geometry
circumcircle
geometric transformation
homothety
ratio
power of a point
radical axis

Let H be the orthocenter of triangle ABC.Let l_a be the line joining the pionts at which the A-excircle touches the lines AB and AC.Define the lines l_b and l_c similarly.Let A_1,B_1,C_1 be the orthogonal projections of H on l_a ,l_b and l_c, respectively.Prove that the lines through A_1,B_1 and C_1 ,perpendicular respectively to BC,CA and AB, pass through the circumcenter of A_1B_1C_1

It follows from the fact that if $\ell{}_a{}_b{}_c$ define the triangle $A'B'C'$, then $H$ is the circumcenter of $A'B'C'$ and $A',B',C'$ lie on $HA,HB,HC$ respectively (I believe this has been posted before and solved here). Now, since $A{}_1B{}_1C{}_1$ is the medial triangle of $A'B'C'$, and the perpendicular through $A_1$ to $BC$, for example (call it $\ell$), is parallel to $A'H$, which passes throughthe circumcenter $H$ of $A'B'C'$, it means that the configuration $A{}_1B{}_1C{}_1,\ell$ is obtained from $A'B'C',A'H$ through a homothety centered at the centroid of $A'B'C'$ of ratio $-\frac 12$, so it passes through the circumcenter of $A{}_1B{}_1C{}_1$, because $A'H$ pases through the circumcenter $H$ of $A'B'C'$.

Sorry, I don't have the time to post the complete proof.

What a nice problem, particularly nice since I wrote a solution half a year ago and now just will copy it from my archive.

[color=blue][b]op02 problem Bulgaria 40 (Bulgarian IMO TSTs 2002):[/b]

Let H be the orthocenter of a triangle ABC. Let $l_a$ be the line joining the points at which the A-excircle of triangle ABC touches the lines CA and AB. Define the lines $l_b$ and $l_c$ similarly. Let $A_1$, $B_1$ and $C_1$ be the orthogonal projections of the point H on the lines $l_a$, $l_b$, $l_c$, respectively. Prove that the lines through the points $A_1$, $B_1$ and $C_1$, perpendicular respectively to the lines BC, CA and AB, pass through the circumcenter of triangle $A_1B_1C_1$.[/color]

[i]Solution.[/i] Let the A-excircle of triangle ABC touch the lines CA and AB at the points $B_a$ and $C_a$, respectively. Similarly define the points $C_b$, $A_b$, $A_c$ and $B_c$. Then, the lines $l_a$, $l_b$, $l_c$ are the lines $B_aC_a$, $C_bA_b$, $A_cB_c$, respectively.

Let the lines $l_b$ and $l_c$ meet at A', let the lines $l_c$ and $l_a$ meet at B', and let the lines $l_a$ and $l_b$ meet at C'. Then, our first lemma will be:

[color=red][b]Lemma 1.[/b][/color] The points A', B', C' lie on the altitudes AH, BH, CH of triangle ABC.

One proof of this lemma was given in the note "Synthetic proof of Paul Yiu's excircles theorem" on my website

  http://de.geocities.com/darij_grinberg/ .

Another (more synthetic) [i]proof of Lemma 1[/i] goes as follows: First denote by (a), (b), (c) the circles with diameters BC, CA, AB, respectively.

If $H_c$ is the foot of the altitude CH of triangle ABC, then $\measuredangle BH_cC=90^{\circ}$ and $\measuredangle CH_cA=90^{\circ}$, so that the point $H_c$ lies on the circles with diameters BC and CA; in other words, the point $H_c$ is a common point of the circles with diameters BC and CA. Clearly, the point C is the other common point of the circles with diameters BC and CA. Thus, the line $CH_c$, joining the two common points of the circles with diameters BC and CA, must be the radical axis of these two circles. But the line $CH_c$, as an altitude in triangle ABC, must coincide with the line CH, and the circles with diameters BC and CA are our circles (a) and (b), respectively. Thus, we can rewrite our result as follows: The line CH is the radical axis of the circles (a) and (b).

Let $I_a$, $I_b$, $I_c$ be the excenters of triangle ABC. Call $X_b$, $X_c$, $Y_c$, $Y_a$, $Z_a$, $Z_b$ the orthogonal projections of the points B, C, C, A, A, B on the lines $I_bI_c$, $I_bI_c$, $I_cI_a$, $I_cI_a$, $I_aI_b$, $I_aI_b$, respectively. (This notation may look strange, but actually it just means that we project each point onto two lines, and onto each line we project two points.) In the MathLinks topic

  http://mathlinks.ro/viewtopic.php?t=5958

I have proved that the points $X_b$, $X_c$, $Y_c$, $Y_a$, $Z_a$, $Z_b$ lie on one circle (k). Since $\measuredangle BY_cC=90^{\circ}$ and $\measuredangle BZ_bC=90^{\circ}$, the points $Y_c$ and $Z_b$ lie on the circle with diameter BC; this is the circle which we denoted by (a). Hence, we can state that the points B, C, $Y_c$ and $Z_b$ lie on one circle; therefore,

  $\measuredangle CY_cZ_b=\measuredangle CBZ_b=90^{\circ}-\measuredangle BCZ_b=90^{\circ}-\measuredangle BCI_a=90^{\circ}-\measuredangle I_aCB_a$,

where, in the last step, we have used the fact that $\measuredangle BCI_a=\measuredangle I_aCB_a$, what is clear, since the line $CI_a$ is the external angle bisector of the angle BCA.

Now, since the A-excircle of triangle ABC has center $I_a$ and touches the line CA at the point $B_a$, we have $I_aB_a\perp CA$, and thus $\measuredangle CB_aI_a=90^{\circ}$. Since we also know that $\measuredangle CY_cI_a=90^{\circ}$, the points $B_a$ and $Y_c$ lie on the circle with diameter $CI_a$, so that

  $\measuredangle CY_cB_a=\measuredangle CI_aB_a=90^{\circ}-\measuredangle I_aCB_a=\measuredangle CY_cZ_b$.

Hence, the points $Y_c$, $Z_b$ and $B_a$ lie on one line. Similarly, the points $Z_b$, $Y_c$ and $C_a$ lie on one line. Altogether, it follows that the four points $Y_c$, $Z_b$, $B_a$ and $C_a$ are collinear; in other words, the line $l_a = B_aC_a$ coincides with the line $Y_cZ_b$. But the points $Y_c$ and $Z_b$ are the common points of the circles (a) and (k); hence, the line $l_a = Y_cZ_b$ is the radical axis of the circles (a) and (k). Similarly, the line $l_b$ is the radical axis of the circles (b) and (k). Thus, the point of intersection C' of the lines $l_a$ and $l_b$ is the point of intersection of the radical axis of the circles (a) and (k) with the radical axis of the circles (b) and (k); in other words, the point C' is the radical center of the circles (a), (b) and (k), and thus it must also lie on the radical axis of the circles (a) and (b). But we know that the radical axis of the circles (a) and (b) is the line CH. Hence, the point C' lies on the line CH. Similarly, the points A' and B' lie on the lines AH and BH, respectively, and thus, Lemma 1 is proven.

Now, being the points of tangency of the C-excircle of triangle ABC with the lines BC and CA, the points $A_c$ and $B_c$ are symmetric to each other with respect to the angle bisector of the angle BCA; hence, $\measuredangle CA_cB_c=\measuredangle CB_cA_c$. In other words, the line $l_c = A_cB_c$ forms equal angles with the lines BC and CA.

The lines AH and BH, being two altitudes of triangle ABC, are perpendicular to the lines BC and CA, respectively. But the line $l_c = A_cB_c$ forms equal angles with the lines BC and CA. Hence, this line $l_c$ also forms equal angles with the lines AH and BH, which are perpendicular to the lines BC and CA, respectively. In other words, we have < HA'B' = < HB'A'. Hence, the triangle A'HB' is isosceles, and HA' = HB'. Similarly HB' = HC'. Consequently, HA' = HB' = HC'. Thus, we have obtained:

[color=red][b]Lemma 2.[/b][/color] The point H is the circumcenter of triangle A'B'C'.

As a consequence of this, the point H, being the circumcenter of triangle A'B'C', must lie on the perpendicular bisector of the side B'C' of this triangle; therefore, the point $A_1$, defined as the orthogonal projection of the point H on the line $l_a = B^{\prime}C^{\prime}$, must be the midpoint of the segment B'C'. Similarly, the points $B_1$ and $C_1$ are the midpoints of the segments C'A' and A'B'. It follows that $B_1C_1\parallel B^{\prime}C^{\prime}$, $C_1A_1\parallel C^{\prime}A^{\prime}$ and $A_1B_1\parallel A^{\prime}B^{\prime}$. Hence, the triangles A'B'C' and $A_1B_1C_1$ are homothetic. [Actually, the triangle $A_1B_1C_1$ is just the medial triangle of the triangle A'B'C'.]

After Lemma 2, the point H is the circumcenter of triangle A'B'C'. Now let's call S the circumcenter of triangle $A_1B_1C_1$. Since the triangles A'B'C' and $A_1B_1C_1$ are homothetic, there exists a homothety mapping the triangle A'B'C' to the triangle $A_1B_1C_1$; this homothety must also map the circumcenter H of triangle A'B'C' to the circumcenter S of triangle $A_1B_1C_1$. Since a homothety maps any line to a parallel line, we thus have $A_1S\parallel A^{\prime}H$. But the line A'H coincides with the line AH and thus is an altitude of triangle ABC; hence, $A^{\prime}H\perp BC$. Hence, $A_1S\perp BC$. In other words, the perpendicular from the point $A_1$ to the line BC passes through the circumcenter S of triangle $A_1B_1C_1$. Similarly, the perpendiculars from the points $B_1$ and $C_1$ to the lines CA and AB also pass through S. So we have proven that the perpendiculars from the points $A_1$, $B_1$ and $C_1$ to the lines BC, CA and AB all pass through the circumcenter S of triangle $A_1B_1C_1$. This completes the solution of the problem.

  darij


Can someone prove the following?

1.Let for a sequence of random variables $ X_1, X_2,...$  there exists a constant $ C <\infty$ so that $ E(X_n^2)<C$ for each n.Then the sequence is uniformly ntegrable.

2. Let $ X_1, X_2, ...$: be a random sequence with $ |X_n|\le K <\infty$ for every $ n$.  Is the sequence $ S_n\equal{}\sum_{k\equal{}1}^nX_k$ Uniformly integrable?

1: you might try to prove that if the family $ \mathcal F \equal{} \{X_i\}_i$ is bounded in $ L^p$ with $ p>1$, then this family is uniformly integrable (can you see why this is false for $ p\equal{}1$). A useful criterion is that for every $ \epsilon>0$ there exists $ \eta>0$ such that for every event $ A$ satisfying $ P[A] < \eta$ then $ E[ |X| 1_A] <\epsilon$ for all $ X \in \mathcal F$.

2: what if $ X_n \equal{} 1$ for all $ n$ ?

Yes you are right alekk.

Is the following correct for (1)?

Let $ \epsilon >0$
$ E[|X_n|.I(|X_n|>\frac {C}{\epsilon})]\le \frac {\epsilon}{C}.E[X_n^2.I{(|X_n|>\frac {C}{\epsilon})]\le \frac {\epsilon}{C}. C\le \epsilon}$ for all $ n$


projective geometry
geometry theorems
geometry

Who can prove Pappus theorem only using the Menelaus Theorem?

You can find it in Praslov. If you don't have it I can rewrite the proof here.

Please, rewrite that solution for me. I do a solution only using Menelaus but very long. If you can, send me "Praslov"

http://www.cut-the-knot.org/pythagoras/Pappus.shtml
I think it's similar to Praslov's solution (maybe even the same). If you want Praslov send e-mail in PM.


limit
integration
trigonometry
calculus
calculus computations

I'd probably rewrite it as $ \lim_{y\to\infty}y\int_y^{\infty}\frac{\cos t}{t^2}\,dt,$ not that that matters.

My hint: try integrating by parts, integrating the cosine and differentiating the $ \frac1{t^2}.$ Then estimate. What does that get you?


parameterization
geometry
power of a point
radical axis

[quote="anandghegde"]Co-axial system of circles- A system(or a family)of circles every pair of which have the same radical axis are called co-axial cirlces.

if S=0 and P=0 are teh equations of the circle and radical axis, then S+$ \lambda$P =0 is the equation of the co-axial system.

my doubt- Does this equation represents a pair of cicles or a single circle? :maybe:[/quote]

As you keep on varying the parameter $ \lambda$, you'll get different circles. SO this represents the whole family of circles.

for a particular value of $ \lambda$ we get a single cirlce
what is meant by radical axis for a single cicle? :maybe:

No.. For 1 value of lambda, we get one circle. As we keep varying the parameter, we get many circles. Just take ANY two of the circles and find their radical axis. As you can obviously see its the given Line.

So for the whole system of circles, any two you take has same radical axis.

i hope this helps to solve your doubt

suppose in a question you see that there are two circles given which are coaxial and also the equation of radical axis is given.
but equation of one of the circles is not given, but the other one 's is given

then to find the eqn of unknown circle --

use S + lambda L = 0 as the eqn of that circle

if we have the value of any point through which the circle passes then we can use it in the above equation to get the equation of that circle

thus the above equation represents a system of circles
S + lambda L = 0 is a general equation but u do need another equation to get the definite equation of circle

thanks shreyas :)

edit:thank you slogger too... :)


For $x_i$ with $i=1,..,n$ positive real numbers less than $1$ and $n\geq 2$ prove

$\frac {\sum_{i=1,n} x_i}{ n} \leq \frac{1}{1+n(1-x_1)(1-x_2)....(1-x_n)}$


[quote="frankvista"]谁能翻译一下这题的证明(Who can translate it?)[/quote]

It's french so...

