x)}{\ln x}\right)$

$ = \frac{1-\ln(\ln x)}{x\ln^2 x}$[/quote]

You are right.

5)

After differentiating, we have

$\sqrt{x^2+3} + \frac{a+x^2}{\sqrt{x^2+3}} = b\sqrt{x^2+3}$

$\frac{2x^2+a+3}{\sqrt{x^2+3}} = b\sqrt{x^2+3} \Rightarrow 2x^2+a+3=bx^2+3b$

So, now we have $b=2$, and $a+3=3b\Rightarrow a = 3$.


linear algebra
matrix
linear algebra unsolved

Let $ A\in \mathcal{M}_n(\mathbb{R})$ prove that :

1) if $ A\equal{}A^T$ then $ \lambda_k \in \mathbb{R}$, for $ k\equal{}\overline{1,n}$
2)if $ A\equal{}\minus{}A^T$ then $ \lambda_k \in \mathbb{C\minus{}R}$ ,for $ k\equal{}\overline{1,n}$

Note : $ A^T$ is the transpose of the matrix $ A$ ,and $ \lambda_k ,k\equal{}\overline{1,n}$ are its eigenvalues.

You can use $ <AX,Y>$=$ <X,A^*Y>$ where $ X,Y\in{M_{n,1}(C)}$ , $ A^*$ is the adjunct matrix(the transpose of the conjugate), and$ <X,Y>$ is the scalar product of $ 2$ column vectors.
In the both case for the demonstration $ Y=X$

2) should say that if $ A\equal{}\minus{}A^T,$ then $ \lambda\equal{}i\mathbb{R}$ - in other words, all of the eigenvalues are pure imaginary.

Saying $ \mathbb{C}\setminus\mathbb{R}$ is simultaneously too weak - "pure imaginary" is (for the most part) a much stronger condition than "not real" - while at the same time being incorrect, since an antisymmetric matrix could have $ 0$ as an eigenvalue, and $ 0$ is real.

Another approach would be the following :

[hide="Notations"]
Let $ z \equal{} a \plus{} ib \in \mathbb{C}$ , $ a,b \in \mathbb{R}$ .I denote $ \overline{z} \equal{} a \minus{} ib$.
If $ X \in \mathcal{M}_n(\mathbb{C})$ , $ X \equal{} (x_{i,j})_{1 \leq i,j \leq n }$ .By $ \overline{X}$, I understand $ \overline{X} \equal{} ( \overline{x}_{i,j})_{1 \leq i,j \leq n}$
[/hide]

1.

Let $ \lambda$ be an eigenvalue of $ A$ and $ X \in \mathcal{M}_{n,1}(\mathbb{C})$ be the corresponding eigenvector.
From $ AX \equal{} \lambda X$ we get $ \overline{A}\cdot \overline{X} \equal{} \overline{\lambda}\cdot \overline{X}$ $ \Rightarrow$
$ \overline{X}^T \cdot \overline{A}^T \equal{} \overline{\lambda}\cdot \overline{X}^T$
But because $ A\in \mathcal{M}_n(\mathbb{R})$ we have that $ \overline{A} \equal{} A$ ,therefore $ \overline{A}^T \equal{} A^T \equal{} A$ .(the last equality is from our hypothesis i.e. $ A^T \equal{} A$)
So $ \overline{X}^T \cdot A \equal{} \overline{\lambda}\cdot \overline{X}^T$ $ \Rightarrow$
$ \overline{X}^T \cdot AX \equal{} \overline{\lambda}\cdot \overline{X}^TX$ $ \Leftrightarrow$
$ \overline{X}^T \cdot \lambda X \equal{} \overline{\lambda}\cdot \overline{X}^TX$ and from here we conclude that  $ \lambda \equal{} \overline{\lambda}$ $ \Rightarrow$ $ \lambda \in \mathbb{R}$

We can prove the second part analogously.


Prove that cos 36 - cos 72 = 0.5, and find the numerical values of cos 36 and cos 72. I couldn't solve this problem without looking at the answers.  :wink:

[quote=vishalarul]Prove that cos 36 - cos 72 = 0.5[/quote]

Maybe use complex numbers.
 :)

[hide]
Consider the 5th roots of unity. They are of the form $\cos{\frac{2k\pi}{5}}+i\sin{\frac{2k\pi}{5}}$, and we're looking for $\cos{\frac{2\pi}{5}}$.

We find them by solving $x^{5}-1=0$. Find, divide by $x-1$ to get $x^{4}+x^{3}+x^{2}+x+1=0$. Then, divide by $x^{2}$ to get $x^{2}+x+1+\frac{1}{x}+\frac{1}{x^{2}}=0$. Let $w: =x+\frac{1}{x}$ This becomes $w^{2}+w-1=0$, so $w=\frac{-1\pm\sqrt{5}}{2}=-\phi,\frac{1}{\phi}$.

If we want x, we know that $x^{2}-wx+1=0$ or $x=\frac{w \pm \sqrt{w^{2}-4}}{2}$. The real part, which is $\cos{\frac{2k\pi}{5}}$ (depending on the root), is $\frac{w}{2}$. If we want $\cos{72}$, we use the positive value for w, so $\cos{72}=\frac{\sqrt{5}-1}{4}=\frac{1}{2\phi}$.

We know that $\cos{36}$ is a positive number such that $\cos^{2}{36}-1=\cos{72}=\frac{1}{2\phi}$.

We see that $\cos{72}+\frac{1}{2}= \frac{1}{2\phi}+\frac{1}{2}=\frac{1}{2}(\frac{1}{\phi}+1)=\frac{\phi}{2}$.
Then, $2(\cos{72}+\frac{1}{2})^{2}-1=2\frac{\phi}{2}^{2}-1=\frac{\phi^{2}}{2}-1=\frac{\phi+1}{2}-1=\frac{\phi-1}{2}=\cos{72}$.
So $\cos{36}-\cos{72}=\frac{1}{2}$.
[/hide]

Nice solutions. The first solution was an ingenious example of using the roots of unity, while the other was similar to the solution in the answer key. Thanks.  :lol:


search
number theory theorems
number theory

Could you tell me if $ \left(a,b,c\right)=\left(m^2 + n^2 - p^2 - q^2,2\left(mp + nq\right),2\left(mp - nq\right)\right)$ and any cyclyc permutation and $ d=m^2 + n^2 + p^2 + q^2$ for integer $ m,n,p,q$ are all integer solutions of $ a^2 + b^2+ c^2 = d^2$?

Yes.  See for example [url=http://en.wikipedia.org/wiki/Pythagorean_quadruple]the Wikipedia article[/url] (interestingly, the Mathworld article does not contain the complete parameterization).  I gave a sketch of one proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=729475814&t=176802]here[/url] and scorpius119 gave another which is quite nice.


What percent of the volume of a $ 10 \times 10 \times 10$ box can be filled with $ 4 \times 4 \times 4$ wooden cubes? Express your answer as a decimal to the nearest tenth.

Does this have to be as many as possible?  Or is this just $ \frac{4^3}{10^3}$?  If the latter is correct, then the answer is $ \boxed{25.6}$.

Think about it. With 4x4x4, you can fill it to axbxc , but a b and c must be multiples of 4.

Mew's right.  You can definitely fit 8x8x8 into a 10x10x10x :P

[quote="AIME15"][size=200][b]Mew's right.[/size][/b]  You can definitely fit 8x8x8 into a 10x10x10x :P[/quote]

Now that's not something you see every day....

:spam: mew....does post count really matter that much?  And yes, you don't see that every day, especially from MOI

Ernie, the reason why your wrong is it says What percent of the volume of a 10x10x10 box CAN BE FILLED, so it implies it's looking for the MAXXXXX (too much caffeine!)

I don't suppose you guys read the stickies too often, do you? I suggest you do so.

Locked.


Thank you, VASC,

By my result post in your topic

http://www.mathlinks.ro/Forum/viewtopic.php?t=144140

I refer that it suffices to consider the inequality in case $\min(a,b,c)=0$. Moreover, I already proved that for all $k\ge 1$ and $a\ge b\ge c$ then

\[\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge \frac{3}{k+1}.\]

However, I will try to solve your problem without my result. Thank you again,

[quote="hungkhtn"] I already proved that for all $k\ge 1$ and $a\ge b\ge c$ then
\[\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge \frac{3}{k+1}. \]
However, I will try to solve your problem without my result. Thank you again,[/quote]

The inequality is not true for $k=1$. See Iran inequality.

I am sorry. I misunderstand at the first time. I did prove that for all $k\ge 1$ and $a\le b\le c$ then
\[\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge \min\left(\frac{3}{k+1},\frac{1}{k\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{(k+1)\sqrt{3}}\right). \]
The equality must holds for $a=b=c=1$ or $a=0,b=c=\sqrt{3}$. It is my implication.

Notice that for $k=2$, we have

\[\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}+\frac{1}{3\sqrt{3}}=1.05>1.\]

So your problem is really sharp, VASC.

There is a nice solution to the original inequality. Who find it?

[quote="Vasc"]If $a,b,c$ are nonnegative numbers such that $a\le b\le c$ and $ab+bc+ca=3$, then

$\frac 1{2a+b}+\frac 1{2b+c}+\frac 1{2c+a}\ge 1$.[/quote]
Using the known $(x+y+z)^{2}\ge 3(xy+yz+zx)$, it suffices to prove that
\[\sum\frac{1}{(2a+b)(2b+c)}\ge \frac{1}{ab+bc+ca}\]
Or
\[\frac{3(a+b+c)}{(2a+b)(2b+c)(2c+a)}\ge \frac{1}{ab+bc+ca}\]
Or
\[ab^{2}+bc^{2}+ca^{2}\ge a^{2}b+b^{2}c+c^{2}a \]
It is trivial since $a \le b \le c$.
We are done. :)

Nice (only nice and not very nice, because this is also my solution  :lol: ).

