[hide]$3s+7c+d=3.15\\ 4s+10c+d=4.20$  Subtracting the first equation from the second gives $s+3c=1.05$  subtracting the new equation from the first one twice gives $s+c+d=3.15-2.10=\boxed{1.05}$[/hide]

yeah...the solutions are supposed to be readable

[quote="anirudh"][quote="i_like_pie"]The cost of 3 sandwiches, 7 cups of coffee and 1 donut is $\$3.15.$ The cost of of 4 sandwiches, 10 cups of coffee and 1 donut is $\$4.20.$ How much will 1 sandwich, 1 cup of coffee and 1 donut cost?[/quote]

[hide="because people think it is not good to not post a solution, I shall post one"]

[I handwrote it, so I just scanned it...]

[/hide][/quote]
Why don't you do the entire problem on the computer? When I do a problem I do the entire thing in latex without paper (particularly on easy questions...)...

[quote="ajai"]use matrices.....i don't want to bother actually doing the work, i'm sure someone else will... :D[/quote]

can you do it with 3 variables and 2 equations?


This is one of the few problems that has continued to stump me. Is there any mathematical way to do it???

A knight is placed on the corner of a four by four grid. He moves two spaces in one direction and one space perpendicular to that direction. What is the maximum number of squares he can travel including the starting square?

Do you have any solution at all? If you don't I can give you one. The solution I have is logical, but involves trial and error. And there's nothing wrong with trial and error; many MATHCOUNTS problems from any level involve trial and error.

yea i tried for 16 and then for 15 and 15 works

[quote="Magixmaniak"]i guessed and checked systematically and found the answer to be 15.[/quote]

Yes, the answer is 15. I remember reading a similar thread.

15 is correct, and trial and error is the only method I can do. I was hoping for something more mathematical. This was problem 30 on the '95 state sprint (yes, state!). How are you supposed to figure it out in 4/3 of a minute?

Well it comes down to the corners. There are 2 black and 2 white corners. A Knight when it moves always has to go to a square of the oppisite color so therefore when its at a white corner it must go to a black square in the center. When the knight is going to the 3rd corner if it reachs it all the middle squares will be filled up by other moves and because the knight cannot visit a square twice it cannot get to one corner.

This was an active topic just last month.  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=181683#p181683]Click Here.[/url]

You can get 15, then prove that 16 is impossible.


geometry
ratio
rectangle
geometry proposed

Let $ M$ be a point interior to the acute-angled triangle $ ABC$ such that $ \angle MBA \equal{} \angle MCA$. Let $ K, L$ be the feet of perpendiculars from $ M$ to $ AB, AC$ respectively. Prove that $ K, L$ are equidistant from the midpoint of $ BC$ and the median from $ M$ of the triangle $ MKL$ has a fixed point when $ M$ varies in the interior of triangle $ ABC$.

--- Denote P,Q,I be the midpoint of MB,MC,BC. $ \triangle IPK\equal{}\triangle LQI$, thus IL=IK
--- Easy to show that the median from M of triangle MKL passes through the orthocenter of triangle ABC.

[quote="kaka__2004"][quote="mr.danh"]--- Denote P,Q,I be the midpoint of MB,MC,BC. $ \triangle IPK \equal{} \triangle LQI$, thus IL=IK
--- Easy to show that the median from M of triangle MKL passes through the orthocenter of triangle ABC.[/quote]
how to prove it?
how to prove "--- Easy to show that the median from M of triangle MKL passes through the orthocenter of triangle ABC"
thank you in advance![/quote]
[b][size=100]PROOF  1.[/size][/b] - Let $ N$ be, the midpoint of the segment $ KL$ and we denote as $ K',\ L',$ the orthogonal projections of $ N,$ on $ AB,\ AC,$ respectively.

It is well known ( and easy to prove by [b][size=100]Thales theorem[/size][/b] ), that if the orthogonal ( or parallel ) projections of three points on two lines intersect each other, form segments with the same ratio, then these points are collinear, with the same ratio. The inverse is also true.

So, in our configuration, it is enough to prove that $ \frac {EL}{LL'} \equal{} \frac {FK}{KK'}$ $ ,(1)$

From the similar right triangles $ \bigtriangleup L'LN,\ \bigtriangleup KMA$ $ \Longrightarrow$ $ \frac {LL'}{MK} \equal{} \frac {NL}{AM}$ $ ,(2)$

From the similar right triangles $ \bigtriangleup K'KN,\ \bigtriangleup LMA$ $ \Longrightarrow$ $ \frac {KK'}{ML} \equal{} \frac {KN}{AM}$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {LL'}{MK} \equal{} \frac {KK'}{ML}$ $ \Longrightarrow$ $ \frac {LL'}{KK'} \equal{} \frac {MK}{ML}$ $ ,(4)$

$ \bullet$ Let be the points $ E\equiv AC\cap BH$ and $ F\equiv AB\cap CH,$ where $ H$ is the orthocenter of $ \bigtriangleup ABC.$

From $ \angle E'BF' \equal{} \angle EBA \minus{} \angle MBA \equal{} \angle FCA \minus{} \angle MCA \equal{} \angle F'CE'$ we conclude that the quadrilateral $ BF'E'C$ is cyclic, where $ E'\equiv MC\cap BH$ and $ F'\equiv MB\cap CH.$

So, we have that $ (ME')\cdot (MC) \equal{} (MF')\cdot (MB)$ $ \Longrightarrow$ $ \frac {ME'}{MF'} \equal{} \frac {MB}{MC}$ $ ,(5)$

But, $ \frac {ME'}{MF'} \equal{} \frac {EL}{FK}$ $ ,(6)$ and $ \frac {MB}{MC} \equal{} \frac {MK}{ML}$ $ ,(7)$

Hence, from $ (5),$ $ (6),$ $ (7)$ $ \Longrightarrow$ $ \frac {EL}{FK} \equal{} \frac {MK}{ML}$ $ ,(8)$

From $ (4),$ $ (8)$ $ \Longrightarrow$ $ \frac {EL}{FK} \equal{} \frac {LL'}{KK'}$ $ \Longrightarrow$ $ \frac {EL}{LL'} \equal{} \frac {FK}{KK'}$ $ ,(9)$ and the proof is completed.

Kostas Vittas.

[b][size=100]PROOF  2.[/size][/b] – We draw the rectangles $ BKMB',\ CLMC'$ and then, we have the configuration of the triangle $ \bigtriangleup MB'C',$ with these similar rectangles erected on its side-segments $ MB',\ MC'$ respectively, outwardly to it.

It is well known ( and easy to prove ), that the line segment $ MN,$ where $ N$ is the midpoint of the segment $ KL,$ is perpendicular to $ B'C'$ and let be the point $ N'\equiv B'C'\cap MN.$

Let $ H$ be, the orthocenter of $ \bigtriangleup ABC$ and we denote the points $ X\equiv MB'\cap BH$ and $ Y\equiv MC'\cap CH.$

From the similar right triangles $ \bigtriangleup B'BX,\ \bigtriangleup C'CY$ $ \Longrightarrow$ $ \frac {B'X}{C'Y} \equal{} \frac {B'B}{C'C} \equal{} \frac {MB'}{MC'}$ $ ,(1)$

Let $ H'$ be, the orthocenter of $ \bigtriangleup MB'C',$ lies on the line segment $ NN'\perp B'C'.$

From $ (1)$ and from $ XH\parallel B'H'$ and $ YH\parallel CH',$ we conclude that the points $ M,\ H,\ H',$ are collinear ( easy to prove by [b][size=100]Thales theorem[/size][/b] ).

Hence, we conclude that the orthocenter $ H$ of $ \bigtriangleup ABC,$ lies on the line segment $ NN'\equiv MN$ and the proof is completed.

Kostas Vittas.


Compute the exact value of $ \cos(2\pi/7) \plus{} \cos(4\pi/7) \minus{} \cos(\pi/7)$.

[hide="Solution"]It is well known that $ \cos\theta\equal{}\frac{e^{i\theta}\plus{}e^{\minus{}i\theta}}{2}$, which transforms this into:
\[ \frac{1}{2}(e^{2\pi i/7}\plus{}e^{12\pi i/7}\plus{}e^{4\pi i/7}\plus{}e^{10\pi i/7}\minus{}e^{\pi i/7}\minus{}e^{13\pi i/7})\]

Let $ \zeta\equal{}e^{\pi i/7}$, a 14th root of unity. Then we have
\[ \frac{1}{2}(\zeta^2\plus{}\zeta^{12}\plus{}\zeta^{4}\plus{}\zeta^{10}\minus{}\zeta\minus{}\zeta^{13})\]

Recognize that $ \zeta^{7}\equal{}\minus{}1$. Therefore, we have
\[ \frac{1}{2}(\zeta^2\minus{}\zeta^5\plus{}\zeta^4\minus{}\zeta^3\minus{}\zeta\plus{}\zeta^6)\]

$ (\zeta^7\plus{}1)\equal{}0\equal{}(\zeta\plus{}1)(\zeta^6\minus{}\zeta^5\plus{}\cdots\minus{}\zeta\plus{}1)\implies(\zeta^6\minus{}\zeta^5\plus{}\cdots\minus{}\zeta\plus{}1)\equal{}0$, so this simply reduces to $ \boxed{\minus{}\frac{1}{2}}$.[/hide]


Sorry, but can you pls translate it? I dunno chemistry in Chinese since I learnt most of them in English. Sorry about that. :D

I'm sorry I cannot translate them into English either :blush: for I've just learnt how to express them in Chinese.
Who can help me to translate them into Enlish?

However,I'll try to retell them in a type of math language(It's actually a math problem) so that you can understand it well.But I haven't found a good way to do that

[quote="wuhao_jshj"]分子式为\[ C_nH_{2n+2}  \]的烷烃共有\[ A_n  \]种结构，试写出数列\[ A_n  \]的通项[/quote]
An alkane with molecular formula\[ C_nH_{2n+2}  \]has\[ A_n  \]different structures. Find the general formula of\[ A_n  \].

P.S. I only know little about chemistry. I found the translation of some term(s) on the Internet, so don't ask me to solve it! :P

请问一下，这种烷烃的结构式是只包括链式结构呢，还是连环式结构也包括进去？

[quote="martha"][quote="dingdongdog"]请问一下，这种烷烃的结构式是只包括链式结构呢，还是连环式结构也包括进去？[/quote]
我不太懂这些，但这样的分子式不是只会出现链式结构吗？
不知道说得对不对。[/quote]
I think so.

应当是可算的(好象是要用什么分子对称群之类,具体的我也不会 :P )
化学书上就是算出来的:

[quote="martha"][quote="dingdongdog"]请问一下，这种烷烃的结构式是只包括链式结构呢，还是连环式结构也包括进去？[/quote]
我不太懂这些，但这样的分子式不是只会出现链式结构吗？
不知道说得对不对。[/quote]
 :blush: sorry,I made a mistake...

[quote="martha"]I am not sure how alkane is defined,  :oops: so i am having problems here: does this question include alkanes with side chains? or is it just asking for single chain alkanes?

For example: is "methyl propane" counted for C4H10 ?[/quote]
Of course

My chem teacher told us that there doesn't exist a formula for this. It is too complex to include all the conditions.


Let $\large 2 \leq\ n \in N$; $\large x_{i} \geq\ 0$. Prove that:
$\large \sum ix_{i} \leq\ \frac{n(n-1)}{2} + \sum x^{i}_{i}$.

all we need to prove is that
$x^n + n \geq\ nx$
which is obvious by bernouli inequality


If $ a,b,c$ are nonnegative real numbers, then

$ a^3\plus{}b^3\plus{}c^3\plus{}12abc \le \sum a^2\sqrt{a^2\plus{}24bc}$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 12abc \le \sum a^2\sqrt {a^2 \plus{} 24bc}$.[/quote]
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 12abc \le \sum_{cyc} a^2\sqrt {a^2 \plus{} 24bc}\Leftrightarrow$ ( after squaring )
$ \Leftrightarrow\sum_{cyc}(a^3b^3\plus{}24a^2b^2c^2)\leq\sum_{cyc}a^2b^2\sqrt{(a^2\plus{}24bc)(b^2\plus{}24ac)}.$
But $ \sqrt{(a^2\plus{}24bc)(b^2\plus{}24ac)}\geq ab\plus{}24c\sqrt{ab}.$
Thus, it remains to prove that $ \sum_{cyc}(ab\sqrt{ab}\minus{}abc)\geq0,$ which is obvious.
Quit easy. :maybe:

$ \left(a^2 \plus{} 24bc\right)\left[7\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 8(bc \plus{} ca \plus{} ab)\right]^2$

$ \minus{} \left[7a^3 \plus{} 8a^2(b \plus{} c) \plus{} 7a\left(b^2 \plus{} c^2\right) \plus{} 92abc \plus{} 48bc(b \plus{} c)\right]^2$

$ \equal{} 24bc\big[(b \minus{} c)^2\left(109a^2 \plus{} 77ab \plus{} 77ac \plus{} 49b^2 \plus{} 89bc \plus{} 49c^2\right)$

$ \plus{} (b \plus{} c \minus{} 2a)^2(25bc \plus{} 7ab\plus{}7ca )\big]\geq0\Longrightarrow$

$ \sum a^2\sqrt {a^2 \plus{} 24bc}$

$ \geq\sum{\frac {a^2[7a^3 \plus{} 8a^2(b \plus{} c) \plus{} 7a(b^2 \plus{} c^2) \plus{} 92abc \plus{} 48bc(b \plus{} c)]}{7(a^2 \plus{} b^2 \plus{} c^2) \plus{} 8(bc \plus{} ca \plus{} ab)}}$

$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 12abc.$

[quote="MathUniverse"][quote="Vasc"]Easy, indeed, Arqady!
I think the following generalization holds:

If $ x_1,x_2,...,x_n$ are positive real numbers such that $ x_1x_2...x_n\equal{}1$, then

$ \sum \frac 1{1\plus{}\sqrt{1\plus{}4n(n\minus{}1)x_1}} \ge \frac 1{2}$.[/quote]

But can't we prove it using RCF theorem?[/quote]

try to prove the contrary! :)

[quote="red3"]we  can prove the general one by only AM-GM! :D[/quote]

It would be nice if you wrote your solution.. Everyone can write that... -.-


inequalities
trigonometry
geometry proposed
geometry

[quote="d.edgar"]Substitute $a=2RsinA$ and the analogues. After some computations, the inequality is equivalent to $(sin^{2}A+sin^{2}B)(cos^{2}{A}+cos^{2}B)\ge sin^{2}C$ which is true from Cauchy-Scwarz ($sin(C)=sin(A+B)=...$).[/quote]sorry can you post a full solution please?

