If $ a,b,c$ are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, then
$ 15(a \plus{} b \plus{} c) \plus{} \frac 1{abc} \geq 18\sqrt {3}$.

The problem is equivalent to the following:
[i]Alternative formulation:[/i] If $ a,b,c$ are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$, then
$ 5(a \plus{} b \plus{} c) \plus{} \frac 3{abc} \geq 18$.

I think using MV isn’t necessary.
Maybe I got another solution, it’s simpler. But let me check it.

I think Muirhead may work if we homogenize it. But MV is a powerful technique I've seen it can solve a large number of challenging problems so far.

m.c.

[quote="bodan"]Try \[{ (\sum_{\mathrm{cyclic}}a+\cdots+a})+\frac{1}{18abc}+\cdots+\frac{1}{18abc}\geq 63\sqrt[63]{\frac{1}{(abc)^3\cdot3^{36}2^{18}}}\geq63\sqrt[14]{\frac{1}{3^72^4}}=  \] \[ =18\sqrt{3}\cdot\frac{7}{\sqrt[14]{3^{14}2^4}}  \] and the inequality follows because $7^{14}\geq3^{14}\cdot 2^4.$  :)[/quote]
Sorry,I think you must be wrong.
because the inquality can be hold... ;)

Sorry to bump such an old one.

The Am-Gm serves as a lower bound because the equality case $\implies a=b=c=\frac{1}{18abc}$ while we have $\sum a^{2}=1\implies a=\sqrt{\frac13}$ but clearly $a\not = \frac{1}{18a^{3}}\iff a\not = \sqrt[4]{18}$ which is obvious. So there is no equality case :)


Show that for any integer m, and $ n \geq 4$ ,

the sum  $ \sum_{k\equal{}1}^nk! \neq m^2$.

oh yeah 3 .... too simple a problem ! i overlooked it...

Thanks, hsiljak !

[quote="bboypa"]Find all x,y,z positive integer with min{x,y,z}>1 such that 1!+2!+3!+...+z!=x^y.  :D

(also very simple..)[/quote]

umm... is it just $ (x,y,z) \equiv (3,2,3)$   :?:

sum of the lhs ends with a 3 therefore it cannot be the square of some number.

[quote="Nora.91"]Show that for any integer m, and $ n \geq 4$ ,

the sum  $ \sum_{k \equal{} 1}^nk! \neq m^2$.[/quote]

Also, $ \text{LHS}\equiv 1\plus{}2\plus{}1\minus{}1\equiv 3\bmod 5$, which is even quicker.

[i]Thank You![/i]

What about [b]bboypa[/b]'s Question, are there more pairs of (x, y, z)  :?:


MATHCOUNTS
analytic geometry
geometry
articles
LaTeX
counting
distinguishability

Welcome to the Tip of the Week!

Before posting, please read the guidelines. Thanks.

[hide="Guidelines"]

1) In this topic, every weekend, I will post a tip(s) that helps in taking a MathCounts test. The tips will range from School level to Nationals level. The tips will be about topics and concepts as well as strategy, stupid mistakes, speed, pressure, etc. The strategy, stupid mistakes, speed, pressure, etc tips will probably be posted a few at a time, but they will be related, because one isn't enough for one week's discussion.

2) In this topic, you can discuss the tip for the week. You can solve problems, use it to solve harder and more complex problems, prove it, and anything else that is related to the tip. If the discussion is not finished in a week, I may either extend the tip for another week or post it in a new topic.

3) Every week, a group of helpers will help with the tip. There is a limit of 10 per week. Please sign up only if you are willing to devote some time (not much, but at least something). Sign up for helpers will start one week before the tip and will end at the time I post the tip. [b]To sign up, please PM me, don’t post in the topic.[/b]

4) Helpers will have to help moderate the topic. By this I mean they will have to help answer questions, post problems, and keep the discussion on topic. They will also help me with finding good tips for future weeks.

5) Every weekend, I will post a new tip. It will include the following:
 - The tip, either explained or a link to an explanation
 - A problem that can be solved using the tip
 - The list of helpers helping with that week’s tip
 - The tip for the next week
 - A reminder saying that sign ups for the next week starts then
 - Any other announcements that may be necessary

If you have any suggestions or requests (for tips), [b]please PM me, don’t post on the topic.[/b] I want to keep the topic discussion about the tip.
[/hide]

I hope you enjoy this.

Im pretty sure you said this week was the shoelace theorem, right?

[size=150][b]Tip of the Week 1[/b]
Shoelace Theorem[/size]

[url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]Here[/url] is an explanation from AoPS Wiki.
[hide="Example Problem"] (maybe a little hard)
Consider the set of all triangles $ OPQ$, where $ O$ is the origin and $ P$ and $ Q$ are distinct points in the plane with non negative integer coordinates $ (x,y)$ such that $ 41x \plus{} y \equal{} 2009$. Find the number of such distinct triangles whose area is a positive integer.
[/hide]
[hide="Helpers (10)"]
HiDN428
pythag011
Ihatepie
nikeballa96
1=2
mathgenius1
Mewto55555
batteredbutnotdefeated
james4l
RoFlLoLcOpT
[/hide]

[b]Next week's topic: Heron's Formula[/b]

Sign ups for helpers for Week 2 have started.
[hide="Week 2 helpers so far (5)"]
FlyAgaric
Thunder365
Sabien5
pytheagle
BOGTRO
[/hide]

The Shoelace Theorem can give you the area for any n-sided polygon with points on the coordinate plane. The wiki article seemed a little confusing, so here is my explanation.

[hide="Easy Explanation of Shoelace Theorem"]
1.)Write all the coordinates going in order by clockwise or counterclockwise, and copy the first one at the end. Here is my example:
$ (2,1)$
$ (3,2)$
$ ( \minus{} 2,3)$
$ ( \minus{} 2, \minus{} 2)$
$ (2,1)$ [b]*Note how this one is copied from the top.[/b]
2)Draw diagonal lines connecting a y coordinate to the x coordinate that is below and to the left of it. Multiply.
..........$ X,Y$
..........$ (2,1)$
....$ 3 \equal{} (3,2)$Product of 1 and 3
$ \minus{} 4 \equal{} ( \minus{} 2,3)$Product of 2 and -2
$ \minus{} 6 \equal{} ( \minus{} 2, \minus{} 2)$Product of 3 and -2
...$ \minus{} 4 \equal{} (2,1)$Product of -2 and 2
I didn't draw the lines because I can't.
3) Now do the same thing on the other side by multiplying each x coordinate with the y coordinate below and to the right of it.
$ X,Y$
$ (2,1)$
$ (3,2) \equal{} 4$
$ ( \minus{} 2,3) \equal{} 9$
$ ( \minus{} 2, \minus{} 2) \equal{} 4$
$ (2,1) \equal{} \minus{} 2$
4) Your lines should now look like a laced shoe. Add both columns separately. The left column is $ \minus{} 11$, and the right one is $ 15$.
5) Take the absolute value of the difference between the two columns, which is $ |\minus{}26|$, and divide by $ 2$, which gives $ \boxed{\boxed{\boxed{13}}}$ in this example[/hide]

To convince yourself that this method works, try this method with easy #'s like $ (1,1), ( \minus{} 1,1), ( \minus{} 1, \minus{} 1),$ and $ (1, \minus{} 1)$, which we know is a square with side 2 and area of 4.

Is it possible to prove it? (I don't know how)

A step by step explanation:

1) write the coordinates underneath each other in clockwise or couterclockwise order
2) multiply diagonal from the top right to the bottom left
3) multiply diagonal from the top left to the bottom right
4) find the positive difference between what you got in steps 2) and 3)
5) find half of what you got in step 4)

What you get in step 5) is the area!

WHOA thats cool! I didnt even know that. OK, I tried it with the number thunder recommended and found out it worked!

I also have a question. What if the question were: Find all values of X if the points 0,1....2,1.....3,5.....and X,4 formed a quadrilateral with area of 20 (this is hypothetical). Would the order of the numbers make a difference in the values for X?

Shoelace method can also be programmed. I use this on a ti-84.
The following site has everething you'll need to know about this method. You can find the program on the bottom.
[url]http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/url]

Some more information can be found [url=http://www.mathreference.com/la-det,shoe.html]here[/url].

To prove it,

a) Prove it for a triangle.
b) Prove it for a quadrilateral by breaking it up into triangles.
c) Prove it for an n-gon by breaking it up into triangles.

I will post a new problem to start some discussion:

A quadrilateral has an area of $ 60$. The four vertices have coordinates of $ (\minus{}4,5), (\minus{}4,\minus{}6), (x, \minus{}3), (x,1)$. Find the positive value of $ x$ using the Shoelace Theorem.

Assuming that's the order that the points go, here's what you do.

Put out the numbers while repeating the first one at the end.

(-4,5)
(-4,-6)
(x,-3)
(x,1)
(-4,5)

Now "cross multiply" like a shoelace.

First the top left to bottom right diagonals (it doesn't matter which one you start on.)
(-4)(-6)+(-4)(-3)+(1)(x)+(5)(x) This sums up to 36+6x.

Then the top right to bottom left diagonals.
(5)(-4)+(-6)(x)+(-3)(x)+(1)(-4) This sums up to -24-9x. Now minus one from the other. You get either 60+15x, or -60-15x. the final equation is |S|/2=A.(where s is the solution to what I just did) Since the Area is 60, |S|=120. So this leaves us with either 60+15x=120, or -60-15x=120. Solving both gives you x=4, or x=-12. Since you want the positive answer, you get x=4.

To prove it for a triangle, don't you use the distance formula, then Heron's formula, then square both the shoelace formula and your result to see if they match up?

Also, to cut up an n-gon into triangles, you first pick a point A, then pick all pairs of two adjacent points $ B_iB_j$ that do not include A. To find the area: for each pair of points $ B_iB_j$, you find the area of $ AB_iB_j$, and sum them all up.

Well, we'll technically going to learn Heron's Formula next week....... :P so...
Anyway, this technique is really cool and handy.

[hide="List of coordinates"]
If the polygon has $ k$ sides, list the coordinates such that they are in clockwise or counterclockwise direction: \[ (x_1,y_1)\newline(x_2,y_2)\newline(x_3,y_3)\newline\vdots\newline(x_k,y_k)\newline(x_1,y_1).\] Most people forget to add $ (x_1,y_1)$ at the end, and get a wrong area.[/hide]
[hide="General Shoelace Theorem"]
"String" $ x_1$ with $ y_2,$ "string" $ y_2$ with $ x_3$ and so on until you "string" $ x_k$ with $ y_1$ or $ y_k$ with $ x_1.$. This will be \[ x_1y_2\plus{}x_3y_2\plus{}x_3y_4\plus{}\ldots\plus{}x_ky_1\] or \[ x_1y_2\plus{}x_3y_2\plus{}x_3y_4\plus{}\ldots\plus{}x_1y_k.\]
Then "string" $ y_1$ with $ x_2,$ "string" $ y_3$ with $ x_2$ and so on until you "string" $ x_1$ with $ y_k$ or $ x_k$ with $ y_1.$ This will be \[ x_2y_1\plus{}x_2y_3\plus{}x_4y_3\plus{}\ldots\plus{}x_1y_k\] or \[ x_2y_1\plus{}x_2y_3\plus{}x_4y_3\plus{}\ldots\plus{}x_ky_1.\]
If you actually drew the "strings" they would look like shoelaces, which is how this theorem got its name.
The Shoelace Theorem states that the area of a polygon with coordinates \[ (x_1,y_1),(x_2,y_2),\ldots,(x_k.y_k,),\] the area is \[ \frac{1}{2}\left|(x_1y_2\plus{}x_3y_2\plus{}x_3y_4\plus{}\ldots\plus{}x_ky_1)\minus{}(x_2y_1\plus{}x_2y_3\plus{}x_4y_3\plus{}\ldots\plus{}x_1y_k)\right|.\][/hide]
[hide="Practice Problems"]
Difficulty is shown by the stars (out of 5).
$ (\star\star)$
1. Find the area of a pentagon with coordinates \[ (3,\minus{}5),(2,3),(0,7),(3,\minus{}4),(1,5).\]

$ (\star\star\star)$
2. Find a positive value of $ x$ such that the quadrilateral with coordinates \[ (x,5),(3,6),(2,4),(\minus{}1,\minus{}5)\] has area $ 30.$

$ (\star\star\star\star)$
3. Let triangle $ ABC$ be a right triangle in the $ xy$-plane with right angle at $ C.$ Given that the length of the hypotenuse $ AB$ is $ 60,$ and that the medians through $ A$ and $ B$ lie along the lines $ y\equal{}x\plus{}3$ and $ y\equal{}2x\plus{}4$ respectively, find the area of triangle $ ABC.$
[hide="hint"]
This doesn't look like it uses the Shoelace Theorem, but it does! Basically, it's the last step and the most crucial.[/hide][/hide]

[size=150][b]Tip of the Week 2[/b]
Heron's Formula and Triangles[/size]
Note: The Shoelace Theorem's discussion will be continued this week. A good place to start is RoFlLoLcOpT's practice problems. Heron's formula will be added to the discussion. Also, the topic of triangles in general can be discussed, like special triangles and similar triangles.

