analytic geometry
graphing lines
slope
calculus
derivative
induction
function

A smooth curve has the property that for all x the value of its slope at 2x is twice the value of its slope at x. What is the slope at 0?

So I set up the equation f'(2x) = 2f'(x)
Then by the chain rule f'(2x) = 2*f'(x)

So now I have f'(2x)=f'(2x). No new information. I can't figure anything out from this except f'(0)=f'(0).

Any help would be appreciated!

There is a calculus forum. Calculus is not appropriate for high school basics.

$ f'(2x)\equal{}2f'(x)\implies f'(0)\equal{}2f'(0)\implies f'(0)\equal{}0$?

[quote="redcomet46"]So I set up the equation f'(2x) = 2f'(x)
Then by the chain rule f'(2x) = 2*f'(x)[/quote]

That is not the correct application of the Chain Rule. $ f'(2x)$ does not mean $ \frac {d}{dx}(f(2x))$

grn_trtle's solution is correct.

alternatively, you can notice that f(x)=n*x^2 for any n works for your conditions, since f'(x)=n*2x. plugging in f'(0) gives 0.

Alternatively say the slope is $ m$. Since $ 2\cdot 0 \equal{} 0$, we know $ m \equal{} 2m$ which gives $ m \equal{} 0$. This isn't exactly calc, by the way.

grn_trtl probably made a typo, but to correct:

$ f'(2x) \equal{} 2f'(2x)$ not $ 2f'(x)$.

You're incorrect, An0maly.

[quote="redcomet46"]A smooth curve has the property that for all x the value of its slope at 2x is [b]twice the value of its slope at x[/b]. What is the slope at 0?[/quote]

What you said implies that $ f'(x)$ is always 0, which means $ f$ is constant, which isn't necessarily true.

But:

$ \frac{d}{dx}f(u) \equal{} f'(u) \frac{du}{dx}$

So $ f'(2x) \equal{} 2f'(2x)$

right?

$ f(x)\equal{}x^2$.

$ f'(x)\equal{}2x$

$ f'(2x)\equal{}4x$

$ 2f'(2x) \equal{} 8x$.

Last time I checked, $ 4x\ne 8x$ for all $ x$.

[hide="Why?"]
Well, $ f'(2x)$ is $ f'(x)$ evaluated at $ 2x$. You took the derivative of $ f(x)$ and then you plugged in $ 2x$.

$ \frac{d}{dx}f(2x)\equal{}2f'(2x)$ by the chain rule, because you're [i]taking[/i] the derivative, not plugging stuff in after you've taken the derivative.

In the previous example, $ f(x)\equal{}x^2$, so $ f(2x)\equal{}(2x)^2 \equal{} 4x^2$, and $ \frac{d}{dx}f(2x)\equal{}8x\equal{}2f'(2x)$.
[/hide]

You're all trying to make something out the chain rule that's simply not there. (grn_trtle correctly understands that.) Back up and start over.

Let $ g(x)\equal{}f'(x).$ Then our condition is that $ g(2x)\equal{}2g(x).$ Basically, that's all we have.

We can turn that around to $ g\left(\frac x2\right)\equal{}\frac12g(x),$ which we can extend by induction to $ g(2^{\minus{}k})\equal{}2^{\minus{}k}g(x).$ Take the limit of that as $ k\to\infty$ and use that $ g$ is assumed to be continuous at zero to see that $ g(0)\equal{}0.$

So $ f'(0)\equal{}0$ and that answers the original question.

Now, what such functions might there be? Yes, $ f(x)\equal{}ax^2$ for arbitrary $ a$ satisfies the equation: $ g(x)\equal{}2ax$ and $ g(2x)\equal{}4ax\equal{}2g(x).$

But can there be anything else? That depends to some extent on how rigorously you enforce the meaning of "smooth." For instance:

$ f'(x)\equal{}g(x)\equal{}x\sin(2\pi\log_2x)$ satisfies the condition for $ x>0.$ We can set $ g(0)$ to be $ 0$ and extend $ g$ to be an odd function and we would have a continuous $ g$ and hence $ C^1$ function $ f$ that satisfies the original condition - but it wouldn't be $ C^{\infty}$ at zero.

[quote="grn_trtle"]Last time I checked, $ 4x\ne 8x$ for all $ x$.[/quote](Except 0 :P)

I'm sorry if I came off as arrogant or something. I just reread the thread and concluded that I must've really been out of it when I posted that.


<<  Is it true that between $n + 1$ positive integers taken from  the numbers  $1 , 2 , 3 , \cdot \cdot \cdot , 2n$  we can find two, such that one of them divides the other ? >>
If yes , which is the simplest explanation?

[u]Babis[/u]

have a look at :
http://www.mathlinks.ro/Forum/viewtopic.php?t=61363

Pierre.

Thanks Pierre!

 I was just wondering if we could make $n$ pairs (couples)  $(a , b)$where $a, b$  are from  $1$ to $2n$  and  $a / b$. But I can not find this algorythm. In that case two of the  $n+1$ numbers belong to the same pair .

[u]Babis[/u]


[quote="Eragon"]I found these on the New Jersey forum(Please, don't try to cheat):

John has 40% fewer marbles than Alicia. Miguel has 40% more marbles than Alicia. If John, Alicia, and Miguel have a total of 180 marbles, how many marbles does Miguel have?If you find the answers,pm me.

Jae was  years old  years ago and will be  years old,  years from now. How old is Jae now? Assume  is a positive integer?[/quote]

Apparently $ \text{\LaTeX}$ doesn't copy onto things, cause the second one makes absolutely no sense

[hide]
So we have $ J=.6A$ and $ M=1.4A$.  So the sum is 3A and it's 180, so A=60 and Miguel has $ \boxed{84}$.
[/hide]

Only #1.. Couldn't understand #2.
[hide]let John=$ J$,Miguel= $ M$,Alicia=$ A$

$ J\equal{}\frac{3A}5$

$ M\equal{}\frac{7A}5$

$ \frac{3A}5\plus{}\frac{7A}5\plus{}A\equal{}180$

$ A\equal{}60$

$ M\equal{}12\times7\equal{}\boxed{84}$[/hide]

[quote="Eragon"]Jae wasa x years old 2x+2 years ago and will be x^2+4 years, 2x+2 years from  now. How old is Jae now? Assume that x is a positive integer.[/quote]

[hide]
So we have $ y$=age now.
$ y\minus{}(2x\plus{}2)\equal{}x$
$ y\plus{}2x\plus{}2\equal{}x^{2}\plus{}4$
Using the first equation, we get $ y\equal{}3x\plus{}2$, so substituting into the second equation, we get
$ 3x\plus{}2\plus{}2x\plus{}2\equal{}x^{2}\plus{}4$
$ 5x\plus{}4\equal{}x^{2}\plus{}4$
$ 5x\equal{}x^{2}$.

So clearly $ x\equal{}5,0$ and since $ x$ is positive, $ x\equal{}5$ and $ y\equal{}3(5)\plus{}2\equal{}17$.
[b]17[/b][/hide]

[quote="alanchou"][quote="Eragon"]Jae wasa x years old 2x+2 years ago and will be x^2+4 years, 2x+2 years from  now. How old is Jae now? Assume that x is a positive integer.[/quote]

[hide]
So we have $ y$=age now.
$ y \minus{} (2x \plus{} 2) \equal{} x$
$ y \plus{} 2x \plus{} 2 \equal{} x^{2} \plus{} 4$
Using the first equation, we get $ y \equal{} 3x \plus{} 2$, so substituting into the second equation, we get
$ 3x \plus{} 2 \plus{} 2x \plus{} 2 \equal{} x^{2} \plus{} 4$
$ 5x \plus{} 4 \equal{} x^{2} \plus{} 4$
$ 5x \equal{} x^{2}$.

So clearly $ x \equal{} 5,0$ and since $ x$ is positive, $ x \equal{} 5$ and $ y \equal{} 3(5) \plus{} 2 \equal{} 17$.
[b]17[/b][/hide][/quote]
Correct!


group theory
abstract algebra
linear algebra
linear algebra unsolved

Given is a finite group. Prove that there are only finitely many conjugacy classes of subgroups of invertible complex $nxn$ matrices that are isomorphic with the given group.

This follows immediately from the fact that in the given setting, every group representation breaks up into a direct sum of irreducible representations (Maschke's Theorem), and there are only finitely many irreducible representations.


1.
 x(x+1)(x+2)....(x+k-1)=y^m (y v stepeni m) gde x,y,k,m prinadlezhat N; k>=4, m>=2, x, y-?
Hotya bi 4astnie slu4ai: a) k=m ili k=4, m-lyuboe>2

 2.
ta zhe zada4a:
a)pri m=2, k-4etnoe, re6enii (x,y) kone4noe koli4estvo


inequalities
integration
inequalities unsolved

Prove that:
$\sum_{i=1}^n\sum_{j=i}^n\frac{a_ia_j}{\binom{k+i+j}{k+2}}\ge 0$.
Where $n,k$ is nonnegative integer; $n>1$ ;{$a_1,a_2,...,a_n$}is any register of real

[quote="keira_khtn"]@Tkhtn :This is a new problem in Math And Young magazine and plz don't ask it here??
@Silouan:i think you must wait for discussing it in one more month,I think!![/quote]
Hi,do you how can I to see Math And Young magazine?

To Silouan,is the inequality should be \[ \sum_{i=1}^n\sum_{j=1}^n\frac{a_ia_j}{\binom{k+i+j}{k+2}}\ge 0 \]
,I just to clear the inequality. :)

[quote="keira_khtn"]@Tkhtn :This is a new problem in Math And Young magazine and plz don't ask it here??
@Silouan:i think you must wait for discussing it in one more month,I think!![/quote]

Sorry boys I didn't know that. In fact Tkhtn send this problem to solve. I can't solve so I posted here .If the problem is really fram Math And Young magazine  I want from the moderators to lock the topic

Keira khtn!
This problem was old.
I think we have could do it one month ago.
Zhaobin,can you post your solution,thank!
(At Math and Young in VN , this problem was done.)

[quote="tkhtn"]
Zhaobin,can you post your solution,thank!
(At Math and Young in VN , this problem was done.)[/quote]
Ok, but can you make sure the inequality first?

[quote="tkhtn"]I don't understand what you say.Sorry(My English is bad).[/quote]
See post#3,i wrote a little different inequality from silouan wrote.I want to ask is the oringinal one is silouan's or mine?

What do you say? :huh: .
I see your problem and siloun'problem is same"same". :o

This inequality was proposed by Han Ngoc Duc in the M and Y Math. Mag. Duc told me that he had a simple and elementary solution.

Now you can discuss about this nice inequality.


logarithms
geometry
perimeter
floor function
function
trigonometry
probability

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[/hide]

[b]
Round 1, Due 9:00 P.M. EDT Sunday 6/21[/b]
1. Let $ a_{t \plus{} 1} \equal{} \frac {1}{3 \minus{} a_{t}}$. Given $ a_1 \equal{} 2009$, $ a_{\infty}$ can be expressed in the form $ \frac {x \plus{} \sqrt {y}}z$, where x, y, and z are positive integers. Evaluate $ x \plus{} y \plus{} z$. (3 pts)

2. If $ \displaystyle{\sum_{k \equal{} 2}^{1003}} \ {2007\choose k}$ can be expressed as $ a^b \minus{} c$, where a, b, and c are (not necessarily distinct) positive integers, find $ a \minus{} b \plus{} c$. (3 pts)

3. Suppose that the solution to: $ (5^{\log x})^2 \plus{} 4x^{\log 5} \minus{} 32 \equal{} 0$ is written as $ k^{\log_i j}$, where k, i, and j are positive integers. Compute $ k \plus{} i \plus{} j$. (4 pts)

4. Every week, a "white and nerdy" kid's 27 friends "beg for his top 8 spaces" on his "totally pimped out" Myspace page. In this week's update, the kid plans to include Caelestor and gfour84 on his top 8. The list members and subsequent rankings is usually chosen randomly; this week, however, the kid prefers gfour84 over the slightly arrogant Caelestor. Thus, the only provision is that Caelestor will come after gfour84 in the list. Let x be the amount of possible combinations (remember, order matters in this list!) that can result. What is the sum of the digits of x? (4 pts)

5. The "white and nerdy" kid is now running Mathematica. He plots a regular hexagon with a perimeter of $ 12\sqrt2$ units that lies in the Cartesian plane in such a way that its center is on the origin, two of the vertices lie on the x-axis, and the midpoints of two of its sides lie on the y-axis. The kid revolves the portion of the hexagon that lies in Quadrant I completely around the x-axis, thus forming a solid whose volume is m cubic units. If that same portion is completely revolved around the y-axis, a solid with a volume of n cubic units results. Suppose that $ \left(\frac {m}n\right)^2 \equal{} \frac {G}T$, where G and T are relatively prime. Demonstrate that you, on the other hand, don't need Mathematica's assistance and evaluate $ G \plus{} T$. (6 pts)

[b]Good luck to all and have fun![/b]

Hello all,
The deadline has passed for round 1. [b] If you still want to join, you may still submit your solutions until the deadline of Round 2, but your maximum score will only be a 15/20.[/b] Solutions and answers posted in next round's results.

Current scores:
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Submissions: 5
Mean Score: 16.4/20
Perfect scores: 3
100 point club: Not yet achieved (Achieve to unlock next ranking).

