linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

I am trying to solve problem 34 in Ch2 of Roger A. Horn's "Matrix Analysis". The problem is to prove "A is normal if and only if every eigenvector of A is also an eigenvetor of A* (A* is Hermitian of A)".

Knowing A is normal, it is easy to show A and A* share every eigenvector. But without assuming A is diagonizable, how can we prove A is normal?

I tried the following:
(1) AQ = QD (A is NxN, Q is NxM with each column as the eigenvector of A, D is MxM diagonal , M<=N)
(2) A*Q = QD'
(1) ==> A*AQ = A*QD = QD'D
(2) ==> AA*Q = AQD' = QDD' = QD'D (by D and D' commute)
Now we have (A*A-AA*)Q = 0 ,or, Q*(A*A-AA*) = 0. If rank(Q) is N, we can have A*A=AA*. But what if M<N?

Also, what can we say about eigenvalues of A and A*?

Any help will be appreciated!!

You will have to prove that $ A$ is (unitarily) diagonalizable; that's equivalent to being normal, after all. This will include finding the eigenvalues.

A hint: can you prove it in the case that $ A$ is upper triangular?

Thanks for the hint. If we have distinct diagonal elements, it is straightforward. But the problem is an upper-triangular matrix could be not diagonizable:
[1 1]
[0 1] has only one eigenvector [1 ; 0].
Can we prove any non-diagonizable matrice A cannot share every eigenvector with its Hermitian A*?

By the way, about eigenvalues of A and A*, by inspecting the characteristic polynomial det(A- \lambda I), we can show they are conjugates. So if A is real, then they have the same eigenvalues.

Can we use SVD decomposition to prove it?

The point is to show directly that an upper-triangular matrix with the same eigenvectors as its Hermitian adjoint is diagonal. The first standard basis vector is always an eigenvector...

As for why this is good enough? Schur's lemma.


If $ a\plus{}b\equal{}12$ and $ a\minus{}b$=6, what is $ a^2 \plus{} 2a \plus{} 2b \plus{} b^2$?

Adding the two equations together, we get $ 2a\equal{}18\Rightarrow a\equal{}9$, so $ b\equal{}3$.  Then, \[ 9^2\plus{}2(9)\plus{}2(3)\plus{}3^2\equal{}\boxed{114}\]

/bump
Is there a faster way to do this without finding out the values?


abstract algebra
calculus
integration
algebra
function
domain
polynomial

1. Let K be the field of fractions of an integral domain R. Show that K is not a Noetherian R-module.

2. Give an example of an integral domain which is not a Noetherian ring.

1. is wrong if $R$ is a field ;)
Otherwise we have a non-invertible element $a \in R$. Then the submodules $a^{-n}R$ are all different in $K$.

2.: $\mathbb{Z}[x_{1},x_{2},...]$, so the integral polynomial ring with $\infty$ (choose a cardinality yourself  :P ) many variables.

I left out something in 1. Sorry about that.

1. Let K be the field of fractions of an integral domain R. Show that K is not a Noetherian R-module [u](except when K=R)[/u].


trigonometry
inequalities
pigeonhole principle

Just was playin around, and came up with this, didnt sovle yet:

Prove true or false:

For every e > 0 there exists a natural n such that 0 < sin n < e.

[quote="radagast"]For every $e > 0$, there exists a positive integer $n$ such that $0 < \sin n < e$.[/quote]

Yes, this is true.

For the proof, I will use the following lemma:

[b]Lemma 1.[/b] For any two positive real numbers $x$ and $f$, there exists a positive integer $n$ such that the fractional part of $nx$ is smaller than $f$.

Here, the [i]fractional part[/i] of a real number $s$ means the number $s - \left[s\right]$, where $\left[s\right]$ is the integer part of $s$ (i. e. the greatest integer less or equal to $s$).

Now, let $A$ be such a number that $\displaystyle A < \frac{\pi}{2}$ and $0 < \sin A < e$ (such a number $A$ always exists, since $e > 0$). Also, define $\displaystyle x=\frac{1}{2\pi}$ and $\displaystyle f=\frac{A}{2\pi}$. Then, by Lemma 1, there exists a positive integer $n$ such that the fractional part of $nx$ is smaller than $f$. In other words, if $u = \left[nx\right]$ is the integer part of $nx$, then we have $u \leq nx < u+f$. In other words, $\displaystyle u \leq n\cdot \frac{1}{2\pi} < u+\frac{A}{2\pi}$. Multiplying this with $2\pi$, we get

$2\pi u \leq n < 2\pi u + A$.

Now, $u$ is an integer, $n$ is also an integer, while $2\pi$ is an irrational number. Therefore, we can never have $2\pi u = n$, so that we can sharpen the inequality $2\pi u \leq n$ to $2\pi u < n$. Thus, $2\pi u < n < 2\pi u + A$, and

$0 < n - 2\pi u < A$.

Now, since $u$ is an integer, the number $2\pi u$ is an multiple of $2\pi$, so that $\sin \left(n - 2\pi u\right) = \sin n$. But from $0 < n - 2\pi u < A$ (and $\displaystyle A < \frac{\pi}{2}$), it follows that $0 < \sin \left(n - 2\pi u\right) < \sin A$. In other words, $0 < \sin n < \sin A$. Finally, since $\sin A < e$, it follows that $0 < \sin n < e$, which yields the existence of a positive integer $n$ as you wished.

In order to complete the proof it just remains to prove Lemma 1. But Lemma 1 is a pretty well-known result and can be proven using the pigeonhole principle. (See, e.g., https://math.stackexchange.com/questions/843763/for-x-in-mathbb-r-setminus-mathbb-q-the-set-nx-lfloor-nx-rfloor-n-in-ma for the proof.)

[[b]EDIT:[/b] Lemma 1 now appears as Exercise 6.3.9 [b](c)[/b] in [url=http://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf]my [i]Math 235: Mathematical Problem Solving[/i] notes[/url] (the notes are still unfinished, so the numbering can changed; but you can find this exercise if you search for the words "fractional part" or "subdivide the half-open interval" in the notes).]

  Darij


[hide="  "]just spamming[/hide]

dang it, my poll got deleted....mine's missuri  :D

ARGH...MY POLL IS RIGGED!!!  just post on teh first occurance.

What happened to teh poll? Remake this thread, no one cares if you "spam".

If you are the type of person that thinks Bay Area is the best forum, then you have obviously not seen the MO forum because it is the same except its even "spammier".

[quote="serialk11r"][quote="007math"]Bay Area = Missouri[/quote]
That is clearly false. I do not see 66 USAMO qualifiers and several previous USAMO winners/HM from Missouri.[/quote]

Ahhh, but I do: The Ince twins a few years ago.

[quote="yaofan"][quote="serialk11r"]
That is clearly false. I do not see 66 USAMO qualifiers and several previous USAMO winners/HM from Missouri.[/quote]
but how many times bigger is california than missouri?[/quote]
66 from the Bay Area, about 60 from Santa Clara County. Santa Clara County is how many times smaller than Missouri?

Okay so count only the urban parts of Missouri. I'm sure Missouri has more than one county.

There's only one city in MO that matters: St. Louis.

Bay Area Combined Statistical Area: 6,985,411
St. Louis-St. Charles-Farmington Combined Statistical Area: 2,866,517

Two to one population ratio: Only about two times.

what happened to all the awesome shpam in bay area forum  :(

Missouri came, saw, and conquered.

All we have left is statistics.

And I'm imbuing that into the Missouri forum[hide=" "] so we can reconquer our rank of superiority! MWAHAHAHAHA![/hide].

Missouri is the gateway to the west. People came, they saw, they left for better kuso.

Man james4l, you need to stop saying "Bay Area" and just exclude all those useless San Francisco, Marin, San Mateo, Santa Cruz, etc. losers who are taking up precious population count and not giving results. It's called SANTA CLARA COUNTY (MSJ counts too, again). Population around 1 million, probably less.

Cupertino's population is 50k - Primary School Source: Homestead
Fremont's population is 210k - Primary School Source: MSJ
Palo Alto's population is 60k - Primary School Source: Paly/Gunn
San Jose's population is 940k - Primary School Source: Lynbrook/Harker

Fine, RECTIFIED population: 1.26 million


Does anyone know how to prove the Descartes Circle Theorem:

For 4 mutually tangent circles, then twice the sum of the squares of their curvatures* is equal to the square of the sum of their curvatures

*The curvature of a circle is equal to the inverse of its radius, and if the largest  circle encloses all the others, then its curvature is negative.

I'm sure this has been discussed on the forum before. I suggest you look for the topic(s). I remember givin a proof based on inversion a very long time ago.

[color=red][Moderator edit: Indeed: http://www.mathlinks.ro/Forum/viewtopic.php?t=6483 and http://www.mathlinks.ro/Forum/viewtopic.php?t=352 .][/color]


Turning everything into simplified fractions usually makes taking square roots etc. easier. Doing so, \[ \sqrt{\frac{16}{25}}-\sqrt[3]{-\frac{1}{125}})+\sqrt[4]{\frac{16}{81}}+\sqrt[5]{32}=\frac{4}{5}-(-\frac{1}{5}+\frac{2}{3}+2.\] Simplifying again and adding the fractions gives \[ 1+2+\frac{2}{3}=\boxed{3\frac{2}{3}}.\]


are $s,t$ variabls or fixed as $a,b,c$ :!:  :maybe:


?知是我自己的问题还是论?的问题,?过在我的version里,出现了很严?的乱?!

Well, still, it does not functioning!
Chinese Communities is turning into an all-English forum!


I had a weird idea as I was waking up one morning, and I would appreciate some feedback on it. Please bear with me here as I explain my train of thought.

Say you have a cup of a certain, predicted mass sitting on a scale. You can only look at this cup from the side, and it's made of opaque ceramic so there is no way to see through it. Upon looking at the scale, you see that the cup is more massive than it should be; you know the mass of the ceramic, but the scale is behaving as if the cup has more mass than what you had predicted.

The cup is actually filled with coffee, but there's no way for you to realize this because you can't see inside the cup; you can only view it in two dimensions. Therefore, you theorize that there is something in the ceramic itself that is giving the cup more mass than is observable. However, if you could see the cup in three dimensions, it would be easy to observe the cup from such an angle that you could look at the coffee inside it.

I thought of this as an explanation for the dark matter in the universe. Maybe the reason we can't see dark matter directly is because it's in a dimension we can't observe, like the third dimension in my analogy. Modern superstring theory states there must be ten, eleven, or twenty-six dimensions, yet we can only observe four: three of space (x, y, z) and one of time (t). What if all of this extra, unpredicted mass that causes things such as galaxies to behave the way they do--“dark matter”--isn't in the four dimensions we usually observe? There are at least six more dimensions out there; what if there's matter in these that affects the observable universe in the way we think dark matter in x,y,z and t should?

Anyway, that's just my idea. Feel free to add to it, disprove it, agree with it, debate about it, etc., and thanks for taking the time to read it. I'm sorry that the analogy's confusing...I can try to explain it better if someone doesn't understand it.

Interesting view on the dark matter contraversy :) .
According to some other prevailing theories, there is no "extra" matter but instead there is just more gravitational strength than expected.  It is from this "extra" gravity that some assumed that the must be the extra mass, though other presume the Newton's Laws of Gravitation do not work on larger scales.  Things to look up for the other side of the story are MOND and the Pioneer Anomaly.
Otherwise, you hypothesis is interesting.


geometry
projective geometry
geometry unsolved

[quote="K®izalid"][color=darkred]I need urgent help with this problem.

Consider a semicircle $w$ of diameter $[AB]$ with $P$ and $Q$ two different points which lie on $w$.
Denote $T\in PP\cap QQ$ and $S\in AP\cap BQ$. Prove that $ST\perp AB$.

[b]Note.[/b] Order of the points on semicircle from left to right is following : $A,\,P,\,Q,\,B$.
Please, can anyone give me a hand, at least an address of topic or site ? Thanks a lot ![/color][/quote]

[hide="Hint."] Denote $O$ is the midpoint of $AB$ and $R=PQ\cap AB$
Thus $ST$ is the polar of $R$ w.r.t. $(O)\implies ST\perp OP$ [/hide]

[color=blue]O.K., [b]April[/b] ! You can prove without pol-polar theory (the [b]Hire[/b]'s theorems) ?! [b]K®izalid[/b] doesn't know about it !![/color]
[quote][color=darkred][b]Hire's theorem.[/b] Let $w$ be a circle and $P,Q,R$ be three points which don't belong to the circle $w$.
Denote the polar $x$ of the point $X$ w.r.t. the circle $w$. Then : $\{\begin{array}{cccc}1\blacktriangleright & Q\in p & \Longleftrightarrow & P\in q\\\ 2\blacktriangleright & R\in p\cap q & \Longleftrightarrow & r\equiv PQ\end{array}$.[/color][/quote]

[color=darkblue]The previous proofs (April's and Butterfly's) are very nicely.
O.K., [b]Butterfly[/b] ! You can prove without the [b]Pascal[/b]'s theorems ?! [b]K®izalid[/b] doesn't know about it !![/color]
[quote][color=darkred][b]Pascal's theorem.[/b] $\left\{\begin{array}{c}\{A,B,C,D,E,F\}\subset w\mathrm{\--\ circle}\\\ X\in AB\cap DE\\\ Y\in BC\cap EF\\\ Z\in CD\cap FA\end{array}\right\|$ $\implies$ $Z\in XY$.[/color][/quote]

[quote="Butterfly"]I give another proof:

Denote $R=AQ\cap BP$

Hexagon $APPBQQ$ is cyclic with regard to $w$,and hence the three points of intersection $S,T$and$R$ lie on the same line according to Pascal theorem,so $ST\perp OP$.[/quote]

I think that by [b][size=100]Pascal's theorem[/size][/b], we have only the collinearity of the points $R,$ $S,$ $T.$

The wanted conclusion of $ST\perp AB,$ is consequent from the fact that the point $R\equiv AQ\cap BP,$ is the orthocenter of the triangle $\bigtriangleup SAB,$ because of $AQ\perp SB$ and $BP\perp SA.$

Kostas Vittas.

