trigonometry algebra function domain inequalities

1 Find the range of $y=\cos{(\sin{x})}$

2 The product of the first $n$ terms of ${a_n}$ is $P_n=(-\frac12)^{\frac{n^2+n-10}{2}}$, then find $a_2$ and the sum of the first $5$ terms.

3 The domain of odd function $f(x)$ is $R$. It is increasing in $[0,+\infty)$. If $f(1)<f(|x-1|)$, find $x$.

4 Let the max of $f(x)=|x^2+bx+c|$ ($x\in[0,2]$) be $t$, and $f(1)=0, b>0$, let t be a function of $g(b)$, find $g(b)$.

5 Let $f_1(x)=\frac{2}{x+1}$, and $f_{n+1}(x)=f_1[f_n(x)]$, n is a positive integer. And let $a_n=\frac{f_n(2)-1}{f_n(2)+2}$, then find $a_{99}$

6 The domain of function $y=f(x)$ is $(0,+\infty)$ while the range is $(-1,4)$, for $x_1$ and $x_2$($x_2$ dose not equal to $x_1$) there is always $f(x_1)+f(x_2)<2f(\frac{x_1+x_2}{2})$, please give two different forms of the function.

[hide="Solution to 1"]
The range of $sin(x)$ which is $x\in(-1,1)$ is also the domain of $cos(sin(x))$.  Therefore, the range of $cos(sin(x))$ is $y\in(cos(1),1)$. :D
[/hide]

[hide="Solution to 2"]
The $n^th$ term be expressed as $\frac{P_0P_n}{P_{n-1}}=(-\frac{1}{2})^{n-5}$ Therefore $a_2=-8$ and $\Sigma^5_ia_i=16\frac{\frac{33}{32}}{\frac{3}{2}}=11$ :D
[/hide]

[hide="Solution to 3"]
$x<0$ and $x>2$. f(|x-1|) forms a pseudo-parabolic shape with its psedo vertex at $(1,f(1))$.  Since $f(|2-1|)=f(|0-1|)$ and on either side of these points the graph is steadily increasing, the only portion of the range less than or equal to  $f(1)$ is on the domain $0\le x\le 2$. From this, I drew my solution.  :D
[/hide]

[hide="Solution to 4"]
$f(1)=0$ gives $c=-2$ and $b>0$ tells that the vertex $-\frac{b}{2a}$ cannot lie in the middle of the interval.  This proves that t will exist either at $x=0$ or at $x=2$ Plugging these x-values in $t=g(b)=max(2, 2+4b)$.  :D
[/hide]

I'll get back with the others

i am expecting the answer of 5 and 6. I am interested in problem 6 and i am really looking forward to see what u think about it.

[quote="JBL"]For number 6, does the question simply ask for two such functions?  Because that's pretty easy to find.  For example, [hide]$f(x) = \frac{-1}{x+\frac{1}{5}} + 4$ is one.[/hide][/quote]

The question is simply asking for 2 such functions but can you give some more general expressions of the functions which satisfy the conditions?

[quote="siuhochung"]any concave function is ok.[/quote]
What is a concave function??

A concave function is one which satisfies the inequality requirement $\displaystyle f(x_1)+f(x_2)<2f(\frac{x_1+x_2}{2})$.  (They can also be categorized a number o f different ways.)  Basically, though, if we talk about continuous functions, it has to approach 4 as x->0 and -1 as x->infinity and be concave in between.  It's hard to come up with a "general" form because there are lots and lots of functions that do this.

On an unrelated note of curiousity, does that concavity condition imply continuity, or can you find non-continuous functions by breaking the reals up into lots of countable dense subsets or something?

[quote="siuhochung"][quote="shobber"]Convex functions...Does it have anything to do with the Jensen's inequality?[/quote]

jensen's inequality is an inequality related to the convexity of a function.[/quote]
Confused......Can you post something about it?

[quote="JBL"]A concave function is one which satisfies the inequality requirement $\displaystyle f(x_1)+f(x_2)<2f(\frac{x_1+x_2}{2})$.

On an unrelated note of curiousity, does that concavity condition imply continuity, or can you find non-continuous functions by breaking the reals up into lots of countable dense subsets or something?[/quote]

A concave function is continuous, except possibly on the boundary of its domain.

For a proof, Google for "convex function is continuous" and there are a lot of hits.  Of course, $f$ is concave iff $-f$ is convex, but it seems everyone likes to state it for convex functions.

Yakko

[quote="shobber"]
2 The product of the first $n$ terms of ${a_n}$ is $P_n=(-\frac12)^{\frac{n^2+n-10}{2}}$, then find $a_2$ and the sum of the first $5$ terms.
[/quote]

Can someone tell me what I'm doing wrong?
I first put 1 in for n and got 16 so the first term is 16
then I put 2 in for n and got 4 so shouldn't the second term be 4 / 16 = 4??

[quote="JBL"]The only references I find on Google that I understand (I have no idea what non-linear programming is) either don't give a definition of convexity or use a more general definition of convexity.  Does our definition here imply $f(tx_1) + f((1-t)x_2) < 2f(tx_1 + (1-t)x_2)$?  (It's clear the reverse implication is true, by setting t = 1/2.  But it's not obvious to me that the case t = 1/2 generalizes to 0 < t < 1.  That's what I'm interested in.)[/quote]

I think the more general definition of convexity you want is $t f(x_1) + (1-t) f(x_2) < f(t x_1 + (1-t)x_2)$.

Unfortunately, this definition not quite implied by the case $t = 1/2$ (the definition given previously in this forum).  See [url=http://groups-beta.google.com/group/sci.math/browse_thread/thread/454ae49eb0392148/b2b2d9e9396f0f2d?q=%22convex+function%22+group:sci.math&rnum=4#b2b2d9e9396f0f2d]this[/url]  thread from sci.math, in particular, the response by David Ullrich.

If $f$ is continuous, then the implication does hold.  Considering the nature of the counterexample given, I suspect that you could replace continuity by a weaker assumption.

Yakko

Yep, put some t's inside that shouldn't have been.  Anyhow, the counterexample he names was what I was worried about -- the inequality in shobber's question isn't strong enough to impose continuity on the function.

Thanks!

We covered it in the forum a few weeks back.

http://www.mathlinks.ro/Forum/viewtopic.php?p=191473#p191473


AMC
AIME
geometry
circumcircle
rhombus
area of a triangle

Source: Alternate AIME 2003

ABCD is a rhombus. THe circumradii of ABD, ACD are 12.5 and 25, respectively. Find the area of the rhombus.

Quote:ABCD is a rhombus. THe circumradii of ABD, ACD are 12.5 and 25, respectively. Find the area of the rhombus.

Solution in spoiler:

[hide]The area of a triangle in terms of its circumradius is abc/4R, where a, b, and c are the lengths of the sides and R is the circumradius.

Let a and b equal half the length of diagonals AC and BD, respectively (so that 2a and 2b are the lengths of the two diagonals).  Then the length of the rhombus' side is sqrt(a:^2: + b:^2:)

Therefore, A(ABD) = [sqrt(a:^2: + b:^2:):^2: * 2b]/(4*12.5)

and A(ACD) = [sqrt(a:^2: + b:^2:):^2: * 2a]/(4*25)

Since each of these triangles is simply half the rhombus, their areas are equal and

[(a:^2: + b:^2:) * 2a]/(4*25) = [(a:^2: + b:^2:) * 2b]/(4*12.5)

Which simplifies to

2a/100 = 2b/50

a/b = 2

Hence, a is twice as long as b.  Let a = 2x and b = x.

Then the length of the rhombus' side is x:rt5:

Again, we'll express the areas in two different ways:

A = 1/2 * base * height

A = abc/4R

ABD has a base of 2x and a height of 2x.  Therefore its area is 1/2 * 2x * 2x = 2x:^2:

Since its sides are x:rt5:, x:rt5:, and 2x, and its circumradius is 12.5, its area is also (x:rt5: * x:rt5: * 2x)/(4*12.5) = 10x:^3:/50 = x:^3:/5

But we're talking about the exact same triangle, so the area is the same.  Therefore

2x:^2: = x:^3:/5

2 = x/5

x = 10

The area of the rhombus is diag1*diag2/2 = 4x*2x/2 = 4x:^2: = 400

[/hide]

Am I right?  Am I right?  Ooh...

Yep that's what I got! =)

I didn't think of it that way though - I used similar right triangles (I don't have the proof here with me so I don't actually remember exactly which ones i used) and got something like AD^2 = 25AC/2 and CD^2 = 25BD. From there it was easy.

Goody.

My work always looks really long, but it doesn't take that long in real life, I just usually show all the steps....

yes, you guys are both right. Congratulations! :D

Though I did it a rather "easier" way, if anyone is interested in my methods at all.....

Yes, yes, what's your way? Easier is good =).

well, you basically use the fact that the area of a triangle can be expressed as abc/4R, where R is the circumradius. Then, you can get 2 linear equations and them solve them to find the diagonals.

The circumradius of a triangle is the radius of the circle that would be circumscribed around it.  Circumradii is the plural form.

Note that the length of the circumradius is equal to the distance between the circumcenter (the point where the three perpendicular bisectors meet) and any vertex.

I think i need to get myself these AoP books and take a few classes here :\

Shore. And yes, the AoPS books are very helpful.


Evaluate the infinite sum \[\sum_{n \equal{} 0}^\infty \binom{2n}{n}\frac {1}{5^n}.\]

[hide="Hint"]Find a different way of expressing $ 2n \choose n$[/hide]

Yeah, the official solution involves generating functions as well as using an integral to help you.

[color=darkred][b]How to solve?[/b][/color] :)

Sorry for reviving but this might help:
$\binom{2n}{n}=(2^n)\left(\frac{1\cdot2\cdot5\cdots(2n-1)}{n!}\right)=\sum_{i=0}^{n}\binom{n}{k}=\left(\frac2n\right)\sum_{i=0}^{n}i\binom{n}{i}^2$

Sorry for reviving, but in the interest of the Contests page I figured i'd post the solutions given by the HMMT staff themselves.  (I'm also doing this because I find these solutions pretty awesome.)

[hide="Solution 1"]
Note that \begin{align*}\binom{2n}n&=\dfrac{(2n)!}{n!\cdot n!}=\dfrac{(2n)(2n-2)(2n-4)\cdots(2)}{n!}\cdot\dfrac{(2n-1)(2n-3)(2n-5)\cdots(1)}{n!}\\&=2^n\cdot\dfrac{(-2)^n}{n!}\left(-\dfrac12\right)\left(-\dfrac12-1\right)\left(-\dfrac12-2\right)\cdots\left(-\dfrac12-n+1\right)\\&=(-4)^n\dbinom{-\frac12}n.\end{align*} Then, by the binomial theorem, for any real $x$ with $|x|<\tfrac14$, we have \[(1-4x)^{-1/2}=\sum_{n=0}^\infty\dbinom{-\frac12}n=\sum_{n=0}^\infty\dbinom{2n}nx^n.\] Therefore, \[\sum_{n=0}^\infty\dbinom{2n}n\left(\dfrac15\right)^n=\dfrac1{\sqrt{1-\frac45}}=\sqrt{5}.\][/hide]
[hide="Solution 2"]
Consider the generating function \[f(x)=\sum_{n=0}^\infty\dbinom{2n}nx^n.\] It has formal integral given by \[g(x)=I(f(x))=\sum_{n=0}^\infty\dfrac1{n+1}\binom{2n}nx^{n+1}=\sum_{n=0}^\infty C_nx^{n+1}=x\sum_{n=0}^\infty C_nx^n,\] where $C_n=\tfrac1{n+1}\textstyle\binom{2n}n$ is the $n$th Catalan number.  Let $h(x)=\sum_{n=0}^\infty C_nx^n$; it suffices to compute this generating function.  Note that \[1+xh(x)^2=1+x\sum_{i,j\geq 0}C_iC_jx^{i+j}=1+x\sum_{k\geq 0}\left(\sum_{i=0}^kC_iC_{k-i}\right)=1+\sum_{k\geq 0}X_{k+1}x^{k+1}=h(x),\] where we're used the recurrence relation for the Catalan numbers.  We now solve for $h(x)$ with the quadratic equation to obtain \[h(x)=\dfrac{1/x\pm\sqrt{1/x^2-4/x}}2=\dfrac{1\pm\sqrt{1-4x}}{2x}.\] Note that we must choose the - sign in the $\pm$, since the $+$ would lead to a leading term of $\tfrac1x$ for $h$ (by expanding $\sqrt{1-4x}$ into a power series).  Therefore, we see that \[f(x)=D(g(x))=D(xh(x))=X\left(\dfrac{1-\sqrt{1-4x}}2\right)=\dfrac{1}{\sqrt{1-4x}}\] and our answer is hence $f(1/5)=\sqrt5$.

$\binom{2n}{n} = (-4)^n \binom{-\frac{1}{2}}{n}$
then use binomial theorem

[hide=Another way (basically equivalent to the above solutions)]
Square the sum and use the identity $\sum_{k=0}^n\binom{2k}{k}\binom{2(n-k)}{n-k}=4^n$.

If we square the sum $S$, then we get
\[S^2=\frac{4^0}{5^0}+\frac{4^1}{5^1}+\frac{4^2}{5^2}+\cdots=\frac{1}{1-\frac{4}{5}}=5,\]
hence $S=\sqrt{5}$ (as $S>0$).
[/hide]

[quote=CantonMathGuy][hide=Another way (basically equivalent to the above solutions)]
Square the sum and use $\sum_{k=0}^n\binom{2k}{k}\binom{2(n-k)}{n-k}=4^n$.

If we square the sum $S$, then we get
\[S^2=\frac{4^0}{5^0}+\frac{4^1}{5^1}+\frac{4^2}{5^2}+\cdots=\frac{1}{1-\frac{4}{5}}=5,\]
hence $S=\sqrt{5}$ (as $S>0$)
[/hide][/quote]

But how do you prove your claimed identity? Using this problem of course! :P


abstract algebra
vector
invariant
group theory
superior algebra
superior algebra unsolved

Show that each irreducible representation of an abelian group has dimension 1 by using Schur's lemma.

