geometry
3D geometry
sphere
conics
ellipse

This weekend there is suppose to be a meteor shower caused by the Comet Kiess but, astronomers have different views on to how the shower will be or if it will occur at all.
Since I always quote SpaceWeather.com :lol:
[quote="SpaceWeather.com"]
WEEKEND METEORS:  On Saturday morning, September 1st, a flurry of bright and colorful meteors might come streaming out of the constellation Auriga.  The source of the shower is ancient Comet Kiess, which has laid down a trail of debris that Earth will cross this weekend.

But will a shower really materialize?  Forecasters are divided. Some expect a brief but beautiful display rivaling the Perseids.  Others say the debris stream is too empty for significant fireworks.  Either way, the peak is due around 4:30 a.m. PDT (11:30 UT) on Saturday morning.  This timing strongly favors sky watchers in Hawaii and western parts of North America all the way from Mexico to California to Alaska.

Spaceweather's full coverage of the Aurigid meteor shower begins Friday, Aug. 31, with links to live audio from a meteor radar, which will monitor activity throughout the weekend. Sky maps and more information are available now at  http://spaceweather.com.
[/quote]
It says it favors the Western part of the U.S. but, viewing of the rest of the continental U.S. should not be too bad.

[hide="Sky Map with Radiant"]
[img]http://science.nasa.gov/headlines/y2007/images/aurigids/skymap_north.gif[/img]
[/hide]

[url=http://science.nasa.gov/headlines/y2007/08aug_aurigids.htm]More information from NASA[/url].

I'll be in San Diego then, so I'll try to get up early and take a look. I might even take some pictures, and I'll post them if I do. :)

There was a 2-hour traffic jam in Los Angeles, so we'll be arriving in San Diego around 3 am. I probably won't go to sleep and instead I'll just stay up and watch the sky (if I see anything). Pictures probably won't work out because my dad didn't bring his tripod and stuff, but I'll see what I can do.

2:00 am: About one hour from San Diego now...

3:00 am: Just arrived at hotel, I'm going to see what's in the sky right now.

3:30 am: Nothing in the sky, just the usual stars and the moon. I can't get a good shot of anything, and there's nothing to shoot anyway. I'll try again in an hour...

Apparently the Aurigids were predicted correctly from reports of meteor shower activity this morning.  :)
[quote="SpaceWeather.com"]
Congratulations to Peter Jenniskens of the SETI Institute and colleagues who correctly predicted this morning's outburst of Aurigid meteors from ancient Comet Kiess. The shower materialized on schedule and put on a vivid display, including many bright fireballs.  So far, positive reports of meteor activity have been received from Arkansas, Arizona, Colorado, California, Ohio, Oregon, Texas and western Canada.  This list will grow; photos, videos and eyewitness reports are accumulating rapidly.  Please visit http://spaceweather.com to experience the sights and sounds of the Aurigids.
[/quote]

i_like_pie , did you see any? I don't know if the light pollution in San Diego helped any but, it should have been a pretty good sight.

Luckily our hotel was off to the east of downtown, so the sky wasn't too bad, but still not so great. I saw 3 shooting stars within 5 seconds, and then another one about 10 minutes later. Several websites predicted an average of 100/hour, so the 3 in 5 seconds were very surprising, but the total of 4 in 15 minutes was way too little.

It's almost impossible to take a picture of a shooting star, so I just tried capturing some bright stars in the sky. Out of 10 or so pictures, 9 turned out slightly blurry and the last one had only a couple faint stars in it. If I had a tripod with me, I could've taken some very high-quality pictures, but it still would've been nearly impossible to capture a shooting star.

Next time, I'll definitely be more prepared, and maybe even drive to an area with very clear skies.

That sounds very cool i_like_pie :).  Too bad about the pictures though but, what can you do? :wink: Post them if you would like.

Hint about take pictures of meteors, long-exposure times.  As in leaving the camera's shutter open for a while.  This is also a good way to get better pictures of the stars, even ones you can not see with you naked eye.  Though if you were to leave it open for a [i]long[/i] period of time, like an hour or two, you might star to get visible streaks, or "star trails", in the picture from the celestial sphere having rotated.

I deleted the pictures as soon as I saw how blurry they were. :P

My exposure times were around a minute or two as I didn't want to wait any longer, but it wasn't worth it anyway. :(

Some sites are now predicting another shower around Thursday and Friday. I'm not sure I want to stay up late on Thursday, but I'll definitely stay up on Friday, and I will definitely not come back empty-handed. :bruce:

:lol: About the pictures.

What meteor shower is it on Thursday and Friday?  Is it just another run of the Aurigids?

Yes, I think it's them again.

There's a lot of info on them [url=http://aurigid.seti.org/]here[/url]. Somewhere on the page there is a quote of somebody seeing 3 in 10 seconds, which were probably the same 3 I saw. :) There's also a graph of the distribution of activity throughout September, peaking around the second week.

I realized that taking pictures of them is pretty pointless. To take any picture of the sky, you need a long exposure time. However, to capture a shooting star/asteroid/whatever, you need a very short exposure time. If it shows up at all on a picture with a long exposure time, it'll be way too faint. So as I said before, I'll try, but I'm most likely not going to get anything good.

Just try, my pictures of the eclipse weren't great, but they were better than nothing.

Nice site you linked to i_like_pie :) , especially those very helpful graphs.

About the long exposure pictures most of the meteors would probably come out faint but, it is that one random one the makes it worth the while. :) The best spot would the radiant of course even though a few of the meteor appear away from the radiant.  If you try it, it will be very interesting to see the results.

Did you try again yesterday night i_like_pie?  :) Was it any good?

Unfortunately the sky was very cloudy (although it was clear throughout the day), and I couldn't see anything besides a faint blob that was probably the moon. Even worse, I won't be able to stay up late tonight as I have to get up at 8 tomorrow.

On the "bright" side, the charts in the link I posted indicate that next weekend might be even better, so I'll try again next Friday.

i_like_pie , if you need an hour-by-hour forecast that is quite accurate try a [url=http://www.mathlinks.ro/viewtopic.php?t=157030]Clear Sky Clock[/url] for your area.  Though the forecast does not past go a day and a half.  It also gives other handy things such as atmospheric transparency.  :wink:


For the last one see [url]http://www.mathlinks.ro/viewtopic.php?t=302086[/url]

Dear luofangxiang,

very nice your proof.

 I tried to prove $\sum ab^2 \sum a^2 \geq \sum a^2b \sum ab $ using only Cauchy-Schwartz but I failed.

Have you a solution by only C-S similar to this one?

$a,b,c \geq 0$ prove that $\sum a^3 \geq \sum a^2b$

I proved it by CS this way:

1) $ (ab^2+bc^2+ca^2)(a^3+b^3+c^3) \geq (a^2b+b^2c+c^2a)^2$
2) $(a^2b +b^2c+c^2a)( b^3+c^3+a^3) \geq (ab^2+bc^2+ca^2)^2$

We set  $A=\sum ab^2$  ,  $B=\sum a^2b$   and $C=\sum a^3$

Now if $A \geq B$ we use 2)   while if $A \leq B$  we have from 1) $ AC\geq B^2 \geq A^2$

Thank you very much.

[quote=can_hang2007]Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{a^2b}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^2c}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^2a}{c^2\plus{}ca\plus{}a^2} \le \frac{ab^2\plus{}bc^2\plus{}ca^2}{ab\plus{}bc\plus{}ca}.$[/quote]
$\sum_{cyc}\frac{a^2b}{a^2+ab+b^2}\leq\frac{ab^2+bc^2+ca^2}{ab+bc+ca}$
$\iff(ab+bc+ca)\sum_{cyc}\frac{a^2b}{a^2+ab+b^2}\leq ab^2+bc^2+ca^2$
$\iff\sum_{cyc}\frac{a^2b(a+b)(a+b+c)}{a^2+ab+b^2}\leq\sum_{sym}a^2b$
$\iff\sum_{cyc}\left(ab-\frac{a^2b(a+b)}{a^2+ab+b^2}\right)\geq ab+bc+ca-\frac{\sum\limits_{sym}a^2b}{a+b+c}$
$\iff\sum_{cyc}\frac{ab^3}{a^2+ab+b^2}\geq\frac{3abc}{a+b+c}$
$\iff\sum_{cyc}\frac{b^2}{c(a^2+ab+b^2)}\geq\frac{3}{a+b+c}$
By CS,
$\sum_{cyc}\frac{b^2}{c(a^2+ab+b^2)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}c(a^2+ab+b^2)}=\frac{a+b+c}{ab+bc+ca}\geq\frac{3}{a+b+c}$

[quote=zdyzhj]can's ineqaulity can be proved with one or two lines[/quote]
You always say "it can be proved by one or two lines" without a proof. That doesn't make sense to any mathlinker.


When $ 200^6$ is converted to base 4, how many zeroes does it have on the end?

200^6=(2^3 * 5^2)^6 = 2^18 * 5^12 = (2*5)^12 * 2^6 =10^12 * 2^6
   ...12 zeros

 Sorry , I'm n00b at LaTeX
   I don't know I can use the LaTeX Code directly , or I should necesary use the program
 By the way help me ! :blush:

The question asked for when $ 200^6$ was in base 4. :wink:

Oh, and to make $ \text{\LaTeX}$ code, put dollar signs($ \$$) around your code. For example:

[code]
$200^{6}$
[/code]

shows up as $ 200^{6}$.

There's a good $ \text{\LaTeX}$ tutorial on the this site. Look at the navigation bar on the left and click where it says LaTeX.

No izzy, read the problem. it asked foro the number of zeroes.

No AIME, read the problem. A zero in base 4 is a 4, not a 10 (izzy was correcting efejzullahu).


geometry
circumcircle
trapezoid
geometry proposed

$\triangle ABC$ is acute with 9-point center $N$ and $\triangle XYZ$ is the pedal triangle of $N$ with respect to $\triangle ABC.$ $R,r,r_0$ stand for the radii of the circumcircle, incircle and incircle of the orthic triangle of $\triangle ABC.$ Show that:

\[NX+NY+NY=\frac{1}{2} \left ( R+3r+r_0+\frac{r^2}{R} \right) \]

$O,H$ are the circumcenter and orthocenter of $\triangle ABC.$ $M_a,M_b,M_c$ denote the midpoints of $BC,CA,AB$ and $H_a,H_b,H_c$ denote the feet of the A-,B- and C- altitudes. Since $PX$ becomes the median of the right trapezoid $OHH_aM_a,$ we get $NX=\frac{_1}{^2}(OM_a+HH_a).$ Thus, adding the symmetric expressions together gives

$NX+NY+NZ=\frac{_1}{^2}( OM_a+OM_b+OM_c+HH_a+HH_b+HH_c)=$

$=\frac{_1}{^2}[R+r+h_a+h_b+h_c-(HA+HB+HC)]=\frac{_1}{^2}(h_a+h_b+h_c-R-r)$

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=259185]this topic[/url], we have $h_a+h_b+h_c=2R+4r+r_0+\frac{r^2}{R}.$ Thus

$NX+NY+NZ=\frac{1}{2} \left ( R+3r+r_0+\frac{r^2}{R} \right).$


Trick or Treat Part 2

We learn the importance of distinguishability in counting problems by counting the number of ways to distribute candy to children at Halloween.

