$ ABCD$ is a square. $ M$ and $ N$ are midpoints of $ \overline{AB}$ and $ \overline{BC}$, respectively. Find the ratio, expressed as a common fraction, of the area of $ \Delta BMN$ to the area of $ ABCD$.

My answer fo thsi was 1/8.  I went to review and it said the answer was [code]$\frac{1}{8}$[/code] whcih is just teh latexed way of what I put, but i still got it wrong..


Positive numbers $ x_1,x_2,...,x_n$ are such that

$ \frac{1}{1\plus{}x_1}\plus{}\frac{1}{1\plus{}x_2}\plus{}...\plus{}\frac{1}{1\plus{}x_n}\equal{}1$.

Prove that

$ \sqrt{x_1}\plus{}\sqrt{x_2}\plus{}...\plus{}\sqrt{x_n}\geq (n\minus{}1)(\frac{1}{\sqrt{x_1}}\plus{}\frac{1}{\sqrt{x_2}}\plus{}...\plus{}\frac{1}{\sqrt{x_n}})$

[quote="indybar"]Positive numbers $ x_1,x_2,...,x_n$ are such that

$ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_2} \plus{} ... \plus{} \frac {1}{1 \plus{} x_n} \equal{} 1$.

Prove that

$ \sqrt {x_1} \plus{} \sqrt {x_2} \plus{} ... \plus{} \sqrt {x_n}\geq (n \minus{} 1)(\frac {1}{\sqrt {x_1}} \plus{} \frac {1}{\sqrt {x_2}} \plus{} ... \plus{} \frac {1}{\sqrt {x_n}})$[/quote]

[hide="hint"]
Let $ a_i\equal{}\frac{1}{1\plus{}x_i}(\Leftrightarrow x_i\equal{}\frac{1\minus{}a_i}{a_i})$
and using this lemmas :

1)$ \sqrt{x_1\plus{}x_2\plus{}\cdots\plus{}x_n}\ge\frac{\sqrt{x_1}\plus{}\sqrt{x_2}\plus{}\cdots\plus{}\sqrt{x_n}}{\sqrt{n}}$

2)$ \frac{1}{\sqrt{x_1}}\plus{}\frac{1}{\sqrt{x_2}}\plus{}\cdots\plus{}\frac{1}{\sqrt{x_n}}\ge n\sqrt{\frac{n}{x_1\plus{}x_2\plus{}\cdots\plus{}x_n}}$
[/hide]


inequalities algebra polynomial integration logarithms topology geometry

Here are two classical inequalities concerning polynomials. Although these useful inequalities are very well-known, this post is addressed to high-school olympians to whom these problems would be a good exercise. The proofs are fully accessible to high-schoolers.

Problem. Let
\[
f(x) = x^n + a_{n-1}x^n  + \cdots + a_0 \in \mathbb{C}[x]
\]
be a monic polynomial with roots $\alpha_1,\ldots,\alpha_n$. Show that:

\[
2^{-n}(1+|a_0|+\cdots + |a_{n-1}|) \leq \prod_{1 \leq j \leq n} \max (|\alpha_j|,1)  \leq \sqrt{1 + |a_0|^2 + \cdots + |a_{n-1}|^2}.
\]

Well, I am not in high-school and sorry, but I could not find elementary solutions for this problem. It is true that the left inequality is pretty clear, but the second one is very tough for me. I will post my solution, which is unfortunately non-elementary. I am not sure about it and I don't know if it is known or not, but here it goes:
  First of all, we have $\prod_{j=1,n}{max(1,|x_j|)}=e^{\int_{0}^{1}{ln|f(e^{2i \pi t})|dt}}$.  I will let to the reader the pleasure of finding a proof for this relation. I am really not sure about my proof and in fact I am not sure about its veridicity, but something tells me it is true. Next, using Parseval's formula, we have $ 1+\sum_{i=0,n-1}{a_i^2}=\frac{\int_{-\pi}^{\pi}{|f(e^{it})|^2dt}}{2\pi}$. From Jensen's inequality we have $ e^{\int_{0}^{1}{ln|f(e^{2i\pi t})|dt}}\leq \int_{0}^{1}{|f(e^{2i\pi t})|dt}$ and from Cauchy's inequality we have $\frac{\int_{-\pi}^{\pi}{|f(e^{it})|^2dt}}{2\pi}\geq (\frac{\int_{-\pi}^{\pi}{|f(e^{it})|dt}}{2\pi})^2$. Combining these two inequalities and using a change of variable gives the result. Sorry for writing non-senses, I am not used with these integrals. It would be so nice to be true, but....

If that formula from my "solution" is true, than we get another nice result:
  $\sup_{|z|=1}{|f(z)|}\geq\prod_{j=1,n}{max(1,|x_j|)}$ where $x_j$ are the roots of f.
  This is stronger than the classical $\sup_{|z|=1}{|f(z)|}\geq 1$.

[quote="harazi"]It is true that the left inequality is pretty clear[/quote]
It took two minutes from me to realize that it is really pretty clear.  :)

Myth, but what is your opinion about the "solution" for the right inequality? Does it have any chance to be made good or is it doomed forever?

I think your solution is right, though I am too lazy to compute integrals.

Yes, the l.h.s. is totally obvious. And, yes, the relation $\prod \max \{1,|\alpha_j|\} = e^{\int_0^1 \ln |f(e^{2\pi i t})|}$ is true, and the solution is correct. This, however, is not to say that an elementary olympiad solution does not exist. Here is a solution accessible to high-school students:

For $h(x) = \sum {b_jx^j} \in \mathbb{C}[x]$, define $||h||:= \sqrt{\sum {|b_i|^2}}$ to be the square root of the sum of the square of the absolute value of its coefficients  ;) .  Consider the polynomial
\[
\widehat{f}(x) := \prod_{|\alpha_i| \geq 1} (x - \alpha_i) \prod_{|\alpha_i|<1} (\overline{\alpha_i}x - 1).
\]
It is an easy, straightforward exercise to check that $||f(x)|| = ||\widehat{f}(x)||$.
Now, the leading coefficient of $\widehat{f}$ is equal to $a:=\prod_{|\alpha_i|<1}\overline{\alpha_i}$, and the free term is equal to $A:=\prod_{|\alpha_i| \geq 1} \alpha_i$; therefore, the sum of the squares of the absolute values of the coefficients of $\widehat{f}$ is at least $|a|^2+|A|^2$. Thus,
\[
||f(x)||^2 = ||\widehat{f}(x)||^2 = |A|^2+|a|^2 \geq |A|^2,
\]
giving the desired inequality
\[
||f(x)||  \geq |A| = \prod_{|\alpha_i|>1} |\alpha_i|.
\]

It's not very difficult, is it?

Just unnecessary remark.
In great book "Polynomials" by Prasolov V.V. I found both harazi's and vess's approaches and many others interesing things!

Classical theorems are CLASSICAL theorems!

[quote="Myth"].
In great book "Polynomials" by Prasolov V.V. I found both harazi's and vess's approaches and many others interesing things!

[/quote]

Yes, it is a pretty good book (lucky me that I can read in Russian). An excellent book ... apart from the "proof" of Schur's theorem that $1+x\cdots+x^n/n!$ is irreducible, of course.  :P  :P  :P

Did you find this bug too?  :P  :P  :P
You know, I looked through new edition - mentioned "proof" is still unchanged.

And today I bought new interesting Prasolov's book "Elements of combinatorial and differential topology"!

Myth, do you know whether this book is also available in english? Or french?  ;)  This seems to be extraordinary if both you and vess say it is nice.

Myth, could you be so kind: hint me the page where is it in this book

@harazi. I am sure there are no translations to other languages.

@eugene. See "Chapter 17. Polynomial's height and other norms"

Myth

 Prasolov.Polynomial , no problem if it is written in russian I will manage to read it.
Is there a web page where it is available ?
----------------------------
It seems Prasolov has publish many good math problem books.
Can you tell us more ?

thank you

What do you mean when say available?

And mentioned books are not problem books. "Polynomials" contains huge amount facts about polynomials, and "Elements of combinatorial and differential topology" contains facts from topology.
I like these books, because all included theorems have nice readable proofs.

Both these books have edition in 1000 copy only.
See http://www.fizmatkniga.ru. Price is about 2*5 euro + 13 (shipment) = about 25 euro.

Thank you a lot Myth :D - after few months I eventually spotted this topic :lol:- it's beautiful coincidence cuz I was planning to buy books of Prasolov soon and now I can easily so thank you again for link to this bookstore  :)

P.S. As books are quite cheap so can you give my some titles of good books with olympiad stuff //inequalities [H-L-P "Inequalities" ?]


trigonometry
quadratics
algebra
system of equations
number theory unsolved
number theory

[quote="t0rajir0u"]I don't understand.  What do you mean by $5, 7$?[/quote]
I think in USA it means $5.7$
To t0rajir0u:  But I can't understand, what you want to say by your problem?

[quote="pavel kozlov"]But I can't understand, what you want to say by your problem?[/quote]
I guess he means: "Solve the following system of equations:
$\begin{cases} x+y+z=5.7\\ xyz=5.7 \end{cases}$
"

And of course it has infinite solutions  :huh:

Let $a = \tan \alpha, b = \tan \beta, c = \tan (\pi - \alpha - \beta)$.  A unique value of $\alpha$ generates a unique solution.

t0rajiru you are fantastic...can you explain your solution ?

$\alpha + \beta + \gamma = \pi \implies \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma$

But it doesn't even need to be that complicated.

$a + b + c = 5.7$
$abc = 5.7$
$c = 5.7 - (a + b)$
$ab(5.7 - (a + b)) = 5.7$

And if $a$ is fixed it can be solved as a quadratic for $b$, determining $c$.  Clearly infinite solutions.

(There must be another condition.  I think $a, b, c \in \mathbb{Q}$ might be it.)


A mixture of an acid anhydride A and mono basic acid B on heating produces another monobasic acid C of equivalent mass 74 and an anhydride D. The acid and anhydride remain in equilibrium. The anhydride D contains two identical fluoroalkyl groups. The acid B contains a trifluoromethyl grp and has an equivalent mass 128. Find A to D with proper reasoning.

Is this possible?  :maybe:
$ A \equal{} (CH_3CO)_2O$
$ B \equal{} CF_3CH_2COOH$
$ C \equal{} CH_3CH_2COOH$
$ D \equal{} (CF_3CH_2CO)_2O$


Well tomorrow is the SHSAT {Specialized high schools entrance exam mentioned in my other post a whilllle ago}. Any last minute tips? Experiences with tests with VERY slim chances that you overcame? Anything will help.  :)

This should be moved to FF  :wink: Good luck, anyway!

While we're on the subject of "luck" : Good Luck to all the participants in the JHMT tommorow!


geometry
rectangle
circumcircle
geometric transformation
reflection
homothety
vector

[quote="paul_mathematics"]Let $ABC$ be a triangle, and erect three rectangles $ABB_1A_2$, $BCC_1B_2$, $CAA_1C_2$ externally on its sides $AB$, $BC$, $CA$, respectively. Prove that the perpendicular bisectors of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent.[/quote]

I remember having seen various solutions of this nice problem. Here is the one from Arthur Engel, [i]Problem-Solving Strategies[/i], 1998, chapter 12.1, problem 3, which I think is the most beautiful one (in the following, I show a slightly modified version of this solution, but the essential idea is from Engel):

[b]Lemma 1.[/b] If ABCD is an arbitrary rectangle and P is a point in its plane, then $AP^2+CP^2=BP^2+DP^2$.

[i]Proof.[/i] Since the quadrilateral ABCD is a rectangle, its diagonals AC and BD are equal in length and intersect at a point M which bisects both of them. In other words, we have AC = BD, and the point M is the midpoint of both segments AC and BD.

Since the point M is the midpoint of the segment AC, the segment PM is a median in triangle APC, and thus, after the formula for the length of a median in a triangle, we have $4\cdot PM^2 = 2\cdot\left(AP^2+CP^2\right)-AC^2$. Thus, $4\cdot PM^2+AC^2 = 2\cdot\left(AP^2+CP^2\right)$. Similarly, $4\cdot PM^2+BD^2 = 2\cdot\left(BP^2+DP^2\right)$. Since AC = BD, we thus have $2\cdot\left(AP^2+CP^2\right)=2\cdot\left(BP^2+DP^2\right)$, thus $AP^2+CP^2=BP^2+DP^2$, and Lemma 1 is proven.

Note that Lemma 1 has a neat converse: If A, B, C, D are four points in the plane such that for any other point P in the plane, we have $AP^2+CP^2=BP^2+DP^2$, then the quadrilateral ABCD is a rectangle. You can prove this for yourself; I won't need this fact in my further solution.

Now let's solve the actual problem, to show that the perpendicular bisectors of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent, where the quadrilaterals $ABB_1A_2$, $BCC_1B_2$, $CAA_1C_2$ are arbitrary rectangles.

In fact, if the perpendicular bisectors of the segments $A_1A_2$ and $B_1B_2$ intersect each other at a point P, then $A_1P=A_2P$ and $B_1P=B_2P$ because of the main property of a perpendicular bisector. If we now succeed to prove that $C_1P=C_2P$, then it will follow that this point P also lies on the perpendicular bisector of the segment $C_1C_2$, so that the perpendicular bisectors of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ will concur at the point P, and the problem will be solved.

Hence, it is enough to prove $C_1P=C_2P$. Well, since the quadrilaterals $ABB_1A_2$, $BCC_1B_2$, $CAA_1C_2$ are rectangles, Lemma 1 yields

$AP^2+B_1P^2=BP^2+A_2P^2$;
$BP^2+C_1P^2=CP^2+B_2P^2$;
$CP^2+A_1P^2=AP^2+C_2P^2$.

Adding these equations together, we get

$AP^2+B_1P^2+BP^2+C_1P^2+CP^2+A_1P^2$
$=BP^2+A_2P^2+CP^2+B_2P^2+AP^2+C_2P^2$.

After subtraction of $AP^2+BP^2+CP^2$, this becomes

$B_1P^2+C_1P^2+A_1P^2=A_2P^2+B_2P^2+C_2P^2$.