Yea, this shouldn't be in this subfolder....... I don't think many Chinese people can read French, I know I can't...

Put this in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=280]France forum[/url].


You, in your new red Porsche, decide to make a 160km trip to Carolina Beach, travelling at 80km/hr. You make the return trip travelling at a rate of 48km/hr. What was your average speed for the entire trip?

[hide]
total distance: 320 miles
time to go: 2 hours
time to come back: 3 hours and 20 minutes
$\frac{320}{\frac{16}{3}} = 320 * \frac{3}{16} = 20 * 3 = \framebox{60mph}$[/hide]


Functional Analysis
real analysis
real analysis unsolved

Let X,Y be Hilbert spaces. Consider $ L(X,Y)$, the space of bounded linear functional $ X \to Y$. This space is Banach under the usual operator norm. Is it Hilbert (i.e. is the operator norm necessarily induced by an inner product)?

No. The operator norm involves $ \sup$, which pretty much kills any hope of the space being Hilbert, or "smooth" in any way.

Concrete example: $ X \equal{} Y \equal{} \mathbb R^2$ (or $ \mathbb C^2$ if you wish). Then $ L(X,Y)$ is the space of all $ 2\times 2$ matrices, with the operator norm (i.e. the norm given by the maximal singular value). To make this even simpler, consider the subspace $ D\subset L(X,Y)$ of diagonal matrices. This is a two-dimensional space, and you should be able to see that its norm is the $ \ell^{\infty}$ norm. [Of course, any subspace of a Hilbert space would be Hilbert too.]

There is a natural inner product of operators, namely $ \langle A,B\rangle\equal{}\mathrm{Tr}(AB^*)$. This is defined on Hilbert-Schmidt operators, not on the entire $ L(X,Y)$.


A two-person game begins with the random selection of an integer
from the set {1, 2, 3}, each choice being equally likely. Then two
players, not knowing the value of x, simultaneously select integers
from {1, 2, 3}. . Each player’s objective is to select an integer x. If
only one player is successful, he wins. If both players are successful,
and their choices are not the same, the winner is the player who
chose the smaller integer. In all other cases, the game is a draw.
The winner gets $ \$$1 from the loser. Describe the pure strategies
for each player, and the payoffs. . Show that the game has a
unique equilibrium strategy solution (this means that if one player
consistently plays this strategy, there is no advantage to the other
in changing their strategy).

How both player may be successful and their choices are not the same?

I assume you mean, "If both players are [b]unsuccessful[/b], and their choices are not the same, the winner is the player who chose the smaller integer."

In that case, I believe the answer is:

[hide]Both players should play a pure strategy of choose 1 every time. Because if the opponent deviates from this strategy q percent of the time while you maintain it, you will win outright $ \frac {1}{3}q$ percent of the time, and you will win by having the lower number $ \frac {1}{3}q$ percent of the time. So you will win a total of $ \frac {2}{3}q$ percent of the time.

But you will only lose when he guesses the right number which will happen $ \frac {1}{3}q$ percent of the time. Giving you an expected gain of $ \frac {1}{3}q$. The only way the opponent can avoid this expected loss is to choose 1 every time also. So "choose 1" is an equilibrium strategy for both players and the expected payoff is 0 to both players.[/hide]

A slight variation of the game would be: the player who uniquely guesses the selected number wins that number of points. And if both players guess incorrectly, the one with the lower number wins one point. Now what is the equilibrium strategy/strategies?


I have tried every method I know of to differentiate the function $ f(x)\equal{}Ke^x\minus{}Vx^2\plus{}F$ and I get the solution below however I am told that is incorrect. Any advice?
[hide]$ Ke^x$[/hide]

You seemed to forget to differentiate $ \minus{}Vx^2$. :lol:

$ f'(x)\equal{}Ke^x\minus{}2Vx$.

As kunny mentioned, you forgot to differentiate $ \minus{}Vx^2$  :wink:


Let $ ABC$ be a triangle, and let $ O$ be the circumcenter of the triangle $ ABC$. The lines $ AB$ and $ AC$ meet the circumcircle of triangle $ BOC$ at two points $ B_{1}$ and $ C_{1}$ (apart from $ B$ and $ C$), respectively. Let $ D$ be the point of intersection of the lines $ BC$ and $ B_{1}C_{1}$. Show that the circle tangent to the line $ AD$ at $ A$ and having its center on the line $ B_{1}C_{1}$ is orthogonal to the circle with diameter $ OD$.

I'll let you prove that the conclusion is equivalent to $AO\perp B_1C_1$. This last assertion follows from the fact that $O$ is the orthocenter of $AB_1C_1$. We can show $OB_1\perp AC_1,\ OC_1\perp AB_1$ by a simple angle chase.

Does someone have a solution with Inversion? :blush:

I finish the solution:
[hide]Let $ C_{1}$ the circle with diameter $ OD$, let $ R$ a point on $ B_{1}C{1}$ and $ RA$ perpendicular to $ AD$, $ C$ is the circle with center $ R$ and radius $ RA$, we shall prove that $ C$ and $ C_{1}$ are orthogonal.
First note that $ < OAC \equal{} < OCA \equal{} 90\minus{}< B$, and $ B_{1}CC_{1}B$ is cyclic, then $ < B_{1}C_{1}C \equal{} < ABC \equal{} < B$, let $ D'$ the intersection of the lines $ AO$ and $ B_{1}C_{1}$, in the triangle $ AD'C_{1}$ we have $ < AD'C_{1}\equal{} 90$, equivalent to: $ OD'D \equal{} 90$, then $ D'$ is on $ C_{1}$.
Now apply an inversion $ P*$ with center $ R$, and radius $ RA$ (circle $ C$), we have
$ (RD')(RD) \equal{} (RA)^{2}$, then $ P*(D) \equal{} D'$.
Finally, the circle $ C$ intersect to $ C_{1}$ at $ X,Y$, then $ P* (X) \equal{} X$, and $ P*(Y) \equal{} Y$
so $ P*(C_{1}) \equal{} C_{1}$, the conclusion follows[/hide]
Can somebody provide another solution?

we know that $ AO\perp B_{1}C_{1}$ , let $ AO\cap B_{1}C_{1}\equiv H$ and $ AO$ intersect circumcircles of $ ABC$ and $ BOC$ at $ A'$ and $ O'$ in order.and let the circle with center on $ B_{1}C_{1}$ and tangent to AD at A be $ \omega$
cause $ \angle H =\angle A'CA = 90^{\circ}$ hence $ AH.AA' = 2R.AH = AC_{1}.AC$ also we have $ AO.AO' = R.AO' = AC_{1}.AC$ hence $ 2AH = AO'$ therefore $ H$ is midpoint of AO' hence $ DO'$ is tangent to the $ \omega$.
let $ DO$ intersect $ \omega$ at $ X,Y$ we have $ (DO,XY) =-1$ hence $ RO^{2}= RX.RY$

Another solution :Use Inversion

The statement is equivalent to the fact that $O$ lies on the polar of $D$ w.r.t. the other circle. but that polar goes through $A$ and is perpendicular to $B_1C_1$, so it remains to prove $AO\perp B_1C_1$.

But this follows immediately from the fact that $B_1C_1$ is antiparallel to $BC$ and $AO$ is isogonal to the $A$-altitude of $BC$, which are orthogonal.


probability
combinatorics unsolved
combinatorics

Of the 2n people in a given collection of n couples. exactly m die (m is fixed). Assuming that the m people have been picked at random, find the mean number of surviving couples

$X_{i}$ - is indicator of $i$ couple survival (1 if yes, 0 if no)
$X$ - the total number of survivals
$X=X_{1}+...+X_{n}$
$E(X)=E(X_{1})+...+E(X_{n})$
$P(X_{i}=1)=(1-\frac{m}{2n})(1-\frac{m}{2n-1})$
$E(X)=nP(X_{i}=1)=n(1-\frac{m}{2n})(1-\frac{m}{2n-1})$
i have done the $P(X_{i}=1)$ quickly so there may be a mistake


1. The smallest case of 5 consecutive odd integers whose sum is a perfect square is 1, 3, 5, 7, 9. (1+3+5+7+9=25.) Find the median of the next larger set of 5 consecutive odd integers whose sum is a perfect square.

I can find the answer, but is there a quick way?

2. A man is running through a train tunnel. When he is 2/5 of the way through, he hears a train that is approaching the tunnel from behind him a speed of 60 mph. Whether he runs ahead or back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate).

Need help with the setup.

KSUNA wrote:Find the median of the next larger set of 5 consecutive odd integers whose sum is a perfect square.

[hide]

If 2n+1 is the first of serie :

The sul of the five consecutive  is 10*n+25=5(2n+5), so 2n+5=5*k, n = (5*k-5)/2. k must be odd, k = 1 is the first case, k = 3 the second.[/hide]

KSUNA wrote:2. A man is running through a train tunnel. When he is 2/5 of the way through, he hears a train that is approaching the tunnel from behind him a speed of 60 mph. Whether he runs ahead or back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate)..

[hide]

3/5 = (V+v).t

2/5 = (V-v).t[/hide]

[quote="JBL"]Belenos, this is not a forum for you to answer challenges on! :)[/quote]

Oh, I'm sorry, I thought this was a need for help and not a challenge. Feel free to remove my posts if you think they are inappropriate.

Oh, actually, you're right.  My fault.  Still, you should be in Advanced :)

Thanks for the help. I needed assistance. They were not meant as challenge questions.


function
limit
real analysis
real analysis unsolved

[b]Let the Rolle's function [/b]$f : [0,1]\rightarrow R$ [b]for which[/b] $f(1)=0.$ [b]Prove[/b] [b]that[/b]
$\forall m\ne 0,\ \exists\ c_m\in (0,1)$ [b]such that[/b] $c^2_m\cdot f'(c_m)+|m|\cdot f(c_m)=0.$

A Rolle's function is a function which is continue on [a,b] and derivated on (a,b).

Let the function  $G:[0,1]\rightarrow R,\ G(0)=0,\ G(x)=f(x)\cdot e^{-\frac{|m|}{x}},\ x\in (0,1]\Longrightarrow.$
$G(0+0)=\lim_{x\searrow 0} f(x)\cdot e^{-\frac{|m|}{x}}=0.$ Thus the function G is a Rolle's function and $G(0)=G(1)=0.$ Results that there is $c\in (0,1)$ such that $G'(c)=0$, i.e. $c^2\cdot f'(c)+|m|\cdot f(c)=0.$
In my opinion, this own problem is very nice.

I see. Thank you. I didn't try to put $x$ in the denominator.

What time is it thereaout ? here time is 00/30 a.m.


I don't really get the idea behind using modular arithmetic :D . Here's a problem:

Four whole numbers, when added three at a time, give the sums 180, 197, 208, 220. What is the largest of the four numbers?

Adding all the numbers gives us:

3a+3b+3c+3d = 180+197+208+220

with a < b < c < d

Now:
220-208 = d-c
220-197 = d-b
220-180 = d-a

Now substitute everything in and you're done.

(Well, since you know the total sum, you can then just subtract each of those 4 sums and you have the 4 numbers).

But anyway - this is nothing to do with modular arithmetic. Modular arithmetic will arise when you are dealing with divisions and remainders.


[quote="jli"]Evaluate

$\sum_{i=1}^{100}\frac{1}{x^{2}+x}$

[hide="hide"]use partial sums[/hide][/quote]
[hide="Elaborating on jli hint"]$\frac{1}{x^{2}+x}= \frac{1}{x}-\frac{1}{x+1}$

Now the series telescopes, and it is easy to see that the answer is $\frac{1}{1}-\frac{1}{101}= \frac{100}{101}$[/hide]


consider asquare (a b c d ) hose side length is 6, p  is a point in side the square with the poperty that
$ph^{2}+pb^{2}=68$
prove that
$20 \leq  pc^{2}+pd^{2}\leq  44$

[hide="Solution"] Draw altitudes down from $P$ to all four sides.  They will divide opposite sides into segments of some certain lengths.  Say that $AB, CD$ have been divided into segments of length $m, 6-m$ and that $BC, DA$ have been divided into segments of length $n, 6-n$ such that

$PA^{2}= m^{2}+n^{2}$
$PB^{2}= (6-m)^{2}+n^{2}$
$PC^{2}= (6-m)^{2}+(6-n)^{2}$
$PD^{2}= m^{2}+(6-n)^{2}$

This allows us to prove what I'm told is called the British Flag Theorem, which states that

$PA^{2}+PC^{2}= PB^{2}+PD^{2}$

But we are not interested in these quantities.  Now, we are given that

$PA^{2}+PB^{2}= 2m^{2}+2n^{2}-12m+36 = 68$

And we want to show that

$20 \le PC^{2}+PD^{2}= 2m^{2}+2n^{2}-12m-24n+108 \le 44$

Well, subtracting $68$ from everything, we want to show that

$-48 \le-24n+72 \le-24$
$5 \ge n \ge 4$

Being lazy, I will just prove this by contradiction.  If $n > 5$ then

$2m^{2}+2n^{2}-12m+36 > 2(m-3)^{2}+18+50 \ge 68$

And if $n < 4$ then

$2m^{2}+2n^{2}-12m+36 < 2(m-3)^{2}+18+32 \le 2 \cdot 3^{2}+50 = 68$

And we are done. [/hide]


determine all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that \[ f(x\minus{}f(y))\equal{}f(f(y))\plus{}xf(y)\plus{}f(x)\minus{}1\].

