rotation
geometry
geometric transformation
complex numbers

q1) A triangle with vertices at $ 3 + i$ , $ i(1 + \sqrt3)$ and $ i(1 - \sqrt3)$ is rotated about its circumcentre through an angle $ \pi/3$ in the counter clockwise direction. FInd the new position of its vertices.

q2) For all complex no's $ z1$ and $ z2$ satisfying $ |z1|=12$ and $ |z2-3-4i|=5$ prove that minimum value of $ |z1-z2|$ is 2.

q3) The affix of $ k{}^t{}^h$ vertex of a polygon is given by $ zk=1+z0+z0^2+.......z0{}^k{}^-{}^1$ of some $ z0$ such that $ |z0|<1$, show that origin lies outside the polygon.

q4) $ z1,z2,z3$ are 3 pair wise distinct complex numbers and $ t1,t2,t3$ are non negative real numbers such that their sum is 1.
PRove that the complex no. $ z=z1t1+z2t2+z3t3$ lies inside a triangle with vertices $ z1,z2,z3$ or on its boundary.

1) Rotation corresponds to being multiplied by another complex number.  The angle gives you one part of the exponential, and in this case, its radius is 1 since it's just pure rotation.  So you're multiplying by the complex number $ z\equal{}e^{i\pi/3}$.

[quote="JRav"]1) Rotation corresponds to being multiplied by another complex number.  The angle gives you one part of the exponential, and in this case, its radius is 1 since it's just pure rotation.  So you're multiplying by the complex number $ z \equal{} e^{i\pi/3}$.[/quote]
 Hey , thanks for the explaination but i have put them up as i saw they are some different types of questions other for beginners , other than find the conjugate and all :lol:

Do try them  :)

2) I found that $ |z_1| \equal{} 12$ represents a circle centered at $ (0,0)$ with radius $ 12$ and $ |z_2 \minus{} 3 \minus{} 4i| \equal{} 5$ represents a circle centered at $ (3,4)$ with radius $ 5$, but I don't know what $ |z_1  \minus{} z_2|$ has to do with them.  :maybe:

[quote="JRav"]1) Rotation corresponds to being multiplied by another complex number.  The angle gives you one part of the exponential, and in this case, its radius is 1 since it's just pure rotation.  So you're multiplying by the complex number $ z \equal{} e^{i\pi/3}$.[/quote]

Of course after translating the circumcenter.  :wink:

@modularmarc101, $ |z_1 \minus{} z_2|$ is the distance between $ z_1$ and $ z_2$ (on the complex plane).


Is it possible to make a theoretical model to describe the behavior of a turbulent fluid (in particular, its internal structures)?
 ;)
 :rotfl:

In don't think so. We have only done a little on it but I think the behavior can be described as chaos. So you can predict at what point the flow of fluid will become turbulent, but after that you can't..

I think a similar case is with convection... at the moment there is no physical model for convection currents, so (as Feynmann said in his lectures) you can predict exactly the point at which convection currents will begin to flow in the star, and after that you can't predict anything. I think this is why weather forcasting is so difficult.


In a magic square, the sum of the entries of each row, column, and diagonal is the same. Find the value of $ B$ in the magic square pictured.
[asy]size(75); defaultpen(linewidth(0.8));
void drawSquare(int side = 1, pair A){
 draw(A--(A.x + side, A.y)--(A.x + side, A.y + side)--(A.x, A.y + side)--cycle);
}
for(int i = 0; i < 3; ++i){
 for(int j = 0; j < 3; ++j){
 drawSquare(1,(i,j));
 }
}
label("$6$",(0.5,2.5)); label("$8$",(2.5,2.5)); label("$5$",(1.5,1.5)); label("$B$",(1.5,0.5)); label("$4$",(2.5,0.5));[/asy]

The sum of each row, column, and diagonal is $ 6 \plus{} 5 \plus{} 4 \equal{} 15$. The first square in the third row must then have a value of $ 15 \minus{}  (8 \plus{} 5) \equal{} 2$, so B must equal $ 15 \minus{} (4 \plus{} 2) \equal{} \boxed{9}$.


Let $a;b;c$ be real numbers.Find min:
$M=\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{b+a}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}$

Chien than, why did you post this Inequality? What do you want?   (http://www.diendan3t.net/forum/showthread.php?t=1330&page=2) :ninja:
I will post the solution. $Min M=3\sqrt{2}+\frac{3\sqrt[4}{8}}{8}$ , but you mustn't post it in [url=http://www.diendan3t.net/forum]3T forum[/url]


algebra
polynomial
binomial theorem
Divisibility Theory

[quote="Peter"]Prove that the number \[\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\] is not divisible by $5$ for any integer $n\geq 0$.[/quote]

[b][Observation][/b] We evaluate the the [b]combinatorial[/b] sum
\[B_{n}= \sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\]
by finding a [b]non-cominatorial[/b] expression. By Binomial Theorem, setting

\[P_{n}= \sum_{0 \leq i \leq 2n+1, \; i \tex{: \;odd}}\binom{2n+1}{i}\left( 2^{\frac{3}{2}}\right)^{i}\;\; \text{and}\;\; G_{n}= \sum_{0 \leq k \leq 2n+1, \; i \tex{: \;even}}\binom{2n+1}{i}\left( 2^{\frac{3}{2}}\right)^{i},\]
we obtain

\[\left( 2^{\frac{3}{2}}+1 \right)^{2n+1}= P_{n}+G_{n}= 2^{\frac{3}{2}}B_{n}+G_{n}\]
and
\[\left( 2^{\frac{3}{2}}-1 \right)^{2n+1}=P_{n}-G_{n}=2^{\frac{3}{2}}B_{n}-G_{n}.\]
Adding these two identities and multiplying $2^{-\frac{5}{2}}$, we obtain

\[B_{n}= 2^{-\frac{5}{2}}\left(\left( 2^{\frac{3}{2}}+1 \right)^{2n+1}+\left( 2^{\frac{3}{2}}-1 \right)^{2n+1}\right).\]

The right hand side is the desired non-combinatorial expression.

[b][Solution][/b]  Our aim is to show that $B_{n}$ is not congruent to $0$ modulo $5$. We need his 'integral' girl friend $G_{n}$. Multiplying the above two indentities, we obtain

\[7^{2n+1}=\left( 2^{\frac{3}{2}}+1 \right)^{2n+1}\left( 2^{\frac{3}{2}}-1 \right)^{2n+1}= 8{B_{n}}^{2}-{G_{n}}^{2}.\]

Our next job is, of course, to read the above equation modulo $5$. Since $7^{2n+1}\equiv (7^{2})^{n}\cdot 7 \equiv (-1)^{n}\cdot 2 \; (mod \; 5)$, we obtain

\[\pm 2 \equiv 8{B_{n}}^{2}-{G_{n}}^{2}\; (mod \; 5).\]

From this, we see that $B_{n}\not\equiv 0 \; (mod \; 5)$. Indeed, $B_{n}\equiv 0 \;(mod \; 5)$ and the above congruence imply that

\[\pm 2 \equiv-{G_{n}}^{2}(mod \; 5).\]
However, this is a contradiction. Even [i]Homer Simpson[/i] can check that $l^{2}\equiv-1, 0, 1 (mod \; 5)$ holds for all $l \in \mathbf{Z}$. :grin:

Suppose for the sake of contradiction that our sum is a multiple of 5 for some $ n$. Then it is 0 in the finite field $ F_{25}$. Considering our equation in $ F_{25}$, we have  $ \sum_{k\equal{}0}^n \binom{2n\plus{}1}{2k\plus{}1} 8^k \equal{} 0$. $ 8 \equal{} 3$ in $ F_{25}$, so $ \sum_{k\equal{}0}^n \binom{2n \plus{} 1}{2k \plus{} 1} 3^k \equal{} 0$. Multiplying both sides by $ \sqrt{3}$, we get that $ \sum_{k\equal{}0}^n \binom{2n \plus{} 1}{2k \plus{} 1} (\sqrt{3})^{2k \plus{} 1}$. Let $ P(x) \equal{} (1 \plus{} x)^{2n \plus{} 1} \equal{} \sum_{k\equal{}0}^{2n\plus{}1} \binom{2n \plus{} 1}{k} x^k$. Then $ \frac{P(x) \minus{} P(\minus{}x)}{2} \equal{} \sum_{k\equal{}0}^n \binom{2n \plus{} 1}{2k \plus{} 1} x^{2k \plus{} 1}$. Thus, if $ \sum_{k\equal{}0}^n \binom{2n \plus{} 1}{2k \plus{} 1} (\sqrt{3})^{2k \plus{} 1}$, then $ \frac{P(\sqrt{3}) \minus{} P(\minus{}\sqrt{3})}{2} \equal{} 0$, that is, $ (1 \plus{} \sqrt{3})^n \equal{} (1 \minus{} \sqrt{3})^n$. Since $ 1 \minus{} \sqrt{3}$ is nonzero, we may multiply both sides by the inverse of $ (1 \minus{} \sqrt{3})^n$, to get $ (\minus{}2 \minus{} \sqrt{3})^n \equal{} 0$. But since $ \minus{}2 \minus{} \sqrt{3}$ is nonzero in $ F_{25}$, we arrive at a contradiction.

I'm still a beginner with finite fields (and I think this is essentially the same as ideahitme's solution), so if anyone could verify this I'd appreciate it a lot.

My idea was to obtain a linear recursion formula, which is pretty straightforward:
$ 2^{3k} \equal{} \frac{1}{\sqrt{8}}(\sqrt{8})^{2k \plus{} 1}$ Which implies the formula:
$ \sum_{k\equal{}0}^{n}\binom{2n\plus{}1}{2k\plus{}1}2^{3k} \equal{} \frac{1}{2\sqrt{8}}\left((1 \plus{} \sqrt{8})^{2n \plus{} 1} \minus{} (1 \minus{} \sqrt{8})^{2n\plus{}1}\right)$
Since $ 1 \plus{} \sqrt{8}$ and $ 1 \minus{} \sqrt{8}$ are the roots of the polynomial $ x^2 \minus{} 2x \minus{} 7$ we will need to consider the sequence $ a_n \equal{} 2a_{n\minus{}1} \plus{} 7a_{n\minus{}2}$, for which the above formula produces the odd terms. Computing shows $ a_0 \equal{} 0$ and $ a_1 \equal{} 1$. The sequence must obviously be periodic modulo 5, and straight computing shows that the sequence looks like this modulo 5:
$ 0,1,2,1,4,0,3,1,3,3,2,0,4,3,4,4,1,0,2,4,2,2,3,0,1$
Since none of the odd terms are 0 the proof is finished.

first we need to define some terms:

$a+b\sqrt{3} \equiv x \mod 5 \iff a \equiv x$ and

$a+b\sqrt{3} \equiv x \mod 5_{\sqrt{3}} \iff b \equiv x$ where $a,b \in Z$

now let $p_n(x)=(1+x)^{2n+1} \implies \frac{p_n(x)-p_n(-x)}{2}=\sum_{k=0}^n \dbinom{2n+1}{2k+1} x^{2k+1}=x(\sum_{k=0}^n \dbinom{2n+1}{2k+1} x^{2k})$

which implies $\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k} \equiv \sum_{k=0}^{n}\binom{2n+1}{2k+1}3^{k} \mod 5$

$\sum_{k=0}^{n}\binom{2n+1}{2k+1}3^{k} = \frac{p_n(\sqrt{3})-p_n(-\sqrt{3})}{2\sqrt{3}}$

now it only takes to prove: $p_n(\sqrt{3})-p_n(-\sqrt{3}) \not\equiv 0 \mod 5_{\sqrt{3}}$

because $p_n(\sqrt{3})-p_n(-\sqrt{3})$ is in form of $n\sqrt{3}$ where $n \in N$

$p_n(\sqrt{3}) \equiv -p_n(-\sqrt{3}) \mod 5_{\sqrt{3}} \implies p_n(\sqrt{3})-p_n(-\sqrt{3})\equiv 0 \mod 5_{\sqrt{3}} \iff p_n(\sqrt{3}) \equiv 0 \mod 5_{\sqrt{3}}$

which is easy to prove $p_{2n+1}(\sqrt{3}) \equiv p_{2n+25}(\sqrt{3}) \mod 5_{\sqrt{3}}$ and because

