MATHCOUNTS
ratio
number theory
Pascal\u0027s Triangle

Hi folks, just found this board today, looks very cool.

I have an odd set of questions.  I've been out of college now for about 6 years, and I'm suffocating to death in a job I can't stand.  I initially majored in Art Illustration as a college freshman before flunking out due to excessive social activities and general burn out.  I went back to school (only after flunking out again), and earned a Bachelors in Communication (emphasis in PR and also Graphic Design).  Looking back, I took that route because it was safe and easy, and I certainly didn't have to extend myself much.  I now have a job that I can't stand, and I need to make a change.

As a younger student I was always considered 'gifted' in Math subjects.  I don't mean John Nash gifted or anything like that, but I tested very well and won little math contests, stuff like that.  School came pretty easy, which was fortunate because all I did was draw all the time.  Eventually, as I placed more (undue, even) priority on partying, social acceptance, and art classes, Math because a turnoff because it was, well, hard.  I also had concentration issues, which made me a poor student.

I apologize for the longish background.  My point is that I am now 30 years old and desperately need a change.  I am still fascinated by mathematics, and I am contemplating going back to school to get a degree in math for the purpose of becoming a math teacher.  If my aptitude and work product allows, I'd want to pursue advanced degrees as well.

My questions are these: how feasible/realistic is this?  How would I go about reintroducing myself to mathematical studies?  I freely admit that my current recollection of formulas and problem solving skills are poor at best, but I would be willing to fix that.  I still have raw tools to work with.

I will depend heavily on loans, so I'd also be interested in relatively inexpensive ways to pursue my next stage of education.  Any recommendations in Illinois?  I'm about 90 minutes south of Chicago, and will more than likely need to hold down a full time job while I pursue a degree.  Any help you can give me would be greatly appreciated.

Thanks for reading this!

It's always feasible to learn math if you are willing to work on your concentration problems.

Here is one suggestion: Our schools need more middle school and high school teachers who are capable of teaching problem solving mathematics.  This doesn't necessarily meaning learning so much math that you could teach calculus.  You might want to check out the MOEMS and MATHCOUNTS curriculum.  You'll probably struggle with some of the MATHCOUNTS material at first as you rae coming back to math after a hiatus and because some of the curriculum is not ordinarily taught in schools (basic number theory and combinatorics), but is not much more difficult than algebra I.

I imagine a teacher capable of running a MATHCOUNTS program would be in even more demand than a teacher simply wishing to teach a regular Algebra I class.

Also, don't focus on math in a boring, "this is the 1-2-3 step process for getting the answer to this kind of problem."  Focus on the exploration process.  Do a bit of reading on Pascal's Triangle, Fibonacci numbers, and other "extra-curricular" topics that good middle school students can start to grasp.

I'm sure that you could receive a dozen completely different thoughts on a teaching career, but I like to promote the idea of new teachers taking on their profession as something more interesting than what most students get in the classroom.  At its heart, mathematics is an art, which should fit right in with your background.

As for the process of looking into loans for going back to school to become a teacher, I know little.  However, I imagine that potential teachers can find specific programs that are built to help bring more teachers into the classroom -- particularly those interested in learning more about mathematics because they are harder to recruit.

Thanks for the reply...it's funny you mention the extra-curricular reading, because I just finished a book called "the Golden Ratio" by Mario Livio, which was an interesting history of the discovery and natural occurrences of phi.  Links to Fibonacci numbers were discussed, and it dipped lightly into fractals by the end.  Great read.  I definitely would want to pursue teaching math as long as I could also use it as a vehicle to study further into interesting concepts afterwards, just as you mention.  I won't consider myself 'done' once I land a teaching gig, by any means.

I'll look more into the math programs you mentioned, as I am not familiar with them.  Thanks again for replying!

From his signature, floydmoseby lives in Illinois. If another reader has a story like his but lives in California, contact me by private message and we'll talk about feasibility.


Let $S$ be a list of positive integers - not necessarily distinct - in which the number 68 appears.  The average (arithmetic mean) of the numbers in $S$ is 56.  However, if 68 is removed, the average of the remaining numbers drops to 55.  What is the largest number that can appear in $S$?

[hide="Solution"]Letting $n$ be the number of terms in $S$, $\frac{55n}{n}=\frac{56n-68}{n-1}$. Simplifying, $n=13$, and the number of terms in the sequence where $68$ is omitted is $12$. Thus, the sum of all terms of $S$ excluding $68$ is $55\cdot12=660$. The largest element appears in $S$ when all the other elements are $1$, so the our answer is $660-11=\boxed{649}$. [/hide]

so let amount of number be x so then sum is 56x
u can write equation
56x-68=55(x-1) so there is 13 number

so sum is 56*13=728
there has to be a 68
so then make 78 and then 11 ones
so then 728-78-11 so answer is 649

[hide=Solution]Let the amount of integers be $a$. The sum is then $56a$. Removing $68$ gives a sum of $56a-68$, which is equal to the average of the remaining $a-1$ numbers, said to be $55$. Therefore, $56a-68=55(a-1)$, or $56a-68=55a-55 \implies a = 13$. The sum of all numbers is $13 \cdot 56=728$ and removing $68$ gives $660$. Then let the remaining $11$ integers equal $1$ to get an answer of $660-11 \cdot 1 = \boxed{649}$.[/hide]


what is the remainder when the 100th number in the fibonacci sequence is diveded by 3? :cursing:

[color=blue]Moderator kyyuanmathcount's note: Sources please![/color]

[quote="pgpatel"]Umm isn't it divided by [b]4[/b] not 3.[/b][/quote]

oh sorry, yes its 4 :blush:

[quote="pgpatel"][hide]First of all the answer could be 0,1,2, or 3. But if you want to be sure look for a pattern. 1,1,2,3,5,8,13,21,34,55,89,144... Let's look at the remainder of them. 1,1,2,3,1,0,1,1,2,3,1,0. It repeats every six terms so we need the fourth. It's [u]3[/u][/hide][/quote]

oh man, that makes me an official idiot!      :ewpu:

[hide]
Look at the pattern of numbers, modulo 4 (that means the remainder when you divide by 4). The first one is 1, the second one is 1, and the third one is 1+1=2. The fourth is 1+2=3. Fifth, 2+3=1 (remember, it's modulo 4). Sixth is 3+1=0. Next comes 1+0=1, 0+1=1, 1+1=2... there's a loop. The number for #100 is the same, then, as the number for #4, which is 3.
[/hide]

Wow!  I remember this problem from last year.  I couldn't figure it out, but I knew the answers were 0, 1, 2, or 3.  I knew it wasn't going to be 0, instinctively.  So I guessed 3, and I got it right!

[hide]You have to look at numbers of the sequence in $(\mod4)$
The pattern goes thus-$1,1,2,3,1,0,1,1,2,3,1,0,1 (\mod4)$  So we use the repeating part of the pattern, $1,1,2,3,1,0 (\mod4)$  And get the residue from $100=4 (\mod6)$   We walk along the pattern 4 numbers, 1,1,2,[b]3[/b]  And get our answer $\fbox{3}$[/hide]


geometry
3D geometry
sphere
algebra unsolved
algebra

Suppose we have $ m$ points in a sphere of radius $ 1$. The minimal distance between two of this points is $ \delta$. Show that $ m < (\dfrac{4}{\delta})^{2}$.


Let 12 pupils,assumed that in any 9 pupils there are always 5 pupils in which each two of them are close friends,prove that there are 6 pupils each 2 of them are close friend...........[b] :blush:  :roll:  :oops:  :P  :D  :? [/b]


Oh man! I know I've seen something like this, but I just can't remember...

:?  :blush: I don't know it is confusing

Oh, god, I remember this prob, I just can't remember the answer...
[hide="Its something like"]Is this not the way to truthtown[/hide] or something...

The sorry thing is that you can only ask one question

[quote="solafidefarms"]Ask something that you can easily verify, as in 'Is it raining?'

Billy[/quote]

Then, you wouldn't be able to ask him the question you need to.

i know that this sounds totally lame but ive watched yugioh and it was on there.

[hide="dont hurt me"]
well if u could ask the man something, u would ask.... can u take me to truth town on the wrong route? if the man is a liar, he would take u on the path to truthtown to liar town, if this is the case, youll see truthtown, go there. if the person tells the truth, go with them. lame huh?
[/hide]

[quote="mysmartmouth"]Maybe this isn't math, but I thought it would be fun:

You are visiting your aunt Ida.  She lives in Truthtown, where the people can only tell the truth.  You come to a fork in the road.  You know one road leads to Truthtown, and one road leads to Liartown, where the people can only tell lies.  You see a man sitting on a bench at the fork in the road.  You are not sure if he is from Truthtown or Liartown.  If you need to know which fork leads to Truthtown, and you can only ask him 1 question, what do you ask him?

(This is like a riddle, but it is similar to a logic problem I once saw on MathCounts, so I thought I would post it)[/quote]

[hide=" Easy..."]You ask the man to point to the town where he's from. The truth guy will point to truth and the lairs guys will also point to truth.  :D  :D [/hide]

[quote="mathgeek2006"]Then you wouldn't be able to tell where he was from, anyway. . .[/quote]
You dont need to know where he's from...you just need to know the way to truthtown

Oh, now I get it.  I was getting a little confused there.

[quote="solafidefarms"]Ask something that you can easily verify, as in 'Is it raining?'

Billy[/quote]

But you only have one question, and now you have wasted it....

[quote="GoBraves"][hide=" Easy..."]You ask the man to point to the town where he's from. The truth guy will point to truth and the lairs guys will also point to truth.  :D  :D [/hide][/quote]
ding ding!!  :P

My logic isn't that great..., I would've never thought about pointing somewhere.

[quote="rcv"][hide="Alternate solution"]"I see you are a resident of Truthtown or Liartown."

"If I were to ask a resident of the other community to tell me the way to Truthtown, which direction would he send me?"

Thank him kindly, but go the opposite way.[/hide][/quote]

I guess that works too.  This question was from Ask Marylin (spelling?) in Parade Magazine.  It comes in the paper every Sunday, and her column has some good math and logic problems every week.  Or check out her site: [url]http://www.marilynvossavant.com/[/url].  She supposedly has the highest documented IQ ever!

grrr... :(

HOLD ON!!!!
Hes taking you to a place. He's ACTING!!!
When you lie, you give a false ANSWER to a question!!!

He'll just say yes, sure and take you to lietown!!!

HA!!!

[hide=" Easy..."]es
[hide=" Easy..."]You ask the man to point to the town where he's from. The truth guy will point to truth and the lairs guys will also point to truth.  :D  :D [/hide][/hide][/quote]
now that's a solution I can do!

But then, oh, no! My other post can apply to this one too!
why, oh WHY did I have to say that? That screws everything up now!
Grrr...

[quote="mathlover"]im sry. i was totally wrong. this is correct [hide="haha"] ask the man to take him to his town. if he's from truthville, he takes u there. if he's form liartown, he'll have to take u to truthville cuase he has to lie. I DID IT!!!!

woohoo[/hide][/quote]
Wrong.

[hide]When he lies, it doesn't mean physically, just verbally.  With your way, he wouldn't have to speak, allowing the liar to take you to Liartown.  So you have to ask for directions to his village is.  Then, you would be able to tell for sure.[/hide]

I think that's nitpicking. We don't have to get philosophical and define a lie to answer this problem. I say it's a fair answer.

[quote="dwx314"]I like mathgeeks answer better too. But, you know, welcome to the world with lawyers, wordplay, technicalities, etc.[/quote]

...loopholes...yeah.


inequalities
calculus
derivative
inequalities unsolved

Prove the ineq for a,b,c≥0.
$ (a\plus{}b\plus{}c)(\frac{1}{\sqrt{a^2\plus{}ab\plus{}b^2}}\plus{}\frac{1}{\sqrt{b^2\plus{}bc\plus{}c^2}}\plus{}\frac{1}{\sqrt{c^2\plus{}ca\plus{}a^2}})$≥$ 4\plus{}\frac{2}{\sqrt3}$

[quote="SUPERMAN2"]Prove the ineq for a,b,c≥0.
$ (a \plus{} b \plus{} c)(\frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}} \plus{} \frac {1}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt {c^2 \plus{} ca \plus{} a^2}})$≥$ 4 \plus{} \frac {2}{\sqrt3}$[/quote]
we may consider $ a \plus{} b \plus{} c \equal{} 3$
let$ p \equal{} \frac {9 \plus{} a^2 \plus{} b^2 \plus{} c^2}{6}$
the inequality becomes
  $ \frac {1}{\sqrt {p \minus{} a}} \plus{} \frac {1}{\sqrt {p \minus{} b}} \plus{} \frac {1}{\sqrt {p \minus{} c}} \geq \frac {4}{\sqrt {3}} \plus{} \frac {2}{3}$ (1)
 let $ f(x) \equal{} \frac {1}{\sqrt {p \minus{} x}}$,$ 0 \leq x < p$
 we have $ g(x) \equal{} f'(x) \equal{} \frac {1}{2(p \minus{} x)^\frac {3}{2}}$ and $ g''(x) \equal{} \frac {15}{8}(p \minus{} x)^\frac { \minus{} 7}{2} > 0$
therefore,g(x)is strictly convex for $ 0 \leq x < p$
according to EV-theorem
  $ 0 \leq a \leq b \leq c$ such that $ a \plus{} b \plus{} c \equal{} 3$and $ a^2 \plus{} b^2 \plus{} c^2$=constant,then (1) is minnimal when either a=0 or $ 0 < a \leq b \equal{} c$
-case $ a \equal{} 0$.the original inequality becomes
  $ \frac {1}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq \frac {4}{3} \plus{} \frac {2}{3\sqrt {3}}$
or $ \frac {3}{bc} \plus{} \frac {1}{\sqrt {9 \minus{} bc}} \geq \frac {4}{3} \plus{} \frac {2}{3\sqrt {3}}$   ($ 0 \leq bc \leq \frac {9}{4}$)

 easy find minimal (1) is $ \frac {4}{\sqrt {3}} \plus{} \frac {2}{3}$ by derivative
-case $ 0 < a \leq b \equal{} c$
original ineqality becomes
$ \frac {2}{\sqrt {a^2 \plus{} ba \plus{} b^2}} \plus{} \frac {1}{\sqrt {3b^2}} \geq \frac {4}{3} \plus{} \frac {2}{3\sqrt {3}}$
or $ \frac {2}{\sqrt {b^2 \minus{} 3b \plus{} 3}} \plus{} \frac {1}{b} \geq \frac {4}{\sqrt {3}} \plus{} \frac {2}{3}$   ($ 1 \leq b \leq \frac {3}{2}$)
(easy prove it by derivative)
 equality is $ a \equal{} 0$,$ b \equal{} c > 0$

I sthere any proof without calculus? :blush:


There are $ n \geq 5$ pairwise different points in the plane. For every point, there are just four points whose distance from which is $ 1$. Find the maximum value of $ n$.