[quote="henry23"]determine all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x \minus{} f(y)) \equal{} f(f(y)) \plus{} xf(y) \plus{} f(x) \minus{} 1
\]
.[/quote]
[hide] $ x\rightarrow f(y)$ will kill it :wink: [/hide]

[quote="MellowMelon"]$ 2f(f(y)) \plus{} (f(y))^2 \equal{} f(0) \plus{} 1$

Not seeing the death here.

I'm pretty sure this was a 3 or 6 on some old IMO.[/quote]
We will prove that for all x there are some y1, such that f(y1)=x and we will have y-y1: f(0)=f(x)+x^2+f(x)-1


How many solutions does the equation $ tanx\minus{}x\equal{}1$ have on the interval $ [\minus{}2\pi,4\pi]$

[hide]The minimal derivative of $ \tan x$ is $ 1$ (equality only at multiples of pi), and the derivative of $ x$ is $ 1$. So on any interval of continuity, $ \tan x \minus{} x$ is monotone increasing, and can have at most one root. Since it goes from $ \minus{} \infty$ to $ \infty$ over such an interval, it must have a root by the Intermediate Value Theorem, and thus each interval of continuity must have [i]exactly[/i] one root.

Intervals of continuity are between consecutive odd multiples of pi/2. We can break each such interval into two halves -- $ \left( \frac {2k \minus{} 1}{2} \pi, \; k \pi \right]$ and $ \left[ k \pi , \frac {2k \plus{} 1}{2} \pi \right)$.

If $ x > 0$, then $ \tan x \minus{} x < 0$ at $ x \equal{} k \pi$, and thus the root on the interval of continuity must fall in the latter half.
If $ x < 0$, then $ \tan x \minus{} x > 0$ at $ x \equal{} k \pi$, and thus the root on the interval of continuity must occur in the former half.
If $ x \equal{} 0$, then $ \tan x \minus{} x \equal{} 0$ at $ x \equal{} 0$.

We have 4 entire intervals of continuity. In multiples of pi:
-3/2...-1/2, -1/2...1/2, 1/2...3/2, 3/2...5/2, 5/2...7/2
Each has one root.

We have two additional half-intervals:
-2...-3/2 and 7/2...4

-2...-3/2 is the second half of -5/2...-3/2. Since x<0 here, the root occurs on the first half of this continuity interval, so there is no root in -2...3/2.

7/2...4 is the first half of 7/2...9/2. Since x>0 here, the root occurs in the second half of this continuity interval, so there is no root in 7/2...4.

This makes 5 roots total.[/hide]

hello, you can compute the solutions of the equation $ \tan(x)\minus{}x\minus{}1\equal{}0$ by a numerical way. See here the so called Newtons method http://mathworld.wolfram.com/NewtonsMethod.html.
Sonnhard.


Ring Theory
advanced fields
advanced fields unsolved

prove the ring of integers of Q(sqrt(-14)) is not a UFD.

Since -14 is 2 mod4, the ring of integers is Z(sqrt(-14)

sqrt(-14)*sqrt(-14)=-14=7*-2

showing that  sqrt(-14), 7, and -2 are all irreducible

Norm(sqrt(-14))= 14
suppose  sqrt(-14)= ab
N(sqrt(-14)=N(a)N(b)
suppose N(a)=7, then

a^2+14b^2=7 which is impossible.

and similarly 7 and -2 are irreducible.

is this correct?


Let $f(n)$ denote the number of permutations $(a_{1}, a_{2}, \ldots ,a_{n})$ of the set $\{1,2,\ldots,n\}$, which satisfy the conditions: $a_{1}=1$ and $|a_{i}-a_{i+1}|\leq2$, for any $i=1,2,\ldots,n-1$. Prove that $f(2006)$ is divisible by 3.

Just a scketch:

Remark: If $|a_i-a_{i+1}|=|a_{i+1}-a_{i+2}|=2$ for some $i$ then, take the first psition when this situation occurs. The sequence will have a prescribed form for the positions $i,...,n$: $k,k+2,...,n-1,n,n-2,n-4,...,k+1$ (depending slightly on the parity of $n$). And the tems $a_1,...a_{i-1}$ are a permutation of $\{1,...i-1\}$.

Hence, the number of permutations with desired property is the same with the total number of permutations of $1,...i-1$ , for all $i$, with the desired property and does not have three consecutive terms of difference 2.
It is easy to prove that if we have $a_{i+1}-a_i=2$ then $a_{i+2}$ is $a_i+1$ and $a_{i+3}=a_i+3$. Hence $a_i,a_{i+1},a_{i+2},a_{i+3}=k,k+2,k+1,k+3$, Hence the sequence $a_n$ is obtained from the identical permutation $1,2,..i-1$ by taking some non-neibourgh (in the sequence) sets of two consecutive elements and intercharging them.

One can compute now $f(2006)$ and check it's residue mod 3. Hope it works  :D

The number of ways to chose some non-neibourgh, two sets with consecutive elements from $1,...,i-1$ is equal to the number of ways to chose some non-overlaping sets with three consecutive consec elements.

If we denote by $c_n$ the number of ways to choose such triples from $1,..,n$ s.t. the last three elements are in a set. we find the following reccurence relation: $c_n=c_{n-3}+c_{n-4}+c_{n-5}$. And hence the number of ways to chose some three consecutive elements sets from $1,..n$ is $d_n=\sum c_n$. And $f(n)=\sum d_n$.

I payed no attention to details but the solution works, I think. I foud prety difficult to compute $f(n)$ but we can work mod 3 to get the result not so hard. What is the official solution?

Here is the solution which I found:
Let $B_n$ is number of the desired permutations in which $n$ ends.Then we have :$f(n+1)=f(n)+b_{n-1}+b_{n-2}$
Otherwise:$b_n=b_{n-1}+b_{n-2}$ for all $n\geq 4$.And it becomes easy!


There are 25 people sitting around a table, and each person has two cards. One of the numbers $ 1, 2, 3, . . . , 25$ is written on each card, and each number occurs on exactly two cards. At a signal, each person passes one of her cards-the one with the smaller number-to her right-hand neighbor. Prove that, sooner or later, one of the players will have two cards with the same numbers.

I have an interesting solution to this one, but I am unsure if it is correct. Post your own solutions too.

[hide="My Solution"]

We can basically "set" the cards and create a contradiction. Imagine the players as having no cards. Now, we are going to place the cards labelled 25. There are two cases.

Case 1: One person has both 25's

Since this obviously contradicts the given conditions, we cannot use this case.

Case 2: Two separate people are dealt the 25's

This must be our desired case.

Now, we make the observation that, after dealing the 25s, placing the 24's leads to only one conclusion: two separate people have the 24's. This is because if we give any or both of the 24's to the people who already have 25's, after a finite number of moves, the 24's will, worst case scenario, "separate" out to separate players (there is a case where they can go to the same player, but again, this contradicts the given provisions; this is a central idea in the proof. We are looking for the worst-case scenario and proving that eventually, the worst-case scenario will sort out into a player having two of one card).

Similarly, assuming all cases with dual cards are ignored, the cards with numbers 14-23 can be placed as well, and as such. Now there are 24 people with cards. By way of the reasoning described above, the last person must be given a 13. Now, to place the other 13, we must either give it to the player with the other 13 card, or we can give it to any other player and let it through a finite number of moves, pass to the player with the other 13 card.

Thus, in the quest to find a case where there will be no player with double cards, we used rigid "if/else" cases, intellectually selecting only those that would not yield a double card. However, when trying to do so, we failed at the end to prove the existence of the case because of a contradiction. Therefore, we have proved what we have been asked.

[/hide]

Hmm, that kind of sounds really informal...

[hide]
let the cards $ \{1,1,2,...,13,13\}$ be called small, and the cards $ \{14,14,...,25\}$ be large.
A large card will only be passed to the right if the person passing the card is holding two large cards. So after $ k$ transformations we will say there are $ M_k$ people holding two large cards.

Originally there are $ M_0$ people holding two large cards $ (M_0 \le 12)$. If every person the the right of the $ M_0$ people is also holding a large card then $ M_1\equal{}M_0$ otherwise we will have $ M_1 < M_0$.

However since there are $ 25$ people and only $ 24$ large cards, we will eventually have a large card passed to a person holding a small card. That is $ M_{k\plus{}24}<M_k$. So we have a decreasing sequence of integers bounded below by zero. So after at most $ 24\cdot 12 \equal{} 288$ moves (very weak bound) all the large cards will be seperate and stationary.

the rest of the proof is essentially the same as yours, veezbo [/hide]


trigonometry
search
inequalities unsolved
inequalities

Prove that in every acute-angled triangle $ ABC$,
$ (\cos A \plus{}\cos B )^2\plus{}(\cos B\plus{}\cos C)^2\plus{} (\cos A\plus{}\cos C)^2 \leq 3$

oke,
the idea is
there exists three positive numbers a,b,c such that
 cos A=$ \frac{a}{ \sqrt{(a\plus{}b)(a\plus{}c)}}$;cos B=...
and now, it is easy!

[quote="kihe_freety5"]oke,
the idea is
there exists three positive numbers a,b,c such that
 cos A=$ \frac {a}{\sqrt {(a \plus{} b)(a \plus{} c)}}$;cos B=...
and now, it is easy![/quote]
You had using substitutions $ xy\plus{}yz\plus{}zx\plus{}2xyz\equal{}1$ (same my idea) but I want know many nice proof.

See here

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=73381531&t=49903


linear algebra
matrix
algebra
polynomial
absolute value
linear algebra theorems

I was wondering if someone could guide me through a proof of the following theorem:

if $ A_{1}$ is a square matrix with integer coefficients, then $ A_{1} \equal{} B_{1} B_{2} B_{3}$, where $ B_{1}$ and $ B_{3}$ are square matrices (of same size) with determinant absolute value $ 1$, and $ B_{2}$ is a diagonal matrix, with the diagonal entries forming a sequence $ b_1 | b_2 | b_3 | ... | b_n$; and that $ b_i$ are uniquely determined.

This is similar, but not the same as the definition of diagonalizability.  If a matrix $ A$ is diagonalizable, then $ A\equal{}QDQ^{\minus{}1}$ where $ Q$ is a matrix of eigenvectors and $ D$ is the diagonal matrix of eigenvalues.

So in your case, if we let $ B_{i}$ be how I defined it above, then it's true for a certain set of matrices.  If the characteristic polynomial of the matrix splits, then you can extend this to a larger class in which $ D$ is the Jordan Canonical form and $ Q$ consists of generalized eigenvectors.


A bag containing some widgets weighs 81 kg.  When 2 of the widgets are removed, the weight of the bag of widgets drops to 73 kg.  The bag weighs 1 kg when empty.  How many widgets are left in the bag?

The two widgets that are removed weigh:

81 kg - 73 kg = 8 kg

so one widget weighs:

8 kg / 2 = 4 kg

and since the bag weighs 1 kg when empty, we subtract 1 kg from the remaining weight of the bag when two widgets are removed, which is 73 kg, so we get that the remaining widgets weigh 72 kg.

Since we know that each widget weighs 4 kg, we divide:

72 kg / 4 kg = 18

Thus we have 18 widgets remaining.

I am not very familiar with LATEX, so I am sorry if this post was not very readable.


inequalities
complex numbers
triangle inequality
inequalities proposed

Let $a, b, c$ be the sides of a triangle, $m_a,m_b,m_c$ be the corresponding medians. Prove that

$m_abc+m_bca + m_cab \geq 4m_am_bm_c$.

Pretty straightforward in complex numbers: let $a,b,c$ be the complex numbers corresponding to the vertices of $ABC$ and let $z_1,z_2,z_3$ be $a-b,b-c,c-a$ respectively. We need to show that $\sum |z_1||z_2||z_1-z_2|\ge |(z_1-z_2)(z_2-z_3)(z_3-z_1)|$, and this is just the triangle inequality ($|x|+|x|\ge|x+y|$) applied for the three complex numbers $z_iz_{i+1}(z_i-z_{i+1})$.

You  can also  prove  it  by  applying  $ \frac{m_am_b}{ab}+\frac{m_bm_c}{bc}+\frac{m_cm_a}{ca}\geq\frac{9}{4}  $  http://www.mathlinks.ro/Forum/viewtopic.php?t=3171  to  the  triangle  wich  has  sides  the medians  of  ABC.


function
logarithms
induction
limit
algebra unsolved
algebra

Sorry, I should have posted this in the open question section, but does anyone happen to know nice results on multiplicative functions f such that $ f(n+1)-f(n)$ is bounded? Or anything like this. This started from a super nice problem given in an oral examination (I thing it is Erdos who found it) which asked to find all increasing multiplicative functions. Thanks in advance.

Don't put it to 'Open Questions!' The only (edit: [b]unbounded[/b]) solutions to this problem are $f(n) = n^c$, $c \leq 1$. In fact, a result of Erdos states that if $f(n)$ is a multiplicative function such that $f(n+1)/f(n) \to 1$ as $n \to \infty$, then $f(n) = n^c$ for some $c$.
In 1970 Wirsing proved the conjecture of Erdos that if $f$ is a multiplicative function, and if $f(n+1)/f(n)$ is bounded, then there is a constant $c$ such that $f(n)/n^c$ is bounded:

E.Wirsing, [i]A charactarization of $\log{n}$ as an additive arithmetic function,[/i] Sympos. Math. Vol IV (INDAM, Rome, 1968/69) Academic press, London, 1970, pp. 45--57.

Could you please send me this article? Do you have it? Thank you very much and sorry for my ignorance.

[quote="harazi"]This started from a super nice problem given in an oral examination (I thing it is Erdos who found it) which asked to find all increasing multiplicative functions.[/quote]

Yes, this is one from Erdös  :)

Pierre.