My thoughts are:
Let $ V$ be a vector space of dimension 1 and let $ \rho^1: G \longrightarrow GL(V)$ be the representation such that $ \rho^1_s$ is the identity automorphism on $ V$ for all $ s \in G$.
Now we let $ \rho^2: G \longrightarrow GL(W)$ be any irreducible representation of $ G$. My task is, I think, to find a linear map $ f: V \longrightarrow W$ which is not equal to zero and which commutes with the action of the group. If this is correct it follows from Schur's lemma that $ f$ is an isomorphism and that $ \dim(W) \equal{} \dim(V) \equal{} 1$.

Is this correct? And how do I find $ f$?

Let $ W$ be an irreducible $ \mathbb{C}G$-module with $ \sigma: G\to GL_\mathbb{C}(W)$ the corresponding representation.
For $ g\in G$ let $ \lambda\in\mathbb{C}$ be an eigenvalue of $ \sigma(g)$.
Then we get the non-invertible $ \mathbb{C}G$-homo $ \sigma(g)\minus{}\lambda \text{id}_W: W\to W$,
and by Schur's lemma we get $ \sigma(g)\equal{}\lambda \text{id}_W\in\mathbb{C}\text{id}_W$ for all $ g\in G$.
So $ \sigma(\mathbb{C}G)\subseteq\mathbb{C}\text{id}_W$, and $ W$ irreducible gives $ \dim_\mathbb{C}W\equal{}1$.

Hi olorin. What does $ \mathbb{C}G$ stand for? Isn't there an easier proof not dealing with modules (I have not encountered them yet)?

$ \mathbb{C}G$ means the group ring, i.e. the formal combination of group elements with coefficients $ \mathbb{C}$, $ \{c_1 g_1 \plus{} c_2 g_2 \plus{} \cdots \plus{} c_ng_n : g_i \in G, c_i \in \mathbb{C} \}$ carrying a natural componentwise addition and multiplication.

If you don't want to use the language of modules, you can look at it like this: (this is the same as olorin's argument)
Consider an irreducible representaiton $ \sigma: G \rightarrow GL(V)$.
$ \sigma(g): V \rightarrow V$ for arbitrary $ g \in G$ is a $ G$-linear endomorphism of $ V$.
Consider an eigenvalue $ \lambda$ of $ \sigma(g)$, then $ \sigma(g) \minus{} \lambda \mathrm{id}$ is a G-linear map with nontrivial kernel, hence it is the zero map. (Schur's lemma used here) Thus, $ \sigma (g) \equal{} \lambda \mathrm{id}$ for some $ \lambda$.

Now consider an arbitrary $ v \in V$, and let $ W$ be the span of $ v$. You see that $ W$ is a G-invariant subspace of $ V$. But $ V$ is irreducible, so $ W \equal{} V$, i.e. V is of one-dimensional.


Congratulations to Lucy for making the state team.

Who else made it?  What were the places of the individuals and schools?

ugh... how did you know I made it?

oh well, thank you.

1st place - Stephen Kwang - Randolph
2nd place - not sure name - Randolph
3rd place - not sure name - Challenger
4th place - Lucy Xie

Randolph got 1st on team. Simmons Middle School (Hoover) got 2nd. We got 3rd :(

we did bad this year.

I was fifth before countdown, and beat this other person from Simmons in countdown. :)

I knew you made it because I saw you post in the MATHCOUNTS thread about AoPSers going to nationals.

I hope Dave Patrick's book is helping you with your studies.  Good luck at nationals.


geometry
angle bisector
geometry unsolved

ABC is an triangle and AA1,BB1 and CC1 are bisectors of ABC triangles.AA1 intersects B1C1 at point L and CC1 intersects A1B1 at point K.
prove that BB1 is the bisector of angle LBK.

Apply the transversal theorem to the triangle ABC, cevian AA_1 and transversal C_1LB_1 and get AL/LA_1 = (b+c)/2a ( 1 ). Call M the intersection of BL and AC, apply Menelaus to the triangle ACA_1 and the transversal BLM and get AM/MC = c/2a ( 2 ). Same way we shall get PC/AP = a/2c ( 3 ), P being the intersection of BK and AC. Dividing (2) and (3) we get (AM/MC)*(AP/PC) = (c/a)^2, which, according to Steiner, shows that BM and BP are symmetrical about the angle bisector of B.

Best regards,
sunken rock


Without a calculator, which of the numbers in the set $\{\log{2^{2005}}, \log{2005^{200}}, \log{314^{271}, \log{271^{371}} \}}$ is the largest?

What's the most efficient way to approach this problem (is this too easy for this forum)?

[hide]

Use logarithm properties and approximations:

$\log 2^{2005} =2005 \log 2$

$\log 2005^{200} =200 \log 2005 \approx 200 \log 2+600$

$\log 314^{271} = 271 \log314 \approx 1355\log 2 + 271$

$\log 271^{371} = 371 \log 271 \approx 2968 \log 2$

Let's approximate $\log 2$ as 0.3.  The three values thus become 601.5, 660, 677.5, and 890.4.  The last one is definitely the largest.

[/hide]

After you simplify the logs to
$2005\log2$
$200\log2005$
$271\log314$
$371\log271$,
you either have to know approximately what the logs are, or just use logical reasoning to figure out the largest.  No matter how you do it, you have to know a fairly close estimate of what each log is.  $\log2\approx0.3$, making $2005\log2\approx601$.  You know $\log2005>\log1000$, and thus > 3.  Because $\log2\approx0.3$, you can also assume that the $\log2005\approx3.3$, making $200\log2005\approx660$.  On the last 2, $371\log271$ appears to be much larger, because $\log314$ is barely larger than $\log271$, but 371 is much larger than 271.  $\log271>2$, so $371\log271>742$ and therefore is the largest.


limit
probability
number theory proposed
number theory

The world-renowned Marxist theorist [i]Joric[/i] is obsessed with both mathematics and social egalitarianism. Therefore, for any decimal representation of a positive integer $n$, he tries to partition its digits into two groups, such that the difference between the sums of the digits in each group be as small as possible. Joric calls this difference the [i]defect[/i] of the number $n$. Determine the average value of the defect (over all positive integers), that is, if we denote by $\delta(n)$ the defect of $n$, compute \[\lim_{n \rightarrow \infty}\frac{\sum_{k = 1}^{n}\delta(k)}{n}.\]
[i]Iurie Boreico[/i]

It is easy to observe that $\delta(n)$ is always less than $10$. If you partition the digits in two groups with difference of sums $\delta(n)$ and $\delta(n)\geq 10$ then move one of the digits from the group with bigger sum to the other group. Hence $\delta(n)$ is bounded by $10$.
I claim that the average is $\frac{1}{2}$. It can be seen that this number is bigger than or equal to $\frac{1}{2}$, because if the sum of the digits of $n$ is odd then $\delta(n)\geq 1$, and the probability of having an odd digital sum is $\frac{1}{2}$.
One can also see that the probability of a number to have at least $10$ digits of $1$, is $1$ (Really easy to show). Now the average of $\delta(n)$ over all numbers will be equal to the average over the special numbers I mentioned (The ones having at least ten digits of $1$)
For each of these numbers like $n$, just remove the first ten digits of $1$ to obtain $m$. Then we know that $\delta(m)<10$. Now using the extra $1$ digits we have at hand we can distribute them among the two groups we have for $\delta(m)$ to decrease the difference between sums to either $0$ or $1$. If the sum of digits of $m$ is odd we can reduce $\delta(m)$ to $1$ and in the other case to $0$. One can again easily observe that the probability of $m$ having an odd digital sum is $\frac{1}{2}$. So for half of the numbers (with respect to probability) we have $\delta(n)=0$ and for another half (disjoint from the last half) we have $\delta(n)\leq 1$. So the average we are trying to find is less than or equal to $\frac{1}{2}$ and hence my claim is proved.

I find

$ \displaystyle\lim_{N \rightarrow \infty}\displaystyle\frac{\displaystyle\sum_{k = 1}^{10^N}\delta(k)}{10^N} $.

Notice that $\forall k$, $\delta(k) \le 9$, since if it is more than $9$, and the digits of $k$ are partitioned into $X \cup Y$, then we can let $X' = X \setminus \{a\}, Y'=Y \setminus \{a\}$ for any $a \in X$, and the defect will be smaller.

Notice that, out of the $10^N$ numbers from $1$ to $10^N$, at most $9\cdot 9^{N}$ of them don't have all the digits. For these numbers, I use the bound $\delta(k) \le 9$.

For any number $k$ that does have all the digits, I prove $\delta(k) \le 1$, by contradiction, assuming $\delta(k) \ge 2$. Let $X \cup Y$ be an optimal partition, with $\sum X - \sum Y = \delta(k) \ge 2$.

If $1 \in X$, then we can put 1 in $Y$ instead of $X$, thus decreasing the defect by $1$. Since the defect was at least 2, I have strictly decreased the defect, contradiction. But since $1$ appears in $k$, then $1 \in Y$. If $2 \in X$, I can interchange 1 and 2 in $X$ and $Y$, and the defect will decrease by 1, contradiction. But since $2$ appears in $k$, then $2 \in Y$. I can continue and I will get that $1,2,...,9$ don't appear in $X$ but they do appear in $Y$. From this, I get that $\sum X=0, \sum Y >0$, and thus $\delta(k) = \sum X - \sum Y $ is impossible! Thus I have proven $\delta(k) \le 1$.

Also, if the sum of digits of $k$ is odd, then clearly $\delta(k)=1$ and if it is even, then $\delta(k)=0$, by parity. Notice that half of the numbers satisfy the first condition and half satisfy the second condition. Therefore,

$\displaystyle\frac{\displaystyle\sum_{k = 1}^{10^N}\delta(k)}{10^N} \le \left( \displaystyle\frac{1}{10^N} \right) \cdot \left( 9*(9^{N}) + 0*\frac{10^N-9^N}{2}+1*\frac{10^N-9^N}{2} \right) $,

and, since $\delta(k) \ge 1$ if the sum of digits of $k$ is odd, which occurs for half of the numbers,

$\displaystyle\frac{\displaystyle\sum_{k = 1}^{10^N}\delta(k)}{10^N} \ge \left( \displaystyle\frac{1}{10^N} \right) \cdot \left(1*\frac{10^N}{2} \right) $,

both of which tend to $1/2$, so we're done.


when you see a problem like this
Block of molding clay are 6 inches by 4 inches by 2 inches. How many whole blocks are needed to mold a cylindrical sculpture 16 inches high and 10 inches in diameter?
when it says "Whole blocks", would we include or exclude the fractional bllock?

I would round up.

Basically the question asks "how many whole blocks should I buy if..."


salaam
I have a nice problem for you!
prove this inquality
if $x_1x_2\dots x_n=1$
then
x1^3+...x4^3>=Max{x1+...+x4,1/x1+...+1/x4}
Fi aman allah

[quote="altusi"]i wonder if they are positives ,  :?
and about their number. is it 4? :)[/quote]
yes they are positive and n=4!

let' solve it: :)

if the maimum is (1/x1+...1/x4) then:
by IGA we have ,
 x1^3+...x3^3>=3x1.x2.x3=3/x4 , we do the same for the three other triplets and we add up the three inegualities, the conclusion follows.

if the maximum is (x1+...+x4) then:
by cauchy we have ,
(x1^3+...x4^3)(1/x1+...1/x4)>=(x1+...+x4)²>=(1/x1+...1/x4)² ,and we are done! ;)


Um...
1.) He adds force to the pedal which accelerates the pedal and thus increases the velocity. The friction between the pedal axle and the chain allows the motion to be transported to the wheels, which turn and increases the velocity of the person and the entire bicycle.
2.) Even though his mass is the same on the moon, the gravity is much weaker. Thus the force acting upon him downwards are much less and therefore he have less weight.

or something like that.

The answers to the question are quite simple as you said.\
1) I will give you the frition part that you couldn't get is-the friction is much greater because the cyclist does not have any momentum yet and there for must work harder against the frition to get started.
2)The westler will way less because there is less gravitational pull on the him/her.

By the way I was supposed to give you the wrong answers someone thought it would be funny and that person will remain unmetioned but I can't give the wrong answer to science questions I love science.

The static friction increases as the cyclist strains against the pedals until the maximum s friction is reached.

Or -- the friction of the ground against the wheels prevents them from spinning freely, forcing the wheels to roll forwards (instead of just spinning in place).  This makes the bicyclist move.

As for the other, weight is a measure of force.  Force is acceleration times mass.  The mass of the wrestler is constant, but his acceleration (due to gravity) is much smaller, so his weight is less.


I have an interesting math limerick:

(12+144+20+3 :sqrt: 4)/7 + 5*11 = 9 2 +0

why is this a limerick?

A dozen, a gross, a score,
Plus three times the square root of four,
Divided by seven,
Plus five times eleven,
Is nine squared and not a bit more.

It's pretty cool, don't you think?  :D

There once was a man from Peru,
Whose limericks always ended on line two. :D

That math one is really cool.


modular arithmetic
number theory
greatest common divisor
number theory unsolved

Sorry if the title is bad ,couldnt think of anything else :(

Find all positive integers $ n$ for which $ a^{12} \equiv 1 \bmod{n}$ for [b]all[/b] integers $ a \ge 1$ which are co-prime to $ n$ .

