AMC
AIME
USA(J)MO
USAMO
AMC 12
geometry
email
\/closed

All classes of AMC/AIME level are good. It's not so much that you can't do these problems. But these classes will dramatically increase your score.

Take just a few and making USAMO is almost guaranteed.

[quote="xirtamax"]I got 110 on AMC 12 and 6 on AIME B
What courses should I take here???[/quote]

I suggest to anyone who scores between 100 and 130 on the AMC12 take that class if they can.

I suggest to anyone who can make the AIME but hasn't made the USAMO take the AIME class we'll offer starting in the fall.  Some students who attend MOP took the AMC/AIME class that is now being split in two.  Those students like using those classes to fill little gaps they might have in knowledge.

Personally, I think our subject courses are best for overall mathematical development and the problem series courses are great for problem practice and broad topic coverage.  The subject courses simply go deeper into each area of mathematics.

[quote="xirtamax"]I got 110 on AMC 12 and 6 on AIME B
What courses should I take here???[/quote]

The courses this summer you should consider are the AMC-12 Problem Series, AMC/AIME Algebra, and the AIME/Olympiad Counting.  The latter two are more challenging than the first.  You can take a look at the diagnostic tests by [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]clicking here[/url].

If you are considering the Independent Study in the fall and are unsure which to take, send us an email at classes@artofproblemsolving.com and we'll help you decide.

[quote="Rep123max"]Will you be offering another AMC 12 series in the winter right before the AMC?[/quote]

Yes, we currently plan on holding the AMC12 class twice a year -- once during the summer and once starting around October (the date for the second AMC12 class has not yet been scheduled).  We will also run an AIME class during the winter.

[quote="joml88"]What other types of problem series do you offer?[/quote]

Currently we offer MATHCOUNTS, AMC12, and AIME.  We also hold an AIME seminar that is a 2-day, 6 hour preparation course held just before the AIME.  Descriptions of the classes can be found here: [url]http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php[/url]

We will be adding an AMC10 course designed for students trying to get over the hump of making it to the AIME.

Some of our problem series will eventually have multiple levels or else staggered problem sets so that students can take problem series courses at the same level twice and work completely new problems.

Question for all students: Are there any problem series that you would like to see that we are not offering?

[quote="Rep123max"]Well once I get better, USAMO? Also, maybe you could have one "Other" which could be ARML, Mandelbrot, other contests.[/quote]

I don't think we'll ever offer a USAMO problem series because that's essentially what our subject courses are for.  Our Inequalities and Olympiad Geometry courses are entirely Olympiad level problems.  The number theory and high level counting and probability courses also go through some Olympiad material.  The next number theory course will go through even more.

It would be impossible to pack the topics of Olympiad mathematics into a single course.  This is why I encourage the most motivated students to take the subject courses (or the independent studies).  They go much deeper into each subject and are better for overall development.  The problem series courses are great for getting a fair amount of practice on a broad array of topics and for filling gaps in knowledge.

[quote="Rep123max"]Well once I get better, USAMO? Also, maybe you could have one "Other" which could be ARML, Mandelbrot, other contests.[/quote]

One thing I did not mention is that we include many problems similar to those at ARML and on the Mandelbrot within all our classes, though perhaps we will one day run a problem series for "various competitions" including those two, the USAMTS, Tournament of Towns, and other competitions.

Even if we do it will not yet be a high priority for us as we are working to create a number of courses that I believe will be more beneficial for students.

[quote="Rep123max"]Ya, I was just suggesting since you asked. I think you are covering most of the subject courses (I cant wait for trig) and what exactly will be covered in the independnet class (not the two courses, i mean the other part)[/quote]

The independent study class will include many exploratory problems -- much like USAMTS problems or Mandelbrot team round problems.  These problems will help students develop an even more solid foundation of the all the kinds of mathematics leading up to the Olympiads (and some other subjects not typically included in the Olympiads).

The independent study courses will also help guide students along texts such as [u]Problem Solving Strategies[/u] and [u]The Art of Craft of Problem Solving[/u] (and the AoPS books for the younger students).  Much of the value of the independent study is that we schedule office hours to work with students and help them get over any stumbling block that could keep a student from developing a complete understanding of an area of mathematics.

Part of the goal is to bring students along the path of research.  I think the program we are developing should be particularly helpful to students interested in research careers in math, science, economics or any quantitative field.  Students will write a paper in each of our independent study programs. (don't take that as an intimidating task.  We are not asking for Westinghouse projects, though our hope is that a few such larger projects might develop out of the papers written.)

[quote="Rep123max"]Also, do you think that with school we will be able to handle the independnet study?[/quote]

I suppose that depends on how much time school takes up.  The independent study (along with the 2 classes) should provide a student with around 15 hours of work per week.  This could take the place of other math team studies and in some cases students are opting out of their regular school math class to set aside time during school to work on the independent study.

I am also of the opinion that students who study math to a high degree have a much easier time with other quantitative courses such as chemistry, physics, and economics.  Students who solidify their mathematics backgrounds typically see much of the material in quantitative subjects as being old material (in terms of thinking and problem solving) in a new framework with a new vocabulary whereas most students find the quantitative side of science more difficult.  This is part of the reason I often suggest to students interested in the sciences that they study mathematics first and foremost (at least until college -- but sometimes even then).

So those 15 hours might replace x hours of regular math team study (which should look much easier after our courses), y hours of time spent learning how to relate math to science, and sometimes z hours of an in-school class.  15 - (x + y + z) might wind up being a relatively small investment of time.


i think we are missing contextual information


algebra
polynomial
complex numbers
number theory
superior algebra
superior algebra unsolved

Suppose $ \alpha$ is algebraic over $ \mathbb{Q}$. Let $ f$ be the minimal polynomial of $ \alpha$ over $ \mathbb{Q}$ of degree n, and let $ F$ to be its splitting field. Let the conjugates of $ \alpha$ be $ \alpha_1 \equal{} \alpha, \alpha_2, ..., \alpha_n$. It is claimed that this polynomial $ g(x) \equal{} \prod_{i\equal{}1}^n (2x \minus{} \alpha_i \minus{} \bar{\alpha_i})$ lies in $ \mathbb{Q}[x]$. ($ \bar{\alpha_i}$ is complex conjugate)

The argument in the book is that all the $ \mathbb{Q}$-homomorphisms of $ F$ would fix the coefficients of the polynomial. I guess he's trying to claim that any such homomorphisms actually send $ \alpha_i \plus{} \bar{\alpha_i}$ to something of this form as well, but I can't see why this is true (why would it preserve the complex conjugate?)

Thanks!

You would need that conjugation commutes with the other elements of the Galois group. Take roots of $ x^3\minus{}2$ as a counterexample.

Thanks! So does it mean that the proof of Liouville's theorem in Esmonde and Ram Murty's book is actually wrong? (p.35)

http://books.google.com.hk/books?id=kOyhgwTqQ0YC&pg=PP1&dq=problems+in+algebraic+number+theory#v=onepage&q=&f=false

Looks that way. I remember taking a short look at that book  some time ago, and I found it strange that they make any difference between real and complex roots (the proof of 3.2.1 should not need $ \alpha$ to be real) or why they do not consider non-real $ \alpha$ to be trivial (there's a canonical constant lower bound for how close a rational can come to $ \alpha$: it's called the imaginary part).

By the way, the first line of page 30 should be "If $ p/q \leq 2M$, then".

Thanks! I have an older version of the book with me where that "p/q \le 2M" problem doesn't exist. Anyway it is really strange for the authors to disgintuish the cases between real and complex roots.


topology
real analysis
real analysis theorems

State the definition of complete metric space.

[quote="Kouichi Nakagawa"]metric space in which every Cauchy sequence is convergent.[/quote]

A metric space in which every Cauchy sequence converges within the space.


I have a solution using inversion.  However, I'd like to see if there is any elementary construction to this problem.

Note that if a center of inversion $ O$ and the radius of inversion $ r$ is given, then the image of a point $ P$ under inversion $ i_O^r$ is constructible.  Therefore, the image of inversion of any circle or any straight line is constructible.

Now, consider an arbitrary inversion $ i$ about $ C$.  Let $ \Gamma_1$ be the circle with center $ C_1$.  $ \Gamma_2$ is defined similarly.  Moreover, let $ \ell$ be the straight line passing through $ A$ and $ B$.  Note that $ i\left(\Gamma_1\right)$ and $ i\left(\Gamma_2\right)$ are two parallel straight lines, and $ i\left(\ell\right)$ is a circle passing through $ C$ and touching both lines.   (I should not have to explain how to construct the two parallel lines, should I?)

The construction of $ i\left(\ell\right)$ is simple.  Let $ m$ be the straight line that is midway between  $ i\left(\Gamma_1\right)$ and $ i\left(\Gamma_2\right)$.  and $ d$ the distance from $ m$ to one of these two lines.  Now, draw a circle centered at $ C$ with radius $ d$.  This circle should cut $ m$ in two points.  The point that is closer to $ \ell$ is the center of $ i\left(\ell\right)$.

Now, it is easy to construct two circles that are tangent to the parallel lines ($ i\left(\Gamma_1\right)$ and $ i\left(\Gamma_2\right)$ ) and the circle $ i\left(\ell\right)$.  Call them $ \omega_1$ and $ \omega_2$.  Do the inversion on these circles.  The circles $ i\left(\omega_1\right)$ and $ i\left(\omega_2\right)$ are what we want.

[quote="jrrbc"]  :(  :(  :(  [color=darkred][b]Japanese Sangaku[/b]  [/color]:?: [/quote]
 [quote="Batominovski"] [color=darkred]I have a solution using inversion.  However, [b]I'd like to see if there is any elementary construction to this problem[/b].[/color] [/quote]
[size=117][color=darkred][b]Yes, exists an elementary construction ! See[/b][/color][/size] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216952][color=red][b]here[/b][/color][/url] [size=117][color=darkred][b].[/b][/color][/size]


geometry
circumcircle
perpendicular bisector
IMO Shortlist
Angle Chasing
httpbit.ly3vjUcdg

Lemma: $ CC_{1}\perp DJ$.

Proof: Let $ w_{A}$ be the $ A$-excircle of $ \triangle ABC$, Let $ w_{C}$ be the circle diameter $ CJ$, let $ w$ be the circle diameter $ C_{1}J$. It is evident that: $ A_{1}$ and $ B_{1}$ lie on $ w_{C}$. Let $ w\cap w_{C}=J,E'$. Note that $ C_{1}C_{1}$, $ E'J$, $ A_{1}B_{1}$ are concurrent; they are radical axes of the three circles I mentioned earlier. But $ \angle CE'J=90$ from $ w$, and $ \angle CE'J=90$ from $ w_{C}$, so $ C_{1}E'C$ is a line; $ E$ and $ E'$ are equivalent since they are both the point that lies on $ CC_{1}$ and $ DJ$.

End Lemma

Let $ \angle A_{1}BD=x$
-$ \angle BA_{1}D=90-x$ since $ BD\perp DA_{1}$.
-$ \angle CA_{1}B=\angle CBA_{1}=90-x$ by vertical angles and $ A_{1}C=A_{1}B$.
-$ \angle JCA_{1}=x$, $ CA_{1}=CB_{1}$ and $ JA_{1}=JB_{1}$ so $ JC$ is the perpendicular bisector of $ A_{1}B_{1}$.
-$ \angle A_{1}JC=\angle CEA_{1}=90-x$ since $ A_{1}C\perp A_{1}J$ and $ A_{1}EJC$ is cyclic.
-$ \angle DEA_{1}=x$ Since $ DE\perp EC$.

Hence $ DBEA_{1}$ is cyclic and $ \boxed{\angle BEA_{1}=90^\circ}$

Now another long angle chase.

-$ \angle A_{1}JC=90-x=\angle CJB_{1}$ since $ A_{1}JC\cong B_{1}JC$ by SSS
-$ \angle CJB_{1}=\angle CEB_{1}=90-x$ since $ CEJB_{1}$ is cyclic.
-$ \angle DEB_{1}=90+x$ by addition.
-$ \angle BCA=180-2x$ since $ \angle A_{1}CJ=\angle B_{1}CJ=x$
-$ \angle BAC=x$ by the sum of the angles in a triangle.
Hence $ \angle DAB_{1}+\angle DEB_{1}=x+180-x=180$ so $ ADEB_{1}$ is cyclic.
It follows that $ \angle ADB_{1}=\boxed{\angle AEB_{1}=90^\circ}$.

Phew that was a lot of angle chasing.

it's obvious that $ AC=BC$.

the polar of $ C$ pass through $ D$ so the polar of $ D$ pass through $ C$,polar of $ D$ also pass through $ C_{1}$ and is perpendicualr to $ I_{a}D$ hence $ C_{1}E$ pass through $ C$.
$ \angle A_{1}I_{a}B_{1}=\angle ACB$ and $ \angle I_{a}EC=\angle CB_{1}I_{a}=90^{\circ}$ hence five points $ A_{1},C,B_{1},I_{a},E$ are on a circle .
$ CI_{a}\parallel AB$ hence $ \angle CB_{1}E=\angle CI_{a}E=\angle I_{a}DC_{1}$ hence $ ADEB_{1}$ is cyclic also you can prove that $ DBEA_{1}$ is cyclic,too.

hence $ \angle AEB_{1}=\angle BEA_{1}=90^{\circ}$

[b]Lemma 1:[/b] $AC=BC$.

It is known that $A_1 B_1$ is parallel to the bisector of $\angle{C}$. Since $A_1 B_1 \perp AB$, the bisector of $\angle{C}$ is perpendicular to $AB$. Hence $AC=BC$.

[b]Lemma 2:[/b] $BA_1 EJ$ is cyclic.

Since $JC_1 \perp DC_1$ and $C_1 E \perp DJ$, $\triangle{DEC_1} \sim \triangle{DC_1 J}$. This yields that $DE \cdot DJ=DC_1^2$. By power of point $D$ with respect to the $A$-excircle,
\[DE \cdot DJ=DC_1^2=DA_1 \cdot DB_1\]
Hence $BA_1 EJ$ is cyclic by converse power of a point.

