geometry
circumcircle
geometric transformation
reflection
similar triangles
geometry unsolved

Let two points $ A, B$ lie on the line $ d$. Let circle $ \omega$ and circle $ \omega$' tangent with line $ d$ at $ A, B$ respectively, circle $ \omega$ and circle $ \omega$' be orthogona at $ M$ and $ N$. Find locus of point $ M$ and find locus of point $ N$.

Inversion with center $ A$ and arbitrary power $ r^2$ takes the line $ d$ passing through the inversion center into itself, the circle $ \omega$ passing through the inversion center and tangent to the line $ d$ into a line $ \omega_i$ parallel to $ d$ and the circle $ \omega' \perp \omega$ tangent to $ d$ at $ B$ into a circle $ \omega'_i$ tangent to $ d$ at the image $ B_i$ of $ B$ and perpendicular to the line $ \omega_i,$ i.e., centered on the line $ \omega_i.$ Loci of the intersections $ M_i, N_i$ of $ \omega_i \cap \omega'_i$ are 2 lines $ l_{Mi}, l_{Ni}$ passing through $ B_i$ and forming angles $ \pm 45^\circ$ with $ d$. As inversion preserves angles between curves, loci of the intersections $ M, N$ of $ \omega \cap \omega'$ are 2 circles $ (L_M), (L_N)$ passing through $ A, B$ and forming angles $ \pm 45^\circ$ with $ d,$ i.e., circumcircles of the 2 squares build on the segment $ AB.$

Another inversive solution...

Lemma: Consider two points $ A$, $ B$.  Let $ P$ and $ P'$ be inverses with respect to the circle with diameter $ AB$.  Then $ m\angle{APB}\plus{}m\angle{AP'B}\equal{}180^{\circ}$.
Proof: Let $ O$ be the center of the circle.  By similar triangles, we have $ m\angle{OAP}\equal{}m\angle{OP'A}$, and similarly on the other side, this proves the lemma.

First note that clearly the loci of $ M$ and $ N$ are reflections of each other over $ d$.  Let these two loci be $ L_M$ and $ L_N$, respectively.  Let $ \Omega$ be the circle with diameter $ AB$.  Both $ \omega$ and $ \omega'$ are orthogonal to $ \Omega$: hence, an inversion in $ \Omega$ must take $ M$ to $ N$.  Therefore, an inversion in $ \Omega$ takes $ L_M$ to $ L_N$.

Now assume $ N$ is further from the line than $ M$ on one side, and closer on the other.  Notice \[ m\angle{ANB}\equal{}\frac{1}{2}(m\widehat{AM}\plus{}m\widehat{MB})\]
\[ m\angle{AMB}\equal{}\frac{1}{2}(m\widehat{AN}\plus{}m\widehat{NB})\equal{}90^{\circ}\plus{}\frac{1}{2}(m\widehat{AM}\plus{}m\widehat{MB})\]
So
\[ 180^{\circ}m\angle{AMB}\plus{}m\angle{ANB}\equal{}90^{\circ}\plus{}m\widehat{AM}\plus{}m\widehat{MB}\equal{}90^{\circ}\plus{}2m\angle{ANB}\]
So $ m\angle{ANB}\equal{}45^{\circ}$, and so $ L_N$ is a circle which passes through $ A$ and $ B$, and makes a $ 45^{\circ}$ angle with the line.  So is $ L_M$.

[quote="thanhnam2902"][color=darkred]Let $ d$ be a fixed line and let $ A$, $ B$ be two its fixed points. Let $ C(O_1)$ , $ C(O_2)$ be the variable circles which are tangent to the line $ d$ at the points $ A$, $ B$ respectively and which are  orthogonally at the their common points $ M$ and $ N$. Find the geometrical locus of the point $ M$ and the geometrical  locus of the point $ N$ .[/color][/quote]

[color=darkblue][u][b]A simple proof[/u][/b] (similarly with [b]K8107[/b]'s). Suppose w.l.o.g. that the line $ O_1O_2$ doesn't separate the points $ A$ , $ M$ .

Denote $ \{\begin{array}{c} m(\widehat {AO_1M}) = 2x \\
 \\
m(\widehat {BO_2M}) = 2y\end{array}$ . Observe that $ \{\begin{array}{c} m(\widehat {BAM}) = m(\widehat {ANM}) = x \\
 \\
m(\widehat {ABM}) = m(\widehat {ANM}) = y\end{array}$

and $ m(\widehat {AMB}) = 180^{\circ} - (x + y)$ . Since $ MO_1\perp MO_2$ obtain $ m(\widehat {AMB}) =$

$ 270^{\circ} - m(\widehat {AMO_1}) - m(\widehat {BMO_2}) =$ $ 270^{\circ} - (90^{\circ} - x) - (90^{\circ} - y) =$ $ 90^{\circ} + (x + y)$ .

Therefore, $ 180^{\circ} - (x + y) = 90^{\circ} + (x + y)$ , i.e. $ \boxed {\ x + y = 45^{\circ}\ }$ and $ m(\widehat {AMB}) = 135^{\circ}$ ([b]constant[/b]).

Obtain analogously that $ m(\widehat {ANB}) = m(\widehat {ANM}) + m(\widehat {BNM}) = x + y$ , i.e. $ m(\widehat {ANB}) = 45^{\circ}$ ([b]constant[/b]).

In conclusion, the geometrical locus of the point $ M$ is the circle $ w$ (for any point $ X\in w$ , $ m(\widehat {AXB}) = 135^{\circ}$ ) and

the geometrical locus of the point $ N$ is the circle $ w'$ (for any point $ Y\in w'$ , $ m(\widehat {AYB}) = 45^{\circ}$ ) .

The circles $ w$ and $ w'$ are symmetrically w.r.t. the its common chord $ [AB]$ .[/color]

Thank yetti, K81o7, Virgil Nicula for your post.


Let $ABCD$ be a trapezium $AB // CD,$ $M$ and $N$ are fixed points on $AB,$  $P$ is a variable point on $CD$. $E = DN \cap AP$, $F = DN \cap MC$, $G = MC \cap PB$, $DP = \lambda \cdot CD$. Find the value of $\lambda$ for which the area of quadrilateral $PEFG$ is maximum.

it can be a nice and a dirty(!) one also


probability
combinatorics unsolved
combinatorics

Hi can people please respond to my lottery probability post.. Its on another forum in here.. I'm actually playing devils advocate on the note to prove to everyone that if you buy 2 tickets on a 135 mil to one lottery you have a 1-67.5 mil chance to win but no one here believes me..

its under the category "other problem solving"


can any one tell me the weight, in pounds, of a kids pool 77" round by 18" deep, when it is filled with water?

Volume of a right cylinder: $\pi r^2 h$

Radius: $\frac{c}{2\pi}=r$

Cubic Inches to Pounds conversion factor for water is.... $0.036127 \frac{lbs.}{in.^3}$


geometry
circumcircle
incenter
geometric transformation
homothety
reflection
angle bisector

The incircle of a triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $D'$ be the midpoint of the arc $BC$ (of the circumcircle of $ABC$) which contains $A$, and define two points $E'$, $F'$ similarly. Show that the lines $DD'$, $EE'$, $FF'$ are concurrent.