$p_{2n+1}  \not\equiv 0 \mod 5_{\sqrt{3}}$ for $1\le 2n+1 \le 25$ which implies  $\forall n: p_n(\sqrt{3}) \not\equiv 0 \mod 5_{\sqrt{3}}$

[quote=ideahitme][quote="Peter"]Prove that the number \[\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\] is not divisible by $5$ for any integer $n\geq 0$.[/quote]

[b][Observation][/b] We evaluate the the [b]combinatorial[/b] sum
\[B_{n}= \sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\]
by finding a [b]non-combinatorial[/b] expression. By Binomial Theorem, setting

\[P_{n}= \sum_{0 \leq i \leq 2n+1, \; i \text{:\;odd}}\binom{2n+1}{i}\left( 2^{\frac{3}{2}}\right)^{i}\;\; \text{and}\;\; G_{n}= \sum_{0 \leq i \leq 2n+1, \; i \text{:\;even}}\binom{2n+1}{i}\left( 2^{\frac{3}{2}}\right)^{i},\]
we obtain

\[\left( 2^{\frac{3}{2}}+1 \right)^{2n+1}= P_{n}+G_{n}= 2^{\frac{3}{2}}B_{n}+G_{n}\]
and
\[\left( 2^{\frac{3}{2}}-1 \right)^{2n+1}=P_{n}-G_{n}=2^{\frac{3}{2}}B_{n}-G_{n}.\]
Adding these two identities and multiplying $2^{-\frac{5}{2}}$, we obtain

\[B_{n}= 2^{-\frac{5}{2}}\left(\left( 2^{\frac{3}{2}}+1 \right)^{2n+1}+\left( 2^{\frac{3}{2}}-1 \right)^{2n+1}\right).\]

The right hand side is the desired non-combinatorial expression.

[b][Solution][/b]  Our aim is to show that $B_{n}$ is not congruent to $0$ modulo $5$. We need his 'integral' girl friend $G_{n}$. Multiplying the above two indentities, we obtain

\[7^{2n+1}=\left( 2^{\frac{3}{2}}+1 \right)^{2n+1}\left( 2^{\frac{3}{2}}-1 \right)^{2n+1}= 8{B_{n}}^{2}-{G_{n}}^{2}.\]

Our next job is, of course, to read the above equation modulo $5$. Since $7^{2n+1}\equiv (7^{2})^{n}\cdot 7 \equiv (-1)^{n}\cdot 2 \; \pmod{5}$, we obtain

\[\pm 2 \equiv 8{B_{n}}^{2}-{G_{n}}^{2}\; \pmod{5}.\]

From this, we see that $B_{n}\not\equiv 0 \; \pmod{5}$. Indeed, $B_{n}\equiv 0 \;\pmod{5}$ and the above congruence imply that

\[\pm 2 \equiv-{G_{n}}^{2}\pmod{5}.\]
However, this is a contradiction. Even [i]Homer Simpson[/i] can check that $l^{2}\equiv-1, 0, 1 \pmod{5}$ holds for all $l \in \mathbb{Z}$. :grin:[/quote]

[quote=BanziC]My idea was to obtain a linear recursion formula, which is pretty straightforward:
$ 2^{3k} \equal{} \frac{1}{\sqrt{8}}(\sqrt{8})^{2k \plus{} 1}$ Which implies the formula:
$ \sum_{k\equal{}0}^{n}\binom{2n\plus{}1}{2k\plus{}1}2^{3k} \equal{} \frac{1}{2\sqrt{8}}\left((1 \plus{} \sqrt{8})^{2n \plus{} 1} \minus{} (1 \minus{} \sqrt{8})^{2n\plus{}1}\right)$
Since $ 1 \plus{} \sqrt{8}$ and $ 1 \minus{} \sqrt{8}$ are the roots of the polynomial $ x^2 \minus{} 2x \minus{} 7$ we will need to consider the sequence $ a_n \equal{} 2a_{n\minus{}1} \plus{} 7a_{n\minus{}2}$, for which the above formula produces the odd terms. Computing shows $ a_0 \equal{} 0$ and $ a_1 \equal{} 1$. The sequence must obviously be periodic modulo 5, and straight computing shows that the sequence looks like this modulo 5:
$ 0,1,2,1,$[size=150][color=#c00]1[/color][/size]$,4,0,3,1,3,3,2,0,4,3,4,4,1,0,2,4,2,2,3,0,1$
Since none of the odd terms are 0 the proof is finished.[/quote]

[hide=PS]corrected [code]\tex{}[/code] to [code]\text{}  [/code],[code]mod 5[/code] to [code]\pmod{5}[/code] [code]cominatorial[/code] to [code]combinatorial[/code] and a slight error in $G_n$'s definition [code]index k[/code] to [code]index i[/code]

There was a $1$ missing(typo) in the second solution, which was messing the parity up :(
[/hide]


Let $ M$ be the mid point of $ BC$ in $ \Delta ABC$.Let $ \omega$ be the circle with diameter $ AM$.Let $ D \equal{} AB \cap \omega$ and $ E \equal{} AC \cap \omega$.Let $ l_{1}$ and $ l_{2}$ be the tangents at $ D$ and $ E$ to $ \omega$ repectively .Let $ P \equal{} l_{1} \cap l_{2}$.Prove that $ PB \equal{} PC$

[hide="hint"]Consider a projection of $ D$ and $ E$ about $ A$ to $ BC$ :wink: [/hide]

ok since no one's trying this(apparently)  let me post the solution
note that $ (DE,MM') \equal{} \minus{}1$
extend $ PM$ to meet the circle at $ M'$ project $ D \rightarrow B$ and $ E \rightarrow C$, $ M \rightarrow M$,$ M^{'} \rightarrow P$ through $ A$  since $ BM \equal{} CM$, $ P \rightarrow \infty$ so $ AM' \parallel{} BC$.
but $ AM'M \equal{} \frac {\pi}{2}$ so $ PM \perp BC$ so $ PB \equal{} PC$ :wink:

no fair....you will use projective in inmo and it will make things easier

[quote="pardesi"]i didn't use though i used inversion :D[/quote]

i saw the proof

but projective,et al was not really needed this year,eh??

but do learn it if u r going to the TC it will be really helpful :)

Too much work da Pardesi.  A simple solution:
Consider the midpoint of $ AM$ to be $ K$. Easily observe that $ KDPE$ is cyclic with diamter $ KP$. Consider midpoint of $ KP$. Let it be $ L$> Easily prove that $ LBP$ and $ LCP$ are congruent. ($ L$ is center of the cyclic quad, so we have one equality ready : $ LB \equal{} LP$). Finish the remaining with the given data ($ BM\equal{}MC$)  and other congruences to arrive at this.


AMC
AMC 10
quadratics
Pythagorean Theorem
geometry

does anybody remember any of the following problems?

2, 8, 9, 14, 15, 18

Do you want me to post the problems, and my solutions?
I will.
[hide="Problem 2"]
Define the operation * by $a*b = (a+b)b$. What is (3 * 5) - (5 * 3)
A) -16
B) -8
C) 0
D) 8
E) 16
[/hide]
[hide="Solution 2"]
Plug the numbers in:
$(3+5)(5)-(3+5)(3) = \boxed{16}= \boxed{E}$
[/hide]
[hide="Problem 8"]
On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form bbcac, where $0 \leq a < b < c \leq 9$, and $b$ was the average of $a$ and $c$. How many five digit numbers satisfy all these properties?
A) 12
B) 16
C) 18
D) 20
E) 24
[/hide]
[hide="Solution 8"]
When we pick an A and a C, we will only ever have one B.
If we fix A, we can list all the possibilities for C, and count those as possible numbers.
When A = 0;
C = 2, 4, 6, 8; (four possibilities for B to be an integer)
When A = 2;
C = 4, 6, 8; (three possibilities for B to be an integer)
When A = 4;
C = 6, 8 (two possibilities for B to be an integer)
Finally, when A = 6;
C = 8 (one possibility for B to be an integer)

The cases are symmetric when A is odd. Therefore there are $2(4+3+2+1) = 20$ numbers that satisfy the conditions, or choice $\boxed{D}$
[/hide]
[hide="Problem 9"]
A cryptographic code is designed as follows. The first time a letter appears in a given message it is replaced by the letter that is one place to the right in the alphabet (assume that A is one right of Z). The second time this same letter appears in the given message, it is replaced by the letter that is 1 + 2 places to the right, the third time it is replaced by the letter that is 1 + 2 + 3 places to the right, and so on. For example, with this code, the word BANANA becomes CBODQG. What letter will replace the last letter s in the message

"Lee's sis is a Mississippi miss, Chriss!"?
A)g
B)h
C)o
D)s
E)t
[/hide]
[hide="Solution 9"]
The letter S occurs 12 times, so it is moved 1 + 2 + ... + 12 letters to the right, or 78 letters. $26|78$, so the letter moves three full alphabets over, and stops where it was, on letter S, or choice $\boxed{D}$
[/hide]
[hide="Problem 14"]
Some boys and girls are having a car wash to raise money for a class trip to China. Initially, 40% of the group are girls. Shortly thereafter, two girls leave and two boys arrive, and then 30% of the group are girls. How many girls were initially in the group?
A) 4
B) 6
C) 8
D) 10
E) 12

[/hide]
[hide="Solution 14"]
The group size remains the same after the change, so we can say that the two girls that left accounted for (40 - 30) = 10% of the overall group.
This means that the original group had (2)(10) = 20 people, and  40% of 20 is $\boxed{8}$ girls, giving us choice $\boxed{C}$
[/hide]
[hide="Problem 15"]
The angles of quadrilateral ABCD satisfy $A = 2B = 3C = 4D$. What is the measure of angle A, to the nearest whole number?
A) 125
B) 144
C) 153
D) 173
E) 180
[/hide]
[hide="Solution 15"]
We rewrite the other angles in terms of A:
$B = \frac{A}{2}$, $C = \frac{A}{3}$, $D = \frac{A}{4}$.
The angles of a quadrilateral always add up to 360, so we have
$A+\frac{A}{2}+\frac{A}{3}+\frac{A}{4}= 360$
$\frac{12A+6A+4A+3A}{12}= 360$
$25A = 4320$
$A = 172.8$
The closest whole number is then $\boxed{173}$ degrees, or choice $\boxed{D}$
[/hide]
[hide="Problem 18"]
A circle of radius 1 is surrounded by four circles of radius r as shown (sorry, no image). What is r?
--------------------------------
I'll describe the image.
It's a circle of radius one tangent to four circles of a bigger radius, and the big circles are arranged in a square around the smaller one.
--------------------------------
A) $\sqrt{2}$
B) $1+\sqrt{2}$
C) $\sqrt{6}$
D) $3$
E) $2+\sqrt{2}$
[/hide]
[hide="Solution 18"]
Connect the centers of the larger circles to form a square.
We can see that the side length of the square is $2r$, and the diagonal is $2r+2$. By the Pythagorean Theorem, $(2r)^{2}+(2r)^{2}= (2r+2)^{2}$
From this, we get
$8r^{2}= 4r^{2}+8r+4$
Solving this quadratic, we get
$r = \boxed{1+\sqrt{2}}$, or choice $\boxed{B}$
[/hide]

No problem. Ask me over Pm if you have any questions about the solutions.


[quote="Michael Niland"]If $x\sqrt{a^{2}-y^{2}}$ +$y\sqrt{a^{2}-x^{2}}$ =$a^{2}$

Prove $x^{2}$ +$y^{2}$=$a^{2}$

$x,y$ and $a$ are real numbers[/quote]

[hide="solution"]The case $a=0$ is trivial. If $a<0$ replace $a$ with $-a$ hence we may assume that $a > 0$
As both $x$ and $y$ belong to $[-a, a]$ we can choose $\alpha$ and $\beta$ belonging to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $x=a\sin \alpha , y=a\sin \beta$
This gives $a^{2}(\sin \alpha \cos \beta+\cos \alpha \sin \beta) =a^{2}\sin (\alpha+\beta )=a^{2}$ and so $\sin (\alpha+\beta )=1$
Since $\alpha+\beta \in [-\pi , \pi]$ there is only one possibility:
$\alpha+\beta = \frac{\pi}{2}$ which implies the desired equality.[/hide]


trigonometry
integration
inequalities proposed
inequalities

Let  $a_1,....,a_{20}$  real numbers such that for each $x\in R$  holds
$\left|a_1sinx+a_2sin2x+.....+a_{20}sin20x\right|\leq \left|sinx\right|$.