So what was the problem, exactly? Finding the increasing multiplicative functions from $\mathbb N^*$ to $\mathbb N^*$? By multiplicative I mean $f(mn)=f(m)f(n)$ for all coprime $m,n$, of course.

This one I can solve. The problems Vess talks about I cannot solve.

Proving that the increasing multiplicative functions $f : \mathbb{N} \to \mathbb{N}$ take the form $f(n) = n^c$  :) . Of course, multiplicativity means $f(mn) = f(m)f(n)$ for $(m,n) = 1$ --- otherwise it's 100% trivial  :D .

@ Harazi: Unfortunately, I haven't seen the article ... I read about it in a great memorial article about Erdos from the Notices that can be found [url=http://www.ams.org/notices/199801/vertesi.pdf]here[/url].

Somewhat related problem http://www.mathlinks.ro/Forum/viewtopic.php?t=16728

[quote="vess"]Proving that the increasing multiplicative functions $f : \mathbb{N} \to \mathbb{N}$ take the form $f(n) = n^c$  :) [/quote]
I am thinking about it  :? Is it hard? No one answered to alekk's question.

I do not find it easy at all. But this is just my humble opinion.

Ok, let's go for it :
Let $f$ be a non-decreasing and multiplicative function from $\mathbb {N}^*$ onto $\mathbb {R}.$
First the function $f(x) = 0$ is a solution, so we assume that $f$ is not the zero-function. It esaily follows that $f(1)=1.$ Thus $f(n) \geq 1$ for all $n$.

For $a \geq 3$ and $k \geq 1$, let $U_k = 1 +a + \cdots + a^k$ and $V_k = a^k - (1+a+ \cdots + a^{k-1}.$
We have $V_1 = a-1 \leq a < 1 + a = U_1$ and $\gcd (a,U_k) = 1 = \gcd ( a, V_k).$
Moreover $U_{k+1} - 1 = aU_k$ and $V_{k+1} + 1 = aV_k.$
Thus $f(U_{k+1}) \geq f(a)f(U_k)$ and $f(V_{k+1}) \leq f(a)f(V_k).$
Thus, by an easy induction : $f(V_k) \leq (f(a))^k \leq f(U_k).$

Let $a,b \geq 3$ and $n \geq b^2$. Let $p = [ \frac {\ln (n)} {\ln (a) } ]$ and $q = [ \frac {\ln (n)} {\ln (b) } ].$
Note that $q \geq 2$ and we have :
$V_{p+2} \geq a^{p+2} - 2a^{p+1} \geq a^{p+1} \geq a^{\frac {\ln (n) } {\ln (a) } } = n.$
Thus $f(n) \leq f(V_{p+2}) \leq (f(a))^{p+2}.$
Similarly $f(n) \geq f(U_{q-1}) \geq (f(b))^{q-1}.$

Then, for all $a,b \geq 3$ and $n \geq b^2$ we have $(f(b))^{\frac {q-1} {\ln (n)}} \leq (f(n))^{\frac 1 {\ln (n)}} \leq (f(a))^{\frac {p+2} {\ln (n) }}.$

But $\frac 1 {\ln ( a)} + \frac 1 {\ln (n)} \leq \frac {p+2} {\ln (n)} \leq \frac 1 {\ln (a)} + \frac 1 {\ln (n)}.$ Thus $\lim_{n \rightarrow +\infty} \frac {p+2} {\ln (n)} = \frac 1 {\ln (a)}.$
Similarly $\lim_{n \rightarrow +\infty} \frac {q-1} {\ln (n)} = \frac 1 {\ln (b)}.$
Therefore : $(f(b))^{\frac 1 {\ln (b)}} \leq (f(a))^{\frac 1 {\ln (a)}}$ for all $a,b \geq 3.$
Since $a,b$ play symmetric parts, we deduce that $(f(b))^{\frac 1 {\ln (b)}} = (f(a))^{\frac 1 {\ln (a)}}$ for all $a,b \geq 3.$

It means that there exists a constant $c$ such that for all $n \geq 3$ we have $(f(n))^{\frac 1  {\ln (n)}} = c$. Since $f(n) \geq 1$ we deduce that $c \geq 1$ so that $f(n) = n^u$ where $u = \ln (c) \geq 0.$ (1)
It is easy to verify that since $f(2) = \frac {f(6)} {f(3)}$, the equality (1) holds for all $n \geq 1.$

Pierre.


hi guys,

anyone knows how to find the partial quotients of $a_m = \sum_{k=0}^{m} \frac{1}{10^{2^k}}$? and also of $lim_{m\rightarrow \infty a_m}$?

Thanks,

Rosa

Sorry, but what is the definition of "partial quotients"? :blush:

Mathworld answers you :) : http://mathworld.wolfram.com/PartialQuotient.html


geometry
circumcircle
Euler
inequalities
geometry solved

let R denote the circumradius of triangle ABC ; a,b,c the sides opposite BC,CA,AB and r_a,r_b,r_c the exradii opposite A,B,C respectively
.if 2R  \leq  r_a prove that
i)  a > max{b,c}
ii) 2R > max {r_b,r_c}

Everywhere below, p is the semiperimeter of the triangle and S is the area.

i) From Euler we have r_a >= 2R >= 4r ===> S/(p-a) >= 4S/p ==>
   p>= 4(p-a)  ===> a+b+c >= 4(b+c-a) ===> 5a >= 3(b+c)
   If a<= min{b,c} it's easy to see that we have a contradiction.

   Assume now that min{b,c} <= a <= max{b,c} ===> (a-b)(a-c) <= 0
   ===> a^2 + bc <= ab + ac (*)

   After a few simple computations we get :
   2R <= r_a <===> 2 abc/4S <= S/(p-a) <===> abc(p-a) <= 2S^2
   <===> abc(p-a) <= 2 p(p-a)(p-b)(p-c)
   <===> 4abc <= (a+b+c)(a+b-c)(a+c-b)
   <===> 2abc <= bc(b+c) + ab(a-b) + ac(a-c) + a^3 - b^3 - c^3 (**)
   (*) ===> (b+c)(a^2+bc) <= (b+c)(ab + ac) (***)

    So using (***) in (**) we get that
    2abc <= 2abc + a^3 - b^3 - c^3 ===> a^3 >= b^3 + c^3
    But the last relation is equivalent to :
    (a-b-c)(a^2+b^2+c^2+ab+ac-bc)>=0, which is obviously false
    because a<b+c

 ii) Let x=p-a,y=p-b,z=p-c. The hypothesis becomes
     x(x+y)(x+y)(y+z) <= 2xyz(x+y+z) <===>
     (x+y)(x+z)(y+z) <= 2yz(x+y+z) <====>
     xy(x+y) + xz(x+z) + yz(y+z) <= 2yz(y+z) <===>
     yz(y+z) >= xy(x+y) + xz(x+z) (****)

     Let us prove for example that 2R > r_b. The other inequality is
     done in a similar manner.
     2R > r_b <===> y(x+y)(x+z)(y+z) > 2xyz(x+y+z) <===>
     (x+y)(x+z)(y+z) > 2xz(x+y+z) <===>
      xy(x+y) + xz(x+z) + yz(y+z) > 2xz(x+z) <===>
      xy(x+y) + yz(y+z) > xz(x+z), which is obvious by (****)

nice job andrei.i gave the same solution


this problem is reached this link
http://www.kms.or.kr/competitions/apmo/data/06APMO-SOL.pdf
when i was  working to understand there solutions.i saw so explanations:
" Since the line $l$ is tangent to the circle $O_{1}$ at $X$, we have
$\angle SPX=90^\circ-\angle XSP=90^\circ-\angle APM=\angle PAM$ "
is that right? for to be true this equation, $SP$ must be a diameter but we don't know this. please somebody help me.

SP is the diameter. Since P is the midpoint of chord AB and O is the centre of the big circle, so $OP \perp AB$. But since AB is tangent to the smaller circle, so OP passes through the centre of the smaller circle and the tangency point of the two circles. Therefore SP is the diameter.


in a 3 by 4 array of evenly spaced dots, how many different squares can be formed by using four of the evenly-spaced dots below as verticies of the square?

[quote="jhredsox"]is the answer
[hide]
8[/hide]???

there should be[hide]
-6 two by two squares (we're not counting one by one squares, as he said 4 dots :lol: )

-2 three by three squares[/hide]

-jorian[/quote]
No, we are counting one by ones since there are four dots (the vertices in all squares)
Also, you are missing the squares with edge length $\sqrt{2}$ and so on.

[quote="jhredsox"]i thought the dots were integer amounts away from eachother

if not...im confused...someone want to help please?

thanks,
jorian[/quote]
Okay, in a 3 by 3 grid the middle dot in the top row, the right most dot in the middle row, the left most dot in the middle row and the middle dot in the bottom row make a square that is $\sqrt{2}$ for each side. It meets all the requirements as well.

oh, i forgot about those, heres a picture of what i disregarded

btw:I JUST SAW MY DAD CONFIRM THE ORDER FOR MY INTRO BOOKS THAT ARE FOR X-MAS!!!

-jorian :)

edit:it wont lemme load a "paint" file, so i'll describe it

                         rephrase them as...:
O O O O              1    2    3    4
O O O O              5    6    7    8
O O O O              9    10  11   12

and connecting the [color=red]2,5,10, and 7[/color]

[hide]$10$

6 1 by 1 squares
2 2 by 2 squares
2 square root of 2 squares
$6+2+2=10$[/hide]

[hide]There are 10, because of the types of squares. The squares are illustrated below:

[img]http://i96.photobucket.com/albums/l167/adasarathy/linedsquares.gif[/img][/hide]

[quote="frost13"]WHile it deosn't apply to this problem, here is a formula for similar problems :

The number of squares formed by an n by n array of dots where all four vertices are among those dots is

$\frac{n^{2}(n^{2}-1)}{12}$[/quote]
Very nice. How would you derive that?

I'll leave that to you :lol:

[hide]I don't actually know, have never tried :( [/hide]

[quote="13375P34K43V312"]If you have something on paint, convert it to .gif (I think that's what you're supposed to do), maybe it's .jpg or .png.[/quote]

how do you convert it

i could probably find out if i played around with it, but its just easier to ask :)

:P
-jorian hoover


number theory
greatest common divisor
algorithm

If GCD($x,y$) denotes the greatest common divisor of integers $x$ and $y$, find GCD($F_{2006},F_{2007}$), where $F_{n}$ denotes the $n$th Fibonacci number. $(F_{0}=F_{1}=1, F_{n}=F_{n-1}+F_{n-2}$ for $n\geq2)$

[hide="Lemma"] $gcd(F_{k}, F_{k+1}) = 1$

The proof is very simple.  $gcd(F_{k}, F_{k+1}) = gcd(F_{k}, F_{k-1})$ by Euclid's Algorithm, and inductively this is equal to $1$ always. [/hide]

that is correct,

but people that have not seen this problem before may not see why the equation holds (even though it is trivial when you rewrite things).

i didn't know that you look around these state forums.

I "View unanswered posts" frequently.   :P


Just had an exam where one of the questions was to characterize the spectrum of the operator

$ Mf(x) \equal{} xf\left(\frac {x^2}{2}\right)$

acting on $ C([0,1])$. I played around with trying to directly determine the spectrum, until I gave up and reluctantly used the gelfand formula for the spectral radius, which luckily turned out to be 0. Is there another argument?

They stripped the problem of most of its beauty :( (but made it reasonably easy to solve, which is more important for an exam :P).

OK, if you want to find another approach to any question or to get convinced that what you did is what had to be done, try to modify it so that your original method doesn't work anymore. The most general problem here would be to determine the spectrum of $ M_{u,v}f(x)\equal{}u(x)f(v(x))$ where $ u\in C[0,1]$ and  $ v: [0,1\to[0,1]$ is any continuous map. But shooting at this at once may be too much, so let's start with keeping one of the features of your original map: assume that $ v(x)$ is increasing and $ v(x)<x$ for all $ x\in(0,1]$ ($ u$ is completely arbitrary). Can you determine the spectrum under this assumption?

No, I can't. The spectral radius could be anything now, so something else is needed.

Have you tried to construct an eigenfunction "by hands"? (this approach will fall apart at the next step but let's see if it survives for now).


An urn contains 16 balls of two colors, red and blue. For balls are drawn from the urn without replacement. The probability of getting exactly 2 red balls and 2 blue balls is 1/20. The urn contains no more red than blue balls. How many blue balls were originally in the urn?

[hide]x=number of blue balls

(x/16)*((x-1)/15)*((16-x)/14)*((15-x)/13)*2*3=1/20. We can do this because the order of the denominators dont matter and we need all combinations so we multiply by 6.

so,

x(x-1)(16-x)(15-x)=2*7*13*2

since, it has only 4 prime factors we must assign one value to each of them. We quickly find that two of them cannot be equal. We then try 2*14*13*1 because it contains two pairs of consecutives. Thus, there are 14 blue balls.[/hide]


function
number theory open
number theory

find all $f: N^*\to N^*$ such that $f(m+\phi(n))=f(m) \mod n$

Somehow, I don't understand the question correctly or I am making a terrible mistake... If you set n=1, then $f(m) \mod n$ is always 0 and then $f(m+\phi(n))=0$... but that isn't allowed because of $f: N^*\to N^*$ hence there is no such function ... where is my mistake?


ABCDEFG is a regular heptagon inscribed in a unit circle centered at O. L
is the line tangent to the circum-circle of ABCDEFG at A, and P is a point
on L such that DAOP is isosceles. Let p denote value of
AP·BP·CP·DP·EP·FP·GP. Determine the value of p^2.