[i]Alternative formulation.[/i] Assume the incircle of a triangle $ABC$ touches $BC$, $CA$, $AB$ in $D$, $E$, $F$. Assume also that the external bisectors of $\angle A$, $\angle B$, $\angle C$ meet the circumcircle $(ABC)$ again in $D'$, $E'$, $F'$. Show that the lines $DD'$, $EE'$, $FF'$ concur.

Here is my solution (not the shortest one, but I had the first part already written down and thus it was easier to use it than to write a completely new solution):

[quote="grobber"]The incircle of a triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $D'$ be the midpoint of the arc $BC$ (of the circumcircle of $ABC$) which contains $A$, and define two points $E'$, $F'$ similarly. Show that the lines $DD'$, $EE'$, $FF'$ are concurrent.[/quote]

Call D", E", F" the midpoints of the arcs BC, CA, AB on the circumcircle of triangle ABC [i]not[/i] containing the points A, B, C. Then we will begin with proving that the triangles DEF and D"E"F" are homothetic.

If I is the incenter of triangle ABC, then the line EF is perpendicular to AI (since the points E and F are symmetric with respect to the line AI). Since the point E" is the midpoint of the arc CA on the circumcircle of triangle ABC, the chords CE" and AE" are equal; but equal chords in a circle have equal chordal angles, and thus we have < CBE" = < ABE". Thus, the point E" lies on the angle bisector of the angle ABC, i. e. on the line BI. Similarly, the point F" lies on the line CI. If the line E"F" meets the line AI at X, then

  < AXE" = 180° - < AE"X - < XAE"
       = 180° - < AE"F" - (< IAC + < CAE")
       = 180° - < ACF" - (< IAC + < CBE")         (cyclic)
       = 180° - < ACI - (< IAC + < CBI)
       = 180° - C/2 - (A/2 + B/2)
       = 180° - (A/2 + B/2 + C/2) = 180° - 90° = 90°,

and thus, the line E"F" is also perpendicular to the line AI. Hence, E"F" || EF. Similarly, F"D" || FD and D"E" || DE. Thus, the triangles DEF and D"E"F" are homothetic.

Note that this homothety yields that the lines DD", EE" and FF" concur; to prove this was the task of the geo problem 191 = Ukraine 2 in the op02 project. Now let's solve the problem Grobber posted:

Since the point D' is the midpoint of the arc BC containing the vertex A, while the point D" is the midpoint of the arc BC not containing the vertex A, the points D' and D" are diametrically opposite points on the circumcircle of triangle ABC. In other words, if O is the circumcenter of triangle ABC, then the point D' is the reflection of the point D" in the point O. Similarly, the point E' is the reflection of the point E" in the point O. Thus, the line D'E' is the reflection of the line D"E" in the point O. This implies D'E' || D"E". Together with D"E" || DE, it follows that D'E' || DE. Similarly, E'F' || EF and F'D' || FD. Thus, the triangles D'E'F' and DEF are homothetic, and consequently, the lines DD', EE' and FF' concur. Proof complete.

[quote="grobber"][i]Alternative formulation.[/i] Assume the incircle of a triangle $ABC$ touches $BC$, $CA$, $AB$ in $D$, $E$, $F$. Assume also that the external bisectors of $\angle A$, $\angle B$, $\angle C$ meet the circumcircle $(ABC)$ again in $D'$, $E'$, $F'$. Show that the lines $DD'$, $EE'$, $FF'$ concur.[/quote]

Just in order to show that this problem is indeed an alternative formulation of the first one:
Since the point D' lies on the external angle bisector of the angle A, we have < D'AC = 180° - < D'AB. But < D'AC is the acute or obtuse chordal angle of the chord D'C in the circumcircle of triangle ABC, and 180° - < D'AB is the acute or obtuse chordal angle of the chord D'B in the circumcircle of triangle ABC. If two chords have equal chordal angles, then their lengths are equal; thus, D'C = D'B, and it follows that the point D' is the midpoint of the arc BC on the circumcircle of triangle ABC which contains the point A. Similarly, the point E' is the midpoint of the arc CA which contains the point B, and the point F' is the midpoint of the arc AB which contains the point C. Thus, the problem is equivalent to the problem I solved above.

  Darij

Just a little bit different way.

If I' is the circumcenter of ABC, then  I'E' and I'F'  are the perpendicular bisectors of AC and AB respectively, therefore they are respectively parallel to IF and IE. from This we have that the two triangles isosceles I'F'E' and IFE are homothetic and II',EE', FF' are concurrent. Similarly one can prove the others concurrences.

A different solution:

Let $D''=OI\cap DD'$. The triangles $D''ID,D''OD'$ are similar, so $\frac{ID''}{D''O}=\frac{ID}{OD'}=\frac{r}{R}$. This means that $DD'$ passes through the internal center of similarity of the incircle and the circumcircle, and the same holds for $EE',FF'$, of course.

This is a very easy problem.

Because of $ID\parallel OD^{\prime}$, where $O,$ $I,$ the circumcenter and incenter of $ABC$ respectively, the line $DD^{\prime}$ intersects the segment $IO$ at a point, so be it $P,$ such that $\frac{OP}{PI}= \frac{R}{r},$ where $R,$ $r,$ are the radii of circumcircle and incircle, respectively. Similarly for the lines $EE^{\prime}$, $FF^{\prime}$.

Kostas Vittas.


$16$ tennis players are in a tournament.with how many ways can we arrange the first turn of the games?(it means each player,play exactly one game)

Can someone move this to Getting Started or some other "easy" forum?

i thinks it was better if you solved it and then talk about it?
(if you think it is easy you don'ht have to think about it) :D

Pick a player - there are 15 possible opponents.
We have 14 players now - pick one, and there will be 13 possible opponents.
We have 12 players now - pick one, and there will be 11 possible opponents, etc
So that gives 15 * 13 * 11 * ... * 3 * 1 unless I'm missing something.


In the non-decreasing sequence of odd integers $ \{A_1, A_2, A_3, ...\}\equal{}\{1,3,3,3,5,5,5,5,5,...\}$ each odd positive integer $ k$ appears $ k$ times. Find a general formula for $ A_n$, where $ n$ is a positive integer.

[hide="hint"] The formula for $ A_i$ where $ i$ is a positive perfect square is easier to find.  Generalize it to any positive integer.[/hide]

First off, the term $ 1\plus{}3\plus{}5\plus{}\cdot \plus{}2i\minus{}1$th term is the last term that is $ 2i\minus{}1$, and the $ 1\plus{}3\plus{}\cdots\plus{}2i\minus{}3 \plus{}1$th term is the first term that is $ 2i\minus{}1$.  Now $ \sum_{i\equal{}1}^k 2i\minus{}1 \equal{} k^2$.  So we know that all $ A_{(k\minus{}1)^2\plus{}1}, A_{(k\minus{}1)^2\plus{}2} \ldots A_{k^2} \equal{} 2k\minus{}1$.  So we have $ A_n \equal{} 2\lceil \sqrt{n} \rceil\minus{}1$


function
integration
limit
probability and stats

$ X_1,X_2,...$ are a sequence of random variables on $ (\Omega,\mathcal{F},P)$. Suppose there exists a function $ p(x)$ with
$ \int_{ \minus{} \infty}^\infty p(x)dx \equal{} 1$ and that satisfies $ \lim\limits_{n\to\infty}P\{X_n\in(x \minus{} \frac {1}{2},x \plus{} \frac {1}{2}]\} \equal{} p(x)$ for any $ x\in\mathbb{R}$. Let $ F_n$ denote the cumulative distribution function of $ X_n$. Find $ \lim\limits_{n\to\infty} F_n$.

if the $ X_i$ are identically distributed, $ P\{X_n\in(x\minus{}\frac{1}{2},x\plus{}\frac{1}{2}]\}$ does not depend on $ n$  :maybe:

[quote="alekk"]if the $ X_i$ are identically distributed, $ P\{X_n\in(x \minus{} \frac {1}{2},x \plus{} \frac {1}{2}]\}$ does not depend on $ n$  :maybe:[/quote]
oops :oops:
You have a sharp eyes.
$ X_1,X_2,...$ - only a "sequence of rv's".