[quote="Virgil Nicula"][color=blue]O.K., [b]April[/b] ! You can prove without pol-polar theory (the [b]Hire[/b]'s theorems) ?! [b]K®izalid[/b] doesn't know about it !![/color] [/quote]
[color=darkblue]Yes, my proof is as same as your, [b]Mr. Virgil Nicula. :lol:[/b] But you were faster than me :)  [/color]

[quote="Virgil Nicula"][color=darkblue][b]Proof.[/b] Denote $\{\begin{array}{c}R\in AQ\cap BP\\\\ M\in (SR)\ ,\ MS=MR\\\\ T_{1}\in PP\cap SR\ ,\ T_{2}\in QQ\cap SR\end{array}$. Observe that the point $R$ is the orthocenter of the triangle $ASB$. Thus,  $SR\perp AB$ and

$\{\begin{array}{ccccc}\widehat{SPT_{1}}\equiv \widehat{ABP}\equiv \widehat{PSR}& \Longrightarrow & \widehat{SPT_{1}}\equiv\widehat{PSR}& \Longrightarrow & T_{1}\equiv M\\\\ \widehat{SQT_{2}}\equiv\widehat{BAQ}\equiv\widehat{QSR}& \Longrightarrow & \widehat{SQT_{2}}\equiv\widehat{QSR}& \Longrightarrow & T_{2}\equiv M\end{array}\|$ $\implies$ $T_{1}\equiv T_{2}\equiv T$ $\implies$ $\{\begin{array}{c}T\in (SR)\\\\ TS=TR\end{array}$ and $ST\perp AB$.

[b]Remark.[/b] $\{\begin{array}{c}\triangle ABC\ ,\ AB\perp AC\\\ M\in (BC)\ ,\ AM=MC\end{array}\|$ $\implies$ $MB=MC$.[/color][/quote]

Very good. Some times, the basic geometric tools, work better than the special ones ( [b][size=100] Pascal, Desqrques, Polars,[/size][/b] etc ).

Kostas Vittas.

Let: arc $AP=2a$, arc $BQ=2b$ $\Rightarrow$ $\measuredangle SPT = a$, $\measuredangle SQT=b$, $\measuredangle ASB=a+b$

Thus $\measuredangle PSQ = 2\measuredangle PTQ$, also we have $TP=TQ$. So $T$ is the circuncentre of $PSQ$

Then $\measuredangle PST = \measuredangle SPT = a$ but also $\measuredangle SAB = 90^\circ-a$

$ST \perp AB$

[color=darkred]Yes, Vittasko ! Among other things, sometimes, when you know too much you can fail the beauty of the particular situations.[/color]

[color=darkblue][b]Here is other proof.[/b][/color]
[quote][color=darkred]Denote $\{\begin{array}{c}R\in AQ\cap BP\\\\ O\in (AB)\ ,\ OA=OB\\\\ T'\in (SR)\ ,\ T'S=T'R\end{array}$. Observe that $SR\perp AB$ and $P,Q$ belong to the [b]Euler's circle[/b] of $\triangle ASB$

with diameter $[T'O]$ $\implies$ $\{\begin{array}{c}PO\perp PT'\\\\ QO\perp QT'\end{array}$ $\implies$ $T'\equiv t$ $\implies$ $T\in SR$ $\implies$ $TP=TQ$ and $ST\perp AB$.[/color][/quote]
[color=darkblue][b]Here is an another elementary proof of one from the my students.[/b][/color]
[quote][color=darkred]Denote the intersection $R\in AQ\cap BP$ and the middlepoint $T'$ of the segment $[SR]$. The points $P,Q$ belong to the circle with the diameter $[SR]$. Thus, $m(\widehat{SPT'})=90^{\circ}-m(\widehat{SQP})$ and $m(\widehat{SQP})=m(\widehat{SAB})=90^{\circ}-m(\widehat{ABP})$. Therefore, $\widehat{SPT'}\equiv\widehat{ABP}$, i.e. the line $PT'$ is tangent to the circle with the diameter $[AB]$. Analogously the line $QT'$ is tangent to the same circle. In conclusion, $T'\equiv T$, i.e. $TS=TR$ and $ST\perp AB$.[/color][/quote]
[color=darkblue][b][u]K®izalid[/u], now are you pleased with this topic and the its messages ? Where are you and the your answer ?[/b][/color]

[quote="Virgil Nicula"]From "K®izalid". I think this problem appeared before on ML.[/quote]
Indeed Virgil Nicula! It was a question in the Pan African Olympiad, 2002.
You can find it here: http://www.mathlinks.ro/Forum/viewtopic.php?t=49092


The classic rule game, you must make a rule without breaking any of the previous rules. Let me start:

1. All rules must contain at least 1 number.

1. All rules must contain at least 1 number.
2. Rule #5 and every multiple of 5 onward must begin with the word all.
3. Rules that have the word "all" must have all the letters a, s, d, f somewhere in the post.

Rule 3 broke rule 1.

Restart.

Rule 1: All rules must include the number 1337.

Rule 1: All rules must include the number 1337.
Rule 2: All rules must contain at least one word in igpay atinlay, this will go on till rule 1337.
Rule 3: All rules ollowingfay this one, must use no S's, unless you are using 1337 of them.
Rule 4: Every rule that=an ultiplemay of 4 or 1337 will have one emoticon.  :D
Rule 5: \$\text{Every rule will have to be written in a kind of language other than the one originating from the Angles. Until you hit rule 1337, all rulez afteray cannot repeat a language that was occupied in a prev. rule}\$

Rule 5 breaks Rule 3.  ("Angles")

Rule 1:  All rules must contain exactly seven words.

Rule 1: All rules must contain exactly seven words.
Rule 2: All prime rules -> contain a mathematical notation.
Rule 3: All even rules -> contain their post count -1.
Rule 4: All rules must have 124 or more letters.

I suppose rule three wasn't really fair because your post count goes up as you post more. Oops. Should we keep going?

Yes, the post number in the upper right hand corner of the post box.

Rule 1: All rules must contain exactly seven words.
Rule 2: All prime rules -> contain a mathematical notation.
Rule 3: All even rules -> contain their post count -1.
Rule 4: All rules must have 124 or more letters
Rule 5: All rules must be .... wait.....   Can you have seven words and 124 letters?  I think not.

Rule 1: All guidelines must be lengthier than the previous restriction.

Rule 2: Oops, I just broke the last rule.

New Rule 1: Each rule must have one [b]bolded[/b] word.

[hide="Rule 1:"]
Rules must be hidden.
[/hide]
[hide="Rule 2:"][color=darkblue]
Rules should be in darkblue[/color]
[/hide]
[hide="Rule 3:"][color=darkblue]Rules should following this rule must include the phrase: "I luv pi(e)."[/color][/hide]
[hide="Rule 4:"][color=darkblue]
Since I luv pi, although i prefer e, rules should contain a giant O at the end.[size=200]O[/size][/color]
[/hide]
[hide="Rule 5:"]
[color=darkblue]
34CH |2UL3 MU57 83 w|21773n 1n 1337... 51nC3 1 LUv p1(3) 0.[size=200]0[/size][/color]
[/hide]
The 1337 O is 0 :D
So I just pwnt 15!!!'s rule... :evilgrin:

[hide][color=darkblue]you broke it meewhee009. you did not post in complete darkblue[/color][/hide]

[b]Rule 1:[/b][b] All rules must be bolded.[/b]

[b][color=darkred]Rule 2: all rules must be in dark red.[/color][/b]

You forgot rule 1, start over.

[color=violet][size=200][b]Rule 1: All rules must be bolded, in huge font, and must be in color violet.[/b][/size][/color]

[b][color=violet][size=200]Rule 1: All rules must be bolded, in huge font, and must be in color violet.[/size][/color]
[color=violet][size=200]|2UL3 2: 4LL |2UL35 MU57 83 w|21773n 1n 1337 [/size][/color][/b]

k00lperson, if you quated my rule you would see that it was bolded.

musicnmath, I meant you forgot to like put
Rule 1: blah blah
Then you do rule 2..
see what I mean?
um
[color=violet][size=200][b]Rule 1: All rules must be bolded, in huge font, and must be in color violet.
|2UL3 2: 4LL |2UL35 MU57 83 w|21773n 1n 1337
Rule 3: All rules must have a smiley, and be awesome like 1337 :P[/b][/size][/color]

k00lperson, rule three breaks rule 2 because rule 2 says that all rules must be written in "the font" 1337.

Rule 1: all rules must have one smiley or say I luv pi(e). :rotfl:

Rule 1: all rules must have one smiley or say I luv pi(e). :rotfl:
Rule 2: All rules must speak of chicken.  :roll:
Chicken 3: Last word of previous rule will substitute "Rule" preceding rule number.  :)
Chicken 4: All rules must worship the all powerful Mr. Green... since I luv pi(e).

BAD. You were supposed to say "Number 4: (as "number" is the last word in "Chicken 3")

Rule 1: Eyeu must rheapalase alternating wardz with feuneytik approximations oer equivalencies.

Rule 1: Eyeu must rheapalase alternating wardz with feuneytik approximations oer equivalencies.
rule 2: oll rules nead to bee written en lowercase.
rule 3: et shall beh known taht rules muhst have (rool number) laytars in teh rule (maud seven).

that's including "rule"


inequalities
trigonometry
inequalities proposed

Prove that in any triangle with angles $ \alpha ,\beta , \gamma$ exists the relation :
\[ Sin(\frac{\alpha}{2})+Sin(\frac{\beta}{2})+Sin(\frac{\gamma}{2}) \ge 1 + \frac{r}{R}
\]

[quote="Sergic Primazon"]Prove that in any triangle with angles $ \alpha ,\beta , \gamma$ exists the relation :
\[ Sin(\frac {\alpha}{2}) \plus{} Sin(\frac {\beta}{2}) \plus{} Sin(\frac {\gamma}{2}) \ge 1 \plus{} \frac {r}{R}
\]
[/quote]
let $ Sin(\frac {\alpha}{2}) \equal{} x,....$
=>$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$
we have to prove :
$ 1 \plus{} 4xyz \leq x \plus{} y \plus{} z$ ,$ f(x,y,z)\equal{}x\plus{}y\plus{}z\minus{}4xyz\plus{}k(x^2\plus{}y^2\plus{}z^2\plus{}2xyz\minus{}1)$
use lagrange multipliers,we get this:f(x,y,z) min when x=y=z
so $ x \plus{} y \plus{} z \minus{} 4xyz\geq 1$

We can solve it easily by Am-Gm, my friend.
Let $ x \equal{} \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}}$...
Square, we only need to prove: $ xy \plus{} yz \plus{} zx \geq\ 5xyz \plus{} 8x^2y^2z^2$
We have: RHS $ \leq\ 6xyz \leq\ xy \plus{} yz \plus{} zx$
<=> $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq\ 6$
<=> $ \sum\ \sqrt {\frac {(a \plus{} c)(b \plus{} c)}{ac}} \geq\ 6$
(obviously trues again by Am-Gm)
 :)

mm, it can be done much easier... recall that $ 1\plus{}\frac{r}{R}\equal{}\cos A\plus{}\cos B\plus{}\cos C$ and note that $ \cos A\plus{}\cos B\equal{}2\cos\left(\frac{A\plus{}B}{2}\right)\cos\left(\frac{A\minus{}B}{2}\right)\leq 2\sin\frac{C}{2}$... take the other sums and we're done  :D


In Parallelogram ABCD, E is a point on AD such that AE is 1/2 of AD and F is the point of intersection of BE and AC. Prove AF is 1/3 of AC.

Here's an outline of my proof:

[hide]Since $\angle{AFE}=\angle{BFC}$ and $\angle{FAE}=\angle{FCB}$, $\triangle{AEF}\sim\triangle{CBF}$, and $\frac{BC}{AE}=\frac{CF}{AF}=2$. Thus $\frac{AF}{AC}=\frac{AF}{AF+FC}=\frac{1}{3}$. [/hide]

I would use mass points... assign a weight of 1 to A, B, D.  Then E has weight 2, F has weight 3, making C have weight 2, and thus AF:FC = 1:2.


$ P \equal{} 2x \plus{} x \plus{} x \plus{} 2x \plus{} x \plus{} x \equal{} 8x \equal{} 8(9) \equal{} \boxed{72}$


induction
combinatorics proposed
combinatorics

Problem number3 Of the last of TST of IRAN.(In an rather old year)
This is my idea of beauty in math.(As you said,grobber)
We call three Sets like $A_1,A_2,A_3$ ,[i]a triangle[/i] if For $1 \le i \le 3$
$A_i \cap A_j \neq \emptyset ,A_1 \cap A_2 \cap A_3 =\emptyset$.
Let $f(n)$ be the smallest natural number such that a set with $f(n)$ number of
subsets of $\{1,2,...,n\}$ contains at least [i]a triangle[/i].
Find a formula for $f(n)$.