Clever - using Halloween as the topic for distinguishability.   :ninja:

[quote="vallon22"]at 10:54 - 10:55, it said something like "as long as these candy bars are separate, they are all different to me."

shouldn't it say that they are the same to him?[/quote]

Yeah, I believe it should, otherwise, it wouldn't make sense.


algebra
polynomial
algorithm
quadratics
AMC
AIME
special factorizations

Pls don't write any questions that need to be proved or shown. Just write questions that have a numeric solution.

problem 1: Find all solutions of the given polynomial equation.
$ x^3\plus{}x^2\plus{}10x \minus{}3 \equal{} 0$

[quote="ertanrock"]Pls don't write any questions that need to be proved or shown. Just write questions that have a numeric solution.

problem 1: Find all solutions of the given polynomial equation.
$ x^3 \plus{} x^2 \plus{} 10x \minus{} 3 \equal{} 0$[/quote]

$ \frac 13\left(\minus{}1\minus{}29 \sqrt[3]{\frac 2{169\plus{}27 \sqrt{173}}}\plus{}\sqrt[3]{\frac{169\plus{}27 \sqrt{173}}2}\right)$ $ \equal{}0.28921621924407...$

[quote="ertanrock"][quote="pco"][quote="ertanrock"]Pls don't write any questions that need to be proved or shown. Just write questions that have a numeric solution.

problem 1: Find all solutions of the given polynomial equation.
$ x^3 \plus{} x^2 \plus{} 10x \minus{} 3 \equal{} 0$[/quote]

$ \frac 13\left( \minus{} 1 \minus{} 29 \sqrt [3]{\frac 2{169 \plus{} 27 \sqrt {173}}} \plus{} \sqrt [3]{\frac {169 \plus{} 27 \sqrt {173}}2}\right)$ $ \equal{} 0.28921621924407...$[/quote]

$ This$ $ is$ $ not$ $ a$ $ solution.$ $ Maybe$ $ this$ $ should$ $ be$ $ an$ $ answer.$  :)[/quote]

The problem stated "no proof", just "numerical result"
I gave a numerical result.
If you want a solution, just use Cardan. That's basic college problem. :)

[quote="pco"][/quote][quote="ertanrock"][quote="pco"][quote="ertanrock"]Pls don't write any questions that need to be proved or shown. Just write questions that have a numeric solution.

problem 1: Find all solutions of the given polynomial equation.
$ x^3 + x^2 + 10x - 3 = 0$[/quote]

$ \frac 13\left( - 1 - 29 \sqrt [3]{\frac 2{169 + 27 \sqrt {173}}} + \sqrt [3]{\frac {169 + 27 \sqrt {173}}2}\right)$ $ = 0.28921621924407...$[/quote]

$ This$ $ is$ $ not$ $ a$ $ solution.$ $ Maybe$ $ this$ $ should$ $ be$ $ an$ $ answer.$  :)[/quote][quote="pco"]

The problem stated "no proof", just "numerical result"
I gave a numerical result.
If you want a solution, just use Cardan. That's basic college problem. :)[/quote]
That doesnt mean we dont need solution. I just wanted to say that you should post problems which have numerical answers. WE NEED SOLUTIONS. PLS post a new problem. Thank you.

For college students, as required by ertanrock :

Original equation : $ x^3\plus{}x^2\plus{}10x\minus{}3\equal{}0$

1) eliminate $ x^2$ term :
================
Let $ x\equal{}y\minus{}\frac 13$ and the equation becomes :

$ y^3\minus{}y^2\plus{}\frac 13y\minus{}\frac 1{27}\plus{}y^2\minus{}\frac 23y\plus{}\frac 19\plus{}10y\minus{}\frac{10}3\minus{}3\equal{}0$

$ \iff$ $ y^3\plus{}\frac {29}3y\minus{}\frac{169}{27}\equal{}0$

2) For easier manipulation, let $ y\equal{}\frac z3$. The equation becomes : $ z^3\plus{}87z\minus{}169\equal{}0$
=================

3) Use Cardan's resolution method (I hope ertanrock consider this is a solution)
=======================
$ \Delta\equal{}q^2\plus{}\frac 4{27}p^3$ $ \equal{}(\minus{}169)^2\plus{}\frac 4{27}87^3$ $ \equal{}\frac{3405159}{27}$ $ \equal{}27^2173>0$
So we have a unique real root $ u\plus{}v$ where $ u\equal{}\sqrt[3]{\frac{\minus{}q\plus{}\sqrt{\Delta}}2}$ and $ v\equal{}\sqrt[3]{\frac{\minus{}q\minus{}\sqrt{\Delta}}2}$

Hence the root $ r\equal{}\sqrt[3]{\frac{169\plus{}27\sqrt{173}}2}$ $ \plus{}\sqrt[3]{\frac{169\minus{}27\sqrt{173}}2}$

$ \iff$ $ r\equal{}\sqrt[3]{\frac{169\plus{}27\sqrt{173}}2}$ $ \minus{}29\sqrt[3]{\frac 2{169\plus{}27\sqrt{173}}}$

4) come back to the original equation :
========================
Since $ r$ is root of $ z^3\plus{}87z\minus{}169\equal{}0$ :

$ \frac r3$ is root of $ y^3\plus{}\frac {29}3y\minus{}\frac{169}{27}\equal{}0$

$ \frac r3\minus{}\frac 13$ is root of original equation.

Hence the result $ \frac 13(r\minus{}1)\equal{}$ $ \boxed{\frac 13\left(\minus{}1\plus{}\sqrt[3]{\frac{169\plus{}27\sqrt{173}}2}\minus{}29\sqrt[3]{\frac 2{169\plus{}27\sqrt{173}}}\right)}$

I hope this will help you.
And I must say that IMHO this is not a correct olympiad problem.
:(

What is the point of posting such useless polynomials to be solved? I am sure that we are not here to get practice on using Cardano's method, but rather to solve interesting problems.

Hi;

What is the difference between a numerical solution gotten by some algorithm or using the Cardano's method ( which could be called an algorithm ). Both of them are just as mind boggling when you just see the answer. Why are the steps used in Cardano's, clearer then the steps used in some iteration?

It is similar to the difference between completing the square and using the quadratic formula.

Hi;

Cardano's method is an algorithm. Newton's iteration is an algorithm. Completing the square is an algorithm. No difference, all can be taught to a computer. There are rules for algebraic manipulation of symbols, arithmetic has rules for the manipulation of symbols, no difference.

[quote="ertanrock"]Pls don't write any questions that need to be proved or shown. Just write questions that have a numeric solution.

problem 1: Find all solutions of the given polynomial equation.
$ x^3 \plus{} x^2 \plus{} 10x \minus{} 3 \equal{} 0$[/quote]

What about the not-real solutions?

[quote="ocha"]The only reason i can see for wanting a marathon with only numerical answers would be to practice for a competition such as the AIME where all answers are integers between $ 0$ and $ 999$. If so, that's where this marathon belongs - in the math competition forum.[/quote]
SURE. That is all I need abou questions.

$ PROBLEM$ $ 2:$ The polynomial $ (x\minus{}a)^3\plus{}b$ is zero at $ x\equal{}1$, and when divided by x, the remainder is $ \minus{}7$. Find all possible values of the pair $ (a,b)$

We have $ (1\minus{}a)^3\plus{}b\equal{}0$ (1). Because $ (x\minus{}a)^3\plus{}b\equal{}x.K(x)\minus{}7$ hence $ b\minus{}a^3\equal{}\minus{}7$ or $ a^3\minus{}b\equal{}7$ (2)

(1)+(2) gives us $ 3a^2\minus{}3a\minus{}6\equal{}0$ or $ 3(a\minus{}2)(a\plus{}1)\equal{}0$ ...

[b]$ PROBLEM$ $ 3:$[/b] None of the numbers $ a,b$ or $ c$ is zero and each is a root of the equation $ x^3\minus{}ax^2\plus{}bx\minus{}c\equal{}0$. Find $ a,b$ and $ c$

Write $ (x\minus{}a)(x\minus{}b)(x\minus{}c) \equal{} x^3\minus{}(a\plus{}b\plus{}c)x^2 \plus{} (ab\plus{}bc\plus{}ca)x \minus{} abc \equal{} x^3\minus{}ax^2\plus{}bx\minus{}c$ and equate coefficients.

$ PROBLEM$ $ 4:$ Let $ P(x)$ be a polynomial of degree $ 3n$ such that \[ P(0)\equal{}P(3)\equal{}...\equal{}P(3n)\equal{}2\] \[ P(1)\equal{}P(4)\equal{}...\equal{}P(3n\minus{}2)\equal{}1\]  \[ P(2)\equal{}P(5)\equal{}...\equal{}P(3n\minus{}1)\equal{}0\] and $ P(3n\plus{}1)\equal{}730$. Determine $ n$


modular arithmetic
number theory unsolved
number theory

[quote="Rushil"]Determine all integers $a$ such that $a^k + 1$ is divisible by $12321$ for some $k$[/quote]

Is $k$ an integer??

 :)

We can get it from $a^9\equiv-1\pmod{37},a^3\equiv-1\pmod{3}$,and take $k=9\times37$,$12321=3^2\times37^2|a^k+1$,for all a that I have given.

Since $a^2\equiv0,1\pmod{3}$,then k must be an odd integer,and $a^k\equiv a\pmod{3}$,so $a\equiv-1 \pmod{3}$,and it's clearly that if $a\equiv-1\pmod{3}$,$a^3\equiv-1\pmod{9}$.
Then for $a^k\equiv-1\pmod{37},a^{2k}\equiv1\pmod{37}$.It's obvious that $\gcd(a,37)=1$.By the Theorem of Fermat,$a^{36}\equiv1\pmod{37}$.
So suppose the least positive integer $d$ satisfied $a^d\equiv1\pmod{37}$,then we must have $d|2k,d\nmid k,d|36$,then we get $d=2t$,t is an odd integer.
$a^t\equiv-1\pmod{37}$,$t|18$,then $t|9$,so $a^9\equiv-1\pmod{37}$.
Suppose that $b=a^9+1,37|b$,\[ a^{9\times37}+1=(b-1)^37+1=b^37-37b^36+\ldots-{37\choose2}b^2+37b- 1+1\equiv0\pmod{37^2} \]
So if $a$ satisfied that $a^9\equiv-1\pmod{37}$,then $a^{9\times37}\equiv-1\pmod{37^2}$
And it's easy to know if $a\equiv-1\pmod{3}$,$a^3\equiv-1\pmod{9}$.then $a^{9\times37}\equiv-1\pmod{9}$.
So we get the necessary and sufficient condition for that there exist integer k such that $12321=3^2\times37^2|a^k+1$ is $a^9\equiv-1\pmod{37}$ and $a\equiv-1\pmod{3}$.
$a^9\equiv-1\pmod{37}\Leftrightarrow a\equiv-1,3,4,-7,-9,-10,11,-12,-16\pmod{37}$,
Then all integers satisfied the conditions are $a\equiv 11,41,62,65,77,95,101,104,110\pmod{111}$.


probability
combinatorics open
combinatorics

A glass strip is dropped down and then it is separated into three parts. What is the probability that all these three small strips can construct a triangle ?

Famous indeed, not quite open, and of course [url=http://www.mathlinks.ro/viewtopic.php?t=115687]posted before.[/url]


I was wondering if you could have an AMC 8 math jam, like the AMC 10 and AMC 12 math jams you already have.  I'm not exactly sure if engouh people would be interested enough to participate though.

Were you thinking of one to help prepare for the AMC-8 or a review of the problems after the contest?

I think both would be good, but I was thinking of a review of problems after the contest.

AMC director Steve Dunbar has graciously offered to send us a copy of the questions from the AMC-8 a few days after the exam.  I will tentatively schedule the AMC-8 Math Jam for Tuesday, November 30th and we will discuss as many of the problems from that exam as we can cover in an hour.  That should be enough time to cover the problems students might have questions about or that might lead students to more interesting concepts.


$ABCDE$ is a regular pentagon, $AP,AQ,$ and $AR$ are the perpendiculars dropped from $A$ onto $CD,CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO+AQ+AR$ equals

$\text{(A)} \ 3 \qquad \text{(B)} \ 1 + \sqrt 5 \qquad \text{(C)} \ 4 \qquad \text{(D)} \ 2 + \sqrt 5 \qquad \text{(E)} \ 5$

[hide]$\overline{OC}$ bisects $\angle BCD=108^\circ\Rightarrow \angle PCO=54^\circ\Rightarrow \angle POC=36^\circ$.

Then $OC=\sec 36^\circ$ and $PC=\tan 36^\circ$.

$AO=OC=\sec 36^\circ$ and $AR=AQ$. From $\angle ABC=108^\circ$, $\angle ABQ=72^\circ \Rightarrow AQ=2\tan 36^\circ \sin 72^\circ=4\sin^2 36^\circ$ from double-angle sine formula.

The sum is $8\sin^2 36^\circ +\sec 36^\circ=4-4\cos 72^\circ +\sec 36^\circ=4+\frac{1-4\cos 72^\circ \cos 36^\circ}{\cos 36^\circ}$.