Prove that  $\frac{a_1}{1}+\frac{a_3}{3}+....+\frac{a_{19}}{19}\leq 1$

Nice inequality. :)
$\left|a_1 \int_0^{\pi}sinxdx+a_2 \int_0^{\pi} sin2xdx+.....+a_{20} \int_0^{ \pi}sin20xdx \right| \leq \int_0^{\pi} \left|sinx\right| dx$
then we have:
$\left|\frac{a_1}1+\frac{a_3}3 +.....+\frac{a_{19}}{19} \right| \leq 1$ :)


i know this site don't discuss much about colleges. but could you read it and give me some advice

like the artofproblem solving people i really like math and i think i'm pretty skilled at it.
but for past 2 years i wasn't encountered with any kind of hard math after the math count. i did pretty good on math count like 1st in the chapter Test of some part of world outside of US. but didn't do so good (like 7~10 when top 5 go to national) on the later part just before national.

now when i'm trying to solve the AMC12 mock questions #1 i only got like 17/25.
for past 2 weeks i've been spending numerous times on studying math. like 3 hours a day which is much more than i spend on school work. for school things i spend like an hour.

i kind of enjoy studying it but it just seems like i wont get anything out of it.

the time when i have to think about colleges is coming up. and even i study for math right now and even score pretty good on AMC, in this phase i wont qualify for USAMO.

and for AMC my junior year is the last chance which will be considered when applying to college. and theres not much math competitions in my region except the math league which i make 1 mistake on every test.

so basically AMC 12 seems like my last chance in math before college.
but does it worth it me studying for it this much. I know its my passion but colleges will be like "ooh this kid is passionate about math. but he didn't win any competitions"

maybe i could qualify for USAMO if i study more. but does it worth it. is qualifying for USAMO very good to top colleges like Ivies?

and do you think i could make it considering that i'm very fast at understanding stuffs. and i do like 2:30 hours a day til the test.

or would you just recommend me studying for SATs + school +APs+Ecs??
the ones i mentioned above seems like things that would come out if i study for it.

i mean is it better to have 5 on 2 self study AP.  2300 SAT instead of 2200. some more minor laundry lists of ECs...
or 110~140 AMC 12. and like 2~7 AIME.

i'm kind of thinking about just not studying for math. I know i am passionate about it but its not a carreer i would choose from. and do you think if i continue like this i would be able to make at least to USAMO.

i'm really discouraged because just months before when i didn't know about these kind of tests, i thought i was really really good even without studying. but when my eyes were opened and saw all these competitions i didn't know and people in AoPS i was like... i SUCK. I need to go study.

so any advice. ?????????/ please..
and can you tell me how much it usually takes you to do one chapter in AoPS volume 2 and how much questions you can do on it and what you got on the last AMC/AIME test compared to your understanding of the book.??

well, i have to say we have different goals for doing math, i do it because i enjoy it

but if you study x hours a day, you wont get that much better in a short amount of time, you can spend a lot of time on math, but you have to let it sink in, i agree that doing everything you can for colleges is good, but i suggest doing a little bit of AMC/AIME math just to improve your problem solving skills

for AoPS Volume 2, i am only 4 chapters in, but it takes maybe 5 hours to get all the problems except 1 or 2, then i move along and come back to them, the thing that i do is whenever i get stuck, i just read ahead in the book (but don't write), and that introduces me to the concepts, so when i actually get to the chapter, i undertand the all the text, i don't know if that is a great idea, but it works for me

thanks.
my sole motivation is not only because of college but its half half with interest.
the problem is that due to this, i became more neglectful of other stuffs. it doesn't mean i'm doing bad at other things but when i start the math solving i just can't stop it.
and i thought i sucked because i couldn't do any of the proof problems+some of actual problems in the volume 2, but i guess everyone does.

This probably should be moved to a different section of the site, but I think that if you know Vol 2 at least half well you should get 9+ on the AIME.  I think it has been said that there is no golden bullet to kill college apps with, so if you are generally strong you should make it to the college of your choice.  I would also like to point out that going from a 2 to a 7 is easier than going from a 7 to a 9, etc. on AIME's, so if you are discouraged now that doesn't bode well for the future practice.  If you have 3 hours to kill everyday, then you can keep up your math work and a few weeks or days before an SAT you should replace math with SAT practice and work on that, likewise for AP's except make few weeks a month at max.  Above all things you shouldn't get discouraged, you should spend less time worrying about things and instead take action.  Good luck in all of your endeavors.

I have the same problem. I am doing about 2 hours a day to math, but i don't win anything(Except Chapter MC) the only difference is that I am in a elemantery (ignore spelling) school.


A 6-inch-by-10-inch rectangle is cut down to a 5-inch-by-8-inch rectangle. What is the percent of decrease in the area? Express your answer as a mixed number percent.

The original area is 60 square inches.  The new area is 40 square inches.  Therefore, 20 square inches are removed, or one third of the original.  As a mixed number, $ 33\frac{1}{3}\%$.

To find the percent of increase (or decrease), there is a formula for you to work with:

\[\dfrac{|\text{original amount}-\text{new amount}|}{\text{original amount}}\]

This should help you cut down confusion most of the time :)


[b]Suppose we form all $ k$-digit palindromes that consist of 8's and 9's such that there is one of each digit. What is the smallest value of $ k$ such that there are at least 2004 numbers in the list?[/b]

If we can use only “one of each digit” how can we go past $ k\equal{}2$? Or am I misunderstanding that bit?

I assume he means at least one.

[hide="solution"]If there are two digits in the palindrome, we have 1 choices for how to choose digits, and similarly for 2 digits. In general, for an k-digit number, there are $ 2^{\lceil{\frac {k}{2}}\rceil}$ palindromes using 8 or 9 as digits. $ 2^{n} > 2004$ for $ n\geq11$, so we must have that $ \lceil{\frac {k}{2}}\rceil \equal{} 11$, which is achieved by $ k \equal{} 21,22$. We want the minimum, so $ k \equal{} \boxed{21}$.[/hide]

If $ k\equal{}2n$, even, then the number of palindromes containing 8's or 9's is $ 2^n$. Since we want to have at least one of each digit, we subtract the two cases where digits are all 8's or all 9's, $ 2^n\minus{}2$.
If $ k\equal{}2n\plus{}1$, odd, then the number of palindromes containing 8's or 9's is $ 2^{n\plus{}1}$. We again subtract the two cases and end up with $ 2^{n\plus{}1}\minus{}2$.
Now, $ 2^m\minus{}2\geq2004\Rightarrow2^{m\minus{}1}\geq1003\Rightarrow m\geq11$
So letting $ n\plus{}1\equal{}m\equal{}11$ will give the lowest value for $ k\equal{}2n\plus{}1$ which is 21.


I am confused as to how to do problems like these. What are they asking for, and how do you find it? The order pairs in sets confuse me.

\[f(x) = {(3, 5), (2, 4), (1, 7)}\]
\[h(x) = {(3, 2), (4, 3), (1, 6)}\]

How would you find... \[(f+h)(1)\] And... \[(f \circ h)(3)\]

I understand how to do it for regular functions like \[\sqrt{x}-2\] and stuff, but not for sets of ordered pairs like these. I have a feeling that it is simple. This question is from a Pre-Calculus level problem.


geometric sequence
real analysis
real analysis solved

0<b<a, x(0)=2, x(n+1)=(ax(n)+b)/(b(x(n)+a)
a) show that x(n) converge, compute lim x(n)
b) y(n)=(x(n)-1)/(x(n)+1) show that y(n) is geometric sequence
c) general term of x(n)
BACAU-2001
-----------------------------------------------------------------------------
We wiil do a) and b) same time
y(n+1)=(x(n+1)-1)/(x(n+1)+1)=(a-b)/(a+b)y(n)  it is geometric sequence
y(0)=1/3 the raison is q=(a-b)/(a+b), with the condition 0<a<b we have
|a-b|/|a+b| <1 so y(n) is convergent lim y(n) = 0 => lim x(n) =1
general term of y(n) is y(n)=y(0)q^n=1/3.((a-b)/(a+b))^n
c) from y(n)=(x(n)-1)/(x(n)+1) we have x(n)=(1+y(n))/(1-y(n))
using the y(n) in term of a,b.

If you have another solution, please post it.

Just a (somewhat useful) observation:

If we have x_(n+1) = f(x_n) with f homographic (f(x) = (ax+b)/(cx+d)) then the sequence a_n = (x_n-u)/(x_n-v) is a geometric progression, where u and v are the 2 fixed points of f.


complex numbers
algebra unsolved
algebra

Find all $a$ such that $-2<a<2$ and $x^{14}-ax^7+1$ is a factor of $x^{154}-ax^{77}+1$

Edited by pbornsztein.

Let $a = 2cos ( \theta)$ for some $\theta \in ] 0 ; \pi [.$

We want that the roots of $P(x) = x^{14} - ax^7 + 1$ are roots of $Q(x)=x^{154} - ax^{77} + 1.$
But the roots of $P$ are the complex numbers $z$ such that $z^7$ is a root of $R(x) = X^2-aX+1 = 0$, that is $z^7 = e^{i theta}.$
Now, such a complex number is a root of $Q$ if and only if $z^{77}$ is also a root of $R.$
This easily leads to $11 \theta  = \pm \theta \mod [2 \pi].$
This means that (hope I didn't miscompute) : $a = 2cos (\theta )$ where $\theta \in \{ \frac {\pi} 5 ,  \frac {2 \pi} 5 ,  \frac {\pi} 6 ,   \frac {\pi} 3 ,  \frac {\pi} 2 ,  \frac {2 \pi} 3 ,  \frac {5 \pi} 6 \}.$

Pierre.


calculus
integration
analytic geometry
function
calculus computations

Consider the solid E given by $ E \equal{} \{ (x,y,z) \in \mathbb{R} ^3: x^2 \plus{} y^2 \leq 1 \wedge 1 \minus{} x^2 \minus{} y^2 \leq z \leq 4 \}$. The density in any point of this solid is proportional to the distance between that point and the axis of the cylinder, with the constant of proportionality being 5. Find the mass of the solid E.

[hide]
This integral is easiest if done with cylindrical coordinates, thus, the density function, $ \rho$, is given by: $ \rho(r,\theta,z) \equal{} 5r$. From the information given,
$ r\in [0,1]$, $ \theta\in [0,2\pi]$, $ z\in [1 \minus{} r^2,4]$ and the mass integral becomes:
$ M \equal{} \int\int\int{\rho dV} \equal{} \int^{2\pi}_{0}\int^{1}_{0}\int^{4}_{1\minus{}r^2}{(5r) r dz dr d\theta}$
= $ \int^{2\pi}_{0}\int^{1}_{0}{5r^2(4 \minus{} (1 \minus{} r^2))drd\theta} \equal{} \int^{2\pi}_{0}\int^{1}_{0}{(15r^2 \plus{} 5r^4) drd\theta} \equal{} \int^{2\pi}_{0}{6d\theta} \equal{} 12\pi$

Edit: sigh, calculation error x_x Thanks for the correction
[/hide]

That's all correct, except on the last line it should be $ (15r^2 \plus{} 5r^4)$.


