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[quote="Raja Oktovin"]Let $ \mathbb{R}^ \plus{}$ be the set of all positive real numbers. Find all functions $ f: \mathbb{R}^ \plus{} \rightarrow \mathbb{R}^ \plus{}$ satisfying
\[ f(x)f(y) \equal{} f(xy) \plus{} f\left(\frac {x}{y}\right)\]
for all positive real numbers $ x$ and $ y$.[/quote]

At least we have the solutions $ f(x)\equal{}2\cosh(g(\ln(x)))$ where $ g(x)$ is any solution of the Cauchy equation $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$

With continuous solutions $ g(x)\equal{}ax$, we get $ x^a\plus{}x^{\minus{}a}$ for any real $ a$

The difficulty (for me) is to see if these are the only solutions or not ...
To be continued.

Hello Raja Oktovin,

Could you now give us your solution, please ?
I give up and it seems no one is searching again.

Thanks.

I can't do this problem so I posted this here...
Actually this topic subject is actually "help me..."
So... someone please...  :(

A reduction:

Let $ f(x) \equal{} g(\log x)$ (doable since $ f$ is defined on $ \mathbb{R}^ \plus{}$), and lettng $ a \equal{} \log x, b \equal{} \log y$,

$ g(a)g(b) \equal{} g(a \plus{} b)g(a \minus{} b).$

Substituting $ b \equal{} 0$, $ g(a)g(b) \equal{} 2g(a)$, so since $ g(a) > 0$, we know $ g(0) \equal{} 2$.

Switching the roles of $ a$ and $ b$, it is evident $ g( \minus{} a) \equal{} g(a)$.

However, here's the big issue: suppose $ \sigma : \mathbb{R} \to \mathbb{R}$ is an additive function,
i.e. one satisfying the Cauchy equation $ \sigma(x \plus{} y) \equal{} \sigma(x) \plus{} \sigma(y)$; then
$ g(\sigma a)g(\sigma b) \equal{} g(\sigma a \plus{} \sigma b) \plus{} g (\sigma a \minus{} \sigma b) \equal{} g(\sigma(a \plus{} b)) \plus{} g(\sigma(a \minus{} b))$.

In other words, if $ g$ is a solution, so is $ g \circ \sigma$; so we're already going to have that amazingly large
degree of freedom.

Not to say it's not solvable (it probably is), but there are definitely an infinitude of ugly solution functions $ g$ if
there are any nontrivial ones.

[quote="pco"][quote="Raja Oktovin"]Let $ \mathbb{R}^ \plus{}$ be the set of all positive real numbers. Find all functions $ f: \mathbb{R}^ \plus{} \rightarrow \mathbb{R}^ \plus{}$ satisfying
\[ f(x)f(y) \equal{} f(xy) \plus{} f\left(\frac {x}{y}\right)\]
for all positive real numbers $ x$ and $ y$.[/quote]

At least we have the solutions $ f(x) \equal{} 2\cosh(g(\ln(x)))$ where $ g(x)$ is any solution of the Cauchy equation $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$

With continuous solutions $ g(x) \equal{} ax$, we get $ x^a \plus{} x^{ \minus{} a}$ for any real $ a$

The difficulty (for me) is to see if these are the only solutions or not ...
To be continued.[/quote]

Actually, can you show that $ f(x) \ge 2$ all the time? And, if so, does your reduction $ f(x) \equal{} 2 \cosh(g \ln(x))$ imply $ g(x\plus{}y) \equal{} g(x) \plus{} g(y)$? If so, that's
a full solution right there.

Actually, might have something from my previous work there...

So, subbing in $ b \equal{} 1$,

$ g(b\plus{}1) \minus{} g(1)g(b) \plus{} g(b\minus{}1) \equal{} 0$,

or $ g(x) \minus{} \alpha g(x\minus{}1) \plus{} g(x\minus{}2) \equal{} 0$, where $ \alpha \equal{} g(1)$. This gives a linear recurrence
with $ g(0) \equal{} 2$ and $ g(1) \equal{} \alpha$, which has solution $ g(n) \equal{} \kappa^n \plus{} \kappa^{\minus{}n}$, where
$ \kappa \plus{} \kappa^{\minus{}1} \equal{} \alpha$. If $ \alpha < 2$, then $ \kappa$ is a unit complex number, and $ \kappa^n \plus{} \kappa^{\minus{}n}$ is eventually negative; this cannot be allowed to happen, so $ \alpha \ge 2$ always.

 Let $ \kappa$ float as a function of $ \alpha$.

 This is necessarily true for any solution $ g$.

Now let $ h(x) : \equal{} g(kx)$; this is also a solution, and so $ h(n) \equal{} \kappa^n \plus{} \kappa^{\minus{}n}$ (where
$ \kappa \plus{} \kappa^{\minus{}1} \equal{} h(1) \equal{} g(k)$) This incidentally shows that $ g(x) \ge 2$ for every $ x$.

Therefore $ g(nk) \equal{} \kappa(g(k))^n \plus{} \kappa(g(k))^{\minus{}n}$ (Here $ \kappa$ is a function -- excuse
the abuse of notation).

So, if we know $ g$ on any value $ k$, we also know it on any of its integral multiples. In fact, it is the case that $ g(k)$ is recoverable from $ g(nk)$ from the above -- the function taking $ \kappa$ to $ \kappa^n \plus{} \kappa^{\minus{}n}$ is invertible modulo the action $ \kappa \mapsto 1/\kappa$ -- from which $ g(qk)$ can be determined for all rational $ q$.

So, we're done... and $ g$ is determined from an arbitrary positive function on basis elements of the space of reals over the rationals.

(If you dig thru the above and actually write out the construction, it's probably something along the lines of what pco wrote above... but what this does do is show we've gotten all the solutions through that line of reasoning.)

@Raja Oktovin :
[quote="Raja Oktovin"]I can't do this problem so I posted this here...
[/quote]
Ok, thanks. Better to post in "Algebra unsolved Problems", next time, IMHO.

@Nukular :
[quote="Nukular"]So, we're done... and  is determined from an arbitrary positive function on basis elements of the space of reals over the rationals.
[/quote]

Thanks a lot for your kind remarks, though I'm not sure to have understood all your conclusions :

From the basis $ g(a)g(b)\equal{}g(a\plus{}b)\plus{}g(a\minus{}b)$, we get that this is the classical D'Alembert equation (without the continuity requirement) and the solutions $ 2\cos(ax)$ and $ 2\cosh(ax)$ for $ x\in\mathbb Q$ are classical results. The first can be eliminated since we are looking for positive solutions.

So $ f(x)\equal{}2\cosh(a\ln(x))$ $ \forall x\in\mathbb Q$ is a normal result deduced from D'Alembert equation.
And $ f(x)\equal{}2\cosh(\sigma(\ln(x)))$ is a family of solution, as I already stated.

But I dont see where (or if) you prove that this is is the unique family, which was my question.

Sorry if I misunderstood or did not see all your demo.

The argument has a couple of key lemmas (and, yes, is missing a last step):

Solutions for $ f(x)f(y) \equal{} f(xy) \plus{} f(x/y)$ are directly in bijective correspondence to positive functions $ g$ for which $ g(x \plus{} y) \plus{} g(x \minus{} y) \equal{} g(x)g(y)$; so I'm
going to stick with that latter equation (*), and solve there. (Eliminates the $ \ln$ term for simplicity).

Let $ \kappa : [2,\infty) \to [1,\infty)$ be defined such that $ \kappa(x) \plus{} \kappa(x)^{ \minus{} 1} \equal{} x$.

Lemma 1) For any solution $ g$, $ g(n) \equal{} \kappa(g(1))^n \plus{} \kappa(g(1))^{ \minus{} n}$ (**) for all integers $ n$. Furthermore, $ g(1) \ge 2$.

I proved this in my earlier post, by considering the recurrence $ g(n \plus{} 2) \minus{} g(1)g(n \plus{} 1) \plus{} g(n) \equal{} 0$.

Lemma 2) If $ \sigma$ is a solution to the Cauchy equation, then $ g \circ \sigma$ is a solution to (*) whenever $ g$ is a solution.

Lemma 3) If $ g$ is a solution to (*), then $ g(kn) \equal{} \kappa(g(k))^n \plus{} \kappa(g(k))^{ \minus{} n}$ (***) for all reals $ k$; furthermore, $ g(k) \ge 2$ for all real $ k$.
Proof: $ \sigma(x) \equal{} kx$ is a solution to the Cauchy equation; letting $ h \equal{} g \circ \sigma$ in equation (**) gives equation (***).

Lemma 4) $ g(p/q)$ is determined completely by $ g(1)$, for all integers $ p,q$.
Proof: We have $ \kappa(g(p/q))^q \plus{} \kappa(g(p/q))^{ \minus{} q} \equal{} g(p) \equal{} \kappa(g(1))^p \plus{} \kappa(g(1))^{ \minus{} p}$. There is only one solution for $ g(p/q)$ from this
equation.

Lemma 5) $ g(kx)$ is determined uniquely, given $ g(x)$, for any rational $ k$ and any real $ x$.
Proof: Same trick as before, apply Lemma 4 to $ g \circ (y \mapsto xy)$.

------------

Yeah, and I have to still add a lemma which gives $ g(x \plus{} y)$ from $ g(x)$ and $ g(y)$, provided $ x$ and $ y$ are independent over $ \mathbb{Q}$. Will
work on that next.

EDIT: Was not aware of this being a classical D'Alembert equation; so most of this is probably just reproducing what you already know. But, at least
it's true that we know $ g$ is at least determined by it's footprint on $ \mathbb{Q} \minus{}$projective equivalence classes in $ \mathbb{R}$.

EDIT 2: Yes, sadly, I believe you get a single additional degree of freedom when you add up two reals from differing rational lines... might be a
ton more solutions; though hopefully not :).

EDIT 3: This can probably be solved if you consider $ g$ to be a positive function from $ \mathbb{Q}^2 \to \mathbb{R}$; if you get more than two degrees of freedom of solutions, then we're going to have a ton here.

Heh: or, you could notice

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

and

$ g(a)g(b) \equal{} g(a\plus{}b)\plus{}g(a\minus{}b)$.

Which should let you get values for both $ g(a\plus{}b)$ and $ g(a\minus{}b)$. So you probably can build it up once you know $ g$ on a basis.

Back to this when I've had some sleep.

Or not:

Sketch:

Use the equations $ g(a)g(b) \equal{} g(a\plus{}b)g(a\minus{}b)$ and $ g(a\plus{}b)g(a\minus{}b) \equal{} g(2a) \plus{} g(2b)$ to lift $ g$ from disjoint
a vector subspace $ V \subset \mathbb{R}$ to $ V \oplus \mathbb{Q}b$. However, this requires a choice, since
unless one of $ a,b$ is zero, then $ g(a\plus{}b)$ and $ g(a\minus{}b)$ are different, and switching the
values of $ g(a\plus{}b)$ and $ g(a\minus{}b)$ can be accomplished by a reflection across $ b^\perp$, without interfering
with other values of $ g$. Once $ g(a\plus{}b)$ is determined, so is $ g(a \plus{} qb)$ for rational $ q$, and therefore so is $ g(ra \plus{} qb)$ for rational $ r$.

So, given the value of $ g$ on each basis element of $ \mathbb{R}$ over $ \mathbb{Q}$,
we must also choose, for each basis element (other than the first), an orientation for $ g$ to grow in --
i.e. whether $ g(a\plus{}b)$ or $ g(a\minus{}b)$ is the larger. This is a complete family of solutions for $ g$.

Yeah, lots of axiom of choice, but I think it works this time around.

[quote="Nukular"] This is a complete family of solutions for $ g$.[/quote]
Hemmm, thanks a lot ...
I'll try to read all your demo when it will be finished ...

Is your family THE complete family of solutions ?

And, if yes, is your family equal to $ \{2\cosh(\sigma(\ln(x)))\}$ where $ \sigma(x)$ is any solution of $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ ?
This was my question and I dont understand if you answered it or not.
Sorry again if I misunderstood.

What I have shown:

$ g$ is completely determined from its values on a basis for $ \mathbb{R}$ over $ \mathbb{Q}$, along with
a direction for all but one basis element; this "direction" tells us whether $ g(a \plus{} b)$ or $ g(a \minus{} b)$ is the larger.
For any values $ \ge 2$ chosen on basis elements, along with choices of direction, a unique $ g$ can be
constructed. So, this is the entire set of such functions $ g$.

However, this "choice of direction" is not substantial, since $ g$ is an even function; so it rather chooses
an automorphism of $ \mathbb{R}$.

So... yes, I your $ 2\cosh(\sigma(\ln x))$ is a complete family of functions, provided that my earlier
work computes $ 2 \cosh (\ln x)$ as the only rational solution (I get a [i] unique [/i] solution, so
if you are correct these would coincide).

Yes, assuming that nice family $ g(x) \equal{} 2\cosh (x)$ for rationals;

$ g(x\plus{}y) \plus{} g(x\minus{}y) \equal{} g(x)g(y) : \equal{} ab$
$ g(x\plus{}y)g(x\minus{}y) \equal{} g(2x)g(2y) : \equal{} a^2 \plus{} b^2 \minus{}4$,

solving gives that

$ g(x\plus{}y) \equal{} \frac{ab \pm \sqrt{(a^2\minus{}4)(b^2\minus{}4)}}{2} \equal{} 2\cosh(x)\cosh(y) \pm 2\sinh(x)\sinh(y) \equal{} 2 \cosh(x \pm y)$,

but that $ \pm$ depends on our choice of $ \sigma$, since $ \cosh$ is an even function.

So, definitively, the solution set is $ f(x) \equal{} 2\cosh(\sigma(\ln x))$.

I appreciate the constant adversarial criticism to make me provide a simpler more constructive solution  :lol:.

[quote="Allnames"]We see that $ f(x) \equal{} f(\frac {1}{x})$; $ f(1) \equal{} 1$;$ f(x^2) \equal{} f^2(x) \minus{} 2$. Thus $ f(x) > \sqrt {2 \plus{} \sqrt {2 \plus{} ...}} \equal{} 2$ when the number of root tends to the infinite . It means $ f(x) > 2$ for all positive numbers $ x$.
Notice that $ f(x^2) \plus{} f(y^2) \equal{} f(xy)f(\frac {x}{y})$. It suffices to write $ x^2 \equal{} xy\frac {x}{y}$ and $ y^2 \equal{} \frac {xy}{\frac {x}{y}}$ thus $ f^2(x) \plus{} f^2(y) \minus{} 4 \equal{} f(x^2) \plus{} f(y^2) \equal{} f(xy)(f(x)(f(y) \minus{} f(xy))$
Now it implies $ f^2(x) \plus{} f^2(y) \plus{} f^2(xy) \equal{} f(x)f(y)f(xy) \plus{} 4$
We know some techniques when proving inequality , now we deduce the existence of continuous function $ g$:$ (0;\infty )\to [1;\infty)$ such that $ f(x) \equal{} g(x) \plus{} \frac {1}{g(x)}$ (since $ f(x) > 2$ above).
The rest is easy enough; $ g(x) \equal{} x^k$ is immediately solution. Thus $ f(x) \equal{} x^k \plus{} \frac {1}{x^k}$[/quote]

Sadly, $ f$ is not required to be continuous -- see the work above for a full solution (modulo
some standard vector space knowledge of $ \mathbb{R}$ over $ \mathbb{Q}$) .

[quote="Erken"]a quick side note: with a condition of continuity the problem appeared in one of the Russian Olympiads. It was either National Olympiad or St. Petersburg olympiad.[/quote]
In my notebook; It appeared as St. Petersburg Olympiad problem. And I posted my solution for it ( without reading carefully the condition). I got a badly mistake!


prove that there is a finit number  of $ (x,y,z)$which they satisfied :
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}\frac{1}{1990}$ and  $ x$ ,$ y$ and $ z$ $ \in N$*


If x0=1 x1=0 x2=2 and x(n+2) = 2x(n) + x(n-1) for n>=1
prove that for any natural m there are two following numbers
that divides by m.

Note that, modulo $ m$, the sequence is periodical, at least from some $ n$, because there are a finite number of possible triples so that at least one will eventually reappear and each term is determined by the three preceeding ones  from the recursive relation. In another hand, the recurrence relation shows that we may define the sequence even for negative subscripts. Thus, the complete sequence (infinite in both directions) is periodical modulo $ m$. Let $ T$ be its period.

But, direct computations show that $ x_{\minus{}1} \equal{} x_{\minus{}2} \equal{} 0$. Thus $ x_{T\minus{}2} \equal{} x_{T\minus{}1} \equal{} 0 \mod [m]$, as desired.

Pierre.


Pick a number from between [b]0[/b] and 20. Whoever picks the second highest number wins! All numbers must be integers.

Just to see what happens, could people please do this and pm me a number? After I get 20 numbers, I will stop. If this has gone for 2 weeks without me getting 20 numbers, I will stop it, unless I am really close. I am going to participate, and to be fair, I have already chosen my number.

Just a note: in the original question, 0 was a legal (though illogical) number to guess.

Oh. Yeah, if you want, you can pick 0.....

I still need 13 more numbers! Please do this, it will be quick and easy for you to do, not a waste of time at all :D

[quote="E^(pi*i)=-1"]Today = Two weeks after the contest started.  What were the results?[/quote]
yeah, mustafa, were are the results? :wink:  :mad:

Well, there weren't as many participants as I would like, but here they are....

Math92:18
Drunner2007:15
Nutz4e:16
Mustafa:15
E^(pi*i):18
Joml88:18
Fireemblem13:15
Deimos:17
Ignite168:18

Congratulations Deimos, winner of.....

[hide]The pride and joy of winning :rotfl: [/hide]


With $ a,b,c > 0$ prove that
  $ \frac {a^2 \plus{} bc}{(a \plus{} b)^3} \plus{} \frac {b^2 \plus{} ca}{(b \plus{} c)^3} \plus{} \frac {c^2 \plus{} ab}{(c\plus{} a)^3} \geq \frac {9}{4(a \plus{} b \plus{} c)}$
I have 2 solutions for this ineq!

[quote="2424"][quote="arqady"][quote="2424"]With $ a,b,c > 0$ prove that
  $ \frac {a^2 \plus{} bc}{(a \plus{} b)^3} \plus{} \frac {b^2 \plus{} ca}{(b \plus{} c)^3} \plus{} \frac {c^2 \plus{} ab}{(a \plus{} b)^3} \geq \frac {9}{4(a \plus{} b \plus{} c)}$
I have 2 solutions for this ineq![/quote]
But it's wrong.   Try $ b\rightarrow \plus{} \infty.$ :wink:[/quote]
Why? :P
If no one can prove this ineq,I will write my solution![/quote]
Your solution is obviously wrong.  :P Try also $ a \equal{} c\rightarrow0^ \plus{} .$ :wink:

[quote="arqady"][/quote][quote="2424"][quote="arqady"][quote="2424"]With $ a,b,c > 0$ prove that
  $ \frac {a^2 + bc}{(a + b)^3} + \frac {b^2 + ca}{(b + c)^3} + \frac {c^2 + ab}{(a + b)^3} \geq \frac {9}{4(a + b + c)}$
I have 2 solutions for this ineq![/quote]
But it's wrong.   Try $ b\rightarrow + \infty.$ :wink:[/quote]
Why? :P
If no one can prove this ineq,I will write my solution![/quote][quote="arqady"]
Your solution is obviously wrong.  :P Try also $ a = c\rightarrow0^ + .$ :wink:[/quote]
Sorry,I edited my ineq.
you can see above! :P
and now,It's true! :lol:

[quote="2424"]With $ a,b,c > 0$ prove that
  $ \frac {a^2 \plus{} bc}{(a \plus{} b)^3} \plus{} \frac {b^2 \plus{} ca}{(b \plus{} c)^3} \plus{} \frac {c^2 \plus{} ab}{(c \plus{} a)^3} \geq \frac {9}{4(a \plus{} b \plus{} c)}$
I have 2 solutions for this ineq![/quote]
We have $ \sum_{cyc}\frac{a}{(a\plus{}b)^2} \geq \frac{9}{4(a\plus{}b\plus{}c)}$ :P

[quote="2424"]
We have $ \sum_{cyc}\frac {a}{(a \plus{} b)^2} \geq \frac {9}{4(a \plus{} b \plus{} c)}$ :P[/quote]
It's not trues, 2424. Try $ (a, b, c) \equal{} (1, 2, 0)$  :wink:

I think that one of $ a,b,c$ can't be equal to $ 0$ since $ a,b,c>0$   :)


There are $ 16$ persons in a company, each of which likes exactly $ 8$ other persons. Show that there exist two persons who like each other.

[hide]
Isn't this just simple pigeonhole?  If the people are nodes and we add a directed edge from $ A$ to $ B$ if $ A$ likes $ B$, there are $ \binom{16}{2}\equal{}120$ pairs of people and $ 16*8\equal{}128$ directed edges, so two edges must come from the same pair . . . [/hide]


linear algebra
matrix
algebra
polynomial

How can I find all possible Jordan forms for a 6x6 matrix A having as minimal polynomial the $ x^{2}(x \minus{} 1)^{2}$

Thanks anyone in advance.

There are five options:

[list]$ J\equal{}J_2(0)\oplus J_2(1) \oplus J_2(0)$[/list]

[list]$ J\equal{}J_2(0)\oplus J_2(1) \oplus J_2(1)$[/list]

[list]$ J\equal{}J_2(0)\oplus J_2(1) \oplus J_1(0) \oplus J_1(0)$[/list]

[list]$ J\equal{}J_2(0)\oplus J_2(1) \oplus J_1(1) \oplus J_1(1)$[/list]

[list]$ J\equal{}J_2(0)\oplus J_2(1) \oplus J_1(0) \oplus J_1(1)$[/list]

Could you explain me how do you come to this result?

Thanks in advance.

Since $ m_A(\lambda)\equal{}x^2(x\minus{}1)^2$, there are two eigenvalues, $ \lambda_1\equal{}0$ and $ \lambda_2\equal{}1$. Further on, because $ A^2$ is minimal, which 'destroys' all jordan blocks of $ \lambda_1$, there is at least one $ 2\times 2$ jordan block for $ \lambda_1$ and also there are no greater blocks for $ \lambda_1$ (otherwise $ A^2$ would not destroy them). The same reasoning for $ \lambda_2$. Now you just list all the possibilities (using the fact, that the sum of dimensions of all blocks is $ 6$).

[quote="tmrfea"]Since $ m_A(\lambda) \equal{} x^2(x \minus{} 1)^2$, there are two eigenvalues, $ \lambda_1 \equal{} 0$ and $ \lambda_2 \equal{} 1$. Further on, because $ A^2$ is minimal, which 'destroys' all jordan blocks of $ \lambda_1$, there is at least one $ 2\times 2$ jordan block for $ \lambda_1$ and also there are no greater blocks for $ \lambda_1$ (otherwise $ A^2$ would not destroy them). The same reasoning for $ \lambda_2$. Now you just list all the possibilities (using the fact, that the sum of dimensions of all blocks is $ 6$).[/quote]

Thank you very much.


This is not High School Basics. Mods, please move this topic.

[quote="Kibu"]Hi! I have a question (propably stupid...):

Why, when each element of group is in order 2 ($ \forall a\in G: a^2 \equal{} 1$), does exist an homomorphism (diffrent from identity)?[/quote]No, if you let $ \varphi$ be a homomorphism, $ 1\equal{}\varphi(1)\equal{}\varphi(a^2)\equal{}\varphi(a\cdot a)\equal{}\varphi(a)\varphi(a)$, which implies that $ \varphi(a)$ is its own inverse for all $ a\in G$.


I've been thinking about this problem with no luck:

Given a polynomial p(x), prove that you can also write an equivalent polynomial in terms of u where u = x + k for any k, where k is an arbitrary real constant.

This proof is trivial if you know how to use Taylor Series from calculus. Can you come up with another proof?

..Maybe I'm missing something, but wouldn't p(x) = p(u-k), and so our proof is complete?

[quote="ComplexZeta"]You should say a little bit more. Perhaps write ``Thus we can replace x by u-k in each term of p(x) and then expand by the binomial theorem to get q(u), a polynomial in u.''[/quote]

Yeah, same deal. But in any case, my point was: the problem is sorta trivial. :D

*smacks forehead*

It's always the simple ones...:(


Two pencils and one pen cost $ 80$ cents, and one pencil and
two pens cost $ \$1.15$. How many cents would three pencils
cost?

Let the cost of a pencil be $ n$, and the cost of a pen by $ m$.  Then:

\begin{align*}2x+y&=80\\x+2y&=115\end{align*}

Multiply the first equation by $ 2$, and subtract the second:

\begin{align*}4x+2y&=160\\x+2y&=115\\3x&=45\end{align*}

Thus, three pencils cost $ 45$ cents.

[quote="AIME15"]Let the cost of a pencil be $ n$, and the cost of a pen by $ m$.  Then:

\begin{align*}2x + y & = 80 \\
x + 2y & = 115\end{align*}

Multiply the first equation by $ 2$, and subtract the second:

\begin{align*}4x + 2y & = 160 \\
x + 2y & = 115 \\
3x & = 45\end{align*}

Thus, three pencils cost $ 45$ cents.[/quote]
ummm do you mean x for n and y for m?

We can also add the 2 equations to get that the cost of 3 pencils and 3 pens is 195 cents. Dividing by 3, one pencil and 1 pen cost 65 cents. Sicne 2 pencils and 1 pen (80) costs 1 pencil(15) more than 1 pencil and 1 pen(65), 1 pencil costs 15, meaning 3 cost 45.


Viet Nam MO (group B)
 let a be a real number # 0 and the sequence of real numbers (x_n) is defined by :
     x1 = 0 , x_(n+1)(x_n+a)=a+1 , n=1,2,3...
 find the x_n ?

The solution from THTT (Mathematics and Youth of Vienamese) is very
 horrible !
I solve it very easily and better than that way.
Do you have nice solution ?

At least it works by induction !

x(2*n+1) = (a+1)a y(n) / z(n)
and
x(2*n) = (a+1)/a z(n) / y(n-1)

with z(n) = (a+1)^(2n) - ((a+1)^(2n)-1) / (a+2)
and y(n) = ((a+1)^(2n)-1) / ((a+1)^2-1)

it can be still be simplified (a bit)  but it is "convenient" on this form to work by induction.

But probably it's smarter to set b=a+1,  then it is immediate to see that

x(n) = b (b^(n-2) - b^(n-3) + ... + (-1)^n) / (b^(n-1) - b^(n-2) + ... - (-1)^n) works  :)

It can again be simplified a bit but it look nice this way.

What's it your method A1lqdSchool ?


[quote="7h3.D3m0n.117"]Bughouse is i think the best strategy game (but it goes in the categories of chess)...anyone else play bughouse?[/quote]

Yeah I have. It's really fun especially when you want to waste time.(usually when you wait for your next match up post-up)

[quote="randomdragoon"]For everyone's information:

Bughouse is a varient of chess played by partnerships of 2 people on two boards. In each partnership, one partner gets white and the other partner gets black. When one partner captures a piece, he passes that piece to the other partner. If you have a piece passed to you, you may drop that piece onto the chessboard instead of making a move.

The only thing you may not do while dropping a piece is dropping a pawn on the first or last ranks.

When one person loses, his entire team loses.

Communication between partners is generally allowed, for example, when one partner tells the other partner to get a knight so that he can go for checkmate in 1 or something. In inexperienced settings, a partner can also give play advice to the other partner.

Typically bughouse games are played 5 min/side (two clocks are needed).
Also, because of the fast time controls, kings can be captured (there would be too much confusion if illegal moves had to be retracted), causing loss of game.

----

Bughouse is similar and different to regular chess. Bughouse pawns and knights are worth a bit more than they are in normal chess. Also, positional weaknesses in bughouse are often exploited much more ruthlessly, as one can just drop pieces in the holes in his opponent's positions rather than needing to maneuver his own pieces into the holes.[/quote]

that was the shortest set of bughouse rules i've ever heard  :lol:

nice job

and besides, the best opening is (assuming black plays e5)

1. e4  e5
2. Bc4 (black does something)
3. Bxf7.

Then you can drop knights, develop your queen, etc. Now white has the tempo and he needs to keep it...so he can sack sometimes rook for checks to keep the pressure on

i always thought bughouse was more elementary. Most people in my area don't play it. They grow out of it after elementary school.
Serious chess is more popular here.

[quote="now a ranger"]
Serious chess is more popular here.[/quote]

chess64 and my chess instructor never played bhouse...he said it was a bad way for rookies to develop strong chess skills (but we always play it after a couple rounds of serious chess)

haha that is true.

I quit mostly because my parents wouldn't let me have a coach because it was expensive.  :(

[quote="now a ranger"]I quit mostly because my parents wouldn't let me have a coach because it was expensive.  :([/quote]

too expensive? wow...chess and i had a science teacher in middle school that had a club that taught anyone who was interested about chess for free...

well our schools don't have clubs. And i looked , and the cheapest coaches were like bad, and they cost 50 + dollars/hour. Anyway, I quit already and at that time I had too much time and didn't do math much. this year i like math a lot, so i literally forgot about chess and haven't played at all since last year.


Just in time for the World Series...

A baseball player has been officially at bat 322 times and has a batting average of .289.  He figures that he will have 53 times at bat for the rest of the season.  (Don't count walks.)  What batting average does he have to maintain for the rest of the season in order to end with a season average of .300?

[hide]He's already batted 322*0.289 or 93 times.

He need to bat 3/10*375 or 113 times to get an average of 0.300

113-93 = 20

So the baseball player need to having a batting average of 0.377 for the rest of the season.[/hide]

Hey baseball fan, batting average is expressed in percent :P .


What scores did you guys get?

My team got 53.25 (which in my bad chapter was 1st)

My team-
Me: 1st-28/16=44
Other guy: 11th-16/14=30
Other guy: 12th-14/16=30
Other guy: 16th-15/14=29
Team:20

64 I think...

Top: 46
Me: 45 BLEGH.
Top's brother: 43
Other person : 42

I think I already posted in the other topic but,
45, 44, 43, 41. Perfect team round, 63.25.
I think the chapter is too easy to tell which team/individual is the best, so I guess we have to wait for states.

well my team got a 62.25 I think we lost to the first place team by about 0.25
someone in our team got one wrong on target :mad:

It was, but recently, MATHCOUNTS posted the tests on their site: http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47

so we can talk about it.

NEWAYS...
46-ME YAY
44-3rd
44-6th
41-10th
team-9 *DONT ASK*
u do the math 4 total *lol*

[quote="happyme"]I wasn't on the team, but I got a 44 (which was like 20th place?)

Our team was in a tie for 5th place, and we had TWO people that got 37's (HORRIBLE!), so we totally failed.

So nobody from our school is going to States.

Oy Miller4Math, are you Jonny (or Johnny), Marcus, that girl, or that tall Asian one?[/quote]

omg, a 44 won't get you into state?!?!?!?!?!?!?!, which state are you in?

Ugh, I got beat my stevenmeow (not a good sign for state!)

Me-45 (1st)
Other ppl: 45 (2nd), 41 (7th), 40 (10th)

Team 9/10=18

Team Total 60.75 ( :huh: poll needs decimals)

[quote="stevenmeow"]Its ok, i've been getting an EXTREMELY long streak of perfects *even @ practices* and im sure to crash someday and get a 42... hopefully ill get a 42 on an AIME-level state test[/quote]

crap....and i havn't made a single perfect since i don't know when....
 :(  :(

45
43
36
34

team round: 8

i think...

[quote="ra5249"]oh crap my team is going to get pwnt xD
me - 46
other ppl - 39, 38, 32[/quote]
Don't worry, my team:
Me - 44 = 3rd place
Other teammates - around 30 : didn't place.
Heh, this is off topic, but for old states, i've gotten:
42
43
45
2008: 46 jk
But I'm improving :lol:

My team was decent, but we screwed up on the team round. People were doing the problems they were weak in, and we missed 2. Careless mistakes killed us too.

Hmm, our team's scores were
46
46
45
34.
And we got 8 on the team round...
So we got 58.75
And we lost to Diamond.
But if I were on the team it would've been 61.25...
Meh, I guess you couldn't really tell, since something may have happened if I were on the team, like me getting a ridiculously low score and causing the team to get a score lower than 8 or something, but you never know...

[quote="Stargirl240"]My team was really, really, really bad!
Me-45, 1st
3rd place, 36
6th place, 30
?th place, 28
Team- 5 (obviously only one person was working *cough*)
Team score=44.75
Yup, we're really bad!  :P But somehow we still got first place.[/quote]
So are we. Lasst year we got 7th place team and this year we were way worse.

This is what we got last year:
Me: 29 (6th)
13th place: 26
VERY BAD PLACE: <20
ANOTHER BAD***: <20
HORRIBLE AT MATH: 11 (he only got 5 sprint and 3 target!)

SUCKERS!!!!! :mad:  :mad:  :mad:

AAAAAAAAAAAAAAARRRRRRRRRRRRRRRRRGGGGGGGGGGGGGGGGGGGGGGGHHHHHHHHHHHHHHHHHHHH!!!!!!!!!

hehe my team didnt do so well...but our chapter phails so we made states!  :D

Me: 43 (1st)
34 (10th)
27
22

4 on the team round... :ninja:  dont ask
39.5 total

wow,
captain:44
me:41
zolojetto:35
other member:31
6 on team...49.75...how skilled are we? :rotfl:

thats not very weak...im in pennsylvannia
44
44
42
41
40
40
38
38
38
38

[quote]I'm on the same team as math154. Our school pwned chapter but this is Missouri so the competition was very weak.

1st-45 (me)
2nd-44(another kid at our school)
3rd-42(math154)
4th-42(kid from another school)
5th-41?(kid from another school)
6th-40?(2nd place's sister)

Team Round: 8 (missed number 10 and a stupid mistake somewhere...)
We beat second place team by like 15.  :rotfl: [/quote]

wiseidiot, i think 5th was also 42. i won the tiebreaker. 6th-8th was 41 i think. i made so many careless mistakes...our team screwed up cause this guy who sucked at geometry was doing 10...and this is STL so the competition is a little harder than in the rest of the state

abacadaea, the reason the scores were so close this year was because of how easy the test was. last year, for example, the state cutoff was about ~30.

and when was the last time Missouri was even in the top 23 at nats???

[quote="math154"]
and when was the last time Missouri was even in the top 23 at nats???[/quote]

Here is some Missouri history :

2006 - 20th at Nats
2005 - 7th at Nats
2005 - 16th at Nats
2004 - 2nd at Nats & 3rd best individual

Looks pretty good to me!!

I don't see how that's bad. It's much better than my state, Wyoming!

[quote="grant.ghs"]At chapter, I got a 44, and our team had 44, 43, 40, 37, with two others not on the official team getting 44 and 40. But first place got a perfect score. You might've heard of him...name's Darryl Wu?[/quote]

lol.

at the museum, bobby and the rest of texas were chasing darryl wanting his autograph.... ra tried to sell me bobbies b4 he lost ..... lol

cd, I would guess?
But wow it seems that Bobby won written, team, AND masters... which... more than makes up for that.[/i]

It makes up for MORE than Darryl! :D

Someday if I made Nats next year I'd ask for Bobby's autograph :P

Yep me too...
Oh wait I got 5th in states in 8th grade.
I can try getting held back tho. I'll still be normally aged, lol.
and btw, reread my other post: [quote="academicsrule"]which... MORE THAN makes up[/quote]

[quote="ra5249"]wait, lost what?
and the entire texas team got darryl's autograph yehhhhh[/quote]
oh ya. that was pathetic lol... You guys were like " SCORE!!! "... -_-...
kemps was fun... ELEVEN.

[quote="dingzhou"]darn, i lost darryl's autograph, i blame it on alanchou.

This has been a public shaming of alan chou, thank you.[/quote]

kinda sad that 8th graders (with exception of bobby) are getting an autograph from a 6th grader. Kinda like the time last year, when someone on one of the ARML teams from indiana saw Kevin chen during arml and asked him for his autograph.

you're just jealous that the tx team didn't ask you for autograph.  :P

deep down, i know you, bobby, kl, ra, darryl, etc. all want my autograph.   :roll:

eh. Ding doesn't want yours. Because you're associated with me. lol

we got a 62.75 team, i think, becuz we had a sucky guy who got a 37

Oh, yeah, on another unrelated thing at nats, Texas [i]half-rhymes[/i] with breakfast


Did anyone every make a mock canadian competition? It might be fun to do something like a mock Euclid. Could someone write it? Also, there shouldn't be any sign-ups, people just PM their answers to the writer (not me). :D

EDIT: all I found in the search was a 'math battle' a while back.

[quote="rzsolt"]If someone else makes one, I'll join. If I get VERY board, I'll try and  make a contest. But I'm never board. ever. Except today.[/quote]
does that mean you're making the contest? :w00t:

[quote="Elyot"]I'll make one.  I've been digging though some old problems and I have tons of gems that I'd love to share with you guys.
Probably I would write something harder than Euclid (at least without as many easy, boring problems as Euclid).  And you know me... I'd put lots of ad hoc puzzles on there rather than hardcore geometry, number theory, or inequalities.[/quote]
Alright! :lol: What are you going to call it? You should probably start a new thread.

why don't you make a canadian open math challenge instead?


logarithms
function
algebra proposed
algebra

a) Prove that there are no one-to-one (injective) functions $f: \mathbb{N} \to \mathbb{N}\cup \{0\}$ such that
\[ f(mn) = f(m)+f(n) , \ \forall \ m,n \in \mathbb{N}.  \]

b) Prove that for all positive integers $k$ there exist one-to-one functions $f: \{1,2,\ldots,k\}\to\mathbb{N}\cup \{0\}$ such that $f(mn) = f(m)+f(n)$ for all $m,n\in \{1,2,\ldots,k\}$ with $mn\leq k$.

[i]Mihai Baluna[/i]

a) It is clear that if such a function does exist then :
1) $f(1)=0$
2) For each prime $p$ and non-negative integer $a$, we have $f(p^a) = af(p).$
3) For pairwise distinct primes $p_1, \cdots, p_k$, and non-negative integers $a_1, \cdots ,a_k$, we have $f(p_1^{a_1} \cdots p_k^{a_k} ) = a_1f(p_1) + \cdots a_kf(p_k).$

Now if $f$ is injective, $f(2)f(3) \neq 0$ (from 1) ), so that $f(2^{f(3)} ) = f(3)f(2) = f(3^{f(2)})$ (from 2) ), which is impossible since $2^{f(3)} \neq 3^{f(2)}$.

b) Let $k$ be given, and let $p$ be a prime greater than all the exponents appearing in the prime decompositions of the numbers $1,2, \cdots, k.$
Let $f(1)=0.$
Let $p_1, \cdots, p_m$ be the primes not greater than $k$. Thus each of the numbers $2,3, \cdots, k$ only decomposes with the $p_i$'s.
For each $i$, let $f(p_i) = p^i.$
Then, for all noin-primes $t \leq k$, define $f(t)$ with 3) above.
It is clear that $f(mn) = f(m)+f(n)$ for all $m,n$ such that $mn \leq k$.
Moreover, using the unicity of the decomposition in base $p$, the equality $f(t) = f(t')$ is equivalent to the fact that $t$ and $t'$ have the same prime decompositions, which is $t=t'$ and $f$ is injective.

Wow! I am now a 2000-man. :) :alien:

Pierre.


[quote="kimby_102"]Is there a limit on the number of people for WOOT?[/quote]

Not officially.  However, if it looked like it were to exceed the very high side of our expectations, we might set a maximum.


Could someone give a brief history of Russia and Europe?  I read the article at Wikipedia but it is too long and involved.  I just need a brief summary of the history.  Thank you.

The Mongols conquered the place, terrorized the people, and took everything of value. The prince of Moscow threw them off, terrorized the people, and took everything of value. A few hundred years later, the Communists threw off the tsar, terrorized the people, and took everything of value.
(Based on what my favorite high school history teacher said)

That's probably not what you really wanted, but what you really wanted probably doesn't exist. If you do want to build an outline of Russian history, start with the Mongolian invasion; the fact that the name "Russia" comes from river vikings a few centuries earlier is a lot less important.

I have a very simply question : when did the territory of what we call Russia today get a Russian - or let's make it simpler - Slavic (relative or absolute) majority? :maybe:

[quote="mathwizarddude"]Could you also summarize the history of Eastern Europe (countries like Croatia, Kosovo, Yugoslavia, etc)?  Thank you.[/quote]

wah wah reading wikipedia is just too much work!

Well, here is the world history in one sentence: [i]people were born, suffered, and died[/i]. (I leave it to you to dig up the source of this quote :P). The history of Russia was once summarized in one word by Karamsin: [i]воруют[/i].

[quote="mathwizarddude"]Could you also summarize the history of Eastern Europe (countries like Croatia, Kosovo, Yugoslavia, etc)?  Thank you.[/quote]

Oh, this could be a good one... Just replace the Mongols, Tsarist Russia and Communists in jmerry's with

-Illyrians
-Romans
-Slavs
-Hungarians
-Turks
-Austro-Hungarians
-Yugoslav royalists
-Yugoslav communists

respectively.

Honestly, what's the use of such condensed history?

Could someone give me a brief summary of calculus?  The wikipedia article was too long and involved.


function
quadratics
algebra
functional equation
algebra solved

Find all functions $f$ from positive integers to positive integers such that, for all positive integers $m$ and $n$, \[(2^m+1)\cdot f(n)\cdot f(2^m n)=2^m\cdot f^2(n)+f^2(2^m n)+(2^m-1)^2 n\]

I can't send solution but answer is
For every odd natural $n$ suppose $s$ is a divisor of $n$.Now for every nonnegative $m$:
$\displaystyle f(2^mn)= 2^ms + \frac ns$

the result of Omid Hatami is right

alternatively, the functional equation can be interpreted as a quadratic equation with variable $f(2^{m}n)$, and the first step is to apply the well-known forula for quadratic equations.

(a very bad functional equation, in my opinion.. i tried to show that $f$ was increasing during the test  ;) )


geometry
perpendicular bisector
angle bisector

In the figure to the right, $ ABC$ is an equilateral triangle. The circles with centers $ B$ and $ C$ both pass through points $ A$ and $ D$. What is the measure of angle $ BAD$?

[asy]import graph;
draw((0,0)--(30,0)--(15,26)--cycle);
draw(Circle((30,0),30));
draw(Circle((15,26),30));
dot((0,0));
dot((15,26));
dot((45,26));
dot((30,0));
label("$A$",(0,0),SW);
label("$B$",(30,0),SE);
label("$C$",(15,26),N);
label("$D$",(45,26),NE);[/asy]

Ah, yes.  Euclid's first construction. (learned that from geometry. :oops:)

[quote="dragon96"]triangle ABC is equilateral because thats how your supposed to construct an equilateral using compass + straightedge. AD is an angle biscetor, so <BAD=30 degrees.[/quote]

I don't really understand. Can you please explain that more clearly? Thanks!

It says in the question ABC is equilateral!

Oh.  :rotfl:
Then ignore the first sentence.

Wow... I didn't notice that. :blush: :blush: :blush:

[quote="ernie"]One can use the compass and straightedge (learned in geometry) to form shapes, such as an equilateral triangle and a perpendicular bisector.  If you look at segment $ CB$, you will see that a circle with radius $ C$ and a circle of radius $ B$ are shown, forming points $ A$ and $ D$.  Triangle $ ABC$ is an equilateral triangle.  If you draw a line from $ A$ to $ D$, you will form segment $ AD$, the perpendicular bisector of $ CB$, and the angle bisector of angle $ BAC$.  Find half of angle $ BAC$, which is $ 60$, and you get your final answer of $ 30$.  Feel free to ask any more questions.[/quote]
so you're bisecting it??

That is correct.  When you take geometry, you'll probably learn this stuff.


[quote="NC7"]Solve this equation
$\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=\sqrt{x+1}$[/quote]
or in other words solve this:
$10x^3+26x^2+18x-7=0$

I always want to find an sencondary-school solution


How many triangular faces does a pyramid with 10 edges have?

[hide]
It's 5. 5 edges on the bottom and 5 edges up the sides.
[/hide]

[hide]
There are 5 edges on the bottom and 5 going up to the top.
So the answer is 5.[/hide]


geometry
calculus
integration
projective geometry
perimeter
calculus computations

i know the formula, but don't know how to set up the integral:

Consider the Torus generated by revolving the circle $ (x\minus{}10)^2\plus{}y^2\equal{}1$ about the $ y$ axis. Compute the area of the surface of revolution located to the right (positive $ x$ axis) of an orthogonal plane to the $ x$ axis and which passes through $ x\equal{}10.$

I think that we can use the PAPPUS theorem.The total surface of the thorus is: 2pir*2piR=4pi^2rR

The Theorem of Pappus gets you the volume, not the surface area.

Let $ y\equal{}\sqrt{1\minus{}(x\minus{}10)^2}.$ Then

$ \frac{dy}{dx}\equal{}\frac{\minus{}(x\minus{}10)}{\sqrt{1\minus{}(x\minus{}10)^2}},$ so

$ \left(\frac{ds}{dx}\right)^2\equal{}\left(\frac{dy}{dx}\right)^2\plus{}1\equal{}\frac{1}{1\minus{}(x\minus{}10)^2}$

The surface area element is $ 2\pi x\,ds\equal{}\frac{2\pi x}{\sqrt{1\minus{}(x\minus{}10)^2}}\,dx.$

Integrate that from $ x\equal{}9$ to $ x\equal{}11.$ That gives you the upper half of the area of the torus, so double it for the whole area.
\[ A\equal{}4\pi\int_9^{11}\frac{x}{\sqrt{1\minus{}(x\minus{}10)^2}}\,dx\]

You are wrong.Pappus gives volume and surface.Volume=area *2pir where r=radius of CG of the area
Surface=perimeter of the area*2pir where r =radius of CG of the perimeter

[quote="Null"]Compute the area of the surface of revolution located to the right (positive $ x$ axis) of an orthogonal plane to the $ x$ axis and which passes through $ x \equal{} 10.$[/quote]

What is this part of the question supposed to mean?  It makes no sense.

Are you asking for the portion of the surface area of the torus for which $ x \ge 10$?  This would require integration.


symmetry
induction
geometric series
geometry proposed
geometry

Let's start with triangle $A_0,B_0,C_0$. If triangle $A_i,B_i,C_i$ is acute then triangle $A_{i+1},B_{i+1},C_{i+1}$ will be orthic triangle of triangle $A_i,B_i,C_i$. For what starting triangles $A_0,B_0,C_0$ is this progression infinite?

The progression is infinite if and only if the triangle $A_0B_0C_0$ is equilateral.

[i]Proof.[/i] We will use:

[b]Lemma 1.[/b] If A, B, C are the angles of an acute triangle ABC, then the angles of the orthic triangle of triangle ABC are 180° - 2A, 180° - 2B, 180° - 2C.

[hide="Proof of Lemma 1"]
[i]Proof of Lemma 1.[/i] The orthic triangle of triangle ABC is the triangle A'B'C', where A', B', C' are the feet of the altitudes of the triangle ABC issuing from the vertices A, B, C, respectively. Since the triangle ABC is acute, these feet A', B', C' lie inside the corresponding sides BC, CA, AB of triangle ABC. Hence, < C'A'B' = 180° - < BA'C' - < CA'B'.

But since < CC'A = 90° and < CA'A = 90°, the points C' and A' lie on the circle with diameter CA. Thus, the quadrilateral CA'C'A is cyclic, so that < C'AC = 180° - < C'A'C. But 180° - < C'A'C = < BA'C'; thus, < C'AC = < BA'C'. In other words, < BA'C' = < C'AC = A. Similarly, < CA'B' = A. Hence, < C'A'B' = 180° - < BA'C' - < CA'B' = 180° - A - A = 180° - 2A. Similarly, < A'B'C' = 180° - 2B and < B'C'A' = 180° - 2C. Hence, the angles of the orthic triangle A'B'C' of triangle ABC are 180° - 2A, 180° - 2B, 180° - 2C. Lemma 1 is proven.[/hide]

Now to solve your problem, we have to prove two assertions:

[i]Assertion 1:[/i] If the triangle $A_0B_0C_0$ is equilateral, then the progression of triangles $A_iB_iC_i$ is infinite.
[i]Assertion 2:[/i] If the progression of triangles $A_iB_iC_i$ is infinite, then the triangle $A_0B_0C_0$ is equilateral.

In fact, [i]Assertion 1[/i] is almost trivial: By symmetry, the orthic triangle of an equilateral triangle is equilateral again, and thus, since the triangle $A_0B_0C_0$ is equilateral, we get by induction that all triangles $A_iB_iC_i$ are equilateral, too, and since equilateral triangles are acute, we can continue our progression infinitely.

Now we will prove [i]Assertion 2[/i]: Assume that the progression of triangles $A_iB_iC_i$ is infinite. Thus, all triangles $A_iB_iC_i$ are acute (else, the sequence would stop at an obtuse triangle and thus be finite). Hence, we can apply Lemma 1, and, remembering that the triangle $A_{i+1}B_{i+1}C_{i+1}$ is the orthic triangle of triangle $A_iB_iC_i$, we see that $A_{i+1}=180^{\circ}-2A_i$, where $A_k$, $B_k$, $C_k$ denote the angles of triangle $A_kB_kC_k$ for any k. Now, the equation $A_{i+1}=180^{\circ}-2A_i$ implies

$60^{\circ}-A_{i+1}=60^{\circ}-\left(180^{\circ}-2A_i\right)=-120^{\circ}+2A_i=-2\left(60^{\circ}-A_i\right)$.

Now, this equation shows that the sequence of angles $60^{\circ}-A_i$ is a geometric series with factor -2; thus, if its first element $60^{\circ}-A_0$ is nonzero, this sequence will be unbounded and thus it will contain some elements which cannot be written in the form $60^{\circ}-\phi$ anymore, where $\phi$ is an angle between 0° and 180°. But all elements of our sequence $60^{\circ}-A_i$ can be written in this form, since $A_i$ is a triangle angle and thus lies between 0° and 180°. Hence, we get a contradiction, which leaves as the only possible case the case when the first element of our sequence, $60^{\circ}-A_0$, is zero. Thus, we must have $60^{\circ}-A_0=0^{\circ}$, so that $A_0=60^{\circ}$. Similarly, we can find $B_0=60^{\circ}$ and $C_0=60^{\circ}$, so that the triangle $A_0B_0C_0$ is equilateral. Hence, Assertion 2 is proven, and the problem is solved.

  Darij


geometry
analytic geometry
rotation
congruent triangles

a point p inside an equilateral triangle  of side s has distances 3,5,4 from the vertice A,B,C respectively.find s?

hello, let $ E,F,D$ be the orthopols form the point $ P$ to the sides $ \overline{AB}$, $ \overline{BC}$ and $ \overline{CA}$ of the given triangle, then we have to solve the following equationsystem
$ u^2\plus{}x^2\equal{}9$
$ (s\minus{}u)^2\plus{}x^2\equal{}25$
$ v^2\plus{}y^2\equal{}25$
$ (s\minus{}v)^2\plus{}y^2\equal{}16$
$ w^2\plus{}z^2\equal{}16$
$ z^2\plus{}(s\minus{}w)^2\equal{}9$
$ x\plus{}y\plus{}z\equal{}\frac{\sqrt{3}}{2}s$
with $ u\equal{}\overline{AE}$, $ v\equal{}\overline{BF}$, $ \equal{}\overline{CD}$ $ x\equal{}\overline{PE}$,$ y\equal{}\overline{PF}$,$ z\equal{}\overline{PD}$ and $ s\equal{}\overline{AB}\equal{}\overline{BC}\equal{}\overline{CA}$.
Sonnhard.

We use [url=http://mathworld.wolfram.com/TripolarCoordinates.html]tripolar coordinates[/url] with respect to $\triangle ABC.$

In the equilateral triangle $ \triangle ABC,$ the tripolar coordinates of $P(x,y,z)$ satisfy the relation:

$ a^2(x^2 \plus{} y^2 \plus{} z^2) \plus{} x^2y^2 \plus{} y^2z^2 \plus{} x^2z^2 \minus{} (x^4 \plus{} y^4 \plus{} z^4) \minus{} a^4 \equal{} 0$

Substituting $(x,y,z)=(3,4,5)$ yields the biquartic equation

$ a^4 \minus{} 50a^2 \plus{} 193 \equal{} 0 \Longrightarrow  \ a^2 \equal{} 25 \plus{} 12\sqrt {3} \Longrightarrow  [\triangle ABC] \equal{} \frac {_{25}}{^4}\sqrt {3} \plus{} 9.$

Let $PA=a, PB=b, PC=c$;rotate $\triangle ABP$ about $B$ with an angle $\alpha=60^\circ$ so that $A$ goes to $C$ and $P$ to a point $X$ and $\triangle BPX$ is equilateral, rotate then $\triangle BCP$ of $60^\circ$ about $C$ so that $B$ goes to $A$ and $P$ to $Y$, with $\triangle CPY$ equilateral, and $\triangle ACP$ of $60^\circ$ about $A$ so that $C$ goes to $B$ and $P$ to $Z$ and $\triangle APZ$ is equilateral.
The hexagon $AZBXCY$ has area twice the area of the $\triangle ABC$ and, joining $P$ with its vertices, we get three congruent triangles $\triangle APY=\triangle PZB=\triangle XCP$ and three equilateral triangles of sides $a, b, c$ respectively.
Consequently, if $l$ is the side of the $\triangle ABC$, and $s$ the area of $\triangle APY$ we get $\frac{l^2\sqrt{3}}{2}=3\cdot s +\frac{\sqrt{3}}{4}\cdot(a^2+b^2+c^2)$.  For the general case, the area $s$ is calculated by Heron Formula, but in our case is about a right-angled triangle, of sides $3,4,5$, hence its area is $6$.

Best regards,
sunken rock


geometry
geometric transformation
homothety
power of a point
radical axis
geometry proposed

Sorry if this was posted before, I searched but didn't find it.

Here's a nice problem.

A circle $\omega$ is tangent to two parallel lines $l_{1}$ and $l_{2}$.
A second circle $\omega_{1}$ is tangent to $l_{1}$ at $A$ and to $\omega$ externally at $C$.
A third circle $\omega_{2}$ is tangent to $l_{2}$ at $B$, to $\omega$ externally at $D$ and to $\omega_{1}$ externally at $E$.

Let $Q$ be the intersection of $AD$ and $BC$.
Prove that $QC=QD=QE$.

Let the centres of circles $\omega, \omega_{1}, \omega_{2}$ be $M, N, K$ respectively, let $\omega$ be tangent to $l_{2}$ at $S$. Let $\angle{NMK}=\alpha, \angle{MNK}=\beta, \angle{MKN}=\gamma, \angle{SMK}=\theta$
We have points $M, C, N$; $N, E, K$; $K,D,M$ collinear(because the three circles are pairwise tangent at $C, D, E$), and also because $M,N,K$ are centres of circles $\omega, \omega_{1}, \omega_{2}$ we have:
$\angle{MCD}=\angle{MDC}=90^{o}-\frac{\alpha}{2}$
$\angle{NCE}=\angle{NEC}=90^{o}-\frac{\beta}{2}$
$\angle{KDE}=\angle{KED}=90^{o}-\frac{\gamma}{2}$
because $MC=MD, CN=EN, DK=EK$
We also have because points $M, C, N$; $N, E, K$; $K,D,M$ are collinear, we get:
$\angle{DCE}=180^{o}-\angle{MCD}-\angle{ECN}= \frac{\alpha+\beta}{2}$
$\angle{CDE}=180^{o}-\angle{MDC}-\angle{EDK}= \frac{\alpha+\gamma}{2}$
$\angle{DCE}=180^{o}-\angle{DEK}-\angle{NEC}= \frac{\gamma+\beta}{2}$
If $Q'$ is the circumcentre of $DCE$, after some angle-chasing using the fact that $DQ'=CQ'=EQ'$ and knowing the angles of $DCE$, we have:
$\angle{DCQ'}=\angle{CDQ'}=\frac{\alpha}{2}$  (1)
Now let's prove that $CB\perp MN$. My part of this solution was wrong, and the only way I could find to prove it is the same as Virgil Nicula's solution. Nice Solution, Virgil Nicula!!!
$BC\perp MN$ so $\angle{MCB}=90^{o}$ so $\angle{DCB}= \angle{DCQ}=90^{o}-\angle{MCD}=\frac{\alpha}{2}$
By analogy, by switching $l_{1}$ with $l_{2}$ we can prove $AD\perp MK$ therefore $\angle{MDA}=90^{o}$ so $\angle{CDQ}=90^{o}-\angle{MDC}=\frac{\alpha}{2}$
But recall (1), therefore:
$\angle{CDQ}=\angle{CDQ'}, \angle{DCQ}=\angle{DCQ'}$ and $DC$ is common so triangles $DCQ'$ and $DCQ$ are congruent and because $Q,Q'$ are on the same side of $CD$ we have $Q=Q'$ so $Q$ is circumcentre of $CED$ so $QD=QC=QE$ QED.

[color=darkblue][b]Proof.[/b] Denote : the circles $w=C(I,r)$, $w_{1}=C(I_{1},r_{1})$, $w_{2}=C(I_{2},r_{2})$ ; the points $M\in l_{1}$, $N\in l_{2}$ so that $I\in MN$ and $MN\perp l_{1}$ ;
the middlepoints $U$, $V$ of the segments $[AM]$, $[BN]$ respectively ; the point $A_{1}\in l_{1}$ for which the line $A_{1}D$ is common tangent for the circles $w$, $w_{2}$ ;
the point $B_{1}\in l_{2}$ for which the line $B_{1}C$ is the common tangent for the circles $w$, $w_{1}$.

$0.\blacktriangleright$ $UA=UM=UC=\sqrt{rr_{1}}$ and $AM=2\sqrt{rr_{1}}$ ; $VB=VN=VD=\sqrt{rr_{2}}$ and $BN=2\sqrt{rr_{2}}$ ; $A_{1}D=A_{1}M$ ; $B_{1}C=B_{1}N\ .$

$1.\blacktriangleright$ The triangle $VIA_{1}$ is $I$- right and $ID\perp VA_{1}\Longrightarrow ID^{2}=DA_{1}\cdot DV\Longrightarrow r^{2}=DA_{1}\cdot \sqrt{rr_{2}}$ and $DA_{1}=A_{1}M$ $\Longrightarrow$ $A_{1}M=\frac{r^{2}}{\sqrt{rr_{2}}}\ \ (1)\ .$

$2.\blacktriangleright$ The triangle $UIB_{1}$ is $I$- right and $IC\perp UB_{1}$ $\Longrightarrow$ $IC^{2}=CB_{1}\cdot CU$ $\Longrightarrow$ $r^{2}=CB_{1}\cdot \sqrt{rr_{1}}$ and $CB_{1}=B_{1}N$ $\Longrightarrow$ $B_{1}N=\frac{r^{2}}{\sqrt{rr_{1}}}\ \ (2)\ .$

$3.\blacktriangleright$ Denote the point $S$ for which $SI_{1}\perp l_{1}$ and $SI_{2}\parallel l_{2}\ .$  $SI^{2}_{1}=I_{1}I^{2}_{2}-(BN-AM)^{2}$ and $MN=AI_{1}+BI_{2}+SI_{1}$ $\Longrightarrow$

$(MN-AI_{1}-BI_{2})^{2}=I_{1}I^{2}_{2}-(BN-AM)^{2}$ $\Longrightarrow$ $(2r-r_{1}-r_{2})^{2}=(r_{1}+r_{2})^{2}-\left(2\sqrt{rr_{2}}-2\sqrt{rr_{1}}\right)^{2}$ $\Longrightarrow$

$4r^{2}-4r(r_{1}+r_{2})=-4\left[r(r_{1}+r_{)}-2r\sqrt{r_{1}r_{2}}\right]$ $\Longrightarrow$ $\boxed{\ r=2\sqrt{r_{1}r_{2}}\ }\ \ (3)\ .$

$(1)\ \ \wedge\ \ (3)$ $\Longrightarrow$ $A_{1}M=\sqrt{\frac{r^{2}\cdot r}{r_{2}}}=$ $\sqrt{\frac{4r_{1}r_{2}\cdot r}{r_{2}}}=2\sqrt{rr_{1}}$  $\Longrightarrow$ $A_{1}M=AM$ $\Longrightarrow$ $A_{1}\equiv A$, i.e. the line $DA$ is radical axis for the circles $w$, $w_{2}\ .$

$(2)\ \ \wedge\ \ (3)\Longrightarrow$ $B_{1}N=\sqrt{\frac{r^{2}\cdot r}{r_{1}}}=$ $\sqrt{\frac{4r_{1}r_{2}\cdot r}{r_{1}}}=2\sqrt{rr_{2}}$ $\Longrightarrow$ $B_{1}N=BN$ $\Longrightarrow$ $B_{1}\equiv B$, i.e. the line $BC$ is the radical axis for the circles $w$, $w_{1}\ .$

For $Q\in AD\cap BC$ the line $EQ$ is the radical axis for the circles $w_{1}$, $w_{2}$. Therefore, $QC=QD=QE$ and the point $Q$ is the radical center for the given circles.[/color]

Virgil Nicula, I like your proof  :P

dear rem, there are some aspects of your proof that aren't quite clear to me. You wrote some things that are not correct (probably typo's), so it's quite hard for me to reconstruct your ideas...
Could you please clear things up a bit?

[quote="rem"]
$\angle{MCD}-\angle{MDC}=90^{o}-\frac{\alpha}{2}$
$\angle{NCE}-\angle{NEC}=90^{o}-\frac{\beta}{2}$
$\angle{KDE}-\angle{KED}=90^{o}-\frac{\gamma}{2}$
[/quote]
Do you mean $\angle{MCD}=\angle{MDC}=90^{o}-\frac{\alpha}{2}$ ?

[quote="rem"]so we have $\angle{CB'S}=180^{o}-\angle{SMC}=180^{o}-\alpha-\theta$. [/quote]
I don't see why, unless you made a typo and you actually want to define $\theta = \angle SMK$.

[quote="rem"]We also have $TB\parallel MS$ [/quote]
Why? This is equal to saying that $MCBS$ is a square, and this doesn't have to be the case.

Perhaps you mean $KB' \parallel MS$, but then you already use what you want to prove, i.e. $B'=B$.
Is my comment incorrect?

Soory, I guess the lat part of my solution was wrong, and the way to prove that $BC\perp MN$ is the same as the one in Virgil Nicula's solution. :blush:

easy problem

I changed the name of points.

notice that $F$ is homothecy center of  $(\omega_{1}\cup \ell_{1})$  and $(\omega_{2}\cup \ell_{2})$  hence $A,F,B$ are collinear  like this we have $A,G,C$ are collinear therefore $\text{Arc}\ AF=\text{arc}\ FB \Longrightarrow \angle AFG=\angle FBC$ therefore quadrilateral $BCFG$  is cyclic hence $AF.AB=AG.AC$ hence $AE$ is tangent to $\omega$ and $\omega_{2}$. like this we have $BG$ is tangent to $\omega$ and $\omega_{1}$ therfore  $O$ is radical center of $\omega\ ,\omega_{1}\ ,\omega_{2}$ hence $OF=OG=OE$


1.$ \frac {5}{x} \equal{} \frac {y}{4} \equal{} \frac {x}{y}$, find $ x^2$

2.[Removed for being inappropriate for Classroom Math]
3.[Removed for being inappropriate for Classroom Math]

[Actually, looking at first problem, It's not really classroom math either.]

YEAH! Problem smackdown!

$ \displaystyle\Omega$

I'm not sure if these problems have real solutions because I made them up... Anyways, give them a shot peoples!

Let us put in $ \frac {a}{10}$ in place of $ \frac {5}{x}$.
\[ 1 \plus{} \frac {a}{10} \equal{} \frac {y}{4}\Rightarrow4 \plus{} \frac {4a}{10} \equal{} y
\]
We plug in $ \frac {5}{x}$ instead of $ \frac {a}{10}$ Now,
\[ \dfrac{\frac {50}{x}}{4 \plus{} \frac {4a}{10}} \equal{} x\Rightarrow\frac {10}{40 \plus{} 4a} \equal{} \frac {50}{x}\Rightarrow\frac {500}{(x)(40 \plus{} 4a)} \equal{} x\Rightarrow\frac {500}{40 \plus{} 4a} \equal{} 1
\]
We find $ a \equal{} 115$. Now, since we have the equation $ \frac {5}{x} \equal{} \frac {a}{10}$, we find $ x \equal{} \frac {10}{23} \Rightarrow x^2 \equal{} \frac {100}{529}$.

If anyone thinks some steps in my solution are unnecessary, please do not post it. I'm just posting the way I did it.

The rest are too advanced for Classroom Math.

Eh... Akhil... No need for all that!

multiplying the last equations is much faster!

which gives you $ \frac{25}{x}\equal{}\frac{x}{4}$

From there you get $ x^2\equal{}100, x\equal{}10$

Proving you incorrect...

dude he said that $ \ x^2$ = 100...
not $ \ x$ =100 :maybe:

Problem 3 is worded quite...badly.

And don't say that problems are your own; I've surely seen problem 2 before.

Yeah... I kind of ripped problem 2 off AoPS volume 1...

[quote="akhil0422"]Let us put in $ \frac {a}{10}$ in place of $ \frac {5}{x}$.[/quote]
Why? How would the new user to classroom math know to do this?

exactly my point levans. Why? Plus Akhil arrived at the wrong answer... :ninja:

OK, since a solution still eludes us...

[hide]Let's "halfway" cross-multiply the first two equations to solve for $ y$

We get $ \frac {20}{x} \equal{} y$

Now, plug this value in for $ y$

We get $ \frac {x^2}{20}$

Now we set this equal to the first equation:

$ \frac {5}{x} \equal{} \frac {x^2}{20}$
$ 100 \equal{} x^3$
$ log 100 \equal{} log x^3$
$ log 100 \equal{} 3 log x$
$ 2 \equal{} 3 log x$
$ \frac {2}{3} \equal{} log x$

Putting this in exponential form, we get:
$ 10^\frac {2}{3} \equal{} x$
$ 10^\frac {4}{3} \equal{} x^2$[/hide]

Giving us a solution...but probably not the right one.
Do you think that anyone can find a mistake in my process?

EDIT: Notice though that this answer is the same as the one batteredbutnotdefeated got.

batteredbutnotdefeated: please explain how cross-multiplying $ \frac {y}{4}$ and $ \frac {x}{y}$ gets you $ x^2 \equal{} 100$. Actually, it gets you $ y^2 \equal{} 4x$.  :wink:

Now I realize the fault in my previous solution, so I shall post another one. If multiply $ \frac{y}{4}$ and $ \frac{x}{y}$, we get $ \frac{x}{4}\equal{}\left(1.\frac{5}{x}\right)^2$. Now, \[ frac{x}{4}\equal{}1\plus{}\frac{1}{x}\plus{}\frac{1}{4x^2}\] Next, after simplification, we get the equation $ \minus{}x^3\plus{}4x^2\plus{}4x\plus{}1\equal{}0$. We get non-integer, but real solutions...

actually it gets you $ x^3\equal{}100$ :wink:

So the problem should have said prove this system can not exist (assuming everyone used the correct steps). Be careful when you are devising your problems! (and Akhil, what he meant was multiplying the last two parts of the equation)

It's not???? I thought the first number was 1 decimal $ \frac{5}{x}$. Then, my answer is what batteredbutnotdefeated and ytrewq got!!

ok just a wild shot in the dark:
if $ \dfrac{5}{x}=\dfrac{y}{4}=\dfrac{x}{y}$, then cross multiply first 2 fractions and then you get $ \dfrac{20}{x}=y$

substitute:
$ \dfrac{y}{4}=\dfrac{x}{y}$ becomes
$ \dfrac{\dfrac{20}{x}}{4}=\dfrac{x}{\dfrac{20}{x}}$
cross multiply:
$ 4x=\dfrac{400}{x^2}$
$ 4x^3=400$
$ x^3=100$
$ x=100^\dfrac{1}{3}$
$ (x)^2=(100^\dfrac{1}{3})^2$
mwahahahahaha :wow: success is mine! lol
so $ x^2=100^\dfrac{2}{3}$
so yeah i agree with ytrewq.

What happened to the 2nd and 3rd probs??

They were "removed for being inappropriate for classroom math".


Here's a problem I saw on Mathlinks:

x_1>1 and x_(n+1)=x_n+(x_n-1)^1/2. What's the limit of x_n/2^n as n->infinity? Does ne1 have any idea? I thought of trigonometric functions at the beginning, but if we put 1/x_n=a_n, getting a_n<1 for any n, and a_n=sinx we get, after the calculations, a_(n+1)=tan(x/2), so we don't get sth with sin, but with tan. Any other ideas?

Are you sure of the terms of the problem  :?
Either I'm completly wrong (becoming an habit  :( ) either it's not too difficult.

If x_(n+1)=x_(n) + (x_(n) - 1)^1/2

Then x(n) is strictly increasing, for n >= n0, x(n) > 4, so sqrt(x(n)-1) <= x(n)/2 and x(n+1) <= 3/2 x(n)

Now this proves x(n)/2^n -> 0 when n ->+oo.

I even think that x(n)/a^n -> 0 when n ->+oo whenever a > 1.

Isn't this the correct version :

a1 > 1, an+1 = an + (an2 - 1)1/2 ?

I saw the following similar problem somewhere but it's much easier :

a1 = 1, an+1 = an + (an2 + 1)1/2.
Compute limn->:inf:an2-n.

In the latter case trig substitution works just fine.
But this is not the case for the problem Grobber posted.

Arne is right, I forgot to write "^2". Indeed Arne, that's the problem I was refering to.


find all functions $ f(x)$ such that it is continuos on $ [0;1]$,differentiable on the open interval $ (0;1)$ and the following two conditions are satisfied:
a)$ 20f(x)\plus{}11f(x)\plus{}2009 \leq 0$ with all $ x \in (0;1)$
b)$ f(0)\equal{}f(1)\equal{}\frac{\minus{}2009}{11}$

I think you have to write the correct statement ... :cool:

[quote="lhplhp"]find all functions $ f(x)$ such that it is continuos on $ [0;1]$,differentiable on the open interval $ (0;1)$ and the following two conditions are satisfied:
a)$ 20f(x) \plus{} 11f(x) \plus{} 2009 \leq 0$ with all $ x \in (0;1)$
b)$ f(0) \equal{} f(1) \equal{} \frac { \minus{} 2009}{11}$[/quote]
The a) condition is equivalent to  $ f(x) \leq \minus{}\frac{2009}{31}$.
What's the interest ? Too many functions saisfay the conditions  :!:
 :cool:

No.No Here is a problem in MYM in my country(T 12/379) so [b]Dont help him [/b]
See here ,please http://www.mathlinks.ro/viewtopic.php?t=168836&start=20


calculus
algebra
polynomial
derivative
function
calculus computations

Calculus problems go in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url] section of the Forum.

The answer to your question can be found in:
1. [url=http://en.wikipedia.org/wiki/Differentiation_rules#The_polynomial_or_elementary_power_rule]Wikipedia: "Differentiation Rules"[/url]
2. [url=http://en.wikipedia.org/wiki/Table_of_derivatives#Derivatives_of_simple_functions]Any table of derivatives[/url]
3. [url=http://en.wikipedia.org/wiki/Calculus_with_polynomials#The_power_rule]Wikipedia: "Calculus with polynomials"[/url]
4. Your calculus book
5. A billion other places


Someday.... somewhere.... the qualifier list is, but here in the real world, everyone is sad; the qualifier list isnae cognizable yet...... (bad adaption of poetry.)

Billy

See they made the final decision at latest at 2.30p.m.(Nebraska zone)
How late or How long will the AMC directors work today? Till the Nebraska time zone 4p.m.? 5p.m.? or 6p.m.? What's the probability they can work out the list today (to corellate the CEEB)?


Let $ ABCD$ be an inscribable and circumscribable quadrilateral. Through the center $ I$ of the incircle of the quadrilateral $ ABCD$ there is constructsd a line parallel to the line $ BC$, which intersects the sides $ AB$ and $ CD$ at the points $ K$ and $ L$, respectively.
Prove that $ KL \equal{} \frac {AB \plus{} BC \plus{} CD \plus{} DA}{4}$.

That sounds interesting!
But, what's so specific about line BC?
[i]As the KL-expression ist symmetric w. r. to all four side-lenghths, we have indeed an even more striking result for the four parallels to the sides going through I.[/i]    :maybe:

$ P$ is tangency point of $ (I)$ with $ BC.$ $ IP$ cuts $ (I)$ again at $ Q.$ Tangent to $ (I)$ at $ Q$ cuts $ AB, CD$ at $ M, N$ and $ MN \parallel BC.$ $ KL$ is midline of the tangential trapezoid $ BCNM,$

$ KL \equal{} \frac {MN \plus{} BC}{2} \equal{} \frac {MN \plus{} BC \plus{} \overline{BM} \plus{} \overline{CN}}{4}.$

$ AB, CD$ intersect at $ X.$ $ ABCD$ is cyclic, $ DA$ is antiparallel to $ BC$ WRT the angle $ \angle BXC.$ Therefore, the $ \triangle AXD \sim \triangle NXM$ are oppositely similar. Since they have common X-excircle $ (I),$ they are oppositely congruent, $ DA \equal{} MN$ and $ \overline{MA} \equal{} \minus{} \overline{ND},$ $ \overline{MA} \plus{} \overline{ND} \equal{} 0.$ Substituting this to the above expression for $ KL,$

$ KL \equal{} \frac {MN \plus{} BC \plus{} \overline{BM} \plus{} \overline{MA} \plus{} \overline{CN} \plus{} \overline{ND}}{4} \equal{} \frac {DA \plus{} BC \plus{} BA \plus{} CD}{4}$


geometry
incenter
analytic geometry
trigonometry
geometry proposed

[color=darkblue]Let $ABC$ be a triangle for which the projection $D$ of the vertex $A$ to the its opposite side-line $BC$ belongs to the line $IG$,
where $I$ is the  incenter and $G$ is the centroid for the given triangle. Ascertain the maximum value of the angle $\widehat{BAC}$.[/color][hide="Here is a direction."]$D\in IG\Longleftrightarrow b+c=p\sqrt 2$, where $2p=a+b+c\ .$[/hide]

Let $A'$ be the orthogonal proyection of $A$ in $\overline{BC}$, so the barycentric coordinates of $A'$ wrt $\triangle ABC$ are $([A'CB]: [A'CA]: [A'AB])$ that is $(0: h_{a}\frac{b.cosC}{2}: h_{a}\frac{c.cosB}{2})=(0: senBcosC: senCcosB)$. As for the condition of $A'\in \overrightarrow{IG}$ then $0=\begin{displaymath}\left| \begin{array}{ccc}0&senBcosC&senCcosB \\ a & b & c \\ 1 & 1 & 1 \end{array}\right|=\left| \begin{array}{ccc}0&senBcosC&senCcosB \\ 0 & b-a & c-a \\ 1 & 1 & 1 \end{array}\right|=\left| \begin{array}{ccc}senBcosC&SenCcosB\\ b-a & c-a \end{array}\right|=(senBcosC)(c-a)-(senCcosB)(b-a)$. And because of $cosB=\frac{a^{2}+c^{2}-b^{2}}{2ac}$ and $cosC=\frac{a^{2}+b^{2}-c^{2}}{2ab}$. That is $\frac{b(c-a)(a^{2}+b^{2}-c^{2})}{2ab}=\frac{c(b-a)(a^{2}+c^{2}-b^{2})}{2ac}$ so $a^{2}c+b^{2}c-c^{3}-ab^{2}+ac^{2}=a^{2}b+bc^{2}-ac^{2}-b^{3}+ab^{2}$ and then $bc(b-c)+(b^{3}-c^{3})=2a(b^{2}-c^{2})+a^{2}(b-c)$ so, a solution is $b=c$. If $b \neq c$ it becomes $b^{2}+c^{2}+2bc=2a(b+c)+a^{2}$ an then $2b^{2}+2c^{2}+4bc=2a(b+c)+a^{2}+b^{2}+c^{2}+2bc$ and then $2(b+c)^{2}=(a+b+c)^{2}=4p^{2}$ and then $(b+c)^{2}=2p^{2}$ thus $b+c=\sqrt[]{2}p$.

It means that $(2-\sqrt[]{2})(b+c)=\sqrt[]{2}a$, that is $(2-\sqrt[]{2})(senB+senC)=\sqrt[]{2}senA$ thus $(2-\sqrt[]{2})(2sen\frac{B+C}{2}cos\frac{B-C}{2})=\sqrt[]{2}senA$ so, making the sustitution $B+C=\pi-A$ it becomes $(2-\sqrt[]{2})sen(B+\frac{A}{2})=\sqrt[]{2}sen\frac{A}{2}$ that is $sen\frac{A}{2}\leq \frac{2-\sqrt[]{2}}{\sqrt[]{2}}$ thus $\frac{A}{2}\leq arcsen\frac{2-\sqrt[]{2}}{\sqrt[]{2}}$ so $A \leq 2arcsen\frac{2-\sqrt[]{2}}{\sqrt[]{2}}$. $\blacksquare$

[quote="Virgil Nicula"][color=darkblue]Let $ABC$ be a triangle for which the projection $D$ of the vertex $A$ in the its opposite side-line $BC$ belongs to the line $IG$,
where $I$ is the  incenter and $G$ is the centroid for the given triangle. Ascertain the maximum value of the angle $\widehat{BAC}$.[/color][/quote]
[color=darkred][b]Huchs[/b], you demonstrated that you know very well the barycentrical coordinates.
In my opinion, is better to work with the barycentrical coordinates than with the triangle complex coordinates (a monstrous tautology !).[/color]

[color=darkblue][b]The my proof.[/b] Suppose $b\ne c$ and denote the projections $T$, $S$ of the points $I$, $G$ respectively in the side $[BC]$.
Prove easily that $GS=\frac{1}{3}\cdot h_{a}$, $\boxed{\ DT=\frac{p-a}{a}\cdot |b-c|\ }$, $MD=\frac{b+c}{2a}\cdot |b-c|$ and $DS=\frac{2}{3}\cdot DM\Longrightarrow \boxed{\ DS=\frac{b+c}{3a}\cdot |b-c|\ }\ .$

$1.\blacktriangleright$ Therefore, $D\in IG$ $\Longleftrightarrow$ $\frac{DT}{DS}=\frac{IT}{GS}$ $\Longleftrightarrow$ $h_{a}(p-a)=r(b+c)$ $\Longleftrightarrow$ $2p(p-a)=a(b+c)$ $\Longleftrightarrow$ $(b+c)^{2}-a^{2}=2a(b+c)$ $\Longleftrightarrow$ $a^{2}+2(b+c)a-(b+c)^{2}=0$ $\Longleftrightarrow$ $a=(\sqrt 2-1)(b+c)$ $\Longleftrightarrow$ $\boxed{\ b+c=p\sqrt 2\ }\ .$

$2.\blacktriangleright$ Denote $\frac{b}{c}=t\ .$ Thus, $a=(\sqrt 2-1)(b+c)$ $\Longleftrightarrow$ $a^{2}=(3-2\sqrt 2)(b+c)^{2}$ $\Longleftrightarrow$ $b^{2}+c^{2}-2bc\cos A=(3-2\sqrt 2)(b^{2}+c^{2}+2bc)$ $\Longleftrightarrow$ $t^{2}-2t\cos A+1=(3-2\sqrt 2)\left(t^{2}+2t+1\right)$ $\Longleftrightarrow$ $(\sqrt 2-1)t^{2}-(\cos A+3-2\sqrt 2)t+(\sqrt 2-1)=0\ \ (1)\ .$  But $b^{2}+c^{2}-a^{2}=$ $b^{2}+c^{2}-(3-2\sqrt 2)(b^{2}+c^{2}+2bc)=$ $2(\sqrt 2-1)\left[b^{2}-(\sqrt 2-1)bc+c^{2}\right]>0$ $\Longleftrightarrow$ $\cos A>0$ $\Longleftrightarrow$ $A<90^{\circ}\ .$ Thus, for the equation $(1)$, $\Delta =(\cos A+3-2\sqrt 2)^{2}-4(\sqrt 2-1)^{2}\ge 0$ $\Longleftrightarrow$ $\cos A+3-2\sqrt 2\ge 2(\sqrt 2-1)$ $\Longleftrightarrow$ $\cos A>4\sqrt 2-5$ $\Longleftrightarrow$ $\boxed{\ A\le \arccos (4\sqrt 2-5)\ }\ .$

[b]Remark.[/b] The Hucht's answer is $A\le 2\arcsin (\sqrt 2-1)$ what is identically with the my answer $A\le\arccos (4\sqrt 2-5)$ because $1-2(\sqrt 2-1)^{2}=4\sqrt 2-5\ .$[/color]

[quote="Virgil Nicula"]
[color=darkred][b]Huchs[/b], you demonstrated that you know very well the barycentrical coordinates.
In my opinion, is better to work with the barycentrical coordinates than with the triangle complex coordinates (a monstrous tautology !).[/color]
[/quote]

LOL thanks!, but I as looking for the use of triangle complex coordinates because I'd like to prove some theorems (like Lester theorem), because when I looked at the Lester's page I got surprised. That's why I'd like to leran it!

Capitan Mandarina!

PS.
[quote="Virgil Nicula"]
[color=darkred][b]Huchs[/b], ...[/color]
[/quote]
It's huch[b]t[/b], with [b]t[/b].


4.
$cosx+\sqrt{3}sinx+1=0$
$cosx+\sqrt{3}sinx=-1$

$(cosx+\sqrt{3}sinx)^2=(-1)^2$
$cos^2x+2\sqrt{3}sinxcosx+3sin^2x=1$
we know that
$sin^2x+cos^2x=1$
$.^..$
$cos^2x+2\sqrt{3}sinxcosx+3sin^2x=sin^2x+cos^2x$
$2\sqrt{3}sinxcosx=-2sin^2x$
$tanx=-\sqrt{3}$

$x=\dfrac{2\pi}{3}+k\pi$$k=1$

[quote="frt"]Your answer for #5 is correct except that $k=0,-1$, not $k=0,1$.
(Watch out for the domain :roll: )

[b]EDIT[/b]- Also, it should be $\dfrac{\pi}{8}+\dfrac{k\pi}{2}$ ($k=0,-1$)

For #4, I think you're on the right track, but your answer is not correct.

I don't get this part:
[quote="felipesa"]
$cos^2x+2\sqrt{3}sinxcosx+3sin^2x=sin^2x+cos^2x$
$2\sqrt{3}sinxcosx=-2sin^2x$
$tanx=-\sqrt{3}$[/quote]
How did you get from the 2nd line to the 3rd line?
Did you divide both sides by $\cos^2 x$? :?[/quote]

NO
I did
$2\sqrt{3}sinxcosx=-2sin^2x$
$\sqrt{3}=\dfrac{-2sin^2x}{2sinxcosx}$
$\sqrt{3}=-\dfrac{sinx}{cosx}$
$tanx=-\sqrt{3}$

wow nobody solved 1. yet, but i ll do it:

First, one should notice that
$4\cos^3 x -3\cos x = \cos 3x$, and
$2\cos^2 x -1 = \cos 2x$

therefore
$\cos 3x+\cos 2x =0\implies \cos 3x = \cos(2x+\pi)$
so actually $3x=2x+\pi +2\pi k ;\ k\in\mathbb Z$ or $2\pi-3x=2x+\pi +2\pi k ;\ k\in\mathbb Z$
but since $x\in[0;\pi)$ we only have
$3x=2x+\pi \implies x=\pi$ (not really a solution because of $x\in[0;\pi)$
$2\pi-3x=2x+\pi \implies x=\dfrac{\pi}5$, or
$2\pi - 3x = 2x-\pi \implies x= \dfrac{3\pi}{5}$

2.

$\dfrac{1}{sinx}+\dfrac{1}{cosx}=2\sqrt2$

$\dfrac{sinx+cosx}{sinxcosx}=2\sqrt2$

$(\dfrac{sinx+cosx}{sinxcosx})^2=(2\sqrt2)^2$

$\dfrac{sin^2x+cos^2x+2sinxcosx}{sin^2xcos^2x}=8$

$\dfrac{1+sin2x}{sen^2xcos^2x}=8$

$1+sin2x=2sinxcosx2sinxcos2$

$1+sin2x=2sin^22x$

$2sin^22x-sin2x-1=0$

$sin2x=1$ or $sin2x=-1/2$

if  $x\in[0,\pi)$

so
$x=\dfrac{\pi}{4}$and$x=\dfrac{11 \pi}{12}$ :)

[quote="riddler"]Can anyone tell me what is the relevance of $\pi$? in angles, what does it stand for?[/quote]

used to measure angles in radians.

Here is a formula $\dfrac{M}{180}=\dfrac{a}{\pi}$

where $M$ is the measure of the angle in degrees and $a$ its measure in radians!
ok? :)

well angles can be expressed in terms of the so called arc length.

the arc length is defined as the length of the arc over a circle  with the radius 1 where the angle goes from the center.

so $2\pi = 360^{\circ}$ because that would be a full circle and the circumfence $c=2\pi r$ and $r=1$. then the arc length is propotional to the angle, hence $\dfrac{\alpha}{360^{\circ}} = \dfrac{x}{2\pi}$ where $\alpha$ is in degree and $x$ in radiants ( a diffrent name for arc length, the more common one actually)

Oh yeah I think we learnt that in physics (angular motion) just forgot

peeta- Yes, that's correct :clap2: There is also a way to solve it without knowing the formula for triple angle.

felipesa- Your answer for #2 is now correct :D

Also, for #4, your work looks fine to me, but somehow it's not quite correct. $\dfrac{2\pi}{3}$ doesn't work, but $\dfrac{5\pi}{3}$ does work. (Does anyone know why?) In addition, there's one more solution...

#3

$sin3x(sin3x-cosx)=sinx(sinx-cos3x)$

$\dfrac{sin3x}{sinx}=\dfrac{sinx-cos3x}{sin3x-cosx}$

$\dfrac{sinx(-1+4cos^2x)}{sinx}=\dfrac{sinx-cosx(1-4sen^2x}{sinx(-1+4cos^2x)-cosx}$

$sin^2x(-1+4cos^2x)^2-sinxcosx(-1+cos^2x)=sin^2x-sinxcosx(1-4sin^2x)$

$sin^2x(-1+4cos^2x)^2-sin^2x=senxcosx(-1+4cos^2x-(1-4sin^2x))$

$-8sin^2xcos^2x+16sin^2xcos^4x=2sinxcosx$

$8sin^2xcos^2x(2cos^2x-1)=2sinxcosx$

$4sen2xcos2x=2$

$8sinxcosx(1-2cos^2x)=2$

$-4sinxcosx+8sinxcos^3x=1$

$sinx(-4cosx+8cos^3x)=1$

$sin4x=1$

$x \in [0,\pi)$

$x=\dfrac{\pi}{8}$
$x=\dfrac{5\pi}{8}$

we can have more two but  in this domain $x \in [0,2\pi)$
$x=\dfrac{9\pi}{8}$
$x=\dfrac{13\pi}{8}$

For this domain $x \in [0,\pi)$  only two solutions are possible.
$x=\dfrac{\pi}{8}$
$x=\dfrac{5\pi}{8}$

Well... you are missing 3 more solutions :(
$\sin 4x=1$ even gives you one of them (because $0\le x<\pi\Longleftrightarrow 0\le 4x<4\pi$)

EDIT (since felipesa edited the answer a bit)-
Close, but it's $x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}$, where $k=0,1$ ;)

Find two more!

EDIT 2- Think about the domain one more time!

Ok, since the posts are getting complicated, I'll clarify which ones we got so far.

peeta solved #1.
felipesa solved #2 and #5.

felipsa tried #3 and #4 as well, but he's still missing some solutions.

[hide="Tips for 3"]
felipesa got $\dfrac{\pi}{8}$ and $\dfrac{5\pi}{8}$, and these are correct. However, there are two more solutions that are NOT in the form of $\dfrac{\pi}{8}+\dfrac{k\pi}{2}$, where$k$ is an integer. Try something different. For example, what if you expand the original equation and simplify it? :) [/hide]

[hide="Tips for 4"]
felipesa gave $\dfrac{2\pi}{3}$ and $\dfrac{5\pi}{3}$ as solutions, but only $\dfrac{5\pi}{3}$ works. In addition, there's one more solution. Think about the following: $\cos x+\sqrt{3}\sin x=a\sin (x-b)$ ;) [/hide]

Anyone else got #3 or #4?

here is 4:

$\cos x + \sqrt 3 \sin x +1 =0 \implies \cos x=- ( \sqrt 3 \sin x +1)$

$\implies \cos ^2 x = 1- \sin ^2x = 3\sin ^2 x + 2\sqrt 3 \sin x +1$
$\implies 4\sin x ( \sin x + \dfrac{\sqrt 3}{2}) = 0$
$\implies \sin x= 0$ or $\sin x = -\dfrac{\sqrt 3}{2}$

because $x\in[0;2\pi)$:
$x_1 = 0$
$x_2 = \pi$
$x_3 = \dfrac{4\pi}{3}$
$x_4 = \dfrac{ 5\pi}{3}$

because of the squaring we have to select the correct solutions, which are after plugging in $x_2$ and $x_4$

#3  :)

$\sin^23x-\sin 3x\cos x=\sin^2x-\sin x\cos 3x$

$\implies (1-\cos 6x)-(\sin 4x+\sin 2x)=(1-\cos 2x)-(\sin 4x-\sin 2x)$

$\implies 2\sin 2x=\cos 2x-\cos 6x$

$\implies 2\sin 2x=2\sin 4x\sin2x$

$\implies \sin 2x(\sin 4x -1)=0$

For $\sin 2x=0$  , $2x=n\pi +(-1)^n\times 0=n\pi \implies x=\dfrac{n\pi}{2}$

So we obtain $\boxed{x=0}$ , $\boxed{\dfrac{\pi}{2}}$

For $\sin 4x=1$ , $4x=n\pi +(-1)^n\times \dfrac{\pi}{2}$

So $\boxed{x=\dfrac{\pi}{8}}$ , $\boxed{\dfrac{5\pi}{8}}$

#4  :)

$\cos x+\sqrt 3\sin x+1=0$

$\implies 2\sin(x+\dfrac{\pi}{6})=-1$

$\implies \sin(x+\dfrac{\pi}{6})=\dfrac{-1}{2}$

$\implies x+\dfrac{\pi}{6}=n\pi+(-1)^n\times \dfrac{-\pi}{6}$

Hence $x=n\pi+(-1)^n\times \dfrac{-\pi}{6}-\dfrac{\pi}{6}$

putting $n=1,2$ and obtain $\boxed{x=\pi}$ , $\boxed{\dfrac{5\pi}{3}}$

[quote="shyong"]
For $\sin 2x=0$  , $2x=n\pi +(-1)^n\times 0=n\pi \implies x=\dfrac{n\pi}{2}$

So we obtain $\boxed{x=0}$ , $\boxed{\dfrac{\pi}{2}}$

For $\sin 4x=1$ , $4x=n\pi +(-1)^n\times \dfrac{\pi}{2}$

So $\boxed{x=\dfrac{\pi}{8}}$ , $\boxed{\dfrac{5\pi}{8}}$[/quote]

Do you know why I dont find $sin2x=0$ in my solution?

thanks

It's because you divided both sides by $\sin{x}\cos{x}(=\dfrac{1}{2}\sin{2x})$ in your solution. To avoid it, you have to factor it out ;)

Anyway, the answers peeta and shyong got were correct :)


[quote="Cezar Lupu"]Let $a, b, c$ three non-negative real numbers. Prove the following inequality

                         \[ \sum_{cyc}\frac{a^{2}(b+c)}{b^{2}+bc+c^{2}}\geq\frac{2}{3}(a+b+c).  \][/quote]

i think that this was posted by darij some time ago and harazi gave a very nice proof!!
 :)

Thank you for reminding me...

http://www.mathlinks.ro/Forum/viewtopic.php?t=14577

  darij

[quote="Vasc"]A stronger inequality is the following
                        $\sum_{cyc}\frac{a^{2}(b+c)}{b^{2}+c^{2}}\geq a+b+c.$[/quote]
NICE INEQUALITY.
And i Sloved by caughy.
${\sum_{cyc}\frac{a^{2}(b+c)}{b^{2}+c^{2}} (\sum a^2(b+c)(b^{2}+c^{2}}) ) \ge (\sum a^2(b+c))^2$

To zhaobin: :) I solved it in the same way as yours.
the Inquality come to be Shur's Ineq. :)

Harazi's solution in the link is also very nice  ;)

hey $\sum_{cyc} \frac{a^2(b+c)}{b^2+c^2}\geq \sum_{cyc} a \leftrightarrow \sum_{sym} a^6 b \geq \sum_{sym} a^5 b^2$

You are right! Check demands no more, than 5 minutes! :D


4 hours to burn the red to the length of the blue one which burned 2.5 hours.  So the red one burned one more inch.

1.5 hours to burn an inch

since the red one burned for 6 hours it had 4 inches

[hide]hm... I got the blue candle burned for 4 hours while the red candle burned for 6 hours

so each cm burns for 2 hours, so the red candle would be 3cm long.... not sure if it's right....[/hide]

[hide]R=B+1
Red burns for 6 hrs
Blue burns for 4 hrs
Red=3 in
Blue=2 in[/hide]
At least, that's what I think. I might be wrong though

I got that too but it doesn't satisfy the condition for when the candles are the same height.

[quote="236factorial"]Grandma bought 2 candles. The red is 1 cm longer than the blue candle. In the afternoon of the Day of Christmas at 17:30 she lit the red candle, at 19:00 she lit the blue candle, also, and let them burn until they were finished. The two candles had the same length at 21:30. The red was finished at 23:30, and the blue was finished at 23:00. How long was the red candle originally?

ABACUS International Math Challenge for 7th and 8th graders[/quote]

[hide]

$r(t)=r_0+\dot{r}t$

$b(\tau)=b_0+\dot{b}\tau$

$r_0 = b_0+1$

$r(4)=b(2.5)$

$r(6)=r_0+6\dot{r} =0 \Rightarrow \dot{r} =\frac{-r_0}{6} =\frac{-(b_0+1)}{6}$

$b(4)=b_0+4\dot{b} =0 \Rightarrow \dot{b} =\frac{-b_0}{4}$

$r(4)=b(2.5) \Rightarrow b_0+1-4\cdot\frac{(b_0+1)}{6} =b_0-2.5\cdot\frac{b_0}{4}$

$\Rightarrow \frac{1}{3}(b_0+1)=\frac{1.5}{4}b_0$

$\frac{1}{3}=b_0(\frac{1.5}{4}-\frac{1}{3})$

$12\times\frac{1}{3}=12\times b_0(\frac{1.5}{4}-\frac{1}{3})$

$4=b_0(4.5-4)$

$4=\frac{b_0}{2} \Rightarrow b_0 = 8$

$r_0=b_0 + 1 = 8 + 1 = 9\text{ cm }\blacksquare$

[/hide]

[quote="236factorial"]Well done, but a little long. Could there possibly be a shorter way to do this?[/quote]

Thanks, everyone for your kind words. I don't know whether there is a shorter method or not. I shall put no restrictions on anyone's cleverness -- if there is a shorter way someone ought to find it.

For myself, I only solved the problem as an example of how to systematically attack such problems and force them to give a correct answer  :D It's the teacher in me I suppose.

[hide="How I did it"]
Blue candle:  2.5 hours   same height    1.5 hours
Red candle:   4 hours      same height    2 hours
They burn at different rates, since after they were the same height, it burned differently.  So:
1.5/2=3/4
2.5/3=.8333333333
Ans x 4 = 3.33333333
4 - Ans = 2/3
6/(2/3)=9
[color=darkblue][u][b][size=200]9[/size][/b][/u][/color][/hide]

[quote="nat mc"]didn't like the numbers, try $\text{\LaTeX}$[/quote]
Blue candle: 2.5 hours....same height.....1.5 hours
Red candle:  4 hours.......same height.....2 hours
They burn at different rates, since after they were the same height, it burned differently. So:
$
1.5 \div 2 \equiv \frac{3}{4}
2.5 \div 3 \equiv .8333333333
 Ans \times 4 \equiv 3.33333333
4 - Ans \equiv \frac{2}{3}
6/(2/3) \equiv 9
Answer: 9$

sorry, but right now I can't remember where those numbers came from....I can't explain it right now.

i think the part that confuses me is that I can't see where the 1 cm longer part went.

[quote="math92"][quote="Dr. No"][quote="nat mc"]So you a teacher?[/quote]

Yes, though I am not currently employed as a teacher. Teaching is the one thing I would do for free though, I love it.[/quote]
Oh that's why you know calculas and all this complicateed math[/quote]

Guilty as charged :) though being a calculus show-off is not at all why I do this.

The one thing I hope students will get from my solutions, is that if you solve the problem in general first (using variables instead of specific numerical values),  and then "plug-in the numbers" after you're done, you'll very seldom go wrong.

[quote="Dr. No"][/quote][quote="math92"][quote="Dr. No"][quote="nat mc"]So you a teacher?[/quote]

Yes, though I am not currently employed as a teacher. Teaching is the one thing I would do for free though, I love it.[/quote]
Oh that's why you know calculas and all this complicateed math[/quote][quote="Dr. No"]

Guilty as charged :) though being a calculus show-off is not at all why I do this.

The one thing I hope students will get from my solutions, is that if you solve the problem in general first (using variables instead of specific numerical values),  and then "plug-in the numbers" after you're done, you'll very seldom go wrong.[/quote]

So, your solution was somewhat calculus?!!
I'm a little too stupid for calculus  :blush:

[quote="236factorial"][quote="4everwise"]I think that I understood his solution. However, it's less beneficial for me to know it thatn to solve it by simpler means.[/quote]

Mainly the symbols threw me off. Sometimes I didn't know what they stood for.[/quote]

The only thing in my solution that required knowledge of anything beyond Algebra I was where I used the symbols $\dot{r}$ or $\dot{b}$. These meant the rates at which the candles, red and blue respectively, were burning. Also $r_0$ and $b_0$ meant the lengths of the candles at time, t = 0. This is all shorthand and shouldn't bother anybody too much.


Who made it to the next round? I didn't, and I got 19/25 on the test...

[size=150][color=blue]
MOD EDIT:I added the Country[/color][/size]

are you sure about that? cuz i got a 14.25 and i made it...you can check your official score on webassign if you took it online...

We dont do webassign, is there a way I can still find out my score?

I'm pretty sure we're talking about the US.

Anyway, yeah, I think we're talking on CC too...(OVB = piccolojunior?)

I think you should check to make sure you got a 19 - they have the answers up. If you want a copy of the test, I can e-mail that to you. (Or, well, once I get your e-mail address, anyway) Otherwise, your teacher got an e-mail with scores. (Or, well, our teacher did).

My teacher only knows who qualify for next round. How can I know my score?

How many perfect scores around the whole country?

Where do you get all this information? I can't find an updated (or detailed) website about the US physics team tests.

I also got through to the next round, however my teacher wasn't able to tell me what my score was. Anyone kows how I can find out my score?

does anyone have an electronic copy of the 2008 F=ma contest (multiple choice as well as free response parts)? could someone possibly PM it to me?

Can anyone forward it to me if they have it too (:
Thanks

I'm curious to see the new format of the preliminary round, as I only have access to previous years when it was not exclusively mechanics.  If any one has a digital copy of this year's F=MA exam (2008), could you please send me a copy at Tommy2030@gmail.com .  It would be greatly appreciated.

-Tommy


function
geometry
3D geometry
logarithms

Well this is something like Kebian's 24 but it is different

Here you have to use all the numbers 0,1,2,3,4,5,6,7,8,9 [u]in the given order only[/u] to form a number n.

Use of all mathematical functions lik log , in , ! etc. is allowed .

You can use square root without using the '2' but for cube roots fourth roots etc. you have to use a '3' or '4' etc.

The first person to answer will get 5 points and the next 4 will get 4,3,2,1 in that order.

In case you post more than one answer you will get a '*' which
will decide who is first in case of a tie.

Today's number is 44.  :huh:

nice problem....i keep getting 42 instead of 41...argggggggggh....

[quote="Klebian"][quote="ch1n353ch3s54a1l"]$1!\cdot 2! \cdot 3! \cdot 4! -5+6-7+8+9$[/quote]

Thats... um... 299...[/quote]

Sorry. *EDITED*

0!-1+2+3+4+5+6+7+8+9

to chinesechess4all, you have to use 0

$(\log(0+1+2+3+4)+5) \times 6 +7 -8 +9$
to vihang, don't I get 5 points because my solution was the first to use 0?

New number is 248.

Will post the scores later.

[quote="Klebian"]Oh, come on, don't be cheap, all of them could just do " 0 +" and be done too....[/quote]
Well maybe if they had read the rules better  :roll:

on the honor code:
$0+(-1+2+34)(5)-6+7+8(9)$
$0+1 \cdot 2^3 \cdot (-4(5)+6(7))+8(9)$
$0+(-1+23+4)(5+6)+7+8+9$
$0+(1+23)(4)+56+7+89$
$0+1 \cdot 2+3 \cdot 4+5!+6 \cdot 7+8 \cdot 9$

[hide="EDIT"]
sorry, but i got in through a link and i didn't realize this was a middle school forum.
i don't think my answers should count because i'm in highshool
 :oops:
[/hide]

[quote="minsoens"]
sorry, but i got in through a link and i didn't realize this was a middle school forum.
i don't think my answers should count because i'm in highshool
 :oops:
[/quote]
Well, try this one then :) : http://www.mathlinks.ro/Forum/viewtopic.php?p=396980#p396980


Suppose $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n \equal{} 2n^2(3 \plus{} a_{n\plus{}1} \plus{} a_{n\plus{}2} \plus{} \ldots \plus{} a_{50})$ holds for $ n \in \{1, 2, \ldots , 50\}.$ Find $ a_1.$

[hide="answer"]
10002[/hide]

How do you do this problem?

[quote="BOGTRO"][quote="Attila the Hun"]$ a_1 \plus{} a_2 \plus{} ..... \plus{} a_n \equal{} (2n^2)(3 \plus{} a_{n \plus{} 1} \plus{} .... \plus{} a_{50})$

=> $ a_1 \equal{} (2) (3 \plus{} a_2 \plus{} \hdots \plus{} a_{50})$                                                            (1)

also,
$ a_1 \plus{} a_2 \plus{} \hdots \plus{} a_{50} \equal{} 3 \cdot 2(50)^2$

=> $ a_2 \plus{} .... \plus{} a_{50} \equal{} 6 * (50)^2 \minus{} a_1$                                                                (2).

substituting (2) in (1),

$ a_1 \equal{} 2 (3 \plus{} 6*(50)^2 \minus{} a_1)$

$ 3a_1 \equal{} 6 \plus{} 12*(50)^2$

$ a_1 \equal{} 2 \plus{} 4*(50)^2$

     = $ 2 \plus{} 10000$

     = $ 10002$. :)[/quote]

Just LaTeXifying (Coining a new word, if I may), for your convieniece.[/quote]

thanks BOGTRO. I don't have LaTeX. :oops:

You don't need to have anything installed to use $ \text{\LaTeX}$. All you need to do is put dollar signs around your math, with [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]commands and other stuff, which are not hard to learn[/url].

[quote="ghjk"][quote="hurdler"]Suppose $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n \equal{} 2n^2(3 \plus{} a_{n \plus{} 1} \plus{} a_{n \plus{} 2} \plus{} \ldots \plus{} a_{50})$ holds for $ n \in \{1, 2, \ldots , 50\}.$ Find $ a_1.$

[hide="answer"]
10002[/hide]

How do you do this problem?[/quote]
How can the last term on the RHS be a_50? It doesn't satisfy the pattern of that sequence(when n=49 or n=50),isn't it?[/quote]

for n=49 or n=50, this problem probably means that no terms exist on the RHS that are of the form a_i, so the right hand side would become $ 2n^2(3)$ when n=49 or n=50.


trigonometry
function
calculus
calculus computations

[quote="red218"]
3.sinx+cosx/sinx-cosx
[/quote]
$ f(x) = \frac{\sin x+\cos x}{\sin x-\cos x}\cdot \frac{\sin x+\cos x}{\sin x+\cos x}$
Thus,
$ f(x) = \frac{\sin^{2}x+\cos^{2}x+2\sin x \cos x}{\sin^{2}x-\cos^{2}x}=-\frac{1+\sin 2x}{\cos 2x}=-\sec 2x-\tan 2x$

Now that is easier to differentiate.


function
LaTeX
real analysis
real analysis unsolved

[code] y y^{'} +y =f
y and f are functions in code  H^s
there is any solution of this ode?
[/code]

hello, can you use $ \text{\LaTeX}$ please?
Sonnhard.


group theory
abstract algebra
search
Ring Theory
superior algebra
superior algebra unsolved

prove that in the ring  $ (\mathbb Q, \plus{} ,.)$ hasn't a maximal ideal .

where $ \plus{} : \text{\ usual addition } ,\cdot: a.b \equal{} 0,\forall a,b\in R$

Also, does anyone know another example ?

 :)

You should say that your "rings" are in fact without unity  :wink:

Well, the only nontrivial ideal condition is that it's closed under addition (the multiplicative part just requires it to contain $ 0$). So you want to show that $ \mathbb Q$ has no maximal proper subgroup.

very nice :D , using the correspondence theorem and $ M$ is maximal subgroup iff $ G/M$ is simple . we get $ \mathbb Q$ hasn't maximal subgroup.

I'm thinking about $ \mathbb{Z}_{p^{\infty}}$ group , with the same multiplication above .(i.e. $ ({\mathbb{Z}_{p^{\infty}}, + ,.)}$ ring

Is there any maximal ideal in it ?

Thank you ZetaX :)

A quick google search turns up [url=http://sierra.nmsu.edu/morandi/notes/NoMaxIdeals.pdf]this note[/url].


Find the minimum area of the triangle whose vertices are $A(-1,1,2)$, $B(1,2,3)$, and $C(t,1,1)$. ($t$ is a real number.)

$\overrightarrow {AB}=(2,1,1), \overrightarrow {AC}=(t+1,0,-1),$ thus

$\triangle {ABC}=\frac{1}{2}\sqrt{|\overrightarrow {AB}|^2|\overrightarrow {AC}|^2-(\overrightarrow{AB}\cdot \overrightarrow {AC})^2}=\frac{1}{2}\sqrt{6\{(t+1)^2+1\}-\{2(t+1)-1\}^2}$

${=\frac{1}{2}\sqrt{2(t+2)^2+3}\geq \frac{\sqrt{3}}{2}},$ the equality holds when $t=-2.$ Therefore desired answer is $\frac{\sqrt{3}}{2}.$


Well, there really haven't been many major math competitions, so until there are, let's talk about some sports teams in Wisconsin. (mustafa I did this for you!)

Bucks Suck

Go Pistons

Brewers Suck

Go Tigers

Packers Suck

So do the Lions

More like

Brewers Suck
Tigers Suck
Go Red Sox (Daisuke Matsuzaka is the shiz)

Packers Suck
Lions Suck
Go Pats

Bucks Suck
Pistons Suck
Oh wait so do the Celtics

Well, the Red Wings are good and there aren't really any hockey teams in Wisconsin.  And the Bruins suck.

Brewers-used to be good, may be good in near future
Bucks-the epitomy of mediocrity
Packers-used to be good, not out of playoff hunt yet
UW Badger Basketball-"the shiz"
Marquette Basketball-almost "the shiz"
Badger Men's Hockey-were "the shiz"
Badger Football- "the super deluxe ultra shiz"

Thanks for letting me use your phrase LEETSPEAK  :rotfl:

[quote="Klebian"]Dang. When I saw this topic, I thought you misspelled Wii sports. Boo.[/quote]Well, WI sports are just as shizzy as Wii sports! :D

[quote="13375P34K43V312"]Well we've owned football for the last six years so i don't care.[/quote]

are you talking about the patriots...its kinda funny what i think

[code]
                   Boston (Ma area)            St. Louis

Baseball         Love them                    Hate them

Football         Hate them                     Love them

[/code]

-jorian

Indeed. I was once a Ram bandwagon fan in the days of Kurt Warner

lol, that was an annoying game against the patriots  :mad: [url]http://en.wikipedia.org/wiki/Super_Bowl_XXXVI[/url]
-jorian

ＢＲＥＴＴ　ＦＡＶＲＥ：　ＲＥＴＩＲＥ．

The Packers are going to win the Super Bowl this year after a magical elf brings Brett Favre a special lotion that makes him younger. :roll:

Well, the Packers are close to the playoffs, the Bucks are going for 6 straight tonight, the Badgers play #12 Arkansas in 5 days, Badger Basketball is #4, and Marquette is #19. Wisconsin sports are looking pretty darn good.

mavericks=awesome
cowboys=almost awesome
stars=close to awesome
rangers=awesome for the first half of the season, then they pretty much die

badgersw win again why am i not surprised FOWIST GUMP

That was far too close for comfort, almost blowing a 12 point lead, but the Badgers prevailed over Ohio State, no matter how good their cafeteria food may be.

[quote="Klebian"]Dang. When I saw this topic, I thought you misspelled Wii sports. Boo.[/quote]

Same with me.


uhh.

I totally want to go to Mathcounts, but my school doesn't have any math team, so I can't go to Mathcounts. (I took amc at some high school :( )

What can I do to go to MC?

Should I register with the private school?

i'm fairly sure you can do mathcounts without a team. The people who do mathcounts without a team are called Individuals. As an individual, the competition is exactly the same, except you dont have a team(which didn't bother me at states; i was an individual) and thus the team round isnt graded. At some places you have an option of taking the team round with other individuals, but it wont really matter. However, many people(i would think) benefit from having a team, since you can interact with your teammates and talk math. But that what i do on AoPS :lol: . So basically, you can still do mathcounts without a team, if that's what you are asking.

You could request a team be made for your school. Ask a math teacher to be the coach, perhaps. You've got a whole year...

I'm lucky to be on a team, and even have someone else on my team to go to nats with. How else would you spend the like 2 HOURS in between team and countdown other than messing around with your team???? It definitely helps me to relax in new surroundings when I have someone else there that is new too. So here's to teams!!! :lol:

p.s.: fantasy lover, what grade and state r u?

yeah i already requested to principal to make a math team but he said that there's too few students who is good at math...

and we don't have any math coach (my school is not very good)

P.S. im in 7th grade, nj.

maybe even you could be the coach! usually the person who wants something has to put more effort than everybody else. And the other people in your school dont have to be good at math, they just have to want to learn and they have to like it. One of the main purposes of mathcounts is to "promote interest in middle school problem solving skill" so if they aren't good at math already, then it's easier to teach them something :lol:

[quote="FantasyLover"]yeah i already requested to principal to make a math team but he said that there's too few students who is good at math...[/quote]

That's too bad. I always geared my teams for those who wanted to participate, not those who were best, or even really good at math. The point is to learn something from it.

The last team I coached really wasn't an official school team. (OK, I've honestly never had an official school team, I've always just made the teams on my own with no support from the school.) It was just several kids who wanted to do it. They weren't even the best kids.  My top participant probably was about 8th best in math at the school.  This out of a school of only 80 7th and 8th graders combined.  Some of the participants I would put near the midline 50th percentile in math ability at the school.  However, they wanted to do it, so that was good enough for me!  Our practices were occasional and very informal, but the students would learn some good math every now and then. Perhaps your school could do something like that.

Now if I could only have my dream of being able to have a fulltime load of daily mathcounts teaching...oh man, that would be great.  I could compete against Boyd and Frost!  Oh well, dream on.

W000000000000T!!!! I MIGHT GO TO MATHCOUNTS!!!!


What is the domain of the function $ f(x) \equal{} (2x\minus{}3)^{x^2 \minus{} 9x \plus{} 20}$ where $ f(x)$ is an exponential function?

The function is defined if $ 2x \minus{} 3 \geq 0 \equal{}> x \geq \frac{3}{2}$, and also $ \forall x \in N$.

[quote="Carcul"]The function is defined if $ 2x \minus{} 3 \geq 0 \equal{} > x \geq \frac {3}{2}$, and also $ \forall x \in N$.[/quote]

Did you mean $ 2x\minus{}3>0$ and $ x>3/2$? because it needs to be an exponential function...

I am not familiar with notations: $ \forall x \in N$, what does it mean?

$ \forall x\in\mathbb{N}$ means "for all $ x$ in $ \mathbb{N}$".

I don't see why $ 2x\minus{}3$ needs to be positive.
My guess is the domain is all reals, where $ x\not\equal{}\frac{3}{2}$ and $ x\not\equal{}2$, because in those values the function becomes linear.

By the definition of exponential function, the base must be positive and not equal to 1, so the domain will be $ {x|x>\frac {3}{2} and x\not=2}$

btw, if a negative number is raised by irrational powers, is it always a complex number? For rational number, this occurs only when it has a even factor in the denominator.

[hide="ANSWER SHEET"] $ x>3/2$ or $ x\in(3,\plus{}\infty)$[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

1. Let M, N be mxn and nxm matrices, respectively. If c is an eigenvalue for MN then it's also an eigenvalue for NM. That's easy to see....but show by example that this isn't necessairly true if c=0.

2. If A is the inverse of I-NM (identity minus NM) and B is the inverse of I-MN then find the expression for B in terms of A, M, N...

i got stuck here...
cheers and thanks for help

[quote="omnyx"]If c is an eigenvalue for MN then it's also an eigenvalue for NM. That's easy to see....but show by example that this isn't necessairly true if c=0.[/quote]

I don't have an example handy, but it's possible to construct $ M, N$ so that $ NM$ is an invertible matrix while $ MN$ is not.

Let $ M\equal{}\begin{bmatrix}1\\1\end{bmatrix}$ and $ N\equal{}\begin{bmatrix}1&1\end{bmatrix}.$

Then $ MN\equal{}\begin{bmatrix}1&1\\1&1\end{bmatrix}$ is not invertible, while $ NM\equal{}[2]$ is invertible.

$ MN$ has characteristic polynomial $ x(x\minus{}2)$ while $ NM$ has characteristic polynomial $ x\minus{}2.$

I don't feel like searching right now, but the proof has appeared on this forum that $ MN$ and $ NM$ have the same characteristic polynomial, except that the one with larger dimension has the polynomial "padded" with enough extra factors of $ x$ to make up the difference in degree.