Much simpler than I though! Thanks for VASC and Can,  :lol:

Toanhocmuonmau, you are really very clever.  :)


1) Outline a synthesis of $ C_6H_5\minus{}C \equiv C\minus{}CH_2OH$.

2) What product will be obtained when the alcohol of 1) is treated first with $ Br_2/CH_2Cl_2$, and then with $ Ph_3P, Br_2$?

Probably start with C6H6:
[hide="click to reveal the hidden content"]1.Gattermann Hydroformylation --> Benzaldehyde
2.Treat with ylide PPh3=CH2 --> Styrene
3.Treat with Br2/CCl4-->dibromo product
4.Treat with alc.NaNH2 to get phenyl ethyne
5.Treat it again with NaNH2 to get the carbanion
6.Treat it with formaldehyde(Nef synthesis)-->Required product[/hide]

My method:
1. Get benzaldehyde from benzene
2. Perkins rn to get cinnamic acid
3. $ \alpha$ halogenation (?doubtful)
4. Alc. $ \ce{KNH2}$
5. Use LAH to reduce the COOH to $ \ce{CH2OH}$
 :)

U can't alpha chlorinate cinnamic acid :)  :)

Chemrock's synthesis is correct, but:

[quote="Chemrock"]alc.NaNH2[/quote]

?

[quote="Chemrock"]Treat it with formaldehyde(Nef synthesis)[/quote]

?

strong base to do 2 dehalogenation to get alkyne.

Nef synthesis is treating metal alknyl with carbonyl compound. It is a nucleophilic addition reaction.

What is alc.KOH then?
It is just that in alcoholic medium it acts as a good base and hence eliminate them.
Probably from benzaldehyde:
[hide="click to reveal the hidden content"]
1.Do claisen smidt reaction with CH3-CHO-->cinnamaldehyde
2.Then brominate and then dehalogenate
3.reduce it with LAH[/hide]

[quote="Chemrock"]What is alc.KOH then?[/quote]

No. Now you are further away from the correct reasoning.

[quote="Carcul"]I know that. But what does it mean "alc.NaNH2"?

And how about a shorter synthesis.[/quote]

Do you want him to say that its a strong base  :maybe:

[quote="chemrock"]U can't alpha chlorinate cinnamic acid [/quote]

hey y cant u alpha chlorinate it??
u can do HVZ rite??

Excuse me the mechanism of HVZ doesn't permit that. :maybe:
[quote="valeriummaximum"]no because it will prefer the benzylic carbon rather than the carbon alpha to teh acid.[/quote]
that also is a sp2-s bond and is stable mr.valerium :huh:

I have given it just above urs.
Here's a better explanation with the mechanism below:
The enol content of cinnamyl chloride is very low so that's why the reaction gives very poor yields

I also do not see the reason...

What you have there is a mechanism for the HVZ reaction.

So does that mean you can alpha chlorinate?  :maybe:   :maybe:

[quote="chemrock"]
The enol content of cinnamyl chloride is very low so that's why the reaction gives very poor yields[/quote]
Here!! :|  :huh:

[quote="Valeriummaximum"]then can u giove the reason why u cant a-chlorinate cinnamic acid...[/quote]

The reason is simple: the alpha carbon of cynnamic acid is a sp2 carbon, not a sp3.

I also gave that before here see:
[quote="chemrock"]It is a sp2-s bond and is stable mr.valerium[/quote]


Let O be the center of a circle on the plane.
We define a fix point M on a circle. We define a fix chord AB.
Let CD be a variable chord (not AB) such that <CMD = < AMB.
Prove that point P = {AB ∩ CD} is a fix point.

Unless I'm misunderstanding something, that can't be true.. What did you really mean?

I also have some troubles taking it in....

I am very sorry Grobber, Sailor and others.
Actually I was thinking about another probelm, which I interpret wrong, because I can't understand it.
This is Fregier theorem :
If through the point M of two fascicols of lines that are in involution we draw a circle, such that lines connect points which are on the circle goes through the fix point. (please explain solution clearly)

Grobber maybe you will explain to everybody involution and omografy.

P.S I hope you will for give me.

I think the best solution is to google "involution", "homography" etc. I'm definitely not an expert in the field, and all I did was apply methods such as the ones you have mentioned to a couple of problems. Besides, I'm really bad at explaining stuff I don't master myself :D.

[url=http://www.euclidraw.com/Eng_fls/EucliDraw_htmls/Vocabulary.htm]Here[/url]'s a very good place to start.

I'd also like to add that I didn't quite understand your formulation of the Fregier theorem.

Many thanks, I think I mustn't to go in the depth of geometry.
By the way just for you again Grobber condition in Romanian:

Daca prin virful M a doua fascicole de drepte in involutie se duce un cerc, dreptele ce unesc punctele situate pe cerc a doua raze omoloage trec printr-un fix punct.


This is a problem taken from kalva, but it 's not easy:
The chord $ CD$ is perpendicular to the diameter $ AB$ and meets it at $ H$.Let $ O$ be the center of the circle. The distances $ AB$ and $ CD$ are integral. The distance $ AB$ has 2 digits and the distance $ CD$ is obtained by reversing the digits of $ AB$. The distance $ OH$ is a non-zero rational. Find $ AB$.

Sorry, i've found the solution in kalva, thanks for paying attention.

just because i want to practice LaTexing, Iwould put the solution.

if $AB = 10a + b$, then $CD = 10b + a$
$(OH)^2 = (CO)^2 - (CH)^2$
$= \left(\frac{1}{2}(AB) \right)^2 - \left(\frac{1}{2}(CD) \right)^2$
$= (5a + \frac{1}{2}b)^2 - (5b + \frac{1}{2}a)^2$
$= \frac{99}{4} \left(a^2 - b^2 \right)$
since OH is a rational, then $a^2 - b^2$ has to be 11
then $a+b = 11, a-b = 1$ so $a = 6, b = 5,$ yielding $AB = 65.$

I ommitted the possibility of $a+b = 1, a-b = 1$, becuase it gives negative values.

THAT WAS A NICE PRACTICE OF LATEXING btw.


If there are n lines that can divide triangle ABC into two parts, so the ratio of the areas of two parts equals to the ratio of the divided perimeter of triangle ABC. (the perimeter doesn't include the line segment that divides the triangle.) prove these n lines pass through the same point.( hint: this same point is G of triangle ABC.)

Consider a arbitrary line cut $ AB,AC$ at $ D,E$. We have:
$ \frac{S_{ADE}}{S_{BDEC}}\equal{}\frac{AD\plus{}AE}{BD\plus{}EC\plus{}CB}\Leftrightarrow \frac{S_{ADE}}{S_{ABC}}\equal{}\frac{AD\plus{}AE}{AB\plus{}BC\plus{}CA}$
$ \Leftrightarrow \frac{AD.AE}{AB.AC}\equal{}\frac{AD\plus{}AE}{AB\plus{}BC\plus{}CA}\Leftrightarrow \frac{1}{AD}\plus{}\frac{1}{AE}\equal{}const$
Let $ DE$ cut bisector of angle $ A$ at $ I$ and construct $ D'E'$ pass through $ I$ ang perpendicular with $ AI$. ($ D',E'$ on $ AB,AC$)
Apply Menelaus theorem for triangle $ AD'E'$ we have:
$ \frac{DA}{DD'}.\frac{ID'}{IE'}.\frac{EE'}{EA}\equal{}1\Leftrightarrow \frac{AE\minus{}AE'}{EA}\equal{}\frac{AD'\minus{}AD}{DA}$
$ \Leftrightarrow \frac{1}{AD}\plus{}\frac{1}{AE}\equal{}\frac{2}{AD'}$ (Because $ ID'\equal{}IE',AD'\equal{}AE'$)
Because $ \frac{1}{AD}\plus{}\frac{1}{AE}\equal{}const$ we have $ AD'\equal{}const$, it mean $ I$ is fix point!
Similarly then all lines pass through incentre $ I$ of triangle $ ABC$!
Recheck we have solution!

good solution! thx.
Sry I gave the wrong hint. It should be the incenter of ABC.


[quote="Loser"]Will there be any results?
Which countries participated?
Can you tell me if possible?[/quote]

The contest is currently running. Problems will be posted in a few days. Russians usually participate in ARO though there might some guests invited, e.g. Chinese contenders.  :)

As I know, Blgarians, Chinese and 1 Ukrainian participate.

Now I know results of SPb guys. They may be found at http://spbolymp.hut.ru forum.

It is very useful info for me! Thank you! You are so kind.

Sorry, but I really do not know any other results. This results are gotten by mobile phone. You may call to your children and ask them aswell, yes?

Finally, you are not the onliest, who is interested in results.

You are not correct here :(
Obviously, the only man here (except you) who partially knows children on ARO and really interested in results is me. For all others, this info is just curious.

I definitely know some people from SPb, who are interested too. Fedor Bakharev, for example. They may be rarely online, but can search some information about ARO here.

It is a strange place for that... Especially for SPb guys.

[quote="Fedor Petrov"]Why? It seems to be the only mathematical olympiads forum, known in Russia. We do not have local forum in SPb, but if we would have, it will be a subforum of Russian subforum of Mathlinks.[/quote]By the way, if you want that I can arrange it for you ;)

[quote="Fedor Petrov"]Thank you. But now even general Russian forum is not active. Thank you in any case, may be somewhen in future, ok?[/quote]

I guess you should tell the SPb math talents about the Russian subforum on ML.  ;)

Well it's never going to be active unless people start using it :D

Finally, I have a file with results. Ask me by PM or e-mail and I may send it.