Now, since $A_1P=A_2P$ and $B_1P=B_2P$, we thus must have $C_1P^2=C_2P^2$, so that $C_1P=C_2P$, and the problem is solved.

  Darij

I am trying to find geometrical solution.

Okay, Myth, if you want, here is a new proof for the new year :D . I hope that you needn't click on the hide tags...

We will use directed angles modulo 180 throughout the solution.

[b]Lemma 1.[/b] Let X, Y and Z be the circumcenters of triangles $AA_1A_2$, $BB_1B_2$ and $CC_1C_2$. Then, the triangles ABC and XYZ are homothetic.

[hide="Proof of Lemma 1"]
[i]Proof of Lemma 1.[/i] Since $ABB_1A_2$ is a rectangle, the perpendicular bisectors of the segments $AA_2$ and $BB_1$ are actually one and the same line g, and this line g is parallel to the line AB. Being the circumcenter of triangle $AA_1A_2$, the point X must lie on the perpendicular bisector of the segment $AA_2$, i. e. on this line g. On the other hand, being the circumcenter of triangle $BB_1B_2$, the point Y must lie on the perpendicular bisector of the segment $BB_1$, i. e. also on the line g. Hence, both points X and Y lie on the line g; this yields that the line XY coincides with the line g. But since g || AB, we thus have XY || AB. Similarly, YZ || BC and ZX || CA. Hence, the corresponding sidelines of triangles ABC and XYZ are parallel to each other; therefore, these triangles are homothetic. Lemma 1 is proven.[/hide]

[b]Lemma 2.[/b] The angle bisector of the angle $A_2AA_1$ is simultaneously the angle bisector of the angle CAB.

[hide="Proof of Lemma 2"]
[i]Proof of Lemma 2.[/i] Since $ABB_1A_2$ is a rectangle, we have $AA_2\perp AB$. Hence, we have $\measuredangle\left(AA_2;\;AB\right)=90^{\circ}$. Similarly, $AA_1\perp CA$ and thus $\measuredangle\left(CA;\;AA_1\right)=90^{\circ}$. Combining, we get $\measuredangle\left(AA_2;\;AB\right)=\measuredangle\left(CA;\;AA_1\right)$.

Now, if u is the angle bisector of the angle $A_2AA_1$, we have $\measuredangle\left(u;\;AA_2\right)=\measuredangle\left(AA_1;\;u\right)$; thus,

  $\measuredangle\left(u;\;AB\right)=\measuredangle\left(u;\;AA_2\right)+\measuredangle\left(AA_2;\;AB\right)$
  $=\measuredangle\left(AA_1;\;u\right)+\measuredangle\left(CA;\;AA_1\right)=\measuredangle\left(CA;\;u\right)$.

This shows that the line u is also the angle bisector of the angle CAB. And Lemma 2 is proven.[/hide]

[b]Lemma 3.[/b] If h is the altitude of triangle ABC issuing from the vertex A, and O is the circumcenter of triangle ABC, then the line h is the reflection of the line AO in the angle bisector of the angle CAB.

[hide="Proof of Lemma 3"]
[i]Proof of Lemma 3.[/i] Since the point O is the circumcenter of triangle ABC, the points A, B and C lie on a circle with center O, and thus, by the central angle theorem, we have < OAB = 90 - < BCA. In other words, < (AO; AB) = 90 - < (BC; CA). On the other hand, as an altitude in triangle ABC, the line h is perpendicular to the side BC, so that we have < (BC; h) = 90, and thus < (CA; h) = < (BC; h) - < (BC; CA) = 90 - < (BC; CA). Comparing this with < (AO; AB) = 90 - < (BC; CA), we see that < (CA; h) = < (AO; AB). If w is the angle bisector of the angle CAB, we have < (CA; w) = < (w; AB); thus,

  < (w; h) = < (CA; h) - < (CA; w) = < (AO; AB) - < (w; AB) = < (AO; w).

Since the line h passes through the point A, which is the point of intersection of the lines w and AO, this equation yields that the line h is the reflection of the line AO in the line w, i. e. in the angle bisector of the angle CAB. This proves Lemma 3.[/hide]

After Lemma 1, the triangles ABC and XYZ are homothetic; thus, the lines AX, BY and CZ concur at a point P.

[b]Lemma 4.[/b] The perpendicular from the point A to the line $A_1A_2$ is the reflection of the line AP in the angle bisector of the angle CAB.

[hide="Proof of Lemma 4"]
[i]Proof of Lemma 4.[/i] The perpendicular from the point A to the line $A_1A_2$ is the altitude of triangle $AA_1A_2$ issuing from the vertex A. The point X is the circumcenter of triangle $AA_1A_2$. Hence, applying Lemma 3 to the triangle $AA_1A_2$, we see that the perpendicular from the point A to the line $A_1A_2$ is the reflection of the line AX in the angle bisector of the angle $A_2AA_1$. But from Lemma 2, the angle bisector of the angle $A_2AA_1$ is simultaneously the angle bisector of the angle CAB; furthermore, the line AX coincides with the line AP. Hence, we can state that the perpendicular from the point A to the line $A_1A_2$ is the reflection of the line AP in the angle bisector of the angle CAB. This proves Lemma 4.[/hide]

[b]Lemma 5 (isogonal conjugates theorem).[/b] If T is an arbitrary point in the plane of a triangle ABC, then the reflections of the lines AT, BT, CT in the angle bisectors of the angles CAB, ABC, BCA concur at one point.

I won't prove this lemma since it is a well-known fact.

[b]Lemma 6.[/b] The perpendiculars from the points A, B, C to the lines $A_1A_2$, $B_1B_2$, $C_1C_2$, respectively, concur at one point.

[hide="Proof of Lemma 6"]
[i]Proof of Lemma 6.[/i] After Lemma 5, applied to the point T = P, the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA concur at one point. Now, after Lemma 4, the reflection of the line AP in the angle bisector of the angle CAB is the perpendicular from the point A to the line $A_1A_2$; similarly, the reflections of the lines BP and CP in the angle bisectors of the angles ABC and BCA are the perpendiculars from the points B and C to the lines $B_1B_2$ and $C_1C_2$. Hence, the perpendiculars from the points A, B, C to the lines $A_1A_2$, $B_1B_2$, $C_1C_2$, respectively, concur at one point, and Lemma 6 is proven.[/hide]

Now we can easily prove the problem statement that the perpendicular bisectors of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ concur.

[hide="Proof of Problem Statement"]
[i]Proof.[/i] After Lemma 6, the perpendiculars from the points A, B, C to the lines $A_1A_2$, $B_1B_2$, $C_1C_2$, respectively, concur at one point. Call this point U.

Since the triangles ABC and XYZ are homothetic, there exists a homothety mapping the triangle ABC to the triangle XYZ. Let V be the image of the point U under this homothety. Then, since a homothety maps any line to a parallel line, we have XV || AU. But since the point U lies on the perpendicular from the point A to the line $A_1A_2$, we have $AU\perp A_1A_2$. Thus, $XV\perp A_1A_2$, so that the point V lies on the perpendicular from the point X to the line $A_1A_2$. But the point X is the circumcenter of the triangle $AA_1A_2$ and therefore lies on the perpendicular bisector of the segment $A_1A_2$; hence, the perpendicular from the point X to the line $A_1A_2$ is nothing but the perpendicular bisector of the segment $A_1A_2$. So we have obtained that the point V lies on the perpendicular bisector of the segment $A_1A_2$. Similarly, the same point V lies on the perpendicular bisectors of the segments $B_1B_2$ and $C_1C_2$. And this proves that the perpendicular bisectors of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ concur.[/hide]

  Darij

[Moderator edit: The following solution is written only for the case when the rectangles $ABB_1A_2$, $BCC_1B_2$, $CAA_1C_2$ are squares; this is due to the fact that the problem was originally stated only for this case. However, you can modify this solution so that it works for the case of general rectangles as well: one has to define the point G not as the centroid of triangle ABC, but as the point where the perpendiculars from the points A, B, C to the lines $A_1A_2$, $B_1B_2$, $C_1C_2$, respectively, concur. This, however, requires first proving that these perpendiculars concur.]

Here's my solution in a nutshell, because I can't write long posts today (very bad hangover from last night; really :)).

By glueing together $AA_1A_2,BB_1B_2,CC_1C_2$ you can obtain another triangle with sides equal to $A_1A_2,B_1B_2,C_1C_1$. Call thir triangle $\Delta$.

Translate $G$ (the centroid of $ABC$) wrt vectors $\overrightarrow{BB_2},\overrightarrow{CC_2},\overrightarrow{AA_2}$ to get $A_3,B_3,C_3$ respectively.

It's not hard to see now that $A_3B_3C_3$ is congruent to $\Delta$, and the perpendicular bisectors of $B_3C_3,C_3A_3,A_3B_3$ are the same as those of $A_1A_2,B_1,B_2,C_1C_2$ respectively. Those perpendicular bisectors must thus be concurrent in the circumcenter of $A_3B_3C_3$.

By congruence of triangles perpendicular bisectors of $A_1A_2$ through by the midpoint of $BC$
Analogously $B_1B_2$ through by the midpoint of  $CA$ and $C_1C_2$ through by the midpoint of $AB$
Thus the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent in the barycenter of $ABC$ $\blacksquare$.


trigonometry
limit
calculus
calculus computations

I'm not sure how this series converges.

(n+2)/((n^3)+sin(n)) (from n=1 to infinity)

I tried using a 1/(n^3) to compare it with it.
Since I know this series converges, because p=3.

But it does not seem to be dominating the (n+2)/((n^3)+sin(n))
series. Can someone help me?

For $n$ large enough
$\frac{n+2}{n^3+\sin n}\leq \frac{2n}{n^3-1}\leq \frac{3}{n^2}$

Or you can try limit test:

$\lim_{n\to\infty} \frac{\frac{n+2}{n^3+\sin n}}{\frac{1}{n^2}}=1$

Hence the convergence of $\sum \frac{n+2}{n^3+\sin n}$ and $\sum \frac{1}{n^2}$ are the same.


Ummm ok there are more Naperville qualifiers than IMSA... something is just wrong with that... also dude on a not totally unrelated note AIME and USAMO qualification in general has gotten pretty messed up as of late... but ummm ok hopefully there will be some red MOPpers (and maybe blue MOPper?) from IL this year

yea...no tony liu? apparently he got owned on the aime I

[quote="Altheman"]Congrats to all who qualified from IL. Many of the names are familiar from ARML, etc.

Winston Luo 11 Glenbrook South HS Glenview IL
Michael Tarczon 10 Glenbrook South HS Glenview IL
Je-ok Choi 11 Illinois Math & Science Academy Aurora IL
Jennifer Iglesias 11 Illinois Math & Science Academy Aurora IL
Arjun Puranik 9 J B Conant HS Hoffman Estates IL
Kevin Hu 8 Kennedy Jr. High Naperville IL
Keshav Puranmalka 11 Naperville Central HS Naperville IL
Kate Rudolph 9 Naperville Central HS Naperville IL
Michael Nasti 12 Naperville North HS Naperville IL
Minseon Shin 10 Naperville North HS Naperville IL
Jeanette Wat 10 Naperville North HS Naperville IL
Benjamin Xie 12 Naperville North HS Naperville IL
John Binder 12 Oak Park & River Forest HS Oak Park IL
Daniel Turcza 12 Oak Park & River Forest HS Oak Park IL
David Woulfe 9 Oak Park & River Forest HS Oak Park IL
Alex Zhai 11 University Laboratory HS Urbana IL
Timothy Berry 11 Walter Payton College Prep Chicago IL
Cory Smith 12 Walter Payton College Prep Chicago IL
Gregory Gauthier 11 Wheaton North HS Wheaton IL
David Woulfe 9 Oak Park & River Forest HS Oak Park IL[/quote]
lol only 11 ppl from IN made it.

i wonder who is going to mop...i know zhai is...

Hi Winston Luo, are you Songjun Luo's son?

[quote="lindachang"]Hi Winton Luo,
May I please have your e-mail address, friends?
thanks
Linda[/quote]

Sure, email me at winluo2@yahoo.com


IF pennies are woth one cent, nickles are worth five cents, dimes are worth ten cents, and quarters are worth 25 cents, how many different combinations of these coins can be used to make 26 cents?

IS THERE anyway to solve this using equations instead of listing all the possiblities?

if you expand the generating function...

$(1+x^{25})(1+x^{10}+x^{20})(1+x^{5}+x^{10}+...+x^{26})(1+x+x^{2}+...+x^{25})$, and find the coefficient of $x^{26}$ in the expansion, then that is your answer

my calculator gives: 13 ways

how does the generating function thing work?

here's a similar problem..its a nice explanation i guess
[url]http://www.cut-the-knot.org/ctk/GeneratingFunctions.shtml[/url]

[quote="Altheman"]if you expand the generating function...

$(1+x^{25})(1+x^{10}+x^{20})(1+x^{5}+x^{10}+...+x^{25})(1+x+x^{2}+...+x^{25})$, and find the coefficient of $x^{26}$ in the expansion, then that is your answer

my calculator gives: 13 ways[/quote]

Shouldn't the last factor have an $x^{26}$ in it?

[quote="SM4RT"]ya..wow i didnt notice that..i think so..because its 26 cents..so that'd make sense..right :maybe:[/quote]

yea...i just typed it up wrong...

How did you make a program for this question?


graph theory
real analysis
real analysis solved

Prove that every sequence has either an ascending either a constant or either a descending sequence using graph teory!

I think that we can construct a graph with the elements of the sequence in this way...
we unit the first element with the second one... the third element will be unit with the first and the second one if he is between them(greater than one of them and smaller than the other one) or we unit it with the first one if it is smaller than it and if the first one is smaller than the second one, and else we unit it with the second ... we proceed like this with all the elements(just that we consider that we can(if we can) unit an element with the k-th and with the j-th element). In this way we obtain a graph with the property that there exists a chain of monotone numbers between any elements of the graph. Now to proove that we have a monotonous sequence we shall choose 2 element and we know that between them exists a chain of monotous elements.

is it correct?

