vector
algebra
polynomial
function
induction
linear algebra
matrix

Let $ A,B \in \mathcal{M}_n(\mathbb{C})$, show that there exist $ P, Q \in \mathrm{GL}_n(\mathbb{C})$ such that $ PAQ$ and $ PBQ$ are upper triangular.

Note: If we added the asumption $ AB \equal{} BA$, this is classical with $ Q \equal{} P^{ \minus{} 1}$ and already on this forum multiple times.

The beginning of the proof:
we must prove there exist 2 basis $ (e_i)_i$ and $ (f_i)_i$ of $ \mathbb{C}^n$ s.t. for all $ k$  $ Ae_k,Be_k$ are in the vector space spanned by $ f_1,\cdots,f_k$.
1) $ k\equal{}1$: we show there exists $ x\not\equal{}0$ s.t. $ Ax,Bx$ are parallel vectors. If $ det(B)\equal{}0$ then we take $ x\in{ker(B)}$ else we consider the polynomial function $ t\rightarrow{det(A\plus{}tB)}\equal{}t^ndet(B)\plus{}\cdots$ where $ t$ is a complex number. There exists $ t_0$ s.t. $ det(A\plus{}t_0B)\not\equal{}0$ and we take $ x\in{ker(A\plus{}t_0B)}$.

[quote="loup blanc"]There exists $ t_0$ s.t. $ det(A \plus{} t_0B)\not \equal{} 0$ and we take $ x\in{ker(A \plus{} t_0B)}$.[/quote]
When you take such $ t_0$, how do you ensure that $ \ker(A\plus{}t_0B)\neq \{0\}$ so that $ x\neq 0$?

Hi JC_math,  lapsus linguae.
There exists $ t_0$ s.t. $ det(A\plus{}t_0B)\equal{}0$,......

Hi loup blanc,

It makes sense.
Is an induction on $ n$ not enough to conclude then?

"lapsus calami" is better.
JC_math, OK for an induction on $ n$:
After 1) $ A\equal{}\begin{pmatrix}a&u\\0&U\end{pmatrix},B\equal{}\begin{pmatrix}b&v\\0&V\end{pmatrix}$ where $ U,V$ are $ (n\minus{}1,n\minus{}1)$ matrices. According to the induction there exist $ P_1,Q_1$ invertible $ (n\minus{}1,n\minus{}1)$ matrices s.t. $ P_1UQ_1,P_1VQ_1$ are upper triangular. Then we choose $ P\equal{}diag(1,P_1),Q\equal{}diag(1,Q_1)$.


Decide whether the following sequence or series are convergent:

1. Xn= (n!)^(1/n)

2. sum of (n!)/(n^n) with n from 1 to infinity

Both are easy using Stirling's approximation

$ n! \approx \sqrt {2 \pi n} \left( \frac {n}{e} \right)^n$


inequalities
function
inequalities unsolved

[b]Problem 1[/b]: Prove that:

$ \sum{\frac {a^{2}}{(b \minus{} c)^{2}}}\geq 2$

[b]Problem 2[/b]: $ 0 \leq x,y,z,t \leq 1$

Prove that: $ \sum{x(1 \minus{} y)}\leq 2$

[quote="Nam Luu"][b]Problem 1[/b]: Prove that:

$ \sum{\frac {a^{2}}{(b \minus{} c)^{2}}}\geq 2$[/quote]
[hide="Solution"]We have $ \frac {a^2}{(b \minus{} c)^2} \plus{} \frac {b^2}{(c \minus{} a)^2}\ge \frac {2ab}{(b \minus{} c)(c \minus{} a)}$ and cyclic.

Adding them up and dividing both sides by 2, we have $ \sum \frac {a^2}{(b \minus{} c)^2}\ge \frac {ab}{(b \minus{} c)(c \minus{} a)} \plus{} \frac {bc}{(c \minus{} a)(a \minus{} b)} \plus{} \frac {ca}{(a \minus{} b)(b \minus{} c)}$.

Hence, it suffices to prove that $ \frac {ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} ca(c \minus{} a)}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}\ge 2.$

Clearing denominator, we want to prove that $ a^2b \plus{} b^2c \plus{} c^2a \minus{} (ab^2 \plus{} bc^2 \plus{} ca^2)\ge 2(ab^2 \plus{} bc^2 \plus{} ca^2) \minus{} 2(a^2b \plus{} b^2c \plus{} c^2a)$$ \implies a^2b \plus{} b^2c \plus{} c^2a\ge ab^2 \plus{} bc^2 \plus{} ca^2$, which is a direct consequence of Muirhead Inequality.[/hide]

[quote="Nam Luu"][b]Problem 2[/b]: $ 0 \leq x,y,z,t \leq 1$

Prove that: $ \sum{x(1 \minus{} y)}\leq 2$[/quote]
[hide="Solution"]The function $ \sum x(1 \minus{} y)$ in $ x, y, z, t$ is convex in $ 0\le x, y, z, t\le 1$.

Since a convex function in closed interval achieves its extrema at end points, we just need to check the values of function when $ x, y, z, t\in \{0,1\}$.

Upon inspection, we find that the maxima of the function is 2, as desired.[/hide]

[quote="Nam Luu"][b]Problem 1[/b]: Prove that:

$ \sum{\frac {a^{2}}{(b \minus{} c)^{2}}}\geq 2$

[/quote]

we have

$ \sum{\frac {a^{2}}{(b \minus{} c)^{2}}} \minus{} 2 \equal{} \frac{\left(\sum{a(a\minus{}b)(a\minus{}c)}\right)^2}{\prod{(a\minus{}b)^2}} \ge 0$

[quote="Nam Luu"][b]Problem 1[/b]: Prove that:

$ \sum{\frac {a^{2}}{(b \minus{} c)^{2}}}\geq 2$

[/quote]

$ \sum \frac{a^2}{(b\minus{}c)^2}\equal{}\left( \sum\frac{a}{b\minus{}c}\right)^2\plus{}2\frac{bc}{(a\minus{}b)(a\minus{}c)}\plus{}2\frac{ac}{(b\minus{}c)(b\minus{}a)}\plus{}2\frac{ab}{(c\minus{}a)(c\minus{}b)}\equal{}\left(\sum\frac{a}{b\minus{}c}\right)^{2}\plus{}2\geq 2$

[quote="FantasyLover"]$ \implies a^2b \plus{} b^2c \plus{} c^2a\ge ab^2 \plus{} bc^2 \plus{} ca^2$, which is a direct consequence of Muirhead Inequality.[/quote]I don't think so... Try $ a\equal{}3,b\equal{}1,c\equal{}2$. Actually, this inequality is equivalent to $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\le0$ which is not necessarily true.

[quote="Johan Gunardi"][quote="FantasyLover"]$ \implies a^2b \plus{} b^2c \plus{} c^2a\ge ab^2 \plus{} bc^2 \plus{} ca^2$, which is a direct consequence of Muirhead Inequality.[/quote]I don't think so... Try $ a \equal{} 3,b \equal{} 1,c \equal{} 2$. Actually, this inequality is equivalent to $ (a \minus{} b)(b \minus{} c)(c \minus{} a)\le0$ which is not necessarily true.[/quote]

but in the beginning, with out loss of generality, we can assume that $ a \ge b \ge c$, thus, the last inequality will holds...

but if we assume that $ a \ge b \ge c$, when we clearing denominator, still get $ a^{2}b\plus{}b^{2}c\plus{}c^{2}a\ge ab^{2}\plus{}bc^{2}\plus{}ca^{2}$ ?

Impossible to show $ \frac{ab(a\minus{}b)\plus{}bc(b\minus{}c)\plus{}ca(c\minus{}a)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\ge 2$ because

$ {ab(a\minus{}b)\plus{}bc(b\minus{}c)\plus{}ca(c\minus{}a)}\equal{}(ab\minus{}ca)(a\minus{}b)\plus{}(bc\minus{}ca)(b\minus{}c)$ $ \equal{}(a\minus{}c)(b\minus{}c)(a\minus{}b)$.

So $ \frac{ab(a\minus{}b)\plus{}bc(b\minus{}c)\plus{}ca(c\minus{}a)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\equal{}\minus{}1$. FantasyLover's  method cannot work, because some of the cyclics of the terms on the RHS of $ \frac{a^{2}}{(b\minus{}c)^{2}}\plus{}\frac{b^{2}}{(c\minus{}a)^{2}}\ge\frac{2ab}{(b\minus{}c)(c\minus{}a)}$ are negative. :(

[quote="spanferkel"]Impossible to show $ \frac {ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} ca(c \minus{} a)}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}\ge 2$ because

$ {ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} ca(c \minus{} a)} \equal{} (ab \minus{} ca)(a \minus{} b) \plus{} (bc \minus{} ca)(b \minus{} c)$ $ \equal{} (a \minus{} c)(b \minus{} c)(a \minus{} b)$.

So $ \frac {ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} ca(c \minus{} a)}{(a \minus{} b)(b \minus{} c)(c \minus{} a)} \equal{} \minus{} 1$. FantasyLover's  method cannot work, because some of the cyclics of the terms on the RHS of $ \frac {a^{2}}{(b \minus{} c)^{2}} \plus{} \frac {b^{2}}{(c \minus{} a)^{2}}\ge\frac {2ab}{(b \minus{} c)(c \minus{} a)}$ are negative. :([/quote]

yeah, I agree with you.

but if use $ \frac {a^{2}}{(b \minus{} c)^{2}} \plus{} \frac {b^{2}}{(c \minus{} a)^{2}}\ge\frac {2|ab|}{|(b \minus{} c)(c \minus{} a)|}$, can prove the original inequality out?


When I typeset my solutions for the USAMTS, I pasted all of the words from fancyhdyr.sty and usamts.tex into the solutions template with my solutions. However, after trying to compile, the computer said there were 41 errors in the file. Is there something wrong with the content in any of these files? If I am doing something wrong, please try to notify me to tell me how to include all of the correct files on the USAMTS Forms page. Thank you in advance for your help.

You're not supposed to copy anything. Just save usamts.tex, fancyhdyr.sty, and solutions.tex in the same folder. Then type your solutions on the file solutions.tex, and compile. :D


geometry
perimeter
circumcircle
Asymptote
trigonometry
AMC 10
perpendicular bisector

Here are a few practice geometry AMC problems (some of the harder AMC 10 and 12)
1. Let $ C_{1}$ and $ C_{2}$ be the circles defined by
$ (x-10)^{2}+y^{2}=36$ and
$ (x+15)^{2}+y^{2}=81$.
What is the length of the shortest line segment XY tangent to the first circle at X and to the second at Y?

2. A regular octagon has side length 2. Find the area of triangle ADG.

3. In triangle XYZ, side XZ and YZ's perpendicular bisector intersect t point M and YM bisects angle XYZ. If XM=9 and MZ=7, what is the area of triangle ABD?

4. How many non congruent triangles with perimeter 7 have positive side lengths?

5.The quadrilateral ABCD has vertices A, B, C, and D. AB=40,CD=5,BC=26, BD=7,and AC=50. Find  AD.

3. What is triangle ABD? Those points are not defined anywhere in the problem.

4. Infinite. Did you mean positive integer side lengths?

5. Uh that diagram is impossible since triangle BDC doesn't even satisfy the triangle inequality... CD=5, BC=26, and BD=7, but 5+7<26...

Okay, I really must have been half-asleep...  :oops:
3. I mean XYZ
4. Yup, integer
5. Hmm... forget about that one then. I might go look up the problem.

[quote="xpmath"]3. In triangle XYZ, side XZ and YZ's perpendicular bisector intersect t point M and YM bisects angle XYZ. If XM=9 and MZ=7, what is the area of triangle ABD? [/quote]

Um, since the intersection of the perpendicular bisectors is supposed to be the circumcenter, $ XM \equal{} MZ \equal{} R$, where $ R$ denotes the circumradius. But you say $ XM\neq MZ$... Am I missing something?

Perhap's I should have worded it better:  the intersection is only of a side and another side's perpendicular bisector.
hey are not both perpendicular bisectors.
Side=XZ, the bisector=YZ

[hide="3"]

[asy]import cse5; size(150,150); pair Y=(0,0),Z=(84,0),T=(42,0); pair X=IP(CR(Y,108),CR(Z,144)); D(MP("X",D(X),N)--MP("Y",D(Y),SW)--MP("Z",D(Z),SE)--X); pair M=IP(X--Z,T--(42,200)); D(MP("T",D(T),S)--MP("M",D(M),NE)--Y);[/asy]

Alright so $ \triangle YMZ$ is isosceles so $ YM \equal{} 7$. By the angle bisector theorem, $ \frac{YX}{YZ}\equal{}\frac{9}{7}$ so let $ YX \equal{} 18x$ and $ YZ \equal{} 14x$. If T is the midpoint of YZ, then $ YT \equal{} 7x$.