I think I have a solution, but I'm not sure it's correct. This is not the first "solution" I'm trying to post :).

Let's rephrase it to make $f(n)$ the smallest $k$ for which any collection consisting of $k$ subsets of $[n]$, [i]each with at least two elements[/i], contains a triangle. This is natural, since the empty set and the singletons cannot be part of any triangle. So from now on, unless I specify it, by "subset of $[n]$" I mean "subset of $[n]$ with at least two elements".

I believe that in this setting, the answer is $2^{n-1}$. First of all, we can find $2^{n-1}-1$ subsets of $[n]$ with no triangles: just take all the subsets containing $n$.

We can prove the other part by induction. Suppose it's proven for numbers less than $n$, and assume we have $2^{n-1}$ subsets with no triangles. Since, by the induction hypothesis, we can't have $2^{n-2}$ subsets of $[n-1]$, it means that at least $2^{n-2}+1$ subsets which contain $n$. Call these sets $A_1,\ldots,A_t$, with $t\ge 2^{n-2}+1$, and for every $i$, let $\tilde A_i=A_i\setminus\{n\}$. Also, let $B_{t+1},\ldots,B_{2^{n-1}}$ be the other sets we have in our collection.

Now, if $\tilde A_i,\tilde A_j$ are complements in $[n-1]$, then all the $B_k$'s have to be subsets either of $\tilde A_i$ or $\tilde A_j$, otherwise $A_i,A_j,B_k$ would be a triangle. If $t=2^{n-2}+a,\ a\ge 1$, we have at least $a$ pairs of complements among the $\tilde A_i$'s. It's easy to see from here that there will be (at least) $a+1$ non-empty subsets of $[n-1]$ (call them $S_1,\ldots,S_{a+1}$) which partition $[n-1]$ and s.t. every $B_k$ is a subset of one of the $S$'s. That's because for each new pair $\tilde A_i,\tilde A_j$ of complements in $[n-1]$, we impose new constraints on the $B$'s, according to the observation in the first sentence of this paragraph.

This means that there are at most $2^{|S_1|}+\ldots+2^{|S_{a+1}|}-n\ B$ 's (the $-n$ comes from the fact that the $B$'s have at least two elements), and this is less than $2^{n-2}-a$, giving a contradiction.

I'm almost certain it's not right, so I apologize in advance :).

i believe you are a real genuis gorbber .
Nice way of thinking.
To be precise the answer is $f(n)=2^{n-1}+n+1$(You said $2^{n-1}$ by decreasing
$\emptyset ,\{1\},...,\{n\}).$
and the proof is exactly by induction but a little different from your.

This one is a Erdös problem from Canadian math journal some years ago... :roll:

Pierre.


In concave hexagon ABCDEF, $ \angle A \equal{} \angle B \equal{} \angle C \equal{} 90^{\circ}, \angle D \equal{} 100^{\circ}, \angle F \equal{} 80^{\circ}$ Also, $ CD\equal{}FA, AB\equal{}7, BC \equal{} 10,$ and $ EF \plus{} DE \equal{} 12$. Compute the area of the hexagon.


limit
calculus
derivative
calculus computations

For the Limit(x,y) -> (0,0) [e^(xy)-1]/[x^2+y^2] is this a correct way to solve the limit?

By setting F(t) = f(tx,ty) = e^[(tx)^2+(ty)^2], you get for the upper bound as:

[F(1)-F(0)]/[1-0]=F(1)= e^[x^2+y^2]. The limit of which is 1..

Is that the right way to do it?

That calculation doesn't make sense. $ f(tx,ty)$ should be $ \frac{e^{tx\cdot ty}\minus{}1}{t^2x^2\plus{}t^2y^2}\equal{}\frac{e^{t^2xy}\minus{}1}{t^2}\cdot \frac1{x^2\plus{}y^2}$; as $ t\to 0$, that tends to $ \frac{xy}{x^2\plus{}y^2}$.

The limit doesn't exist.

But shouldn't after you get:

[e^(xy)-1]/[x^2+y^2] less than or equal to [e^(x^2 + y^2)-e^0]/[x^2+y^2 - 0]

BY paramatizing since the limit is between t = 1 and t =0,
you can do F(t) = f(tx,ty) = e^[(tx)^2+(ty)^2]

then lower bound is less than [F(1)-F(0)]/1

Or am I still mistaken and the limit doesn't exist?

Your first step is a bad idea- it eliminates all of the interesting variation in the function. Saying that $ |f(u)|\le c$ for some nonzero $ c$ doesn't tell you anything about whether $ f$ actually has a limit.

Work with the exact values, and try limits along different paths approaching the origin.

here is one way of looking at it: assume you are approaching $ (0,0)$ from the family of curves (i.e. lines) given by $ y\equal{}mx$

you have $ L\equal{}\lim_{(x,y)\to(0,0)}\;\dfrac{e^{xy}\minus{}1}{x^2\plus{}y^2}\;\equal{}\;\lim_{x\to 0}\;\dfrac{e^{m\,x^2}\minus{}1}{x^2\,(1\plus{}m^2)}$

apply l'hospital

$ L\equal{}\lim_{x\to 0}\;\dfrac{m\,e^{m\,x^2}}{1\plus{}m^2}\;\equal{}\;\frac{m}{1\plus{}m^2}$

since $ L$ is dependent on $ m$, that implies limit doesn't exist.

what it simply means is had the limit $ L$ existed, then it would be [i]independent[/i] of $ m$ $ \Rightarrow$ $ L$ would be the [i]same[/i] no matter how you approach $ (0,0)$.

Okay... I have another question...
I'm totally stuck on this limit:

Limit(x,y) -> (0,0) [e^(xy)-1]/sqrt[x^2+y^2]

How can I solve it? I can't prove it doesn't exist and can't find an upper bound either.... HELP!

WEll, I think this may be a way to do the problem I mentioned in the previous post. Can you tell me if this is right?

[e^(xy)-1]/sqrt[x^2+y^2] less than or equal to [e^(x^2+y^2)-1]/sqrt[x^2+y^2] which is less than or equal to:
Sum from n=1 (z^n), where z = x^2+y^2

Then it divides out with the bottom and get:
Sum from n=1 (z^(n-1/2)) which when evaluated at z=0 is 0.

Your idea is right, but there is no need to use infinite series here. You also need a lower bound, as in
\[ \frac{e^{\minus{}(x^2\plus{}y^2)}\minus{}1}{\sqrt{x^2\plus{}y^2}}\le
\frac{e^{xy}\minus{}1}{\sqrt{x^2\plus{}y^2}}\le \frac{e^{x^2\plus{}y^2}\minus{}1}{\sqrt{x^2\plus{}y^2}}\]
Rewrite this in terms of $ r\equal{}\sqrt{x^2\plus{}y^2}$:
\[ \frac{e^{\minus{}r^2}\minus{}1}{r}\le
\frac{e^{xy}\minus{}1}{\sqrt{x^2\plus{}y^2}}\le \frac{e^{r^2}\minus{}1}{r}\]
As $ r\to 0$, the limit on the right is the derivative of $ e^{r^2}$ at $ r\equal{}0$; the limit on the left is the derivative of $ e^{\minus{}r^2}$ at $ r\equal{}0$. Both derivatives are equal to $ 0$, so by the squeeze theorem the limit in the middle is also $ 0$.

Ah. I got it. YOu just didn't say to use l'hospital's rule. Thanks for the help :)

Actually, I did not use the rule - I used the definition of a derivative instead. Works the same here.


[quote="mr.danh"]Find all real numbers $ x$  satisfying:
$ sin x \plus{} sin 2x \plus{} sin 3x \equal{} 1$[/quote]

But $ \frac{\pi}{2}\minus{}x$ and $ \frac{\pi}{2}\plus{}x$ don't have sum $ \frac{\pi}{2}$ and difference $ \minus{}x$. Finding extraneous solutions just by using identities is a sign something went wrong somewhere. :wink:

[quote="santosguzella"][quote="scorpius119"]But $ \frac {\pi}{2} \minus{} x$ and $ \frac {\pi}{2} \plus{} x$ don't have sum $ \frac {\pi}{2}$ and difference $ \minus{} x$. Finding extraneous solutions just by using identities is a sign something went wrong somewhere. :wink:[/quote]

Where I made a mistake???.. thx! :maybe:[/quote]

So $ \sin \frac{\pi}{2}\minus{}\sin x\neq 2\sin \left( \frac{\pi}{2}\minus{}x\right) \cos \left( \frac{\pi}{2}\plus{}x\right)$.

[quote="scorpius119"][quote="santosguzella"][quote="scorpius119"]But $ \frac {\pi}{2} \minus{} x$ and $ \frac {\pi}{2} \plus{} x$ don't have sum $ \frac {\pi}{2}$ and difference $ \minus{} x$. Finding extraneous solutions just by using identities is a sign something went wrong somewhere. :wink:[/quote]

Where I made a mistake???.. thx! :maybe:[/quote]

So $ \sin \frac {\pi}{2} \minus{} \sin x\neq 2\sin \left( \frac {\pi}{2} \minus{} x\right) \cos \left( \frac {\pi}{2} \plus{} x\right)$.[/quote]

Your're right.. I will correct! Thx again! :blush:
$ \sin \frac {\pi}{2} \minus{} \sin x \equal{} 2\sin \left(\frac{ \frac {\pi}{2} \minus{} x}{2}\right) \cos \left(\frac{ \frac {\pi}{2} \plus{} x}{2}\right)$


why do no one solve it?
is it to hard?
or is it wrong?
 :(

yes ,it is wrong
i have a misstake on some where
please correct it :blush:

[quote="Fermat -Euler"][quote="eagerkill"]x,y,z>0 satisfy $\ zx^2+xy^2>yz^2;xy^2+yz^2>zx^2;yz^2+zx^2>xy^2$

:D[/quote]

[quote="IP"]Oh it is not true . You can take $x=y=z=0,1 \to$ :d[/quote]

 x ,y ,z are >0[/quote]

I means $x=y=z=10^{-1}$


inequalities
complex analysis
complex analysis unsolved

I don't have any clue how to prove this. Will someone help me? Thanks.
If $ f \in  \mathfrak{F_a}$ with $ a>0$, then show that for any positive integer $ n$, $ f^{(n)} \in \mathfrak{F_b}$ where $ 0 \leq b < a$

Hint: Let $ z$ be an element of the infinite strip with height $ b$. Let R be the maximum radius such that $ D_R(z)$ is a a subset of that strip (the open disk centered around $ z$ with radius R). Now apply the inequality: $ |f^{(n)}(z)|\leq \frac{n!}{R^n}sup_{\delta D}|f(z)|$ Combining this with what is given should yield the result.

Thank you for your help. But can you help me how to bound the supremum of $ \frac{1}{1\plus{}z^2}$ where $ z \in \delta D_R$. I think I'm almost done here if I can bound that term.


geometry
circumcircle
concurrency
Triangle
orthocenter
IMO Shortlist
geometry solved

[hide="Solution"]AH,BH,CH meets the circumcircle (O) at A',B'C'.
By figure, we guess $ D\equal{}OA'\cap BC, E\equal{}OB'\cap CA, F\equal{}OC'\cap AB$. The problem will be solved if AD,BE,CF are concurrent. And we make a use of Ceva's theorem.
* First way, $ \frac{BD}{CD}\equal{}\frac{sin2B}{sin2C}$
*Second way, see at [url]http://www.mathlinks.ro/viewtopic.php?search_id=1848850242&t=207305[/url][/hide]

We know that the orthocenter reflects over the sides of the triangle on the circumcircle. Therefore the minimal distance $ OD\plus{}HD$ equals $ R$. Obviously we can achieve this on all sides, so we assume that $ D,E,F$ are the intersection points between $ A',B',C'$ the reflections of $ H$ across $ BC,CA,AB$ respectively. All we have to prove is that $ AD$, $ BE$ and $ CF$ are concurrent.

In order to do that we need the ratios $ \dfrac {BD}{DC}$, $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$, and then we can apply Ceva's theorem.

We know that the triangle $ ABC$ is acute, so $ \angle BAH \equal{} 90^\circ\minus{} \angle B \equal{} \angle OAC$, therefore $ \angle HAO \equal{} |\angle A \minus{} 2(90^\circ \minus{}\angle B)| \equal{} |\angle B\minus{} \angle C|$. In particular this means that $ \angle OA'H \equal{} |\angle B\minus{}\angle C|$. Since $ \angle BA'A \equal{} \angle C$ and $ \angle AA'C \equal{} \angle B$, we have that $ \angle BA'D \equal{} \angle B$ and $ \angle DA'C \equal{} \angle C$.

By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get
\[ \dfrac {BD}{DC} \equal{} \dfrac { \sin B }{\sin C }.\]

Using the similar relationships for  $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$ we get that those three fractions multiply up to 1, and thus by Ceva's, the lines $ AD, BE$ and $ CF$ are concurrent.

[b]Lemma 1:[/b] If an ellipse is inscribed triangle $ABC$, tangent to $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively, then $AD$, $BE$, and $CF$ are concurrent.
[i]Proof:[/i] Scale the ellipse about its major axis so that it becomes a circle; here, $A'D'$, $B'E'$, and $C'F'$ concur at the Gergonne point of $A' B' C'$. Scaling preserves incidence, so $AD$, $BE$, and $CF$ concur.