Denote by Koo the complete graph with a infinite countable number of vertices. There is the following proposition:if we color the sides with r colors, there exists a complete infinite monochromatic subgraph.

Yes, this is the infinite Ramsey's theorem.

Pierre.


about positive numbers a,b,c,x,z,y we know that
1)a<b<c,a<x<y<z<c.
2)abc=xyz,a+b+c=x+y+z.
prove that a=x,b=y,c=z.
help me

Consider two functions
\[ f\left( t \right) \equal{} {t^3} \plus{} p{t^2} \plus{} {q_1}t \plus{} r\]
with roots a, b, c
\[ g\left( t \right) \equal{} {t^3} \plus{} p{t^2} \plus{} {q_2}t \plus{} r\]
with roots x, y, z
We have
\[ f\left( a \right) \minus{} g\left( a \right) > 0 \Rightarrow \left( {{q_1} \minus{} {q_2}} \right)a < 0\]

\[ f\left( c \right) \minus{} g\left( c \right) > 0 \Rightarrow \left( {{q_1} \minus{} {q_2}} \right)c > 0\]
This is a contradiction. The conclusion follows easily


What is the least positive integer divisible by each of the first
eight positive integers?

$ 8\times7\times3\times5 \equal{} \boxed{840}$


Hi,

I was just wondering how you will do a problem like this?

Suppose a standard fair dice is thrown. What is the probability of two consecutive sixes appearing for the first time after the nth roll? (i.e. a 6 on the n-1 roll and then another 6 on the n roll).
How does the answer change if we say any two consecutive numbers instead of two consecutive sixes?

Thanks in advance.

not sure if this is right....

[hide]((5/6)^n)(1/36)

since for the first n times, you must have anything but a 6, so its 5/6*5/6*5/6...and two consecutive sixes is 1/6*1/6.[/hide]

[quote="Klebian"]Hmm... but couldn't you have a 6 as long as you didn't get 2 in a row?[/quote]

Hi Walk Around the River,
Thanks for the suggestion but I agree with Klebian's point. I was having some trouble with that point too. Some of my thoughts on the problem follows:

[hide]Say, if we are just talking about the
Pr(win after n=2 throws) , it'd be 1/6*1/6
Pr(win after n=3 throws), 5/6*1/6*1/6
Pr (win after n=4 throws), 1*5/6*1/6*1/6
[the last two throws are fixed because of the demands of the problem, we can't have the 2nd throw come up 6 since that'll make  two 6s in a row after 3 throws. But if the 2nd throw is demanded to not show a 6, we can have anything appearing on the first throw, rite?]
Pr (win after n=5 throws), ?*?*5/6*1/6*1/6
[Here I got confused. It seems like either one of ? can be 1 as long as the other one is demanded to not show a 6 (i.e. put down as 5/6). But how does I express this as probability?]
Please help. I am trying to solve this problem by first looking at simple cases. Please tell me if this is not the right approach. Thanks.[/hide]

[quote="Klebian"]Hmm... but couldn't you have a 6 as long as you didn't get 2 in a row?[/quote]

good point. the language of the question wasnt very clear.  thats why i wasnt sure.


[quote="jacob8888888"]An ausdal bus from Austin to Dallas leaves every hour on the hour. An Ausdal bus from Dallas to Austin leaves every hour on the half-hour. The trip between the two cities takes three hours.How many Ausdal buses going in the opposite direction does an Ausdal bus from Austin to dallas pass?

Thanks[/quote]

[hide]
I made a diagram and I found it out to be three buses. One more thing: Can Ausdal be Greyhound? It's really annoying to keep on saying Ausdal bus, Ausdal bus in my head.
[/hide]

Please hide your methods for approaching the probelm.[hide]Well since there are already three buses on the other road coming his way. He will pass those. There will aslo be 3 more busses sent within the 3 hour period that he is on the road. I think that is the answer.[/hide]


linear algebra
matrix
search
complex numbers

How can I show that any trace 0 matrix can be written in the form AB-BA for some matrices A,B.

I think I am missing something obvious.

That's not the answer to the question the game asked. Of course the trace of $AB-BA$ is zero. But why can an arbitrary matrix of trace zero be written in that form?

Let, $AB=C=(c_{ij}), BA=D=(d_{ij})$
$\text{tr}(AB-BA)=\text{tr}(C-D)=\sum_{i}(c_{ii}-d_{ii})=\sum_{i}c_{ii}-\sum_{i}d_{ii}$
$=\text{tr}(C)-\text{tr}(D)=\text{tr}(AB)-\text{tr}(BA)$
$\therefore \text{tr}(AB-BA)=\text{tr}(AB)-\text{tr}(BA)$

Let, $A=(a_{ij}), B=(b_{ij})$
$\text{tr}(AB)=\sum_{i}(\sum_{s}a_{is}b_{si}) = \sum_{s}(\sum_{i}b_{si}a_{is})=\text{tr}(BA)$
$\therefore \text{tr}(AB)=\text{tr}(BA)$

Therefore, the following is formed.

[quote="Kouichi Nakagawa"]$\begin{array}{lll}\text{tr}(\text{AB}-\text{BA}) & = & \text{tr}(\text{AB})-\text{tr}(\text{BA}) \\ & = & \text{tr}(\text{AB})-\text{tr}(\text{AB})\\ & =& 0 \end{array}$[/quote]

Kouichi Nakagawa: have you understood the problem?

In fact it is an old result, discussed at
http://www.mathlinks.ro/Forum/viewtopic.php?t=25560
http://www.mathlinks.ro/Forum/viewtopic.php?t=1986
I haven't checked the solution in the second topic as I am still trying to solve the problem.

  darij

Sorry :(
I thought"Proof:tr(AB-BA)=0"

This Problem is Lie Ring?
$\mathfrak{sl}_{n}= \mathfrak{sl}(n;\mathbb{C}) = \{X\in\mathfrak{gl}_{n}\ ; \ Tr(X) = 0\}$

PS Had better two irrelevant articles delete it?

The proof referred to requires the complex numbers (an algebraically closed field could be substituted).
Is this true over an arbitrary field?

I thought of the following approach (over $\mathbb R$): consider the map $F\colon \mathbb R^{n\times n}\oplus \mathbb R^{n\times n}\to \mathbb R^{n\times n}$ defined by $F(A,B)=AB-BA$, and prove that its image is both open and closed in the set $\{A\in \mathbb R^{n\times n}\colon\mathrm{tr}(A)=0\}$. But I never got far.  :(

I want to make a collection of links ;)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=443239&highlight=convex#443239]Link 1[/url] (for the complex case)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=219655&highlight=basis#219655]Link 2[/url] (for the real case)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?&t=86676]Link 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?&t=89027]Link 4[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ens&t=101774]Link 5[/url] (a refinement)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104546]Link 6[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1473841900&t=113155]Link 7[/url]

to which I add the following (thanks to Darij)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25560]Link 8[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1986]Link 9[/url]

but also

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=147182]this thread[/url]

I know what you are all thinking... and you're right :(

Let $f : M(n;K) \rightarrow M(n;K)$be the map dedined by $f(M)=MP-PM$.
$\left(M \in M\in M(n;K) \ \text{be a nonzero matrix whose}\ Tr(M)=0, \ M(n;K)\ni P_{ij}=\left\{\begin{array}{ccc}0 & for & i \not=j \\ i & for & i=j \end{array}\right.\right)$
$f(M)_{ij}=(j-i)M_{ij}, \ f(M)=0 \iff M \ \text{is diagonal}$

Clearly $0 \in Im(f)$.
If $M \not= 0$ has trace $0$, there exists an invertible matrix $Q$ s.t. $Q^{-1}MQ \in Im(f)$.
So there is matrix $C$ s.t. $Q^{-1}MQ=CP-PC$.
Thus, $M=QCPQ^{-1}-QPCQ^{-1}=(QCQ^{-1})(QPQ^{-1})-(QDQ^{-1})(QCQ^{-1})$.
Let $A=QCQ^{-1}, \ B=QPQ^{-1}$.
So, $M=AB-BA$.
Conversely, every matrix in Im(f) is traceless.

For example,
Let $M=\begin{pmatrix}2 &-1 & 2 \\ 1 & 4 &-4 \\-2 & 4 &-6 \end{pmatrix}, \ P=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}.$
So, $Q=\begin{pmatrix}0 & 2 &-3 \\-6 &-5 & 12 \\ 0 &-6 & 12 \end{pmatrix}, \ C=\begin{pmatrix}a & 1 &-1 \\ 0 & b & 3\\ 1 &-1 & c \end{pmatrix}$.
Therefore, $A=QCQ^{-1}=\begin{pmatrix}11+4 b-3 c & \frac{1}{2}& \frac{11}{4}+b-c \\-41-2 a-10 b+12 c &-2+a &-9-\frac{3}{2}a-\frac{5}{2}b+4c \\-38-12 b+12c &-2 &-9-3 b+4 c \end{pmatrix}$,
$B=QPQ^{-1}=\begin{pmatrix}-1 & 0 &-1 \\ 14 & 1 & \frac{11}{2}\\ 12 & 0 & 6 \end{pmatrix}$

$(a, b, c \in K)$

[quote="darij grinberg"][quote="Kouichi Nakagawa"]If $M \not= 0$ has trace $0$, there exists an invertible matrix $Q$ s.t. $Q^{-1}MQ \in Im(f)$.[/quote]

Why?

  dg[/quote]
There exists an invertible matrix $D$ s.t. first column of $D^{-1}MD$ is $\begin{pmatrix}0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$.
Therefore, there exists an invertible matrix $Q$ s.t. all the diagonal entries of $Q^{-1}MQ$ are $0$.


The distance from Goteborg to Jonkiping on a map is 88 cm. The
scale on the map is 1 cm: 15 km. How far is it between the two city centers, in kilometers?

We can set up a proportion.

$ \frac{1}{15} \equal{} \frac{88}{x}$

$ x \equal{} 88 \times 15$

$ x \equal{} \boxed{1320}$

If it's 88cm on a map (big map!), then it is a total of $ 15\times 88\equal{}1320$ km between them in reality.


geometry
incenter
circumcircle
angle bisector
radical axis
geometry proposed

$ \triangle ABC$ is right-angled at $ A.$ $ D$ and $ E$ are the feet of the altitude and internal angle bisector issuing from $A.$ Let $ I_1$ and $ I_2$ be the incenters of $ \triangle ADB$ and $ \triangle ADC.$ Prove the following statements:

[i]1) [/i]If $ \odot(AI_1I_2)$ cuts the circumcircle of $ \triangle ABC$ again at $ A',$ then $ AA',BC, I_1I_2$ concur.

[i]2)[/i] Internal angle bisector of $ \angle DAE$ cuts $ BC$ and $ I_1I_2$ at $ K,P.$ Then $\frac {\overline{PK}}{\overline{PA}} \equal{} \sqrt {2} \minus{} 1$

From $ AI_1 \perp CI_2$ and $ AI_2 \perp BI_1$  we get $ I$, the incenter of $ \triangle ABC$ is the orthocenter of $ \triangle AI_1I_2$, hence $ I_1I_2 \perp AI$, $ I$ being the incenter of $ \triangle ABC$, consequently it's easy to see that $ I_1BCI_2$ is cyclic, so $ BC, I_1I_2$ and $ AA'$ concur at the radical center of the circles $ ABC, AA'BC$ and $ I_2I_1BC$, done.

Further more, see that, if $ F \equal{} BC \cap AI_1$ and $ G\equal{} BC \cap AI_2$, $ \triangle AI_1I_2 \sim  \triangle AGF$, therefore considering $ A'$ the common point of the circles $ (ABC)$ and $ (AFG)$, that would have made the point #1 little bit more difficult, but I thing this should reduce little bit the calculations for #2, i.e. seeing that, $ \frac{AP}{AK}\equal{}\frac{AI_1}{AG}$, but I am sure the proposer has an easier calculation method!

Best regards,
sunken rock

Thanks sunken rock. My proof to 1) is the same as your, so I'll post my solution for 2)

From $ \triangle ADC \sim \triangle BDA$ we have $ \frac {_{DI_1}}{^{DI_2}} \equal{} \frac {_{AB}}{^{AC}}$ $\Longrightarrow$ $ \triangle ABC$ and $ \triangle DI_1I_2$ are inversely similar. Thus if $ X,Y$ are the  projections of $ A$ on $ DI_2$ and $ I_1I_2,$ respectively, we have $ \angle I_1I_2D \equal{} \angle XAY \equal{} \angle C.$ Then

$ \angle XAI_2 \equal{} \angle CAI_2 \minus{} \angle CAX \equal{} \frac {_1}{^2}\angle B \minus{} (\angle B \minus{} 45^{\circ})$

$ \angle XAI_2 \equal{} \frac {_1}{^2}(\angle B \plus{} \angle C \minus{} \angle B) \equal{} \frac {_1}{^2}\angle C \Longrightarrow$ $ AI_2$ bisects $ \angle XAY$

Since $ \triangle AXD$ is isosceles right at vertex $ X,$ we get $ AY \equal{} AX \equal{} \frac {\sqrt {2}}{2}AD$

Let the perpendicular from $ K$ to $ \ell$ meet $ \ell$ and $ I_1I_2$ at $ M,N,$ respectively. Then

$ \angle MAK \equal{} 45^{\circ} \plus{} \angle C \plus{} \frac {_1}{^4}(\angle B \minus{} \angle C) \equal{} 90^{\circ} \minus{} \frac {_1}{^4}(\angle B \minus{} \angle C)$

$ \angle MAK \equal{} 90^{\circ} \minus{} (90^{\circ} \minus{} \angle AKD) \equal{} \angle AKD$

$ \Longrightarrow$ $ \ell,AK$ and $ BC$ bound an isosceles triangle with base $ AK.$ Hence, $ KM \equal{} AD$

$\frac {PK}{PA} \equal{} \frac {KN}{MN} \equal{} \frac {AD \minus{} AY}{AY} \equal{} \frac {AD \minus{} \frac {\sqrt {2}}{2}AD}{\frac {\sqrt {2}}{2}AD} \equal{} \sqrt {2} \minus{} 1.$


topology
calculus
real analysis
\/closed

Where would topology problem and questions go?

