geometry
parallelogram
angle bisector
area of a triangle
Heron\u0027s formula

In triangle $ABC$, $AB=13$, $BC=14$, and $AC=15$. Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$. Which of the following is closest to the area of the triangle $ADE$?

\[ \text {(A) } 2 \qquad \text {(B) } 2.5 \qquad \text {(C) } 3 \qquad \text {(D) } 3.5 \qquad \text {(E) } 4  \]

[hide="Solution"]
Extend $AD$ to $F$ so that we have a parallelogram $ABFC$, with diagonals $BC$ and $AF$ intersecting at $D$. Since the sum of the squares of the sides of a parallelogram is equal to sum of the squares of the diagonals, we have \[ 15^2+15^2+13^2+13^2 = 14^2+(2AD)^2,  \] so $AD = 2\sqrt{37}$.

Now, by the Angle Bisector Theorem, we have \[ \frac{13}{BE} = \frac{15}{14-BE}, \] and $BE = 6.5$. By Stewart's Theorem, we have \[ (13^2)(6.5)+(15^2)(7.5) = (AE)^2(14)+(6.5)(7.5)(14),  \] and $AE = \frac{\sqrt{585}}{2}$.

Finally, $DE=14-7-6.5=.5$.

Now applying Heron's Formula to $\triangle ADE$, we get $[ADE]=3 \implies C$. (The work is really messy, so it's best to do it on your calculator.)
[/hide]

[hide]well, i think some of those things are a bit unnecessary...

the area of that triangle is $1/2(14)*h$ where $h$ is the altitude from A to BC

we know that the area of the triangle is 84 (i have this memorized...seeing it so many times now) anyway, h=12

to find DE, make BE=x, and apply the angle biscector theorem (and using proportion properties), to get

$x/13=(14-x)/15=14/28 \to  x=13/2$

so $DE=BD-x=7-13/2=1/2$, so $1/2(1/2)12=3$

so $C$ is our answer[/hide]


inequalities
trigonometry
inequalities unsolved

[quote="tdl"]Let $ x,y,z\geq 0: xy \plus{} yz \plus{} zx \equal{} 1$. Prove that: $ \frac {1}{x \plus{} yz} \plus{} \frac {1}{y \plus{} zx} \plus{} \frac {1}{z \plus{} xy}\geq 3$[/quote] Substituting $ x\equal{}\tan \frac A2$ etc. we get that, $ \sum \tan \frac A2 \tan \frac B2 \equal{}1$
From Cauchy, \[ \sum \frac{1}{x\plus{}yz} \geq \frac{9}{x\plus{}y\plus{}z \plus{}1} \iff \sum x \leq 2 \iff   \sum \tan \frac A2 \leq 2 \]
But my solution works only if the triangle is acute angled. :(

Because from Majorization inequality we get, $ \sum \tan \frac A2 \leq 2$

I think that there must be some way to solve this using this substitution.  :maybe:

Thank for your reply. I also solve it in the case triangle $ ABC$ is acute like Moonmathpi496 and I think I can solve it by mixing variable but I false!
By the way if I true then I think that it is from a math contest of Russia 2005, but it isn't national olympiad.
Origin problem: Let $ a,b,c > 0: a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that: $ \frac {a}{a^3 \plus{} bc} \plus{} \frac {b}{b^3 \plus{} ca} \plus{} \frac {c}{c^3 \plus{} ab}\geq 3$
With $ x \equal{} \frac {bc}{a},y \equal{} \frac {ca}{b},z \equal{} \frac {ab}{c}$ we have our problem
Please check and solve it, especially member from Russia!

I do not think this problem can be solved so
can be proven:$ f(x,y,z)\geq f(x,t,t)$ where t such that:$ t^2\plus{}2tx\equal{}1$.
or contract provisions
but I still have not found a suitable solution

[quote="tdl"]Let $ x,y,z\geq 0: xy \plus{} yz \plus{} zx \equal{} 1$. Prove that: $ \frac {1}{x \plus{} yz} \plus{} \frac {1}{y \plus{} zx} \plus{} \frac {1}{z \plus{} xy}\geq 3$[/quote]
If someone wants to prove it when $ x\equal{}y$: By the uvw-method it is almost obvious that we only have to prove it when $ x\equal{}y$ and when $ z\equal{}0$ - the latter being trivial.

[quote="Moonmathpi496"][quote="Mathias_DK"]If someone wants to prove it when $ x \equal{} y$: By the uvw-method it is almost obvious that we only have to prove it when $ x \equal{} y$ and when $ z \equal{} 0$ - the latter being trivial.[/quote]
Sorry, I don't understand your solution/outline. Could you please explain it a bit more?[/quote]
I didn't explain it very well.
First have a look at: http://www.mathlinks.ro/viewtopic.php?t=278791

Then write everything in terms of $ u,v^2,w^3$. Then it is concave in $ w^3$(To me it is obvious :P) and the conclusion follows.

My solution,it might be quite complex :blush: .
We have prove when $ x \plus{} y \plus{} z\leq 2$.We need prove in case $ x \plus{} y \plus{} z\geq 2$.
We have:
 $ \frac {1}{x \plus{} yz} \plus{} \frac {1}{y \plus{} zx} \plus{} \frac {1}{z \plus{} xy}\geq 3$
$ \Leftrightarrow \sum (x \plus{} yz)(y \plus{} zx)\geq \prod (x \plus{} yz)$
$ \Leftrightarrow (x \plus{} y \plus{} z) \plus{} 7xyz(x \plus{} y \plus{} z)\geq 3x^2y^2z^2 \plus{} 3xyz(x \plus{} y \plus{} z)^2 \plus{} 2$
Ineq is true because $ x \plus{} y \plus{} z \plus{} 6xyz(x \plus{} y \plus{} z)\geq 2 \plus{} 3xyz(x \plus{} y \plus{} z)^2$(because $ x \plus{} y \plus{} z\geq 2$)
 and $ xyz(x \plus{} y \plus{} z)\geq 3x^2y^2z^2$

[quote="Mathias_DK"][quote="math10"]I really do not like to use methods not Elementary
I really want to see a nice solution for a nice problem[/quote]
What do you consider to be beautiful? Are EMV,SMV,SOS and such techniques beautiful? Is a solution using only AM-GM or only Cauchy-Schwartz more beautiful than one with Jensen? What if the one with Jensen is much shorter?[/quote]
I always priority the method  Elementary.For me the EV,SMV method method is too strong, not used in the exams
, SOS is complicated.So I I  little use those methods.
Jensen is one of the priorities my.

In fact, with this problem, I had a solution with abc technique of [b]Nguyen Anh Cuong[/b]. But I hope to see another solution, simplier and elementery. And I see a solution of [b]math10[/b]. It is the solution which I'm waiting!

[quote="math10"]Ineq is true because $ x \plus{} y \plus{} z \plus{} 6xyz(x \plus{} y \plus{} z)\geq 2 \plus{} 3xyz(x \plus{} y \plus{} z)^2$(because $ x \plus{} y \plus{} z\geq 2$)
 and $ xyz(x \plus{} y \plus{} z)\geq 3x^2y^2z^2$[/quote]
Sorry, I don't understand why these inequalities are true? :|
For the first one with $ x\plus{}y\plus{}z\geq 2$, I got $ 6xyz(x\plus{}y\plus{}z) \geq 3xyz(x\plus{}y\plus{}z)^2 \iff 2 \geq x\plus{}y\plus{}z$!!!

[quote="Moonmathpi496"][quote="math10"]Ineq is true because $ x \plus{} y \plus{} z \plus{} 6xyz(x \plus{} y \plus{} z)\geq 2 \plus{} 3xyz(x \plus{} y \plus{} z)^2$(because $ x \plus{} y \plus{} z\geq 2$)
 and $ xyz(x \plus{} y \plus{} z)\geq 3x^2y^2z^2$[/quote]
Sorry, I don't understand why these inequalities are true? :|
For the first one with $ x \plus{} y \plus{} z\geq 2$, I got $ 6xyz(x \plus{} y \plus{} z) \geq 3xyz(x \plus{} y \plus{} z)^2 \iff 2 \geq x \plus{} y \plus{} z$!!![/quote]
ineq is true because:$ x \plus{} y \plus{} z \plus{} 6xyz(x \plus{} y \plus{} z)\geq 2 \plus{} 3xyz(x \plus{} y \plus{} z)^2$
$ \Leftrightarrow (x\plus{}y\plus{}z\minus{}2)((zy\plus{}yz\plus{}zx)^2\minus{}3xyz(x\plus{}y\plus{}z))\geq 0$

[quote="math10"]$ \Leftrightarrow (x \plus{} y \plus{} z \minus{} 2)((zy \plus{} yz \plus{} zx)^2 \minus{} 3xyz(x \plus{} y \plus{} z))\geq 0$[/quote]
With $ xy\plus{}yz\plus{}zx\equal{}1$, we don't have $ x\plus{}y\plus{}z \geq\ 2$, math10. Try $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt{3}}$  :wink:

we easily prove when $ x\plus{}y\plus{}z\leq 2$.So we need prove in case $ x\plus{}y\plus{}z\geq 2$

[quote="tdl"]Let $ x,y,z\geq 0: xy\plus{}yz\plus{}zx\equal{}1$. Prove that: $ \frac{1}{x\plus{}yz}\plus{}\frac{1}{y\plus{}zx}\plus{}\frac{1}{z\plus{}xy}\geq 3$[/quote]
Let $ x,y,z\geq 0: xy\plus{}yz\plus{}zx\equal{}1$. Prove that:
\[\frac{1}{x+yz}+\frac{1}{y+xz}+\frac{1}{z+xy}-3 \geq \frac{81}{2}(-5+3\sqrt{3})x^2y^2z^2\]

BQ


Έστω $ x,y\geq 0$ και $ z,t>0$ ώστε $ z\minus{}x\geq y\minus{}t$. Να αποδείξετε ότι $ \left(\frac{z}{t\plus{}y}\plus{}\frac{t}{x\plus{}z}\right)_{min}\equal{}\sqrt{2}\minus{}\frac{1}{2}$.

P.S Η άσκηση προτάθηκε απο τον xp

[quote="Γιώργος"]Έστω $ x,y\geq 0$ και $ z,t > 0$ ώστε $ z \minus{} x\geq y \minus{} t$. Να αποδείξετε ότι $ \left(\frac {z}{t \plus{} y} \plus{} \frac {t}{x \plus{} z}\right)_{min} \equal{} \sqrt {2} \minus{} \frac {1}{2}$.

P.S Η άσκηση προτάθηκε απο τον xp[/quote]

Αν και έχει παραμείνει άλυτη για μέρες βάζω τη λύση σε hide γιατι κάποιοι μου είπαν πως την προσπαθούν ακόμα..

[hide] WLOG έχουμε ότι $ x\geq y$ και $ z\geq t$ και $ z\plus{}t\geq x\plus{}y\Longrightarrow 2(z\plus{}t)\geq x\plus{}y\plus{}z\plus{}t\Longrightarrow z\plus{}t\geq \frac{x\plus{}y\plus{}z\plus{}t}{2}$. Άρα

$ \frac{z}{t\plus{}y}\plus{}\frac{t}{x\plus{}z}\equal{}\frac{z\plus{}t}{t\plus{}y}\minus{}t\left( \frac{1}{t\plus{}y}\minus{}\frac{1}{x\plus{}z}\right)\geq$

$ \geq \frac{1}{2}\cdot \frac{x\plus{}y\plus{}z\plus{}t}{t\plus{}y}\minus{}(t\plus{}y)\left( \frac{1}{t\plus{}y}\minus{}\frac{1}{x\plus{}z}\right)\equal{}\frac{1}{2}\frac{x\plus{}y\plus{}z\plus{}t}{t\plus{}y}\minus{}1\plus{}\frac{t\plus{}y}{x\plus{}z}\equal{}\frac{1}{2}\cdot 1 \plus{}\frac{1}{2}\cdot \frac{x\plus{}z}{t\plus{}y}\minus{}1\plus{}\frac{t\plus{}y}{x\plus{}z}\equal{}\minus{}\frac{1}{2}\plus{}\frac{1}{2}\cdot \frac{x\plus{}z}{t\plus{}y}\plus{}\frac{t\plus{}y}{x\plus{}z}\geq \sqrt{2}\minus{}\frac{1}{2}$[/hide]


Let $ f(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$
and  $ 4b \plus{} 8d \plus{} 1 \equal{} 0$ and $ f(x)\ge 0$ for $ x \in [ \minus{} 1,1]$. Find $ b$ .

$ b\equal{}\frac{\minus{}1}{4}\minus{}2d$. Because $ (f(x)\plus{}f(\minus{}x))/2\equal{}x^4\plus{}bx^2\plus{}d\ge 0$ in $ x^2\in [0,1]$ we get $ \frac{\sqrt 2 \minus{}1)^2}{8}\le d\le \frac{\sqrt 2 \plus{}1)^2}{8}.$
And any mean for $ d$ is well, when $ a\equal{}c\equal{}0$. Therefore we can not define some $ b$.


http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_12A_Problems/Problem_18
http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_12A_Problems

It says in the problem that there are 168 primes under 1000 and in the solution that there are 165. Is it just me, and is my reasoning just off?
How many primes under 1000 really are there?

I got a 112.5 on that one :P. Brute forced me way through.


Well, it is easy to get:
$\log_{m} {(x^4+m^4)} \geq \log_{m}{(\frac{x^2}{4}+m^4)}$

Since $m \in (0,1)$, so $f(x)=\log_m x $ is a [b]decreasing function[/b]. Hence we have:
$x^4+m^4 \leq \frac{x^2}{4}+m^4$

$x \in (-\frac12, \frac12)$.

[quote="shobber"][quote="paladin8"]Hmm, the problem changed after I posted a solution... now my post seems to be nonsense... :blush:[/quote]
lol.
Can you please tell us the original problem?[/quote]

I think it was let $f(x) = 1-\frac{1}{1-x}$. Find $f^{1998}(3)$.


How does one specify the text alignment of blocks within an align environment?

For example,

\begin{align*} 1 & 22 & 333 & 4444 \\
5555 & 666 & 77 & 8\end{align*}
[code]\begin{align*} 1 & 22 & 333 & 4444 \\
5555 & 666 & 77 & 8\end{align*}[/code]

What if I wanted the $ 333$ / $ 77$ column to be aligned left and the $ 22$ / $ 666$ column centered, for example?

align is based on an array with specifications {rl rl rl ...}. If you don't want r followed by l then you can add further ampersands (&), so for example

\begin{align*}
& 1 && 22 && 333 && 4444 \\
& 5555 && 666 && 77 && 8
\end{align*}

is completely left-aligned.

Align is meant for equations which have an = in them (hence r followed by l) so centering is not an option. For centering of numbers use an array:

$ \begin{array}{lclr}
1 & 22 & 333 & 4444 \\
5555 & 666 & 77 & 8
\end{array}$

with specifications {l c l r}.


