geometry
circumcircle
invariant
perpendicular bisector
geometry solved

Let $O$  be the center of the circumcircle of triangle $ABC$;
$M,N,P$  be the midpoints of $BC,CA,AB$, respectively,
$O_1,O_2,O_3$ be the centers of the circumcircle of triagles $OAM,OBN,OCP$, respectively.
Prove that $O_1,O_2,O_3$ are collinear. Sorry if this problem was posted before!

Since the three circles have a common point, showing that their centers are collinear is equivalent to showing that they have another common point. Alternatively, we can consider an inversion of pole $O$ and show that the lines which are the images of the three circles are concurrent.

Consider the inversion of pole $O$ and power $R^2$ (i.e. the inversion wrt the circumcircle). $A,B,C$ are fixed by this inversion, and $M,N,P$ turn into the vertices of the triangle formed by the tangents through $A,B,C$ respectively to the circumcircle. These vertices lie on the symmedians of $A,B,C$, so the three lines (the images of the circles $(AOM),(BON),(COP)$) are concurrent (they intersect in the symmedian point, $K$), Q.E.D.

Okay, it was clear that Grobber would be quicker, and it was also clear that my solution would be the same as his one, but at least, I have written it in a slightly more detailed way, so that maybe it will be easier to follow.

[In the following, I will say "circle $P_1P_2P_3$" for the circle through three given points $P_1$, $P_2$, $P_3$.]

Since the points $O_1$, $O_2$, $O_3$ are the centers of the circles OAM, OBN, OCP, it is clear that, in order to prove that the points $O_1$, $O_2$, $O_3$ are collinear, it is enough to show that the circles OAM, OBN, OCP are coaxal.

Consider the inversion with respect to the circle ABC. The point O is the center of this inversion. Furthermore, this inversion clearly leaves the points A, B, C invariant (since they lie on the circle ABC). Now, let XYZ be the triangle enclosed by the tangents to the circle ABC at the points A, B, C (in other words, let XYZ be the tangential triangle of triangle ABC). Since the two tangents from a point to a circle have equal length, we have XB = XC, and thus the point X lies on the perpendicular bisector of the segment BC. The circumcenter O of triangle ABC also lies on this perpendicular bisector, and so does the midpoint M of the segment BC. Hence, the point M lies on the line OX. We have < OMB = 90 (since $OM\perp BC$ by the definition of a perpendicular bisector); on the other hand, < OBX = 90 (since $BX\perp OB$, as a tangent to a circle is perpendicular to the corresponding radius). Hence, < OMB = < OBX. Since we also clearly have < MOB = < BOX, this yields that the triangles OMB and OBX are similar, and thus OM : OB = OB : OX, and thus $OM\cdot OX = OB^2$. Since O is the center and OB is the radius of the circle ABC, and since the point M lies on the line OX, this equation shows that the point M is the inverse of the point X in the circle ABC. Similarly, the points N and P are the inverses of the point Y and Z in this circle.

Now, the triangle XYZ is the tangential triangle of triangle ABC, and thus, the lines AX, BY and CZ concur at one point K (the symmedian point of triangle ABC). If K' is the inverse of this point K in the circle ABC, then, since the point K lies on the line AX, it follows that the point K' lies on the inverse of the line AX in the circle ABC. But the inverse of the line AX in the circle ABC is a circle through O (since O is the center of our inversion); this circle passes through the point A (since the line AX passes through A, and since A is invariant under our inversion), and this circle passes through the point M (since the line AX passes through X, and since M is the inverse of X). Hence, the inverse of the line AX is the circle OAM. And thus, the point K' lies on the circle OAM. Analogous reasoning shows that the same point K' lies on the circles OBN and OCP; hence, the circles OAM, OBN and OCP have two distinct common points, namely, the points O and K', and therefore must be coaxal. $\blacksquare$

  Darij


Let $a,b,c$ be positive reals. Prove that$(1+a^3)(1+b^3)(1+c^3)\geqq(1+abc)^3$

Yes.

Multiplying out, we get:

$1+[3,0,0]+[3,3,0]+[3,3,3]\geq 1+3[1,1,1]+3[2,2,2]+[3,3,3]$, which follows immediately from AM-GM or Muirhead.

Is AM-GM enough?

$a^3+b^3+c^3\geq 3abc$

and so on.  That's all I mean by writing $[3,0,0]\geq 3[1,1,1]$


Solve the diophantine equation: $x^y-y^x=1$ $(x,y>1)$

Indeed, this one is easy and very well known. In fact, I think we have already discussed it several times.


Quote:Math club membership grew from 75 to 99 members at West Junior High.
The increase represents a growth of 25% for girls and 40% for boys.
How many girls were in the math club last year?[hide]

Let G = number of girls last year.

Let B = number of boys last year.

 . . . Then: G + B = 75

This year there are 25% more girls, or 1.25G girls.

And there are 40% more boys, or 1.40B boys.

. . . Then: 1.25G + 1.40B = 99

Multiply the first equation by -1.40 and add the second equation:

. . . .  -1.40G - 1.40B = -105

.. . . . 1.25G + 1.40B = 99

We get: -0.15G = -6  --->  G = 40

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

Last year, there were 40 girls and 35 boys.

This year, the girls increased by 25%: 40 + 10 = 50

and the boy increased by 40%: 35 + 14 = 49

And this year's membership is 99 . . . check![/hide]

[hide="answer o_o"]
[color=darkred]
b = boys
g = girls

b + g = 75
25b + 25g = 1875

1.4b + 1.25g = 99
28b + 25g = 2475

3b = 600
b = 200

Tell me what I did wrong >_< .
[/hide]
[/color]

[quote="Treething"]
[color=darkred]
1.4b + 1.25g = 99
28b + 25g = 2475

Tell me what I did wrong >_< .
[/color][/quote]

Your post appeared at a wrong [hide="time."]  multiplication
28b + 25g = 1980[/hide] :D


Given isosceles right triangle $ ABC$ with leg of 12. If a new, smaller similar triangle is constructed with each corresponding side one-half as long as each side of triangle $ ABC$, the area of the new triangle is what fraction times the original?

[hide="fast way to do it"]
The ratio of the areas is just $ (\text{the ratio of the side lengths})^2.$ In this case it would be $ \left(\frac{1}{2}\right)^2\equal{}\boxed{\frac{1}{4}}.$[/hide]

[hide="slower way to do it"]
The smaller triangle has legs of length $ 6.$ The area of the larger triangle is $ \frac{12\times12}{2}\equal{}72$ while the area of the smaller triangle is $ \frac{6\times6}{2}\equal{}18.$ The ratio of the areas is $ \frac{18}{72}\equal{}\boxed{\frac{1}{4}}.$[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

$A\in M_3(R)$ such that $A+I_3$ and $A^2-A+I_3$ are not invertible

Compute $det(A^2+d.A^{-1})$  where $d=det(A)$

I think this should be right...

First we find the eigenvalues of A -- we know that $\det (A + I) = 0 \; \Rightarrow \lambda_1 = -1$ is an eigenvalue. Similarly $\det (A^2 - A  + I) = 0$, so $A^2 - A$ has eigenvalue -1. Solving $\lambda^2 - \lambda = -1$ we get $\lambda_2 , \lambda_3 = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i$ (since the matrix is real we must have both roots as eigenvalues)

Now write $\det (A^2 + \det(A) A^{-1} ) = \det (A^{-1}) \det (A^3 + \det(A) I) = \frac{1}{\det(A)} P_{A^3}( - \det A ) $ where $P_{A^3}(x)$ is the characteristic polynomial of $A^3$. Then  use that $P_{A^3}(x) = (\lambda_1^3 - x)(\lambda_2^3 - x)(\lambda_3^3 - x)$, since $A^3$ just has the same eigenvalues cubed, and that $\det (A) = \lambda_1 \lambda_2 \lambda_3$, to obtain the expression $(\lambda_3^2 + \lambda_1 \lambda_2 )(\lambda_2^2 + \lambda_1 \lambda_3 )(\lambda_1^2 + \lambda_2 \lambda_3 ) = 5$ after some calculating I don't have time to verify  :blush:


How do you find the sum of the roots of $4ix^2-x+2i-1$ without factoring?

The roots of $ix^2-x+2i=0$ are (ix+1) and (x+2i)  ,so the sum of the roots is i-2i=-i .

  but the coefficient of -x in $ix^2-x+2i=0$  is -1, not -i.  shouldn't the coefficient of -x be the sum of the roots?

 as in $(a+b)(a+c)= a^2+ac+ba+bc= a^2+a(b+c)+bc$

thanks

The sum of the roots of a given quadratic is $\frac{-b}{a}$.

The two roots are $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$. Adding we have $\frac{-2b}{2a}$, which simplifies to $\frac{-b}{a}$

[hide="answer"]so the answer is $\frac{1}{4i}$[/hide]

[quote="myc"]but the coefficient of -x in $ix^2-x+2i=0$  is -1, not -i.  shouldn't the coefficient of -x be the sum of the roots?
[/quote] Only if the leading coefficient is 1.


geometry
geometric transformation
reflection
geometry unsolved

Let $ ABC$ be an acute triangle. Draw the Apollonius circle, i.e. the locus of points $ X$ such that $ \frac{AX}{XB}\equal{}\frac{AC}{CB}$. Let $ P$ be a point on the circle, interior of triangle $ ABC$, and let $ P'$ be the reflection of $ P$ in $ AB$. Prove that $ \angle ACP'\equal{}\angle BCP$.

[quote="jbmorgan"]Let $ ABC$ be an acute triangle. Draw the Apollonius circle, i.e. the locus of points $ X$ such that $ \frac {AX}{XB} \equal{} \frac {AC}{CB}$. Let $ P$ be a point on the circle, interior of triangle $ ABC$, and let $ P'$ be the reflection of $ P$ in $ AB$. Prove that $ \angle ACP' \equal{} \angle BCP$.[/quote]
We have $ P'$ lie in Apollonius circle and chord $ PX$ equal chord $ P'X$ ; $ AX$ bisect $ \angle ACB$
So $ \angle ACP' \equal{} \angle BCP$


Let a_1,a_2, . . .,a_n be an arbitrary rearrangement of 1,2, . . .,n. Prove that if n is odd, then (a_1−1)(a_2−2) . . . (a_n−n) is even.

Let $ n=2p+1$. There are $ p$ even numbers and $ p+1$ odd numbers.
In the factors $ a_{1}-1,a_{3}-3,\ldots ,a_{2p+1}-(2p+1)$ exists $ j$ such that $ a_{j}$ is odd, so the factor $ a_{j}-j$ is even and the product is even.


function
induction
limit
algebra unsolved
algebra

Let $F$ is set consist all fuction $f(x): R^{+}\to R^{+}$ such that
           $f(3x)\geq f(f(2x))+x$ .Find real number $a$ is greatest satisfying $f(x)\geq ax$

$1_{man}+1{woman}=1_{baby}\Rightarrow 1+1=1$

noone? :P. it  have a solution be very nice

it  is very easy see $f(x)=1/2$ belong to $F$.
$\Rightarrow a\leq \frac{1}{2}$.
 Suppose $f$ is any fuction in $F$.
From problem $\Rightarrow f(x)>\frac{1}{3}$.
  Suppose  sequence $(a_n)$ is defined  $a_1=1/3, a_{n+1}=\frac{2a_{n}^2+1}{3}$.
By induction ,I have $f(x)\geq a_n.x$  $\forall n$
Then,I have $\lim_{n\to\infty}{a_n}=1/2$  $\Rightarrow a_{max}=1/2$


[quote="Jonhs"]Solve the equation:
$\sqrt[3]{x+\frac{1}{2}}=16x^{3}-1$[/quote]
$\sqrt[3]{x+\frac{1}{2}}=16x^{3}-1\Leftrightarrow\sqrt[3]{2\sqrt2x+\sqrt2}=16\sqrt2x^{3}-\sqrt2\Leftrightarrow$
$\Leftrightarrow\sqrt[3]{t+\sqrt2}=t^{3}-\sqrt2,$ where $t=2\sqrt2x.$
Let $f(x)=\sqrt[3]{x+\sqrt2}$ and $g(x)=x^{3}-\sqrt2.$
Then $f(g(x))=g(f(x))=x$ and $f$ is increasing function. Hence, $f(x)=g(x)\Leftrightarrow f(x)=x.$
Thus, $\sqrt[3]{t+\sqrt2}=t^{3}-\sqrt2\Leftrightarrow t^{3}-\sqrt2=t\Leftrightarrow t=\sqrt2.$
Id est, $x=\frac{1}{2}.$ :)

[quote="VANCHANH123"]Solve:
$16x^{3}-1=\sqrt[4]{x+\frac{1}{2}}$[/quote]

Let $f(x)=16x^{3}-1$ and $g(x)=\sqrt[4]{x+\frac{1}{2}}$

For $x\leq 0$ $f(x)\leq 0$ and $g(x)>0$ $\Rightarrow $ no solution.
For $x>0$, $f(0)<g(0)$, $f(x)$ convex, $g(x)$ concave and $f(1)>g(1)$ $\Rightarrow $ a unique real root

Obviously, $\frac{1}{2}$ is a root, so it is the unique root.