If you received a perfect score, good for you. If not, you may appeal before the deadline of next round's submissions.
[hide]
New Participants:
438436 +20
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[/hide]
[/hide]

[b] Check back tomorrow for a new round of the BAME! Good job to all who participated![/b] :D

Welcome back! Please take a look at the FAQ again, some things may have changed:

[b]Basic FAQ:[/b]
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- [b]Initial Entry form[/b]: If this is your first submission, please choose a name that you would like to go by (it can be your actual username or just a random five-digit number, you decide). This is the name you will go by when the scoreboard is updated. Please also tell us whether or not you would like your solutions posted.
- [b]What does BAME stand for?[/b] Biweekly Aptitude in Mathematics Examination.
- [b]What is the scoring system?[/b] The typical round consists of 5 problems, each worth a different amount of points (ranging from 2-6, depending on difficulty). The answer alone to a question is worth at most 1 point, so please attach a solution! You have plenty of time to do these problems (shouldn't take more than half an hour).
- [b]I don't understand this problem![/b] If you have issues with the wording, please post a response on this topic thread and the authors will get back to you. If you just don't know how to do this problem, then we suggest you do some studying.
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[/hide]

[b]Round 2, Due 12:00 P.M. EDT Friday 6/26:[/b]
1. Find the last three digits of the sum $ \displaystyle\sum_{i \equal{} 09}^{62} {( \minus{} 1)^{\lfloor \frac {i}{22} \rfloor} 622_i}$, where $ \lfloor x \rfloor$ is the floor function and the subscript next to the 622 represents the current base (i.e. $ 622_{9} \equal{} 506$ in base 10). [b]You are allowed a calculator (so don't miss this one!), but do not use brute force![/b] (2 pts)

2. Consider the function $ f(x) \equal{} x^3 \minus{} 3x^2 \plus{} 12x \minus{} 4$ whose roots are a, b, and c. Determine $ \frac {a}{c} \plus{} \frac {b}{c} \plus{} \frac {a}{b} \plus{} \frac {c}{b} \plus{} \frac {b}{a} \plus{} \frac {c}{a}$. (3 pts)

3. If $ \sin(\omega) \equal{} 3\cos(\omega)$ for $ 0\le \omega \le \frac {\pi}{2}$, then $ x \equal{} \displaystyle{\sqrt {\frac {2\sin(\omega) \minus{} \cos(\omega)\sin(2\omega)}{\sin(2\omega)\sec(\omega)}}}$. $ x^4$ can be expressed as a fraction $ \frac {m}n$, where m and n are relative prime. Evaluate $ |m \minus{} n|$ [b] without a calculator![/b] (5 pts)

4. An army plans to execute a prisoner donning red armor today. He is told to drink 3 out of 12 bottles given to him in rapid succession. There are 6 bottles containing 0.5 moles of poison each, and 2 bottles containing 1 mole of poison each. Hidden among them are also an antidote that can be taken at any time without harm (specifically it will cure someone who has ingested 0.5 moles of poison), 1 harmless bottle of wine, and 2 extremely potent antidotes that will cure someone who has ingested [b]only (no more, no less)[/b] 1 mole of poison (otherwise, the prisoner will still perish from the potency). ([b]Note that the antidotes are not vaccines and cannot be used to prevent death from poison[/b], and that the poison requires an hour to take effect.) If the prisoner is still alive after one day, then he will be freed. The probability of his survival is $ \frac {r}s$, where r and s are relatively prime. What is $ r \plus{} s$? (4 pts)

5. Against all odds, the prisoner and 4 of his cohorts (also wearing red) are freed. Whilst in the jungle, they are challenged to a faceoff by a platoon of 5 troopers wearing white and black. The two groups will participate in an RPG-style battle, in which each team will alternate "moves." During each "move," one team member will fire one shot at the opposing team. A team wins if they outnumber the other by 2 or more (or eliminate all 5 of the opposing team). If the troopers go first, the probability that red wins with 2 of 5 people still surviving can be expressed as $ \frac {v}w$, where v and w are relatively prime integers. Given that a "red shirt" has an accuracy of 60% and a "storm trooper" has an accuracy of 75%, find $ v \plus{} w$ (6 pts).

[b]More FAQ on Submitting:[/b]
[hide]
- [b]If you are PMing something, please PM both Caelestor and gfour84 for quicker grading/response. PLEASE do not post your solutions on this thread. If you have a general comment, question, or discussion, feel free to post.[/b]
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- [b]What do I get for doing these problems? First of all, these questions are great practice for late AMC 12-mid AIME. Second, I'm sure you would like to ace all of these problems or compare your aptitude with others, right? The problems given let you practice without the temptation of the help of others' solutions. Third, the answers and solutions will only be initially provided to those who submit answers.
- [b]Did you come up with these problems?[/b] Caelestor created problems 1, 4, and 5, while gfour84 provided questions 2 and 3. Some questions may have been adapted from others.
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[b]Good luck and have fun to all![/b]

One more clarification. The first antidote always eliminates 0.5 mols, regardless of how much poison is in the body.
Also note that #3 is actually only worth 4 points and #4 is worth 5 (error on my part, sorry).


Let $(a_{n})$ and $(b_{n})$ be two sequences such that $a = b_{n+1}-n$ and if lim $(b_{n})$ exists in $R$ then $\sum_{b=1}^{n+1}a_{n}$ converges.

I can't even figure out what you probably mean to say. Could you please carefully reread your source and proofread what you've written?


Stie cineva cum as putea sa cumpar carti de la editura s.s.m.r?

suni la numarul de telefon pe care il gasesti pe gazeta, iti dai datele si platesti cartea prin mandat. o vei primi apoi acasa  :wink:


Do you have an order, or metric, or topology or something? I'm not sure how to define bounded on an arbitrary set.

Suppose that $ F \equal{} \{x_1,\dots,x_n\}$ is a finite subset of $ \mathbb R$. Let $ m \equal{} \min(x_1,\dots,x_n)$ and $ M \equal{} \max(x_1,\dots,x_n)$. Then $ F\subset [m,M]$, which means that $ F$ is bounded.

In a more general metric space $ (M,d)$ a subset is called bounded if it is contained in some ball. In this case we can let $ R \equal{} \max(d(x_2,x_1),\dots,d(x_n,x_1))$ and observe that $ F\subset \overline{B}(x_1,R)$.

Try to find a more descriptive title next time you post.


Given five positive reals $ a,b,c,d,e$ such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\leq 1$. Prove that

$ \frac{1}{1\plus{}ab}\plus{}\frac{1}{1\plus{}bc}\plus{}\frac{1}{1\plus{}cd}\plus{}\frac{1}{1\plus{}de}\plus{}\frac{1}{1\plus{}ea}\geq \frac{25}{6}$

[quote="outback"]Given five positive reals $ a,b,c,d,e$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2\leq 1$. Prove that

$ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} cd} \plus{} \frac {1}{1 \plus{} de} \plus{} \frac {1}{1 \plus{} ea}\geq \frac {25}{6}$[/quote] Using Jensen for $ \frac{1}{1\plus{}x}$ we get
 $ \frac{1}{1\plus{}ab}\plus{}\frac{1}{1\plus{}bc}\plus{}\frac{1}{1\plus{}cd}\plus{}\frac{1}{1\plus{}de}\plus{}\frac{1}{1\plus{}ea}\ge \frac{5}{1\plus{}\frac{ab\plus{}bc\plus{}cd\plus{}de\plus{}ea}{5}}\ge \frac{5}{1\plus{}\frac{a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2}{5}}\ge\frac{5}{1\plus{}\frac{1}{5}}\equal{}\frac{25}{6}$


geometry
quadrilateral
rotation
perpendicular
IMO Shortlist

The diagonals of a quadrilateral $ ABCD$ are perpendicular: $ AC \perp BD.$ Four squares, $ ABEF,BCGH,CDIJ,DAKL,$ are erected externally on its sides. The intersection points of the pairs of straight lines $ CL, DF, AH, BJ$ are denoted by $ P_1,Q_1,R_1, S_1,$ respectively (left figure), and the intersection points of the pairs of straight lines $ AI, BK, CE DG$ are denoted by $ P_2,Q_2,R_2, S_2,$ respectively (right figure). Prove that $ P_1Q_1R_1S_1 \cong P_2Q_2R_2S_2$ where $ P_1,Q_1,R_1, S_1$ and $ P_2,Q_2,R_2, S_2$ are the two quadrilaterals.

[i]Alternative formulation:[/i] Outside a convex quadrilateral $ ABCD$ with perpendicular diagonals, four squares $ AEFB, BGHC, CIJD, DKLA,$ are constructed (vertices are given in counterclockwise order). Prove that the quadrilaterals $ Q_1$ and $ Q_2$ formed by the lines $ AG, BI, CK, DE$ and $ AJ, BL, CF, DH,$ respectively, are congruent.

It is clear that a counterclockwise rotation of $90 ^\circ$ transforms $\overline {R_1Q_1}$ into $\overline{R_2Q_2}$. (In this post $\overline {R_1Q_1}$ will mean the extended line through $R_1$ and $Q_1$.)  Let $\overline{R_1Q_1} \cap \overline{R_2Q_2}=B_1$, and define $A_1,C_1,D_1$ similarly. Now $\angle AB_1C = 90^\circ$ and $\overline{BB_1}$ bisects this angle.  In fact a counterclockwise rotation about any point on $\overline{B_1B}$of $90 ^\circ$ transforms $\overline {R_1Q_1}$ into $\overline{R_2Q_2}$.  If we can prove that $\overline{AA_1},\overline{BB_1},\overline{CC_1},\overline{DD_1}$ at some point $X$, then the counterclockwise rotation of $90 ^\circ$ about $X$ will transform $P_{1}Q_{1}R_{1}S_{1}$ into $P_{2}Q_{2}R_{2}S_{2}$.

Construct $B_2$ on the opposite side of $AC$ from $B$ so that $AB_2C$ is an isosceles right triangle with vertex $B_2$.  $AB_1CB_2$ is cyclic (two opposite right angles), so $\overline {B_1B_2}$ bisects $\angle AB_1C$ as well.  Thus $B_1,B,B_2$ are collinear.  Defining $A_2,C_2,D_2$ similarly, we have that because $\overline {AC}$ and $\overline{BD}$ are perpendicular, the squares $AB_2CD_2$ and $A_2BC_2D$ have parallel sides (and diagonals of course).  Hence the lines connecting corresponding vertices are concurrent.


Wat will u get on treating acetone and cinnamaldehyde in basic medium ?
 :D

What basic medium are we talking about? And how do we mix the reagents? The product might depend on that.

i mean medium of OH- .
And if many rns may be expected can you highlight the various rns? :D  :D  :D

i was thinking of aldol reaction. but since it is in basic medium, aldol product wont be isolated. i think the final product will be a cylic hemiacetal.

tell me the product and i'll see then  :D

initially you get the aldol product. but it doesnt get dehydrated as there is no acid or heating. in presence of base, oh group acts as acid and o- is formed which attacks the carbonyl carbon of acetone to give a hemiacetal which has another oh group and hence cyclizes to give a cyclic hemiacteal

In that case, two reactions are possible: aldol condensation and Michael addition, although I would probably vote on the first one.

No VARUNARASU, even in basic conditions dehydration occurs. Why? Because, the product formed has an extended pi system, which is quite favourable.

then how do you ever isolate an aldol product in good yield

By first converting one component to the enolate with a very strong base like LDA, and then adding the other component, followed by carefull hydrolysis. Lower temperatures also prevent dehydration.

why do u consider aldol superior to micheal addn?  :huh:

Because the carbonyl carbon is stereochemically more available than the beta carbon; also, Michael addition would destroy extended conjugation. So in my opinion the final product would be Ph-CH=CH-CH=CH-C(=O)-CH3. But of course, I can be wrong.

but i think a micheal and an intramolecular aldol are more favourable right?
Atleast that's what Majid's rule says  :D

Or the reverse: aldol followed by intramolecular Michael addition. Unless you are referring to something like Robinson Annulation, which is commonly used with uncrowded unsaturated carbonyls.

And what is "Majid's rule"?

I think that it is perfectly right according to majid's rule.
cinnamaldehyde doesnt undergo aldol and so michael followed by intramolecular aldol is the correct anwer.

[quote="madness"]Prof Majid of crap and crap has said that " Always try the Micheal first " bcos an aldol first might be unfavourable and a Micheal foll by an aldol (intramol) is favourable .
And btw does cinnamaldehyde undergo aldol???[/quote]
and why shouldnt it undergo aldol, it can act as the acceptor.
and for michael reaction to take place you need a more reactive donor, michael with acetone is not practically possible unless you use a very strong base( not with oh-)

I would advise Professor Majid, and the rest of you, to take a look [url=http://jchemed.chem.wisc.edu/jcesoft/CCA/CCA5/MAIN/1ORGANIC/ORG12/TRAM12/F/0410121/MOVIE.HTM]here[/url].

That's perfect. :thumbup:  :clap2:
But what is your take on majid's rule for other situations. :maybe:

The most important of the other situations is Robinson Annulation.

I accept that Majid's Rule is not very theoretical it is like markownikoff's rule to find you the product but it failed this time


linear algebra
matrix
linear algebra unsolved

Suppose $ A$ are $ n\times n$ matrix (over$ \mathbb{R}$ or $ \mathbb{C}$) such that $ tr(A) \equal{} 0$. Is there matrix $ B$ with zero diagonal entries such that $ B$ is similar to $ A$?

For $ n\equal{}2$, take $ A\equal{}\left(\begin{array}{cc}1&0\\0&\minus{}1\end{array}\right)$ and $ B\equal{}\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$. Clearly they can’t be similar because $ A$ is invertible while $ B$ isn’t.