I need your help to know how to do it,
but i kno the answer

[hide="ANSWER"]113[/hide]

this problem doesn't make sense, DAOP seems to form a concave quadrilateral. what do you mean by isoceles?

It was a typo, see #10 on wiki - [[Mock AIME 1 Pre 2005 Problems]] (or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14682]here[/url]).

[hide="Hint"]
Complex numbers. Let $ P \equal{} 0 \plus{} 0i, D \equal{} 0 \plus{} i, O \equal{} \minus{}1 \plus{} i$, then vertices of heptagon are given by $ (z \minus{} (\minus{}1 \plus{} i))^7 \equal{} 1$.
[/hide]


algebra
difference of squares
special factorizations

Okay. Use the difference of square thing to get:
$ (2^{512}\plus{}1)(2^{256}\plus{}1)(2^{128}\plus{}1)(2^{64}\plus{}1)(2^{32}\plus{}1)(2^{16}\plus{}1)(2^{8}\plus{}1)(2^{4}\plus{}1)(2^{2}\plus{}1)(2\plus{}1)(2\minus{}1)$. The smallest primes here are 3 and 5, which, coincidentally are the smallest primes (other than 2, but 2 can't be a factor), so $ 3*5 \equal{} 15$.

I think one exclamation point is enough. :D

[quote="12markkram34"]Can someone please tell me how to solve this:

26. What is the product of the two smallest prime factors of 2^1024 - 1?

Thanks.
-12markkram34[/quote]

Ugly, but difference of squares gives us

$ (2^512\plus{}1)(2^256\plus{}1)(2^128\plus{}1)(2^64\plus{}1)(2^32\plus{}1)(2^16\plus{}1)(2^8\plus{}1)(2^4\plus{}1)(2^2\plus{}1)(2\minus{}1)$.

The primes are $ 3, 5$, so $ \boxed{15}$

BTW, Next year at MC (My first year)

I was confident I'd make Nats, then choke, but if these are

Nats questions, then I seriously doubt I won't make Nats CD

at least.


You can find this problem and solution for it in Solved problems. It was posted by [u]hxtung[/u] and solved by
[u] grobber[/u]


if a,b,c,d are real numbers all greater than1 then prove that 8(abcd+1) > (1+a)(1+b)(1+c)(1+d)

hello, please use $ \text{\LaTeX}$ in your post, is this your problem?
If $ a,b,c,d$ are real numbers all greater than1 then prove that $ 8(abcd+1) > (1+a)(1+b)(1+c)(1+d)$.
Sonnhard.

Yes that is right. How do we attach our Latex file to a post. Do I post it as an image?

hello, no you must enclose the variables and numbers in the dollar sign and type $ a$, this is an $ \$ a\$$.
Sonnhard.

Very nice!  :D
We have
$ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1)(d \minus{} 1) > 0$, so $ abcd \plus{} ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \plus{} 1 > abc \plus{} abd \plus{} bcd \plus{} acd \plus{} a \plus{} b \plus{} c \plus{} d$
Expanding the original equation and simplfying, we get
$ 3abcd \plus{} 3 > ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd$
But $ (ab \minus{} 1)(cd \minus{} 1) > 0$, so $ abcd \plus{} 1 > ab \plus{} cd$
Adding this cyclically gives the desired result.

Or what you could do is take $ (a\minus{}1)(bcd\minus{}1)>0$ and cyclics and $ (ab\minus{}1)(cd\minus{}1)>0$ and cyclics and add them, also solves the problem.

Let $ w\equal{}a\minus{}1$, $ x\equal{}b\minus{}1$, $ y\equal{}c\minus{}1$, and $ z\equal{}d\minus{}1$. Then $ w,x,y,z>0$.

The condition is:

$ 8( (w\plus{}1)(x\plus{}1)(y\plus{}1)(z\plus{}1)\plus{}1)>(w\plus{}2)(x\plus{}2)(y\plus{}2)(z\plus{}2)$

But this is equivalent to $ 4 w x \plus{} 4 w y \plus{} 4 x y \plus{} 6 w x y \plus{} 4 w z \plus{} 4 x z \plus{} 6 w x z \plus{} 4 y z \plus{}
 6 w y z \plus{} 6 x y z \plus{} 7 w x y z>0$ which is obvious.

[quote="enndb0x"]I was wondering for a solution of this problem using AM-GM
 Anyone ?[/quote]Sure, have you tried solving it using AM-GM? Try and find out if it works...

[quote="enndb0x"]Of course I tried ,but I couldn't do anything using that inequality.I am working in AM-GM and I was wondering how this problem can be solved using AM-GM.[/quote]The problem is 'symmetric' with respect to the variables, meaning, you can shuffle them around and still be the same. Now, a common result from AM-GM is that equality holds when the terms are "equal"... Are there equal quantities to be found? I haven't found any yet... or say what happens if $ a\equal{}b\equal{}c\equal{}d$? Then, $ 8(a^4\plus{}1)\ge(1\plus{}a)^4$? If we use AM-GM, it would be bashing it with strict inequalities...

The inequality, after dividing both sides by $ \sqrt{abcd}$ we get it equivalent to with:
$ 8(\sqrt{abcd}\plus{}\frac{1}{\sqrt{abcd}})>\prod _{cyclic} (\frac{1}{\sqrt{a}}\plus{}\sqrt{a})$

But it doesn't work :blush:


do you know how much are laptops in vietnam? are there soooo cheap?

I'm a Vietnamese.Can you comment what 's laptops? If I'm true,I thinks it's very cheap.
Sorry,are you a member of a IMO team which will go to Ha Noi,Viet Nam next month?

Yes I'm a polish team member. hmm laptop well it's notebook in english i think. :), the cheapest notebook

LOL, yeah stuff should be cheap in Vietnam I think. Laptop -- a portable computer. Can you just post the price of the laptops in Vietnam?

I don't get it ... why would laptops be cheaper in Vietnam than in EU?

lol....
I dont know why u all r going to buy laptops in Vietnam :rotfl:
anyway....If anything [which is small and can be taken easily to home :P ] is cheap in Vietnam please mention....i am going to buy them :blush: ...

Ok ,portable computer,notebook,MP4 or IPOD in VietNam are very cheap
portable computer:about 926$(pretty good) ,and about 1852$ (so good)
notebook:so cheap that I don't want to write here
MP4 or IPOD: about 61$(not much fuction),and about not more than 300$ (good)
but I think that you should buy in EU although I'mVietnamese

Don't believe him, of course everything in VN is cheaper because of our neighbor Chineses :blush: . :d But I am not sure about their quality. If you live in VN, that is a profite, you can bring it to the store for reparing (normally in VN, they give us a brandnew in case of bad-quality. If not, you only could cry when it is broken. So when you want to buy some thing, ask them if it is world-wide insurance. :wink:
Anyways, you should buy clothes and  shoes in VN, it cost almost-freely  :D

[quote="nmtruong1986"]
Anyways, you should buy clothes and  shoes in VN, it cost almost-freely  :D[/quote]
then isnt it a nice idea to go to vietnam without shoes and buy one from there...as it ias almost-free :wink:
But about cloth its ur own choice.... :rotfl:

I dont think bying Notebook in Vietnam is the good idea.

My friends always bring Notebook from:
1) US
2) Japan
3) Singapore

Of course, there are cheap Notebook in Vietnam, but they are
1) Second hand
2) Of not very wellknown producers

For reference, I bought my IBM T60 for 1.400 USD. It not very good price, but it is price in Vietnam, I cant avoid it.

Namdung

But the Price for a IBM T60-CE4 is just 1,310USD and IBM ThinkPad T60 1951 CR9 is  1,179 USD. [url=http://www.thegioididong.com/Laptop/ProductDetail.aspx?IDBranch=1&IDIndex=1032&IDProduct=6431]take a look here[/url] . What's the kind of your notebook?


its the same as to questions like "how many 3 digit numbers can be made without using a 3 or 4" that you see on mathcounts tests.  first, draw 3 blanks.  one for the shirts, one for the pants, and one for the shoes.  now write in each blank the number of choices you have for that item.  now multiply the numbers to get ur answer.  i dont know why this works, but i think it has to do with combinations.  to solve the answer above about the 3 digit number, first, write the numbers in each blank that can be put there.  for the first one, you can have a 1,2,5,6,7,8,9, but NOT 0, or else it wouldn't be a 3 digit number.  for the 2nd and 3rd, u can hav all of the other numbers and 0.  so the answer is $7*8*8=448$.  so there are $448$ 3 digit numbers that can be made without using a 3 or 4.

i think you mean independent.

(and i think that you cannot add if you are using the indifference principle, which is a horrible principle anyways)

[hide]
you choose:

5 for the first
3 for the second
4 for the third

5x4x3=60

5+4+3->never add in counting/choosing
12c3 -> You may pick 3 shirts
12p3 -> you're not ordering them
[/hide]

[quote="alanchou"][quote="kohjhsd"][quote="sharkman"]Costello purchased 5 shirts, 3 pairs of pants and 4 pairs of shoes.
How many different outfits consistings of one shirt, 1 pair of pants and 1 pair of shoes can Costello make?

The answer is 5 x 4 x 3.

I want to know what makes the above answer correct.

I also want to know why the other 3 choices are not the correct answers.

OTHER CHOICES:

(a) 5 + 4 + 3

(b) 12C3

(c) 12P3[/quote]

[hide]5 ways to choose a shirt.
3 ways to choose pants.
4 ways to choose shoes.
These are exclusive, which gives $3 \cdot 4 \cdot 5 = 60$.

5+4+3: We never add in counting except when we're joining things together.
12C3: You can pick 3 shirts, which will be useless if you don't have your pants or shoes. :P
12P3: You are just picking 3 things again and ordering them, which is not quite right.[/hide][/quote]

Wait a sec, in exclusive events we add them.  I think you mean independant events.[/quote]

 :blush: Right. I forgot those two words are independent.


geometry perimeter trigonometry inradius

Find the perimeter and area of the regular polygon.
[img]http://img221.exs.cx/img221/6418/pa6qh.jpg[/img]
There's no examples of these in my textbook. Is there a formula or something? Please help.  :(
Thank you.

[quote="tarquin"][hide]$\frac {nx^2}{\tan \frac {180}{n}}$[/hide][/quote]

What's that for? The area or perimeter? and is n for the number of sides right? and x would be the number given???  :?

Area of any regular polygon given the measure of the side $x$ and the number of sides $n$.

i plugged it in. doesn't seem right. :(

3(10) 2
______= 937.4
tan60

><

Here's the proof.
[hide]A regular polygon with $n$ sides each of measure $x$ can be split into $n$ congruent isoscoles triangles with vertex angles equal to $\frac {360}{n}$. We can draw the altitude $y$ to the longest side of one of the isoscoles triangles creating two congruent right triangles and splitting the vertex angle in two. $\frac {\frac {1}{2}x}{y} = \tan \frac {180}{n}$, so $y = \frac {\frac {1}{2}x}{\tan \frac {180}{n}}$. Knowing this, the area of one of the isoscoles triangles is $\frac {x^2}{4 \tan \frac {180}{n}}$. Since there are $n$ congruent isoscoles triangles in the polygon, the area of the polygon is therefore $\frac {nx^2}{4 \tan \frac {180}{n}}$.[/hide]

orenji, I might have made a mistake typing the formula the first time. I think you need to divide by 4.
edit- Actually I think equilateral triangles are, for some weird reason, an exeption. Everything else should work.

Tarquin, it does work for equilateral triangles. It works for any regular polygon.

The first figure:
[hide]Is the first one supposed to be an equilateral triangle? If so, you would know the height is 20. You can make this into a 30, 60, 90 triangle. The length of x, the base of the piece, would be 20/:rt3: , and the hypotenuse would be 2 times the base, so 40/:rt3: .

That would give you a perimeter of 40/:rt3:  x 3 = 120/ :rt3:  = 120 :rt3:/3 = 40 :rt3: .

The area would be 20/ :rt3:  x 20 = 400/ :rt3:  = 40 :rt3: /3.[/hide]
The second figure:
[hide]If it's supposed to be a square, the length of the diagonal is 8.
a 2 + b 2 = c 2

So, since the lengths of the sides are the same, you can really say, a 2  + a 2 = 64.

Half of 64 is 32, so a  2 must equal 32. Therefore, the length of a must be 4 :rt2: .

The perimeter would be 4 :rt2:  x 4, giving you 16 :rt2: .

The area would then be (4 :rt2:) 2  which gives you 32. [/hide]

Hopefully this is what you wanted, and sorry if it isn't. I don't know any cool formulas, just how to find the area and  perimeter of the regular, basic figures.  :)

You could also say that generally: $[A] = \frac{mrn}{2}$, where $m$ is the side length, $r$ is the inradius, and $n$ is the number of sides.

$r = \frac{m}{2} \tan{\frac{90(n-2)}{n}}$

This also gives you the same result.

[quote="Zyloch"]Why not just use $\frac{1}{2}ap$ where $a$ is the apothem and $p$ is the perimeter?

Then all it amounts to is finding the apothem and the sides, which is relatively easy for all of them with a bit of trig sometimes.[/quote]

That's basically what tarquin did, except that he solved for the apothem with trig. and plugged it in so that all you'd need is the number of sides and the side length.

How about the formula for finding the perimeter?


For each positive rational number a let the numbers (a+1) and a/(a+1) be the children of a. Prove that each positive rational number, being not equal to 1, is a uniquely defined descendant of 1.

[i]German IMO Selection Test 1982, problem 5[/i]

First let us prove that any positive rational number is a descendant of 1.