[hide="Herons Formula"]
The Heron's Formula finds the area of a triangle with sides $ a$, $ b$, and $ c$, where $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$ (the semiperimeter). Heron's formula tells us:

$ Area \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
[/hide]

[hide="Problem"]
Let $ A$ be the area of a triangle with side lengths $ 5$, $ 6$, and $ 7$. Find $ A^2$.
[/hide]

[hide="Helpers (6)"]
FlyAgaric
Thunder365
Sabien5
pytheagle
BOGTRO
alex960531
[/hide]

[b]Next week's topic: How to Count Arrangements[/b]

Sign ups for helpers for Week 3 have started.
Please PM me to sign up for helpers.

[hide="Comment"]
Week 1 turned out be a bad week. I will try to make the topics more general, so that there are more areas to talk about.

If you have any suggestions to increase activity, please PM me.
[/hide]

I'll post a problem using the Binomial Theorem.

Find, to the nearest integer, $ (4 \plus{} \sqrt {2})^4$ without a calculator.

[hide="Solution to Problem"] $ (4 \plus{} \sqrt{2}) ^ 4 \equal{} 452 \plus{} 288\sqrt{2}$, which to the nearest integer is $ 452 \plus{} 407 \equal{} 859$. [/hide]

[hide="Another problem using the binomial theorem"]What is the constant term in $ (3x^3\plus{}2\sqrt{x})^7$?[/hide]

Wow, yeah there is no constant term I meant:

What is the constant term in $ (3x^2\plus{}\frac{2}{x})^9$?

[hide="Solution"] You have to find the term in which the powers of $ x$ are equal. Since the $ 3x^2$ has an exponent that is twice the exponent of $ \frac {2}{x}$, we need the exponents to be $ (3x^2)^y * (\frac {2}{x})^{2y}$. In this case, $ y \equal{} 3$. We get $ 27x^6 * \frac {64}{x^6} \equal{} \frac {27\times64\times{x^6}}{x^6}$. This simplifies to $ 27\times64 \equal{} \textbf{1728}$.[/hide]

Also, try this question with the Binomial Theorem from MCNats, 07 Sprint #29:

How many terms are in the expansion of the expression $ [(3x\plus{}2y)^2(3x\minus{}2y)^2]^3$ after it is simplified to lowest terms?

You're missing the main binomial theorem part.

You need to multiply by $ \binom{9}{3}$ by that answer, because there are that many ways to pick the terms.

[size=150][b]Week 6[/b]
Number Thoery[/size]

Anything related to number theory. I tried to find some problems, but I couldn't. No helpers either.

[hide="Helpers (0)"]
None  :(
[/hide]

I don't know if this will continue if no one posts.

Finding the amount of zeros at the end of factorial.

I'm not good at explainin.

Since to get a zero there needs to be a multiple of ten we need the prime factorization of ten
$ 10 \equal{} 2\times5$
We obviously know there are more 2's then 5's (try it out for your self) so we just count how many 5's. But some 5's have to be counted more than once because they have more than 1 5 in the prime factorization so the formula is for $ n!$ the amount of zeros are:
$ \frac{n}{5} \plus{} \frac{n}{5^2}\plus{}\frac{n}{5^3}\plus{}\frac{n}{5^4}\plus{}\frac{n}{5^5}...$

The formula should actually be \[ \left\lfloor \frac{n}{5}\right\rfloor \plus{}\left\lfloor \frac{n}{25}\right\rfloor \plus{}\cdots\] otherwise it would simplify to $ n/4$, which fails for at least $ 75\%$ of the possible values of $ n$.

The binomial expression:
$ \ 10_b\plus{}1\equal{}(b\plus{}1)^{2}\equal{}b^{2}\plus{}2b\plus{}1$


The inequality does not hold:try $ a\equal{}1,b\equal{}2,c\equal{}\frac{5}{4}$.But we have

  $ \sqrt{\frac{a^{2}}{a^{2}\plus{}(b\plus{}c)^{2}}}\plus{}\sqrt{\frac{b^{2}}{b^{2}\plus{}(c\plus{}a)^{2}}}\plus{}\sqrt{\frac{c^{2}}{c^{2}\plus{}(a\plus{}b)^{2}}}\leq 3\sqrt{\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{5(ab\plus{}bc\plus{}ca)}}$.

A related one is
\[ 1\leq\sqrt{\frac{a^{2}}{b^{2}\plus{}(c\plus{}a)^{2}}}\plus{}\sqrt{\frac{b^{2}}{c^{2}\plus{}(a\plus{}b)^{2}}}\plus{}\sqrt{\frac{c^{2}}{a^{2}\plus{}(b\plus{}c)^{2}}}\leq\frac{3}{\sqrt5}.\]
(Problem 6.9 page 257 in Bat Dang thuc, SL & KP.)

I wish to find $ p$ such that
\[ \sum_{\rm cyclic}\sqrt{\frac{a^{3}}{a^{3}\plus{}(b\plus{}c)^{3}}}\leq\left(\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}\right)^{p}.\]

$ (a,b,c)\equal{}(1,1,0)$, the inequality holds: $ \sqrt2\leq 3/2$.

[quote="tanpham90"]To fjwxcsl : Try this one : $ (1,1,0)$  :wink:
And the following is true :
Prove that if $ a,b,c\geq0$, then
$ 1\leq\ \sqrt{\frac{a^{3}}{a^{3}\plus{}(b\plus{}c)^{3}}}\plus{}\sqrt{\frac{b^{3}}{b^{3}\plus{}(a\plus{}c)^{3}}}\plus{}\sqrt{\frac{c^{3}}{c^{3}\plus{}(b\plus{}a)^{3}}}\leq\ \sqrt{2}$
Apply Holder's Inequality in Special Way and we will kill it ( very easy ) , Mr Thuan  :lol: ![/quote]
The Vasc's proof for the left part:
$ \sum_{cyc}\sqrt{\frac{a^{3}}{a^{3}\plus{}(b\plus{}c)^{3}}}\geq\sum_{cyc}\frac{2a^{2}}{2a^{2}\plus{}(b\plus{}c)^{2}}\geq\sum_{cyc}\frac{a^{2}}{a^{2}\plus{}b^{2}\plus{}c^{2}}\equal{}1.$


geometry
3D geometry
sphere
advanced fields
advanced fields solved

Suppose that $X$ is the union of a sphere (subset of $\mathbb{R}^{3}$) and one of its diameter. Find a universal covering of $X$ ? (I can draw an horrible picture of such a covering, but I'm quite sure there is a nice way to describe it ? isn't it ?)

If you turn the space inside out, it becomes a sphere hanging on a circle (like a bead on a string that is tied into a loop). The universal covering space is obtained by untying the string: it is a line with infinitely many spherical beads.


pigeonhole principle
combinatorics unsolved
combinatorics

Prove that some power of 2 has a decimal expansion that begins with 1999...
[hide]Hint: pigeonhole principle[/hide]


When Susan took her first job as a biologist, she received a salary of $ \$28,\!000$ and was promised a raise of at least $ \$1,\!400$ after each year of service. What is the maximum number of years it could take for her salary to reach $ \$35,\!000$?

$ 35000\minus{}28000\equal{}7000$ is the amount she needs to go up.  $ \frac{7000}{1400}\equal{}5$, so it will take at most 5 years.


trigonometry
Trigonometric Equations
number theory
Diophantine equation
IMO Shortlist
IMO Longlist

Suppose $\tan \alpha = \dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Prove that the number $\tan \beta$ for which $\tan {2 \beta} = \tan {3 \alpha}$ is rational only when $p^2 + q^2$ is the square of an integer.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

This is just some calculations!

we can easily show that $ \tan3\alpha\equal{}\frac{3\tan\alpha\minus{}\tan^{3}\alpha}{1\minus{}3\tan^{2}\alpha}\equal{}\frac{3\times\frac{p}{q}\minus{}\frac{p^{3}}{q^{3}}}{1\minus{}3\times\frac{p^{2}}{q^{2}}}\equal{}\frac{p\left(3q^{2}\minus{}p^{2}\right)}{q\left(q^{2}\minus{}3p^{2}\right)}$

now let $ \frac{1}{A}\equal{}\frac{p\left(3q^{2}\minus{}p^{2}\right)}{q\left(q^{2}\minus{}3p^{2}\right)}$

and as said in the problem:
$ \tan2\beta\equal{}\frac{2\tan\beta}{1\minus{}tan^{2}\beta}\equal{}\tan3\alpha\equal{}\frac{1}{A}\Longrightarrow\tan^{2}\beta\plus{}2A\tan\beta\minus{}1\equal{}0\Longrightarrow\tan\beta\equal{}\minus{}A\plus{}(\text{or }\minus{})\sqrt{A^{2}\plus{}1}$

so $ \tan\beta$ is rational if and only if $ A^{2}\plus{}1$ is perfect square!

$ A^{2}\plus{}1\equal{}\frac{q^{2}\left(q^{2}\minus{}3p^{2}\right)^{2}\plus{}p^{2}\left(3q^{2}\minus{}p^{2}\right)^{2}}{p^{2}\left(3q^{2}\minus{}p^{2}\right)^{2}}$

so its sufficient that $ q^{2}\left(q^{2}\minus{}3p^{2}\right)^{2}\plus{}p^{2}\left(3q^{2}\minus{}p^{2}\right)^{2}\equal{}\left(p^{2}\plus{}q^{2}\right)^{3}$ is perfect square but this is only when $ p^{2}\plus{}q^{2}$ is a perfect square and we're done.


greatest common divisor
MATHCOUNTS
analytic geometry
graphing lines
slope
number theory
relatively prime

[b]A 24-foot by 72-foot rectangular dance floor is completely tiled with 1-foot by 1-foot square tiles. Two opposite corners of the dance floor are connected by a diagonal. This diagonal passes through the interior of how many tiles?
[/b]

My original answer was 95 because of m+n-1, but I now realize it doesn't work when the side lengths are not relatively prime. The answer is 72, which is the length of one of the sides. Is there an add on to m+n-1 when m and n share factors?

Also, I've been wondering about how to solve the problem if the problem stated the same thing but with two diagonals connecting two corners of the dance floor.

[b]A 24-foot by 72-foot rectangular dance floor is completely tiled with 1-foot by 1-foot square tiles. Two opposite corners of the dance floor are connected by a diagonal, and another two opposite corners are connected by another diagonal . These diagonals pass through the interior of how many tiles?
[/b]

Thanks

I forgot, is there a formula similar to $m+n-GCF(m,n)$ for this problem?

For the first problem, the formula is
$m+n-\text{gcf}(m,n)$
$24+72-24=72$
So the formula does work.
For the second problem, you double the amount the first diagonal passes through and cubtract one if they intersect in the middle of a tile.
If they intersect on an edge of corner, you don't have to subtract 1.
The answer to the second one is 144 because they intersect at a vertex.

Well, if they aren't relatively prime, you can simplify it.  So, a $72 \times 24$ is comprised of 24 $3 \times 1$.  Since every $3 \times 1$ has a diagonal going through all three squares, the answer would be [hide]72[/hide]

EDIT.

I calculated the answer for one diagonal.  Since none of the diagonals intersect, the answer is 144.

so the formula always works for natural numbers, right?

The $m+n-\text{GCF}(m,n)$, yes.

As for the second problem, notice that the diagonals do not intersect in the interior of any tile. Therefore, no tiles are affected by more than one diagonal, so you just double the answer.

[quote="probability1.01"]Whoa, but what if you had two diagonals with a 17x39? Unless I am extremely mathcounts-rusty, the answer is not 2*17 + 2*39 - 3 = 109.[/quote]

Wait, where did you get 2*17 + 2*39 - 3 from?  :?

[quote="JesusFreak197"]Two diagonals with a 17x39 should be $2(17+39-1)-1=109$, I'm pretty sure. Why wouldn't it be?[/quote]

Oh. Thanks.  I just didn't recognize the expression 2*17+2*39-3 until after it was simplified when you gave the first step.  :fool:

TY all -- I realized there was something wrong with them not being relatively prime, but I forgot.

Also, thanks for the help on the second problem too.

[quote="JesusFreak197"]Two diagonals with a 17x39 should be $2(17+39-1)-1=109$, I'm pretty sure. Why wouldn't it be?[/quote]

What about a 1x3? Is it $2(1+3-1)-1 = 5$? Didn't think so.

Hmm, right. In that case, you just need to find a way to factor slope into it. Basically, it seems like if the dimensions are odd, you subtract the ceiling value of the larger over the smaller; if both even, subtract none. If one is odd and one is even, subtract two.

Applying this to your example of a 1x3, we do get the correct answer of $3$. ;) So, in the case of the 17x39, I think it would be $107$ instead. Anyone want to draw it out and check? :P


If the famous baseball player for the San Francisco Midgets, Larry Ponds, has a $ \frac{2}{5}$ chance of earning a walk on each plate appearance, what is the probability that he will earn a walk exactly once in his next two plate appearances? Express your answer as a common fraction.

The probability that he walks the first time but not the second is $ \frac{2}{5}\cdot\left(1\minus{}\frac{2}{5}\right)\equal{}\frac{2}{5}\cdot\frac{3}{5}\equal{}\frac{6}{25}$.  The probability that he walks the second time but not he first is $ \frac{3}{5}\cdot\frac{2}{5}\equal{}\frac{6}{25}$ as well.  We add these together, so our final answer is $ \boxed{\frac{12}{25}}$

[hide="My slightly shorter solution"]
The cases are symmetric, so you do not need to do casework.