Given a triangle with acute angle $ \angle BEC,$ let $ E$ be the midpoint of $ AB.$ Point $ M$ is chosen on the opposite ray of $ EC$ such that $ \angle BME \equal{} \angle ECA.$ Denote by $ \theta$ the measure of angle $ \angle BEC.$ Evaluate $ \frac{MC}{AB}$ in terms of $ \theta.$

[hide="Solution"][quote="orl"]Given a triangle with acute angle $ \angle BEC,$ let $ E$ be the midpoint of $ AB.$ Point $ M$ is chosen on the opposite ray of $ EC$ such that $ \angle BME = \angle ECA.$ Denote by $ \theta$ the measure of angle $ \angle BEC.$ Evaluate $ \frac {MC}{AB}$ in terms of $ \theta.$[/quote]
Let F on EM such that $ BF=BE=AE$. Since angle BEC is acute, $ \angle BEC=\angle BFE$
$ \Rightarrow\angle AEC=\angle BFM\Rightarrow \triangle AEC=\angle BFM\Rightarrow CE=FM$
$ \Rightarrow MC=EF, \frac{MC}{AB}=\frac{EF}{2BE}=cos\theta$[/hide]

shouldn't it be "on the opposite ray of $ CE$"?

There is a small mistypo. Edited problem:
[quote="mr.danh"]Given a triangle with acute angle $ \angle BEC,$ let $ E$ be the midpoint of $ AB.$ Point $ M$ is chosen on the ray of $ EC$ (or "opposite ray of CE") such that $ \angle BME \equal{} \angle ECA.$ Denote by $ \theta$ the measure of angle $ \angle BEC.$ Evaluate $ \frac {MC}{AB}$ in terms of $ \theta.$[/quote]

let $ \alpha\equal{}\angle ACE$ and $ \beta\equal{}\angle BCE$.

we have that $ \frac{BM}{BE}\equal{}\frac{\sin \theta}{\sin\alpha}\equal{}\frac{AC}{AE}$, so $ BM\equal{}b$.

note that $ \cos \theta\equal{}\frac{b^2\minus{}a^2}{2c\cdot m_c}>0$, so $ b>a$. and note that $ \frac{\sin\alpha}{\sin\beta}\equal{}\frac{a}{b}<1$, so $ \alpha<\beta$, which implies that the order of the points on ray $ EC$ is $ E\minus{}C\minus{}M$...

we have also that $ \cos\alpha\equal{}\dfrac{b^2\plus{}\frac{a^2\plus{}b^2\minus{}c^2}{2}}{2b\cdot m_c}$ and $ \cos\beta\equal{}\dfrac{a^2\plus{}\dfrac{a^2\plus{}b^2\minus{}c^2}{2}}{2a\cdot m_c}$...

note that $ \angle BCM\equal{}180\minus{}\beta$, so we have $ MC\equal{}BM\cos\alpha\plus{}BC\cos(180\minus{}\beta)\equal{}\dfrac{b^2\minus{}a^2}{2m_c}\equal{}c\cos \theta$, and we're done.  :D

Let $ \angle ACE = \angle BME = \beta$
$ \frac {MC}{AB} = \frac {CF}{AB} - \frac {ME}{AB} = \frac {1}{2}.(\frac {CE}{AE} - \frac {ME}{EB}) = \frac {1}{2}.(\frac {sin(\alpha - \beta)}{sin\beta} - \frac {sin(\pi - \alpha - \beta)}{sin\beta}) = cos\alpha$


Let $ x^2\plus{}2y^2\equal{}1$.
Find the max value of $ P\equal{}\sqrt{x}\plus{}\sqrt{3y}$

hello, you can use the method of Lagrange Multipliers, consider the function
$ F(x,y,\lambda) \equal{} \sqrt {x} \plus{} \sqrt {3y} \plus{} \lambda(x^2 \minus{} 2y^2 \minus{} 1)$.
Now you must calculate the partial derivatives of F with respect to $ x,y,\lambda$ and solve the equation system
$ F_x \equal{} 0$
$ F_y \equal{} 0$
$ F_\lambda \equal{} 0$
Sonnhard.

[quote="Kyoshiro"]Let $ x^2 \plus{} 2y^2 \equal{} 1$.
Find the max value of $ P \equal{} \sqrt {x} \plus{} \sqrt {3y}$[/quote]
We can use CS 2 times $ P\le \sqrt{2}\sqrt{x\plus{}3y}\le\sqrt{2}\sqrt{\sqrt{1\plus{}\frac{9}{4}}\sqrt{x^2\plus{}2y^2}}\equal{}13^{\frac{1}{4}}$

[quote="dima ukraine"][quote="Kyoshiro"]Let $ x^2 \plus{} 2y^2 \equal{} 1$.
Find the max value of $ P \equal{} \sqrt {x} \plus{} \sqrt {3y}$[/quote]
We can use CS 2 times $ P\le \sqrt {2}\sqrt {x \plus{} 3y}\le\sqrt {2}\sqrt {\sqrt {1 \plus{} \frac {9}{4}}\sqrt {x^2 \plus{} 2y^2}} \equal{} 13^{\frac {1}{4}}$[/quote]
Your solution is wrong.
$ x\equal{}?, y\equal{}?$ :huh:


let's share our photos!!

mine are in
[url=http://www.esnips.com/web/frengoIMO2006photos]http://www.esnips.com/web/frengoIMO2006photos[/url]

(i'm francesco cavazzani, silver medalist at 26 points)

my friend simone's (simo_the _wolf,gold at 28) ones are in
[url=http://www.esnips.com/web/Videoimo06]http://www.esnips.com/web/Videoimo06[/url]
[url=http://www.esnips.com/web/sthewphotoIMO06]http://www.esnips.com/web/sthewphotoIMO06[/url]

there are many photos also in
[url=http://imo2006.dmfa.si/images.html]http://imo2006.dmfa.si/images.html[/url] and [url=http://www.dic.si/galerija/thumbnails.php?album=34]http://www.dic.si/galerija/thumbnails.php?album=34[/url]

i hope many of you will reply...

bye bye

EDIT3:thanks to the admins... :oops:

Sorry to tell this, but I have to...

I was stolen my camera the last day, after the closing ceremony (during the dinner). I just want to let my mail if someone have some informations... : boitemail75@yahoo.fr .

I had good pictures, some "file photos" as we say in france. So, Help me lol.

It was a Canon Ixus 50 with a part of my phone number written on it.
Thank you

PS : I am french, so excuse my language lol.

Hi!

I have uploaded my photos on my blog. Feel free to visit it.
[url]http://imo2006.blogspot.com[/url]
This is the story of IMO 2006 from the perspective of the Singapore team, as well as from the perspective of an Observer B.
Enjoy the pictures and story, and remember to tag! Do include your name/nickname and country you are from so that I know who you are.

Pls PM me if you see any errors on my blog.

Thanks!


I don't understand the question but I hope
\[ 2207\minus{}\frac{1}{2207\minus{}\frac{1}{2207\minus{}\cdots}}\equal{}\left(\frac{3\pm\sqrt{5}}{2}\right)^8\]
helps  :P

[quote="Albanian Eagle"]I don't understand the question but I hope
\[ 2207 \minus{} \frac {1}{2207 \minus{} \frac {1}{2207 \minus{} \cdots}} \equal{} \left(\frac {3\pm\sqrt {5}}{2}\right)^8\]
helps  :P[/quote]

right this gives the result but How can I achieve this result?

[quote="epsilon07"]\sqrt[8]{2207-\frac{1}{2207-\frac{1}{.\frac{.}{....}}}}=\frac{a+b\sqrt{c}}{d}

a,b,c and d are real numbers. Find the minimum value of a + b + c + d[/code][/quote]

There is no response to this problem since we can find $ a,b,c,d$ such that $ a \plus{} b \plus{} c \plus{} d$ is as little as we want :

Let $ LHS \equal{} u > 0$. Let $ a \equal{} 1$ and $ c \equal{} 0$ and $ d \equal{} \frac 1u$. We have $ u \equal{} \frac {a \plus{} b\sqrt {c}}{d}$ and $ a \plus{} b \plus{} c \plus{} d \equal{} b \plus{} 1 \plus{} \frac 1u$ and we just have to take $ b\to \minus{} \infty$ to get the value as little as we want.


Using pennies, nickels, dimes, quarters, and half-dollars, how many ways are there to make $ 50n$ cents, where $ n$ is a fixed integer?

[hide="Hint"]Suppose you have 49 pennies, 9 nickels, 4 dimes, 1 quarter, and 0 half-dollars.  Then there is 1 way to make 0 cents, 45 ways to make 50 cents, 52 ways to make 100 cents, and 2 ways to make 150 cents.[/hide]

Can u use a diophantine equation and find how many tuple..

hmm i dont think diophantine equations are the way to go here. I was thinking of generating functions but im not sure how to find the coefficients of all $ x^{50n}$ for

[hide]
\[ \frac{1}{(1-x^{5})(1-x^{5})(1-x^{10})(1-x^{25})(1-x^{50})}\]

Note that the pennies act like nickels since having a number of pennies not equivalent to a multiple of $ 5$ produces amounts never equivalent to $ 50n$.
[/hide]

You can make that generating function argument work.  One key idea is to notice that
[hide]all the factors in the denominator of your generating function are divisors of $ 1-x^{50}$.[/hide]

I think it is easier to solve the problem without generating functions, however.  Here is a second hint that builds on the hint in the original post.

[hide="Hint_2"]Imagine putting all the pennies you use into coin wrappers of 50 pennies each, leaving at most 49 pennies loose.  Similarly, put all the nickels you use into wrappers of 10 nickels each, leaving at most 9 nickels loose.  Do something similar for dimes, quarters, and half-dollars, making each wrapper have value 50 cents.[/hide]

I think generating functions are the approach over here. This problem appears in Zeitz page 142 on the section on generating functions. But he does a simpler problem with $ n=1$ (dollar).

Hmm, no one seemed to give a complete solution, so I will present one without generating functions.  It helps to read my Hints 1 and 2 above.

[hide="Solution"]As Hint 2 suggested, given a way to make $ 50 n$ cents, imagine wrapping the pennies into groups of 50, the nickels into groups of 10, the dimes into groups of 5, the quarters into groups of 2, and the half-dollars each in their own group.  So each coin wrapper is worth 50 cents.  There may be some loose coins: at most 49 (actually, 45) pennies, 9 nickels, 4 dimes, and 1 quarter.  So the loose change is at most 159 cents.  Because we made $ 50 n$ cents, and each wrapper was worth 50 cents, the loose change must be a multiple of 50 cents.  So there are only four cases for the loose change: 0 cents, 50 cents, 100 cents, or 150 cents.  We will handle each case separately:

[b]0 cents in loose change:[/b] We must have $ n$ wrappers, each labeled penny, nickel, dime, quarter, or half-dollar, with order not counting.  By a usual "balls and walls" argument, there are $ \binom{n+4}{4}$ such possibilities.  (Think of the $ n$ wrappers as the balls, with 4 walls to separate them into 5 coin types.)

[b]50 cents in loose change:[/b] Here we have $ n-1$ wrappers.  So there are $ \binom{n+3}{4}$ ways to label the wrappers.  By Hint 1, there are also 45 ways to make the loose change add up to 50 cents.  Thus we have $ 45 \binom{n+3}{4}$ possibilities for this case.

[b]100 cents in loose change:[/b] Here we have $ n-2$ wrappers, so there are $ \binom{n+2}{4}$ ways to label the wrappers.  Again, by Hint 1, there are 52 ways to make the loose change add up to 100 cents.  Thus we have $ 52 \binom{n+2}{4}$ possibilities in this case.

[b]150 cents in loose change:[/b] Here we have $ n-3$ wrappers, so there are  $ \binom{n+1}{4}$ ways to label the wrappers.  There are only 2 ways to make the loose change add up to 150 cents.  Thus we have $ 2 \binom{n+1}{4}$ possibilities in this case.

Adding up all the cases, the number of ways to make $ 50 n$ cents is
\[ \binom{n+4}{4}+45 \binom{n+3}{4}+52 \binom{n+2}{4}+2 \binom{n+1}{4}\, . \]
[/hide]


function
calculus
derivative
calculus computations

The height of water during the rise and fall of a tide is recorded.
when the water height h(in meters) at time t(in hours) starting when the level is equal to the mean level of 3 meters and has an amplitude of 2.5 meters and a period of 15 hours.

this is a function of sinusoidal  :maybe:

the formula is y = a +b sin (Kx + c )

y = 3 + 2.5 sin ( (2pi/15)*x) ,  since c = 0 because no phase shift .

how i can Find  the highest and lowest values of h and the times at which they are taken.

Isn't it just $ \minus{}1 \leq sin(x) \leq 1$ for all $ x \in \mathbb{R}$? Then:

\[ 3 \minus{} 2.5 \leq y \leq 3 \plus{} 2.5\]

etc.

In general case, you need to use derivative.

technically your function should be in the form $ h(t)$.

Anyway, I'd go with mathangel's idea.  Then you need to find the solutions of $ \sin(2\pi t/15) \equal{} \pm 1$.  Simple enough.

IF you are supposed to use the calculus, what you should do is take the derivative $ h'(t)$ and find the points times where it is $ 0$ or undefined (the critical points).  Test these points to figure out which are mins, maxs, or neither (1st derivative test works well - remember you are looking for the derivative to change sign.  positve to negative gives a max and the opposite gives a min.)


quadratics
modular arithmetic
number theory
relatively prime

Does there exist a couple $ (x,y)$ in rationals such that $ x^2\plus{}y^2\equal{}3^{2009}$　$ ?$

[hide="hint"]
Note the quadratic residues mod $ 4$.

[/hide]

[hide="Solution"]
[b]Lemma:[/b] Let $ a$ and $ b$ be integers that are relatively prime to an odd prime, $ p$.  Then, $ a^2\equiv \minus{} b^2\bmod p$ is possible only if $ p\equiv 1\bmod 4$.
[i]Proof:[/i] We have that $ a^2\equiv \minus{} b^2\bmod p$, so $ (\frac {a}{b})^2\equiv \minus{} 1\bmod p$, meaning that this is equivalent to saying that $ \minus{} 1$ is a quadratic residue $ \bmod p$.  Let $ \frac {a}{b} \equiv c\bmod p$ and let $ k$ be the smallest positive integer so that $ c^k\equiv 1\bmod p$.  We have that $ c^2\equiv \minus{} 1\bmod p$, so $ c^4\equiv 1\bmod p$.  $ k$ must be a factor of $ 4$, so $ k \equal{} 1,2,4$.  If $ k \equal{} 1$ or $ k \equal{} 2$, then $ c^2\equiv 1\bmod p$, contradicting the fact that $ c\equiv \minus{} 1\bmod p$.  Then, $ k \equal{} 4$.  Yet, $ k|\phi (p)$, so $ k|(p \minus{} 1)$, so $ 4|(p \minus{} 1)$, so $ p\equiv 1\bmod 4$.

Now for the problem.  Assume solutions exist.  Let $ x \equal{} \frac {a}{b}$ and $ y \equal{} \frac {c}{d}$.  Then, this becomes $ (ad)^2 \plus{} (bc)^2 \equal{} 3^{2009}b^2d^2$.  By the Lemma, if $ ad$ and $ bc$ do not share a factor, with $ 3$, then $ (ad)^2 \plus{} (bc)^2$ cannot be a multiple of $ 3$, meaning that $ 3|ad$ and $ 3|bc$, so we can divide off $ 9$ and get that $ 3^k|[(\frac {ad}{3})^2 \plus{} (\frac {bc}{3})^2]$, where $ k$ is an odd integer (note that $ k$ is odd since $ b^2d^2$ is a square, so $ b^2d^2$ has an even number of powers of $ 3$ dividing it.)  We keep on repeating the process, and we will eventually get $ 3\parallel{}[(\frac {ad}{3^l})^2 \plus{} (\frac {bc}{3^l})^2]$ for some integer $ l$, which is a contradiction, since if $ 3|x, y$, so $ 9|(x^2 \plus{} y^2)$.  Then, no rational solutions exist.  [/hide]

Please correct me if I am wrong but if you clear denominators, the equation becomes

$ x_1^2\plus{}x_2^2 \equal{} 3^{2009}x_3^2$ where $ x_1,x_2.x_3 \in \mathbb{Z}$

It is easily proved that$ 3|x_1$ and $ 3|x_2$

An even power of 3 divides LHS and an odd power of 3 divides RHS which is a contradiction and hence there are no solutions in rationals.

