[quote="skcd7"]
$ a>0,$ $ b>0,$ $ c>0,$ $ a\plus{}b\plus{}c\equal{}1.$ Prove that:
\[ \sum_{cyc}\sqrt{a^2\plus{}abc}\geq\frac{2}{3}\left(\sqrt a\plus{}\sqrt b\plus{}\sqrt c\right)\]
[/quote]
$ \left(\sum_{cyc}\sqrt{a^2\plus{}abc}\right)^2\equal{}\sum_{cyc}\left(a^2\plus{}abc\plus{}2\sqrt{(a^2\plus{}abc)(b^2\plus{}abc)}\right)\geq$
$ \geq\sum_{cyc}(a^2\plus{}2ab\plus{}3abc).$
Hence, it remains to prove that
$ (a^2\plus{}b^2\plus{}c^2)^3\plus{}9a^2b^2c^2\geq\frac{4}{9}(a\plus{}b\plus{}c)^2(a^2\plus{}b^2\plus{}c^2)^2,$ which is obvious.

$ a,b,c\geq 0$

prove that

$ \frac {abc}{a^3 \plus{} b^3 \plus{} c^3} \plus{} \frac {2}{3}\geq \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$

$ \Longleftrightarrow \sum_{sym} (a^2 \minus{} ab) \sum_{sym}(a^3 \minus{} a^2b)\geq 0$

$ a \plus{} b \plus{} c\geq 3$

prove that $ (a^2 \plus{} k)(b^2 \plus{} k)(c^2 \plus{} k)\geq (1 \plus{} k)^3$

try $ k \equal{} \frac {1}{75},a \equal{} 17,b \equal{} \frac {1}{8},c \equal{} \frac {1}{4}.$

$ a,b,c\geq 0$

prove that
$ 4(a \plus{} b \plus{} c)^3\geq 27(ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc)........(*)$

assume $ a \equal{} max{(a,b,c)}$

【1】$ a\geq b\geq c$

substitution $ a \equal{} x \plus{} y \plus{} c,b \equal{} y \plus{} c$($ x,y\geq 0$)

 $ (*)\Longleftrightarrow (9y^2 \plus{} 9x^2 \plus{} 9xy)c \plus{} (5y \plus{} x)(2x \plus{} y)^2\geq 0$

【2】$ a\geq c\geq b$

substitution $ a \equal{} x \plus{} y \plus{} b,c \equal{} y \plus{} b$($ x,y\geq 0$)

 $ (*)\Longleftrightarrow (9y^2 \plus{} 9x^2 \plus{} 9xy)b \plus{} (5y \plus{} 4x)(y \minus{} x)^2\geq 0$

[quote]$ a,b,c\geq 0$

prove that

$ \frac {abc}{a^3 \plus{} b^3 \plus{} c^3} \plus{} \frac {2}{3}\geq \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$.
[/quote]

$ \frac {abc}{a^3 \plus{} b^3 \plus{} c^3} \plus{} \frac {2}{3}\geq \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$

$ \iff \frac {3abc \minus{} (a^3 \plus{} b^3 \plus{} c^3)}{3(a^3 \plus{} b^3 \plus{} c^3)}\ge \frac {ab \plus{} bc \plus{} ca \minus{} (a^2 \plus{} b^2 \plus{} c^2)}{a^2 \plus{} b^2 \plus{} c^2}$

$ \iff \frac {(a \plus{} b \plus{} c)\sum{(a \minus{} b)^2}}{6(a^3 \plus{} b^3 \plus{} c^3)} \le \frac {\sum{(a \minus{} b)^2}}{2(a^2 \plus{} b^2 \plus{} c^2)}$

$ \iff \sum{\left( \frac {1}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {a \plus{} b \plus{} c}{3(a^3 \plus{} b^3 \plus{} c^3)}\right)(a \minus{} b)^2} \ge 0$

$ \iff \sum{\frac {\sum{(a \plus{} b)(a \minus{} b)^2}}{3(a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3)}(a\minus{}b)^2} \ge 0.$


In trapezoid ABDC, AB=2 BC = 6. AD IS parallel to BC. and BAD is 60 degrees. Find the area of the trapezoid.

I did this and i kept getting 8 radical 3, but it's wrong , someone helpme :)

ABCD not ABDC oops :X

And no AB and CD arent parallel

[hide]Set up a system of Equations using Pythags and the length the segments get cutoff.[/hide]

[quote="snipefreak"][quote="funcia"]do you mean abcd or abdc?
oh and are ab and cd parallel too...[/quote]
They wouldn't be it is a trapezoid:

A        D
--------
\[color=white]........[/color]/
 ------ Kinda like that.  :wink:
B       C[/quote]
thanks i just didnt read the question thoroughly :(

In trapezoid ABCD, AB=2 BC = 6. AD IS parallel to BC. Angle BAD is 60 degrees. Find the area of the trapezoid.

If AD // BC and angle BAD = 60, then angle ABC = 120.  Draw triangle ABC with angle ABC = 120, BA = 2 and BC =6.  Draw ray AP parallel to BC so that angle BAP = 60.  Note that D may be placed anywhere on ray AP, thus there are infinite possibilities.

What’s the source of this problem?  Has something been omitted from the statement of the problem?

I got an answer!

Attached with picture.

[quote="anirudh"]I got an answer!

Attached with picture.[/quote]
[hide]
what makes FD equal 1?
[/hide]

I dunno what FD is so he might want to find out what it is. THere is something missing.

You assumed it was isosceles.  That wasn't part of the given. The fact that no information was given to determine the position of D is the flaw in the problem.

Your answer is correct for the assumption you made.

[quote="ProtestanT"]assume FD is 0 and it is shaped liek this:
..___
./.....|
/___|
wouldn't that be easier?[/quote]

Thats what I did when I started on the problem but I realized that it never said the trapezoid was right.

but its easier wen the trapezoid is right :D


IMO Shortlist
number theory
prime numbers
prime factorization

[hide="Lemmas"]
[b]Lemma 1[/b]: There are infinitely many highly divisible numbers.
[i]Proof:[/i] Suppose for the sake of contradiction that there are only finitely many highly divisible numbers. Then the sequence $d(1), d(2), \ldots$ will be bounded, which is impossible since any $n$ distinct primes $p_1, p_2, \ldots, p_n$ satisfy $d(p_1 p_2 \ldots p_n) = 2^n$, which grows arbitrarily large.

[b]Lemma 2[/b]: Suppose $n = p_1^{e_1} p_2^{e_2} \cdots p_n^{e_n}$ is highly divisible. Then $e_1 \geq e_2 \geq \cdots \geq e_n$.
[i]Proof:[/i] Suppose that $e_1 \geq e_2 \geq \cdots \geq e_n$ does not hold; let $f_1, f_2, \ldots, f_n$ be the permutation of $e_1, e_2, \ldots, e_n$ for which $f_1 \geq f_2 \geq \cdots \geq f_n$. Then $n' = p_1^{f_1} p_2^{f_2} \cdots p_n^{f_n} < n$, while $d(n') = d(n)$, which is a contradiction.

[b]Lemma 3[/b]: For any prime $q$ and integer $k$, all sufficiently large highly divisible $n$ satisfy $q^k | n$.
[i]Proof:[/i] We will first prove this when $k=1$. Suppose for the sake of contradiction that the result is false when $k=1$. Lemmas 1 and 2 imply that $q \neq 2$, so we may let $q = p_{m+1}$ be the $m+1$st prime, where $m \geq 1$. If ouy result is false, then there are infinitely many highly divisible numbers of the form $p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}$. This combined with the fact that $e_1 \geq e_2, e_3, \ldots, e_m$ shows that $e_1$ can attain arbitrarily large values. Hence, we can find some highly divisible $n = \prod_{i=1}^m p_i^{e_i}$ with $\frac{e_1 + 1}{e_1 - \lceil \log_2 p_{m+1} \rceil + 1} < 2$ (since the fraction goes to 1 as $e_1$ approaches infinity).

We have
\begin{align*}
d(n)
&= d \left (\frac{2^{e_1} n}{2^{e_1}} \right) \\
&= \frac{(e_1 + 1)}{e_1 - \lceil \log_2 p_{m+1} \rceil + 1} d \left (\frac{n}{2^{e_1}} 2^{e_1 - \lceil \log_2 p_{m+1} \rceil} \right) \\
&< 2d \left (\frac{n}{2^{\lceil \log_2 p_{m+1} \rceil}} \right) \\
&= d \left (\frac{p_{m+1} n}{2^{\lceil \log_2 p_{m+1} \rceil}} \right),
\end{align*}
while $n = 2^{\lceil \log_2 p_{m+1} \rceil} \frac{n}{2^{\lceil \log_2 p_{m+1} \rceil}} \&> \frac{p_{m+1} n}{2^{\lceil \log_2 p_{m+1} \rceil}}$, implying that $n$ is not highly divisible, a contradiction.

For $k > 1$, suppose for the sake of contradiction that the reuslt is false. Let $s  = k+2$. Then $\frac{d(q^{k+s})}{d(q^k)} = \frac{2k+3}{k+1} > 2$. Combining the $k=1$ case with our assumption that this is false implies the existence of a highly divisible $n$ with $q^k \not | n$ for which some prime $r > p^s$ divides $n$. Note that
\begin{align*}
d(n)
&= d \left (\frac{n}{r} r \right) \\
&\leq 2 d \left (\frac{n}{r} \right ) \\
&< \frac{2k+3}{k+1} d \left (\frac{n}{r} \right) \\
&\leq d \left (\frac{n p^s}{r} \right ),
\end{align*}
while $\frac{n p^s}{r} < n$, so we have reached a contradiction. [/hide]

[hide="Part a"]
By lemma 3, all sufficiently large highly divisible $n$ are divisible by 16. Furthermore, the fact that 4 is the only highly divisible power of 2 (for $n > 2$, $3 \cdot 2^{n-2} < 2^n$ while $d(3 \cdot 2^{n-1}) = 2n > n+1 = d(2^n)$) combined with lemma 2 tells us that all sufficiently large highly divisible $n$ are divisible by 48. I claim that for any highly divisible $n$ divisible by 48, there exists a highly divisible number between $n$ and $2n$, from which the result follows.

Let $n = 2^a 3^b m$, where $2, 3 \not | m$. If $a \geq b + 2$, then $a(b+2) > (a+1)(b+1)$. Setting $k_1 = \frac{3n}{2}$ gives
\begin{align*}
d(k_1)
&= d \left( \frac{3n}{2} \right) \\
&= d(2^{a-1} 3^{b+1} m) \\
&= a(b+2) d(m) \\
&> (a+1)(b+1) d(m) \\
&= d(2^a 3^b m) \\
&= d(n).
\end{align*}
Hence, we can find some highly divisible $k_1'$ such that $n < k_1' \leq k_1$ and $d(k_1) \leq d(k_1')$. $n < k_1' < 2n$, so we are done in the case where $a \geq b+2$.

Recalling that $a \geq b$, we see that $a \leq b+1$ yields $b = a$ or $b=a+1$. Since $48 | n$, $a > 3$, so $(a+3)b > (a+1)(b+1)$. Setting $k_2 = \frac{4n}{3}$ gives
\begin{align*}
d(k_2)
&= d \left( \frac{4n}{3} \right) \\
&= d (2^{a+2} 3^{b-1} m) \\
&= (a+3)b \, d(m) \\
&> (a+1)(b+1) d(m) \\
&= d(2^a 3^b m) \\
&= d(n).
\end{align*}
Hence, we can find some highly divisible $k_2'$ such that $n < k_2' \leq k_2$ and $d(k_2) \leq d(k_2')$. $n < k_2' < 2n$, so our proof of part a is complete.
[/hide]

[hide="Part b"]
Let $k_0$ be any positive integer. Let $n$ be the smallest highly divisible number for which $p^{k_0} | n$, and let $p^k$ fully divide $n$. I claim that $n/p$ is highly divisible as well. If $n/p$ is not highly divisible, then we can find some highly divisible $a$ less than $n/p$ for which $d(n/p) \leq d(a)$. Let $p^j$ fully divide $a$. We must have that $j \leq k-1$ (by the minimality of $n$. Furthermore, $p^{k-1} || n/p$. Note that
\begin{align*}
d(n)
&= d(p^k n/p^k) \\
&= (k+1) d(n/p^k) \\
&= \frac{k+1}{k} d(p^{k-1} n/p^k) \\
&= \frac{k+1}{k} d(n/p^{k-1}).
\end{align*}
We can deduce that $d(pa) = \frac{j+2}{j+1} d(a)$ in a similar way. Note that $j \leq k-1$ implies $\frac{k+1}{k} \leq \frac{j+2}{j+1}$, yielding
$d(n) = \frac{k+1}{k} d(n/p) \leq \frac{k+1}{k} d(a) \leq \frac{j+2}{j+1} d(a) = d(pa)$. On the other hand, since $n$ is highly divisible and $n > pa$, we must have $d(n) > d(pa)$, so we have reached a contradiction.[/hide]

From now on, I will call highly divisible integers "highlies", for convenience. It is easy to see that the sequence of highlies is infinite and divisible by any number for sufficiently big highlies. To see this, we see $p^k | n$ for a sufficiently big highly $n$, where $p$ is prime. If $k=1$, then $n$ must be a product of a limited number of primes. Say $n=q_1^{e_1}*...*q_l^{e_l}$, where $q_1=2$. We know $e_1$ is bounded, otherwise we can decrease it and multiply $n$ by $q_{l+1}^e$ for an adequate $e$. Therefore $e_i$ is bounded for all $i$, so $n$ is bounded. For $k > 1$, take a big enough $n$, and we will have an arbitrarily large prime divisor of $n$. Divide $n$ by that prime divisor and multiply it by $p^V$ for a big $V$. It is easy to see this new number is smaller than $n$ but has more divisors. So we're done.

