geometry
perimeter
trigonometry
inequalities
trig identities
Law of Sines
angle bisector

In triangle $\, ABC, \,$ angle $\,A\,$ is twice angle $\,B,\,$ angle $\,C\,$ is obtuse, and the three side lengths $\,a,b,c\,$ are integers.  Determine, with proof, the minimum possible perimeter

[hide="partial solution, then stuck"]

let $m\angle B=x, m\angle A=2x, m\angle C=180-3x$.
$180-3x>90\implies x<30$

Let a be side opposite angle A, b be side opposite angle B, and c be side opposite angle C. By the law of sines:

$\frac{a}{\sin x}=\frac{b}{\sin 2x}=\frac{c}{\sin(180-3x)}$
$\frac{a}{\sin x}=\frac{b}{\sin 2x}=\frac{c}{\sin 3x}$
$\frac{a}{\sin x}=\frac{b}{2\sin x\cos x}=\frac{c}{3\sin x-4\sin^{3}x}$
$a=\frac{b}{2\cos x}=\frac{c}{3-4\sin^{2}x}$
$a=\frac{b}{2\cos x}=\frac{c}{3-4(1-\cos^{2}x)}$
$a=\frac{b}{2\cos x}=\frac{c}{4\cos^{2}x-1}$
Let $y=2\cos x$
$a=\frac{b}{y}=\frac{c}{y^{2}-1}$

$0<x<30\implies\sqrt{3}<y< 2$. We also know that $y$ is rational (otherwise the the expression $\frac{b}{y}$ couldn't be an integer, which we know it is). So I need to find the rational value of $y$ within that interval that would minimize the $a+b+c$. Does anybody have any idea how to do this? Should I try a different approach?
[/hide]

I have the equation:BC(2cosB+1)!So, will you continue to find the minimum tjhance? :blush: I think you can give BC=1 and then find cosB(cosB must be an integer!) :)

Okay I found a nicer way to do this problem with a nice identity I recently learned.

[hide]
The identity is:

If $ \angle A \equal{} 2\angle B$ then

$ BC^{2}\equal{} CA^{2}\plus{}CA\bullet AB$
or
$ a^{2}\equal{} b^{2}\plus{}bc$

[b]Proof[/b]

Let the angle bisector of A meet CB at D. The $ \triangle ADB$ is isosceles. Let AD=BD=x. By the angle bisector theorem, $ CD \equal{}\frac{bx}{c}$ and $ a \equal{} x\left(1\plus{}\frac{b}{c}\right)$

By stewart's theorem:

$ \frac{bx}{c}(c^{2})\plus{}xb^{2}\equal{}\left(\frac{bx}{c}\plus{}x\right)x^{2}\plus{}\frac{bx}{c}(x)\left(\frac{bx}{c}\plus{}x\right)$
$ bcx\plus{}b^{2}x \equal{}\frac{bx^{3}}{c}\plus{}x^{3}\plus{}\left(\frac{bx}{c}\right)^{2}x\plus{}\frac{bx^{3}}{c}$
$ bcx\plus{}b^{2}x \equal{} x^{3}\left(1\plus{}2\frac{b}{c}\plus{}\left(\frac{b}{c}\right)^{2}\right)$
$ bc\plus{}b^{2}\equal{} x^{2}\left(1\plus{}\frac{b}{c}\right)^{2}$
$ bc\plus{}b^{2}\equal{}\left(x\left(1\plus{}\frac{b}{c}\right)\right)^{2}$
$ bc\plus{}b^{2}\equal{} a^{2}$

Okay so now that I've established that I can get on with the problem. I basically need to find the minimal solution to the equation $ bc\plus{}b^{2}\equal{} a^{2}$

With $ \angle C$ obtuse, we know c is the greatest. Also, because $ \angle A >\angle B$, $ a > b$. So we have $ c > a > b$. Also $ a\ne b\ne c$. Now it shouldn't be difficult to find the minimum from here. But it's getting kind of late so I'm just going to leave it here for now... The rest is just number theory.

EDIT- okay so I'm ready to finish this up.

$ a^{2}\equal{} b(b\plus{}c)$

If $ \gcd(b,c) > 1$ this this common factor divides a as well. Then we can divide a, b, and c by this factor to acheive a smaller solution, so we know $ gcd(b,c) \equal{} 1$

Because b and b+c are relatively prime, and their product is a square, b and b+c are both squares. First we try $ b \equal{} 1$

$ a^{2}\equal{} 1\plus{}c$
for 1+c to be a perfect square, $ c \equal{} 0$ which is impossible.

Now try $ b \equal{} 4$
Letting $ c \equal{} 5$, we get $ a \equal{} 6$ but then a>c which we don't want. We can't have c=12, since $ \gcd(b,c) \equal{} 1$ but if c=21 then a=10 which satisfies $ c > a > b$. Unfortunately it doesn't satisfy the triangle inequality.

Try $ b \equal{} 9$
We can't have $ c \equal{} 7$, since 7<9 and the next possible value of c is 16. Then $ a \equal{} 15$ but this give perimeter of $ 40$. But (15,9,40) doesn't satisfy the triangle inequality either.

Well I really don't feel like going through all this but you guys get the idea  :D

[/hide]

since no ones completed this old problem (good study for USAMO)...ill get started on it with some other method
[hide="ummm"]
In a triangle, call the side $ a$ that is across from angle $ A$, $ b$ that is across from $ B$, and $ c$ that is across from $ C$.

Clearly, $ \frac {1}{2}ab\sin{C} = \frac {1}{2}ac\sin{B} = \frac {1}{2}bc\sin{A}$,
If $ A = 2B$, and $ C > 90$, then clearly $ A + B < {90}$, so $ B < {30}$, so $ \cos{B} > \frac {\sqrt {3}}{2}$, and obviously less than $ 1$

The three equalities means that $ ab\sin{(3B)} = ac\sin{B} = bc\sin{(2B)}$, since $ \sin{(180 - 3B)} = \sin{(B)}$ and the $ \frac {1}{2}$ cancel out

using those 3 equations, you can conclude that (just by moving them around a bit)
$ b = \frac {a\sin{B}}{\sin{(2B)}} = \frac {a}{2\cos{B}}$
$ c = \frac {a\sin{(3B)}}{\sin{(2B)}} = [\frac {4{\cos^2{B}} - 1}{2\cos{B}}]a$

its not obvious from here, but $ \cos{B}$ has to be a fraction (rational number) inorder for those $ 3$ to even have to be a chance to be integers, and since $ \cos$ is being squared in the $ c$ equation, you have to choose the cosine with the smallest number+denominator (hehe this will probably lose me some points), such that $ \cos{B} > {\frac {\sqrt {3}}{2}}$ which is around $ 0.86$ so the denominator has to be $ \geq{8}$

so using my "assumption" you pick $ \frac {7}{8}$ (which satisfies what I said), youll get two equations with just an $ a$, then choose $ a$ to be as smallest possible and youll get the minimum perimeter

umm so basically if you want to finish it off
plug in that $ \frac {7}{8}$ for $ \cos{B}$
you get that $ b = \frac {4}{7}a$ and $ c = \frac {33}{28}a$ and for $ b$ and $ c$ to be minimum possible integers as well as $ a$, clearly $ a = 28$, so we have $ b = 16$ and $ c = 33$ .  It satisfies the triangle inequality and $ 16 + 28 + 33 = 77$

Of course this isnt a $ 7$ solution because of that "assumption" (but it might be trivial so i dont know...)
[/hide]


What is the least possible positive integer-value of $ n$ such that $ \sqrt{18\cdot n \cdot 34}$ is an integer?

This problem is very simple

You first look at this....

$ \sqrt{18\cdot n \cdot 34}$

it may look hard at first but you can simplify it to make it easier...

$ \sqrt{18\cdot 2\cdot 17\cdot n}$

Does it seem easier now?

$ \sqrt{36\cdot 17\cdot n}$

Well you know that $ \sqrt{36}$ $ \equal{}$ $ 6$
so we are not worried about that, but there is no
$ \sqrt{17}$, so in this case you will need
to get another $ 17$ so $ 17$ can get out of the
square root sign.

SO here is the final step:

$ \sqrt{36\cdot 17\cdot 17}$

which equals

$ 6\sqrt{17\cdot 17}$

which equals

$ 6\cdot 17$

so now we know that $ {n\equal{}17}$

(by the way, $ 6\cdot 17$ $ \equal{}$ $ 102$ if you were curious :lol: )


If anyone interest in participating following weekend's mathematical practice through AIM, please sign up in this topic.
On the practice, we'll talk about serious mathematical topic so if you have any idea please post your idea. I hope this weekend we will start up with little bit easy stuffs so I meant no stuffs like trig or log right now. If you have idea or want to sign up, please write your AIM sn in this topic.

Thank you.

[i]There is always road to the people who seek the road.[/i]

Good point. I haven't decide yet.  I hope we can cover little bit higher than algebra II because that's class I'm taking now..

[quote="Silverfalcon"]To AMC10 'cause I might take that test this year[/quote]

You can practice old AMC10 tests any time, but you can only take the real test in February. So unless you have a very unusual school year schedule, you CAN'T take the AMC 10 this year as to school year, nor can you take it this year as to calendar year (on the Gregorian calendar). Full details about the AMC 10 test can be found on the [url=http://www.unl.edu/amc/e-exams/e5-amc10/amc10.html]AMC 10 Web page[/url], including links to ordering old tests for practice.

Hope this helps! Have fun with the practice AMC10 tests; they have some interesting math in them.

[quote="Silverfalcon"]To AMC10 'cause I might take that test this year[/quote]

But the answers are already out...and they were a long time ago in fact...

You mean next year right? It's easy for kids to get confused when they're doing they're electives and scheduling for the next school year...

Okay.. Um.. Then are there any tests in upcoming Fall, or Winter?
And where can you see the old AMC tests?

Sorry but I just had surprised news from school!!
I have a lot test today so I don't think any discussino through AIM would be possible this weekend..
Sorry..

I like the idea of discussing math through aim.  It's too bad that you can't do it this weekend Silverfalcon but maybe some other people might want to get together and get something started.  I'm pretty busy also this weekend but I'll be able to be on at night on saturday but i'm not sure about sunday.  my aim sn is BaseballJoe2002  Feel free to im me.  So anyone whos interested post or im.

Saturday, 7:00 P.M. EST today.. I'm possible just that time...
And please post the topic to discuss.

Thank you.

Time schedule for mathematic discussion:

6:00 EST on Friday, 4/10/04

People who are so far available:
MithsApprentice - Probably helping out the other people =)
Rep123max
Me

Anyone wants to join may sign up with their AIM...

is this on aim? what ur sn?  i might be able to do it but im not sure.
thursday would be better

ieatrice89
is there any way u will be able to do it on thursday?

i will sign up to do some math, on thursday or whatever day is convenient.

My bad on that...

I forgot what I learned in here:

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Questions.php

And for NAS, I'll postpone the time to Saturday

And one more thing to remind:  I know you're in university, NAS, but this discussion will be cover maximum level in AMC12.  NO AIME or further ones..

Correct me if Im wrong but I think NAS lives in the same town as U Washington but is a HS student, and his HS is connected w/the university.

Why dont we have 2 math chats, one on friday and one on saturday?

I'm not from Washington and yes, we might have two ones but then we might need more people on that.. So far, we only have four people:

Tare
You
Me
NAS

Okay, you also have joml88.

I think you misunderstood me.  I was explaining that NAS was not a college student.

Why don't you PM people who were in the last chat and aks them if they want to be in this one?

1.  Yes I'm a Junoir in HS (not Collage)
2.  Yes our scedule (everyone's life) is influenced by Washington State University (not UW) that's why are break is earlier (sorry for the confusion).
3.  Anyhow I was just saying that if this was a regular occurance (over the course of weeks) that it would be nice if we could have some on Saturday.

I'd be interested in participating in one on Saturday, at least this week (as long as nothing comes up..)
screen name: loro46mpnorp

So far people joining:

Caroline
NAS
JS1527
joml88
Tare
and Me

Even we only have six people, please identify which day you are available:

For Caroline and NAS, you two are only available in Saturday, right?

And JS1527, you're available in both days, right?

For others, please tell... It's really important for me to keep the track of people since I'm conductiong this.

Thanks.

Silverfalcon

[quote="Silverfalcon"]For Caroline and NAS, you two are only available in Saturday, right?[/quote]

Correct, for me at least.

The fourth discussion:

Friday 4/23 6:00 EST

Discussion I'm involve will last one hour.

For the fourth disscussion, please sign up here.

Thanks you

I'll probably do it.. if I'm online, I'll be there.

I'll try, I probably have the same scheduling aka school conflict, but I could be there by 4:00 PDT.

I will try to be there but I am not sure.  I'll make the chatroom if I am online but if not just invite yourself to MathChat. =)


function
real analysis
real analysis unsolved

Construct a continuous function $ f: \mathbb{R} \to \mathbb{R}$, which takes every real value exactly three times.  $ ( \Longleftrightarrow f(x) \equal{} c$ has tree real solutions for every $ c \in \mathbb{R} )$

On the first page of this forum right now (posted 3 days ago) is the following thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261571


geometry
quadratics
angle bisector
Pythagorean Theorem
geometry proposed

Let $\Omega$ be a semicircumference with diameter $\overline{BC}$ and $A$ is a variable point on $\Omega.$ The medians $BM_b$ and $CM_c$ of $\triangle ABC$ issuing from $B$ and $C$ and the internal angle bisector of $ \angle BAC$ bound a triangle $\triangle XYZ.$ Prove that the area of $\triangle XYZ$ is maximum when the altitude from $A$ to $BC$ equals the fifth part of $\overline{BC}.$

[quote][color=darkred][b][u]Lemma[/u].[/b] In $ \triangle ABC$ consider the points $ \left\|\begin{array}{ccc} M\in (BC) & ; & \frac {MB}{MC} \equal{} m \\
 \\
N\in (CA) & ; & \frac {NC}{NA} \equal{} n \\
 \\
P\in (AB) & ; & \frac {PA}{PB} \equal{} p\end{array}\right\|$ for which denote $ \left\|\begin{array}{c} X\in BN\cap CP \\
 \\
Y\in CP\cap AM \\
 \\
Z\in AM\cap BN\end{array}\right\|$ .