Use your judgment.

If it's a basic topology problem, you can probably put it in the calculus / real analysis forum, since most basic topology is useful in analysis.

If it's algebraic topology, you can probably put it in either algebra or advanced fields.

If it's general topology but advanced enough as to be pretty much inapplicable to analysis, it's probably safe to put it in advanced fields.

As far as I can tell, there's probably not yet enough discussion of topology in the College Playground to merit a whole subforum for it.


abstract algebra
group theory
superior algebra
superior algebra unsolved

As a public service, here are the questions. [b]nivinjill[/b], please let us know what you've done already and where you get stuck - folks here are (justifiably) reluctant to just do someone's homework for them. Both problems are fairly straightforward applications of the definitions.

1. Let $ R$ be a unitary ring and $ M$ an abelian group. Prove that $ M$ is an $ R$-module if and only if if there is a unit-preserving ring morphism $ f: R \to \mbox{End}(M)$.

2. Prove that the ring of endomorphisms of the abelian group $ \mathbb{Z}$ is isomorphic to the ring $ \mathbb{Z}$ and that the ring of endomorphism of the abelian group $ \mathbb{Q}$ is isomorphic to the ring $ \mathbb{Q}$.


[quote="duythuc_lqd"]Help me!!!
[url=http://img266.imageshack.us/my.php?image=97371736.png][img]http://img266.imageshack.us/img266/2874/97371736.png[/img][/url][/quote]

The general solution for second problem can be found at http://www.mathlinks.ro/viewtopic.php?p=1360482#1360482
Just applied $ m_1 \equal{}\minus{}1 , m_2\equal{}0, m_3\equal{}1$ to the lemma.

I've find the solution for the first problem...but the book was writen in Vietnamese ...
It is such a hard inequlity...

:blush: I have to post the solution by two times...

ok.I'm Vietnamese.And author of the inequality is Vo Quoc Ba Can.Thanks for answering me.


function
integration
limit
real analysis
advanced fields
advanced fields unsolved

[b]Suppose $ f$ is an integrable function such that $ f \geq 0$.    Then, the function

\[ F(x)\equal{}\int_{\minus{}\infty}^{x} f(t) dt\]

is continuous.

HINT: Use the Monotone Convergence Theorem. [/b]

I was reading a post here and the guy who proved this used the Dominated Convergence Theorem, but I was asked not to use this, but the Monotone Conv. Thm.

I have been thinking very hard about this exercise.  For example, I was able to see that if $ f$ is bounded, then the statement is certainly true, but I don't know how to generalize this (Perhaps I can use the fact  every function can be approximated by simple functions, which are bounded  :maybe: )

Thank you in advance

If $ (X,\mu)$ is a measure space and $ f$ is integrable then $ \nu(E) \equal{} \int_E \, f \,d\mu$ is a (complex) measure on $ X$. This can be proven, as you suggest by approximating with simple functions. Furthermore, this measure is absolutely continuous in the sense that $ |\mu|(E) \equal{} 0$ implies $ |\nu|(E) \equal{} 0$. By measure theory we then have:

1) If $ E_n$ are measurable sets with $ \lim E_n \equal{} E$ (in the sense of liminf and limsup) then $ \lim \nu(E_n) \equal{} \nu(E)$
2) For every $ \epsilon$ there is $ \delta$ such that $ |\mu|(E) < \delta$ implies $ |\nu|(E) < \epsilon$

The second property may also be proven directly by considering simple functions, again as you suggested.

It follows then that your function $ F$ is absolutely continuous, in particular continuous.

Proving these things is a first step towards the bounded convergence theorem.

Another approach to the bounded convergence theorem is via Fatou's lemma/monotone convergence theorem. In this case, let $ x_n$ be a "monotone" sequence tending towards $ x$. $ F(x) \minus{} F(x_n) \equal{} \int_{x_n}^x f(t) \, dt$. Let $ g_n \equal{} \chi_{(x_n,x)}f$ Then monotone convergence can be applied to $ f \minus{} g_n$. We get $ \lim F(x) \minus{} F(x_n) \equal{} \lim \int g_n \equal{} 0$. This type of argument can be extended to prove the bounded convergence theorem.

Wrote kind of too much I guess  :P But the point is whatever you do it amounts to bounded convergence, and also  that $ F$ is much more than continuous.

Thanks Kalle.  I am not familiar with measure spaces, so I cannot use the first part of your proof.   However, I read your explanation using the Monotone Convergence Theorem, but I don't see why you can pick just a monotone convergent sequence of real numbers.   In order to prove continuity, I should pick [b]any[/b] convergent sequence, and prove that the images of the points of the sequence converge to the image of the limit point.

Thank you in advance

It follows for any sequence from the monotone sequences. Do a proof by contradiction. If there is a sequence that converges but the sequence of images don't converge try to extract a monotone subsequence whose images don't converge

I think what you want to say is that in general it is enough to show that if for any monotone convergent sequence, the images converge then the function is continuous.  However, I don't see it and in fact I don't see how to get a contradiction in what you say.

Thanks Kalle

P.S.: at least, by drawing some sequences, it looks plausible what you say but I can't prove it  :wink:

If $ x_n \to x$ but $ f(x_n)$ does not converge to $ f(x)$ then, we can extract the two subsequences $ y_n$ and $ z_n$ of those $ x_n$ to the left of $ x$ and those to the right, respectively. Without loss of generality, suppose that $ f(y_n)$ does not converge to $ f(x)$. Then there is an $ \epsilon > 0$ and a subsequence $ a_n$ of $ y_n$ such that $ |f(a_n) \minus{} f(x)| \geq \epsilon$ for all $ n$. Of course, $ a_n \to x$ still.Then choose $ n_1 \equal{} 1$. Given $ n_k$ choose $ n_{k\plus{}1}$ such that $ n_{k\plus{}1} > n_k$ and $ a_{n_k} < a_{n_{k\plus{}1}}$

$ a_{n_k}$ is a subsequence of $ x_n$, strictly increasing. $ a_{n_k}$ converges to $ x$ of course, but  $ |f(a_{n_k}) \minus{} f(x)| \geq \epsilon$ for all $ k$ so it does not converge to $ f(x)$.


I thought up of the following proof, any comments?: No two matter (defined as something with mass and volume) can exist in the exact same place. Assume the empty space we consider to be empty is filled with matter. Then by the pigeonhole principle space is infinite.

Course there is the big problem of "Is empty space filled with something we can not observe by the science we have now?" but other than that it seems fine to me...

According to Quantum mechanics, the position of an observable is only defined in terms of probablity. Two pigeons may be at the same place at the same time. In theory at least

>> Assume the empty space we consider to be empty is filled with matter

(false => true) is true

whatever you're taking about (tautology). This is much used by lawyers (and by my girl-friend when she wants to have the last word [hide](i.e always)[/hide]  :mrgreen:)

[quote="Magnara"]Well, that's only if the pigeons are the size of electrons....[/quote]
It's regardless of size; it just becomes more uncertain when things are little.


integration calculus geometry parallelogram function algebra domain

Do you guys know another method to prove that $ \int_0^1 {\int_0^1 {\frac{{dx\,dy}}{{1 \minus{} xy}}} }  \equal{} \frac{{\pi ^2 }}{6}$ with out applying the fact $ \frac{1}{{1 \minus{} xy}} \equal{} \sum\limits_{k \equal{} 0}^\infty  {(xy)^k }$?

See Proof 1 in this paper by Robin Chapman, which contains 14 proofs of zeta(2):
[url]http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf[/url].

Yes, I've seen that solution, but I have several doubts:

[list][*]When setting $ (x,y) \equal{} (u \minus{} v,u \plus{} v)$ we require the Jacobian, isn't it?
[*]How do we know that $ S$ is the square with vertices $ (0,0),\,\left( {\frac {1}{2}, \minus{} \frac {1}{2}} \right),\,(1,0)$ and $ \left( {\frac {1}{2}, \frac {1}{2}} \right)$?
[*]Before reversing integration order, how was $ \zeta (2) \equal{} 2\iint_S {\frac {{du\,dv}}{{1 \minus{} u^2 \plus{} v^2 }}}$? (With integration limits.)
[*]By reversing integration order how he got $ \zeta (2) \equal{} 4\int_0^{1/2} {\int_0^u {\frac {{dv\,du}}{{1 \minus{} u^2 \plus{} v^2 }}} } \plus{} 4\int_{1/2}^1 {\int_0^{1 \minus{} u} {\frac {{dv\,du}}{{1 \minus{} u^2 \plus{} v^2 }}} }$?[/list]
Thanks.

Yes, that transformation has a Jacobian. The Jacobian is a constant; we multiply the integrand by $ 2$. Draw the picture- this is an affine transformation, so our square becomes a parallelogram, and vertices go to vertices. Knowing that this integral is $ \zeta(2)$ requires the power series, but we're not interested in that question right now. As for the integral calculations- work the limits out yourself, and note that we have even functions on symmetric domains in the inner integral.

[quote="Field"]Do you guys know another method to prove that $ \int_0^1 {\int_0^1 {\frac {{dx\,dy}}{{1 \minus{} xy}}} } \equal{} \frac {{\pi ^2 }}{6}$ with out applying the fact $ \frac {1}{{1 \minus{} xy}} \equal{} \sum\limits_{k \equal{} 0}^\infty {(xy)^k }$?[/quote]

Here is how to do the substitution. First let $ \bold{g}: \mathbb{R}^2\mapsto \mathbb{R}^2$ be a transformation. Then (without any conditions) $ \iint_D f(x,y) ~dA \equal{} \iint_{\bold{g}^{ \minus{} 1}(D)} f(\bold{g}(x,y)) \cdot | \text{J}(\bold{g})(x,y) | dA$. Where the set $ \bold{g}^{ \minus{} 1}(D) \equal{} \{ (x,y) | \bold{g}(x,y) \in D \}$.

Now, apply that formula above. We are given $ D$ to be the the unit square $ [0,1]\times [0,1]$. Let $ \bold{g}(x,y) \equal{} (x \plus{} y,x \minus{} y)$ and $ f(x,y) \equal{} 1/(1 \minus{} xy)$. Then $ f(\bold{g}(x,y)) \equal{} 1/(1 \minus{} x^2 \plus{} y^2)$. Now if $ (x,y) \in \bold{g}^{ \minus{} 1}(D)$ then it means $ \bold{g}(x,y) \in D \implies (x \plus{} y,x \minus{} y) \in [0,1]\times [0,1]$, i.e. $ 0\leq x \plus{} y \leq 1$ and $ 0\leq x \minus{} y\leq 1$, and this is a tilted square as jmerry was talking about. Now $ \text{J}(\bold{g}(x,y)) \equal{} 2$ by just computing the determinant, and the rest follows.


Given,

$ 4\cdot 10^n \equiv 1 p mod{13}$

How many values of "n" between $ 0 and 10^{2007}$ satisfy this equality?

I am more interested in the approach and to see if we can tackle all such problems the same way

$ 4\times10^n\equiv 40\times10^{n \minus{} 1}\equiv 10^{n \minus{} 1}\equiv 1\pmod{13}$

A bit of trial and error shows that $ 10^6\equiv 1\pmod{13}$, so $ 10^{6k}\equiv 1\pmod{13}$.
This means that $ n\minus{}1$ has to be a multiple of $ 6$, or $ n\equal{}6m\plus{}1$, for some integer $ m$.

$ 0 < n < 2007$
$ 0 < 6m \plus{} 1 < 2007$
$ \minus{} 1 < 6m < 2006$
$ \minus{} .166 < m < 334.333$
Since $ m$ is an integer, it can be $ 0,1,2,3...,334$. Thus, $ 335$ values of $ n$ satisfy the congruency.

4 x 10^n  = 1(mod 13)
40 x 10^(n-1) = 1(mod 13)
[13(3) + 1] x 10^(n-1) = 1(mod 13)
1 x 10^(n-1) = 1(mod 13)
10^(n-1) = 1(mod 13)

When you get to this point, is there any other way of solving other than trial and error?

If $ 10^{n \minus{} 1}\equiv1\pmod{13}$, then n=13 is a solution immediately from Fermat's Little Theorem. Any other values of n-1 have to divide 12, so you can use trial and error.
PS. Just look it up in the Wiki if you don't know what it is. Or search google.

actually i thinkthe inequality is $ 0<6m\plus{}1<\boxed{10^{2007}}$
making the answer $ \frac{10^{2007}\plus{}2}{6}$
otherwise, the approach is good

Wasn't the exponent already brought down when making the inequality?


group theory
combinatorics proposed
combinatorics

Let $S(n)$ be the set of permutation of the set $\{\ 1,2,...,n\}\ $.for every $\pi \in S(n)$. put$A(\pi)=\{\ \eta \in S(n) ,\eta \circ \pi=\pi \circ \eta \}\ $.let $\Phi(\pi)$ be the number of $A(\pi)$ find all values of $\Phi(\pi)$.

What do you mean by "the number of $A(\pi)$" ? Does it mean the number of elements of $A(\pi)$?

Consider the action by conjugation performed by $G=S_n$ on itself. We know that $|G|=|Gx|\cdot|Stab_G(x)|$ for any $x\in G$. What we want is to find the possible values for $|Stab_G(x)|$, and these are the possible values of $\displaystyle\frac {|G|}{|Gx|}=\frac{n!}{|Gx|}$. We are now looking for the possible values of $|Gx|$, that is the possible numbers of elements in a conjugacy class.

It is well-known (any group theory course which deals with permutations) that two permutations are conjugations of one another iff they have the same type of decomposition, i.e. iff when they are decomposed in disjoint cycles (every permutation has a unique such decomposition), they have the same numbers of cycles of length $k,\ \forall k\in\overline{1,n}$. This means that the number of elements in a conjugation class is number of the type $\frac {n!}{k_1!\ldots k_n!1^{k_1}\ldots n^{k_n}}$, where $k_i\in\mathbb N$ s.t. $\displaystyle\sum_{i=1}^n k_ii=n$. In turn, this tells us that the numbers we were initially looking for are $k_1!\ldots k_n!1^{k_1}\ldots n^{k_n}$ with $k_i\in\mathbb N$.

