number theory proposed
number theory
IMO
IMO 1964

For both parts we will consider the cases $ n\equal{}3k, 3k\plus{}1, 3k\plus{}2$. Working in $ mod 7$ we have $ 2^{3} \equiv 1 (mod 7)$ hence the following:

$ 2^{3k} \equiv 1 (mod7)$
$ 2^{3k\plus{}1} \equiv 2 (mod7)$
$ 2^{3k\plus{}2} \equiv 4(mod 7)$

Now the rest is easy.

a)

$ 2^{3k}\minus{}1 \equiv 0 (mod7)$
$ 2^{3k\plus{}1}\minus{}1 \equiv 1 (mod7)$
$ 2^{3k\plus{}2}\minus{}1 \equiv 3(mod 7)$

So only for $ n\equal{}3k$ where $ k\in\mathbb{Z}^{\plus{}}$ does $ 7|2^{n}\minus{}1$.

b)

$ 2^{3k}\plus{}1 \equiv 2 (mod7)$
$ 2^{3k\plus{}1}\plus{}1 \equiv 3 (mod7)$
$ 2^{3k\plus{}2}\plus{}1 \equiv 5(mod 7)$

So there is no value of $ n\in\mathbb{Z}^{\plus{}}$ such that $ 7|2^{n}\plus{}1$.

Here is my solution for part (a) (wow this thread is 12 years old!)

Let $k$ be any number.\begin{align} \nonumber7\equiv 0\pmod 7\\ \nonumber (7+1)\equiv 1\pmod 7\\ \nonumber(7+1)^k\equiv 1\pmod 7 \\ \nonumber8^k\equiv1\pmod7\\ \nonumber2^{3k}\equiv 1\pmod7\\ \nonumber2^{3k}-1\equiv 0\pmod7\end{align}We also know that $2^n-1\equiv 0\pmod 7$. Thus, $n=3k$, which means that $n$ must be a multiple of $3$.

There are $3$ possible cases:
$\implies n\equiv0\pmod3$
$\implies n\equiv1\pmod3$
$\implies n\equiv2\pmod3$

If $n\equiv0\pmod3$, we can express $n$ as $3y$, where $y\in \mathbb{Z}^+$.

If $n\equiv1\pmod3$, we can express $n$ as $3y + 1$, where $y\in \mathbb{Z}^+$.

If $n\equiv2\pmod3$, we can express $n$ as $3y + 2$, where $y\in \mathbb{Z}^+$.

(Note that $y$ in these 3 lines must not be the same constant.)

Note that:
$\implies 2^{3y}\equiv1\pmod7$
$\implies 2^{3y + 1}\equiv2\pmod7$
$\implies 2^{3y + 2}\equiv4\pmod7$

(a): $7|(2^n - 1)$ iff (if and only if) $2^n \equiv 1\pmod7$.

We check our cases and see that this is only true when $n$ can be expressed as $3y$.

This implies that $n = 3y$ where $y \in \mathbb{Z}^+$.

(b): $7|(2^n + 1)$ iff (if and only if) $2^n \equiv 1\pmod7\equiv6\pmod7$. We check our cases to see that no such $n$ satisfies this.

Credits to my blog

[hide=Solution](a) Note that $n = 3a \implies n \equiv 0 \pmod 3$ always works, because $2^{3a}-1 \equiv 8^a-1 \equiv 0 \pmod 7$. Our other two cases are $n = 3a+1$ and $n=3a+2$, the first implying $n \equiv 1 \pmod 7$ and the second implying $n \equiv 3 \pmod 7$. Therefore, neither of these work, so the solutions are $\boxed{n \equiv 0 \pmod 3}$.

(b) This would mean that $2^n \equiv 6 \pmod 7$. As can be seen via a quick check of $n \equiv 0,1,2,3,4,5,6 \pmod 7$ (which give $1,2,4,1,2,4,1$ respectively) $2^n \equiv 6 \pmod 7$ is impossible for any number that's a residue $\pmod 7$. All numbers can be simplified down to residues $\pmod 7$, so this is impossible. $\blacksquare$[/hide]

By Euler's Theorem, $2^6\equiv1\pmod7$, so we only need to consider $n\in[0,6]$, since it cycles.
If $n\pmod7\in\{0,3,6\}$, then $2^n-1\equiv0\pmod7$ and $2^n+1\equiv0\pmod7$.
If $n\pmod7\in\{1,4\}$, then $2^n-1\equiv1\pmod7$ and $2^n+1\equiv3\pmod7$.
If $n\pmod7\in\{2,5\}$, then $2^n-1\equiv3\pmod7$ and $2^n+1\equiv5\pmod7$.
So $\textbf{(A)}~\boxed{n\in3\mathbb N}$ and $\textbf{(B)}~\square$.

[quote=Wikipedian1337](a) Find all positive integers $ n$ for which $ 2^n\minus{}1$ is divisible by $ 7$.
(b) Prove that there is no positive integer $ n$ for which $ 2^n\plus{}1$ is divisible by $ 7$.[/quote]

For part (a) Note that:
$$2^n-1 \equiv 0 \pmod 7 \implies 2^n \equiv 1 \pmod 7 \implies n=3k \; \forall k \in \mathbb N$$
That becuase $2^3 \equiv 1 \pmod 7$ and $1^k=1$ for all $k \in \mathbb N$.
For part (b) use case work: We know that $n$ can be $3i$ , $3i+1$ and $3i+2$ so:
Case 1.- $n=3i$
$$2^{3i} \equiv -1 \pmod 7 \implies 1 \equiv -1 \pmod 7 \implies \; \text{contradiction!!}$$
Case 2.- $n=3i+1$
$$2^{3i+1} \equiv -1 \pmod 7 \implies 2 \equiv -1 \pmod 7 \implies \; \text{contradiction!!}$$
Case 3.- $n=3i+2$
$$2^{3i+2} \equiv -1 \pmod 7 \implies 4 \equiv -1 \pmod 7 \implies \; \text{contradiction!!}$$
So we proved that no such $n$ for part (b).
Thus we are done  :blush:

[hide=Sol]For (a), I claim the answer is all $n$ such that $n$ is divisible by $3$. We see that the order of 2 modulo 7 is 3. This means that $2^{1\pmod3}\equiv2\pmod7, 2^{2\pmod3}\equiv4\pmod7, \text{and } 2^{0\pmod3}\equiv1\pmod7$. Since $2^n$ must be $1 \pmod 7$ for (a) to be true, our claim is true. For (b), we see that $2^n$ must be $6 \pmod 7$, which will never be true because a power of $2$ is always 1, 2, or 4 $\pmod 7$. [/hide]

(a) We have $2^3\equiv 1\text{(mod 7)}$, but $2^1\equiv 2\text{(mod 7)}$, so $\text{ord}_72=3$, and hence $3|n$.

(b) Notice that $2^n$ is a QR mod $7$, but $-1$ isn't, contradiction.

[hide=Solution](a) We claim that $n\equiv 0\pmod{3}$ are the only solutions. If $n\equiv r\pmod{7},$ where $0\le r<3,$ we know $$2^n-1\equiv 2^{3k+r}-1\equiv 2^{3k}\cdot 2^r-1\equiv 8^k\cdot 2^r-1\equiv 2^r-1\pmod{7}.$$ We test $r=0,1,2$ and find that $2^n-1=0,1,3$ respectively. Since we want $2^n-1\equiv 0\pmod{7},$ our claim is proven.

(b) Similarly from (a), we find that if $n\equiv r\pmod{3},$ then $2^n+1\equiv 2^r+1\pmod{7}.$ This is equal to $2,3,5$ when $r=0,1,2$ respectively so there are no solutions.[/hide]

[hide=my sol for part a]The order of 2 mod 7 is 3, so n must be a multiple of 3.[/hide]

[hide=my sol for part b]Since 2 is a quadratic residue mod 7, we have a contradiction.[/hide]

Don't forget to upvote my post if you like my solution! :D

Let's bash it a bit :P
[hide=part a]$-1\equiv -1\text{(mod 7)}$
$2\equiv 2\text{(mod 7)}$
$2^2\equiv 4\text{(mod 7)}$
$2^3\equiv 1\text{(mod 7)}$
After this due to cyclicity of 2 the remainders start repeating, hence $2^4\equiv 2\text{(mod 7)}$, $2^5\equiv 4\text{(mod 7)}$ and such. The only power of 2 which satisfies the condition for becoming $ 1\text {(mod 7)} $ is $2^3\equiv 1\text{(mod 7)}$. Hence all [b]n's[/b] which are divisble by 3 are a solution to $2^n - 1\equiv 0\text{(mod 7)}$[/hide]
[hide=part b]$1\equiv 1\text{(mod 7)}$
There is no +ve [b]n[/b] which satisfies becoming   $ -1\text{(mod 7)}$ to make $2^n +1\equiv 0\text{(mod 7)}$ Hence there is no +ve integer [b] n [/b] that satsifies $2^n +1\equiv 0\text{(mod 7)}$[/hide]

[hide=Sol]
[hide=part a] $ord_7(2)=3$,there fore $n=3k$[/hide]
[hide=part b]case bash :P,too lazy to post [/hide]

a) Fermat's Little Theorem gives $2^6\equiv 1 \pmod 7.$ But since $7|(2^3-1)(2^3+1)$ we have $2^3\equiv 1\pmod 7.$ So all $n$ divisible by $3$ satisfy.

----------
b) Let $n=3r+s,$ where $s\in \{1,2\}.$ Then $$2^n\equiv (2^3)^r\cdot 2^s \equiv 2\text{ or } 4 \not\equiv 1 \pmod 7.$$ $\text{Q.E.D.}$

Observe that since $2^3=8\equiv1\pmod{7}$, $2^n\pmod{7}$ is cyclic with period $3$: $1,2,4,1,2,4,\dots$. Thus, for (a), the residues of $2^n-1$ modulo $7$ are $0$, $1$, and $3$ when $n\equiv0,1,2\pmod{3}$, in that order, so $\boxed{\text{all }n\text{ s.t. }3|n}$ is the requested answer. Similarly, for (b), the residues of $2^n+1$ modulo $7$ are $2$, $3$, and $5$, and in particular, $0$ is not one such residue. Hence proven. $\blacksquare$

We split into cases.

Case 1: $n \equiv 0 \pmod 3$
\[2^{3k}-1 \equiv 8^k-1 \equiv 1-1 \equiv 0 \pmod 3\]
\[2^{3k}+1 \equiv 8^k+1\equiv 1+1 \equiv 2 \pmod 3\]

Case 2: $n \equiv 1 \pmod 3$
\[2^{3k+1}-1 \equiv 2\cdot8^k-1 \equiv 2-1 \equiv 1 \pmod 3\]
\[2^{3k+1}+1 \equiv 2\cdot8^k+1\equiv 2+1 \equiv 3 \pmod 3\]

Case 3: $n \equiv 2 \pmod 3$
\[2^{3k+2}-1 \equiv 4\cdot8^k-1 \equiv 4-1 \equiv 3 \pmod 3\]
\[2^{3k+2}+1 \equiv 4\cdot8^k+1\equiv 4+1 \equiv 5 \pmod 3\]

Thus, the only case that satisfies $2^n-1$ is divisible by $7$ is when $n \equiv 0\mod3$ or $n$ is divisible by $3$ and no cases satisfy $7$ dividing $2^n+1$.

[b](a)[/b]$7 | 2^n-1 \implies 2^n \equiv 1 \pmod 7 \implies n=3k  \forall k \in \mathbb N$ since $\text{ord}_7(2) = 3.$
[b](b)[/b] $7 | 2^n+1 \implies 2^n \equiv 6 \pmod 7.$ However, the residues of $2^n \pmod 7 \forall n \in \mathbb N$ is $1, 2,$ or $4.$ Hence, there is no positive integer $ n$ for which $ 2^n+1$ is divisible by $ 7$.

(a) We claim that the answer is when $n$ is a multiple of 3. We will use proof by induction. The base case is when $n=3$, which works, now $n=3k$ is 1 mod 7, $n=3k+3$, is 2^3 mod 7 which is 1 mod 7, hence, proved.

(b) $2^n$ needs to be 6 mod 7, but it can only be 1 mod 7, 2 mod 7, 4 mod 7, hence, proved.


There are $1985$ participants to an international meeting. In any group of three participants there are at least two who speak the same language. It is known that each participant speaks at most five languages. Prove that there exist at least $200$ participans who speak the same language.

I think this is just basic pigeonhole.
If there aren't two persons who speak different languages , then any two persons that we take speak a common language, thus for a certain person $X$ all of the remaining $1984$ persons speak at least one of the languages that he speaks. Since $X$ speaks at most $5$ languages there are at least $397$persons who speak a  common language.
If there exist two persons who don't speak a common language we shall denote them by $A$ and $B$. By taking groups of three which consist of $A$,$B$ and one at the time from the remaining $1983$ persons it gives us that at least $992$ persons speak a common language with $A$ or with $B$. Since $A$ and $B$ speak at most 5 languages and $992=198\cdot 5+2$ there are at least $199$ persons who speak a common language with $A$ or $B$ which solves the problem.

Hello.

My solution (I hope it's correct).

Let $G$ be a graph with $1985$ vertices,each belonging to one participant.We connect two vertices if and only if,the corresponding participants don't speak the same language.From the statement of the problem,we obtain that $G$ has no triangles,thus from Turan's theorem,it has at most $\frac{1985^2}{4}$ edges.Hence,there exists a vertex with degree at most $\lfloor \frac{1985}{2}\rfloor =992$.Then,the corresponding participant has common languages with at least $993$ other participants.Since,that participant speaks at most $5$ languages,there is one language spoken by at least $\lfloor \frac{993}{5}\rfloor +1=199$ other participants,which means that this language is spoken by at least $200$ participants in total,and we are done.