For part (a), assume there exists infinitely many highlies $H$ such that $kH$ is the next highly, for some $H$. Note that the next highly is simply the least integer $m$ such that $d(m) \ge d(H)$. Clearly $m=2H$ satisfies this. Therefore $k=2$. Now, suppose $H=\prod_{i=1}^k p_i^{e_i}$ is the factorization of $H$. It is easy to see that $p_1=2$, $p_2=3$, etc and also $e_1 \ge e_2 \ge ... \ge e_k$. Now take $m=H\frac{q_{x+1}}{q_x}$ where $q_x$ and $q_{x+1}$ are consecutive prime numbers, and $e_x \textgreater e_{x+1} +1 \ge 1$. By Bertrand's Postulate, $m \textless 2H$. But also $d(m) = d(H) \frac{(e_{x+1}+2)(e_x)}{(e_x+1)(e_{x+1}+1)} \textgreater d(H)$ since $e_x \textgreater e_{x+1}+1$. Therefore, $2H$ isn't the least integer $z$ with $d(z) \textgreater d(H)$, since $m$ also satisfies. So we must have $e_x \le e_{x+1}+1$ for all $x$.  If we take $m=4H/3$ it will satisfy, and we only have to care about $6^{V} | H$ where $V$ is a big number I don't want to compute. But that divisibility is true for sufficiently big $H$'s. So we're done.

For part (b), choose a $k$ and take a highly $H$ such that $p^k || H$ and WLOG $H$ is the first highly to be divisible by $p^k$. Now suppose $H/p$ isn't highly. Then there exists a highly $x$ such that $d(x) \ge d(H/p)$ and $x \textless H/p$. Now suppose $p^k$ doesn't divide $x$, so say $v_p(x)=q<k$. Then $d(xp) = d(x) \frac{q+2}{q+1} \ge d(H/p)\frac{q+2}{q+1} \ge d(H/p)\frac{k+1}{k} = d(H)$ since $v_p(H/p)=k-1$. So $x$ is a highly divisible by $p^k$ and less than $H$, contradicting the minimality of $H$. So we were wrong and $H/p$ is in fact highly and we're done, since we can choose $k$ from an infinite subset of the naturals.

Here's my solution for part (a). I highly suspect that this might be wrong, so I would be grateful if someone points out any flaws.

[hide=" Part (a)"]

Oops incorrect, please ignore.

[/hide]

Amazing problem! :)

[hide=for fun]My reaction on solving: "To think you'd go this far, in your factorisation". (Kudos if you get the reference.)[/hide]

[quote=Yimin Ge]Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$.
Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ satisfying $m < s < n$.

(a) Show that there are only finitely many pairs of consecutive highly divisible
integers of the form $(a, b)$ with $a\mid b$.

(b) Show that for every prime number $p$ there exist infinitely many positive highly divisible integers $r$ such that $pr$ is also highly divisible.[/quote]

For brevity, replace the term "highly divisible" with "peak".

First note that infinitely many peak numbers exist, otherwise $(d(n))_{n \ge 1}$ is bounded above, which is clearly not the case. Let $q_1<q_2<\dots$ be the sequence of primes. Now we give three important claims.

[b]Lemma.[/b] All peaks consist of a contiguous block of the first few primes in their factorisation.

(Proof) Let $n=\prod^m_{j=1} p_j^{\alpha_j}$ be a peak. Then $d(a)=d(n)$ where $a=\prod^m_{j=1} q_j^{\alpha_j}$ and $a \le n$. So $a=n$ and the claim is true. $\blacksquare$

[b]Lemma.[/b] Exponents of these primes in the peak are in decreasing order.

(Proof) Suppose $p^a || n$ and $q^b || n$ for $p<q, a<b$ and a peak $n$. Then $d(n)=d\left(\left(\tfrac{p}{q}\right)^{a-b}n\right)$ while the latter is smaller. $\blacksquare$

[b]Lemma.[/b] For any prime $q$ and integer $k>0$, all sufficiently large peaks are divisible by $q^k$.

(Proof) For $k=1$, suppose for infinitely many peaks $n$ we have $p_{m+1} \nmid n$. Then for a large such $n$ we can find $1 \le s \le m$ with $p_s^{\alpha_s}>\sqrt[m]{n}$. Now we can conveniently drop a few factor of $p_s$ in favour of $p_{m+1}$ to get $n'<n$ but $d(n')>d(n)$, contradiction!

Suppose $q=2$ fails for some $k>1$. Then all exponents are small, so we can drop a few factors at the end and append a suitable number of $2$'s.

Let $q_{m+1}$ for $m \ge 1$ be the smallest prime that fails for some fixed $k>1$. Then for all peaks $n$ sufficiently large, $v_2(n), v_{p_m}(n)$ exceed a constant $M>0$. Since $q_{m+1}<2q_m$, we see $d\left(\frac{q_{m+1}}{2q_m}n\right)>d(n)$ a contradiction! Hence the lemma is true. $\blacksquare$

Now pick a sufficiently large peak $n$. It is easy to check $\operatorname{max}\left( d(\tfrac{4}{3}n), d(\tfrac{3}{2}n) \right)>d(n)$ so we have a peak $n<k<2n$ solving part I.

For part II, fix any large $k$ and let $r$ be the largest peak with $v_p(r)=k$. Let $j<pr$ be the largest peak (so $j>r$). Let $s=v_p(j)>k$. Clearly, $$d(j) \ge d(pr)=\left(\frac{k+2}{k+1}\right)d(r)>\left(\frac{k+2}{k+1}\right)d\left(\frac{j}{p}\right)=\left(\frac{k+2}{k+1}\right)\left(\frac{s}{s+1}\right)d(j)$$ hence $(k+2)s<(k+1)(s+1)$ leading to $s<k+1$, that is clearly false! Hence $pr$ is a peak two! $\blacksquare$

Since no one has pointed it out yet, the complete list for part (a) is 1, 2, 6, 12, 60, 360, 2520. http://oeis.org/A072938

Solid problem. Before we go into each part, we first prove a bunch of facts about highly divisible numbers. First, note that there are clearly infinitely many of these numbers, as $d(n)$ can be arbitrarily large, and therefore there are infinitely many peaks. Let $p_1<p_2<\cdots$ be the prime numbers.

[b]Lemma:[/b]  Consider a highly divisible number $n$, and let its prime factorization be $n=\prod_i p_i^{e_i}$, where $e_i=0$ for sufficiently large $i$. We have the following two results. [list]     [*](a) We have $e_1\ge e_2\ge e_3\ge\cdots$.     [*](b) For fixed $t$ and $i$, there exists some $N$ such that $n\ge N\implies e_i\ge t$. [/list]

[i]Proof:[/i]  Note that permuting the exponents in the prime factorization does not change the value of $d(n)$, so if the $e_i$s were non increasing, we could permute them to be non increasing, decreasing the value of $n$ while maintaining the value of $d(n)$, thus contradicting highly divisible. This proves (a).

Before proving (b), we first show it with $t=1$ (i.e. for fixed $i$, all sufficiently large highly divisible numbers contain $p_i$). Suppose that this was not the case, and that there was some $k$ such that all highly divisible numbers are only divisible by $p_1,\ldots,p_k$. For sufficiently large highly divisible $n=p_1^{e_1}\cdots p_k^{e_k}$, we would have $e_1$ arbitrarily large, and in particular such that \[\log_{p_1}(p_{k+1})<\frac{e_1}{100}.\] Now, pick a positive integer $x$ such that \[\log_{p_1}(p_{k+1})<x<\frac{e_1+1}{2}.\] Now, let $m=\tfrac{n}{p_1^x}p_{k+1}$. We see that $m<n$, and \[\frac{d(m)}{d(n)} = \frac{e_1-x+1}{e_1+1}\ge 1,\] so $n$ is not highly divisible. This proves (b) for $t=1$.

Now we'll show (b) for arbitrary $t$. Suppose that there did not exist such $N$, and that we have $e_i<t$ for all highly divisible $n$. Let $u$ be the largest value such that $e_i=u$ for infinitely many highly divisible $n$. Indeed, suppose we have such $n=p_1^{e_1}\cdots p_j^{e_j}$ (with $e_j\ge 1$), so $e_i=u$. Let $m=\tfrac{n}{p_j}\cdot p_i^{2u}$, so $m<n$ (as long as $n$ is large enough, so that $p_j$ is large enough). We see that \[\frac{d(m)}{d(n)} = \frac{e_j}{e_j+1}\cdot\frac{3u+1}{u+1}\ge 1,\] so $n$ is not highly divisible. This is the desired contradiction, so (b) is proved. $\blacksquare$

We'll now solve the original problem.

(a) Suppose that $n=p_1^{e_1}\cdots p_k^{e_k}$ is highly divisible, and is sufficiently large. Suppose we have $e_i\ge e_{i+1}+2$ for some $i$. Let $m=\tfrac{n}{p_i}\cdot p_{i+1}$. Then,  \[\frac{d(m)}{d(n)} = \frac{e_i}{e_i+1}\cdot\frac{e_{i+1}+2}{e_{i+1}+1}\ge\frac{e_{i+1}+2}{e_{i+1}+3}\cdot\frac{e_{i+1}+2}{e_{i+1}+1}>1,\] and $m<2n$ (by Bertrand). Thus, we have $n<m<2n$ such that $d(m)>d(n)$, so the next highly divisible number will be less than $2n$, so not a multiple of $n$.

Now, we have $e_{i+1}\in\{e_i,e_i-1\}$. The idea now is to focus on $i=1$, so $e_2\in\{e_1,e_1-1\}$ (and recall that $e_1$ is sufficiently large by the second part of the lemma). Now, let $m=\tfrac{4n}{3}$, and now \[\frac{d(m)}{d(n)} = \frac{e_1+3}{e_1+1}\cdot\frac{e_2}{e_2+1}\in\{\frac{(e_1+3)(e_1-1)}{e_1(e_1+1)},\frac{e_1(e_1+3)}{(e_1+1)^2}\},\] both of which are greater than $1$, so we finish by the same logic as before.

(b) Note that $v_p(n)$ is unbounded as $n$ ranges over highly divisible positive integers (by the second part of the lemma). The key idea is to pick highly divisible $n$ such that $v_p(n)$ is at a peak, or that $v_p(m)<v_p(n)$ for all highly divisible $m<n$. We now show that $r=n/p$ is in fact highly divisible, which finishes since there are infinitely many peaks.

Suppose $r$ is not highly divisible. Let $x$ be the largest highly divisible number less than $r$, and since $r$ isn't highly divisible, we must have $d(x)\ge d(r)$. Since $pr$ is highly divisible, we have $d(px)<d(pr)$.

Now, we have \[\frac{v_p(x)+2}{v_p(x)+1}d(x) = d(px) < d(pr) = \frac{v_p(r)+2}{v_p(r)+1}d(r),\] so \[\frac{\frac{v_p(r)+2}{v_p(r)+1}}{\frac{v_p(x)+2}{v_p(x)+1}}>1,\] or $v_p(x)>v_p(r)$. Thus, \[v_p(x)\ge v_p(pr)=v_p(n),\] which violates the peak $v_p$ assumption on $n$. This solves (b).

[quote=Yimin Ge]Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$.
Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ satisfying $m < s < n$.

(a) Show that there are only finitely many pairs of consecutive highly divisible
integers of the form $(a, b)$ with $a\mid b$.

(b) Show that for every prime number $p$ there exist infinitely many positive highly divisible integers $r$ such that $pr$ is also highly divisible.[/quote]

Highly divisible is too long and boring so call them dead integers.
[b][color=#f00]Claim:[/color][/b] If $n=\prod_i p_i^{e_i}$ is an dead integer then:-
(a) We have $e_1\ge e_2\ge e_3\ge\cdots$.
(b) $n$ must be divisible by the first $k$ primes.
[b][color=#f00]Proof:[/color][/b] Note that permuting the exponents in the prime factorization does not change the value of $d(n)$, so if the $e_i$'s were non increasing, we could permute them to be non increasing, decreasing the value of $n$ while maintaining the value of $d(n)$,which is alive a contradiction to the graveyard.
For part(b),Let $n=\prod^m_{j=1} p_j^{\alpha_j}$ be dead. Then $d(a)=d(n)$ where $a=\prod^m_{j=1} q_j^{\alpha_j}$ and $a \le n$. So $a=n$ and the claim is true. $\blacksquare$

For part a)Choose a large number which is dead.
Then $\operatorname{max}\left( d(\tfrac{4}{3}n), d(\tfrac{3}{2}n) \right)>d(n)$ so there is another dead integer $X$ which is $n<X<2n$,so this is proved.
For part b)Choose a large monster $F$ with $\nu_p(n)=F$
Now suppose that $pn$ isn't dead and let $X$ be the smallest dead number which is less than $X<pn$
Now clearly,$$d(j) \ge d(pr)=\left(\frac{k+2}{k+1}\right)d(r)>\left(\frac{k+2}{k+1}\right)d\left(\frac{j}{p}\right)=\left(\frac{k+2}{k+1}\right)\left(\frac{s}{s+1}\right)d(j)$$,so $(k+2)s<(k+1)(s+1)$ or $k>s+1$,another contradiction,so we are done.$\blacksquare$

We first prove a few generic properties of highly divisible numbers.