--
Patrick

[quote="VANCHANH123"]Solve:
$16x^{3}-1=\sqrt[4]{x+\frac{1}{2}}$[/quote]
Still $x=\frac{1}{2}.$ :D
But another reasoning. :wink:

[quote="Jonhs"]Solve the equation $\sqrt[3]{x+\frac{1}{2}}=16x^{3}-1$.[/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $\frac{2x+1}{2}=t^{3}$, i.e. $x=\frac{2t^{3}-1}{2}$, where $t\in \mathcal R$. The equation becomes $t+1=2\left(2t^{3}-1\right)^{3}$ $\Longleftrightarrow$ $\frac{t-1}{2}=\left(2t^{3}-1\right)^{3}-1$ $\Longleftrightarrow$ $t-1=2\left[\left(2t^{3}-1\right)-1\right]\left[\left(2t^{3}-1\right)^{2}+\left(2t^{3}-1\right)+1\right]$ $\Longleftrightarrow$ $t-1=4\left(t^{3}-1\right)\left[\left(2t^{3}-1\right)^{2}+2t^{3}\right]$ $\Longleftrightarrow$ $t-1=4(t-1)(t^{2}+t+1)(4t^{6}-2t^{3}+1)$. Suppose $t\ne 1$. In this case the equation becomes $E\equiv 4(t^{2}+t+1)(4t^{6}-2t^{3}+1)=1$.  But $t^{2}+t+1\ge \frac{3}{4}$ and $4(t^{3})^{2}-2(t^{3})+1\ge \frac{3}{4}$. Therefore, $E\ge 4\cdot\frac{3}{4}\cdot\frac{3}{4}=$ $\frac{9}{4}> 1$ $\Longrightarrow$ $E\ne 1$. In conclusion, the given equation has only the zero $t=1$, i.e. $x=\frac{1}{2}$.

[b][u]Remark.[/u][/b] I think that we can solve similarly and the general equation $\sqrt[n]{x+\frac{1}{2}}=16x^{3}-1$, where $n\in N\ ,\ n\ge 2$.

.............$t\ne 1\Longrightarrow E\equiv 4(t^{n-1}+t^{n-2}+\ldots+t^{2}+t+1)(4t^{2n}-2t^{n}+1)=1\Longrightarrow t\in\emptyset$ ..........................[/color]

[quote="Jonhs"][color=darkred]Solve the equation:[/color]
\[\sqrt[3]{x+\frac{1}{2}}=16x^{3}-1 \]
[/quote]
[color=darkblue][b]Solution.[/b] Denote $\sqrt[3]{x+\frac{1}{2}}=t,$ we have: $\begin{cases}16x^{3}=t+1\\ 2t^{3}=2x+1\end{cases}$
Assume that: $2x\ge t\iff 16x^{3}\ge 2t^{3}\iff t+1\ge 2x+1\iff t\ge 2x$
Thus $t=2x,$ so $16x^{3}-2x-1=0\iff (2x-1)(8x^{2}+4x+1)=0\iff x=\frac{1}{2}$[/color]

[color=darkred][b]Similar problem.[/b] Solve the equation:[/color]
\[\sqrt[3]{\frac{x+1}{2}}=2x^{3}-1 \]

[quote="Jonhs"]Solve the equation $\sqrt[3]{x+\frac{1}{2}}=16x^{3}-1$.[/quote]
[color=darkblue][b]Another idea[/b] : denote $f : \mathcal R\rightarrow\mathcal R$, where $f(x)=\sqrt[3]{x+\frac{1}{2}}$. Thus the equation becomes $f(x)=2f^{-1}(2x)$, where the function $f$ is increasing a.s.o.[/color]


suppose that N is a fixed point.prove that:

the locus of points like M inwich the sum of MN is fixed,is a circle with the centerN.and the radius MN.

ohh...thanks for the diagram ;) .but i think its not clear... :(

i think we can prove it like this:
first we should prove that each point on the circle is a good point and each point out of circle is not good. and for this each point on circle has exactly the fixed length. and each point out of the circle has longer or shorter length (i think it is easy to prove it :lol: )
anyway easy but useful question. :D

[quote="ashegh"]suppose that N is a fixed point.prove that:

the locus of points like M inwich the sum of MN is fixed,is a circle with the centerN.and the diameter MN.[/quote]

let $N(a,b)$ and $M(x,y)$ and $MN = r >0$
we have $MN=r$
then
$MN^2=r^2$
then
 $(x-a)^2+(y-b)^2=r^2$
then the locus of  $M$ ,is a circle with the center $N(a,b)$ and the diameter $r$

[quote="frt"][quote="ashegh"]suppose that N is a fixed point.prove that:

the locus of points like M inwich the sum of MN is fixed,is a circle with the centerN.and the [b]diameter[/b] MN.[/quote]
Shouldn't it be "radius"? ;)
(Or am I misinterpreting the problem?)[/quote]
no you are right,and i think ashegh mean exactly what you mean but may be there is a little mistake in typing. :D


geometry
geometric transformation
reflection
USAMTS
AMC
USA(J)MO
USAMO

Does anyone know when the results for MAML II come out?

... and ... what's a good score on this thing?  :maybe:

Thanks,
~tZf

If I remember from last year, they take quite some time. I don't remember getting the results until mid-April.

Generally the scores are pretty high, with several 100s, but last year was an obvious exception.

eh can we discuss problems/projected scores yet?

Better wait. I'm not sure all schools have taken it yet.

One of the weakness of the MAML I think is that the weighting just does not make any sense. There's no reason why problems such as 2a and 4a should be weighted as much as 6b and more than 7b. The parts should be weighted according to difficulty, not with the goal of making each question worth the same number of points. The current system just does not really reflect mathematical ability.

hmm I believe question 2 is wrong. The problem didn't state that the big circle has to intersect the base of the equilateral triangle at its midpoint. So basically r_1 and r_2 varies depending on where the circle intersects the base.

Is anyone protesting #2? Did it go through?

Level 2 results are online: http://www.wocomal.org/statistics/olympiad/200607/200607o_level2.html

Rick Yanco
Worcester Academy

does 5th place get any money?  :(

Stephen beat me by 1 point on the level I too...  :o

Yeah, I believe all the places get money. I do know that 3rd place gets $\$$150 and 5th through 10th places get $\$$75. (So I get $\$$75 yay! Time to buy Engel...)

Yay! 14th place!

Do you think we will get our graded solutions back?

Well we certainly didn't get anything back last year. I would be interested in knowing where I lost points though.

[quote="diophantient"][quote="13375P34K43V312"]I doubt it, but this is my first time doing it; I'm not sure. I've never heard of a proof contest that gives you your solutions back, but for USAMTS they do give you like 4-5 word comments[/quote]
with USAMO you get to keep your solutions, don't you?[/quote]

Yeah, you keep your original and your test center coordinator faxes copies


Let a regular pentegon ABCDE with all vertices lie on a circle. M is a point lying on the minor arc AB of that circle.
Prove that MA+MB+MD=MC+ME
[hide="Extension"]A regular polygon $ A_1A_2...A_n(n\ge 3)$ with all vertices lie on a circle. M is a point lying on the minor arc $ A_1A_n$ of that circle. Prove that $ MA_1\plus{}MA_3\plus{}..\plus{}MA_n\equal{}MA_2\plus{}MA_4\plus{}...\plus{}MA_{n\minus{}1}$ if $ n$ is odd.  [/hide]

$ MA \cdot CB\plus{}MB \cdot CA\equal{}AB \cdot CM$
$ BM \cdot EA\plus{}EB \cdot MA\equal{}EM \cdot AB$
$ DB \cdot MA\plus{}DA \cdot MB\equal{}DM \cdot AB$

Let $ AB\equal{}a$ and $ CM\equal{}b$.

$ \Rightarrow MD\equal{}\frac {b} {a} \cdot (MA\plus{}MB)$
$ \Rightarrow (2b\plus{}a)(MA\plus{}MB)\equal{}a(MD\plus{}MC\plus{}ME)$

$ MC\plus{}ME\equal{}\frac {2b\plus{}a} {a} \cdot (MA\plus{}MB)\minus{}MD\equal{}\frac {2b\plus{}a} {a} \cdot (MA\plus{}MB)\minus{}\frac {b} {a} \cdot (MA\plus{}MB)$
$ \equal{}\frac {a\plus{}b} {a} \cdot (MA\plus{}MB)$

$ MA\plus{}MB\plus{}MD\equal{}(MA\plus{}MB)\plus{}\frac {b} {a} \cdot (MA\plus{}MB)\equal{}\frac {a\plus{}b} {a} \cdot (MA\plus{}MB)$

$ \Rightarrow MA\plus{}MB\plus{}MD\equal{}MC\plus{}ME$.


are there commands in LATEX that formats computer code for you??

I want to use it for school paper because I'm too lazy to format it using Word.

How about [url=http://www.andre-simon.de/index.html]Highlight[/url] which will output your code in LaTeX (and other formats), beautifully coloured, indented, numbered etc?


OK if we are given the following first order rxn:

$ 2N_2O_5 \minus{} > 4NO_2 \plus{} O_2$

The rate constant is $ k$. Now how would we express the rate law for the decomposition of $ N_2O_5$.

$ \minus{} d[N_2O_5]/dt \equal{} k[N_2O_5]$

OR

$ \minus{} d[N_2O_5]/dt \equal{} 2k[N_2O_5]$

I think it is the latter but I see all books use the former.zzz

Its urgent.

Let's take your reaction for example:

$ 2 \ce{N2O5} \rightarrow 4\ce{NO2} \plus{} \ce{O2}$

$ \text{ Rate of Reaction} \equal{} \minus{}\frac{1}{2}\frac{d[\ce{N2O5}]}{dt}\equal{} \frac{1}{4}\frac{d[\ce{NO2}]}{dt} \equal{} \frac{d[\ce{O2}]}{dt}$

In this case since you said the reaction is of first order then we have:

$ \text{Rate of Reaction} \equal{} k[\ce{N2O5}]$.

rate=k[N205]= -0.5 d/dt[N2O5]=0.5b[N205]
HENCE RATE CONSTANT=k=b/2
or b=2k


function
calculus
derivative
linear algebra
matrix
trigonometry
vector

What is the minimal value of $k$ ( depending on $n$) such that one can find a $C^{\infty}$ periodical function $f$ from $R^n$ to $R^k$ such that if $D_i f$ are the partial derivatives of $f$ one has $<D_i f(x), D_j f(x)> = d_{ij}$ where $<>$ is the scalar product and $D_{ij}$ is the Kronecker symbol.

Clearly, $k\ge n$. It was shown on the forum that if $f: \mathbb R^n\to\mathbb R^n$ is $C^1$ and its Jacobian matrix is orthogonal, then $f$ is an isometry, hence it couldn't be periodic. Finally, $k=n+1$ works: we just take $f(x_1,\ldots,x_n)=(\cos x_1,\sin x_1,x_2,\ldots,x_n)$ (it has period $(2\pi,0,\ldots,0)$). This means that the answer is $\min k=n+1$.

Ooppss, sorry, I forgot a small, but important detail: the period must be $(2\cdot\pi,...,2\cdot\pi)$. Sorry again.

Unless I'm missing something obvious, that makes no difference whatsoever: let $Q: \mathbb R^n\to\mathbb R^n$ be an orthogonal transformation sending $(2\pi,2\pi,\ldots,2\pi)$ onto $(2\pi,0,\ldots,0)$, and let $f$ be the function I used in my previous post. Then $f\circ Q: \mathbb R^n\to\mathbb R^{n+1}$ satisfies the required condition.

I suspect you might have meant that $f$ is supposed to be $2\pi$-periodic in each variable, i.e. if $e_i,\ i=\overline{1,n}$ is the vector having $1$ in position $i$ and $0$ everywhere else, then $2\pi e_i$ is a period of $f$ (for all $i$).

Indeed, I did not express corectly the periodicity: such an $f$ should verify $f(x+y)=f(x)$ for all vector $x$ and all vector $y$ such that all components of $y$ are in $2\cdot\pi\cdot Z$. Sorry again.  In this case, for example, for $n=2$ the answer is 4.


polynomial
number theory proposed
number theory

Find all pairs of positive integers $ x$ and $ y$ that satisfy equation:

$ x^{2} \plus{} xy \plus{} y^{2} \equal{} 2005(x \plus{} y)$

2005?  I would think the motivation behind these problems is too well-known for this to be a good Olympiad problem...

First write $ (3x)^2 \plus{} (3x)(3y) \plus{} (3y)^2 \equal{} 3 \cdot 2005(3x \plus{} 3y)$ and let $ 3x \equal{} u \plus{} 2005, 3y \equal{} v \plus{} 2005$.  This gives

$ u^2 \plus{} uv \plus{} v^2 \plus{} 3 \cdot 2005^2 \equal{} 6 \cdot 2005^2 \implies$
$ u^2 \plus{} uv \plus{} v^2 \equal{} 3 \cdot 2005^2$.

Over $ \mathbb{Z}[\omega]$, the [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integers[/url], $ 5$ and $ 401$ are both Eisenstein primes, so integer solutions are possible if and only if $ 2005 | u, v$.  In other words, the only solutions occur when $ u \equal{} v \equal{} \pm 2005$, but then $ x, y$ are not positive integers, so there are no solutions.


If I have a question about the clearness of a problem where should I ask it?

Email the USAMTS staff. You can go to contacts on the usamts website and it'll give you an email.

I did just that today, but unfortunately, they couldn't tell me what I needed to know.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=233202]this thread[/url]: we don't give out "clarifications" of problems while a round is in progress.

If I base my solution on an interpretation of the problem, and I get the problem right based on that interpretation, but that interpretation is not correct, would I still be able to get full credit?

It depends on whether the graders feel that the interpretation is reasonable and/or if it trivializes the problem.  It's hard for me to give an example.

There are two interpretations that I have that are equally likely to be correct. They make the problem drastically different from each other. It's a possible typo.