I think you can use highlight:

http://www.tex.ac.uk/tex-archive/support/highlight/README

thanks plinius, but I finally don't need it. I will keep the link, just in case.

bye

[quote="krümel"]Hi, I would like to include in my latex document the content of an html document, but the pages of the html document should keep the numeration and header of the latex document. Is this possible???

thanks in advance!![/quote]

use the latex2html package


calculus
integration
geometry
perimeter
algorithm
ceiling function
congruent triangles

Let $ T(n)$ be the number of incongruent triangles with integral sides and perimeter n greater than or equal to 6.. Prove that $ T(n) = T(n-3)$ if n is even, or disprove by a counterexample

[hide]This is true.
n-3 is odd and we will devise an algorithm to find all possible pairs.
a+b+c=k
a+b>c
a+c>b
b+c>a.
We will choose one of the variables to vary from 1 to [n/3]-1.  Without loss of generality, assume that is a.
a=1, b=c=(n-4)/2
a=2, $ b=\frac{n-4}{2}-1$, c=(n-4)/2.  Again, this is without loss of generality and since we want non-congruent triangles, we cannot switch the values for c with that of b.
a=3, b=c=(n-6)/2 and b=$ \frac{n-6}{2}-1$, c=$ \frac{n-6}{2}+1$.  We can keep adding +n to b and -n to c in increments of 1 until the triangle created no longer satisfies the triangle inequality.  This increases the difference between b and c by two for each new triangle with a fixed "a".  When a is even, the difference between b and c starts at 1 while when a is odd, the difference between b and c starts at 0, which is why any even a and a-1 have the same number of triangles.  From this, the amount obtained for a given "a" besides that one is $ \lceil{a/2}\rceil$
This continues until [n/3]-1.  At that point, one can only obtain one more because any more would recreate a variation that has already been seen.
WE don't need to evaluate T(n-3); we just need to show it equal to T(n).
First, we observe that for T(n), a>1 since n is even and n-1 is odd which would force b and c to differ by 1 and violate the triangle inequality.
Also, if a=2, there's only one possibility because +1/-1 to b and c causes a difference of 2, which would violate the triangle inequality.  The same argument would suffice for a=3.  In other words, this is T(n-3) but the a for T(n-3) is a+1 for T(n).  Similarly, the maximum a is also shifted up one to [a/3] since the -1 from T(n-3) resulted from [(n-3)/3].  Thus, we can conclude that T(n-3)=T(n).[/hide]


A sphere with diameter 10 units fits exactly inside a cube with side length 10 units. How many cubic units are in the volume of the largest cube that can be placed in one of the corners of the box so that one vertex just touches the sphere? Express your answer to the nearest tenth of a cubic unit.

[hide]The length of the space diagonal of the big cube is $10\sqrt{3}.$ That minus the diameter of the sphere is twice the length of the space diagonal of the smaller cube. So the length of the space diagonal of the smaller cube is $\frac{10\sqrt{3}-10}{2},$ or $5\sqrt{3}-5.$ The side length is $\frac{5\sqrt{3}-5}{\sqrt{3}},$ or $\frac{15-5\sqrt{3}}{3}.$ That is approximately equal to $2.113.$ The volume of the smaller cube, to the nearest tenth of a unit, is $\boxed{9.4}.$[/hide]


Let $ a,b,c$≥0.Prove the ineq:
$ \sum_{cyc}\frac{a(b\plus{}c\minus{}a)}{a^2\plus{}2bc}$≥$ 0$

[quote="SUPERMAN2"]Let $ a,b,c$≥0.Prove the ineq:
$ \sum_{cyc}\frac {a(b \plus{} c \minus{} a)}{a^2 \plus{} 2bc}$≥$ 0$[/quote]
The following Hang's inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:
\[ \sum_{cyc}\frac {a(b \plus{} c \minus{} a)}{a^2 \plus{} 2bc}\geq\frac{81abc(ab\plus{}ac\plus{}bc)}{(a\plus{}b\plus{}c)^5}\]

[quote="SUPERMAN2"]But how can I prove it?[/quote]
After expanding you can obtain:
$ \sum_{cyc}\frac{a(b\plus{}c\minus{}a)}{a^2\plus{}2bc}\equal{}\frac{2(a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}3abc(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)}{(a^2\plus{}2bc)(b^2\plus{}2ac)(c^2\plus{}2ab)}.$

[quote="arqady"][quote="SUPERMAN2"]But how can I prove it?[/quote]
After expanding you can obtain:
$ \sum_{cyc}\frac {a(b \plus{} c \minus{} a)}{a^2 \plus{} 2bc} \equal{} \frac {2(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 \plus{} 3abc(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}{(a^2 \plus{} 2bc)(b^2 \plus{} 2ac)(c^2 \plus{} 2ab)}.$[/quote]

 :o I tried to expand it this morning,but I gave up finally....How can you get this fantasic formula!!!
argady...Could you help me solve my problem which to find the max of cos2a+cos2b+cos2c in the forum...


Odd question :)

Determinant is a number, hence draw your own conclusion.

ops I thought determinant is matrix...anyway, what is singular non singular..?

Singular matrix is a matrix whose determinant is zero.

Non-singular matrix is a matrix which is not singular :)

And you could've found that out by yourself if you read what's written under the matrix.

[quote="bpms"][quote="srulikbd"]what is the meaning of "odd determinant"?[/quote]
Is this from IMTS/USAMTS?[/quote]

odd determinant- probably a determinant that is odd  :lol:

I agree that seems likely and I haven't seen anything like that in my work....odd det. i think i'll google it

as will i.. matrices seem hard... and i have no idea what those columns mean.

I think what they're trying to get at is [hide] that if the determinant is $\equiv 1 \mod 2$, then it can't be zero. Just use modular multiplication (look at each term in the matrix in $\mod 2$, and then calculate the determinant in $\mod 2$)[/hide]


There is a rocket flying 8500 miles per hour and a rocket flying 3500 miles per hour. They are flying towards each other. How far apart are they 1 minute before they collide?

I'm not sure how to solve this. Please explain this problem to me. Thank you.

Since it asks for minutes, it might be a good idea to first convert the velocities into miles per minute.  Call the two velocities $ v_{1}$ and $ v_{2}$.  Next, work backwards.  At time $ t\equal{}1\ \text{minute}$, the distance between them is 0 miles.  1 minute prior, the first plane is $ d_{1}\equal{}v_{1}t\equal{}v_{1}\cdot 1\ \text{min}$ miles to the left, and the second plane is $ d_{2}\equal{}v_{2}t\equal{}v_{2}\cdot 1\ \text{min}$ miles to the right.  Then the total distance between them is $ d_{1}\plus{}d_{2}$.

Because they are going to collide, it means that they are flying toward each other. Then relative velocity is $ v \equal{} v1 \plus{} v2 \equal{} 12000 mph$.
At the time is 1 minute, the distance is $ 1/60$ of the distance in 1 hour or $ 12000/60 \equal{} 200 miles$.

I will note that thdanh90 and I worked in two different reference frames yet still obtained the same answer.  This is frequently the case.


Problema suna in felul urmator : fie planul "alfa" dat si pctele $A$ si $B$ ce nu apartin planului. Determinati un punct $M$ din planul "alfa" astfel incat $MA^2 + MB^2$ sa fie minim. Iar rezolvarea mea s-a folosit de inegalitatea dintre media patratica si media aritmetica  pentru numerele $MA$ si $MB$,astfel ca am ajuns la $2(MA^2) + 2(MB^2)\le (MA+MB)^2$  deci min$(MA^2+MB^2)=(MA+MB)^2$ /2  daca si numai daca MA=MB.  AM vazut celelalte rezolvari si am vazut ca s-a gasit alta pozitie a punctului M si am inteles rezolvarea..dar nu reusesc sa inteleg unde e scaparea acestei rezolvari,caci pe aici pe undeva am gresit ceva..va multumesc anticipat si imi cer scuze pentru modul in care am explicat problema (vezi "^2","/"),inca invat latex..

Pai $MA+MB$ nu are valoare constanta. Daca avea valoare constanta atunci solutia ta era corecta.

Rescriind problema cu $\LaTeX$:

[i]Fie planul $\alpha$ si punctele $A,B \not\in \alpha$. Determinati un punct $M \in \alpha$ astfel incat $MA^2 + MB^2$ sa fie minima.[/i]

Folosind geometrie analitica problema este usoara: fie $A(0,0,t)$, $B(u,v,w)$, $M(x,y,z)$; avem $MA^2+MB^2=2(x^2-xu)+2(y^2-yv)+t^2+u^2+v^2+w^2$; prima paranteza e minima pt $x = \frac{u}{2}$, iar a doua pentru $y = \frac{v}{2}$; asadar, $MA^2+MB^2$ e minim pt. $M=$ mijlocul segmentului determinat de proiectiile lui $A,B$ pe $\alpha$.

Daca notam mijlocul $C$ al segmentului $AB$, atunci incearca sa aplici teorema medianei $[\ \"4m^2_a= 2(b^2+c^2)-a^2\"\ ]$ din varful $M$ in triunghiul $AMB$.


Find the sum

1). $ \sum\limits_{n\equal{}1}^{\infty}\frac{1}{\sqrt{n}(n\plus{}1)\plus{}\sqrt{n\plus{}1}n}.$

2). Prove that

$ \sum\limits_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\minus{}1}}{\sqrt{n}(n\plus{}1)\plus{}\sqrt{n\plus{}1}n}\equal{}2(1\minus{}\sqrt{2})\cdot s\minus{}1,$

where

$ s\equal{}\lim\limits_{n\rightarrow \infty}\left(\sum\limits_{k\equal{}1}^{n}\frac{1}{\sqrt{k}}\minus{}2\sqrt{n}\right),$ denotes the $ \textbf{Ioachimescu's}$ constant.

First one:

$ \frac{1}{\sqrt{n} (n\plus{}1)\plus{}n \sqrt{n\plus{}1}}\cdot \frac{\sqrt{n} (n\plus{}1)\minus{}n \sqrt{n\plus{}1}}{\sqrt{n} (n\plus{}1)\minus{}n \sqrt{n\plus{}1}} \equal{}$

$ \equal{}\frac{\sqrt{n} (n\plus{}1)\minus{}n \sqrt{n\plus{}1}}{\left(\sqrt{n} (n\plus{}1)\right)^2\minus{} \left(n \sqrt{n\plus{}1}\right)^2} \equal{}$

$ \equal{}\frac{\sqrt{n} (n\plus{}1)\minus{}n \sqrt{n\plus{}1}}{n(n\plus{}1)} \equal{}$

$ \equal{}\frac{1}{\sqrt{n}}\minus{}\frac{1}{\sqrt{n\plus{}1}}$

Therefor:

$ \sum^\infty_{n\equal{}1}\frac{1}{\sqrt{n} (n\plus{}1)\plus{}n \sqrt{n\plus{}1}}\equal{}\sum^\infty_{n\equal{}1}\left(\frac{1}{\sqrt{n}}\minus{}\frac{1}{\sqrt{n\plus{}1}}\right) \equal{} 1$


algebra
polynomial
function
superior algebra
superior algebra unsolved

A, a ring with n elements and x^2 = x. Find the number of polynomial functions defined on A with values in A.

it is clear that every polynomial function $ A \to A$ is linear, i.e. has the form $ x \mapsto ax \plus{} b$. it is easy to see that the function determines $ a,b$ uniquely.


function
integration
calculus
real analysis
real analysis unsolved

Let f be a derivable function with  $ \displaystyle \int_{a}^b f(x) dx \equal{}(b\minus{}a)f(a)$ . Prove that exist $ c \in (a,b)$ s.t. $ f(c)\minus{}f(a)\equal{}f'c)$

[hide="Proof sketch"]Suppose $ f$ is continuously differentiable and there is no $ c\in(a,b)$ such that $ f(c) \minus{} f(a) \equal{} f'(c)$.  Then since $ f(x) \minus{} f'(x)$ is continuous, we must have either (1) $ f'(x) > f(x) \minus{} f(a)$ or (2) $ f'(x) < f(x) \minus{} f(a)$ for all $ x\in (a,b)$.  Let $ g(x) \equal{} f(x) \minus{} f(a)$.

