[quote="MithsApprentice"]Request for ARML problem series next spring if possible :-D[/quote]

An ARML problem series is not currently a high priority for us.  All of the topics covered would be redundant to the classes we already offer.  The AMC-12 and AIME problem series would cover material similar to the Individual and Team questions.  Our subject courses cover problems of a similar depth to the Power Question.

Students interested in learning to work the Power Questions better might consider taking our Olympiad Problem Solving course which covers the most common kinds of Power Question-ish tools such as induction.

[quote="Kent Merryfield"]The best thing to do is to contact the person or persons who organize the ARML team that serves your area. Find out if there are practice sessions or problem solving activities for that team and find out if you can attend them.[/quote]

Unfortunately, we don't have that in Texas. We have school until just right before ARML and if we were to meet, there would be team members driving at least 7 hours to attend the meeting no matter where it's held...


Begin with the $n$-term sequence $1, 2, 3, \ldots, n$. Form the $(n-1)$-term sequence $3, 5, 7, \ldots, 2n-1$ from the sums of consecutive pairs of elements of the original sequence. If we keep doing this until we have a single term, what is its value?

[quote="t0rajir0u"]Maybe a little easy for Pre-Olympiad?  :|

[hide="Solution"] It's not hard to observe and then prove by induction that our final number is

${n-1 \choose 0}1+{n-1 \choose 1}2+...+{n-1 \choose n-1}n$.

Let this sum be $S$.  Because binomial coefficients are symmetric,

$2S = \left({n-1 \choose 0}+...+{n-1 \choose n-1}\right) (n+1) = 2^{n-1}(n+1)$

So

$S = \boxed{ 2^{n-2}(n+1) }$ [/hide][/quote]

Maybe. I rejected it as problem for #24 on my Mock AMC because it was too easy (to guess, at least) :P.

This is basically pascal's triangle. Instead of a single $1$ for the row 1, we have a line of invisible $1$'s and then it's not hard to see that everything behaves the same way as in pascal's triangle.


Two deck of cards are arranged in the following way.
The larger Joker is on top, followed by the smaller Joker. The four Aces are next followed by the four 2's and the four 3's ... and the four Queens and the four Kings. In each type of card they are arranged with Spades first, then Hearts, then Diamonds, then Clubs.
One deck is put on top of the other deck.

The dealer removes the cards in the following way.
The first card he removes off the top, he says the number 1. The next card he removes off the top he says the number 2. The next he says 3 and so on.
Whenever he says an odd number he throws the card away. Whenever he says an even number, he puts the card on the bottom of the deck. Eventually, there will only be one card left. What is the suit and name of the card?

After cycling through the deck once, shouldn't the only numbers remaining be even?

[hide="Solution"]Number the cards from 1 to 108. After iteration 1, where he goes through all the cards exactly once, only the multiples of 2 are left (2, 4, 6, ..., 108), and the dealer's count is even. After iteration 2, where he goes through all the remaining 54 cards exactly once, only the multiples of 4 are left (4, 8, 12, ..., 108), and the dealer's count is even. After iteration 3, where he goes through the remaining 27 cards, only the multiples of 8 are left (8, 16, 24, ..., 104), and the dealer's count is now odd.

Since the dealer's count is odd, he will keep only the odd number-positioned cards, leaving: (8, 24, 40, 56, 72, 88, 104). He has went through another 104/8=13 cards, so his count is now even. He will go through 7 cards, leaving: (24, 56, 88) and making his count odd again. Then he will go through 3 cards, leaving: (24, 88) and making his count even. Then the last card is the 88th card.

Cards 1 to 54 are from the top deck, while cards 55 to 108 are from the bottom deck. Cards 55 and 56 are the jokers, so Aces are 57, 58, 59, 60, 2s are 61, 62, 63, 64 and so forth. Subtracting 56 from everything, Aces are 1, 2, 3, 4, 2s are 5, 6, 7, 8 and so forth, and the last card is 32. Therefore the last card is the last suit of 8, which is the 8 of Clubs.[/hide]


vector
geometry
geometric transformation
rotation

On the sides of an arbitrary convex quadrilateral $ABCD$, squares are constructed, all lying external to the quadrilateral, with centers $M_{1}, M_{2}, M_{3}, M_{4}$. Show that $M_{1}M_{3}=M_{2}M_{4}$ ad that $M_{1}M_{3}$ is perpendicular to $M_{2}M_{4}$.

The outline of complex number solution is on [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116980[/url]

I'd also be interested if there was a euclidean solution...I have a
[hide="vector solution"]
First, we will use the notation $| \stackrel{\rightharpoonup}{AB}$ to denote a $90^{\circ}$ clockwise rotation of the vector $\stackrel{\rightharpoonup}{AB}$.

Now, by the basic properties of vector addition, etc.
$\stackrel{\rightharpoonup}{M_{1}}= \frac{1}{2}\left( \stackrel{\rightharpoonup}{B}+\stackrel{\rightharpoonup}{A}\right)+\frac{1}{2}|(\stackrel{\rightharpoonup}{A}-\stackrel{\rightharpoonup}{B})$

$\stackrel{\rightharpoonup}{M_{2}}= \frac{1}{2}\left( \stackrel{\rightharpoonup}{C}+\stackrel{\rightharpoonup}{B}\right)+\frac{1}{2}|(\stackrel{\rightharpoonup}{C}-\stackrel{\rightharpoonup}{D})$

$\stackrel{\rightharpoonup}{M_{3}}= \frac{1}{2}\left( \stackrel{\rightharpoonup}{D}+\stackrel{\rightharpoonup}{C}\right)+\frac{1}{2}|(\stackrel{\rightharpoonup}{C}-\stackrel{\rightharpoonup}{D})$

$\stackrel{\rightharpoonup}{M_{4}}= \frac{1}{2}\left( \stackrel{\rightharpoonup}{A}+\stackrel{\rightharpoonup}{D}\right)+\frac{1}{2}|(\stackrel{\rightharpoonup}{D}-\stackrel{\rightharpoonup}{A})$

Therefore

$M_{1}M_{3}= M_{3}-M_{1}= \frac{1}{2}\left(\stackrel{\rightharpoonup}{C}+\stackrel{\rightharpoonup}{D}-\stackrel{\rightharpoonup}{B}-\stackrel{\rightharpoonup}{A}\right)+\frac{1}{2}| \left(\stackrel{\rightharpoonup}{C}+\stackrel{\rightharpoonup}{B}-\stackrel{\rightharpoonup}{D}-\stackrel{\rightharpoonup}{A}\right)$

$M_{2}M_{4}= M_{4}-M_{2}= \frac{1}{2}\left( \stackrel{\rightharpoonup}{D}+\stackrel{\rightharpoonup}{A}-\stackrel{\rightharpoonup}{C}-\stackrel{\rightharpoonup}{B}\right)+\frac{1}{2}| \left( \stackrel{\rightharpoonup}{C}+\stackrel{\rightharpoonup}{D}-\stackrel{\rightharpoonup}{B}-\stackrel{\rightharpoonup}{A}\right)$

Now it is clear that $M_{2}M_{4}= | M_{1}M_{3}$, implying that they are orthogonal and equal in length.
[/hide]

It is called Aubel's theorem. The only solution I know is one that uses rotations.

Complex and vector solutions are completely identical, short of notation. (Rotating a vector $90^\circ$ CCW is the same as multiplying a complex number by $i$.)


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ l_1, l_2, l_3$ be arbitrary lines in $ \mathbb{Q}^3$.
Show there exists a nonzero polynomial $ f \in \mathbb{Q}[x, y, z]$ of degree $ \le 2$ such that vanishes on $ l_1, l_2$ and $ l_3$.

Let $ P(x,y,z)\equal{}0,\,Q(x,y,z)\equal{}0,\,R(x,y,z)\equal{}0$ be three planes that are parallel two $ l_1$ and $ l_2$, $ l_2$ and $ l_3$, $ l_3$and $ l_1$.
Then $ l_1$ would be the equations $ P\equal{}a_1,\, R\equal{}c_2$ , $ l_2$ would satisfy the equation $ Q\equal{}b_1,\, P\equal{}a_2$, and $ l_3$
 would be the equations $ R\equal{}c_1,\,Q\equal{}b_2$
Consider now the polynomial $ (P\minus{}a_1)(Q\minus{}b_1)(R\minus{}c_1)\minus{}(P\minus{}a_2)(Q\minus{}b_2)(R\minus{}c_2)$  :wink:


This is my first time and i would love some help!? i have been trying to figure this question out for three days now and i have given up!?

Students Who Tell the Truth

  A school has 200 students. These are special students: some of them ALWAYS tell the truth, and the rest of them, alas, NEVER tell the truth. Among the subject areas of math, science, and social studies, each student has one favorite. A survey was conducted, each student was asked three yes or no questions: “Do you like math the most?” “Do you like science the most?” “Do you like social studies the most?”

The results were as follows:

104 students said “yes,” they like math the most.
86 students said “yes,” they like science the most.
60 students said “yes,” they like social studies the most.

How many students tell the truth, and how many do not?

if anyone could help me it would be great!!!! THANK YOU!!!

Obviously, those who said they like science and social studies the most are lying. :D

Um, I think this is a solution, but I'm not sure:

Out of three subjects, you can only have one favorite.  Obviously, the thruthful people just respond truthfully, so the each will get "one subject" (that contributed to the tally)

The Liars will say they don't like their fav subject, and that the like their 2 least fav's.  So, they each get "two subjects."  One more then they should.

Those three (104+86+60)=250  So, there are 50 "extra subjects", which means 50 of the students are liars.

150 tell the truth
50 don't

That makes since!! thank you so much for your help!!! have a great day!!


Let $w=C(I,r)$ be the incircle for the triangle $ABC$. The circle $w$ touches the sides $[BC]$, $[CA]$, $[AB]$ in the points $D$, $E$, $F$ respectively. I note the middlepoints $M$, $N$ of the sides $[BC]$, $[CA]$ and the intersection $T\in EF\cap BI$. Prove that $TB\perp TC$ and $T\in MN$.

This is exactly the Extended Problem from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187743#p187743]http://www.mathlinks.ro/Forum/viewtopic.php?t=23677 post #7[/url].

  darij


trigonometry
parameterization
parametric equation
calculus
calculus computations

Hi guys.

I was just hoping to verify if I got my answers right to the following problems.

1) Is the equation of the tangent plane of $z = e^{x^{2}+y^{2}}$ at $(1,-1,1)$ equal to $z = 2x+2y+1$ ?

2) Is the parametric equation of the curve of intersection of $z = x^{2}$ and $x^{2}+y^{2}= 1$ equal to $x = \cos t$ ; $y = \sin t$ ; $z = \cos^{2}t$?

Thanks for the help.

1) There is no tangent plane, since the surface doesn't go through that point.
2) That is an equation for the curve, if you choose some interval of length $2\pi$ for $t$. There are infinitely many others.

Sir thanks for your answer. I made a mistake regarding question no. 1. It should be this:

1) Is the equation of the tangent plane of $z = e^{x^{2}-y^{2}}$ at $(1,-1,1)$ equal to $z = 2x+2y+1$ ?

My mistake on the sign of the exponent of $e$. It should really be minus. Thanks again for the verification sir. Hope you could get back to this question again.


calculus
integration
geometry
3D geometry
modular arithmetic
similar triangles

How do you solve these problems?

1. Two similar triangles with integral sides have two of their sides the same. The third sides differ by 387. What are the lengths of the sides?

Source : A puzzle book

2. Find the largest prime divisor of $ 6^7\plus{}37$.

Source : Mandelbrot

[hide="A Start for 1"]
Let the sides of the triangles be $ c\leq a\leq b$ and $ a\leq b\leq c\plus{}387$ (the most plausible set of triangles  :wink: ).

Then we have $ r\equal{}\frac ca\equal{}\frac ab\equal{}\frac b{c\plus{}387}\equal{}\frac{a\plus{}b\plus{}c}{a\plus{}b\plus{}c\plus{}387}$, where $ a,b,c,a\plus{}b\plus{}c\in\mathbb Z^\plus{}$ and $ 0<r<1$.
[/hide]
[hide="2"]
$ 6^7 \plus{} 37 \equal{} 6^7 \plus{} 6^2 \plus{} 1$

$ \equal{} 6^7 \minus{} 6^4 \plus{} 6^4 \plus{} 6^3 \plus{} 6^2 \minus{} 6^3 \plus{} 1$

$ \equal{} 6^4(6^3 \minus{} 1) \plus{} 6^2(6^2 \plus{} 6 \plus{} 1) \minus{} (6^3 \minus{} 1)$

$ \equal{} (6^5 \minus{} 6^4)(6^2 \plus{} 6 \plus{} 1) \plus{} 6^2(6^2 \plus{} 6 \plus{} 1) \plus{} (1 \minus{} 6)(6^2 \plus{} 6 \plus{} 1)$

$ \equal{} (6^2 \plus{} 6 \plus{} 1)(6^5 \minus{} 6^4 \plus{} 6^2 \minus{} 6 \plus{} 1)$

$ \equal{} 43\cdot6511 \equal{} 43\cdot17\cdot\boxed{383}$.

Not exactly satisfying, but...
[/hide]

[hide="1 and proof of uniqueness"]If the sides of the triangle are $ a,b,c$ and $ b,c,a\plus{}387$, then $ \frac{b}{a} \equal{} \frac{c}{b} \equal{} \frac{a\plus{}387}{c}$. Multiplying these three together gets that $ \frac{a\plus{}387}{a}$ is the cube of a rational number. Note that $ 387 \equal{} 3^2 \cdot 43$, with divisors $ 1,3,9,43,129,387$. The fraction $ \frac{a\plus{}387}{a}$ can have a reduced form $ \frac{k\plus{}d}{k}$ for any $ d \in \{1,3,9,43,129,387\}$ and some divisor $ k$ of $ a$. So $ \frac{k\plus{}d}{k} \equal{} \frac{p^3}{q^3}$, and since both are in reduced form $ k\plus{}d \equal{} p^3, k \equal{} q^3$. Any solution to $ p^3 \minus{} q^3 \equal{} d$ with $ p,q > 0$ gives a pair of triangles.