[quote="JC_math"][quote="payman_pm"]Suppose $ A,B$ are $ n\times n$ matrices (over$ \mathbb{R}$ or $ \mathbb{C}$) such that $ tr(A) \equal{} 0$ and diagonal entries of $ B$ are zero.Is $ B$ similar to $ A$?[/quote]
I guess you question might be slightly different.
Suppose $ A$ is an $ n\times n$ matrix (over$ \mathbb{R}$ or $ \mathbb{C}$) such that $ tr(A) \equal{} 0$, does there exist $ B$ whith diagonal entries zero such that $ B$ is similar to $ A$?[/quote]
Sorry for the mistake.I corrected that!

Then the answer is "yes", and I think it has been posted on this forum already by alekk or harazi.

In fact you can show that there exist $ B$ with zero diagonal entries such that $ B$ unitarily equivalent to $ A$.


If a, b, c and d are four multidigit numbers and

if

a + b + c + d, a + b, a + c, a + d, b + c, b + d and c + d are all square numbers,

find minimum values of a, b, c and d.

If "multidigit" means that all four must be larger than 9, I'll go searching again.  Otherwise, the smallest solution is [hide]$\{a, b, c, d\} = \{4, 87, 121, 191\}$.[/hide]

i meant all numbers greater than 9.. but thats good too.

can u tell how did u found out.. or just crude search?

thanks,

{a, b, c, d} = {4, 87, 121, 191}.

how is "a+b" a square number?

am i missing something?

"If a, b, c and d are four multidigit numbers and

if

a + b + c + d, a + b, a + c, a + d, b + c, b + d and c + d are all square numbers"

I think this is incorrect as none of the above is a square number with the given set of {4, 87, 121, 191}.

 :(  :(  :(

Hmm, that's an excellent point.  I'm not quite sure what I was doing -- let me get back to you on this.

The method I was following:  we need a square, $a + b + c + d$, which can be expressed as a sum of squares in at least 3 different ways.  Thus, if we look at primitive Pythagorean triples, it has to be a multiple of three distinct hypotenuses.  I neglected to put the squares in everywhere they should have been  :blush:

When I actually solve for my smallest solution, I get $\{a, b, c, d\} = \{-21847, 30128, 37472, 59872\}$ which obviously has a negative member, which is too bad.  I don't know any other way of doing this than picking three Pythagorean triples and hoping the result gives you all positive $a, b, c, d$.  (The triples I used for this case were (4, 3, 5), (12, 5, 13) and (24, 7, 25).)

That example I gave was pretty badly non-minimal.  Here are other, smaller (but still not meeting the conditions) solutions for $a + b + c + d = 4225 = 65^2$:
$\{-495,1120,1584,2016\}$, $\{-88,344,1177,2792\}$, $\{-104,360,729,3240\}$ and $\{-320,576,945,3024\}$ all have one negative element, while $\{\frac{193}2,\frac{1057}2,\frac{1985}2,\frac{5215}2\}$ is half-integers.  In fact, I can multiply this last example by 4 and I'll get a perfectly acceptable (but larger) solution.  It's possible there are other smaller solutions than this one -- in particular, taking $a + b + c + d = 85^2$ seems promising to me.  (125 could also work.)

Edit added later:
85 yields 4 solutions with one negative element plus one solution of positive half-integers.  125 yields only solutions with a negative element.  The result is that $\{386, 2114, 3970, 10430\}$ is the solution with the smallest value of $a + b + c + d$, and that we have another solution with $a + b + c +d = 170^2$, but this is not necessarily the next smallest.  In addition, we have a number of solutions in which exactly one of the numbers is negative.

Thaks a lot !! Now let me understand this.

 :)

EDIT:
   [b]The method I was following: we need a square, (a+b+c+d), which can be expressed as a sum of squares in at least 3 different ways. Thus, if we look at primitive Pythagorean triples, it has to be a multiple of three distinct hypotenuses. [/b]

   Can you explain this in more detail... especially this part.. [u][i]"which can be expressed as a sum of squares in at least 3 different ways. Thus, if we look at primitive Pythagorean triples, it has to be a multiple of three distinct hypotenuses"[/i][/u]

   Apologies for asking such primitive question  :?:  :?:  :?:

No worries -- that was rather light on explanation :-).
A number $c$ is the hypotenuse of a right triangle with integer sides if we can write $c^2 = a^2 + b^2$ for positive integers $a, b$.  We say that $(a, b, c)$ is a Pythagorean triple.  If $(a, b) = 1$ (that is, $a$ and $b$ have no common factor), we say that $(a, b, c)$ is a primitive Pythagorean triple.
We have that $a + b + c + d = (a + b) + (c + d) = (a + c) + (b + d) = (a + d) + (b + c)$.  Thus, the number $a + b + c + d$ is a square which can be expressed as a sum of squares in three distinct ways.  So, we need three Pythagorean triples, all with the same hypotenuse.  Not all of these triples need to be primitive: if we only wanted two expressions, we could use $(15, 20, 25)$ and $(7, 24, 25)$.  Thus, we need a number $c$ which is the multiple of the hypotenuse of 3 distinct Pythagorean triples.  One can find a list of the hypotenuses of primitive triples in the [url=http://www.research.att.com/~njas/sequences/A020882]Encyclopaedia of Integer Sequences[/url], or use the [url=http://mathworld.wolfram.com/PythagoreanTriple.html]formula[/url] for such triples (see equation (11) at that link).  (Actually, what we really want are members of [url=http://www.research.att.com/~njas/sequences/A009000]this sequence[/url] which appear at least 3 times, but it doesn't have enough terms to be really useful.)  Then we find all the Pythagorean triples associated with this number and solve the equations which result from assigning our various sums ($a + b, a + c, \ldots$) to the squares of the legs in an appropriate fashion.

Out of curiousity, where did this question come from?

Thanks !! Its much clearer now.

One of colleagues at work asked me this question... probably he got it in a mailing list of something.
 :)


darkquantum wrote:Consider the equation x^2+y^2=z^2+18, where x,y,z are positive integers.

Prove there are no solutions where z=y+5

[hide]substituting in, we get x^2=10y+43. The right hand side will always end in a 3, which is not possible. (the quadratic residues mod 10 are 1, 4, 5, 6, and 9)[/hide]

Yesiree.

By the way, Mystic, do you particularly like number theory?

definitely. how'd u guess? jk

though some meanie stole the last problem!!! jk

Do you want me to put up some more Number theory ones?


[quote="mathelete"]The list contains:
1. mathelete
2. kldaace
A third player must sign up. When I say so, I go first with a circle, kldaace goes second with an X, and the last player to on the list puts a triangle. You must get four in a row. We have a 7X7 board, and you have to give me the x and y coordinates. Remember, it has not started, so there is time for sign up.[/quote]
Okay, the list is almost final (we can still accept replacements, just in case):
1. mathelete
2. kldaace
3. AIME15

[quote="iphailedusamo"]If you need an ALTERNATE, PM me.[/quote]
You are already in.
Circle at (4,4)

Hang on, you are probably confused.
The list now contains:
mathelete
iphailedusamo
AIME15

As you see, I already went, so it is ipailedusamo's turn.


[quote="thanhvien"]Solve the equation (x):

(x^3+a^3):((x+a)^3)+(x^3+b^3)):((x+b)^3)+(x^3+c^3):((x+c)^3)=3/2 (1- [(x-a):(x+a)][(x-b):(x+b)][(x-c):(x+c)]
  (a,b,c of R)[/quote]

Don't have a clue what you mean, I tried to edit it but I really don't understand what you are trying to say with this equation.

(..) : (..) = (..) / (..) probably. It would make sense if the poster re-read the post and removed the smilies. :)

So, it should be this:

\[\frac{x^3+a^3}{(x+a)^3} + \frac{x^3+b^3}{(x+b)^3} + \frac{x^3+c^3)}{(x+c)^3} = \frac{3}{2} \left(1- \frac{x - a}{x + a} \frac{x - b}{x + b} \frac{x - c}{x + c}\right).\]

Correct, thanhvien?


algebra
polynomial
geometric sequence
geometric series

Find factorised form of

$x^n-y^n$  and  $x^n+y^n$ :D

Thats not the total factorisation, e.g. $x^4-y^4 = (x-y)(x+y)(x^2+y^2)$ or $(x^6-y^6)=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)$.
There will be as many factors as there are positive divisors of $n$ for $x^n-y^n$.

For $x^n+y^n$, just note that $x^{2n}-y^{2n} = (x^n-y^n)(x^n+y^n)$, thus it can be reduced to the other case.

In order to factor something like that we may want to know if $n$ is even or if it is odd.
If it's odd, then we use the formula used by molina, but if it's odd then we must start taking factors of two until we find a prime factor as Zeta X showed

To give the factorisations:
\[ x^n-y^n = \prod_{d|n} \Phi_d(x,y) \]
\[ x^n+y^n = \prod_{d|2n \ , \ d \nmid n} \Phi_d(x,y) \]
where $\Phi_k(x) = \prod_{s \in \mathbb{Z}/ k \mathbb{Z}^*} (x-\zeta^s y)$ with $\zeta$ being a primitive $k$-th root of unity (e.g. $\zeta = cos\left(\frac{2\pi}{k}\right) + i \cdot sin\left(\frac{2\pi}{k}\right)$).

Showing that the $\Phi_k(x,y)$ can't be factored anymore is not that easy indeed. Thus I think it's inappropriate for Intermediate (probably also for Pre Olympiad).

wouldn't the solutions just be the nth roots of y?

so for eample for n=4 and y=16 it would be
(x-2(1))(x-2(i))(x-2(-1))(x-2(-i))

it doesn't seem that hard.

Depends on if you want a complex or a rational factorisation. The problem isn't very clear about that; I assumed a rational one since it's what most problems of that kind concern with. The complex one is in my opinion sensless since it doesn't contain any information and clearly exists by the fundamental theorem of algebra. But if you want the complexe one, see the further factorisation of the $\Phi_k(x,y)$ above.

But you can't set in special values for $y$ and then factor ;)

I was just giving an example. y can be anything

[quote="shadysaysurspammed"]By find i meant "find",like molina, show how you got those.[/quote]
 :?
What factorisation now¿ And how should be clear: just put some of the factors $(x-\zeta^s y)$ together such that the polynomials get integral.

well actually i maent prove that $x^n-y^n=(x-y)(......)$the thing which molina said,ofcourse are are right about further factorisation thing but leave that,prove the thing given by molina in her first post,simple =)

[quote="shadysaysurspammed"][quote="molina"]$\large{x^n-y^n\,=\,(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...+y^{n-1})}$

And :

$\large{x^n+x^n\,=\,(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-...+-1)}$[/quote]

I'm asking to prove that(i remember kunny proving this once but i searched all over and couldn't find it,so i'm not sure if any one proved it)[/quote]

That's not really a "factoring." It's just insight that you know.. Like $x^3 - y^3 = (x-y)(x^2+xy+y^2)$. You know that if one product is $(x-y)$, there should be $x^2$ to have $x^3$ and $y^2$ to have $y^3$ etc.. It's just looking at the numbers and seeing it.

Zetax is the one who really "factored" it. I wouldn't really call yours factoring but if you have to, I guess it's OK.

[quote="shadysaysurspammed"][quote="molina"]$\large{x^n-y^n\,=\,(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...+y^{n-1})}$

And :

$\large{x^n+x^n\,=\,(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-...+-1)}$[/quote]

I'm asking to prove that(i remember kunny proving this once but i searched all over and couldn't find it,so i'm not sure if any one proved it)[/quote]

Just multiply those together and it is easy to see that they are true. All the middle terms cancel out.

[hide="Constructive proof"]
Consider the geometric progression

$G = 1 + \frac{x}{y} + \frac{x^2}{y^2} + ... +\frac{x^{n-1}}{y^{n-1}}$.  I hope everyone here knows how to find the sum of a geometric series, so:

$G = \frac{ \frac{x^{n}}{y^{n}} - 1}{ \frac{x}{y} - 1 } = \frac{ \frac{x^{n}}{y^{n-1}} - y}{x - y}$

Now, notice that

$y^{n-1} G = y^{n-1} + xy^{n-2} + x^2 y^{n-3} + ... + x^{n-1}$

But that, from our summation,

$y^{n-1} G = \frac{ x^{n} - y^{n} }{x - y}$

And hence

$x^{n} - y^{n} = (x-y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})$

The second case can be found by substituting $y \to -y$.
[/hide]


When you do hide and stuff you just click on the little hide button creating:
[code][hide][/hide][/code]
But in asy you have to manually write:
[code][asy][/asy][/code]
Why not make a little button for asy?

That question has been asked long ago. Here is the "official answer":
[quote="Valentin Vornicu"][quote="math92"]Are there going to be asy tags up soon?[/quote]No they will not appear on the list of tags. If you cannot write the tags manually then it's doubtful you'll be able to write any good code anyway :)[/quote]
As to my own opinion, I do not really care, I have never used any of the tag buttons anyway when posting: the time it takes to take the hands off the keypad and operate the mouse is hardly any less than the time it takes to type 5 symbols...

My answer from back then still stands today :)


modular arithmetic
AwesomeMath
summer program
number theory proposed
number theory

Find all positive integers $m$ and $n$ such that both $3^{m}+1$ and $3^{n}+1$ are divisible by $mn$.