So the probability for a certain order is $\frac{3}{5} \times \frac{2}{5}$, and there are $2$ possible orders.

The answer is $\frac{12}{25}$.

For this problem, it is perfectly fine to do casework, but with more complicated problems, the method mentioned above works much better.

By the way, the method is called Binomial Probability, and here is a [url=http://stattrek.com/probability-distributions/binomial.aspx]very technical explanation[/url] of it.
[/hide]

[hide="Alternatively"]
The probability that he walks neither time is $\frac35\cdot\frac35=\frac{9}{25}$, and the probability that he walks both times is $\frac25\cdot\frac25=\frac{4}{25}$, so the probability that he walks 1 time is $1-\frac{9}{25}-\frac{4}{25}=\boxed{\frac{12}{25}}$
[/hide]

$8$ year bump, woohoo! :D

[hide=Solution]
There are $2$ ways to choose the plate appearance that earns a walk. There's a $\frac{2}{5}$ probability that he earns the walk on one plate appearance, and there's a $\frac{3}{5}$ probability that he doesn't earn the walk on the next plate appearance. $2 \cdot \frac{2}{5} \cdot \frac{3}{5} = \boxed{\frac{12}{25}}$.
[/hide]

Edit: I just realized that I posted a different solution from Alcumus and everyone else-

[hide=solution]The answer is 12/25, because he can either be walked on the first appearence or the second. In both of those cases, there is a 2/5 probability that he is walked, and then a 3/5 probability that he is not walked. So, the final answer is (3/5*2/5)+(3/5*2/5) = 12/25.


For a number $ A\equal{}x^2\minus{}1002000y^2$, where $ x,y\in \mathbb{Z}$, it holds that $ A>0$ and $ A$ is not a perfect square. Find the least possible value of $ A$.

We can simplify $ 1002000 \equal{} 2^4\cdot 3\cdot 5^3\cdot 167 \equal{} 3\cdot 5\cdot 167\cdot (2^2\cdot 5)^{2} \equal{} 2505\cdot (20)^2$
\[ x^{2} \minus{} 2505(20y)^2 \equal{} A
\]
where $ A\geq 2$ and $ A\ne n^2\;\forall n\in \mathbb{N}$.

If $ A \equal{} y^2$ then pair $ (x,y) \equal{} (1001y,y)$ is a solution and we may build a sequence of solutions for any this $ A$. But for another $ A$ i can't find any solution. I cheched this by cimputer and get nothing. That's why i think that there dos not exist such $ A$ not a perfect square that equation has solutions, but we should prove this.


let a,b,c >0 and abc=1 .
 find the minimal of expression P
       $(\frac{a}{(a+1)(b+1)}+1)( \frac{b}{(b+1)(c+1)} +1)( \frac{c}{(c+1)(a+1)}+1)$

It 's very easy. I 've found a very special solution.

First we can find
$a=\frac{x}{y} ; b=\frac{y}{z} ; c=\frac{z}{x}$
then the expression become
$P = \prod (\frac{xz}{(y+x)(y+z)} + 1)$
We need to prove that
$P \geq (\frac{5}{3})^3$
This problem is quite easy.


quadratics
modular arithmetic
absolute value
number theory unsolved
number theory

group A is consisted of integers where the absolute value is under 111 (i.e. A={-111,-110,...,110,111})

and  Set S ={$ (a,b)| x^2\minus{}y^2\minus{}1 is multiple of 223$and x and y are in A}

If we define $ \bar{x}$ as the remainder when we devide x with 223

Then find the n(T) where T={$ \bar{x^2}$|(x,y)$ \in$S}.

Note that $ (x,y)\in S$ if and only if $ y^2$ and $ x^2$ represent a pair of consecutive quadratic residues modulo $ 223$ (i.e., $ x^2\equiv y^2 \plus{} 1\pmod{223}$).
The number of such pairs is $ \frac {223 \minus{} 3}{4} \equal{} 55$ (see [url=http://en.wikipedia.org/wiki/Quadratic_residue#Pairs_of_residues_and_nonresidues]this link[/url] for more info) and and each such pair gives an unique value of $ x^2$ modulo $ 223$. Therefore, $ |T|\equal{}55$.


limit
function
calculus
calculus computations

Evaluate :       $\lim_{x\rightarrow 0}\ \frac{sin(tanx)-tan(sinx)}{sin^{-1}(tan^{-1}x)-tan^{-1}(sin^{-1}x)}$

How to go about the denominator part...the inverse functions ?

It's been posted before. Who can find the reference?


$1993$ points are arranged in a circle. At time $0$ each point is arbitrarily labeled $+1$ or $-1$. At times $n = 1, 2, 3, ...$ the vertices are relabeled. At time $n$ a vertex is given the label $+1$ if its two neighbours had the same label at time $n-1$, and it is given the label $-1$ if its two neighbours had different labels at time $n-1$. Show that for some time $n > 1$ the labeling will be the same as at time $1.$

Let $a(n, t)$ be the labeling of vertex $n$ at time $t$.  Then by the givens $a(n, t+1) = a(n+1, t)\cdot a(n-1, t)$.

Now, I claim $a(n, t+2^{k}) = a(n+2^{k}, t)\cdot a(n-2^{k}, t)$ for all $k \geq 0$.  We've shown the case $k = 0$.  Assume it true for $k = i$.  Then $a(n, t+2^{i+1}) = a(n, (t+2^{i})+2^{i}) = a(n+2^{i}, t+2^{i})\cdot a(n-2^{i}, t+2^{i}) = a(n+2^{i+1}, t) \cdot a(n, t)^{2}\cdot a(n-2^{i+1}, t) = a(n+2^{i+1}, t) \cdot a(n-2^{i+1}, t)$, so we have the claim by induction.

Now, 1993 is prime, so $2^{1992}\equiv 1 \mod 1993$.  Thus $a(n, 2^{1992}) = a(n+2^{1992},0) \cdot a(n-2^{1992}, 0) = a(n+1, 0) \cdot a(n-1, 0) = a(n, 1)$, and we're done.

Wow, nice solution. I think you are true. Thanks!  :wink:


trigonometry
inequalities unsolved
inequalities

Show that for x in (0,pi/2) we have (cos(x))^(1/3)<sin(x)/x

a ugly solution with complicated computations:

it is easy to prove that sin x > x - x^3/6 and cos x < 1 - x^2/2 + x^4/24

(cos x)^(1/3) < sin x / x
<==>
x^3 cos x  < (sin x)^3

it is sufficient to prove
x^3(1 - x^2/2 + x^4/24) < (x - x^3/6)^3
<==>
0 < (x - x^3/6)^3 - x^3(1 - x^2/2 + x^4/24)
expand the right side  :( and we finally get
0 < x^7(9-x^2)/216
which holds when x\in (0,\pi/2).


Find all functions $f : Z \rightarrow Z$ which satisfy $f(m+ f(n)) = f(m) + n$ for all$m, n \in Z$.

Suppose $f(a)=f(b)$.  Then $a+f(m)=f(m+f(a))=f(m+f(b))=b+f(m)$, and hence $a=b$, so $f$ is injective.

$m=n=0$ gives $f(0)=0$.

$m=0$ gives $f(f(n))=n$, so $f$ is surjective, and hence bijective.

Let $n=f(p)$. Then $f(m+f(n))=f(m+f(f(p)))=f(m+p)=f(m)+f(p)$, so $f$ is additive, and hence $f(k)=ck$.

$f(f(k))=c^2k=k$, so $c=\pm1$, and $f(x)=\pm x$.

Yeah I did pretty much the same thing too.
    Ashwath

i proved that $f(f(n))=n,f(kn)=kf(n)$ and i proved that this function is an one - to - one function and $f(n)$ can be any integer number and i think it is proved but i have no more time. :D


Right now two clocks agree but the ﬁrst clock loses $ 3.4$ seconds every hour and the second gains $ 2.6$ seconds every hour. How many hours will it be until the clocks again show the same time?

The clocks have an effective difference of six seconds per hour. It would take 7200 hours for the difference to reach 12 hours, so the clocks would then show the same time.

[hide="Solution"]The first rate is $ \minus{} 3.4$ seconds per hour. The second is $ 2.6$ seconds per hour. After $ h$ hours, we need $ \minus{} 3.4h\equiv 2.6h\bmod 12\cdot 60\cdot 60$.

That is, $ 6h\equiv 0\bmod 43200\implies h\equiv 0\bmod 7200$, so the smallest positive $ h$ is $ \boxed{7200}$.[/hide]

Edit: lol, multiplied wrong.

I don't think grn_trtle's answer is correct, unless the two clocks show only minutes.


Two radii OA and OB of a circle c with midpoint O are perpendicular. Another circle touches c in point Q and the radii in points C and D, respectively. Determine $ \angle{AQC}$.

$ 45$(if $ C$ lies on $ OA$) or $ 90$ (if $ C$ lies on $ OB$)

Label the center of the circle touching circle c and the two radii as P.
Then it immediately follows that OCPD is a square.
Thus $ \angle CQD \equal{} \frac{1}{2} \angle CPD \equal{} 45$.
By symmetry we get that $ \angle AQC \equal{} \angle BQD$.
Also $ \angle AQB \equal{} 180 \minus{} \frac{1}{2} \angle AOB \equal{} 135$.
Thus $ \angle AQC \equal{} \frac{1}{2} (135 \minus{} 45) \equal{} 45$ if $ C$ lies on $ OA$
or $ \angle AQC \equal{} 45\plus{}45 \equal{} 90$ if $ C$ lies on $ OB$


calculus
integration
trigonometry
calculus computations

More specifically, the Lagrange remainder term. My teacher showed us a few examples where he could approximate the values of integrals using this. For example, in integrating $sin(x^2)$ the easiest approach is to taylor expand $sin(u)$ with Lagrange remainder term $R(u)$. Then, for $sin(x^2)$ it's possible to use $R(x^2)$ as the remainder. My problem is then this: This seems to imply that $R(x^2)$ is indeed a correct remainder term for the maclaurin expansion of $sin(x^2)$, however, it looks nothing like the remainder term you would get if you instead differentiated $sin(x^2)$ however many times.

However, since all other terms in the expansions are the same, this seems to imply that $R(x^2)$ IS the lagrange remainder term. No doubt, this is because we don't have a certain value for $\xi$, but this still confuses me. Please, someone tell me what's up.

Thanks!

I think you have answer your own answer. the value of $\xi$ is the case. And also if you differentiated $\sin x^2$ instead of using $\sin u$ expansion. you will get different remainder. If you taking the limit $n \rightarrow \infty$. they are not approaching the limit with similar way (hope you understand that  :) ). Sorry if my english isn't  good .


inequalities
logarithms
integration
absolute value
abstract algebra
complex analysis
complex analysis unsolved

This is an exercise from Garnett. It asks to prove Hardy's inequality
$ \sum \frac{|a_n|}{n\plus{}1} \leq C \lVert f \rVert_{H^1}$
by "using duality".

Well, the obvious thing I see is that $ \sum \frac{a_n}{n\plus{}1}$ is the "inner" product between $ f$ and a logarithm. Since the logarithm is $ BMO$, we get that $ \lvert\sum \frac{a_n}{n\plus{}1} \rvert \leq  C \lVert f \rVert_{H^1}$. However, I cannot really figure out how to move the absolute value inside the sum.

What is the relation between $ a_n$ and $ f$? This doesn't make sense without that.

Well, obviously I intended that $ f(z) \equal{} \sum a_n z^n$. I've seen a couple of other proofs of this inequality, so I am specifically interested in a proof relying on that we know the dual of the Hardy space $ H^1$.

I figured out how to do this, if anyone cares. We have to lift the problem to $ H^2$ where putting absolute values on the coefficients in the series expansion is a continuous operation. If $ f \equal{} \sum a_n z^n$, let $ f_* \equal{} \sum |a_n| z^n$. So if we factor $ f \equal{} gh$ where $ \lVert g \rVert_{H^2}^2 \equal{} \lVert h \rVert_{H^2}^2 \equal{} \lVert f \rVert_{H^1}$ , then $ F \equal{} g_*h_* \equal{} \sum A_n z^n$ has smaller $ H^1$-norm than $ f$. Thus
$ \sum \frac{|a_n|}{n\plus{}1} \leq \sum \frac{A_n}{n\plus{}1} \leq C \lVert F \rVert_{H^1} \leq C \lVert f \rVert_{H^1}$

Yes, this is the classical factorization proof. It is interesting that you can do it in a different way, almost exactly as you started. All we need is to show that $ g(z)\equal{}\sum_{k\ge 0} b_k z^k$ is in BMO when $ |b_k|\le\frac 1{k\plus{}1}$. Take any arc $ A$ of length $ \ell$, let $ N\approx \ell^{\minus{}1}$ and split $ g$ into the sum $ p\plus{}h$ where $ p(z)\equal{}\sum_{k\equal{}0}^N b_k z^k$ and $ h$ is the tail. Note that $ \int_A|h|^2\le\int_{\mathbb T}|h|^2 \le C\ell$ and $ \int_A |p'|\le N\ell\le C$, which is enough to ensure the BMO property.