If $ x$ and $ y$ are positive integers satisfying $ 2(x\minus{}y)^2$ $ \plus{}$ $ 4y^2$ = $ 54$, find $ x$

[hide]divide both sides by 2 to get $ (x\minus{}y)^2\plus{}y^2\plus{}y^2\equal{}27$.
Thus 27 is written as the sum of 3 squares.  Reducing mod 8, we can see that $ x\minus{}y, y, y$ are all odd.  Testing $ y$ being $ 1$ and $ 3$, we get $ x\minus{}y\equal{}5$ and $ 3$ respectively, and $ x\equal{}6$ and $ 6$.  So either way, $ x\equal{}\boxed{6}$[/hide]


Set $ A$ has 5 members and Set $ B$ has 7 members. $ A\cap B$ has 4 members. How many members are in $ A\cup B$?

$ \#(A \cup B) \equal{} \#(A) \plus{} \#(B) \minus{} \#(A \cap B)$

$ \#(A \cup B) \equal{} 5 \plus{} 7 \minus{} 4 \equal{} \boxed{8}$


Give all solutions of the equation:
$sin(9x)+sin(5x)+2sin^{2}x=1$


Sorry, don't know latex yet.

Prove: The mth fibonacci number, for any m, divides the (m times n)th fibonacci number, for any n.

[b]Problem:[/b]
[color=blue]Let $F_n$ be a Fibonacci sequence. If $n|m$, then $F_n|F_m$[/color]

[b]Solution:[/b]
$n|m \iff m=nk$
I will use inducion on $k$
For $k=1$ it's obvious.
$F_n|F_{kn} \Rightarrow F_n|F_{kn+n}$

Formula: $F_{n+m}=F_{n-1}F_m+F_{n}F_{m+1}$
By the formula we have:
$F_{kn+n}=F_{kn-1}F_n+F_{kn}F_{n+1}\Rightarrow F_n|F_{kn+n}$


logarithms
function
real analysis
real analysis unsolved

Does anyone know whether
$ \sum\limits_{\gamma} \frac{1}{\left| \gamma \right|}$
is convergent?

$ \gamma$ runs over the nontrivial zeros of $ \zeta(s)$

Certainly divergent. They behave like $ n/\log n$ asymptotically, if I remember it right.

The number of zeroes of $ \zeta(\sigma\plus{}it)$ in the region $ 0<\sigma<1$, $ 0<t<T$ is $ \frac{T}{2\pi}\log \frac{T}{2\pi}\minus{}\frac{T}{2\pi}\plus{}O(\log T)$. This was conjectured by Riemann (1859 ...of course) and proved by von Mangoldt more than 30 years later. [Source: Ivic's book.]

holy smokes, is that so ?  :wacko:  guess i have to read up on the book by harold m edwards.  :oops:  i am sorta rusty when it comes to riemann.

http://www.amazon.com/Riemanns-Zeta-Function-Harold-Edwards/dp/0486417409

Merci :)

My main problem is to show
$ \sum\limits_{\gamma} \frac{x^{\gamma}}{\gamma} \equal{} \hbox{o}(x)$

According to the author of the script, it's obviously. Who can give me a hint to see the obviousness? :lol:
I'm not sure if I can check the limit of every single summand.

As $ x\to\infty$? Is $ x$ positive, negative, imaginary?

Hm. Even convergence of the series looks like a deep result about the distribution of zeros. As we know, the series does not converge absolutely. Assuming RH, we can write the series as $ x^{1/2}\sum \frac {\exp(i t_n x)}{\frac12 \plus{} it_n}$. This is convergent if there is some uniformity in the distribution of $ t_n$. It's known that $ t_{n \plus{} 1} \minus{} t_n \equal{} O(t_n^{\theta})$ with $ \theta < 0.156$ (Ivic), but even this - rather non-obvious - fact does not imply that the series is convergent. For instance, one must rule out large clusters of zeros.

And if you don't assume RH...  :ninja:

May I ask who is the author of the "script"?

[quote]May I ask who is the author[/quote]
I don't know. I printed out the last three pages only.

I found two quotations in different books

[quote]$ \sum\limits_{\gamma} \frac {x^{\gamma}}{\gamma} \equal{} \hbox{o}(x),$ because the limit can be evaluated termwise and $ \forall \gamma: x^{\gamma \minus{} 1} \to 0 \, \, \, \, \, \, (x \to \infty).$[/quote]

[quote]$ \sum\limits_{\gamma} .. \equal{} \hbox{o}(x)$ is a result of the PNT proof by Hadamard and Poussin.[/quote]

:huh: Trivial? Hmm..


logarithms
number theory
prime numbers
number theory unsolved

Let $ k$ be a positive integer, we write k as a product of prime numbers (not necessarily distinct, par example 18=2.3.3). Let $ T(k)$ be the sum of all factors in the factorization above. Find the largest constant $ C$ such that $ T(k) \geq C.lnk$ for all positive integer k.

Let $ k \equal{} p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$. Then $ T(k) \equal{} a_1p_1 \plus{} a_2p_2 \plus{} \dots \plus{} a_np_n$. Hence $ T(k)\ge C\ln k$ is equivalent to $ a_1p_1 \plus{} a_2p_2 \plus{} \dots \plus{} a_np_n\ge C(a_1\ln p_1 \plus{} a_2\ln p_2 \plus{} \dots \plus{} a_n\ln p_n)$ $ \iff \sum a_i(p_i \minus{} C\ln p_i)\ge 0$. Therefore $ C\le \min\left\{\frac {p_i}{\ln p_i}\right\}$. But since $ \frac {3}{\ln 3} < \frac {2}{\ln 2}$ and $ \frac {x}{\ln x}$ is increasing for $ x\ge 3$ we get $ \max C \equal{} \frac {3}{\ln 3}$. Equality holds when $ k \equal{} 3^m$.

:mad: What's the use of posting M&Y problems here and using other people's solutions?

I really don't understand.

Please friends, don't help the cheaters.

Sounds familiar?
[url]http://www.mathlinks.ro/viewtopic.php?t=188838[/url]


geometry
limit
integration
function
calculus
calculus computations

Find the area bounded by $ y \equal{} f(x)$ and the curve $ y \equal{} \frac{2}{1 \plus{} x^2 }$ satisfying the conditions  $ f(x) f(y) \equal{} f(xy)$ x , y belongs to R and $ f'(1) \equal{} 1$ .

[quote="vijaymenon"]Find the area bounded by $ y \equal{} f(x)$ and the curve $ y \equal{} \frac {2}{1 \plus{} x^2 }$ satisfying the conditions  $ f(x) f(y) \equal{} f(xy)$ x , y belongs to R and $ f'(1) \equal{} 1$ .[/quote]

$ f'(x)\equal{}\lim_{h\to 0}\frac{f(x\plus{}h)\minus{}f(x)}{h}\equal{}\lim_{h\to 0}\frac{f(x(1\plus{}\frac{h}{x}))\minus{}f(x.1)}{h}$

$ f'(x)\equal{}\lim_{h\to 0}\frac{f(x)f(1\plus{}\frac{h}{x}))\minus{}f(x)f(1)}{h}$

$ f'(x)\equal{}xf(x)\lim_{h\to 0}\frac{f(1\plus{}\frac{h}{x}))\minus{}f(1)}{\frac{h}{x}}$

$ f'(x)\equal{}xf(x)f'(1)$

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

$ \int \frac{f'(x)}{f(x)}dx\equal{}\frac{x^2}{2}\plus{}c$

$ f(x)\equal{}ke^{\frac{x^2}{2}}$

Thus $ f(x)\equal{}0$ or $ f(x)\equal{}e^{\frac{x^2}{2}}$

f(x) cant be constant since f΄(1)=1
For x=x and y=1 at $ f(x) f(y) \equal{} f(xy)$ we get f(x)[f(1)-1]=0 for every x real number that means f(1)=1
For x=x and y=0 and similar way we get f(0)=0
Lets assume that exists $ k \ne 0$ with f(k)=0 then putting x=x and y=k we get f(kx)=0 cant happened since f(x) isnt constant. So f(x)=0 only when x=0

Working similar to pankajsinha we have
$ \begin{array}{l} \mathop {\lim }\limits_{x \to {x_0}} \frac {{f\left( x \right) \minus{} f\left( {{x_0}} \right)}}{{x \minus{} {x_0}}}\mathop \equal{} \limits_{{x_0} \ne 0}^{x \equal{} {x_0} \cdot h} \mathop {\lim }\limits_{h \to 1} \frac {{f\left( {{x_0} \cdot h} \right) \minus{} f\left( {{x_0}} \right)}}{{{x_0} \cdot h \minus{} {x_0}}} \equal{} \frac {1}{{{x_0}}}\mathop {\lim }\limits_{h \to 1} \frac {{f\left( {{x_0}} \right)f\left( h \right) \minus{} f\left( {{x_0}} \right)}}{{h \minus{} 1}} \equal{} \\
\equal{} \frac {{f\left( {{x_0}} \right)}}{{{x_0}}}\mathop {\lim }\limits_{h \to 1} \frac {{f\left( h \right) \minus{} 1}}{{h \minus{} 1}} \equal{} \frac {{f\left( {{x_0}} \right)}}{{{x_0}}}\mathop {\lim }\limits_{h \to 1} \frac {{f\left( h \right) \minus{} f\left( 1 \right)}}{{h \minus{} 1}} \equal{} \frac {{f\left( {{x_0}} \right)}}{{{x_0}}} \cdot f'\left( 1 \right) \equal{} \frac {{f\left( {{x_0}} \right)}}{{{x_0}}} \\
\end{array}$
That means
 $ \begin{array}{l} f'\left( x \right) \equal{} \frac {{f\left( x \right)}}{x},x \ne 0 \Leftrightarrow \\
\Leftrightarrow xf'\left( x \right) \minus{} {\left( x \right)^\prime }f\left( x \right) \equal{} 0 \Leftrightarrow \\
\Leftrightarrow {\left[ {\frac {{f\left( x \right)}}{x}} \right]^\prime } \equal{} 0 \\
\end{array}$
From the above we get $ f\left( x \right) \equal{} \left\{ {\begin{array}{*{20}{c}} {{c_1}x,x > 0} \\
{0,x \equal{} 0} \\
{{c_2}x,x < 0} \\
\end{array}} \right.$

Since f΄(1)=1 we get $ f\left( x \right) \equal{} \left\{ {\begin{array}{*{20}{c}} {x,x > 0} \\
{0,x \equal{} 0} \\
{{c_2}x,x < 0} \\
\end{array}} \right.$

 For x,y negative numbers and from the relation $ f(x) f(y) \equal{} f(xy)$ we get  $ {c_2} \equal{} 1 \vee {c_2} \equal{} \minus{} 1$
So f can be written as
$ f\left( x \right) \equal{} x \vee f\left( x \right) \equal{} \left\{ {\begin{array}{*{20}{c}} {x,x \ge 0} \\
{ \minus{} x,x < 0} \\
\end{array}} \right.$
For the area we know from ypothesis that is bounded and that can be provide when $ f\left( x \right) \equal{} \left\{\begin{array}{l} x,x \ge 0 \\
\minus{} x,x < 0 \\
\end{array} \right.$

and since $ \frac {2}{{1 \plus{} {x^2}}} \minus{} f\left( x \right)$
 is an odd function we can calculate
\[ \int\limits_0^1 {\left( {\frac {2}{{1 \plus{} {x^2}}} \minus{} x} \right)dx \equal{} 2\left( {\frac {\pi }{4} \minus{} 0} \right)} \minus{} \left( {\frac {1}{2} \minus{} 0} \right) \equal{} \frac {{\pi \minus{} 1}}{2}\]
Final result π-1


You are going to play a board game four times (in succession).

Against opponent A, you win $ \frac23$ of the time playing this board game,
and against opponent B, you win $ \frac13$ of the time playing this board game.
Your goal is to win at least two games in a row in the set of four games.

You are given the option to play against these opponents in this order:

A - A - B - A.

Or, instead, you can elect to play against these opponents in this order:

B - B - A - B.

Is there an advantage to doing one of the above options over the other
option in order to try to win at least two games in a row throughout the
set of four games?

Is there something wrong here, or have I misunderstood the question?

It seems to me that, since you are twice as likely to win against $ A$, you will want to play $ A$ as often as possible.

There is only one way of getting $ 4$ victories:

$ WWWW$, with probability $ (\frac{2}{3})^{4} \equal{}\frac{16}{81}$.

There are $ 4$ ways of getting $ 3$ victories, all of which entail two wins in a row:

$ WWWL, WWLW, WLWW, LWWW$, with probability $ 4(\frac{2}{3})^{3}(\frac{1}{3})\equal{}\frac{32}{81}$.

There are $ 6$ ways of securing $ 2$ victories, $ 3$ of which entail two victories in a row:

$ WWLL, LWWL, LLWW$, with probability $ 3(\frac{1}{3})^{2}(\frac{1}{3})^{2} \equal{} \frac{12}{81}$.

Therefore the $ A\minus{}A\minus{}B\minus{}A$ strategy has a probability of $ \frac{60}{81}\equal{}\frac{15}{27}$.

The other strategy involves interchanging $ \frac{2}{3}$ with $ \frac{1}{3}$,  giving $ \frac{7}{27}$.

umm, I prob misunderstood it, but couldn't there also be orders like WLWW and WWLW? that would still give 2 consecutive wins... :huh:

Anytime, Arrange your tan  :D

Uhh, you kinda lost me in the fractions(not really, I'm just too lazy to read all of it, but I see your thinking), in the end, it [u]IS[/u] more beneficial to play person A, is that correct?

Yes, it is more beneficial to play person A.

With my errors, it showed me that I wanted to put
different conditions on my question, after all ,to
achieve a different result.


Solve $ \frac{1}{y}\frac{dy}{dx} \equal{} xy \plus{} \frac{x}{1\plus{}x^{2}}$ where $ y\equal{}1$ when $ x\equal{}0$.