But $(4\cos 72^\circ \cos 36^\circ)(\sin 36^\circ)=2\sin 72^\circ \cos 72^\circ=\sin 144^\circ=\sin 36^\circ$

$\Rightarrow 4\cos 72^\circ \cos 36^\circ=1$.

The fraction is then 0 so the sum is just 4 $\Rightarrow \boxed{C}$.[/hide]

This is perfectly fine. However, can anyone find a method that does not use trig at all?

[hide="Hint"]
Let's think about splitting. What can we do to this pentagon to write area of it in say... more than 1 way? :D [/hide]


This year, we're hosting the HCSSiM Math Jam on April 1st, starting at 7:30p.m. EDT.  We'll have at least a couple of alumns on hand, and will answer questions about the program.  Additionally, I'll pose an Interesting Test problem (from several years ago) and we'll discuss the solution.

Everyone is invited!

It's here:  http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=232
so feel free to take a look.


HMMT Harvard college topology geometry MIT function

With this year's HMMT fresh in people's minds, I'd like to hear your comments and suggestions for next year.

What went well?  What went poorly?  Which mini-events were good and bad?  What would you like to see next year?

How bad was space in the team round?  It'll be slightly better at Harvard, but we'll still have to trade off between rejecting teams and cramping them.  In response to this, try to include your room size or number.

And if you're a senior, it's time to start writing problems for HMMT 2008!

I was a volunteer, and it appeared to me that HMMT ran substantially more smoothly than it has the previous two years.  (Well, excepting the loss of two trophies.)  Nice work, organizers!

I didn't go myself, but I heard from our state team members that it was very well organized (aside from the minor facilities problem). As a matter of fact, some of them described it as the best experience of their lives. Good job to the organizers.

A few things:

- The team round actually wasn't cramped for us, we shared one room with another team, but that's still only 16 people, and we did fine
- The Guts Round, on the other hand, was insanely crowded (we were in the A room)

Mini Events:
- The Useless Math Olympiad was ok, but some of the events were kind of lame (random guessing that wasn't even funny)...I was hoping it would be fun because it's a MOP event

Problems:
- The individual rounds were pretty good
- The team round...only an hour for proofs?....a bit more would be nicer (and more problems), I liked the idea to use advanced NT though, and topology (Mildorf mentioned this?) would have been nice
- Similarly, more time on the GUTS round might be more fun (the guts round is really fun)

Lunch:
Eww...the lines were so crowded, and it was hard to find an area to eat. You know what...they should really close the MIT student center to students during HMMT...it was waaay too crowded, the mini events had already started by time I ate!

Scoring:
- I don't like the way they do it now because general is actually a lot worse than the others - you see, they weight it based on how people do AND they do the 1.6 thing - I think just the weighting based on how people do would be enough.

[quote="xxazurewrathxx"]Well aren't the general tests usually easier than the topic tests (correct me if I'm wrong - but I'm assuming that's how it was intended)? Judging from past results, it doesn't seem like the weighting system has had any severe flaws, people who took the general test were still able to place high in individual rankings.[/quote]
Not this year, this year first in general had 5 pts more than 2nd but he wasn't in the top 10 since the average score on general is much higher than with the subject tests (I would think)

Oh yeah I just noticed that. Hmm... wouldn't it be more fair to do it like that though? To compensate for the difficulty of the tests, that is. What exactly is the new scale anyway?

I was under the impression that the point of general was to be easy. I don't think there should necessarily be an expectation that the easier test will be considered equally with the harder tests. If you are scoring at the top of general, in a sense you should be taking one of the other, harder tests. If you aren't, then the relatively smaller deviation in results (in the other contests, many people are getting single digit scores while the winners get 30-40 points) means that your normalized score is higher than it would've been in the other contests. I personally think it would be fair either way, though the 1.6 thing does make some sense.

It would be nice if you listed more than the top 10 for overall, since you did give stuff to the top 22 or so. Maybe go to top 25 or something?

I agree that guts was a bit crowded. I think most teams learned to cope with it, but more space would be nice, if it is actually possible.

The useless math olympiad was pretty cool, but I was disappointed that there was no dumbassing competition or one of those other funny things that we did at MOP. I didn't mind it though, it was still pretty cool, but I just think that more diverse stuff could be added.

I also think that the power round subject was a bit unfair this year. A lot of people already knew a lot about multiplicative functions, specifically our team and I'm guessing the WOOTlings, based on the fact that they had MOPpers and we learned about it at MOP.

I guess -- Geometry test was too easy, based on the scores? But you probably already knew that.

[quote="JSteinhardt"]
I guess -- Geometry test was too easy, based on the scores? But you probably already knew that.[/quote]
No kidding, my score was 2.5 times what I was getting on practices.

Do you think that you could move HMMT to President's Day Weekend next year? That way, more teams would be able to go.

[quote="chess64"][quote="JSteinhardt"]I also think that the power round subject was a bit unfair this year. A lot of people already knew a lot about multiplicative functions, specifically our team and I'm guessing the WOOTlings, based on the fact that they had MOPpers and we learned about it at MOP.[/quote]

Actually, we only had one MOPper (Jeremy), but yeah the rest of us already knew about all those functions too.[/quote]

Did we learned about it in WOOT??

well it is definitely in ACoPS, and I think all of us had it... :rotfl:

man...i was reading that section....and i was like OMG those are multiplicative...thats cool...but i did not do any of the problems...i was literally like...this is cool and all, but it will not be useful in the near future

Here's my take.

Problems-great. except for some repeats, they were great.

breakfast area- eh. I mean it was good enough, but it was kinda dark...

individual rooms-Pretty good. The desks were a little small, but it did fine.

team rooms-I would rather be in a single, smaller room and working around a table than what we had, but it was good.

lunch-well we saw the student center, but due to my uber knowledge of the MIT campus, me and 5 others went to a small (hot) sandwich shop down the street on Mass Ave.

guts-well it would have been better if EVERYONE was allowed to split their team into two rows, we were at the back so we couldn't. I don't see a clear solution to make it less crowded, maybe there could be stations in the infinite corridor and we could be running out of classrooms...of course then real time scoring would go away. I don't know.

I like the individual scoring system better this year. But I think people taking general should not be eligible for individual awards.

When do we find out our team rankings? I mean I came with my school and we have not found out how we placed since we were not in the top 10.

oh dawg yes. cause then i could take: CALCULUS AND NUMBER THEORY. my two greatest passions. that'd be pretty hot.

seriously, who cares about combinatorics? or if you're really cool, you'll just add it to the bill and not leave out combinatorics.

oh dude and bringing back advanced topics would be SWEET

except they should be like actually advanced topics

[quote="bpms"]It'd also be very cool if you guys added a number theory section.[/quote]

agreed

Then maybe I wouldn't have done general.

no take out calculus, and add number theory

and also advanced topics. i'm sure mildorf would be happy to write some inequalities to put on there :P

damn son, inequalities are not advanced

complex analysis is more advanced

(I was sort of perturbed the limits of integration on calc #10 were 0 to infty - if it were -infty to infty, I think complex analysis works?)

It's hard to write non-guessable inequality questions with numerical answers. (Unless you make them extraordinarily hard.)

Right, HMMT individual rounds should emulate IMO number 3s.   :roll:


geometry
geometric transformation
reflection
complex analysis

I was going through some of the exercises from chapter $1$ of "Complex Analysis" (4th edition) by Serge Lang, and the first two questions just didn't make sense.

Exercise $(1)$ basically says $f(z)=\frac1{z}$ describes a map called [b]inversion[/b] through the unit circle, and exercise $(2)$ says $f(z) = \frac1{\overline{z}}$ describes a map called [b]reflection[/b] through the unit circle.

Now, I am pretty sure that $f(z) = \frac1{\overline{z}}$ describes [b]inversion[/b] and [b]not reflection[/b].

Am I correct or is the book right?

To me, "reflection in a circle/sphere" = "inversion". And this is what $1/\bar z$ does.
The map $f(z)=1/z$ is an inversion followed by reflection in the real axis. Or the other way around.

$f(z) = \frac 1{\overline z}$ is geometric inversion, but $g(z) = \frac{1}{z}$ is multiplicative inversion.  It's a collision of names, and Lang just decided to resolve it in his particular way.  Notation can't be "wrong," just unconventional.


geometry
circumcircle
Euler
geometry unsolved

Let $ O$ the circumcenter of a triangle $ ABC$ and $ A_1,B_1, C_1$ the midpoints of the sides $ BC,CA , AB$

 respectively. We consinder the points $ A_2,B_2,C_2$ such that :

$ \vec {OA_2} \equal{} \lambda \cdot \vec {OA_1}$ , $ \vec {OB_2} \equal{} \lambda \cdot \vec {OB_1}$ ,  $ \vec {OC_2} \equal{} \lambda \cdot \vec {OC_1}$ ,

 where $ \lambda > 0$ . Prove that lines $ AA_2, BB_2 , CC_2$ are concurent .

 Babis

$ A_2B_2 \parallel A_1B_1 \parallel AB$ then $ \triangle A_2B_2C_2$ is homothetic to $ \triangle ABC$. We can also prove taht the locus of the points of concurrence when k change is the Euler Line of ABC.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]$ A_2B_2 \parallel A_1B_1 \parallel AB$ then $ \triangle A_2B_2C_2$ is homothetic to $ \triangle ABC$. [/quote]
Is the concurrency trivial after finding that $ A_2B_2C_2$? I mean is it true for any homothetic triangle in general? I would like to see a complete proof.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]We can also prove taht the locus of the points of concurrence when k change is the Euler Line of ABC.[/quote]
And how do you proof this remark?

[quote="math_07"]...I mean is it true for any homothetic triangle in general? I would like to see a complete proof.?[/quote]
Suppose $ A'B' \parallel AB, B'C' \parallel BC, C'A' \parallel CA.$ The triangles $ \triangle A'B'C' \sim \triangle ABC$ are similar by ASA, having equal angles $ \Longrightarrow$ $ \frac{B'C'}{BC} \equal{} \frac{A'B'}{AB}$. Let $ A'A, B'B$ meet at $ S.$ $ \triangle A'B'S \sim \triangle ABS$ are similar by ASA, having equal angles $ \Longrightarrow$ $ \frac{B'S}{BS} \equal{} \frac{A'B'}{AB}.$ Combining, $ \frac{B'C}{BC} \equal{} \frac{B'S}{BS}.$ Then $ \triangle SBC \sim \triangle SBC$ by SAS $ \Longrightarrow$ $ \angle C'SB' \equal{} \angle CSB$ $ \Longrightarrow$ $ S, C', C$ are collinear. $ S$ is similarity center of $ \triangle A'B'C' \sim \triangle ABC.$

[quote="math_07"][quote="¬[ƒ(Gabriel)³²¹º]¼"]We can also prove that the locus of the points of concurrence when k change is the Euler Line of ABC.[/quote]
And how do you proof this remark?[/quote]
$ O$ is the similarity center of $ \triangle A_1B_1C_1 \sim \triangle A_2B_2C_2$ and $ G$ is similarity center of $ \triangle A_1B_1C_1 \sim \triangle ABC$ $ \Longrightarrow$ similarity center $ S$ of $ \triangle A_2B_2C_2 \sim \triangle ABC$ is on the line $ OG$ (d'Alembert theorem).


function
calculus
integration
real analysis
real analysis unsolved

May anyone tell me that
 F(5/2,5/2,1,z^2)= ?
 F(11/2,11/2,1,Z^2)= ?
I have looked up a handbook to find out how they are presented,
but that book didn't show the style above.
For example F(1,1,2,Z)=(1/z)ln(1-z)

Hello! I recommend that you use the handbook by Prudnikov, Brychkov and Marichev, it is called
Tables of Series and Integrals, published by Gordon and Breach. Look up the volume on "More Special Functions".
In this volume, there is a section which lists down many many special cases of the hypergeometric function.


conics
ratio
trigonometry
geometry
projective geometry
geometry unsolved

In triangle $ABC$, let $D$ be on $BC$ such that $AD$ bisects $\angle BAC$. $M$ is a point on $AD$. $E$=$BM\cap AC, F=CM\cap AB, J=FD\cap BE, K=ED\cap CF, T=AK\cap BC$. Show that $F,M,K,C$ form a harmonic division, and $B,D,T,C$ also form a harmonic division.