[b]Lemma 2:[/b] Let $\ell$ be any line, and let $X$ and $Y$ be any points on the same side of $\ell$. Let $Z$ be the reflection of $X$ across $\ell$, and let $W$ be the intersection of $YZ$ and $\ell$. Then the ellipse with foci $X$ and $Y$ that passes through $W$ is tangent to $\ell$.
[i]Proof:[/i] Suppose that the ellipse intersected line $\ell$ at another point, $W'$. Then $XW' + YW' = YW' + ZW' > YZ$ by the triangle inequality, since $W'$, $Y$, and $Z$ are not collinear. On the other hand, by definition of ellipse, $XW' + YW' = XW + YW = ZW + YW = YZ$, so $YZ > YZ$, which is a contradiction.

By lemma 1, it is sufficient to show that there exists an ellipse with foci $O$ and $H$ that are tangent to the sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.

Reflect $H$ across $BC$, $CA$, and $AC$ to get $A'$, $B'$, and $C'$, respectively, and let $OA'$ hit $BC$ at $D$, $OB'$ hit $CA$ at $E$, and $OC'$ hit $AB$ at $F$. . $A',B',C'$ lie on the circumcircle of $ABC$, so $R = OA' = OB' = OC' = HD + DO = HE + EO = HF + FO$, where $R$ is the circumradius of $\triangle ABC$ (since $A', B', C'$ are the reflections of $H$ across the sides of the triangle.)

Consider the ellipse with foci $O$ and $H$ with a major axis of length $R$. By definition, the ellipse passes through $D$, $E$, and $F$. However, by lemma 2, it is tangent to sides $BC$, $CA$, and $AB$, so by lemma 1, we are done.

[hide="Some motivation"]The problem statement implies that $D$, $E$, and $F$ are intersection points of the ellipse with the sides of the triangle. If the ellipse is not tangent to a side, then it intersects it twice, which leaves some "confusion" as to which intersection point to choose. It is therefore natural to expect the ellipse to be tangent to the sides of the triangle. (Lemma 1 also suggests that this is true.)[/hide]

@Zhero: Instead of scaling, it is more elegant to just use Brainchon's theorem.

Cheers,

Rofler

Probably. I believe we could also just centrally project it to a circle as well (my first approach). :P I chose to scale because I felt it was the most elementary solution; anyone who knows anything about ellipses should be able to understand the solution I gave.

since $H$ and $O$ are two isogonal conjugate points, there exists an ellipse which these two points are its foci and its tangent to sides of the triangle.

[quote="Valentin Vornicu"]

By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get
\[ \dfrac {BD}{DC} \equal{} \dfrac { \sin B }{\sin C }.\]

[/quote]

By the law of sines for triangles $~$ $ \triangle{BA'D} $ $~$ and $~$ $ \triangle{DA'C} $ $~$ we get:
\[ \frac { BD } { \sin B } = \frac { A'D } { \sin \angle{DBA'} } \quad \wedge \quad \frac { DC } { \sin C } = \frac { A'D } { \sin \angle{DCA'} } \]
So for:
\[ \dfrac {BD}{DC} \equal{} \dfrac { \sin B }{\sin C }.\]
to be true, the following:
\[ \sin \angle{DBA'} = \sin \angle{DCA'} \]
should be true as well. I don't think that's correct. :maybe:

[quote="Virgil Nicula"]You are right. See [url=http://www.artofproblemsolving.com/Forum/blog.php?u=9882&b=53379][color=darkred][b]here[/b][/color][/url] [b]PP8[/b].[/quote]

Wow you had a math blog!!!  :w00t: I unfortunately can't read the entire problems. Only half of the picture is displayed. Do you know why? :(

[hide="Solution"]
Let $D',E',F'$ be the reflections of $H$ across $BC,AC,AB$ respectively. Let $D$ be the intersection of $OD'$ and $BC$. Define $E$ and $F$ similarly. Let $R$ be the radius of Circle $O$. It is well known that $D',E',F'$ lie on the circumcircle of $\triangle{ABC}$. Thus, $DD'=DH$, so $OD+DH=OD+DD'=OD'=R$. Likewise, $OE+EH=OF+FH=R$, so $D,E,F$ are the points we are looking for.

Let line $OD'$ intersect Circle $O$ again at $A'$. We have $\angle{A'AD'}=90$, so $AA'\perp AD'$. But $AD'\perp BC$, so $AA'||BC$. It follows that $AA'BC$ is an isosceles trapezoid, implying that $\widehat{A'B}=\widehat{AC}$ and $\widehat{AB}=\widehat{A'C}$. Hence, $\angle{DD'B}=B$ and $\angle{DD'C}=C$. But $ABD'C$ is cyclic, so $\angle{DBD'}=\angle{CAH}=90-C$ and $\angle{DCD'}=\angle{BAH}=90-B$.

By Law of Sines on $\triangle{BDD'}$ and ${CDD'}$, we have
\[ \frac{BD}{DC}=\frac{\frac{DD'\sin\angle{DD'B}}{\sin\angle{DBD'}}}{\frac{DD'\sin\angle{DD'C}}{\sin\angle{DCD'}}}=\frac{\sin B\cos B}{\sin C\cos C}=\frac{\sin2B}{\sin2C}\]
By similar reasoning, $CE/EA=\sin2C/\sin2A$ and $AF/FB=\sin2A/\sin2B$. As a result, by Ceva's Theorem,
\[\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{\sin2B}{\sin2C}\cdot\frac{\sin2C}{\sin2A}\cdot\frac{\sin2A}{\sin2B}=1,\]
so $AD,BE,CF$ are concurrent. // [/hide]

[hide="Sol"] Clearly, the condition implies that $D, E, F$ lie on an ellipse with focus $O, H$. Then, notice that for any conic that is tangent to the sides of the triangle, the point of tangency are concurrent with the opposite vertex (so see this, just use brianchon (on degenerate $AZBXCY$ is $X, Y, Z$ are points of tangency) or use a projective transformation taking the conic to a circle. Note that tangency is preserved for the latter since it is one-to-one). Now, since $O, H$ are isogonal conjugates, we know that the ellipse is indeed tangent to the sides of the triangle. Therefore, take $D, E, F$ as the point of tangency.[/hide]

[quote="orl"]Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.[/quote]

This problem also appeared in one of the Indian TSTs!!!

Take $H_a, H_b, H_c$ be the mirror images of $H$ about $BC, AC, $ and $AB$ respectively. They lie on the circumcircle of $\Delta ABC$. Now, take $\{D\} = OH_a \cap BC$, $\{E\} = OH_b \cap AC$ and $\{F\} = OH_c \cap AB$. Now, using Pool's Theorem, we get that $OD + DH = OE + EH = OF + FH$. Moreover, in this case the sum $OD+DH$ is minimal, so the points $D, E, F$ are situated on an ellipse of foci $O$ and $H$, having the constant sum equal to $R$ (the ray of the circumcircle of $\Delta ABC$), and which is tangent to the edges of the triangle. Now, this is just Gergonne's theorem, under a projective transformation.

Here is a bashing proof.

$A',B',C'$ are the reflection point of $H$ with respect to $BC,CA,AB$ respectively
$D,E,F$ are the intersection point of $OA',OB',OC'$ with $BC,CA,AB$ respectively

Let

$ S_A : = (b^2 + c^2 - a^2) [ b^2 (c^2 + a^2 - b^2 )^2 + c^2 (a^2 +b^2 - c^2 )^2 ] $
$ S_B : = (c^2 + a^2 - b^2 ) [ c^2 (a^2 + b^2 - c^2 )^2 + a^2 (b^2 +c^2 - a^2 )^2 ]$
$ S_C : = (a^2 + b^2 - c^2 ) [ a^2 (b^2 +c^2 -a^2)^2 + b^2 (c^2 +a^2 -b^2 )^2 ] $

 One can calculate the barycentric cooridnates of $D,E,,F$ respectively to be

$ D = (0 : \frac{1}{S_B} : \frac{1}{S_C} ) $
$ E = (\frac{1}{S_A} : 0 : \frac{1}{S_C} ) $
$ F = (\frac{1}{S_A} : \frac{1}{S_B} : 0 ) $

 It is straightforward to see that $AD,BE,CF$ are concurrent.

[hide=Solution]
Let $H_A$ be the reflection of $H$ about $BC$ and define $H_B, H_C$ similarly. Choose $D=OH_A \cap BC, E=OH_B \cap CA, F=OH_C \cap AB$. We claim that these are the $D, E, F$ we are looking for. By properties of reflections $OD+DH=OE+EH=OF+FH$.

Now, $\angle BOH_A=2(90-B)$ and similarly so by Ratio Lemma $\frac{BD}{CD}=\frac{2 \cos B \sin B}{2 \cos C \sin C}$ and similar cyclic variations for $\frac{CE}{AE}, \frac{AF}{BF}$. Thus Ceva yields $AD, BE, CF$ concur as desired.
[/hide]

[quote=zuss77]So, let's play with it...[/quote]

How did you do that?

[quote=orl]Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$.  Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.[/quote]
[hide=Solution]Let $H_a,H_b,H_c$ be reflection of $H$ about $BC,CA,AB$.Let [list][*]$D=OH_a\cap BC$ \\ [*]$E=OH_b\cap CA$ \\ [*] $F=OH_c\cap AB$[/list]
Clearly  $OD+DH=OE+EH=OF+FH=R$.Now Since $(O,H)$ are isogonal conjugate so there is an inconic tangent to $AB,BC,CA$ at $D,E,F$.Hence $AD,BE,CF$ concur at the perspector of this Conic hence we are done.
[/hide]

I don't think anyone has mentioned this yet, but [hide=note]the concurrency point is actually just the isotomic conjugate of the circumcenter[/hide]

we can draw an $elipse$ with focuses $O$ and $H$ that is tangent to the sides of $ABC$ in $D,E,F$ then from $brianchon's theorem AD,BE,CF$ concur. $\blacksquare$

When I solved this I thought of the following problem: Let $A,B$  be points in the plane and $l$ be a line in that plane. Find the locus of the point(s) $S$ such that $AS+BS$ is minimal.
This is solved by Euclid's lemma and reflecting $B$ or $A$ over $l$.

Let $R$ be the circumradius of $ABC$, $AH, BH, CH$ meet $(ABC)$ again at $H_a, H_b, H_c$ respectively, and $D = BC \cap OH_a$, $E = CA \cap OH_b$, $F = AB \cap OH_c$. By Orthocenter Reflections, it's clear that $$OD + DH, OE + EH, OF + FH = R.$$
It's well-known that $O$ and $H$ are isogonal conjugates. Hence, the Ratio Lemma and LoS implies $$\frac{BD}{DC} = \frac{H_aB}{H_aC} \cdot \frac{\sin BH_aD}{\sin CH_aD} = \frac{\sin H_aCB}{\sin H_aBC} \cdot \frac{\sin CH_aA}{\sin BH_aA} = \frac{\sin H_aAB}{\sin H_aAC} \cdot \frac{\sin CBA}{\sin BCA}$$ $$= \frac{\sin (90^{\circ} - B)}{\sin (90^{\circ} - C)} \cdot \frac{\sin B}{\sin C} = \frac{\cos B \sin B}{\cos C \sin C}.$$ Now, finding $\frac{CE}{EA}, \frac{AF}{FB}$ via analogous processes and applying Ceva's finishes easily. $\blacksquare$

[b]Remark:[/b] This solution is clearly motivated by the Orthocenter Reflection Lemma.

Step1 : Determine where are $D,E,F$.
Let $H1$ be reflection of $H$ across $BC$ and let $OH_1$ meet $BC$ at $D$. Use same trick to define $E$ and $F$ now note that:
$DH + DO = DO + DH_1 = OH_1$ , $EH + EO = EO + EH_2 = OH_2$ , $FH + FO = FO + FH_3 = OH_3$ and $OH_1 = OH_2 = OH_3 = R$.

Step2 : proving $AD,BE$ and $CF$ are concurrent.
By Ceva's Theorem we only need to prove $\frac{AF}{FB} . \frac{BD}{DC} . \frac{CE}{EA} = 1$.
$\frac{BD}{DC} = \frac{\sin{BCD}}{\sin{DBC}} . \frac{\sin{DH_1B}}{\sin{CH_1D}}$ and $\angle DH_1B = \angle HAO + \angle C = \angle B - \angle C + \angle C = \angle B$ and $\angle CH_1D = \angle B + \angle C - \angle B = \angle C$.
now if we multiple all we will have 1 so $AD,BE$ and $CF$ are concurrent.

Let $H_A,H_B,$ and $H_C$ be the reflections of $H$ over $\overline{BC},\overline{AC},$ and $\overline{AB},$ respectively. Let $D=\overline{OH_A}\cap\overline{BC}$ and define $E$ and $F$ similarly. Notice $$DO+DH=DO+DH_A=OH_A=r$$ where $r $ is the radius of $(ABC).$ Similarly, $EO+EH=FO+FH=r.$ We see $$\angle HAO=\angle A-2(90-\angle B)=\angle A+2\angle B-180.$$ Hence, by the Law of Sines, $$\frac{BD}{CD}=\frac{\sin\angle BH_AD\cdot DH_A/\sin\angle H_ABC}{\sin\angle CH_AD\cdot DH_A/\sin\angle H_ACB}=\frac{\sin(\angle C+\angle HAO)\cos\angle B}{\sin(\angle B-\angle HAO)\cos\angle C}=\frac{\sin\angle B\cos\angle B}{\sin\angle C\cos\angle C}$$ and Ceva finishes. $\square$

What is branchion's.