[b][color=#00f]Property 1:[/color][/b] There are infinitely many highly divisible numbers.
[color=#00f]Proof:[/color] $d(i)$ is unbounded. $\blacksquare$

[b][color=#00f]Property 2:[/color][/b] If $n=p_1^{k_1}\ldots p_n^{k_n}$ is highly divisible, then $k_1 \geq \cdots \geq k_n$.
[color=#00f]Proof:[/color] If we had $k_i<k_{i+1}$ for some $i$, then increase $k_i$ by $1$ and decrease $k_{i+1}$ by $1$, which decreases $n$. Further, this multiplies $d(n)$ by
$$\frac{k_i+2}{k_i+1}\cdot \frac{k_{i+1}}{k_{i+1}+1},$$
which is at least $1$ as
$$\frac{k_i+2}{k_i+1}\cdot \frac{k_{i+1}}{k_{i+1}+1}\geq 1 \iff k_ik_{i+1}+2k_{i+1}\geq k_ik_{i+1}+k_{i+1}+k_i+1 \iff k_{i+1} \geq k_i+1,$$
which is true. $\blacksquare$

[b][color=#00f]Property 3:[/color][/b] For any prime $p$ and $k\geq 1$, all large enough highly divisible $n$ are divisible by $p^k$.
[color=#00f]Proof:[/color] First, note that if the exponent of some prime $p$ is unbounded, then the exponent of every prime $q \neq p$ is unbounded as well. This is because if the exponent of $q$ is at most $m-1$, we may pick some highly divisible $n$ such that $p^{1001qm} \mid n$. Since $(p^{1000q+q})^{m}>p^{1000qm}q^{m}$, $\frac{nq^m}{p^{qm}}<n$. On the other hand, $\frac{nq^m}{p^{qm}}$ has at least
$$\frac{2m}{m}\cdot \frac{1000qm+1}{1000qm+qm+1}>1$$
times as many divisors as $n$, hence $n$ is not highly divisible: contradiction.
However, if the exponent of every prime $p$ is bounded, then arbitrarily large highly divisible $n$ must be divisible by arbitrarily large primes. But if the exponent of $2$ is bounded by $m$, simply take some prime $p>2^m$ that divides some highly divisible $n$. Then $\frac{n2^m}{p}<n$, yet $\frac{n2^m}{p}$ has at least
$$\frac{2m-1}{m-1}\cdot \frac{1}{2}>1$$
times as many divisors as $n$: another contradiction. $\blacksquare$

We are now ready to do part (a). The idea here is that if $a,b$ are highly consecutive and $a \mid b$, then $4a/3$ must have more divisors than $a$ if $a$ is large enough. I claim that we have $\nu_2(n)\leq \tfrac{5}{3}\nu_3(n)+\tfrac{16}{3}$ for large highly divisible $n$. Indeed, if this was not the case (so $\nu_2(n)>\tfrac{5}{3}\nu_3(n)+\tfrac{17}{3}$), we could multiply $n$ by $\tfrac{27}{32}$. Letting $\nu_2(n)=x,\nu_3(n)=y$, this multiplies the divisor count of $n$ by
$$\frac{x-4}{x+1}\cdot \frac{y+4}{y+1},$$
which is at least one, as
$$(x-4)(y+4)\geq (x+1)(y+1) \iff xy-4y+4x-16\geq xy+x+y+1 \iff 3x \geq 5y+17$$
Then, let $\nu_2(a)=x,\nu_3(b)=y$, so $4a/3$ has
$$\frac{x+3}{x+1}\cdot \frac{y}{y+1}$$
times as many divisors as $a$. This is greater than $1$ if
$$(x+3)y\geq (x+1)(y+1)+1 \iff xy+3y \geq xy+x+y+2 \iff x \leq 2y-2,$$
which holds for all $x$ large enough, as $\tfrac{5}{3}y+\tfrac{16}{3}\leq 2y-2$. $\blacksquare$

Now for part (b). By property 3, for any given prime $p$ there exist infinitely many highly divisible $n$ such that $\nu_p(n)<\nu_p(m)$ for all $m>n$ and $m$ highly divisible—call such $n$ [i]extinct[/i]. I claim that for all extinct $n$, $pn$ is also highly divisible. Indeed, suppose $pn$ wasn't, so there exists some rational $r<1$ such that $rpn$ is highly divisible and $d(rpn)>d(pn)$. It is clear that we require $rpn>n$.
Suppose $\nu_p(r)=k$, so let $r=p^kr'$. If multiplying $n$ by $r'$ multiplies $d(n)$ by some rational $t$, then multiplying $pn$ by $r'$ does the same. Thus,
$$\frac{d(rpn)}{d(pn)}=t\cdot \frac{k+\nu_p(n)+2}{\nu_p(n)+2}>1.$$
On the other hand,
$$\frac{d(rn)}{d(n)}=t\cdot \frac{k+\nu_p(n)+1}{\nu_p(n)+1}.$$
If $k \geq 0$, this is at least $t\cdot \frac{k+\nu_p(n)+2}{\nu_p(n)+2}>1$, which contradicts the fact that $n$ is highly divisible. Hence $k<0$, but then $\nu_p(rpn)\leq \nu_p(n)$ and $rpn>n$, hence $n$ cannot be extinct (by taking $m=rpn$ in the definition): contradiction. $\blacksquare$

Wow, surprisingly tough for a 2005 ISL.
Let $P = \{2= p_1 < p_2 < p_3 < \ldots\}$ be the set of all primes

(a) Let $n$ be some highly divisible number. We can see that $n$ can be represented in the form of $p_1^{e_1}p_2^{e_2} \ldots p_k^{e_k}$ where $e_1 \geq e_2 \geq \ldots \geq e_k$

Claim) For all $l \in \mathbb{N}$, there exists some constant $N_l$ such that all highly divisible numbers $n \geq N_l$ satisfies $v_2(n) > l$
[hide=Claim Proof]
Although the wording makes it seem complicated, the fundamental intuition is that we want to bound $v_2(n)$ from below
It suffices to provide an upper bound for highly divisible numbers with $v_2(n) \leq l$.
FTSOC assume that no such upper bound exists, and let $n$ be an arbitrarily large highly divisible number with $v_2(n) \leq l$
Note that $e_1 \geq e_2 \geq \ldots$, and thus $n \leq p_1^l p_2^l \ldots p_k^l$
Since $n$ is arbitrarily large, assume that $k$ is also arbitrarily large, such that $p_k >> 2^{100l}$
Note that $n > \frac{n}{p_k} \cdot 2^{100l}$, but $d(RHS) > d(n)$, creating a contradiction.
[/hide]

Claim 2) For a sufficiently large highly divisible integer $n$, $v_2(n) - v_3(n) \geq 2$
[hide= Claim 2 Proof]
By Claim 1 we can obtain a lower bound for $v_2(n)$. Assume that $v_2(n)$ is arbitrarily large.
Let $k = v_3(n)$
Case 1: $v_2(n) = k$
$n > \frac{8}{9} \cdot n$, but $(k+1)^2 < (k+4)(k-1) \implies d(n) < d(\frac{8n}{9})$
Case 2: $v_2(n) = k+1$
$n > \frac{8}{9} \cdot n$, but $(k+1)(k+2) < (k+5)(k-1) \implies d(n) < d(\frac{8n}{9})$
Both cases create a contradiction, and thus Claim 2 is proven
[/hide]

We can now finish the problem.
FTSOC assume there are infinitely many consecutive highly divisible numbers of the form $(n,kn)$
We can see that $d(2n) > d(n)$, and thus $k = 2$.

Let $n$ be an arbitrarily large number such that $(n,2n)$ are consecutive highly divisible numbers
Note that $\frac{3}{2}n < 2n$, but $(v_2(n))(v_3(n)+2) > (v_2(n)+1)(v_3(n)+1) \implies d(\frac{3}{2}n) > d(n)$
Thus, there must exist a highly divisible number between $n$ and $\frac{3}{2}n$, causing a contradiction

(b)
We can obviously generalize the first claim to all primes
[hide=Explanation]
Hint: It's the exact same logic but with $p$ instead of $2$
[/hide]

Let $h_1$, $h_2$, $\ldots$ be the sequence of all highly divisible numbers, and let $p$ be an arbitrary prime
By the generalized Claim 1, it is possible to take infinitely many $h_i$ such that $v_p(h_i) < v_p(h_j)$ for all $j > i$

We claim that for all such $i$, $p \cdot h_i$ is also highly divisible
[hide=Proof]
Since $d(p \cdot h_i) > d(h_i)$, it is enough to show that $d(p \cdot h_i) > d(s)$ for all highly divisble numbers $s$ satisfying $h_i < s < h_i \cdot p$
Note that $v_p(h_i) < v_p(s)$, and thus $v_p(h_i) + 1 \leq v_p(s)$
Furthermore, we can trivially see that $d(h_i) > d(\frac{s}{p})$
$\therefore d(p \cdot h_i) = d(h_i) \frac{v_p(h_i) + 2}{v_p(h_i) + 1} > d(\frac{s}{p}) \frac{v_p(s)+1}{v_p(s)} = d(s)$
[/hide]


How do you do Graphics & GUI programming in C++ on a mac?  I can't really make console programs forever!   :D

First off, I'm not all that familiar with graphics and GUI, but I do use a mac.  I think for graphics, OpenGL is compatible with macs.  Needless to say, DirectX is out of the picture.  For C++ GUIs, I'm not sure what is out there.  However, you could learn Java (which is more like C++ than you might think), since its portable and GUI is pretty much built-in.

Isn't OpenGL just for 3-D stuff?  And I perfer C++ to Java.

[quote="nr1337"]Isn't OpenGL just for 3-D stuff?[/quote]
According to [url=http://www.opengl.org/about/overview.html]OpenGL Overview[/url] you can also do 2D graphics with OpenGL, but I think it is mostly used for 3D stuff.


In the MathLinks board, there are links to "Profile" and "Preferences". If you click on them, they go to the exact same page. Is this the same for everyone? Shouldn't the "Preferences" link go to Preferences? Or the "Profile" link go to the place where you can see the number of posts, posts per day, interests, occupation, ...?

In the top right hand corner, the "News" link doesn't work either. This is all in the MathLinks board style.

I don't know, but when I tried Mathlinks, every link I clicked went to the same page, except on AoPS instead, but the URL is still mathlinks.ro, I know it's practically the same site, but its supposed to be a different skin, not just a different index page.


Let $ x;y;z$ be real numbers such that $ 0<z \leq y \leq z \leq 8$ and $ 3x\plus{}4y \geq max\{xy;\frac{1}{2}xyz\minus{}8z\}$
Find the maximum value of $ A\equal{}x^5\plus{}y^5\plus{}z^5$

[quote="pro_math"]Let $ x;y;z$ be real numbers such that $ 0 < z \leq y \leq z \leq 8$ and $ 3x \plus{} 4y \geq max\{xy;\frac {1}{2}xyz \minus{} 8z\}$
Find the maximum value of $ A \equal{} x^5 \plus{} y^5 \plus{} z^5$[/quote]
Delete it,here is a problem from MYM contest see here   http://www.mathlinks.ro/viewtopic.php?t=168836


Show that there are infinitely many pairs of positive integers $(m, n)$
such that $\frac{m+1}{n}+\frac{n+1}{m}$ is a positive integer.

Let $m=2d(d+a),n=2d(d-a)$, then $\frac{m+1}{n}+\frac{n+1}{m}=\frac{2d^{2}+2a^{2}+1}{d^{2}-a^{2}}$. It is equal 3 if $d^{2}-5a^{2}=1$. Pele's equation had infinetely many solutions: $d+a\sqrt 5 =(9+4\sqrt 5 )^{k}, \ k\in N$.


linear algebra
matrix
combinatorics proposed
combinatorics

This is pretty cute :).

Let $A=(a_{ij})$ be a real $m\times n$ matrix with at least one non-zero entry. For every $i\in\overline{1,m}$, let $R_i$ be the sum of the entries in the $i$'th row, and for every $j\in\overline{1,m}$, let $C_j$ be the sum of the entries in the $j$'th column.

Prove that we can find $i,j$ s.t. $a_{ij}>0,\ R_i\ge 0,\ C_j\ge 0$ or $a_{ij}<0,\ R_i\le 0,\ C_j\le 0$.

nice and easy, Call zone $A$ the union of squares that are the intersection of a positive row and positive column, Zone $B$ the union of the squares in the intersection of a negative row and a negative column, Zone $Y$ the union of squares of the intersection of a negative Row and a Positive column.

Let $S(T)$ be the sum of elements in a region $T$.

If we cant find neither of the situations we must have: $S(A)<0$ and $S(B)>0$ but also $S(A)+S(Y)>0$ (sum of positive columns) and $S(B)+S(Y)<0$ (Sum of negative Rows) which implies $S(A)>S(B)$ contradiction!

(to visualize it in a nicer way, permute rows and columns to order them acording to their sums in increasing order)


How many four-letter "words" are there with vowels(not 'y') in the middle two places and a consonant at the end?

Ok, so, for the first letter : 26 possibilities
The two middle letters : 5 possibilities
The last one : 26-5 = 21 possibilities
So, I'd tend to say : 26*5*5*21 = 26*25*21 = 13650

[hide]I assume the first letter can be either a vowel or a consonant.
So there are 26 different letter the first letter can be
2nd letter 5 (only vowels)
3rd letter-5 (only vowels)
4th letter 26-5=21 (only consonants)

The answer is \[26\times5\times5\times21=13650\][/hide]

[hide]any letter for the first, 26.
vowels for 2nd and 3rd, 5 each.
Last letter, consonant, 21.

26*5*5*21 = 13650[/hide]


inequalities
function
calculus
derivative
inequalities unsolved

Let $ x,y,z>0$ and $ x\plus{}y\plus{}z\le 1$.
Find the maximum value of $ P\equal{}(x^2\plus{}y^2\plus{}z^2)\sqrt{xyz}$

It is obvious that the maximum value is taken when $ x\plus{}y\plus{}z\equal{}1$.
The problem is equivalent to:
Find the maximum value of $ P(x,y,z)\equal{}(x^2 \plus{} y^2 \plus{}z^2 )\sqrt{xyz}$, where x,y,z are positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}1$.

[Claim 1] $ P(x,y,z) \leq P( \frac{x\plus{}y}{2}, \frac{x\plus{}y}{2}, z)$.

[hide="Proof"] The inequality is equivalent to $ (x^2 \plus{} y^2 \plus{}z^2 ) \sqrt{xyz} \leq ( (\frac{x\plus{}y}{2})^2 \plus{} (\frac{x\plus{}y}{2})^2 \plus{} z^2 )\sqrt{(\frac{x\plus{}y}{2})(\frac{x\plus{}y}{2})z}$

<=> $ (x^2 \plus{} y^2 \plus{} z^2 )\sqrt{xy} \leq ( \frac{(x\plus{}y)^2}{2} \plus{} z^2 )\frac{x\plus{}y}{2}$
<=> $ z^2 (\frac{x\plus{}y}{2} \minus{} \sqrt{xy} ) \plus{} ( (\frac{(x\plus{}y)^2}{2}) (\frac{x\plus{}y}{2}) \minus{} (x^2 \plus{} y^2 )\sqrt{xy} ) \geq 0$

It is trivial that $ z^2 (\frac{x\plus{}y}{2} \minus{} \sqrt{xy} ) \geq 0$.