This can hardly be called a "problem". The only reason why one would not be able to solve this, is because he/she doesn't know what harmonic division is.

"Solution": $AK$ is the polar of $B$ wrt to $AC,AD$.
Therefore, $AC,AT,AD,AB$ is a harmonic bundle. The conclusion follows.

[quote="Jan"]

"Solution": $AK$ is the polar of $B$ wrt to $AC,AD$.
[/quote]

What do you mean by that? A polar is with respect to a circle..

Not necessarily. A polar is defined as the locus of all points that are harmonically conjugated to a given point P wrt the intersection points of a variable line through P and a given conic section. (Probably there is even a more general definition of a polar)

Two intersecting lines are also a conic section (degenerate) !

See http://mathworld.wolfram.com/Polar.html

Well, I don't understand much there. Can you give your definition of polar with two intersecting lines?

(I'm just putting the image... I first drew it and then discovered that this problem is very easy if you know something about cross ratios)

This is exactly what I meant in my first post. The only reason why one wouldn't be able to solve this problem is because of unfamiliarity with polar lines.

A polar line of a point P wrt two interesecting lines (degenerate conic) is the locus of all points Q such that P and Q divide the intersection points of the degenerate conic with PQ harmonically.

We can even change this degenerate conic to a general conic, but since you were asking for an explanation in the case of a degenerate conic, I gave this specific example.
Anyway, Google is a very friendly person, why not ask him (or should I say her)?  :wink:

Hey, are you sure the problem is correct? By simple trig bashing (looking at $\frac{FM}{MK}=\frac{FC}{CK}\leftrightarrow \frac{\frac{FM}{MA}}{\frac{MK}{MA}}=\frac{\frac{FC}{CA}}{\frac{CK}{CA}}$..., I got that $F,M,K,C$ form a harmonic division iff $\tan BAD=3\tan DAK$, which is clearly untrue!

umm...what? the angle thing is not necessary

T, B, D, C form a harmonic progression by definition [i.e. the way T was constructed]

then note that FT, MB, KD, are concurrent at E...so an easy lemma shows that [use sine area formula?]

F, M, K, C are harmonic

[if you have 4 lines, then a 5th line interesects those 4 so you get a harmonic progression, then any other line that intersects the initial 4 lines makes a harmonic progression]


[quote="limes123"]Find all positive integers $ n$ with the following property: for any real numbers $ a_1, a_2, . . . , a_n$, knowing the numbers $ a_i \plus{} a_j$ , $ i < j$, determines $ a_1, a_2, . . . , a_n$ uniquely.[/quote]

Clearly $ n \ge 3$. Then

for $ i<j<k$, we have $ a_i\plus{}a_j.......(1)$, $ a_j\plus{}a_k.......(2)$ , $ a_k\plus{}a_i.........(3)$ are known.

$ \Rightarrow a_i\equal{}\frac{(1)\plus{}(3)\minus{}(2)}{2}, a_j\equal{}\frac{(1)\plus{}(2)\minus{}(3)}{2}, a_k\equal{}\frac{(2)\plus{}(3)\minus{}(1)}{2}$ are uniquely defined.

Now for any $ n \ge 3$, we put $ i\equal{}1,j\equal{}2,k\equal{}3,4,...,n$ we get $ a_1,a_2,...,a_n$ are uniquely defined.

So all $ n \ge 3$ are the answers.

I'm thinking of 4 numbers, $ a_1, a_2, a_3, a_4$.  The set of all $ a_j\plus{}a_k$ with $ 1 \le j < k \le 4$ is this set $ S \equal{} \{13, 17, 18, 19, 23\}$.  I sorted these outcomes so you won't know which one corresponds to any given $ j,k$.  (Since there are only $ 5$ items in $ S$ rather than $ 6$, one of the sums $ a_j\plus{}a_k$ must be equal to another such sum.) Find my set $ \{a_1, a_2, a_3, a_4\}$.  Can you do it?


Is anyone familiar with the Exeter Math Institute or the Summer Academy held there?


floor function
number theory proposed
number theory

hi,
Let $ p$ be a prime and $ a \in \mathbb{R}$ s.t. $ pa^{2} < \frac {1}{4}$. Prove:
For $ n \geq \lfloor \frac {a}{\sqrt {1 \minus{} 2a\sqrt {p}}} \rfloor \plus{} 1$ we have
\[ \lfloor n\sqrt {p} \minus{} \frac {a}{n} \rfloor \equal{} \lfloor n\sqrt {p} \plus{} \frac {a}{n} \rfloor.
\]

still no idea ?
can somebody give me a hint please ???


Let $X$ be a non-empty set of positive integers which satisfies the following: [list] [*] if $x \in X$, then $4x \in X$, [*] if $x \in X$, then $\lfloor \sqrt{x}\rfloor \in X$. [/list] Prove that $X=\mathbb{N}$.

Let $ m\in X$ the smallest element of $ X$. Then $ m\le[\sqrt m]$, from where $ m\equal{}1$. Then $ 4^k\in X$, $ \forall k\in\mathbb{N}_0$, from where $ 2^k\equal{}[\sqrt{4^k}]\in X$, $ \forall k\in\mathbb{N}_0$. All powers of $ 2$ are in $ X$.
Now suppose some $ t\notin X$. Then the equation ($ n$ is the unknown) $ t\equal{}[\sqrt n]$ has no solution $ n\in X$. This means that $ \forall m$, $ t^2\le m<(t\plus{}1)^2$, $ m\notin X$. We prove by the induction on $ k\ge1$ that if $ t^{2^k}\le m<(t\plus{}1)^{2^k}$, then $ m\notin X$. Base case $ k\equal{}1$ is proved.
Suppose for some $ k$ we have proved this claim. Now take $ m$ such that $ t^{2^{k\plus{}1}}\le m<(t\plus{}1)^{2^{k\plus{}1}}$. Suppose $ m\in X$. Then $ [\sqrt m]\in X$. But $ t^{2^k}\le[\sqrt m]\le\sqrt{m}<(t\plus{}1)^{2^k}$, contradicting the induction hypothesis.
Now suppose some $ X\neq \mathbb{N}$. Assume some $ l\notin X$. Then $ l\ge3$. Let $ x\equal{}\log_2l$, $ y\equal{}\log_2(l\plus{}1)$. As we have proved, for any $ k\ge2$, no positive integer in $ [l^{2^k},(l\plus{}1)^{2^k})$ is in $ X$, or no positive integer in $ [2^{2^kx},2^{2^ky})$ is in $ X$. Because we have proved that all powers of $ 2$ are in $ X$, this means that no positive integer is in the interval $ [2^kx,2^ky)$ for any $ k$, implying that $ 2^k(y\minus{}x)<1$. However $ y\minus{}x>0$ is a fixed positive real number, while $ k$ is as big as we want, so $ 2^k(y\minus{}x)$ is not bounded. Contradiction. Thus $ X\equal{}\mathbb{N}$.

Very nice solution :) And I had another approach.
Step 1$ [\sqrt {\sqrt {[x]}}]=[\sqrt {\sqrt {x}}]$
Step 2:$ 1\in X$
Step 3:$ [2^{\frac{n}{2^{m-1}}}]\in X$for all positive integers $ m,n$
Step 4:For a certain $ n \in N,n\ge 2$Choose $ m$ sufficiently large so that $ 2^{\frac{1}{2^{m-1}}}$$ \le \frac{n}{n-1}$ Then $ n\in X$
QED


$ 1\leq a,b,c,d \leq 2$  prove:

$ \frac{a\plus{}b}{b\plus{}c}\plus{}\frac{c\plus{}d}{d\plus{}a}\leq\frac{4(a\plus{}c)}{b\plus{}c}$

it is simple zaya_yc
$ \frac{{3c}}{{b \plus{} c}} \ge 1$
$ \frac{{2c}}{{b \plus{} c}} \plus{} \frac{{3a}}{{b \plus{} c}} \ge \frac{{c \plus{} d}}{{d \plus{} a}}$
Equality does not hold


If g represents the numerically smallest generator mod p, is g also a generator mod
p^n for every value of n? Can you find a counterexample?

Can you explain generator? Do you mean that $ g$ is the smallest number such that the numbers $ g^1$ through $ g^{p \minus{} 1}$ iterate over the values $ 1$ to $ p \minus{} 1$ modulo $ p$ in some order?

Sorry, for purposes of this question g is a generator mod p if the set of number of the form g^n for n between 1 and p-1, will contain all of the units mod p.

Your "generators" are known as primitive roots. [url=http://mathworld.wolfram.com/PrimitiveRoot.html]MathWorld[/url] and [url=http://en.wikipedia.org/wiki/Primitive_root_modulo_n]Wikipedia[/url] have articles about them.

If I remember correctly, if $ g$ is a primitive root for $ p$ and $ p^2$, it is a primitive root for all $ p^n$ with $ n$ positive. However, $ g$ being a primitive root for $ p$ does not necessarily imply $ g$ being a primitive root for $ p^2$. However, either $ g$ or $ g \plus{} p$ is a primitive root for $ p^2$.


Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\frac{N}{11}$ is equal to the sum of the squares of the digits of $N$.


Αυτή την ανισότητα την βρήκα στον απειροστικό 1. Την έλυσα με επαγωγή. Υπάρχει κάποιος άλλος τρόπος εκτός αυτού?

Να δειχθεί ότι $ \frac{a^{n}\plus{}b^{n}}{2}\geq (\frac{a\plus{}b}{2})^{n}$ με $ a,b>0$

Θέσε $ x\equal{}\frac{a}{b}$ και μελέτησε τη συνάρτηση που προκύπτει.

Durandal 1707

Η επαγωγή ισχύει για $ n \equal{}$ φυσικό. Θα δείξω τη γενικότερη περίπτωση: ότι ισχύει για $ n \equal{}$ πραγματικό $ \ge 1$.

Εφαρμόζοντας την ανισότητα των δυνάμεων είναι για κάθε πραγματικό $ n \ge 1$:

$ (\frac {a_n \plus{} b_n}{2})^{\frac {1}{n}} \ge (\frac {a^1 \plus{} b^1}{2})^{\frac {1}{1}} \Leftrightarrow$

$ \frac {a^n \plus{} b^n}{2} \ge (\frac {a \plus{} b}{2})^n$

[quote="Γιώργος"]Αυτή την ανισότητα την βρήκα στον απειροστικό 1. Την έλυσα με επαγωγή. Υπάρχει κάποιος άλλος τρόπος εκτός αυτού?

Να δειχθεί ότι $ \frac {a^{n} \plus{} b^{n}}{2}\geq (\frac {a \plus{} b}{2})^{n}$ με $ a,b > 0$[/quote]
Με Jensen για την κυρτή $ x^n$ στο $ [a,b]$ δες αν θέλεις το ποστ 15 Ιουλιου Γεωμετρικές συνθήκες κυρτότητας...

Υπάρχει μια  απόδειξη παρόμοια με του Μανώλη χρησιμοποιώντας την ΑΜ-ΓΜ
Θέτεις $ P\equal{}\frac{a^{n}\plus{}b^{n}}{2}$ άρα η ανισότητα παίρνει τη μορφή 1/(2P^(1/n))+b/(2P^(1/n))<=1
Παρατηρούμε ότι a/(2P^(1/n))=(a^v/2P)^(1/n)*(1/2)^((v-1)/n)<=1/n*(b^n)/2P+(n-1)/n*1/2
παρατηρούμε το ιδιο και για το b/(2P^(1/n)) προσθέτουμε κατα μέλη τις 2 τελευταιες ανισότητες και η απόδειξη τελειωσε :D

Συγγνωμη που δεν τα έγραψα ολα σε latex γιατί βιαζόμουν.