[quote="d.edgar"]$a, b,c$ are the sides of a triangle, $R$ is the circumradius. This problem is from the Baltic Way 1998.

Substitute $a=2RsinA$ and the analogues. After some computations, the inequality is equivalent to $(sin^{2}A+sin^{2}B)(cos^{2}{A}+cos^{2}B)\ge sin^{2}C$ which is true from Cauchy-Scwarz ($sin(C)=sin(A+B)=...$).[/quote]
The inequality is equivalent to \[4(\sin^{2}a+\sin^{2}b)\leq 4\sqrt{2\sin^{2}a+2\sin^{2}b-\sin^{2}c}\]
\[(\sin^{2}a+\sin^{2}b)^{2}+\sin^{2}c\leq 2(\sin^{2}a+\sin^{2}b)\]
Now notice that \begin{eqnarray*}&&(\sin^{2}a+\sin^{2}b)^{2}+(\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b)\\&=&(\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b+\sin^{2}a+\sin^{2}b)\\&=&2(\sin^{2}a+\sin^{2}b)\\ \implies (\sin^{2}a+\sin^{2}b)^{2}+\sin^{2}c\leq 2(\sin^{2}a+\sin^{2}b)&\iff& \sin^{2}c\leq  (\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b)\end{eqnarray*}
Recognize $\sin c=\sin a\cos b+\cos a\sin b$ and the inequality is equivalent to:\[(\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b)\geq (\sin a\cos b+\cos a\sin b)^{2}\]
Obvious from CS

[quote="me@home"]Now notice that
\begin{eqnarray*}&&(\sin^{2}a+\sin^{2}b)^{2}+(\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b)\\ &=&(\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b+\sin^{2}a+\sin^{2}b)\\ &=&2(\sin^{2}a+\sin^{2}b)\\ \implies (\sin^{2}a+\sin^{2}b)^{2}+\sin^{2}c\leq 2(\sin^{2}a+\sin^{2}b)&\iff& \sin^{2}c\leq (\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b) \end{eqnarray*}
Recognize $\sin c=\sin a\cos b+\cos a\sin b$ and the inequality is equivalent to:
\[(\sin^{2}a+\sin^{2}b)(\cos^{2}{a}+\cos^{2}b)\geq (\sin a\cos b+\cos a\sin b)^{2}\]
Obvious from CS[/quote]ah i see.. very nice and thanks!


superior algebra
superior algebra unsolved

if p is prime that include I  show that $ %Error. "display" is a bad command.
\sqrt {I}$  $ %Error. "display" is a bad command.
\subseteq$P.now use from this to proove:the ideal(x,4) in$ %Error. "display" is a bad command.
\mathbb{Z}$[x] is primary but is not the potent of a prime ideal.$ %Error. "display" is a bad command.
\sqrt {I}$={x$ %Error. "display" is a bad command.
\in$ R $ %Error. "display" is a bad command.
\mid$ exist  m $ %Error. "display" is a bad command.
\in$ $ %Error. "display" is a bad command.
\mathbb{N}$ $ %Error. "display" is a bad command.
\ x^{m}$ $ %Error. "display" is a bad command.
\in$ I}.

The first part is easy working with definitions.
Check that $ \sqrt{(x,a)}\equal{}(x,2)$, this is a prime ideal. But $ (x,2)^2\equal{}(x^2,4) \subsetneq (x,4) \subsetneq (x,2)$.


[quote="HelloWorld"]What is Stevanovic Circle ?  :)[/quote]

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


John saw an ad for tomato soup at 24 cents per can. At the price advertised, what is the greatest number of whole cans of soup John could buy for $ \$$1.00?


calculus
integration
trigonometry
calculus computations

Splendid! xufrog.

Your solution is correct. :)

kunny


Okay, I'm working with a question that asks for the equation of a parabola.  The vertex is given as (-2,4).  The parabola passes through the point (-1,8).  I'm uncertain how to solve this.

I'm familiar with the equation y=a(x-h)^2+k, but I don't think it can be applied here.  It's for when a parabola's been moved...

[quote="Silis"]I'm familiar with the equation y=a(x-h)^2+k, but I don't think it can be applied here.  [/quote]

Why not?  The vertex is $(h, k)$.  So all you need to do now is find $a$.

When a parabola is "moved", its vertex is also "moved", and as t0rajir0u said $(h,k)$ is the vertex and is those are the parts which are changed to "move" the parabola.
[hide]
$y = a (x-h)^{2}+k$
Where $h=-2$ and $k=4$.
To find $a$, let $x=-1$ and $y=8$.
Then $a = \frac{8-4}{(-1+2)^{2}}= 4$.
[/hide]


MATHCOUNTS
FTW
probability
geometry
3D geometry
search
AoPS Books

What's up ppls! I need help w/ my MathCounts. Does anyone no a great study technique for MC?

Do problems, and don't play games.
I guess that's why I'm not improving much...

If you can, read AoPS intro books/ Volume 1, strengthen your weak points, practice countdown w/ FTW, read problems carefully, and read the rules carefully (i know that sounds stupid, but if you put your answer in incorrect form......)

Read AoPS books. Try taking easy-to-refer notes so that you will be able to easily look over them if you're ever unsure. Also, don't fool around, and start getting serious.....
 :lol:

rihim

Print out some old target rounds and all those other things from the MATHCOUNTS website and time youself while doing them.

No offense to anyone, but I kinda asked for help on MathCounts :) , not some random aops advertisement. Can someone tell me a good way to improve me on my targets, etc.?
That would be a great help.
Once Again, thank you all that have replied.

Well I mainly think that target is pretty easy because there is more time per problem. I don't really know a way to improve, but there will definitely be a least one geometry or probability problem.

Sprint Round: The sprint has problems increasing in difficulty. Thus, when taking it, instead of going through all problems too fast, try getting the easier problems correct, and save the hard ones until the last, when you have more time. Also, try leaving some time in the end to check over your answers, so you can avoid any easy mistakes.

Target Round: Since you have more time in the target round, read the problems first, and pick the easier problem to start with. Check over your calculations with your calculator.

Actually, they look at the back problems of the sprint when breaking ties, so it's very important to get the back problems correct as well as the easy ones.

[quote="cognos599"]dont just read them, do all of the problems in them. And try to take old mathcounts sprints in 30-35 minutes instead of taking all 40, it helps with speed.[/quote]

hey u, cognos, you saved any "old mathcounts sprints?" cuz in mathcounts.org they only gots 2007 ones and there ain't no 2006, 2005...... and stuff.

Hey Cognos,
In my opinion, it isn't about working [u]just[/u] the old ones in 30-35 minutes. It's about working all the problems at a great speed with accuracy.
I need a lot of help with accuracy practices other than the MathCounts workbook things. Does anyone have any helpful sites, books (excluding the aops one's because I have them already), etc. that can help me in getting accuracy in Targets.
Cause I have plenty of time to work the problems, I just need to know how to do them.
You understand?
I welcome all suggestions.

you can find 2007 sprint, target, team, and countdown rounds and solutions on the mathcounts website(http://www.mathcounts.org).otherwise, i cant help u. sorry

Work on accuracy and understanding the problems.  Then find tricks that will make a hard question as simple as $ \pi$.
AoPS intro series are the place to start for Mathcounts.  After just halfway through the book I can see I've drastically improved.
Again, get the problems right other than putting answers down for each problem and get only 2 right.
:)

one thing that I do, is skip lengthly problems. I put a box around all of the problems that I didn't do, and then a circle around all of the ones I'm not sure if I got the right answer for. This really helps for accuracy.

and always learn how to do the problems you missed or figure out why you missed them.

Also, try taking previous competition rounds. You can order them from the MATHCOUNTS store at [url]http://www.mathcounts.org[/url]

Hey guys, I still need help! Is there any more advise?

For mathcounts, speed is really important.
Push yourself in doing the problems at Sprint 1 minute per problem so that u have extra time for the hard ones at the end. As for the Target Round, i recommend just practicing them from past tests....u should finish all the previous tests from 1992-2007 school, chapter, state, or even nationals.  In Mathcounts, there are some concepts you should practice: probability, counting, a bit of number theory, geometry, and rate problems. Those are like the basics. I'm not trying to "advertise" anything, but u should definitely do AoPs book 1 and buy some Mathcounts Handbooks. After doing each tests or warm-up or watever it is, u should always know why u made the mistake.  Try seeing the type of problems you tend to get wrong and work on them so that the next time, u make fewer mistakes (duh).
And also, u should know some really basic geometry, coutning, algebra 1 and 2 formulas, and the squares and cubes of number 1-25 at least. These are going to be ur lifesavers for Countdown Round or anything. I do believe that in the Internet, if u search for "Mathcounts Toolbox" on the Internet, there is like this whole list of formulas and good stuff like that.  It might come in handy. Good luck!

You should practice EXPLICITLY for mathcounts, I made some massive mistakes, and I hope that I can prevent some of you from making them.
In 7th grade I made nationals, got top 30.
Between nationals in 7th and 8th grade, I spent nearly all my time practicing for AMC/AIME, and assumed that it would also aid my mathcounts ability.
During the year, I did literally 4 sprint/target practices outside of the competitions themselves, and got bored with the problems. Nationals was a bit of a slap in the face for me, as my ability to do mathcounts problems had waned despite me being much much better at math, resulting in my rank dropping massively (more than doubling).
So if you want to do well in mathcounts, go through each and every practice test you can get at least twice, even more for safety.

Is it wise to study MC in fifth grade, before you reach the qualifying age for MC (in sixth grade)? Even if I don't know most of the content, and I'm not very good at problem solving (it will change later in the year), do you think it is advisable to study my brother's AoPS books volume 1-2? I barely understand pre-algebra as it is, though...

[quote="D3m0n Shad0w"]I barely understand pre-algebra as it is, though...[/quote]

Pre-alg is basically finding $ x$. Example: $ 2x\plus{}1\equal{}5$

Starting early is a [i]really[/i] good idea for MathCounts. It gives you the most important thing you need in prepping for a math competition: time/

Ah, I forgot to mention...I have Science Fair? I could always just work faster, but still...

[quote]Pre-alg is basically finding $ x$. Example: $ 2x \plus{} 1 \equal{} 5$[/quote]

Oh. It's that? Well...
2x + 1 = 5
2x + 1 - 1 = 5 - 1
2x= 4
2x/2=4/2
x=2

Like that?  :wink: Is MC about that, except in word problems?

actually, mc is much harder  :)

for example, here is a state problem (isn't that hard actually.)

Even though the snapshot doesn't work, I can still see how hard it is. I just looked at MOEMS, and it is pretty hard.

Well anyways, here is a problem that isn't too hard from the State round.

28. The Fibonacci sequence is the sequence 1,1,2,3,5, ... where each term is the sum of the previous two terms. What is the remainder when the 100th term of the sequence is divided by 4?

[quote="7h3.D3m0n.117"]
Starting early is a [i]really[/i] good idea for MathCounts. It gives you the most important thing you need in prepping for a math competition: time/[/quote]

The funny thing with this is I just learned about it in December, and I'm in 8th grade :)

But, those aops intro books are crp expensive, where else can i get material?

As I stated elsewhere, the AoPS books are actually quite cheap as compared to any other math textbook you wish to buy.


Hi, I am in 6th grade and I am starting a chemistry class. If anyone has anything taht I could learn, could you please post it with every step. Thanks :)

Whoa, 6th grade learning chemistry is a little early. It's usually a high school course.

Do you know what atoms, molecules, mixtures, compounds, and solutions are? I'm sure that others will explain it fully to you (I'm just starting chemistry too but I'm entering 8th  :blush: ).

[quote="amirhtlusa"]i started chemistry in 6th grade also ;)[/quote]

Oh yeah I was reading in 4th... oh well.

When is "real chemistry" (like actually dedicating a year or semester to chemistry)?

[quote="amirhtlusa"]yeah my Prepration for Chemistry olympiad started in 6th Grade.

i had 2hrs classes 3 days a week, every week  ;)

that is more that one semester or anything else like that...[/quote]

I'm learning chemistry by myself, because my school is so far behind and we "started" chemistry and haven't learned anything about electron orbitals or molecular mass yet... what's the difference between chemistry olympiad and normal class chemistry (a lot of hard concepts, etc)?

i guess something like regular math class and a olympiad math class...


find ${ \left({\prod_{k = 0}^{499}2k + 1}\right)}$ divided by $ {\left({\prod_{k = 1}^{499}2k}\right)}$ cubed to the nearest integer. For some reason I can't figure out how to put the two big products over a fraction bar in LATEX.