PS: This forum is great.


MATHCOUNTS
geometry
perimeter
rhombus
percent
probability
analytic geometry

Every week, I'll post three problems on this thread. Only middle schoolers should posts actual solutions to the problems. The rules are simple.

The problems will start off very easy and progress as the year goes on.

Participation is voluntary. It's always good to be the first on answering though  :D

Solutions will come when the problems from the following week are posted (that is, if there's a huge confusion with the problems).

[size=200]Week 1[/size]:

1. What is the positive difference between $\frac{1}{5}$ and $\frac{5}{7}$? Express your answer as a common fraction.

2. What is the sum of the first 7 prime numbers?

3. How many numbers between 1 and 100 have all even digits?

I hope that i'm qualified even though i'm taking math at a high school (i'm still officially in 8th grade and MATHCOUNTS elligible).

[hide="1"]
we have a common demoninator of 35 and $\frac 57$ is larger so we have:

$\frac {25}{35} - \frac 7{35} = \boxed{\frac{18}{35}}$

[/hide]

[hide="2"]
$2+3+5+11+13+17 = \boxed{58}$

[/hide]

[hide="3"]
There are 5 choices for the tens digit (0, 2, 4, 6, or 8) and 5 choices for the ones digit (also 0, 2, 4, 6, or 8), so the total number is $5\cdot5 - 1(for 00) = \boxed{24}$

[/hide]

[quote="236factorial"]Every week, I'll post three problems on this thread. Only middle schoolers should posts actual solutions to the problems. The rules are simple.

The problems will start off very easy and progress as the year goes on.

Participation is voluntary. It's always good to be the first on answering though  :D

Solutions will come when the problems from the following week are posted.

[size=200]Week 1[/size]:

1. What is the positive difference between $\frac{1}{5}$ and $\frac{5}{7}$? Express your answer as a common fraction.

2. What is the sum of the first 7 prime numbers?

3. How many numbers between 1 and 100 have all even digits?[/quote]
 :idea:
[hide="answer1"]$\frac{5}{7}$ - $\frac{1}{5}$ = $\frac{25}{35}$ - $\frac{7}{35}$ = $\frac{18}{35}$[/hide]

[hide="answer2"]2+3+5+7+11+13+17=19+16*2+7=58[/hide]

[hide="answer3"]between 1,10  are 4. 20~30, 5, so on 40~50, 60~70, 80~90=
24 numbers. [/hide]

[quote="Hamster1800"]I hope that i'm qualified even though i'm taking math at a high school (i'm still officially in 8th grade and MATHCOUNTS elligible).
[/quote]

Well, since you are Mathcounts eligible, you can still answer questions.

[quote="math92"]do we need to post solutions or just answers?[/quote]

Solutions preferred (and i'm sure some will yell at you if you don't explain  :P ).

This isn't a competition, so have fun!

[quote="236factorial"][hide="solutions to week 1"]
1. Simply subtract the fractions to get $\frac{18}{35}$.

2. 2+3+5+7+11+13+17=58

3. There are 5 even numbers from 0-9. We want an even number for each digit, so there are 5 choices for each. 5*5=25 (one digit numbers are assumed like 02, 04, etc.).

However, 00 isn't a choice, so there are actually only 24 choices. [/hide]

[size=200]Week 2[/size]

1. Through what quadrants does the line $y=x+1$ go through?

2. If the diagonal of a square is $6\sqrt2$, what is the square's perimeter?

3. In which types of special quadrilaterals are the diagonals perpendicular to each other?[/quote]

[hide="1"]
Graph it to see that it goes through I, II, III (assuming I remember the quadrant numbers right)
[/hide]

[hide="2"]
Say each side of the square is x. Then, by the pythagorean theorem, the diagonal is $\sqrt{x^2 + x^2} = \sqrt{2x^2} = \sqrt{2}\sqrt{x^2} = x\sqrt{2}$. Since $x\sqrt{2} = 6\sqrt{2}$, $x = 6$. So, the perimeter is $\boxed{24}$

[/hide]

[hide="3"]
[b]kites[/b]. Please don't forget kites. Rhombi are kites and lozenges are rhombi so you don't need to mention them.

[/hide]

[url=http://mathworld.wolfram.com/Lozenge.html]Lozenge from mathworld[/url]

basically, it's a rhombus with angles measuring 45 degrees and 135 degrees

[hide="2"]
since the diagonal is $6\sqrt{2}$ the side must be 6. 6*4=24
[/hide]

[hide="3"]
um, i think diamonds could be the answer.[/hide]

[quote="236factorial"]Wait a second.... what are the rules for this topic now???

I have simply neglected this topic for two months (just one or two posts, no problems) because I thought no one cared about this anymore  :D[/quote]
Why would we turn down more problems? Right now I'm at summer camp for TIP :(

[quote="236factorial"]I believe MCrawford said that there shall be no marathons of any kind, and structure of this topic is starting to remind me of a marathon.  :wink:[/quote]
He didn't say that there shall not be any marathons, I think he was referring to the marathon in the basics, that it was too high level and that many marathons were becoming too high level for their forum.
At any rate, the Mathcounts marathon would've been locked if "there shall be no marathons of any kind", but it's running strong.

I posted 33 weeks, with three problems each. That's 99 problems. So here's

[color=blue][i][size=150]The final problem[/size][/i][/color] (multistep)

100. a) In trapezoid ABCD, BC=10, leg CD=5.5, angle ABC=50 degrees, angle ADC=100. Find AD-1.5, and let this value be S.

b) If $M=(1+\sqrt{S})^{2002}-2(1+\sqrt{S})^{2001}-2(1+\sqrt{S})^{2000}$, find M.

c) What is $M^{3}+63M+12$? Let this value be T.

d) Let $R=\sqrt{4+\sqrt{T}}+\sqrt{4-\sqrt{T}}$. Find $R^{2}$.

e) The temperature shown on my thermometer on my window (in the sun) is $(R^{2}+4)(R^{2}-4)-4$. What does my thermometer read?

A calculator is not necessary.