What can one say when we replace 12 in the problem by a fixed $ k \in \mathbb{N}$  and have it as $ a^{k} \equiv 1 \bmod{n}$ ?

we use the same idea mentioned by ZetaX [url=http://www.mathlinks.ro/viewtopic.php?t=150516]here[/url]
in fact we have $ 12 \times 3 \plus{} 1 \equal{} 37$
but $ 37$ is a prime and so it has a primitive root(exactly $ 12$ primitive root) .take one of those primitive root let's say $ r$.
now we have to check two case :
1-$ 37 \not | n$ which implies that $ n |37^{12} \minus{} 1$.the rest is to check different cases.(in order to save time we can consider the fact that if $ n$ is odd then $ n | 2^{12} \minus{} 1$).
2-$ 37 | n$
now let $ n \equal{} 37^{\beta}p_1^{a_1}...p_k^{a_k}$
now supose that $ t$ is the smallest answer to
$ 37t \equiv \minus{} r \pmod{p_1...p_k}$
in $ N$
the case $ k \equal{} 1,2$ are obvious since we can have just $ 12$ different values for $ r$ and so we can determine the value of $ 37t \plus{} r$ and then determine the value of $ p_j$.
so let's suppose that $ k \neq 1,2$
now consider the number $ 37(t \minus{} 1) \plus{} r$ then we have $ \gcd(37(t \minus{} 1) \plus{} r,n) \equal{} 1$
because we have $ p_j | 37t \plus{} r$ and if the gcd isn't one then there exists at least one prime like $ p_s$ for which we have
$ p_j | 37(t \minus{} 1) \plus{} r$ which implies that $ p_j | 37$
which is false.
and obviously $ 37 \nmid 37(t \minus{} 1) \plus{} r$
so we must have $ (37(t \minus{} 1) \plus{} r)^{12} \equiv 1 \pmod{n}$ but we had $ 37 \mid n$
which implies $ (37(t \minus{} 1) \plus{} r)^{12} \equiv 1 \pmod{37}$ or $ r^{12} \equiv 1 \pmod{37}$
which is false since $ r$ is a primitive root.
(also we can choose the prime $ 11$ instead of $ 37$)
so we r done.
i did the same for the generalization and i got some conditions for $ n$ which was similar to what has been mentioned.
nothing more unfortunately :(

@shoki - very nice :)

As regards to the generalisation , we can reciprocate the same method done by shoki if we know any value of $ x$ , (and also we have a highly advanced scientific calculator and patience :D) .But we have to "generalise" ,so we should get a general expression for $ n$ , like say $ n \equal{} f(x)$ ,such that given $ x$ we can find all possible values for $ n$ .

[quote="bboypa"]Let $ k \in \mathbb{N}_0$ fixed and define $ S_k: \equal{} \{n \in \mathbb{N}_0: n \mid x^k \minus{} 1\text{ for all }x\in (\mathbb{Z}/n\mathbb{Z})^*\}$ then:

i) if $ \upsilon_2(k) \equal{} 0$ then $ S_k \equal{} \{1,2\}$;

ii) if $ \upsilon_2(k) \in \mathbb{N}\setminus\{0\}$ and $ n\in S_k$ then $ \upsilon_2(n) \le \upsilon_2(k) \plus{} 2$;

iii) if $ \upsilon_2(k) \in \mathbb{N}\setminus\{0\}$ and $ n\in S_k$ then $ \text{gpf}(n) \le k \plus{} 1$;

iv) if $ \upsilon_2(k) \in \mathbb{N}\setminus\{0\}$ and $ n\in S_k$ then $ \upsilon_2(p \minus{} 1) \le \upsilon_2(k)$ for all $ p \in \mathbb{P} \setminus\{2\}$ such
that $ p\mid n$;

v) if $ \upsilon_2(k) \in \mathbb{N}\setminus\{0\}$ and $ n\in S_k$ then $ \text{rad}(p \minus{} 1) \mid k$ for all $ p \in \mathbb{P} \setminus\{2\}$ such that $ p\mid n$;

vi) if $ \upsilon_2(k) \in \mathbb{N}\setminus\{0\}$ and $ n\in S_k$ then $ \upsilon_p(n) \le \upsilon_p(k) \plus{} 1$ for all $ p \in \mathbb{P} \setminus\{2\}$ such that $ p\mid n$;

vii) if $ \upsilon_2(k) \in \mathbb{N}\setminus\{0\}$ and $ n\in S_k$ then $ \text{lpf}(n\cdot 2^{ \minus{} \upsilon_2(n)})$ is a Fermat prime;

viii) Deduce that we can costructe $ S_k$ by a finite number of steps.[/quote]
er..bobypa ...can you help us by explaining the notations you have used  :blush:


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inequalities unsolved
inequalities

[quote="Stephen"]$ a_1, a_2, ...$ is a sequence of positive real numbers.
And $ b_n$ is a arithmetic mean of $ a_1, a_2, ..., a_n$.
Prove that

$ \sum_{n \equal{} 1}^{N}b_n^2\le4\sum_{n \equal{} 1}^{N}a_n^2$ for all positive intergers $ N$.[/quote]
http://www.mathlinks.ro/viewtopic.php?search_id=1472742605&t=148513


Prove that there exist 10000 successive positive integers, exactly 150 of which are primes.

As far as I can tell, this isn't a number theory problem at all. Anyway, I like it, so I want to post a solution:

We can agree that among the positive integers from 1 to 10000, there are more than 150 primes. We can also agree that among the positive integers from 10001!+2 to 10001!+10001, there are 0 primes. Now consider every time we take the integers from n to n+9999 and move to n+1 to n+10000. There are three options for what our prime number count can do: 1) it can increase by 1; 2) it can remain the same; 3) it can decrease by 1. Thus we will get each number from 0 to pi(10000) primes in some range of 10000 (and maybe some others too) if we start somewhere between 1 and 10001!+2. One of those will be 150, as desired.

Why do you think this is no number theory ?

It just has to do with ``continuity,'' so I don't really consider it number theory. Is my solution the same as yours?

Of course it is the same.

But well ... this is a problem concerning primes ... so for me this is NT.


If a class is made up of 23 students, what is the probability that at least 2 students will have the same birth date? (excluding the use of leap years)

(If I remembered the question correctly, Bill Nye said the answer is $\frac{1}{2}$. I think it's a little too high.)

No, I think that Bill Nye is absolutely right...
Consider the following (pun intended  :D ):
[hide]If we look at the complement, i.e. everyone has different birthdates,
The first person has 365 days to choose from
the second has 364 to choose from
the third has 363 to choose from... ... ...
the nth person has 366 - n days to choose from then

So,
if we multiply this out, for n persons,
[P(n) denotes probability]
$P(n) =(\frac{365}{365})(\frac{364}{365})(\frac{363}{365})...(\frac{366-n}{365}) = \frac{365!}{(365-n)!(365^{n})}$
Thus, just sub in 23 for n, and you should get 0.49... but we must remember to do 1- P(n), as we took the complement, and this yields about 0.51, which is 51%.
[/hide]
Very interesting problem!

- Nawahd Werdna

Isn't the probability some really small number since the problem never said that everyone was born in the same year?

just saying

and $0.51 \neq \frac{1}{2}$

Well more children could tend to be conceived on certain dates...that's as far as I'll go explaining that

[quote="me@home"]In this case they are - they are uniform random variables.

And this is very well known and called the "23" theorem. The general solution (of when there will be x probability instead of 1/2=x) is a very complicated expression that is difficult to solve for.[/quote]

Do you know where I could find out more about the [b]$23$ therom[/b]?

Thanks to everyone for their help!

[quote="vamathletes"]There's a formula here
http://www.madsci.org/posts/archives/aug97/869256748.Ot.r.html[/quote]

 :blush: \[P(n)=1-\frac{{365 \choose n}}{365^{n}}\]


trigonometry
modular arithmetic
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algebra
polynomial
Vieta
ARML

Sorry for no pic.

So basically, using Law of cosines, you get the product of the segment lengths from each A and B is $ \sqrt{8^6(1\plus{}\cos{\frac{\pi}{7}})(1\minus{}\cos{\frac{\pi}{7}})(1\plus{}\cos{\frac{2\pi}{7}})(1\minus{}\cos{\frac{2\pi}{7}})(1\plus{}\cos{\frac{3\pi}{7}})(1\minus{}\cos{3\frac{\pi}{7}})}$ so the desired answer is just $ 8^6(1\plus{}\cos{\frac{\pi}{7}})(1\minus{}\cos{\frac{\pi}{7}})(1\plus{}\cos{\frac{2\pi}{7}})(1\minus{}\cos{\frac{2\pi}{7}})(1\plus{}\cos{\frac{3\pi}{7}})(1\minus{}\cos{3\frac{\pi}{7}}) \equal{} 8^6\cos^2{\frac{\pi}{7}}\cos^2{\frac{2\pi}{7}}\cos^2{\frac{3\pi}{7}}$ so basically after uglifying the product to sum things you get the answer to end in $ \boxed{672}$ i missed this in the actual contest cuz i, for some reason, put the answer to be $ 8^6(\cos^2{\frac{\pi}{7}}\plus{}\cos^2{\frac{2\pi}{7}}\plus{}\cos^2{\frac{3\pi}{7}})$ giving me an answer of 752. so i basically hate myself.

[quote="cognos599"]Sorry for no pic.

So basically, using Law of cosines, you get the product of the segment lengths from each A and B is $ \sqrt {8^6(1 \plus{} \cos{\frac {\pi}{7}})(1 \minus{} \cos{\frac {\pi}{7}})(1 \plus{} \cos{\frac {2\pi}{7}})(1 \minus{} \cos{\frac {2\pi}{7}})(1 \plus{} \cos{\frac {3\pi}{7}})(1 \minus{} \cos{3\frac {\pi}{7}})}$ so the desired answer is just $ 8^6(1 \plus{} \cos{\frac {\pi}{7}})(1 \minus{} \cos{\frac {\pi}{7}})(1 \plus{} \cos{\frac {2\pi}{7}})(1 \minus{} \cos{\frac {2\pi}{7}})(1 \plus{} \cos{\frac {3\pi}{7}})(1 \minus{} \cos{3\frac {\pi}{7}}) \equal{} 8^6\cos^2{\frac {\pi}{7}}\cos^2{\frac {2\pi}{7}}\cos^2{\frac {3\pi}{7}}$ so basically after uglifying the product to sum things you get the answer to end in $ \boxed{672}$ i missed this in the actual contest cuz i, for some reason, put the answer to be $ 8^6(\cos^2{\frac {\pi}{7}} \plus{} \cos^2{\frac {2\pi}{7}} \plus{} \cos^2{\frac {3\pi}{7}})$ giving me an answer of 752. so i basically hate myself.[/quote]
I managed to simplify it to something like that, but how do you compute the answer from that? How do you find the cosine or sine of a multiple of $ \frac {\pi}{7}$?

It should be $ \left(2^9\sin\frac\pi7\sin\frac {2\pi}7\sin\frac {3\pi}7\right)^2 \equal{} 2^{18}\sin\frac\pi7\sin\frac {2\pi}7\sin\frac {3\pi}7\sin\frac {4\pi}7\sin\frac {5\pi}7\sin\frac {6\pi}7$
$ \equal{} 2^{18}\cdot\frac7{2^6} \equal{} 2^{12}\cdot7\equiv\boxed{672}\pmod{1000}$, not what cognos599 had.

Refer to (24) [url=http://mathworld.wolfram.com/TrigonometryAngles.html]here[/url] and problem 2 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1343563925&t=12775]here[/url].

[quote="cognos599"]$ 8^6\cos^2{\frac {\pi}{7}}\cos^2{\frac {2\pi}{7}}\cos^2{\frac {3\pi}{7}}$[/quote]
[quote="math154"]The expression is equal to $ \left(2^9\sin\frac\pi7\sin\frac {2\pi}7\sin\frac {3\pi}7\right)^2$, not what cognos599 had.

Refer to (24) [url=http://mathworld.wolfram.com/TrigonometryAngles.html]here[/url] and problem 2 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1343563925&t=12775]here[/url].[/quote]
Are the two expressions equal to each other?

Is the problem statement supposed to have $ \overline{BC_i}$?

And cognos599, that product evaluates to $ 4096$? Edit: and math154, that evaluates to $ 3136$?

I get $ 28672$ when I evaluate it...

[b]ProtestanT:[/b] They are not equal. I just realized my wording might have been confusing: I'm talking about what it should be.

[hide="Solution"][asy]pair rt (int n){return 2 * expi(n * pi / 7);}

size(100); pathpen = linewidth(0.7);

pair A=MP("A",(2,0)), B=MP("B",(-2,0)); D(arc((0,0),2,0,180)); D(A--B);
for(int i = 1; i < 7; ++i) D(A--D(rt(i))--B);[/asy] It is easy to verify that the length of a chord of a circle with radius $ r$ with central angle $ \theta$ is $ 2r \sin \frac {\theta}2$. Thus, the answer is given by $ \left[\prod _{i = 1}^6 4 \sin \frac {i\pi}{14}\right]^2$. Note that
\begin{align*}\frac {\sin 14x}{\sin x} = \frac {\text{Im} \left(\left(\cos x + i\sin x\right)^{14}\right)}{\sin x} & = {14 \choose 1}\sin^{12}x - {14 \choose 3} \sin^{10} x \cos^2 x + {14 \choose 5} \sin^{8} x \cos^4 x - \cdots + {14 \choose 13} \cos^{12} x \\
& = \left[{14 \choose 1} + {14 \choose 3} + \cdots + {14 \choose 13}\right]\sin^{12} x + T(\sin x) \sin x + 14 \\
\end{align*} using $ \cos^2 x = 1 - \sin^2 x$, where $ T(x)$ is a polynomial of degree $ 10$. Note that the roots of this $ 12$th degree polynomial with leading coefficient $ (1+1)^{14}/2 = 2^{13}$ are given by $ \sin \frac {k\pi}{14}, k \equiv \pm 1,2,3,4,5,6$ (because they satisfy $ \sin 14x = 0$). Thus, by Vieta's, \[ \prod _{i = 1}^6 \sin \frac {i\pi}{14} = \sqrt{\prod _{0 \neq i = -6}^6 \sin \frac {i\pi}{14}} = \sqrt {\frac {14}{2^{13}}} = \frac {\sqrt {7}}{2^6}.\] Therefore, we need to find $ \left(2^6 \cdot \sqrt {7}\right)^2 = 2^{12} \cdot 7 \equiv 1024 \cdot 28 \equiv \boxed{672} \pmod{1000}$.
[/hide]

[hide="Fastest, I think"]We scale the entire thing down by $ 2$, so we will multiply by $ 2^{12}$ at the end. Let $ \zeta=e^{\pi i/7}$ be the first primitive fourteenth root of unity. If we place our semicircle in the complex plane with $ A=1$ and $ B=-1$, $ C_k=z^k$. The length of a segment $ AC_k$ is $ |1-z^k|$ and similarly $ BC_k=|-1-z^k|=|1+z^k|$. Note that absolute value distributes over multiplication; that is, $ |ab|=|a\parallel{}b|$. Our product is now:
\begin{eqnarray*}
P&=&\left(\prod_{k=1}^6 \left|1-\zeta^k\right|\right)\left(\prod_{k=1}^6 \left|1+\zeta^k\right|\right)\\
&=&\prod_{k=1}^6 \left|1-\zeta^k\right|\left|1+\zeta^k\right|\\
&=&\prod_{k=1}^6 \left|1-\left(\zeta^2\right)^k\right|
\end{eqnarray*}$ \zeta^2$ is a primitive seventh root of unity, so its first through sixth powers are all the primitive seventh roots of unity. This suggests using a polynomial with exactly these roots, the 7th cyclotomic polynomial:
\[ \Phi_7(x)=\frac{x^7-1}{x-1}=\sum_{k=0}^6 x^k=\prod_{k=1}^6 \left(x-\left(\zeta^2\right)^k\right)
\]$ P=|\Phi_7(1)|=7$. We must now scale this by $ 2^{12}$: $ 2^{12}P=28672\equiv\boxed{672}\pmod{1000}$.[/hide]

I'm surprised this was on the test; there's an ARML problem almost exactly like this.