Then the area of $ \triangle XYZ$ is given by the relation $ [XYZ] \equal{} \frac {(1 \minus{} mnp)^2}{(1 \plus{} m \plus{} mn)(1 \plus{} n \plus{} np)(1 \plus{} p \plus{} pm)}\cdot [ABC]$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b]  Apply the [u][b]Menelaus[/b]' theorem[/u] to the following transversals in the mentioned triangles :

$ \left\|\begin {array}{cccc} \overline {BZN}/ACM\ : & \frac {BM}{BC}\cdot\frac {NC}{NA}\cdot\frac {ZA}{ZM} \equal{} 1 & \Longleftrightarrow & \frac {ZA}{m \plus{} 1} \equal{} \frac {ZM}{mn} \equal{} \frac {AM}{1 \plus{} m \plus{} mn} \\
 \\
\overline {CXP}/BAN\ : & \frac {CN}{CA}\cdot\frac {PA}{PB}\cdot\frac {XB}{XN} \equal{} 1 & \Longleftrightarrow & \frac {XB}{n \plus{} 1} \equal{} \frac {XN}{np} \equal{} \frac {BN}{1 \plus{} n \plus{} np} \\
 \\
\ \overline {AYM}/CBP\ : & \frac {AP}{AB}\cdot\frac {MB}{MC}\cdot\frac {YC}{YP} \equal{} 1 & \Longleftrightarrow & \frac {YC}{p \plus{} 1} \equal{} \frac {YP}{pm} \equal{} \frac {CP}{1 \plus{} p \plus{} pm}\end{array}\right\|$ . Prove easily that

$ [XYZ] \equal{} [ABC] \minus{} [ABM] \minus{} [BCN] \minus{} [CAP] \plus{} [CXN] \plus{} [AYP] \plus{} [BZM]$ $ \implies$

$ \frac {[XYZ]}{[ABC]} \equal{} 1 \minus{} \frac {BM}{BC} \minus{} \frac {CN}{CA} \minus{} \frac {AP}{AB} \plus{} \frac {MB}{BC}\cdot\frac {ZM}{AM} \plus{} \frac {NC}{CA}\cdot\frac {XN}{BN} \plus{} \frac {PA}{AB}\cdot\frac {YP}{CP}$ $ \implies$

$ \frac {[XYZ]}{[ABC]} \equal{} 1 \minus{} \sum\frac {BM}{BC}\cdot\left(1 \minus{} \frac {ZM}{AM}\right) \equal{}$ $ 1 \minus{} \sum\frac {BM}{BC}\cdot\frac {ZA}{AM} \equal{}$ $ 1 \minus{} \sum\frac {m}{m \plus{} 1}\cdot\frac {m \plus{} 1}{1 \plus{} m \plus{} mn} \equal{}$

$ 1 \minus{} \sum\frac {m}{1 \plus{} m \plus{} mn}$ $ \implies$ $ \frac {[XYZ]}{[ABC]} \equal{} 1 \minus{} \frac {m}{1 \plus{} m \plus{} mn} \minus{} \frac {n}{1 \plus{} n \plus{} np} \minus{} \frac {p}{1 \plus{} p \plus{} pm}$ $ \implies$

$ \ldots\ldots$ $ \implies$ $ \boxed {\ [XYZ] \equal{} \frac {(1 \minus{} mnp)^2}{(1 \plus{} m \plus{} mn)(1 \plus{} n \plus{} np)(1 \plus{} p \plus{} pm)}\cdot [ABC]\ }$ .[/color]

[quote="Luis González"] [color=darkred] Let $ ABC$ be a $ A$ - right triangle with two fixed  vertices $ B$ , $ C$ and a mobile vertex $ A$  on the its circumcircle . Denote

the midpoints $ N$ , $ P$ of the sides $ [CA]$ , $ [AB]$ respectively and the point $ M\in (BC)$ for which $ \widehat {MAB}\equiv\widehat {MAC}$ .

The lines $ AM$ , $ BN$ , $ CP$ delimit a triangle $ XYZ$ . Prove that the area of $ XYZ$ is maximum if and only if $ \boxed {\ h_a \equal{} \frac a5\ }$ .[/color] [/quote]
[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $ a \equal{} 1$ , i.e. $ b^2 \plus{} c^2 \equal{} 1$ and $ x \equal{} \frac cb\ge 1$ . Observe that $ [ABC] \equal{} \frac {bc}{2(b^2 \plus{} c^2)} \equal{} \frac {x}{2(x^2 \plus{} 1)}$ .

Apply the upper lemma for $ n \equal{} p \equal{} 1$ , $ m \equal{} x$ and  obtain $ [XYZ] \equal{} \frac {x(1 \minus{} x)^2}{6(m \plus{} 2)(2m \plus{} 1)(m^2 \plus{} 1)}$ , i.e. $ [XYZ] \equal{} \frac 16\cdot f(x)$ ,

where $ \boxed {\ fx) \equal{} \frac {x(1 \minus{} x)^2}{(2x^2 \plus{} 5x \plus{} 2)(x^2 \plus{} 1)}\ ,\ x\ge 1\ }$ . Thus, $ [XYZ]$ is maximum iff $ f(x)$ is maximum on $ [1,\infty )$ . Prove easily that

$ f'(x)\ .s.s.\ \minus{} 2(x \minus{} 1)\left(x^5 \minus{} 3x^4 \minus{} 7x^3 \minus{} 7x^2 \minus{} 3x \plus{} 1\right) \equal{}$ $ \minus{} 2(x \minus{} 1)(x \plus{} 1)\left(x^4 \minus{} 4x^3 \minus{} 3x^2 \minus{} 4x \plus{} 1\right)\ .s.s.$

$ \minus{} \left(x^2 \plus{} x \plus{} 1\right)\left(x^2 \minus{} 5x \plus{} 1\right)\ .s.s.\ \minus{} \left(x^2 \minus{} 5x \plus{} 1\right)\ .$ Therefore, $ f(x)\ ,\ x\ge 1$ is maximum iff $ \boxed {\ x: \equal{} x_{\mathrm{max}} \equal{} \frac {5 \plus{} \sqrt {21}}{2}\ }\ .$

On other hand, $ 5h_a \equal{} a$ $ \Longleftrightarrow$ $ 5ah_a \equal{} a^2$ $ \Longleftrightarrow$ $ 5bc \equal{} b^2 \plus{} c^2\Longleftrightarrow x^2 \minus{} 5x \plus{} 1 \equal{} 0$ , i.e. $ x \equal{} \frac {5 \plus{} \sqrt {21}}{2}\ .$ In conclusion,

the area of the triangle $ [XYZ]$ is maximum $ \Longleftrightarrow$ $ 5h_a \equal{} a$ $ \Longleftrightarrow$ $ \frac cb \equal{} \frac {5 \plus{} \sqrt {21}}{2}$ $ \Longleftrightarrow$ $ \boxed {\ \frac {a}{2\sqrt 5} \equal{} \frac {b}{\sqrt 7 \minus{} \sqrt 3} \equal{} \frac {c}{\sqrt 7 \plus{} \sqrt 3}\ }\ .$[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle with two fixed  vertices $ B$ , $ C$ and a mobile vertex $ A$  on the its circumcircle, where $ A\le 90^{\circ}$ .

Denote the midpoints $ N$ , $ P$ of the sides $ [CA]$ , $ [AB]$ respectively and the point $ M\in (BC)$ for which $ \widehat {MAB}\equiv\widehat {MAC}$ . The lines $ AM$ ,

$ BN$ , $ CP$ delimit a triangle $ XYZ$ . Prove that the area of $ XYZ$ is maximum if and only if $ \boxed {\ \left(3\sqrt 2 \plus{} 4\sin\frac A2\right)\cdot h_a \equal{} 2\cos\frac A2\cdot a\ }$ .[/color] [/quote]

Now, I would like to present my solution. By Routh theorem for the triangle $\triangle GMN$ bounded by $BM_b,$ $CM_c$ and the A-internal bisector, we have:

$\frac{|\triangle GMN|}{|\triangle ABC|}=\frac{( \frac{c}{b}-1)^2}{(1+\frac{c}{b}+\frac{c}{b})(1+1+\frac{c}{b})(1+1+1)}=\frac{b^2+c^2-2bc}{6(b^2+c^2)+15bc}$

Substituting $a^2=b^2+c^2$ (Pythagorean theorem) and $2|\triangle ABC|=a \cdot h_a= bc$ yields

$|\triangle GMN|= f(h_a)=\frac{a^2h_a-2a{h_a}^2}{12a+30h_a} \Longrightarrow f'(h_a)=\frac{60a{h_a}^2+48h_aa^2-12a^3}{(12a+30h_a)^2}.$

Solving the quadratic $f'(h_a)=0$ reveals the maximum of $f(h_a)$ at $h_a=\frac{_1}{^5}a.$

Thank you. I didn't know it.
See above my proof of your nice problem.
The proof of the general case is similarly.
In the case $ A > 90^{\circ}$ is a bit of studing, but isn't hardly.
I am in Florida. Here is the hour 6:50 PM.
And in your country ([b]Venezuela[/b], I think !)) is same time ?!


trigonometry
function
integration
algebra
polynomial
algebra unsolved

sum of tangents
solve:

1) (tg1)^2 + (tg2)^2 + ... + (tg89)^2

2) (tg1)^2 + (tg3)^2 + ... + (tg89)^2 Only odd

I need the solution( with complex, and the solution with trigonometry) , thanks!

I'm not so inspired now, but you might use tg(x) = e^ix - e^-ix/(e^ix + e^-ix)i.

$ tg^2x\plus{}tg^2(90\minus{}x)\equal{}\frac{4}{sin^22x}\equal{}\frac{4}{1\minus{}cos^22x}\equal{}4(1\plus{}cos^22x\plus{}cos^42x\plus{}...).$
Therefore \[ \sum_{k\equal{}1}^{89}tg^2k\equal{}A_{90}\equal{}\sum_{k\equal{}1}^{90\minus{}1}tg^2{\frac{\pi k}{2*90}}\equal{}\sum_{m\equal{}0}^{\infty}S_m(90).\]
Were $ S_m(n)\equal{}2(\sum_{k\equal{}1}^n cos^{2m}\frac{\pi k}{n}  \minus{}1)$.
Consider function  \[ f_m(x)\equal{}cos^{2m}x\equal{}\sum_{k\equal{}0}^{\infty} a_k(m)cos kx, a_0(m)\equal{}\frac{1}{\pi}\int_{0}^{\pi}f_m(x)dx\equal{}\frac{(2m\minus{}1)!!}{(2m)!!},\]
\[ a_k\equal{}\frac{2}{\pi}\int_0^{\pi}f_m(x)coskx dx.\]
Therefore $ S_m(n)\equal{}2\sum_{k\equal{}1}^n f_m(\frac{k\pi }{n}) \minus{}2\equal{}2n(a_0(m)\plus{}a_{2n}(m)\plus{}a_{4n}(m)\plus{}...)\minus{}2$

The answer for sum of 1 to 89 is  4005   u_u

i dont  understand.

and  the sum of  only odd???


trigonometry
inequalities
rearrangement inequality
inequalities proposed

real numbers a,b,c
$ \sum_{cyc}a^4\plus{}\sum_{cyc}ab^3 \geq 2\sum_{cyc}a^3b$
hold for $ (a,b,c)\equal{}(1\plus{}2\cos{\frac{\pi}{9}},1\plus{}2\cos{\frac{2\pi}{9}},\minus{}1)$

we have

$ \sum{a^4}\plus{}\sum{a^2b^2}\geq2\sum{a^3b} \Leftrightarrow \sum a^2(a\minus{}b)^2\geq0$ which is true

and we have by the rearangment inequality

$ \sum ab^3\geq \sum a^2b^2$

and so the inequality follows :)

Are you sure ? We don't have the condition that we can apply the rearrangement inequality
The solution follows immediately from a nice identity
\[ \sum {{a^4}}  \plus{} \sum {a{b^3}}  \minus{} 2\sum {{a^3}b}  \equal{} \sum {{{\left( {{a^2} \minus{}ab \plus{} ac \minus{} {c^2}} \right)}^2}}\]

[quote="Evariste-Galois"]we have

$ \sum{a^4} \plus{} \sum{a^2b^2}\geq2\sum{a^3b} \Leftrightarrow \sum a^2(a \minus{} b)^2\geq0$ which is true

and we have by the rearangment inequality

$ \sum ab^3\geq \sum a^2b^2$

and so the inequality follows :)[/quote]
here are  real numbers

[quote="dvtvd"]Are you sure ? We don't have the condition that we can apply the rearrangement inequality
The solution follows immediately from a nice identity
\[ \sum {{a^4}} \plus{} \sum {a{b^3}} \minus{} 2\sum {{a^3}b} \equal{} \sum {{{\left( {{a^2} \minus{} ab \plus{} ac \minus{} {c^2}} \right)}^2}}\]
[/quote]
yes ,the thought is right ,but your identity is wrong

[quote="ye109"]real numbers a,b,c
$ \sum_{cyc}a^4 \plus{} \sum_{cyc}ab^3 \geq 2\sum_{cyc}a^3b$
hold for $ (a,b,c) \equal{} (1 \plus{} 2\cos{\frac {\pi}{9}},1 \plus{} 2\cos{\frac {2\pi}{9}}, \minus{} 1)$[/quote]

$ \sum{{a^{4}}}\plus{}\sum{a{b^{3}}}\minus{}2\sum{{a^{3}}b}\equal{}\frac{1}{2}\sum{{{\left({{a^{2}}\minus{}b^2\minus{}ab\plus{}bc}\right)}^{2}}} \ge 0$

Equality holdes for :  $ (a\equal{}b\equal{}c)$   or  $ (a\equal{} \minus{}1$ , $ b\equal{} 1 \plus{} 2\cos{\frac {\pi}{9}}$ , $ c\equal{} 2\cos{\frac {2\pi}{9}})$

[quote="ye109"]real numbers a,b,c
$ \sum_{cyc}a^4 \plus{} \sum_{cyc}ab^3 \geq 2\sum_{cyc}a^3b$
hold for $ (a,b,c) \equal{} (1 \plus{} 2\cos{\frac {\pi}{9}},1 \plus{} 2\cos{\frac {2\pi}{9}}, \minus{} 1)$[/quote]

This problem was first seen in gazeta mathematica by Vasc. He uses the transformation $ b \equal{} a \plus{} p$, $ c \equal{} a \plus{} q$ etc...


can anyone prove or disprove that -

In a cyclic polygon, for any triangulation, the sum of inradii of each of the triangles in a specific triangulation is constant.

it's true and [url=http://www.mathlinks.ro/viewtopic.php?p=329668#p329668]here[/url] is the proof


arithmetic sequence
combinatorics proposed
combinatorics

Each point of a circle is painted with one of three colors. Prove that there
exist three points of the same color on the circle which are vertices of an
isosceles triangle.