[quote="236factorial"]I posted 33 weeks, with three problems each. That's 99 problems. So here's

[color=blue][i][size=150]The final problem[/size][/i][/color] (multistep)

100. a) In trapezoid ABCD, BC=10, CD=5.5, angle ABC=50 degrees, angle ADC=100. Find AD-1.5, and let this value be S.

b) If $M=(1+\sqrt{S})^{2002}-2(1+\sqrt{S})^{2001}-2(1+\sqrt{S})^{2000}$, find M.

c) What is $M^{3}+63M+12$? Let this value be T.

d) Let $R=\sqrt{4+\sqrt{T}}+\sqrt{4-\sqrt{T}}$. Find $R^{2}$.

e) The temperature shown on my thermometer on my window (in the sun) is $(R^{2}+4)(R^{2}-4)-4$. What does my thermometer read?

A calculator is not necessary.[/quote]

[hide]a) $AD=\boxed{4.5}$   $S=\boxed{3}$

b) $M=(1+\sqrt{3})^{2002}-2(1+\sqrt{3})^{2001}-2(1+\sqrt{3})^{2000}=(1+\sqrt{3})^{2000}((1+\sqrt{3})^{2}-2(1+\sqrt{3})-(1+\sqrt{3}))=(1+\sqrt{3})^{2000}(0)=\boxed{0}$

c) $T=\boxed{12}$

d) $R=\sqrt{4+\sqrt{12}}+\sqrt{4-\sqrt{12}}$
$R^{2}=4+\sqrt{12}+4-\sqrt{12}+2\sqrt{4^{2}-12}=\boxed{12}$

e) $(12+4)(12-4)-4=144-16-4=\boxed{\boxed{\boxed{124}}}$
[/hide]

Is the topic locked now?

While MCrawford's comments [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98833](see this)[/url] pertained specifically to the Classroom Math Forum, his reasoning applies equally well to this Forum and the marathons herein.  [Compare the creep in difficulty from this thread's "Week 1" to the "Final Problem", for example.]

I'm not going to ban marathons at this time.  But I will happily lock this thread on 236factorial's request.  I would also encourage future posts be more to individual threads, and less to "marathon" threads.

I thought that only applied to marathons in the classroom forum.

[quote="SplashD"][quote="lotrgreengrapes7926"]Could you please post the solution to A? I must be missing something.  :fool:[/quote]

I actually used trig for #1 and didn't want to post it in a MATHCOUNTS forum. I couldn't find a better way. Could you?[/quote]
And it's not very basic trig either. :huh:

[quote="SplashD"]basic sine, cosine, and tangent actually, nothing fancy.[/quote]
Really? This is how I did it: [hide] Draw altitudes from $D$ and $A$ to $BC$, intersecting at $X$ and $Y$, respectively. $CX=5.5\sin10^\circ$. $DX=5.5\cos10^\circ$ so $AY=5.5\cos10^\circ$, therefore, $BY=5.5\cos10^\circ\cot50^\circ$.

So, $AD=10-5.5\sin10^\circ-5.5\cos10^\circ\cot50^\circ=10-5.5(\sin10^\circ+\frac{\cos10^\circ\cos50^\circ}{\sin50^\circ})=10-5.5(\frac{\sin10^\circ\sin50^\circ+\cos10^\circ\cos50^\circ}{\sin50^\circ})$.

Using the cosine subtraction identity, $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$, we get $10-5.5(\frac{\cos(50-10)^\circ}{\sin50^\circ})=10-5.5(\frac{\cos40^\circ}{\sin50^\circ})$.

Then, from the identity $\cos(90-\theta)^\circ=\sin\theta^\circ$, we get $10-5.5(\frac{\sin40^\circ}{\sin40^\circ})=10-5.5=\boxed{4.5}$. [/hide]

Wow! That's looks MUCH more complicated than what I did.

[hide="No Trigonometry allowed"]The "final problem" has crept up in difficulty.  Some of you seem to think the problem requires trigonometry.  (Of course, trig isn't required, or 236factorial wouldn't have posted the problem here.)

The above trigonometry discussion illustrates why MCrawford banned marathons from the Classroom Forum.  Trigonometry is forbidden in this forum!  As a moderator, I have few good choices on what to do.  I can leave the trig (in violation of this forum's rules) or I can delete the trig (we mods like to avoid deleting posts) or I can split the trig into another post (without any context of what question was being answered).[/hide]

Part A of the final problem is slightly underspecified, as there are two possible trapezoids that meet the conditions of the problem.  Choosing $AD\parallel BC$ eliminates the ambiguity.  Attached is a drawing, with one extra line segment that provides a big hint.

The first problem was supposed to have "leg CD=5.5" instead of "CD=5.5". I have changed that. rcv's picture is right, and the hint is very good  :D


hey, ok...i'm having trouble with my type two gr 8 math equations, and i tried so many tmes but nevere get them right./..... what wuould you do for this...

>>   4/5x=60

>>   x/7-8=10

If by what you wrote, you mean:

$\frac{4}{5} x = 60$
$\frac{x}{7} - 8 = 10$

Then, you are posting on the wrong forum.  You should post on beginner topics or less.

For solving these types of problems you have a few goals:

1) getting the variable on one side of the equation (in the first problem you don't have to bother)
2) getting the variable to be in terms of $1x$ by division or multiplication by an appropriate number

First question:

Step 1 is done.

You can divide by $\frac{4}{5}$ on both sides of the equation to ascertain the value of $x$.
Thus, $x = \frac{60}{\frac{4}{5}} = \frac{60}{\frac{4}{5}} \frac{\frac{5}{4}}{\frac{5}{4}} = 60 \cdot \frac{5}{4} = 15 \cdot 5 = 75$

Note: dividing by $\frac{a}{b}$ is equivalent to multiplying by $\frac{b}{a}$

Second question:

Step 1: $\frac{x}{7} - 8 + 8 = 10 + 8 \rightarrow \frac{x}{7} = 18$

Step 2: Multiply by $7$ or divide by $\frac{1}{7}$

$x = 18 \cdot 7 = 126$


Let $a_n,a_{n+1}$, which are two terms of the sequence $a_1,a_2, \cdots\cdots a_n, \cdots\cdots$, be the roots of the quadratic equation $x^2+3nx+C_n=0$.

If $a_1=1$, find $\sum_{n=1}^{2p} C_n$

[quote="Singular"]$a_n + a_{n+1} = 3n$ :(  [/quote]

That is a typo.  All the other stuff is correct, however.


inequalities
trigonometry
inequalities proposed

[quote="Lovasz"]Prove the inequality for some positive reals:
\[{a \over b}+{b \over c}+{c \over a}+{3(ab+bc+ca) \over a^{2}+b^{2}+c^{2}}\geq 6. \]
[/quote]

I think it is not correct. Check for $a=2$, $b=2$, $c=3$. The inequality does not hold because the constant term for $\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}$ is too large.

Sorry. That seems my ugly solution had a mistake.

[quote="weiquan"][quote="Lovasz"]Prove the inequality for some positive reals:
\[{a \over b}+{b \over c}+{c \over a}+{3(ab+bc+ca) \over a^{2}+b^{2}+c^{2}}\geq 6. \]
[/quote]

I think it is not correct.[/quote]

It is correct. Read more carefully. It was asked to prove for some positive reals. Now the inequality is correct if $a=b=c=1$.