[b]Lemma 3:[/b] $A_1 EBD$ is cyclic.

Note that since $B_1 J \perp AB_1$,
\[\angle{A_1 ED}=\angle{JB_1 D}=90-\angle{AB_1 D}=\angle{BAC}=\angle{A_1 BD}\]
Hence $A_1 EBD$ is cyclic.

[b]Lemma 4:[/b] $ADEB_1$ is cyclic.

Note that
\[\angle{B_1 ED}=\angle{A_1 ED}+\angle{B_1 EA_1}=\angle{CAB}+\angle{B_1 JA_1}=\angle{CAB} + 180-2\angle{CAB}=180-\angle{CAB}\]
Hence $ADEB_1$ is cyclic.

Therefore $\angle{AEB_{1}}=\angle{BEA_{1}}=90$.

[geogebra]4a3df5a7b7dfb5852c91d817df7006087e5e29e3[/geogebra] A solution by inversion!!

[u]Invert in the [b]excircle[/b][/u]

Let $AJ\cap C_1B_1=A^'$

Consider the quadrilateral $AEA^'B_1$

Under inversion,
$A \rightarrow A^'$ . . . ($\Delta JC_1B$ is right and $C_1E \bot DJ$)

$E \rightarrow D$ . . . ($\Delta JC_1B$ is right and $C_1E \bot DJ$)

$A^' \rightarrow A$

$B_1 \rightarrow B_1$

$ADA^'B_1$ is cyclic and as circles remain circles under inversion,
$AEA^'B_1$ is cyclic

Therefore, $\angle AEB_1= \angle AA^'B_1 = 90$

Show the case of the other angle similiarly.

Sorry for reviving an old thread, I just felt the urge to post this solution, and by the looks of things, it hasn't been posted in this thread. So here goes:

Let $\gamma_1$ denote the circumcircle of triangle $A_1BD$, centred at the midpoint $M$ of $A_1B$ with radius $MB=MA_1$, since $\angle A_1DB=90^0$. Now, note that $JC_1E \sim JDC_1$ by trivial angle chasing, hence $JC_1^2=JE \cdot JD$. We note that the power of $J$ with respect to $\gamma_1$ is:

$JM^2-MB^2=JA_1^2+A_1M^2-MB^2=JA_1^2=JC_1^2=JD \cdot JE$

implying that $E$ lies on $\gamma_1$, or $A_1EBD$ is cyclic, which readily implies that $\angle BEA_1=180^0-\angle BDA_1=\boxed{90^0}$.

Similarly, let $\gamma_2$ denote the circumcircle of triangle $ADB_1$, again, since $\angle ADB_1=90^0$, we have that $\gamma_2$ is centred at $N$, the midpoint of $AB_1$, with radius $NA=NB_1$. Again, we have that the power of $J$ with respect to $\gamma_2$ is:

$JN^2-NA^2=JB_1^2+B_1N^2-NA^2=JB_1^2=JC_1^2=JD \cdot JE$

implying that $E$ also lies on $\gamma_2$ i.e. $ADEB_1$ is cyclic i.e. $\angle AEB_1=\angle ADB_1=\boxed{90^0}$.

Done!

We use directed angles modulo $180^{\circ}$.

$A_1B_1\perp AB$ implies that $CA = CB$; let $M$ be the midpoint of $AB$, then let $AB = 2a, CM = b, AC = c$, so that $a^2+b^2 = c^2$. Observe that $DB = \frac ac(c-a)$, since $BA_1 = c-a$, hence $MD = \frac {a^2}{c}$, and $DC_1 = \frac{(c+a)(c-a)}{c} = \frac{b^2}{c}$. Hence, $$CD^2 - DC_1^2 = b^2 + \frac{a^4}{c^2} - \frac{b^4}{c^2} = b^2 + (a^2-b^2) = a^2 = c^2-b^2 = CJ^2 - C_1J^2$$ And so $DJ\perp CC_1$.

Thus $E$ is also the foot from $C$ to $DJ$. By Miquel's Theorem, $(CAB_1), (BDA_1), (ADB_1)$ concur at some point $E'$. Observe that $(CAB_1)$ is the circle with diameter $CJ$. We see that since $$\measuredangle A_1E'J = \measuredangle A_1CJ = \measuredangle CBA = \measuredangle A_1BD = \measuredangle A_1E'D$$Hence $D,J, E'$ are collinear. But $E'$ is on the circle with diameter $CJ$, so $CE'\perp E'J = DJ$, so $E'$ is the foot from $C$ to $DJ$, so $E' = E$. Therefore, since $E'$ is on $(A_1DB) = (A_1B)$, and $(ADB_1) = (AB_1)$, we see that $\measuredangle BEA_1 = \measuredangle AEB_1 = 90^{\circ}$.

My solution:
Let $\angle BCA=2x,$ then $\angle AB_1D=x=\angle CA_1B_1=90-\angle BAC=90-\angle A_1B_1J.$
$1:$ $AC=BC.$
Proof: We know $\angle BAC=90-x$ and $\angle ACB=2x\to \angle ABC=180-\angle ACB-\angle BAC=90-x,$ as desired.
$2:$ $EJB_1A_1$ is cyclic.
Proof: We know $DC_1$ tangent to $(C_1EJ),$ since $\angle JC_1D=C_1EJ=90.$
Then $DC_1^2=DE\cdot DJ,(i)$
Also we have from PoP $DC_1^2=DA_1\cdot DB_1 (ii).$
From $(i),(ii)$ we have $DE\cdot DJ=DA_1\cdot DB_1\to EJB_1A_1$ is cyclic.
Claim1: $BDA_1E$ is cyclic.
Proof: $$\angle DBA_1=\angle ABC=^{1} \angle BAC=90-\angle AB_1D=\angle JB_1A_1=^{2} \angle DEA_1\to BDA_1E$$ is cyclic.Then $\angle BEA_1+\angle BDA_1=\angle BEA_1+90=180\to \angle BEA_1=90.$
Claim 2: $ADA_1B_1$ is cyclic.
Proof: $$\angle BAC=90-\angle DB_1A=\angle A_1B_1J=\angle DEA_1=^{2} 90-x,$$ also $\angle A_1EB_1=2x\to \angle DEB_1=90+x,$ Also $\angle DAB_1=90-x\to \angle DAB_1+\angle DEB_1=180.$ Then $\angle ADB_1=\angle AEB_1=90.$
Then $\angle AEB_1=\angle BEA_1=90.$

[hide="length bash + spiral sim"]

We show that $\angle BEA_1 = 90$ by proving that there is a spiral similarity at $E$ taking $\triangle EA_1B$ to $\triangle EDC_1$.

Equivalently, we show that there is a spiral similarity at $E$ taking $\triangle EA_1D$ to $\triangle EBC_1$. Because $A_1B_1 \perp AB$ (this condition also implies $\angle A = \angle B$), we have

$$\angle A_1DE = 90 - \angle EDC_1 = \angle EC_1D.$$

It remains to show $\frac{ED}{DA_1} = \frac{EC_1}{C_1B} \iff \frac{BC_1}{DA_1} = \frac{EC_1}{ED} = \frac{C_1J}{DC_1}$. If the midpoint of $AB$ is $M$, then by similar triangle $\triangle BMC \sim \triangle BDA_1$, we compute $DA_1 = R_a \left( \frac{s-c}{a} \right)$ where $R_a$ is the A-exradius. Similarly, we find $BD = \frac{c}{2} - \frac{c^2}{4a}$, so $DC_1 = DB + BC_1 = \left( \frac{c}{2} - \frac{c^2}{4a} \right) + \left( \frac{a + a - c}{2} \right) = a - \frac{c^2}{4a}$.

Finally, we have

$$\frac{BC_1}{DA_1} = \frac{s-c}{R_a\left( \frac{s-c}{a}\right) }, \:\:\:\: \frac{C_1J}{DC_1} = \frac{R_a}{a - \frac{c^2}{4a}}$$

Equating the above yields $R_a^2 = a^2 - \frac{c^2}{4}$, true by Pythagorean Theorem on $\triangle CMB$ (where the height to $AB$ has length $R_a$).

To show $\angle AEB_1 = 90$, repeat the above arguments to show there is a spiral similarity at $E$ taking $\triangle EB_1A$ to $\triangle EDC_1$.

[/hide]

[hide=Solution]
[quote=ISL 2006 G5]In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. [/quote]
[b][color=#000]Solution:[/color][/b] Let $BB_1 \cap (J)$ $=$ $Z$ $\implies$ $B_1A_1ZC_1$ is Harmonic. $$-1=(B_1,Z;A_1,C_1) \stackrel{B_1}{=} (A,B;D,C_1)$$
Since, $\angle DEC_1=90^{\circ}$ $\implies$ $DE$ bisects $\angle AEB$.
-----------------------------------------------------------------------
 Also, Since, $A_1B_1 \perp AB$ $\implies$ $CJ||AB$ $$\implies \angle ABC=\angle BCJ=90^{\circ}-\tfrac{1}{2} \angle ACB \implies \angle ABC=\angle BAC$$
---------------------------------------------------------------------
By LaHire's Theorem, $D$ lies on Polar of $C$ $\implies$ $C$ lies on Polar of $D$ $\implies$ $C$ $\in$ $C_1E$
$$\implies -1 = (C_1,CC_1 ~ \cap ~ (J) ; A_1,B_1) \stackrel{C_1}{=} ( D, EC ~ \cap ~ A_1B_1 ; A_1,B_1 )$$
Hence, $EC$ bisects $\angle A_1EB_1$
--------------------------------------------------------------
It follows that, $CA_1EJB_1$ is cyclic $\implies$ $\angle BDE=\angle EJC=\angle BA_1E$ $\implies$ $BDA_1E$ is cyclic. Hence, $\angle BEA_1=90^{\circ}$ and since, $\overline{BD}$ $\stackrel{E}{\mapsto}$ $\overline{CB_1}$ $\implies$ $\angle AEB_1$ $=$ $90^{\circ}$ $\qquad \blacksquare$
[/hide]

[asy]
size(8cm);
defaultpen(fontsize(10pt));

pair A1 = dir(15), B1 = dir(345), C1 = dir(90), J = origin, C = extension(A1,A1+dir(foot(B1,A1,J)--B1),J,J+dir(0)), B = extension(A1,C,C1,C1+dir(0)), A = extension(B,C1,C,B1), D = extension(B,C1,A1,B1), E = foot(C1,D,J);

dot("$A_1$", A1, dir(15));
dot("$B_1$", B1, dir(345));
dot("$C_1$", C1, dir(90));
dot("$J$", J, dir(225));
dot("$C$", C, dir(330));
dot("$B$", B, dir(90));
dot("$A$", A, dir(45));
dot("$D$", D, dir(90));
dot("$E$", E, dir(180));

draw(arc(J,1,315,480));
draw(A--J--B1--A--C1--J--D--B1^^C1--E^^B--C--J);
draw(B--E--A1^^A--E--B1, dashed);
[/asy]
We claim that $\boxed{\angle BEA_1 = \angle AEB_1 = 90^\circ}$.

First, note that $$\angle A + \dfrac{1}{2}\angle C = \angle B_1AD + \angle AB_1D = \angle A_1BD + \angle BA_1D = \angle B + \dfrac{1}{2}\angle C,$$ so $AC = BC$, or $\overline{AB} \parallel \overline{CJ}$.

Now we use complex numbers. Set the $A$-excircle as the unit circle (so $J$ is the origin) with $c_1 = i$. Since $C$ lies on the real axis, $b_1 = \overline{a_1}$. It follows from the tangent formula that $$a = \dfrac{2b_1c_1}{b_1+c_1} = \dfrac{2i \cdot \tfrac{1}{a_1}}{\tfrac{1}{a_1}+i} = \dfrac{2i}{ia_1+1} = \dfrac{2}{a_1-i}$$ and $$b = \dfrac{2a_1c_1}{a_1+c_1} = \dfrac{2ia_1}{a_1+i}.$$ To solve for $d$, note that $$d+\overline{d} = 2\text{Re}(d) = 2\text{Re}(a_1) = a_1 + \dfrac{1}{a_1}$$ and $$d-\overline{d} = 2\text{Im}(d) = 2i.$$ Adding gives $$2d = a_1 + \dfrac{1}{a_1} + 2i \implies d = \dfrac{a_1^2+2ia_1+1}{2a_1}.$$ We can also calculate $$\overline{d} = \dfrac{(\tfrac{1}{a_1})^2-\tfrac{2i}{a_1}+1}{\tfrac{2}{a_1}} = \dfrac{a_1^2-2ia_1+1}{2a_1}.$$ It remains to find $e$. Since $E \in \overline{DJ}$, we can let $e = kd$ for some $k \in \mathbb{R}$. $\overline{C_1E} \perp \overline{DJ}$ implies \begin{align*} 0 &= \dfrac{c_1-e}{d-j} + \overline{\left(\dfrac{c_1-e}{d-j}\right)} \\ &= \dfrac{i-kd}{d} + \dfrac{-i-k\overline{d}}{\overline{d}} = \dfrac{i}{d} - k-\dfrac{i}{\overline{d}}- k \\ \implies 2k &= \dfrac{i}{d} - \dfrac{i}{\overline{d}} \\ \implies e = kd &= \dfrac{i}{2}\left(1-\dfrac{d}{\overline{d}}\right) \\ &= \dfrac{i}{2}\left(1-\dfrac{a_1^2+2ia_1+1}{a_1^2-2ia_1+1}\right) \\ &= \dfrac{2a_1}{a_1^2-2ia_1+1}.\end{align*} Finally, we verify \begin{align*} \dfrac{b-e}{a_1-e} &= \dfrac{\tfrac{2ia_1}{a_1+i}-\tfrac{2a_1}{a_1^2-2ia_1+1}}{a_1-\tfrac{2a_1}{a_1^2-2ia_1+1}} \\ &= \dfrac{2i(a_1^2-2ia_1+1)-2(a_1+i)}{(a_1+i)(a_1^2-2ia_1+1)-2(a_1+i)} \\ &= \dfrac{2a_1(ia_1+1)}{(a_1^2+1)(a_1-i)} \\ &= \dfrac{2ia_1}{a_1^2+1} = -\dfrac{\tfrac{-2i}{a_1}}{\tfrac{1}{a_1^2}+1} \\ &= -\overline{\left(\dfrac{b-e}{a_1-e}\right)}\end{align*} and \begin{align*} \dfrac{a-e}{b_1-e} &= \dfrac{\tfrac{2}{a_1-i} - \tfrac{2a_1}{a_1^2-2ia_1+1}}{\tfrac{1}{a_1} - \tfrac{2a_1}{a_1^2-2ia_1+1}} \\ &= \dfrac{2a_1(a_1^2-2ia_1+1)-2a_1^2(a_1-i)}{(a_1-i)(a_1^2-2ia_1+1)-2a_1^2(a_1-i)} \\ &= \dfrac{2a_1(-ia_1+1)}{-(a_1-i)(a_1^2+2ia_1-1)} = \dfrac{-2ia_1(a_1+i)}{-(a_1-i)(a_1+i)^2} \\ &= \dfrac{2ia_1}{a_1^2+1} = -\dfrac{\tfrac{-2i}{a_1}}{\tfrac{1}{a_1^2}+1} \\ &= -\overline{\left(\dfrac{a-e}{b_1-e}\right)}.\end{align*} The proof is now complete. $\blacksquare$

[quote=pohoatza]In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$.