Protestant, i wrote cos instead of sin. sorry, in my actual calc., i used sin, i just didn't want to retype everything so i was copying and pasting and i forgot to change from cos to sin.

I'm surprised this was on the AIME...too well-known.

Also, AIME II seems to have more complex number problems than AIME I in the recent two years. Coincidence?

I think that this is the same as Ebean's, but I don't understand the math in his. Using simple trig, but not proven yet. Try this with a diameter of 2, and only one point.
The product is 2. With 2 points, The product is 3. With three points the product is 4. With 6 points assume the product is 7. Each of the 12 chords should be twice as long, so $ 2^12*7\equal{}28672$

[quote="rrusczyk"]Hint: Last week, I wrote a problem and solution in the Precalculus book I'm writing that is essentially equivalent to this problem.  (Too bad I didn't finish the book earlier!)

The problem is in a chapter on complex numbers.[/quote]

So is there a more straightforward, non-complex numbers way of solving this?

This question is trivial after you watch this video by Dr Peyam: [url]https://www.youtube.com/watch?v=hQ532SjbxNo[/url]

[quote=worthawholebean]Let $ A$ and $ B$ be the endpoints of a semicircular arc of radius $ 2$. The arc is divided into seven congruent arcs by six equally spaced points $ C_1,C_2,\ldots,C_6$. All chords of the form $ \overline{AC_i}$ or $ \overline{BC_i}$ are drawn. Let $ n$ be the product of the lengths of these twelve chords. Find the remainder when $ n$ is divided by $ 1000$.[/quote]

Progress:

All angles are in degrees.

First, note that $AC_i = BC_{7-i}$ by symmetry. Next, notice that $$AC_1 = 4\sin\left(\frac{90}{7}\right), AC_2 = 4\sin\left(\frac{180}{7}\right), AC_3 = 4\sin\left(\frac{270}{7}\right), AC_4 = 4\sin\left(\frac{360}{7}\right), AC_5 = 4\sin\left(\frac{540}{7}\right), AC_6 = 4\sin\left(\frac{720}{7}\right).$$ Also, we have $\sin p \sin(90-p) = \sin p \cos p = \frac{\sin(2p)}{2}$ for any $p$, so we can rewrite $AC_1 \cdot AC_2 \cdot AC_3 \cdot AC_4 \cdot AC_5 \cdot AC_6$ as $$\frac{4^6}{2^3}\sin\left(\frac{180}{7}\right)\sin\left(\frac{360}{7}\right)\sin\left(\frac{540}{7}\right).$$

How can I evaluate this expression?

[url=https://cdn.artofproblemsolving.com/attachments/0/c/3d45222c7fc8cd7562d0a3d3134c1042882c36.pdf]This handout might help[/url] (Exercise 2.2.1 is very similar)

@below By browsing the forums enough lol

[quote=samrocksnature][url=https://cdn.artofproblemsolving.com/attachments/0/c/3d45222c7fc8cd7562d0a3d3134c1042882c36.pdf]This handout might help[/url] (Exercise 2.2.1 is very similar)[/quote]

where do you find so many handouts?

How can I express the sine of something in terms of roots of unity? I'll use radians for an example: If I wanted to do something with $\sin\left(\frac{\pi}{7}\right)$, how could I do that? I know that $e^{\frac{\pi}{7}} = \cos\left(\frac{\pi}{7}\right)+i\sin\left(\frac{\pi}{7}\right)$, but how do I just get the sine term?


function limit calculus derivative algebra polynomial induction

Please tell me the taylor series expansion of the following...

                 $ e^{\frac{\minus{}1}{x^2}}$

You know that $ e^x\equal{}\sum_{k\equal{}0}^\infty\frac{x^k}{k!}.$ Proceed from there.

Please tell the whole procedure that how will you proceed from this.

Think of it as a variable substitution.

$ e^u\equal{}\sum_{n\equal{}0}^\infty \frac{u^n}{n!}$, you have $ u\equal{}\frac{\minus{}1}{x^2}$, so

$ e^{\minus{}1/x^2}\equal{}\sum_{n\equal{}0}^\infty \frac{(\minus{}1/x^2)^n}{n!}$

You can just do some algebra to clean that up. When I started these, I always used $ u$ in the definition of the power series instead of $ x$ to not get confused.

The original poster left out one crucial piece of information: Taylor series [b]centered at what point?[/b]

The series HilbertThm90 is writing is not a Taylor series at all, since it contains negative powers of $ x$ rather than positive powers of $ x.$ It is what is known as a Laurent series centered at zero.

If we were to ask for the Taylor series of the original function centered at some $ x_0\ne 0,$ that could be obtained, although it would be algebraically messy. I suspect that's not the question.

I'm going to engage in some mind-reading, and deal with what I think might have been the original question. We're dealing with functions defined on the real line.

Let $ f(x)\equal{}\begin{cases}e^{\minus{}\frac1{x^2}},&x\ne0\\ 0,&x\equal{}0\end{cases}.$

We can show that $ f$ is continuous at zero, but we can do more than that:

$ f'(0)\equal{}\lim_{h\to0}\frac1he^{\minus{}\frac1{h^2}}\equal{}0,$ since the exponential goes to zero faster than any power of the argument.

We can extend this by induction. We can compute the $ n$th derivative:

$ f^{[n]}(x)\equal{}\begin{cases}P_n \left(\frac1x\right) e^{\minus{}\frac1{x^2}},&x\ne0\\0,&x\equal{}0\end{cases},$ where $ P_n$ is a polynomial.

Dividing by $ x$ and letting $ x\to0,$ we see that $ \frac1xP_n \left(\frac1x\right)$ is just another polynomial in $ \frac1x,$ and the exponential still goes to zero faster than it grows. Hence, $ f^{[n\plus{}1]}(0)\equal{}0$ and the induction continues.

We have that $ f\in C^{\infty}(\mathbb{R}).$ We also have that $ f^{[n]}(0)\equal{}0$ for all $ n.$ That gives us the Taylor series for $ f$ centered at zero:

The Taylor series for $ f$ centered at zero is

$ \sum_{n\equal{}0}^{\infty}0\cdot x^n\equal{}0.$

Which clearly does not equal the function. Which is the point of the exercise - not all $ C^{\infty}$ functions have Taylor series, or equal their Taylor series if they do.

Woah. Sorry. I need to get my head out of complex analysis if I am to post here.  :blush: That's no excuse, since it says Taylor series, and what I wrote isn't a Taylor series in any type of analysis.


[quote="Davron"]This one is good inequality problem i dont know even its true or not please help to solve it :
$5(a^2+b^2+c^2)+4(xy+yz+zx)$ $\geq$ $9(x+y+z)$
Show that inequality holds for $xyz=1$...
Davron[/quote]
I guess that $a,b,c$ are positive real numbers and $abc=1.$
Prove that $5(a^2+b^2+c^2)+4(ab+ac+bc)\geq9(a+b+c).$
Proof:
$5(a^2+b^2+c^2)+4(ab+ac+bc)=\sum_{cyc}(5a^2+ab+ac+2bc)\geq\sum_{cyc}9\sqrt[9]{a^{12}b^3c^3}=$
$=\sum_{cyc}9\sqrt[9]{a^9}=9(a+b+c).$ :)

Ohhh i am sorry with that actually we have a=x b=y c=z sorry guys now can you prove that ...

Davron

Perfect arqady ... how you come with this inequality ...? Please say by PM or just write here ....  :)  Perfect !

Davron

[quote]I guess that $a,b,c$ are positive real numbers and $abc=1.$
Prove that $5(a^2+b^2+c^2)+4(ab+ac+bc)\geq9(a+b+c).$
[/quote]

From AM-GM: $a+b+c \ge 3\sqrt[3]{abc}=3$.

Since $a^2+b^2+c^2\ge ab+ac+bc$, we have:

$5(a^2+b^2+c^2)+4(ab+ac+bc)\ge3(a^2+b^2+c^2)+6(ab+ac+bc)$

$=3(a+b+c)^2\geq9(a+b+c)$.


n people lined up. And they lined up again. How many ways to line up again s.t no one stands behind the same person.

We can restate the question, "How many permutations $ \pi\in S_{n}$ are there such that there does not exist $ i$ satisfying $ \pi(i\plus{}1) \equal{}\pi(i)\plus{}1$?"

The sequence is [url=http://www.research.att.com/~njas/sequences/A000255]A000255[/url] in the OEIS.  It satisfies the recursion $ a_{n\plus{}1}\equal{} n a_{n}\plus{}(n\minus{}1)a_{n\minus{}1}$ for $ n\geq 3$ with $ a_{1}\equal{} a_{2}\equal{} 1$ because in every good permutation of $ \{1,\ldots, n\plus{}1\}$, either:
1) $ n\plus{}1$ lies between two nonconsecutive numbers, in which case we can remove it to give a good permutation of $ \{1,\ldots, n\}$, and this map is $ n$-to-1, or
2) $ n\plus{}1$ lies between two consecutive numbers $ i$ and $ i\plus{}1$, in which case we can remove both $ n\plus{}1$ and $ i\plus{}1$ from the permutation and replace each value $ k > i\plus{}1$ in the result with $ k\minus{}1$.  This gives a good permutation of $ \{1,\ldots, n\minus{}1\}$, and this map is $ n\minus{}1$-to-1.

Without the condition we have n! way(s) to line up.
In n! way(s), one stands behind the same person in (n-1)! time(s). We all have n people so the answer should be n!/((n-1)!*n)=1

 :?:  :huh:


$Question 1$ In the class the professor asked one student to write down his Room No; a three digit number.
He asked him to write it again after that; making that a six digit one ; ( like for $123$ ; it would be $123123$ ..) and give it to the second student .

He asked the second student to divide it by a number $a$ .
" Don't worry abt the remainder ; there will be none ; give the quotient to the third . "

The third was asked to divide by $b$ ; again told there will be no remainder & asked to give the quotient to the fourth .

The fourth was told to divide by $c$ , again being told of no remainder ;and was asked to hand over the quotient to the Prof .

The Prof studied that and correctly predicted his Room number .

What were $a$, $b$ & $c$?

$Question 2$ A professor asked me about the membership strength of a group I am part of; and I said more than $400$.
With a chuckle he gave this proposition :

Choose some $n$ friends from yr group .
I'll draw an Isosceles triangle $ABC$ , $BC$ being the base ; and $a$ being the equal sides .

Each member in turn can extend the base by an arbitrary length ( Ob'sly < $2a$ ) and make another isosceles triangle with the same equal sides $a$ on his extension .

Thus there could be $n$ such triangles ; all isosceles with different bases , but equal sides .

Now , using two adjacent vetices as diagonals ; a Rhombous should be formed ; thus abt $n-1$ Rhombi .

Repeat the process again and again ; until you get a single Rhombous at the top .
The final figure may look like an uneven ragged ridge or a mountain .

Call this peak point as $P$ .

The idea is to choose such entensions that this $P$ is Non Equidistant from $B$ & $C$ .

Is it possible ???

If yes, how?
If not, why not?

1) If the room number is $N$, then the six digit number is $1001N$. Since $1001=7\cdot 11\cdot 13$, the numbers $a,b,c$ are obviously $7,11,13$ and the final quotient is the room number itself.

Or $(a, b, c)=(1001, 1, 1)$ or any permutation.  :P
There are a few others, too. :D

It's not obvious what the meaning of the process you've described in question 2 is.  I have isosceles triangle ABC and I extend it to rhombus ABDC?  Or some other way?  How do I combine the various rhombi into this big picture?  Which points B, C are we talking about at the end (because we have more than 400 different such pairs)?


trigonometry
algebra
polynomial
limit
product of roots

1. let a, b, c, d be rational numbers, a^2+b^2+c^2+d^2<=1, prove:
(a+b)^4 + (a+c)^4+(a+d)^4+(b+c)^4+(b+d)^4+(c+d)^4<=6

2.cos6cos42cos66cos78?  the numbers are all in degrees

[hide="1"] Square the given inequality.  Now multiply by $ 6$ and apply AM-GM a couple of times.[/hide]

[hide="2"]
That's equivalent to $ \cos(6*1)\cdot\cos(6*7)\cdot\cos(6*11)\cdot\cos(6*13)$
See if you can try to find a formula for $ \cos(6\theta)$.. I'm not sure how helpful this is.
[/hide]

[hide="2"]
The equation $ 2\cos 5\theta \equal{}\sqrt{3}$ holds for five values of $ \cos\theta$, namely $ \cos 6^{\circ}$, $ \cos 78^{\circ}$, $ \cos 150^{\circ}\equal{}\minus{}\frac{\sqrt{3}}{2}$, $ \cos 222^{\circ}\equal{}\minus{}\cos 42^{\circ}$, and $ \cos 294^{\circ}\equal{}\cos 66^{\circ}$.