$ d \equal{} 1,3,9$: Too small. These can be brute forced.
$ d \equal{} 43$: Cubes mod 7 can only be $ 0,1,\minus{}1$. We have $ p^3 \minus{} q^3 \equiv 1 \pmod{7}$, so either $ p$ or $ q$ is divisible by $ 7$. Then one is at least $ 7$, which is too high to get a difference of $ 43$ ($ 7^3$ is more than $ 43$ away from any other cube).
$ d \equal{} 129$: Mod 7 again. $ p^3 \minus{} q^3 \equiv 3 \pmod{7}$. No solutions.
$ d \equal{} 387$: $ p \equal{} 8, q \equal{} 5$ is the smallest by brute force, giving $ a \equal{} 125$. Note $ (p\minus{}q)(p^2 \plus{} pq \plus{} q^2) \equal{} 387$. We have a solution for $ p \minus{} q \equal{} 3$, and since $ p^2 \plus{} pq \plus{} q^2$ is monotonically increasing there are no others for it. The only divisor of $ 387$ smaller than $ 3$ is $ 1$. If $ p\minus{}q \equal{} 1$, this reduces to $ 3p^2 \minus{} 3p \plus{} 1 \equal{} 387$. But the right is divisible by $ 3$ and the left is not, contradiction.

The only triangles are $ 125,200,320$ and $ 200,320,512$.[/hide]

A swifter approach for 2), but essentially the same idea:

After realizing that $ 6^{7}\plus{}37\equal{}6^{7}\plus{}6^{2}\plus{}1$, one can note that the third root of unity, i.e. $ \omega^{3}\equal{}1$, shows us that $ \omega^7 \plus{}\omega^2 \plus{}1 \equal{} \omega^2 \plus{}\omega \plus{}1\equal{}0$. Hence, we immediately have that $ (6^{2}\plus{}6\plus{}1)$ is a factor of the desired expression and the result follows as above.


[quote="Linkin_Park"]1. Let $a,b,c \in [1,3]$ and $a+b+c=6$
Prove that $a^{2}+b^{2}+c^{2}\le 14$
[/quote]
Let $D=\{(a,b,c)\in [1,3]^{3}|a+b+c=6\}$ then $D$ is polygon.
For any $M(a,b,c)$ we have $OM=\sqrt{a^{2}+b^{2}+c^{2}}$.
Now, use lemma : let three points $A,B,C$ , we have for any $M\in [AB]$ then $CM\leq\max(CA,CB)$.

1) The solutions to $a+b+c=6$, are $a=b=c=2$ and $a=1,b=2,c=3$, and the sum of the squares of this solutions is $\leq 14$.

In the same way you solve the second one.

I assumed $a,b,c \in N$. I guess he should have pointed out that $a,b,c \in R$, then.

Of course, a,b,c are real numbers :D
If they are $\in N$, there's no more to say

[quote="Linkin_Park"]
2. Let $a,b,c \in [0,2]$ and $a+b+c=3$
Prove that $a^{3}+b^{3}+c^{3}\le 9$[/quote]

Also 1) Denote $D=\{(a,b,c)\in [0,2]^{3}|a+b+c=3\}$ then $D$ is convex polygon.

For any $M(a,b,c)\in D$ let $f(M)=a^{3}+b^{3}+c^{3}$. We have $f$ is convex on $D$,...

[hide="2"]
WLOG let $a\geq b\geq c$. Just use
\[b^{3}+c^{3}\leq (b+c)^{3}\]
so it just remains to prove
\[a^{3}+(3-a)^{3}\leq 9\]
But we have
\[9-a^{3}-(3-a)^{3}=9(a-1)(2-a)\geq 0\]
because we are given $a\leq 2$, and $3a\geq a+b+c=3\Rightarrow a\geq 1$.
[/hide]

[hide="1"]
Substitute $a=x+1$ etc. so $x,y,z$ lie in the interval $[0,2]$, $x+y+z=3$, and we need to prove
\[x^{2}+y^{2}+z^{2}\leq 5\]
This problem dies using the exact same method as before: WLOG let $x\geq y\geq z$, then $x\geq 1$, and we use
\[y^{2}+z^{2}\leq (y+z)^{2}=(3-x)^{2}\]
The remaining inequality
\[x^{2}+(3-x)^{2}\leq 5\]
follows from
\[5-x^{2}-(3-x)^{2}=2(x-1)(2-x)\geq 0\]
[/hide]


geometry
inradius
trigonometry
3D geometry
area of a triangle
Heron\u0027s formula
trig identities

It would be really helpful if I had a formula for the third side of a triangle, c, given the other two sides, a and b,and the inradius, r.  If you try using just Heron's Formula, you get a cubic in c, which seems to proffer no reasonable formula.

There is probably a nicer expression but it would take some time to come up with. Try playing with angles using the Law of Sines and the fact that they are supplementary.

[quote="ineedmathhelp"]I think you can use Heron's formula and set it equal to the inradius multiplied by the semiperimeter since they are both areas of the same triangle. Since you have the values of a, b, and r, plugging in shouldn't be very messy.[/quote]

[quote="juggle7"]If you try using just Heron's Formula, you get a cubic in c, which seems to proffer no reasonable formula.[/quote]

Depends on what you call a "reasonable formula". There is no expression in radicals, since the problem of constructing a triangle from a, b and r is not solvable using compass and ruler, so that your cubic is, indeed, the best expression one can give for the problem. See http://www.mathlinks.ro/Forum/viewtopic.php?t=17413 .

  darij

[quote="darij grinberg"]There is no expression in radicals, since the problem of constructing a triangle from a, b and r is not solvable using compass and ruler, so that your cubic is, indeed, the best expression one can give for the problem.[/quote]

I guess you meant square roots.  Theoretically, I could use my cubic formula and find an expression for c in radicals (including cube roots).

Yes, I meant square radicals, sorry.

  Darij


Fermat was able to show that $26$ is the only number sandwiched between a square number and a cubed number $25=5^{2}$, $27=3^{3}$. How did he do it?

From all I know you just show that the Elliptic Curve

$y^{2}=x^{3}-2$ has only one pair of integer solution $(x,y)=(3,5)$.

http://mathforum.org/library/drmath/view/66130.html

I really don't know if the solution is the same as Fermat's, but probably he used similar idea.

[quote="Franz Joseph"]$y^{2}=x^{3}-2$ [/quote]

There is a good way to do this using the arithmetic of $\mathbb{Z}[ \sqrt{-2}]$.  :)

Of course, this theory wasn't developed during Fermat's time, so he wouldn't have been able to access this proof.

Two pairs of integer solutions; also $(3,-5)$.  This problem has been posted many many times.  It is an elementary unique factorization problem in $\mathbb{Z}[\sqrt{-2}]$ as stated by t0rajir0u.  However, I have not even considered Fermat's solution.

Masoud Zargar

$\mathbb{Z}[\sqrt{-2}]$ What's this :?: . Could you post a link and your solution please?

$y^{2}+2=x^{3}\implies (y+i\sqrt{2})(y-i\sqrt{2})=x^{3}$.  Since we are dealing with cubes, unique factorization yields $y+i\sqrt{2}=(a+ib)^{3}=a^{3}+3a^{2}bi-3ab^{2}-ib^{3}$.  Equating real and imaginary components gives us $y=a^{3}-3ab^{2}=a(a^{2}-3b^{2})$ and $\sqrt{2}=3a^{2}b-b^{3}=b(3a^{2}-b^{2})$.  From them, we get $b=\sqrt{2}\implies a=\pm 1$.  These give us $y=\pm 5\implies x=3$.  Thus, all solutions are $(x,y)=(3,\pm 5)$.

Masoud Zargar

EDIT: This topic is not really suitable in this forum. :wink:


limit
real analysis
real analysis unsolved

Let $(a_n)$ be a sequence such that : $\lim\limits_{n\to \infty}(a_{n+2}-a_n)=0$
  Prove that : $\lim\limits_{n\to \infty}\dfrac{a_{n+1}-a_n}{n}=0$

$\lim\limits_{n\to \infty}(a_{n+2}-a_n)=0$
So we've got:
$\lim\limits_{n\to \infty}{a_{2n}} = g_1$
and
$\lim\limits_{n\to \infty}{a_{2n+1}} = g_2$
Lets take $c = |g_1 - g_2| + \epsilon$
$0 < \dfrac{a_{n+1}-a_n}{n} < \dfrac{c}{n}$ for n big enough
$0 < lim_{n\to \infty}{\dfrac{a_{n+1}-a_n}{n}} < lim_{n\to \infty}{\dfrac{c}{n}} = 0$

[quote]So we've got:
$\lim\limits_{n\to \infty}a_{2n} = g_1$
and
$\lim\limits_{n\to \infty}a_{2n+1} = g_2$ [/quote]
No, we don't. Consider $a_n=(-1)^n\sqrt{n}$ as an example.

I think it is trivial.
We have $\lim\limits_{n \to \infty}a_n = L \in {\mathbb R}$ so $\lim\limits_{n \to \infty}\dfrac{a_{n+1}-a_n}{n}=\lim\limits_{n \to \infty}\left ({a_{n+1}-a_n}\right ) \lim\limits_{n \to \infty}{\dfrac{1}{n}} =(L-L)0=0$

Not true. The sequence 1, 2, 1, 2, ... doesn't have a limit but satisfies the conditions.

(It is not given that the sequence is convergent! It doesn't need to be convergent.)

yes arne....sorry for my stupid mistake  :blush:

[quote="liyi"]Stolz Theorem.  :P[/quote]

   How can I apply Stolz theorem here ?


I am trying to get an opinion on how important it is to continue pursuing anvanced English placement for a Math oriented student with pretty good, but not perfect English skills (SAT 650 in 7th grade).
Advanced English this year in his school is a disaster, teacher bragging about nobody getting As, answering all questions with "Writers make choices", tons of homework and poor grades.
Of course, it could be just random luck this year, but the general question is:
Is it worth taking a chance next year in high school with advanced English or it would be more prudent to qualify, say "Thank you, but No" and save some time and energy for Math and Science. After all, one can't be perfect in everything and sometimes choices have to me made. A top scoring Math student preparing for Math competitions etc. can't compete equally well in English with students who are going to be English majors and only pursue English/Humanities?
Will a B in Advanced English ruin his GPA compared with A in regular English?
Or will the colleges frown for not taking the most challenging English course while taking the most challenging Math/Science courses?

Well I have heard colleges look for well-rounded people. :maybe:

[quote="dkushnir"]It is a well known general answer, I am asking about specifics and how the AoPSers balance this aspect for themselves. I am not talking about flunking English, just how much English is good enough for a non-English major kid who enjoys Math and Math competitions and not English?[/quote]

Colleges look for a [i]rigorous academic record[/i]. Last time I attended a Stanford information session someone asked the following question.

PERSON: Would it be better to take a regular course and get an A or take an advanced course and get a B?

STANFORD PERSON: Take the advanced course and get an A.

Of course if nobody is getting an A in that class the schools you apply to will know (or you let them know). Taking advanced courses always looks good to colleges; and I mean not just looks good, but actually is good in reflecting your own abilities. But don't commit academic suicide either, taking too many hard courses at once, because having B's across the board probably won't look too great.

If you know you don't want to take that course and that you won't do well in it, however, there's really no reason to take it unless it somehow demonstrates a below average English level, which I'm sure it doesn't. I took the fairly standard honors/AP English track at my school which is probably above my English level, while some students were a year ahead of me; it hasn't made a difference thus far in college applications.

EDIT: Disclaimer - I may have contradicted myself in my opinion, but I basically wrote whatever thoughts came to me at the time...


analytic geometry
graphing lines
slope
geometry
geometric transformation
reflection

Point (6,3) is reflected over line $ \textit{l}$ to (2,5). What is the slope of line $ \textit{l}$?

draw a line that connects point (6,3) and (2,5).

The line is perpendicular with line l.

so the slope of line l * the slope of a line that connects point (6,3) and (2,5) = -1.

the slope of a line that connects point (6,3) and (2,5) is $ \minus{}\frac{1}{2}.$

so $ \minus{}\frac{1}{2}$ * l = -1.

the slope of line l is 2.

answer : 2

why would you find the slope of the line perpendicular to point (6,3) and (2,5)

never mind
reflections are off perpendicular lines


A rectangular box has sides of length x, x-2, and x-4 . Write a function that determines the volume of the box in terms of x.

Can I just multiply them all like L x W x H?

I got this:  V(x) = x(x -2)( x - 4).

Is this correct?

Yes. It is correct.

EDIT: All right, it is correct for $ x\in[4,\plus{}\infty)$

[quote="firecricket91"]Actually that is not correct.

You didnt put any restrictions on $ x$, i mean can x be 1?[/quote]
Good observation.  We still have to define the domain of the function.


I just had 57 posts and posted another one, but it didn't change.  Why?

Some forums, notably including GFF, are not included in post count.

Yes, I just checked your recent posts, and most of them have been in G&FF, where posts do not count.

Other forums include,

Test Forum
Community Forums

Lately they have been counting Community forums.

But GFF, Polls and Test are the ones I know off that don't count.


MATHCOUNTS geometry probability \/closed

So I convinced my sister to do mathcounts this year and my mom wants to let me know what class she should take.
She's in 8th grade and taking geometry at school. She has no mathcounts experience, obviously, and when she took the 2009 chapter sprint round she got 18/30. She is a pretty fast learner. What class should she take?
The Basic or Advanced?