EDIT: srry i messed up...lol..btw t0rajir0u's answer is correct for the previous problem i posted

So you want to find $ (\frac{1*3*5*\cdots*999}{2*4*6*8*\cdots *998})^3$

We have $ \frac{3}{2}*\frac{5}{4}*\frac{7}{6}*\cdots*\frac{999}{998}\equal{}\sqrt[3]{x}$

And we want to find x.

.... you go get your calculator, hit the on button, and *bursts out laughing* :rotfl:

Seriously, I can't find a pretty way to do this.

[hide="Well,"]
We can impose some weak bounds by telescoping;
\[ \frac {4}{3} \cdot \frac {6}{5} \cdot \frac {8}{7} \cdots \frac {1000}{990} < \frac {3}{2} \cdot \frac {5}{4} \cdot \frac {7}{6} \cdots \frac {999}{998} < \frac {2}{1} \cdot \frac {4}{3} \cdot \frac {6}{5} \cdots \frac {998}{997}
\]
Multiplying all three quantities by the middle term and then taking the square root,
\[ \sqrt {\frac {1000}{2}} \approx 22.36 < \frac {3}{2} \cdot \frac {5}{4} \cdot \frac 76 \cdots \frac {999}{998} < \sqrt {999} \approx 31.607
\]
We can tighten the above bounds in particular by replacing some of the fractions, eg on the rightmost term we could change $ \frac {2}{1}$ with $ \frac 32$ and get a slightly better upper bound of $ \approx 27.37$. Of course, since you're cubing the result, this wouldn't get you anywhere close to an exact integer value.

If only that cube was 'cube root', then the answer would be $ 3$ (wishful thinking  :roll:). Even better, if that said 'the denominator cubed' (which is what I thought it first) the answer would come out to be close to $ 0$.

I brute-forced this and the actual answer comes out to be $ \approx 25.225^3 \approx \boxed{16051}$, which seems nearly impossible by hand; also the product doesn't seem to converge to any value if we change $ 499$ to something higher.

Edit: In fact, your result before cubing can only be on the scale of $ .0004 \%$ off before you get a different answer ... thus the problem is probably asking for something else.
[/hide]


inequalities trigonometry AMC AIME MATHCOUNTS function algebra

pythag011 has asked me to host this.

16 people to start, single-elimination tournament, based on iTest's ToC.  pythag011 will be writing the problems.  In each round, there will be five questions, worth 2, 3, 5, 7, and 8 points, respectively.  The answers will be integers from 1-999.  The person with the highest score wins.  In case of a tie, the first person to submit wins.  You have one week to submit each round.

[color=darkred][size=75]Extra thread removed.
-Neal (nebula42)[/size][/color]

Btw you double posted...

The main reason I asked ernie to host this is that I realized that I am a very very very very bad host.

[quote="Dojo"]Join based on the level. (If its easy I join, if its pythag level, I dont :P )[/quote]

the 2-pointer-easy mc
the 3-pointer-medium mc
the 5-pointer-hard mc
the 7-pointer-aime medium level
the 8-pointer-easy olympiad level

[quote="AIME15"]Translated to what it'll really be :D

2 pointer - medium-hard mc
3 pointer -  hard mc - medium-hard amc
5 pointer - aime level
7 pointer - easy olympiad
8 pointer - medium-hard olympiad

:([/quote]

I promise you this will not happen...

what i ssaid is first round.

2nd round is same.

3rd is

2-pointer medium mc
3-pointer hard mc
5-pointer medium aime
7-pointer very easy olympiad (this one is easier than the easy olympiad)
8-pointer easy olympiad

4th is what aime said  :)

[quote="ernie"]Can I join. :D

I just host, I don't get the problems... do I?[/quote]

No joining ernie... but you can take each round unofficially :P

join.

[hide=".."]not long enough.[/hide]

[hide][hide][hide] arent you bored yet?
I join![/hide][/hide][/hide]

[hide="click it"][hide="wait, no, this one"][hide="actually, this one"]DAFJBOMB SPAMS! :spam: [/hide][/hide][/hide]
btw, (to make this post NOT spam), when are we starting?

[quote="AIME15"]I submitted (via AIM) on time.[/quote]

You said you thought that the answers might be those, but you didn't officially submit them.

[quote="dafjbomb"]hehe, i guess that makes me the only one to submit on time![/quote]
Either do I:wink: .

Hey! I thought I submitted on the last day!  :mad:

I have no clue who submitted or not, since i got so many pms that a lot of my inbox was deleted :(

I'm just using my memory.

Anyways, round 2:

1. Bananas cost 3 dollars. Apples cost 4 dollars. Pineapples cost 5 dollars. Joe buys some fruit. Joe spent 20 dollars. Find the number of ways he could have bought those (i.e. 5 apples is different from 4 pineapples, etc.)

2. We have:

x+2y+3z = 100
y+2z+3w = 200
z+2w+3x = 250
w+2x+3y = 150

Find (x+z)(y+w).

3. How many ways are there to put 1,2,3,4,5,6,7,8,9 in a 3x3 grid so that every 2x2 subsquare has an even sum?

4. ABC is a triangle with angle A = 60. If AB = 2008 and all sides have positive integer lengths, find the product of the possible values of side AC. (Give me the answer 0 if you believe that there are no such triangles).

5. P and Q are polynomials such that if a and b are complex numbers such that if P(a) = P(b) then Q(a) = Q(b), prove that there exists some polynomial H with P(x) = H(Q(x))

A bracket has already been made, and you are not in it.

You are first on the replacement list though; you can submit, and if it turns out that i need a replacement, you'll be in.

[quote="modularmarc101"]Oh okay. So are the problems for this round the one pythag011 posted?[/quote]

Yes. When you are done, submit them to ernie though  :)

Points of each problem pleased! I assumed that points increased when # of problems increased(so, #5 has the largest point).


The probability of getting a blue marble on the first try

is $ \frac{1}{3}$ then the probability of a yellow is

$ \frac{1}{2}$ so

$ \frac{1}{6}$ is the answer.

[quote=Possible][quote=GameBot]A red marble, a blue marble and a yellow marble are placed in a bag. A marble is drawn and not returned to the bag. A second marble is then drawn. What is the probability that the first marble is blue and the second marble is yellow? Express your answer as a common fraction.[/quote]

The first solution is incorrect.
The actual answer is 1/2.[/quote]

The first solution looks right to me. How did you get 1/2?


Let $ C$ be a circle of radius $ 1$ and $ D$ be a diameter of $ C$.  Let $ P$ denote the set of points in $ C$ whose distance from $ D$ is less than their distance from the boundary of $ C$.

(formally, $ P\equal{}\{x\in C: \text{dist}(x,\partial C)>\text{dist}(x,D)\}$)

What is the area of $ P$?

I could have sworn that I posted this in Calculus Computations and Tutorials.  Did someone move it?


I'm having some trouble with probability questions.

9. A cafeteria offers vanilla and chocolate ice cream, with or without fudge sauce.  The manager kept records on the last 200 customers who ordered ice cream.  What's the probability of the choices below?   (I don't understand the difference between some of the letters, for example, a, and b.  Are they not the same thing?)

                   Fudge Sauce     No fudge sauce       Total
Vanilla                64                         68                 132
Chocolate           41                        27                   68
Total                  105                       95                 200

a. vanilla icre cream with no fudge sauce
b. vanilla icre cream, given that it does not include fudge sauce
c. no fudge saurce, given that it is for vanilla ice cream
d. vanilla ice cream
e. chocolate ice cream, given that it includes fudge sauce
f. fudge sauce, given that it is for chocolate ice cream

10. Are the events "an order is for chocolate ice cream", and "an order includes fudge sauce" independent?

11. Given a family with two children.
a. Find the sample space.
b. If it is known that at least one of the children is a girl, find the probabilit that both are girls.
c. If the older child is a girl, find the probability that both children are girls.

12. A single card is drawn from a standard 52 card deck.  Let B denote the event that the card is black, and let D denote the event that it is a diamond.
a. Are events B and D independent?
b. Are events B and D mutually exclusive?

Thanks!

Probablility is hard, but hopefully I'll make some progress.  Right now, I have to study for a school test pretty soon, so I'm not really going in depth, mostly focusing on getting the stuff I need to know for the test down...:)

To understand 9A vs 9B, you need to understand the concept and language of "conditional probability" which then leads to the definition of "independent evets".   9A. is an "intersection" or "and" problem whereas 9B is "conditional probability"

9A.  Of all 200 customers, 68 had 'vanilla and no fudge'. P=68/200

9B.  Of the 95 'no fudge' customers, 68 chose vanilla:  P=68/95

10.  Independence is basically asking, "Does knowing the probability that the customer chose fudge sauce alter the probability that the customer chose chocolate?"  If "no", then the events are independent.  In this case, compare:  "prob of choosing chocolate" = 68/200 = 17/50 (34%)" to "prob of choosing chocolate if you know they chose fudge sauce" = 41/105 (about 39%).  Not equal, thus Dependent events.  That is, learning that they chose fudge sauce increased the probablity that they also chose Chocolate.

Hope this helped in time for test. :)


let $p$ be a prime and odd number and $A,B$ be two sets such that:$A\cap B=\emptyset \ \ ,\ \ A\cup B=\{1,2,\dots,p-1\} $
and forall$a,b \in A\ ,\ c,d\in B$we have $a\times b\in A\ ,\ c\times d\in A\ ,\ \ \ a\times c,a\times d,b\times c,b\times d \in B$ Prove that:

$A=\left\{\begin aaa: \left(\begin {\frac{a}{p}}\right)=1 \ ,\ 1\leq a\leq p-1 \right\}$

$B=\left\{\begin bb: \left(\begin  {\frac{b}{p} }\right)=1 \ ,\ 1\leq b\leq p-1 \right\}$

Well, I think what you want is:
Show that the only surjective homomorphism from $\mathbb{Z}_p^*$ to $\mathbb{Z}_2$ is the Legendre-symbol.
But this follows eysily from the fact that $\mathbb{Z}_p^*$ is cyclic, so isomorphic to $\mathbb{Z}_{p-1}$.

Actually I don't know what is $\mathbb{Z}_p^*$
would U please solve it easier

$\mathbb{Z}_p^*$ is meant to be the prime residue classes $\mod p$, so all different from $0$.

And the rest follows from the existence of a primitive root $\zeta \mod p$:
If $\zeta \in A$, all residue classes would be in $A$, which I think is excluded as trivial case.
So $\zeta \in B$, so all even powers of $\zeta$ are in $A$ and the odd ones in $B$.


ceiling function
ratio
combinatorics unsolved
combinatorics

A necklace consists of 100 blue and several red beads. It is known that every segment of the necklace containing 8 blue beads contain also at least 5 red beads. What minimum number of red beads can be in the necklace?

[i]Proposed by A. Golovanov[/i]

If the necklace is linear (and not circular), I claim that the answer is 70. An alternating pattern of 7 blue beads and 5 red beads achieves this number, and this clearly satisfies the condition since any segment that contains 8 blue beads will have to contain blue beads from more than one of these sections, and thus contain the 5 red beads in between.

To show that this is sufficient, let's just consider the first 99 blue beads. Consider the segments from the 1st to the 8th, the 8th to the 15th, the 15th to the 22nd, and in general the $ 7k\plus{}1$st to the $ 7k\plus{}8$th bead. There will be 14 of these segments, and each needs to contain 5 red beads. However, a red bead can only lie in one of these segments since they each overlap in exactly one blue bead and no other beads. Thus, 70 red beads are required.

For a circular necklace, this requires some fixing, which I'll think about later.

[quote="Hamster1800"]If the necklace is linear (and not circular), I claim that the answer is 70.
For a circular necklace, this requires some fixing, which I'll think about later.[/quote]

   It is circular. First, I believe a necklace must be on somebody's neck (beautiful girls are always preferred), and second, for a string (as I think a linear necklace is properly called) the problem is way too simple.

For each blue bead give it a value of $ 1$ and each red bead give it a value of $ -1$ and for any segment $ s$, $ f(s)$ is the sum of values of these beads. The problem tells us for any segment $ x$ containing $ 13$ beads we know $ f(x) \le 3$. Let there be $ r$ red beads.
Now summing over all segment of length $ 13$ we have
$ 13(100-r)=%Error. "sumf" is a bad command.
(x) \le 3(100+r)$ so $ 14r\ge 1000$ and $ r\ge72$. I am not sure if this bound is the best one though. :)

[quote="Ahwingsecretagent"]
$ 13(100 - r) = %Error. "sumf" is a bad command.
(x) \le 3(100 + r)$ so $ 14r\ge 1000$ and $ r\ge72$.[/quote]

Is not $ 16r \ge 1000$? then $ r \ge 63$

If we, instead of considering segments of 13 beads, consider segments containing exactly 8 blue beads, I believe we can obtain a bound of 72. This is because each of the 100 segments requires 5 red beads, but each red bead will be in 7 segments, so we need at least $ \lceil\frac{500}{7}\rceil \equal{} 72$ red beads. It feels like this should be a tight bound (since a 7-5 ratio works for necklaces with $ 7k$ blue beads for some $ k$), but it's not as immediate as it would be if it were 98 blue beads, for example.

Sorry I didn't check my previous post. It should have been
$ 13(100\minus{}r) \equal{} \sum_{x}f(x)\le 3(100\plus{}r)$. And yes it should have been
$ 16r\ge 1000$ so $ r\ge 63$. Thank you for correcting me mszew. I think this is the correct bound.