[quote="Rust"]Let $p$ is minimal prime divisor of m and $T_{3}(p)$ is semiperiod ($T_{3}(p)$ is minimal k, suth that $p|3^{k}+1$).
Then $p|3^{m}+1\to T_{3}(m)|m\to T_{3}(m)=1\to p=2\to 3^{m}+1=2(mod \ 8)$.
Therefore n odd and $n=1$. It give solutions (m,n) $(1,1), \ (2,1), \ (1,2), \ (4,1), \ (1,4)$.[/quote]
What is $T_{3}(m)$?

@Please post concrete your solution at all! If not, I will must send to you questions only one line as above!

[quote="N.T.TUAN"]What is $T_{3}(m)$?
[/quote]
$T_{3}(p)$ is minimal number k, suth tat $p|3^{k}+1$, then $p|3^{m}+1\Longrightarrow T_{3}(p)|m$, because $2T_{3}(p)|m-T_{3}(p)$.
$T_{3}(p)\le p-1$, because p minimal prime divisor $T_{3}(p)=1\to p=2$. If n odd, it give n=1.

[quote="N.T.TUAN"]Find all positive integers $m$ and $n$ such that both $3^{m}+1$ and $3^{n}+1$ are
divisible by $mn.$[/quote]
If $m=1$ or $n=1$ then easy.
If $m>1$ and $n>1$ then one of $m,n$ is odd because $3^{2k}+1\equiv 2\pmod{4}$. WLOG, assume that $m$ is odd, because $m|3^{m}+1$ we have $\gcd (3,m)=1$. Called $p$ is minimum prime divisor of $m$ ($p$ exist by $m>1$)  then $\gcd (9,p)=1$ and $9^{p-1}\equiv 1\pmod{p},9^{m}\equiv 1\pmod{p}$ therefore $\gcd (p-1,m)>1$ (in fact, $\text{ord}_{p}(9)>1$), so $m$ has prime divisor $q<p-1<p$, contradiction.

I've just found in a half-forgotten folder a solution of this by prowler on the Awesomemath forum (back before it was hacked):

[quote="prowler"]Just an idea, obviously $3 \not | m, n$.

Suppose $p$ is the smallest prime number such that $p | m$ and $m = p\cdot l $
$ 3^{p\cdot l}\equiv -1(mod p), 3^{2p\cdot l}\equiv 1(mod p)$
and $ 3^{p-1}\equiv 1(mod p)$

Let $d$ is the smallest power such that $3^{d}\equiv 1(mod p)$.

Then the least common divisor for $(2p\cdot l, p-1)= k\cdot d$, $k\in N$,
but (l, p-1) = 1 because l has prime divisor greater or equal $p$
and (p,p-1)=1, then (p-1, 2) = 1,2 =k \cdot d.

$d = 1 or d = 2$ then $p | 2 $ or $p | 8 $ , $p = 2$.
If $ 2 | m$ and $2 | n$ then $4 | (3^{2\cdot l}+1)$,but

$3^{2\cdot l}+1\equiv (-1)^{2\cdot l}+1\equiv 2 (mod 4)$ - contradiction.

Now let $m = 1$ then $n =2$ or $n=4$, checking just pairs $(2,1)$ and $(1,2)$ are good.[/quote]


Okay I havent been able to figure this problem out, so I figured some good geometer's from AoPS could help me:

There is an equilateral triangle ABC with interior point P.  if AP = 57, BP = 73, and CP = 65, then what are the sides of ABC?

I realized that (57+73)/2 = 65. But I don't think that helps any  :( . Any suggestions?

i'm not sure how to use the diagram to my advantage Palytoxin  :(

you spent 4 minutes into my drawing expecting answer right away? hmm, think hardly my friend. remember equilateral has 60 degree angles. $CD$ is a side of triangle...

[quote="Iversonfan2005"]Please someone get the answer soon, I really need to learn how to do these kind of problems :? There's one just like this recently on the Intermediate section.[/quote]

Let PA=57, PB=65, PC=73. Construct QAB(Q lies outside triangle) such that QAB is congruent to PCB. Thus, we get <QPB=60deg. and QP=65; now find <QPA using cosine rule.
And the angle APB is 60deg. + inv_cos(11/38).

you can get the side AB using cosine rule again.

What I tried to do was that I tried to do it algabraicly with two variables, it took a while and then I think I got each side, but I don't know if it's right though. I guess you $need$ to do trigonometry don't you?

If $P$ lies on a median there is a really neat formula. Let $D, E$, and $F$ be the points where where lines $PA, PB,$ and $PC$ intersect sides $BC$, $CA$, and $AB$ respectively. Then we know that

\[[PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]\]

can you be sure P lies on a median though?


inequalities
quadratics
calculus
derivative
inequalities proposed

arqady, what inspired you in putting a $\frac{1}{6}(x^2-1)$ term?

Secondly, can the inequality be generalized?

[quote="perfect_radio"]arqady, what inspired you in putting a $\frac{1}{6}(x^2-1)$ term?[/quote]
Let $f(x)=\frac{1}{x^5-x^2+3}+\lambda(x^2-1)-\frac{1}{3}.$
Then $f'(1)=0$ and $\lambda=\frac{1}{6}.$ :)
[quote="perfect_radio"]
Secondly, can the inequality be generalized?[/quote]
I would not do it. ;)

[quote="arqady"][quote="Vasc"]$\sum\frac{1}{x^5-x^2+3}\leq1$:)[/quote][/quote]
it is will be easier
if we can prove:$(x^5-x^2+1)(y^5-y^2+1)(z^5-z^2+1) \ge 1$
(which is a sepical case I mentioned [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=62775]here[/url],nobody proved or disproved it.)
And then use the famous inequality:
$abc=1,a,b,c>0$
$\sum \frac 1{2+a} \le 1$

[quote="perfect_radio"] Secondly, can the inequality be generalized?[/quote]
Long time ago I conjectured that:
[i]If $x_1,x_2,...,x_n$ are positive numbers such that $x_1+x_2+...+x_n \geq n$, then for $p \geq 1$ the inequality holds:
$\frac{1}{x_1^p+x_2+...+x_n}+\frac{1}{x_1+x_2^p+...+x_n}...+\frac{1}{x_1+...+x_{n-1}+x_n^p} \leq 1$.[/i]

I proved this conjecture only for $p=2$ and $p=3$.
On the other hand, it is very nice and interesting the zhaobin's inequality:
$(1-a+a^p)(1-b+b^p)(1-c+c^p) \ge 1$
for $a,b,c$ positive numbers such that $a+b+c=3$.
I proved the similar inequality
$(1-a_1+a_1^2)(1-a_2+a_2^2)...(1-a_n+a_n^2) \geq 1$
for $a_i$ positive numbers such that $a_1+a_2+...+a_n=n$, and $n \leq 10$.
However, it seems that this ineq. is valid for $n \leq 13$.

[quote="Vasc"]I proved the similar inequality
$(1-a_1+a_1^2)(1-a_2+a_2^2)...(1-a_n+a_n^2) \geq 1$
for $a_i$ positive numbers such that $a_1+a_2+...+a_n=n$, and $n \leq 10$.
However, it seems that this ineq. is valid for $n \leq 13$.[/quote]

[b]What follows has a critical sign error, as pointed out by Vasc in the next post.[/b]

If you have proved it for the cases $n \le 10$, then I believe it is true for all positive integers $n$. I'll outline the solution here.

Note that $\frac{d^2}{dx^2}\left[\ln(x^2-x+1)\right] = \frac{1+2x-2x^2}{(1-x+x^2)^2}$ has a single positive zero, i.e. $f(x) = \ln(x^2-x+1)$ has a single inflexion point. By my version of the $n-1$ equal value principle (illustrated in example 21, page 16 of http://web.mit.edu/~tmildorf/www/Inequalities.pdf ) we can write $x_1 = \cdots = x_{n-1} = a$ and $x_n = n-an+a$. It is then left to see that $g(a) = (n-1)\ln(a^2-a+1) + \ln((n-an+a)^2-(n-an+a)+1) \ge 0$. For $n \ge 10$, this single variable inequality in $a$ falls to the same lines of argumentation illustrated in http://web.mit.edu/~tmildorf/www/ASharpBound.pdf . $g(1)=0, g'(1)=0, g''(1) > 0$, the inequality holds as $a \rightarrow \pm \infty$, so if false there must be a root of $g'(a) = 0$ other than $a=1$.  Removing nonzero junk factors of $g'(a)$, it must be that

$h(a) = g'(a) * (a^2-a+1)*(1-a+a^2-n+3an-2a^2n+n^2-2an^2+a^2n^2)/(n(n-1))$
$= (a-1)(1+2a-2a^2+n-3an+2a^2n)$

has root $h(a)=0$ other than 1. Then by Rolle's theorem there must be some root of $h'(a) = -1+8a-6a^2+4n-10an+6a^2n = 0$. The minimum of $h'(a)$, a quadratic in $a$, occurs where $h''(a) = 8-12a-10n+12an=0$ or $a = (5n-4)/(6n-6)$, but we have

$h'((5n-4)/(6n-6)) = (n^2-10n+10)/(6(n-1)) > 0$

for $n\ge 10$. Thus there can be no solution to $g(a) = 0$ other than $a=1$.

[b]In the above work, I dropped a minus sign at the very last step. It is actually:

$h'((5n-4)/(6n-6)) = -(n^2-10n+10)/(6(n-1)) > 0$ for $n < 9$.[/b]

Darn!

You are right, I have dropped a minus sign in my computations.

An even more straightforward, brutally computational solution is possible. First a clarification. $\ln(x^2-x+1)$ has two inflexion points over the reals, convex on the bounded interval. By Karamata, we can spread values on the concave intervals so that $n-2$ values are on the convex interval. But since $x^2-x+1$ is minimized in the convex interval, we can push the two values on the concave intervals together so that at least one can be combined with the $n-2$ in the convex interval, all the while decreasing the logarithmic sum in question. Having equated $n-1$ variables, the derivative with respect to $a$ is 0 only when $a=1$ or $a = (-2+3n\pm\sqrt{12-12n+n^2})/(2(-2+2n))$. $a=1$ is the trivial equality case. The two nontrivial critical points arise when $n \ge 11$.

\[ g\left((-2+3n-\sqrt{12-12n+n^2})/(2(-2+2n))\right) \qquad (*)  \]

is the interesting critical value. One easily checks that it $g$ positive at this value for $11 \le n \le 13$. $g$ at the other critical value is also positive for these values of $n$. What seems apparent from the graph of (*) is that it is negative for $n \ge 14$, which would mean 13 is the best possible. I will check for this later.

Of course 13 is the best bound! The same computations lead to the counterexample $a_1=\cdots=a_{13} = 0.6476047054\dots = \frac{20-\sqrt{10}}{26}$ and $a_{14} = 5.581138830\dots = 4+\sqrt{5/2}$. In this case,

\[ \prod_{i=1}^{14} (a_i^2-a_i+1) = 0.9158544491\dots  \]

Finally, take $a_{15} = \cdots = a_{n} = 1$, completing the problem.


Find the sum $\sum_{n=1}^\infty \frac{n^{4}}{3^{n}}$.

thats a lot of bad computation...the idea is set it equal to S, then multiply by 3, and take 3S-S, you get the degree of the top is 1 less, rinse and repeat until it becomes a geometric series,

can anyone find some sleek way of doing this? the answer should be 15... it is too nice to involve lengthy calculations.

What a criterion - "too nice for ugly calculations"... Are you prone to the idea that, for example, beautiful people can't be criminals because they're just too beautiful for that? :)

On a more mathemathical basis: Einstein's $E=mc^{2}$ is even more beautiful, but guess what? It's preceded by a 30-page paper.

Or on an even more mathemathical basis: Confer Wiles's proof of Fermat's Last Theorem, which is monstrously complicated for such an elegant and seemingly innocent statement.


[quote="GameBot"]Suppose that we win $ \$2$ if we flip a heads on a coin toss, but lose $ \$1$ if we flip tails.  What is the expected value, in dollars, of our winnings after one flip?[/quote]

$ \frac{1}{2}(2\minus{}1)\equal{}.5$.


At the midpoint of line segment $ AB$ which is $ p$ units long, a perpendicular $ MR$ is erected with length $ q$ units.  An arc is described from $ R$ with a radius equal to $ \frac{1}{2}AB$, meeting $ AB$ at $ T$.  Then $ AT$ and $ TB$ are the roots of:

$ \textbf{(A)}\ x^2\plus{}px\plus{}q^2\equal{}0 \\
\textbf{(B)}\ x^2\minus{}px\plus{}q^2\equal{}0 \\
\textbf{(C)}\ x^2\plus{}px\minus{}q^2\equal{}0 \\
\textbf{(D)}\ x^2\minus{}px\minus{}q^2\equal{}0 \\
\textbf{(E)}\ x^2\minus{}px\plus{}q\equal{}0$

[hide="Click for solution"]
Since $ MT\equal{}\frac{\sqrt{p^2\minus{}4q^2}}{2}$, $ AT\equal{}AM\plus{}MT\equal{}\frac{p\plus{}\sqrt{p^2\minus{}4q^2}}{2}$ and $ TB\equal{}AB\minus{}AT\equal{}\frac{p\minus{}\sqrt{p^2\minus{}4q^2}}{2}$.  The required equation is thus $ x^2\minus{}px\plus{}q^2\equal{}0$ or $ \boxed{\textbf{(B)}}$.
[/hide]

[quote="mathwizarddude"][quote="AIME15"]An arc is described from $ R$ with a radius equal to $ \frac {1}{2}AB$, meeting $ AB$ at $ T$.  [/quote]
What does "is described" mean here?[/quote]It means that the description of the 'arc' is part of a circle whose radius is specified.