$ \cos\angle XYM \equal{}\cos\angle MYT \equal{}\frac{7x}{7}\equal{} x$

By the law of cosines on triangle XYM,

$ (18x)^{2}\plus{}7^{2}\minus{}2(18x)(7)\cos\angle XYM \equal{} 9^{2}$
$ (18x)^{2}\plus{}7^{2}\minus{}2(18x)(7)x \equal{} 9^{2}$

We get $ x \equal{}\frac{2}{3}$ by solving, so the sides of the triangle are $ XY \equal{} 18\left(\frac{2}{3}\right) \equal{} 12$ and $ YZ \equal{} 14\left(\frac{2}{3}\right) \equal{}\frac{28}{3}$ and $ XZ \equal{} 9\plus{}7 \equal{} 16$.

Now I found Heron's formula on the entire triangle too messy, so instead I'll just do it on $ \triangle XYM$ and then add it to the area of $ \triangle YMZ$.

For XYM, $ s \equal{} 14$ and $ \sqrt{s(s\minus{}12)(s\minus{}7)(s\minus{}9)}\equal{} 14\sqrt{5}$

For YMZ, the altitude MT, by the pythagorean theorem, is $ \frac{7}{3}\sqrt{5}$ so the area is $ \frac{14}{3}\left(\frac{7}{3}\sqrt{5}\right) \equal{}\frac{98}{9}\sqrt{5}$

Adding these up gives $ \frac{224}{9}\sqrt{5}$

EDIT- in fact, i think the easiest way is with the formula $ A \equal{}\frac{1}{2}\sin\theta ab$

$ [XYZ] \equal{}\frac{1}{2}\sin\angle XZY (16)\left(\frac{28}{3}\right) \equal{}\frac{224}{3}\sin\angle MYZ \equal{}\\ \frac{224}{3}\sqrt{1\minus{}\cos^{2}\angle MYZ}\equal{}\frac{224}{3}\sqrt{1\minus{}\left(\frac{2}{3}\right)^{2}}\equal{}\frac{224}{9}\sqrt{5}$

[/hide]


Emrooz sobh talahaye sale 1387 ham maloom shodan :
1-Mahed Abroshan
2-Ehsan Azarmsa
3-Mohammadjavad Akbari
4-Farshid Bateni
5-Khashayar Khosravi
6-Hossein Dabirian
7-Khashayar Sartipi
8-Sepehr Ghazinezami
9-Ehsan Gera'eli
10-Masood Givechi
11-Mohiedin Motevassel
12-Soheil Nemati
13-Masood Noorbakhsh
14-Pooya Vahidi Ferdosi

khashayar behet tabrik migam az samime ghalb omidvaram beterekooni , dar zemn noghre yek ki shod ? tala yek ki shod ? (noghre yek ham sahmie kamel migire dige na ?)


Let ABCDE be a pentagon. Suppose AB meets CD at F, and BC meets DE at G. Let the circumcircles of triangles BCF and CDG meet again at A' (different from C). Define points B', C', D', E' likewise. Show that pentagon A'B'C'D'E' is cyclic.  :D

It's called Clifford's circle, I think. It's all angle chasing+Miquel's theorem.

I'll change some notations first. Assume $AB\cap CD=E_1$, for instance, and define $A_1,B_1,C_1,D_1$ similarly. Then assume circles $(BCE_1),(CDA_1)$ intersect at $C,C'$, and, similarly, define $A',B',D',E'$.

It suffices to show that $A'B'C'D'$ is cyclic. From Miquel's Theorem we know that $A',D',A_1,D_1,E$ are concyclic. Now the angle chase can start :): $\angle B'A'D'=\angle D_1A'D'-\angle D_1A'B'=\pi-\angle D'A_1D_1-\angle D_1BB'$, which is equal to $\pi-\angle D'C'C-\angle CC'B'=\pi-\angle D'C'B'$, Q.E.D.

Yeah, I didn't think this problem would go past you, grobber, except your solution is about 1/5 of the length of mine!!! (but yes, it was all angle-chasing, which is my favorite... :D )

This is indeed a well-known result. See also

[url=http://www.math.ust.hk/excalibur/v6_n1.pdf]Kin Y. Li, [i]Concyclic Problems[/i], Mathematical Excalibur 1/2001 (6) p. 1-2[/url].

  Darij

Thanks darij.  :)  (For a second there, I thought you understood chinese!  :D )


(a) Given real numbers a; b; c, with a + b + c = 0, prove that [tex]a^3 + b^3 + c^3 > 0[/tex] if and only if [tex]a^5 + b^5 + c^5 > 0[/tex]:
(b) Given real numbers a; b; c; d, with a+b+c+d = 0, prove that [tex]a^3+b^3+c^3+d^3 > 0[/tex] if and only if [tex]a^5+b^5+c^5+d^5>0[/tex].

http://www.mathlinks.ro/Forum/viewtopic.php?t=25457 .

  darij


The product of three consecutive positive integers is $ 8$ times their sum. What is the sum of their squares?

$ \textbf{(A)}\ 50 \qquad
\textbf{(B)}\ 77 \qquad
\textbf{(C)}\ 110 \qquad
\textbf{(D)}\ 149 \qquad
\textbf{(E)}\ 194$

If you follow the directions, you'll eventually end up with a cubic: $ x^{3}\plus{}3x^{2}\minus{}22x\minus{}24\equal{}0$ which you need to solve.  It's not hard to see that $ x\equal{}4$ satisfies the equation.  Then the numbers are $ 4,5,6$ and from there it is easy.

yes so $ 4^2 \plus{} 5^2 \plus{} 6^2 \equal{} 77 \equal{} \boxed{B}$

[hide="easier than a cubic"]Let the middle term be x, so you have (x-1)(x)(x+1)=24x. Simplify to find $ x^3\minus{}25x\equal{}0$, since x is not 0, we find $ x^2\minus{}25\equal{}0$ and x=5. Simple after that.[/hide]

its even more simpler
suppose the nos to be n-1,n, n+1
then, (n^2-1).n=8*3n
this gives, n=5
hence , answer c :D


Given that kids require time and money,

[size=150]Kids=Time*Money[/size]

We all know that time is money so:

[size=200]Kids=Money*Money=Money^2[/size]

And because money is the root of evil:

[size=150]Money=sqrt(Evil)[/size]

Therefore:

[size=150]Kids=(sqrt(Evil))^2[/size]

and so we get that:

[size=150]Kids=Evil[/size]

oops...sorry I pressed submit but I wanted to change the title so I went back and now there are 2 of the same post :blush:

[quote="jli"]oops...sorry I pressed submit but I wanted to change the title so I went back and now there are 2 of the same post :blush:[/quote]Not anymore :P

:rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl::rotfl: :rotfl::rotfl:

[quote="tetrahedr0n"]We've already had this same joke on the forum, except with girls instead of kids  :D .[/quote]

oops...my dad told me it :blush:

I love it when u can use math to prove weird things.

[quote]We've already had this same joke on the forum, except with girls instead of kids Mr. Green .[/quote]

I feel so offended by that!!!!!
don't be a sexist!!!!

guys and girls are both equal!!!

It's not that I'm sexist... that is just the way to original joke posted here went.

Nothing is better than eternal happiness does not mean that literally... Its funny anyways.

[quote="djimenez"][quote="shobber"]Who can give some other identities?[/quote]

Well, what about this one: $\$1 is better than eternal happiness.
Let's start sayint that we know that $\$1 is better than nothing!, everybody agrees up to here. On the other hand, we all know deep in our heart that nothing is better than eternal happiness. Therefore by simple transitivity, $\$1 is better than eternal happiness!, am I right?

Best,[/quote]

ha ha ha! that's funny but nothing has more than one meaning

[quote="peter"][quote="Valentin Vornicu"][quote="A+MATH"]sin(x)/n = 6       si(x) = 6  six = 6[/quote]Actually 6 six = $6 \cdot 6 $ = 36 ;)[/quote]

I think he was clarifying how he went from sin(x)/n=6 to 6, not multiplying.[/quote]Yes I know and I proved him wrong, that is actually sin(x)/n=36 not 6 ;) :D :D

[quote="Valentin Vornicu"][/quote][quote="peter"][quote="Valentin Vornicu"][quote="A+MATH"]sin(x)/n = 6       si(x) = 6  six = 6[/quote]Actually 6 six = $6 \cdot 6 $ = 36 ;)[/quote]

I think he was clarifying how he went from sin(x)/n=6 to 6, not multiplying.[/quote][quote="Valentin Vornicu"]Yes I know and I proved him wrong, that is actually sin(x)/n=36 not 6 ;) :D :D[/quote]

Not at all...to find sin(x)/n, cancel the n's and get six.  (I've seen this before)

[quote="peter"]Not at all...to find sin(x)/n, cancel the n's and get six.  (I've seen this before)[/quote]Are you trying to prove that my proof of sin(x)/n = 36 based on the proof of sin(x)/n = 6 is wrong?  :lol:  :lol:  :D

[quote="solafidefarms"]DJiminez, I was going to post that joke![/quote]

Sorry man, I also was there when Lew told that joke  :D , (one of the few good jokes that day  ;)  :D )

Best,

OK, Let's get this thread doing for a while.  Here is one from http://www.workjoke.com:

One day a mathematician decides that he is sick of math. So, he walks down to the fire department and announces that he wants to become a fireman.
The fire chief says, "Well, you look like a good guy. I'd be glad to hire you, but first I have to give you a little test."

The firechief takes the mathematcian to the alley behind the fire department which contains a dumpster, a spigot, and a hose. The chief then says, "OK, you're walking in the alley and you see the dumpster here is on fire. What do you do?"
The mathematician replies, "Well, I hook up the hose to the spigot, turn the water on, and put out the fire."

The chief says, "That's great... perfect. Now I have to ask you just one more question. What do you do if you're walking down the alley and you see the dumpster is not on fire?"
The mathematician puzzles over the question for awhile and he finally says, "I light the dumpster on fire."
The chief yells, "What? That's horrible! Why would you light the dumpster on fire?"
The mathematician replies, "Well, that way I reduce the problem to one I've already solved."

Here is another one from http://www.workjoke.com:

Top ln(e^10) reasons why e is better than pi

10) e is easier to spell than pi.
9) pi ~= 3.14 while e ~=2.718281828459045.
8) The character for e can be found on a keyboard, but pi sure can't.
7) Everybody fights for their piece of the pie.
6) ln(pi^1) is a really nasty number, but ln(e^1) = 1.
5) e is used in calculus while pi is used in baby geometry.
4) 'e' is the most commonly picked vowel in Wheel of Fortune.
3) e stands for Euler's Number, pi doesn't stand for squat.
2) You don't need to know Greek to be able to use e.
1) You can't confuse e with a food product.

[quote="mcalderbank"]that is exactly why it is funny ;)

I just know the basic 2 = 1 'proof'

a = b
2a = a+b
2a - 2b = a - b
2(a - b) = a - b
2 = 1[/quote]
THE SECOND LAST STEP TO THE LAST STEP IS TRUE WHILE a!=b
therefore 1!=2
it's obvious... :blush:

[quote="zmli"]8) The character for e can be found on a keyboard, but pi sure can't.[/quote]What about in a greek computer's keyboard :D

[quote="zmli"]5) e is used in calculus while pi is used in baby geometry.[/quote]... and in Fourier Analysis, and in calculus too, and in many other areas of mathematics.

[quote="zmli"]3) e stands for Euler's Number, pi doesn't stand for squat.[/quote]Actually, according to what I know, it comes from exponential, not from Euler. Also, I think it is known more as the Napier's number than for the Euler's number.


geometry
3D geometry
sphere
tetrahedron
analytic geometry
symmetry
trigonometry

Given a tetrahedron such that  $ AB \equal{} \sqrt {3},\ AC \equal{} AD \equal{} BC \equal{} BD \equal{} CD \equal{} 2$ on a sphere with a radius $ r$. Find the value of $ r$.

[hide="With Coordinates"][color=darkblue]Let the four points be $ C(1,0,0)$, $ D ( \minus{} 1,0,0)$, $ B(0, \sqrt {3}, 0)$, and $ A \left( 0, \frac {\sqrt {3}}{2}, \frac {3}{2} \right)$.