[img width=50]https://media.discordapp.net/attachments/925784397469331477/950082397293150208/Screen_Shot_2022-03-06_at_10.27.55_AM.png?width=1314&height=1169[/img]

Let $H_1,H_2,H_3$ be reflections of $H$ over $BC,AC,AB$ respectively. Let $OH_1,OH_2,OH_3$ intersect $BC,AC,AB$ at $E,F,D$ respectively. Let $r$ be the circumradius. Note that $(OD+DH,OE+EH,OF+FH)=(OD+DH_3,OE+EH_1,OF+FH_2)=(r,r,r).$

Finally, note that $D,F,E$ are collectively on an ellipse with foci $O,H.$ Since $D,E,F$ minimize the distance $OD+DH,OE+EH,OF+FH$ respectively, this ellipse must be tangent to $\triangle ABC'$s sides. Then we can do a scaling in the direction of $OH$ that makes this ellipse into a circle. This transformation does not disrupt intersections, so this ellipse is still tangent to $\triangle ABC'$s sides, and it preserves concurrentness. This ellipse is now an incircle so $AE,BF,CD$ are concurrent. I know I messed up point labeling so just substitute $D$ in for $E,$ $F$ in for $E$ and $D$ in for $F$ to get the result.

[hide=Solution]
First we will prove that there exists points $D,E,$ and $F$ that satisfy the conditions in the problem Reflecting $H$ about $BC,$ we find that $$OD+DH=OD+DH_A.$$ Suppose that $D$ lies on $OH_A.$ Thus, $$OD+DH=OH_A.$$ Definined $E,F$ similarly, we find that $$OD+OH=OE+EG=OF+FH=\text{Radius of (ABC)}.$$ Now, we will prove that such points $D,E,$ and $F$ will result in cevians $AD,BE,$ and $CF$ that are concurrent. From the Law of Sines on $\triangle{BOD}$ and $\triangle{COD},$ we find that $$\frac{\sin{\angle{BOD}}}{BD}=\frac{\sin{\angle{OBD}}}{OD}\text{ and }\frac{\sin{\angle{COD}}}{CD}=\frac{\sin{\angle{OCB}}}{OD}.$$ Since $OB=OC,$ we must have $\angle{OCB}=\angle{OBC}.$ This means that $$\frac{CD}{BD}=\frac{\sin{\angle{COD}}}{\sin{\angle{BOD}}}\cdot\frac{\sin{\angle{OBD}}}{\sin{\angle{OCB}}}=\frac{\sin{\angle{COD}}}{\sin{\angle{BOD}}}.$$ Now,$$\frac{\sin{\angle{COD}}}{\sin{\angle{BOD}}}=\frac{\sin{\angle{COH_A}}}{\sin{\angle{BOH_A}}}=\frac{\sin{(2\sin{\angle{H_AAC}})}}{\sin{(2\sin{\angle{H_AAB}})}}=$$$$\frac{\sin{(2(90-\angle{C}))}}{\sin{(2(90-\angle{B}))}}=\frac{\sin{(180-2\angle{C})}}{\sin{(180-2\angle{B})}}=\frac{\sin{(2\angle{B}})}{\sin{(2\sin{\angle{C}})}}.$$ Similarly, we find that $$\frac{BF}{FA}=\frac{\sin{(2\angle{B})}}{\sin{(2\angle{C})}}\text{ and }\frac{AE}{EC}=\frac{\sin{(2\angle{A})}}{\sin{(2\angle{C})}}.$$ Thus, $$\frac{CD}{BD}\cdot\frac{BF}{FA}\cdot\frac{AE}{EC}=\frac{\sin{(2\angle{C})}}{\sin{(2\angle{B})}}\cdot\frac{\sin{(2\angle{B})}}{\sin{(2\angle{A})}}\cdot\frac{\sin(2\angle{A})}{\sin{(2\angle{C})}}.$$ In conclusion, by Ceva's theorem, $AD,BE,$ and $CF$ are concurrent, as desired.
[/hide]

[hide=Storage]
Consider an inconic with focii as $O$ and $H$ (since circumcentre and orthocentre are isogonal conjugates).

For $DH+OD=R$, consider the segment joining reflection of $H$ in $BC$ and $O$.
$BC \cap OH_{BC}$ is the required point on $BC$, ie $D$

Now we finish with Brianchon's theorem.[/hide]

We claim that the conditions are satisfied when $D=OH_A \cap BC$ and define $E,F$ symmetrically, where $H_A$ is the reflection of $H$ over $BC$. By the reflecting the orthocenter lemma,  have that $OD+DH=OD+DH_A=R$,  where $R$ is the circumradius of $ABC$, so the first condition holds true. We now prove the concurrency using complex numbers. By the reflection formula, we get that $h_a=\frac{-bc}{a}$. Next using the intersection of line formulas we have $d=\frac{bc(b+c)}{a^2+bc}$. We have $\frac{BD}{DC} = \frac{\lambda}{1-\lambda}$ where $\lambda$ satisfies $\lambda b + (1-\lambda ) c = \frac{bc(b+c)}{a^2+bc}$. This gives $\lambda=\frac{c(b^2-a^2)}{(b-c)(a^2+bc)}$. Thus by Ceva's Theorem, the problem is equivalent to $$\prod_{\text{cyc}} \frac{\frac{c(b^2-a^2)}{(b-c)(a^2+bc)}}{1-\frac{c(b^2-a^2)}{(b-c)(a^2+bc)}} = \prod_{\text{cyc}} \frac{c}{b} \cdot \frac{b^2-a^2}{a^2-c^2} =1,$$ but this is obvious so we are done.

Reflect $H$ over the sides $BC,AC,AB$ of the triangle to get the points $H_A,H_B,H_C$, which must lie on $(ABC)$. Note that if we let $D$ be the intersection of lines $OH_A$ and $BC$, and similarly for $E$ and $F$, then the $OH+DH=OE+EH=OF+FH$ condition is satisfied.

    We now apply law of sines on $\triangle BH_AD$ and $\triangle CH_AD$ and then angle chase to find that $$\frac{BD}{DC}=\frac{\sin{\angle BH_AD}}{\sin{\angle CH_AD}}\cdot \frac{\sin{\angle H_ABD}}{\sin{\angle H_ACD}}=\frac{\sin{\angle B}}{\sin{\angle C}}\cdot \frac{90^\circ-\sin{\angle B}}{90^\circ-\sin{\angle C}}.$$ We get similar results for $\tfrac{CE}{EA}$ and $\tfrac{AF}{FB}$, so $\tfrac{BD}{DC}\cdot\tfrac{CE}{EA}\cdot \tfrac{AF}{FB}=1$, and we are done.

Let the [url=https://mathworld.wolfram.com/MacBeathInconic.html]MacBeath inconic[/url] touch $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at $D$, $E$, and $F$, respectively. By Brianchon on $AFBDCE$, lines $AD$, $BE$, and $CF$ concur.

Let $H_1,H_2,H_3$ be the reflections of $H$ over $BC,CA,AB$ respectively. Let $OH_1,OH_2,OH_3$ intersect $BC,CA,AB$ at $D,E,F$ respectively.\\

We claim that these points $D,E,F$ are the desired points in the problem.\\

Let $\angle A=\alpha,\angle B=\beta,\angle C=\gamma$.\\

Note that $\angle BAH_1=90-\beta$, so $\angle BOH_1=180-2\beta$. Since $OB=OH_1$, $\angle DH_1B=\beta$. Similarly, $\angle DH_1C=\gamma.$\\

Additionally, note that $\triangle ABH_1$ and $\triangle ACH_1$ have the same circumradius, so $$\frac{BH_1}{CH_1}=\frac{\sin(90-\beta)}{\sin(90-\gamma)}.$$ Then, by Ratio Lemma, $$\frac{BD}{CD}=\frac{BH_1}{CH_1}\cdot \frac{\sin \beta}{\sin\gamma}=\frac{\sin(90-\beta)\sin\beta}{\sin(90-\gamma)\sin\gamma}.$$ Taking the cyclic product of this over $\alpha,\beta,\gamma$ yields 1 as everything cancels, so by Ceva's, $AD,BE,CF$ concur so we are done.

We look for an obvious placement for D,E,F, because this condition is not so convenient. First reflect H to obtain H_a, H_b, H_c. Then the condition is that H_aD+DO=..., so we make this a straight line by placing D as the intersection of H_aO and BC, and cyclically, so that the sum is equal to the circumradius. Now employ Ceva's to get the desired result.
We can express these lengths as law of sines. We have <H_aBD=<H_aBC=<H_aAC=90-<C, and <DH_aB=<OH_aB=90-<H_aOB/2=90-H_aAB=<B. Then BD/DH_a (by law of sines) = sinB/cosC, and an analogous argument gives CD/DH_a=sinC/cosB. Dividing, we have BD/CD=sinBcosB/sinCcosC and cyclically. Then by Ceva's we're done $\blacksquare$

[hide=extra sol for storage]BD/R=sin<BOD/sin<BDO=sin(2<BAH_a)/sin<BDO=sin2B/sin<BDO and analogously for CD/R=sin2C/sin<CDO, dividing gives BD/CD=sin2B/sin2C and analogously, so we're done multiplying cyclically. Also note that 2sinBcosB/2sinCcosC=sin2B/sin2C so the earlier sol was indeed equivalent. [/hide]

inellipse!
[hide = solution]
Let $H_A$ be the reflection of $H$ about $BC$ and define $D = OH_A \cap BC$. Define $E, F$ analogously. Now observe that we obviously have $OD+DH = OH_A$, which is constant across similar quantities for $E, F$.  Therefore we need only show that $AD, BE, CF$ concur. For now, let $\mathcal E$ be the ellipse with focii at $O, H$ passing through $H_A, H_B, H_C$. As $\angle CDO = \angle BDH_A = \angle BDH$, it follows that $BC$ externally bisects $\angle ODH$. This implies that $BC$ is tangent to $\mathcal E$. From similar implications we infer that $\mathcal E$ is an inellipse of $\triangle ABC$. Hence by Brianchon on circumscribing (degenerate) hexagaon $AFBDCE$ we obtain that $AD, BE, CF$ concur.
[/hide]
[hide = remark]
The above proof works for any pair of isogonal conjugates $\{P, Q\}$ in place of $\{H, O\}$; since the only property of $H, O$ we used is that $O$ is equidistant from the reflections of $H$ about the sides of the triangle; a property which is true for any arbitrary pair of isogonal conjugates. Thus we obtain that for any pair of isogonal conjugates, there is an inellipse of $\triangle ABC$ with the two as focii. The converse can also be proven similarily.
[/hide]

Note that by the existence of the MacBeath inconic, there is an ellipse with foci at $O$ and $H$ inscribed in $\triangle  ABC$. Take $D, E, F$ to be the tangency points of the ellipse with sides $BC, AC, AB$, respectively. By Brianchon theorem, $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ concur. By ellipse properties $OD + DH = OE + EH = OF + FH$, so we are done.

To be fair, this is more or less cheating. However, proving that the MacBeath inconic exists can be done easily using phantom point + angle chase, but I guess I am too lazy to write it up.

If $H_A,H_B,H_C$ are the reflections of $H$ about $BC,CA,AB$, $(D,E,F):=(OH_A\cap BC,OH_B\cap CA,OH_C\cap AB)$ satisfy the length condition with a constant sum of $R_{(ABC)}$. We prove concurrency by Ceva's: by the Law of Sines,
\begin{align*}
BD=\sin\angle BOH_A\cdot\tfrac{OD}{\sin\angle OBC}&=OD\cdot\tfrac{\cos2B}{\cos A}\\
CD=\sin\angle COH_A\cdot\tfrac{OD}{\sin\angle OCB}&=OD\cdot\tfrac{\cos2C}{\cos A}.\\
\end{align*}
So $\tfrac{BD}{CD}=\tfrac{\cos2B}{\cos2C}$ and analogously $(\tfrac{CE}{AE},\tfrac{AF}{BF})=(\tfrac{\cos2C}{\cos2A},\tfrac{\cos2A}{\cos2B})$, whence multiplying gives $\tfrac{BD\cdot CE\cdot AF}{CD\cdot AE\cdot BF}=1$ as desired. $\square$

Let $A'$ be the reflection of $H$ across $BC$ and let $D$ be $OA'\cap BC$. Construct $E$ and $F$ similarly. Note that by orthocenter properties, we have that $A'\in (ABC)$, so we have that
\[OD+DH=OD+DA'=OA'=R,\]
and similarly, we also find that $OE+EH=OF+FH=R$ also. Now we must prove that the $AD$, $BE$, and $CF$ are concurrent. \\\\
Through angle chasing, we get that
\[\angle DHA'=\angle OAH=\angle BAC-\angle BAH-\angle OAC=a-(90-b)-(90-b)=180-(2b+a)=b-c,\]
by orthocenter properties. Additionally, we also get that $\angle BCH=90-a$, which gives that
\[\angle DHC=90-\angle DHA'-\angle DCH=c.\]
Now using Law of Sines, we get that
\[BD=DH*\frac{\sin b}{\cos c},\]
and
\[CD=DH*\frac{\sin c}{\cos b},\]
which gives that
\[\frac{BD}{CD}=\frac{\sin b\cos b}{\sin c\cos c}=\frac{2\sin b\cos b}{2\sin c\cos c}=\frac{\sin 2b}{\sin 2c}.\]
Similarly, we get that $\frac{CE}{EA}=\frac{\sin 2c}{\sin 2a}$ and $\frac{AF}{FB}=\frac{\sin 2a}{\sin 2b}$. This gives us that
\[\frac{BD}{CD}*\frac{CE}{EA}*\frac{AF}{FB}=1,\]
which means that $AD$, $BE$, and $CF$ are concurrent by Ceva's Theorem, finishing the problem.