[quote="hsbhatt"]Please correct me if I am wrong but if you clear denominators, the equation becomes

$ x_1^2 \plus{} x_2^2 \equal{} 3^{2009}x_3^2$ where $ x_1,x_2.x_3 \in \mathbb{Z}$

It is easily proved that$ 3|x_1$ and $ 3|x_2$

An even power of 3 divides LHS and an odd power of 3 divides RHS which is a contradiction and hence there are no solutions in rationals.[/quote]

Oh,I haven't noticed such a clear solution!Thank you hsbhatt!

[hide]$ x^2\plus{}y^2\equal{}3^{2009} \implies x^2\plus{}y^2 \equiv 3^{2009} \text{ (mod 4) }$

$ 3^{2009} \equiv 3  \text{ (mod 4)  , and}$

$ \text {for any odd no. 2k\plus{}1 and any even no. 2k ,}$

$ (2k\plus{}1)^2 \equiv 1  \text{ (mod 4) } , (2k)^2 \equiv 0  \text{ (mod 4) }$

$ (2k\plus{}1)^2\plus{}(2k)^2 \equiv 1 \text{ (mod 4)}$

$ \text{x and y should be an odd no. and an even no. respectively , and vice versa .}$

$ \text{But  } 3^{2009} \not \equiv (2k\plus{}1)^2\plus{}(2k)^2  \text{ (mod 4)}$

$ \text{It is a contradiction.}$
[/hide]

[quote="TZF"][b]@ vinskman:[/b] I believe that doesn't apply, because the question was to find the solution set in $ \mathbb{Q}$ and not in $ \mathbb{Z}$[/quote][hide="Fixed"]$ \frac{p_1}{q}$ and $ \frac{p_2}{q}$ are solutions: \[ p_1^2\plus{}p_2^2\equal{}3^{2009}q^2\]where $ (p_1,q)\equal{}(p_2,q)\equal{}1$. We have that $ 3^{2009}\equiv3\pmod4$, which is impossible unless $ q^2\equiv0\pmod4$. However, this makes $ p_1$, $ p_2$, and $ q$ even, a contradiction.[/hide]

[hide]You don't necessarily have (p1, q) = 1 and (p2, q) = 1, but then again the solution can be easily fixed (We can assume (p1, p2, q) = 1 or else we can reduce the fractions, thus they cannot share a factor of 2...).[/hide]


Is it possible to make a numbered list start with 2. instead of 1. ?

In case you were wondering: i was working on a test, and needed to start answering questions starting with 2.

After \begin{enumerate} use  \setcounter{enumi}{1}
(enumii, enumiii, enumiv would deal with counters for sublists, subsublists etc)

if enumi is for lists,
is there something similar for sections?
i tried:

\setcounter{section}{1} but that didn't work

It should work unless you have another system involved eg chapters.
[code]\documentclass[12pt]{article}
\begin{document}
\setcounter{section}{1}
\section{this is a section}
\end{document}[/code]
gives
2 this is a section

So give a minimal document showing where it doesn't work.

my bad... i did \setcounter after \section

thanks


Find positive real numbers x,y,z satisfying x+y+z=12 and xyz=2x+5y+10z

ongoing problem !!!!!!!!! MOD plz lock this topic !!


Find all pairs $ (m,n)$ of positive integer such that

$ n!\plus{}3\equal{}m^2$

[hide] For all n that are greater than 3, $ n!\equiv0\pmod{4}$. As all squares leave a remainder of 0 or 1 when divided by four, the left hand side would then be 3 (mod 4) while the right hand side would be 0 or 1. Thus, n must be 1, 2, or 3.

For n=1, we have m=2.

For n=2, we get no solution.

For n=3, we find m=3.

Thus, all solutions are (3,3) and (2,1).[/hide]


Maths is fun![hide]

There are 15 that aren't married, 30 without a cell, 25 without a car, and 20 without a house. 15+30+25+20=90, so 10 have to have/do all four by pigeonhole.

Hmm. That may have been a bit easy for intermediate.[/hide]

[hide]This problem isn't too bad, but I think a little hard to put in the Beginner's Forum. Especially if you've never seen something like this before.[/hide]

And it naturally leads to other variations, like what's the least number men who could have exactly 3 of the 4 properties? Or exactly 2 of the 4 properties? How can you solve these types of problems in general?

gauss202 wrote:In a certain town, there live exactly 100 men. Exactly 85
of these men are married, exactly 70 have a cell phone, exactly 75 own a car, and exactly 80 own a house. What is the smallest possible number of men in this town who do all four (i.e., are married, have a cell phone, own a car, and own a house)?

See if you can find the general way to solve problems like this.

(Taken from the 2002 University of Tennessee FERMAT contest)

[hide]Looking at the numbers the "opposite" way, we can say that

15 men aren't married

30 don't have a cell phone

25 don't own a car

20 don't own a house.

To find the minimum number that do have all four, we only need to find the maximum number of people that satisfy one of the four "opposite" properties, and then subtract from 100.

The maximum occurs if the four "opposite" groups contain nobody in common.  15+30+25+20 = 90.  Therefore, at minimum, 10 people do have all four things.[/hide]

To generalize, [hide]the number of people who have all i characteristics, where there are a1 people who have characteristic 1, a2 people who have characteristic 2, and ai people who have characteristic i out of a total pool of n people, there are a1 + a2 + ... + ai - n(i-1).[/hide]


Harvard college Princeton Columbia Yale geometry MIT

Is there any way to check how many math majors a college has, such as some sort of database?  I've browsed through some websites such as Harvard, Princeton, Columbia, Yale, etc., and the only one that gave a number was Princeton - about 34 (I believe this includes both juniors and seniors since they're the ones that have declared their majors).

E-mail the person in the math department whose job is to answer that sort of question.  You should be able to find such a person from the websites.  Be aware that different colleges may count differently: at U. Chicago, for instance, there is no distinction between an "applied math" and "pure math" degree while at Harvard there is, so the two numbers aren't directly comparable to each other.

And just want to ask a question about math major. Which univerisities do you think are prestigious at Applied Mathematics Major (Combining with Science, Engineering, Finance, etc.), since I really want to major in that area? Thanks!

i think the US news and world report listing has MIT first and NYU second for applied math.


An $ f$ einai sinexhs sto $ R$ (me times sto $ R$) kai $ f(x) \equal{} \int_o^xf(x \plus{} t)dt$  gia kathe pragmatiko $ x$ tote na deiksete oti i $ f$ midenizetai apeires fores.

Με την αλλαγή μεταβλητής $ u\equal{}x\plus{}t$ έχουμε $ f(x)\equal{}\int_x^{2x} f(u)du$ και αν $ F$ αρχική της $ f$ τότε $ F'$ συνεχής με $ F'(x)\equal{}F(2x)\minus{}F(x)$
Mε Θ.Μ.Τ σε καθένα από τα διαστήματα $ [\minus{}2,\minus{}1],[\minus{}4,\minus{}2],[\minus{}8,\minus{}4],...$ υπάρχουν $ t_1 , t_2 ,t_3 ,...$ στα αντίστοιχα ανοικτά διαστήματα ώστε $ F'(\minus{}1)F'(t_1) \le 0 ,F'(\minus{}2)F'(t_2) \le 0 ,F'(\minus{}4)F'(t_3) \le 0 ,...$ Με Bolzano στα διαστήματα $ [t_1,\minus{}1],[t_2,\minus{}2],[t_3,\minus{}4],...$ προκύπτει το ζητούμενο

Αλλάζοντας τα $ \minus{}1,\minus{}2,\minus{}4,...$ σε $ \minus{}1,\minus{}\frac{1}{2},\minus{}\frac{1}{4},...$ βρίσκεις άπειρες ρίζες στο $ [\minus{}1,0]$ και βέβαια όχι μόνο εκεί.
Επίσης είναι άπειρες φορές παραγωγίσιμη, γιατί το ολοκλήρωμα συνεχούς είναι παραγωγίσιμη συνάρτηση κ.ο.κ επαγωγικά (τα όρια ολοκλήρωσης είναι παραγωγίσιμα ως πολυωνυμικά)
Εύκολα επίσης δείχνουμε ότι $ f^{(n)}(0)\equal{}0$ για κάθε $ n\in N$ παραγωγίζοντας και θέτοντας όπου $ x\equal{}0$ σε κάθε σχέση
Πραγματικά ενδιαφέρουσα άσκηση

Euxaristw poli gia tin lisi sou... :)  :)

Otan tha exw ligo xrono tha grapsw kati parapanw gia autin... kathos kai mia enallaktiki lisi (sinoptika: pollaplasiase me tin $ e^x$ tin sxesh afou paragogiseis kai meletas tin sinartisi $ e^xf(x)$.

Kai pali thanks

[quote="staretak"]Thanks gia tin lisi sou...  :)  :)

Px mia poli wraia paratirisi gia tin Analysi tis askisi tha itan na doume an i f einai smooth mono (exei oles tis paragwgous- olwn twn taksewn) i einai kai analytiki. Arxika loipon dinw to parakatw, pou kanei tin askisi topologika pio endiaferousa kai pou paramenei se theoria Likiou (genika otan kataskeuaza tin askisi ithela na stirizetai mono se theoria Likiou):

An i f ikanopoiei tis sinthikes tis askishs tote na deithei oti i f exei apeires rizes pou oles exoun apoliti timi mikroteri tou 2008.[/quote]

Quick note: αν η $ f$ της παραπάνω άσκησης είναι αναλυτική, τότε είναι ταυτοτικά $ 0$ (άπειροι μηδενισμοί σε συμπαγές διάστημα).

Για μένα, η ενδιαφέρουσα ερώτηση είναι: [i]μπορεί να κατασκευαστεί ομαλή συνάρτηση (παράγωγοι κάθε τάξης) που να μην είναι ταυτοτικά $ 0$ και να ικανοποιεί την εκφώνηση της άσκησης[/i]?

Μόλις βρω χρόνο θα γράψω κάτι και γι'αυτό και για τις διάφορες ασκήσεις που χρωστάω λύσεις στο forum. Sorry, απλώς είμαι σε πίεση τον τελευταίο καιρό... :(

Cheerio,

Durandal 1707

Edit: καλά, τώρα είδα το post του r_boris... Ninja'ed  :ninja:

Δεν βλέπω τι κερδίζεις Staretak πολλαπλασιάζοντας με $ e^x$ απλώς καταλήγεις να μελετάς την εξίσου δύσκολη εντελώς μη γραμμική Δ.Ε $ g'(x)\equal{}e^{\minus{}x}g(2x)$ αν θέλεις γράψε τι έχεις βρεί
Παναγιώτη προς το παρόν εγώ δεν βρήκα τίποτα αλλά η μελέτη είναι σύντομη και πρόχειρη, έχω   την εντύπωση ότι κάτι καλό βρίσκεται εδώ!!

[quote="r_boris"]Δεν βλέπω τι κερδίζεις Staretak πολλαπλασιάζοντας με $ e^x$ απλώς καταλήγεις να μελετάς την εξίσου δύσκολη εντελώς μη γραμμική Δ.Ε $ g'(x) \equal{} e^{ \minus{} x}g(2x)$ αν θέλεις γράψε τι έχεις βρεί
Παναγιώτη προς το παρόν εγώ δεν βρήκα τίποτα αλλά η μελέτη είναι σύντομη και πρόχειρη, έχω   την εντύπωση ότι κάτι καλό βρίσκεται εδώ!![/quote]

Basika douleuoume eite me tin $ f$ eite me tin $ F$. Kai gia tis dio to idio epixeirima douleuei(kai malista kai gia mia terastia oikogenia sinartisewn-pou akoma den exw prosdiorisei plhrws).Loipon, opws eipes, douleuontas gia tin $ F$ thetontas $ e^xF(x) \equal{} g(x)$  pernoume:
$ g'(x) \equal{} e^{ \minus{} x}g(2x)$ (*)

Twra an thewriseis oti i $ g$ (pou einai smooth) oti exei peperasmeno plithos lisewn sto [-1.0] (profanws $ F(0) \equal{} 0$, opote $ g(0) \equal{} 0$) tote briskeis ena h  arnitiko wste sto [h,0) i $ g$  den exei rizes kai ara diatirei prosimo. Omos an perioristeis tora sto [h/2,0) ekei to $ g(2x)$ diatirei prosimo kai logw tis (*) exoume kai antistoixi monotonia. Auto, omws, einai poli eukolo na dei kaneis oti einai atopo (dioti $ g(0) \equal{} 0$)

Twra eipa auto to parapanw me tis fragmenes rizes, dioti (opos proanaferthike) exoume rizes me accumulation point, opote an exoume tin analytikotita tote exoume kateuthian oti i $ f$ einai pantou 0 sto sinektiko $ R$. Oso afora gia ti periptwsi pou exoume mono smoothness, anti gia kati allo, tha grapsw to eksis:

Blepoume oti an exoume analytikotita tote poli eukola prosdiorizoume tin $ f$. Wstoso gia tin smooth case i katastasi den einai akrivws etsi. Gia auto to logo i analytikotita einai ena para poli simantiko thema sta Mathimatika. Iparxoun theorimata, opws i Elliptic regularity pou lene oti an exoume elliptic PDE kai lusi tis pou einai smooth tote einai kai analytiki. I ligo pio geometrika exei px apodeixthei i analytiki eikasia Caratheodori alla oxi i geniki. I eikasia leei oti kathe epifaneia pou einai omoiomorfi tis sfairas exei dio toulaxiston omfalika simeia. Genika i idea gia tin proseggisi autis einai na kataskeuastei katalliko dianismatiko pedio pou tha midenizetai mono sta omfalika (opoia kai an einai auta). An deixthei oti o index ton krisimon simion einai to poli ena, tote apo to theorima Poincare-Hopf exoume teliwsei afou i sfaira exei xaraktiristiki Euler 2. WSTOSO, i anisotita gia ton index mporei na deixthei mono sti periptwsi pou exoume analytiki epifania (plus kapoia kurtotita) se ena sunolo peripou 300 selidwn..heh. I geniki periptwsi einai akoma open...

PS. 1)Apla na anaferw oti poli tis modas einai ta generic thewrhmata. Etsi poli prosfata apodeixtike i generic eikasia Caratheodori apo ton Feldmann.
2)Ta teleutaia den einai kai toso ''Gia to Likio''...

[quote="r_boris"]Παναγιώτη προς το παρόν εγώ δεν βρήκα τίποτα αλλά η μελέτη είναι σύντομη και πρόχειρη, έχω   την εντύπωση ότι κάτι καλό βρίσκεται εδώ!![/quote]

Sorry, είδα εντελώς επιπόλαια το πρόβλημα και είχα κάτι εντελώς διαφορετικό στο μυαλό μου όταν είπα ότι η απάντηση είναι εύκολη. Άκυρο το σχόλιο.

