induction
counting
distinguishability
combinatorics proposed
combinatorics

Here is a problem created I think by a Romanian student in informatics:
We are given $n$ coins. They are of 3 types, A, B, and C. We know that $\displaystyle |C|>\frac{n}{2}$. At one step, we are allowed to select some pairs of coins and compare them (we will find out if they are different or equal). A coin can take part in only one pair during a step. Find the minimum numer of steps required to find a coin of type C.

Well, Dust said me the answer is $3$.
Let's prove by induction on $n$ the number of coins that we can find in at most $3$ steps all the coins, moreover that any coin we wish will be weighted at most $2$ times.
base is obvious now induction step:
Suppose for $1,\cdots,n-1$ we proved now we prove for $n$. Let $C$ be our coin desired to be weighted $2$ times.
First case. $n$ is even. Then divide the coins into $\frac n2$ pairs and weight them. Since some kind of coins will be more than half,  in these pairs some coins will be equal. Let's unite them into bigger coins. It's obvious this bigger coins again satisfy the condition of the problem, moreover a coin of dominant type in this set would lead to a coin of dominant type in the original set of coins. By induction, we can find a coin of dominant type in the new set by three steps.
But since every 'big' coin contains two smaller coins, we can unite two steps into one.
Now let $D$ be the coin $C$ was matched with. If we got $C\neq D$ then $C$ would no more be weighted so at most one time. If $C=D$ they were united in the new set. Hovewer, in this new set they were weighted $3$ times and since they are no longer distinguishable, we may suppose that in the last step(when just one coin was weighted) the weighted coin was $D$ so $C$ was weighted at most $2$ times, as desired.
Second case. $n$ is odd. Divide the coins into $\frac {n-1}2$ coins and let $C$ be unweighted (so already $C$ can be weighted at most $2$ times). If no match occurs, then $C$ is of dominant type. If matches occur, then again united them into bigger coins and add $C$ to the set. By induction hypothesis, we can find a coin of dominant type in this new set s.t. $C$ is weighted at most two times. Now we can again unite two steps if in one of them $C$ is not weighted (this is true). Hence this case is also handled.

To prove two steps don't suffice, use the very same induction idea. Suppose that at first weighting all coins were matched. Now we unite the matched coins into 'bigger' ones and the left step would transform into two steps in the new set, so it's not possible to find a coin of dominant type in this new set, and hence in the original one too.
QED


Let $ ABC$ be an acutable triangle. Prove or disprove that
$ 4\cos(\frac {A}{2})\cos(\frac {B}{2})\cos(\frac {C}{2})\ge\sin(A \plus{} \frac{B}{2} \plus{} \frac{C}{3})$ .

___________
 :idea:

$ LHS \equal{} \frac{p}{R}$ as well-know and clear $ p \ge R$ when the triangle is acutangle.


Why is 1 to the infinity considered an indeterminate form?

limn-->infinity(1+1/n)n=e.

limn-->infinity(1-1/n)n=1/e.

Both are 1infinity.

Its basically because anything over 1 to the power of infinity goes to infinity, anything under 1 goes to 0. So you can't tell when it goes towards 1 to infinity..

Taking log may help to see why :

if u(n) = a(n)b(n) is of the form 1(+oo) then (assuming everything is > 0), log(u(n))=b(n)log(a(n)) is of the form +oo x 0 which is a well known indeterminate form.

I am afraid just saying something is of the form +inf * 0 doesnt mean its indeterminate....

for example lim x->+inf 1 = lim x->+inf x * 1/x which is +inf * 0... but the limit is clearly one.

Indeterminate means 'which can't be determinated by a single rule', not that the limit never exists (or is not finite).

+oo x 0 :
. can be anything you want : a*n * 1/n with a  IR
. may not exist : n * ((-1)^n / n)
. can be +oo : n * 1/n


A box contains both chewy-center and soft-center pieces of candy.  There are 8 more chewy-center pieces than soft-center pieces.  When a piece of candy is chosen at random, the probability that it has a chewy center is 3/5.  How many pieces of candy are in the box?


Solve the equation:
$x^{3}-\sqrt[3]{6+\sqrt[3]{x+6}}=6$

Bad title.

http://www.mathlinks.ro/Forum/viewtopic.php?t=144404


In triangle ACE, B is on AC and D is on CE, and BE meets AD at F. EB bisects angle AEC. If AB = 3, BC = 4, AE = 6, and F is the midpoint of BE, find CD.

[hide] Angle bisector says "CE = 8."
Mass points say "CD/DE = 7/3"
Thus, CD = 5.6?[/hide]

[quote="pontios"]djshowdown2,

mass point of which triangle?

and why it is $\frac{CD}{DE}=\frac{7}{3}?$[/quote]

The whole triangle, centered around F.  So masses on (A,B,C,D,E,F) are (4,7,3,10,7,14) respectively.

How I went about assigning - from the given lengths, put 4 on A and 3 on C so a total of 7 on B.  Since F is the midpoint of BE, E is also 7, so there's 14 on F, making 10 on D.  If you wanna extend CF to meet AE at G to check, then G has 11 and everything balances.  So since there's 3 on C and 7 on E, CD/DE = 7/3

[hide="Another way"]
After we find $CE=8$ as djshowdown2 said, take $P$ on $BC$ such that $PF\parallel CE$.

Since $BF=FE$ and $PF\parallel CE$, we have $PF=\frac12CE=4$ and $BP=PC=2$.
Then, since $PF\parallel CD$, $\frac{PF}{CD}=\frac{AP}{AC}=\frac57$
Hence $CD=4\times\frac75=\boxed{\frac{28}{5}}$

[/hide]


1-digit - 2 ways
2-digit - 2 choices for first digit, 3 for second: 6 ways
3-digit - 2 choices for first digit, 3 for second, 3 for third: 18 ways.

Summing these yiels 2+6+18=26

uh just 3*3*3(which is the 3 digit choices)-1 (000 is not positive)=26 seems to work quite well....

Use policecaps way, it's much much faster.

[quote="ajain"]How many positive integers less than 1000 can be written using only the digits 0, 1, 2?

I was one off on this one, and I want to know how to solve this problem faster.

Thankx.[/quote]

Use combinatorics. Once you get good at it, you can get really fast at these problems when you recognize it's that type of problem. All you have to do is believe  :). Forget I said that...

....so?

He was emphasizing the 'YOU HAFTA BELIEVE" part of it :D


induction
number theory unsolved
number theory

let $ A \subset N$ prove that there exist $ B$ such that :
i $ A \subseteq B$
ii the sum of the elements of $ B$is divisible by all elements of $ B$

[quote="pco"][quote="mathnice1"]let $ A \subset N$ prove that there exist $ B$ such that :
i $ A \subseteq B$
ii the sum of the elements of B is divisible by all elements of B[/quote]

That's wrong : Take as $ A$ any infinite subset, then the sum of elements of $ B$ is not defined.[/quote]
$ A$ is not infinite  :)

Of course, it was self-understood that $ A$ and $ B$ ought to be finite.

If $ | A | \equal{} 1$ one may take $ B \equal{} A$.
If $ A_n \equal{} \{ 1, n \}$, with $ n > 1$, we will prove it by induction on $ n$. Denote a set $ B$, built to the requirements, by $ B(A)$, and the sum of its elements by $ \sigma(B)$. One may clearly take $ B(A_2) \equal{} \{ 1, 2, 3 \}$; notice that $ 3 \ | \ 6 \equal{} \sigma(B(A_2))$.
Assume it proven for all $ k \leq n\minus{}1$, with the extra property that $ k\plus{}1 \ | \ \sigma(B(A_k))$. Let's take $ B(A_n) \equal{} A_n \cup \{ b(n\plus{}1) \; ; \; b \in B(A_{n\minus{}1}) \setminus \{ 1 \} \}$.
But $ \sigma(B(A_n)) \equal{} \sigma(B(A_{n\minus{}1}))(n\plus{}1)$. It follows that $ n \ | \ \sigma(B(A_n))$, since $ n \ | \ \sigma(B(A_{n\minus{}1}))$ by induction hypothesis,
and $ b(n\plus{}1) \ | \ \sigma(B(A_n))$ for all $ b \in B(A_{n\minus{}1})$, since $ b \ | \ \sigma(B(A_{n\minus{}1}))$ by induction hypothesis, including the case $ b \equal{} 1$. Induction is thus completed.

Consider now $ A \equal{} \{ x, y \}$, with $ 1 < x < y$. Take $ z \equal{} 2xy \minus{} x \minus{} y$ and $ B(A) \equal{} A \cup \{ z \} \cup \{ 2xyb \; ; \; b \in B(A_z) \setminus \{ 1 \} \}$. Then $ \sigma(B(A)) \equal{} 2xy\sigma(B(A_z))$, so $ B(A)$ is clearly built to the requirements.

Moreover, we have up to now built $ B(A)$ such that $ b > a$ for all $ a \in A$ and $ b \in B(A) \setminus A$. We will assume this property in the induction on $ | A |$, proven in the above for $ | A | \leq 2$. For $ | A | \geq 3$, take $ C \equal{} B(A \setminus \{ \min A \})$, and $ D \equal{} B(\{ \min A, \sigma(C) \})$. Finally, take $ B(A) \equal{} (D \setminus \{ \sigma(C) \}) \cup C$. Clearly $ \sigma(B(A)) \equal{} \sigma(D)$, and $ c \ | \ \sigma(C) \ | \ \sigma(D)$ for all $ c \in C$, while $ d \ | \ \sigma(D)$ for all $ d \in D$, therefore $ B(A)$ is built to the  requirements, and this concludes the problem.


calculus
Sequence
floor function
Summation
number theory
IMO Shortlist

[hide]By $a_{n}=\frac{d(1)+d(2)+...+d(n)}{n}$, where $d(n)=\sum_{k|n}1$ is number of positive divisors of $n$.

We have $a_{n+1}-a_{n}=\frac{nd(n+1)-d(1)-d(2)-...-d(n)}{n(n+1)}$

a)Choose $n=...$ , can you find?  :maybe:
b)Choose $n=p-1$ with $p$ is prime[/hide]

As Mr.Tuan said, for $(b)$ we can choose $n=p-1$ where $p \geq 7$. To prove part $(a)$ first observe that
\begin{align*}
  a_{2n}&=\frac{1}{2n}\left(\left\lfloor\frac{2n}{1}\right\rfloor+\left\lfloor\frac{2n}{2}\right\rfloor+\cdots+\left\lfloor\frac{2n}{n}\right\rfloor+\left\lfloor\frac{2n}{n+1}\right\rfloor+\cdots+\left\lfloor\frac{2n}{2n}\right\rfloor \right)\\
&>\frac{1}{2n}\left(\left\lfloor\frac{2n}{1}\right\rfloor+\left\lfloor\frac{2n}{2}\right\rfloor+\cdots+\left\lfloor\frac{2n}{n}\right\rfloor\right)\\
& \geq \frac{1}{2n}\left(2\left\lfloor\frac{n}{1}\right\rfloor+2\left\lfloor\frac{n}{2}\right\rfloor+\cdots +2\left\lfloor\frac{n}{n}\right\rfloor\right)\\
&=\frac{1}{n}\left(\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\cdots+\left\lfloor\frac{n}{n}\right\rfloor\right)\\
&=a_{n}
\end{align*}
where we used the fact that $\lfloor 2x \rfloor \geq 2 \lfloor x \rfloor $ for any real number $x$.\\\\
Now, suppose there exists only finitely many $n$ such that $a_{n+1}> a_n$ and let $m$ be largest such $n$ .Then
$$a_{m+1}\geq a_{m+2}\geq a_{m+3} \geq \dots \geq a_{2m+2}$$
which is a contradiction.

[quote="bodom"]for part a) you can also put n= 2^k -1 for k big enough:dyou just use some easy analysis to prove that:D[/quote]
Please post concrete!

You are true, i think so. But
[quote="bodom"]$ lim[ (1+1/2+.....+1/2^{k})/k]=ln2$ [/quote]
It is wrong!  :D

:oops: Sorry, I mean that $ 1+\frac{1}{2}+\frac{1}{2^{2}}+...$ . I am wrong!

Actually, it's not necessary to use a specific sequence like $ n \equal{} 2^{k}\minus{}1$. Since $ d(n)$ is unbounded, you can find infinitely many values of n such that $ d(n\plus{}1) > d(1), d(2),\ldots, d(n)$, and such an $ n$ does the job.

Another way to treat (a): Assume the contrary. Then $ a_{n}$ has to be bounded above. But

\[ a_{n}\geq\frac{1}{n}\left(\frac{n}{1}\minus{}1\plus{}\frac{n}{2}\minus{}1\plus{}\ldots\plus{}\frac{n}{n}\minus{}1\right) \equal{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\ldots\plus{}\frac{1}{n},\]
which is known to diverge, contradiction.

[quote="mattilgale"]We define a sequence $ \left(a_{1},a_{2},a_{3},...\right)$ by setting
\[ a_{n} \equal{} \frac {1}{n}\left(\left[\frac {n}{1}\right] \plus{} \left[\frac {n}{2}\right] \plus{} \cdots \plus{} \left[\frac {n}{n}\right]\right)\]
for every positive integer $ n$. Hereby, for every real $ x$, we denote by $ \left[x\right]$ the integral part of $ x$ (this is the greatest integer which is $ \leq x$).

[b]a)[/b] Prove that there is an infinite number of positive integers $ n$ such that $ a_{n \plus{} 1} > a_{n}$.
[b]b)[/b] Prove that there is an infinite number of positive integers $ n$ such that $ a_{n \plus{} 1} < a_{n}$.[/quote]
First a lemma:
$ \sum_{k\equal{}0}^{n\plus{}1} \left \lfloor \frac{n\plus{}1}{k} \right \rfloor \minus{} \left \lfloor \frac{n}{k} \right \rfloor \equal{} \tau(n\plus{}1)$, $ \tau(n)$ is the number of divisors of $ n$. The proof follows from the fact that $ \left \lfloor \frac{n\plus{}1}{k} \right \rfloor \minus{} \left \lfloor \frac{n}{k} \right \rfloor \equal{} 1 \iff k \mid n\plus{}1$.
So:
$ (n\plus{}1)(a_{n\plus{}1}\minus{}a_n) \equal{} \sum_{k\equal{}0}^{n\plus{}1} \left ( \left \lfloor \frac{n\plus{}1}{k} \right \rfloor \minus{} \left \lfloor \frac{n}{k} \right \rfloor \right )\minus{}\frac{1}{n} \sum_{k\equal{}0}^{n} \left \lfloor \frac{n}{k} \right \rfloor  \equal{} \tau(n\plus{}1)\minus{}a_n$

Assume that for some $ N$ $ a_{n\plus{}1} \ge a_n$ for all $ n \ge N$. Then $ \tau(n\plus{}1) \ge a_n$, and $ 2 \equal{} \tau(p\minus{}1\plus{}1) \ge a_{p\minus{}1}$ for all primes and hence $ a_n \le 2$ for all $ n \ge N$. But $ a_n$ is unbounded. Contradiction.
Assume that for some $ N$ $ a_{n\plus{}1} \le a_n$ for all $ n \ge N$. Then $ a_n$ is decreasing. But $ a_n$ is unbounded. Contradiction.