Case (1):  $ g'(x) > g(x)$ and $ g(a) \equal{} 0$ implies $ g(x) > 0\Leftrightarrow f(x) > f(a)$ for all $ x\in (a,b)$.
Case (2):  $ g'(x) < g(x)$ and $ g(a) \equal{} 0$ implies $ g(x) < 0\Leftrightarrow f(x) < f(a)$ for all $ x\in (a,b)$
(This assertion requires some more justification, hence 'sketch' rather than proof)

In either case we have a contradiction, since MVT applied to $ F(x) \equal{} \int_a^xf(t)dt$ guarantees the existence of $ \xi \in (a,b)$ satisfying $ F'(\xi) \equal{} f(\xi) \equal{} f(a)$.[/hide]

Let $ g(x) \equal{} f(x) \minus{} f(a)$. Then $ \textstyle \int_{a}^{b} g(x) \, dx \equal{} 0$. In this case, we have to prove that there exists $ c \in (a, b)$ such that $ g(c) \equal{} g'(c)$.

By mean value theorem of integral, there is $ \xi \in (a, b)$ such that $ g(\xi) \equal{} 0$.

Now let $ h(x) \equal{} e^{\minus{}x} g(x)$. Since $ h(a) \equal{} h(\xi) \equal{} 0$, $ h$ is continuous on $ [a, \xi]$ and differentiable on $ (a, \xi)$, there is some $ c \in (a, \xi)$ such that $ h'(c) \equal{} 0$.

But since $ h'(x) \equal{} e^{\minus{}x} (g'(x) \minus{} g(x))$, we have $ g(c) \equal{} g'(c)$ as desired.


How many odd perfect squares are between 5 and 211?

$ 3^2$ , $ 5^2$ , $ 7^2$ , $ 9^2$ , $ 11^2$ , $ 13^2$ , $ 14^2$ $ \equal{}>$ $ 7$ odd perfect squares.

However, apparently the correct answer is $ 6$. I'm confused. I need help.


Prove that if $ ABCD$ is an inscribed quadrilateral , then the sum of the inradii of triangles $ ABC$ and $ ACD$ is equal to the sum of the radii inscribed in triangles $ BCD$ and $ BAD$.

In an inscribed quadrangle $ ABCD$, $ h_a,h_b,h_c,h_d$ are the distances of the centre of the circumscribed circle to the corresponding sides{$ a,b,c,d$ where {$ a$ is opposite to $ c$}.
Now prove that if the centre of the circumscribed circle lies inside the quadrangle then
$ a h_a \plus{} c h_a \equal{} b h_d \plus{} d h_a$.

Mathlinkers, please post a solution for these 2 nice sums.
Thanks.,


(a) By expressing $ cos4\theta, sin4\theta$ in terms of powers of $ cos\theta$ and $ sin\theta$, show that $ tan4\theta \equal{}\frac{4tan\theta\minus{}4tan^{3}\theta}{1\minus{}6tan^{2}\theta\plus{}tan^{4}\theta}$

(b) Hence solve the equation $ t^{4}\minus{}4t^{3}\minus{}6t^{2}\minus{}4t\plus{}1$

---------------------
I did part (a) but how do I do part (b)

The book has it like this

Let $ t\equal{}tan\theta$. Then $ tan4\theta$ = 1 $ \equal{}\Rightarrow $ $ \frac{4t\minus{}4t^{3}}{1\minus{}6t^{2}\plus{}t^{4}}$

How did they get $ tan4\theta$ = 1?

[quote="MellowMelon"]So first of all, the problem should be $ \plus{}4t^{3}$, not $ \minus{}4t^{3}$, or else this method is not as pretty.

Anyways, they explained things in a strange order. This should make it more clear.

$ t^{4}\plus{}4t^{3}\minus{}6t^{2}\minus{}4t\plus{}1 \equal{} 0$
$ 4t^{3}\minus{}4t \equal{}\minus{}t^{4}\plus{}6t^{2}\minus{}1$
$ \frac{4t^{3}\minus{}4t}{\minus{}t^{4}\plus{}6t^{2}\minus{}1}\equal{}\frac{4t\minus{}4t^{3}}{t^{4}\minus{}6t^{2}\plus{}1}\equal{} 1$
Put $ \tan\theta \equal{} t$
$ \frac{4\tan\theta\minus{}4\tan^{3}\theta}{\tan^{4}\theta\minus{}6\tan^{2}\theta\plus{}1}\equal{} 1$
Use the identity you derived in part a.
$ \tan 4\theta \equal{} 1$
This is much, much easier to solve, of course.[/quote]

I guess that's what the question was intended to be, indeed. One could solve $ t^{4}\minus{}4t^{3}\minus{}6t^{2}\minus{}4t\plus{}1 \equal{} 0$ by noticing that $ t$ is a solution if and only if $ t^{\minus{}1}$ is a solution. Therefore, there exist real numbers $ a$ and $ b$ such that $ t^{4}\minus{}4t^{3}\minus{}6t^{2}\minus{}4t\plus{}1 \equal{} (t^{2}\plus{}at\plus{}1)(t^{2}\plus{}bt\plus{}1)$. By expanding, one finds $ a\plus{}b \equal{}\minus{}4$ and $ ab \equal{}\minus{}8$, so (wlog) $ a \equal{} 2\sqrt{3}\minus{}2$ and $ b \equal{}\minus{}2\sqrt{3}\minus{}2$. It's easy to finish it off now.


i didn`t know where to place this, so here it goes:
is every linear transformation T:Rm-------->Rn open?

No. Consider the case in which the image is a proper subspace.

if it is surjective, it is OK. Try ''open mapping theorem'" on google.


Prove that there is a function $f$ from the set of all natural numbers into itself such that $f(f(n))=n^2$  for all $n \in \mathbb{N}$.

First, let $ f(1) \equal{} 1$.

Let $ S\subseteq\mathbb{N}$ be the set of numbers that aren't perfect squares. Take the two smallest elements of $ S$, and remove them. Call them $ a,b$. Define for all nonnegative integers $ n$, $ f(a^{2^{n}}) \equal{} b^{2^{n}}$, and $ f(b^{2^{n}}) \equal{} a^{2^{n\plus{}1}}$. Thus $ f(f(a^{2^{n}})) \equal{} a^{2^{n\plus{}1}}$, and $ f(f(b^{2^{n}})) \equal{} b^{2^{n\plus{}1}}$ as desired. Continue this procedure of removing the two smallest elements of $ S$ and pairing them up in this way ad infinitum.

By examining the exponents in prime factorizations, we can see that for any two elements $ a,b\in S$, we have that $ a^{2^{n}}\neq b^{2^{k}}$ regardless of $ n$ and $ k$, so there won't be any multiple outputs for this function. Additionally, $ \{a^{2^{n}}| a\in S, n\in\mathbb{N}\cup\{0\}\}\equal{}\mathbb{N}$, so we have defined the function for all natural numbers.

Correct, except that I think that you should remove all powers of $ a$ and $ b$ in the next $ S$. :)

None of the numbers of the form $ a^{2^{n}}, b^{2^{n}}$ are in $ S$ if $ n\geq 1$, so I think we only have to remove $ a$ and $ b$. For instance, if one of $ a$ and $ b$ is $ 2$, we don't want to remove $ 8$, as $ f(8)$ has yet to be defined.

[i]This problem is very easy

   Frist , choose an arbitrary odd prime $ p$ . Define the function $ g : \mathbb{N} \to \mathbb{N}$ as follows :

  $ g(n) \ \equal{} \ \frac {n}{p}$ if $ 2 \not |v_p(n)$ and $ g(n) \ \equal{} \ 2np$ if $ 2 | v_p(n)$

    An easy check can verify that : $ g(g(n)) \ \equal{} \ 2n \ \forall n \ \in \ \mathbb{N}$

   Then , we can construct the function $ f$ as follows :

   $ f(1) \ \equal{} \ 1 \ \equal{} \ 1^2$

 For the positive integer $ n > 1$ , assume that it's prime fractorization is $ p^{\alpha_1}_1 \cdot p^{\alpha_2}_2 \cdots p^{\alpha_k}_k$

 With $ p_1 ; p_2 ; ...; p_k$ are distinct pimes , $ \alpha_1 ; \alpha_2 ; ...; \alpha_k$ are positive integers .

   Then , we can choose  $ f(n) \ \equal{} \ p^{g(\alpha_1)}_1 \cdot p^{g(\alpha_2)}_2 \cdots p^{g(\alpha_k)}_k$

  So that $ f(f(n)) \ \equal{} \ p^{g(g(\alpha_1))}_1 \cdot p^{g(g(\alpha_2))}_2 \cdots p^{g(g(\alpha_k))}_k$

   $ \rightarrow f(f(n)) \ \equal{} p^{2\alpha_1}_1 \cdot p^{2\alpha_2}_2 \cdots p^{2\alpha_k}_k \ \equal{} \ n^2$

  In conclusion , $ f(n) \equal{} n^2 \ \forall \ n \ \in \ \mathbb{N}$ , Done  :lol: [/i]

I had the same solution as MellowMelon, but also proved that the function is unique by choice (you can arbitrarily pair the nonsquare numbers up, but that's it). Here it goes:

First note that f(1) = 1. For the rest of the problem, none of the variables equal 1.

Suppose that f(n) = k. It follows that $k \neq n$ since that would imply f(f(n)) = n, a contradiction.

Iterating the function shows us that $f(n^{2^m}) = k^{2^m}$ and $f(k^{2^m}) = n^{2^{m+1}}$. We can prove this inductively. Moreover, group all such elements in sets $S_{n,k}$ = {$f^m(n), f^m(k)$} of basis f(n) = k.

Now suppose $f(n^2) = k$ and f(n) = p. Iterating the function gives $f(n^2) = p^2$. Hence, $k = p^2$, and we can write the basis for this pair as f(n) = p. Running this procedure until completion eventually results in n not being a perfect square. Hence all $S_{n,k}$ must be derived from a nonsquare n (regarding basis f(n) = k).

Now suppose $f(n) = k^2$. Since f(f(k)) = $k^2$, we have f(n) = f(f(k)). Since f is injective, it follows that f(k) = n. Hence, we can rewrite the set $S_{n,k}$ in terms of f(k) = n. Applying the above procedure yields that k is nonsquare in basis f(n) = k.

Hence all $S_{n,k}$ must have bases that satisfy n, k are not perfect square numbers. Since $\mathbb{N}$ is partitioned into sets $S_{n,k}$ by arbitrarily pairing nonsquare n, k, this function is unique. (MellowMelon showed this through uniqueness of prime factorization).


The sum of three binomials each of the form ma+nb, where m and n are natural numbers, is 4a+8b. How many distinct sets of such three binomials exist?

 :help:

[hide="Hint"]Find the number of distinct ways that $ m_1,m_2,m_3$ can add to 4, then find the number of ways that $ n_1,n_2,n_3$ can add up to 8.[/hide]

[hide="Solution"]We have $ m_1\plus{}m_2\plus{}m_3\equal{}4$. This can be done in $ \dbinom{3}{2}\equal{}3$ ways.
We also have $ n_1\plus{}n_2\plus{}n_3\equal{}8$. This can be done in $ \dbinom{7}{2}\equal{}21$ ways.
The total number of sets is thus $ 3\cdot21\equal{}\boxed{63}$.[/hide]

Googling natural numbers, theres apparently a big controversy about whether 0 should be included or not. So if the definition of natural numbers means numbers of the set {0,1,2...}, then our answer would actually be $ \binom{6}{2} \binom{10}{2} \equal{} 675$ by the same method that bbgun used. The difference between our two solutions was that he said $ m,n \neq 0$.

 This method is known as [url=http://www.artofproblemsolving.com/Wiki/index.php/Distinguishability]balls and urns[/url].

Assuming this was indeed from mathcounts, i think they would stick to the original convention of natural numbers; that is, all the positive integers. Otherwise, it might be unsure, although i think if they meant nonnegative integers they would specify it. oh well.  :roll:

I don't think natural numbers include 0.


A number $n$ is called a Niven number, named for Ivan Niven, if it is divisible by the sum of its digits. For example, $24$ is a Niven number. Show that it is not possible to have more than $20$ consecutive Niven numbers.