If you think there's a typo, PM me.  I'll tell you if it's a typo or not.


geometry
parallelogram
circumcircle
geometry proposed

[quote="Fang-jh"]Let $ O$ be the circumcenter of acute triangle $ ABC$. Point $ P$ is in the interior of triangle $ AOB$. Let $ D,E,F$ be the projections of $ P$ on the sides $ BC,CA,AB$, respectively. Prove that the parallelogram consisting of $ FE$ and $ FD$ as its adjacent sides lies inside triangle $ ABC$.[/quote]

[hide="Solution"]Let
$ \angle BAC\equal{}\alpha$
$ \angle ABC\equal{}\beta$
$ \angle ACB\equal{}\gamma$
Let $ P$ is forth point of parallelogram $ FEKD$.
It's enough to prove that
$ \angle AEF\plus{}\angle FEK<180^o$ and
$ \angle BDF\plus{}\angle FDK<180^o$

Let's prove it:
We have:
$ 90^o\plus{}\angle OBP>90^o$
                $ <\equal{}>$
$ \gamma\plus{}\angle PAF\plus{}\angle OAP\plus{}\angle OBP>90^o$
    So
$ \angle EAP\plus{}\angle PBD\equal{}\gamma\plus{}\angle OAP\plus{}\angle OBP>90^o\minus{}\angle PAF\equal{}\angle APF\equal{}\angle AEF$
                 $ <\equal{}>$
$ \angle EFD>\angle AEF$
                 $ <\equal{}>$
$ \angle AEF\plus{}180^o\minus{}\angle EFD<180^o$
                 $ <\equal{}>$
$ \angle AEF\plus{}\angle FEK<180^o$

Similiarly we achieve that
$ \angle BDF\plus{}\angle FDK<180^o$
So parallelogram $ FEKD$ lies inside $ \triangle ABC$[/hide]


calculus
integration
calculus computations

generated by the following parametric components:
x=e^t * cos(t)
y=e^t * sin(t)
t: [0,pi]
I've never been good with parametrics. Can someone help explain to me how this problem needs to be setup?

This should be in your book.  The arc length integral is

$ \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 \plus{} \left( \frac{dy}{dt} \right)^2} \, dt$

oh ok, thank you
i didnt have my book with me sorry ><

Only in rare and exceptional cases is it possible to compute an arclength in closed form. This happens to be one of those rare and exceptional cases.

This also works well expressed in polar coordinates.


quadratics
number theory unsolved
number theory

Hi please help have exams in few days .

ii) If p is a prime number , prove that $(p-1)! \equiv -1 mod p$.  [b]Solved [/b]

iii) If , in addition to (ii) , $p=4n+1$ for some n , prove that the congruence

$T^2 \equiv -1 mod p$

has the solution $T=(2n)!$ [i]( Dont know how to solve)[/i]

Thanks

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46300 , Post 5, Lemma 1.

$4n\equiv -1 (mod 4n+1)$
$4n-1\equiv -2 (mod 4n+1)$
.
.
.
$2n+2\equiv -(2n-1)(mod 4n+1)$
$2n+1\equiv -(2n)(mod 4n+1)$

[2n+1,4n]  <- in this interval there is an even amount of numbers
...................actually there are 4n-2n-1+1=2n numbers
so $\prod_{i=2n+1}^{4n} i \equiv(-1)^{2n}\prod_{i=1}^{2n} i\equiv (2n)!$
now because $(4n)!=(2n)! \cdot \prod_{i=2n+1}^{4n} i \equiv (2n)!\cdot (2n)!$*
from wilsons theorem you have that $(4n)!\equiv -1 (mod 4n+1)$
using * you have that $(2n)!^2\equiv -1 (mod 4n+1)$

[quote="bomb"]You could extend this result to other primes = 1 mod n by quadratic forms

Bomb[/quote]
The direct extension is that one to arbitrary fields which can be proven by totally the same way. But what are you speaking of¿


The sum of a certain number of positive integers is 31. What is the biggest their product can be?

(1) 55296 (2) 78732 (3) 118098 (4) 49152
(5) None of the above

[hide="blaah"]
I can tell you the steps, but you solve it from there.
x+y=31
Try middle numbers..multiply them.your answer![/hide]

[hide="solution"]
Start testing out numbers from 2. We could have 2^15*1, however we want to make it as big as possible, so 2^14*3=49152.

This seems reasonable but we keep on going...we try 3^10*1, but we could break this down into 3^9*4=78732.

we try 5 now (4 would yeild same ans as first one)...and we get an answer of 5^5*6=18750 or something..

the correct answer is [b](B) 78732[/b][/hide]


When the first 50 counting numbers are arranged in columns, as shown, what is the sum of all of the counting numbers in column E?

                               A         B       C      D       E
                               1         2       3      4        5
                              10         9      8      7        6
                              11        12     13     14      15
                              20        19     18     17      16
                               .          .        .      .        .
                               .          .        .      .        .
                               .          .        .      .        .

[hide]50 will end in column A. The pattern in E is that it contains the numbers that end it 5 and 6, so the sum is

$ (5*5\plus{}10\plus{}20\plus{}30\plus{}40)\plus{}(5*6\plus{}10\plus{}20\plus{}30\plus{}40)\equal{}\boxed{255}$[/hide]

There are two sequences of numbers here:
5+15+...+35+45 =125  (use the sum formula)
6+16+...+36+46 =130
125+130=  [u][b]255[/b][/u]

Or...
Since each term in the sequence 6+...+46 is one more than each term in 5+...+45 and there are 5 terms, then
((5+15...+45)*2)+5
250+5=  [u][b]255[/b][/u]


By number, its level is Pre-Olympiad although I solved this in ten minutes :D .

Number 4

Suppose that $f$ satisfies the functional equation

\[2f(x)  + 3f(\frac{2x+29}{x-2}) = 100x+80\]

Find $f(3)$.

Enjoy!

solution
[hide]
$2f(3)=380-3f(35)$

$2f(35)=3580-3f(3)$

combining this two equations, we get that $760=10740-5f(3) \Rightarrow f(3)=1996$

not that hard indeed... (if i didn't make any stupid calculating mistakes  :? )[/hide]

Haha..

You're right for the answer and the year. :)


a. Unlimited pairs of parentheses
b. One "^" for exponentiation
c. 14 sets of "-1" (where they can be read as "subtract one" or
"negative one," depending on where each is placed)

Each "-1" group must be kept together as a group/pair. Parentheses
may be placed in front of, between, and/or behind any or all of the
"-1" pairs. The minus signs are not allowed to be split up, nor are the
ones allowed to be split up, or regrouped with other ones.

You must use ALL 14 "-1" pairs, and you must use the "^"
symbol, except you can avoid the latter and use Latex to
show a quantity being raised to another quantity, and then
in that case you will not use the "^" symbol.

I am looking for an expression in these symbols that gives the
largest positive value.
I am not asking for it to be simplified.

I think this is what you mean...
For a maximum to occur, we need the exponent to be positive so after multiplying $ (-1-1-1...)$ by $ (-1)$ we find that a miximum value will occur when;

$ (-1-1-1...-1)^{(-1)(-1-1...-1)}$

So the question really asks: what is the maximum of $ a^b$ given that $ a+b = 13$
(Note that for the expression to be positive, either $ a$ or $ b$ must be even which is always the case.)

$ \Rightarrow \Max : (13-a)^a$, obviously $ a$ will be and intiger somewhere in the middle of $ (0,13)$ so we only need to check a couple of values

$ 10\log_e 3 \approx 10.986...$
$ 9\log_e 4 \approx  12.476...$
$ 8\log_e 5 \approx 12.875...$
$ 7\log_e 6 \approx 11.6755...$

And this is all we need, so the maximum is $ 5^8$

$ \Rightarrow \box{(-1-1-1-1-1)^{(-1)(-1-1-1-1-1-1-1-1)}}$

[quote="ocha"]

And this is all we need, so the maximum is $ 5^8$

$ \Rightarrow \box{( - 1 - 1 - 1 - 1 - 1)^{( - 1)( - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1)}}$[/quote]
You have an extra open parenthesis that isn't matched with anything.

Open hint:  You can also have formations such as:

(-1 - 1)(- 1 - 1 - 1) = 6, just for a subpart of the expression.

So your search would have to be extended...

(Nice description/work.)

Everyone, the previous attempt is NOT the answer.


Most of the national communities are based on languages, and the Africans speak languages that are already well represented.

[quote="erdos"]I just was wondering why there isn't any african community in the National and Local Communities Round Table. As I know there are many african countries which participate in the IMO : South Africa, Bangladesh, Morocco, Tunisia ....
I know that a few African people are on the forum, but it would be great if the admins could create african communities. and Talking about my country, now there are many moroccans who know the forum and there are many that are on it (even my leader and deputy  :lol: ) So I think that a moroccan community can attract more moroccans to the forum. What do the others think about that ??[/quote]

just a small correction: Bangladesh is not in Africa.

The African countries which take part regularly at IMO are: Algeria (since 1977), Morocco (since 1983), Tunisia (since 1981) and South Africa (since 1992). Since one of the official languages in SA is english, it would be a nice idea if the webmaster could create a north african forum for the three countries Morocco, Algeria and Tunisia and the languages can be Arabic, French and English.

What do you think Valentin?

I'm tying to get a mathlinks-style site [url=http://www.mathspage.co.za]www.mathspage.co.za[/url] for the South African olympiad community going and some stuff for Africa.  I don't want to duplicate what mathlinks is doing very well and we encourage our students to use mathlinks, but I'd like to develop our site into a targetted community that supports and integrates with our programs especially for student that are starting out while our top students would interact and benefit from mathslinks.  I think it would be great to get more African guys on mathlinks, and they would probably need their own forum, not around language as much the level of expertise and difficulty of the problems.  The top students from SA, Morroco, Tunisia and Algeria who are at IMO level are fine but there are a number of African countries that are keen to take part in olympiad maths but don't know where to start - there literally is no one in their country who can do this stuff and very few resources.  Take a look at this years PAMO problems that i've posted to get an idea of the level of competition, there are a number of countries that go to the PAMO that don't get anywhere on those problems, just because they haven't got anyone with olympiad experience in their countries, some of the countries select their teams for a competition like this just by taking the top students from their school exams and sending them here.  I know i'm waffling on a bit but my point is that a lot of countries in Africa could and really need to benefit from  resource like this but they have very specific need and probably need a lot of support along the way.

I agree, I would really like to see an African forum. It would encourage more Africans to become MathLinkers.

Hm, interesting, as a South African IMO coach I would like to add a few things, too :)

Well, first, something I experienced myself in South Africa: a lot of South Africans do NOT have a good or fast internet connection (of course there are exceptions) and many of them just don't have a clue that Mathlinks exists because they are just never surfing on the web, or something.

Then, Brad, [quote="bradpreston"]I think it would be great to get more African guys on mathlinks, and they would probably need their own forum, not around language as much the level of expertise and difficulty of the problems.[/quote]

Well, don't forget that there are forums on Mathlinks like "intermediate", "pre-olympiad" etc, so that these forums could be a good starting point. I like the idea of "targetted community" but I have seen that there is hardly any activity on the South African forum you created [the majority of the posts is made by the coaches...], so I'm afraid the "community" you are aiming at is just too small. I mean, there are like 40 guys each year who do some IMO/PAMO preparation camp, and let's say that at most 20 (half of them) have a good internet connection, are busy on the web quite often, and are interested in discussing things on a math forum. I don't think it will be easy to get the forum going, you see - although it's definitely worth a try.

So, some African communities on Mathlinks *could* be a solution for the small number of guys who are interested in a forum like this one, since there are plenty of other guys around here, from all possible nationalities, and plenty of possibilities (because of the different forums and levels here, etcetera).

Hmm, so you guys are suggesting that I create one forum for Africa?  :?

Yeah, we are. I mean, you might as well try it. If it works, it works, if it doesn't work, it doesn't.

I don't know, I am neutral :D

Maybe some young African mathematicians will find it easier to enter the Mathlinks communities if there is some African forum? I don't know.


P is a point such that PA=3, PB=5 and PC=7. Prove that P is the incenter of triangle ABC, if ABC has the maximal perimeter.


function
integration
probability
complex analysis
topology
advanced fields
advanced fields unsolved

Pretty standard easy problem, can be solved in a few ways:

Suppose that $f$ is continuous on $\mathbb{R}$. Show that $f$ and $\hat{f}$ cannot both be compactly supported unless $f=0$. This can be vewed in the same spirit as the uncertainty principle.

The Fourier Transform is linear, and $f = \alpha\hat{\hat f}$ where $\alpha>0$ is a real constant, and depends only on the exact definition of the Fourier Transform you are using (usually the definition is taken in such a way that $\alpha=1$), then it is enough to prove that if $f$ is a compactly supported function, then $\hat f$ is not. I am particularly lazy right now to finish (I haven't lunch yet, I'm hungry), but it is at least a start.

Here's the real argument: If $f$ is compactly supported, then $\hat{f}$ is entire. Here's why:

$\hat{f}(\xi)=\int_{-R}^Rf(x)e^{-2\pi i\xi x}dx=\int_{-R}^Rf(x)\sum_{n=0}^{\infty} \frac{(-2\pi i)^n\xi^nx^n}{n!}\,dx$

$=\sum_{n=0}^{\infty}\frac{(-2\pi i)^n}{n!}\left(\int_{-R}^Rf(x)x^n\,dx\right)\xi^n=\sum _{n=0}^{\infty}c_n\xi^n$

where $|c_n|\le\frac{2(2\pi R||f||_{\infty})^n}{n!}$ so the power series has an infinte radius of convergence.

By analytic continuation, a nontrivial analytic function (and an entire function is certainly analytic) cannot be compactly supported. If it's zero on an open set, then it's zero everywhere.

Moreover, $f$ and $\widehat f$ cannot be simultaneously supported by a set of finite measure (unless $f=0$, of course). If you know (or can solve) this, try to show also that $f$ and $\widehat f$ cannot be simultaneously supported by $E\cup (0,+\infty)$ where $E$ is some set of finite measure. The only proof of that I know is quite complicated (too complicated to post here anyway). Maybe somebody will come up with some simple idea?  ;)

Here's the uncertainty inequality. Define the Fourier transform as it is here (with $2\pi$ in the exponent of the exponential). Assume that $f\in L^2(\mathbb{R})$ with $||f||_2=1.$ By Plancherel, this also means that $||\hat{f}||_2=1.$ Interpret $|f(x)|^2dx$ as a probability density for $x$ and $|\hat{f}(\xi)|^2d\xi$ as a probability density for $\xi.$ Let $\sigma_x$ be the standard deviation of the first probability density and $\sigma_{\xi}$ the standard deviation for the second probability density. (Either or both of the standard deviations might be infinity.) Then,

$\sigma_x\cdot\sigma_{\xi}\ge4\pi.$

If $f$ is a Gaussian ([i]i.e.[/i], a normal distribution), equality is achieved.