[quote="Valentin Vornicu"]Fedor, why don't you attach the file with the results here? :? :)[/quote]

YEs,I wanna know what our team did there.I've heard they have participated

[quote="Fedor Petrov"]OK. It is.[/quote]
10x again.


geometry
parallelogram
ratio
angle bisector
exterior angle
geometry proposed

Let $ ABC$ be a triangle.
The $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$.
The $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$.
The $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$.
The lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$.
Prove that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram.

[i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$.
Similarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined.

[hide="Source of the problem"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]

$ O_aB \parallel A_bC_b,$ both perpendicular to the bisector $ BO_b$ of the $ \angle B,$ and $ O_aC \parallel A_cB_c,$ both perpendicular to the bisector $ CO_c$ of the $ \angle C.$ The $ \triangle O_aBC \sim \triangle A_bA_cX$ are centrally similar, having parallel sides. (Since $ BA_c \equal{} s \minus{} a \equal{} CA_b$ and $ A_bA_c \equal{} b \plus{} c,$ the similarity center is the common midpoint $ A'$ of their corresponding sides $ BC \sim A_bA_c$ and the similarity coefficient $ \frac {a}{b \plus{} c}.$) Let $ D \in BC$ be foot of the A-altitude and let $ K \in BC$ be foot of the $ \angle A$ bisector $ AO_a$ of the $ \triangle ABC.$ A is the internal similarity center of the excircles $ (O_b),\ (O_c),$ $ \frac {AO_b}{AO_c} \equal{} \frac {r_b}{r_c} \equal{} \frac {DA_b}{DA_c}.$ The points $ A_a,\ K$ divide the side $ BC$ of the $ \triangle O_aBC$ in the ratios

$ \frac {A_aB}{A_aC} \equal{} \frac {s \minus{} c}{s \minus{} b},\ \ \frac {KB}{KC} \equal{} \frac {c}{b},$

while the points $ D,\ A_a$ divide the corresponding side $ A_bA_c$ of the $ \triangle XA_bA_c$ in the ratios

$ \frac {DA_b}{DA_c} \equal{} \frac {r_b}{r_c} \equal{} \frac {s \minus{} c}{s \minus{} b},\ \ \frac {A_aA_b}{A_aA_c} \equal{} \frac {s \minus{} a \plus{} s \minus{} b}{s \minus{} a \plus{} s \minus{} c} \equal{} \frac {c}{b}.$

Thus $ A_a \sim D,\ K \sim A_a$ are corresponding points of the centrally similar $ \triangle O_aBC \sim \triangle A_bA_cX,$ therefore $ O_aA_a \parallel XD,\ O_aK \parallel XA_a.$ $ A \in O_aK$ by definition, $ O_aA \equiv O_aK.$ $ A \in XD,$ $ XA \equiv XD,$ because $ AD \perp BC$ by definition and $ O_aA_a \perp BC,$ hence $ XD \perp A_bA_c \equiv BC.$

[quote="yetti"]$ O_aB \parallel A_bC_b,$ both perpendicular to the bisector $ BO_b$ of the $ \angle B,$ and $ O_aC \parallel A_cB_c,$ both perpendicular to the bisector $ CO_c$ of the $ \angle C.$ [/quote]
Another solution of using this.
Denot $ P,Q\equal{}l\cap XA_{b},XA_{c}$ where $ l$ is line that pass $ A$ and parallel to $ BC$. Then triangle $ O_{a}BC$ and $ XPQ$ are homothety. $ BA_{b}\equal{}BC_{b}$ imply $ AP\equal{}AC_{b}\equal{}A_{a}B$, and similarly $ AQ\equal{}A_{a}C$. Hence $ PQ\equal{}BC$. Therefore, triangle $ O_{a}BC$ and $ XPQ$ are congruent and homothety.Since $ AP\equal{}A_{a}B$ and $ AQ\equal{}A_{a}C, A_{a},A$ are corresponding points.So $ XA\equal{}O_{a}A_{a}$ and $ XA\parallel O_{a}A_{a}$.Hence $ AO_{a}A_{a}X$ is parallelogram.


inequalities
linear algebra
linear algebra unsolved

Let $X, Y\in M_{2}(\mathbb{C})$ such that \[ det(X^{2}+Y^{2})+det(XY+YX)=\frac{1}{2}. \]
  Prove that
                                            \[ det(X+Y)+det(X-Y)\leq\sqrt{2}. \]

i'll make it short because i'm in a rush. i have a biology test paper tomorrow. i will explain tomorrow more, if necesarry. here we go . let's calculate

$det{(X+Y)^2}=det(X^2+Y^2+XY+YX)=det(X^2+Y^2)+det(XY+YX)+\alpha=\frac 12 + \alpha$ , and also

$det{(X-Y)^2}=det(X^2+Y^2-(XY+YX))=det(X^2+Y^2)+det(XY+YX)-\alpha =\frac 12 - \alpha$

add them , and u get $det{(X+Y)^2}+det{(X-Y)^2}=1$ , so ${[det(X-Y)]}^2+{[det(X+Y)]}^2=1$

now we'll apply that inequality $\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2}$ for $det(X+Y)$ and $det(X-Y)$

so u get that $\sqrt{\frac 12} \geq \frac{det(X+Y)+det(X-Y)}{2}$ from which u get the conclusion

i'd like to thank my fans, appologyse for the awful writing, and admire Cezar Lupu cause he's cool and does Math in the same time


problem 1

$ \sqrt {x^{\rm{2}}  \minus{} 16}  \minus{} \left( {x \minus{} 4} \right) \equal{} \sqrt {x^{\rm{2}}  \minus{} 5x \plus{} 4}$

problem2

\[ 2\left( {x^2  \plus{} \frac{1}{{x^2 }}} \right) \minus{} 9\left( {x \plus{} \frac{1}{x}} \right) \plus{} 14 \equal{} 0
\]

members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com

[hide="#1"]
$ \sqrt{(x\minus{}4)(x\plus{}4)}\minus{}\sqrt{(x\minus{}4)(x\minus{}1)}\equal{}x\minus{}4$
$ x^2\minus{}16$ is concave upwards with solutions of $ x\equal{}\pm4$, thus $ x\le\minus{}4$ & $ x\ge4$
Clearly, $ \sqrt{0}\minus{}\sqrt{0}\equal{}0\Longrightarrow\boxed{x\equal{}4}$[/hide]

[hide="Another Solution for 1"]$ y \equal{} x \minus{} 4$
[list]
$ \sqrt {y(x \plus{} 4)} \minus{} y \equal{} \sqrt {y(x \minus{} 1)} \implies \sqrt {x \plus{} 4} \minus{} \sqrt {x \minus{} 1} \equal{} \sqrt {y}$

$ y \equal{} (x \plus{} 4) \minus{} 2\sqrt {(x \plus{} 4)(x \minus{} 1)} \plus{} (x \minus{} 1) \implies$

$ (x \minus{} 4) \equal{} (2x \plus{} 3) \minus{} 2\sqrt {(x \plus{} 4)(x \minus{} 1)}$

$ 2\sqrt {(x \plus{} 4)(x \minus{} 1)} \equal{} x \plus{} 7 \implies 4(x \plus{} 4)(x \minus{} 1) \equal{} x^2 \plus{} 14x \plus{} 49$

$ 3x^2 \minus{} 2x \minus{} 65 \equal{} 0 \implies x \equal{} \frac {1\pm14}{3} \implies x \equal{} \boxed{5}, x\not\equal{}\minus{}\frac{13}{3}$

[/list]
[/hide]

[hide="2"]$ 2\left(x^2 \plus{} \frac {1}{x^2}\right) \minus{} 9\left(x \plus{} \frac {1}{x}\right) \plus{} 14 \equal{} 0 \implies 2x^4 \minus{} 9x^3 \plus{} 14x^2 \minus{} 9x \plus{} 2 \equal{} 0$

$ \equal{} (x \minus{} 1)^2(2x \minus{} 1)(x \minus{} 2)$

So $ x \equal{} \boxed{1, 1, \frac {1}{2}, 2}$[/hide]


First equation: $3(m+n)=24$
Second equation: $m-n=12$

Now:

$m-n = 12 \implies m+n = 12+2n \implies 12+2n = 8 \implies n =-2 \implies m = 10 \implies m \cdot n =-20$

 :mad:

Here is a worse solution:
$3(m+n)=24$, so $m+n=8$ and we are given that $m-n=12$.  Therefore, $(m+n)^{2}-(m-n)^{2}=4mn=8^{2}-12^{2}=64-144=-80$.  Therefore, $4mn=-80\Rightarrow mn=-20$

[quote="xxxxx"]Find mn if 3 times the sum of m and n is 24 and the difference of m and n is 12.[/quote]
[hide]$3(m+n)=24;\,m-n=12$

$m+n=8;\,m-n=12$

$2m=20\Rightarrow m=10$

$10+n=8\Rightarrow n=-2$

$mn=\boxed{-20}$[/hide]


Prove that given a circle and a tangent to that circle, the tangent line and the radius that meets the tangent line form a right angle.

[quote="arsemesomething"]can we use the fact that angles in a semicircle are 90 degress?[/quote]

I thought there were $180^\circ$ in a semicircle, which isn't that just gotten from the fact that there are $360^\circ$ in a circle?

i meant that if you have a semicircle, and a point on its arc, the angle at that point subtended by the diameter is 90 degrees.

can the tangen theorem be considered a special case of the chord theorem, where the two ends of the chord is the same point?


modular arithmetic
algebra proposed
algebra

Find alls integers n and k :
$k^n\equiv 0\pmod{n^k +1}$

$(2,3)$ is the only positive solution. This is because for any $n>2, n^k > k^n$ if and only if $k > n$. So $n^k - k^n$ would be smallest if $k = n + 1$, and $n^{(n+1)} - (n+1)^n$ increases with $n$.

$(0,k)$ and $(n,0)$ work for all positive $n$ and $k$. $(0,0)$ doesn't work. These are trivial.