Is it correct (terribly sorry if it’s all nonsense :D)?


calculus
integration
logarithms
function
calculus computations

Evaluate $ \int_0^1 \frac{x\minus{}e^{2x}}{x^2\minus{}e^{2x}}dx$.

$ \ln \left(e^{2}\minus{}1 \right)^{\frac{1}{2}}$      am i right?

Right!

The problem was very easy! :blush:

YA, RIGHT.  on differentiating the  denominator , we get twice the numerator. :lol:

Could someone post the work for this? I'm in AP Cal and I'm really not sure how to solve this.

Edit: Googled a bit, and it appears to be an application of Leibniz integral rule. How can that be, though, since it's not a function in terms of f(x, y) ?

[quote="The Wicker Man"]$ (x^2 - e^{2x})' = 2(x - e^{2x})$, so

$ \int \frac {x - e^{2x}}{x^2 - e^{2x}}dx = \frac {1}{2}\int \frac {2(x - e^{2x})}{x^2 - e^{2x}}dx = \frac {1}{2}\ln{|{x^2 - e^{2x}|} + c}$.[/quote]

Thanks. Where does the dx term go though? If you say that $ y = x^2 - e^{2x}$, then $ \frac {1}{2}\int \frac {2(x - e^{2x})}{x^2 - e^{2x}}dx = \int{(dy / y)} dx$ and the dx is left over, you would need integration by parts right? Or am I making a fundamental mistake?

The latter: it's $ \int \frac{\frac{dy}{dx}}{y} \, dx$.

[quote="JBL"]The latter: it's $ \int \frac {\frac {dy}{dx}}{y} \, dx$.[/quote]

Yeah, when I woke up this morning I realized my error lol.


algebra
function
domain
combinatorics proposed
combinatorics

A plane is divided by n straight lines, no three are concurrent and no two are parallel.

Find the maximum and minimum number of angular domains in this configuration.

What is your definition of an angular domain?

Pierre.

I believe angular domain is an infinite section of the plane which is bounded by exactly 2 lines.. I hope I got the problem right..

Sorry about not being explicit.. :blush:

I meant : May the angular domain contain other regions of the division by the lines?

Pierre.


kk thats what i thought...thanks so much

[quote="ProtestanT"][quote="linzk3"]sqrt(x+6)=x-3 how can i find x? please and thanks[/quote]
[hide]
square both sides. $x\ge6$
$x+6=x^{2}-6x+9$
$x^{2}-7x+3=0$
$(x-\frac{7}{2})^{2}-\frac{37}{4}=0$
$x-\frac{7}{2}=+/-\frac{\sqrt{37}}{2}$
$x=\frac{7+/-\sqrt{37}}{2}$
$\frac{7-\sqrt{37}}{2}$ is less than 6, so it doesn't work
$x=\frac{7+\sqrt{37}}{2}$
test it in the original equation to see if it works
[/hide][/quote]

Your restriction on $x$ is off. For [i]real[/i] $x$ we want $x+6 \geq 0 \ \wedge \ x-3 \geq 0 \implies x \geq 3$. But the smaller solution will give a negative $\text{RHS}$, which is impossible for square roots on real values. (With the primary root.)

[quote="linzk3"]sqrt(x+6)=x-3 how can i find x? please and thanks[/quote]
[hide]$\sqrt{x+6}=x-3$

$x+6=x^{2}-6x+9$

$x^{2}-7x+3=0$

$x=\frac{7\pm\sqrt{49-12}}{2}$

The lesser value will make the square root negative, so...

$x=\frac{7+\sqrt{37}}{2}$[/hide]
@ ProtestanT - You can use $\pm$ for plus/minus sign.


There are three roots which are all real numbers of the equation $ x^{3} \plus{} px^{2} \plus{} qx \plus{} 1 \equal{} 0$
$ p > 0, q > 0$  Prove that $ pq \geq 9$.

Since all the coefficients of the polynomial are positive what can you say about the real roots?  Now call the roots $ \minus{} r_1, \minus{} r_2, \minus{} r_3$, and express both $ p$ and $ q$ in terms of them.  You should conclude that $ pq \equal{} 3 \plus{} \left(\frac {r_1}{r_2} \plus{} \frac {r_2}{r_1}\right) \plus{} \left(\frac {r_1}{r_3} \plus{} \frac {r_3}{r_1}\right) \plus{} \left(\frac {r_2}{r_3} \plus{} \frac {r_3}{r_2}\right)$ (hint: don't multiply things out right away, simplify first using $ r_1r_2r_3\equal{}1$).  When $ x$ is a positive number, what is the smallest $ x \plus{} \frac {1}{x}$ can be?

No, in my post above $ r_1r_2r_3 \equal{} 1$, because I defined the roots as $ \minus{} r_1, \minus{} r_2, \minus{} r_3$ to avoid sign difficulties in expressing $ p$ and $ q$.  It is true that the roots are negative, and as far as I can tell the rest of the argument is also correct.


A circle is centered at $A$ with a radius of $1$ and a circle centered at $B$ with a radius of $4$ are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is:

A. $\frac{1}{3}$

B. $\frac{2}{5}$

C. $\frac{5}{12}$

D. $\frac{4}{9}$

E. $\frac{1}{2}$

[img]7033[/img]

[hide="Hint"]
Lots of triangles and rectangles!!   :P  [/hide]

Let $C$ be the radius of third circle.

Sangaku problem 3 says:

\[\frac{1}{{\sqrt C }}= \frac{1}{{\sqrt A }}+\frac{1}{{\sqrt B }}\]

Choice [b]d)[/b].

Using that formula, I believe you miscalculated.  Plugging in 1 and 4, we have $\frac{1}{\sqrt{C}}=\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{1}}\Rightarrow \frac{1}{\sqrt{C}}=1+\frac{1}{2}\Rightarrow \frac{1}{\sqrt{C}}=\frac{3}{2}\Rightarrow \sqrt{C}=\frac{2}{3}\Rightarrow C=\frac{4}{9}$.

Why you say I mistook?, I've got the same answer.

here's a different way

[hide]Let $r$ denote the radius of the third circle.

Plot a point $D$ such that $\angle{ADB}=90^\circ$ and it is on the right of line $\overline{AB}$

Similarly Plot a point $E$ below point $A$ such that $\angle{AEC}=90^\circ$ and $F$ below point $B$ such that $\angle{BFC}=90^\circ$

We quickly see that:
$\overline{AB}=4+1=5$
$\overline{BD}=4-1=3$
$\overline{AD}=\sqrt{25-9}=4$

Also note that $\overline{EC}+\overline{CF}=\overline{AD}=4$

In $\triangle{AEC}$
$\overline{AE}=1-r$
$\overline{AC}=1+r$
$\overline{EC}=\sqrt{(1+r)^{2}-(1-r)^{2}}\implies 2\sqrt{r}$

In $\triangle{BFC}$
$\overline{BF}=4-r$
$\overline{BC}=4+r$
$\overline{CF}=\sqrt{(4+r)^{2}-(4-r)^{2}}\implies 4\sqrt{r}$

So,
$4\sqrt{r}+2\sqrt{r}=4\implies r=\boxed{\frac{4}{9}}$
[/hide]


[quote="goldendomer"]Hi everybody. Haven't posted for a LONG time. Here are two questions I need help in. Please give solutions.

14. $Find$ $the$ $simplest$ $common$ $fraction$
$equivalent$ $to$ $.0377777777777777777........$

22. $Express$ $the$ $sum$ $as$ $a$ $base$ $ten$ $numeral:$
11 base 2 + 11 base 3 + 11 base 4 +...+ 11 base 12.[/quote]

hints

14. that thing = $\frac{3}{100}+\frac{7}{900}$

22. calculate 11 base 2 then 11 base 3 then 11 base 4 then tell me if you see a pattern

Or notice that 11 base n = n+1.

For the first question, use infinite geometric series formula or multiply it by 10.

[quote="goldendomer"]Hi everybody. Haven't posted for a LONG time. Here are two questions I need help in. Please give solutions.

14. Find the simplest common fraction equivalent to $0.03777\dots$

22. Express the sum as a base ten numeral:
$11_{2}+11_{3}+11_{4}+...+11_{12}$.[/quote]

use the way specified in the mathcounts toolbox:

14 is $\frac{17}{450}$


Given three distinct unit circles, each of which is tangent to the other two, find the radii of the circles which are tangent to all three circles.

There's another one between the three circles, which its radius can easility be found using the cosine law.

[quote="seungjun"]Actually, the two circles' radii are $(2\surd3\pm3)/3$. Actually, the 2004 AMC10B problem #16 asks for the larger circle radius.[/quote]

I agree with your answer for both radii.  This was from 2001 MMPC and 2004 AMC10.

it makes since since the difference between the big and little should be twice the radius, or the diameter of the unit circles and that is satisfied.  I think using equilateral triangles is the most straightfoward way of solving this problem.  using the ratio of the sides you can easily find the $\sqrt{3}$

[quote="seungjun"]What is the MMPC?
And mna851, if you want to know how I got the larger radius, connect the three unit circles' radii to form an equilateral triangle. The large radius is (draw a diagram on your own) 1 + 2/3*(altitude of the equilateral triangle).[/quote]

MMPC is a Michigan Mathematics Prize Competition.

Actually, there's something called Descartes Circle Theorem:

Assume that the three mutually tangent circles have radius of $\frac{1}{a_{1}}$, $\frac{1}{a_{2}}$, and $\frac{1}{a_{3}}$.  There exists 2 circles that are tangent to the other three circles, one in the "interior" of the configuration, and one that encloses all of the original three  (outside).  Let the radius of the tangnet circle in question be $\frac{1}{a_{4}}$.

$2(a^{2}_{1}+a^{2}_{2}+a^{2}_{3}+a^{2}_{4})=(a_{1}+a_{2}+a_{3}+a_{4})^{2}$

Solving for $a_{4}$ yields 2 values, the smaller one being the smaller tangent and the larger one being the larger tangent circle.


TV screens are described by the lengths of their diagonals.
A 19'' TV has a rectangular screen with a diagonal length of
19 inches. The screen of a 20'' TV is 12 inches tall. How many
inches wide is the screen?

it said TVs are discribed by the lengths of their diagonals

therefore, we know that we have a right triangle with hypotenus 20 and one leg 12
thus, from our pythagorean triple, we know that the screen is $ \boxed{16}$ inches wide

We can use 3-4-5 triangles and so we see that 20=5times 4 so we are looking for 4times 4=16.


The number 2004 has 6 as the sum of its digits. How many positive integers betwenn 1 and 9999 has 6 as the sum of their digits?

[quote="Moonmathpi496"]Here is the solution

[hide="hint"]
actually its the number of integer solution the equation
$ x_{1}+x_{2}+x_{3}+x_{4}=6$ has
and the ans is $ 9C3=84$.
[/hide][/quote]

Your answer is correct, but can you explain why it is $ \binom{9}{3}$?

It's simple balls and urns, search it
Try [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904611533&t=145047]this[/url]

[quote="Ritty"]Can you explain why it is $ \binom{9}{3}$?[/quote]
[hide]We break up the $ 6$ into six "ones".
111111  Then we divide the ones into four piles using 3 dividers:
1|11|1|11  This represents 1212
||111111|  This represents 0060
Since we have $ 9$ places to put $ 3$ dividers, the answer is $ \binom{9}{3}$.[/hide]


In LaTeX:

$ A_1 \equal{} 1$
$ A_{n \plus{} 1} \equal{} 1 \plus{} \prod_{k \equal{} 1}^n A_k$

Find $ S \equal{} \sum_{k \equal{} 1}^{\infty} \frac {1}{A_k}$

[hide="Possible Start"]$ A_{n \plus{} 1} \equal{} 1 \plus{} \prod_{k \equal{} 1}^n A_k$
$ A_{n \plus{} 1} \equal{} 1 \plus{} A_n \prod_{k \equal{} 1}^{n \minus{} 1} A_k$
$ A_{n \plus{} 1} \equal{} 1 \plus{} A_n (A_n \minus{} 1) \equal{} A_n ^2 \minus{} A_n \plus{} 1$[/hide]

[hide]
Notice that from $ A_{n\plus{}1}\equal{}A_n^2\minus{}A_n\plus{}1$, we obtain $ A_{n\plus{}1}\minus{}1\equal{}A_n^2\minus{}A_n\equal{}A_n(A_n\minus{}1)$. This implies that $ \frac 1{A_{n\plus{}1}\minus{}1}\equal{}\frac 1{A_n(A_n\minus{}1)}\equal{}\frac 1{A_n\minus{}1}\minus{}\frac 1{A_n}$ by partial fraction decomposition. This rearranges to $ \frac 1{A_n}\equal{}\frac 1{A_n\minus{}1}\minus{}\frac 1{A_{n\plus{}1}\minus{}1}$.

The convenient thing about the above rearrangement is that we can now get this sum to telescope. We have:
\[ \frac 1{A_2}\plus{}\frac 1{A_3}\plus{}\dots\plus{}\frac 1{A_n}\equal{}\frac 1{A_2\minus{}1}\minus{}\frac 1{A_{n\plus{}1}\minus{}1}.\]
The reason we start at $ A_2$ instead of $ A_1$ is because the recursion does not hold for $ A_1$, so we have to treat it separately. This leads to
\[ \frac 1{A_1}\plus{}\frac 1{A_2}\plus{}\dots\plus{}\frac 1{A_n}\equal{}1\plus{}\frac 1{A_2\minus{}1}\minus{}\frac 1{A_{n\plus{}1}\minus{}1}.\]

Now the proof is straightforward. As $ n$ approaches infinity, the last term approaches $ 0$, so the infinite sum is $ 1\plus{}\frac 1{2\minus{}1}\equal{}2$.[/hide]


geometry
circumcircle
geometric transformation
reflection
parallelogram
ratio
analytic geometry

Let $\omega$ be the circumcircle of $\triangle{ABC}$. $P$ is an interior point of $\triangle{ABC}$. $A_{1}, B_{1}, C_{1}$ are the intersections of $AP, BP, CP$ respectively and $A_{2}, B_{2}, C_{2}$ are the symmetrical points of $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of side $BC, CA, AB$.
Show that the circumcircle of $\triangle{A_{2}B_{2}C_{2}}$ passes through the orthocentre of $\triangle{ABC}$.