It is well known that $\displaystyle \sum_{i=1}^{n} \left \lfloor \dfrac{n}{i} \right \rfloor = \displaystyle \sum_{i=1}^{n} d(i)$ where $d(i)$ denotes the number of natural divisors of $i$ (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3379845&sid=f64a70ac8438b1455bb9ee7efd90751f#p3379845]INMO 2014 P2[/url]). Now,
\[a_{n+1} > a_n \iff n \displaystyle \sum_{i=1}^{n+1} d(i) > (n+1) \displaystyle \sum_{i=1}^{n} d(n) \\
\iff d(n+1) > \dfrac{1}{n}\displaystyle \sum_{i=1}^{n} d(i) = a_n.\]
Similarly, $a_n < a_{n+1}$ iff $d(n+1) < a_n.$

For the first part, just let $n+1$ be a prime. For the second part, choose a $n$ such that $d(n+1) > d(i)$ for all $1 \leq i \leq n$ (if only finitely many such $n$ existed, there would exist a $k$ such that $d(i) < k$ for all $i \in \mathbb{N}$ which is obviously impossible because for any prime $p,$ $d(p^k) = k+1$).

For a) just notice that $a(2n)>a(n)$ and this is true cause $[2n/t]>=[n/t]+[n/t]$ so we are finished.
For b),just pick $n=p-1$,where $p$ is a prime.Now,cause it is obvios that $[p/k]=[p-1/k]$,for all k non equal $p$ or $1$,so we just need $a(p-1)>2$ and this is true for all $p>=7$,so we are finished(there are infinitely many primes).

a)Assume the opposite and proceed.
Clearly this implies that the sequence is bounded.
However $a_n=\sum_{i=1}^n \lfloor{\frac{n}{i}}\rfloor>\sum_{i=1}^n \frac{1}{i}-1$ which is unbounded a contradiction so we are done.
b)Take $n=p-1$ and so it suffices to show $1+\sum_{k=1}^n (\lfloor{\frac{n+1}{k} \rfloor -\lfloor \frac{n}{k} }\rfloor)=1+\sum_{k \le n,k|n+1}<a_n$,as required(it implies $a_{p}>2$ so just choose $p$ to be large enough and since the sequence diverges we are done)

The first time I have succeeded in solving an N3 :D
[quote=mattilgale]We define a sequence $ \left(a_{1},a_{2},a_{3},\ldots \right)$ by
\[ a_{n} \equal{} \frac {1}{n}\left(\left\lfloor\frac {n}{1}\right\rfloor \plus{} \left\lfloor\frac {n}{2}\right\rfloor \plus{} \cdots \plus{} \left\lfloor\frac {n}{n}\right\rfloor\right),
\] where $\lfloor x\rfloor$ denotes the integer part of $x$.

[b]a)[/b] Prove that $a_{n+1}>a_n$ infinitely often.
[b]b)[/b] Prove that $a_{n+1}<a_n$ infinitely often.

[i]Proposed by Johan Meyer, South Africa[/i][/quote]
[b]LEMMA[/b]
$$[\frac{n+1}{k}]-[\frac{n}{k}]=1(\text {if k divides n+1})=0(\text {otherwise})$$
See that
$$(n+1)a_{n+1}=n.a_n + d(n+1)$$
We will prove part $b$ first :)
---------------------
[b]CLAIM[/b]
$1+\frac{1}{2}+\cdots \frac{1}{2^n}>\frac{n}{2}$
Proof:
$$1+\frac{1}{2}>1$$
$$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=
\frac{1}{2}$$
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
Similarly continuing with break  at every power of $2$ and summing gives us the stated inequality.
-----------------------
So there will exist some $N$ for which $\forall n\ge N$
$$1+\frac{1}{2}+\cdots \frac{1}{n}>3$$
Or using $[x] >x-1$ ,
$$a_n>2$$
Take $n+1$ to be a prime $>N$
Thus
$$(n+1)a_{n+1}=n.a_n + 2>(n+1)a_n$$
-----------------------
Now part (a)
Integrate the function $f(x)=\frac{1}{x}$ and get the inequality $$1+\frac{1}{2}+\cdots \frac{1}{n}< \ln(n+1)$$
So using $[x]\leq x$,
$$a_n\leq  \ln(n+1) $$
Also , $\forall k>1, 2^{k-1}>k$
Set $n+1=2^k$
Use $e>2$ and get $ \ln(2^k)<k$
and get $$a_n<d(n+1)\Rightarrow a_{n+1}>a_n$$

By counting in two ways the number of pairs $(a,b)$  with $a,b \le n$ and $a\mid b$, we get that
\[a_n = \frac{1}{n}\sum_{i=1}^{n} d(i).\]
[b]a)[/b]
Let $n+1$ be a highly composite number (a number such that for all $k<n+1$, $d(k)< d(n+1)$), of which there exists infinitely many positive integers, because $d(n)$ is unbounded.
Then, since $d(n+1)$ is bigger than any preceding $d(k)$, it is bigger than their average, and hence its average with them must be bigger than their average, that is, $a_{n+1}>a_n$.

[b]b)[/b]
Take $p\ge 11$ prime, $n=p-1$. Then, if $s = na_n$,
\[s = d(1)+d(2)+d(3)+d(4)+d(5)+d(6) + \sum_{i = 7}^{p-1}d(i)\ge 1+2+2+3+2+4+\sum_{i = 7}^{p-1}d(i)>2p-2\]
Hence,
\[\frac{s}{p-1} > \frac{s+2}{p},\]
i.e, $a_{n+1}<a_n$.

[quote=mattilgale]We define a sequence $ \left(a_{1},a_{2},a_{3},\ldots \right)$ by
\[ a_{n} \equal{} \frac {1}{n}\left(\left\lfloor\frac {n}{1}\right\rfloor \plus{} \left\lfloor\frac {n}{2}\right\rfloor \plus{} \cdots \plus{} \left\lfloor\frac {n}{n}\right\rfloor\right),
\] where $\lfloor x\rfloor$ denotes the integer part of $x$.

[b]a)[/b] Prove that $a_{n+1}>a_n$ infinitely often.
[b]b)[/b] Prove that $a_{n+1}<a_n$ infinitely often.

[i]Proposed by Johan Meyer, South Africa[/i][/quote]
Nice Problem. bit easy for N3
Btw here's how I thought, Posting it for storage.
For any $n$ let $d(n)$ count the number of all divisors. Then we can easily see if $i\nmid n+1$ then $\lfloor \frac{n+1}{i}\rfloor =\lfloor \frac{n}{i}\rfloor $ and if $i|n+1$ then $\lfloor \frac{n+1}{i}\rfloor +1 =\lfloor \frac{n}{i}\rfloor $
Hence to get $a_{n+1}>\text{or} < a_n$ it's only the divisor points of $n+1$ which will make the difference. So clearly $\sum_{i=1}^{n+1} \lfloor \frac{n+1}{i}\rfloor =\sum_{i=1}^{n} \lfloor \frac{n}{i}\rfloor +d(n+1)$ to get $a_{n+1}<a_n$ we must have $nd(n+1)<\sum_{i=1}^{n} \lfloor \frac{n}{i}\rfloor$ the smallest possible value of $d(n+1)$ we can choose is $2$ So let's do that and if we get $2n<\sum_{i=1}^{n} \lfloor \frac{n}{i}\rfloor$ which is true in fact we can get after some $n_0\in N$ we have $2n<n+\frac{n}{2}+\frac{n}{3}+\frac{n}{4}-3<\sum_{i=1}^{n} \lfloor \frac{n}{i}\rfloor$ $\forall n\geq n_0$  Hence for $(b)$ if we take $n+1=\text{prime}$, we are done as there are infinitely many primes.
Now we are left to prove $a_{n+1}>a_{n}$ also holds for infinitely many $n$ so we need $nd(n+1)>\sum_{i=1}^{n} \lfloor \frac{n}{i}\rfloor$ but wait isn't $b_n=\sum_{i=1}^{n} \frac{1}{i}$ is just a big motivation to use cauchy Condensation bound i.e we always have $\sum_{k=0}^{m} 2^{k}\frac{1}{2^{k}}>b_{2^m}$ which can be proven easily by just using the fact $\frac{1}{n}$ is decreasing. Hence if we choose $n+1=2^m$ then we get that $d(n+1)=m+1$ and $m+1>b_{2^m}>b_{2^{m}-1}>\frac{\sum_{i=1}^{2^m-1} \lfloor \frac{2^m-1}{i}\rfloor}{2^m-1}$
Hence as $\sum_{i=1}^{2^m} \lfloor \frac{2^m}{i}\rfloor =\sum_{i=1}^{2^m-1} \lfloor \frac{2^m-1}{i}\rfloor +m+1$ So $a_{2^m}>a_{2^m-1}$

First, define $d(n)$ as the number of positive divisors of $n$. We note that $\left\lfloor\frac {n}{k}\right\rfloor$ is the number of positive multiples of $k$ less than or equal to $n$, so summing these over $k=1$ to $k=n$ is also just the sum of the number of divisors of each positive integer less than or equal to $n$. This means that $$a_n=\frac1n\sum_{k=1}^{n}d(k).$$ It is now clear that part (a) happens for $n+1$ being highly composite(well known that there are infinitely many $n$ that does this), and part (b) happens for $n+1$ being every prime greater than or equal to $7$(also clear that there are infinitely many $n$ that does this).

It is well known that $$\sum_{i=1}^{n}\left \lfloor \frac{n}{i}\right \rfloor = \sum_{i=1}^{n}\tau(i),$$ for positive integers $n \geq 1$ (which can be proven by considering $n\rightarrow n+1$ and noticing only divisors of $n+1$ get affected). We will use this many times throughout our proof.

Note that $$na_n = \left \lfloor \frac{n}{1}\right \rfloor + \left \lfloor \frac{n}{2}\right \rfloor + \cdots + \left \lfloor \frac{n}{n}\right \rfloor = \sum_{i=1}^{n}\tau(i)$$ and $$(n+1)a_{n+1} = \left \lfloor \frac{n+1}{1} \right \rfloor + \left \lfloor \frac{n+1}{2}\right \rfloor + \cdots + \left \lfloor \frac{n+1}{n+1}\right \rfloor = \sum_{i=1}^{n+1}\tau(i),$$ so $$(n+1)a_{n+1} - na_n = \tau(n+1).$$ We will use this to prove both parts.

[b][color=#0ff]Part a):[/color][/b] Assume for the sake of contradiction that $a_{n+1} > a_n$ finitely often. Then, there exists a positive integer $N$ such that for all $n > N$, we have $a_{n+1} < a_n$ (equivalently $a_{n+1} < a_n$ for sufficiently large $n$). Then, for all $n > N$, we have $$a_n = (n+1)a_n - na_n > (n+1)a_{n+1} - na_n = \tau(n+1).$$ Now, take an integer $n > N$ such that $n+1$ has more divisors than any of $1, 2, \cdots, n$. This $n$ exists since if it did not, $\max(\tau(1), \tau(2), \cdots, \tau(n+1))$ would be bounded for all $n$, so $\tau(n)$ would always be bounded, which is clearly false (taking $n = 2^k$ gives $\tau(n) = k+1$ which can grow arbitrarily large). Then, for this particular $n$, $$n\tau(n+1) < na_n = \left \lfloor \frac{n}{1}\right \rfloor + \cdots + \left \lfloor \frac{n}{n}\right \rfloor = \tau(1) + \tau(2) + \cdots + \tau(n) < \underbrace{\tau(n+1) + \tau(n+1) + \cdots + \tau(n+1)}_{n \text{ times}} = n\tau(n+1),$$ a contradiction, as desired.

[b][color=#0ff]Part b): [/color][/b] Assume for the sake of contradiction that $a_{n+1} < a_n$ finitely often. Then, there exists a positive integer $N$ such that for all $n > N$, we have $a_{n+1} > a_n$ (equivalently $a_{n+1} > a_n$ for sufficiently large $n$). Then, for all $n > N$, we have $$a_n = (n+1)a_n - na_n < (n+1)a_{n+1} - na_n = \tau(n+1).$$ Now, take $n = p-1$, where $p > \max(49, N+1)$ is a prime. Then, $a_{p-1} < \tau(p) = 2$. Thus, $$2(p-1) > (p-1)a_{p-1} = \left \lfloor \frac{p-1}{1}\right \rfloor + \cdots + \left \lfloor \frac{p-1}{p-1}\right \rfloor \geq \left \lfloor \frac{p-1}{1}\right \rfloor + \left \lfloor \frac{p-1}{2}\right \rfloor + \left \lfloor \frac{p-1}{3}\right \rfloor + \left \lfloor \frac{p-1}{4}\right \rfloor.$$ Note that $\left\lfloor \frac{a}{b}\right\rfloor = \frac{a}{b} - \left \{ \frac{a}{b}\right \} \geq \frac{a}{b} - 1,$ so $$2(p-1) > \frac{p-1}{1} + \frac{p-1}{2} + \frac{p-1}{3} + \frac{p-1}{4} - 4 = \frac{25}{12}(p-1) - 4 \iff 4 > \frac{1}{12}(p-1)\iff 48 > p-1 > 49 -1 = 48,$$ a contradiction, as desired.

I really like this problem, as it uses two very neat key observations.

[hide=Observation 1]
First, we [b]claim:[/b]
$$\sum_{k=1}^n \left\lfloor \frac nk \right\rfloor = \sum_{k=1}^n \tau(k).$$
While others have given an iterative explanation, here's a more combinatorial approach.

Consider an algorithm to compute the sum of divisors of all numbers up to $n$:

[quote=Algorithm]
Define $d$ to be a list of $n$ zeros, indexed from $1$ to $n$.
For each $i$ from $1$ to $n$:
$\texttt{    }$ For each $1 \leq j \leq n$ that is a multiple of $i$:
$\texttt{    }$ $\texttt{    }$ Add $1$ to $d_j$.
[/quote]

Once this algorithm completes, $d_k = \tau(k)$ for each $1 \leq k \leq n$.
For each $i$, a $1$ is added exactly $\lfloor n/i \rfloor$ times. $\square$
[/hide]

[hide=Observation 2]
Next, note that $a_n$ is the average of the number of divisors of the numbers from $1$ to $n$. Thus, if $\tau(n+1) > a_n$, then (a) is satisfied; and if $\tau(n+1) < a_n$, then (b) is satisfied. If $n+1$ is a highly composite number then (a) is done; there are infinitely many of these (as otherwise $\tau(n)$ is required to be bounded). If $n+1$ is prime then (b) is done.
[/hide]

Set $$na_n=f(n)=\sum_{i=1}^{n} \lfloor \dfrac{n}{i} \rfloor $$ and $f(0)=0.$ Then we have $f(n)-f(n-1)=\tau (n)$ and hence $a_n= \tfrac{\tau (1)+\dots +\tau (n)}{n}.$ Clearly $a_n > \tau (n+1)$ for $n+1$ prime, which proves $(a).$ And $(b)$ otherwise.