Cheerio,

Durandal 1707

BTW, μία μικρή διόρθωση γιατί έχει ξεφύγει ένα $ 2$ από την παραγώγιση του ολ/τος με πάνω όριο το $ 2x$: η συναρτησιακή που μου προκύπτει είναι $ F'(x) \equal{} \mathbf{2} F(2x) \minus{} F(x)$ και, αναλόγως: $ g'(x) \equal{} 2e^{ \minus{} x}g(2x)$.

Durandal 1707

[quote="Durandal"]BTW, μία μικρή διόρθωση γιατί έχει ξεφύγει ένα $ 2$ από την παραγώγιση του ολ/τος με πάνω όριο το $ 2x$: η συναρτησιακή που προκύπτει είναι $ F'(x) \equal{} \mathbf{2} F(2x) \minus{} F(x)$ και, αναλόγως: $ g'(x) \equal{} 2e^{ \minus{} x}g(2x)$.

Durandal 1707[/quote]

O r_boris apo oti katalava anaferetai stin paragousa tis $ f$ pou einai i $ F$ kai oxi stin idia tin $ f$. Etsi gia tin $ f$ isxuei oti $ f'(x) \equal{} 2f(2x) \minus{} f(x)$ kai gia tin $ F$ isxuei $ F'(x) \equal{} F(2x) \minus{} F(x)$. H idia analysi douleuei kai gia tin $ F$ kai gia tin $ F' \equal{} f$ kai gia tin $ f'$ ktlp...

[quote="staretak"][quote="Durandal"]BTW, μία μικρή διόρθωση γιατί έχει ξεφύγει ένα $ 2$ από την παραγώγιση του ολ/τος με πάνω όριο το $ 2x$: η συναρτησιακή που προκύπτει είναι $ F'(x) \equal{} \mathbf{2} F(2x) \minus{} F(x)$ και, αναλόγως: $ g'(x) \equal{} 2e^{ \minus{} x}g(2x)$.

Durandal 1707[/quote]

O r_boris apo oti katalava anaferetai stin paragousa tis $ f$ pou einai i $ F$ kai oxi stin idia tin $ f$. Etsi gia tin $ f$ isxuei oti $ f'(x) \equal{} 2f(2x) \minus{} f(x)$ kai gia tin $ F$ isxuei $ F'(x) \equal{} F(2x) \minus{} F(x)$.[/quote]

Ωωωωωωωωωωωωωωωωωωωχ, ΜΑ ΑΝ ΕΙΝΑΙ ΠΟΤΕ ΔΥΝΑΤΟΝ...... :( :( :(

Ok, guys, παραιτούμαι, προφανώς δε με θέλει το άθλημα στο συγκεκριμένο πρόβλημα. 2/2 βλακείες λέω επειδή είμαι απρόσεχτος και διαβάζω διαγώνια το τί λέγεται και δεν κάθομαι να προσέξω σε τί αναφέρεται ο καθένας...

Τί να πω, sorry, το μυαλό μου είναι αλλού απ'ό,τι φαίνεται :( Cheerio,

Durandal 1707

Γεια σας,

Βάζω μία άσκηση που βρήκα πριν λίγο στο βοήθημα του Μπαϊλάκη, "ΘΕΜΑΤΑ ΜΑΘΗΜΑΤΙΚΩΝ", σελίδα 104, άσκηση 119. Μοιάζει (τουλάχιστον οπτικά), με αυτή του staretak. Νομίζω πως η άσκηση του staretak είναι σαφώς πιο δύσκολη, αλλά παραθέτω και αυτή για να τη δείτε:

Δίνεται συνάρτηση $ f: \mathbf{R}\rightarrow\mathbf{R}$, δύο φορές παρ/σιμη στο $ R$, για την οποία ισχύει:

$ f(x) \equal{} \int_{0}^{1}f(x \plus{} t)dt$

Να δείξετε ότι η $ f''$ τέμνει τον $ x'x$ άπειρες φορές.

Autin tin askisi tin eixa dei prin kanw tin parapanw...euxaristw pou tin anafereis...wstoso otan to pano orio einai 1 tote einai poli pio eukoli...

Αφού παραγωγίσεις την $ f(x)\equal{}\int_x^{x\plus{}1}f(u)du$ ΘΜΤ για την $ f$ σε διαστήματα $ [k,k\plus{}1] ,k \in Z$ και μετά τα αντίστοιχα Rolle για την $ f'$

Mias kai egrapse o Stefanos gia elliptic regularity se P.D.E's sas grafo tin parakato askisi, i opoia einai basika to elliptic regularity se 1 diastasi, alla den xreiazete na gnorizei kaneis p.d.e's gia na tin lisei :

Na brethoun oles oi sinexeis sinartiseis $ f: R \rightarrow R$ etsi oste $ \int_{\minus{}\infty}^{\plus{}\infty}f\cdot g''dx\equal{}0$ gia kathe smooth $ g: R \rightarrow R$ i opoia einai isi me miden ekso apo kapoio fragmeno kleisto diastima.

Osoi asxolithoun me to problima tha exoun parei mia geusi pros tin apodeiksi orismenon themeliodon theorimaton stin theoria ton merikon diaforikon eksisoseon.

Stergios

[quote="Anto"]Mias kai egrapse o Stefanos gia elliptic regularity se P.D.E's sas grafo tin parakato askisi, i opoia einai basika to elliptic regularity se 1 diastasi, alla den xreiazete na gnorizei kaneis p.d.e's gia na tin lisei :

Na brethoun oles oi sinexeis sinartiseis $ f: R \rightarrow R$ etsi oste $ \int_{ \minus{} \infty}^{ \plus{} \infty}f\cdot g''dx \equal{} 0$ gia kathe smooth $ g: R \rightarrow R$ i opoia einai isi me miden ekso apo kapoio fragmeno kleisto diastima.

Osoi asxolithoun me to problima tha exoun parei mia geusi pros tin apodeiksi orismenon themeliodon theorimaton stin theoria ton merikon diaforikon eksisoseon.

Stergios[/quote]
Περιμένουμε την λύση αν και υποψιάζομαι ότι $ f\equal{}0$?

Ipodeiksi :

Mia proti skepsi einai na doume ti simbainei ean i f einai smooth, i esto paragogisimi. Eufkola tha doume oti tote exoume polles liseis (opote i apantisi den einai f=0)
I Elliptic regularity leei ligo poli oti i f prepei na einai isi me mia smooth sinartisi (ektos enos sinolou metrou miden), opote exoume brei oles tis liseis.

Pos omos mporoume na to apodeiksoume auto ?

Stergios


When sad, I usually listen to sad songs... and you?

I do too. It always makes you feel better.

i listen to metal instead of your depeche mode/three days grace/KRAWWLLINNGG

[quote="bubka"]I study - because when I am sad, it is generally because I have just sat down to study. :([/quote]
LOL

and yea, go metal
metal is good mental food.

You should have broken knuckles, by now.

When I'm sad, I do what I do when I'm angry: hold it in as long as possible. When I can bear it no longer, I whisper one little word, and I don't want to say it here.

[quote="1=2"]You should have broken knuckles, by now.
[/quote]

Not really, lol

WhenI'm sad I do math, generally a olympiad problem because you have concentrate hard on the problem, and forget that your sad.(I normally end up getting mad at the fact that I cant do anything in the problem though lol)

When I'm sad and not studying and not at my computer (the last one is just a formal inclusion, that i am sad generally impies that i am not at my computer), i sit or lie down somewhere alone (and preferably a bit cold) and think, i feel sadder, i enjoy the sensation, the slight shivers ...

I cut pneguinintegral's wrist with a spork.

[quote="Treething"][quote="PenguinIntegral"]I cut treething's wrist with a razor.[/quote]

man so that's where those cuts came from[/quote]
dude haven't you ever played Dungeons and Dragons

under ingredients for fireball it says "blood of mathcounts champion"

BTW, [url=http://www.adamchance.com/funny.htm]I put on my robe and wizard hat... [inappropriate][/url]

Other things I do are read, write, or just lay there

I watch Pokemon  & Digimon, and then write analytical essays about them.

Part of my statement was false.


fractional part
modular arithmetic
geometry
algebra
polynomial
USAMO

Let $p > 2$ be a prime and let $a,b,c,d$ be integers not divisible by $p$, such that
\[ \left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\} + \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2  \]
for any integer $r$ not divisible by $p$. Prove that at least two of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$ are divisible by $p$.
(Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)

[quote="spdf"]There are a solution not using primitive root of unity appear in a book of Zuming Feng[/quote]

Can you tell the book please? Thanks! :)

Define R(x) be the remainder obtained upon dividing x by p.

Let a' be positive number such that aa' = 1 (mod p), multiply a, b, c, d by a' we have aa', ba', ca', da'. Let $ R(ba') \equal{} k_{1}, R(ca') \equal{} p\minus{}k_{2}, R(da') \equal{} p\minus{}k_{3}$, WLOG let $ R(ba')\leq R(ca')\leq R(da')$, or $ k_{1}\leq p\minus{}k_{2}\leq p\minus{}k_{3}$.

so $ 1\plus{}k_{1}\equal{} k_{2}\plus{}k_{3}$--------------------**

Now for the sake of contradiction, assume $ k_{3}> 1$.

Therefore from ** we see that $ k_{2}, k_{3}< k_{1}$, now WLOG let
$ k_{3}\leq k_{2}< k_{1}$

Lemma: $ k_{1}< (p\plus{}1)/2$

Proof: By definition $ k_{1}$ is the smallest among $ (k_{1}, p\minus{}k_{2}, p\minus{}k_{3})$. Assume
$ k_{1}\geq (p\plus{}1)/2$, then
$ R(2k_{1})\leq R(k_{1})\minus{}1$, (If$ 2k_{1}\minus{}p>k_{1}$, then$ k_{1}>p$ to contradiction)
$ R(2(p\minus{}k_{2}))\leq R(k_{2})\minus{}1$, etc...  So let $ r \equal{} 2$,

$ 2\plus{}R(2k_{1})\plus{}R(2(p\minus{}k_{2}))\plus{}R(2(p\minus{}k_{3})) < R(1)\plus{}R(k_{1})\plus{}R(p\minus{}k_{2})\plus{}R(p\minus{}k_{3}) \equal{} 2p$, contradiction. Hence the lemma follows.

Note that when we multiply 1, k_1, p-k_2, p-k_3 by positive integer r<p, then
$ R(rk_{1}) \equal{} rk_{1}\minus{}[rk_{1}/p]p$

$ R(r(p\minus{}k_{2})) \equal{} ([rk_{2}/p]\plus{}1)p\minus{}rk_{2}$
$ R(r(p\minus{}k_{3})) \equal{} ([rk_{3}/p]\plus{}1)p\minus{}rk_{3}$

$ R(r)\plus{}R(rk_{1})\plus{}R(r(p\minus{}k_{2}))\plus{}R(r(p\minus{}k_{3})) \equal{} 2p$, from ** we deduce

 $ [rk_{2}/p]\plus{}[rk_{3}/p] \equal{} [rk_{1}/p]$-----------------(*)

Let $ a$ be the smallest positive integer satisfying $ [ak_{1}/p] \equal{} 1$, and simultaneously $ [ak_{2}/p]$ must also be 1 to satisfy (*) (since $ k_{2}\geq k_{3}$).

If $ 0 < s < k_{1}$, then either $ a$ or $ a\plus{}1$ must be the minimum value for $ m$ so that
 $ [(s\plus{}mk_{1})/p] \equal{} 1$ (This is trivial. since $ s < k_{1}$, the minimum for $ m$ is at least $ a$, and, $ s\plus{}k_{1}> k_{1}$, so the minimum is at most $ a\plus{}1$).

Similarly, If $ 0 < s < k_{2}$, then either $ a$ or $ a\plus{}1$ must be the minimum value for m so that $ [(s\plus{}mk_{2})/p] \equal{} 1$.

Now let $ m$ be the largest integer value for $ [rk_{1}/p]$, where $ r\leq [p/k_{3}]$. Let $ n$ be the smallest integer sastifying $ [nk_{1}/p] \equal{} m$. Since $ n\leq [p/k_{3}]$, $ [nk_{3}/p] \equal{} 0$ so by (*), $ [nk_{2}/p] \equal{} m$.

Claim: If $ t$ is the smallest integer so that $ [tk_{1}/p] \equal{} m\plus{}1$, then $ t$ is either $ n\plus{}a$ or $ n\plus{}a\plus{}1$.

Proof: Note that $ R(nk_{1}) < k_{1}$ (else $ R(nk_{1})\minus{}k_{1}\geq 0$ and contradict well-ordering principle ). $ nk_{2}\equal{} mp\plus{}R(nk_{1})$, so
$ [(nk_{2}\plus{}(t\minus{}n)k_{2})/p] \equal{} m\plus{}[(R(nk_{1})\plus{}(t\minus{}n)k_{2})/p]$, minimum for $ t\minus{}n$ is either $ a$ or $ a\plus{}1$, so miminum for $ t$ is either $ a$ or $ a\plus{}1$.

Similarly the smallest integer $ t$ satisfying $ [tk_{2}/p] \equal{} m\plus{}1$ is either $ n\plus{}a$ or $ n\plus{}a\plus{}1$.

Now consider setting $ r$ to be $ [p/k_{3}]\plus{}1$ (since $ k_{3}> 1$ by assumption, $ [p/k_{3}]\plus{}1 < p$.)

So $ [k_{3}r/p] \equal{} 1$, and the only way to satisfy (*) is if $ [k_{1}r/p] \equal{} m\plus{}1$, and
$ [k_{2}r/p] \equal{} m$, which implies $ r \equal{} n\plus{}a$ (if $ r \equal{} n\plus{}a\plus{}1$, both $ [k_{1}r/p], [k_{2}r/p] \equal{} m\plus{}1$ and (*) is violated).
Consider $ r \equal{} [p/k_{3}]\plus{}2 \equal{} n\plus{}a\plus{}1$, $ [rk_{2}/p] \equal{} m\plus{}1$, and
By Lemma $ R((n\plus{}a)k_{1})\plus{}k_{1}< 2k_{1}< p$, so
$ [(n\plus{}a\plus{}1)k_{1}/p] \equal{} [m\plus{}1\plus{}(R((n\plus{}a)k_{1})\plus{}k_{1})/p ] \equal{} m\plus{}1$,

Similarly, $ [rk_{3}/p] \equal{} [1\plus{}(R((n\plus{}a)k_{3})\plus{}k_{3})/p ] \equal{} 1$ ,

and (*) is violated, contradiction. It follows $ k_{3}\equal{} 1$, so
$ R(da') \equal{} p\minus{}1$ => $ R(da')\plus{}R(aa') \equal{} p$ => $ p|(a'd\plus{}aa')$ => $ p|(a\plus{}d)$

 QED

[hide="Idea for another solution"]Say $ g$ is a generator $ \pmod p$. Consider $ \frac {1}{p}$ mod $ g$; it repeats with period $ g \minus{} 1$. Let the repeating unit, an integer, be $ n$. Multiplying the fraction by $ a$, $ b$, $ c$, and $ d$ is equivalent to rotation of the digits of $ n$, since these are all powers of $ g$ mod $ n$. If we add the four fractions together, they must sum to $ 2$; we can analyze the first digit and last digit of the repeating unit to see that every four corresponding digits add to $ 2g \minus{} 2$; the multiplication by $ r$ is also equivalent to simultaneous rotation of all four fractions' repeating units, so anything true about the last digit or first digit must be true about all digits.[/hide]

I have no idea if this is useful.

Solution by dysfunctionalequations and me.