Therefore to evaluate
\[ \cos 6^{\circ}\cos 42^{\circ}\cos 66^{\circ}\cos 78^{\circ}\]
we only need to find the product of roots of $ 2P(x)\minus{}\sqrt{3}$ (where $ P$ is the fifth degree polynomial such that $ P(\cos\theta )\equal{}\cos 5\theta$) and divide by $ \frac{\sqrt{3}}{2}$.

Now $ P(x)$ has constant term $ \cos\frac{5\pi}{2}\equal{}0$ and fifth-power term 16 (a cheap way to see this is to note that $ \cos ix\equal{}\frac{e^{x}\plus{}e^{\minus{}x}}{2}$ and then note that the leading coefficient of $ P$ is
\[ \lim_{\cos\theta\rightarrow\infty}\frac{\cos 5\theta}{\cos^{5}\theta}\equal{}\lim_{x\rightarrow\infty}\frac{16(e^{5x}\plus{}e^{\minus{}5x})}{(e^{x}\plus{}e^{\minus{}x})^{5}}\equal{}16\lim_{x\rightarrow\infty}\frac{1\plus{}e^{\minus{}10x}}{(1\plus{}e^{\minus{}2x})^{5}}\equal{}16\]
but this can certainly be done by determining the $ \cos^{5}\theta$ term after expanding $ (\cos\theta\plus{}i\sin\theta )^{5}$ and writing everything in terms of cosine), so the product of the roots of $ 2P(x)\minus{}\sqrt{3}$ is $ \frac{\sqrt{3}}{32}$, making the desired product $ \boxed{\frac{1}{16}}$.
[/hide]

The problem is to find the product: $ \sin 84\sin 48\sin 24\sin 12$

We can use $ z\equal{}\text{cis}\,\frac{2\pi}{30}$ to do this...

$ \frac{1}{16}*(z^{7}\minus{}z^{\minus{}7})*(z^{4}\minus{}z^{\minus{}4})*(z^{2}\minus{}z^{\minus{}2})*(z^{1}\minus{}z^{\minus{}1})\equal{}$
$ \frac{1}{16}*(1\plus{}(z^{8}\plus{}z^{7}\minus{}z^{5}\minus{}z^{4}\minus{}z^{3}\plus{}z\plus{}1)(z^{6}\minus{}z^{5}\plus{}z^{3}\minus{}2z^{2}\plus{}2z\minus{}z^{\minus{}1}\plus{}z^{\minus{}2}\minus{}z^{\minus{}3}\plus{}z^{\minus{}4}\minus{}z^{\minus{}5}\plus{}z^{\minus{}6}\minus{}z^{\minus{}7}\minus{}z^{\minus{}8}\plus{}2z^{\minus{}9}\minus{}2z^{\minus{}10}\plus{}z^{\minus{}11}\minus{}z^{\minus{}13}\plus{}z^{\minus{}14}\minus{}1))$

$ \equal{}\frac{1}{16}$ since $ (z^{8}\plus{}z^{7}\minus{}z^{5}\minus{}z^{4}\minus{}z^{3}\plus{}z\plus{}1)$ is the 30th cyclotomic polynomial.


superior algebra
superior algebra unsolved

let R be a ring, I and J are ideals in R with the intersection I ∩ J = {0}. Suppose there are x ∈ I and y ∈ J with x+y = 1. Prove that x and y are idempotent.

$ xy$ is in $ I$ and $ J$, thus $ xy\equal{}0$. Multiplying $ x\plus{}y\equal{}1$ with $ x$ gives $ x^2\plus{}xy\equal{}x$, giving $ x^2\equal{}x$. Analogously, $ y^2\equal{}y$.


analytic geometry
geometry
graphing lines
slope
function
calculus
factorial

(Not sure if this topic is  belongs here..)

As the title says, throw some questions for me to solve! I need to study for my math midterms for chapters 1 - 4 (grade ten, Canadian education system.) This includes..
[hide="Click for Table of Contents from my textbook (Chapters 1- 4)"]
Chapter 1: Number Sequences
- Arithmetic Growth
- General Term of Aithmetic Sequence
- Sum of Arithmetic Series
- Geometric Growth
- Exponent Laws
- Data in Tables
- Sequences in Tables

Chapter 2: Real Nmbers
- Radicals
- Extending Exponent Laws to Rational Exponents
- Applications of Rational Exponents
- Pythagorean Theorem
- Irrational Numbers
- Relating the Sides of Special Triangles
- Multiplying Radicals
- Dividing Radicals
- Adding, Subtracting Radicals
- Combined Operations with Radicals

Chapter 3/4: Coordinate Geometry
- Length of Line Segment
- Midpoint of Line Segment
- Slope of Line Segment
- Slopes of Parrallel Line Segments
- Slopes of Perpendicular Line Segments
- Using Equation to Draw a Graph
- Slope of Line
- Equation of Line
- Interpreting Equation Ax + By + C = 0
[/hide]
Thank you! =)

[quote="jnsoccer66"]O.K.  Simplify:
$\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{6^{2}}+...$.
 :lol:[/quote]

Ooh stumped on the first question! How does that work? =/

[quote="adalton"]JNSoccer was just trying to be mean, lol.

That's the Zeta Function with n=2, so it simplifies to $\frac{\pi^2}{6}$[/quote]

Lol.. I so don't get that.. hah.

[quote="jnsoccer66"]What class is this?  It looks like a combination of algebra and geometry.  In my :censored: school, we never have two subjects in one year.  Its just ..., Algebra, Geometry, Algebra II, Pre-Calc, Calc.[/quote]

In my province in Canada, geometry, algebra, whatnot are all combined. For high school, our mathematics classes are math 8, 9, 10, 11, 12 with honours for every grade, calculas and calculas AP. It's the same with science up untill grade 10.. and then you can pick chemistry, physics, biology or earth sciences.

[quote="orl"]Are discussing the curriculum? Or are you looking for problems to be solved?  :D[/quote]
Looking for problems to solve =)

[quote="ch*yeuk"][quote="t0rajir0u"]What an odd curriculum.

Problem:  What is the area of an isosceles triangle with side lengths $17, 17, 30$?[/quote]

Aw man.. radicals.. I stopped midway, didn't know what to do

[hide]
$\\ x = \sqrt{17^2 + 15^2} = \sqrt{289 + 225} = \sqrt{514} \\ a = \frac{30 \sqrt{514}}{2}$
[/hide][/quote]

Perhaps you mean $\sqrt{17^2 - 15^2}$

[quote="b-flat"]OK, find the equation of a line perpendicular to the line that goes through (3,4) and (-2,-6).  The perpendicular line goes through (0,0).  Write your answer in the form y=Ax+B

Hope this helps, I'll post more if I can think of some.

By the way (I dislike using "ChatSpeak"), I made that up off the top of my head, so I don't know if it works out nicely.[/quote]

[hide]
$\\ \frac{-6 - 4}{-2 - 3} = \frac{10}{5} = 2 \\ -\frac{1}{2} = \frac{y}{x} \\ -2y = x \\ y = -\frac{1}{2}x$
Oh wow, how did my answer end up to be like that? =/
[/hide]

[quote="prototypex24"][quote="ch*yeuk"][quote="t0rajir0u"]What an odd curriculum.

Problem:  What is the area of an isosceles triangle with side lengths $17, 17, 30$?[/quote]

Aw man.. radicals.. I stopped midway, didn't know what to do

[hide]
$\\ x = \sqrt{17^2 + 15^2} = \sqrt{289 + 225} = \sqrt{514} \\ a = \frac{30 \sqrt{514}}{2}$
[/hide][/quote]

[hide="Answer"]0.5*30*8=120[/hide][/quote]
Hmm how did you get those numbers?

*runs in*

*gets hit by problem*

Ow! *rubs head*

[quote="ch*yeuk"][/quote][quote="prototypex24"][quote="ch*yeuk"][quote="t0rajir0u"]What an odd curriculum.

Problem:  What is the area of an isosceles triangle with side lengths $17, 17, 30$?[/quote]

Aw man.. radicals.. I stopped midway, didn't know what to do

[hide]
$\\ x = \sqrt{17^2 + 15^2} = \sqrt{289 + 225} = \sqrt{514} \\ a = \frac{30 \sqrt{514}}{2}$
[/hide][/quote]

[hide="Answer"]0.5*30*8=120[/hide][/quote][quote="ch*yeuk"]
Hmm how did you get those numbers?[/quote]
He used $A=\frac12 bh$.
If we regard the base as the side whose length is $30$, then the height is...[hide]by Pythagorean Theorem, $\sqrt{17^2-15^2}=8$
(It's minus, not plus)
[/hide]
And the rest is... :D

So, here are some more questions, arranged (roughly) in order of difficulty.  Truthfully, the last two will probably be harder than the midterm.

[hide="Problem 1"]
Simplify $128^{ \frac{10}{7} }$.
[/hide]

[hide="Problem 2"]
$\triangle ABC$ is a triangle with $\angle B = 60^{\circ}, \angle C = 90^{\circ}$ and $AB = 15$.  A triangle $\triangle ACD$ is drawn such that $\angle ACD = 60^{\circ}, \angle ADC = 90^{\circ}$.  What is the length of $AD$?
[/hide]

[hide="Problem 3"]
Evaluate the sum $1 + 7 + 4 + 5 + 7 + 3 + 10 + 1 + 13 - 1 + 16 - 3 + 19 - 5 + 22 - 7 +-... + 1000 - 659$.  :)
[hide="Hint"]
See what happens when you pair the terms up.  Alternately, see what happens when you only look at the even-numbered terms, and then only look at the odd-numbered terms.
[/hide]
[/hide]

[hide="Problem 4"]
What is the area of a triangle with side lengths $13, 37, 40$?  (It is not a right triangle :)  There is a formula for this.  Don't use it.)
[hide="Hint"]
Pythagoras and algebra are still your friends.
[/hide]
[/hide]

let an arithmetic sequence of 10 terms have a sum of x. Let the 4th term be y. Let the 6th term be z.  Find the 7th term of the sequence in terms of x, y, and z

[quote="ch*yeuk"][quote="frt"]
He used $A=\frac12 bh$.
If we regard the base as the side whose length is $30$, then the height is...[hide]by Pythagorean Theorem, $\sqrt{17^2-15^2}=8$
(It's minus, not plus)
[/hide]
And the rest is... :D[/quote]

Ooh how silly of me to mix up the formula! Oh well.. we get a formula cheat sheet heh. I get it noww though thanks =)

[quote="t0rajir0u"]
[hide="Problem 1"]
Simplify $128^{ \frac{10}{7} }.$
[/hide]
[/quote]

[hide]
$\\ 7\sqrt{128^{10}}\\ = 28\sqrt{8^{10}}\\ = 56\sqrt{2^{10}}\\ = 56\sqrt{1024}\\ = 224\sqrt{64}\\ = 8$
[/hide][/quote]

Simplify $128^{ \frac{10}{7} }.$
$128 = 2^7$
$((2^7)^\frac{1}{7})^{10}$
$2^{10} = 1024$

[quote="shadysaysurspammed"]This might be a little hard.....
Prove that $n^2-1$(n)(3n+2) is divisible by 24 for even and odd n.No its easy :)[/quote]

That's my question...

[quote="ch*yeuk"]
[hide]
$\\ 7\sqrt{128^{10}}\\ = 28\sqrt{8^{10}}\\ = 56\sqrt{2^{10}}\\ = 56\sqrt{1024}\\ = 224\sqrt{64}\\ = 8$
[/hide][/quote]

Sorry, the first step and the last step are both wrong :(  That first step should be $\sqrt[7]{128^{10}}$ which is rather a different quantity, and $224 \sqrt{64} = 224 \times 8 \neq 8$ in any case.

[hide]1) 1024 or two to the power of ten
2) 11.25
3)58283
4) 240
[/hide]

nice problems I enjoyed working on them.

Why is infiniti factorial equal to the square root of two pi?

[quote="dimak11214"]Why is infiniti factorial equal to the square root of two pi?[/quote]

Long story :|

It involves the extension of factorial to the reals and the complex numbers, and then a renormalized product of some sort...

http://mathworld.wolfram.com/GammaFunction.html

[quote="t0rajir0u"]What an odd curriculum.

Problem:  What is the area of an isosceles triangle with side lengths $17, 17, 30$?[/quote]

Area=\boxed{120}

You have to put that in latex, in between $\$$'s.  Also, you should edit your post by clicking the "edit" button at the bottom right rather than making a new post each time.


For what value of $ x$ does $ 3^{2x^{2}\minus{}5x\plus{}2} \equal{} 3^{2x^{2}\plus{}7x\minus{}4}$? Express your answer as a common fraction.

[quote="GameBot"]For what value of $ x$ does $ 3^{2x^{2} \minus{} 5x \plus{} 2} \equal{} 3^{2x^{2} \plus{} 7x \minus{} 4}$? Express your answer as a common fraction.[/quote]

There exists a certain exponent property that if:
$ a^x \equal{} a^y$

It holds that:
$ x \equal{} y$

$ 3^{2x^{2} \minus{} 5x \plus{} 2} \equal{} 3^{2x^{2} \plus{} 7x \minus{} 4}$

$ 2x^{2} \minus{} 5x \plus{} 2\equal{}2x^{2} \plus{} 7x \minus{} 4$

$ 6\equal{}12x$

$ \boxed{x\equal{}\dfrac{1}{2}}$


A quadrilateral has sides length $24, 25, 26,$ and $27$, and area $180x$, where $x$ is a real number. Find $x^{2}$.

If it was cyclic x would be 13..but it never says its cyclic

The area of an arbitrary quadrilateral cannot be determined by its side lengths alone.

