Does anyone here actually play viola?
It's funny, everyone plays cello and violin, and there are barely any violists out there...
(I'm a violist, by the way.)
So... anyone?

I'm sure there are plenty, actually. Violas and math go together very well.

Oh yeah- I'm a violist.

Violas are pointless.  They're just obese violins.

Sopranos, altos, tenors, and basses all contribute to making beautiful vocal music, and similarly violas help make beautiful string music.

For math92's sake we need more pieces like Brandenburg Concerto #6 - no violins allowed.

Really? There are three times as many violas as cellos in my school, sometimes I'm lonely..  :wink:  (I'm a cellist)

No offense, but here's some famous jokes: http://www.mit.edu/~jcb/jokes/viola.html

No offense taken- who do you think tells most viola jokes?

The greatest composer used to play the viola in his early years as an orchestral musician.

What do you mean, "one of"? J.S. Bach usually played viola in ensemble settings (and we thank him for Brandenburg #6). Mozart as a child prodigy and as a teenager in Salzburg played violin, but when he was living in Vienna and writing string quartets (and soloing on piano), he sat in as a violist in quartets. Haydn played viola. Brahms? Maybe. But in all of these cases, it's not in their "early years."

If you're looking for the common thread that unites the largest number of composers, it's that they were keyboard players (organ, harpsichord, piano). But viola isn't that uncommon. Playing viola puts you in the middle of the sound and the middle of the orchestra - you hear everyone, and you hear how it's related.

Thanks for the support...
Because at my school, there are like five times more violinists than violists (literally, probably more =( )
So yeah...

2nd graders don't play viola - they play violin. Little bitty violins - 1/4 size or so. And even 5th graders who ever thought about viola might wind up playing something like a violin restrung and retuned as a viola (an abomination of a sound!). To have viola players someone has to persuade middle schoolers to change instruments. It sounds like the middle school teachers in your system weren't all that persuasive. (Do you get pieces with "third violin" parts? Yecchh.)

Are there more celli or violas at your school?

[quote="Kent Merryfield"]... balanced mixture of instrumentation.[/quote]
With very large size- the main orchestra at the high school in question has over 100 players, and is limited by the physical size of the stage they perform on. I was one of about 15 violas, and the numbers are similar now.

http://www.viola-in-music.com/

http://music.lib.byu.edu/piva/VnV.html

Good material here for those interested in the viola.

[quote="Kent Merryfield"]Was that addressed to me? If so, spoken like someone who doesn't know my age. I was a violinist from elementary school into college. The schools my children have attended are pretty good at getting a balanced mixture of instrumentation. My older child - jmerry - became a violist starting in the sixth grade. It was with his instrument in the house that I gradually became a violist myself. I was in my 40's when I started playing it. My younger child has always been strictly a violinist.[/quote]
I was addressing ra, actually, but never mind.

wow, an orchestra with more violas than cellos? i can't imagine that. in my school orchestra there are usually about 8-10 cellos, and maybe half that in violas. and about 30 violins. i started viola in 4th grade, incidentally, you don't have to be older to start. can't you play a 13 inch viola or something? i started with that size. i don't know if they have smaller sizes, though. and as an aside: have you ever taken a look at the pellegrina viola? it looks so... unorthodox. i wonder what it sounds like, though. here's a link: http://www.rivinus-instruments.com/Pellegrina.htm

At one time in my orchestra, i was the only viola.it was really sad  i used to play violin when i was little then i quit. Now i hate violins for some weird reason.

idk in our school orchestra we have about 10 violas but only like the first 3 chairs play above a pianissimo. in an orchestra i'm in outside of school we have about 6 violist. they all play really well though so i works out against the 30 or so violins =] (i'm a violinist) o and some people at my school switch to viola cuz their arms r too long 4 violin...(weirdos :D ) no offense to anyone :maybe:

[quote="Kent Merryfield"]And a plague be upon arrangers who insert completely unnecessary treble clefs into orchestral viola parts.[/quote]

Well I play cello and piano and also compose...
I usually write a lot of the viola parts in treble clef just becuase 1) Its easier for me.  :blush: and 2) I don't like having like 4~5 ledger lines for the viola parts.

You're reading the message of a violist and pianist. I can play in 3 clefs, while all my friends can only read 1. Sucks to be them.

I play violin and piano, but in my school's orchestra I often (almost always, actually) fill in the viola parts, so I can read the alto clef as well. It's pretty easy, actually. I enjoy playing the viola part because it's helped me gain a better understanding of how the parts in an orchestra fit together.


Please, an easy proof of this one.

For a,b,c positive real numbers such that a+b+c=1 and a \geq b \geq c

prove that a(1-a) \leq b(1-b) \leq c(1-c)

The problem is incorrectly stated. It is a(1 - a) \geq b(1 - b) \geq c(1 - c).

This is really too easy.

Sorry, you are right.

I have made a huge error. The inequality is very easy.

If a=x/(x+y+z)  b=y/(x+y+z) c=z/(x+y+z) we obtain

 a(1-a)=x(y+z)   and  b(1-b)=y(x+z)  so x(y+z) \geq y(x+z) and

 finally xz \geq yz.


Solve this equation: $ 2^{[\sqrt[3]{x} ]}\equal{}x$. I see x=2 is a solution, but i don't know how to prove that the equation has only one solution :maybe:

That's because it doesn't have only one solution. $ x \equal{} 1024$ is also a solution.

[hide]Now, you can say $ 2^{\lfloor \sqrt [3]{x} \rfloor } > 2^{\sqrt [3]{x} \minus{} 1 }$, and find that after a certain point, the latter becomes strictly greater than $ x$. If this value is $ p$, then it is clear that $ RHS > LHS$ for all $ x\geq p$.

For $ x$ less than that, you can note that $ 2^{\lfloor \sqrt [3]{x} \rfloor }$ is an integer power of $ 2$, so by the equation, $ x$ must be this integer power of $ 2$. Then, there are relatively few cases to try out.[/hide]

[quote="ely_en"]Solve this equation: $ 2^{[\sqrt [3]{x} ]} \equal{} x$. I see x=2 is a solution, but i don't know how to prove that the equation has only one solution :maybe:[/quote]

Shouldn't the answer be the cube root of 2?

no, because $ [\sqrt[3]{2}]\equal{}1$ and $ 2^1$ isnt't equal with $ \sqrt[3]{2}$

[quote="ely_en"]no, because $ [\sqrt [3]{2}] \equal{} 1$ and $ 2^1$ isnt't equal with $ \sqrt [3]{2}$[/quote]

I still don't understand the symbol you used here $ [\sqrt [3]{2}] \equal{} 1$

Greatest integer function. In other words, the greatest integer less than the cube root of 2 (in that case).

I always find it easier to write as $ \lfloor x \rfloor$


1) $75 is invested at 7.5% compounded continuously. When will the amount in the bank be quadrupled? Answer correct to the nearest hundredth of a year.

2) How many negative roots do you expect for the equation
8x^6-3x^4+2x^2-4=0.

3)Solve for x if 1+i is one of the roots of x^3-5x^2+8x-6=0.

4) If lnA=x, lnB=y, and lnC=w, then ln(B^4C^2/A^(1/2))=?.

How do you solve them all?

first of all, these are definetly homework problems so post what youve done so far and then i might help you

second, you already posted this in the"other problem solving topics" forum.  no double posting!!!

for the first one... i dunno the formula to use, so im stuck there.
2nd question, i was able to find 3 negative roots; but the answer was 1 or 3 negative roots. Which has lost me.
3rd, i just dunno how to start.
4th, i have no clue either

and sorry about double posting.

[hide]1) If an amount is compounded continuously the equation is
$P=P_{0}e^{rt}$ Where $P$ is final principle, $P_{0}$ is the intial, $r$ is the rate and $t$ is the time in years. So $300=75e^{.075t}$. Solving this for $t$, We get $t=\frac{\ln4}{.075}$ or $t=18.48$ years

3) If $1+i$ is a solution to the equation, then the conjugate must be a solution as well, so two solutions are $1+i$ and $1-i$. Then, since the sum of the roots is $5$, the last root must be $3$. Therefore, $x=1+i,1-i,3$

4) If $\lnA=x$ then $e^{x}=A$, using the same logic $e^{y}=B$ and $e^{w}=C$. Plugging these values into the original equation, you find  \[\ln{\frac{e^{4y}e^{2w}}{e^{\frac{x}{2}}}}\] or by exponent rules $\ln{e^{4y+2w-\frac{x}{2}}=4y+2w-\frac{x}{2}}$
[/hide]

[quote="DownBringer"]how do u find the sum of the roots and the product of the roots??[/quote]

Alright, if you take a cubic in the form $(x-a)(x-b)(x-c)$, when you expand it you find that it is of the form, $x^{3}-(a+b+c)x^{2}+(ab+bc+ac)x-abc$
So, applying what we have just done to the equation $Ax^{3}+Bx^{2}+Cx+D$ we find the generalization, that sum of the roots is $\frac{-B}{A}$, and the product of the roots is $\frac{-D}{A}$

You can apply this same process to find the generalization for quadratics and quartics, etc.

[quote="DownBringer"]how do u find the sum of the roots and the product of the roots??[/quote]

Use Vieta's formulae.

For #2, use Decartes' [url=http://en.wikipedia.org/wiki/Rule_of_signs]Rule of Signs[/url].

[quote="DownBringer"]#3 is still confusing to me. Because the sum is 5, then the last number is 3?? Im confused there...[/quote]

We had the roots as $1+i, 1-i$, adding these up gives us $2+i-i$ or $2$. So, since the sum of the roots is 5, and 2 of the 3 roots add up to 2, then the last root has to be $3$

Ex. $a+b+c=5$, then $1+i+1-i+c=5$, so $2+c=5$ or $c=3$


Show that any number of the form
$2^{k}(2^{k+1}-1)$
is perfect if $2^{k+1}-1$ is prime.

A perfect number is a number $n$ such that the sum of all its divisors equal $2n$.

I remember answering this before. You should use the search function. :)

[hide] The sum of the factors of $2^{k}(2^{k+1}-1)$ is $(1+2+4+...+2^{k})(1+2^{k+1}-1)=(2^{k+1}-1)(2^{k+1})=2\cdot2^{k}(2^{k+1}-1)$.
QED [/hide]

[hide]also, if the perfect  number is even, it has to be of the form given above.
see http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=number+perfect&t=103013[/hide]


geometry
incenter
circumcircle
geometry proposed

Prove, that in non-isosceles triangle $ IO \bot E_LE_R$, where:
$ I$ - incenter;
$ O$ - circumcenter;
$ E_L$ - left Jerabek point:
$ A_1  \in BC$, $ B_1  \in AC$, and $ C_1  \in AB$, $ E_LA_1\parallel CA$, $ E_LB_1\parallel AB$, $ E_LC_1\parallel BC$  and  $ E_LA_1 \equal{} E_LB_1 \equal{} E_LC_1$;
$ E_R$ - right Jerabek point:
$ A_2  \in BC$, $ B_2  \in AC$, and $ C_2  \in AB$, $ E_RA_2\parallel AB$, $ E_RB_2\parallel BC$, $ E_RC_2\parallel CA$  and  $ E_RA_2 \equal{} E_RB_2 \equal{} E_RC_2$.

Is it known $ E_L \equal{} (1/b, 1/c, 1/a)$ and $ E_R \equal{} (1/c, 1/a, 1/b)$, so line $ E_LE_N$ has equation $ \sum(a\minus{}\frac{bc}{a})x\equal{}0$ so it has infinite point $ (a(b\minus{}c)(ab\plus{}ac\minus{}bc): ...: ...)$ and OI- line has infinite point $ (ca(c\minus{}a)(s\minus{}b)\minus{}ab(a\minus{}b)(s\minus{}c): ...: ...)$ so we need to prove

$ \sum (b^2\plus{}c^2\minus{}a^2)[ca(c\minus{}a)(s\minus{}b)\minus{}ab(a\minus{}b)(s\minus{}c)][a(b\minus{}c)(ab\plus{}ac\minus{}bc)]\equal{}0$

which is true  :P !


function
MATHCOUNTS
trigonometry
AMC
USA(J)MO
USAMO
number theory

$1.$ Create the number 24 using the numbers 3, 3, 7, 7, and the following functions: $+-\times \div.$ This is not a trick question and does not involve a number system other than base 10. Decimal points are not allowed.

$2.$ 5 cows eat 2 acres of grass in 10 days. 7 cows eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. How many days will it take 16 cows to eat 7 acres of grass?

$3.$ The front wheels of a wagon measure 3.5 feet in diameter. The real wheels measure 4.25 feet in diameter. While the wagon is stopped somebody makes a chalk mark on both a front and rear wheel. How far must the wagon travel before both chalk marks return to their original positions at the same time?

$4.$ You have nine pearls: eight are real and one is fake. All the real ones weigh the same and the fake weighs a little more than the real ones. Using a balance scale twice, how can you find out which one is fake?