[i]Proposed by Dimitris Kontogiannis, Greece[/i][/quote]
[asy]
import olympiad;
import geometry;
size(12cm);
pair O=origin;
pair I,A,B,C,B1,A1,E,J,C1,D,M;
A=dir(210);
B=dir(330);
C=dir(90);
M=dir(30);
I=incenter(A,B,C);
J=2*M-I;
B1=foot(J,A,C);
C1=foot(J,A,B);
A1=foot(J,B,C);
D=foot(B1,A,B);
E=foot(C1,D,J);
draw(segment(A,B1),magenta);
draw(segment(A,C1),magenta);
draw(segment(B,C),magenta);
draw(segment(J,D),magenta);
draw(segment(J,B1),magenta);
draw(segment(J,C1),magenta);
draw(segment(J,A1),magenta);
draw(segment(B1,D),magenta);
filldraw(circumcircle(A,B,C),invisible,purple);
filldraw(circumcircle(E,A1,C),invisible,dashed);
filldraw(circumcircle(E,D,B),invisible,dotted);
filldraw(circumcircle(A,D,E),invisible,dotted);
filldraw(circumcircle(A1,B1,C1),invisible,purple);
dot(A^^B^^C^^J^^A1^^B1^^C1^^E^^D);
label("$A$",A,NW);
label("$B$",B,NW);
label("$C$",C,NW);
label("$D$",D,NW);
label("$E$",E,NW);
label("$A_1$",A1,SW);
label("$B_1$",B1,N);
label("$C_1$",C1,SW);
label("$J$",J,NE);
[/asy]
[b]Claim  :[/b] $ABC$ is an isosceles triangle
[b]Proof : [/b]  $A_1B_1\perp AB\implies 90^{\circ}=\angle{ABC}+\angle{BA_1D}=\angle{ABC}+\angle{B_1A_1C}=\angle{ABC}+\tfrac{\angle{ACB}}{2}\implies \angle{BAC}=\angle{ABC}\square$
 ----------------
Now we have $JB_1^2=JC_1^2=JE\cdot JD\implies JB_1$ is tangent to $\odot{B_1ED}$.Similarly $JA_1$ is also tangent to $\odot{A_1ED}$.Thus \[\angle{JA_1E}=\angle{JDA_1}=\angle{JDB}=\angle{JB_1E} \implies \odot{JB_1A_1E}\].Thus $\angle{A_1BD}=\angle{ABC}=\angle{JB_1A_1}=\angle{A_1ED}\implies E\in \odot{A_1DB}$ Similarly $E\in \odot{ADB_1}$
Thus $E : A_1B\to B_1A$.Thus $\angle{BEA_1}=\angle{AEB_1}=90^{\circ} \; \blacksquare$

Solution with Jeffrey Kwan:

We first show $\angle BEA_1 = 90^{\circ}$. Note that $E$ is the inverse of $D$ with respect to $(J)$. Let $B^\ast$ and $A^\ast$ denote the inverses of $B$ and $A$. Since $\measuredangle BDA_1 = \measuredangle BB^\ast A_1 = 90^{\circ}$, those four points are cyclic. Since $JB^\ast \cdot JB = JE \cdot JD$, point $E$ lies on that circle too. Hence $\measuredangle BEA_1 = 90^{\circ}$. The proof that $\angle B_1EA = 90^{\circ}$ is similar; note that $\measuredangle ADB_1 = \measuredangle AEB_1 = 90^{\circ}$ and $JA \cdot JA^\ast = JE \cdot JD$.

[asy] size(7cm);

pair J = origin; pair C_1 = dir(180); pair A_1 = dir(110); pair B_1 = dir(70); pair D = extension(A_1, B_1, C_1, C_1+N); pair E = foot(C_1, D, J); pair B = 2*A_1*C_1/(A_1+C_1); pair A = 2*B_1*C_1/(B_1+C_1); pair C = 2*A_1*B_1/(A_1+B_1); pair T = -C_1; pair F = D+T-C_1; draw(A--B_1, lightred); draw(B--C, lightred); draw(D--A, lightred);

draw(D--J, blue); filldraw(unitcircle, invisible, blue); filldraw(C_1--D--F--T--cycle, invisible, blue); draw(C_1--E, lightblue);

draw(B--E--A_1, deepgreen+dashed); pair Bs = midpoint(A_1--C_1); draw(A_1--C_1, grey+dotted); draw(B--J, grey+dotted);

dot("$J$", J, dir(270)); dot("$C_1$", C_1, dir(C_1)); dot("$A_1$", A_1, dir(A_1)); dot("$B_1$", B_1, dir(B_1)); dot("$D$", D, dir(D)); dot("$E$", E, dir(270)); dot("$B$", B, dir(B)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$B^\ast$", Bs, dir(190));

/* TSQ Source:

!size(10cm);

J = origin R270 C_1 = dir 180 A_1 = dir 110 B_1 = dir 70 D = extension A_1 B_1 C_1 C_1+N E = foot C_1 D J R270 B = 2*A_1*C_1/(A_1+C_1) A = 2*B_1*C_1/(B_1+C_1) C = 2*A_1*B_1/(A_1+B_1) T = -C_1 F = D+T-C_1 A--B_1 lightred B--C lightred D--A lightred

D--J blue unitcircle 0.1 yellow / blue C_1--D--F--T--cycle 0.1 yellow / blue C_1--E lightblue

// First batch B--E--A_1 deepgreen dashed B* = midpoint A_1--C_1 R190 A_1--C_1 grey dotted B--J grey dotted

// Second batch A--E--B_1 deepgreen dashed A* = midpoint B_1--C_1 R0 B_1--C_1 grey dotted A--J grey dotted

*/ [/asy]

Nice and easy :)
2006 G5
[hide=Solution]First note that $A_1B_1 \perp AB \implies A+ \frac C2 = \frac {\pi}{2} \implies B+ \frac C2 =\frac {\pi}{2}$ So we have that if $C^*$ is the midpoint of $A_1B_1$, then $DC_1JC^*$ is a rectangle.  Now let $B^* = BJ \cap A_1C_1$ and define $A*$ similarly . Note that since $E$ is the inverse of $D$ in $\odot(J)$ , hence $BB^*DE$ is a cyclic quadrilateral . Now note that $$\measuredangle{BDA_1}=\measuredangle{BB^*A_1}= \frac{ \pi}{2} \implies A_1 \in (BB^*DJ)$$ . So we have that $\measuredangle{BEA_1}=\measuredangle{BDA_1}=\frac{ \pi}{2}$
Similarly we can prove $\measuredangle{AEB_1}= \frac { \pi}{2}$[/hide]

First of all, we know that the polar of $C$ with respect to the excircle is $DA_1B_1$, hence $C$ lies on the polar of $D$. It is easy to see that $C_1$ and $E$ also lie on the polar of $D$. Hence, we find that $\measuredangle CEJ = 90^\circ$, which implies $A_1CB_1EJ$ is cyclic with diameter $CJ$.

Now, note that $\overline{DA_1B_1} \parallel \overline{C_1J}$, so $\overline{C_1J}$ is tangent to $(A_1CB_1EJ)$. Additionally, if $I$ is the incenter, then we know $\measuredangle ICJ = 90^\circ$, and so combined with our previous result we find that $AC = BC$. Now, evaluating the power at $C_1$ gives $$C_1A \cdot C_1B = s(s - c) = \frac{s(s - b)(s - c)}{(s - a)} = r^2 \cdot \frac{s^2}{(s - a)^2} = R_A^2 = C_1J^2 = C_1E \cdot C_1C,$$ hence $E$ lies on the circumcircle.

Next, we claim that $DBA_1E$ is cyclic, which quickly implies $\measuredangle BEA_1 = 90^\circ$. We proceed with a directed angle chase: $$   \measuredangle DBE = \measuredangle ABE = \measuredangle ACE = \measuredangle B_1CE = \measuredangle B_1A_1E = \measuredangle DA_1E.$$ In the same vein, we can show that $ADEB_1$ is cyclic which implies $\measuredangle AEB_1 = 90^\circ$.

Note that $JA_1^2 = JB_1^2 = JC_1^2 = JD \cdot JE$, so $\overline{JA_1}$ and $\overline{JB_1}$ are tangent to $(A_1DE)$ and $(B_1DE)$, respectively. Now $\angle JA_1E = \angle A_1DE = \angle B_1DE = \angle JB_1E$, so $E$ lies on $(JA_1CB_1)$. Remark that $ACB_1E$ is an isosceles trapezoid, so $\angle B_1DE = \angle JB_1E = \angle JA_1E = 90^\circ - \angle A_1EB_1 = 90^\circ - \angle CB_1E = 90^\circ - \angle AB_1E = \angle B_1AE$, and thus $ADEB_1$ is cyclic.

Now $\angle BAE = \angle DAE = \angle DB_1E = \angle A_1B_1E = \angle A_1CE = \angle BCE$, so $ABEC$ is cyclic. Finally, $\angle EDA_1 = \angle EDB_1 = \angle EAB_1 =  \angle EAC = \angle EBC = \angle EBA_1$, so $EBDA_1$ is cyclic as well. Now an angle extraction on cyclic quadrilaterals $BDA_1E$ and $ADEB_1$ yields $\angle AEB_1 = 90^\circ$ and $\angle BEA_1 = 90^\circ$.

Can someone write a barycentric Solution, please because it looks good in bary but I can't finish my answer
Thanks

[asy]     size(10cm); defaultpen(fontsize(10pt)); pair A = dir(120);     pair B = dir(200);     pair C = dir(340);     pair J = 2 * dir(270) - incenter(A, B, C);     pair A1 = foot(J, B, C);     pair B1 = foot(J, A, C);     pair C1 = foot(J, B, A);     pair D = extension(A, B, A1, B1);     pair E = foot(C1, D, J);
draw(circumcircle(A, B, C));     draw(B1--A--C1);     draw(B--C);     draw(CP(J, A1));     draw(B1--D);     draw(D--J);     draw(C1 -- E);     draw(C--J);     draw(circumcircle(D,B,A1), dashed);     draw(circumcircle(D, E, B1), dashed);     dot('$A$', A, dir(A));     dot('$B$', B, dir(B));     dot('$C$', C, dir(C));     dot('$J$', J, dir(J));     dot('$A_1$', A1, dir(A1));     dot('$B_1$', B1, dir(B1));     dot('$C_1$', C1, dir(C1));     dot('$E$', E, dir(E));     dot('$D$', D, dir(D));   [/asy]

[b][color=red]Claim:[/color][/b]    $A_1, E, B, D$ is cyclic.
[i]Proof.[/i]    Since $\angle BDA_1 = 90^{\circ},$ we have that the circumcenter $O$   of $\triangle BDA_1$ is on $B A_1,$ so then $\angle OA_1J = \angle   BA_1J = 90^{\circ},$ implying that the two circles are orthogonal.   Finally, $E, D$ are swapped by the inversion because $D$ is the pole   of $C_1E,$ so $A_1, E, B, D$ is cyclic as claimed. $\blacksquare$
By symmetry, we also have $ABEB_1$ cyclic. Now, we know that $BDA_1, ADB_1 = 90^{\circ},$ so then $\angle BEA_1 = \angle AEB_1 = 90^{\circ}.$

Because of the perpendicularity, $(A_1DB)$ and $(B_1DA)$ are tangent to lines $A_1J$ and $B_1J$ respectively. Since $\triangle{JC_1D}$ is right, $JE \cdot JD = JC_1^2 = JB_1^2 = JA_1^2$, so by Power of a Point $A_1DBE$ and $B_1ADE$ are cyclic, which implies that $\angle{BEA_1} = \angle{A_EB_1} = 90^{\circ}$.

[asy]
size(250);
pair A = (2, 6), B = origin, C = (9.84, 0);
pair I = incenter(A, B, C);
pair X1 = extension(circumcenter(A, B, C), (B+C)/2, (50, -50), (100, -50));
pair X2 [] = intersectionpoints(circumcenter(A, B, C)--X1, circumcircle(A, B, C));
pair L = X2[0];
pair J = 2*L - I;
pair A1 = foot(J, B, C), B1 = foot(J, A, C), C1 = foot(J, A, B);
pair D = foot(A1, A, B);
pair EE = foot(C1, D, J);
draw(A--B--C--cycle, red);
draw(C1--A--B1, red);
draw(C--C1, blue);
draw(circumcircle(A1, B1, C1), magenta);
draw(D--J, blue);
draw(B1--D, orange);
draw(circumcircle(B, EE, A1), green);
draw(circumcircle(A1, B1, EE), green);
draw(circumcircle(A, D, EE), green);
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, NE);
dot("$J$", J, S);
dot("$A_1$", A1, S);
dot("$B_1$", B1, NE);
dot("$C_1$", C1, NW);
dot("$E$", EE, S);
dot("$D$", D, NW);
[/asy]

The answer is $90^{\circ}$ for both parts. Indeed, the entire problem is just equivalent to the three following claims:

[b][color=#f00]Claim.[/b][/color] $\triangle ABC$ is isosceles.