Since $\sum_{j=1}^m\left\lfloor\frac{m}{j}\right\rfloor=\sum_{j=1}^m\tau(j)$ and $\sum_{j=1}^m\left\lfloor\frac{m}{j}\right\rfloor-\left\lfloor\frac{m-1}{j}\right\rfloor=\tau(m),$ we see \begin{align*}a_{n+1}-a_n&=\frac{1}{n+1}\sum_{i=1}^{n+1}\left\lfloor\frac{n+1}{i}\right\rfloor-\frac{1}{n}\sum_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor\\&=\frac{1}{n(n+1)}\left(n\left(\sum_{i=1}^{n+1}\left\lfloor\frac{n+1}{i}\right\rfloor-\left\lfloor\frac{n}{i}\right\rfloor\right)-\sum_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor\right)\\&=\frac{1}{n(n+1)}\left(n\tau(n+1)-\sum_{i=1}^n\tau(i)\right).\end{align*} If $n+1$ is prime and $n\ge 6,$ then $$\sum_{i=1}^n\tau(i)=1+2+2+3+2+4+\sum_{i=7}^n\tau(i)>2n>n\tau(n+1)$$ since $\tau(m)\ge 2$ if $m>1.$ This shows (b). Then, we claim by induction that there exist infinite $n$ such that $\tau(n+1)>\tau(i)$ for $1\le i\le n,$ which would prove (a). For our base case, take $n=5,$ noting we have $\tau(6)>\tau(i)$ for $1\le i\le 5.$ For the inductive step, suppose $\tau(k+1)>\tau(i)$ for $1\le i\le k.$ Then, choose the smallest $\ell$ such that $\ell>k+1$ and $\tau(\ell)>\tau(k+1).$ Clearly $\ell$ must exist since $\tau(m)$ can grow arbitrarily large. Then, $n=\ell-1$ works and our induction is complete. $\square$

Note that \[\frac {1}{n}\left(\left\lfloor\frac {n}{1}\right\rfloor + \left\lfloor\frac {n}{2}\right\rfloor + \cdots + \left\lfloor\frac {n}{n}\right\rfloor\right)=\frac1n (\tau(1)+\tau(2)+\dots+\tau(n))\] which means that $a_{n+1}>a_n$ if and only if $n\tau(n+1)>\tau(1)+\tau(2)+\dots+\tau(n)$ and $a_{n+1}<a_n$ if the inequality is flipped. Since $\tau$ function is unbounded, there must be infinitely many times when $\tau(n+1)>\tau(1),\dots,\tau(n)$ and $a_{n+1} > a_n$ in this case, and since $\tau(p)=2$, for all large enough primes, $a_{p}<a_{p-1}$.

We have the equivalence
\[a_n=\frac{\tau(1)+\tau(2)+\dots+\tau(n)}{n}\]
and now by the unboundedness of $\tau$ and the fact that primes have $2$ factors (which is the minimum, other than $1$), we are done with both parts. $\blacksquare$

Note that \[\frac {1}{n}\left(\left\lfloor\frac {n}{1}\right\rfloor + \left\lfloor\frac {n}{2}\right\rfloor + \cdots + \left\lfloor\frac {n}{n}\right\rfloor\right)=\frac1n (\tau(1)+\tau(2)+\dots+\tau(n))\] from an elementary divisor counting argument. Thus, $a_{n+1}>a_n$ if and only if $n\tau(n+1)>\tau(1)+\tau(2)+\dots+\tau(n)$. By the unboundedness of $\tau$, there are infinitely many $n$ with $\tau(n+1)>\max\{\tau(1),\ldots,\tau(n)\}$, so $a_{n+1}>a_n$ for infinitely many $n$. Also, $a_{n+1}<a_n$ if and only if $n\tau(n+1)<\tau(1)+\tau(2)+\dots+\tau(n)$; choosing sufficiently large $n$ such that $n+1$ is prime completes part (b). We are done.

The following claim trivializes the problem:

[color=#f00][b]Claim:[/b][/color] \[a_n = \frac{\sum_{a=1}^n \tau(a)}{n},\] where $\tau(n)$ is the divisor function.

[i]Proof: [/i]Note that $\left\lfloor\frac{n}{1}\right\rfloor$ counts the number of multiples of $1$ up to $n$, $\left\lfloor\frac{n}{2}\right\rfloor$ counts the number of multiples of $2$ less than $n$, etc. Therefore, for some number $1 \geq k \geq n$, each divisor $d$ of $k$ is counted once once and only once, in the term $\left\lfloor\frac{n}{d}\right\rfloor$. By this argument for all $k$, we obtain that \[ \sum_{a=1}^n \tau(a) = \sum_{a=1}^n \left\lfloor\frac{n}{a}\right\rfloor,\] from which the claim easily follows.

Now the result is clear - for part a) take some number $n$ such that $\tau(n+1) > \tau(1), \tau(2), ... \tau(n)$, and for part b) take $n = p-1$ for a prime $p$. It is easy to see with our claim that this works.

It is obvious to see that
\[
a_n = \frac 1n \sum_{i=1}^n d(n)
\]
[b]a)[/b] It remains to find $n$ such that
\[
d(n+1) > \frac{\displaystyle\sum_{i=1}^n d(n)}{n}
\]
Let $n+1$ be a highly composite number, this finishes.

[b]b)[/b] We claim primes minus 1 that are $1$ mod $6$ work (an infinite number of these exist by Dirichlet). It remains to show that $\frac{a_{p-1}+2}{p} < \frac{a_{p-1}}{p-1}$ or $a_{p-1} > 2p-2$. If $ 6\mid p-1$, then $a_{p-1} > \frac{p-1}{1}+\frac{p-1}{2}+\frac{p-1}{3} + \frac{p-1}6 = 2(p-1)$, as desired.

storage from a while ago

It's easy to see that $a_n=\frac{\sum d_i}{n}$ where d is the divisor function (for 1s, it's a factor floor[n/1] times, then 2s its a factor [n/2] times etc.). Then, for part b) take an odd prime p, which works since the sum in numerator increases only by 1 (from p/p) but denominator is also increased by 1, so it goes to $\frac{x+1}{y+1}<\frac{x}{y}\text{ }\forall x>y$. For part a), there are infinitely many times where $d(n+1)>d(1),d(2),\dots,d(n)$, which suffices because this implies $n(d(1)+d(2)+\dots+d(n+1))>(n+1)(d(1)+\dots+d(n))$, and dividing on both sides by n(n+1) finishes. $\blacksquare$


Bobby is given a question on what he wants his salary to be from his chores.
He is given 2, 5, and 6.
He fits the numbers in $?+?\cdot?$
What is the largest amount of money he can get?

[hide]You can get the most from multiplying 2 big numbers together, so he would want the numbers to be $2+5\cdot6=\boxed{\$32}$.  The 2 is low so you want to add it, not multiply.[/hide]

Accorind to the so-called "BEDMAS," brackets, exponents, division/multiplacation, addition/subraction occur first, respectively, therefore, $2+5/cdot 6=32\neq 42$.

Masoud Zargar

[quote="R49917S"][hide="B FLats answer"]Bflat: Wouldn't that be 42 since there are not parentistheis (i cnt spell that). Because $2+5=7$ and $7\times6=42$[/hide][/quote]

Well, there are no parenthesis. You have to follow the order of operations, which say that multiplication comes before addition  ;)

O yeah!! I forgot

P.S. Did they all of a sudden decide to redo teh look of the forum


Russian Roulette with cash, salaries, health, armor, and different bullet types.

Each person has a gun with six slots, as before, and you load as you wish. PM me your shot configuration, please
[I will post list of all bullet types & armor soon]

Each round, you target somebody. They will choose a number from 1-6; in return you will choose a number from 1-6. This will determine which bullet you will get shot with and they will get shot with.

Obviously if you are hit with a bullet you lose health (unless you have armor; armor could neutralize all damage)

You have 100% health at the beginning, heal 25% every turn, and armor can decrease the amount of damage taken for a hit. Let the games begin!

P.S. There's really less luck and more strategy here.

SIGNUPS:
1. dragon96
2. pacman2812
3. budi713
4. Brut3Forc3
5. cf249

Answering questions:

cf249:
[b]You can have over 100% health, right?[/b]
Well, no.

dragon96:
[b]Will there be alliances?[/b]
That kinda screws up the game, so no.

1=2:

[b]1. What are the bullet types?[/b]
That has been posted
[b]
2. I don't get the armor. When you neutralize damage, say 750%, what exactly does that do? Also, is the neutralization for one time only?[/b]
An example:
Say someone attacks you with a Railgun [White]. You have Pierce-Free Armor.
A diceroll makes the DMG of the railgun 245%. Your armor makes that 45%.
[b]
3. How much money do you have originally, how do you get money, and can you lose money without buying anything?[/b]
That has been posted

ARMOR

$0 Normal Clothes: no armor protection
$2 Wool Clothes: neutralize 10% damage
$6 Heavy Padded Clothes: neutralize 20% damage

$9 Leather Armor: neutralize 25% dmg
$11 Bronze Armor: neutralize 30% dmg

$18 Iron Armor: neutralize 45% dmg
$30 Titanium Armor: neutralize 65% dmg

$55 Ceramic Armor: neutralize 90% dmg, all descriptions with "bullet": -90% atk
$85 Kevlar Armor: neutralize 140% dmg, all descriptions with "bullet": -90% atk

$110 Pierce-Free Armor: neutralize 200% dmg, all descriptions with "dart": -90% atk
$200 Mithril Armor: neutralize 300% dmg, all "poison" weapons: no poison

$500 Imbued Armor: neutralize 750% dmg
$1000 Epic Armor: immunity to all bullets except "Epic Bullet" - Epic Bullets do 25% dmg
HEALING

$10 Shaman's Robe: increase HP regen rate to 32%
$20 Alchemist's Veil: increase HP regen rate to 40%

$35 Healer Stuff: regen = 45%
$70 Field Doctor's Uniform: regen = 50%

$100 Professional Doctor's Suit: regen = 60%
$250 Quantum Physicist = regen = 80%

$500 Epic Heal : regen = 100%

no going over 100% hp

you may have 1 armor and 1 healing at a time. Old suits will be sold for 50% orig value.

1. What are the bullet types?
2. I don't get the armor. When you neutralize damage, say 750%, what exactly does that do? Also, is the neutralization for one time only?
3. How much money do you have originally, how do you get money, and can you lose money without buying anything?

/IN

The message is too small. Please make the message longer before submitting.

This is kinda complicated

I guess i will join for now

JOBS

day = real game day
round = game round

Early Militiaperson: Salary of $1/day, + 25% ATK. Allows you to wear up to "Iron Armor" Alchemist: Salary of
$1/day, +10% HP Regen. Allows you to wear up to "Leather Armor" and "Alchemist Robes".

Pay $7 to get lv 2 job [$4 to switch]
Soldier: Salary of $3/day, + 50% ATK. Allows you to wear up to "Kevlar Armor" Professional Doctor: Salary of
$5/day, +33% HP Regen. Allows you to wear up to "Titanium Armor" and "Professional Doctor Uniform"
Scientist: Salary of $6/day, bonus of$10/round. Neutralize 30% dmg. Allows you to wear "Quantum Physicist's Uniform" but no armor

Lv 3 jobs [pay $50,$10 to switch]
Modern Infantry: Salary of $16/day, + 75% ATK. Allows you to wear up to "Imbued Armor" Neurosurgeon: Salary of
$22/day, +40% HP Regen. Allows you to wear up to "Ceramic Armor" and "Professional Doctor Uniform"
Tycoon: Salary of $30/day, bonus of$45/round. Allows you to wear up to"Ceramic Armor".
Quantum Physicist: Salary of $28/day, bonus of$40/round. Neutralize 60% dmg. Allows you to wear "Quantum Physicist's Uniform" but no armor
Bulletproof Guy: No salary. Get $25 dollars for every bullet level shot at you. Neutralize 500% dmg. No armor allowed. Lv 4 jobs [pay
$250, $75 to switch] General: Salary of$35/day, +100% ATK. Allows you to wear any armor, "Field Doctor Uniform"
Entrepreneur: Salary of $65/day, bonus of$100/round. Up to "Mithril Armor"

Lv 5 jobs [pay $1000] Epic Creator: Wear anything. + 100% ATK. Salary of
$100/day

BULLETS

$ 0 - Thin Air - No damage at all.$1 - Basic Bullet - 25% ATK
$ 1 - Smoke Bomb - Target only has 75% chance of attacking this round. Not cumulative.
$2 - Heavy Bullet - 40% ATK
$ 3 - Grenade - 30-70% ATK
$3 - Smoke Grenade - Target only has 50% chance of attacking this round. Not cumulative.
$ 4 - Poison Dart - 20% ATK, for rest of game, 5% ATK unless cured
$4 - Weakening Dart - Makes target take 150% damage
$ 4 - Weighted Bullet - 55% ATK
$5 - Cash Stealer - You get $ 7 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per
$1.
$ 6 - Poison Spray - 20% ATK, for rest of game, 8% ATK unless cured
$6 - Glaive Dart - 5% ATK for all players, another 35% ATK on target
$ 7 - Massive Bullet - 75% ATK
$7 - Power Grenade - 55-90% ATK
$ 7 - Smoke Screen - Target only has 25% chance of attacking this round. Not cumulative.
$7 - Piercing Dart - 65% ATK, armor has no effect
$ 8 - Plague Dart - 30% ATK, for rest of game, 10% ATK unless cured
$8 - Muter - Target only does half damage for two rounds
$ 9 - Rockets - 45% ATK, If target has armor, rocket does double damage
$10 - Cash Stealer v2 - You get $ 15 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per
$1.
$ 11 - Railgun [Red] - 75-120% ATK
$12 - Smoke Guard - Target only has 10% chance of attacking this round. Not cumulative.
$ 13 - Reflector - Target has 50% chance of attacking self. Not cumulative.
$14 - Super Bullet - 130% ATK
$ 15 - Armor Stealer - Downgrades target armor by 1 level.$16 - Plague Bullet - 45% ATK, for rest of game, 25% ATK unless cured
$ 17 - Silencer - Target only does 1/3 damage for three rounds$19 - Railgun [Yellow] - 110-190% ATK
$ 21 - Shining Reflector - Target has 75% chance of attacking self. Not cumulative.
$23 - Mega Bullet - 180% ATK
$ 25 - Cash Stealer v3 - You get$40 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per $ 1.$28 - Powerful Armor Stealer - Downgrades target armor by 2 levels.
$ 32 - Railgun [White] - 200-360% ATK
$36 - Ultra Smoke Guard - Target has no chance of attacking this round. Not cumulative.
$ 40 - Subduer - Target only does 1/4 damage for four rounds$45 - Great Reflector - Target has 90% chance of attacking self. Not cumulative.
$ 50 - Giga Bullet - 400% ATK
$55 - Cash Stealer vFinal - You get $ 100 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per
$1.
$ 60 - Railgun [Blue] - 400-750% ATK
$65 - Pandemic Bullet - 150% ATK, for rest of game, 70% ATK unless cured
$ 70 - Great Armor Stealer - Downgrades target armor by 5 levels.$75 - Hyper Bullet - 700% ATK
$ 80 - Railgun [Purple] - 550-1000% ATK
$85 - Subduer Plus - Target does no damage for six rounds
$ 90 - Perfect Reflector - Target has 100% chance of attacking self. Not cumulative.
$95 - Armor Killer - Destroys all target armor
$ 100 - KO Bullet - autokills target$125 - Epic Bullet - autokills target, penetrates Epic Armor

