ratio
geometry
circumcircle
geometric transformation
reflection
rotation
trapezoid

http://mcis.jsu.edu/mathcontest/geo03.pdf

I did 2003 and 2002 tests yesterday. I guess I'm having a huge learning day. I've been reading a lot of the solutions... well all of them ..  :lol:

I need help on $7,13,11,16,27,31,32,43,44,45,46,48,50$

Due to guessing on these ones I didn't know how to do, and narrowing down as many choices as I could I managed to get 5 of these right by guessing.

So I got a 42/50.

Could someone show the elegant way of doing these problems?  :lol:

[hide="#7"]
The two lines may only intersect at $1$ point.

To maximize the circle-to-circle intersections, each circle should intersect the other two at 2 points to give $2+2+2=6$ points.

To maximize the line-to-circle intersections, each line should intersect each circle at 2 points to give $2\cdot 2\cdot 3=12$ points.

The answer is $19$, or $\boxed{\text{C}}$.
 [/hide]

Could you please provide a link?

I think it just disappeared.

oops I guess I accidentally erased when editing  :blush:

K i provided it

[hide="#16"]
Draw the medians of the equilateral triangle. They intersect at the centroid, which is also the circumcenter.

The centroid divides the medians into a $2: 1$ ratio, so the shorter segment of each median is $\frac{x\sqrt{3}}{6}$.

By the Pythagorean Theorem, you get the circumradius to be $R=\frac{x}{\sqrt{3}}$.

The area, then, is $\frac{\pi x^{2}}{3}$, or $\boxed{D}$.[/hide]

[hide="#50"]
The Angle Bisector Theorem states that each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides, which in this case is $6: 4=3: 2$.

We see that $AG: GC=5.4: 3.6=3: 2$, so $\overline{BG}$ is the angle bisector of $\angle{ABC}$, or $\boxed{D}$.[/hide]

[hide="43"]draw a hexagon with vertices as the centers of the smaller circles. now call the length of the smaller circle r. connecting the radii of the larger circle to a vertex of the hexagon, we have it as length 18-r. but this also equals the side length of the hexagon which is 2r. solving for r yields r=6. so the area of the larger circle is 324pi and the smaller circles have area of 216pi. the answer is C, 108pi[/hide]

[hide="44"]the shaded region can be found by subtracting the areas of triangles ABE and BHE from the whole rectangle. since AE=12 and AB=8-7=1, BE=$\sqrt{145}$ and using pythag again, BH=9. so [ABE]=12(1)/2=6 and [BHE]=8(9)/2=36 so the answer is 54, B[/hide]

[hide="13"]recall the centroid divides the median in a 2:1 ratio. calling DB=x and AE=y, we have

$(\frac{1}{3}x)^{2}+(\frac{2}{3}y)^{2}=324\rightarrow \frac{1}{9}x^{2}+\frac{4}{9}y^{2}=324$
and similarly
$\frac{4}{9}x^{2}+\frac{1}{9}y^{2}=576$
adding them together yields
$\frac{5}{9}(x^{2}+y^{2})=900$
since $AB^{2}=\frac{4}{9}(x^{2}+y^{2})$,
we can do some rearanging to find that $AB=\sqrt{720}=12\sqrt{5}$[/hide]

POST #100!!!
 :winner_first:  :clap2:  :clap:  :omighty:  :harhar:  :icecream:  :10:  :heli:  :laugh: [hide="11."]Draw a line connecting G and D. The top triangle is $\frac{1}{4}$ the area of the big triangle, so it is $27$. Reflect across $\overline{GD}$ and you have a quad. with area $54$. Cut it out and you are left with $54$. Notice that the remaining gray triangles [b]do not[/b] have the same base and height. Their combined bases are equal to each of the white triangle bases. So the area of the remaining gray bases are $\frac{1}{3}*54=18$. Subtract $108-18=90$ and you get $\rightarrow\boxed{E}$.[/hide]

[hide="48."]Write the degrees of each into ratios. The ratios are $\frac{4}{3}: 1: \frac{4}{5}$. Let's find out the number of degrees required of middle to turn the top $180^{o}$. Do the math and it's $135^{o}$. So we know that the middle has to turn $(135+k270)^{o}$, for integer k, in order to result in the prescribed position. Expiriment with a few k's and you get $\longrightarrow\boxed{E}$.[/hide]

[hide="45"]let the semicircles be tangent at $T$, and the large semicircle intersect DC at $P$ and $Q$ then $EC=ET=x$, then $x^{2}=EP*EQ$ by power of a point, $PQ=2\sqrt{45}$, it is easy to get $EP=9-\sqrt{45}-x$ and $EQ=9+\sqrt{45}-x$, use the equation to solve: $x=2\implies \boxed{E}$[/hide]

[hide="46"]this is a good ol' circle packing problem...i think the best arrangement is:

X
.  X
X
.  X
X

where the left row is tangent, you can easily get $2+\sqrt{3}\implies \boxed{B}$[/hide]

[hide="48"]let us suppose the middle gear rotates x degrees, then the top gear rotates 8x/6 degrees, and the bottom gear rotates 8x/10 degrees

we need the smallest x such that both gears rotate a 180+360n degrees, it is easy to verify that $\boxed{E}$ is the answer...[/hide]

[quote="bigboy82892"]POST #100!!!
Draw a line connecting G and D. The top triangle is $\frac{1}{4}$ the area of the big triangle, so it is $27$. Reflect across $\overline{GD}$ and you have a quad. with area $54$. Cut it out and you are left with $54$. [b]Notice that the remaining gray triangles have the same base and height.[/b] Their combined bases are equal to each of the white triangle bases. So the area of the remaining gray bases are $\frac{1}{3}*54=18$. Subtract $108-18=90$ and you get $\rightarrow\boxed{E}$.[/quote]
I don't think that assumption is correct.

Sorry for triple posting,  :oops:  :blush: , but this will be the second-to last.
[hide="46."]We know we can fit $3$ circles on the side, giving us width $2$ so far. Now, we need to fit in $2$ now circles. The best way is to fit it tangent to $2$ circles. Draw lines connecting the centers of the circles that look like 8o, and you have an equilateral triangle whose base is $1$cm away from the side and the opposite angle that is $1$cm away from the other side. The height of this triangle is $\sqrt{3}$ since the sides are length $3$, so the answer is ${\rightarrow\boxed{B}}$.[/hide]
[hide="31."]Draw lines from the centers to their respective tangents and to each other. You'll get two trapezoids, but let's focus on one. It's a right trapezoid. Draw a perpendicular line from the center of the smaller circle to the line connecting the tangent to the center of the other. The resulting right triangle has hypotenuse $8$ and leg $4$. We want to find the other leg so do the Pythagorean Theorem and you get $4\sqrt{3}$. Multiply by $2$ so you have $8\sqrt{3}$. The only answer choice that has "$+8\sqrt{3}$" is $B$ so the answer is $\longrightarrow\boxed{B}$.[/hide]
EDIT(To save post count):@rnwang:OOPS! :oops:  :blush: I'll change it.

These are the answers I got tell me if they're right:
[hide]
1. E
2. E
3. B
4. D
5. D
6. D
7. C
8. D
9. C
10. C
11. D
12. C
13. A
14. D
15. B
16. D
17. E
18. B
19. E
20. C
21. D
22. D
23. D
24. A
25. The question doesn't make sense.
26. A
27. B
28. C
29. A
30. D
31. B
32. Too hard for my puny brain.
33. A
34. B
35. E
36. C
37. D
38. B
39. A
40. D
41. A
42. C
43. C
44. B
[/hide]

I'll do the rest tomorrow.

So 27 and 32 are the only problems that haven't been done.

I'm trying to figure out 32.

edit:

[hide="partial solution, you may look in here"]We can use the triangle law to figure out if ABC is right, obtuse, or acute:

10^2+6^2=136<144, so the triangle is obtuse. Therefore, $\angle EBA<90$

Since $\angle ABC$ is obtuse, then $\angle BAC<90$ and $\angle BAD>90$ So we are dealing with two obtuse triangles here:

BAD and CAB.[/hide]

32
[hide="hint"]draw a parallel through D to BC, extend lines, and use similar triangles, (that are created not only by the parallel line, but using the congruent angles)[/hide]

27
[hide="hint"]just use teh sine area formula, and some similar triangles...[/hide]

Wow thanks guys. I printed your solutions by cutting and pasting onto a word document and located AoPS Nice Helper's Corrections that I need to Look at that are Elegant.  :lol:

I finished 2002 and 2005 and 2006 Statewide corrections. I guess I will do this 2003 one tomorrow.  :lol:

[hide="progress on #27"]ABC is similar to AEF. AEF's sides are 1/2 of ABC's sides, so the area of AEF is 30.[/hide]

[hide="27"]
We know that HIJ and EFJ are similar. Since EFJ's base is 1/2 of ABC's, and HIJ's base is 1/3 of ABC's, the ratio of EFJ's sides to HIJ's is 3:2. We also know that the height of both triangles add together to 1/2 of ABC's height. Since HIJ's height is 3/5 of HIJ's + EFJ's height, it is 3/10 of ABC's height. It's base is 1/2 of ABC's base. Therefore its area is 3/10 * 1/2 of ABC's area. 120 * 3/20 is 18.[/hide]


Prove $ x^2 \plus{} 5 \equal{} y^3$ has no integer solutions.

this problem was proposed on bulgaria

so it has surely a simple solution.


After a short discussion on another thread, I began wondering... what is the political swing of this forum? So, I'm creating a poll just to see. It has five options, which are to be seen as gradients from most liberal to most conservative. Please choose the option that represents your political views [b]overall[/b]. Post if you would like to elaborate on why you chose a certain view. Thanks!

I tend to swing to the left.  Often far to the left.

Take the political compass http://www.politicalcompass.org/

I'm economically in the middle and socially libertarian.

According to the compass thing, I'm pretty economically left and somewhat libertarian.

[quote="shobber"]How do you define those different views??[/quote]

What are your views on the war? Gun control? Homosexual marriage? Abortion? Education... how should it be funded? Homeland security? Budget control? Immigration? Welfare? Tax reform? What are your principles and values that you bring into a political arena?

that's a cool test. according to it, i am a leftist libertarian. but i already knew that.

I think it is wrong to be right or left. It is correct to be right in your decisions for each issue, you cant jsut be right or left. I think in most decisions, there is a right answer. So it occurs that one side is always wrong made not in all instances but a lot of them and your decision souhld be based on the topic. Sorry for typing errors, I am lazy to correct them.

Hmm... I agree that one should not be a puppet--just because one is a liberal doesn't mean he/she has to have liberal views on every issue.

From the test, I found out (but I already knew) that I am a hard-core liberal.  Still, I don't have liberal views on all subejcts.  For example, I'm pro-death penalty in theory.  I'm anti-death penalty in practice because in the South, where most of the nation's executions take place, many people are sentenced to death because of their race.  This is not a stereotype or an insult--I have nothing against Southerners at all--they're my fellow countrymen--it's just that it's a known fact that the death penalty is applied unfairly.

that's why the words "tend" and "overall" are capitalized and bolded...................

Libertarian, according to the political compass test. But in the middle for the "right - left" scale.

It's kind of weird, cause I put "agree" and "disagree" for most, but then for certain things, especially regarding religion or certain popular "phrases", I would put "strongly disagree". Interesting test though...

According to the test, I'm a leftist libertarian.  What a surprise.

hmm... leftist libertarian, just like everybody else on this forum, it appears. Almost exactly at the same spot as the Dalai Lama and directly opposite... Bush  :furious: .

Can somebody please explain why anyone would be proud to be on the same side as Hitler and exactly opposite from Gandhi??? It just doesn't make sense--don't they see something wrong with their reasoning... killing tons of people with different opinions than yourself and trying to conquer the world versus trying to free your country from colonization. I'm proud to be a leftist libertarian.

[quote="jb05"]Almost exactly at the same spot as the Dalai Lama and directly opposite... Bush  :furious: .[/quote]

 :rotfl:

Well, not everyone who thinks like Hitler will do what he did... ;)

What happened to all of the Bush supporters? Come now, I have seen a LOT of conservatives on this site... mostly tweener's from Texas... I think a bunch of them go to school together... where are they?

Perhaps they don't want to post where everyone else is saying how they are all leftist libertarians.

Although there are many, many conservatives on this site, I'd say that there are more people with liberal views than conservative views.

Keep in mind that this political compass test is [i]not[/i] scientific. I just thought that it was interesting.

I'm
Economic Left/Right: -6.25
Social Libertarian/Authoritarian: -5.79

[quote="MathFiend"][quote="political compass"]Your political compass
Economic Left/Right: -4.38
Social Libertarian/Authoritarian: -0.87 [/quote]

now....could someone please explain that to me? i'm not big on politics and am just a little bit confused... :)[/quote]

The page they gave you when it showed you your results had explanation, but I can do a little for you:
On the Social Libertarian vs. Authoritarian, the relevant question is "How much control over the individual lives of people should the state excercize?"  So, for example, my -5.79 to your -0.87 shows that I am much more libertarian than you, at least according to this test: you probably are less worried about the government impinging on your civil rights, for example, and more likely to feel pornography should be illegal than I am.  (I'm just picking example questions that would have moved us along that particular axis.)