It remains to prove that $ ( (\frac{(x\plus{}y)^2}{2}) (\frac{x\plus{}y}{2}) \minus{} (x^2 \plus{} y^2 )\sqrt{xy} ) \geq 0$, or equivalently,

$ (x\plus{}y)^3 \geq 4(x^2 \plus{}y^2 )\sqrt{xy}$.
Substituting $ x\equal{}a^2$ and $ y\equal{}b^2$, this inequality is equivalent to
$ (a^2 \plus{} b^2 )^3 \minus{} 4(a^4 \plus{}b^4 )ab \geq 0$
<=>$ (a\minus{}b)^4 (a^2 \plus{}b^2 ) \geq 0$ which is trivially true.

[/hide]

Due to the claim, for any $ (x,y,z)$ s.t. $ x\plus{}y\plus{}z\equal{}1$,

$ P(x,y,z) \leq P(\frac{x\plus{}y}{2} , \frac{x\plus{}y}{2} , z)$.

Substituting $ \frac{x\plus{}y}{2} \equal{} t$, $ z\equal{}1\minus{}2t$

Therefore $ P(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},z) \equal{}P(t,t,1\minus{}2t)$

This is a one-variable function on t -> use differentiation to find the max.value!

[quote="qwerty414"]
Substituting $ x \equal{} a^2$ and $ y \equal{} b^2$, this inequality is equivalent to
$ (a^2 \plus{} b^2 )^3 \minus{} 4(a^4 \plus{} b^4 )ab \geq 0$
[/quote]
which is wrong. Try $ a \equal{}7$ and $ b \equal{} 3.$ :wink:

qwerty414: Proving that we only have to consider the case $ x\equal{}y$ is the easy part of the problem... ;)

We have to find the max of $ f(t) \equal{} t^2(1\minus{}2t)(t^2\plus{}t^2\plus{}(1\minus{}2t)^2)^2 \equal{} \minus{}72t^7\plus{}132t^6\minus{}104t^5\plus{}44t^4\minus{}10t^3\plus{}t^2$, when $ 0 \le t \le \frac{1}{2}$.
Then $ f'(t) \equal{} \minus{}2t(3t\minus{}1)^2(28t^3\minus{}25t^2\plus{}9t\minus{}1)$. (This took really long time to calculate. :P )
$ t \to 28t^3\minus{}25t^2\plus{}9t\minus{}1$ is increasing, so it has at most one root. Let $ \alpha$ be that root. (We easily see that $ 0 < \alpha < \frac{1}{3}$). So $ 28t^3\minus{}25t^2\plus{}9t\minus{}1 \equal{} (t\minus{}\alpha)g(t)$, where $ g(t) > 0 \forall t \in \mathbb{R}$ Then $ f'(t) \equal{} \minus{}2t(3t\minus{}1)^2(t\minus{}\alpha)g(t)$. So when $ t \in [0;\alpha]$ we have $ f'(t) \ge 0$. When $ t \in [\alpha;\frac{1}{2}]$ we have $ f'(t) \le 0$.
So the max of $ P$ is $ \sqrt{f(\alpha)}$.. I'm still thinking about, how to get an exact value of it..

Sometimes wikipedia is useful^^
[url]http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method[/url]

[quote="Mathias_DK"]qwerty414: Proving that we only have to consider the case $ x \equal{} y$ is the easy part of the problem... ;)

We have to find the max of $ f(t) \equal{} t^2(1 \minus{} 2t)(t^2 \plus{} t^2 \plus{} (1 \minus{} 2t)^2)^2 \equal{} \minus{} 72t^7 \plus{} 132t^6 \minus{} 104t^5 \plus{} 44t^4 \minus{} 10t^3 \plus{} t^2$, when $ 0 \le t \le \frac {1}{2}$.
Then $ f'(t) \equal{} \minus{} 2t(3t \minus{} 1)^2(28t^3 \minus{} 25t^2 \plus{} 9t \minus{} 1)$. (This took really long time to calculate. :P )
[/quote]

You were wrong in calculation of f'. Actually

       $ f'(t) \equal{} \minus{}2t(3t\minus{}1)(6t^2\minus{}4t\plus{}1)(14t^2\minus{}8t\plus{}1)$

(Of course, f'(t) should be divided by $ 6t^2 \minus{} 4t \plus{} 1$!)

so, the maximum is achieved at $ x \equal{} \frac{4\minus{}\sqrt{2}}{14}$.

This maximum equal $ \frac{1}{2401}(10\plus{}\sqrt{2})(4\minus{}\sqrt{2})\sqrt{21\plus{}7\sqrt{2}}$ ~ 0.06833.


[color=cyan]How many pairs of non-negative integers (m, n) each sum to 1492 without any carries? [/color]

[hide]There are 3 possibilities for the units digit of the numbers- (2,0), (1,1) and (0,2) (We're looking for the number of pairs of digits that add up to the digit in the units place). Similarly we can count 10 sets for the tens digit, 5 for the hundreds digit and 2 for the thousands digit. Total number of ways : 5*2*3*10= 300. Divide by two to account for repeats = 150. [/hide]

[hide]your answer was fine at 300, it had already accounted for the repeats me thinks[/hide]

Nope -- there are no repeats, because it asks for ordered pairs.

lol, yeah, so the answer is [hide]300[/hide] right?

hrm, i think it is

I think [hide]300.[/hide] is correct.

Oops.

[hide]Let the two numbers be abcd and efgh.  Since there are no carries, a+e=1, b+f=4, c+g=9, d+h=2.

a and e can be (0,1) or (1,0), so two options here.

b can be 0 through 4 (with f as 4 through 0), so five options here.

c can be 0 through 9 (with g as 9 through 0), so 10 options here.

d can be 0 through 2 ("    " ) so 3 options here.

2*5*10*3 = 300.[/hide]

Amiright? Amiright?  *checks* My answer is consistent with Jen's.

Im sorry but I dont understand wat this question is asking.

How many pairs of non-negative integers (m, n) each sum to 1492 without any carries?

This isnt the way the problem was worded originally was it? Because when I read this problem, I immediately think that m + n = 1492 and to solve for how many pairs are possiblehere,which the answer is definiely not 300 or 150. So um, someone decipher please?

I'm not quite sure what you mean either Jelyman :) but maybe you missed the "no carries" bit? Ie which means you when actually add them up by hand you don't use any carries..

[color=blue]i think you answered the question TripleM.  like 6+5 carries over, but were trying to find the numbers which dont. thats why its not up there in the 700's as the answer[/color]

I wish problems this easy popped up more often, especially on AIMEs...  Well, this problem was easy if you knew what they meant by the word "carries".

u mean solutions like (1100, 300) is ok... but a solution like (700, 792) isnt ok because to add the 7's u gotta carry the ones to an extra tens place?

Yes, except for they typo (1100 + 300 isn't 1492).

eh... was just typing in haste. But im all good now, thanx.


function
geometric series
algebra
rational function
number theory proposed
number theory

Here's something fun... Let me know if there is already information available on this subject.

Given a character $ \chi$ and a complex number $ |s| < 1$, define the corresponding Dirichlet G series in the following way.

$ G(s, \chi) \equal{} \sum_{n\equal{}1}^{\infty} \chi(n)s^{n}$

How does this function behave on the unit disk? How can it be used?  :)

[quote="t0rajir0u"]In the literature, these are referred to as [url=http://mathworld.wolfram.com/DirichletL-Series.html]Dirichlet L-series[/url] and have a variety of uses, most notably to prove Dirichlet's theorem on arithmetic progressions.[/quote]

Yes, yes, I love L series tremendously, but look closely. You will see that I have swapped the base and exponent!  :)

[quote="t0rajir0u"]My apologies.  As you have written it, $ G(s, \chi)$ is a sum of several geometric series, hence a rational function, so I can't imagine the theory behind such sums is too involved.[/quote]

Very good point. So now we have each Dirichlet character corresponding to a rational function in this way. It kind of vaguely reminds me of this handy tool for exponential sums.

$ \tau(\chi) \equal{} \sum_{k \equal{} 1}^{n} \chi(k)e^{2\pi i k \over n}$

UPDATE

OK, we have:

$ G(s, \chi) \equal{} \sum_{k \equal{} 1}^{n} \chi(k){s^{k} \over 1 \minus{} s^{n}}$

Does this mean a G series is defined everywhere on the unit circle except for at the nth roots of unity, where n is the conductor of the character? I wonder if this is actually an analytic continuation to more than just the unit disk.

Additionally, we have:

$ {1 \over \phi(n)}\sum_{\chi \, (mod \, n)} G(s, \chi) \equal{} {s \over 1 \minus{} s^n}$


inequalities
integration
inequalities unsolved

Suppose $ n \in \mathbb{N}$ , $ x_0\equal{}0, x_i>0$ for $ i\equal{}1,2,...,n$, and $ \displaystyle\sum_{i\equal{}1}^{n}x_i\equal{}1$. Prove that:

$ 1 \le \displaystyle\sum_{i\equal{}1}^{n}\frac{x_i}{\sqrt{1\plus{}x_0\plus{}...\plus{}x_{i\minus{}1}}\cdot \sqrt{x_i\plus{}...\plus{}x_n}}<\frac{\pi}{2}$

This is an old problem:D
Lemma *with $ 0<sinx<siny<1$ we have (y>x>0)
$ \frac{siny-sinx}{cosx}\le y-x$
Let
$ x_{0}+...+x_{i-1}=sin(a_{i-1})$   (remember that   :$ 0<a_{i}<\frac{\pi}{2}$)
so we have:
$ f(x)=\sum \frac{sin(a_{i})-sin(a_{i-1})}{cos(a_{i-1})}$
using lemma *
 so we have
$ f(x)\le a_{n}-a_{0=\frac{\pi}{2}}$
the other ineq is simple(we can use AMGM here)

I have my own solution which uses different technique.

[b]Proof.[/b] Suppose that $ n\in\mathbb{N}$, $ x_0: = 0$ and $ x_i > 0$ for $ i\in\{ 1,2,\dots ,n\}$ with $ \sum_{i = 1}^{n}x_i$. Let us denote $ T_i: = \sum_{k = 0}^{i - 1}x_k$ for $ i\in\{ 1,2,\dots ,n\}$. This implies that $ T_i = 1 - \sum_{k = i}^{n}x_k\in [0,1)$. On the other hand, we can rewrite the original inequality in the form
$ 1\le\sum_{i = 1}^{n}\frac {x_i}{\sqrt {1 + T_i}\cdot\sqrt {1 - T_i}} < \frac {\pi}{2}.$

(1) First, we prove the left inequality. It is obvious that
$ x_i\le\frac {x_i}{\sqrt {1 - T_i^2}} = \frac {x_i}{\sqrt {1 + T_i}\cdot\sqrt {1 - T_i}},\qquad 1\le i\le n$.
Hence,
$ 1 = \sum_{i = 1}^{n}x_i\le\sum_{i = 1}^{n}\frac {x_i}{\sqrt {1 + T_i}\cdot\sqrt {1 - T_i}}$.
The proof of the inequality on the left side is done.

(2) For the second part we define the intervals $ J_i: = [T_i,T_{i + 1})$ for $ 1\le i\le n - 1$ and $ J_n: = [T_n,1]$. It is obvios that $ \bigcup_{i = 1}^{n}J_i = [0,1]$ and $ |J_i|=x_i$. Note also that
${ \frac {x_i}{\sqrt {1 + T_i}\cdot\sqrt {1 - T_i}} < \int_{J_i}}\frac {\mathrm{d}x}{\sqrt {1 - x^2}}.$
Thus
$ \sum_{i = 1}^{n}\frac {x_i}{\sqrt {1 + T_i}\cdot\sqrt {1 - T_i}} < \int_{0}^{1}\frac {\mathrm{d}x}{\sqrt {1 - x^2}} = \frac {\pi}{2}$.

The proof is complete.


Let $ A$ be a closed set in $ \mathbb{R}^n$ and smooth function $ f: A \rightarrow \mathbb{R}$. Does there exist a open set $ U$ s.t. $ A\subseteq U$ and a smooth function $ \widetilde{f}: U\rightarrow R$ s.t. $ \widetilde{f}(x)\equal{}f(x)$ for any $ x \in A$ ?

Define "smooth in $ A$" for closed $ A$. There are several valid definitions.

Only after posting it i realized that i didn't pay attention to the definition of the smoothness on some non-open set. So, $ f$ is smooth on $ A$ iff $ \forall a\in A: \exists V \in \mathcal{V}(a),\exists \widetilde{f}_{a}: V\rightarrow\mathbb{R}$ smooth and $ \widetilde{f}_{a}(x)\equal{}f(x),\forall x \in A\cap V$. I also found the answer to my question: using the partition of the unity the answer is yes(the fact that $ A$ is closed is very important).


LaTeX
number theory unsolved
number theory

k is a number more than 1 and p=6k+1 ,p is prime and m=2^p - 1
Prove that (2^(m-1) - 1)/127m is an integer.

Bad tile, latex missing... :mad:

Pierre.


Sequence
number theory
Divisibility
IMO Longlist
IMO Shortlist

Given a finite sequence of integers $a_{1},$ $a_{2},$ $...,$ $a_{n}$ for $n\geq 2.$ Show that there exists a subsequence $a_{k_{1}},$ $a_{k_{2}},$ $...,$ $a_{k_{m}},$ where $1\leq k_{1}\leq k_{2}\leq...\leq k_{m}\leq n,$ such that the number $a_{k_{1}}^{2}+a_{k_{2}}^{2}+...+a_{k_{m}}^{2}$ is divisible by
$n.$

[b]Note by Darij:[/b] Of course, the $1\leq k_{1}\leq k_{2}\leq ...\leq k_{m}\leq n$ should be understood as $1\leq k_{1}<k_{2}<...<k_{m}\leq n;$ else, we could take $m=n$ and $k_{1}=k_{2}=...=k_{m},$ so that the number $a_{k_{1}}^{2}+a_{k_{2}}^{2}+...+a_{k_{m}}^{2}=n^{2}a_{k_{1}}^{2}$ will surely be divisible by $n.$

A trivial one?
For $p=1, \cdots, n$, let $\displaystyle S_p = \sum_{i=1}^p a_i^2$.
If one of the $S_p$ is 0 modulo $n$ we are done. Otherwise, from the pigeon-hole principle, there exist $p<q$ such that $S_p = S_q$ mod[$n$]. Then, $\displaystyle  S_q - S_p = \sum_{i=p}^q a_i^2 = 0$ mod[$n$], and we are done.