If $n<0$, the LHS is a fraction, and a fraction mod an integer is a fraction (if you even consider it to exist), so that doesn't work.

If $k<0$, you've got fractions again, so if you want to worry about fractional mods, it seems like there ought to be a solution. Namely, $(2,-3)$. For positive $n$, that should be the only solution, by the same argument as above.

For both $n$ and $k$ less than 0, we've got $(-2,-1)$. I believe that's the only one, but I'm tired and I don't want to think about how to prove that.


If we count by 3's starting with 1, the following sequence is obtained: 1, 4, 7, 10,...
What is the 100th number in the sequence?

i think that's wrong because the sequence starts with 1.

Forgot to subtract 2  :(
[hide="Corrected answer"]$298$[/hide]

298. We see that the 100th number is obviously 3x100 which is 300. Subtracting 2 by the first and last digits according to a formula which I forget what it is.

pplz u just revived so many old topic.
plz stop as nick42 has already mentioned


inequalities
analytic geometry
trigonometry
Pythagorean Theorem
geometry
triangle inequality

Given:  isosceles triangle $ PAB$, with $ |AP|\equal{}|PB|$ and $ C$ on side $ PB$, strictly between $ P$ and $ B$.

Prove:  $ |CB|<|AC|.$

On the coordinate plane, let $ A \equal{} ( \minus{} m,0), B \equal{} (m,0), P \equal{} (0,n), C \equal{} (x,y)$. Drop a perpendicular from $ C$ onto $ AB$ and call the intersection point at $ (x,0)$ $ D$. Since $ C$ is on side $ PB$, strictly between $ P$ and $ B, 0 < x < m$. Therefore, $ AD > DB$ and from the Pythagorean theorem, it follows that $ AC > BC$.

[quote="gooeymeister"]On the coordinate plane, let $ A \equal{} ( \minus{} m,0), B \equal{} (m,0), P \equal{} (0,n), C \equal{} (x,y)$. Drop a perpendicular from $ P$ onto $ AB$ and call the intersection point $ D$. Since $ C$ is on side $ PB$, strictly between $ P$ and $ B, 0 < x < m$. Therefore, $ AD > DB$ and from the Pythagorean theorem, it follows that $ AC > BC$.[/quote]

I'm not following this -- could you clarify?  Specifically:

First, since the triangle is isosceles with apex $ P$, $ D$ is the midpoint of $ A$ and $ B$, and is thus the origin.

Second, for the same reason, $ AD \equal{} DB$, not $ AD > DB$.

Third, in your last sentence, if the proof is still viable despite the first two comments, to what triangle are you applying Pythagoras's theorem?

(Note that my intuition is that there must be a way to solve this through repeated application of the triangle inequality, but I'm not seeing it.)

[quote="Peterhi"]Given:  isosceles triangle $ PAB$, with $ |AP| \equal{} |PB|$ and $ C$ on side $ PB$, strictly between $ P$ and $ B$.

Prove:  $ |CB| < |AC|.$[/quote]

It is clear that $ \angle CAB < \angle PAB$ , because $ AC$ is an internal ray to the angle $ \angle PAB$ but $ \angle PAB \equal{} \angle PBA \longrightarrow \angle PBA > \angle CAB \longrightarrow CA > CB$ as desired  :)

[quote="gooeymeister"][quote]First, since the triangle is isosceles with apex P, D is the midpoint of A and B, and is thus the origin. [/quote]

Sorry, I miswrote $ C$ as $ P$ and have now corrected it.[/quote]

Gooey, could you explain your use of the Pythagoras's theorem in the last point of your original post?

Thanks.

Since $ AD \bot CD$, $ AD^2\plus{}CD^2\equal{}AC^2$. Similarly, $ BD^2\plus{}CD^2\equal{}BC^2$. Therefore, $ AD>BD \Rightarrow AC^2>BC^2 \Leftrightarrow AC>BC$.

[quote="gooeymeister"]Since $ AD \bot CD$, $ AD^2 \plus{} CD^2 \equal{} AC^2$. Similarly, $ BD^2 \plus{} CD^2 \equal{} BC^2$. Therefore, $ AD > BD \Rightarrow AC^2 > BC^2 \Leftrightarrow AC > BC$.[/quote]

Very nice, thanks.

Trigonometric Solution:

 [geogebra]dc33a2b03ce2ab5b74b9c06b2776cc4dc4452079[/geogebra]

By the Sine Law, it follows that $ \frac{CB}{AC}\equal{}\frac{\sin{\angle{CAB}}}{\sin{\angle{CBA}}}$. Since $ \angle{CAB} < \angle{CBA} < 90$, and because $ \angle{CAB}$ increases so does $ \sin{\angle{CAB}}$ (by the graph of sine) it also follows that $ \sin{\angle{CAB}} < \sin{\angle{CBA}}$ therefore $ \frac{\sin{\angle{CAB}}}{\sin{\angle{CBA}}}\equal{}\frac{CB}{AC}<1$ and $ CB<AC$.


number theory
greatest common divisor
modular arithmetic
factorial
induction
relatively prime
number theory unsolved

Let $ n$ and $ q$ be positive integers. Prove that the sum

$ \sum_{k \equal{} 0}^{n \minus{} 1} \binom{\left(n,k\right)q}{\left(n,k\right)}$

is divisible by $ nq$.

Here, for any two integers $ x$ and $ y$, the term $ \left(x, y\right)$ denotes the greatest common divisor of $ x$ and $ y$, and the term $ \binom{x}{y}$ denotes the binomial coefficient $ \frac {x\left(x \minus{} 1\right)...\left(x \minus{} y \plus{} 1\right)}{y!}$.

[quote="snake_pissken"]Let $ n$ and $ q$ be positive integers. Prove that

the sum of binomial coefficients

$ \sum_{k \equal{} 0}^{n \minus{} 1} \binom{(n,k)q}{(n,k)}$

is divisible by $ nq$, where $ (x, y)$ denotes the greatest common divisor of $ x$ and $ y$.[/quote]

The sum can be easily rewritten as $ \sum_{d|n}\phi(\frac nd)\binom{dq}d \equal{} q\sum_{d|n}\phi(\frac nd)\binom{dq \minus{} 1}{d \minus{} 1}$, so all we have to do is to prove that $ n|\sum_{d|n}\phi(\frac nd)\binom{dq \minus{} 1}{d \minus{} 1}$. Now, using the well known identity $ \phi(m) \equal{} \sum_{d|m}\mu(d)\frac md$ we can rewrite again the sum as $ \sum_{m|n}\frac nm\sum_{d|m}\mu(d)\binom{\frac mdq \minus{} 1}{\frac md \minus{} 1}$, so that to prove that it is congruent to $ 0$ modulo $ n$ it is enough to prove that
\[ m|\sum_{d|m}\mu(d)\binom{\frac mdq \minus{} 1}{\frac md \minus{} 1}\]
but this last is a trivial task, because if $ p^e$ is a prime power factor of $ m$, you can group the summands $ \binom{\frac mdq \minus{} 1}{\frac md \minus{} 1}$ and $ \binom{\frac m{d'}q \minus{} 1}{\frac m{d'} \minus{} 1}$ in couples such that $ \frac md$ and $ \frac m{d'}$ differ in a power of $ p$, and they are congruent modulo $ p^e$.

NOTE: I will clarify the last assertion with an example. Suppose we want to prove that $ 300$ divides $ \sum_{d|300}\mu(d)\binom{\frac {300}dq \minus{} 1}{\frac {300}d \minus{} 1} \equal{} \binom{300q \minus{} 1}{300 \minus{} 1} \minus{} \binom{150q \minus{} 1}{150 \minus{} 1} \minus{} \binom{100q \minus{} 1}{100 \minus{} 1} \minus{} \binom{60q \minus{} 1}{60 \minus{} 1} \plus{} \binom{50q \minus{} 1}{50 \minus{} 1} \plus{} \binom{30q \minus{} 1}{30 \minus{} 1} \plus{} \binom{20q \minus{} 1}{20 \minus{} 1} \minus{} \binom{10q \minus{} 1}{10 \minus{} 1}$, then it suffices to prove that it is a multiple of $ 4$, of $ 3$ and of $ 25$.
To prove that $ 25$ divides the above sum it suffices to prove that $ 25$ divides $ \binom{300q \minus{} 1}{300 \minus{} 1} \minus{} \binom{60q \minus{} 1}{60 \minus{} 1}$, $ \binom{150q \minus{} 1}{150 \minus{} 1} \minus{} \binom{30q \minus{} 1}{30 \minus{} 1}$, and so on, grouping together the summands who have coefficients as $ 300, 60$ and $ 150, 30$ differing in a $ 5$ factor.
Now, as an example, we prove that $ 25$ divides $ \binom{300q \minus{} 1}{300 \minus{} 1} \minus{} \binom{60q \minus{} 1}{60 \minus{} 1}$, what is inmediate because
\[ \binom{300q \minus{} 1}{300 \minus{} 1}\equiv\frac {300q \minus{} 1}{300 \minus{} 1}\frac {300q \minus{} 2}{300 \minus{} 2}\frac {300q \minus{} 3}{300 \minus{} 3} \frac {300q \minus{} 4}{300 \minus{} 4}\frac {300q \minus{} 5}{300 \minus{} 5}\cdots\equiv\frac {300q \minus{} 5}{300 \minus{} 5} \frac {300q \minus{} 10}{300 \minus{} 10}\frac {300q \minus{} 15}{300 \minus{} 15}\cdots\equiv\frac {60q \minus{} 1}{60 \minus{} 1} \frac {60q \minus{} 2}{60 \minus{} 2}\frac {60q \minus{} 3}{60 \minus{} 3}\cdots\equiv\binom{60q \minus{} 1}{60 \minus{} 1}\pmod{25}\]

Maybe you should rename the subject "quite fun".  I liked this problem.