$5.$ I am as old as John will be when I am twice as old as John was when my age was half the sum of our present ages. John is as old as I was when John was half the age he will be 10 years from now. How old am I now?

$6.$ Four people need to cross a bridge at night. The bridge is only strong enough to hold at most two people at once. Since it is dark, a flashlight must be used for all crossings (there is only 1 flashlight). It takes person A ten minutes to cross, person B five minutes, person C two minutes, and person D one minute. If two cross at the same time they must walk at the slower man's pace. How can you get everyone across in 17 minutes?

$7.$ What is the height of a regular pentagon with side length 1? Answers may not be expressed in trigonometric functions.

$8.$ What are the solutions for $a+b= 25$ where $a$ and $b$ are prime numbers? Also solve for 50 and 100.

[color=red][b]SOLVED: 1, 2, 3, 4, 5, 6, 8
UNSOLVED: 7[/b][/color]

[hide="1"]$7\cdot(\frac{3}{7}+3)=24$[/hide]
[hide="3"]The 2 tires circumferences are $4.25\pi$ and $3.5\pi$.  In $59.5\pi$ feet they will both be in their original positions.[/hide]
[hide="4"]First put 3 pearls on the left side of the scale, 3 pearls on the right side of the scale, and 3 pearls off the scale.  If the scale is heavier on the left side, the fake is on the left side.  If the scale is heavier on the right side, the fake is on the right side.  If both sides of the scale weigh the same, the pearl is off the scale.  Now that the pearl is narrowed down to 1 of 3, put 1 pearl on the left side, 1 pearl on the right side, and 1 pearl off the scale.  If the scale is heavier on the left side, the pearl is the one on the left side.  If the scale is heavier on the right side, the pearl is the one on the right side.  If the scale weighs the same amount on both sides, the pearl is the one off the scale.[/hide]

I don't understand #2.  Does every cow eat the same amount per acre per day?  Thats what I understood from the problem, but then the 2 equations don't have the same amount of cows per acre per day.

Yes, they eat at constant rates.

[hide="Hint"]Consider the amount of grass which grows in one acre per day, which is also constant.[/hide]
[hide="Another Hint"]Also consider the initial amount of grass in an acre.[/hide]

[quote="i_like_pie"]Yes, I did mean you. The problem says create the number 24 using 3, 3, 7, 7, and basic symbols. You just put 24.  :?:[/quote]

It's because he has a habit of looking at other people's answers and then making a post containing solely the answer to increase his post count. However, I guess he didn't even read the question this time!

:wallbash_red:
that's not true, klebian
but i solved th eproblem in a piece of paper, and just wrote 24 :blush:

You didn't post the answer, but I guess it was correct since you got 24. Unless what Klebian said was true, which I doubt 100%.

I added several problems yesterday and I'm thinking of another one right now.

Is #7 considered Mathcounts level?

[quote="anirudh"]yes, because you said that trig is not allowed :D[/quote]

If I gave you an USAMO number theory problem and said trig isn't allowed, that doesn't mean it's MathCounts level.

I also don't think #7 has enough information, is the pentagon regular as well?

[hide]6. first the 2 min guy and the 1 min guy go across, then the 1 min comes back.
           then the 5min and 10min guy go and the 2 min guy comes back.
            then the 2min guy and 1min guy go across to finish
2+1+10+2+2=17[/hide]

[hide="6"]first, d goes with c and d brings back the flashlight.  then a goes with b and c brings back the flashlight.  then d and c go back[/hide]

[quote="ProtestanT"][quote="i_like_pie"]
$2.$ 5 cows eat 2 acres of grass in 10 days. 7 cows eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. How many days will it take 16 cows to eat 7 acres of grass?
[/quote]
[hide="2"]
Let $x$ = acres of grass a cow eats per day
Let $y$ = days it takes 16 cows to eat 7 acres of grass
Let $z$ = rate @ which grass grows (in grass-acres-per-day if its a measure)
$2+20z-50x=0$
$2+20z=50x$
$x=\frac{1}{25}+\frac{2z}5$
$210x=3+90z$
$\frac{42}5+84z=3+90z$
$\frac{27}5=6z$
$z=\frac{9}{10}$
$x=\frac{1}{25}+\frac{18}{50}$
$x=\frac{2}{5}$
$16yx=7+7yz$
$\frac{32y}{5}=7+\frac{63y}{10}$
$\frac{y}{10}=7$
$y=70 days$

If I made a mistake, please tell me if its logical or arithmetic.
[/hide][/quote]
Correct! :D
The same answer was posted in the new topic, "Some Problems 2." The pentagon height problem has not been solved yet.

[quote="ProtestanT"]5,6,and 8 are solved on the other post[/quote]
Yep. However, nobody has correctly calculated the height of the pentagon yet.

[quote="i_like_pie"]$1.$ Create the number 24 using the numbers 3, 3, 7, 7, and the following functions: $+-\times \div.$ This is not a trick question and does not involve a number system other than base 10. Decimal points are not allowed.

$2.$ 5 cows eat 2 acres of grass in 10 days. 7 cows eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. How many days will it take 16 cows to eat 7 acres of grass?

$3.$ The front wheels of a wagon measure 3.5 feet in diameter. The real wheels measure 4.25 feet in diameter. While the wagon is stopped somebody makes a chalk mark on both a front and rear wheel. How far must the wagon travel before both chalk marks return to their original positions at the same time?

$4.$ You have nine pearls: eight are real and one is fake. All the real ones weigh the same and the fake weighs a little more than the real ones. Using a balance scale twice, how can you find out which one is fake?

$5.$ I am as old as John will be when I am twice as old as John was when my age was half the sum of our present ages. John is as old as I was when John was half the age he will be 10 years from now. How old am I now?

$6.$ Four people need to cross a bridge at night. The bridge is only strong enough to hold at most two people at once. Since it is dark, a flashlight must be used for all crossings (there is only 1 flashlight). It takes person A ten minutes to cross, person B five minutes, person C two minutes, and person D one minute. If two cross at the same time they must walk at the slower man's pace. How can you get everyone across in 17 minutes?

$7.$ What is the height of a regular pentagon with side length 1? Answers may not be expressed in trigonometric functions.

$8.$ What are the solutions for $a+b= 25$ where $a$ and $b$ are prime numbers? Also solve for 50 and 100.

[color=red][b]SOLVED: 1, 3, 4
UNSOLVED: 2, 5, 6, 7, 8[/b][/color][/quote]

Question #2 has been solved here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110953 :)


1)IN a tournament , the RED and the BLUE play a best four out of seven series, i.e teams play at most seven games and the series ends when the first team reaches four wins. Suppose both teams have 50% chance of winning any game they play. What is the probability that the REDS will win the series in the seventh game? In how many different win-loss sequences can the RED win the series?

2) A square is tiled by 24 congruent rectangles. Points A and B are the midpoints of a pair of opposite sides of the square. On each side of line AB, four rectangles are chosen at random and colored black. The square is then folded over line AB. Compute the probability that exactly one pair of black rectangles is coincident.

[hide="the most likely incorrect answer to 1a"]If RED wins in the seventh game, then each team has won 3 games in the first 6 games. Thus the number of possible win-loss combinations for all seven games is $ \binom{6}{3} \cdot 1$, and the total number of win-loss combinations for 7 games is $ 2^7$ (it doesn't matter that the game might end earlier; each situation is counted an appropriate number of times, since this is probability). Thus the probability that RED will win in the seventh game is $ \frac{\binom{6}{3}}{128} \equal{} \boxed{\frac{5}{32}}$.[/hide]

[hide="the most likely incorrect answer to 1b"]If RED wins in the fourth game, there is one way. If RED wins in the fifth game, there are $ \binom{4}{1}$ ways for the BLUE to make its one win. If RED wins in the sixth game, there are $ \binom{5}{2}$ ways for BLUE to make its one win. Similarly, if RED wins in the seventh game, there are $ \binom{6}{3}$ ways. Summing these, we find that there are $ \boxed{35}$ ways for RED to win.

More simply, consider the number of ways RED could win four games out of seven, which is $ \binom{7}{4} \equal{} 35$.[/hide]

comment to second post:
some threads take much longer to be answered; your thread took less than 36 hours!

[hide="hint"]
find the umber of ways to assign the black rectagles.
then find the umber of arrangements that satisfy the given condition.
[/hide]
[hide="my solution to number 2"]
there are a total of $ \binom{12}{4} \cdot \binom{12}{4}$ ways to assign the black rectangles.
the number of arrangements that saatisfy the condition in the question is
$ 12 \cdot \binom{11}{3} \cdot \binom{8}{3}$
the first 12 is to denote the number of ways to choose the coincident rectangle, the $ \binom{11}{3}$ denotes the number of ways to assign the other 3 black rectangles to the first rectangle, and $ \binom{8}{3}$ denotes the number of ways to assign the black rectangles to the other side. the top number is less because there are less places to choose from since no more rectangles can be coincident.
the answer is then $ \frac{12 \cdot 165 \cdot 56}{495 \cdot 495}\equal{}\frac{224}{495}$
my accuracy is no better than leoxnlin's though
[/hide]

In question 2, when you say the rectangles are congruent that means they are the same size and shape, but does it also imply that they are set out uniformly in grid like pattern?

For example, what if rectangles are tiled horizontally in one half of the square and vertically in the other. Then it is impossible to get coincident black rectangles ??


draw AP and PB. ADP and EBP are similar triangles. But even AEP and PCB are. So PE:DP=PC:PE.


F:N :arrow: N
and F(F(m)+F(n))=m+n (m and n are natural numbers)
find F(n).


Find the largest real number $k$ such that:
$a^3+b^3+c^3+kabc\ge \frac{1}{9}+\frac{k}{27}$ $\forall a,b,c\ge 0$ and $a+b+c=1$

This seems t be a stupid question: what's the basic of choosing $a=0,b=c=\frac12 \implies k\le \frac{15}{4}$ (this variable is true), while $a=b=0,c=1 \implies k \le 54$, - false?

This is old inequality and easy problem
$k=\frac{15}{4}$

[quote="yptsoi"]Have you solved the n-variables case?[/quote]
Have you asked me ? ;)


given that $ \lambda \equal{} 3$ is an eigenvalue of

$ A \equal{} \left( \begin{array}{ccccc} \minus{} 3 & 2 & 1 \\
\minus{} 1 & 4 & 1 \\
\minus{} 4 & 0 & 2 \\
\end{array} \right)$

find a corresponding eigen vector

What would be the quickest way to solve this?

I done like this $ (A\minus{}I\lambda)X\equal{}0$

$ A \equal{} \left( \begin{array}{ccccc} \minus{} 6 & 2 & 1 & | & 0 \\
\minus{} 1 & 1 & 1 & | & 0 \\
\minus{} 4 & 0 & \minus{} 1 & | & 0 \\
\end{array} \right)$

and then tried to use Gaussian reduction

Where do you get stuck? Is it because you have a rank 2 matrix, and thus don't have a unique solution? Remember, you can take any multiple of an eigenvector and it is still an eigenvector.

I get something along the lines of $ \left(\begin{matrix} 4 & 0 & 1 \\
0 & 4 & 5 \\
0 & 0 & 0 \\
\end{matrix}\right)$ (your's probably doesn't look exactly like that since I usually scale to get all integer entries and I swapped two rows at one step).

Now I usually interpret this as $ 4x \plus{} 1z \equal{} 0$ and $ 4y \plus{} 5z \equal{} 0$. I need a vector $ (x,y,z)$ that will satisfy these conditions. So I stick a $ 1$ in the first spot and then fiddle. That forces a $ \minus{} 4$ in the third spot, which forces a $ 5$ in the second spot giving $ (1,5, \minus{} 4)$.


[quote="catcurio"]Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?

$\text{(A)}\ \frac15 \qquad\text{(B)}\ \frac13 \qquad\text{(C)}\ \frac25 \qquad\text{(D)}\ \frac23 \qquad\text{(E)}\ \frac45$[/quote]

[hide]$1-3\cdot\frac{1}{5}=\frac{2}{5}=C$[/hide]

[hide]If she used $\frac15$ of her money for $\frac13$ of the CD's then she used $\frac35$ of her money for all of the CD's.

$1-\frac35=\frac25$.