In the economically left-right portion, we're both rather far left.  We both probably feel that individuals trump multinational corporations and that the social safety net is a good thing, for example.  I imagine that being protectionist would push us left, although that's somewhat questionable as generally protectionists fall in both camps.  (Pat Buchannan, for example.)

Oh, one other thing: all of you who labelled yourself as "Left," what exactly do you imagine "Left" means?  Ted Kennedy?  Bill Clinton?  For most of the twentieth century, "left" referred to a place well on the far side of those two, much more along the lines of Socialism or Marxism.

Hmm...I'm in the first quadrant. :-) A lot closer to the x-axis than the y-axis though. :-)

[quote="shobber"]How do you define those different views??[/quote]

You can use an easy way to see if you are left or right. We often definite communist as far left, see how many of your viewpoints is similar with communist , then you can know your political views.

According to the survey I'm almost exactly in the center. i suppose I must be moderate then. Put I tend not to like anyone politicians, becuase barely any politicians today are well educated. Politics used to be about who was best for the job but now its just about money and influence making whoever runs our country incapable to to so. Personaly I would opt for a deferent set of guidelines to those who run for president. Something along the lines of education and things like that.

I have a lot of strong liberal views. I often though find "liberals" to be complete idiots in their thinking. I have a lot of strong conservative views. I often though find "conservatives" to be complete idiots in their thinking. Should I take a mathematical average and say I'm moderate? (even though I have not many moderate views). I typically make up my mind based on the issue at hand, as I think everyone should.

Well I think this explains it:

Economic Left/Right: 0.00
Social Libertarian/Authoritarian: -0.05

I'm nothing! Yay, right where I want to be.

Economic Left/Right: -4.25
Social Libertarian/Authoritarian: 0.82

I stand on the far right side except economy(far left)

Economic Left/Right: -3.00
Social Libertarian/Authoritarian: 1.74

im very far to the right on the economic scale and a little more authoratarian than libertarian on the social scale. I consider myself to be mostly 'right', but im left on some things.
EDIT: so according to that thing, im authoritarian left.


real analysis
algebra
polynomial
integration
function
advanced fields
advanced fields unsolved

Let $ \mu$ be a finite (in general, complex-valued) Borel measure on the unit circle such that $ d\mu \perp d\theta$ ($ d\theta$ lebesgue measure on unit circle). Then there are polynomials $ p_n$ such that $ \sup_{\overline{\mathbb{D}}}{|p_n|} \leq 1$, $ \int p_n \, d\mu \to \parallel{}\mu\parallel{}$ and $ p_n \to 0$ $ d\theta$-a.e.

There is a hint outline that I do not understand how to follow, probably because I am not well versed enough in measure/integration. However, I will sketch a proof I have constructed when $ \mu$ is [b]real[/b]-valued. I do not understand how to carry it over to the complex case though, and that's where I need help.

Take a Hahn decomposition $ A \cup B$ for $ \mu$. Since $ d\mu \perp d\theta$ we can find $ A_1 \subset A_2 \subset \cdots \subset A$ and $ B_1 \subset B_2 \subset \cdots \subset B$ such that $ A_n$, $ B_n$ are compact and of lebesgue measure zero, such that $ \mu$ is supported on the union of all $ A_n$ and $ B_n$. By the construction of Fatou/F & M Riesz there are functions $ f_n$ and $ g_n$ analytic in the disc, continuous in the closure, and $ f_n$ is equal to 1 on $ A_n$ and $ g_n$ is equal to -1 on $ B_n$ and both have modulus strictly less than 1 [b]everywhere else[/b]. Then if we take some high odd powers $ f_n^k$, $ g_n^k$, then $ \int f_n^k \plus{} g_n^k \, d\mu$ will be close to $ |\mu|(A_n \cup B_n)$. Approximating this continuous function $ f_n^k \plus{} g_n^k$ with polynomials I think we'll be almost done.

As I said, I don't understand what to do when $ d\mu \equal{} h \,d|\mu|$ and $ h$ assumes complex values not just $ \pm 1$. I mean, I see one approach in having a "Hahn decomposition" with infinitely many components, corresponding to preimages of $ h$, but this seems messy and difficult to carry out...

To complete my solution above, one has to pass to a subsequence to get convergence to zero a.e.

Anyway, for the general case, it turns out there is a theorem by Rudin and Carleson that given a continuous function on any compact subset of the unit circle with measure zero there is an analytic function in $ \mathbb{D}$ extending continuously to the boundary with exactly the values of the continuous function on the compact subset. The proof is not too hard given the functions I used in the proof above, but still too tricky for me to find on my own.


As dori sa stiu daca vor mai fi unul sau doua teste de selectie la juniori. De asemenea as dori o confirmare a perioadei de pregatire (3-10 iunie).

Din ce am auzit eu, vor mai fi inca 2 teste, iar in legatura cu data pregatirii ai dreptate.

[color=blue]Se mai dau inca 2 baraje pentru Juniori... Perioada e cea pe care ati zis-o voi... Succes![/color]


If $ X \sim \Gamma (\alpha, \lambda)$, then show $ cX \sim \Gamma (\alpha, \frac {\lambda}{c})$

Here is what I did:
\[ P(X < \equal{} x) \equal{} \int_0^t \lambda e^{ \minus{} \lambda t} (\lambda t)^{\alpha \minus{} 1} / \Gamma (\alpha) dt\]

\[ P(Y < \equal{} y) \equal{} \int_0^y \lambda e^{ \minus{} \lambda y} (\lambda y^{\alpha \minus{} 1} / \Gamma (\alpha) dy\]
Let Y = cX

If
\[ cX \sim \Gamma (\alpha, \frac {\lambda}{c})\]
then
\[ P(cX < \equal{} y) \equal{} \int_0^y \lambda e^{ \minus{} \lambda t/c} \frac {\lambda t}{c}^{\alpha \minus{} 1} / c \Gamma (\alpha) dt\]
and
\[ P(X < \equal{} \frac {y}{c}) \equal{} \int_0^{y/c} \lambda e^{ \minus{} \lambda t} (\lambda t)^{\alpha \minus{} 1} / \Gamma (\alpha)\]
Where did I make a mistake? They're not equal; the first statement has an extra 1\c

So suppose I did a change of variable.
\[ \tau \equal{} ct, t \minus{} > \tau/c, d\tau \equal{} cdt\]
$ P(X < \equal{} \frac {y}{c}) \equal{} \int_0^{y} \lambda e^{ \minus{} \lambda \frac {\tau}{c}} (\frac {\lambda \tau}{c})^{\alpha \minus{} 1} / c \Gamma (\alpha) d\tau$

I think it is because you were to sloppy when you said if $ Y= cX\sim\Gamma (\alpha,\frac{\lambda}{c}) ...$ you should get by plugging in $ \lambda/c$ for $ \lambda$ in $ P(X \leq x)$

\[{ \frac{1}{\Gamma(\alpha)} \int_0^y  \frac{ \lambda }{ c } e^{\frac{-\lambda t}{c}} (\frac{\lambda t}{c}})^{\alpha - 1} dt\]

Which is precisely what you get for your change of variables.

This is probably easiest to do via the MGF:

Let $ Y \equal{} cX$
$ M_Y(t) \equal{} E[e^{Yt}] \equal{} E[e^{ctX}] \equal{} M_X(ct) \equal{} \left(\dfrac{\lambda}{\lambda \minus{} ct}\right)^\alpha
\equal{} \left(\dfrac{\lambda/c}{\lambda/c \minus{} t}\right)^\alpha$

and this is the MGF of a Gamma$ \left(\alpha, \frac{\lambda}{c}\right)$ random variable


Does anyone have the answer key to the 2004-2005 MAML answer key? Thanks!

you can find them on the wocomal website, but for some reason, that set does not have an answer key as the last page


Prove that the product of four consecutive integers is one less than a perfect square.

[hide="Solution"]
Let the smallest integer be $ x \minus{} 1$. So that, the product is:
\[ (x \minus{} 1)(x)(x \plus{} 1)(x \plus{} 2) \equal{} (x^2 \minus{} 1)(x^2 \plus{} 2x) \equal{} (x^2 \minus{} 1)((x \plus{} 1)^2 \minus{} 1)\]
Then,
\[ (x^2 \minus{} 1)((x \plus{} 1)^2 \minus{} 1) \equal{} (x^2 \plus{} x \minus{} 1)^2 \minus{} 1\]
Thus we are done. :ninja:

Edit: Sorry for being vague in my solution, I was in a bit of a hurry when I wrote the solution, I thought about it and I think this way is better:
\[ (x \minus{} 1)(x)(x \plus{} 1)(x \plus{} 2)\equal{}(x^2\plus{}x\minus{}2)(x^2\plus{}x)\],
Let $ a\equal{}x^2\plus{}x$, so we have to show that $ (a\minus{}2)a$ is one less then a perfect square or in other words show that $ (a\minus{}2)a\plus{}1$ is a perfect square,
\[ (a\minus{}2)a\plus{}1\equal{}a^2\minus{}2a\plus{}1\equal{}(a\minus{}1)^2\]
And substituting $ x^2\plus{}x$ for $ a$ we get,
\[ (x^2\plus{}x\minus{}1)^2\]
Hope that helps
[/hide]

How do you get from $ (x^2\minus{}1)((x\plus{}1)^2\minus{}1)$ to $ (x^2\plus{}x\minus{}1)^2\minus{}1$ without intermediate steps?


inequalities
geometry
geometric transformation
reflection
algebra
polynomial

[quote="sinajackson"]Let $ a,b,c$ be triangle sides. Prove that:

$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\geq k\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-3\right)$

where $ k=1-\frac{2}{(2\sqrt{2}-1)\sqrt{5+4\sqrt{2}}+1}$ :huh:[/quote]

Actually, this is not so strange for me since this is my own problem. I first posted it in Mathlink

http://www.mathlinks.ro/Forum/viewtopic.php?t=108053

Dear Jackson: What is this magazine?

Oh :huh:  cool!
well this is The BEst MAthematics Magazine in Iran :lol:  there was a part teachin the method of Mixing Variables for solving Inequalities and this problem waz in its Excercises :)
and the link you gave has solved it in this way

EDIT: This problem is purposed by (hungkhtn) in Mathematical reflections article about Mixing variables (Part1) and in Dastchin this paper was translated

By this method (mixing all variables), we can also prove the following inequality

http://www.mathlinks.ro/Forum/viewtopic.php?t=153536

It is the same as Jack, but unsolved in Mathlinks.

[quote="navid"]This problem is purposed by (hungkhtn) in Mathematical reflections article about Mixing variables (Part1) and in Dastchin this paper was translated ,Please modify your post Sina.[/quote]

Ok! :wink:


geometry
Stanford
college
calculus
MATHCOUNTS
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blogs

1st period, 8th grade PE: Oh yes I love PE. Walking into the boys locker room gives the pleasant aroma of the football players. *sniff* ahh Febreze!!! Thank god I didn't end up with a locker near them... Oh boy I forgot my locker combo. I have to wait outside the "man-shed" for one of the teachers to unglue their eyes from the computer screens... I do my cruel 20 push ups (with my PE teacher, I would round it up to more like one million) rush to my locker throw on my clothes and head up to the weight room to watch the announcements so poorly done by our students. Then down the stairs to run in an endless circle around the gym, great hulking bodies of the 8th graders shoving me back and forth. After about 30 minutes of running and stretching, out to the field to play the [sarcasm] great game of lacross. [/sarcasm] Then the loud shriek of the whistle blows us in to the locker room's toxic odors of the locker room. Then out to science.

2nd period, 7th science: Oh boy this is one of the okay periods. A fun teacher who likes to show us a picture of something that looks like a black blueberry with a black head... the picture says Why Not to Fart in a Wetsuit... and after a couple courtsy laughs, on to cells. *snort* like I need to know what a Ribosome is... (It makes protien for cells) the bell rings, then off to LA/

3rd period, 7th LA: Oh yes I love Language Arts. We get to talk about our feelings of some boring story written by some boring guy who lived in an cave his whole life, writing that book. Also my teacher like to talk on and on and on about how generous she was to give her ex the boat, or her sons going to college or her dog Vinny... ding ding ding, period over.

4th period, 7th SS: Dang, same teacher... even the same class to talk to... more comments on her dog Vinny, and we do our LA stuff in SS!!! Of course there's the occasional who exported more wheat? USA or China packets... Exciting.

Lunch: Yes I love lunch. I can't turn my head for more than half a second without my lunch bieng dismanted and rationed out to everyone in the table. Then you have to negotiate and soon you come with a sort of depleted lunch. Yum.