By Van der Waerdens theorem, there exists a V s.t. no matter how you color V consecutive integers with 3 colors, there will be 3 of the same color in arithmetic progression. Consider a regular V-gon in the circle. 3 in arithmetic progression implies isoceles triangle. Yay.

Cheers,

Rofler


could people from different colleges post which subjects are best at thier respective colleges?
it will be a great help
thanks

You're really going to have to be more specific than that.  What are you interested in?  And you have to look at more than "Which school has the best this?"  Top tier schools in a field are comparable, so depending on who you ask, you may get very different answers.

for instance which collegeis good fr geometry and numbertheory

Are you applying as an undergraduate or a grad student?  If you're applying as an undergraduate, that's not really that important since you have to get down the basics-vector calculus, linear algebra, analysis, algebra before you can move on to the more advanced stuff.

[quote="JRav"]Again, not really that important.  You're not going to be taking courses in geometry until you get through the basics first.  Look for a school that suits your needs as a student, has a good student body, and you're comfortable with the program.

For example, the school I go to is not world-famous for it's math program (still good, just not Harvard/Princeton/MIT-good).  But I really like the program, it gives me a lot of flexibility, it lets me take graduate courses as soon as I'm ready, and is small so I can get to know the professors.[/quote]
where do u go?

I'm going to chance a guess, Johns Hopkins?

Indeed.  I'm pretty sure I've mentioned it at some point.  Want more information?

And I'm glad you remembered the "s" at the end of Johns!


geometry
Harvard
college
Support
3D geometry
sphere
inequalities

i am a student pursuing ms in geology from iit kharagpur,india.however i am facing a dilemma:i have a deep interest in mathematics and i want to pursue INDEPENDENT research in the subject.the problem is thati don't know how to begin.Here,in india,our primary approach while learning a subject is problem solving.in mathematics ,for instance,we may sometimes soar to olympiad level problems,never beyond that.I don't think that research in this subject is primarily concerned with solving problems.more important is to find beauty in equations.Can you please elucidate to what extent i am true?Also ,i must have something to think upon.I feel that mathematics  today hs developed to a level where no  individual can proclaim mastery over every field of this subject.If that is the case how am i supposed to  learn about every development in this field?Also,i am currently studying quantum mechanics.Please suggest some topic which would be relevant in this field.
thank you

As far as I know, [url=http://en.wikipedia.org/wiki/Operator_algebra]Operator Algebras[/url] is a mathematical area that is somewhat relevant to quantum mechanics. (Did I mention that Texas A&M has a strong research program in this area?) However, I suggest that you precede your attempts at independent research in mathematics by several years of formal schooling.

[quote="_anant"]I don't think that research in this subject is primarily concerned with solving problems.more important is to find beauty in equations.[/quote]

Half of mathematics research is finding the right problems to try to solve; in other words, asking the right questions.  It takes creativity to do this, as well as a deep understanding of the field in which you're working.

There is also a big difference between Olympiad problem-solving and research:  you know that every Olympiad problem has a solution, and moreover you know that that solution involves known techniques.  Mathematical research may require that you invent entirely new techniques; that is part of the challenge!

[quote="_anant"]I feel that mathematics  today hs developed to a level where no  individual can proclaim mastery over every field of this subject.[/quote]

I think this was already true at least half a century ago.  The fields of mathematics are extraordinarily diverse.

[quote="_anant"]If that is the case how am i supposed to  learn about every development in this field?[/quote]

Well... you aren't.  As you progress in your mathematical research, you'll tend towards a specialty.  For example, my research mentor at RSI is a graduate student at Harvard specializing in algebraic geometry, which, albeit a vast and exciting subject, is still only one part (albeit a central part) of modern mathematics.

If you haven't already, refer to [url=http://en.wikipedia.org/wiki/Mathematical_formulation_of_quantum_mechanics]this discussion of the mathematical formulation of quantum mechanics[/url].

[quote](albeit a central part)[/quote]
It would be interesting to know which parts of modern mathematics are central and which are not.

I freely admit to not having enough of an understanding of the state of modern mathematics to actually support that statement with any authority; that comment was largely based off of the relevant quote

[quote="en.wikipedia.org"]Algebraic geometry is a branch of mathematics which, as the name suggests, combines techniques of abstract algebra, especially commutative algebra, with the language and the problematics of geometry. It occupies a central place in modern mathematics and has multiple conceptual connections with such diverse fields as complex analysis, topology and number theory.[/quote]
which is supported by my own experience with algebraic geometry.

[quote="mlok"]
It would be interesting to know which parts of modern mathematics are central and which are not.[/quote]
Just think of mathematics as a surface of a sphere on which one stays somewhere and can see about 1 mile in each direction from his position beyond which everything seems foggy and unclear. What will he think about where the center of the known universe is? :P

As to the differences between olympiad problem solving and mathematical research, the only ones I am really sure about is that you have much more relaxed time constraints when you are doing research and that it is no longer necessary to solve most of the problems to get a high score :lol:.

[quote="t0rajir0u"]  Half of mathematics research is finding the right problems to try to solve; in other words, asking the right questions. [/quote]
Following mlok I am tempted to say "It would be interesting to know how to distinguish between a "right" and a "wrong" mathematical question" (or why it is "half" and not 45.239%). In my humble opinion, every mathematical question nobody knows how to answer is a good one, and we have never had a shortage in people asking questions while people being able to answer them are still badly missing.

What one should really understand about mathematical research is that it is an enterprise in which you fail miserably 9 out of 10 times if you are a genius and 99 out of 100 times if you  are at the average level. You'll have just to get used to it or to quit in despair.

As to "finding beauty in equations", I'm not really sure what you are asking about here. There are equations in mathematics and there is a concept of beauty in it, of course, but the main goal still remains to solve problems.

At last, I do not really understand the word INDEPENDENT in _anant's post. There is just no such thing: you'll work on questions posed by other people using techniques invented by other people; your solution if you find one will be looked at by other people, so where is the independence here? Well, when you start flying high enough, you'll find one though it will be not quite what you might currently think it is but asking for it in the very beginning is the same as asking for independence in your way to play chess before you even know how to position the pieces on the board. No wonder that the result is that you don't know how to start. Fortunately, the way out is simple: forget about independence until better times and go through a few years of standard apprenticeship.

As to mastering all the mathematics, you cannot do this, indeed, but the good news is that nothing prevents you from mastering any particular part of it you might get interested in now  or later. Learning [b]every[/b] development in the field is no more necessary for becoming a good mathematician than reading every book ever written to become a good writer, though you may correctly speculate that [b]some[/b] amount of reading will be quite useful in the latter case.

[quote="fedja"][quote="t0rajir0u"]  Half of mathematics research is finding the right problems to try to solve; in other words, asking the right questions. [/quote]
Following mlok I am tempted to say "It would be interesting to know how to distinguish between a "right" and a "wrong" mathematical question" (or why it is "half" and not 45.239%). [/quote]

"Half" is rhetorical, of course.  When I say "right" question, I mean a question that leads to interesting ideas most readily.  It's not that there are "wrong" questions so much as there are more and less enlightening ones.

[quote="fedja"][quote="mlok"]

At last, I do not really understand the word INDEPENDENT in _anant's post. There is just no such thing: you'll work on questions posed by other people using techniques invented by other people; your solution if you find one will be looked at by other people, so where is the independence here? Well, when you start flying high enough, you'll find one though it will be not quite what you might currently think it is but asking for it in the very beginning is the same as asking for independence in your way to play chess before you even know how to position the pieces on the board. No wonder that the result is that you don't know how to start. Fortunately, the way out is simple: forget about independence until better times and go through a few years of standard apprenticeship.

.[/quote][/quote]
by independent research,i meant that i find my own questions to think upon,not by enrolling myself into a standard phd course,  and to find solutions to them all by myself.Of course, i will need help of internet and other material.Please don't exaggerate the meaning of the phrase.

[quote="_anant"]
by independent research,i meant that i find my own questions to think upon,not by enrolling myself into a standard phd course,  and to find solutions to them all by myself.[/quote]
Well, that also is hardly possible to do in the beginning. I know a few people who would, probably, be capable of that but even they followed the standard path of enrolling into a PhD program, having an advisor, etc. and it didn't hurt them in the slightest, quite the opposite. The reason is that you need to know the terrain well enough to be your own guide. The hardest part of it is to pick a problem that matches your strength. The difficulty is that an attempt to solve some "real" question will, probably, be futile in the beginning. What many people usually start with are some little things like proving some inequality or lemma or whatever for some project of their advisors. In many cases, the advisor knows how to do it himself and can provide hints along the way. He also can see from the way you work on the problem what your srtong and weak sides are, what are the gaps in your education, etc. and can give you some good recommendations. Basically, an ideal way to start is solving the problem that is already solved (or almost solved) by someone else who can give you help along the way when you get stuck. Next time you may try to use less help; eventually you'll be working almost entirely on your own, but to attempt to do it all by youself from the very beginning is the same as an attempt to climb a high mountain without any previous experience, with no equipment and no guidance. Getting trained for mathematical research is no different from getting trained for any other craft: you get somebody who has the skill and start with patiently learning what he can teach you. When you see that you can do all he can do, you find some better craftsman to learn from and so on. Eventually you'll get to the point when other people will learn from you too but you'll never stop learning from them. At least that's the standard way of becoming a master and I do not see why one should reject  it :). As I said, my advice is to forget about independence in the beginning no matter what you mean by it; it just doesn't exist in the sense you want it.

_anant just got completely blew away by this post. a good lesson learned for anant. be more knowledgeable next time before making a post.

lenny, not tht he got blew away or anything. If he was knowledgeable about this why would ask for help? He was not sure so that is why he has asked.


Given three positive numbers. Prove that:
\[{ab \over c(a+b)^{2}}+{bc \over a(b+c)^{2}}+{ca \over b(c+a)^{2}}\geq{9 \over 4(a+b+c)}. \]

write our inequality like this :

$\Longleftrightarrow abc( \sum{\frac{1}{(ac+bc)^{2}}}) \geq  \frac{9}{4\sum{a}}$

but :$\sum{\frac{1}{(ac+bc)^{2}}}\geq  \frac{9}{4abc \cdot \sum{a}}$  :wink:

[quote="Lovasz"][quote="pohoatza"]
but :$\sum{\frac{1}{(ac+bc)^{2}}}\geq \frac{9}{4abc \cdot \sum{a}}$  :wink:[/quote]
What's your [b] but [/b]?[/quote]

Let $ab=z$ $bc=y$ $ca=x$, we can get the Iran 96 inequality


analytic geometry
geometry unsolved
geometry

Given a square $ABCD$ and any two points $E,F$ lying on $BC$ and $CD$ such that $DF$ and $BE$ are not equal. Now let $AE$ and $AF$ intersect diagonal $BD$ at $P,Q$ respectively. If we are given $BP.CE = DQ.CF$ prove that the points $C,P,Q,E,F$ are all concyclic.
 AR

Put the origin of Cartesian plane in $A$ such that
$A(0,0),B(1,0),C(1,1),D(0,1)$ and $E(1,e)$ and $F(f,1)$ where
$e,f\in (0,1)$. Now, it isn't hard to calculate that
$P(\frac1{e+1},\frac{e}{e+1})$ and $Q(\frac{f}{f+1},\frac1{f+1})$.
Condition $BP\cdot CE=DQ \cdot CF$ now becomes
\[\frac{e}{e+1}(1-e)=\frac{f}{f+1}(1-f).\] That is equivalent with
$(f-e)(ef+e+f-1)=0$, but $e\not= f$ so $ef+e+f-1=0$.