Note that $D$, $E$, and $F$ must be the tangency points of $\triangle ABC$ and the Macbeath ellipse, since the ellipse has foci $O$ and $H$. Then Brianchon's finishes.

I feel actually guilty for writing this up because I didn't know the existence of this conic before searching it up lol
but the important thing is figuring out that the ellipse must have foci $O$ and $H$.

We claim finding the smallest possible value of the quantity $OD+DH$ works. It's a classic result that this point $D$ is just $OH_A \cap BC$, where $H_A$ is the reflection of $H$ over $BC$, so
\[OD+DH = OD+DH_A = R,\]

which is obviously cyclically symmetric. Ceva's gives confirms the final requirement, as
\[\prod \frac{BD}{DC} = \prod \frac{BH_A \cdot \sin \angle BH_AD}{CH_A \cdot \sin \angle CH_AD} = \prod \frac{\sin
B \cos B}{\sin C \cos C} = 1. \quad \blacksquare\]


Two roots of the real quartic $ x^4 \minus{} 18x^3 \plus{} ax^2 \plus{} 200x \minus{} 1984 \equal{} 0$ have product -32. Find $ a$.

[hide="Solution, please criticize"]Let the two roots be solutions to the equation $ x^2 \plus{} mx \minus{} 32$ for some complex m. other two roots of the quartic have product 62 by Vieta's Formula. We can then factor the quartic into two quadratics, $ (x^2 \plus{} mx \minus{} 32)(x^2 \plus{} nx \plus{} 62)$. Expanding this expression, we have that it is

$ x^4 \plus{} (m \plus{} n)x^3 \plus{} (mn \plus{} 30)x^2 \plus{} (62m \minus{} 32n)x \minus{} 1984$

We can set coefficients of corresponding $ x^i$ equal to each other. We thus have two equations,

$ m \plus{} n \equal{} 18$
$ 62m \minus{} 32n \equal{} 200$

Solving this system, we see that $ (m,n) \equal{} ( \minus{} 4, \minus{} 14)$.

$ mn \plus{} 30$, the $ x^2$ term in our quartic that was expressed in terms of m and n, corresponds to a. Substituting values for m and n, $ a \equal{} mn \plus{} 30 \equal{} 86$.[/hide]

It's a pretty solid solution.
The only thing that I would expand on would be your claim that you can factor the quartic into two quadratics: $ (x^{2}\plus{}mx\minus{}32)(x^{2}\plus{}nx\plus{}62)$.  I realize it's decently obvious, but since the foundation of  your proof lies on this statement you should certainly explain it a little more.

I think that is pretty obvious... the first quadratic has the roots that have the given product and the other one is the quadratic with the remaining roots. The constant term of the second quadratic is just found using the product of all four divided by the product of the first two

[quote="nittanylion"]It's a pretty solid solution.
The only thing that I would expand on would be your claim that you can factor the quartic into two quadratics: $ (x^{2} \plus{} mx \minus{} 32)(x^{2} \plus{} nx \plus{} 62)$.  I realize it's decently obvious, but since the foundation of  your proof lies on this statement you should certainly explain it a little more.[/quote]

Hmm, like for example actually go through explaining Vieta's?


1. A series is given such that
pi=3*3^(1/2)+24{(1/12-1/(5*2^5)-1/(28*2^7)-1/(72*2^9)-......}
how many term is needed to obtain the value of pi to sixteen decimal place?
2.Another series of pi
pi=6{1/2+1/(2*3*2^3)+1*3/(2*4*5*2^5)+.....}
how many term is needed to obtain the value of pi to ten decimal place?
3. What is the remainder when 1+x^1111+x^2222+.............x^8888+x^9999 is divided by 1+x^1+x^2+.............x^7+x^8+x^9?
4. Which one is greater 99^n +100^n or 101^n? given that n>50 and n is an integer.
5. When x^1951-1 is divided by x^4+x^3+2x^2+x+1 what is the co-efficient of x^14 ?

[quote="jli"][quote="13375P34K43V312"]These seem too easy for HSB. Could a mod kindly move this to intermediate?

And also please delete this post.[/quote]

intermediate is harder that HSB[/quote]

I think he meant that the question(s) are too hard for HSB.  :wink:

i hope he did-also this is way too hard not in latex

[quote="now a ranger"][quote="Ahmed Abdullah"]1. A series is given such that
$\pi=3*3^{(}1/2)+24{(1/12-1/(5*2^{5})-1/(28*2^{7})-1/(72*2^{9})-......}$
how many term is needed to obtain the value of pi to sixteen decimal place?
2.Another series of pi
$\pi=6{1/2+1/(2*3*2^{3})+1*3/(2*4*5*2^{5})+.....}$
how many term is needed to obtain the value of pi to ten decimal place?
3. What is the remainder when $1+x^{1111}+x^{2222}+.............x^{8888}+x^{9999}is divided by 1+x^{1}+x^{2}+.............x^{7}+x^{8}+x^{9}?$
4. Which one is greater $99^{n}+100^{n}or 101^{n}?$ given that $n>50$ and $n$ is an integer.
5. When $x^{1951}$$-1$ $\div$ $x^{4}+x^{3}+2x^{2}+x+1$ what is the co-efficient of $x^{14}$ ?[/quote][/quote]
[hide="4"]
101^n because 9^n+10^n<11^n
or we could think
x^n+(x+1)^n<(x+2)^n[/hide]
put = instead of<

If a problem needs to be moved, one post is plenty enough to tell me.

Moved to Intermediate.

It's generally not advisable to post multiple problems in one post.

[quote="funcia"]9^n+10^n<11^n[/quote]

False for $n < 5$.  Please confirm your conjectures numerically before you post them.

[hide="3"] $\bmod x^{10}-1$ it is clear that the remainder is $\boxed{0}$. [/hide]
[hide="4"] For $n \ge 50$ we have

$(100+1)^{n}-(100-1)^{n}> 2{50 \choose 1}100^{n-1}= 100^{n}$

So clearly $101^{n}> 100^{n}+99^{n}$.  [/hide]
[hide="5"] There is no coefficient of $x^{14}$ since the result upon division by a quartic must be cubic or smaller.  What did you mean by this problem?  [/hide]

t0rajir0u, he meant the coefficient of $x^{14}$ in the quotient upon dividing $x^{1951}-1$ by $x^{4}+x^{3}+2x^{2}+x+1$


integration
calculus
calculus computations

The title of your post should be instead "Not hard but maybe not nice". This is not difficult to compute, its just quite boring.

I'm what's called a lazy person.
Take $ f(x)\equal{}\frac{1}{1\minus{}x}\equal{}1\plus{}x\plus{}x^2\plus{}x^3\plus{}\ldots$ \[ \int f(\minus{}x^8)\text{ d}x\equal{}\int (1\minus{}x^8\plus{}\plus{}x^{16}\plus{}\ldots\plus{}(\minus{}1)^nx^{8n})\text{ d}x\equal{}C\plus{}\sum^{\infty}_{k\equal{}1} (\minus{}1)^n\frac{x^{8k\plus{}1}}{8k\plus{}1}\] PNC the lims of integration.. \[ \displaystyle \sum^{\infty}_{k\equal{}0} (\minus{}1)^k\frac{1}{8k\plus{}1}\equal{}\text{answer}\]
I'm quitting here. Gonna go out. Have fun evaluating that  :)


AMC USA(J)MO USAMO geometry calculus probability vector

I would have to say that almost every homework assignment I get in my high school IS a waste of time, and I qualified for USAMO freshman year...

Mostly because all my homework assignments are repetition to such excess that doing more than 5 of them on a problem set of about 20 is so tedious that you can't believe you are sitting there doing it.

Therefore I tend to not do my homework. However, I sitll get A's in my classes, even with a 3-4 point deduction for not doing homework. I think homework should be completely optional. If you don't do it and you fail the test... thats your problem. But if you feel you have a mastery over the topic area, get 100+ on the test, and don't do your homework because it IS a waste of your time which could be spent on more constructive studies, you should not be penalized.

(After edit by moderator:) This thread was split off from an earlier thread that was moved to this College Forum from the AMC Forum. Now I'll try to spice up the discussion a little by adding a poll. The original text of this message is intact as originally posted.

I [i][b]DEFINITELY[/b][/i] agree with you about that.  Sadly, I get more points off for not doing my homework than you do though, but homework should be optional like you said.  I mean, instead of grading a person on their homework, they could just even offer a homework quiz at the least.

I disagree that homework should be optional.

First of all, if homework were optional, practically no one would do it, and poor students would almost certainly not do the harder problems.  At least in my experience, students will blaze through the easy problems, then sort of fiddle with the harder problems, and maybe ask someone else for help.  If homework were optional, these people probably wouldn't do any of the problems, and if they did, they wouldn't bother trying the harder ones (say, they'd give up after about 30 seconds).  As a result, they wouldn't be capable of passing quizzes or tests.

You can't exactly say "oh, they don't matter" because that's the whole point of school: people should be educated.  The sad fact that many teachers aren't very effective in teaching means that if you were to merely assign quizzes, almost everyone would fail because there is virtually no lesson and no practice problems.   If students are forced to do homework, they might at least do some work.

Second of all, grades are partially based on effort: if you won't put in the 20 minutes each day to show that you care about the topic at hand, why should the teacher consider you an "outstanding" student? (Recall that an "A" is defined loosely as "outstanding".)  Part of being an "outstanding" student is working hard, regardless of whether or not it's all busywork.

I also disagree with the statement that bright students should be kept out of high school as much as possible.  Although I almost certainly represent the minority of students, since I attend a private school, I think that participating in school is an important part of life, and colleges will look for that.  Besides, school should be easy for a bright student, who will still have time to learn other things, right?

If I were homeschooled, I would most likely focus on math and science, since these are my main interests.  By being forced to take other subjects - English, History, Spanish, Art, Drama, Music, P.E., etc. - I think that my interests have been broadened.  For instance, I've taken more of an interest in literature thanks to my English teacher, who is excellent.

Just some thoughts.

Try to look at someone else like me.  I already understand the homework and how to do it, so why bother doing it?  Homework should be for those people that are [b][i]struggling [/i][/b]with it.

[quote="MysticTerminator"]ahem. school is easy but it's still a huge waste of time.[/quote]

Do you think you would be the same person, with the same abilities, if you never went to school? The thing about school is that most people learn without really knowing they're learning. But if you take a minute to look at all the things you you've learned through your education - you would be amazed.

Long division? That's something people don't just pick up on their own. Chemistry? Molecular Biology? Poetry? Foreign Languages?.. These are things that most people probably wouldn't learn anything about if they didn't go to school. Certainly not all of these subjects.

The problem with school is that as a teacher I can't just teach to the pace of the smartest individuals in the class. I wouldn't be getting the best result that way. I have to teach somewhere towards the slowest students so that everyone will learn. That's why it's good to try group students as close in ability as possible though.

But the purpose of homework isn't just to force students to spend some time with the material. Its also to demonstrate discipline and responsibility when they're asked to do something - whether they like the task or not. I required all students to do homework. If they already understand, then it shouldn't take that long. And even if they do, the repetition should reinforce and make the subject second nature to them. It's just like being an athlete - even if you're the best on the team you can still get something from routine practice.

I always try to throw in a few exercises that are out of the ordinary though, to keep the homework interesting. Perhaps something about in interesting application or new idea, or about something that they probably haven't considered before. But some of the questions just have to be to show me that they understand the material.

[quote="mathfanatic"]I disagree that homework should be optional.

First of all, if homework were optional, practically no one would do it, and poor students would almost certainly not do the harder problems.  At least in my experience, students will blaze through the easy problems, then sort of fiddle with the harder problems, and maybe ask someone else for help.  If homework were optional, these people probably wouldn't do any of the problems, and if they did, they wouldn't bother trying the harder ones (say, they'd give up after about 30 seconds).  As a result, they wouldn't be capable of passing quizzes or tests.
[/quote]

But that is their own fault and they should take responsibility for their actions. If the teacher isn't motivating them enough to want to do their homework, the teacher probably isn't that good either.

[quote]
You can't exactly say "oh, they don't matter" because that's the whole point of school: people should be educated.  The sad fact that many teachers aren't very effective in teaching means that if you were to merely assign quizzes, almost everyone would fail because there is virtually no lesson and no practice problems.   If students are forced to do homework, they might at least do some work.
[/quote]

Actually, a lot of people just copy the answers down if they don't want to do it. That's never going to change. It just makes it a little annoying to get out of doing your work, but still much easier than doing it.