5th period, 8th Math.: Oh yes another great class with 8th graders that are approximately 1 foot taller than me. Of course there are a couple migits but oh well. I love our course. We get to talk and disscuss and on and on and on. And oh yeah, we get to make cute litte bar graphs about automobile ratings... then we dissucss more and more and talk more and more in our cute little groups where we are assigned roles... :|

6th period, 7th Orchestra: Dang I was supposed to have 8th grade orchestra but whatever... our tuning process takes about 30 minutes and whoever uses the school cello before me needs to learn not to basically take the C string off every day... and oh yes Wednsdays. Sectionals. I have to teach about the most talkative kids in the whole school how to play a sting instrument... not fun. :(

After school, Cross Country: Hey not so bad. I feel my endurance raising and actually got 2nd place last race... no applause there. Just run and do core excersises and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run and run.
Love it.

half day, saturday schedule. which means ACHE is the order. ><

A block - SLEEP IN!!!! woohoo!!!
C block - turns out freshman PE was cancelled because it's saturday schedule!!!! yay! no. i practiced violin for my violin lesson.
BREAK - violin lesson. oh goody. it was interesting. not to mention that i'm not very good.
H block - math. made the stupidest mistakes anyone could ever make in the entire world. *shoots*
E block - photography. today we were in the ceramics studio making our Artist Trading Cards. we played with stamps and all sorts of fun artsy stuff!! XD

Our school was flooded but they didn't call it off. So we spent half our morning in the cafeteria as our alternate classroom.

Power outage from random power failure. We sat in first hour texting...best hour ever! :D

1st period- acc. bio- tests back. yuck 91%
2nd- french 3- sophs thought i was a soph dressed as frosh... but i'm really a frosh... anyway... that's what happens if ur in a soph french class...
3rd- frosh girls pe- we went outside and ran and power walked. it was COLD!!!
4th-lunch- lunchy
5th- acc. eng- uhh media center. worked on speech.
6th- art- creepy sr. stalker!! worked on shading project
7th- acc. geometry- errrrr i don't remember. boring ness
8th- dorkestra- teacher not there. led 1st violin sectionals. go section leaders!!

hmm after school i went 2 french club. it was fun we worked on homecoming golf carts =]

Yippy, fun fun fun.

AB Block (don't ask why): English - Yay! At least the English teacher was $ x$ times better than last year's (8th grade) English teacher... with $ 1000 < x < \infty$.

B Block: French - Fun! I get to complan why my teacher doesn't teach relative pronouns already! ("La salle [b]dans laquelle[/b] j'ai fait mes devoirs...", "Le poisson [b]auquel[/b] j'ai donné le poison..."). Instead, I get endless pictures of people who "s'habille" and "se déshabille" and who "se lève" and "se réveille"... and my homework largely consists of copying words three times in a row.

But a great improvement over middle school French!

C Block: JAVA Programming! Except the teacher couldn't find the test that he meant to give me, and I had already completed all the exercises up to the middle of Quarter 2... so yeah, I'm not really doing Java anymore. I'm doing C++/UNIX/Researching wobbly tables.

Lunch: Nothing to say here.

CD Block (again, don't ask me why): AP Physics! The first actual challenging course! Everything's good :)

E Block: History. I get nine concurrent assignments that are due over the next five weeks. Big deal.

Other days:

A Block: Fitness concepts - Stretches + mile except on Fridays, where we play a game. Actually involves math! How many times do we get to play a fun game during the year? And we also did the target heart rate too. (Not that that was fun... simple arithmetic... fun to watch other people struggle xD)

Other A Block: Film analysis! Watching movies! Yay! :D

D Block: Online, M52A on Stanford = Multivariable Differential Calculus. I'm in the computer lab with 5 others who are taking this course as well - 4 seniors and 1 junior, and a freshman (guess who :D)

X Block: Robotics half the time and Chess the other half.

hmm my school has block schedule so i have odd classes on one day, and even the other day..
todayy wasss oddd.

1st- open period. stayed in library. goofed on aops website for a little. then studied for bigg ass honors spanish 4 test.

tutorial: such useless time but whatevs. practiced violin for orchestra seating auditionssss..

3rd: span 4...shareddd our original story [which had to be put together in a book form..] freakin 8 PAGES LONG@!!@@ AHHH. crazy hardcore class mannn. soo hardddd. our class average right now is like a D. i swear.

lunchh: went off campus to chipotle...thenn violin practice again.

5th: philharmonic orchestra. practiced. auditioned. BLEW MY AUDITION. so im assuming that ill get like last chair as opposed to...5th last year. whatever. talked texted prankcalled my boyfriennndd and left a really weird voicemail xD thenn took gazillion pictures with my frienndss

7th: math EW EW EW. i hate school math. so easy though. i guess that's good?
lol

today was horrible. i didnt sleep last night to finish all my workkk
:x yikkes

Block schedule, periods 1-4.

Period 1: Spanish-- We finished basic verbs and we're starting to go into the culture of Spanish-speaking countries. We have to learn the capitals of 20 countries.

Period 2: Algebra II-- Sometimes taking math online can be a bit of a nuisance, especially when you're using the English room and there is a substitute today. I had to explain EVERYTHING to him before I could do any math.

Period 3: PE-- Gah I HATE PE (except on Mondays, where we get to play  :ddr: ) because I'm not good at anything (except  :ddr: ) First we ran .75 miles, then we played volleyball. My team got CRUSHED.

Lunch: Nothing much to say here.

Period 4: Name That Book-- Well, only here because I have to (no other good elective that is not full.) Just reading, reading, and more reading for 90 mins.

Homeroom: Just a study period.

DISMISSAL WOOHOO!

At least not much homework.

Best day ever today:

Period 1 (Precalc): Our math class is still flooded so we went to an alternate classroom and I got to finish my chem notebook that was due. We had math homework but I didn't do - she never collected it though so I was saved.

Period 2 (English): Quiz and fun crossword.

Period 3 (AP World History): Didn't do my document questions but thankfully she didn't check them.

Period 4 and 5 (AP Chemistry): Went over a packet but had 30 min of free time at the end to talk with the ladies  :P

Period 6 (AP Stat): I just sat there and pretended to listen.

Period 7 (Spanish): We had homework due today but we had a sub of a sub (our usual sub is another spanish teacher but she also wasn't here so we actually had a sub) and the person couldn't find that directions and we just had a free period at the end of the day.

Man I got really lucky. The three assignments that I didn't do weren't checked!

After school: Call of Duty 4 on the Xbox and I got my Red Tiger Camo for my AK-74u  :) . Next up is the G36C and the Barrett 0.50 cal sniper!

Also, I'm going to move this to the Games forum, since this topic isn't that bad for the Fun Factory.

1st period: Design and Tech. People in my class don't understand concepts at all, and we are maybe a week behind other Design and Tech classes.

2nd period: English. BORING and UNEVENTFUL.

3rd period: Biology. Basically like English, but easier.

4th period: PE. I don't like PE at all. We run around in circles. We also play field hockey. With mouth guards. I hate mouth guards. And if the choice is between inhaling noxious fumes than and getting hit in the face with a stick, then why do we even play sports at all?

5th period: AP Calculus BC. I'm in a class with only one other 9th grader, and everyone else besides us is getting D's and stuff.

6th period: AP Computer Science. Basically just programming. I'm 2 programs ahead of everyone else in my class. I would actually be 5 programs ahead, but A LOGIC ERROR prevented that...

7th period: Latin 2. Really strange people in this class...

8th period: Well, at our school, we sign up for activities for 8th period. I usually go to Math Team/Certamen (Latin for "contest" or "struggle") or Computer Team/Electronics Club or World Domination Club (No, we don't actually dominate the world, we actually play the board game Diplomacy)

Oh, and don't ask me why I'm in Calculus or AP Computer Science.

(Yongyi781, how are you in AP Physics????????????)

dang

A: HAH THERE WAS NO A-BLOCK TODAY = NO ENGLISH

B: Orchestra. So I brought my violin home on a weekday for the first time all year yesterday to practice for a quiz on a part that I had read like once. Turns out I didn't actually need it. So the way quizzes work is that you listen to one row at a time. I'm in the second row of second violins, with 2 stands (actually I normally sit farther back because I suck but just for this piece). One of the kids is really bad and didn't know it at all, my stand partner was absent, so it was only two of us playing. Then the teacher said we were playing too softly (I think?). Fun start to the day.

C: Spanish. Dang I've never seen our teacher in such a good mood. He basically passed back quizzes for the first half hour of the block. I think my grades were 94, 100, B-, 100, 83, 85. The fourth was a trivial thing that everyone got 100 on, and the last two was some ridiculous thing that everyone failed. And we went over that. And we argued over answers. Fun stuff.

D: AP Biology. I love this class. So we had this quiz where you draw a fat (as in a glycogen + 3 fatty acids). That was pretty fun. Then we talked about some stuff that I don't remember. But yesterday's class was particularly memorable as we spent the last 20 minutes or so discussing steroids. Woo.

E: Study. I read my history textbook.

Lunch: I ate lunch? Dang I need to find sit with different people...lunches suck this year. All the people whom I sit with, who aren't actually my friends (my actual friends are in the Asian cafeteria which I've never eaten in for some weird reason) talk about drugs. Good stuff.

F: Another study. I read my history textbook.

G: AP Calc. Test on...derivatives? I finished around 20 minutes early. Then I read my history textbook. AND I STILL HAVE LIKE 3 PAGES LEFT IN THE CHAPTER WTF.

H: AP World History. We discussed Palin-Biden for some weird reason for like 15 minutes. Then we quickly did a review of River Valley Civs. I think? I wasn't really listening. It was the last block of the week.

that was pretty pointless...

1st - 3rd period:

BLOCK WITH MY EVIL BLOCK TEACHER WHO DECIDES TO TAKE AWAY OUR BREAKS 999/1000 OF THE TIME!  NOT FAIR

Rating: 1/10.  Woulda been 0 if i didnt have friends.

4th period;

LUNCH

Rating: 8/10, dont have that many freinds in 4th lunch, most are in 5th

5th period:

spanish quiz

Rating: 4/10, easy quiz rofl but stupid teacher who doesnt really know spanish

6th

science
we had a sub for our uber great and funny teacher buti got a laugh out of two people getting referaals

6/10

7th

pe rofl thats fun!

7/10

8th
boring math, but we have mathcounts today!

9/10

OOH I PREDICT THIS WONT MAKE ME FEEL BETTER AT ALL.

1st hour: 8th grade PE!!! Dress out. I have a CORNER PE LOCKER THE WORST THING THAT CAN HAPPEN TO YOU. I have a pe locker (maybe 10 in x 10 in x 1 foot. Size 14 mens dont fit well in these things.) with about 6 people within 2 feet of me (not joking) and we have nearly taken out eyeballs dressing out. To top it off, the guy directly next to me is SUPER sweaty. Like even before pe he is dripping, im certain he has a medical issue. After pe?  **cringes** i dont want to talk about it. Dodgeball= Schweet its time to hang out with my emo friends in the back of the gym and talk about anything. we're doing a dub on saturday i might give you guys the link. I havent sweated in PE since 6th grade :D

2nd hour: The TV show. Woot script today. This is my fave class because my teacher is like a 60 year old lady who will talk "Lip Gloss" for us and lately we've made it our mission to get her to say the word "crack". Mission failed today. Also every one in this class is HILARIOUS and so fun.

Assembly: WE ARE SO PROUD OF YOU FOR DOING THIS FUNDRAISER WE ARE GOING TO MAKE YOU DO 2 MORE FUNDRAISERS. (I never do fundraisers.) They stretched this between 6 speakers for 1 hour.

3rd hour:  ADV Science: we were SUPPOSED to have a debate on drilling in alaska today, but the assembly actually ran over... we did nothing but talked.

4th hour: Tech Ed. Graphic Communications Module. Hehe my friend gave me a Nintendo DS lite today for an early bday gift. My mom will make me give it back.

5th hour: Alg. II. I actually took the official Mercury RCML yesterday and today i took the Discovery for fun :D

Lunch! Pizza!

6th Hour: ADV SS. Worked on a Mixbook. [url=http://www.mixbook.com/books?bid=2480009]HEHE[/url]

7th hour: ADV english Worked on a photostory.

Wow. I feel better.
AW MAN ITS FALL BREAK! I WANT TO DO THIS AGAIN AND I WONT GET TO UNTIL NEXT WEEK!

I ultimately feel worse.

1st- bio- lab on metric system
2nd-french3- quiz on subjunctives
3rd- gym- human Foosball!!! very fun =]
4th- lunch- uhh sandwich/ go 2 practice room to practice violin
5th- eng- uhh worked on speech for outline and got beaten up by my eng teacher cuz i told him he wasn't important XD
6th- art- cut out stuff with an x-acto knife
7th- geometry- ezpz math quiz
8th- dorkestra- play stuff
pep assembly- very loud screamy stuff. powder puff cheerleaders were hilarious

after school- pit dorkestra- eh. boring. made me too tired 2 go to homecoming football game

tomorrow- HOMECOMING DANCE!!!

for gum, certain teachers care. most don't.
but they DO yell at you if you have sweatshirts on. -_- (its not dress code b/c it's "distracting")

yesterday was an okay day. I screwed up on my photography flipbook so I have to do it again.
Our Junior Studies Presentation uber failed.
I screwed up in math class.
Physics project an A- because one girl didn't do anything so she decided to lower the whole class's grade instead of just giving her a D.