It probably depends I bet.  What state does she live in?  Is her goal to make nats or just make states?  I'm taking advanced.  It'll probably be pretty tricky for me, but it's probably a good idea.
If you live in a competitive state or have a killer chapter, then go advanced.  Also, is the 18/30 due to careless mistakes or conceptual errors?

I would recommend doing the basics class first, and if your sister does well, then maybe trying the advanced class. Like maybach said, it also depends on whether the mistakes she made were careless, or she didn't know how to solve the problem. Try pinpointing which problems she's having the most difficulty in (ex: algebra, geometry, probablility), and take a course on those too. Good luck!!

[quote="isabella2296"]18/30 on chapter sprint is fairly low and definitely wouldn't make states (at least in my chapter), so I would recommend the basics. She might also want to look into the introduction series.[/quote]

Well in my chapter, all you need is about 15 on sprint and 5 on target.......not joking, Calumet Chapter is pretty pathetic.

But probably basics is the way to go, but if you think you can make states with that score, than go advanced.

[quote="maybach"]But probably basics is the way to go, but if you think you can make states with that score, than go advanced.[/quote]

You should really do basics first, because the advanced class is harder. If your sister got 18/30 on chapter sprint then she should do basics before advanced.

Another option is to take [b]both[/b], namely, the September 14 Basics class and the November 11 Advanced class.

In my chapter (Santa Clara, California), a 27 in the sprint and perfect in target would be needed to make states, but it really depends on what chapter and state you are in.

Well, if your sister learns fast, she can try the Basics first, and quit after 2 classes if it is clear that it is too easy.

i would say basics for 2 classes and if that's easy than advance

Another choice I would suggest is to study Introduction to C&P book and Introduction to Geometry textbooks and then take the Advanced class. The books cover almost everything that is needed for success in MathCounts, while the problem series classes cover only the important materials and strategies (probably due to the limited time). That way, your sister can learn a variety of topics on the book and then do lots of problems in the class.

If she's good about asking questions, I'd recommend advanced.  I'd probably recommend the Intro C&P class over either of them if she has the time.


Hello,
I'm very hesitated which courses level should I take? If I know the concepts in the introductory level I get hesitated but is this means I'm ready for the middle once?

For me I can adapt if it's adaptable I get more challenging stuff than math in my life and pass it.

Please past AMSP participants show us your experience about this issue.

Intermediate.

You're there to challenge yourself ;]] especially if you already have a good background.

I'm ready for challenges and I like it. Thank you for the post.


linear algebra
matrix
algebra
polynomial
real analysis
real analysis unsolved

Find $ e^{tA}$ when
$ A\equal{}$\[ \left(\begin{array}{ccc}\lambda & 1 \\0 & \lambda \end{array}\right)\]

$ A$ is Jordan normal form.
Thus, $ e^{tA}\equal{}\begin{pmatrix}e^{t\lambda} & te^{t\lambda} \\ 0 & e^{t\lambda} \end{pmatrix}$

Justifying that claim:

$ A \equal{} \lambda I \plus{} B$ where $ B \equal{} \begin{bmatrix}0 & 1 \\
0 & 0\end{bmatrix}.$

Note that $ B^2 \equal{} 0$ and that $ I$ commutes with $ B.$

We can show that if $ A$ and $ B$ commute (that is, $ AB \equal{} BA$) then $ e^{A \plus{} B} \equal{} e^Ae^B.$ The proof is essentially the same as that for the the scalar complex exponential based on the power series.

Hence $ e^{t(\lambda I \plus{} B)} \equal{} e^{\lambda tI}e^{tB}.$

Any polynomial or power series in $ I$ is simply a scalar matrix (a scalar multiple of $ I$). In particular, $ e^{\lambda tI} \equal{} e^{\lambda t}I.$

Then $ e^{tB} \equal{} \sum_{k \equal{} 0}^{\infty}\frac1{k!}t^kB^k \equal{} I \plus{} tB$ since $ B^2 \equal{} 0.$

So $ e^{tA} \equal{} e^{\lambda t}(I \plus{} tB).$ Multiply that out to get what Kouichi Nakagawa said.

(Either that, or use the formula in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177357]other topic.)[/url]


A person walks down the street from work. The buses that come from behind him pass him every 6 minutes. The buses going towards him pass him every 3 minutes. If the buses depart consistently then
1) at what interval do the buses leave the bus station?
2) the ratio of the man's speed to the speed of the buses if both have constant speeds?

distance between consecutive buses is constant, so if $ m$ denotes the man's speed and $ b$ the speed of a bus, we have $ \frac{b\plus{}m}{b\minus{}m} \equal{} \frac{2}{1}$, so $ b: m \equal{} 3: 1$. since buses that come from behind the man pass him every 6 minutes, and they go 3 times as fast as he does, they must be leaving the station once every $ 6\cdot \frac{2}{3} \equal{} 4$ minutes.

[quote="pleurestique"]$ \frac {b \plus{} m}{b \minus{} m} \equal{} \frac {2}{1}$, so $ b: m \equal{} 3: 1$.[/quote]

huh? I don't understand what the fraction means or how u got to 3:1 from 2/1. help please.


Jamie counted the number of edges of a cube, Jimmy counted
the number of corners, and Judy counted the number of faces.  They
then added the three numbers.  What was the resulting sum?

Basically it is edges+corners+faces.
edges=12
corners=8
faces=6
12+8+6=26

Alternatively, you have a 3x3x3 cube, every cube (except the center) uniquely represents either an edge, corner, or side. Thus, 3^3-1=26.


I know that Shobber is always looking for Chinese old problems, and here I found some: http://www.imo.org.yu/index.php?options=oth|Chn|Ne|Ne&p=M8571

[quote="Inspired By Nature"]someone change the title to 'for shobber and 4everwise and IBN (Inspired by Nature)'! :D[/quote]

Jajaajaj!!! I'm waiting you Shobber to post, ajajajja lol

Thanks you very much, Jose, for providing the link!

But the all the past problems of the Chinese Team Selecting Test can be found here: http://www.mathlinks.ro/Forum/resources.php?c=37&cid=47, which is more complete than the one in the IMO Compendium.

The Chinese Team Selecting Test has two parts. One is the Test, which contains 6 problems and 21 points for each. They can be found in the link above. There are also some other problems that are called "Quizzes". They are worth 7 points each. There were 6 sets of 18 problems of quizzes in 2003, 2004, 2005, 2006, and all of them can be found in the National Olympiad Forum. It is the Quizzes that are very very difficult to find. I have the Quizzes from 2001 to 2006 and 14.5 problems of 2000. But I can't find the problems before 2000 anywhere! Maybe the only think I can do now is to wait for other people to find the problems and pos them online......  :(

Anyway, I still thank you so much for the link and for your help!  :)

I'd like to have them to. If you need my e-mail, ask me by PM. Thanks.

[quote="shobber"][quote="mecrazywong"]Well, I have the "quizzes" of the year 1994.[/quote]
Where did you get the quizzes of 1994? Is is ok to send it to me?[/quote]
It's contained in a book called "Math Olympiad Collection 1994" published by the IMO 1994 Chinese Team leader. All the training materials, including notes and TST are included in the book.
For the quizzes, I can PM them to you one by one (it's a really harsh job if I am to finish translating and typing them within a short period of time).
To Jose: I will also PM the problems to you.

Hi Mecrazywong! Thank you very much for the problems! If you do not have time translating them, you can just send them to me ( a GIF figure is also ok ) and I shall try to do the translation!

thank you for sharing,would you like send it to me?thank you.


AMC
AIME
ratio
geometric sequence
geometric series

log6 a means "log base 6 of a"

log6 a + log6 b + log6 c = 6

a,b, and c are positive integers that form an increeasing geometric sequence (could you also define this, the exact definition slips my mind)

b-a is the square of an integer.

find a+b+c

Thanks for any help you can give me.

If a, b, and c are in a geometric progression, then b/a=c/b. So let's rewrite a, b, and c as b/r, b, and rb. Then we get log6 a + log6 b + log6 c = log6 b^3 = 6. Thus b=36. If b-a is a perfect square, then a can be 11, which will make c not an integer, or a=20, which will also make c not an integer, or a=27, which makes c=48, or a=32, which will make c not an integer, or a=35, which will make c not an integer. Thus (a, b, c)=(27, 36, 48), so the sum is 111.

Here's my solution (in spoiler):

[hide] Since a,b, and c form a geometric series, we can write the three out as a, ar, and ar^2.  (That's what geometric means, btw.)  From that we get log6 (a)+log6 (ar)+log6 (ar^2)=6, which reduces down to log6 (a^3*r^3)=6.  We can reduce this to ar=6^2=36.  We are given that ar-a=a(r-1) is a perfect square also.  The factors of 36 are 1,2,3,4,6,9,12,18,36.  Testing a few, we see that a=36 and r=1.  This seems a bit trivial, but I'll go on anyways.  Thus a+b+c=36+36*1+36*1^2=108.[/hide]

Not sure if it's right though...

-interesting_move

wow, complexzeta sure posts quickly...I guess my answer was wrong afterall.

-interesting_move

interesting_move, the problem states that it is an "increeasing geometric sequence". Yours isn't increasing since r=1.

Fierytycoon

Here's my solution too! I guess i didn't solve it quickly enough to post first lol.

[hide]

Using log properties, we get abc=6^6.

Now, using r as the common ratio and setting a=a, ar=b, ar^2=c(a geometric series is defined by a, ar, ar^2, ar^3,..., where a and r are real numbers), we get a^3r^3= 6^6

ar=36

ar-a=x^2

The only possible pairs (a,r) are (11, 36/11) , (20,9/5) (27,4/3) (32,9/8) (35,36/35). However, the only pair for which ar^2=c is an integer is (27, 4/3).

Then a=27, b=36, and c=48, and a+b+c= 111[/hide][/hide]

Ah..I missed that part.  So would my answer be right if it didn't specify it?

-interesting_move

Thanks for the help. I guess I need to make sure I know definitions of maths terms like that.

[quote]So would my answer be right if it didn't specify it? [/quote]

I think so.


If $ ( a_n )$ is a sequence of positive real numbers such that $ \frac{a_{n\plus{}1}}{a_n}$ converges to $ \alpha$, then $ \sqrt[n]{a_n}$ also converges to $ \alpha$. You may use this fact.

Here's a direct argument:

Choose an arbitrary $ k\in \mathbb{N}.$ For $ n> k,$ we have $ n!\ge k!\cdot k^{n\minus{}k}.$

Then, for $ n>k,\ \left(\frac1{n!}\right)^{\frac1n}\le\frac1{(k!)^{1/n}}\cdot \frac1{k^{1\minus{}k/n}}.$

But as $ n\to\infty,$ $ (k!)^{1/n}\to 1$ and $ k^{1\minus{}k/n}\to k.$

Hence, $ \limsup_{n\to\infty} \left(\frac1{n!}\right)^{\frac1n}\le \frac1k.$

Since $ k$ was arbitrary and since this is a positive sequence, we conclude that
\[ \lim_{n\to\infty}\left(\frac1{n!}\right)^{\frac1n}\equal{}0.\]

[quote="scatteredclouds"]$ lim_{n \minus{} > \infty}{(1/n!)^{1/n}}$ = ?

I would guess it goes to 0. But don't no how to prove?
any help pls? :)[/quote]
You can use that $ \sum_{n \equal{} 0}^{\infty}a_nx^n$ has radius of covergenes $ R\equal{}\frac {1}{\lim_{n\rightarrow\infty}{^n\sqrt {a_n}}}$.

Note that there "lim..." is the highest possible sub-limit in denominator in this formula.

Put $ a_n \equal{} \frac{1}{n!}$ and you will see that $ \sum_{n \equal{} 0}^{\infty}\frac {x^n}{n!} \equal{} e^x$ has radius of covergenes $ \infty$, so your the highest sub-limit is 0.
Try to prove that this limit is exists (it means, that the highest sub-limit is only one).

Here is a simple and elementary solution:

If $ n \equal{} 2k$, then $ n! \equal{} [(2k) \cdots (k \plus{} 1)]\cdot [k \cdots 1] \geq (n/2)^{n/2}$.
If $ n \equal{} 2k \minus{} 1$, then $ n! \equal{} [(2k \minus{} 1) \cdots k]\cdot [(k \minus{} 1) \cdots 1] \geq (n/2)^{n/2}$.

So we have $ (n/2)^{n/2} \leq n!$. Then

$ 0 \leq \frac {1}{\sqrt [n]{n!}} \leq \frac {1}{\sqrt {n/2}}$

and this immediately yields the desired result by squeezing theorem.

Another simple solution (but requires some knowledges):
Use that $ \sqrt {2\pi n}(\frac {n}{e})^n < n! < \sqrt {2\pi n}(\frac {n}{e})^n e^{1/12n}$.


Hey guys! I have a question and need an explanation for that. Thanks!

First, 4 is subtracted from x and the square root of the difference is taken. Then, 7 is added to the result, giving a final result of 9. What is the value of x?

Try putting the words into an equation in $x$ and then solving for $x$.

[hide="The equation would be"]$\sqrt{x-4}+7=9$[/hide]

I understand how you converted from the words to the symbols and numbers. Unfortunately, I have never seen anything like that before because I have never tried it before. So I need your help to explain how to get the answer for that question. Thanks!

[hide="Solution"]So we have $\sqrt{x-4}+7=9$
$\Rightarrow\sqrt{x-4}=2$     subtracting 7 from both sides
$\Rightarrow x-4=4$     squaring both sides to get rid of the square root
$\Rightarrow x=8$     adding 4 to both sides
[/hide]

I have understood those steps, which you have shown. But there is only one problem that I don't understand is "squaring both sides to get rid of the square root." Could you show me how to do this? Besides, I do not know a lot of information about square roots. Thank you!

you know how you can add a number to both sides to undo subtraction ($x-4=6$ so $x=10$) and can multiply both sides by a number to undo divison ($\frac{x}{3}=4$ so $x=12$)? That is because addition and subtraction are inverse functions, and so are multiplication and division. In the same way, squaring both sides and taking the square root of both sides are inverse functions. For example, if you had $\sqrt{x}=7$ you'd just know that $x=49$, but the math reason for that is that you squared both sides.