[quote="puuhikki"][quote="The soul of rock"]Don't be silly![/quote]
I'm not. I'm exact.[/quote]

"[i]Some positive reals[/i]" in [b]Mathematics[/b] always means [i]any[/i] given positive reals...

My friend told me the best constant for the inequality
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k\cdot\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\ge 3+k \]
is $k=2\sqrt{2}$. But i think it is not true, see $a=1.48,b=1.1,c=1$. Please check it for me, thanks!!!!! :P

Can you post your solution? It seem to be very hard


calculus
number theory proposed
number theory

Find all natural numbers with the property that, when the first digit is moved to the end, the resulting number is $\dfrac{7}{2}$ times the original one.

I don't know if it's true but...:

Let $n$ the number.

So, we can write $n=A+10^kb$ where $b$ is a digit, $k+1$ the number of digit of $n$ and $A$ a number.

So, we have : $\frac{7}{2} (A+10^kb)=10A+b$

And finally: $13A=b(7*10^k-2)$

So, $13|7*10^k-2$ because $b$ is a digit.

After calculus, $k=-1 (mod 6)$

So, you can find $n$ when $k=-1(mod6)$ and you take :

$A=\frac{b*(7*10^k-2)}{13}$ with $b$ such that $A$ has $k$ digit.

All these problems emanate, more or less, from the classical $\dfrac {1} {7} = 0.(142857)$. The least such number is $153846$, the period of $\dfrac {2} {13}$.

$\frac{7}{2}(10^na+b)=10b+a$ with $(a,b,n)\in \mathbb{N}_0^3$, $a\in\{1,2,\ldots,9\}$, $b\in\{1,2,\ldots,10^n-1$ implies $b=\frac{7\cdot 10^n-2}{13}a$ implies $n=6k-1$ for some $k\in \mathbb{N}_0$. If $a\ge 2$ then $10^{6k-1}-1\ge b = \frac{7\cdot 10^n-2}{13}a \ge$ $\frac{14\cdot 10^{6k-1}-4}{13}$, that is false: then $a=1$ (and all contraints are satisfied). All solutions are of the form $(a,b)=(1,\frac{7\cdot 10^{6k-1}-2}{13})$. []

Here's another solution.

Let the desired number be N and let the digits of the desired number be $a_1a_2a_3 \ldots a_n$.
Form a new number $x = 0.\overline{a_1a_2a_3 \ldots a_n}$ as a repeating decimal formed from the digits of N.

Then by the conditions of the problem, $\frac{7}{2}x = 0.\overline{a_2a_3a_4 \ldots a_1}$
But this number is also the same as $10x - a_1$
Therefore, $\frac{7}{2}x = 10x - a_1 \implies x = \frac{2a_1}{13}$

So the repeating part of any of the fractions $\frac{2k}{13}$ whose first digit is less than 3 will satisfy the conditions of the desired number (notice that we need the first digit to be less than 3 so that when it is multiplies by 7/2 it will still have the same number of digits).

The only fraction with this property is $\frac{2}{13} = \overline{.153846}$, so possible solutions are N = 153846, N = 153846153846, etc.

[quote=Labiliau

After calculus, $k=-1 (mod 6)$

So, you can find $n$ when $k=-1(mod6)$.[/quote]
How do you get this?


geometry
geometric transformation
reflection
calculus
derivative
function
inequalities

The minimal value of $ f(x) \equal{} \sqrt{a^2 \plus{} x^2} \plus{} \sqrt{(x\minus{}b)^2 \plus{} c^2}$ is

A. $ a\plus{}b\plus{}c$
B. $ \sqrt{a^2 \plus{} (b \plus{} c)^2}$
C. $ \sqrt{b^2 \plus{} (a\plus{}c)^2}$
D. $ \sqrt{(a\plus{}b)^2 \plus{} c^2}$
E. None of these

[hide="Im not sure but .."]
Consider three points $ P\ (x,0),\ A\ (0,a),\ B\ (b,c)$. Then $ f(x) \equal{} PA \plus{} PB$.

If $ a, b, c > 0$, then we wish to minimize the length of two segments joined at a common endpoint. Using the well-known geometric technique, reflect $ B$ across the x-axis to $ B'\ (b, \minus{} c)$; then $ PB \equal{} PB'$ and $ PA \plus{} PB'$ is minimized when $ P$ lies on $ AB'$. Thus $ f(x) \equal{} AB' \equal{} \sqrt {b^2 \plus{} (a \plus{} c)^2} \Rightarrow \mathrm{(C)}$.

Unfortunately this argument doesn't work when one of $ a, c < 0$, because then $ P$ can lie on $ AB$ and the answer would be $ \sqrt {b^2 \plus{} (c \minus{} a)^2}$.

So, the answer is both C and E.
[/hide]

[hide="Does this work?"]
Not that it solves the problem, but it gets it down to two options:
Consider the case where $ b \equal{} 0$. Then the minimum value is at $ x \equal{} 0$, so $ f(x) \equal{} a \plus{} c$.
This only leaves options A and E open.
Then if you consider the case where a and c are 0, the minimum value is b... so option A seems to fit (But it could still be E).

You could also try taking the derivative of the function, setting it equal to 0, solving for x, and plugging it back into the function.  I got $ \frac { \minus{} a^2b \pm abc}{c^2 \minus{} a^2}$ for x, but that might be wrong and I'm too lazy to plug it back in and simplify. It'll probably get messy.
[/hide]

[hide="Solution"]Letting $ b\equal{}0$, we see that the answer must be one of A, C, and E.

Note that negating $ a$ does not change $ f(x)$, yet it changes the expressions in A and E, hence neither is the answer. Hence, the answer is [b]E. None of the above[/b].[/hide]

[quote="t0rajir0u"]The problem only really make sense if we assume $ a, b, c$ are non-negative.  azjps is correct.[/quote]
But it still can't need it! How you can state like this? Minxcopxki inequality doesn't need the restriction of a,b,c. If a,b,c is non-negative, there's nothing to say more! :wink:

[quote="azjps"][quote="imMAW"][hide="Does this work?"]
You could also try taking the derivative of the function, setting it equal to 0, solving for x, and plugging it back into the function.  I got $ \frac { - a^2b \pm abc}{c^2 - a^2}$
[/hide][/quote]

Indeed this simplifies to $ \frac {ab(c - a)}{(c + a)(c - a)} = \frac {ab}{a + c}$. (It is $ \pm$ because if one of $ a,c < 0$ we want the negative root, and through some algebra we get the other solution I noted above).