All this is some sort of cheating without a non-factorization proof of the $ H^1$-BMO duality, of course, but, fortunately, such proofs do exist.

Thank you very much. Those estimates have quite a clever feel to me, I had trouble bounding the norm of $ g$ uniformly. I need to learn how to work with concrete information.

About the classical proof, I do not think I know it. One of the proofs I have seen depends on theory of Hankel matrices/operators, and the other was some clever trick with integrating over halves of a circle and along the real axis. For the classical proof, how are you meant to deduce that $ \sum \frac {A_n}{n \plus{} 1} \leq C \lVert F \rVert_{H^1}$ (in the notation I introduced)? I saw this inequality as a consequence of what you wrote with $ b_k \equal{} \frac {1}{k \plus{} 1}$.

About your last comment, I think the factorization $ H^1 \equal{} H^2 \cdot H^2$ actually follows from the $ H^1 \minus{} BMO$ duality, at the very least as a sort of weak factorization (like a tensor product of two copies of $ H^2$).


Let $ a$, $ b$ and $ c$ be positive real numbers not greater than $ 1$.

Prove that the three numbers $ a(1 \minus{} b)(2 \minus{} 3c)$, $ b(1 \minus{} c)(2 \minus{} 3a)$, $ c(1 \minus{} a)(2 \minus{} 3b)$ cannot all be greater than $ \dfrac{1}{8}$.

:)

[hide]Assume the contrary.

$ \frac{9\minus{}3(a\plus{}b\plus{}c)}{9}\ge (abc(1\minus{}a)(1\minus{}b)(1\minus{}c)(2\minus{}3a)(2\minus{}3b)(2\minus{}3c))^{\frac{1}{9}}$

$ 1\minus{}\frac{1}{3}(a\plus{}b\plus{}c) \ge (abc(1\minus{}a)(1\minus{}b)(1\minus{}c)(2\minus{}3a)(2\minus{}3b)(2\minus{}3c))^{\frac{1}{9}}$

But the RHS $ > 2$

Therefore the assumption is false. [/hide]

Oops.. sorry my solution is wrong  :blush:

Sorry, the inequality isn't true. Try $ a\equal{}b\equal{}c\equal{}\dfrac{3}{8}$.

:)


[hide]OK... first, factor the equation:

$ 3x^3 \plus{} 2x^2 \plus{} x \equal{} 0 \implies x(3x^2 \plus{} 2x \plus{} 1) \equal{} 0$

We immediately know that one of the roots is $ 0$.

Now we find the roots of $ 3x^2 \plus{} 2x \plus{} 1 \equal{} 0$ with the quadratic formula...

$ \frac { \minus{} 2 \pm \sqrt {4 \minus{} 12}}{6} \equal{} \frac { \minus{} 2 \pm 2\sqrt 2i}6$

Therefore, the roots are...

$ x \equal{} \left\{0, \frac { \minus{} 1 \plus{} \sqrt 2i}3, \frac { \minus{} 1 \minus{} \sqrt 2i}3\right\}$

The only real root, however, is $ 0$.[/hide]

Edit: simplified!

Could you not simplify the unreal roots? Id do it, but I don't know latex.

hey actually the real root is 0;
coz complex roots are unimplified in mathematics

[hide]$ 3x^3 \plus{} 2x^2 \plus{} x \equal{} 0$ can be factorized and written as $ x(3x^2 \plus{} 2x \plus{}1)\equal{}0$.  From this we can tell that one possible value of X is 0.  We can use the quadratic equation to solve for the other term, assuming that it is equal to 0.  $ \dfrac{\minus{}2\pm{\sqrt{2^2\minus{}12}}}{6}\equal{}\dfrac{\minus{}2\pm{\sqrt{8i}}}{6}$.  Therefore the roots are...

0,$ \dfrac{\minus{}2\plus{}\sqrt{8i}}{6}$, $ \dfrac{\minus{}2\minus{}\sqrt{8i}}{6}$[/hide]

hello, write your equation in the form $ x(3x^2\plus{}2x\plus{}1)\equal{}0$ and we have
$ x_1\equal{}0$ or $ 3x^2\plus{}2x\plus{}1\equal{}0$, this implies $ x_{2,3}\equal{}\minus{}\frac {1}{3}\pm\frac{\sqrt{2}}{3}i$.
Sonnhard.


Let $ ABC$ be a triangle with $ A\equal{}120^{\circ}$ and $ m\left(\widehat {OI,BC}\right)\equal{}45^{\circ}\ .$

Ascertain $ B$ and $ C$ ($ O$ is circumcenter and $ I$ is incenter of the given triangle).


greatest common divisor
number theory
geometry
algorithm
least common multiple
analytic geometry
Euclidean algorithm

I noticed many people used "greatest common divisor" and not "greatest common factor" so if you could vote that'll help me decide which I should use for my problems. Can you also post which you use and roughly where you live (West Coast, Bay Area, Mid-West, etc.)? Thanks.

East Coast.  GCF.  But I have no objections to GCD.

So far a tie...this isn't helping you guys :lol:
Guess we need more votes.

Also am I being to rash to conclude that in East Coast you say "greatest common factor" and in the West Coast yoy say "greates common divisor"?

east, gcd i think, but i dont really pay attention, so not sure.

doesn't GCD mean "greatest common [i]denominator[/i]?"

Posted by MrMcGeek

GCD.  (From East Coast.)  When I was a researcher in computer science, we would always talk about the Euclidean algorithm for gcd, never gcf.

Mmm, haven't thought of that...although even if that do exist it won't really mean much since it'll always equal :inf: and the fraction won't be a fraction anymore (it'll always be 1; or 0 if the numerator is also :inf:)
LCD will mean something but then it'll be just the same thing as LCM of the denominators so I don't think the term "greatest common denominator" don't exist.

I definitely use GCD.. I think its more standard really? I haven't seen gcf used anywhere, although some textbooks might use hcf for highest common factor. I think the majority of websites and books and stuff would use gcd..

I think Calculators also use gcd, im pretty sure the ti-89 does, but i forget about ti-83.  i cant check though, my 83 broke for some reason after not using it for a while.

"If you don't put enough love into it it won't love you back" some quote from some person (I even forgot their gender) about "relationships" :lol:
btw it also uses "gcd"

So should I go with "GCD"?

I vote GCD (East).   It seems to be more standard (except in high school texts), but the main reason is a bit more complex.  When you say [i]factor[/i], you are usually talking about a specific factorization of a number (perhaps prime factorization, or something else), whereas [i]divisor[/i] just refers to a number that divides evenly into the one we're talking about.  Hence greatest common [i]divisor[/i] makes more sense to me.  Of course I may be wrong about this logic, but it fits with the way I've usually seen these terms ("factor" vs "divisor") used.

I was going to write a response to the "denominator" post, but then I realized that Tare already had.  :P  Textbooks often use "least (or lowest) common denominator" = "LCD", which is of course the LCM.  Stupid textbooks.  "LCD" is "liquid crystal display".

*giggle* The denominator can be as big as necessary, so there is no GCD.  Just like there's no LCF.

I vote with GCF, but it's probably just because I was raised that way.

Oh, when I was just 2 months old,
My mammy an' my pappy, they did told
Me how to use the GCF
(Not GCD or LCF!)
So now I'm old and wizened too,
But I still know my numbers true
And if you ask me what to do,
Then I'll say GCF!

P.S. West Coast, if you can call it that.  Generally, the desert isn't a [i]coast[/i].

[quote="ComplexZeta"]I use GCD sometimes, but more often I'll use (a,b).[/quote]

...which unfortunately looks like both the open interval a < x < b and the cartesian coordinate that's a units right and b units above the origin.

It's usually rather clear which one is being used.

[quote="mathfanatic"]Oh, when I was just 2 months old,
My mammy an' my pappy, they did told
Me how to use the GCF
(Not GCD or LCF!)
So now I'm old and wizened too,
But I still know my numbers true
And if you ask me what to do,
Then I'll say GCF![/quote]
"Oh you must've been a brilliant baaaaby, cause baby, look at you now!" :lol:

Even though I'm in the East Coast I learned English in the West Coast so I guess now I could conclude that
West Coast="GCF"          Central="GCF"?          East Coast"GCD"

I still can't decide on which to use... :|

sigh. they'll know what you're talking about...

as a side note, my dad (learned math in india) uses HCF (highest common factor).

Yeah, TripleM (from New Zealand) also mentioned HCF -- I've never seen it before this thread, though.

For what it's worth, I did some google searches.  "Greatest common divisor" had 34,400 hits, which beat "greatest common factor" (13,700), which beat "highest common factor" (3,620).  Majority rules?! :)

theres actually a website googlefight.com i think where it compares the number of sites on google for any given keywords that you enter.  I would think its just as easy to put them into google though.

[quote="ComplexZeta"]I disagree that West Coast people use GCF in general. I think SFBAers use GCD mostly.[/quote]
Mmm, I was in Bay Area...I still use GCF though, mmm...

By the way way before I noticed someone mention textbooks but all of mine (all three by Forester) uses GCF...this is getting annoying, it's like split into half and half except GCD's slightly winning...

i think a couple people already said this, but it really doesnt matter; if you say either one people will know what you're saying.  You're not going to get mugged for saying 'gcf' when you're on the east coast.

[quote="Tare"]By the way way before I noticed someone mention textbooks but all of mine (all three by Forester) uses GCF...this is getting annoying, it's like split into half and half except GCD's slightly winning...[/quote]

I'm pretty sure the distinction is partly between SCHOOL mathematics textbooks, and mathematics books for undergraduates and professional mathematicians. If you wanted to write problems for people familiar with the higher-level literature, you would write gcd, for the reasons mentioned by other participants. And even the younger people here should be ready for the terminology in the higher-level literature, because they'll all be reading it in a few years at most, right?

I recognize both the gcf and gcd terminology, and grew up in the upper Midwest (that is, between the two coasts of the United States). I have traveled EVERYWHERE in the United States but usually wasn't talking about math in my travels. The hcf terminology appears to be British (or British colonial) school mathematics terminology. Certainly any British mathematician would recognize the gcd terminology.

Just my humble opinions and observations,

Okay then.  It looks like it's Tare, me and GCF against the world.  Rock on!

Aaah, that cleared the confusion. Thanks tokenadult.

So GCD's more sophisticated than GCF...hmm, but GCF still has the upper hand as to the commoner's vote..."It's six of one and a half dozen of the other."


Anyone else from the New York area doing the Columbia Science Honors Program? I'm taking Exploring Black Holes.

If you are in Columbia SHP, can you go there in the fall and take a break in the spring and go there again next fall?

I was in "Nano-From Science to Technology." Now, that is sort of irrelevant . . . .

spse, I am pretty sure you can't. You could always ask, though. There is always somebody in that office on the seventh(?) floor . . . . But, again, that is irrelevant. You could always call. I am just not sure what the phone number is.


algebra
polynomial
algorithm
calculus
calculus open

Let $ f(x,y)$ be polynomial of degree 3 with real coefficients, which is irreducible over $ R$. Its curve (e.g. set of points $ (x,y) \in R^2$ for which $ f(x,y)\equal{}0$) sometimes contains an [url=http://en.wikipedia.org/wiki/Acnode]Acnode[/url] (sometimes called "isolated point" or "hermit point"). For instance, if $ f(x,y)\equal{}y^2\plus{}x^3\plus{}x^2$, it's curve contains isolated point $ (0,0)$:

[img]http://upload.wikimedia.org/wikipedia/en/7/75/Acnode.gif[/img]

For this particular polynomial it's easy to prove the existence of isolated point. However, there are much more complex cases. Here is the curve for $ f(x,y) \equal{} 3\,{x}^{3}\minus{}5\,{y}^{2}x\minus{}4\,{x}^{2}\minus{}10\,yx\plus{}10\,{y}^{2}\minus{}6\,x\plus{}20\,y\plus{}12$ ([url=http://staff.jccc.net/swilson/planecurves/cubics.htm]source[/url]).

[img]http://staff.jccc.net/swilson/planecurves/imagescubic/cubic27.gif[/img]

My question is: how to check, whether given polynomial's curve has an isolated point? E.g. for which real $ a,b,c,d,e,f,g,h,i,g$ the polynomial
\[ f(x,y)\equal{}ax^3\plus{}by^3\plus{}cx^2y\plus{}dxy^2\plus{}ex^2\plus{}fy^2\plus{}gxy\plus{}hx\plus{}iy\plus{}g\]
has an isolated point? I expect some semialgebraic conditions for the polynomial's coefficiens. If anybody knows the solution for this problem, or has any suggestions, please, write a note.

Thank you!
Sergey Markelov
markelov@mccme.ru

Sergey, the following paper presents an algorithm that finds isolated points on a real algebraic curve:
http://www.nd.edu/~dbates1/preprints/real_curves.pdf


Support
geometry
geometric transformation
\/closed

I see that nobody updates the "Resources" section of this forum of important math contest like Greece, Moldova, etc. ... this makes me so unhappy, because it was a great collection of olympiad problems!

Maybe the people who used to update it, now have no more time to do this work... in this case I (and many users of this forum, I suppose) would be pleased to do this!

I'm preparing a new version of the problem introduction tool. For now that is restricted so that people don't post wrong problems (or edit/delete good ones). So only people that I trust are able to do that for now.