[hide]In the context of this proof, for any integer $x$, let $[x]$ denote the remainder when $x$ is divided by $p$ (so $0 \leq [x] < p$.) The condition of our problem is equivalent to $[ra] + [rb] + [rc] + [rd] = 2p$ for $r$ between 1 and $p-1$. Call any 4-tuple $(w,x,y,z)$ [i]good[/i] if it satisfies the conditions of this problem. We observe the following four facts:

[list][*] $p | [a] + [b] + [c] + [d]$. This follows from setting $r=1$.
[*] If $p | a + b \iff p | c+d$, $p | a + c \iff p | b+d$, and $p | a+d \iff p | b + c$, i.e., it is sufficient to show that $p | a + b, a + c$, or $a+d$. This follows from the first observation.
[*] $(w,x,y,z)$ is good if and only if $(rw, rx, ry, rz)$ is good, for any $p \not | r$.
[*] If $(w,x,y,z)$ is good, then exactly two of $[w], [x], [y], [z]$ are larger than $(p-1)/2$. To show this, we first let $Q(t) = 0$ if $1 \leq [t] \leq (p-1)/2$ and $Q(t) = 1$ if $(p-1)/2 < [t] < p$, and we note that for any $t$, $[2t] - [t] = [t] - Q(t)p$. Now,
\begin{align*}
0
&= (2p - 2p) \\
&= ([2w] + [2x] + [2y] + [2z]) - ([w] + [x] + [y] + [z]) \\
&= ([2w] - [w]) + ([2x] - [x]) + ([2y] - [y]) + ([2z] - [z]) \\
&= [w] + [x] + [y] + [z] - p(Q(w) + Q(x) + Q(y) + Q(z)) \\
&= 2p - p(Q(w) + Q(x) + Q(y) + Q(z)),
\end{align*}
whence $Q(w) + Q(x) + Q(y) + Q(z) = 2$. The result follows. [/list]

Combining the last two observations, we see that for $r$ between 1 and $p-1$, $Q(ra) + Q(rb) + Q(rc) + Q(rd) = 2$. Taking this mod $p$, we have that $Q(ra) + Q(rb) + Q(rc) + Q(rd) \equiv 2 \pmod{p}$ for all $r \not \equiv 0 \pmod{p}$.

By the Lagrange interpolation theorem, there exists a polynomial $P(x)$ of degree at most $p-2$ such that $P(x) \equiv Q(x) \pmod{p}$ for all $x \not \equiv 0 \pmod{p}$. Let $R(x) = P(x+1) - P(x)$. We have that $R(x) \equiv 0 \pmod{p}$ when $x \equiv 0, 1, \ldots, \frac{p-3}{2}, \frac{p+1}{2}, \ldots, p-2$ and $R((p-1)/2) \not \equiv 0 \pmod{p}$, so the degree of $R$ is exactly $p-3$. Hence, the degree of $P(x)$ is exactly $p-2$.

The polynomial $S(r) = P(ra) + P(rb) + P(rc) + P(rd)$ is congruent to 2 mod $p$ for $p-1$ values. Since the degree of $S$ is $p-2$, $S-2$ must be the zero polynomial mod $p$. Since $S$ has degree $p-2$, the coefficient of $r^{p-2}$in $P(r)$ is nonzero mod $p$. Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$, or $\frac{1}{a} + \frac{1}{b} \equiv -\frac{1}{c} - \frac{1}{d} \pmod{p}$.

From the first observation, we have $a+b \equiv -(c+d) \pmod{p}$. If $p | a + b$, we are done, so we may suppose that $a+b \not \equiv 0 \pmod{p}$. Hence, from $\frac{1}{a} + \frac{1}{b} \equiv -\frac{1}{c} - \frac{1}{d} \pmod{p}$, we get $ab \equiv (-c)(-d) \pmod{p}$. Therefore, $a$ and $b$ are both the roots of the polynomial $(x+c)(x+d)$ mod $p$, so $p | a + c$ or $p | a + d$, which completes our proof.[/hide]

My solution to this problem as follows:

From the assumption we have: $\frac{r(a+b+c+d)}{p}=2+[\frac{ra}{p}]+[\frac{rb}{p}]+[\frac{rc}{p}]+[\frac{rd}{p}]$ is an integer.

And we get $a+b+c+d$ is divisible by $p$.

[u]Now, we work with modulo $p$[/u]. Let $a+b=x,a+c=y,a+d=z$ leads $c+d=-x,b+d=-y,b+c=-z$

We have: $2xyz=(a+b)(a+c)(a+d)+(b+c)(c+d)(d+b)$

or $2xyz=a^2(a+b+c+d)+a(bc+cd+db)+bcd+(b+c)(c+d)(d+b)$

But $a=-(b+c+d)$ so $a(bc+cd+db)+bcd+(b+c)(c+d)(d+b)=0$.

From which we get $p$ divides $2xyz$ but $p>2$ is prime, so:  or $p|x$ or $p|y$ or $p|z$

And we assume $p|x$. hence $a+b$ and $c+d$ are divisible by $p$. Q.E.D

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

With the solution above, we can replace 2 by 1 or 3. A very nice problem! :lol:  :lol:

[quote="LoveMath4ever"]

[u]Now, we work with modulo $p$[/u]. Let $a+b=x,a+c=y,a+d=z$ leads $c+d=-x,b+d=-y,b+c=-z$

We have: $2xyz=(a+b)(a+c)(a+d)+(b+c)(c+d)(d+b)$
[/quote]

That's not true, because LHS is actually 0?

@bomb Your complex solution is very very beautiful.But I still have something to understand.I can't come up with such beautiful complex solution,Idon't know how it was thinked about.Can you say something about your thought when you were solving the problem?

[quote=bomb]Courtesy of Kalva...

n denote the residue of n mod p, so n = 0, 1, 2, ... , or p-1. Thus {na/p} = na/p, and we have that na + nb + nc + nd = 2p for n not a multiple of p.

Let ω be a complex pth root of 1. We show first that ω + 2ω2 + 3ω3 + ... + (p-1)ωp-1 = p/(ω - 1). Suppose the sum is S. Then (1 - 2ω + ω2)S = ω - pωp + (p-1)ωp+1 (we need only look at the two lowest and two highest powers - the others all cancel because k - 2(k-1) + (k-2) = 0) = p(ω - 1). Hence S = p/(ω - 1).

Take residues a', b', c', d', so that aa' = bb' = cc' = dd' = 1 mod p. Then for any integers m, n we have mnaa' = mn mod p. Hence -ma' na = -mn mod p. Hence ω-ma' na = ω-mn. So na ω-ma' na = na ω-mn. Similarly for b, c, d. Take n not a multiple of p and add the four equations to get: na ω-ma' na + nb ω-mb' nb + nc ω-mc' nc + nd ω-md' nd = 2p ω-mn.

As n runs through 1, 2, ... , p-1, each of na, nb, nc, nd runs through a complete set of non-zero residues. If we take m to be not a multiple of p, then so does -mn, so adding the equations for n = 1, 2, ... , p-1, we get na ∑ kω-a'mk + ∑ k ω-b'mk + ∑ kω-c'mk + ∑ kω-d'mk = 2p(ω + ω2 + ... + ωp-1) = -2p.

Since ω-a'm is also a complex pth root of 1, we have ∑ kω-a'mk = p/(ω-a'm - 1) and similarly for the other terms. So the equation becomes: 1/(ω-a'm - 1) + 1/(ω-b'm - 1) + 1/(ω-c'm - 1) + 1/(ω-d'm - 1) = -2.

Multiplying through by (ω-a'm - 1)(ω-b'm - 1)(ω-c'm - 1)(ω-d'm - 1), expanding and simplifying, we get 2 + ω-a'm-b'm-c'm + ω-a'm-b'm-d'm + ω-a'm-c'm-d'm + ω-b'm-c'm-d'm = ω-a'm + ω-b'm + ω-c'm + ω-d'm + 2ω-a'm-b'm-c'm-d'm (*). Note that this equation is also true for m = 0 (when it is just 5 = 5). Now sum the equations for m = 0, 1, 2, ... , p-1. We have ∑ ωk = p if k is a multiple of p, and 0 otherwise. Obviously the first term ∑ 2 gives 2p and the other terms on the lhs give non-negative sums, so ∑ lhs is at least 2p. The first four terms on the rhs all have zero sum, so the last term must have sum 2p, so a' + b' + c' + d' must be a multiple of p. Thus (*) becomes ωa'm + ωb'm + ωc'm + ωd'm = ω-a'm + ω-b'm + ω-c'm + ω-d'm. Multiply through by ω-a'm and sum for m = 0, 1, 2, ... , p-1. The first term on the lhs has sum p and the others have non-negative sum. The first term on the rhs has zero sum, so one of the others must have positive sum. Hence p divides at least one of (a'+b'), (a'+c'), (a'+d'). Without loss of generality it divides a' + b'. In other words a' + b' = 0 mod p. Multiplying by ab, we get a + b = 0 mod p.

Bomb[/quote]

https://mks.mff.cuni.cz/kalva/usa/usoln/usol993.html

[quote=Delray][quote=Zhero]
The polynomial $S(r) = P(ra) + P(rb) + P(rc) + P(rd)$ is congruent to 2 mod $p$ for $p-1$ values. Since the degree of $S$ is $p-2$, $S-2$ must be the zero polynomial mod $p$. Since $S$ has degree $p-2$, the coefficient of $r^{p-2}$in $P(r)$ is nonzero mod $p$. Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$, or $\frac{1}{a} + \frac{1}{b} \equiv -\frac{1}{c} - \frac{1}{d} \pmod{p}$.
[/quote]

In particular, it isn't clear why this is true:
[quote=Zhero]
Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$
[/quote][/quote]

collab with janabel

Actually the fact that S(r) is the zero polynomial means that S(r) = 0, which means everything about it equals 0. Also since it is a p-2 degree polynomial, we can just take the p-2 degree terms and have that ra^{p-2} + rb^{p-2} + rc^{p-2} + rd^{p-2} = 0. Setting r =1 results in the desired equation.

[quote=vsathiam]Actually the fact that S(r) is the zero polynomial means that S(r) = 0, which means everything about it equals 0.[/quote]
Wow much rigorous.

Solution from [i]Twitch Solves ISL[/i]:

First of all, we apparently have $r(a+b+c+d) \equiv 0 \pmod p$ for every prime $p$, so it automatically follows that $a+b+c+d \equiv 0 \pmod p$. By scaling appropriately, and also replacing each number with its remainder modulo  $p$, we are going to assume that \[ 1 = a \le b \le c \le d < p. \] We are going to prove that $d = p-1$, which will solve the problem.

[b][color=red]Claim:[/color][/b]  	For each integer $r = 1, 2, \dots, p-1$ we have 	\[ 2(r-1) = \left\lfloor \frac{rb}{p} \right\rfloor 		+ \left\lfloor \frac{rc}{p} \right\rfloor 		+ \left\lfloor \frac{rd}{p} \right\rfloor. \]
[i]Proof.[/i]  	By plugging in $r=1$ to the given we have $a+b+c+d=2p$. 	Now, we have 	\[ 2 = \sum_{\text{cyc}} \left( \frac{ra}{p} 		- \left\lfloor \frac{ra}{p} \right\rfloor \right) \] 	and since $a+b+c+d=2p$ the conclusion follows. $\blacksquare$

We vaguely outline the approach now, before giving a formalization. Imagine the interval $[0,1]$. One by one, for each $r = 1, 2, 3, \dots, p-1$, we mark the fractions with denominator $r$ on this number line; the resulting pictures may be better known as [i]Farey fractions[/i]. At each step, we can place the three numbers $b/p$, $c/p$, $d/p$ into one of the resulting sub-intervals. Our goal is to show that $d/p$ is always in the rightmost interval, while $b/p$ and $c/p$ are always to the right of symmetrically mirrored points. An example of a possible diagram is shown below (not to scale).
[asy]
	unitsize(11cm);
	usepackage("amsmath");
	real h = 0.05;
	draw( (0,0)--(1,0) );
	void tick(string s, real x, real t, pen p) {
		label(s, (x,-t), dir(-90), p);
		draw((x,-t)--(x,t), p);
	}
	tick("$0$", 0, h, black);
	tick("$1$", 1, h, black);
	tick("$\frac{1}{2}$", 1/2, h, black);
	tick("$\frac{1}{3}$", 1/3, 0.7*h, blue);
	tick("$\frac{2}{3}$", 2/3, 0.7*h, blue);
	tick("$\frac{1}{4}$", 1/4, 0.5*h, deepgreen);
	tick("$\frac{3}{4}$", 3/4, 0.5*h, deepgreen);
	tick("$\frac{1}{5}$", 1/6, 0.3*h, grey);
	tick("$\frac{2}{5}$", 2/5, 0.3*h, grey);
	tick("$\frac{3}{5}$", 3/5, 0.3*h, grey);
	tick("$\frac{4}{5}$", 5/6, 0.3*h, grey);
	dot("$\boxed{\frac{b}{p}}$", (0.3,0), dir(90), red);
	dot("$\boxed{\frac{c}{p}}$", (0.72,0), dir(90), red);
	dot("$\boxed{\frac{d}{p}}$", (0.95,0), dir(90), red);
[/asy]
In symbols, It will be enough to prove the following.
[b][color=red]Claim:[/color][/b]  	For each $r = 1, 2, \dots, p-2$ we have 	$\frac{r-1}{r} < \frac dp < 1$.
Equivalently, for each $r = 1, 2, \dots, p-2$ 	we have $\left\lfloor \frac{rb}{p} \right\rfloor 	+ \left\lfloor \frac{rc}{p} \right\rfloor = r-1$.
[i]Proof.[/i]  	Assume this is not true and take the minimal counterexample $r > 1$. 	Then evidently 	\[ r-1 > \left\lfloor \frac{rd}{p} \right\rfloor 		\ge \left\lfloor \frac{(r-1)d}{p} \right\rfloor 		= r-2.  \] 	Now, we have that 	\[ 2(r-1) = \left\lfloor \frac{rb}{p} \right\rfloor 		+ \left\lfloor \frac{rc}{p} \right\rfloor 		+ \underbrace{\left\lfloor \frac{rd}{p} \right\rfloor}_{= r-2}.  \] 	Thus 	$\left\lfloor \frac{rb}{p} \right\rfloor >  	\left\lfloor \frac{(r-1)b}{p} \right\rfloor$, 	and $\left\lfloor \frac{rc}{p} \right\rfloor > 	\left\lfloor \frac{(r-1)b}{p} \right\rfloor$. 	An example of this situation is illustrated below with $r = 7$ 	(not to scale).