Pierre.

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


inequalities
geometry
3D geometry
sphere

On a horizontal table with friction a ping-pong ball rolls purely towards a totaly smooth wall, perpendicularly to the plane of the wall. Can it happen that the ball collides with the wall exactly two times? (the time of the collisions are negligible, and the collisions are not perfectly elastic)

Is there friction between the wall and the ball :?:

It should be possible if the $v_{\text{after}}/v_{\text{before}}> 1/2$ inequality holds, according to my calculations. Is that ok? These are the translational speeds of the ball right before and right after the collision with the ball.

Good to see you again Djole :D
No your calculations are wrong then, since it cant happen that it collides two times. But we you have to prove that somehow :)

Just one question: what is the moment of inertia of the ball? I took it to be $mR^{2}$, which it obviously can't be  :mad:  Maybe that's why I'm wrong...

The ping-pong ball is a hollow sphere with a very thin border. So calculate the moment of inertia of a hollow sphere with very thin wall. :wink:

So it's $2mR^{2}/3$ (experience, eh ;)).

Ok, this is how I did it. Before hitting the wall, the ball will have an angular speed $\omega$ and a translational speed $v = \omega R$. Now, after the collision, suppose the ball has the translational velocity $v' = \eta v$, $\eta < 1$, but retains its angular velocity. Now, the friction will slow down both velocities until one of them changes direction and the equation $v' = \omega' R$ is reached. By Newton's 2nd law, we get $\omega' = \frac{2\omega-v'/R}{5}$, which is $>0$ when $2\omega R/3 > v'$, which may still hold, implying the condition $\eta < 2/3$.

Now, the ball will again collide with the wall, and its velocity will be $v'' = \eta v'$. I got that $15\eta^{2}+12\eta-4 < 0$ must hold for the ball to hit the wall again. This may hold for $\eta \lnsim 0.25$, which is still ok... I don't know, maybe I've screwed it up somewhere, I'm too tired to check... What's your solution?

Assuming the coefficient of restitution, $\eta$ following Djole's symbolism, less than one, as usual, we conclude that  we'll have  only one collision, if   $\,{2\over3}\,<\,\eta\,<\,1\,$,  or more than two, if   $\,\eta\,<\,{2\over3}\,$ .

dude. a simple arguments hold...

the velocity of the ball  before it hits the wall is $v>0$
after it hits the wall is $u <0$

it changed the signal. same for angular velocity, the signal changes.

So? There'll be sliding after the collision, so it's not that simple...


geometry
ratio
trigonometry
trapezoid
Pythagorean Theorem

Here's  the question(i dont know if its below intermediate level or something)

Find the height of a regular pentagon of side 1km.(answer not to be expressed in t-ratios,no trigono) :D

well, without trig it's hard to get an exact answer. i got about 1.54km as the height after dividing the pentagon into a trapezoid an triangle and working my way from there, probably plus minus 0.05km. does anyone know an exact answer without trig?

i DO know the solution,but i'm waiting for u guys to solve the problem. :lol:
i didnt get the question myself.

The length of a diagonal is $\frac{1+\sqrt{5}}{2}$, which can be found by using similar triangles.
[hide="how to find it"]Let's name the 5 vertices of the pentagon $A$, $B$, $C$, $D$, $E$.
Draw two diagonals $AC$ and $BD$, and let $P$ be the intersection.

Then we see that $\triangle BPC$ and $\triangle BCD$ are similar.
Also, $CD=PD$ since $\angle CPD=\angle PCD=72^\circ$.

Let $BP=a$. Then, $BD=BP+PD=BP+CD=a+1$
Since $\triangle BPC\sim\triangle BCD$,
we have $\frac{1}{a}=\frac{a+1}{1}\Longleftrightarrow a^2+a-1=0$

Solving for $a$, we get $a=\frac{-1+\sqrt{5}}{2}$.
So $BD=a+1=\frac{1+\sqrt{5}}{2}$ [/hide]

Then, using Pythagorean theorem, we can calculate the height:
$h=\sqrt{\left(\frac{1+\sqrt{5}}{2}\right)^2-\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3+\sqrt{5}}{2}-\frac{1}{4}}=\frac{\sqrt{5+2\sqrt{5}}}{2}\approx 1.54$

that's pretty much what i did too. i also used some formulas i found on math world. :D

I think I did it differently maybe wrong but my method is short.
extend the pentagon to make an isoscleses triangle. so then we have 2 equations.
then you have to equate the equations to form an equation.
then you can get the value of height

[quote="shadysaysurspammed"]ur  right  frt. :lol:

how do u understand what construction n stuff to make in such questions? :blush:[/quote]
well I think that is what geometry is all about :P  ....good guesswork :D


So who came this year? What did you think? I liked the problem sets this year (afterall, I did write several of the problems), and I think the rounds were really interesting. I had fun grading papers this year, although it is a lot of work.

So what were your thoughts?

You might want to check out the discussion of JHMT in the Maryland forum:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229571


[color=darkblue] Let $ x,y,a,b\in \mathbb{R}_ \plus{}$, $ x \not \equal{} y$, $ x \not \equal{} 2y$, $ y \not \equal{} 2x$, $ a \not \equal{} 3b$ and $ \dfrac{2x \minus{} y}{2y \minus{} x} \equal{} \dfrac{a \plus{} 3b}{a \minus{} 3b}$. Prove that $ \dfrac{x^2 \plus{} y^2}{x^2 \minus{} y^2}\ge 1$[/color]

Oops...
Sorry.... Move it please to ineq...

The inequality follows if $ x>y$. But it's given that $ \frac{2x\minus{}y}{2y\minus{}x}\equal{}\frac{a\plus{}3b}{a\minus{}3b}\Leftrightarrow 2xa\minus{}6xb\minus{}ya\plus{}3yb\equal{}2ya\minus{}ax\plus{}6yb\minus{}3xb$ $ \Leftrightarrow xa\equal{}ya\plus{}yb\plus{}xb\Leftrightarrow a(x\minus{}y)\equal{}(y\plus{}x)b>0$ $ \Rightarrow x>y$.


The sum of the squares of 5 consecutive positive integers is 750. Find the middle integer.

There is no solution for $ x\in\mathbb{Z}$.

[hide="Solution"]$ (x\minus{}2)^2\plus{}(x\minus{}1)^2\plus{}x^2\plus{}(x\plus{}1)^2\plus{}(x\plus{}2)^2\equal{}750$

$ \left(x^2\minus{}4x\plus{}4\right)\plus{}\left(x^2\minus{}2x\plus{}1\right)\plus{}x^2\plus{}\left(x^2\plus{}2x\plus{}1\right)\plus{}\left(x^2\plus{}4x\plus{}4\right)\equal{}750$

$ 5x^2\plus{}10\equal{}750$

$ x^2\equal{}148$

$ x\equal{}\boxed{\pm2\sqrt{37}}$[/hide]

[hide="solution"]
let the middle integer be $ x$. now write the other integers in terms of $ x$. you get the following algebraic equation:
$ (x\minus{}2)^{2} \plus{}(x\minus{}1)^{2} \plus{} x^2 \plus{} (x\plus{}1)^{2} \plus{} (x\plus{}2)^{2}\equal{}750$
evaluating the binomial expressions and gathering terms we get
$ 5x^2 \plus{}10\equal{}750 \Longrightarrow x^2\equal{}148$?????
that doesn't make sense because $ x$ will not be an integer, whereas the problem says it is.
what mistake did i make?[/hide]

is there a way to combine $ (x\minus{}1)^2$ and $ (x\plus{}1)^2$, and $ (x\minus{}2)^2$ and $ (x\plus{}2)^2$?

[quote="abacadaea"]is there a way to combine $ (x \minus{} 1)^2$ and $ (x \plus{} 1)^2$, and $ (x \minus{} 2)^2$ and $ (x \plus{} 2)^2$?[/quote]

Umm...both of the posts above you did that already, just an FYI.


The polynomial    $p(x)=ax^2+bx+c=a(x-x_1)(x-x_2) \; ,\; ( a > 0)$ , has properties :
1) $a , b ,c$ are integers ,
2) $0 < x_1 < x_2 < 1$ ,
3) $|p(x)| \le 1 \;$  for all $x$  in $[0,1]$ .
[color=green] [b]Find  all polynomials , of degree two,   having properties 1)--3).  [/b] [/color]

[hide="Beginning"]The following conditions must be fulfilled:

1. $|f(0)|\le 1$
2. $|f(1)|\le 1$
3. $|{D\over{4a}}|\le 1$
4. $D\ge 0$

Hence $|c|\le 1, |a+b+c|\le 1,|{b^2\over {4a}}-c|\le 1, b^2\ge 4ac$.

Since $a,b,c$ are integers, $c$ can be only $0$ or $\pm1$ and $a+b+c$ can be only $0$ or $\pm 1$. And now it comes down to casework... (for which i'm lazy right now).[/hide]

[quote="Farenhajt"][hide="Beginning"]The following conditions must be fulfilled........ Sinnce $a,b,c$ are integers, $c$ can be only....[/hide][/quote] [hide]The solution is not complete ![/hide]


An AIME question asked a question that involved finding the roots of z :^6: +z :^4: +z :^3: +z :^2: +1=0. First of all, how would I have just seen to factor it to (z :^2: -z+1)(z :^4: +z :^3: +z :^2: +z+1), and second, did anybody else know the trick involving roots of polynomials with all coefficients the same? Are there any other tricks that I should know?

There's a great deal of symmetry, so divide by z3 and make it a polynomial in z+1/z. That's the easiest thing to do. You might just find the factorization if you play around with that.

The solution to this question was in a topic called "AIME Polynomials" or something like that in the intermediate topics, so check that out.


superior algebra
superior algebra unsolved

Show that the intersection of all maximal left ideals is a 2-sided ideal. in a ring R

ooops... sorry for my ignorance.
Thanks for the link though, quite instructive.  :oops:


Determine the prime factorization of $ 5^5\minus{}5^3$ through factorizing the expression.

hello, we get $ 5^5\minus{}5^3\equal{}3000\equal{}2^3\cdot3\cdot5^3$.
Sonnhard.

Just to let you know, please don't revive threads unless you have a good reason :)

[hide=hint]use the prime factorization after you solved $5^5 - 5^3$[/hide]


It has been discussed many times before...$0.99999\cdots=1$


LaTeX
quadratics
geometry
3D geometry
algebra
quadratic formula

These are two challenge problems that I'm not totally sure how to solve. I'd like some help on them  :P

find all real numbers r that satisfy the equation below.

(1/r3 + 7) − 7 = −r3/r3 + 7.

and this one,which is pretty similar but also involves radicals...

find all values of x that satisfy

3sqrt/(x + 5) − sqrt(x + 5) = 2x + 1/ sqrt(x + 5)

sorry my $ \text{\LaTeX}$ is really bad so I didn't put the problems in that form... It would probably just be confusing. i think you'll be able to understand what I wrote though. Thanks for the help. :D

Could you put this in latex please? Its hard to understand your plain text.

Whoops: sorry :blush: didn't see your last two sentences. And sorry, but i can't understand your plain text.

What does r3 mean?  Does it mean $ r^{3}$?  And what are the denominators of both fractions?

In the second problem, is it $ 3\sqrt{x\plus{}5}\minus{}\sqrt{x\plus{}5}\equal{}2x\plus{}\frac{1}{\sqrt{x\plus{}5}}$, or is that first part something different?

It's hard to solve your problems if we can't understand them.

yeah. I think that's what he means for the second problem. Anyways, if its not, then too bad. Here's my solution:

Assume $ \sqrt{x\plus{}5}$ is $ y$. The equation then becomes:
$ 3y\minus{}y\equal{}2(y^2\minus{}5)\plus{}\frac{1}{y}$
$ 2y\equal{}2y^2\minus{}10\plus{}\frac{1}{y}$
$ 2y^2\equal{}2y^3\minus{}10\plus{}1$
$ 2y^3\minus{}2y^2\minus{}9\equal{}0$
This cannot be factored, and we got ourselves a complicated problem here.

You could try replacing $ y^2$ with z and then solving for z using the quadratic formula, then finding y, and then finding x.

[quote="Math Champion"]his cannot be factored, and we got ourselves a complicated problem here.[/quote]
It can be, just not necessarily in terms of real numbers.  See [url=http://en.wikipedia.org/wiki/Cubic_equation#General_formula_of_roots]here[/url].  Not pretty, but it exists.

Um. Yeah. Wow. I'm not gonna even go that far... but actually that's not my question. I understand that my expressions were confusing and I'll have a proper $ \text{\LaTeX}$ version as soon as I can... sorry about that guys...

Ok this is the second problem in $ \text{\LaTeX}$ form.

$ \frac{3}{\sqrt{x+5}}-\sqrt{x+5}=2x+\frac{1}{\sqrt{x+5}}$

Yeah. I got it right... :lol:

The first one...

$ \frac{1}{r^3+7}-7=\frac{r^3}{r^3+7}$

That's it. Sorry for any confusion. If you're not mad at me or anything then I would like your help. Thanks!

[hide="solution"]
To stay away from cubed numbers until the end, let $ x=r^3.$ The equation becomes \[ \frac{1}{x+7}-7=\frac{x}{x+7}\Rightarrow1-7x-49=x^2+7x\Rightarrow x^2+14x+48=0.\] The solutions to the quadratic are $ x\in\{-8,-6\}.$ Since $ x=r^3,$ the solutions for $ r$ are ${ r\in\{-2,\sqrt[3]{-6}\}}.$ Checking, only $ \sqrt[3]{-6}$ works, so $ -2$ is extraneous. That means the only solution for $ r$ is $ \boxed{\sqrt[3]{-6}}.$
(I think I did something wrong because I don't think $ -2$ should be extraneous.)[/hide]

(Sharp intake of breath) Bonnie, I thought you were better than this! Anyway, RoFlLoLcOpT, -2 should not be included (plug it into the equation) and therefore cube root of -6 is the answer.