For the other post also,solving the general $ \frac{y'(x)}{y(x)}\equal{}y(x) \alpha (x)\plus{}\beta (x)$.
Substituing $ y \equal{} z^{\minus{}1}$
$ z'(x)\plus{}z(x) \beta (x)\plus{}\alpha (x)\equal{}0$
Using [url=http://en.wikipedia.org/wiki/Integrating_factor]Integrating Factor[/url] with the original equation
$ z(x)\equal{}\frac{A}{\sqrt{x^2\plus{}1}}\minus{}\frac{1}{3} \left(x^2\plus{}1\right)$
$ y(x)\equal{}\frac{1}{\frac{4}{3 \sqrt{x^2\plus{}1}}\minus{}\frac{1}{3} \left(x^2\plus{}1\right)}$


I've heard the Kosnita's theorem these days.But I haven't found any proof of it?
Anyone can show me a proof of the Kosnita's theorem?
Thanks a lot.

See [url]http://pagesperso-orange.fr/jl.ayme/[/url] Volume 1 (2007) Le point de Kosnitza est l'isogonal du centre du cercle d'Euler
Your problem should be put in [url]http://www.mathlinks.ro/Forum/index.php?f=50[/url]


summer program
Mathcamp
Stanford
college
AwesomeMath
email

I have been accepted into Mathcamp 2006, but I'm still not sure how much portion of the camp fee could be subsidized... :(

By the way... I was also admitted by Stanford University Mathematics Camp (SUMaC). They've promised me full scholarship :roll: !

Anyway, since the date of the two programs overlap with each other, I would only attend one.

Congratulations to all successful applicants this year ! :rotfl: Moreover, also good luck to those on the waiting list. You may gain a vacancy from me because ultimately I have to forfeit either of the two camps :D

It seems that the admissions process was done fairly quickly this year... do you think there were less applicants?

Yeah, I was amazed that it only took a week after the deadline.  Anyway, I was accepted, so I'll see everyone there!

Yep, I'm in too, so I hope to see you guys there. :D

I was accepted but I  probably(like 99%) can't go. Do you know if  they let you defer acceptances for another year?

I got accepted.  But my first priority is probably PROMYS...

Got admitted. Excited! (they emailed me at 11:40 p.m., they work that hard? till...?)
  But I am still thinking about going there or not, because I will apply college next year, and this summer seems the golden time for that. I don't want to spend that long time on things other than application.
  What do you think of it?

  Also, I think, one of the reasons that they offer us so quickly is that they try to avoid the competition between Mathcamp and Awesome, since the Awesome still has 3rd round left. Mathcamp wants us to drop that idea...  :D  kidding.

First of all: if anyone's trying to decide if they want to go to camp or not, please feel free to contact me.  I'm both a student alum of the camp and a staff member there; there are also a ton of other alums floating around here!

To answer a few questions:

1) We'll have scholarship information out soon.  We should be able to do pretty well by everyone; we definitely don't want you to be kept away because of money.
2) 11:40pm is definitely not even close to the latest we were up, but it was important to us to get these out early (reasons unconnected to AwesomeMath, since I didn't even know about the status of their admissions --- just because it's better for everyone to know what's up).  We were also helped by our new and improved database, and because the quiz was harder this year --- scores were down across the board.  But that makes it easier to grade, I suppose.
3) We can't let you defer your admission for a year, although you could always apply next year and would probably have a good chance of getting in since you got in this year (and you'll only get better at math, right?).
4) Waiting list people will know their status as soon as we get decisions from everyone else.  We'll let someone new in each time someone else declines a spot.  We'll do our best to make this as quick as possible!
5) You should all come!  Camp is great!

Congratulations to everyone who got in.  This is shaping up to be a really good year for the camp.

Congratulations to everyone admitted to MathCamp 2006.

Thank you, Danz, for the information. And sorry for guessing the reason connected to Awesome Math.  :blush:
By the way, can we know some more details about the selection of this year? like, how many questions correct are solid to get in, how much you focus on things other than quiz, how many people applied...
Also when you mail me the info for registration, can I know the details for admission of myself, which part did well, which part needs improvement?... (because I want to gain some experience for applying college in the future through this admission)

The solutions of quiz will be given during the camp, right?

Thank you.

I will see you in Seattle!

Heh -- I'm spending this semester in Germany, so if you think that Mira and Dan were working late, I was at those meetings too. (-:

[quote="Go Beyond"]
By the way, can we know some more details about the selection of this year? like, how many questions correct are solid to get in, how much you focus on things other than quiz, how many people applied...
[/quote]

I can answer some parts of this directly, some parts vaguely, and other parts not at all.  We had in the vicinity of 150 applications this year -- just enough fewer than last year that we were able to cram all of the grading into one weekend, rather than having it spill over into a second weekend (since we all have other jobs to do during the week).  That's part of the reason that we were able to get decisions out so quickly.

We don't disclose how well one has to do on the quiz to get in. There are several good reasons for this.  The main reason is that we don't want applicants to be shooting for a target -- we want applicants to do as well as they can on the quiz.   Another reason, while not going into too much detail, is that on many of the problems it's not so easy to get a perfect score in our grading scheme.   So "number of problems solved" is not something we can meaningfully tally, and even if we tried, it would be misleading.  No one gets a score equivalent to solving k problems perfectly by solving k problems perfectly and nothing else -- always by submitting solutions to more than k problems and earning partial credit.

Along the same lines, many applicants think they've solved more problems than they actually have.  If we announced that solving n problems was enough to get in, we would have to deal with a few dozen angry emails and phone calls from applicants claiming to have solved n+1 problems, but who in fact have correct solutions for n/2 problems.

As a rule of thumb, 80-90% of spots at camp are determined by the quiz (that is, we'll accept everyone who scores above suitably chosen lines on the quiz, as long as there's nothing else in the application that causes concern for us, which happens on average less than once a year).     Near the line, the essay and recs become much more important: we're trying to put together a community, and while doing well on the quiz is the surest way to get in, there are other ways.  It is a fact that there are people who we have rejected (not waitlisted) this year who scored higher on the quiz than people we have accepted.

[quote]
Also when you mail me the info for registration, can I know the details for admission of myself, which part did well, which part needs improvement?...
[/quote]

We're pretty busy right now, but corner any of me, Mira, or Dan during camp, and we'll be happy to talk with you then.

[quote]
The solutions of quiz will be given during the camp, right?
[/quote]

Yes.

I got through!!!!!!!!!

By the way, I am Akash, 15 years old, from India. :)

I'd like to know how many others have qualified from India. How many from this part of the world? [because we could plan to go together].


calculus
integration
trigonometry
calculus computations

[img]http://img236.imageshack.us/img236/9905/integralsom6.jpg[/img]

Please help me how to solve these two integrals. On the 2nd one, I am unsure if it's either 1/2arcsec x^2 + C or 1/2arccsc x^2 + C.

Thanks. :)

For the first:
write it in the form $\frac{ f'(x) }{ \sqrt{1-f^{2}(x) }}$

For the second set  $x = \frac{1}{ \sqrt{ \sin{x}}}$.

#2: Since arcsec and arccsc differ only by a constant, it's both- whichever you like better.

the fisrt one $\int\frac{cos(x)}{\sqrt{9-sin^{2}(x)}}=\int\frac{cos(x)}{3\sqrt{1-(\frac{sin(x)}{3})^{2}}}$

take $u=\frac{sin(x)}{3}$ then $du=\frac{cos(x)}{3}$

so $\int\frac{1}{\sqrt{1-u^{2}}}$


Let random variable Z obbey $ \mathcal N (0,1)$ and $ X\equal{} \alpha\plus{}\beta Z\plus{} \sigma Z^2$ ($ \alpha, \gamma, \sigma$  is constants.)
Caculate $ E(X)$ and $ V(X)$.

$ E(X)$ is simple enough; we don't even need to know the specific distribution of $ Z,$ just that $ E(Z)\equal{}0$ and $ \text{Var}(Z)\equal{}1.$ Note that $ E(Z^2)\equal{}(E(Z))^2\plus{}\text{Var}\,(Z)\equal{}1.$ Hence $ E(\alpha\plus{}\beta Z\plus{}\sigma Z^2)\equal{}\alpha\cdot 1\plus{}\beta\cdot 0\plus{}\sigma\cdot 1\equal{}\alpha\plus{}\sigma.$

But to get $ \text{Var}(X),$ we will need more details.

$ E(X^2)\equal{}E(\alpha^2\plus{}2\alpha\beta Z\plus{}(\beta^2\plus{}2\alpha\sigma)Z^2\plus{}2\beta\sigma Z^3\plus{}\sigma^2Z^4)$

$ \equal{}\alpha^2\plus{}\beta^2\plus{}2\alpha\sigma\plus{}2\beta\sigma E(Z^3)\plus{}\sigma^2E(Z^4).$

The moment generating function for $ Z$ is $ e^{t^2/2}\equal{}1\plus{}\frac{t^2}2\plus{}\frac{t^4}{8}\plus{}\cdots.$

We can read off from that that $ E(Z^3)\equal{}0$ and $ E(Z^4)\equal{}\frac{4!}{8}\equal{}3.$

So $ E(X^2)\equal{}\alpha^2\plus{}\beta^2\plus{}2\alpha\sigma\plus{}3\sigma^2.$

Subtract off $ (E(X))^2\equal{}\alpha^2\plus{}2\alpha\sigma\plus{}\sigma^2$ to get

$ \text{Var}(X)\equal{}\beta^2\plus{}2\sigma^2.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

show that any group of order $ 4n\plus{}2$ has a subgroup of index $ 2$.

$ G$ acts on itself by left multiplication.  An element of order 2 gives rise to a permutation that is the product of $ 2n\plus{}1$ transpositions, in particular it is odd.  If this permutation representation has any odd permutations, exactly half of the permutations are odd, so the set of even permutations forms a group of index 2.


trigonometry
calculus
function
algebra
polynomial
logarithms
calculus computations

How do I use Taylor approx. to estimate sin 0.3 with less than 0.01 error. Im guessing i have to use that Rn(x) error thingy but don't know how. Can anyone help?

Also, may seem silly, but do you think 0.3 refers to degrees or radians. I'm never quite sure unless they specify.

Thankyou x

Denote f(x)=sin x. $f(x)=f(0)+f'(0)x+...+\frac{f^{(n-1)}(0)}{(n-1)!}x^{n-1}+\frac{f^{(n)}(\theta x)}{n!}x^{n}$, $0<\theta<1$ In our case $|f^{(n)}(y)|\le1$ for all y, x=0,3. I f we take n such that $\frac{0.3^{n}}{n!}<0.01$ then we shall have what we need($\frac{f^{(n)}(\theta x)}{n!}x^{n}$ is error)
P.S. 0.3 refers to radians, of course

In this case you don't even need the Taylor remainder estimate: $\sin 0.3=\sum_{n=0}^{\infty}(-1)^{n}\frac{0.3^{2n+1}}{(2n+1)!}$
For alternating series, the error does not exceed the first discarded term (in absolute value). Since $0.3^{3}/3!=0.0045<0.01$, just one term suffices: we have $\sin 0.3\approx 0.3$ with required precision.

[quote]Also, may seem silly, but do you think 0.3 refers to degrees or radians. I'm never quite sure unless they specify. [/quote]
In calculus - and in differential equations, which is essentially calculus continued - trigonometric functions are [b]always[/b] in radians.

Even if you were asked, for instance, for the sine of $32^{\circ}$ (approximated by a Taylor polynomial), the very first thing you'd do is to convert those degrees into radians; instead of $32^{\circ}$ make that $\frac{\pi}6+2\cdot\frac{\pi}{180}.$

There's a similar calculus-based assumption about logarithms: if you're not explicitly told what the base of the logarithm is, assume it's the natural logarithm.


On the bottom of the forums page you have users guest, etc .
What are the hiddens?

This indicates the number of registered users that are online but have chosen to not have their user name listed as being online. This can be set at:

Profile->Preferences->Privacy Choices

Hidden users, are usually in itallic, because they don't want us to know if they're online or offline. Also, you can't find out in the "online list" where they are right now :)


I have a question about situations that involve overlapping in the solutions using the UC method.  For example, suppose we had:
$y''-2y'+y = xe^{x}$, the auxiliary equation is $m^{2}-2m+1$ which gives that there is a repeated root of $m=1$. Then $y_{c}= c_{1}e^{x}+c_{2}xe^{x}$.
Now for the particular solution we begin with:
$(Ax+b)e^{x}$ and we see that there is an overlap with $xe^{x}$ and $e^{x}$ so we multiply by $x^{n}$ where $n$ is the smallest integer that eliminates the overlap, in this case 2.  My question is: Do we multiply $x^{2}$ into the existing $(Ax+b)$  and use $(Ax^{3}+Bx^{2})$, or do we multiply it into the $xe^{x}$ and then use $(Ax^{3}+Bx^{2}+Cx+D)$?

You should use $(Ax^{3}+Bx^{2})$, because the purpose is to get something linearly independent. The $C$ and $D$ terms in the second thing are linearly dependent with the complementary solution (or whatever you guys call the I guess "general" solution since you use particular).

Ok I see, I didn't realize that $C$ and $D$ would be linearly dependent with $y_{c}$.  As a side note, I was taught the the general solution for a non-homogeneos equation is $y_{c}+y_{p}$ where $y_{c}$, the complementary function, is the general solution to the corresponding homogeneous equation and $y_{p}$ is a particular solution to the non-homogeneous equation.

It doesn't matter which you use.  You have free control over the linear and constant parts because of the $c_{1}$ and $c_{2}$ in the homogeneous solution, so there is no need to duplicate that freedom in the particular solution.  Any function you could express using $Ax^{3}+Bx^{2}+Cx+D$ you also could express using just $Ax^{3}+Bx^{2}$ by tinkering with $c_{1}, c_{2}$.


projective geometry
geometry solved
geometry

Let $D$ be a point on side $BC$ of triangle $ABC$ and point $P$ on segment $AD$. A line through $D$ intersects segments $AB$, $PB$ at $M$, $E$ respectively, also intersects the extensions of segments $AC$, $PC$ at $F$, $N$ respectively.
If $DE = DF$, prove that $DM =DN$.

Simple exercise in projective geometry.