[quote="vangelismak"]Πολύ σωστά κύριε Μπόρη :)   Λογικά για να είναι σε βιβλίο με Απειροστικό λογισμό θα θέλει την απόδειξη με Jenssen :)[/quote]

Όχι απαραίτητα... αν και στοιχειώδης, η Jensen χρησιμοποιείται κυρίως σε πιθανότητες και συναρτησιακή ανάλυση σε πανεπιστημιακό επίπεδο. Τώρα βέβαια, τα πάντα εξαρτώνται από το βιβλίο, eh? ;)

Ok, σταματάω το spamming :P Cheerio,

Durandal 1707

Just of curiosity:Μιας και είπαμε για απειροστικό 1 ποιοι από το forum ειναι φοιτητές του μαθηματικού?

[quote="kunny"]Induction.[/quote]

We solved this inequality with induction. Now we're looking for other beautiful solutions. Can you suggest anything except of course induction? :)

1.Compare $ 2(a^{m + n} + b^{m + n})$ with $ (a^m + b^m)(a^n + b^n)$  for $ a,\ b > 0,\ m,\　n\in\mathbb{N}$.

2. $ \frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$ for $ a>0,\ b>0.\ n\in\mathbb{N}$

$ \Longleftrightarrow \left(\frac{a}{a+b}\right)^n+\left(\frac{b}{a+b}\right)^n\geq \frac{1}{2^{n-1}}.$

We can set $ x=\frac{a}{a+b},$ the inequality becames $ x^n+(1-x)^n\geq \frac{1}{2^{n-1}}$ for $ 0<x<1.$


Let $ a,b,c$ positive reals such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ abc\plus{}\frac{15}{ab\plus{}bc\plus{}ca}\geq6$

[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ abc \plus{} \frac {15}{ab \plus{} bc \plus{} ca}\geq6$[/quote]

$ abc \plus{} \frac {15}{ab \plus{} bc \plus{} ca} \minus{} 6 \equal{} \frac {15}{ab \plus{} bc \plus{} ca} \plus{} \frac {5(ab \plus{} bc \plus{} ca)}{3} \minus{} 10$

$ \plus{} abc \minus{} \frac {5(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)}{9} \plus{} \frac {4(a \plus{} b \plus{} c )^3}{27}$

$ \equal{} (\frac {15}{ab \plus{} bc \plus{} ca} \plus{} \frac {5(ab \plus{} bc \plus{} ca)}{3} \minus{} 10 )\plus{} 3(\sum_{\text{cyc}}a^{3} \minus{} \sum_{\text{sym}}ba^{2} \plus{} 3abc)$

$ \plus{} (a^{3} \plus{} b^{3} \plus{} c^{3} \minus{} 3abc) \geq 0$

this is my solution:
firstly I put, $ p\equal{}a\plus{}b\plus{}c$ , $ q\equal{}ab\plus{}bc\plus{}ca$ and $ r\equal{}abc$
then from Schur, $ r\geq \frac{4}{9}(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\minus{}\frac{1}{9}(a\plus{}b\plus{}c)^3\equal{}\frac{4q}{3}\minus{}3$
then it's suffice to prove that, $ 4q\plus{}\frac{45}{q}\geq 27$
$ \Leftrightarrow 4\left(q\plus{}\frac{9}{q}\right) \plus{} \frac{9}{q}$
wish is obvious , because by AM-GM, we have $ q\leq \frac{p^2}{3}\equal{}3$ and $ q\plus{}\frac{9}{q}\geq 6$
equality holds when $ a\equal{}b\equal{}c\equal{}1$

[quote="peine"]this is my solution:
firstly I put, $ p \equal{} a \plus{} b \plus{} c$ , $ q \equal{} ab \plus{} bc \plus{} ca$ and $ r \equal{} abc$
then from Schur, $ r\geq \frac {4}{9}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \minus{} \frac {1}{9}(a \plus{} b \plus{} c)^3 \equal{} \frac {4q}{3} \minus{} 3$
then it's suffice to prove that, $ 4q \plus{} \frac {45}{q}\geq 27$
$ \Leftrightarrow 4\left(q \plus{} \frac {9}{q}\right) \plus{} \frac {9}{q}$
wish is obvious , because by AM-GM, we have $ q\leq \frac {p^2}{3} \equal{} 3$ and $ q \plus{} \frac {9}{q}\geq 6$
equality holds when $ a \equal{} b \equal{} c \equal{} 1$[/quote]

My Friends Thank you very much.... :)

[quote=sqing]Let $ a,b,c$ be positive reals such that $ abc(a\plus{}b\plus{}c)\equal{}3$. Prove that\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{18}{(a+b+c)^2}\ge 5,\]
[/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers such that $abc(a\plus{}b\plus{}c)\equal{}3$. Prove that:
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{21}{(a+b+c)^2}\ge \frac{16}{3}\]

[quote=sqing]Let $ a,b,c$ be positive reals such that $ abc(a\plus{}b\plus{}c)\equal{}3$. Prove that
\[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}\ge 11\][/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$.
In another hand, $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}=\frac{9v^4-6uw^3}{w^6}+\frac{8}{u^2}$ is an increasing function of $v^2$.
Thus, by $uvw$ it remains to check one case only: $b=a$ and $c=\sqrt{a^2+\frac{3}{a^2}}-a$, which gives that our inequality is true.


[quote="anirudh"]An equilateral triangle and a square have the same perimeter of 12 inches. What is the ratio of the side length of the triangle to the side length of the square? Express your answer as a common fraction.[/quote]
[hide]$\frac{4}{3}$[/hide]

[hide]a side length of a triangle is 4 and a side length of a square is 3, so $\boxed{\frac{4}{3}}$[/hide]

[hide] each side of the equilateral has the same length and the equilateral has 3 sides so a side length of the equilateral triangle is
$12: 3=4$
 and the square has 4 sides so the side length of the square is
$12: 4=3$
=> the ratio is $4/3$ [/hide]

[hide]I got 4/3.

Each side of the equilateral triangle is 4.
Each side of the square is 3.

So 4/3[/hide]

[hide="solution"]The side length of the triangle is 4 and the square is 3. The ratio is four thirds.[/hide]

[quote="moogra"][hide]An equilateral triangle and a square have the same perimeter of 12 inches. What is the ratio of the side length of the triangle to the side length of the square? Express your answer as a common fraction.

4 inches = triangle
3 inches = square

3/4[/hide][/quote]

Read the question again  :wink:

[hide]The square has sides that are 3 inches and the triangle has sides that are 4 inches. The ratio is 4/3.[/hide]

[quote="i_like_pie"]Don't revive old topics, especially ones with questions that have been answered multiple times already.[/quote]
Also he never posts solutions and use hide tags.

My answer
[hide]Each side of the triangle is 4 inches and each side of a square is 3 inches. Therefore, the answer is $\frac{4}{3}$.[/hide]


Hello, all.  I am starting another tourney that can hopefully grab some attention, the Countdown Cup.  It will be a min. 28, max. 32 player cd tournament, single elimination.  Seeding will be based on highest rating (I give full credit to the AoPS user name of "mewto55555").  The first 32 signed up will get top priority.  If we have, say 20 people, the Countdown Cup will be postponed until more interest is generated.  I would like to also introduce new members to start joining and making tourneys (just look at me, I made 2 in the past week! :D ).  Along with your post, please post your highest rating.

Thank You.

gotta put this at the tippy-top  :D  :D  :D

I'll join!  I'll get owned though, my highest rating is a 1270.

why not ill die but thats ok :rotfl:  1357

I shall join
oh yes highest rating:1292 (pathetic)

[quote="mathemagician1729"]I shall join[/quote]

and ur highest rating is?

u gotta read all the junk at the tippy-top! :lol:

:D  1239

I am not so great

I will have a go anyway

[quote="ash-Pokemon"]my rating is dead, iwas almost muted, and now im going to lose another tournament!!!!!!!;(  :)  :([/quote]

READ THE JUNK AT THE TOP!

oh btw, i no wat ur highest is (1351)  :D  :D  :D

I will join but i know I will get murdered in 59.684 seconds flat. (My highest rating is a 1414 which is pathetic)

I just started playing today. Highest rating: 1272(I think)

Fantasy forgot to mention his highest rating which is 1502.

I'll join too. My highest rating is 1563.

AIME15 will sign up

Highest: 1354

I SUCK!

1 superllamel (1582)
2 mewto55555 (1529)
3 ra5249 (1517)
4 FantasyLover (1502)
5 alexhhmun? (1485)
6 Yongyi781 (1467)
7 vallon22 (1457)
8 Jongao (1452)
9 ernie (1451)
10 RomanianGenius (1438)
11 BOGTRO (1425)
12 lotsofmath (1422)
13 Brut3Forc3/winner of coin toss (1414)
14 skaldjfhrulez (1414)
15 kl2836/winner of coin toss (1412)
16 budi713 (1412)
17 hurdler (1411)
18 violin321 (1367)
19 mathblitz (1357)
20 bob einstein (1356)
21 ash-Pokemon (1351)
22 thdanh90/winner of coin toss (1349)
23 isabella2296 (1349)
24 AIME15 (1340)
25 kookamunga (1335)
26 nikeballa96 (1323)
27 PWNER Express (1316)
28 mathemagician1729 (1292)
29 vamathletes (1280)
30 Meta Knight (1247)
31 Mind Storm (1241)
32 blue1421 (1239)

NO MORE CAN JOIN!

superllamel's new rating is 1600 :huh:  :huh:

[quote="BOGTRO"]*EDITEED*

my highest rating is 1439. Currently 1369... well it was 1397,

my lagging comp wouldn't load the problems and I lost 30 pts, won two 15-player games,

and gained a point each for both of them.[/quote]

srry BOGTRO- but matches hav already begun

vallon22 (7) defeats nikeballa96 (26)

1st game: 4-0, vallon22
2nd game: 4-0, vallon22
3rd game: 4-0, vallon22

[quote="isabella2296"]Wait -- so who am I playing?[/quote]

ur playing RomanianGenius

[quote="isabella2296"]Wait -- so who am I playing?[/quote]

i made a new thread for the games that can take place currently.

yes u r, plz check other thread for all match-ups


What are the products when $ \text{2\minus{}methylcyclohexanone}$ is treated with:

a) $ \ce{NaOH}$ in acetone

b) $ \ce{NaEtO}$ in water

Which one is the kinetic product and which one is the thermodynamic product? The products are structural isomers and both sodium salts.

Shouldn't it be $ \ce{NaOEt}$ in ethanol?

That's the problem I was wondering about that too! The problem stated water.

So yeah I am not actually too sure of what the isomers are.

Two isomeric enolate ions will form: the thermodynamic product will result from the weakest base, and the kinetic product from the strongest base. Haven't we discussed this already before?

Yeah, just want to confirm that the sodium ethoxide should be in ethanol thanks.

A question why wouldn't we get aldol products for the first one?

Because with a weak base as NaHO, the equilibrium

$ 2\minus{}Methylcyclohexanone \plus{} Acetone \rightleftharpoons aldol\, product$

lies to the left.

Wikipedia says that the formation of an enolate must be kept in strictly Bronsted Acid free conditions. So what would actually happen if water was present rather than ethanol? Would it add protonate the enolate?

[quote="BanishedTraitor"]Wikipedia says that the formation of an enolate must be kept in strictly Bronsted Acid free conditions.[/quote]

Of course! Enolate anions are also bases.

What species are present in a NaOEt/water mixture?

$ \ce{Na \plus{}} , \ce{H2O}, \ce{OEt \minus{}} , \ce{EtOH}, \ce{OH \minus{}} , \ce{H3O \plus{} }$ I think those are all the species.

[quote="Carcul"]Because with a weak base as NaHO, the equilibrium

$ 2 \minus{} Methylcyclohexanone \plus{} Acetone \rightleftharpoons aldol\, product$

lies to the left.[/quote]
i know the steps of aldol condensation are equilibria but only the fist step lies in unfavourable equilibrium,doesnt it as enolate is a stronger base. so if the enolate is already formed, i dont see why you cant get an aldol product.
also to get an enolate with naoh, you will have to use excess of the alkali to shift the equilibrium to the right, so in the other steps too, the equilibrium has to shift to the right, shouldnt it?

[quote="VARUNARASU"]i know the steps of aldol condensation are equilibria but only the fist step lies in unfavourable equilibrium[/quote]

That depends on the base and on the carbonyl compound(s).

with naoh, the equilibrium lies definitely to the left.

then how is the enolate formed? coz the first step is the formation of enolate.