----------------------------------------------------------------------------------------------------------
Visit the High School Forum!

You're supposed to keep Challenge Set discussion to the message board :wink:

I don't know, these are a little too easy to be Challenge Problems by middle school standards... I don't think they should be moved unless you're referring to something else :wink: )

These are Challenge Set Problems from the Algebra 1 class :wink:

You know what guys? I figured this out a while ago...sorry lol. RoFlLoLcOpT, -2 isn't extraneous, if you think it is just plug it into the original equation. it's one of the answers...and so is cube root of -6.

This is what I have:

you start with

$ \frac {1}{r^{3} \plus{} 7} \minus{} 7 \equal{} \frac {r^{3}}{r^{3} \plus{} 7}$

then you go on to add seven and subtract the $ \frac { \minus{} r^3}{r^{3} \plus{} 7}$ so that you have the whole number on one side and the fractions combine on the other side. It's pretty easy from there. Cross multiply...

$ \frac {1 \plus{} r^{3}}{r^{3} \plus{} 7} \equal{} \frac {7}{1}$

and then you can use algebra to find r, which in this case is $ r^3 \equal{} \minus{}8$ and then $ r \equal{} \minus{} 2$. nemobengalII is right too...

you know what RoFlLoLcOpT? Actually, the equation doesn't look like this. Maybe I missed it when I wrote the equation the first time, but I rechecked and it looks like this. It's a very small thing but it changes things.
This is what it's supposed to be.

$ \frac{1}{r^{3}+7}-7 =\frac{-r^{3}}{r^{3}+7}$

what happens when you plug -2 in for r now?

$ \frac{1}{(-2)^{3}+7}-7 =\frac{-(-2)^{3}}{(-2)^{3}+7}$

${ \frac{1}{-1}-7=\frac{-(-8)}{-1}}$

And that turns out to be $ -8=-8$

Sorry for not putting that earlier... must have missed that little subtraction sign when putting the equation in $ LaTeX$.

Set a common denominator to get
$ 1\minus{}7r^{3}\minus{}49\equal{}\minus{}r^{3}\Rightarrow \minus{}48\equal{}6r^{3}\Rightarrow \minus{}8\equal{}r^{3}\Rightarrow r\equal{}\sqrt[3]{\minus{}8}$.  This has three solutions by the fundamental theorem of algebra, which are $ r\equal{}\minus{}2,1\plus{}i\sqrt{3},1\minus{}i\sqrt{3}$.

I think it's unlikely that they would talk about complex no.s at the middle school class level... and the problem said reals.

[quote="SuperNerd123"]I think all the are looking for is -2 seeing as they don't go over imaginary numbers till like 8th grade or something[/quote]

Well in my school the advanced math class learns it in 7th. Also, the reason you got -2 as extraneous was that the equation was written wrong.


The integer $ a$ is the sum of three consecutive positive integers and is divisible by 5. What is the least possible value for $ a$?

$ a\equal{}(k\minus{}1)\plus{}k\plus{}(k\plus{}1)\equal{}3k$ then the minimum $ a$ is with $ k\equal{}$ the minimum multiple of $ 5$ then $ a\equal{}15$


function
calculus
derivative
real analysis
real analysis unsolved

What is wrong with the following argument?

Suppose $w=f(x,y,z)$ and $z=g(x,y)$.

By the chain rule,

$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}+\frac{\partial w}{\partial y} \frac{\partial y}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x}$.

Hence, $0=\frac{\partial w}{\partial z} \frac{\partial z}{\partial x}$, and so $\frac{\partial w}{\partial z}=0$ or $\frac{\partial z}{\partial x}=0$.

the problem is "What is wrong with the following argument"  :D

take $f(x,y,z)=x^3+y^2-z$ and try what u did

[quote="Chang Woo-JIn"]What is wrong with the following argument?

Suppose $w=f(x,y,z)$ and $z=g(x,y)$.

By the chain rule,

$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}+\frac{\partial w}{\partial y} \frac{\partial y}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x}$.[/quote]

This part doesn't look right  :huh:

The chain rule is dangerous if you don't distinguish functions and variables.

Those derivatives you have labeled "$\frac{\partial w}{\partial x}$" don't mean the same thing as each other.

Yes, you are right jmerry.

$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}+\frac{\partial w}{\partial y} \frac{\partial y}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x}$ is right.

however, we cannot say $\frac{\partial w}{\partial x}$ on left side and $\frac{\partial w}{\partial x}$ on right side are equal.

They are different because $\partial x$ on left side mean just a variable which is related with $y,z$ in $f$.

and $\partial x$ on right side mean a variable of $f$ which is not related with $y,z$ in $f$.  :D  :D

It was a comic problem.  :lol:


Is AP Physics equivalent to the college Physics 1?

There are several separate issues here.

The first issue is that "AP Physics" is not a unique identifier. There are two separate AP Physics tracks (B and C) leading to three exams. The second issue is that "college Physics 1" is also not a unique identifier - there can be multiple physics tracks at colleges, as will be explained below.

AP Physics B does not make use of calculus. The material covered in the AP Physics B exam most closely resembles a year-long sequence with a title like "General Physics" that either has no calculus prerequisite, or possibly is meant to be taken with a "Survey of Calculus" or "Short Calculus" course. Such courses may satisfy the major requirements for biological science degrees and may satisfy the physics requirement for medical schools or similar health professional schools. Such a course [b]does not count[/b] towards a degree in engineering or the physical sciences (physics, chemistry, geology).

AP Physics C does make use of calculus - you should not be in an AP Physics C course unless you're also in AP Calculus, preferably Calculus BC. The course leads to two separate exams: Physics C Mechanics, and Physics C Electricity and Magnetism. The material covered in these two exams most closely resembles the first two semesters or first two quarters of the sequence of physics courses that [b]does[/b] count towards a degree in engineering or physical science.

Then there's the third issue: each college and each department in each college decides for itself what credit should be given for AP exams. Physics departments are often among the most skeptical of AP credits, and there are departments that don't accept any of them, even for the purposes of placement. You would need to check that with the institution you propose to attend.


inequalities
integration
function
calculus
analytic geometry
graphing lines
slope

First, some notation.
Suppose that $\mu$ is $\sigma$-finite measure on the space $X$. Suppose that $T$ is a measure-preserving operator on $X$.
For $f \in L^{1}(\mu)$, let
$A_{n}[f](x) = \frac{1}{n}\sum_{k=0}^{n-1}f(T^{k}x)$
Finally, define
$\mathcal{M}[f](x) = \sup_{n \in \mathbb{N}}A_{n}[f](x)$

The maximal inequality or maximal ergodic theorem states that for such $f$,
$\mu(\{x : \mathcal{M}[f](x) > \lambda\}) \le \frac{C\|f\|_{1}}{\lambda}$

In my analysis class, we discussed this inequality for proving the pointwise ergodic theorem (Birkhoff's one). In class we proved that
[hide="the constant is le 2"] ---------------------
The proof is actually somewhat complicated. First, we prove it in the case that $X = \mathbb{Z}$ and $\mu$ is the counting measure $\#$.

For given $\lambda$, let $\Omega$ be the set of all $n$ such that $\mathcal{M}[f](n) > \lambda$.

In this case, let $F$ be the collection of all intervals over which the sum is taken for given $n$. That is, $I_{j}= [n_{j}, n_{j}+1, ..., n_{j}+k_{j}-1]$, where $\frac{1}{k_{j}}\int_{n_{j}}^{n_{j}+k_{j}-1}f(x)d\#$ is the maximal average. It is not too hard to see why all of these intervals are of finite length, and there are only finitely many intervals in $F$. We claim that there exists a subcollection $G \subset F$ of disjoint intervals with the property that
${|\bigcup\{I \in G\}| \ge \frac{1}{2}|\bigcup_{n_{j}\in \Omega}I_{j}}\ge \frac{1}{2}|\Omega|$
The second inequality is obvious from the fact that each $n_{j}\in I_{j}$. The first requires some proof.

Note that if any three intervals intersect, then one of them must be contained within the union of the other two. So discard all of the intervals in $F$ that are thus contained, so that any point $m$ is in at most two intervals. Now we can divide the remaining intervals into two disjoint subcollections. The union of these still covers the union of everything in $F$, so at least one of them must cover half of the points of $F$. This is the subcollection we want.
[This is basically the Besicovitch covering theorem.]

Now suppose $J_{k}$ are the intervals in $G$.
Then we have
$\|f\|_{1}= \sum_{n}f(n) \ge \sum_{k}\sum_{n \in J_{k}}f(n) \ge \sum_{k}|J_{k}|\lambda \ge \frac{\lambda}{2}|\Omega|$
whence
$\#(\Omega) \le \frac{2\|f\|_{1}}{\lambda}$
as desired.

We prove the general case by defining the "partial max" $\mathcal{M}_{n}[f](x)$ by the max over the first $n$ ergodic averages and proving that the measure of the set of $x$ where that thing is greater than $\lambda$ is uniformly bounded wrt $n$ by the same bound as before.

A sketch of this (it's not important to my question) is as follows:

Define a function on $\{0, 1, \ldots, N\}$ ($N$ is arbitrary and large) $\mathcal{M}_{n}[f](T^{k}x)$ for fixed $x$. Then we apply Fubini's theorem to find the measure of the set in the original question as the limit as $n \to \infty$ of the integral wrt $x$ of the measure of the same set except for fixed $x$ as a function of $k$. We can apply the thing I just proved.
----------------[/hide]

However, the optimal constant is actually 1, and I don't see how the above proof can be modified to improve the constant any further. Can anyone suggest a different way to prove this inequality that will show $C=1$ suffices?

Suppose $\mu$ is the counting measure on $\mathbb Z$. Without loss of generality $f\ge 0$. The inequality holds with constant $1$ if $f$ is supported at one point. Now look at what happens when you add two (or more) nonnegative functions together: the $L^{1}$ norms are simply added, what about the measures of upper level sets of the maximal function? *

(*) Disclaimer: I do not claim that this leads anywhere.

I tried your idea, but I can't see how to go with it. I come to the following problem:

Let $f(n) = f_{1}(n)+f_{2}(n)$ where $f_{1}$ and $f_{2}$ are supported on sets of size less than $k$. Suppose that the inequality is true for these functions and constant 1. We want to prove that the same holds for $f$. Then we win, right?

Hau-kay. So we might make the estimate $\mathcal{M}[f](n) \le \mathcal{M}[f_{1}](n)+\mathcal{M}[f_{2}](n)$ -- obviously equality holds if the sup for $f_{1}$ and $f_{2}$ occur at the same length average for every $n$ and if that's not the case the inequality is strict. But then we have this problem,

$\#\{n | \mathcal{M}[f_{1}](n) > \lambda \vee \mathcal{M}[f_{2}](n) > \lambda\}\le \#\{n | \mathcal{M}[f_{1}](n)+\mathcal{M}[f_{2}](n) > \lambda\}$

$\#\{n | \mathcal{M}[f_{1}](n) > \lambda \vee \mathcal{M}[f_{2}](n) > \lambda\}\le \#\{n | \mathcal{M}[f_{1}](n) > \lambda\}$
$+\#\{n | \mathcal{M}[f_{2}](n) > \lambda\}$*

They are both the same way so I can't conclude anything!

But I don't know how to make a better estimate of $\mathcal{M}[f]$ in terms of $\mathcal{M}[f_{1}]$ and $\mathcal{M}[f_{2}]$ because we can find $f_{1}$ and $f_{2}$ which do not satisfy $f_{1}= kf_{2}$ for a constant $k$ but do satisfy $\mathcal{M}[f] = \mathcal{M}[f_{1}]+\mathcal{M}[f_{2}]$! For example,

$f_{1}(n) = \begin{cases}1 \text{ if }n = 0 \\ 1 \text{ if }n = 6 \\ 0 \text{ otherwise}\end{cases}$
$f_{2}(n) = \begin{cases}1 \text{ if }n = 0 \\ \frac{11}{10}\text{ if }n = 6 \\ 0 \text{ otherwise}\end{cases}$

I know this isn't the simplest example, but it's the one I happened to come up with. So, what am I missing? Or does this approach seem a dead end?

*By the way, does anyone know how to make a long equation spill over to a new line in a nice way?

Well, I was sleepy when I posted that thing.  :blush:

This is a rising sun argument due to F.Riesz. Let $F$ be an antiderivative of $f$, that is, $F(n)=\sum_{k< n}f(k)$. Then $\Omega=\{n\colon \exists k\ F(n+k)-F(n)>\lambda k\}$. Imagine sunrays coming from the right with slope $\lambda$. The set $\Omega$ is the part of the graph of $F$ that lies in the shadow. It is a union of intervals $[a,b]$ such that neither $a-1$ nor $b+1$ is in $\Omega$. Since $a$ is shadowed but $b+1$ is not, it follows that $F(b+1)-F(a)>\lambda(b+1-a)=\lambda\#[a,b]$. (Draw a picture). Finally, take the sum over all such (disjoint!) intervals.

That rising sun trick is so cool! We used it last semester in analysis proving something about functions of bounded variation being differentiable a.e.

Thanks for your help. We didn't actually need this but the prof mentioned that a better constant could be found, so I was wondering how. :D


sorry,first post just wanted to try & find out how to solve my problem

*cough* How can we colve your problem without knowing it?


a,b,c - lengths of triangle sides.

Prove that $ a^2b(a\minus{}b)\plus{}b^2c(b\minus{}c)\plus{}c^2a(c\minus{}a)\ge0$

[hide="Incorrect"]
By Muirhead, $ a^3b \plus{} b^3c \plus{} c^3a\geq a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$, so

$ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a)\geq0$.
[/hide]

Edit: Sorry. :oops:

[quote="math154"]By Muirhead, $ a^3b \plus{} b^3c \plus{} c^3a\geq a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$, so

$ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a)\geq0$.[/quote]

A question about Muirhead:
Is $ \displaystyle\sum_{sym} a^2b^2c^0 \equal{} \frac {2}{3!}(a^2b^2 \plus{} b^2c^2 \plus{} a^2c^2)$ or just $ \frac{1}{3!}a^2b^2 \plus{} b^2c^2 \plus{} a^2c^2$
That is, when you count the permutation $ a^2b^2c^0$ do you also count $ b^2a^2c^0$??