I seem to be getting $ MT\equal{}\sqrt{(p/2)^2\minus{}(p\minus{}q)^2}$ instead if I'm understanding the problem correct.
By the way, is M taken to be the midpoint of line segment AB?

[quote="mathwizarddude"]I seem to be getting $ MT \equal{} \sqrt {(p/2)^2 \minus{} (p \minus{} q)^2}$ instead if I'm understanding the problem correct.
By the way, is M taken to be the midpoint of line segment AB?[/quote]Yeah, and I think the problem was a bit vague about that since it doesn't specifically say M is the midpoint. $ MR$ is perpendicular to the secant line $ AB$... and yea $ MR\equal{}q$.


Elementary question:

How would one insert a brace with a text caption above or below content?

$ \overbrace{x + y}^{\text{integer}}$

\overbrace{x+y}^{\text{integer}}

$ (a + b)^2\ne\underbrace{a^2 + b^2}_{\text{obvious}{\text{, isn't it?}}}$

(a+b)^2\ne\underbrace{a^2+b^2}_{\text{obvious}{\text{, isn't it?}}

Thanks a lot! Is there a way to get rid of the margin?

[code](a+b)^2\ne\underbrace{a^2+b^2}_{\substack{\text{obvious,}\\ \text{isn't it?}}}[/code]
$ (a\plus{}b)^2\ne\underbrace{a^2\plus{}b^2}_{\substack{\text{obvious,}\\ \text{isn't it?}}}$


A cube is increased to form a new cube so that the surface area of the new cube is double that of the original cube. By what factor is the volume of the cube increased? Express your answer in simplified radical form.

Setting up equations, we have $ 6s_1^2 \equal{} 12s_2^2$. Obviously $ s_2$ is $ \sqrt {2}$ times $ s_1$. So the volume is increased by a factor of $ 2\sqrt {2}$. I hope....  :maybe:


geometry
circumcircle
parallelogram
geometry proposed

[quote="Virgil Nicula"]Let $ ABCD$ be a convex quadrilateral for which $ A \equal{} 30^{\circ}$ , $ AC\perp BD$ and

$ m(\widehat {BAC}) \plus{} m(\widehat {CBD}) \equal{} 60^{\circ}$ . Prove that $ m(\widehat {ACD}) \equal{} 3\cdot m(\widehat {CAD})$ .

See [url=http://www.mathlinks.ro/viewtopic.php?t=199367][size=117][color=red][b]here[/b][/color][/size][/url] a trigonometrical proof. I wish a synthetical proof, if it is possibly ![/quote]

[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. - A triangle $ \bigtriangleup ABC$ with $ \angle A \equal{} 60^{o}$ is given and let $ M$ be, the midpoint of the arc $ BC$ of its circumcircle $ (O),$ not containing the point $ A.$ The $ A$-altitude $ AD,$ intersects $ (O)$ at one point so be it $ E$ and let be the point $ F\equiv BC\cap AM.$ Prove that $ \angle FEA \equal{} 3\angle FAE$. [/color][/size][/b]

[b][size=100]PROOF[/size][/b]. – Let $ H,\ O$ be, the orthocenter and circumcenter respectively of $ \bigtriangleup ABC.$

It is easy to show that the sideline $ BC,$ is the midperpendicular of the segments $ HE,\ OM.$

Also, we can easy to show that the line segment $ AM,$ is the nidperpendicular of the segment $ OH$ $ ($ From $ AH\parallel OM',$ where $ M'$ is the antidiametric point of $ M$ and $ \angle HOM \equal{} \angle EMO \equiv \angle EMM' \equal{} \angle AM'M,$ we conclude that $ AHOM'$ is a parallelogram and so, from $ AH \equal{} OM' \equal{} AO$ and from $ \angle HAM \equal{} \angle AMO \equal{} \angle OAM,$ we are done $ ).$

So, we have $ \angle FEA \equal{} \angle FHE \equal{} \angle EAF \plus{} \angle HFA \equal{} \angle EAF \plus{} \angle OFA \equal{} \angle EAF \plus{} 2\angle FMO$ $ \Longrightarrow$ $ \angle FEA \equal{} 3\angle EAF$ and the proof is completed.

Kostas Vittas.

PS. If we consider the tangent of $ (O)$ at vertex $ B,$ it is easy to show that in the quadrilateral $ ABEF,$ except $ AE\perp BF,$ we have also $ \angle BAF \equal{} 30^{o}$ and $ \angle EAB \plus{} \angle EBF \equal{} 60^{o}.$


Thanks but I am sure there is formula for sum of this series.

The sum telescopes.  Factor $ x^4 \plus{} x^2 \plus{} 1$.

I do not think so . This sumation is useful when each term can be written in form $ \frac{1}{A}\minus{}\frac{1}{B}$ which is not case here.

[quote="canada"] ... summation is useful when each term can be written in the form $ \frac {1}{A} \minus{} \frac {1}{B}$, which is not the case here.[/quote]

Well,

$ \frac {1}{x^{4} \plus{} x^{2} \plus{} 1} \equal{} \frac {1 \minus{} x}{2(1 \minus{} x \plus{} x^{2})} \plus{} \frac {1 \plus{} x}{2(1 \plus{} x \plus{} x^{2})}.$

Hence, the sum of the given series is

$ \frac{1}{2}\left(\frac{1}{1\plus{}1\plus{}1^{2}} \plus{} \frac{1}{1\plus{}2\plus{}2^{2}} \plus{} \frac{1}{1\plus{}3\plus{}3^{2}} \plus{} \ldots \right)$


calculus
integration
function
LaTeX
floor function
ceiling function
algebra

If
   $ \ f(x) \equal{} \minus{} 1$

Then evaluate [size=200] [b]$ \int_{ \minus{} 1}^{0}\frac { [ x ] dx }{ [\frac { x \minus{} f(x) \plus{} n \minus{} 1}{n} ] \plus{} [\frac {x \minus{} f(x) \plus{} n}{n}] \plus{} [\frac { x \minus{} f(x) \plus{} n \plus{} 1}{n} ] \plus{} [\frac { x \minus{} f(x) \plus{} n \plus{} 2 }{n} ] }$[/b][/size]

Where [ ] denotes the greatest integer function

You forgot $ dx$ I guess, but $ [x] \equal{} 0$ for $ x \in [ \minus{} 1,0]$?

no actually $ [x] \equal{} \minus{}1$ for $ x\in [\minus{}1,0]$. and ya i forgot dx now changed

Notation/$ \text{\LaTeX}$ hint:

I know that $ [\cdot]$ is moderately well established as the greatest integer function, but brackets also have many other uses, including being a synonym for parenthesis. I feel like campaigning for an end to the bracket notation in favor of a notation that is also well-established, completely unambiguous, and visually evocative: Refer to it as the "floor" function and use the symbol $ \lfloor\cdot\rfloor$.

The $ \text{\LaTeX}$ part of this comment: the two required symbols are given by \lfloor and \rfloor. That is, \lfloor x\rfloor becomes $ \lfloor x\rfloor.$

Furthermore, these symbols can be adaptively sized with \left and \right:
\left\lfloor \frac{n}{2}\right\rfloor for $ \left\lfloor \frac{n}{2}\right\rfloor.$

With that background, you can probably guess the meaning of $ \lceil x\rceil$ without having to look it up.

is  f(x)=-1  for all reals?   i  mean is it  a constant function ?  then  what is the purpose of giving  it in the  q ?

ya f(x) = 1 for x>= 0
           = -1 for x<0

. . . try to be as precise as possible and specify everything without unnecessary symbols.  From what I can interpret above, looks like you have:

\[ \minus{}\int_{\minus{}1}^0\frac{1}{\lfloor{\frac{x\plus{}n}{n}}\rfloor\plus{}\lfloor{\frac{x\plus{}n\plus{}1}{n}}\rfloor\plus{}\lfloor{\frac{x\plus{}n\plus{}2}{n}}\rfloor\plus{}\lfloor{\frac{x\plus{}n\plus{}3}{n}}\rfloor}dx,\quad n\in \mathbb{N}^{\plus{}}\]


so i solved the cube couple of times with the original method...

but it took me like 5 to 10 minutes...

but i saw people doing it in like 30 seconds...

is there faster way to solve the cube?

Well using the basic "booklet" method I can get under 40 seconds easily.

Hmm...lube your cube and get very familiarized with the basic method before moving on.  Then, try the Fridrich method.

Memorize algorithms. Easiest solution. Don't just memorize and nit practice though.

I know a person that knows the same amount of algorithms as I do, but I do it twice as fast somehow.

most people can go sub-minute using the beginner layer method without much dexterity, sub40 with. i don't know much about other methods, but if you go with fridrich, start by learning cross OLL's and F2L.


[quote="abacadaea"]let sum $ S$ equal $ 1+2+.......5687$. Find $ n-k$ if $ S=\binom{n}{k}$.[/quote]

actually it isnt that hard if you write it as the sum of an arimetic progression.

Its just $ \frac{(1+5687)(5687)}{2}$.  that should get you to the answer...(dont try to multiply that out) but there are actually two answers..I dont know if you want the bigger or smaller one.

im going to get this problem SOOOO wrong..
but at least ill give it a try
[hide]setting up the equation for sums
so. $ (5687*5688)/2=n!/k!(n-k)!$ so we're looking for n-k. i thought it was okay to assume that 5687 is $ n$ and $ 5687*5688$ was the simplified form, when $ (n-k)!$ is taken out of the numerator. therefore $ n-k$ =5686?????[/hide]

dont look at the solution if you havent done the problem yet.(P.S. they aren't very good)

[hide="solutions"][hide="1"] we know that $ \binom{n}{k}=\frac{n(n-1)....(n-k)}{k}$ and we set that equal to $ \frac{n(n-1)}{2}$ where n is 5686 instead of 5687. Note the left side had 2 monomials multiplied together on the top and a 2 for the denominator. therefore $ k$ must be $ 2$.and the answer is $ 5686$[/hide][hide="2"]if we say it were $ 1+2+3....n=S$ then $ S=\frac{n(n+1)}{2}$ We know $ \binom{n}{k}=\frac{n!}{k!(n-k)!}$. Now we need to solve for $ k$ in the equation $ \binom{n}{k}=S=\frac{n(n+1)}{2}$. multiplying the left side by $ \frac{n+1}{n+1}$ gets us $ \frac{(n+1)!}{k(n-k)!(n+1)}=\frac{n(n+1)}{2}$. Dividing both side by $ S$ gets us $ \frac{2(n-1)!}{k(n-k)!(n+1)}$. Multiplying both sides by $ \frac{k(n-k)!(n+1)}{(n-1)!}$ gets us $ k(k-1)!=2=k!$ therefore $ k=2$ and $ 5688-2=5686$[/hide][/hide]
[hide="answer"]5686[/hide]

actually, how would you get that.
abacadaea posted the answers and they are both 5686..how did you get 2?

This is because 5688C2 is the same as 5688C5686. Therefore the answer can be
[hide]5688-2 = 5686 or 5688-5686 = 2[/hide]

oh yeah, i forgot to consider that :blush:

[quote="ordinary boy"][quote="mihail911"]Yay!!!
i can't believe i got it right!!!
 :coolspeak:  :trampoline:  :first:
good problem, abacadaea!!![/quote]

You are quite aware the answer can be $ 2$ also right?[/quote]

yes i am aware of that because $ \binom{n}{k}=\binom{n}{n-k}$


modular arithmetic
quadratics
number theory unsolved
number theory

Positive integer $n$ cannot be divided by $2$ and $3$, there are no nonnegative integers $a$ and $b$ such that $|2^a-3^b|=n$. Find the minimum value of $n$.

This is old but since there is no solution, I will add mine.

It is easy to check that all numbers below 35 satisfying the condition have solution.

Consider now $ |2^a \minus{} 3^b| \equal{} 35$.
The first case is that of $ 2^a \minus{} 3^b \equal{} 35$.  It is easy to see that if a<4, there is no solution - now take modulo 16, and note that $ \minus{} 3^b \equal{} \equal{} 3 \pmod{16}$ has no solution.

The second case is $ 3^b \minus{} 2^a \equal{} 35$.  By mod 4, b is odd, and by mod 3, a is even.  Thus, we have an equation of the form $ 3x^2 \minus{} y^2 \equal{} 35$.  Consider this modulo 7.  Note that 3 is not a quadratic residue, so we are done.


trigonometry
MATHCOUNTS
geometry
LaTeX
rich tag

Lets see what the average grade for this forum is...

Now, be honest..

what do you mean? I don't understand..

maybe I'm missing something lol  :blush:

hmm...i cant seem to get the poll to work...

nvm i got it..

*

plz state your grade...(unless yoiur embarassed then please dont)

worst for me is a middle A

I think a 96 in Algebra 1 in 2005.

I got a 70.0 in Algebra I...

100% Test Average
0% Homework Average
0% Bring A Reading Book Average
0% Return Progress Reports

somehow thats a 70.

I'm homeschooled, so.. Mathcounts test??? ok. Here's mine:

.6304347826 on my first state competition in 6th grade.

umm i got a B in reading in fourth grade cuz my teacher was racist but ur talking about math now i see...

never gotten lower than a 93 ever in math (93 was because i was too lazy and "failed" some quizes)

I cant remember far back, lol. But the worst I can remember was a 85 this year in grade 9. (I'm realy in grade 7, so I have a excuse)

Let's see, through now, A, 95. However, I didn't feel like showing ALL my work on my alg. 2 midterm, so i colud dip to a...oh...nvm...99.