[quote="james4l"]BULLETS

$ \$$0 - Thin Air - No damage at all.
$ \$$1 - Basic Bullet - 25% ATK
$ \$$1 - Smoke Bomb - Target only has 75% chance of attacking this round. Not cumulative.
$ \$$2 - Heavy Bullet - 40% ATK
$ \$$3 - Grenade - 30 - 70% ATK
$ \$$3 - Smoke Grenade - Target only has 50% chance of attacking this round. Not cumulative.
$ \$$4 - Poison Dart - 20% ATK, for rest of game, 5% ATK unless cured
$ \$$4 - Weakening Dart - Makes target take 150% damage
$ \$$4 - Weighted Bullet - 55% ATK
$ \$$5 - Cash Stealer - You get $ \$$7 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per $ \$$1.
$ \$$6 - Poison Spray - 20% ATK, for rest of game, 8% ATK unless cured
$ \$$6 - Glaive Dart - 5% ATK for all players, another 35% ATK on target
$ \$$7 - Massive Bullet - 75% ATK
$ \$$7 - Power Grenade - 55-90% ATK
$ \$$7 - Smoke Screen - Target only has 25% chance of attacking this round. Not cumulative.
$ \$$7 - Piercing Dart - 65% ATK, armor has no effect
$ \$$8 - Plague Dart - 30% ATK, for rest of game, 10% ATK unless cured
$ \$$8 - Muter - Target only does half damage for two rounds
$ \$$9 - Rockets - 45% ATK, If target has armor, rocket does double damage
$ \$$10 - Cash Stealer v2 - You get $ \$$15 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per$ \$$1.
$ \$$11 - Railgun [Red] - 75 - 120% ATK
$ \$$12 - Smoke Guard - Target only has 10% chance of attacking this round. Not cumulative.
$ \$$13 - Reflector - Target has 50% chance of attacking self. Not cumulative.
$ \$$14 - Super Bullet - 130% ATK
$ \$$15 - Armor Stealer - Downgrades target armor by 1 level.
$ \$$16 - Plague Bullet - 45% ATK, for rest of game, 25% ATK unless cured
$ \$$17 - Silencer - Target only does 1/3 damage for three rounds
$ \$$19 - Railgun [Yellow] - 110-190% ATK
$ \$$21 - Shining Reflector - Target has 75% chance of attacking self. Not cumulative.
$ \$$23 - Mega Bullet - 180% ATK
$ \$$25 - Cash Stealer v3 - You get $ \$$40 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per $ \$$1.
$ \$$28 - Powerful Armor Stealer - Downgrades target armor by 2 levels.
$ \$$32 - Railgun [White] - 200 - 360% ATK
$ \$$36 - Ultra Smoke Guard - Target has no chance of attacking this round. Not cumulative.
$ \$$40 - Subduer - Target only does 1/4 damage for four rounds
$ \$$45 - Great Reflector - Target has 90% chance of attacking self. Not cumulative.
$ \$$50 - Giga Bullet - 400% ATK
$ \$$55 - Cash Stealer vFinal - You get $ \$$100 from target. If the target does not have the cash, target must go to the blood drive and lose 5% per $ \$$1.
$ \$$60 - Railgun [Blue] - 400 - 750% ATK
$ \$$65 - Pandemic Bullet - 150% ATK, for rest of game, 70% ATK unless cured
$ \$$70 - Great Armor Stealer - Downgrades target armor by 5 levels.
$ \$$75 - Hyper Bullet - 700% ATK
$ \$$80 - Railgun [Purple] - 550 - 1000% ATK
$ \$$85 - Subduer Plus - Target does no damage for six rounds
$ \$$90 - Perfect Reflector - Target has 100% chance of attacking self. Not cumulative.
$ \$$95 - Armor Killer - Destroys all target armor
$ \$$100 - KO Bullet - autokills target
$ \$$125 - Epic Bullet - autokills target, penetrates Epic Armor[/quote]
Wow. Dang. Uh... /in....?

I'm in.

Question:  You can have over 100% health, right?

BONUS ITEMS

Antidote: [2 dollars per potence] Heals poison
What do I mean by potence? For the poisonous stuff they have "x% ATK unless cured", well x is potence.

More confusing than I thought it would be. I buy infinite bullets while the game hasn't started and prices are 0. :D

ok Iguess I'll join. If there is massive inactivity *cough mafia 33*cough* then I will be out of here breaking the speed of light barrier

[quote]Each person has a gun with six slots, as before, and you load as you wish. PM me your shot configuration, please
[I will post list of all bullet types & armor soon]

Each round, you target somebody. They will choose a number from 1-6; in return you will choose a number from 1-6. This will determine which bullet you will get shot with and they will get shot with. [/quote]
Questions:
1-How many bullets can we load in all? How many guns are there in all?
2-If there is more than 1 gun, how will the gun to be used be chosen?
3-Will we determine which slot the bullets are in, or do we only determine the number of bullets in each gun?

SIGNUPS:
1. dragon96
2. pacman2812
3. budi713
4. Brut3Forc3
5. cf249
6. 1=2
7. lotsofmath

Answering questions:

cf249:
[b]You can have over 100% health, right?[/b]
Well, no.

dragon96:
[b]Will there be alliances?[/b]
That kinda screws up the game, so no.

1=2:

[b]1. What are the bullet types?[/b]
That has been posted
[b]
2. I don't get the armor. When you neutralize damage, say 750%, what exactly does that do? Also, is the neutralization for one time only?[/b]
An example:
Say someone attacks you with a Railgun [White]. You have Pierce-Free Armor.
A diceroll makes the DMG of the railgun 245%. Your armor makes that 45%.
[b]
3. How much money do you have originally, how do you get money, and can you lose money without buying anything?[/b]
That has been posted

ProtestanT:
[b]
1-How many bullets can we load in all? How many guns are there in all?[/b]
You have 1 gun with 6 bullet spaces
[b]
2-If there is more than 1 gun, how will the gun to be used be chosen?[/b]
There is one gun.
let me explain:

You have a gun. The person you target also has a gun. You type a number, they also type in a number. This determines the slot number.

[b]3-Will we determine which slot the bullets are in, or do we only determine the number of bullets in each gun?[/b]
Only the number. Random generated slots.

Also: after each duel those two guns are re-scrambled minus the bullet just fired

So, in the first duel, both people are guarenteed to die? (there's 6 bullets, and 6 slots in each gun)

Or do we have to buy bullets to stuff into the guns?

No. For one, you don't have the money to get 6 bullets (you start off with 5 dollars). Second, one dollar bullets do like a total of 25% damage.

Yes, you buy the bullets = huge customization

[quote="james4l"]SIGNUPS:
1. dragon96
2. pacman2812
3. budi713
4. Brut3Forc3
5. cf249
6. 1=2

[/quote]

What about me?

SIGNUPS:
1. dragon96
2. pacman2812
3. budi713
4. Brut3Forc3
5. cf249
6. 1=2
7. lotsamath

One more signup, and we're ready to go.

You can start PMing your bullet choices, job choice, and armor choices

[BTW, Job income starts when all have pm'ed me bullet choices!]

Get ready for a rocky ride  :D

[b]
Prod Mails: Sunday, October 12, 2008

Bullet Choices are Due: Wednesday, October 15, 2008[/b]

Um... I assume it will start shortly after the deadline.  :wink:

DAY BEGINS... NOW!

To those who haven't submitted, too bad, you are now "CPU-controlled" for this round.

Brut3forc3, it is now your turn.

Pick someone to attack.

List:
Brut3forc3
cf249
BoesFX
dragon96
lotsofmath
pacman2812
budi713 [CPU 1-1]
1=2 [CPU 1-2]

I'll just attack1=2, the CPU.
I think making each of us choose numbers is useless since it's completely random.

You can still send your bullet choices...

but they will be null for this "sub-round" only

My Deadline was intended to be a prod

[b]*THIS IS ALSO A PROD FOR BUDI713*[/b]

2 days worth of salary to everyone! WOOHOO!

Anyways, the story:

Brut3forc3 decided to aim at 1=2. He aimed, fired, and...

Well, you see, he forgot his bullets at home. Oops.
1=2 decided to be sneaky and wait outside of Brut3forc3's house. When Brut3forc3 came back home, 1=2 aimed, fired, and...

Oops, he also forgot his bullets at home  :roll:

[b]So that's that. If you need to buy more bullets, now's the time, as with job switches and stuff.[/b]

1=2 and Brut3forc3 are immune this round; they cannot be aimed at.

Brut3forc3
cf249 < On deck
BoesFX < Get ready
dragon96
lotsofmath
pacman2812
budi713 [CPU 1-1]
1=2 [CPU 1-2]


complex analysis
complex analysis unsolved

Find all the roots of the complex equation
$ e^z\equal{}i$

By roots do you mean all $ z$ such that $ e^z\minus{}i\equal{}0$?

Then $ z\equal{}\log(e^z)\equal{}\log(i)\equal{}\ln|i|\plus{}iArg(i)\equal{}i(2n\pi \plus{}\frac{\pi}{2})$ for all integers $ n$.


The first term of an arithmetic sequence is 1, another term of the sequence is 91 and all of the terms of the sequence are integers. How many distinct arithmetic sequences meet these three conditions?

The common difference must be an integer factor of $ 91\minus{}1\equal{}90\equal{}2^1 \cdot 3^2 \cdot 5^1$ Thus there are $ (1\plus{}1)(1\plus{}2)(1\plus{}1)\equal{}\boxed{12}$ sequences.


Suppose there is a right** cylindrical can containing both its top and bottom sections.  If its radius is increased by 2 units, the volume will double.  If the height of this same can is increased by 6 units, its total surface area will double.

Find the exact dimensions of the original can.

** To be clear, this is a usual can where the section of the can between the top part and the bottom part meets each of them at right angles, hence the name "right cylindrical" can.

Are you sure this wasn't recently posted on these forums, because I think I just saw it somewhere in high school recently...

I couldn't be sure that someone else posted it and then deleted their post.  I have never seen it outside of my own.

However, I created my math problem.

You "think" you saw it in high school.

Even if that were true, logically, it has no bearing on it showing up in this forum.  There would not be a logical connection of
likelihood.

You might be suffering from "false memory syndrome."

All you have to do with this is set up two proportions. The rest is just easy algebra.   :)

cincodemayo5590,

yours is a relatively unhelpful post to me.

Please do not bother to post your alleged nonspecific method/suggestions for help and to not show the supporting work and the required solution, as in "The rest..."

Also, the words "easy algebra" are meaningless because they're subjective.


Consider $n$ point from the plane ($n \geq 5$) such that there are no $3$ colinear points from these points.
Show that there exist at least $C_{n-3}^{2}$ convex quadrilateral whose heads are from these points .

"Heads" means : if we have a quadrilateral $ABCD$ the heads are the points $A$ and $B$ ..., Sorry but I don't know what to call that   :blush:  :blush:   :lol:  :lol:

I think you mean 'vertices' (and 'vertex' if there is only one...).
In the other hand, this is the rest of the statement that I'm not sure to understand : It is '3 of the points are collinear' or 'no 3 of the points are collinear' or 'exactly 3 of the points are collinear'??

Pierre.

Thanks for clearing that.  :lol:
For the statement : [b]"no 3 of the points are collinear"[/b]

Thus, the bound is not really good.
For any $5$ of the points, the famous first case of the Erdös-Szekeres problem states that among them we may find $4$ which are the vertices of a convex quadrilateral. In another hand, if $P$ is one of these convex quadrilateral, then $P$ appears at most $n-4$ times in the above reasoning.
Therefore, the number of convex quadrilaterals is at least $\frac {C_n^5} {n-4} = \frac {n(n-1)(n-2)(n-3)} {120} = \frac {n(n-1)(n-2)} {60(n-4)} C_{n-3}^2.$
But it is easy to verify that $n(n-1)(n-2) - 60(n-4) = (n-5)(n-6)(n+8) \geq 0$ for $n \geq 5$.
Thus the above bound is better than desired (it is the same for $n = 5,6$).

Pierre.


[hide]

By definition

$10^1 = x^2 - 3x + 6$

Looks like we got a quadratic on our hands, so let's move everything to the side where $x^2$ is positive

$0 = x^2 - 3x - 4$

Something that multiplies to -4 but adds to -3.. perhaps -4 and 1

$0 = (x - 4)(x + 1)$

sooo $x = 4$ or $-1$

And these don't violate any rules about logarithms so it's $(D)$[/hide]

[quote="MCrawford"]If $\log_{10}{(x^2 - 3x + 6) = 1}$, the value of $x$ is:
\[ \text{(A)}\ 10 \text{ or } 2 \qquad \text{(B)}\ 4 \text{ or } -2 \qquad \text{(C)}\ 3 \text{ or } -1 \qquad \text{(D)}\ 4 \text{ or } -1 \]
\[ \text{(E) none of these} \][/quote]

[hide]$(D)$[/hide]

All you have to do is solve the quadratic equation because 10^0=1, so the answer will be D, wait wouldn't it be just 4, because negatives are extranneous.

[quote="WindSlicer"]No, you can plug the -1 in and see that it's not extranneous.[/quote]

Oh yeah, never mind, because it will still equal 0, allright


If $x,x'$ be the roots of equation $y^2+y-1=0$ and $x>x'$ then find the value of  $5x^2+3x'^2+(7-\sqrt{5})(x+x')$ without solving the equation.


A couple of questions here -- They look similar but I'll prefer it if you can give an individual solution for each problem --

1) $A$ normally comes to the hotel everyday at 6.00p.m , where he is picked up from here by his wife. One day, he arrived 1 hour earlier and so decided to walk home.On the way, he meets his wife,  and they go back together in the car.If they reach home 10 minutes early that day , find the ratio of $A$'s walking speed to that of his wife's driving speed.

2 ) $B$ arrives at a hotel everyday at 6.00p.m , where he is picked up by his wife.It normally takes them one hour to get home .One day , he arrived an hour early and decides to walk home. On the way , he meets his wife , and they reach home 10 minutes earlier than usual.Find the time for which $B$ walked.

Er....could someone move this to a higher level ?? I dont seem to be getting any reply in getting started.... :(


quadratics
modular arithmetic
Congruences

Let $p$ be an odd prime of the form $p=4n+1$. [list=a][*] Show that $n$ is a quadratic residue $\pmod{p}$. [*] Calculate the value $n^{n}$  $\pmod{p}$. [/list]

Recall that $a$ is a $k$-th power iff $a^\frac{p-1}{\gcd(p-1,k)} \equiv 1 \mod p$.

We show that $n$ is a fourth power $\mod p$:
We have $16n \equiv -4 \equiv (2i)^2 \mod p$, thus we want that $2i$ is a square $\mod p$ (here $i$ is such that $i^2 \equiv -1 \mod p$, existing by $p \equiv 1 \mod 4$).
- If $p \equiv 1 \mod 8$, then $2$ is a quadratic residue and $i$ is, too, the latter by $i^{p-1}2 = (-1)^\frac{p-1}4 = 1 \mod p$.
- If $p \equiv 5 \mod 8$, then $2$ is nonsquare $\mod p$ and $i$ is, too, the latter by $i^\frac{p-1}2 \equiv (-1)^\frac{p-1}4 = -1 \not\equiv 1 \mod p$.
In both cases, $2i$ is a quadratic residue, thus there is an $a$ with $a^4 \equiv  n \mod p$.

Or shorter: $(1+i)^4 \equiv (2i)^2 \equiv -4 \mod p$, giving the same result with $a=1+i$.

This solves directly a. (and shows more), but also b. follows: $n^n \equiv (a^4)^\frac{p-1}4 = a^{p-1} \equiv 1 \mod p$.


ratio
geometry
circumcircle
geometry unsolved

In the following, the point of intersection of two lines $ g$ and $ h$ will be abbreviated as $ g\cap h$.

Suppose $ ABC$ is a triangle in which $ \angle A \equal{} 90^{\circ}$ and $ \angle B > \angle C$. Let $ O$ be the circumcircle of the triangle $ ABC$. Let $ l_{A}$ and $ l_{B}$ be the tangents to the circle $ O$ at $ A$ and $ B$, respectively.