On the other hand, $Q,E,C,F$ are concyclic iff $EQ\bot FQ$ which is
equivalent with \[\frac{e-\frac1{1+f}}{1-\frac{f}{f+1}}\cdot \frac1{f}=-1.\] The last equality stands iff $ef+e+f-1=0$ and that's
it :)

Bye

Does anybody know a solution without coordinates?


trigonometry
geometry
inequalities
inequalities proposed

Let  ABC be a triangle, and  A,B  ,C  its angles. Prove
$ 6\sqrt{3tan(A/2).tan(B/2)\plus{}3}\plus{}4\sqrt[3]{3tan(B/2).tan(C/2)\plus{}7}\plus{}3\sqrt[4]{3tan(A/2).tan(C/2)\plus{}80}\le\ 29$

You can write $ \tan(\frac{A}{2})$, etc. in terms of the lengths of the tangents to the incircle from the vertices in order to turn the inequality into a 'normal' algebraic inequality.


[quote="wengzilu"]A circle with points (-2,-2), (1,4) and (x,2006) can exist if x =?[/quote]
For any three points on the plane there exists a circle which contains them, so just $x\in\mathbb{R}$.  :maybe:

[quote="lordWings"]Except when [hide]$x=1002$, then they form a line $y=2x+2$[/hide]:P[/quote]
Sorry.  :blush:
But maybe $\text{wengzilu}$ forgot about the fourth point? Then the problem is a bit more difficult.


Find all positive numbers  $\; p\;  $ for which the inequality
\[ \left(\frac{h_a}{a}\right)^{p}+ \left(\frac{h_b}{b}\right)^p +
\left(\frac{h_c}{c} \right)^{p} \ge  \frac{ 3^{1 + \frac{p}{2}
}}{2^p }
\] is satisfied in any acute triangle.

For $p=2$ the inequality is easy.Well,is equivalent with the one proposed at iran 1996 for which I gave a new proof. ;)

$p \ge 2$ follows from $p = 2$, by the Power-Mean inequality, since

$\left(\frac{\left(\frac{h_a}{a}\right)^p + \left(\frac{h_b}{b}\right)^p + \left(\frac{h_c}{c}\right)^p }{3}\right)^{\frac{1}{p}} \ge \left(\frac{\left(\frac{h_a}{a}\right)^2 + \left(\frac{h_b}{b}\right)^2 + \left(\frac{h_c}{c}\right)^2 }{3}\right)^{\frac{1}{2}} \ge \frac{\sqrt{3}}{2}$

[quote="Mildorf"]$p \ge 2$ follows from $p = 2$, by the Power-Mean inequality, since ...[/quote]
[b]What about the case $ p \in (0,2)  $ ?? [/b]


ABC  is  triangle .Show  that
[tex] R\Sigma(\frac{m_a}{a})^2 \geq 3\sqrt(\Sigma a^2)\  [/tex]

there must be 4 missing on the left. else the inequality fails for the equilateral triangle if i didn't miscalculate.

Peter

We know m_a<=(b+c)/2.
Now we know a,b,c are medians in the triangle with medians as length.
So we should get a<=k(m_b+m_c).
Now replace and we get sum{ma/a}>=c( a constant)
Now use 9R^2>=a^2+b^2+c^2 multiply and Bingo!

can you give your proof for $\sum (\frac{m_a}{a})^2\geq k$ more explicitly? and can you derive the result $\sum (\frac{m_a}{a})^2\geq \frac{9}{4}$? i think your bounds $m_a\leq \frac{b+c}{2}$ don't give a result that is sharp enough to proof the inequality. it is only true if $a=0$ i think.

Peter

[Edit: typo corrected]

Yes Peter I'm wrong. i thought m_a=(b+c) can happen when a=b=c and hence mistake.
But is it true that sum{m_a/a}>=3sqrt(3)/2 ?

now, we want to proof that $\sum (\frac{m_a}{a})^2\geq \frac{9}{4}$. this is quite easily done: we have $m_a^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}$(really easy to proof), so adding $\frac{3}{4}$ on both sides and multiplying with 2, we need to proof that $\sum \frac{a^2}{b^2}\geq 6$, where the sum is symmetric. but that is clear since $\frac{a^2}{b^2}+\frac{b^2}{a^2}\geq 2$ by AM-GM. so the inequality follows since $9R^2\geq a^2+b^2+c^2$, which isn't hard as well.

Peter

[Edit: $ was missing, typo corrected]

Not surprisingly :-), this can be proved by the inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558

By taking $x=1/a^2, y=1/b^2, z=1/c^2$ and M the centroid of the triangle, we get

\[
\left(\frac{m_a}{a}\right)^2 + \left(\frac{m_b}{b}\right)^2 + \left(\frac{m_c}{c}\right)^2 \geq \frac{9}{4}\cdot \frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2} \geq  \frac{9}{4}.
\]


Find the largest interval, $ (a, b]$, such that for $ a < k \leq b$, the limit exists and is finite.

$ \lim_{x \rightarrow \infty}\frac {7^{kx} \plus{} 8}{7^{5x} \plus{} 6}$

? $ < k \leq$ ?

If $ k>5$, then the limit blows up because the numerator will eventually grow larger than the denominator.  If $ k\leq 5$, then the limit will exist.  If $ k\equal{}5$, then the limit is 1, and if $ k<5$, then the limit is 0 since the denominator will eventually grow much more rapidly than the numerator.  Let's suppose $ k$ is negative.  Then we get
$ \lim_{x\rightarrow\infty}\frac{7^{\minus{}kx}\plus{}8}{7^{5x}\plus{}6}\equal{}\lim_{x\rightarrow\infty}\frac{\frac{1}{7^{kx}}\plus{}8}{7^{5x}\plus{}6}\equal{}\lim_{x\rightarrow\infty}\frac{1\plus{}8\cdot 7^{kx}}{7^{kx}(7^{5x}\plus{}6)}\equal{}\lim_{x\rightarrow\infty}\frac{1\plus{}8\cdot 7^{kx}}{7^{(k\plus{}5)x}\plus{}6\cdot 7^{kx}}\equal{}0$
Hence I would say that the largest such interval is $ (\minus{}\infty,5]$.

[quote="ben_b"]Find the largest interval, $ (a, b]$, such that for $ a < k \leq b$, the limit exists and [b]is finite[/b].

$ \lim_{x \rightarrow \infty}\frac {7^{kx} \plus{} 8}{7^{5x} \plus{} 6}$

? $ < k \leq$ ?[/quote]

I'm pretty sure that both 0 and 1 are finite numbers.  It asks that the limits be finite, not the interval.


At a local gymnastic competition, five judges gave points for a routine by one gymnast. Disregarding the highest and lowest scores produced an average score of 9.46. If you only ignore the highest score, then the average would be 9.40 points. Ignoring the lowest score instead produces a score of 9.50. What is the difference between the highest and lowest scores given by the judges?

Our total score is 47.3 (without high or low score). Disregarding the highest score, the total is 37.6. Thus the high score was 9.7. Disregarding the lowest score, the total is 38. Thus the low score was 9.3. The difference is then $ \frac{2}{5}$.


Hello everybody !

I have installed Asymptote (it works in interactive mode !) but I have a message labeled Xasy in the black window wich I write here:

/cygdrive/C//Program Files/Asymptote/xasy.py:1.1:invalid token '#'
/cygdrive/C//Program Files/Asymptote/xasy.py:1.3:sintax error
error: could not load module 'cygdrive/C//Program Files/Asymptote/xasy.py'

I don't know what is that. Can anybody help me please ? thanks a lot !

Do you get this message all the time or just for one particular file? If the latter is the case, what is the file name and what is the code in that file. In particular, is there [b]import xasy;[/b] line anywhere? If not, what exactly are the names of packages and other files that you import (if any)?

It seems to me that Asymptote tries to import either xasy or xasy.py as an .asy file and fails to realize that # stands for the comment symbol there. This is not what it should normally do but you can force such behavior, so at this point I suspect the problem is with your file, not with Asymptote itself...


[quote="Imperial Effect"][quote="ashrafmod"]$\sqrt{x-1}-\sqrt{x-\sqrt{3x}}=\frac{1}{2}\sqrt{x}$[/quote]
Is it 4?  :huh:[/quote]

Yes 4 is the solution to $\sqrt{x-1}-\sqrt{x-\sqrt{3x}}=\frac{1}{2}\sqrt{x}\iff 225x^{4}-1872x^{3}+4192x^{2}-1280x+256=0$


I'm trying to apply to some academic programs that ask me to fax or mail test results for certain tests, and I think I might be able to turn in my AIME or AMC 10 score for part of the applications.

Is there anywhere that I can find my score on some official-looking piece of paper that has my AIME or AMC score?

Hmm... I don't know such place but can't you just use your AIME qualification certificate? I think that should suffice... Of course, this is assuming that you have qualified to AIME.

Your teacher will have a copy of the "official" score report.


Let $ e$ and $ b_{n}$ a decreasing sequence in$ (0,1)$ . The sequence $ a_{n}$ satisfies

$ e+a_{n+1}\le a_{n}(1+\frac{b_{n}}{n})$

Prove that $ lim$$ inf$ $ a_{n}\gneq 0$ :D


What is the most efficient way to allocate memory for big matrices in the heap?

I use this method: I allocate memory for $n \times n$ vector then store the appropriate pointers in a vector of pointers (*s[n]). Then I access the data via s[i][j]. (this seems to work, but I used it just once, during the olympiad that took place this morning)

I also tried to use something like: we take a vector of pointers *s[n] then for each i, s[i]=new type[n]. But that causes errors. (I think that happens because it doesn't use memory properly when allocating)

Anyway, do you know any better ways?

Yes, use C++ with STL and rely on the implemented automatic memory management. :) It works OK and takes no time at all to use in your code. And I'm serious. If you know how to use STL you have a big advantage.

But back to your question - the first method is fine, I'd do it like this too. The second one is harder to code but has one advantage - you can resize the matrices. I see no reason why it wouldn't work although I can't say I've tried it :). But as long as allocating one n x n vector works for me I would stick with it. (If your question was if it works - allocating n * n memory and then accessing it as s[i][j] - then the answer is it does.)


How many cubic yards of gravel will be required for a walk surrounding a rectangular lawn 200 yards long and 100 yards wide if the walk is to be 3 feet wide and 3 inches deep? Express your answer as a mixed number.

We will find the area above the ground and then multiply by the depth. We have $ (200 \plus{} 2\cdot1)(100 \plus{} 2\cdot1) \minus{} 200\cdot100 \equal{} 202\cdot102 \minus{} 200\cdot100 \equal{}$
$ 4(101\cdot51 \minus{} 100\cdot50) \equal{} 4(100 \plus{} 50 \plus{} 1) \equal{} 604$.
Multiplying by the depth of $ \frac1{12}$ yards, we come up with a final answer of $ \frac {151}3 \equal{} \boxed{50\frac13}$.

Wait, I'm not understanding where the extra 2 from $(200+2)(100+2)$ came from...

You're adding 3 extra ft onto both sides of the lengths and the widths. Since 3ft = 1 yd, you add 2 yds to each side.  :)


[quote="mdk"]A triangle has lattice points as vertices and contains no other lattice points. Prove that its area is 1/2.[/quote]
Picks theorem,
$ A=\frac{1}{2}\cdot 3+0-1$


Hello,

1) Is scalar multiplication commutative?

Say S * [b]V [/b]is a scalar multiplication of vector V by a scalar S.

Is it equivalent to [b]V [/b]* S ?

Does such a expression exist?

2) Can a vector be multiplied (normal multiplication not dot or cross product) by another vector?

say [b]V1 [/b]* [b]V2[/b]

If yes, what is the result?

I am confused because a book I’ studying says:

A dot B dot C is not defined but (A dot B) C and A (B dot C) are nothing more than multiplying a vector by a scalar. I don’t see any scalar in the two expressions.

3) The book also says at another place:

The projection of [b]A [/b]onto [b]B[/b] is the same as projection of [b]B[/b] onto [b]A[/b].

Using AutoCAD I drew two intersecting lines and projected each line onto another. The length of projections were not same.

(see attached file VectorProjection.jpeg)

What wrong am I doing? Or is the statement wrong?

Please help

[quote="Ganitopasak"]

A dot B dot C is not defined but (A dot B) C and A (B dot C) are nothing more than multiplying a vector by a scalar. I don’t see any scalar in the two expressions.

[/quote]

I think it's because after you take (A dot B) you have a scalar quantity right? so multiplying by a vector C is basically multiplying a vector by a scalar. The same applies for A(B dot C) I hope that helps a little :)

My Answers in italics....

1) Is scalar multiplication commutative?

Say S * V is a scalar multiplication of vector V by a scalar S.

Is it equivalent to V * S ?

Does such a expression exist?

[i]Bear in mind that a scalar multiple S[b]V[/b] of a vector [b]V[/b] is a vector in the same or opp. direction of [b]V[/b]
depending on whether S is +ve or -ve. Though I 've never seen an expression like [b]V[/b] * S,u can define it if u wish  ;) [/i]

2) Can a vector be multiplied (normal multiplication not dot or cross product) by another vector?

say V1 * V2

If yes, what is the result?

I am confused because a book I’ studying says:

A dot B dot C is not defined but (A dot B) C and A (B dot C) are nothing more than multiplying a vector by a scalar. I don’t see any scalar in the two expressions.

[i]There are two types of Vector products . Scalar or Dot  and Vector or Cross.

The scalar(dot) product [b]A[/b].[b]B[/b]  is defined as

[b]A[/b].[b]B[/b]  = ABcos :theta:  where  :theta: is the angle between [b]A[/b] and [b]B[/b]  .

A = magnitude of [b]A[/b]
B = magnitude of [b]B[/b]

The vector(cross) product [b]A[/b]X[b]B[/b]  is a vector as defined below.

[b]A[/b]X[b]B[/b]  = ABsin :theta:[b]n[/b]

where [b]n[/b] is a unit vector in the direction of the normal to the plane containing  [b]A[/b] and [b]B[/b] .