The enolate is formed by reaction of 2-methylcyclohexanone with NaHO.

i know dat. but my doubt was this- for aldol condensation in this case, the major hurdle wud b the formation of enolate, but once the enolate is formed, why dont you get an aldol product? :D


trigonometry
inequalities unsolved
inequalities

# generated from Geo#

$ B$ , $ A$ acute angles
Prove:
$ \tan \frac {B \minus{} A} {3} \ge \frac {\sin B \minus{} \sin A} {\cos A \plus{} \cos B \plus{} 1}$

Taking $ A \equal{} \frac {\pi}{4}$ and $ B \equal{} 0$, we obtain a contradiction.


geometry
power of a point
radical axis
projective geometry
geometry unsolved

With your naming of points, we shall prove that actually OE and FP are perpendicular.
Name Q the intersection of tangents at A and C to the given circle, apply Pascal to the "hexagon" ABCCDA and get P, Q and F collinear, name R the intersection of the tangents at D and B, apply Pascal to ABBCDDA, get P, F and R collinear.
Now see that O and E have equal powers w.r.t. the circles (Q,QA) and (R,RB) [OA, OB are equal tangents to both circles, and AE*CE = BE*DE from ABCD], hence OE is the radical axis of the 2 circles.

Best regards,
sunken rock

[quote="Litlle 1000t"][quote="FOURRIER"]Let $ A,B,C,D$ be 4 points in a circle of centre $ O$ and:

$ F \equal{}$ Intersection of $ AD$ and $ BC$
$ E \equal{}$ Intersection of $ AC$ and $ BD$
$ P \equal{}$ Intersection of $ AB$ and $ CD$

Then prove $ EF$ Perpendicular to $ PO$.

Can some one help, I need it.[/quote]

$ EF$ is polar line of $ P$  :P[/quote]

whats that?

for a point $ P$,we know that the polar of point $ P$ wrt circle $ (O)$ is perpendicular to $ PO$...
here are two useful links about poles and polars:
[url]http://www.mathlinks.ro/viewtopic.php?p=957064#957064[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=5501[/url]

The radiacl axis OE is perpendicular onto QR, the line joining the centers of these circles, and P and P belong to this line.
See drawing, attached.

Best regards,
sunken rock

I have heard that this is called "Brokard" bla bla~.Oh,I make my first reply to other's post. :blush:

[quote="FOURRIER"]Let $ A,B,C,D$ be 4 points in a circle of centre $ O$ and:

$ F \equal{}$ Intersection of $ AD$ and $ BC$
$ E \equal{}$ Intersection of $ AC$ and $ BD$
$ P \equal{}$ Intersection of $ AB$ and $ CD$

Then prove $ EF$ Perpendicular to $ PO$.

Can some one help, I need it.[/quote]

 Well , during the  Balkan MO 2008 , a greek student and MLinker (Nick Rapanos) used this Lemma in a clever way and gave a nice solution to the first geometry problem. This geve him one more medal by the competition !As you can see , it is a very important proposition.

 Babis Stergiou


Ross Mathematics Program
calculus
topology
geometry
summer program
Mathcamp
abstract algebra

Hi,

I am an Australian and i am going to try out for a few camps next year and the year after. I will be trying out for Ross, ISSYP, SSP and mathcamp. I just have a few questions regarding mathcamp.

I have looked at the 2009 classes and found there are extremely interesting classes such as multivariable calculus, Elliptic curves, algebraic topology, ring theory and Galois theory. I was wondering what is the level of difficulty of these classes, and how deep do they go, is it mathematically intense or do they just use words to summarize the topics?

Also, I will most likely be starting university mathematics through distance education while still in high school (i will be in my second last high school year). Is there any rule stating that i can not go because of that (but let me convey to you that i will still be a high school student when i apply for the camps).

Thank you for helping  :lol:

[quote="archimedes1"]No classes are just summaries; there is real mathematics going on in each one.  For quick overviews and brief tastes of different areas, Mathcamp has colloquia on weekends.

Many campers take courses with a college during the school year, so it shouldn't prevent you from coming.[/quote]

Thankyou  :lol:

The answer is very much that real mathematics does go on. We rigorously prove everything in these classes, and the problems are as difficult as one might expect in a normal college math class. E.g., last year in algebraic topology, we rigorously proved homotopy lifting and the existence of a universal cover, or this year in ring theory, we proved Wedderburn's theorem on semisimple rings and the classification of modules over a PID. Usually, the class must go quite fast and/or not cover everything that a full college class might cover, but we are doing serious mathematics.

[quote="calc rulz"]the problems are as difficult as one might expect in a normal college math class.[/quote]

I greatly envy you if you can honestly say that. My college math classes have been moving slower than all but the slowest Mathcamp classes. Classes taught by David Roe, who taught Elliptic Curves and the six-hour-a-day Modular forms class, are especially notorious for insane difficulty. It's a bit hard to accurately convey the difficulty of the courses (after all, my adviser told me I'd be swamped in work if I took the classes I'm taking now, but I've found my load extremely light [I'm a freshman at CMU]), but I'll try me best by saying this: It is expected that, after completing Elliptic Curves and Modular Forms, you will be able to follow a sketch of a proof of Fermat's Last Theorem.

Ehhhhhhh okay so by sketch of the proof of Fermat's Last Theorem you actually mean sketchiest of sketchiness sketchier sketchy proofs. The first three days required complex analysis and group theory, and we legitimately proved dimension formulas, the Fourier expansion, and some facts about Hecke operators.

The last two days, we learned Galois representations under the assumptions that if you know what a Galois group is, what a homomorphism is, and what a vector space is, then you know what a Galois representation is ;). Now just assume a significant graduate background in algebraic geometry and homological algebra, and you can prove Fermat's Last Theorem.

Back to Darmani's point, it is true that Mathcamp classes often move faster than college because you only have a week or two to cover a large amount of material.


Let $ a$ and $ b$ be integers such that the equation $ x^3\minus{}ax^2\minus{}b\equal{}0$ has three integer roots. Prove that $ b\equal{}dk^2$, where $ d$ and $ k$ are integers and $ d$ divides $ a$.

[quote="Johan Gunardi"]Let $ a$ and $ b$ be integers such that the equation $ x^3 \minus{} ax^2 \minus{} b \equal{} 0$ has three integer roots. Prove that $ b \equal{} dk^2$, where $ d$ and $ k$ are integers and $ d$ divides $ a$.[/quote]
Let $ u,v,w$ be the three integer roots. We get :
$ (1)$ : $ u\plus{}v\plus{}w\equal{}a$
$ (2)$ : $ uv\plus{}uw\plus{}vw\equal{}0$
$ (3)$ : $ uvw\equal{}b$

If $ b\equal{}0$, the required property is true since $ b\equal{}0\equal{}1\cdot 0^2$ with $ 1|a$ so we'll now consider $ b\ne 0$ and so $ u,v,w\ne 0$

Let $ u\equal{}u'c$ and $ v\equal{}v'c$ with $ \gcd(u,v)\equal{}c>0$

$ (2)$ $ \implies$ $ u'v'c^2\plus{}w(u'\plus{}v')c\equal{}0$ $ \implies$ $ u'v'c\plus{}w(u'\plus{}v')\equal{}0$ $ \implies$ $ u'\plus{}v'|c$ (since $ \gcd(u'v',u'\plus{}v')\equal{}1$

So $ c\equal{}(u'\plus{}v')k$ and $ u'v'c\plus{}w(u'\plus{}v')\equal{}0$ becomes $ w\equal{}\minus{}u'v'k$. So :

$ u\equal{}u'(u'\plus{}v')k$
$ v\equal{}v'(u'\plus{}v')k$
$ w\equal{}\minus{}u'v'k$

Then $ (3)$ $ \implies$ $ b\equal{}\minus{}(u'v'(u'\plus{}v')k)^2k$ and $ (1)$ $ \implies$ $ k|a$

Hence the result.


group theory
abstract algebra
advanced fields
advanced fields solved

Prove that a non-trivial normal subgroup with infinite index of a free group on finitely many generators cannot be free on finitely many generators.

(It's well-knwon that a subgroup of a free group is free, so the subgroup in the problem is indeed free; the issue is proving that it can't be finitely generated)

It's classical, but cute. Enjoy it :).

[Edited]

I think the whole thing's wrong.

If the free group has only one generator, this is obviously false; the normal subgroups of $\mathbb{Z}$ are isomorphic to $\mathbb{Z}$.

Here's a counter example:
Let the free group $G$ be generated by $a,b$, and let $N$ be the set of all elements which are products of an even number of terms $a,b,a^{-1},b^{-1}$.
$N$ is obviously generated by $a^2,b^2,ab,ab^{-1},a^{-1}b,a^{-1}b^{-1}$, but of course is not free on these generators. We can reduce the generating set to $a^2,b^2,ab$, and I think it is free on those three.

Oh, how terribly sorry I am! :) The normal subgroup is supposed to have infinite index. I've edited it.


A landscape designer is designing a triangular garden with two sides that are 4 feet and 6 feet, respectively. The angle opposite the 4-foot side is 30°. How many distinct triangular gardens can the designer make using these measurements?

He can make two distinct triangular gardens
One with side (4, 6, 3. (root(3)) + root(7)
Other with side (4, 6, 3. (root(3)) - root(7)

Thanks,
http://www.teacherone.com

I thank you but what are the steps to help me find the answers on my own?


calculus integration function LaTeX logarithms trigonometry algebra

Evaluate $ \int_{\minus{}1}^1 \frac{2x^4\minus{}x^3\minus{}2x^2\plus{}1}{x^3\minus{}x\plus{}\sqrt{1\minus{}x^2}}\ dx$.

Is this an ordinary Riemann's integral or another type e.g. a Lebesgue's integral ?As a Riemann's type the integrand is not defined in -1 and  it is not possible to calculate this integral .

You can take limit as x approaches 1 from the left of the same integral from -x to x. If the function is nice in a certain way, then this will converge. For example, we can use a similar method to evaluate the integral of $ x^{\minus{}2}$ from 0 to 1.

hes talking about the -1 term i think... only the +1 term seems like it would work your way

(pi)(sqrt.(3)-2)
Replace x by(-x) and then add the two integrals.After this put x=cosz.
Would be extremely grateful if somebody puts it up using latex
Thanks

Computer generated.....Maple 11.

$ \minus{}2\minus{}\frac{2}{3}\,\sqrt {3}\pi \plus{}2\,\pi \plus{}\frac{1}{3}\,\sqrt {3}\ln  \left( 2\plus{}\sqrt {3}
 \right) \minus{}\frac{1}{3}\,\sqrt {3}\ln  \left( 2\minus{}\sqrt {3} \right)$

[quote="Mathematica Integrator"]Mathematica could not find a formula for your integral. Most likely this means that no formula exists.[/quote]

I'll keep trying though  :play_ball:

hello, i have
$ \minus{}2\plus{}2\pi\plus{}\frac{2arcosh(2)}{\sqrt3}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have
$ \minus{} 2 \plus{} 2\pi \plus{} \frac {2arcosh(2)}{\sqrt3}$.
Sonnhard.[/quote]
yes, $ sint\equal{}x$

[quote="kunny"]I can't receive the right answer from anyone. :([/quote]

oh ! don't feel bad, i actually did it on the  same day that you originally posted the question, i just didn't feel like typing at the time. [img]http://forum.ah.fm/images/smilies/console.gif[/img]

$ I\equal{}\int_{\minus{}1}^1 \frac{2x^4\minus{}x^3\minus{}2x^2\plus{}1}{x^3\minus{}x\plus{}\sqrt{1\minus{}x^2}}\;\; dx$

i want to get rid of that annoying $ \ldots \plus{} \sqrt {1\minus{}x^2}$ in the denominator, so i will use the substitution

$ x\equal{}\frac{2t}{1\plus{}t^2}$ to get

$ I\equal{}2\int_{\minus{}1}^1  \frac{ 2\left( \frac{2t}{1\plus{}t^2}\right)^4 \minus{} \left( \frac{2t}{1\plus{}t^2}\right)^3 \minus{}2\left( \frac{2t}{1\plus{}t^2}\right)^2 \plus{} 1} {  \left( \frac{2t}{1\plus{}t^2}\right)^3 \minus{}  \frac{2t}{1\plus{}t^2} \plus{}  \frac{1\minus{}t^2}{1\plus{}t^2} } \; \;  \frac{1\minus{}t^2}{(1\plus{}t^2)^2}\;\; dt$