As stated, the infinite square lattice is such a set.  What's it supposed to say?

Minimum. See Zhaoli's pdf in http://www.mathlinks.ro/viewtopic.php?t=5283

The minimum value of $n$ is $\boxed{9}.$

First let us construct this. Throw the points onto the complex plane. Let $a, b$ denote the complex numbers $e^{i \sqrt 2}, e^{i \sqrt 3},$ respectively. Let $c, d$ denote the complex numbers $a \cdot e^{i \frac{\pi}{3}}, b \cdot e^{i \frac{\pi}{3}}.$ Then, we claim that the set of $9$ points corresponding to the complex numbers in the set $\{0, a, c\} + \{0, b, d\}$ works. This is easily verified.

Let us now show that $9$ is indeed optimal. Consider the graph on the given $n$ points where we connect two points iff their distance is $1.$ It's easy to see that a $K_4$ is impossible, and so this easily proves $n \ge 6.$

Suppose that $n = 6.$ Take two points $A, B$ which are connected. As there are four other points, and each of $A, B$ is connected to three of them, there exists a point $C$ so that $\triangle ABC$ is equilateral. Now, there's three other points $D, E, F$. Each of $A, B, C$ is connected to two of them, so it follows that $\triangle DEF$ is the anticomplementary triangle of $\triangle ABC.$ These six points clearly do not work.

Suppose that $n = 7.$ Again, we can force an equilateral $\triangle ABC.$ Let $D, E, F, G$ be the other four points. Each of $A, B, C$ is connected to exactly two of them. Furthermore, each of $D, E, F, G$ is connected to at most $2$ of $A, B, C.$ As there are a total of $3 \cdot 2 = 6$ connections between $\{A, B, C\}$ and $\{D, E, F, G\}$, we can WLOG assume that $D$ is connected to $A, B$ and $E$ is connected to $B, C.$ Up to this point, $D$ is connected to only $A, B$, and so thus it must be connected to both $F$ and $G$. Analogously, so must $E$. However, as $D, E$ are a distance two apart, this is clearly absurd.

There are essentially two cases to be considered when $n =8.$

[b] Case 1. [/b] There is an equilateral triangle

Let's again suppose that $\triangle ABC$ is equilateral. As there are five other points, and again $3 \cdot 2 = 6$ connects between $A, B, C$ and the others, we know that there exists $D$ which is connected to $A, B$, WLOG. Observe that $C, D$ each have only two connections up to this point ($A$ and $B$). Furthermore, every point which isn't $A$ or $B$ is connected to at most one of $C, D.$ Hence, among the other four points $E, F, G, H$, we know that exactly two of them are connected to $D$, and exactly two are connected to $C.$ In other words, if we let $\omega_c, \omega_d$ be the circles centered at $C, D$ with radius $1$, two of $E, F, G, H$ lie on $\omega_c$ and two on $\omega_d.$ Observe that $A$ is connected only to $A, B, C$ at this point, and so WLOG $E$ is connected to $A$. By symmetry, we may also assume that $E \in \omega_c.$ This implies that $\triangle ACE$ is equilateral. However, notice that $E$ can now be connected to at most $2$ points on $\omega_c$ ($A$ and possibly another point) and no points on $\omega_d$ which are not $A$. Hence, $E$ is connected to at most three points: $A, C,$ and potentially the other point on $\omega_c$. Thus, this case is a failure.

[b] Case 2. [/b] There is no equilateral triangle.

Let $A, B$ be two connected points. Let $\omega_a, \omega_b$ be the circles centered at $A, B$ with radius $1.$ Since no points are on both $\omega_a, \omega_b$ (else an equilateral triangle is formed), each point is on precisely one of $\omega_a, \omega_b.$ Let $P \neq A$ be a point on $\omega_a.$ Then, it can't be connected to any other point on $\omega_a$ (else an equilateral triangle is formed), and is connected to at most two points on $\omega_b.$ This means that $P$ is connected to at most two points, and so this case is also a failure.

As we've exhausted all cases when $n \le 8$ and concluded that no configurations of $n \le 8$ points work, we conclude that the minimum value of $n$ is $\boxed{9}.$


(1)
$2^{30} - 1 = (2^{15} - 1)(2^{15} + 1) = (2^5 - 1)(2^{10} + 2^5 + 1)(2^5 + 1)(2^{10} - 2^5 + 1) = = 31 \cdot 1057 \cdot 33 \cdot 993$
Now numbers are better.
$2^{30} - 1 = 3^2 \cdot 7 \cdot 11 \cdot 31 \cdot 151 \cdot 331$
We can delete and $331$ immediately and try (brute) with the others.
So divisors in $\left[50,200\right]$ are $63, 77, 93, 99, 151$ and they are five.

(2)
The only prime divisor in the interval is $151$

Work:[hide]Let's factor this number out.  2^29+...+1=2^30-1.  This is binary representation.  Now, notice that 1+2^1 is divisible by 3 while 1+2^1+2^2 isn't.  But adding 1+2^1+2^2+2^3 is.  This is because powers of 2 are either one or 2 mod 3 and they start on 2^0 or 1 mod 3.  2 is 2 mod 3.  So when added they equal 0 mod three.  Since the last added power was an odd power, this number is divisible by 3.  Notice the numbers that come out when we divide by 3 is 1,5,21,85,...  In other words, we start with one.  We multiply by 4 and add 1.  Obviously, it's futile to do this 28 times to get our number.  However, let's try to factor out what we can.  Notice that the third number is divisible by three.  If you calculate further, the 6th number is divisible by 3 and the 9th.  1-29 including the odds only is 15 so the last or 15th number is divisible by 3.  Now we divide by 3 and remove the parts of the series that's irrelevant.  7,455,29127,...  Each term is divisible by seven because it starts with 7 and the formula that generates the next number is 64x+7 where x is the previous number.  1,65,4161,266305,17043521.  (The of the previous few steps wasn't just to find factors, it was to shrink the number.)  Now, we focus on finding more factors of that number.  1+0+3+2-7-4-5-1=-11 so 17043521 is divisible by 11.  We get 1549411.  But notice back in our original series every 5th number was divisible by 31.  This is because looking at this mod 31 we have 1,5,21,23,0,... and the cycle repeats.  Since 15 is divisible by five and the 15th number previously is now the number we're considering, 31 divides 1549411.  We get 49981.  I had to use a calculator to find that 49981=151*331.  The number's prime factorization is 3^2*7*31*11*151*331.  Does anyone know of a faster (less tedious or not brute force) way to find the last two factors?[/hide]
The answers: Part 1: [hide]63,93,99,77,151.  5 factors[/hide]
Part 2: [hide]One prime divisor in there and that's 151 so that's the sum.[/hide]

dengmi, can you $\LaTeX$ your solution and also create more space between lines? It's like all jumbled up and hard to follow...


Have you stopped to plug $x=\pm 1$ into the initial equation?  :)

(BTW, $x^{4}+x^{4}+x^{4}+1\geqslant 4x^{3}$ by AM-GM.)

I forgot that AM-GM gives you a coefficient...  :wallbash_red: sorry guys

Then it probably just goes back to the well-known fact that you can write $x^{n}+\frac{1}{x^{n}}$ as a polynomial in $x+\frac{1}{x}$. :wink:

If the equation above is right then that one hasn't a root. Because if $x>0$ then applying Cauchy's ine, we have the right hand is larger than left hand. If $x<0$ then the left hand < 1, right hand >1. Hence the equation has no root. I'm sorry my mistake typing. The problem i want to everybody discus in this forum that is:
[b]Solve the equation[/b]: $\left(x^{3}+\frac{1}{x^{3}}+1\right)^{4}=3\left(x^{4}+\frac{1}{x^{4}}+1\right)^{3}$

sorry -  the original message deleted to remove confusion ...

Is it possible to find the non-real solutions?

I have a thing want someone anwser to me. Is there a solution which is different with [b] Farenhajt'[/b] s solution? It means that only use Cauchy's inequality or Cauchy - schawtz' one in proving inequality bewteen two hands.


analytic geometry
trigonometry
function
calculus
derivative
quadratics

What is the maximum value of the angle of projection of a particle so that the displacement of the particle from the point of projection is continuously increasing?

Another nice problem :D

Let $ y$ be the vertical coordinate pointing upwards, and $ x$ be the horizontal coordinate
The equations of motion:

\[ \ddot y = -g \;\;\;\;\;\; \ddot x=0\]

Integrating:

\[ y(t)=-\frac{g}{2}t^2+At+B\]

and

\[ x(t)=Ct+D\]

The initial conditions: $ x(0)=y(0)=0$ and $ \dot x(0)=v_0\cos\varphi$ and $ \dot y(0) = v_0\sin\varphi$
Where $ v_0$ is the initial velocity.With these we have for the constants: $ B=D=0$ and $ C=\v_0\cos\varphi$ and $ D=v_0\sin\varphi$.

So the time functions:

\[ x(t)=v_0t\cos\varphi\]
\[ y(t)=v_0t\sin\varphi-\frac{g}{2}t^2\]

Now the square distance from the point of projection is $ x^2(t)+y^2(t)$. Now we want this to be, growing monotonic function of time, that is we want it to increase continuously as time passes by, Hence its derivative with request to time must by greater than (or equal to) zero, for all $ t$.

So we have:

\[ \frac{d}{dt}(x^2(t)+y^2(t))\geq 0\]

for all $ t$.
calculating the derivative we have:

\[ t(g^2t^2-3v_0g\sin\varphi \, t +2v_0^2 ) \geq 0\]

time is always nonnegative, hence the quadratic term in the brackets has to be nonnegative for all $ t$. This will only happen if its discriminant is smaller than or equal to zero, that is:

\[ 9v_0^2g^2\sin^2\varphi - 8v_0^2g^2 \leq 0\]

So from here:

\[ \sin^2\varphi \leq \frac{8}{9}\]

So the maximum value of the angle will be:

\[ \varphi_{max}=\text{arcsin}\sqrt{\frac{8}{9}}\]

Its interesting how the sine and cosine of $ \sqrt{\frac89}$ appear in different parts of physics. For example the cosine appears in the study of the waves induced by a ship travelling in a fluid. It can be derived very lengthily that this is the angle (the cosine part) enclosed by the line of waves, and the direction the ship is heading. Try to measure it, and you will always get about $ 20$ degrees regardless of the velocity of the ship..

[quote="Thaakisfox"]
So the maximum value of the angle will be:
$ varphi_{max} \equal{} \text{arcsin}\sqrt {\frac {8}{9}}$

Its interesting how the sine and cosine of $ \sqrt {\frac89}$ appear in different parts of physics. For example the cosine appears in the study of the waves induced by a ship travelling in a fluid. It can be derived very lengthily that this is the angle (the cosine part) enclosed by the line of waves, and the direction the ship is heading. Try to measure it, and you will always get about $ 20$ degrees regardless of the velocity of the ship..[/quote]
Thanks for the solution and the last bit information too!  :D


13. In the middle of a vast prairie, a fire truck is stationed at the intersection of two perpendicular straight highwas. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the fire truck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

You're posting these for the contests section, right?  One thing: please put just the problem in the post (i.e. no problem number).  The source, of course, is ok (as well as very helpful :P ).

Thanks


hi everyone, this question has almost nothing to do bout maths, but here it goes:
nearly every topic that i click on contains a question on sequences and functional equations, so i'm wondering what other problem would fall under algebra, because they seem quite seldom round here!

Equations, systems of equations, quadratics or anything that resumes to that.


number theory
greatest common divisor
calculus
integration
algebra
polynomial
quadratics

Hide please, but that is correct, anyone have a solution?  I ended up leaving this one blank.

