geometry
3D geometry
sphere
Harvard
college
MIT
prism

Source: Harvard-MIT Math Tournament

A cube of sidelength $2$ is inscribed in a sphere, then a second cube with face parallel to the first is inscribed between the sphere and one of the faces of the first cube. What is the length of a side of the second cube?

$ \frac{1}{\sqrt2} $ ???? seems wrong!!!

[quote="Rushil"]$ frac{1}{\sqrt2} $ ???? seems wrong!!![/quote]

It is wrong...next time show your work ;)

[quote="Rushil"]Hey is it correct now???[/quote]

No....how did you get that answer?

[quote="Rushil"]$ \frac{4}{10} $[/quote]

No...please show your work ;)

[quote="Rushil"]Let the side of the cube be $x$
Now if I consider the planar view and apply pythagoras thm. onb a triangle whose verices are centre of sphere, edge of small cube and Mid point of small cube.The radius of sphere is $ \sqrt2 $ form given

Thus $ ( \frac{x}{2} )^2 +\ (1+x)^2 = 2 $

This gives $0.4$ as the answer.
?????[/quote]

No, the radius of the sphere is not in $\sqrt{2}$ form, think about it, how would you express the diagonal of a cube?

[hide]
First note the radius of the sphere is half the space diagonal of the cube.  The diagonal of a cube is $s \sqrt 3$ where $s$ is the cube's side, so the sphere's radius is $\sqrt 3$.
Now make a rectanglular prism with diagonal points the center of the sphere and a vertex of the smaller cube touching the sphere.  The prism has sides length $\frac s 2$, $\frac s 2$, and $1 + \frac s 2$ and space diagonal $\sqrt 3$.  Using the formula for space diagonal of rectangular prism ($l^2 + w^2 + h^2 = d^2$), we get
$\frac{s^2 + s^2 + (s+3)^2}{4} = 3$
$s^2 + 2s - 1 = 0$
$s = -1 \pm \sqrt 2$
$s = \sqrt 2 - 1$
[/hide]
I hope that is right, I am not sure.

Rushil, you should not just post a bunch of answers, however confident you are with them.  At least post a solution with them so we can point out your mistakes and you can learn from them.

So my radius calculation is wrong. Plz correct me fast!!!!

$\frac{2}{3}$ is what i got.  Here's my solution...
[hide]Let the center of the sphere be $O$.  Now let one of the four vertices of the smaller square (on the smaller cube) which touch the sphere be $M$.  Draw $OM$.  Now take the face of the smaller cube which is touching the larger cube.  Take the vertex here which shares an edge with $M$, and call it $N$.  Draw $MN$.  Now draw the perpendicular from $O$ to $MN$.  Let it intersect at $P$.  Then we have a right triangle with hypotenuse the radius of the sphere, $\sqrt{3}$, and legs $x+1$ and $\frac{x}{\sqrt{2}}$, where $x$ is a side of the square.  So
$\frac{x^2}{2}+x^2+2x+1=3$
$3x^2+4x-4=0$
$x=\frac{-4+\sqrt{16+48}}{6}=\frac{-4+8}{6}=\frac{2}{3}$[/hide]
Rushil, you were on the right track, you just have to remember that you're dealing with a sphere, not a circle.

No, both of you are wrong...I don't want to give away the solution yet, because I think you guys should be able to get this, think about the hints that I'm giving, to get the radius of the sphere, you can use the diagonal of the cube, then form a right triangle using the side lengths of the two cubes as well as the diagonals of both cubes.

[quote="K81o7"]$\frac{2}{3}$ is what i got.  Here's my solution...
[hide]Let the center of the sphere be $O$.  Now let one of the four vertices of the smaller square (on the smaller cube) which touch the sphere be $M$.  Draw $OM$.  Now take the face of the smaller cube which is touching the larger cube.  Take the vertex here which shares an edge with $M$, and call it $N$.  Draw $MN$.  Now draw the perpendicular from $O$ to $MN$.  Let it intersect at $P$.  Then we have a right triangle with hypotenuse the radius of the sphere, $\sqrt{3}$, and legs $x+1$ and $\frac{x}{\sqrt{2}}$, where $x$ is a side of the square.  So
$\frac{x^2}{2}+x^2+2x+1=3$
$3x^2+4x-4=0$
$x=\frac{-4+\sqrt{16+48}}{6}=\frac{-4+8}{6}=\frac{2}{3}$[/hide][/quote]

 :)  :D  :lol: :first:  :w00t:  :10:  :clap2:  :clap:

damn I was a fool . yes the radius is $ \sqrt3 $ !!!!! :wallbash_red:  :wallbash:  :stretcher:

[quote="Rushil"]- I was a fool . yes the radius is $ \sqrt3 $ !!!!! :wallbash_red:  :wallbash:  :stretcher:[/quote]

Yes, but you still would have gotten it wrong because you didn't see that the triangle has one side that is half the diagonal of the face of the smaller cube....see above, you thought the base of the triangle was $\frac{x^2}{4}$ not $\frac{x^2}{2}$.

Gah.  I just had a calculation error with the quadratic formula... oh well.  Good solution.


[quote="bigman"]How many real solutions does the following system of equations have? Justify

x+y=3
3xy-z^2=9[/quote]
we have: $ (x\plus{}y)^2 \geq 4xy$ thus, $ xy \leq \frac{9}{4}$
 From the equation (2), we have: $ 9\equal{}3xy\minus{}z^2 \leq \frac{27}{4} \minus{}z^2 \leq \frac{27}{4} < 9$

 not exist real roots for system of equations.


Prove that the equation  $ x^{2}\minus{}16nx\plus{}7^{5}\equal{}0$ equation $ n\in N$ has no integer solutions.
 :(

 _____________________________________
Azerbaijan Land of the Fire

[quote="Pirkuliyev Rovsen"]Prove that the equation  $ x^{2} \minus{} 16nx \plus{} 7^{5} \equal{} 0$ equation $ n\in N$ has no integer solutions.
 :(
[/quote]

So $ 7^5\equal{}x(16n\minus{}x)$ and so $ 12$ possibilities :

$ x\equal{}1$ and $ 16n\minus{}x\equal{}7^5$ and so $ n\equal{}\frac{7^5\plus{}1}{16}\notin N$

$ x\equal{}7$ and $ 16n\minus{}x\equal{}7^4$ and so $ n\equal{}\frac{7^4\plus{}7}{16}\notin N$

$ x\equal{}7^2$ and $ 16n\minus{}x\equal{}7^3$ and so $ n\equal{}\frac{7^3\plus{}7^2}{16}\notin N$

$ x\equal{}7^3$ and $ 16n\minus{}x\equal{}7^2$ and so $ n\equal{}\frac{7^2\plus{}7^3}{16}\notin N$

$ x\equal{}7^4$ and $ 16n\minus{}x\equal{}7^1$ and so $ n\equal{}\frac{7\plus{}7^4}{16}\notin N$

$ x\equal{}7^5$ and $ 16n\minus{}x\equal{}1$ and so $ n\equal{}\frac{1\plus{}7^5}{16}\notin N$

$ x\equal{}\minus{}1$ and $ 16n\minus{}x\equal{}\minus{}7^5$ and so $ n\equal{}\minus{}\frac{7^5\plus{}1}{16}\notin N$

$ x\equal{}\minus{}7$ and $ 16n\minus{}x\equal{}\minus{}7^4$ and so $ n\equal{}\minus{}\frac{7^4\plus{}7}{16}\notin N$

$ x\equal{}\minus{}7^2$ and $ 16n\minus{}x\equal{}\minus{}7^3$ and so $ n\equal{}\minus{}\frac{7^3\plus{}7^2}{16}\notin N$

$ x\equal{}\minus{}7^3$ and $ 16n\minus{}x\equal{}\minus{}7^2$ and so $ n\equal{}\minus{}\frac{7^2\plus{}7^3}{16}\notin N$

$ x\equal{}7^4$ and $ 16n\minus{}x\equal{}\minus{}7^1$ and so $ n\equal{}\minus{}\frac{7\plus{}7^4}{16}\notin N$

$ x\equal{}\minus{}7^5$ and $ 16n\minus{}x\equal{}\minus{}1$ and so $ n\equal{}\minus{}\frac{1\plus{}7^5}{16}\notin N$

Hence the result


calculus
integration
limit
logarithms
calculus computations

Find $ \lim_{a\to\infty} \frac {1}{a^2}\int_0^a \log (1 \plus{} e^x)\ dx.$

By l'Hospital's rule, I get that the integral diverges:

\[ \lim_{a\rightarrow 0} \frac {\int_0^a \log (1 \plus{} e^x)\ dx}{a^2} \equal{} \lim_{a\rightarrow 0} \frac {\log (1 \plus{} e^a)}{2a}\rightarrow \frac {\log 2}{0}\rightarrow \pm\infty\]

If $ \lim_{x\to0^ \plus{} }f(x) \equal{} L\ne0,$ then we must have $ \lim_{a\to0^ \plus{} }\frac1{a^2}\int_0^af(x)\,dx$ diverges.

In fact, $ \int_0^af(x)\,dx\equal{}aL\plus{}o(a),$ so that $ \frac1{a^2}\int_0^af(x)\,dx\equal{} \frac {L}{a} \plus{} o(a^{ \minus{} 1}).$

Sorry, I made a typo. I will edit the problem.

kunny

Note that $ \ln(1\plus{}e^x)\equal{}\ln\left(e^x\cdot\frac{1\plus{}e^x}{e^x}\right)\equal{}
x\plus{}\ln(1\plus{}e^{\minus{}x}).$

Hence $ \int_0^a\ln(1\plus{}e^x)\,dx\equal{}\frac{a^2}2\plus{}\int_0^a\ln(1\plus{}e^{\minus{}x})\,dx.$

But note that $ \ln(1\plus{}e^{\minus{}x})$ is comparable to $ e^{\minus{}x}$ for large $ x.$ Hence, that second integral is bounded for all $ a.$

$ \int_0^a\ln(1\plus{}e^x)\,dx\equal{}\frac{a^2}2\plus{}O(1)$ as $ a\to\infty.$

$ \frac1{a^2}\int_0^a\ln(1\plus{}e^x)\,dx\equal{}\frac12\plus{}O(a^{\minus{}2}).$

The answer to kunny's corrected problem is $ \frac12.$


search
complex numbers
algebra unsolved
algebra

Let $a$, $b$, $c$ be complex numbers such that:
(i) $|a| = |b| = |c|$,
(ii) $a + b + c = a^4 + b^4 + c^4$,
(iii) $a^2 + b^2 + c^2 = a^8 + b^8 + c^8$.

Prove that $a$, $b$, $c$ form the vertices of an equilateral triangle.

We note $S$ the triplets verifying the three conditions.
We have for all complex $a, (a,a.e^{i\frac{2\Pi}{3}},a.e^{i\frac{4\Pi}{3}})\in S$
We have just to see that $(a,b,c)\in S, (a,b',c')\in S \Rightarrow (b,c)=(b',c')\textnormal{ or }(b,c)=(c',b')$
For that, we use $(i)$ and $(ii)$ to have $\frac{b^4-b'^4}{b-b'}=\frac{c^4-c'^4}{c-c'}$ iff $b\neq b'$ and $c\neq c'$ (so we have to see that $b'=c$ and $c'=b$).
I continue to search


Euler
vector
projective geometry
geometry solved
geometry

How can i demonstrate Euler's line whit the help of vectors

hello Campulung Muscel!!! :D :D

You must first prove that HA+HB+HC=2HO and OA+OB+OC=OH then you get that GH=-2GO (all the vectors are oriented)

Lets consider that M is the middle point of BC and we get that 2OM=AH and 2HM=HB+HC. Then HA+HB+HC=HA+2HM=2(MO+HM)=2HO.
Lets consider N and P the middle points of AC and AB then 2OM=OB+OC, 2ON=OC+OA and 2OP=OA+OB then you get that: 2(OM+ON+OP)=2(OA+OB+OC) and 2(OM+ON+OP)=AH+BH+CH=-2HO=2OH from which we get that OA+OB+OC=OH.

In the triangle AHM, we have that G is on AH and so we get that |GA|/|GM|=2 so HG=1/3*HA+2/3HM= (HA+HB+HC)/3=2/3HO.
In the same way you get that in OAM(triangle) you have OG=1/3OA+2/3OM=1/3OA+(OB+OC)/3=(OA+OB+OC)/3=1/3OH so we get that GH=2/3OH and that GO=-1/3HO so GH=-2GO (relation which shows the colinearity between O,G and H) all done!
hope it is Okay! :D :D
cheers! :D :D :D

And how can i demonstrate Euler's line whitout vectors, pls :D  :D

       Traieasca C-lung Muscel :D   :)  :D  :)  :D  :)

Try to prove that if M is the midpt of BC OM=HA/2. Then the triangles AHG and MOG are similar.