[quote="Hamster1800"]If we, instead of considering segments of 13 beads, consider segments containing exactly 8 blue beads, I believe we can obtain a bound of 72. This is because each of the 100 segments requires 5 red beads, but each red bead will be in 7 segments, so we need at least $ \lceil\frac {500}{7}\rceil \equal{} 72$ red beads. It feels like this should be a tight bound (since a 7-5 ratio works for necklaces with $ 7k$ blue beads for some $ k$), but it's not as immediate as it would be if it were 98 blue beads, for example.[/quote]

One example with $ 72$ is:
$ RBRB$ and $ 12$ consecutive blocks of $ BRBRBRBBRBRB$


[b]Rules for the Collection[/b]
[u]What belongs here?[/u]
Any interesting or noteworthy chemical fact that you want to share with others.  Ideally, at some point we'll end up categorizing the Collection, but for now let's just start building it up.
[u]What does not belong here?[/u]
Off-topic discussion.  Corrections to existing content are acceptable (and indeed encouraged); however, corrective posts will be removed after the relevant error is fixed.

(This sticky is an implementation of an idea due to pardesi.)

At temperatures approaching, absolute zero, odd things happen due to Bose-Einstein condensates.

Another one: 1 amu is defined to be the mass of $ \frac{1}{12}$ of an atom of C-12 in its unbound ground state.

The relevant posts from the old Collection can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179750]here[/url].

If you have a copy of the AP Chemistry Reference Tables, that's full of equations.

Sodium Acetate is a supersaturated liquid at room temperature. When force is transferred into it, it crystalizes, which is substantially exothermic. When it is heated up to a bit over 100F/40C, it turns back to a liquid. It is a byproduct of baking soda and vinegar reactions, and is used to flavor salt and vinegar potato chips.

[quote="greenpepper9999"]Radon glows when you freeze it.[/quote]
It's a noble gas, which doesn't exist as a solid or liquid. It glows when it has an electric charge running through it.

Silicon Aerogel is the lightest solid at STP currently known to man. It is also the best insulator, it can support thousands of times it's weight, is made out of the same stuff as glass, and holds 15 world records.

Radon has a freezing and melting point. Also, aerogel is not the lightest solid known, certain carbon structures are.

"STP" stands for standard temperature and pressure. When something is at STP, it is commonly denoted by a degree symbol.
e.g. If you want the heat of fusion of something at STP, it would be $ \triangle H_f ^\circ $

Hmm, there are only one metal and one nonmetal that are liquids at room temperature.

Metal-mercury
Nonmetal-bromine

The word "element" is missing. Also, this depends on pressure; and room temperature ;)
My gallium is regularily melting on hot summer days.

Oh yeah, you're right.  :oops:

Here's a good one.
Metallic bonds form when metallic cations are attracted to the free valence electrons in the electron cloud get attracted to the nucleus.

[quote="JigglykongisFUM16"]
Metallic bonds form when metallic cations are attracted to the free valence electrons in the electron cloud get attracted to the nucleus.[/quote]
This sentence doesn't make sense...

Let me try again.

Metallic bonds form when metallic cations are attracted to the free valence electrons in the electron clouds of other metallic atoms, which then get attracted to the atomic nuclei of the cations.

(Usually transition metals)

The mass of 6 deuterium atoms is greater than the mass of 1 atom of Carbon12, despite the fact that both sets have the same number of protons, neutrons, and electrons. This difference in mass can be determined using $E=mc^2$.

Edit: Ok, well the difference in mass is from the nuclear force and if given the energy of that nuclear force, it would be possible to calculate the actual mass of 12C.

[quote="NT2048"]The mass of 6 deuterium atoms is greater than the mass of 1 atom of Carbon12, despite the fact that both sets have the same number of protons, neutrons, and electrons. This difference in mass can be determined using $E=mc^2$.[/quote]
I want you to demonstrate to me how this single equation implies anything on their masses ;)
What you actually want to say is, that like for any other nuclear reaction, the difference of masses is released (or consumed) as energy according to the formula, which by the way is true for any case where energy is involved: the masses change accordingly. Thus two $H_2$ and and $O_2$ weigh more than two $H_2 O$.

If you cool mercury with liquid helium to below -452°F, it lost all electrical resistance and became an ideal conductor. It would be like cooling and iPod down to hundreds of degree below zero and finding out that the battery remain fully charged no matter how long you use it. From The Disappearing Spoon by Sam Kean.

Bricks are funny

But in all seriousness
Stereoisomers are molecules that have chirality centers and are chiral.
Molecules with two or more chirality centers but are achiral are called meso compounds.
Enantiomers are compounds that are mirror images of each other.
Diastereomers are stereoisomers that are not enantiomers.

Why can't' people get serious talk about chemistry instead of bricks? :bruce:

As this stood around without correction for quite some time now:
[quote="ScienceLover969"]The Big Bang theory is proven to bee false beacuse you cannot fuse elements heavier than nickel and according to the big bang elements fused together to get all of our current ones.[/quote]
That's just complete nonsense. About everything in it is wrong, so I won't even start fixing every detail. Don't post such crap if you obviously have no clue about the facts (your nickname is chosen very poorly...).

Just let it be said that the Big Bang mostly produced hydrogen and helium, stars produce everything up to iron, and supernovae produce everything.


geometry
trapezoid
geometric transformation
homothety
geometry proposed

Consider an isosceles trapezoid $ ABCD$ with $ AD$ parallel to $ BC$. In $ \triangle ABD$, a circle tangent to $ AB$ and $ BD$ intersects $ AD$ at $ M$ and $ N$. Now let $ X$ and $ Y$ be intersection of $ CM$ and $ CN$ with the incircle of $ \triangle ACD$ closer to $ C$. Prove that $ XY$ is parallel to $ AD$.

Suppose $ BA$ and $ BD$ are tangent to the said circle at $ U,V$.
Let $ U',V'$ be points on $ CA$ produced and $ CD$ produced such that $ AU \equal{} AU'$ and $ DU \equal{} DU'$.
Let $ (C)$ be the circle tangent to $ CU'$ and $ CV'$ at $ U',V'$ respectively.
Let $ (C)\cap AB \equal{} \{M',N'\}$
$ AM'\cdot AN' \equal{} AU'^2 \equal{} AU^2 \equal{} AM\cdot AN$
$ DN'\cdot DM' \equal{} DV'^2 \equal{} DV^2 \equal{} DN\cdot DN$
$ M \equal{} M'$ and $ N \equal{} N'$
Since both $ (C)$ and the incircle of $ CAD$ touch $ CA$ and $ CD$, the two circles are homothetic.
This homothety maps $ M,N$ to $ X,Y$. Hence $ XY\parallel{}MN$


integration
complex analysis
complex analysis unsolved

Compute the following integral.

$ \int^{\infty}_{0} \frac{x^2}{x^8\plus{}1}dx.$

I think that I should get $ \frac{1}{4} \sqrt{1\minus{}\frac{1}{\sqrt 2}} \cdot \pi$. I see we know this is even. Then there is work to show that
$ \int^{\infty}_{\minus{}\infty}$ is$ 2\pi i$ times the sum of the residues of $ \frac{z^2}{z^8\plus{}1}$ in the upper half plane. Then I found that

$ c_k \equal{} \text{exp}[i(\frac{\pi}{8}\plus{}\frac{2k\pi}{8}]$ where $ k\equal{}0,1, \ldots, 7.$ Then the zeros in the upper half plane occur at $ k\equal{}0,1,2,3$.

Then I used

$ \text{Res}_{z\equal{}c_k}\equal{}\frac{c_k^2}{8c_k^7}\equal{}\frac{1}{8c_k^5}\equal{}\frac{1}{8}c_k^{\minus{}5}$ where $ k\equal{}0,1,2,3.$ Then when I add these residues up I did not get the answer above, I got $ \frac{\sqrt{\sqrt{2}\plus{}2} \cdot \pi}{16} \minus{} \frac{\sqrt{2\minus{}\sqrt{2}} \cdot \pi }{16}i.$

I agree with you until you say "when I add these residues up" ... your answer seems to be wrong.  I get the one you expect.


Usually I am good at the geometry problems, but this one I got stuck on. I just want a hint, so hide the solution if you give it.

You have triangle ABC with AB=5, AC=6, and BC=7. Point X is on BC so that the sum of the areas of the circumcircles of ABX and ACX is minimized. What is BX?

[hide="hint"]
the area of a circle that goes through two points is minimized when the two points are on the circle's diameter.

[/hide]

Thanks, Circlesquared. So I guess the altitude gives the minimum. By herons, the area of the whole triangle is $ 6\sqrt {6}$, so the height is $ 12\sqrt {6}/7$ and by Pythag we have $ BX \equal{} 19/7$, which is correct according to the solution's manual. They give a whole long explanation on why the altitude is the key instead of just saying what you said.

$ \frac{a}{\sin A}\equal{}2R$. In this case, we must minimize \[ \frac{b^2}{4\sin^2 \theta}\plus{}\frac{c^2}{4\sin^2\pi\minus{}\theta}\equal{}\frac{b^2\plus{}c^2}{4\sin^2\theta} \] where $ \theta\equal{}\angle ADC$. It is obvious that this is minimized when $ \theta\equal{}\frac{\pi}{2}$.

[quote="aidan"]Just another question, what if you want to maximize/minimize (I don't know which it should be, will work that out later) the areas of the [b]incircles[/b] of the triangles $ ACX$ and $ BCX$?[/quote]

Minimize it by putting X = C. :P

[quote="aidan"]And then maximizing it?  :D[/quote]

Think it would be BX = 0 or CX = 0.

But $ X \equal{} C$ means $ CX \equal{} 0$ and $ X \equal{} B$ means $ BX \equal{} 0$  :maybe:


Hello,

As I usually hear that people from Pro-Olympiad should learn to demonstrate, here it goes an easy one to practice.

Let $m$ and $n$ two positive integers such that $m|n$ and $mn$ is a perfect square. Prove that $\frac{m}{n}$ is also a perfect square.

It is easy, but try to write it very very carefully (and elegant!). It shouldn't take more than a few lines (for sure less than ten, even very well written)

Best,

[quote="probability1.01"][hide]Since $m | n$, $n = km$ with $k \in \mathbb{N}$

$\sqrt{mn} = \sqrt{km^2} = m\sqrt{k}$ is a natural number.  $\therefore \sqrt{k} \in \mathbb{N} \implies k$ is a square.  $k = \frac{n}{m}$ is therefore also a square.[/hide][/quote]

Correct :first:
I will see if something else similar comes to my mind soon.

Although its basically the same thing as probability 1.01 did, you can also use factorization and prove that the powers of all the prime factors of both $mn$ and $\frac m n$ are even.


geometry
3D geometry
tetrahedron
trigonometry
AMC
AIME

In tetrahedron $ABCD$, edge $AB$ has length 3 cm.  The area of face $ABC$ is 15 $\text{cm}^2$ and the area of face $ABD$ is 12 $\text{cm}^2$.  These two faces meet each other at a $30^\circ$ angle.  Find the volume of the tetrahedron in $\text{cm}^3$.

Still needs some corrections, I think. :P

[hide]The height of $ABD$ is $2\left(\frac{12}{3}\right) = 8$
Assume WLOG that $ABC$ is the base of the tetrahedron.
Since they are "hinged" $30^\circ$ apart, the height of the tetrahedron is $8\sin{30^\circ}=4$
$V = \frac{bh}{3} = \frac{15*4}{3} = \boxed{20}$[/hide]

What is edge ABC?
How can you have an edge with 3 vertices?

The problem should read:

In tetrahedron $ ABCD$, edge [color=red]AB[/color] has length 3 cm. The area of face [color=red]ABC[/color] is $ 15\text { cm}^2$ and the area of face $ ABD$ is $ 12 \text { cm}^2$. These two faces meet each other at a $ 30^\circ$ angle. Find the volume of the tetrahedron in $ \text {cm}^3$.

[quote="alan"][hide]The height of $ ABD$ is $ 2\left(\frac {12}{3}\right) \equal{} 8$
Assume WLOG that $ ABC$ is the base of the tetrahedron.
Since they are "hinged" $ 30^\circ$ apart, the height of the tetrahedron is $ 8\sin{30^\circ} \equal{} 4$
$ V \equal{} \frac {bh}{3} \equal{} \frac {15*4}{3} \equal{} \boxed{20}$[/hide][/quote]

How did you transform two faces which hinged 30 degrees apart to a right triangle?

This is almost the same problems as AIME 1992 #7!!! :ninja:

[quote="TZF"]There is a unique line through D that is parallel to AB. If D is allowed to vary along this line, then neither the area of ABD nor the volume of ABCD will change (why?). [/quote]

I see why the area doesn't change but why doesn't the volume change?  I mean I get it intuitively, I'm just not sure that what I'm thinking is rigorous.   :(  Help please?

[quote="cdymdcool"][quote="TZF"]There is a unique line through D that is parallel to AB. If D is allowed to vary along this line, then neither the area of ABD nor the volume of ABCD will change (why?).

So, let's move it to a more convenient place. In fact, let's move it so that the projection of D onto AB is the same point as the projection of C onto AB; suppose we call this point P.

If we do this, we get a very nice property -- the projection of D onto plane ABC is on segment PC! Let this point be Q.

Because Q is the projection of D onto plane ABC, QD is an altitude of tetrahedron ABCD. But because Q is the projection of D onto PC, QD is also an altitude of triangle DPC. So, if we want to compute the height of ABCD, we can just compute the height of triangle CPD.

We know CP is an altitude of ABC so we can compute CP. And we know CP and DP are separated by 30 degrees, so we can compute QD within triangle CPD. Then we can multiply this by 1/3 * Area(ABC), which is given.[/quote]

I'm sorry but whats "the projection of D onto AB is the same point as the projection of C onto AB; suppose we call this point P."  What do you mean by projection?[/quote]

I was also wondering this.