[quote="ghjk"]This problem is kind of weird because there isn't restriction of a,b,c!This is called:" Minxcopski's inequality"(maybe I misspell it) and we just need to consider whether the equality happens or not to find out C or E is the correct answer!
The equality happens when a(b-x)=cx, which means x=ab/(a+c).Now if a=-c then we can't find the value of x! Therefore, I think E is correct![/quote]

[hide]
Again, as t0rajir0u stated we assume $ a,c > 0$ so $ a + c \neq 0$. Thus
\begin{align*} f\left(\frac {ab}{a + c}\right) & = \sqrt {a^2 + \left(\frac {ab}{a + c}\right)^2} + \sqrt {\left(\frac {ab}{a + c} - \frac {b(a + c)}{a + c}\right)^2 + c^2} \\
& = \frac {\sqrt {a^2(a + c)^2 + a^2b^2}}{a + c} + \frac {\sqrt {( - bc)^2 + c^2(a + c)^2}}{a + c} \\
& = \frac {a\sqrt {b^2 + (a + c)^2} + c\sqrt {b^2 + (a + c)^2}}{a + c} \\
\text{min}(f(x)) & = \boxed{\sqrt {b^2 + (a + c)^2}} \end{align*}
Which agrees with the geometrical interpretation solution.
[/hide][/quote]
What do you meant when rephrasing your solution? I pointed out the ambiguous restriction of this problem! This is exactly what I will do if a,c>0! :)

Oh okay, I misunderstood your comment.  :oops:

I think we can agree though that the answer is
[hide]
\begin{align*}
\text{min}\ (f(x)) =\left\{
\begin{array}{lr}
\sqrt{b^2 + (a+c)^2} & ,\ ac \ge 0 \\
\sqrt{b^2 + (a-c)^2} & ,\ ac < 0
\end{array}
\right.
\end{align*}
[/hide]

Given such, I would just assume that the test-makers meant $ ac \ge 0$ on a contest. That's what I don't like about having a "none of the above" option.

The problem is indeed weird. I didn't make this so don't attack me but the answers they gave is both C and E. What azjps had about Pythagorean Theorem was correct. I'll use that as the official solution on the Resources section.


The number $121_b$, written in the integral base $b$, is the square of an integer for
(A) $b=10$, only
(B) $b=10$ and $b=5$, only
(C) $2\leq b\leq 10$
(D) $b>2$
(E) no value of $b$

[hide="solution"]
$121_b = b^2 + 2b + 1 = (b+1)^2$

So $b + 1$ must be an integer.  That is always true because $b$ is an integer.  However, the number has 2 for one of its digits, meaning the base must be greater than 2.

Answer: (D) $b > 2$[/hide]

[hide]You can write $121_b$ as $b^2+2b+1 = (b+1)^2$.  So the answer is d.[/hide]

[hide="my answer"]It's D because 121b = (b+1)^2.  Therefore, b must be greater than 2.[/hide]

Sorry h_s_potter2002, I didn't see your post.


Here's a variation of a problem I posted in the marathon. It's a target problem.
$8^{x}-1=x^{4}$
If the product of all possible values of x is y and the sum is z, then find $yz$ as a decimal to the nearest tenth.

[quote="SplashD"][hide]That's easy, the answer is [hide="."]$\boxed{0}$[/hide][/hide][/quote]
correct.

[quote="SplashD"]This is nothing like the problem you posted on the marathon, it's actually much easier.[/quote]
[hide]Only if you see that 0 is a possibility. Not everyone would[/hide]

[hide]On any problem like this, you should always check 0 and 1 to see if they work.  $8^{0}=1-1=0$ and $0^{4}=0$, so 0 is a solution.  If 0 is a solution, then 0 is the product of all solutions because it doesn't matter what the other solutions are:  it's still 0.  And, the sum doesn't matter because of the same thing, making the answer, incredibly, $\boxed{0}$[/hide]

I actually spent the first few seconds looking for an integer solution when I saw the 1 and the answer hit me.

[quote="b-flat"]Umm, 0 is an integer...[/quote]

I mean an integer like 16 or something.

[quote="b-flat"]Yes, I know what you mean  :D

When I first saw the problem, I was like wow, that will take some time, but then, a few seconds later, I realized that 0 as a solution would be really really nice.  And what do you know??  0 is a solution  :o

BTW, does anyone know any of the other solution?  I stopped looking as soon as I realized 0 was a solution.  :)[/quote]

What's the point, you got the answer.

Using a graphing calculator, there are no other solutions besides 0, since $8^{x}$ grows far too quickly.
But, I wonder if there are any complex solutions. :D

[quote="nebula42"][quote="b-flat"]I would be curious to know what other types of solutions there are to a problem like that, but that would probably be way above mathcounts.[/quote]
The only general "solution" I know would be to graph the equation on a graphing calculator and find the zeroes of the function  :P . Logarithms are a good way to find general solutions to exponential equations, but they don't work in cases like this one because terms like $8^{x}$ and $x^{4}$ in the same equation stop you from being able to use them.[/quote]
I just added as a decimal to the nearest tenth so the new answer would be 0.0 as well as serving the purpose of throwing off people who found decimal answers and wonder why doesn't it say as a decimal.

Wow, you are right  :lol:

$8^{x}\equiv0\pmod{8}$, unless $x=0$ and $-1\equiv7\pmod{8}$

4th powers
$0^{4}\equiv0\pmod{8}$
$1^{4}\equiv1\pmod{8}$
$2^{4}\equiv0\pmod{8}$
$3^{4}\equiv1\pmod{8}$
$4^{4}\equiv0\pmod{8}$
$5^{4}\equiv1\pmod{8}$
$6^{4}\equiv0\pmod{8}$
$7^{4}\equiv1\pmod{8}$

This clearly eliminates every integer, except 0.

[quote="solafidefarms"]There are no other solutions save 0; this can be proven using modular arithmetic. Mod 8 works quite well for 4th powers; this problem is trivial in mod 8.[/quote]
This is indeed a very good method  :lol: , but unfortunately, this only shows that there are no [i]integer[/i] solutions, which means we won't be done yet unless the problem asks for integer solutions only.

[hide]I see the answer is 0, since a solution is 0 and hence the answer has to be zero since y is 0 and the product of y and z must be  0.[/hide]


Find the smallest positive integer,$x$, such that all of the below fractions are written in their lowest form.

$\frac{x+9}{8}, \frac{x+10}{9},........ \frac{x+49}{48}$.

[hide="Solution"]Rewriting those as $1+\frac{x+1}{8},1+\frac{x+1}{9},\dots,1+\frac{x+1}{48}$, we see that $x+1$ mustn't have any of the prime factors of the numbers $8,9,\dots,48$. The smallest such number is the first next prime greater than $48$, namely $53$, which gives $x=52$.[/hide]

Extension:

Find the smallest positive integer,$x$, such that all of the below fractions are integers.

$\frac{x+9}{8}, \frac{x+10}{9},........ \frac{x+49}{48}$.

[quote="ritchjp"]When you put the series of periods after the second fraction are you implying that there is a series of fractions, or are we dealing with just the three fractions that are shown?[/quote]He may have put too many periods (3 periods suffice) but whenever you see 3 periods it usually means "and so on". So in this problem, we are dealing with 41 fractions in total.

[quote="srulikbd"]series. gcd(8...48)-1[/quote]

I got lcm(8,9...48)-1

[quote="srulikbd"]series. gcd(8...48)-1[/quote]

This would be equal to $0$, since, for example, the sequence contains both $11$ and $13$  ;)

[quote="SplashD"]I got lcm(8,9...48)-1[/quote]

Better :)

Fix the hide tag in your signature, it hid the quote button.