$\boxed{\text{C}}$[/hide]

[hide]Supposing X is the money she has and Y is the amount of cds;
1/5X=1/3Y
3/5X=Y
Thus she has 2/5 of her money left (1-3/5)
[b](C)[/b][/hide]

[hide]
If she spends $\frac 15$ of her money on $\frac 13$ of the CD's, she will spend $\frac 35$ of the money on all of the CDs.
Therefore she has $1-\frac 35=\frac 25$ of her money left.
$\text{C}$
[/hide]

[hide]x=her money
y=number of CDs
1/5x=1/3y
3/5x=y
So, 2/5
or C[/hide]

[hide]Simple. All you need to do is subtract 3/5 from 1(1-3/5). 1-3/5 equals 2/5. 2/5=[size=150]C[/size][/hide]

[hide]
She uses 1/5 of the money to buy 1/3 of the CD's, which means she uses 3/5 of the money to buy the CD's. Thus, she has 2/5 left.
[/hide]

[hide]If 1/3 of the CDs costs 1/5 of the money, then 3(1/3) of the CDs costs 3(1/5)=3/5 of the money. Brianna will have 2/5 of her money left. (answer C).[/hide]

[quote="catcurio"]

Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?

$\text{(A)}\ \frac15 \qquad\text{(B)}\ \frac13 \qquad\text{(C)}\ \frac25 \qquad\text{(D)}\ \frac23 \qquad\text{(E)}\ \frac45$

[/quote]

[hide]
2/5[/hide]


[quote="SpaceWeather.com"]
ERUPTING COMET:  Astronomers in Japan and Europe report that Comet 17P/Holmes is undergoing a spectacular eruption.  The 17th-magnitude comet has brightened by a factor of five hundred thousand or more during the past 24 hours becoming a naked eye object in the evening sky. This may signify a breakup of the comet's core or a rich vein of ice suddenly exposed to sunlight--no one knows. Look for a yellow 2.5th-magnitude fuzzball in the constellation Perseus after sunset. ("2.5th magnitude" means a little dimmer than the stars of the Big Dipper.) At present the comet looks more like a star than a comet; it does not have a discernable tail, but it might grow one as the outburst continues.  Visit http://spaceweather.com for a sky map, photos and updates.
[/quote]

Have not looked yet for the comet yet, it is cloudy and is supposed  to be for the next couple of days.

Also, the upcoming full moon with be the "biggest" moon of 2007.  On the nights of October 25 and 26 the moon will not be an illusion, it will actually appear to be bigger than normal (14% more) because it is at perigee, the closest point on it reaches to the Earth.

Saw it. Looks big. Looking forward to the moon.

Not a great combination is it? Biggest full moon and a comet? :(
I'll try tonight though
[b]@hunter34[/b] did you see it with naked eye?

The comet is or, now, was $ 2.5$ in magnitude, which is quite right.  Unless there is alot of light pollution besides the mooon, it should be easily seen.  :wink:

Saw it. Star like with naked eye. Fabulous through $ 10\times 50$, like a [i]very[/i] bright and slightly assymetrical globular   :lol:
Did anyone see greenish-yellow tinge?

There are STILL clouds here. :mad:

shreyas_patankar , you never said you had a pair of binoculars in the thread about it.

 :wink: If anyone else looks, post about it.

[quote="AstroPhys"]shreyas_patankar , you never said you had a pair of binoculars in the thread about it.[/quote]
Sorry :oops:
I voted, but somehow forgot to post.
I've got a great Olympus $ 10\times 50$ (hand held). I use it mostly for globulars etc. I've also seen 73P Schwassmann-Waschmann (if that's how you spell it) last year through the same instument.

It is fine that you did not pot. :wink:
Nice binoculars.

[quote=SpaceWeather.com"]
Last week, Comet 17P/Holmes shocked sky watchers around the world with a sudden million-fold increase in brightness.  It literally exploded into view, rapidly becoming a naked eye "star" in the constellation Perseus.  Since then the comet has expanded dramatically. It is now physically larger than the planet Jupiter and subtends an angle in the night sky similar to the Moon's Sea of Tranquility, the right eye of the "Man in the Moon."  Photographers, this amazing comet is an excellent target for off-the-shelf digital cameras and backyard telescopes.  It grows visibly from night to night and no one knows how large it will become.  Stay tuned to http://spaceweather.com for sky maps and the latest images.
[/quote]

Here is a image of what it looked like.

[img]http://www.astronomy.com/asy/objects/images/comet_17p_wjt_1107_500.jpg[/img]

That must have looked nice. Have not looked yet, houses are blocking Perseus.  :(

[quote="SpaceWeather.com"]
COMET TAIL: Exploding Comet 17P/Holmes continues to amaze onlookers. On Nov. 8th and 9th part of the comet's blue tail broke away in view of many backyard telescopes.  This event does not necessarily signal a new outburst of Comet Holmes. Comet tails can be disconnected by gusts of solar wind which trigger magnetic storms around the comet akin to geomagnetic storms which fuel auroras on Earth.

TAURID METEOR SHOWER: The annual Northern Taurid meteor shower caused by Comet Encke peaks on Nov. 12th. Although Taurid rates are normally low, only about five meteors per hour, those five can be doozies. The shower is a well-known producer of slow, bright fireballs visible at all hours of the night. If you're outside after dark this weekend, be alert for Taurids.
[/quote]

Instead of making a new topic for the meteor shower.  :wink:

[quote="SpaceWeather.com"]
GIANT COMET:  University of Hawaii astronomers have measured the diameter of Comet 17P/Holmes: 1.4 million kilometers.  This makes the exploding comet bigger than the sun and now the largest object in the solar system. Not surprisingly, the comet is visible to the naked eye; with only a backyard telescope you can watch its gigantic debris cloud expand from night to night. Nov. 19th is an especially good night to look: Comet Holmes will glide by Mirfak, the brightest star in the constellation Perseus, and appear to swallow it.

(Note: The sun remains by far the most massive object in the solar system.  Comet 17P/Holmes' diaphanous atmosphere of dust and gas, which is what the astronomers measured, contains less mass than a typical asteroid. In spite of its great size, Comet Holmes is a lightweight that won't be deflecting the orbits of planets or causing any other such catastrophes.)
[/quote]

Very cool.  :) Though I have not seen it yet.

any explanation of what caused this dramatic explosion?

[quote="AstroPhys"]
This event does not necessarily signal a new outburst of Comet Holmes. Comet tails can be disconnected by gusts of solar wind which trigger magnetic storms around the comet akin to geomagnetic storms which fuel auroras on Earth.
[/quote]

That is one possible explaination.  :wink:

[quote="SpaceWeather.com"]COMET 17P/HOLMES UPDATE:  Now that the full Moon has left the evening sky, Comet Holmes is visible again. The comet is not as bright as it was when it first exploded in late October. Most people now have trouble finding it with the naked eye. A quick sweep through Perseus with a pair of binoculars, however, will pinpoint the comet while the smallest of telescopes reveals it to be a truly impressive object.   Comet Holmes now occupies about 10 times the volume of the Sun and it is developing a hint of a blue-green tail.  A similar tail was observed in early November, but it broke off during an apparent magnetic storm.  Now the tail is growing back. Visit http://spaceweather.com for sky maps and the latest images.[/quote]

 :) Is there anyone still observing this comet?

No, or at least I am not. I can't find it.

Yes I do it is:
[img]http://sci.esa.int/science-e-media/img/16/17P-position410b.jpg[/img]

Good.  :)
Especially since it has become diffuse, it is harder to see in urbanized areas.  :(

A comet is an icy small Solar System body that, when passing close to the Sun, warms and begins to release gases, a process called outgassing. This produces a visible atmosphere or coma, and sometimes also a tail. These phenomena are due to the effects of solar radiation and the solar wind acting upon the nucleus of the comet.

The Moon is an astronomical body that orbits planet Earth, being Earth's only permanent natural satellite. It is the fifth-largest natural satellite in the Solar System, and the largest among planetary satellites relative to the size of the planet that it orbits


LaTeX
analytic geometry
MATHCOUNTS
AMC
AIME
geometry
probability
\/closed

Is the classroom moderated?...or like a chatroom?

It's a chatroom, but we will be able to monitor transcripts (read: do not abuse this privilege - those who do will be banned from this free time in the Classroom).

[quote="towersfreak2006"]When you say enrolled in a course, does this exclude auditing students?[/quote]

Nah, I'm audited and I've been able to access it tonight.

[quote="jen7"]How do we access the open classroom time?  I thought we could just click "Classroom," but when I clicked it, I couldn't get in even though it was 12:17p.m. EST when I tried.[/quote]Oops -- we messed up.   :blush:  We had a bug in the classroom software and it didn't open this morning like it should have.

It's been fixed and it should work from now on.

What about students enrolled in problem series? What level can they access? Can they even access the chatroom at those times?

The classroom doesn't seem to have been opening this weekend at the right times.

Nuts, sorry about that.  Fixed the weekend times now.

Problem Series students can access the rooms, too.

MATHCOUNTS, AMC-10: Intro
AMC-12, AIME: Intermediate.

We highly recommend that you use the message board to coordinate times to get together in the classroom - then you'll be much more likely to find someone to work with.

[quote="MysticTerminator"]just out of curiosity, is there a benefit to the classroom versus AIM?[/quote]

$\LaTeX$?

I think it's also beneficial to have a regular place students can go without even having to organize a chatroom.  It also seems natural to include everyone with a common curriculum.

Of course, the virtual classroom doesn't prevent you from chatting elsewhere.

I am new at this so be patient with me.

I am not sure how to navigate the system.  Where and how go I get to the open classroom ?

If you are currently enrolled in a class at AoPS (or have enrolled in one in the past 6 months or so), you can access the classroom during the aforementioned open times by doing the following:

1) Log in to the forum
2) Click the Classroom tab on the left side of the page

If you have never been to the classroom before, a box will open asking you if you want to run the classroom applet - say 'Always' or 'Yes' and the classroom will load automatically.

isnt there supposed to be a classroom open at 7pmET today? it is not happening.

it opened eventually, although it took about 15-20min after the supposedly opening time. But thanks anyways.

was there office hour for intro IS B today tuesday?  I wasn't able to go.  when I went to my class, it did not list my class.  It looked different.

What exactly happened and what time did you show up?  There was an office hours at 3 PM Pacific time.  Maybe you showed up a little after 4?  Then, you would not have been able to get in the office hours since it was over by then.

[quote="rrusczyk"]We are expanding our services to students enrolled in our classes by allowing them access to the online classroom at specific times.  The times at which the classroom will be open are:

Weekdays

11AM - 1PM ET
5PM - 7PM ET
9PM - 11PM ET

Weekends

11AM - 1PM ET
3PM - 5PM ET
7PM - 11PM ET

There will be 3 different rooms, one each for Intro level, Intermediate level, and Olympiad level.  Any student who has enrolled in a course that started in the last six months will be able to access the rooms corresponding to their course level at the times stated above.

A note to students using the rooms: please keep the math conversation at the appropriate level.  Also, if possible please do not put large blocks of LaTeX in the classroom.

We recommend students use the message board to coordinate meeting times.[/quote]

I cant get in. Its not letting me, even though that i have java.

[quote="WindSlicer"]I think you're going by the forum times?

ET is the forum times + 3 hours. So when you posted that message it was 2:38 pm, which meant the room should've been closed.[/quote]

There aren't designated "forum times" I don't think... Mine are set to PST because I'm in WA.

[quote="WindSlicer"]I think there's a default forum time though, I've seen people have similar time problems before. And plus it makes sense - the person was posting during a time that the classroom wasn't open, yet they thought it was.[/quote]

I think you're right.  Spencer, be sure to correct for time zone differences.

yeah, i'm having trouble opening classroom

if you read my other post, then sorry, but i got it to open the after i posted my reply

[quote="EmoryABrown"]I just tried to log on at 8:28 eastern time, and it did not work!!!  and geometry starts in two days going on at this time :!:[/quote]The open classroom is not open at 8:28 Eastern Time.  The classroom will also be open during classtimes.  There is no class at 8:28 Eastern on Tuesdays, and it was not one of the open times, so that's why you couldn't get it.

The open times are:

Weekdays

11AM - 1PM ET
5PM - 7PM ET
9PM - 11PM ET

Weekends

11AM - 1PM ET
3PM - 5PM ET
7PM - 11PM ET

Typically, students in a given class arrange a meeting time.  If you just go at random, it will almost certainly be empty.

does abusing the privlige disclude talking about things like school or asking where people live and what grade they are in, and other general chat

Socializing with your classmates in open classroom time is fine.

so now we can actually like make friends here and stuff like that. yay  :D  so were like aloud to tell people where we are and stuffs like that? what school were in?

When i logged in around 10:43 central it said that the room was still moderated and nobody was in it. it let me in though.

ya sometimes they're moderated. mostly not.

[quote="Scary math"]
And also if there would be any classes on probability or set theory in near future.
[/quote]
There is Introduction to Counting and Probabily, along with the Intermidiate one.

[quote="Scary math"]Hi
I have been an AoPSian for 5-6 weeks now and I wish to know if non-US citizens can take online classes,and if so,then details regarding timings,payment,etc.
And also if there would be any classes on probability or set theory in near future.
Thank u.[/quote]

Yes, non-US citizens can enroll.  You can find all the details on timing and cost, as well as what classes we offer, on the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]class list and enrollment page[/url].