I want to give a little motivation to the solution I am posting.  There seemed to be two parts of the problem to me, one of them was how to control the gcd thingy and the other was how to finish it after you can phrase that question in the right way.  To provide some insight into how the second part would work out, I tried the simpler cases of $n = p^{k}$, where it is clear exactly how the gcds will work out.  The trick that fell out from there ended up being the final trick in the full solution.  And it was because that trick existed, then I decided to consider one prime power at a time and to provide all the conditions necessary for that one final trick.

Here is the actual solution after that introductory junk:

We first clean up a little clutter and restate the problem.  To get rid of $q$ we remember that $\binom{aq}{a}= q \binom{aq-1}{a-1}$.  Secondly, in order to more clearly understand what is happening, we sum in a different way.  We consider all of the possible $k$ for a given $(n,k)$ and realize that we are going to have $\phi(n/(n,k))$ duplicates.  Thus we can rewrite the entire problem as: \[n|\sum_{d|n}\phi(\frac{n}{d})\binom{dq-1}{d-1}\] Now, we consider one prime power at a time.  That is to say we let $n = p^{k}t$ where $(p,t) =1$ \[p^{k}|\sum_{d|p^{k}t}\phi(\frac{p^{k}t}{d})\binom{dq-1}{d-1}\] Now is the big trick.  We hope that this problem is not that bad and hope that this big sum actually decomposes into many smaller reasonable sums that are also all zero in mod $p^{k}$. In fact, this doesn't require that much faith after having checked the cases when $n= p^{k}$.  We now consider divisors $d'$ of $t$   and for each divisor we notice that we have the divisors $d', d'p, \cdots, d'p^{k}$ of $n$.  So we only consider one of these sums at a time: \[\sum_{a=0}^{k}\phi(\frac{p^{k-a}t}{d'})\binom{p^{a}d'q-1}{p^{a}d'-1}\] We now remember that $\phi(x)$ is multiplicative so we factor out the useless (it turns out) $\phi(\frac{t}{d'})$  So we have: \[(1)\binom{p^{k}d'q-1}{p^{k}d'-1}+(p-1)\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}+\cdots+(p^{k}-p^{k-1})\binom{d'q-1}{d'-1}\] Now remember that we need $p^{k}$ to divide this junk.  Our first step is to prove that: \[\binom{p^{k}d'q-1}{p^{k}d'-1}\equiv \binom{p^{k-1}d'q-1}{p^{k-1}d'-1}\pmod{p^{k}}\] But if you just write out the terms in the factorials for the LHS, you notice that all the terms on the top and on the bottom are the same in mod $p^{k}$.  Thus we expect it to be 1. However, we must be careful that some of the terms are not relatively prime to $p$, so division is a little tricky.  We save all of those leftover terms and cancel a factor of $p$ from each of them.  What we have left over is exactly equal to the RHS.
So now all we need is that: \[p^{k}|(p)\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}+(p^{2}-p)\binom{p^{k-2}d'q-1}{p^{k-2}d'-1}+\cdots+(p^{k}-p^{k-1})\binom{d'q-1}{d'-1}\] But we can know divide out one factor of $p$ and continue the process through induction or whatever. \[p^{k-1}|(1)\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}+(p-1)\binom{p^{k-2}d'q-1}{p^{k-2}d'-1}+\cdots+(p^{k-1}-p^{k-2})\binom{d'q-1}{d'-1}\] At the very end we will just have that we need: \[1|\binom{d'q-1}{d'-1}\] which is obvious.  This works for all the prime power divisors of $n$, so we are done by CRT or whatever.  WOOHOO!

I think you are expert at Number Theory. Thank you for your solution.

Wow, I'm flattered.  You give me too much credit, sometimes I solve the problems, other times I don't.  All I can say is that I really like number theory.  That's why it was no problem at all for me to post this solution, it's fun for me.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Prove that: $\exists A \in M_n (R) $ invertible, such that \[ :A + A^{ - 1}  = I_n  \Leftrightarrow n = 2k,k \in N. \]

We need an invertible matrix $A$ with minimal polynomial $x^2-x+1$. The characteristic polynomial has the roots $\omega,\bar{\omega}$, and they must have the same multiplicity, because the polynomial is real. This means that the degree is even.

If $n$ is even, we must find an example. I bet that's not hard to do :).

:) indead

by using companion matrics :
consider A=  [0   -1
                     1    1]

its inverse is [1    1
                     -1   0]

now just put these matrices as blocks on the diagonal of the 2n dimensional matrix and u have a good A

fredbel6's $2\times2$ blocks:

$A=\begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$

$A^{-1}=\begin{bmatrix} 1 & 1 \\ -1 & 0 \end{bmatrix}$

Of course, any matrix that satisfies the condition of this problem also satisfies $A^3=-I$ and thus also $A^6=I.$


Hey!

I know the AOPS books are great for AMC and AIME, but what are some other good books

and is Book 2 AOPS good for the AIME

What are some good books for AMC AND AIME??

AoPS books seem to be the "standard" books for preparing for the AMC and AIME.  If AoPS Volume 1 and 2 aren't helping you for one reason or another, you could check out the AoPS Intro and Intermediate books that are geared toward specific subjects.  Most other books that are geared toward contest math are closer to USAMO/IMO level.  One book that many people recommend, although I don't like it that much, is Art and Craft of Problem Solving, which is somewhere between AIME and USAMO level.

I suggest you stick with the AoPS Volume 1 and 2 books for learning new concepts, and focus on solving a lot of problems.

[quote="mathswimmer"]You should master the 4 introductory books and have a good understanding of the new intermediate books coming out this year.[/quote]This is not good advice. Besides being prohibitively expensive, buying *all* the AoPS books is unnecessary. I would be very surprised if *anyone* had the time to really master all the books. I think they're best if you are weak in a particular subject and/or actually do need an introduction to it.

So what did people use to prepare for the amc's (or ahsme) before aops came out?

I think that's why AoPS has been so successful as *the* introductory math problem solving books, there really wasn't anything comparable before them. Does it matter what people used before? All I could think of is past tests and math circles (if you're lucky in that regard).

[quote]What are some good books for AMC AND AIME??[/quote]

[quote]what did people use to prepare for the amc's (or ahsme) before aops came out[/quote]

Consider the Contest Problem Books I-IX and all the other books in the Problem Books category available from the
MAA on-line bookstore and major on-line book retailers.

Steve Dunbar
MAA Director, American Mathematics Competitions


[quote="mathclass"]Is it sufficient to find integers (c1, c2, c3, c4, c5, c6) that satisfy the conditions and prove it? Do we have to prove that it is the only such representation, or anything like that?[/quote]
Does the problem ask for that? If you are uncertain, reread the problem. Solutions are also subject to Rule 1 on the first page ("1. You must show your work and prove your answers on all problems.")

Erin Schram
USAMTS grader


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a group such that $ G'$ is abelian and each normal and abelian subgroup of $ G$ is finite. Prove that $ G$ is finite.

[quote="Ahwingsecretagent"]Sorry what is $ G'$?[/quote]

[url=http://en.wikipedia.org/wiki/Commutator_subgroup]here[/url], note that this is fairly difficult to deal with.

Isn't $ \mathbb{Z}/p^{\infty}$ a counterexample? Is the problem supposed to be worded differently?

If $ G$ is itself abelian, it's one of those subgroups that has to be finite.

Suppose G is a metabelian group in which every abelian normal subgroup is finite.

Since G'=[G,G] is abelian it is finite.  By Zorn's lemma, choose some maximal abelian normal subgroup N containing [G,G].  Consider z in the centralizer of N. Since G/N is abelian, the subgroup <z,N> is normal, and since z centralizes itself and N, <z,N> is abelian normal.  Since N is maximal, <z,N>=N, and N is its own centralizer. Hence the map from G into Aut(N) has kernel N, and G is an extension of the finite abelian group N by a finite abelian subgroup of the finite group Aut(N).


Fie $f: R\to R$, $f\neq 0$ o functie continua, periodica, F o primitiva a sa si g un polinom de grad cel putin 2. Sa se arate ca $Fog$ nu este periodica.
-----
Am reusit sa rezum problema la cazul F periodica. Altfel zis, cum pot demonstra ca daca compunem o periodica cu o polinomiala de grad cel putin 2, nu mai obtinem o periodica?

La asta am dat o demonstratie pe forumu de calculus. Era un post a lu perfect-radio. Vezi si tu daca e buna solutia.  Poate gasesti una mai simpla  :)


The product of the base seven numbers $ 24_7$ and $ 30_7$ is expressed in base seven. What is the base seven sum of the digits of this product?

Converting to base 10, $ 18\cdot{21}\equal{}378$ which is $ 1050$ in base 7, so $ 6$

Or, you can say
[code]
 1
  24
 x30
-----
1050
[/code]


probability quadratics MATHCOUNTS AMC USA(J)MO USAMO algebra

This round isnt going to be hard, for this is the first of the series of competition and I wanted to open it up to even the least skilled people (no offence :))

OK, Question 1:
Correct Answer: 1 points
Each different ways of arriving at the answer: 1/4 points

What is the probability of rolling 2 dices and getting (2^(1/3)+1)(4^(1/3)-2^(1/3)+1) as the sum of the two numbers rolled?

Question 2:
Correct Answer: 2 points
Each different ways of arriving at the answer: 1/4 points

Find the equation of a line going through the origin and (2, (5^(1/3)+25^(1/3)-(2*40^(1/3))/(5^(1/3)-1)) in point-intercept form

Question 3:
Correct Answer: 3 points
Each different ways of arriving at the answer: 1/2 points

Find all quadratic equation which goes through (a,b), (b,c), and (c,b) where as a=(5^(1/3)+25^(1/3)-(2*40^(1/3))/(5^(1/3)-1), b=8^(1/3)+64^(1/3), and c=((((8/27)^(-1/3))^(3/2))/(3/2)^(-1/2))^1/2

Enjoy! and remember, use spoilers!