By symmetry, the center of the sphere is on the segment connecting the midpoint of $ AB$ and the midpoint of $ CD$, which we will parametrize as $ k (0, \sqrt {3} , 1)$. [i]Note that $ k \equal{} 3/4$ and $ k \equal{} 0$ give our endpoints.[/i]

So, [b][u]equation of sphere is:[/u][/b] $ x^2 \plus{} (y \minus{} \sqrt {3} k)^2 \plus{} (z \minus{} k)^2 \equal{} r^2$.

$ (1,0,0)$ is a solution, $ \implies 1 \plus{} 4k^2 \equal{} r^2$.
$ (0, \sqrt {3}, 0)$ is a solution, $ \implies 3 (1 \minus{} k)^2 \plus{} k^2 \equal{} r^2$
$ \therefore 1 \plus{} 4k^2 \equal{} k^2 \plus{} 3(1 \minus{} k)^2 \implies k \equal{} \frac {1}{3}$

Then, $ 1 \plus{} 4 \left( \frac {1}{3} \right) ^2 \equal{} r^2 \implies \mathbf{ r \equal{} \frac {\sqrt {13}}{3}}$[/color][/hide]

That's the correct answer.  :)

Can anyone solve the problem by other method?

[hide="Without Coordinates"]

Let M be the midpoint of AB, N be the midpoint of CD, O be the center of the circumsphere, and O' be the projection of O onto the plane containing triangle ACD. Observe that O' is the centroid of ACD.

We know $ AM = \frac {AB}{2} = \frac {\sqrt {3}}{2}$, that $ AN = \cos 30 ^{\circ} \cdot AC = \sqrt {3}$, and that $ \angle M$ is right. Therefore, $ \text{m} \angle ANM = 30 ^{\circ}$. It follows that $ OO' = \frac {1}{\sqrt {3}} O'N = \frac {1}{\sqrt {3}} \left( \frac {1}{3} AN \right) = \frac {1}{3}$.

Finally, $ r = OA = \sqrt {OO' ^2 + O'A ^2} = \sqrt {OO' ^2 + (2O'N) ^2} = \sqrt {\left( \frac {1}{3} \right) ^2 + \left( \frac {2}{\sqrt {3}} \right) ^2} = \frac {\sqrt {13}}{3}$[/hide]


calculus
integration
trigonometry
algebra
partial fractions
real analysis
real analysis unsolved

hello, do you mean $ \int\sqrt[6]{\tan(x)}\,dx$?
Sonnhard.

Substitute $ t\equal{}(\tan x)^{1/6}$, $ x\equal{}\arctan(t^6)$. The integral becomes $ \int\frac{6t^6}{1\plus{}t^{12}}\,dt$. Now you have a complicated mess of a partial fractions problem. Have fun.

I did this only,but the solution became complicated..I am looking for a shorter solution or can someone post the complete solution to this messi question


inequalities
trigonometry
inequalities unsolved

I am trying to solve this problem

In an acute-angled triangle ABC find the best value of K>1 such that

SinA +SinB +SinC >= K( CosA +CosB +CosC)

Thank you for your help!

Dear Michael Niland,

This is not the complete proof but maybe it can be helpful for you.

First we have 2 indentities
  sinA + sinB + sinC = p/R
  cosA + cosB + cosC = 1 + R/r

We have to find k such that
 p >= k(R+r)

But in acute triangle we have the inequality
  p > 2R+r

Apply this one and the well known R >= 2r  we can find k >=5/3

The equality case is impossible. Now I do not know if 5/3 is the best or not. I still try to improve it.

I think the best value is $ k = \frac{1}{\sqrt{2}} + 1$ if you let ABC be an acute issoceles triangle at A and let $\measuredangle{BAC}$ get closer and closer to $90^0$
   You can prove this inequality by let A = max {A,B,C} and using derivation to prove this inequality.
   The equality holds if and only if it is a right issoceles triangle
    I 'll give you my solution later (maybe in my next post)


inequalities
function
calculus
derivative
inequalities proposed

[quote="cefer"]
$x^{4}+y^{4}+z^{4}+6 \geq 3(x^{3}+y^{3}+z^{3})$
[/quote]
This inequality is not true.

We can prove it by Cauchy-Schwarz Inequality

$\sum \frac{x^{2}}{1+y^{3}+z^{3}}\geq \frac{(x^{2}+y^{2}+z^{2})^{2}}{\sum x^{2}(1+y^{3}+z^{3})}\geq 1$.

But the proof of the right inequality is not very easy. :)

This form is correct, and suggests the nice inequality

$a^{5}+b^{5}+c^{5}+abc(ab+bc+ca) \geq \sum a^{2}b^{2}(a+b)$.

But I don't think this inequality help you to prove the inequality above. :P

[quote="nayel"][quote="tripleth"]Let $x,y,z$ be positive numbers satisfying $x+y+z=3$. Prove the inequality:
\[{x^{2}\over 1+y^{3}+z^{3}}+{y^{2}\over 1+z^{3}+x^{3}}+{z^{2}\over 1+x^{3}+y^{3}}\geq 1. \]
[/quote]
can we say that $x^{3}+y^{3}+z^{3}\le 3$? if we can, then i might be able to give a proof.[/quote]

Problem is, $x^{3}+y^{3}+z^{3}\geq 3$.

sorry we can't since $3(a^{3}+b^{3}+c^{3})\ge (a^{2}+b^{2}+c^{2})(a+b+c)\ge 9$. i felt too lazy to check it so haded it to you guys.

[quote="silouan"]The last inequality mr Vasc wrote is equivalent to
$x^{4}+y^{4}+z^{4}+xyz(xy+zy+zx)\geq x^{2}y^{2}+z^{2}y^{2}+z^{2}x^{2}+x^{2}+y^{2}+z^{2}$[/quote]
We can prove it using the topic "Schur's Inequalities"
http://www.mathlinks.ro/Forum/viewtopic.php?t=124095

It helps that $9+3abc\geq 4(ab+bc+ca)$   ?

Only for $ab+bc+ca\leq 2$, when the third degree Schur's Inequality is stronger than the fourth degree Schur's Inequality.

$9+3abc\geq 4(ab+bc+ca)$  comes from
$(a+b+c)^{3}+9abc\geq 4(a+b+c)(ab+bc+ca)$ .
But coulf this help ? :maybe:
If no , mr Vasc could you post your solution?

I believe that this is correct.
[b]Solution.[/b]

Let $f(a,b,c)=\frac{a^{2}}{1+b^{3}+c^{3}}$.  Note that $f$ is convex.

Using the multivariable Jensen's Inequality,

\[\sum_{cyc}f(a,b,c)\geq 3 f\left(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3}\right)=3f(1,1,1)=1,\]
as required.$\blacksquare$

Masoud Zargar

[quote="Gibbenergy"][quote="boxedexe"]I believe that this is correct.
[b]Solution.[/b]

Let $f(a,b,c)=\frac{a^{2}}{1+b^{3}+c^{3}}$.  Note that $f$ is convex.

Using the multivariable Jensen's Inequality,
\[\sum_{cyc}f(a,b,c)\geq 3 f\left(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3}\right)=3f(1,1,1)=1, \]
as required.$\blacksquare$

Masoud Zargar[/quote]

 I am curious to know what is Mutivariable Jensen's inequatity. What is it?[/quote]

It is Jensen's Inequality for a function with more than one variable. For the function to be convex it has to be convex for each variable.

[quote="weiquan"][quote="Gibbenergy"][quote="boxedexe"]I believe that this is correct.
[b]Solution.[/b]

Let $f(a,b,c)=\frac{a^{2}}{1+b^{3}+c^{3}}$.  Note that $f$ is convex.

Using the multivariable Jensen's Inequality,
\[\sum_{cyc}f(a,b,c)\geq 3 f\left(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3}\right)=3f(1,1,1)=1, \]
as required.$\blacksquare$

Masoud Zargar[/quote]
/quote]

It is Jensen's Inequality for a function with more than one variable. For the function to be convex it has to be convex for each variable.[/quote][/quote]

  You mean the second partial derivative of the function with respect to each variable must be positive? Then, in the solution above, he must differentiate the term in the denominator too.  $\frac{c^{2}}{1+a^{3}+b^{3}}$:D
  Can anyone give me a reference for Jensen's multiple variable inequalities? Thank you a lot.

[quote="Lovasz"][quote="boxedexe"]I believe that this is correct.
[b]Solution.[/b]

Let $f(a,b,c)=\frac{a^{2}}{1+b^{3}+c^{3}}$.  Note that $f$ is convex.

Using the multivariable Jensen's Inequality,
\[\sum_{cyc}f(a,b,c)\geq 3 f\left(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3}\right)=3f(1,1,1)=1, \]
as required.$\blacksquare$

Masoud Zargar[/quote]

That seems this solution is wrong. Suppose the powers in denominators is 9, then $f(a,b,c)={a^{2}\over 1+b^{9}+c^{9}}$ is convex, but the inequality $\sum f(a,b,c) \geq 1$ is wrong![/quote]

Actually you are right, this solution is wrong, as Boredexe later said, but so is your claim :P. $f(a,b,c)={a^{2}\over 1+b^{9}+c^{9}}$ is neither convex nor concave!

[quote="tripleth"][quote="tripleth"]Let $x,y,z$ be positive numbers satisfying $x+y+z=3$. Prove the inequality:
\[{x^{2}\over 1+y^{r}+z^{r}}+{y^{2}\over 1+z^{r}+x^{r}}+{z^{2}\over 1+x^{r}+y^{r}}\geq 1. \]
[/quote]
I guess the inequality is still true if we put $r=4,5,6,...$ Did someone pay attention to this point?[/quote]
I think, you are wrong. For $n\geq4$ try $x=y\rightarrow1.5^-$ and $z\rightarrow0^+.$


quadratics
function
algebra
polynomial
Vieta
number theory solved
number theory

Markov style, again :D

Key is to write in terms of x.

(x + y + z)^2 = kxyz. (*)

If there are integer solutions to *, then

x^2 + x(2y + 2z - kyz) + (y + z)^2 = 0 has integer solutions.

Suppose there does not exist a family of solutions (x, y, z) with x \geq y + z.
Then x < y + z for all solutions (x, y, z).

By Viete's formula, (x', y, z) must also satisfy *, where x' = (y + z)^2 / x, clearly an integer because x | (y + z)^2 simply from observing *.
But x < y + z, so x` > y + z. Contradiction.

So let us look at the family of solutions (x, y, z) with x \geq y + z.
Let x = x_0 be minimal.

By Viete's formula, (kyz - 2y - 2z - x_0, y, z) must also satisfy *.

But x_0 is minimal, so kyz - 2y - 2z - x_0 \geq x_0.

Then (y + z)^2 = x_0(kyz - 2y - 2z - x_0) \geq x_0^2 implies

y + z \geq x_0. But x_0 \geq y + z, so x_0 = y + z.

Then 4y + 4z = kyz after substitution. WLOG, y \geq z.

If k > 8, kyz > 8yz \geq 8y \geq 4y + 4z, a contradiction.

Thus, for * to have solutions, it is necessary to have k \leq 8. But k cannot equal 7, for otherwise 4y + 4z = 7yz implies (7y - 4)(7z - 4) = 16. It is easy to check that all other k (i.e. 1 \leq k \leq 8, k is not 7) generate solutions for *.

The source of the problem , please! I know it from AMM (as far as I remember, in 1994) or from Mongolian Olympiad (I don't know the year, but I believe is 2000 or 2001).

I think it's both Mongolia 2001 (just remebered this one) and AMM 1994.

[quote="StRyKeR"]
So let us look at the family of solutions (x, y, z) with x \geq y + z.
Let x = x_0 be minimal.

By Viete's formula, (kyz - 2y - 2z - x_0, y, z) must also satisfy *.

But x_0 is minimal, so kyz - 2y - 2z - x_0 \geq x_0.
[/quote]

Can someone explain to me why we must have kyz - 2y - 2z - x_0 \geq x_0? I thought x_0 is minimal only for solutions in which x \geq y + z. Wouldn't the solution (kyz-2y-2z-x_0, y, z) statisfy kyz-2y-2z-x_0 <= y+z?