Assume we have a 21 consecutive Niven numbers, with the smallest one being $ a$. Let $ s(x)$ denote the sum of digits of $ x$. Now if we have two odd Niven numbers $ b$ and $ b\plus{}10$ whose tens digits differ by one (as opposed to them being 9 and 0), then $ s(b\plus{}10)\equal{}s(b)\plus{}1$. Therefore one of $ s(b),s(b\plus{}10)$ is even, which is a contradiction since $ b$ and $ b\plus{}10$ are odd.

Such a situation clearly arises if no number in our list in divisible by $ 100$; we could take the pair $ (a,a\plus{}10)$ or $ (a\plus{}1,a\plus{}11)$. Therefore some number in our list is divisible by $ 100$; call it $ c$. The list cannot have both $ c\plus{}1$ and $ c\plus{}11$, so $ c\plus{}11$ is not in the list. Similarly, we cannot have both $ c\minus{}1$ and $ c\minus{}11$, so $ c\minus{}11$ is not in the list. This restricts our list to be $ c\minus{}10$ through $ c\plus{}10$.

Now for $ 0\le i\le 9$, we have $ s(c\plus{}i)\equal{}s(c)\plus{}i$. Then one of $ s(c),s(c\plus{}1),...,s(c\plus{}9)$ is divisible by 10. This number must be $ s(c)$ since none of $ c\plus{}1,...,c\plus{}9$ are divisible by $ 10$. Since $ s(c)$ cannot be $ 0$ (as $ 0$ dividing a number is not well-defined), we must have that $ s(c)\ge 10$.

Also note that $ s(c\plus{}1)\equal{}s(c\plus{}10)$. Then $ s(c\plus{}1)|c\plus{}1$ and $ s(c\plus{}1)|c\plus{}10$, so

$ s(c\plus{}1)|(c\plus{}10)\minus{}(c\plus{}1)\equal{}9$ which means that $ s(c\plus{}1)\le 9$.

However, this means that

$ 9\ge s(c\plus{}1)\equal{}s(c)\plus{}1\ge 10\plus{}1\equal{}11$

which is a contradiction.


Does anyone have any books (besides the ones on this website, of course,) to recommend to someone who is almost, but not quite, on the intermediate level. Just for clarification, I mean books like AoPS and [u]Advanced Mathematics[/u] by Brown.

There are probably already a bunch of threads just like this one.  Try searching this site.  Also you can look at the resources tab on the left side of the web site.  It has a few.

ya i saw the ones on the resources tab, im just wondering if anbody has any [b]other[/b] books to recommend.

[quote="joml88"]There are probably already a bunch of threads just like this one.[/quote]

Yes, the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]readable books thread[/url] opened by mathnovice includes several interesting suggestions, including some I posted.

Hope this helps!

I recommend a book for precalc (not problem-solving, but the knowledge can help).  It's called Precalculus with Limits by Larson, Hostetler, and Edwards.  The second edition is the one I have and the one I think looks best.  It might take some googling/amazon searching to find the right book.


linear algebra
matrix
linear algebra unsolved

Let $A_1,...,A_k$ be some idempotent $nxn$ matrices with complex elements such that their sum is also idempotent. Then prove that $A_i A_j=0$ whenever $i,j$ are different.

[quote="harazi"]...that their sum is also idempotent. [/quote]

You mean identity matrix for the sum $A_1+...+A_k=I_n$ but $A_i^2=A_i$

$a_i$ morphism of $C^n$ associate to the matrix $A_i$

 Prove that (1) $C^n=\oplus_{i=1}^{k}Im(a_i)$ where $ker(a_j)=\oplus_{i\neq j}Im(a_i)$

To finish : use a basis wich is union of basis of $Im(a_i)$

No, Moubi, I refer that the sum of these matrices is idempotent and not the identity matrix.

So if $A_1+..+A_k$ is a projector then $I-(A_1+...+A_k)$ is also a projector say $A_{k+1}$

$A_1+....+A_k+A_{k+1}=Id$ apply post 2

I think it's easier to deal with the initial problem, as Harazi posted it, without making that extra complication.

We have $r(\sum A_i)=tr(\sum A_i)=\sum tr(A_i)=\sum r(A_i)$, so the dimension of $Im(\sum A_i)$ is equal to $\sum dim(Im(A_i))$, and the dimension of $\sum Im(A_i)$ lies between those two, so it's also equal to both of them. From here we find that the sum of the images of $A_i$ is direct, and the conclusion follows (we get, for $i\ne j$, $Im(A_i)\subseteq \ker(A_j)$).


Find all pairs $ f(x),g(x)$ of polynomials such that

$ f(g(x))\equal{}x^{2007}\plus{}2x\plus{}1$

aren't there an infinite amount of them, as the function f can add or subtract any thing added by g(x)?

Well, sure, but are there any other solutions?


calculus
integration
trigonometry
logarithms
calculus computations

Compute $ \int_0^{\infty}\frac{\sin^2\left(\frac{\omega x}{2}\right)}{x(e^x\minus{}1)}\,dx$, where $ \omega \neq 0$.

is it $ \frac {1}{4}\log(\prod_{k \equal{} 1}^\infty \frac {k^2 \plus{} \omega^2}{k^2})$?

I also obtained $ \frac {1}{4} \log \left( \frac {\sinh (\pi \omega)}{\pi \omega} \right)$, although my method seems somewhat indirect.

[hide="My approach"]
\begin{eqnarray*} \int_{0}^{\infty} \frac {\sin^2 \left( \frac {\omega x}{2}\right)}{x (e^x - 1)} \, dx & = & \sum_{n = 1}^{\infty} \int_{0}^{\infty} \frac {\sin^2 \left( \frac {\omega x}{2}\right)}{x} e^{ - nx} \, dx \\
& = & \sum_{n = 1}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} \sin^2 \left( \frac {\omega x}{2}\right) e^{ - nx} e^{ - xt} \, dt dx \\
& = & \sum_{n = 1}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} \frac {1 - \cos(\omega x)}{2} \, e^{ - (t + n)x} \, dx dt \\
& = & \sum_{n = 1}^{\infty} \int_{0}^{\infty} \frac {1}{2}\left( \frac {1}{t + n} - \frac {t + n}{(t + n)^2 + \omega^2} \right) \, dt \\
& = & \sum_{n = 1}^{\infty} \frac {1}{4} \ln \left( 1 + \frac {\omega^2}{n^2} \right) \\
& = & \frac {1}{4} \ln \prod_{n = 1}^{\infty} \left( 1 + \frac {\omega^2}{n^2} \right) \\
& = & \frac {1}{4} \ln \left( \frac {\sinh (\pi \omega)}{\pi \omega} \right) \end{eqnarray*}[/hide]


let $ f : R \rightarrow R$ be a function uniforme continuous ,such that for every real number $ x>0$, the sequence $ (f(nx)_{n \in N}$ converge .
what we can say about $ f$ ?

Look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178178]this thread[/url] and at the references there. On the other hand, the [b]uniform[/b] continuity assumption makes the problem much simpler, so much that I even won't move it to Unsolved and Proposed section. Can anybody see why? ;)


A die is tossed.   It is a fair, unbiased die.   List the outcomes for each event below and the probability of the event.

a.  A number less than or equal to 3 appears.

b.  An odd number is tossed.

c.  A number greater than 2 appears.

[hide="a"]Thats just $\frac{1}{2}$ because there are six numbers and three numbers that work. 1,2, and 3.[/hide]

[hide="b"]There are 3 odd numbers so it is $\frac{1}{2}$. They are 1,3, and 5. [/hide]

[hide="c"]All numbers greater than 2 are 3,4,5,and 6 so it is $\frac{2}{3}$.[/hide]

All great answers and great service.

Thanks!


logarithms
limit
real analysis
real analysis unsolved

Do you happen to know whether the series

$ \sum\limits_{n\equal{}2}^{\infty}(\minus{}1)^{n}(\sqrt[n]{n}\minus{}1)$,

can be calculated in terms of well known constants?

Perhaps by using some nonstandard techniques...

it's very interesting for me.  :!:  therefore I will try to post some idea:
$ \sum_{n \equal{} 1} ( \minus{} 1)^{n}(n^{1/n} \minus{} 1) \equal{} \sum_{n \equal{} 1}( \minus{} 1)^{n}(\sum_{k \equal{} 1}\frac {\ln^{k}n}{k!n}) \equal{} \sum_{k \equal{} 1}\frac {1}{k!}\sum_{n \equal{} 1}( \minus{} 1)^{n}\frac {\ln^{k}n}{n}$=
$ \sum_{k \equal{} 1}\frac {1}{k!}\lim_{s \minus{} > 0^{ \minus{} }}(2^{s}\zeta(1 \minus{} s) \minus{} \zeta(1 \minus{} s))^{(k)}$
let $ \alpha(s) \equal{} (2^{s}\zeta(1 \minus{} s) \minus{} \zeta(1 \minus{} s))^{(k)}$

hence $ ... \equal{} \sum_{k \equal{} 1}\frac {\lim_{s \minus{} > 0^{ \minus{} }} (\alpha(s))^{(k)}}{k!} \equal{} \lim_{s \minus{} > 0^{ \minus{} }}\sum_{k \equal{} 1}\frac {(\alpha(s))^{(k)}}{k!} \equal{} \alpha(1) \minus{} \lim_{s \minus{} > 0^{ \minus{} }}\alpha(s)$ by taylor expansion.
if I have not mistake in my calculation then last I think can be calculated and we have exact value of your sum.

Using this technique I got that

$ \sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}(\sqrt[n]{n}\minus{}1)\equal{}\gamma\ln 2\minus{}\frac{1}{2}\ln^{2}2\plus{}\sum\limits_{k\equal{}2}^{\infty}\frac{(\minus{}1)^{k}}{k!}\left((2^{1\minus{}s}\minus{}1)\zeta(s)\right)^{(k)}\biggm|_{s\equal{}k}$

$ \equal{}\gamma \ln 2\minus{}\frac{1}{2}\ln^{2}2 \plus{}\sum\limits_{k\equal{}2}^{\infty}\frac{(\minus{}1)^{k}}{k!}f^{(k)}(k)$,

where $ f(x)\equal{}(2^{1\minus{}x}\minus{}1)\zeta(x)$,

and I don't think that the last sum can be calculated exactly.

didilica where is from this problem? do you know solution?

This is one of the problems I was thinking of for quite some time, I put it aside and then I was thinking of posting it on the net  hoping that a solution can be developed. I do not have a solution for this problem and I feel, I may be wrong, that the series cannot be expressed in terms of well-known constants.


I heard that too. Nice job, BTW, to your whole team. I didn't win anything, as you know. I made so many careless mistakes, and on a lot of problems I changed my answer from right to wrong. :wallbash_red:

That was one heck of a nice job.  You done good.  :)

(Okay, my math is better than my grammar :D).

[quote="E^(pi*i)=-1"]So . . . how did everyone do at JHMT?  This is going to sound like bragging, but:

First Place Algebra
Second Place Combinatorics
First Place Individual
First Place Team Round
First Place Scientific Math
First Place Overall Team

I have a lot of awards. :D  I hope I heard right that AoPS is going to give gift certificates to the winners.[/quote]


linear algebra
matrix
calculus
integration
inequalities
Functional Analysis
real analysis

I posted this a long time ago, but no one provided a solution, so I'm posting it again :)

Let $ (\alpha_n)_{n\in\mathbb{N}}$ be a sequence of real numbers.  Prove that

\[ \sum_{i,j\geq 1} \frac{\alpha_i\alpha_j}{i\plus{}j} \leq \pi\sum^\infty_{k\equal{}1} \alpha^2_k
\]

You might as well phrase that as
\[ \sum_{i,j\ge1}\frac{\alpha_i\beta_j}{i\plus{}j}\le\pi\left(\sum_{k\ge 1}\alpha_k^2\right)^{\frac12}\left(\sum_{k\ge 1}\beta_k^2\right)^{\frac12}\]
In other words, the infinite Hilbert matrix generates a bounded linear transformation from $ \ell^2$ to $ \ell^2,$ and we give a specific bound for its norm.

I know I've posted the integral analogue of this problem, including the $ L^p$ versions for $ 1<p<\infty.$ The integral version is easier, because there are more tools to use - but the same truth is there for both integral and sum.