[hide]Combine to get $\frac{9a^{2}+14b^{2}}{9ab}$. Now if a prime $p$ other than 9 or 14 divides $a$ or $b$, we know that $p$ does not divide the other one, so it would divide the denominator of $\frac{9a^{2}+14b^{2}}{9ab}$ but not the numerator. However, we could have a divisor of 14 divide $a$, as the 14 times $b^{2}$ on the top would take care of that, meaning $a$ can be $1,2,7,14$. Furthermore, to take care of the $9$ on the bottom, $b$ must be a multiple of $3$. We can see that $b=9$ does not work as the numerator would not be a multiple of $81$, but if $b=3$, the top is a multiple of $9$ as well as one extra factor of $a^{2}+14=15=3(5)$ because $a^{2}$ must be congruent to 1 mod 3. Thus, there is one possibility for $b$ and four for $a$, producing $\boxed{4}$ solutions.[/hide]

[hide]I let a/b=r, and the integer the expression equals be x.

So x=r+14/(9r)
9r^2 - 9xr + 14 =0

Since x is an integer and r is rational, the expression on the LHS can be factored. The possibilities are:

(3r-1)(3r-14)
(3r-2)(3r-7)
(r-1)(9r-14)
(r-14)(9r-1)
(r-2)(9r-7)
(r-7)(9r-2)

We want the coefficient of the 9xr term to be divisible by 9. (9x has to be divisible by 9 if x is an integer.)

For the first, 9x=3*1+3*14 = 45, which is divisible by 9. For the second, 9x=3*2+3*7 = 27, which is also divisible by 9. The last four have 9xr terms with coefficients that are not divisible by 9 because you have 9*something + something that's not divisible by 9.

Therefore there are only four possibilities for r, which come from setting the first two expressions equal to zero:
(3r-1)(3r-14) = 0
(3r-2)(3r-7) = 0
r = a/b = 1/3, 14/3, 2/3, 7/3.

4 [b](A)[/b]
Sorry that was kind of long >>;[/hide]

I don't know how I got the right answer with this shabby proof, but it works  :D
[hide]Simplify to 9a^2+14b^2=9nab. Then

14b^2=9a(bn-a)
(14b^2)/(9a)=bn-a

Since bn-a is an integer and gcd(a,b)=1, 9 must divide b^2 --> 3 divides b. So let b=3k, and then

(14k^2)/(a)=3kn-a

Similarly, gcd(a,k)=1 so a must divide 14 --> a=1,2,7,14. Then there are 4 cases:

7k^2-3kn+2=0
2k^2-3kn+7=0
14k^2-3kn+1=0
k^2-3kn+14=0

And given ak^2+bk+c, the only positive integral solutions for k are when b=a+c or b=ac (in other words, n=3, 5), so the integral solutions for k are 1,2,7,14 which gives 4 values for b and 4 positive values for a.[/hide]

[quote="kyyuanmathcount"]How many pairs of positive integers $(a,b)$ are there such that $\text{gcd}(a,b)=1$ and $\nonumber \begin{eqn}\frac{a}{b}+\frac{14b}{9a}\end{eqn}$ is an integer?

Choices were (A) 4      (B) 6   (C) 9   (D)  12   (E) infinitely many[/quote]
[hide="fast solution"]To be an integer, both denometers have to be the same. so b has to equal 9a or an simplified version. working out all the possibilities of the denometers, only b=3 works since a and b are relatively prime. then you can find 4 possibilities for a by brute force.[/hide]

[hide="Yet one more argument"]

$9a^{2}+14b^{2}=9kab$ for some $k\in\mathbb{N}.$

First: look at that mod 3 and realize that $b$ must be divisible by 3. Write $b=3y.$ We now have

$a^{2}-3kay+14y^{2}=0.$

Now let $q=\frac{a}{y}.$ $q$ must be rational:

$q^{2}-3kq+14=0.$

At this point, invoke the rational root theorem: the only rational roots of an integer-coefficient polynomial have numerators that divide the constant term and denominators that divide the lead coefficient. That means that the only possible positive values for $q$ are $1,2,7,$ and $14.$ All four of these work, so those are our four answers.[/hide]

[hide]$\frac{9a^{2}+14b^{2}}{9ab}=k\Rightarrow9|14b^{2}\Rightarrow3|b$ Let $b=3j$. $\frac{a^{2}+14j^{2}}{3aj}=k\Rightarrow j|a^{2}$, $(a,j)=1\Rightarrow j=1\Rightarrow b=3$.  $\frac{a^{2}+14}{3a}=k\Rightarrow a|14\Rightarrow a=1,2,7,14$. Since none are divisible by $3$, all $\boxed4$ work.[/hide]

overview of yet another method
[hide]
Here's the overview of another method that hasn't been discussed, directly.

$\frac{a}{b}+\frac{14b}{9a}=n$

1.  Substitute $x=\frac{a}{b}$.  This yields a quadratic, in $x$.

2.  Apply the quadratic formula.  Pull any fractions out of the radicand.  Note that the radicand must be a non-negative perfect square for $x$ to be rational.

3.  My radicand was $(3n)^{2}-56.$  So, the difference of two perfect squares must be 56.  There are several ways to solve.  I found $n=3$ or $n=5$.

4.  Plug the known values of $n$ back into the quadratic formula.  Each one generates a distinct pair of rational values for $x$, which means there are four ordered pairs, $(a, b).$
[/hide]


I'm thinking of running AMC Week Day.

I personally don't like AMC Day anymore since it fills my inbox too quickly. So, here's a new thing that will help me (since I have to find the problems) and you (since you will be solving them) for upcoming AMC. We all want to get good score on AMC so let's try to do that.

Rule:

If enough people are interested, I will post 7 problems in one day. You'll have a week to do these problems. IMPORTANT RULE: DO NOT SEND THE EACH PROBLEMS IN DIFFERENT PRIVATE MESSAGE! Also, if you had a mistake and want to send again, well, try not to send it again. Do it if you only have to.

I had one message like "ooops, my answer is 6" and then came "no, it's 3" and "oh, it's 4." It's very annoying and I hate it.

The level of problem will be strictly AMC-12 but some may apply to AMC-10.

Anybody interested?

A week to do 7 AMC? Isn't that too much...

Oh, I wasn't on the forums when the AMC Day was around.

I know it says I registered early, but I wasn't on these forums till this summer <== really bad decision on my part, heh

[quote="joml88"]It's basically the same concept of one problem a day...[/quote]

Yup.  :)

I just like week day better than posting 7 different problems everyday. Nothing wrong with that but it's just more tiring work. I rather go hard on one day and rest for other 6 days. :D


quadratics
modular arithmetic
number theory proposed
number theory

Let $p=4k+3$ be a prime number. Find the number of different residues mod p of $(x^{2}+y^{2})^{2}$ where $(x,p)=(y,p)=1.$

[quote="bilarev"] Find the number of residues mod p of $(x^{2}+y^{2})^{2}$ [/quote]
I don't understand  :maybe: This problem has form beautifuly.

$x$ and $y$ runs over all residues mod $p$ except 0.

You not understand what I said :) Ok, problem said that: Find $0\leq r<p$ such that $(x^{2}+y^{2})^{2}=p\cdot q+r$, for some $q\in\mathbb{N}$?

Sorry for this inexactness ... I think that now it's clearer.

If p=2 all residues can be write in this way.
If $x=y=0$, then $r=0$. If one of $x,y$ not equal 0, then $(x^{2}+y^{2})^{2}=x^{4}z^{2}\in Z_{p}^{2},z=1+t^{2}, t=y/x$.
If $p=3(mod 4)$ all quadratic residue can be write in this way. Number of residius is (p+1)/2.
If $p=1(mod 4)$ sufficiently consider z. If exist t, suth that $(\frac{1+t^{2}}{p})=-1$, then all quadratic residues can be write in these way, else only residues from $Z_{p}^{4}$.
Because for all $p=1(mod 4)$ exist t, suth that $(\frac{1+t^{2}}{p})=-1$ all quadratic residues can be write in these way.

We need use following lemmas
[b]lemma 1[/b]. If $p=4k+3$ is prime and $\gcd (x,p)=\gcd (y,p)=1$ then $x^{2}+y^{2}\not \equiv 0\pmod{p}$.
[i]Proof.[/i] Trivial.
[b]lemma 2[/b]. If $p=4k+3$ is prime then there exists $x_{0},y_{0}\in\mathbb{N}$ such that $x_{0}^{2}+y_{0}^{2}+1\equiv 0\pmod{p}$ and $\gcd (x_{0},p)=\gcd (y_{0},p)=1$.
[i]Proof.[/i] Theorem 87 in Hardy and Wright.

If $0\leq r<p$ such that for some $x,y$ :$\gcd (x,p)=\gcd (y,p)=1$ and $(x^{2}+y^{2})^{2}\equiv r\pmod{p}$ then by lemma 1, $r$ must is quadratic residue of $p$, therefore there at most $\frac{p-1}{2}$ of $r$.

Now, by lemma 2 we have $(x_{0}^{2}+y_{0}^{2})^{2}\equiv 1\pmod{p}$ therefore $((p_{i}x_{0})^{2}+(p_{i}y_{0})^{2})^{2}\equiv q_{i}^{2}\pmod{p}$ for $i=1,2,...,\frac{p-1}{2}$ and $q_{1},q_{2},...,q_{\frac{p-1}{2}}$ are all quadratic residues and $q_{i}\equiv p_{i}^{2}\pmod{p}(i=1,2,...,\frac{p-1}{2}$), but numbers $q_{1}^{2},q_{2}^{2},...,q_{\frac{p-1}{2}}^{2}$ are all incongruent, we have at least $\frac{p-1}{2}$ of $r$.

Final, answer $\frac{p-1}{2}$.

The $\frac{p+1}{2}$ numbers $x^{2}(x=0,1,...,\frac{p-1}{2}).$ are incongruent and so are the $\frac{p+1}{2}$ numbers $-1-y^{2}(y=0,1,...,\frac{p-1}{2}).$ But there $p+1$ numbers in the two sets together, therefore... :)

[quote="ciprian"]Can you post a proof for lemma 2 N.T.TUAN? (I do not have that book:)
Edit: Ok I have an idea but I am not sure:
Let $r_{1},...r_{\frac{p-1}{2}}$ the quadratic residues. So there is enough to find a quadratic residue r such that $r+1$ is a nonquadratic residue. If we assume that this does not hold, then all the residues are quadratic, a contradiction :)[/quote]
Sufficiently proof, that exist t, suth that $(\frac{1+t^{2}}{p})=-1$. If $(\frac{1+t^{2}}{p})\ge 0 \ \forall t$, then $r=1+(r-1)=1+1+(r-2)=...=1+1+...+1$ and $(\frac{r}{p})\ge 0 \ \forall r$. It give contadition for p>2.

Here's another solution, copied from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=587&t=290862]here.[/url] The answer is $\frac{p-1}{2}$. Let $S$ be the set of residues $\pmod{p}$ ($0$ is not a residue here). I'll show that for each number $k \not\equiv 0 \pmod{p}$, either $k$ or $p-k$ is in $S+S$. (*) Since $p \equiv 3 \pmod{4}$, $0 \not\in S+S$. But by Cauchy-Davenport, $|S+S| \ge 2|S|-1 = p-2$. So $S+S$ can be missing at most one other element other than $0$. So if $k$ is missing, then $p-k$ can't be missing. So (*) is proven and the result trivially follows.

Obviously zero can't be represented as that would imply that $\left(\frac{-1}{p}\right)=1$ contradiction with the first condition.So let's show that all QR-s can be represented and this will kill it as this is obviously necessary.
$p\equiv_4 3$  $\implies$ at least one of $\{a,-a\}$ is a QR and so it suffices to show that for every $z$ $x^2+y^2=z^2$ has non-trivial solution in $\mathbb{F}_p$.But we could just pick
$(x,y)=\left(\frac{s-1}{s+1}\cdot z,\frac{2s}{s+1}\cdot z\right)$ for some $s\not =-1$ and we,re done.

[hide = Umm...]

Just take $x = y$, so that $(x^2 + y^2)^2 = 4x^4$. Since $x^4$ covers all the quadratic residues as $x$ is nonzero, we're done. The answer is that all nonzero quadratic residues can be obtained, for a total of $\boxed{\frac{p-1}{2}}.$

[/hide]

I claim that all nonzero quadratic residues modulo $p$ work, hence there are $\frac{p-1}2$ solutions.

For Fermat Christmas reasons, $p \mid x^2+y^2$ is impossible. On the other hand, it is clear that $r$ must be quadratic, so we show that all remaining QR's work. This is simply because as $p \equiv 3 \pmod 4$, one of $a$ and $-a$ is a NQR mod $p$ by multiplicativity. Hence for each $a$, there exists a pair $(x, y)$ such that among the two equations
\begin{align*}
x^2+y^2 &\equiv a \\
x^2+y^2 &\equiv -a
\end{align*}
one has at least $p-1$ solutions with $p \nmid x, y$. (It is well-known that an $x^2+y^2 \equiv a$ type equation has at least $p-1$ solutions in general.) This is enough to conclude.