I have been coming in introchat at 6:30EST weekdays, and have been able to stay in the classroom past 7:00. Is this supposed to happen?


geometry
circumcircle
parallelogram
complex numbers
geometry proposed

A line $ AD$ through the veertex $ A$ meets the circumcircle of trinagle $ ABC$ in$ D$.If $ U,V$ are the orthocenters of the triangles $ ABD,ACD$ , prove that $ UV$ is equal and parallel to side $ BC$

 [u]babis[/u]

It is nice and quite easy .
Clearly we have that $ CV//UB$ (they are both perpendicular to $ AD$ . Also we have that

$ CV\equal{}2OM\equal{}UB$ where $ M$ it the midpoint of $ AD$ . So we have that $ BUVC$ is a parallelogram so $ BC\equal{}UV$ and

$ BC//UV$  .

(Also we could solve this problem using complex numbers to obtain a cute solution )


algebra
polynomial
Rational Root Theorem

Only possible rational solutions are:

$ \pm 1,\pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}$
Just plugging some of these numbers in, we find that $ \frac{1}{2}$ works.
So the equation now becomes

$ \displaystyle\left(a\minus{}\frac{1}{2}\displaystyle\right)(8a^2\minus{}12a\plus{}2)\equal{}0$
So the solutions are
$ \frac{1}{2}, \frac{3 \pm \sqrt{5}}{4}$

bah i hate those equations. just learned them a week ago.

@mathcountsfreak  i assume your getting these from descartes rule of signs, right???

No, descarte's rule of signs finds how many positive/negative real roots there are.
I used the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational root theorem[/url]


limit
inequalities
calculus
topology
calculus computations

Can anyone help me find $ \lim_{x \to 0}{\frac{x}{1+e^{\frac{1}{x}}}}$ Thanks!

[quote="Zakary"]I think that it is $ 0$.

The numerator goes to $ 0$.  The denominator goes to $ \infty$ from the right-hand side and goes to $ 1$ from the left-hand side.

Am I wrong in my reasoning?[/quote]

Well, your reasoning is very informal.

[hide="Click here for a rigorous proof"]Let $ \epsilon>0$. Let $ \delta = \epsilon$.

Note the inequality
$ |\frac{x}{1+e^{1/x}}| < |x|$
(this is very easy to prove, all you have to do is note that $ e^{1/x}>0$)

If $ |x-0|<\delta$, then:

$ |\frac{x}{1+e^{1/x}}| < |x| <\delta = \epsilon$

In other words,

$ |\frac{x}{1+e^{1/x}}-0| <\epsilon.$

Therefore, $ \lim_{x \to 0}{\frac{x}{1+e^{\frac{1}{x}}}}$ by the $ \epsilon-\delta$ definition of a limit.[/hide]

It may be informal, but it's good enough. I would certainly accept that reasoning from a calculus student.

Yeah, unfortunately, I don't know the epsilon-delta definition of limits -- all I know is L'Hopital's Rule, algebraic manipulation, and "informal" reasoning, etc.

I've never really "understood" the epsilon-delta definition when I've looked it up by myself -- I've never been taught in class because I've only taken AP Calculus as my only Calculus class.  Could epsilon and delta be called by different names?  Why are they called "epsilon" and "delta"?

Thanks.

The specific letters $ \epsilon$ and $ \delta$ are just there by tradition; the quantifiers (for every  ___, there exists ___) are the important part.
A version of the definition of a limit using words:
If $ f(x)$ can be made arbitrarily close to $ L$ by choosing $ x$ sufficiently close to, but not equal to $ a$, then $ \lim_{x\to a}f(x)=L$.

You can go two directions from there; on the real line or in a metric space, we get $ \epsilon$ to indicate "arbitrarily close" and $ \delta$ to indicate "sufficiently close". In a more general topological context, we have the concept of neighborhoods; for every neighborhood of $ L$, there exists a punctured neighborhood of $ x$ satisfying the condition.


Well, I've been thinking about this: Right now, Harry Potter is so popular that people have threatened to commit suicide just because of this book. But will Harry Potter survive the test of time? Will Harry Potter still be considered a great work of literature 300 years later? Post your opinions here.

PS Although I know this is the Games and Fun Factory, there usually aren't Harry Potter related topics on the Round Table. I would still like reasonable and honest votes  :)  (if your a Harry Potter fan, don't just automatically vote for "the rest of humanity").

oh uhh, my mom snuck up on the computer when i was playing piano and voted what she thought, thats why it says 11 - 50. I think maybe 100 - 151.

umm its going to be somewhat like Lord of the Rings series..

People are going to remember it for a while because they'll remember how popular it was. But I don't think they'll remember it for the literature value. Part of the reason it was so popular was the constant hype about the next book. Once the excitement over the 7th movie dies out, there won't be any more of that driven enthusiasm that it used to enjoy, which was one of the things that would get new readers started.

People will probably remember Harry Potter for a while, but I don't think they'll remember it 50 years from now.  Like A M Y said, there won't be as much excitement about it now, since the 7th book is already out.  Sure, there are two movies left, but after those come out, it'll probably just die away and people will move on.  It's a pretty safe bet that there will be a new series of books that caused as much hype as the Harry Potter series has.

There a bunch of books from the 1800's that are still famous, but not exactly popular. Harry Potter might become one like that.

yeah, vamathletes interpreted my question well: it doesn't neccesarily have to be popular, just well known.


there was a question about box-and-whiskers plot on a math tournament, and none of us knew what it was. the teacher barely knew anything about it... so can someone explain what a box and whiskes plot is?

Sorry, but I'm not going to draw a sample plot.

There are three vertical lines in a box and whiskers plot. The boundaries between the quartiles are marked by the the lines.  The numbers inside the box are called the innerquartile. The difference between the two furthest lines is the innerquartile range. 3/2 times the innerquartile range is the farthest a number can be from the ends of the box before it's an outlier. Outliers may be denoted by asterisks. The highest non-outlying number and the lowest non-outlying number form the ends of the whiskers.

Note that the box and whisker plot may have vertical whiskers, with the three vertical lines above being horizontal lines.

Ex. If the data set was {1,2,3,4,5,6,7}, the median is 4, the upper quartile is 6, the lower quartile is 2, and the ends of the whiskers are 1 and 7.

[quote="LuCy4EvA"]I'm not exactly getting it... I now that you divide the set of numbers by 4, and you get like 4 quartiles, and two of them is inside the "box" and the other two are the whisker part that's not in the box[/quote]

The median is the middle number when you arrange the numbers in order from least to greatest. It's the middle line.

The difference between the two numbers that determine the box is the innerquartile range. You multiply the innerquartile range to get some nameless number. You add the nameless number to the greater boundary of the box to get another number. Any number greater than the sum is an outlier. You subtract the nameless number from the smaller boundary of the box to get some number. That number is greater than any outlier.

http://mathworld.wolfram.com/Box-and-WhiskerPlot.html has examples of box and whisker plots.


I use following code for including graphics and I would like to add an box of 1 pt around the picture. Could you tell me how to do that?

When I use the commands [keepaspectratio=true,width=360pt] I assure the picture will have the same width as the text and has an appropriate height.  I prefer to add a box without interfering with these commands.

\usepackage{graphicx}
\begin{document}
\begin{figure}[htbp]
	\centering
		\includegraphics[keepaspectratio=true,width=360pt]{grexp3.jpg}
	\caption{Experiment 3}
	\label{fig:gr_exp3}
\end{figure}
\end{document}

\fbox is an easy way to put a box around anything:
\[ \text{\fbox{anything}} \]
\setlength{\fboxrule}{1pt} will change the width of the lines to 1pt.
\[ \text{\setlength{\fboxrule}{1pt}\fbox{anything}} \]


[hide]First draw an altitude from C to AB and call that X.  That length is $\sqrt{48}$ from the Pythagorean Theorem.

Then apply the Pythagorean Theorem to triangle CXD.  You get the equation:

$64-48=(x+1)^2$.

Solving gives $x=3$?[/hide]

From Stewart's Theorem...

$c*n*c + b*m*b = m*a*n + d*a*d$

When you draw out the triangle, you get $BC$ to be a cevian; therefore...

$c = 7, m = 2, d = 7, b = 8, n = ??$

Where we are trying to solve for $??$. Note that this formula is extremely useful, and is a good idea to memorize. (Understand why it works though... AoPs Volume 2 >> More Triangles!!)

We'll use $n$ instead of $??$...

Plugging in and simplifying, we get $n^2 +2n-15=0$. Factor to get $(n+5)(n-3)=0$. Since $n>0, n = \boxed3.$

[hide]Call the midpoint of AB E. Triangle CEB and triangle CED are right. $EB=1$.
$CE^2=49-1=48$.
$ED^2=64-48=16$
$ED=4$
$BD=ED-EB=4-1=3$

$\boxed{A}$[/hide]

[quote="catcurio"]Source: 2005 AMC-10B #10

In $\triangle ABC$, we have $AC=BC=7$ and $AB=2$. Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$. What is $BD$?

$\text{(A)}\ 3 \qquad\text{(B)}\ 2\sqrt3 \qquad\text{(C)}\ 4 \qquad\text{(D)}\ 5 \qquad\text{(E)}\ 4\sqrt2$[/quote]

Denote O such that CO┴AB. Then, CO=√48.

Then, OD=√(64-48)=4.

So, BD=OD-OB=4-1=3.

So, BD=3 (A).


trigonometry
quadratics
geometry unsolved
geometry

Let $A,B$ be two points on a give circle, and $M$ be the midpoint of one of the arcs $AB$ . Point $C$ is the orthogonal projection of $B$ onto the tangent $l$ to the circle at $A$. The tangent at $M$ to the circle meets $AC,BC$ at $A',B'$ respectively. Prove that if $\hat{BAC}<\frac{\pi}{8}$ then $S_{ABC}<2S_{A'B'C'}$.

Assuming that $C'\equiv C,$ and setting $\angle BAC=\angle AOM=\theta,$ we have: $AC=R\sin(2\theta),$ and $AA'=R\tan\frac{\theta}{2},$ where $R$ is the radius of the circle. Then $f(\theta)=\frac{[ABC]}{[A'B'C]}=\frac{AC^2}{A'C^2}=\frac{\sin^2(2\theta)}{\left(\sin(2\theta)-\tan\frac{\theta}{2}\right)^2}.$ To simplify set $x=\tan\frac{\theta}{2},$ and rewrite $f(x)=\frac{16(1-x^2)^2}{(-x^4-6x^2+3)^2}.$ Since $x>0,$ it is easy to show that as long as $x<\sqrt{2\sqrt{3}-3}$ (or $\theta<68.53^\circ$), the quadratic in $x^2$ in the denominator is positive, and therefore we can write $\sqrt{f(x)}=\frac{4(1-x^2)}{3-6x^2-x^4}.$ We are asked to show that $\sqrt{f(x)}<\sqrt{2},$ which is obvious by solving the resulting quadratic in $x^2.$

[quote=Prasanthinalluri]I'm sorry this might be a stupid question.
But I'm from the US and have never seen notation like this.
What does the $S_{ABC}$ notation mean in this context.[/quote]

Area of $ABC$.


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

Let $ ABC$ be a triangle, $ P,Q$ the isogonal conjugate points, $ O$ the circumcenter, $ L$ the Anti-Steiner point of the line $ OQ$ wrt median triangle of $ ABC$. It is knwon that $ L$ lie on pedal circle of $ P$. Prove that Simson line of $ L$ wrt pedal triangle of $ P$ is parallel to $ OQ$.