Denote the line through the points D, M, E, F and N by d.

The Desargues involution theorem states that, if X, Y, Z, W are any four points and g is any line in the plane, then there exists an involution on the line g such that the pairs of points $\left(XY\cap g;\;ZW\cap g\right)$, $\left(YZ\cap g;\;WX\cap g\right)$ and $\left(XZ\cap g;\;YW\cap g\right)$ are conjugate pairs in this involution. Applying this to the points X = A, Y = B, Z = C and W = P and the line g = d, we conclude that there exists an involution i on the line d such that the pairs of points $\left(AB\cap d;\;CP\cap d\right)$, $\left(BC\cap d;\;PA\cap d\right)$ and $\left(AC\cap d;\;BP\cap d\right)$ are conjugate pairs in this involution i. But $AB\cap d=M$, $CP\cap d=N$, $BC\cap d=D$, $PA\cap d=D$, $AC\cap d=F$ and $BP\cap d=E$. Hence, we obtain that the pairs of points (M; N), (D; D) and (F; E) are conjugate pairs in this involution i. In other words, the points M and N are conjugate to each other in the involution i, the point D is conjugate to itself in the involution i, and the points F and E are conjugate to each other in the involution i.

Of course, the result that the point D is conjugate to itself in the involution i simply means that the point D is a fixed point of the involution i. Thus, the involution i has a fixed point, i. e. it is a hyperbolic or parabolic involution. Let D' be the second fixed point of this involution i (in the degenerate case when the involution i is parabolic, i. e. has only one fixed point, set D' = D). It is well-known that any two conjugate points in an involution are harmonic conjugates with respect to the segment joining the two fixed points of the involution. Since the two fixed points of our involution i are D and D', and since the points M and N are conjugate to each other in the involution i, we thus can conclude that the points M and N divide the segment DD' harmonically. Hence, as a consequence, the points D and D' divide the segment MN harmonically. Similarly, since the points F and E are conjugate to each other in the involution i, the points D and D' divide the segment FE harmonically. In other words, the point D' is the harmonic conjugate of the point D with respect to the segment FE.

But since DE = DF, the point D is the midpoint of the segment FE, and thus, its harmonic conjugate D' with respect to the segment FE is an infinite point. But since the points D and D' divide the segment MN harmonically, the point D is the harmonic conjugate of the point D' with respect to the segment MN; since the point D' is an infinite point, its harmonic conjugate D must therefore be the midpoint of the segment MN. Thus, DM = DN. And the problem is solved.

  Darij

Applying menelaus theorem three times.
\[
\frac{{DP}}{{PA}}\frac{{AB}}{{BM}}\frac{{ME}}{{ED}} = 1
\]
\[
\frac{{AP}}{{PD}}\frac{{DN}}{{NF}}\frac{{FC}}{{CA}} = 1
\]
\[
\frac{{FC}}{{CA}}\frac{{AB}}{{BM}}\frac{{MD}}{{DF}} = 1
\]
Combining them we have
\[
\frac{{DN}}{{DM}} = \frac{{NF}}{{MD}}
\]
which gives the result.


I recently read that "a container of water can be 'heaped over.'" I've heard of other properties such as strong resistance to heat and cohesion but I've never heard or know what this mean. Anyone has any good idea?

This is actually a really good question, and thus worthy of my first ever post:

As you well know, water is a big factor in biological systems (most if not all reactions in a biological system happen in an aqueous environment)... and as with everything, there is an underlying reason.

Now, the main point of your question was concerning something known as surface tension--or, more simply, the ability to fill a container past the brim with water.  Hydrogen bonding is the source of this particular characteristic.  When you have hydrogen paired with a particularly electronegative heteroatom, the tendency exists for the electronegativity of the heteroatom (Oxygen, Nitrogen, or Sulfur) to decentralize the electron cloud of the hydrogen, creating a slight, though relatively dense, charge capable of acting as an intermolecular force.

Think about the structure of water.  It contains hydrogen and oxygen in abundance, with oxygen being a highly electronegative atom.  So, it is only logical that we'd see the formation of hydrogen bonds that, though individually much weaker than a covalent or ionic bond, could be quite strong en masse.  The fact that water molecules form these hydrogen bonds between each other is what cause them to be 'cohesive'.  So, when you fill a container above the rim, the cohesive forces of the hydrogen bonds are strong enough to keep it from spilling over.

More fascinating from a biological standpoint is the role that water via hydrogen bonds plays in mediating temperature (essential in a biological system to prevent the denaturing of proteins) and even in plants.  Water's cohesive properties are part of the reason it flows [i]up[/i] the stem, along with adhesion.

This is just a summary introduction, but there should be a lot of literature available online about this sort of thing.  Hydrogen bonding is a very important fundamental principle in biology.

    Hope I was helpful,
          Matt Corley

You can see this more in containers with a small diameter.

You can also enhanse the property if you add energy enhansing surfacants.

Another way to do this would be to use the principle of freezing, but be ready to have the heaping run out when it melts, just like a particulate heaped might flow if vibration started the particles flowing (past their immediate angle of repose).

If one is talking about biology, one can moderately over-hydrate a life form without damage.

My last, and most extreme example, is to change the force of gravity - short time while the continer is falling to remove/reduce the force against which surfact tension must act.  (In orbit, the limiting condition, one can heap a container with a lot of water.)

Thanks for an interesting idea.


algorithm
combinatorics proposed
combinatorics

At a math contest there are $2n$ students participating. Each of them submits a problem to the jury, which thereafter gives each students one of the $2n$ problems submitted. One says that the contest is [i]fair[/i] is there are $n$ participants which receive their problems from the other $n$ participants.

Prove that the number of distributions of the problems in order to obtain a fair contest is a perfect square.

i think i might actually have a bijective solution here....
firstly rephrasing in terms of permutations, the number of 'good' distributions is simply the same as saying that if we consider the digraph of the corresponding permutation, then the undirected graph that results from this digraph is bipartite with partition sets of equal size. Since the corresponding graph has to be bipartite, it means that every cycle in the cyclic representation is even. Conversely, starting with a permutation that has only even cycles say $C_1,C_2,\ldots,C_r$ of sizes $2k_1,2k_2\ldots,2k_r$ resp., it is easy to see that if we consider a proper coloring of this graph, then both the color classes are equal-that has to be the case, anyway, since this is a permutation.
so what we need to count is simply the number of permutations that have all cycles even.I claim that this number is $1^2\cdot 3^2\cdots(2n-1)^2$ and so in particular is even.
this can be bijectively seen as follows: Consider all ordered tuples of partitions $(\Pi,\Psi)$ of $X=\{1,2,\ldots,2n\}$ of type $2^n$(i.e., all parts have size $2$). The number of such tuples is of course, the reqd number. Wha we now need is a bijection between these and the permutations with only even cycles.
What we can do is, firstly retain(and remove) all the common sets in both partitions. Call the new set again as $X$(after deleting the elements of the common partitions). Now consider the two partitions and the parts of $\Pi$(resp. $\Psi$) that are not yet removed. Now starting from the least element of $x\in X$ and its corresponding partition, (say $\{x,y\}$),start with the word $xy$ and proceed to the other partition to that part that contains $y$.Call it $z$. enlarge the word to $xyz$ and again go to the part of the other partition that contains $z$ and keep proceeding till we hit an element that was already covered in the word.since both are partitions, one can argue that when we reach an already registered entry it has to be $x$ again. And since there are pairs counted at all times except the beginning and the end we reach a word of even length. This is one of the cycles of the reqd permutation. Again delete these elements from $X$ and iterate. This algorithm is easily inverted and accomplishes the desired result.

[quote="seshadri"]we consider the digraph of the corresponding permutation, then the undirected graph that results from this digraph is bipartite with partition sets of equal size.[/quote]

I couldn't get how did you obtain this undirected graph?

I couldn't understand, can a question be given more than one student?

We claim the answer is $f(n) = (2n-1)!!^2$, assuming that no two students receive the same problem.

First, the re-distribution of problems can be modeled by a permutation $\sigma$ of $[2n]$ (the first $2n$ positive integers). Write the permutation in cycle notation. Suppose we have a fair contest; without loss of generality, let contestants $1, 2, 3, \dots, n$ receive their problems from contestants $n+1, n+2, \dots, 2n$. This easily implies $n+1, n+2, \dots, 2n$ receive their problems from contestants $1, 2, 3, \dots, n$. Let $A = \{1, 2, \dots, n\}$ and $B = \{n+1, n+2, \dots, 2n\}$. Then $1 \in A$, so $\sigma(1) \in B, \sigma(\sigma(1)) \in A$, $\dots$. This implies the cycle containing 1 has even length. By a similar argument, all cycles of the permutation representing the fair contest have even length. Conversely, if all cycles have even length, then in each cycle color the numbers red and blue, such that no two consecutive numbers in the cycle are the same color. Then the red contestants receive their problems from the blue contestants, and vice versa. Therefore, it suffices to count the number of permutations of $[2n]$ with all cycles being even length. There are two ways to do this.

[b]Solution 1. (HMMT-style bash)[/b] Let $f(n)$ denote this quantity for a given $n$. Then by considering the length of the cycle $1$ is part of, we derive the recurrence $f(0) = 1$, and for $n \ge 1$,
$$f(n) = \sum_{k=0}^{n-1} (2n-1)(2n-2)\dots(2k+1) f(k).$$
Let $f(n) = (2n)! g(n)$; then we get $g(0) = 1$ and $2n g(n) = g(n-1) + g(n-2) + \dots + g(0)$. Thus $2n g(n) = g(n-1) + 2(n-1) g(n-1) = (2n-1) g(n-1)$, so $g(n) = \frac{2n-1}{2n} \cdot g(n-1)$. By induction we get $g(n) = \frac{(2n-1)!!}{(2n)!!}$, so $f(n) = (2n-1)!!^2$, as claimed.

[b]Solution 2. (Bijection)[/b] We claim that $f(n) = g(n)^2$, where $g(n)$ is the number of ways to partition $[2n]$ into $n$ pairs. (It is well-known that $g(n) = (2n-1)!!$.) Indeed, it suffices to establish a bijection between permutations $P$ of $[2n]$ with all cycles of even length, and ordered pairs of permutations $(A, B)$ of $[2n]$ with all cycles of length $2$.

For a given $P$, write it in cycle notation. In each cycle, color the smallest number red and alternate coloring numbers red and blue such that no two consecutive numbers are the same color. Then we can let $A, B$ be the set of pairs $(x, P(x))$ for all red, blue $x$ respectively. Conversely, suppose we are given $A, B$. They determine $n$ red and $n$ blue edges on a graph with vertices at $1, 2, 3, \dots, 2n$. Forgetting about the colors for a moment, we see that each vertex has degree $2$, so the graph is a disjoint union of cycles. Now let's make everything colorful again. The edges of each cycle alternate red, blue, red, blue, .... We assign the cycle a direction by considering the vertex $v$ (of the cycle) with smallest number, and pointing the cycle in the direction of the red edge emanating from $v$. Thus, by piecing together all directed cycles, we obtain $P$ from $A, B$. Thus, our claimed bijection is invertible, proving that $f(n) = g(n)^2$.

[quote=Jettofaiyafukushireikan][quote=seshadri]This algorithm is easily inverted and accomplishes the desired result.[/quote]

you mean that, the number of the desired number is the square of the number of the all possible matchings in a 2n-complete graph.
this, however, can not be inverted
for exanple,
Π={1,2}{3,4}{5,6}{7,8} Ψ={1,3}{2,4}{5,7}{6,8}
and
Π={1,2}{3,4}{5,7}{6,8} Ψ={1,3}{2,4}{5,6}{7,8}[/quote]

Using my algorithm, it can be inverted. The first one is $(1243)(5687)$ and the second is $(1243)(5786)$. I hope this example better illustrates my algorithm and proves the correctness of the solution.


I've just solved this poblem:If $ n$ is composite then $ 2^{n}\minus{}1$ is also composite. Solution in the book was different from mine. It says that proof follows from the fact that $ a\minus{}b|a^{n}\minus{}b^{n}$, and I don't understand why is that. Someone help me?

[hide="Method 1"] $ a^{n}\minus{}b^{n}\equal{} (a\minus{}b)(a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}a^{n\minus{}3}b^{2}\plus{}...\plus{}ab^{n\minus{}2}\plus{}b^{n\minus{}1})$ [/hide]
[hide="Method 2"] [b]Lemma:[/b]  $ a\equiv b\bmod m, c\equiv d\bmod m\implies ac\equiv bd\bmod m$

[b]Corollary:[/b]  $ a\equiv b\bmod m\implies a^{k}\equiv b^{k}\bmod m$

Because $ a\equiv b\bmod (a\minus{}b)$ trivially, we conclude $ a^{k}\equiv b^{k}\bmod (a\minus{}b)\Leftrightarrow a\minus{}b | a^{k}\minus{}b^{k}$. [/hide]

I now that $ a\minus{}b|a^{n}\minus{}b^{n}$, but I want to know why solution follows from that fact.

If $ n \equal{} pq$ (a composite number), then $ 2^{p}\minus{}1|(2^{p})^{q}\minus{}1^{q}$ and you have the claim.

That's the way I solved it, but I'm interested in understanding the solution in the book which says that it follows from $ a\minus{}b|a^{n}\minus{}b^{n}$.

That's exactly what I've written down. Put $ a\equal{}2^{p}, b\equal{}1, n\equal{}q$

I have found also this solution, suppose $ n$ is composite and $ 2^{n}\minus{}1$ is prime, then for $ k$ with $ (2^{n}\minus{}1,k)\equal{}1$ we can say that $ k^{2^{n}\minus{}2}\equal{}1 mod 2^{n}\minus{}1$.

Then $ k^{2^{n\minus{}1}\minus{}1}\equal{}1 mod 2^{n}\minus{}1$ so $ 2^{n\minus{}1}\minus{}1\equal{}2^{n}\minus{}2$ and the only solution for this is $ n\equal{}1$ but $ 2^{n}\minus{}1$ would not be a prime then. Contradiction ?