[quote="brconterato"]$ \displaystyle\sum_{sym} a^2b^2c^0 \equal{} a^2b^2c^0 \plus{} a^2c^2b^0 \plus{} b^2a^2c^0 \plus{} b^2c^2a^0 \plus{} c^2a^2b^0 \plus{} c^2b^2a^0$.

In general,
$ \displaystyle\sum_{sym} F(x,y,z) \equal{} F(x,y,z) \plus{} F(x,z,y) \plus{} F(y,x,z) \plus{} F(y,z,x) \plus{} F(z,x,y) \plus{} F(z,y,x)$[/quote]

Okay, thanks

[hide]Hint:
Let a=x+y, b=y+z and c=z+x with x,y,z are three lengths that are divided by the inscribed circle in the triangle with three sides a,b,c.
By the way, it's the IMO problem!(very famous. There's a 1-line solution with assumption c=min(a,b,c)) :) [/hide]

Thanks ghjk, but your hint is not enough for me :(

I tried it as well, and it didn't work (Rearrangement on $ a,b,c$ and $ a^2b,b^2c,c^2a$--note that if $ a\geq b\geq c$, we don't have to have $ a^2b\geq b^2c\geq c^2a$; $ a^2b\geq c^2a\geq b^2c$ is possible as well).

You have to use the restraint that $ a$, $ b$ and $ c$ are sides of a triangle. I think rearrangement works after you use the substitution ghjk suggested -- but I have seen it done by Cauchy.

[hide="The relevant inequality is..."]

$ \frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x} \ge x \plus{} y \plus{} z$, which follows either by Cauchy:

$ (x \plus{} y \plus{} z)(\frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x}) \ge (x \plus{} y \plus{} z)^2$

or rearrangement, with the assumption that $ x \le y \le z$.

[/hide]

[quote="ghjk"][quote="tim26270746"]This problem is quite easy for an IMO problem.
I used rearrangement inequalities.[/quote]
Can you show your proof? I don't think it works though. :roll:[/quote]

I think I got a step wrong... sorry... I'll check it again


group theory
abstract algebra
linear algebra
matrix
logarithms
superior algebra
superior algebra solved

[quote="alekk"]prove that $exp(Mn(C))=Gln(C)$.[/quote]

$G=Gln(\mathbb{C})$ is a connected algebraic group and $H = exp(Mn(\mathbb{C}))$ is
an algebraic subgroup of $G$. Check that the tangent space at $Id$ is the same - as Lie algebras - then check the dimensions, the result follows.

Fabrice

[quote="alekk"]well... it's far too complicated for me  :D  do you have a nice proof
with local inversion or something like that ? I'm sure there is one such proof.
(mine is with matrix logarithm but is quite ugly !)[/quote]

What we loose by using less thoery we have
to recover it by calculation...
Thus your matrix logarithm is one of the
"straightforward" tool you'll have to use
if you don't any additional structure.

Fabrice

of course  :D
but my teacher told me there was a proof with local inversion but didn't give that proof.
Do you knwo this proof ?
thanx


inequalities
function
inequalities proposed

Hang on a second: what if $n=2, a_1=a_2=0.5$? Then we need $\sqrt{0.5+0.5} \leq (\sqrt 2)(0.5)$, i.e. $1 \leq \frac{1}{\sqrt 2}$   :?

Did you mean $\sqrt{k(a_k-a_{k+1})}$ for the RHS terms? If so, I think I have a solution:

The result is clearly true if $a_i=0$ for all $i$. Now we perform a number of "steps": in each step, we increase $a_i$ by $x$ for each $i \leq k$ for some natural $k \leq n$ and real $x$. It is clear that we can reach any configuration by the right combination of steps. It remains to show that each step preserves the inequality.

A step increases the square of the LHS by $kx$. The only term on the RHS to change is $\sqrt{k(a_k-a_{k+1})}$. This term started at zero and we only need one step for each $k$, so we can assume that its value before the step is zero. So the step increases the RHS by $\sqrt{kx}$.

Now let $y$ and $z$ be the total values of the RHS and LHS before the step. The values of their squares after the step are $y^2+kx$ and $(z+\sqrt{kx})^2$. But now $(z+\sqrt{kx})^2 \geq z^2 + kx \geq y^2+kx$.

ooops, i made a terrible mistake.... I have edited it now....

Why everyone that tries to post this problem makes a mistake? I think i have a much more nicer solution than the official one, i ll post it in a while

[quote="darij grinberg"]In fact, this was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=6584 (a really funny discussion, by the way).

  Darij[/quote]

I remember at that time it was a little more frustrating than amusing :) .

here is the solution i wanted to share and i like it because it uses weighted jensen, wich i think is the first time i use it.

Let b_k= :( :sqrt:  k- :sqrt: (k-1))/ :sqrt: n. then clearly, the right side is:

  :sqrt: n*:Sigma:  :( :sqrt:  a_k)*b_k. Now,  :Sigma: b_k=1 (it telescopes) So we are going to do jensen with the function  :sqrt: x and weights b_1,b_2....b_n.

We obtain: RHS>=  (:sqrt: n)(:sqrt: ( :Sigma: a_k*b_k)) (*)

Now b_1>b_2>b_3...>0 again by the convexity of  :sqrt: x and also:
       a_1>=a_2>=....>=a_n>=0. So:

:Sigma: a_k*b_k>= ( :Sigma: a_k)(  :Sigma: b_k)/n= ( :Sigma: a_k)/n. By tshevichev.

 Replacing this in (*) we get the desired inequality.

Well, well...I like the fact that [u]pascual's[/u] solution seemed a little too sad with the sad faces intermingled with the formulas  :D  :D  :D

[quote="pascual2004"]We obtain: RHS>=  (:sqrt: n)(:sqrt: ( :Sigma: a_k*b_k)) (*)[/quote]

Since $\sqrt x$ is a concave function, the inequality should be the other way round.

So that makes my solution totally wrong.....thanks anyway


AMC
AMC 10
AMC 12
AIME
ARML
AIME I
AIME II

High School Indiana Math League Dates

Oct. 26
Nov. 30
Jan 11
Feb. 8
March 8
April 5
[url]http://www.mathleague.com/reports/2004_05/IN1.HTM[/url]

AMC10 and AMC12
Feb. 1 or Feb. 16

AIME I
March 8

Aime II
March 22

Huntington College Invitational
April 13

ICTM
April 23

ARML
June 5
[url]http://www.rose-hulman.edu/~rickert/ARML/IN/[/url]

Pistons vs. Pacers
Dec. 25

If you have any other dates that you would like me to add, send me a message.

It is, and of course, this forum is dead.


calculus
integration
vector
function
real analysis
trigonometry
advanced fields

$u,v$ vectors of norm $1$ in $R^n$

Find $A(n)$ such that

$\int_{R^n}sgn(u|x).sgn(v|x)e^{\frac{-||x||^2}{2}}dx =A(n)\arcsin((u|v))$

where $(..|..)$ is scalar product, $sgn$ is sign function, $dx$ Lebesgue
measure on $R^n$

$A(n)=2^{n/2+1}\pi^{n/2-1}$?

It is clear that we only need to solve problem in case $n=2$, $u=(1,0)$, $v=(\cos\phi,\sin\phi)$. So we obtain a couple of nice integrals...

You get the right expression of $A(n)$ Myth :)


$ BCDEF$ is a cyclic pentagon.$ CB$ and $ EF$ are produced to meet at $ A$.

If $ \angle EAC\equal{} 30^o,\angle EDF\equal{} 36^o,\angle BDC\equal{} 40^o\angle DBC\equal{} x^o$and $ \widehat{DE}\equal{}3\widehat{DC}$, find $ x$.
. :gathering:

This is definitely not Pre-Olympiad material (probably HSB).

Anyway,...

[hide="Solution"]
$ \angle C = 140^\circ - x^\circ$ and $ \angle E = 144^\circ - 3x^\circ$, so

$ 2(144^\circ - 3x^\circ) + 2(140^\circ - x^\circ) - 360^\circ = 208^\circ - 8x^\circ = \widehat{BF}$.

$ \widehat{CE} = \widehat{CD} + \widehat{DE} = 2x^\circ + 6x^\circ = 8x^\circ$, and since

$ \angle A = 30^\circ = \frac {\overarc{CE} - \overarc{BF}}2$, we find that

$ 60^\circ = 16x^\circ - 208^\circ\implies x = \boxed{16.75 = \frac {67}4}$.
[/hide]

Edit: Yes, it was a typo. :)

Edit2: Again, a typo. :huh:

[quote="math154"]This is definitely not Pre-Olympiad material (probably HSB).

Anyway,...

[hide="Solution"]
$ \angle C = 144^\circ - 3x^\circ$ and $ \angle E = 140^\circ - x^\circ$, so

$ 2(144^\circ - 3x^\circ) + 2(140^\circ - x^\circ) - 360^\circ = 208^\circ - 8x^\circ = \widehat{BF}$.

$ \widehat{CE} = \widehat{CD} + \widehat{CE} = 2x^\circ + 6x^\circ = 8x^\circ$, and since

$ \angle A = 30^\circ = \frac {\overarc{CE} - \overarc{BF}}2$, we find that

$ 60^\circ = 16x^\circ - 208^\circ\implies x = \boxed{16.75 = \frac {67}4}$.
[/hide][/quote]

$ \boxed{\angle C = 144^\circ - 3x^\circ ,\angle E = 140^\circ - x^\circ}$

Is typo error?

[quote="math154"]This is definitely not Pre-Olympiad material (probably HSB).

Anyway,...

[hide="Solution"]
$ \angle C = 140^\circ - x^\circ$ and $ \angle E = 144^\circ - 3x^\circ$, so

$ 2(144^\circ - 3x^\circ) + 2(140^\circ - x^\circ) - 360^\circ = 208^\circ - 8x^\circ = \widehat{BF}$.

$ \widehat{CE} = \widehat{CD} + \widehat{CE} = 2x^\circ + 6x^\circ = 8x^\circ$, and since

$ \angle A = 30^\circ = \frac {\overarc{CE} - \overarc{BF}}2$, we find that

$ 60^\circ = 16x^\circ - 208^\circ\implies x = \boxed{16.75 = \frac {67}4}$.
[/hide]

Edit: Yes, it was a typo. :)[/quote]

May I be stupid ,would you please offer details from:

$ \boxed{2(144^\circ - 3x^\circ) + 2(140^\circ - x^\circ) - 360^\circ = 208^\circ - 8x^\circ = \widehat{BF}}$.

$ \boxed{\widehat{CE} = \widehat{CD} + \widehat{CE} = 2x^\circ + 6x^\circ = 8x^\circ}$ ------Is it a typo?

 :D  :o

Since $ \angle BCD,\angle DEF$ are [url=http://en.wikipedia.org/wiki/Inscribed_angle]inscribed angles[/url], we can see that $ 2\angle BCD\equal{}\widehat{BF}\plus{}\widehat{FE}\plus{}\widehat{ED}$ and $ 2\angle DEF\equal{}\widehat{BF}\plus{}\widehat{BC}\plus{}\widehat{CD}$.

Thus, $ 2\angle BCD\plus{}2\angle DEF\equal{}2\widehat{BF}\plus{}\widehat{FE}\plus{}\widehat{ED}\plus{}\widehat{BC}\plus{}\widehat{CD}$, and since $ \widehat{BF}\plus{}\widehat{FE}\plus{}\widehat{ED}\plus{}\widehat{DC}\plus{}\widehat{CB}\equal{}360^\circ$, this is equivalent to $ 2\angle BCD\plus{}2\angle DEF\equal{}\widehat{BF}\plus{}360^\circ$, or $ \widehat{BF}\equal{}2\angle BCD\plus{}2\angle DEF\minus{}360^\circ$, which in this problem translates to

$ \widehat{BF}\equal{}2(140^\circ \minus{} x^\circ)\plus{}2(144^\circ \minus{} 3x^\circ) \minus{} 360^\circ \equal{} 208^\circ \minus{} 8x^\circ$.

[quote="math154"]Since $ \angle BCD,\angle DEF$ are [url=http://en.wikipedia.org/wiki/Inscribed_angle]inscribed angles[/url], we can see that $ 2\angle BCD \equal{} \widehat{BF} \plus{} \widehat{FE} \plus{} \widehat{ED}$ and $ 2\angle DEF \equal{} \widehat{BF} \plus{} \widehat{BC} \plus{} \widehat{CD}$.

Thus, $ 2\angle BCD \plus{} 2\angle DEF \equal{} 2\widehat{BF} \plus{} \widehat{FE} \plus{} \widehat{ED} \plus{} \widehat{BC} \plus{} \widehat{CD}$, and since $ \widehat{BF} \plus{} \widehat{FE} \plus{} \widehat{ED} \plus{} \widehat{DC} \plus{} \widehat{CB} \equal{} 360^\circ$, this is equivalent to $ 2\angle BCD \plus{} 2\angle DEF \equal{} \widehat{BF} \plus{} 360^\circ$, or $ \widehat{BF} \equal{} 2\angle BCD \plus{} 2\angle DEF \minus{} 360^\circ$, which in this problem translates to

$ \widehat{BF} \equal{} 2(140^\circ \minus{} x^\circ) \plus{} 2(144^\circ \minus{} 3x^\circ) \minus{} 360^\circ \equal{} 208^\circ \minus{} 8x^\circ$.[/quote]

Thanks very  much, you are nice and great.... :clap2:  :D


When I was at camp, I was homesick, but now I [b]REALLY[/b] wish I was back at camp.  :( I didn't realize just how much I enjoyed MathPath until it was too late. Anyone else?

Well, I wasn't homesick while I was at camp, but ya, I miss it a lot. So many cool people there this past summer.

That's the only bad thing about camp: you realize how much you miss it when you go home.