Cylic quads, yes, see Art of Owna's post.


geometry
geometric transformation
homothety
geometry unsolved

Let $ABC$ is a triangle and $M \in [AB], N \in [BC] ,L \in [CA]$.  Can be found every time $M,N,L$ points that  $MNL$ is a equilateral triangle? What are the conditions?

I'm almost sure of it  such points can be found.

Choose two points $M',N'$ on $AB$ and $BC$ such that there is a point $L'$ internal to the triangle $ABC$ such that $M'N'L'$ is equilateral (moving $M',N'$ toward $B$ you'll finally find such a point $L'$).
Now, consider the line parallel to $AC$ passing through $L'$ and let $h$ be the homotety which maps that line to $AC$, then $M=h(M')$, $N=h(N')$, $L=h(L')$ are the vertices of an equilateral triangle and lay on $AB$, $BC$, $CA$.


algebra
polynomial
induction
algebra unsolved

i have read  a book about polinomials (author:mr adeel khan) .there,i read about Newton's indentities .this is very useful for computation but i can't prove this.HELP ME!

[quote="N.T.TUAN"]What is newton's indentities? Please post it! We will help you, if We can  :wink:[/quote]

newton's identities are as follows

if $P(x)$ $=$ $\sum_{i=0}^{n}a_{i}x^{i}$
be a polynomial of degree n and $\alpha_{1},\alpha_{2},\dots\alpha_{n}$
be the n roots of the polynomial then newton's identities are

$\sum\alpha_{i}$ $=$ $(-1)^{n}\frac{a_{n-1}}{a_{n}}$
.....

good friend u can prove this by induction on n :D

[quote="N.T.TUAN"]pardesi,
No, I don't think so. It is Viete's theorem.[/quote]

yes this is viete's theorem :D
but newton's identities are quite related to them
see
[url]http://en.wikipedia.org/wiki/Newton's_identities[/url]

PLEASE GIVE ME A PROOF OF THEOREM .I TRIED TO SOLVE BUT I CAN'T FIND A SOLUTION FOR THIS


Solve in $Z$ the equation $3^x=x^2+3x+1$.

For $x<0$ LHS isn't integer.

For $x\geqslant 3$ we can use induction to prove $3^x>x^2+3x+1$:

$1^\circ\quad$ For $x=3$ we have $3^3>3^2+3\cdot 3+1$ which is true.

$2^\circ\quad$ Inductive step:

\begin{eqnarray*}x\geqslant 3 &\implies& (x+1)^2-2>0\\ &\implies& 2x^2+4x-2>0\\ &\implies& 3x^2+9x+3>x^2+5x+5\\ &\implies& 3(x^2+3x+1)>(x+1)^2+3(x+1)+1\end{eqnarray*}

Hence $3^x>x^2+3x+1\implies 3^{x+1}>3(x^2+3x+1)>(x+1)^2+3(x+1)+1$

Therefore, we have to check $x=0,1,2$ and we find that the unique solution is $x=0.$

Or you can use a simple graphing calculator  :P

I like the $\mod3$ solution more.

[hide]
With the exception of the case $x=0$,
$x^2 + 3x + 1 \equiv x^2 + 1 \equiv 0\mod3$
Since the only quadratic residues in $\mod3$ are 0 and 1, the only two values that $x^2 + 1\mod 3$ can produce are 1 and 2, and will never be equal to zero.  Therefore there are no solutions except for $x=0$.
[/hide]


Sorry about posting in the Advanced section. I can't figure out this one:

Suppose you have 19 computer chips. 9 of these chips have 16 prongs each, 7 have 32 prongs and 3 have 128 prongs each. We are going to place these chips on a computer board with 750 sockets. (A socket holds 1 prong). Will all 19 chips fit on the computer board? Briefly explain why or why not?

Are we just adding up all the prongs on the chips which comes out to be 752 which means all 19 chips can't fit. It just seems too simple to be correct.

Well I am not sure how to state it mathematically but each chip has a number of prongs that is a multiple of the smallest chip.  Restated the \[GCD ( 16, 32, 128) =16\] Now since 16 does not go into 750 evenly 750 = 14 mod 16  then there is no way all the chips will fit.

Yeah that was supposed to be greatest common factor, not denominator.

I'm pretty sure what you did (adding them up) was correct. There aren't enough sockets for all the prongs, so the question was pretty easy it seems.


Determine the limit of  the sequence a_n = (1+(1+(1 +...+ 1^0.5)^0.5)^0.5)^0.5 which has n 1's.

Very famous one,
a(n+1)= \sqrt (1+a(n))
the sequence a(n) is convergent to a limit k
k^2=1+k =>consider the positive root k=(1+ \sqrt 5)/2
Moubinool Omarjee


Explain why the compound 1,2,3,4-dibenzocyclooctatetraene is chiral. See the attached file for the structure.

Cyclooctatetraene is non-aromatic and conforms itself, and the molecule is asymmetric

Can someone please elaborate??
I cant understand the answer....
thanks a lot....

That's right.

Adwaitvedant Mathkar, the structure of this compound is the following:


You have until 7:15.

Regular rules apply. If there are questions contact me at whitehorseking88 for AIM.

Times up. Send ur answers please.

Here they are. Lag caused it to be posted 10 minutes late. Everyone has until 7:30.

BTW: Here's the PDF file.

[quote="white_horse_king88"]Here they are. Lag caused it to be posted 10 minutes late. Everyone has until 7:30.[/quote]

EDIT: i didnt see your post :P

Sure everyone. The deadline is 7:30. For everybody. 75 minutes from now.

Hey white_horse_king, I sent you my answers.  Can you grade it now?

It seems I made this way too easy. But anyways, it was meant to be an AMC 10, so I guess the results show that it was relatively easy AMC 10.

5 MINS LEFT.

Can you post the questions here now that time is up, preferably separate from solutions? I didn't get to my computer soon enough to know you were going to do this. otherwise, I would have participated.

I thought it was pretty typical of an AMC-10, actually, and I actually omitted more questions than I usually do on actual old tests. (Though that might be because every time I do a real AMC test I see one or two questions I've seen previously on this forum or in an AoPS class midterm/final challenge. I don't mean to but it skews my results.) Good job for making a true AMC-10 instead of one of those weird 10/12 hybrids.

I tried to make this as real of an AMC 10 as possible. I didn't include too much difficult stuff. I'll post statistics soon.

it was great! i loved being able to solve the questions..
it was a lot easier than other mock AMCs (for me, at least) considering my score improved 60+ points from the last one i tried.

ah, i should really calm down before tuesday comes.

can someone imform me how we join these mock AMCs and submit answers and can you guys inform me by emailing me at kaycubs@gmail.com when the next mock AMC is so i can participate, sorry i am new here

Just a note, kaycub. Try not to revive very old threads.  I have noticed you have done this several times already.  Usually if there is a Mock AMC the writer will post on this forum when it will be and post the questions.  Usually the people taking it will private message the writer their answers.

There may not be another mock AMC for a while since many people are gunning for the USAMO now. However, you can still test yourself with old ones by taking the time just to look over the problems and then comparing to the given answers. You might also want to try the mock AIMEs. Most of these mock contests have discussions and solutions that you could look over if you have any issues.

Here is mathfanatic's directory:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15513


[hide="Hint"]Denote $ AF\equal{}y,BD\equal{}z$, then $ x(x\plus{}1)\equal{}1\plus{}y,y(y\plus{}1)\equal{}1\plus{}z$ and $ z(z\plus{}1)\equal{}x\plus{}1$.[/hide]

[quote="N.T.TUAN"]$ \|\begin{array}{c} D\in (AB)\ ,\ E\in (BC)\ ,\ F\in (CA) \\
 \\
\boxed {\ CD\perp AB\ ,\ AE\perp BC\ ,\ BF\perp AC } \\
 \\
AD = BE = CF = 1 \\
 \\
AF = y\ ,\ BD = z\ ,\ CE = x\end{array}\|$ $ \implies$ $ \|\begin{array}{c} x(x + 1) = y + 1\ . \\
 \\
y(y + 1) = z + 1\ . \\
 \\
z(z + 1) = x + 1\ .\end{array}$[/quote]
[color=darkblue][b]Good hint, N.T.Tuan ![/b] I think that you applied the power of the point. For example, the power of the point $ A$ w.r.t. the circumcircle of the quadrilateral $ BDFC$ : $ AD\cdot AB = AF\cdot AC$ $ \implies$ $ y(y + 1) = z + 1$ a.s.o. I"ll continue from here.

[b]Method I.[/b] Observe that $ xyz = 1$ and $ \sum x^2 = 3\ .$ Thus,

$ 1 = \frac 13\cdot\sum x^2\ge \sqrt [3] {(xyz)^2} = 1$ $ \implies$ $ x^2 = y^2 = z^2$ $ \implies$ $ x = y = z = 1\ .$

[b]Method II.[/b] The functions $ f(x) = x^2 + x$, $ g(x) = x + 1$ are strict increasing over $ \mathcal R^*_ + \ .$

The our system becomes $ \|\begin{array}{c} f(x) = g(y) \\
\ f(y) = g(z) \\
\ f(z) = g(x)\end{array}\ .$ Therefore, $ \boxed {x\le y}\implies$ $ f(x)\le f(y)\implies$

$ g(y)\le g(z)\implies$ $ \boxed {y\le z}\implies$ $ f(y)\le f(z)$ $ \implies$ $ g(z)\le g(x)$ $ \implies$ $ \boxed {z\le x}\ .$

Show that analogously and in the case $ x\ge y\ .$ In conclusion $ x = y = z = 1\ .$

[b]An easy extension.[/b][/color]

 [quote="Virgil Nicula"]$ \|\begin{array}{c} D\in (AB)\ ,\ E\in (BC)\ ,\ F\in (CA) \\
 \\
\boxed {\widehat {ADC}\equiv\widehat {BEA}\equiv\widehat {CFB}} \\
 \\
AD = BE = CF = 1 \\
 \\
AF = y\ ,\ BD = z\ ,\ CE = x\end{array}\|$ $ \implies$ $ \|\begin{array}{c} x(x + 1) = y + 1\ . \\
 \\
y(y + 1) = z + 1\ . \\
 \\
z(z + 1) = x + 1\ .\end{array}$[/quote]
[hide="Hint."]Denote the common value $ m(\widehat {ADC})=\phi\ .$ Observe that the quadrilaterals $ ADOF$, $ BEOD$, $ CFOE$ are cyclically. Thus $ AD\cdot AB=AO\cdot AE=AF\cdot AC\implies$ $ AD\cdot AB=AF\cdot AC$ $ \implies$ the quadrilateral $ BCFD$ is cyclically $ \implies$ $ \phi=90^{\circ}$ a.s.o.[/hide]


Let A,a,B,b,C,c,D,d, be primes not necessarilly distinct such that:

-Aa+Bb=Cc+Dd.
-ABCD is even.
-ABCD + x^2=y^2 for x , y integers.
-For all letter M, M>m.

a) Show that ABCD and abcd are both perfect squares.
b) Is there a condition redundant? That is, is there a condition of the four above that can be removed and the statement a) in the problem is still true?

In the problem there was a condition that stated:

- For all letter M, M>m.

It's wrong. The correct one is:

-For all letter M, M>=m.

Enjoy!!


function
calculus
derivative
inequalities
calculus computations

Let g be a real function that has a continuous first derivative g' on R.Suppose further that g(0)=0 and |g'(x)|<=|g(x)| for all x in R.Prove that g vanishes identically.

[quote="pankajsinha"]I just got it
|g'(x)|<=|g(x)|<=0

[/quote]

how did u get that

That doesn't work. You seem to be saying that since $f'(x)=0$ when $f(x)=0$, $f$ is identically zero. That isn't true; we do need the stronger condition.

This is related to uniqueness theorems for differential equations. A standard uniqueness theorem:
Let $I$ be an interval, and let $f$ be a function defined on $I\times \mathbb{R}$ such that $f$ is continuous and $|f(t,y_{1})-f(t,y_{2})|\le L|y_{1}-y_{2}|$ ($f$ is Lipschitz in $y$). Suppose $t_{0}\in I$ and $y_{0}\in \mathbb{R}$. Then there is at most one solution $y$ defined on $I$ to the differential equation $y'(t)=f(t,y(t)),y(t_{0})=y_{0}$.
If you can solve this thread's problem, try to prove that theorem.

That uniqueness Theorem you stated is just a stronger version of Picard's Theorem. Where you have $f$ satisfying [i]Lipschitz condition[/i], Picard's version has $f_{y}$ continuous. The proof of Picard's Theorem of differential equations that I know is fairly long (but straight forward).

Continuity certainly can't be enough to guarantee uniqueness.

Yes, continuity of partial derivatives is stronger than merely satisfying Lipschitz condition. The continuity of $f_{y}$ in Picard's Theorem is only used to create the inequality, namely, Lipschitz condition, which is why it makes sense to just replace the continuity of partial derivative with Lipschitz condition, which I think make this theorem a little stronger because many functions do satisfy Lipschitz condition without having continuous partial derivatives.

To Kalle: Continuity of $f$ isn't enough, but continuity of the partial derivative is plenty.

I took the statement of that uniqueness theorem from my current class- actually, the version proved there had $f$ mapping $I\times\mathbb{R}^{n}$ to $\mathbb{R}^{n}$.

I should probably repeat my initial response to pankajsinha: the argument in the second post is not correct.

Well,I imagined the continuous function y=g(x) passing through (0,0) and then plotted y=|g(x)|
and then since it is given that |g'(x)|<=|g(x)| for all x in R and therefore i assumed that
|g'(x)|<=minimum of |g(x)|=0      (since g(o)=0)

|g'(x)|<=0

Hence |g'(x)|=0

Therefore,g'(x)=0 due to which g(x)=constant=0

Does there exist a counter example to this argument

[quote="pankajsinha"]Well,I imagined the continuous function y=g(x) passing through (0,0) and then plotted y=|g(x)|
and then since it is given that |g'(x)|<=|g(x)| for all x in R and therefore i assumed that
|g'(x)|<=minimum of |g(x)|=0      (since g(o)=0)

|g'(x)|<=0

[/quote]

well that's not a valid argument because it may so happen that the minimum os $g(x)$ be less than that of $g^{'}(x)$ well this arguement of ur only hold for those $x$ where $g(x)$ $=$ $0$ .