Let $S$ be the set of reduced nonzero quadratic residues modulo $p$, so that $S$ contains exactly the reduced residues of $i^2$, where $i\in\left\{1,2,\cdots,\frac{p-1}2\right\}.$ Letting $S'$ be the set of possible reduced residues attainable by $(x^2+y^2)^2$ for integers $x$ and $y$ relatively prime to $p,$ we claim that $S=S'.$

First, we note that $(x^2+y^2)^2$ must clearly be a quadratic residue modulo $p$, implying that all quadratic nonresidues modulo $p$ are not in $S'.$ Additionally, note that having $(x^2+y^2)^2\equiv0\pmod{p}$ is equivalent to having $$x^2+y^2\equiv0\pmod{p},$$ which is impossible since $p\equiv3\pmod4$, due to Fermat's Christmas Theorem. Hence $0\notin S'$, thus ruling out all quadratic residues and zeroes modulo $p$. It follows that $S'\subseteq S$; to show that $S=S'$, it suffices to prove that there exist $x$ and $y$ such that $x^2+y^2\equiv a\pmod{p}$ for all $a\in\{1,2,\cdots,p-1\}$.

Let $N$ be some solutions to the equation $x^2+y^2\equiv a\pmod{p}$, where $(x,y)\in\{1,2,\cdots,p-1\}^2.$ Fixing $y$, we find that $$y^2\equiv a-x^2\pmod{p},$$ and it is well known that there are $\left(\frac{a-x^2}{p}\right)+1$ solutions $y\in\{0,1,\cdots,p-1\}$ to the above equation. However, note that we must have $y\not\equiv0\pmod{p}$ in our original setup; if such a case happens, there must exist some $x$ such that $a-x^2\equiv0\pmod{p}$, which happens iff $a$ is a quadratic residue modulo $p$, corresponding to $\left(\frac{a}{p}\right)+1$ such values of $x$, each of which correspond to $1$ bad value of $y=0.$ Thus, as $x$ ranges from $1$ to $p-1$, we find that $$N=\sum_{x=1}^{p-1}\left(\left(\frac{a-x^2}p\right)+1\right)-\left(\left(\frac{a}{p}\right)+1\right),$$ where the terms $\left(\frac{a-x^2}p\right)$ and $\left(\frac{a}{p}\right)+1$ respectively correspond to the number of pairs $(x,y)$ for some fixed $x$ and the number of bad pairs $(x,0)$.

We then find that
\begin{align*}
N&=\sum_{x=1}^{p-1}\left(\left(\frac{a-x^2}p\right)+1\right)-\left(\left(\frac{a}{p}\right)+1\right)\\
&=p+\sum_{x=0}^{p-1}\left(\left(\frac{a-x^2}p\right)\right)-\left(\left(\frac{a-0^2}{p}\right)-1\right)-\left(\left(\frac{a}{p}\right)+1\right)\\
&=p+\sum_{x=0}^{p-1}\left(\left(\frac{a-x^2}p\right)\right)-2\left(\left(\frac{a}{p}\right)+1\right)\\
&=(p-2)+\sum_{x=0}^{p-1}\left(\left(\frac{a-x^2}p\right)\right)-2\left(\frac{a}{p}\right).
\end{align*}
As it is well known that $$\sum_{x=0}^{p-1}\left(\left(\frac{a-x^2}p\right)\right)=(-1)^{\frac{p+1}2}$$ for $\gcd(a,p)=1$, as in our case, it follows that, since $p=4k+3$,
\begin{align*}
N&=(p-2)+\sum_{x=0}^{p-1}\left(\left(\frac{a-x^2}p\right)\right)-2\left(\frac{a}{p}\right)\\
&=(p-2)+(-1)^{\frac{p+1}2}-2\left(\frac{a}{p}\right)\\
&=(p-2)+(-1)^{2(k+1)}-2\left(\frac{a}{p}\right)\\
&\ge p-2+1-2\\
&=p-3.
\end{align*}
Thus, for $p>3$, we have $N\ge1$, so there must some $(x,y)$ such that $x^2+y^2\equiv a\pmod{p}$. It therefore follows that $S=S'$ for all $p>3.$

Now consider the case were $p=3.$ Since the only nonzero quadratic residue modulo $1$ and both $S$ and $S'$ are nonempty, it follows that $S=S'=\{1\}.$

We therefore conclude that $S=S'$ for all primes $p=4k+3$. Our final answer is therefore
\begin{align*}
|S'|&=|S|\\
&=\frac{p-1}2.
\end{align*}
$\blacksquare$

- Jörg


function
inequalities
logarithms
number theory
totient function
number theory unsolved

Let $ \phi$ be the [b][u]Euler's totient function[/b][/u]. Given any two real numbers $ \alpha$ and $ \beta$ with $ 0\leq \alpha<\beta\leq 1,$ prove that there exists a natural number $ m$ such that:

\[ \alpha<\dfrac{\phi(m)}{m}<\beta.\]

First estblish a lemma

Let $ a_n$ is an increasing sequence which diverges, but for which $ a_n \minus{} a_{n\minus{}1}$ is decreasing and convergent to zero.

Then for any reals $ A>0$ and $ B>A$ we can find $ n$ and $ k> n$ such that $ A < a_n \minus{} a_k < B$

Proof: We can take $ k>1$ such that $ a_k \minus{} a_{k \minus{} 1}$ < $ B \minus{} A$
Since $ a_n$ diverges there is a smallest $ n$ such that $ a_n \minus{} a_k > A$ but then $ a_{n\minus{}1} \minus{} a_k \le A$
Since $ n > k$, and the differences decrease, $ a_n \minus{} a_{n\minus{}1} < B \minus{} A$ adding the two inequalities gives $ a_n \minus{} a_k < B$ which completes the lemma

Now let $ p_i$ be the ith prime number and take $ m \equal{} \prod_{i \equal{} x}^{y} p_i$ for some positive integers $ x$ and $ y > x$

Then $ \frac{\phi(m)}{m} \equal{} \prod_{i \equal{} x}^{y} (1 \minus{} \frac{1}{p_i})$

$ \minus{}\ln( \frac{\phi(m)}{m}) \equal{} \sum_{i \equal{} x}^{y} \ln(1 \plus{} \frac{1}{p_i\minus{}1})$

Take the sequence $ a_n \equal{} \sum_{i \equal{} 1}^{n} \ln(1 \plus{} \frac{1}{p_i\minus{}1})  \implies \minus{}\ln(\frac{\phi(m)}{m}) \equal{} a_y \minus{} a_x$

This series is known to diverge while $ a_n \minus{} a_{n\minus{}1}$ is decreasing and convergent to zero, so it satisfies the conditions of the lemma

Take $ A \equal{} \minus{}ln(\beta)$ and $ B \equal{} \minus{}ln(\alpha)$ and apply the lemma to get the required result.


Evaluate:
$ \sum_{n\equal{}1}^\infty \frac{n}{4n^{4}\plus{}1}$

[hide="Solution"]By Sophie Germain Identity, we have $ 4n^4\plus{}1\equal{}(2n^2\minus{}2n\plus{}1)(2n^2\plus{}2n\plus{}1)$.

Since $ \frac{1}{2n^2\minus{}2n\plus{}1}\minus{}\frac{1}{2n^2\plus{}2n\plus{}1}\equal{}\frac{4n}{(2n^2\minus{}2n\plus{}1)(2n^2\plus{}2n\plus{}1)}$, the given summation is equal to $ \frac{1}{4}\sum_{n\equal{}1}^\infty\left(\frac{1}{2n^2\minus{}2n\plus{}1}\minus{}\frac{1}{2n^2\plus{}2n\plus{}1}\right)$.

Now, noting that $ 2(n\plus{}1)^2\plus{}2(n\plus{}1)\plus{}1\equal{}2n^2\plus{}2n\plus{}1$, the summation telescopes.

The given summation is $ \frac{1}{4}\left(\frac{1}{1}\minus{}\frac{1}{5}\plus{}\frac{1}{5}\minus{}\frac{1}{13}\plus{}\frac{1}{13}\minus{}\cdots\right)\equal{}\frac{1}{4}$.

Answer: $ \frac{1}{4}$[/hide]


Duke
college
trigonometry
calculus
MATHCOUNTS
AMC
AMC 8

well, i just got it today. i'm in the seventh grade. NOT satisfied.
CR - 610 BLEH.
Math - 790
Writing - 690.

How about you guys????

I think this is the wrong forum for SAT scores...Put this in the college forum...

If you want middle schooler's scores, then maybe Middle School math would work better.

I took the ACT in 7th, and I will take the SAT in 8th.

I did really well on the ACT.

I took the SAT in 7th..I'm in 8th right now
But i wasn't satisfied with my score

I think i got 690 reading or something

how do you take it? i wanna take it but i dont know where to go

i took it december of 2009 and im waiting for my results.....
btw i'm in seventh grade too

I'm in eighth grade this year, and I took it last year.  I got a really bad score, about 30 points lower than mathfreak in every category  :(

[quote="pinkmuskrat"]Wait, sorry, but why are non-juniors and seniors taking the SAT? Is this like a practice program?[/quote]

Kinda, Duke TIP allows Gifted and Talented students to take the SAT in 7th grade (it might be 8th or 9th too, i'm not sure).  I guess you can call it practice for the real thing a few years later, but to some people it does matter a lot (not to me).

Many schools let middle schoolers through sophomores take the PSAT even though it doesn't count until Junior year.  Its the same as the SAT except without the essay section.

I don't now about now, but about 8-9 years ago, I know Johns Hopkins' CTY program had middle schoolers take the SAT's for admission into their program through the "CTY Talent Search".

For any middle schoolers that are not satisfied with their score -- I fell asleep during the SAT's in 7th grade and got a 480 in CR, and I'm fine, so no worries :)

For any 7th graders not satisfied with their score, I did not take the SAT until my junior year in high school.

[quote="TZF"]I fell asleep during the SAT's in 7th grade and got a 480 in CR, and I'm fine, so no worries[/quote]

 :)

Hi! I'm taking the SAT's too (I'm in 7th grade) in a few weeks. I heard from some people at my school who took it in september that there was a few trigonometry and calculus questions. I'm not very sure about this, can anyone tell me is that is true or give me an overview of what is on it? In SAT prep books they never mention trigonometry or calculus. If you have taken the SAT's this year (2009-2010 school year) could you tell me? Thanks! It would be appreciated a lot!

I'm pretty sure there is some trig but no calculus. The trig is not hard, if there is any at all, you should be fine just knowing the angle/side relationships (sin, cos, tan).
If you know people who said there was calculus, there is DEFINITELY a non-calculus (and easier) way to approach the problem.

[url]http://www.collegeboard.com/student/testing/sat/about/sat/math.html[/url]
Unless I'm mistaken, I don't think there is any trigonometry. I've never seen any on practice questions. I'm pretty sure the ACT has trigonometry, though.

ofcourse there was no calculus on the test. you shouldn't be a little sheep and believe everything you hear. They are not allowed to just put calc. on the test w/o telling students, there would be a major uproar. I took it on Dec 5, 2009.

[quote="Watermelon99"]MathWise, yes I have never seen a trigonometry or calculus on any practice book test. However the book 'Master Math For The SAT' covers trig, and I know some people who have encountered a couple in the past. I do know a bit of trig (just the sine, cosine, tangent) but I'm sort of scared of the calculus. Several people at my school said that this year (2009-2010 school year) the SAT has calculus (or maybe hard math they have never seen before and they just called it calculus because it looks hard and reminds them of calculus?) and some other scary math they have never seen before. Actually one of the people who said that there was calculus is on the school MathCounts team and got a perfect on the AMC8 this year. This actually may be the case because every year too many people get a 800 since SAT math has always been so easy and they want fewer people to get 800's so the SAT will continue to be called a "reliable" test. Anyone who took it on December 5th 2009 (MathFreak77, do you know?)... is this true? I really want to get to the bottom of this!  :lol:[/quote]

No chance there is calculus - the highest level math you really need is about 8th-9th grade.

Not even the SAT II has calculus, if I'm not mistaken.

Got my scores.
Math- 570
Reading - 540
Writing- to fail to post.
7th grader
Maybe compared to you guys I failed but I scored higher than the average SAT taker (compared to last years scores).

I would be able to get my sat scores online now but since I'm not 13 til' tomorrow I don't get to know my scores yet.  WHY DO I HAVE TO NOT BE 13 YET.  SO UNFAIR. :bomb:  :bomb:  :wallbash_red:

Hmm...

800/700/660 for math/critical reading/writing.  Much better than last year.  I'm happy with this.

[quote="buzzer11"]Why is a 12 year old taking the SATs.... What are you, in sixth grade? CTY?[/quote]

I'm in 7th grade btw.  Today's my birthday so I have my scores:
Math:  710
CR:  630
Writing:  600

sigh.....

I took SAT 2 years ago in 7th grade - 790 math 730 reading 670 writing wasn't bad, considering that I was only 11 at the time.

Wow, congratulations! I took my first SAT recently and got a 2340: 790 reading, 800 math, 750 writing (11th grade). I got only an 8 out of 12 on the essay, so I guess I need to try to do better on that part. My overall score was better than I expected though, since I got only a 227 on the PSAT (but I didn't take that test very well). I didn't use my calculator at all on the math section; it almost seems to help if I don't.

@MayBach,

I just added another year to my birthday, when I change thirteen I will edit (you are allowed to) back to normal.  :D  :lol:

I took it in 7th grade and I failed...