What about a monochromatic shirt? Do they yell at you for wearing that?

To tell you the truth my tie is more distracting than my sweatshirts except for one.

Wait, you wear a uniform?

We can wear w/e we want pretty much.

Well... We have a dress code. We have to wear a formal-ish shirt. a shirt with no writing. or a polo. I wear a polo every day.
you can wear a jacket, but no sweatshirts. which is just stupid. I don't understand why sweatshirts are distracting.
and you can wear fingertip length skirts and jeans and flip flops and stuff like that...

but i feel bad for the guys. They have to wear a button-up shirt with a tie every day. and nice pants. :D

lol...

my school is straight out stupid. here's my day:
Math- slept
Social Studies: was blowing my nose like half the time (i was sick yesterday) so i didn't listen to the video
Science: Stupid idiotic memorization quiz
Lunch: Sat there and talked to my few (unmathy as possible) friends
Language Arts: OMG MEMORIZATION QUIZ
P.E.: stupid baseball with a batting order that no one ever follows
Health: filled with stupid people who don't the meaning of "replicate"

French - Ew, test. First page was easy, second page was easy, third page was sort of easy, last page was murder. I'm pretty sure I got a 100 though. Finished with 10 minutes to spare, so I did my french homework.

Social Studies - Yuck, another test. It was really easy, so that was good. I got to read for about 25 minutes.

Computers - Worked on my Photoshop project. Amazingly, it's actually turning out pretty good.

English - Yay, I got a 100 on my test. It was uber-easy...my teacher gave you like 6 lines to write a whole story. I used 6 pages(front AND back!). Then we worked on our essays. I finished my rough draft and started my english homework.

Lunch - Did my math homework, did my english homework, ate, read.

Science - Test again. All easy, except for one question. I don't know if I got it right, but even if I got it wrong I still get a 97, so it's okay.

Math - Doodled a lot. Finished this really easy worksheet in like, 10 minutes, so I got to read a bunch too.

Drama - Hate this class. We had to watch this black-and-white movie called "It's A Wonderful Life." I was bored to tears. We had a sub, so my friend and I snuck in the back of the room and talked.

Woodshop - Finished my project. It stinks. My average is only a B+, and I need at least an A- to stay on high honor roll.

[quote="AIME15"]Oh boy...failed the Pacer test today.[/quote]

I did that a week ago. How many did you get? I got 12. :(  :(  :(

12? wow. it's okay tho. I know people who got to 8. -_-

anywayy..

today was sorta messed up...

A free. sleep in. ^^
B spanish. Got tests back. made stupid mistakes.
C D free because we dont' have PE.
Lunch - Violin lesson. so i ate lunch during D.
E photography. woww.. i hate my teacher...
F junior studies. class outside. worked on narratives.
G physics. got 19/20 on a lab because we didn't center the cup right. (Roll a ball down the ramp and place a cup so that the ball will fall through the cup.)
H math. talked about quadratic equations and stuff of that sort.

me get 61 on pacer test
thats all we have to do

bio- cell test that i studied a lot for so it was ez
french- test back. i made one stupid spelling error  :mad:
gym- power walking and some weird game where we all lay on the floor of a tennis court and there's a LOT of yucky goose poop there
lunch-eat lunch and get ppls asking me questions about who i went to homecoming with...which was like a MONTH ago
english- continued getting questions about homecoming... and watched raisin in the sun
art- started my final shading project
math- umm something... i don't remember
orchestra- played stuff.

after school- seating audition for orchestra. fail.
after seating audition- post this post  :P

[quote="AIME15"]I got 100 on pacer :(

Some kid tied with me...epic phail.  For a guy who runs cross country.[/quote]

How long is each lap? 5 meters? Even the most athletic kid in my school only got around to 70. :|
But our lap is 15 meters, I think. (Yeah, ebilness)

i used to be considered very athletic in india where many people are vegetarian. But, once i moved here, everybody ate meat and got more nutrition so i fell behind in athletics and took up math as an activity (i was good a academics too) and that's how i got started at math. :)

Our lap is almost one basketball court long.


In a multiple choice test there are 3 questions: For the first question there are 3 answers. For the second question there are 5 answers and for the third question there are 4 answers. Suppose that the candidate answers the 3 questions (without knowing the correct answer). Find the value of E(X) of the correct answers.

[hide]Let $u(x)$ be the probability that an event $x$ happens and $f(x)$ be the event (in this case, picking one right answer).  Then

$E(X)=\sum_{i}u(x_{i})f(x_{i})=\frac{1}{3}\cdot 1+\frac{1}{5}\cdot 1+\frac{1}{4}\cdot 1= \boxed{\frac{47}{60}}$[/hide]

$\textrm{elementary probability}\not\subset \textrm{Advanced fields}$.  This belongs in one of the high school forums (getting started or intermediate).


number theory
prime factorization
prime numbers

A positive integer $x$ is [i]tivisible[/i] by a positive integer $y$ if the number of factors of $x$ (including 1 and itself) is divisible by the number of factors of $y$ (including 1 and itselft). How many positive integers less than $1000$ is $1000$ tivisible by?

hey but $1000=(5^{3})(2^{3})$ so the number of divisors is $16$ and 2 divides 16 and each prime has exactly 2 divisors. So we have to count the number of primes less than 1000???

this is one of those problems where you can use lots of resources.

You have to count all numbers that have 1, 2, 4, 8, or 16 factors. Quite a few.

Yeah. I was hoping someone would see it and come up with a nice way to do it.

Let $p_{i}, p_{j}, ...$ be distinct primes.
1 factor: none.
2 factors: all primes ($p_{i}$).
4 factors: all numbers in the form $p_{i}^{3}$, or $p_{i}p_{j}$.
8 factors: all number in the form $p_{i}^{7}$, or $p_{i}p_{j}^{3}$, or $p_{i}p_{j}p_{k}$.
16 factors: all numbers in the form $p_{i}^{15}$, or $p_{i}p_{j}^{7}$, or $p_{i}^{3}p_{j}^{3}$, or $p_{i}p_{j}p_{k}^{3}$, or $p_{i}p_{j}p_{k}p_{l}$. :D

[quote="lotrgreengrapes7926"]1 factor: none.[/quote]Did you count 1?  :P

[hide="partial solution"]The prime factorization of $1000=2^{3}*5^{3}$. There are 16 factors in 1000. We are looking for:
a) 1
b) prime numbers less than 1000
c) numbers in the form of $a^{3}$ or $a*b$
d) numbers in the form of $a^{7}$, $a^{3}*b$ or $a*b*c$
e) numbers in the form of $a^{15}$, $a^{7}*b$, $a^{4}*b^{4}$, $a^{4}*b*c$ or $a*b*c*d$[/hide]

That's as far as I can go without getting into a muddle of casework. Sorry, guys.

Yeah, it takes forever with all the casework.  Is there someway you could do it with a combination?  I was thinking that would be easiest once you found the number of primes, but I can't figure out how to limit it to results less than 1000...

I don't think there's an easy way to do it. I think drunner just wanted us to figure out how many factors it can have and just state "find those numbers".  :P

I agree with Hokkage. If it were something simple, like 300, it would be much easier.


What day of the week will it be 40 days from Tuesday?

[hide="solution"]
$ 40\equiv5\pmod{7},$ so the day it will be 40 days after Tuesday is the day 5 days after Tuesday, which is $ \fbox{Sunday}.$[/hide]

In 6 weeks (42 days) it will be Tuesday.
Subtract 2 days.
So in 40 days it will be Sunday.


function
real analysis
limit
real analysis unsolved

8.2 Theorem Suppose $ \sum^{\infty}_{n \equal{} 0} {c_n}$ converges.
and $ f(x) \equal{} \sum^{\infty}_{n \equal{} 0} {c_n x^n} ( \minus{} 1 < x < 1)$

$ Then \lim_{x \minus{} > 1} f(x) \equal{} \sum^{\infty}_{n \equal{} 0} {c_n x^n}$

$ Pf: Let S_n \equal{} c_0 \plus{} c_1 \plus{} c_2 \plus{} .. \plus{} c_n,$
$ \sum^{m}_{n \equal{} 0} {c_n x^n} \equal{} \sum^{m}_{n \equal{} 0} {(s_n \minus{} s_{n \minus{} 1})x^n} \equal{} (1 \minus{} x)\sum^{m \minus{} 1}_{n \equal{} 0} {s_n x^n} \plus{} s_m x^m$

$ For |x| < 1,$ we let $ m \minus{} > \infty$ and obtain $ f(x) \equal{} (1 \minus{} x)\sum^{\infty}_{n \equal{} 0} {s_n x^n}$ (I DON'T get this line=( )

So can some1 pls explain the last line of the pf, why when m goes to infinity we only count the first term, is there any theorem to refer to? thx a lot  :)

Please use complete words. You have enough time to write them out; this is not the sort of instant communication in which abbreviations make sense.

This is Abel's theorem. The technique in the proof is summation by parts, which is just an algebraic identity. What do you know about the $ s_n$* that you can use to estimate this sum, first as $ m\to\infty$ and then as $ x\to 1$?

*You should be consistent about case. In mathematics, we treat $ s$ and $ S$ as different symbols, which can be used to mean different things.


find the sum of the squares of the first 32 natural numbers.

is there an easy way to do this??

in fact there is a formula,
$ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} \ldots n^2 \equal{} \frac {(n)(n \plus{} 1)(2n\plus{}1}{6}$

EDIT: oops ignore me
but there is still a proof of this that may show up on the left hand panel

the problem is 'find the sum of the squares of the first 32 natural numbers. '

which is $ 1^2\plus{}2^2\plus{}...\plus{}32^2$

formula : $ 1^2\plus{}2^2\plus{}...\plus{}n^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$

$ n\equal{}32 \implies \frac{32*33*63}{6}\equal{}11088$

answer : 11088


Let $ ABC$ be a triangle with circumcircle $ (O;R)$.let $ E$ be the midpoint of $ AB$ and donote by $ F$ the point on $ AC$ such that $ \frac{AF}{AC}=\frac{1}{3}$.Construct a parallelogram $ AEMF$.Prove that $ MA+MB+MA \leq \sqrt{11(R^{2}-OM^{2})}$
Now suppose that $ (O;R)$ is fixed.Construct a traingle $ ABC$ ,inscribed in $ (O;R)$ such that
$ MA+MB+MC=\sqrt{11(R^{2}-OM^{2})}$

You've already posted this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159283[/url]

Do not double post.  Find the right forum, and then post only once in that forum.


look at this link image:

http://img203.imageshack.us/img203/6439/area.gif

ABCD is a trapezoid

ABE area = 256 m^2
EDC area = 25 m^2

How is area of ADE and BEC ?

ABE and EDC are similar by AA (the bases are parallel, so angle CDE is congruent to angle ABD, for example); thus, the ratio of their side lengths equals the square root of the ratio of their areas; the ratio of their areas is 256:25, so the ratio of the side lengths is 16:5. Therefore, DE/EB=5/16, CE/EA=5/16. Triangles ADE and BEC both share altitudes with triangle DEC, so the ratio of the areas of the two triangles equals the ratio of the bases of the triangles. The ratio of the bases (DE to EB, CE to EA) are both 5/16, and it can be easily calculated that the area of both triangles is [b]80[/b].

[quote="randomguy64"] Triangles ADE and BEC both share altitudes with triangle DEC, so the ratio of the areas of the two triangles equals the ratio of the bases of the triangles. The ratio of the bases (DE to EB, CE to EA) are both 5/16, and it can be easily calculated that the area of both triangles is [b]80[/b].[/quote]

Um maybe I'm just not getting this but which altitudes do they share with DEC?

The perpendiculars from $ D$ to $ AC$ and from $ C$ to $ DB$.

By the way, randomguy64 and I have already posted solutions to this problem here: http://www.mathlinks.ro/viewtopic.php?t=309263

Oh, I see. That's what confused me. I didn't notice the double thread, and in different forums.


In searching for a proof of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14522]Tung Lam's problem[/url], I found the following (maybe it's known before).

[b]Exponential Rarrangement Inequality[/b]

Let $a_1\ge a_2 \ge \cdots \ge a_n\ge 1$ and $b_1\ge b_2 \ge \cdots \ge b_n \ge 0$. For any permutation $\sigma$ of $\{1,2,\ldots,n\}$, we have

\[ a_1^{b_1} + a_2^{b_2}+\cdots+a_n^{b_n} \geq a_1^{b_{\sigma(1)}} + a_2^{b_{\sigma(2)}}+\cdots+a_n^{b_{\sigma(n)}} \geq a_1^{b_n} + a_2^{b_{n-1}}+\cdots+a_n^{b_1}.  \]

:D  ... A very well known method ...:D
Let x,y,u,v be real numbers such that x $\geq$ y $\geq$ 1 and  p $\geq$ q $\geq$ 0 . It's easy to see that  xp+ yq $\geq$ xq+ yp.
So for all  "elementary inversion" in {$b_i$}, the value of the Sum doesnt increase .
But all permutation  $\sigma$ of $\{1,2,\ldots,n\}$  is a composition of elementary inversions, so we have immediately the asked inequalities .
:cool:


counting
distinguishability
number theory
prime numbers

For how many three-element sets of positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$?