[quote="Truffles"]I found it a much easier solution to just find the height using a 30-60-90 triangle.  So let X be the foot of the perpendicular from D to AB.  $ DX \equal{} 2 \cdot \frac{[ABD]}{AB} \equal{} 8$  The height from D to face ABC is part of 30-60-90 triangle including DX so the height is 4.  The area of the tetrahedron is $ \frac{1}{3} \cdot 15 \cdot 4 \equal{} 20$

[wow I have the last 4 posts on this page xD sorry about that][/quote]
I still don't understand how you got that 30-60-90 triangle; Maybe I need to be more imaginative XD

here you go (you've probably left the forum now)


function
integration
calculus
limit
real analysis
absolute value
real analysis unsolved

Assume that $f\in L_\text{loc}^{1}(\mathbb{R}^{n})$ and $x\in \mathbb{R}^{n}$. Show that
\[t\mapsto m(B(x,t))^{-1}\int_{B(x,t)}f(y)\text{ d}y \]
is continous as  $t\in (0,\infty)$.

The measure of the ball is just a constant multiple of $t^{n}$; dividing by $t^{n}$ won't affect continuity. That leaves the integral; we need to show that $\lim_{z\to 0^{+}}\int_{B(x,t+z)-B(x,t)}f=0$ and $\lim_{z\to 0^{+}}\int_{B(x,t)-B(x,t+z)}f=0$.

More generally, if $S_{n}$ are nested sets ($S_{n+1}\subset S_{n}$) with $\cap S_{n}$ empty and $f$ is any function integrable on $S_{1}$, $\lim \int_{S_{n}}|f|=0$. This is a basic fact about Lebesgue integration, closely related to the convergence theorems (it's a special case of the dominated convergence theorem). By countable additivity, $\int_{S_{1}}|f|=\sum_{n}\int_{S_{n}-S_{n+1}}|f|$; from convergence, we get the desired result.

This is a convolution. (The title of the topic is a little misleading. This convolution is not itself the maximal function; the maximal function would be a supremum over $t>0$ of integrals like these, only with an absolute value somewhere). The maximal function itself won't be continuous, although it will be ____ - semi-continuous. (I'll let you decide what word we need to fill in the blank.)

Here's a different "more generally" than the one jmerry stated:

If $f\in L^{1}(\mathbb{R}^{n})$ and $g\in L^{\infty}(\mathbb{R}^{n}),$ then $f*g$ is continuous. In fact, we can say more: $f*g$ is uniformly continuous and tends to zero at infinity, and $|f*g(x)|\le \|f\|_{1}\|g\|_{\infty}$ for all $x.$

That doesn't quite apply here, because we were told that $f$ was locally $L^{1}$ but not globally $L^{1}.$ But $g,$ which was the normalized characteristic function of a ball, was compactly supported. Hence, $f*g(x)$ does not depend on all values of $f$ but only on those on some bounded set, and our desired conclusion - that $f*g$ is continuous - is a local conclusion.

We also have that if $f\in L^{p},\ g\in L^{p'},$ where $\frac1p+\frac1{p'}=1,\ 1\le p\le\infty,$ then $f*g\in C_{0}$ (the continuous functions that tend to zero at infinity) and $\|f*g\|_{\infty}\le\|f\|_{p}\|g\|_{p'}.$

talking about maximal function (this one with the sup, and the Lebesgue measure). Show that if $f \in L^{1}_{loc}(\mathbb{R})$ is continuous at $x_{0}$, then $Mf(x_{0})$ is also continuous at $x_{0}$.


algebra
polynomial
quadratics
geometry
3D geometry
inequalities
AMC

here are a couple from a sample test on the website:

1) find the real solution to the equation x^3 + 3x^2 + 3x = 3. and, aside, what are some typical strategies to find real, but irrational, solutions to polynomials of deg > 2 when you cannot find enough rational roots to hack down the original polynomial to a quadratic (by dividing)?

2) what is the minimum possible value of (a/2b) + (b/4c) + (c/8a), where a, b, c are positive real numbers?

Thx.

Fierytycoon

I can answer #1 pretty quickly.  Although I'm only using a special case to solve this one, there is usually some hidden "trick" that you must observe to solve problems like these.

[quote]1) find the real solution to the equation x^3 + 3x^2 + 3x = 3[/quote]

I couldn't help but notice that x^3 + 3x^2 + 3x is a lot like the expansion of (x+1):^3:.  If you add a 1 to both sides, you get:
x:^3: + 3x:^2: + 3x + 1 = 4
(x+1)^3 = 4
x + 1 = 4^(1/3)
x = -1 + 4^(1/3)

So in this case, I just completed the cube, in a way similar to the method of solving quadratic equations by completing the square.  You could find the imaginary solutions (there are 2) by dividing the original polynomial by (x+1-4^(1/3)).

-mathfanatic

For the second question, we use the case n=3 of the arithmetic mean-geometric mean inequality. In this case we have (x+y+z)/3>=(xyz)^(1/3), where x,y,z are all positive. So here we have (a/2b+b/4c+c/8a)/3>=(1/64)^(1/3), or a/2b+b/4c+c/8a>=3/4. It's always a good idea with these sorts of problems to make sure that it is possible to achieve the minimum value (3/4). If this is the case, we need a/2b=b/4c=c/8a=1/4, since the only equality case of the AM-GM inequality occurs when x=y=z. We can easily check that a=1, b=2, c=2 satisfies this condition, and thus 3/4 is the minimum value for this expression.

Thanks for the solutions. One quick thing though:

[quote]. It's always a good idea with these sorts of problems to make sure that it is possible to achieve the minimum value (3/4). [/quote]

Why do you say that? Does it often so occur that, for some reason, it is not possible to achieve some min/max value in a problem?

Fierytycoon

It can be that the minimum is not achievable. For problems like this one the minimum is generally achievable, but sometimes we bound things especially with more complicated expressions with methods like this, and maybe we cannot reach the bound. At the very least, on a competition like the USAMO, you won't get full credit if you don't check.

hmm, do you happen to know a problem off the top of your head in which what you said is true?

Fierytycoon

[quote="ComplexZeta"]Someone posted this problem to the MOP 2002 email list. It doesn't have to do with this directly, but I dared to assume some equality cases of the AM-GM and got horrible contradictions. I didn't imagine it would be sound anyway, but I felt I had nothing to lose by trying it. Incidentally, no one posted an acceptable answer to this post, so I'd be interested to know if anyone can solve it.

Let ABC be a triangle with angle A=60 degrees. Let P be a point in the plane of ABC such that AP=1, BP=2, and CP=3. Find the maximum area of triangle ABC.[/quote]

This seems to be a particularly difficult problem. I can turn it into a max-min problem and solve it numerically, but I can't seem to find a closed form solution to it.

What I did was first to show that P must be interior to the triangle. That shouldn't be too hard to see why. Then I called the angle CAP theta, and I dropped perpendiculars from P to AC and from P to AB. The lengths of these perpendiculars are sin (theta) and sin (60-theta) respectively. So from there you can see AC = cos (theta) + sqrt(9 - sin^2 (theta)) and AB = cos (60-theta) + sqrt(4 - sin^2 (theta)).

The Area of the Triangle is 1/2*AB*AC*sin(60) = sqrt3/4*AB*AC. So you have a function of a single variable, theta, that you want to maximize. I get something like 4.6644101 happening at theta = 110/3 degrees. I don't know what the closed form of that number is though.

[quote="ComplexZeta"]
Let ABC be a triangle with angle A=60 degrees. Let P be a point in the plane of ABC such that AP=1, BP=2, and CP=3. Find the maximum area of triangle ABC.[/quote]

I was never able to solve this in closed form, but I found a solution here: http://www.mathlinks.ro/phpBB/viewtopic.php?t=48

But in any case, if a problem asks you to find the minimum or maximum value, and you prove that it is greater (or less than) 3/4, or whatever, you haven't actually solved the problem. You would lose marks if you don't actually *show* its the max/min.


Let $n$ be a positive integer and $a, b$ integers such that

$gcd(a, b, n) = 1$

Prove that there exist integers $a_{1}, b_{1}$ with

$a_{1}\equiv a \bmod n$
$b_{1}\equiv b \bmod n$
$gcd(a_{1}, b_{1}) = 1$

[hide="Hint"] This is a little easy if you're familiar with Bezout's Lemma, which you can apply repeatedly.  If you're not, it might be hard, which is why I put it here. [/hide]

[b]t0rajir0u[/b], does the lemma also hold if we have $(a,b,c)$

[hide="That is to say"] $(a,b,c)=ax+by+cz$[/hide]

If so then this proof might work (I have no idea as I am not good with number theory or anything along those lines).

[hide="My feeble attempt"] $(a,b,n)=ax+by+nz=1$. Now let: $a_{1}=kn+a$ and $b_{1}=cn+b$. We have that $(a_{1},b_{1})=a_{1}x_{1}+b_{1}y_{1}=(kn+a)x_{1}+(cn+b)y_{1}=ax_{1}+by_{1}+n(kx_{1}+cy_{1})$. If we let $x_{1}=x$ and $y_{1}=y$ then select appropriate values $a_{1}$ and $b_{1}$ so that $kx+cy=z$.[/hide]

I probably made some stupid mistake somewhere.

[quote="BanishedTraitor"][b]t0rajir0u[/b], does the lemma also hold if we have $(a,b,c)$

[hide="That is to say"] $(a,b,c)=ax+by+cz$[/hide][/quote]

[hide="Well..."] Yes.  And you can prove it directly by the first lemma once you realize that

$(a, b, c) = ( (a, b), c)$ [/hide]

[quote="BanishedTraitor"][hide="My feeble attempt"] $(a,b,n)=ax+by+cz=1$. Now let: $a_{1}=kn+a$ and $b_{1}=cn+b$. We have that $(a_{1},b_{1})=a_{1}x_{1}+b_{1}y_{1}=(kn+a)x_{1}+(cn+b)y_{1}=ax_{1}+by_{1}+n(kx_{1}+cy_{1})$. If we let $x_{1}=x$ and $y_{1}=y$ then select appropriate values $a_{1}$ and $b_{1}$ so that $kx+cy=z$.[/hide][/quote]

I don't know why you have both $c$ and $n$ in your proof... anyway, you're close.  However, [hide="you havent justified..."] That you can find $x, y$ such that $kx+cy = z$. [/hide]

Any other attempts? Either to solve the problem or complete where I left off?


Hi I am monkeygirl12 I decide to start a topic very similar to a post from about 5 years ago. Here are the rules no insults, or inapporite content and remember have fun. Please List how you got your user name, and an interesting fact about you
Here an example.

Hi am Monkeygirl12
I am 12 years old. I choose my username because I act like a monkey and have an iq of an monkey. An interesting fact about me is I can read 2,000 words a minute.

Okay, okay... I'll give it a go.

Hello I'm meewhee009, I'm 11 years old. I chose my username for various reasons (the good times my friend). An interesting fact about me is... erm.... I can jump using a jump rope 42 times per minute while eating a banana.  :rotfl:  It's a talent that comes in handy every now and then.

Hi, I'm Bob and I'm 12 (one of only two true facts in this post :P ) I didn't get to choose a username because an rabid lab rat incapacitated me, and then registered for me. :( An interesting fact is that ferrets attack more people than grizzly bears (according to a Mountain Dew commercial.)

Hi I'm Roland, I'm 10+pi years old, my username was the result of a cataclysmic nuclear meltdown, and I hate stalkers.

I was hoping repliers could be a tad more realitic

It [i]is[/i] the fun factory. Besides, the information I omitted can easily be gathered from my username. I wonder what my name is and why my username is what it is. :D *nudges r15, who has no excuse*

ok I will play along
Hi i am monkeygirl12 and I am a real Monkey. I got my username because I am a monkey and my iq is 12.  2+2=Monkey

That is creative. I applaud your creativity  :clap:

Okay, fine, I'll get into the details.

My name is Roland Ylxylynyien Zambordo, I chose my username because I didn't want to spend as much time typing it as most other usernames, and I am an idiot.

My names maybach I'm 12 and my brain died about 7 years ago.....................

My IQ is probably around 2 pi...................

i'm really jgyao04, but that's ok. i'm 10 yrs old, and well, a tree fell on me when i went to Whipple hill(note that all these are true.)


MATHCOUNTS
geometry
geometric transformation
reflection
blogs

ummmm do you mean practice or the real thing, or either?

practice:
chapter sprint-28
chapter target-16(8)
chapter overall-41(thats 2 test from the same year)

state sprint-16
state target-6
state overall-NA

real thing:
chapter sprint-18
chapter target-14(7)
chapter overall-32

state sprint-16
state target-8(4)
state overall-24

I am currently an 8th grader in Oklahoma.

I don't know what you mean by practice, but I went up to last year's state(and I only know my state score..Sorry).

I am not that good, but I achieved a score of 21(13 in sprint, 8 in target).

That's not that good but I somehow got 16th in the state... :roll:

my best practice scores:

nats: 38 24/14

states:  43 27/16

chapter: 46  30/16

but my real scores at the competitions havn't come close to that, so these practice scores don't reflect how good i am.