I got like a 760 on math and a 540 on critical reading cause I stink at reading.

the weird thing is that I love to read.


function
trigonometry
algebra solved
algebra

We define a fuction $ f(x)$ circulate on R if there exists $ a>0$ :
          \[ f(x+a)=f(x)\]
We define a fuction $ f(x)$ anti-circulate on R if there exists $ a>0$ :
          \[ f(x+a)= -f(x)\]
 $ a$ is a cycle of the circulate or anti-circulate function $ f(x)$
We define $ a_0$ the basic cycle of the circulate or anti-circulate function $ f(x)$ if $ a_0=min(a)$
     Is this clause right or wrong?
          "If $ f(x)$ is anti-circulate on R with the basic cycle $ b$ then $ f(x)$ is circulate on R with the basic cycle $ 2b$"

The answer is right
Since the function is anti-circulate with cycle b we have that:

\[f(x+2b)=f(x+b+b)=-f(x+b)=f(x)\]
By applying the anti-circulate condition twice, giving that the function is circulate with cycle $2b$

This is if you assume that the same having $a$ basic cycle of $b$ implies this holds for all $x$, while it is also allowed to hold for smaller values than b at some locations - take $y=\sin x$ at $\pi$ versus $\frac{\pi}{2}$.


find all positive integerx,y,z satisfy: 3^x+4^y=7^z

[quote="eryaman"]I don't think this should be in getting started.[/quote]
why, i think thi pro is very interesting,
if it easy, can you solve equation $3^x+5^y=19^z$  :D

[quote="DPopov"]I can...but I don't think many people who view this forum can...I'd post in the intermediate ;)[/quote]
oh, and youcan solve $x^3+y^3=54z^3$- thanks :D

well...hmm

Solve or rather prove

$x^n+y^n=z^n$

has no nonzero solutions for $n\geq3$  :P  :P  :P

there's a diaphontine equation for you :D

The problem (or problem[b][u]s[/u][/b]) posted in this thread is(are) not fair to the reader's of this forum. Basically you cannot expect an average math student about the grade level of midschool to solve some tricky exponential diophantine equations, unless you have moved on pretty advanced (that's an exception, you're a genius). If any of you think this is fair, it's because you do not belong to 'Getting Started' and you should move on. Thank you.

i agree...as i stated above...move to intermediate or pre olympiad


analytic geometry
vector
geometry
geometric transformation
rotation

Professional recording studios often use reel-to-reel tape decks. On such a tape deck, the tape is pulled past the playback head at a constant linear speed of 0.381 m/s. (a) Using the data in part a of the drawing, find the angular speed of the take-up reel. (b) After 2.40 × $ 10^3$ s, the takeup reel is almost full, as part b of the drawing indicates. Find the average angular acceleration of the reel and specify whether the acceleration indicates an increasing or decreasing angular velocity.

[img]http://univirtual.info/elementaryphysics/art/images/cutnell3158c08/image_n/ngr014.gif[/img]

I know the definitions of displacement, velocity, and acceleration.  I also know the four kinematics equations.  Are those enough to solve this problem?  I can't seem to find a way to relate all of the given information together.

Thanks!

Do you know these equations?

$ l\equal{}r\theta$
$ v\equal{}r\omega$
$ a\equal{}r\alpha$

These are the equations that relate linear position/velocity/acceleration to their angular counterparts.

Yes, I know those formulas. :) So I found that part (a) is 7.5 rad/s.  For part (b), I know that $ \alpha \equal{} \Delta \omega / \Delta t$.  Is $ \Delta \omega$ positive or negative?  Also, how do I tell if the average ang. acc. means an increasing or decreasing velocity.  Thanks!

The sign depends on which way you orient your coordinate system.  But the rate, and sign of angular acceleration, can tell you a lot about velocity.  A high acceleration means that velocity is increasing quite fast, while an opposite sign will tell you the object is slowing down.

[quote="JRav"]The sign depends on which way you orient your coordinate system.  But the rate, and sign of angular acceleration, can tell you a lot about velocity.[/quote]

Could you please explain what you mean by orienting the coordinate system?  How would that apply to this question?

[quote="JRav"]A high acceleration means that velocity is increasing quite fast, while an opposite sign will tell you the object is slowing down.[/quote]

I don't get this... If a negative sign means that the object is slowing down, then how can sign depend on how I orient my coordinate system?  By the way, I'm getting an acceleration of 0.0017 (I don't know if it should be positive or negative.)

Thanks!

The standard example or coordinate orientation is when you drop an object.  You can calculate the speed from the equations of motion-that's easy.  Velocity is a vector so you need direction.  Since the object is going down, I would establish down to be the positive direction and up to be the negative direction.  So if a mass hit the floor with speed 5 m/s, the sign would be positive because I established down to mean positive.  But if the mass now bounces up at 3 m/s, the sign is negative, because I established up as the negative direction.  This is what I mean when I say that direction determines sign.

Suppose a ball is rolling on a surface to the right.  I might say that right is the positive direction.  If the acceleration is negative (to the left), it means that acceleration is in the opposite direction to the motion of the ball, so it would be slowing down.  This is actually what it means to be slowing down, that acceleration is in the opposite direction of motion.  (I could decide that right is negative and left is positive, so the acceleration would be positive and the ball would be slowing down, but such an orientation is silly).

Ok, but in this problem, there is a spinning motion.  Do I define my direction in terms of rotations?  Would something like counterclockwise=positive and clockwise=negative make sense?


The product of two numbers is 10,000. However, the two numbers do not have a 0 as any of their digits. What are the digits?

[hide]Only way that can happen is if none of the factors have both 2s and 5s
625*16
>>1,2,5,6<<[/hide]

[hide]
The prime factorization of 10,000 is $2^5\times5^5$.
For a factor to not have a 0, it can't have both a 2 and 5 in the prime factorization.
Therefore the factors are $2^5=32$ and $5^5=3125$
[/hide]

I think you mean that the prime factoriation of 10000 is 2^4*5^4

:oops:
then the answer is [hide]$2^4=16\times5^4=625$[/hide]

[hide]10000=2^4x5^4
since the numbers have no zeroes, they can't have both a 2 and a 5
2^4=16 and 5^4=625

1,6,2,5[/hide]

[hide]10000 equals 2^4 times 5^4.
Thus it is 16 and 625.[hide][/hide][/hide]


Let $x,y,z$ be positive real numbers such that $xy+yz+zx=1$. Prove that
\[\frac{x}{\sqrt{2x+y}}+\frac{y}{\sqrt{2y+z}}+\frac{z}{\sqrt{2z+x}}\ge \sqrt[4]{3}.\]

I think I can reduce to the much easier inequality: \[9\sum \frac{x^{4}}{(2x+y)^{2}}\ge \sum xy \]

I'm sorry, mine is much weaker and can't be used in any way to help this problem, in my solution I had a similarly sorted expression when I needed it oppositely sorted.

I know why it's so hard... it's completely wrong!!!
In fact, no result holds when exponent r is between 3 and 5.
For r<=3 your inequality holds, and for r>=5 it is flipped.

[img]8023[/img]
[img]8024[/img]

Sorry...

Continue then, mehome. I think you can solve this.  :)

I'm not quite sure what you mean...
do you mean the general inequality:
\[\sqrt[r]{\frac{x^{4}}{(2x+y)^{2}}}+\sqrt[r]{\frac{y^{4}}{(2y+z)^{2}}}+\sqrt[r]{\frac{z^{4}}{(2z+x)^{2}}}\ge \sqrt[r]{3}\qquad \bigcap \qquad xy+yz+zx=1\]
Oh by the way here is the $r=1$ proof:
Let $S_{a}=\frac{x^{2}}{2x+y}$. Then we must prove: \[\sum S_{a}^{2}\ge3\sum xy \]
But, \[\sum S_{a}^{2}\ge \frac{\left(\sum S_{a}\right)^{2}}{3}\]
Also, \[\sum S_{a}=\sum\frac{x^{2}}{2x+y}\ge \frac{\left(\sum x\right)^{2}}{3\sum x}=\frac{\sum x}{3}\]

Thus, \[\sum S_{a}^{2}\ge \frac{\left(\sum S_{a}\right)^{2}}{3}\ge \frac13 \cdot \left(\frac{\sum x}{3}\right)^{2}=\frac{\left(\sum x\right)^{2}}{27}\ge \frac{\sum xy}{3}\]

I don't usually bump old topics, but I had the occasion to run across this one and I would like to clear up the confusion.

This particular problem is not true, but it probably can be salvaged somehow.

Take $(x,y,z)=(0,1,1)$; the inequality becomes $\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{2}}\geq \sqrt[4]{3}$.
However, we may see that $\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{2}}- \sqrt[4]{3} = -0.031616962576319171909225759190192560227630901247221264124...$.


I know I've seen it before but is there a way to find what the following sequence converges to or simplify it?

$ a_{1}\equal{}1$

$ a_{2}\equal{}\sqrt{3a_{1}\plus{}1}\equal{}2$
$ a_{3}\equal{}\sqrt{3a_{2}\plus{}1}\equal{}\sqrt{7}$
$ a_{4}\equal{}\sqrt{3a_{3}\plus{}1}\equal{}\sqrt{3\sqrt{7}\plus{}1}$

I wasn't sure where this problem should go.

[hide="Convergence and limit"]1. The sequence is bounded above.

We show that $ a_i < {3 \plus{} \sqrt {13}\over 2}$, inductively. For $ i \equal{} 1$ it's obvious, and $ a_{i \plus{} 1} \equal{} \sqrt {3a_i \plus{} 1} < \sqrt {{11 \plus{} 3\sqrt {13}\over 2}} \equal{} \sqrt {{22 \plus{} 6\sqrt {13}\over 4}} \equal{} {3 \plus{} \sqrt {13}\over 2}$

2. The sequence is strictly increasing.

The function $ f(x) \equal{} x^2 \minus{} 3x \minus{} 1$ is negative in $ \left(0,{3 \plus{} \sqrt {13}\over 2}\right)$ - actually, it's negative in a wider interval, but this reduced one is of our interest. Since all $ a_i$ lie in that interval (see 1.), we have $ a_i^2 \minus{} 3a_i \minus{} 1 < 0\iff a_i < \sqrt {3a_i \plus{} 1}\iff a_i < a_{i \plus{} 1}$

From 1. and 2. it follows that $ a_i$ is convergent. Its limit $ L$ must satisfy $ L \equal{} \sqrt {3L \plus{} 1}\iff L \equal{} {3 \plus{} \sqrt {13}\over 2}$[/hide]

[quote="Farenhajt"]We show that $ a_i < {3 \plus{} \sqrt {13}\over 2}$, inductively.[/quote]
How did you come up with this? I see why its true but how did you find specifically $ {3 \plus{} \sqrt {13}\over 2}$?  Just through observation?  What induction did you use?

Also, what if $ a_{1} \equal{} 2$?  Would you have to find another value that $ a_{i}$ is less than?

Thanks for your help, it is greatly appreciated.

Well, when doing exercise problems like this, it's usually easiest to "assume" the sequence is converging and to find its limit - in the given case, by solving $ L \equal{} \sqrt {3L \plus{} 1}$. Then it's easier to construct a proof of convergence using some of the basic criteria.

As for the other question, you accomplish nothing by choosing $ a_1 \equal{} 2$, since $ a_2$ is already $ 2$ :) But, if you chose $ a_1 \equal{} 4$, for example, then you could argue that the sequence is bounded BELOW and strictly DECREASING, which I leave for you to do as an exercise.

NOTE: I could just as well use the estimation $ a_i < 5\implies a_{i \plus{} 1} < \sqrt {3\cdot 5 \plus{} 1} < 5$. The exact boundary isn't that important, what's important is that there IS one. However, the value I gave was important for the part 2.


As a beginning 'mathematical enthusiast i began to
examined the solve "Fermat's Last Theorem." Fortunately,the said problem that baffled great mathematical minds for the past
350 years finally ended in 1995."Andrew Wiles" did solved the
riddle. I have few easy questions related to 'Riemannian Hypothesis:"

             1)What really makes 'riemannian hypothesis' unique,and very different?
             2)What are the distinctive qualities do possessed  by the said hypothesis that makes it a seeminly hard problem to crack?
             3)What commonality,and differences exixst between
the 'riemannian hypothesis,' and 'fermat's last theorem?'

                      Before i embark into these mathematical
mazes(Riemannian Hypothesis),i just want to get few advices.mathematically.Any "INTELLECTUALLY BRAVE SOUL'S"
comment?

                                                Respectfully,
                                              A.Eddington

How much do you already know about the Riemann Hypothesis? I'd imagine a quick google of "Riemann Hypothesis" would fill you in on all your questions, although I'll offer quick answers.

1) It's unique because it is. There is no other theorem that is the same.

2) It's a seemingly hard problem to crack because there does not appear to exist a moderately easy-to-find (even in scales of years) proof.

3) They are similar probably only to the extent that they are both in the field of Number Theory. However, they are otherwise very different, as googling will inform you.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14074[/img]

 :(

$ \angle T_aIT_c \equal{} 90 \plus{} \frac{\beta}{2}$, so we can costruct a angle equal to $ \beta$, let e point P (interior the convess angle $ \angle T_aIT_c$) such that $ \angle T_aPT_c \equal{} \beta$. Call $ \Gamma$ the circumcircle of $ \triangle T_aPT_c$. Then B stay on $ \Gamma$. Let M be the midpoint of the arch $ T_aT_c$ that not contain B. Then IM meet $ \Gamma$ at B and $ BT_a$ meet $ IT_c$ at C and $ BT_c$ meet $ IT_a$ at A.