Let $ BC \cap l_{A} \equal{} S$ and $ AC \cap l_{B} \equal{} D$. Furthermore, let $ AB \cap DS \equal{} E$, and let $ CE \cap l_{A} \equal{} T$. Denote by $ P$ the foot of the perpendicular from $ E$ on $ l_{A}$. Denote by $ Q$ the point of intersection of the line $ CP$ with the circle $ O$ (different from $ C$). Denote by $ R$ be the point of intersection of the line $ QT$ with the circle $ O$ (different from $ Q$). Finally, define $ U \equal{} BR \cap l_{A}$. Prove that
\[ \frac {SU \cdot SP}{TU \cdot TP} \equal{} \frac {SA^{2}}{TA^{2}}.
\]

My Sketch showed that his problem is wrong.

Mathx, are you sure that the problem is wrong? A credible friend of min claims that he has proven it. Could you please post the picture (figure) of the counterexample? :wallbash:

[quote="pardesi"]i suppose the biggest problem here is drawing the figure itself :rotfl:[/quote]
agree,the pictureis really too complex,It hard too draw it down and make it  clear

But it can be drawn... how to solve it?  :D

But then again... who cares?  :D  :D  :D

  dg

The conclusion is true,
here is a way to prove:
let the tagent line pass through $ {C}$ neet $ {AB}$ at point $ {E'}$ ,by Pascal Theory we get that :$ {E',D,S}$ are collinear.  Thus ,$ {E \equal{} SD\cap AB \equal{} E'}$, So we get that: $ {EC}$ is tagent to $ {\odot O}$,
[b][1][/b]
$ {TA^2 \equal{} TR*TQ}$;$ {SA^2 \equal{} SB*SC}$ $ {\rightarrow}$ $ {\frac {SA^2}{TA^2} \equal{} \frac {SB*SC}{TR*TQ}}$;

Let $ {RQ\cap CB \equal{} X}$
[b][2][/b]
By Melelaus :
 $ {\frac {XR}{RT}*\frac {TU}{SU}*\frac {SB}{CX} \equal{} 1}$;$ {\frac {XQ}{TQ}*\frac {PT}{PS}*\frac {SC}{XC} \equal{} 1}$;
[b][3][/b]
EASILY GET: $ {CX*XB \equal{} XR*XQ}$;

By {[1][2][3]} $ {\Longrightarrow}$
$ {\frac {SP}{TP}*\frac {SU}{TU} \equal{} (\frac {XR}{TR}*\frac {SB}{XB})*(\frac {XQ}{QT}*\frac {SC}{XC}) \equal{} \frac {SB*SC}{TR*TQ} \equal{} \frac {SA^2}{TA^2}}$  QED. :wink:

It's very simple!

Observe that $SA^2$ = $SB\cdot SC$
and $TA^2$ = $TQ\cdot TR$.
Now, simply apply sine law with some angle chasing to get
$\frac{SU\cdot SP}{TU\cdot TP}$ = $\frac{SB\cdot SC}{TQ\cdot TR}$.
This proves the result.


Find a solution of the simultaneous equations

$ xy\plus{}yz\plus{}zx\equal{}12$

$ xyz\equal{} 2\plus{}x\plus{}y\plus{}z$

in which all of $ x, y, z$ are positive and prove that it is the only such solution.

Show that a solution exists in which $ x, y, z$ are real and distinct.

hello, using the AM-GM inequality for three positive real numbers we get
$ \frac{xy\plus{}yz\plus{}zx}{3}\ge \sqrt[3]{(xyz)^2}$
from here we get with
$ xy\plus{}yz\plus{}zx\equal{}12$
the inequality $ 8\geq xyz$.
From the second equation we get
$ \frac{xyz}{4}\equal{}\frac{2\plus{}x\plus{}y\plus{}z}{4}\geq\sqrt[4]{2xyz}$
after a little algebra we get
$ xyz\geq 8$ therefore is $ x\equal{}y\equal{}z\equal{}2$ the unique positive solution of our system.
Sonnhard.


conics
ellipse
geometry proposed
geometry

Let $ ABB',ACC'$ collinear and denote $ D\equal{}BC\cap B'C'$. Prove that:
$ AB\plus{}BD\equal{}AC'\plus{}C'D\Leftrightarrow AB'\plus{}B'D\equal{}AC\plus{}CD$

It is very easy to check that: In a triangle with side $ a,b,c$ and angle $ A,B,C$ we have: $ \frac{a\plus{}b\plus{}c}{a}\equal{}\frac{2}{1\minus{}tan\frac{B}{2}.tan\frac{C}{2}}$
Now:
$ AB\plus{}BD\equal{}AC'\plus{}C'D\Leftrightarrow \frac{AB\plus{}BD}{AD}\equal{}\frac{AC'\plus{}C'D}{AD}\Leftrightarrow tan\frac{\widehat{ADB}}{2}.tan\frac{\widehat{DAB}}{2}\equal{}tan\frac{\widehat{ADC'}}{2}.tan\frac{\widehat{DAC'}}{2}$
And:
$ AB'\plus{}B'D\equal{}AC\plus{}CD\Leftrightarrow \frac{AB'\plus{}B'D}{AD}\equal{}\frac{AC\plus{}CD}{AD}\Leftrightarrow tan\frac{\widehat{ADB'}}{2}.tan\frac{\widehat{DAB'}}{2}\equal{}tan\frac{\widehat{ADC}}{2}.tan\frac{\widehat{DAC}}{2}$
$ \Leftrightarrow cot\frac{\widehat{ADC'}}{2}.tan\frac{\widehat{DAB}}{2}\equal{}cot\frac{\widehat{ADB}}{2}.tan\frac{\widehat{DAC'}}{2}$
It is OK!

[quote="mihai miculita"]$ nktp$ $ wrote$:
If $ {BB'}\cap{CC'} \equal{} \{A\}$ and $ AB \plus{} BD \equal{} AC' \plus{} C'D$ then: $ AB' \plus{} B'D \equal{} AC \plus{} CD$. ($ FALSE!$)
$ HINT$:
$ AB \plus{} BD \equal{} AC' \plus{} C'D$ $ \Longleftrightarrow$ $ {points}$ $ B$ and $ C'$ $ \in$ $ {ellipse1}$($ A$; $ D$); and
$ AB' \plus{} B'D \equal{} AC \plus{} CD$ $ \Longleftrightarrow$ $ {points}$ $ B'$ and $ C$ $ \in$ $ {ellipse2}$($ A$; $ D$).[/quote]

Please recheck your idea! This problem is complete true!

$ {Let}$     ($ P_4$): $ lines$ $ BB'$, $ CC'$, $ BC$ and $ B'C'$ $ are$ $ tangent$ $ to$ $ the$ $ same$ $ circle$; $ then$:
($ P_1$) $ \Longleftrightarrow$ ($ P_4$) $ \Longleftrightarrow$ ($ P_2$)
$ {and}$
($ P_4$) $ \Longleftrightarrow$ ($ P_3$).

From $ AB \plus{} BD \equal{} AC' \plus{} C'D$ $ \Longrightarrow$ $ AB \minus{} C'D \equal{} AC' \minus{} BD$ $ ,(1)$

From $ (1),$ we conclude that the convex quadrilateral $ ABDC',$ has an excircle, so be it $ (O').$

Let $ K,$ $ L,$ $ M,$ $ N,$ the tangent points of $ (O'),$ to the sidelines $ AB,$ $ C'D,$ $ BD,$ $ AC',$ respectively.

It is easy to show now, that $ AB' \plus{} B'D \equal{} AB' \plus{} B'K \plus{} LD$ $ ,(2)$ and $ AC \plus{} CD \equal{} AC \plus{} CN \plus{} MD$ $ ,(3)$

Because of $ AB' \plus{} B'K \equal{} AK \equal{} AN \equal{} AC \plus{} CN$ and $ LD \equal{} MD,$ from $ (2),$ $ (3),$ we conclude that $ AB' \plus{} B'D \equal{} AC \plus{} CD$ and the proof is completed.

Kostas Vittas.


geometry
trigonometry
trig identities
Law of Sines
Law of Cosines
area of a triangle
Heron\u0027s formula

Find the area of the triangle which has the side lengths $\sin \theta ,\ \sin 2\theta , \sin 3\theta \ (0\leq \theta <2\pi).$

[hide="hint"]
$0<\sin{\theta}$, $\sin{2\theta}$, $\sin{3\theta}$
the first restricts to $0<\theta<\pi$
the second restricts to $0<\theta<\frac{\pi}{2}$
the third restricts to $0<\theta<\frac{\pi}{3}$
So, $0<\theta<\frac{\pi}{3}$
$0<\sin{\theta}<\frac{\sqrt{3}}{2}$
I think this is correct. I don't how to do the problem and this is as far as I can get. :( You can also draw the triangle and try to force out the area using Law of Sines and Law of Cosines.[/hide]

[quote="lotrgreengrapes7926"][quote="t0rajir0u"][hide="Bigger Hint"]
Extended Law of Sines
[/hide][/quote]
[hide]I don't know the names of anything. Is that $Area=\frac{1}{2}ab\sin{C}$[/hide][/quote]

[hide="The Extended Law of Sines"]
$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} = 2R$, where $R$ is the circumradius.[/hide]
[hide="Why this is such a big hint"]
Set $R = \frac{1}{2}$.  ;) [/hide]

Why can you set $\ R= \frac{1}{2}$ and Heron just becomes messy.

[quote="mna851"]Why can you set $\ R= \frac{1}{2}$[/quote]

It's actually necessary to look at tummykung's solution for a rigorous proof of why this must be.  But solving in this way is so elegant that I cannot resist it  :D

[quote="tummykung"][hide="My Solution"]
let A,B,C is the angle opposite side lenghts $sin 3\theta ,2\theta ,\theta$ respectively

I use Law of Cosines
$\sin^23\theta = \sin^2\theta+\sin^22\theta-2sin\theta\sin2\theta\cos A$
$(3-4\sin^2\theta)^2=1+4\cos^2\theta-4\cos\theta\cos A$
$(4\cos^3\theta-3\cos \theta)(4\cos \theta)=4\cos\theta\cos A \qquad(\because 3-4\sin^2\theta = 4\cos^2\theta-1)$
$\cos A =4\cos^3\theta -3cos\theta=\cos 3\theta$
from condition of $\theta$ then I have $A=3\theta$

then use Law of Sines
\[ \frac{\sin 3\theta}{\sin A}=\frac{\sin 2\theta}{\sin B}=\frac{\sin \theta}{\sin C}  \]
hence $B=2\theta$ and $C=\theta$
sum of angle in triangle is $\pi$
so $\theta = \frac{\pi}6 \therefore A,B,C\ is\ \frac{\pi}2,\frac{\pi}3,\frac{\pi}6$ respectively
hence the area is $\frac 12 \big(\frac 12\big)\big(\frac{\sqrt 3}2\big)=\boxed{ \frac{\sqrt 3}{8}}$
[/hide]
ps.sorry for my bad english  :P[/quote]
Oh nooo... I find my errors
first [quote]$(4\cos^3\theta-3\cos \theta)(4\cos \theta)=4\cos\theta\cos A \qquad(\because 3-4\sin^2\theta = 4\cos^2\theta-1)$[/quote]
It shoud be $(4\cos^3\theta-3\cos \theta)(4\cos \theta)=\mathbb{-}4\cos\theta\cos A$
then $\cos A =\mathbb{-}\cos 3\theta$
and then I have $A = \pi -3\theta$

second, I can't conclude that $X=Y$ if $\sin X=\sin Y$ (X,Y is angle in triangle)
if I have $\sin X=\sin Y$ (X,Y is angle in triangle) I can only know $X=Y,\pi-Y$

$\therefore B = 2\theta ,\pi-2\theta$
and $C = \theta ,\pi-\theta$

Oh...I don't sure to select one of two choises
I force them by Law of Cosines again (with other side)
Ugh!... so now I can conclude that $B=2\theta$ and $C=\theta$

hence the Area of triangle is \[ \frac{1}{2}\sin\theta \sin 2\theta\sin 3\theta\qquad,0<\theta<\frac{\pi}3 \]
here is a graph of $y=\frac{1}{2}\sin\theta \sin 2\theta\sin 3\theta\qquad,0<\theta<\frac{\pi}3$
[img]http://img269.imageshack.us/img269/8724/until5hx.jpg[/img]

upper bound is about 0.27?
what is real value of its?  :rotfl:

That's right. The answer is $\frac{1}{2}\sin \theta \sin 2\theta \sin 3\theta \ \left(0<\theta <\frac{\pi}{3}\right).$

[quote="luimichael"]Can it be solved by using Heron's Formula ?[/quote]
like this?
\[ Area =\frac{\sqrt{(\sin\theta+\sin2\theta+\sin3\theta)(-\sin\theta+\sin2\theta+\sin3\theta)(\sin\theta-\sin2\theta+\sin3\theta)(\sin\theta+\sin2\theta-\sin3\theta)}}{4} \]

[b]Direct Calculation[/b]
Let $a = \sin \theta, b = \sin 2\theta, c = \sin\ 3\theta$.
Then $\cos C$
$=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{\sin^2\theta -(\sin^2 3\theta-\sin^2 2 \theta)}{2 \sin \theta \sin 2 \theta}$
$=\frac{ \sin^2 \theta -( \sin 3 \theta + \sin 2 \theta)( \sin 3 \theta - \sin 2 \theta)}{2 \sin \theta sin 2 \theta}$
$=\frac{\sin^2 \theta -2 \sin \frac{5 \theta}2 \cos \frac{\theta}2 \cdot 2 \cos \frac{5 \theta} 2 \sin \frac{\theta}2}{2 \sin \theta \sin 2 \theta}$
$=\frac{\sin^2 \theta - \sin 5 \theta \sin \theta}{2 \sin \theta \sin 2 \theta}$
$=\frac{\sin 5 \theta - \sin \theta}{2 \sin 2 \theta}$
$=\frac{2 \cos 3 \theta \sin 2 \theta}{2 \sin 2\theta}= \cos 3 \theta$

Hence,
area of the triangle = $\frac 1 2 ab \sin C$ =$\frac 1 2 \sin \theta \sin 2 \theta \sin 3 \theta$.

Note:
$\sin C = \sqrt{1- \cos^2 C} = \sqrt{ 1- \cos^2 3 \theta} = \sqrt{\sin^2 3 \theta} = \sin 3 \theta$ because both $\sin C$ and $\sin 3 \theta$ are positive.

[b]By Heron's Formula[/b]:
In the following discussion, I would replace $\theta$ by $x$ for easy typing.
The required area
$=\frac 1 4 \sqrt{(\sin x + \sin 2x + \sin 3x)(-\sin x + \sin 2x + \sin 3x)(\sin x - \sin 2x + \sin 3x)(\sin x + \sin 2x - \sin 3x)}$
$=\frac 1 4 \sqrt{(2 \sin x \cos x + \sin 2x)(2 \cos x \sin x + \sin 2x)(2 \sin 2x \cos x - \sin 2x)(-2 \cos 2x \sin x + \sin 2x)}$
$=\frac 1 4 \sqrt{ \sin^2 {2x} (2 \cos x +1)( 2 \cos x -1) \cdot 4 \sin^2 x (\cos 2x + \cos x)(- \cos 2x + \cos x)}$
$= \frac 1 2 \sin x \sin 2x \sqrt{(2 \cos x +1)(2 \cos x -1)(\cos 2x + \cos x)(-\cos 2x+\cos x) }$
$=\frac 1 2 \sin x \sin 2x \sqrt{(4 \cos^2 x -1)(2 \sin \frac{3x}{2} \sin \frac{x}{2})(2 \cos \frac{3x}{2} \cos \frac{x}{2})}$
$=\frac 1 2 \sin x \cdot \sin 2x \sqrt{(4 \cos^ 2x-1) \sin x \cdot \sin 3x}$
$=\frac 1 2 \sin x \cdot \sin 2x \sqrt{(3 - 4 \sin^ 2 x) \sin x \cdot \sin 3x}$
$=\frac 1 2 \sin x \cdot\sin 2x \sqrt{(3 \sin x - 4 \sin^3 x) \sin 3x}$
$=\frac 1 2 \sin x \cdot \sin 2x \sqrt{\sin 3x \sin 3x}$
$=\frac 1 2 {\sin x}\cdot{ \sin 2x }\cdot{\sin 3x}$    Q.E.D.


induction
trigonometry
MATHCOUNTS
SFFT
special factorizations
geometry
trig identities

[hide]This must be in the form $x\sqrt{2}+y\sqrt{3}$ or $x+y\sqrt{6}$.