Practice:
Chapter:46 (on atleast 2 of them)
State:42 (on 2003)

Real:
Chapter (37)
State (28...ouch)

xpmath are you in 7th or 8th grade? and the states one-not much of an ouch. i got 24, maybe even less. im pretty sure thats what it is though

I'm in 7th grade now. Maybe, but I made so many silly mistakes on target (I solved it all right, but messed up righting the numbers down). NOT COMPLAINING THOUGH

lol. i do that more on the sprint round and i dont read the question is my problem. also, i cant read them very fast

I got the two silliest (not really) sprints wrong:
On the how many knights and  whatever there are around a table, I got that wrong.
And on the moon,earth, and satellite one, I thought it said distance from moon to satellite over moon to earth.

what place did you get at states?

the knights one, it was just logic, was the satalite one with scientific notation.

one problem i have is skipping so many questions because i think they will take to long. i do the ones i can solve realy fast first. the longer ones may be easy to get right, but they just take too much time. i just end up timing out(states only) and i dont have the time to get the simple arithmatic ones.

I got 22nd :P
I know, the knights one was easy, I just messed up big.
I can't exactly remember about the satellite...

8th grade

chapter
23/16 = 39 (11th)

state
18 ( :ninja: )/14 = 32 (hehe, easy state, 7th written, 4th overall)
this was a 14 pt improvement from 7th, so remember you can always improve

nats
15/10 = 25 (114th)

the knights/knaves problem was really easy...

Anyways, heres mine in da real thing (i never made it to state)
chapter sprint: 19
chapter target: 12 (6)
chapter score: 31

[quote="mathcrazed"]
$ 34$ in Indiana would have been like 6th or 7th lol... scary... at least, I think so.[/quote]
thats state right? in pennsylvannia, it would take a 40 to get 1st i think. something like that

34 got 8th and 9th.

trust me lyndon, i would know. -_-

lol... hahahahah... Edit out the name plz... :D Feel sry for you

[quote="abacadaea"]are you kdding?!??!?!?! you litterally need a perfect score to get in the countdown round. do you live in california?
either way, thats crazy. i wonder why they still cant win nats....[/quote]\
isn't a 46 a perfect score?

[quote="abacadaea"][quote="mamathbug"]Practice
---------------------------
Chapter

Sprint 30 - Target 8

State

Sprint 30 - Target 8

Nationals

Sprint 29 - Target 8

don't want to give away real (sorry guys, don't want you to track me down - i got a perfect on chapter, i'll say that)[/quote]
did yoiu time yourself?[/quote]

Yeah, and if you want the year I believe it was 1998.

To all the coaches out there, what we did is we had some Saturdays where we would go to someone's house, and do a chapter target/sprint then a state target/sprint and then a national target/sprint. This worked really well since all the kids would know at which point they were missing problems. And by the way, there *was* a time where I could get those scores on nationals, but that was my highest (never managed to get a 46, however).

Ok, hope that helps!

All states r different, and have remarkably different score ranges.  For eg, in my state, a 28 was enough to get u on the nats team, but in other states, u would need 40+ to make the nats team

[quote="diophantient"][quote="13375P34K43V312"][quote="diophantient"]My best mathcounts score is a 24 at chapters.[/quote]

PFT YEAH OK[/quote]
What's that supposed to mean?[/quote]
Shut Up

I'm from Georgia, and frankly, the competition in middle school nowadays isn't that big.  Here are my best scores:

School: dunno
Chapter: dunno
State: 33- 19/14 (I know, I'm steeeupid)
Nats: 24- 18ish/6 (again, I'm steeeeupid)

My best finish (last year), suprisingly:
Chapter: 1st
State: 1st (I know, right?)
Nats: 99th (Well my goal was to get a double digit placing)

But I think my math's a bit better now thanks to AOPS  :lol:

[quote="Math Geek"]But I think my math's a bit better now thanks to AOPS  :lol:[/quote]
what grade are you in? and yes thats the same case with me :)

I'm in 8th grade, though I guess it wasn't really a good idea to skip 3rd grade, because the competition in 8th is sooooo much greater than that in 7th grade. :(

Guys don't regret skipping a grade 5 years ago because a) it probably wasn't your choice and b) it doesn't matter - since you would have gotten through the *exact* same thing...

It's not so much your age as what grade you're in...


If a stack of eight quarters is exactly one-half inch high, how many quarters will be needed to make a stack one foot high?

$ 8*2\equal{}16$ for 1 inch
now foot=12 inch
so $ 16*12\equal{}64*3\equal{}192$ for a foot


Two different base 120 units digits are chosen randomly and their product is written in base 120. What is the probability that the units digit of the base 120 product is 1??

Im just wondering if for example the numbers 1B 2B 3B(11 21 31) are considered to have an  units digit of one :huh:  :maybe:

[quote="t0rajir0u"]$ 1B_{120} \equal{} 131$, not $ 11$.  Are you sure you understand what base $ 120$ is?  That number has a units digit of $ B$.[/quote]

OH, so how can u express the number lets say.. 19 in base 120?? would it be J?? OR how about a number out of the alphabet?

Well, you need $ 120$ different symbols.  For that reason, base $ 120$ isn't very practical to write out.  That doesn't matter.  What matters is this:  a number has a units digit of $ 1$ in base $ 120$ if and only if it is congruent to $ 1 \bmod 120$.

Hmm ic....but since 120 is a multiple of 10 then only numbers such as(A1 base 120, 91 base 120......)(numbers that end in 1 both when expressed in base 10 and 120) are the numbers that leave a rem of 1 right?So lets say that  we were to count these units digits 1 by one one without using "inclusion-exclusion" they would have to be pairs that end in the units digit (3,7) (1,1) or (9,9) or stuff like (B,B)aka(11,11)

Is this correct?

PS:I could rewrite this if its not clear :oops:  :blush:  :D

Rem=remainder

This is what the question is asking:

Pick two integers $ a, b \in [1, 120]$.  What is the probability that $ ab \equiv 1 \bmod 120$?

Think about it this way:

Given an integer $ a \in [1, 120]$, how many integers $ b \in [1, 120]$ have the property that $ ab \equiv 1 \bmod 120$?

[hide="Hint"] There are two cases: $ \gcd(a, 120) \equal{} 1$ and $ \gcd(a, 120) > 1$. [/hide]

[quote="t0rajir0u"]This is what the question is asking:

Pick two integers $ a, b \in [1, 120]$.  What is the probability that $ ab \equiv 1 \bmod 120$.

Think about it this way:

Given an integer $ a \in [1, 120]$, how many integers $ b \in [1, 120]$ have the property that $ ab \equiv 1 \bmod 120$?

[hide="Hint"] There are two cases: $ \gcd(a, 120) \equal{} 1$ and $ \gcd(a, 120) > 1$. [/hide][/quote]

Thanks alot...i just needed to understand what the question was asking(i have a bad tendency of misinterpreting questions about base numbers  :rotfl: )


Theorem:

Let f be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a) = f(b)$ then $\exists c \in (a,b)$ such that $f'(c) = 0$. (c may not be unique)

Proof: If f is constant, then any point will do.
          If if is not constant, blah blah  blah...

What does it mean, when it says, $f$ is constant?

$f$ is constant means $f(x) = k$ for all $x$. Obviously, then $f'(x) = 0$ for all $x$.


modular arithmetic
algebra
polynomial
calculus
derivative
binomial theorem
number theory

I came across the following nice identity, but don't know how to prove it
(other than expanding both sides, using the binomial theorem for instance)

           (X+Y)^7-X^7-Y^7=7XY(X+Y)(X^2+XY+X^2)^2

Does anyone know how to prove it?

you mean ?
$ (x \plus{} y)^7 \minus{} x^7 \minus{} y^7 \equal{} 7xy(x \plus{} y)(x^2 \plus{} xy \plus{} x^2)^2$

ops Kouichi Nakagawa post the answer before me

As
$ (x\plus{}y)^3\minus{}x^3\minus{}y^3\equal{}3xy(x\plus{}y)$
$ (x\plus{}y)^5\minus{}x^5\minus{}y^5\equal{}5xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)$
$ (x\plus{}y)^7\minus{}x^7\minus{}y^7\equal{}7xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)^2$

can it be generalized?

[b]generalization[/b]

$ k\ge 3$

$ \begin{array}{l}
\left \{ \begin{array}{ll}xy(x\plus{}y) | (x\plus{}y)^k \minus{} x^k \minus{}y^k & (k\equal{}3\plus{}6n) \\
xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2) | (x\plus{}y)^k \minus{} x^k \minus{}y^k & (k\equal{}5\plus{}6n) \\
xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)^2 | (x\plus{}y)^k \minus{} x^k \minus{}y^k &  (k\equal{}7\plus{}6n) \\
\end{array} \right. \\
\\
\cdot \ \ \ k | (x\plus{}y)^k \minus{} x^k \minus{}y^k \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k \in \mathrm{Prime})
\end{array}$

More specific:
If $ n$ is even, then it is divisible by $ x,y$, but not by any higher power of these or by $ x\plus{}y,x^2\plus{}xy\plus{}y^2$.
If $ n \equiv 1 \mod 6$, then it is divisible by $ x,y,x\plus{}y,(x^2\plus{}xy\plus{}y^2)^2$, but not by any higher power of these.
If $ n \equiv 3 \mod 6$, then it is divisible by $ x,y,x\plus{}y,$, but not by any higher power of these or by $ x^2\plus{}xy\plus{}y^2$.
If $ n \equiv 5 \mod 6$, then it is divisible by $ x,y,x\plus{}y,x^2\plus{}xy\plus{}y^2$, but not by any higher power of these.

Try to prove the generalization without expanding anything!

[quote="josejacobi"]well...actually what I meant was if it is possible to factor the left side ((X+Y)^7-X^7-Y^7) to get the right side (7XY(X+Y)(X^2+XY+X^2)^2 ) without applying the binomial formula (which is what  Kouichi Nakagawa  did). Thanks for replying my post anyway...[/quote]

[b]Solution : Modular Arithmetic[/b]

if $ k\equal{}7\plus{}6n \Rightarrow xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)^2 | (x\plus{}y)^7\minus{}x^7\minus{}y^7$
So,
$ (x\plus{}y)^7\minus{}x^7\minus{}y^7 \equal{} kxy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)^2$
$ \iff k \equal{} \frac{(x\plus{}y)^7\minus{}x^7\minus{}y^7}{xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)^2} \equal{} 7$
(dimension about $ x(or \ y)$ is $ 6$ together.)
Thus, $ (x\plus{}y)^7\minus{}x^7\minus{}y^7 \equal{} 7xy(x\plus{}y)(x^2\plus{}xy\plus{}y^2)^2$

My solution is different: consider the roots of $ (x\plus{}1)^n\minus{}x^n\minus{}1$.

Thanks a lot for the solutions... I came across this identity while working on the following problem:

Let a, b, c be complex numbers, prove that if (a-b)^7+(b-c)^7+(c-a)^7=0 then a, b, c are the vertices of an equilateral triangle.

By substituting x=a-b, y=b-c and z=c-a we get x^7+y^7+z^7=0. Notice also that x+y+z=0. Thus, we get (x+y)^7-x^7-y^7=0. Using the identity it is easy to get the result

[quote="Kouichi Nakagawa"][quote="josejacobi"]well...actually what I meant was if it is possible to factor the left side ((X+Y)^7-X^7-Y^7) to get the right side (7XY(X+Y)(X^2+XY+X^2)^2 ) without applying the binomial formula (which is what  Kouichi Nakagawa  did). Thanks for replying my post anyway...[/quote]

[b]Solution : Modular Arithmetic[/b]

if $ k \equal{} 7 \plus{} 6n \Rightarrow xy(x \plus{} y)(x^2 \plus{} xy \plus{} y^2)^2 | (x \plus{} y)^7 \minus{} x^7 \minus{} y^7$
So,
$ (x \plus{} y)^7 \minus{} x^7 \minus{} y^7 \equal{} kxy(x \plus{} y)(x^2 \plus{} xy \plus{} y^2)^2$
$ \iff k \equal{} \frac {(x \plus{} y)^7 \minus{} x^7 \minus{} y^7}{xy(x \plus{} y)(x^2 \plus{} xy \plus{} y^2)^2} \equal{} 7$
(dimension about $ x(or \ y)$ is $ 6$ together.)
Thus, $ (x \plus{} y)^7 \minus{} x^7 \minus{} y^7 \equal{} 7xy(x \plus{} y)(x^2 \plus{} xy \plus{} y^2)^2$[/quote]

[quote="ZetaX"]More specific:
If $ n$ is even, then it is divisible by $ x,y$, but not by any higher power of these or by $ x \plus{} y,x^2 \plus{} xy \plus{} y^2$.
If $ n \equiv 1 \mod 6$, then it is divisible by $ x,y,x \plus{} y,(x^2 \plus{} xy \plus{} y^2)^2$, but not by any higher power of these.
If $ n \equiv 3 \mod 6$, then it is divisible by $ x,y,x \plus{} y,$, but not by any higher power of these or by $ x^2 \plus{} xy \plus{} y^2$.
If $ n \equiv 5 \mod 6$, then it is divisible by $ x,y,x \plus{} y,x^2 \plus{} xy \plus{} y^2$, but not by any higher power of these.[/quote]

Let $ f(x)\equal{}(x\plus{}1)^n\minus{}x^n\minus{}1$... let $ \omega$ be a root of $ x^2\plus{}x\plus{}1$.

consider $ f'(x)\equal{}n(x\plus{}1)^{n\minus{}1}\minus{}nx^{n\minus{}1}$ and $ f''(x)\equal{}(n)(n\minus{}1)(x\plus{}1)^{n\minus{}2}\minus{}(n)(n\minus{}1)(x)^{n\minus{}2}$

We note that $ 1\plus{}\omega\equal{}\minus{}\omega^2$...also $ \omega^3\equal{}1$. That lets us evaluate the above polynomials completely. We consider the derivatives to show the multiplicities...


Hello!