[i]Proof.[/i] Simply notice that $$90^{\circ} - A = \angle CB_1D = \frac C2. \ \blacksquare$$

[b][color=#f00]Claim.[/b][/color] $DBEA_1$ is cyclic.

[i]Proof.[/i] We will show that $\angle DEA_1 = \angle B$. First, verify that quadrilateral $CA_1EJ$ is cyclic because $\angle CEJ = \angle CA_1J$, while $\triangle CA_1B_1$ is isosceles, so $$\angle DEA_1 = 90^{\circ} - \angle A_1EC = 90^{\circ} - \angle A_1B_1C = 90^{\circ} - \frac C2 = B,$$ proving the claim. $\blacksquare$

[b][color=#f00]Claim.[/b][/color] $ADEB_1$ is cyclic.

[i]Proof.[/i] We will show that $\angle DEB_1 = 180^{\circ} - \angle A$. Observe that this is equivalent to $$\angle CEB_1 = 90^{\circ} - A \iff \angle CA_1B_1 = 90^{\circ} - A,$$ which is obvious. $\blacksquare$

It is not hard to see that from these two claims, we must have $\angle BEA_1 = \angle AEB_1 = 90^{\circ}$, as desired. $\square$

[b][color=#9a00ff]Remark.[/color][/b] I believe there are no nontrivial configuration issues in this problem -- if there are, they should not be hard to clean up.

miQuel

Note that $\angle C_1JC=\angle C_1JA_1+\angle A_1JC=\angle DBA_1+\angle CA_1B_1=90^\circ.$ Let $F$ be $CJ\cup A_1B_1$ then $C_1DFJ$ rectangle.

$\angle EJF=\angle DJF=\angle C_1DJ=\angle EC_1J$ so $C_1,E,C$ collinear. $\angle JEC=90^\circ.$ Thus, $EA_1CB_1J$ is cyclic.

$\angle BAC=\angle JCB_1=\angle JEB_1$ so $ADEB_1$ is cyclic. Thus, $E$ is the miquel point of $ADA_1C$ which implies $\angle BEA_1=\angle AEB_1=90^\circ.$

Just a remark, is it just a coincidence or did ABC always seem isosceles? IDK but that looked like the only possible diagram
Also a very trivial problem, not sure why G5 :P
Anyways, here are two incredibly quick solutions I found in <15 minutes (even though they're short, it's easily motivated and also I think the first one is unique):
1. DEB_1=90+CEB_1=90+CJB_1=180-JCB_1. This is equal to 180-A iff CJ parallel to AB, which is evident since CJ perp. B_1A_1 perp. AB. Hence ADEB_1 is cyclic and AEB_1=90 degrees. Then from BA_1E=180-EA_1C=DJC=BDE, BDA_1E is cyclic so BEA_1=90 degrees as well. $\blacksquare$
2. We have that =JB_1^2=JA_1^2=JC_1^2=JE*JD (similarity of right triangles and cross multiplying). Noting that JA_1 and JB_1 are tangents to A_1DB and ADB_1, this implies E lies on (A_1DB) and (ADB_1), so the supplement of right angles is the answer, as desired. $\blacksquare$

Note that the condition essentially says that the external bisector of $\angle C$ is parallel to $AB$, which is equivalent to $AC=BC$, so $\triangle ABC$ is isosceles.\\

Claim; $C$, $E$, and $C_1$ are collinear. Note that $C_1E$ is the polar of $D$ with respect to the excircle, since $C_1$ clearly lies on the excircle and $E$ also does since it is the foot from $C_1$ to $DJ$. However, $D$ lies on the polar of $C$ (which is $A_1B_1$), so $C$ lies on the polar of $D$, showing the claim.\\

Hence, since $$\angle CEJ=\angle CA_1J=90,$$ $CA_1EJ$ is cyclic. Then, since $$\angle BA_1E=\angle EJC=\angle BDE,$$ $BDA_1E$ is cyclic as well, so $\angle BEA_1=90$.\\

Note that $B_1$ also lies on $(A_1CJE)$ since $\angle JB_1C=90$. Thus, $$\angle AB_1E=\angle BA_1E=\angle BDE,$$ so $ADEB_1$ is cyclic and thus $\angle AEB_1=90$ as well.

[b][color=#f00]remark[/color][/b]: lesson here, actually draw the circles :skull: solved it pretty quickly after i drew the excircle

ok i really need to :sleeeeeeeeppppppp: now


Is $ {n^2\plus{}3n\plus{}5}$ ever disible by $ {121}$ for any $ {n}$?.
What is infinite descent with respect to unique techniques of problem solving?

[hide]We need to show that if n^2+3n+5 is divisible by 11, then it is not divisible by 121.

n^2+3n+5=n^2-8n+16 mod11 = (n-4)^2 mod11

This means that 11 divides n^2+3n+5 if n=11k+4. Substituting this into our original expression we obtain:

(11k+4)^2 + 3(11k+4) + 5
121k^2 +88k +16 +33k +12 +5
121k^2+121k+33
121k(k+1) + 33 = 33mod121

Therefore 121 never divides n^2+3n+5.[/hide]

And you can read about infinite descent here: http://en.wikipedia.org/wiki/Infinite_descent.


real analysis
calculus
integration
function
real analysis solved

The Vitali's Covering Theorem on my textbook is:

$E\in \mathbb{R}$ and $m^*(E) < \infty$. $\Gamma$ is a Vitali's covering of $E$, then for any $\varepsilon > 0$, there exist finite disjoint $I_j \in \Gamma (j=1,2,\dots,n)$ such that $m^*\left( E \setminus \bigcup\limits_{j=1}^n I_j\right) < \varepsilon$

My question is:

Prove that the conclusion can be modified as : there exists countably many $\{I_j\}$ such that $m^*\left( E \setminus \bigcup\limits_{j\geq 1} I_j\right) = 0$

liyi, did you by any chance post, some time ago, the following problem?

Given a finite set of intervals $I_1,I_2,\ldots,I_n$, there is a subset of pairwisely disjoint intervals $I_{k_1},\ldots,I_{k_m}$ s.t. $m(\bigcup_{j=1}^m I_{k_j})\ge\frac 13\cdot m(\bigcup_{t=1}^n I_t)$.

I think I can remember it being solved by Myth, and I also think (not sure about this, I have to check some details) this problem here can be solved by employing the one I mentioned above (actually, while searching for "Vitali covering" with Google, I found a page which presented this lemma in the same article as the theorem, so I assumed they used it in the proof; they also had proofs, but I didn't read those :)).

Yes, I remember something like that.
I even drew a special figure for this problem. :)

Ok, I'll assume your $\Gamma$ is a sequence of intervals and $E$ is an interval (for simplicity; I don't know if it's used :)), since this is part of the statement of the Vitali Theorem I know.

First of all, given a sequence of intervals $(I_n)$ which covers an interval $E$, it's easy to show that for each $\varepsilon>0$, we can find $n_{\varepsilon}$ s.t. $m(I_1\cup \ldots\cup I_{n_{\varepsilon}})>m(E)-\varepsilon$ (we can use the countable additivity of $m$).

Now take a finite set of intervals covering a subset of $E_1$ (I renamed $E$) with measure $>m-\varepsilon_1$ ($m$ is the measure of $E$), and extract from it a disjoint subset $\mathcal S_1$ covering a set of measure $>\frac{m-\varepsilon_1}3$. We now have an uncovered set of measure $<\frac 23\cdot m+\frac{\varepsilon_1}3$, and a sequence of intervals which Vitali covers it, namely those intervals of the initial Vitali cover which are disjoint from the intervals in $\mathcal S_1$. We can repeat the process above, choosing a finite set of intervals $\mathcal S_2$ which covers "almost" $\frac 13\cdot$ (the measure of $E_2=E_1\setminus\bigcup _{I\in\mathcal S_1}I$), and so on.

This proves the first part of the theorem, because the measure covered by the intervals in $\mathcal S_1\cup\mathcal S_2\cup\ldots\cup \mathcal S_n$ for some $n$ can be made arbitrarily close to $m\cdot(\frac 13+\frac 2{3^2}+\frac{2^2}{3^3}+\ldots+\frac{2^{n-1}}{3^n})$, which goes to $m$ as $n\to\infty$. However, this construction has an important feature: not only does it show that we can choose finite sets of disjoint intervals so that they cover a measure arbitrarily close to $m$, but it also shows that we can find increasing sets of such intervals, meaning that each set of intervals includes the preceding set in our construction.

To get from here to the actual theorem we want to prove requires a tiny bit more work and I'm still not sure about it, but I think you can see what I mean.

Sorry if it eventually turns out that this doesn't prove the theorem..

Excuse me for my question. Is it slightly modified proof of Vitali's theorem?

I think so :). I haven't seen a proof of Vitali's theorem, and, frankly, I don't even know which one of these is considered to be Vitali's theorem :). liyi says it's the first one, involving finite sets of intervals. I looked it up with Google, and I found that the second one is usually referred to as being Vitali's Covering Theorem, so I'm a bit confused here.. :) Anyway, I'm sure this imitates a standard proof.

When I want to prove the Lebesgue theorem on the differentiaion of the integral by way of proving that the Hardy-Littlewood maximal function is weak type 1,1, the covering lemma I need is the one grobber wrote in #2: you can, from the collection of covering intervals, select a pairwise disjoint family that covers at least a fixed fraction of the original set. Grobber quoted that fraction as 1/3; I've quoted and used the weaker 1/5. That doesn't really matter. The same thing can be done for balls in $\mathbb{R}^n.$

Now, out of all the covering lemmas, which one is Vitali, and which one is Wiener, and which one goes by some other name - that I'm not so sure of.

Yes. Most books says the conclusion is the second one...
However, they don't have big difference in the proofs.

[quote="liyi"][quote="grobber"]liyi, did you by any chance post, some time ago, the following problem?

Given a finite set of intervals $I_1,I_2,\ldots,I_n$, there is a subset of pairwisely disjoint intervals $I_{k_1},\ldots,I_{k_m}$ s.t. $m(\bigcup_{j=1}^m I_{k_j})\ge\frac 13\cdot m(\bigcup_{t=1}^n I_t)$.

[/quote]

hehe, you don't remember it correctly.
I think what I said is
Given a finite set of open intervals $I_1,\dots,I_n$, and $m(I_k\cap E)\geq \frac{2}{3}m(I_k)$. Show that $m(E\cap(\bigcup_{k=1}^n I_k) \geq \frac{1}{3}m(\bigcup_{k=1}^n I_k).$[/quote]
I think it is exactly our case, since we may assume $I_k\subset E$, i.e. $m(I_k\cap E)= m(I_k)$.

I don't know.. In this case, that problem (the one liyi posted some time ago) doesn't say anything, since $m(E\cap(\bigcup_{k=1}^n I_k))=m(\bigcup_{k=1}^n I_k)$, so it's trivially $\ge\frac 13\cdot m(\bigcup_{k=1}^n I_k)$.

Maybe I misunderstood what you meant, Myth? :?

Oh, you are right. But I remember my proof. If we have the finite system of segments $\{I_k\}$, $I_k\subset E$ and $\cup I_k=E$, then we can extract subsystem $\{I_{k_n}\}$ s.t. $m(\cup I_{k_n})\geq\frac{1}{2}m(E)$.
Do you need this fact, don't you?

With $\frac 13$ instead of $\frac 12$ (I don't know if it's true with $\frac 12$). Actually, a Kent Meryfield said, it's enough to show that we can extract a disjoint family of intervals that cover $\alpha\cdot m(I_1\cup\ldots\cup I_n)$, where $0<\alpha<1$ is some fixed real. For the one with $\frac 13$, it can be done by a greedy argument (select the longest interval, then the longest interval that doesn't intersect the first one, and so on).

I know that it is sufficient to do it with constant $\alpha$...
Anyway, I think everyone got the idea.


i solved this question the following way.....
the inner radius of cross section of the refill is 1mm
height upto which ink is filled is about 11cm
vol=pi r^2 h

width of line is about 0.4 mm
length of wire = x
height upto which ink rises on paper is about 0.001 mm
vol = length x breadth x height

equating these 2 volumes we get x= 830m or so

i wrote 820 m
I assumed Aniket to be between 15-20 years of age....so that he holds the pen hard and so the ink coming out of the pen in drawing a dot is say $ 0.1 ml$....
Also Sometimes his hand may tremble(Not due to fear :D ) and he actually may end up drawing a curve which we cannot include in the straight line but the ink is wasted say he draws a semi circle of radius $ r\equal{}0.2 mm$...and this happens for every 0.5 m he draws a line
Also the paper quality may be bad he may have the ink may blot at some places and he may have to use more ink say sucha thing happens every 1 m  of drawing a line.....and he uses extra $ 10^{\minus{}6} ml$
Now let's concentrate on the thickness of the line he draws....Say it's on a normal $ 0.1 mm$...but sometimes due to bad quality of paper or due to irritation on his part he may end up doing it more thick say he does so every 1.5 m he draws a line and say in such a case the height is increased by $ 10^{\minus{}6}m$...
Using this form a equation :)

Why don't you merge this with the other thread moderator?

bcos my dear cyber boy our moderator is busy wid many other things (like for instance pming and things like that)   :D   :wink:

dei venna da ...mude rohit  :P
shreyas ok  i forgot to do so this morn was busy for FIITJEE


Hey guys, go ahead and check out

http://iSabio.com

for fully animated solutions to this year's AMC 10 problems.