[quote="D.P.L"]why minus one?[/quote]

it's for the extension, because in the extension you gotta make them integers


let m \geq 2 be an integer.find the smallest integer n \geq m such that for any partition of the set
{m,m+1,m+2,.....,n}  into 2 subsets,atleast one of the subsets contains three numbers a,b,c (not necessarily distinct) such that a^b= c
help
 regards

the solution is n=m^(m^m+2)
I think if you this answer the proof is reachable

was that awesome??
dude,u throw such a thunderbolt and then u tell me that the proof is reachable. :D i couldn't have reached this answer(not in this life time)
don't know how people create such beatiful mathematical problems.

well it's not fully original. I have seen something very similar with ab=c and also the answer is similar. I really belive it was adapted, but the fact is that the problem is very nice :)

at least try to prove that you can partitionate the set with that mm^m+2 -1 such that no two elements can be found with ab = c.

I can't understand. Consider two sets $\{m,m+1,...,m^m-1\}$ and $\{m^m,m^m+1,...,\left(m^m\right)^{m^m}-1\}$.
It is impossible to satisfy equlity $a^b=c$ if $a$, $b$ and $c$ lie in the same set. And $n=m^{m^{m+1}}-1>m^{m^m+2}-1$.
What is wrong?
Maybe you meant $n=m^{m^{m+2}}$?

No one wants to answer me... :(
Ok. Let's prove estimate for $n$.

Suppose $n\geq m^{m^{m+2}}$.
We can write following chain
$m\in A$ $\Longrightarrow$ $m^m\in B$ $\Longrightarrow$ $\left(m^m\right)^{m^m}=m^{m^{m+1}}\in A$ $\Longrightarrow$ $m^{m+1}\in B$ and $\left(m^{m^{m+1}}\right)^m=m^{m^{m+2}}\in B$.
So $m^m$, $m^{m+1}$ and $m^{m^{m+2}}$ lie in $B$ -- contradiction.

And last step. Required partition for $n=m^{m^{m+2}}-1$.
Let
\[A=\{m,m+1,m+2,...,m^m-1,m^{m^{m+1}},m^{m^{m+1}}+1,...,m^{m^{m+2}}-1\}\]
and
\[B=\{m^m,m^m+1,m^m+2,...,m^{m^{m+1}}-1\}.\]

Actually, problem is not that hard.
I can estimate it as ARO round 3 or round 4 for 9-th grade (ARO has 5 rounds).

Hi Myth,

Sorry I cannot answer to you as quickly as possible on all the threads :)
Indeed I ment $m^{m^{m+2}}$ not $m^{m^m+2}$ :)


Let a,b,c be positive real numbers.prove that:
$ {a^2}\plus{}{b^2}\plus{}{c^2}\leq {\sqrt{b^2\minus{}bc\plus{}c^2}}{\sqrt{c^2\minus{}ca\plus{}a^2}}\plus{}{\sqrt{c^2\minus{}ca\plus{}a^2}}{\sqrt{a^2\minus{}ab\plus{}b^2}}\plus{}{\sqrt{a^2\minus{}ab\plus{}b^2}}{\sqrt{b^2\minus{}bc\plus{}c^2}}$

[quote="xiaohui"]Let a,b,c be positive real numbers.prove that:
$ {a^2} \plus{} {b^2} \plus{} {c^2}\leq {\sqrt {b^2 \minus{} bc \plus{} c^2}}{\sqrt {c^2 \minus{} ca \plus{} a^2}} \plus{} {\sqrt {c^2 \minus{} ca \plus{} a^2}}{\sqrt {a^2 \minus{} ab \plus{} b^2}} \plus{} {\sqrt {a^2 \minus{} ab \plus{} b^2}}{\sqrt {b^2 \minus{} bc \plus{} c^2}}$[/quote]
Yes, of cause!
$ \sum_{cyc}\sqrt{(a^2\minus{}ab\plus{}b^2)(a^2\minus{}ac\plus{}c^2)}\equal{}$
$ \equal{}\sum_{cyc}\sqrt{\left(\left(a\minus{}\frac{b}{2}\right)^2\plus{}\frac{3b^2}{4}\right)\left(\left(a\minus{}\frac{c}{2}\right)^2\plus{}\frac{3c^2}{4}\right)}\geq$
$ \geq\sum_{cyc}\left(|\left(a\minus{}\frac{b}{2}\right)\left(a\minus{}\frac{c}{2}\right)|\plus{}\frac{3bc}{4}\right)\geq$
$ \geq\sum_{cyc}\left(\left(a\minus{}\frac{b}{2}\right)\left(a\minus{}\frac{c}{2}\right)\plus{}\frac{3bc}{4}\right)\equal{}a^2\plus{}b^2\plus{}c^2.$

It is old and easy.....
Give me your e-mail ,  I can sent you many inequality problems and their solutions... :rotfl:


Is there any way or trick to search mathlinks  for posts containing certain expressions written in $\LaTeX$  ?
Form of ASCII search can't do much very often.

I think if you search the functions it works fine (i.e. frac, sqrt, and such), but it's difficult to do things like specific $\LaTeX$ code because it filters out $\{$ and $\}$ characters afaik.

Hi,
I tried.Sometimes it works ,sometimes not.
Do you suggest to type in expressions as they are written in LaTeX code (on this forum) and the search will return results?

Just put the $ signs in. It works for me that way.  :)


inequalities
calculus
integration
Cauchy Inequality
inequalities proposed

Find the best constant k such that for any n and any nonnegative real numbers $x_1,...x_n$ we have $ (x_1+2x_2...+nx_n)(x_1^2+...+x_n^2)\geq k(x_1+...+x_n)^3$.

$ k = \frac{2}{3\pi} $. I dont know is there the elementary proof for this result, but there is a classical result of B.Nagy:

Let $ 1 < s \le 2, 1 \le t \le 2, q = 1 + \frac{s}{t(s-1)}, a_1, ..., a_n > 0 $ then

$(\sum_{k=1}^n a_k)^q + |(\sum_{k=1}^n (-1)^ka_k)|^q \le \frac{q\pi}{2}(\sum_{k=1}^n a_k^s)^{1/t(s-1)}(\sum_{k=1}^n n^ta_k^t)^{1/t}$

Harazi's question is for s = 2, t = 1, q = 3.

Note that Carlson inequality (posted by Harazi some days ago) follows from this inequality. Unfortunately, the proof is not elementary.

Namdung: can you show this non-elementary solution ?

I think I've got a proof. Hope there is no mistake.
Set $x_1=n-1,x_2=n-2,...,x_n=0$. By taking n->infinity,k->4/9.
Now we want to prove that
$
(x_1^2+...+x_n^2)(x_1+...+nx_n)\geq\frac{4}{9}(x_1+...+x_n)^3
$
which is equivalent to
$
(3x_1^2+...+3x_n^2)(6x_1+...+6nx_n)\geq(2x_1+...+2x_n)^3
$
We may assume that $3x_1^2+...+3x_n^2=6x_1+...+6nx_n$. Now we just have to show that
$
(\frac{3x_1^2+...+3x_n^2+6x_1+...+6nx_n}{2})^2\geq(2x_1+...+2x_n)^3
$
which is equivalen to
$
(\frac{3(x_1+1)^2+3(x_2+2)^2+...+3(x_n+n)^2-3(1+2^2+...+n^2)}{2})^2
$
$
\geq(2x_1+...+2x_n)^3
$
Now note that the left side of the inequality becomes smallest when $x_1+1=...=x_n+n$.
Again we may assume that $a_1+...+a_n\geq 1+...+(n-1)=\frac{n(n-1)}{2}$. Because if not, we can have $x_n=0$ (this makes the left side smaller) and then n is reduced. Now set $k=\frac{(x_1+...+x_n)-\frac{n(n-1)}{2}}{n}$. Now it is sufficient to prove that
$
(\frac{3n(n+k)^2-\frac{n(n+1)(2n+1)}{2}}{2})^2\geq(n(n-1)+2nk)^3
$
this is equivalent to
$
(\frac{(2n^2-\frac{3}{2}n-\frac{1}{2})n+6n^2k+3nk^2}{2})^2\geq(n(n-1)+2nk)^3
$
which is equivalent to
$
((n+\frac{1}{4})(n-1)n+3n^2k+\frac{3}{2}nk^2)^2\geq(n(n-1)+2nk)^3
$
This one can be easily verified when $k\geq 0$.