But a new system is underway (it's being tested as we speak) and soon anyone will be able to contribute (there will be a checking between the submission and the actual appearance of a problem in the resources section - but it will be much faster than it is today - and hopefully more accurate).

I will make a general announcement when the tool is ready. It will include translation versions (so that people can translate the problems in their own language - for now I'm thinking just Spanish, Romanian, German, French, Dutch, Chinese, Vietnamese and Iranian as these are the countries with the most users on the Forum).

[quote="Valentin Vornicu"][quote="edriv"]any news?  :maybe:[/quote]The tool is somewhat ready, we are doing different tests right now, and we have other stuff preparing as well. I hope everything will be in place and ready before the IMO :)[/quote]

Then you don't have too much time left :P

[quote="kdano"]Then you don't have too much time left :P[/quote]It's in place but there are some other things to take care of right at this moment :)


please help me with this question!(please show work)

1.A new high school has just been completed.There are 1000 lockers in the long hall of the school and they have been numbered from 1 to 1000. during lunch, the 1000 students decide to try an experiment

_ The first student, student 1, runs down the row of lockers and opens every door
_Student 2 closes the doors of lockers 2,4,6,8 and so on to the end of the line
_student 3 changes the state of the doors of lockers 3,6,9,12 and so on to the end of the line.(the student opens the door if it is closed and closes the door if it is open)
_ Student 4 changes the state of doors 4,8,12,16 and so on.
_Student 5 changes the state of every fifth door
_student 6 changes the state of every sixth door,and so on until all 1000 students had a turn.

when the students are finished, which locker doors are open?

is it the same when my problem ask which locker doors are open and theirs ask what is the last door open?

Read the first part of the question, it asks "Which locker doors are open?"


calculus
algebra
binomial theorem
calculus computations

Hey! Can anyone please tell me how to simplify:

(3+h)^4 - 81
___________
       h

I don't understand how to expand something thts to more than power of 2

Thanks  :)

[quote="ROLEX4life"]Or you could use the binomial theorem, even though fieryhydra's solution was pry a bit easier.[/quote]
Lahnals mentioned he/she cannot expand terms that have powers more than 2. Thus, keeping that in mind, it is not a wise thing to talk about Binomial Theorem at this point in time. But, of course, your proposed solution is a correct one too!

The point is the subtraction of the squared numbers is the product of the addition and subtraction.


Let $a_{n}> 0$, and given that the series $\sum a_{n}$ diverges. Determine the convergence of \[\sum \frac{a_{n}}{1+n a_{n}}\]

I think if $a_{n}= n$ then we have
$\sum_{n=0}^\infty \frac{n}{1+n^{2}}$
which clearly diverges by comparison to $\frac{1}{2n}$ (ignoring the first term which is 0 anyway).

edit: maybe you actually want to prove that it always must diverge though. But we can write it as $\frac{1}{1/(a_{n})+n}$, so unless $a_{n}\to 0$, then it must diverge.

But the $n$ can't matter. If it does (i.e., $1/(a_{n})$ is not big compared to it) then the series diverges by a comparison like above. So we ignore then $n$ and we're just looking at $\sum a_{n}$ again.

Clearly my proof is not rigorous but this should give the idea.

The issue isn't that it sometimes diverges; the question is whether it always diverges.

Hint: $\frac{a_{n}}{1+na_{n}}\ge\frac12\min\left(a_{n},\frac1n\right)$

If we had the condition that $a_{n}$ was monotone decreasing, we would be able to prove the the series in question diverges. This fails if we don't restrict to the monotone case. For that, consider this example:

$a_{n}=\begin{cases}\frac1{n^{2}}&n\text{ not a perfect square}\\ 1&n=k^{2}\text{ for some }k\end{cases}$

$\sum_{n=1}^{\infty}a_{n}=\infty$ since the terms do not tend to zero.

However, $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ converges.

Whoa, thanks. Although it's obvious now, I didn't realize as I was writing my post that I implicitly assumed monotone (or at least, some form of "nice" behavior). Cool example (I wanted to think of one like that, but then I "proved" that it always diverges...)


Let $ ABC$ be a triangle. $ D$ is the midpoint of $ BC$. $ AD$ intersects the circumcircle at $ E$. Let $ H$ be the orthocentre, $ O$ be the circumcentre. Let the circle with diameter $ AD$ intersect the circumcircle again (apart from $ A$) at $ P$. Prove: $ AP, OH, BC$ are concurrent iff $ AH\equal{}HE$.

it was not quite easy for me :blush:

I'm too busy to write all the proof now......(sorry :D )
So I will write some main ideas of the proof. Later, I will post the full.

[hide]
1. $ AH \equal{} HE \leftrightarrow B,C,G,H$ are concyclic.($ G$ is the centroid of triangle.)
2. $ D, H, P$ are collinear.
3. $ AP, OH, BC$ meet at one point $ \leftrightarrow B,C,G,H$ are concyclic

with these 3 truths, problem can be solved
[/hide]


Solve for x algebraically (or any method, as long as it doesnt involve visual aids.. e.g. graphs):

3x^2 - 2 + 2sinx = 0

[hide:a start]$ x\equal{}arcsin(sin x)$[/hide]

i do not think that would help simplify or solve the equation
but there may be a more subtle trick...

by the way: its [hide=...]

[hide="real start"]$ 3x^2\plus{}2\sin x\equal{}2 \implies 2\sin x\equal{}2\minus{}3x^2 \implies \sin x\equal{}1\minus{}\frac{3}{2}x^2$

since $ \minus{}1\le \sin x \le 1$, $ \minus{}1\le 1\minus{}\frac{3}{2}x^2\le 1 \implies \minus{}2 \le \minus{}\frac{3}{2}x^2\le 0 \implies 4\ge 3x^2 \ge 0$

therefore, $ \frac{4}{3}\ge x^2\ge 0$[/hide]

well, that's a lot better than my idea! :D

You can't solve this one algebraically.

You can use the Taylor cubic approximation $ 2 \sin x \approx 2x \minus{} \frac{x^3}{3}$. There are better cubic approximations for $ \sin$, but they involve the Graham-Schmidt process, and you'd probably need a computational device to find them. In any case, the Taylor approximation will give you the lower root within 1/200 and the higher root within 1/5000, so it works fine. The cubic will also give you a third root, but you can just ignore that because it does not satisfy $ x^2 \leq 4/3$.

So anyway, you have the approximation $ x^3\minus{}9x^2\minus{}6x\plus{}x \equal{} 0$ (I multiplied by -3). Then you can use the cubic formula, or alternately, the [url=http://en.wikipedia.org/wiki/Newton%27s_method]Newton-Raphson method[/url].

The cubic formula will be exact but will involve complex numbers and gross computation.

The Newton-Raphson method involves simple calculus and straightforward computations, and will be more and more accurate depending on how many recursions you use. You probably won't need too many to get a pretty accurate root. And the roots of the cubic aren't exactly the roots of the original equation, so there's nothing holy about finding an exact root to the cubic here.

hello, you can use this method:
http://mathworld.wolfram.com/NewtonsMethod.html
Sonnhard.


Let p be a prime number such that $ p\equiv3 [4]$ and a,b,c,d positive integers such that $ a^{2p}\plus{}b^{2p}\plus{}c^{2p}\equal{}d^{2p}$
Prove that $ p|abcd$

See [url=http://www.mathlinks.ro/viewtopic.php?p=934245#934245]this post[/url]for a stronger result.


The formula for finidng volume of pyramid is 1/3 * b * h.  Find the length of a diagonal of base in pyramid if volume of a pyramid is 216 and height is 18.  Write your answer in simplest radical form.

[hide]

216=1/3*b*h

b=36

Since pyramid is a square base pyramid...

Length of square side is 6

Diagonal =  :sqrt: (6 :^2: +6 :^2: ) = 6 :sqrt: 2

[/hide]

I might be wrong.  Tell me though.

It's correct, though..
But I want to add that without doing that 6 :^2: + 6 :^2:  or something you did.. Here's something that explains the way:

Diagonal of Square: rt 2 * length a

length a is the base or height in right triangle with 45-45-90 degrees where base and height has same length.  And diagonal in square is same as hypotenuse in the right triangle, therefore, it's just rt 2 * length a

Good Job!

Thx

I never knew that.

I was just using Pythagorus to find the diagonal.

i get it.... but what does rt 2 stand for ??????

2 right triangles????

[hide]oh, but i thought that the square root would be expressed as

[sqrt][/hide]

as long as you understand what it means it doesn't matter how you write it... I saw many people wrote rt 2 as sqrt 2 because sqrt and rt means same thing.. But, I'll try to write more appropriately..

oh, thanks

i got confused because i thought that "sqrt" was formal on the computer like the + sign or = sign or * for multi.

oh, thanks

i got confused because i thought that "sqrt" was formal on the computer like the + sign or = sign or * for multiplication


function
real analysis
real analysis unsolved

Let $f:[0,1]\to[0,1]$ be a continuous function s.t. $f(0)=0$ with the property that for each $x\in[0,1]$ there is a positive integer $n=n(x)$ s.t. $f^{(n)}(x)=x$, where $f^{(n)}$ means $f$ composed with itself $n$ times. Show that $f(x)=x,\ \forall x\in[0,1]$.

*$f$ is surjective
*$f$ is injective: if $f(x)=f(y)$ then
$f^{(n_xn_y)}(x)=f^{(n_x)}(y)=f^{(n_x)}(x)=x$ and $f^{(n_xn_y)}(x)=f^{(n_y)}(x)=f^{(n_y)}(y)=y$
*$f(0)=0$, $f$ is continuous,$f:[0,1] \to [0,1]$ . So $f$ is strictly increasing.
Suppose that $f(x)>x$(resp. $f(x)>x$) then $f^{(n)}(x)>x$(resp. $f^{(n)}(x)<x$)
for $n \geq 1$ -contradiction-


Anyway, I made another mock aime for this year, and I was wondering if anybody would take it any time soon.

I have all the problems written, but there are a few that i still need to get an exact answer for. I would like some second opinions on some of the problems though. There were a few that i got fishy answers, and there was no really good way to check them. Please p.m. me if you want to check over a few or all the problems, i will give outline solutions to look over.

My guess is that I will hold the test over the next few weeks, and you can send answers whenever you want. But if too many people are busy, then I can either postpone the start or deadline. The problems should be reasonable, but I really don't know since they all seem clear to me since i made them...

I'm going to be very busy over the next couple of weeks, but don't let that stop you if most other people are free.

yay! another mock AIME!

i'll take it as long as there isn't [b]too[/b] much emphasis on geometry...

[Recalls how a certain Mock AIME went.] :rotfl:

i tried to make the problem distribution as even as possible, but we will see what you guys think, some of them seem like one topic, but they really are another, i have everything done, and there are a few people who will look over the problems some time soon, so i guess that i will wait for them, and there are a few problems that seem a little too easy, i might throw in a few more tricks...

i was thinking about posting it soon, how are people doing with finals and stuff, personally, i am done, except i have my math final tomorrow...but i don't need to worry...i could fail my final and still get an A, also the people who were going to look over never did...the only thing is that i caught a few mistakes, so i think the test is done more or less

I'll be done this week, so I should be able to take it anytime this weekend.

It would be perfect timing for me if you posted it at the beginning of next week, since that's when I have my semester break, and my play will be done.


The participants to an international conference are native and foreign scientist. Each native scientist sends a message to a foreign scientist and each foreign scientist sends a message to a native scientist. There are native scientists who did not receive a message.

Prove that there exists a set $S$ of native scientists such that the outer $S$ scientists are exactly those who received messages from those foreign scientists who did not receive messages from scientists belonging to $S$.

[i]Radu Niculescu[/i]

I always had a problem with the way this is formulated :). For instance, the set $S$ could be taken to be empty if there were no native scientists which did not receive any messages, so this condition doesn't significantly change the problem, I think :?.

Anyway, consider a maximal (with respect to inclusion) set $S$ of native scientists, which contains the set of native scientists who did not receive any messages (so that they are non-empty), and with the property that the set $S'$ of native scientists who received messages from those foreigner scientists who did not receive messages from the scientists in $S$ is disjoint from $S$. There exists at least one such set, namely the set of natives who did not receive messages itself. Let's show that such a set $S$, together with $S'$, covers the whole set $N$ of native scientists.

Suppose we can find a native scientist $x_1$ in $N$, but outside both $S$ and $S'$. If he sends a message to a foreigner scientist who also receives a message from $S$, we can safely add $x_1$ to $S$ and leave $S'$ unchanged, thus contradicting the maximality of $S$. If, on the other hand, $x_1$ sends a message to some $y_1$ who does not receive any messages from $S$, two things ca happen: either $y_1$ sends a message to $x_2$, who also receives messages from other foreigners who do not receive messages from $S$, in which case we add $x_1$ to $S$ and leave $S'$ unchanged, or $y_1$ sends a message to $x_2$, and it's the only foreigner who does not receive messages from $S$ to do so. In this case, we add $x_1$ to $S$ and take $x_2$ out of $S'$. Either way, the existence of $S\cup\{x_1\}$ contradicts the maximality of $S$, and we have a contradiction.

I hope it's correct.