We know that $ OQ$ is the [b]Steiner[/b] line of triangle $ A_1B_1C_1$ wrt L and it parallel to [b]Simson[/b] line wrt L. So we prove that [b]Simson[/b] line of triangle $ A_1B_1C_1$ parallel to [b]Simson[/b] line of triangle $ A_2B_2C_2$ wrt L.
$ \Leftrightarrow \angle A_2RZ\equal{}\angle A_1WT \Leftrightarrow \angle RA_2Z\plus{}\angle RZA_2\equal{}\angle A_1TW\plus{}\angle TA_1W$
$ \Leftrightarrow \angle A_2B_2C\plus{}90^o\minus{}\angle C_2A_2L\equal{}\angle B\plus{}90^o\minus{}\angle LA_1C_1$
$ \Leftrightarrow \angle A_2C_2C\equal{}\angle B\plus{}\angle C_2A_2L\minus{}\angle LA_1C_1\equal{}\angle C_1B_1A_1\minus{}\angle LB_1C_1\plus{}\angle C_2L'L\equal{}\angle C_2L'A_1$
Let $ U\equal{}A_1L'\cap (A_2B_2C_2), A_3\equal{}(A_2B_2C_2)\cap BC$ we have to show that $ \angle C_2L'U\equal{}\angle A_3C_2B_2$
$ \Leftrightarrow C_2B_2UA_3$ is a trapezoid.
Applying the first [b]Fontene[/b]'s theorem we have $ C_2A_2, C_1A_1, B_2L'$ are concurrent at point B'
Let $ G\equal{}C_2U\cap B_2A_3$, applying [b]Pascal[/b]'s theorem for hexagon $ C_2A_3A_2UL'B_2$ we get G lies on $ C_1A_1$
Finally we will prove that $ GC_2\equal{}GB_2$ and complete the proof.
Let J be the projection of A onto BP, we will show that J lies on $ A_3B_2$.
Since $ C_2B_2PJ$ is cyclic quadrilateral we get $ \angle JB_2C_2\equal{}\angle JPC_2\equal{}\angle C_2A_2A_3\equal{}\angle A_3B_2C_2$ so $ A_3,J,B_2$ are collinear.
Let D be the midpoint of $ AP$, it's easy to see that $ DB_2\equal{}DC_2$ so we prove $ D,O,G$ are collinear.
Because $ AC_2JB_2$ is cyclic and $ GC_1//AB_2$ then $ C_2JGC_1$ is cyclic.
$ \angle C_1C_2J\equal{}\angle AB_2J\equal{}\angle APJ\equal{}\angle DJP\equal{}\angle C_1DJ$ (because $ C_1D//BP)$ then $ C_1C_2DJ$ is cyclic.
Therefore $ C_1,C_2,D,J,G$ are concylic $ \Rightarrow \angle C_1DG\equal{}\angle C_1JG\equal{}180^o\minus{}\angle C_1JB_2 (1)$
On the other side, $ \angle AJC_1\equal{}\angle C_1AJ\equal{}\angle C_2B_2J$ then $ \angle AEC_2\equal{}180^o\minus{}\angle C_2JB_2 (2)$
But $ AJ\perp C_1D, OD\perp C_2B_2$ so $ \angle AEC_2\equal{}\angle ODC_1 (3)$
From $ (1)(2)(3)$ we have $ \angle ODC_1\equal{}\angle C_1DG$
Thus $ O,D,G$ are collinear.
We are done!

Another proof is base on [b]Yetti[/b]'s nice proof at here:
http://www.mathlinks.ro/viewtopic.php?search_id=308431607&t=173756

Let $ \triangle P_AP_BP_C, $ $ \triangle Q_AQ_BQ_C $ be the pedal triangle of $ P, $ $ Q $ WRT $ \triangle ABC $, respectively. Let $ T, $ $ H $ be the projection of $ A $ on $ OQ, $ $ QQ_A $, respectively. Since the 9-point circle of $ \triangle ABC $ and $ \odot (AOT) $ are symmetry WRT A-midline of $ \triangle ABC $, so $ L $ and $ T $ are symmetry WRT A-midline of $ \triangle ABC $ $ \Longrightarrow $ $ Q_AHLT $ is an isosceles trapezoid.

Let $ V $ $ \in $ $ \odot (P_AP_BP_C) $ be the point such that $ LV $ $ \perp $ $ P_BP_C $. From $ \measuredangle P_AVL $ $ = $ $ \measuredangle P_AQ_AL $ $ = $ $ \measuredangle THA $ $ = $ $ \measuredangle TQA $ and $ AQ $ $ \parallel $ $ LV $ $ \Longrightarrow $ $ P_AV $ $ \parallel $ $ OQ $, so notice $ P_AV $ is parallel to the Simson line of $ L $ WRT $ \triangle P_AP_BP_C $ (well-known) we conclude that the Simson line of $ L $ WRT $ \triangle P_AP_BP_C $ is parallel to $ OQ $.

Let $\triangle P_AP_BP_C,\triangle Q_AQ_BQ_C$ be the pedal triangles of $P,Q$ WRT $\triangle ABC$ and let $\triangle DEF$ be the medial triangle of $\triangle ABC.$ Parallel from $A$ to $BC$ cuts $QQ_A$ at $J$ and $R$ is the projection of $A$ on $OQ.$ In this configuration [url=https://artofproblemsolving.com/community/c6h414593]it's already known[/url] that $J,L,R,Q_A$ are concyclic and $A,R,Q,J$ are obviously concyclic. Thus simple angle chase gives:

$\angle LP_AP_B=\angle LQ_AJ-\angle P_BQ_AQ=\angle RAQ-\angle P_BQ_AQ=\angle (OQ,P_BP_C)-(90^{\circ}-\angle P_BQ_AP_A) =$

$=\angle (OQ,P_BP_C)+90^{\circ}-\angle P_AP_CP_B=90^{\circ}-\angle (OQ,P_AP_C).$

This latter expression means that the isogonal of $P_AL$ WRT $\angle P_BP_AP_C$ is perpendicular to $OQ,$ i.e. the Simson line of $L$ WRT $\triangle P_AP_BP_C$ is parallel to $OQ.$


algebra
polynomial
function
domain
real analysis
real analysis unsolved

Let $f:[0,2\cdot\pi)->[-1,1]$ be a real trygonometric polynomial of degree at most $n$ such that $|f(x)|=1$ for exactly $2n$ values in its domain of definition. Then find $f$.

Do we also have the condition $|f(x)|\le 1$ for all $x$ ? If not, we have many rather uninteresting solutions exemplified by the following case: let $n=4$ and let $f(x)=a\cos(2x)$ for some $a>1.$

I specified in the statement that the polynomial takes values in $[-1,1]$. By the way, this is really a super-nice problem from Putnam.


Consider the triangle $ ABC$ with $ m(\angle C) > 90^\circ$ and the point $ D$ on the side $ (BC)$ such that $ AC \plus{} AB \equal{} 2AD$.
The median $ CM$, with $ M$ on $ AB$, intersects the line $ AD$ at the point $ N$.Prove that $ AN\leq 2ND$.

[b]BTW, please, everyone, try to solve the question [url]http://www.mathlinks.ro/viewtopic.php?t=200023[/url][/b]
[b]Thanks.[/b]

[color=darkblue]By [b][u]Menelaus theorem[/u][/b] in $ \triangle{ADB}$ for transversal $ \overline{CNM}$ we have: $ \dfrac{AM}{MB} \cdot \dfrac{CB}{CD} \cdot \dfrac{ND}{AN}\equal{}1 \Longleftrightarrow \dfrac{AN}{ND}\equal{}\dfrac{CB}{CD}\equal{}1\plus{}\dfrac{BD}{CD}$.

So we need to prove that $ 2 > \dfrac{AN}{ND} \equal{}1\plus{}\dfrac{BD}{CD} \Longleftrightarrow  \dfrac{BD}{CD} < 1\Longleftrightarrow  BD < CD$.

Let $ L$ be midpoint of $ BC$. We have $ AD\equal{}\dfrac{AB\plus{}AC}{2}>AL$. Because $ \angle{C}>90^\circ$ points $ C,L,D,B$ lie on $ BC$ in this order, so $ CD> \dfrac{BC}{2}>BD$, $ \mathrm{Q.E.D.}$[/color]


Captain Sugar posts 7 posts a day, unless the internet is down. he posts between 7-8 a.m. and 6-8 p.m. every day (unless the internet is down.) how many posts will Captain Sugar post in 29 2/3 days if the internet is down 7 of those days?

edit: round down to the nearest whole number, since you can't have a fraction of a post.

it depends because of the 2/3 days

you have to know when he startss posting

i think...

[edit]  it is now 8.14 posts per day :D

assume the day starts at 12.01 a.m.

[hide="answer"] 156.  2/3 of a day falls at 4:00 p.m, when he has only posted at 7-8 a.m.  22 days (because of the week the internet was down) *7 = 154 posts, plus the 2.3 posts in the morning rounded down is 156.[/hide]

[hide]
ok so 7 of the days the internet is down
Thus there are $19\frac 23$ days
For the 19 part, there are $19\cdot7=140-7=133$
For the $\frac 23$ part you can't tell because it depends how many times he posts in each time period. $\frac 23$ of the day is 16 hours=12am-4pm
If they were distributed evenly, then he would get in half of his days posts, 3.5
Thus the answer would be 138.5[/hide]

sorry, biffanddoc, not quite, don't forget to round down because Captain Sugar can't post half a post!

Boy, it's fun to talk about yourself in third person  :lol:

[hide]29 2/3 - 7 = 22 2/3 days, not 19 2/3! also, since 2/3 of the day has passed, he would have posted only during the 7-8 a.m. period, where he would post 1/3 of his daily posts, assuming he posts evenly. so, on the 2/3 day, he would post 2 posts, because 7/3 = 2.33333 repeating, rounded down to 2.[/hide]


induction
arithmetic sequence
number theory unsolved
number theory

Define a sequence $a_n$ such that $a_1=1$, $a_n=a_{n-1}+(a_{n-1} mod n)^2$, where $a_{n-1} mod n$ is the remainder. How many consecutive identical terms can there be at most?

It is easy to prove (by induction) $a_n=n$.

It's not easy to prove since it's wrong. ($a_3=6$)
Note that the square is outside of $\mod$...

I am sorry, I solve $a_n=a_{n-1}+(a_{n-1}^2 mod \ n).$

Another solution:

We have $a_1=1, a_2=2, a_3=6, a_4=a_5=10, a_6=26, \ldots, a_{21}=1085=49\cdot22+7.$ And here an arithmetic progression begins!!! :lol:

Really, if $a_n=49(n+1)+7$ then $a_n mod (n+1)=7$ and $a_{n+1}=a_n+7^2=49(n+2)+7.$ Hence by induction we have $a_n=49(n+1)+7, n\ge21.$  :) :lol:
Therefore  the alone consecutive identical terms of our sequence are $a_4=a_5=10$ and $a_{10}=a_{11}=a_{12}=132.$

[quote="rogue"]Another solution:

We have $a_1=1, a_2=2, a_3=6, a_4=a_5=10, a_6=26, \ldots, a_{21}=1085=49\cdot22+7.$ And here an arithmetic progression begins!!! :lol:
[/quote]

What's the motivation behind this?

[quote="Rust"]It is remind interest problem: $a_1\in N, a_n=a_{n-1}+(a_{n-1} mod \ n)$. Prove that exist m,k: $a_n=k(n+2)$ for all n>m.
Generelized: $a_1 \in N,a_n=a_{n-1}+(a_{n-1} mod \ n)^l$. Prove that exist m,k: $a_n=k^l(n+1)+k$ for all n>m.[/quote]

Or disprove that ... :?


algebra
polynomial
function
national olympiad

VietnamTST2000
                                                  Day1
1.Two circles $C_1$and$C_2$ intersect at points $P$and$Q$.Their common tangent,closer to $P$ than to $Q$ ,touches $C_1$ at A and $C_2$ at B .the tangents to $C_1$ and $C_2$ at $P$ meet the orther  circle at points $E$ # $P$ and $F$#$P$ ,respectively.Let $H$ and $K$ be the points on th rays $AF$ and $BE$ respectively such that  $AH=AP$and $BK=BP$.Prove that $A,H,Q,K,B$ lie on a circle.
2.Let $k$ be a given postive  integer.Define $x_1=1$ and for each $n>1$ ,set$x_{n+1}$ to be the smallest positive integer not belong to the set   {$x_i,x_i+ik/ i=1,2,...n$}.
Prove that there is a real number a such that $x_n=[an]$ for all $n$ belong to $N$.
3.Twos players alternately replace the stars in the expression
                            $x^{2000}+ x^{1999}+..+x+1$
By real numbers.the player who makes the last move lose if the resulting polynomial has a real  root $t$ with $|t|<1$, and wins ortherwise.Give a winning  strategy for one of the players.
                       Day2
4.Let $a,b,c$ be pairwise coprime natural numbers.$A$  positive integer $n$ is said to be stubborn  if it cannot be written in the form
                                                  $n=bcx+cay+abz$ for some $x,y,z$be long to$N$
  Determine th number of  stubborn  numbers.
5.Let $a>1$ and $r>1$ be real numbers.
(a) Prove  that if $f: R^{+}\to R^{+}$ is a fuction satisfying the conditions
        (i)$f(x)^2\leq ax^r.f(\frac{x}{a})$ for all $x>0$
        (ii)$f(x)<2^{2000}$ for all $x<\frac{1}{2^{2000}}$
    Then  $f(x)\leq x^r.a^{1-r}$ for all $x>0$.
(b) Contruct a fuction $f: R^{+}\to R^{+}$  satisfying condition (i) such that for all $x>0,f(x)>x^r.a^{1-r}$.
6.A collection of $2000$ congruent circle is given on the plane such that no two circles are tangent and each circle meets at least two orther circles.Let $N$ be the number of points that belong to at least two of the circles.Find the smallest possible value of $N$.


Cevians dividing triangle sides proportionate to the angles
    at these sides are concurrent. Name the triangle center.

     Col. Pestich.