$ \equal{} 2 \int_{\minus{}1}^1  \frac{ \left[ 2(2t)^4 \minus{} (2t)^3 (1\plus{}t^2) \minus{} 2(2t)^2 (1\plus{}t^2)^2 \plus{} (1\plus{}t^2)^4\right]\;(1\minus{}t^2)} { (1\plus{}t^2)^3 \left[(2t)^3 \minus{} 2t (1\plus{}t^2)^2 \plus{} (1\minus{}t^2)(1\plus{}t^2)^2 \right] }  \;\; dt$

first simplify, and then break it into partial fractions to get

$ I\equal{}6 \int_{\minus{}1}^1 \frac{1} {1\plus{}t^2} \; dt \minus{}4 \int_{\minus{}1}^1 \frac{2t\plus{}1}  {(1\plus{}t^2)^2} \plus{}16 \int_{\minus{}1}^1 \frac{t}  {(1\plus{}t^2)^3} \minus{}4 \int_{\minus{}1}^1  \frac{t(t\plus{}1)} {t^4\plus{}2t^3\plus{}2t^2\minus{}2t\plus{}1}$

and if you are smart enough, you can also break the last one to get

$ I\equal{}6 \int_{\minus{}1}^1 \frac{1} {1\plus{}t^2} \; dt \minus{}4 \int_{\minus{}1}^1 \frac{2t\plus{}1}  {(1\plus{}t^2)^2} \plus{}16 \int_{\minus{}1}^1 \frac{t}  {(1\plus{}t^2)^3}\plus{}\frac{2}{\sqrt 3}  \int_{\minus{}1}^1\frac{t} {t^2\plus{}(\sqrt 3\plus{}1) t \plus{} (\sqrt 3 \plus{}2) } \; dt\minus{}\frac{2}{\sqrt 3}  \int_{\minus{}1}^1  \frac{t} {t^2 \minus{}(\sqrt 3\minus{}1) t\minus{}(\sqrt 3 \minus{}2) }\; dt$

$ \equal{}\left[6 \tan^{\minus{}1} t \right]\; \bigg| _{\minus{}1}^1 \minus{}\left[ 2\left( \tan^{\minus{}1} t\plus{}\frac{ t\minus{}2}{1\plus{}t^2}\right) \right]\; \bigg| _{\minus{}1}^1\minus{}\left[ \frac{4}{(1\plus{}t^2)^2} \right]\; \bigg| _{\minus{}1}^1 \minus{} \left[ \frac{\sqrt 2(\sqrt 3 \plus{}1)}{\sqrt {3(2\plus{}\sqrt 3 )}}\tan^{\minus{}1} \left( \frac{2t \plus{} (\sqrt 3 \plus{} 1)}{\sqrt {2(2\plus{}\sqrt 3) }}\right) \minus{}\frac{1}{\sqrt 3} \ln\left( t^2 \plus{} (\sqrt 3 \plus{}1) t \plus{} ( \sqrt 3 \plus{} 2) \right)\right]\; \bigg| _{\minus{}1}^1
\minus{}\left[ \frac{\sqrt 2(\sqrt 3 \minus{}1)}{\sqrt {3(2\minus{}\sqrt 3 )}}\tan^{\minus{}1} \left( \frac{2t \minus{} (\sqrt 3 \minus{} 1)}{\sqrt {2(2\minus{}\sqrt 3 ) }}\right) \plus{}\frac{1}{\sqrt 3} \ln\left( \minus{}t^2 \plus{} (\sqrt 3 \minus{}1) t \plus{} ( \sqrt 3 \minus{} 2) \right)\right]\; \bigg| _{\minus{}1}^1$

which then yields

$ I \equal{} \boxed{2\left[ \minus{}1\plus{}\pi\left(1\minus{} \frac{1 } { \sqrt 3 }\right) \plus{}\frac{\ln(2\plus{}\sqrt 3)}{\sqrt 3}\right]}$

At last I could get the right answer.  :lol:

Sorry for that [quote="Euclid's Turtle"]Computer generated.....Maple 11.

$ \minus{} 2 \minus{} \frac {2}{3}\,\sqrt {3}\pi \plus{} 2\,\pi \plus{} \frac {1}{3}\,\sqrt {3}\ln \left( 2 \plus{} \sqrt {3} \right) \minus{} \frac {1}{3}\,\sqrt {3}\ln \left( 2 \minus{} \sqrt {3} \right)$[/quote] was right.  :blush:

kunny

Can someone show me how to get the antiderivative that the integrator has got?
[url]http://integrals.wolfram.com/index.jsp?expr=%282x%5E4-x%5E3-2x%5E2%2B1%29%2F%28x%5E3-x%2Bsqrt%7B1-x%5E2%7D%29&random=false[/url]

It seems much more complicated than misan's and with the imaginary unit introduced.


search
analytic geometry
trigonometry
geometry
angle bisector
Pythagorean Theorem
geometry solved

It's a problem that I search the solution sinds a long time!

$D, E$ et $F$ are points on $BC$ such that $AD, AE$ and $AF$ are respectively an altitude, an angle-bisector and a median of triangle $ABC$. If $AD=12$, $AE=13$ and $AF=m$, determine the values of m for which angle $BAC$ is
     1)  an acute angle
     2)  a right angle
     3)  an obtuse angle

Put the coordinate origin at the vertex $A$ and the positive x-axis along the ray $AD$, so that the coordinates of the points $A, D$ are $A = (0, 0)$, $D = (12, 0)$. The line $BC$ is perpendicular to the altitude $AD$ (the x -axis). Using Pythagorean theorem for the right angle triangle $\triangle ADF$,

$DE = \sqrt{AE^2 - AD^2} = \sqrt{13^2 - 12^2} = \pm 5$

Hence, the point $E$ has coordinates $E = (12, 5)$ or $E = (12, -5)$. Consider the 1st case (the 2nd one does not lead to anything different). The angle bisector $AE$ has an equation

$y = x \tan \phi = \frac{y_E}{x_E}\ x = \frac{5}{12}\ x$

If the angle $\angle BAC$ is right, the lines $AB, AC$ form angles $\pm 45^o$ with this angle bisector. Hence, their equations are

$AB:\ \ y = x \tan{(\phi + 45^o)} = x\ \frac{\tan \phi + \tan{45^o}}{1 - \tan \phi \tan{45^o}} = x\ \frac{\frac{5}{12} + 1}{1 - \frac{5}{12}} = \frac{17}{7}\ x$

$AC:\ \ y = x \tan{(\phi - 45^o)} = x\ \frac{\tan \phi - \tan{45^o}}{1 + \tan \phi \tan{45^o}} = x\ \frac{\frac{5}{12} - 1}{1 + \frac{5}{12}} = -\frac{7}{17}\ x$

These 2 lines intersect the line $DE$ with an equation $x = 12$ at points $B, C$ with coordinates

$B = \left(12, \frac{12 \cdot 17}{7}\right),\ \ C = \left(12, -\frac{12 \cdot 7}{17}\right)$

The y-coordinate of the midpoint $F$ of the segment $BC$ is then

$y_F = \frac{y_B + y_C}{2} = 6 \left(\frac{17}{7} - \frac{7}{17}\right) = \frac{1440}{119} \doteq  12.1008$

and the median length $m = AF$ is

$AF = \sqrt{x_F^2 + y_F^2} = \sqrt{12^2 + \left(\frac{1440}{119}\right)^2} = \frac{12}{119}\sqrt{13^4} = \frac{12 \cdot 169}{119} \doteq 17.0420$

Obviously, $AF > AE$, otherwise the angle would be $\angle BAC \le 0$ and the triangle $\triangle ABC$ would not exist. Denote $m_0 = \frac{12 \cdot 169}{119} \doteq 17.0420$.  As a result,

$0 < \angle BAC < 90^o$ [color=white].[/color] for [color=white].[/color] $13 < AF < m_0$

$\angle BAC = 90^o$ [color=white].[/color] for [color=white].[/color] $AF = m_0$

$\angle BAC > 90^o$ [color=white].[/color] for [color=white].[/color] $AF > m_0$


A B C D  a quadruple ,
A B=A C ,
< adb = <bdc=60
prove that  ab=bc=ac
and  ad+dc=bd

[color=darkred][b]Proposed problem.[/b] Let $ABCD$ be a convex quadrilateral for which $AB=AC$ and $m(\widehat{ADB})=m(\widehat{CDB})=60^{\circ}\ .$
Prove that $AB=BC=AC$ and $DA+DC=DB\ .$[/color]

[color=darkblue][b]Proof.[/b] Denote the second intersection $B'$ of the line $BD$ with the circumcircle $w$ of the triangle $ACD\ .$
Prove easily that $B'A=B'C=AC\ .$ But $B\in BD$ and $AB=AC\ .$ Thus, $AB=AB'$ $\Longrightarrow$ $B\equiv B'\in w$ , i.e. $AB=BC=CA\ .$
The second part of the conclusion it is well-known (apply the [u]Ptolemeu's theorem[/u] in the quadrilateral $ABCD$).   [/color]


geometry
trapezoid
homothety
IMO Shortlist
DDIT

The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$.

[i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

[quote="Sunkern_sunflora"]Nice short solution , plane geometry!
Will you please resend your soln. with a diagram. Moreover, I didn't understand what you wanted to mean by [b]A ->C B ->D Q ->Q’ [/b]. proabably you meant constructions!!! :huh:[/quote]
Sorry，I didn't say it clearly
A ->C  means A,C are homothety w.r.t P

[quote="Sunkern_sunflora"]plane geometry wrote:
[quote]Sorry，I didn't say it clearly
A ->C means A,C are homothety w.r.t P[/quote]

What on the earth does [b][size=150]homothety[/size][/b] mean???? :wacko:[/quote]
SEE HERE
http://mathworld.wolfram.com/Homothecy.html

Thank you for providing me with that link plane geometry!
I will be very obliged if you kindly clarify , cause I couldn't make out much about [b]homothecy [/b]from your link.

I have a nice and easy solution with trigonometry.

 :P

Have fun!

[quote="plane geometry"]Using homothety,P is the center,A ->C B ->D Q ->Q’
Then we have Q’,P,Q are collinear  by homothety ∠AQ’D= ∠CQB=∠AQD => AQ’QD is concyclic => ∠DAQ=∠DQ’Q=∠DQ’P=∠BQP[/quote]
that is actually the official solution, isn't it?

I found a solution without even mentioning the word homothety, but it is a bit unnatural.

[quote="April"]The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$.

[i]Author: unknown author, Ukraine[/i][/quote]

___________________
My answer without words  :lol:

Just out of curiosity, what was the motivation behind this solution? How would one know to construct such a point Q'?

Hi, does anyone have any non-homothetic solutions to this problem? will be greatly appreciaited.

Best wishes,

manifest  desniyeeeeeeee

[hide]Let the line parallel to QD intersect line QP at X.

Since we have XB || DQ, BC || AD, we have ∠XBC = ∠QDA.
Since XB || DQ, we have ▲XBP ~ ▲QPD. Therefore we have XB / DQ = BP / DP = BC / AD. Therefore ▲XBC ~ ▲QDA.
Since ∠BXC = ∠AQD = ∠CQB, we have XBCQ is cyclic. Therefore, ∠BQX = ∠BCX = ∠DAQ. [/hide]

I found pretty weird solution using fourth lemma from here: http://sem.edu.pl/materialy/sem-plakat3.pdf
We apply this lemma to triangle $BAQ$ (constructing earlier new point) and solution from now is obvious. But unfortunately this lemma isn't so obvious. I'm not aware of its proof without using trigonometric Ceva's theorem and law of sines :/.