[quote="perfect628"]However, I didn't feel like showing ALL my work on my alg. 2 midterm, so i colud dip to a...oh...nvm...99.[/quote]

yeah i hate how teachers take off points for stuff you can do mentally in your head

I got a B on my final last year.  And a couple of B's in 5th grade.

I don't think I've ever gotten more than a 92 or a 93 since 8th grade though, which seems really low compared to everyone else here.

I don't remember but I'm pretty sure I never got below an A on a report card, though I used to get some poor grades on a test occasionally.

Now I try never to miss a single point on a test, and I get a bunch of extra credit, so my grade is usually above 100%  :D

[quote="white_horse_king88"]I think I got a 93 in sixth grade. Because all I did was play chess and eat candy during class. And I didn't do homework. Rofl...
Glad I don't do that anymore.
I have a 100 in Calc. Actually more like 110... but too bad the extra points don't count.[/quote]

wow if i didn't do my homework my maximum grade (assuming i missed a week's worth of hw) would be like an extremely low B

probably a lowwwwwwwww A in 6th grade (worst teacher ever)

Wow.  The results clearly show that I'm on a math forum right now! :)

About a 4.6 in calculus. (On the normal 4-point GPA scale)

My lowest grade is a 90.9 (A-) (in math)
I made a silly mistake so I got a 10/11

OOOOPs clicked the wrong buttun!!!!!!!!!!!!!!!!

[quote=n0vad3m0n][quote="Thales418"]how can i type infinity in latex?[/quote]

completely not related to this topic unless your thinking that your grade is $\infty$...[/quote]

Dang, 2007 pretty old


algebra polynomial trigonometry function vector induction calculus

If a function $f: U \subseteq E \rightarrow F$ (where $U$ is an open subset of a normed vector space $E$, and $F$ another normed vector space) is $n$ times differentiable on $U$, then you have the approximation: $f(x+h)=\sum_{k=0}^{n}{\frac{1}{k!}D^{k}f(x)(h^{k})}+o(h^{n})$, where $h^{k}=(h,\dots,h)$ $k$ times, whenever, it is understood, $x \in U$ and $h$ is small enough.
A proof of this is easy by induction, using the mean value theorem (inequality).

[quote="shyong"]Find the Taylor Polynomial of

$f(x,y)=e^{x}\cos y$ about $(0,0)$ .

I know how to do Taylor Polynomial with 1 variable , but how to deal with 2 variables ?What is the general pattern for 2 variables taylor polynomial?[/quote]

I look up on the Internet and find it in Wiki:
  http://en.wikipedia.org/wiki/Taylor_series
Is it what you are looking for?

The general Taylorseries expression for several variables is given by: \[T_{N}(x) \&= \sum_{i_{1}=0}^\infty \cdots \sum_{i_{n}=0}^\infty \frac{f^{(i_{1},i_{2},\ldots,i_{n})}(0)}{i_{1}!i_{2}!\cdots i_{n}!}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}\,, \quad f^{(i_{1},i_{2},\ldots,i_{n})}(0) = \frac{\partial^{i_{1}+\dotsb+i_{n}}f}{\partial x_{1}^{i_{1}}\cdots \partial x_{n}^{i_{n}}}\mid_{x=0}\,, \\ T_{2}(x) \&= \sum_{i+j = 1}^\infty \frac{f^{(i, j)}(0)}{i!j!}x_{1}^{i}x_{2}^{j}= \sum_{\abs{I}= 1}^\infty \frac{1}{I!}\frac{\partial^{\abs{I}}f}{\partial x^{I}}\mid_{x=0}\, x^{I}\,.\] (For simplicity, convergence centrered at $a= 0$.)

For example, the second-degree Taylor polynomial for $f(x,y) = e^{x}\cos(y)$
centered about $(0,0)$
is \[( 1+x+1/2\,{x}^{2}+1/6\,{x}^{3}+1/24\,{x}^{4}+{\frac{1}{120}}\,{x}^{5}+O(x^{6})) ( 1-1/2\,{y}^{2}+1/24\,{y}^{4}+O(y^{6})) \\ = 1-1/2\,{y}^{2}+1/24\,{y}^{4}+x-1/2\,x{y}^{2}+\ldots\] found by just multiplying the two-one variable Taylor series together keeping the degrees lower than 3. This procedure works because of the uniqueness characterization of Taylor polynomials.

[quote="shyong"]isn't that just Taylor Polynomial for 1 variable ? Im asking for 2 variable though ...[/quote]

Not it is not, and it is more general than the all-made formula given by Ottem. The point is, working in a general normed vector space $E$, nothing prevents you from taking $E=\mathbb{R}^{n}$ for instance, and you will work with $n$ variables.


algebra
polynomial
number theory
greatest common divisor
least common multiple
vector
Ring Theory

Hi,

that's not from any contest. It occurred to me when
reading a paper and I cannot solve it:

Let $ \mathbb{C}[x_1,x_2]$ be the polynomial ring in two
variables over $ \mathbb{C}$. Let $ 0\leq k_1 < k_2 < k_3$
be integers. Does there exist a polynomial $ f\in \mathbb{C}[t_1,t_2,t_3]$,
such that $ f(x_1^{k_1}x_2, x_1^{k_2}x_2, x_1^{k_3}x_2) \equal{} 0$ ?
Does somebody know about this?

I assume you mean nonzero polynomials.  Take $ f(x, y, z) \equal{} x^{k_3 \minus{} k_2} y^{k_1 \minus{} k_3} \minus{} z^{k_3 \minus{} k_1}$.

Thanks for your answer.

Yes, the non-triviality condition $ 0\neq f$ should be added.

However, since $ k_1 < k_3 \Leftrightarrow k_1 \minus{} k_3 < 0$, you have not given a polynomial at all!

If I take your suggestion to be
$ f : \equal{} t_1^{k_3\minus{}k_2} t_2^{k_3\minus{}k_1} \minus{} t_3^{k_3\minus{}k_1}$
then

$ f(x_1^{k_1}x_2,x_1^{k_2}x_2,x_1^{k_3}x_2)\equal{}$
$ (x_1^{k_1}x_2)^{k_3\minus{}k_2}\cdot (x_1^{k_2} x_2)^{k_3\minus{}k_1}\minus{}(x_1^{k_3}x_2)^{k_3\minus{}k_1}\equal{}$
$ x_1^{k_1k_3\minus{}k_1k_2}x_2^{k_3\minus{}k_2}x_1^{k_2k_3\minus{}k_1k_2}x_2^{k_3\minus{}k_1}\minus{}x_1^{k_3k_3\minus{}k_1k_3}x_2^{k_3\minus{}k_1}$

and without further simplification it is obvious that this is not zero. So
unfortunately, this is not a polynomial of the desired kind.

[quote="t0rajir0u"]My apologies; I intended $ f(x, y, z) \equal{} x^{k_2 \minus{} k_1} z^{k_3 \minus{} k_2} \minus{} y^{k_3 \minus{} k_1}$.[/quote]

Hi t0rajir0u,

thank you very much for your suggestion!

with this $ f$ we get

$ f (x_1^{k_1} x_2, x_1^{k_2} x_2, x_1^{k_3} x_2) \equal{}$
$ (x_1^{k_1} x_2)^{k_2 \minus{} k_1} \cdot (x_1^{k_3} x_2)^{k_3 \minus{} k_2} \minus{} (x_1^{k_2} x_2)^{k_3 \minus{} k_1}$

Now just look at $ x_1$. On the left side of the minus
sign, there appears an $ x_1^{k_1k_2 \minus{} k_1^2 \plus{} k_3^2 \minus{} k_2k_3}$ within a monomial. On the right-hand side
of the minus sign, there merely appears an $ x_1^{k_2k_3 \minus{} k_1k_2}$, so these sub-monomials, when added (which
they of course aren't when expanding out your suggestion, but their vanishing is necessary for
such a polynomial to work and here it shows that the suggested one does not work),
still leave the positive power $ x_1^{k_3^2 \minus{} k_1^2}$, so for this reason alone, that does not work!

Why do you think it works?

I tried many things more or less similar to your
attempt. I restricted myself to polynomials having
up to four monomials, and got a bunch of ugly
systems of linear equations. In all cases, if I could
show solvability, then the solutions corresponded
to exponents in $ f$ which were partly $ \emph{negative}$,
so all these tries did not give me a polynomial.

:oops:

I'm going to check my work one more time.  Take $ f(x, y, z) \equal{} x^{k_3 \minus{} k_2} z^{k_2 \minus{} k_1} \minus{} y^{k_3 \minus{} k_1}$.  Then (letting $ x \equal{} x_1$ and $ y \equal{} x_2$)

$ f(x^{k_1} y, x^{k_2} y, x^{k_3} y) \equal{} x^{k_3 k_1 \minus{} k_2 k_1} y^{k_3 \minus{} k_2} x^{k_3 k_2 \minus{} k_3 k_1} y^{k_2 \minus{} k_1} \minus{} x^{k_3 k_2 \minus{} k_2 k_1} y^{k_3 \minus{} k_1}$

and $ (k_3 k_1 \minus{} k_2 k_1) \plus{} (k_3 k_2 \minus{} k_3 k_1) \equal{} k_3 k_2 \minus{} k_2 k_1$ and $ (k_3 \minus{} k_2) \plus{} (k_2 \minus{} k_1) \equal{} k_3 \minus{} k_1$ as desired.

Perhaps I should explain the motivation:  $ (k_2, 1)$ is a vector on the line between $ (k_1, 1)$ and $ (k_3, 1)$, so some weighted average of the latter two is equal to the former.  Since $ k_1, k_2, k_3$ are positive integers, the weights are positive rationals, and so we can clear the denominators and get integer weights.


A man requests something and gets it, but he never intended to use it.

What happened¿

yes/no questions, right?
I'll start with something really bizzare: did he have counterferit money, and bought something cheap (that he would never use) to get a lot of real money in change?

[hide]Is he physically handicapped in some way?

Edit: I was thinking of the riddle where a wheelchair guy adds a 2nd floor section to his house for guests to stay in.
[/hide]

did he have counterferit money, and bought something cheap (that he would never use) to get a lot of real money in change? - no

Did he want it? - yes, in some sense

Is he physically handicapped in some way? - no

[hide="Is he..."]...buying a coffin for a deceased person?[/hide]

that's what i was gonna say..
a [hide="........."]coffin for a dead man[/hide]

Is he buying a coffin for a deceased person? - no

maybe it's a

[hide]fire alarm or security system?[/hide]

[hide="How about..."] Insurance? [/hide]

Good idea, but no. In fact it's a rather long riddle, so you better ask questions (yes-no/finite given answers).

What form did the 'request' take? In person? Telephone? E-mail? (Does it matter?)
Does the fact that it is a 'man' alter the outcome? Could it instead be a woman or a child?

Are the preparations to make the object reusable expensive? requiring an expert? undoable by the man alone?

Did the man need the object before he met the person who gave him the object?
Did he need it one hour before that?
Two hours?
8?
12?
18?
a day?
two days?
4?
7?
28?

Are the objects sold in a store or something?

What form did the 'request' take? In person? Telephone? E-mail? - none of the mentioned
(Does it matter? - a bit)
Does the fact that it is a 'man' alter the outcome? no
Could it instead be a woman or a child? - woman yes, child not

Are the preparations to make the object reusable expensive? - no
requiring an expert? - yes
undoable by the man alone? - unknown/irrelevant

Did the man need the object before he met the person who gave him the object? - no

Are the objects sold in a store or something? - special stores only

I don't really understand how "Are the preparations to make the object reusable undoable by the man alone?" could be irrelevant...

However, what the penultimate (I'm bored) question says appears to be pretty important... my guess: the other man poisoned the man and offered him the antidote for a bunch of money.

Did something like that, with "poisoned" probably different, happen?

"but he never intended to use it..."

Tough.

Is it some kind of armor? (like bulletproof vest or something else in that venue)?
Does he need it to survive under some circumstances that he hopes will never really happen (like, say, nuclear war)?
Is it a measuring device of some sort (like a Geiger counter or a carbon monoxide detector)?

I don't really understand how "Are the preparations to make the object reusable undoable by the man alone?" could be irrelevant... - let's say: it's irrelevant for the story itself, and I don't think it helps you to find out what it is.

However, what the penultimate (I'm bored) question says appears to be pretty important... my guess: the other man poisoned the man and offered him the antidote for a bunch of money. - no

Did something like that, with "poisoned" probably different, happen? - no

Is it some kind of armor? (like bulletproof vest or something else in that venue)? - no
Does he need it to survive under some circumstances that he hopes will never really happen (like, say, nuclear war)? - no
Is it a measuring device of some sort (like a Geiger counter or a carbon monoxide detector)? -no

This reminds me of an old story I read when I was a child... Let's see if that is it. Did he request several objects because, despite he needed only one he was afraid that if he asked for just one, he would be given a faulty one? - yes

Oh, I think I got it!
I'm going to ask a question to check.
Was the man a terrorist (or something similar)?

Probably a total spoiler:
[hide]Oh, was it that one with a guy hijacking a plane and asking for two parachutes, so people would think he was going to take a hostage , and they didn't want to endanger the hostage's life, so they had to give him two good ones, but he didn't take a hostage after all?[/hide]

Was the man a terrorist (or something similar)? - I wouldn't call this a terrorist...