If there are solutions to
$a^{3}+b^{3}+c^{3}=x^{3}$
for $a,b,c,x\in\mathbb{N}$
find a way to generate all solution.
However, if there are no solutions prove that none exist.

[quote="maokid7"]However, if there are no solutions prove that none exist.[/quote]

 :lol:

i read in a number theory book that the proof of this is too complex for even undergrads.
may be the mods can help better.


Prove that $ n^4 \plus{} 4^n$ is composite $ \forall n \in \mathbb{N}$.

EDIT: Oops, it should have been $ \mathbb{N} \minus{} \{1\}$.

It's false when $ n\equal{}1$ because $ 5$ is prime.if you let $ n>1$...

[hide="solution"]
It's obvious when $ n$ is even,so let's check in the case $ n$ is odd.
By Fermat't Little Theorem,we have $ n^4\equiv 1\pmod 5$,then
$ n^4 \plus{} 4^n\equiv 1 \plus{} ( \minus{} 1)^n \equal{} 0\pmod 5$,so it's dividible by $ 5$ and greater than 5 when $ n > 1$,so $ n^4 \plus{} 4^n$ is composite.[/hide]

[hide="hint"]
sophie germain identity
$ a^4\plus{}4b^4\equal{}\left(a^2\plus{}2ab\plus{}2b^2\right) \left(a^2\minus{}2ab\plus{}2b^2\right)$
[/hide]

[hide="hint"]
sophie germain identity
$ a^4 \plus{} 4b^4 \equal{} \left(a^2 \plus{} 2ab \plus{} 2b^2\right) \left(a^2 \minus{} 2ab \plus{} 2b^2\right)$
[/hide]

EDIT: double post... oops
please delete...

[quote="uglysolutions"]You're missing the case $ 5|n$.  :wink:[/quote]

oops,thank you for pointing it out :lol:

[quote="stevenmeow"][hide="hint"]
sophie germain identity
$ a^4 \plus{} 4b^4 \equal{} \left(a^2 \plus{} 2ab \plus{} 2b^2\right) \left(a^2 \minus{} 2ab \plus{} 2b^2\right)$
[/hide][/quote]
I think this identity helps when $ n \equiv 1\mod 4$ (such that $ n^4 \plus{} 4^n \equal{} n^4 \plus{} 4*(4^k)^4$), but what about other cases?
(btw, there's a delete button at the bottom of one's post - except for the 1st post)

$ 4$ is itself a square, so the identity works just fine when $ n \equiv 3 \bmod 4$ as well.  And the expression is clearly composite when $ n$ is even.


playing with Mathematica I discovered a nice property of Fibonacci rests which are zeros mod powers of 10:

$ 10|f_i\iff 15|i$

$ 100|f_i\iff 150|i$

$ 1000|f_i\iff 750|i$

is there any general formula for this ?

-is it known or not ?

Principially, there are some standard effects that give these and more general effects:
Let $ p$ be a prime and $ m,n$ integers.

Let $ a_n$ be the smallest positive integer such that $ n|F_{a_n}$.
Then $ n|F_m \implies a_n|m$. Additionally, $ p|F_{p\minus{}\left(\frac p5\right)}$ (where $ \left(\frac p5\right)$ is the Legendre symbol), thus $ a_p | p\minus{}\left(\frac p5\right)$.
For $ p \neq 2,5$, we have $ a_p \equal{} \mathrm{ord}_p\left(\minus{}\frac{3\plus{}\sqrt 5}{2}\right)$ (where $ \mathrm{ord}_p(a)$ is the smallest $ k$ such that $ a^k \equiv 1 \mod p$; we subtly work in algebraic extensions to get that $ \sqrt 5$); this holds for integers $ p$ (not only primes) coprime to $ 10$ as well.
$ p^s|F_n \implies p^{s\plus{}1} | F_{pn}$ for $ s>0$; we have the inverse direction if $ p \neq 2$ or $ s>1$. So in that cases, $ p^s\parallel{}F_n \implies p^{s\plus{}1} \parallel{} F_{pn}$.
We now got everything necessary to describe those $ n$ such that a given prime power $ p^s$ divides $ F_n$. Even more, by the first property, we can extend it to arbitrary divisors.

Do you know [b]Fibonacci entry points[/b]?

Cf. [url=http://www.research.att.com/~njas/sequences/A001177]A001177[/url] - OEIS


articles
geometry
geometric transformation

I wanna see what ppl primarily think about Hinduism with this question. It's not really a poll, but still. I know the "accepted" answer but what do you think?
BTW: Polytheistic means we believe in many gods.
Monotheistic means we believe in one god.

It's a monotheistic religion, but the god has a lot of different reincarnations. (24..I think..not sure)

Hinduism is essentially polytheistic. It accepts the existence of three male gods - Brahma (the creator), Vishnu (the preserver), Maheshwara (the destroyer) - and a female mother goddess, Shakti, among many others. It is similar to the Greek, Roman, Egyptian or Norse civilizations.

@ nonie: You are alking of the avatars of Vishnu - matsyo, kurmo, boraho, nrisingho, bamon, krishno, ram, buddho, poroshuram, kolki - these are the ten avatars of Vishnu.
[url]http://en.wikipedia.org/wiki/Vishnu#Avatars[/url]

However, there are several monotheistic sects among hindus:
1) 'Brahmo's (worshipper of Brahma)
2) 'Vaishnava's (worshipper of Vishnu)
3) 'Shaiva's (worshipper of Shiva)
4) 'Shakto's (worshipper of Shakti)
among others.

Information on all of them is available at Wikipedia.
http://en.wikipedia.org/wiki/Brahma
http://en.wikipedia.org/wiki/Vishnu
http://en.wikipedia.org/wiki/Shiva
http://en.wikipedia.org/wiki/Shakti

[hide="NOTE:"]I am a radical atheist.[/hide]

In 7th grade we learned that Hinduism has several gods. Therefore it is polytheistic.

Our textbook might be wrong.

It's not the first time a textbook has been wrong. Recently there was a textbook incident in California where Hindu students there questioned the validity of over 40% of the facts presented about Hinduism in the textbook. The textbooks there were flatout wrong, and it's not uncommon. In Hinduism, we believe in one Brahman(not to confuse with Brahma) which is the "supreme being." The different gods you know of are as bubka said, different sects of Hinduism. From the Vedic texts, "Ekah sat vipraa bahuda vadanti, " God is one, but ppl call him by different names.

Sorry fireemeblem, but normal Hinduism has the Trimurty concept at the heart of it. Hindu sacred sriptures and myhtologies all speak 300 million gods. Hinduism never speaks of different forms of one God. The monotheism concept is much too modern.

There is nothing like one Brahman either. Why, I am a Brahman by birth. If I had had the sacred thread ceremony I would have become a real Brahman. But as both my parents are agnotstic, we never thought of the ST ceremony.

Brahman is at the top of the Varnas. This system say that all Hindus are divided into four categories ('Varna's) which are, in order of social prestige:
1) Brahman - the priests
2) Kshatriya - the warriors and the monarchs
3) Vaishya - the bourgeoisie
4) Shudra - the proletariat

no, no, no. you're getting the two mixed up. The caste "Brahman" that you said in your previous post is "Brahmin." I'm a Brahmin too as are most Hindus in teh US. I'm talking about Brahman. Brahman is not in any form, and there are 3 different aspects of Brahman-creation, sustenance, and destruction. This is what you know as Trimurty. last, Hinduism is not a mythology, despite what label westerners have applied.

yea, doesnt it have 64 billion?

thats what i learned

to bubka,
thats the caste system

to fireemblem13,
there r two: brahman and atma

atma is like the soul, it is everywhere but no one can see it
brahman is what u worship (shiva, vishnu, etc

a good metaphor for god is that god is "behind a curtain"

@ fireemblem:

I have heard of that somewhere. But are you trying to say that the Hindu holy books do not accept the existence of many gods, different from one another in identity?

@ akalra1:

Another good metaphor is God is as real as $\sqrt{-1}$

I beleive hinduism is polytheistic because there are many gods.  I don't see how it can be monotheistic.  :maybe:

[quote="bubka"]
Another good metaphor is God is as real as $\sqrt{-1}$
[/quote]

That is a very good metaphor.

Here are some hindu gods -

Indra, the King of Heaven
Varuna, the God of Wind
Agni, the God of Fire
The Ashwinikumar duo, the Gods of Medicine
Surya, the Sun god

and the list goes on.

[quote="bubka"]Here are some hindu gods -

Indra, the King of Heaven
Varuna, the God of Wind
[b]Agni, the God of Fire[/b]
The Ashwinikumar duo, the Gods of Medicine
Surya, the Sun god

and the list goes on.[/quote]

btw, before "god" was worshiped, fire was worshipped (true vedic worship is worship to the fire). back when people didnt have stoves, heaters, etc, fire was really useful. so they worshipped fire  because it was very important in their daily lives. but as the human mind developed, people wanted a more humanlike god, so they eventually worshipped humanlike gods like Durga, Vishnu, Shiva, etc. but one god remained: agni (i think agni means fire in sanskrit). modern pujas involve fire worship and "god" worship (u offer things like ghee into the fire, etc). if u have ever seen an indian wedding, the bride and teh groom walk around the fire, and that blesses their relationship.

I shall prove that there are infinitely many forms of God. Assume for contradiction that there are a finite number of Gods. Now, we know, God is omniscient, right? well, in hinduism, it's true. God is omniscient and omnipotent. Since God is omniscient, God is in each and every person. This is also true described in Hinduism. Well, if God is finite, then God can only exist in a finite number of people, which is a contradiction. Therefore there are infinitely many Gods, but this cannot be. Therefore, there must be infinitely many [b]FORMS[/b] of God. QED.

@fireemblem13:
there are only finitely many people, which destroys your proof.

@akalra1:
I have seen like 20 odd indian weddings! I have grown up in India, in Bengal.

[quote="fireemblem13"][quote="bubka"]@fireemblem13:
there are only finitely many people, which destroys your proof.[/quote]
good point. lol. well then forget just ppl and consider the entire planet. If God were to inhabit every corner of the world, God would have to be infinite. But really, God is not measurable. He is not a "person" you can visualize like Krishna or Rama. God is the ultimate Moksha, the goal in life, realization, and liberation from the cycle of truth and untruth. Everything you see around is a form of god. Not just the "gods" we know of but EVERYTHING is a form of god. It's much more abstract.[/quote]

@fireemblem:
but hey, take this for example

Laxmi is the god of wealth. wealth is god then, and should not be abused. bill gates has infinite money, therefore god is infinite

[quote="bubka"]
@akalra1:
I have seen like 20 odd indian weddings! I have grown up in India, in Bengal.[/quote]

what do u mean "odd"
have u ever seen any indian movies?

[quote="akalra1"]have u ever seen any indian movies?[/quote]
i haven't seen any recently, but i definitely used to see a lot of them. they're not really worth watching anymore unless you're really bored w/ the exception of a couple movies. instead of wasting my time watching one of them, i'd rather do math.

[quote="fireemblem13"][quote="bubka"]@fireemblem13:
there are only finitely many people, which destroys your proof.[/quote]
good point. lol. well then forget just ppl and consider the entire planet. If God were to inhabit every corner of the world, God would have to be infinite. But really, God is not measurable. He is not a "person" you can visualize like Krishna or Rama. God is the ultimate Moksha, the goal in life, realization, and liberation from the cycle of birth and rebirth. Everything you see around is a form of god. Not just the "gods" we know of but EVERYTHING is a form of god. It's much more abstract.[/quote]

You are not talking hinduism there.

@ akalra1:
20 odd means 20 plus. And I have seen many indian movies. Bollywood is the world's largest movie industry.

[quote="bubka"]You are not talking hinduism there. [/quote]
[b][size=200]WHAT!!![/size][/b] what do u mean im not talking hinduism. hinduism isn't just what you know: ramayana, mahabharata, etc. It's a lot more. have you heard of the Purushartas, Dharma, Aarth, Kama, and Moksha. Liberation from the cycle of birth and rebirth comes straight from this. I have no idea what u meant to say when you said "You are not talking hinduism there" but hinduism is a lot more vast than the stories we all know about. You need to do your research. As for me, I've researched, written essays, written articles, given speeches, and read books on Hinduism. Please excuse me if i were a bit harsh when i wrote this. it was not intended.

i have reached moksha

*fades into heaven*

jk

hey guys, r u hindu?

btw, moksha is the eternal goal of life

[quote="fireemblem13"][quote="akalra1"]hey guys, r u hindu?[/quote]
yes, i'm a very devout hindu. moksha is indeed the goal of life, but unfortunately, today, most hindus are completely oblivious to that and not only to that but hinduism in general. in the vedic times, this was not so. moksha guided everyones' lives and thus everyone acted with dharma in order to achieve moksha. Sadly, ppl today are all materialistic and have no idea about moksha or dharma.[/quote]
It is a very similar situation between muslims. Many seem to have forgotten the true meanings and teachings of islam. That is why we have all these wierd so-called islamic laws that give women a lower status. That is not written anywhere in the quran, in fact it gives women and men equal status. And there are ofcourse a lot other misinterpretations. I apologize, this topic is about the hindu religion, please continue. Just wanted to point out this similarity.