I fully understand your explanation now. By the way, the problem is solved. Thanks again, helpers!


I am really bad at number theory.
Where should i start (any books)?

I already have intro to NT. Are there any other books?

I don't know where to post this actually.  :blush:

I would recommend that you start with Introduction to Number Theory, as it really is a good introduction.

[u]Other random books/problem sources.[/u]

Other contests
[url]http://www.artofproblemsolving.com/Wiki/index.php/Number_theory#Books[/url]
[url]http://www.artofproblemsolving.com/Wiki/index.php/Category:Introductory_Number_Theory_Problems[/url]
[url]http://www.artofproblemsolving.com/Wiki/index.php/Category:Intermediate_Number_Theory_Problems[/url]
Hardy+Wright: An Introduction to the Theory Of Numbers
Andrews: Number Theory
[url]http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf[/url]


if you see this message, post your e-mail address. i got some question to ask.

would I get spammed if I post my e-mail address here?  :maybe: I won't risk that. Anyhow, I send you a private messages.


A positive integer has 1000 digits (decimal system), all but at most one of them being the digit $5$. Show that this number isn't a perfect square.

[hide="Solution"]
If the last two digits are 55, then it can't be a perfect square. So one of those digits has to be the different one. Mod 20, a perfect square can only be 0, 1, 4, 5, 9, or 16. This means if we change the units digit, we must change it to a 6. If we change the tens digit, note that any square ending in a 5 must end in a 25. So we must change it to a 2.

That gives us two possibilities: $5555\ldots556$ and $5555\ldots525$. The former's digits sum to a multiple of 3 that is not a multiple of 9, meaning it can't be a perfect square. The latter's square root must be of the form $10k+5$, so we have $100k^{2}+100k+25$. But $100k^{2}+100k$ must be a multiple of 200, and $5555\ldots5500$ is not a multiple of 200. So that's not a perfect square either. And we're done.[/hide]


inequalities
analytic geometry
trigonometry
geometry
function
triangle inequality

The Triangle Inequality is pretty trivial, but could someone post a rigorous proof of it?

consider a coordinate system, with the verticies of the triangle as the coordinates $(0,0)$, $(a,b)$, and $(c,d)$

$\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}\ge \sqrt{(a-c)^{2}+(b-d)^{2}}$

$\sqrt{a^{2}+b^{2}}\sqrt{c^{2}+d^{2}}\ge ac+bd$

$a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}\ge a^{2}c^{2}+b^{2}d^{2}+2abcd$

$(ad-bc)^{2}\ge 0$

which is true, all steps are equivalences so the first statement is true, note that since i used an arbitrary triangle, i proved all 3 triangle inequality statements

[quote="13375P34K43V312"]Couldn't it be somewhat simpler if we used (0,0), (a,0), and (c,d)? Also, I don't understand what you're doing between the first and second lines.[/quote]

The first equation is the triangle inequality. He squared it to get the second inequality.

Do we need to use coordinates? Let $ABC$ be a triangle. Then

$AB+BC \ge AB \cos{(\angle CAB)}+BC \cos{(\angle ACB)}= AC$.

Yeah, that's much simpler. Most geometry problems can be done without analytic geometry.

You can generalies it to any biquadratic form in N dimensions aswell. (Biquadratic form basically meaning distance function)

Edit: Think it might be bilinear forms


function
calculus
derivative
logarithms
trigonometry
limit
calculus computations

Among the following function,
-one is continuous everywhere,but not differentiable at 0
-one is differentiable,but derivative is discontinuous at 0
-one is differentiable and has continuous derivative

1)$ f(x) = \{\begin{array}{cc}\ln (1 + x^3)\sin\frac {1}{x},x > 0 \\
0,x\leq0\end{array}$

2)$ g(x)=\{\begin{array}{cc}\ln^2(1+x)\sin\frac {1}{x},x>0\\0,x\leq0\end{array}$

3)$ h(x)=\{\begin{array}{cc}\ln (1+\frac{\sin x}{2}),x>0\\x\leq0\end{array}$

Solution:

1)$ \lim_{x\rightarrow 0}\frac{\ln (1 + x^3)\sin\frac {1}{x}}{x}=\lim_{x\rightarrow 0}\frac{(x^3+0{x^6})\sin\frac {1}{x}}{x}=0$

differentiable

$ f'(x)=\frac{3x^2\sin\frac {1}{x}}{1+x^3}-\frac{\cos\frac {1}{x}\ln (1+x^3)}{x^2}$

2)$ \lim_{x\rightarrow 0}\frac{(x^2+o(x^4))\sin\frac {1}{x}}{x}=0$ differentiable

$ g'(x)=\frac{2\ln (1+x)\sin\frac {1}{x}}{1+x}-\frac{\cos\frac {1}{x}\ln^2 (1+x)}{x^2}$

3)$ \lim_{x\rightarrow 0}\frac{(\frac{\sin x}{2}+0(\sin^2x))\sin\frac {1}{x}}{x}=\lim_{x\rightarrow 0}\sin\frac {1}{x}$ not differentiable

I couldn't find which function f'(x) or g'(x) is not differentiable at 0,which one continuous.

[hide="f"]
When $ x>0$, derivative is $ \frac{3x^2\sin \frac{1}{x}}{1\plus{}x^3}\minus{}\frac{\ln (1\plus{}x^3)\cos \frac{1}{x}}{x^2}$. When $ x<0$, derivative is 0. Check for derivative at $ x\equal{}0$: the limit $ \lim_{h\rightarrow 0^\minus{}} \frac{f(h)}{h}\equal{}0$ for the negative part. For the positive part:
\[ \lim_{h\rightarrow 0^\plus{}} \frac{f(h)}{h}\equal{}\lim_{h\rightarrow 0^\plus{}} \frac{\ln (1\plus{}h^3)\sin \frac{1}{h}}{h}\]
\[ \equal{}\lim_{h\rightarrow 0^\plus{}} h^2\cdot \frac{\ln (1\plus{}h^3)}{h^3}\cdot \sin \frac{1}{h}\]
There is a product of three terms: the first tends to 0, the second tends to 1, the third is bounded between -1 and 1. As a result, the limit is 0.

We then get
\[ f^\prime (x)\equal{}\left\{
\begin{array}{rl}
\frac{3x^2\sin \frac{1}{x}}{1\plus{}x^3}\minus{}\frac{\ln (1\plus{}x^3)\cos \frac{1}{x}}{x^2},x>0\\
0,x\leq 0\\
\end{array}\right.\]
It remains to determine $ \lim_{x\rightarrow 0^\plus{}} f^\prime (x)$. This is 0 since we have a sum of two terms tending to 0, by similar reasoning as showing that $ f^\prime (0)\equal{}0$.

So $ f$ has a continuous derivative.
[/hide]
[hide="g"]
When $ x>0$, the derivative is $ \frac{2\ln (1\plus{}x)\sin \frac{1}{x}}{1\plus{}x}\minus{}\frac{\ln (1\plus{}x)^2\cos \frac{1}{x}}{x^2}$. When $ x<0$, the derivative is 0. Now to compute the derivative, we need $ \lim_{h\rightarrow 0^\plus{}} \frac{f(h)}{h}$:
\[ \lim_{h\rightarrow 0^\plus{}} \frac{\ln (1\plus{}h)^2\sin \frac{1}{h}}{h}\equal{}\lim_{h\rightarrow 0^\plus{}}h\cdot \left( \frac{\ln(1\plus{}h)}{h}\right)^2\cdot \sin \frac{1}{h}\]
which is 0. Also, $ \lim_{h\rightarrow 0^\minus{}} \frac{f(h)}{h}\equal{}0$ so $ f^\prime (0)\equal{}0$.

However, $ \lim_{x\rightarrow 0^\plus{}} f^\prime (x)$ doesn't exist! This derivative is the difference of two terms. The first tends to 0, but the second can be expressed as
\[ \left( \frac{\ln (1\plus{}x)}{x}\right)^2\cos \frac{1}{x}\]
This term can be made greater than 1/2 when $ x\equal{}\frac{1}{2n\pi}$ for sufficiently large $ n$, and less than -1/2 when $ x\equal{}\frac{1}{(2m\plus{}1)\pi}$ for sufficiently large $ m$. So the limit cannot exist.

We conclude $ g$ is differentiable, but its derivative has a discontinuity at 0.
[/hide]


Find $ n\in \mathbb{N}\ ,\ n\ge 2$ such that $ \frac{1}{1\plus{}\sqrt[n]{n\plus{}1}\plus{}\sqrt[n]{(n\plus{}1)^2}}\equal{}\frac{\sqrt[n]{n\plus{}1}\minus{}1}{n}$ .

That's just another way of rewriting

$ n\equal{}\sqrt[n]{(n\plus{}1)^3}\minus{}1$

$ n\equal{}1$ does not work, $ n\equal{}2$ does not, and $ n\equal{}3$ works.

If $ n\ge 4$ then $ \sqrt[n]{(n\plus{}1)^3}\minus{}1<(n\plus{}1)\minus{}1\equal{}n$

Hence $ n\equal{}3$ is the only solution.


Problem one:
Rewrite it: $|(x+2)^2-1|+2x+5=0$

If $(x+2)^2 \geq 1$, then $x \geq -1$ or $x \leq -3$. Now the equation becomes:
\[x^2+6x+8=0\]
Hence $x=-2$, $x=-4$.
Only $x=-4$ satisfy the inequality above.

If $(x+2)^2 \leq 1$ then $-3 \leq x \leq -1$. The equation becomes:
\[x^2+2x-2=0\]
$x=-1+\sqrt{3}$, $x=-1-\sqrt{3}$.
Similarly, we get $x=-1-\sqrt{3}$.

Therefore: $x=-4$, $x=-1-\sqrt{3}$.


I've always read studies on how classical music is good for your brain and enhances some features of it (I don't know the science behind it or exactly how it works).

I'm wondering, what type of music do some of you listen to? When you work on math, do you prefer silence or some sort of music/noise?

For me, I can't read when there is a lot of noise or some sort of music I do not like. But when I'm doing math, I find a basic rhythm to be helpful. I often listen to hip hop instrumentals just for the drumkicks...

If any of you play the piano, you understand that music is a lot of math. But playing music and me is comparable to vegetable oil and water.

Anyways, back to the original question. What/When are your music preferences and why?

I love classical music because it is so much more complex, interesting, and beautiful than other popular forms of music. As far as helping me with math, though, there is no way I can listen to nice music and do math at the same time. I can't think about the music and put all my effort into a math problem at the same time.

.I am also a huge fan of what most people call "classical" music, but my specific affinities lie more towards the Romantic and Twentieth century spectrum.  I bet quite a few people on this forum would enjoy listening to some of the newer stuff out there, especially minimalism, considering its highly mathematical structure  :D .
Check out some John Adams, Eric Whitacre, Cindy McTee, Richard Wagner, Johannes Brahms, Philip Glass, Sergei Rachmaninoff, Prokofiev, Copland, Frank Ticheli and such to hear some great music.

miraculouspostmaster      , I like a lot minimalists composers and post modern composers.. like philip glass, steve reich , john adams..
I don't like john cage, but i dont considere him a minimalist, I consider him as an analaogue of the neo dadaist movement in painting.
Also  I like a lot Erik Satie he is great and Debussy  of the romantics..
Bela Bartok is another composer that I admire  a lot.

BTW, philip glass studied mathematics and philosophy in NYU!

When I'm alone listening to music of any kind, I end up humming.. or sometimes singing along.
So I've given up listening to music when I'm trying to concentrate.

whenever I take a test or concentrate on math, there is always a song in my head, even as I type the song that I have in my head is "Highway to Hell"

try listening to "fractals" by mannheim steamroller sometime

[quote="Magnara"]I agree about the beat thing.  The Postal Service works really well for me.[/quote]

The Postal Service is pure awesomeness.

[quote]I find if i really need to study, classical music, really helps. To help me concentrate I chose something that's really beautiful, but not full of surprises, if you know what I mean...(Like no surprise cymbol crashes or sudden tempo changes) That way I end up tuning it out. I love to listen to other types of music, rock mostly, but it distracts me sometimes as I insist on singing the lyrics out to all my favorite songs, hehehe
[/quote]

I would highly recommend the following (for study music):
Any of Eric Whitacre's choral works (especially Water Night and Sleep)
Richard Wagner's Overture to Lohengrin
2nd Movement of Rachmoninoff's 2nd Piano Concerto
John Adams' Christian Zeal and Activity (if you can manage to ignore the sermon in the middle)
Almost any of the Brahms intermezzi - (Op. 109, No.2 is fantastic)


Blue... no, wait... green!

AAAAAAHHHHHHHHHH!!!!

Edit: Actually, I voted for red.

[quote="Hexaditidom"]"How do you know so much about swallows?"
"You have to know these things when you're a king..."

 :D[/quote]

haha...

green

at first I was going to google it.

but, yeah, voted for green.

[quote="eryaman"]what's with the name of this topic?[/quote]

It's a reference to the movie [i]Monty Python and the Holy Grail[/i].

"Wait, wait.  What if we make this giant wooden badger..."

blue; there are so many different shades that there's one for every single mood. like a "teal"ish for happy, and then deep depressing navy, and everything inbetween. i love blue; it's absolutely perfect

orange, cuz if its a gloomy, dark day, you can wear a bright orange shirt, therefore reducing the rates of suicide...

really, it's true

[quote="white_horse_king88"]Yellow, no Blu - argh............!! :D

Is the actual quote I think...[/quote]

No, it's "Blue.  No, gr-aaaagh".  I saw it again a couple of days ago.

red
actually, arrest-me-ultra-red to be specific  :D

Blue ......just because it's my favorite. purple is good too

Im sort of like all colours but my favorite from that list would be white, what? It is sign of power.