$\textbf{(A)}\ 32 \qquad
\textbf{(B)}\ 36 \qquad
\textbf{(C)}\ 40 \qquad
\textbf{(D)}\ 43 \qquad
\textbf{(E)}\ 45$

[quote="Elemennop"]Wouldn't all permutations of those sets also work? That would make the answer far larger than any of them given.[/quote]
The question asks for the number of sets of $\{a,b,c\}$, but not the number of ordered pairs of $(a,b,c)$.

The official solution (which I really didn't get much out of) uses non-brute forcing method. I'm not sure how to do since I'm bad at counting, however.

Basically, you have $32$ divisors of $2310$, so there should be $16$ values that begin with a $1$, as you're just matching up two divisors at a time. For $2$, you remove $2$ as a possible factor and so are left with $16$ divisors, then you remove $1$ and $\frac{2310}{2}=1155$ because you've already counted that set, so you should have $7$. Same thing with $3$, except you also remove the single $2$ and $\frac{2310}{2\cdot 3}=385$, so you have $6$. Keep going to $11$, and you should have $16+7+6+5+4+3=41$, which is what I got. You don't need to do any that begin with $2$ factors, because the closest possible distribution of $5$ integers is $(1,2,2)$, so everything is covered by beginning with each of the $1$ possibilities.

I know I am reviving a VERY old topic, but I don't think a repost is appropriate here.  Anyways:

[hide="Method 1"]I also got 41, but then realized my mistake:
We factor $2310 = 2\cdot3\cdot5\cdot7\cdot11$.  So we have 5 distinguishable balls into 3 indistinguishable boxes.  We use a recurrence relation (see bottom of [url=http://mcraefamily.com/MathHelp/CountingObjectsInBoxes.htm]this[/url] page) and see that the answer is clearly $1+15+25$, because boxes can be empty (you really need to visit the link and scroll to the bottom of the page if you have no idea what I am talking about).

BUT WAIT!

What if two boxes were empty?  The we would have the set (1, 1, 2310}.  However, this is a set of TWO elements, and the problem requires 3 elements.  So we must subtract a set, giving $41-1 = \boxed{40}$.

Answer: C[/hide]

If you don't understand the recursion, you can do it Nathan's way, which is clarified below.  (DP so I could add it to the AoPS Resources/Contests section).

$2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$.  It thus has 32 divisors.  Let us start counting sets now:

Starting with the set {1, x, y}, we see that xy will have to equal 2310.  Since 2310 has 32 factors, we can have 32 different values of x.  However, since order doesn't matter, we need to divide this by 2 (i.e. the set with x = 2; y = 1135 is the same as x = 1135, y=2, etc.).  So we have $\boxed{16}$ sets here.

Next, we consider the sets {2, x, y}.  Obviously, xy must multiply to 1135, which has 16 divisors.  Thus, we can have 16 different values of x.  But order doesn't matter, so we divide by 2 and find 8 different sets that work.  However, one of these sets will be {2, 1, 1135} which we counted earlier.  So we simply subtract that set, leaving $\boxed{7}$.

Next, we consider the sets {3, x, y}.  Obviously, xy must multiply to 770, which has 16 divisors.  Since 770 has 32 factors, we can have 32 different values of x.  However, since order doesn't matter, we need to divide this by 2, leaving 8 different sets.  However, the sets {3, 1, 770} and {3, 2, 385} are counted here, and were counted earlier too, so we need to subtract those two, leaving $\boxed{6}$ sets.

In a similar fashion, there will be:
$\boxed{5}$ sets {5, x, y}
$\boxed{4}$ sets {7, x, y}
$\boxed{3}$ sets {11, x, y}

To find the total number of sets, we add all the boxed numbers, and get $16+7+6+5+4+3= 41$.  However, we overlooked something.  All members of a set must be distinct, and the problem asked for 3-element sets.  Thus, the set {1, 1, 2310} from the first box does not work, because it is only has two elements.  Thus, we subtract that set from our 41, leaving $\boxed{\boxed{40}}$ sets that work.

Answer: C

$2310=2(3)(5)(7)(11)$.  We want to either partition these 5 prime numbers in groups of 3 or groups of 2 with the other third number 1.  Note that $5=1+4=2+3$ and $5=1+1+3=1+2+2$.  As a result, the answer is
\[\binom{5}{4}\binom{1}{1}+\binom{5}{2}\binom{3}{3}+\frac{1}{2}\binom{5}{3}\binom{2}{1}\binom{1}{1}+\frac{1}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}=40. \]
Masoud Zargar

hello [hide] Heres a non-genius solution. 1)Prime factor the number to find 5 numbers. My understanding of the problem is that we have to make it so that these 5 numbers turn into 3 bigger numbers that multiply to 2310. One way to think of it is in grouping of possibilities logically.First we see the number of ways the 5 digits can be separated in 3 groups. Namely, we can have a max of a group of 3 in a group, and we can only have 3's with 1's. This creates group A)1,1,3 and B)2,2,1. For group A, it's simply a matter of looking at the individual 1's as two such that we just group in two's >we have AB,AC,AD,AE,CB,BD,BE,CD,CE,DF - 10 -(threes's can be thought of as a dependent variable and neglected).
B)Perhaps more of the same. One way to do it would be to take look at the one first as it has the least possibilities. We clearly have 5. Next we look at each letter(the one I am referring to) and with 4 letters left(since we have one already) we group with two, creating AB AC AD BC BD CD for example 6 for the individual letter E. Then multiply by 5 for 30. Now for the hardest part 10+30=40 lol. [/hide] I think this may be what boxedexe did.

[hide]
$ 2310\equal{}2\cdot 3\cdot 5\cdot 7\cdot 11$. The number of ordered triples $ (x,y,z)$ with $ xyz\equal{}2310$ is therefore $ 3^5$, since each prime dividing 2310 divides exactly one of $ x,y,z$.

Three of these triples have two of $ x,y,z$ equal (namely when one is 2310 and the other two are 1). So there are $ 3^5\minus{}3$ with $ x,y,z$ distinct.

The number of sets of distinct integers $ \{ a,b,c\}$ such that $ abc\equal{}2310$ is therefore $ \frac{3^5\minus{}3}{6}$ (accounting for rearrangement), or $ \boxed{40}$.
[/hide]

Hmmm I don't understand why my solution is incorrect. so what i did was this:

2310 = 7 * 11 * 3 * 2 * 5

Since it is asking for how many a,b,c, satisfy this... then why can't you just do stars and stripes? there are 5 objects and 2 dividers so do (7 choose 2)...?? but then add 3 back because a,b,c, are positive??... or am I just plain wrong.. sorry

I still don't understand why $\binom{7}{2}=21$ doesn't work.

[quote=afroromanian]Because the 5 prime factors are distinguishable. Say you list them as {2, 3, 5, 7, 11} and you divide them as {2, 3}, {5} and {7, 11} What if you want to switch the 2 and 11? The divider method won't count this possibility: {11, 3}, {5}, {7,2} which is clearly valid. Why doesn't it count it? Because the five divisors are distinguishable so you can't use that method. So $\binom{7}{2}$ is undercounting the total by quite a bit.[/quote]

I suppose that makes sense. But if you're putting distinct objects into boxes, wouldn't the answer just be $P(9,3)$?

No, you're assigning each of 5 distinct object to 3 distinct boxes, that would give $3^5$ ways. However, you're not allowed to assign all 5 to one box, so subtract three to get $240$. But now remember that we are making *sets*, so order doesn't matter, (e.g {2, 3}, {5, 7}, {11} is the same as {5,7}, {2,3}, {11}), so you need to divide by $3!$ to get a total of $\boxed{40}$.

[quote=afroromanian]However, you're not allowed to assign all 5 to one box.[/quote]
Sorry, why is this?

Because the problem states that you are looking for all 3-element *sets* {a, b, c} of positive integers, such that $a \times b \times c = 2310$. So if you put all 5 numbers in one box (set), then the other two boxes would have 1, and 1, but {1, 1, 2310} isn't a valid set, as by definition, a set cannot have repeated elements.

Okay. Thanks! So it's sort of like permutations with repetitions?

Sure, I guess.. basically you're thinking of $a, b, c$ in $abc = 2310$ as 3 distinct boxes in which you put the 5 distinct factors of 2310. But you realize that you have too many permutations (you're overcounting) so you have to correct for that. That's the strategy to the *pretty* solution.

The straightforward solution in the above thread of breaking it down into cases isn't bad either (see @boxedexe solution), in fact that's how I first did it before I saw this much more elegant solution.

$2310 = 2 \cdot 3 \cdot 5\cdot 7 \cdot 11$
So for the first number you can pick $2\cdot5$ numbers,but some of them are same like $2^0=11^0$, so there is 6 different solutions,  for the second one you can pick eather $5$ or $6$, and last but not least depends on first two. So final nubmer of solutions is: $5\cdot5+6$ or $31$ and permutations.

[quote=CantonMathGuy][hide=The good solution]
$2310=2\cdot3\cdot5\cdot7\cdot11$. There are three ways to assign each prime factor to one of $a,b,c$, for a total of $3^5=243$. Our three numbers must be pairwise distinct. The only way two of them could be equal is when two of them are equal to $1$, which can happen in three ways. Thus, $240$ of the ways have three distinct numbers. Finally, we divide by $3!=6$ because we can permute $a,b,c$ for an answer of $\boxed{40}$.
[/hide][/quote]

this solution doesn't count (1, 1, 2310), right? Why isn't the answer 40 + 1 = 41?

[quote=Designerd][quote=Boi115]The set $\{1,~1,~2310\}$ is equal to $\{1,~2310\},$ which makes it a two-element set.[/quote]

so what you're saying is that a set has to have distinct numbers?[/quote]

Yes, that's part of the definition of a set.


function
integration
real analysis
real analysis unsolved

Let $u(x)=x+\frac{x^3}{1\cdot3}+\frac{x^5}{1\cdot3\cdot5}+\cdots$

Then $u$ satisfies the differential equation $u'-xu=1;\,u(0)=0.$

We can then calculate the formula $u(x)=e^{x^2/2}\int_0^xe^{-t^2/2}dt.$

From this, it's clear that Moubinool's answer is $D(x)=e^{x^2/2}.$


geometry
3D geometry
sphere
calculus
search
real analysis
real analysis unsolved

Tell the volume of infinite dimensional sphere.

I dont know this topic belongs to calculus but this question was asked me in the calculus class.

The question doesn't really belong anywhere because it doesn't make sense. What do you mean by "infinite dimensional sphere"? Worse, in this context, what do you mean by "volume"?

The question "what is the $ n$-dimensional measure of the unit ball in $ \mathbb{R}^n$?" makes perfect sense, and if we take the limit of that as $ n\to\infty,$ then this "volume" tends to zero.

Ok then tell me the volume of n dimensional sphere

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1131143527&t=45050]this topic[/url], and also follow the link in my last post there.


number theory
greatest common divisor
number theory unsolved

For each positive integer $ n$, find the greatest common divisor of $ n!\plus{}1$ and $ (n\plus{}1)!$.

The $ gcd \equal{}1$ if $ n\plus{}1$ is composite and $ \equal{}n\plus{}1$ if $ n\plus{}1$ is prime. Clearly any divisor of $ (n\plus{}1)!$ which is not coprime to $ n!$ cannot divide both. If $ n\plus{}1$ is composite then any divisor of $ (n\plus{}1)!$ is not coprime to $ n!$, so the $ gcd\equal{}1$. Otherwise $ n\plus{}1$ is prime and $ n\plus{}1|n!\plus{}1$ by Wilson's theorem and $ n\plus{}1|(n\plus{}1)!$ but no other number divides both.


algebra
binomial theorem
combinatorics proposed
combinatorics

prove that the number of distinct terms in the expansion of $ /(x_{1}+x_{2}+\ldots+x_{m})^{n}$ is given by $ H^{m}_{n}=/(^{n+m-1}_{n})$.

We know $ (x_{1}\plus{}x_{2}\plus{}...\plus{}x_{m})^{n}\equal{}\sum\frac{(y_{1}\plus{}y_{2}\plus{}...\plus{}y_{m})!}{y_{1}!y_{2}!...y_{m}!}x_{1}^{y_{1}}... x_{m}^{y_{m}}$ and  $ y_{1}\plus{}...\plus{}y_{m}\equal{}n$, ($ y_{i}\geq 0$) and the number of solutions of this equality are $ \binom{n\plus{}m\minus{}1}{m\minus{}1}\equal{}\binom{n\plus{}m\minus{}1}{n}$.