Another proof Euler's line

A',B'  the midle of [BC], [AC]
the triangle ABH and A'B'O their sides are paralel in pair meet at infiny

By Desargues theorem the lines AA',B4,OH meet
O,G,H are on a line

There are tons of proofs. I'm sure you can find some on the net!

Okay .. here is a full solution:

Let A' be the middle of BC and A'' the oposite(diametrally) of A.
Because BH is paralel with CA" and CH with BA" then we haver that BHCA" is a paralelogram.
A' is the middle of HA" so OA' is paralel with AH and OA'=1/2AH. Let G be the intersection of HO and AA'. It is obious that the triangles A'OG and AHG are alike with AH=2OA', so we have that AG=2GA' from which we have that G is the weight point of ABC. So we have the colinearity of the points.

cheers! :D :D

another full solution for this one is:

Lets consider that A' is the middle of BC and A1 the projection of A on BC. The same you do to construct B1. The triangles A'O'B' and AHB are alike. Because we have that AB/A'B'=2 we get that AH/OA'=2.

Now all you do is prove that m(HGA)=m(OGA') and we see that AG=2GA' and that m(GA'O)=m(HAG)  so the triangles GA'O and AHG are alike. So we get the colinearity of O,G and H.

I think that there are many proofs for these problem(like grobber said) and all these are enough! :D

cheers! :D :D


trigonometry
geometry
percent
probability
geometric series

Hi all, I'm not sure if this is the place to ask this but I lost the link to one of the big TI calc forums.

I'm going to take the ACT in April, but I have a TI-89, which they banned...I don't understand why they would allow it on the SAT but not the ACT..

Anyway, I was wondering which calculator to buy. On the description the TI-84 plus has stuff like statistics, graphing, trig, etc. But on the TI-84 plus Silver Edition it has "pre-calc, pre-algebra, geometry" that the regular 84plus doesn't have, and it costs $ 20 more. I was wondering if it's worth the$20 to pay for them or am I able to download them off the internet?
Thanks.

The Silver Edition only has a lot of extra RAM, and a lot of extra applications added. If you're going to use it only for the ACT, there's probably no point in buying that over a normal 84 plus.

I understand, but I just meant like will the programs under "pre-calc, pre-algebra, and geometry" help me on the ACT or can I just download those programs

You can get these programs online, provided you have a cable connecting the calculator's small USB thingy to a computer's bigger USB port. I think this cable comes with it, so that should be no problem. Look [url=http://education.ti.com/educationportal/sites/US/productDetail/us_ti84p.html?bid=7]here[/url] for the applications.

I wouldn't get anything better than a Ti-84 for a couple different reasons: my ti-84 served me fine on my ACT, and will do so until I need something with 3D capabilities. Secondly, [i]you're probably not going to need your calculator for the majority of the ACT[/i]. This may sound crazy, but believe me, its true. Most of the problems require little more than a couple scratchy lines of algebra, and the ones that deal with the intersection of lines wont just give you one or two; you'll most likely end up with a linear system of 4 variables or so, which is a case where I would probably bust out my Ti and solve using matrices and reduced row echelon form. You don't even need a Ti-84 to do that thought: Ti-83's will suffice for brute forcing linear systems.

I will admit, that the 84 family has a much improved display over the 83's. They are also faster, and as someone who knows how to write in basic on Ti's, That is quite an advantage. I can't really say whether or not the SE will be considerably more powerful, but it also won't make much of a difference.

My suggestion to you is know your formulas: midpoints, distance, arithmetic + geometric series (and the sums thereof), it probably wouldn't hurt to review trig, special triangles, averages, percents, probability, and your basic quadratics. Aside from one or two problems, the most complicated thing I was using my Ti-84 for was to simplify fractions. The calculator is the probably the least of your worries though: The ACT is a 4-part test, your English and Sciences count too!

Sincerely,
Grant H.
High School Senior

[quote="iamhe"]Since we are talking about calculators here, where can I get a cheap and good deal on a ti 89 titanium?

Anybody know?

Thanks a lot.[/quote]

Ebay is good. They're around $ \$50$

To me, the Silver Edition lags (probably because of it's excess applications). Get the regular 84+.

If you're just going to use it on the ACT, then I suggest you go to EBay and spend nothing over 25. You're not going to need the calculator much on the ACT. Even a regular scientific calculator will probably be enough.

Moogra,
I looked at ebay. None of them were even close to 50 bucks. Are you sure that ebay is the place? Or is it just my impatience?

Thanks

eBay might not be the [i]best[/i] but it will certainly be the cheapest. You can buy a regular TI-84+ for maybe around 100 USD and significantly less during the end of the summer.

You can also think about the TI-83+ if you're partial to the cost, which is only around 70 USD for a brand new one. It has pretty much everything you need except for memory.


You see, stupid mistakes are just as important if not more important than "real" mistakes. If everyone cut their stupid mistakes "in half", then we'd have had a lot of people with 40+ points. Lyndon made dumb errors on that test. So did Jeffery. So did Youkow [youkow made A LOT of dumb errors]. So did Michael Mardini.

Do SO MANY problems so that you can sorta tell when, where, how, under what conditions, why, how many, etc. you'll make in a given test. Sometimes as you're reading it's key to take special note of a word and think "I'm not making a dumb error on this".

Anyhow there have been 1908573219875239857 threads in the psat on stupid mistakes... so you might either get A) a lot of responses, or B) not very at all from people being impatient

Stop watching football. I'm not kidding.

Before (in like November) when I was practising for MathCounts, and I watched football (quite a lot actully) I used to consistently get 33s. Then my dad got pissed and cut our cable and now I get 45/46s (on the large part) since I have nothing better to do than do math.

But I guess since Footbal season is over, that doesn't really apply  :wink: .

I only have one other tip, once you finish the problem, just look over the problem and your work, and make sure your answer is the one they are looking for (i.e. you put gallons used instead of amount of dollars spent on gallons of gas), and it is in the correct (I cannot tell you how annoyed I am for putting the answer to this certain problem in decimal form instead of fraction).

Well, I'm off to finish the AoPS: Basics book  :huh:

when you finish your problem, check what the problem is asking for... then check if your answer makes sense by estimation. make sure it's reduced to the simplest fraction, simplest radical, to the nearest integer, etc. Also make sure the decimal is in the right place, by estimation.

often, in the last few problems of the sprint, I do some serious calculations in the second step or something, and I forget that I have one more step to do. I think that occurs most likely in complex geometry problems. That messed me up last year.

in response to kevinchen, if u do all that, ur gonna run out of time~!

I wouldn't have any tips...  I make too many careless errors..

@footballrocks: Then how did you get to nationals.

You know what really bothered me? The guys who read the problem out loud to us thinking we can't understand it ourselves in CountDown. It really messes me up and I know it would be so much simpler if everything was quiet and he'd just let US look over it.

What score earns 3rd in Wisconsin?  And where is the board telling what last year's cutoffs were?

[quote="indianamath"]I feel that if i reduced my reading/computational erros by half i very well could have made Nats. Now i'm debating whether i should slow down, or should i practice doing problems fast and so i can be accurate when i do them slowly(slower, not definitely slowly). if you have any additional suggestions they would be welcolmed and appreciated.[/quote]
I definitely agree, 100% of my incorrect questions on last year's AMC/AIME were due to carelessness

I guess I'm just lucky...  Careless Errors was to explain for a 21 at Chapter... :blush:

Well, one thing that Mr. Fischer told us one day was that he knew a student who underlined problems as he/she read them to make sure they understood the question.
Also, make sure the graders can read your handwriting, because it seems like that kind of mistake appears at the worst possible time. Imagine knowing you were one point away from nationals or your team need an extra .25 on their team score to win, and you missed one because you had bad handwriting.

i think i missed one to handwriting in 2006 lol.


Find the value(s) of $ x$ for which the following is true:

\[ 862.5 \div 2.3 \equal{} 3.75 \cdot 10^x\]

Find the value of $ 862.5 \div 23$.  This value is $ 375$.

This can also be expressed in scientific notation as $ 3.75 \times 10^2$, so our answer is $ \boxed{2}$.

[quote=econaxis]...but how can we find it without a calculator and fast[/quote]

[hide]You don't.
Assume $x$ is an integer because if it weren't, $x$ would be a weird irrational number. Now you can see the answer is more than $1$ and less than $3$.


calculus
geometry
geometric transformation
reflection

Random question, why is your aops screenname your screenname?

well : I am Frédéric , I was BEL6 IMO2002 contestant
therefore fredbel6

I validate this name very much
It was established in order to keep in touch with people and my own family during the IMO (it's not that i really am so obsessed about being an IMO contestant)

I am also the Wikipedian evilbu, if you are wikipedian, do come greet me if you like (and I participate in the mathproject)
evilbu =evil buu (if you know Dragonball you understand this)

I made up my name by throwing together some letters. :) It is also the name of one of my characters.

I just took every other letter of my last name, and that's my username :)

ch1n353ch3s54a1l=chinese chess for all, back when i thought 1=i and i was addicted to chinese chess. Yes, very unoriginal.  :P

My name has a bit of a story behind it.  I play the trombone in my school's band, and a few months ago, we had solo festival.  As a result of this, our lesson teacher told us that we probably wouldn't do anything productive that day.  One of my fellow trombonists suggested that we play baritone, and we did.  That, combined with the fact that about halfway through the lesson the power to our school went out, made for a very interesting lesson.  Soon, the baritone became a recurring joke among the trombone section.  During another lesson, a few weeks after the baritone episode, we saw a trombone quartet called "T-Bone Party."  Colin (the person who had suggested that we play the baritone in the first place) immediately switched the first two letters to B-tone party, (Baritone Party,) and the rest is history.

mine is a bit cheeky.....(i'm in england on vacation so i thought i would talk like they do in england...)
i like math so i had math in my name and i was born in 92

My pet name is bubka!

When I was born in 88, Sergei Bubka was in his golden days. Reportedly, I jumped around too much in my bed, and got the name (which is a bit ironical, as since then I have turned out to be as unsportsmanlike as I possibly could!)

My Favorite Game!!! I think Fire Emblem is a great game and i got addicted to it the first time i played it although some other people think its boring.

I am named jhollenbeck because I took my first name's initial, Jonathan, and my last name.  now I have jhollenbeck!  So original :D  :rotfl:

Just a juxtaposition of two seemingly unrelated words...
Perhaps I oversee a miraculous post office.

My name is Andrei Ciupan..figure it out :P

[quote="dogseatcheese"][quote="daine"]Daine is a character in some books I read (Tamora Pierce!) so I just chose it.[/quote]

Which book was it? I haven't refreshed my memory on her books for a long time. Is it like wolf speaker or sumthing like taht.[/quote]

She's the main character in the Immortals quartet. Wolf Speaker is the second book.

[quote="daine"]Daine is a character in some books I read (Tamora Pierce!) so I just chose it.[/quote]

Guess where I got my screenname :D . The Protector of the Small quartet is my favorite, and L.K. happens to be my first two initials, so I picked this name. The book itself isn't my favorite. I like Squire the best. The Immortal quartet is really good, too. I've been rereading it recently.

[quote="spix"]"Spix" is derived from the name of a nuthouse in my hometown. Now everyone calls me that.[/quote]

Where's that ?  :P

I don't know what slizzel means. I was a random word in a song. Once I saw a movie and it was translated like "subtile".. However it's just a username  :roll:

3 letters from my name and three from my last name
dad uses something better but i do not want to have it with Jr :mad:

I like the sixers and I joined in '05. That's pretty much it. ;)

Captain Sugar is the main character in a book i'm writing. he's a minor hero, not a superhero, who doesn't have any super powers. he would use cool gadgets, but spiderman and batman took all of the good ones, so all he gets are the novelty gadgets.  his archenemies are the ones the superheroes are too lazy to take down.

[quote="t0rajir0u"]Torajirou was [url=http://senseis.xmp.net/?Shusaku]Honinbo Shusaku[/url]'s given name.  I picked it up from watching Hikaru no Go, liked the name, so it stuck and I use it for pretty much everything.   :D[/quote]

I was wondering what your name kept reminded me from as soon as I saw your username.

My screen name means... :| ...dunno.....I like eminem(known as slim shady) and before i joined the site i visited a few pages and evrywhere people were writing SPAM SPAM SPAM,so there u go.

Its pretty cool thiough when people call me SHADY,yeahhh,i think shady means someone with a bad character.

[quote="deimos"]some greek mithology name; one of the martian moons

[quote="fredbel6"]
I am also the Wikipedian evilbu, if you are wikipedian, do come greet me if you like (and I participate in the mathproject)
[quote]

what math project? i'm ganimed.[/quote][/quote][/quote]

What do you mean?
[url]http://en.wikipedia.org/wiki/User:ganimed[/url]
Do you mean you are on the English wikipedia the user known as ganimed?

My name is pretty self-explanatory. It reflects the deep mysteries of my uncreativeness (is that a word?).