Hehe, quite interesting... I thought it was easier :blush: . It seems it is the key point for the Heisenberg Uncertainty Principle.


For how many positive integer values of $ n$ is $ 3^n$ a factor of $ 15!$?

$ \left\lfloor \frac {15}{3} \right\rfloor \plus{} \left\lfloor \frac {15}{3^2}\right\rfloor \equal{} 5 \plus{} 1 \equal{} \boxed{6}$.


calculus integration trigonometry logarithms function Princeton college

hello there,   folks. :D

  show
(1) $ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\;\minus{}\;\cos\,(\beta\,x)}{x}\;\textbf dx\;\equal{}\;\boxed{\ln\left(\dfrac{\beta}{\alpha}\right)}$

then use it to show
(2)  $ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\;\minus{}\;\cos\,(\beta\,x)}{x^2}\;\textbf dx\;\equal{}\;\boxed{\dfrac{\pi}{2}(\beta\minus{}\alpha)}$

now use (2) to compute
(3) $ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\;\minus{}\;\cos\,(\beta\,x)}{x^3}\;\textbf dx\;$  :lol:

I am not familiar with the "cosine integral" that is yielded by the first integral.

The first is Frullani's Integral.
I leave (b) and (c) to someone else.

[quote="The Wicker Man"]The first is Frullani's Integral.[/quote]
But how can you prove that $ \cos (0) \minus{} \cos (\plus{}\infty) \equal{} 1$ :huh:

As for #3:

If $ \alpha\ne\beta,\quad\int_0^{\infty}\frac{\cos\alpha x\minus{}\cos\beta x}{x^3}\,dx$ diverges,

since the integrand behaves like a constant times $ \frac1x$ as $ x\to0.$

The misan principal value still exists by formal calculation though  :P

$ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\; \minus{} \;\cos\,(\beta\,x)}{x}\;\textbf dx\;\equal{}\int_0^{\infty}\mathcal{L} \{\,\cos{\alpha s}\minus{}\cos{\beta s}\}\;\textbf ds\;\equal{}\int_0^{\infty}\left(\frac{s}{s^2\plus{}\alpha^2}\minus{}\frac{s}{s^2\plus{}\beta^2}\right)\;\textbf ds\;\equal{}\ln{\frac{\beta}{\alpha}}$.

$ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\; \minus{} \;\cos\,(\beta\,x)}{x^2}\;\textbf dx\; \equal{}\int_0^\infty\;\dfrac{\beta\sin\,(\beta\,x)\; \minus{} \alpha\;\sin\,(\alpha\,x)}{x}\;\textbf dx\; \equal{}\int_0^\infty\;\mathcal{L}\{\,\beta\sin\,(\beta\,s)\; \minus{} \;\alpha\sin\,(\alpha\,s)\}\textbf ds\; \equal{}\int_0^\infty\;\left(\frac{\beta^2}{s^2\plus{}\beta^2}\minus{}\frac{\alpha^2}{s^2\plus{}\alpha^2}\right)\;\textbf ds\;\equal{}\frac{\pi}{2}(\beta\minus{}\alpha)$.

Another solution to the first part:

We start with:
\[ \log (t)\equal{}\int_0^\infty \frac{e^{\minus{}x}\minus{}e^{\minus{}tx}}{x}dx\]
Use cauchy's theorem to turn the line of integration by $ \pi /2$, so
\[ \log (t)\equal{}\int_0^\infty \frac{e^{\minus{}ix}\minus{}e^{\minus{}itx}}{x}dx\]
\[ \equal{}\int_0^\infty \frac{\cos{x}\minus{}\cos{tx}}{x}dx\minus{}i\int_0^\infty \frac{\sin{x}\minus{}\sin{tx}}{x}dx\]
If $ t$ is real (and positive), then the imaginary part of $ log(t)$ is zero, so we have
\[ \int_0^\infty \frac{\cos{x}\minus{}\cos{tx}}{x}dx\equal{}\log(t)\]
Finally,
\[ \int_0^\infty \frac{\cos{\alpha x}\minus{}\cos{\beta x}}{x}dx\equal{}\int_0^\infty \frac{\cos{x}\minus{}\cos{\beta x}}{x}dx\minus{}\int_0^\infty \frac{\cos{x}\minus{}\cos{\alpha x}}{x}dx\]
\[ \equal{}\log(\beta)\minus{}\log(\alpha)\equal{}\log(\frac{\beta}{\alpha})\]

[quote="misan"]
  show
(1) $ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\; - \;\cos\,(\beta\,x)}{x}\;\textbf dx\; = \;\boxed{\ln\left(\dfrac{\beta}{\alpha}\right)}$[/quote]
\begin{eqnarray*}
  \int_0^\infty  {\frac{{\cos (\alpha x) - \cos (\beta x)}}
{x}\,dx}  &=& \int_0^\infty \! {\int_0^\infty  {\bigg\{ {e^{ - xy} \cos (\alpha x) - e^{ - xy} \cos (\beta x)} \bigg\}\,dx} \,dy}  \hfill \\
   &=& \int_0^\infty  {\left\{ {\frac{y}
{{\alpha ^2  + y^2 }} - \frac{y}
{{\beta ^2  + y^2 }}} \right\}\,dy}  \hfill \\
   &=& \frac{1}
{2}\ln \frac{{\alpha ^2  + y^2 }}
{{\beta ^2  + y^2 }} \Bigg|_0^\infty   \hfill \\
   &=& \ln \frac{\beta }
{\alpha }.\quad\blacksquare
\end{eqnarray*}

[quote="misan"]show
(2)  $ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\; - \;\cos\,(\beta\,x)}{x^2}\;\textbf dx\; = \;\boxed{\dfrac{\pi}{2}(\beta - \alpha)}$[/quote]

\begin{eqnarray*}
  \int_0^\infty  {\frac{{\cos (\alpha x) - \cos (\beta x)}}
{{x^2 }}\,dx}  &=& \int_0^\infty \! {\int_\alpha ^\beta  {\frac{{\sin (xy)}}
{x}\,dy} \,dx}  \hfill \\
   &=& \int_\alpha ^\beta \! {\underbrace {\int_0^\infty  {\frac{{\sin (xy)}}
{x}\,dx} }_{\frac{\pi }
{2}}\,dy}  \hfill \\
   &=& \frac{\pi }{2}(\beta  - \alpha ).\quad\blacksquare
\end{eqnarray*}

Is this stuff one would learn in analysis?

What's up JRav? You like misan's problems or not? I really like his(er) questions, most of them are pretty hard and interesting.

[quote="JRav"]In what class would one learn this?[/quote]
I dunno exactly, I think one sees it when covering Calculus III. (Here in Chile we see multiple integration on Calculus II.) Anyway one have to consider several facts to solve hard problems, one of them is practice.

There're lots of hard integrals which can be solved by applyin' double integration or series, it's pretty cool.

so Black, using a similar technique, can you show ( for $ \alpha, \beta > 0$ )

$ \int_0^\infty\dfrac{\tan^{ \minus{} 1}\,(\alpha\, x) \minus{} \tan^{ \minus{} 1}\,(\beta\,x)}{x}\;\textbf dx\; \equal{} \;\boxed{\dfrac{\pi}{2}\;\ln\,\left(\dfrac{\alpha}{\beta}\right)}$

don't use frullani.  :lol:

[ PS: edited: "upper limit" for the above integral   edited as per Black's post below.   :blush:  ]

Of course the upper integration limit should be $ \infty,$ in that case makes sense to talk about Fullani's Integral.

We start with a similar fashion: we're going to create a double integral, and the only reason to do this is to reverse integration order with Fubini-type theorem.

\begin{eqnarray*}
  \int_0^\infty  {\frac{{\arctan (\alpha x) - \arctan (\beta x)}}
{x}\,dx}  &=& \int_0^\infty \! {\int_\beta ^\alpha  {\frac{{dy\,dx}}
{{1 + x^2 y^2 }}} }  \hfill \\
   &=& \int_\beta ^\alpha \! {\underbrace {\int_0^\infty  {\frac{{dx}}
{{1 + x^2 y^2 }}} }_{\frac{\pi }
{{2y}}}\,dy}  \hfill \\
   &=& \frac{\pi }
{2}\ln \frac{\alpha }
{\beta }.\quad\blacksquare
\end{eqnarray*}

[quote="The Wicker Man"]$ \int_0^\infty\;\dfrac{\cos\,(\alpha\,x)\; \minus{} \;\cos\,(\beta\,x)}{x}\;\textbf dx\; \equal{} \int_0^{\infty}\mathcal{L} \{\,\cos{\alpha s} \minus{} \cos{\beta s}\}\;\textbf ds$[/quote]

I don't see that, could anybody explain me wich property was aplied here?.

Have you ever seen Laplace Transform? That's was applied there.

Can you also show

$ \int_0^\infty \frac{|\cos \alpha x|\minus{}|\cos \beta x|}{x}\, dx\equal{}\left(1\minus{}\frac{2}{\pi}\right)\,\ln\left(\frac{\beta}{\alpha}\right)$

and

$ \int_0^\infty \frac{|\cos \alpha x|\minus{}|\cos \beta x|}{x^2}\, dx\equal{}\beta\minus{}\alpha \qquad\qquad ?$

[quote="JRav"]I think they're interesting but I have no clue how to solve them.  In what class would one learn this?[/quote]

You might learn some simple versions of these techniques in a Complex Analysis class or a college Advanced Calculus course. Some complex analysis classes, however, spend more time of integration techniques and special functions than others.

In general a focus on teaching complicated integration techniques, which was once popular in the British University system (see old Cambridge Tripos Exams), has diminished greatly from what it was, say 100-150 years ago.

A classic book that presents some of these types of integrals and special functions is [u]A Course of Modern Analysis[/u] by Whittaker and Watson. A lot of the exercises are old Tripos Exam questions. Although the title says "Modern" keep in mind that it was written in around 1902. It was, however, still being used for the second semester Complex Analysis class at Princeton in the 1990s. I believe a more modern book is now being used for their complex analysis class and I doubt that many American universities still use Whittaker and Watson nowadays. I could be wrong about that though.

A good book to get you introduced to the advanced integration techniques used in this thread is [u]Schaum's Outline of Advanced Calculus[/u] by Robert Wrede and Murray Spiegel. Its very readable by a high school student with a BC Calculus background.

We know $ \int_0^\infty \frac{\cos \alpha x\minus{}\cos \beta x}{x}\, dx\equal{}\log\frac{\beta}{\alpha}$

and $ \int_0^\infty \frac{\cos\alpha x\minus{}\cos\beta x}{x^2}\, dx\equal{}\frac{\pi}{2}(\beta\minus{}\alpha)$ .

And we know the Fourier series $ |\sin \alpha x|\equal{}\frac{2}{\pi}\plus{}\frac{2}{\pi}\sum_{n\equal{}1}^\infty \left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)\cos 2n\alpha x$

and $ |\cos \alpha x|\equal{}\frac{2}{\pi}\plus{}\frac{2}{\pi}\sum_{n\equal{}1}^\infty (\minus{}1)^n\left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)\cos 2n\alpha x$ .

Therefore $ \frac{|\cos\alpha x|\minus{}|\cos \beta x|}{x}\equal{}\frac{2}{\pi} \sum_{n\equal{}1}^\infty (\minus{}1)^n\left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)\, \frac{\cos 2n\alpha x\minus{}\cos 2n\beta x}{x}$ .

And this implies $ \int_0^\infty \frac{|\cos \alpha x|\minus{}|\cos \beta x|}{x}\, dx$

$ \equal{}\frac{2}{\pi} \underbrace{\sum_{n\equal{}1}^\infty (\minus{}1)^n \left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)}_{\equal{}\frac{\pi}{2}\minus{}1}\, \log\frac{\beta}{\alpha}\equal{}\left(1\minus{}\frac{2}{\pi}\right)\log\frac{\beta}{\alpha}$ .

Analogous $ \int_0^\infty \frac{|\cos \alpha x|\minus{}|\cos \beta x|}{x^2}\, dx$

$ \equal{}\frac{2}{\pi}\sum_{n\equal{}1}^\infty (\minus{}1)^n \left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)\underbrace{\int_0^\infty \frac{\cos 2n\alpha x\minus{}\cos 2n\beta x}{x^2}\, dx}_{\equal{}\frac{\pi}{2}(2n\beta\minus{}2n\alpha)}$

$ \equal{}\underbrace{\sum_{n\equal{}1}^\infty (\minus{}1)^n\, 2n\, \left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)}_{\equal{}1}\, (\beta\minus{}\alpha)\equal{}\beta\minus{}\alpha$ .

And $ \int_0^\infty \frac{|\sin \alpha x|\minus{}|\sin \beta x|}{x}\, dx$

$ \equal{}\frac{2}{\pi}\sum_{n\equal{}1}^\infty \left(\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\minus{}1}\right)\,\int_0^\infty \frac{\cos 2n\alpha x\minus{}\cos 2n\beta x}{x}\, dx\equal{}\minus{}\frac{2}{\pi}\log\frac{\beta}{\alpha}$ .