Well, I know a solution for the general version, but I was wondering if there was perhaps a 'nice' one for this specific case.

The Schur test with $ A(i)\equal{}B(i)\equal{}\frac 1{\sqrt i}$ certainly does it (the sums are less than the integrals and the integrals give you $ \pi$ in the continuous version of the problem).

yes, that was the solution (as I recall, it's a (solved) exercise in 'A Hilbert Space Problem Book' by Halmos).  maybe it was silly of me to expect that there might be a particularly elementary solution.


limit
trigonometry
integration
calculus
calculus computations


Given one hundred positive real numbers such that: $\sum_{i=1}^{100} a_i = 300$ and $\sum_{i=1}^{100} a_i^2 > 10000$. Show that there exist three numbers with sum more that $100$.

i think something must be wrong in the q

take ai=3  for i=1 till 90
a91=21  and all others as 1

so sum(ai)=90*3 + 21 + 1*9=300

and sum(ai^2)>10000

but no 3 satisfy the condition

 :oops: sorry i looked it up as 1000

i think i have got the soln
it is very long!! there must be a shorter one

let a_1>=a_2>=.......................a_100

now sum(ai)=300.......................................1

sum(ai^2)>10000........................................2

let k=100/3

so sum(ai^2)>10000=300k=k*sum(ai)

hence sum (ai(ai-k))>0....................................3

which gives  a_1>k

so let a_1=k+a

now let a_1+a_2<2k               ie i am assuming side by side that a_1<2k

ie a_2<k-a      which gives a<k
also   a_n<=a_2<k-a    for n>2

thus a(n)-k<-a

putting these valuse in 3

a*a_1 -a(a_2 +a_3........a_100)>sum(ai(ai-k))>0
ie a(2a_1 -300)>0

ie a_1>150>2k
hence the contradiction simply gives that a_1+a_2>=2k or a_1>=2k

ie a_2>k-a

so now let a_1=k+a
                a_2=k-a+b
                a_3=k-b-c

so that a_1+a_2+a_3 <3k
also a_2>a_3 gives
a<2b+c

putting these vals in 3 similarly as previous case
a_1*a + (a_2)(b-a) -(b+c)(300-a1-a2)>0

now observe a_1*a + (a_2)(b-a) < 2(b+c)(a_1+a_2)
can be easily proved using a<2b+c

thus (b+c)(3a1+3a2-300)>0
ie. a1+a2>100
which completes the proof


limit
superior algebra
superior algebra unsolved

Consider $ \hat{\mathbb{ Z}}$ the (profinite) completion of $ \mathbb Z$ and $ P$ the set of primes. Prove that the invertible elements of $ \hat{\mathbb Z}$ are precisely the elements of $ \overline P\minus{}P$.

$ \hat{\mathbb Z}\equal{}\lim_{\leftarrow}\mathbb Z/n\mathbb Z$

proof (by ZetaX): let $ x \equal{} (\overline{x_{n}})_{n}$ be a unit in $ \hat{\mathbb{Z}}$. then $ gcd(x_{n},n) \equal{} 1$ for all $ n$. for every $ k$, put $ m \equal{} lcm(1,...,k)$ and choose, according to dirichlet's theorem, some prime $ p_{k}$ s.t. $ p_{k}\equiv x_{m}$ mod $ m$. in particular, $ p_{k}\equiv x_{i}$ mod $ i$ for every $ i\leq m$. thus $ p_{k}$ converges to $ x$. if $ x$ is a prime $ p$, then we get a contradiction $ x_{p}\equiv 0$ mod $ p$. thus, $ x\in\overline{\mathbb{P}}\minus{}\mathbb{P}$.

conversely, assume $ x\in\overline{\mathbb{P}}\minus{}\mathbb{P}$. then there is a sequence $ p_{k}$ of primes s.t. for every $ n$ there is some index $ k_{n}$ s.t. for all $ k\geq k_{n}$ we have $ x_{n}\equiv p_{k}$ mod $ n$. so if for some $ n$, $ x_{n}$ is not invertible mod $ n$, then $ p_{k}$ divides $ n$ for all $ k\geq k_{n}$. in particular, one prime $ p$ occurs infinitely often in the sequence of the $ p_{k}$. but then $ x_{m}\equiv p$ mod $ m$ for all $ m$, i.e. $ x\in\mathbb{P}$, a contradiction. thus, for all $ n$, $ x_{n}$ is invertible mod $ n$. clearly, the sequence of the inverse elements also lies in the projective limit, so that $ x$ is invertible.


For $a,b,c$ positive real numbers such that $abc=3$ prove that

$  \frac{1}{\sqrt {a(1+b)}} + \frac{1}{\sqrt{b(1+c)}} + \frac{1}{\sqrt{c(1+a)}} > \frac{3}{2} $

This inequality is without equality, isn't it ?

You are right, Vasc.
Sorry a lot for this terrible mistake.

If $x,y,z$ be positive real numbers such that $xyz=3$, then
\[\frac{1}{\sqrt{x(1+y)}}+\frac{1}{\sqrt{y(1+z)}}+\frac{1}{\sqrt{z(1+x)}}>\frac{2}{\sqrt[4]{3}},\]
the infimum $\frac{2}{\sqrt[4]{3}}=1.5196713713031850946623\cdots$ be obtained when $x=\frac{\sqrt{3}}{y}\rightarrow 0$.


[hide]The square's area is 16 sq.in. The triangle's area is 60. 60+16=76 or d. [/hide]

[hide="my answer"][quote="anirudh"]A square has a perimeter of  16  in. A triangle has sides with lengths  17,8 and 15  in. What is the sum of the areas of these two figures, in square inches?
a. 	136
b. 	134
c. 	79
d. 	76
e. 	74[/quote]

The triangles area is 8/2 *15 so 60
th area of the square can be 16,15,12,7
The only one that fites is 16 so the area is d. 76[/hide]


If x and y are positive numbers, prove the inequality
$\frac{x}{x^{4}+y^{2}}+\frac{y}{x^{2}+y^{4}}\le \frac{1}{xy}$


[quote="nebula42"]How many factors of $2006!$ are greater than $2005!$?[/quote]

[hide]$2006!$ can be missing any factor except 2006, so the answer is $\boxed{2005}$[/hide]

[hide]When $2006!$ is divided by any thing from $1$ to $2005$, the result will always be greater than $2005!$, because when it is divided by 2006, it will equal $2005!$.  Therefore, the numbers $1$ to $2005$ is $\boxed{2005}$ numbers.[/hide]

Yes, you're both correct (again  :)  :P ).

It is just any number that is multiplied $2005!*2006/n$, where n is between 1 and 2005, inclusive.

$2005$

[quote="nebula42"]How many factors of $2006!$ are greater than $2005!$?[/quote]
[hide]$\boxed{2005}$ because all numbers from 1 to 2005 are factors of 2006! that can be multiplied to give 2006.[/hide]


A ''superball" of mass m bounces back and forth between two surfaces with speed vo, Gravity is neglected and collisions are perfectly elastic.
b. If one surface is slowly moved toward the other with speed V<< v, the bounce rate will increase due to the shorter distance between collisions, and because the ball's speed increases when it bounces from the moving surface. Find F in terms of the seperation of the surfaces, x.

I've already worked out the answer, but there's a step I still have some doubts. My TA told me after each collision, the ball gains 2V speed. so it's v + 2V. But how to you justify 2V, the rate of change in speed?? I'm really confused cuz momentum is conserved and kinetic energy's conserved after each collision,and with (v+2V) as the after speed these two equations don't really hold.  :(

thxx

Hey buddy,

Not to be confused u hv to first consider the surfaces to have mass M,then write energy and momentum conservation eqn's: mv-MV=-mv'-MV' , I take down to be +ve,up -ve.Just for now let's assume the velocity of surface has changed after impact.Rearrange it as m(v+v')=-M(V'-V)       EQN 1
Now write energy conservation:

        (mv^2)/2+(MV^2)/2=(mv'^2)/2+(MV'^2)/2 rearrange it as m(v+v')(v-v')=M(V'-V)(V'+V)  EQN 2

Using eqn1 in eqn2 and we get  V'={V(1-m/M)-2vm/M}/[1+m/M]   and  v'=v+V+V'

Now, in our real problem M>>m thus we get V'=V  and  v'=v+2V

As you see,the energy and the momentum are really conserved,but there comes a mistake when you directly take the V(surface velocity) to be constant or surface mass to be infinity! First you calculate then take limit.

But the bounce rate problem needs a tedious working.It took my more than one hour.I think we have to calculate it using a ''period'' of ball.I refer to time needed to come to the same point after bouncing 2 times,each from upper and lower surfaces.I took the initial separation to be x0 and the velocity v0.at any instant x=x0-Vt.Time can be expressed as t=(x0-x)/V then for nth period rate is=2/''period''  2 is the number of bouncings in each period.''period''=2x0*v0/{[v0+2V(n-1)]*[v0+2Vn]}  , n=v0*t/{x0-Vt}

We get R=(v0/x0)*{[(2x0/x)-1]^2}.This really needs very tedious work.In my opinion we cannot take the derivative respect to time,time steps are quite big and only 2 bouncing happens in one period.
2V is not the rate of change,is just change of speed in each period.Rate of change of speed is found by dividing the 2V by ''period''.Rate is time related,but you can say 2V is change of speed per 2 bounces(1 from up,1 from down).


point set
combinatorics
Turan s theorem
Extremal Graph Theory
Extremal combinatorics
IMO Shortlist

In the plane we are given a set $ E$ of 1991 points, and certain pairs of these points are joined with a path. We suppose that for every point of $ E,$ there exist at least 1593 other points of $ E$ to which it is joined by a path. Show that there exist six points of $ E$ every pair of which are joined by a path.

[i]Alternative version:[/i] Is it possible to find a set $ E$ of 1991 points in the plane and paths joining certain pairs of the points in $ E$ such that every point of $ E$ is joined with a path to at least 1592 other points of $ E,$ and in every subset of six points of $ E$ there exist at least two points that are not joined?

The wording is a bit iffy for those who know graph theory.  Here, by "path" they mean the graph-theoretic edge - not path.

Let's say we have a point $ Q_{n\plus{}1}$ in a set $ S_n$ of points connected by an edge to points $ Q_1,Q_2,...,Q_n$ and we want to determine if $ Q_{n\plus{}1}$ is connected to another vertex in $ S_n$.  Then we simply have to check whether $ 1991\minus{}|S_n|\minus{}n< 1593$.  Then we are assured that there exists a set $ S_{n\plus{}1}\in S_n$ with $ |S_{n\plus{}1}|\ge|S_n|\plus{}n\plus{}1593\minus{}1991\equal{}|S_n|\plus{}n\minus{}398$ and for all $ P\in S_{n\plus{}1}$, $ P$ is connected by an edge to points $ Q_1,Q_2,...,Q_{n\plus{}1}$.

Since the sum of the degrees of a graph must be even, we must have one vertex with degree at least $ 1594$.  It follows that $ |S_1|\ge 1594$.  Then we have $ |S_2|\ge1197,|S_3|\ge801,|S_4|\ge406,|S_5|\ge12$.  Since $ 12>5$, there exists $ Q_6\in S_5$ so that $ Q_1,Q_2,Q_3,Q_4,Q_5,Q_6$ are our desired points.

 [quote=va2010]We use Zarankiewicz Lemma.

Assume the graph is $6$-free. Then by Zarankiewicz lemma, there exists a vertex with at most degree $[\frac{4}{5}*1991] = 1592$. However, this is a contradiction as all vertices have degree strictly greater than $1592$.[/quote]
In the actual contest though, (Indian National MO (INMO) in my case), would we need to prove the Zarankiewicz Lemma, or would we get full marks for just this much?


Hi everyone,

I am writing a complete mock MathCounts exam for my brother's school. I figured since I am writing one, you all might be interested in taking it as a mock contest here on AoPS.

If there is enough interest, I will finish the exam by this Friday and we can decide on a date to hold the contest. So, please let me know if you are interested.

By the way, I figure the exam will be around a NATIONAL level exam, possibly as hard as some of the hardest Nationals. I will basically be drawing from my experiences, and creating similar problems to the hardest problems I personally faced during my MathCounts years.

Thanks, please let me know if you are interested.

i'm interested...
is it going to be a test compiled from actual MATHCOUNTS nationals, or are you making up national level questions? a good date would be Jan. 17... martin luther king day

He's writing the questions, which means it won't be from past contests.

I'm interested.

Good practice for AMC! :D

[quote="darktreb"]Yes, I'm writing the questions. However, you'll probably notice similarities to many problems you have seen, as should be the case with MathCounts.

Personally, I think it's very possible to know how to do EVERY type of MathCounts problem. To support my believe, that is how many National Champions have said they approached the contest, learning how to do every problem ever written. Each year at Nationals they pull out one or two new ones, but almost every other problem has been used in some similar form before.[/quote]

I think it's certainly true that a student can study every MATHCOUNTS problem type.  When we wrote our MATHCOUNTS Problem Series, we had maximum coverage of those problem types in mind.  When I prepared for MATHCOUNTS I took tests until I felt like I had seen all of the problem types...and then took more tests from harder competitions.  Both were extremely helpful in their own ways.

I'd like to do it. Even though I'm not anything close to national level.

I was national level last year, but I probably forgot so much :P . Ah well. I made the school team, so that's good :P .

I will do it. Thank you for your time and effort! More mock mathcounts are welcome (especially in State level)!!!

I'm not sure when I'm going to hold it. It will probably be this weekend or next. Perhaps I should ask everyone when would be best?

It will have to be some weekend during the afternoon, but which weekend is really up for debate. I will have the test done by this weekend, MOST LIKELY. However that's not a guarantee since I want to proof it with some people first.

[quote="cj"]Please be aware that there is Mathcounts online class here on Sunday.  So do not start too late.[/quote]

What class is it? It doesn't seem to be a mathjam.

[quote="yif man12"]Mock Mathcounts practice thing over AIM.[/quote]

 No . It is a class of Mathcounts problem series.

Oops, spoke too quickly. My bad.  :blush:

i'm interested...a little more practice for me :D  :D  :D

Mock MathCounts will be at 1:30 PM Central Time, tomorrow, January 30th.

How will it happen? Where should we go? What should we do? And that is 2:30 ET, right?

Is it just me, or is there a reason that my target rounds are in my outbox instead of my sentbox?

I don't plan on posting scores publically. I don't plan on posting answers just quite yet either.

what about for people who missed start time... I was at mathcounts practice... will there be a copy of the test made available to us?


a,b,c are real numbers such that   abc greater or equal than a^2+b^2+c^2

prove that abc greater or equal than a+b+c+18[/hide][/list][/code]

[quote="bvarici"]WHY nobody look this problem[/quote]
Because it seems to be incorrect.
Just take $ (a,b,c)\equal{}(0,0,0)$  :wink:

[quote="bvarici"]$ a,b,c$ are real numbers such that $ abc \geq a^2 \plus{} b^2 \plus{} c^2$

Prove that $ abc\geq a \plus{} b \plus{} c \plus{} 18$[/quote]

Maybe is for nonzero real numbers, I'll try it.

[quote="FelixD"]If I haven't made any mistakes, the inequality should be true with the restriction $ abc \ne 0$.[/quote]
I agree with you.

I'm sorry I wrote mistake , equality not hold only bigger
And I solved it it's easy :)


geometry
vector
geometric transformation
homothety
ratio
angle bisector
geometry proposed

I came across this problem while solving another one and used it as a lemma. My solution to this used vectors and some computations so please give a synthetic proof of the following:
Let the incircle of $\triangle ABC$ touch the side $AB$ at $K$ and let $M$ be the midpoint of the line segment $AB$. Prove that $CK$ is parallel to $I_c M$ where $I_c$ is a C-excenter of $\triangle ABC$.

Let F be the foot of the bisector of the angle $\angle C$, L the tangency point of the excircle $(I_c)$ with the side AB and $r, r_c$ radii od the circles $(I), (I_c)$, respectively. The vertex C is the external homothety center of the circles $(I), (I_c)$, while the angle bisector foot F is their internal homothety center (the intersection of the center line with the common internal tangent). Therefore, F divides the segment $II_c$ in the ratio $\frac{KI_c}{KI} = \frac{r_c}{r}$ internally, while C divides this segment in the ratio $\frac{CI_c}{CI} = \frac{r_c}{r}$ externally. Consequently,

$CF = CI + FI = \frac{r \cdot II_c}{r_c - r} + \frac{r \cdot II_c}{r_c + r} = \frac{2r_c r \cdot II_c}{r_c^2 - r^2}$

$I_cF = \frac{r_c \cdot II_c}{r_c + r}$

$\frac{CF}{I_cF} = \frac{2r}{r_c - r}$

The internal homothety center F also divides the segment length KL of the common internal tangent in the ratio $\frac{FL}{FK} = \frac{r_c}{r}$ internally, hence

$FK = \frac{r \cdot KL}{r_c + r},\ \ FL = \frac{r_c \cdot KL}{r_c + r}$

The midpoint M of the side AB is also the midpoint of the segment length KL of the common internal tangent, hence

$FM = FL - ML = \frac{r_c \cdot KL}{r_c + r} - \frac{KL}{2} = \frac{(r_c - r) \cdot KL}{2(r_c + r)}$

$\frac{FK}{FM} = \frac{2r}{r_c - r}$

As a result, the triangles $\triangle FKC \sim \triangle FMI_c$ with equal angles at the common vertex F are centrally similar with the similarity center F by SAS, which means that their corresponding sides $CK \parallel I_cM$ are parallel.

[hide="Here is a short proof."]
[color=red][b]Lemma.[/b] Let $ABC$ be a triangle for which denote the its incircle $w=C(I,r)$ and the its $C\ -$ exincircle $w_c=C(I_c,r_c)\ .$ Define the points : $K\in AB\cap w$, $K_c\in AB\cap w_c\ ;$ $\{K,K'\}\subset IK\cap w$, $\{K_c,K'_c\}\subset I_cK_c\cap w_c$. Then $C\in K_cK'\cap K'_cK\ .$[/color]

[color=blue][b]Proof.[/b] Denote the projection $D$ of the point $C$ on the line $AB\ (CD=h_c)\ .$ Remember the relations $S=\frac{ch_c}{2}=rp=r_c(p-c)$ and $K_cA=KB=p-b$. Prove easy: $KD=\frac{|a-b|(p-c)}{c}$, $KK_c=|a-b|$, $K_cD=\frac{|a-b|p}{c}\ .$ Thus,
$\blacktriangleright C\in K_cK'$ $\Longleftrightarrow$ $\frac{K_cK}{K_cD}=\frac{2r}{h_c}$ $\Longleftrightarrow$ $\frac cp=\frac{2r}{h_c}\ ,$ yes.
$\blacktriangleright C\in K'_cK$ $\Longleftrightarrow$ $\frac{KK_c}{KD}=\frac{2r_c}{h_c}$ $\Longleftrightarrow$ $\frac{c}{p-c}=\frac{2r_c}{h_c}\ ,$ yes.
[/color]
[color=darkblue][b]Proof of the proposed problem.[/b]  The points $M$, $I_c$ are the middlepoints of the segments $K_cK$, $K_cK'_c$ respectively, i.e. the segment $MI_c$ is the $K_c\ -$ middle-line in the triangle $K_cKK'_c\ .$ Therefore, $MI_c\parallel \overline {CKK'_c}\ .$

[b]Remark.[/b] This problem is a nice application of the homotety from the origin $C$ with the ratio $\rho=\frac{r_c}{r}\ :$ $L\rightarrow L'$ $\Longleftrightarrow$ $CL'=\rho\cdot CL\ .$
See the my last reply from[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=82316[/hide]


linear algebra
matrix
calculus
integration
search
linear algebra unsolved

Let $A$ be an $n\times n$ antisymmetry matrix,i.e.$A^{T}=-A$,with all integral entries.Show that $\det A$ is a square. :)

Been posted long before :oops: Thank you alekk :)


algebra
polynomial
calculus
derivative
calculus computations

I am trying to show for $ q(x): \mathbb{R} \rightarrow \mathbb{R}$ is smooth, where:

$ q(x) \equal{} \begin{cases} e^{ \minus{} 1/x} & x> 0 \\
0 & x\le 0\end{cases}$

So I'm trying to take the $ r^{th}$ order Taylor polynomial of  $ q$ at x, but it's coming out to something that's really not nice looking at all (for each higher derivative I calculate). How should I do this to show $ q$ is smooth and analytic?

It's smooth. It's not analytic.

You want to show that all of those derivatives are zero at zero. Apply the limit definition.

I keep trying to find some way to show that $ e^{\minus{}1/x} \rightarrow 0$, but Taylor series, L'Hopital's, nothing will show that it willl.....am I missing something here?

Well, $ e^{\frac1x}$ doesn't go to zero. The quantity you actually care about, $ e^{\minus{}\frac1x}$, isn't even an indeterminate form.


number theory
prime numbers
number theory unsolved

for which values of n, the expression  $n^{n+1} + (n+1)^n$  is a prime?

Hmm.. I stopped after finding two and I don't know how to find others if there are..

[hide="My attempt"]

Since the second part is $(n+1)^{n}$, we don't want $n$ to be negative because that way, we'll get fraction and prime numbers are integers that have two divisors, 1 and itself. So, inspection on $n = 0$ gives $0^{0+1} + (0+1)^0 = 0+1 = 1$ but 1 is not a prime. For $n = 1,2$, the results are 3 and 17, respectively. The next one was 81+64 which ends in 5 so it's not a prime. The next one is 1024+625 = 1649 which is not prime since it's divisible by 17 and 97. I stopped here because the numbers were getting large and I didn't want to do all the tedious work.[/hide]

So, I know there are at least two of them that works but to me, it looks like there aren't anymore. But again, I don't know if this is true and even it is, I don't know how to prove it..  :?


[quote="1=2"]Myself[/quote]
That's what I'm going as, too!

I am dressing up as a Klein Bottle and my imaginary friend (Borat) is dreassing up as a Hypercube.

[quote="pii"]umm, a monk from Monty Python and the Holy Grail. So I can walk around, chant and hit myself on the head. So much fun...AND CANDY![/quote]HILARIOUS IDEA!!

I'm Calvin & Hobbes this year, but I think I have a plan for next year - A Day Late.  The day after halloween, I go around ringing peoples' doorbells and saying "trick or treat!" with no costume.  They say "what are you?"  I say "I'm a Day Late!"

Though it might annoy people.  And it wouldn't get much candy.

nothing this year

next year, im gonna be something(not telling you!)

few more hours before it starts for me (about 5:30 PM right now).

[hide="I had so much fun"]at Panera Bread. I also went to Target to get some new shoes and to Lowes to get a new light.[/hide]

Candy: two pieces of candied ginger(YUMMY) from the candied ginger jar

[quote="mlok"]I taught Calculus until 7pm today. So I was just that, a scary calculus teacher. :diablo:[/quote]

I learned Calculus until 3pm today. So I was just that, an average calculus student.

I was a rather sexy girl.

Eyeliner, lipstick, head decorations, miniskirt, and ooh fairy wings.

Yeah a couple of my friends played dressup on me right before chem class

i had wayyy too much math stuff 2 do so i didnt go trick-or-treating

Went trick-or-treating yesterday with a bunch of friends. It was fun. These two teenagers were chasing us cuz we wouldn't take the candy in under 0.0000001 seconds lol!

so my 3 hours nap has led to 17 hours of sleep in the last 24 because i'm sick

Wow, story from a classmate:

So his parents are real strict and don't celebrate Halloween. So they put up a sign that says "Sorry, we don't celebrate Halloween"

But these kids come anyway and then ring the doorbell.

Classmate: Sorry, we don't celebrate Halloween
Kid (literally): WTF? **@#*$ \$$(@)!@@+#_@)!(\ !  GIVE US CANDY
Doorbell: ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong ding-dong
Classmate:  :maybe:  :huh:

Holy............... That's so .......... (can't find the word)......... Maybe awkward or something.