Make sure to use Internet Explorer because firefox has a couple issues.

The solutions are actually pretty good. The only bad part is the guys jokes. :D

Dr. CHoi always does those... I prefer to ignore it..


Let $A,B,C \in \mathcal l M_3(\Re)$ such that $A^3=B^3=C^3=O_3$ and $A,B,C$ commute one with each other . Prove that $ABC=O_3$.

a generalization has already been posted ..

Sorry  :blush:  could you give the link pls?

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=nilpotent&t=42447
Or if you don't want to use field extension, notice that $r_k=Rank(A_1A_2..A_k)$ is stricly decreasing.


calculus
integration
analytic geometry
symmetry
geometry
LaTeX
calculus computations

In this case,
r=4*cos(3(theta))
How should I set the lower and upper limits? Is it about symmetry?
And in general, how can I determine the lower and upper limits in a faster way?

Thanks for your attention!

================
Sorry for my mistake
Actually, I'm trying to find the area of the region enclosed by this equation

You have not given any explanation of just what you are trying to compute, and without that explanation, no one can help you. Please write a coherent explanation of what the underlying problem is. (And if you do that, there's a chance that you will see the answer yourself.)

In addition try to use Latex - that makes mathematical writing much more amenable.

In this particular case, "the area of the region enclosed by this equation" is somewhat ambiguous: the graph is that of a three-petaled rose, and each petal is traced out twice, once for $ r<0$ and once for $ r>0$, as $ \theta$ goes from $ 0$ to $ 2\pi.$

You need to understand the graph even to understand what you are asking.


function
algebra
floor function
real analysis
real analysis unsolved

For any $n \in {\mathbb N}$ we have $[na^3]+[nb^3]+[nc^3]=[3nabc]$ where $a,b,c$ are reals. Show that $a=b=c$ or $a+b+c=0$.

 :ninja:

[quote="jianuovidiu"]For any $n \in {\mathbb N}$ we have $[na^3]+[nb^3]+[nc^3]=[3nabc]$ where $a,b,c$ are reals. Show that $a=b=c$.

 :ninja:[/quote]

counter example:
$n=1,a=b=0, c=1/2$.  BTW, $[ \quad]$ is a floor function? or something else??

[quote]counter example:
$n=1,a=b=0, c=1/2$.  BTW, $[ \quad]$ is a floor function? or something else??[/quote]

hey read carefuly the problem.
[x] means integer part of x

p.s. i've edited  :D


A [i]permutation[/i] of the integers $1901, 1902, \cdots, 2000$ is a sequence  $a_1, a_2, \cdots, a_{100}$ in which each of those integers appears exactly once. Given such a permutation, we form the sequence of partial sums
\[s_1 = a_1,\;\;s_2 = a_1 + a_2,\;\;s_3 = a_1 + a_2 + a_3, \; \ldots\;, \; s_{100} = a_1 + a_2 + \cdots + a_{100}.\]
How many of these permutations will have no terms of the sequence $s_1, \ldots, s_{100}$ divisible by three?

Among the numbers 1901, 1902, ..., 2000, there are 34 congruent to 2 mod 3 and 33 congruent to 1 and 0 mod 3.  Note that if $s_n$ is congruent to 2, $a_{n + 1}$ must be congruent to 2 or 3, while if $s_n$ is congruent to 1, $a_{n +1}$ must be congruent to 1 or 3.  Note that it doesn't matter at all where in the sequence the numbers congruent to 0 mod 3 go -- if we were to delete all of them, everything would still work out fine.  So, let's do that and add them back in later.  Now, we have a sequence $s_n$ that, reduced mod 3, either goes 1, 2, 1, 2, 1, ... or 2, 1, 2, 1, 2, ... and has a corresponding sequence $a_n$ which either goes 1, 1, 2, 1, 2, 1, 2, ... or 2, 2, 1, 2, 1, ....  Because we have more terms congruent to 2 mod 3, we must be in the second case, so the positions of all the numbers 1 and 2 mod 3 are fixed, giving us 34!*33! possible permutations.  We have to add back in the numbers congruent to 0 mod 3, however.  We now have 67 places they can go (at any point other than being the first term of the sequence).  This gives us a nice factor of ${99 \choose 33}$ multiplied by another 33! for arranging them, giving us a final total of $\frac{99!\cdot 34!\cdot 33!}{66!}$.


1)Let be $ x\in \mathbb{R}$ . Solve this equation :

$ \left\lfloor \sqrt{\frac{x\plus{}7}{2}}\right\rfloor\equal{}\frac{x\plus{}1}{2}$

2)Let be $ a,x\in \mathbb{R}$ . Solve this equation :

$ ax\plus{}a\lfloor x\rfloor\equal{}a$

3)Let be $ x\in \mathbb{R}$ . Solve this equation :

$ \left\lfloor \frac{x\plus{}1}{3}\right\rfloor\equal{}\frac{3x\plus{}5}{7}\plus{}\left\lfloor \frac{1\minus{}2x}{6}\right\rfloor$

4)Solve this equation in $ \mathbb{R}$ :

$ \lfloor x^2\rfloor\plus{}1\equal{}2\lfloor x\rfloor$

Doesn't 2 just become $ x\plus{}\lfloor x \rfloor \equal{} 1$, a fairly trivial equation, for nonzero $ a$?

Yes , you can do that , but can you find a solution ?

It's easy.

$ x\equal{}1\minus{}\lfloor x \rfloor \in \mathbb{Z}$

so

$ \lfloor x \rfloor \plus{}x\equal{}2x\equal{}1$

$ \frac{1}{2} \not \in \mathbb{Z}$

First one will be

$ LHS \in \mathbb{Z} \Rightarrow x\equal{}2k\minus{}1 \quad k\in \mathbb{Z}$
$ x\ge \minus{}7$

Our equation becomes

$ k \le \sqrt{k\plus{}3}<k\plus{}1$

rest is easy.

[hide="#4"]Put $ x\equal{}n\plus{}\alpha$ where $ n\equal{}[x], \alpha\equal{}\{x\}$

Then $ n^2\plus{}[2n\alpha\plus{}\alpha^2]\plus{}1\equal{}2n\quad(*)$

From the given equation, $ 2n\equal{}[x^2]\plus{}1$, hence $ n$ is strictly positive.

Now $ (*)$ becomes $ (n\minus{}1)^2\plus{}[2n\alpha\plus{}\alpha^2]\equal{}0$

Since $ n$ is strictly positive, the expression under the integer part sign is non-negative. Therefore, the only possibility is $ n\equal{}1\land 0\leqslant 2n\alpha\plus{}\alpha^2<1$. Then we get $ 1\leqslant (1\plus{}\alpha)^2<2\iff 1\leqslant 1\plus{}\alpha<\sqrt{2}\iff 0\leqslant\alpha<\sqrt{2}\minus{}1$

That gives the interval of solutions: $ x\equal{}n\plus{}\alpha\in[1,\sqrt{2})$[/hide]

[hide="#3"]$ {3x \plus{} 5\over 7}$ is obviously integer, so $ {3x \plus{} 5\over 7} \equal{} k\implies x \equal{} {7k \minus{} 5\over 3}$. Plugging that into the equation and simplifying, we get

$ \left[{7k \minus{} 2\over 9}\right] \equal{} k \plus{} \left[{13 \minus{} 14k\over 18}\right]\quad(*)$

Now let $ \alpha \equal{} \left\{{7k \minus{} 2\over 9}\right\}, \beta \equal{} \left\{{13 \minus{} 14k\over 18}\right\}$. Then

$ {7k \minus{} 2\over 9} \minus{} \alpha \equal{} k \plus{} {13 \minus{} 14k\over 18} \minus{} \beta$

After rearranging and solving for $ k$, we get

$ k \equal{} {17 \plus{} 18(\alpha \minus{} \beta)\over 10}$

By the definition of $ \alpha,\beta$, we have $ \minus{} 1 < \alpha \minus{} \beta < 1$, hence $ \minus{} {1\over 10} < k < {7\over 2}$, therefore $ k \equal{} 0,1,2,3$. Checking those values in $ (*)$ shows that only $ k \equal{} 1,2$ work. Therefore $ x \equal{} {7k \minus{} 5\over 3}\in\left\{{2\over 3}, 3\right\}$[/hide]

[hide="#1"]$ {x\plus{}1\over 2}\equal{}k\in\mathbb{Z}\implies x\equal{}2k\minus{}1$, so the equation becomes $ [\sqrt{k\plus{}3}]\equal{}k$, and $ k$ is obviously non-negative (because of the root). Let's now check $ k>\sqrt{k\plus{}3}$, because for such values of $ k$ the equation can't possibly be satisfied. We get $ k\geqslant 3$, hence it's sufficient to check $ k\equal{}0,1,2$, and only $ k\equal{}2$ works. Therefore $ x\equal{}2k\minus{}1\equal{}3$ is the only solution.[/hide]

[b]2)[/b]For non-zero $ a$, the equation $ \Leftrightarrow x\plus{}\lfloor x \rfloor\equal{}1$, so $ x$ is an integer or $ 1$ is not an integer, impossible.

So it becomes $ 2x\equal{}1 \Leftrightarrow x\equal{}\frac{1}{2}$, which is not an integer, contradiction, so there is no solution.

[b]4)[/b] We have $ \lfloor x^2 \rfloor \ge (\lfloor x \rfloor)^2............(1)$

So the equation $ \Rightarrow (\lfloor x \rfloor\minus{}1)^2 \le 0$, so $ \lfloor x \rfloor\equal{}1$

Equality of $ (1)$ holds $ \Leftrightarrow 2 >x^2 \ge 1$, together with $ 2>x \ge 1$ gives $ x \in [1, \sqrt{2})$ as solutions.


The coordinates of one of the endpoints of a diagonal of a rectangle are $ (\minus{}4,2)$, and the coordinates of the point of intersection of the diagonals are $ (1,\minus{}1)$. The sides of the rectangle are parallel to the axes. What is the number of square units in the area of the rectangle?

The diagonals of a rectangle bisect each other, meaning their midpoint is the point of intersection.

WE have two corners of the rectangle at (-4,2) and (x,y).

(x-4)/2=1

x=6

(y+2)/2=-1

y=-4

The endpoints of the diagonal of the rectangle are (-4,2) and (6,-4)

It has a width of 6-(-4)=10 and a height of 2-(-4)=6 for an area of 60.


A round hole is to be completely covered with two square boards. The sides of the squares are 1 metre. In what interval may the diameter of the hole vary?

[i]Kömal, problem A.269, September, 2001[/i]
http://komal.elte.hu/verseny/2001-09/A.e.shtml

i believe it is 2- \sqrt 2
but i haven't proved it.

You mean the maximum radius is that, right? I got the same thing, but I have no proof, just some intuitive observations,

First of all, we can "push" the 2 squares s.t. they still cover the disk but 2 sides of each of the squares is tangent to the circle.

I then aligned the 2 squares along a diagonal and I looked for the largest disk which can be covered by such a construction. I found it to be the disk of radius 2- \sqrt 2, as you said, so up until now the answer seems to be (0, 4-2 \sqrt 2] (we're asked for the interval in which the diameter may vary).


integration geometry inequalities calculus function rectangle trapezoid

[img]http://www.xthost.info/skydragun/Picture%20170.jpg[/img]

can anyone solve these?

They're both equal to $\frac32.$ The only thing you really need about the sum is this: For any $p>0$, there is a constant $C$ which depends on $p$ such that

$\sum_{k=1}^mk^p=Cm^{p+1}+O(m^p).$

You don't even need to know what $C$ is for this particular problem - just that it's some constant.

wait can you explain in a little more detail. what rule is this that you are using.

this is part of my cal 2 project, and its just an extra credit problem, so we probably havent covered the concept in class

Zhaobin's comment is essentially what I was using. It's not any more complicated than that. Draw a picture of the sum as the areas of boxes, either below or above the curve $y=\sqrt{x}.$

[quote="zhaobin"]I think you can also use this equality:$\int_1^{2^n}\sqrt{x}dx< \sum_{k=1}^{2^n}\sqrt{k} <1+\int_1^{2^n}\sqrt{x}dx$[/quote]

can you explain me how did you get this ineqs? i'm quite new to calculus...

Zhaobin's estimate is not quite correct; in particular, his right-hand inequality doesn't hold.

Let's consider the integral $\int_0^m\sqrt{x}\,dx.$

The function $f(x)=\sqrt{x}$ is an increasing function. I would strongly suggest drawing a picture. Partitition the interval $[0,m]$ with integers: $\{0,1,2,\dots,m-1,m\}.$

As an increasing function, the integral is greater than its left-endpoint Riemann sum and less than its right-endpoint Riemann sum. (This is where the drawings with the rectangles above and below the graph come in.) Thus,

$\sum_{n=0}^{m-1}\sqrt{n}<\int_0^m\sqrt{x}\,dx<\sum_{n=1}^m\sqrt{n}$

We can rearrange the right hand inequality to $\frac23m^{3/2}<\sum_{n=1}^m\sqrt{n}.$

We deal with the left hand inequality by ignoring the vanishing $n=0$ term and by adding $\sqrt{m}$ to both sides. We get the following:

$\frac23m^{3/2}<\sum_{n=1}^m\sqrt{n}<\frac23m^{3/2}+\sqrt{m}.$

This is more than accurate enough for the purposes of the original problem.

If you want more, you can average the two sides of this estimate. The difference between $\sum_{n=1}^m\sqrt{n}$ and $\frac23m^{3/2}+\frac12\sqrt{m}$ is bounded.

thanks. i understand now how to use the estimates.

but why is the difference in the last paragraph bounded?

and what if $m \to \infty$? is the difference in this case about $0$? (though i don't think so)

Roughly speaking, estimating the error in that last approximation is an error estimate for integration by the trapezoid rule. The magnitude of the error increases with $m$ but remains less than 1 as $m\to\infty.$

With a little numerical extrapolation, I get that for large $m,$

$\sum_{n=1}^m\sqrt{n}\approx \frac23m^{3/2}+\frac12\sqrt{m}-.2079$

This is also overkill; prelewd's original problem does not require this level of precision.