I am an atheist. I give too little importance to Hinuism to do research on that. But I can tell you that the concept of Mukti does not contradict polytheism. In fact it doesn't even contradict atheism, but that's another matter. What you are sayin is essentially along Buddhist lines.

Well, It's every bit hinduism, believe it or not... im not saying buddhism doesn't have similar concepts.

maybe I should post the  4000 word essay i wrote. i think i will. don't read all of it but plz read a section.

So you are called Aman Sharma, are you?

Listen Aman. Are you under the impression that monotheism is better than polytheism?

wtf :rotfl: where'd u get that :| this whole time, i have only been trying to correct a natural misunderstanding about Hinduism. I have no idea what ur talking about. besides, another concept in hinduism is the fact that we don't discriminate betweet other religions and accept all religions as different paths to the same destination.

From what I've read and been told, Hinduism is MONISTIC (meaning only one basic/fundamental reality). Not necessarily monotheistic, nor polytheistic. The impression of polytheism comes from the fact that there are many "spirits" which Hindus worship (devas I think), but others tend to loosely interpret them as "Gods."

So what are gods? These spirits are what we call gods.

To aman: The fact is I cannot understand why you are so hellbent to prove Hinduism monotheistic.

I suppose the closest thing to a translation of "devas" is Gods, but there is a fine line between the two, which is why Hinduism is not polytheistic. There is pretty much NO God also, which is why it is not considered monotheistic either.  However, as I stated earlier, it is categorized as monistic.

well...
[quote]The impression of polytheism comes from the fact that there are many "spirits" which Hindus worship (devas I think), but others tend to loosely interpret them as "Gods."[/quote]
when u say this, i partially agree with you. the only place i disagree about is when you say "spirits" or "devas." I'm not quite sure what you mean by that and I would like a reference if possible. hinduism is a religion which is very open. It is not imperative for one to be hindu, to believe in Krishna or Rama or Brahma or whoever. In fact, it is not even necessary to worship any particular God. That is not what makes one Hindu. Because of the fact that Hinduism is so open, it allows Hindus to worship whoever they wish. I don't quite understand by what you mean when you say Monistic, i've never heard that be4. a reference would be wonderful if possible.

EDIT: TO bubka, plz excuse me if i have been too irrational on this topic, but i am only trying to clear up this misconception of polytheism. i'm sorry, but there were many posts that declared this fact, and i had to respond, i guess. i'm sorry. it was not intended.

guys, remember what the teachings of hinuism r

god is in everyone

god is everywhere

etc etc blah blah blah

Moderatorial duty:  If I see any more chatspeak on this thread, I'm deleting the post.  Write in English, for crying out loud!

The California text book brouhaha was initiated by a Hindu fundamentalist group arguing that only Hindus can legitimately analyze Hinduism and that therefore their views should be substituted for the "standard textbook view" in textbooks.  (This is not to say that the "standard textbook view" was without error, of course.)

I think the question "Is Hinduism monotheistic or polytheistic?" is problematic because the terms "monotheistic" and "polytheistic" are left undefined.  If they had obvious, agreed-upon conventional meanings, the question wouldn't be very interesting at all.  There are certainly complications: after all, some branches of Christianity recognize their god as three separate (but unified) entities, the trinity.  The Catholic and Orthodox churches have long lists of saints.  Some branches of Christianity also provide a significant role to a creature of god-like strength and power (the devil).  Yet we call Christianity monotheistic.  This is somewhat odd.
The problem arises with Hinduism because of its theology, which includes the belief that everything is essentially joined in some way, including the various different gods (each of which has multiple manifestations).  This is the basic root of the argument that Hinduism is monotheistic.

Frankly, though, I don't think the monotheistic-polytheistic dichotomy is very interesting as a dichotomy of theology.  It's much more interesting as a dichotomy of worship practice.  And in this respect, Hinduism is unquestionably polytheistic.

For once, I'm also going to agree with bubka -- it's not clear to me why anyone would care enough to start this argument.

There was also some earlier discussion along these lines in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78038]this post[/url].

[quote="bubka"]You have a point in that we should start with the definitions.

So what exactly is polytheism?[/quote]

from answers.com

pol·y·the·ism
The worship of or belief in more than one god.

mon·o·the·ism

The doctrine or belief that there is only one God.

that much we all know. Then the problem lies in the definition of god.

I would like to get this from aman himself. a clear definition as a student of maths should give.

Sorry I haven't been able to post earlier. Your question isn't as simple as it may seem. You see, God is not perceived by the 5 senses. We know that by meditation to God(realize how difficult that is) we realize God and realize that nothing in this material world matters(realize how difficult that is as well.) [quote="bubka"] a clear definition as a student of maths should give.[/quote]
hmmm... Well, we know that God is omnipotent and omniscient which comes from the Vedic shlok,

Ishavasyamidamsarvam, Yatkinch Jagatyam Jagat
Tenataktena bhunjita, magrida kasya swiddhanam.

Well, we also know that God can be understood in countless forms, Krishna, Rama for different people. What do you think would accurately define God?

I meant what in the definition of god stops Narayana and Shiva from being two different gods? And how is deva different from god?

P.S.: I was asking some of my friends here who have studied in the Rama Krishna Mission schools this question, and they all say that polytheism came much later, but I am still not convinced.

[u]From bubka's Factfile:[/u]

[b]FACT 1:[/b]

In Sid Meyers' turn based strategy game Civilization IV, you can found Hinduism by researching polytheism.

I think Hinduism has one primary belief but we believe in many gods! :)


calculus
integration
conics
ellipse
linear algebra
matrix
calculus computations

Evaluate the double integral of (x^2 - xy + y^2) over D, where D is the region bounded by the ellipse x^2 - xy + y^2 = 2.
By using the transformation x = root(2)u - root(2/3)v ; y=root(2) + root(2/3)v

May I know what D looks like??
And how do I know what are the ranges of u and v after the transformation?
Thanks

I'm guessing you have a typo under the definition of $ y.$ Shouldn't that be $ y\equal{}\sqrt{2}\,u\plus{}\sqrt{\frac23}\,v\,?$

The image of an ellipse or circle under a linear transformation (and this is certainly a linear transformation, given by $ \begin{bmatrix}x\\y\end{bmatrix}\equal{} \begin{bmatrix}\sqrt{2}&\minus{}\sqrt{\frac23}\\ \minus{}\sqrt{2}&\sqrt{\frac23}\end{bmatrix} \begin{bmatrix}u\\v\end{bmatrix}$ ) is an ellipse or a circle. You'd hope it's a circle - there's not much point in the transformation otherwise. But I'll let you find out whether it is.

I'm going to suggest some simple-minded brute force. Just put those definitions of $ x$ and $ y$ into the equation $ x^2\minus{}xy\plus{}y^2\equal{}2$ of the boundary curve of your region, grind through some algebra, and see what you get. It will be the boundary curve of your transformed region.


Let $f(x)=sin^{-1}x$, $g(x)= cos^{-1}x$, and $h(x) = g(x)+f(x)$, when all the angles are measures in radians.

a. Graph all three functions on the same graph, labeling all points of intersection and at least four other points on the graph.
b. What is the domain of $h(x)$?
c. Using the insight gained from the graph, simplify the formula for h(x), Explain why this makes sense.

I need help on [b]C[/b].
Thanks!

Did you do [b]A[/b] and [b]B[/b] yet?  That might be useful.

[hide="You should see that..."] $h(x)$ is a horizontal line, so it is a constant.  Moreover, $h(0) = g(0)+f(0) = 0+\frac{\pi}{2}$, so $h(x) = \frac{\pi}{2}$ for all $x$.

This makes sense because $\cos x = \sin \left( \frac{\pi}{2}-x \right)$. [/hide]

[hide]

$h(x)$ could be simplified to the constant function, $h(x)=\frac{\pi}{2}$

If you look at a triangle, the sine of one angle and the cosine of the other acute angle are the same, so $\sin^{-1}(x)$ and $\cos^{-1}(x)$ would be those two angles. The sum of these two acute angles would be $\frac{\pi}{2}$ radians.

[/hide]


$47)$ At what time between $3$ and $4$ o'clock are the minute and hour hands of a clock together?

$49)$ A tree $123$ feet high breaks off a certain distance up, and the moment the top strikes a stump 15 feet high the broken part points to a spot $108$ feet from the base of the tree. Find the length of the part broken off.

[hide="47"]We write the time as 3+x hours, or 15+5x ticks around the clock for the hour hand, or 60x minutes, which is 60x ticks around the clock for the minute hand. 55x=15, x=3/11 (180/11=16.363636 minutes)[/hide]

49: I can't get a graphical picture at all of this. The tip is 15 feet high. What direction does it fall down? Straight? Or does the breakpoint remain together?[/hide]

On 49, there wasn't a picture. I guess people are supposed to visualize it. I couldn't visualize it either, which is why I didn't work this problem out.

One way of looking at it: h is the height it broke at, x is the distance it fell.

h/108=15/(108+x)

Also: (15-h)^2+x^2=123-h. Solve this system?

I do not know if this helps, but the answer is 78.  :blush:

How I reach that conclusion, I do not know. I have reached an impasse.


if $ a,b,c > 0$ prove that
$ \sum\frac {a \plus{} 2b}{4a \plus{} 7b \plus{} c}\le\frac34$

[quote="Megus"]when you group elements like that:

$\sum \frac{1}{4}-\frac{a+2b}{4a+7b+c}=\sum\frac{c-b}{16a+28b+4c} \geq 0$

it looks quite obvious[/quote]

Is what you write assumed to be a proof, or a hint to a proof, or a part of a proof? If yes, I must say I am completely lost here, since the above may [i]look[/i] obvious but there is no clear reason why it [i]is[/i] so. Anyway, the inequality in question is straightforward. Mainly, one has to substitute x = 4a + 7b + c, y = 4b + 7c + a and z = 4c + 7a + b, to show by a simple calculation that $a+2b=\frac{11x-y-z}{36}$ and cyclically, so that the inequality $\sum \frac{a+2b}{4a+7b+c} \leq \frac34$ can be rewritten in the form $\sum \frac{\left(\frac{11x-y-z}{36}\right)}{x} \leq \frac34$. Now, after multiplication with 36, this becomes $\sum \frac{11x-y-z}{x} \leq 27$; but $\frac{11x-y-z}{x} = 11-\frac{y}{x}-\frac{z}{x}$, so that $\sum \frac{11x-y-z}{x} = 3\cdot 11 - \sum_{\mbox{symmetric}} \frac{x}{y}$, and it remains to prove $3\cdot 11 - \sum_{\mbox{symmetric}} \frac{x}{y} \leq 27$. This can be directly simplified to $\sum_{\mbox{symmetric}} \frac{x}{y} \geq 6$; but this is clear from AM-GM, as $\prod_{\mbox{symmetric}} \frac{x}{y} = 1$.

  Darij

my solution to this one is:
denote by $x=\frac{a+2b}{a+b+c}$, $y=\frac{b+2c}{a+b+c}$ and $z=\frac{c+2a}{a+b+c}$. It is clear now that $x+y+z=3$
and the inequality can be rewritten as
$\sum\frac{x}{3x+1}=\frac13\sum\frac{3x+1-1}{3x+1}=\frac13\sum(1-\frac1{3x+1})=1-\frac13\sum\frac1{1+3x}\le\frac34$ $\Leftrightarrow$ $\frac34\le\sum\frac1{1+3x}$ but from AM-HM results that $\sum\frac1{1+3x}\ge\frac9{3+3\sum x}=\frac9{12}=\frac34$

[quote="Kuba"]if $a,b,c>0$ prove that

$\sum\frac{a+2b}{4a+7b+c}\le\frac34$[/quote]
$\sum {\frac{1}{3}-\frac{a+2b}{4a+7b+c}}=\frac{1}{3}\sum \frac{a+b+c}{4a+7b+2c}$

And we prove:

$(a+b+c)(\frac{1}{4a+7b+c}+\frac{1}{4a+7b+c}+\frac{1}{4a+7b+c}) \ge \frac{3}{4}$
It's too easy!


inequalities
function
real analysis
real analysis solved

Let f:[a,b]-->R be a Rolle function so that f'(x) is not constant on (a,b). Prove that there exists c in (a,b) so that:
|f'(c)|>1/(b-a)|f(b)-f(a)|.

Let
[tex]F(x) = f(x) - f(a) - \frac{f(b)-f(a)}{b-a} (x-a)[/tex]
thus
[tex]F(a)=F(b) = 0[/tex]

From [tex]f'(x)[/tex] is not constant, we know that [tex]F(x)[/tex] is not always zero. Without loss of generality, suppose [tex]f(y) > 0[/tex].

Apply Lagrange's Mean Value theorem to [tex]F(x)[/tex] on [tex][a,y][/tex] and [tex][y,b][/tex],
we have
[tex]F'(x_1) = \frac{F(y)}{y-a}>0\quad x_1\in (a,y) [/tex]
and
[tex]F'(x_2) = \frac{F(y)}{y-b}<0\quad x_2\in (y,b) [/tex]

Since
[tex]F'(x) = f'(x) - \frac{f(b)-f(a)}{b-a}[/tex]

thus
[tex]f'(x_1) > \frac{f(b)-f(a)}{b-a}[/tex] and [tex]f'(x_2) < \frac{f(b)-f(a)}{b-a}[/tex]

If [tex]\frac{f(b)-f(a)}{b-a}\geq 0[/tex] then let [tex]c=x_1[/tex]; if it is smaller than 0 then let [tex]c=x_2[/tex]. In both cases we have [tex]|f'(c)| > \left|\frac{f(b)-f(a)}{b-a}\right|[/tex].