Finally I get an answer!

As for the question, [hide]your correct, but can you explain the (2^1/3+1)(2^2/3-2^1/3+1)=3 part a bit more?[/hide]

You know, I never realized how hard it is to select a phrase or an equation from a Spoiler text

I should give you partial credit because you double-posted, but oh well. Oh, and by the way, [hide]you're correct![/hide]

Yup, that's definitely one of the key points to this problem.

note that I was not competing in the first place...

[quote="Syntax Error"]why does he only get partial credit? he was busy[/quote]
 :?: Why does busy people gets to double-post? I mean, those two things aren't connected...(or so I think? :) )

and MysticTerminator, sorry.

im not participating either, just postin mass amounts of problems

Okey dokey. I'd still love it if you refrain from posting anymore questions until I have sufficient participants for the competition. I mean, your problems can be solved anytime but the competition has to end by September 30th, you know. :)

Am I being too self-centered? :P

Waiit..then who IS going to participate?

i dunno, but ill keep on posting problems cause it draws MORE attention.  :twisted:

um...you realize simon would easily destroy me in such a competition...but yeah, besides that i think that's a good idea

[quote]I should learn how to write problems, so maybe I should write such contests for others rather than solving other problems.[/quote]

Simon, are you majoring in math in college? Anyways, you still have the Putnam.

I wouldn't consider not majoring in math for a second.

Oh, come on, Syntax!
We're just dividing those precious attention amongst ourselves.

Can you do this, as a favor?

[quote="TheSouthpaw"]Personally, I think there should be different divisions for member of the month...

One for the MathCounts participants

One for the AMC people

One for the USAMO making people

and One for Simon and MysticTerminator  :wink:[/quote]
The last one's no good *booing and geering from the audience* but as for the rest, I'll post my "plans" in October. Till then, you'll just have to speculate :)

[quote="MysticTerminator"]i feel confused and betrayed...[/quote]

Again, I publically apologize.
I mean, what else can I do? harakiri? Chinese water torture? The ripply thing!? :)

[hide]1.  (2^(1/3)+1)(4^(1/3)-2^(1/3)+1)=8^(1/3)-4^(1/3)+2^(1/3)+4^(1/3)-2^(1/3)+1=3

Total possiblities of 2 dice = 6*6 = 36

Combinations that yield 3 = 1,2 and 2,1 = 2 combinations

2/36 = 1/18, the answer

2.  The equation of the line will be y = m*x + b.

The line passes through 0,0 so that when x = 0, and y = 0, b must equal 0.  So, 2*m = (5^(1/3) +25^(1/3) - (2*(40^(1/3))))/(5^(1/3)-1)

Simplify

2*m = (25^(1/3)-(3*5^(1/3)))/(5^(1/3)-1)

so, y = 2*(25^(1/3)-(3*5^(1/3)))/(5^(1/3)-1)*x, or

y = 2*(5^(1/3)*(5^(1/3)-3))/(5^(1/3)-1)

3.  You can get three equations, with a, b, and c currently not simplified.

d*a^2+e*a+f = b

d*b^2+e*b+f = c

d*c^2+e*c+f = b

Solving for f in the first equation, pluging it into the second and solving for e, pluging both of them into the third and solving for d gives that:

d = 1/(a-b)

e = -(a+c)/(a-b)

f = (a*(b+c)-b^2)/(a-b)

When simplified, b comes out as 6, and c as 3/2.

a simplifies to (5^(1/3)*(5^(1/3)-3))/(5^(1/3)-1)

Leaving a as a (otherwise the final equation is very long), I get:

y = (2*x^2-(2*a+3)*x+3*(5*a-24))/(2*(a-6))[/hide]

OK,. be patient with me here bookworm, because I'm making some tough decisions to make here...I'll be done in a minute.

Valo, (OK in Spanish, fsi (for somebody's info  :)) here is the official score: 4 1/2 points
Explanation: The first one is correct. 1 and a quarter points.
The second one is correct, but it could be improved upon, so half credit for the answer. However, you're on the right track, so I'll give you the quarter points. Total of 1 and a quarter points.
The third one is the same as the second. Good answer, not the best. 2 points.
Making up the total of 4 and a half points.

'Tis truly sad that there are less people actually participating than the people who said "Count me in!" :(


analytic geometry
graphing lines
slope
AMC
AIME
quadratics
calculus

=) Hello. This is a question from the 2005 Fermat contest, which is written in Ontario, Canada. It was the last, most difficult question, and nobody in the school solved it (although some guessed). Apparently it took the entire math department and a few grade 12's about half an hour working together to solve it (the contest itself was only an hour long). A full solution written for someone at a Grade 11 level (so that I can understand this interesting problem :)) would be greatly, greatly appreciated! An interesting problem for all, I hope ^_^

A 'triline' is a line with the property that three times its slope is equal to the sum of its $x$-intercept and its $y$-intercept. For how many integers $q$ with 1<= q <= 10,000 is there at least one integer $p$ so that there is exactly one triline through $(p,q)$?

A) 60
B) 57
C) 58
D) 61
E) 59

btw, Fermat is written in more than just Ontario. Students from all over Canada write it.

Hey... My Solution was actually correct.. That's surprising..

Now I should just hope that I didn't do anything silly..  :D

By the way, I am in grade 11, and blahblahblah's solution is exactly the same as mine..

[quote="blahblahblah"]I'll sketch a solution, the problem isn't that hard (if I'm reading it correctly).

Consider the line through $p,q$ with slope $m$.  By the triline equation, we get:

$3m=q-pm+p-q/m\Rightarrow$
$(p+3)m^2-(p+q)m+q=0\Rightarrow$ two solutions unless $\triangle=0$, therefore
$(p+q)^2-4pq-12q=0\Rightarrow$
$(p-q)^2=12q$

This is equivalent to the number of integers between $1,346$ divisible by $6$; the answer is thus $57$.[/quote]

I'm not sure I understand what you did here -- could you expand on your solution a bit more for a relative beginner?  :lol:

I especially don't understand your last part... "the number of integers between $1,346$ divisible by $6$"... did you mean the number of integers between $1,346$ *and* $1,000$? If not, then there are more than 57 integers divisible by six between 1-1,346. If so, where'd that come from?

Sorry for all these questions  :blush: I'd just like to understand what you did here....

In general, I never use commas when writing out numbers, so that should be read as the number of integers between $1$ AND $346$.

You should be able to follow the first part pretty easily, for the last part, notice that $12\vert p^2$ if and only if $6\vert p$.

I'd say this is a pretty difficult problem for a high school competition.  I wouldn't expect to see it before the end of the AIME, and maybe not even there.

good AIME problems are much harder. This is a very straightforward problem. I'll expand slightly on blah^3's soln (which is in fact the way I worked it, and really seems to be the most logical way to work it)

So we are given this fairly random condition for trilines. Let's try to quantify this. First, we know any line can be written in the form $y = mx + b$, where m is the slope and b is the y-intercept.  It's also pretty easy to find the x-intercept in this general case - simply plug 0 in for y and solve for x to get $-\frac{b}{m}$. So, the condition for "trilinity" is that thrice the slope is the sum of the intercepts, i.e. $3m = b + -\frac{b}{m}$. Cleaning this up a little, we have $b = \frac{3m^2}{m-1}$, so the general form of a triline is $y = mx + \frac{3m^2}{m-1}$. Now, we are interested in the trilines that pass through some point $(p,q)$. In specific, we want there to be a single solution for $m$ (the slope). So, plugging $(p,q)$ in for $(x,y)$, we have $q = mp + \frac{3m^2}{m-1}$. We want to solve for $m$ (as we have a condition on $m$), so doing so, we find $(p+3)m^2 - (p+q)m + q = 0$. So, as this is a quadratic in $m$, if we want this to have a single solution, the discriminant must be $0$. So, we have $(p+q)^2 - 4(p+3)q = 0$. We want to solve this for $p$ (as we have a condition for $p$ in terms of a specific $q$). Doing so, we find $(p-q)^2 = 12q$. As $q$ is an integer, there will be an integral solution for $p$ if and only if $12q$ is a perfect square. We can make this a bit easier on ourselves by noting $12q$ will be a perfect square if and only if $q$ is $3$ times a perfect square. From here, all we have to note is that $3 = 3 \cdot 1^2$ is the smallest integer of this form in the given range, and $9747 = 3 \cdot 57^2$ is the largest integer of this form in the range, for a total of $57$ eligible values of $q$.

MysticTerminator, you have made my day. I fully understand that solution, and now I see where blahblahblah is coming from :P Thank you so much for your patience with that solution!
I love this place  :lol:

How do you know that this:

[quote="MysticTerminator"]From here, all we have to note is that $3 = 3 \cdot 1^2$ is the smallest integer of this form in the given range, and $9747 = 3 \cdot 57^2$ is the largest integer of this form in the range, for a total of $57$ eligible values of $q$.[/quote]

is the same as this:

[quote="blahblahblah"]This is equivalent to the number of integers between $1,346$ divisible by $6$; the answer is thus $57$.[/quote]

Okay, so $346$ is the largest integer whose square is less than $120000$

We want integers whose square is divisible by $12$.  $12=2^3\cdot 3$, and the extra factor of $2$ doesn't help.  So we're just looking for integers divisible by $6$.