My favorite color is purple. I like green, too.

Aww... and here I was hoping that this would be a trivia topic so I could answer Nineveh.

haha Lucky . . . you and your trivia! :)
How's the AoPS trivia challenge been going, anyway? Are you still running it?

AOPS Trivia Challenge has kinda died.

AOPS Trivia Invitational on the other hand may start up again soon with a tournament. Dunno when though. Not now that's for sure.

[quote="Lil_mad_dog"]Someone:An Asian swallow or a European one?
Gatekeeper:Well, I don't know that... Aiggh![/quote]
Actually, [i]King Arthur [/i]said, "Is it an African swallow or a European one?"  The [i]Bridge[/i]keeper said "What?  I dunno that.  AAAAAGGGHHH!"

UNC BLUE [color=cyan]CLOSEST I COULD GET[/color]

I unclog my nose in your direction, you son of a window dresser!

how is this topic popular?
it hasnt been posted on for more than two weeks

 Blue

black

popular topics mean they have about 2 pages, no matter how old

[quote="kyyuanmathcount"]I liked blue always.

I used to like green, but when I found out it was the colors of our rival sports team, i cut it.[/quote]
How can you always liked blue, but once liked green?  This calls for an existence theorem for colo(u)rs.

Blue. No, yellow...!

Actually, it is blue.

Red is the color of evil, blood, and the shortest color on the list.

What do colors have to do with the capital of Assyria?! :?

You obviously haven't seen Monty Python and the Holy Grail.

Gogo, Holy Hand Grenade of Antioch.
I can't decide between blue and green.

My favorite color is blue.  :D
sort of like the shade of that smiley.

Green is a good color too, but i really like blue.


Find
$ \sum\limits_{i \equal{} 1}^n {i^k }$

[quote="The Enigma"][quote="varunrocks"]If n is 0 mod 4 than the value of the summation would be 0.
if n is 1 mod 4 then the value of the summation would be i.
if n is 2 mod 4 then the value of the summation would be -1+i.
if n is 3 mod 4 then the value of the summation would be -1.
This is because i runs in cycles of 4.
i+-1-i+1=0.[/quote]
you are mistaken he never meant that i denoted complex numbers , he meant that i was any number in the given interval..[/quote]
Yes, what I meant is that just the sum of $ i^k$ with i  is the natural numbers. :)


Demidec needs some additional beta testers for their quizes. Thought you guys might be interested in having a lil fun, especially the science quizzes. The questions are mostly 11-12 standard.  The site is - and you need to mail dan@demidec.com to become a part of the team. Its basically a volunteer job, but you do start getting paid if you do a lot of beta testing.
In case you are wondering, I don't get any referral perks for promoting demidec here  :rotfl: Its just that Dan is a friend of mine and you are the smartest bunch of guys I could get to help him(plus you get additional ques for practice) :)
Also FYI- http://en.wikipedia.org/wiki/DemiDec

@nuclear
Forgive me,but I'm a rural dweller :rotfl:
So mind telling me what "beta testing for demidec" exactly is?
[hide]U may laf at me if u want,but I really cudn't make out what it is[/hide]

Its basically solving quizzes and finding errors in the questions/options before they are released for the actual contests or books.


Prove that if n = 11^m, the number of tens of n is equal to the number of unit of m

[hide] We use induction. The base case, m=1 is obviously true.  Note that the units digit of n will always be 1.  n ends in k1, then 11(n) will end in 11+110k, so the number of tens is k+1.  So the tens place increases by one when m increases by one, so it is always the units digit of m. [/hide]

[hide]
$n=11^m=(10+1)^m$
Using binomial expansion, the coefficient of $10^{1}1^{m-1}$ is $\binom{m}{m-1}=m$.
Since the last term of the expansion (=units digit) is always 1, the only term that determines the tens digit is $10^{1}1^{m-1}$.
Therefore, the units digit of $m$ ($m\pmod{10}$) is equal to the tens digit of $n$ ($m\pmod{10}$).
[/hide]


inequalities
AMC
USA(J)MO
USAMO
rearrangement inequality

im having trouble doing this old usamo problem that goes like this:
prove  a^a*b^b*c^c => abc^((a+b+c)/3).  i got the feeling that i should use AM-GM on itm but i cant figure out how, does anybody have any ideas>

[color=cyan]This is a nice problem.  Nice in that I can figure it out.  AM-GM isn't going to help -- what might help is the rearrangement inequality, which shouldn't be that hard to prove I think.  It says that given two lists,
a1<=a2<=a3<=. . .<=an and b1<=b2<=b3<=. . .<=bn,
if we add pairwise multiples of the sequences, we get

a1*b1+a2*b2+. . .+an*bn >= a1*bi + a2*bj + . . . +an*bk >=
>= a1*bn + a2*b(n-1) + . . . + an*b1

where in the middle term, each of the b terms appears exactly once.  Although how to apply it I leave up to you.[/color]

I came across this cute inequality some time ago I tried to find as many proofs as possible. Here are the ones I found. Any more very welcomed !

Let's prove a more general result :

(*) prod xi^xi >= (prod xi)^(sum xi / n)

1) AM-GM-HM

By weighed geometric - harmonic mean (for instance see http://planetmath.org/encyclopedia/PowerMean.html)

(prod xi^xi)^(1/(sum xi)) >= sum (xi) / sum(xi/xi) = sum xi / n

AM >= GM :

sum xi / n  >= (prod xi)^n

Taking 'power' (sum xi) of last inequality gives (*)

2) Jensen

f(x) = x*ln(x), f''(x)=1/x > 0 so f is strictly convex.

So we can apply Jensen inequality :

1/n (sum f(xi)) >= f (sum xi / n)

Taking 'exp' of last inequality gives (*)

3) Chebyshev

Chebyshev inequality :

if a1 <= a2 ... <= an
if b1 <= b2 ... <= bn

then sum (ai)/n * sum(bi)/n <= sum (ai*bi)/n

(*) being symetric we can suppose wolog x1 <= x2 <= .. <= xn
ln being increasing, ln(x1) <= .. <= ln(xn)
we can apply Chebyshev :

(sum xi)/n * (sum ln(xi))/n <= (sum xi*ln(xi))/n

Taking 'exp' of last inequality gives (*)

4) Rearrangement inequality

See JLB.

5) Trick

for n = 4, wolog a >= b >= c >= d

(a^4a b^4b c^4c d^4d) / (abcd)^(a+b+c+d) =
(a/b)^(a-b) * (a/c)^(a-c) * (a/d)^(a-d)
(b/c)^(a-c) * (b/d)^(b-d) *
(c/d)^(c-d) >= 1  because each term is of the form x^y with x >= 1 et y >= 0.

My favorite is the 5th.


quadratics
number theory unsolved
number theory

Hi, friends! I'm posting MMO 2005 (9th grade questions)...

    9th Grade

   [u][b]Problem1:[/b][/u]   The discrimnants of 3-quadratic equations with non-zero coefficient are 1,4 and 9.  Prove that it's possible to take one of the roots of equestion such that the sum of them is equal to the sum of the rest of the roots.

   [u][b]Problem2: [/b][/u]  Are there 2005-distinct natural numbers such that the sum of any-2004 of  them can be divisible by the rest number.

   [u][b]Problem3: [/b][/u]  Circle O_1 passes through the center of the circle O_2. Two tangent lines drawn from a point C on the cicle O_1 to the circle O_2 cut the circle O_1 at the points A and B. Prove that AB is perpendicular to the line passes through the centers of the circles.

   [u][b]Problem4: [/b][/u] Is it possible to make a square from 1000-pieces obtained by breaking into pieces a triangle.(any triangle)

   [u][b]Problem5: [/b][/u] Given a point P (not intersection of altitudes) in an acute triangle ABC. Prove that the circles pass through the midpoints of the triangles PAB, PAC, PBC, ABC and the circle passes through the projections of O to the sides intersect at one point.

fatih sahin

Solution3: Let O_1 and O_2 be the centers of each circle and the tangency points of the lines drawn from C on CA and CB be X and Y, respectively.
 If AO_2=BO_2 then AO_2BO_1 are kite since AO_1 = BO_1 and AB will be perpendicular to O_1O_2 ...let's prove.
     XO_2= YO_2 so angle XCO_2 = angle YCO_2 and angle AO_1O_2=angle O_2O_1B
so O_1O_2 is the bisector of angle AO_1B and AO_1=BO_1 then AO_2=BO_2...  it's proved.

                                                                                Fatih Sahin

[b][u]Problem6:[/u][/b] There are  n-nonzero digits on the circle. Myth and Pierre both choose n-1 consecutive digits and write them into their sheets, reading digits in clockwise order. It is known that Myth and Pierre chose not the same digits, but n-1 -digit numbers they obtained are the same. Prove that it is possible to partition circle into several parts such that all numbers on the arcs will be the same.

matematikator,
we appreciate you posting these problems, but you should post contests in the National Olympiads section ([url=http://www.mathlinks.ro/Forum/index.php?f=125]www.mathlinks.ro/Forum/index.php?f=125[/url]).


Can the product of three consecutive integers be 111111111? Why or why not?

Not too hard.

[hide]No,at least one of the numbers must be even, and 111111111 is not divisible by 2.[/hide]

Okay -  A slight modification and a little more challenge -  (Still not too hard though)

Prove that 111111111 is not a product of three consecutiveintegers  in ANY base (That is need not be base 10)
[hide](  Annoying fact:  numbers ending in 1 need not be odd in base 3 ) [/hide]

hmm.  maybe it's because the product of three consecutive numbers could never end in a 1?  I can't prove it though.


algebra
polynomial
linear algebra
matrix
algorithm
linear algebra unsolved

$ p(x)\equal{}ax^2\plus{}bx\plus{}c\in\mathbb{Z}[x]$ is an irreducible polynomial of degree $ 2$. Find a necessary and sufficient condition upon $ a,b,c$ so that the equation $ p(X)\equal{}0$, where the unknown is $ X\in\mathcal{M}_2(\mathbb{Z})$, admits an infinity of solutions.

The sufficient and necessary condition is $ a\in\{\minus{}1,1\}$.

Proof: Denote by $ S(a,b,c) \subset M_2(\mathbb Z)$ the set of integer matrix solutions of $ p(M)\equal{}0$.

1 - If $ S(a,b,c)$ is infinite, there is $ M\in S(a,b,c)$ whose minimal polynomial (considered in $ \mathbb Q[x]$) has degree 2:[hide="Proof of 1"]Otherwise the minimal polynomial of every $ M\in S(a,b,c)$ has degree 1. In this case $ M$ is conjugated (over $ \mathbb Q$) to $ \lambda E_2$ with $ \lambda\in\mathbb Q$. This implies $ M\equal{}\lambda E_2$ and in particular $ \lambda\in\mathbb Z$ with $ p(\lambda)\equal{}0$. Hence $ S(a,b,c)$ has 2 elements at most - a contradiction.[/hide]

2 - $ a\equal{}\pm 1$ is a necessary condition:[hide="Proof of 2"]Let $ M\in S(a,b,c)$ with minimal polynomial $ m_M(x)\in\mathbb Q[x]$ of degree 2. Since the minimal polynomial and the characteristic polynomial coincide, we can conclude $ m_M(x)\in\mathbb Z[x]$. Then we get $ p(x)\equal{}a\cdot m_M(x)$, using the division algorithm in $ \mathbb Q[x]$ at first, then stepping down to $ \mathbb Z[x]$ again. Since $ p(x)$ is irreducible (in $ \mathbb Z[x]$ - as I interpret the assumption), $ a$ is a unit in $ \mathbb Z[x]$, i.e. $ a\equal{}\pm 1$.[/hide]

3 - $ a\equal{}\pm 1$ is a sufficient condition:[hide="Proof of 3"]Wlog we have $ p(x)\equal{}x^2\plus{}bx\plus{}c$. Define the companion matrix $ M_0\equal{}\left(\begin{array}{cc}0&\minus{}c\\1&\minus{}b\end{array}\right)$ of $ p$. Then the infinitely many, pairwise different matrices \[ \left(\begin{array}{cc}1&n\\0&1\end{array}\right)^{\minus{}1}M_0\left(\begin{array}{cc}1&n\\0&1\end{array}\right)\] with $ n\in \mathbb Z$ are contained in $ S(1,b,c)$.[/hide]


LaTeX
limit
logarithms
geometry
geometric transformation
real analysis
real analysis unsolved

Salut tout le monde j'ai besoin un aide pour étudier la nature de la série numérique du terme générale donné par
la formule suivante:

       u_n^ =〖(-1)〗^(n+1)/n ∑_(k=1)^n▒〖(-1)〗^k/k

Just to translate into Latex:

$ u_n \equal{} \frac{(\minus{}1)^{n\plus{}1}}{n} \sum_{k\equal{}1}^n \frac{(\minus{}1)^k}{k}$

English version (not so accurate):

"Hi everyone I need some help to study the convergence of the sequence whose general term is given by the following formula:"

just a question: if you are not french, why would you write in french in an english forum ?

Since $ \lim_{n\to\infty}\sum_{k\equal{}1}^n\frac{(\minus{}1)^k}{k}$ converges to a finite real number (to $ \minus{}\ln 2,$ although we don't really need to know that), then if we multiply that by another sequence which tends to zero, the resulting limit is zero.

After all, the limit of the product is the product of the limits, provided both limits exist.