Prove for all $x,y,z>0$ that
\[(x+y+z)^{2}+(x+y+z)\frac{3xyz}{xy+yz+zx}\geq 4(xy+yz+zx)\]

[quote="Tiks"][quote="Yimin Ge"]Prove for all $x,y,z>0$ that
\[(x+y+z)^{2}+(x+y+z)\frac{3xyz}{xy+yz+zx}\geq 4(xy+yz+zx) \]
[/quote]
\[(x+y+z)^{2}+(x+y+z)\frac{3xyz}{xy+yz+zx}\geq 4(xy+yz+zx) \]

\[\Longleftrightarrow \]

\[(x+y+z)^{2}(xy+yz+zx)+3xyz(x+y+z)\geq 4(xy+xz+yz)^{2}\]

\[\Longleftrightarrow \]

\[xy(x^{2}+y^{2})+xz(x^{2}+z^{2})+yz(y^{2}+z^{2})\geq 2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}) \]

\[\Longleftrightarrow \]

\[xy(x-y)^{2}+xz(x-z)^{2}+yz(y-z)^{2}\geq 0 \]
:)[/quote]
nice solutions  :)  :)  :)  and I tried to solve it by mixing variable. It is not hard .

But isn`t this just a weaker form of Schur?
\[2\sum ab-\sum a^{2}\leq\frac{9abc}{a+b+c}\leq \frac{3abc(a+b+c)}{ab+bc+ca}\]


Let $ x\in R $ satisfying :
 $ x^2 - x \in N $
 $ x^n - x \in N $ , $ (n \in N , n>2) $
Prove that : $ x \in N $

This is pretty hard for Intermediate.  I've moved it up a notch.

Dear Moderator ,
 If you this problem is not available for pre-olympiad , please move it to Olym section . Thank you

[hide="Is this right?"]Factor $x^2-x$ to get $x(x-1)$.  Since x-1 is an integer iff x is an integer and the product of two non-integers is a non-integer, we know that x must be an integer in order for $x^2-x$ to be a natural number.

Now, factor $x^3-x$ to get $x(x+1)(x-1)$.  Since, from the first paragraph, we know that x is an integer, we know that x must be at least 2 in order for the expression $x(x+1)(x-1)$ to be positive, so x must be at least 2 in order for $x^3-x$ to be a natural number.

Since we know that x is an integer, and x is at least 2, we can conclude that x is a natural number.[/hide]

The product of two non-integers can be an integer, for example take $x=\frac{\sqrt 5 + 1}{2}$

I assume that $x$ is supposed to be non-negative, am I right?
In that case :

- First we cannot have $0<x<1$ otherwise the number $x,x^2,x^n$ would all belong to $]0;1[$ and then would be equal, which cannot occur.

- Assume that $x>1$. Then $x^n > x^2 > x$ and there exist an integer $a>0$ such that $x^2 = x+a.$
Then $x^3 = x^2 +ax = x(a+1) + a$ and $x^4 = x^2(a+1) + ax = x(2a+1)+(a^2+a).$
And so on...By induction we easily prove that $x^k = u_kx + v_k$ where $u_k > a$ and $v_k$ are integers.
Thus $x^n = u_nx + v_n$ so that $x^n - x = (u_n-1)x+v_n.$ It follows that $(u_n-1)x$ is an integer and then $x$ is rational.

Now assume that $x  = \frac r s$ is not an integer, where $\gcd (r,s) = 1$ and $s>1$. Then there exists a prime $p$ which divides $s$. Let $\alpha$ be its exponent in the prime decomposition of $s$. It follows that $2 \alpha$ is the exponent of $p$ in the prime decomposition of $s^2$ so that $2\alpha$ is the exponent of $p$ which divides the denominator of $x^2-x$, which contradicts that $x^2-x$ is an integer.
Thus $x$ is an integer and we are done.

Pierre.

Okay, wasn't really thinking when I made my last post.

Oh , this is my teacher's problem !
 This problem is simple .
 First , if $ x = \frac{p}{q} $ , we can easily show the contradiction !
 Second , $ x^3 - x^2 = x(x^2-x) \in N $ . So $ x \in Q $ : contradiction

Solution:[hide]
$x^2 - x = n_1$, $x^3-x = n_2$, so $x \cdot n_1 = x(x^2 - x) = x^3 - x^2 = n_2 - n_1$ which gives $x = \frac{n_2-n_1}{n_1} \in \mathbb{Q}$. Let $x = \frac{m}{n}$, and suppose $p$ is some factor of $n$ which does not divide $m$. Then $x^2 - x = \frac{m^2}{n^2} - \frac{m}{n} = \frac{m(m-n)}{n^2} \in \mathbb{N}$ but $p$ does not divide $m$ or $m-n$, contradiction.[/hide]


Prove that for every natural n

sin(pi/2n)*sin(3pi/2n)*...*sin(pi*(2n-1)/2n)>=2^(1-n).

hello, do you mean $ \sin\left(\frac{\pi}{2n}\right)\sin\left(\frac{3\pi}{2n}\right)\cdots\sin\left(\frac{(2n\minus{}1)\pi}{2n}\right)\ge2^{1\minus{}n}$?
Sonnhard.

I think we using complex number.(may using root equation $ w^{n} \equal{} 1$,this result, I do not think clearly)

[quote="Dr Sonnhard Graubner"]hello, do you mean $ \sin\left(\frac {\pi}{2n}\right)\sin\left(\frac {3\pi}{2n}\right)\cdots\sin\left(\frac {(2n \minus{} 1)\pi}{2n}\right)\ge2^{1 \minus{} n}$?
Sonnhard.[/quote]
$ F(k)\equal{}\boxed{\sin\left(\frac{\pi}{k}\right)\sin\left(\frac{2\pi}{k}\right)\cdots\sin\left(\frac{(k\minus{}1)\pi}{k}\right)\equal{} k\cdot 2^{1\minus{}k} }$

$ LHS \equal{} \frac{F(2n)}{F(n)}\equal{}2^{1\minus{}n}$ :)

Why
$ F(k)\equal{}\boxed{\sin\left(\frac{\pi}{k}\right)\sin\left(\frac{2\pi}{k}\right)\cdots\sin\left(\frac{(k\minus{}1)\pi}{k}\right)\equal{} k\cdot 2^{1\minus{}k}}$ ?


find all triples(x,y,z) of positive integers

$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}\frac{4}{5}$

This was just posted in the Intermediate forum -- what's the source?  Are you cheating off the same test, or what?  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275221

JBL, I don't see how those are the same...


geometry
3D geometry
icosahedron
geometry unsolved

Suppose an icosahedron is centered at the origin and three of its twelve vertices are at
(0, 0, 1), (0, 0,−1), and (a, 0, b). What are the values of a and b?

If you choose two pairs of opposite vertices on a regular icosahedron, then they form a golden rectangle. Then $(0,1)$ and $(0,-1)$ are the opposite vertices of a golden rectangle. So $|b|=\frac{2\sqrt5}5$ and $|a|=\frac{\sqrt{5}}5$


1) A block starting from rest slides down a frictionless plane of length $L$.  When the block has attained one-half it's final speed, what is the distance it has traveled along the plane?
2) A ball starts from rest and rolls without slipping down a plane.  What is the ratio of the angular velocity of the ball at the end of the plane to the angular velocity at the center of the plane?
3) Suppose the force required to tow a canal bridge is directly proportional to its speed.  If it takes $4$ horsepower to tow the barge at a speed of $2$ miles per hour, what horsepower is required to move the barge at a speed of $6$ miles per hour?

[hide="Answers"]
1) $\frac{L}{4}$

2) $\sqrt{2}$

3) $36$[/hide]

Heh, I missed #3 on a multiple choice section earlier this year.
If force is proportional to velocity, then F=k*v for some constant k.
Power=Force*Velocity
Power=k*Velocity*Velocity
4 hp=k*2 m/s*2m/s, so k=1 (and some units)

Then Power=1*6 m/s*6 m/s--> 36 hp.

Thanks, I forgot that power=force * velocity, any idea about the others?

The first one:

E=KE+PE
KE=0 at the beginnning
PE=0 at the end
When the block has 1/2 of its final velocity, it has (1/2)^2=1/4 of its final kinetic energy, so potential energy is 3/4 of the original, so it has only traveled 1/4 of the way.

another solutioin to problem 1 follows the Newton's Second Law.

We know that $v_{t}^2-v_{0}^2 = 2as$. In this one, $v_{0}=0$. So we have $v_{t}^2=2as$.
Since the ratio of the two velocities is $\frac12$, therefore $\frac{s}{L}={(\frac12)}^2$, which implies $s=\frac14 L$.

By the way, what does AP physics C mean?? Is that mean the easiest test in AP class?

[quote="shobber"]
By the way, what does AP physics C mean?? Is that mean the easiest test in AP class?[/quote]

No, the only two AP physics tests offered are B and C, C is the harder of the two.

I finally figured out the second one:

Let point $A$ represent the top of the incline, let point $C$ represent the center point, and let point $B$ represent the end, we then have

$E_A = E_C = E_B$

$mgh = \frac{1}{2}mgh + \frac{1}{2}mv_C^2 + \frac{1}{2}I\omega_C^2 = \frac{1}{2}mv_B^2 + \frac{1}{2}I\omega_B^2$

The moment of inertia for a solid ball around its center is $\frac{2}{5}mr^2$, so now we have

$mgh = \frac{1}{2}mgh+\frac{1}{2}mr^2\omega_C^2 + \frac{1}{2}\cdot \frac{2}{5}mr^2\omega_C^2 = \frac{1}{2}mr^2\omega_B^2 + \frac{1}{2}\cdot \frac{2}{5}mr^2\omega_B^2$

$mgh = \frac{1}{2}mgh + \frac{7}{10}mr^2\omega_C^2$
$\frac{1}{2}mgh = \frac{7}{10}mr^2\omega_c^2$
$\frac{5}{7}gh = r^2\omega_C^2$
$\omega_C = \sqrt{\frac{5gh}{7r^2}}$

$mgh = \frac{7}{10}r^2\omega_B^2$
$\frac{10}{7}gh = r^2\omega_B^2$
$\omega_B = \sqrt{\frac{10gh}{7r^2}}$

$\frac{\omega_B}{\omega_C} = \sqrt{2}$

Is it just me or does that seem like too much work for a multiple choice problem?

[quote="shobber"]For problem two there is a hint.
If a ball roll with out any sliding, we will have: $v_c=v=\omega r$ where $v_c$ is the velocity of the centre of the ball.
So follow this one and Newton's Second Law, we will get $\sqrt 2 $.

I don't know what you learn in AP classes, but for problem 1, it is one of the easiest quiz in our textbook.[/quote]

This is also a very easy problem here, it's just that I haven't done any mechanics in about 4 months.


Al Khawarizmi:algorithmic resolution of second degree equation with one unknown, a general solution .The "son"  of the indian father of algebra.The son who got beyond his father.
Omar khayyami:classification and resolution of third degree equation using conics.How to, geometrically, solve third degree equations.The equivalence btween fifth postula of Euclid and the existence of rectangles.
Ibn al Haytham:Dynamic geometry.Segment and point moving through lines and curves.
Al tusi:Taylor formula.Approximation of polynomial roots
AL mutaman ibn hud :ceva theorem
When Arab and Muslims inherited the greek science of mathematics , they gave this inheritance their fingerprint , say that they translated it algebrically and transmitted it to europe,even if Algbra has his root in india.
Many others not in my mind right now.

What's the point in posting this in the Career forum? :?

I don't get your point.  It is widely acknowledged that algebra has its foundations in India.


Show that there is no number x fror which x^2=3 mod7

[quote="math154"]Um, testing, we see that $ \{0^2,(\pm1)^2,(\pm2)^2,(\pm3)^2\}\pmod7 \equal{} \{0,1,2,4\}\pmod7$.[/quote]

But why did you stop at x=positive and negative 3?

Because the set of integers $ \{0,\pm1,\pm2,\pm3\}\pmod7$ includes all integers (i.e. all integers can be expressed as $ 7m,7m\pm1,7m\pm2,7m\pm3$).

cdymdcool, math154 used the well known fact that

a^k (mod n) = (a+bn) ^k (mod n) for a,b,n,k, positive integers.

Which is true because in the expansion of $ (a\plus{}bn)^k$

$ \equal{}a^k\plus{}\dbinom k1a^{k\minus{}1}(bn)\plus{}\dbinom k2a^{k\minus{}2}(bn)^2\plus{}\cdots$, all terms are divisible by $ n$ (i.e. $ \equiv0\pmod n$) except for $ a^k$.


Let $ a,\ b,\ c$ be positive real numbers such that $ \frac{a^3\plus{}2a}{a^2\plus{}1}\plus{}\frac{b^3\plus{}2b}{b^2\plus{}1}\plus{}\frac{c^3\plus{}2c}{c^2\plus{}1}\equal{}\frac{9}{2}$.

Prove that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq 3$.