By the way, dr. rice mentioned that there's something funny about this problem.
(not to worry, the answer is 57) anybody sees anything interesting or extremely funny?


vector
linear algebra
linear algebra unsolved

Let $V$ be a 10-dimensional real vector space and $U_1,U_2$ two linear subspaces such that $U_1 \subseteq U_2, \dim U_1 =3, \dim U_2=6$. Let $\varepsilon$ be the set of all linear maps $T: V\rightarrow V$ which have $T(U_1)\subseteq U_1, T(U_2)\subseteq U_2$. Calculate the dimension of $\varepsilon$. (again, all as real vector spaces)

Trivial number 1 as always: pick a basis of $U_1$, extend it to a basis of $U_2$, extend that to a basis of $V$. Then we have fixed $3(10-3)+(6-3)(10-6)=33$ elements, and the rest can be chosen completely free. So $\dim\varepsilon = 100-33=67$.


email AMC AIME USA(J)MO USAMO LaTeX geometry

I emailed a school in Kobe, Japan (I'm moving there by early February) asking if someone could proctor the AMCs for me to take. Here is the reply:

Dear Soo-Hyun Yoo,

I am the high school counselor at Canadian Academy on Rokko Island in
Kobe, Japan.  Our headmaster forwarded your request for a test proctor
to me.   Before I can look for a proctor I will need more information.
 How much does the test cost to take and how will we get it?  Are
students such as yourself allowed to take the test independent of
attending our school?  I haven't had time yet to look at the websites.
 Please make sure that you have reviewed the conditions necessary for
you to take the test as an individual and then let me know exactly
what you will need from us.

[hide="Original email"]Dear Mr. Wesson,

I am a high school freshman (9th grade in high school) currently living in the United States of America. I will be moving to Kobe, Japan by early February.

I have regularly taken the American Mathematics Competitions (AMC) since 6th grade. This is a huge contest in which over 250,000 students participate annually. I am seeking a place to take this year's AMCs and subsequent tests in Kobe, and a proctor to administer the test.

Test format:
AMC 10/12 - taken by over 250,000 students - high scorers qualify to take the AIME
AIME - score calculated with AMC score - qualification step for USAMO
USAMO - taken by only about 500 U.S. residents, qualification step for International Math Olympiad team

Could you please see if one of the teachers at your school can offer his/her time to proctor the AMC? A quiet classroom after school is sufficient. Even a corner in the school office where I can be monitored (to ensure the authenticity of my answers) is a fine place to take the test. I would greatly appreciate it.

More information on the AMC, offered by the MAA, can be found on their website: http://www.unl.edu/amc/

Information for the AMC 10/12, as well as the order forms, can be found in their information packet: http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf

Thank you so much for taking the time to read this. Please reply as soon as possible.

Sincerely,
Soo-Hyun Yoo
[/hide]

I'm aware that there is a \$40 registration/shipping fee per school with each bundle of 10 exams sold at \$14 or \$16 for the AMC 10 and 12, respectively.

The overnight shipping cost is \$60.

The completed order form needs to be faxed to the AMC office by the deadline.

[b]However,[/b] I am not sure if I am allowed to take the test officially, registered as a student at a school that I don't attend.

Also, how should I pay the fees? Is mailing cash okay?

Could someone confirm the conditions necessary and any additional information I should provide for the school?

If you aren't sure, just take the AMC B, since by that time you'll be in Japan.  I think you can register online, but I'm not sure if you can pay by credit card or not.  If you want to ask the MAA if they will allow your condition, you should try to contact them directly.

By the way, to use Latex, surround your code with dollar symbols.  For example, instead of "x^2+z \$10" you should type
[code]$x^2+z$ $\$10$[/code]
which will give you $ x^2\plus{}z$ $ \$10$.

It wasn't if if or not I could take the A test that I wasn't sure about - it's if or not I can take the tests officially registered as a school I don't attend. I'll be homeschooled - but registered as a student at a public school.

I've already contacted the MAA weeks ago, but I haven't gotten a reply yet. Should I email them again?

Thanks though.

EDIT: I've just emailed to amcinfo@maa.org my original post above.

The AMC staff member who formerly answered email at amcinfo@maa.org retired effective December 22, then the AMC office was closed for the holidays.  Then the new person who answers amcinfo@maa.org was at the Joint Math Meetings in San Diego almost immediately after returning.  That probably accounts for the delay in answering.  As of late last week, information requests were caught up again.

Please email amcinfo@maa.org and state your question simply and clearly.  From the posts above I do not understand your request.

Further pursuit of the question here on the Forum will probably not help, your request appears to be too specfiic to your individual situation to make general advice possible.

Steve Dunbar
MAA Director, American Mathematics Competitions

My school only offers the AMC A tests. Would it be possible to take the B test at another school in the area (a school that I am not enrolled in)?

You can take the B tests at a local university or higher education site:

[url]http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-CU-list.shtml[/url]

provided that you are lucky enough to be near one.


abstract algebra
group theory
Galois Theory
superior algebra
superior algebra unsolved

I thought this was a nice problem.

[b]Problem. [/b] A [i]minimal non-abelian group[/i] $G$ is a group such that $G$ is not abelian, but every proper subgroup of $G$ is abelian. Likewise, a minimal non-nilpotent group is a group that is not nilpotent, but every proper subgroup is nilpotent.

(a) Prove that a minimal non-abelian group cannot be simple;
(b) Prove that a minimal non-nilpotent group cannot be simple;
(c)* Can a minimal non-solvable group be simple?

[I have solved (a) and (b), but not (c). ]

--Vesselin

:D  Just in case ... (c) is a joke  ;) .

First (c):

$A_{5}$ is minimal non-solvable and it is simple. I believe this is what you meant when you said that it was a joke :).

(a)

In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=abelian&t=54910&sid=5bc9072c1c1cc41498c0e44e1a303d6d]this[/url] other thread there is a proof (which can be simplified) of the stronger fact that if a group has a maximal subgroup which is abelian then it is solvable. Here's a proof of the statement (stronger than the problem (a) but weaker than the problem in that other topic):

If a finite group (non-abelian) $G$ has a maximal subgroup $A$ which is abelian, then $G$ cannot be simple.

Assume $G$ is simple. Let $g$ be an element outside $A$. If we were to have an element $1\ne x\in A\cap gAg^{-1}$, then the centralizer of $x$ in $G$ would be a subgroup of $G$ containing the maximal subgroup $A$ strictly, so $x$ would be central in $G$, which contradicts the simplicity of $G$. This means that $gAg^{-1}\cap A=\{1\}$ whenever $g\in G\setminus A$. I claim that this implies $(|A|,[G: A])=1\ (*)$.

Make $A$ act by multiplication to the left on the set of $[G: A]$ left cosets of $A$. Let $a\in A$ be an element of prime order $p\ |\ |A|$. $a$ fixes $A$ (regarded as the coset of $1$), and if $g\not\in A$, then, because $g^{-1}ag\not\in A$, we have $agA\ne gA$ (i.e. $a$ does not fix $gA$). This means that the cycle structure of $a$ as a permutation of the left cosets of $A$ consists of one cycle of length $1$, and several cycles of length $p$. This means that $p|[G: A]-1$, which proves $(*)$.

Now let $P$ be a Sylow subgroup of $A$. By $(*),\ P$ is  Sylow subgroup of $G$. It’s not normal in $G$ because we assumed $G$ to be simple, so the normalizer of $P$ is $A$. Clearly, it’s also its centralizer, so $P$ is a Sylow subgroup of $G$ contained in the center of its normalizer. By Burnside’s Normal Complement theorem (discussed on the forum several times), $P$ has a normal complement in $G$. In particular, $G$ is not simple, and we have our contradiction.

(b)

I don’t really have a proof to this, except for one (which sort of feels like cheating :)) using the following theorem of Hall (which I’ve also cited in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47434&sid=807169e25f4b3b935f52c90249cadc15]this[/url] other thread started by vess):

Let $G$ be a finite group, and let $P$ be a Sylow $p$-subgroup of $G$ with the property that for every subgroup $Q$ of $P$, every element of $G$ of order prime to $p$ which normalizes $Q$ centralizes $Q$. Then $G$ has a normal complement of $P$.

Using this, the problem becomes easy. Let $P$ be a Sylow $p$-subgroup of our minimal non-nilpotent group $G$, assumed simple, and let $Q\le P$. $Q$ is not normal, so its normalizer is contained in some nilpotent subgroup $N$ of $G$. $N$ is the direct product of its Sylow subgroups, so every element in $G$ with order prime to $p$ which normalizes $Q$ certainly centralizes $Q$. The hypotheses of the theorem above are satisfied, so $G$ has a normal subgroup, contradicting the assumption that it’s simple.


trigonometry
algebra
polynomial
complex numbers
algebra unsolved

:arrow: [color=red]Let k be a positive integer. Prove that sqrt(k+1)-sqrt(k) is not the real part of the complex number z with z^n=1 for some positive integer n.[/color]

If $\cos{\alpha}=\sqrt{k+1}-\sqrt{k}$, here $z=\cos{\alpha}+i\sin{\alpha}$ and $z^{n}=1$. Then $\cos{n\alpha}=1$ therefore $\sqrt{k+1}-\sqrt{k}$ is a root of $g(x)-1$, here $g(\cos{\alpha})=\cos{n\alpha}$. But minimal polynomial of $\sqrt{k+1}-\sqrt{k}$ is $f(x)=x^{4}-(4k+2)x^{2}+1$, so $f(x)|[g(x)-1]$ contradiction because product of moduls of all roots of $f(x)$ is equal to $1$.

And now try this problem :
http://www.mathlinks.ro/Forum/viewtopic.php?t=113831


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14613[/img]

 :(

You posted this problem at [url]http://www.mathlinks.ro/viewtopic.php?t=205642[/url]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14623[/img]

 :thumbup:


are you sure about the second equation :?:

i think the second equation must be y^2 + xy = b^2
then it reduces to the most trivial thing ..
just add (i) and (ii) .. get 2xy first add and then subtract and get x+y , x-y ... solve them :)[/img]


search
number theory unsolved
number theory

Show that for any $ a > 0$ there exist infinitely many $ n\in N$ such that $ n^2 \plus{} 1$ divides $ [an]!$.
(where $ [x]$ is integer part of x.)