I'm pondering about generating all numbers $ x$ so that each of the digits from 0 to 9 appears exactly once in either $ x$ or $ 2x$. It's easy to see that 48651 satisfies this criterion since all the digits 0-9 occur exactly once in either 48651 or 97302.

From the above facts we can conclude that what I'm actually looking for is a number x $ x * (10^5 \plus{} 2) \equal{} x * (100002) \equal{} p$ so that $ p$ is a pandigital number.

That way I'd only have to check for numbers $ 10000 < \equal{} x < \equal{} 99999$ and see if $ x*(100002)$ contains all the digits from 0..9. Now, it's easy to check that $ 9|p$ and as 100002 is only divisible by 3 $ x$ has to be a multiple of 3, therefore I'd only have to check one third of the numbers from the specifed interval. This is quite fast if done by a computer but I belive there are even more optimisations that could be done and wondering if someone see other paterns $ x$ should satisfy.

Thanks

I have some thoughts about the numbers $ x$ and $ 2x$
Then, we can apply them in the number $ p$

$ \blacksquare$ The number $ x$ shouldn't start with $ 5$ or more, because then $ 2x$ will have 6 digits

Why we don' t want a 6-digit number:
The number of digits of $ 2x$ is at most one more than the number of digits of $ x$
So $ x$ and $ 2x$ should have 5 digits each

$ \blacksquare$ One other fact is this:

$ 2x$ should end at [color=red]0[/color],2,4,6 or 8

$ \star$ So we know that $ p$ should start with $ 0,1,2,3,4$ and should end at $ 0,2,4,6,8$

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

But, since you want to use a computer, I'd prefer another way:

Take all the permutations for number $ x$. We have:
$ 5$ choices for the 1st digit
$ 9$ choices for the 2nd digit
$ 8$ choices for the 3rd digit
$ 7$ choices for the 4th digit
$ 6$ choices for the 5th digit

So you have $ 5 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \equal{} 15120$ permutations to check

Ok, by computer it can be solved, I wonder if there is some other way to do this..

[color=blue][b]P.S.[/b] I found that there are 60 solutions[/color]

[color=red]EDIT: I removed the condition "no zero at the end", because when x ends at 5 then 2x ends at 0[/color]


It's tomorrow and Saturday. Good luck to everyone going.

I'll be at San Jose State for southern California. Anyone else going to be there? :D

I'll also be at San Jose tomorrow. West side!

[size=200]WASHINGTON STATE GOT FIRST PLACE[/size][size=59] in division B[/size][size=200]WOOHOO!!![/size]

Y'all better watch out for us next year. We have experience now, and word on the street is we're actually going to practice.

Glad to meet you, rrusczyk, complexzeta, paladin8 (though I don't think I talked to you specifically) and others.

Argh ARML sounds like fun but my school doesn't do it :-(  I think I can enter by myself somehow but I'll have to figure all that out over the summer.

Both Southern California A and B got "Site Honors," which is more or less the "you-suck-but-not-that-bad-so-here-have-a-nice-little-book" award.

Oh well. Watch out next year :D.

Oh yeah, in case anybody here is on an SC team, I'm Roger Lee and not Roger Li (sat next to and roomed with Macky-the-Rubik's-Kid on the bus).

ARML was really great though. As much as I'd like to say that the math was the best part, the people really made it what it is. From mafia to stealing everything in sight to inflating condoms to showering in the women's bathroom, ARML is awesome. God willing, I'm going next year.

And, yet again, those kids at Thomas Jefferson, VA, beat everybody in the nation. For the [b]3rd[/b] year in a row.

I was on South Carolina B, and we tied for 59th out of 63!

YAAAY GEORGIA ARML OWNS(most of <-- later added)YOU

YAAAAY GEORGIA A 4TH PLACE BABY WE RULE!! how did people do invidually?? i got a 5...sad we're losing all of our good seniors this year though so we're gonna suck next year...i was so happy when ricky biggs didnt win, i guess i was kinda biased because he owned me during the final round of the lightning round at the duke competition earlier this year

I got a 6... bleh. I was going strong right until the last set of questions, and the pressure all came in on me and I just couldn't reason at all. I did both problems in about four minutes after the contest when my mind was back in order... I've got another shot, I suppose.

It's hard to keep concentrating after getting the first six. There was a guy on my relay team (on SFBA B) in 2001 who got the first six and then missed the last two even though #7 in 2001 was the easiest problem on the contest. (I think only one other person on SFBA B missed it.) There were also three people on SFBA A 2004 who started out with 6/6, but then only one ended up getting 8, even though all were clearly capable of doing the problems.

The same thing happened to me. After getting the first 6, it took me a couple minutes just to get back on track during 7 and 8. By the time I had solved #7 (and caught my error in taking phi(2004) instead of d(2004)), I had about 2 minutes for #8... not enough time for the necessary casework.

Last year apix solved the first 6, made a stupid mistake on 7, and spent all his time trying to figure out what was wrong with it.

Getting an 8 is hard... I know, but I won't do my "I wanted that 8" routine.


integration
calculus
calculus computations

hello, setting $ t\equal{}\sqrt{x\plus{}b}$, then we get $ dx\equal{}2tdt$ and our integral is
$ 2\int\frac{1}{t^2\plus{}a\minus{}b}\,dt\equal{}\frac{2}{\sqrt{a\minus{}b}}\arctan\left(\frac{t}{\sqrt{a\minus{}b}}\right)\plus{}C$.
Sonnhard.


[hide="1"]1. Multiply both sides by $5^{x-1}$. Use the Laws of Exponents to simplify the RHS. Then, take the log of both sides to get $x\log3=x-1$ using Laws of Logarithms. Then, subract each side by x and undistribute. Isolate x to receive the answer.[/hide]

How does the Law of Exponents help when 2,3,5 are different bases?


need help with this problem fast:

x^2 + y^2 = 25
x - y + 1 = 0

thanks...

?!? What's the problem?

Just substitute from the second one into the first:

$x-y+1=0\Rightarrow x=y-1\Rightarrow (y-1)^2+y^2=2y^2-2y+1=25$
$\Rightarrow y=\frac{1\pm\sqrt{1^2+4\cdot12}}{2}=-3$ or $4$.

So $(x,y)=(-4,-3)$ or $(3,4)$.

thanks a lot... i really appreciate it.

by the way, that was FAST. thanks again.

[quote="anirudh"][hide="my sol"]$x=y-1
\\
\\
(y-1)^2+y^2=25
\\
\\
y^2+1+2y+y^2
\\
\\
2y^2+2y+1=25
\\
\\
2y^2+2y-24
\\
\\
\frac{-b \pm \sqrt{b^2-4ac}}{2a}
\\
\\
\frac{-(-2)\pm \sqrt{(-2)^2-4(2)(-24)}}{4}
\\
\\
\frac{2\pm14}{4}
\\
\\
\textbf{The two answers for y this way are 4 and -3. Therefore, the answers for x must be -3 and 4}$[/hide][/quote]
First of all, the equation is not $2y^2+2y-24=0$, but it's $2y^2-2y-24=0$.
Watch for the sign ;)

From $2y^2-2y-24=0$, you can factor like this:

$y^2-y-12=0\Longleftrightarrow (y+3)(y-4)=0\ \therefore y=-3,4$

$\therefore x=-4, 3$

$(x,y)=\boxed{(-4,-3),(3,4)}$


trigonometry
geometry
trig identities
Law of Sines

hi!
I have a book that refers the equallity of triangles  conditions and next refers that A-S-S does not guarantee triangle equallity (give a counterexample)
unless the angle is grater than or equal to 90 or the second side is longer than the first... :maybe:
[color=green]1)can you find the counterexample reqested :?: [/color]
[color=darkred]2)can you proved the second part othe prepasition(that A-S-S holds if the angle is grater than or equal to 90 or the second side is longer than the first) :?: [/color]

That's kind of strange. For any triangle which doesn't satisfy the 2 conditions, using law of sines should give you an A-A-S triangle congruence (but I haven't seen A-S-S in any regular geometry book  :oops: ). If the book is correct, (which it probably is) you would need to find a triangle such that Law of Sines is not permissable (Wait, there might be a typo: do you mean that the counter-example is provided when the 2 conditions are satisfied because the sines of obtuse angles are negative, and therefore law of sines can't be used?)

Make it, 30º, 4,3. The 3 can go inwards (towards the 4) or outwards (away from the 4) to get different sides.
The thing is, The two (possible) new angles are of the form 90+x and 90-x, so if the first angle is 90 or greater, you can only choose the 90-x option.

I  write what exactly the book (art and craft of problem solving p.260) writes:
[quote]There are several congruence conditions:
S-A-S,  A-S-A,  S-S-S,  A-A-S
Note that A-A-S does not guarantee congruence (give a counter example),unless the angle is greater than or equal to 90 or the second side is longer than the first[/quote]


Friends... I have  the following problem. Please, if possible, try to give a reasonably lengthy hint or the solution.

Consider two integer-valued random variables $ X$ and $ Y$ that are independent. The problem is to express the CDF of the random variable $ Z\equal{}X^2\plus{}Y^2$ in terms of the pmfs of $ X$ and $ Y$.

Let's assume that $ X$ and $ Y$ are discrete random variables and $ X(\Omega)=\{a_n: n\in\mathbb{N}\}, Y(\Omega)=\{b_n: n\in\mathbb{N}\}$ where $ a_n,b_n$ are some positive real numbers. First, it would be useful to determine $ Z(\Omega)$, and this is of course the set $ \{a_n^2+b_n^2: n\in\mathbb{N}\}$.
Although the problem is to find $ \mathbb{F}_{Z}$, it's enough to find $ \mathbb{P}_{Z}$ ( this is probability induced by random variable $ Z$), cause $ \mathbb{P}_Z\{c_n\}=\mathbb{F}_Z(c_n)-\mathbb{F}_Z(c_n-)$ ( this, of course wouldn't work if $ X$ and $ Y$ were continuous random variables ), and then one has $ \mathbb{F}_Z (x)=\sum_{c_n\leq x}\mathbb{P}\{Z=c_n\}$, where $ c_n\in Z(\Omega)$. The next procedure is almost classic in these kind of problems; first we can agree that $ \mathbb{P}(X\in \{a_n: n\in\mathbb{N}\})=1$, so we move to the next step;
$ \mathbb{P}\{Z=c_n\}=\mathbb{P}(\{X^2+Y^2=c_n\}\cap \{X\in \{a_k: k\in\mathbb{N}\}\})=\sum_k\mathbb{P}(X^2+Y^2=c_n,X=a_k)=\sum_k\mathbb{P}(Y^2=c_n-X^2,X=a_k)=\sum_n\mathbb{P}(Y^2=c_n-a_k^2,X=a_k)=\sum_k\mathbb{P}(Y=\sqrt{c_n-a_k^2},X=a_k)$, now we use that $ X,Y$ are independent, and we have $ \mathbb{P}(Z=c_n)=\sum_k \mathbb{P}(Y=\sqrt{c_n-a_k^2})\mathbb{P}(X=a_k)$, and notice that the sum is much "smaller" then it seems, it goes only for $ k$ such that $ \sqrt{c_n-a_k^2}=b_l$ where $ b_l\in Y(\Omega)$


inequalities
inequalities unsolved
algebra

[hide=Thanks.][quote=vietmath]The solution is in the attached.[/quote]
[/hide]


[problem]

what is the remainder when $ 5^{30}$ was divided by 7?

i hate mod problems... plz help me...

yea i found that pattern...

but it takes too long time...

you need to do 125/7, 625/7, 3125/7, 15625/7...

is there any other way to do this kind of problem?

Lovely application, Brut3. I humbly thank you. :D


function
linear algebra
matrix
number theory
prime factorization
algebra proposed
algebra

Consider $ f: \mathbb{N}^* \rightarrow \mathbb{N}$ such that $ f(n)\equal{}$ the number of the pairs $ (x,y) \in \mathbb{Z}^2$ with the property $ |x^2\minus{}y^2|\equal{}n$.

$ a)$ Find $ f(2009)$ and $ f(2010)$

$ b)$ Determine $ f(n)$

We have $ |x^2 - y^2| = n$   ($ n\in\mathbb{N}$), and we want to find the number of pairs $ (x;y) \in \mathbb{Z}^2$ which satisfy the equation.

We can rewrite the equation in this way:

$ |(x + y)(x - y)| = n$
$ (x + y)(x - y) = \pm n$

Now, it's clear that $ (x + y)$ and $ (x - y)$ have to be divisors of $ n$.

Let's say $ (x + y) = k$ and $ (x - y) = h$

$ \eft \{{\begin{matrix} x + y = k \\
x - y = h\end{matrix}}$

which leads:
$ x = \frac {k + h}{2}$, and
$ y = \frac {k - h}{2}$.

So, $ k$ and $ h$ have to be both even or both odd.

Now, let's write $ n$ as its prime factorization:

$ n = d_1^{\alpha_1}\cdot d_2^{\alpha_2}\cdot ... \cdot d_n^{\alpha_n}$

Now there are two cases:

1) $ n$ is even ($ \Rightarrow$ $ n = 2^{\alpha_1}\cdot d_2^{\alpha_2}\cdot ... \cdot d_n^{\alpha_n}$)
2) $ n$ is odd

If $ n$ is odd, all right: the number of divisors will be exactly
4 times $ \prod_{k = 1}^n {(\alpha_k + 1)}$
(this is because we have to consider all the combinations of $ \pm k$ and $ \pm h$).