Funcia the (former)PW to the F.U.N. of the C.I.A.'s secret hideout. The new one is "mary had a little rattlesnake". kFrivolous Undercover Ninnies of Criminal Iditarod Assasinators...I'm rushing so my spelling suckls......BYE

[quote="solafidefarms"][quote="Xantos C. Guin"]Xantos is a name from somewhere, can't remember where though.
C. is a random initial that is supposed to stand for something but I can't remember what.
Guin is the last sylable of penguin.[/quote]

If I remember correctly Xantos is a corruption of your first name. C is for your last name, right?

Sola Fide means "by faith alone", one of the watchphrases of the Protestant Reformation. 'solafidefarms' is the name of my family's farm.

Billy[/quote]

Xantos was meant to be totally random. I didn't notice the similarity between my real name and Xantos for 6 months after I started using it.

So that's why your farm is called Sola Fide. I was wondering that.


n, x, y &  z are  integers such that : $x^n +y^n = z^n$

1- Show that : $z^n z^n-4x^n y^n$  is a perfect square.

2- Prove that for all n>5, there are no integers  x, y & z such that:

 $z^n z^n-4x^n y^n$  is a perfect square

3- So...

It is

(x^n+y^n)²-(x^n-y^n)²=z^2n-(z^n-2y^n)²

is equivalent to

z^2n-4x^ny^n=(z^n-2y^n)²

The assertion 2 is not true.

[quote="pco"][quote="amparvardi"]Any solutions ?[/quote]

It's obviously a joke.

Question 1 is college-level.

Giving an olympiad-level answer to the question 2 is equivalent to give an olympiad-level proof for Wiles theorem,

and so ... no olympiad-level answer to this question.[/quote]

Sure, but I thought this one will solve by olympiad-level solutions and not college-level.
It didn't solve for a long while...


ARML
AMC
AIME
USA(J)MO
USAMO
AMC 12
AwesomeMath

I'm going to be new to ARML, can someone explain to me the level of problems, how they select teams, how often it is, who from AoPS is going, and any other things I should know for the upcoming year.
-MrRubix

The problems are somewhere around AIME level, but less computationally intensive, as they give you 5 minutes per problem instead of 12.  I'd say that of the people there, probably around 80% solve the first problem, and lately, the last problem has been hard enough that fewer than 10 people (<1%) solve it.

Practices are from 2-5 on Sundays throughout April and May.  You're not required to go to every practice, but team selection is usually set up in such a way that the more you go to, the more chance you have of being on a higher team.  Team selection is based on your performance at practices, your USAMO index (with a bonus if you take AMC12), past ARML scores, and possibly other factors.  This year's exact selection methodology hasn't been decided yet, but it will probably be announced at some point.

A lot of AoPSers go, some of whom will probably respond to this below me.  I usually go as a coach/chaperone, but I'm studying abroad this spring, so I can't.

Great to hear that you're joining the team!  ARML is a blast; I have the time of my life there every year.

i thought arml was the best thing in this world until i went to awesomemath

[quote="junggi"][quote="Sly Si"]We usually take a bus down on Thursday night, arriving Friday morning.
...
We bus back Saturday afternoon and arrive sometime late that night.[/quote]
guys dont let this scare you the bus is the best part of the trip[/quote]
HEAR HEAR

 :o Silas youre not coming this year????? nooooooooooo

what!??
silas isnt coming?
nooooo!

wtf whos replacing?

omg the bus trip is fun
[sarcasm]
we hold a really loud party on teh way back so whoever wants to sleep cant and theyr too sleepy and full of math to beat us up lol
[/sarcasm]

WAIT WHOA WHOA I JUST SAW THAT PART ABOUT NO SILAS
dude what that takes away about half the awesomeness out of our team
=(

Yeah I'm not going this year, I don't want to hold the A team back since we have tons of new awesome people.

Yep, sadly, I'm not going this year.  I think other people will take my place.  Nathan Pflueger wanted to go last year, but broke his foot right before ARML and so had to cancel; I think he'll be going this year.  You'll like him.  A couple of other people also have talked to me about going, so hopefully we might actually have more than one Stanford (or Berkeley) student going this year.  I'll miss ARML a lot, but I couldn't pass up the chance to spend a quarter abroad.

Ubemaya, go to ARML.

Oh, and I'll be back next year.  Even if I have to schedule my final exams around ARML, I'm going in 2009.

Err.
When/where/etc are practices?

Nvm, I'm blind.

Can we get Evan O'Dorney to join the SFBA ARML team this year?

That will definitely boost our team performance.

[quote="CircleSquared"][quote="Sly Si"]Practices are at Stanford.  The meet itself is May 30th and 31st (Friday and Saturday) in Las Vegas.  We usually take a bus down on Thursday night, arriving Friday morning.  The contest starts Friday night, and continues Saturday morning.  We bus back Saturday afternoon and arrive sometime late that night.[/quote]

I am suddenly extremely interested.

Where exactly is it (is it the same place as the math circles?)[/quote]

*cough*

yes fyi
even the same basement classroom

we usually split up in the middle, and we may get lost while walking in building 380 but its no problem really


real analysis
rudin
topology
advanced fields
advanced fields unsolved

Before heading back to school, I decided to review the first couple of chapters of Baby Rudin, and got stuck rather quickly.

Show that if $ X$ is a connected metric space with at least two points, then $ X$ contains uncountably many points.

Guess: Suppose $ X$ is a connected metric space with only countably many points. Then you can construct a disconnection somehow vaguely related to picking a point $ x_{0}\in X$ and a $ \delta$ such that every point of the space is in $ B_\delta(x_{0})$ or is in $ \{x : d(x_{0}, x) > \delta \}$. If this is truly how it is done, then how do we know that such a $ \delta$ exists? Something to do with listing every distance between $ x_{0}$ and every other point, then picking a real that isn't one of those? Countable sets vs. uncountable gets a little muddled in what is allowed in my mind sometimes.

You have the right idea. To prove that $ X$ is uncountable, it is enough to produce a surjective map $ f\colon X\to Y$ where $ Y$ is known to be uncountable. In this case, $ f(x)=\mathrm{dist}(x,x_{0})$, and $ Y$ is an interval of $ \mathbb R$. (The image of a connected set under a continuous map is connected.)

As for the full problem; take an $ x$ and a $ y$ in the set. Let $ d(x,y)=\alpha$. There are only countably many distances in the set and there are uncountably many reals strictly less than $ \alpha$.So take an element less than alpha that is distinct from all distances and call it $ r$.Define $ A=\{u;d(u,x)>r\}$ and $ D=\{u;d(u,x)<r\}$. $ A$ is non-empty - it contains $ y$. $ D$ is non-empty as it contains $ x$ and the union of $ A$ and $ D$ is $ X$. I think this more or less does it.
[b]Edit;[/b] mlok beat me to it - I take ages writing anything in even the most basic Latex.

Ah, clearly. I hate it when things seem so easy after you see them.  :)


[hide]Let $x$ be the first of the 100 consecutive integers.  $x+(x+1)+(x+2)... (x+99)$ is $100x+4950$.  Since $x$ is an integer, the correct answer has to end in _____50.  Thus, the answer is $(a)$.[/hide]

[hide="Answer"]The sum of $100$ consecutive integers is $n+(n+1)+...+(n+98)+(n+99)=100n+4950$. Therefore, the number must end with $50$, and the only one that does that is $\boxed{A}$.[/hide]


Find the number of solutions satisfying the inequality:

$0$ is less than or equal to $x+y+z$ is less than or equal to $30$

Where $x$, $y$, and $z$ are non-negative integers.

[hide="nothing to do with the problem"]What does "have been around alcohol" mean, in your signature???[/hide]

This is probably wrong but...
[hide]
$\displaystyle\sum_{n=2}^{32}\frac{n!}{(n-2)!2!}=\boxed{5456}$?
[/hide]

Yay partitions!
[hide]$\binom{33}{3} = \boxed{5456}$[/hide]

Ah, you create three partitions with four spaces so you can keep some off to the side and sum to every number. thanks (no crazy summations).


Prove that a list can be made of all the subsets of a finite set in a way that

(i) the empty set is first in the list,
(ii) each subset occurs exactly once, and
(iii) each subset in the list is obtained either by adding one element to the preceding subset or by deleting one element of the preceding subset.

[hide="Induction"]

Without loss of generality, label the elements of the set $ 1,2,\cdots, n$

Claim: We can satisfy the given conditions of the set $ 1,2,\cdots,k$. In addition, we can make the last member of the list  be $ k$.

Empty set clearly works. Base case of one element clearly works. Now assume this works for the set with elements $ 1,2,\cdots,n \minus{} 1$. Now attach $ n$ to this set.

Label the $ i$th member of the list $ A_i$. Complete the rest of the list by letting $ A_{2^{n \minus{} 1} \plus{} r} \equal{} A_{2^{n \minus{} 1} \minus{} r \plus{} 1}\cup\{n\}$ for $ 0 < r \leq 2^{n \minus{} 1}$. We see that the conditions are satisfied.

[/hide]

[quote="rofler"]Gray Code <--My shortest Solution yet! 2 words, hehe. Doesn't even need to be explained, it's so apparent.

Q.E.D.[/quote]Yea, but not every reader of this forum knows what 'Gray Code' is. Thus, your solution fails to be apparent.
Here is a general info about [url=http://en.wikipedia.org/wiki/Gray_code]Gray Code[/url]

Some other questions to think about:

1) What subsets can be the final subset in such a list?

2) What about the set of finite subsets of $ \mathbb{N}$?  (Obviously, our list will have to be infinite.)

3) What about the set of [i]all[/i] subsets of $ \mathbb{N}$?

4) What about if we start our list with $ \mathbb{N}$ instead of $ \emptyset$ -- can we get all the infinite subsets of $ \mathbb{N}$ in this way?

Well, obviously those readers that don't know what gray code is would look it up, now wouldn't they :P

[quote="rofler"]Well, [b]obviously[/b] those readers that don't know what gray code is would look it up, now wouldn't they :P[/quote]Yea and the [b]obvious[/b] thing as well is to provide a link ;)

@JBL:
1) singletons? So that it is 'cycles' back to the empty set
2) I'm not too sure, maybe there is a countable basis...
3) Power set of $ \mathbb{N}$ is uncountable...
4) This one is very tricky, we can't just take the complement of part 2...

[quote="10000th User"]1) singletons? So that it is 'cycles' back to the empty set[/quote] Okay, and Eternica's inductive solution essentially gives us that we can get all singletons.  If you want it to cycle back (which the original problem didn't ask for) then you are certainly limited to the singletons.  But if we don't impose this restriction, then there might be other possibilities -- e.g., with a 3-element set we can get $ \emptyset \to \{1\} \to \{1,2\} \to\{2\} \to \{2,3\} \to \{3\} \to \{1,3\} \to \{1,2,3\}.$  So, I want to know what these other possibilities are.
[quote="10000th User"]2) I'm not too sure, maybe there is a countable basis... [/quote] "Countable basis?"
[quote="10000th User"]3) Power set of $ \mathbb{N}$ is uncountable...[/quote]  Okay, this is more sophisticated than I was looking for -- the simpler argument is just to note that [hide][hide="Are you sure you want to peek?"]you can't get any infinite set in finitely many steps.[/hide][/hide]
[quote="10000th User"]4) This one is very tricky, we can't just take the complement of part 2...[/quote]
[hide]Once you've done part 3 your way, this shouldn't be too hard ...[/hide]

1) Just using your example, I'm unable to have a 2-element subset as the final subset in the list... So many ideas: it seems like a simple graph with 2 odd degrees (Eulerian?), or a Pascal's triangle related, or as simple as parity... any hints JBL?  :blush:

4) Adding and removing an element from an infinite set is still an infinite set... but is that allowed too because there is no way to get all infinite subsets?

My hint for number 1 is that to rule out subsets of size 2 it is sufficient to use one of the three things you named.  (This will rule out a lot of possible ending sets in the general case, but it won't show that the other sets can be the ending set.  That will take another argument.)

For number 4, the set of finite subsets of $ \bf N$ is countable and the set of all subsets of $ \bf N$ is uncountable, as you noted.  So, the set of infinite subsets of $ \bf N$ is uncountable, and we certainly can't get all of them by a listing procedure.

Problem 2 is still open (and I have no idea what was meant by "countable basis").


Does anyone have AIME 2004 B????

Thanks,

If you are in Mathlinks mode, click "Resources" at the top.  If you are in AoPS mode, click "Contests" at the top.


I can't think of any off the top of my head (from the AMC 10 - AMC 12 [b]definitely[/b] has log problems, see this year's). But prepare yourself for those anyway.

Frequently Asked Question AMC 10/12 #5
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Since logarithms are not usually considered part of the grades 9 and 10 mathematics curriculum, and are usually considered part of advanced algebra, logarithms do not appear on the AMC 10.

You can also get the AMC 10/12 Math Club Package, look in the index of problem types under "logarithm" and find the problems on the AMC 12 since about 2002 that did involve logarithms, along with the fact that no AMC 10 problems involved logarithms.