[quote="Valentin Vornicu"]For which positive integer $k$ there exist positive integers $x,y,z$ such that $(x+y+z)^2= kxyz$ ?[/quote]
k=1,2,3,4,5,6,8,9

First we prove that $k\le 9$:

Let $(a,b,c)$ be a natural solution to  $(x+y+z)^2=kxyz...  ...(1)$  so that sum $a+b+c$ is minimal.  Then

$a^2+(2b+2c-kbc)a+(b+c)^2=0$

so we see that $a $ is a natural solution of quadratic function:

$f(x)=x^2+(2b+2c-kbc)x+(b+c)^2$

wlog assume that $a\ge b\ge c$.

Then let $a_1$ be a second solution of $f(x)$. Then by vieta theorem we have that

$a+a_1 = kbc-2b-2c$

$aa_1 = (b+c)^2$

from these we can see that $a_1\in N$.  So $ (a_1,b,c) $ is also a natural solution of (1) . So $a+b+c\le a_1+b+c$ equivalen to   $a_1\ge a $ .So since $a_1\ge a\ge b$ we have $f(b)=(b-a)(b-a_1)\ge 0$ so

$f(b)=4b^2+c^2+4bc-kb^2c \ge 0$

if we had that $k>9$
then  $4b^2+c^2+4bc \ge kb^2c>9b^2c\ge 4b^2+c^2+4bc$

contradiction. So

We have $k\le 9$.

for $k= 1,2,3,4,5,6,8,9$ there are example solutions. But for $k=7$  there are no solution.
To prove it we must prove that $(x+y+1)^2=7xy$ has no solution in N. It can be proven by same vieta method. Then after asuuming that $x,y,z>1$ we can prove that (1) has no sol. for $k=7$ by same method as proven above.  So finally

$k=1,2,3,4,5,6,8,9$. is answer.

For $k=7$

$ 4b^2+c^2+4bc\ge 7b^2c$

$\implies 9b^2\ge 4b^2+c^2+4bc\ge7b^2c$

$\implies c\le \frac 97<2\implies c=1$

$ \implies 4b^2+1+4b\ge 7b^2$

$\implies 2b+1\ge \sqrt 7 b\implies b=1$

Hence $7=\frac{(a+1+1)^2}{a\times 1\times1}\implies a^2-3a+4=0\implies $ No Solution Done!


AMC
AMC 12
AMC 10
USA(J)MO
USAJMO
AIME
USAMO

[quote="batteredbutnotdefeated"]ummm... Let's say I score higher on an AMC12 than on an AMC10... How would I remedy such a problem? I'm stronger in advanced concepts than I am at the basic ones...[/quote]

I guess if you score higher on AMC 12 than you do on AMC 10, you can choose to use your AMC 12 score to participate in the USAMO...

but I want to be ready for [b]everything[/b]...

including AMC10. Would doing tons of AMC10s help?

You can never be ready for everything, since there are infinitely many things to be ready for. So you have to choose what you're going to prepare for. I think choosing to be ready for both the AMC10 and AMC12 is just an arbitrary choice. You're free to make it, but since it seems unlikely to benefit you in any way (unless this USAJMO thing somehow benefits you for taking AMC10), it seems silly to bother preparing for the AMC10. I personally didn't prepare for the AMC12 even, focusing instead on AIME, since that seemed to be more important in terms of making the USAMO, plus if you can do well on the AIME you will probably do moderately well (i.e. 120+) on the AMC12.

Perhaps you mean that you think it's bad that you can't solve problems on "basic" topics? Do you actually mean that you have a lower score on the AMC10? This is probably just random fluctuation, since the AMC10 is usually the AMC12 minus the 5 hardest problems (with 5 easier problems substituted).

[quote="batteredbutnotdefeated"]I find myself over thinking AMC10 problems... I think that if I keep reviewing, then I can remedy this...

Yes, I am trying to make USAMO in a couple different ways:

AMC12 index
AMC10 index(does this make USAMO?)
AIME score of 11[/quote]

If you want to achieve [i]that [/i]AIME score, you need to forget about AMC 12 and start doing AIME and oly problems...

Just focus on getting the Introduction books and not making many mistakes...strange problem though, haven't seen that before

Trying to be ready in [b]everything[/b] is not only foolish, but unrealistic.

First of all, what do you mean by scoring better on AMC 12? Are we talking about 140 on AMC 12 and 120 on AMC 10? Are you getting most of your differences in trig, complex numbers, and logarithm questions? Are you losing points on AMC 10 because of geometry?

In many cases, students can score well on AMC 10 if they do well on AMC 12. The only exception is geometry. You need to be more specific on your questions, and honestly, if you are worried about AMC 10 (which is really not needed) and want to attain 11 on AIME, you need to come back to reality and ask yourself, "Where am I at?" Before you set the goals, make sure understand where you are at.

If your scores are like 140 on AMC 12, 120 on AMC 10, and 8 on AIME, then yes, it's possible. But, even then, two-digit on AIME is very hard and like someone already said, you should be spending time to do AIME problems..

that DOES sound rather odd, but i'll offer what advice i have for studying amcs, which is just to do as many amc problems as possible. since many of the amc 10 and 12 problems overlap, i'd do the amc12 problems.

maybe if you're overthinking it, try to just do the problem fast and see what the first intuitive answer you get should be. you're probably on the right track anyway.

[quote="batteredbutnotdefeated"]ummm... Let's say I score higher on an AMC12 than on an AMC10... How would I remedy such a problem? I'm stronger in advanced concepts than I am at the basic ones...[/quote]

If you really want to remedy that problem, do mathcouns porblems.

[quote="dysfunctionalequations"]You do know that when Ubemaya says "Do MATHCOUNTS problems" in every single thread, he's joking?[/quote]

yes, i know. but mathcounts porblems are always fun.

[quote="xpmath"]Perhaps you are applying the fun method of "overkill" to each problem.

First try using the basic stuff before going on to more advanced things that you're learning. Could you give an example of a problem that you couldn't solve on the AMC 10 and what you tried on it?[/quote]

actually, here's a problem that I [b]did[/b] solve on the AMC10:

$ 4^a\equal{}5$, $ 5^b\equal{}6$, $ 6^c\equal{}7$, and $ 7^d\equal{}8$. Find abcd.

I solved it using the chain rule for logarithms where as the actual solution didn't even use log...

umm that actually seems like a good method to solve that problem with and would take a short amount of time to do (that is assuming the answer is 1.5 and I didnt mess up).

So I am assuming that it took you a long time to do that problem which is why you brought it up. Its good that you spotted a great method to solve the problem,but in that case you just need to make sure that you fully understand the method you are using before you get too deep into the question. Eg, if you use some obscure theorem that you dont really understand to solve a problem, you should stop yourself and look for other ways to solve the problem.

[quote="batteredbutnotdefeated"][quote="xpmath"]Could you give an example of a problem that you couldn't solve on the AMC 10 and what you tried on it?[/quote]

actually, here's a problem that I [b]did[/b] solve on the AMC10:

$ 4^a \equal{} 5$, $ 5^b \equal{} 6$, $ 6^c \equal{} 7$, and $ 7^d \equal{} 8$. Find abcd.

I solved it using the chain rule for logarithms where as the actual solution didn't even use log...[/quote]

So usually, if there is a way to do something with logs, there's a way to do it with exponents.  In this case, we can see that by the way the exponent equations are set up, we may have a link between the $ 8$ and the $ 4$ after we manipulate exponents.  So for this problem, here's a:
[hide="solution without logs"]
$ 8\equal{}7^{d}\equal{}6^{cd}\equal{}5^{bcd}\equal{}4^{abcd}\Rightarrow abcd\equal{}\frac{3}{2}$
[/hide]
This would work the same way if you took logs, multiplied by the next exponent, and then undid the log.  So it seems that your problem here was that you went one level too deep into the problem, when really there was an analogous solution on a simpler level.  Like the others said, look for an elementary solution first; there probably is one :)

I think it would still be beneficial if you gave us an example of a problem on the AMC10 that you couldn't solve, as other have suggested.

What are you scoring on the two? Because if you are scoring a 150 on the AMC 12 and like a 144 on the AMC 10, there is really nothing to worry about (your careless error could have just as easily been made on the AMC 12). The same goes for the case in which your scores differ by 12 or less. But if your scores are like AMC 12: 130 and AMC 10: 90, then I can see how you could be concerned.

[quote="batteredbutnotdefeated"]ummm... Let's say I score higher on an AMC12 than on an AMC10... How would I remedy such a problem? I'm stronger in advanced concepts than I am at the basic ones...[/quote]

Haha, that was me in freshman year^v^ I got 112.5 on AMC 12B that year but <100 (not sure about what I got) on AMC 10A...
It's not too bad.
That was also the only time I have taken AMC 10

[quote="batteredbutnotdefeated"]ummm... Let's say I score higher on an AMC12 than on an AMC10... How would I remedy such a [b]problem?[/b][/quote]

Exactly why is this a problem? Maybe you're prone to stupid mistakes and get some of the really easy questions on the AMC 10 wrong. Maybe you mis-bubbled on the AMC 10 but not the AMC 12. Maybe your weakest subject was all over the AMC 10 but your strongest was all over the AMC 12.

Also, if you can score more on the AMC 12, then by all means take the AMC 12, study for the AMC 12, etc.
Besides the later problems on the AMC 12 are AIME-worthy. The last two problems on the AMC 10 are AIME #1 or #2 worthy only. AMC 12 is IMHO the better gateway to the AIME.

Maybe you're just working off of a small sample size.


Find $ \lim_{x-->\infty} \frac {x^k}{e^x} $ with k = constant

We can use De L'Hopital rule again and again anmd get the limit is 0.


inequalities
geometry
circumcircle
geometry solved

We have a triangle with incribed circle of radius $r$. Prove that for circumradius $R$ of this triangle we have\[ R\geq2r  \]

[color=red][Moderator edit: Discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3414 .][/color]

ok sorry for this easy problem... now I know the solution..please don't solve it!


probability
combinatorics unsolved
combinatorics

The problem is.

A company has 12 distinct motors, two of these motors are from supplier A.  The motors must be divided into three lines, each line consisting of four motors.  Prove that the probability that both motors from supplier A will be on the same line is .091

I got this problem from a printout given to us in class, I think its a typo, because I keep getting .019 NOT .091
A = motor from supplier A
B = other motors
P(A) = 1/6
P(B) = 5/6

Taking 4 motors the chances that the two A motors will be together is:
(1/6)*(1/6)*(5/6)*(5/6) = .019

Does everyone else agree that is must be a typo?

I think neither probability is true.  The probability that the three other people in the line with an A motor are not the other A motor is 10/11*9/10*8/9=8/11, so the probability that the two A motors are together is 3/11.


Sign ups:

Players.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

Set up:

Rounds:
1. Mafia Day I
2. Mafia Day II
3. Mafia Day III
4. Elect the executioner Day I
5. Elect the executioner Day II
6. Elect the executioner Day III
7. Egg Toss Day I
8. Egg Toss Day II
9. Egg Toss Day III

I need 10 players.

/in

The message is too small. Please make the message longer before submitting.

Dojo is this another attempt at BTM? we already have 2.... anyway unless you plan only 1 kill for each round  you'll need more than 10.... w8 you do only need ten, but you had better plan mafia REALLY carefully. I can't join because I hav too much I already have to keep up with.

It's not BTM.

My cuz wanted to start using this forum so I chose a set of games open to newbs to learn some of the games played here.

One person will be killed per round.

1. hsp2014
2. Super Math Boy
3. eagylv999
4.
5.
6.
7.
8.
9.
10.

This could be nice :)  I will join this.

ok i think i will join.

I want to get familiarized with games other than mafia.

Just to tell you, having 3 days of mafia will be a very bad idea.

And you could use some more games. Like... meh, ready aim fire?

1. hsp2014
2. SuperMathBoy
3. eggylv999
4. AIME15
5. #5-1Mole
6. crazychinesemaniac
7. hedgehog12
8. kimmystar94
9.
10.

What's wrong with 3 days of mafia?

this will be short mafia, with only 10 people. and, egg toss is no fun with 2-4 people.

Ur at L-2.

[b]The Everyone's gone votecount 2!!! [/b]
zapi 2007: (1) Ernie
hsp2014: (1) kimmystar94
SuperMathBoy: (4) zapi2007, ernie, AIME15, hedgehog12

[i]With 10 players, it takes 6 to lynch.[/i]
not voting (everyone else.)

Prods going out.
If they do not respond. The replacements, in order are:

1. jjx1
2. Bugi
3. mathelete

[quote="Prod PM"]If you do not post on the game marathon: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246794[/url] by December 9th, you will be replaced.