Let p0/q0, with gcd(p_0,q_0)=1 be a rational positive number. If p_0=q_0, we are done. Otherwise, define the sequence p_n/q_n so that for all n>=1 , p_{n-1}/q_{n-1} is  the "child" of p_n/q_n.

If p_n>q_n , then p_{n+1}=p_n-q_n , q_{n+1}=q_n
If p_n<q_n , then p_{n+1}=p_n, q_{n+1}=q_n-p_n

In both cases, p_{n+1} + q_{n+1} < p_n + q_n, if p_n is not equal to q_n. Clearly this construction cannot go on forever, because otherwise we have found an infinite strictly decreasing sequence of natural numbers. Therefore, we must reach 1. Also from this construction it's obvious that there are not two paths from 1 to p0/q0. Otherwise the elements before p0/q0 must coincide, and we proceed to show by induction that the k'th elements before p0/q0 in the two sequences coincide. It's clear that they reach 1 the same time (or otherwise contradiction) so we are done.


Why must $ 4\minus{}7b^2$ be rational if $ b$ is rational?

The sum, product, and negative of two rational numbers is rational.


Two-thirds of the people in a room are seated in three-fourths of the
chairs. The rest of the people are standing. If there are 6 empty chairs,
how many people are in the room?

If there are 6 empty chairs, there are 18 taken chairs. 18 people is 2/3 of the people, so the total number of people is 27 people.


calculus
integration
LaTeX
calculus computations

[quote="lianafan"]The question is to integrate by parts: the integral of 2x sqrt2x-3 dx; dv =sqrt2x-3

I started to solve the problem but i keep getting stuck. Can you show me what I am doing wrong??

u=2x     dv=sqrt2x-3
du=2     v=2/3(2x-3)^3/2

uv-integral vdu= 2x*2/3(2x-3)^3/2-integral 2*2/3(2x-3)^3/2
After integrating, I got:
4x/3(2x-3)^3/2-8/15(2x-3)^5/2

Now where do i go from here?[/quote]
try the substitution $2x-3=t^2$
So, $dx=t dt$
which will make the integrand as $t^2(t^2+3)$ which can be easily integrated

the question says to integrate by parts not by substitution.

The question is to integrate by parts: the integral of 2x sqrt2x-3 dx; dv =sqrt2x-3

I started to solve the problem but i keep getting stuck. Can you show me what I am doing wrong??

u=2x dv=sqrt2x-3
du=2 v=2/3(2x-3)^3/2

uv-integral vdu= 2x*2/3(2x-3)^3/2-integral 2*2/3(2x-3)^3/2
After integrating, I got:
4x/3(2x-3)^3/2-8/15(2x-3)^5/2

first
if u = 2x , du = 2 dx, not 2

dv = sqrt(2x-3)dx
v = 1/3 (2x-3)^(3/2)

int udv = 2/3x(2x-3)^(3/2) - int 2/3 (2x-3)^(3/2) dx
=  2/3x(2x-3)^(3/2) -2/15 (2x-3)^(5/2) +C

[quote="lianafan"]the integral of 2x sqrt2x-3 dx; dv =sqrt2x-3[/quote]
Two comments:

1. $\text{\LaTeX}$ isn't all that hard to use. Seing something like your statement makes me less likely to read it and hence less likely to respond.

It's mostly pretty intuitive. Just write:

$\$ 2x\sqrt{2x-2}dx;dv=\sqrt{2x-3} $\$

and it becomes

$ 2x\sqrt{2x-2}dx;dv=\sqrt{2x-3 }$

(There are ways to fix the spacing - put a \, in where you want a space - but that's already a big improvement.)

2. Parentheses always matter! And they matter even more when you're trying to write in-line straight-text mathematics. From the context, I guessed that you meant $2x\sqrt{2x-3}$ but that's not what you said. You said $2x\sqrt{2x}\,-3,$ which is a completely different expression.

One thing about trying to enter formulas into spreadsheets and sophisticated calculators - either that will teach you to mind your parentheses, or you'll be getting a lot of wrong answers.

Note that isaacchao, in answering you, did use parentheses. I'd rather he used $TeX,$ but it is an improvement.

Thanks for showing me how to use the code. I could never figure it out before. sorry about how bad it looked before. Next time, it will be much better! :)


analytic geometry
vector
superior algebra
superior algebra unsolved

The question said : Let S5 act on (F5)^5 be permuting the coordinates, and form the semi-direct product G=    (F5)^5 [i][u]semi[/u][/i] S5                  ...

Where [i][u]semi[/u][/i] is that semi-direct product sign..

My question is, what is (F5)^5???? is it a number from F5 to the 5th exponent??? Or something else? And What does it mean by permuting the coordinates???

It's the $ 5$ dimensional vectors space over the field with $ 5$ elements. Thus the space of $ 5$-tuples from this field.

o...so an element in (F5)^5 would be like (a,b,c,d,e)  where a,b,c,d,e, each in F5??


[hide]If $ n$ is a mod 5 and b mod 4, then the term $ n^n$ will be equivalent to $ a^b$ mod 5. Thus, n's residue mod 20 will uniquely determine what it is mod 5. But from 1 to 100, every residue mod 20 is repeated 5 times, so $ 1^1\plus{}2^2\plus{}\cdots\plus{}100^{100}\equiv 5(1^1\plus{}2^2\plus{}\cdots\plus{}20^{20})\equiv 0 \pmod 5$.[/hide]

[quote="Brut3Forc3"]If $ n$ is a mod 5 and b mod 4, then the term $ n^n$ will be equivalent to $ a^b$ mod 5. Thus, n's residue mod 20 will uniquely determine what it is mod 5. .[/quote]

Hey that's clever!  :winner_first:

[quote="Brut3Forc3"][hide]If $ n$ is a mod 5 and b mod 4, then the term $ n^n$ will be equivalent to $ a^b$ mod 5. Thus, n's residue mod 20 will uniquely determine what it is mod 5. But from 1 to 100, every residue mod 20 is repeated 5 times, so $ 1^1 \plus{} 2^2 \plus{} \cdots \plus{} 100^{100}\equiv 5(1^1 \plus{} 2^2 \plus{} \cdots \plus{} 20^{20})\equiv 0 \pmod 5$.[/hide][/quote]

Beautiful work!! :trampoline:


How many six tuples (a,b,c,d,e,f) of odd positive integers satisfy 1/a+1/b+1/c+1/d+1/e+1/f =1?

[hide="I know that this may not be entirely rigorous, but here goes..."]Lets re-write this as:

$\frac{bcdef+acdef+abdef+abcef+abcdf+abcde}{abcdef}= 1$

Since these are all odd, and we know that the product of odd integers is odd, and the sum of odd and odd is even, and the sum of even and even is even, this is what we really have:

$\frac{ \textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}}{ \textrm{odd}}= 1$

$\frac{ \textrm{even}+\textrm{even}+\textrm{even}}{ \textrm{odd}}= 1$

$\frac{ \textrm{even}}{ \textrm{odd}}= 1$

Cross multiplying results in:

$\textrm{even}$ $=$ $\textrm{odd}$

This is clearly impossible!

So there are no solutions to the original equation. That restriction "odd positive integers" made all the difference.

There are $0$ solutions to the question.[/hide](Edit: Thanks for pointing that out!)

[quote="vishalarul"][hide="I know that this may not be entirely rigorous, but here goes..."]Lets re-write this as:

$\frac{bcdef+acdef+abdef+abcef+abcdf+abcde}{abcdef}= \frac{1}{6}$

Since these are all odd, and we know that the product of odd integers is odd, and the sum of odd and odd is even, and the sum of even and even is even, this is what we really have:

$\frac{ \textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}}{ \textrm{odd}}= \frac{1}{6}$

$\frac{ \textrm{even}+\textrm{even}+\textrm{even}}{ \textrm{odd}}= \frac{1}{6}$

$\frac{ \textrm{even}}{ \textrm{odd}}= \frac{1}{6}$

Cross multiplying results in:

$6$ $\cdot$ $\textrm{even}$ $=$ $\textrm{odd}$

Since $6$ is even,
$\textrm{even}$ $\cdot$ $\textrm{even}$ $=$ $\textrm{odd}$

This is clearly impossible!

So there are no solutions to the original equation. That restriction "odd positive integers" made all the difference.

There are $0$ solutions to the question.[/hide][/quote]
That is rigorous, it is called parity.

[quote="lotrgreengrapes7926"][quote="vishalarul"][hide="I know that this may not be entirely rigorous, but here goes..."]Lets re-write this as:

$\frac{bcdef+acdef+abdef+abcef+abcdf+abcde}{abcdef}= \frac{1}{6}$

Since these are all odd, and we know that the product of odd integers is odd, and the sum of odd and odd is even, and the sum of even and even is even, this is what we really have:

$\frac{ \textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}+\textrm{odd}}{ \textrm{odd}}= \frac{1}{6}$

$\frac{ \textrm{even}+\textrm{even}+\textrm{even}}{ \textrm{odd}}= \frac{1}{6}$

$\frac{ \textrm{even}}{ \textrm{odd}}= \frac{1}{6}$

Cross multiplying results in:

$6$ $\cdot$ $\textrm{even}$ $=$ $\textrm{odd}$

Since $6$ is even,
$\textrm{even}$ $\cdot$ $\textrm{even}$ $=$ $\textrm{odd}$

This is clearly impossible!

So there are no solutions to the original equation. That restriction "odd positive integers" made all the difference.

There are $0$ solutions to the question.[/hide][/quote]
Why is it equal to $\frac{1}{6}$, shouldn't it $=1$? Nice solution!! You could also multiply both sides by $abcdef$.[/quote]

yes...i think so but it wont make a difference because its still an oodd*even=odd which is a contradiction

[quote="mathgeniuse^ln(x)"]Hah, this was a problem on chess64's mock AMC.

I liked that question[/quote]

me too.


geometry
circumcircle
geometric transformation
Euler
power of a point
radical axis
geometry proposed

Let $(I),(O)$ be the incircle and the circumcircle of the triangle $ABC$. Prove that
the line $IO$ is orthogonal to the Monge's Line of the excircles of $ABC$.

Please, what is the Monge's line of the circle... ?

I think it's the line which is a result of this theorem:

http://mathworld.wolfram.com/MongesCircleTheorem.html

Yes, the Monge's line of the excircles of $ABC$ is the line through the exsimilicenters of the pairs of excircles $[(Ia),(Ib)]; [(Ib),(Ic)]; [(Ia),(Ic)]$.

The orthic axis and the Euler line intersect orthogonally

   at point X(468).

    T.Y.

    M.T.

Yes, this is derived from a well-known fact:

 [i]Fact[/i]  Let $ABC$ be a triangle and $A'B'C'$ be the orthic triangle. Then the axis of perspective of $ABC$ and $A'B'C'$ is the radical axis of the nine point circle and the circumcircle of $ABC$ and is, therefore, perpendicular to the Euler line of $ABC$.

 Back to the problem the Monge line $l$ of the excircles is the axis of perspector of $I_aI_bI_c$ and $ABC$. Notice that $ABC$ is the orthic triangle of $I_aI_bI_c$ and as a result, $l$ is perpedicular to the Euler line of $I_aI_bI_c$ which is the line $OI$.

 From this problem, we obtain the following corollary:

 [i]Corollary[/i]  If we call $W$ the nine point center, and $S$ is the center of the circle that intouches the excircles, then $SW$ is parallel to $OI$.


Hi I hope this is the right place to ask the following question. I'm going into a non-elite university this year and was wondering if it was worth while for a math major to pursue a second major in physics. Should math majors seeking to go to graduate school do a second major in physics or just do as many upper level math courses as possible with the time? Hope you can help as I don't know any pure maths majors personally. Thank you!

A lot of math people pursue majors in physics as well.  It's very common.

It's worthwhile if you are interested in physics, and it certainly won't hurt to know something other than math.


algebra
polynomial
calculus
integration
quadratics
modular arithmetic
algebra unsolved

Prove that the Polynomial $ f(x) \equal{} x^{4} \plus{} 26x^{3} \plus{} 56x^{2} \plus{} 78x \plus{} 1989$ can't be expressed as a product $ f(x) \equal{} p(x)q(x)$ , where $ p(x)$ and $ q(x)$ are both polynomial with integral coefficients and with degree at least $ 1$.

Apply Eisenstein's criterion to $ f(x\plus{}1)$ by choosing $ p\equal{}2$.

following solution is proposed by earldbest (Poincare Conjecture)

I'm not sure, but...

consider f(x + 1) = x^{4} + 30x^{3} + 140x^{2} + 272x + 2150, which is irreducible by Eisenstein's Irreducibility Criterion

[quote="makar"]following solution is proposed by earldbest

consider $ f(x \plus{} 1) \equal{} x^{4} \plus{} 30x^{3} \plus{} 140x^{2} \plus{} 272x \plus{} 2150$, which is irreducible by Eisenstein's Irreducibility Criterion[/quote]

this is the same as [b]MeKnowsNothing[/b]'s solution :D

We can directly apply extended Eisenstein criterion to the polynomial as each term other than the first is divisible by 13

[quote="Pascal96"]each term other than the first is divisible by 13[/quote]

$13$ does not divide $56$.  So, it is not too simple if you are gonna use this prime, as you can only conclude that the polynomial is either irreducible, or divisible by some quadratic factor.  This gives us basically nothing.

Oops, I think I read 56 as 65 by mistake.. Sorry about that!