Steve Dunbar
MAA Director, American Mathematics Competitions


logarithms
number theory
algebra unsolved
algebra

Let $ (\log_2(x))^2 \minus{} 4 \cdot \log_2(x) \minus{} m^2 \minus{} 2m \minus{} 13 \equal{} 0$ be an equation in $ x.$ Prove:

[b](a)[/b] For any real value of $ m$ the equation has two distinct solutions.
[b](b)[/b] The product of the solutions of the equation does not depend on $ m.$
[b](c)[/b] One of the solutions of the equation is less than 1, while the other solution is greater than 1.

Find the minimum value of the larger solution and the maximum value of the smaller solution.

Before I need to move even more of these things away from Number Theory: such problems definitely don't belong to Number Theory!
By the way: practically all of them have been posted before...


Hi guys,
    If anyone is interested, my IMO pictures are at yahoo.photos.com/pengrocks1. Have a great summer,

Peng Shi

The actual link is photos.yahoo.com/pengrocks1
nice swap of words there, peng  :)


The problem with our inequalities is the following:
   We have great teachers who propose extremely interesting problems, but unfortunately only this kind of problems appear in contests. And this is so bad. But it's the reality. And yes, Myth, this really is a joke. Again, unfortunately.

another one in the solved section and no solution attached!!  :(  :(


U, W, W' is a subspace of V(vector space over a field F) such that U+W=U+W' and U∩W= U∩W' = {0}, but W≠W'

$ V$ is the plane, and $ U,W,W'$ are three distinct lines through the origin.


What does the notation $ \left( \binom{n}{k} \right)$ mean, and what is it called?

Here is the context in which I saw it:

[quote="HMMT 2007 - Combinatorics #9"]Let $ S$ denote the set of all triples $ (i,j,k)$ of positive integers where $ i + j + k = 17$. Compute $ \sum_{(i,j,k) \in S} ijk$. [/quote]
[quote="2nd Official Solution"]Let $ s_n = \sum_{i + j + k = n} ijk$. Then $ \sum_{n \geq 0} s_n x^n = \left( \sum_{n \geq 0} n x^n \right) ^3 = \left( \frac {x}{(1 - x)^2} \right) ^3 = \frac {x^3}{(1 - x)^6}$. So, ${ s_n = \boxed{ \left( \binom{6}{n - 3} \right) }} = \binom{n + 2}{5}$, yielding $ s_{17} = \binom{19}{5}$[/quote]

[i]-- Thanks!  :oops: [/i]

Looks like "multichoose" -- it's the number of $ k$-element [[multiset]]s, each of whose elements belongs to an $ n$-element set.


[quote="amd2357"]For $r$ an odd positive integer there exist positive integers $a$ and $b$ such that
$(2+\sqrt{3})^r=a+b\sqrt{3}$
Prove that $a-1$ is a perfect square.[/quote]

Note from the binomial expansion:
$a=2^r+\binom{r}{2}\left(2^{r-2}\right)(3)+...+\binom{r}{r-3}\left(2^3\right)\left(3^\frac{r-3}{2}\right)+\binom{r}{r-1}\left(2\right)\left(3^\frac{r-1}{2}\right)$
Hence, $a\equiv2\bmod{3}$ since $r$ is odd, and $(a+1)|3$.
Furthermore, $a\equiv0\bmod{2}$.

If $(2+\sqrt{3})^r=a+b\sqrt{3}$, then $(2-\sqrt{3})^r=a-b\sqrt{3}$.
Then $(2+\sqrt{3})^r(2-\sqrt{3})^r=(a+b\sqrt{3})(a-b\sqrt{3})=a^2-3b^2$
At the same time $(2+\sqrt{3})^r(2-\sqrt{3})^r=\left((2+\sqrt{3})(2-\sqrt{3})\right)^r=1^r=1$
$a^2-3b^2=1$
$(a+1)(a-1)=3b^2$
Since $(a+1)|3$, if $\frac{a+1}{3}$ is a perfect square, then $(a-1)$ is a perfect square.

Recall that $a$ is even so $a+1$ and $a-1$ are relatively prime. Then $\frac{a+1}{3}$ and $(a-1)$ are relatively prime, so each must be a perfect square since their product is a perfect square.

Nice and short.  Good job Towersfreak2006


LaTeX
exterior angle
number theory
prime factorization

I can't believe this account is still active but I'm too lazy to create a new one so please ignore the username please :blush:

AoPS 1-10.5 has a proof for "The measure of an angle formed by two secant which intersect outside the circle is half the difference of the intercepted arcs" but not one for a tangent and a secant or two tangents. I figured it out for two tangents but couldn't do so for tangent-secant and no matter how hard I look online there just isn't a proof out there.

If someone here can post a proof I'll greatly appreciate it.

:rotfl: proving the tangent-tangent theorem is twice as hard as the tangent-secant theorem.

As you'll probably kick yourself if I give the proof away, I'll just give you scrappy clues.

[hide="hints"]
1) Make 2 additional lines that don't pass through the center
2) use the remote exterior angle theorem
3) apply tangent-chord theorem once
4) look for inscribed angles and supplementary angles
[/hide]

I'm not that great with names, so could you post the theorems, or a link to a place that describes them. (I could look them up myself, but posting within the topic will be a convenience for others plus im lazy  :blush: )

If you insist.

[hide="solution"]
[asy]unitsize(60); pair A,B,C,D,F,G,H,I; A=origin; B=(-1.5,0.2); C=(-0.9991095,0.04219142); D=1.2*dir(-30); F=1.2*dir(40); G=(0.84286033,-0.5381324); H=dir(120); draw(unitcircle); draw((B--D),EndArrow); draw(C--H,dashed); draw(B--H); draw(G--H,dashed); label("\small { $O$} ",origin,E); dot(origin); label("\small { $B$} ",B,W); label("\small { $C$} ",C,SW); label("\small { $D$} ",G,S); label("\small { $E$} ",H,N); draw(anglemark(B,H,C,26,28)); draw(anglemark(H,G,C,26,28)); draw(anglemark(G,B,H,20,20)); draw(anglemark(G,C,H,12,14,16));[/asy]

$ \angle EBC \cong \angle EBD$ (shared angle)
$ \angle BDE = \frac{1}{2}(\stackrel{\frown}{CE})$ (inscribed angle theorem)
$ \angle CEB = \frac{1}{2}(\stackrel{\frown}{CE})$ (tangent-chord theorem) $ \therefore$
$ \angle BDE \cong \angle CEB$
$ \triangle BEC \sim \triangle BDE (A.A.)$
$ \frac{\overline{BE}}{\overline{BD}}= \frac{\overline{BC}}{\overline{BE}}$
$ \boxed{(\overline{BE})^{2}= \overline{BC}\times \overline{BD}}$
$ \angle DBE+\angle BEC = \angle DCE$  (remote exterior angle)}
$ \angle DBE = \angle DCE-\angle BEC$ (EQUATION 1)
$ \angle DCE = \frac{1}{2}(\stackrel{\frown}{DE})$ (inscribed angle theorem)

Substitute $ \angle DCE$ by $ \frac{1}{2}(\stackrel{\frown}{DE})$ and $ \angle BEC$ by $ \frac{1}{2}(\stackrel{\frown}{CE})$ into EQUATION 1 to get:

$ \boxed{\angle DBE= \frac{1}{2}(\stackrel{\frown}{DE}-\stackrel{\frown}{CE})}$

[/hide]

Edit: Thanks to minsoens for fixing the picture. Anyway,

In case you don't understand the theorems I used, look below:

Remote exterior theorem (for triangles): The sum of any 2 angles in a triangle equals its remote exterior angle.

Inscribed angle theorerm: An angle inscribed in a circle equals half the measure of its intercepted arc.

Tangent-chord theorem (again, it applies to circles): An angle between a tangent and a chord equals half the measure of the arc that the chord intercepts.

[quote="Prime factorization"]Edit: Sorry my picture's not working.[/quote]hi, i think this is your picture:[asy]unitsize(60); pair A,B,C,D,F,G,H,I; A=origin; B=(-1.5,0.2); C=(-0.9991095,0.04219142); D=1.2*dir(-30); F=1.2*dir(40); G=(0.84286033,-0.5381324); H=dir(120); draw(unitcircle); draw((B--D),EndArrow); draw(C--H,dashed); draw(B--H); draw(G--H,dashed); label("\small{$O$}",origin,E); dot(origin); label("\small{$B$}",B,W); label("\small{$C$}",C,SW); label("\small{$D$}",G,S); label("\small{$E$}",H,N); draw(anglemark(B,H,C,26,28)); draw(anglemark(H,G,C,26,28)); draw(anglemark(G,B,H,20,20)); draw(anglemark(G,C,H,12,14,16));[/asy]i don't know what's wrong either, but i just changed [code]draw(anglemark(G,B,H,20));[/code] to [code]draw(anglemark(G,B,H,20,20));[/code] and it worked. :huh:

First off I just want to thank Prime factorization and minsoens for the clear, consise answer I needed. I can't believe you can use LATEX on this forum... :coolspeak: The community sure has grown since (way) before

So yeah, I just want to say thank you before saying the following

If you have AoPS 1 (fifth ed.) it's on pg 116 (Chap 10.5) and the proof is in the end of the chapter. In the proof for secant-chord theorem, the book uses secant-tangent theorem, meaning for the proof of the secant-tangent theorem you can't use the secant-chord theorem (which is why I got stuck).

Is there a way to prove either secant-chord or secant-tangent theorem without using the other in the proof?

(Heck if you have sixth/seventh ed. of the text, maybe it says it there. I bought this book like three and a half years ago...)

Go here [url]http://www.mathlinks.ro/viewtopic.php?t=160871[/url] and download the geometry.pdf file. In it, about halfway through, is all the proofs of the angle theorems related to chords, secants and tangents.

Ok now I see why it couldn't be solved; the proof in the book for the tangent-chord theorem was wrong (used tangent-secant). I vaguely remember dividing the tangent-chord cases into three parts and proving each separately (as in your proofs)

Thanks for the help :D


What is the minimum vertical distance between the graphs of 2 + sin(x) and cos(x)?

[hide="Elaborated"]The first graph is strictly above the second graph, so it suffices to minimize $ 2\plus{} \sin x \minus{} \cos x$. You can then use the angle-subtraction formula of Sine to find that the amplitutde of $ \sin x \minus{} \cos x$ is $ \sqrt{2}$, so the answer is $ 2\minus{} \sqrt{2}$[/hide]

take derivative of sinx - cos x and set = 0. so cos x = - sin x. Then minimum occurs when x = -pi/4.


If b>1, sin x>0, cos x>0 and $\log_b sin x=a$, then $\log_b cos x$ equals what?

If $\log_7(\log_3(log_2 x))= 0$, then the negitive square root of x equals what?

Suppose p and q ar positive numbers for which $\log_9 p=\log_{12} q=\log_{16} (p+q).$
What is the value p/q?

1
[hide]
Knowing the pythagorean identity, we can easily see that the expression we want is equal to $\frac 1 2 \log_b 1-b^{2a}$, which I think can simplify, but I can't seem to do it.
[/hide]

2
Solafide, you cannot just divide out a log statement like that.
[hide="hint"]
Try cancelling out the logs there by raising both sides to the power of the base.
[/hide]
[hide="ans"]
Using my hint over and over, we get $x=8$, so the answer we want is $-2\sqrt 2$
[/hide]

i hate logs!...now why did I do what I did instead of IntrepidMath's way?! :oops:  :blush:

[quote="solafidefarms"]2.[hide]Let's add 1 to both sides: $\log_7(\log_3(\log_2 x))+\log_7 7= \log_7 7$. Then we can take out the log_7's: $\log_3(\log_2 x)= 0$. Add 1 again, take the log, take out the log_3: $\log_2x=0. Log it again: we get x+1=1. So x=0, so the negative square root of x does not exist. Am I correct?[/hide]

3 I did in the AoPS book 2 yesterday, so it isn't fair for me to answer it ;)[/quote]Actually would you mind posting your solution for 3? At almost 3:00 AM I'm not sure how I would go about solving that one.  :D

[hide="3"]

Say $p=9^k$ ; $q=12^k$ ; $p+q=16^k$ for some real $k$

Then $p+q = 9^k+12^k=16^k$ . Divide both side by $16^k$

$\left(\frac{3}{4}\right)^{2k}+\left(\frac{3}{4}\right)^k=1$

Now letting $\left(\frac{3}{4}\right)^k=x$ , you will get a quadratic equation

$x^2+x-1=0$  :arrow:  $x=\frac{\sqrt{5}-1}{2}$  (because exponent is positive)

But $\frac{p}{q}=\left(\frac{3}{4}\right)^{k}=x=\frac{\sqrt{5}-1}{2}$

 :) [/hide]


Suppose that we have a sequence $\{a_{k}\}_{k=1}^{\infty}$ so that $\sum_{d|n}a_{d}=2^{n}$. Prove that for every natural number like $n$, $n|a_{n}$.