Thank you.[/quote]

hahaha

[b]Unvote, Vote:SuperMathBoy[/b]

I am out of this game. I got the message and decided not to play.

Ernie is on the vote count twice. why is that?

[b]VOTE: SUPERMATHBOY[/b]
i don't know my role.

You're dead anyway, doesn't matter xD

Twilight talk...

jjx, that's a really dumb move.  I mean, REALLY dumb.  What if you're scum, and he's your scum-mate?  Seriously, that's dumb.  I won't mind it if he's scum, but still, a really dumb move...

AIME15 you take this game to seriouslt. I'm doing this for fun. geez. I don't care if its a dumb move. ANd wouldn't I know if I'm not a scum. :wink:

[quote="Dojo"]Okay effective imediately:

jjx1 replaces #5-1 mole
Bugi replaces eggylv999
mathelete replaces hsp2014

Roles going out

EDIT: Sorry, somehow my files like are like gone. Can everyone pm me their roles?[/quote]

HUH???

 :mad:  :mad:  :mad: WHAT DID I DO???

[quote="Dojo"]
EDIT: Sorry, somehow my files like are like gone. Can everyone pm me their roles?[/quote]

That's fail.


trigonometry
Pythagorean Theorem
geometry
power of a point

Triangle $ABC$ is right angled at $A$. The circle with center $A$ and radius $AB$ cuts $BC$ and $AC$ internally at $D$ and $E$ respectively. If $BD = 20$ and $DC = 16$, determine $AC^2$.

[hide]Let $AC=a$ and $AB=r$. By Pythagorean Theorem, $a^2+r^2=36^2$.

Extending $\overline{CA}$ to intersect the circle again at $X$, $CX=a+r$. So by Power of a Point,

$CE*CX=CD*CB\Rightarrow (a+r)(a-r)=a^2-r^2=16*36$.

$a^2=\frac{(a^2+r^2)+(a^2-r^2)}{2}=\frac{36(16+36)}{2}=36*26=31^2-5^2=\boxed{936}$.[/hide]

use pythagorous theorem on ABC. cosine rule on ADB and ACD.

[b]The point $E$ is in [u]addition[/u]! [/b] Here are two proofs without the power of a point w.r.t. a circle.

[hide="Proof 1"]

[b][u]Proof 1[/u].[/b] I"ll use the [u][b]Stewart[/b]'s relation[/u]: $20\cdot b^2+16\cdot c^2=36\cdot c^2+20\cdot 16\cdot 36\iff$ $20\cdot b^2=20\cdot c^2+20\cdot 16\cdot 36\iff$

 $b^2=c^2+16\cdot 36\implies$ $\left\{\begin{array}{cc}
b^2+c^2=36\cdot 36\\\\
b^2-c^2=16\cdot 36\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b^2=26\cdot 36\\\\
c=6\sqrt{10}\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b=6\sqrt{26}\cdot 36\\\\
c^2=10\cdot 36\end{array}\right\|\ .$[/hide]

[hide="Proof 2"]

[b][u]Proof 2[/u].[/b] Let $K\in BC$ so that $AK\perp BC$ , i.e. $AB=AD\iff KB=KD=10$ . If denote $AK=x$ , then  $KB=10$ ,

$KC=26$ and $\left\{\begin{array}{c}
b^2=x^2+26^2\\\\
c^2=x^2+10^2\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b^2+c^2=36\cdot 36\\\\
b^2-c^2=16\cdot 36\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b^2=26\cdot 36\\\\
c=6\sqrt{10}\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b=6\sqrt{26}\cdot 36\\\\
c^2=10\cdot 36\end{array}\right\|\ .$

[b][u]Remark[/u].[/b] You can use directly the equivalence $XY\perp AB\iff XA^2-XB^2=YA^2-YB^2$ .[/hide]


A domino is a rectangular tile composed of two squares. An integer is represented on both squares, and each integer 0-9 is paired with every integer 0-9 exactly once to form a complete set. A [i]double[/i] is a domino which has the same integer on both its squares. What is the probability that a domino randomly selected from a set will be a double? Express your answer as a common fraction.

[hide]
For the first number of dots,
There are 10 choices to go with it.
For the second, there are 9 left (because we already paired it with the first number)
For the third, 8
All of the way down to 1
$10+9+8+\cdots1=55$
There are 10 doubles (0-0,1-1,...9-9)
Thus the probability is $\frac{10}{55}=\frac{2}{11}$[/hide]


Euler
function
number theory unsolved
number theory

Can anyone help me with this question, I don't know where to start?

If d = gcd(a,b). Show that phi(d)*phi(a*b) = d*phi(a)*phi(b)

I was thinking along the lines of unique factorization of a, b and d into primes, then somehow show that the two equations equal. The problem is I'm unsure how i should go about factorizing them to make it work out in the end. Any help is appreciated. thanks

Just divide the equation by $dab$ and use the fact that $\frac{\phi(n)}n=(1-\frac 1{p_1})(1-\frac 1{p_2})\ldots(1-\frac 1{p_k})$, where $p_i$ are the prime factors of $n$. It should be pretty obvious.. :?

You are right. That's how it could be done.
Let $p$ be any prime dividing $a$ or $b$.

- If it divides only one of them, say $a$ with exponent $\alpha$, then it does not divide $d$ so that its contribution in both sides of the desired equality is $(p-1)p^{\alpha - 1}$.

- If it divides both $a$ and $b$, with respective exponent $\alpha$ and $\beta$
Wlog, we may assume that $\alpha \leq \beta$ so that $p^{\alpha}$ is the contribution of $p$ into the prime decomposition of $d$.
Thus, the contribution of $p$ on the lhs is $(p-1)p^{\alpha - 1} \times (p-1)p^{\alpha + \beta -1}$, whereas its contribution in the rhs is $p^{\alpha} \times (p-1)p^{\alpha - 1} \times (p-1) p^{\beta -1}$, hence result.

Pierre.


I have the following word problem that I just can’t figure out.
Austin, Devin, Olivia and William want to make an ice-cream. Each one has different amounts of milk (three, seven, five and four quarts). A container of ice cream needs 3 cups of milk and 2 cups of sugar. Determine the amount of milk and sugar that each person has.

1.	The person that has 10 cups of sugar has 2 cups less sugar than milk.
2.	If Olivia has 6 times more milk, she would have 10 gallons and 2 quarts of milk.
3.	If Devin has 7 cups more sugar, Devin would have a total of 5 pints of sugar.
4.	Olivia could, at the least, make a container of ice cream.
5.	The person that has 3 cups of sugar, has 17 cups less sugar than cups of milk.
6.	If Austin has 7 times more milk, she would have 7 gallons of milk.
7.	Austin could make at least 2 containers of ice cream.
8.	If Olivia had 11 cups more sugar, she would have a total of 9 pints of sugar.

Ok since a gallon has 4 quarts in it Olivia would have 7 quarts of milk and at least 2 cups of sugarand Austin has 4 quarts and at least 4 cups of sugar, since you need 2 cups. Devin has 3 cups of sugar (10-7=3), and 5 quarts. Thats beacuse he has 3 cups of sugar which is 17 less cups than he has milk, so he has 20 cups of milk which makes 10 pints. since 2 pints make a quart, Devin has 5 quarts. Process of elemination means William would have three quarts of milk. Back to Olivia for the actual sugar. she has 7 cups of sugar, because she has 11 cups less than 9 pints, or 18 cups. William has 10 cups of sugar, beacuse he has 3 quarts of milk, whcih comes from 12 cups, which is 2 more than 10 cups of sugar.

So, all in all:
Olivia: 7 quarts, 7 cups
William: 3 quarts, 10 cups
Devin: 5 quarts, 3 cups
Austin: 4 quarts, at least 4 cups

I hope this helps you.


geometry
incenter
circumcircle
geometry proposed

Given triangle $ABC$ with incenter $I$, circumcenter $O$,nine pointcenter $E$,circumradius $R$, prove that $OI=2\sqrt{IE\cdot R}$

Use Euler's fomular, it is trivial  :wink:

Euler's formula ?  :rotfl: ok! $OI^{2}=R^{2}-2Rr$ but what does $IE$ equal?

[quote="skywalker"]Euler's formula ?  ok! $OI^{2}=R^{2}-2Rr$ but what does $IE$ equal?[/quote]
And  Feuerbach 's theorem, now it is trivial.


Calculate: $ \displaystyle \frac{2 \plus{} \frac 23}{2 \minus{} \frac 23}$.


Each of the five numbers 1, 5, 9, 13 and 17 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column.  What is the largest possible sum of the three numbers in the horizontal row?
[asy]size(70);
draw((10,0)--(10,30)--(20,30)--(20,0)--(10,0)); draw((0,10)--(0,20)--(30,20)--(30,10)--(0,10));[/asy]

Since we want to maximize the sum, it'd be nice to have the largest possible number in the middle. Suppose we put 17 there. It then follows that one of the columns/rows has a 5 and 9 and the other has a 1 and 13.

Thus, our answer is $ 31$.


geometry
circumcircle
geometric transformation
reflection
ratio
geometry unsolved

The triangle $ ABC$ with the circumcircle $ k(U,r)$ is given. On the extension of the radii $ UA$ a point $ P$ is chosen. The reflection of the line $ PB$ on the line $ BA$ is called $ g$. Likewise the reflection of the line $ PC$ on the line $ CA$ is called $ h$. The intersection of $ g$ and $ h$ is called $ Q$.
Find the geometric location of all possible intersections $ Q$, while $ P$ passes through the extension of the radii $ UA$.

Denote P1,P2 are reflections of P  w.r.t  AB,AC  respectively
Q1,Q2 are intersections of UA∩BP1  UA∩BP2  respectively
PP1⊥AB AB⊥BD   =>  BD∥PP1
Analogously, CD∥PP2
From what P1,P2 is defined we know A is the circumcenter of △PP1P2
Notice that ∠P1AP2=2∠A  => ∠P2PP1=180-∠A=∠BDC
∠PP1P2=1/2∠PAP2=∠CAD=∠DBC    ∠PP2P1=1/2∠PAP1=∠BAD=∠BCD
=> △DBC∽△PP1P2   => CD/PP2=BD/PP1
Combined with the fact that CD/PP2=DQ2/DP  BD/PP1=DQ1/DP
We conclude DQ1=DQ2  =>  BP1,CP2,AU are concurrent at Q
So Q lies on segment UA

$ DA$ is a diameter of $ k(U, r).$ Let $ g, h$ cut $ DA$ at $ Q_1, Q_2.$ $ BA, CA$ bisect the angles $ \angle PBQ_1, \angle PCQ_2$ internally and $ BD \perp BA, CD \perp CA$ externally $ \Longrightarrow$ the cross ratios $ (D, A, Q_1, P) \equal{} (D, A, Q_2, P) \equal{} \minus{} 1$ are both harmonic $ \Longrightarrow$ $ Q_1 \equiv Q_2 \equiv Q.$ When $ P \longrightarrow \infty,$ then $ Q$ goes to the midpoint $ U$ of $ DA.$


Determine the smallest positive decimal integer N such that, N is formed entirely by 1’s and 0’s and is divisible by 225.

Note:  N cannot have any leading zero.

Set $ N\equal{}\sum_{k\equal{}0}^ma_k10^k$ with $ m\geq 2$ and $ a_k\in\{0,1\},a_m\equal{}1$ and note that $ 225\equal{}9\cdot 25$ holds. Now $ N\equiv 10a_1\plus{}a_0\bmod{25}$ requires $ a_0\equal{}a_1\equal{}0$. From $ N\equiv\sum_{k\equal{}0}^ma_k\equal{}\sum_{k\equal{}2}^ma_k\bmod{9}$ we conclude that the smallest $ N$ is given by $ N\equal{}\sum_{k\equal{}2}^{10}10^k$

Basically, the answer is [b]11,111,111,100[/b].


calculus
derivative
real analysis
real analysis unsolved

find all critical points of
f(x, y)=ex(1-cos y)

and classify these critical points.

Is that meant to read f(x,y)=e^x.(1-cosy) ?

In which case, find the partial derivatives, equate to zero for max/min/stationary point and look at the second derivatives to determine the character of the point.