Indeed there's more of products viz. scalar triple products and vector triple products. But I believe those are a bit advanced for u now..
first get ur fundas strong on elementary products of vectors(the ones I described above)

Also as somebody has pointed out, (A dot B) C and A (B dot C) are indeed nothing more than multiplication of a vector by a scalar because (A dot B) and (B dot C) are scalar quantities as defined above
[/i]

3) The book also says at another place:

The projection of A onto B is the same as projection of B onto A.

 [i]  U have read the book incorrectly..
In general,the projection of A onto B is NOT the same as projection of B onto A.
They are same only when their magnitudes are equal

Actually the dot product can be defined,based on the defnition of projections, as

[b]A[/b].[b]B[/b] = A*proj of [b]B[/b] on [b]A[/b] =  B*proj of [b]A[/b] on [b]B[/b]

NB: proj of [b]B[/b] on [b]A [/b]= Bcos :theta:

[/i]

1. Yes, I believe scalar multiplication is commutative, because it is simply expanding or contracting the vector.

2. I don't think the "regular product" of two vectors is defined, but I'm not sure. $A(B \cdot C)$ is defined because $B \cdot C$ is a scalar by the definition you gave, and so you can multiply a scalar by a vector. Same with $(A \cdot B)C$. However $A \cdot B \cdot C$ is not defined at all.

3. I'm not really sure...

[quote="blahblahblah"]$A(B\cdot C)\not=A\cdot(B\cdot C)$...[/quote]

Yeah... that's what I meant by it's not associative...

?!  $A\cdot (B\cdot C)$ isn't even defined.

[quote="paladin8"]
2. I don't think the "regular product" of two vectors is defined, but I'm not sure. $A \cdot (B \cdot C)$ is defined because $B \cdot C$ is a scalar by the definition you gave, and so you can multiply a scalar by a vector. Same with $(A \cdot B) \cdot C$. However $A \cdot B \cdot C$ is not defined because you don't know which of the previous two it is (intuitively) and the dot product is a binary operator (I think that's the word) that is not associative.
[/quote]

But it doesn't make sense to find the dot product between a scalar and a vector. Scalar multiplication (dot product) is not the same as multiplying by a scalar. So none of $A \cdot (B \cdot C)$, $(A \cdot B) \cdot C$, $A \cdot B \cdot C$ are meaningful.

And what blahblahblah said.

Hehe. Whoops, I was writing it wrong... I meant $A(B \cdot C)$...


Let $a_1,a_2,...,a_n$ and $b_1,b_2,...b_n$ be real numbers between 1001 and 2002 inclusive. Suppose $\displaystyle \sum^n_{i=1} {a_i}^2 = \sum^n_{i=1} {b_i}^2$
Prove that $\displaystyle \sum^n_{i=1}\frac{{a_i}^3}{b_i} \leq \frac{17}{10} \sum^n_{i=1} {a_i}^2$

Determine when equality holds.

I am sure that this inequality has been posted before.
A recent one is http://www.mathlinks.ro/Forum/viewtopic.php?t=33023

liyi, this was from the link you provided.

[quote="liyi"]$\frac 12\leq \frac{a_i}{b_i}\leq 2$
$\left(\frac{a_i}{b_i}-\frac{1}{2}\right)\left(\frac{a_i}{b_i}-2\right)\leq 0$
$a_i^2+b_i^2\leq \frac{5}{2}a_i b_i$ (*)
On one hand, we have from (*) that $\sum a_i^2\leq \frac{5}{4}\sum a_ib_i$
On the other hand (from (*) again) , we have $\frac{a_i^3}{b_i}+a_ib_i\leq \frac{5}{2}a_i^2$, thus $\sum\frac{a_i^3}{b_i}+\sum a_ib_i\leq \frac{5}{2}\sum a_i^2$.
Combine them, we obtain that $\sum\frac{a_i^3}{b_i} \leq \frac{17}{10}\sum a_i^2$[/quote]

How to you transform (*) into two following inequalities, which you said were from "on one hand" and "on the other hand"? I am confused.

Taking sum of both sides of (*), we get
$\sum a_i^2 + \sum b_i^2 \leq \frac{5}{2}\sum a_ib_i$
Then apply $\sum a_i^2 = \sum b_i^2$ to it.

Multiply $a_i$ and divide $b_i$ in both sides of (*) we get $\frac{a_i^3}{b_i}+a_ib_i\leq \frac{5}{2}a_i^2$

From, $a_i^2+b_i^2 \le \frac{5}{2}a_ib_i$, you can get

$\frac{a_i^3}{b_i}+a_ib_i \le \frac{5}{2}a_i^2$ from multiplying through by $\frac{a_i}{b_i}$
$\frac{b_i^3}{a_i}+a_ib_i \le \frac{5}{2}b_i^2$ from multiplying through by $\frac{b_i}{a_i}$

Combining those two we get

$\sum \frac{a_i^3}{b_i}+2a_ib_i \le \frac{5}{2}\sum a_i^2$.

We can replace $2a_ib_i \ge \frac{4}{5}\sum a_i^2$ from our original inequality. So

$\sum \frac{a_i^3}{b_i}+\frac{4}{5}\sum a_i^2 \le \frac{5}{2}\sum a_i^2$

and finally

$\sum \frac{a_i^3}{b_i} \le \frac{17}{10}\sum a_i^2$.

This is a very nice inequality!

EDIT: liyi beat me to it


ABCD is a cyclic quadrilateral with side AD = d, where d is the diameter of the circle. AB = a, BC = a, and CD = b. If a, b, and d are integers with a not equal to b,

a) Prove that d cannot be a prime number.
b) Determine the minimum value of d.

a) Let $O$ be the center of the circle, $M$ be the mid point of the cord $AC$. Since $O,\ M,\ B$ are collinear and $OM=\frac{1}{2}CD=\frac{1}{2}b.$ By Pythagorian's theorem for $\triangle{ACD}$, we have that $AM=\frac{1}{2}AC=\frac{1}{2}\sqrt{d^{2}-b^{2}}$. Use this theorem again for $\triangle{AMB}$, we have that $\left(\frac{\sqrt{d^{2}-b^{2}}}{2}\right)^{2}+\left(d-\frac{b}{2}\right)^{2}=a^{2}\Longleftrightarrow 5d^{2}=4(a^{2}+bd).$ Since 5 and 4 are coprime, there exists some positive number $m$ such that $d^{2}=4m,\ a^{2}+bd=5m.$


10. Write $ \sqrt{5\minus{}2\sqrt{6}}$ as the difference of the square roots of two integers.

[hide="Answer"]$ \sqrt{3}\minus{}\sqrt{2}$[/hide]

[quote="illusiat"]for these, u just find 2 numbers whose sum is the 1st term and product is the 2nd term(inside the square root). in this case, 2 and 3. Then, write them from largest to smallest, and square root of largest minus square root of smallest.[/quote]
Oh.. Okay. I think I have it. Thanks.

[quote="DonkeyKong"]10. Write $ \sqrt {5 \minus{} 2\sqrt {6}}$ as the difference of the square roots of two integers.

[hide="Answer"]$ \sqrt {3} \minus{} \sqrt {2}$[/hide][/quote]
Look, you just want $ \sqrt {5 \minus{} 2\sqrt {6}}$ to be in the form $ \sqrt {x^2 \minus{} 2xy \plus{} y^2}$
Seeing as you have $ 2\sqrt {6}$, $ x$ and $ y$ must have a product of $ \sqrt {6}$
Using this, we find that $ \sqrt {2}$ and $ \sqrt {3}$ Works.
Therefore, we can factor $ \sqrt {5 \minus{} 2\sqrt {6}}$ as $ \sqrt {(\sqrt {2})^2 \minus{} 2\sqrt {2}\sqrt {3} \plus{} (\sqrt {3})^2} \equal{} \sqrt {(\sqrt {3} \minus{} \sqrt {2})^2} \equal{} \sqrt {3} \minus{} \sqrt {2}$

[quote="ZhangPeijin"][quote="DonkeyKong"]10. Write $ \sqrt {5 \minus{} 2\sqrt {6}}$ as the difference of the square roots of two integers.

[hide="Answer"]$ \sqrt {3} \minus{} \sqrt {2}$[/hide][/quote]
Look, you just want $ \sqrt {5 \minus{} 2\sqrt {6}}$ to be in the form $ \sqrt {x^2 \minus{} 2xy \plus{} y^2}$
Seeing as you have $ 2\sqrt {6}$, $ x$ and $ y$ must have a product of $ \sqrt {6}$
Using this, we find that $ \sqrt {2}$ and $ \sqrt {3}$ Works.
Therefore, we can factor $ \sqrt {5 \minus{} 2\sqrt {6}}$ as $ \sqrt {(\sqrt {2})^2 \plus{} 2\sqrt {2}\sqrt {3} \plus{} (\sqrt {3})^2} \equal{} \sqrt {(\sqrt {3} \minus{} \sqrt {2})^2} \equal{} \sqrt {3} \minus{} \sqrt {2}$[/quote]
Wow, Thanks. I get it now..


induction
number theory theorems
number theory

for given $ n\in\mathbb{N}$ describe all $ k\in\mathbb{Z}$ which can be represented as $ e_1\cdot 1\plus{}e_2\cdot 2\plus{}\dots\plus{}e_n\cdot n$ where $ e_i\in\{\pm 1\}$ for $ i\equal{}1,\dots,n$.

-is there any reasonably nice characterisation ?

Let $ S_n\equal{}1\plus{}2\plus{}\cdots\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$

By induction all the numbers $ \{\minus{}S_n,\minus{}S_n\plus{}2,\cdots,S_n\plus{}2,S_n\}$ can be expressed.

Case $ n\equal{}1$ Trivial

Case $ n\equal{}2$, $ \{\minus{}3,\minus{}1,1,3\}\equal{}\{\minus{}1\minus{}2,\plus{}1\minus{}2,\minus{}1\plus{}2,\plus{}1\plus{}2\}$

Suppose it is true for $ n$

There are two possibilities for $ e_{n\plus{}1}$

a) $ \{\minus{}S_n,\minus{}S_n\plus{}2,\cdots,S_n\plus{}2,S_n\}\minus{}(n\plus{}1)$
$ \{\minus{}S_{n\plus{}1},\minus{}S_{n\plus{}1}\plus{}2,\cdots, S_n\plus{}2\minus{}(n\plus{}1),S_n\minus{}(n\plus{}1)\}$
 and $ S_n\minus{}(n\plus{}1)>0$

b) $ \{\minus{}S_n,\minus{}S_n\plus{}2,\cdots,S_n\plus{}2,S_n\}\plus{}(n\plus{}1)$
$ \{\minus{}S_n\plus{}(n\plus{}1),\minus{}S_n\plus{}2\plus{}(n\plus{}1),\cdots, S_{n\plus{}1}\plus{}2,S_{n\plus{}1}\}$
and $ \minus{}S_n\plus{}(n\plus{}1)<0$

Then there is overlap between both subsets. Then it is true for $ n\plus{}1$


Show that the equation $a^2 + b^2 = c^2 + 3$ has infinetely many triples of integers $a, b, c$ that are solutions.

We take two numbers, say $2a$ and $2a+1$. Then we search the value of $x$ in the following equation: $x+4a^2 = (2a+1)^2+3$

\begin{eqnarray} x&=& (2a+1)^2-4a^2+3\\ &=& 4a^2+4a+1-4a^2+3\\ &=& 4(a+1) \end{eqnarray}
$x$ is a perfect square if and only if $(a+1)$ is a perfect square.
There are infinitely many values for which $a+1$ is a perfect square, so there are also infinitely many values so that $x$ is a perfect square.
\[ a^2+b^2=c^2+3 \]
has infinitely many solutions, QED.

Nice Jan :D But I'm sure you've seen the problem before, haven't you? ;)

Well, I must confess it looked rather familiar...
But I still found the solution myself, so why wouldn't I post it?  :D


Okay...here is the problem, I looked at the solution but I don' t understand it.

In an algebra class, half the students are boys.  One-third of the students are wearing glasses.  Half the boys are wearing glasses.  What fraction of the girls is wearing glasses?  Express your answer as a common fraction.

[hide]I am not going to tell the answer because some people just say the answer even though it's written right here!!!! *cough cough* <insert name here> *cough cough*[/hide]

Easiest way- say there are 12 people
so 6 boys 6 girls
4 people wearing glasses
3 boys are wearing glasses
therefore 1 girl is wearing glasses.
so there is 1 girl out of 6 girls wearing glasses or 1/6

I [i]would[/i] have done it that way, but I was trying to do it using algebra, so I got it wrong.  Using a random number would have gotten the answer faster.... Thanks!!!

Also, seeing that we have fraction slike 1/2 and 1/3 , we want a number that is (ideally) divisible by 2 and 3 a couple times, like 12.

Using algebra, you get 1/3 of the students wearing glasses. To find out how many girls wear glasses, you first have to find out how many people wear glasses that are boys, and subtract that from 1/3. The number of boys is 1/2 and the number of glasses-wearing boys is 1/2 of 1/2, or 1/4. So, we have 1/3-1/4=1/12.
This means that 1/12 of all the people in class are girls that wear classes. But we want just the fraction of the girls, not the fraction over the total number of people in the class.

So... We know that 1/2 of the students are boys, that implies the other 1/2 of the students are girls. So if 1/2 of the class are girls, we set up the equation $ g \equal{} \frac {1}{2}t$, where t is the total number of students and g is the number of girls in the class. We also know that $ f \equal{} \frac {1}{12}t$, where f is the number of girls wearing glasses. That must mean $ f \equal{} 2 \cdot \frac {1}{12}g$, so $ f \equal{} \frac {1}{6}g$. The answer is 1/6.