It is the Multinomial Theorem: it follows immediately from repreated applications of the Binomial Theorem, i.e. write $ x_{1}\plus{}\ldots\plus{}x_{n}\equal{} (x_{1}\plus{}\ldots\plus{}x_{n\minus{}1})\plus{}x_{n}$ and then expand with the Binomial Theorem.


probability
AMC
AIME
number theory
\/closed

In which order should I take these classes?

Algebra 3
Intermediate Number Theory
Intermediate Counting and Probability
Precalculus
AIME Problem Series (A or B)

Thanks so much.

I recommend Algebra 3 first.  The others can really be taken in any order.

Also know that you need a strong base (introductory level) in all subjects to take the AIME classes.


geometry
geometric transformation
reflection
combinatorics proposed
combinatorics

Here's my problem from the first issue of Mathematical Reflections (the second issue is online, so it's time to post the problems here). By the way, I advise you to visit the journal, I like the problems there.

Given is a positive integere $n$ with $(n,6)=1$.
We have a $k\times l$ table. In each cell of the table there is a bulb. One can take $n$ consecutive bulbs on a row, column, or diagonal, and switch all of them.

Initially all the bulbs are ON. Prove that one can switch all of them OFF if and only if $n|k$ or $n|l$.

The problem might be well-known, but I didn't meet it before.

By "diagonal" you mean both diagonal directions are allowed? Right now, I can prove that $n|k$ or $n|l$ if only one diagonal direction is allowed, but in this case I'm only using $(n,2)=1$.

Yes, both diagonals.
This is why I needed $(n,6)=1$. If there was just one diagonal, $(n,2)$ would be enough for me, too.

I apologise in advance if I am being extremely dense, but why do you need the condition that $n$ is odd (whether or not both diagonal directions are allowed)? :oops:

Let $k = an + p$, $l = bn + q$, where $a, b \in \mathbb{Z}$, $0 \leq p, q < n$.  Colour the cells of the table with the numbers $0, 1, 2 \ldots, n - 1$ in the following manner:

1   2   3   ...   0   1   2   ... q
3   4   5   ...   2   3   4   ...   (q+2)
5   6   7   ...   4   5   6   ...   (q+4)
...
0   1   2   ...   -1   0   1   ...   (q-1)
2   3   4   ...   1   2   3   ...   (q+1)
4   5   6   ...   3   4   5   ...   (q+3)
...
-1   0   1   ...   -2   -1   0   ...   (q-2)
1   2   3   ...   0   1   2   ... q
3   4   5   ...   2   3   4   ...   (q+2)

(2p-1)   2p   (2p+1)   ...   (2p-1)   (2p-1)   2p   ...   (q+2p-2)

where the colour of the cell in row $i$ and column $j$ is $2i +j - 2$, taken modulo $n$.  Throughout the rest of this argument, all colours will be considered modulo $n$.

Since $\gcd(n, 6) = 1$, we observe that any $n$ consecutive horizontal, vertical or diagonal entries have exactly one cell of each colour.  This means that at any moment, the number of bulbs of colour $i$ which are off has the same parity as the number of times an operation has been carried out.  Let $a_i$ denote the number of cells coloured with colour $i$.  To prove that one can eventually turn all the bulbs off only if $n$ divides $k$ or $n$ divides $l$, it suffices to prove that if $n$ does not divide both $k$ and $l$, there exist $i$ and $j$ such that $a_i - a_j$ is odd.

Suppose on the contrary that $n$ does not divide both $k$ and $l$ and $a_i - a_j$ is even for all $1 \leq i, j \leq n$.  In particular, this implies that $a_i - a_{i-1}$ is even for all $i$.  Clearly, it suffices to consider the $p$ by $q$ sub-table at the bottom-right corner. Observe that colours $i$ and $i - 1$ both do or do not appear once in a row unless the first cell is of colour $i$ or the last cell is of colour $i - 1$ (but not both).  Since the first column consists of the numbers $\lbrace 1, 3, 5, \ldots, 2p - 1 \rbrace$, the last column must consist of the numbers $\lbrace 0, 2, 4, \ldots, 2p - 2 \rbrace$ for our supposition to hold.  But the last column consists of the numbers $\lbrace q, q + 2, q + 4, \ldots, q + 2p - 2 \rbrace$.  This is possible only if $q = 0$, a contradiction.  Therefore, we conclude that $n$ divides $k$ or $n$ divides $l$.

Now suppose that $n$ divides $k$ or $n$ divides $l$.  We may assume WLOG that $n$ divides $k$.  By dividing the $k$ by $l$ table into $\frac{kl}{n}$ subtables of dimensions $n$ by $1$, and changing the state of the bulbs in each of those subtables once, all the bulbs will be switched off.

Wow nice solution Valiowk, I couldn't have thought of it. :)

To iura: Nice, thank you for providing the new [i]Mathematical Reflections[/i]. It is surely a great addition to my Favorites Links ;)


In the United States, Congress has tried to amend the Constitution to prohibit flag burning 6 times. Each time, it passed in the house with a supermajority as required, but three times it failed in the Senate. The most recent of those times it failed by only 1 vote.

Source: http://en.wikipedia.org/wiki/Flag_Desecration_Amendment#Congressional_votes

This raises the question, SHOULD flag burning be illegal? Some say it is part of free speech, other's ask what does flag burning say flag burning isn't speech at all, more like spitting on a country, what is your opinion on this matter?

I think that flag burning should be illegal- not because it is 'spitting on a country' but because it incites violence amongst others. This would leave the possibility of only banning flag burning in most public places, but allowing people to burn flags in the privacy of their own homes.

Could you remove the 'did you hear' part of the topic title: it is clearly there just to incite people, given fredbel's recent comments.

[quote="tim1234133"]because it incites violence amongst others.[/quote]

Freedom of assembly and freedom of the press provide an easy way to incite violence among others.  Should they also be illegal?

Burning a flag should only be illegal when it is dangerous to people or property.
Messing it up is perfectly fine!!

It's done all the time here, like using it to wipe your feet : :)
[img]http://www.willemcoppenolle.be/nieuws/05-08-23.jpg[/img]

[quote="t0rajir0u"]
Freedom of assembly and freedom of the press provide an easy way to incite violence among others.  Should they also be illegal?[/quote]

Fair point, though it is more the short term thing - 'I see a flag burning, lets go riot and maybe burn some buildings/people too' - that bothers me.

Even if a law was passed, how well would it be enforced?

I think making flag burning illegal is a weird kind of idolatry -- what's so special about the flag? Are we going to make it illegal to burn effigies of political leaders too? What about burning things that have red, white, and blue on them? What about burning things that were made in the USA?

I don't want to burn any flags, but as a general principle I don't like Congress telling me what I can and cannot do. They need to convince me that they know better, and in this case they've failed miserably to convince me why there's anything wrong with burning a flag (except, as fredbel6 said, if it's potentially harmful to other property, but there's already a law for that: arson).


Given that $ a,b,c \geq 0$ and satisfies: $ a^3 \plus{} b^3 \plus{} c^3 \geq 6abc$. Prove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \left(\frac {9abc}{a \plus{} b \plus{} c}\right) \cdot \left(\frac {a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}\right) \geq 2(ab \plus{} bc \plus{} ca)$

 :)

Why no one come to solve it?
We just need to use a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
after some calculations,we can turn it into schur inequality,sorry for my poor English

I see, very few people are interested. So, here is the solution with identity:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \left(\frac {9abc}{a \plus{} b \plus{} c}\right) \cdot \left(\frac {a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}\right) \minus{} 2(ab \plus{} bc \plus{} ca)$

$ \equal{} \frac {(a^3 \plus{} b^3 \plus{} c^3 \minus{} 6abc)(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \minus{} a^2(b \plus{} c) \minus{} b^2(c \plus{} a) \minus{} c^2(a \plus{} b))}{(a \plus{} b \plus{} c)(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)}$

 :)


Are there infinitely many sets of 3 consecutive primes in arithmetic progression.

Yes, but you wont be able to prove it easily.
It can be thought of as a corollory of the Grenn-Tao Theorem, that fro any $ k$ there are $ k$ primes in arithmetic order.

http://mathworld.wolfram.com/Green-TaoTheorem.html


geometry
trigonometry
geometric transformation
rotation

Prove that, if triangle $ ABC$ is equilateral and $ M$ is variable between $ \stackrel{\frown}{BC}$, $ \overline{MA} \equal{} \overline{MB} \plus{} \overline{MC}$.

you can apply straightforward Potolemy's theorem to get :
AB.MC+AC.BM=AM.BC
since ABC is a equilateral triangle , AB=AC=BC
we obtain MC+MB=MA

Nice HTA. I did a very ugly solution using law of sines in triangles $ ABM$ and $ AMC$.

[hide="rotation"]draw MB and MC. we know that $ \angle{ABM}$ is supp to $ \angle{ ACM}$ rotate the quadrilateral until AB coincides with AC and so we know that since $ \angle{ABM}$ is supp to $ \angle{ ACM}$ the points M C and M' are collinear  since AM=AM' and MAM'=60 we now that  MAM' is equilateral therefore AM=BM+CM. [/hide]


At a unit circle, we construct chords $PQ$ and $MN$ parallel to the radius $OR$. If $MP=PQ=NR=s$ and $MN=d$, are these statements below correct ?
(i)$d-s=1$
(ii)$ds=1$
(iii)$d^2-s^2=\sqrt{5}$

For the first one, should it be |d-s|=1?  Because it is possible that d<s


[quote="Derek"]Hi,

Evaluate: $ \displaystyle \sum_{n=4} ^\infty \frac {5}{(n^2-n-6)} $

Can you show me step by step on how to solve this? Thanks.[/quote]
the bottyom is (n+2)(n-3)
the whole thing is
 :Sigma: 1/(n-3)-1/(n+2)
now if you write out a few terms, a whole bunch of them cancel out
so is=ts 1-1/6+1/2-1/7+...+1/6(cancels out with first 1/6)+1/11 +1/7(cancels out with first 1/7)+1/12, etc
since everything cancels out except 1+1/2+1/3+1/4+1/5, (and the last 5 terms which are infinitely small) that should be your answer

Okay, so we want $\displaystyle\sum_{4}^{\infty}\frac{5}{(n-3)(n-2)}$.  Let's try to use partial fraction decomposition, to see if the series will telescope.

So we say that $\frac{5}{(n-3)(n+2)}=\frac{A}{n-3}+\frac{B}{n+2}$.  Multiplying both sides by $(n-3)(n+2)$, we see that $5=A(n+2)+B(n-3)$.  Letting $n=-2$ gives us $5=A(0)+B(-5)$, so $B=-1$.  Letting $n=3$ gives us $5=A(5)+B(0)$, so $A=1$.  That means we can rewrite $\frac{5}{(n-3)(n+2)}$ as $\frac{1}{n-3}-\frac{1}{n+2}$.

So now we want to find the sum $\displaystyle\sum_{4}^{\infty}\frac{1}{n-3}-\frac{1}{n+2}$.  Plugging in some numbers, we find the first few terms of the series are:
$\frac{1}{1}-\frac{1}{6}+\frac{1}{2}-\frac{1}{7}+\frac{1}{3}-\frac{1}{8}...$  Notice that every term will cancel, since the denominator of the postive terms increases by one each time, and the denominator of the negative terms also increases by one.  That means everything will cancel out except the positive terms with no negative additive inverse in the series.  There are the first 5 positive terms, since the first negative one is $-\frac{1}{6}$.  So the sum is equal to $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{137}{60}$.

Oh ok, the part I didn't get was the partial fractio decomposition. (Now I sort of get it... I think)

Thank you for your help(s).


ratio
trigonometry
geometry unsolved
geometry

in the triangle $ \triangle ABC$ below , if you have that :

$ \frac{|AM|}{|MB|} \equal{} \frac{2}{3}$ ,,, $ \frac{|BN|}{|NC|} \equal{} \frac{1}{4}$

find : $ \frac{|OC|}{|MO|}$ .

$ \frac {OC}{MO} \equal{} \frac {AC \cdot \sin{\angle CAO}}{MA \cdot \sin{\angle MAO}} \equal{} \frac {b \cdot \frac {\sin{\gamma} \cdot CN}{AN}}{\frac {2}{5}c \cdot \frac {\sin{\beta} \cdot BN}{AN}} \equal{} \frac {5}{2} \cdot 4 \equal{} 10$

You can apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {AON}$ of the triangle $ BCM$ :

$ \frac {AM}{AB}\cdot\frac {NB}{NC}\cdot\frac {OC}{OM} \equal{} 1$ $ \implies$ $ \frac {OC}{OM} \equal{} \frac {AB}{AM}\cdot\frac {NC}{NB} \equal{} \frac 52\cdot\frac 41$ $ \implies$ $ \frac {OC}{OM} \equal{} 10$ . [b]Otherwise[/b],

denote $ P\in AN$ for which $ MP\parallel BC$ . Thus, $ \frac {OC}{OM} \equal{}$ $ \frac {NC}{MP} \equal{}$ $ \frac {NC}{NB}\cdot\frac {NB}{MP} \equal{}$ $ 4\cdot\frac {AB}{AM} \equal{}$ $ 4\cdot\frac 52$ $ \implies$ $ \frac {OC}{OM} \equal{} 10$ .