Was he intending to use his object for military purposes?
Educational?
Business?
- None

Probably a total spoiler: yes (four, not two)

[url=http://en.wikipedia.org/wiki/D._B._Cooper]Here[/url] is the full story I was thinking of (it is amazing that you can find almost everything online nowdays :))

Well, he was "something similar", I guess, since he was hijacking a plane.


It's $sin$. In spanish is $sen$ so i got confused.


Find the least positive integer with $ 100$ divisors.

This is what I did:

Since we are trying to find the smallest integer with 100 divisors we could think of finding something in the form of $ 2^x3^y$.  The amount of divisors would be $ (x \plus{} 1)(y \plus{} 1) \minus{} 1 \equal{} 100 \implies (x \plus{} 1)(y \plus{} 1) \equal{} 101$.  $ 101$ is a prime number though.  So one of either x+1 or y+1 must equal 1 and that means one of x or y must be 0.  To make this the smallest possible y = 0.  So our number is $ 2^{100}$.

(you can also see that having more than two prime factors is impossible since 101 only has one factor other than itself)

[quote="Truffles"]This is what I did:

The amount of divisors would be $ (x \plus{} 1)(y \plus{} 1) \minus{} 1 \equal{} 100$ [/quote]
If you had read the post above you....you would have realised that you were wrong here....

Yeah, that's exactly what I thought tomekk had done....but he made a mistake in the last step....


On a calculator screen, what year since 1795 looks the same backward, forward, and upside down?

The numbers 1 and 8 look the same upside down. This year number must be a palindrome, and must be greater than 1795, so the answer is $ \boxed{1881}$.

I'm going to assume something- though on a calculator, you want look "backwards", like behind the calc, I believe you need to in this prob. If you assume backwards as reading the number backwards, there are multiple solutions.

We find, through looking at all 10 digits, 0, 1, and 8 are the only ones which look the same in every way. We need to make a number from these digits that is from 1795 to 2009 and is a palindrome. This is $ \boxed{1881}$.

- I'm sending this only to clarify backwards. Small correction- Actually, 0, 1, 2, 5, and 8 look the same upside down. The way you did it IS the ftwer's quick thinking way though.

It would help if they reworded the problem a bit.  It was kind of confusing.

Isn't this problem a bit out of date? Because 2002 works as well.

Not exactly...

If you look closely, there is a curve at the top of the $2$.

If you flip the $2$, the top would become flat.

Not a really big difference...but possibly...

But the problem says "on a calculator screen" so the 2 would look the same upside down.

On a Scientific TI-30 calculator, the 2 has a curve on the top. So yes, my reasoning is still correct.

You guys are missing the condition "backward".
2002 backwards looks more like 5005, but maybe that's just me.

I guess you can go with that.

but it says that the numbers are to be equal.
[quote="GameBot"]On a calculator screen, what year since 1795 [b]looks the same[/b] backward, forward, and upside down?[/quote]

Is the question for the least year since 1795?

Cause I just put 11111 and it was wrong...


algebra
polynomial
calculus
integration
modular arithmetic
quadratics
algebra unsolved

Find all integers $n$ such that the polynomial $ p(x)=x^5-nx-n-2$ can be written as product  of two non-constant polynomials with integral coefficients.

[quote="math10"]Find all integer  $ n$ such that the polynomial $ p(x) \equal{} x^5 \minus{} nx \minus{} n \minus{} 2$ can be written as product  of two no_constant polynomials  with integral coefficients.[/quote]

[b]Case 1:[/b] One of the two polynomials is of degree $ 1$, i.e. $ (x\minus{}a)$ for some integer $ a$ . Then by Gauss's lemma we have $ a|\minus{}2 \Rightarrow$ possible value of $ a\equal{}1,\minus{}1,2,\minus{}2$.
Put $ x\equal{}1,\minus{}1,2,\minus{}2$ into the equation $ P(x)\equal{}0$ we get $ n\equal{}10, 34$ as solution for this case.

[b]Case 2:[/b] One of the two polynomials is of degree $ 2$, i.e. $ (x^2\plus{}ax\plus{}b)$, then by Gauss's Lemma we have possible values for $ b\equal{}1,\minus{}1,2,\minus{}2$.

Now $ P(x) \equal{}(x^2)^2x\minus{}nx\minus{}n\minus{}2 \equiv (ax\plus{}b)^2x\minus{}nx\minus{}n\minus{}2$
$ \equiv [(2ab\minus{}a^3)x\plus{}(b^2\minus{}a^2b)]x\minus{}nx\minus{}n\minus{}2 \equiv (a^4\minus{}3a^2b\plus{}b^2\minus{}n)x\plus{}(\minus{}2ab^2\plus{}a^3b\minus{}n\minus{}2) \equiv 0 \pmod{x^2\plus{}ax\plus{}b}$

$ \Leftrightarrow a^4\minus{}2a^2b\plus{}b^2\minus{}a^2b\equal{}\minus{}2ab^2\plus{}a^3b\minus{}2\equal{}n$

$ \Rightarrow (a^2\minus{}2b)(a^2\minus{}ab\minus{}b)\equal{}b^2\minus{}2$

[b](i)[/b] $ b\equal{}1$, then $ (a^2\minus{}2)(a^2\minus{}a\minus{}1)\equal{}\minus{}1 \Leftrightarrow a\equal{}\minus{}1$
[b](ii)[/b] $ b\equal{}\minus{}1$, then $ (a^2\plus{}2)(a^2\plus{}a\plus{}1)\equal{}\minus{}1$, no solution.
[b](iii)[/b] $ b\equal{}2$, then $ (a^2\minus{}4)(a^2\minus{}2a\minus{}2)\equal{}2$, no solution.
[b](iv)[/b] $ b\equal{}\minus{}2$, then $ (a^2\plus{}4)(a^2\plus{}2a\plus{}2)\equal{}\minus{}2$, no solution.

$ (a,b)\equal{}(\minus{}1,1) \Rightarrow n\equal{}\minus{}2(\minus{}1)(1)^2\plus{}(\minus{}1)^3(1)\minus{}2 \Rightarrow n\equal{}\minus{}1$

So $ n\equal{}\minus{}1,10,34$ are the only possible $ n$s

Note: stevencheng's solution appears to erroneously treat $ \minus{} 2$ as the constant term of $ p(x)$ instead of $ \minus{} n \minus{} 2$, which makes the problem much easier, and misses some solutions.

[hide="Solution"]Suppose that $ p(x)$ factors into a first-degree and a fourth-degree polynomial--then we have a factor $ (x \minus{} x_0)$ and $ x_0^5 \minus{} n(x_0 \plus{} 1) \minus{} 2 \equal{} 0\Leftrightarrow n \equal{} \frac {x_0^5 \minus{} 2}{x_0 \plus{} 1} \equal{} x_0^4 \minus{} x_0^3 \plus{} x_0^2 \minus{} x_0 \plus{} 1 \minus{} \frac {3}{x_0 \plus{} 1}$ for an $ x_0\in \mathbb{Z}$. This yields $ x_0 \plus{} 1\mid 3\Leftrightarrow x_0\in \{2,0, \minus{} 2, \minus{} 4\}$. These values of $ x_0$ correspond to $ n\in \{10, \minus{} 2,34,342\}$. (Note that we can discount $ x_0 \equal{} \minus{} 1$ because $ p( \minus{} 1) \equal{} \minus{} 3\neq 0, \forall n$.)

Now suppose that $ p(x)$ factors into the product of a cubic polynomial and a quadratic, both of which must be monic and have integral coefficients, say $ x^3 \plus{} ax^2 \plus{} bx \plus{} c$ and $ x^2 \plus{} dx \plus{} e$. Expanding the product and equating terms with $ x^5 \minus{} nx \minus{} n \minus{} 2$, we have:
[list][*]$ a \plus{} d \equal{} 0$
[*]$ e \plus{} ad \plus{} b \equal{} 0$
[*]$ ae \plus{} bd \plus{} c \equal{} 0$
[*]$ be \plus{} cd \equal{} \minus{} n$
[*]$ ce \equal{} \minus{} n \minus{} 2$[/list]
Then we have $ d \equal{} \minus{} a$, $ e \equal{} a^2 \minus{} b$, and $ c \equal{} 2ab \minus{} a^3$. The last two equations specify $ be \plus{} cd \minus{} 2 \equal{} ce\Leftrightarrow (2a \minus{} 1)b^2 \minus{} (a^2 \plus{} 3a^3)b \plus{} a^5 \plus{} a^4 \minus{} 2\equal{}0$ (1). This is a quadratic in $ b$, with discriminant $ D \equal{} (a^2 \plus{} 3a^3)^2 \minus{} 4(2a \minus{} 1)(a^5 \plus{} a^4 \minus{} 2) \equal{} a^6 \plus{} 2a^5 \plus{} 5a^4 \plus{} 16a \minus{} 8$. As $ a$ and $ b$ must be integral, $ D$ must be a perfect square (this is a necessary but not necessarily sufficient condition for $ b$ to be integral). But for $ a\geq 3$ and $ a\leq \minus{} 4$ respectively, we have:
[list][*]$ (a^3 \plus{} a^2 \plus{} 2a \minus{} 2)^2 < a^6 \plus{} 2a^5 \plus{} 5a^4 \plus{} 16a \minus{} 8 < (a^3 \plus{} a^2 \plus{} 2a \minus{} 1)^2$
[*]$ (a^3 \plus{} a^2 \plus{} 2a \minus{} 1)^2 < a^6 \plus{} 2a^5 \plus{} 5a^4 \plus{} 16a \minus{} 8 < (a^3 \plus{} a^2 \plus{} 2a \minus{} 2)^2$[/list]
This can be verified easily using the expansions $ (a^3 \plus{} a^2 \plus{} 2a \minus{} 2)^2 \equal{} a^6 \plus{} 2a^5 \plus{} 5a^4 \minus{} 8a \plus{} 4$ and $ (a^3 \plus{} a^2 \plus{} 2a \minus{} 1)^2 \equal{} a^6 \plus{} 2a^5 \plus{} 5a^4 \plus{} 2a^3 \plus{} 2a^2 \minus{} 4a \plus{} 1$.

This means that for $ a > 2$ and $ a < \minus{} 3$, $ D$ cannot be a perfect square (as it is bounded by two consecutive squares), so $ b$ cannot be an integer. Testing $ a\in \{2,1,0, \minus{} 1, \minus{} 2, \minus{} 3\}$ shows that $ D$ is a perfect square for $ a \equal{} 1$ only ($ D \equal{} 4$) and this corresponds to $ b\in \{4,0\}$ when solving (1). Finally, setting $ a \equal{} 1$ and using $ b \equal{} 4$ and $ b \equal{} 0$, respectively, yields the factorizations $ (x^3 \plus{} x^2 \plus{} 4x \plus{} 7)(x^2 \minus{} x \minus{} 3) \equal{} x^5 \minus{} 19x \minus{} 21$ ($ k \equal{} 19$) and $ (x^3 \plus{} x^2 \minus{} 1)(x^2 \minus{} x \plus{} 1) \equal{} x^5 \plus{} x \minus{} 1$ ($ k \equal{} \minus{} 1$).

Therefore, $ p(x) \equal{} x^5 \minus{} nx \minus{} n \minus{} 2$ is reducible over $ \mathbb{Z}[X]$ only for $ n\in \{10, \minus{} 2,34,342,19, \minus{} 1\}$.[/hide]


superior algebra
superior algebra unsolved

Does there exist a set containing all other sets?

That set would contain itself, thus contradicting axioms of set theory (the axiom of regularity, to be exact).

Yes, but what is the original paradox that arises from considering such a set?

There is no "original" paradox. Maybe you talk about Russel's paradox, but that's somewhat different (and in fact independent).

usually you say that the sets of subsets of a given set is bigger than the set itself, and this leads to a contradiction (without appealing to regularity).

More rigorously, considering $ S$ the set containing all sets we could argue that the set of all subsets of $ S$ would have a bigger cardinality than $ S$ thus making it as ma_go said "bigger" than $ S$.
Although there are axiomatic set theories where there exists a universal set (see http://en.wikipedia.org/wiki/Universal_set).


Would anybody be interested in competing in a Mathcounts Tournament or Contest? This would probably be kind of like the Getting Started Tournament level and may be good practice for those competing in Mathcounts next year.

can non mathcount age qualifyers join the tournament?

sure, mathcounts alums and those who are too young to participate can be in this too.

I want to compete too. Never can have too much competition in your life.

era

I'll join.  I need to get out all my not-knowing-how-to-read and not-knowing-how-to-add before the real competition.

(I seem to be joining every competition I see  :D )

[quote="tandjsnell"]Great! I really need practice on sprint round questions.
Will you be doing this just like a regular mathcounts competition or did you have a different format in mind?[/quote]

I was thinking of making some regular season Sprints and Targets, just having 40 minutes to do 30 sprint problems and 25 minutes to do 8 target problems. (Yes, I know it's supposed to be 24, but I wanted it to be a rounder number.) I haven't really thought of the format, but I've got a 16 hour plane ride upcoming, so I'll think of something.  :lol:

[quote="h_s_potter2002"]I'll be in.

It's weird how there are so many contests and stuff all of sudden.  When I first joined, there was only Rep's getting started contests.[/quote]

Yeah, I got the idea from everyone else's contests so they deserve some of the credit.

These kinds of contests are good for practicing for everybody and just doing them for fun to remember for the older people here. A while ago, there was the UNL math day contests, which went on for a rather long time with many, many questions.