Otherwise, if $ n$ is even, we have to find in how many ways we can "split" $ \alpha_1$ 2-factors and combine them with the other factors. This number is given by:

$ 4 \cdot {(\alpha_1 - 1)} \cdot \prod_{k = 2}^n {(\alpha_k + 1)}$

So... here is my answer:

$ n = d_1^{\alpha_1}\cdot d_2^{\alpha_2}\cdot ... \cdot d_n^{\alpha_n}$

If $ 2\mid n$, $ \alpha_1 =$number of 2-factors

$ f(n) : = \{\begin{matrix} 4 \cdot {(\alpha_1 - 1)} \cdot \prod_{k = 2}^n {(\alpha_k + 1)}, & \mbox{if }2\mid{x} \\
 \\
4 \cdot \prod_{k = 1}^n {(\alpha_k + 1)}, & \mbox{if }2\nmid x \end{matrix}$

So,
$ f(2009) = f(7^2 \cdot 41) = 4\cdot 3\cdot 2 = 24$

and

$ f(2010) = f(2\cdot 3\cdot 5\cdot 67) = 4\cdot 0\cdot 2^3 = 0$
Is this solution correct?... great problem, Moldovan...  :lol:

Gabriele  :)


.......and difficult for me  :blush:

Let M a point inside a triangle ABC and [color=blue][b]AB=AC[/b][/color], we put:
$ (AB)\bigcap {(CM) \equal{} \{ E\} }$
and:
$ (BM)\bigcap {(AC) \equal{} \{ F\} }$

we suppose that  AEFM and BMC have the same area , Prove that:
$ < BMC \equal{} \frac{{2\pi }}{3}$

$ S(AEMF)\equal{}S(BMC) \Longrightarrow S(AEMF)\plus{}S(EMB)\equal{}S(BMC)\plus{}S(EMB)$

$ \Longrightarrow S(ABF)\equal{}S(CBE)$

consider$ BH,CH'$az the altitudes of triangle $ ABC$.

$ \Longrightarrow BH.AF\equal{}CH'.BE$

and because of $ BH\equal{}CH'$$ \Longrightarrow AF\equal{}BE$

its clear that we also have:$ CF\equal{}AE$

dear,we might add another hypothesis to this problem,

we should have $ ABC$ is equilateral triangle

because in the limited case,if u make $ F$,closer to$ C$,and when the distance of$ CF$ closes to $ 0$,

angle $ BMC$,closes to the angle betwin ray$ AC$,and ray$ CB$.

but the angle betwin ray$ AC$ and ray$ CB$,depends on $ \angle BAC$,and is equal to $ 90\plus{}\frac{\angle BAC}{2}$

if $ 90\plus{}\frac{\angle BAC}{2}\equal{}2\frac{\pi}{3} \Longrightarrow \angle BAC\equal{}\frac{\pi}{3}$

it shows that the hypothesises in the problem arnt complete,and we should have $ ABC$is equilateral.

in this case,triangles$ BFC$and $ AEC$ are equal to eachother,and:

$ \angle MBC\equal{}\angle ACM$

$ \angle ACM\plus{}\angle BCM\equal{}\frac{\pi}{3}  \Longrightarrow \angle MBC\plus{}\angle BCM\equal{}\frac{\pi}{3}$

$ \Longrightarrow 180\minus{}(\angle MBC\plus{}\angle BCM)\equal{}\angle BMC\equal{}2\frac{\pi}{3}$


prove for all $ n\in N$*
$ 2^n>1\plus{}n\sqrt{2^{n\minus{}1}}$

$ 2^n>1+n\sqrt{2^{n-1}}$
$ \Leftrightarrow 2^{n-1}+2^{n-1}-2\frac{n}{2}\sqrt{2^{n-1}+\frac{n^2}{4}> 1+\frac{n^2+4}}$
$ \Leftrightarrow (\sqrt{2^{n-1}}-\frac{n}{2})^2+2^{n-1}>1+\frac{n^2}{4}$
$ \Leftarrow 2^{n-1}>1+\frac{n^2}{4}$
$ \Leftrightarrow 2^{n+1}>4+n^2$  (*)
For $ n=1$, $ 2^n=2, 1+n\sqrt{2^{n-1}}=2$ so have equality in the original inequality $ 2^n>1+n\sqrt{2^{n-1}}$. For $ n=2$, $ 2^n=4, 1+n\sqrt{2^{n-1}=1+2\sqrt{2}}$ and $ 4>1+2\sqrt{2}\Leftrightarrow 3>2\sqrt{2}\Leftrightarrow 9>8$ which is true. If $ n=3$, in (*), $ LS=16, RS=13$ so (*) is true and the original inequality is true. For $ n>3$, $ (2^{n+1}-4-n^2)'=(n+1)ln2-2\ge 3ln2-2>0$ so $ f(n)=2^{n+1}-4-n^2$ is increasing on $ [3,\infty)$ so for $ n>3$, $ 2^{n+1}>4+n^2$ and hence the original inequality is true.
So the original inequality is true for all $ n\in N$ with equality when $ n=1$.

[quote="greatestmaths"]prove for all $ n\in N$*
$ 2^n > 1 \plus{} n\sqrt {2^{n \minus{} 1}}$[/quote]
This ineq true with $ n \leq 3$
With $ n \geq 3$,suppose that this ineq true with $ n\equal{}k$,$ 2^k > 1 \plus{} k\sqrt {2^{k \minus{} 1}}$
We will prove that it true with $ n\equal{}k\plus{}1$
We have:
$ 2^{k\plus{}1}\equal{}2.2^k > 2 \plus{} 2k\sqrt {2^{k \minus{} 1}}$
=$ 2\plus{}\sqrt{2}k.\sqrt{2^k}\equal{}1\plus{}(k\plus{}1)\sqrt{2^k}\plus{}1\plus{}\sqrt{2^k}(\sqrt{2}k\minus{}k\minus{}1)$
$ >1\plus{}(k\plus{}1)\sqrt{2^k}$
Done!


prove that there is exactly 1 pair r and s such that $ 45/61 > r/s > 59/80$

you must have forgotten to write some condition because as written the problem is obviously false, between any two rational numbers exist infinitely more rational numbers


geometry
circumcircle
Euler
geometry proposed

Given triangle $ ABC$ and the circumcenter $ O. D$ is an arbitrary point on $ BC. A$ circle which has diameter $ AD$ intersects the Nine-point circle of triangle $ ABC$ at $ E$. Prove that $ \angle ADE\equal{}\angle ODC$

Denote A’ is the intersection of altitude through A and BC
Then A’ is on both ⊙N’ and ⊙O’
K is the midpoint of HA ,also the intersection of AA’ and  ⊙N’
∠ELA’=∠EKA’  ∠EAK=∠EDL  => ⊿LED∽⊿KEA  => KE/AK=LE/LD  KE/AE=LE/ED   =>
⊿KEL∽⊿AED  OKE/OL=LE/LD  =>  ⊿OLD∽⊿KEL  => ⊿OLD∽ADE =>  ∠ODL=∠ODC=∠ADE

$ (ADE) \cap ME \equal{} S,E$.
$ (ADE)\cap BC \equal{} D,F$
Denote by$ M$ the midpoint of$ BC$.
Denote by$ I$ the$ Euler$ center of $ ABC.$
From the $ Reim's$theorem,applied to circles$ (ADE)$ and $ (MEF)$ ,we have :
$ IM\perp DS \equal{} > AO\perp DS$
But $ AS \perp DS \equal{} > A,O,S$ are collinear $ \equal{} > OSMD$ is cyclic.
We have :
$ \angle {ADE} \equal{} \angle{OSE} \equal{} \angle{ODC},$
which completes the proof.
 :)


geometry
perimeter
algorithm
combinatorics proposed
combinatorics

A magician should determine the area of a hidden convex $ 2008$-gon $ A_{1}A_{2}\cdots A_{2008}$. In each step he chooses two points on the perimeter, whereas the chosen points can be vertices or points dividing selected sides in selected ratios. Then his helper divides the polygon into two parts by the line through these two points and announces the area of the smaller of the two parts. Show that the magician can find the area of the polygon in $ 2006$ steps.

Let our polygon be $ AA_1...A_n$ (that is for fixing a vertex) and $ M(A,B)$ be the midpoint of segment $ AB$.
1)Select $ A$ and $ M(A_1,A_2)$ points. Helper gives the area of $ AA_1M(A_1,A_2)$ so the area of $ AA_1A_2$. Similarly we can learn the area of $ AA_nA_{n \minus{} 1}$.
2)After that assume that we know the areas of $ AA_1...A_i$ and $ AA_n...A_j$ ($ i < j$). If the area of $ AA_1...A_i$ is smaller than that of $ AA_n...A_j$, select the points $ A$ and $ M(A_i,A_{i \plus{} 1})$. Helper gives the area of $ AA_1...A_iM(A_i,A_{i \plus{} 1})$, so the area of $ AA_iM(A_i,A_{i \plus{} 1})$ and $ AA_iA_{i \plus{} 1}$ then $ AA_1...A_{i \plus{} 1}$. Otherwise do the same for learning the area of $ AA_n...A_{j \minus{} 1}$.
With this algorithm we can find the whole area with $ n \minus{} 2$ questions

I think that the number of steps can be reduced to 23.

We consider the polygon as a 2050-gon with some vertex forming 180º angles. We choose a vertex and name it A, and the rest of them $B_{0}$, $B_{1}$, ..., $B_{2048}$. First we need to find the side on which there is point P such that AP divides the polygon in two parts of the same area. Then we ask the area for $AB_{1024}$, $AB_{512}$ and $AB_{1536}$. If the first is the largest, then P is between $B_{512}$ and $B_{1536}$. If the second is the largest, then P is between $B_{0}$ and $B_{1024}$, and if the third is the largest, then P is between $B_{1024}$ and $B_{2048}$. In each case we reduced the number of sides in half. Now we divide the two sections we have in half, and again find the areas and reduce the number of sides. After 21 steps we only have two sides left, and we know the areas from A to their three vertex. We now ask for the areas for the midpoints of those sides. If one of the midpoints has a larger area than the central vertex, then P is on that side, and we can find the area of the triangle formed by that side and A (it's twice the smallest difference with one of the neighboring vertex). If the central vertex has the largest area, then the area of each of the triangles is twice the difference of the areas for the not central vertex and the midpoint of the side). In any case we can find the area of both sides of the polygon, and the area of the two triangles. This gives us the total area.


Find all real solutions to the system of equations:

$2x+\sqrt y=1$
$2y+\sqrt z=1$
$2z+\sqrt x=1$

you can see $(1/4, 1/4, 1/4)$ is a solution.  now suppose wlog that to get another solution, you increase $x$ by some value, starting from $(x,y,z)=(1/4,1/4,1/4).$  Then the first equation implies that $y$ must decrease.  Then the third equation imples that $z$ must also decrease.  But then the second equation can't hold.  A similar argument holds for decreasing $x$.  So this is the only solution.

[quote="Farenhajt"]Put $f(t)=\frac{\sqrt{t}-1}{2}$. Then we have $x=f(y), y=f(z), z=f(x)$, which gives $x=y=z$. Hence $2x+\sqrt{x}-1=0$, which has the unique real solution $x={1\over 4}$. Therefore the unique solution is $({1\over 4},{1\over 4},{1\over 4})$[/quote]
From the above, $x=f(y) \Rightarrow x=\frac{\sqrt{y}-1}{2}\Rightarrow 2x+1=\sqrt{y},$  which isn't what the problem says.

[quote="FieryHydra"][quote="Farenhajt"]Wasn't paying attention... Thanks :)

The function should be $f(t)=\frac{1-\sqrt{t}}{2}$, but the point remains the same, though.[/quote]
Are you using the idea that if $x=f(y), y=f(z), z=f(x)$ and $f$ is either increasing or decreasing, then $x=y=z$ is the only possible solution?

I remember seeing this in one of the posts earlier but don't recall which. Could you please write down the exact idea? Thanks in advance.[/quote]

I saw it [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=447041#447041]here[/url]. The idea is that if - for example - $f$ is an increasing function on an arbitrary interval and $x, y, z$ are in that interval such that $f(x) = y$, $f(y) = z$ and $f(z) = x$, then $x=y=z$.

Proof: Suppose this isn't true. The statement is cyclic, so suppose WLOG that $z = \min(x, y, z)$. Clearly, if two of $x, y, z$ are equal, say $x$ and $y$, then $y = f(x) = f(y) = z$, so they're all equal then and the proof is done. There remain two cases: $x>y>z$ and $y>x>z$. First case: $y = f(x) > f(z) = x$ because $f$ is increasing and $x>z$: contradiction. Second case: $x = f(z) < f(y) = z$ because $f$ is increasing and $y>z$: contradiction. And we're done.


I got as far as [hide](-1+:sqrt:53)/2[/hide]. Am I right so far?

Hmm, I got [hide](3  +/- 2 Sqrt(3) ) / 2[/hide] instead.

spoiler

[hide] x=3+1/(4+1/x)

x= 3 + x/(4x+1)

x= (13x+3)/(4x+1)

4X^2-12X-3= 0

Which I believe comes out to Fierytycoon's answer.

[/hide]

Can you explain that solution a bit, I don't understand the x=part.. I am very tired tonight...

Well we define x as "3+1/(4+1/(3+1/(4+1/(3+1/(4+1/(3+1/(4+1/(3+......)))))))"
Since the pattern continues forever, you get 3+1/(4+1/x) which is equal to x. Thus 3+1/(4+1/x)=x, and you keep rolling from there.


Don't know, if the inequality is true, at least I'm not able to prove it

$x,y,z>0$

\[ x^2+y^2+z^2+27\cdot\left(\frac{xyz}{xy+yz+zx}\right)^2 \geq 2(xy+yz+zx) \]

wrong, try x = 2, y = 2, z = 1.

  darij