I remember this problem was proposed in the TST of Sankt-Peterburg(Russia) by A.Smirnov two or tree years ago,I solved this problem with using complex numbers,I will post my solution,if noone find batter solution :) .

Lemma: Let $P,Q,R$ be points on the circumcircles of $BHC,CHA,AHB$. Then $H,P,Q,R$ are concyclic (or collinear) if and only if $\frac{BP}{CP} \frac{CQ}{AQ} \frac{BR}{AR}=-1$, where the ratio is taken positive if if e.g. $HBPC$ is convex and negative otherwise.

Proof: Invert wrt $H$ if $A,B,C,P,Q,R$ go to $A',B',C',P',Q',R'$ we must have $P',Q',R'$ collinear which by Menelaos is possible if an only if $\frac{B'P'}{C'P'} \frac{C'Q'}{A'Q'} \frac{B'R'}{A'R'}=-1$. However $\frac{B'P'}{C'P'}=\frac{BP}{CP} \frac{HB'}{HC'}$. By multiplying these relations we get the desired result. Note that $H$ is not necessarily the orthocenter.

The problem follows clearly from this Lemma, since from Hamilton Theorem the symmetric of circumcircle of $ABC$ wrt $BC$ is the circumcircle of $HBC$ and the needed product of ratios is $1$ by Trigonometric Ceva Theorem.

look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=114907, there was part of a TST peru, too.

[quote="Cezar Lupu"]Yes, Tiks, i'm interrested in your solution with complex numbers. I remeber that I have created a similar problem for RMNO, but this problem is nicer. :)[/quote]
Only for you dear Cezar!  :wink:

Let that the circumcircle of the triangle $ABC$ is a unit one.

And let's take complex plane with center in the circumcentr.

I will write the small letter $x$ as a complex coordinate of the a point $X$.
We have that $A_{1}\in{(O)}$ and $P\in{AA_{1}}$=> \[a_{1}\overline{a_{1}}=1\] and \[\frac{a-p}{\overline{a}-\overline{p}}=\frac{a-a_{1}}{\overline{a}-\overline{a_{1}}}=-aa_{1}\] => \[a_{1}=\frac{a-p}{a\overline{p}-1}\] Now as we have that $A_{2}$ is simetric of $A_{1}$ with respect to $BC$ then $a_{1}+a_{2}=b+c$=> \[a_{2}=b+c+\frac{p-a}{a\overline{p}-1}\] Similary \[b_{2}=a+c+\frac{p-b}{b\overline{p}-1}\]  \[c_{2}=a+b+\frac{p-c}{c\overline{p}-1}\] But we also know that $h=a+b+c$(H is the orthocenter)
So finally we just need to check that $\lambda$ is a real number,that is \[\lambda=\overline{\lambda}\] where \[\lambda=\frac{(a_{2}-c_{2})(b_{2}-h)}{(b_{2}-c_{2})(a_{2}-h)}\] :wink:

[quote="Tiks"]So finally we just need to check that $ \lambda$ is a real number,that is
\[ \lambda \equal{} \overline{\lambda}
\]
where
\[ \lambda \equal{} \frac {(a_{2} \minus{} c_{2})(b_{2} \minus{} h)}{(b_{2} \minus{} c_{2})(a_{2} \minus{} h)}
\]
:wink:[/quote]
Could you or somebody else explain this step in details?

[b]Lemma[/b]: Let $W$, $X$, $Y$, and $Z$, be any points. Then if the reflection of $Y$ across the midpoint of $WZ$ is reflected across the midpoint of $XZ$, the point obtained is the same as the reflection of $W$ across the midpoint of $XY$.
[i]Proof[/i]: Let $w, x, y, z$ be the vectors corresponding to $W$, $X$, $Y$, and $Z$. For any vectors $x$, $p$, and $q$, the reflection of $x$ across the midpoint of $p$ and $q$ is given by $p+q-x$, since $\frac{(p+q-x)+x}{2} = \frac{p+q}{2}$. Hence, the reflection of $Y$ across the midpoint of $WZ$ is $w+z-y$, and the reflection of this point across the midpoint of $XZ$ is $x+z-(w+z-y) = x+y-w$, which is indeed the reflection of $W$ across the midpoint of $X$ and $Y$.

Let $M_a$ be the midpoint of $AA_1$, $M_b$ be the midpoint of $BB_1$, $M_c$ be the midpoint of $CC_1$, and let $O$ be the circumcenter of $\triangle ABC$. $M_a$, $M_b$, and $M_c$ lie on the circle with diameter $OP$. Let the homothety centered at $A$ with factor 2 map $M_a$ to $R_a$, $M_b$ to $R_c$, and $O$ to $A'$. $M_a$ is mapped to $A_1$, so we can conclude that $R_b R_c A_1 A'$ is cyclic. If we reflect these points across the midpoint of $BC$, we find that $A_1$ is mapped to $A_2$, $A'$ is mapped to $H$ (a result of trivial angle chasing), and by the lemma, $R_b$ is mapped to $B_2$ and $R_c$ is mapped to $B_3$, so $A_2 B_2 C_2 H$ is indeed cyclic.

[quote="treegoner"]Let me first rewrite the problem

[i] Problem [/i] Let $ABC$ be a triangle and $A_1B_1C_1$ be the circumcevian triangle of a point $P$ wrt $ABC$. Let $A', B',C'$ be the reflections of $A_1, B_1, C_1$ through $BC, CA, AB$ respectively. Then the circumcircle of $A'B'C'$ passes through $H$.

[i] Proof [/i]

Let $A_2B_2C_2$ be the circumcevian triangle of the isogonal conjugate $P'$ of $P$ wrt $ABC$. Let $D, E, F$ be the midpoints of $BC, CA, AB$. Call $P'_a, P'_b, P'_c$ the reflections of $A, B, C$ through $P$.

Let $K$ be the reflection of $P'_a$ through $D$. Then $BP'_aCK$ is a parallelogram. Since $ABP'_aP'_b$ and $ACP'_aP'_c$ are parallelograms, it follows that $AP'_bCK$ and $AP'_cBK$ are parallelograms. This implies that $K$ is also the reflection of $P'_b$ through $E$, of $P'_c$ through $F$.

Under the reflection through the midpoint $D$ of $BC$, $H$ is mapped the antipode $A_3$ of $A$ wrt circumcircle of $ABC$, $A'$ is mapped to $A_2$, $P'_a$ is mapped to $K$. Thus, $\measuredangle{HA'K} = \measuredangle{A_3A_2P'_a} = \measuredangle{A_3A_2A} = 90^0$. This yields that $A'$ lies on the circle with diameter $HK$.

By similar argument , we obtain $B', C'$ also lie on the circle with diameter $HK$. Therefore, the circumcircle of $A'B'C'$ passes through $H$. And we are done.[/quote]

CAN SOMEBODY EXPLAIN WHY $\measuredangle{A_3A_2P'_a} = \measuredangle{A_3A_2A}$?

I have nice approach.

I restate the problem like this, using Ceva:

Let $\omega$ be a circle through $H$, the orthocenter of $ABC$. Let $\Omega$ be the circumcircle of $ABC$, and let $\Omega_A, \Omega_B, \Omega_C$ be the reflections of $\Omega$ through sides $BC, CA, AB$, respectively, with centers $O_A, O_B, O_C$, respectively. Let $\omega$ intersect these three circles in $H_A,H_B,H_C$, respectively. Prove that:

$\left( \frac{BH_A}{H_AC} \right) \left( \frac{CH_B}{H_BA} \right) \left( \frac{AH_C}{H_CB} \right) = 1$.

PROOF: Let $O$ be the center of $\omega$. I prove that

$\displaystyle\prod_{cyc} \left( \frac{BH_A}{BH} \right) \left( \frac{CH}{CH_A} \right) = 1$.

I use Sine Law in triangles $BHH_A$, $CHH_A$. Let $\alpha(a,c)$ be the sine of the angle formed by side $a$ (namely $BC$) with line $OO_C$, and define $\alpha(b,a), \alpha(a,b)$ and so on similarly. Then the problem reduces to proving

$\alpha(a,b) \alpha(b,c) \alpha(c,a) = \alpha(b,a) \alpha(c,b) \alpha(a,c) $

Let $X$ be a point such that $XA || OO_A$. Notice that $\frac{ \alpha(b,a)}{ \alpha(c,a)} = \frac{ sinCAX}{sin XAB}$.

Therefore, by trigonometric Ceva, the problem reduces to proving that the parallel lines to $OO_A,OO_B,OO_C$ through $A,B,C$ (respectively) concur.

But this is trivial since triangles $ABC$ and $O_AO_BO_C$ are homothetic (the center is the 9-point center). :)

Dear Mathlinkers,
this is the P-circle of Hagge.
For a synthetic proof and more,
http://perso.orange.fr/jl.ayme  vol. 5 le P-cercle de Hagge.
Sincerely
Jean-Louis

This is a very nice generalization of Fuhrmann circle

Let $ H $ be the orthocenter of $ \triangle ABC $ .
Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ .
Let $ A_2, B_2, C_2 $ be the reflection of $ A_1, B_1, C_1 $ in $ BC, CA, AB $ .
Let  $ A_3=AH \cap (A_2B_2C_2), B_3=BH \cap (A_2B_2C_2), C_3=CH \cap (A_2B_2C_2) $ .

Then
(1) $ A_2, B_2, C_2, H $ are concyclic ( $ P $-Hagge circle ) .
(2) $ \triangle ABC \cap \triangle A_1B_1C_1 \cap P $ ~ $ \triangle A_3B_3C_3 \cap \triangle A_2B_2C_2 \cap P $ .

( See [url=http://www.artofproblemsolving.com/community/c6t319f6h624925_some_properties_of_hagge_circle]Some properties of Hagge circle[/url] )

After inversion about $H$ with radius $1$, it would suffice to show that
\[
\prod \frac{A_2'B'}{A_2'C'}=1
\]
Now $A_2'B'=\frac{A_2B}{HA_2\cdot HB}$, but all factors $HA,HB,HC,HA_2,HB_2,HC_2$ cancel out in the product. So it suffices to show
\[
\prod \frac{A_2B}{A_2C}=1
\]
But of course $\frac{A_2B}{A_2C}=\frac{A_1B}{A_1C}$, and $\prod \frac{A_1B}{A_1C}=1$ iff $AA_1,BB_1,CC_1$ concur. So we are done.

i am sorry for reviving this thread .
[hide=solution using dilation]Notice $ABC , AA_1A_2,BB_1B_2,CC_1C_2$ have the same Centroid ( $G$ ).
Now consider a dilation with center $G$ and scale factor $-1/2$.
Notice that the dilation sends $A_2$ to $A'$ where $A'$ is the midpoint $ AA_1$ ,
$B_2$ to $B'$ where $B'$ is the midpoint $ BB_1$ and $C_2$ to $C'$ where $C'$ is the midpoint $ CC_1$
Also notice that $H$ is sent $O $( the circumcenter of $ABC$ ).
So it remains to show that $A'B'C'O $ is concyclic.
to show the above statement we just need to notice that $OA' \perp AA_1$ $OB' \perp BB_1$ and $OC' \perp CC_1$
Or $\angle OA' P = \angle OB' P = \angle OC' P = 90 ^{\circ}$ , which shows that A'B'C'O is concyclic.

[/hide]
P.S.this actually proves the existence of Hagge Circle.(if the solution is not wrong )


trigonometry
calculus
calculus computations

hello there :D

show

$ \sum_{n\equal{}1} ^\infty \; \frac{\theta}{2^n}\;\tan\; \frac{\theta}{2^n} \;\equal{}\; \boxed{1 \minus{} \theta \;\cot\;\theta}$

$ \displaystyle{\sum_{r\equal{}1}^n \frac{1}{2^r} \tan \frac{x}{2^r} \equal{} \frac{1}{2^n} \cot \frac{x}{2^n} \minus{} \cot x}$ :D


search
AoPSwiki
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AMC 8

Well there's old AMC 12's at http://www.math.ksu.edu/main/handbook/ProblemSets/AMC12Problems. I think if you navigate more around there you can find old AMC 8s

You know, never mind.

Thanks #H34N1, the Kansas site works...
Saved me a lot of typing on my part... Bookmark time

Repetitive link
http://www.math.ksu.edu/main/handbook/ProblemSets

I had a similar problem to isabella2916. Thanks so much! this is a great website :)

Yes, we are cheating AMC people out of their money, and yes, I do feel guilty about it. Just like getting all the past AMC's from my coach and storing them in my backpack... Just like going on AoPSWiki and doing past AIME's... I feel so guilty.

I have that one too! :D Very helpful resource :)

Only problem is, i've done all of the AMC's so i remember questions if i take them again.  But i think you can buy all the test back to 1980 something.

There is also a new CD available that is all of the AMC8 (and its predecessor the AJHSME) since the contest inception.  I don't believe it is listed on our website, as our web developer has been extraordinarily busy with some other projects this summer.  If you email amcinfo@maa.org, you can request more information.  The advantage of purchasing directly from us is you know it will be accurate since there have been no transcriptions.  AoPS and the KSU site do have our permission for reproduction as we are close partners with them.

There are plans for similar AMC10/12 and AIME/USAMO disks, but no definite timeline has been determined for their production.

Elizabeth Claassen
AMC Senior Accounting Clerk

[quote="solafidefarms"]The AMC people gave permission to AoPS people to post AMCs/AIMEs/USAMOs, thus you needn't feel too guilty :).[/quote]

if this is so, why are there no AMC 8 problems in the resources section??? i'm not trying to be a jerk, i am just curious why that is so.