[quote="The Nguyen"]Find $ f: \mathbb R^ \plus{} \to \mathbb R^ \plus{}$ such that $ xf(xf(y)) \equal{} f(f(y))$[/quote]
Let $ x \equal{} \frac{z}{f(1)}, y \equal{} 1$:
$ \frac {z}{f(1)} f(z) \equal{} f(f(1)) \iff f(z) \equal{} \frac{f(f(1)) \cdot f(1)}{z}$
Hence $ f(x) \equal{} \frac {a}{x}, a > 0$ is the only solution, and it is easily seen to fit the given equation :)

[quote="The Nguyen"]Find $ f: \mathbb R^ \plus{} \to \mathbb R^ \plus{}$ such that $ xf(xf(y)) \equal{} f(f(y))$[/quote]

Let $ P(x,y)$ the $ xf(xf(y)) \equal{} f(f(y))$ $ \forall x,y >0 \qquad (*)$

$ P(x,1)$ $ \implies$ $ xf(xf(1))\equal{}f(f(1))$ $ ,\qquad \forall x>0$

Let $ f(1)\equal{}c$ with $ c>0$ we have $ f(cx)\equal{}\frac{a}{x}$ where $ a\equal{}f(c)>0$

$ P(\frac{x}{c},1)$ $ \implies$ $ f(x)\equal{}\frac{ac}{x}$ $ ,\qquad \forall x>0$ and $ a>0,c>0$

$ f(1)\equal{}c$ $ \implies$ $ ac\equal{}c$ $ \implies$ $ a\equal{}1$ $ \implies$ $ f(x)\equal{}\frac{c}{x}$ where $ c,x>0$

This is indeed a solution of $ (*)$. Hence the solution is $ \boxed{f(x)\equal{}\frac{c}{x},\qquad x \in \mathbb R^ \plus{}, c>0 }$


Solve these equations with x,y,z belong to Z
a) $\ 2^x=1+3^y.7^z$
b) $\ x^3+y^6+27^z=1996^2$ :harhar:


I usally don't have problems using sin, cos and tan on a calculator, but now i've got a TI-84 plus. If we take a triangle ABC where C=90, a=5cm and b=12cm.

sinA=a/c
sinA=5/13
sinA=0,38

now what buttons do i hit on my calculator? I thought it was as simple like i just had to hit sin(0,38) or sin^-1(0,38) but then i get these results:

A=0,3752
or
A=0,3948

What do i do wrong?

Well, arcussine of 5/13 is approximately 0.395 radians. Maybe you want your answer as degrees? Then you should multiply the radian answer by $180/\pi$.

Well, I do not have a TI 84, but a 83, so maybe...

Probably you're in mode "Radian", while you should be in "Degree". Change it with the "MODE"-button.

Ah thx, this is really great. I recall that i did change it to degree once, but maybe i didn't. It has been changed now, so no problem there (unless it changes again :) ). Thx a lot to both of you ;)


limit
logarithms
calculus
calculus computations

Fix $ p,q \in \mathbb N$.  Let $ a_{n}= (-1)^{n-1}\frac{1}{n}$ for $ n \geq 1$.  Rearrange $ a_{n}$ to form a new sequence $ b_{n}$ by taking the first $ p$ positive terms of $ a_{n}$, followed by the first $ q$ negative terms, followed by the next $ p$ positive terms, and so on.  Show that $ \sum b_{n}$ is convergent and find its sum.

Let $ H_{n}= \sum_{k=1}^{n}\frac{1}{k}$. If we let $ s_{n}$ be the partial sum of the rearranged series, then

\begin{eqnarray*}s_{(p+q)n}& = & \left( \frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{2p-1}\right)-\left( \frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2q}\right)+\cdots\\ && \cdots+\left( \frac{1}{2(n-1)p+1}+\frac{1}{2(n-1)p+3}+\cdots+\frac{1}{2np-1}\right)-\left( \frac{1}{2(n-1)q+2}+\frac{1}{2(n-1)q+4}+\cdots+\frac{1}{2nq}\right)\\ & = & \sum_{k=1}^{np}\frac{1}{2k-1}-\sum_{k=1}^{nq}\frac{1}{2k}\\ & = & \left( H_{2np}-\frac{1}{2}H_{np}\right)-\frac{1}{2}H_{nq}\\ & = & \frac{1}{2}\left( H_{2np}-H_{np}\right)+\frac{1}{2}\left(H_{2np}-H_{nq}\right) \end{eqnarray*}

But one can show thaht $ \lim_{n\to\infty}H_{pn}-H_{qn}= \ln \frac{p}{q}$ for positive integers $ p, q$ by changing the limit into an integral. Since $ s_{n}$ converges, we have

$ \lim_{n\to\infty}s_{n}= \lim_{n\to\infty}s_{(p+q)n}= \frac{1}{2}\ln \frac{4p}{q}$


Let $F_{i}$ denote the $i$th Fibonacci number.  Prove that $F_{n+2}=\Sigma^{n}_{i=0}F_{i}+1$

Induction works:

It's obviously true for $F_{3}$.

Now suppose we proved it for $F_{n+2}$, then we will prove it for $F_{n+3}$: \[F_{n+3}= F_{n+2}+F_{n+1}\]  \[= \sum_{i=0}^{n}{F_{i}}+1+F_{n+1}\]  \[=  \left(\sum_{i=0}^{n}{F_{i}}+F_{n+1}\right)+1\]  \[=  \sum_{i=0}^{n+1}{F_{i}}+1\] So it is true for all $n$.


Peresejo has $ 4$ enumerated baskets and he places balls of $ 6$ colors, and the white color is one of those, inside the baskets, no more than one ball for canast and he always places a quantity of balls greater or equal than $ 2$, but less or equal than $ 4$. Peresejo says that a $ tuanis$ colocation occurs when two white ball are in adyacent baskets. How many $ tuanis$ colocations can Peresejo form?
Note: $ tuanis$ is a regular word in Costa Rica which means nice  :)

there must be at least 2 white balls. if there are exactly 2 whites, there are $ 3\cdot 6^2$ ways to have 2 adjacent whites. (3 ways to pick two adjacent baskets to contain the two whites, and 6 choices for each of the remaining empty squares -- empty, or one of the other 5 colors.) if there are exactly 3 whites, then 2 of them will necessarily be adjacent. there are $ 4\cdot 6$ ways in this case: 4 ways to pick the 3 baskets containing a white, and 6 choices for the remaining basket. if there are 4 whites, the number of ways is clearly 1. so the answer is $ 3\cdot 6^2\plus{}4\cdot 6\plus{}1\equal{}133$.

Nice solution, I proposed this problem in my country olympiad :lol: , I did it with the $ PIE$ principle. There are three ways of placing the white balls in adyacent baskets, placing the two white balls in the first two basket, in the second and third baskets and finally in the two last baskets. Then each way has $ 7^{2}$ for accomodation taking in count that there can be an empty basket, but each way has common accomodations so the way $ 1$ and way $ 2$ has $ 7$ and way $ 2$ and $ 3$ have $ 7$ common accomodations and way $ 1$ and $ 3$ has $ 1$ common accomodation, and the three ways have $ 1$ common accomodation with all the $ 4$ baskets with white balls. Then by the $ PIE$ the number of $ tuanis$ accomodations are $ 7^{2}\minus{}2\cdot 7\minus{}1\plus{}1\equal{}133$


If x² is congruent to a mod p then x² is congruent to a mod p^j where j is any integer
 Show that the equation X²+Y²=qZ, where q is any integer number and consider that q is congruent to 3 mod4, has solutions only if x,y,z are integer numbers

The first line as problem was already posted...
But nonetheless, the second line makes no sense to me. Please restate it (should be something like: "the only solution is $x=y=z=0$"). But also posted before ;)
Oh, and please use $\LaTeX$ and post only one problem per thread.

The second part is the following: considering the equation X²+Y²=qZ, where q is congruent to 3 mod4, then thereis a solution for this equation in x,y,z only if these numbers are integers

I can read, you just repeated.
But that makes less than no sense!

if i understand you in the rigth way then $X=1, Y=\sqrt{2}, Z=1$ isn't a solution for $Q=3$?  :o

Naphthalin


Unless I have done something very stupid, this is a very short angle-chasing exercise. It's better suited for Getting Started.

Care to show how you chased the angles? Maybe I'm just being incredibly stupid as well but I can't see it.  :?

I'm pretty sure it's no angle chase. It's one of those nasty problems which can be solved either by a smart construction or by trigonometry.

$\frac{\sin x}{\sin12}=\frac{CD}{DA}=\frac{\sin30}{\sin84}\frac{\sin(x+36)}{\sin42}$ so $\frac{2\sin84\sin42}{\sin12}=\frac{\sin42}{\sin6}=\cos36+\cot x\sin36$. Since $\sin42=\cos6\sin36+\sin6\cos36$, we get $\cot6\sin36+\cos36=\cos36+\cot x\sin36$, so $x=\boxed6$.


If 3 is divided by (3+x) and absolute value of x is less than 3, the result is a+bx+cx^2+dx^3+ex^4+......
Find the value of a+b+c+d Express answer as a common fraction reduced to lowest terms.

This looks a good sum where did you get it from can you tell me?
I live in England aswell where abouts do you live, I live in Hampshire.

I see...never though about using the sum formula backwards....
Thanks a lot


modular arithmetic
number theory unsolved
number theory

How many different ways are there to write 2004 as a sum of one or more positive integers which are all "aproximately equal" to each other? Two numbers are called aproximately equal if their difference is at most 1. The order of terms does not matter: two ways which only differ in the order of terms are not considered different.

[quote="paganinio"]"aproximately equal" to each other?
then
[hide]
it must be like n, n+1, n, n+1, ...
so that n*m+(n+1)*m=2004
m(2n+1)=2004 (*)
list all possible 'm's, you will got the result.

then we look at how many 'm's there are.
2004=2*2*3*167
m could be 1, 2, 3, 167, 4, 6, 2*167, 3*167, 12, 2*2*167, 6*167, 2004
then put them into (*) one by one, see if n and n+1 are both positive integers.
[/hide][/quote]

here's a serious mistake
[hide]
2n+1 is an odd number
so (*) should be n*m+(n+1)*(m+1)=2004
therefore 2mn+m+n=2003
then find integer solutions for this equation.
[/hide]

Why do you assume, that there are [tex]m[/tex] terms equal to [tex]n[/tex] in the sum and [tex]m+1[/tex] (or [tex]m[/tex] in your first post) terms equal to [tex]n+1[/tex]? Can't there be [tex]m[/tex] terms equal to [tex]n[/tex] and [tex]m+k[/tex] terms equal to [tex]n+1[/tex] for some integer [tex]k[/tex], and then [tex]nm + (n+1)(m+k)=2004[/tex] ? Sorry, if I missed something trivial.

Lets say there are K terms, with a being equal to n+1 and the rest equal to n.

We can assume there must be at least one term equal to n (otherwise we count n+n+n.. again as (n+1)+(n+1) + ..)

Our sum is $nK+a=2004$.

So for any given K,  $0\le a<K$, but also $a\equiv 2004 \pmod{K}$, so there is exactly one value for a. Thus, since K can be from 1 to 2004, there are 2004 solutions.


A normal alkane, $ C_{n}H_{2n\plus{}2}$ has a vapor density of 3.66 mg/mL at 200 degrees C and 1 atm. Using the ideal gas law, determine the structure of the alkane.

Here, I manipulated the ideal gas equation $ PV \equal{} nRT$ to yield that the ratio of the vapor densities of two gases equals the ratio of their molecular weights. So if I did this right, all I need is the vapor density of air at 200 degrees, but how can I find this value (either by calculation or table - but I can't find a table!)

Thanks

No no. All you have to do is to manipulate the ideal gas law only for the alkane.

[hide="Answer"]We have $ d \equal{} \frac{m}{V}$ and $ n \equal{} \frac{m}{M}$. Substituting this in the ideal gas law we get

$ M \equal{} \frac{dRT}{P}$

Substituting, $ M \equal{} \frac{3.66\,g/L \times 0.08206\,LatmK^{\minus{}1}mol^{\minus{}1} \times 473.15\,K}{1\,atm} \equal{} 142.1\,g/mol$.

But the molar mass is given by $ M \equal{} n \cdot 12.011 \plus{} (2n\plus{}2) \cdot 1.0079$,

and substituting and solving for n we get n = 10, and so our alkane is decane.[/hide]


I'd like to set up a Mu Alpha Theta chapter in my school, but I'm not sure how to go about doing that. Truth be told, I'm not too sure about what Mu Alpha Theta [i] does[/i]. Can anyone help me/give me information?

Thanks,
sran

Thanks for the info.

Some questions:

What kinds of activities should be done at the regular meetings? Practice? Promotional? What?