Sorry I was unclear at any part.


inequalities
logarithms
function
algebra proposed
algebra

Let $ x,y \in (0,\infty)$ with $ x\plus{}y\equal{}1$. Prove that $ x\ln x\plus{} y\ln y \geq \ln \frac{1}{2}$

Let:

$ f(x) \equal{} x \ln(x) \plus{} (1 \minus{} x) \ln(1 \minus{} x)$

Using derivations we get the local extrema for this function at:

$ x \equal{} \frac {1}{2}$

We see this is a minimum. So

$ x \ln(x) \plus{} y \ln(y)$

has it's local extrema (minimum) at $ x \equal{} y \equal{} \frac {1}{2}$ and is equals:

$ \ln\frac {1}{2}$

so the inequality holds.

[quote="DocEmBr"]Let $ x,y \in (0,\infty)$ with $ x \plus{} y \equal{} 1$. Prove that $ x\ln x \plus{} y\ln y \geq \ln \frac {1}{2}$[/quote]

Let $ f: (0,\infty)\minus{}>\mathbb{R}$ , $ f(x)\equal{}\ln \left( \frac{1}{x}\right)$ $ f$ convex.
And by Jensen 's inequality for convex function we have:

$ xf\left(\frac{1}{x}\right)\plus{}yf \left (\frac{1}{y}\right) \geq f \left( x \cdot\frac{1}{x}\plus{}y \cdot \frac{1}{y}\right)\equal{}\ln\frac{1}{2}$

Let $ f(x)\equal{}x\ln x$ then $ f''(x)\equal{}\frac 1x>0$ hence $ f$ is convex. Thus $ f(x)\plus{}f(y)\ge 2f\left(\frac{x\plus{}y}{2}\right)\equal{}\ln\frac 12$.


quadratics algebra polynomial function geometry conics parabola

At the Pummill Math Relays, a recent math competition me and some guys at my school went to, a set of 8 challenging problems was given out to each school to be solved within 2.5 hours.  Our school got first, solving 7/8 of the problems! Let's see how you guys do.

1. The average of [i]n[/i] consecutive integers is 75. The product of the smallest and largest of these integers is 3600. Find [i]n[/i].

2. How many 8-letter words can be made from the letters in MISSOURI?

3. If P(x) is a quadratic polynomial such that when the remainder of P(x) is divided by x-1 is 3, when divided by x+1 is -3, and when divided by x-2 is 12, then what is the remainder when P(x) is divided by x+2?

4. Three circles of radius 1 are all tagent to one another. A fourth smaller circle is tangent to the other three circles. Find the radius of the smaller circle.

5. Find all x, - :pi: < x < :pi: , such that cos x + sin x = 1/2
[For full credit, give your answer in terms of inverse trigonometric functions]

6. Four quarter circles are centered at the vertices of a square whose sides have length 1. Find the area of the region where they overlap.

7. A line is normal (i.e. perpendicular) to the parabola y = x^2 at the point (a, a^2) != (not equal) (0,0). The line meets the parabola at a second point. Find the coordinates of that point.

8. The function floor(x) gives the greatest integer that is less than or equal to x. For example, floor( :pi: ) = 3,  floor(2) = 2, and floor(-1.2) = -2. Sketch the region of the plane defined by the equation floor(x)^2 + floor(y)^2 = 1.
[NOTE: You guys may just describe this graph, if you'd like.]

Good luck!

Ok, here goes:

1.

[hide]

3600 factors out to 3^2*5^2*2^4 you want a sum of (75)*2=150 so you know you most likey there's going to be a five and a two in each factor a short amont of expermentation shows that 2^3*3*5+3*2*5=150, so your factors are 120, 30 so you have 120-30+1=91 consecutive integers.

[/hide]

2.

[hide]

you know you have 8 letters with 2 sets of double letters for 8!/(2*2)=10080 permetations.

[/hide]

3.

Quote:such that when the remainder of P(x) is divided by x-1 is 3

huh?

I'll leave the rest to someone else.[/hide]

NAS wrote:[hide]so you have 120-30=90 consecutive integers.[/hide]

How many days are there in January, given that the first day of January is January 1 and the last day of January is January 31?

I think an easier method for number one than what NAS posted is the following: [hide](75-(n/2))(75+(n/2))=3600; solving this quadratic yields n/2=45, so n=90.[/hide]

Yes, your method is nicer, but it still got you the wrong answer :)

Number 6 needs a picture or we need to know the length of the radii.

Good point - my answer doesn't include the 75.  So, [hide]91[/hide]... I hope.  Otherwise I've just made a total fool of myself

mpn wrote:3. If P(x) is a quadratic polynomial such that when the remainder of P(x) is divided by x-1 is 3, when divided by x+1 is -3, and when divided by x-2 is 12, then what is the remainder when P(x) is divided by x+2?

I'll take a stab at this..

[hide]Let P(x)=ax^2+bx+c.  Dividing this by x-1, we get a remainder of a+b+c=3.  Similarly, a-b+c=-3, and 4a+2b+c=12.  Solving this system, we get a=3, b=3, c=-3.  Dividng 3x^2+3x-3 by x+2 yields a remainder of 3.[/b][/hide]

Find all x, -  pi < x <  pi, such that cos x + sin x = 1/2

[hide]

(cos x + sinx)(cos x + sinx) = 1/4

cos^2(x) + sin^2(x) + 2sinxcosx= 1/4

1 + 2sinxcosx = 1/4

1+sin(2x) = 1/4

sin(2x) = -3/4

Let y=2x

sin y = -3/4

y=-arcsine(3/4)+2kpi or y = (pi +arcsine(3/4)) + 2kpi

2x = arcsine(3/4) + 2kpi or 2x = pi + arcsine(3/4) + 2kpi

x=-arcsine(3/4)/2 + kpi or x = pi/2 + arcsine(3/4)/2 + kpi

x = -arcsine(3/4)/2, pi/2 + arcsine(3/4)/2

[/hide]

Edit: Thanks JBL

1/4 - 1 = -3/4, not 3/4 :)

Many silly mistakes on this thread, folks :)

[hide]4.  First draw the lines connecting all the centers.  This is an equilateral triangle.  Then draw the lines connecting the center of the smaller circle to the other centers.  If you draw the line from the center of the small circle to the point of tangency of any of the bigger circles you will have a 30-60-90 triangle(because the big triangle is equilateral).  The lenght of the side opposite the 60 is 1 so since the lenght of the the radius of the big circle plus the radius of the small circle is the hypotenuse it is equal to (2/ :rt3: )-1 which when simplified and rationalized equals (2 :rt3: -3)/3[/hide]


I wont add a poll cuz there are too many options.
Just post what you like.

Mine is Ride of Valkyries.

I don't have a favorite piece, and if I tried to answer this, my answer would change every 10 minutes. But just at the moment that I saw this, the tune that was running around in my imagination happened to be the slow movement from Beethoven's Symphony #7. (Which I have played - that helps hold it in my head.)

My favorite is Chopin's [i]"Raindrop" Prelude in D-Flat Major[/i]  :D

[quote="IrinaSharova"][quote="jeez123"]Hmm... Liszt's Liebestraum (hope I spell it correctly). It is a bit impossible for me to learn the original version.... so I opt for arrangements now :D .[/quote]

Liszt's Liebestraum is indeed very elegent and very difficult to perform, it is one of my favourite as well. :)[/quote]

It's not that hard compared to other pieces like Mephisto Waltz, Tarantella, Transcendental Study etc.

Nearly any of Bach's violin concertos are my favorites. I also like Beethoven's 5th, 7th, and 9th symphonies.

Currently, I'm really big on Liszt's La Campanella, though I do not listen to a lot of classical music, for the most part.  Mostly just whatever I play or hear other people play.

I particularly like Brandenburg Concerto #5, but I haven't really heard a wide variety of classical music.

Wieniawski - Scherzo Tarantelle
Bach - Double Concerto in D
Scarlatti's Sonata in D Major (435)
Paganini's 24th Concerto
Tchaikovsky Piano Concerto No.1
Brandenburg's 5th Concerto (Harpsichord is amazing)
Beethoven's 5th Symphony :D

Among many, many others.

Beethoven Appasionata Mvmt. 1
La Campanella
Variations on a theme of paganini...
what else?
Well, the appasionata is the only one I play currently.

I pretty much like all classic music.

But I think my favorite is 'Amadeus' by Mozart.
The songs title is his middle name.. isn't that cool?
Anyways, we also played that in school today.
It was awesome, even though we didn't play as well and it wasn't the original..
Original's too hard...but oh well.. :lol:

The Blue Danube - Strauss
The Goldberg Variations - Bach
L'isle Joyeuse - Debussy
The Planets (esp. Mars, Saturn, Neptune) - Holst

...and of course, I could listen to this one for hours...

Canon in D Major - Pachelbel

That is a very hard question to answer, though I will to the best of my ability.
Last month it was Vivaldi's Recorder Concerto in C major, RV558
The week before it was Handel's Water Music Suites
Last week my favorite was Bach's Brandenburg Concertos
Today I like Beethoven's "Moonlight"(Piano Sonata No. 14 in C Sharp Minor) and "Appassionata"(Piano Sonata No. 23 in F Minor)

[quote="Kent Merryfield"]But just at the moment that I saw this, the tune that was running around in my imagination happened to be the slow movement from Beethoven's Symphony #7. (Which I have played - that helps hold it in my head.)[/quote]

Yes, it seems that my favorite piece is usually the one that I'm currently playing/ogling. I really like the movement you're talking about too; however, my current favorite is Mozart's G minor Viola Quintet (K 516). It's actually been one of the few works that remain on my "favorites list," as it were, for more than a month or so.

I recently began admiring Mahler for his 2nd symphony because it mirrors our world so well.

I also like most songs by Vivaldi, some by Mozart, Beethoven (even though I'm the only one who doesn't know much music by him), Holst, Bach and Mendelssohn.

Hmmm...Liebestraum isn't hard...cadenzas are trickyish but everything else is pretty simple.
I don't really listen to much music, but here goes my long preferences description:
Liszt is really great, Mazeppa is really good (transcendental no.4), as well as a few others. Mephisto Valse 1, Un Sospiro (concert etude 3) sounds really good too even if its not incredibly hard. All his paganini etudes are pretty good. Chopin's first ballade is definitely one of my favorites, but I don't really like his other ballades. All his scherzos are pretty good, haven't listened to his sonatas. His etudes are sorta too short for me to want to listen to them as music. Chopin is just pretty awesome in general.
Rach I don't like that much, except for his concertos, which are awesome.
As for more modern people, I liked all the prokofiev stuff I've ever listened to but its sorta wierd hehe, so I like older stuff.

Mozart, Bach, etc. I absolutely hate. Mozart stuff is way too simple sounding, and lacks the complexity of Romantic era compositions. Beethoven isn't bad, but I still prefer later romantic era (since he's like, beginning of romantic era). I don't like Schubert or Schumann all that much, neither do I like Grieg.

[quote="davidyko"]Hehe, you guys are such old farts. 20th century faves:
Mahler: Symphony no. 6
Stravinsky: Le sacre du printemps (The Rite of Spring)
Scriabin: Piano Sonata no. 5
Debussy: La mer
Shostakovich: Symphony no. 10
Ravel: Piano Concerto in G
Prokofiev: Piano concerto no. 3
Ligeti: L'escalier du diable (Piano etude no. 13)[/quote]
Meh I don't like 20th century besides prokofiev and rach if he counts, they all sound too...wierd for me.

hahaha i dont like classical much but this song might become a rock classic!!

[youtube]WcwweoJDkV0[/youtube]

watch it!!! lol

I like that symphony that that S.H.E. song was based on!!!!  :rotfl:

But, seriously, I like half the gavottes I've ever played on violin, especially the last one in the 3rd Suzuki book. They're all pretty easy though, I think. (Will anyone help me find classical sheet music for clarinet???  :oops: )

But I think by classical I can say things that aren't western, right???  :o  b/c everyone else thought that, even tho the real definition of the word, I think, is Greek/Roman.... So yay! I like A Bing's Er Quan Ying Yue. Is that classical? And I have like 3 CD's of the greatest surviving Chinese classical music ever, but idk the composers. lol er hu and any Chinese flute that isn't shui huo can make any song but the worst sound really good. And they make good songs sound like magic.

EDIT: And other instruments like guqing, pipa, xiao (wait that's a flute right?), etc. too. And I mean rap, pop, classical, etc. Like that 12-yr-old's playing of Qing Hua Ci on Youtube...[youtube]PvfNHT1lJFU[/youtube]
(No that's not it...)

BWV 1,4,12,21,36,61,62,63,78,82,93,102,103,106,131,198,211,212,213,227,229,232,235,243,244,245,248,404,539,542,578,996-1006a,1017,1043,1065,1079,1080 and some I forgot
Monteverdi, L'orfeo
Gluck, Orfeo ed Euridice, Che faro ...
Wagner, Der Fliegende Hollander, Senta
Vivaldi, Stabat mater
Cage, 4"33'' (the best piece of the 20th century!)
Schubert, Erlkonig, Winterreise
Handel, Messiah
Missa salisburgensis
Paganini, Caprice No. 24

Hmmm... I like:
     Mendellessohn, Hunting Song
     Chopin, Minute Waltz
    uhh.. can't think of any others right now.

[youtube]Z67z1EypsRk[/youtube]

^ beasts playing a beastly piece :)

the handel-halvorsen is the only piece that I can listen to for more than ten consecutive times.

favorites:

Rosin eating zombies from Outer Space

Night Shift

Waltz of the flowers

Santa's Classical Christmas

Ukrainian Bell Carols

12/24 Sargevo

Uh...rosin eating zombies from outer space?