$P$ is the inner center of similitude of the circles $QAD$ and $QBC$. After noticing this fact, the rest part of the problem becomes very easy. This proof is almost same as the previous proof by homothety. But I posted this one because viewing from this perspective provides a better excuse to extend $QP$. (The motivation of what SnowEverywhere was talking.)
[quote="SnowEverywhere"]Just out of curiosity, what was the motivation behind this solution? How would one know to construct such a point Q' ?[/quote]

By isogonal theorem $QP,Q\infty_{AD}$ are isogonal in angle $AQB,$ implying $\angle (QB,QP)=\angle (Q\infty_{AD},QA)=\angle (AD,AQ).$

G3?

[img width=25]https://media.discordapp.net/attachments/986049049834713139/1004381934312509490/Screen_Shot_2022-08-03_at_9.35.05_AM.png?width=1102&height=1026[/img]

Let $PQ$ intersect $(ADQ)$ again $Q'.$ Let $\mathcal{H}$ be the homothety about $P$ taking $BC$ to $AD$. Since $\angle AQ'D=\angle BQC$, $Q$ is taken to $Q'$

Thus, $BQ\parallel Q'D$ which implies $\angle BQQ'=\angle QQ'D=\angle DAQ$ as desired.

Solved with [b][url=https://artofproblemsolving.com/community/user/603664]SatisfiedMagma[/url][/b].

[b][color=blue]Lemma: [/b][/color] Suppose that $\overline{AP}$ and $\overline{AQ}$ are isogonal with respect to $\angle A$. Let $X = \overline{PB} \cap \overline{QC}$ and $Y = \overline{PC}\cap \overline{QB}$. Then $\overline{AX}$ and $\overline{AY}$ are isogonal with respect to $\angle A$.

The above lemma works projectively as well! Clearly $AQ$ and $QC$ are isogonal to each other in $\triangle QCD$ with respect to $\angle DQC$. Apply the lemma twice with points $A$ and $B$. One application will give you that $QP$ is the angle bisector of $\angle DQC$. The second one will tell us that $AD \cap BC = \infty \cap P$ is also isogonal with respect to $Q$. Therefore $PQ \parallel AD \parallel BC$. Finally
    \[ \angle PQB = \angle PQA = \angle DAQ \]
and we're done. $\blacksquare$

Let $K$ be the point on $PQ$ such that $BQ \parallel DK$. It's easy to see that $ADK$ and $CBQ$ are homothetic at $P$, implying $$\angle AKD = \angle CQB = \angle AQD$$ which means $ADQK$ is cyclic. Hence, $$\angle DAQ = \angle DKQ = \angle BQP$$ as desired. $\blacksquare$

[b]Remark:[/b] I've encountered the Forgotten Isogonal Line Lemma in a [hide=few places]see [i]USA TST 2012/2.2[/i] and [i]ISL 2012/G2[/i][/hide] over the past couple of days. Thus, it's quite strange that I didn't manage to spot it here.

[asy]
        /*
            Converted from GeoGebra by User:Azjps using Evan's magic cleaner
            https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
        */
        /*
            A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings.
        */
        pair A = (-34.42819,67.89340);
        pair B = (-74.88348,-31.82015);
        pair C = (25.84231,-32.05941);
        pair D = (8.33576,67.79182);
        pair Q = (74.46684,-13.21240);
        pair P = (-16.46590,38.10467);
        pair X = (100.00792,-32.23557);

        import graph;
        size(15cm);
        pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);

        pen wwzzqq = rgb(0.4,0.6,0);
        draw(arc(Q,17.37791,129.22787,143.32099)--Q--cycle, linewidth(0.75) + blue);
        draw(arc(Q,17.37791,-172.89803,-158.81352)--Q--cycle, linewidth(0.75) + blue);
        draw(arc(A,21.23966,-36.67900,-0.13609)--A--cycle, linewidth(0.75) + red);
        draw(arc(Q,21.23966,150.56218,187.10196)--Q--cycle, linewidth(0.75) + red);
        draw(arc(X,21.23966,143.32099,179.86390)--X--cycle, linewidth(0.75) + red);
        draw(A--B, linewidth(0.5));
        draw(B--C, linewidth(0.5));
        draw(C--D, linewidth(0.5));
        draw(D--A, linewidth(0.5));
        draw(B--D, linewidth(0.5));
        draw(A--C, linewidth(0.5));
        draw(D--Q, linewidth(0.5) + red);
        draw(Q--C, linewidth(0.5) + red);
        draw(A--Q, linewidth(0.5) + wwzzqq);
        draw(Q--B, linewidth(0.5) + wwzzqq);
        draw(Q--P, linewidth(0.5) + blue);
        draw(arc(A,21.23966,-36.67900,-0.13609), linewidth(0.75) + red);
        draw(arc(A,19.11570,-36.67900,-0.13609), linewidth(0.75) + red);
        draw(arc(Q,21.23966,150.56218,187.10196), linewidth(0.75) + red);
        draw(arc(Q,19.11570,150.56218,187.10196), linewidth(0.75) + red);
        draw(Q--X, linewidth(0.5));
        draw(X--C, linewidth(0.5));
        draw(arc(X,21.23966,143.32099,179.86390), linewidth(0.75) + red);
        draw(arc(X,19.11570,143.32099,179.86390), linewidth(0.75) + red);
        draw((-106.73860,-12.78198)--(143.20593,-13.37567), linewidth(0.5) + blue);

        dot("$A$", A, NW);
        dot("$B$", B, SW);
        dot("$C$", C, dir(270));
        dot("$D$", D, NE);
        dot("$Q$", Q, NE);
        dot("$P$", P, dir(180));
        dot("$X$", X, SE);
[/asy]

Let $X=PQ\cap BC$. Also $\ell$ denote the line through $Q$ parallel to $BC$.

Now by DDIT on quadrangle $ACBD$ w.r.t. $Q$, we get that $\left\{(QP,Q\infty_{BC}),(QA,QB),(QC,QD)\right\}$ are pairs under some Involution. Now due to our given angle condition, we get that $\angle AQB$ and $\angle DQC$ share the same angle-bisectors. This gives us that $\left\{(QD,QC),(QA,QB)\right\}$ are reciprocal pairs under the Involution of isogonal conjugation w.r.t. the bisector of $\angle QAB$. This thus also means that $QP$ and $\ell$ are isogonal w.r.t. the angle-bisector of $\angle QAB$.

Thus to finish, we have that,
\begin{align*}
    \measuredangle DAQ&=\measuredangle DAX\\
    &=\measuredangle CXA\\
    &=\measuredangle (\ell,QA)\\
    &=\measuredangle (\ell,QP)+\measuredangle PQA\\
    &=\measuredangle (\ell,QP)+\measuredangle (BQ,\ell)\\
    &=\measuredangle BQP
,\end{align*}
and we are done. :)

This is a boring problem :unamused: it's one construction which trivializes, nothing to learn from it.
Let the line through D parallel to BQ intersect PQ at R. It's obvious by sim triangle ratio of sides that ADR and CBQ are homothetic at P (RP/RQ=RD/RB=RA/RC). Then ARD=CQB=AQD, meaning R lies on the circumcircle of ADQ. Then DAQ=DRQ=BQP, which finishes. $\blacksquare$

Cute!

Suppose the homothety at $P$ sending $BC$ to $DA$ takes $Q$ to point $R.$ Then we have $BQ \parallel DR,$ and $\angle ARD=\angle CQB=\angle AQD,$ so $ARQD$ is cyclic. Then, we have $\angle BQP=\angle BQR=\angle DRQ=\angle DAQ$ and we are done.

Let $X$ be the second intersection of line $PQ$ with $(BQC)$. Clearly $P$ is the center of the homothety from $(BQC)$ to $(AQD)$. This homothety sends arc $BX$ in $(BQC)$ to arc $QD$ in $(AQD)$, hence $\angle BQX = \angle DAQ$.

Extend $QP$ to meet the circumcircle of $ADQ$ at a point $R$. Then it is easy to see from the angle condition that there exists a homothety taking $\triangle BCQ \mapsto \triangle DAR$ so we have $RD \parallel BQ$ from which we finish as $\angle QAD = \angle QRD = \angle BQP$.

Take a homothety centered at $P$ that maps $C$ to $A$ and $B$ to $D$, because they are similar triangles. Let us say this homothety sends $\triangle{BCQ}$ to $\triangle{DAT}$. Then:
\[\angle DTA=\angle BQC=\angle AQD,\]
so $ADQT$ is cyclic.\\

We then angle chase to see:
\[\angle BQP=\angle DTQ=\angle DAQ,\]
as desired $\square$

ok heres th e dumb solution first i guess

By $DDIT$ involution swaps $(QA,QB),(QC,QD),(QP,Q\infty_{AD})$ so we're done by angle chase.

Homothety at $P$ sending $AD$ to $CB$ will also send $Q$ to a point $R$ such that $AQ \parallel CR$ and $BQ \parallel DR$. Thus
\[\angle DRA = \angle BQC = \angle DQA \implies ADQR \text{ cyclic} \implies \angle BQP = \angle QRD = \angle QAD. \quad \blacksquare\]

Assume WLOG that $\frac{AD}{BC} = r \le 1,$ and let $Q'$ be the image of $Q$ under a homothety $h$ centered at $P$ with scale $-r.$ Then $h(A) = C$ and $h(D) = B,$ so the given condition is equivalent to $ADQQ'$ being cyclic. This means that $\angle DAQ = \angle DQ'Q.$ Therefore, we'd like to show that $\angle BQQ' = \angle DQ'Q,$ or $DQ' \parallel BQ.$ This is evident since $Q \to Q'$ and $D \to B.$

Let $\mathcal{H}$ be the homothety sending $AD \to BC$, centered at $P$. Then let $\mathcal{H}(Q) = X$. Notice that $\triangle{BCQ} \sim \triangle DAX$, by homothety. It follows that $\angle{AQD} = \angle BQC = \angle{DXA}$, so $ADQX$ is cyclic. From our homothety, we also know that $BQ \parallel XD \implies \angle DAQ = \angle QXD = \angle BQP$, so we are done.
Nice problem. I think the solution makes it look easy but it's a bit hard to spot the homothety centered at $P$.

If $Q'$ is the image of $Q$ under the homothety (centered at $P$) sending $\overline{AD}$ to $\overline{CB}$ then the angle condition implies $BCQQ'$ cyclic and $\measuredangle BQP=\measuredangle BQQ'=\measuredangle BCQ'=\measuredangle DAQ$. $\blacksquare$

[b]Remark:[/b] I think this problem becomes 10x (not an exaggeration) easier if you make $ABCD$ a [b]self-intersecting trapezoid[/b], i.e. such that $ADBC$ is convex instead of $ABCD$. If we translate the given angle conditions to the appropriate directed angle equalities (by drawing the diagram for when $ABCD$ is convex and inspecting orientations), the problem should still be true when $ADBC$ is convex. It's generally much easier to spot positive homotheties than negative ones, and in this case it becomes much easier (maybe even obvious) to notice that $\overline{AD} \mapsto \overline{CB}$ gives you a cyclic quadrilateral. In fact, when $P$ goes to infinity in this case I think the resulting problem is quite well-known (and older than this).


Find all functions $f$ such that
$f(f(x)+y)=f(x^{2}-y)+4f(x)y$,
for all reals $x;y$.

$y=x^{2}$ $\to$ $f(f(x)+x^{2})=f(0)+4f(x)x^{2}$,
$y=-f(x)$ $\to$ $f(0)=f(x^{2}+f(x))-4f^{2}(x)$.
Combining the last two statements,
$f^{2}(x)=f(x)x^{2}$ $\to$ $f(x)(f(x)-x^{2})=0$
Therefore, for each separate $x$, $f(x)=0$ or $f(x)=x^{2}$:
$f(x)=0$ if $x\in Z$,
$f(x)=x^{2}$ if $x\not\in Z$,
where $Z$ is the set of zeroes of $f$.
Usually it's easy to prove that either $Z\equiv\mathbb{R}$ or $Z\equiv\{0\}$. I'll try later.


Find integer solution:
$ x^4 \minus{} y^4 \plus{} z^4 \plus{} 2x^2z^2 \plus{} 3x^2 \plus{} 4z^2 \plus{} 1\equal{}0$

[quote="kunny"]No integer solutions.[/quote]

huh ?  at least $ (x,y,z)\equal{}(0,\pm 1,0)$