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14095[/img]

 :thumbup:


geometry
parallelogram
geometric transformation
rotation
rectangle

In parallelogram ABCD, |AB|>|BC| let K,L,M,N be points of touch of incircles of triangles ACD, BCD, ABC, ABD (respectively) to one of diagonals (so K is the point where incircle of ACD touches AC). Prove that KLMN is rectangle.

First, it is a parallelogram:
[hide]Draw the incircles of triangles ABD and CDB.  We see that the resulting diagram is symmetric in a 180 degree rotation around the center of the parallelogram O and so it follows that LO and NO are equal.  Similarly, MO and KO are equal.  Thus, the diagonals of KLMN bisect each other and KLMN is a parallelogram.[/hide]
Second, it is a rectangle:
[hide]We can calculate the length MO.  Let |DA| = a > b = |CD| and let |AC| = c.  Then AO = c/2 and AM = (a - b + c)/2 (from the fact that the tangents to a circle from a point are equal) so that MO = (a - b)/2.  Similarly LO = (a - b)/2.  Thus the diagonals are of equal length and so the figure is a rectangle. [/hide]


Does anyone know how to do a piecewise function?  You know with the curly braces.

Like this?

\[
|x| = \left\{ \begin{array}{l}
x\; \mathrm{if}\, x \geq 0 \\
-x\; \mathrm{if}\, x < 0
\end{array}\right.
\]

[code]$$
|x| = \left\{ \begin{array}{l}
x\; \mathrm{if}\, x \geq 0 \\
-x\; \mathrm{if}\, x < 0
\end{array}\right.
$$[/code]

Another way:

[tex]s\left(x,y\right) = \left\{ \begin{array}{cc}0,&\mbox{ if } x \neq y\\1,&\mbox{ if } x=y\end{array}\right.[/tex].

[code]s\left(x,y\right) = \left\{ \begin{array}{cc}0,&\mbox{ if } x \neq y\\1,&\mbox{ if } x=y\end{array}\right.[/code]

I have learnt this method from Valentin.

  Darij

Ahh, thanks. I should have thought of using an array.

Eum, darij, why is that another method? :?

That's the same but without extra enters, isn't it? Or did I miss something obvious?

Yes, sorry, I see it's the same, I just used centers and \mbox instead of \mathrm.

  Darij

The way I learned to do it is like this:

\[f(x)=\begin{cases}1&x\in\mathbb{Q},\\0&x\in\mathbb{R}\setminus\mathbb{Q}.\end{cases}\]

I think this looks a bit better since it doesn't have such a large space after the curly braces.


limit integration trigonometry logarithms


At the topic [b]Hard one [/b](which is locked  for REPLY  ;) ), Harazi wrote:
[quote]Prove that for all x_1,...,x_n>0 with product 1 and for all n>4 we have
     1/sqrt(1+x_1)+1/sqrt(1+x_2)+....+1/sqrt(1+x_n)<n-1.
I will bring this one to your attention since it is a very interesting and hard inequality. I hope that someone will be so smart to prove it. I find it one of the hardest inequalities.[/quote]
Then Namdung wrote:
[quote]So, all we have to do is to prove it for n = 4.
I think someone in our site can help me in finding solution for this one.[/quote]
About this inequality I can say the following:
Last summer (in June), I sent this inequality to Gazeta Matematica B, but.
In the same day, I sent to Gazeta Matematica A the inequality
     $\frac{1}{\sqrt{4+5x_1}}+...+\frac{1}{\sqrt{4+5x_n}}\leq\frac{n-1}{2}$,
where $n\geq3$ and $ x_1...x_n=1.$
This last inequality was posted in GM-A no.1, 2004.
For this reason, I cant post now my solution to the first inequality. I will do it after some months.

[quote="harazi"]I agree with Vasc this is a very hard problem. But, Vasc, please look at the sixth round of the mathlinks contest for an even harder problem! I have a very very nice solution to your problem![/quote]
Hi,Harazi,can you post your solution for Vasc's last inequality when $n=3$,thanks.

Put $f(x) = \frac{1}{\sqrt{4+5x}}$

To important steps:

1) Use logconcavity to show that if
  $x, y \le 2.5$ then $f(x) + f(y) \le 2f(\sqrt{xy})$

2) Prove that for all $t \ge 0$

   $\frac{2}{\sqrt{4+5t}} + \frac{t}{\sqrt{4t^2+5}} \le 1$

the equality holds when t = 0 or t = 1.

Use IGI theorem or UMV theorem we can replace $4/5$ by an abiraty $k$.

What do you means, Hung? Could you formulate your general statement?

Namdung

[quote="hungkhtn"]Yes, It's in my book I sent to you. IGI and UMV theorem.[/quote]
Ok :D
To Hungkhtn, Can you sent it to me? :P
UMV is U.. Mixing Variable. I think it very interesting and I want see SMV (stronger mixing variable), too :D

[quote="hungkhtn"]Yes, It's in my book I sent to you. IGI and UMV theorem.[/quote]

Dear Hung, I didnt ask about IGI or SMV but about generalization of Vasc result.

Namdung

OK, By IGI or UMV we can find the maximum of
\[ \frac{1}{\sqrt{1+kx_1}}+\frac{1}{\sqrt{1+kx_2}}+...+\frac{1}{\sqrt{1+kx_n}} \]

I know we can, but I ask about the result, Hung. Could you formulate it?

IGI or UMV show that we only need to consider the problem in case $n-1$ variable equals.

Dear Hung,

I did know about that. But what is the maximum in general case? Generalization is quite complicated work, we should be carefull.

Namdung


well those aime problems were eaten up quickly:

Note that 6! = 8910. What is the largest n such that n! is a product of n-3 consecutive positive integers.

Here is a theorem I concocted on my own:

1!*3!*5!*...*(2n-1)!=(n(n+1)/2)!

You never know how often come across triangle numbers in some math problems.

The only problem is that the theorem isn't true.  For example, n = 5 is a counterexample.

Oh! :cry:

There were a lot of correct answers for there to be exceptions.  It works if n=1,2,3,4 and 5.  I got this from a Mathcounts problem.  I can't remember the year this problem was from.

[hide]OK, first find what the product of n-3 consecutive integers is:

(n-3)!, but (n-3)! < n! no matter what so we try to change the form.

(n-3)! = (n-2)!/1!. However this is still too small. Keep going:

(n-2)!/1! = (n-1)!/2!. Nope. Keep going:

n!/3!... n! never equals 3!*n! so obviously no.

(n+1)!/4!... aha! if (n+1)!/4! = n!....

then (n+1)! = 24(n!)

which means that if n is greater than 23 it will not work. But obviously if you plug 23 back into that you get 24! = 24*23! which is true. Therefore, n = 23.

Sorry if this isn't a great explanation, but its the best i could do.

[/hide]

[quote="paladin8"](n+1)!/4!... aha! if (n+1)!/4! = n!....

then (n+1)! = 24(n!)
which means that if n is greater than 23 it will not work[/quote]
I am not sure that I understand this part. How do you know that for $k>1$ the equation $(n+k)!=(k+3)!\cdot n!$ would not have the solution for $n$ greater than $23$?

[quote="frodo"]Here is a theorem I concocted on my own:

1!*3!*5!*...*(2n-1)!=(n(n+1)/2)![/quote]
The reason why I am reopening this tread is very interesting frodo's "theorem". I have found a very nice proof that this equality holds only for $n\in\{1,2,3,4\}$.

For $n=5$ or $n=6$ the right side is divisible by $13$ while the left side is not. Suppose that $n\geqslant 7$. We have \[n^2-7n\geqslant 0\iff 4n\leqslant\frac{n(n+1)}2\] By Bertrand's postulate there is a prime number $p$ such that $2n\leqslant p<4n$ and thus the problem is solved.

So you're saying we want to find facorials where n!=(n-3+k)!/k!.  If we divide by n!, we get 1=(n-3+k)!/(k!n!).  Now, k>3 or we have 6n!=n! or 6=1.  If k=4, (n+1)!/24n!=1 so n+1=24 and n=23.  For k=5, (n+2)!/120!n! and we get n+2=120 and n=118.  This should extend infinitely for identifying which numbers satisfy this condition.


Find all positive triple $ (t,a,b)$ with $ a,b>1$ such that $ 2^t \equal{} a^b \pm 1$

Did you mean $ a,b>1$, or indeed $ a,b\geqslant 1$? In the first case, there is only a solution $ (3,3,2)$, by Catalan's conjecture (now theorem). In the second case, add $ (1,1,r)$ and $ (r,2^r\mp 1,1)$ for all $ r\in\mathbb{N}$.

[quote="Bojan Basic"]Did you mean $ a,b > 1$, or indeed $ a,b\geqslant 1$? In the first case, there is only a solution $ (3,3,2)$, by Catalan's conjecture (now theorem). In the second case, add $ (1,1,r)$ and $ (r,2^r\mp 1,1)$ for all $ r\in\mathbb{N}$.[/quote]
Any solution without Mihailescu Theorem?
(That's too hard theorem!)

[quote="Bojan Basic"]I have no idea, I don't really like putting unnecessary constraints to oneself (like "no using hard theorems"—whatever that means, "no using computers" etc.). To me, there is only a correct and incorrect solution. :) Therefore, I would probably not think about another approach to this problem, and you’ll have to wait for someone else.[/quote]
My idea is the opposite of you [b]Bojan Basic[/b]  :lol:
what i think is that using only elementary method to kill the problem is the best way to do Mathematics problem...
There aren't many people who know a lot of theorem like you [b]Bojan Basic[/b] !! I wanna be a Math teacher who can make every students understand the way to solve the solution easily...And here is an elementary solution for this one...
[u]case1[/u]
$ 2^t \equal{} a^b \minus{} 1$
for $ t \equal{} 0,1$ is trivial we have no answer here
for $ t\geq 2$ since [b]a[/b] must be odd from
$ 2^t \equal{} (a \minus{} 1)(a^{b \minus{} 1} \plus{} a^{b \minus{} 2} \plus{} ... \plus{} a \plus{} 1)$ but from$ (a^{b \minus{} 1} \plus{} a^{b \minus{} 2} \plus{} ... \plus{} a \plus{} 1) \equal{} b (mod 2)$we got that [b]b[/b] must be even let $ b \equal{} 2w$ for some $ w\in N$
$ 2^t \equal{} (a^w \minus{} 1)(a^w \plus{} 1)$ but $ (a^w \minus{} 1,a^w \plus{} 1) \equal{} (2,a^w \plus{} 1) \equal{} 2$ since [b]a[/b] is odd
so $ a^w \minus{} 1 \equal{} 2$ then $ (t,a,b) \equal{} (3,3,2)$ is also an answer so
$ (t,a,b) \equal{} (3,3,2)$ for this one
[u]case2[/u]
$ 2^t \equal{} a^b \plus{} 1$
for $ t \equal{} 0,1$ is trivial we have no answer here
for $ t\geq 2$ since [b]a[/b] must be odd from $ a^{2l} \equal{} 1 (mod 4)$ for all $ l\in N$ out [b]b[/b] must be odd let $ b \equal{} 2w \plus{} 1$
$ 2^t \equal{} (a \plus{} 1)(a^{2w} \minus{} a^{2w \minus{} 1} \plus{} ... \plus{} a^2 \minus{} a \plus{} 1)$ but $ (a^{2w} \minus{} a^{2w \minus{} 1} \plus{} ... \plus{} a^2 \minus{} a \plus{} 1)$ is odd so $ a \equal{} 0$
no answer of this one


MATHCOUNTS
geometry
perimeter
ratio
percent
number theory
least common multiple

How many two-digit numbers less than 50 have an odd number of factors?

What is the smallest positive integer greater than 1 such that division by each of 4, 5, 6, 9, and 10 gives a remainder of 1?

What is the positive difference between the two greatest prime factors of 2730?

How many positive integers less than 100 have reciprocals with terminating decimal representations?

What is the sum of the least and the greatest three-digit prime numbers that are less than 200?

Zeno had to paint a 15 x 15 square floor.  He decided that each day he would paint half of the unpainted part from the previous day until there was only one square foot or less left, in which case he would stay and finish the job that day.  Using this strategy, how many days did it take Zeno to paint the entire floor?

The lengths of two sides of a triangle are 7 cm and 3 cm.  If the number of centimeters in the perimeter is a whole number, what is the number of centimeters in the positive difference between the greatest and least possible perimeters?

What is the units digit of $1998^{1998}$?

The perimeter of an isosceles triangle is 38 centimeters and two sides of the triangle are whole numbers in the ratio 3:8.  What is the number of centimeters in the length of the shortest side?

The lengths of two sides of an isosceles triangle are 10 centimeters and 12 centimeters.  Find the number of square centimeters in the positive difference between the greatest and least possible areas for this triangle.  Express your answer as a decimal to the nearest hundredth.

There are 16 boys and 24 girls in an algebra class.  What percent of the class is girls?

Seven points are marked on the circumference of a circle.  How many different chords can be drawn by connecting two of these seven points?

1. [hide]The numbers are written as axby. X and Y must both be even in order for there to be an odd number of factors, (since and even+1=odd, and the number of factors is (x+1)(y+1)...). So, prime numbers are out. The first even number besides 0 is 2, and so all perfect squares are in (4-7).

And I think that's all. There's a 99% chance I'm wrong.