What exactly is the National Conference? I really could not figure it out from the website, it seems designed for someone who already knows what it is. What do you do there? What incidental expenses should one expect? etc...

National Converntion you mean??

if it is National convention, then I know a little about it.
it is just like any other contest, except it is a week-long contest.

you are tested in a lot of tests:
Individual, Ciphering, Topic Tests(3 rounds), Hustle, etc.

you can see the schedule in the Mu Alpha Theta website.
It is in Hawaii this year, so it will be great. the cost is also on the website. You can expect to pay that amount plus your plane ticket.

Chapters aren't required to attend the National Convention.  For example, my school doesn't go, this is probably because my school isn't exactly a math school and going would be way to expensive.


function
real analysis
real analysis unsolved

[img]http://img223.imageshack.us/img223/5952/26547437.jpg[/img]

This function seems not tend to zero if x is infinity??

[img]http://img5.imageshack.us/img5/4933/19014970.jpg[/img]

Why the terms of the square of e disappear?

[img]http://img84.imageshack.us/img84/8489/92862099.jpg[/img]
[img]http://img147.imageshack.us/img147/6716/78163006.jpg[/img]

The assumption of t and the term of t in big-o notation is different??

[quote="kit.kit44"]This function seems not tend to zero if x is infinity??[/quote]
The limit you underlined just shows that the terms in the sum tend to zero as $ n \rightarrow \infty$. You might be thinking that it says the entire sum approaches zero as $ x \rightarrow \infty$, which is not true and not what the limit means.

[quote="kit.kit44"]Why the terms of the square of e disappear?[/quote]
The big-oh notation ignores constants. $ K \cdot O(f(x)) \equal{} O(f(x))$ for any constant $ K$.

[quote="kit.kit44"]The assumption of t and the term of t in big-o notation is different??[/quote]
Once again, the big-oh notation ignores constants. $ O\left(\left(\frac {x \minus{} e}{e}\right)^4\right) \equal{} O\left(\frac {(x \minus{} e)^4}{e^4}\right) \equal{} O\left(\frac {1}{e^4}(x \minus{} e)^4\right)$, and since $ 1/e^4$ is a constant, we can ignore it and just write $ O\left(\frac {1}{e^4}(x \minus{} e)^4\right) \equal{} O((x \minus{} e)^4)$.


Prove

$\displaystyle { \lim_{x\to a}{cf(x)}=cL}$

$\displaystyle { \lim_{x\to a}{(f/g)(x)}=L/M}$

c=constant

As written, the post doesn't make much sense: it is completely unclear what is $L$, what is $M$ (also I suspect that "Proof" should be "Prove"). I suspect however that you refer here to the standard properties of limits and your post should look something like this:

Assume that $\lim_{x\to a}f(x)=L$ and $c$ is an arbitrary constant. Then $\lim_{x\to a}cf(x)=cL$.

Assume that $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M\neq 0$. Then $\lim_{x\to a}f(x)/g(x)=L/M$

The proofs of both properties can be found in any decent Calculus textbook. Try to read them there (of course, I can rewrite them here word by word, but I don't think it is a great idea). If you have trouble understanding the proofs, you are always welcome to ask questions, but try to be as clear and specific as possible (it is impossible to answer a question when it is unclear what is asked).


geometry
trigonometry
circumcircle
incenter
analytic geometry

Let $\Delta ABC$ be an acute angled triangle. Proof that the baricenter of the system $(A,\tan A), (B,\tan B), (C,\tan C)$ is the ortocenter of $\Delta ABC$.

Is this intermediade? In my math book is the seven problem :)

[quote="probability1.01"]Define baricenter and I'll try it :D.[/quote]

It's the centroid, i.e. intersection of medians

http://mathforum.org/library/drmath/view/57768.html

Hey! The centroid is a particulaty case of barycenter I guess  :?

What do you mean by (A,tanA), (B,tanB), (C,tanC)?

If I remember correctly from geometry, the orthocenter only equals the circumcenter in an equilateral triangle.  (That's when all four points of concurrency meet--the incenter, circumcenter, orthocenter, and centroid).

Proving this for an equilateral triangle wouldn't be very difficult.

$A,B$ and $C$ are the vertex of $\Delta ABC$ triangle and $\tan A, \tan B$ and $\tan C$ are their weights.

To ilustrate this i'll give an exemple:

We have a segment $AB$ and point $A$ has the weight $m_{1}$ and $B$ hat it's weight $m_{2}$. So it's barycenter has the coordinates $x_{barycenter}=\frac{x_{A} \cdot m_{1}+x_{B} \cdot m_{2}}{m_{1}+m_{2}}$. If $m_{1}=m_{2}$ than the coordinates of barycenter are the same with the centroid of the segment.
I conssidered 2 points, with 2 weights but we can have this for $n$ points and weights $x_{baricenter}=\frac{1}{m_{t}}\sum_{k=1}^n{x_{i}}$ and $m_{t}=\sum_{k=1}^n{m_{i}}$. Also for $y_{barycenter}, z_{barycenter}$

First, you should prove that for example $\frac{m_A}{m_B} = \frac{S_{BPC}}{S_{APC}}$.


inequalities
algebra
polynomial
inequalities unsolved

This is so nice that I think it  deserves an own  topic.

For $a,b$ positive reals prove that

[quote="Vasc"]

$\frac{1}{1+a+a^2+a^3}+\frac{1}{1+b+b^2+b^3}\geq \frac{1}{1+(ab)^{3/2}}$
[/quote]

No idea about this one !!!!!!!!

Very hard for me  :(

For ab ≥ 1 it is easy:
1/(1+a)(1+a2) +1/(1+b)(1+b2) ≥ 2/(1+√ab)(1+ab) ≥ 1/(1+ab√ab).
Another story when ab < 1.
I will write my solution believing that you will read it to the end.
(because I spent on it three hours). :)
Our inequality is equalent to
(1+ab√ab)(2+a+a2+a^3 +b+b2+b^3) ≥ (1+a+a2+a^3)(1+b+b2+b^3)

 1 + 2ab√ab + ab√ab(a+b) + ab√ab(a2+b2) + ab√ab(a^3+b^3) ≥
≥  ab + a2b2 + ab(a+b) + ab(a2+b2) + a2b2(a+b) + a^3b^3.

1  ab - a2b2 + a^3b^3 + ab√ab(a2 + b2 - √ab(a+b)) + ab√ab(a^3 + b^3 - 2ab√ab)
≥  ab (a + b - 2√ab) + ab(a2 + b2 - √ab(a+b))

(1-ab)2(1+ab) + ab√ab(√a -√b)2(a+√ab+b) + ab√ab(√a -√b)2(a+√ab+b)2
≥  ab(√a - √b)2 + ab(√a -√b)2(a+√ab+b)

(1-ab)2(1+ab) + ab√ab(√a - √b)2(a+√ab+b)(1+a+√ab+b)
≥  ab(√a - √b)2 (1+a+√ab+b)

 ab(√a - √b)2 (1+a+√ab+b)(1-√ab(a+√ab+b)) ≤(1-ab)2(1+ab).

We denote [b]t [/b]= 1 - √ab(a+√ab+b) > 0    (it [b] t [/b]< 0 then we are done)

√ab(√a - √b)2 = √ab (a+√ab+b) -3ab = (1-[b]t[/b]-3ab)
√ab (1+a+√ab+b) =√ab +√ab (a+√ab+b) = (√ab +1-[b]t[/b])

Hence we have to demonstrate:
[b]t[/b] * (√ab +1-[b]t[/b]) * (1-[b]t[/b]-3ab) ≤ (1-ab)2(1+ab).

3/2 [b]t[/b] * (√ab +1-[b]t[/b])/2 * (1-[b]t[/b]-3ab) ≤  (1-ab)2(1+ab).

By AM-GM inequality: 3/2 [b]t[/b] * (√ab +1-[b]t[/b])/2 * (1-[b]t[/b]-3ab) ≤ (3/2 +√ab /2 -3ab)^3 / 27.

We will show   (3/2+√ab /2 -3ab)^3 / 27 ≤  (1-ab)2(1+ab). Noting ab= r we get :

(3 +√r - 6r)^3 / 8*27 ≤  (1-r)2(1+r).

We use (3+√r - 6r)^3 / 8*27 ≤ (3+3√r - 6r)^3 / 8*27 = (1+√r-2r)^3 /8 ≤  (1-r)2(1+r).

(1-√r)^3 * (1+2√r)^3 ≤ 6(1-√r)2(1+√r)2(1+r)

(1-√r)(1+2√r)^3 ≤ 6(1+√r)2(1+r)

1+2√r +4r +8r√r -√r-2r-4r√r-8r2 ≤ 6 +6r+12√r +12r√r+6r2+6r^3         q.e.d

Thats all, huh !!!    :)  :)  :)

I am interesting in Vasc solution or another one easier.
I think this is very unbreakable inequality, because I couldn't solve it
using simple methods.  :D

Thank you very much, Prowler.
I admire your great mental effort and I will read very carefully your solution to the end.

Thank you very much again.

Dear Prowler,

I checked two times your computations and in my opinion they are CORRECT.

Only two little misprints

[quote]ab(√a - √b)2 (1+a+√ab+b)(1-√ab(1+a+√ab+b)) ≤(1-ab)2(1+ab).

We denote t = 1 - √ab(1+a+√ab+b) > 0 (it t < 0 then we are done) [/quote]

should be

[quote]ab(√a - √b)2 (1+a+√ab+b)(1-√ab(a+√ab+b)) ≤(1-ab)2(1+ab).

We denote t = 1 - √ab(a+√ab+b) > 0 (it t < 0 then we are done) [/quote]

and in the last line

[quote]1+2√r +4r +8r√r -√r-2r-4r√r-8r^2 ≤ 6 +6r+12√r +12r√r+6r^2+6r^3  [/quote]

Your proof is very very hard to find so I congratulate with you for your great reasoning.

Now I am very curious to see if Vasc's proof is easier but I don't think.

Thank you very much again.

Thank you manlio !!!
You was so kind to check it.
So we are waiting for Vasc's solution or others one. :)

Thank you very much, Darkseer.

I will study it carefully.

Thank you very much, Darkseer.

All that is very interesting and new for me.

My solution is similar to Prowler' solution.
First part is identical and the second part is alike.

You are very kind, Harazi.
But I forgot my old response concerning this cyclic ineq.
My present response is the following: mainly, by intuition and by pleasure to find
nice inequalities - simple as form and hard as proving.
And a non-stupid question: How do you find your hard and nice inequalities?

[quote="harazi"]Just a stupid question: how on earth do you find these inequalities? Do not give me the same answer as for the cyclic inequality $ 3(a^3b+b^3c+c^3a)\leq (a^2+b^2+c^2)^2$  ;)[/quote]
Do you means mine or Vasc's?
However, both are not stupid at all. If you mean mine, That's I just looked at my picture and tried to make it simplier and simplier.
If you mean Vasc's, then I think that's a good question. I am also wondering how Vasc find such a nice lemma. And an interesting thing is, the lemma seemed to be harder than the original problem!
I also wonder if the lemma can be generalized. For example, is it true that
$
\frac{1}{1+a+a^2+a^3+a^4+a^5}+\frac{1}{1+b+b^2+b^3+b^4+b^5}\geq \frac{1}{1+(ab)^{\frac{5}{4}}+ab^{\frac{5}{2}}}
$

Sorry, Vasc, probably you misunderstood me because of my poor english. What I wanted to say is that is very very remarkable how you find these inequalities that do not have a direct solution, but rather a very long and complicated one. I think that intuition  is the answer, isn't it? In this case, congratulations and my respect. And about my "hard and nice "inequalities, they are not that hard nor that nice as yours and certainly many are not the result of intuition, although the most difficult were indeed found by intuition. Sorry again. As about Darksheer's methods, what can I say except for beautiful and difficult?

[quote="Darkseer"]I also wonder if the lemma can be generalized. For example, is it true that
$
\frac{1}{1+a+a^2+a^3+a^4+a^5}+\frac{1}{1+b+b^2+b^3+b^4+b^5}\geq \frac{1}{1+(ab)^{\frac{5}{4}}+ab^{\frac{5}{2}}}
$[/quote]
I propose to start with the inequality:

If $a,b$ are positive numbers such that $r=\sqrt{ab}\geq1$, then
$\frac{1}{1+a+a^2+...+a^n}+\frac{1}{1+b+b^2+...+b^n}\geq\frac{2}{1+r+r^2+...+r^n}$.

Find somebody a nice solution ?

i have a strong feeling that we can use $: a+b >= 2\sqrt ab$
this gives $ \frac{1}{b}+\frac{1}{a} >= \frac{2}{\sqrt ab} $

sorry, i have now no time to check it..maybe later

i think it will be more interesting and nice , if we open tis one for 3 numbers


calculus
integration
logarithms
search
real analysis
real analysis unsolved

Find the value of the integral
$\int^1_0 \frac{x-1}{\ln x} \,dx$

It must have been discussed before.
Try to search on the forum.