I have a vague sense that this might not have been the question intended, but I'm not enough of a mind-reader to come up with what the more interesting question might have been.

in fact the question asked is to study:
$ \sum_n \frac {( \minus{} 1)^{n \plus{} 1}}{n} \sum_{k \equal{} 1}^n \frac {( \minus{} 1)^k}{k}$

I feel that this is a particular case of a more general problem.

Assume that $ S\equal{}\sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}x_{n}$ and $ T\equal{}\sum\limits_{n\equal{}1}^{\infty}x_{n}^{2}$ both converge.

Find

$ \sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}x_{n}\left(\sum\limits_{k\equal{}1}^{n}(\minus{}1)^{k}x_{k}\right)$.

And the series can be expressed, via Abel summation by parts formula, in terms of $ T$ and $ S$.

Ah, so now it makes sense. The misunderstanding wasn't in the translation from French to English but in the translation from whatever that was (which doesn't show properly on my screen) to $ \text{\LaTeX}.$

Here's a low-tech argument: by the Alternating Series Test, $ \sum_{k=1}^n\frac{(-1)^n}{k}=L+(-1)^ny_n$ where $ L$ is the limit of the (convergent) infinite series and $ 0<y_n<\frac1{n+1}.$

Then the whole sum is $ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(L+(-1)^ny_n)$

Break this into two sums. $ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\cdot L$ converges (to $ -L^2$) by the Alternating Series Test. The other term is

$ \sum_{n=1}^{\infty}\frac{-1}{n}\cdot y_n.$ This series is negative and converges by comparison to $ \sum_{n=1}^{\infty}\frac1{n(n+1)}=1.$

We get that the whole original series converges to a number between $ -L^2$ and $ -L^2-1.$ And we also note that $ L=-\ln 2$ so the series converges to a number in the interval $ (-(\ln 2)^2-1,-(\ln 2)^2).$

I suspect that it's likely that we can compute this limit in closed form, but I have a class to teach right now.

Assuming that there are no errors in my calculations I got that

the value of the series equals

$ \minus{}\frac{\ln^{2}2}{2}\minus{}\frac{\pi^{2}}{12}$,

which happens to agree with the numerical calculations.

[quote="didilica"]I feel that this is a particular case of a more general problem.

Assume that $ S \equal{} \sum\limits_{n \equal{} 1}^{\infty}( \minus{} 1)^{n}x_{n}$ and $ T \equal{} \sum\limits_{n \equal{} 1}^{\infty}x_{n}^{2}$ both converge.

Find

$ \sum\limits_{n \equal{} 1}^{\infty}( \minus{} 1)^{n}x_{n}\left(\sum\limits_{k \equal{} 1}^{n}( \minus{} 1)^{k}x_{k}\right)$.

And the series can be expressed, via Abel summation by parts formula, in terms of $ T$ and $ S$.[/quote]

$ \sum\limits_{n \equal{} 1}^\infty  {\left[ {( \minus{} 1)^n x_n \sum\limits_{k \equal{} 1}^n {( \minus{} 1)^k x_k } } \right]}  \equal{} \frac{1}{2}\left( {S^2  \plus{} T} \right)$


For sequence of matrix ($ A_m$) such that :
   $ A_0\equal{}A \in S_n(R)$,$ A_1 \in S^{\plus{}\plus{}}_n(R)$ and : $ A_m\equal{}2A_{m\minus{}1}\minus{}2009A^{2}_{m\minus{}1}$
Prove that : $ \lim_{m \to \plus{}\infty} {A_m}\equal{}\frac {I_n}{2009}$
Here $ S^{\plus{}\plus{}}_n(R)$ is Of $ S^{\plus{}}_{n}$ and $ GL_n(R)$
__________________
 :pilot:

We have :  $ I_n\minus{}2009.A_m\equal{}I_n\minus{}2.2009A_{m\minus{}1}\plus{}2009^2.A^{2}_{m\minus{}1}\equal{}(I_n\minus{}2009.A_{m\minus{}1})^2$
from here by induction : $ I_n\minus{}2009.A_m\equal{}(I_n\minus{}2009.A)^{2^m}$
 you use : $ A_1\equal{}A(2.I_n\minus{}2009A)>0$ therefore : $ \minus{}I_n<(I_n\minus{}2009.A)<I_n$ and $ (I_n\minus{}2009.A)^{2^m} \to 0$  and we have prove .


Find all $ f: R\rightarrow R$ such that $ f(x\plus{}f(x)f(y))\equal{}f(x)\plus{}xf(y)$ for all $ x,y\in R.$

there is my solution  :blush: :

for $ x\equal{}y$ we have:
$ f(x\plus{}(f(x))^{2})\equal{}f(x)(x\plus{}1)$
Let :$ f(x\plus{}(f(x))^{2})\equal{}f(y\plus{}(f(y))^{2})$
so:
$ f(x)(x\plus{}1)\equal{}f(y)(y\plus{}1)$ $ (1)$
Let $ f(x)\equal{}f(y)$
using $ (1)$ we have $ x\equal{}y$ so f is injective.
Let $ f(0)\equal{}a$
$ f(a^{2})\equal{}a$
and let $ x\equal{}0$
$ f(af(y))\equal{}a\equal{}f(a^{2})$
so $ af(y)\equal{}a^{2}$
if $ a\neq0$ :
$ f(y)\equal{}a$ and it's a contradiction because f is not a constante function.
so $ a\equal{}0\equal{}f(0)$
we have: $ f(x\plus{}(f(x))^{2})\equal{}f(x)(x\plus{}1)$
for $ x\equal{}\minus{}1$ : $ (f(\minus{}1))^{2}\equal{}1$
$ \mid f(\minus{}1) \mid\equal{}1$
so $ f(f(x)\plus{}x)\equal{}f(x)\plus{}x$ or $ f(x\minus{}f(x))\equal{}\minus{}(x\minus{}f(x))$
===>
-$ f(x)\equal{}0$
-$ f(x)\equal{}x$
-$ f(x)\equal{}\minus{}x$

[quote="reason"]there is my solution  :blush: :

for $ x \equal{} y$ we have:
$ f(x \plus{} (f(x))^{2}) \equal{} f(x)(x \plus{} 1)$
Let :$ f(x \plus{} (f(x))^{2}) \equal{} f(y \plus{} (f(y))^{2})$
so:
$ f(x)(x \plus{} 1) \equal{} f(y)(y \plus{} 1)$ $ (1)$
Let $ f(x) \equal{} f(y)$
using $ (1)$ we have $ x \equal{} y$ so f is injective.
[/quote]
Sorry, but you're wrong.

$ (1)$ is true only under your hypothesis $ f(x \plus{} (f(x))^{2}) \equal{} f(y \plus{} (f(y))^{2})$, and not for any $ x,y$. So what you have shown is :

If $ f(x)\equal{}f(y)$ [u][b]and [/b][/u]$ f(x \plus{} (f(x))^{2}) \equal{} f(y \plus{} (f(y))^{2})$, then $ x\equal{}y$. So you did not show that $ f(x)$ is injective (and, btw, $ f(x)$ is not injective since, for example, $ f(x)\equal{}0$ $ \forall x$ is a non injective solution).


probability quadratics geometry parallelogram trapezoid AMC

Please show your work in order to recieve full credit.

Source: Columbus State University Math Tournament, 2003 Individual Round
Difficulty: Easy

i)  If 1/a - 1/b = 1, find the value of (a-2ab-b)/(2a+3ab-2b).

ii) If a + b = 1 and a:^2: + b:^2: = 85, find a:^3: + b:^3:.
Note: if you don't know how to solve this, refer to AoPS Vol 1, p. 93

iii) N consecutive integers sum to 2003, where N is an odd number greater than three.  Determine the smallest of these integers.

iv) A point P(x,y) is randomly selected from the region |x| + |y|< 1.  What is the probability that 2x + 3y < 1?

ii) [hide]

a+b=1  (given)

a:^2:+b:^2: = 85 (given)

manipulate the first eqn to get a=1-b, and substitute into the 2nd eqn.

(1-b):^2:+b:^2:=85

expand and simplify

1-2b+2b^2=85

0=2b^2-2b-84

Use the quadratic eqn to solve for b. You'll get b=6 or 7.

But the sign of b has been lost because of the :^2: in the second eqn. Looking at the first eqn, it's easy to see that a=7, b=-6, and this is easily checked for the second eqn (and it holds true).

Final answer: 7:^3:+(-6):^3: = 127.

[/hide]

i) [hide]

Commentary provided for those that might not understand the algebra.

1/a - 1/b = 1

1/a = 1+ 1/b

1/a = (b+1)/b  (common denominator)

a=b/(b+1) (inverse)

Substitute into other equation.

  (b/(b+1)) - 2b*(b/(b+1)) - b

2(b/(b+1)) + 3b *(b/(b+1)) - 2b

(that's one equation, but would have been really messy on one line. The underline is division)

Factor out a b from the numerator and denominator and cancel it (not shown). Then rewrite the 1 in the end of the numerator as (b+1)/(b+1) and the 2 in the end of the denominator as 2*(b+1)/(b+1). Combine.

(1-2b-b-1)/(b+1)

(2+3b-2b-2)/(b+1)

Cancel the (b+1) s, combine terms, and do the suitable algebra.

-3b/b = -3

[/hide]

...right?

edit: that's all for me for now... I have to read the AP US History book. (Although I'd prefer to keep doing math)

[hide]iii) -1000

since 2003 is prime and that the series has to have more than 3 numbers... than the median or the 1002nd term is 2003/2003 = 1 so the 1st term is 1-1001 = -1000[/hide]

i think. im never sure of myself... especially not today... lol... if you don't get it... go to the parallelograms help thing in getting started.

Hmm... [hide]number 3 could also be -2002.  As for numbers 2 and 1:  you solved the quadratic wrong, that's why your sign got messed up confuted.  There are two solutions, (a, b) = (7, -6) or (-6, 7), which both give your same correct answer.  For number 1, you are really much better off not solving for a or b.  Instead:

a - b = -ab  (multiplying through by -ab), so

(a - 2ab - b)/(2a + 3ab - 2b) = (-3ab)/(ab) = -3[/hide]

[hide]

Haha, indeed I did solve the quadratic wrong. lol.

And I like your way of solving 1 better  Oh well, at least mine worked. How did you know to multiply by -ab? Did you just "know," or something else?[/hide]

Well, at first I just collected them into one fraction, (b - a)/ab = 1.  Then I looked at the expression I was trying to find and noticed that it only included pieces that looked like (a - b) and ab.  So I created an (a - b) by multiplying through by -ab, and then substituted it in.  (I also could have multiplied through by just plain old ab, of course.  And I also could have made the substitution the other way, putting in (b - a) for ab instead of the other way around.)  Generally speaking, I look at questions like that and make the assumtion that I don't have to actually solve for the variables.  This is very often the case in problems of this type -- number 2 can also be done without explicitly finding either a or b.  So I guess it's drilled-into-me intuition, built up after nearly 10 years of math competitions.

Oof.

JBL is right in saying that -2002 is the lowest number possible in the original question.  It occurred to me, however, that the question actually used in the competition required that the number of integers was an odd number, so if this were the case, then Chinaboy is right.

Now I feel stupid.  Why do I make so many miscopies?

Number 4 is an excellent question.  Not exactly easy, but very good.

#4-

[hide] k, just draw |x|+|y|<1, you get a square with side of  :rt2: , area 2. then just plot 2x+3y<1. and find where it hits y=x+1 and y=x-1. then because the intersection of the former two ineqs is the diamond, find the area of the trapezoid thus formed and substract it from 2, then divide by 2 and simplify[/hide]

[quote="JBL"]And when you do all of that, the answer you get is?[/quote]

i think i got 7/10.


algebra
polynomial
quadratics
Rational Root Theorem

For some real number $r$, the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$. Which of the following numbers is closest to $r$?

$\text{(A)} \ 1.22 \qquad \text{(B)} \ 1.32 \qquad \text{(C)} \ 1.42 \qquad \text{(D)} \ 1.52 \qquad \text{(E)} \ 1.62$

[hide]Let $y=2x$. The polynomial becomes $y^3-y^2-21y+45$. Testing out choices for $y$ using Rational Root Theorem, find that $y=3$ works.

The remaining quadratic is $y^2+2y-15=(y-3)(y+5)$. The repeated value of $y$ is 3 $\Rightarrow r=1.5\Rightarrow \boxed{D}$.[/hide]

Using Ruffini's twice to divide $8x^3-4x^2-42x+45$ by $x-r$, we get that $12r^2-4r-21=0$.

So $r=1'5$, and the needed answer is $\boxed{D}$.

Excuse me, but this you just revived a year-old thread

Huh, is Ruffini the same thing as [url=http://www.physics.arizona.edu/~restrepo/475A/Notes/sourcea/node24.html]Horner's Method[/url]?

Here's a method without guess and check:
Note that if a polynomial $ P(x)$ has repeated root $ r$ of multiplicity $ m$, then $ P(x) \equal{} (x \minus{} r)^m Q(x)$; but $ P'(x) \equal{} m(x \minus{} r)^{m \minus{} 1}Q(x) \plus{} (x \minus{} r)^mQ'(x)$ and so $ P'(r) \equal{} 0$.

So after differentiating $ 8x^3 \minus{} 4x^2 \minus{} 42x \plus{} 45$, we get $ 24x^2 \minus{} 8x \minus{} 42 \equal{} 0$ and so $ x \equal{} \frac {1\pm8}6$.

By the way, does anyone have a proof for synthetic division?

a proof for synthetic division?  it's just a special case of polynomial long division.

Yeah, all you're doing is just like you're dividing polynomials, except that you take the variables out of it.
If you just look at a synthetic division problem you'll see why it works.


linear algebra
matrix
linear algebra unsolved

Let V=R_(nxn ) be the space (over R)of all nxn matrices with real entries and let us denote by M={A∈V| A^t=A} and N={B∈V| B^t=-B} the subspaces of all symmetric and antisymmetric matrices,respectively.
What is dim(M),dim(N),and how to prove that M+N=V?