Posted a [i]very[/i] long time ago:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mobius&t=2461[/url]


function
complex analysis
complex analysis unsolved

let $ f: G\rightarrow \mathbb{C}$ is analytic and one-to-one, where $ G$ is open subset of $ \mathbb{C}$. Prove that $ f'(z)\neq 0 \forall z\in G$

[hide="Hint"]If $ n\equal{}ord(f;z_0)>0$ for some $ z_0\in G$, thenthere is an open neighbourhood $ z_0\in U\subset G$ and an analytic function $ h: U\to \mathbb C$ such that $ f(z)\equal{}h(z)^n$ for all $ z\in U$.[/hide]


conics
ellipse
geometry
geometric transformation
homothety

Suppose you have the length of the major and minor axes of an ellipse. Now suppose you inscribe a triangle in the ellipse with area $k$. How can you find the maximum value of $k$?

[quote="robinhe"][hide]I very interesting trick taught to me by someone :ninja:

Actually I don't know 100% but I think it should work.

First find the max area of the triangle inscribed in a circle (unit)

And then just "stretch" the circle into the ellipse by multiplying by the major and minor axis (like how you would find the area of an ellipse compared to a unit circle).

It should work.[/hide][/quote]

[hide]Yep that is how you should do it.

You have to make sure you are multiplying by 1/2 the major axis and 1/2 the minor axis, not the whole things to get the area of the triangle.[/hide]

If you used a different triangle, wouldn't there be a bigger area, If you could make a new triangle at a different angle, wouldn't the area be bigger?

wait wait, so would it work if lets say, you take a circle, draw an equilateral triangle, cuz thats the biggest triangle for a circle, then stretch the whole circle with the triangle inside so that it creates an ellipse?

[quote="deej21"]Hm, so you are using homothecy.  Would a circle and ellipse be homothetic?[/quote]

no, homothecy (homothety) is dialation, i.e. expanding, the above method is stretching (there might be another word for it like orthogonal projection)

You can convert the given ellipse into a circle by linear transformation.


topology
real analysis
real analysis unsolved

Does there exist a continuous (/differentiable) bijection from the rationals to the algebraic numbers?

I think that any two countable dense subsets of $ \mathbb R$ can be put into bijection by means of a homeomorphism of $ \mathbb R$ onto itself. WLOG you may assume that one of these subsets is the set of dyadic rationals, which has an appealing tree structure. Try to mimic this structure in the other set.

What about diffeomorphic?

I am principally interested in whether there is an analytic bijection (with analytic inverse).

If by "analytic" you mean "real-analytic", the answer is "yes". Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59853]here[/url] and follow the links in grobber's post.


algebra
polynomial
inequalities
LaTeX
inequalities unsolved

[quote="dominicleejun"]isn't this true only for a triangle with sides a,b,c?[/quote]
It's true for all non-negative $a,$ $b$ and $c:$
Let $a\geq b\geq c.$ Then $a+b-c\geq0,$ $a+c-b\geq0$ and $(a-c)^{4}\geq(b-c)^{4}.$ Hence,
$\sum_{cyc}(a-b)^{4}(a+b-c)\geq(a-b)^{4}(a+b-c)+(b-c)^{4}(a+c-b)+(b-c)^{4}(b+c-a)=$
$=(a-b)^{4}(a+b-c)+2c(b-c)^{4}\geq0.$


hello I have a mid term exam tomorow and was wondering if anyone could help me out. I am in grade 10.

how do you figure out  the answer to

7(squareroot) x 5(squareroot)

I'm not sure quite what you mean. to display the square root of x, type this
\sqrt{x}
and put a dollar sign on either side of it.
it will look like this
$\sqrt{x}$
click preview before you post to see if it came out right  :wink:

It might help you to know that when you multiply 2 radicals of the same degree (they're both square roots, both cube roots, etc.) then you can just multiply the two numbers inside the radicals, and put that under the radical.
[hide]
If you meant to say
$\sqrt{7}*\sqrt{5}$
the answer would be
$\sqrt{35}$
since there are no squares under the radical, it does not simplify. [/hide]

Do you mean $\sqrt{7}\cdot \sqrt{5}$ ?

$\sqrt{7}\cdot \sqrt{5}= \sqrt{35}$

Just multiply what's inside the root sign (called the radical), and then put the root back on.

EDIT: Kate beat me to it!

i do this in 7th grade   :D .
It is really easy:
For any [u]positiv[/u]e a and b we have:$\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$
That would give us:$\sqrt{35}$


Compute the area and perimeter of a regular hexagon inscribed in a circle of radius 2.

[hide="hint"]Divide the hexagon into six equilateral triangles.[/hide]

[hide="Click here to reveal hidden content"]

Separate the hexagon into 6 different congruent triangles, using 6 radii. The triangles are all isosceles, because radii are congruent. Thus the vertex angle is 360/6=60 degrees. Because the sum of the angles in a triangle is 180, 60 + x + x=180, so x=60 degrees. Therefore all the angles in the triangle are 60 degrees, and it is equilateral. The area of an equilateral triangle is [sqrt(3)*r^2]/4, where r is the side of the triangle.(See if you can prove this). r=2, so the area of the triangle is [4*sqrt(3)]/4. Since there are 6 triangles, the area of the triangle is [b]6*sqrt(3)[/b]. Also, since the triangle is equilateral, each of the sides is 2, and 6*2=[b]12[/b] is the perimeter.[/hide]


inequalities
Extremal combinatorics
graph theory
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in $\LaTeX$. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

This promblem seems to be very difficult and beautiful!!! :!: Does anybody know some solution???

answer: 3/32 is best possible.

proof:
 let v1,v2,...,vn be vertices of graph g and E be set of its edges.for i=1,2,...,nlet A i be set of vertices connectedto vi by an edge,Gi subgraph of g whose set of vertices is AiandEi set of edges of Gi.aNd vi,ei,ti=f(Gi) be no. of vertices,edges and tri.  in Gi res.
note,no. of tetrashedra and tri. one of whose vertices is Vi are res. ti,ei.so,summationi=1ton of vi=2cardinality of E, summationi=1ton of ei=3f(G) ,summationi=1ton of ti=4g(G)
note ei<=vi(vi-1)/2 and ei<=cardinality of E,so ei62,=vi^2*cardinality of E/2,i.e.ei<=vi*cardinality of E/261/2.summing  over alli f(G)^2<=2(cardinality of E)^3/9 so, ti=f(Gi)<=(2/9)^1/3 f(G)61/38e
summing all such inequalities we are done.and 3/32is best possible taking complete graph and limit. :) i

[hide]
Let $e(G)$ denote the number of edges in a graph, and $|G|$ denote its size. To prove

(1) $32g(G)^3 \leq 3f(G)^4$

we apply induction to reduce it to

(2) $9f(G)^2 \leq 2 e(G)^3$

which in turn we can reduce by induction to $2e(G) \leq |G|^2$, which is trivial, since $e(G) \leq \binom{|G|}{2}$.

Clearly for small cases (1) holds. Assume by induction that we know the statement is true for graphs of size up to $n$. Now, suppose we have a graph $G$ of $n+1$ vertices. Choose a vertex $v$, and let $G'$ denote the subgraph with all vertices of $G$ except $v$. Also let $G_0$ denote the subgraph consisting of all vertices adjacent to $v$. Then, we want to show

$32g(G')^3 \leq 3f(G')^4$

But we have $g(G') = g(G) + f(G_0)$, because the number of tetrahedra not containing $v$ is $g(G_0)$ and the number containing $v$ is $f(G_0)$, and $f(G') = f(G) + e(G_0)$, so we can expand the previous inequality and use the inductive hypothesis to reduce the result to

$96g(G)^2 f(G_0) + 96g(G) f(G_0)^2 + 32f(G_0)^3 \leq 12f(G)^3 e(G_0) + 16 f(G)^2 e(G_0)^2 + 12 f(G) e(G_0)^3$

The last term on the right was dropped because it isn't needed. Splitting this inequality into its three component inequalities (i.e. comparing the first terms on both sides, the second terms, and the last terms), and using the facts that $g(G) \leq (\frac{3}{32} f(G)^4) ^{\frac{1}{3}}$ and $f(G) \geq f(G_0)$, we find that the inequalities can be reduced to

$2e(G_0)^3 \geq 9f(G_0)^2$

$3e(G_0)^3 \geq 8\sqrt{2}f(G_0)^2$

$3e(G_0)^3 \geq 8f(G_0)^2$

which are progressively weaker and are all true if (2) is true for all graphs. Hence it remains to prove (2). We make a basically identical induction for (2), and show that it reduces to

$18f(G)e(G_0) + 9e(G_0)^2 \leq 6e(G)^2|G_0| + 6e(G)^2|G_0|^2$

and the two components of this are both true because $2e(G) \leq |G|^2$ for all graphs.
[/hide]

First a lemma

A graph has $f$ triangles and $h$ edges, then $f^2 \le (2/9)h^3$.
Proof: Let $d_i$ be degree of vertex $i$. Then that vertex is in at most $h$ triangles, but also at most $d_i^2/2$ triangles. Thus,

$3f \le \sum \sqrt{hd_i^2/2} = \sqrt{h/2} \sum d_i = \sqrt{2h^3}$, which implies result.

Now back to problem. Graph has $f$ triangles and $g$ K4s. Let vertex $i$ be in $f_i$ triangles and $g_i$ K4s. Then consider graph of neighbours of vertex $i$. Applying lemma to it, since it has $f_i$ edges and $g_i$ triangles, then $g_i^2 \le (2/9) f_i^3$. But also $g_i \le f$ trivially. So we get

$ 4g = \sum g_i \le \sum ((2/9)f_i^3f)^{1/3} = (2f/9)^{1/3} * \sum f_i = (2f/9)^{1/3} * (3f)$

and so $ 64g^3 \le (2f/9) * (27f^3) = (2*27)f^4$ and so $g^3 \le (3/32)f^4$.

Equality is for big cliques. Answer is $3/32$.

A Generalization from a book - let $N_t$ be the number of $K_t$s present in $G$.

Prove that $\frac{N^t_{t+1}}{N^{t+1}_t} \le \frac{t!}{(t+1)^t}$. For $t=3$ we have this problem.

@below Indeed, the book does not mention the equality at all - it just says that the proof of this inequality is similar to the ISL problem itself.

@below again Nope, it's from a Korean combo book.

Then again the same argument proves that $c \ge \frac{t!}{(t+1)^t}$ but proving equality seems hard.

@above Is this from "Straight from the Book"? :D

[hide=Solution]Let $V_1,V_2,...,V_n$ be the vertices of $G$ and $E$ be its set of edges.

For each $i\in\{1,2,...,n\}$, let $A_i$ denotes set of vertices connected to $V_i$ by an edge, $G_i$ is a subgraph of $G$ admitting $A_i$ its set of vertices. Let $E_i$ be the set of edges of $G_i$. Let $v_i,e_i$ and $t_i=f(G_i)$ be the number of vertices, edges and triangles in $G_i$ respectively. Then,
$$
\begin{aligned}
\sum_{i=1}^nv_i&=2|E|,\\
\sum_{i=1}^ne_i&=3f(G),\\
\sum_{i=1}^nt_i&=4g(G).
\end{aligned}
$$
We have $e_i\le\dbinom {v_i}{2}\le\dfrac{v_i^2}{2},\quad e_i\le |E|$, whence
$$
\begin{aligned}
e_i^2&\le\frac{v_i^2\cdot |E|}{2}\\
\iff e_i&\le v_i\sqrt{\frac{|E|}{2}}\\
\iff 3f(G)&\le 2|E|\sqrt{\frac{|E|}{2}}\\
\iff f(G)^2&\le\frac{2|E|^3}{9}.
\end{aligned}
$$
Thus,
$$
\begin{aligned}
t_i=f(G_i)&=f(G_i)^{\frac 13}f(G_i)^{\frac 23}\\
&\le\left(\frac 29\right)^{\frac 13}e_if(G_i)^{\frac 13}\\
\implies 4g(G)&\le 3\left(\frac 29\right)^{\frac 13}f(G)^{\frac 43}\\
\iff g(G)^3&\le\frac{3}{32}f(G)^4.
\end{aligned}
$$
Hence, we conclude that $c_{\min}=\dfrac{3}{32}$, with equality occurs if and only if $G$ is a completed graph.
[/hide]

[quote = rkm0959]A Generalization from a book - let $N_t$ be the number of $K_t$'s present in $G$.

Prove that $\frac{N^t_{t+1}}{N^{t+1}_t} \le \frac{t!}{(t+1)^t}$[/quote]

Proof is by induction:

[b]Base Case[/b]: For $t = 1$, the conclusion is clear. We wish to show $|E| \le \frac{1} {2} n^2$, where $n$ is the number of $K_1$'s, i.e. vertices, in $G$ (assuming $G$ has $n$ vertices).
$|E| \le {{n} \choose {2}} \le \frac{1}{2} n^2$, with equality occurring at $K_n$ for sufficiently large $n$, where $\frac {n^2 - n} {2}$ behaves like $\frac {n^2} {2}$.