A uniform cylinder of mass M and radius R is at rest on a block of mass m, which in turn rests on a horizontal, frictionless table.  If a horizontal force F is applied to the block, it accelerates and the cylinder rolls without slipping.  Find the acceleration of the block.

[hide="solution"] it is easy to derive by the fact that there is no slipping betwen the two surface sthat
$ a_{0} \equal{} a \plus{} \alpha R$ where $ a_{o}$ is the accelartaion of block and $ a$ the accelaration of centre o f mass of the clinder in earth frame
also we have if $ f$ is the griction then $ f R \equal{} \frac {MR^{2}\alpha}{2}$ so $ f \equal{} \frac {M \alpha R}{2}$
also we have $ f \equal{} Ma$ also $ F \minus{} f \equal{} ma_{0}$
so we have $ \frac {F \minus{} f}{m} \equal{} \frac {f}{M} \plus{} \frac {2f}{M} \Rightarrow f \equal{} \frac {MF}{3m \plus{} M}$
so we have $ a_{0} \equal{} \frac {3F}{3m \plus{} M}$
[/hide]


There is a 20*20 array of dots where each dot is coloured blue or red. Its given that there are totally 219 red dots out of which 39 are on the border but not in the corners. Any 2 adjacent red/blue dots are joined by red/blue lines resp. and 2 adjacent dots of different colours (one red, one blue) are joined by a black line.
Its given that there are 237 black lines. Find the no. of blue lines.[/b]

Let me add another similar problem (B) which i think is similar:

There are 2004 points $P_{1}, P_{2}, \ldots, P_{2004}$ on a circle, and these points are labelled with the integers from 1 to 2004 inclusive in a certain way such that there are:
( I) 600 groups of three adjacent points which consist of all even integers
(II) 500 groups of three adjacent points which consist of 2 even integers and one odd integer,
with the understanding that point $P_{1}$ and $P_{2004}$ are adjacent.

Find:
(III) The number of groups of three adjacent points which consist of 1 even integer and 2 odd integers
(IV) The number of groups of three adjacent points which consist of all odd integers.

HINT [hide]Maybe we should use Fubini's principle to solve the problem[/hide]
MY ANSWER [hide]263[/hide][/b]

[hide="Solution to original question"]Consider \[\sum_{\textrm{segments}}\textrm{number of red endpoints}.\]  This is equal to \[2\cdot (\textrm{number of red edges})+\textrm{number of black edges}.\]  However, it's also equal to \[\sum_{\textrm{red vertices}}\textrm{degree of vertex},\] which is \[39\cdot 3+(219-39)\cdot 4 = 837.\]  Since there are 237 black lines, there are 300 red lines and thus $760-300-237 = 223$ blue lines.[/hide]

THERE ARE IN FACT 800 TOTAL NUMBER OF LINES NOT 760.
THE REST OF THE SOLUTION IS PERFECT.

You are wrong.  Each of 20 columns has 20 vertices and thus 19 edges, thus $20\cdot19 = 380$ vertical edges.  Similarly 380 horizontal edges, so 760 total edges.


Given a regular tetrahedron $ OABC$ with side length 1. Let $ P,\ Q$ and $ R$ be the internally dividing points such that : $ OP: PA \equal{} 2: 1,\ OQ: QB \equal{} 1: 2$ and $ OR: RC \equal{} t: (1 \minus{} t)$ for $ 0 < t < 1$. Find the value of $ t$ for which the area of $ \triangle{PQR}$ is minimized.

It might help if you tell us what the vertexes of the tetrahedron are and stuff  :wink:

Sorry, I meant like, what the names of the vertexes are or something, or else I have no idea what any of the points mean


Find all polynomials $f(x)$ with real coefficients for which
$f(x).f(2x^{2})=f(2x^{3}+x)$


Looking at college apps and scholarships, I have found that employment comes up more than I expected it to.  To get into highly selective colleges and receive some decent outside scholarships, how important is work experience? And to add, how does work experience measure up to volunteer activities?  I have several volunteer activities, many of which could have been paid but I chose to go pro bono.  I just can't imagine how highly selective colleges and scholarships could expect kids to put in so much time into school & academics and WORK too.

Sadly, I have seen work experience counts for very little- on the same level as "minority status" in some cases. This, to me, is very disappointing, as I have held a steady job since I was 14. I wish that, at least, colleges would take into account the [i]type[/i] of work done, as mine has consistently been within my chosen field (vet med). If you go to College Board's website, you will see how important different things are for different schools. Under their "college finder," where they detail every college you could ever think of, there is a section where colleges rate how important different factors (test scores, essays, work exp, minority status, and so on). That should give you a better idea.

huh?  in most cases i would say a job counts like any other extra-curricular.  an instance where it can/should help a bit more is if it has to do with your interests, i.e if math is your passion, then a job teaching math would count for significantly more than, say, a job bagging groceries at a local store.

There is no one thing that can cause rejection or acceptance. Work experience is not a vital part of the application. They ask just to see what you do with your time and to see what kind of responsibilities you've handled before. The colleges and scholarship people want to get a really good picture of what kind of person you are, and asking such questions help them.
If you have nothing to put in work experience, you shouldn't try to get work just to have something to fill in. Just continue doing what you're interested in.

Well, of course they're going to want someone who spends more time on extracurriculars (school activity) than those who are constantly working off campus (not productive for the school!). They don't want to admit you to their school and then end up seeing that you're working during your free time instead of participating.

[quote="bleumoose"]Sadly, I have seen work experience counts for very little- on the same level as "minority status" in some cases.[/quote]

You have never worked inside an admissions office, I take it. The kind of material you quote in your next post, also in this thread, should not be taken as any kind of mathematical description of how an admissions office weighs different factors in deciding on admissions cases.

Your quoted paragraph above looks odd to me because usually minority status is a HUGE factor in college admission, having more actual real-world effect than a 100-point difference between two applicants on their SAT aggregate scores on the "old" SAT. If colleges don't admit that minority status counts for a lot in actual decision-making, that merely illustrates that you shouldn't take lists like that too seriously.

Colleges always say that school grades count for a lot--they have to say that to please the constituency of schools that refer students to one college or another. But of course perfect grades at MOST schools counts for very little indeed for getting into the top colleges, because there are 38,000 high schools in the United States, and tens of thousands more high schools in other countries where students apply to United States colleges, but only a few thousand places in the entering classes of the most desired colleges each year. There isn't even room in the top colleges for all of the high school students who are valedictorians, so high school grades really only act as a threshold factor, and other factors distinguish which applicants get in and which don't from a pool of applicants who all have fairly high grades.

Returning to the OP's question, work experience is good, because it shows using one's time responsibly. A person who does work for free ("pro bono") is showing the same thing, and needn't worry about not working for money. A person who doesn't work at all, but who merely hangs out at the mall, OUGHT TO look less desirable than a student from a poor family who works to help his parents buy groceries. Colleges are asking more about work experience on their application forms and in their interviews recently because they found out that former admissions practices admitted too many rich slackers and excluded too many of the smart but poor students who had to work at high school age.

So wait a second- are you basically saying that (by only these factors) as a white person who has been working diligently to save some money (FOR college) since the age of 14, I am less desirable to a college than a minority who has not held a job? This seems incredibly disappointing to me.

Also, I wonder about what WindSlicer said. Our school is notably lacking in "extracurriculars" because of its nature. We have no athletic teams, and no clubs outside of MAO, NHS, Math team, Science Olympiad, Heritage Panel, and SGA. Math team falls off by the end of tenth grade because of lack of interest. The Heritage Panel and SGA I really had no passion for (politics aren't my thing, and a series of disappointing runs for SGA when less qualified candidates were chosen made me jaded). I am in NHS and MAO, but other than that, I have no extracurriculars. My job, to me, IS my extracurricular activity (as well as my volunteer ops). I  enjoy it to no end and feel I am learning a lot. Yes, I get paid, but not as much as many of my friends who work in retail, etc. I am saving the money to pay for college as well as for a backpacking trip to Europe I hope to take next summer (if my mother is not too terrified over me being abducted). Is there any way to explain this to a college? I am spending my time wisely, I believe, and building up some money so that I can pay for my own education (i.e. I'm not buying a bunch of CDs with the money).

Also... why would College Board say that (about the important factors, see above) if it's not true? I'm guessing this is what the colleges themselves are reporting TO College Board, but if that's so- why would they not tell the whole truth? I realize I'm asking for clarity in something that no one has ever been able to clarify, but... could anyone give me some reasoning on this? Thanks.

Well, of course if your school doesn't offer many extracurricular activities you can't participate in them. You just have to make sure that they know that you are working to save up and pay for college (I'm sure you could sneak that in a college essay or two). It's all about the different situations. And don't forget that you can always start up a club that you are interested in, not only will it give you something to do, but I would think it looks good if you start a club (although I don't know how much free time you have with the work, school, current clubs you are in). But either way, I dont' think it will be a deciding factor in whether you get accepted or not.

[quote]you can always start up a club that you are interested in[/quote]

*laughs* why do you think we don't have any clubs? all attempts have been smashed- even those for CHESS and helping the environment and so forth. apparently, by law, if one school club is allowed to start, all must be allowed. the forming of some such clubs would take away from the time we are supposed to be concentrating on our specialties, therefore, no clubs allowed, outside of already formed clubs. GR.

but i digress. good idea, sneaking it into essays. danke!

where do you live?  you have to realize everything's relative... so long as you make the best use of the resources given to you, you'll be attractive to colleges.  If you have a genuine passion for your job, you're most likely ahead of a lot of applicants who participate in lots of different activities but don't really have a strong interest in any of them.

[quote="bleumoose"]extracurriculars count for more than work experience  AND volunteer work.[/quote]

Work that requires skills and selection, such as a programming job, is a whole other animal.

[quote="bleumoose"]MIT's Admission Factors (From College Board website)

Very important admission factors:
Character/Personal Qualities
Secondary School Record

Important admission factors:
Class Rank
Extracurricular Activities
Interview
Recommendations
Standardized Test Scores
Talent/Ability

Considered:
Alumni/AE Relation
Essay
Geographical Residence
Minority Status
Volunteer Work
Work Experience

As you can see, extracurriculars count for more than work experience  AND volunteer work.[/quote]

That list is absurd. Let me assure you there is no set list. On any given day, depending on the admissions person's mood, they might choose to consider one more than the other. In general I would say that, especially at MIT who places a bit TOO MUCH emphasis on minority status, that list is nothing more than PR.

[quote="tokenadult"][quote="bleumoose"]Sadly, I have seen work experience counts for very little- on the same level as "minority status" in some cases.[/quote]

You have never worked inside an admissions office, I take it. The kind of material you quote in your next post, also in this thread, should not be taken as any kind of mathematical description of how an admissions office weighs different factors in deciding on admissions cases.

Your quoted paragraph above looks odd to me because usually minority status is a HUGE factor in college admission, having more actual real-world effect than a 100-point difference between two applicants on their SAT aggregate scores on the "old" SAT. If colleges don't admit that minority status counts for a lot in actual decision-making, that merely illustrates that you shouldn't take lists like that too seriously.
[/quote]

As usual, tokenadult is on top of things and got here first. Well said.

[quote="bleumoose"]So wait a second- are you basically saying that (by only these factors) as a white person who has been working diligently to save some money (FOR college) since the age of 14, I am less desirable to a college than a minority who has not held a job? This seems incredibly disappointing to me.[/quote]

Whoa whoa now... let's be correct, a white female IS a minority (at MIT), as a white male IS NOT. "White" simply doesn't cut it. Since you are male, I'll assume you're were relating the question to yourself. I think it's a hueristic that, all other things aside, a minority student is more desirable than a non minority. That's not to say that your work experience could not make up for it, or any other of your attributes for that matter. If I had to give you a yes/no answer to this question though, I'd have to say yes, and I too find it disappointing. But there is some good to come of this, life is disappointing, why not get used to it now.

[quote="bleumoose"]
Also... why would College Board say that (about the important factors, see above) if it's not true? I'm guessing this is what the colleges themselves are reporting TO College Board, but if that's so- why would they not tell the whole truth? I realize I'm asking for clarity in something that no one has ever been able to clarify, but... could anyone give me some reasoning on this? Thanks.[/quote]

School's like MIT care a lot of about how they look to the public. Like any business, they give what they want the general public to see and not necessarily the truth or at least the whole truth.

I will give you one example of this. Have you ever sat and wondered how some schools like MIT retain 99% of their students? Considering how "hard" MIT is, don't you think that quite a few can't handle it or maybe can no longer afford it or simply don't like it? Here is a little unspoken trick that they use. They don't allow students without some very serious reason to "withdraw" from the school. If you go and decide to leave, they will place you on "temporary leave". Technically you are still a student there, and can return up to two years later and face no penalty. You get placed on "temporary leave" even if you and MIT both know fully well you aren't going to return. Why? Because they can then record you as still being a student, even if you left. Now here's the kicker, if you then at the end of your leave decide not to come back, they aren't required to count you against their retention rate. Go figure. So basically they can fudge their numbers to make it look like everyone stays when infact this is not true.

I know this because I personally know (know well for that matter) people who have left MIT for various reasons who were all explained this process and who were all put on "leave". When I found this out I was a bit disappointed with MIT.

The more I learn, the more I realize that MIT and other like schools are like Super Models, that is they only look good when wearing their makeup.

Edit: I want to clarify that my words are specifically to MIT's undergrad program, and don't represent my feelings about their graduate programs, as they are completely separate and indeed different animals.


Could sombody be so kind as to walk me through the basics of polar function calculus?

Thanks, ive been googling and reading wikipedia so i have some knowledge, i just need you super smart forum members to help me consolidate it in my head!

This should go in the Calculus Computation and Tutorial Forum I think.