It is not difficult to see that the subspace of symmetric matrices M has a basis consisting of matrices A such that $a_{ij}=a_{ji}=1$ for and 0's everywhere else, fo i,j=1,...,n. The subspace of antysymmetric matrices N has a similar basis: $a_{ij}=1=-a_{ij}$ and 0's everywhere else, for i,j=1,...,n , but $i \neq j$. From this we can find the dimensions of both subspaces.

As for the sum, it is sufficient to prove that dimM+dimN=dimV and $M \cap N =0$. That dimM+dimN=dimV can be shown from the previous point, and it's obvious that if $A \in M$ and $A \in N$ (since it would have to be symmetric and antysymmetric), then A=0, QED.

Alternately, you can just write the explicit decomposition: $A=\frac12(A+A^{T})+\frac12(A-A^{T})$. This requires that our field is not of characteristic 2; that's the only role your assumption that the field is $\mathbb{R}$ plays.


Determine all integer solutions (a,b,c) with $c\leq 94$ for which:
$(a+\sqrt c)^2+(b+\sqrt c)^2 = 60 + 20\sqrt c$

Well I'm not completely clear on your meaning of natural solutions.
If you mean that a,b,c are all natural numbers then we have a simple solution procress.

[tex](a+\sqrt{c})^2+(b+\sqrt{c})^2=60+20\sqrt{c}[/tex]
[tex]a^2+2a\sqrt{c}+c+b^2+2b\sqrt{c}+c=60+20\sqrt{c}[/tex]

Since a,b,c are all natural we have that:
[tex]a^2+b^2+2c=60[/tex]
[tex]2(a+b)\sqrt{c}=20\sqrt{c}[/tex] or [tex]a+b=10[/tex]

Now we use the fact [tex]a^2+b^2 < 60[/tex] to produce the [tex](a,b,c)[/tex] triples:
[tex] (3,7,1), (4,6,4), (5,5,5), (6,4,4), (7,3,1)[/tex]

ps - the tex code for [tex]\sqrt{c}[/tex] is "\sqrt{c}"

[quote="tom11784"]ps - the tex code for [tex]\sqrt{c}[/tex] is "\sqrt{c}"[/quote]

I know, just forgot to close the dollar signs. :)

[quote="tom11784"][tex](a+\sqrt{c})^2+(b+\sqrt{c})^2=60+20\sqrt{c}[/tex]
[tex]a^2+2a\sqrt{c}+c+b^2+2b\sqrt{c}+c=60+20\sqrt{c}[/tex]

Since a,b,c are all natural we have that:
[tex]a^2+b^2+2c=60[/tex]
[tex]2(a+b)\sqrt{c}=20\sqrt{c}[/tex] or [tex]a+b=10[/tex]

Now we use the fact [tex]a^2+b^2 < 60[/tex] to produce the [tex](a,b,c)[/tex] triples:
[tex] (3,7,1), (4,6,4), (5,5,5), (6,4,4), (7,3,1)[/tex][/quote]

Ever thought that $c$ can be a perfect square? :D

hmm... well here 0 is possible too for naturals so that adds two more couples
anyway a little ugly

[quote="tom11784"]ps - what $'s?[/quote]
same as [tex]


LaTeX
real analysis
real analysis unsolved

two parts:

prove the following for A,B being bounded sets of the real numbers.

part a: if A+B is defined as the set of all a + b such that a is in A and b is in B

then

LUB(A + B) = LUB(A) + LUB(B)

and the second part:

part b: if every element in A,B is positive, and

A*B is defined as the set of all a*b such that a is in A and b is in B

then

LUB(A*B) = LUB(A) * LUB(B)

And what have you tried?

By the way, for communicating in mathematics it's very helpful to know some [[LaTeX]] -- see that link and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

so for part a my thoughts were that you need to prove that
LUB(A+B)>= LUB(A) + LUB(B)
which is actually easy, I have that part. And then also prove,

LUB(A) + LUB(B) >= LUB(A+B)  in order to prove equality
I'm really at a loss for this part; you can't use the fact that the maximal element is the LUB, because the sets don't necessarily have a max element, they're only bounded.

let $ \sup A \plus{} B \equal{} u, \sup A \equal{} a, \sup B \equal{} b$

to show $ u\leq a \plus{} b$, just note that $ \forall x\in A,y\in B$, we have $ x \plus{} y\leq a \plus{} b$ then use the supremum property of $ u$.

i think the tricky part is $ u\geq a\plus{}b$, where some epsilon argument is used. Since you are ok with it, I ll not demonstrate.

As for part b, it is merely the same thing, by changing the addition to multiplication ...

sorry I guess I mixed it up...what I meant was that I got the direction you proved it, the easy one, but did not get the hard direction with epsilons.

we use the supremum property of $ a,b$.

We know that for any $ \epsilon>0$ there exist $ x\in A,y\in B$ such that

$ a\minus{}\epsilon<x$ and $ b\minus{}\epsilon<y$

So that $ a\plus{}b\minus{}2\epsilon<x\plus{}y\leq u$

Since we can make epsilon arbitrary small, we ll obtain $ a\plus{}b\leq u$


You're looking for a way to take out all of the i's from the denominator,
so multiplying by 4-2i will allow you to cancel out the i's since i^2=-1,
and since (a-b)(a+b)=a^2-b^2, it will take away the possibility of a linear
expression in the denominator containing i.

what's going on? what are you talking about? did you make a new thread instead of replying to one?

Yes, I believe he did. I'm guessing this should be closed, or merged with the other thread.


geometry
circumcircle
quadratics
trigonometry
geometric transformation
homothety
reflection

Let a square  $ABCD$  and a point $E$  on side $BC$   such that $CE = 2EB$  . The line $AE$   meets the circumcircle of the square at point $M$  . If the line $BM$  meets line $DC$  at $N$  , prove that  $AB = 2CN$ .

  [i]I think I have seen a trig.solution.A synthetic one would be better.[/i]
 [u]babis[/u]

Let $a=AB.$ Draw the perpendicular $MT$ from $M$ to $BC$. Also take the center of the square $O$ and draw the perpendicular $OP$ to $BC.$ If $MT=x$, then $\triangle MTE\sim\triangle ABE\implies TE=\frac{AB}{BE}\cdot MT={x\over 3}.$ Since $PB={a\over 2}$ and $BE={a\over 3}$, we have $PT=PB-BE-ET={a\over 6}-{x\over 3}.$ If we draw the perpendicular $MF$ to $OP$, we have $MF=PT$ and $PF=MT$, and Pythagoras gives $OM^2=MF^2+OF^2.$ Now

$OM=\frac{a\sqrt{2}}{2}$
$MF=PT={a\over 6}-{x\over 3}$
$OF=OP+PF=OP+MT={a\over 2}+x$

Substituting this into $OM^2=MF^2+OF^2$ we get

\begin{eqnarray*}{a^2\over 2}={a^2\over 36}-{ax\over 9}+{x^2\over 9}+{a^2\over 4}+ax+x^2 &\implies& \frac{10x^2}{9}+\frac{8ax}{9}-\frac{2a^2}{9}=0\\ &\implies& 5x^2+4ax-a^2=0\\ &\implies& 5x^2+5ax-ax-a^2=0\\ &\implies& (5x-a)(x+a)=0\end{eqnarray*}

Since the solution must be positive, we take $x={a\over 5}$

Now $BT=BE+ET={a\over 3}+{x\over 3}={a\over 3}+{a\over 15}={2a\over 5}$

Therefore $BT=2TM$, and since $\triangle BTM\sim\triangle BCN$, we get $BC=2CN$

PS. I'm not sure if solving a quadratic is "synthetic enough", but it sure seems simpler than a trig solution.

This a metric solution and of course synthetic. There are many nice problems which are solved only metric. For example  THALES THEOREM ,  similarity , pythagorean theorem ,homothety , inversion , power of a point   ... , are metric aproaches.
Your solution is pure geometric and I' ll read it carefully.I doubt , if we can find a not metric solution , but in maths everything is .... possible.I' ll try it for a little too, just when I have some free time.

 Tomorrow we have our Olympiad! Silouan is a contestand. We'll see. I'll send the problems  as well as other  greek members in the olympiad !

  [u]babis[/u]

Let $CM \cap AB = L$. Since $\angle{ABE}=\angle{CBL}$, $\angle{BAE}=\angle{BCL}$, $AB=CB$, thus $\triangle{ABE} \cong \triangle{CBL}$, thus $LB=BE$. Therefore $AB=3BL$, $\frac{LA}{BA}= \frac{4}{3}$. In $\triangle{CLB}$ and line $MEA$, use Menelaus:
\[ \frac{CM}{ML}\cdot \frac{LA}{BA} \cdot \frac{BE}{EC}=1 \]
\[ \frac{CM}{ML}=\frac{3}{2} \]
Since $CN || BL$, so $\frac{CN}{BL}=\frac{3}{2}$. Because $BL=BE= \frac{1}{3} \cdot AB$, so $\frac{CN}{AB}=\frac{1}{2}$.

Note: A picture is added to my solution.

Dear Mathlinkers,
by considering the outer square AUVB of ABCD and the tangent to the circumcircle od ABCD, we applie the Pascal’s theorem (U, E, N collinear) and finish with the two similar triangles CNE and EUV….

Sincerely
Jean-Louis

Let $G$ be the centroid of $\triangle BCD$; easily $EG\bot AC$ and $ABEG$ is cyclic, hence $\angle EAG=\angle GBC$. From the cyclic $ABMC$ we get $\angle MBC=\angle MAC$, hence $\angle NBC=\angle GBC$; as $BG$ is a median of $\triangle BCD$, we infer that $N$ is the reflection of midpoint of $CD$ in $C$ and we are done.

Best regards,
sunken rock

$\begin{gathered}
  {\text{Let's}}\mathop {}\limits_{}^{} {\text{ look }}\mathop {}\limits_{}^{} {\text{at }}\mathop {}\limits_{} {\text{this}}\mathop {}\limits^{} {\text{, }}\mathop {}\nolimits_{} {\text{the }}\mathop {}\nolimits^{} {\text{inverse}}\mathop {}\nolimits^{} : \hfill \\
  \left. \begin{gathered}
  AB = BK\mathop {}\limits_{}  \hfill \\
  BZ = ZC \hfill \\
\end{gathered}  \right\}\mathop {}\limits_{}^{}  \Rightarrow \mathop {}\limits_{}^{} tr.KAD \approx tr.BCN\mathop {}\limits_{}^{}  \Rightarrow \mathop {}\limits_{}^{} BC = 2CN \hfill \\
  \mathop {}\limits_{}^{} and \hfill \\
  A\hat BM = A\hat MD = D\hat MC = C\hat MN = {45^o} \hfill \\
  In\mathop {}\limits_{}^{} tr.BZM\mathop {}\nolimits_{}^{} (th.\mathop {}\limits_{}^{} {\text{bisectors}})\mathop {}\limits_{}^{}  \Rightarrow  \hfill \\
  \frac{1}{2} = \frac{{CZ}}{{CB}} = \frac{{EZ}}{{EB}}\mathop {}\limits_{}^{}  \Rightarrow \mathop {}\nolimits_{}^{} EC = 2EB \hfill \\
\end{gathered} $


[quote="itazzirun"]let a, b any two positive integers, show that square root of 2 is always lies between a/b and (a+2b)/(a+b).[/quote]
Let's notice that $\frac{a+2b}{a+b}= 1+\frac{b}{a+b}= 1+\frac{1}{\frac{a}{b}+1}$, so setting $x=\frac{a}{b}$ we have to prove that $\sqrt{2}$ lies between $x$ and $1+\frac{1}{1+x}$. If $x\leqslant \sqrt{2}$, then $1+\frac{1}{1+x}\geqslant 1+\frac{1}{1+\sqrt{2}}= 1+\sqrt{2}-1= \sqrt{2}$. Analogously we prove that if $x\geqslant \sqrt{2}$, then $1+\frac{1}{1+x}\leqslant 1+\frac{1}{1+\sqrt{2}}= 1+\sqrt{2}-1= \sqrt{2}$ and we are done.


function
induction
real analysis
real analysis solved

let a_n be a sequence defined in this way:
a_1=1 and a_n= 1+an/2 if n is even and 1/a_(n-1) if n is odd.
a)find n if we know that a_n=19/99.
b)prove that the function f:N*->Q*+,f(n)=a_n is surjective.

okey... seems that nobody wants to solve this one
i have a nice solution but i used continuous fractions and this is not in the school program so i'm interested if someone know.. or have a proof without this...
i'll give now only some hints because i don't have enough time to write the entire solution.. (if someone really wants then i'll write it)

for n=2^p,p>=0 => a_n=a_(2^p)=1+a_(2^(p-1))=...=p-1+a_1=p for any natural p.
n=2^p+1=>a_n=1/p.
n=2^p+2=>a_n=1+a_(2^p+1)=1+1/p.
.....
Now we know that any number N can be write like this:
N=2^x_1+2^x_2+...+...+2^x_k, with (x_k) a strictly increasing sequence of natural numbers.
we denote x_k-x_(k-1)=d_k

and now we shall write N in function of d_i, i=1,..,k.

we should get a continuous fraction with d_i.
example:
n=2^3+2^2+1=>a_n=1/a_(n-1)=1/(2+a_(2+1))=1/(2+1/a_2)=1/(2+1/2).
n=2^5+2^4+2^2+1=>a_n=1/a_(n-1)=1/(2+a_(2^3+2^2+1))=1/(2+1/(2+a_(2+1)))=1/(2+1/(2+1/2))
(these are some simple examples but u can figure out how to write N in function of d_i,i=1,..,k and x_1).

enjoy!

oh! I enjoy incompletely  :( please post complete solution if possible :D
thanx

the solution
i hope it's ok... actually it's 2:30 AM and i'm very tired... hope i didn't make a mistake(or too many :D )

More general, let f,u,v  be  three functions  f:N*-->Q*+ , u:N*-->N*, v:N*-->N* such that f(u(n)) = 1+f(n) , f(v(n)) = 1/(1+f(n))
We prove that there is g(a/b) such that f(g(a/b))=a/b  by using induction on m = a+b .
For m=2 , so a/b = 1/1 = 1,  we have f(1)=1, so g(1)=1
Suppose that for each a/b  such that a,b are natural and a+b =< m,  there is g(a/b) such that f(g(a/b)) = a/b
Let a/b such that a,b are natural and  a+b= m+1,  :
a > b ===> a-b + b =< m ===> there is g((a/b-1) such that f(g(a/b-1)) = (a-b)/b ===> f(u(g(a/b -1))) = 1 + f(g(a/b-1)) = 1 + (a-b)/b = a/b ==> g(a/b) = u(g(a/b -1))
a < b ===> a + b-a =< m ===> there is g((b/a-1) such that f(g(b/a-1)) = (b-a)/a ===> f(v(g(b/a -1))) = 1/(1 + f(g(a/b-1))) =1/(1 + (b-a)/a) = a/b ==> g(a/b) = v(g(b/a -1))
....(q.e.d) ...


geometry
circumcircle
geometric transformation
homothety
ratio
reflection
geometry proposed

Let c be the circumcircle of the triangle ABC. If P and Q are two points of c collinear with the center of c (  :) I don't know how you say in english"diametralmente opposti") find the locus of the intersection of the lines of Simpson related to P and Q wrt ABC.