Let $ a \geq b$ be real number such that $ a^2\plus{}b^2 \equal{} 31$ and $ ab \equal{} 3.$ Then $ a\minus{}b$ equals

$ \text{(A)}\ 5 \qquad \text{(B)}\ \frac{31}{6} \qquad \text{(C)}\ 2 \sqrt{6} \qquad \text{(D)}\ \frac{5}{6} \sqrt{31} \qquad \text{(E)}\ \frac{5}{6} \sqrt{37}$

[hide="Solution"]
$ a^2\plus{}b^2\equal{}31$
$ ab\equal{}3$
$ a^2\plus{}b^2\minus{}2ab\equal{}31\minus{}2(3)$
$ (a\minus{}b)^2\equal{}25$
$ a\minus{}b\equal{}\pm 5$
Since $ a\geq b$, $ a\minus{}b$ must be either positive or zero.  Therefore, the answer is $ \boxed{\text{(A)}\ 5}$. [/hide]


geometry
3D geometry
number theory unsolved
number theory

Prove that there are infinitely many positive integers $n$ such that $n \times n \times n$ can not be filled completely with 2 x 2 x 2 and 3 x 3 x 3 solid cubes.

[quote="rgep"]Consider one face of the cube.[/quote]

I'm not sure how this fact helps. Can you give further explanation?


I did honors chemistry this year and next year will be taking AP Chemistry using the 11th edition of Brown. Will I be prepared for USACO to get by the first round if I do that?

Yes, totally http://en.wikipedia.org/wiki/USACO

Also, you might want to try some practice problems: http://train.usaco.org/usacogate You may also consider taking AP Computer Science A and learning Java, Python, or another programming language to help you with the USA Computing Olympiad if you have not done so.

[hide]The first round can be done without AP Chemistry material so you are all set. Also, if you wish to go to the upper levels of USNCO, you may not be able to go to MOP. [/hide]

lol ummm i think you USNCO.  USACO refers to the Computer Science Olympaid and I don't see how taking Honors Chemistry will prepare you for that :P.  And yes, if you take AP Chemistry you'll be prepared to get by the first round of USNCO.

I believe USACO was the thread's subject. I see that you are a PROMYS alumnus. Attending both MOP and the Chemistry Olympiad camp is not allowed while going to USACO and MOP has been done many times in the past.

Last year's IOI team (like IMO except for computers) had at least two MOP participants that I know of: a blue MOP student and a black one.

haha sorry no I meant USNCO  :blush:
[hide]i doubt i will make blue so its not a big deal  :lol: [/hide]

What is the typical minimum score on the local exam required to get past the first round of the UNSCO?  How about on the national exam to get to the summer camp?

^For the local exam, our section was pretty easy so it was only about 45/60, but I know that for some sections, it can be like 58/60.  I don't know the other question because we never got our National scores back.


Given the series $ 2\plus{}1\plus{}\frac {1}{2}\plus{}\frac {1}{4}\plus{}...$ and the following five statements:
(1) the sum increases without limit
(2) the sum decreases without limit
(3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small
(4) the difference between the sum and 4 can be made less than any positive quantity no matter how small
(5) the sum approaches a limit
Of these statments, the correct ones are:

$\textbf{(A)}\ \text{Only }3 \text{ and }4\qquad
\textbf{(B)}\ \text{Only }5 \qquad
\textbf{(C)}\ \text{Only }2\text{ and }4 \qquad
\textbf{(D)}\ \text{Only }2,3\text{ and }4 \qquad
\textbf{(E)}\ \text{Only }4\text{ and }5$

[hide]On quick inspection, (1) and (2) are not true and (5) is true. I don't really understand what becomes very small but if it means teh positive quantity being minutely small, (4) is true, whichimplies that the answer is $ \boxed{E}$.[/hide]

What does statement 4 mean?
Confusing....

In other words, as the number of terms approaches infinity, the difference between the finite series and 4 approaches 0.


Hey,
Could you guys delete the blogs that are marked for deletion?

I was hoping to remake mine, but cannot because it is still awaiting deletion.

Thanks

Do you want yours deleted, or would you like your old one revived?

If an admin does delete for good, is it possible to revive?

[quote="chess64"]Can someone revive my old one?[/quote]

Revived.

As an answer to 236factorial, if we nuke it for good, it's gone.

Is there a way to revive my first blog that I deleted?


Suppose $a = \lg z + \lg (x(yz)^{-1}+1)$, $b = \lg x^{-1} + \lg(xyz+1)$, $c=\lg y+\lg ((xyz)^{-1}+1)$. Let $M$ be the maximum value of $a,b,c$. Find the min value of $M$.

[i]source : China Provincial Test 1996, the original problem is a filling-in-blank problem.[/i]

[quote="digitalboy1"]Sorry, but I quite don't understand the problem, what do you mean by "M is the [i]maximum[/i] value of a,b,c. Find the [i]minimum[/i] value of M' ?[/quote]
for example: if $a<b<c$ then $M=c$. Find the minimum of $c$ if it is greater than $a$ and $b$.

any solution??? any restiction on $x , y, z $


nC0+nC1+nC2+nC3+nC4+nc5------------nCn=2^n
pl explain this result logically in a easy manner.i know one proof by binomial therom .explain other logical way.
_________________

Hmmm.... it seems to me that nC0.... nCn represents all the different ways you can make selections from the set of n.  I think it's obvious that it covers every possible selection exactly once.

There are 2 possibilities for every person.  The person CAN be selected or CANNOT be selected.  If you understand the concept of multiplying along independent pathways, just skip straight to the conclusion that the selection of a given person is independent of the selection of others in the set of people and thus it is 2*2*2... n times.

If this doesn't make sense, try this.  Imagine a tree-like diagram with branches sprouting from the previous levels.  Think of the selection of the first person as the first level of branching.  We have 2 options for the first guy (selected or not selected).  From EACH of these two options sprout 2 more branches representing possible outcomes of the selection of the second person.  From each of the 4 resulting branches 2 more can sprout making the total 8 for branching level 3...  as you see, independent pathways multiply.

[geogebra]1f4ed9841c0cec7cdd7ed177d10a7f4c06966407[/geogebra]

This tree diagram visually shows how this technique exactly covers every outcome once, just like the nC0 + nC1.... + nCn technique.

I hope this helps and you like my beautiful diagram.

Aha!  I remember a combinatorial problem that involves this concept.

"Let S be a set with six elements.  In how many different ways can one select two not necessarily distinct subsets of S so that the union of the two subsets is S?  The order of selection does not matter; for example the pair of subsets (a,c) (b,c,d,e,f) represents the same selection as the pair (b,c,d,e,f) (a,c)."

The not necessarily distinct means that there can be overlap between the sets, like with c in the example the problem gave.


algebra
polynomial
calculus
integration
linear algebra

Find p in P_3 (R) such that p(0) = 0 and p'(0) = 0, and integral from 0 to 1 of |2+3x - p(x)|^2 is as small as possible

This is orthogonal projection in $ L^2[0,1]$. You want to find the projection of $ 2\plus{}3x$ onto the subspace spanned by $ x^2$ and $ x^3$.


LaTeX
algebra
polynomial
geometric series
probability
expected value

[quote="sunnyboy780"]No latex, since I'm new.

Let x = 3 + (3+d)/4 + (3+2d)/16 ... (your sum).  Then, 4x = 12 + 3+d + (3+2d)/4 + (3+3d)/16 ...

Subtract the two, and you get 3x = 9 + d + d/4 + d/16...

It should be obvious from here, the sum of d + d/4 + d/16... is 4d/3 by the infinate geometric series sum formula.

3x = 9 + 4d/3, x = 3 + 4d/9[/quote]

Isnt there any other way? I got this using method of difference that is same as above , and if i am correct this is an AGP??
No other summations? Like using the general term and then using sigma from 1 to infinity??

There are a couple of general ways to compute the sum $ \sum_{k\equal{}0}^{\infty} P(k) x^k$ for a polynomial $ P$; one is by repeatedly differentiating $ \frac{1}{1 \minus{} x}$ and summing, for example (equivalently, you can raise this series to various powers).  Here in the linear case there is also a probabilistic method:  $ \sum_{k\equal{}0}^{\infty} \frac{k}{r^{k\plus{}1}}$ is the expected number of times you need to toss an $ r$-sided die before you get any particular face, which is $ r$.

[quote="t0rajir0u"]There are a couple of general ways to compute the sum $ \sum_{k \equal{} 0}^{\infty} P(k) x^k$ for a polynomial $ P$; one is by repeatedly differentiating $ \frac {1}{1 \minus{} x}$ and summing, for example (equivalently, you can raise this series to various powers).  Here in the linear case there is also a probabilistic method:  $ \sum_{k \equal{} 0}^{\infty} \frac {k}{r^{k \plus{} 1}}$ is the expected number of times you need to toss an $ r$-sided die before you get any particular face, which is $ r$.[/quote]
 I got the probabilistic method, thanks!


Fie $A_{1}, A_{2}, \ldots, A_{n}$ $n$ puncte, nu toate pe aceeaşi dreaptă.

Demonstraţi că ele formează cel puţin $n$ drepte distincte.

In esenta cazul cand se obtine nr.minim de drepte este atunci cand n-1 puncte coliniare si atunci se formeaza n drepte.In alte situatii evident ca se formeza un nr de drepte>n.
De aici concluzia problemei.

As sugera sa te gandesti de mai multe ori inainte de a vorbi de "cazul cel mai nefavorabil"...

Pe lângă soluţia prin inducţie (cred că e în Iaglom) sun posibile şi altele, de exemplu:
Fie $P_{i},i=1,\ldots ,n$ punctele. Să presupunem că ele determină $m$ drepte, $d_{j},j=1,2,\ldots,m.$ Considerăm matricea $A=(a_{ij})\in \cal{M}\it_{n,m}(\mathbb{R}),$ unde \[a_{ij}=\left\{ \begin{array}{ll}1, & P_{i}\in d_{j}\\ 0, & P_{i}\notin d_{j}\\ \end{array}\right.\] Atunci matricea $B=A\cdot^{t}A$ are ordinul $n$ şi $\det{B}\ne 0,$ deci $n\le m.$
Daca $B=(b_{ij}),$ sa observam ca $b_{ij}$ reprezinta numarul de drepte care trec prin punctele $P_{i}$ si $P_{j}$. De aceea, pentru $i\ne j$ avem $b_{ij}=1$ iar in rest $b_{ii}\ge 2.$


Let $ n$ be a natural number. How many set $ A\subset\{1,2,...,2n\}$ satisfy $ \nexists x,y\in A: x\plus{}y\equal{}2n\plus{}1$

Partition $ \{1,2,\dots,2n\}$ into $ \{1,2n\}$, $ \{2,2n\minus{}1\}$, ..., $ \{n,n\plus{}1\}$.  Clearly, $ x\plus{}y \equal{} 2n\plus{}1$ iff they both belong to the same subset.  Hence, there are $ 2^{n}$ such sets (since you may choose an element from each subset arbitrarily).

In fact, there are $ 3^{n}$ such subsets: from each member of the partition, you may choose the left element, the right element, or neither element.


From what I know (which may not be correct), MIT has a lot of people in math major, and Princeton has only a few (but they are all good).  Princeton is another good school to consider, but you may not find too much people doing the same thing as you do.

I don't know about caltech (but it cannot be bad), but the weather is good.  The only bad thing you might say about the campus is that it's really small....

How good is Harvard's math department?  I hear of many people associated with math things who go there.

If anyone has specific questions about the Princeton math department, I will be happy to answer. If you have specific questions about MIT, try joml88. For Harvard, JBL. I'm not sure if there's an obvious choice of whom to ask about Chicago or Caltech; there are definitely several people doing math at Stanford and Harvey Mudd who read this forum.

I'll be going to Caltech, so I can probably help people with questions sometimes next year.

[quote="pagene"]I'm applying to colleges in the fall and want to be a math major. I'm interested in pure maths much more than applied maths. I know Caltech and MIT have really good math departments (especially MIT) but they seem more focused on applied math. My question is, if I'm interested in pure math, would I find a better math department elsewhere (like Princeton or Chicago)? Thanks[/quote]

The math department at MIT is very large.  Having a good applied program doesn't necessarily imply and lesser theoretical program.  If you look at just about all of the other majors at MIT, they're very math oriented.  Thus it makes sense to place an emphasis on having a strong applied math program and, conversely, for the math program to become stronger as a result of the other departments relying on applied math.  That said, the theoretical math program has a lot to offer.  It, also, is top notch.

One way for you to do your own research into this is to look at the course catalogs for the schools you're looking at.  This usually is a fairly good indication of what the program will be like.  Also, talking to people at that school in that major is probably the best way to get information.

It's also important to consider whether the all-star faculty is accessible to students. At small universities (especially ones with small grad schools too) this is likely to be the case. Ask students at the university.

Some schools have a great reputation for math, but really it's mainly the graduate program that's good because the best professors spend most of their time with the grad students.

Of course, I do have an ulterior motive in pointing this out -- Princeton is particularly notable for the accessibility of the faculty to undergrads. There are nearly twice as many math faculty as undergraduate majors, and the number of grad students is around the number of undergrad majors. So the profs really don't need to divide their attention too much. And the famous profs don't use their status to hide away from the undergrads, either. Most of them regularly teach undergrad classes, and you can do seminars or reading courses with them too once you are a math major.

I think that all of the schools which you have listed have good enough (pure) math programs that your decision should mostly be based on other criteria. Consider life, other academics, campus, location, finaid...