[hide]$ Q(h,k)\equal{}\frac{\partial^2 f}{\partial x^2}(a,b)h^2\plus{}2\frac{\partial^2 f}{\partial x\partial y}(a,b)hk\plus{}\frac{\partial^2 f}{\partial y^2}(a,b)k^2$ where (a,b) is the point in question
If Q(h,k) is negative for all (h,k) not (0,0) there is a local max in (a,b)
If Q(h,k) is positive for all (h,k) not (0,0) there is a local min in (a,b)
If Q(h,k) takes both positive and negative values then it is a saddle point
For other cases (e.g. Q(h,k)>=0) this test is inconclusive and you need to do something else to determine what character the point has.[/hide]


function
analytic geometry
graphing lines
slope
combinatorics open
combinatorics

on a table some $n$ numbers are written. one operation consists of replacing two numbers $a,b$ by a single number, namely $\displaystyle \frac {a+b} 4 $.

they asked to prove is that if we begin with all the $n$ numbers equal with 1, at the end of $n-1$ operations, the remaining number is at least $\displaystyle \frac 1n$.

that is easy to prove, as  \[ \frac 1a + \frac 1b \geq \frac 4{ a+b} \textrm{ from Cauchy} \] thus the sum of the reversed numbers is always decreasing, so \[ n \geq \frac 1k \Rightarrow k \geq \frac 1n \] where $k$ is the remaining number.

The question is, what exactly is the closed formula of the sequence $a_n$, where $a_n$ is the smallest number that can be obtained starting, as above, with $n$ one-s?

[quote]one operation consists of replacing two numbers [tex]a,b[/tex]  by a single number, namely[tex]\frac{a+b}2[/tex] .
[/quote]
Perhaps you wanted to say [tex]\frac{a+b}4[/tex](if it is [tex]\frac{a+b}2[/tex] you can get only 1s  :)  )
Recurrence for  [tex]a_n[/tex] is
[tex]a_n=\min\frac{a_i+a_{n-i}}4[/tex], [tex]a_1=1[/tex]
I believe the solution is as follows:
Let [tex]2^k\leq n < 2^{k+1}[/tex]
Then [tex]a_n=\frac{3\cdot2^k-n}{2^{2k+1}}[/tex]

yes cotan, you are right. I made a slight typo, and I have corrected it.
what are your arguments for the minimal value?

Here's my proof.
Recurrence for [tex]a_n[/tex] is pretty obvious - on the last step there are two numbers, one of which is made of  [tex]i[/tex] 1's and other from [tex]n-i[/tex]. To obtain the minimum we should take minimal possible values of these, i.e. [tex]a_i[/tex] and [tex]a_{n-i}[/tex]
Now the solution for more general version of recurrence [tex]\displaystyle{a_n=\min_{1\leq i <n} \frac{a_i+a_{n-i}}A}
[/tex], [tex]a_1=1[/tex],[tex]A\geq2[/tex] is [tex]a_n=\frac{(A-1)2^{m+1}-(A-2)n}{A^{m+1}}, 2^m\leq n<2^{m+1}[/tex]. We prove this by induction. Consider function [tex]f(x)=\frac{(A-1)2^{m+1}-(A-2)x}{A^{m+1}}, 2^m\leq x<2^{m+1}, x>1[/tex], then we want to prove [tex]a_n=f(n)[/tex]. Notice that f(x) is continuous and convex. (On each of the intervals [tex][2^m,2^{m+1}][/tex] it is linear, and its slope increases when we move to the next interval). So [tex]a_i+a_{n-i}=f(i)+f(n-i)\geq2f(\frac{i+n-i}2)=2f(\frac n2)[/tex], and we have minimum at [tex]i=[\frac n2][/tex], that is [tex]i=n-i=\frac n2[/tex] for even n, and [tex]i=[\frac n2],n-i=[\frac n2]+1[/tex] for odd n. Now it is easy to prove that [tex]f(x)=\frac 2A f(\frac x2)[/tex] so if this assumtion is true for [tex]k<n[/tex] then it is true for [tex]k=n[/tex]


[hide]
$\displaystyle \left(\sec x - \tan x \right) \displaystyle \left(\sec x + \tan x \right) = \sec^2 x - \tan^2 x$

Using the identity $\tan^2 x + 1 = \sec^2 x$,

$\sec^2 x - \tan^2 x = 1$

We know that $\sec x - \tan x =2$, so:

$(2)(\sec x + \tan x) = 1$
$\sec x + \tan x = \frac{1}{2}$
[/hide]

$\frac{1-cos(x)}{sin(x)}=2$
$1-sin(x)=2cos(x)$
$1-4sin(x)+sin^2(x)=4cos^2(x)$
 After solving this equation
$sec(x)+tan(x)=\frac{1}{2}$

[quote="inom"]$\frac{1-cos(x)}{sin(x)}=2$
$1-sin(x)=2cos(x)$
$1-4sin(x)+4sin^2(x)=4cos^2(x)$
 After solving this equation
$sec(x)+tan(x)=\frac{1}{2}$[/quote]

[quote="kimby_102"]Source: AHSME 1999
15. Let $x$ be a real number such that $\sec x-\tan x =2$. Evaluate $\sec x + \tan x$.[/quote]

[hide]By the identity, $\sec^2 x-\tan^2 x=1$, we shall divide by $\sec x-\tan x=2$ to get $\sec x+\tan x$.  Hence, $\sec x+\tan x=\frac{1}{2}$.

$\frac{1}{2}$[/hide]


What is the numerical value of k if the roots of the 2nd degree polynomial x² - 3(a-2)x + a² - 2a + 2k are rational?
How do you solve this one? The information above are the only ones given.

Is $ a$ an arbitrary rational number?  Use the fact that the roots of a quadratic polynomial are rational if and only if its discriminant is the square of a rational.


linear algebra
matrix
linear algebra unsolved

The following matrix satisfies the property that the sum of the squares of any 2 adjacent numbers (horizontally and vertically) is also a square.
Can this 4x4 array be extended to a 5x5 array?
If so, what is the 5x5 array you get?

[1836 105 .252 . 735]
[1248 100 .240 . 700]
[.936  .75 . 180 . 525 ]
[.273 560 1344 3920]

More generally, does there exist a 5x5 Pythagorean matrix?

Note: the .'s are just for spacing; not decimal points :!:

The legs of every Pythagorean triple can be written in the form $ k(2mn), k(m^2 \minus{} n^2)$ where $ (m, n) \equal{} 1, m \not \equiv n \bmod 2$.  This representation is [b]unique[/b], so assign to each entry the designation "$ 2mn$" or "$ m^2 \minus{} n^2$"; the only requirement then is that entries of the first form are adjacent to entries of the second.  Then it's a matter of finding numbers with enough factors.  This should be possible with an $ n \times n$ matrix.


inequalities
number theory
number theory unsolved

Determine all triples $ (x, y, p)$ with $ x, y$ positive integers and $ p$ a prime number verifying the equation $ p^x \minus{} y^p \equal{} 1$.

by Fermat's little theorem it follows that
$ y\equal{}kp\minus{}1$,for some natural $ k$.
if $ x\equal{}1$ and $ y>1$,then there is no solution as $ y^p$ would be greater than $ p$.
if $ x\equal{}y\equal{}1$,then $ p\equal{}2$.
now assume that $ x>1$.
it means that $ y^{p}\minus{}1$ is divisible by $ p^x$.
on the other hand by Newton's binom:
$ y^{p}\minus{}1\equal{}(kp)^{p}\minus{}\dots\plus{}(kp)^2\binom{p}{2}\minus{}kp^2$.
therefore,$ k$ is divisible by $ p^{x\minus{}2}$.
therefore,$ y\geq p^{x\minus{}1}\minus{}1$.
let's solve the following inequality:
$ p^{x}>(p^{x\minus{}1}\minus{}1)^{p}$.
after some accurately made calculations and after considering several cases we conclude that $ x<3$ and $ p\leq 3$.
thus another solution is $ (2,2,3)$.recalling the initial one,all the solutions are $ (2,2,3),(1,1,2)$

[quote="outback"][quote="broniran"]The only solution is $ (2,2,3)$[/quote]

Proof, please...[/quote] It follows from the Catalan conjecture. It says that the only two consecutive powers of natural numbers are 8,9  :wink:  It was proved in 2003 ( I'm not sure about the year...) :roll:


Dear friend,
Thank you for your time and kindness.
To offer an apology, I have got that it is to keep strict in math as important as to keep promise and faith.
To offer a thanks, I am selfish to ask you read a manuscipt even I can find out the mistakes after some days.
To offer a note, this is my last try in some years and I can't help continuing my crazy work if I try to check it right now.
To hear from you is my honor.
Thank you!

Consider $ k(2k\plus{}3x^n)^2\plus{}k(2k\plus{}3y^n)^2\equal{}(2k\plus{}z^n)^3$.
Then If there is no solution of $ 5z^{2n}\minus{}18(xy)^n\equal{}m^2\plus{}n^2$.
then it follows Fermat's last theorem.

[quote="Taoli"]Consider $ k(2k \plus{} 3x^n)^2 \plus{} k(2k \plus{} 3y^n)^2 \equal{} (2k \plus{} z^n)^3$.
Then If there is no solution of $ 5z^{2n} \minus{} 18(xy)^n \equal{} m^2 \plus{} n^2$.
then it follows Fermat's last theorem.[/quote]

$ z \equal{} 3$, $ xy \equal{} 2$, $ n \equal{} 3$ yields:
$ 5z^{2n} \minus{} 18(xy)^n \equal{} 3^4 \cdot 41 \equal{} 45^2 \plus{} 36^2$

The formula solution of $ k(2k \plus{} 3x^n)^2 \plus{} k(2k \plus{} 3y^n)^2 \equal{} (2k \plus{} z^n)^3$ yields $ (m, n)\equal{}1$.
Can you check it through computer?

[quote="Mathias_DK"]
$ z \equal{} 3$, $ xy \equal{} 2$, $ n \equal{} 3$ yields:
$ 5z^{2n} \minus{} 18(xy)^n \equal{} 3^4 \cdot 41 \equal{} 45^2 \plus{} 36^2$[/quote]

Does it yields $ 5z^{2n} \minus{} 18(xy)^n \equal{} 5\cdot729\minus{}18\cdot8 \equal{} 3501$ which is not $ 45^2 \plus{} 36^2?$

[quote="greentreeroad"][quote="Mathias_DK"]
$ z \equal{} 3$, $ xy \equal{} 2$, $ n \equal{} 3$ yields:
$ 5z^{2n} \minus{} 18(xy)^n \equal{} 3^4 \cdot 41 \equal{} 45^2 \plus{} 36^2$[/quote]

Does it yields $ 5z^{2n} \minus{} 18(xy)^n \equal{} 5\cdot729 \minus{} 18\cdot8 \equal{} 3501$ which is not $ 45^2 \plus{} 36^2?$[/quote]
$ 3501 \equal{} 3^2 \cdot 389$, which can obvouisly by the sum of two squares.


calculus
integration
calculus computations

For a real number $ a$, evaluate $ \int_0^a |x(x \minus{} a^2)|\ dx$.

[quote="kunny"]For a real number $ a$, evaluate $ \int_0^a |x(x^2 \minus{} a)|\ dx$.[/quote]
if $ a > 0$ then $ \int_0^a |x(x^2 \minus{} a)|\ dx \equal{} \int\limits_{0}^{\sqrt {a}}x(a \minus{} x^2)dx \plus{} \int\limits_{0}^{\sqrt {a}} \minus{} x(a \minus{} x^2)dx \equal{} \frac {a^4}{4} \minus{} \frac {a^3}{2}$

I'm very sorry, I have made a typo.
 :blush:
I will edit soon.

kunny

Let $ \ I(a) \equal{} \int_0^a |f(x)| \cdot |g(x)| \ dx,\ f(x) \equal{} x,\ g(x) \equal{} x \minus{} a^2$

If $ a \geq 1,\ f(x) > 0,\ g(x) \leq 0, \forall x \in (0, a)$

If $ a < 0,\ f(x) < 0,\ g(x) < 0,\ \forall x \in (a, 0)$

If $ 0 \leq a < 1,\ f(x) > 0,$ however $ g(x)$ changes sign at $ x \equal{} a^2$

Hence, accounting for these signs (and notably the change of sign of g(x) in case 3), we get:

$ I(a) \equal{} \begin{cases}
\frac{1}{6} \left(2a^3 \minus{} 3a^4\right) & \mbox{if } a < 0 \mbox{,} \\
\frac{1}{6} \left(2a^3 \minus{} 3a^4 \plus{} 2a^6\right) & \mbox{if } 0 \leq a < 1 \mbox{,} \\
\minus{}\frac{1}{6} \left(2a^3 \minus{} 3a^4\right) & \mbox{if } a \geq 1 \mbox{.} \\
\end{cases}$


"If a = b (mod m) and p = q (mod m), then:

(a+b) = (p+q) (mod m) & ab = pq (mod m)"

(note I am using = instead of the three bars sign because I cannot latex)

Are the properties listed above true?
[b]If so, can someone please write up a (formal) proof of it?[/b]

Simple substitution.