Solution: Go $\pmod 2$. We have $f(x)\equiv x^4+1\pmod 2$. Now $(x+1)^4+1=x^4+4x^3+6x^2+4x+2$, and we are done by Eisenstien's. $\blacksquare$

[quote=makar]Can someone give detailed solution if possible alternate approach.
Sorry to say but i have no idea about Eisenstein's criterion.please explain it a little or provide me some link in which it explained in detail using examples[/quote]

Assume that the roots of the polynomial $f(x)=x^4+26x^3+56x^2+78x+1989$ are $\alpha,\beta,\gamma,\delta.$ By Vieta's Relation we get, $$(\alpha+\beta)+(\gamma+\delta)=-26$$ $$\alpha\beta+\gamma\delta+(\alpha+\beta)(\gamma+\delta)=56$$ $$\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)=-78$$ $$(\alpha\beta)(\gamma\delta)=1989$$
Assume for contradiction!! Let's Assume that $f(x)$ can be factored as a product of two non-constant polynomial $p(x)$ and $q(x)$ with integer coefficient and degree at least $1$. $\implies \boxed{f(x)=p(x)q(x)}$
Since $f(x)$ is a monic Quartic polynomial therefore it can be break into two monic polynomials!!
Therefore, $p(x)$ and $q(x)$ both are monic polynomials!! Therefore only three cases arise..
[color=#f00]case1[/color]:- $p(x)$ and $q(x)$ both are Quadratic!!
[color=#00f]case 2[/color]:- $p(x)$ have degree $3$ and $q(x)$ have degree $1$
[color=#0f0]case 3[/color]:- $p(x)$ have degree $1$ and $q(x)$ have degree $3$
First Look for Case 1:- Assume $\boxed{p(x)=x^2+b_1x+c_1}$ and $\boxed{q(x)=x^2+b_2x+c_2}$ where $b_1,b_2,c_1,c_2$ are integers!!!
Therefore,  $\boxed{f(x)=(x^2+b_1x+c_1)(x^2+b_2x+c_2)}\implies x^{4} + 26x^{3} + 56x^{2} + 78x + 1989=(x^2+b_1x+c_1)(x^2+b_2x+c_2)\implies\boxed{c_1.c_2=1989}$
We get, $$\boxed{x^{4} + 26x^{3} + 56x^{2} + 78x + 1989=x^4+(b_1+b_2)x^3+(c_1+c_2+b_1b_2)x^2+(b_1c_2+b_2c_1)x+c_1c_2}$$
Since $c_1c_2=1989$ it states that there are $3$ pairs of $\boxed{(c_1,c_2)=(1,1989),(9,221),(3,663)}$
 Let's check for $\boxed{c_1=1}$ and $\boxed{c_2=1989}$ Therefore we get, $\boxed{b_1+b_2=26}$ and $\boxed{b_1b_2+1990=56}$ and $\boxed{1989b_1+b_2=78}$ Third equation only satisfies when $b_1=0$ and $b_2=78$ for integers!! but it cant satisfied the first equation therefore we get a contradiction!!

Now Let's check for $\boxed{c_1=9}$ and $\boxed{c_2=221}$ get, $\boxed{b_1+b_2=26}$ , $\boxed{b_1b_2+230=56}$ and $\boxed{221b_1+9b_2=78}$ get from first equation $b_1=26-b_2$ putting this value on third we get a contradiction since $b_1$ and $b_2$ are integers!!

Lastly we check for $\boxed{c_1=3}$ and $\boxed{c_2=663}$ get, $\boxed{b_1+b_2=26}\implies b_1=26-b_2$ , $\boxed{b_1b_2+666=56}$ and $\boxed{3b_2+663b_1=78}$ we get contradiction since $b_1$ and $b_2$ are integers!!
Therefore Case 1 can't happen!!

We also get, $\boxed{f(0)=p(0)q(0)}$

Now let's explore Case 2:- Assume $\boxed{p(x)=x+a}$ and $\boxed{q(x)=x^3+bx^2+cx+d}$ where $a,b,c,d$ are integers!!!
Therefore we get, $$x^4+26x^3+56x^2+78x+1989=(x+a)(x^3+bx^2+cx+d)$$
We Get, $\boxed{ad=1989}$ similarly there are only $3$ pairs which satisfied this!! $\boxed{(a,d)=(1,1989),(9,221),(3,663)}$
Therefore we get, $$\boxed{x^4+26x^3+56x^2+78x+1989=x^4+(a+b)x^3+(ab+c)x^2+(ab+c+d)x+ad}$$

Now, First check for $\boxed{a=1}$ and $\boxed{d=1989}$ get $\boxed{b=25}$ and $ab+c=56\implies \boxed{c=31}$ $\implies$ $25+31+1989=78$ gives us the contradiction!!

Check for $\boxed{a=9}$ and $\boxed{d=221}$   get $\boxed{b=15}$ and $\boxed{c=-79}$  similarly the third case doesn't satisfied gives us the contradiction !!

Similarly the last case also doesn't satisfied Since $a,b,c,d$ are integers!!
which gives the contradiction!!
Therefore $f(x)$ can't be expressed into two non constant polynomials $p(x)$ and $q(x)$ with integer coefficient and degree at least $1$.
$$\boxed{f(x)\ne p(x)q(x)}$$


a).Let $(a_n)$ be an increasing sequence of positive integers and let $S_n=a_1+a_2+...+a_n.$Show that if $\displaystyle \liminf \frac{a_n}{n}>2+\sqrt2$, then for all n sufficiently large there exists a perfect square between $S_n$ and $S_{n+1}$.
b). If $\displaystyle \liminf \frac{a_n}{n}=2+\sqrt2$ and the above conclusion fails, the show that $\displaystyle \limsup \frac{a_n}{n}=\infty$

weird !
one knows that there exists some $\epsilon>0$ such that for $n$ large, $\frac{a_n}{n}>C+\epsilon$. (here $C=2 + \sqrt{2}$).
From here it immediately follows that for $n$ large, there exists some $\mu > 0$ such that $\frac{S_n}{n^2}> \frac{C}{2} + \mu$

But there is always a perfect square between $n$ and $n+2\sqrt{n}+1$. Hence, because $S_{n+1}-S_n=a_{n+1}$, one just have to prove that for $n$ large, $\frac{a_n}{n} > \frac{2\sqrt{\(frac{C}{2} + \mu)}n+1}{n}$.
But for $n$ large, $\frac{a_n}{n} > C + \epsilon$ and $\frac{2\sqrt{(\frac{C}{2} + \mu)}n+1}{n}$ tends towards $2\sqrt{\frac{C}{2} + \mu}$. And because $\mu$ can be chosen as small as one wants, the conclusion is trivial because for all $x>2$, $x>2\sqrt{\frac{x}{2}}$. So I made a big mistake somewhere but I cannot see where ?

[quote="alekk"]weird !
one knows that there exists some $\epsilon>0$ such that for $n$ large, $\frac{a_n}{n}>C+\epsilon$. (here $C=2 + \sqrt{2}$).
From here it immediately follows that for $n$ large, there exists some $\mu > 0$ such that $\frac{S_n}{n^2}> \frac{C}{2} + \mu$
[/quote]
So, from here you have $\sqrt{S_n}> (\sqrt{\frac{C}{2}+\mu})n$ (*).
[quote="alekk"]
But there is always a perfect square between $n$ and $n+2\sqrt{n}+1$. Hence, because $S_{n+1}-S_n=a_{n+1}$, one just have to prove that for $n$ large, $\frac{a_n}{n} > \frac{2\sqrt{\(frac{C}{2} + \mu)}n+1}{n}$.
[/quote]
I think here is the mistake. So, as we want to prove that there is a square between $S_n$ and $S_{n+1}$, it would be enough to prove that $a_{n+1}=S_{n+1}-S_n>2\sqrt{S_n}+1$ (***).But (*) says that $2\sqrt{S_n}+1> 2(\sqrt{\frac{C}{2}+\mu})n+1$ and below you prove that $a_{n+1}>2(\sqrt{\frac{C}{2}+\mu})n+1$, but this doesn't give us (***). If i'm not wrong you give a bound for two numbers below and it doesn't give us any comparision between them.
[quote="alekk"]
But for $n$ large, $\frac{a_n}{n} > C + \epsilon$ and $\frac{2\sqrt{(\frac{C}{2} + \mu)}n+1}{n}$ tends towards $2\sqrt{\frac{C}{2} + \mu}$. And because $\mu$ can be chosen as small as one wants, the conclusion is trivial because for all $x>2$, $x>2\sqrt{\frac{x}{2}}$. So I made a big mistake somewhere but I cannot see where ?[/quote]

stupid me  :blush:  you're right Eugene  :lol:


integration
function
algebra
rational function
partial fractions
calculus
calculus computations

$ \int \frac {x^2 \plus{} 5x \minus{} 4}{(x^5 \plus{} 1)^3} dx$

any hints?

Can this be done in closed form? The answer is yes. It's a rational function. The only possible obstacle to expressing the antiderivative of a rational function in closed form is the question of whether we can factor the denominator. But we know that fifth roots of unity (actually, because of the $ \plus{}$ sign, tenth roots of unity) can be expressed using square roots (which might be nested). So we can factor that 15th degree denominator, perform partial fractions, and proceed.

Am I willing to attempt to do that by hand? Not a chance.

Is there a gimmick that makes things shorter? Maybe. I haven't seen one yet.

This might be a good test case for the capacities and limitations of computer algebra systems. (I don't routinely use any such system myself.)

[url]http://www.integrals.com[/url] is free.  Its answer to this problem is explicit but ugly.

I looked at that. All I can say is that I actually expected uglier - somehow, some of the possible terms must have combined together. I do see those fifth roots of unity in there.


algebra polynomial calculus integration induction algorithm

prove that
\[ P(x) \equiv 0(\bmod p).
\]
has at most $ n$ solutions where $ n$ is the degree of the the polynomial $ P$ with integral coefficients and $ p$ is a prime number.

Prove this without induction or the use of determinants.

Well, just apply the division algorithm (which is possible since the inverses all exist mod p except for 0, which is trivial) to each of the roots one by one.

Then the result follows.

 :)

Induction and determinants?  What proofs have you seen?  Once you've verified the field axioms for $ \mathbb{F}_p$ the rest follows by the divison algorithm in $ F[x]$ for any field $ F$.


Actually, I think it's $(x-6)(x^{2}-4x+5)$.

The Factor Thm: If C is a root of f(x), (x - C) is a factor of f(x)
Therefore, we can construct a polynomial with the roots given.

The Fundamental Thm. of Algebra: if f(x) is a polynomial of degree N, f(x) has N roots (either complex or real).

We are given three roots, so N must be at least degree three.

$(x-6)$ is a factor.
So is $(x-(2+i))$ and $(x-(2-i))$
The polynomial of least degree with these factors, is necessarily
$(x-6)(x-(2+i))(x-(2-i)$,
or
$(x-6)(x^{2}-(2x+ix)-(2x-ix)+5)$
$(x-6)(x^{2}-4x+5)$
$x^{3}-4x^{2}+5x-6x^{2}+24x-30$
$\boxed{x^{3}-10x^{2}+29x-30}$

[quote="Algebra 1 ©"]What is the polynomial of least degree that has the zeros 6,2+ i, and 2-i?

I got this as the answer:
 $x^{3}+6x^{2}-3x-18+(-4x-24)i$

Can anyone confirm that this is correct?[/quote]

Graargh, we just spent last chapter doing that in Precalc.

Actually, you could arrive at the same answer using only the zeros 6 and 2+i, since by the conjugate thingy, 2-i also has to be a zero.

DiscreetFourierTransform appears to be correct.


[quote="jhcreinhardt"]Hint:
[hide]
what does the left side remind you of?[/hide][/quote]

Of algebra class.

Joke!

So, the products of the real roots are a factor of 2005.

Then, using these factors, we find which produce 2005=2005 when plugged into the equation.

Then, 2005/(reals) is the answer rt?

Please don't make fun of me...

[quote="13375P34K43V312"]Let $P$ be the product of the non-real roots of $x^{4}-4x^{3}+6x^{2}-4x=2005$. Find floor(P).[/quote]
[hide]
Add 1 to both sides
$(x-1)^{4}=2006$
The two non-real roots are $\pm\sqrt[4]{2006}+1$ where $\sqrt[4]{2006}$ are not real
$(i\sqrt[4]{2006}+1)(-i\sqrt[4]{2006}+1)=\sqrt{2006}+1$, floor of that is $\boxed{45}$[/hide]

edit: a little late  :huh:

OOOOH

My bad. I forgot we were dealing with irrationals as well.

Stupid me!


trigonometry
inequalities
algebra unsolved
algebra

Solve  equation :)
$ tgx \minus{} \sqrt {3}cotgx \minus{} sinx \plus{} \sqrt {3}cosx \equal{}\sqrt{3} \minus{} 1$

[quote="nkht-tk14"]Solve  equation :)
$ \tan{x} - \sqrt{3}\cot{x} - \sin{x} + \sqrt {3}\,\cos{x} = \sqrt {3} - 1$[/quote]

[hide="Here is my solution."]
Hi nkht-tk14!