By inspection, we find that $\sqrt{2}+\sqrt{3}$ works.

We could also have set it equal to one of those and solved.  [/hide]

[quote="anirudh"][hide]This problem is in the form of $a\sqrt{2}+\sqrt{3}$. Now, time for guess and check. You get $(1, 1)$ for $a, b$.

Also, one could do it as:

$\left(a\sqrt{2}+b\sqrt{3}\right)=\sqrt{5+2\sqrt{6}}$.

By squaring it becomes $2x^{2}+3y^{2}+2xy\sqrt{6}=5+2\sqrt{6}$

$2x^{2}+3y^{2}+2\sqrt{6}(xy+1)=0$

Simon's Favorite Factoring Trick and you're done and get the same answer. :) [/hide]

EXTENSION:

Simplify: ${\sqrt{\sqrt{\sqrt{2+\sqrt{2}}}}}$[/quote]
Simon's favorite factoring trick is overkill and might not even work in this case. It's like proving the Egyptian Fractions with induction, using the law of sines in a right triangle, or proving Fermat's Little Theorem with Euler's Totient Theorem. Here's a hint, equate the rationals and the radicals and you will get an easily solvable system of equations. Also, you assumed it was in the form
$a \sqrt{3}+b \sqrt{2}$ but it could be $a+b \sqrt{6}$.

We can always set the part with the radical equal and the part w/o equal...

[quote="anirudh"][hide]This problem is in the form of $a\sqrt{2}+\sqrt{3}$. Now, time for guess and check. You get $(1, 1)$ for $a, b$.

Also, one could do it as:

$\left(a\sqrt{2}+b\sqrt{3}\right)=\sqrt{5+2\sqrt{6}}$.

By squaring it becomes $2x^{2}+3y^{2}+2xy\sqrt{6}=5+2\sqrt{6}$

$2x^{2}+3y^{2}+2\sqrt{6}(xy+1)=0$

Simon's Favorite Factoring Trick and you're done and get the same answer. :) [/hide]

EXTENSION:

Simplify: ${\sqrt{\sqrt{\sqrt{2+\sqrt{2}}}}}$[/quote]

Is there an easy way to do this question, Anirudh.

Then do you want to supply us with a solution to your problem?

Well, I know one thing.

$\sqrt{2+\sqrt{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}-\sqrt{\frac{2-\sqrt{2}}{2}}$

Well we could write it as $(2+\sqrt{2})^{1/8}$ then use the binomial theorem but that would be both messy and above mathcounts.

That wouldn't really help since it won't break it into a finite number of square roots, I don't think.

[quote="anirudh"]No. I don't know how to do it. That's why I posted it.[/quote]

Ah, so THAT's why you called it an extension, and somehow KNEW it must be an extension of our current topic.

I believe that this idea is above MATHCOUNTS level.  Locked until somebody pms me a solution within MC.


geometry
rectangle
incenter
parallelogram
geometry proposed

Let a quadrangle $ ABCD$ so that radius of incircle of triangles $ ABC,BCD,CDA,DAB$ are equal. Prove that: $ ABCD$ is a rectangle!

Let $ I_A,\ I_B,\ I_C,\ I_D$ be incenters and $ r_A,\ r_B,\ r_C,\ r_D$ inradii of the $ \triangle ABD,\ \triangle BCA,\ \triangle CDB,\ \triangle DAC.$

The proposition "$ ABCD$ is cyclic $ \Longrightarrow\ r_A \plus{} r_C \equal{} r_B \plus{} r_D$" was proved in the thread [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=39268[/url] . The converse proposition "$ r_A \plus{} r_C \equal{} r_B \plus{} r_D$ $ \Longrightarrow\ ABCD$ is cyclic" was proved in the thread [url]http://www.mathlinks.ro/viewtopic.php?t=77259[/url] (the thread has a misleading title). Thus $ r_A \equal{} r_B \equal{} r_C \equal{} r_D$ makes $ ABCD$ at least cyclic.

The proposition "$ ABCD$ is cyclic $ \Longrightarrow I_AI_BI_CI_D$ is a rectangle" is well known and a stronger converse proposition "$ ABCD$ is not cyclic $ \Longrightarrow I_AI_BI_CI_D$ is not cyclic" was proved in the thread [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=23192[/url], post 2, step 7. Combining "$ I_AI_BI_CI_D$ is a rectangle" and $ r_A \equal{} r_B \equal{} r_C \equal{} r_D$ makes $ ABCD$ a rectangle.

Another solution: (may be better I think  :P )
Construct parallelograms $ ABB'C,ACD'D$ we have:
$ \Delta ABC \equal{} \Delta B'CB,\Delta ADC \equal{} \Delta D'CD,\Delta ABD \equal{} \Delta CB'D'$
Then $ r_{BCD} \equal{} r_{CDD'} \equal{} r_{CD'D} \equal{} r_{B'CB}$ with $ BDD'B'$ is a parallelograms!
Then easy to check that: $ C \equal{} BD'\cap B'D$
It mean $ ABCD$ is a parallelograms!
Then it's OK! (Clearly $ AC \equal{} BD$ by formula $ r \equal{} \frac {S}{p}$)


I posted this in Pre-Olympiad and they told me to put it here
So here goes

Problem 1:

Unit square $S_{1}$, whose vertices are $A_{1},B_{1},C_{1},D_{1}$, is drawn in a plane.
Square $S_{2}$, whose vertices are $A_{2},B_{2},C_{2},D_{2}$, is drawn with its vertices on $S_{1}$ such that $A_{1}A_{2}=A_{2}B_{1}$.
An infinite number of more squares are drawn in a similar fashion:
Square $S_{x+1}$, whose vertices are $A_{x+1},B_{x+1},C_{x+1},D_{x+1}$, is drawn with its vertices on square $S_{x}$ such that $A_{x}A_{x+1}=(2^{x}-1)A_{x+1}B_{x}$.
Determine the area of $S_{\infty}$.

[hide="as far as Ive gotten to finding the answer"]
The area of $S_{\infty}$ can be expressed as
$\prod_{k=1}^{\infty}\frac{(2^{k}-1)^{2}+1}{2^{2k}}$
[/hide]

Problem 2:

Four unit squares are placed together to form a $2\times2$ square. One point is randomly selected from each unit square, and line segments are drawn in to create the quadrilateral that has these 4 points as vertices. What is the probability that this quadrilateral's area is greater than 1?

wow - are these problems really that hard?


Did anyone participate this year?  :)

My school had two junior varsity teams and one varsity.  The varsity team got 48 of the 80 multiple choice in Part I correct.


when i typed the answer in it said it was wrong... :mad:

Theprodigy, please do not go on the alcumus problem forum just to search for answers, as this is considered cheating.

@ shangdevin i totally agree with you.
@the prodigy it is fixed now

[quote=The Archer][quote="GameBot"]Automobile license plates for a state consist of four letters followed by a dash and two single digits. How many different license plate combinations are possible if exactly one letter is repeated exactly once, but digits cannot be repeated?
[asy]size(150);
draw((0,0)--(0,5)--(10,5)--(10,0)--cycle);
label("\Huge{CHIC - 03}",(1,3)--(9,3),S);
label("\small\emph{State of Excellence}",(1,1)--(9,1),S);
draw((0.5,3.5)--(0.5,4.5)--(2,4.5)--(2,3.5)--cycle);
label("\footnotesize 5-03",(1.25,4));
draw((9.5,3.5)--(9.5,4.5)--(8,4.5)--(8,3.5)--cycle);
label("\footnotesize FX",(8.75,4));[/asy][/quote]

I dissected this problem into two different categories: choices for letter combinations and digits. By the Fundamental Counting Principle, we should be able to multiply these two numbers to get the total combinations.

[b]Letter Combinations:[/b]
The positions the repeated letter could take number $ \binom{4}{2}$ (choosing two slots out of a total of 4) and there are 26 choices for what exactly the letter is. The remaining two have 25 and 24 choices respectively. (Note: letters can't be repeated.)

$ 6*26*25*24 \equal{} 93600$
[b]
Digit Combinations:[/b]
10 choices for the first, and 9 for the second accounting for no repetition of the digits.

$ 9*10 \equal{} 90$

In total:

$ 93600(90) \equal{} \boxed{8424000 \mbox{ combinations}}$[/quote]

What about the permutation of the first 3 letters you chose?? :(

[hide=Alternate Solution]
You can choose the first three letters in 26 choose 3 ways. The repeated letter can be chosen in 3 ways(you can repeat any of the first three letters). Since order does matter we need to take care of permutations we can permute these letters in 12 ways(there would be 4! ways to choose them if they were different but since one letter is repeated we have to divide by 2!).

We can arrange the number in 90 ways(10*9)

So the total number of arrangements is( 26 choose 3*3)12*90 which is equal to 8,424,000
[/hide]

Why can't I do
[hide]10 * 9 = 90 for the possibilities for the numbers. For the letters we can do 26 choose 3 and then multiply that by 3 to choose the letter which will repeat. Then, to order the letters, we can do 4! * 26 choose 3 * 3 = 24 * 2600 * 3 = 187200. 187200 * 90 = [b]16848000[/b].[/hide]
Why is that wrong?

@above, yeah, that was what I answered too
this problem is kind of unclear

[hide=Solution]There are $26$ options for the repeated letter and $\binom{25}{2}$ ways to pick the other two letters. There are $12$ arrangements. There are $10 \cdot 9$ ways to pick the digits, so our answer is $26 \cdot \binom{25}{2} \cdot 12 \cdot 10 \cdot 9 = \boxed{8424000}$.[/hide]

Just deal with the digits first(10x9, this is not too hard to see), and then you want to go ahead and deal with the letters, by picking out the same ones first, with 26 choices and arranging them in 4C2 ways, and then the 2 different out of the remaining 25, and getting 25C2, and the 2 different can be arranged in 2! ways.


trigonometry
LaTeX
Pythagorean Theorem
geometry

[hide="Pythagorean"]
$\cos^{2}x+\sin^{2}x = 1 \Rightarrow \frac{\cos^{2}x }{\sin^{2}x}+1 = \frac{1}{\sin^{2}x}\Rightarrow \cot^{2}x+1 = \csc^{2}x \ \ \ \text{QED}$
If that is what you meant
[/hide]

[quote="cdymdcool"]How to prove 1+cotX^2=cscx^2[/quote]

[hide]take the identity $\cos^{2}x+\sin^{2}x=1$  and divide everything by $\sin^{2}x$
And by the way, use $\LaTeX$ next time, it makes it much easier to read equations.[/hide]

[hide="Easier Solution"]Or, if you're new to trig and you don't understand the proof of the Pythagorean identities, we can do a proof of it from our right triangle trig.
Let us have a right triangle with legs A and B, and hypotenuse C.
We know that $A^{2}+B^{2}= C^{2}$.
Now, we need to represent these as trigonometric functions.
Without Loss of Generality, let angle $\alpha$ be the angle between A and C.
Then, we have $\sin \alpha = \frac{B}{C}$ and $\cos^{2}\alpha = \frac{A}{C}$
To show the identity, we square each of these.
$\sin^{2}\alpha = \frac{B^{2}}{C^{2}}$ and $\cos^{2}\alpha = \frac{A^{2}}{C^{2}}$
Then add them:
$\sin^{2}\alpha+\cos^{2}\alpha = \frac{A^{2}+B^{2}}{C^{2}}$

However, we know that from the Pythagorean theorem, the numerator $A^{2}+B^{2}$ is equal to $C^{2}$, which makes the right hand side = $\frac{C^{2}}{C^{2}}= 1$
Therefore,
$\sin^{2}\alpha+\cos^{2}\alpha = 1$
From here, it is a matter of dividing by $\sin^{2}\alpha$
[/hide]

[quote="DiscreetFourierTransform"]Or, if you're new to trig and you don't understand the proof of the Pythagorean identities, we can do a proof of it from our right triangle trig.
Let us have a right triangle with legs A and B, and hypotenuse C.
We know that $A^{2}+B^{2}= C^{2}$.
Now, we need to represent these as trigonometric functions.
Without Loss of Generality, let angle $\alpha$ be the angle between A and C.
Then, we have $\sin \alpha = \frac{B}{C}$ and $\cos^{2}\alpha = \frac{A}{C}$
To show the identity, we square each of these.
$\sin^{2}\alpha = \frac{B^{2}}{C^{2}}$ and $\cos \alpha = \frac{A^{2}}{C^{2}}$
Then add them:
$\sin^{2}\alpha+\cos^{2}\alpha = \frac{A^{2}+B^{2}}{C^{2}}$

However, we know that from the Pythagorean theorem, the numerator $A^{2}+B^{2}$ is equal to $C^{2}$, which makes the right hand side = $\frac{C^{2}}{C^{2}}= 1$
Therefore,
$\sin^{2}\alpha+\cos^{2}\alpha = 1$
From here, it is a matter of dividing by $\sin^{2}\alpha$[/quote]

Hide, please.
[hide="Slightly faster"]Read DFT's definitions.
We know:
\[\cot(\alpha) = \frac{A}{B}\]

\[\csc{\alpha}= \frac{C}{B}\]

(these are the definitions)

From the Pythagorean Theorem,

\[A^{2}+B^{2}= C^{2}\]

\[\left(\frac{A}{B}\right)^{2}+1 = \left(\frac{C}{B}\right)^{2}\],

where the last equality comes from dividing by $B^{2}$.

Substituting, we get:

\[\cot^{2}(\alpha)+1 = \csc^{2}(\alpha)\]

QED.[/hide]


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Since the deadline for submitting solutions to Mathematical Reflection third issuse 's problem column has passed, I can post this problem for discussion.

Given $ABC$ be a triangle, and $P$ be a point inside the triangle. Let $A', B', C'$ be the intersections of $AP, BP, CP$ with the triangle's sides. Through $P$, we draw a perpendicular to $PA$ that intersects $BC$ at $A_{1}$. Define $B_{1}, C_{1}$ analogously. Let $P^{*}$ be the isogonal conjugate of $P$ with respect to $A'B'C'$. Show that $A_{1}, B_{1}, C_{1}$ lie on a line that is perpendicular to $PP^{*}$.