$ a\equiv b\pmod m$
$ a\plus{}p\equiv b\plus{}p\pmod m$
$ p\equiv q\pmod m$
$ \therefore a\plus{}p\equiv b\plus{}p\equiv b\plus{}q\pmod m$

$ a\equiv b\pmod m$
$ a\cdot p\equiv b\cdot p\pmod m$
$ p\equiv q\pmod m$
$ \therefore a\cdot p\equiv b\cdot p\equiv b\cdot q\pmod m$.


1. A stock market investor owns shares in stock whose present value is 25. She has decided that she must sell her stock if it either goes down to 10 or up to 40. If each change of price is either up 1 point with probability .55 or down 1 point with probability .45, and the successive changes are independt, what is the probability that the investor retires a winner?

Define a [b]path[/b] of the price the sequence of ups and downs and let u represent up and d represent down.
Then to every [b]path[/b] of the change of price so that it ends with 40 there corresponds one and only one [b]path[/b] of change of price so it ends with 10 -- by switching every "u" in the path a "d" and every "d" with a "u". This is always possible because 25 is same distance from 40 as it is from 10. In the same way we prove to every [b]path[/b] of the change of price so that it ends with 10 there corresponds one and only one [b]path[/b] of change of price so it ends with 40.
The probability of a probability when price reaches 40 is $0.55^{k}*0.45^{l}$ where $k,l$ is number of times price goes up or down, respectively. Then, as price goes from 25 to 40, $k=15+l$. The case when price goes to 10 that corresponds to this case has probability $0.55^{l}*0.45^{k}$ (as we replace every u with d and every d with u). The ratio of the first and second probability is always $\frac{0.55^{15}}{0.45^{15}}$. But this holds for every case, so the ratio of probability that stock goes up to 40 to probability that it goes down to 10 is also $\frac{0.55^{15}}{0.45^{15}}$. But they must add to 1.
Let probability of going to $40$ be $P$. Then probability of going to 10 is $1-P$. Then:
$\frac{P}{1-P}=\frac{0.55^{15}}{0.45^{15}}$
So, $P=\frac{0.55^{15}}{0.55^{15}+0.45^{15}}$
Nice problem :D


Hello all,

The 1st problem is
"A box with height (x+5) cm has a square base with side x cm. A second box with height (x+2)cm has a square base with side (x+1) cm.
If the tow boxes have he same volume, find the value of x.

The 2nd problem is
"A box with a square base and no lid is to be made from a square  piece of metal by cutting squares from the corners and folding up the sides. The cut-off squares are 5 cm on a side. If the volume of the box is
100 cm to the power of 3, find the dimensions of the original piece of metal.

Note: I really don't understand word problem, is kinda hard for me. Please explain this to me, thanks a lot.

We need length x width x height for the volume of a box.  Since the base of each box is a square we have length = width for each box.  So we have:
$x^{2}(x+5) = x^{3}+5x^{2}$ for the first box and,
$(x+1)^{2}(x+2) = x^{3}+4x^{2}+5x+2$ for the second box.
Since their volumes are equal set the two equations equal to each other:
$x^{3}+5x^{2}= x^{3}+4x^{2}+5x+2$. Simplify the equation and move everything over to the left.  That leaves a quadratic:
$x^{2}-5x-2 = 0$.  This cannot be factored so we need to use the quadratic formula:
$ax^{2}+bx+c = 0 \rightarrow \frac{-b \pm \sqrt{b^{2}-(4ac)}}{2a}$
This gives $\frac{5 \pm \sqrt{33}}{2}$. The volume cannot be negative and so $x \approx 5.37$ .

Edited: I made an error earlier.

illcrowflu, thanks a lot for the explanation but dunno if your answer is wrong or correct, the book says that the answer is x= 5.37(the first problem), but I need the procedure to get this.


trigonometry
geometry
angle bisector
trig identities
Law of Sines

Prove that, in any triangle $ ABC$, $ \sin(\frac{A}{2}) \le \frac{a}{b\plus{}c}$.

Let the bisector of angle $ A$ meet $ BC$ at $ D$, and let $ m\angle ADB \equal{} \theta$. By Law of Sines, $ \sin \frac {A}{2} \equal{} \frac {BD}{AB}\sin \theta\leq \frac {BD}{AB} \equal{} \frac {a}{b \plus{} c}$, the last equality following by the angle bisector theorem. Equality occurs iff $ AB\equal{}AC$.


How many ways 6 students can be divided into two teams containing 3 students each?

There are 6! ways to arrange all students in a line abcdef
Then we group them into such groups: abc|def
But abc|def is the same as acb|def or abc|dfe or def|abc
Since there are 3! ways to arrange abc, 3! ways to arrange def and 2 ways to arrange two teams, we have the answer equal to $ \frac{6!}{3!3!2}\equal{}10$


Why is it believed one bad apple in a basket spoils the rest. Is there any reason? :D  :rotfl:  :rotfl:

[quote="skand"]yes,Its true :D

It is because the rotten apple releases ethylene gas that spoils the rest(I think ethylene is a phytohormone :) )

In my opinion it is the right answer.Or wait for ritu to come and give better answer(he is the only one here with bio as a subject)[/quote]
it is spot on as usual :)

ethylene is a plant growth inhibitor and it is resp for spoiling  :wink:
and am i the only 1 with bio really? :rotfl:

then who else is?????
I suppose Hell,Da vinci and slogger are not bio students
and count me out
I don't know when i'll opt for Physical education(while taking admission,I asked about this permission to choose physical as my subject whenever I wish  :rotfl: )

well there is no need to tell
in my school ,every month one or the other CSE student drops to PE :rotfl:  :rotfl:


Here is the IMO shortlist problems 1986, I got it from internet.

Here is a new version of the above International Mathematial Olympiad (IMO) Longlist 1986 attachment which is a bit more comprehensive. For Windows there is a [url=http://www.stdutility.com/download/stduviewer.exe]STDU viewer.[/url]

[b][size=150]IMO Long List 1986 Problems:[/size][/b]

Problem 1   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=364109

Problem 2   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364111

Problem 3   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364112

Problem 4   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=364113

Problem 5   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364116

Problem 6   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364117

Problem 7   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=364118

Problem 8   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364119

Problem 9   : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364120

Problem 10 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364121

Problem 11 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364122

Problem 12 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364124

Problem 13 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=364125

Problem 14 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=364127

Problem 15 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364128

Problem 16 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=364130

Problem 17 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364196

Problem 18 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364197

Problem 19 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=364198

Problem 20 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364199

Problem 21 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364201

Problem 22 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=364202

Problem 23 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364204

Problem 24 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364205

Problem 25 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=364207

Problem 26 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364208

Problem 27 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364209

Problem 28 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364210

Problem 29 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364212

Problem 30 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364213

Problem 31 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364214

Problem 32 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=364215

Problem 33 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364218

Problem 34 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=364219

Problem 35 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=364220

Problem 36 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364221

Problem 37 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364222

Problem 38 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364225

Problem 39 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364226

Problem 40 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364227

Problem 41 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364228

Problem 42 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364229

Problem 43 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364230

Problem 44 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=364231

Problem 45 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=364234

Problem 46 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=364235

Problem 47 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364236

Problem 48 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364237

Problem 49 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364238

Problem 50 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364239

Problem 51 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364240

Problem 52 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364241

Problem 53 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=364242

Problem 54 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364243

Problem 55 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=364246

Problem 56 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364247

Problem 57 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364248

Problem 58 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=364249

Problem 59 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364250

Problem 60 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=364252

Problem 61 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364253

Problem 62 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=364254

Problem 63 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364255

Problem 64 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=364257

Problem 65 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364258

Problem 66 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364259

Problem 67 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364260

Problem 68 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364261

Problem 69 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=364262

Problem 70 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364263

Problem 71 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364264

Problem 72 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=364265

Problem 73 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364266

Problem 74 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364267

Problem 75 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364268

Problem 76 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364269

Problem 77 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=363943

Problem 78 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364270

Problem 79 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364271

Problem 80 : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364272


P is a point on the graph of the equation $y=x^{2}+2x-15$. The tangent to the curve at P is parallel to the $x$ axis. Find the equation of the tangent line at point P.

[hide]
To find the bottom of the parabola, which is point $P$, you have to find the value of $x$ that makes y as small as possible.
If $x=0$, then $y=15$.
If $x=1$, then $y=16$.
If $x=2$, then $y=15$.
So, the answer is $y=-16$.[/hide]


algebra
polynomial
linear algebra
linear algebra unsolved

Determine the matrices are similar to each other.

[0,-1,1;1,2,-1;0,0,1]  and
[1,0,0;2,0,-1;-2,1,2]

same characteristic polynomial: $P(x)=(x-1)(x+1)^{2}$. So one can look if there are diagonalisable : I don't know. But if one is, and the other is not, this is finished: they are not similar. If the two are diagonalisable, it is also finished: the two are similar. And if the two are not diagonalisable: they have the same Jordan form thus are similar over $\mathbb{C}$ .. just look what is the appropriated case  :)

I agree that they have the same characteristic polynomial, but isn't it $(x-1)^{3}$?

yep, you're right  :blush:  here, for example, it suffices to look at the rank of $M-Id$. The matrices are similar iff they have the same Jordan blocks, and in this case, iff $Rank(M_{1}-Id)=Rank(M_{2}-Id)$


Hmm,
are there a lot?

Seems to only work with primes whose n-1 is some power of 2.

[quote="SnowStorm"]
Seems to only work with primes whose n-1 is some power of 2.[/quote]
why? :?

Is there any way you can do it without guessing and checking, but then again you won't find all of them. Hmmmm.... Does anyone know how to do this?

[quote="chess64"][quote="warut_suk"]Find all primes $p,q$ such that $p-1$ divides $3q-1$ and $q-1$ divides $3p-1$[/quote]

Hmmm.... well $(p,q)=(3,3)$ works.

$3q-1 \equiv 0 \mod {p-1} \ \ (1)$ and $3q-1 \equiv 0 \mod {q-1} \ \ (2)$. Does that help any? Probably not....

$p-1$ and $q-1$ are always going to be even unless $p=2$ or $q=2$. That means that $3q$ and $3p$ are odd. So $q \neq 2$ and $p \neq 2$.

Hmm... this is difficult.[/quote]

i dont understand why $p\neq 2$ and $q\neq 2$. because $p-1=q-1=1$ 1 divides any integer, so why is $p=2$,$q=2$ excluded?
the thing i found out is that they both cannot have different parity because if say $p$ is odd and $q$ is even, then $p-1$ is even but $3q-1$ is odd and $p-1$ does not divide $3q-1$

this problem is very difficult, im stuck. im able to prove that the primes cannot be of form $6m+1$. only if i can do the same with primes of the form $6m-1$...

[quote="Palytoxin"][quote="chess64"][quote="warut_suk"]Find all primes $p,q$ such that $p-1$ divides $3q-1$ and $q-1$ divides $3p-1$[/quote]

Hmmm.... well $(p,q)=(3,3)$ works.

$3q-1 \equiv 0 \mod {p-1} \ \ (1)$ and $3q-1 \equiv 0 \mod {q-1} \ \ (2)$. Does that help any? Probably not....

$p-1$ and $q-1$ are always going to be even unless $p=2$ or $q=2$. That means that $3q$ and $3p$ are odd. So $q \neq 2$ and $p \neq 2$.

Hmm... this is difficult.[/quote]

i dont understand why $p\neq 2$ and $q\neq 2$.[/quote]

Sorry, I made an incorrect assumption.  :blush:

so far i got $(p,q)=(2,2),(3,3)$ im unable to prove that there are no more other solutions... does anyone know how to proceed? pls give hints if you know, ty

If you express the primes in the form $2x+1$ then we can test for values that work.

For example if we choose some prime, say 5, we have $2x|14$ and $4|6x+2$ and of course the only possible values of $x$ are $1$ or $7$, both which work so plugging into $2x+1$ we have $3$ and $15$, the latter which we can throw out.