How to prove these inequalities without using college knowledge , because I'm just a high-school student

$a_{i}>0 \ \ , \ i=1 \ , \ 2 \ , \ 3 \ , \cdots \ n$

$\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}}\geqslant\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\geqslant\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\geqslant\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}$

for the first one, apply Svac.
the second one, using Cauchy.
the third, Cauchy again (with the denominator)

Sorry, I don't have much time to type it out, hope you understand  :D )


[quote="Master"]it is new in 'M and Y'
[find min of A
A=sum((a^6)/(a^3+b^3))
when a,b,c >0 and a+b+c=1]
it is easy to prove that A>=1/18
but  with one more condition ,we can prove that A>=1/4
can anyone find this? :D[/quote]
We have $ \sum\frac{a^{6}}{a^{3}\plus{}b^{3}}\geq\frac{(a^{3}\plus{}b^{3}\plus{}c^{3})^{2}}{2(a^{3}\plus{}b^{3}\plus{}c^{3})}$
If you want to prove $ A\geq\frac{1}{4}$,we need prove:
$ 2(a^{3}\plus{}b^{3}\plus{}c^{3})^{2}\geq a^{3}\plus{}b^{3}\plus{}c^{3}$
This inequality is true because $ a;b;c > 0$!Done! :blush:


a) Rezolvati ecuatia:
$\frac{a}{b}+\frac{b}{a}=2007$. Generalizare.
b) Rezolvati ecuatia:
$\frac{a_{1}}{a_{2}}+...+\frac{a_{2007}}{a_{1}}=2007$. Generalizare.

in ce multime trebuie rezolvate ecuaţiile?


calculus
integration
function
calculus open

I have heard many times that the integral of $ e^{x^2}$ cannot be expressed in terms of elementary functions.   Of course, I agree about it, but how can I prove that?

Thanks

P.S.: To be honest, some guy asked me about that and I said "Well, I am pretty sure about that is true, but I don't even know how to prove it  :huh: "

I couldn't find a proof online but if you can get a hold of this book, you'll get a proof:
[i]The Integration of Functions of a Single Variable[/i] (2nd edition), GH Hardy; Cambridge University Press 1958

http://math.hunter.cuny.edu/ksda/papers/churchill.pdf

Corollary 5.5


superior algebra
superior algebra solved

1. Let (A,+,*) be a ring with 8 elements and characteristic 2. Prove that A is a field iff the eqn a^3+a+1=0 has at least a soln in A.

2. Let (A,+,*) be a ring. a is a non-invertible element in A. Let Xa={x in A|ax=1}. Prove that if Xa has at least an element then it has infinitely many elements.

2)

Let b in Xa fixed, f : Xa -> Xa , f(x) = xa + b - 1
af(x) = axa + ab - a = ab = 1
f(x) = f(y) => xa + b - 1 = ya + b - 1 => xa = ya => xab = yab => x=y

Assume Xa is finite => f is injective, thus bijective. Since b is in Xa, there is c such that f(c) = b => ca + b - 1 = b => ca = 1

cab = (ca)b = b = c(ab) = c

Therefore c=b and ba=ab=1, so a is invertible, contradiction.

So Xa cannot be finite.

1)

First of all let us establish the implication A is a field => the equation a^3 + a + 1 = 0 has at least one root.

If A is a field, then for every x in A\{0}, x^7 = 1 => since the 7 elements in A\{0} are distinct, this equation of degree 7 has 7 roots over A. Since it's a field, unique factorisation works in A[X] so if we prove that x^3+x+1 divides x^7-1, given that char(A)=2, we are done.
That's easy since (x^3+x+1)(x^4+x^2+x+1)=x^7-1
Therefore the equation has at least three roots (so in particular, one)

If the equation a^3+a+1=0 has one root in the ring A=> a!=0 or else 0=1, so the ring has one element.
(a^3+a+1)(a^4+a^2+a+1)=a^7-1=0, because char(A)=2
So a^7=1, therefore the elements of {1,a,a^2,..,a^6} are all roots of a^7=1. So it's clear now that A = {0} U <a> and thus every element in A\{0} is invertible. Therefore, it's a field.

In particular, we notice that we get that it's commutative for free, no need to use Weddeburn's theorem.


inequalities
symmetry
inequalities unsolved

If [tex]a\ge b \ge c \ge0[/tex] and [tex]a+b+c=3[/tex],then prove that [tex]ab^2+bc^2+ca^2\le\frac{27}{8}.[/tex]
Sonnhard.

(a+b+c)^3 = 27 = a^3 + b^3 + c^3 + 6abc + 3(LHS) + 3(a^2b + b^2c + c^2a)

We apply Schur's inequality,

a^3 + b^3 + c^3 + 3abc >= ab(a+b) + ac(a+c) + bc(b+c)

to obtain:

27 >= 3abc + 4(LHS) + 4(a^2b + b^2c + c^2a)

We now need to show that:

4(a^2b + b^2c + c^2a) + 3abc >= 4(ab^2 + bc^2 + ca^2)

But this is true because:

(a^2b + b^2c + c^2a) - (ab^2 + bc^2 + ca^2) = (a-b)(b-c)(a-c), which is clearly nonnegative.

Equality holds if a=b=3/2, c=0.

Thanks for the assist, fuzzy.

[quote="blahblahblah"] now need to show that:

4(a^2b + b^2c + c^2a) + 3abc >= 4(ab^2 + bc^2 + ca^2)
[/quote]

You are on the right track, to finish the proof, note that

$(a^2b + b^2c + c^2a)  - (ab^2 + bc^2 + ca^2) = (a-b)(b-c)(a-c) \ge0$.

And here's another similar solution:

With f(a,b,c) = ab^2 + bc^2 + ca^2

Looking at the equality case, we consider:

f([a+b+c]/2,[a+b+c]/2,0) - f(a,b,c) =

(a+b+c)^3/8 - ab^2 + bc^2 + ca^2

We pair off terms with Schur and use the earlier factorization to obtain that f([a+b+c]/2,[a+b+c]/2,0) - f(a,b,c) >= 0 as desired.

This seems to be a similar problem to the CMO #5, 1999:
http://www.cms.math.ca/Competitions/CMO/solutions/sol_1999.pdf

However, I believe that CMO #5, 1999 is weaker because of the way the two sequences are sorted.

[quote="blahblahblah"][quote="siuhochung"]hi, i'm not familiar with schur, can you tell me what schur inequality is?[/quote]

Schur (for t=1) states that a^3 + b^3 + c^3 + 3abc >= ab(a+b) +bc(b+c) +ac(a+c)

The proof follows from considering:

By symmetry, x >= y >=z
x(x-y)(x-z) >= y(y-x)(y-z) and z(z-x)(z-y) >=0

Adding and expanding, we get the desired result.[/quote]

thanks. :)


linear algebra
matrix
LaTeX
linear algebra solved

Let $A\in M_{n}\mathbb{(C)}$ a non-singular matrix such that $3ABA^{-1}+A=2A^{-1}BA$.
 prove that $det(ABA^{-1}-A^{-1}BA)=0$ ;)

This is my first post giving a solution to a problem. Please feel free to give comments.

By the hypothesis, we have
\[3ABA^{-1}+3A=2A^{-1}BA+2A\]
\[3A(BA^{-1}+I)=2(A^{-1}B+I)A\]
\[3A(B+A)A^{-1}=2A^{-1}(B+A)A\]
\[\Rightarrow \det(A+B)=0\]
\[\Rightarrow \det(BA^{-1}+I)=\det(A+B)\det(A^{-1})=0\]

Let $M=ABA^{-1}-A^{-1}BA$. Then $AM=A^2BA^{-1}-BA\Rightarrow BA=A^2BA^{-1}-AM$. Putting this to the equality in the hypothesis,
\[3ABA^{-1}+A=2A^{-1}(A^2BA^{-1}-AM)\]
\[A(BA^{-1}+I)=-2M\]
\[\Rightarrow \det(M)=0\]

Very good, you can also complete you prove using  A(BA^-1-I)=-2M and you finish it
very quickly  :P

sorry , the - shold be read +, all i want to say that the formula is direct without manipulation  :blush:

sorry i don`t know latex yet but  i will be short....  Let S be the sum for which we must prove that det(S)=0. Simply calculate 3S and 2S from the initial condition and you find that their determinants are equal ;)


It is a classical and beauiful problem to find all real addtivie functions with the property that $ f(x^n)=f^n(x)$ for all $x$, where $n$ is given. The problem is: can we make weaker the assumption, that is, what happens if we only ask that for any real $x$ there is an integer $n$ such that $ f(x^n)=f^n(x)$?

harazi , after 4 days of struggleing i have reached this :
Nothing , :blush:  :blush:  :blush:  at all.
Do you have any suggestion for the beginning ?

P.S:Can you give a solution for that classical problem,$f(x^n)=f^n(x)$?


function
invariant
number theory proposed
number theory

Define f(x) to be what we will call the "eventual sum" of x, which is an iterating function until it reaches a single digit. For example, f(4435672) goes first to 4+4+3+5+6+7+2 = 31, and since 31 is not a single digit we f(31) which gives 3+1 = 4, which is the single digit we're looking for, so f(4435672)=4.

My idea: prove that for any integer a, there exists some b such that a-100<b<a+100 and f(1),f(2),...,f(b) yields the exact same number of 1's, 2's, ..., 9's. Else: provide a counterexample for a.

Don't really know where to go with this one, but I'm pretty sure it's true. You can sort of see it in the graph of f(x).

The "eventual sum" of a number n will simply be the remainder of n upon division by 9 (where we use 9 instead of 0) since digit sums are invariant mod 9. Thus, if b is divisible by 9, the statement holds, and there is certainly a value of b divisible by 9 in the given interval.


I've been learning logarithms, but run into a problem.

I just don't understand base conversion.  Could some please derive the formula for me and explain it?

Here is what I got:

log2^20=  (log4^20)/(log4^2)

How would you use that formula to solve a common problem?

[quote="Vaele"]I've been learning logarithms, but run into a problem.

I just don't understand base conversion.  Could some please derive the formula for me and explain it?

Here is what I got:

[b]log2^20=  (log4^20)/(log4^2)[/b]
How would you use that formula to solve a common problem?[/quote]
Well itll be better to tell us what the problem is?  because that problem that you have has no variables to solve for nor does the equality even hold.

are you trying to say: $\log_{2}20$,
then that would become $\frac{\log{20}}{\log{2}}$

Basically, if you've got a log, like $\log_{2}5$, for example, you can rewrite it as $\frac{\log 5}{\log 2}$. The base goes in the numerator and the part in the log goes in the numerator. It doesn't matter what the base of the logs in the fraction is, as long as the numerator and denominator are the same. So you can say

$\log_{2}5 = \frac{\log 5}{\log 2}= \frac{\log_{7}5}{\log_{7}2}= \frac{\log_\pi 5}{\log_\pi 2}= \frac{\ln 5}{\ln 2}= \frac{\log_{983456}5}{\log_{983456}2}$

See? It doesn't matter what the base is that you're converting to.

Ok, I understand that now, but

1) How does this work?
2) What are some uses of it?

Here is a sample problem btw (sorry I didn't post one earlier):

Given log10^2 = a, and and log10^3 = b, write the following in terms of a and b.

log18^8/9 =

It works like this:

$\log_{a}b = c$ means that $a^{c}= b$.  Taking a logarithm in a base $d$ gives $\log_{d}a^{c}= \log_{d}b$ and so $c\log_{d}a = \log_{d}b$ and $c = \frac{\log_{d}b}{\log_{d}a}$.

As a step in there, I used the fact that $\log_{d}a^{c}= c\log_{d}a$.  We can prove this as follows:
If $k = \log_{d}a^{c}$ then $d^{k}= a^{c}$ and so $d^{\frac{k}{c}}= a$ and $\frac{k}{c}= \log_{d}a$ and $k = c\log_{d}a$.

oh ok, that makes sense

Thanks so much!


Obviously, this is a quadratic equation on $ x$. Therefore, we have

$ x_{1,2}\equal{}\frac{4cos\frac{\pi}{12} \pm \sqrt{16(cos\frac{\pi}{12})^2\minus{}24((cos\frac{\pi}{8})^4\minus{}(cos\frac{\pi}{12})^4})}{12}$

Now, it's a simple matter of calculating $ cos\frac{\pi}{12}$ and $ cos\frac{\pi}{8}$.
$ cos\frac{\pi}{6}\equal{}2(cos\frac{\pi}{12})^2\minus{}1$, and since $ cos\frac{\pi}{6}\equal{}\frac{\sqrt{3}}{2}$, we get that $ cos\frac{\pi}{12}\equal{}\frac{\sqrt{\sqrt{3}\plus{}2}}{2}$.
Similarly, we have $ cos\frac{\pi}{4}\equal{}2(cos\frac{\pi}{8})^2\minus{}1$, and since $ cos\frac{\pi}{4}\equal{}\frac{\sqrt{2}}{2}$, we get that $ cos\frac{\pi}{8}\equal{}\frac{\sqrt{\sqrt{2}\plus{}2}}{2}$.

So, we have $ cos\frac{\pi}{12}\equal{}\frac{\sqrt{\sqrt{3}\plus{}2}}{2}$, from which we get $ (cos\frac{\pi}{12})^2\equal{}\frac{\sqrt{3}\plus{}2}{4}$ and $ (cos\frac{\pi}{12})^4\equal{}\frac{7\plus{}4\sqrt{3}}{16}$.
We also have $ cos\frac{\pi}{8}\equal{}\frac{\sqrt{\sqrt{2}\plus{}2}}{2}$, from which we get $ (cos\frac{\pi}{8})^2\equal{}\frac{\sqrt{2}\plus{}2}{4}$ and $ (cos\frac{\pi}{8})^4\equal{}\frac{6\plus{}4\sqrt{2}}{16}$.

Now, we just use it in the original equation:
$ x_{1,2}\equal{}\frac{2\sqrt{\sqrt{3}\plus{}2} \pm \sqrt{4(\sqrt{3}\plus{}2)\minus{}24(\frac{6\plus{}4\sqrt{2}}{16}\minus{}\frac{7\plus{}4\sqrt{3}}{16}})}{12}$.
That's it, maybe you can simplify it some more, but I think this will do (also, in decimal notation the solutions can be approximated to $ \minus{}0.0349$ and $ 0.6788$).

Now, excuse me, but what I don't see is the point of this problem. If I haven't misread something, this is pure calculation and not nice whatsoever. There is nothing which requires you to think in this problem - you just need to be careful while calculating.