[asy]
unitsize(11cm);
	usepackage("amsmath");
		real h = 0.05;
		draw( (0,0)--(1,0) );
		void tick(string s, real x, real t, pen p) {
			label(s, (x,-t), dir(-90), p);
			draw((x,-t)--(x,t), p);
		}
		tick("$0$", 0, h, black);
		tick("$1$", 1, h, black);
		tick("$\frac{1}{2}$", 1/2, h, black);
		tick("$\frac{1}{3}$", 1/3, 0.7*h, blue);
		tick("$\frac{2}{3}$", 2/3, 0.7*h, blue);
		tick("$\frac{1}{4}$", 1/4, 0.5*h, deepgreen);
		tick("$\frac{3}{4}$", 3/4, 0.5*h, deepgreen);
		tick("$\frac{1}{5}$", 1/6, 0.3*h, grey);
		tick("$\frac{2}{5}$", 2/5, 0.3*h, grey);
		tick("$\frac{3}{5}$", 3/5, 0.3*h, grey);
		tick("$\frac{4}{5}$", 5/6, 0.3*h, grey);
		tick("$\frac{1}{6}$", 1/8, 0.3*h, grey);
		tick("$\frac{5}{6}$", 7/8, 0.3*h, grey);
		dot("$\boxed{\frac{b}{p}}$", (0.31,0), dir(90), red);
		dot("$\boxed{\frac{c}{p}}$", (0.725,0), dir(90), red);
		dot("$\boxed{\frac{d}{p}}$", (0.92,0), dir(90), red);
		tick("$\dfrac{2}{7}$", 2/7, 1.5*h, red+1);
		tick("$\dfrac{5}{7}$", 0.7, 1.5*h, red+1);
		tick("$\dfrac{6}{7}$", 0.9, 1.5*h, red+1);

		tick("$\dfrac{7}{8}$", 0.95, 2*h, orange+1);
		tick("$\dfrac{3}{8}$", 0.375, 2*h, orange+1);
		tick("$\dfrac{6}{8}$", 0.75, 2*h, orange+1);
[/asy]
	Right now, $\frac bp$ and $\frac cp$ 	are just to the right of 	$\frac ur$ and $\frac vr$ for some $u$ and $v$ with $u+v=r$. 	The issue is that the there is some fraction just to the right of 	$\frac bp$ and $\frac cp$ from an earlier value of $r$, 	and by hypothesis its denominator is going to be strictly greater than $1$.
It is at this point we are going to use the properties of Farey sequences. 	When we consider the fractions with denominator $r+1$, 	they are going to lie outside of the interval 	they we have constrained $\frac bp$ and $\frac cp$ to lie in. 	Indeed, if right now $\frac ur < \frac bp < \frac st$, 	then Farey theory says the next fraction 	inside the interval $[\frac uv, \frac st]$ 	is $\frac{u+s}{v+t}$, and since $t > 1$, we have $v+t > r+1$. 	In other words, we get an inequality of the form 	\[ \frac ur < \frac bp < \text{something} \le \frac{u+1}{r+1}. \] 	The same holds for $\frac cp$ as 	\[ \frac vr < \frac cp < \text{something} \le \frac{v+1}{r+1}. \] 	Finally, 	\[ \frac dp < \frac{r-1}{r} < \frac{r}{r+1}. \] 	So now we have that 	\[ \left\lfloor \frac{(r+1)b}{p} \right\rfloor 		+ \left\lfloor \frac{(r+1)c}{p} \right\rfloor 		+ \left\lfloor \frac{(r+1)d}{p} \right\rfloor 		\le u + v + (r-1) = 2r-1 \] 	which is a contradiction. $\blacksquare$
Now, since \[ \frac{p-3}{p-2} < \frac dp 	\implies d > \frac{p(p-3)}{p-2} = p - 1 - \frac{2}{p-2} \] which for $p > 2$ gives $d = p-1$.

This was fun... I didn't really use anything besides floors

[hide=Solution]
WLOG $a=1$ (by rescaling $r$ appropriately). Also WLOG $b = \max(b,c,d), c = \min(b,c,d)$; we will show $b=p-1$. The equation turns into $\lfloor \frac{br}{p} \rfloor + \lfloor \frac{cr}{p} \rfloor + \lfloor \frac{dr}{p} \rfloor = 2r-2$ for all $1 \leq r \leq p-1$. Let $f_b(r) = \lfloor \frac{b(r+1)}{p} \rfloor-\lfloor \frac{br}{p} \rfloor$, and define $f_c(r)$ and $f_d(r)$ analogously; these take on $0$ or $1$ as values, and we aim to show given $f_b(r)+f_c(r)+f_d(r) = 2$ for all $1 \leq r \leq p-2$ that $f_b(r) = 1$ for all these $r$.

Assume FTSOC; let $r_0$ be the point closest to $\frac{p-1}{2}$ where $f_b(r) = 0$; call any point $r'$ where $f_b(r') = 0$ a "hole".

Claim: $c < \frac{p}{2}$

Proof: This follows from plugging in $r=1$ into our equation and using the min; as the increment doesn't happen, $c$ doesn't exceed $\frac{p}{2}$.

We require $f_c(r_0) = f_d(r_0) = 1$ now. However, $f_c(r_0-1), f_c(r_0+1) \neq 1$ by the claim, so $f_d(r_0-1) = f_d(r_0+1) = 1$. Due to having $3$ consecutive increments, this implies that $d(r_0+2)-d(r_0-1)>2$, or $d>\frac{2p}{3}$.

Claim: $c < \frac{p}{3}$

Proof: Plug in $r=2$ into our equation now, using $b,d > \frac{2p}{3}$ to see $f_b(2)=f_d(2) = 1$.

Now $f_c(r_0-2),f_c(r_0-1),f_c(r_0+1),f_c(r_0-2) = 0$ by this size bound, so $f_d$ is $1$ from $r_0-2$ to $r_0+2$. This implies $d > \frac{4p}{5}$ by a similar argument.

Now we can perform the same steps above to show $c < \frac{p}{5}$, then $f_d$ is $1$ from $r_0-4$ to $r_0+4$, etc. (widening the hole, sort of), getting $d > \frac{8p}{9}$, and so on; ultimately our streak of $1$'s for $d$ gets longer than $\frac{p}{2}$, forcing $d = p-1$ so $b=p-1$ for size reasons, contradicting the existence of a hole in $f_b$ at $r_0$. Thus, no holes exist and $f_b(r) = 1$ for all $r$, and $b=p-1$. Q.E.D.
[/hide]


ARML
probability
modular arithmetic
counting
distinguishability
linear algebra
matrix

42. Find the remainder when 87! is divided by 89.

A. 0   B. 1    C. 44   D. 87   E. 88

48. (Free Response) A circular dartboard with radius 4 inches has an equilateral triangluar bullseye with edge length of 2 in. What is the probability that 5 out of 50 darts thrown will hit the bullseye? (round your answer to four decimal places)

[I've got as far as the probability for one dart is (Sqrt 3)/(16pi) but I don't know what to do after that.]

36. Ten points P,Q,R,S,T,U,V,W,X, and Y are equally spaced around a circle of radius 1. What is the difference in the lengths of the lines PQ and PS?

A. (1/2)(sqrt(5)+1)    B. 1   C. sqrt(5)-1    D. (1/4)(sqrt(5)+sqrt(2))    E. sqrt(5) - sqrt(2)

35. If A, B, and C are each invertable n-square matrices, then
[img]http://mars.walagata.com/w/bthings-2000/Matrices.JPG[/img]

[hide="42"]
By Wilson's Theorem, $88!\equiv -1 \pmod {89}$. Let $87!\equiv k\pmod {89}$. Then $88!\equiv 88k\equiv -k\pmod {89}$. So $k=\boxed {1}$.
[/hide]

[hide="36"]The answer is 1 (B).  If you want to see how I got it, I'll post that in a little bit, once I have a good diagram to use to show you...[/hide]

[hide="48"]
isn't it just
$\binom{50}{5}\left(\frac{\sqrt{3}}{16\pi}\right)^5\left(\frac{16\pi-\sqrt{3}}{16\pi}\right)^{45}$
because you have to choose 5 distinguishable darts out of 50?
 :|
[/hide]

[hide="35"]B.  The inverse of a matrix is such that putting it on the left of the original matrix multiplies to an identity.  Thus, the inverse of $AC^{-1}=CA^{-1}$, because first A and its inverse cancel, leaving C and its inverse to cancel.  Using this idea, we chug through the expression to get an answer.[/hide]

[hide="48"]
The probability of hitting the dartboard for a given dart is $\frac{\sqrt{3}}{16}$, and the probability of not hitting is $1-\frac{\sqrt{3}}{16}$.  We must arrange 5 "hits" and 45 "misses" in a row to represent the occurrences of the darts.  One ordering represents one possibility.  There are $\binom{50}{5}$ possible permutations.  Thus, the answer is $(\frac{\sqrt{3}}{16})^5(1-\frac{\sqrt{3}}{16})^{45}\binom{50}{5}$[/hide]

For #36, just
[spoiler]

They are the ten roots of unity, so use law of cosines on a 36-72-72 triangle for the shortest side.

[hide="35"]
Just assume $A = [2]$, $B=[3]$, and $C=[5]$ then multiply them like scalars. You end up with $1.5$, which can only be choice B.
[/hide]


Prove that the ortocentre of a triangle whose vertices are $ (a \cos \alpha, a \sin \alpha)$, $ (a \cos \beta, a \sin \beta)$, $ (a \cos \gamma, a \sin \gamma)$ is
$ (a \cos \alpha \plus{} a \cos \beta \plus{} a \cos \gamma, a \sin \alpha \plus{} a \sin \beta \plus{} a \sin \gamma)$

Call $ A(a \cos \alpha, a \sin \alpha), B(a \cos \beta, a \sin \beta), C(a \cos \gamma, a \sin \gamma)$ and $ H(a \cos \alpha \plus{} a \cos \beta \plus{} a \cos \gamma, a \sin \alpha \plus{} a \sin \beta \plus{} a \sin \gamma)$ we have
$ \overrightarrow{AH}(a \cos \beta \plus{} a \cos \gamma,a \sin \beta \plus{} a \sin \gamma)$ and $ \overrightarrow{BC}( \minus{} a \cos \beta \plus{} a \cos \gamma, \minus{} a \sin \beta \plus{} a \sin \gamma)$ then
$ \overrightarrow{AH}.\overrightarrow{BC} \equal{} (a \cos \beta \plus{} a \cos \gamma)( \minus{} a \cos \beta \plus{} a \cos \gamma)$
$ \plus{} (a \sin \beta \plus{} a \sin \gamma)( \minus{} a \sin \beta \plus{} a \sin \gamma)$
$ \equal{} \minus{} a^2 \cos^2 \beta \plus{} a^2 \cos^2 \gamma \minus{} a^2 \sin^2 \beta \plus{} a^2 \sin^2 \gamma \equal{} 0$
Similarly then OK!


How many strings of A,B,C are there that satisfies:
1. There are exactly $ n$ A's
2. There are exactly $ 2n$ total of B's and C's
3. No two adjacent letters are the same

Lets first keep all the A's in a line. We shall insert D's to the right and left of A's and in between them giving rise to many configurations, then color every D either as B or C without violating the adjacency condition.

(i) Space to the left of A's is empty and Space to the right of A's is empty

So we distribute $ 2n$ D's into the gaps between A's so that there is at least one D between any two A's. For a bunch of D's within two A's, they have to be alternately colored C and D or D and C. Hence it depends on the coloring of the first element within the bunch. This can be done in $ 2^{n\minus{}1}{(n\plus{}1)\plus{}(n\minus{}1)\minus{}1\choose n\minus{}1\minus{}1}$ ways.

Continuing mechanically similarly,

(ii) Space left of A's is filled and Space left of A's is filled

We get $ 2^{n\plus{}1}{(n\plus{}1)\plus{}(n\minus{}1)\minus{}1\choose n\minus{}1\minus{}1}$

(iii) Either the left or the right is empty but not both

Adding two symmetric cases, we get $ 2^{n\plus{}1}{n\plus{}n\minus{}1\choose n\minus{}1}$


Let $m$ be a positive integer. Prove that there are positive integers $a,b,k$, so that $a,b$ are odd numbers, $k\geq 0$, and $2m=a^{19}+b^{99}+k.2^{1999}$

Um ... what if m = 2?  How do I express 4 in that form, anyone?

Also, digitalboy, could you please post sources for your problems?

This is not possible. Like JBL said, for $m=2$, we have to have $4=a^{19}+b^{99}+k \cdot 2^{1999}$ - not possible. Perhaps digitalboy1 meant $a^{19}+b^{99}+2 \cdot k^{1999}$. Then it would be possible - $k=a=b=1$.

but then what if $m=3$ ?

$6=a^{19} + b^{99} + 2*k^{1999}$ has solutions for integers?  What are they?

well...for one thing...he said k greater than or equal to zero, yet he also said that k is a positive integer...


LaTeX
limit
calculus
calculus computations

Find lim n -sqrt(n+a)sqrt(n+b) as n tend to infinity. ive narrowed it down for it to be in (0,1). but cant find a better bound. im guessing the value is cero.

The limit of what you wrote down is $ \minus{} \infty$.  Did you mean something else?  Also, what are $ a$ and $ b$?

It isn't $ \minus{} \infty$, in fact it is $ \frac { \minus{} a \minus{} b}{2}$ (if $ a$ and $ b$ are constants).

[hide="Solution"]
$ n \minus{} \sqrt {n \plus{} a}\sqrt {n \plus{} b} \equal{} \frac {n^2 \minus{} (n \plus{} a)(n \plus{} b)}{n \plus{} \sqrt {(n \plus{} a)(n \plus{} b}} \equal{} \frac {( \minus{} a \minus{} b)n \minus{} ab}{n \plus{} \sqrt {n^2 \plus{} (a \plus{} b)n \plus{} ab}} \equal{} \frac { \minus{} a \minus{} b \minus{} \frac {ab}{n}}{1 \plus{} \sqrt {1 \plus{} \frac {a \plus{} b}{n} \plus{} \frac {ab}{n^2}}}$

and taking limits, you get the answer I said above. My steps, in case it isn't clear, is I multiplied and divided by the conjugate, simplified a bit, then divided everything by $ n$ and $ n^2$ inside the radical.
[/hide]

Sorry, I didn't know how to interpret that hanging $ n$.  I have trouble reading jumbled equations.

Please use Latex or learn the basics of it - your topic is quite incomprehensible as it is.
Did you mean $ \displaystyle \lim_{n\rightarrow \plus{}\infty} n\minus{}\sqrt{n\plus{}a} \cdot \sqrt{n\plus{}b}$ ?
If it is so, then you find that it tends to [hide]$ \minus{} \frac {a\plus{}b}{2}$[/hide], but now it's too late (and I'm too tired) to show how to demonstrate it. :sleeping:

Bye
GM

P.S. I see only now caffeineboy's solution and it's the same as mine  :lol:

We can write:

$ \lim_{n\rightarrow \infty}  \left(n\minus{}\sqrt{n\plus{}a} \cdot \sqrt{n\plus{}b}\right)\equal{}$

$ \equal{}\lim_{n \rightarrow \infty} \left(n\minus{} \sqrt{n^2 \plus{} n(a\plus{}b) \plus{} ab}\right)\equal{}$

$ \equal{}\lim_{n \rightarrow \infty} n\left(1\minus{} \sqrt{1 \plus{} \frac{a\plus{}b}{n} \plus{} \frac{ab}{n^2}}\right)\equal{}$

$ \equal{}\lim_{n \rightarrow 0} \frac{1}{n}\left(1\minus{} \sqrt{1 \plus{} (a\plus{}b)\,n  \plus{} a\,b\,n^2 }\right)\equal{}^H$

$ \equal{}\lim_{n \rightarrow 0} \minus{}\frac{a\plus{}b\plus{}2\,a\,b\,n}{2\sqrt{1 \plus{} (a\plus{}b)\,n  \plus{} a\,b\,n^2 }}\equal{}$

$ \equal{}\lim_{n \rightarrow 0} \minus{}\frac{a\plus{}b\plus{}2\,a\,b\cdot 0}{2\sqrt{1 \plus{} (a\plus{}b)\cdot 0  \plus{} a\,b\cdot 0^2 }}\equal{}$

$ \equal{}\minus{}\frac{a\plus{}b}{2}$

where $ \equal{}^H$ denotes the usage of the L'Hopital's rule.


function
AMC
AIME
combinatorics unsolved
combinatorics

The number of partitions of 2007 that have an even number of even parts can be expressed as $ a^b$, where $ a$ and $ b$ are positive integers and $ a$ is prime. Find the sum of the digits of $ a \plus{} b$.