$ \begin{cases}\minus{}11.7\cos(x) \plus{} 11.4\cos(y) \equal{} 0\\
11.7\sin(x) \plus{} 11.4\sin(y) \equal{} 4.9\end{cases}$

Squaring both sides and adding them up:

$ 11.7^2\plus{}11.4^2\minus{}2\cdot 11.4\cdot11.7\cdot\cos(x\plus{}y)\equal{}4.9^2$

$ \cos(x\plus{}y)\equal{}\frac{11.7^2\plus{}11.4^2\minus{}4.9^2}{2\cdot 11.4\cdot11.7}\simeq 0.91$

$ \therefore\;x\plus{}y\equal{}\cos^{\minus{}1}0.91$

Substituting in the first equation:

$ 11.4\cos y\equal{}11.7\cdot \cos(\cos^{\minus{}1}0.91\minus{}y)$

$ 11.4\cos y\equal{}11.7\cdot(0.91\cdot \cos y\plus{}0.41\cdot \sin y)$

$ 0.64\cos y\equal{}4.797 \sin y$

$ 0.41(1\minus{}\sin^2y)\equal{}23\sin^2y$

$ \sin y\equal{}0.13\;\Rightarrow\;y\simeq 7.6^\circ$

And so on...
This is ugly, hope it is correct.

[quote="triplebig"]$ \begin{cases} \minus{} 11.7\cos(x) \plus{} 11.4\cos(y) \equal{} 0 \\
11.7\sin(x) \plus{} 11.4\sin(y) \equal{} 4.9\end{cases}$

Squaring both sides and adding them up:

$ 11.7^2 \plus{} 11.4^2 \minus{} 2\cdot 11.4\cdot11.7\cdot\cos(x \plus{} y) \equal{} 4.9^2$

$ \cos(x \plus{} y) \equal{} \frac {11.7^2 \plus{} 11.4^2 \minus{} 4.9^2}{2\cdot 11.4\cdot11.7}\simeq 0.91$

$ \therefore\;x \plus{} y \equal{} \cos^{ \minus{} 1}0.91$

Substituting in the first equation:

$ 11.4\cos y \equal{} 11.7\cdot \cos(\cos^{ \minus{} 1}0.91 \minus{} y)$

$ 11.4\cos y \equal{} 11.7\cdot(0.91\cdot \cos y \plus{} 0.41\cdot \sin y)$

$ 0.64\cos y \equal{} 4.797 \sin y$

$ 0.41(1 \minus{} \sin^2y) \equal{} 23\sin^2y$

$ \sin y \equal{} 0.13\;\Rightarrow\;y\simeq 7.6^\circ$

And so on...
This is ugly, hope it is correct.[/quote]

 thanks for the reply, I think it should be cos(x-y) , x-y = cos^-1 (0.91)
was there anything wrong with my method?

 thank you

No problem.

Unless I am doing something wrong, when you add them up you get the middle term:

$ (\sin x\sin y \minus{} \cos x\cos y) \equal{} \minus{} (\cos x \cos y \minus{} \sin x \sin y) \equal{} \minus{}\cos (x \plus{} y)$

Remember the signs are switched in the cosine formulas.

[quote]$ ( \minus{} 11.7)^2 \cos(x)^2 \plus{} (11.4)^2 \cos(y)^2 \equal{} 0$[/quote]

Why is this? Shouldn't there be a $ 2\cdot 11.7\cdot 11.4\cdot \cos x\cos y$ in the middle?

[quote="triplebig"]No problem.

Unless I am doing something wrong, when you add them up you get the middle term:

$ (\sin x\sin y \minus{} \cos x\cos y) \equal{} \minus{} (\cos x \cos y \minus{} \sin x \sin y) \equal{} \minus{} \cos (x \plus{} y)$

Remember the signs are switched in the cosine formulas.

[quote]$ ( \minus{} 11.7)^2 \cos(x)^2 \plus{} (11.4)^2 \cos(y)^2 \equal{} 0$[/quote]

Why is this? Shouldn't there be a $ 2\cdot 11.7\cdot 11.4\cdot \cos x\cos y$ in the middle?[/quote]

-11.7cos(x) = -11.4cos(y)
-11.7cos(x) + 11.4cos(y) = 0
(-11.7cos(x))^2 = (-11.4cos(y))^2
(-11.7cos(x))^2 + (11.4cos(y))^2 = 0 ?

Oh, I did something different, after squaring and combining the equations, 117.76sinasinb = 117.76cos(a-b) where 117.76cosacosb = 0 and 117.76sinasinb = 0.91

would this method work out in the end? it seems like it should.. thanks a lot :)

hello, rewriting your system in the form
$ \minus{}a\cos(x)\plus{}b\cos(y)\equal{}0$
$ a\sin(x)\plus{}b\sin(y)\equal{}c$
so we get
$ \minus{}a\cos(x)\equal{}\minus{}b\cos(y)$
$ a\sin(x)\equal{}c\minus{}b\sin(y)$
adding and squaring we get
$ a^2\equal{}b^2\plus{}c^2\minus{}2bc\sin(y)$
therefore
$ \sin(y)\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}$
Sonnhard.


inequalities
function
inequalities proposed

Let $a,b,c,d$ be positive reals whose sum is 4. Prove that \[\frac{1}{\sqrt{a+2b+c}}+\frac{1}{\sqrt{b+2c+d}}+\frac{1}{\sqrt{c+2d+a}}+\frac{1}{\sqrt{d+2a+b}}\geq2.\]

Alas, by Cauchy-Schwarz (I hope I spell it right) inequality: \[LHS(\sqrt{a+2b+c}+...+\sqrt{d+2a+b})\geq 16\] so if we prove that $16\geq 2 (\sqrt{a+2b+c}+...+\sqrt{d+2a+b})$ that's it, and the last
inequality comes directly from AM - QM inequality.
[edit] The equality obviously holds for $a=b=c=d=1$

this is very easy! :|

use jensen for the convex function $\frac{1}{x}$ and you get :
$\sum_{cyc}\frac{1}{\sqrt{a+2b+c}}\geq \frac{4}{\sqrt{\frac{4(a+b+c+d)}{4}}}=2$

Why must you use Convex function? It's more complex than AM-GM. And product of two sums with n terms not less than square of n, is very familiar.


Let $n$ be the natural number whose decimal representation is just composed from $k$ ones, e.g. $\underbrace{11 \ldots 11}_{k \quad digits}$ .

a.) In dependance of $k$, analyse how many digits $n^2$ has and with which digit it begins.

b.) We notice: $11^2=121$ and $111^2=12321.$ How many such symmetric numbers, palindromes, are there. For which $k$'s is $n^2$ a symmetric number ?


geometry
Putnam
HMMT
geometric transformation
reflection
number theory

[b]DIVISION A RESULTS[/b]

Power Test (out of 65 points)
1. Quagga with 58
2. Pig, aka Wheelchair Epidemic, with 50
3. Colbert with 49
4. Yellow Pigs with 47
5. Thomas Jefferson A with 43

Team Test (out of 40 points)
1. Thomas Jefferson A with 35
2.5. Quagga with 31
[b]2.5. AoPS with 31[/b] Note: AoPS was erroneously left out during the awards ceremony.
4. North Carolina with 30

[b]Individual Finals[/b]
1. Shaunak Kishore
2. Bryce Taylor
3. Matt Superdock
4. John Berman
5. Gaku Liu
6. Jacob Steinhardt
7. Wenyu Cao
8. Haitao Mao
9. Andrew Ardito
10. Chris Olund
(I'm not 100% sure of that ordering; I don't have the records of the tiebreaking, so I tried to construct it from memory and people's comments.)

Algebra (out of 40 points)
1. Palmer Mebane from Quagga with 20
2. Wenyu Cao from Quagga with 18
3. Hyunji Bae from North Carolina with 17
4. Neel Patel from South Brunswick Duckies with 16
5. Tao Ran Chen from Colbert with 15
7.5. Brian Hamrick from TJ A with 12
7.5. Ian Askins from AAST 1 with 12
7.5 Dan Li from TJ A with 12
7.5 Kam Chuen Chan from Colbert with 12

Geometry (out of 40 points)
1.5. Shaunak Kishore from Quagga with 28
1.5. Haitao Mao from TJ A with 28
3. Alex Anderson from Quagga with 22
4. Kevin Lee from AAST 1 with 21
5.5. Tao Ran Chen from Colbert with 20
5.5. Wenyu Cao from Quagga with 20
7. Damien Jiang from North Carolina with 18
8.5 Dan Li from TJ A with 16
8.5 Ian Frankel from Aztec Gold with 16

Number Theory (out of 40 points)
1. Jacob Steinhardt from TJ A with 22
2. Andrew Ardito from AAMC A with 19
3. John Berman from North Carolina with 16
4.5 Chris Olund from TJ B with 15
4.5 Bryce Taylor from North Carolina with 15
7.5 Palmer Mebane from Quagga with 14
7.5 Eric Morphis from AoPS with 14
7.5 Krishanu Sankar from Colbert with 14
7.5 David Bieber from AAMC A with 14
10. Taylor Sutton from AoPS with 13

Combinatorics (out of 40)
1. Jacob Steinhardt from TJ A with 23
2. Matt Superdock from Quagga with 22
3. Kevin Lee from AAST 1 with 19
4. Michael Wu from Aztec Gold with 18
5.5. Gaku Liu from Quagga with 16
5.5. Haitao Mao from TJ A with 16
7. John Berman from North Carolina with 15
10. Louis Evans from Goat (aka Gladiator) with 14
10. Justin Kopinsky from Yellow Pigs with 14
10. Adam Sealfon from Pig (aka Wheelchair Epidemic) with 14
10. Schuyler Smith from AAMC A with 14
10. Andrew Ardito from AAMC A with 14

Sweepstakes score was calculated as follows:
4 * Power Test score + 5 * Team Test score + top 4 scores from each individual test

Sweepstakes (out of 1100)
1. Quagga with 606
2. Thomas Jefferson A with 516
3. North Carolina with 468
4. Colbert with 446
5. AAMC A with 411

[b]DIVISION B RESULTS[/b]

Power Test (out of 65 points)
1. Florida B with 40
2. AAST 4 with 36
3. White Station with 27
4. AAST 5 with 25
5. South Pirates (aka Pirate's Cove?) with 24

Team Test (out of 40 points)
1. Florida B with 26
2.5 AAST 4 with 23
2.5 South Pirates (aka Pirate's Cove?) with 23
4. AAST 6 with 16
5. Worcester Academy B with 15

[b]Individual Finals[/b]
1. Preet Patel
2. John Chiarelli
3. Gene Katsevich
4. Troy Wu
5. Zev Rosengarten
6. Jeffrey Chao
7. Shawn Pan
8. Ed Zheng
9. Matt Myers
10. Tong Chen or Raja Rajan Vel (I can't tell which from the score sheet I have)

Algebra (out of 40 points)
1. Gene Katsevich from Florida B with 25
2.5. Raja Rajan Vel from Charter School of Wilmington with 18
2.5. Jeffrey Chao from Charter School of Wilmington with 18
7.5 Taylor Clarke from White Station with 13
7.5 Jeffrey Cai from AAST 4 with 13
7.5 John Chiarelli from AAST 4 with 13
7.5 Peter Kwon from AAST 6 with 13
7.5 David Huang from Charter School of Wilmington with 13
7.5 Annie Huang from Florida B with 13
7.5 Nikhil Narula from AAST 4 with 13
7.5 John Song from White Station with 13

Geometry (out of 40 points)
1. Gene Katsevich from Florida B with 23
2. Tong Chen from White Station with 16
3. Rui Hu fom Charter School of Wilmington with 14
5. Ian Harding from St. Andrews Saints with 12
5. Kevin Dong from Florida B with 12
5. Surya Viswanathan from Worcester Academy B with 12
7. Anna Lin from Florida B with 11
8. Jeffrey Fan from Lawrenceville with 10
9. Mengfei Yang from Charter School of Wilmington with 9

Number Theory (out of 40 points)
1. Zev Rosengarten from Lawrenceville with 28
2. Shawn Pan from White Station with 19
3.5. Troy Wu from White Station with 18
3.5. Preet Patel from Florida B with 18
5. Jeffrey Chao from Charter School of Wilmington with 16
6. Hannah Gramling from Howard County Math Team with 15
8. David Huang from Charter School of Wilmington with 14
8. Nikhil Narula from AAST 4 with 14
8. Raja Rajan Vel from Charter School of Wilmington with 14

Combinatorics (out of 40 points)
1. Preet Patel from Florida B with 24
2. Ed Zheng from Lawrenceville with 20
3. Shawn Pan from White Station with 16
4.5. Matt Mayers from AAST 4 with 15
4.5. John Chiarelli from AAST 4 with 15
6.5. Tommy Gill from Charter School of Wilmington with 11
6.5. Michael Tan from AAST 5 with 15
9.5. Theo DuBose from St. Andrews Saints with 8
9.5. Paul Hwang from AAST 5 with 8
9.5. Yang Tang from Florida B with 8
9.5. Zev Rosengarten from Lawrenceville with 8

Sweepstakes score was calculated as follows:
4 * Power Test score + 5 * Team Test score + top 4 scores from each individual test

Sweepstakes (out of 1100)
1. Florida B with 479
2. AAST 4 with 372
3. Charter School of Wilmington with 291
4. White Station with 262
5. AAST 5 with 261

Sorry for the poor formatting. It was already a lot of work to type all of that; making it into a nice table is just too much.

Three notes:
1) These results, and possibly more, as well as statistics, will be posted on our website soon. If I have time I might go through the tests and collect statistics for each problem too.

2) I did the rankings Putnam-style, i.e. if there was a tie, I averaged all of the places that get used up rather than giving everyone the first of the places that's covered. This should explain why people did/didn't make the finals. On our website, when the results are posted, we will show the more common (and more flattering) way of ranking.