[quote="yif man12"]Does anyone want to do a mock Chapter, or should we skip right to State level or National level?

Also, I was hoping for a little more participation and will probably do it if we can get at least 10 people. Right now, we're at 7 or 8.[/quote]

If we only have 8 people then we should skip all the way to National level.  They have 12 people there, I don't see why we should make our number even lower than it already is.

I m in high school but I wont mind getting beat by middle schoolers if you need more people..

Wednesdays are equally bad. Mondays,Thursdays,Fridays, Saturdays, and Sundays are good days for me.

Anytime during the weekdays before 3:30 is not good for me. School already started.

Well, I haven't looked at the AMC forum in a long time, because I don't have any time to do many problems, but I'm sure a time can be worked out so that everybody can take it.

By the way, this will be a bit harder than an average State competition. I kind of went crazy making it, so don't get too frustrated on it.

The AMC D has been moved to 5:00 pm. Could I take it in the morning around 10:00 am EST? I feel that I need time to get ready for the Mock AMC

If anyone wants to take the contest earlier, pm me, and I shall get back to you.

Is the test in Latex?

4 and 19 didn't show up on my browser.

Aren't the problems supposed to be posted yet?  Sorry, I'm impatient.

GAH!!!  I'm so sorry.  I misread my co-mod's post and assumed it was 3pm CST, when it was actually EST.

I think we can do this tomorrow or something.  I'm really sorry.

Actually, I'm going to post the questions in a few minutes and leave the contest open for a day or two.  I think we can rely on the honor system.  Rice University does :cool:

Are the teams up yet, or is this going to be an individula thing, like the mock AMC's?

And how hard are these MATHCOUNTS questions, chapter, state, or national level?

[quote="Syntax Error"][quote="h_s_potter2002"]And how hard are these MATHCOUNTS questions, chapter, state, or national level?[/quote]

Personally, I think this test is above national level.  The questions are pretty much all national level, but some are too hard for certain questions yada yada.  Yet, I didn't compile this test.  Don't worry, if you think it's hard, it's hard for everyone - at least how my dad used to tell me when I competed.[/quote]

Hehe,  sorry about making the test a little on the harder side. There may be another one, except much easier this time, one actually around the state level.  :)

Good thing nobody here knows where I live, or else I may be getting some not-so-happy visitors.  :lol:  ;)

Again, please stop bumping threads from over 15 years ago.  The announcements there referencnig have most likely been deleted.


function
number theory unsolved
number theory

Let $n$  be natural number. If $2n$ has $28$ positive devisors and $3n$ has $30$ how many does $6n$ have?
I came to the conclusion if $n=2k$ then $6n$ has $56$ devisors and if $n=3t$ then $6n$ has $60$ devisors

Not sure if this is the correct answer :(
Tell me if it's right and if its wrong please tell me the right way to solve it :)

Nope... $6n$ should have $35$ positive divisors.

We make use of the multiplicative property of the $\tau (x)$ function, where $\tau (x)$ is defined as the number of positive divisors of $x$.

Let $n = 2^{\alpha_1}3^{\alpha_2}n_0$, where $\gcd (2, n_0) = \gcd (3, n_0) = 1$.
Number of positive divisors of $2n = (\alpha_1 + 2)(\alpha_2 + 1) \tau (n_0) = 28$ (Eqn 1)
Number of positive divisors of $3n = (\alpha_1 + 1)(\alpha_2 + 2) \tau (n_0) = 30$ (Eqn 2)
Number of positive divisors of $6n = (\alpha_1 + 2)(\alpha_2 + 2) \tau (n_0)$
Subtracting (Eqn 1) from (Eqn 2), we have $(\alpha_1 - \alpha_2) \tau (n_0) = 2$.

Case 1: $\tau (n_0) = 1, (\alpha_1 - \alpha_2) = 2$
$\Leftrightarrow \alpha_1 = \alpha_2 + 2$
$\Leftrightarrow (\alpha_1 + 2)(\alpha_2 + 1) \tau (n_0) = (\alpha_2 + 4)(\alpha_2 + 1) = 28$, $(\alpha_1 + 1)(\alpha_2 + 2) \tau (n_0) = (\alpha_2 + 3)(\alpha_2 + 2) = 30$.
By exhaustion of the finite cases, we have $\alpha_2 = 3$ as the only possibility, thus number of positive divisors of $6n = (\alpha_1 + 2)(\alpha_2 + 2) \tau (n_0) = 7 \cdot 5 = 35$.

Case 2: $\tau (n_0) = 2, (\alpha_1 - \alpha_2) = 1$
$\Leftrightarrow (\alpha_1 + 2)(\alpha_2 + 1) \tau (n_0) = 2(\alpha_2 + 3)(\alpha_2 + 1) = 28$,
$(\alpha_1 + 1)(\alpha_2 + 2) \tau (n_0) = 2(\alpha_2 + 2)(\alpha_2 + 2) = 30$ (Eqn 3)
From (Eqn 3), ${(\alpha_2 + 2)}^2 \mid 15$, yet $15$ is square-free (Contradiction)

$\therefore$, from the $2$ cases, we conclude that $6n$ has $35$ divisors.


geometry
parallelogram
power of a point
geometry unsolved

A circle k through vertices A and B of a triangle ABC intersects the
sides AC and BC at points L and N, respectively. Let M be the
midpoint of arc LN lying in the triangle.
NOTATION: () means INTERSECTION
Let D = AM()BL, F = AM()BN, G = BM()AL, E = BM()AN. Prove that
(a) DE is parallel to FG;
(b) if DEFG is a parallelogram, then it is a rhombus.

a)
$ ALNB$ is cyclic. Then $ \angle ABC \equal{}\angle CLN$.
Since $ M$ is the midpoint of arc LN, then $ \angle GAM \equal{}\angle FBM$ and by the converse of power-of-a-point theorem $ AGFB$ is cyclic.
Then $ \angle CGF \equal{}\angle ABC \equal{}\angle CLN$, and $ GF\parallel LN$.
Let $ AN\cap BL \equal{} K$. Then $ \Delta ADK$ and $ \Delta BKE$ are similar and so $ ADEB$ is cyclic. Then $ \angle ABD \equal{}\angle AED \equal{}\angle ANL$ and $ DE\parallel LN$, from which $ GF\parallel DE$.

b)
If $ GFED$ is a parallelogram, it has perpendicular diagonals, and so it is a rhombus.

Hold on, why must GFED have perpendicular diagonals if it's a parallellogram? I don't know if I understand you correctly. THe first part of the proof is very nice. But what assumption have you made in the 2nd half, I donot understand that.
Can someone please explain me? Much appreciated.

THanK YoU

Sorry I was a bit unclear.  :blush:
GFED has perpendicular diagonals. Now if it is a parallelogram, it is a parallelogram with perpendicular diagonals, and so it is a rhombus.

How so? How do we know that the diagonals are orthogonal? The condition for that is that AB must be a diameter of the circle. The problem does not state that. So I am missing something maybe here. Can you prove the diagonals are orthogonal anyways?

THanK YoU

[quote="chessfreak"][color=darkred][b]A circle through vertices $ A$ and $ B$ of a triangle $ ABC$ intersects the sides $ [AC]$ and $ [BC]$ at points $ L$ and $ N$ respectively.

Let $ M$ be the midpoint of arc $ LN$ lying in the triangle.  Let $ \{\begin{array}{c}D\in AM\cap BL\ ,\ F\in AM\cap BN\\\\ G\in BM\cap AL\ ,\ E\in BM\cap AN\end{array}$. Prove that :

$ 1\blacktriangleright DE\parallel FG$.

$ 2\blacktriangleright DEFG$ is a parallelogram $ \implies$ $ DEFG$ is a rhombus[/b].[/color][/quote]

Hold on, am I missing something important here? The diagonals are obviously not necessarily orthogonal. There is a vital part of the proof I am missing here. As I earlier mentioned, the diagonals are perpendicular to each other if and only if the chord AB is actually a diameter.

So the problem part ii now reduces to:

Prove that DG parallel to EF if and only if AB is a diameter of the circle k.

Where has this been proved? Or maybe Sepp I am a morone and cannot get it.

Anyone please help?

THanK YoU

Since $ DEFG$ is a parallelogram, $ GM \equal{} ME$ and $ \angle EFA \equal{}\angle FDG$. Consider a point $ P$ on the line $ AD$ such that $ \angle GPD \equal{}\angle EAM \equal{}\angle GAM$.
Then by ASA $ \Delta GPM\equiv\Delta EAM$. This means $ AM \equal{} MP$, and since $ \Delta GAP$ is isosceles, $ GM\bot AP$.

chessfreak, once more, you aren't supposed to ask anyone for help in this forum. You have now posted four of the problems. Please don't post further problems.

As for your combinatorics proof - I will post it on the forum once the deadline is over. That will be the 12th of this month.


Before anyone else posts this, please note that any collection of successive terms in a sequence can be represented by any one of an infinite collection of polynomials, so therefore there is no distinct next term. You could make an infinite number of polynomials of degree 8 to give you any integral next term you want.

forget the infinite just write one of those polynomial that works.

[quote="unknown_pattern18"]find the nth term for the sequence
1, 2, 10, 67, 467, 3268, 22876[/quote]
Can you say it more strictly? Every sequnce $\{a_n\}$ satisfying $a_1=1,a_2=2,a_3=10,a_4=67,a_5=467, a_6=3268,a_7=22876$ is an answer for your question, for example $\{a_n\}$ defined as $a_1=1,a_2=2,a_3=10,a_4=67,a_5=467, a_6=3268,a_7=22876$ and $a_n=\frac{\pi^{n}}{e}$ for $n\geq 8$. I hope that you can write version of your problem which doesn't allow sequence defined above.

It was discovered in a previous topic that the sequence satisfies a recurrence; specifically,

$a_1 = 1$
$a_n = 7(a_{n-1} - 1) + n$


Rules of Craps:
Two standard, six-sided dice are rolled.
If the total is 2, 3, or 12, all bets on the PASS line lose and all bets on the DON'T PASS bar win.
If the total is 7 or 11, all bets on the PASS line win and all bets on the DON'T PASS bar lose.
If the total is anything else, that number becomes the point.  The dice are rolled more times until the point or a 7 shows up.  If the point shows up, all bets on the PASS line win and all bets on the DON'T PASS bar lose.  If a 7 shows up, all bets on the PASS line lose and all bets on the DON'T PASS bar win.

The odds for the PASS line is 1:1.  If you bet one, you will either lose that 1 or win 1 more.
The odds for the DON'T PASS bar is 1:1.  If you bet one, you will either lose that 1 or win 1 more.

Is it better to bet on PASS or DON'T PASS?  Prove your answer.

the probability of getting a 2, 3, or a 12 is $\frac19$
the probability of getting a 7 or an 11 is a $\frac29$

pull it from there

Your odds make no sense. Don't you mean 2:1 ?

Ignite168, that's how it is.  If you bet $x$ and the odds are $w: x$, you will either lose your $x$ or keep that $x$ and win $w$.

[quote="i_like_pie"][url=http://img151.imageshack.us/img151/5718/clockformulayo8.gif]Clock Formula[/url]

[quote="Treething"]it's a [b]math[/b] problem, it doesn't have to make any sense[/quote][/quote]

Uh, why did you just post a link to the clock formula? If I'm not mistaken, it has nothing to do with this problem.

[quote="mathnerd314"]Uh, why did you just post a link to the clock formula? If I'm not mistaken, it has nothing to do with this problem.[/quote]
Mistake. Long story. Sorry. :blush:


$ a,b,c > 0$, $ a\plus{}b\plus{}c\equal{}1$ Then prove:

$ \sqrt {\frac {ab}{c \plus{} ab}} \plus{} \sqrt {\frac {bc}{b \plus{} ac}} \plus{} \sqrt {\frac {ac}{b \plus{} ac}} \le 3/2$

$ \sum\sqrt {\frac {ab}{c \plus{} ab}} \equal{} \sum\sqrt {\frac {ab}{(b \plus{} c)(c \plus{} a)}}\le \frac 12\sum\left(\frac {a}{c \plus{} a} \plus{} \frac {b}{b \plus{} c}\right) \equal{} \frac 32$

Similar but not straightforward: show that $ \sum\sqrt {\frac {ab}{b \plus{} ca}}\le \frac 32$

Have fun! :P


(3^n - 1)(2^n - 1)=m^2
Solve for natural numbers m,n
It is APMO or China question

i think this is what you mean
$ (3^n \minus{} 1)(2^n \minus{} 1)\equal{}m^2$.. solve for naturals $ m$ and $ n$

[quote="Altunshukurlu"]Yes,thank you
I dont know how to write it. :blush:[/quote]
use this symbole $ \$$ before and after the code


So the question is plot the graph of the function in the indicated domain.Indentify the range.
8.y=3x-5;(reals)
10.y=2x^2;(non-positive reals)
12.y=-2/x;(positive reals)
13.y=3/x;(.5 greater than or equal to x less than or equal to x 4 )
15.y=|x|-5
17.y=x^2+2;(1,2,3,4,5)

So how is this trig?

[hide="Solution"]
1. Y can be anything.
2. Since $ x^{2}$ is always non-negative, the range is all non-negative numbers.
3. $ x$ can be infinitesimally small or large so $ \minus{}\frac{2}{x}$ can be any negative number.
4.Just $ \frac{3}{4}$ to $ 6$.
5.$ |x|$ must be positive so it is any number greater than or equal to $ \minus{}5$
6. Just try out the numbers to get $ 3,6,11,18,27$.
[/hide]