Here's mine:

[url]http://www.youtube.com/watch?v=oHg5SJYRHA0[/url]

Guys, I memorized the Rickroll watch ID on Youtube. Not effective.

The Protoss soundtrack is probably the closest you can get to being "classical" in Starcraft. Hmmm.

1812 overture for the win! Egmont overture, and butterfly legend (chinese song)

sorry, i was just browsing after a long time off aops, i 4got to check when was last post. My bads.  :oops:

[quote="Richard_Min]Mod: Please Lock

Joe, please don't revive old threads.[/quote]

But a perfectly legitimate thread.

Could the mod instead just move this to the Music and Art thread? (That thread needs some love).

Hi, please do not revive old threads prematurely (already said).

Additionally, do not continue posting if you know it's a revival, unless you have something important to say.

Even if it's a legitimate thread, the Games/FF policy is that you create a "revival thread" (a new thread about the same topic) to continue the topic, and hopefully post a link to the old thread within the first post.

I feel that the generality of the topic makes it okay for this forum, but I will lock this thread for being a revival.

I tend to unlock it because I see no reason to forbid revivals. It may be usefull in case of games (at least in most cases), but this isn't one. In general, application of rules should be replaced by application of the ideas behind them (here: a lot of revivals create a mess and especially for games make not much sense in most cases; but again, a bigger mess is caused by having a new thread on the same general topic every week).

For clarification on my part, any mod can lock any thread, even if Phire wasn't a mod in games/ff?

Mine would have to be all of Chopin's wonderful etudes for the piano  :lol:

Please do not revive dead threads.

I made that mistake...

[size=84][color=red]Technically this isn't a revival. It does make a point that it should be avoided, though. --PhireKaLk6781[/color][/size]

Sorry, but I am new here. What is so bad about reviving dead posts? Isn't that good?

No, revivals bring up dead threads, cluttering up space and squeezing out current discussion. For more information, or if you have future questions like this, private-message me (or dragon96, or the active moderator for the forum you're posting in) rather than post on the thread.


MATHCOUNTS
geometry
rhombus
FTW
summer program
MathPath
Duke

OK.  If you want to take it, PM me for your answers and the highest score wins!  Good luck.  (No need to sign up).  Here's MY mock mathcounts test (first seen in the algebra 1 forum archive)

So far:

1st:  ilovepink - 41/46 (27 sprint, 14 target)
2nd: Magic Bard - 39/46(23 sprint, 16 target)

gj!  Please, more ppl PM answers.

OK, more:
1st:  ilovepink - 41
2nd:  Magic Bard - 39
3rd:  math rulz - 33(23 sprint, 5 target)
4th:  Scamper - 19(19 sprint, yet to turn in target)

More updates:
1st:  ilovepink - 41
2nd:  the cliu - 40(24 sprint 16 target)
3rd:  Magic Bard - 39
4th:  math rulz- 33
5th:  Scamper - 19

Great test.  I PMed you my answers.  It was perfect, other than that I think the target test was a little easier than it should have been to be honest.  Well, it's only school round.  I don't know.  Great job!

Are you going to continue this with qualifications for mock chapter, state, and then national?

Yes,  I will make a chapter, state, and a national test, but I'll need some help with state and nats, otherwise it'll be too easy(ahem pythag011)

Anyways, new placings:
1st:  AdithyaGanesh-42(26 sprint, 16 target)
2nd:  ilovepink-41
3rd:  the cliu-40
4th:  Magic Bard-39
5th:  math rulz-33
6th:  Scamper-19

imo the target WAS too easy but I was kinda lazy since it took awhile to make the sprint and I KNEW ppl were gonna mess up on sprint so I thought target should be ez......

Sure.  I'd need some help with states and A TON of help with nats(only have 2 nats test, don't really know the difficulty that well of it)

Do scores from this round carry over?  They don't right?  And is there going to be a CD?

Once I see how many ppl submitted answers, I'll make a cutoff score.

Note:  The cutoff score, under no circumstances, will be more than 42.

Any thoughts about the test?  Was it original?

I won't have time to turn in the test until Friday evening. When is the school round due?

THE SCHOOL ROUND IS DUE FRIDAY, DECEMBER 11

I can help write for later rounds if you need it.

[quote="Maybach"]Any thoughts about the test?  Was it original?[/quote]

Not very original...just old problems with different numbers and wording.  Similar to the other mock test - very routine problems.  I understand that this is intended to be a school level competition, but they do have some original problems.

UPDATE:

1st: yaofan-42(28 sprint, 14 target, won tiebraker due to higher sprint score)
2nd: AdithyaGanesh-42
3rd: ilovepink-41
4th: the cliu-40
5th: Magic Bard-39
6th: math rulz-33
7th:  Scamper-19

what is ur label policy.
on number 21 it does not specify whether the answer should be in minutes or in hours.

Honestly, these protests don't really matter.  It won't change who qualifies.  This is practically unofficial.  worry about state and nats.

[quote="ytrewq"]I would like to protest some of the questions.

[hide="Sprint 24"]
Between $ 1$ and $ 30$ does not include $ 30$ so the answer should be $ 30$ less or $ \boxed{60}$.
[/hide]

[hide="Sprint 25"]
$ x^2 \plus{} (a \plus{} b)x \plus{} ab \equal{} x^2 \minus{} 5x \plus{} 10$
$ ab \equal{} 10$
$ a \plus{} b \equal{} \minus{} 5$
$ ab \minus{} (a \plus{} b) \equal{} \boxed{15}$
[/hide]

[hide="Sprint 29"]
$ \frac {x}{360}\pi 12^2 \equal{} 3\pi$
$ \frac {x}{360}144 \equal{} 3$
$ x \equal{} 3*360/144$
$ x \equal{} \boxed{7.5}$
[/hide]

[hide="Target 5"]
Could someone explain what this problem means?
[/hide][/quote]

i can understand your claim for 24 and 25, but for 29 the problem says round to the nearest hundredth, u have it to the tenths.

5 target means that if 1/2 and 3/2 were on a number line, there are 3 numbers in between them that split that section into 4 equal parts. you have to find the sum of those 3 numbers.

i hope i helped

yes i agree with dafjbomb....but it really doesnt matter cuz all 3 of us qualify 4 state!  :D

[quote="dafjbomb"]

i can understand your claim for 24 and 25, but for 29 the problem says round to the nearest hundredth, u have it to the tenths.

5 target means that if 1/2 and 3/2 were on a number line, there are 3 numbers in between them that split that section into 4 equal parts. you have to find the sum of those 3 numbers.

i hope i helped[/quote]

I looked over the protests ... I think they are all legitimate (@ Maybach ... though they might not make a difference in the standings, they do make a difference in the perception of the mock competition and should probably be addressed).

In 29, the answer was 7.5, so rounding to the hundredth seemed a little odd ... 7.50 I guess?

Thanks for explaining the Target question ... I didn't understand it either.

OK.  I grant all protests, but somne people did get ALL of these problems right, so plelase PM me if you missed points on it so I can readjust the scores.

[quote="CRICKET229"]Maybach...do you have any updates for State yet?  :D[/quote]

No updates yet.

[quote="dafjbomb"]yea, it felt odd for me too when i had to write 7.50 instead of 7.5 (which really surprises me b/c i never pay attention to small detail like that and end up getting it wrong when it actually counts on like an amc or aime or something)[/quote]

See 2004 Chapter Sprint #9:

"A pizza costing \$17.60 is cut into 16 equal pieces.  Brad eats six pieces.  In dollars, how much was the portion of the pizza that Brad ate?  Express your answer as a decimal to the nearest [b]hundredth[/b]."

Answer is \$6.60 :)

Hello people!

Round 1 (page 2) and Round 2 (Page 1) of my ___tourney are out!!!! Click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319772]here[/url] to go there!!!!!!

The rounds are due in the middle of February!!!!!

[quote="TheMan1998"]I think it's different because money is money but this is a decimal number of degrees...why would you need the extra zero? :)[/quote]

m'bad.

What happened was I thought the answer was 3.75 before I checked.  Then I realized it was 7.5 and I forgot to change it.


[quote="Mitochondria"]Hi I'm studying some topics in math, and over the past few days I was reading up some basics of set theory.  I tried many of the proofs and I came across this one which involves proving a biconditional statement.  I was wondering if someone here could please help me.  Thanks!

Question: Let $A$ and $B$ be subsets of a universal set $U$.  Prove that $A\subseteq B\Leftrightarrow A\cap B=A$[/quote]

From $A\subseteq B$, we know that set $A$ is entirely inside of set $B$.  Since set $A$ is inside of set $B$, the intersection of the two sets must be the set $A$.


An interesting observation:

[color=blue]Problem[/color].Let
\[1+{ 1 \over 2 }+{ 1\over 3}+{ 1 \over 4}+...+{ 1 \over n}= \frac{a_{n}}{b_{n}}\]
where $(a_{n},b_{n})=1$

Show that there are infinitely many integers like $n$ that
\[b_{n+1}< b_{n}\]

We have $\frac{a_{n+1}}{b_{n+1}}= \frac{a_{n}}{b_{n}}+\frac{1}{n+1}$

If $n+1 = p^{a}$ ($p$ is prime and $a \geq 1$). Let $p^{k}| b_{n}$ and $p^{k+1}\not| b)n$
Then, $a_{n+1}= a_{n}\cdot p^{a-k}+\frac{b_{n}}{p_{k}}, b_{n+1}= b_{n}\cdot p^{a-k}$ with $0 \leq k \leq a$ and $(p^{k},b_{n}) = 1$.
It's clear that $(a_{n+1},b_{n+1}) = 1$. In this case, we have $b_{n+1}\geq b_{n}$.

Thus, $n+1$ must be composite with at least two prime factors.

Why $b_{2 \cdot 3^{k}}\not\equiv 0(mod 3^{k})$ ?  :maybe:
Can you explain more completely about that ?

First comments: the issue of establishing a ($\textbf{surefire}$) bound on two consecutive denominators turned out to be a hard task, especially if I wasn't sure about the particular $n$ that I need to shoot for. \\

In short, this Solution will analyze the $v_p$ of $b_n$ for each individual $p$ prime, while forcefully ignoring the simultaneous side effects of other primes which divide $n+1$.

Then, after securing the value of $b_n$ for large $n$, that is
\[ b_n = \text{LCM}(1,2,\ldots,n); \]
we will show that this definition contradicts itself for some $n$. \\

Before we analyze the size of $b_n$, let's establish some almost-certain-properties of denominators when adding fractions: \\
$\color{green} \rule{4.4cm}{2pt}$ \\
$\color{green} \diamondsuit$ $\boxed{\textbf{A Bound on LCM.}}$ $\color{green} \diamondsuit$ \\
$\color{green} \rule{4.4cm}{2pt}$ \\
Let we have
\[ \dfrac{a}{b}+\dfrac{c}{d} = \dfrac{e}{f} \]
with $(a,b) = (c,d) = (e,f) = 1$. Then, $f \mid \text{LCM}(b,d)$. \\
$\color{green} \rule{4.4cm}{0.2pt}$ \\
This should have been taught in any primary school: we can just pick $f = \text{LCM}(b,d)$ as the common denominator and adjust the new numerator of $a/b$ and $c/d$ accordingly. \\

Alternatively, consider all primes $p$ dividing $b$ or $d$. Notice that if $v_p(b) = b_p$ and $v_p(d) = d_p$,
\[ v_p(f) \leq \max{(b_p,d_p)} \]
after simplification. \\
$\color{green} \rule{25cm}{0.2pt}$ \\
Next, we replace each letter $a$ to $f$ with its respective variable $a_n,b_n,1,n+1,a_{n+1},b_{n+1}$ and specify the relation between $b,d,f$ in some different cases. \\
$\color{green} \rule{6cm}{2pt}$ \\
$\color{green} \clubsuit$ $\boxed{\textbf{Each Prime's Increase Rate.}}$ $\color{green} \clubsuit$ \\
$\color{green} \rule{6cm}{2pt}$ \\
For each $(n+1,p)$, define $p$ to $\textit{drop}$ in $n+1$ if $v_p(b_{n+1}) < v_p(b_n)$. If the sign is $=$ or $>$, define it to $\textit{stagnate}$ or $\textit{expand}$. \\

(This should serve a clue on why the config is in the form $\color{green} \textbf{\textit{something significant}} - 1$). \\

Then, if
[list = disc] [*] $v_p(n+1) > v_p(b_n)$, then $p$ expands; [*] $v_p(n+1) < v_p(b_n)$, then $p$ stagnates; [*] $v_p(n+1) = v_p(b_n)$, then $p$ stagnates or drops. [/list]
Particularly, it takes an equal power for $p$ between the summand's denominators to drop. \\
$\color{green} \rule{25cm}{0.2pt}$ \\
$\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ \\
Just look at the $v_p$ of the numerator and denominator of $\dfrac{a_n (n+1) + b_n}{(n+1)b_n}$.
$\color{red} \rule{25cm}{1pt}$ \\
Now we get to actually analyzing the composition of $b_n$. For starters, assume that there exists some number $N$ so that for every $n \geq N$, $b_{n+1} \geq b_n$. \\
$\color{red} \rule{5.2cm}{2pt}$ \\
$\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Restrictive Expansions.}}$ $\color{red} \clubsuit$ \\
$\color{red} \rule{5.2cm}{2pt}$ \\
Fix a prime $p$, and let we have $p^{a} \geq N+1$ for some $a \in \mathbb{N}$. Then, we Claim that for $n+1 \geq p^{a}$, $p$ expands on $n+1$ if and only if $n+1$ is a power of $p$. \\
$\color{red} \rule{25cm}{0.2pt}$ \\
$\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ \\
First, note that this is obvious for $n+1 = p^{a}$ (by $\color{green} \clubsuit$ $\boxed{\textbf{Each Prime's Increase Rate}}$ $\color{green} \clubsuit$ and noting that $v_p(b_n) \leq a-1$). \\