Let's substitute $ x=2\,\arctan(y)$ to the equation and remember that $ \tan(x)=\frac{2\,y}{1-y^2}$, $ \sin(x)=\frac{2\,y}{1+y^2}$ and $ \cos(x)=\frac{1-t^2}{1+t^2}$. So we will have:

\[ {\frac {2y}{1-{y}^{2}}}-\frac{\sqrt{3}\,(1-y^2)}{2\,y}-\frac{2\,y}{1+y^2}+\frac{\sqrt{3}\,(1-y^2)}{1+y^2}-\sqrt{3}+1=0\]
\[ \left( {y}^{4}-4\,{y}^{3}-1 \right)  \left( y-\frac{\sqrt{3}}{3}\right)  \left( y+\sqrt {3} \right) = 0\]

Now let us solve the equation $ y^4-4\,y^3-1=0$.

\[ y^4-4\,y^3-1 =  \left( {y}^{2}-2\,y-1 \right) ^{2} + \left({y}^{4}-4\,{y}^{3}-1- \left( {y}^{2}-2\,y-1 \right) ^{2}\right)=\\
=\left( {y}^{2}-2\,y-1 \right) ^{2} -2\, \left( y+1 \right) ^{2} = (y^2-(2+\sqrt{2})\,y-1-\sqrt{2})\,(y^2-(2-\sqrt{2})\,y-1+\sqrt{2})\]

First equation $ y^2-(2+\sqrt{2})\,y-1-\sqrt{2}$ has solutions $ y_1=1+\frac{\sqrt {2}}{2}+\frac{\sqrt {10+8\,\sqrt {2}}}{2}$ and $ y_2=1+\frac{\sqrt {2}}{2}-\frac{\sqrt {10+8\,\sqrt {2}}}{2}$. Second equation $ y^2-(2-\sqrt{2})\,y-1+\sqrt{2}$ has discriminant $ 10-8\,\sqrt {2}=\sqrt{100}-\sqrt{128}\le0$, so it has no real solutions.

Now have found all values for $ y$: $ \frac{\sqrt{3}}{3}, -\sqrt{3}, 1+\frac{\sqrt {2}}{2}+\frac{\sqrt {10+8\,\sqrt {2}}}{2}$ and $ 1+\frac{\sqrt {2}}{2}-\frac{\sqrt {10+8\,\sqrt {2}}}{2}$

So we can write the solutions of the original equation:
\[ x=\frac{\pi}{3}+2\,k\,\pi\\
x=-\frac{2\,\pi}{3}+2\,k\,\pi\\
x=2\,\arctan(1+\frac{\sqrt {2}}{2}+\frac{\sqrt {10+8\,\sqrt {2}}}{2}) +2\,k\,\pi\\
x=2\,\arctan(1+\frac{\sqrt {2}}{2}-\frac{\sqrt {10+8\,\sqrt {2}}}{2}) +2\,k\,\pi\]

First 2 solutions can be written as one series $ x=\frac{\pi}{3}+k\,\pi$.

Also we need to substitute $ x=k\,\pi$ manually into the original equation to make sure that solutions have not been lost (since $ \tan{\frac{x}{2}}$ is not defined at these points). No of them are solutions of the original inequality, because $ cot(x)$ is also not defined at these points. This finish the solution.

Sergey Markelov
markelov@mccme.ru[/hide]

and this is mine :D :
[hide]
$ tg(x) \minus{} \frac {\sqrt {3}}{tg(x)} \minus{} sin(x) \plus{} \sqrt {3} cos(x) \equal{} \sqrt {3} \minus{} 1$
$ \Leftrightarrow \frac {sin(x)}{cos(x)} \minus{} \frac {\sqrt {3}cos(x)}{sin(x)} \minus{} sin(x) \plus{} \sqrt {3} cos(x) \equal{} \sqrt {3} \minus{} 1$
For computations simplification set $ a \equal{} sin(x)$, $ b \equal{} cos(x)$ then:
$ \Leftrightarrow \frac {a}{b} \minus{} \frac {\sqrt {3}b}{a} \minus{} a \plus{} \sqrt {3}b \equal{} \sqrt {3} \minus{} 1$
$ \Leftrightarrow a^2 \minus{} \sqrt {3}b^2 \minus{} a^{2}b \plus{} \sqrt {3}ab^2 \equal{} (\sqrt {3} \minus{} 1)ab$
$ \Leftrightarrow (a \plus{} b \minus{} ab)(a \minus{} \sqrt {3}b) \equal{} 0$
$ \Leftrightarrow a \plus{} b \minus{} ab \equal{} 0$ or $ a \minus{} \sqrt {3}b \equal{} 0$
Now the second equation can be solved easily , for the first we can solve it by squaring and verifying and I found 3 solutions:
$ \frac {\pi}{3} \plus{} k \pi$
$ \frac {\arcsin(2 \minus{} 2\sqrt {2})}{2} \plus{} k \pi$
$ \frac{\pi}{2} \minus{} \frac {\arcsin(2 \minus{} 2\sqrt {2})}{2} \plus{} k \pi$
[/hide]


geometry
circumcircle
trapezoid
geometry proposed

Let $a$ be the side length of a regular 14-gon. Show synthetically that the radius $R$ of its circumcircle  is a real positive solution of the equation
\[ (a \plus{} R)^3 \equal{} 4a^2R \plus{} 5aR^2\]

Let $O$ be the center of the 14-gon and $ B,C$ two consecutive vertices. Thus $\angle BOC \equal{} \frac {_{180^{\circ}}}{^{7}}.$ There exists two points $ P,Q$ on  $OC,OB$ such that $ BP \equal{} PQ \equal{} QO=a.$ Draw  parallels $ QT \equal{} x$ and $ PS \equal{} y$ to $ BC.$ Then $ \triangle  CBP$ and $ \triangle QOT$ are congruent $\Longrightarrow$ $ PC \equal{} QT \equal{} x.$  But $\triangle BCP$ and $\triangle OBC$ are similar

$\Longrightarrow \ \frac {PC}{BC} \equal{} \frac {BC}{R} \Longrightarrow \ x \equal{} \frac {a^2}{R} \ (1)$

$QTPS$ is a trapezoid with $ PS \equal{} QS \equal{} y$ and since $ \triangle OSP \sim  \triangle OBC,$ we get

$ \frac {SP}{BC} \equal{} \frac {OS}{OB} \ \Longrightarrow \ \frac {y}{a} \equal{} \frac {y \plus{} a}{R} \Longrightarrow \ y \equal{} \frac {a^2}{R \minus{} a} \ (2)$

$ QS \equal{} TP \equal{} y \Longrightarrow \ TP \plus{} PC \equal{} OC \minus{} OT \Longrightarrow \ y \plus{} x \equal{} R \minus{} a \ (3)$

Combining $(1),(2)$ and $(3)$ yields

$ \frac {a^2}{R \minus{} a} \plus{} \frac {a^2}{R} \equal{} R \minus{} a \ \Longrightarrow \  R^3 \plus{} a^3 \equal{} a^2R \plus{} 2aR^2$

$\Longrightarrow \ (a \plus{} R)^3 \equal{} 4a^2R \plus{} 5aR^2.$


function
real analysis
real analysis theorems

I've some difficulties whit the little-o notation. I understand it but cannot work with it good. Do you have some file than can help me ?

thanks

thanks.

I was not clear enough, where can I find a proof (with landau notation) of the fact that the product of two totally differentiable functions is also total differentiabel.

Suppose that $ f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $ g: \mathbb{R}^n \rightarrow \mathbb{R}^m$.
If we know that $ f(x) \equal{} f(x_0) \plus{} A(x \minus{} x_0) \plus{} o(|x \minus{} x_0|) \ x \rightarrow x_0$ and that
$ g(x) \equal{} g(x_0) \plus{} B(x \minus{} x_0) \plus{} o(|x \minus{} x_0|) \ x \rightarrow x_0$.
$ A,B$ are matrices.

How can I proof that $ f(x)g(x) \equal{} f(x_0)g(x_0) \plus{} A(x \minus{} x_0) g(x_0) \plus{} f(x_0)B(x \minus{} x_0) \plus{} o(|x \minus{} x_0|) \ x \rightarrow x_0$.
Can you give me a push how to start proving this ?


Here are some problems that I thought were interesting:

4.40

Ammonia is burned to form nitric oxide as per the reaction

$ 4 \text{NH}_3 \plus{} 5 \text{O}_2 \leftrightarrow 4 \text{NO} \plus{} 6 \text{H}_2\text{O}$

a)  Calculate the net rate of generation of all species as a ration of that for $ \text{NH}_3$.

b)  If ammonia is fed into a continuous flow reactor at the flow rate of 100 mol/h and air is used at a flow rate corresponding to $ 40.0$ % stoichiometric excess of oxygen, calculate the molar flow rates and mole fractions of all species exiting the reactor, assuming complete conversion ($ X \equal{} 1$) for ammonia.

4.42

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous flow reactor to which ethylene and hydrogen bromide are fed.  The product stream is analyzed and found to contain $ 51.7$ mol % $ \text{C}_2\text{H}_5\text{Br}$ and $ 17.3$ mol % $ \text{HBr}$.  Which is the limiting reactant ($ \text{A}$), which reactant ($ \text{B}$) is in stoichiometric excess?  Calculate the fractional conversion $ X$ of the limiting reactant and the molar feed ratio $ \Theta_B$ of the reactant that is in stoichiometric excess.

Speaking of which, is anyone else doing anything related to chemical enineering?

[quote="Zakary"]Speaking of which, is anyone else doing anything related to chemical enineering?[/quote]

Some time ago I have teached "Energy and Mass Balances", which is a topic belonging to Chemical Engineering and Environmental Engineering courses.

I'm a Chemical Engineering major, and do research in Catalysis.


Does anyone have any Mandelbrot problems and solutions? If so please post or PM me.

[quote="dondigo"]What are Mandelbrot problems :?:[/quote]

I believe he's referring to problems from the [url=http://www.mandelbrot.org/welcome.html]Mandelbrot Contest[/url], am I right?

Are you referring specifically to team Rounds or individual problems, or problems in general?

You can get a few problems from the current year from http://www.mandelbrot.com I think  :D


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let k[x, y] be the polynomial ring in two variables x and y over a field k.
Prove that any ideal of the form (x − a, y − b) for a, b belongs to k is a maximal
ideal of this ring

[quote="ZetaX"]The ideal $ \mathfrak m$ is maximal in the ring $ R$ iff $ R/\mathfrak m$ is a field.[/quote]

Hmm, not that I am trying to insult you, but dont we need to know that $ R$ is a commutative unitary ring?
Meaning, An ideal $ I$ is maximial in a commutative unitary ring iff $ R/I$ is a field.

yes, and $ k[x,y]$ is commutative and unitary ... and in many contexts: ring = commutative unitary ring.

by the way: $ I$ is a maximal ideal in a ring $ R$ if and only if $ R/I$ is simple.


[hide]
i dont think this is right, but is goes like $\frac{1979}{1980}*\frac{1980}{1981}$ so i think it is $\frac{1}{2003}$[/hide]

[hide]As we evaluate each separate term, we start seeing a pattern:$1979/1980*1980/1981*1981/1982...$All the numerators and denominators cross-cancel except for the first and last, which leaves us with an answer of $1979/2005$.[/hide]

oh darnit i messed up the numerator
Answer is[hide]Easy as pi's, $\frac{1879}{2003}$[/hide]

this is a really old topic.
And learn to edit your posts instead of making new ones.


Given the acuted-angle triangle $ ABC$. Let $ w_1$ be the circle with diameter $ AB$ and $ w_2$ be the circle with diameter $ AC$. Denote that the line through $ C$ and perpendicular to $ AB$ intersect $ AB$ at $ C_1$ and $ w_1$ at $ C_2$ ( $ C_2$ is outside the triangle). Define $ B_1$ and $ B_2$ similarly. Let $ I$ be the intersection of the line $ B_1C_2$ and the line $ B_2C_1$. Prove or disprove that $ AI$, $ BC_2$, $ CB_2$ are concurrent.

A bit computation solution:  :blush:
Denote $ \angle{AB_1I} \equal{} \angle{ABC_2} \equal{} a$ we have:
$ \frac {sinC_1B_1I}{sinAB_1I}.\frac {sinCBC_2}{sinABC_2} \equal{} \frac {sin(B \minus{} a).sin(B \plus{} a)}{sin^2a} \equal{} \frac {cos2a \minus{} cos2B}{1 \minus{} cos2B} \equal{} 1 \minus{} \frac {1 \minus{} cos2a}{1 \minus{} cos2B} \equal{} 1 \minus{} \frac {sin^2a}{sin^2B} \equal{} 1 \minus{} \frac {AC_2^2.4R^2}{AB^2.AC^2}$
Similarly and use $ AC_2^2 \equal{} AC_1.AB \equal{} AB_1.AC \equal{} AB_2^2\Leftrightarrow AB_2 \equal{} AC_2$ we have:
$ \frac {sinC_1B_1I}{sinAB_1I}.\frac {sinCBC_2}{sinABC_2} \equal{} \frac {sinB_1C_1I}{sinAC_1I}.\frac {sinBCB_2}{sinACB_2}$
Now use Xeva sines theorem for two triangle $ ABC,AB_1C_1$ we have solution!


basics
(1)why is magnetic field generated when current is passed through a conductor?
and (2)why is current induced in a conductor by change in magnetic flux?     :?:

At a fundamental level, the best physics can do is mathematically describe natural phenomenon - it can't answer the "why," since that is in the realm of philosophy.

Ampere's law/the Biot-Savart law and Faraday's law explain those effects you mentioned above, but I didn't really answer your question because those laws come from empirical observations.  I don't believe anyone can tell you the "why" behind physical laws.

[quote="MysticTerminator"]one possible answer (to the first question) is relativistic effects applied to the electric field. if you appropriately shift your frame of reference in the first example, try to see why the magnetic force in the original frame is actually just an electric force in this new frame[/quote]

That is a fascinating point.  Do you mean that not only are electric and magnetic fields intimately related, but they are actually the same?  Is this a principle of relativistic electrodynamics or something?

yup. If you have Griffiths' book on electromagnetism (because where else would you learn the subject from?), he gives a good introduction in the last chapter. It's not very mathematically deep, but going further usually uses some basic differential geometry, at least.