[quote="mathemonster"]youve done all the AIMES?[/quote]

No just the AMC's.  Like 5 aimes.

http://mathjunk.googlepages.com/amc8

Someone gave it to me.

[quote="AIME15"][quote="isabella2296"]I tried to find some old AMC 8's to practice, but after a thorough web search, I can't find anything but a few scattered problems. Does anyone have one they could private message me/post here? Thanks in advance![/quote]
On the AMC website you can purchase previous AMC's.  I bought a CD with all the AMC 8, AMC 10, AMC 12, AIME, and USAMO tests from 2001-2007.  It also had the MOSP tests for 2007.[/quote]
what is the link to buy that cd and how much did it cost?

Don't revive.

Order [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/Pub,Revision.pdf]here[/url].

Number 51.  They changed it, so you buy all the amc 8's w/o the other stuff.

My mom bought them on a disk.  Not sure where she got them but it's a great way to practice!!
So if your willing to buy them that's an option!

Yeah the disk with all AMC 8's from 1985-1999 came yesterday!

Also AHSME's from 1950-1999


Let $A_{1},...A_{n}$ be $n$ distinct (2 by 2) complex numbers. For $i<n$ let
$U_{i}=\frac{1}{A_{i+1}-A_{i}}$ and $U_{n}=\frac{1}{A_{1}-A_{n}}$. Show the following equivalence:
for all $j$ in $[1,...,n-2]$ is $\sum_{i=1}^{n}U_{i}^{j}=0$
iff
$A_{1},...,A_{n}$ are (perhaps in disorder) the  vertices of a regular n-gone.

Rust, really, read the problems more carefully; and check your answers twice!

If $A_{k}=A+aexp(\frac{2\pi k}{n}),k=1,2,..,n$, were A and a complex const it is true. Because in this case $U_{k}=cq^{k}, \ c=\frac{1}{a(1-q)}, \ q=exp(-\frac{2\pi }{n}).$
It is obviosly: \[\sum_{k=1}^{n}U_{k}^{j}=c^{j}\sum_{k=1}^{n}q^{jk}=0\] for $j=1,2,...,n-1.$

Hi Rust,
your post relates only to the part:" ii) implies i) " of the proposed equivalence. Moreover your answer is very partial because, as said in the question, the points A1,...,An are not necessarily successive vertices of the regular n-gone.

For $i)\to ii)$ it is nesessary $\sum_{k}U_{k}^{j}=0 \ \forall j=1,2,...,n-1.$
For example let let $Uj$ are roots of equation $x^{n}=ax+b$, then
$\sigma_{j}(U_{i})=0, j=1,2,...,n-2$, therefore $s_{j}(U_{i})=0,j=1,2,...,n-2$.
But $A_{i}=A_{1}+\sum_{j=1}^{i-1}U_{j}^{-1}$ don't belongs to regular n-gone, when $a\not =0$.

Hi Rust,
you are a good mathematician but you don't work carefully.
What is the a's value?


i'm just testing the percent sign because i can't get the expression x% to work in the TeXer.  please respond to this post (as defined by a reply with helpful suggestions) iff the percent sign does not show up below..

$x\%$

don't you see the symbol? here, it _does_ show op

yes, but I wondered why you posted it then :?

intelligence tests are not for in this section. please post them in the correct places. this is for people who need help

and if you just wanted to do the test, there is something like a "Preview" button when writing a post ;)

[quote="nr1337"][quote="farmertim"]to test your powers of deductive reasoning.  3712 people passed.[/quote]

How did you get that number?[/quote]
I think he got 3712 from the number of current registered users (3713) minus 1 user (Peter VDD).

"iff" means "if and only if", right... I'm just a getting started person

Edit: I'm assuming this is right, since no ones answering

[quote="farmertim"]i'm just testing the percent sign because i can't get the expression x% to work in the TeXer.
[/quote]

If you type what you typed there (x\%) into the texer, you should get

$x\%
$

But if you type [quote="farmertim"]the expression x%[/quote]

$x%
$

That's what he did :)

But instead of using the test button (to see it only for yourself) he apparently rather posted it publicly on the forums :)

oh well perhaps it saves someone else the work to test it :D


ARML
trigonometry
geometry
AMC
USA(J)MO
USAMO
group theory

okay, so how did everybody do?

i got a 5, which is awesome cause i was expecting something lower
here's my problem breakdown:
1:  trivial, got it in about a minute, but its a #1 so its okay
2:  same as above, but its a #2 so thats okay, maybe it shoulda been a bit harder
3:  i liked this problem, there were many ways to do it, i did this thing with substitution and the difference of squares, it was fun
4:  good problem, i used the law of cosines to find the big side length and the equation worked out nicely.  however, I used the fact that 90+60=120.  so i got it wrong.  which sucks.  bleargh
5:  this one was okay, i got $1\le x<1.5$ and with about 30 seconds left realized that maybe it could be higher, but i was finishing up 6, so i didnt get it.  i still havent worked through it yet, but it doesn't seem extremely hard to get, but maybe kinda computational
6:  this was a good problem, it required some problem solving skillz
7:  easy, but thats okay considering how hard #8 was.  i personally solved it in 3 ways, one of which was assuming all points were equal :D
8:  no idea

1.triv
2.triv
3.um this wasn't that good of a problem. straightforward expanding plus use the given
4.my solution was cool. using law of cosines is like bashing. 3 triangles on the side are of area ab/4, given says, after splitting some areas, that 2 of those thin ones=other one+33 so solve easily
5. obviously 0<x<2pi -> 0<2x<2pi so i got it wrong
6. i almost solved it, but im like oh nos im will not do casework even though i only needed to consider like 2-3 cases and i iddnt get it
7. dot product its like that one problem we did in practice. @eric: assuming points are equal isn't "solving" it =P
8. um what?

anyways, ARML was awesomely fun regardless of how bad we failed
thanks to all you guys for making this an awesome arml

We'll do better next year, and even better the year after, etc.
Our team was really young this year, which is probably the reason why we sucked. Lots of juniors and seniors on A team = win, looking at past years' results.

EDIT: [b][u][size=150]OMG.[/size][/u][/b]

I just realized we got the same score as last year on the individual and team rounds. Just the relay and the power went to hell:
Individual: 74 > 74
Relay: 28 > 14
Team: 28 > 28
Power: 39 > 21
Total: 169 > 137
So yeah.

Well this year's individual was probably a bit easier than last year's individual so bad (especially since tons of B and C team members beat the A team). The team was pretty bad - we should have *at least* had 8, if not 9 *cough* Vishal *cough* #4 *cough*. The relay is supposedly okay because Silas said usually it's about this level, last year's was much easier (he said that most years you needed 10+ for a respectable score, but last year you needed 20+). The power round definitely went to hell.

I think that one problem is that for team selection, we mostly concentrate on individuals - not teamwork or proof solving/writing. I mean, a couple people (not saying who) didn't really do that much on the team and power rounds. We could sacrifice a couple individual points for not burning up on those rounds.

Also, our relay teams were definitely stacked. It was nice and all that one team got an 8, but come on, two zeroes? We could have broken up that team and put one or two strong individuals on each team. For example I was a #1 because I was with two better people but I could have been put on #2 or #3 on another team and put someone else on the #1 slot.

We also didn't have enough practices and we didn't really practice "correctly", in my opinion. First, team selection took up almost half of all the practices and we needed more where we already knew the teams. And we definitely needed more relay practice.

Just my two cents... but no one has to take heed because I'm one of the worst on the team.

EDIT: Yay 900th post!

[quote="Ubemaya"]
I think that one problem is that for team selection, we mostly concentrate on individuals - not teamwork or proof solving/writing. I mean, a couple people (not saying who) didn't really do that much on the team and power rounds. We could sacrifice a couple individual points for not burning up on those rounds.
[/quote]
yea, proof writing skills are good and we focus only on indivs
having all the usamo qualifies or usamo high scorers or some other proof contest high scorers automatically into A might help

[quote]We also didn't have enough practices and we didn't really practice "correctly", in my opinion. First, team selection took up almost half of all the practices and we needed more where we already knew the teams. And we definitely needed more relay practice.
[/quote]
also, we'll probably we better off if we did ARML problems instead of the easier NYSMLs.
we were all like "whoa we aced the NYSMLs, yay!"
but got owned at ARMLs

[quote="Ubemaya"]Except my relay team.[/quote]
and um what the hell is your point?

[quote="Ubemaya"]To contradict you.[/quote]
except it did NOT contradict me
by "we", i meant we as the whole 15 person team
SADLY YOU 3 DO NOT MAKE UP THE ENTIRE TEAM
SORRY ALBERT
SORRY

[quote="Ubemaya"]Exactly... the whole 15 person team did *not* fail because a subgroup of it did *not* fail at one particular thing. Actually since all three of us got a 6 on the individual we didn't fail in general.[/quote]
19th place is not failing?
anyway i was talking about practice armls so shut up now please

[quote="bookaholic"][quote]19th place is not failing?[/quote]
Well, if you were competing with teams from Exeter and places like that...[/quote]
WELL, BUT THE A TEAM LAST YEAR JUST SO HAPPENED TO GET 6TH PLACE OR SOMETHING LAST YEAR
HMMMMMMMMMM
MAYBE ITS JUST THAT WE SUCK
i lost

You'll do better next year, don't worry  :D
Just trying to be optimistic.

On the topic of next year, can I join the ARML team next year?

Comment for whoever likes rating my posts 1s:
If you think my 900th post is downright spammy enough to be a 1, you have issues.

Then again I might be wrong, can I have a general consensus on this?

I rated a 5 so that it doesn't get classified as spamzorz.

Please don't rate this post lol...

Dude Patrick you'd better join ARML next year to make up for tards like me on the team.

Although I'm not quite sure if I want to join... I think I might be kicked out. :(

About the relay team selection:

When I was competing, we essentially made up our own relay teams for each practice.  The coaches forced us to change teams every couple of questions so that we tried out a variety of configurations.  Then they would crunch the numbers from practice and put us into teams.  Usually, this resulted in one stacked team (like ours this year) as well as another surprisingly good team that came together randomly during practice.  This didn't seem to come at the cost of three weak teams; the others would be decent.  I decided to try to adapt this approach to our A team.  That's why I had you guys make your own teams and switch up a lot.

Then last week, something fell into place and all four teams came out really well, just like I had hoped for.  I figured we should just keep those, rather than trying to fix what wasn't broken.  The stacked team happened to consist of the three people who weren't there.

My original strategy was to take four of the top students and put them in 2nd and 3rd on two teams, and then pair them with two of the younger students in 1st, so the stacked team probably would have ended up together anyway.

What the heck, who got an 8?

[hide="I lost "]Me get 5.[/hide]

I got my ARML book from the USAMTS on Saturday...
lol, that sucks.

I might join the team next year, but you'd probably be better off w/o me... I really suck. :(

I'll be lucky to be allowed to go on C team, I think... :(

I think I might join the fabled Z team everyone seemed so excited to be on the start of this thing...

 :D


inequalities
geometry
similar triangles
combinatorics proposed
combinatorics

For those who like challenges, here is a truly amazing problem formulated by the Russian mathematician V. Proizvolov:

[b]Problem. [/b] Consider a convex $n$-gon $\mathcal{P}=P_1P_2 \ldots P_n$ and a set $A=\{A_1,A_2,\ldots,A_n\}$ of $n$ points in the plane. Prove that there exists a permutation $\pi$ of $1,\ldots,n$ such that the triangles $P_iA_{\pi (i)}P_{i+1}$ completely cover the interior of $\mathcal{P}$. In other words, one may assign a point to each side of $\mathcal{P}$ such that the resulting $n$ triangles form a cover of $\mathcal{P}$.

Enjoy!

--Vesselin

Oh, It's really hard problem, I think.

Let $\Gamma_1, \ldots, \Gamma_n$ be the sides of our polygon $\mathcal{P}$. Clearly we have to find a permutation $\pi=(i_1,\ldots,i_k)$ of the set $\{1,\ldots,n\}$ such that the triangles $A_k\Gamma_{i_k}$ cover our polygon $\mathcal{P}$.

Let $h_{ij}$ be the length of an altitude from $A_i$ to $\Gamma_j$.

For every permutation $\pi$ let us consider the quality $F(\pi)=\prod_{i=1}^{n}{h_{i\pi(i)}}$ of permutation $\pi$.

And now we are going to show that if permutation $\pi$ is not desired (that is the triangles $A_i\Gamma_{\pi(i)}$ don't cover $\mathcal{P}$), then we can find a permutation $\pi'$ with better quality (that is $F(\pi')>F(\pi)$).

If we are done with this, then we may consider "the best" permutation (with maximal quality) and it will be desired. Indeed, if it is not desired, we may choose a permutation with better quality --- contradiction.

So, we are going to learn how to improve the quality of a permutation.

Clearly, it is sufficient to improve the quality of the identity permutation (since the enumeration of $\Gamma_i$ was arbitrary). So we have triangles $A_i\Gamma_i$ and at least one point $A$ inside $\mathcal{P}$ which is not covered by any triangle $A_i\Gamma_i$. Clearly there is a side $\Gamma_{i_1}$ such that the triangle $A_1\Gamma_{i_1}$ contains $A$. Then there is a side $\Gamma_{i_2}$ such that the triangle $A_{i_1}\Gamma_{i_2}$ contains $A$. Continue the process like this until we obtain a triangle $A_{i_k}\Gamma_{i_p}$, where $p<k$. Then we will do the following. The first $p$ triangles we will replace back (that is to $A_i\Gamma_i$) and the remaining triangles $A_{i_p}\Gamma_{i_{p+1}}$, ..., $A_{i_{k-1}}\Gamma_{i_k}$, $A_{i_k}\Gamma_{i_p}$ we will keep.

So we have a permutation $\sigma=(i_p \to i_{p+1} \to \ldots \to i_{k} \to i_{p})$.

[b]Claim.[/b] The quality of $\sigma$ is better than the quality of the identity permutation $\epsilon$, that is $F(\sigma)>F(\epsilon)$.

To prove this Claim we need a simple lemma.