You can't leap to "less than the minimum"- it simply isn't true. Suppose instead we had the inequality $|y'|\le 2|y|^{\frac12}$, $y(0)=0$.
There are nonzero solutions to that, such as $y(x)=\begin{cases}0& x\le 1\\ (x-1)^{2}&x>1\end{cases}$.

thanks for clearing my concepts.Now i will try  to arrive at the correct proof


how many solutions $(x_{1},x_{2},...,x_{n})$ in nonnegative integers are there to the equation

$x_{1}+x_{2}+....+x_{n}=10$?

I think its:
[hide]C ( 9 + n, n - 1)[/hide]

i thought it would be

$\binom{n+9}{10}$

My reasoning is that the variables are the urns, and 10=1+1+1+1+1+1+1+1+1+1, each of the "1" is a ball.. so 10 balls and n urns.

And the balls-and-urn argument, if i remember correctly, is C ( balls + urns - 1, urns - 1), which is C (10 + n - 1, n - 1) = C (9 + n, n - 1).

Sorry for not knowing Latex..

Soo...ten balls, $n$ urns...

$\dbinom{b+u-1}{u-1}$=$\dbinom{n+9}{n-1}$.

If it were for positive integers, then we would have $\dbinom{9}{n-1}$.


Let be given a parallelepiped $ ABCD.A'B'C'D'$. Show that if a line $ \Delta$ intersects three of the lines $ AB'$, $ BC'$, $ CD'$, $ DA'$, then it intersects also the fourth line.

$ \mbox{The lines A'B, B'C, C'D and D'A met all four skew lines AB', BC', CD' and DA'}\Rightarrow$
$ \Rightarrow\mbox{the four lines: AB', BC', CD' and DA' is four "hiperbolic lines"}\Rightarrow \\
\Rightarrow \mbox{all line }\triangle \mbox{ intersects tree of the lines, intersects also the fourth line.}$


calculus
derivative
function
trigonometry
LaTeX
calculus computations

Caculate derivative of this function
$ y\equal{}sinx.|cosx|$

consider all such $ x$ for which $ \cos x\geq 0\rightarrow y =\sinh\cos x$
now consider those for which $ \cos x < 0$

$ \text{\LaTeX}$ tip;Use \cos x instead of cos x

Solution

$ y \equal{}\sin{x}\,|\cos{x}|$

$ \Rightarrow y \equal{}\sin{x}\cos{x}\,sgn(\cos{x}) \equal{}\frac{1}{2}\sin{2x}\,(H(\cos{x})\minus{}H(\minus{}\cos{x}))$

$ \Rightarrow\frac{dy}{dx}\equal{}\cos{2x}\,(H(\cos{x})\minus{}H(\minus{}\cos{x}))\plus{}\frac{1}{2}\sin{2x}\,(\minus{}\sin{x}\,\delta(\cos{x})\minus{}\sin{x}\,\delta(\minus{}\cos{x}))$

$ \equal{}\cos{2x}\,sgn(\cos{x})\plus{}\frac{1}{2}\sin{2x}\sin{x}\,(\delta(\cos{x})\plus{}\delta(\minus{}\cos{x}))$


Hi!
I am very glad that IMO is taking place in my country.
Slovenia being such a small and unnoticed country,
I've been wondering if you know anything about it.

Pero please can you send all Math Olympiads of your own country ...?
 I think the problems of the Long Listed part will be selected by the professors of the host country ... so that : slovenian's will select this part and hence we can say that Solenian MO's will be usefulll...  ;)

Davron

I guess this but i would like to know the problem types ... of the Slovenian MO's
did you get me ?

I love Slovenia!!!
Its name is so romantic...there would not be such an attractive country to hold an IMO in my high scholl life...It is really a lifetime pity to lose the chance to enter 06 China Team. (I failed at the very first step...)
I hope that I will be able to attend a worldwide mathematics meeting in Slovenia after I become a mathematician...in the future...

[quote="José"]This maybe help anyone to know a bit more:

http://en.wikipedia.org/wiki/Slovenia[/quote]I think Pero wanted to know what people knew about Slovenia before this IMO. I think this is not such a good idea. Perhaps even have a topic after the IMO to see what people have learned about the country :)

I personally do not know much besides some geographical facts (like it's inside EU, near Italy, has nice beaches, smooth weather, etc.) and I hope to learn more when I get there.

Dober dan!

Slovenia is my favorite country in the world :)

I have been 4 times: when I was an infant, when I was 4 years old, a few summers ago and just this past year for a few days while I was studying in Hungary.  The whole country and the capital Ljubljana are small and beautiful.  I guess what you find interesting about a place depends on what you have in your own country, but the limestone caves are quite incredible to me.  Bled is gorgeous, a town on a lake surrounded by Alps.  Also, they have a mathematician on their currency!

Limestone caves? I gotta see those :) ... although I'm not a big fan of caves since I've seen [url=http://www.imdb.com/title/tt0435625/]The Descent[/url]  :?

[quote="Valentin Vornicu"]Limestone caves? I gotta see those :)[/quote]

The last time I visitted I went to the caves at Postojna, which are very commercialized (tours in many languages, a train-ride to the inside of the cave, very expensive to get in, lots of tourists, and quite amazing) but I think that there are manymany of them throughout the country -- someone who knows better can probably say more.

I'm glad to hear this. Yes, there are many other caves (many for Slovenia, of course  ;) ). For example,[url=http://www.park-skocjanske-jame.si/Eng/a-Skocjan%20Caves%20Regional%20Park.htm]"Škocjanske jame"[/url] are very nice. There are a lot of other interesting things, too (castles, vale [url=http://www.logarska.si/ang/index_a.html]"Logarska dolina"[/url]...). Welcome  :)

Well, Yes, most of those European countries that joined EU  in 2004 have been shown in media.

In my opinion, Slovenia is like a rather succesful story in a dangerous region : it got out of Yugoslavia relatively easy, the most prosperous of all ten nations that joined in 2004, and the most prosperous part of former Yugoslavia (is that correct?).
Edit :Recently, the country reached headlines again as it annouced it will have EUros in 2007 as currency, making it the thirteenth country to adopt it, and the first since the start of the Euro.

Question : is it true that about every man in Slovenia is called Janez, preferably Janez Novak? :D

Of course not, women are called Marija  :lol:

(it's really off topic but we can make a new one for questions like this  ;) )

[quote="spider_boy"][quote="Davron"]
 ... so that : slovenian's will select this part and hence we can say that Solenian MO's will be usefulll...  ;)

Davron[/quote]

Davron, I think Slovenian NMO's are a little easy.[/quote]

proud???????????

[quote="spider_boy"]Maybe this is off topic :!: I have heard a lot, so I want to ask:
 Is there a relation between slovenian language and russian language? I mean do slovenian people understand  in russian?[/quote]
They are both Slavic languages, but I think they are quite far apart.  Slovene is very close to Serbo-Croatian and probably some of the other "near-by" Slavic languages.

[quote]the most prosperous part of former Yugoslavia (is that correct?).[/quote]
Yes -- Croatia is next, and the other parts are substantially poorer.  I don't know about Montenegro, now that it is going to be its own nation.

[quote]Of course not, women are called Marija :lol:[/quote]
The reason I have been to Slovenia is because my mother's closest friend is Slovene.  Of course, she (the friend) is named Marija.

[quote="JBL"][quote="spider_boy"]Maybe this is off topic :!: I have heard a lot, so I want to ask:
 Is there a relation between slovenian language and russian language? I mean do slovenian people understand  in russian?[/quote]
They are both Slavic languages, but I think they are quite far apart.  Slovene is very close to Serbo-Croatian and probably some of the other "near-by" Slavic languages.
[/quote]

There is no such thing as Serbo-Croatian language. Serbian and Croatian are two different languages, altough very similar (Serbs and Croats can understand each other completely). Serbo-Croatian and Croato-Serbian were "made out". http://en.wikipedia.org/wiki/Serbo-croatian#Croatian_linguists

And what about alphabet then?   It came to my attention that in Serbian, the Cyrillic alphabet is used, while Croatian is written in Latin alphabet?

With Serbs and Croats etc living together in the past with the same Yugoslav nationaly, wasn't that confusing?

You speak of the distinction between roman Catholic and orthodox for Croats and Serbs respectively, but where is Slovenia in that? Isn't Slovenia mostly catholic as well? If so why is Croatia sometimes described as 'Pope John Paul's Poland next door"

[quote="fredbel6"]With Serbs and Croats etc living together in the past with the same Yugoslav nationaly, wasn't that confusing?[/quote]
The official languages of Yugoslavia were Serbo-Croatian, Slovene and Macedonian, I believe.  Specifically, Yugoslavia did [i]not[/i] recognize two distinct languages of Serbian and Croatian.

[quote="fredbel6"]You speak of the distinction between roman Catholic and orthodox for Croats and Serbs respectively, but where is Slovenia in that? Isn't Slovenia mostly catholic as well? If so why is Croatia sometimes described as 'Pope John Paul's Poland next door"[/quote]
As I said, Slovenes are mostly Catholic and use the Latin alphabet.  Croatians identify more heavily as Catholic than Slovenes do (88% to 58%).  In Croatia, "none" is the second-most practiced religion, 5%.  In Slovenia, "other or unspecified" is second, with a whopping 23%, and "none" is third with 10%.
All these numbers are drawn from the
[url=http://www.cia.gov/cia/publications/factbook/index.html]CIA World Factbook.[/url]

[quote="fredbel6"]

Question : is it true that about every man in Slovenia is called Janez, preferably Janez Novak? :D[/quote]

 :D  :D  No! There are also Seads, Sanels, Ermins, and esspecialy Mujos, Suljos and Hasos  :rotfl:  I guess that only people from ex-Yugoslavia will understand what I wanted to say, but what the h*ll...

[quote="hsiljak"]

 :D  :D  No! There are also Seads, Sanels, Ermins, and esspecialy Mujos, Suljos and Hasos [/quote]

Only Mujo and Haso are known in Slovenia...  But we have our Janezek (=little Janez)  ;)

(for the others: There are really a lot of jokes about Janezek, Mujo&Haso.)

[quote="vedran6"][quote="hsiljak"][quote="JBL"]The distinction between the Serbian and Croatian languages, like that between Czech and Slovak, is much more a political distinction than a linguistic distinction.  If America had divided from Britain in the late twentieth century, we might equally well insist that American and English were two different languages (both of which, of course, must never be confused with Canadian and Australian).  Bosnia lays claim to the distinct (and essentially identical) language of Bosnian, as well.  In any event, no slights or insults are intended by my use of the term.

Serbs, who for the most part belong to the Orthodox church, write in the Cyrillic alphabet while Croats and Slovenes, for the most part Catholic, write in the Latin alphabet.  But all the languages are Slavic and all are extremely closely related and mutually intelligible.  I don't know what the official alphabet of Bosnian is -- the country is plurality Moslem, so the basic "which branch of Christianity?" test doesn't work.[/quote]

Bosnian official alphabets are both Cyrillic and Latin. In the beginning of the 20th century, Arab alphabet was used too, but when Austro-Hungary annexed Bosnia (1908), Cyrillic and Latin alphabet became the only official alphabets.
Latin alphabet is used in the Federation of BiH, and the Cyrillic alphabet is used in the other entity of Bosnia, so-called RS. The authorities in "RS" use Cyrilic alphabet even in writing foreign languages (i.e. English). it sounds dumb-well, it is...
Bosnia is a strange country...[/quote]
       I think that people in RS dont use cyrilic when they are writing in foreign languuges.
  btw.when we are speaking about langauges Bosnia and Herzegowina is there most complicated with three equal langauges  :P[/quote]

Well, they don't do it anymore... But in the years after the war, I had a chance to talk to a student frm Teslic, who had to write his English papers in Cyrillic (in the year 1999).

[hide](E moj Vedrane, cjepidlako... Nisam ni mislio da to i danas rade... pogresno sam se izrazio ;) )[/hide]

Since there are a lot of jokes in Slovenia mentioning "two stupid Bosnians called "Mujo i Haso" ", there was try in Bosnia to launch new series of jokes mentioning "two Slovenians called "Mujo in Haso" ".

Regarding this discussion on languages, official language of former Yugoslavia was, besides Slovenian and Macedonian, a language called "Serbo-Croat", "Croato-Serbian", "Serbian" and "Croatian", depending on region. When I was much younger, I have learned a class called "Serbo-Croat/Croato-Serbian Language and Literature", and so was written in my class report.

Bosnian language was mentioned regularly in documents since XI century. Basicly, language is same as Croatian and Serbian, so we in Bosnia are, at least, trilingual.  :) Every person has a right to call his/her language as he/she wants. Bosnian constitution points out, that official languages of State of Bosnia and Herzegovina are "Bosnian, Serbian and Croatian", and alphabets are latin and cyrillic, with equal rights to be used. All three languages can be written in both latin and cyrillic, although it is rarely done.

[quote="Mojca"]Well, I thought that the problems on IMO sites were a joke. The following languages are available:
- Croatian
- Bosnian
- Serbian
- Serbian (BIH)
Do people select Bosnian as the first language and Serbian (BIH) as the second one?

There's even no British ...[/quote]
        Someone choses Bosnian someone Serbian as first langauge,that is only the thing of your personal choice.They are very similar so it is not so intelligent to select other for second langauge so usually English is second langauge.  :P


Are you sure you'd have to register an extra team?  I took a brief look at some of the forms on the website and couldn't find anything to that effect.  Strangely, I also couldn't find anything about paying extra to send alternates, so based on that you might be right.

I'd e-mail someone at ARML as soon as you can.  Their e-mail addresses are on the ARML site.

Good luck.