How fast can you say "the nine-point circle" ? :)

First of all, let's show that the two Simson lines are perpendicular. I won't write the whole proof here, just the main idea I used. If $P',Q'$ are the points where the perpendiculars from $P,Q$ to $BC$ cut the circumcircle for the second time, then $AP',AQ'$ are parallel to the Simson lines of $P,Q$ respectively, so the Simson lines are perpendicular, since we obviously have $AP'\perp AQ'$.

For the rest of the proof I'll use another well-known result: the Simson line of $P$ (or $Q$, of course), passes through the midpoint of $PH$ ($QH$), where $H$ is the orthocenter. Let $P'',Q''$ be the midpts of $PH,QH$. It's now clear that $P'',Q''$ are antipodes on the nine-point circle, since this circle is the image of the circumcircle through the homothety of center $H$ and ratio $\frac 12$, so the Simson lines are perpendicular and pass through two antipodes on the nine-point circle, meaning that their point of intersection lies on this circle.

We should also show that the entire circle is part of the locus, but I haven't done that yet.

[quote="grobber"]How fast can you say "the nine-point circle" ? :)

For the rest of the proof I'll use another well-known result: the Simson line of $P$ (or $Q$, of course), passes through the midpoint of $PH$ ($QH$), where $H$ is the orthocenter.

[/quote]

The same  way I did, substantially. The only difference being that this "well-known" fact was not at all known by me.

Infact I named that ("discovery"  :) )  LemmaRL151004.

Could you, please,  give one o more proof of that?

I am very sorry but more or less everything we are discussing in this thread is well-known...

If you can read Dutch, I could refer to http://www.pandd.demon.nl/simson2.htm . If you have H. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967, I would point at chapter 2, 7. In any case, the story is shortly mentioned at http://www.cut-the-knot.org/Curriculum/Geometry/CircumSimson.shtml .

  Darij

As I have already mentioned, we still have to prove that the entire nine-point circle is included in the locus. I haven't proven this rigorously, but I think that when $P$ moves on the circle in a direction (clockwise, for example), the point of intersection of the Simson lines moves on the nine-point circle in an opposite direction.

[quote="grobber"]As I have already mentioned, we still have to prove that the entire nine-point circle is included in the locus.[/quote]

You know, questions like this are often very aggravating, especially when tracking back the construction requires solving a 9th degree equation...  :D  So when I ask for loci I usually ignore this part...

But in the case of our problem, it's quite easy: It can be readily shown that if P and Q are any two diametrically opposite points on the circumcircle of triangle ABC, and if the Simson lines of these points P and Q intersect at J, then the Simson line of the point J [b]with respect to the medial triangle of triangle ABC[/b] is parallel to the line PQ. Even more is true: The line PQ is the Steiner line of the point J with respect to the medial triangle of triangle ABC. [The [i]Steiner line[/i] of a point which lies on the circumcircle of a triangle with respect to that triangle is defined as the line passing through the reflections of the point in the three sidelines of the triangle.] Hence, having an arbitrary point J on the nine-point circle, we can easily track back the line PQ with the property that the Simson lines of the points P and Q intersect at J.

  Darij


group theory
abstract algebra
superior algebra
superior algebra unsolved

I am having difficulty proving that the rationals under addition are not isomorphic to any proper subgroup of itself :(
Hints would be more appreciated, if possible. Thanks =)

Edit: Im thinking along lines of there will be some infinite sum of a proper subgroup (of infinite order) which will equal one of the "omitted" elements of the original. Is this along a good line?

Characterize all homomorphisms out of the rationals to another abelian group.

So I just found that $ \phi(\frac{a}{b}) \equal{} \frac{a}{b}\phi{(1)}$. Which... disproves it I believe. =)

Why do you believe that?  (We have $ r \cdot {\bf Q} \equal{} \bf Q$ for every $ r \in {\bf Q}\smallsetminus \{0\}$.)


induction
combinatorics unsolved
combinatorics

In a plane, consider a finite set of identical squares whose sides are parallel. For any k+1 squares, at last two of them contain points in common. Prove that this set can be divided into no more than 2k-1 subsets so that in each subset all squares have a common point !

Do you know solution?
Jury's solution span almost whole page of text.

It is time to raise this problem again!  ;)
Please, do not send official solution.  :P

Myth, what do you mean?
The official solution is quite short and easy:
Take the leftmost square $A$, then apply induction on $k$ for the squares not intersecting $A$!

You are right!
But one year ago, when I saw my archive, I discovered that official solution (which was given to all contestants after contest) are too long. Unfortunately I don't remember main idea.
After olympiad ending official solution can be changed, because it is highly possible that students invented more nice solution than author.

Thus we can move it to Solved section.


function
calculus
derivative
algebra unsolved
algebra

what is the smallest value that

$ \sqrt{49\plus{}a^{2}\minus{}7a\sqrt{2}}\plus{}\sqrt{a^{2}\plus{}b^{2}\minus{}ab\sqrt{2}}\plus{}\sqrt{50\plus{}b^{2}\minus{}10b}$

can have for positive real numbers a and b?

please help!

[quote="wailam010"]what is the smallest value that

$ \sqrt {49 + a^{2} - 7a\sqrt {2}} + \sqrt {a^{2} + b^{2} - ab\sqrt {2}} + \sqrt {50 + b^{2} - 10b}$

can have for positive real numbers a and b?

please help![/quote]
I hope this is a real problem given in olympiad or olympiad preparation and not just an invented one.
I hope, as a consequence, there is a clever solution for this problem. I did not find it  :blush: and used a heavy basic method :

Let $ f(x,y) = \sqrt {49 + x^2 - 7x\sqrt 2}$ $ + \sqrt {x^2 + y^2 - xy\sqrt 2} + \sqrt {50 + y^2 - 10y}$

$ f(x,y)$ is a positive function, defined over $ \mathbb R^2$ and whose derivative is defined over $ \mathbb R^2\backslash\{(0,0)\}$.
Since $ 49 + x^2 - 7x\sqrt 2 > (x - 7\frac {\sqrt 2}2)^2$ and $ 50 + y^2 - 10y > (y - 5)^2$, we get that $ f(x,y) > |x - 7\frac {\sqrt 2}2| + |y - 5|$

So $ f(x,y) > |x| + |y| - 5 - 7\frac {\sqrt 2}2$ and $ f(x,y)$ has an absolute minimum value somewhere in the interior of the square $ |x| + |y| = f(0,0) + 5 + 7\frac {\sqrt 2}2$

As a consequence, the absolute minimum is either at $ (0,0)$ (where derivatives are not defined), either at $ (u,v)$ such that $ f'_x(u,v) = f'_y(u,v) = 0$. let us now solve this equation :

1) $ f'_x(x,y) = 0$

$ \iff$ $ \frac {2x - 7\sqrt 2}{2\sqrt {49 + x^2 - 7x\sqrt 2}}$ $ + \frac {2x - y\sqrt 2}{2\sqrt {x^2 + y^2 - xy\sqrt 2}} = 0$

$ \iff$ $ (2x - y\sqrt 2)\sqrt {49 + x^2 - 7x\sqrt 2}$ $ + (2x - 7\sqrt 2)\sqrt {x^2 + y^2 - xy\sqrt 2}) = 0$

$ \iff$ $ (2x - y\sqrt 2)^2(49 + x^2 - 7x\sqrt 2)$ ${ = (2x - 7\sqrt 2)^2(x^2 + y^2 - xy\sqrt 2})$ and $ 2x - y\sqrt 2$ and $ 2x - 7\sqrt 2$ with opposite signs.

$ \iff$ (after simplifications) $ x(xy + 7x - 7y\sqrt 2)(y - 7) = 0$ and $ 2x - y\sqrt 2$ and $ 2x - 7\sqrt 2$ with opposite signs.

2) $ f'_y(x,y) = 0$

$ \iff$ $ \frac {2y - x\sqrt 2}{2\sqrt { x^2 +y^2- xy\sqrt 2}}$ $ + \frac {2y - 10}{2\sqrt {50 + y^2 - 10y}} = 0$

$ \iff$ $ (2y - 10)\sqrt {x^2 +y^2 - xy\sqrt 2}$ $ + (2y - x\sqrt 2)\sqrt {50 + y^2 - 10y} = 0$

$ \iff$ $ (2y - 10)^2(x^2 + y^2 - xy\sqrt 2)$ $ = (2y - x\sqrt 2)^2(50 + y^2 - 10y) = 0$ and $ 2y - 10$ and $ 2y - x\sqrt 2$ with opposite signs.

$ \iff$ (after simplifications) $ y(xy + 5y\sqrt 2 - 10x)(x - 5\sqrt 2) = 0$ and $ 2y - 10$ and $ 2y - x\sqrt 2$ with opposite signs.

The system
$ x(xy + 7x - 7y\sqrt 2)(y - 7) = 0$
$ y(xy + 5y\sqrt 2 - 10x)(x - 5\sqrt 2) = 0$

Gives quickly the set of solutions $ (0,0)$, $ (5\sqrt 2,7)$, $ (5\sqrt 2,\frac {35}2)$, ${ (\frac {35\sqrt 2}3},7)$ and ${ (\frac {35\sqrt 2}{17},\frac {35}{12}})$

The conditions "$ 2y - 10$ and $ 2y - x\sqrt 2$ with opposite signs" and "$ 2x - y\sqrt 2$ and $ 2x - 7\sqrt 2$ with opposite signs" imply then that the absolute minimum is either at $ (0,0)$, either at ${ (\frac {35\sqrt 2}{17},\frac {35}{12}})$.

And, since ${ f(0,0) = 7 + 5\sqrt 2 > 13 = f(\frac {35\sqrt 2}{17},\frac {35}{12}})$, we get that the minimum value of $ f(x,y)$ is $ 13$

[quote="nsato"][hide="big hint"]
[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/e4f129bf1be8687f547a79b007a33676.png[/img]
[/hide][/quote]

Mwaaaaahhhhhhaahhhaaa !  :lol:

What a giant hint !

Very very nice, indeed,
I'm quite ashame with my long ugly demo  :blush: (but at least my result is good since $ \sqrt {7^2 \plus{} (5\sqrt 2)^2 \plus{} 7\cdot 5\sqrt 2\cdot \sqrt 2} \equal{} 13$)

Congrats!

Actually, this problem has been posted about $ 5$ times, or even more often^^


calculus
integration
algebra proposed
algebra

IF WE HAVE (a,b)=(c,d)=1, a non equal b (mod 2) c non equal d (mod 2) and c^2+d^2 is a multiple of a^2+b^2 ten (c^2+d^2)/(a^2+b^2)=x^2+y^2 for some integers x and y.

If you consider that every prime of the shape 4k+1 can be writen as sum of two perfect squares this is obvius that is why i need a solution where don´t need that.

thanks

Why not just proving it for those primes¿
I don't see any approach not using harder methods...

SORRY BUT I TINK THAT IS A HARDER PROBLEM AND I TOOK A LOT OF TIME TO THE MATHEMATICS OF THE FERMAT GENERATION.

CAN YOU BE MORE ESPECIFIC ABOUT THE SOLUTION YOU GIVE

Shouting will not higher the number of people willing to answer...

You said:
[quote="Chaos_Angel"]If you consider that every prime of the shape 4k+1 can be writen as sum of two perfect squares this is obvius that is why i need a solution where don´t need that.[/quote]
so: why not just proving that¿ (it seems the easiest way)

i didn´t get mad about your topic but that is my opinion about the dificult of this problems, i am not trying to scream or anything and anyway that was not a helpful coment.

i am not trying to ofend you so don´t get mad.

THANKS FOR YOUR ANSWER

My post was just a question ([u]not[/u] a comment) cause I wanted to know why you can't use it (at least it's possible that it shall be shown by just using some specific theorems (e.g. known at this time); at least your posts aren't very specific about that).

I can offer:
- a proof that primes of type $4k+1$ are sum of two squares (which, as you said, gives the result).
- a proof that when integral $n$ can be written as sum of two rational squares, then also as sum of integral squares (which gives by considering $\frac{a^2+b^2}{c^2+d^2} = \left( \frac{ac-bd}{c^2+d^2} \right)^2 + \left( \frac{ad+bc}{c^2+d^2} \right)^2$ the result).