Excuse me once more if I've misread the problem or falsely calculated something, so I mistakenly fail to see the point behind this. (Is there some nice property of this equation, maybe? Some nice way to solve it?)

[quote="mornik"]\[ x_{1,2} \equal{} \frac {2\sqrt {\sqrt {3} \plus{} 2} \pm \sqrt {4(\sqrt {3} \plus{} 2) \minus{} 24(\frac {6 \plus{} 4\sqrt {2}}{16} \minus{} \frac {7 \plus{} 4\sqrt {3}}{16}})}{12}
\]
[/quote]
This can be "simplified" to
\[ x_{1, 2} \equal{} \frac{\sqrt{2} \left( \sqrt{3} \plus{} 1 \right)}{12} \pm \frac{1}{24} \sqrt{40\sqrt{3} \minus{} 24\sqrt{2} \plus{} 38}
\textrm{,}
\]
because of
\[ \sqrt{2 \plus{} \sqrt{3}} \equal{} \sqrt{\frac{\left( \sqrt{3} \plus{} 1 \right)^{2}}{2}} \equal{} \frac{\sqrt{2} \left( \sqrt{3} \plus{} 1 \right)}{2}
\textrm{.}
\]


[quote="conejita"]Demuestra que para todos los numeros naturales N existe un UNICO entero formado por N digitos del conjunto (1, 2, 3, 4, 5)  el cual es divisible por 5^N.

Es divertido este problema!!!

Espero soluciones!!

 :rotfl:[/quote]
Por inducción:
Para N=1: 5 es el unico entero de un digito del conjunto {1, 2, 3, 4, 5}, que es div. por 5.
suponiendo que es cierto para N=k.
Ahora veamos la cantidad de soluciones de:
$5^{k+1}$ divide a $a_{1}a_{2}\ldots a_{k+1}$ ; $a_{i}\in{1, 2, 3, 4, 5}$
Entonces 5^k divide a $a_{2}\ldots a_{k+1}$
De la hipotesis inductiva $a_{2}\ldots a_{k+1}$ es único.
luego al dividirlo entre $5^{k}$ es entero. Si este entero es de la forma 5p+t, $t\in{1, 2, 3, 4, 5}$
Sea s el resto que deja $2^{k}$ al dividirlo entre 5, el cual es diferente de 0, pues 2 y 5 son coprimos. Entonces  $5^{k+1}$ divide a $a_{1}.10^{k}+5^{k}(5p+t)$. Entonces 5 divide a $a_{1}(s)+t$ Como s es 1,2,3 o 4, se concluye que a_1 es unico en modulo 5. Es decir a_1 toma exactamente uno de los valores: 1,2,3,4,5.(a_1 es único).
Con lo que completamos la induccion.


Hello. I am a relative of someone who was at the Penn State ARML - I don't want to reveal too much info about what team they were on, but suffice it to say that about a dozen or more students on the teams within their region (and those in the region already know the region I'm referring to) came down with some sort of Flu and/or bad virus within 1 - 6 days of coming home. The symptoms most prevalent were REALLY high fevers (of like 103 degrees) and many also had respiratory symptoms such as sore throats, runny noses, and cold-like symptoms.  I am trying to find out if this Flu was contained to only one region (maybe they spread on their bus or within their dorms?) or if some students (or adults) in other/all regions at Penn State have been similarly sick?  If you are among them or know someone ill, please state your symptoms, and if you saw a doctor, what did he diagnose you with, and what medications did he prescribe or suggest? It's sort of scary how contact with others for only 24 or so hours (since most participants who got sick did NOT go to the same school), spread SO quickly to make so many so sick.  And for those of you who did get "hit" by illness, I wish you all a very speedy recovery! Oh, so far, no one has revealed having the "Swine Flu" - just "general flu" - but clearly, if someone was diagnosed with the "Swine Flu", or something more severe, it would be really helpful to other participants to have this information. You need not reveal your identity, obviously. Stay healthy!!

One of my teammates (Western PA) was diagnosed with Influenza A yesterday! And his parents believe it happened after he got back from ARML!  :o  Gee, I hope everyone is OK. I hope your relative's team gets better soon  :(

I, too came down with the flu, right after ARML. I think this was because I was sleeping in front of the air conditioner the whole time....

OR MAYBE IT'S a PANDEMIC!!!

This is scary...

West Virginia competitors showed no symptoms at the competition.  I had a stuffy nose for two days IMMEDIATELY following the competition, but I disregarded it as my sinuses acting up because of the severe change in temperature around my area earlier this week.  It's just nerve-racking to think that the virus could have ravaged around Penn State and possibly even some of the competitors.  Please inform all of us any details you find out regarding this.

worthawholebean - you guys (gals) are lucky! being sick is no fun.  :mad:   Thanks for posting though and letting all know.

There were probably 500 people from all over the northeastern quadrant of the country in relatively close contact for two days - I don't find it that surprising that something was going around.

Edit: Think about lunch on Saturday. We were packed really tightly into that cafeteria and expecially in the lines to enter it. I wouldn't even be that surprised if a few people had gotten H1N1 - it's contagious, yes, but not really any more dangerous than the normal flu.

//according to the media//

The media tends to blow things out of proportion.

[quote="gauss1181"]I thought, according to the media, that swine flu is significantly more serious than normal flu...? Or was that being said just to spread propaganda/make ppl more aware?[/quote] The swine flu is actually a bit [i]less[/i] dangerous than the normal flu. The media just had nothing to report on (the people were getting tired about the economy) and invented a new scare.

I actually got pretty sick the week after getting back from ARML.  I didn't have a fever, but I had other really bad cold-like symptoms. I'm taking medicine right now to get over it. That's in Virginia  :maybe:

I got a bad throat infection/cough (accompanied by a mild fever/congestion) but I started taking amoxicillin immediately, which helped and basically cured it so it appears it was bacterial and not caused by a virus, which Influenza would be caused by.

nine34 (and everyone else) - not sure what region you are from, but I can share the following: - a state that had multiple teams attending Penn State ARML from their region had [i]many[/i] participants report to me privately.  Approximately 10 students communicated with ME (so likely there were many more who have not communicated but no way of knowing these #'s - hence, why, in part, I started this post). What they said was this - virtually all experienced HIGH fevers when returning from ARML - ranging from a couple days to a week after attending - reported fevers usually hovered around 103 degrees.  There were also reports of a lot of upper respiratory symptoms - ie: lots of coughs, sore throats, leaky noses, etc., as well as headaches (usually at the start).  I haven't heard much about "body aches" though I think a couple of people might have mentioned.  Several told me they went to doctors who diagnosed as flu, but not sure if this was based on clinical tests or just "general knowledge/guess of physician".  One participant DID reveal they were being tested to see if the Flu was the "Swine Flu" in the news - but didn't receive the results as of yet.  Several told me they were either on Flu meds or antibiotics (I guess to treat the symptoms/infection).  Though I didn't think antibiotics were usually prescribed for flu, many people told me that's what their doctors gave them.

 Again, I am trying to ascertain just HOW widespread at ARML this particular flu or viral or bacterial infection has gotten, as people are STILL reporting either being sick or recently becoming ill.  Personally, I don't think anyone's illness was caused by a change in temperature or an open window. In fact, I think "being chilled" causing illness is a fallacy that has been largely disproved by medical science. And, no, I don't believe it's "normal" to have SO many people getting SO ill (with such high fevers and lingering symptoms) in such a short period of being together. Again, I will state I am NOT a physician nor a statistician, so I could be wrong. But again, I don't know how many people were infected. I hope this info helps someone, and PLEASE share if you or anyone you know (no names, of course) got sick after Penn State ARML.

people I communicated with indicated their docs said they had "flu" and gave them antibiotics - I question this practice myself, as I thought the flu was different than a bacterial infection requiring antibiotics.  I don't know if or why any drs prescribed antibiotics for what they thought were viral infections.


IMO
polynomial
roots
minimum value
algebra
IMO 1973
coefficients

Determine the minimum value of $a^{2} + b^{2}$ when $(a,b)$ traverses all the pairs of real numbers for which the equation \[ x^{4} + ax^{3} + bx^{2} + ax + 1 = 0 \] has at least one real root.

It is also necessary to prove that if $a^2+b^2\ge\frac45$ then the given equation has at least one real root.

Since $x=0$ isn't the solution of the given equation, we can rewrite as follows.

$x^{4}+ax^{3}+bx^{2}+ax+1=0\Longleftrightarrow x^{2}+ax+b+\frac{a}{x}+\frac{1}{x^{2}}=0\ (x\neq 0).$

By $\left|x+\frac{1}{x}\right|=|x|+\frac{1}{|x|}\geq 2\sqrt{|x|\cdot\frac{1}{|x|}}=2$ when the eqality holds when $|x|=\frac{1}{|x|}\ (x\neq 0)\Longleftrightarrow x=\pm 1.$

Let $x+\frac{1}{x}=t,$

we are to find the condition for which the equation with respect to $t, \ t^{2}+at+b-2=0$ has at least one real root at $t\leq-2\ or\ t\geq 2.$

The condition for which $a,\ b$ should satisfy this is given by $2|a|-2\leq b\leq \frac{1}{4}a^{2}+2$ and $|a|\geq 4\ \cdots [A],\ |b+2|\geq2|a|\ \cdots [B].$

Sketching the distribution satisfying $[A]$ or $[B]$ on $a-b$ plane gives the answer, $a^{2}+b^{2}$ has the minimum value of $\frac{4}{5}$ when $(a,\ b)=\left(\pm \frac{4}{5},\ \frac{2}{5}\right)$ which are the intersection points between two graphs of $b=2|a|-2$ and $b=-|a|.$

[color=darkblue][b]Lemma.[/b] For $k>0$ the function $f: [k,\infty)\rightarrow R\ ,\ f(x)=x+\frac{k^{2}}{x}$ is increasing. Particularly, $0<k\le a\le x\Longrightarrow a+\frac{k^{2}}{a}\le x+\frac{k^{2}}{x}$ with equality iff $x=a\ .$
Indeed, for $0\le k\le u<v$ we have $f(u)-f(v)=\frac{(u-v)(uv-k^{2})}{uv}<0$, i.e. $f(u)<f(v)$ because $uv>0$, $u-v<0$ and $u\ge k>0\ ,\ v>k>0\Longrightarrow uv>k^{2}\ .$

[b]Example.[/b] $m=\min_{x\ne 0}\frac{\left(x^{4}+1\right)^{2}}{x^{2}\left(x^{4}+3x^{2}+1\right)}=$ $\min_{y>0}\frac{\left(y^{2}+1\right)^{2}}{y\left(y^{2}+3y+1\right)}=$ $\min_{y>0}\frac{\left(y+\frac{1}{y}\right)^{2}}{\left(y+\frac{1}{y}\right)+3}=$ $\min_{z\ge 2}\frac{z^{2}}{z+3}=$ $\min_{t\ge 5}\frac{(t-3)^{2}}{t}$,
where $x\ne 0$ $\Longleftrightarrow$ $y=x^{2}>0$ $\Longleftrightarrow$ $z=y+\frac{1}{y}\ge 2$ $\Longleftrightarrow$ $t=z+3\ge 5\ .$ Thus, $m=\min_{t\ge 5}\left(t+\frac{9}{t}-6\right)=-6+\min_{t\ge 5}\left(t+\frac{9}{t}\right)\ .$ Using the above lemma $(3=k<a=5\le t)$, obtain $m=-6+\left(5+\frac{9}{5}\right)=\frac{4}{5}$ and the minimum points are $x\in \{\pm 1\}$ because $t=5\Longleftrightarrow z=2\Longleftrightarrow y=1\Longleftrightarrow x\in \{\pm\ 1\}\ .$

[b]The proof of the proposed problem.[/b] Can consider that the relation $x(x^{2}+1)\cdot \underline a+x^{2}\cdot \underline b+(x^{4}+1)=0$, $x\in R^{*}$ is the equation of a mobile line $d$in the analytical plane $aOb\ .$ Thus, square of the distance $PO$, where $P(a,b)\in d\ ,\ OP\perp d$ is $a^{2}+b^{2}=\frac{(x^{4}+1)^{2}}{x^{2}(x^{2}+1)^{2}+(x^{2})^{2}}=\frac{(x^{4}+1)^{2}}{x^{2}(x^{4}+3x^{2}+1)}\equiv f(x)\ .$ Therefore, $m=\min_{P(a,b)\in d}\{a^{2}+b^{2}\}=\min_{x\ne 0}f(x)=\frac{4}{5}$ and $x\in \{\pm 1\}$ (from the above example), i.e. the minimum points are the solutions of the systems $2\epsilon a+b+2=0\ ,\ a-2\epsilon b=0$, where $\epsilon^{2}=1\ .$  In conclusion, $a_{\mathrm{min}}=\frac{4\epsilon}{5}$ and $b_{\mathrm{min}}=-\frac{2}{5}$, where $\epsilon^{2}=1\ .$[/color]

[color=indigo][b]Proof for the proposed problem.[/b] Since $x=0$ isn't the solution of the given equation, we can rewrite as follows.

$x^{4}+ax^{3}+bx^{2}+ax+1=0\Longleftrightarrow x^{2}+ax+b+\frac{a}{x}+\frac{1}{x^{2}}=0\ (x\neq 0).$

By $\left|x+\frac{1}{x}\right|=|x|+\frac{1}{|x|}\geq 2\sqrt{|x|\cdot\frac{1}{|x|}}=2$ when the eqality holds when $|x|=\frac{1}{|x|}\ (x\neq 0)\Longleftrightarrow x=\pm 1.$

Let $x+\frac{1}{x}=t,$ then $|t|\ge 2$ and the given equation becomes: $t^{2}+at-b-2=0\iff t^{2}-2=-at-b$

We have: $(a^{2}+b^{2})(t^{2}+1)\ge (at+b)^{2}=(t^{-}2)^{2}\implies a^{2}+b^{2}\ge\frac{t^{4}-4t^{2}+4}{t^{2}+1}\quad (1)$

On the otherhand, \[\frac{t^{4}-4t^{2}+4}{t^{2}+1}-\frac{4}{5}=\frac{(5t^{2}-4)(t^{2}-4)}{5(t^{2}+1)}\ge 0\quad (2)\]

From $(1)$ and $(2)$ we have: $a^{2}+b^{2}\ge\frac{4}{5}$ [/color]

a nice solution pco but i am asking too about [the equation of a mobile line]i need more informations because it s the first time time that i heard about that.