[quote="pestich"]Cevians dividing triangle sides proportionate to the angles
    at these sides are concurrent. Name the triangle center.[/quote]

If your cevians divide the triangle sides in the ratios of the [b]opposite angles of the adjacent sides[/b] (this means that, e. g., the cevian AA' is such that BA' : A'C = C : B), then the cevians concur at the Hofstadter Zero Point $X_{360}$ of triangle ABC. But if your cevians divide the triangle sides in the ratios of the [b]adjacent angles[/b] (this means that, e. g., the cevian AA' is such that BA' : A'C = B : C), then the cevians concur at a point which is not in the ETC yet. This point is the isotomic conjugate of the Hofstadter Zero Point $X_{360}$, of course.

  Darij


When logged into [url]http://www.artofproblemsolving.com[/url], I often click the "View posts since last visit" button to give me a page of new posts.  On that page, there are occasionally links to portal posts that Valentin or someone added to the MathLinks home page.  For example, Valentin recently added a portal post about "New search feature added".  On the AoPS page of new posts, the corresponding link was to [b]www.artofproblemsolving.com[/b]/Forum/viewtopic.php?t=226526.  When I click on that link, I get sent to the Forum index page, not the portal post.  In contrast, if I were to manually type the address [b]www.mathlinks.ro[/b]/Forum/viewtopic.php?t=226526, I get sent to the correct portal post.

Is that the way the links are supposed to work?


AMC
AIME
geometry
perimeter
analytic geometry
area of a triangle
similar triangles

So you have two circles, one internally tangest to the other. Their point of tangency is T. There are two points on the larger circle, namely A and B, so that AB is tangent to the smaller circle at T'. Given that AT' = 7 and T'B = 56, maximize the area of ATB.

Hey Kamil! This is Matt from HCSSiM.

I'll post what I can think of.

The area of a triangle (in terms of it's sides) can be given by Heron's formula.
Let S be the semi-perimeter of the triangle ATB $ S = \frac {1}{2}(AT + TB + BA)$
We know that $ BA = AT' + T'B = 7 + 56 = 63$ from the given information.

Then the area of the triangle ATB is $ \sqrt {S(S - AT)(S - TB)(S - BA)}$ =
$ \sqrt {\frac {AT + TB + 63}{2} \cdot (\frac {AT + TB + 63}{2} - AT) \cdot (\frac {AT + TB + 63}{2} - TB) \cdot (\frac {AT + TB + 63}{2} - 63)}$

Let $ m$ be the segment $ TB$, and let $ n$ be the segment $ AT$.

Cleanup: through algebraic simplification, our expression is
$ \frac {1}{4} \sqrt {(63 + m - n) (63 - m + n) ( - 63 + m + n) (63 + m + n)}$

All we need to do now is bound $ m$ and $ n$, and then we can attack this with a Lagrangian-multiplier esque solution. I've got to get back to English work now, but I think this is on the right track.

I don't see how that uses the hypothesis that the circles are tangent. Here is a hint:

[hide="hint"]what is the locus of point T?[/hide]

If I'm not mistaken: [hide="The locus of T is"]a circle.[/hide]

[hide="My First Approach"]Let $ C$ and $ D$ be the centers of the large and the small circles, respectively.

We can use the following system of equations:
1. $ AC \equal{} CT$
2. $ \overline{DT'} \perp {AB} \implies T'D \equal{} DT$
3. $ D \in \overline{CT}$

If we fix $ A$ and $ B$ and apply coordinates, we can obtain equations relating $ C$ and $ D$ with the coordinates of $ T$ using the first two equations above. Then, we can use similar triangles to apply the third equation as a condition.

From here, it's fairly light algebra.[/hide]


Misha’s favorite number is even. Her favorite number is also non-negative and less than 1. What is Misha’s favorite number?

[hide]$0$[/hide]
It is the only possibility.  :D

mmm...
[hide]0
easy :first:  :icecream: [/hide]

Was it...? [url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=54[/url] I don't know where to look!
[hide] Well since we know she is wearing a pink dress and her hair is brunette and she is eating popcorn the answer is 0[/hide]
If this was on MC would someone PM me and tell me which round?

[quote="Bluestalk"]Wasn't that like.....#1 in Chapter Round of Mathcounts last year?[/quote]

I am inclined to think that is was #1 on the School countdown round..... naah still too easy

[hide="By common sense,"]Misha's favorite number is 0.[/hide]

[quote]I am inclined to think that is was #1 on the School countdown round..... naah still too easy[/quote]

NO WAIT. I think it was in Chapter Countdown #1, because I seem to remember that we were laughing at the first question because it was so dumb. Might be a faulty memory though.

Here is the link.
http://www.mathcounts.org/webarticles/articlefiles/510-06CCountdown.pdf
Yes, it was from a countdown

Waaah! No one PMed me. Umm. just wondering but does MC always have some really easy ones...?

My chapter has some extremes. Usually it is really crappy and it took 5-10 seconds for the people up there to get thos one right. My teammate made official countdown with a 38. However, last year four of the people in countdown at state were from my chapter, and two of the people on national team were from there.

MY chapter from what I here is really mean but I'm not sure about smartness...(ok maybe they aren't mean i don't know....i wasn't there)

EDIT: I meant by that that they sound dumb. Picking on little girls. Well it is tempting...JK it isn't really for me.

You guys can discuss mathcounts in the appropriate forum.

Since the problem has been solved, this topic is locked.


Let  $ h_a, h_b, h_c$ be the heights,  $ r_a, r_b, r_c$ be the exradii of an triangle $ ABC$. Let  $ P$ be any point inside triangle $ ABC$. Prove or disprove that

($ \frac{h_a}{d_a} \plus{} \frac{h_a}{d_a} \plus{} \frac{h_a}{d_a})(\frac{h_b \plus{} h_c}{r_a} \plus{} \frac{h_c \plus{} h_a}{r_b} \plus{} \frac{h_a \plus{} h_b}{r_c}) \ge 54$,

where  $ d_a, d_b, d_c$ are the distances from  $ P$ to the sides  $ BC, CA$ and $ AB$ respectively.

___________
 :?:  :)


These aren't a question from the Math Team meet (I thought up of these during the meet; refering to [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1687]Math Team Meet 2 Question 3[/url]
), so these might be ridiculously hard, but...

What is 1/(1+1/(2+1/(3+1/4+...)))
What is 1/(2/(3/(4/(5/...))))

If you made this problem up it's probably impossible to evaluate in closed form. I can see no way of evaluating it.

[hide]1/1+(1/2+(1/3+(1/4+...= 2, cuz you have one already and the more you add, the closer you get to 2, so if it goes on forever it will equal 2[/hide]

I don't think so. It seems to be converging somewhere around .69777. Unfortunately, that isn't .6931, which would be really beautiful.

I believe this is solvable, just find the pattern...

I have no idea what your problem is meant to be, but as it is written 1/1 + 1/2 + 1/3 + .. diverges. And 1/1 + (1/2 + (1/3...) is exactly the same thing. I think you need to put the brackets in different places, maybe you mean 1/(1+1/(2+1/(3+..)?

As for the second one.. you are gonna needa put brackets in there too, for example there is no such thing as 1/2/3, it could be 1/(2/3) = 3/2, or (1/2)/3 = 1/6. Neither are solvable as is.

He meant 1/(1+1/(2+1/(3+1/4+...))), if I understood correctly. (Tare, is that right?)

Yup, I thought I put them in correctly...
As for number 2 I already clarified that point by saying / denotes fraction, not division, but I'll go fix that too.

Well, a fraction is the same as division, so that didn't really make sense.

A division: A/B/C=(A/B)/C=A/BC
A fraction: A/B/C= A/(B/C)=AC/B

This is because if all the functions are divisions, you do them from the first one while in a fraction you do them from the last...I think...I hope :lol:

Nah fraction and division are exactly the same thing. Theres no set order of doing either, you have to put in brackets in either case.

okeydokes. Anyone solved the problems yet?

Because it doesn't converge, it diverges

Well, that's not really fair. It converges extremely quickly, so I can give you any degree of accuracy you want. The number is transcendental for sure since it has such terrific rational approximations. It doesn't seem to have anything to do with e or pi, but then again, the continued fraction representation for e is rather enlightening, to say the least.

Food for thought: What if it reaches infinity? What then?

Oops. I started writing my post before confuted posted. It doesn't diverge. It converges to somewhere around .69777.


topology
advanced fields
advanced fields unsolved

Does anyone know of an example of spaces $ X$ and $ Y$ that are homeomorphic and subsets $ A \subset X$ and $ B \subset Y$ that are also homeomorphic but where $ X \minus{} A$ and $ Y \minus{} B$ are not homeomorphic?

Sure. Take $ X \equal{} Y \equal{} [0,1]$, $ A \equal{} [0,1/2]$ and $ B \equal{} [1/4,3/4]$.

In general, the homeomorphism type of a complement of $ A$ depends not only on the ambient space $ X$ and the homeomorphism type of $ A$, but also on "how" $ A$ sits within $ X$. For a more interesting example, consider a (planar) circle $ A$ and a [url=http://en.wikipedia.org/wiki/Trefoil_knot]trefoil knot[/url] $ B$ in $ X \equal{} Y \equal{} \mathbb{R}^3$ (it's not quite trivial to show that the complements are not homeomorphic, but in fact, they're not).

actually, i seem to remember that knot complements determines its isotopy type (at least in dimension three).

That is correct, but it's a highly sophisticated result. Showing that the unknot and the trefoil have non-homeomorphic complements is much easier, of course, but still not quite trivial (the only way I can think of requires the fundamental group).


Can anybody please help me with a problem I'm having with transcripts? I need to know how to download transcripts where the LaTex and the pictures will show up on Microsoft Word.

Well, I'm not a very computer trained person but looking at proprties it says "Microsoft Office 2000 SR-1 Professional." I hope that helps.

Hurm.  If you just try copy-paste, it doesn't work?  Do you have any other editors?

i dont think i have any other editors, microsoft word is the only one i have  :(

Hey I have the same version of word, and here copy-paste does work... did you try that? Perhaps you should try again, your memory may have been full or so :)

( select the text -- ctrl+c -- goto word -- ctrl+v )


it's even $\geq 2$ and was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=34079 (some of the posts at the nend of the thread).

  dg


There are 35 objects that need to be carried away which in total weigh 18 pounds. A spaceship can carry away up to a total of three pounds per trip. Show that if the spaceship can carry away any combination of 34 of the objects in 7 trips, then it can carry away all 35 of the objects in 7 trips.

This is a very nice problem, try to see if you can solve it. You can post any progress you've made.
[hide="Hint"]Consider the object of smallest weight[/hide]

[hide="Partial solution"]If the sum of weights of 35 objects is 18 pounds, then the smallest object weighs less than or equal to $ \frac{18}{35}$ pounds. This is easy to prove, since if all the objects exceeded this weight, then the total weight would exceed 18 pounds. Call this weight $ a$

Now consider the combination of 34 objects which does not include the smallest weight. The total combined weight of these objects is equal to $ 18\minus{}a$, so using a similar technique, one of the 7 trips must have a total weight of at most $ \frac{18\minus{}a}{7}$. Now consider adding $ a$ to this smallest trip (without exceeding 3 pounds).

$ \frac{18}{7}\minus{}\frac{a}{7}\plus{}a\equal{}\frac{6(a\plus{}3)}{7}$

Set this less than or equal to 3 and we find that $ a \le \frac{1}{2}$ so for any $ a$ up to $ \frac{1}{2}$ then it is possible to add $ a$ to one of the trips, and therefore carry all 35 objects in 7 trips.

Now for the cases where $ \frac{18}{35} \ge a > \frac{1}{2}$ which is where I am stuck.[/hide]

Yes, you have got the right idea of breaking it up into $ a \ge \frac {1}{2}$ and $ a \le \frac {1}{2}$ Now for the second case...

[hide=" full solution"]
label the objects $ a_i$ with $ a_1 \ge a_2 \ge...\ge a_{35}$

If $ 34$ objects can be moved in $ 7$ trips, and each trip can take at most $ 3$ pounds, then there is a total of $ 3\cdot 7 \equal{} 21$ pounds available among all $ 7$ trips. Consider the first $ 34$ objects, which we know can be taken in $ 7$ trips. We have $ a_1 \plus{} ... \plus{} a_{34} \equal{} 18 \minus{} a_{35}$

So by the box principle there is atleast one trip with $ \frac {21 \minus{} (18 \minus{} a_{35})}{7}$ pounds of free space.
If $ \frac {21 \minus{} (18 \minus{} a_{35})}{7} \ge a_{35} \Leftrightarrow \frac {1}{2} \ge a_{35}$ then we are done.

So assume $ a_{35} > \frac {1}{2}$

Since $ 34$ objects (all more than $ \frac {1}{2}$) can be taken in $ 7$ trips, we must have each trip taking exactly $ 5$ objects except one trip which takes $ 4$ objects.

Consider the trip which takes $ 4$ objects, the worst case is that the four objects are the heaviest.