[quote]
Second of all, grades are partially based on effort: if you won't put in the 20 minutes each day to show that you care about the topic at hand, why should the teacher consider you an "outstanding" student? (Recall that an "A" is defined loosely as "outstanding".)  Part of being an "outstanding" student is working hard, regardless of whether or not it's all busywork.
[/quote]

Well if it were limited to 20 minutes, I would be fine with that. But usually I get about an hour and a half of busywork, and thats not combined either. Thats not good. And a good student doesn't have to work hard, he or she has to work efficiently. I think part of learning to work is to learn what things are most important, worthy of your time, and constructive, and to do these things first and foremost.

[quote]
I also disagree with the statement that bright students should be kept out of high school as much as possible.  Although I almost certainly represent the minority of students, since I attend a private school, I think that participating in school is an important part of life, and colleges will look for that.  Besides, school should be easy for a bright student, who will still have time to learn other things, right?
[/quote]

Though it remains easy... if you do nothing but sit there for hours on end bored waiting just to get out of school because you know everything thats going on, just to have an hour and a half of busywork, you are wasting a lot of time that could be spent on more constructive endeavors. Any waste of time is a bad one.

Thing is, they don't want to group bright kids with bright kids. My school is a magnet school so its kind of weird. But we have ONE top math section where we do BC calc w/ proofs. The section has about 16-20 students. It SHOULD have about 6, to make it a constructive class where everyone is top top and can follow a highly accelerated theoretical abstract curriculum with proofs. But instead we have people asking the same question for about 2 weeks because they can't follow the class and don't want to take the time to work on their own. Now thats a waste of my time once again. And the class gets "dumbed down" to going from proof-based to endless drills, and absurd amounts of calculator work. That just makes me disappointed.

[quote="gauss202"]

I do try to always throw in a few exercises that are out of the ordinary though, to keep the homework interesting. Perhaps something about in interesting application or new idea, or about something that they haven't considered before.[/quote]

Well at least you do that. If my teachers did that I would have SOMETHING new to work on

My Calculus teacher uses the following, quite effective homework system: he checks the homework, but no points are deducted from a student's average if his/her test average is above a 90. For students with test averages 85-89, they lose a half-point for every missing homework assignment, students with test averages 80-84 lose one point for every missing assignment, etc. Thus homework is optional for the students who know what they're doing, but those with poorer performances are punished for not taking the time to improve their understanding of the subject.

As to the original topic- there are definitely USAMO quals who were rejected from colleges, especially the top colleges. I've heard of one student who, in addition to being a multiple-time USAMO qualifier was also an Intel semifinalist and had performed well in the other science olympiads- he didn't get into his first choice college.

Nukular: What kind of math education did you have to qualify for USAMO as a freshman?

Optional homework does not work. I know because our AP physics homework is completely optional. The class is far too large; usually it's just a senior course, but the teacher (who is also the geometry math team coach) allowed sophomores who had been on the math team to take AP Physics. So the class this year has about 15 sophomores and 10 seniors, and honestly, only me and one other sophomore care. There are only about 4 sophomores who do their homework, and the ones who don't fail...horribly. We take tests about every 3 weeks, and the lowest grade on the last test was a 2...yes, a 2 out of 100. But this kid is not stupid, he just doesn't do his homework. And it's not just him, it's a trend. The teacher gives you lots of freedom and is really good at explaining things, but the kids who don't do their homework do awful in his class.

I agree with you all; most of school is really easy and some is boring. I'm taking algebra 2 this year, but I already completed the book it 8th grade so everything is a review....and the school won't let me move up to precal (even though i make 100 on everything in the class). I used to be a big history nerd, so I already know most of that class. And we could go on and on. But school and homework are absolutely necessary for a well-rounded education. And don't forget, most of the people in this forum are very bright and are at the top of their class, so some of school obviously seems like a waste of time because we like to learn outside of the classroom.

I know most effort is outside of school but Nukular, what math class were you taking freshman year?  I'm just wondering.

Well its called "Advanced Analysis II" but it's really just a precalculus course.

I'm in 8th grade, but i'm takin precal right now
but what you guys are discussing in some of the more advanced problems are way over my head, how should i study for those kind of tests?

I opted out of Algebra 1 because I was in som accerlated program in 6th grade and  2 because I did every problem in the Algebra 2 book over the summer.

(After edit:) I was going to start a new thread here about what approach to take to studying secondary math to maximize

1) learning math,

and

2) chance of college/university admission,

and now that [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9596]the old thread[/url] has had some recent posts on that issue, I have split off from it the posts that pertain most to the issue of what courses to take in high school, how hard to study in high school, how to go beyond the high school curriculum, and other how-to-study-math-in-high-school posts. Feel free to deepen and expand the discussion here in the new thread.

As I noted in a reply to the original thread, I homeschool my children precisely because there is NO school in my state that has the sustained commitment to high-level development of math ability in young people that would meet my oldest son's (and possibly his younger sibling's) needs. There are some good schools here in MN, and some good mathletes who come out of them, but those with the best quality mostly have high expense, and those with low expense mostly don't have minimally adequate quality. The majority of the young people who post on this AoPS Web site are in average schools, with only meager opportunities to go beyond the school curriculum in math unless they make those opportunities themselves. Let's talk about what forms of self-help we can engage in so that mediocre expectations from our local public school don't hose our chances for challenging study in higher education.

The best advice I could give someone in preparing for college mathematics is to learn how to read from the textbook. In (most) secondary schools, math classes meet 5 times a week for about 9 months. In college, the same amount of material will be covered in 3 classes a week for about 8 months (if not less). So what that means is that you have to do a lot of learning on your own. When I got to college I noticed that a lot of students had a hard time doing this. So even if you have a good teacher, read you book too. It can often be a really valuable resource.

As for the courses people should take, there's really no mystery. The basic courses of Geometry, Algebra, Trigonometry/Pre-Cal, and Calculus are good. I don't think enough time is spent on Probability and Combinatorics, at least not where I live - so those might have to be studied on your own.

I would encourage more students to take Multivariable Calculus and Linear Algebra if they get the chance. I know this is a very controvertial idea, as many people feel that too many students are rushing to calculus these days. In general I agree with that, but for students who have solid foundations in algebra, geometry, and trig. I think it's a good idea to have some exposure to these subjects before college. Because it makes other subjects (like Physics, Chemistry, and Economics) so much easier in college if you have seem MVC and LA before. But unfortunately many schools don't offer those courses at the high school level, or even if they do they're so watered down that you'll just have to take them again in college.

Good Books that I know of:

Trigonometry/Pre-Cal - [u]Advanced Mathematics: A Precalculus Course[/u] by Brown and Robbins
Calculus - [u]Calculus, 7th edition[/u] by Larson, Hostetler, and Edwards + a Barron's AP Prep Book
Multivariable Calculus - [u]Vector Calculus[/u] by Marsden and Tromba
Linear Algebra - [u]Introduction to Linear Algebra[/u] by Strang

[quote="gauss202"]The best advice I could give someone in prepare for college mathematics is to learn how to read from the textbook. In (most) secondary schools, math classes meet 5 times a week for about 9 months. In college, the same amount of material will be covered in 3 classes a week for about 8 months (if not less). So what that means is that you have to do a lot of learning on your own. When I got to college I noticed that a lot of students had a hard time doing that. So even if you have a good teacher, read you book too. They are often really valuable resources.[/quote]

Wow, that is so true.  I took a course at NYU senior year, and did it ever kick my butt!  (It only met once a week, too, so it was even worse.)

I get homework points as long as the assignment is done, it doesn't even need to be  right.

Same here.

I like the setup in our A.P. Physics class. Homework is optional and non-punitive, but helps your grade (sometimes a lot) if you do it. However, half of the class is failing, and half of the class has solid A's. I wish they would just kick out the failing people.

yeah...a lot of people in my honors/ap classes are failing...they shouldn't even be there. I wonder how they managed to get in the course in the first place...

[quote="hikari"]yeah...a lot of people in my honors/ap classes are failing...they shouldn't even be there.[/quote]

So they bomb the tests? Do you think that slows down the pace of the course for the more able students, or do the teachers still keep up the level of the class?

It depends on the class. Some teachers slow down the pace while others keep the pace the same. And yes, I think having people in harder classes that can't really handle it keeps more abled students from learning what they want. Honors/AP classes are for people who can learn harder things faster and for people who really want to be there.

In my physics class, homework is usually about 30%, and if you don't have it in class when he collects it, it's a 0, even say, if you did it but left it in your locker... and 0 points if you bring it in the next day.  I think that's a little excessive though.  And its not graded on correctness or anything, just if you have physics stuff written down.


A truth table is always a valid way to do these proofs, but here's more intuitive ways of seeing it.

$ (A\Rightarrow C )\wedge (B\Rightarrow C)$ is equivalent to $ (A\vee B)\Rightarrow C$. But we have $ A\vee B$. By the properties of $ \Rightarrow$ (modus ponens), we can then conclude $ C$.

I can't figure out what the second one means. What is $ J\minus{}C$ supposed to mean? And what's the order of operations after that? Does the "and" go with the statement before the first arrow or the second? Latex is helpful, by the way.

MellowMelon, thanks. I have trouble with Latex. I tried to follow the steps, but i wasn't able to download this software. I'm missing something, but i can't figure out what it is. Pls. help.

That was my problem for a while too...until I realized you don't need any software, just enclose the Latex code with dollar signs...

for example, to write 2+2=4 in latex, just write $ \$2\plus{}2\equal{}4\$$

For more complicated things, you can look in the AOPS wiki. I don't know about drawing diagrams though...

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php

then click on symbols or commands for complicated symbols etc. like integration

and all you need is question mark


Can you find intersection points for vectors?

In other words,
If there are vectors $w,x,y,z$ to points $W,X,Y,Z$ and assuming that $WY$ and $XZ$ intersect at $A$, can you express the vector to $A$ in terms of $w,x,y,z$?

[hide="The ugly duckling"] Let $w = (a, b), x = (c, d), y = (e, f), z = (g, h)$.  Then the point $A$ is the intersection of the lines

$y-b = \frac{f-b}{e-a}(x-a)$
$y-d = \frac{h-d}{g-c}(x-c)$

Which is, of course, incredibly ugly. [/hide]


[b][color=darkred]A)[/color][/b]
Εστω $A=2a^3-6a^2+11$
και    $B=3a^3-9a^2+17$  οπου $a \in \mathbb{N}$

Aν Μ το ΕΚΠ των Α,Β                [u]ΝΔΟ[/u]

                                 \[ \frac{2}{A}-\frac{3}{B}=\frac{1}{M}  \]

[b][color=darkred]B)[/color][/b]
Nα προσδιορισετε τα στοιχεια του συνολου $M=\{x \in Q \mid x \ne -3 ,\ \frac{x^2+2x+9}{x+3} \in \mathbb{Z}\}$

 :)

Να μια λύση για την  1). Βρίσκουμε ότι (Α,Β)=(33,34)=1 .
Αντικαθιστούμε σε αυτή που θέλουμε να αποδείξουμε και βρίσκουμε 0=0 ισχύει .

Να ένα πλάνο λύσης για την  2)
Θέτουμε $x=\frac{k}{l}$ με $(k,l)=1$  παίρνουμε ότι πρέπει ο αριθμός

$Α=\frac{k^2+2kl+9l^2}{l(k+3l)}$ να είναι ακέραιος.
Είναι $Α=\frac{k^2+6kl-4kl+9l^2}{l(k+3l)}=\frac{(k+3l)^2-4kl}{l(k+3l)}$  ακέραιος

κτλ

Ένας άλλος τρόπος για το Β είναι ο παρακάτω:

$\frac{x^2+2x+9}{x+3}=k \in \mathbb{Z}$ άρα η διακρίνουσα του τριώνυμου

$x^2+(2-k)x+9-k$ πρέπει να είναι τέλειο τετράγωνο $D=k^2+8k-32 = m^2 , m\in \mathbb{Z}$.

Τώρα βρίσκουμε τα $k,m$ από τη σχέση $(k+4-m)(k+4+m)=48$.

Άν έκανα σωστά τις πράξεις $x \in \{-15,-9,-7,-6,-5,-2,-1,0,1,3,9\}$  :?


yeah... haven't really used anything much, so i was wondering how to add contributers to my blog (check it out by clicking on Mr. Green in my signature!). its kinda smallish, so i was wondering if i could have more people than me to contribute to the blog.


function
logarithms
algebra
domain
complex analysis
complex analysis unsolved

Let $C_1=\{z=e^{i\phi}, 0\leq \phi \leq \pi \}$,  $C_1=\{z=e^{i\phi}, \pi \leq \phi \leq 2\pi \}$- the upper and lower semicirlces of the unit complex circle. Let $f_1(z)$ be any regular branch of the function $\log \frac{1-z}{1+z}$ in $\mathbb{\bar{\mathbb{C}}} \setminus C_1$, and $f_2(z)$ be any regular branch of the function $\log \frac{1+z}{1-z}$ in $\mathbb{\bar{\mathbb{C}}} \setminus C_2$. Is it true that $f_1(z)=f_2(z)$ or $f_1(z)=-f_2(z)$ for $z \in \mathbb{C} \setminus C_1 \cup C_2$.

I'm having difficulty with understanding regular branches. Maybe you could give some kind of relation between $f_1(z)$ and $f_2(z)$. It just seemed that one of the $f_1(z)=f_2(z)$,$f_1(z)=-f_2(z)$ should be true because of the relation in the usual sense of real logarithm that $\log \frac{1-z}{1+z} =- \log \frac{1+z}{1-z}$, but i don't know what happens in this case.