[b]Lemma.[/b] Let M and A be arbitrary poits inside a convex polygon with sides $\Gamma_1$, ..., $\Gamma_n$ such that the triangle $M\Gamma_1$ contains the point $A$. Let $h_1$, ..., $h_n$ be M-altitudes of the triangles $M\Gamma_1$, ..., $M\Gamma_n$ and $r_1$, ..., $r_n$ be altitudes of the triangles $A\Gamma_1$, ..., $A\Gamma_n$. Then for every $i=2,3,\ldots,n$ the inequality $\frac{h_1}{r_1}>\frac{h_i}{r_i}$ holds. (See the picture.)

[i]Proof of the Lemma.[/i] Since $A \in \triangle{M\Gamma_1}$ then the triangle $M\Gamma_i$ doesn't contain the point $A$. Hence the ray $\overrightarrow{MA}$ meets $\Gamma_1$ and doesn't meet $\Gamma_i$ (see the picture). Let $B_1$ be the point of intersection $\overrightarrow{MA}$ with $\Gamma_1$; and $M_1$ and $A_1$, $M_i$ and $A_i$ be the feet of perpendiculars from $M$ and $A$ to $\Gamma_1$ and $\Gamma_i$ respectively. Since $\triangle{MM_1B_1}$ similar to $\triangle{AA_1B_1}$ we have $\frac{h_1}{r_1}=\frac{MB_1}{AB_1}=1+\frac{MA}{AB_1}$. If $\overrightarrow{MA} \cap l_i = \emptyset$ (where $l_i$ is a line containing the side $\Gamma_i$) then there are two cases: 1) $\overrightarrow{MA} \parallel \Gamma_i$ or 2) $\overrightarrow{AM}$ meets $l_i$. In the first case $\frac{h_i}{r_i}=1$ and we are done. And in the second case we have $\frac{h_i}{r_i}<1$ and we are also done (since $\frac{h_1}{r_1}>1$). And if $\overrightarrow{MA} \cap l_i \neq \emptyset$ then let $B_i=\overrightarrow{MA} \cap l_i$. It is easy to see that $AB_i>AB_1$. Then from the pair of similar triangles $\triangle{MM_iB_i}$ and $\triangle{AA_iB_i}$ we have $\frac{h_i}{r_i}=\frac{MB_i}{AB_i}=1+\frac{MA}{AB_i}$. Therefore $\frac{h_i}{r_i}=1+\frac{MA}{AB_i}<1+\frac{MA}{AB_1}=\frac{h_1}{r_1}$ and we are done. The lemma is proven.

[i]Proof of the Claim.[/i] Since the triangles $A_{i_p}\Gamma_{i_{p+1}}$, ..., $A_{i_k}\Gamma_{i_p}$ contain $A$ and the triangles $A_p\Gamma_{i_p}$, ..., $A_k\Gamma_{i_{k}}$ don't contain $A$, we may apply our Lemma and write the inequalities $\frac{h_{i_pi_{p+1}}}{r_{i_{p+1}}(A)} > \frac{h_{i_pi_p}}{r_{i_p}(A)}$, $\frac{h_{i_{p+1}i_{p+2}}}{r_{i_{p+2}}(A)} > \frac{h_{i_{p+1}i_{p+1}}}{r_{i_{p+1}}(A)}$, ..., $\frac{h_{i_ki_p}}{r_{i_p}(A)} > \frac{h_{i_ki_k}}{r_{i_k}(A)}$. Multiplying all this inequalities and additing the same vaules to left and right sides (for other sides of our polygon) we will have $\frac{F(\sigma)}{\prod_{k=1}^{n}{r_{i_k}(A)}} > \frac{F(\epsilon)}{\prod_{k=1}^{n}{r_{i_k}(A)}}$. And since the denominators are the same, we finally obtain $F(\sigma)>F(\epsilon)$. The Claim is completely proven, and the problem is solved.

[b]Finally observation.[/b] Using the Lemma we essentially use the fact that all $A_i$ are inside of $\mathcal{P}$. But if it is not true, then we may consider not all permutations but anly the set of permutations $\Omega$ containing only the permutations $\pi$ which have minimal triangles $A_i\Gamma_{\pi(i)}$ not-intersecting $\mathcal{P}$ at all. The remaining is easy.


solve in $ N$ the equation $ (1+a)(1+b)+(1+c)=2abc$

There shouldn't be anything amazing about this problem.  Note that $ c=\frac{(a+1)(b+1)+1}{2ab-1}$ and so if $ c\ge 1$, we know that $ (a-1)(b-1)\le 4$.  From here, just straight case crunching gets you three unique solutions up to permutation: $ (1,1,5);(2,5,1);(3,3,1)$.


suppose we are give z=n(mod x) and z=f(mod y), is there a way i can express z in mod(xy)?

Your question is closely related to the Chinese Remainder Theorem. Find more information here:

http://mathworld.wolfram.com/ChineseRemainderTheorem.html

Although I think the theorem only shows the existence of such a z, there is a way to find this z.

Check out page 21 of these useful lecture notes from Hong Kong:

http://www.math.ust.hk/faculty/makyli/MATH190/lect-notes.pdf


induction
superior algebra
superior algebra unsolved

find a smith normal form of the A=(aij) E Mn(Z) where aij=j^i

i think a solution is:[hide] you can make induction to the dimension and then using det vandermonde compute the last element of the diagonal. if you want i can be more specific let me know[/hide]

in the solution i sketched there is a point that i havent proved..  i guessed too the smith normal form is that but, i dont see the proof of that  is  abvious...


Let
\[ S\equal{}\{(x,y)\mid y\equal{}\sin^2x,\,0\le x\le\pi /2\}.\]
Find all sets $ \{A,B,C\}\subset S$ such that $ B$ is the midpoint of line segment $ AC$.

Let $ A(a,\sin^2{a})$ and $ C(c,\sin^2{c})$ be two points with $ 0\leq a < c\leq\frac {\pi}{2}$
The midpoint $ B$ of line segment $ AC$ is $ B(\frac {a \plus{} c}{2},\frac {\sin^2{a} \plus{} \sin^2{c}}{2})$
Since $ B\in S$, $ \therefore\frac {\sin^2{a} \plus{} \sin^2{c}}{2} \equal{} \sin^2{\frac {(a \plus{} c)}{2}}$
$ 1 \minus{} \frac {\cos{2a} \plus{} \cos{2c}}{2} \equal{} 1 \minus{} \cos{(a \plus{} c)}$
$ \cos{(a \plus{} c)}(\cos{(a \minus{} c)} \minus{} 1) \equal{} 0$

Suppose $ \cos{(a \minus{} c)} \equal{} 1$. From $ |a \minus{} c|\leq\frac {\pi}{2}$
$ \therefore |a \minus{} c| \equal{} 0$ which means $ a \equal{} c$, contradicting with $ a < c$

$ \therefore\cos{(a \plus{} c)} \equal{} 0$
From $ 0 < a \plus{} c < \pi$ $ \therefore a \plus{} c \equal{} \frac {\pi}{2}$

$ \therefore$ All sets $ \{A,B,C\}\in S$ such that $ B$ is the midpoint of line segment $ AC$ are $ A(a,\sin^2{a}),B(\frac {\pi}{4},\frac {1}{2}),C(\frac {\pi}{2} \minus{} a,\cos^2{a})$ for all $ a\in [0,\frac {\pi}{2}] \minus{} \{\frac {\pi}{4}\}$


linear algebra
matrix
symmetry
vector
invariant
analytic geometry
geometry

Given $ p\plus{}q\plus{}r\equal{}0$, find $ x$:

$ \begin{vmatrix} p \minus{} x & r & q \\
r & q \minus{} x & p \\
q & p & r \minus{} x \end{vmatrix} \equal{} 0$

You can end up with a messy cubic if you don't see the significance of this equation.
[b]Edit:[/b] [i]Edited accordingly[/i]

[hide="A Start"]Notice that this is asking for the eigenvalues of the matrix $ M = \left( \begin{array}{ccc} p & r & q & r & q & p & q & p & r \end{array} \right)$

Notice that the columns are cyclic permutations of each other, so by symmetry, we expect $ \left( \begin{array}{c} 1 & 1 & 1 \end{array} \right)$ to be an eigenvector. Indeed:

$ \left( \begin{array}{ccc} p & r & q & r & q & p & q & p & r \end{array} \right) \left( \begin{array}{c} 1 & 1 & 1 \end{array} \right) = \left( \begin{array}{c} p + q + r & p + q + r & p + q + r \end{array} \right) = \left( \begin{array}{c} 0 & 0 & 0 \end{array} \right)$

So, the associated eigenvalue is $ 0$, and thus this is also a root of the original equation.[/hide]

So now, how do we get the other roots? Think about the image space of the transformation represented by $ M$, and see if you can figure out what kind of transformation it does to vectors within this image space!

I just noticed that I missed a critical part of the problem in the original post - it was: $ p\plus{}q\plus{}r\equal{}0$

use any of the properties of matrices and use 0 matrices!

Well, obviously you can crank out the cubic. That would make this a pretty useless and uninteresting problem -- certainly tedious!

Having noticed the eigenvalue in my previous post, we now have a pointer as to how to factor the cubic. Or, we can choose to avoid computing the determinant altogether if we can see what's:

[hide="going on here"]We see that the plane perpendicular to $ (1, 1, 1)^T$ is going to be the image space, and so this space is going to be invariant under $ M$. That is, we'll get our other two eigenvalues from here. And what is this space? The set of vectors perpendicular to $ (1,1,1)^T$, or, in your usual 3D coordinates, $ X \plus{} Y \plus{} Z \equal{} 0$ (where I used capitals to distinguish the coordinates from the $ x$ in the original equation).

You can see that this transformation is going to be a scale in all directions followed by a length-preserving transform by symmetry, so we only need to figure out if it's a rotation (imaginary eigenvalues) or a reflection (real eigenvalues). This is great, because it means we can find out a lot about $ M$'s eigenvalues by looking at it's effect on any random vector in this plane!

Note that $ (1,0, \minus{} 1)^T$ is in the plane and has length $ \sqrt {2}$. $ M(1,0, \minus{} 1)^T \equal{} (p \minus{} q,r \minus{} p,q \minus{} r)^T$, which has length:
$ \sqrt {(p \minus{} q)^2 \plus{} (r \minus{} p)^2 \plus{} (q \minus{} r)^2}$
$ \equal{} \sqrt {2p^2 \plus{} 2q^2 \plus{} 2r^2 \minus{} 2pq \minus{} 2qr \minus{} 2pr}$
$ \equal{} \sqrt {3(p^2 \plus{} q^2 \plus{} r^2) \minus{} (p \plus{} q \plus{} r)^2}$
$ \equal{} \sqrt {3(p^2 \plus{} q^2 \plus{} r^2)}$

Thus, every vector in the plane increases in length by the ratio of these two lengths -- namely, $ \sqrt {\frac {3}{2}(p^2 \plus{} q^2 \plus{} r^2)}$

Any guesses as to what the resulting eigenvalues might be?

[b]Hint:[/b] We first need to figure out if this is a scaled rotation or a scaled reflection!

[hide="Answer"]A translation. Hurrah! No complex numbers here. Just notice that the columns permute in the opposite direction that they do in the identity matrix $ I$, so it's not a rotation around $ (1,1,1)^T$.

If you don't "see" how it works, pick p=2, q=1, r=0 and draw it in your head.

In a scaled reflection, if each vector's length increases by a factor of $ c$, the eigenvalues are $ c$ and $ \minus{} c$. (why?)
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Thus, the other two roots are: $ \pm \sqrt {\frac {3}{2}(p^2 \plus{} q^2 \plus{} r^2)}$[/hide]

If you don't know anything about eigenvectors and eigenvalues, the point of my post is to show you that these seemingly tedious determinant equations can have meaning in a larger context. Check out linear algebra when you have the time. It's very fundamental and interesting, and will be a pleasant break from trying to blast ahead through calculus and multivariable calculus and analysis and etc.


Let $q$ be a prime $\geq 5$ and let $p$ be an odd prime such that $p|q+1$. Prove that:

$p | \sum_{i=1}^p (-1)^{i+1}\left (\sum_{m=1}^q m^i \right )$

It is easy to show, that $p |\sum_{m=1}^q m^i, \ \ if \ (p-1)\not|i$. Therefore
$\sum_{i=1}^p (-1)^{i+1}\left (\sum_{m=1}^q m^i \right )=\frac{q+1}{p} (mod \ p).$

[quote="{x}"][quote="Rust"]It is easy to show, that $p |\sum_{m=1}^q m^i, \ \ if \ (p-1)\not|i$.[/quote]

Nope :(. This is unfortunately false.
Take $q = 7$, $p = 5$ and $i = 1$.
Then $p - 1 = 4 \nmid i = 1$.
But
\[ \sum_{m = 1}^7 m^1 = 28  \]
And $5 \nmid 28$.
You probably confused it with
\[ p | \sum_{m = 1}^p m^i \Longleftrightarrow ( p - 1 ) \nmid i  \][/quote]

I assume that one of the conditions of the problem is $p|q+1$. But $5\nmid7+1$.
Also since $p|q+1$ you can simply write $q=pk-1$ take these sums and you will see Rust's arguments are true
\[ \sum_{m=1}^pm^i=0 (mod p) \]
\[ \sum_{m=p+1}^{2p}m^i=0 (mod p) \]
\[ \vdots \]
\[ \sum_{m=p(k-1)+1}^{pk}m^i=\sum_{m=p(k-1)+1}^{pk-1}m^i=0 (mod p) \]
adding these sums the result is obvious.


0=6t^4-8t^3-3t^2-3
a question i got for assignment can anyone show me a way to solve this? thx

Investigation of the rational roots tells us all the roots are irrational.  It might be the product of two quadratics or something.  I don't really think there's a nice way to do this by hand.

Sure, there are some less-than-nice ways...  :lol:

:rotfl: From Mathematica, the solutions are attatched. (there's one more, but you get the idea)

[quote="me@home"]With mathematica, can you get it to output the generalized formula for a quartic? That would be fun to put on a t-shirt... :rotfl:[/quote]

 :spam:  :spam: