If $ Y$ is the concentration of alcohol in a person's blood, then the rate at which it is removed by the liver obeys the following differential equation  (assuming no further drinking):
\[ \frac{dY}{dt}   \equal{}  \minus{}  \frac{V_{\mathrm{max}}Y}{K_{\mathrm{M}} \plus{} Y}\]
where $ K_{\mathrm{M}}$ and $ V_{\mathrm{max}}$ are constants which may vary from person to person.

[b]a)[/b] There are two extreme situations for which an approximation may be in order: first, what if $ Y >> K_{\mathrm{M}}?$ Then the differential equation may be approximated as:
\[ \frac{dY}{dt}   \equal{}  \minus{} V_{\mathrm{max}}\]
What are the solutions of this approximate equation?  Draw a rough sketch of the family of solutions.

[b]b)[/b] Second, what if $ Y << K_{\mathrm{M}}?$ Then the differential equation may be approximated as:
\[ \frac{dY}{dt}   \equal{}  \minus{} \left[\frac{V_{\mathrm{max}}}{K_{\mathrm{M}}}\right]\, Y\]
What are the solutions of this approximate equation? Draw a rough sketch of the family of solutions.

[b]c)[/b] Solve the full original equation (it is separable). Your answer cannot be solved explicitly in closed form for $ Y$ in terms of $ t$  but (as a hint for later) notice that you can solve for $ t$ in terms of $ Y.$

[b]d)[/b] Suppose that for a particular person, we know that $ K_{\mathrm{M}} \equal{}  0.02$ (units)  and that $ V_{\mathrm{max}} \equal{} 0.025$ (units)/hour. This person gets drunk: the initial condition is $ Y(0)  \equal{}  0.13$ (units). Make an approximate prediction, based on whichever of  a) or b)  is applicable, as to how long it will take for this person's blood alcohol to drop to the legal  $ 0.08$ (units).

[b]e)[/b] Find the exact solution of this initial value problem.

[b]f)[/b] Use the exact solution to answer these two questions: how long will it take for the concentration to reach $ 0.08$ (units)? How long will it take for the concentration to reach $ 0.005$ (unit) ?  [The unit is % by volume - not a standard scientific unit. Don’t unnecessarily confuse yourself by trying to think of is as % - it’s just some unit.]

Carefully drawn graphs might be useful.

Some comments about this problem:

This is a favorite homework exercise of mine for the first weeks of a differential equation course. I'm bringing it up now in the last weeks of a Calculus II course because we're doing a smidgen of differential equations there.

My wife has taught general biochemistry many times and an advanced enzyme kinetics course several times. The rate law here is a simple version of the Michaelis-Menten rate law for enzyme-catalyzed reactions. The terminology used - $ V_{\mathrm{max}}$ and $ K_{\mathrm{M}}$ - is standard in that context. (The "M" in the subscript stands for "Michaelis.")

In fact, the removal of ethanol from the bloodstream is primarily through an enzyme-catalyzed reaction. The reaction is an oxidation, with NAD+ as the oxidizing agent, that transforms it into acetaldehyde. The acetaldehyde is then further oxidized to acetyl-coenzyme A, which is then a substrate for the citric acid cycle.

The Michaelis-Menten law would apply not just to enzymes, but to any catalyst-mediated chemical reaction in which there's a relatively small amount of the catalyst compared to the concentrations of the reactants.

The actual numbers I used for $ V_{\mathrm{max}}$ and $ K_{\mathrm{M}}$ were pulled out of thin air, just to make the homework problem interesting. But I do understand that $ K_{\mathrm{M}}$ for alcohol dehydrogenase is small compared to physiologically meaningful concentrations of ethanol.

[hide="Is this it?"]
$ \frac {dY}{dt} \equal{} \minus{} \frac {VY}{K \plus{} Y}$

$ \frac {dt}{dY} \equal{} \minus{} \frac {K \plus{} Y}{VY}$

$ \frac {dt}{dY} \equal{} \minus{} \frac {K}{VY} \minus{} \frac {1}{V}$

$ t(Y) \equal{} \minus{} \frac {Y \plus{} KlnY}{V}\plus{}C$[/hide]

That's a portion of part c). You haven't yet made use of initial conditions, nor have you yet addressed any of the other portions of the project.

well once you have the solution to the equation, plugging in the given numbers isn't too hard

a)$ Y\equal{}\minus{}Vt\plus{}C$
b)$ Y\equal{}Ce^{\minus{}\frac{V}{K}}$
d)$ 2$
e)$ t\equal{}.8(ln(.13)\minus{}ln(Y))\plus{}40(.13\minus{}Y)$
f)
$ 2\plus{}.8ln(1.625)$
$ 5\plus{}.8ln(26)$

b) would be $ Ce^{\minus{}\frac{V}{K}t}.$ Exponential decay, but only for small values of $ Y.$

The other answers are probably correct, although I'll expect my students to use a calculator and get me approximate numerical values for those answers on f).


Let $ a,b,c$≥$ 0$ satisfy $ ab\plus{}bc\plus{}ca\equal{}1$.Prove that
$ \frac{ab\plus{}1}{a\plus{}b}\plus{}\frac{bc\plus{}1}{b\plus{}c}\plus{}\frac{ca\plus{}1}{c\plus{}a}$≥$ 3$

[quote="drnohad"]I think that ab+bc+ca=3[/quote]
I think the ineq is right.The equality occurs when $ a\equal{}b\equal{}1,c\equal{}0$

[quote="SUPERMAN2"]Let $ a,b,c$≥$ 0$ satisfy $ ab + bc + ca = 1$.Prove that
$ \frac {ab + 1}{a + b} + \frac {bc + 1}{b + c} + \frac {ca + 1}{c + a}$≥$ 3$[/quote]

$ q = ab+bc+ca =1$ , $ p = a+b+c$ , $ r = abc$

$ \Leftrightarrow p^2-3p+2+r(p+3) \ge 0$

1. for $  0 < p \le 1 $ or $ p \ge 2$ it is obviously .

2. $ 1 \le p \le 2$
By Shur : $ r\ge \frac{1}{9}p(2-p)(2+p)$
$ \Leftrightarrow \frac{1}{9}(2-p)(p^3+5p^2-3p+9) \ge 0$

Equality holdes for $ a=b=1$ , $ c=0$

[quote="nguoivn"][quote="SUPERMAN2"]Let $ a,b,c$≥$ 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 1$.Prove that
$ \frac {ab \plus{} 1}{a \plus{} b} \plus{} \frac {bc \plus{} 1}{b \plus{} c} \plus{} \frac {ca \plus{} 1}{c \plus{} a}$≥$ 3$[/quote]
I had 2 proofs by AM-GM for it  :)[/quote]
Prove it  :lol:
Em rất muốn xem lời giải Am-Gm của anh  :)

We need prove:
   $ \sum(\frac {ab\plus{}1}{a\plus{}b}\plus{}1)\geq 6$
$ \Leftrightarrow \sum\frac {(a\plus{}1)(b\plus{}1)}{a\plus{}b}\geq 6$
Using ineq AM-GM,we have:
    $ \sum\frac {(a\plus{}1)(b\plus{}1)}{a\plus{}b}\geq 3 \sqrt[3]{\frac {\prod (a\plus{}1)^{2}}{\prod (a\plus{}b)}}$
We need prove:
 $ \prod (a\plus{}1)^{2}\geq 8\prod (a\plus{}b)$
$ \Longleftrightarrow ((\sum a)\plus{}2\plus{}abc)^2\geq 8((a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc)$
we have inqe is true because $ ((\sum a)\plus{}2\plus{}abc)^2\geq (2\plus{}\sum a)^2\geq 8\sum a\geq 8((a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc)$

[quote="math10"]We need prove:
   $ \sum(\frac {ab \plus{} 1}{a \plus{} b} \plus{} 1)\geq 6$
$ \Leftrightarrow \sum\frac {(a \plus{} 1)(b \plus{} 1)}{a \plus{} b}\geq 6$
Using ineq AM-GM,we have:
    $ \sum\frac {(a \plus{} 1)(b \plus{} 1)}{a \plus{} b}\geq 3 \sqrt [3]{\frac {\prod (a \plus{} 1)^{2}}{\prod (a \plus{} b)}}$
We need prove:
 $ \prod (a \plus{} 1)^{2}\geq 8\prod (a \plus{} b)$
$ \Longleftrightarrow ((\sum a) \plus{} 2 \plus{} abc)^2\geq 8((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc)$
we have inqe is true because $ ((\sum a) \plus{} 2 \plus{} abc)^2\geq (2 \plus{} \sum a)^2\geq 8\sum a\geq 8((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc)$[/quote]
Nice proof, math10. And here is one of my proofs :
Using Am-Gm with the form: $ (x\plus{}y\plus{}z)^2 \geq\ 3(xy\plus{}yz\plus{}zx)$ we only need to prove:
$ \sum\ \frac{(1\plus{}ab)(1\plus{}bc)}{(a\plus{}b)(b\plus{}c)} \geq\ 3$ <=> $ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}a)\plus{}abc \geq\ ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)$
(obviously trues)  :)

[quote="math10"]can you  clarify your solutions?
I do not think it can be written as simple.(I do not want them to write the check :blush: )
Ps caubetoanhoc 94:Em có tham gia kì thi VIC không?
Bài này hình như tương tự bài 1 của nó.[/quote]

Nice;math10 :)
Em có tham gia VIC mà :)..Nhưng ở VIC em expand :)..Mà anh có tham gia ko :-?? Tên anh gì vậy :P

perhaps the solution of nguoivn is true.
Ps caubetoanhoc94:anh cũng đăng kí tham gia nhưng sau đó lại thôi,bởi bây giờ đang cố gắng hết tuầnn này phải có 225 bài trên mathlink.Anh tên Lê Trọng Cường,nick:ltcuongdhqg.

[quote="nguoivn"][quote="math10"]We need prove:
   $ \sum(\frac {ab \plus{} 1}{a \plus{} b} \plus{} 1)\geq 6$
$ \Leftrightarrow \sum\frac {(a \plus{} 1)(b \plus{} 1)}{a \plus{} b}\geq 6$
Using ineq AM-GM,we have:
    $ \sum\frac {(a \plus{} 1)(b \plus{} 1)}{a \plus{} b}\geq 3 \sqrt [3]{\frac {\prod (a \plus{} 1)^{2}}{\prod (a \plus{} b)}}$
We need prove:
 $ \prod (a \plus{} 1)^{2}\geq 8\prod (a \plus{} b)$
$ \Longleftrightarrow ((\sum a) \plus{} 2 \plus{} abc)^2\geq 8((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc)$
we have inqe is true because $ ((\sum a) \plus{} 2 \plus{} abc)^2\geq (2 \plus{} \sum a)^2\geq 8\sum a\geq 8((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc)$[/quote]
Nice proof, math10. And here is one of my proofs :
Using Am-Gm with the form: $ (x \plus{} y \plus{} z)^2 \geq\ 3(xy \plus{} yz \plus{} zx)$ we only need to prove:
$ \sum\ \frac {(1 \plus{} ab)(1 \plus{} bc)}{(a \plus{} b)(b \plus{} c)} \geq\ 3$ <=> $ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} a) \plus{} abc \geq\ ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)$
(obviously trues)  :)[/quote]
Very nice solution! :)


geometry
vector
function
3D geometry
tetrahedron
trigonometry
geometry proposed

[u][b]The author of this posting is : Piotr Achinger[/b][/u]
____________________________________________________________________

Given a convex polyhedron with $ n$ faces $ S_1, S_2, \ldots, S_n$.
We denote the area of the face $ S_i$ by $ P_i$ for every $ i\in \left\{1,2,\ldots,n\right\}$.

Consider a set of vectors $ v_1, v_2, \ldots, v_n$, where the vector $ v_i$ is perpendicular to $ S_i$, pointing outside the polyhedron, and its length is equal to $ P_i$ for every $ i\in \left\{1,2,\ldots,n\right\}$.

Prove that $ v_1 \plus{} v_2 \plus{} \ldots \plus{} v_n \equal{} 0$.

This problem has a simple physical interpretation:
Just imagine that the polyhedron is filled with gas that puts an equal pressure on each face; then the resultant force is O.

This is a very cool problem.

It's enough to prove it for tetrahedra, since every polyhedron can be decomposed into tetrahedra, and it's easy to see that this function is "additive" (for example, if two tetrahedra have a common face, then the two vectors corresponding to that face cancel out). On the other hand, for a tetrahedron we only need to play with some cross products. The details are really easy to fill in.

[quote="Piotr Achinger"]Given a convex polyhedron with $ n$ faces $ S_1, S_2, \ldots, S_n$.
We denote the area of the face $ S_i$ by $ P_i$ for every $ i\in \left\{1,2,\ldots,n\right\}$.

Consider a set of vectors $ v_1, v_2, \ldots, v_n$, where the vector $ v_i$ is perpendicular to $ S_i$, pointing outside the polyhedron, and its length is equal to $ P_i$ for every $ i\in \left\{1,2,\ldots,n\right\}$.

Prove that $ v_1 \plus{} v_2 \plus{} \ldots \plus{} v_n \equal{} 0$.[/quote]

Here is a solution found by Peter Scholze and me when discussing the problem with the Iranian IMO team in Athens.

Let g an [i]arbitrary[/i] line somewhere in space, and let p be a plane perpendicular to g.

For every $ i\in\left\{1;\;2;\;...;\;n\right\}$, let $ u_i$ be the orthogonal projection of the vector $ v_i$ on the line g (note that $ u_i$ is just a directed segment, not a real vector), and let $ R_i$ be the [i]directed[/i] area of the orthogonal projection of the polygon $ S_i$ on the plane p. Then, $ u_i \equal{} \left|v_i\right|\cdot\cos\measuredangle\left(v_i;\;g\right)$ and $ R_i \equal{} P_i\cdot\cos\measuredangle\left(S_i;\;p\right)$ (note that $ S_i$ stands both for the i-th face of our polyhedron and for the plane containing this face). Since the length $ \left|v_i\right|$ of the vector $ v_i$ equals to $ P_i$, we have $ u_i \equal{} \left|v_i\right|\cdot\cos\measuredangle\left(v_i;\;g\right) \equal{} P_i\cdot\cos\measuredangle\left(v_i;\;g\right)$.

But since $ g\perp p$, we have $ \measuredangle\left(v_i;\;g\right) \equal{} 90^{\circ} \minus{} \measuredangle\left(v_i;\;p\right)$, and since $ v_i\perp S_i$, we have $ \measuredangle\left(S_i;\;p\right) \equal{} 90^{\circ} \minus{} \measuredangle\left(v_i;\;p\right)$. Thus, $ \measuredangle\left(v_i;\;g\right) \equal{} \measuredangle\left(S_i;\;p\right)$. Consequently,

$ u_i \equal{} P_i\cdot\cos\measuredangle\left(v_i;\;g\right) \equal{} P_i\cdot\cos\measuredangle\left(S_i;\;p\right) \equal{} R_i$.

So we have shown that $ u_i \equal{} R_i$ holds for every $ i\in\left\{1;\;2;\;...;\;n\right\}$.

But, if we orthogonally project the faces of a convex polyhedron on a plane, then the sum of the directed areas of the projections is 0. Thus, $ R_1 \plus{} R_2 \plus{} ... \plus{} R_n \equal{} 0$. Since $ u_i \equal{} R_i$ holds for every $ i\in\left\{1;\;2;\;...;\;n\right\}$, this becomes $ u_1 \plus{} u_2 \plus{} ... \plus{} u_n \equal{} 0$. But for every $ i\in\left\{1;\;2;\;...;\;n\right\}$, the number $ u_i$ is the orthogonal projection of the vector $ v_i$ on the line g; thus, the number $ u_1 \plus{} u_2 \plus{} ... \plus{} u_n$ is the orthogonal projection of the vector $ v_1 \plus{} v_2 \plus{} ... \plus{} v_n$ on the line g.

Hence, the equation $ u_1 \plus{} u_2 \plus{} ... \plus{} u_n \equal{} 0$ signifies that the orthogonal projection of the vector $ v_1 \plus{} v_2 \plus{} ... \plus{} v_n$ on the line g is 0. Now note that this holds not just for one special line g; this holds for [i]every[/i] arbitrary line g in space! Hence, the vector $ v_1 \plus{} v_2 \plus{} ... \plus{} v_n$ itself must be the zero vector. $ \blacksquare$

Hope that the above solution was halfways intelligible...

  Darij


In a triangle $ ABC$
$ D\in [AC]$ , $ E\in [BC]$
$ |AD|\equal{}15$ , $ |CD|\equal{}40$ , $ |CE|\equal{}32$ , $ |EB|\equal{}18$ and $ |ED|\equal{}12$
then
$ \frac{m( \widehat A)}{m( \widehat B)}\equal{}?$

It really depends on the time given for the problem - Requires here 10 minutes for the calculation.
[hide]I could find angle C using the law of cosine,
Then AB using law of cos
And the angles using law of sins.
[/hide]
But is there a quicker way to do it?

yes there is a much more simple way by SAS congruent but i dont know the solution yet :)


[quote=TaiPan~SP!]Let $n$ be a positive integer and $\alpha$ and $x$ be real numbers. Let also $0 < \alpha <1$ and $\alpha^{n+1}\leq x \leq 1$.

Prove that

$P(x)=\frac{x-\alpha}{x+\alpha}.\frac{x-\alpha^{2}}{x+\alpha^{2}}.\frac{x-\alpha^{3}}{x+\alpha^{3}}.........\frac{x-\alpha^{n}}{x+\alpha^{n}}\leq P$

where

$P=P(1)$.[/quote]
[url]https://artofproblemsolving.com/community/c6h1344193p7312528[/url]


The volume of a cylinder is $ 54\pi$ $ \text{cm}^3$. How many cubic centimeters are in the volume of a cone with the same radius and height as the cylinder? Express your answer in terms of $ \pi$.
[asy]import solids; currentprojection=orthographic(0,100,25); defaultpen(linewidth(0.8));
revolution cyl = cylinder((5,0,0),1,5,Z);
revolution cone = cone((0,0,0),1,5,Z);
cyl.draw(backpen=dashed);
cone.draw(backpen=dashed);[/asy]

The volume of a cylinder is $ \pi r^2 h$ and the volume of a cone is $ \dfrac{\pi r^2 h}{3}$ so the volume of the cylinder divided by three is the volume of the cone. $ \dfrac{54 \pi}{3}\equal{} \boxed{27 \pi}$

You mean $ \frac{54\pi}{3}\equal{}\boxed{18\pi}$  :)


For a tetrahedron $ABCD$, with vertex $D$, find a formula for its volume if you are given the side lengths $a,b,c$ of triangle $ABC$ and lengths $d_{a}, d_{b}, d_{c}$ as edges $DA,DB,DC$.

i think we need f2 solve for this one...there might be some fancy formulas on wolfram...but thats not that reasonable...just coordinatize, and solve for D, there are certain cases where the projection of D onto plane ABC is easy to compute....

If you want the fancy formula, look up the Cayley-Menger formula on MathWorld.  It's a 5x5 determinant.


in most book we can see this example to define work done

that is take a piston and apply a pressure $P$ on it  and then move the piston by a small distance $dx$ then the work done is $Fdx=PAdx=PdV$...

but my doubt is u r applying the force on the piston so the work is done on the piston,but the piston not necessarily applies the same force on the gas then why is
work done on the gas inside
$W=\int_{V_{i}}^{V_{f}}PdV$

[quote="Pardesi"]but the piston not necessarily applies the same force on the gas[/quote]

Why not? In that derivation (and the usual work computations envolving pistons) its usually considered that there is no friction between the piston and walls of the container.

[quote="Carcul"]b its usually considered that there is no friction between the piston and walls of the container.[/quote]
how does taht ensure that the force exerted by the piston on the gas is same as tha force on the piston


integration
calculus
calculus computations

compute $ \int \frac{xe^x}{(x\plus{}1)^2}$

Indochina  u bet me again as  usual ah....... :wallbash_red:       :P . jus joking ,good solution  :P

any other solution like substitution maybe?
i think, maybe i can't always reform that problem like indochina solution.. :)

You might be able to get it integrating by parts.

On the other hand, this is precisely rigged to work; any slight change would make the antiderivative not an elementary function. Systematic methods don't like that sort of thing.


Let P(x)=$x^3+a.x^2+b.x+c$ and P(x) has three real roots.Let Q(x)=$x^2+2.x+2005$ and P(Q(x)) has no real  roots.Show that P(2005)>1.


AMC USA(J)MO USAMO USAMTS geometry inequalities vector

I'm happy to have (probably) qualified for USAMO, but I haven't been preparing at all for USAMO. I know I can't do too well, but I'd like to do ok, and in particular I may be able to get into freshman MOSP.

Yes, I do know what a proof is, and I indeed have proof-writing experience from USAMTS, but haven't yet tackled problems at the USAMO level.

So, what should my strategy be? How much time should I spend trying to solve problems vs. reading solutions, or even learning new theorems/tools? Which subjects should I focus on the most?

Regarding geometry, what are some important theorems I should know?

Regarding inequalities, should I try to learn anything about them? If so, could someone list the major theorems/inequalities/ideas/tools I should learn? (I've seen AM-GM, WLOG, and Cauchy mentioned, but I forget what else).

And are there any other theorems etc I should know? (ie in other topics)

[quote="calc rulz"](I've seen AM-GM, WLOG, and Cauchy mentioned, but I forget what else)[/quote]
WLOG isn't an inequality theorem, its an acronym menaing "without loss of generality"

Some general strategic advice:
On each half, it's likely that the first problem will be the easiest; spend some time on it. On the other hand, you shouldn't ignore the other problems- there's always a chance that you happen to know something useful.
You have more than four hours for each half, so you can afford to spend some time on each problem just playing around and looking for something interesting. If you get stuck, look at something else for a while.
There's usually one or two problems that can be approached with no technical background. That doesn't necessarily make them easy, but it's certainly worth trying them.

If you want practice, check out the Olympiad section on this site- there are lots of problems there, from all over the world. Most national olympiads are easier than the USAMO, but the style of problems are similar.

Forget AM-GM, you're better off learning the full QM-AM-GM-HM (for nonnegative $a, b$):

$\sqrt{ \frac{a^2 + b^2}{2} } \ge \frac{a + b}{2} \ge \sqrt{ab} \ge \frac{2ab}{a+b}$

The common AM-GM statement is an extension of the middle two:

$\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 a_3 ... a_n}$

As for Cauchy, for nonnegative sequences $a_1, ... a_n, b_1, ... b_n$:

$\left( a_1^2 + a_2^2 + ... \right) \left( b_1^2 + b_2^2 + ... \right) \ge \left( a_1 b_1 + a_2 b_2 + ... \right)^2$

In vector form, it becomes simply

$| \vec{v} | | \vec{u} | \ge | \vec{v} \cdot \vec{u} |$

Which is how I remember the other form in the first place.  :P

Any other important results are things you're better off not memorizing now because they won't stick.  If you can, try to do a lot of practice problems - that's usually the best way to familiarize yourself with new concepts.

And yes, as was previously stated, WLOG is an acronym meaning "Without Loss of Generality."  For example, if one were trying to prove the inequality

$a^2 + b^2 + c^2 \ge ab + bc + ca$

You could realize that this inequality is "symmetric" with respect to $a, b, c$, which means that $f(a, b, c) = a^2 + b^2 + c^2 - ab - bc - ca = f(b, c, a) = f(a, c, b) = ...$, so you can assume, Without Loss of Generality, that $a \ge b \ge c$ (since you can just pick them in that order).

Just some tips on what *I'm* doing (i'm in the same boat as you are, except I got a borderline 8 on the AIME I (after stupid mistakes I won't get into here) - trying to sneak into MOP as a freshman). I am going through the Art and Craft of Problem Solving (often abbreviated ACoPS), which has a good chunk of chapters about actually writing good, convincing proofs (methods of writing proofs, common tools of writing proofs, things to look for, etc.) and a section of book covering some advanced topics and ideas.

I have also been googling and, especially looking at DPopov's thread in this forum, there are some users that posted a list of common theorems/formulas/inequalities you should know (I can't recall the actual title of this thread, but you should be able to find it).

Also, I have heard *many* times that simply working past USAMO problems for some time like the actual time you are allowed, and then working through the suggested solutions to find what you were doing correctly, what you were doing incorrectly, and what you obviously need to start looking for/improve on. I have also heard Titu Andreescu's 2000-2001 Olympiad Problem book cited a few times as a useful resource, but I cannot confirm this, as I do not have the book myself.

(is the power mean inequality useful? what about Bernoulli's inequality? chebychev's?)

Anyway, that is all I have really learned from many hours of Googling/reading. If anybody has any additions to my ideas above, *please* share them for the sake of all of us youngsters (and others) that are preparing for the USAMO. Thanks!

Alex

Good luck on the USAMO!

Calcrulz said he got an 8 on the USAMO too.

Inequalities is so much about having experience...there are tons of techniques and substitutions and everything that one normally won't have the insight to use unless you've seen it before. Say you have the condition $ab+ac+bc+2abc=1$...looks fairly unassuming, but you can substitute $a=\frac{x}{y+z}$ and so on, which trivializes alot of ineq...also, you can do $a=\cos\alpha$ where the three angles sum to $180^{\circ}$, again helping alot. And this is just a fraction of what you can do that really needs experience to work with.

The other stuff isn't nearly as bad, though.

I have Art and Craft of Problem Chapters, which sections do you recommend focusing on? (as the book is too large to read all or even most of it in less than a month)

Regarding other olympiads, does anyone know any specific countries that have relatively easier ones than USAMO?

Also, wouldn't it help to say, know something about inequalities so as to get partial credit on such a problem? And in general, should I always write up something for each problem so as to get some partial credit?

Phelpedo, I got an 8 on the AIME II, not USAMO.

Yeah, I meant AIME, not USAMO.

I wouldn't say 3 weeks before the USAMO is too late to prepare. I started preparing for the AIME II hardly a week before the test and was able to raise my average score up 5 points; I did the same last year before MathCounts States to move up from 5th chapter to 4th state. I would caution you not to do too much math and burn yourself out though. Unfortunately I board at a school and on top of homework&sports, there isn't much time to do extracurricular math, so I'm not able to really start preparing until vacation.

[quote="probability1.01"]Ehhh, the only thing that you should focus on is being good at writing proofs. Your USAMO score is going to be what it's going to be, and practice at this point isn't going to really help you. Just don't get intimidated, but also don't over-stretch yourself.[/quote]
That might be true for people of your calibur, but a lot of others can easily make a big improvement in three weeks.

Unless it isn't true for you either, in which case I need to question why you even posted this comment.

[quote="Phelpedo"][quote="calc rulz"]Phelpedo, the USAMO is different; to truely prepare you need months of preparation. And probability, I know that. I'm just asking what the best thing to do is given the amount of time I have.[/quote]

Yes but that isn't to say that you can't improve in these few weeks. And since you're in 9th grade, you can always look ahead to next year's USAMO. It's never to early to start.[/quote]

I know, so I'm asking what preparation I should do. I was just saying you can't do the same kind of hardcore preparation that people who want to do really well do.

Also, I'm particularly focused on this year because of freshman MOP (they take approximately 30 ninth graders to MOP).

[quote="amitavadhar"]I'm a freshman and i think i got into to USAMO.  Can anyone please tell me how they select freshman MOSP. And if its by USAMO score, what you should shoot for to get in.[/quote]
Try to get one question and bs a whole bunch of other questions, that should be enough. Though I don't know with the expansion, it might be higher.

[quote="amitavadhar"]I'm a freshman and i think i got into to USAMO.  Can anyone please tell me how they select freshman MOSP. And if its by USAMO score, what you should shoot for to get in.[/quote]

They select based on ur usamo score and shoot for at least completely solving one problem.

Just a quick update:

calc rulz, sorry you didn't make USAMO =(. I heard you had the 8 on AIME, but only a 122 on the AMC12, so you did not make the floor ...

also, congrats Gaurav! Who else from ... your school... qualified? (I was the only one from Klein; Akash actually only got a 6 on aime! and our sophmore dude owned himself ... I won't even get into that =D)

Anyway, congrats to all others that made USAMO (AIME score of 8 (with at least a 212 index) for us young people, index of 217 for you older people), and best of luck. I only hope that people will continue to expand this thread to help new test-takers out.

So I really did qualify, now I will have to start preparing for USAMO. So here is my to do list:

Just go and look at some important inequality, geometry, and/or any other theorems/tools/tactics I should know (then should I do problems specifically related to these theorems, does anyone know where I can find such problems as such?)

Next I will look through AaCoPS, picking out tools, problems, and some other advice (does anyone recommend any particular sections?)

Finally I will start to go through USAMO problems, contemplating them until I'm not getting anywhere, then looking at the solution and seeing what I was doing wrong and doing right (I heard other olympiads are easier and are better for someone starting to do olympiad-level problems, does anyone recommend any particular country olympiads?)

I would appreciate comments on this plan, including adding things to the plan, taking things off the plan, editing parts of the plan, or a comment that it is a good plan if it is.

[quote="calc rulz"]

Finally I will start to go through USAMO problems, contemplating them until I'm not getting anywhere, then looking at the solution and seeing what I was doing wrong and doing right (I heard other olympiads are easier and are better for someone starting to do olympiad-level problems, does anyone recommend any particular country olympiads?)[/quote]

Well...no, but:

The non-calculus parts of the Putnam are designed to be solved in a far shorter time period (approx. 30-40 minutes a problem IIRC - 6 problems in 3 hours, although it's rare that anyone makes a perfect score) than those on most national Olympiads. When it comes to just getting into the proper state of mind to take an Olympiad, I'd highly recommend Putnam problems and USAMOs from before (I believe) 1996. You don't have to contemplate those for as long.

Personally, I go for the full experience - shut myself away for 3 to 4.5 hours and do a full USAMO/IMO. That's what I should be doing now, but I'm taking a short break 2 hours in  :blush:

Also remember that this is asking what I should do in a couple weeks, not preparation for next year.

[quote="amitavadhar"]Does anybody also know what a good score to aim for is to get into freshman mop.[/quote]

Solve at least one problem.

I think last year it was around 7.  But since there's more people this year, I'm thinking at least a 10 maybe?

I dunno cause I've never had any type of erperience with USAMO though.

It will probably be something like 7. Last year's USAMO was particularly easy.

Is it generally easier to do well on the usamo the second time doing it than the first?  Will the experience in the actual thing be replaceable by doing past problems?

The reason I'm asking is that I think I was capable of doing quite well as usamo this year, but I bombed with lots of careless mistakes on AIME, putting me one below the floor.  And I'm a sophomore.  So, will having not made usamo this year affect how well I would do at usamo next year?


inequalities
geometry
circumcircle
trigonometry
function
triangle inequality
geometry proposed

Let $ABC$ be a triangle and $M,N,P$ be the midpoints of sides $BC, CA, AB$. The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$. Show that the area of triangle $ABC$ is at most the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$.

We have  $\frac{[A_{1}BC]}{[ABC]}=\frac{A_{1}M}{AM},...$ etc, therefore we only need prove \[\frac{A_{1}M}{AM}+\frac{B_{1}N}{BN}+\frac{C_{1}M}{CM}\geq 1.\]

But this statement is easy, i think so :D

I'm going to use the fact that the symmetric of the ortocentre wrt. a side lies on the circumcircle.

Let's call $H_{A},H_{B},H_{C}$ the intersections of the altitudes with the circumference; $D,E,F$ the intersections of the altitudes with the sides.

Since the baricentre lies between the ortocentre and the circumcentre, and the last ones are isogonal conjugate,  we have that $C_{1}$ is further than $H_{C}$ from the line $AB$. Then $(ABC_{1}) \ge (ABH_{C}) = (ABH)$.
Summing up, we get:
$(ABC_{1})+(BCA_{1})+(CAB_{1}) \ge (ABH)+(BCH)+(CAH) = (ABC)$
QED   :D

@ n.t.tuan: could you explaing why your inequality is easy?

very nice solution  :)

[quote="edriv"]@ n.t.tuan: could you explaing why your inequality is easy?[/quote]

If $ABC$ has an obtuse angle, it is obvious.
Otherwise if you call $A_{0}$ the midpoint of the smallest arc BC and $A'$ the midpoint of the greatest arc BC, then you have $\frac{A_{1}M}{AM}\ge \frac{A_{0}M}{A'M}= tan^{2}\frac{\alpha}{2}$.
Since tan² is convex we have $tan^{2}\frac{\alpha}{2}+tan^{2}\frac{\beta}{2}+tan^{2}\frac{\gamma}{2}\ge 3 tan^{2}30 = 1$, qed.

Benjamin

[quote="iura"][color=darkred]Let $ABC$ be a triangle and $M,N,P$ be th midpoints of sides $BC, CA, AB$. The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$. Show that the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$ is at least equally to the area of the triangle $ABC$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Suppose that the triangle $ABC$ hasn't an obtuse angle. Denote the circumcircle $w=C(O,R)$ of the given triangle, the area $S=[ABC]$ and the points $A'\in w\cap(OM$, $B'\in w\cap (ON$, $C'\in w\cap (OP$. Observe that $MA_{1}\ge MA'$ and $MA'=OA'-OM=R(1-\cos A)$ a.s.o. Therefore, $\sum\frac{[BA_{1}C]}{[ABC]}=$ $\sum\frac{MA_{1}}{MA}=$ $\sum\frac{MA^{2}_{1}}{MA\cdot MA_{1}}\ge$ $\sum\frac{A'M^{2}}{MB\cdot MC}=$ $\sum \frac{4R^{2}(1-\cos A)^{2}}{a^{2}}=$ $\sum\frac{4R^{2}(1-\cos A)^{2}}{4R^{2}(1-\cos^{2}A)}=$ $\sum\frac{1-\cos A}{1+\cos A}=$ $\sum f(A)\ge 3\cdot f\left(\frac{A+B+C}{3}\right)=$ $3\cdot f\left(\frac{\pi}{3}\right)=$ $3\cdot\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=1$, where $f: \left(0,\frac{\pi}{2}\right]\rightarrow \mathrm R$, $f(x)=\frac{1-\cos x}{1+\cos x}=\tan^{2}\frac{x}{2}$ is a convex function (prove easily) and I applied the Jensen's inequality. In conclusion, $\sum [BA_{1}C]\ge [ABC]$.[/color]

[quote="barasawala"][quote="Virgil Nicula"] Observe that $MA_{1}\ge MA'$ [/quote]

Why is this?[/quote]

Given a circle $k$ and a point $P$, how would you find the point on $k$ which lies nearest to $P$?

If you don't find this obvious, here is a more formal argument: $A^{\prime}\in w\cap(OM$ yields $OM+MA^{\prime}=OA^{\prime}$. Since both points $A^{\prime}$ and $A_{1}$ lie on the circumcircle of triangle $ABC$, and the center of this circumcircle is $O$, we have $OA^{\prime}=OA_{1}$. Thus, $OM+MA^{\prime}=OA_{1}$. But the triangle inequality in $\triangle OMA_{1}$ yields $OM+MA_{1}\geq OA_{1}$. Thus, $OM+MA_{1}\geq OM+MA^{\prime}$, so that $MA_{1}\geq MA^{\prime}$, qed.

  darij


Prove that
a/b +b/c +c/a >= (a+c)/(b+c)+ (b+a)/(c+a) + (c+b)/(a+b)
for all positive real a,b,c.
I have nice solution, and you????

Go to Mathlinks archieve and you will find there 2 solution.

Hey, I've read these soluion, but I've nice solution, different them

Hey, this is my solution:
Put x=a/b, y= b/c, z= c/a, then x,y,z>0 and xyz=1.
 inequaly is equivalent: x+y+z >= (1+xz)/(1+x) + (1+xy)/(1+y) + (1+yz)/(1+z).
 after a few trivial computations, we get x2 + y2 + z2 + x2z + y2x + z2y >= 3 + x+ y +z (*)
 Applying Cauchy's, we get x2z + y2x + z2y >=3xyz=3 (1)
 x2 +1 >=2x, y2 +1 >= 2y, z2 +1 >= 2z, then x2 + y2 + z2 + 3 >= 2(x+y+z)
 On the other hand: x+ y +z >=3(xyz)1/3=3, then x2 + y2 + z2 >= x+ y +z (2)
 Combine (1),(2), then (*) is true


if a square matrix has all entries 0 or 1 then the determinant is -1, 0 or 1. (In the exam it was not writen square, just matrix, maybe it was soposed to be false becaause of that) :roll:


I'm going to type up the solutions to the admission test in LaTeX, and was wondering if there was an official LaTeX submission form somewhere, like there is for the USAMTS.

I actually used the USAMTS template for it, by modifying the document "usamts" that's all like DO NOT MODIFY.


I want to get 120+. I got a 106.5 on the AMC 12 A  :oops:
Hope I do well.

144 AMC10B

because that's what I unofficially got on the 10A

I expect to get a zero on the 10B, that's what I'm aiming for.

And... the same amount of effort to get a 0 could be tweaked slightly to get a 150... cough cough.

Not really, since you only have to find an answer that is wrong, so for example for questions that ask "how many blah" if you find $2$ examples, you could circle $1$ instead of having to prove that there can't be $3$.

[quote="ehehheehee"]Or you could circle all of the answers, fill in  A, B, C, D and E.  That should give you a 0.[/quote]

Or you could do what my little brother does on my math notebook:

Shade in the entire page.

Expected score on AMC B: 0

Reason: not taking it.

Yes a 150 is easy to get on the 10 unless you get stumped by one annoying geometry problem and don't guess on it. That's what I did last year.

If I could take it, probably mid or high 120s, but as it is, a 0 (see Xantos C. Guin's post).  I really want to take it, because my 112.5 on the A involved solving about 4 more problems within 5 minutes after the test was over, but my school doesn't offer it, and I wouldn't have any transportation for another location...

133.5 on 12A, aiming for 150 on 12B. *hopes*

[quote="Ignite168"][quote="worthawholebean"][quote="Ignite168"]0 on B, school doesn't offer it, nearest University that does is over an hour away, not much reason too.[/quote]

No it isn't It's like a half hour. :D

I'm not taking the B for the same reason as Ignite. I'm pretty sure I got a 103.5 on the 12A.[/quote]

You don't know how Asians drive.[/quote]
Asians drive?  :huh:

I think I have a chance of getting a 150 unless I get stuck on an annyoing geometry problem.

I'm hoping for the same score on 12B as 10A (aka not great, not bad)

I'm shooting for 135+ so that I only have to get like an 8 on AIME to make USAMO.


logarithms
LaTeX
integration
algebra
linear equation
calculus
calculus computations

Hi I was wondering if anyone can help me with this problem.  Thanks.

Question: Solve $ \displaystyle xy' \equal{} y\ln(xy)$.

Start with the substitution $ y\equal{}e^z$. The equation becomes $ x e^z z'\equal{}e^z(\ln x\plus{}z)$, or $ xz'\equal{}z\plus{}\ln x$. That's a linear equation, which you should know how to handle.

Let $ u \equal{} \ln(xy)$

$ \frac {du}{dx} \equal{} \frac {xy' \plus{} y}{xy}$

$ xy' \plus{} y \equal{} xy\frac {du}{dx}$

$ xy' \equal{} xy\frac {du}{dx} \minus{} y$

$ y\ln(xy) \equal{} xy\frac {du}{dx} \minus{} y$

$ x\frac {du}{dx} \minus{} 1 \equal{} u$

$ x\frac {du}{dx} \equal{} u \plus{} 1$

$ \frac {du}{u \plus{} 1} \equal{} \frac {1}{x}\, dx$

$ \ln(u \plus{} 1) \equal{} \ln x \plus{} \ln A$

$ u \plus{} 1 \equal{} Ax$

$ \ln(xy) \equal{} Ax \minus{} 1$

$ xy \equal{} e^{Ax \minus{} 1}$

$ y \equal{} \frac {e^{Ax \minus{} 1}}{x}$.

We can check this by differentiating.

$ y' \equal{} \frac {xAe^{Ax \minus{} 1} \minus{} e^{Ax \minus{} 1}}{x^{2}}$

$ xy' \equal{} \frac {(Ax \minus{} 1)e^{Ax \minus{} 1}}{x}$

and

$ y \equal{} \frac {e^{Ax \minus{} 1}}{x}$

so

$ xy \equal{} e^{Ax \minus{} 1}$

and

$ xy' \equal{} y\ln(xy)$

(I must practise so I can LaTeX faster.)

[quote="jmerry"]Start with the substitution $ y \equal{} e^z$. The equation becomes $ x e^z z' \equal{} e^z(\ln x \plus{} z)$, or $ xz' \equal{} z \plus{} \ln x$. That's a linear equation, which you should know how to handle.[/quote]

$ xz' \minus{} z \equal{} \ln x\Longrightarrow \left(\frac {z}{x}\right)' \equal{} \frac {\ln x}{x^2}$

$ \therefore \frac {z}{x} \equal{} \int \frac {\ln x}{x^2}\ dx \equal{} \minus{} \frac {1}{x}\ln x \plus{} \int \frac {1}{x}\cdot \frac {1}{x}\ dx$

$ \equal{} \minus{} \frac {\ln x}{x} \minus{} \frac {1}{x} \plus{} A$, yielding $ z \equal{} \minus{} (1 \plus{} \ln x) \plus{} Ax$.

The answer is $ y \equal{} e^z \equal{} \boxed{\frac {e^{Ax} \minus{} 1}{x}}$ which is AndrewTom's result.


let $ a_{n}$ a sequence such that:
$ a_{1}\equal{}1$ and  $ a_{n\plus{}1}\equal{}\frac{1}{16}(1\plus{}4a_{n}\plus{}\sqrt{1\plus{}24a_{n})}$

find $ a_{n}$ in function of $ n$.

[hide]you can use a sequence $ b_{n}$ such that: $ b_{n}^2\equal{}1\plus{}24a_{n}$ and $ b_{n}\geq 0$ [/hide]


What is the smallest integer k, so that $ (k^2 \plus{} 5k \minus{} 25)/ (k\plus{}1)$ is also an integer?

[hide="hint"]
rewrite the expression as a polynomial plus a fraction with denominator x+1
[/hide]

[hide="solution"]$ \frac {k^2 \plus{} 5k \minus{} 25}{k \plus{} 1} \equal{} \frac {k^2 \plus{} 5k \plus{} 4 \minus{} 29}{k \plus{} 1} \equal{} k \plus{} 4 \plus{} \frac { \minus{} 29}{k \plus{} 1}$in order for the original expression to be an integer, the the fraction $ \frac { \minus{} 29}{k \plus{} 1}$ must be an integer, meaning that k+1 must divide -29.
the factors of -29 are -29, -1, 1, and 29; and the respective values of k are -30, -2, 0, and 28
$ \boxed{ \minus{} 30}$ is the smallest of these[/hide]


Let $ a,b,c$ be the lengths of the sides of a triangle.
Prove that $ \frac{a}{\sqrt{2b^2\plus{}2c^2\minus{}a^2}}\plus{}\frac{b}{\sqrt{2c^2\plus{}2a^2\minus{}b^2}}\plus{}\frac{c}{\sqrt{2a^2\plus{}2b^2\minus{}c^2}}\ge \sqrt{3}$

hint :F(x)=1/(x)^1/2 is convex and let a+b+c=1 :wink:

I think, the following reasoning is better:
$ \sum_{cyc}\frac{a}{\sqrt{2b^2\plus{}2c^2\minus{}a^2}}\equal{}\sum\frac{\sqrt3}{3\sqrt{1\cdot\left(\frac{2(a^2\plus{}b^2\plus{}c^2)}{3a^2}\minus{}1\right)}}\geq$
$ \geq\sum_{cyc}\frac{\sqrt3\cdot2}{3\left(1\plus{}\frac{2(a^2\plus{}b^2\plus{}c^2)}{3a^2}\minus{}1\right)}\equal{}\sum_{cyc}\frac{\sqrt3 a^2}{a^2\plus{}b^2\plus{}c^2}\equal{}\sqrt{3}.$


geometry
perimeter
analytic geometry
function
inequalities
circumcircle
geometric transformation

Prove that among all the triangles on the same base and between the same parallels an isosceles triangle has the least perimetre.

 :( can anyone help me out? :(

Without loss of generality, let these lines be described by the equations $y = 0$ and $y = 1$

WLOG, Label points $(-1, 0)$ and $(1,0)$ A and B, respectively.
We'll use the distance formula for this proof.

The distance from A to B is clearly two.
Now pick a point on the line $y = 1$ and call it C. Clearly, it's coordinates will be $(x,1)$.

The perimeter of the triangle will be the sum of the distance between these points.

The distance between A and C is $\sqrt{1+( X-(-1))^{2}}$
or
$\sqrt{x^{2}+2x+2}$

The distance between B and C is $\sqrt{1+(x-1)^{2}}$
or
$\sqrt{x^{2}-2x+2}$

The total perimeter is then
$2+\sqrt{x^{2}+2x+2}+\sqrt{x^{2}-2x+2}$
We wish to minimize this.
There are a few ways to do this, but the easiest is with logic.
Note that the minimum for this particular triangle will be at least two, because the square root function always returns a non-negative value. In order to minimize the perimeter, we need to try and minimize $\sqrt{x^{2}+2x+2}+\sqrt{x^{2}-2x+2}$.

This is minimized when $x = 0$ (trivial).
Therefore, the perimeter is minimized when point C lies on the Y axis.
This is halfway between points A and B. Because AC = BC,  triangle ABC must be isosceles. Therefore, the minimum possible perimeter for a triangle between two parallels with at least one point on each, and a common base, is when the triangle is isosceles.

:maybe: Can anyone solve it by pure geometry without using coordinates? :maybe:

[hide="definitely overkill, but it works anyway"]
Let the base be $\overline{AB}$ and the parallel line be $l$. Suppose the perpendicular bisector of $\overline{AB}$ intersects $l$ at $C$ and the circumcircle of $\triangle ABC$ at $C$ and $D$.

For any point $P$ on $l$, Ptolemy's Inequality on quadrilateral $PADB$ gives
\[PA\cdot BD+PB\cdot AD\geq AB\cdot PD \]
and since $AD=BD$, this reduces to $PA+PB\geq \frac{AB\cdot PD}{AD}$. But $PD=\sqrt{PC^{2}+CD^{2}}\geq CD$, so
\[PA+PB\geq \frac{AB\cdot CD}{AD}\]
Finally, using Ptolemy's on quadrilateral $CADB$ (this time it's an equality because the quadrilateral is cyclic), the right hand side of the above inequality is equal to $CA+CB$. Therefore $PA+PB+AB\geq CA+CB+AB$, so perimeter minimized at $C$, where $CA=CB$.
[/hide]

[size=59]edit: of course it's overkill, considering what tools i used[/size]

[quote="t0rajir0u"][quote="parijat"]on the same base and between the same parallels[/quote]Not really sure what you mean here.  Could you elaborate?[/quote]i think he means "all triangles that have the same length base and the last point on a line parallel to the base"

[hide="okay found a better solution"]
I keep the same $A,B,C,l$ as above, and let $X$ be the reflection of $B$ over $l$. Then for any point $P$ on $l$, triangle inequality gives
\[PA+PB=PA+PX\geq AX=CA+CB\]
and the problem is solved.
[/hide]

Yay, that's the solution that Minkowski's suggests.  :)

well [b]I really appreciate all the approaches to the problem[/b].
I found the sum in the chapter related to circumcircle, incircle and related topics. So, I personally feel the 2nd solution is perhaps what the authors wanted.  :|

Here's an approach I thought of, but I can't seem to finish it off nicely.

Let the base be $\overline{BC}$ and the parallel line be $l.$ $A$ is the point on $l$ such that triangle $ABC$ is isosceles and $A'$ is some other point on $l.$

Since triangles $ABC$ and $A'BC$ have the same area, we mus have $r_{1}s_{1}=r_{2}s_{2}.$ So, if $ABC$ has the smallest perimeter ($2s$) of all triangles $A'BC,$ it must have the largest inradius.

Now how do I show that it has the largest inradius?  :P

consider an ellipse...it is the locus of points such that the sum is constant, if we intersect this with the parallel lines, we get the last vertex of our triangle...we want to make the sum as small as possible...so we shrink the ellipse...eventually it is tangent to the line, due to symmetry, the triangle is isosceles

this is the same as the second solution...using the reflective properties of an ellipse, etc

[quote="t0rajir0u"][quote="parijat"]on the same base and between the same parallels[/quote]

Not really sure what you mean here.  Could you elaborate?

Edit:  [b]DFT:[/b]

$\sqrt{(1+x)^{2}+1^{2}}+\sqrt{(1-x)^{2}+1^{2}}\ge \sqrt{2^{2}+2^{2}}= 2 \sqrt{2}$

By Minkowski's inequality (also known as the Triangle Inequality... hint, hint.)[/quote]

Which is attained when [b]x = 0[/b];
then $y= 2\sqrt{2}$.

How is what I said incorrect?

It wasn't incorrect, but you didn't support your claim that the minimum occurs when $x = 0$.  Phrasing it in terms of Minkowski's inequality also strongly suggests scorpius' second solution.

Fair enough.

And the Minkowski's solution works well too.
Good point with the ellipse, Al. I think I'll use that one on some other problems too.

I think these two probs which were sent by u r same.
http://www.mathlinks.ro/Forum/viewtopic.php?p=790905#790905#790905#790905#790905
just add base=BC
then u'll get perimeter.


calculus
integration
function
geometry
trigonometry
search
calculus computations

Even function, might as well integrate from $ 0$ to $ a$ and double it.

Question: is that outer exponent supposed to be $ \frac{2}3$ or should it really be $ \frac{3}2$ ?

[quote="Kent Merryfield"]Even function, might as well integrate from $ 0$ to $ a$ and double it.

Question: is that outer exponent supposed to be $ \frac{2}{3}$ or should it really be $ \frac{3}{2}$ ?[/quote]

its $ \frac{2}{3}$.. thx anyway!

[quote="Carcul"]This integral seems to be related to the area of the astroide, and in that case the exponent should be 3/2. Considering that, then make the substitutions $ x \equal{}\frac{a}{(1\plus{}t^{2})^{3/2}}$ and then $ t \equal{}\tan u$, [hide="and your integral should become"]$ I \equal{} 6a^{2}\int^{\pi /2}_{0}\sin^{4}u\cos^{2}u\,du$.[/hide][/quote]
its exactly whats that about.. thx a lot!

[quote="Kent Merryfield"]Even function, might as well integrate from $ 0$ to $ a$ and double it.

Question: is that outer exponent supposed to be $ \frac{2}{3}$ or should it really be $ \frac{3}{2}$ ?[/quote]

im the most stupid man in the world! is $ \frac{3}{2}$.. sorry!

Look at the curve parameterized by $ \begin{cases}x \equal{} a\cos^{3}\theta\\ y \equal{} a\sin^{3}\theta\end{cases}$ for $ \theta\in[0,2\pi].$

This is a simple closed curve symmetric about both axes. In the first quadrant, we have $ x^{\frac{2}{3}}\plus{}y^{\frac{2}{3}}\equal{} a^{\frac{2}{3}}$ or $ y \equal{}\left(a^{\frac{2}{3}}\minus{}x^{\frac{2}{3}}\right)^{\frac{3}{2}}.$

Your integral $ I$ is the top half of the area enclosed by this curve; therefore if we can compute the area enclosed by the curve, we need only divide by $ 2$ to obtain $ I.$

If $ C$ is a simple closed counterclockwise curve, one formula for the area enclosed by it is

$ \text{area}\equal{}\frac{1}{2}\int_{C}(\minus{}y\,dx\plus{}x\,dy).$

(The formula can be verified either by Green's Theorem or by more elementary and tedious methods.)

In this case $ \minus{}y\,dx\plus{}x\,dy \equal{} 3a^{2}(\cos^{2}\theta\sin^{4}\theta\plus{}\cos^{4}\theta\sin^{2}\theta)\,d\theta,$ which simplifies to $ 3a^{2}\cos^{2}\theta\sin^{2}\theta\,d\theta.$

Hence, $ 2I \equal{}\frac{3a^{2}}{2}\int_{0}^{2\pi}\cos^{2}\theta\sin^{2}\theta\,d\theta \equal{}\frac{3a^{2}}{8}\int_{0}^{2\pi}\sin^{2}(2\theta)\,d\theta$

By the "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=598460340&t=79074]yin-yang argument[/url]", the integral of the square of a sine or cosine over an interval that includes only whole quadrants is one half the length of the interval. Hence:

$ 2I \equal{}\frac{3a^{2}\pi}{8}$ or $ I \equal{}\frac{3a^{2}\pi}{16}.$

Another way to express this is that the area of the astroid is $ \frac{3}{8}$ of the area of the circle it is inscribed in.


what is the easiest and most simple way of finding the formula for Summation 1^n +2^n +3^n + . . . + k^n =??. What does this mean?

"The sum of kth powers of first n natural number fk(n) is a polynomial in n of degree k + 1. For each polynomial fk(n), the sum of its coefficients equals 1."

How could i explain that in other words??
Thanks

Ben

The summation basically asks for the sum of all the numbers to some power (is this what you wanted). For example, the sum of $1+2+3+...+n = \frac{(n+1)(n)}{2}$ (in this case n = 1).

I'm not sure about your second question also, are you asking for a proof? I'll give you an example of what it means.

For a summation when k = 5 for example, the closed form summation would be of the form $Ax^{6}+Bx^{5}+Cx^{4}+Dx^{3}+Ex^{2}+Fx$ (There is no constant term, which your description omits). Also, $A+B+C+D+E+F = 1$ There are a few more rules in finding these kinds of summations. For example, $A = \frac{1}{k+1}$

"the closed form summation would be of the form"

I learned this stuff in another language, thus i dont really know what  closed form means.. could someone explain? I live in Chile,South America :P
Thanks

Ben

Yea my wording wasn't perfect. The closed form of a summation is basically an easy formula you can obtain that, instead of having to add all the individual terms, you can just put in the number for n.

For example, the closed form of the summation $1+2+3+...+n$ is $\frac{(n+1)n}{2}$ Hopefully that was more clear.

thanks man.. i got it right now... you know im checking the telescopical way of getting the algebraic expresion for this summation... is this a good method???


Determine all functions $f : \mathbb N_0 \rightarrow \mathbb N_0 - \{1\}$ such that
\[f(n + 1) + f(n + 3) = f(n + 5)f(n + 7) - 1375, \qquad \forall n \in \mathbb N.\]

See a similar problem [url=http://www.mathlinks.ro/viewtopic.php?t=169128]here[/url].

We have $ f(n\plus{}1)\plus{}f(n\plus{}3)\equal{}f(n\plus{}5)f(n\plus{}7)\minus{}1375$ and $ f(n\plus{}3)\plus{}f(n\plus{}5)\equal{}f(n\plus{}7)f(n\plus{}9)\minus{}1375$. Thus $ f(n\plus{}5)\minus{}f(n\plus{}1)\equal{}f(n\plus{}7)(f(n\plus{}9)\minus{}f(n\plus{}5))$. But $ f(n\plus{}9)\minus{}f(n\plus{}5)\equal{}f(n\plus{}11)(f(n\plus{}13)\minus{}f(n\plus{}9))$ and hence $ f(n\plus{}5)\minus{}f(n\plus{}1)\equal{}f(n\plus{}7)f(n\plus{}11)(f(n\plus{}13)\minus{}f(n\plus{}9)$. Continuing this, we have
\[ f(n\plus{}5)\minus{}f(n\plus{}1)\equal{}f(n\plus{}7)f(n\plus{}11)\ldots f(n\plus{}4k\minus{}1)(f(n\plus{}4k\plus{}1)\minus{}f(n\plus{}4k\minus{}3)),
\]for any positive integer $ k$. If $ f(n\plus{}5)\minus{}f(n\plus{}1)$ does not equal $ 0$, then none of $ f(n\plus{}7),f(n\plus{}11),\ldots,f(n\plus{}4k\minus{}1),(f(n\plus{}4k\plus{}1)\minus{}f(n\plus{}4k\minus{}3))$ may equal $ 0$. But they do not equal to $ 1$, and hence $ f(n\plus{}7),f(n\plus{}11),\ldots,f(n\plus{}4k\minus{}1)\ge2,f(n\plus{}4k\plus{}1)\minus{}f(n\minus{}4k\minus{}3)\ge1$. So $ f(n\plus{}5)\minus{}f(n\plus{}1)\ge2^{k\minus{}1}$ for any positive integer $ k$, which is impossible. Hence $ f(n\plus{}5)\minus{}f(n\plus{}1)$ must equal to $ 0$, i.e., $ f(n\plus{}5)\equal{}f(n\plus{}1)$.

So we have $ f(4)\equal{}f(8)\equal{}\ldots$, $ f(5)\equal{}f(9)\equal{}\ldots$, $ f(2)\equal{}f(6)\equal{}f(10)\equal{}\ldots$, $ f(3)\equal{}f(7)\equal{}f(11)\equal{}\ldots$. Let $ f(4)\equal{}a,f(5)\equal{}b,f(2)\equal{}c,f(3)\equal{}d$. We have $ a\plus{}c\equal{}ac\minus{}1375$ and $ b\plus{}d\equal{}bd\minus{}1375$. It is easy to solve these equations. $ f(0)$ and $ f(1)$ can be any non negative integer unequal to $ 1$.


limit geometry parallelogram function circumcircle induction symmetry

I hope you enjoy this
1) Find the smallest integer number n>1 such that:
	$ M \equal{} \frac{{1^2  \plus{} 2^2  \plus{} ... \plus{} n^2 }}{n}$ is a perfect square

2) Let sequence  $ \{ u_n \}$: $ u_1  \equal{} a > 1$ ; $ u_{n \plus{} 1}  \equal{} u_n^2  \minus{} u_n  \plus{} 1,\forall n \ge 1$
Find:
$ \lim (\frac{1}{{u_1 }} \plus{} \frac{1}{{u_2 }} \plus{} ... \plus{} \frac{1}{{u_n }})$

3) Let $ \Delta ABC$: AB=AC. M is a point in $ \Delta ABC$ such that  $ \angle BMC \equal{} 90^0  \plus{} \frac{{\widehat A}}{2}$. Construct the paralellogram MKBD and MHCE; K in AB, H in AC; D,E in BC. N is the intersection point of KD and HE. Find the locus of N?

4) $ a,b,c \in R$ and  $ \left\{ \begin{array}{l}
 a^2  \plus{} c^2  \equal{} 10 \\
 b^2  \plus{} 2b(a \plus{} c) \equal{} 60 \\
 \end{array} \right.$
A=b(c-a)
Find Max and Min of A

5)Find all the finited set have at least 2 elements with all the element is the positive integers.And if a,b are the two element of A(b>a),  $ \frac{{a^2 }}{{b \minus{} a}}$ is also the element of A.

6)
Find all the function  $ f: R \to R$ such that
$ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(f(y)) \minus{} x)\forall x,y \in R$

How you can solve Problem 3 Hung ? :blush:  :blush:
This is the second day, post the first day if you have. I hope MathVnpro would write hís solution. :)

I'll post the solution but not now.Let's wait for everyone sovle it.

[hide="one"]

$ \frac {(n \plus{} 1)(2n \plus{} 1)}{6} \equal{} k^2$, $ k\in \mathbb{N}$

Let $ n \plus{} 1 \equal{} m$ and we get $ 2m^2 \minus{} m \equal{} 6k^2 \qquad(1)$

Since $ m$ is an integer, the discriminant of (1) is a perfect square
$ \therefore 1 \plus{} 48k^2 \equal{} a^2 \Leftrightarrow a^2 \minus{} 48k^2 \equal{} 1$

We have a pell equation, and the smallest solution is clearly $ a \equal{} 7, k \equal{} 1$

$ (7 \plus{} \sqrt {48})^2 \equal{} 97 \plus{} 14\sqrt {48}$, but this gives $ m \equal{} \frac {1 \pm a}{4} \equal{} \frac {49}{2}$

$ (7 \plus{} \sqrt {48})^3 \equal{} 1351 \plus{} 260\sqrt {48}$ giving $ m \equal{} \frac {1 \plus{} 1351}{4} \equal{} 338$

So $ n \equal{} 337$ and $ \frac {(337 \plus{} 1)(2\cdot 337 \plus{} 1)}{6} \equal{} 195^2$ which works

$ n \equal{} \boxed{337}$

[/hide]

[hide="Hints for 3"]It's the circumcircle
[hide="More hints"]A, M, and N are collinear[/hide][/hide]

[hide="another similar solution for 1"]
${ \frac {(n + 1)(2n + 1)}{6} = k^2}$
since $ (n + 1)(2n + 1)$ are coprime we get,
$ n + 1 = 2a^{2}$ and $ 2n + 1 = 3b^{2}$
if $ 2a = k$ eliminating $ n$ we get,
$ k^2 - 3b^2 = 1$
solving, $ (2,1)(7,4)(26,15)$
since $ b$ is odd, we take $ 26,15$ to get solution $ 337$

The other case :
$ n + 1 = 6a^2$
$ 2n + 1 = b^2$
It gives $ 12a^{2} = b^{2} + 1$
This
has no solutions since there is no square leaving a remainder of 11 on division with 12.
[/hide]

5. Partial, I might add more as I figure out more about the problem.
[hide]
Let $ A$ be the set of all sets $ S$ with the properties listed.
(It has at least 2 elements, all of its elements are positive integers, and if $ a,b\in S$ and $ b > a$, then $ \frac {a^2}{b \minus{} a}\in S$).
Part 1: All sets of the form $ \{k, 2k\}$, $ k\in \mathbb{N}$, are in $ A$.
Proof: Since we are working with positive integers, $ 2k > k$. Then the only possible choices of $ a$ and $ b$ are $ a \equal{} k$ and $ b \equal{} 2k$. Then $ \frac {a^2}{b \minus{} a} \equal{} \frac {k^2}{2k \minus{} k} \equal{} k$, which is already in the set, so the set has all the properties.
[/hide]

[hide="number 2"]
$ u_{n \plus{} 1} \equal{} u_n^2 \minus{} u_n \plus{} 1$

$ u_{n \plus{} 1} \minus{} 1 \equal{} u_n(u_n \minus{} 1)$

$ \frac {1}{u_{n \plus{} 1} \minus{} 1} \equal{} \frac {1}{u_n(u_n \minus{} 1)} \equal{} \frac {1}{u_n \minus{} 1} \minus{} \frac {1}{u_n}$

$ \therefore \frac {1}{u_n} \equal{} \frac {1}{u_n \minus{} 1} \minus{} \frac {1}{u_{n \plus{} 1} \minus{} 1}$

So by telescoping series

$ \sum_{k \equal{} 1}^{\infty} \frac {1}{u_k} \equal{} \sum_{k \equal{} 1}^{\infty} \frac {1}{u_k \minus{} 1} \minus{} \frac {1}{u_{k \plus{} 1} \minus{} 1} \equal{} \frac {1}{u_1 \minus{} 1}$[/hide]

[hide="number 6"]
$ f(f(x)\plus{}y) \equal{} 2x\plus{}f(f(f(y))\minus{}x) \qquad(1)$

[hide="Proof that f is surjective"]
let $ y\equal{}\minus{}f(x)$ and rearrange to get

$ f(0)\minus{}2x \equal{} f[ f(f(\minus{}f(x)))\minus{}x]$

Since $ f(0)$ is fixed, we can choose $ x$ such that $ f(0)\minus{}2x$ is any real number, hence $ f$ is surjective
[/hide]

[hide="proof that f is injective"]
assume $ f(a)\equal{}f(b) \Longrightarrow f[f(f(a))\minus{}x]\plus{}2x \equal{} f[f(f(b))\minus{}x]\plus{}2x$

$ \therefore f(f(x)\plus{}a)\equal{}f(f(x)\plus{}b$, but since $ f$ is surjective, let $ f(x)\equal{}y\minus{}a$

$ \therefore f(y) \equal{} f(y\plus{}b\minus{}a)$ and by induction $ f(y)\equal{}f(y\plus{}n(b\minus{}a))$

Now plugging into $ (1)$ gives
$ f(f(x)\plus{}y)\equal{}f(f(x\plus{}b\minus{}a)\plus{}y) \equal{} f[f(f(y))\plus{}x\plus{}b\minus{}a] \plus{} 2x \plus{} 2(b\minus{}a) \equal{} f[f(f(y))\plus{}x]\plus{}2x \plus{} 2(b\minus{}a)$

So $ f(f(x)\plus{}y) \equal{} f[f(f(y))\minus{}x] \plus{} 2x \plus{} 2(b\minus{}a)$, but comparing this with $ (1)$ shows that $ 2(b\minus{}a)\equal{}0$

So $ f(a)\equal{}f(b)\Longrightarrow a\equal{}b$ and $ f$ in injective
[/hide]

[hide="proof that f(x)=x"]
Letting $ x\equal{}y\equal{}0$ gives $ f(f(0))\equal{}f(f(f(0)))$, but fince $ f$ is injective $ f(0)\equal{}f(f(0))\Longrightarrow f(0)\equal{}0$

Letting $ x\equal{}0 \Longrightarrow f(y)\equal{}f(f(f(y)))\Longrightarrow f(f(y))\equal{}y \qquad(2)$

Letting $ y\equal{}0 \Longrightarrow f(f(x))\equal{}f(\minus{}x)\plus{}2x \qquad(3)$

From $ (2)$ and $ (3)$, $ x\equal{}f(f(x))\equal{}f(\minus{}x)\plus{}2x \Longrightarrow f(\minus{}x)\equal{}\minus{}x$

And $ \boxed{f(x)\equal{}x}$ is the only solution
[/hide][/hide]

[hide="number 4"]
$ \left\{\begin{array}{l}a^{2} \plus{} c^{2} \equal{} 10 \\
b^{2} \plus{} 2b(a \plus{} c) \equal{} 60 \\
\end{array}\right.$

Putting the equations together
$ (b \plus{} a \plus{} c)^2 \minus{} (a \plus{} c)^2 \equal{} 60 \Longrightarrow b \equal{} \minus{} (a \plus{} c) \pm \sqrt {60 \plus{} (a \plus{} c)^2}$

$ A \equal{} b(a \minus{} c) \equal{} c^2 \minus{} a^2 \pm (c \minus{} a)\sqrt {60 \plus{} (a \plus{} c)^2}$

Now set $ c \minus{} a \equal{} x, c \plus{} a \equal{} y$ and $ (x \plus{} y)^2 \plus{} (x \minus{} y)^2 \equal{} 40 \Longrightarrow x^2 \plus{} y^2 \equal{} 20$

$ A \equal{} x(y \pm \sqrt {60 \plus{} y^2})$

First consider the the case where the $ \pm$ is $ \plus{}$
By AM-QM,
$ \sqrt {y^2} \plus{} 4\sqrt {\frac {60 \plus{} y^2}{16}} \le \sqrt {5}\sqrt {y^2 \plus{} \frac {60 \plus{} y^2}{4}} \equal{} \frac {5}{2}\sqrt {12 \plus{} y^2} \qquad(1)$

From $ (1)$ we have
$ A \equal{} \frac {5}{2}\sqrt {20 \minus{} y^2}\left(y \plus{} \sqrt {60 \plus{} y^2}\right) \le \frac {5}{2}\sqrt {12 \plus{} y^2}\sqrt {20 \minus{} y^2}$

By AM-GM

$ \frac {5}{2}\sqrt {12 \plus{} y^2}\sqrt {20 \minus{} y^2} \le \frac {5}{2}\frac {12 \plus{} y^2 \plus{} 20 \minus{} y^2}{2} \equal{} 40$

Equality when $ y \equal{} \pm 2$

So the Maximum is $ \boxed{40}$ equality when $ (a,b,c) \equal{} (1, \minus{} 3,10), ( \minus{} 1,3, \minus{} 10)$

Now consider the case where $ \pm$ is $ \minus{}$
$ A \equal{} x(y \minus{} \sqrt {60 \plus{} y^2})$

But from symmetry we only have to negate our previous solutions

Minimum $ \equal{} \boxed{ \minus{} 40}$ when $ (a,b,c) \equal{} ( \minus{} 3,1,10), (3, \minus{} 1, \minus{} 10)$

[/hide]

[hide="and number 5"]Let $ x \equal{} \text{min}\,(S)$, and let $ y \in S$ with $ y \neq x$. Then $ z : \equal{} \frac {x^2}{y \minus{} x} \in S$, and $ z \equal{} \frac {x^2}{y \minus{} x} \ge x \Longrightarrow y \le 2x,\ \forall y \in S$. Hence $ z \equal{} \frac {x^2}{y \minus{} x} \le 2x \Longrightarrow y \ge \frac {3x}{2}$. Suppose $ z \neq x$; then the same bounds for $ y$ apply to $ z$. If $ z \neq y$, wlog $ y < z$; then $ \frac {3x}{2} \le y < z \le 2x$, and $ 2x \equal{} \frac {x^2}{x/2} < \frac {y^2}{z \minus{} y} \in S$, contradiction. Thus either $ y \equal{} z \equal{} \frac {x^2}{y \minus{} x} \Longrightarrow y^2 \minus{} yx \minus{} x^2 \equal{} 0 \Longrightarrow (2y \minus{} x)^2 \equal{} 5x^2$, infinite descent gives no solutions, or $ x \equal{} z \equal{} \frac {x^2}{y \minus{} x} \Longrightarrow y \equal{} 2x$. Hence the set of solutions is $ \boxed{\{n,2n\},\, \forall n \in \mathbb{Z}_{ > 0}},$ which we easily verify work.  [/hide]

Here is the solution for 3
It is easy to prove
$ \angle KBM \equal{} \angle MCB \equal{} \angle HMC$
then
$ \begin{array}{l} \Delta BKM \sim \Delta MHC \\
\Rightarrow BK*HC \equal{} KM*MH \\
\Rightarrow \Delta KBD \sim \Delta ECH \\
\Rightarrow \angle KDB \equal{} \angle EHC \\
\Rightarrow \angle DKM \equal{} \angle MEH \\
\end{array}$
Then MKNB can be inscribed in a circle
$ \angle KNM \equal{} \angle KEM \equal{} \angle KBM$
Then KMNB can be inscribed in a circle
$ \begin{array}{l} \angle MNB \equal{} \angle AKH \\
\Rightarrow \angle BNC \plus{} \angle BAC \equal{} 180^0 \\
\end{array}$
Can you give any comment about this test?Which is the hardest to you?
@to Ocha how do you have$ \therefore f(f(x)\plus{}a)\equal{}f(f(x)\plus{}b)$ ?


ratio limit geometry perimeter integration

Um, didn't you already post something like this in Other Problem Solving Topics?

isn't $\pi$ derived from the ratio of the circumference to the diameter?

Also,
$\pi$ /4 = 1 - 1/3 + 1/5 - 1/7 + ....
$\pi$ /2 = 2/1 x 2/3 x 4/3 x 4/5 x 6/5 x 6/7 x 8/7 x 8/9

[quote="amirhtlusa"]the following was derived by Rumanujan:
$ \frac{1}{\pi} = \frac{\sqrt{8}}{9801} \sum_{n=0}^\infty \frac{(4n)!(1103 + 26390n)}{(n!)^4396^{4n}} $[/quote]

amazing...  :P

[quote="NoSoupForYou"]Is $\pi=4\displaystyle\sum_{i=1}^{\infty}\frac{1}{2i-1}(-1)^{i+1}$?[/quote]

Yeah, I think this is the same as the first series in my post:
$\displaystyle \frac tion {\pi} / {4}= 1-\frac tion {1}/{3}+\frac tion {1}/{5}-\frac tion{1}/{7}... $

P.S. This series converges [i]real[/i] slowly, for 10 decimals you'd have to calculate $i$=166

Well, you would simply find the $\lim{l}_n$ of the perimeter of a regular $2^n\text{-gon}$ (because these are easy to construct) and divide it by twice the radius.  :D  Actually that's how the circumference (and hence $\pi$) is defined.

Also, $\pi=\left(\int_{-\infty}^{\infty} e^{-x^2} \ dx \right)^2$


If I have two summations in a fraction (one as the numerator and one as the denominator), how do I make the upper and lower and upper bounds appear below and above the Sigma, rather than to the side?

Use \sum\limits - $ \frac {\sum\limits_{n \equal{} 1}^{\infty}2^{ \minus{} n}}{\sum\limits_{n \equal{} 1}^{\infty}2^{ \minus{} n}}$ . Also works with integrals: $ \frac{\int\limits_0^1 x\,dx}{\int\limits_0^1 x\,dx}$


How many distinct permutations can be formed from the letters in the name NATHAN?

There are $ \frac{6!}{2!2!}\equal{}\boxed{180}$ ways to arrange the letters in NATHAN.


I was wondering whether anybody here reads the wall street journal? If so, what sections do you read and how often? Do they help you?

I have a subscription, but only get to read it as much as I'd like about 2-3 days a week.  Its a fantastic source of information on the corporate world.

When I was in high school, I had a couple of friends who read it on a regular basis.  I wish I had been reading it back then.  If not that newspaper, then find one you enjoy...just a great habit to get into.

And on a math-related note, there have been some great pieces in WSJ recently (past few months) on the overall decline of mathematics education across the country.

I have seen some of those. When you read the journal, do you read anything in particular?

I read wall street journal europe and financial times. It seems that politics and economy are the most important parts of journals in general and they depend on each other quite heavily.  :)

Did anyone see the article about drexel and its president. Its the college I go to.

typically, i read the front page and personal journal about once every 2 weeks. :) i usually read BBC news on the internet.


algebra
polynomial
coefficients
IMO Shortlist

Let $ P(x)$ be a polynomial with real coefficients such that $ P(x) > 0$ for all $ x \geq 0.$ Prove that there exists a positive integer n such that $ (1 \plus{} x)^n \cdot P(x)$ is a polynomial with nonnegative coefficients.

[hide="proof idea"]The statement clearly holds for linear polynomials (trivially with $ n \equal{} 0$).  Next we prove it for quadratic polynomials of the form $ Q(x) \equal{} x^2 \minus{} bx \plus{} c$, $ b < 2\sqrt {c}$, $ c > 0$.  Let $ a_k$ denote the coefficient of the $ k$th power of $ (1 \plus{} x)^n Q(x)$, then
\[ a_0 \equal{} c \\
a_1 \equal{} nc \minus{} b \\
a_{n \plus{} 1} \equal{} n \minus{} b \\
a_{n \plus{} 2} \equal{} 1 \\
a_k \equal{} {n \choose {k \minus{} 2}} \minus{} b{n \choose {k \minus{} 1}} \plus{} c{n \choose k} \ \ \ \ \text{for } 2\leq k \leq n
\]
now if we can prove there is an $ n$ so that $ a_k > 0$ for each $ 0\leq k\leq n \plus{} 2$, we're done.  This can be done, for example, by expressing $ {n \choose {k \minus{} 2}}$ and $ {n \choose {k \minus{} 1}}$ as $ {n \choose k}$ multiplied by the appropriate coefficient, and using the fact that $ b < 2\sqrt {c}$.  The resulting inequality, though messy, is manageable.  Maybe there's a cleaner way to do this step, which I suppose is where the meat of the proof is.

Now having proved it for the quadratics above, it follows immediately for arbitrary polynomials with real coefficients, since any real polynomial satisfying $ P(x) > 0$ for $ x\geq 0$ may be expressed as the product of linear polynomials and quadratic polynomials of the given form. [/hide]

[b]Solution:[/b]
Obviously,coefficient in front of $ x^n$,where $ n$ is degree of $ P$,is positive,otherwise when $ x\rightarrow\infty$ our polynomial $ P\rightarrow \minus{} \infty$.So we can divide all polynomial to this coefficient and we may assume that $ P(x)$ is of the form:
\[ x^{n} \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \dots \plus{} a_0
\]
We can proceed by induction on the degree of $ P(x)$ to prove the statement.
We will check the base later,let's at first show how can induction be used here:
Suppose that we have proved the problem for all quadratic and linear polynomials and for all polynomials of degree $ \leq n$,satisfying to the necessary conditions.We need to prove the statement for $ n \plus{} 1$.

It is well-known fact that $ P(x)$ can be represented in the following form:
\[ P(x) \equal{} (x \plus{} \alpha_1)(x \plus{} \alpha_2)\dots(x \plus{} \alpha_k)(x^2 \minus{} p_1x \plus{} q_1)\dots(x^2 \minus{} p_l \plus{} q_l)
\]
,where $ k \plus{} 2l \equal{} n \plus{} 1$ and polynomials $ x^2 \minus{} p_ix \plus{} q_i$ are always positive.
Denote
\[ Q(x) \equal{} (x \plus{} \alpha_1)(x \plus{} \alpha_2)\dots(x \plus{} \alpha_k)
\]

\[ R(x) \equal{} (x^2 \minus{} p_1x \plus{} q_1)\dots(x^2 \minus{} p_l \plus{} q_l)
\]
Since $ x^2 \minus{} p_ix \plus{} q_i > 0$ for all $ x\in\mathbb{R}$,it follows that there exist some $ m_i$ for each $ x^2 \plus{} p_ix \plus{} q_i$,such that polynomial
\[ (1 \plus{} x)^{m_i}(x^2 \plus{} p_ix \plus{} q_i)
\]
has nonnegative coefficients.
On the other hand,since $ P$ has no positive roots it follows that $ \alpha_1,\alpha_2,\dots,a_k > 0$,so $ Q(x)$ has nonnegative coefficients,finally $ (1 \plus{} x)^{m_1 \plus{} m_2 \plus{} \dots \plus{} m_l}\cdot P(x)$ has nonnegative coefficients.

The last step is to show why statement true for quadratic trinomials.
Let $ P(x) \equal{} x^2 \minus{} bx \plus{} c$,where $ b,c > 0$.We may also assume that $ P$ has no roots,because if it has then they are negative and $ P$ has nonnegative coefficients and we will have nothing to prove.

Consider polynomial $ F(x) \equal{} (1 \plus{} x)^{n}P(x)$.We can represent $ F(x)$ as follows:

$ F(x) \equal{} x^{n \plus{} 2} \plus{} x^{n \plus{} 1}\cdot\left(\binom{n}{1} \minus{} b\right) \plus{} x^{n}\left(\binom{n}{2} \minus{} b\cdot\binom{n}{1} \plus{} c\right) \plus{} \dots \plus{} x^{n \minus{} k}\left(\binom{n}{k \plus{} 2} \minus{} b\cdot\binom{n}{k \plus{} 1} \plus{} c\cdot\binom{n}{k}\right) \plus{} x\left(c\cdot\binom{n}{1} \minus{} b\right) \plus{} c$

But for enough large value of $ n$ all the coefficients will be positive.So the base is checked and the proof is complete.

P.S:
I didn't read [b]JoeBlow[/b]'s post above and I am sorry if has the same idea I used in my proof...

Since both of above proofs are not complete how to manage quadratic polynomial case, I'll prove this in detail here.\\\\
Consider
$$(1+x)^{m}(x^2 + bx + c)$$
where $b^2 - 4c < 0$. Note that we have $c > 0$ and $|b| < 2\sqrt{c}$. Observe that
$$(1 + x)^m(x^2 + bx + c)= \sum_{k=0}^{m+2}\left(\binom{m}{k-2} + \binom{m}{k-1}b + \binom{m}{k}c\right)x^k=\sum_{k=0}^{m+2}\frac{m!}{k!(m-k+2)!}f(k)x^k$$
where
$$f(k) = (1+c - b)\cdot k^2 - \big(m(2c - b)+3c -2b + 1\big)\cdot k+(m^2 +3m + 2)c.$$
We need to show that there exists $m$ such that $f(k) \geq  0$ for all $0 \leq k \leq m + 2$. Observe that
$$\bigtriangleup_f = \big( m(2c - b) + 3c - 2b + 1\big)^2 - 4(1 + c - b)(m^2 + 3m + 2)c = (b^2 -4c)m^2 + A m + B$$
So
$$\bigtriangleup_f \to -\infty \,\, \text{ as }\,\, m \to +\infty.$$
In particular, there exists $m$ such that $\bigtriangleup_f < 0$. Since coefficient of $k^2$ in $f(k)$ is
	$$1 + c - b > 1+c- 2\sqrt{c} = (1 - \sqrt{c})^2 \geq 0$$
we deduce that $f(k) > 0$ for all $k$.

[hide=Solution]
It suffices to show this for all quadratic polynomials with nonreal roots and real coefficients since the product of polynomials with nonnegative coefficient has nonnegative coefficients. Let the roots be $a$ and $b$, where $a$ and $b$ are conjugates. Then, it suffices to show that $(x-a)(x-b)(x+1)^{n-2}$ has nonnegative coefficients for some $n\geq2$. This is equivalent to showing that there exists an $n$ such that the $k$th symmetric sum of $-a$, $-b$, and $1$ appearing $n-2$ times is positive for all $1\leq k\leq n$. The $k$th symmetric sum of these numbers is equal to $\binom{n-2}k-a\binom{n-2}{k-1}-b\binom{n-2}{k-1}+ab\binom{n-2}{k-2}$. This is equal to
\begin{align*}
\binom{n-2}k-a\binom{n-2}{k-1}-b\binom{n-2}{k-1}+ab\binom{n-2}{k-2}&=\frac{(n-2)!}{(n-2-k)!k!}-(a+b)\frac{(n-2)!}{(n-1-k)!(k-1)!}+ab\frac{(n-2)!}{(n-k)!(k-2)!}\\
&=\frac{(n-2)!}{k!(n-k)!}((n-k)(n-k-1)-(a+b)(n-k)k+abk(k-1)).
\end{align*}

Now, let $p=a+b$ and $q=ab$. Since $a$ and $b$ are nonreal, we must have $p^2<4q$. The expression is positive if and only if $(n-k)(n-k-1)-pk(n-k)+qk(k-1)>0$. This is equivalent to $k^2(p+q+1)-k(2n-1+pn+q)+n^2-n>0$, or $k^2(p+q+1)-kn(p+2)+n^2-n>k(q+1)$. The minimum value of the left hand side occurs when $k=\frac{p+2}{2p+2q+2}n$, so the minimum value is $n^2-n-\frac{p^2+4p+4}{4p+4q+4}n^2$. Since $p^2<4q$, the minimum value of the left hand side is $\frac{4q-p^2}{4p+4q+4}n^2-n$, which is $\theta(n^2)$. Therefore, since the right hand side is $\theta(k)\leq\theta(n)$, this means that for all sufficiently large $n$, we have that $(x-a)(x-b)(x+1)^{n-2}$ has all nonnegative coefficients.

Therefore, for all polynomials $P$ that satisfy the conditions in the problem, there exists a positive integer $n$ such that $(x+1)^nP(x)$ is a polynomial with nonnegative coefficients.

The crux of the problem is the following reduction.

[b][color=#f00]Claim.[/b][/color] It suffices to prove the claim for all quadratic polynomials with nonreal roots.

Factor $P(x)$ into quadratics and linear terms with real coefficients by grouping its complex roots into conjugate pairs. Because the product of two polynomials with nonnegative coefficients has nonnegative coefficients, it suffices to group the $n$ extra terms throughout the quadratics. If we can guarantee that each of the resulting products has nonnegative coefficients, we are done because $P(x)$ has no nonnegative real roots. $\blacksquare$

Now the problem is a matter of computation. Scale down by the leading coefficient, which we know is positive, to obtain some quadratic $x^2+ax+b$ with $a^2>4b$. Observe $$(x+1)^n(x^2+ax+b) = \left(x^n+nx^{n-1}+{n \choose 2}x^{n-1}+\cdots+{n \choose 2}x^2+nx+1\right)(x^2+ax+b),$$ and in particular the coefficient of $x^k$ for $2 \leq k \leq n$ is given by $$\left[{n \choose n-k+2} + {n \choose n-k+1} a + {n \choose n-k}b\right] = \left[{n \choose k-2} + {n \choose k-1}a + {n \choose k} b\right].$$ This coefficient equals some positive factor of $$(n-k+1)(n-k+2)+bk(k-1)+ak(n-k+2)$$ after expansion. From here set $a$ to be its negative, so it suffices to prove that $$ak(n-k+2) < (n-k+1)(n-k+2)+bk(k-1)$$ for all $k$. This is a quadratic in $k$, and in particular, its discriminant $$\Delta = (2a+na+2n+b+3)^2-4(a+b+1)(n^2+3n+2)$$ has quadratic term $$4n^2+a^2n^2-4an^2+4an^2-4bn^2-4n^2 = (a^2-4b)n^2$$ has a negative coefficient. This means that by picking $n$ sufficiently large, the expression will be positive for all $k$, because the quadratic term of $k$, $a+b+1$ is positive by the $a^2>4b$ condition.

Finally, the edge cases also work out; the coefficient of $x^{n+2}$ is 1, and the coefficient of $x^{n+1}$ is $n+a$ which can be positive by taking sufficiently large $n$, and similarly the constant term is $b$ and the $x$-coefficient is $bn+a$, again positive after taking sufficiently large $n$. Hence, we are done by the first claim.

[hide=Solution]Let the roots of $P(x)$ be $r_1,r_2,\ldots,r_j.$ This means that all roots are either nonreal or negative. Assume that $r_1,r2,\ldots,r_i$ are negative. Now, let $x_1,x_2\ldots,x_i$ be the absolute value of those roots. This means that we can factor out $(x-r_1)(x-r_2)\ldots(x-r_i)=(x+x_1)\ldots(x+x_i),$ for $x_1,x_2\ldots,x_n>1.$ Taking this out, it wouldn't make any more negative coefficients. Now, we only have to consider the case such that $r_1,r_2,\ldots,r_j$ are all nonreals. We know that the nonreal roots come in conjugate pairs. Now, we can arrange them like this: $$(x-(a_1+b_1i))(x-(a_1-b_1i))\ldots(x-(a_j+b_ji))(x-(a_j-b_ji)).$$ It suffices to prove that $(x-(a+bi))(x-(a-bi))(x+1)^{k}=(x^2-2ax+(a^2+b^2))(x+1)^{n}$ has nonnegative coefficients, for some integer $k.$ We claim that all terms must be positive. Indeed, when combining, we find that $\binom{n}{k}+\binom{n}{k-1}2a+\binom{n}{k-2}(a^2+b^2)>0,$ for all $n\geq k.$ This is because after expanding, we find that $$\binom{n}{k}+\binom{n}{k-1}2a+\binom{n}{k-2}(a^2+b^2)>0=$$$$2+4ak+a^2k+b^2k-2ak^2+a^2k^2+b^2k^2+2akn-3k+3n+k^2-2kn+n^2\geq$$$$2+4ak+2ak(n-k)+a^2k+b^2k+a^2k^2+b^2k^2+3(n-k)+(n-k)^2\geq0,$$ since $n\geq k.$ Now, it suffices to prove that there exists an integer $n$ such that $$(a^2+b^2)n-2a>0,a^2+b^2>0,1>0,\text{ and }n-2a>0.$$ The second and third statements are obvious. The fourth requires a larger value of $n.$ Finally, the first statement just requires a large enough $n,$ after $a$ and $b$ are fixed. We conclude that there always exists an integer $n$ such that $(x-(a+bi))(x-(a-bi))(x+1)^n$ has only nonnegative coefficients. Combining this with the first statements, we conclude that there exists $n$ such that $(x+1)^nP(x)$ consists of only nonnegative coefficients, if $P(x)>0$ for all $x\geq0.$
[/hide]


algebra
polynomial
vector
advanced fields
advanced fields solved

[color=blue][b] Nice lemma.[/b]

  Prove that any affine hyperplane in $\mathbb{R^{d}}$ cuts the curve $c(t)=(t, t^{2},\ldots ,t^{d})$ in at most $d$ points.

   [/color]

This is just a very fancy way to say that a polynomial of degree not exceeding $d$ has at most $d$ real roots...

Yes, but can you explain this rigourous? :)

I try to explain what fedja said:

Each linear subspace $U$ can be defined by: $\{x|(x,v)=0\}$ with the usual scalar product. If you consider the affine subspace $a+U$ you receive the equation \[ ((t,...,t^d)-a,v)=0\Leftrightarrow\sum_{i=1}^d(t^i-a_i)v_i=0. \] Here the left-hand side is polynomial of degree at most $d$ which has at most $d$ zeros.

Misha

Addendum: since $v$ is not the zero vector, the polynomial is not identically zero.

The choice of field is irrelevant.


calculus
derivative
integration
geometric sequence

Compute the sum:
S=1/2tan(x/2)+1/2^2*tan(x/2^2)+..+1/2^n*tan(x/2^n).
thank you!

Tiger Sandre,
Germany.

Are the tangents in the numerators or denominators?  Using parentheses and spaces makes what you write much more readable.  I am also (somewhat) skeptical that this belongs in this forum, either.

Where do you get these problems from?  what are they for?

Well my maths teacher gave them to me, he put it under the indermediate topic and perhaps it is harder.. but I don't really know how to solve it!

well the sum is like this: sum (1/2^n)*tan(x/2^n)

thank you very much
Tiger Sandre,
Germany

If you can sum the tangents alone, you can consider S/2 and take S-S/2 and end up with something involving only the tangents. I don't know if you can do that though.

So, I have no idea if this is valid or not.  But, anyhow:
Let me try a power series:
tan(x) = a0 + a1*x + a2*x:^2: + ...

general term = ai*x^i

General term of
tan(x/2^n)/2^n = ai*(x/2^n)^i / 2^n = ai*x^i / 2^((i + 1)*n).
We want to sum this over n = 1 to infinity.  This involves adding
2^(-i - 1) + 2^(-2i - 2) + 2^(-3i - 3) + ... which is a geometric sequence with sum

2^(-i - 1) / (1 - 2^(-i - 1)) = 1/(2^(i+1) - 1)

So f(x) = sum ai * x^i / (2^(i + 1) - 1)

This does not look too promising.  Oh well.  Chug along.  tan(x) has zero even derivatives -- so a0, a2, a4, etc. are all 0.  Thus, we really "only" need the sum
f(x) = :sigma: ai * x^(2i - 1) / (2^(2i) - 1)  for i from 1 to infinity
Now, the denominator factors so you can split this into a sum and difference-- I'm not going to write any more, because if this works (a) it's too ugly to be the best solution and (b) I've reached a point on my scrap paper I can't get past.  Keep following this trail if you want.  Part of this evaluates to the integral of tan(x)/(2x).  The other part I don't really know, but it should be something vaguely similar, I hope.

Anyhow, there is for sure a better way.  Especially because tan has such an unpleasant Taylor series.


function
limit
real analysis
real analysis unsolved

Let $f:\mathbb{R}\to\mathbb{R}$ be a derivable function.
Find $\lim_{x\to\infty}\frac{xf(x)}{f'(x)}$.

Are there any additional conditions on this problem? As it stands I don't see anything keeping this limit from doing just about anything.

no... no conditions ... I haven't done it either
it is a problem from our school olympiad (today) ...


linear algebra
matrix
linear algebra unsolved

Hello,

I would need some guideline on how to solve the following matrix equation:

$ \mathbf{C(I \plus{} DC)^{ \minus{} 2}} \circ \mathbf{I} \equal{} k\mathbf{I}$.

$ \mathbf{C}$ is a positive definite hermitian matrix. $ \mathbf{D}$ is a diagonal matrix with positive and real diagonal entries. $ \mathbf{I}$ is the identity matrix and $ k$ is a real number > 0.

I would like to solve this equation for $ \mathbf{D}$. Any ideas?

$ C$ is diagonalizable and, if a solution of the form you seek exists, then $ (I \plus{} DC)$ is invertible.  So we just calculate,
\[ C(I \plus{} DC)^{ \minus{} 2} \equal{} kI \\
\Rightarrow (I \plus{} DC)^{ \minus{} 2} \equal{} kC^{ \minus{} 1} \\
\Rightarrow (I \plus{} DC)^2 \equal{} \frac {1}{k}C \\
\Rightarrow I \plus{} DC \equal{} \frac {1}{\sqrt {k}}C^{\frac {1}{2}} \\
\Rightarrow D \equal{} \left(\frac {1}{\sqrt {k}}C^{\frac {1}{2}} \minus{} I\right)C^{ \minus{} 1}\]
where $ C^{\frac {1}{2}} \equal{} VA^{\frac {1}{2}}V^{ \minus{} 1}$ and $ C \equal{} VAV^{ \minus{} 1}$ for some diagonal matrix $ A$.  Of course this is just a symbolic solution, if you're planning on computing this numerically you wouldn't want to compute $ C^{ \minus{} 1}$, but rather just solve $ DC \equal{} \frac {1}{\sqrt {k}}C^{\frac {1}{2}} \minus{} I$ directly, and find $ C^{\frac {1}{2}}$ by diagonalizing or an iterative method like Denman-Beavers.

Sorry, I forgot to say what $ \circ$ stands for. It stands for Hadamard product, that is, elementwise multiplication of matrices. So what the equation means is that the diagonal elements of $ \mathbf{C(I \plus{} DC)^{ \minus{} 2}}$ should be equal. Now solving as you did JoeBlow doesn't guarantee that $ \mathbf{D}$ is diagonal, right?

Okey, I can reformulate the problem in the following way, which might be more easy to deal with?
Find a diagonal matrix $ D$ so that the following equation is fullfilled:

$ L^{\ast}(LDL^{\ast} \plus{} I)^{ \minus{} 2}L \circ I \equal{} kI$.

All of the matrices are square matrices. $ L$ is a known lower-triangular matrix with positive and real main diagonal. $ I$ denotes the identity matrix. $ k > 0$ is a positive real number. $ \circ$ denotes Hadamard product and $ \ast$ denotes Hermitian transpose. How to solve the above equation for $ D$, any hints?

It seems that the fixed point iteration for that equation works partiallly. So I just multiply by D on both sides and iterate.
However, one interesting thing can be noted. In cases when the iteration does not converge to the optimum, it converges to the optimal D but reversed! So basically, by reversing the main diagonal in D, it is correct! So the iteration converges to a solution that is reversed. And in other cases it converges to the right one. Anyone noticed something like this before, how to guard oneself against this reversed solution? No way to do that but just iterating more with different starting points?
Grateful for any insight.

Okey, it seems that my fixed point iteration method works great (by reformulating my equation in a nicer way), so I consider it to be solved (algorithmically at least) :)


I need help in proving the following for $ x>0$

1) $ \left(\sqrt{2}\minus{}x\right)e^{2\sqrt{2}x}\minus{}\left(x\plus{}\sqrt{2}\right) \le \frac{e^{2\sqrt{2}x}\minus{}1}{\sqrt{2}e^{x^2/2}}$

2) $ \left(\sqrt{3}\minus{}x\right)e^{2\sqrt{3}x}\minus{}\left(x\plus{}\sqrt{3}\right) \le \frac{e^{2\sqrt{3}x}\minus{}1}{\sqrt{3}e^{x^2/2}}$

Sorry,
The RHS of 2) should be:    $ 2\left(\frac{e^{2\sqrt{3}x}\minus{}1}{\sqrt{3}e^{x^2/2}}\right)$


calculus
function
integration
calculus computations

[quote="Antonio1965"]$ f(x) \equal{} xsen({x^2} \plus{} 2),{\rm{ x }} \in {\rm{ R}}$

Antonio[/quote]

hmm.... What is the sen function?

Okay then ...

I = $ \int x \ sin(x^2 \plus{} 2) dx$

$ x^2 \plus{} 2 \equal{} t \ \Rightarrow \ 2x dx \equal{} dt$

I = $ \frac{1}{2} \int sin(t) \  dt \equal{} \frac{\minus{}1}{2} cos(t) \plus{} C$

= $ \frac{\minus{}1}{2} cos(x^2 \plus{} 2) \plus{} C$

Thank you,
when I look for your solution is easy, :oops:
Regards,
Antonio


linear algebra
superior algebra
superior algebra solved

Calculate

|(a0+b0)n  (a0+b1)n (a0+b2)n ... (a0+bn)n|
|(a1+b0)n  (a1+b1)n (a1+b2)n ... (a1+bn)n|
|(a2+b0)n  (a2+b1)n (a2+b2)n ... (a2+bn)n|
|............................................................|
|(an+b0)n  (an+b1)n (an+b2)n ... (an+bn)n|

The one PiDeltaPhi posted is related to this. Actually, this problem is the previous problem of PiDeltaPhi's on the linear algebra problem set book.

And I will bring the same idea as before (slightly modified :D) :
take P(X0,...Xn,Y0,...Yn)=detM, where M=m_ij=Xi+Yj.

Note that P has degree at most n in each variable.
Also P(Xi,X2...Xn,Y0...Yn)=0 for i=1..n so (X0-X1)(X0-X2)...(X0-Xn)|P.

Analogously we ger P=c* \pii<j (Xi-Xj)(Yi-Yj).
The question is how to determine c, which is a constant.
An idea is to make Xi=Yi=i+1 and use the result in the previous determinant topic to find c. But this means that the idea does not work for that problem too.

Smth else. This problem fits best into the Math for College section. Have you ever seen an IMO problem with determinants?
There was a problem proposed for the IMO that had a quick determinants solution, but this was the reason the problem only reamained proposed and not actually given in contest.

And smth else: My last year algebra teacher recommended "Proskuriakov - Problems in linear algebra". Have you heard of it?

I have the complete solution to this determinant (and the one i posted).I will post it,if the professor who made it agrees to it,so bear with me.

[quote="amfulger"]And smth else: My last year algebra teacher recommended "Proskuriakov - Problems in linear algebra". Have you heard of it?[/quote]
that is exactly the book i mentioned.

i don't know why the post was deleted... so i post it here again.

The solution to this problem can be found on my site:

[url]http://www.blacksun.de.tc/math/determinant.pdf[/url]

Here is a texed web version of the determinant:

[url]http://www.sosmath.com/matrix/determ3/determ3.html[/url]


The best way to play “Fur Elise” and “Moonlight sonata” by Beethoven.

My performing and teaching work is based on accurate indication of Note Strength. Without playing, just by looking at any score for the first time, any of my students or choir members (my other work – chorus master at opera) can say about each note - whether it should be played louder or softer in relation to the previous and the following note, and by how much.

This is an extremely important knowledge because the right or wrong stressing and softening of notes is the only difference between the best and the worst pianist in the world. All of us play the same notes on the same beat, and speed (in spite of mass obsession) eventually does not matter (listen to Horowitz).

We just follow some simple and easy “Dynamics Rules” and it saves a lot of time and energy. In one lesson’s time I can teach anybody to play a piece (which s/he is able to play, of course) as the best professionals play. All the students of any level receive inevitably excellent marks (“for exceptional musicality”) at any exam or competition, as soon as they have learned 25-30 basic rules. Quite often mere following these rules improves technique dramatically. (This happened recently to my student in 2nd Scherzo by Chopin. She amazed adjudicators with “ultra fine finger work” in the C# minor episode, though at the beginning she could not play the 2nd voice at all). A majority of these rules are well known and followed by all good musicians, but the problem is that the rules have never been published altogether in one book, and they are still scattered in many sources. Therefore it is very important and useful to exchange our knowledge with other musicians.

I am gathering and testing these rules for more than 40 years (the test for a rule is considered “passed” if nobody can show me the music in which this rule is not applicable). Unfortunately, after I moved to America, I can not find here anybody with whom I can discuss my concerns regarding dynamics, phrasing, and articulation, etc. People around me say that they have never heard about any musical rules and laws. Instead, they just “self-express themselves in the way they want and feel” (could you imagine – they do it in the music of Bach or Mozart?). Judging from my experience as an adjudicator, this “self-expression” instead of knowledge of the basics of music is a real problem for teaching nowadays. Sometimes I even hear the proud words: “I think in phrases in music and I teach to think in phrases. I do not care what they are made of”. Would you like to learn English from a teacher who “thinks in phrases” because he does not know the words? If you are serious, will you rely on the phrasebook instead of the dictionary?

Proper stressing /softening in music is as important as the proper stressing in the words of any language. Nobody will understand even very familiar words pronounced with the wrong stresses, e.g., AmerIca , TorontO, SonAtina, etc. Similarly, no one will understand and enjoy the music played with wrong stressing and softening. Music is just one of human languages, and for any language, the same rules apply.

 The Scientific Acoustic Research Laboratory of Moscow Conservatory does not work at this time, and I do not know which Western software can be used to display or print (in Disklavier 124 degrees scale) the Note Strength (volume) of each particular note. This is a very effective and convenient way to discover “artistic secrets” of your favourite pianists, if visual information about timing and strength of each note in their recording is available. (Fortunately, almost everything in the Piano Repertoire is recorded today digitally by “Disklavier” and the best performers, so we have a lot to choose from).

I will be very glad to hear from or about somebody who knows “what musical phrases are made of” and wants to share or exchange with me useful rules or can suggest the best ways to perform. I will appreciate the information regarding software as well.

Today my concern is in the very first bars of “Moonlight sonata” and “Fur Elise” by Beethoven.
Which notes of the melody G# G#G#   G#G#G#   A G#F# B E   and  of the melody E-D#-E-D#-E-B-D-C-A      C-E-A-B       E-G#-B-C     should be stressed, i.e. played stronger than regular ones?
(Stressed notes can be marked with “+” or ”++ “or “+++”, the more pluses – the louder).

Which ones should be softened (marks can be “-“, “- - “ , or “---“) ?

Which notes are just regular (not stressed, not softened – no mark needed or mark “0” can be used)? The differences between + and ++ between - and -- , between  0 and +   equal 1 degree. (One degree is THE SMALLEST difference in volume between two notes that we can hear).

For example, G#, G#--, G#++ ,      G#+, G#, G+++    A,  G#++,  F#++, B, E+++ (“Moonlight”)

Another example: E+,  D# , E+++, D#--, E+, B---, D+, C-, A+++ (“Fur Elise”).

 ( I hope that nobody plays like this, of course).

I will appreciate any opinion expressed.

  Vladimir.

I think in Fur Elise, I hear people go like

E--  D#-  E  D#+  E++  B++  D+ C  A++ ...etc.

With a slight (and very slight) ritard at the end of the phrase.

In Fur Elise I play it like this:

E+, D#, E++, D#+, E+++ (++?), B++, D++, C+, A

I think, I don't understand the method of +'s and -'s... E+ is played a little louder than [i]pp[/i] and D# is played [i]pp[/i] and the notes are gradually and subtly growing louder until you get to B. And then A played just so you can hear it over the left hand played [i]pp[/i] .[/i]

I'm not sure if you're supposed to play like that (although I'm sure not many will mind), because the phrasing doesn't indicate loud-soft pattern for individual  notes.

I think that everyone has their distinctive playing style, and they should express it.  Of course, the notes should flow evenly and not consist of randomly assigned volumes.

I agree! With everything you just said actually... And I play it even, one thing that I have seen to be suggested is playing Fur Elise with the Una Corda pedal. Well, on grand pianos that is... There's the left pedal that brings the hammers closer to the strings, but the action is a bit weird, and it's harder to play.

Una corda pedal?! I don't think I've ever heard of anyone playing with that..

I have a book that says to [i]practice[/i] with the Una Corda pedal...


In rectangle CDEF, CD=32 and DE=24 units. In rectangle DFGH, FG = 12. What is the intersection of the interiors of the two rectangles?

there isn't enough info. Are points H and G on CD and EF respectively?

I got [hide]330[/hide] after setting up ratios with the similar triangles. I'm checking right now...

can someone else check too?

EDIT: dingzhou, do you know the answer?

[hide]
Umm... intersection is the area common to the two rectangles, which is just half of $ CDEF$. Union is the area of the region bounded by the whole "thing". Using intersection, the answer is $ \boxed{384}$. Using union, we would require more info. Are the rectangles similar?[/hide]

Ummmm....
An intersection is a set of points, not an area...

[quote="wiseidiot"]Yongyi, your solution assumes that the second rectangle intersects the vertex of the 24x32 rectangle, which is not the case in this problem.[/quote]

Even if it extends outside, the area of the [i]intersection[/i] is still the area of that triangle. And you still haven't specified the width of $ DFGH$!!

[quote="dingzhou"]FG = 12[/quote]

looks like the width to me

EDIT: also, the area turns out not to be the triangle...

Oops... I'm getting worse and worse at reading... :blush:

And I got [hide]330[/hide] also (the triangles turned out to be a bunch of 3-4-5's)!

well, similar to, but... we get the idea :)


find all perfect squares which differ by 2 from a prime from either end?

(example: 7+2= 3^2=11-2)

[hide="Hint"]Primes bigger than 3 are $ 6k\plus{}1$ or $ 6k\plus{}5$, so our square is like $ 6k\plus{}3$.
$ 9\mid 6k\plus{}3$ so $ t^2\equal{}18l\plus{}9$ where $ 6k\plus{}3\equal{}18l\plus{}9$.
$ t^2\equal{}9(2l\plus{}1)$[/hide]


[hide]I'll assume you mean integral dimensions...

Since x^2 + y^2 + z^2 = diagonal^2 (because of two obvious Pythag. theorems) we just have to find x, y, and z which sum to 24 and whose squares sum to 200, and it took a few seconds to find that they are 6, 8, and 10.

The surface area is, I believe, [b]376[/b]?[/hide]

[quote="espark52"]The dimensions of a rectangular box are x, y and z cm. The sum $x+y+z$ equals 24 cm and the product of two sqace diagonals is 200 square centimeters. What is the number of square centimeters in the surface area of the box?[/quote]
[hide]We know $x^{2}+y^{2}+z^{2}=d^{2}$ and $d^{2}=200,$ so $x^{2}+y^{2}+z^{2}=200.$ The dimensions are $6\times8\times10,$ so the surface area is $2(6\cdot8)+2(8\cdot10)+2(6\cdot10),$ or $\boxed{376\text{ cm}^{2}}.$[/hide]

[quote="espark52"]The dimensions of a rectangular box are x, y and z cm. The sum $x+y+z$ equals 24 cm and the product of two sqace diagonals is 200 square centimeters. What is the number of square centimeters in the surface area of the box?[/quote]

[hide]Notice that $(x+y+z)^{2}-x^{2}-y^{2}-z^{2}=2(xy+xz+yz)$.

So the answer is just 376

$376$[/hide]


[quote="autre"]hmm... I read in old post that english-speaking canadian could understand the written french so I think it's useless to translate my initial post.[/quote]

Hmm, je m'en doute.  When in doubt, they don't, that's my rule.

[quote]Je suis français et je suis admis dans quelques camps d'été aux Etats-Unis. Je voudrias savoir s'il y aura des participants canadians et francophones. Je précise que je n'ai rien du tout contre les canadiens anglophones. Simplement, je suis évidemment plus à l'aise pour parler le français que l'anglais.[/quote]

Bon : "I am French and I have been allowed into some summer camps in the United States.  I wanted to know whether or not there will be Canadian and French-speaking participants.  I stress that I have no problem at all with English-speaking Canadians.  Simply : I would feel more at ease speaking French than English."

At your service! :ninja:

Je ne suis pas sûr, mais normalement les gens aux Etats-Unis ne parlent qu'anglais, et parfois on peut même se douter de cela aussi ;)  :o


I don't understand what you mean when you say "particular". A square is a polygon with 4 sides of equal length and 4 right angles

Can you show the problem that it was in?

If you have multiple squares, and someone is talking about only one of them, you can say that they're referring to a "particular square."

Well, let us say that you have a certain number of squares

Which there are infinate. So, a particular square would mean

they are talking about a square.

IN A GEOMETRY CONTEXT

example: A particular square has the same perimeter as its area.

Find the side length.

IN AN ALGEBRA CONTEXT

A particular positive perfect square has the property that when

subtracted by 56,

you get the square root of the number. Find the square.

[hide="Answers"] Call the side length $ s$. Then, $ s^2\equal{}4s$.

Moving the $ 4s$ over, we get $ s(s\minus{}4)\equal{}0$. So $ s\equal{}0,4$. Since it

can't be 0, the answer would be 4.

-----------------
Call the number $ x$. Then, $ x\minus{}56\equal{}\sqrt{x}$, or, if we call

$ \sqrt{x}\equal{}y$, $ y^2\minus{}56\equal{}y$, or $ y\equal{}8$. So, $ 64$.  [/hide]

That would be what it means... If you showed the problem where

you saw it, it would help.


Three numbers $N,n,r$ are such that the digits of $N,n,r$ taken together are formed by $1,2,3,4,5,6,7,8,9$ without repetition.
If $N = n^2 - r$, find all possible combinations of $N,n,r$.

Sorry for reviving this thread
But what is the solution?

@IMI-Mathboy,i don't understand anything of your solution...explain please..... :(


With $ (F_n)$ is Fibonacci sequence: $ F_1\equal{}F_2\equal{}1,F_{n\plus{}1}\equal{}F_{n}\plus{}F_{n\minus{}1}$ then we have $ f(n)\equal{}\frac{F_{n\plus{}1}}{F_n}$. (Easy to prove it by induction)

[quote="tdl"]With $ (F_n)$ is Fibonacci sequence: $ F_1 \equal{} F_2 \equal{} 1,F_{n \plus{} 1} \equal{} F_{n} \plus{} F_{n \minus{} 1}$ then we have $ f(n) \equal{} \frac {F_{n \plus{} 1}}{F_n}$. (Easy to prove it by induction)[/quote]

thank you.


Two times the complement of an angle is forty degrees less than the supplement of an angle. Find the measure of degrees in the measure of the angle.


I have been going to the University of Utah Math Circles (http://www.math.utah.edu/mathcircle/) for along time. Although the topic discussed has nothing to do with mathcontests, it is really interesting, and there is free food, toys (mathematical like rubik's cubes), and prizes for contest winners . Today I finally won a monthly contest YEAHHHH!.  It's prteetyy funny because when the professor asked me how i did it i told him i brute forced some them. So do you guys have mathcircles and can you tell about it and give the website.

I go to a generic state university.  I tried to start a math club but no one was interested.  I there might be people that would be interested but its hard to find them.

What was math like in high school for you?  I wasn't exposed to anyhting.  They wouldn't let me take calculus because my grades weren't up to par in stuff like English class.  High school was the biggest waste of 4 years.  None of the math teachers knew about any interesting math, they just needed a job somewhere.

But, it's also extremely rare that they get students that would give enough of a crap to study on their own in any subject, much less math.

is a math circle just a math club?  why arent they called hexagons? or triangles? or some other weird polygon that mathematicians can think of?

i guess irregular funny looking ones.. or the ones that have so many sides that they are impossible to draw by hand? lol.. i can never seem to draw heptagons.. hehe

That sounds sweet, I wish there was a professor/teacher with that kind of motivation and dedication where I live. Free food plus free math, awesome.


Let $a_0\neq 0$ and $a_{n+3} = 2a_{n+2}+5a_{n+1}-6a_n$ for all $n\geq 0$. Find all possible values for $\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}$

Do you know about linear recursive sequences and the characteristic equation?

[hide]The characteristic equation is $x^3 - 2x^2 - 5x + 6 = 0$, with roots 1, -2 and 3.

So there exist constants $\alpha$, $\beta$ and $\gamma$ such that \[a_n = \alpha + \beta \cdot \left(-2\right)^n + \gamma \cdot 3^n\] and it's easy from there.[/hide]

For example, if $\alpha,\beta,\gamma \neq 0$ then \[\lim_{n \rightarrow \infty}\frac{a_{n + 1}}{a_n} = \lim_{n \rightarrow \infty}\frac{\gamma \cdot 3^{n + 1} + \beta \cdot (-2)^{n + 1} + \alpha}{\gamma \cdot 3^n + \beta \cdot (-2)^n + \alpha}\] wil be equal to 3, do you see that?