Yeah, I feel that way every time I leave camp. But when you go to camp (any camp) next summer, you'll feel better. I've done it before.

lol you have been to mathpath right?
yes im overly campsick. and the-people-there-sick too

dayum, I miss all the teachers, students (MathPath 2009, like pinkmuskrat), pool tables, and those professional chefs in Colorado..(loved all 4 categories)


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

$G$ is a finite group
$\forall g\in G $    $  g^2=1$
Prove that $G$ has $2^m$ elements

Indead cauchy is the best way to tackle this, however it is overkill, i think it is possible without!

first of all note that this group is abelian

because if a^2=1
then a=a^(-1)

but if (ab)^2=1
then (ab)^(-1)=a*b
or b^(-1)a*^(-1)=a*b
or b*a=a*b

now let a1...am be a minimal set of generators of the group  (meaning that every element of G can be written as a product of a finite amount of combinations of those am and their inverses, and no subset of my bunch of am's does that)
note that for all ai , ai^2=1

now, notice that because this group is abelian , every element can be written as
a1^t1*....am^tm
where all t's are zero or one (so you have some sort of binary notation)

now are all these elements different, sure!
suppose t1...tm and s1...sm are not all equal , but that a1^t1*...am^tm=a1^s1*....am^sm
then at least one ti would be zero and the corresponding si one, or the other way around,
but that would mean that that ai can be expressed as a product of the others an that my set of a's wasn't minimal
(you can compare this with the fact that all vectors in a basis of a vector space are not a linear combination of the others)

so there you have it : there are 2^m elements in your group

i hope you understand my proof and that it is correct
if you have questions, please ask

as a matter of fact, you can prove that if you have an abelian group ,and all elements have the same order k
that that order k must be prime and that your group is isomorphic to Ck*....*Ck

however you have to say 'abelian and all elements have same order' to get that
you can't say ' all elements have same order' because than you can have counterexamples

again , if you want proof or examples i'll give them if you ask

:) Thanks! I think I have fully understood your beutiful proof fredbel and thanks julien for the theorems. I didn't know them because i've just started reading about groups :)

[quote="julien_santini"]It is elementary if you know Cauchy theorem - or even more powerful: Sylow theorem -. If card(G) is not of the form 2^m, then there will be a prime number p <> 2 such that p divides card(G), and hence there will exist a subgroup of G with cardinal p (Cauchy theorem, which also follows immediately from the existence of a p-Sylow subgroup). But then some element g of G has period p and hence g^2 <>e (the neuter element).[/quote]

Sorry for the question but i'm not sure.... in your last sentence i think you mean : $\exists g\in G$   $g^p=1$
If so could you prove it.

And to fredbel:
That fact abaout abelian group and elements which have the same order... i'm not sure if i understand properly what is Ck

Easier: $G$ is abelian and has structure of vector space over $Z_2$. As such it is isomorphic with the vector space $Z_2^k$, for some $k$. Hence, $|G|=2^k$.

[quote="fredbel6"]Ck is 'a' notation (there are others) for the additive group modulo k
it always has k elements and it is always abelian

for instance C3 contains three classes {0 mod 3,1 mod 3,2 mod 3}
and now   0 mod 3+1 mod 3=1 mod 3
2 mod 3 + 2 mod 3=1 mod 3

it is in a sense the most basic group of order k[/quote]

I have always used $Z_k$ for that :)

yes that is a notation too,
in my course of group theory we always used Ck

perhaps that was to avoid confusion between the ring and the group modulo k?


function
induction
inequalities
algebra unsolved
algebra

The functions $f_0, f_1, f_2, ...$ are defined on the reals by $f_0(x) = 8$ for all $x$, $f_{n+1}(x) = \sqrt{x^2 + 6f_n(x)}$. For all $n$ solve the equation $f_n(x) = 2x$.

Notice that if $x$ is a solution, then we must have $x \geq 0$.

I will prove by induction on $n$ that
(a) if $x > 4$, then $f_n(x) < 2x$ for all positive integers $n$, and
(b) if $x < 4$, then $f_n(x) > 2x$ for all positive integers $n$.

Suppose that $x > 4$. It is obvious that $f_0(x) = 8 < 2x$ and also \[ f_1(x) = \sqrt{x^2 + 48} < 2x \] since \[ \sqrt{x^2 + 48} < 2x \Leftrightarrow x^2 + 48 < 4x^2 \Leftrightarrow x^2 > 16 \Leftrightarrow x > 4. \] Now, for the induction hypothesis, suppose that $f_n(x) < 2x$ for some positive integer $n$. Then we get \[ f_{n + 1}(x) = \sqrt{x^2 + 6f_n(x)} < \sqrt{x^2 + 12x} < 2x \] where the last inequality holds because $x > 4$. This completes the induction.

Similarly one can prove that if $x < 4$, then $f_n(x) > 2x$ for all positive integers $n$.
(One just needs to reverse all inequality signs in the proof for (a).)

Finally, it is easy to see that $x = 4$ satisfies $f_n(x) = 2x$ for all $n$.

Hence the only solution is $x = 4$, for all positive integers $n$.

As a matter of fact this result holds only for $n = 1$.  For higher values, x becomes complex.

Hello

This is USAMO 1990: https://artofproblemsolving.com/community/c6h58270p356624


my cousin is studying for the Army test and i was helping him. this was his problem and he has no answr keys and he is a bit slwo right now l0l.
this is the problem
2-1/8n=16+3/4n
i got -12.25 is that right?

No.

2 - 1/8n = 16 + 3/4n which is more readily seen as

2 - (1/8)n = 16 + (3/4)n does not have your solution.

And if you typed it as

2 - 1/(8n) = 16 + 3/(4n),

then that also would not have your solution.


If \[ \sum_{i\equal{}1}^n a_i\geq \sqrt{(n\minus{}1)\sum_{i\equal{}1}^na_i^2}\], prove that $ a_i \geq 0$


can anyone help me to solve this
triangle ABC, M is the point inside the triangle.
let MA=x, MB=y, MC=z
prove that x/a + y/b + z/c  >=  :sqrt: 3 [/quote]

This problem was posted on ML before !
  You can use Leibnitz's formula to solve it .

Look at http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 post #14.

  Darij

A stronger one is true.

$\frac{xy}{ab} + \frac{yz}{bc} +\frac{zx}{ca}  \ge 1$


integration
logarithms
trigonometry
calculus
calculus computations

Compute:

$ \int_{0}^{\frac {\pi}{2}} \ln(\sin(t)) dt$  EDITED

I have a nice solution for it.

It looks like your answer involves Catalan's constant.

I don't think so, I'd like to see your solution. Hide it.

[quote="FOURRIER"]no its very elementary.[/quote]
[b]Very[/b] elementary is $ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin ( \ln x ) \, \mathbf{d}x$ not the one you've posted :huh: .

[quote="FOURRIER"]no its very elementary.[/quote]

no, it isn't.   :oops:  if you mentioned the solution is elementary   to my students, they would   humiliate you so fast you wouldn't know what hit you.   :D

like Black   said, it involves   catalan's constant. it is   well  known  that

$ \int_0^{\dfrac{\pi}{2}}\;\ln\,\cos\,t\;\textbf dt\; \equal{} \; \minus{} \dfrac{\pi}{2}\;\ln\,2\qquad \ldots (1)$

and

$ \int_0^{\dfrac{\pi}{4}}\;\ln\,\cos\,t\;\textbf dt\; \equal{} \; \minus{} \dfrac{\pi}{4}\;\ln\,2 \plus{} \dfrac{\beta(2)}{2}\qquad \ldots (2)$

and i know quite a few things about $ \zeta(3)$ and $ \beta(2)$, so i know what i am talking about.  :P

the integral you have posted is essentially $ (2)$

I had a mistake, its actually :

$ \int_{0}^{\frac {\pi}{2}} ln(sin(t)) dt$

Now it's corrected

and this one has surely an elemtary solution.

[hide="solution"]
$ \underbrace{\int_{0}^{\frac {\pi}{2}} \ln (\sin t) dt}_{\textbf{I}} \stackrel{2x \equal{} t}{ \equal{} } 2 \int_0^{\frac {\pi}{4}} \ln 2 \, dx \plus{} 2 \int_0^{\frac {\pi}{4}} \ln ( \sin x ) \, dx \plus{} 2 \int_0^{\frac {\pi}{4}} \ln ( \cos x ) \, dx \equal{}$
$ \equal{} \frac {\pi}{2} \ln 2 \plus{} 2 \int_0^{\frac {\pi}{4}} \ln ( \sin x ) \, dx \plus{} 2 \int_0^{\frac {\pi}{4}} \ln \left( \sin \left( \frac {\pi}{2} \minus{} x \right) \right) \, dx \equal{}$
$ \equal{} \frac {\pi}{2} \ln 2 \plus{} \underbrace{2 \int_0^{\frac {\pi}{4}} \ln ( \sin x ) \, dx \plus{} 2 \int_{\frac {\pi}{4}}^{{\frac {\pi}{2}}} \ln ( \sin x ) \, dx}_{2 \textbf{I}}$
hence $ \textbf{I} \equal{} \int_{0}^{\frac {\pi}{2}} \ln(\sin t) dt \equal{} \boxed{ \minus{} \frac {\pi}{2} \ln 2}$
[/hide]

[hide="hint using the definition of the definite integral"]
for $ m \equal{} 2,3,4, \ldots$

$ sin\frac {\pi}{m} sin\frac {2\pi}{m} sin\frac {3\pi}{m} \ldots sin\frac {(m \minus{} 1)\pi}{m} \equal{} \frac {m}{2^{m \minus{} 1}} \ldots (*)$

Take logarithms of $ (*)$ and write the limit as $ m \to \infty$ as a definite integral.

[/hide]


Let $ x_1,x_2,\dots,x_n$ be real numbers such that $ x_i \ge 1$ for $ i\equal{}1,2,3,\dots,n$ and $ x_1\plus{}x_2\plus{}\dots\plus{}x_n\equal{}2n$. Find the maximum value of \[ \frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}\dots\plus{}\frac{1}{x_n}.\]

Let $ x_i\equal{}y_i\plus{}1$
$ \implies \sum{y_i}\equal{}n$
And we have to find the maximum of that:
$ \sum{(y_i\plus{}1)^{\minus{}1}}$
Let $ f(x)\equal{}(x\plus{}1)^{\minus{}1}$
$ f$ is concave because $ f"(x)\equal{}\minus{}1$
Then by Jensen inequality:

$ \sum{(y_i\plus{}1)^{\minus{}1}}\le n\frac{1}{1\plus{}\frac{\sum{y_i}}{n}}\equal{}\frac{n}{2}$

[quote="Raja Oktovin"]Let $ x_1,x_2,\dots,x_n$ be real numbers such that $ x_i \ge 1$ for $ i \equal{} 1,2,3,\dots,n$ and $ x_1 \plus{} x_2 \plus{} \dots \plus{} x_n \equal{} 2n$. Find the maximum value of
\[ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \dots \plus{} \frac {1}{x_n}.
\]
[/quote]
$ (1,1,\dots,1,n \plus{} 1) \succeq (x_1, x_2, \dots, x_n) \Rightarrow$
$ \Rightarrow \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \dots \plus{} \frac {1}{x_n} \leq n \minus{} 1 \plus{} \frac {1}{n \plus{} 1}$ by Karamata inequality

[quote="gilcu3"]Let $ x_i \equal{} y_i \plus{} 1$
$ \implies \sum{y_i} \equal{} n$
And we have to find the maximum of that:
$ \sum{(y_i \plus{} 1)^{ \minus{} 1}}$
Let $ f(x) \equal{} (x \plus{} 1)^{ \minus{} 1}$
$ f$ is concave because $ f"(x) \equal{} \minus{} 1$
Then by Jensen inequality:

$ \sum{(y_i \plus{} 1)^{ \minus{} 1}}\le n\frac {1}{1 \plus{} \frac {\sum{y_i}}{n}} \equal{} \frac {n}{2}$[/quote]
you wrong used Jensen inequality becouse $ \sum{(y_i \plus{} 1)^{ \minus{} 1}}\geq n\frac {1}{1 \plus{} \frac {\sum{y_i}}{n}} \equal{} \frac {n}{2}$

[quote="Raja Oktovin"]is karamata inequality the best approach to this problem?
i've never heard it and i think here, the students should use well-known theorem...[/quote]
see here [url]http://www.mathlinks.ro/viewtopic.php?t=14975[/url].


Is nobody interested in this problem?

I solved it with vectors, and I enjoyed it very much  :)

The equations just scream "vectors":

[hide]If $\vec x=\langle x_{1},x_{2},x_{3}\rangle,\vec y=\langle y_{1},y_{2},y_{3}\rangle,\vec u=\langle 1,1,1\rangle$, then $\vec u\cdot\vec x=\vec u\cdot\vec y=\vec x\cdot\vec y=0$, hence $\vec u,\vec x,\vec y$ are pairwise perpendicular. The desired expression reduces to $\cos^{2}\alpha_{1}+\cos^{2}\beta_{1}$, where $\alpha_{1},\beta_{1}$ are the angles between $_{1}\text{-axis}$ and $\vec x,\vec y$ respectively. If we observe the vectors in $\vec x,\vec y,\vec u$ coordinate system, then the unit vector of the "old" $_{1}\text{-axis}$ will be $\vec v=\langle \sin\phi_{0}\cos\theta,\sin\phi_{0}\sin\theta,\cos\phi_{0}\rangle$ where $\phi_{0}=\arctan\sqrt{2}$ (i.e. the angle between $\vec u$ and "old" $_{1}\text{-axis}$) and $\theta$ is a free variable. Therefore $\cos\alpha_{1}=\sin\phi_{0}\cos\theta,\cos\beta_{1}=\sin\phi_{0}\sin\theta$, giving $\cos^{2}\alpha_{1}+\cos^{2}\beta_{1}=\sin^{2}\phi_{0}={\tan^{2}\phi_{0}\over 1+\tan^{2}\phi_{0}}={2\over 3}$[/hide]

Ok, nice.  :)

I found a cool algebraïc solution this morning. I'll post it soon.

Notice that if we can prove that \[\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}+\frac{y_{1}^{2}}{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}\]
is a constant, then it is equal to $\frac23$
because after replacing $x_{1},y_{1}$ with the other variables and adding cyclicly, we find that three times the constant equals two.