After making the substitution t=x+1/x, analyse the 3 different cases. Either both roots of the quadratic are greater than 2 in absolute value and have same sign or both are greater than 2 in absolute value and have opposite sign or there is exactly 1 root between -2 and 2. Using properties of quadratic expressions we get the desired result.


[quote="Cube"](1) Danny's strategy for his first marathon ( 26.6 miles ) was to run 2 miles, walk 1 mile, run 2 miles, walk 1 mile, and continue this pattern until he completed the race. Danny's average running pace is 6 minutes per mile and his average walking pace is 12 minutes per mile. How many miles will it take Danny to complete the marathon? Express your answer as a decimal to the nearest tenth.

ANSWER: [hide]211.2[/hide]

(2) Let  ABCD  be a rectangle such that  AB=2BC=6  . Let  O  be a semicircle of diameter  AB  that is tangent to  CD  at  M and let  U  be a semicircle of diameter  CD  that is tangent to  AB  at  N . Find the number of square units in the area that inside the two semicircles. Express your answer as a decimal rounded to the nearest hundredth. Use  Pi=3.14.

ANSWER: [hide]11.05[/hide][/quote]
[hide]i am pretty sure you mean minutes in #1
he is going 24 minutes per 3 miles. that is 192 minutes. for the last 2.6 miles, he runs the first 2. thats 12 more minutes. 204 minutes total. the last .6 is 12 mph so multiply 12*.6=7.2 so we have his marathon time is 211.2[/hide]


Find all values of $ a$ such that the system
$ x^2\plus{}y^2\plus{}z^2\equal{}a$
$ |x\plus{}y|\plus{}|x\minus{}y|\plus{}|y\plus{}z|\plus{}|y\minus{}z|\plus{}|z\plus{}x|\plus{}|z\minus{}x|\equal{}6$
has 8 solutions.

If you negate any combination of $ x$, $ y$ and $ z$, you get a new solution.  Also if you permute the variables at all, you get new solutions.  For any positive $ x$, $ y$ and $ z$, if $ x\neq y\neq z\neq x$, then there are 6 positive permutations.  If one pair is equal, then there are 3.  Thus since 8 isn't divisible by 3 but both 6 and 3 are, all three must be equal.  Hence $ x\equal{}y\equal{}z$ and the bottom system turns into:

$ |2x|\plus{}|2x|\plus{}|2x|\equal{}6|x|\equal{}6$ so $ |x|\equal{}1$.  Hence $ x^2\plus{}y^2\plus{}z^2\equal{}3x^2\equal{}3\equal{}a$.

We can verify that this does have 8 solutions: $ (1, 1, 1), (1, 1, \minus{}1), (1, \minus{}1, 1), (1, \minus{}1, \minus{}1), (\minus{}1, 1, 1), (\minus{}1, 1, \minus{}1), (\minus{}1, \minus{}1, 1), (\minus{}1, \minus{}1, \minus{}1)$.

Your method is simple and good :wink:
Also it's easy to find $ a$ if we know that $ x^2\plus{}y^2\plus{}z^2 \equal{} a$ is the function of sphere, which has the radius $ a$ and center $ (0, 0, 0)$ and $ |x\plus{}y|\plus{}|x\minus{}y|\plus{}|y\plus{}z|\plus{}|y\minus{}z|\plus{}|z\plus{}x|\plus{}|z\minus{}x|\equal{}6$ is the function going on edges of $ 2*2*2$ cube with vertices $ (1, 1, 1), (1, 1,\minus{}1), (1,\minus{}1, 1), (1,\minus{}1,\minus{}1), (\minus{}1, 1, 1), (\minus{}1, 1,\minus{}1), (\minus{}1,\minus{}1, 1), (\minus{}1,\minus{}1,\minus{}1)$ 8 intersections can be only in these vertices so the radius of sphere is $ \sqrt{3}$, and the answer is 3.


geometry
geometric transformation
reflection

A bear is chasing a mn and the man falls into a trench with maximun acceleration.What is the colour of the bear?  :rotfl:  :D (seriously,this is a good physics question)

if it is really physics question then please tell me the answer with correct explanation

Yes it is a physics question which has appeared in IIT.Sounded really corny to me when i my friend first asked it.

I will tell the solution after everyone tries it.Any random guesses?

I dunno,give me a reason :D  Its not a panda bear or something,just a BEAR.Read the question properly

white because of reflection of colors made the mans eye go crazy and he stumbled and fell.

He just went running and fell into it,nothing like he was blinded :D The explanation is wrong.No optics at all.REad the question is all that i can say....

the bear is white .

The man fell with max acceleration , and this can only occurs in North or South Pole where $g=9.81m/s^2$ . Since there are no bear in South , we conclude that the bear is from North. And we know that only Polar Bear (only type of bear ) lives in North Pole . :D

Yeas thats right shyong :D   :D   See,it was a perfect physics question

wow.. that is awesome.. g is gravity? or gravitational acceleration?

the answer is white     since  u r sayin that g is max (probably 9.84 m/s2)
 :D                 havent heard bout any other bear  up or down there :D

This, of course is very old story and there are many such funny question, but lest some one take it seriously ... I will make a comment..

FWIW -  Seriously, I doubt, this could be a IIT type question,  or even a  serious physics type question... may be okay as a joke but not to be taken seriously.

For perspective, the effect of  air resistance on the  acceleration of the falling  man  (which BTW would depend more on what is his position /cross section area in horizontal plane - then the   value of "g" due to geographic location)  would be much more than the variation due to the value of g.

My friends gave me this question and they told me it was an old IIT screening queston(which is quiote possible).

I'm sorry i didn't mention to neglect air resistance+falls verticallly and stuff,this is how the question ws given to me,i didn't see teh original one.

Some of the questions in Polya's "How to Solve It" are similar.
Still, it would be weird if they ask that on an IIT entrance exam.


is there an elementary way to show that $ \pi$ is irrational?

well there is a way which is not elementary but still a cool way to prove that...  :lol:
first they prove that $ \pi^{2}$ is irrational.

There is a proof which uses only a bit of calculus, cfr. the attachment.

It is in Dutch but it should be possible to just read the formulas and understand the proof.

Also some online translator could help (if it is really necessery) e.g. http://babelfish.altavista.com

then probably I should post the solution I was talking about too...
[quote="Proofs from THE BOOK"]
[b][u]lemma[/u].for some fixed $ n\geq 1$ let

$ f(x)=\frac{x^{n}(1-x)^{n}}{n!}$

$ (i)$the function $ f(x)$ is a polynomial of the form $ f(x)=\frac{1}{n!}\sum_{i=n}^{\infty}c_{i}x^{i}$ where the coefficients $ c_{i}$ are integers.

$ (ii)$for $ 0<x<1$ we have $ 0<f(x)<\frac{1}{n!}$

$ (iii)$the derivatives $ f^{(k)}(0)$ and $ f^{(k)}(1)$ are integers for all $ k\geq 0$[/b]

[hide="proof"]
parts $ (i)$ and $ (ii)$ are clear.

for $ (iii)$ note that by $ (i)$ the $ k$-th derivative $ f^{(k)}$

vanishes at $ x=0$ unless $ n\leq k\leq 2n$,and in this range $ f^{(k)}(0)=\frac{k!}{n!}c_{k}$ is an integer.from $ f(x)=f(1-x)$ we
get $ f^{(k)}(x)=(-1)^{k}f^{(k)}(1-x)$ for all $ x$,and hence $ f^{(k)}(1)=(-1)^{k}f^{(k)}(0)$ which is an integer.[/hide]

[b]theorem.$ \pi^{2}$ is irrational[/b]

[i]proof[/i].assume that $ \pi^{2}=\frac{a}{b}$ for integers $ a,b>0$.we now use the polynomial

$ F(x): =b^{n}(\pi^{2n}f(x)-\pi^{2n-2}f^{(2)}(x)+\pi^{2n-4}f^{(4)}(x) \mp ...)$

which satisfies $ F''(x)=-\pi^{2}F(x)+b^{n}\pi^{2n+2}f(x)$.

from part $ (iii)$ of the lemma we get that $ F(0)$ and $ F(1)$ are integers.

elementary differentiation rules yield

$ \frac{d}{dx}[F'(x)\sin \pi x-\pi F(x)\cos \pi x]=(F''(x)+\pi^{2}F(x))\sin \pi x$

$ =b^{n}\pi^{2n+2}f(x)\sin \pi x$

$ =\pi^{2}a^{n}f(x)\sin \pi x$

and thus we obtain

$ N: =\pi \int_{0}^{1}a^{n}f(x)\sin \pi x dx=[\frac{1}{\pi}F'(x)\sin \pi x-F(x)\cos \pi x]_{0}^{1}$

$ =F(0)+F(1)$

which is an integer.furthermore $ N$ is positive since it is defined as the integral of function that is positive (except on the boundary).however,if we choose $ n$ so large that $ \frac{\pi a^{n}}{n!}<1$,then from part $ (ii)$ of the lemma we obtain

$ 0<N=\pi\int_{0}^{1}a^{n}f(x)\sin \pi x dx<\frac{\pi a^{n}}{n!}<1$

a contradiction.[/quote]


geometry
parallelogram
circumcircle
geometric transformation
reflection
conics
angle bisector

Let $l,m$ be two parallel lines in the plane.
Let $P$ be a fixed point between them.
Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$.
(By angle $EPF$ we mean the directed angle)
Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.

Let $n$ be the line through $P$ which is parallel to $\ell,m$, and let $S,T$ be points on $\ell,m$ respectively such that $\angle SPT=2\angle EPF\ (1)$. Finally, let $Q$ be the fourth vertex of the parallelogram $SPTQ$. We'll prove that the point we're looking for is $Q$, and that $\angle EQF=\pi-\angle EPF\ (*)$.

Since $QF\mapsto QE$ is a projective transformation, and so is the map sending a line through $Q$ to the line which makes an angle of $\pi-\angle EPF$ with it (such that the two lines have the appropriate orientation, i.e. the same as the orientation of the angle $\angle EPF$), it suffices to check $(*)$ for three particular positions of $E,F$ on $\ell,m$. We'll take the first two to be $E\to\infty,F\to\infty$, where the conclusion is easy to draw, and we'll take the third one such that $PF$ is the angle bisector of $\angle QPT$. Because of $(1),\ PE$ is the angle bisector of $\angle QPS$. Since $SE$ is the external angle bisector of $\angle PSQ$, it means that $E$ is the $P$-excenter of $PSQ$, so $QE$ must be the external angle bisector of $\angle PQS$, and thus orthogonal to the angle bisector of $\angle PQS$. This latter line is parallel to $PF$, the angle bisector of $\angle QPT$, so we have $QE\perp PF$. In the same way we show that $QF\perp PE$, so $Q$ is the orthocenter of $PEF$, and $(*)$ folows.

[quote="amir2"]let circumcircle of $PEF$ intersect $\ell$ at $Q$ so we have:$\angle EQF=\alpha$

therfore all $QF$ lines are parallel to each other ,hence the locus of circumcenter will be a line,hence circumcircle pass thorugh the reflection of $P$ wrt locus of circumcenter.

So We are done[/quote]
You are completely wrong!
Just tell me why you conclude from $QF$ moving parallel that the locus of circumcenter is a line?
And of course your result is wrong for $0<\alpha<\frac{\pi}2$ because just see the two cases where one of $E,F$ goes to infinity, then the circumcenter will be the point of infinity that is perpendicular to $PF$ or $PE$ (The one that has no infinity point)
But if these two points are not equal (which occurs when $0<\alpha<\frac{\pi}2$) then the line you mentioned shall be the infinity line! But of course it is not!

Take $Q$ as in grobber's post. It is enough to show that the triangle $\triangle XQE$ is similar to $\triangle YFQ$ (play with angles around $Q$  to show that $\angle EQF$ is $\pi -\alpha$).  To prove the similarity notice that $\angle EXQ =\angle FYQ=\alpha$ and that the triangles $\triangle PXE$ and $\triangle FPY$ are similar. From the latter we have $XE/PY=XP/YF\Rightarrow XE/QX=QY/YF$.

I don't know if following is correct.

Map $E\mapsto F$ is projective from line $l$ to line $m$. So envelope $EF$ is some conic. Since it is bounded betwen parallel lines $m$ and $l$ it is not parabola.  It is quite obviously that $m$ and $l$ are tangents to this conic.

Now, following is well known:

Let $l$ and $m$ be tangents (they don't have to be parallel) on conic. If $F_{i}$ is focus of this conic then directed angle $\angle XF_{i}Y$ is constant, where $X$ and $Y$ are variable points on $l$ and $m$ such that $XY$ is also tangent on this conic.

Useing this one of focus $F_{1}$ and $F_{2}$ is for sure not equal $P$ wich proves our statement.

[b] Note:[/b] From this we see that point $P$ need not to be betwen lines $l$ and $m$ since in this case envelope $EF$ is hyperbola.

Please, give some comments. Thanks!

initialy you can prove that the locus of H(Orthocenter of triangle EPF) is a line perpendicular to lines l and m named it n.now assume that the intersecion point of circumcircle of EHF and line n is T.now prove that the distance between T and line l is equal with distance between P and line m.consequently T is a point that sees EF with angle 180-EPF.