In fact, one can prove a stronger statement: if $A''B''C''$ be the pedal triangle of $P$ with respect to $A'B'C'$, then the line $A_{1}, B_{1}, C_{1}$ is the polar of the point $P$ with respect to the circumcircle of triangle $A''B''C''$.  :)

We prove the stronger version of the problem:

$\triangle A'B'C'$ is the cevian triangle of $P$ WRT $\triangle ABC.$ $X,Y,Z$ are the projections of $P$ on $B'C',$ $C'A',$ $A'B'.$ Perpendiculars to $PA,PB,PC$ at $P$ cut $BC,CA,AB,$ at $A_1,B_1,C_1,$ respectively. It's known that these points are collinear (orthopolar $\tau$ of P WRT ABC). Then $\tau$ is the polar of $P$ WRT $\odot(XYZ).$

[b]Proof:[/b] Let the perpendicular to $PA$ at $P$ cut $A'B'$ and $A'C'$ at $N,M.$ $U \equiv MN \cap YZ$ and $A_2$ is the reflection of $P$ about $U.$ $\angle MNZ=\angle ZPA'=\angle ZYA'$ $\Longrightarrow$ $MNZY$ is cyclic and since $UP$ is clearly tangent to $\odot(PYA'Z),$ then we have ${UA_2}^2=UP^2=UY \cdot UZ=UM \cdot UN$ $\Longrightarrow$ circle $(U)$ with diameter $\overline{PA_2}$ is orthogonal to both $\odot(XYZ)$ and $\odot(MNZY)$ $\Longrightarrow$ $P,A_2$ are conjugate points WRT $\odot(XYZ)$ $\Longrightarrow$ $A_2$ is on the polar of $P$ WRT $\odot(XYZ).$ It also follows that $(P,A_2,M,N)=-1$ $\Longrightarrow$ $A'(P,A_2,C',B')=-1,$ but $A'(P,A_1,C',B')-1$ $\Longrightarrow$ $A'A_1 \equiv A'A_2 \equiv BC$ $\Longrightarrow$ $A_1 \equiv A_2,$ hence $A_1$ lies on the polar of $P$ WRT $\odot(XYZ).$ Analogously, $B_1$ and $C_1$ lie on the polar of $P$ WRT $\odot (XYZ)$ $\Longrightarrow$ $\tau \equiv A_1B_1C_1$ is the polar of $P$ WRT $\odot(XYZ).$

Dear Mathlinkers,
the reference in question
Nguyen K. L., 	O13, Mathematical Reflections 3 (2006) ; http://reflections.awesomemath.org/
Sincerely
Jean-Louis


Determine all continuous functions $ f: \mathbb {R}\to\mathbb {R}$ such that:
\[ f'\left(\frac {x\plus{}y}{2}\right) \equal{}\frac {f(x)\minus{}f(y)}{x\minus{}y}\]
for all reals $x$ and $y$.

$ f'(\frac {x \plus{} y}{2})$ $ \equal{} \frac {f(x) \minus{} f(y)}{x \minus{} y}$

$ x \equal{} 1 \plus{} z, y \equal{} 1 \minus{} z \implies 2zf'(1) \equal{} f(1 \plus{} z) \minus{} f(1 \minus{} z)$
$ x \equal{} 1 \plus{} z, y \equal{} z \minus{} 1 \implies 2f'(z) \equal{} f(1 \plus{} z) \minus{} f(z \minus{} 1)$
$ x \equal{} 1 \minus{} z, y \equal{} z \minus{} 1 \implies 2(1 \minus{} z)f'(0) \equal{} f(1 \minus{} z) \minus{} f(z \minus{} 1)$

$ \therefore f'(z) \equal{} zf'(1) \plus{} (1\minus{}z)f'(0)  \implies f'(z) \equal{} 2az \plus{} b$ for constants $ 2a \equal{} f'(1) \minus{}f'(0), b \equal{} f'(0)$
$ \implies \boxed{f(z) \equal{} az^2 \plus{} bz \plus{} c}$

It can be checked by substitution that this form satisfies the equation for any constants $ a$, $ b$ and $ c$

[quote="PhilG"]$ f'(\frac {x \plus{} y}{2})$ $ \equal{} \frac {f(x) \minus{} f(y)}{x \minus{} y}$

$ x \equal{} 1 \plus{} z, y \equal{} 1 \minus{} z \implies 2zf'(1) \equal{} f(1 \plus{} z) \minus{} f(1 \minus{} z)$
$ x \equal{} 1 \plus{} z, y \equal{} z \minus{} 1 \implies 2f'(z) \equal{} f(1 \plus{} z) \minus{} f(z \minus{} 1)$
$ x \equal{} 1 \minus{} z, y \equal{} z \minus{} 1 \implies 2(1 \minus{} z)f'(0) \equal{} f(1 \minus{} z) \minus{} f(z \minus{} 1)$

$ \therefore f'(z) \equal{} zf'(1) \plus{} (1 \minus{} z)f'(0) \implies f'(z) \equal{} 2az \plus{} b$ for constants $ 2a \equal{} f'(1) \minus{} f'(0), b \equal{} f'(0)$
$ \implies \boxed{f(z) \equal{} az^2 \plus{} bz \plus{} c}$

It can be checked by substitution that this form satisfies the equation for any constants $ a$, $ b$ and $ c$[/quote]

Very short and nice, congrats !


A cylindrical can that is six inches high has a label that is $ 60\pi$ square inches in area and exactly covers the outside of the can excluding the top and bottom lids. What is the radius of the can, in inches?

A cylindrical can would take the shape of a [b]cylinder[/b].

The formula for the lateral surface area of a cylinder (which would be the area of the label only, not the top and bottom sides) is $ 2\pi r h$.

We know that the l. surface area of the cylinder is $ 60\pi$.

We also know that the cylindrical can is 6 inches tall.

Thus we can say that

$ 2\pi r *6 \equal{} 60\pi$

We can get rid of the $ \pi$ on both sides.

We are left with $ 12 * r \equal{} 60$

Thus, r = $ 5$.


geometry
analytic geometry
area of a triangle
Heron\u0027s formula
Pythagorean Theorem

I believe heron's formula is

$A = \sqrt{(s)(s-a)(s-b)(s-c)}$

where s is 1/2perimeter, a, b, and c, are the three sides of a triangle, and A is the area of the triangle.

why is this so (proving involved)?
Does it involve the pythagorean theorem?

If the proof involves strange high-level math, feel free to move this to another forum.

I will look it up for you. I think one of my books has it in there.

I think it's also called the shoelace method.

Look at this proof http://en.wikipedia.org/wiki/Heron%27s_formula#Proof it uses law of cosines.

I proved it a while ago
http://www.livejournal.com/users/killergerbah/4472.html

Interesting proof tarquin, and it doesn't use the law of cosines.

What's the shoelace method?

About the "shoelace method", I googled and found an explanation of it on[url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]this page[/url]. It's something about matrices. In the unfortunate situation that you don't have Adobe Acrobat Reader, you can download it off the internet for free somewhere.

With a triangle with coordinates $(0,0)$, $(h,j)$, and $(k,l)$, the area is $A=\frac{|hl-jk|}{2}$.

The length of one side, from (0,0) to (h,j) is $\sqrt{h^2+j^2}$. Call that $a$. The side from (0,0) to (k,l) is $\sqrt{k^2+l^2}$. Call that $b$. The remaining side is $\sqrt{(h-k)^2+(j-l)^2}$. Call that $c$.

From Heron's formula, we have $A=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$. I'll finish later...

You'll end up with Heron's no matter what If you just solve for the height in terms of the sides and then use $\frac {bh}{2}$

I proved it for a isosceles triangle a while ago. http://artofproblemsolving.com/Forum/weblog_entry.php?e=545

At one of our Nationals practices, Boy Soprano tried to prove it.
He got a huge jumble and none of us knew what to do with it.
The paper he did the work on ended up as a paper hat.  :D


modular arithmetic
number theory proposed
number theory

Find all positive integers (m,n) that  12^m= 7+ 5^n

[hide="Somewhat Tedious Solution"]If $ m\equal{}1$ then the equation becomes $ 12 \equal{} 7\plus{}5^n \Rightarrow n\equal{}1$. Thus, $ (m,n)\equal{}(1,1)$ is a solution.

If $ m \ge 2$, then $ 7\plus{}5^n \equal{} 12^m \equiv 0\pmod{144}$. Therefore, $ 5^n \equiv \minus{}7 \equiv 137 \pmod{144}$.

Calculating the possible residues of $ 5^n$ in $ \pmod{144}$:
$ 5^1 \equiv 5 \pmod{144}$
$ 5^2 \equiv 25 \pmod{144}$
$ 5^3 \equiv 125 \pmod{144}$
$ 5^4 \equiv 49 \pmod{144}$
$ 5^5 \equiv 101 \pmod{144}$
$ 5^6 \equiv 73 \pmod{144}$
$ 5^7 \equiv 77 \pmod{144}$
$ 5^8 \equiv 97 \pmod{144}$
$ 5^9 \equiv 53 \pmod{144}$
$ 5^{10} \equiv 121 \pmod{144}$
$ 5^{11} \equiv 29 \pmod{144}$
$ 5^{12} \equiv 1 \pmod{144}$

The possible residues of $ 5^n$ in $ \pmod{144}$ are: $ \{ 5,25,125,49,101,73,77,97,53,121,29,1 \}$.

Therefore, $ 5^n \not \equiv 137 \pmod{144}$, and thus, there are no solutions with $ m \ge 2$.

So, the only solution is $ (m,n) \equal{} (1,1)$. [/hide]

If $ n$ is odd, say $ n \equal{} 2k \plus{} 1,$ then $ 5^n \plus{} 7 \equal{} 5(24 \plus{} 1)^k \plus{} 7 \equal{} 5\left[(24 \plus{} 1)^k \minus{} 1\right] \plus{} 12$ is divisible by 12 since $ \left[(24 \plus{} 1)^k \minus{} 1\right]$ is divisible by 12. In fact, $ (24 \plus{} 1)^k \minus{} 1$ is divisible by 24 and so $ \frac {(24 \plus{} 1)^k \minus{} 1}{12}$ is an even integer. Hence $ 5\cdot\frac {(24 \plus{} 1)^k \minus{} 1}{12} \plus{} 1$ is odd and so $ 5^n \plus{} 7$ is a product of 12 with an odd integer; hence it cannot be equal to $ 12^m$ for $ m > 1.$ Hence, if $ n$ is odd, the only solution is $ (m,n) \equal{} (1,1).$

If $ n \equal{} 2k$ is even, $ 5^n \plus{} 7 \equal{} (24 \plus{} 1)^k \plus{} 7\equiv8\pmod{12}$ and so cannot be a positive power of 12.

Another approach is suppose that $ m\ge 2$ and see that $ 7\plus{} 5^n\equiv 0  (mod 8)$, so $ n$ is even. But, on the other hand, we got $ 7\plus{} 5^n\equiv 0  (mod 3)$, so n is odd, a contradiction. Therefore, $ m\equal{}n\equal{}1$, and this is the unique solution.


trigonometry
IMO
geometry
perimeter
IMO 1988

Consider 2 concentric circle radii $ R$ and  $ r$ ($ R > r$) with centre $ O.$ Fix $ P$ on the small circle and consider the variable chord $ PA$ of the small circle. Points $ B$ and $ C$ lie on the large circle; $ B,P,C$ are collinear and $ BC$ is perpendicular to $ AP.$

[b]i.)[/b] For which values of $ \angle OPA$ is the sum $ BC^2 \plus{} CA^2 \plus{} AB^2$ extremal?

[b]ii.)[/b] What are the possible positions of the midpoints $ U$ of $ BA$ and $ V$ of $ AC$ as $ \angle OPA$ varies?

[b](i)[/b] Let K the midpoint of the chord BC and $\theta = \angle OPA$. Using Pythagorean theorem for the right angle triangles $\triangle ABP, \triangle ACP$, we have

$AB^2 + AC^2 + BC^2 = 2\ PA^2 + PB^2 + PC^2 + BC^2$

$PB^2 + PC^2 = \left(\frac{BC}{2} + PK\right)\left(\frac{BC}{2} - PK\right) = \frac{BC^2}{2} + 2\ PK^2$

$AB^2 + AC^2 + BC^2 = 2 (PA^2 + PK^2) + \frac 3 2\ BC^2$

The power of the point P to the larger circle (O, R) is

$PB \cdot PC = (R + r)(R - r) = R^2 - r^2$

On the other hand,

$PB \cdot PC = \left(\frac{BC}{2} + PK\right)\left(\frac{BC}{2} - PK\right) = \frac{BC^2}{4} - PK^2$

so that

$\frac{BC^2}{4} - PK^2 = R^2 - r^2$

$BC^2 = 4(R^2 - r^2) + 4\ PK^2$

$AB^2 + AC^2 + BC^2 = 2 (PA^2 + PK^2) + 6(R^2 - r^2) + 6\ PK^2 =$

$= 2\ PA^2 + 8\ PK^2 + 6(R^2 - r^2)$

Obviously,

$PA = 2r \cos \theta$

According to the solution of the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=54056]PQR similar to DGH[/url], the locus of midpoints K is a circle (Q, r/2) with the diameter OP = r. The argument used to solve the proposition (ii) later on could be used as well. Since $PK \perp PA$,

$PK = r \cos (90^\circ - \theta) = r \sin \theta$

Substituting for PA, PK into the last formula for $AB^2 + AC^2 + BC^2$,

$AB^2 + AC^2 + BC^2 = 8r^2 \cos^2 \theta + 8 r^2 \sin^2 \theta + 6(R^2 - r^2) = 6R^2 + 2r^2$

Thus the sum of squares $AB^2 + AC^2 + BC^2$ does not depend on the position of the point A on the smaller circle (O, r).

[b](ii)[/b] The triangle $\triangle KLM$ is the medial triangle of the triangle $\triangle ABC$. Since the vertex A and the vertices B, C are on concentric circles (O, r), (O, R), respectively, the midpoint K and the midpoints L, M are also on concentric circles (Q, r/2), (Q, R/2), respectively.

Comment: The point P just provides an [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=anker&p=224443#p224443]anker[/url] (sic) of this fine configuration concentric circles :rotfl:.

[quote="armpist"]It was a very interesting comment Yetti put at the end of his post, and here is the

 reason why I consider it interesting.  As it is a case with most Olympiad problems,

 there usually exists a deeper current in them,  and this one is a good example of

 such problems.   Yetti delivered a reasonably good solution, taking the path that

 the problem composer wanted a problem-solver to take.   But his comment at the

 end about anchor was highly unusual, because he was obviously making a joke

 that made him laugh in a horizontal position.?[/quote]
You are a magnificently split personality, which is why I put the comment there.

[quote="armpist"]
 Somewhere else on this Geometry forum (Find Perimeter problem) Yetti asked me

 for explanation of some 3 points  collinearity. ... I hoped that Yetti will realize

 the simple reasons of questioned collinearity and edit his question out. But no!

 He insists, claiming that without my personal proof the points are not collinear,

 and that's it.   I trully belive that I can not upstage Darij in such explanation, so I

 decided ( que sera sera) that the points should stay triangular, I will not bring

 them in line. On top of that I disagree with the entire solution I gave. It is free for

 grabs, I'm not returning to that thread.[/quote]
The reasons for collinearity of the said points are indead simple, Ceva's theorem and the ratio of sides $1 : \sqrt 3$. I calculated the ratio and deduced the collinearity. You postulated the collinearity and used it to get the ratio. Thus you did not have even a sketch of a solution, as you claimed, not to mention the desired result in a closed form, regardless of the fact that the said points indeed are collinear.

[quote="armpist"]
  Thank you.

   M.T.[/quote]
Does sfwc know that you are using his signature ?

Yetti

[quote="yetti"]The reasons for collinearity of the said points are indead simple, Ceva's theorem and the ratio of sides $1 : \sqrt 3$. I calculated the ratio and deduced the collinearity. You postulated the collinearity and used it to get the ratio. [/quote]


Let $ n$ be an integer greater than three, then prove that;

$ 1!\plus{}2!\plus{}3!...\plus{}n!$  cannot be a perfect power

This is a well known problem. Firstly, note that it cannot be a perfect square by considering its last digit. If $ n\ge8$, then the expression is divisible by $ 9$ but not $ 27$, and hence cannot be a perfect power. It remains to consider the case for $ x<8$.