[b]Inductive Step[/b]: Assume that $\frac{N^{t-1}_{t}}{N^{t}_{t-1}} \le \frac{(t-1)!}{t^{t-1}}$ holds for $t-1$. We will now show that the inequality holds for $t$.

Indeed, consider the graph $G$ with vertices $v_1, v_2, ..., v_n$ and fix vertex $v_i$. Let $A_i$ be the number of $K_{t+1}$'s that $v_i$ belongs to. We can bound this number in two ways.
On one hand, consider the subgraph induced by the neighbor set of $v_i$. $B_i$, the number of $K_{t}$'s contained in this subgraph, is also the number of $K_{t+1}$'s $v_i$ belongs to.
Let $C_i$ be the number of $K_{t-1}$'s contained in the subgraph. By the inductive hypothesis, $ \frac{B^{t-1}_i}{C^t_i} \le \frac{(t-1)!}{t^{t-1}}$.
It follows that $A^{t-1}_i \le \frac{(t-1)!}{t^{t-1}} \cdot C^t_i$

The second way of bounding $A_i$ is simpler. $A_i \le \frac {\sum_{v_j \perp v_i} B_j} {t} \le \frac {\sum_{v_i} B_i} {t} = \frac{t \cdot N_t} {t} = N_t$, since for each neighbor the number of $K_t$'s it is in is counted $t$ times, and the sum of $B_i$'s count $N_t$ $t$ times.

Combining our bounds, $A^t_i \le \frac{(t-1)!}{t^{t-1}} \cdot C^t_i \cdot N_t$. Then $A_i \le C_i \cdot \sqrt [t] {\frac{(t-1)!}{t^{t-1}} \cdot N_t}$.

Thus, $N_{t+1} = \frac {1} {t+1} \cdot {\sum_{v_i} A_i} \le \frac {1} {t+1} \cdot \sqrt [t] {\frac{(t-1)!}{t^{t-1}} \cdot N_t} \cdot {\sum_{v_i} C_i}$.

But notice that ${\sum_{v_i} C_i} = t \cdot N_t$, since the number of $K_t$'s $v_i$ is part of is $C_i$, so summing over all vertices and multiplying by $t$ since each $K_t$ is counted $t$ times gives us this relation.

Then $N_{t+1} \le \frac {1} {t+1} \cdot t \cdot N_t \cdot \sqrt [t] {\frac{(t-1)!}{t^{t-1}} \cdot N_t}$. Raising both sides to the $t$th power, we obtain:

$N^t_{t+1} \le \frac{(t-1)!}{t^{t-1}} \cdot N^{t+1}_t \cdot \frac {t^t} {(t+1)^t} = \frac {t!} {(t+1)^t} \cdot N^{t+1}_t$, as desired.

[b]Conclusion[/b]: $N^t_{t+1} \le \frac {t!} {(t+1)^t} \cdot N^{t+1}_t$ for all $t$. To show this constant is optimal, simply take $K_N$ for $N$ sufficiently large, and note equality occurs.

[quote=orl]For a finite graph $G$, let $f(G)$ be the number of triangles and $g(G)$ the number of tetrahedra formed by edges of $G$. Find the least constant $c$ such that \[g(G)^3\le c\cdot f(G)^4\] for every graph $G$.

[i]Proposed by Marcin Kuczma, Poland [/i][/quote]

Let $V, E$ denote the vertex set and edge set of graph $G$. Let $\nu_i$ denote the $i^{th}$ vertex. Let $G_i$ be the subgraph induced by $\nu_i$ and let $V_i, E_i$ be it’s vertex set and edge set. Denote by $T_i$ to be the number of $3$-cycles in $G_i$ and let the total number of triangles and tetrahedrons in $G$ be $X, Y$ respectively.

[b]Claim :[/b] $X \leq \frac{2 \cdot |E|^{3/2}}{3 \sqrt{2}}$

[i]Proof.[/I] Note that $|E_i| \leq \tbinom{|V_i|}{2} \leq \tfrac{|V_i|^{2}}{2}$. Also note that $\sum{|E_i|} = 3 \cdot X$. So, we have $$3 \cdot X = \sum{|E_i|} = \sum{\sqrt{|E_i|} \cdot \sqrt{|E_i|}} \leq |E|^{1/2} \cdot \sum{\tfrac{|V_i|}{\sqrt{2}}} = \tfrac{2 \cdot |E|^{3/2}}{\sqrt{2}}$$ $\square$

By [b]Claim[/b] in graph $G_i$, we obtain that $T_i \leq \tfrac{2 \cdot |E_i|^{3/2}}{\sqrt{2}}$. So, we have $T_i = T_i^{1/3} \cdot T_i^{2/3} \leq \tfrac{2^{1/3} \cdot |E_i| \cdot X^{1/3}}{3^{2/3}}$. Also note that we have $\sum{T_i} = 4Y$ and $\sum{|E_i|} = 3X$. So, we have $4Y = \sum{T_i} \leq \tfrac{3 \cdot 2^{1/3} \cdot X^{4/3}}{3^{2/3}}$ which in turn rearranges to $X^3 \leq \frac{3}{32} \cdot Y^4$.

We claim that $\mathcal{C} = \tfrac{3}{32}$ is the smallest possible one. Indeed note that the value of $\tfrac{X^3}{Y^4}$ for $K_n$, the $n$-vertex complete graph, as $n \mapsto \infty$ “approaches” $\frac{3}{32}$. $\blacksquare$

[hide=fake]We claim that $g^3\le\frac{3}{32}f^4$ for any graph $G=(V,E)$. The constant $\frac{3}{32}$ cannot be improved due to the asymptotically tight equality case of $K_n$.

We first prove a lower level analogue of the problem which will act as an aid in the main solution.

[b]Lemma:[/b]  For any graph $G=(V,E)$ with $f$ triangles, we have \[f^2\le\frac{2}{9}|E|^3.\]

[i]Proof:[/i]  Note that \begin{align*} 3f &\le \sum_{w\in V}\min\left\{\binom{\mathrm{deg}(v)}{2},|E|\right\} \\ &\le \sum_{w\in V}\sqrt{\binom{\mathrm{deg}(v)}{2}|E|} \\ &\le |E|^{1/2}|V|\sqrt{\binom{2|E|/|V|}{2}} \\ &\le |E|^{1/2}|V|\frac{\sqrt{2}|E|}{|V|} \\ &= \sqrt{2}|E|^{3/2}, \end{align*} as desired. $\blacksquare$

Let's return to the problem at hand now. For any $w\in V$, let $G_w=(V_w,E_w)$ be the induced subgraph with vertices given by the neighbors of $w$ in $G$. Note that the number of triangles in $G$ with vertex $v$ is $|E_w|$ and the number of tetrahedra with vertex $v$ is $f(G_w)$. Thus, \[3f=\sum_{w\in V}|E_w|\] and \[4g=\sum_{w\in V}f(G_w)\le\frac{\sqrt{2}}{3}\sum_{w\in V}|E_w|^{3/2}.\] We also have that \[f\le\binom{|V|}{3}\le\frac{|V|^3}{6}\implies |V|\ge (6f)^{1/3}\] From this, we see that \begin{align*} g &\le \frac{\sqrt{2}}{12}\sum_{w\in V}|E_w|^{3/2} \\ &\le \frac{\sqrt{2}}{12}|V|\left(\frac{3f}{|V|}\right)^{3/2} \\ &= \frac{\sqrt{6}}{4}|V|^{-1/2}f^{3/2} \\ &\le \frac{\sqrt{6}}{4}(6f)^{-1/6}f^{3/2} \\ &= \sqrt[3]{\frac{3}{32}f^4}, \end{align*} as desired.[/hide]

[quote=ayan.nmath][quote=yayups]
\begin{align*} g &\le \frac{\sqrt{2}}{12}\sum_{w\in V}|E_w|^{3/2} \\
 &\textcolor{red}{\le \frac{\sqrt{2}}{12}|V|\left(\frac{3f}{|V|}\right)^{3/2}} \\
 &\vdots\end{align*}
[/quote]
Can you explain this step? I think the reverse inequality is true.[/quote]

oops yea, i think my solution is fake :P
I'll fix it at some point.

The answer is \(\tfrac3{32}\), attained by arbitrarily large cliques. Let the vertices of \(G\) be \(v_1\), \(\ldots\), \(v_n\), and let \(e_i\) edges contain \(v_i\), let \(f_i\) triangles contain \(v_i\), and let \(g_i\) tetrahedrons contain \(v_i\). Also let \(G\) contain \(E\) edges, \(F\) triangles, and \(G\) tetrahedrons, so that \(2E_i=\sum e_i\), \(3F_i=\sum f_i\), \(4G_i=\sum g_i\).

[color=red][b]Claim:[/b][/color]      \(F^2\le\tfrac29E^3\).

[i]Proof.[/i]      Note that \(f_i\le E\) and \(f_i\le\tbinom{e_i}2\). Then \[3F=\sum_{i=1}^nf_i\le\sum_{i=1}^n\sqrt{E\binom{e_i}2}<\sqrt{\frac E2}\sum_{i=1}^ne_i=\sqrt2E^{3/2},\]     and the claim follows. \(\blacksquare\)

Note the trivial bound \(g_i\le F\). Consider the subgraph of \(G\) comprised of the neighbors of \(v_i\) (but not \(v_i\)). It contains \(f_i\) edges and \(g_i\) triangles, so \(g_i^2\le\tfrac29f_i^3\). Hence \[4G=\sum_{i=1}^ng_i\le\sum_{i=1}^n\sqrt[3]{F\cdot\frac29f_i^3}=\sqrt[3]{\frac{2F}9}\sum_{i=1}^nf_i=3\sqrt[3]{\frac29}F^{4/3},\] and the desired bound follows.

[color=red][b]Remark:[/b][/color]      Repeating the argument presented above gives the following generalization: if \(A_t\) is the number of \(K_t\)'s formed by edges of \(G\), then \[\frac{A_{t+1}^t}{A_t^{t+1}}\le\frac{t!}{(t+1)^t}.\]

We claim the best bound is $\frac{3}{32}$, which can be easily attained by large cliques.

First, define $e(G)$ to be the number of edges. We claim that $f(G)^2\le \frac{2}{9}e(G)^3$

We prove this with induction. Consider adding a vertex, $v$, to $G$ which already satisfies the condition and the induced subgraph $H$ which consists of vertices incident to an edge through $v$. The addition of $v$ adds $v(H)$ edges at at most $e(H)$ triangles, so we wish to show that $$(f(G)+e(H))^2\le \frac{2}{9}(e(G)+v(H))^3$$
Of course, it suffices to assume $f(G)=\sqrt{\frac{2}{9}e(G)^3}$, and $v(H)=\sqrt{2e(H)}$ to stricten the inequality. Now, if we let $x=e(G)$, $y=e(H)$, we wish to show $\frac{2}{9}\left(x+\sqrt{2y}\right)^3\ge\left(\sqrt{\frac{2}{9}x^3}+y\right)^2\iff 3x^2\sqrt{2y}+6xy+2y\sqrt{2y}\ge 3xy\sqrt{2x}+\frac{9}{2}y^2$. However, as $x\ge y$, $3x^2\sqrt{2y}\ge 3xy\sqrt{2x}$ and $6xy>\frac{9}{2}y^2$, so we immediately get the desired result.

Now, we proceed in the exact same fashion. Suppose that graph $G$ already satisfies the inequality, and consider adding a vertex $v$, incident upon an induced subgraph $H$. We add at most $e(H)$ new triangles and $f(H)$ tetrahedra, so we wish to show $$\left(\sqrt[3]{\frac{3}{32}x^4}+y\right)^3\le \frac{3}{32}\left(x+\sqrt[3]{\frac{9}{2}y^2}\right)^4$$
where $x=f(G)$ and $y=f(H)$. Expanding once again, this becomes $$3y\sqrt[3]{\frac{9}{1024}x^8}+3y^2\sqrt[3]{\frac{3}{32}x^4}+y^3\le \frac{3}{8}x^3\sqrt[3]{\frac{9}{2}y^2}+\frac{9}{16}x^2\sqrt[3]{\frac{81}{4}y^4}+\frac{27}{16}xy^2+\frac{27}{64}y^2\sqrt[3]{\frac{9}{2}y^2}$$
This is true because $\frac{3}{8}x^3\sqrt[3]{\frac{9}{2}y^2}\ge 3y\sqrt[3]{\frac{9}{1024}x^8}$, $\frac{9}{16}x^2\sqrt[3]{\frac{81}{4}y^4}>3y^2\sqrt[3]{\frac{3}{32}x^4}$, and $\sqrt{\frac{27}{16}}xy^2>y^3$

Use $A$ and $B$ in place of $f(G)$ and $g(G)$ respectively.