Posted before.
I will solve it again :
Consider the Pell's equation :
$ x^2\plus{}1\equal{}(d^2\plus{}1)y^2$ where $ d>0$
It has smallest  positive solution :
$ (x,y)\equal{}(d,1)$
So it has infinite solution :
$ y_1\equal{}$
$ y_{n\plus{}2}\equal{}2(d^2\plus{}1)y_{n\plus{}1}\minus{}y_n$
This is a sequence of root of this equation.
So $ \gcd(y_{2n\plus{}1},d^2\plus{}1)\equal{}1$
Consider the sequence $ \{y_{2n\plus{}1}\}$
Chose d,n large enough we have $ ax_{2n\plus{}1}>2y_{2n\plus{}1},\forall n\geq n_0$
So $ [ax_{2n\plus{}1}]\geq y_{2n\plus{}1}$ and $ d^2\plus{}1<[ax_n\plus{}1]$ for all $ n\geq n_0$
So $ y_{2n\plus{}1}^2|[ax_{2n\plus{}1}]!$ and $ d^2\plus{}1|[ax_n]!$
But from $ \gcd(y_{2n\plus{}1},d^2\plus{}1)\equal{}1$ so
$ x_{2n\plus{}1}^2\plus{}1|[ax_{2n\plus{}1}]!$
Proof is complete .
Note : please use search to find the general problem I have post .
Exist infinite $ x$ such that
$ ax^2\plus{}b|[cx]!$ :wink:


Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$ f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$ for all $ x,y \in \mathbb{R}.$

$ P(x,0): (f(x) \minus{} 2)f(0) \equal{} 0$
If $ f(0)\ne0$, we get the first solution is $ f(x) \equal{} 2$.
Now $ f(0) \equal{} 0$.
$ P(2,2): f(2)^2 \equal{} 2f(2)$. So $ f(2) \equal{} 0$ or $ f(2) \equal{} 2$.
$ P(1,1): f(1)^2 \minus{} 3f(1) \plus{} f(2) \equal{} 0$. We can solve $ f(1)$ for all possible $ f(2)$ and there are four cases.

Case 1) $ f(2) \equal{} 2,f(1) \equal{} 2$
$ P(x,1): f(x \plus{} 1) \equal{} 2$, No solution.

Case 2) $ f(2) \equal{} 0,f(1) \equal{} 3$
$ P(x,1): f(x \plus{} 1) \equal{} 3 \minus{} f(x)\Rightarrow f(x \plus{} 2) \equal{} f(x)$
$ P(x,2): f(2x) \equal{} f(x \plus{} 2) \minus{} f(x) \equal{} 0$. No solution

Case 3) $ f(2) \equal{} 0,f(1) \equal{} 0$
$ P(x,1): f(x \plus{} 1) \equal{} 2f(x)\Rightarrow f(x \plus{} n) \equal{} 2^nf(x)$, where $ n$ can be any integer.
We also see $ f(n) \equal{} 0$
$ P(x,n): f(nx) \equal{} (2^n \minus{} 1)f(x)$
$ P(nx,2x)$ implies $ f(x^2) \equal{} \frac3{2^n \minus{} 1}(f(x)^2 \plus{} f(x))$ for any integer $ n$.
We can let $ n$ be arbitrarily large and we get the second solution $ f(x) \equal{} 0$.

Case 4) $ f(2) \equal{} 2,f(1) \equal{} 1$
$ P(x,1): f(x \plus{} 1) \equal{} f(x) \plus{} 1\Rightarrow f(x \plus{} n) \equal{} f(x) \plus{} n$ and $ f(n) \equal{} n$.
$ P(x,n): f(nx) \equal{} nf(x)\Rightarrow f(rx) \equal{} rf(x),\forall r\in \mathbb Q$ and $ f(r) \equal{} r$
$ P(x,r): f(x \plus{} r) \equal{} f(x) \plus{} r$
$ P(x,x): f(x^2) \equal{} f(x)^2\ge0$
For any $ x$, we find rational sequences $ r_n\uparrow x,R_n\downarrow x$.
$ f(x) \equal{} f(x \minus{} r_n) \plus{} r_n\ge r_n$ and $ f(x) \equal{} R_n \minus{} f(R_n \minus{} x)\le R_n$.
So $ f(x) \equal{} x$.

Conclusion: $ f(x) \equal{} x$, $ f(x) \equal{} 0$ and $ f(x) \equal{} 2$.


Suppose that $ f$  $ : (0,\infty) \to (0,\infty)  $ is continuous differentiable and   strict monotonic  on  $ (0,\infty) ,$ such that $ \displaystyle f^{-1}(x)=\displaystyle f^{\prime}(x) \; ,\; \forall x\in (0,\infty) .$
[b]  If $ \varphi:=\displaystyle \frac{1+\sqrt{5}}{2} , $  find $ \displaystyle  f(\varphi)\; .$ [/b]
[b]Remark : [/b] My solution need some inequalities.

I just love this problem. I saw it for the first time at oral ENS and after more than one month I finally posted a solution on the analysis forum. Fedja gave a more elegant solution. It is a very very difficult problem, I think.


vector
analytic geometry
Functional Analysis
advanced fields
advanced fields unsolved

Let $ a,b$ be unit vectors in real Banach space such that $ \parallel{}2a \plus{} b\parallel{} \equal{} \parallel{}a \minus{} 2b\parallel{} \equal{} 3$. prove that there exists continuous linear functionals $ f,g$ with the norm $ 1$ such that $ f(a) \equal{} f(b) \equal{} 1$, $ g(a) \equal{} \minus{} g(b) \equal{} 1$.

[quote="mlok"]Are the functionals required to have unit norm too? Otherwise the problem is easy: just check that $ a$ and $ b$ cannot be collinear.[/quote]

Yes,yes, sorry. they are both required to have norm $ 1$.

Actually, such a and b cannot exist. Mark the points (1,0), (0, 1), (2,1), (1,-2) on a coordinate plane, then mark the points symmetric to them about (0,0). There is no convex region that contains all these points on the boundary.

Once again, sorry, i've edited the problem conditions: there must be $ \parallel{}2a \plus{} b\parallel{} \equal{} \parallel{}a \minus{} 2b\parallel{} \equal{} 3$.

Mark the points (1,0), (0, 1), (2/3,1/3), (1/3,-2/3) on a coordinate plane, then mark the points symmetric to them about (0,0). The only convex region that contains all these points on the boundary is the square with vertices $ (\pm 1,0)$, $ (0,\pm 1)$. So the norm on the span of $ a,b$ is the $ \ell_1$ norm, and the required functionals can be given explicitly (and extended to the entire space by Hahn-Banach).


inequalities
calculus
function
inequalities unsolved

Find the largest value for real $k$ such that $a^2+b^2+c^2+3k\geq (k+1)(a+b+c)$ for all $a,b,c\in \mathbb{R^{+}}$ satisfy $abc=1$.
(Clearly $k\geq 1$)

And clearly $k<3$, because if we set $c$ close to $1$, and $a=b$, we get the inequality become false for $k=3$.
Can anyone help me?

very nice problem.
I guess the largest value of $k$ is $\frac 8 3$ ;)
but have no proof  :(

Nice problem !  :)
 We use mixing variables and calculus to solved this problem :
 The ineq is equivalent to : $\frac{a^2+b^2+c^2-a-b-c}{a+b+c-3} \geq k$
 Let $f(a,b,c)= \frac{a^2+b^2+c^2-a-b-c}{a+b+c-3}$
 [b]Step 1:[/b] prove  $f(a,b,c) \geq f(a,\sqrt{bc},\sqrt{bc})$ , where $a= min (a,b,c)$
  Indeed ,
$f(a,b,c) - f(a,\sqrt{bc},\sqrt{bc}) =(\sqrt{b}-\sqrt{c})^2[3-3(\sqrt{b}+\sqrt{c})^2+a(\sqrt{b}+\sqrt{c})^2-a^2+2\sqrt{bc}(b+c+\sqrt{bc})] \geq (a+2\sqrt{bc})(\sqrt{b}+\sqrt{c})^2-3(\sqrt{b}+\sqrt{c})^2+3-a-2\sqrt{bc} =(a+2\sqrt{bc}-3)((\sqrt{b}+\sqrt{c})^2-1) \geq 0$

 [b]Step 2:[/b] we find the minimum value of $f(a,t,t)$ , where $at^2=1$ and $t\geq 1$
 We have :
 $f(a,t,t)=g(t)=\frac{2t^6-2t^5-t^2+1}{2t^5-3t^4+t^2}$
 $g'(t)=\frac{2t(t+1)(t-1)^4(t^2+t+1)(2t^2-2t-1)}{(2t^5-3t^4+t^2)^2}$
 $g'(t)=0 \iff t=\frac{1+\sqrt{3}}{2}$
 So $g(t) \geq g(\frac{1+\sqrt{3}}{2}) = 2,803847578$
 So the maximum value of $k$ is $g(\frac{1+\sqrt{3}}{2}) = 2,803847578$

Hmm but the solution using calculus bothers me. Is there a solution without using calculus?
Because the function simplifies to $g(t) = (t+\frac{1}{2}) + \frac{\frac{3}{4}}{t+\frac{1}{2}} + \frac{1}{t^2}$
And also, if $x = \frac{\sqrt{3}+1}{2}$ then $x^2 = x+\frac{1}{2}$
So maybe there is a no-calculus way to find the minimum of $g(t)$?

Btw, I calculated that $g(\frac{\sqrt{3}+1}{2}) = 8 - 3\sqrt{3}$


logarithms
calculus
calculus computations

Compute  $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\plus{}1}n^2}{n^3\plus{}1}$