*prod* Is this problem solvable? Here's my progress:

Let $\mathcal{P}$ be the set of all such partitions $P$ of 2007. We wish to find $|\mathcal{P}|$. For a given odd number $o$ and a given even number $e$, let $\mathcal{O}_o$ be the set of all partitions $O$ of $o$ into odd parts, and let $\mathcal{E}_e$ be the set of all partitions $E$ of $e$ into an even number of even parts, with no odd parts. Every partition of 2007 must have an odd number of odd parts, so every such partition $P$ of 2007 corresponds uniquely to an ordered pair $(O,E)$, where $O\in\mathcal{O}_o$ and $E\in\mathcal{E}_{2007-o}$, for some positive odd $o\leq 2007$. In fact, this correspondence is a bijection. It therefore suffices to find
\[\sum_{i=0}^{1003}|\mathcal{O}_{2i+1}|\cdot |\mathcal{E}_{2006-2i}|.\]

More ramblings: The number of ways to partition an integer $n$ into odd parts is equal to the number of ways to partition an integer $n$ into [i]distinct[/i] parts from a special case of Glaisher's theorem, and the generating function for the number of partitions into distinct parts $q(n)$ is
\[\sum_{n=0}^{\infty} q(n)x^n=\prod_{k=1}^{\infty} (1+x^k).\]

I don't think the problem is solvable (as an AIME problem) as stated either. It is if it meant ordered partitions (compositions) instead of unordered partitions, in which case it should be $2^{2005}$ with a final answer of 9.

By the way, "even number of even parts" is a red herring. It's equivalent to "odd number of parts" when you remember the sum of all of the parts has to be odd.

Yes, I realized that as soon as I went to dinner. Thank you. I guess I'll go about finding $A_{000009}(2007)$ from OEIS.


Show that $ [(4 \plus{} \sqrt {11})^n]$ is odd for any positive integer $ n$.

[hide][quote="apratimdefermat"]Show that [$ (4 + √11)^n]$ is odd for any positive integer n.[/quote]
I believe you mean [(4+sqrt{11})^n] as in your topic subject

we denote $ x_1 = 4 + sqrt{11}$
and $ x^2 = 4 - sqrt{11}$
So by Vi-et theorem , $ x_1$ and $ x_2$ are two roots of : $ X^2 - 8X + 5 = 0$
now set $ S_n = x_1^n + x_2^n$
note that  $ S_{n}.S_1 = S_{n + 1} + x_1x_2.S_{n - 1}$ which means $ 8.S_n = S_{n + 1} + 5 S_{n - 1}$(*)
$ S_1 = 8 , S_2 = 54$ therefore S_3 must be even (due to (*) ) and since $ S_2$ and $ S_3$ are even , $ S_4$ must be even (due to (*)) ...etc ... We have S_n is even for all positive integer n

note that $ 0 < (4 - sqrt{11})^n < 1$ so $ S_n - 1 < (4 + sqrt{11})^n < S_n$
so$ [(4 + sqrt{11})^n] = S_n - 1$ and since $ S_n$ is even for all positive integer we conclude that$ [(4 + sqrt{11})^n]$ is odd for all n integer[/hide]


Given a right angle triangle with the hypotenuse a, and the two legs b and c, I need to the lengths of b and c in terms of a.

This is the solution I was told:
[quote]
We can find the angle of this triangle by using tan^{-1}(\frac{1}{2})

Now, since we have the angle, we can use it to find the length of the sides using sine and cosine.
c=acos(tan^{-1}(\frac{1}{2}))=\frac{2a}{\sqrt{5}}

b=asin(tan^{-1}(\frac{1}{2}))=\frac{a}{\sqrt{5}}
[/quote]
Could somebody here please break this down for me and further explain it?

In order to determine $a$ and $b$, we need one further piece of information, $\theta$, the angle opposite $a$.  Now,
\begin{align*}sin\theta &=\frac{a}{c}\\ csin\theta &=a\\ \\ cos\theta &=\frac{b}{c}\\ ccos\theta &=b\\ \end{align*}

Thus, your formulae are $a=csin\theta$ and $b=ccos\theta$.

In general, you need to know $3$ of the aspects of a triangle in order to determine the other aspects.


Prove that for any natural number $ n > 1$

\[ \frac{1}{2}<\frac{1}{n^{2}\plus{}1}\plus{}\frac{2}{n^{2}\plus{}2}\plus{}\ldots\plus{}\frac{n}{n^{2}\plus{}n}<\frac{1}{2}\plus{}\frac{1}{2n}.\]

Please search before posting.........

http://www.mathlinks.ro/viewtopic.php?p=356525#356525

The soln given in the above link is not complete. Could anyone provide a completed one :P .


limit
integration
logarithms
real analysis
real analysis unsolved

Ascertain $ \lim_{x\to\infty}\left(x^3\cdot\int_{\minus{}\frac 1x}^{\frac 1x}\frac {\ln (1\plus{}t^2)}{1\plus{}e^t}\ \mathrm {dt}\right)$ .


How does AMC notify you if you make AIME?

They don't notify you, they notify your organizer at school.

You only get notified for USAMO.

I guess ask the college professor that distributed it. AMC sends emails of the student scores to the teacher so if you know the distributor's email then you can just ask him to forward it instead of taking a trip to the local college.

For the AMC 10 and AMC 12, the AMC office sends two copies of the School Results Report.  The first copy of the report is sent by email to the email address supplied on the school registration form.  The second copy is a paper copy sent by postal mail to the address supplied on the school registration form.  Both copies are addressed  to the School Contest Manager.  Usually, the school contest manager is your math teacher, but it may be a Math Chair, a gifted-and-talented coordinator, a guidance counselor, or even a principal.  Typically, the School Contest Manager is the person who registers the school and takes the responsibility to administer the contests on the day of the contests.

The "they" in your question is the School Contest Manager.  The contest manager has complete details about everything related to your contest and your score.  Talk to your school contest manager first about your score or your AIME qualification.

You can read more details in the AMC 10/AMC 12 Teachers' Manual, Section V, page 6:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote]What if we need a notify or certificate to apply to college or other programs?[/quote]

I have never encountered a situation where a student needed a certificate from the AMC to apply for college or other programs.  There are some college applications, and perhaps some program applications that may ask for self-reporting of an AMC or AIME score if available.

Once again, we have anticipated this question, and detailed instructions to teachers can be found in the AIME Teachers' Manual, Section V. page 5, second paragraph.

Steve Dunbar
MAA Director, American Mathematics Competitions

Wait, why would you take the AIME in college?
Isn't it only for people below 12th grade?

I'm pretty sure that "grade 13" only has meaning in Canada.

[quote="ZzZzZzZzZzZz"]If your school doesn't offer the AMC, you can take it at a nearby university, and I think they also have an option for grade "13".[/quote]
That doesn't mean university students can take it. It's an option for students in other countries.

"Option" is the keyword here. Remember, this is the [b]American[/b] Math Competition. International students can use the AMC as a practice, but I don't believe they can take it officially.

I'm in Japan with a Korean citizenship, but I still took it officially :D

going to take the AIME too.. it's just the USAMO you can't take internationally since it's the qualifying test for the US's IMO team.


In a 100-yard dash, Bob came in second and Albatross came in next to last. What is the least number of students in the race?

2..

assuming that next to last meaning before the last...

ya
next to last means before the last

btw you're correct


inequalities
symmetry
geometry
geometric transformation
reflection

$ ABCD$ is a $ 4\times 4$ square. $ E$ is the midpoint of $ \left[AB\right]$. $ M$ is an arbitrary point on $ \left[AC\right]$.  How many different points $ M$ are there such that $ \left|EM\right|\plus{}\left|MB\right|$ is an integer?

$\textbf{(A)}\ 2  \qquad\textbf{(B)}\ 3  \qquad\textbf{(C)}\ 4  \qquad\textbf{(D)}\ 5  \qquad\textbf{(E)}\ 6$

[hide="Solution"]Let $ E'$ be the reflection of $ E$ over $ AC$.  Let $ E'B \cap AC \equal{} \{M'\}$
$ |E'M| \plus{} |MB|$ varies on the interval $ [4,4\plus{}2\sqrt5]$. From $ A$ to $ M'$, it is $ [4,2\sqrt5]$. From $ M'$ to $ C$, it is $ [2\sqrt5,4\plus{}2\sqrt5]$.
The integer solutions are $ 4,4,5,6,7,8$.
$ \boxed D$.[/hide]


Positive reals $a,b,c$ satisfy $ab+bc+ca=abc$. Prove that:

$\frac{a^4+b^4}{ab(a^3+b^3)} + \frac{b^4+c^4}{bc(b^3+c^3)}+\frac{c^4+a^4}{ca(c^3+a^3)} \geq 1$

It was quite easy problem, my solution:
$ab + bc + ca = abc$ can be rewritten as: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=1$
Standart substitution: $x=\frac{1}{a}$ etc.
So we have $x+y+z=1$ and we want to prove that:
$\sum \frac{x^4+y^4}{x^3+y^3} \geq 1 = x+y+z$
And now it's easy because the inequality:
$\frac{x^4+y^4}{x^3+y^3} \geq \frac{x+y}{2}$ is equivalent to
$(x-y)^2(x^2+xy+y^2) \geq 0$
We write two similar inequalities and then just sum up to finish, QED.

Indeed it was easy though I did it other way (brutally  :D ) - I've just multiplied all out - a lot of terms reduce and we are left with symmetric inequality which instantly follows from Muirhead  :)

We have $ab+ac+bc=abc$ which is equivalent to $\frac1a+\frac1b+\frac1c=1$ Let $x=\frac1a, y=\frac1b, z=\frac1c$
Then the inequality becomes $\displaystyle\sum_{cycl}\frac{x^4+y^4}{x^3+y^3}\geq 1$.

Now, using Cauchy-Schwarz, we have
$(x^{n+1}+y^{n+1})(x^{n-1}+y^{n-1})\geq (x^n+y^n)^2$
which is the same as $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}\geq \frac{x^n+y^n}{x^{n-1}+y^{n-1}}$

hence
$\frac{x^4+y^4}{x^3+y^3}\geq\frac{x^3+y^3}{x^2+y^2}\geq \frac{x^2+y^2}{x+y}\geq \frac{x+y}2$

Do this for the two other fractions as well, sum up, and the inequality is proven
qed

//Edit: Oh no, TomciO was faster  :blush:

Even without the substitution, it's a one-liner  :) (Sums are cyclic)

\[ \frac{abc}{ab+bc+ca}\sum \frac{a^4+b^4}{ab(a^3+b^3)} = \sum \frac{a^4+b^4}{a^3+b^3}\frac{c}{ab+bc+ca} \geq \sum \frac{a+b}{2} \frac{c}{(ab+bc+ca)} \]

which is just equal to 1!

This solution uses [b]Soarer[/b]'s method from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=107998[/url] (I posted there my inequality, similar to this).
At first, lets notice that Chebyshev gives: \[\frac{a^{3}+b^{3}}{2}\cdot\frac{a+b}{2}\leq\frac{a^{4}+b^{4}}{2}\iff\]  \[\iff\frac{a^{4}+b^{4}}{a^{3}+b^{3}}\geq\frac{a+b}{2},\] so we have \[\sum\frac{a^{4}+b^{4}}{ab(a^{3}+b^{3})}\geq\sum\frac{a+b}{2ab}= \frac12\cdot\sum\frac1a+\frac1b= \sum\frac1a= 1.\] BTW, does there exist any non-uglification (i.e. without expanding) solution which doesn't use $\frac{p^{4}+q^{4}}{p^{3}+q^{3}}\geq\frac{p+q}{2}$ (my solution was proving $f(m)\geq f(m-1)$ for $f(m)= \frac{a^{m}+b^{m}}{a^{m-1}+b^{m-1}}$ by Schwarz or $\frac{a}{b}+\frac{b}{a}\geq 2$)?

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=480817&hilit=Poland+inequality:
 $\frac{a^4+b^4}{a^3+b^3} \geq \frac{a+b}{2}$ ,thus it is reduced to
$LHS \geq \frac{1}{2}\left[\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}\right] =\frac{ab+bc+ca}{abc} =1.$

The inequality is equivalent to

$\sum_{cyc}\frac{a^4c+b^4c}{(ab+bc+ca)(a^3+b^3)}\geq 1$

Clearing the denominator, we get

$\sum_{cyc}a^7b^4+\sum_{cyc}a^7c^4\geq \sum_{cyc}a^6b^4c+\sum_{cyc}a^6bc^4$

This holds by Muirhead inequality :D

I shall prove the generalized form:
Let $S_n=\sum\frac{a^n + b^n}{ab(a^{n-1} + b^{n-1})}$ then for all $n > 2$ we have $S_n \geq S_{n-1}$ and $S_2 \geq 1$
We prove the claim by induction on $n$ for $n=2$ we easily have(by C-S)
$$(a^2 + b^2)(a^2 b^2 +b^2 a^2) \geq (ab(a+b))^2 \rightarrow \frac{a^2 + b^2}{ab(a+b)} \geq \frac{1}{2a} + \frac{1}{2b}$$
summing the same inequality for other variables,we reach the desired result.
now assume that claim is true for $n-1$ we will prove it for $n$.by C-S we have:
$$(a^n + b^n)(a^n b^2 + b^n a^2) \geq (ab(a^{n-1} + b^{n-1}))^2 \rightarrow \frac{a^n + b^n}{ab(a^{n-1} + b^{n-1})} \geq \frac{a^{n-1} + b^{n-1}}{ab(a^{n-2} + b^{n-2})} \rightarrow S_n \geq S_{n-1}$$
the last result comes from summing the inequality for other variables.
so we have proved that for all $n \geq 2 , S_n \geq 1$ this also includes $S_4$ which is this problem.


If we had only signs + the limit would be the positive solution of $x^2-x-8=0$, or rather $\frac{\sqrt{33}+1}{2}$. Since the signs have a period (3: - + -), we can square 'n ordining three times and solve the equation, isn't it?

[quote="JohnConstantine"] Since the signs have a period (3: - + -), we can square 'n ordining three times and solve the equation, isn't it?[/quote]
That is to say solve the equation $\sqrt{8-\sqrt{8+\sqrt{8-x}}}=x?$ Try!(providet the
limit is existing.)  :)

i think that we should try to see that the number is a solution to the equation and then viewing it as a function to show that that's maybe the only positive solution. :ninja:
what do you mean by unsolved?

[quote="Albanian Eagle"]i think that we should try to see that the number is a solution to the equation and then viewing it as a function to show that that's maybe the only positive solution. [/quote]
You are right.
[quote="Albanian Eagle"]
what do you mean by unsolved?[/quote]
Actually, I don't know, what do I mean. Sorry. :)

Our job should be the same as here http://www.mathlinks.ro/Forum/viewtopic.php?t=59094


n \geq 2 , x1,x2,...,xn  are  reals  with -1<x1<x2<...<xn<1  and
Σxi^13 = Σxi . Show  that

x1^13*y1 + x2^13*y2+...+xn^13*yn < x1*y1 + x2*y2 + ...+xn*yn
for  any  reals  y1<y2<...<yn

Im  looking  if  there is  a  different  solution , thanks!