[color=#f00][b]Lemma:[/b][/color] Show that in a graph with $m>0$ edges, there are fewer than $\frac{\sqrt{2}}{3}m^{3/2}$ triangles.

[i]Solution:[/i] The number of triangles is at most
\[\frac{1}{3}\sum_v \min\left(\tbinom{\deg v}{2},m\right) < \frac{1}{3}\sum_v \deg v\sqrt{m/2} = \frac{\sqrt{2}}{3}m^{3/2}.\]
Now, onto the problem itself. We claim $c=\tfrac{3}{32}$ does the job. Taking large cliques $K_n$, $B^3$ is asymptotically $\frac{n^{12}}{24^3}$ and $A^4$ is asymptotically $\frac{n^{12}}{6^4}$, so a lower constant than $\tfrac{3}{32}$ cannot work.

Now it remains to show
\[B^3\le \frac{3A^4}{32}.\]
If $A>0$, we actually claim
\[B<\frac{1}{4}\sqrt[3]{6}A^{4/3}.\]
Let $f(v)$ for a vertex $v$ denote the total number of times the vertex is contained in a triangle, so $\displaystyle\sum_v f(v)=3A$. By the lemma, a vertex $v$ is part of at most
\[\min\left(\frac{\sqrt{2}}{3}m^{3/2},A\right)\]
tetrahedrons. Thus, we have
\[B\le \frac{1}{4}\sum_v \min\left(\frac{\sqrt{2}}{3}m^{3/2},A\right) < \frac{1}{4}\sum_v \sqrt[3]{\frac{2}{9}}m\cdot A^{1/3} = 3A^{4/3}\cdot \frac{1}{4}\cdot \sqrt[3]{\frac 29} = \frac 14\sqrt[3]{6}A^{4/3},\]
as desired.

I claim our answer is $c = \tfrac {3}{32}$, which can be achieved when we let $G$ be a complete graph $K_n$ and take $n \to \infty$. Indeed, note\[\frac{g(G)^3}{f(G)^4} \approx \frac{(n^2/24)^3}{(n^3/6)^4} = \frac{3}{32}\]for super large $n$. Let $G$ consist of $n$ vertices $\{v_i\}_{i = 1}^n$. Each vertex has degree $d_i$.

First, note that $e(G)$, the number of edges, always must satisfy $f(G)^2 \leq \tfrac29e(G)^3$. Indeed, for every vertex, if it is in $t_i$ triangles, then\[f(G) = \frac13\sum t_i\]where clearly $t_i \leq e(G) - d_i$ and $\tbinom{d_i}{2}$. Hence, we may stupidly bound by $t_i \leq \sqrt{(e(G) - d_i)\tbinom{d_i}{2}} \leq d_i\sqrt{\tfrac{e(G)}{2}}$. It follows that\[f(G) = \frac13 \sum t_i \leq \frac 13\sqrt{\frac{e(G)}{2}}\sum d_i = \sqrt{\frac29 e(G)^3}\]since $\sum d_i = 2e(G)$ is known. Squaring proves our result, which holds for any graph $G$. We can more or less, iterate this to get a similar result for $f(G), g(G)$.

In our original graph $G$, say vertex $v_i$ is part of $t_i$ triangles and $T_i$ copies of $K_4$. Then\[g(G) = \frac14\sum T_i\]For each vertex $v_i$, examine the graph $G_i = G \backslash v_i$. Note $t_i$ counts the number of edges in $G_i$, and $T_i$ the number of triangles. Hence it follows that $T_i^2 \leq \tfrac29 t_i^3$. Clearly $T_i = f(G\backslash v_i) \leq f(G)$ so note $T_i^3 \leq \tfrac29 t_i^3f(G)$. Thus,\[g(G) = \frac14\sum T_i \leq \frac14\sqrt[3]{\frac{2f(G)}{9}}\sum t_i = \sqrt[3]{\frac{3}{32}f(G)^4}\]which directly leads to our desired result.

[color=#00f][b]Remark:[/b][/color] In fact, continuously iterating like this actually yields that $C_{n+1}(G)^n \leq \tfrac{n!}{(n+1)^n}C_n(G)^{n+1}$ where $C_i(G)$ counts the number of copies of $K_i$ in $G$.

[quote=Stronger C8]Given a graph $G$ and an integer $t\geqslant 3,$ let $K_t(G)$ denote the number of copies of $K_t$ in $G.$ Fix a positive integer $n$. Find the least constant $C_n$ such that for any graph $G$ we have \[K_{n+1}(G)^n\leqslant C_n\cdot K_n(G)^{n+1}.\][/quote]

We will prove that $C_n=n!/(n+1)^n.$ To show that the inequality holds, we will proceed by induction. Let's firstly prove the base case, $n=2.$

[b]Lemma.[/b] Let $G$ be a graph with $e$ edges and $t$ triangles. Then, $T^2\leqslant 2/9\cdot E^3.$

[b]Proof.[/b] For every vertex $v_i,$ let $T_i$ denote the number of triangles containing $i$ and $E_i$ denote the number of edges in the neighbourhood of $v_i.$ Consequently, \[T_i=E_i=\sqrt{E_i\cdot E_i}\leqslant \sqrt{E\cdot \binom{\deg(v_i)}{2}}\leqslant\sqrt{\frac{E}{2}}\cdot \deg(v_i).\]By summing the latter over all vertices $v_i,$ it follows that\[3T=\sum_{i}T_i\leqslant\sqrt{\frac{E}{2}}\cdot \sum_{i}\deg(v_i)=\sqrt{\frac{E}{2}}\cdot 2E\]and thus, $T^2\leqslant 2/9\cdot E^3,$ as desired.

Now, assume that the inequality holds for $n-1.$ For every vertex $v_i,$ let $K_t(v_i)$ denote the number of $K_t$'s which contain $v_i.$ Furthermore, let $N(v_i)$ denote the neighbourhood of $v_i.$ Then, \[K_{n+1}(v_i)=K_n(N(v_i))=\sqrt[n]{K_n(N(v_i))\cdot K_n(N(v_i))^{n-1}}\leqslant \sqrt[n]{K_n(G)\cdot C_{n-1}}\cdot K_{n-1}(N(v_i))\]As seen previously, by summing the latter over all vertices $v_i$ it follows that \begin{align*}(n+1)\cdot K_{n+1}(G) &=\sum_i K_{n+1}(v_i)\leqslant \sqrt[n]{K_n(G)\cdot C_{n-1}}\cdot \sum_iK_{n-1}(N(v_i)) \\ &=\sqrt[n]{K_n(G)\cdot C_{n-1}}\cdot n\cdot K_n(G)\end{align*}And, by substituting $C_{n-1}=(n-1)!/n^{n-1},$ we finally get that \[K_{n+1}(G)^n\leqslant \frac{n^n\cdot C_{n-1}}{(n+1)^n}\cdot K_n(G)^{n+1}=\frac{n!}{(n+1)^n}\cdot K_n(G)^{n+1}.\]Therefore, the inequality also holds for $n,$ completing the induction. Now, we shall prove that $C_n=n!/(n+1)^n$ is indeed optimal. To do so, notice that \[\lim_{t\to\infty}\frac{K_{n+1}(K_t)^n}{K_n(K_t)^{n+1}}=\lim_{t\to\infty}\binom{t}{n+1}^n\binom{t}{n}^{-(n+1)}=\frac{n!^{n+1}}{(n+1)!^n}=\frac{n!}{(n+1)^n}.\]

I will prove just the normal version. The generalization is basically the same proof.

[b]Notation:[/b] for any discrete random variable $X$, denote its entropy by
$$H(X) := -\sum_{s} \mathbb P(X=s)\log\mathbb P(X=s).$$

The answer is $\boxed{3/32}$, achieved by taking an arbitrarily large clique. To prove the bound, take a uniform (ordered) $K_4$ subgraph $WXYZ$ of $G$. Then, using [url=https://en.wikipedia.org/wiki/Shearer%27s_inequality]Shearer's Lemma[/url], we get
\begin{align*}
3\log(24g(G)) &= 3H(W,X,Y,Z) \\
&\leq H(W,X,Y) + H(X,Y,Z) + H(Y,Z,X) + H(Z,X,W) & \text{(Shearer)}\\
&= 4H(X,Y,Z) \\
&\leq 4\log(6f(G)), &\text{(uniform bound)}
\end{align*}
giving $(24g(G))^3 \leq (6f(G))^4$, proving the bound for $c=6^4/24^3 = 3/32$.

Wow, such an impressive application of Shearer's lemma - never would have thought of this!

[hide = Solution]
The answer is $\tfrac{3}{32}$. To show we must have $c \geq \tfrac{3}{32}$, for each integer $n \geq 4$ the graph $K_n$ has $\tbinom{n}{3}$ triangles and $\tbinom{n}{4}$ tetrahedra, so \[\lim_{n \to \infty} \frac{g(K_n)^3}{f(K_n)^4} = \lim_{n \to \infty} \frac{\binom{n}{4}^3}{\binom{n}{3}^4} = \frac{(3!)^4}{(4!)^3} = \frac{3}{32}.\] To prove that the statement is true with $c = \tfrac{3}{32}$, let $f^*(G)$ and $g^*(G)$ denote the number of [i]ordered[/i] triangles and tetrahedra in $G$, so $f^*(G) = 6f(G)$ and $g^*(G) = 24g(G)$. We will show that $g^*(G)^3 \leq f^*(G)^4$, which will give the desired conclusion.

We use [hide = entropy]
The entropy of a random variable $X$, denoted $H(X)$, is defined as \[H(X) = \sum_s -p_s\log_2 p_s,\] where $s$ ranges over the possible values of $X$ and $p_s$ denotes the probability that $X$ takes the value $s$. We will use the fact that if $X$ takes on only values in a set $S$, then $H(X) \leq \log_2 |S|$. Equality holds if $X$ is uniform. (This is called the [i]uniform bound[/i].) We will also use [url = https://en.wikipedia.org/wiki/Shearer%27s_inequality]Shearer's lemma[/url]. One reference is [url]https://yufeizhao.com/pm/probmethod_notes.pdf[/url].
[/hide] --- let $H(X)$ denote the entropy of a random variable $X$. Let $(v_1, v_2, v_3, v_4)$ be an ordered tetrahedron in $G$ chosen uniformly at random (so $v_1$, $v_2$, $v_3$, and $v_4$ are all random variables); then \[H(v_1, v_2, v_3, v_4) = \log_2g^*(G).\] Now by Shearer's lemma we have \[3H(v_1, v_2, v_3, v_4) \leq H(v_1, v_2, v_3) + H(v_1, v_2, v_4) + H(v_1, v_3, v_4) + H(v_2, v_3, v_4).\] But if $(v_1, v_2, v_3, v_4)$ is an ordered tetrahedron, then $(v_1, v_2, v_3)$ is an ordered triangle (though not necessarily one chosen uniformly), so by the uniform bound \[H(v_1, v_2, v_3) \leq \log_2f^*(G).\] The same is true for the other three terms, so we obtain \[3\log_2g^*(G) \leq 4\log_2f^*(G),\] which means $g^*(G)^3 \leq f^*(G)^4$.
[/hide]

Let $f_r(G)$ denote the number of $K_r$'s in a graph $G$. We prove a more general statement: for all graphs $G$,
\[
f_r(G)\leq c_r f_{r-1}(G)^\frac{r}{r-1}
\]
where
\[
c_r=\frac{(r-1)!^\frac{r}{r-1}}{r!}.
\]

We proceed by induction on $r$. For $r=2$, the statement is
\[
||G||\leq\frac{1}{2}|G|^2
\]
which is true.

Assume the statement for $r$. Let $G$ be an arbitrary graph. For every vertex $v$, let $t_r(v)$ denote the number of $K_r$'s containing $v$. By the induction hypothesis,
\[
t_r(v)=f_{r-1}(G[N(v)])\leq c_{r-1}f_{r-2}(G[N(v)])^\frac{r-1}{r-2}=c_{r-1}t_{r-1}(v)^\frac{r-1}{r-2}.
\]
We also trivially have
\[
t_r(v)=f_{r-1}(G[N(v)])\leq f_{r-1}(G).
\]
It follows that
\[
t_r(v)\leq\min\left\{c_{r-1}t_{r-1}(v)^\frac{r-1}{r-2},f_{r-1}(G)\right\}\leq\sqrt[r-1]{c_{r-1}^{r-2}t_{r-1}(v)^{r-1}f_{r-1}(G)}
\]
so
\begin{align*}
f_r(G)&=\frac{1}{r}\sum_{v\in V}t_r(v)\\
&\leq\frac{1}{r}\sum_{v\in V}\sqrt[r-1]{c_{r-1}^{r-2}t_{r-1}(v)^{r-1}f_{r-1}(G)}\\
&=\frac{1}{r}c_{r-1}^{r-2}f_{r-1}(G)^\frac{1}{r-1}\sum_{v\in V}t_{r-1}(v)\\
&=\frac{r-1}{r}c_{r-1}^\frac{r-2}{r-1}f_{r-1}(G)^\frac{r}{r-1}\\
&=c_r f_{r-1}(G)^\frac{r}{r-1}.
\end{align*}
The induction step is complete.