3) If you are wondering how you or your team would have done in the other division, remember that:
a) We didn't grade Power Test as leniently in A as we did in B. So if a team moved up to B the Power Test score would probably go down.
b) The team tests were the same.
c) There were 15 questions for each individual subject area. The first 10 were on the B division test and the last 10 were on the A division test.

My name is Eric Morphis. I am 7.5th place in number theory and would like to protest my score in number theory. I believe my actual score was seventeen. There is, of course, a possibility that I mis-marked my answers, but if possible, I would like you to review my number theory test. Thanks.

Edit: Ignore the last edit.

[quote="Twiz"]My name is Eric Morphis. I believe my actual score was seventeen.[/quote]
I'll check.

If anyone else thinks they should have a different score, let me know. I'm probably going to go through the tests to compile stats about each question, so I can check the grading at the same time. After I do this, we will mail the graded answer sheets and tests back to your advisors, along with results and certificates and suchlike.

[quote="jb05"]Can you post results/scores for the individual finals? Also, how did you break ties for it?[/quote]
I don't have them on my computer but when I can get them I'll post them.

We broke ties by direct comparison of individual test results. If two people qualified from the same test and tied in the finals, the tie was broken by their scores in the qualifying test. I wasn't in the grading room, so I don't know if that broke all ties. If not, the next comparison was that if one person qualified for finals in two tests, and the person he tied with only qualified in one test, the first person was placed higher. So for example Jacob Steinhardt was guaranteed to win any tie in the finals. This was enough to determine the top 10 in each division with no ties.

Thanks for getting these results up so fast!

Also I thought my score on Combinatorics was 26, but I might have copied answers wrong as I usually do.  If you could check, that would be great.

-Matt Superdock

A lot of scores that were reported by people once the contest was done, such as Matt's above, and mine in Algebra (I thought it was 25) and in Number Theory (thought it was 18), as well as some others, seem to be lower. Maybe the answers were graded harshly or something? I figured the 5 point difference in my case is the fact that, for Algebra A 10, instead of writing $ \frac {(1 \plus{} \sqrt {5})}{2}^{1/5} \plus{} \frac {(1 \minus{} \sqrt {5})}{2}^{1/5}$, I wrote $ \sqrt [5]{\phi} \minus{} \frac {1}{\sqrt [5]{\phi}}$ and included that $ \phi \equal{} \frac {1 \plus{} \sqrt {5}}{2}$. Those are the same answer, but I didn't know if my form would be accepted.

EDIT: Mistake I just noticed. Haitao Mao, one of the Geometry test winners, is from TJ A, not Quagga. I didn't think we had swept the top 3 of that test...

hm it saids aops placed 2.5 and tied with Quagga, but at the awards ceremony, the places were:

1) TJ A
2) Quagga
3) North Carolina

so we were never included  :huh:  Was that just a mistake?

[quote="MellowMelon"]A lot of scores that were reported by people once the contest was done, such as Matt's above, and mine in Algebra (I thought it was 25) and in Number Theory (thought it was 18), as well as some others, seem to be lower. Maybe the answers were graded harshly or something? I figured the 5 point difference in my case is the fact that, for Algebra A 10, instead of writing $ \frac {(1 \plus{} \sqrt {5})}{2}^{1/5} \plus{} \frac {(1 \minus{} \sqrt {5})}{2}^{1/5}$, I wrote $ \sqrt [5]{\phi} \minus{} \frac {1}{\sqrt [5]{\phi}}$ and included that $ \phi \equal{} \frac {1 \plus{} \sqrt {5}}{2}$. Those are the same answer, but I didn't know if my form would be accepted.[/quote]
In your particular case, that's likely. I don't know exactly how strict the standards were for each question in the grading room. Although I personally probably would have accepted your answer, because it's actually right, when you write your answers you should consider that the people grading might not be able or willing to mentally check every single answer that differs from what's on the answer sheet. So you need to write the answer in the most obvious possible form.

Also, if you notice that the answer on the answer sheet is the same as yours but in a different form, you might want to come to the challenge/appeal room and ask whether your form will be accepted. Unfortunately it's now a bit late for that, but we'll make sure it's clear that you are allowed (and encouraged) to do this next year.

Tenoreoz: What I posted is correct. We made a mistake during the awards ceremony. I'll indicate this in the original post.

Wow, huge gaps in the sweepstakes top 3.

My score actually went up (8 -> 12, not very good still). I must have missed noticing that I got the problem that I was actually confident about on algebra.

Thanks for posting these! Do you know about when the tests will be available online?

[quote="Hamster1800"]Thanks for posting these! Do you know about when the tests will be available online?[/quote]
I'll do it tonight if none of the other officers beat me to it.

[quote="Xevarion"]Also, if you notice that the answer on the answer sheet is the same as yours but in a different form, you might want to come to the challenge/appeal room and ask whether your form will be accepted. Unfortunately it's now a bit late for that, but we'll make sure it's clear that you are allowed (and encouraged) to do this next year.[/quote]
Maybe also you could announce the individual finals cut-offs so that if someone thinks they should have made it, they can appeal?

In my case, my Combinatorics score of 23 was higher than the expected score of 22. Not that I'm complaining, but this seems to indicate that either lots of us fail at adding, or some scores were incorrectly recorded.

[quote="Twiz"]Maybe also you could announce the individual finals cut-offs so that if someone thinks they should have made it, they can appeal?[/quote]
Actually, we meant to print out the top 10 in each test and each division (like they are listed above) and post/distribute those, but that didn't end up happening. We'll be sure to do something to this effect in the future.

Regarding incorrectly recorded scores: It is entirely possible that this happens. When people are adding up more than 500 scores and then entering them into a database, it's nearly inevitable that there will be some errors. If you have a suggestion for how we could reduce this problem (besides computer grading, which seems difficult to implement), I would be delighted to hear it.

Actually, to partially answer my own question, in the future we might re-check all of the top 10 scores. We probably won't have the manpower to introduce much more redundancy into the grading of ALL tests, but we could at least add in a triple-check for the top scores to improve the likelihood that everything is on the straight and narrow for individual finals.

edit: This brings up an interesting question. Suppose that your grade on the test is the score you actually should get, plus a random variable $ X$ which is symmetric about 0 and has small standard deviation. (This is a good model of what happens in real life.) It seems like you'd all agree that this grading system is much less fair than perfectly accurate grading. This suggests that we should be able to devise some operator $ U(X)$ which maps random variables to the positive reals as a measure of unfairness. Of course $ U(0) \equal{} 0$, but what else can we say about $ U$? Does it matter if the mean of $ X$ is not 0? (It seems like it shouldn't; if $ X$ is constant, then the ordering is preserved, so changing $ X$ by a constant shouldn't change the unfairness of $ X$.) If we develop a good theory of unfairness here, PUMaC can try to create a grading system that will minimize unfairness in a rigorously demonstrable fashion!

[quote="Xevarion"]Twiz, MellowMelon, JSteinhardt, matt276eagles: your tests all appeared to have been graded correctly. If you still have/remember your answers, you can refer to the tests with solutions that I'm about to post. If not, you can wait until we mail you your stuff, which will be some time this week probably.[/quote]

You're right, I rechecked the answers I wrote down and my test was graded correctly after all.  Thanks for going through the trouble of rechecking that.

[quote="Xevarion"][quote="ProtestanT"]Jus wondering, for the #7 in the team test, would it have been marked correct if it was in degrees instead of radians?[/quote]
I would think that both should be accepted. However I wasn't in the grading room so I don't know what they decided to do. Sometimes, if a particular notation is used in the problem, the same notation is expected in the answer. I can't remember the wording of the problem so I don't remember if degrees were used...[/quote]

Funny story about that: one coach wanted to accept $ 60^\circ$ for $ \frac {2\pi}{3}$ before I assured him that they weren't, in fact, equal...

Have the packets with our results been mailed out yet????

No mail on Sundays... they'll be mailed tomorrow.

And how long do you expect it will be until the results are online? It has been weeks since the actual competition already...

Well there won't be anything different from what's in this thread. I'll bug the webmaster about it, though; thanks for reminding me.

I believe they were not in fact all from the same high school. We made an exception for that team because they told us everyone on the team was grade 9 or younger. It's clear that it was our mistake. In the future, assuming the nature of divisions does not change substantially, that rule will be enforced strictly.

I understand.

What a huge difference between Division A and Division B.  I put my top eight students in Division A, only to have them be blown away; while I put eight regular students in Division B, and they did reasonably well (although I don't know how many Division B teams there were, so maybe they were 10th of 12).

The A/B separation is brilliant, however, allowing "regular" high schools to face off against one another (and AAST) with easier questions while Quagga and TJ and the HCSSiM/PROMYS mixed teams can square off with harder questions.  The aforementioned rule 10 is pure genius -- it may not be perfect in all cases, but it does a good job of separating teams by ability.

The organizer of Florida B was not even aware of that rule when he made the registration. I assure you he meant no malice (he saved that for Division A).

The reason he had put that team in Division B in the first place is because he thought it was not going to even be a full team (2 people at first), but then he found another team to merge it with. In hindsight, Gene and Preet, both on Florida B, belonged in Division A. One of them has already expressed his disappointment on this forum that he couldn't take the more challenging Divison A tests.

In most contests with divisions and such, though, this almost always happens. The lower divisions are won by someone who should have been moved up.

[quote="rjyanco"]Thanks for the results.
Just to clarify, rule 10 (shown on the PUMaC website) says "Teams that do not come from a single high school must compete in the A Division," while the Division B results you listed show[quote="Xevarion"]Sweepstakes (out of 1100)
[b]1. Florida B with 479[/b]
2. AAST 4 with 372
3. Charter School of Wilmington with 291
4. White Station with 262
5. AAST 5 with 261[/quote] Were all eight students on Florida B from the same high school?  If so, congratulations to that high school![/quote]
Listen, we didn't register in the B division hoping to trump everyone unfairly.
Nearly all of the people who were supposed to be on Florida B (excluding Preet and Gene) dropped out, leaving the organizer with only two of the eight people needed. They merged with 6 people from NY, who preferred to be in the B division.

I think the 6 people out of 8 from NY were all from Stuy. My bad guys.

I didn't expect them to win quite like that.

a 9th and 10th grader plus 6 kids from the same school. I also did not know that the school was Stuy.

By the way, #8 on the B standings-his name is Matt Mayers, not Matt Myers (I'm a pretty reliable source seeing that I am his little brother :) )


Sorry if this has been asked before -- I ran a search but couldn't find anything.

What editions of the texts and solutions are you guys presently selling? (i.e. If I order a complete set in the next few weeks, will they all be the same size/style?) I noticed on Dave's blog that there's a new cover for Vol. 1, but the bookstore page is still displaying the old one...

We have sold out of the older (6th) edition of the Volume 2's.  All orders are getting the new editions.

For the Volume 1's, we still have the older editions in stock, and those are the ones being shipped unless someone specifically requests the newer edition.

The content of the editions are the same.  The only differences are the size, the cover, and the layout.

What about the answer keys? Do they have a corresponding new cover? If not, are they the same size/dimensions as the new texts?

The new solution manuals are the same as the new texts (8.5 x 11, new color covers).

No, but there are [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php]lots of other books that we recommend[/url].

[quote="daermon"]is there going to be a vol. 3 for supergeniuses?  :rotfl:[/quote]

Well, "The Art and Craft of Problem Solving" by Paul Zeitz is nicknamed volume 3.  :D


A chemistry student decomposes a 78.0 gram sample of potassium chlorate, using manganese dioxide as a catalyst. Her effort yields 24.247 grams of oxygen. What is her percent yield?

[hide] First, find the mol of potassium chlorate KClO3.
Equation: 2KClO3 = 2KCl + 3O2.
Then find the mol of oxygen produced based on the balanced equation.
Find the mass of oxygen produced.
Finally, find the yield.
I got 79.3 % for my final answer.[/hide]


If five geometric means are inserted between 8 and 5832, the fifth term in the geometric series:

$\textbf{(A)}\ 648 \qquad
\textbf{(B)}\ 832 \qquad
\textbf{(C)}\ 1168 \qquad
\textbf{(D)}\ 1944 \qquad
\textbf{(E)}\ \text{None of these}$

If there are 5 terms in between $ 8$ and $ 5832$, we multiply by the common rate six times.

$ 8r^6\equal{}5832$
$ r^6\equal{}729$
$ r\equal{} \pm 3$

The fifth term of the series is $ 8\cdot (\pm 3)^4\equal{} 648$, so A.

How in the world is someone supposed to know that
$ 3^6\equal{}729$  ?!

$ 3^6 \equal{} 9^3 \equal{} 729$

Thats how I would do it as I know my cubes until 12.

[quote="Truffles"][quote="azn6021023"]If there are 5 terms in between $ 8$ and $ 5832$, we multiply by the common rate six times.

$ 8r^6 \equal{} 5832$
$ r^6 \equal{} 729$
$ r \equal{} \pm 3$

The fifth term of the series is $ 8\cdot (\pm 3)^4 \equal{} 648$, so A.[/quote]

If there are five terms [i]in between[/i] 8 and 5832 wouldn't it be $ 8 \cdot 3^5 \equal{} 1944$?[/quote]

Letting the common ratio be r, we have

$ 8, 8r, 8r^2, 8r^3, 8r^4, 8r^5, 8r^6$. The five terms in between are $ 8r, 8r^2, 8r^3, 8r^4, 8r^4$. Thus we have $ 8r^6\equal{}5832$, and we continue from there.

This is poorly written, as 8 is not said to be the first term (Usually I don't complain about things that should be obvious, but I might mess that up)

[hide=Solution]
Let the common ratio be $r$.