$ 1!\plus{}2!\plus{}3!\plus{}4!\equiv3\bmod{5}$   Since $ n!$ is divisible by $ 5$ for every $ n>4$ we conclude that every term ($ n>3$) is equal to $ 3 (mod 5)$ thus it can't be a perfect square

There are countless ways to attack this problem,this seems an easy one, yet I think it has been posted before

Daniel


The sequence $ (a_n)$ is given by $ a_1 \equal{} 4, a_2 \equal{} 6$ and $ a_{n\plus{}2}a_n \equal{} a^2_{n\plus{}1} \plus{} 8$ for $ n\geq 1$. Prove that $ 9a_n^2\minus{}128$ is a square of a rational number for all $ n$.


calculus
integration
real analysis
limit
calculus computations

The whole problem is from the financial/actuarial mathematics but the exercice is typically from analysis.

The whole problem can be reduced to this one:
I have to show from definition (!!!) of definite integeal (Riemann integral or if you want we can treat it as Lebesgue integral) that:

$\bar{a}_{n]}={\int_{0}}^{n}e^{-\delta t}dt=\frac{1}{\delta}(1-e^{-\delta n})$.

where

$\delta =ln(1+i)$ where $i$ means interests (but if you want you can use $\delta$, not $i$). $n$ is any natural

$\bar{a}_{n]}$ means continuous annuity , well, doesn't matter.

The problem is to show by definition that ${\int_{0}}^{n}e^{-\delta t}dt=\frac{1}{\delta}(1-e^{-\delta n})$

I'd be very grateful for help.

Use the equal-widths partition. You'll have to evaluate a geometric sum, and then apply the basic limit $\lim_{s\to0}\frac{e^{s}-1}{s}=1$.


geometry
calculus
derivative
trigonometry
perpendicular bisector
geometry proposed

Ascertain the convex pentagon $ABCDE,\ AC\perp BE$ with the maximum area  inscribed in a given circle.

Assume that the pentagon is inscribed in the circle $(O,R)$ with $R=1$

I note the angles:

${\phi}_1 = \widehat {AOB}$
${\phi}_2 = \widehat {BOC}$
${\phi}_3 = \widehat {COD}$
${\phi}_4 = \widehat {DOE}$
${\phi}_5 = \widehat {EOA}$

The pentagon is divided by $CE$ to the quadrilateral $ABCE$ and the triangle $CDE$

If $P = AC \cap BE$ then from the right triangle $APB$ we have

$\widehat{PAB}+\widehat{PBA} = \frac{\pi}{2} \Rightarrow$

$\widehat{CAB}+\widehat{EBA} = \frac{\pi}{2} \Rightarrow$

arc $(BC)$ + arc $(AE)=\pi$

and similarly arc $(AB)$ + arc $(CE)=\pi$

so we have : ${\phi}_1 + {\phi}_3 + {\phi}_4 = {\phi}_2 + {\phi}_5 = \pi \ \ \ (1) \Rightarrow$

$sin{\phi}_1 = sin({\phi}_3 + {\phi}_4)$ and $sin{\phi}_2 = sin{\phi}_5$

$(OAB) = \frac{1}{2} sin{\phi}_1$
$(OBC) = \frac{1}{2} sin{\phi}_2$
$(OCE) = \frac{1}{2} sin({\phi}_3+{\phi}_4)$
$(OEA) = \frac{1}{2} sin{\phi}_5$

$(OAB)+(OCE) = \frac{1}{2} sin{\phi}_1 + \frac{1}{2} sin({\phi}_3+{\phi}_4) = \frac{1}{2} sin{\phi}_1 + \frac{1}{2} sin{\phi}_1 = sin{\phi}_1$

$(OBC)+(OAE) = \frac{1}{2} sin{\phi}_2 + \frac{1}{2} sin{\phi}_5 = \frac{1}{2} sin{\phi}_2 + \frac{1}{2} sin{\phi}_2 = sin{\phi}_2$

The sum $E_2 = (OBC)+(OAE) = sin{\phi}_2$ depends only on ${\phi}_2$, and it is independent of the other angles.

Also, the other sum $E_1 = (OAB)+(OCE)+(CDE) = sin{\phi}_1 + (CDE)$ is independent of ${\phi}_2$.

So, we can maximize one of the sums and then maximize the other severally.

Of course all the angles are greater than $0$.

So from ${\phi}_2 + {\phi}_5 = \pi$ we get that ${\phi}_2,{\phi}_5 < \pi$

Also, ${\phi}_1 + ({\phi}_3+{\phi}_4) = \pi \Rightarrow {\phi}_1 , ({\phi}_3+{\phi}_4) < \pi$

So, all the angles are in $(0,\pi)$.

[hide="The area of the triangle $CDE$:"]
If $DK$ is the altitude, then $(CDE) = \frac{1}{2} CE \cdot DK$ so, given the length of $CE$ the area is maximized when $DK$ is the perpendicular bisector of $CE$

So, let $K$ be the midpoint of the segment $CE$.

$\widehat{COE} < \pi \Rightarrow K$ is between $O,D$.

[color=blue]As the length of $CE$ increases, $DK$ also increases, so the area $(CDE) = \frac{1}{2} CE \cdot DK$ increases.

So, we [u]wish[/u] that $CE$ is the longest side of $ABCE$ (if it is not, then we can rename the vertices of $ACDE$, so that $CE$ is the longest side
$\Rightarrow {\phi}_3+{\phi}_4 =max \{ {\phi}_1,{\phi}_2,{\phi}_3+{\phi}_4,{\phi}_5 \}$).[/color]

$CE = 2 \cdot KE = 2 \cdot R \cdot sin{\phi}_3 = 2sin{\phi}_3$

The altitude $DK$ of $\triangle CDE$ is $DK = OD-OK = R-Rcos{\phi}_3 = 1-cos{\phi}_3$

So $(CDE) = \frac{1}{2} CE \cdot DK = \frac{1}{2} 2sin{\phi}_3(1-cos{\phi}_3) = sin{\phi}_3(1-cos{\phi}_3) \Rightarrow$[/hide]

$(CDE) = sin{\phi}_3-sin{\phi}_3cos{\phi}_3$

$E_1 = (OAB)+(OCE)+(CDE) = sin{\phi}_1 + sin{\phi}_3-sin{\phi}_3cos{\phi}_3$

$sin{\phi}_1 = sin({\phi}_3 + {\phi}_4) = sin(2{\phi}_3) \Rightarrow$
$sin{\phi}_1 = 2 sin{\phi}_3cos{\phi}_3 \Rightarrow$

$E_1 = 2 sin{\phi}_3cos{\phi}_3 + sin{\phi}_3-sin{\phi}_3cos{\phi}_3$

$E_1 = sin{\phi}_3cos{\phi}_3 + sin{\phi}_3$

[hide="maximize $E_1$"]
$f(x) = sinxcosx+sinx$
$f'(x) = cos^2x-sin^2x + cosx$
$f'(x) = 2cos^2x-1 +cosx$

$f'(x)$ is factorized  to $f'(x) = (2cosx-1)(cosx+1)$
$\widehat{COE} < \pi \Rightarrow {\phi}_3 < \frac{\pi}{2} \Rightarrow cosx+1>0$

So we have $f'(x)=0$ only for $2cosx=1 \Rightarrow x=\frac{\pi}{3}$ (only positive values are accepted)

So for $x=\frac{\pi}{3}$ we have a maximum or minimum

using the second derivative $f''(x) = -sinx(4cosx+1)<0$ we see that $f(\frac{\pi}{3})$ is a maximum

So we have: [/hide]

${\phi}_3 = {\phi}_4 = \frac{\pi}{3} \Rightarrow {\phi}_1 = \pi-2{\phi}_3 = \frac{\pi}{3}$

[hide="maximize $E_2$"]
It is more easy

$E_2 = (OBC)+(OAE) = sin{\phi}_2$

so we have a maximum for $E_2$ when $sin{\phi}_2$ is maximum $\Rightarrow sin{\phi}_2=1\Rightarrow$[/hide]
${\phi}_2 = {\phi}_5 = \frac{\pi}{2}$

Applying the values of the angles we have:

$E_2 = sin \frac{\pi}{2}=1$

$E_1 = sin \frac{\pi}{3}(1+cos \frac{\pi}{3}) = \frac{\sqrt{3}}{2}(1+\frac{1}{2}) = \frac{3\sqrt{3}}{4}$

so, the max area is $E_1+E_2 = 1+\frac{3\sqrt{3}}{4}$, about $73,2 \%$ of the circle's area.

[quote="Virgil Nicula"][color=darkred]Let $ ABCDE$ be a convex pentagon inscribed in the given circle $ w = C(O,R)$

for which $ AC\perp BE$. Establish when the its area $ S = [ABCDE]$ is maximum.[/color][/quote]
[color=darkblue][b]Proof I (metric).[/b] Denote $ \{\begin{array}{ccc} M\in (CE) & , & MC = ME \\
\ AB = x & , & CE = y\ . \\
\ BC = u & , & AE = v\ .\end{array}$

Suppose w.l.o.g. that $ OD\perp CE$. Thus, $ S = [ABCE] + [CDE]\ .$

Since $ AC\perp BE$ prove easily that the line $ CE$ separates the points $ O$ and $ D\ !$ Hence,

$ AC\perp BE$ $ \implies$ $ \{\begin{array}{c} x^2 + y^2 = u^2 + v^2 = 4R^2\ . \\
 \\
\boxed {OM = \frac x2}\ ,\ MD = R - \frac x2\ . \\
 \\
xy + uv = AC\cdot BE = 2\cdot [ABCE]\ .\end{array}$ $ \Longrightarrow$

$ S = \frac 12\cdot AC\cdot BE + \frac 12\cdot CE\cdot DM =$ $ \frac {xy + uv}{2} + \frac y2\cdot (R - \frac x2)\ .$

Therefore, $ S\mathrm {\ - \ max}$ $ \Longleftrightarrow$ $ 2uv + y(x + 2R)\mathrm {\ - \ max}\ .$

Since $ u^2 + v^2 = 4R^2$, then the product $ uv$ is maximum iff $ \boxed {u = v = R\sqrt 2}\ .$

The product $ y(x + 2R)\mathrm {\ - \ max}$ $ \Longleftrightarrow$ $ y^2(x + 2R)^2\mathrm {\ - \ max}$ $ \Longleftrightarrow$

$ (4R^2 - x^2)(x + 2R)^2\mathrm {\ - \ max}$ $ \Longleftrightarrow$  $ (2R - x)(2R + x)^3\mathrm {\ - \ max}$  $ \Longleftrightarrow$

$ (2R - x)(\frac {2R + x}{3})^3\mathrm {\ - \ max}\ .$ Thus, $ (2R - x) + 3\cdot(\frac {2R + x}{3}) = 4R$ $ \Longleftrightarrow$

$ y(2R + x)$ is maximum iff $ 2R - x = \frac {2R + x}{3} = R$,  i.e. $ \boxed {x = R\ ,\ y = R\sqrt 3}\ .$

In conclusion, $ \boxed {\begin{array}{c} AB = CD = DE = R \\
 \\
BC = AE = R\sqrt 2\end{array}}$ and in this case, $ \boxed {S_{\mathrm {max}} = \frac {4 + 3\sqrt 3}{4}\cdot R^2}\ .$[/color]

[color=darkblue][b][u]$ 1\blacktriangleright$ A similar proposed problem[/u][/b] (it is a bit more easily ! ).[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABCDEF$ be a convex hexagon inscribed in the given circle $ w = C(O,R)$

for which $ AD\perp CF$. Establish when the its area $ S = [ABCDEF]$ is maximum.[/color][/quote]

[color=darkblue][b][u]$ 2\blacktriangleright$ A similar proposed problem[/u][/b] (it is a bit more difficulty ! ).[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABCDEFG$ be a convex heptagon inscribed in the given circle $ w = C(O,R)$

for which $ AD\perp BF$. Establish when the its area $ S = [ABCDEFG]$ is maximum.[/color][/quote]

[color=darkblue][b]Proof II (trigonometric).[/b] Suppose w.l.o.g. $ R = 1$ and $ OD\perp CE\ .$ Denote

$ \{\begin{array}{c} m(\widehat {ACE}) = y\ ,\ m(\widehat {BEC}) = z\ . \\
 \\
m(\widehat {DCE}) = m(\widehat {CED}) = x\ .\end{array}$ Since $ AC\perp BE$, then $ \{\begin{array}{c} x < 45^{\circ}\ . \\
 \\
y + z = 90^{\circ}\ .\end{array}$

Show easily $ \{\begin{array}{c} m(\widehat {CDE}) = 180^{\circ} - 2x\ ,\ m(\widehat {ADB}) = 90^{\circ} - 2x \\
 \\
BE = 2\sin (2x + z)\ ,\ AC = 2\sin (2x + y) \\
 \\
DC = DE = 2\sin x\end{array}\ .$ Therefore,

$ S = [ABCE] + [CDE] = \frac 12\cdot AC\cdot BE + \frac 12\cdot DC^2\cdot \sin\widehat {CDE} =$

$ 2\sin (2x + z)\sin (2x + y) + 2\sin^2x\sin 2x$ $ \Longrightarrow$ $ \underline {S = \cos (y - z) + \sin 2x(1 + \cos 2x)}$

$ \blacktriangleright \cos (y - z)\mathrm {\ - \ max}$ $ \Longleftrightarrow$ $ \boxed {y = z = 45^{\circ}}\ .$

$ \blacktriangleright \sin 2x(1 + \cos 2x)\mathrm {\ - \ max}$ $ \Longleftrightarrow$ $ \sin^22x(1 + \cos 2x)^2\mathrm {\ - \ max}$ $ \Longleftrightarrow$

$ (1 - \cos 2x)(\frac {1 + \cos 2x}{3})^3\mathrm {\ - \ max}\ .$ Observe that $ (1 - \cos 2x) + 3\cdot\frac {1 + \cos 2x}{3} = 2$

(constant). $ S\mathrm {\ - \ max}$ $ \Longleftrightarrow$ $ 1 - \cos 2x = \frac {1 + \cos 2x}{3} = \frac 24$, i.e. $ \boxed {x = 30^{\circ}}\ .$

In conclusion, $ \{\begin{array}{c} AB = DC = DE = 1 \\
 \\
AE = BC = \sqrt 2\end{array}$ and $ \boxed {S_{\mathrm {max}} = 1 + \frac {3\sqrt 3}{4}}\ .$[/color]

[color=darkred][b]Attention ! [/b] In the last two replies I presented still two similar problems
and a second proof (with [u]trigonometry[/u]) of the initial proposed problem.[/color]


Does anyone know an interesting proof for irrationality of $\pi$.

Take a look at this.  [url=http://www.mcs.csuhayward.edu/~malek/Mathlinks/Pi.html]www.mcs.csuhayward.edu/~malek/Mathlinks/Pi.html[/url]
I believe this is the most simple proof ever.
Johaan Heinrich Lambert proved that if $\theta$ is rational, then $\tan\theta$ is irrational.  As a result, he took $\tan\frac{\pi}{4}=1$.  Since $\frac{\pi}{4}$ is irrational, $\pi$ must be irrational.

EDIT:  Johaan Heinrich Lambert method also works vice versa, so it is now clear.

Masoud Zargar