After that, we prove this for $n+1$ a power of $p$. Again, just use the same method as the previous paragraph. \\

We get to the not-so-obvious part: let $n+1$ be the first drop of $p$. Assuming that
\[ p^{a} < n+1 < p^{a+1}, \]
$v_p(n+1)$ must be $a$, since we know that $v_p(b_n) = a$ (observe that $p$ has not dropped before $n+1$). Write $n$ as $p^{a} \cdot b$ with $b < p$. \\

This part is by far the most magical part (arguably) of this proof: by $\color{green} \diamondsuit$ $\boxed{\textbf{A Bound on LCM}}$ $\color{green} \diamondsuit$ we know that
\[ b_{n+1} \mid \text{LCM}(b_n,p^{a} \cdot b) \Rightarrow b_{n+1} \leq p^{a-1} \cdot \text{LCM}(b_n,b) \]
so we obtain that
\[ b_{n+1} \leq \dfrac{b_n \cdot b}{p} < b_n.  \]
Note that the first equality is valid since $v_p(b_{n+1}) \leq a-1$ as a consequence of $p$ dropping. \\

This will imply that $p$ will stagnate on all $n+1$ between $p^{a}$ and $p^{a+1}$, so no expansions take place. \\

For powers higher than $a$, use the same method, as in the second paragraph of this proof. $\blacksquare$
--------
And now we go to the promised exact value of $b_n$: \\
$\color{cyan} \rule{5.8cm}{2pt}$ \\
$\color{cyan} \diamondsuit$ $\boxed{\textbf{Surprisingly Quick Result.}}$ $\color{cyan} \diamondsuit$ \\
$\color{cyan} \rule{5.8cm}{2pt}$ \\
We prove the Claim in the beginning for $n > N^2$. \\

For primes $p < N$, there must exist a prime power $p^{a}$ between $N$ and $N^2$; so $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Restrictive Expansions}}$ $\color{red} \clubsuit$ will ensure that
\[ v_p(b_{n}) = \max{(v_p(1),v_p(2),\ldots,v_p(n))} = v_p(\text{LCM}(1,2,\ldots,n)) \]
for $n > N^2$. \\

Moreover, for $p > N$, the first prime dividing $n+1$ greater than $N$ must be $p$ itself; and we can directly apply  $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Restrictive Expansions}}$ $\color{red} \clubsuit$ to ensure the same result for $n > N$ (a stronger bound, given that we do not have to pinpoint a specific $a$ for large $p$.) \\

Since we have guaranteed equality for all $p$, the conclusion immediately follows. $\blacksquare$ $\blacksquare$ \\
--------
Now we break the overwhelming symmetry on $b_n$, establishing that the numerator does not want $b_n$ to act in that fashion. \\
$\color{blue} \rule{7.4cm}{2pt}$ \\
$\color{blue} \clubsuit$ $\color{blue} \boxed{\textbf{Finishing: The Numerator's Revolt.}}$ $\color{blue} \clubsuit$ \\
$\color{blue} \rule{7.4cm}{2pt}$ \\
We Claim that $n = 2 \cdot 3^k$ for $n+1 > N^2$ yields $(a_n,b_n) \ne 1$. \\
$\color{blue} \rule{25cm}{0.2pt}$ \\
$\color{blue} \spadesuit$ $\color{blue} \boxed{\textbf{Proof.}}$ $\color{blue} \spadesuit$ \\
Obviously, $3 \mid b_n$. We prove that $3$ also divides $a_n$. \\

Since
\[ \dfrac{a_n}{b_n} = \sum_{i=1}^{n} \dfrac{1}{i} \Rightarrow a_n = \text{RHS} \, \cdot b_n, \]
we get
\[ a_n = 2 \cdot 3^k \left( \dfrac{1}{1}+\ldots+\dfrac{1}{3^k-1} + \dfrac{1}{3^k+1} + \ldots + \dfrac{1}{2\cdot 3^k-1} \right) + \dfrac{2 \cdot 3^k}{3^k} + \dfrac{2 \cdot 3^k}{2\cdot 3^k} \]
which is divisible by $3$. A contradiction. \\

We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$ \\

[hide = Notes]
Alternatively, one may see this as $a_{2 \cdot 3^k}/b_{2 \cdot 3^k}$ having to be simplified, implying this denominator to be smaller than the pre-established $b_{2\cdot 3^k-1}$ --- prompting $n = 2\cdot 3^k-1$ to be the desired $n$ that satisfies the problem. \\

Also, there are a variety of $n$ that can yield a contradiction (in this method of pinpointing $b_n$); as long as $(a_n,b_n) \ne 1$ for that $n$ we're good to go (see last bit of Motivation for an elaboration.)
[/hide]
--------
[hide = Motivation: Eliminating Wild Expansions and Drops.]

A story on solving this: in my training camp for the 2021 IMO, the team were asked to pick some of their own problems (preferably with solutions) and solve them within one designated hour. I picked this, and awfully enough, I realized that I fakesolved this problem when I had to submit my Solution. \\

My original way was claiming that $b_n \mid b_{n+1}$ if $b_{n+1} \geq b_n$, but this need not be true. As an example, pick $n+1$ so that some big prime experiences an expansion, and force the other primes dividing $n+1$ to have equal $v_p$ with $b_n$. \\

This will ensure some (small) prime to drop by a small margin. \\

The hardest step in this proof was to prove that $b_n$ needs to behave $LCM$-ly, and to ensure that, the other components of $n+1$ needs to be put on hold to facilitate the stagnancy of $p^{a}$. \\

$\color{green} \rule{25cm}{2pt}$ \\
$\color{red} \text{Sec 1: Arbitrary Substractions, and Unharmonious LCMs.}$ Direct computation of $\dfrac{a_n}{b_n}$ shows that $n = 6$ is the only $n$ that satisfies $b_{n+1} < b_n$ from $1 \leq n \leq 9$. (By this point I re-read the problem to avoid doing a misread problem.) \\

The next obvious way to proceed is to consider the fraction in $\color{green} \clubsuit$ $\boxed{\textbf{Each Prime's Increase Rate}}$ $\color{green} \clubsuit$'s proof. Here I noted that we can divide $(n+1)b_n$ by the common divisor of $n+1$ and $b_n$, yielding that the new denominator must be smaller than their LCM (as the simplification in primary school shows you.) \\

Then, I tried to claim for $n = 2p, p-1$ or $p^2-1$, only to find out that we need to guarantee the size ratio of $b_{p^2-1}$ and $b_{p^2}$ to win. Then I am back to considering
\[ \dfrac{a_n}{b_n}- \dfrac{a_{n-1}}{b_{n-1}} = \dfrac{1}{n} \]
and realizing that the LHS' denominator, $b_{n-1}b_n$, to be absurdly large. \\

By now, it seems that the only way to move forward is to let $n+1 = p^{a} y$ where $y$ is to be customised later. Some further manipulations conviced me that $p^a$ is less important than $y$ --- while $p^a$ needs to be harmonious with $b_n$ so that $p$ can drop, $y$ also needs to cooperate with the previous numerator and/or fractions. \\

$\color{green} \rule{25cm}{2pt}$ \\
$\color{red} \text{Sec 2: The Spin on Variety.}$ This is the catalyst for isolating the $v_p$ of each prime: the denominators being $\color{red} \textbf{\textit{harmonious}}$ implies their $v_p$ to be equal for some $p$. In the exact opposite, if
\[ v_p(b_n) \ne v_p(n+1) \]
whether the sign is $<$ or $>$, we can easily predict the standing of $b_{n+1}$ as in the first Claim. Consequently, this implies that every $\textbf{drop}$ needs to be propelled by an $\textbf{expansion}$. \\

At first glance I already felt a suspicion towards this statement --- an expansion can only occur with LARGE exponents!

$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{blue} \text{Sec 2.5: The Circular Circumventing + Interval Smash.}$ Writing this is the hardest part. While an artificially created expansion can be forged if $v_p(b_n)$ is still low, there's no surefire way to ensure a drop to instantly win. So now I focused on the expansion's behavior. \\

This is the place where I first tried to use a $n > N$ bound; and here I noted that the first $\textbf{\textit{lone}}$ expansion isn't very far --- just take $n+1 = p^a$! The next issue is showing that $p^{a'} \cdot b'$ cannot be an expansion for $a' < a$; or even $a' = a$. \\

Namely, I wanted the value of $v_p(b_n)$ to be somewhat stagnated after it is fortified by $n+1 = p^a$ --- otherwise its drop can create a leeway for the expansion criteria. This seemed to be trivial for $p = 2$ or $p = 3$; in the latter case the only danger to drop is for $n = 2 \cdot 3^a$ where the $2$ can create an expansion and $3$ could drop. \\

This is negated easily, as $\dfrac{2}{3} < 1$. I tried to induct this upwards for larger $p$, prompting larger $p$ (and even $p > N$) to be fortified this way, and I realized that there's no need for circular induction by smashing the large interval $[N, \infty)$ into
\[ [p^{a}, p^{a+1}], [p^{a+1},p^{a+2}], \ldots \]
and tackling each case separately, as it is easily done for $p = 2$ or $p = 3$. \\

$\color{green} \rule{25cm}{2pt}$ \\
$\color{red} \text{Sec 3: Mechanical Finish.}$ As $b_n$'s position is already set in place, I wondered how $a_n$ need not be accounted this far in regards to establishing $b_n$'s position. \\

After seeing that $a_n$'s exact value can be obtained by simple multiplication, this problem still proves to be a bit stubborn --- not all $n$ can directly yield a contradiction. For example, to prove that
\[ \gcd{(a_n,b_n)} \ne 1 \]
or $p \mid a_n,b_n$, $p$ needs to be less than $n$ (as $b_n$ is the LCM of numbers between $1$ and $n$) while $a_n$ supposedly has an reciprocal of a large prime power. (Take $n = 2^k+4$, for example. While $v_2$ of the LCM is $k$, the $v_2$ of the harmonic sum is also $-k$, as the term $1/2^k$ creates a large valley) \\

To combat this, one would need a lot of equal-sized valleys (read: harmonious denominators) to give a prime-divisor output. It turns out that a quick experimentation on $p = 3$ yields the result, as $1+2 \equiv 3$. \\
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
[/hide]
(Note: The Finishing could also be done with $1+2+3+\ldots+(p-1) \equiv 0 \mod p$ as in the official solution attached)

Surprisingly ugly solutions to this beatiful problem...

I might be missing something.
First note that sum 1/i, i=1 to (p-1)(p) = a/b, then p doesn’t divide b.
But if sum 1/i, i=1 to ((p-1)p) - 1= x/y, then p divides y.
Obviously b<y in this example. This works for every prime > 2


algorithm
combinatorics unsolved
combinatorics

In a graph, every vertex has degree $ k$. Prove that there is a cycle of length at least $ k \plus{} 1$.

A trivial note is that "graph" here means finite, simple, and loopless.  Consider the following algorithm:  choose a vertex $ v_0$ of the graph to start.  At step $ i \plus{} 1$, choose a vertex $ v_{i \plus{} 1}$ such that $ v_{i \plus{} 1}$ is a neighbor of $ v_{i}$ and $ v_{i \plus{} 1} \not \in \{v_{i \minus{} 1}, v_{i \minus{} 2}, \ldots, v_{i \minus{} k \plus{} 1}\}$.  Stop if we ever have $ v_i \equal{} v_j$ for $ i \neq j$.  Note that we can always make the required choice by the degree condition, and that when we halt the cycle $ v_j \to v_{j \plus{} 1} \to \ldots \to v_{i\minus{} 1} \equal{} v_j$ has length at least $ k \plus{} 1$ by the choice of $ v_i$.


geometry
geometric transformation
reflection

$ \left|x \plus{} 2\right| \plus{} \left|y \minus{} 3\right| \equal{} 1$ is an equation for a square.  How many units are in the lengths of its diagonals?

[hide]2[/hide]

It may be easier to imagine $ \left|x\right| \plus{} \left|y\right| \equal{} 1$, then translate it $ 2$ units left and $ 3$ units down (Although you don't need this translation to determine the length of the diagonal.)

For any solution $ (x,y)$ of the graph $ \left|x\right| \plus{} \left|y\right| \equal{} 1$, $ ( \minus{} x,y), (x, \minus{} y),$ and $ ( \minus{} x, \minus{} y)$ are also solutions. So you can assume that $ x,y > 0$, giving the line $ x \plus{} y \equal{} 1$ in the first quadrant only. Then place the 3 reflections of that line to make a square in a sort of "diamond" shape, with diagonal length of $ 2$.

Sorry that I left this question out for a long time.  However, I still don't get it.  Could someone explain it a little more, please?  Thanks.

Translation doesn't change anything, so we make it $ |x|\plus{}|y|\equal{}1$, and we get a quadrilateral with vertices at points (0,1),(0,-1),(1,0), and (-1,0).


modular arithmetic
number theory
algebra unsolved
algebra

Determine the last two digits of the sum
$7^{1}+7^{2}+...+7^{2005}.$.

[hide="Another solution"]
This is a geometric series. The sum is:
$\frac{7^{2007}-7}{6}$
$\phi{(100)}=16$
The sum is congruent to $\frac{7^{7}-7}{6}$, which reduces to $\boxed{7}$ so the last digits are 07.
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Shouldn't this be in the Number Theory Forum?