Note that
\[
\lim_{n\rightarrow\infty}\frac{f_r(K_n)}{f_{r-1}(K_n)^\frac{r}{r-1}}=c_r
\]
so this constant cannot be improved.

The key lemma is the following:

[b][color=#ff9a00]Lemma.[/color][/b] In a graph with $m$ edges, there are at most $\frac{\sqrt 2}3 m^{3/2}$ triangles.

Notice that at some vertex $v$, at most $\min\left\{{\deg v \choose 2}, m\right\}$ triangles emanate from $v$. Thus we have the bound $$T \leq \frac 13 \sum_{v \in G}\min\left\{{\deg v \choose 2}, m\right\} < \frac{\sqrt 2}3 m^{3/2}$$ triangles, as needed. $\blacksquare$

Now pick some vertex $v$ and consider all the neighbors of $v$. There are at most $f(v) = \frac{\sqrt 2}3 (T(v))^{3/2}$ triangles among these neighbors, where $T(v)$ is the number of triangles at $v$; as a result, there are $\min(A, f(v))$ tetrahedrons at $v$. Then, $$4B \leq A^{\frac 13} \sum_{v \in G} \sqrt[3]{\frac 29} T(v) =\frac{\sqrt[3]6}4 A^{4/3},$$ which proves the result.


[color=blue]Hace un mes, se me ocurrió un método para demostrar que existen tantos números reales entre $0$ y $1$ como entre $1$ y $+\infty$ (por el momento, excluyamos a los extremos).

Antes de publicar mi demostración, quisiera que me digan si están de acuerdo con dicha posibilidad, puesto que en el foro de mi país no creen que sea posible (sucede que tampoco les mostré la demostración para saber cómo reaccionarían…).

Saludos.[/color]

Es conocido que existe una biyección entre $(0,1)$ y $\mathbb{R}$, también una biyección entre $(0,1)$ y $(1,+\infty)$ (basta tomar la función inversa). En general, se prueba que para dos intervalos cualesquiera (ya sean acotados o no) de $\mathbb{R}$ existe una biyección entre ellos.

$Tipe$

[color=blue]No había pensado que podía explicarse con las definiciones que ha presentado, pero mi demostración coincide con las características de las mismas. ¿Quiere que envíe mi demostración, para comprobarlo?[/color]

[color=blue]¡Vaya! Te adelantaste, Amilcar. Aunque, para verificar la biyección, hay que tomar la misma función para $f: (1;+\infty)\rightarrow\Re^{+}$.

Es interesante, que todo número real tiene su recíproco. ¿Qué sucede con los números complejos?[/color]

[quote="aev5peru"]ya ps mandala.
Tipe, te refieres a esta funcion.
$f(x)=1/x$, donde $f: (0,1)\to R^{+}$
chau salu2.[/quote]
plop!!, me confundi al escribir, jeje yo queria poner lo siguiente.
$f(x)=1/x$, donde $f: R^{+}\to (0,1)$, y pues claro la biyeccion que tiene esta funcion como indica R.G.A.M.
chau salu2.

Bueno, no es exactamente esa pues la "función" que indicas no está bien definida($f(1/2) \notin (0,1)$), me referia específicamente a $f: (0,1) \to (1,+\infty)$ dada por $f(x)=1/x$.

$Tipe$

[color=blue]Cierto es y, para demostrar la biyección, habría que asumir $f: (1;+\infty)\rightarrow(0;1)$ dada por $f(x)=\frac{1}{x}$, ¿no?

¿Existe alguna propiedad similar en los números complejos? No es tan simple, a mi parecer, definir dos conjuntos que cumplan una propiedad similar.[/color]

Lo que sucede con los numeros complejos es que son pares ordenados por tanto es un $R^{2}$ y en general entre todos los conjuntos acotados o no de $R^{n}$ siempre existe una biyeccion en particular hay tantos numeros reales como numeros complejos

Me corrijo yo solo antes de que lo haga otro :

[quote="DarthTenebrus"] y en general entre todos los conjuntos acotados o no de $R^{n}$ siempre existe una biyeccion[/quote]

Obviamente me falto aclarar que me referia a los conjuntos infinitos no numerables


geometry
circumcircle
ratio
trigonometry
angle bisector
geometry proposed

This has the same setting as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32558]this problem[/url].

Use the observation in post number $6$ in that thread to show that the power of $N$ wrt $(ABC)$ is $NB\cdot NC$. On the other hand, since the distance from $A$ to $BN$ is equal to the distance from $B$ to $AM$, and the distance from $A$ to $CN$ equals the distance from $C$ to $BM$, and those are equal, it means that $A$ lies on the angle bisector of $\angle BNC$. From this we also deduce that $AN\perp ON$ ($O$ is the circumcenter of $ABC$), so the power of $N$ wrt $(ABC)$ is also $NA^2$.

From all that we have $NA^2=NB\cdot NC$ and $\angle CNA=\angle ANB$. This shows that $CNA,ANB$ are similar, so $\angle BAN=\angle NCA=\angle CAM$.

[quote="payman_pm"]Let $ABC$ be an acute triangle, and let $M$ be the midpoint of its side $BC$. Suppose that a point $N$ lies inside the triangle $ABC$ and satisfies $\measuredangle ABN=\measuredangle BAM$ and $\measuredangle ACN=\measuredangle CAM$.

Prove that $\measuredangle BAN=\measuredangle CAM$.[/quote]

This problem becomes much more familiar if one works backwards:

Let the circle passing through the points A and C and touching the line AB at A and the circle passing through the points A and B and touching the line AC at A meet at a point W (apart from the point A, of course). Then, I will show that < ABW = < BAM, < ACW = < CAM and < BAW = < CAM. But there can exist only one point X satisfying < ABX = < BAM and < ACX = < CAM (in fact, the first of these two equations puts the point X on a fixed line through the point B, and the second of these two equations puts the point X on a fixed line through the point C; these two lines can have only one common point); hence, knowing that < ABW = < BAM and < ACW = < CAM on the one hand, while < ABN = < BAM and < ACN = < CAM on the other, we will thus conclude that W = N, so that the equation < BAW = < CAM will become < BAN = < CAM, and the problem will be solved.

Hence, in order to solve the problem, it is enough to show that < ABW = < BAM, < ACW = < CAM and < BAW = < CAM.

Well, let Y and Z be the orthogonal projections of the point W on the sides CA and AB of triangle ABC, and let E and F be the orthogonal projections of the point M on the sides CA and AB of triangle ABC.

Since the circumcircle of triangle ACW touches the line AB at A, the tangent-chordal angle theorem yields < BAW = < WCA. Similarly, < CAW = < WBA. The latter equation can be also rewritten in the form < WBA = < CAW. Now, from < BAW = < WCA and < WBA = < CAW, it follows that the triangles AWB and CWA are similar. Therefore, the ratio of the corresponding sides AB and CA of these triangles equals the ratio of the altitudes to these sides; in other words, $\frac{AB}{CA}=\frac{WZ}{WY}$ (in fact, since Y and Z are the orthogonal projections of the point W on the lines CA and AB, the segments WY and WZ are the altitudes of triangles CWA and AWB to the sides CA and AB, respectively). So we have $\frac{WZ}{WY}=\frac{AB}{CA}=\frac{c}{b}$.

On the other hand, in the right-angled triangle CME, we have $ME=CM\cdot\sin\measuredangle MCE=CM\cdot\sin C$. Similarly, $MF=BM\cdot\sin B$. But since the point M is the midpoint of the segment BC, we have CM = BM, and thus $\frac{ME}{MF}=\frac{\sin C}{\sin B}$. By the sine law in triangle ABC, we have $\frac{\sin C}{\sin B}=\frac{c}{b}$. Thus, $\frac{ME}{MF}=\frac{c}{b}$. Comparing this with $\frac{WZ}{WY}=\frac{c}{b}$, we get $\frac{ME}{MF}=\frac{WZ}{WY}$.

Since < AYW = 90° and < AZW = 90°, the points Y and Z lie on the circle with diameter AW. Thus, the quadrilateral AYWZ is cyclic. From this, it follows that < YWZ = 180° - < YAZ on one hand; on the other hand, by the chordal angle theorem, we obtain < WYZ = < WAZ.

Similarly, we get < FME = 180° - < FAE and < MFE = < MAE.

From < FME = 180° - < FAE and < YWZ = 180° - < YAZ, it follows, noting that < FAE = < YAZ, that < FME = < YWZ. This, together with $\frac{ME}{MF}=\frac{WZ}{WY}$, implies that the triangles FME and YWZ are similar. Hence, < WYZ = < MFE. But we know that < WYZ = < WAZ and < MFE = < MAE. Thus, < WAZ = < MAE. In other words, < BAW = < CAM. This is already one of the three equalities we have to prove.

Remains to verify < ABW = < BAM and < ACW = < CAM.

Actually, in the above, we have showed < BAW = < WCA and < BAW = < CAM. Thus, < WCA = < CAM. In other words, < ACW = < CAM. Similarly, one can show < ABW = < BAM. And the proof is complete.

By the way, the point W = N is the A-vertex of the second Brocard triangle of triangle ABC. The equation < BAW = < CAM means nothing but just the (well-known) fact that this vertex lies on the A-symmedian of triangle ABC.

  Darij

Here's yet another solution:

Consider the inversion of pole $M$ and power $MB^2$. It turns $A$ into a point $P$ on the median $AM$. We have $\angle MCP=\angle MAC=\angle ACN$. In the same way we show that $\angle MBP=\angle ABN$, so $P$ is the isogonal conjugate of $N$. Since it lies on the $a$-median of $ABC$, $N$ must lie on the $a$-symmedian of $ABC$, Q.E.D.

Let $x = \measuredangle BAM, y = \measuredangle CAM, \beta = \measuredangle ABC, \gamma = \measuredangle ACB,$ $b = AC, c = AB, a = BC, \phi = \measuredangle AMC$. Then, $b \sin y = CM \sin \measuredangle AMC = BM \sin \measuredangle AMB = c \sin x (*)$. Thus, as $c^2 + \frac{a^2+b^2-c^2}{2} = \frac{a^2+b^2+c^2}{2} = b^2 + \frac{a^2 - b^2 + c^2}{2}$, we have $b \sin y (c^2 + \frac{a^2+b^2-c^2}{2}) = c \sin x (b^2 + \frac{a^2-b^2+c^2}{2}) \Rightarrow b \sin y (c^2 + ab \cos \gamma) =$ $c \sin x (b^2 + ac \cos \beta) \Rightarrow c^2 \cdot \frac{2b \sin y}{a} + 2b^2 \sin y \cos \gamma$ $= b^2 \cdot \frac{2c \sin x}{a} + 2c^2 \sin x \cos \beta$. Now, $\frac{2b \sin y}{a} = \sin \phi = \frac{2c \sin x}{a}$, so keeping in mind that $\sin \phi =$ $\sin (\gamma + y) = \sin (\beta + x)$, we rewrite the above as $c^2 \sin \phi - 2c^2 \sin x \cos \beta = b^2$ $\sin \phi - 2b^2 \sin y \cos \gamma \Rightarrow c^2$ $(\sin \beta \cos x - \sin x \cos \beta) = b^2 (\sin \gamma \cos y  - \sin y \cos \gamma)$ $\Rightarrow \frac{c^2 \sin (\beta - x)}{b^2 \sin (\gamma - y)} = 1$, so by $(*)$ and the givens, we have $\frac{\sin y}{\sin x} \cdot \frac{\sin y}{\sin (\gamma - y)} \cdot \frac{\sin (\beta - x)}{\sin x} = 1 \Rightarrow \frac{\sin y}{\sin x} \cdot$ $\frac{\sin \measuredangle ACN}{\sin \measuredangle NCB} \cdot \frac{\sin \measuredangle CBN}{\sin \measuredangle NBA} = 1$. But if we apply Trig Ceva to rays $AN, BN, CN$, with the use of the above expression, we find that $\frac{\sin \measuredangle BAN}{\sin \measuredangle CAN} = \frac{\sin y}{\sin x}$. Now, as $\measuredangle BAC$ is acute, $f(t) = \frac{\sin t}{\sin (\measuredangle BAC - t)}$ is strictly increasing and can thus have only one solution, so as $x + y = \measuredangle BAC$, we find that $\measuredangle BAN = y$, as desired.


in barname ro natonestam download konam. age mishe rahnemayim konin. :(  :?:

mansooret barnameye "Cave" hast ke "ashegh" gozoshte?

are man eshtebahi be jaye javab ye topic jadid baz kardam. bebakhshid