Harvard
college
advanced fields
advanced fields unsolved

[quote="Moubinool"]Qualifying examination Harvard University
problem 3/Day 2/September 2004

a) Show that every linear fractional transformation wich maps the upper half  to the plane onto itself is of the form
$F(z)=\frac{az+b}{cz+d}$ with $a,b,c,d \in \mathbb R \; ad-bc=1$

b) Show that every linear fractional transformation wich maps the unit disk onto
itself is of the form
$F(z)=\frac{az+b}{\overline{b}z+\overline{a}}$ with $a,b\in \mathbb C \; |a|^2 -|b|^2 =1$[/quote]

You can have the answer of this question here :
http://www.dptmaths.ens-cachan.fr/IMG/pdf/cours_ana_compl_desvillettes.pdf

Enjoy reading.

--
Fabrice


The Olympiad section is for Olympiad problems, not homework.  This belongs in [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]HSB[/url].

Consult a list of trigonometric identities.

[hide]Use the double angle formula:

$ \cos(2a) \equal{} \cos^2 a \minus{} \sin^2 a$[/hide]

[hide]Use the double angle formula: cos 2x=(cos x)^2-(sin x)^2

So (cos (x/5))^2-(sin (x/5))^2=cos (2x/5)[/hide]


ok Dude! solve it. I [b]killed[/b] myself but I couldn't...

Isn't $\frac{a^{2}}{2b^{2}}+\frac{1}{2}\ge \frac{a}{b}$? Maybe there is typo in the problem?

no there is no typo
the $\sum$ is a cyclic sum I suppose.
I think this problem was solved by arqady on the forum.

[quote="VIETNAM"]Another problem.It 's also simple.
a,b,c,d are real positive numbers.Prove that:
$\sqrt[n]{\frac{a}{b+c+d}}$+$\sqrt[n]{\frac{b}{a+c+d}}$+$\sqrt[n]{\frac{c}{a+b+d}}$+$\sqrt[n]{\frac{d}{a+b+c}}$ $>2$[/quote]

For different problems please use different threads.

conceive

[quote="VIETNAM"]For any a,b,c>0.Prove:
 $\sum$ $\frac{a^{2}}{b^{2}}$+3 $\geq$ $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$.[/quote]
 $\text{Appling AM-GM inequality then}$
 $\frac{a^{2}}{2b^{2}}+1\geq \frac{a}{b}\text{ and}$
 $\frac{a^{2}}{2b^{2}}+\frac{b^{2}}{2c^{2}}\geq \frac{a}{c}$
 $\Rightarrow \text{ As desire}$ :)  :)


the source may be irritating, but this is a numerical corollary from a theorem proven there which is nontrivial.

let $a,b_1,...,b_6$ be positive real numbers. assume that

$a>b_i+b_j$ for all distinct $i,j$,
$2a>\sum_{j\neq i} b_j$ for all $i$.

then prove the inequality $a^2>\sum b_i^2$.

and 10 euro for the guy who guesses the theorem it comes from without looking it up :D

This is a really beautiful inequality! I am wondering about possible extensions to more than 6 numbers $b_1$, $b_2$, ..., $b_n$. Somehow it looks like you are doing a new kind of inequalities... Anyway, to give a halfways traditional proof, I am posting one that was found by a rather promiment olympiad mathematician at the Nuremberg seminar which just ended this morning.

At first we can WLOG assume that $b_1\geq b_2\geq b_3\geq b_4\geq b_5\geq b_6$. Then, of course, the only conditions worth thinking about are $a>b_1+b_2$ and $2a>b_1+b_2+b_3+b_4+b_5$. We have to prove the inequality

$aa>b_1b_1+b_2b_2+b_3b_3+b_4b_4+b_5b_5+b_6b_6$.

Well, at first, since $b_5\geq b_6$, it is clearly enough to prove

$aa>b_1b_1+b_2b_2+b_3b_3+b_4b_4+b_5b_5+b_5b_5$.

If we now call $b=b_1+b_2-b_3$, then

$\left(bb+b_3b_3\right)-\left(b_1b_1+b_2b_2\right)=2\left(b_1-b_3\right)\left(b_2-b_3\right)$,

what is $\geq 0$ since $b_1\geq b_3$ and $b_2\geq b_3$. Thus, $bb+b_3b_3 \geq b_1b_1+b_2b_2$, and instead of proving

$aa>b_1b_1+b_2b_2+b_3b_3+b_4b_4+b_5b_5+b_5b_5$,

it is enough to prove

$aa>bb+b_3b_3+b_3b_3+b_4b_4+b_5b_5+b_5b_5$.

In other words, we have to prove

$aa>bb+2b_3b_3+b_4b_4+2b_5b_5$.

Note that we have $b\geq b_3$ since $b=b_1+b_2-b_3\geq b_1$ (that's because $b_2\geq b_3$) and $b_1\geq b_3$. If we now denote $b_3=c$, we thus have $b\geq c$. Also, clearly, denoting $b_4=d$ and $b_5=e$, we have $c\geq d\geq e$. Combining, we get $b\geq c\geq d\geq e$. Now, using the notations c, d, e, the inequality in question rewrites as

$aa>bb+2cc+dd+2ee$.

On the other hand, the conditions $a>b_1+b_2$ and $2a>b_1+b_2+b_3+b_4+b_5$ rewrite as $a>b+c$ and $2a>b+2c+d+e$ (in fact, since $b=b_1+b_2-b_3$, we have $b_1+b_2=b+b_3=b+c$, thus $\underbrace{b_1+b_2}_{=b+c}+\underbrace{b_3}_{=c}+\underbrace{b_4}_{=d}+\underbrace{b_5}_{=e}=b+2c+d+e$).

Now we distinguish between two cases:

[b]Case 1.[/b] We have $b\geq d+e$. Then, $b-\left(d+e\right)\geq 0$. But since $a>b+c$, we have

$aa - \left(bb+2cc+dd+2ee\right) > \left(b+c\right)^2 - \left(bb+2cc+dd+2ee\right)$
$= \left(c-d\right)\left(2b-c-d\right)+2\underbrace{\left(b-\left(d+e\right)\right)}_{\geq 0}d+2\left(d-e\right)e \geq 0$,

so that $aa>bb+2cc+dd+2ee$, as desired.

[b]Case 2.[/b] We have $b\leq d+e$. Then, $d+e-b\geq 0$. But since $2a>b+2c+d+e$, we have

$4\left(aa - \left(bb+2cc+dd+2ee\right)\right) = \left(2a\right)^2 - 4\left(bb+2cc+dd+2ee\right)$
$> \left(b+2c+d+e\right)^2 - 4\left(bb+2cc+dd+2ee\right)$
$=4\left(c-d\right)\left(b-c+e\right)+\left(3b-3d+e\right)\underbrace{\left(d+e-b\right)}_{\geq 0}+8\left(d-e\right)e \geq 0$,

so that $aa > bb+2cc+dd+2ee$ again, and the proof is complete.

  Darij

I've just taken a new look at this problem and figured I'd do what anyone would do nowadays: Rewrite it in terms of polyhedra. Let $P$ be the polyhedron in $\mathbb{R}^6$ consisting of all $6$-tuples $\left(b_1, b_2, \ldots, b_6\right)$ of nonnegative reals satisfying $b_i + b_j \leq 1$ for all $i \neq j$ and $\sum_{j \neq i} b_j \leq 2$ for each $i$. Then, the inequality in question is just saying that this polyhedron $P$ is a subset of the closed ball with center $0$ and radius $1$. So I've [url=https://sagecell.sagemath.org/?z=eJydkLFuwjAQhnc_xY-62EqLkgxsGfIGlRgRQgGc9pBzR-0UibfHiYmIGBiYTvpkfZ_vGBVWqvXSgXrrexEXQN1ZfI-DdHvipifhoBTZv5DH15uNLkAtCFWFU1bAumCRG7TiIySGb_jHas4Ksx3haQbNdhQVSfSVTMNgrJNpZs8n9wv5eoDzn-qp9InyXivHWvnsnfJYvLdIHbXf4q6_9uiF9VCKJN0pQ1ozzdKoD2dZ18tLvDEdbNDGqHp5_G_c7vxQmBvqLXmO&lang=sage&interacts=eJyLjgUAARUAuQ==]implemented this polytope in SageMath[/url]. It is a polytope with $99$ vertices and $27$ facets; each facet has $26$ vertices. This looks suspiciously like a dual of [url=https://en.wikipedia.org/wiki/2_21_polytope]the $2_{21}$-polytope[/url]. Is it isomorphic to the latter (in some reasonable way -- ideally affinely equivalent)?

This wouldn't be a complete coincidence, of course -- both the original problem and the $2_{21}$-polytope are connected to the $27$ lines on a cubic surface and thus to the Coxeter group $E_6$.


geometry
analytic geometry
area of a triangle

[hide] Notice how 15-20-25 is a multiple of the primitive Pythagorean triple 3-4-5. Hence, our triangle is right. One way to find the area is just base*height/2, which is 15*20/2=150.

Now, it makes sense that the smallest altitude is opposite the longest side, as they are inversely proportional (since their product is constant). Letting the altitude be x, we have 25x=150*2, so x=12.[/hide]

[quote="shortstack28"]The sides of a triangle have lengths of 15, 20, and 25. Find the length of the shortest altitude.[/quote]

If you want to solve problems like these, all you need to do is draw the right triangle. Let the coordinates be (0,x), (0,0), and (y,0). Then you know that two of the altitudes are 'x' and 'y', but the shortest is always the third one. You find this altitude by drawing a line from the origin (the right angle) that is perpendicular to the hypotenuse. You can say that the length of the hypotenuse is 'z'. Then you say that z = w + m. w is the distance between the point where the altitude intersects the hypotenuse and (0,x), and m is the distance between the point where the altitude intersects the hypotenuse and (y,0). Notice thta if z = w + m, then z - w = m. Also notice that if this altitude is perpendicular to the hypotenuse, then you have just broke the triangle into two smaller right triangles. Let the shortest altitude be 'a'. For one of the triangles, m is the short side, a is the long side, and y is the hypotenuse of this triangle. In the other triangle the two sides are a and w, and the hypotenuse is 'x'. You can substitute (z-w) for m, and then you just apply the pythagorean theorum to get two equations with the two smaller triangles, and then you solve for w, and then for 'a'.

I hope that helped.

[quote="shortstack28"]The sides of a triangle have lengths of 15, 20, and 25. Find the length of the shortest altitude.[/quote]

In a triangle, the shortest altitude is the altitude to the hypotenuse (if not a right triangle, it will be the altitude to the longest side)... Here, we have a right triangle where 15 and 20 are the legs and 25 is the hypotenuse...

Now we know that the area of a triangle has the formula A = bh/2...

but, there are 3 possible bases and three possible altitudes... first, we can consider 15 as the base, so the altitude will be 20...

Then A = 15(20)/2 = 150...

Another way of computing for the area is by cosidering the hypotenuse as the base... By this we will have

A = 25h/2...

so we can form the equation

25h/2 = 150
25h = 300
h = 12...


If four circles are all internal touch a fifth circle, then denote the length of the outer common tangent line of the first and second circle $l_{12}$. Similarly we define the other common tangent lines, and the first 4 circles lie clockwise. Show that if we use $l_{12}, l_{23}, l_{34}, l_{41}$ to construct a quadrilateral, then the four vertices are concyclic.

It is the consequence of Casey theorem. :)


function
logarithms
calculus
calculus computations

Hi, I am in the middle of solving this problem:
Let f(x) = 3 +x^2 + tan (πx/2), where -1 < x < 1.
a.) Find the(inverse) f^–1  (3).
b.) Find f(f^-1(5)).

I know the steps on how to find the inverse equation, but I do not know how to solve for x, in order to get it alone, because of the tan[(pi)(x/2)]. I do not remember how to get x alone when it is in sin (x), tan (x), etc. Do you multiply everything by tan x in order to cancel it out?

I'm pretty sure the second part requires a cancellation equation, but then again I'm stuck on the tan issue. Trigonemtric functions confuse me  :( .

Thank you so much for your help!!  :)

$ f(x)$ does not have an elementary inverse (I don't think it's even clear that the inverse is unique).  However, note that $ f(0) \equal{} 3 \plus{} 0 \plus{} 0 \equal{} 3$.

As for the second problem, what is the definition of an inverse function?

$ f$ is strictly increasing on its interval of definition, so it has a unique inverse. That inverse is defined on all of $ \mathbb{R}$, and does not have a simple formula.


function
LaTeX
algebra
functional equation

Λοιπόν, έπεσα πάνω σε αυτό το διαμάντι (το λέω έτσι επειδή είναι λίγο unusual στο tackling) οπότε καλή δουλειά!!! :wink:

Να βρεθούν όλες οι $ f:$ $ R\minus{}\minus{}\minus{}\minus{}>R$ που ικανοποιούν:

$ f(x\plus{}yf(x))\equal{}f(x)\plus{}xf(y)$

για κάθε $ x,y$ στο $ R$.

Πολύ ωραία άσκηση Κώστα! Δεν έχω χρόνο για να βάλω αναλυτική λύση για αυτό θα βάλω απλά ένα hint...

[hide]Αφού δείξετε ότι $ f(0) \equal{} 0$, θεωρήστε την $ g(x) \equal{} \minus{} \frac {x}{f(x)}$ (για κάθε χ διάφορο του 0) και δείξτε ότι αυτή κάνει πάντα 1! Για ευκολία θέστε όπου $ g(x)$ to $ u$, και μετά στην αρχική βάλτε όπου $ x \minus{} > u$[/hide]

[quote="NickNafplio"]Δημητρη αυτο εκανα χτες που την κοιταξα λιγο. Με την αντικατασταση που λες προκειπτει οτι υπαρχουν $ u \equal{} \frac { \minus{} x}{f(x)}$ τετοιο ωστε $ f(u) \equal{} \frac {1}{u}$ και προσπαθησα μετα κανοντας διαφορα τεχνασματα (αντικαταστασεις, επαγωγη, ορια κ.α) να δειξω οτι $ u \equal{} \minus{} 1$, αλλα δεν καταφερα να ολοκληρωσω τη λυση... βρηκα αρκετα τεχνασματα τα οποια οδηγουσαν στη λυση αλλα για ολα επρεπε να ξερουμε οτι η συναρτηση ειναι συνεχης... ενω δεν ξερουμε τιποτα τετοιο

ΥΓ: Καλη επιτυχια στον Εσωτερικο[/quote]

Εγώ δούλεψα με αντικαταστάσεις!
Προσπάθησε να βρεις κάποια σχέση για $ f(u^2)$, απέδειξε επίσης ότι είναι $ 1\minus{}1$ και μετά (τουλάχιστον εμένα έτσι μου βγήκε) δείξε ότι $ f(u^2\plus{}u)\equal{}f(u\plus{}1)$. Α επίσης είναι σημαντική η σχέση $ f(u\plus{}1)\equal{}f(u)\plus{}1$ (η οποία ισχύει και για κάθε $ x$!).