$ 18 \equal{} \sum a_i \ge (a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_{35}) \plus{} 30a_{35} > (a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_{35}) \plus{} \frac {30}{2} \Longrightarrow a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_{35} < 3$

Since $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_{35} < 3$, we can send them together and we are done  :)
[/hide]


Yeah, so here's a problem that I debated with my teacher, but it's really controversial and stuff.

Sorry, my typing skills with numbers are bad:

Square root of (2m^2-9) = m

I got 3 and -3 as solutions, and when I plug them in they both work.  (i.e. square root of 9 is equal to both 3 and -3).

However, my teacher doesn't seem to think so.  She says that the answer should be 3, and -3 is extraneous, which I don't understand why...

I can't tell if this is sarcastic or not.

Anyway, the square root of a number is defined to be the positive square root if the sign is already in the problem. So, $ \sqrt{9}\equal{}3$.

Watch when this is just a spam thread.

[quote]Anyway, the square root of a number is defined to be the positive square root if the sign is already in the problem. So, . [/quote]

Sorry, but what sign...?

Dude wth? PLug it in.

You get rt(18-9)=-3

rt9=-3

However, sqrt(x) implies the positive sqrt!

[quote]However, sqrt(x) implies the positive sqrt![/quote]

I knew how to plug it in.  I was confused whether it was positive or both positive or negative sqrt.

Thanks guys.


Let be $ a,b,c>0$ such that $ abc\equal{}1$ . Show that :

$ \frac{2a}{ab\plus{}1}\plus{}\frac{2b}{bc\plus{}1}\plus{}\frac{2c}{ca\plus{}1}\le a\sqrt{c}\plus{}b\sqrt{a}\plus{}c\sqrt{b}$

It is nice. We may use amgm or...
[hide]
$ a\sqrt {c} \minus{} \dfrac{2a}{ab \plus{} 1} \equal{} a\sqrt {c} \minus{} \dfrac{2ac}{abc \plus{} c} \equal{} a\sqrt {c} \minus{} \dfrac{2ac}{1 \plus{} c} \equal{}$

$ a\sqrt {c}\left(1 \minus{} \dfrac{2\sqrt {c}}{1 \plus{} c}\right) \equal{} \dfrac{a\sqrt {c}(1 \minus{} \sqrt {c})^2}{1 \plus{} c}\ge 0$

Then $ a\sqrt {c}\ge\dfrac{2a}{ab \plus{} 1}$ with equality only when $ c \equal{} 1$.

Cycling the variables in the last inequality and summing up gives the desired inequality. Equality happens only when $ a \equal{} b \equal{} c \equal{} 1$.
[/hide]

:)

Because of $ abc \equal{} 1$ we have  $ a \equal{} \frac {1}{bc}$ , $ b \equal{} \frac {1}{ac}$, $ c \equal{} \frac {1}{ab}$
By replaceing each \sqrt{variable} our inequality can be transformed in :

$ \frac {a}{\sqrt {ab}} \plus{} \frac {b}{\sqrt {bc}} \plus{} \frac {c}{\sqrt {ac}}\ge\frac {2a}{ab \plus{} 1} \plus{} \frac {2b}{bc \plus{} 1} \plus{} \frac {2c}{ac \plus{} 1}$

From AM-GM the following ineqaulities hold

 (1)        $ \frac {a}{\sqrt {ab}}\ge \frac {2a}{ab \plus{} 1}$

 (2)        $ \frac {b}{\sqrt {bc}} \ge \frac {2b}{bc \plus{} 1}$

 (3)        $ \frac {c}{\sqrt {ac}} \ge \frac {2c}{ac \plus{} 1}$

The sum (1)+(2)+(3) gives the desired inequality

Equality holds for $ a \equal{} b \equal{} c \equal{} 1$


how do you do problems with 3 variables in proportions. Example provided below

Four typists can type a total of 600 memos in three days. At that rate, how many memos can three typists type in one day?

(From Chapter Round 2006-2007 CTD #12)

So... [hide]

4 typists -- 600 memos -- 3 days.
To get to 3 typists, multiply the first and second by $ \frac{3}{4}$  We do not need to multiply the last one as that would result in the exact same proportion.

3 typists -- 450 memos --3 days.
3 typists -- 150 memos -- 1 day.

For such problems with

x -- y -- z and asking for a -- b -- ?, try to find a formula for the missing variable.

So:

x -- y -- z $ \rightarrow$ a -- $ \frac{ya}{x}$ -- z $ \rightarrow$ a --  b -- $ \frac{zbx}{ya}$

or something of the sort if my calculations were correct.[/hide]

[quote="ngade"][quote="Ihatepie"]So first change it from

four typists-600 memos- three days

to four typists- 200 memos - 1 day.

So make a proportion

4/200=3/x so 4x=600 so x is 150. So 150 memos.[/quote]

you could do that, or you could simply divide 600 by four to get the unitary value:

1 typist-150 memos 3 days
1 typist-50 memos 1 day
3 typist-150 memos 1 day

its count down so this should be pretty quick mental math

btw, for 3 variable problems- ignore one of the variables each time
make each calculation with 2 variables it makes life easier  :lol:[/quote]

So you can just divide it by each other until you get a unitary value? Hmmm that would make it a whole lot easier

[code]
Typists      Memos       Days
4            600         3
4            200         1
1            50          1
3            150         1
[/code]

Well, to can set up the proportion like this.....

4 to 600 to 3
=
3 to x to 1

Convert the first proportion to a unit rate, comparing the # of memos made in one day (divide by 3)

the proportions can then be set up as..

4 to 200
=
3 to x

All you need to do is cross multiply.

600=4x

x=150 memos.

Er.... I think this is what rihim ment, but you could just combine proportions....
because typings and days share indirect proportionality with each other and direct proportionality with memos. you could have:
(4)(3)/600 = (3)(1)/x
12x  =  1800
x = 150


Besides, APMO what other international math olympiads (excluding the obvious IMO) does the US take part in?

And how are these teams selected?

I don't believe we participate in any others.  There are some international regional math olympiads for Europe and elsewhere (e.g. Balkan Math Olympiad), but we just do IMO and APMO.  You already know about IMO selection, and you're probably heard that the people who take the IMO are past (top 30) MOPpers and AMC perfect scorers.  (It's unlikely they'll let the former freshman MOPpers take APMO next year, because the test is only supposed to be given to ~30 students.)

This year all AMC perfect scorers were invited to take part in APMO.


function
real analysis
combinatorics unsolved
combinatorics

Any ideas how to evaluate the summation using generating functions:

$ \sum_{k \equal{} 0}^n ( \minus{} 1)^k \binom{n}{k}\binom{m \minus{} k}{r}$

where $ m\geq n$
?

Can you make it a convolution? Please just give hints.

Note that $ k \plus{} (m \minus{} k) \equal{} m$, so the sum is $ \sum_{i \plus{} j \equal{} m} ( \minus{} 1)^i {n \choose i} {j \choose r}$.


How would I enter this into my calculator.

cos 28º 15’ (where the ' symbol means minutes)

Would I first convert to radians by taking (28.15/60) x (п / 180º)
[/code]

hello, which calculator do you have?
Sonnhard.

TI 83

I would have to have it in radian mode correct. I feel like I am making this harder than it should be.

hello, with my TI 83 plus
Angle->Degree
type $ \cos(28 \plus{} \frac {15}{60}) \approx 0.9496991262$
Sonnhard.

why would i use 18 instead of the 28?  or was it a typo?

hello, it was a typo, i have corrected it, sorry.
Sonnhard.

Under the Angle menu (2nd Matrix, I think) there is a minute symbol.


The distribution of random sample means has the same mean as for the mean of individual values, but the standard deviation is smaller.

Is this true or false or are we unable to tell?

Can someone please explain the answer to me?

Thanks

This is true, and is one of the fundamental concepts ofstatistics.

Conceptually, think of it this way: when you pick any 1 value, it is quite likely that it is NOT equal to the mean. After all, you've got a large range of values, and the mean is only 1 of those values. So you have a certain deviation.

Now, as you pick more and more values, the mean of that sample gets closer and closer to the true mean, because some of the low values will be balanced out by the high values, etc. Thus the deviation for samples decreases as the sample size increases, until you "sample" is the entire population and deviation = 0.

Hope that made sense.

I think lasagnaman only touched upon the answer, but what he said was completely true.

The distribution of random sample means is the distribution of the means of several random samples. For example, if you distribute surveys to several schools and ask for the schools to return the mean value of the surveys, you have a sample of random sample means. (Not really random, but still a sample of means.)

The answer to the OP's question is true.
First the means are the same. Out of a population of size N, consider all possible samples of size n. Call the mean of the individual values $\mu$. Then the sum of the individual values is $N\mu$. The mean of random sample means is the sum of all values divided by the number of samples. This is simply the sum of all the values in each sample divided by $n$. Each item is counted ${N-1\choose n-1}$ times, and there is a total of ${N\choose n}$ samples. So the mean of random sample values is $\frac{\frac{1}{n}N\mu{N-1\choose n-1}}{{N\choose n}}=\mu$.
The standard deviation is smaller because the effects of the extreme values (the smallest and largest) are diminished because the means of the samples are considered. The means of the samples are never as low as the smallest value and never as high as the greatest value. The extremes "cancel out" in the samples and give a mean that is closer to the population mean. Since more of the sample means are distributed near the mean of the population, the standard deviation is smaller.


I think you mean $x$, $y$, $z \ge 0$!

Expanding, this is equivalent to
\[x^{3}+y^{3}+z^{3}+3xyz \ge x^{2}y+xy^{2}+x^{2}z+xz^{2}+y^{2}z+yz^{2},\]
which is Schur's.

This particular phrasing appeared on [url=http://www.kalva.demon.co.uk/canada/can92.html]Canada, 1992[/url].


3 blue marbles and 4 red marbles are placed in an opaque bag. Bob takes one marble out at a time until he has taken out an equal number of red and blue marbles, or has taken all the marbles. Find the probability that Bob takes all of the marbles.

[hide]
Note that Bob can take out equal number of both marbles only if he took out even number of marbles.  We'll count the number of ways to take equal number of marbles and subtract.

case 1: $n=2$
we have 2 ways, $ab$ and $ba$, to take out the marbles

case 2: $n = 4$
we have 2 ways also, $aabb$ and $bbaa$. (because if it starts with $ab$ or $ba$, by case 2 it'll end already).

case 3: $n = 6$
we need to start with $aaba$ $aaab$, or the complement, then end with two of the same.  So we have 4 ways to do it.

So we get:
$1-\left(2*\frac{3*4}{7*6}+2*\frac{3*2*4*3}{7*6*5*4}+4*\frac{3*2*1*4*3*2}{7*6*5*4*3*2}\right) = \frac{1}{7}$
[/hide]


As for the second question, the code in old mathcounts tests was pretty self-explanatory.  There was nothing in those questions that you should need special programming knowledge to know.  It just provided a convenient method for doing algorithms.

you might know but just telling for just in case you don't know:
You could get all these in Mathcounts.org
Also, for 2002, just change the address on 2003  that has number 2003 to 2002.. It's simple but it took me long time to get it...
I had to go Minnesota Mathcounts site to get it..


quadratics
number theory open
number theory

Dear friends,

I have two questions which i don't know how the results will be:

Problem: Let a,b,c,d are integers .We contruct the sequence
    u1=c ; u2=d ; u(n+2)= au(n+1)+ bu(n)
Put delta:= a^2 +4b and assume that delta is NON QR modulo p , where p is an odd prime number.

1) Prove or disprove that if delta <0 then the sequence is period ie... there exists and positive integer m such that
             u(n+m)= u(n) for all n is a positiove integer
2) Prove or disprove 1) when we omit the hypothesis delta <0 (ie... delta may be is a positive integer)

Plz help me to answer the question

Many thanks,

[u][b]The author of this posting is : pluricomplex[/b][/u]
____________________________________________________________________

[quote="grobber"]I think it means that it isn't a quadratic residue modulo p.[/quote]

Yes this's what i mean but note gcd(delta,p)=1


The base angles in an isosceles trapezoid are 30 degrees. What is the measure of the the greatest angle in this trapezoid?

hope this isn't too hard.

[hide="myanswer"]I think its 360-60=300/2=150[/hide]

[hide]divide the trapezoid into 2 right triangles and a rectangle. if the base angles are 30 degrees, the top angles of the triangles must be 60 degrees. add that to the 90 degrees of the rectangle and we get 150 degrees.[/hide]

correct (360-2(30)=300, 300/2=150, since it's a iscosceles trapezoid).


geometry
circumcircle
angle bisector
geometry unsolved

Suppose that $ABCD$ is a quadrilateral such that $A,B,C$ and $D$ are on a circle. Prove that there exist exactly four points like $M$ such that $\frac{MA}{MB}=\frac{MD}{MC}$. Also, prove that the diameters of the quadrilateral consisting of these four points are perpendicular.