So we have one pair that works, $(p,q)=(3,5)$

Testing with 2 and 3 we get two more that work $(p,q) = (2,2), (3,3)$

7 is impossible because $7-1=6$ can never divide $6x+2$ which shows that the prime can never be 1 mod 6.

11 gives $(p,q) = (11,17)$

This is as far as I have gotten I cannot seem to find anymore...

hey someone solve or warut post solution please, what is answer? i can get as far as tarquin did  :(

Well, assume $p \leq q$. We've checked the case $p=2$ and $p=q$ already, so assume $p \leq q-2$. Now $\frac{3p-1}{q-1}$ must be an integer, but $\frac{3p-1}{q-1} \leq \frac{3q-7}{q-1} \leq 3$...

[quote="warut_suk"]Find all primes $p,q$ such that $p-1$ divides $3q-1$ and $q-1$ divides $3p-1$[/quote]
i think i got full solution by now, this very hard problem but i learned a lot
[hide="my full solution"]ok we start off with assuming $p\leq q$ (this makes things easier by symmetry).

they must have same parity because if, say, $p$ is odd and $q$ is even, then $p-1$ is even but $3q-1$ is odd and $p-1$ does not divide $3q-1$.

if $p=2$, $q-1|5$, so either $q-1=1$ or $q-1=5$. but $q=6$ is not prime, we get one answer $(p,q)=(2,2)$

if $p=q$, $p-1|3p-1$. since $3(p-1)=3p-3<3p-1$, $4p-4=3p-1$ and get $(p,q)=(3,3)$. also, $5p-5=3p-1$ yields above answer $p=q=2$. there are no more because $np-n=3p-1$ solving we get $p=\frac{n-1}{n-3}$ no longer integer for $n\geq 6$

now we can assume $3\leq p<q$, both odd primes. as warut_suk said, $p-1|3q-1$ no restriction because $p<q$, so we work with $q-1|3p-1$ and let $m$ be some integer such that $(q-1)m=3p-1$.
\[m=\frac{3p-1}{q-1}<\frac{3q-1}{q-1}=3+\frac2{q-1}\]
because $p<q$. but $\frac2{q-1}<1$ if $q>3$, so $m\leq 3$.

suppose $m=3$. we get $3q-3=3p-1$, $3q-3p=2$, no solution because 2 is not divisible by 3

suppose $m=2$. we get $2q-2=3p-1$, $q=\frac{3p+1}2$ and this is possible because $p$ is odd. let $k$ be some integer such that $(p-1)k=3q-1$. plug $q$ into new equation and solve for $k$:
\[
k = \frac{9p+1}{2p-2}=\frac{8p-8}{2p-2}+\frac{p+9}{2p-2}=4+\frac12\left(\frac{p+9}{p-1}\right)=4+\frac12\left(1+\frac{10}{p-1}\right)
\]
and so $\frac{10}{p-1}$ must be an odd integer. this happens when $p-1=2,10$

if $p-1=2$, $p=3$ is prime, $q=\frac{3p+1}2=5$ also prime, and $k=7$

if $p-1=10$, $p=11$ is prime, $q=17$ also prime, and $k=5$

suppose $m=1$. we get $q-1=3p-1$, $q=3p$, impossible because $q$ must be prime

altogether, all solutions are $(p,q)=(2,2),(3,3),(3,5),(11,17)$[/hide]

warut_suk, please check if correct...

[quote="warut_suk"]Yeah.. checking briefly, I think so  :)[/quote]
ty warut_suk because i have never spent that long on a problem, it was hard. but i guess this is nothing compared to real olympiad problems, which are insane, in my opinion

[quote="warut_suk"]Indeed... actually I took it from part of a real olympiad problem.[/quote]

 :wow:  :wow:  :wow:  :jump:  w00t w00t i did an olympiad problem?  :clap:  im happy. and guess what? number theory, inequalities and combinatorics are my weakest subjects now, i have improved my number theory, yay!

[quote="Palytoxin"][quote="warut_suk"]Indeed... actually I took it from [size=150][i][u][b]part[/b] [/u][/i][/size]of a real olympiad problem.[/quote]

 :wow:  :wow:  :wow:  :jump:  w00t w00t i did an olympiad problem?  :clap:  im happy. and guess what? number theory, inequalities and combinatorics are my weakest subjects now, i have improved my number theory, yay![/quote]

Just kidding, great job!  :) You were the only one to get it. That's because you persevered and didn't give up. I'm lazy, lol.


calculus
derivative
integration
function
real analysis
limit
topology

try to make sure there's no calculus on the test

Definitely not, cuz I don't even know calculus, except maybe like derivatives or something.

[quote="me@home"]Pssh! "I don't even know calculus except maybe derivatives"... lol
that's all you ever use it for in any high school contest problem!!!![/quote]
Oh yeah, well then there won't be any of those either.

How's this, [size=150]NO CALCULUS ON THE TEST![/size]

well yes you are right; i meant that if you could take the derivative of most simple functions then you can logically integrate some easy integrals, in fact all integrals seen onsuch a test :)

[quote="Xevarion"][quote="me@home"]When I say derivatives, that includes integrals (deriv's give you all the tools you need for integrals...)[/quote]
AIIIEEEE!!!! Derivatives follow from integrals! Not the other way! They only teach you that way in school because calculating particular derivatives is easier than calculating particular integrals and they have some aversion to starting with harder stuff and some obsession with calculating.[/quote]

huh? how do derivatives follow from integrals?

Hmmm, this isn't even a calculus forum, let alone a calculus topic.  Talk about this somewhere else.

Isn't an integral a reverse derivative?  So why does the derivative come from the integral?  (I haven't taken calculus yet  :roll: ).

EDIT:  I just realized this is my 1337th post! 1337ness!

[quote="Xevarion"]You are thinking of the integral as something you "calculate." The action of producing a closed form for the integral of some function is indeed pretty much the reverse of doing a derivative.

However, from a theoretical point of view, it is more natural to define the integral first, and then actually define the derivative on integrals (i.e., functions that are integrals of other functions) as the limit of the integral over a small region as the size of the region goes to 0.

The reason for this is that integration is a smoothing process and is therefore more general than differentiation. A differentiable function is always integrable; an integrable function is not necessarily differentiable.

Any more details and I'm going to have to start explaining measure theory (although I think Arnav already knows that).[/quote]

Derivatives can also be approached from a measure theoretic standpoint? I was previously only aware of the Fréchet derivative, which (I think) is unrelated... furthermore, how do we define functions that are integrals of other functions? If we take $\nu(A) = \int_{A}f \, d\mu$, the measure theory definition provides a new measure on the space, but not so much a function; do we differentiate the measure in some sense or is there some other function we define?

If $f$ is Lebesgue integrable on $\mathbb{R}^{d}$,
$\lim_{m(B) \to 0, x \in B}\frac{1}{m(B)}\int_{B}f(y)dy = f(x)$ for a.e. $x$.

Sorry, I guess what I said before was kinda messy and unclear. So that's the Lebesgue differentiation theorem or something. But anyway, I guess the way I was thinking of it was if we want the FTC
$F(b)-F(a) = \int_{a}^{b}F'(x) dx$
then we need to know what kind of integral we're talking about. For the Lebesgue integral it works whenever $F$ is of b.v..

But it's true that a function that's not of b.v. can still have a derivative a.e.... I don't know what I was thinking now...  :|

Oh yeah also, you can define integration on any measure space, even if it's not a metric space (even if it's not a topological space!), and I don't really know how you'd define a derivative in such a space except as the inverse function for integration. (Maybe that's because I'm a noob and don't really understand those very general derivatives on Wikipedia.)


How difficult would it be to get in though,
do they look at grades, competitions, ideas, reccomendations?
How much?

I think that the main thing that they are looking for your [i]attitude[/i] towards math.

The six track system offers a wide range of difficulty so your abilities don't worry that much about your math ability. I suppose that you would do best if you had a level of understanding (like able to solve the first fifteen to twenty problems on an AMC 10 given unlimited time...I don't have an official measure, but I think that is a reasonable start).

How do you study math? Books? Old contests? AoPS Forum? Math circle? How much time do you spend working on problems? What is your attitude towards problem solving? What is your reaction to doing math for 6+ hours a day? What types of problems do you like? Anecdotes about what you do can serve to show your enthusiasm.

From this regard, I think that personal statements and recommendations convey your attitudes best. Competition results also serve as evidence that you work hard at math (although a lack of stellar results in contests won't hurt you).

It is nice to do well on the practice tests as well.

Of course, all of these comments are merely my impressions as a person who went there. I have not participated in the reading of applications.

[quote]How do you study math? Books? Old contests? AoPS Forum? Math circle? How much time do you spend working on problems? What is your attitude towards problem solving? What is your reaction to doing math for 6+ hours a day? What types of problems do you like? Anecdotes about what you do can serve to show your enthusiasm[/quote]

........... well ive already been to a math camp where we did 9 hrs, so im used to it. I like proofs, and problems i have to think about. I spend about 2/3.5 hrs a day, i use all except mathcircle

[quote]From this regard, I think that personal statements and recommendations convey your attitudes best. Competition results also serve as evidence that you work hard at math (although a lack of stellar results in contests won't hurt you).[/quote]

 i have placed @ tamu, and several other major stuff.

neways, is it very difficult to be accepted?

THX for ur help

Here's an acceptance poll from last year:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142965
Most AoPSers who applied got in.

korammath, I think that you'd have no problem getting into AMSP. Remember, what I said is not checklist. The point of your application should be to convey your enthusiasm in math, not to answer each question that I posed systematically. You should include specific evidence (of what you do, like, etc), but they all serve for to argue a larger point about your attitude.

[quote="korammath"]So when would i need to apply?
early, regular or late. im assuming not late but i dk....

thx altheman.

btw fightthetide says ur awesome at math[/quote]

The early application is posted; see another topic.

And yes, Alex is.

[quote="korammath"]OK, missed early cuz of MATHCOUNTS, so shud i do reg, and how many shud i get rite??[/quote]

Uhh...Reread what I wrote above. Pure scores or accomplishments are secondary considerations. It is important to convey a positive attitude towards math.


A ball is attached to a string with a length of 15 cm. the string is attached to a vertical pole radius 1 such that the center of the pole, the point where the string is attached to the pole, and the ball are colinear. The ball is swung around the pole at a constant rate, traveling in a spiral path. As the ball moves, the string is being wrapped around the pole. The ball comes to a complete stop when the length of the string not wrapped around the pole is 0. how far has the ball traveled?

This would have been the arc length of a polar curve had you not added gravity, but if this is the case, the distance traveled is highly dependent up the initial rotational speed, which was not given.

Also, this would be better placed in the physics forum.

the ball is on the ground, and the string is parallel to the ground.

there shouldn't be physics involved...i don't think atleast


I've been wondering how to solve problems such as..

Given that today is Friday, find the day of the week YY^ZZ days from now.

Can anyone take me through one of these problems? Thanks.

You need to find that number mod 7, then add to Friday and you got the result

Example:

$ XX\equal{}22,YY\equal{}55$

$ 22^{55}\equiv 1^55\equiv 1\pmod 7$, so that day will be Saturday.

Ooo thanks. Though now I have another question..

How can you quickly find the mods of huge powers? IE 38^39 or 49^56, etc?

Well you can use [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem]Euler's Totient Theorem[/url]

Example:

$ 38^{39} \equiv 3^{39} \pmod 7$ because $ 3 \equiv 38$

As $ \phi(7)\equal{}6$ , $ 3^{39} \equiv 3^3 \equiv 27 \equiv 6 \pmod 7$ because $ 39 \equiv 3 \pmod 6$


I have a suggestion that may have already been done, but i think we should delete all the multi's of one user and let him/her keep one account.
Thanks

In the case of a multi that you want to delete, you should contact an admin.

Some actually have relatives on the same IP, not multis so be careful

In addition this would raise the issue with if they actually POSTED with that account.

What's the problem with multis? Maybe I make 5-10 xD

The only problems I see are: games forum, FTW
Add if there is more.

Yes thats true. That's the reason people CREATE multis usually. Otherwise they "lose" all the time on FTW against the real account.

And the search function is the part of the site...

http://www.mathlinks.ro/viewtopic.php?search_id=1552566030&t=251127


geometry
3D geometry
induction
combinatorics unsolved
combinatorics

A brick has the shape of a cube of size $2$ with one corner unit cube removed. Given a cube of side $2^{n}$ divided into unit cubes from which an arbitrary unit cube is removed, show that the remaining figure can be built using the described bricks.