Υ.Γ. Thanks! Καλή επιτυχία σε όλους όσους δίνουν αύριο και τα λέμε παιδιά στο Χημείο!

[quote="bilot4"][quote="NickNafplio"]Δημητρη αυτο εκανα χτες που την κοιταξα λιγο. Με την αντικατασταση που λες προκειπτει οτι υπαρχουν $ u \equal{} \frac { \minus{} x}{f(x)}$ τετοιο ωστε $ f(u) \equal{} \frac {1}{u}$ και προσπαθησα μετα κανοντας διαφορα τεχνασματα (αντικαταστασεις, επαγωγη, ορια κ.α) να δειξω οτι $ u \equal{} \minus{} 1$, αλλα δεν καταφερα να ολοκληρωσω τη λυση... βρηκα αρκετα τεχνασματα τα οποια οδηγουσαν στη λυση αλλα για ολα επρεπε να ξερουμε οτι η συναρτηση ειναι συνεχης... ενω δεν ξερουμε τιποτα τετοιο

ΥΓ: Καλη επιτυχια στον Εσωτερικο[/quote]

Εγώ δούλεψα με αντικαταστάσεις!
Προσπάθησε να βρεις κάποια σχέση για $ f(u^2)$, απέδειξε επίσης ότι είναι $ 1 \minus{} 1$ και μετά (τουλάχιστον εμένα έτσι μου βγήκε) δείξε ότι $ f(u^2 \plus{} u) \equal{} f(u \plus{} 1)$. Α επίσης είναι σημαντική η σχέση $ f(u \plus{} 1) \equal{} f(u) \plus{} 1$ (η οποία ισχύει και για κάθε $ x$!).

Υ.Γ. Thanks! Καλή επιτυχία σε όλους όσους δίνουν αύριο και τα λέμε παιδιά στο Χημείο![/quote]

Ναι πραγματι ετσι λυνεται. Με καποιες αντικαταστασεις ειχα βρει το 1-1 και τη 2η απο τις 2 σχεσεις που λες και ειχα βρει και το $ f(u^2) \equal{} f^2(u)$ αλλα δεν καταφερα να τα συνδιασω

mim vale th lysh s. Allh exw yp opsin!!! Einai ontws wraia askhsh. Vevaia ayth h anitkatastash kane rule out thn profanh lysh f(x)=0 alla kala ayto einai piece of cake. Thelw na dw th lysh s.

Kalh epityxia avrio sto Xhmeio. Cu :wink:

Χρόνια πολλά σε όλους. Παιδιά συγγνώμη. Ξέχασα να αναρτήσω τη λύση που υποσχέθηκα. Τη γράφω στα αγγλικά για να μην αλλάζω language συνεχώς. Here it goes:

It is easy to check that both $ f(x)\equal{}0$ and $ f(x)\equal{}x$ are trivial solutions. Our target therefore reduces to showing that these two functions are the only solutions to our problem.

For this reason suppose that $ f$ is a solution that is not the zero function. All we have to show is that $ f(x)\equal{}x$ for all $ x$.

First, setting $ y\equal{}0$ and $ x\equal{}1$ we get $ f(0)\equal{}0$. If now $ f(x)\equal{}0$ then $ 0\equal{}xf(y)$ for all $ y$, which directly implies (as $ f$ is not the zero function) that $ x\equal{}0$. So $ f(x)\equal{}0$ if and only if $ x\equal{}0$.

Then, putting $ x\equal{}1$ we get the equation $ f(1\plus{}yf(1))\equal{}f(1)\plus{}f(y)$ for all $ y$. Let's call this useful equation (1). If $ f(1)$ is not equal to $ 1$, then putting $ y\equal{}\frac{1}{1\minus{}f(1)}$ in (1) yields $ f(y)\equal{}f(1)\plus{}f(y)$, resulting in $ f(1)\equal{}0$, which is absurd by the statement in the previous paragraph. Consequently, we must have $ f(1)\equal{}1$ and (1) becomes $ f(1\plus{}y)\equal{}f(1)\plus{}f(y)$, which we know (by a simple inductive argument) that is equivalent to $ f(n)\equal{}n$ for every integer $ n$.

For integer $ n$, real $ z$, plugging in the original functional equation $ x\equal{}n$, $ y\equal{}z\minus{}1$, we arrive at: $ f(nz)\equal{}f(n\plus{}(z\minus{}1)f(n))\equal{}n\plus{}nf(z\minus{}1)\equal{}nf(z)$

If $ a\equal{}\minus{}b$, then $ f(a)\equal{}f(\minus{}b)\equal{}\minus{}f(b)$ (we can easily see this with $ n\equal{}\minus{}1$), means that $ f(a)\plus{}f(b)\equal{}0\equal{}f(a\plus{}b)$. If $ a$ is different from $ \minus{}b$, then $ a\plus{}b$ is not $ 0$ and thus (recall our first paragraph) $ f(a\plus{}b)$ is not $ 0$. Setting $ x\equal{}\frac{a\plus{}b}{2}$, $ y\equal{}(\plus{}\minus{})\frac{a\minus{}b}{2f(\frac{a\plus{}b}{2})}$ we get respectively:
$ f(a)\equal{}f(\frac{a\plus{}b}{2}\plus{}\frac{a\minus{}b}{2f(\frac{a\plus{}b}{2})}f(\frac{a\plus{}b}{2}))\equal{}f(\frac{a\plus{}b}{2})\plus{}\frac{a\plus{}b}{2}f(\frac{a\minus{}b}{2f(\frac{a\plus{}b}{2}}))$

$ f(b)\equal{}f(\frac{a\plus{}b}{2}\plus{}\frac{b\minus{}a}{2f(\frac{a\plus{}b}{2})}f(\frac{a\plus{}b}{2}))\equal{}f(\frac{a\plus{}b}{2})\plus{}\frac{a\plus{}b}{2}f(\frac{b\minus{}a}{2f(\frac{a\plus{}b}{2}}))$

Adding these, we get $ f(a)\plus{}f(b)\equal{}2f(\frac{a\plus{}b}{2})\equal{}f(a\plus{}b)$ (by our result for the integers)

Now, applying what we just found to the original functional equation we get the useful equation (2) which is: $ f(yf(x))\equal{}xf(y)$. Setting $ y\equal{}1$ we get $ f(f(x))\equal{}x$, which makes clear the fact that our $ f$ is bijective. Setting $ z\equal{}f(x)$ in (2), we have $ f(yz)\equal{}f(y)f(z)$ for all $ y,z$

We are almost done. Put $ z\equal{}y$ in the last equation which gives $ f(y^2)\equal{}f(y)^2\geq{0}$
Put $ z\equal{}\minus{}y$ in the same equation which gives $ f(\minus{}y^2)\equal{}\minus{}f(y^2)\equal{}\minus{}f(y)^2\leq{0}$ So we conclude that $ f(a)$ is greater than $ 0$ if and only if $ a$ is greater than $ 0$.

Our last step consists of setting $ y\equal{}\minus{}1$ in the original equation. Indeed, we get $ f(x\minus{}f(x))\equal{}f(x)\minus{}x$. Since $ x\minus{}f(x)$ and $ f(x)\minus{}x$ are of opposite signs, by our last observation we must have $ x\minus{}f(x)\equal{}0$ for all $ x$, which implies $ f(x)\equal{}x$ for all $ x$.

Μετά από όλη την ώρα που γράφω Sil δε σε παίρνει πλέον να λες τίποτα για το Latex μου... :wink:

na po ego gia to latex sou..
gia na sou vgainoun oraies oi parentheseis xrisimopoiise tin entoli \displaymath{............}


function
integration
calculus
derivative
limit
algebra
functional equation

Find all continuous functions $f : \mathbb R \to \mathbb R$ such that for all $x \in \mathbb R$ and for all $n \in \mathbb N^{\ast}$ we have \[ n^2 \int_{x}^{x + \frac{1}{n}} f(t) \, dt = n f(x) + \frac12 .  \]

[i]Mihai Piticari[/i]

For every $a$  we have.
$\int_0^af(x)dx=\sum_{k=0}^{n-1} \int_{a\frac k n}^{a\frac{k+1}n}=\frac 1 n \sum_{k=0}^{n-1} f(a\frac k n) + \frac 1 n f(a)-\frac 1 n f(a)+\frac{1} {2n} =$
$\frac 1 n \sum_{k=0}^{n} f(a\frac k n)-\frac 1 n f(a)+\frac{1} {2n}$
But we know that :
$\frac{1-2f(a)}{2n}=\int_0^af(x)dx-\frac 1 n \sum_{k=0}^n f(a\frac k n)=\frac 1 n (f'(a)+f'(0))+o(\frac 1 n)$ (I'm not sure whether this is the correct form or the RHS must be divided by 2 :D )

Hence we have $1-2f(a)= f'(a)+f'(0)$ Which can be easily solved by multiplying with $e^{2a}$.

Here's the correct lemma:

[color=blue][b]Lemma 1.[/b] Let $f : \left[ a, b \right] \to \mathbb R$ be a differentiable function whose derivative is Riemann integrable. Consider the sequence $\left\{ a_{n}\right\}_{n \geq 1}$ given by:
\[a_{n}= \int_{a}^{b}f(t) \, dt-\frac{b-a}{n}\sum_{k=1}^{n}f \left( a+\left( k-\lambda \right) \cdot \frac{b-a}{n}\right) , \]
where $\lambda \in \left[ 0, 1 \right]$. Then, we have that $\lim_{n \to \infty}n a_{n}= \left( \lambda-\frac12 \right) (b-a) \left[ f(b)-f(a) \right]$.[/color]

I'm not going to write its proof... It's kind of long.

[i]Solution to the problem from RMO 2004.[/i] Fix a real number $x \in \mathbb R$ and $\frac{p}{q}\in \mathbb Q^{+}$, where $p, q \in \mathbb N^\ast$ and $\gcd (p,q) = 1$.
For all integers $t > 0$ we have that
\begin{eqnarray*}\int_{x}^{x+\frac{tp}{tq}}f(t) \, dt &=& \sum_{k=0}^{tp-1}\int_{x+\frac{k}{tq}}^{x+\frac{k+1}{tq}}f(t) \, dt \\ \ &=& \sum_{k=0}^{tp-1}\left( \frac{f \left( x+\frac{k}{tq}\right)}{tq}+\frac1{2 t^{2}q^{2}}\right) \\ \ &=& \frac{p}{2 t q^{2}}+\sum_{k=0}^{tp-1}\frac{f \left( x+\frac{k}{tq}\right)}{tq},\end{eqnarray*}
i.e. $tp \cdot \left( \int_{x}^{x+\frac{p}{q}}f(t) \, dt-\sum_{k=0}^{tp-1}\frac{f \left( x+\frac{k}{tq}\right)}{tq}\right) = \frac{p^{2}}{2 q^{2}}$.
It's easy to deduce that $f$ has a continuous derivative, so by taking $\lambda = 1$ in Lemma 1 we get that
\[\lim_{t \to \infty}\left[ tp \cdot \left( \int_{x}^{x+\frac{p}{q}}f(t) \, dt-\sum_{k=0}^{tp-1}\frac{f \left( x+\frac{k}{tq}\right)}{tq}\right) \right] = \frac{p}{q}\cdot \frac{f \left( x+\frac{p}{q}\right)-f(x)}2 . \]
Hence, $f \left( x+\frac{p}{q}\right) = f(x)+\frac{p}{q}$.
Now, if $x_{0}= 0$, then $f(x) = f(0)+x$ for all rational $x \geq 0$.
By continuity, $f(x) = f(0)+x$ for all $x \geq 0$.
For any $x < 0$ we can find $r \in \mathbb Q^{+}$ such that $x+r > 0$, so $f(x) = f(x+r)-r = f(0)+x+r-r = f(0)+x$.
Conversely, every function of the form $f(x) = x+c$ with $c \in \mathbb R$ verifies the given conditions.

An other one ..
$f(x)$ is continuous, so $n^{2}\int_{x}^{x+\frac{1}{n}}f(t)= n f(x)+\frac12$ is derivable, so $f(x)$ is dérivable, and I can write :
$n(f(x+\frac{1}{n})-f(x))= f'(x)$ so $\forall x,n$ there is $u_{n}(x)\in ]x,x+\frac{1}{n}[$ such that $f'(u_{n}(x))=f'(x)$
Let $a,b$ be such that $a < b$.  I denote $A = \{t \parallel a \leq t \leq b ,\ f'(t)=f'(a)\}$
Manifestly $A$ is non empty because $a\in A$. I denote $S_{a}=sup(A)= superior\_{b}ound\_{o}f\_{A}$.
We have $f'(S_{a})=f'(a)$ because of $f'$ is continuous.
If $S_{a}< b$ then there is a natural $n_{a}$ such that $S_{a}+\frac 1{n_{a}}< b$, so there is $u\in ]S_{a},S_{a}+\frac 1{n_{a}}[$
such that $f'(u)=f(S_{a})=f(a)$, so $S_{a}$ is not the superior bound of $A$ (contradiction !!).
So we have to have $S_{a}=b$, so $f'(b)=f'(a)$. Then we obtain that  $f'$ is a constant function $f'(x) = k\Longrightarrow f(x)=kx+c$
Finally I replace in the initial functional equation and I obtain $f(x)=x+c$
 :cool:

Another aproach is: let $g(x)=f(x)-x$, then we get \[\int_{x}^{x+\frac{1}{n}}g(t)dt=\frac{1}{n}g(x)\] From this problem ( http://www.mathlinks.ro/Forum/viewtopic.php?highlight=shortlist&t=84732 ) we know that $g(x)=k$ and $f(x)=x+k$.