[/hide]

2. [hide]So, x=1mod4, 1mod5, 1mod6, 1mod9, and 1mod10. Take the one over, and x=1mod(LCM (4, 5, 6, 9, 10))...x=1mod180, so x is at least 181.

[/hide]

3.

[hide]2730=(2)(3)(5)(7)(13), so the answer is 6.

[/hide]

5.

[hide]101+199=300??? Haha...[/hide]

I have to go buy some school supplies. Be back later.

1.
[hide]For a number to have an odd number of factors, it has to be a perfect square.  so all the two digit numbers less than 50 that's perfect square are 16, 25, 36, and 49.
So the answer is 4[/hide]

2.
[hide]For number 2, I think it would just be the LCM of all the numbers and u add a 1 to it. The LCM of 4, 5, 6, 9, and 10 is 180. So the answer would be 181[/hide]

3.
[hide]The prime factorization is 5*3*2*13*7 so the difference of the the greatest prime factors are 13-7 which is 6[/hide]

5.
[hide]101+199 = 300[/hide]

6.
[hide] 15*15 = 225
so ...
day one 112.5 left
day two 56.25
day three 28.125
day four 14.0625
day five 7.03125
day six 3.515625
day seven 1.75.....
day eight is the finishing day

If I divided wrong, then the answer would probably be wrong, but I think it's 8[/hide]

7.
[hide]I think the greatest would be 9 and the least would be 5, so the answer should be 4. 3+7+9 = 19 and 3+7+5 = 15[/hide]

8.
[hide] this is the same as calculating the unit digit of 8 to the 1998th power, powers of 8 can end in 8, 4, 2, and 6. 1998/4= 499 R 2, so the answer would be 4.[/hide]

9.
[hide]38 is equal to (3+8+8)x or (3+3+8)x, so it would be either 19x or 14x, side the sides have to be whole numbers, it can't be 14x. x is equal to 2 and the shortest side is 6. [/hide]

11.
[hide]24/40 = 60%[/hide]

12.
[hide]The Combinations of choosing 2 out of 7 is 21[/hide]

So far nobody has written a solution to 10.  First, what are the two possible triangles?  Next, what seems like a reasonable way of calculating the area of each of the possible triangles?

Looking at your clues, I figured this far:
[hide]the length of the triangles can be 12,12,10 or 10,10,12.

 the hight of the 12,12,10 triangle would be 13 and the area would be 65.
 the height of the 10,10,12 triangle would be the square-root-of-136 or 2-times-the-square-root-of-37 and area would be 12-times-the-square-root-of-37

but I don't know what the square-root-of-37 would be so I don't really know the answer...

is my method close?:? [/hide]

When you divide the two triangles up into symmetric right triangles, which length is the hypotenuse on each triangle?

[hide]o, never mind, I figured that you are suppose to subtract instead of add

but on the 12,12,10 triangle the height would be the square-root-of-119, wouldn't it?

the 10,10, 12 triangle would have a height of 8 and have the area of 48.

I still don't know what the square-root-of-119 would be, so I still don't know how to solve the problem

I got a calculator and got that 5-times-the-square-root-of-119 is 54.54356.. and so I think the answer would be 6.54356...[/hide]


I am positively confused by Thermodynamics. So many formulae, derivation etc. And this is only the first law. Any suggestions of what I can do?

i don't know which book u r following...but i would suggest atkins and if u read with positive sense going deep into why such thing was defined so on and so....u will like it
thermodynamics is atleast one of my favorite subject

Actually, you should only memorize a few equations. Read carefully the subject, and try to solve a reasonable number of exercises and problems. Any doubts? Post them here.

Here's an example:

A mole of ammonia gas is condensed at its boiling point at $ 1.01325 bar$ of $ \minus{}33.4^{\circ}C$ by the application of a pressure infinitesimally greater than $ 1 bar$. To exaporate ammonia at its point requires the absorption of $ 23.30kJmol^{\minus{}1}$. Calculate:

a) $ w_{rev}$ b)$ q$ c)$ \Delta H$ d)$ \Delta U$

Phase changes are constant T and P processes. Therefore:

$ q \equal{} q_P \equal{} \Delta H$

And because it's isothermal so $ q_p \equal{} \minus{}w_{rev}$ right?

Wrong., because we cannot consider the behaviour of ammonia as ideal.

Well yes but for the purpose at hand (before April we always consider Ideal Gas properties) we consider it as an ideal gas.

Four months considering ideal-gas behaviour?! What kind of physical-chemistry course is that? Your bad-luck is that I am not your teacher.

That's because in April there is another camp teaching more advanced stuff. :P

[quote="Carcul"] Your bad-luck is that I am not your teacher.[/quote]
that's true...otherwise i hwould have started liking chemistry a long long ago...but i had a wonderful organic techer....initially i thought u were him  :) (may be that's true  :P )

By the way Carcul, didn't you say that you are teaching Physical Chemistry this year or something?

Wondering about what? Yes, I am giving a Physical Chemistry course to Pharmaceutical Sciences students, which, as can be expected, features "only" Thermodynamics and Kinetics.

By the way, I am currently writing a "Collection of Thermodynamics Problems" thread: its far from being finnished, but I can post already what I have (with the condition that no one else post it, at least until I complete it).

I wouldn't mind to wait until you finish. On a different note, do you have any Biochemistry-related problems?

No, I do not have (except a few that are physical-chemistry problems featuring biomolecules).

Never mind then. I wish Carcul was my Physical Chemistry teacher (crash course in Thermodynamics)!

Even i would join the corse if you agree....
i think the course should feature Thermodynamics,Equilibrium,Kineticsand some others which are important..........

[quote="Carcul"]Thanks. Just by curiosity, what are the subjects that your Physical Chemistry course will feature?[/quote]

Thermodynamics, Kinetics, Analytical. They sort of integrate each topic (each is taught by a different tutor).

Analytical is not a part of Physical Chemistry. Anyway, if that's the structure of your PC course then why the worry about Quantum Chemistry?

It's not just a Physical Chemistry course. Quantum Chemistry is part of the programme.


algebra
polynomial
number theory
superior algebra
superior algebra solved

Let F be a field which contains finite elements. It is not difficult to prove that $|F|=p^n$. And it can be shown that for every prime p there are such fields which contains p and $p^2$ elements. Is there a field containing $p^n$ elements, where $n>=3$?

[quote="grobber"]This is far from an open question. For each prime $p$ and each natural $n$ there is exactly one field with $p^n$ elements, and it's the splitting field of the polynomial $x^{p^n}-x$ over the field with $p$ elements.[/quote]
Maybe I've misunderstood you. Buy why such a field exist?

Try google-ing "Galois theory".

Let $F$ be the splitting field of $f(x)=x^{p^n}-x$ over $F_p$. Try to show that the roots of this polynomial are distinct, and form a subfield of $F$, meaning that $F$ is, in fact, formed by these $p^n$ roots.

if you can get your hands on: "A Classical Introduction to Modern Number Theory (Graduate Texts in Mathematics)"

by Ireland and Rosen, they have a nice section on finite fields (although they do not prove that fields of the same size are unique up to isomorphism) which provides a nice proof of the fact you stated (which would be take too much time to post to this board)

Thanks for your informations.
I thought it had something to do with theories...but I was not sure.


summer program
Mathcamp
combinatorics unsolved
combinatorics

Suppose you start with the number 1 and go through a series of steps, where at each step you add a (positive integer) divisor of your current number to the number itself, to get a new number. For instance, the first step is forced: you have to take 1 + 1, so your new number is 2. Now you have two choices; the next number could be 2 + 1 = 3 or 2 + 2 = 4. If you choose 4, the next step after that can take you to 5, 6, or 8.

   1. Show that if N is less than or equal to 2n+1, then you can always reach N using 2n steps or fewer.
   2. Show that if N is larger than 2n and you can reach N in n+1 steps, then N is a sum of two powers of two.

This is Problem #4 on the Mathcamp Qualifying Quiz 2009. [url]http://mathcamp.org/quiz.php[/url]

According to this, applications are accepted until April 25th 2009. [url]http://mathcamp.org/apply.php[/url]


What is the maximum number of points of intersection of two congruent squares that do not share a common line segment?

Tilt one square 45 degrees relative to the other square.  Then they will intersect 8 times.


Hi! I would like to include in my text some html tags, the problem is that if I write <a> this is showned as !a? in the pdf. Any ideas how could I correct this???

thanks in advance!

[quote="krümel"]Hi! I would like to include in my text some html tags, the problem is that if I write  this is showned as !a? in the pdf. Any ideas how could I correct this???

thanks in advance![/quote]

Not all the html tags work in the forum, but you can use the BBCode here that works better. For example, instead of posting

[code]this topic[/code] you can post

[code][url=http://www.mathlinks.ro/Forum/posting.php?mode=quote&p=224175]this topic[/url][/code]It takes a while but eventually you will remember them easely. Just in case, the buttons just over the edition screen, are very useful, and those are the ones you usually need.

[quote="Ledurt"]what is a "html tag" ?
is it some sort of link in the document or a small drawing?

ledurt[/quote]It stand for Hyper Text Markup Language (html), and it is the usual internet exchange text format. If you want, read [url=http://www.cosy.sbg.ac.at/~lendl/tags.html]this list[/url] to understand the sintaxis. This forum uses BBCode that is another version of html (somehow).

[b]BTW[/b], what is this doint in the $\LaTeX$ forum?

djimenez, I think you missunderstood me, I'm not talking about using html tags in this forum, I'm talking about including html tags in my latex file  :?

thanks in advance!

OK, I got it. Now, do you want to write them down or you want to actually use them? Actually, it shouldn't be !a? but ¡a¿, and that is because those signs (used mainly in spanish, as you should now) are not usually finded on a regular keybord, therefore $\text{\TeX}$ replace < and > by them (because they are no usually part of a text unless it is math, and in that case, it would be within an equation enviroment). If you want to include html code to be written in the text, if it is short, I would recomend

[code]\verb|
the text you want
|[/code]

If it is a lot longer (probably several lines) I would recommend the use of the full enviroment tool

[code]\begin{verbatim}
    show(the=0; text < you ; want++){
        to.show();
    }
    and(after,that);
    close(the);
    brackets[right];
\begin{verbatim}[/code]

I hope it helps!

Ok! thanks David, it worked! I use your first solution because I just need to add a few tags separately, I'm not going to add html code.

Muchísimas gracias!! ;)

If you're going to be doing a lot of this sort of thing (showing computer code such as HTML in a LaTeX document), I suggest using a specialized package, such as the [b]listings[/b] package, available at:
[url]http://www.ctan.org/tex-archive/help/Catalogue/entries/listings.html[/url]

thank you DPatrick. I don't think I'm going to include any code, but it's good to know it, just in case in the future I need it.

thanks!


Show that the arithmetic sequence $2,5,8,11,14,...,2+3n$ may or may not contain one perfect squares.

well, i know u said to prove it, and i will, but the answer is no.

[b]Do not answer this question. It is too closely related to a USAMTS problem.[/b]

o, gosh, im sorry, i forgot...damn...i am sorry....i should have taken a better look at the USAMTS problems....damn...


vector
linear algebra
linear algebra unsolved

Prove that any linearly independent subset of the span of any $ r$ vectors is has size less than or equal to $ r$.

Attempt:  Suppose $ w_{1},\dots ,w_{s}\in\textrm{Span}\left(v_{1},\dots ,v_{r}\right)$ so that for each $ j\in\left[1,s\right]$, we can write

\[ w_{j}=a_{1j}v_{1}+\cdots +a_{rj}v_{r}.\]
Now if $ x_{1}w_{1}+\cdots +x_{s}w_{s}=0$, then $ x_{1}=\cdots =x_{s}=0$ so

\begin{align*}x_{1}w_{1}+\cdots +x_{s}w_{s}= &\left(a_{11}x_{1}+\cdots +a_{1s}x_{s}\right)v_{1}+ \\ &\cdots +\left(a_{r1}x_{1}+\cdots +a_{rs}x_{s}\right)v_{r}.\end{align*}
Where do we go from here?

$ A$ with entries $ a_{ij}$ is a matrix.

What size is it? What happens if you row eliminate it (or column eliminate it)? How many pivot elements can it have, at most? (That is, what is its maximal rank.) Does it have a nontrivial null space?


geometry rhombus conics ellipse rotation

x,y is real numbers

|x|/2 + |y| = 1

find the max,min of
f(x,y) = x^2 - xy + y^2

[/code][/u]

[quote="AndrewTom"]I'm not sure about this, but I think:

$ \frac {\mid x \mid}{2} \plus{} \mid y \mid \equal{} 1$ defines a rhombus with vertices at $ (2, 0), (0, 1), ( \minus{} 2, 0)$ and $ (0, \minus{} 2)$.

$ x^{2} \minus{} xy \plus{} y^{2} \equal{} 4$ is an ellipse passing through these four points, so the maximum value is $ 4$ and the minimum value (at the origin) is $ 0$.[/quote]

how do you know that f(x,y) = x^2 -xy + y^2 is a ellipse?