Hey,it is the problem in 'Messy but easy' in Proposed & Own Problems'.However,I have done it,so I will write down my soluiton here.
We can be easily to prove that: \sum{i=1 to k} x^13> \sum{i=1 to k} x where k \leq n.
Then the inequality is equal to.
A= \sum (x^13-x)*y <0.And by Abel formular,
A= \sum{i=1 to n}  [(\sum{j=1 to i} (x^13-x))(y_i-y_i+1)]<0.
So we have what we need now. :D .
Hope this is different from your solution.... :maybe:

I allreaddy  knew  that , thank  you!!
I was  just  wondering  if there  is something else...
Thanks again!

You have allready known that solution,so this is the second one.However,this is not mine:
Let t_i=x_i^13-x_i.Note that if -1<x_i \leq 0 then t_i \geq 0,else t_i \leq 0.
The inequality is equal to:
 \sum t_i*y_i<0.We can see that \sum (y_i+m)t_i= \sum y_i*t_i.So we can let y_i \geq 0.
Let k is the number where x_k \leq 0;x_k+1>0.So t_1,....,t_k \geq 0 and
t_k+1,...,t_n<0.We have
 \sum t*y<y_k* \sum{i=1 to k} t_i+y_k+1 \sum{j=k+1 to n} t_j<y_k+1 \sum t=0.
Don't tell me you have known that too. :)


trigonometry algebra polynomial BritishMathematicalOlympiad

Determine the sequences $ a_0 , a_1 , a_2, \ldots$ which satisfy all of the following conditions:

a) $ a_{n \plus{} 1} \equal{} 2a^2_n\minus{}1$ for every integer $ n \ge 0$
b) $ a_0$ is a rational number and
c) $ a_i \equal{} a_j$ for some $ i, j$ with $ i \not \equal{} j$.

[hide]
For $ x > 1$ then $ f(x) \equal{} 2x^2 \minus{} 1 > x$, so if $ a_0 > 1$ then $ a_{n}$ is a strictly increasing sequence and hence fails (c). If $ a_0 < \minus{} 1$, then $ a_1 > 1$ and again fails (c).

So $ \minus{} 1 \le a_0 \le 1$. Let $ a_0 \equal{} \cos \theta$, then the recursion yields $ a_n \equal{} \cos 2^n \theta$. To satisfy (c), we require $ \cos 2^i \theta \equal{} \cos 2^j \theta \Longrightarrow 2^i \theta \equal{} 2^j \theta \plus{} 2\pi n$ or $ 2^i \theta \equal{} (2n \plus{} 1) \pi \minus{} 2^j \theta$. Solving gives $ \theta \equal{} \frac {2n}{2^i \minus{} 2^j} \pi$ or $ \theta \equal{} \frac {2n \plus{} 1}{2^i \plus{} 2^j} \pi$.

We require $ a_0 \equal{} \cos \theta \in \mathbb{Q}$; its well-known that the only rational multiple of $ \pi$ whose cosine is rational are $ 0, \frac {\pi}{3}, \frac{\pi}{2}, \pi, \frac {4\pi}{3}$ (and equivalent), and all of these can be expressed in the desired form, so our solution set is $ a_0 \equal{} \minus{} 1, \frac { \minus{} 1}{2}, 0, \frac 12, 1$.

[/hide]

I'm not sure but when I plug $ a_0\equal{}(\minus{}1, 0,\frac{1}{2})$ in $ a_1\equal{}2a_0 \minus{}1$ I don't get $ a_0\equal{}a_1$.
My solution is
Because $ a_i\equal{}a_j \equal{}> a_i\equal{}a_0$.
$ 2x_0^2\minus{}1\equal{}x_0$
$ \equal{}>2x_0^n\minus{}x_0\minus{}1\equal{}0$
And I get $ x_0$ is $ 1$ and $ \frac{\minus{}1}{2}$

They only have to be equal for at least one i and j, not all i and j.

a0 is rational so a0 = a/q with a, q coprime and q odd OR a0 = a/(2q) with a, 2q coprime.
Suppose a0 = a/q where q is odd
Then a1 = 2a^2/q^2 - 1 = (2a^2 - q^2)/q^2
Since a is coprime to q and q is odd, 2a^2 - q^2 is coprime to q^2 therefore current expression for a1 will not cancel i.e. a1 = b/q^2
Carrying on like this you get a0 = a/q, a1 = b/q^2, a2 = c/q^4, ... all in their lowest terms
If any of these is equal we require q = 1 or q = -1 i.e. a0 is an integer
For a0>1 or a0<-1 sequence diverges so a0 = -1, 0, or 1.

Suppose a0 = a/2q
Then a1 = 2a^2/4q^2 - 1 = (a^2 - 2q^2)/(2q^2)
Since a is coprime to 2q again this is in its lowest terms i.e. a1 = b/2q^2 etc. etc. as above
Hence again we need q = 1 i.e a0 = 1/2 or a0 = -1/2

How does one prove that the only rational multiples of pi whose cosine is rational are 0, pi/3, pi/2, pi, 4pi/3 (and equivalents)?

Show that $ 2\cos \theta$ is an [url=http://en.wikipedia.org/wiki/Algebraic_integer]algebraic integer[/url], in other words, the root of a monic polynomial with integer coefficients. It follows (by RRT) that if $ 2 \cos \theta$ is rational, it must be an integer, and then we must have $ \cos \theta \equal{} \pm 0, \frac{1}{2}, 1$.

we may prove that an= an+1 for any n positive integer ( using the fact and ai = aj then ai+1=aj+1 , and WLOG we put that i>j then all ( a1 a2 a3 a4 .... an ) are equal )
Can we ?


$ a,b,c>0$. Prove that:

$ \frac{a^4\plus{}b^4\plus{}c^4}{a^2\plus{}b^2\plus{}c^2}\plus{}ab\plus{}bc\plus{}ca\geq \frac{4}{3}\left ( a^2\plus{}b^2\plus{}c^2 \right )$

that's wrong. Take $ a,b \to 0$, $ c\equal{}1$.


analytic geometry
complex numbers
geometry proposed
geometry

Prove that $S_{IOH}=\frac{|(a-b)(b-c)(c-a)|}{8r}$.
 :lol:

I have a solution with complex numbers and coordinate method, but I haven't any ideas to solve this problem with geometrical ways.

Have you any Geometrical solutions ?  :wink:

[quote="vardanch"]Have you any Geometrical solutions ?  :wink:[/quote]
No !/Alex


superior algebra
superior algebra unsolved

$n$ is a given natural number. Prove that the number of groups like $G$ which have exactly $n$ conjugate classes is finite.

[Mod.: edited]

[quote="payman_pm"]$n$ is a given natural number. prove that there are finte group like $G$ which has exactly $n$ conjugate class.[/quote]
..
$G=S_{n}$

This doesn't work, $S_{4}$ has exactly $5$ conjugacy classes (represented by $(1),(12),(123),(1234),(12)(34)$.

But take $\mathbb{Z}/ n \mathbb{Z}$.

@sylow:
No, thats wrong. Take S_4 than there are 5 conjugacy classes, not 4.

I dont know why you make examples.$n$ is only a given natural number and it is not the $G$'s size.please someone explain more!!! :maybe:

You solved another problem!
I mean:
$n$ is a given natural number.prove that the number of groups like $G$ which has exactly $n$ conjugate classes is finite!!!!! :blush:

Opps,
S is next to the Z on keyboard  :rotfl:

I wrote the the first post fast and maybe it has chaneged the meaninig of the actual problem.I tried to edit my first post but i could't
(i dont know why? :|  :!: )and i want to know why one of my post is deleted on this topic?

The first is just a matter of fact and belongs to all normal users; I've changed the first post now.

And I don't like posts that are pure dumping after small periods of time, this is why I deleted that post (two times in fact).
And I don't like reposting posts, either, if someone had a reason to delete them...


Given that the perimeter of an isosceles triangle is 36 units and the altitude to the base is 12 units, what is the number of square units in its area?

Using a little guess and check and knowledge of pythagorean triples, we have the base is $ 10$ and the congruent sides are each $ 13$. The area is $ \frac{(10)(12)}{2}\equal{}\boxed{60}$.

the base of this triangle have 13units and we have 65 units for area of this triangle

13*13=169 so tha base of triangle is 10units and each side of this triangle have 13 units and area will be 65 units


Each letter represents a non-zero digit. What is the value of $t$?
$c+o=u$
$u+n=t$
$t+c=s$
$o+n+s=12$

[hide]Substitute t+n for s: $o+n+t+c=12$.
Substitute u for c+o: $t+u+n=12$.
Substitute t for u+n: $2t=12$.
$t=\boxed{6}$.[/hide]


[quote="Ligouras"]Let $ a,b,c$ be positive reals. Prove that

$ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)\geq abc(a \plus{} b \plus{} c)\sqrt {3abc(a \plus{} b \plus{} c)}$[/quote]

From Holder inequality

$ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)(1 \plus{} 1 \plus{} 1) \ge (ab \plus{} bc \plus{} ca)^3 \equal{} (ab \plus{} bc \plus{} ca)(ab \plus{} bc \plus{} ca)^2 \ge 3abc(a \plus{} b \plus{} c)\sqrt {3abc(a \plus{} b \plus{} c)}$

  :P

Done .

[quote="Pain rinnegan"][quote="Ligouras"]Let $ a,b,c$ be positive reals. Prove that

$ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)\geq abc(a \plus{} b \plus{} c)\sqrt {3abc(a \plus{} b \plus{} c)}$[/quote]

It's not your inequality, it's only manlio's inequality weakened[/quote]

Dear [b]PAIN[/b] [color=red][b]Manlio Marangelli [/b][/color]is

$ 3(a^2b\plus{}b^2c\plus{}c^2a)(ab^2\plus{}bc^2\plus{}ca^2)\geq abc(a\plus{}b\plus{}c)^3$

OK?


inequalities
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Guess what! It begins with Cauchy and ends with Schwarz!

Rearrangement can do so much with something so intuitive  :)

Holder's Inequality is my favorite :lol:

Yeah, I like $x^{2}\ge 0$ much. I also like $A_{m}-G_{m}$ but I not often use it  :D
P/s: This topic is for spam?

I really like Jensen, and Cauchy (Andreescu's form)

I think it's the same as Engel form.

By the way, this is a spammy thread.

I like Cauchy-Schwarz because it is true in arbitrary vector space with inner product. :D Also minkowski because it is an important result about $L^{p}$ spaces. But for algebraic ineqs the ineq of Newton seems fun (you can get lots of others from it).

[quote="stancioiu sorin"]I agree with you![/quote]
What do you mean about "stronger"?

[quote="Arne"]Bergström = Cauchy :roll:[/quote] like [b]"Titu Andreescu "and "Engel" too[/b]

My favourite is $Chebishev$ and $Muirhead$

hey I like $x^{2}\ge 0$  :D
but in geometry I think $AM-GM$ is most useful !

[quote="Leao"][quote="me@home"]Andreescu's Form???[/quote]
$\frac{a^{2}}{x}+\frac{b^{2}}{y}\geq \frac{(a+b)^{2}}{x+y}$ for $a,b>0$ and $x, y$ real and not equal to zero. Equality holds iff $ay=bx$. You can use induction to show that it holds for n elements.[/quote]

Hello!I would like to add the proof   :).Really easy
From this inequality we get:$(x+y)(a^{2}y+b^{2}x)\geq(a+b)^{2}xy\Rightarrow a^{2}y^{2}+b^{2}x^{2}\geq2abxy\implies(ay-bx)^{2}\geq0$

I like $x^{2}\geq0$   :P (altough i don't know how to prove it completely)(the (-)(-)=(+) part)

[quote="alex__ib"]I like $x^{2}\geq0$   :P (altough i don't know how to prove it completely)(the (-)(-)=(+) part)[/quote]
Prove that $(-1)(-1)=1$. Then use the fact that if $x\geq 0$ then $x^{2}\geq 0$.


this is a basic precal question.

The four sides of a skew quadrangle and the two segments joining the midpoints of the opposite sides are realized by rigid bars. The bars are linked by hinges. Prove that this apparatus is not rigid.

EDIT:

This has already been posted:

[I had at least one error in a character for the link here.]

Then it was corrected by the next user.

That should be http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229135


Let a,b,c be non-negative numbers such that $ (abc)^3 \equal{} 2$. Prove that

     $ \frac {1}{a^3(1 \plus{} b^3)} \plus{} \frac {1}{b^3(1 \plus{} c^3)} \plus{} \frac {1}{c^3(1 \plus{} a^3)}\geq 1$

    [i]Marcel Chirita / Bucharest[/i]

let $ a^3\equal{}x,b^3\equal{}y ,c^3\equal{}z$ and see here : http://www.mathlinks.ro/Forum/viewtopic.php?p=1205#1205


Is there a mathematical formula that can be used to determine the number of triangles in problems like this? <I can count # of triangles to be 24, but looking for a smarter way.>

It says "Howmany Triangles are in this figure?" I am not sure how will I put that figure here (I have a paper copy). It is a triangle with the three lines of angle bisectors.

hmmm. I'm having a problem here. :maybe:

I count 16.

[hide]
[img]7051[/img]

Just do organization. Startlike this:

There are 6 traingles with 1 mini triangle (1, 2, 3, 4, 5, 6)
3 Semi (1+2, 4+6, 3+5)

and so on and so forth and you should get at your answer.
[/hide]

As far as I know, there s now formula, as there are far too many conditions.

Can't you take the number of intersection points choose 3 and subtract one for every set of 3 collinear points?

EDIT: Oops, this is wrong. Ignore it.

The number of triangles is the number of interior angles. (Only if the triangles are regular. As in they don't make weird shapes that can't create triangles.)

NO, its the interior angles times two because around each angle there are 2-3 triangles.


If I understand correctly, the sequence is $ 1,0,\minus{}1,1,0,\minus{}1,...,1$.  Since, the first three terms cancel out, and so do the next three, etc., they leave only one term, $ \boxed{1}$.

[quote="sri340"]1, 0, –1, +1, 0, –1, +1, 0, –1 .  .  .  . and so on, where the last number is +1, has a sum of

(A) +1   (B)  –1   (C) +2   (D) 0   (E) –2[/quote]

Because the sequence repeats, we can just let $ 1, 0 , \minus{}1, 1$ be the sequence.  $ 1 \plus{} 0 \minus{}1 \plus{} 1 \equal{} 1$.


[quote="zerog"]Hi there,, I found another challenge that I would like to share,, I am feeling confused with that. If some one could solve it I will be very happy to know,

Find the values of 'm' and 'c' if the straight line y=mx+c passes throught the given points!!!

(2,8) and (8,20)

(4,6) and (10, 3)[/quote]

[hide]You know since the line passes through the points that
$8=2m+c$
$20=8m+c$

Solve this system of equations.[/hide]

The same method is used to solve the other equation.


James was conversing with his friend Peter, his nephew Danny, and Danny's friend Nellie recently. James possessed  two tickets to a new movie in his pocket that he had just purchased, and James mentioned this and noted that there were two four-digit numbers on the tickets and that the sum of all 8 digits was 25. Peter asked if any digit appeared more than twice out of the 8, which James answered, and then Danny asked if the sum of the digits of either ticket was equal to 13, which James answered also. Much to James’ astonishment Nellie immediately told him what the two numbers were.
What were they?

The answers  to the 2 questions have to be no and yes respectively. So have:
a + b + c + d = 13
e + f + g + h = 12
The digits are 1,2,4,5 on one ticket and 1,3,4,5 on the other.