[quote="Virgil Nicula"]Prove that for $ \{a,b,c\}\subset (0,\infty )\ \ ,\ \ \left\{\begin{array}{c}ax+by=(x-y)^{2}\\\ by+cz=(y-z)^{2}\\\ cz+ax=(z-x)^{2}\end{array}\right\|$ $ \implies$ $ x=y=z=0$.[/quote]
[b]Proof.[/b]
We have $ ax+by+cz=\sum x^{2}-\sum yz$
$ \Rightarrow ax+(y-z)^{2}=\sum x^{2}-\sum yz\Rightarrow ax=x^{2}-xy-xz$
Similarly, $ by=y^{2}-yz-yx$
Thus, $ ax+by=x^{2}-xy-xz+y^{2}-yz-yx$
$ \Rightarrow (x-y)^{2}=(x-y)^{2}-z(x+y)\Rightarrow z(x+y)=0$

Similarly, we also have $ x(y+z)=0$ and $ y(z+x)=0$
And now we can easily to show that $ x=y=z=0$


It takes Mary $ 30$ minutes to walk uphill $ 1$ km from her home to school, but it takes her only $ 10$ minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 3.125 \qquad
\textbf{(C)}\ 3.5 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 4.5$

it takes her 40 min to walk uphill 2 km so average speed is $ \frac{2}{\frac{2}{3}}\equal{}3\frac{km}{h}$

[quote=Zanderman]Can someone tell me why this is wrong?

By $d=rt$ on the way there, $1 = \frac{1}{2}r \rightarrow r = 2$ (km/hr).

On the way there: $1 = \frac{1}{6}r \rightarrow r = 6$ (km/hr).

So then we average: $\frac{2+6}{2} = 4$.[/quote]

Since d = rt, r = d/t
So average rate = total distance/total time

[hide=Sol]

Total distance is $2$ km.

Total time is $\frac12+\frac16=\frac23$ hours.

Average speed is $\frac2{\frac23}=3\frac{\text{km}}{\text{hour}}$

[quote]Since d = rt, r = d/t
So average rate = total distance/total time[/quote]

That doesn't really answer my question though... Why can't you figure out the rates of uphill and downhill then average them?


geometry
circumcircle
Euler
geometric transformation
reflection
homothety
perpendicular bisector

Figure, $ ABC$ is a triangle, its sides tangent the inscribed circle $ (I)$ at $ D,E,F$ . $ EF$ meet $ BC$ at $ L$, $ AI$ intersect the circumcircle of △$ ADL$ at $ J$, $ DM//AI$, $ M$ is on $ LJ$ .
$ (w_1)$ is the nine point circle of △$ DML$, similarly delimit $ (w_2),(w_3)$ . Prove that $ (w_1),(w_2),(w_3)$ are coaxial .

Moreover we can prove that the diacentral line IO is their common radical axis.

[b]Lemma:[/b] Let $ M_a,M_b,M_c$ be the midpoints of the sides $ BC,CA,AB$ of $\triangle ABC$ and $ P_a$ denotes the intersection of the tangent of its circumcircle $ (O)$ at $ A$ with the sideline $ BC.$ Points $ P_b,P_c$ are defined similarly. Then 9-points circles of $ \triangle AP_aM_a,$ $ \triangle BP_bM_b$ and $ \triangle CP_cM_c$ are coaxal having the Euler line of $ \triangle ABC$ as their common radical axis.

Let $ D$ be the foot of the A-altitude on $ BC$ and define $ M'$ and $ A_1$ as the midpoints of $ AM_a,$ i.e $ M_bM_c$ and $ AP_a.$ From the fact that the the orthocenter $ H$ is the center of the negative inversion mapping the 9-point circle $ (N)$ of $ \triangle ABC$ into its circumcircle $ (O),$ it follows that the angles formed by $ AD$ and the tangents to $ (N),(O)$ at $ D,A$ are equal and oppositely directed. Since $ \triangle AM_bM_c$ and $ \triangle DM_bM_c$ are symmetric about the midline $ M_bM_c,$ we have that the tangent to $ (N)$ at $ D$ passes through $ A_1$ $ \Longrightarrow$ $ N$ lies on 9-point circle $ (N_1)$ of $ \triangle AP_aM_a$ due to $ NM' \perp M_bM_c.$ Similarly, 9-point circles $ (N_2)$ and $ (N_3)$ of $ \triangle BP_bM_b$ and $ \triangle CP_cM_c$ pass through $ N ,$ $(\star).$

Since $ \angle NM'A_1'$ is right, we get that $ N_1$ is the midpoint of $ NA_1$ and since $ \triangle M_aM_bM_c$ and $ \triangle DM_cM_c$ are symmetric about the perpendicular bisector of $ M_bM_c,$ we have that $ A_1$ is the reflection of the intersection $ K_a$ of the tangent to $ (N)$ at $ M_a$ with $ M_bM_c$ about $ M'.$ $ K_a$ lies on the Lemoine axis of $ \triangle M_bM_cM_c$ (trilinear polar of its symmedian point K') $ \Longrightarrow$ $ A_1$ lies on the trilinear polar of the isotomic conjugate of $ K'$ WRT $ \triangle M_aM_bM_c,$ i.e. the trilinear polar $ \ell$ of its 3rd Brocard point, which is orthogonal to its Euler line. Thus, $ N_1,N_2,N_3$ lie on the line $ \ell'$ centrally similar to $ \ell$ through the homothety $(N,\frac {_1}{^2})$ and hence orthogonal to the Euler line of $ \triangle ABC.$ From this fact together with $ (\star)$ the proof  is ended.
__________________________________________

From $ DM \parallel AI$ we easily see that $ MD \perp \overline{LFE}.$ Let $ A'$ be the midpoint of $ EF$ and $ T \equiv EF \cap AD.$ Since $ AD$ is the polar of $ L$ WRT $ (I)$ $ \Longrightarrow$ $ (E,F,T,L) \equal{} \minus{} 1$ $ \Longrightarrow$ $ LF \cdot LE \equal{} LT \cdot LA' \equal{} LD^2.$ This implies that $ \angle LTD \equal{} \angle LDA',$ but from the cyclic quadrilateral $ AJDL$ we get $ \angle DAJ \equal{} \angle DLJ.$ Therefore, if $ R \equiv LJ \cap AD',$ we have $ \angle LRD \equal{} \angle LA'J \equal{} 90^{\circ}$ $ \Longrightarrow$ $ M$ is the orthocenter of $ \triangle A'LD.$ The 9-point circle $ \omega_1$  of $ \triangle MLD$ is then identical with the 9-point circle of $ \triangle A'LD$  and the same happens for the 9-points circles $ \omega_2$ and $ \omega_3.$  Using the previous lemma for $ \triangle DEF,$ we obtain that $ \omega_1 , \omega_2 ,\omega_3$ are coaxal having its Euler line $ IO$  as their common radical axis.

wow, I think [b]luis[/b] is super shrewd . Good job .
your method is similar with me, next I 'll show how I found this problem .

Very nice solution dear [b]Luis[/b]! :)
This is another proof for your lemma:
Let $ w_a, w_b, w_c$ be three 9-point circles of triangle $ AP_aM_a, BP_bM_b, CP_cM_c$, respectively. Let $ E$ be the center of $ w_a, H_a, H_b, H_c$ be the projections of $ A,B,C$ onto $ BC,CA,AB$, respectively.
We have $ \angle EM_aH_a\equal{}90^o\minus{}\angle EM_aO\equal{}90^o\minus{}|\angle M_aM_bM_c\minus{}\angle M_aM_cM_b|\equal{}90^o\minus{}|\angle B\minus{}\angle C|$.
But $ \angle AP_aM_a\equal{}\angle B\minus{}\angle P_aAB|\equal{}|\angle B\minus{}\angle C|$ so $ \angle EM_aH_a\plus{}\angle AP_aM_a\equal{}90^o$.
Therefore $ M_aE\perp AP_a$. Let $ T_a\equal{}M_aE\cap AH_a$ then $ T_a$ is the orthocenter of triangle $ AP_aM_a$. Because $ E$ is the midpoint of $ M_aT_a$ then $ E$ lies on $ w_a$. Similarly we get $ E$ is the intersection of $ w_a$ and $ w_b$. $ (1)$
On the other hand, let $ X_a$ be the intersection of $ w_a$ and $ AH_a$ then $ X_a$ is the midpoint of $ AT_a$. But $ T_a$ is the midpoint of $ AH$ thus $ HX_a.HH_a\equal{}\frac{3}{4}HA.HH_a\equal{}\frac{3}{4}HB.HH_b\equal{}HX_b.HH_b$. Therefore $ H$ lies on the radical axis of $ w_a$ and $ w_b. (2)$
From $ (1)$ and $ (2)$ we get Euler line of triangle $ ABC$ is the radical axis of $ w_a, w_b$. Similarly we are done.


geometry
geometric transformation
trigonometry
ratio
trig identities
Law of Sines
Law of Cosines

Source: A book on geometry in Swedish

In a circle, a chord is being cut in the relation 1:1:1 of an angle of a sector, which equals 45 degrees. Find the relationship between chord and diameter.

I can post a picture if my crappy translation isn't readable.

A picture would be useful, I'm not exactly sure what the problem means.

Perhaps not the best picture but...
edit: Woops, it was much crappier than I thought. I'll render a new one.

I think this is what he means

$AB:BC:CD=1:1:1$

Sorry I am really dumb but is the
[hide="answer"]
$\boxed{\frac{W}{D}=0.385}$
I used mainly $2$ equations to get the answer,
$(\frac{3k}{2})^2+h^2=r^2$
where $3k$ is the lenght of the chord,
and $h$ is the height of the perpendicular drawn from the centre to the chord,
and $r$ is the radius of the circle.
we get that $r*sin67.5=h$
solving this by a calculator :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:
we get the answer $\boxed{\frac{W}{D}=0.385}$
[/hide]

Thx. But I was looking for an non-trig solution. Mainly because the book hasn't introduced it yet.  :D

Sorry I had to delete the pic, I reached my maimum upload quota   :oops:

[quote="riddler"]anyway I used a calculator, so it is not proper.
Is my answer right??????????????????????? :?  :([/quote]
I dunno. Probably. I have no solution. (And that you're telling me now?!!);) But your reasoning seems correct.

Here, I think I hvae it.
[hide]

Make the radius $1$, and one third of the chord $x$, so we are trying to find $\frac{3x}2$.  Now call the distance from the center to an intersection on the chord $y$.  It is obvious that the triangle that is made from first drawing the picture is isosceles with sides $y$, $y$, and $x$, with vertex angle $45$, and base angles $\frac{180-45}{2}=67.5$   By law of sines, we get $\frac{y}x=\frac{\sin{67.5}}{\sin{45}}$ or $y=\frac{\sin{67.5}}{\sin{45}}x$.   Now we draw the radii to the endpoints of the chord to make triangles with sides $1$, $x$, and $y$, with an angle $112.5$ opposite the side length $1$.   Now we use law of cosines ot get $1^2=x^2+y^2-2xy\cos{112.5}$.  We subsitute for $y$, to get
$1=x^2+\left(\frac{\sin{67.5}}{\sin{45}}\right)^2x^2-x^2\frac{2\sin{67.5}\cos{112.5}}{\sin{45}}$.   TO find $\sin{67.5}$, we use the fact that it is $\sin{\frac{135}{2}}$, so $\sin{67.5}=\sqrt{\frac{1-\cos{135}}2}$   $\sin{67.5}=\sqrt{\frac{1+\frac{\sqrt2}2}2}=\frac{\sqrt{2+\sqrt2}}2$, and we know that $\sin{45}=\frac{\sqrt2}2$, so $\left(\frac{\sin{67.5}}{\sin{45}}\right)^2=\left(\frac{\frac{\sqrt{2+\sqrt2}}2}{\frac{\sqrt2}2}\right)^2=\left(\frac{\sqrt{2+\sqrt2}}{\sqrt2}\right)^2=\frac{2+\sqrt2}2=1+\frac{\sqrt2}2$
Now we have $1=x^2+x^2(1+\frac{\sqrt2}2)-x^2\frac{2\sin{67.5}\sin{112.5}}{\sin{45}}$

We know that $\sin{67.5}=\sin{180-67.5}=\sin{112.5}$, we subsitute in to get

$1=x^2+x^2(1+\frac{\sqrt2}2)-x^2\frac{2\sin{112.5}\cos{112.5}}{\sin{45}}$

We use the fact that $2\sin{\theta}\cos{\theta}=\sin{2\theta}$, and apply here to get

$1=x^2+x^2(1+\frac{\sqrt2}2)-x^2\frac{\sin{225}}{\sin{45}}$

$\sin{225}=-\sin{45}$, so

$1=x^2+x^2(1+\frac{\sqrt2}2)+x^2$

$1=x^2(3+\frac{\sqrt2}2)$

$\frac1{3+\frac{\sqrt2}2}=x^2$

$x=\frac1{\sqrt{3+\frac{\sqrt2}2}}$

$3x=\frac3{\sqrt{3+\frac{\sqrt2}2}}$

$\boxed{\frac{3x}2=\frac{3}{\sqrt{12+2\sqrt2}}\approx{0.7791}}$[/hide]
you pretty much have to use trig

it seems right.... better than mine I used a calculator.

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(If this is a load of unintelligible crap, ignore this and ask someone else.)

Umm... You mean that all lengths in the figure are the same?

no, the radii are the same.  Then from the center of the circle to the two intersection points on the chord, those are the same.  And then the chord, as given was split into a 1:1:1 ratio so each of the partial segments on the chord are the same.