[quote="Amir.S"]it was posted before in ML but I couldn't find it.

Hint:Use bc=2R.h_a[/quote]
How do I prove that the diameters of the quadrilateral consisting of these four points are perpendicular?

[quote="probablity"]Are you sure this is correct? Seems like there should be infinitely many points like M. Does M need to be on the circumcircle or something?[/quote]
yes,$M$ is on circumcircle

[quote="Matematica"]How do I prove that the diameters of the quadrilateral consisting of these four points are perpendicular?[/quote]

 let $H_{1},H_{2}$ be the feet of perpendicular from $M$ to $AC$ and $BD$ so we have $MA.MC=MB.MD=2R.MH_{1}=2R.MH_{2}$ hence $MH_{1}=MH_{2}$

therefore $M$ is on internal and external angle bisector of intersction of $AC$ and $BD$ hence there are four points with that property and the diagonals of that quadrilateral are perpendicular


In general, $ 2^{n + 1}$ is $ 2^n$ more than $ 2^n$.  As everyone else said, the answer is $ \boxed{2^{2000}.}$


I am stood in the middle of a forest on $ R^{2}$ and cannot see anything but trees in every direction. Is it necessary possible to remove all but finitely many trees so that I still can't see out?

If trees are open sets: yes, since $ S^1$ is compact, and the projection $ \mathbb{R}^2 \setminus \{0\} \to S^1$ is a quotient map. Otherwise, no: if trees can be single points, for example, then you need infinitely many to obscure your view.

I don't think you have to assume that trees are open sets; I think it's more natural to just assume that their projections are beefy enough to contain an open interval. Trees that are closed disks, for instance, are fine as well.

No. You can easily cover the circle by countably many non-degenerate closed intervals so that there is no finite subcover. For instance, take the covering $ (0,1]\equal{}\bigcup_{j\ge 1}[\tfrac 1{j\plus{}1},\tfrac 1j]$ and wrap $ (0,1]$  the usual way to make it a circle.

Oops  :oops: ! Of course. I somehow convinced myself that it's enough to look at the interiors of the closed intervals, but that's not necessarily a covering at all. Sorry!


Find the first three positive values of x for which y=5
$y = 1+5 sin \frac{\pi}{2}(x+0.7)$

I get everything up to here...

[quote="lotrgreengrapes7926"]
$-0.7-(\frac{2}{\pi}\arcsin\frac{4}{5}-0.7)+1.3=\boxed{1.3-\frac{2}{\pi}\arcsin\frac{4}{5}}=-0.7+0.11+1.3=\boxed{0.71}$
$\frac{2}{\pi}\arcsin\frac{4}{5}-0.7+4=\boxed{\frac{2}{\pi}\arcsin\frac{4}{5}+3.3}=-0.11+4=\boxed{3.89}$
$-0.7-(\frac{2}{\pi}\arcsin\frac{4}{5}-0.7)+1.3+4=\boxed{5.3-\frac{2}{\pi}\arcsin\frac{4}{5}}=-0.7+0.11+1.3+4=\boxed{4.71}$ quote]

where did you get 1.3 from and why do you start adding?
 I accidentally posted this here. It should probably be in getting started...(sorry)
oh, and sorry about the latex....

[quote="xenon_nightmare152"]I get everything up to here...

[quote="lotrgreengrapes7926"][hide]
$-0.7-(\frac{2}{\pi}\arcsin\frac{4}{5}-0.7)+1.3=\boxed{1.3-\frac{2}{\pi}\arcsin\frac{4}{5}}=-0.7+0.11+1.3=\boxed{0.71}$
$\frac{2}{\pi}\arcsin\frac{4}{5}-0.7+4=\boxed{\frac{2}{\pi}\arcsin\frac{4}{5}+3.3}=-0.11+4=\boxed{3.89}$
$-0.7-(\frac{2}{\pi}\arcsin\frac{4}{5}-0.7)+1.3+4=\boxed{5.3-\frac{2}{\pi}\arcsin\frac{4}{5}}=-0.7+0.11+1.3+4=\boxed{4.71}$ [/hide][/quote]

where did you get 1.3 from and why do you start adding?
 I accidentally posted this here. It should probably be in getting started...(sorry)
oh, and sorry about the latex....[/quote]
Well, I kinda graphed it.
Let's call $\frac{2}{\pi}\arcsin\frac{4}{5}-0.7\approx-0.11=x$.
Then, you can see that the distance from $x$ to $-0.7$ is the same as the distance from the next solution to $1.3$. So, we have $1.3-(x-(-0.7))=2-x$. The next solution is $4$ more than $x$, since the period of $\sin\frac{\pi}{2}(x+0.7)$ is $4$. And the last one is $4$ more than $2-x$, which we calculated above.


search inequalities ratio induction AMC USA(J)MO USAMO

Source: Pomona-Wisconsin Math Talent Search, 2003/1/3

Let F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, and in general, F_n = F_(n-1) + F_(n-2) for all n :ge: 3.   Show that F_n/F_(n-1) < 1.7 for all n :ge: 4.

well, it should be pretty easy to show that as n-> infinity, F_n/F_(n-1) approaches (1 + sqrt(5))/2...and F_n/F_(n-1) is decreasing. The inequality should then hold (i haven't actually worked this out) for n=4 and therefore all n>4.

Actually it's not decreasing. Look more closely. It's a bit trickier than that.

Yea, it's not decreasing. You can expand the ratio of Fibonacci numbers as a continued fraction, which alternates up and down. I forgot wat the infinit expansion will approach, but it's like phi or something, i gorget.

[quote="froufrou4189"]Sorry!

I still haven't worked it out, but then it seems that the <i>absolute value</i> of the difference between F_n/F(n-1) and phi should be decreasing as n gets larger.[/quote]

No, it won't. Since F_n/F_(n-1) will alternate, jumping above and below what the infinite continued fraction expansion will aproach, the absolute value will not change. Besides, since we are dealing with positive numbers here, absolute value makes no sense.

froufrou4189 is correct -- the absolute value of the difference is decreasing. That's one way of solving the problem. I personally wouldn't solve it exactly like that, but there's nothing wrong with it at all.

Oh!
Oh!
My bad, I didn't see the "phi" part of it in there.

Yes, you are correct. The absolute will decrease, since the fractions are approaching phi.

Right. The terms are the convergents of the continued fraction representation of phi: 1+1/(1+1/(1+...)). These convergents can be proved to be the best rational approximations to phi given their denominators. (That is, there is no better approximation to phi with denominator less than 21 than 34/21, for example.)

Could someone please post a solution for this problem because I didn't understand anything from the reply posts.

Induction works fine here.  The base cases n = 4 and n = 5 are easy to check.  For a particular n :ge: 6, assume by induction that the inequality holds for n - 1 and n - 2.  Then
Fn = Fn-1 + Fn-2 < 1.7 Fn-2 + 1.7 Fn-3 = 1.7 Fn-1.

Thanks Ravi. Now I partially understand how the USAMO differs from the usual problems that I do on the AMCs. I guess I better be at the USAMO math jam to learn more fully about induction and theorems!

Unfortunately, you won't learn a whole lot of mathematical approaches in the upcoming USAMO Math Jam - we'll primarily be talking about strategy.  We will breeze through the kinds of approaches you might try, but we won't be doing much actual math.  The goal of the Math Jam is to talk about general strategies for being organized on the test, writing proofs, optimizing your performance, what to do when you're stuck, that sort of thing.

By the way, if you don't know or like induction, here is another approach.  First, define F0 = 0 and F-1 = 1.

By plugging the recursion for Fibonacci numbers into itself two times, we see that Fn = 3Fn-3 + 2Fn-4 (for n :ge: 3).  Plugging in one more time gives Fn = 5Fn-4 + 3Fn-5 (for n :ge: 4).  Thus, for n :ge: 4, we have 5Fn-1 - 3Fn = 5(3Fn-4 + 2Fn-5) - 3(5Fn-4 + 3Fn-5) = Fn-5 :ge: 0.  Hence for n :ge: 4, we have Fn / Fn-1 :le: 5/3 < 1.7.

This proof assumes that the Fibonacci numbers are nonnegative.  If you forced me to prove that, then I guess I would have to use induction.


algebra
function
domain
advanced fields
advanced fields unsolved

For some real number $\lambda$ consider an equation $\triangle u+\lambda u = 0$ on an open domain
$V \subseteq \mathbb{R}^{2}$. Is it true that each such a solution is real-analytic and if it is, why?

If $u\in C^{2}$ (classical solution), then so is $\Delta u$, and therefore (by Green's formula), $u\in C^{3}$ with estimates.  By iteration this gives $u\in C^{\infty}$ with $D^{n}u=O(C^{n})$. Thus, the remainder of the Taylor series goes to zero.

[url=http://eqworld.ipmnet.ru/en/solutions/lpde/lpdetoc3.htm]Eqworld[/url]  has a bunch of explicit solutions.


modular arithmetic
number theory unsolved
number theory

Find all positive intergers $ m$ and $ n$ such that $ 3^m\plus{}1$ and $ 3^n\plus{}1$ are divisible by $ mn$.

[quote="Stephen"]Find all positive intergers $ m$ and $ n$ such that $ 3^m \plus{} 1$ and $ 3^n \plus{} 1$ are divisible by $ mn$.[/quote]

Nobody's interested in it? :maybe:

if $ m$ or $ n$ is odd then the problem will be obvious(we can easily find all odd numbers for which $ n |3^n\plus{}1$)
suppose that both of them are even and so we have $ 4 | mn | 3^m\plus{}1$
but we know that $ m$ is even so we get $ 3^m\plus{}1 \equiv 2 \pmod {4}$
which contradicts $ 4 | 3^m\plus{}1$.
we r done!

let p is the smallest prime divisor of m, we have $ p|3^m \plus{}1$ infer that $ p|3^{2m} \minus{}1$ we have $ p|3^{p\minus{}1}\minus{}1|$
so $ p|3^{gcd(2m,p\minus{}1)}\minus{}1$ so $ p|3^2 \minus{}1\equal{}8$ so we attain p=2
similarly we also attains n is even;
$ 3^m\plus{}1$ is divisible by 2 but not 4 but m.n is divisible 4
so, there doesn't exist (m,n)


Which graphing software do the people who teach the classes use to make those graphs?

There are two usable formats for graphs and the such: [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]Asymptote[/url] and Geogebra (found in the message body tag options).

However, [b]in the classes[/b], Asymptote is primarily used.  In the Introduction to Geometry class, some diagrams are made with The Geometer's Sketchpad.


Euler
combinatorics unsolved
combinatorics

How many integer solution of this equation: $ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\equal{} k$ with satisfy $ 0\le\ x_{i}\le\ m$ for $ i\equal{}1,2,...n$   ($ k,m,n$ are integer and $ k\le\ mn$ )

Please explain your result, how you can get it ehsan2004.
Thanks.

Let $ y_{i}\equal{}m\minus{}x_{i}$, then we can transform it to well-know problem (Euler candies divide problem)
Find number of solution of this equation:$ y_{1}\plus{}y_{2}\plus{}...\plus{}y_{n}\equal{}mn\minus{}k$ where $ 0\le\ x_{i}\le\ m$
Then, we have the result:
\[ C_{mn\plus{}n\minus{}k\minus{}1}^{n\minus{}1}\]
We can use bijection method to count it.

By using the principle of inclusion and exclusion you can get the result and we know that the number of the solutions without any condition are $ \binom{k\plus{}n\minus{}1}{n\minus{}1}$ then the number of solutions such that  $ x_{i}\geq m\plus{}1$  (By substituting $ y_{i}\equal{}x_{i}\minus{}(m\plus{}1)\geq 0$) are $ \binom{k\minus{}(m\plus{}1)\plus{}n\minus{}1}{n\minus{}1}$ and so on.  :)


A school store sells 7 pencils and 8 notebooks for $ \$4.15$. It also sells 5 pencils and 3 notebooks for $ \$1.77$. How much do 16 pencils and 10 notebooks cost?

$ \textbf{(A)}\ \$1.76 \qquad \textbf{(B)}\ \$5.84 \qquad \textbf{(C)}\ \$6.00 \qquad \textbf{(D)}\ \$6.16 \qquad \textbf{(E)}\ \$6.32$

[hide] Say that notebooks are sold for $n$ dollars each and pencils are sold for $p$ dollars each.  Thus, we have that
\[7p+8n=4.15\Rightarrow 35p+40n=20.75 \]
and
\[5p+3n=1.77\Rightarrow 35p+21n=12.39 \]
Thus, we can subtract our equations to have that $19n=8.36\Rightarrow n=.44$.  Plug n in to get that
\[p=.09\Rightarrow 16p+10n=4.4+1.44=5.84 \]
So, the answer is $B$. [/hide]

You can use substitution, but this can get on your nerves to solve the problems, because it uses cents