The ellipse is rotated in relation to the axis, of course.

how do you know that the ellipse is rotated?

if there is an XY term the conic is rotated


function
inequalities proposed
inequalities

[quote="Marius Mainea"][quote="Marius Mainea"][quote="crystaljewel3"]:blush: $ \frac {(b \plus{} c \minus{} a)^{2}}{(b \plus{} c)^{2} \plus{} a^{2}} \plus{} \frac {(c \plus{} a \minus{} b)^{2}}{(c \plus{} a)^{2} \plus{} a^{2}} \plus{} \frac {(a \plus{} b \minus{} c)^{2}}{(a \plus{} b)^{2} \plus{} c^{2}}\geq \frac {3}{5}$. :?:[/quote]

  Sorry.[/quote]

 Maybe  $ \frac {(b \plus{} c \minus{} a)^{2}}{(b \plus{} c)^{2} \plus{} a^{2}} \plus{} \frac {(c \plus{} a \minus{} b)^{2}}{(c \plus{} a)^{2} \plus{} a^{2}} \plus{} \frac {(a \plus{} b \minus{} c)^{2}}{(a \plus{} b)^{2} \plus{} c^{2}}\le \frac {3}{5}$[/quote]
No! It can be $ \frac {(b \plus{} c \minus{} a)^{2}}{(b \plus{} c)^{2} \plus{} a^{2}} \plus{} \frac {(c \plus{} a \minus{} b)^{2}}{(c \plus{} a)^{2} \plus{} a^{2}} \plus{} \frac {(a \plus{} b \minus{} c)^{2}}{(a \plus{} b)^{2} \plus{} c^{2}}\geq \frac {3}{5}$
I have found a nice proof:
$ \sum_{cyc}\frac {(b \plus{} c \minus{} a)^{2}}{(b \plus{} c)^{2} \plus{} a^{2}} \equal{} \sum_{cyc}\frac {(b^2 \plus{} bc \minus{} ab)^2}{a^2b^2 \plus{} (b^2 \plus{} bc)^2}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2b^2 \plus{} (b^2 \plus{} bc)^2)}.$
Id est, it remains to prove that $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2b^2 \plus{} (b^2 \plus{} bc)^2)}\geq\frac {3}{5},$ which is equivalent to
$ (a^2 \plus{} b^2 \plus{} c^2)^2\geq3(a^3b \plus{} b^3c \plus{} c^3a).$ :lol:

Arqady, nice use of Cauchy lemma or Cauchy Schwarz !!!!!
Had exactly the same solution in my mind.

Anyway, for those who might have problems with the finishing of Arqady's solution: It is just an application of Cauchy Schwarz.

Can someone give a proof using Convexity? I am trying for one but till now haven't got it.

Why don't you try to set $ a \plus{} b \plus{} c \equal{} 3$ and give it a shot with convexity.

Because the function wouldn't be convex? :roll:


geometry AMC AIME USA(J)MO USAMO HMMT ARML

With the Intermediate Geometry book from AoPS still in the works, does anybody know of a well-written, comprehensive geometry book AIME #5-15 level?

Here is how I would recommend anyone beyond plain geometry after Volume I.

[url=http://www.amazon.com/Challenging-Problems-Geometry-Alfred-Posamentier/dp/0486691543]Challenging Problems in Geometry[/url]
- I have this book and it is pretty nice. It does not have competition-like format as AoPS Volume I does but nevertheless, it is one of the often mentioned books for AIME geometry. It covers several important theorems that only appear in Volume II.

[url=http://www.amazon.com/Geometry-Revisited-New-Mathematical-Library/dp/0883856190/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1203030327&sr=1-1]Geometry Revisited[/url]
- I also have this book but I have not gotten chance to read it much mainly because it is very difficult. The book is best for very difficult AIME geometry and probably, more of USAMO and IMO preparation. The book has problems but a lot of things is not really for AIME/AMC like nine-point circle and butterfly problems.

My best suggestion is that if you want to master AIME #5-15, do all the geometry problems in [url=http://web.mit.edu/hmmt/www/]HMMT[/url] first. I repeat: Do these problems first. Then, or in process of while doing most of them, you feel like you need more help beyond Volume II, buy the first book I mentioned. Then, work through the rest of the problems.

I know some might argue that Geometry Revisited works fine for AIME, and I won't argue against that. But it is just my understanding that the student can learn more from doing problems from previous years of AIME (which has TONS of geometry), ARML, and HMMT for practices solely on AIME.

Geometry Revisited presents a very large collection of theorems. In general, its problems are pretty easy, if you have read the proofs of the theorems carefully (which are far from easy). It doesn't really have olympiad problems and such (that's not its purpose), as far as I know (maybe I should just look more closely at the book).

Challenging Problems in Geometry is almost the complete opposite. It proves nothing for you, and it poses big theorems with nontrivial proofs (Desargues, Menelaus, etc.) as problems. It is a book of problems only.

Geometry Revisited is good for learning facts; Challenging Problems is good for problems. So get Geo Revisited first, read it like, a lot, and then work your way through Challenging Problems. Actually, I think that only about half the problems in Challenging Problems are above AIME level... but oh well.


inequalities
calculus
derivative
function
inequalities proposed

Second part of inequality:$ \frac{y}{x}>\frac{y^{x}}{x^{y}}$
If $ x,y > 1$
Then:$ yx^{y}>xy^{x}$,
$ ln(y)+yln(x)>ln(x)+xln(y)$,
$ (y-1)ln(x)>(x-1)ln(y)$,
or: $ \frac{y-1}{ln(y)}>\frac{x-1}{ln(x)}$.
This can be proven by the differentiation of the function $ (\frac{z-1}{ln(z)})^{\prime}>0$ if $ z>1$.
if $ x,y<1$ we have:
$ \frac{y-1}{ln(y)}<\frac{x-1}{ln(x)}$ and the same case.

[quote="N.T.TUAN"]What is solution of 1st inequality? :D[/quote]
 :D   :blush:


Solve $(2^{n}-1)(3^{n}-1)=m^{2}$ with non-negative integer $m, n$.

It works mods 12. All numbers are equaivalent 6 mods 12. There is no such $m^{2}$

What if n is even then $(2^{n}-1)(3^{n}-1) \equiv 0(mod 12)$
and we clearly have m such that $m^{2}=0(mod 12)$

It is obvious that $(m, n)=(0, 0)$ is one of the solutions, but is there any others?

this was posted not long before
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=chinese&t=127680


i want to show, by using AM > GM, that in a triangle

sinA + sinB + sinC =< 3(sqrt3)/2

(I have to use AM>GM to do this, so I can't use concavity of sine)

See [url=http://www.mathlinks.ro/viewtopic.php?t=209782]here[/url]  :wink:

[quote="miff"][quote="dgreenb801"]I heard this can be proven directly with Jensen's inequality. Can someone explain Jensen's inequality, and how it solves this problem, in addition to the AM-GM proof?[/quote]
By Jensen's inequality

we have (sinA + sinB + sinC) /3 =< sin ( (A+B+C)/3 ) =< sin 60              (because A+B+C = 180)[/quote]

That's not true. $ \sin(x)$ is not convex in the interval $ (0,\pi)$. Plus, your inequality sign points in the wrong direction relative to Jensen's. Instead, we use $ f(x) \equal{} \minus{} \sin(x)$.

By Jensen's, we have $ \frac { \minus{} (\sin(A) \plus{} \sin(B) \plus{} \sin(C))}{3} \geq \minus{} \sin( \frac {A \plus{} B \plus{} C}{3}) \equal{} \minus{} \frac {\sqrt {3}}{2}$

Multiplying out the negative, we switch the inequality direction,
$ \sin(A) \plus{} \sin(B) \plus{} \sin(C) \leq \frac {3 \sqrt {3}}{2}$

$ \sin(x)$ is indeed concave on $ (0,\pi)$. Also, if you use Jensen for concave functions the inequality sign is in the right direction. So, while your method of transforming it into a convex function isn't wrong, it's not neeed at all.

Oh, ok. Thanks.
I didn't realize there was an an analogous Jensen's inequality for concave functions, even though I inadvertently proved it.


f: \mathbb{R^{+}}\rightarrow\mathbb{R}

and are greated in

show that is greated too

NB: can be note derivative

[url=http://www.math123xyz.com]Multimidea Math Resources[/url]


conics
ellipse
geometry
geometric transformation
homothety
analytic geometry
trigonometry

Se considera o elipsa si doua drepte perpendiculare variabile d si d', ce trec prin originea elipsei. Fie M si N ( respectiv M' si N') punctele de intersectie ale dreptei d(respectiv d') cu elipsa.
Sa se arate ca suma inverselor patratelor lungimilor segmentelor MN si M'N' este constanta

let there be a fixed ellipse and two lines $d$ and $d'$ such that $d\bot d'$, and $d\cap d'=\left\{O\right\}$, where $O$ is the center of the ellipse. Let $M$ and $N$ be the intersections of $d$ with the ellipse, and $M'$ and $N'$ the intersections of $d'$ with the ellipse.
show that $\frac{1}{MN^{2}}+\frac{1}{M'N'^{2}}$ is constant.

and the
[hide="solution"]
first, assume that the ellipse is given by
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
it is obvious that a homothety of center $O$ does not affect the problem, so let $b=1$.
next, let $M$ have coordinates $(a \cos t,\, b\sin t)$, and $M'$ have coordinates $(a \cos u,\, b\cos u)$.
the fact that $d\bot d'$ is equivalent with $\tan t \tan u=-a^{2}$.
all that`s left is to prove :
$\frac{1}{a^{2}\cos^{2}t+\sin^{2}t}+\frac{1}{a^{2}\cos^{2}u+\sin^{2}u}=1+\frac{1}{a^{2}}$ (*) (the right hand side is the value of the sum for the case when $d$ and $d'$ are the axis of the ellipse)
for easier work, let $\sin t=p$, $\cos t=x$, $\sin u=q$, $\cos u=y$.
let`s substitute $a^{2}$ in (*) with $-\frac{pq}{xy}$, and denote by $S$ the sum above :
$S=\frac{1}{-\frac{pqx}{y}+p^{2}}+\frac{1}{-\frac{pqy}{x}+q^{2}}$,
that is,
$S=\frac{1}{pq\cdot\left(\frac{p}{q}-\frac{x}{y}\right)}+\frac{1}{pq\cdot\left(\frac{q}{p}-\frac{y}{x}\right)}$,
that is
$S=\frac{1}{py-qx}\cdot\bigg(\frac{y}{p}-\frac{x}{q}\bigg)$,
that is
$S=\frac{qy-px}{pq(py-qx)}$,
that is
$S=\frac{qy-px}{p^{2}qy-q^{2}px}$,
that is (using that $p^{2}+x^{2}=1$)
$S=\frac{p^{2}qy-q^{2}px+x^{2}qy-y^{2}px}{p^{2}qy-q^{2}px}$,
that is
$S=1+\frac{xy(xq-yp)}{pq(py-qx)}$,
that is (recall that $pq=-xya^{2}$)
$S=1+\frac{1}{a^{2}}$, exactly what we had to prove.[/hide]

Mai bine scriai solutia in romana, ca poate unii de pe forumul romanesc nu stiu engleza. :)


Q1 Four whole number s are taken at random and multiplied together.Find the probability that last digit in product is 1,3,7,9 .
Q2 n dice are rolled .The probability that at least one of the dice shows even number is 15/16.Find value of n
Q3 A and B throw a dice each.Find probability that A's throw is not greater than B's throw
[hide]
Ans1  16/125
Ans2  1/6
Ans3  7/12

[/hide]

Hi jackie-chan01. You give the answer for Q2 as $ \frac {1}{6}$ which cannot be the value of $ n$.

The probability that at least one is even equals one minus the probability that  they are all odd. Therefore

$ 1 \minus{} (\frac {1}{2})^{n} \equal{} \frac {1}{16}$.

So $ n \equal{} 4$.

For Q3, we can draw a two-way table or list the $ 21$ ordered pairs $ (A, B)$ for which A's score is less than or equal to B's score out of the total of $ 36$ possible pairs, giving a probability of $ \frac {21}{36} \equal{} \frac {7}{12}$

I've got the wrong answer for Q1:

if the last digit of the product is $ 1, 3, 7$ or $ 9$, then each number is odd and its last digit is not $ 5$. So the last digit of each number is $ 1, 3, 7$ or $ 9$, giving the probability as $ \frac{4^{4}}{10^{4}} \equal{} (\frac{2}{5})^{4} \equal{} \frac{16}{625}$.

Hope this helps.

All answers are right.. :) I mixed up the answers while typing.Sorry for that :(


I was wondering if anyone knew of any good statistics books that are calculus-based and prove everything. It is driving me nuts right now as I follow through this book for AP Stats and it always floats away from the rigorous proofs of a lot of the important theorems and formulas.

As for Linear Algebra, I have a question. Would Apostol Vol. 1 & 2 suffice for a theoretical linear algebra course? Also is it sufficient for probability & statistics or is it better to get another book?

There might be a bit of linear algebra you're missing from Apostol. I'd suggest Linear Algebra Done Right by Axler.


When each side of a square is decreased by $ 25%
$, the area of the square decreases by $ 7$ square meters. What is the number of square meters in the area of the original square.

I think the question has an error. According to the title, the sides were decreased by $ 25\%$, but the question shows simply $ 25$, which, I think, is an invalid case.

The correct answer, if anybody wants to know, is $16$
It was a 4x4, then decreased by 25% makes it a 3x3 which does have 7 as the difference


Kiesha's average for ten tests in math class this term is 92. When Mrs. Foley ignores the lowest score, Kiesha's average is 94. What was Kiesha's lowest math test score this term?