Thanks

Well- we know that $\exp(f_1(z)+f_2(z))=1$, so $f_1(z)+f_2(z)=2\pi ig(z)$, where $g$ is an integer-valued continuous function. This makes $g$ constant on each component of its domain, but there are two components, and the values will be different.

There's no way to have $f_1=f_2$ except at isolated points.

[quote="jmerry"]Well- we know that $\exp(f_{1}(z)+f_{2}(z))=1$, so $f_{1}(z)+f_{2}(z)=2\pi ig(z)$, where $g$ is an integer-valued continuous function. This makes $g$ constant on each component of its domain, but there are two components, and the values will be different.
[/quote]

Sorry, for reviving this old thread, but i have a question.
I must be misunderstanding something, but can't we take $f_{1}$ and $f_{2}$ to be such regular branches that $f_{1}(0) = 0, f_{2}(0) = 0$ and then $f_{1}(z) =-f_{2}(z)$ for all $z$ and $g(z) = 0$ $\forall z \in \mathbb{C}\setminus C_{1},C_{2}$.
Jmerry, did you really mean that $g$ must be constant on each of components with two [b]different[/b] values or it can be constant, say, $0$ in the whole complex plane?

The transformation $w=\frac{1-z}{1+z}$ turns $\mathbb{C}\setminus C_{j}$ into $\mathbb{C}\setminus L_{j}$, where $L_{j}$ is a halfline connecting $0$ to $\infty$. Also, $f_{1}(w)=\log w$ and $f_{2}(w)=-\log w$. Now, there are multiple branches of the logarithm, so in general $f_{1}(w)+f_{2}(w)=2\pi i m$ for some integer $m$. But if you know that $f_{1}(w)+f_{2}(w)=0$ at some point, then you know $m=0$ (in the connected component of that point).

(Edit: you can have two different constants $m$ in two components. For simplicity consider $\arg w$ and take its branches with $\arg 1=0$. Then $\arg (-1)=\pi$ for one branch and $\arg (-1)=-\pi$ for the other.

When we cross $C_{1}$, $f_{1}$ changes by about $2\pi i$ and $f_{2}$ changes by a tiny amount. The sum $f_{1}+f_{2}$ must change by $2\pi i$ when crossing the unit circle. and the two values of $f_{1}+f_{2}$ are different.


group theory
abstract algebra
induction
set theory
superior algebra
superior algebra unsolved

I have been struggling with group presentations for quite some times. I understand the definition of a group presentation but I am just so bad at doing problems involving presentations. Anyway, please help me with these:

1) Let F be the free group on {a,b} and N the normal subgroup generated by $ \{a^4,a^2 b^{ \minus{} 2},abab^{ \minus{} 1}\}$. Show that every element of F/N is of the form $ a^i b^j N$ with $ 0 \leq i \leq 3$ and $ j \equal{} 0,1$.

2) Prove that the cyclic group of order 6 is the group defined by the generators a,b and relations $ a^2 \equal{} b^3 \equal{} a^{ \minus{} 1}b^{ \minus{} 1} ab \equal{} e$.

Attempted solutions

1) I could do the rest if we could somehow show that all elements were of the form $ a^n b^m$ with $ m,n \in \mathbb{Z}$. But without knowing that, I don't even see how to show the group is finite...

2) So, the way to do this is to find an explicit isomorphism $ \phi$ from the presented group to $ Z_6$ right? So I think we want $ \phi(a) \equal{} 3$ and $ \phi(b) \equal{} 2$ because of the first two relations. Because a and b generate the group, that means the homomorphism is completely specified, right? I think I can prove this is onto since $ ab^ {\minus{} 1}$ goes to 1 so $ (ab^{ \minus{} 1})^n$ must go to n mod 6. But then how do I show injectivity...show the kernel is trivial? How do I use the commutativity ab = ba here... ?

1) Actually, you don't need to show that every element is of the form $ a^n b^m$, just that the group is generated by these elements.  So if you can show that they form a group, you're done. (you can think of a group generated by a set as the smallest possible group that contains that set, so if it's already a group...)

2) This is a little tricky.  Usually a really good first step for a problem like this is to make sure that the result is [i]intuitively[/i] obvious to you.  Why is the problem false without $ ab \equal{} ba$, for example?

Well, notice that(like the first problem), each element of the presented group can be written $ a^i b^j$ with $ 0\leq i \leq 1$, $ 0\leq j\leq 2$.  If $ \phi(a^i b^j) \equal{} 0$, then $ 3i \plus{} 2j \equal{} 0$, and you can convince yourself that this means $ i \equal{} j \equal{} 0$.

An interesting question is: how do you know that this homomorphism $ \phi$ exists?  It would be impossible to write down such a map, for example, if the relations implied that $ a\equal{}b$, for example.  I wouldn't worry too much about it, but there is a deep idea at work here. (the universal property of free groups)

1) Sorry, I don't know what you mean. The problem wants me to find an upper limit on the number of elements in the group F/N. When you say "the group" in the first sentence, you are referring to F/N not N right? Why would showing that F/N is generated by the elements $ a^n b^m$ help me at all? First, isn't that obvious because we could take n = 1, m = 0 and then n=0, m = 1 to get the generators of the original free group... Second, I said I could finish the problem knowing that ALL of the elements in F/N were of the form a^n b^mN with n,m in Z not only that the generators had that form...

2) Let's deal with 1) first.

BTW, thanks for helping me.

1) Sorry, I believe that I read what you wrote in reverse. (so I only solved the part that you solved already)

The other part is to show that any product of elements $ a^n b^m a^{n'} b^{m'}$ is of the form $ a^{n''}b^{n''}$.  You can do this using induction and the commutativity relation.  Once you know that, you are done. (this is because you have constructed a group that generates $ F/N$, so it must equal $ F/N$)

1. I see! $ a^n b^m a^{n'} b^{m'} \equal{} a^{n \minus{} 1} b^{m' \minus{} 1} a^{n' \minus{} 1} b^{m' \plus{} 1}$ and if we keep doing that eventually we will get to $ a^{n''} b^{m''}$. Then we just use $ a^4N \equal{} N$ and $ b^2N \equal{} a^2N$

2.
[quote="LydianRain"]
This is a little tricky.  Usually a really good first step for a problem like this is to make sure that the result is [i]intuitively[/i] obvious to you.  Why is the problem false without $ ab \equal{} ba$, for example?
[/quote]
Because its clear that the other relations do not imply that relation and we need abN = baN to hold true since Z_6 is cyclic hence abelian.

[quote="LydianRain"]
Well, notice that(like the first problem), each element of the presented group can be written $ a^i b^j$ with $ 0\leq i \leq 1$, $ 0\leq j\leq 2$.  If $ \phi(a^i b^j) \equal{} 0$, then $ 3i \plus{} 2j \equal{} 0$,
[/quote]

I see!

[quote="LydianRain"]
and you can convince yourself that this means $ i \equal{} j \equal{} 0$.
[/quote]
Hmm...you mean a^i = b^j = 1 in the presented group, right? That is, because 3i = -2j implies that i is congruent to 0 mod 2 and j is congruent to 0 mod 3 (this is just number theory).

[quote]
An interesting question is: how do you know that this homomorphism $ \phi$ exists?  It would be impossible to write down such a map, for example, if the relations implied that $ a \equal{} b$, for example.  I wouldn't worry too much about it, but there is a deep idea at work here. (the universal property of free groups)[/quote]

That is something else I am struggling to understand. Is this what the "universal property of free groups" means to you:

If $ F$ is a free group on the set $ X$ and $ \iota: X \to F$ the inclusion map, then if $ G$ is any group and $ f: X \to G$ a map of sets, then there exists a unique homomorphism $ \bar{f}: F \to G$ such that $ \bar{f} \iota \equal{} f$. (Hungerford Theorem 9.2)

Are you saying that I used this unwittingly in my proof? Can you elaborate?

Oh by the way, it occurred to me that a possibly better way to show that $ \phi$ is an isomorphism is to explicitly construct the inverse map.

[quote]Are you saying that I used this unwittingly in my proof? Can you elaborate?[/quote]
Yes, we're talking about the same universal property.  I think that you used this property implicitly by assuming that $ \phi$ exists.  Specifically, here is how I would write the proof, if I were being very rigorous:

1) Define a map of sets $ \psi: \{a,b\} \to Z_6$ by $ \psi(a)\equal{}3$, $ \psi(b)\equal{}2$. (basic set theory)

2) Then there is some $ \overline{\psi}: F \to G$ with $ \overline{\psi}(a)\equal{}3$ and $ \overline{\psi}(b)\equal{}2$. (universal property of free groups)

3) Since $ N\subset ker\ \overline{\psi}$, this induces a map $ \phi: F/N \to G$ that does what we want. (universal property of quotient groups)

If this proof seems like overkill to you, remember that in order for $ \phi$ to be well-defined, we have to verify that there is no way to deduce [i]any[/i] relation like, for example, $ a\equal{}b$, which would imply $ 2\equiv 3 (mod\ 6)$.

Universal properties are very hard to use correctly when you first see them, so don't worry if it's a little confusing.  Anyway, you don't [i]need[/i] to use them, you can just observe that your definition of $ \phi$ is $ 0$ on all the generators of $ N$, so it is $ 0$ on all of $ N$, and you can use this to show that it is well-defined.  But you asked, so I included the above. :)

No, the proof definitely doesn't seem like overkill. In fact it seems rather perfect. One small question about step 3:

Don't you actually need $ \ker \bar{\psi} \equal{} N$ also to use the First Isomorphism Theorem? Is the other inclusion obvious for some reason?

We just need $ N\subset ker\ \overline{\psi}$ in order to show that such a map exists. (though this is not enough to show that it is an isomorphism)

Actually, showing that $ \phi$ is injective is exactly the same as showing that $ ker\ \overline{\psi}\equal{}N$. (which we talked about earlier)

(actually, $ ker\ \phi \equal{} ker\ \overline{\psi}\ /\ N$, which is a slightly stronger statement of the first isomorphism theorem)

OK. So, $ N$ is normal in $ \ker \bar{\psi}$ means that the cosets of $ N$ in $ G$ are finer than the cosets of $ \ker \bar{\psi}$ in $ G$ so the induced homomorphism $ F/\ker \bar{\psi} \to G$ can be refined to a homomorphism $ F/N \to G$ by sending aN wherever $ a\ker \bar{\psi}$ goes. So, I see why $ \phi$ exists but where did we discuss $ \phi$ being injective before?

You said, "how do I show injectivity," and my first post answered this question.

Oh yeah...I'm a little forgetful sometimes. I get it all now. That's like the first group presentation problem I [i] completely [/i] understood. Knowing about how to use the universal property of free groups was key.


if $ k$ be a positive integer, then prove that
$ k^3 \plus{} k^2 \plus{} 1$ can be divided into two parts such that if one part is divisible by $ 6$, the other part less $ 1$ is also a factor of the first part.

What we are supposed to prove is that there will be at least one set of combination where if one part is divisible by $ 6$, it will also be divisible by the second part.

For example say with a value of $ k$ is $ 3$ and the given expression equalling $ 37$, the combimnation $ 6$ and $ 31$ will not satisfy the requirement ($ 6$ being not divisible by $ 30$) but the combination of $ 30$ and $ 7$ will satisy ($ 30$ being divisible by $ 6$) or $ 24$ and $ 13$ will satisfy ($ 24$ being divisible by $ 6$ ans $ 12$).

for $ k$ having a value of $ 4$ and the given expression equalling $ 81$, conbination of $ 42$ (divisible by $ 6$) and $ 39$ will not satisfy ($ 42$ being not divisible by $ 38$), but $ 72$ and $ 9$ will satisfy ($ 72$ being divisible by $ 6$ and $ 8$).

you'll have to prove that such combination is possible witth any positive integer value of $ k$.

This sounds like you would want to use inductive reasoning.  Assume a $ k$th case is true, and then show that the $ k\plus{}1$ case is also true.

i gave that as an example. but if it can be solved using induction, then also is fine.


As a sign of my happyness caused by your splendid solution, I will give you a gift:
   Prove that for all a,b,c>0 we have 27+(2+a^2/bc)(2+b^2/ca)(2+c^2/ab)>=6(a+b+c)(1/a+1/b+1/c).

Thank you very much Harazi.
Naturally, as I am not clever to solve inequality, I extend this gift to all members of the forum.

Hi all,

I post no so nice, but effective solution: By expanding two sides, this inequality is equivalent to

2abc(a^3+b^3+c^3+3abc-a^2.b-a^2.c-b^2.a-b^2.c-c^2.a-c^2.b) + (a^3.b^3+b^3.c^3+c^3.a^3+3.a^2.b^2.c^2-a^3.b^2.c-a^3.b.c^2-a.b^3.c^2-a.b^2.c^3-a^2.b^3.c-a^2.b.c^3)

which holds by Schur for a, b, c and ab, bc, ca.

By my opinion, almost symmetric, homogenus inequality of low order can be proof via Schur and Muirhead. (Anybody has contrexample?)

But this approach requires many computation. However, sometimes can use it.

Namdung

It is also my solution. I posted this problem because I would like to see whether there exists a solution without computations. Well, I will have to wait.

I want to help xzlbq, sorry harazi if it is wrong:
\[27+(2+\frac{a^2}{bc})(2+\frac{b^2}{ca})(2+\frac{c^2}{ab})\ge6(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\]