So $8r^6=5832\implies r^6=729\implies r=\pm3$ assuming $r$ is real. So $8r^4=8\cdot81=\boxed{\textbf{(A)}\ 648}$. [/hide]

[quote=kante314]So basically u take the first two digits and divide by 8 and that gives 7 and remainder 2, then 23 divided by 8 is 2 and remainder 7, then 72 divided by 8 is 9, so 729 :P[/quote]

Oh, until now I have totally forgotten division.  :blush:  :wallbash:  :wallbash:

[quote=Jwenslawski]Is their an easier way to calculate $5832/8$ besides using a calculator?[/quote]

5832=5600+232=5600+240-8, so 5832/8=700+30-1=729

[quote=brainomega]It is helpful to mem up to (if not more than): $ 2^{12},3^6,4^6,5^5,$ and a bunch more of cubes and squares.[/quote]
A bit of a side note, but you don't actually need to memorize $4^{n}$ as these can be simply found from powers of $2^{2n}$


rotation
geometry
geometric transformation

The conductor shown in the figure is turned so as to form a middle circle with r radius.It is rotated at f frequence in a magnetic field with B inductivity.Find the amplitude of the voltage and the ampitude of the inducted current in the conductor if the inside resistence of the amperometer A is R and the resistence of the circuit is negligible.

Assuming:
  $ \vert B$ orthogonal to the rotation axis:
   conductor length finite;
  A connected between  the ends of the arc:
  Self inductance negligible;

[hide]and $ \; \omega = 2 \pi f \;,\;i\; = {{B\pi R^2f} \over 2} sin (2\pi ft)$ [/hide]


I am new to latex and i have already downloaded texnic centre ..

i have read many times to make sure that "LATEX>=PDF"
is displayed ...
but where it is displayed and if it is not displayed then what to do....

plz help me  :|

Do you install a $ \text{\LaTeX}$ engine like MiKTeX?
You must do it first, follow this: http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php

Yes i have downloaded texcnic latex editor which is mentioned on the link that u suggested..

 now what to do???

I mean do you download MiKTeX along with TeXnicCenter?

No i havent downloaded miktex ...
does there is a need to download both of them..???

Yes because MiKTeX is the language itself and TexnicCenter is just an interface to you that's why it's size is small.
Now you need to download MiKTeX and when you open TexnicCenter again after installing the MiKTeX don't forget to enter the directory of the MiKTeX to TexnicCenter. You just need to follow the instruction on the past link I gave you.

Note that you don't need to download anything if all you want to do is use LaTeX on the forum -- see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


I live in the U.S., but I will be coming to China with the US IMO team a week or so before the IMO!

So the US team will train with the China team?!?

What do you mean? rem is the first three letters of my last name.

[quote="JJ_JJ_JJ_JJ"]rem: is your name from the death note?  :?:[/quote]

You are in danger bcos you reveal his true identity :D  :D  :D
Heart attack in 40 seconds

bet some of the us's teammember are chinese

haha
i came here last year, and i'm going back to china this sunday
my middle school is in 武钢三中，a pretty good school for math olympiad~ :P

i didn't know this website when i was in china~

haha
This is my first logon.
I am in China. :P

Yunnan province lies in southwest China.

Soy Fukie'nio. Vivo en la ciudad de foochow. Yo voy a estudiar el espanol a  Espana. :blush:

我是lz
夏天了马甲凉快XD

Xi'an Shaanxi Province China
I'll be a high school student next semester
I'm in NPU high school西工大附中

death note is"死亡笔记"  , a cartoon, pretty popular
Don't you hear it?

There'er many math gennius born in China.. :wink:

我怎麼感覺論壇上武漢寧波人特別多
可能那幾個都是教育大省吧

[quote="shobber"]你知道有多少人，人在中，心在美

[quote="Un4give"]人在美 心在中[/quote][/quote]

這一句強大。話說，我們那邊也有相當數量的人也盼望著美軍來佔領我們，推翻天朝統治，或者獨立成國的。我看著心也寒死。

NO MONGALIA?!?!?! YAH, CHINA STOLE A BIG PART OF US, BUT YOU CAN'T JUST IGNORE US LIKE THAT!!!! :bomb:

I'm am now pulling hair out. DON'T TOUCH ME! :bruce:

Arggg all i see is a bunch of boxes because chinese characters don't show up on this computer.

在中国的路过……竟然还有字数限制


How do you "Mass PM"?

Can you put for the Username like infinitypi, Fantasylover, mz94 or does it just mean sending over and over again?

You can't mass PM.  Someone's asked that before.  You have to send over and over again.  Which sucks when sending reminder PM's for tourneys.

[quote="AIME15"]You can't mass PM.  Someone's asked that before.  You have to send over and over again.[/quote]
Speaking only for myself (but I suspect the administrators would likely agree), I'd call that a feature, not a bug.

You can't mass PM. However in the future moderators will have this option available, as well as regular users that will probably be able to mass pm their [b]mutual[/b] friends.


if you choose a real number between 0 and 1, what is the possibility that the number you choose is rational?
i think the answer must be 0, but i cant prouve it.
and another question: let you choose a sequence infinite of integer between 1 and 10, what is the possibility that your sequence doesnt have any "1"?

For the first question, the answer depends on the probability distribution you are using and, generally speaking, can be anything between $ 0$ and $ 1$. Assuming that the uniform distribution was meant (which is usually the case when one talks about "random choice on $ [0,1]$" and forgets to mention which probability law he uses), your answer is correct. How to prove it? Well, first show that the probability of evry particular number to be chosen is $ 0$ and then use countable additivity ;)

The second question has the same problem with it: the exact probability law for the choice is missing, and it is important. Again, let us make some reasonable assumption about what might be meant. Assume that all the numbers in the sequence are chosen independently and the probability of each number to be chosen in each position is $ \frac{1}{10}$. Then the first step would be to ask yourself what is the probability that there is no $ 1$ in the first $ n$ positions. Then look at what happens as $ n\to\infty$ :)


I was just wondering that if we can find a rule for this?
for $n$ positive variable like $x_1,x_2,...,x_n$ we show the mean of power $i$ by $A_i$ where $i \in \mathbb{N}$  and we denot this:

$ A_i = \sqrt[i]{\frac{\sum_{j=1}^n (x_j)^i}{n}}$

now, is this true?

$\frac{A_{i+1}}{A_i} \geq \frac{A_i}{A_{i-1}}$

and in the general form if :

$\frac{A_a}{A_c} \geq \frac{A_b}{A_d}$

Do we have any rule between the $a,b,c,d$?

well guys i think this is really beautifull but its wrong (a pitty , a really pitty) :(  :(
and for a solution this is myths one :
 , $Myth $ has said that it is almost wrong cause[quote="Myth"], conjecture is obviously wrong. Indeed,  ineq means that $\ln A_i$  is a convex function of $i$, but we know that $\ln A_i$ is increasing and bounded, it is impossible for convex functions. Also you can easily check the "conjecture" for 2 numbers using a computer to be sure it is wrong. Though I think that the reverse ineq is true for sufficiently large $i$. [/quote]
so i better dont start thinking about making inequalities again :blush:


Can someone explain in a relatively simple manner what a transcendental L-function is and perhaps give examples?  The web as far as I've checked does not provide much help in the matter.

Thanks,
Shaw

Hmmm... I just read about this. I'm guessing you're asking due to [url=http://www.aimath.org/news/gl3/]this[/url] article. A rather great web (though not "simple" per se) is given [url=http://golem.ph.utexas.edu/category/2008/03/the_world_of_l.html#comments]here[/url].

Thanks Hilbert.  Yea, I ran across that one also.  It's tough and I personally cannot accept there is no way to present the matter in a more accessible manner.  I think it would be nice if AoPS could become a resource in such matters but  perhaps that is not what Rusczyk had in mind.


linear algebra
matrix
linear algebra unsolved

Hi! I need help with a problem from Artin's Algebra, that I don't seem to make any progress on.

Set $A^* : = \bar A^t$.

Let $A$ be an arbitrary complex matrix and let I be the identity matrix.
Then the problem wants you to prove that

$I + AA^*$ is a regular matrix, i.e. $\det(I + AA^*) \ne 0$.

The only idea I had, was to use the spectral theorem in some way (that I don't really know, apart from knowing that you can use the spectral theorem to diagonalize $AA^*$) to show that $AA^*$ does not have (-1) as an eigenvalue.

This is [b]not[/b] a homework problem, so I would be very thankful, if you could help me.

The matrix $AA^*$ is a positive semidefinite Hermitian matrix, therefore all its eigenvalues are non-negative.
Here is the proof. Let $v$ be an eigenvector of $AA^*$: $AA^* v = \lambda v$. Then
\[ \lambda = \langle \lambda v, v\rangle = \langle (AA^*) v, v\rangle = v^T \bar A A^T \bar v = (A^* v)^T (\overline{A^* v}) = \langle A^* v, A^* v\rangle = \|A^*v\|^2\geq 0.  \]
So every eigenvalue is non-negative.


post something FUN that happened in your math class today.  :lol:

[quote="lingomaniac88"]How about something funny that happened a while ago?

Teacher: (after working through a math problem with the class) So $\tan x=2$.
Student: But isn't $\tan x$ only $-1$, $0$, $1$, or undefined? (he was thinking about how he graphs $\tan x$, plotting a point at intervals of $\frac\pi4$)
Teacher: No, because...
[hide="Warning, somewhat dirty joke"]as $\tan x$ goes up the asymptote...[/hide]
Teacher: *pause*
Everyone laughs.[/quote]
i dont get it.
how immature of me :(
i have an idea of what it might have to do with, but it doesnt make sense

hmm, something funny that happened in class?
well we were doing this game with dice and at the end i went up to the board to explain a mathematically perfect strategy, and i said that 1+2+3+4+6=21 (just did 6*7/2 and forgot to take away the 5) but nobody noticed
i did after class though


Let a,b,c positive real numbers such that a+b+c=1. Prove that

5(a^2 + b^2 + c^2) \leq 6(a^3 + b^3 +c^3) +1

Homogenizing we want to prove that
5(a + b + c)(a + b + c) \leq 6(a + b + c) + (a + b + c)
or
ab(a + b) + bc(b + c) + ca(c + a) \leq a + b + c + 3abc
which is exactly Schur's inequality.


abstract algebra
complex analysis
real analysis
Functional Analysis

What's the difference between Graduate Math Courses and Undergraduate Math Courses of the same title? For example graduate complex analysis course and undergraduate complex analysis course. What's the difference between the two versions?

I would say that an undergraduate course is more of an introductory course, which introduces various ideas at a basic level. Whereas a postgraduate course, explores the course in further detail and spends more time going into detail.

You could say an undergraduate course is teaching you a language and its rules (grammar, punctuation... ), and a postgraduate course teaches you how to use the language.

Beans' analogy is good.

Part of it also depends on the quality of your university.  For example, at my institution, the undergraduate course in complex analysis is basically a "cookbook" course, where the main goal is to compute lots of complex integrals.  Most of the students who take it are physics or electrical engineering majors.  The graduate course there is the rigorous introduction to complex variable theory.

In some cases, a graduate course in a subject presupposes the undergrad course; for example, the content of a graduate real analysis course (measure theory, L^p spaces, some functional analysis, etc.) is usually disjoint from and dependent on the content of an undergrad real analysis course (e.g. "Baby Rudin").  Other times, it's a lot of the same material, only from a more "mature" or advanced perspective.  I'm thinking of, for instance, algebra, where a lot of graduate courses cover the basic definitions (groups, modules, rings, etc.) but both faster and with more (perhaps category-theoretic?) sophistication.


Person A is swinging a slingshot. He draws the slingshot back with a potential energy of 8 J. He releases the slingshot and the stone fires forward with a kinetic energy of 6 J. What happened to the other 2 J?

Those 2 J correspond to energy dissipated as heat (by friction or otherwise) and sound.


Let $ f(n)$ be the sum of the digits of $ n$. Calculate $ f(f(f(2^{2006})))$.

[hide="classic"]Clearly, $ 2^{2006}$ has less digits than $ 10^{2006}$ which has 2007 digits, so $ f(2^{2006}) < 9\times2007 \equal{} 18063$. Thus, $ f(f(2^{2006})) < f(17999) \equal{} 35$. So, $ f(f(f(2^{2006}))) < f(29) \equal{} 11$. Now, a number is congruent to the sum of its digits mod 9(commonly known). $ 2^6 \equal{} 64\equiv1\mod9$, so $ 2^{2006}\equiv2^2 \equal{} 4\mod9$. The only number less that 11 with this true is 4, so $ f(f(f(2^{2006}))) \equal{} 4$.[/hide]

ah so that's how you're supposed to approach these types of problems. oh btw

[quote="perfect628"][hide="classic"]Clearly, $ 2^{2006}$ has less digits than $ 10^{2006}$ which has 2007 digits, so $ f(2^{2006}) < 9\times2007 \equal{} 18063$. Thus, $ f(f(2^{2006})) < \boxed{f(f(17999)) \equal{} 35}$. So, $ f(f(f(2^{2006}))) < \boxed{f(f(f(29)))}\equal{} 11$. Now, a number is congruent to the sum of its digits mod 9(commonly known). $ 2^6 \equal{} 64\equiv1\mod9$, so $ 2^{2006}\equiv2^2 \equal{} 4\mod9$. The only number less that 11 with this true is 4, so $ f(f(f(2^{2006}))) \equal{} 4$.[/hide][/quote]

Shouldn't that just be $ f(17999)\equal{}35$ and $ f(29)$.


Let $ f(x)\equal{}2009x^5\minus{}x^4\minus{}x^3\minus{}x^2\minus{}2006x\plus{}1$.Prove that $ f(n),f(f(n)),f(f(f(n))),...$ are pairwise coprime for any positive integer $ n$

$ f(n) \equal{} n(n \minus{} 1)Q(n) \plus{} 1$ where $ Q\in \mathbb{Z}[X]$
we take $ f^{0}(n) \equal{} n,f^{k}(n) \equal{} fofo..k\ times..of(n)$.

so $ f^{h \plus{} m}(n) \equal{} 1 \plus{} (f^{m}(n) \minus{} 1)\prod_{i \equal{} 1}^{k}Q(f^{i \minus{} 1}(f^{m}(n)))f^{i \minus{} 1}(f^{m}(n))$
so $ gcd(f^{h \plus{} m}(n),f^{m}(n)) \equal{} 1$