geometry rhombus trigonometry logarithms inequalities parallelogram function

from Robert W Martin's Second Year Algebra: Patterns and Connections

$\star$1.  The sides of a rhombus have length $\sqrt{14}$, and the sum of the lengths of the diagonals is $10$.  Find the area of the rhombus.

$\star$2.  Find the three possible values of $x^{20}$ if $x^{20} + 6x^{10} = 5x^{15}$.

$\star$3.  Graph all $x$ on the number line for which $(x^2-2x)^2 - 11(x^2-2x) + 24 < 0$

$\star$4.  If $0^{\circ} < A < 90^{\circ}$ and $\cos{A} = \frac{3\sqrt{11}}{10}$, find the value of $\log{\sin{A}} + \log{\cos{A}} + \log{\tan{A}}$.

$\star$5.  If $x^2+4x-6=0$ has solution set $\{r,s\}$ while $x^2+ax+b=0$ has solution set $\{r-1,s-1\}$, find the value of $a+b$.

$\star$6.  The sum of two of the solutions of $3x^3-20x^2+kx-10=0$ is $6$.  Find the value of constant $k$.

$\star$7.  Suppose $g(x) = x^2+kx+m$ for some constants $k$ and $m$.  If $g(1-x) = g(x+1)$ [u]for all values of x[/u], find the value of $k$.

$\star$8.  Find the distance from the origin to the triangular region whose vertices correspond to the $x$, $y$, and $z$ intercepts of the plane $x+y+z=2$.

$\star$9.  Simplify: $\prod_{k=4}^{21}{(\frac{1}{k^2}-1)}$

$\star$10.  Find the degree measure of angle $\alpha$ between $0^{\circ}$ and $360^{\circ}$ for which $\tan{\alpha}+\cot{\alpha} = 2\sec{\alpha}$.

good book, I might edit in more as I come across them.
($\star$'s by correctly answer ones, as of last time I checked)

[quote="djshowdown2"]
2.  Find the three possible values of $x^{20}$ if $x^{20} + 6x^{10} = 5x^{15}$.

[/quote]

[hide]- $x=0$ works
- $x$ is different from $0$
$\implies x^{10}-5x^5+6=0$
$\implies (x^5-\frac{5}{2})^2=\frac{1}{4}$
$\implies x^5-\frac{5}{2}=\frac{1}{2}$ or $x^5-\frac{5}{2}=\frac{-1}{2}$
Then $x^5=3$ and $x^5=2$ gives us $x^{20}=81$ and $x^{20}=16$

So three possible values for $x$ are $0,81,16$[/hide]

Watch your area formula on #1.
I've editted in 2 more, an inequality and a nice log/trig problem

3
[hide]
Factor to $(x^2-2x-8)(x^2-2x-3)<0$, then to $(x-4)(x+2)(x-3)(x+1)<0$, using interval testing we get $x=0$ to be greater, so the intervals between the roots alternate pos/neg, so we get $x \in (-2,-1) \cup (3,4)$
[/hide]

5
[hide]
Using viete, we get $r+s=-4$ and $rs=-6$.  We also get $(r-1) + (s-1) = -a$ and $(r-1)(s-1) = rs - (r + s) + 1 = b$.  Plugging in, we get $a = 6$ and $b=-1$, so the solution is $a+b=5$
[/hide]

[quote="djshowdown2"]6.  The sum of two of the solutions of $3x^2-20x^2+kx-10=0$ is $6$.  Find the value of constant $k$.[/quote]
Isn't it $3x^3-20x^2+kx-10=0$?

[quote="frt"]
Isn't it $3x^3-20x^2+kx-10=0$?[/quote]

Yes  :oops:

[quote="djshowdown2"]1.  The sides of a rhombus have length $\sqrt{14}$, and the sum of the lengths of the diagonals is $10$.  Find the area of the rhombus.
[/quote]
[hide]
In a parallelogram, the sum of the squares of the sides is equal to the sum of the squares of the diagonals. The area of a rhombus, a type of parallelogram, is equal to half of the product of the diagonals.

$\sqrt{14}^2 \cdot 4 = d_1^2+d_2^2\\ 56 = (d_1+d_2)^2-2d_1d_2\\ 56=100-2d_1d_2\\ d_1d_2=22\\ K=11$[/hide]

[hide="7"]Since $g(1-x)=g(x+1)$ we also have that $(1-x)^2+k(1-x)+m=(x+1)^2+k(x+1)+m$. The terms cancel to leave us with $-2x-kx=2x+kx$. Divide both sides by x to get $-2-k=2+k$, and $k=-2$. [/hide]

right right, you're all doin well.  A note on #7, thats a fine way to do it, but if the function were more complicated a nice solution comes from choosing a simple value to substitute in, since it must be true for all values.

Also I editted in a couple more problems, 10 might be a good number.

ALSO, I put these problems here based on difficulties of other problems in this forum, but I just read rruscyzk's Cleaning Up thread.  are these a good difficulty for this board, in case a decide to post more some time?  Thanks

#9[hide]
if you set k=4
the first term is $(\frac{1}{k}+1)(\frac{1}{k}-1)=-(\frac{k-1}{k})(\frac{k+1}{k})$
the second term is $(\frac{1}{k+1}+1)(\frac{1}{k+1}-1)=-(\frac{k}{k+1})(\frac{k+2}{k+1})$
the third term is $-(\frac{k+1}{k+2})(\frac{k+3}{k+2})$
.
.
.
the 18th term is $-(\frac{k+16}{k+17})(\frac{k+18}{k+17})$
so-(to make it visually clear)
(since all the terms are negative and there are an even number of terms,  all the minus signs are omitted)
$(\frac{k-1}{k})(\frac{k+1}{k})(\frac{k}{k+1})(\frac{k+2}{k+1})...(\frac{k+15}{k+16})(\frac{k+17}{k+16})(\frac{k+16}{k+17})(\frac{k+18}{k+17})$
the second factor of the first term cancels out with the first factor of the second term,
the second factor of the second term cancels out with the first factor of the third term,
the second factor of the third term cancels out with the first factor of the fourth term,
until only $(\frac{k-1}{k})(\frac{k+18}{k+17})$
k=4, so the total product equals $\frac{11}{14}$.
please correct me if im wrong, im noob

[/hide]

[quote="djshowdown2"]ALSO, I put these problems here based on difficulties of other problems in this forum, but I just read rruscyzk's Cleaning Up thread.  are these a good difficulty for this board, in case a decide to post more some time?  Thanks[/quote]

I'll move this to Intermediate Forum so you don't have to worry about difficulty. :)

[hide="8"]
Focus on the tetrahedron whose vertices are $O(0,0,0)$, $A(2,0,0)$, $B(0,2,0)$, and $C(0,0,2)$.

The area of the tetrahedron is $\frac{1}{3}(2)\left(\frac{1}{2}\cdot 2\cdot 2\right)=\frac{4}{3}$

Let $d$ be the distance from $O$ to $\triangle ABC$. The volume of the tetrahedron can be calculated by $\frac{1}{3}d(\triangle ABC)$, and we know that $\triangle ABC=\frac{(2\sqrt{2})^2\sqrt{3}}{4}=2\sqrt{3}$ (because $\triangle ABC$ is an equilateral triangle with side length $2\sqrt{2}$.)

Therefore, $d=\frac{4}{3}\times 3\times\frac{1}{2\sqrt{3}}=\boxed{\frac{2\sqrt{3}}{3}}$
[/hide]

EDIT-[hide="If you know the formula"]
$d=\frac{|Ax+By+Cz+D|}{\sqrt{A^2+B^2+C^2}}$

$=\frac{|0+0+0+2|}{\sqrt{1^2+1^2+1^2}}$

$=\frac{2}{\sqrt{3}}$

$=\boxed{\frac{2\sqrt{3}}{3}}$[/hide]

[quote="shyong"][hide="question 1"]

as chess64 said , sum of square of four side is equal to sum of square of both diagonal. Hence we have $d_1^2+d_2^2=56$ where $d_1,d_2$ are the diagonal.
Since $d_1+d_2=10$ , we can get $d_1d_2=22$

From cosine rules , we can obtain

$(d_1d_2)^2=28^2(1-\cos^2\theta)$ hence $\sin\theta=\frac{11}{14}$

So the area is $\frac{1}{2}d_1d_2\sin\theta=\frac{121}{14}=8\frac{9}{14}$
 ;)
[/hide][/quote]

$A=(1/2)d_1d_2$

http://mathworld.wolfram.com/Rhombus.html

[quote="djshowdown2"]from Robert W Martin's Second Year Algebra: Patterns and Connections
10.  Find the degree measure of angle $\alpha$ between $0^{\circ}$ and $360^{\circ}$ for which $\tan{\alpha}+\cot{\alpha} = 2\sec{\alpha}$.

good book, I might edit in more as I come across them.
($\star$'s by correctly answer ones, as of last time I checked)[/u][/quote]

$\frac{\sin{\alpha}}{\cos{\alpha}}+\frac{\cos{\alpha}}{\sin{\alpha}} = \frac{2}{\cos{\alpha}}$
$\frac{1}{\sin{\alpha}\cos{\alpha}} = \frac{2}{\cos{\alpha}}$
$\frac{1}{\cos{\alpha}}(\frac{1}{\sin{\alpha}}-2) = 0$, as $\frac{1}{\cos{\alpha}} \not= 0$ for all $\alpha$ we get:
$\sin{\alpha} = \frac{1}{2} \rightarrow \alpha = 30^{\circ}, 150^{\circ}$.

[quote="pontios"][hide="1"]
The diagonals are $x$ and $y$, so $x+y=10$

The area is $x \cdot y$

$(\frac{x}{2})^2+(\frac{y}{2})^2=(\sqrt{14})^2$

$\frac{x^2}{4}+\frac{y^2}{4}=14$

$x^2 + y^2 = 4 \cdot 14$

$x^2 + y^2 =56$

$x^2+y^2+2xy = 56 + 2xy$

$(x+y)^2 = 56+2xy$

$100 = 56+2xy$

$2xy = 44$

$xy = 22$[/hide][/quote]

[quote="djshowdown2"]Yep, that's all of them.  A note on number 1, it really is not meant to be complicated.  You don't need trig, you don't even need to know that the sum of squares of sides = sum of squares of diagonals.  Pontios had a good one, just messed up the area formula, it should have been half that.  Simplest way (for me) to see it is call half of each diagonal $a$ and $b$, so we know $a^2 + b^2 = 14$ and $2a+2b = 10$, trying to find $2ab$  ;)[/quote]

I was wrong. Sorry :oops:

Of course the area is $\frac{1}{2}xy$


calculus
integration
trigonometry
function
logarithms
calculus computations

hello, this integral is not easy, i think the substituion $ z\equal{}\tan\left(\frac{x}{2}\right)$ will help.
Sonnhard.

$ z\equal{}\tan\left(\frac{x}{2}\right)$ doesn't help.

[quote="Xaenir"]
\[ \int {\frac {{dx}} {{\cos ^5 x \plus{} \sin ^5 x}}}
\]
[/quote]
This is same as the integral of algebraic function you posted, :wink:

Hey,　you, Xaenir
Why do you use my avatar? :mad:

Yeah, kunny was right. At first, I did not look at the name but the picture alarmed me that it was kunny. Ha.. It is quite a bit hilarious! :rotfl:
By the way, there is no contemporary method to solve this problem with the exception of setting $ t\equal{} tan(\frac{x}{2})$

[quote="kunny"]
Why do you use my avatar? :mad:

[/quote]

Even i thought it was Kunny!

[quote="ith_power"]
This is same as the integral of algebraic function you posted

[/quote]

Where? I don't see any post like that here!

Indeed. It's 0.2 times that of Integral of Algebraic Function"-13

I have a solution , but its a bit too lengthy,,,anyone knows a shorter method

Xaenir, hey you, You should delete your avatar and apologize for your rudeness. :mad:

What's on Earth, why do you stole my avatar from me? :wink:

\[ \begin{gathered} \cos ^5 x \plus{} \sin ^5 x \hfill \\
\equal{} (\cos ^2 x \plus{} \sin ^2 x)(\cos ^3 x \plus{} \sin ^3 x) \minus{} \cos ^2 x\sin ^2 x(\cos x \plus{} \sin x) \hfill \\
\equal{} (\cos x \plus{} \sin x)(1 \minus{} \cos x\sin x \minus{} \cos ^2 x\sin ^2 x) \hfill \\
I \equal{} \int {\frac {{dx}} {{\cos ^5 x \plus{} \sin ^5 x}}} \equal{} \int {\frac {{dx}} {{(\cos x \plus{} \sin x)(1 \minus{} \cos x\sin x \minus{} \cos ^2 x\sin ^2 x)}}} \hfill \\
\equal{} \int {\frac {{(\cos x \plus{} \sin x)dx}} {{(\cos x \plus{} \sin x)^2 (1 \minus{} \cos x\sin x \minus{} \cos ^2 x\sin ^2 x)}}} \hfill \\
\equal{} \int {\frac {{(\cos x \plus{} \sin x)dx}} {{(1 \plus{} \sin 2x)(1 \minus{} \cos x\sin x \minus{} \cos ^2 x\sin ^2 x)}}} \hfill \\
\end{gathered}
\]
Put
\[ \sin x \minus{} \cos x \equal{} t
\]

\[ \begin{gathered} \therefore (\cos x \plus{} \sin x)dx \equal{} dt \hfill \\
t^2 \equal{} 1 \minus{} \sin 2x \Leftrightarrow 1 \plus{} \sin 2x \equal{} 2 \minus{} t^2 \hfill \\
\therefore I \equal{} \int {\frac {{dt}} {{(2 \minus{} t^2 )\left( {1 \minus{} \frac {{t^2 \minus{} 1}} {2} \minus{} \frac {1} {4}(t^2 \minus{} 1)} \right)}}} \hfill \\
\equal{} 4\int {\frac {{dt}} {{(2 \minus{} t^2 )(5 \minus{} t^4 )}}} \hfill \\
\equal{} 4\int {\left[ {\frac {1} {{2 \minus{} t^2 }} \plus{} \frac {1} {{(2 \minus{} \sqrt 5 )(2\sqrt 5 )(\sqrt 5 \minus{} t^2 )}} \plus{} \frac {1} {{(2 \plus{} \sqrt 5 )(2\sqrt 5 )(\sqrt 5 \plus{} t^2 )}}} \right]} dt \hfill \\
\equal{} \sqrt 2 \log \left| {\frac {{\sqrt 2 \plus{} t}} {{\sqrt 2 \minus{} t}}} \right| \plus{} \frac {1} {{(2 \minus{} \sqrt 5 )5^{\frac {3} {4}} }}\log \left| {\frac {{5^{\frac {1} {4}} \plus{} t}} {{5^{\frac {1} {4}} \minus{} t}}} \right| \plus{} \frac {2} {{(2 \plus{} \sqrt 5 )5^{\frac {3} {4}} }}\tan ^{ \minus{} 1} \left( {\frac {t} {{5^{\frac {1} {4}} }}} \right) \plus{} C \hfill \\
\end{gathered}
\]
Anyone has a shorter one??...
Please reply.

Kunny, come on man how old are you?
Grow up dude, i mean, he uses your avatar.. so what.., then there are two of you... :D

well I changed my decision.... :wink: ..........an avatar is too trivial a thing to deprive MathLinks of the great "kunny".So here is my apology  :P
Now can we be friends again?


why is spam called spam??
is there some connection between the food spam??

I know the food means Shoulder Pork and hAM.


Stie cineva cum se repartizeaza premiile (Isi II) la concursul national de matematica??Va rog sa-mi raspundeti .

da ma, la arhimede cred ca zici ...cum s-au dat si anul trecut
-35.5 - 40 premiu 1
-28,5 sau ceva in genu - 35 premiu 2
-24,5 - 28,5 sau ceva in genu - premiu 3
-18,5 - 24.5 mentiune...parca:|


Hi

My sister made an account here and forgot her username. The site isn't letting her create an account with the same email. What should she do to get that account with the same email?

Email me her email address ( valentin@mathlinks.ro ) and I will email her back the username.


inequalities
function
number theory unsolved
number theory

[hide]Let $ p_n$ be the larger of a pair of twin primes.

$ Q.E.D.$[/hide]

A bit longer than I wanted it...nice problem though.

Wait, am I wrong?  :maybe:

EDIT:  OOPS, twin prime conjecture not yet proven :oops:

The [url=http://mathworld.wolfram.com/TwinPrimeConjecture.html]Twin Prime Conjecture[/url] is still open.  See the link for more details.  I highly suspect that its proof is non-elementary, and so even if this conjecture had been settled, its use here would not be acceptable to bboypa.

Let's see if I can do it right this time...

[hide="Take 2"]Clearly, in order for a given prime $ p_k$ to fit the condition in the problem, we must have

${ p_k-p_{k-1}<p_{k+1}-p_k}$

Assume the statement to be proven is false and that there are only finitely such primes.  This implies that there are an infinite number of primes that do not satisfy this condition and an inifinite number of "prime gaps" that are larger than the next.

Let $ p_k$ be the smallest prime preceding a prime gap of at least $ k$.  Since our assumptions implies there are only a finite number of prime gaps less than the preceding one, there are infinitely many such primes.  By the way we defined these primes, the prime gap preceding any such prime must be less than the prime gap following it, that is

$ p_k-p_{k-1}<p_{k+1}-p_k$

Hence, our assumption was wrong and infinitely many such primes exist.[/hide]

Please say I didn't mess it up this time :maybe:

AMS, I'm not sure what you're doing.  You don't seem to have negated the statement correctly, and you're using $ k$ in two different ways.  You also haven't used any properties specific to the primes.

Suppose $ 2p_n \ge p_{n \minus{} 1} \plus{} p_{n \plus{} 1}$ for all but finitely many $ n$.  Letting $ d_n \equal{} p_n \minus{} p_{n \minus{} 1}$, this is equivalent to the inequality $ d_n \ge d_{n \plus{} 1}$, but this would imply that $ p_n \le cn$ for some constant $ c$.  This contradicts the Prime Number Theorem, although there's an easier way to see it:  for any fixed $ k$, the primes greater than $ p_k$ belong to one of

$ (p_1 \minus{} 1)...(p_k \minus{} 1)$

residue classes $ \bmod p_1 ... p_k$, which implies that

$ c \le \prod_{i \equal{} 1}^{k} \left( 1 \minus{} \frac {1}{p_i} \right)$.

Now it's a standard exercise to show that this product diverges to zero.  (Hint:  its reciprocal "converges" to $ \zeta(1)$.  For this reason, it's slightly more natural to phrase this argument in terms of an asymptotic for the prime number counting function $ \pi(n)$.)

[quote="bboypa"]Show that there exist infinitely many $ n \in \mathbb{N}$ such that $ 2p_n < p_{n \minus{} 1} \plus{} p_{n \plus{} 1}$ where $ p_i$ denote the $ i \minus{}$ th prime (i.e. $ p_1 \equal{} 2,p_2 \equal{} 3,\ldots$).

Ps1. We want [i]only[/i] very elementary proof
Ps2. I hope you like it  :D[/quote]

[hide="My elementary proof"]
Let $ s_n \equal{} p_{n \plus{} 1} \minus{} p_n$

If this sequence is always non increasing from a given point, then it has a limit $ a$ and $ s_n \equal{} a$ $ \forall n\geq N$. Then $ p_{N \plus{} k} \equal{} p_N \plus{} ka$ and so $ p_{N \plus{} p_N} \equal{} p_N(1 \plus{} a)$ and is not prime (I consider $ a>0$, since the case $ a\equal{}0$ is obviously impossible).

So, for any $ N$, it exists $ n\geq N$ such that $ s_{n} > s_{n \minus{} 1}$
Q.E.D
[/hide]

[quote="pco"][quote="bboypa"]Show that there exist infinitely many $ n \in \mathbb{N}$ such that $ 2p_n < p_{n \minus{} 1} \plus{} p_{n \plus{} 1}$ where $ p_i$ denote the $ i \minus{}$ th prime (i.e. $ p_1 \equal{} 2,p_2 \equal{} 3,\ldots$).

Ps1. We want [i]only[/i] very elementary proof
Ps2. I hope you like it  :D[/quote]

[hide="My elementary proof"]
Let $ s_n \equal{} p_{n \plus{} 1} \minus{} p_n$

If this sequence is always non increasing from a given point, then it has a limit $ a$ and $ s_n \equal{} a$ $ \forall n\geq N$. Then $ p_{N \plus{} k} \equal{} p_N \plus{} ka$ and so $ p_{N \plus{} p_N} \equal{} p_N(1 \plus{} a)$ and is not prime (I consider $ a > 0$, since the case $ a \equal{} 0$ is obviously impossible).

So, for any $ N$, it exists $ n\geq N$ such that $ s_{n} > s_{n \minus{} 1}$
Q.E.D
[/hide][/quote]

Yes patrick, the same of mine (however s_n, s_{n+1} and s_{n+2} cannot be equal mod 3..)<---------(eeeeeeeeeehmm  sorry )
Thanks also to t0rajir0u for the answer!

Now the [i]bonus question[/i]:
Show that there exist infinitely many $ n \in \mathbb{N}$ such that $ 2p_n > p_{n \minus{} 1} \plus{} p_{n \plus{} 1}$ where $ p_i$ denote the $ i \minus{}$ th prime (i.e. $ p_1 \equal{} 2,p_2 \equal{} 3,\ldots$).  :D
(Again only elementary proof)

[quote="bboypa"] ...(however s_n, s_{n+1} and s_{n+2} cannot be equal mod 3..)...[/quote]

You're wrong. It's possible to have $ s_n, s_{n\plus{}1}$ and $ s_{n\plus{}2}$ equal mod $ 3$ : take for example $ 251,257,263,269$ and you have $ s_n\equal{}s_{n\plus{}1}\equal{}s_{n\plus{}2}\equal{}6$


You're wrong : $ a\equal{}\minus{}\log_2(\ln 2)$ is the zero of $ f'(x)$ and so you can just say that $ f(x)\equal{}2^x\minus{}x\minus{}1$ is decreasing on $ (\minus{}\infty,a)$ and increasing on $ (a,\plus{}\infty)$

So $ f(x)$ has at most two zeroes.

And since we trivially have $ f(0)\equal{}f(1)\equal{}0$, we know that $ f(x)\equal{}0$ has exactly two roots $ 0$ and $ 1$ and so :

$ f(x)>0$ $ \iff$ $ x\in(\minus{}\infty,0)\cup(1,\plus{}\infty)$

[quote="theSA"]I found this, but I don't know if it is useful.
Set  $ f(x) \equal{} 2^x \minus{} x \minus{} 1$. then using  $ f'(x)$  I find out that critical point is  $ \minus{} \log _2 (\ln 2)$.
So solution can be  $ ( \minus{} \log _2 (\ln 2),\, \plus{} \infty )$.
I think...  :?:[/quote]

Yes I see it is wrong. since  $ x\equal{}\minus{}3$  also an answer.

[b]pco[/b] how you get that it has got ONLY 2 ROOTS?

[quote="theSA"] [b]pco[/b] how you get that it has got ONLY 2 ROOTS?[/quote]

Since $ f(x)$ is strictly decreasing on $ ( \minus{} \infty,a]$, there is at most one root in this interval.
Since $ f(x)$ is strictly increasing on $ [a, \plus{} \infty)$, there is at most one root in this interval.
And so at most two roots in $ \mathbb R$

And since we found two trivial roots, we got the whole set of roots.


When I try to visualize division by a whole number I imagine splitting the dividend into number of compartments indicated by the divisor and then I fill up each compartment with as much numbers as I can.

For example 6 : 3 = 2  -->  six split up in three compartments and each compartment can hold maximum two numbers.

But I am having problems visualizing division by fraction. It confuses me that the resulting quotient is bigger than the dividend.

For example 1 : 1/5 = 5

In this example my method of visualising is nonsense because each compartment can not hold 5 numbers when you have been given just 1 number to work with in the first place. In real world whatever you brake into parts those parts are always smaller than the initial thing but in mathematics it doesn't seam so..

I hope you can see what is confusing me if not I will try to explain again. Could you please share with me some of your methods of visualising division by fraction?

Thank you

Look at it this way: Diving by a fraction is the same as multiplying by its reciprocal. I always switch it to multiplication because it's easier. $2\div\frac{2}{3}=2\cdot\frac{3}{2}$

Try this:

Your compartment idea is good, and will work with fractions also   :)

For example, lets look at $\frac{1}{\frac15}$.

This can also be said in words as "What times $\frac15$ equals 1?"

There are five parts of the whole when using fifths, so look at this picture:
------------
|1|2|3|4|5|
------------

There are five sections, called 1-5.  Each section has an equal area, and they all add up to the entire bar.  Thus, each section is $\frac15$ of the entire area.  Now, can you see that $\frac1{\frac15}=5$?

--------------------------------------------------------------

Another way to look at this:

Write the problem as $1\div\frac15=x$.  Then, multiply by $\frac15$ to eliminate dividing by fractions.

$1\div\frac15\times\frac15=\frac{x}5$

When simplified, this becomes

$1=\frac{x}{5}$

Multiply each side by five to solve for $x$, and get

$5=x$

This is the answer you want from the first equation, so we are done.


I've only been here for a few days; I apologize if there is an answer to the question below somewhere on the site but i haven't found it yet...

I know that you can get exponents etc. under the emoticons, but when I use them they come out in a hideous yellow that I can barely read.

example: x :^2:  :ge: 0 for all real x is very hard for me to make out; I've been typing x^2 [u]>[/u] 0 ... instead, but it takes me longer...

Any suggestions on how to change the color of the math symbols?  FYI, I am using the "large black font" when I log in - the other typeface is less pleasant for me to read.

"hideous"? :)
Anyways, if you highlight the emoticons, it's not [b]that [/b]bad. You'll need to highlight Spoilers anyways so you shouldn't have any problem.

highlighting is fun anyways. so i do it for most posts (lots of people put hidden spoilers)

Really? Some people doesn't like unnecessary Spoilers. In fact, I think you reprimanded me for using too much spoilers when I just began posting on AoPS.

I have highlighted things before (especially Joel's posts in cyan), but when I highlight any of the symbols generated by emoticons all I get in their place is a shaded square.

Maybe the software doesn't like me? :-x  :roll:

[quote="Tare"]"hideous"? :)
[/quote]

I don't mind yellow in general (I'm wearing a yellow tie right now), but since I can't read any of the yellow printing it annoys me, so it must be hideous - QED
(the author makes no warrantee that the above line of reasoning will win any logic awards)

[quote="Tare"]What browser are you using?

Even if it was a shaded square, the yellow should stand out...at least on my computer.[/quote]

I have a laptop supplied by the school where I teach - it has explorer 5.0 (we're not allowed to use anything else on their machines).

...so for you, if a=annoying, than it follows that a=hideous?
Counter example: If your mom was annoying, is she hideous?

and why are you wearing a tie? :)

Getting back on topic, if you put your mouse pointer above the yellow emoticon, then the description would show up. It'll take a long time, but it would work. Try these: :rt2:  :le:  :cong:  :inf:

Unfortunately, we can't make a different set of Emoticons for each style (which would solve the problem).  Sorry...

Maybe when I have time I'll see if I can turn off the math emoticons for the other board (so you'll see the text that generates the math symbol instead of the math symbol).  This will probably have to wait until we port the site to a new server this winter...

I'm still wearing a tie because I came back to school (there was a "fun night" tonight where they set up a kareoke and a jousting game and some other stuff) - male teachers are generally expected to wear a tie (they also want us to wear a suit jacket, but I usually dodge that request).

FYI, I teach at a small, private, religiously-based (although I'm not religious) all-girls school, pre-school through grade 12.

I had already discovered that holding the pointer over the yellow text works, provided that I see it in the first place, but thanks for your suggestion.   I'm hoping for a better answer from "the powers that be" on this board.

in this case, the "powers that be" are the admin...

:?: or it could be any "oldies" on the AoPS like mathfanatic or Syntax Error, no?

...although I think that I mentioned all the possible solutions, though, except for the obvious ones like "Change the style to Normal" and "Chnage display settings"


What is the sum of all possible values of $ k$ for which $ x^2 \plus{} kx \minus{} 9x \plus{} 25$ is the square of a binomial?

We want $ k\minus{}9\equal{}10$ or $ k\minus{}9\equal{}\minus{}10$, so $ k\equal{}19$, or $ k\equal{}\minus{}1$. The sum is $ \boxed{18}$. I think..... :maybe:

that is correct
a slightly more detailed explanation follows

$ x^2\plus{}kx\minus{}9x\plus{}25$

we want a value of $ k$ for which that is the square of a binomial

we see that the first term is a $ x^2$ and the last term is a $ 25$

therefore, we know that the binomial must be $ (x\pm 5)$

that means the middle term must be $ \pm10x$

case 1: the middle term is $ 10x$

that means $ kx\minus{}9x\equal{}10x$

and $ kx\equal{}19x$, so $ k\equal{}19$

case 2: the middle term is $ \minus{}10x$

that means $ kx\minus{}9x\equal{}\minus{}10x$

and $ kx\equal{}\minus{}x$, so $ k\equal{}\minus{}1$

our sum of all possible values of $ k$ is $ 19\plus{}(\minus{}1)\equal{}\boxed{18}$

thats how i did mine i like your way better though, vallon22


Twenty switches in an office computer network are to be connected so that each switch has a direct connection to exactly three other switches. How many connections will be necessary?

The answer is 30. Is there a quick way to figure it out?

Say that a connection occurs at each computer at a port. Then since there are 20 computers and each computer has 3 connections and thus 3 ports, there are 60 ports in total. However, each connection connects the port of one computer to the port of another, so we must divide the number of ports by 2, to get 60/2 = 30 connections.

Quick way ??? What;s the long way ? {Only if you tell us how you've done it can we tell you a {possibly}shorter method of doing it.} :|


Given: $ \frac xy \equal{} \frac 83$ . What is the value of the ratio $ \frac y{2x}$? Express your answer as a common fraction.

hello, from $ \frac{x}{y}\equal{}\frac{8}{3}$ we get $ \frac{y}{x}\equal{}\frac{3}{8}$ multiplying by $ 1/2$ one get
$ \frac{y}{2x}\equal{}\frac{3}{16}$.
Sonnhard.

8y=3x
192y=72x
(5 1/3)y=2x
So the fractions is $ \frac {y}{\frac {16}{3}\times y}$
we simplify to get 3/16.
Ugh, that was ugly


algebra
polynomial
calculus
derivative
number theory
greatest common divisor
relatively prime

There exists a polynomial of degree 5 with the following property: if $ z$ is a complex number such that $ z^5 \plus{} 2004z \equal{} 1$, then $ P(z^2) \equal{} 0$. Calculate the quotient $ \frac{P(1)}{P(\minus{}1)}$.

Not completely confident about this

[hide]
If z is a root of P(z) then,

$ \sqrt{z}^5 \plus{} 2004 \sqrt{z} \equal{} 1 \Rightarrow z(z^2\plus{}2004)^2 \equal{} 1 \Rightarrow P(x) \equal{} z(z^2\plus{}2004)^2 \minus{} 1$

So $ \boxed{\frac{P(1)}{P(\minus{}1)} \equal{} \minus{}\frac{2005^2 \minus{} 1}{2005^2\plus{}1}}$
[/hide]

[hide="Another way."]
=====
Let $ f(z) \equal{} z^5 \plus{} 2004z \minus{} 1$ and let its roots be $ r_1,...,r_5$. Then $ f(z) \equal{} (z \minus{} r_1)\dots (z \minus{} r_5)$ and $ P(z) \equal{} a(z \minus{} r_1^2)\dots (z \minus{} r_5^2) \equal{} a(\sqrt {z} \minus{} r_1)(\sqrt {z} \plus{} r_1)\dots (\sqrt {z} \minus{} r_5)(\sqrt {z} \plus{} r_5) \equal{} af(\sqrt {z})\cdot ( \minus{} 1)^5f( \minus{} \sqrt {z}) \equal{} \minus{} af(\sqrt {z})f( \minus{} \sqrt {z}),\ a\in \mathbb{R}$

Thus:

$ \frac {P(1)}{P( \minus{} 1)} \equal{} \frac { \minus{} af(1)f( \minus{} 1)}{ \minus{} af(i)f( \minus{} i)} \equal{} \frac {2004\cdot \minus{} 2006}{(2005i \minus{} 1)( \minus{} 2005i \minus{} 1)}$
$ \equal{} \minus{} \frac {(2005 \minus{} 1)\cdot (2005 \plus{} 1)}{2005^2 \plus{} 1} \equal{} \boxed{ \minus{} \frac {2005^2 \minus{} 1}{2005^2 \plus{} 1}}$
=====
[/hide]

[quote="aidan"]No doubt then that the answer is correct.. but do we need to verify in the beginning that all roots of $ f(z) \equal{} z^5 \plus{} 2004z \minus{} 1$ are distinct?[/quote]

Why? Which step in our proofs are invalidated in that case?

I also don't think any steps are invalidated.. I was wondering because the official solution begins like this:

Let $ z_1, \ldots, z_5$ be the roots of $ Q(z) \equal{} z^5 \plus{} 2004z \minus{} 1$. We can check these are distinct (by using the fact that there's one in a small neighbourhood (what does this mean?) of each root of $ z^5 \plus{} 2004z$, or by noting that $ Q(z)$ is relatively prime to its derivative (what does this tell us?)). And certainly none of the roots of $ Q$ is the negative of another, since $ z^5 \plus{} 2004z \equal{} 1$ implies that $ (\minus{}z)^5 \plus{} 2004(\minus{}z) \equal{} \minus{}1$, so their squares are distinct as well.

Only after this "introduction" do they proceed with solving the problem.. haha.

[quote="aidan"]I also don't think any steps are invalidated.. I was wondering because the official solution begins like this:

Let $ z_1, \ldots, z_5$ be the roots of $ Q(z) \equal{} z^5 \plus{} 2004z \minus{} 1$. We can check these are distinct (by using the fact that there's one in a small neighbourhood (what does this mean?) of each root of $ z^5 \plus{} 2004z$, or by noting that $ Q(z)$ is relatively prime to its derivative (what does this tell us?)). And certainly none of the roots of $ Q$ is the negative of another, since $ z^5 \plus{} 2004z \equal{} 1$ implies that $ ( \minus{} z)^5 \plus{} 2004( \minus{} z) \equal{} \minus{} 1$, so their squares are distinct as well.

Only after this "introduction" do they proceed with solving the problem.. haha.[/quote]

If $ \mathrm{gcd} ( Q(z), Q'(z) ) \equal{} 1$ then the two equations have no roots in common, which means Q(z) cannot have a repeated root, otherwise it would be a root of Q(z) and Q'(z)


What is       lim            (1 + 2/n)^n.
                 n ->  :inf:

This problem was on my precalculus test recently and there was one problem we were allowed to skip, and everybody skipped this problem, so they had to tackle this long and tedious physics problem instead.  Well...one person got this right :-D.

Hint: [hide]sub in n = 2m[/hide]

Solution:

[hide]limn -> :inf:(1 + 2/n)n

= limm -> :inf:(1 + 1/m)2m

= e2.[/hide]

Good job.  7/7.  That's what I did also.

Because you had to substitute 2m for n everywhere, including the exponent.


geometry
incenter
trigonometry
geometric transformation
homothety

The incircle of triangle $ABC$ touched side $BC$ at $D$. Let the midpoint of $BC$ be $M$. Show that $MI$ bisects $AD$ where $I$ is the incentre of triangle $ABC$.

Consider ABDC as a degenerate tangential quadrilateral with diagonals AD, BC and incircle (I). Newton line of any quadrilateral ABDC connecting midpoints M, N of its diagonals BC, AD is the locus of points P such that area sums $S_{\triangle PAB}+S_{\triangle PDC}= S_{\triangle PBD}+S_{\triangle PCA}$ are equal. If ABDC is tangential with incircle $(I, r),$ then $S_{\triangle IAB}+S_{\triangle IDC}= \frac{r}{2}(AB+DC) = \frac{r}{2}(BD+CA) = S_{\triangle IBD}+S_{\triangle ICA},$ hence $I \in MN.$

[color=darkblue][b]Proof 1.[/b] Suppose w.l.o.g. that $b>c\ .$ Denote the intersections $S\in BC\cap AI$, $N\in AD\cap MI\ .$
Show easily that $MD=\frac{b-c}{2}$, $MS=\frac{a(b-c)}{2(b+c)}$, $SD=\frac{(b-c)(p-a)}{b+c}$ and $\frac{IA}{IS}=\frac{b+c}{a}\ .$
Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{MIN}$ for the triangle $ADS$ :
$\frac{MS}{MD}\cdot \frac{ND}{NA}\cdot\frac{IA}{IS}=1$ $\Longrightarrow$ $\frac{a(b-c)}{2(b+c)}\cdot\frac{2}{b-c}\cdot\frac{ND}{NA}\cdot\frac{b+c}{a}=1$ $\Longrightarrow$ $\boxed{\ NA=ND\ }\ .$
Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{AIS}$ for the triangle $NDM$ :
$\frac{AN}{AD}\cdot \frac{SD}{SM}\cdot\frac{IM}{IN}=1$ $\Longrightarrow$ $\frac{1}{2}\cdot\frac{(b-c)(p-a)}{b+c}\cdot\frac{2(b+c)}{a(b-c)}\cdot\frac{IM}{IN}=1$ $\Longrightarrow$ $\boxed{\ \frac{IN}{IM}=\frac{p-a}{a}\ }\ .$

[b]Remark.[/b] Denote the projection $P$ of the vertex $A$ to the opposite sideline $BC$ and the intersection $R\in AP\cap MI\ .$
Prove easily that $NR=NI$ and $AR=ID$, i.e. $AR=r\ .$
[u]Example.[/u] The orthocenter $H\in MI$ $\Longleftrightarrow$ $H\equiv R$ $\Longleftrightarrow$ $AH=ID$ $\Longleftrightarrow$ $2R|\cos A|=r$ $\Longleftrightarrow$ $|\cos A|=\frac{r}{2R}$ a.s.o.

[b]Lemma (well-known).[/b] Given are two concurrent (in the point $A$) fixed lines $d_{1}$, $d_{2}$ and four fixed points $\{A,B\}\subset d_{1}$, $\{C,D\}\subset d_{2}\ .$
Then the geometrical locus of the point $L$  for which $[LAB]=[LCD]$ is a parallelogram.
[b]Particular case.[/b] Given is a quadrilateral $ABCD$ which is  circumscribed to the circle $w=C(I,r)$ Denote the the middlepoints $M,N$ of the diagonals $AC$, $BD$.
Then $I\in MN$ (the [u]Newton's line[/u]). Indeed, $[IAB]+[ICD]=[MAB]+[MCD]=[NAB]+[NCD]=\frac{1}{2}\cdot[ABCD]\ .$

[b]Proof 2.[/b] The nice above [u]Yetti's demonstration[/u] using the previous lemma which was applied to
a degenerate circumscribed (to the circle $w$) quadrilateral $ABDC$,where the its diagonals are $BC$ and $AD\ .$ [/color]

Let $w=(I,r),w'=(I',r')$ be respectively the incircle and the A-excircle of $\triangle ABC$

Let $DE$ be a diameter of $w$.
The circle $w'$ touches the side $BC$ at $E'$. It's known that the point $M$ is the midpoint of $DE'$

$A$ is the homothety center of the circles $w,w'$.
The directed segments $IE,I'E'$ have the same direction, so the points $A,E,E'$ are collinear.

$I$ is the midpoint of $DE$
$M$ is the midpoint of $DE'$

So $MI \parallel AE' \Rightarrow  MI$ bisects $AD$


[quote="kunny"]Prove that $\sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq \log_{10}\sqrt{xy}\leq \log_{10}(x+y)-\log_{10}2$ for $1\leq x,\ 1\leq y.$

Then for $1\leq x,\ 1\leq y,\ x+y\leq 20,$ find the maximim value of $\left(\log_{10}xy+\log_{10}\frac{x}{y}\right)\left(\log_{10}xy-\log_{10}\frac{x}{y}\right).$[/quote]
[hide]
$\sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq \log_{10}\sqrt{xy}\leq \log_{10}(x+y)-\log_{10}2$
$\iff \sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq \log_{10}\sqrt{xy}\leq \log_{10}(\frac{x+y}{2})$
Then the right inequality is just AM-GM after getting rid of the logs.
Now on to the left inequality
$\sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq \log_{10}\sqrt{xy}$
$\iff \sqrt{\log_{10}x\log_{10}y}\leq \frac{\log_{10}xy}{2}$
$\iff \sqrt{\log_{10}x\log_{10}y}\leq \frac{\log_{10}x+\log_{10}y}{2}$
which is, again, AM-GM

Now the second part:
$\left(\log_{10}xy+\log_{10}\frac{x}{y}\right)\left(\log_{10}xy-\log_{10}\frac{x}{y}\right).$
$=(\log x^{2})(\log y^{2})$
$=4\log x\log y$
So we only need to maximize $\log x\log y.$
Using the inequality proven above,
$\sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq \log_{10}\sqrt{xy}\leq \log_{10}(\frac{x+y}{2})$
The right most part is maximized when $x+y=20,$ so substituting that in, we get
$\sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq \log_{10}\sqrt{xy}\leq 1$
The middle part is of course maximized when $x=y,$ since their sum is fixed. Then $x=y=10,$ and this gives $\sqrt{\log_{10}x}\sqrt{\log_{10}y}=1$
$\log x\log y=1$
This is certainly obtainable when $x=y=10,$ and nothing greater than this value can be obtained, because
$\sqrt{\log_{10}x}\sqrt{\log_{10}y}\leq\log_{10}(\frac{x+y}{2})\leq1.$
So the maximum of
$\left(\log_{10}xy+\log_{10}\frac{x}{y}\right)\left(\log_{10}xy-\log_{10}\frac{x}{y}\right).$
$=4\log x\log y$
is $4\cdot1=\boxed4.$

[/hide]


The natural numbers from 1 to 100 are arranged on a circle with the characteristic that each number is either larger as their two neighbours or smaller than their two neighbours. A pair of neighbouring numbers is called "good", if you cancel such a pair, the above property remains still valid. What is the smallest possible number of good pairs?

Consider five consecutive numbers. There are ways to arrange them:
1) $a_1>a_2<a_3>a_4<a_5$. Then if $a_2>a_4$, then pair $(a_2,a_3)$ is good;
2) $a_1<a_2>a_3<a_4>a_5$. Then if $a_2>a_4$, then pair $(a_3,a_4)$ is good.
It follows that each number is in at least one good pair. So there are at least 50 good pairs.
I am sure that it is far from right answer. It is just first step for so long unsolved problem :)

Hmm... wont' there be at MOST 50 good pairs, since at most 50 pairs are available at all?

???
There are 100 available pairs.
I think, the answer is something like 51 :?

[quote="Myth"]I think, the answer is something like 51 :?[/quote]
Of course, it's 51, Myth!

In the previous post Myth proved that it $\geq 50$. At first we will show that it >50. It means that we are going to prove that it can't be precisely 50.

Suppose there are exactly 50 good pairs. Without loss of generality we may assume that $(a_1,a_2)$, $(a_3,a_4)$, ..., $(a_{99},a_{100})$ are good pairs and $a_1<a_2>a_3<\ldots>a_{99}<a_{100}>a_1$. Since pair $(a_2,a_3)$ is "bad" then $a_1>a_4$. Since pair $(a_4,a_5)$ is bad then $a_3>a_6$. And so on. But then we will have $a_1>a_4>a_3>a_6>a_5>a_8>a_7>\ldots>a_{97}>a_{100}>a_{99}>a_2>a_1$. It is a contradiction, I think.

And now we are gonna give an example of arrangment with exactly 51 good pairs. Here it is:

$1, 3, 2, 5, 4, 7, 6, 9, 8, \ldots, 97, 96, 99, 98, 100$.

It is easy to see that only pairs $(3,2)$, $(5,4)$, ..., $(97, 96)$, $(99, 98)$, $(98, 100)$ and $(1,3)$ are good. The rest pairs are bad.

I think this so long unsolved problem is solved now!


algebra
polynomial
linear algebra
matrix

Let $A=\left(\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ a & b \end{array}\right).$ Determine the value of $a,b$ such that $A^3=A.$

The characteristic polynomial of A is

$p(\lambda) = (\lambda - \textstyle{\frac{1}{2}})(\lambda - b) - \textstyle{\frac{a}{2}} = \lambda^2 - p\lambda + q,$

where

$(1) \; p = b + \textstyle{\frac{1}{2}}$
$(2) \; q = \textstyle{\frac{b - a}{2}}$.

According to Cayley-Hamiltons theorem

$p(A) = A^2 - pA + qI = \bf{0},$

i.e.

$A^2 = pA - qI.$

Multiplying by A, the result is

$A^3 = pA^2 + qA = p(pA - qI) - qA = (p^2 - q)A - pqI.$

The condition $A^3 = A$ gives

$(p^2 - q - 1)A = pqI,$

which is equivalent to

$(p^2 - q - 1) \left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ a & b \end{array} \right) = pq \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$.

This matrix equation is satisfied if and only if $p^2 - q - 1 = pq = 0$. Hence $q = p^2 - 1$ and $p(p^2 - 1) = 0$ giving

$(3) \; (p,q) = (-1,0), \: (0,-1), \: (1,0).$

Futhermore

${(4) \; b = p - \textstyle{\frac{1}{2}}}$
$(5) \;a = p - 2q - \textstyle{\frac{1}{2}}$

by (1) and (2).  Inserting the three solutions (3) in the formulas (4) and (5), we get the three solutions of this matrix problem, namely

$(a,b) = (\textstyle{-\frac{3}{2}}, \textstyle{-\frac{3}{2}}), \: (\textstyle{\frac{3}{2}}, \textstyle{-\frac{1}{2}}), \: (\textstyle{\frac{1}{2}}, \textstyle{\frac{1}{2}}).$

Your answer is correct. We can classify whether there existed $A^{-1}$ or not.


"Fboxing" works when you are working with simple numbers like $\fbox{1},\fbox{1865},$and
$\fbox{85739514}$

But when I try to "Fbox" fractions, it says dangerous text like here:

\fbox{\frac{2}{3}}

Any tips? Or other "boxing" codes for fractions/squareroots?

\boxed{\frac{1}{2}} gives:
$\boxed{\frac{1}{2}}$

I hope that's what you want. :)

Awesome, thanks a lot!

$\boxed{\frac{1}{2}}$
$\boxed{\sqrt{3}}$
$\boxed{3}$
$\fbox{3}$

Looks good to me!


SOHO have discovered a Comet that orbits the sun.

http://www.astronomy.com/asy/default.aspx?c=a&id=6046

Nice article find hunter34.  :)
To clarify, all comets we find have some orbit around the sun but, not that many of found to be in periodic orbits, as the article states.


topology
search
real analysis
real analysis unsolved

Are $ \mathbb{Q}$ and $ \mathbb{Q}^2$ homeomorphic ?

All countable metric spaces with no isolated points are homeomorphic. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1126232041&t=48347]this[/url] (don't let the four long posts scare you; my "proof" is much more complicated than it needed to be :)).


function
floor function
modular arithmetic
number theory unsolved
number theory

posted here [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=474994#p474994]APMO 2004[/url]


Find all three digit number $ abc$ where $ a\plus{}b\plus{}c\equal{}a(b\plus{}\frac{c}{10})$ where $ a,b$ and $ c$ are natural numbers.

[hide="Solution"]

$ 10\mid ac$

1) a=5
$ 5\plus{}b\plus{}c\equal{}5b\plus{}\frac c2$
$ 5\plus{}\frac c2\equal{}4b$, all solutions for (b,c) are (2,6)
526 solution
2) c=0
a+b=ab, then (a-1)(b-1)=1, so a=b=2
220 solution
3) c=5
$ a\plus{}b\plus{}5\equal{}ab\plus{}\frac a2$
$ \frac a2\plus{}b\plus{}5\equal{}ab$
Let a=2k
$ b\equal{}\frac{k\plus{}5}{2k\minus{}1}$ must be integer, so k=1, and a=2, and b=6
265 solution

All solutions: 220,265,526
[/hide]


I can't even find the 2005 school round on mathcounts.org though, so are they gonna post the chapter exam on the site?

[quote="rcv"]If you go to the "Students" section, you'll find a link to last year's Chapter and State competitions.  [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47](Click here.)[/url]

To the best of my knowledge, MATHCOUNTS has never posted the School or National Competitions.[/quote]

Mathcounts is selling the '04 National exam for like $30.

Would you buy it for $30? I'd rather print out the handbook problems, even though they're not as hard.

There are a few old tests around in (illegal) circulation.  Your school or another school nearby might have some.  If anyone asks, I never knew that :D.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

$f:M_4(C) \rightarrow M_4(C)$  defined by

For any $A\in M_4(C)$

A=
( a b c d )
( e f  g h)
(  i j  k l)
( m n o p)

f(A)=

( e a g c)
( f  b h d)
( m i o k)
( n  j p l)

Prove that $f$ is diagonalisable, find the eigenvalues eigenvectors

This problem was trivial. I saw the answer as soon as I saw the problem, and just posted immediately. It didn't really need proof, either; the answer is enough.

$f^{4}(A)=A$

 $f^4-Id_{M_4(C)}=0_{M_4(C)}$

$x^4-1=0$ and $x^4-1=(x-1)(x+1)(x+i)(x-i)$ split with distincts roots

$f$ is diagonalisable

The characteristic polynomial of $f$ is
$\chi_{f}(x)=(x^4-1)^4$ so we can deduce that

$det(f)=1$
$tr(f)=0$
$tr(Com(f))=0$ where $Com(f)$ is the comatrix


calculus
integration
geometry
inequalities
algebra proposed
algebra

Several clocks lie on the table. It is known that at some moment the sum of distances between a point $X$ of the table and the ends of their minute hands is not equal to the sum of distances between $X$ and the ends of their hour hands. Prove that there is a moment when the sum of distances between $X$ and the ends of their minute hands is greater than the sum of distances between $X$ and the ends of their hour hands.

The official solution is very nice, but I wonder if there is a more elementary one.

Let's do it for one clock in a way which will obviously be extendable to more clocks, because it's easier to describe if we don't always add "the sum of the distances" :).

Let $f(t)$ be the distance from $X$ to the tip of the hour hand at time $t$. Let $[0,a]$ be an interval of $12$ hours. The integral from $0$ to $a$ of the difference between the distance from $X$ to the tip of the minute hand and the distance to the hour hand is $\int_0^a [f(12t)-f(t)]dt\ (*)$ (because the minute hand moves twelve times faster). However, $\int^a_0 f(12t)dt=\frac 1{12}\cdot\int^{12a}_0 f(t)dt=\int^a_0f(t)dt$, so the expression $(*)$ equals $0$. Since the quantity we have integrated is not always $0$, it means that there is a $t$ when it's positive (what we want).

Is it Ok?

I did. I really don't think they're supposed to have different lengths in this problem. It would be ridiculous :).

It was absolutely unclear for me. I even couldn't imagine that they have equal lengths.  :?

Grobber's solution is OK, that's the one I have, but I can't believe there is not some which doesn't depend on definite integrals.

Yes, hours and minutes hands are assumed equal, I forgot to mention that because I just retyped the original statement of the problem (which says nothing about that but this fact is used in the solution, so that's probably an oversight).

[quote="_tequilla"]it is from this book "Plane geometry" Victor Prasolov
but i don't remeber solution, maybe tomorrow i will find it[/quote]
You wrong. There is another problem in that book.

[quote="pbornsztein"]Didn't we look at it for WOP-02? [/quote]

What is WOP 2002?

[quote="pbornsztein"]Didn't we look at it for WOP-02?
Pierre.[/quote]
But the solution from WOP is somewhat extra short and doesn't make a sense.

[quote="fedja"][quote="Myth"]Do you assume that hours and minutes hands are equal?[/quote]
It is actually not necessary. What is really needed is that on each clock the hour hand is not longer than the minute hand. All one needs to do then is to check that for each such "normal" clock the average distance from a given point to the end of its minute hand is not smaller than the average distance to the end of the hour hand. Actually, for the clocks whose hour hands are strictly shorter than their minute hands, the inequality will be strict and the additional assumption about the existence of the moment when the sums of the distances are not the same won't be necessary...[/quote]
It was known for me, of course ;) . That's why the problem was so strange for me.


[quote="orl"]
A sequence a_1, a_2, a_3 is defined by

(1) a_1 = 1 and a_2 = 2

(2) a_{n+2} = 5*a_{n+1} - 3*a_{n} if a_{n}*a_{n+1} is even
a_{n+2} = a_{n+1}- a_{n} if a_{n}*a_{n+1} is odd

Prove:

a.) The sequence generates infinitely many positive and negative elements.
b.) No element of the sequence is equal to zero.
c.) If n = 2^k - 1 (k  {2,3,4,...}) then a_{n} is divisible by 7.[/quote]
=================
I suggest the following proof (in brief)

First we note that in modulo 2, the relation is the same: a(n+1) = a(n) + a(n-1) and the sequence is:
1 0 1 1 0 1 1 0 1 1 0 ...
So, if we put  u(n) = a(3n+1), v(n) = a(3n+2) and w(n) = a(3n+3) then
u(n+1) = 5w(n) - 3v(n)
v(n+1) = u(n+1) - w(n)
w(n+1) = 5v(n+1) - 3u(n+1)
From that we can get that u(n), v(n), w(n) all of them fullfils the relation
X(n+1) = 2X(n-1) - 9X(n)  (1)

Now, the problem is nearly obviuos. Note that for solution we must'nt solve the recurence (1).

Namdung


geometry
rectangle
perimeter
combinatorics unsolved
combinatorics

There is a 7x7 square board divided into 49 unit cells, and tiles of three types: 3x1 rectangles,3-unit square corners, and unit square. Tery has infinitely many rectangles and one corners, while Tom has only one square.
(a) Prove that Tom can put his square somewhere on the board (convering exactly one unit cell) in such a way that Jerry cannot tile the rest of the board with his tiles.
(b)Now Jerry is given another corner. Prove that no matter where Tom puts his square (convering exactly one unit cell), Jerry can tile the rest of the board with his tiles.

I'll try. Hope there's no mistake.

a)If we colour the sqare board with numbers from { 0, 1, 2 } giving to square ( i, j ) the colour i+j  (mod 3) [ where 0<=i, j<=6 and we start to count from the top left corner ] we obtain 16 1-coloured cells, 16 2-coloured cells and 17 0-coloured cells. If Tom places his unit square on a k-c. (means k-coloured) cell where k=+-1 (mod 3) we are let with 17 0-c. cells, 15 k-c. cells and 16 (-k)-c. ones. Since any rectangle covers exactly 1 cell of each type, Jerry is forced to use his corner to make the number of uncovered cells of each type equal by covering with it 2 0-cs. and one (-k)-c. If his corner must cover 2 0-c.s it is easyly seen that must be placed pointing perpendiculary to the diagonal of cells which satisfy i+j=6.
Now consider the "symmetric" colouring given by C(i, j)= i-j (mod 3). It has exactly the same properties as the previous one, except for the fact that in the last sentence the diagonal to consider is no longer i+j=6, but i-j=0. If Tom places his unit square on a cell that has colour different from 0 in both colourings ( for instance, (1, 0) ), Jerry's corner, for him to be able to cover the 7*7 square, should be placed pointing in 2 different directions at the same time, which is impossible.

b) Consider all the 4 4*4 squares contained in the main square and in the "corners" of it. Their union is the main square, so the cell where tom places his sqare must be contained in one of them. If we divide such square into four 2*2 smaller squares, it is possible for Jerry to cover with his corner the leftover of the one where Tom's square was placed. Consider one of the other 2*2 squares, adyacent to this one. If Jerry places his second corner inside it in a way that the remaining cell is adyacent to one of the last 2 2*2 squares and has a side in common with the perimeter of the 4*4 square, it is easy to check that he can then cover the rest of the 4*4 square with 1*3 rectangles. What remains of the main square can be divided into a 3*7 and a 3*4 rectangles, which can obviously be tiled with Jerry's rectangles.

Sorry for my English.


Prove that for any positive integer n, the number [(n - 1)!/(n² + n)] is even.

well isn't this just like my old little problem I proposed in the MathLinks Contest 2nd Edition, round 6, problem 1? :) well it certainly has the same idea :P

yeah, but once you've seen that problem, it really comes to you in a jippy :D

he idea is to separate 2 cases:

1: when both n and n+1 are composed numbers.
2: when one of n and n+1 is a prime, and in this case use wilson theorem!


There are 250 hobbits in Shir. Each hobbit lives in his own house. Every evening some hobbits pay a visit to all hobbits which sit in their houses this evening.
a) How many evenings do they need in order each hobbit pays a visit to each other hobbit?
b) How many evenings do they need in order for each two hobbits $A$ and $B$ either $A$ pays a visit to $B$ OR $B$ pays a visit to $A$?

:what?:
I think the statement is unclear for you. Sorry for my english.
Each evening many hobbits may pay visits. For example, 125 hobits pay visits to all 125 remaining hobbits. So totally we will have $125^2$ visits this evening.

a) If the hobbits are $1,2,...,250$ and the evenings are $1,2,...,n$, let us assign to each hobbit the set $A_i$ of nights when he goes out. The condition that hobbit $i$ visits $j$ sometime is that $A_i$ is not included in $A_j$. $n=10$ is enough, because if each $A_i$ were a different 5-element subset of $1,2,...,10$ these $A_i$'s satisfy (amazingly, there are 252 such subsets). $n=9$ is not enough, as it is well-known that the maximum number of subsets of $1,2,...,9$ which do not contain each other is $\dsp {9 \choose 4}=126$, so not enough.

b)I didn't understand this very clearly, but I will suppose that the statement is "either A pays at least a visit to B or B pays at least a visit to A". Then the fact that for every $i$ and $j$ one of them visits the other, but not the other way around, means that for every two $A_i$ and $A_j$ one is strictly contained in the other. So the $A_i$'s are going to form a sequence of subsets of $1,2,...,n$ such that every one of them is contained in the next. It is clear that we need $n\geq 249$, and it's easy to see how to make it work for $249$.

Your solution for a) is correct.

Unfortunately, your interpretation for b) is untrue. For example, for 3 hobbits we can have $A$ pays a visit to $B$, $B$ pays a visit to $C$ and $C$ pays a visit to $A$, and no other visits.

Maybe I misunderstand the problem, but I think RHS is right.
If A pays a visit to C, then on that day if B sits in his home, A does pay a visit to B. Otherwise, B pays a visit to C on that day. So there cannot be a cycle.

Indeed, you are right. It was just silly example. I tried to explain the condition to RHS.

I too believe that RHS is right.

Suppose that for some $i,\ j$ we have $|A_i|=|A_j|$. But since $A_i\subset A_j$ or $A_j\subset A_i$ we find out that $A_i=A_j$. On the other hand, this means $A_i\subset A_j$ and $A_j\subset A_i$ which is a contradiction.

Now we may find strictly increasing sequence of numbers $\{|A_1|,|A_2|,...,|A_{250}|\}$ and the rest is obvious.

Oh, it seems you think that OR in the condition is XOR. It is not true. It is possible that $A$ pays a visit to $B$ and $B$ pays a visit to $A$.

Then by similar argument the answer is 8=the least integer $\geq log_2 250$, isn't it?


Let  ABC a  triangle  and x,y,z  are  reals  with  x+y+z=0.Show  that
a^2*x*y+b^2*y*z+c^2*z*x<=0
10  th  tournament of towns , autumn  1988
thank  you!

I remember to have already solved this one. Maybe I'm getting old, so I will post again that solution:
   There are two numbers between x,y,z that have the same sign. Let them be x and y. The inequality becomes (x+y)(b^2y+c^2x)>=a^2xy. Of course, we can suppose that x,y>=0. Apply Cauchy and find that
 (x+y)(b^2y+c^2x)>=xy(b+c)^2>=xya^2.


trigonometry
geometry
circumcircle
inequalities
inequalities proposed

Prove that:
   $ 1\plus{}cosAcosBcosC\geq\sqrt{3}sinAsinBsinC$ :)

It is well known (or easily provable using the Cosine Law and Hero's formula for the triangle area) that $ 1\plus{}\cos A\cos B\cos C\equal{}\frac{a^2\plus{}b^2\plus{}c^2}{8R^2}$, where $ R$ is the circumradius and $ a,b,c$ are the sidelengths, while $ \sin A\sin B\sin C\equal{}\frac{abc}{8R^3}\equal{}\frac{S}{2R^2}$, where $ S$ is the area.  the inequality is then equivalent to
\[ a^2\plus{}b^2\plus{}c^2\geq4\sqrt{3}S.\]
Squaring and using Hero's formula again, we find this inequality to be equivalent to
\[ a^4\plus{}b^4\plus{}c^4\plus{}2a^2b^2\plus{}2b^2c^2\plus{}2c^2a^2\equal{}3(16S^2)\equal{}6a^2b^2\plus{}6b^2c^2\plus{}6c^2a^2\minus{}3a^4\minus{}3b^4\minus{}3c^4,\]
or
\[ a^4\plus{}b^4\plus{}c^4\geq a^2b^2\plus{}b^2c^2\plus{}c^2a^2,\]
trivially true by Cauchy-Schwarz, with equality iff $ a\equal{}b\equal{}c$.


I think this is too hard for Classroom Math.

[quote="Elio (n)"]Find all the solutions pourque $ 2*(a/a\minus{}2)$ produce entire number[/quote]

Does "entire number" mean integer?

and it'll get moved eventually, no need to repost.

[quote="Elio (n)"]Find all the solutions pourque $ 2*(a/a\minus{}2)$ produce entire number[/quote]
[hide]$ a\minus{}2\mid2a\,\forall\,a\equiv0\bmod2,\,a\ge\minus{}2$

Also, $ a \equal{}\{1,3\}$.[/hide]

[quote="Elio (n)"]$ a(a/a\minus{}2) \equal{} 2a/a\minus{}2 \equal{} (2/a\minus{}2)a$

$ a \equal{}{0,1,3,4}$[/quote]
Ok my solution is incorrect, but what about $ a\equal{}6$?

[quote="i_like_pie"][quote="Elio (n)"]$ a(a/a\minus{}2) \equal{} 2a/a\minus{}2 \equal{} (2/a\minus{}2)a$

$ a \equal{}{0,1,3,4}$[/quote]
Ok my solution is incorrect, but what about $ a \equal{} 6$?[/quote]

[hide="Here is a better solution"]$ 2\left(\frac{a}{a\minus{}2}\right) \equal{} 2\plus{}\frac{4}{a\minus{}2}$.

So $ a\minus{}2$ is a factor of $ 4$. So:

$ a\minus{}2 \equal{}\pm 1,\pm 2,\pm 4$

So, the set of all solutions for $ a$ is:

$ \{\minus{}2, 0, 1, 3, 4, 6\}$[/hide]

vishalarul, nicely done!

This type of idea happens very often in problem. Let's take this problem one step further.

What are all numbers $ x$ such that $ \frac{x}{2}\cdot\frac{x\plus{}2}{3x}$ is a positive integer?

I'm not going to solve this but remember the idea of multiplying and dividing fractions.

Silverfalcon, I might be doing something wrong since it seems so easy...
[hide]
${ \frac{x}{2}}\cdot\frac{x+2}{3x}=\frac{x+2}{6}$

$ x+2$ has to be divisible by 6, so $ x=6k+4$ for any integer $ k$[/hide]

Find all $ x$ such that $ (6x)|(x\plus{}1)(x\plus{}2)$.

[hide]$ 2|(x\plus{}1)(x\plus{}2)$ for all integral $ x$ so we just need $ 3|(x\plus{}1)(x\plus{}2)$ and $ x|(x\plus{}1)(x\plus{}2)$. Clearly, the first is true if and only if $ x\equal{}\pm1\pmod{3}$. Now $ x|(x\plus{}1)(x\plus{}2)\Rightarrow x|(x^{2}\plus{}3x\plus{}2)\Rightarrow x|2$, so $ x\equal{}\pm1,\pm2$, which satisfies the first condition.[/hide]

Yeah sorry about that.. It wasn't supposed to be that easy.

I meant to do something like $ \frac{x^{5}\plus{}4x^{2}\plus{}7}{7x^{3}}$ is an integer.  But you got the idea though.

[hide="solution"]We check 0-6 mod 7, and see that only 0 works. So $ x\equal{}7y$

$ \frac{7^{5}y^{5}\plus{}196y^{2}\plus{}7}{7^{4}y^{3}}$

We factor out a 7 and get

$ \frac{7^{4}y^{5}\plus{}28y^{2}\plus{}1}{7^{3}y^{3}}$

But the numerator does not have a factor of 7 in it. Therefore, for all integers x non-zero, $ \frac{x^{5}\plus{}4x^{2}\plus{}7}{7x^{3}}$ is not an integer.[/hide]


[quote="DPopov"]$\triangle ROT$ is a right isosceles triangle with a right angle at $O$. $MN$, $AC$, and $RT$ are parallel. Prove or disprove the following \[AB^{2}+AC^{2}\geq BM^{2}+MN^{2}+NP^{2}+PC^{2}\] [/quote]What is $P\ ??$


algebra
polynomial
modular arithmetic
number theory proposed
number theory

Polynomial $ f(x)\equal{}x^4\minus{}x^3\plus{}2x^2\plus{}1$
Sequence $ (a_n): a_1\equal{}1, a_{n\plus{}1}\equal{}f(a_n)$
Show that there are infinitely many primes $ p$ that does not divide every terms $ a_n$ of the sequence.

Notice that
\[ f(x) \minus{} x \equal{} (x^4 \plus{} 2x^2 \plus{} 1) \minus{} x(x^2 \plus{} 1) \equal{} (x^2 \plus{} 1)(x^2 \minus{} x \plus{} 1)\]
Thus, if $ p$ is a prime number such that $ p | a_n^2 \plus{} 1$ then $ (a_n,p) \equal{} 1$ and
\[ a_m \equal{} f(f(f...(a_n)...)) \equiv a_n \pmod{p}, \forall m \geq n\]
Therefore $ p \nmid a_m, \forall m \geq n$. If we can choose $ n$ and $ p$ such that $ p$ does not divide any term $ a_k$ for $ k < n$ then $ p$ divides no term of $ \{a_n\}$ as we want. So, we will first prove that there are infinitely many prime number $ p$ such that $ p | a_n^2 \plus{} 1$ for some $ n$. Indeed, we have
\[ (a_{n \plus{} 1}^2 \plus{} 1) \minus{} (a_{n}^2 \plus{} 1) \equal{} (a_{n \plus{} 1} \minus{} a_n)(a_{n \plus{} 1} \plus{} a_n) \equal{} (a_n^2 \plus{} 1)(a_n^2 \minus{} a_n \plus{} 1)(a_{n \plus{} 1} \plus{} a_n) \\
\Rightarrow \dfrac{a_{n \plus{} 1}^2 \plus{} 1}{a_n^2 \plus{} 1} \equal{} 1 \plus{} (a_{n \plus{} 1} \plus{} a_n)(a_n^2 \minus{} a_n \plus{} 1) \equiv 2 \minus{} a_n a_{n \plus{} 1} \pmod{a_n^2 \plus{} 1}\]
We also have
\[ 2 \minus{} a_n a_{n \plus{} 1} \equal{} 2 \minus{} a_n(a_n^4 \minus{} a_n^3 \plus{} 2a_n^2 \plus{} 1) \equiv 3 \pmod{a_n^2 \plus{} 1}\]
So $ a_n^2 \plus{} 1 | a_{n \plus{} 1}^2 \plus{} 1$ and $ \gcd\left(\dfrac{a_{n \plus{} 1}^2 \plus{} 1}{a_n^2 \plus{} 1}, a_n^2 \plus{} 1 \right) \equal{} 1$. On the other hand, since $ \{a_{n}^2 \plus{} 1\}$ is strictly increasing, there exists a prime number $ p$ such that $ p | a_{n \plus{} 1}^2 \plus{} 1$ but $ p \nmid a_n^2 \plus{} 1$ and so $ p$ does not divide any $ a_k^2 \plus{} 1$ for $ k < n$. The claim is proved $ (1)$.
Now we prove that the prime number $ p$ chosen like above does not divide any term $ a_k$ for $ k < n$. Indeed, suppose the contrary. Let $ k$ be the smallest index such that $ p | a_k$ then $ a_{k \plus{} 1} \equal{} f(a_k) \equiv 1 \pmod{p}$. Thus the sequence $ \{a_n \pmod{p}\}$ is periodical. Therefore $ a_{n \plus{} 1} \equiv a_i \pmod{p}$ for an index $ i \leq k < n$. This leads to $ p \nmid a_{n \plus{} 1}^2 \plus{} 1$ due to $ (1)$. The conclusion is then followed.


induction
combinatorics unsolved
combinatorics

Some pairs of towns in a country are joined by roads, so that there is a unique route from any town to another which does not pass through any town twice. Exactly 100 of the towns have only one road. Show that it is possible to construct 50 new roads so that there will still be a route between any two towns even if any one of the roads (old or new) is closed for maintenance.

Hi, I was thinking at this problem last year, which in terms of graphs can be formulated:

Let G be a tree (it is equivalent to have a unique simple path between each pair of nodes). How many edges do you have to add in order to obtain a strongly connected graph (remains connected even if you delete one edge)?

And the answer is [L+1/2], where L is the number of leafs (nodes of degree 1).

It can be proved by induction in the number of leafs, with a bit of care, but is not too difficult. If you want to think the proof by induction, ok, if you want that I post it, please tell me.

bye,
fla

Hi, nice to have a new member....
Please post full solutions to the problems.

Pierre.

Welcome fbonomo, that is a cute avatar!!! Please post your solution so we all can enjoy it  :D

[b]Lemma[/b]
There is vertex $X$ s.t. for any edge $XY$ number of leafs in subtree $Y$ is at most 50 (we consider $X$ as a root).

[i]Proof.[/i]
$\square$ Consider arbitrary vertex $V$. If $V$ is not an apropriate vetrex than there is edge $VV'$ s.t. subtree $V'$ contains at least 51 vertices. Then consider $V'$ as a new root. In this case subtree $V$ contains at most 49 vertices. If $V'$ again is not an appropriate vertex then repeat procedure for $V'$.
Finally, we necessarily will find required vertex $X$. $\blacksquare$

Let $X$ be vertex from lemma.
If $\deg X=2$ then both subtree of $X$ have 50 leafs. Just connect leafs from one subtree to leafs from second subtree.
Suppose $\deg X\geq 3$. Then it is possible to 'join' some of subtrees in such way that joined subtrees again have at most 50 leafs and number of joined subtrees is 3 (see Figure 1). Let $a$, $b$ and $c$ be numbers of leafs in corresponding trees. Then $a+b+c=100$.
Connect
$\frac{a+b-c}{2}$ leafs from $A$ with $\frac{a+b-c}{2}$ leafs from $B$;
$\frac{a-b+c}{2}$ leafs from $A$ with leafs from $C$;
$\frac{-a+b+c}{2}$ leafs from $B$ with leafs from $C$
(each leaf has exactly one connection).
It is clear that removing of arbitrary edge from $A$, $B$ and $C$ doesn't break connectivity.

----------------------

Case of odd number of edges can be done in the same way. Just one leaf we must connect to root $X$.

I forgot to say that I was unable to find solution with induction, and my attemps to do it took whole hour :( . So finally, I found this non-inductive solution :)

Ok, another to be moved  :)
But, not today to let everybody look at your solution.

Pierre.


[hide]
smallest square = 32^2= 1024
largest square   = 99^2= 9801

Since we've found the highest and lowest squares everything between it must also be squares.

99-32= 67 +1 (for "32")= [b] 68[/b]

  What fact did i discover?

[/hide]

The four-digit natural numbers are the square of a number.

[quote="4everwise"]How many four-digit natural numbers are the square of a number?[/quote]

[hide]
$\sqrt{1000} ~ 31.622776601683793319988935444327$
so 32 is the lowest number.
$\sqrt{9999} ~ 99.994999874993749609347654199058$
so 99 is the largest number.

99 - 32 + 1 = [b]68[/b]
[/hide]

I got another one that is quite simple. In a 45, 45, 90 degree triangle ABC, with AC as the hypotenuse, a line is drawn from A to the midpoint of AC (D). Show that AD = 1/2 AC

I dont understand...... First you tell us that D is the midpoint of AC and now you're asking why AD which is half of the line segment, why its half of the line segment....

Oops, sorry. AC is the hypotenuse of the 45, 45, 90 triangle. D is the midpoint of AC. Prove AD = 1/2 AC

lol isnt that the same thing.

D is the midpoint of AC, meaning that AC is now divided into two equal halves.  Therefore, AD is one of the two equal halves, and is half of AC.

Can you use the fact that it is a 45, 45 90 degree triangle and how the angle measures show it?

Connecting the mid-points of each line, and the two in bewtween genereates 4 idenctical right-angle, isosceles triangles?

[quote="4everwise"]How many four-digit natural numbers are the square of a number?[/quote]
Aren't all four-digit natural numbers the square of "a number"?  [See 4everwise's second post in this thread.]

Good Job! :10:
Draw the altitude of the right triangle from the midpoint D. This bisects the right angle, making two 45 degree angles. Angle A and Angle C were both 45 degrees before, and the altitude is always perpendicular to the line drawn.

In a 45, 45, 90 degree triangle, the 2 legs are equal. Since AC, the hypotenuse, consists of the legs of 2 isoceles right triangles and AD is one of the legs,

AD = 1/2AC


number theory
least common multiple
number theory unsolved

A is the subset of {1,2,3,4, ... , n }. For All x,y in A, LCM of x,y is at most n.
 then. Prove that |A| $ < \equal{} 1.9\sqrt {n}\ \plus{} 5$

[quote="gus135791"]A is the subset of {1,2,3,4, ... , n }. For All x,y in A, LCD of x,y is at most n.
 then. Prove that |A| $ < \equal{} 1.9\sqrt {n}\ \plus{} 5$[/quote]
I suppose you are speaking about "least common multiple" when you write "lcd"!!

Let then $ U \equal{} \{A\subseteq\{1,2,...,n\}$ such that $ \forall x,y\in A$, lcm$ (x,y)\leq n\}$
Obviously, $ U\neq\emptyset$ ($ {1}\in U$)

Let $ p \equal{} \max_{A\in U}|A|$ (the problem is to show that $ p\leq1.9\sqrt n \plus{} 5$)

Let $ B\in U$ such that $ |B| \equal{} p$. Then : if $ x\in B$, any divisor $ y$ of $ x$ is in $ B$.
This is easy to see since, else, $ B\cup\{y\}$ would be in $ U$ and would have a greater cardinal than $ p$.

So $ B$ is build with one number $ u \equal{} \max_{x\in B}x$ and all its divisors.

Let $ f(x)$ the number of divisors of $ x$  : $ f(\prod p_i^{n_i}) \equal{} \prod(n_i \plus{} 1)$. We have $ p\equal{}f(u)$ and the problem would be solved if we show that $ f(x)\leq1.9\sqrt x \plus{} 5$

$ \forall$ prime $ p\geq 5$ : $ n \plus{} 1\leq \sqrt {p^n}$ $ \forall n\geq 0$

For prime $ p \equal{} 2$ : $ n \plus{} 1\leq \frac {3}{2}\sqrt {p^n}$ $ \forall n\geq 0$

For prime $ p \equal{} 3$ : $ n \plus{} 1\leq \frac {2\sqrt 3}{3}\sqrt {p^n}$ $ \forall n\geq 0$

So, for any $ x$ : $ f(x)\leq \frac {3}{2}\frac {2\sqrt 3}{3}\sqrt x$ $ \equal{} \sqrt 3\sqrt x < 5 \plus{} 1.9\sqrt x$

Q.E.D.


modular arithmetic
number theory unsolved
number theory

hi
find all prime "p" such that (2^(p-1)-1)/p is a perfect number

$ 2^{p\minus{}1}\minus{}1\equal{}p.a^{2}$

if $ p\equal{}2$  $ \implies$  $ 2\nmid 2^{p\minus{}1}\minus{}1$ Contradiction!

$ \implies$  $ p>2$

Let $ p\equal{}2k\plus{}1$   $ \implies$  $ 2^{2k}\minus{}1\equal{}p.a^{2}$

$ (2^{k}\minus{}1)(2^{k}\plus{}1)\equal{}p.a^{2}$

$ (2^{k}\minus{}1,2^{k}\plus{}1)\equal{}1$   $ \implies$

$ i)2^{k}\minus{}1\equal{}px^{2},2^{k}\plus{}1\equal{}q^{2}$  $ \implies$  $ (q\minus{}1)(q\plus{}1)\equal{}2^{k}$

$ \implies$  $ q\minus{}1\equal{}2^{l},q\plus{}1\equal{}2^{c}$  $ \implies$  $ 2^{c}\minus{}2^{l}\equal{}2$  $ \implies$  $ c\equal{}2,l\equal{}1$

$ \implies$  $ q\equal{}3$  $ \implies$  $ k\equal{}3$  $ \implies$  $ 2^{3}\minus{}1\equal{}px^{2}$  $ \implies$  $ p\equal{}7$

$ p\equal{}7$  $ \implies$  $ \frac{2^{p\minus{}1}\minus{}1}{p}\equal{}9$

$ ii)2^{k}\minus{}1\equal{}q^{2},2^{k}\plus{}1\equal{}px^{2}$

if $ k\geq2$   $ \implies$  $ q^{2}\equiv \minus{}1\pmod{4}$ Contradiction!

$ \implies$  $ k\leq1$  if $ k\equal{}0$  $ \implies$  $ p\equal{}1$ Contradiction!

if $ k\equal{}1$  $ \implies$  $ p\equal{}3$  $ \implies$  $ \frac{2^{p\minus{}1}\minus{}1}{p}\equal{}1$ :wink:

I thought the question was to find the perfect numbers. But then it's a open problem, because we don't know if there exists an odd perfect number.

hi
I dont think t is open question
It's from iranian test in one year and it is solved.
can you explain more Bugi?


poison I think I am going to poison the main character of my story I am writing(sci-fi)...what type of poison should I use? I know it is a little weird...

Ahhh.... Try research. Try finding the wierdest poison available, like mushroom poison. Maybe even clorox mixed with a little cyanide.

[(I have a knack of mixing things!)

-- quote from the move "Aresenic and Old Lace"]

 :rotfl:

If it's scifi, you can make something up yourself (especially if you make it sound like an industrial chemical or something). Many chemicals are very poisonous, you can look on chemistry safety websites. Or you could use a natural poison -- snakes and spiders come to mind, as well as oleander (I don't know much about plants but I read that book). I think sea snakes have the most powerful poisons of all animals, but plants are probably more deadly.

Yes I guess I could always make up something but that sometimes goes a little too sci-fi for my taste of writing...but I could just make up a spider or snake...or a venemous lycanthrope..just wanted to make a point I am not going to use arsenic too original.

if you're killing off a programmer/coder inject him with an overdose of caffeine

hmm....science fiction.......you should probably use some poison that's so well done that it harms only the target and is completely palatable to anyone else.

Use Arsenicous (the gasious form of Arsenic)

[quote="Xevarion"][quote="bubka"]You can make a poisonous robotic snake and get it bite people.[/quote]

ON A PLANE[/quote]

wow that movie was bad, no wonder its box office revenue was a failure and it got 2 stars...

well does it have senseless violence...i'd go see it then

Use orange juice.  As a twist in the story, the main character turns out to be not human and, let's say, boron based or something, and citric acid is poisonous

That's rather good but I've already got a lycanthrope on my mind..maybe..

I'VE HAD IT WITH THIS MOTHER****ING SNAKES ON THIS MOTHER****ING PLANE.

yes lots of people have...did i mention my last main character in this story commited suicide? i love to kill mcs.


inequalities
geometry
inradius
trigonometry
inequalities unsolved

let $ ABC$ a triangle , $ a$ , $ b$ , $ c$ he lenghts of the sides , $ A$ , $ B$ , $ C$ the angles , $ r$ the inradius and $ S$ the area , prove that :

$ \frac{\cos^2\big(\frac{A}{2}\big)}{a}\plus{}\frac{\cos^2\big(\frac{B}{2}\big)}{b}\plus{}\frac{\cos^2\big(\frac{C}{2}\big)}{c} \geq \frac{27r}{8S}$

[quote="y-a-s-s-i-n-e"]let $ ABC$ a triangle , $ a$ , $ b$ , $ c$ he lenghts of the sides , $ A$ , $ B$ , $ C$ the angles , $ r$ the inradius and $ S$ the area , prove that :

$ \frac {\cos^2\big(\frac {A}{2}\big)}{a} \plus{} \frac {\cos^2\big(\frac {B}{2}\big)}{b} \plus{} \frac {\cos^2\big(\frac {C}{2}\big)}{c} \geq \frac {27r}{8S}$[/quote]
$ \frac {\cos^2\big(\frac {A}{2}\big)}{a} \plus{} \frac {\cos^2\big(\frac {B}{2}\big)}{b} \plus{} \frac {\cos^2\big(\frac {C}{2}\big)}{c} \equal{} \frac{p}{4Rr}\equal{}\frac{p^2}{4RS}\geq \frac {27r}{8S}$;$ ( 2p^2 \ge 27Rr)$; :)

$ \cos^2\theta \equal{}\frac{1\plus{}\cos 2\theta}{2}$ and $ \cos A\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}$ so $ \cos^2\frac{A}{2}\equal{}\frac{(a\plus{}b\plus{}c)(b\plus{}c\minus{}a)}{4bc}$. Plugging that in,

$ \sum_{cyc} \frac{\cos^2\frac{A}{2} }{a}\equal{} \sum_{cyc}\frac{(a\plus{}b\plus{}c)(b\plus{}c\minus{}a)}{4abc}\equal{}\frac{a\plus{}b\plus{}c}{4abc}\sum_{cyc} b\plus{}c\minus{}a\equal{}\frac{(a\plus{}b\plus{}c)^2}{4abc}$

The $ 4AR\equal{}abc$, so $ \equal{}\frac{s^2}{4AR}\equal{}\frac{s}{4rR}$ since $ A\equal{}rs$. $ r$ is inradius. $ R$ is circumradius. $ A$ is triangle area. $ s$ is semiperimeter.

[quote="y-a-s-s-i-n-e"]thank"s a lot

can you prove that

$ \frac {\cos^2\big(\frac {A}{2}\big)}{a} \plus{} \frac {\cos^2\big(\frac {B}{2}\big)}{b} \plus{} \frac {\cos^2\big(\frac {C}{2}\big)}{c} \equal{} \frac {p}{4Rr} \equal{} \frac {p^2}{4RS}$

$ 2p^2 \geq 27Rr$

 :lol:[/quote]
$ b^2\plus{}c^2\minus{}2bcCos(A)\equal{}a^2$  ,  $ (cos(A/2))^2\equal{} \frac{1\plus{}cos(A)}{2}\equal{}> (cos(A/2))^2\equal{} \frac{p(p\minus{}a)}{bc}$;
so we have : $ \frac{(cos(A/2))^2}{a}\plus{}\frac{(cos(B/2))^2}{b}\plus{}\frac{(cos(C/2))^2}{c}\equal{} \frac{p(p\minus{}a)}{abc}\plus{}\frac{p(p\minus{}b)}{abc}\plus{}\frac{p(p\minus{}c)}{abc}$ $ \equal{}\frac{p^2}{4RS}$;
$ Rr \equal{}RS/p\equal{}abc/(4p) \le 2/27p^2$ ;(by AM-GM) :)


factorial
number theory
prime factorization
AMC

Let $N$ be the number of consecutive 0's at the right end of the decimal representation of the product $1!\times2!\times3!\times4!\cdots99!\times100!.$  Find the remainder when $N$ is divided by 1000.

[hide]As usual, the number of 0's at the end is the same as the number of 5's in the prime factorization, since there will be more than enough 2's. Since 25's provide two 5's, we count them twice. For the fives, we have $1+1+1+1+1+2+2+2+2+2+3+\cdots+18+19+19+19+19+19+20$, or $5(1+2+\cdots+19)+20=5\cdot190+20=950+20=970$. Since we already counted each of the 25's once, we count them once more: $1+1+\cdots+\text{(25 times)}+1+2+2+\cdots+3+3+4=25(1+2+3)+4=150+4=154$. So $970+154=1124$ and our answer is $\fbox{124}$.[/hide]

[hide]
we must count the number of 5's,
notice that from 5! through 99! the number of 5's are 1 through 22, each repeating 5 times

so the answer is 5(1+2+..+22-5-11-17)+24
(the subtractions are becuase of multiples of 25)
so the answer is 1124 so 124[/hide]

[hide]This problem comes down to counting the number of fives in the expression shown. So start with fives (ignoring 25 for right now):
We see 20 fives in 100! (as 100/5=20), then 19 in 99!-95!, 18 in 94!-90!, etc. so basically we have $20+5\cdot \frac{19\cdot20}{2}= 970$. Then we have to account for all the occurances of multiples of 25. We have 4 additional fives in 100, three in 99-75, 2 in 74-50, etc. This comes out to $4+25\cdot6=154$. $154+970=1124$ Since the answer asks for the remainder when 1124 is divided by 1000 (or the last three digits), the answer is $\fbox{124}$.[/hide]

I'm a newb, thought I'd post something

[hide]I rewrote the expression as 1^100*2^99...100^1, noting that the sum of each number and it's exponent was 101.  as others have noted, the key was the number of 5's.  so we get 5^96*10^91... and used a sumation to put dem babies together, and then double count all of the perfect squares multiples (25, 50, 75, 100).  a friend who also took this screwed up by not double counting the mutiples of 25[/hide]

[quote=naren_pr]15 year bump[/quote]

thats 3 years older than me :o

[quote=RedFireTruck][quote=naren_pr]good solution.

yay, now my number of recieved matches my number of gaven upvotes[/quote]

are you trying to get people to upvote you using reverse psychology lol.[/quote]

[s]you found me out[/s]. no, i actually meant it was a good solution.

[hide=Solution]We do casework. If the number is from $1$ to $25$, then the result is $0+5(1+2+3+4)+6$. If the number is from $26$ to $50$, then the result is $24+5(7+8+9+10)+12$. If the number is from $51$ to $75$, then the result is $48+5(13+14+15+16)+18$. If the number is from $76$ to $100$, then the result is $72+5(19+20+21+22)+24$. Adding gives $1124 \implies \boxed{124}$.[/hide]

quite possibly the worst problem I've ever done :(

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLsclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

By Legendre and a bit of double counting the answer is clearly
$$
\sum_{i=1}^{20} \nu_5(i) \cdot (101-5i) = 24 + \sum_{i=1}^{20} (100-5i) + \sum_{i=1}^{4} (100 - 25i).
$$
This sum evaluates to $24 + 950 + 150 = 1\fbox{124}$.


function
integration
real analysis
real analysis unsolved

Lemma:Suppose that $ \{g_n,n\geq 1\}$ are real valued continuous functions such that $ g_n\to g$ as $ n\to\infty$,where $ g$ is
    continuous in a neighborbood of $ b\in\mathbb{R}$.Then,for every $ \varepsilon >0$,there exists $ h_0>0$,such that
    \[ \limsup_{n\to\infty}\left|\frac{1}{2h}\int_{|x\minus{}b|<h} g_n(x) dx\minus{}g(b)\right|<\varepsilon\quad \mbox{for all} h\in (0,h_0).\]

I wonder if the lemma is true.

Consider:

$ g_n(x) \equal{} 1 \minus{} |1 \minus{} nx|,0 \le x \le \frac {2}n$,otherwise $ g_n(x) \equal{} 0$

Then $ g_n$ converges to $ 0$ obviously.
But:
$ \frac {1}{2h}\int_{ \minus{} h}^hg_n(x)dx \equal{} \frac {1}{2h}\int_{0}^hf(x)dx \equal{} \frac {1}{4}$,for $ h \equal{} \frac {2}n$

Which will contradict with your conclusion.(Set $ b\equal{}0$)

I may follow jmerry's idea [url]http://www.mathlinks.ro/viewtopic.php?t=179914[/url]


[quote="236factorial"]There has been a lot of hard questions lately. Here's an easy sequence.

What are the next two terms in the sequence

2, 4, 7, 11, 16, 22, 29, _, _ ?[/quote]
[hide]
37, 46. Each term is x more than the last, and each term x increases by 1.[/hide]

[hide="Answer"]$37$, $46$; $(n^{th}\; triangular\; number) +1$.[/hide]

[quote="superman"]cmon, don't make it too easy[/quote]

Hey, breaks can't be too hard.  :P

I'll think of a hard sequence in a while.

[quote="4everwise"]Here's the formula for the $n^{th}$ term:  :lol:
[hide]$\frac{n(n+1)}{2}+1$[/hide][/quote]
or $\frac{n^2+n+2}{2}$ :D


Find the number of ordered pairs of integers $ (m,n)$ for which \[ \frac{1}{m}\plus{}\frac{1}{n}\equal{}\frac{1}{2009}\]

actually after

$ 2009n \plus{} 2009m \equal{} nm$

[hide]$ nm \minus{} 2009n \minus{} 2009m \equal{} 0$

$ (n\minus{}2009)(m\minus{}2009) \equal{} 2009^2$

this question then becomes how many factors $ 2009^2$ has

$ 2009^2 \equal{} 7^4 * 41 ^2$

$ 5*3 \equal{} 15$[/hide]

You forgot that both (n-2009) and (m-2009) can both be negative  :roll:. That means you have to multiply the total by 2 to get $ \boxed{30}$ as an answer.

I think.

I think answer is $ \boxed{29}$.
Because, $ (n\minus{}2009, m\minus{}2009)\equal{}(\minus{}2009, \minus{}2009) \ \text{i.e.} \ (n, m)\equal{}(0, 0)$ is not answer.

I think that was the trick of the problem (once you simplify, you have to check that your steps reverse i.e. $ m,n\ne 0$.

Can anybody solve this problem:

$ \frac{1}{m}\plus{}\frac{1}{n}\equal{}\frac{1}{a}$ where $ a$ is a given positive integer.

How many solutions are there in terms of $ a$? (The given problem is $ a\equal{}2009$).

I am assuming m and n are integers as well?

As I showed in my algebraic steps with the original problem, we arive at:

$ an+am=nm$
$ nm-am-an=0$
$ (n-a)(m-a)-a^2=0$
$ (n-a)(m-a)=a^2$
$ xy=a^2$ where there are as many ordered pairs $ (x,y)$ as there are $ (m,n)$.

As snail2 said, now we find the number of factors of $ a^2$. We multiply that by two, since $ x$ and $ y$ have negative possibilities as well. Then we subtract the one case where $ x$ and $ y$ are both $ -a$ ($ m$ and $ n$ are $ 0$). The answer becomes:

$ \boxed{2(\text{\# of factors of }a^2)-1}$


The graph shows six labeled points.
How many distinct circles of radius
2 units are in the coordinate plane
and pass through exactly two of the
labeled points on this graph?
[asy]unitsize(1inch);
draw((-2,0)--(5,0),Arrows);
draw((0,-2)--(0,4),Arrows);
dot((-1,1));
dot((-1,-1));
dot((1,1));
dot((1,-1));
dot((3,1));
dot((3,-1));
label("$(-1,1)$",(-1,1),S);
label("$(-1,-1)$",(-1,-1),S);
label("$(1,1)$",(1,1),S);
label("$(1,-1)$",(1,-1),S);
label("$(3,1)$",(3,1),S);
label("$(3,-1)$",(3,-1),S);[/asy]

I count 18?

In particular any pair of points which are closer than 4 to each other will be struck by two circles of radius 2.

That gives me 6 circles for the vertical pairs and 8 circles for adjacent horizontal pairs.

If you center a circle radius 2 on a corner, it strikes two adjacent points... but the other circle striking those same points will also hit a third point, so I only found 4 this way.

The extreme corners are of course too far away from each other.

These counting problems are a bit depressing unless there's a slick observation to be made...

Couldn't you just do $ \binom{6}{2} \equal{} \boxed{15}$?

No, isabella, because there are two circles of radius 2 that can go through each set of two points (one centered to the left of the points and one centered to the right of the points--or one above, one below).

That said, I didn't know how to get the answer  either :( .  RightAnnihilator's solution makes sense, though.

Oh, I see.

Does anyone happen to know what the answer is? The person who asked this, did you review the questions and get the answer? If so, what is it?

I happen to know the answer is $ \boxed{18}$.

as far as I know, brute force is the only way that works.

I think:

If you connect/group the points, you find $ 7$ "pairs" of adjacent points. Note that the circle can be located on either/both sides of the points (most the circle located to the left, or most of it located to the right). So, we have $ 7 \times 2 \equal{}14$ circles.

Also, $ 4$ more circles can be formed with their centers at each of the points in the corners.

So our answer is $ 14\plus{}4\equal{}\boxed{18}$

the correct answer is 22. i do not know how.

No, even the MATHCOUNTS Answer Key says 18.  So 18 is the correct answer.

Dudes, On the FTW!, the answer says 22. Of course, who's to say who's right?

Well, the MathCounts answer key is the official answer key so I would say 18 is right.

i put 18, but ftw says it's 22!!!! ergh!!

The next time you see that question, you can click Review and Report the problem to prevent future incidences with that problem.


geometry
inequalities
circumcircle
geometry unsolved

Find the largest value of the real number $ \lambda$, such that as long as point $ P$ lies in the acute triangle $ ABC$ satisfying $ \angle{PAB}\equal{}\angle{PBC}\equal{}\angle{PCA}$, and rays $ AP$, $ BP$, $ CP$ intersect the circumcircle of triangles $ PBC$, $ PCA$, $ PAB$ at points $ A_1$, $ B_1$, $ C_1$ respectively, then $ S_{A_1BC}\plus{} S_{B_1CA}\plus{} S_{C_1AB} \geq \lambda S_{ABC}$.

Denote $ \angle{PAB} \equal{} \angle{PBC} \equal{} \angle{PCA} \equal{} \omega$. From $ BPCA_1 \minus{}$cyclic $ \Longrightarrow$ $ \angle AA_1C \equal{} \angle PBC \equal{} \omega$. Hence $ CA_1\parallel{}AB$ $ \Longrightarrow$ $ \frac {S_{A_1BC}}{S_{ABC}} \equal{} \frac {AM}{MA_1} \equal{} \frac {MC}{MB} \equal{} \frac {S_{PAC}}{S_{PAB}}$, where $ M \equal{} BC\cap AA_1$. Hence $ S_{A_1BC} \equal{} \frac {S_{PAC}}{S_{PAB}}.S_{ABC} \equal{} \frac {PA.PC.sinA}{PA.PB.sinB}.S_{ABC} \equal{} \frac {sin\omega .sinA}{sin(C \minus{} \omega).sinB}.S_{ABC}$ $ \equal{} \frac {sin^2A}{sin^2C}.S_{ABC} \equal{} \frac {a^2}{c^2}.S_{ABC}$ (since $ cot\omega \equal{} cotA \plus{} cotB \plus{} cotC$). So the problem reduces to $ \boxed{\frac {a^2}{c^2} \plus{} \frac {c^2}{b^2} \plus{} \frac {b^2}{a^2}\geq \lambda}$. It's easy to see that $ \lambda\geq 3$
Let $ c \equal{} max\{ a,b,c\}$. If we fix $ a \equal{} b$ and assume $ f(c) \equal{} \frac {a^2}{c^2} \plus{} \frac {c^2}{a^2} \plus{} 1$ then $ f'(c) > 0$ and $ f(c)$ is continuous and if we take $ c\to a$ we get that $ \lambda \equal{} 3$

Denote $ \angle{PAB} = \angle{PBC} = \angle{PCA} = \omega$. Suppose that $AA_1$ cuts $BC$ at $A_2$ define $B_2,C_2$ similarly. Note that $$\frac {S_{A_1BC}}{S_{ABC}} = \frac{A_1A_2}{AA_2}=\frac {ABsin(B)}{A_1Bsin(A_2BA_1)}=\frac {sin(PCB)sin(B)}{sin(\omega)sin(A_2BA_1)}=\frac {BPsin(B)}{CPsin(CPA_1)} =\frac {BPsin(B)}{CPsin(A)}.$$
Since the last fraction is cyclic the sum is at least $3$. Taking an equilateral triangle gives a equality case  for $ \lambda = 3.$


group theory
abstract algebra
superior algebra
superior algebra solved

Let $n\geq 5$ and $2<t<n.$ Prove that $S_{n}$ has no any subgroup of index $t.$

I think you mean normal subgroup :maybe:

For normal subgroups, it's easy and well-known. Do you have a counter-example for subgroups?

If there is such a subgroup then you get a non-trivial homomorphism from $S_{n}$ to $S_{t}.$ Now use the fact that $A_{n}$ is the only proper normal subgroup of $S_{n}$ for $n\geq 5.$


[hide]$\frac1a+\frac1b+\frac1c = \frac1{a+b+c}$
$\Rightarrow \frac{(a+b)(b+c)(c+a)}{abc(a+b+c)}=0$ (This isn't hard to show.)
From here, we can derive our results.

One obvious generalisation is we can replace $2007$ with any odd $n$. [/hide]

it's kind of annoying to put "Easy" as the subject/ source of every problem you post, just leaving those fields blank is probablly better.

I agree with [b]cats... [/b]on this one! :)

Is this some kind of commercial to write Easy on every problem you send? And if it is, is it to find sovers or to state how easy it was for you? :mad:
One shud use "easy" and "obvious" very carefully in math! :D


[quote="bane"]is it possible to find any (apart from 2 and 5) prime number in factorization of repunits (numbers that contains only the digit 1, like 11...111)?[/quote]
$ \forall p>5: \ 111..(p\minus{}1\ times)...111\equal{}\frac{10^{p\minus{}1}\minus{}1}{9}\equiv 0\ (mod\ p)$


calculus
integration
function
real analysis
real analysis unsolved

Let $ f: [a,b]\rightarrow\mathbb R$ be a continue function. Prove that exists $ c\in (a,b)$ so that $ \int_a^cf(x)\ \mathrm{dx}\equal{}(b\minus{}c)f(c)$ .

We consider the function $ g: [a,b]\minus{}>R$ $ g(x)\equal{}(b\minus{}x) \int_{a}^{x}f(t)dt$.We have $ g(a)\equal{}g(b)\equal{}0$ , so from Rolle we know that exist' s $ c \in (a,b)$ so that $ g'(c)\equal{}0$.And so we obtain the conclusion.


Se considera multimea $A=\{1, 4, 7,\ldots, 100\}$ astfel incat alegem 19 elemente din multuimea $A$. Sa se arate ca exista doua dintre ele cu suma $104$.

Avem $A = \left\{ \left. k \in \overline{0,33}\right| 3k+1 \right\}$.
Impartim $\left\{ 0, 1, \ldots, 33 \right\}$ in:
$B_{0}= \left\{ 0 \right\}$;
$B_{1}= \left\{ 1, 33 \right\}$;
$B_{2}= \left\{ 2, 32 \right\}$;
$B_{3}= \left\{ 3, 31 \right\}$;
$\ldots$;
$B_{16}= \left\{ 16, 18 \right\}$;
$B_{17}= \left\{ 17 \right\}$.

Daca alegem $19$ numere din $\overline{0,33}$, doua din ele, $m,n$, o sa fie in aceeasi multime $B_{i}$, $i \in \overline{1,16}$.

Suma elementelor corespunzatoare este $(3m+1)+(3n+1) = 3(m+n)+2 = 102+2 = 104$.

observam ca elementele sunt de forma $3k+1$, ele fiind in nr de $34$

Consideram numerele de la $100,97,...55,52$, ele sunt in numar de $17$, iar suma oricaror doua nr dintre acestea este mai mare ca $104$.

[b]obs[/b]: daka luam si nr $49$ sau un alt numar mai mic decat acesta exceptand pe $1$ din multimea $A$, suma de $104$ se realizeaza cu elementul de pe o pozitie simetrica fatza de numarul $52$ din multimea $A$
                (de exemplu [b]55[/b]- 52 - [b]49[/b];
                                   [b]58[/b] - 55 - 52 - 49 - [b]46[/b] etc.)

Astfel putem sa il consiram si pe $1$ in in submultimea celor $17$ intrucat el nu are "simetric" in raport cu $52$ in multimea $A$.Asadar se obtine o submultime de $18$ elemente incat oricum am alege $2$ nu ar da suma $104$.

[b]Concluzie:[/b] Oricum am mai alege un element din $A$ pt a-l "pune" in submultimea noastra, el va da suma $104$ adunat cu "simetricul" sau fatza de $52$.Deci oricum am lua $19$ elemente din $A$ , exista $2$ cu suma $104$

Amin!


:D I am almost finished with AoPS 1 and 2. I want to do well on the AMC-12 and AIME, but I am not sure what books to use to prepare?

Any suggestions?

Are Mathematical Circles, Art of Problem Solving, Geometry Revisited any good? Comments?

Thanks!    :lol:

For another general topic book like AoPS, take a look at "Art and Craft of Problem Solving" by Paul Zeitz.

Fierytycoon

Yeah, Zeitz is considered the classic "after AoPS book" - although I know people (ahem, Simon) who have done Zeitz in between the two books.  However, if you've finished both the books and can do the problems discussed in there quickly, you will probably do really well on the AMC and AIME.  I recommend buying (ahem!) some old exams from the AMC and doing those, if you haven't already.

Actually I confess that I didn't ever read AoPS 1. I didn't know about it when I was at that sort of a level. I skimmed over it a bit last year to see what was in it though.

Oh.  Scratch that from my post... grrr... where's the edit button?

For some reason, you can see what I'm thinking.  Anyway, I guess Simon means that you could read Zeitz before AoPS2, and if you've already finished AoPS2, then you should do fine with Zeitz's book.

[quote]Actually I confess that I didn't ever read AoPS 1.[/quote]

Same here. Hopefully I didn't miss anything important... I understood AoPS 2 without too much trouble though, so I hope I will be ok.

Fierytycoon[/quote]

I was wondering if there is a solutions manual for Art and Craft of Prob Solving?

Not an official one, at least.  I think.  You can probably find an unoffical one...or you could just ask here... dunno... starting to sound like Syntax... too many ellipses...

There is a solution manual for Zeitz, and I have it. He told me that he usually didn't give it to students, but he gave it to me because he liked my website. I'm not sure if I'm supposed to give it to you, but contact me about it if you want it.

Oh...so it's more of a personal thing.  I'd never heard of a solution, so it somewhat surprised me.  But then again, he'd have had to work out all of the problems to make sure they work, and for fun, so he might as well have written them down.

The solution manual is designed for people teaching a problem solving class. I believe the reason that he won't give it to students is that he doesn't want people taking the class to cheat in the class.

Complex Zeta,

How should I contact you?    :roll:

Private messaging is a good way. I would give you my email address, but I don't want everyone to send me random spam (i.e. non-math-related emails), so private message me, and I'll give you my email.

Yeup I talked to Richard and he said that The Art and Craft of Problem Solving was the way to go for a Post-Aops kid...

Enjoy!
IW.OAV


logarithms
modular arithmetic
floor function

1) Find the remainder when $ 3^{1874}$ is divided by 11.

2) How many digits are in the expansion of $ 2^{2004}$ (Log 5 = 0.6990)

3) $ 3^{x \plus{} 2} \plus{} 3^{x \minus{} 1}$ divided by $ 5$ times $ 3^{x \plus{} 1}$ Simplify.

I don't know how to do these.

Hint 1)  3^5 mod 11 = 1. Then write 1874 = 5A+R

Hint 2) Write $ 2^{2004} \equal{} A\cdot 10^n$ where $ 1\le A < 10$ (You want to find $ n$)
	Then $ \log 2^{2004} \equal{} \log A \plus{} n$ where $ \log A < 1$
	Also $ \log 2^{2004} \equal{} 2004\log 2 \equal{} 2004\log (10/5)$

3) Is "5 times $ 3^{x \plus{} 1}$" the denominator? Or just 5?

1
[hide]Using Euler's theorem $ a^{\varphi (m)} \equiv 1\pmod {m}$
remainder of $ \frac {3^{1874}}{11}$
$ (3,11) \equal{} 1\implies 3^{\varphi (11)} \equiv 1\pmod {11}$
$ {\varphi (11)} \equal{} 10\implies3^{10}\equiv 1\pmod {11}$
$ \implies3^{1874}\equiv(3^{10})^{187}\cdot3^4\equiv1^{187}\cdot3^4\equiv 81\equiv4\pmod{11}$
$ \implies 3^{1874}\equiv4\pmod{11}$
therefore the remainder when $ 3^{1874}$ is divided by $ 11$ is $ \boxed4$[/hide]
i will do the rest soon

[quote="Bicameral"]1) Find the remainder when $ 3^{1874}$ is divided by 11.

[/quote]

[hide="solution"]
$ \phi{11} \equal{} 10$
$ 3^{1874 \pmod{10}} \equal{} 3^4 \equal{} 81$
$ 81 \pmod{11} \equal{} \boxed{4}$
[/hide]

[hide="2"]
The amount of base-$ 10$ digits in an integer $ n$ is $ \lfloor \log n \rfloor \plus{} 1$.
$ \lfloor \log {2^{2004}} \rfloor \plus{}1 \equal{} \lfloor \log {\dfrac{10^{2004}}{5^{2004}}} \rfloor \plus{}1 \equal{} \lfloor \log {10^{2004}} \minus{} \log {5^{2004}} \rfloor \plus{}1 \equal{} \lfloor 2004 \minus{} 2004 \log 5\rfloor \plus{} 1 \approx \lfloor 2004 \minus{} 2004(.6990)\rfloor \plus{} 1 \equal{} 603 \plus{} 1 \equal{} 604$
[/hide]

[quote="10000th User"][quote="Crox"]Hi bicameral, here is the 2)

[hide]log (2^2004)=log ((5^2004)(2^2004)/(5^2004))=
=log ((5*2)^2004))-log (5^2004)=2004-2004*log (5)=2004(1-0.6990)=
=(2004*301)/1000=603.204  Then the number of digits it's 603[/hide][/quote]This answer is off by 1 digit. For example, try finding the number of digits of 128 with what you just did in your solution. The answer with this method is 2... which I'm sure you'll know what to do now :)[/quote]

 :blush: lapsus. I have just edit

If you want all 3 in one post:
[hide="hint for 1"]Totient function.[/hide]
[hide="solution"]Since 11 is prime, we have $ \phi{11}\equal{}10$. Thus, $ 3^{10}\equiv1\pmod{11}$. So 81 (mod 11) which is 4.[/hide]
[hide="hint for 2"]The number of digits in $ n^x$ is $ x\lfloor \log{n}\rfloor\plus{}1$. [/hide]
[hide="solution"]So, $ \lfloor2004*{\log10\minus{}\log5}\rfloor\plus{}1\equal{}\lfloor2004(0.301)\rfloor\plus{}1\equal{}604$[/hide]
[hide="hint for 3"]Factoring[/hide]
[hide="solution"]$ \frac{3^{x\plus{}2}\plus{}3^{x\minus{}1}}{5*3^{x\plus{}1}}\equal{}\frac{3^{x\minus{}1}(28)}{45*3^{x\minus{}1}}\equal{}\frac{28}{45}$.[/hide]


algebra
polynomial
modular arithmetic
algebra unsolved

Find all polynomials p(x) with integer coe±cients such that for each positive integer
n, the number 2 n -1 is divisible by p(n).

I am really dumb these days :(
If $q\mid p(n)$ and $q$ is a prime number, then $q\mid p(n+q)$, so $q\mid 2^n-1$ and $q\mid 2^{n+q}-1$. Since, obviously, $q\neq 2$, it follows $q\mid 2^q-1$, but $2^q-1=1\pmod{q}$.
Therefore $p(n)=\pm 1$.