Edit: Seems I didn't read quite that carefully.  On the website, there's a form called "General information for all sites".  On the second page, there's a section on alternates.  It says you have to pay 20 dollars extra for each alternate you plan to bring.  There's no mention of a limit on how many alternates you can take without registering an extra team, so I think you'll be OK!  Just remember to contact ARML to (a) make sure of this just to be safe, and (b) tell them you'll be bringing extra people.

:)

i think the max number of alternates is 14, if you bring 15 people you make a new team. alternate teams join up at the contest to make full teams of 15, so bringing a 15+ alternate team wouldn't quite work


Find the largest two-digit number which is equal to four times the sum of its digits.

hello, let $ z\equal{}10a\plus{}b\equal{}4(a\plus{}b)$ from here we get $ b\equal{}2a$ and if we set $ a\equal{}4$ and $ b\equal{}8$ we have found the largest possible number $ 4z\equal{}48\equal{}4(4\plus{}8)\equal{}48$.
Sonnhard.

This is technically
$ \ 10a\plus{}b\equal{}4(a\plus{}b)$
$ \ 10a\plus{}b\equal{}4a\plus{}4b$
$ \ 6a\equal{}3b$
$ \ 2a\equal{}b$
So giving us 48.
Because the greatest multiple of 2 in single digit numbers is 8.
8=b
so 2a=8 or a=4.


geometry
incenter
LaTeX
area of a triangle
geometry theorems

I have a question regarding barycenters.

On this site: http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

I have that the incenter has the barycentrics  equal to  $ I (a , b , c)$, but i in a book have them equal to $ I(a/2p, b/2p, c/2p)$, where $ p$ is the semiperimeter.

Also the centroid $ G(1,1,1)$ on ETC, but in my book it is $ G(1/3,1/3,1/3)$.

And the list could go on forever.

Why is this?

Can someone please explain?

Thank you very much!

If you want to use LaTeX, please don't disable BBCode in your posts.

As for your question: There is a difference between the notions of normalized and homogeneous barycentrics. The [i]normalized barycentrics[/i] of a point $ P$ with respect to a triangle $ ABC$ are the triple $ \left(\frac{\left[BPC\right]}{\left[ABC\right]},\frac{\left[CPA\right]}{\left[ABC\right]},\frac{\left[APB\right]}{\left[ABC\right]}\right)$, where the abbreviation $ \left[P_{1}P_{2}P_{3}\right]$ stands for the directed area of a triangle $ P_{1}P_{2}P_{3}$. [i]Homogeneous barycentrics[/i] (not: "the" homogeneous barycentrics) of a point $ P$ with respect to a triangle $ ABC$ can be an arbitrary triple $ \left(x,y,z\right)$ of reals $ x$, $ y$, $ z$ which are not all equal to $ 0$ and whose proportion is $ x: y: z=\left[BPC\right]: \left[CPA\right]: \left[APB\right]$.

Of course, the triple of normalized barycentrics of a point $ P$ is a triple of homogeneous barycentrics of $ P$. But the converse doesn't hold, as there are many triples of homogeneous barycentrics of $ P$.

Your book gives normalized barycentrics, while the ETC gives homogeneous barycentrics.

  Darij


calculus
integration
trigonometry
complex numbers
calculus computations

Hi I have another integral question that I need some help with.  I was wondering if anyone might know how to evaluate this.  Thanks.

Question:  Evaluate $ \displaystyle\int_0^\infty x\cos(x^3)dx$

Introducing the substitution t=x^3 transforms the original integral to

$ \frac{1}{3}\int_{0}^{\infty}t^{\minus{}\frac{1}{3}}\cos{t}dt$

It's equivalent to

$ \frac{1}{6}\int_{0}^{\infty}t^{\minus{}\frac{1}{3}}(e^{ix}\plus{}e^{\minus{}ix})dt$

or

$ \frac{1}{6}\int_{0}^{\infty}t^{\minus{}\frac{1}{3}}e^{ix}\plus{}\frac{1}{6}\int_{0}^{\infty}t^{\minus{}\frac{1}{3}}e^{ix}$

Using $ u\equal{}it$ solves both integrals:

$ \frac{1}{6}\Gamma\left(\frac{2}{3}\right)\left(e^{\frac{\pi}{3}i}\minus{}e^{\frac{2\pi}{3}i} \right)$

(the complex numbers in the last bracket appeared as third roots of minus one).

Now, finally we obtain the answer

$ \frac{1}{6}\Gamma\left(\frac{2}{3}\right)$.

Is this ok?

\begin{eqnarray*}
\int_{0}^{\infty} x \cos(x^3) \, dx
& = & \frac{1}{3} \int_{0}^{\infty} \frac{\cos t}{t^{1/3}} \, dt \qquad (\text{where } x^3 = t)\\
& = & \frac{1}{3 \Gamma\left( \tfrac{1}{3} \right)} \int_{0}^{\infty} \cos t \left( \int_{0}^{\infty} u^{-2/3} e^{-tu} \, du \right) \, dt\\
& = & \frac{1}{3 \Gamma\left( \tfrac{1}{3} \right)} \int_{0}^{\infty} \int_{0}^{\infty} u^{-2/3} \cos t \, e^{-tu} \, dt du\\
& = & \frac{1}{3 \Gamma\left( \tfrac{1}{3} \right)} \int_{0}^{\infty} \frac{u^{1/3}}{1 + u^2} \, du\\
& = & \frac{1}{3 \Gamma\left( \tfrac{1}{3} \right)} \int_{0}^{\frac{\pi}{2}} \tan^{1/3} \theta \, d\theta \qquad (\text{where } u = \tan \theta )\\
& = & \frac{1}{6 \Gamma\left( \tfrac{1}{3} \right)} \beta \left( \tfrac{2}{3}, \tfrac{1}{3} \right)\\
& = & \frac{1}{6} \Gamma\left( \tfrac{2}{3} \right)
\end{eqnarray*}

In general, we have

$ \int_0^{\infty}\frac{\cos x}{x^p}\,dx \equal{} \frac{\pi}{2 \Gamma(p)} \sec\left(\frac{p \pi}{2}\right)$, for 0 < p < 1.


I was just discussing this with a friend and thought I could get your opinion as well. Majority of the IIT aspirants who frequent this forum write FIITJEE's All India Test Series. (I think almost 12th student except soumya).

The question I'd like to pose is that. Is AITS the right way to prepare for IIT-JEE. Yes, FIITJEE's question set is good, no doubt, very good. I doubt we have encountered 90% of them before (clearly shows that they have a better collection of books by authors) but the fact is that their physics objectives clearly deserve to be subjectives. Infact I felt that some of the 3 mark objectives were much much tougher than the Numericals, which were dead easy. :(

And FIITJEE's passages are total bull.... :(( Atleast one passage in JEE has some new concept based. FIIJEE asks only theory questions in passage giving some stupid story. And the std of the paper is defn higher than JEE not only that but the solutions are so discouraging. :(((

IITJEE has a good mix of easy, moderate, tough and Superduper tough questions while I don't find this kinda variety in FIITJEE. You may say doing a tougher paper may prepare you well for JEE which is easier but the fact is the results toppled. The ranks gotten by Olympiadians last year was much different from usual. Usually AIR-1 ends up winning a gold in physics (4 times since 2000). :(

What are your comments.

P.S: @Mod: Please let this be a separate thread.

[quote="pardesi"]according  me i pretty much like olympiad kinda physics problems(minus  this years...where their checking s**** and their InPho wasn't any good looking too)
also though FIITJEE should be giving good problems but not tough and long...a tough but tricky problem is adjustable but not a long and sh*** problem...which has been happening...
and those paras are almosta formality.. :P  theydon't give anything new...
i's almos a aloos for people whioc start with the toughest section i di this time with maths....
so good questions but bad place to have them[/quote]

Man, you use 5 letter swear words... Are you suffixing everything with 's'. (like s.h.i.t.s). Anyway well, they are lengthy questions, not really tough. Might probably become less lengthier if you had come after mugging up all the formulas in 120 different IIT books and NCERT. :D. Sums are not tough if you are talking about level of difficulty. they can easily be solved and do not involve difficult ideas except some good conceptual questions. ( I loved that friction question in paper I ans is: Both forward and backward).

P.S: I bet  madness voted "God save the queen". I included that just for him.

Of course  :D  :D  If god doesnt save the queen then the world is doomed  :P

i liked the maths problems this time around...and yes nothing is seemingly tough after PT 2

I hated that Sania Mirza sum to the core. I wasted 10 mins on it cause I wanted to score fully in the probability question, tried thrice and my answers missed the given answer by a constant. I wish they had given in terms of combinations and exponennts

Ofcourse it is also illogical that she would beat sarapova 6-1 but lets not get into that shall.

[quote="achu_182"]i study at supposedly the headquarters of FIITJEE (south delhi), so i get a lot of exposure to such discussions. i know that one of the papers(i think PT3) wa made by our Physics teacher (though i must clear any doubts that that gives any advantage).

our chemistry sir had told us that the FIITJEE AITS difficulty level is planned. the difficulty level increases from PT 1 to around PT3 or FT1 (around which it is mindblowingly high, but in our case, i think the max was at PT2). after that the difficulty level steadily declines and around the Open Test, the difficulty is still higher than IITJEE but reasonable, and is supposedly the best paper in the AITS(best in the sense, that it has the right mix of difficult and easy questions). he also said that the result of AITS has nothing do with ur JEE performance (even though FIITJEE claims the opposite) and u must take the AITS as a chance to create different situations and try out new things, like for example changing the order of attempting questions etc.) and give ur real performance in OT, where there will be the maximum no. of people. never take AITS results to heart( even its solutions). i have also been frustrated many times by solutions, where u have thought abt something extra which deserves to be there as an answer, but due to the solution-makers' shortsightedness (in some cases, blindness!!) he hasnt taken care of that and i get a -1 instead of +3.

the point is that do reasonably well in the OT, and the rest of the AITS results shouldn't bother u much!![/quote]

Wow, well said. It is good exposure and practise. there is no doubt about that.

I think FIITJEE papers prepare us for the worst case scenario so I think and I certainly hope it is a better preclude to IIT-JEE

I voted for God save the Queen, b'cos thats the only option that seemed sane among the rest.. :D

Ah, keep making fun!!! Lets see who has the last laugh mister!!! :D Anyhow, I aint gonna get perturbed at all.. So why bother?

[quote="chemrock"]Yeah for the 'INSANE' the 'INSANE' looks 'SANE' :rotfl:  :rotfl:  :rotfl:[/quote]

Me name is madness so i aint care man  :D  :D   :P

[quote="chemrock"]Yeah for the 'INSANE' the 'INSANE' looks 'SANE' :rotfl:  :rotfl:  :rotfl:[/quote]

Thats the sanest thing I've seen in this thread. :) You get an excellent rating for this. :D


geometry
rectangle
combinatorics proposed
combinatorics

Determine all pairs $(m,n)$ of positive integers such that a $m\times n$ rectangle can be tiled with L-trominoes.


[quote="redsmartie"]$ tan44 \times cot45 \times tan46$
What is the answer?[/quote]

No trig in MathCounts forum.
To a mod: move to HSB.

[hide="Click here for the solution"]We know that:
$ \cot\theta = \tan(90^{\circ}-\theta)$

$ \tan\theta=\frac{1}{\cot\theta}$

So, we may substitute to find:
$ \tan\theta \cdot \tan(90^{\circ}-\theta) = 1$

Hence:

$ \tan 44^{\circ}\cdot \tan 46^{\circ}= 1$

Since $ \cot 45^{\circ}= 1$, we get a final answer of:

$ \tan44 \times \cot45 \times \tan46 = \boxed{1}$[/hide]

maybe most middle schooler hasn't taken precal?

[quote="redsmartie"][b]Why not?[/b]
You could accomplish the questions, the person before me could, and I can. :idea:[/quote]

THERES NO TRIG IN ANY [b]MATHCOUNTS[/b] COMPETITOINS.  AND THIS IS A MATHCOUNTS FORUM.  AGAIN< DONT REVIVE OLD TOPICS, ESPECIALLY TRIG ONES.

Pesky caps lock key sticking again eh?  Give him a break please, he's probably unfamiliar with the rules - MATHCOUNTS does not include trig in it's curriculum list.    Moved to HSB.


Can 2008 be expressed as the sum of 3 squared integers?

Hey Nayel, do you mind my asking how you deduced that? (presuming it wasn't brute force).

Thanks!

Well, first I deduced that $ x,y,z$ must all be even. Thus I considered the equation $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 502$. Noticing (after some failure approaches in proving 502 cannot be expressed as a sum of 3 squares :P ) $ 22^2 \equal{} 484$, I found $ 502 \minus{} 484 \equal{} 18 \equal{} 3^2 \plus{} 3^2$. Hence $ 502 \equal{} 22^2 \plus{} 3^2 \plus{} 3^2\implies 2008 \equal{} 44^2 \plus{} 6^2 \plus{} 6^2$. So it was mostly luck. :)

Ah nice, it's so clear now!

Thanks for that Nayel (and also speed!).

You're welcome. :) And also welcome to AoPS-MathLinks!

[hide="Theorem"]Write $ a\equal{}4^k(8m\plus{}n)$ with $ 4\not| n$. Then $ a$ can be represented as the sum of three perfect squares iff $ n\neq 7$.

Please don't ask for a proof :) If I'll ever be able to prove it I'll make it go into history as Bertram's postulate. ;)[/hide]

[quote="bertram"][hide="Theorem"]Write $ a \equal{} 4^k(8m \plus{} n)$ with $ 4\not| n$. Then $ a$ can be represented as the sum of three perfect squares iff $ n\neq 7$.

Please don't ask for a proof :) If I'll ever be able to prove it I'll make it go into history as Bertram's postulate. ;)[/hide][/quote]
I think that can be easier stated as $ k$ is inexpressible as the sum of three squares iff it is of the form $ 4^{k}(8a\plus{}7)$ where k and a are nonnegative.
I saw this theorem once elsewhere in the forums...

That's right, but you used $ k$ twice in your statement :P

PLease don't send me the link because I've been pondering this problem for about 8 years now and I want to find the solution myself. ;)