So you state that Nagy's result is far from accuracy.

Here is a very nice solution ( Darksheer's method is also very very beautiful!):
  Put $A=\sum_{i=1,n}{a_i^2}, B=\sum_{i=1,n}{ia_i}$ and then take $ u^3=12A^{-1}B^2$ and f(u) the integer part of u. Apply the Cauchy inequality in the form:
  $\sum_{i}{a_i}\leq\sqrt {\sum_{i\leq f(u)}{(1-\frac{i}{y})^2} A}+\frac{B}{y}$. Then use again an integral estimation and you are done. Cauchy roules!

This method of using Cauchy and then an integral estimation is new,for me, and very interesting to apply.

[quote="harazi"]I also think that Namdung's result is false. I obtained I think $\frac{4}{9}$. The solution is based again on Cauchy. I will post it this week-end (hope I will find a place at the Internet club)/[/quote]

I will check again in Polya-Hardy-Littlewood "Inequalities".

Namdung

Dear All,

Harazi's solution is nice and correct. I'd checked, Nagy result is true, but the constant is not the best (except the case s=t=2 - Carlson inequality). So, congratulation to creator and solver of this very nice inequality.


function
geometry
geometric transformation
rotation
analytic geometry
calculus

We form a circle of radius $R$ out of a straight wire. We put a small ring of mass $m$ on it. The coefficient of friction between the ring and the wire is $\mu$. What initial velocity should we give the ring so that it goes round the circle exactly once, and stops just where it was in the beginning?

isn't frictional force F= u*m*g the force that deccelerates constantly the ring?
in this case V=2* sqrt(u*pi*r*g) .what do i need m for?

Is the circle vertical or horizontal? I think it should be vertical, because otherwise it won't stop (its velocity will decay exponentially)...

could you explain me please why exponentially?

You have that the friction force is $F = \mu N$, where $N$ is the reaction force of the circle. If the circle is horizontal, $N = mv^{2}/R$; solving the resulting differential equation you get an exponential decreasing function $v(x)$.

Yes you are right Djole, the circle is horizontal. here is a tip: you dont need the $\dot{\varphi}(t)$ function you only need the $\dot{\varphi}(\varphi)$ function. $\varphi$ is the angle of rotation by the radius drawn to the ring.

Well, yes, I got that $\dot\varphi (\varphi)$ is exponentially decaying...

It's not just as simple like that. yes there are exponential stuff in the solution, but not only $e^{-...}$.

Really? Now I'm confused, I got $\dot\varphi = \dot\varphi_{0}e^{-\mu\varphi}$. Could you post your solution?

the reaction force by the wire is $\vec{T}$. $\varphi$ is the angle of rotation of the radius drawn to the ring. The equations of motion in cyllindrical coordinates:
$mR\ddot{\varphi}=-\mu |\vec{T}|$,
$T_{r}=mR{\dot{\varphi}}^{2}$ and
$T_{z}=mg$
so from these using: $|\vec{T}|=\sqrt{T_{r}^{2}+T_{z}^{2}}$
We get:
$R\ddot{\varphi}=-\mu \sqrt{g^{2}+R^{2}{\dot{\varphi}}^{4}}$
now we dont need the $\dot{\varphi}(t)$ function , but only the $\dot{\varphi}(\varphi)$ function.
lets use the following substitution: $u=\dot{\varphi}$.
so:
$\ddot{\varphi}=\dot{u}=\frac{du}{dt}=\frac{du}{d\varphi}\frac{d\varphi}{dt}=\frac{du}{d\varphi}u=\frac12 \frac{du^{2}}{d\varphi}$
now we get:
$R\frac{1}{2}\frac{du^{2}}{d\varphi}=-\mu \sqrt{g^{2}+R^{2}u^{4}}$
lets use another substitution:
$\xi=u^{2}$
the equation now:
$\frac{d\xi}{\sqrt{g^{2}+R^{2}{\xi}^{2}}}=-\frac{2\mu}{R}d\varphi$
integrating both sides from $0$ to $2\pi$:
$arsinh \frac{R}{g}\xi(2\pi)-arsinh\frac{R}{g}\xi(0)=-4\pi \mu$
we want the ring to stop after one round  : $\xi(2\pi)=0$
So the wanted initial velocity is:
$v(0)=R\dot{\varphi}(0)=R\sqrt{\xi(0)}=\sqrt{Rg \sinh{4\pi \mu}}$

Ah, but there is a gravitational force! Great -- I never would've thought of that! :D

yep gravitational force is there too. :D

Hey,

What kind of physics background do you guys have? I've never seen this sort of stuff in calculus-based books like Tipler. Is this Olympiad-level?

No, it's not in Tipler.  I used Tipler as well, and when the prelims came around, I don't think I was as prepared the open-response as I should of been.  Tipler is much easier.


combinatorics solved
combinatorics
graph theory

There are n girls and n boys at the party. Participants who belong to the same sex do not know each other. Moreover, there cannot be found two girls who know the same two boys. At most how many acquaintances can be among the participants of the party ?

a.) n = 5
b.) n = 7

I proved estimation S \leq n(1+\sqrt (4n-3))/2, where S is a number of acquaintances.

If n=5 then S \leq 12.807... ==> S \leq 12. It is easy to construct corresponding example.

If n=7 then S \leq 21.
Example. Let us to enumerate all boys and girls with numbers 0,1,2,...,n-1. Then girl k must know boys k, k+1 and k+3 - overall 7*3=21 acquaintance.

Estimation for S.
Let xk=(xk0,xk1,...,xk(n-1)), where xkm=1 if girl k knows boy m and xkm=0 otherwise.
It is clear S=\sum xkm.
Condition " there cannot be found two girls who know the same two boys" has equivalent (xk,xt) \leq 1 for k<>t.
So (x0+x1+...+x(n-1))2 \leq \sum xkm2 + n(n-1) = (trick!) \sum xkm + n(n-1) = S + n(n-1), but
x0+x1+...+x(n-1) = ((x00+x10+...),(x10+x11+...),...) ==>
(x0+x1+...+x(n-1))2 = \sum (x0m+x1m+...)2 \geq (\sum xkm)2/n=S2/n.

So S2-nS-n2(n-1) \leq 0 !!!!!

I got the same results, but I examined a few cases to get them (cases like: the maximum degree of a vertex in the corresponding graph is 3,4,5 etc.) so it's impossible to apply my soln to the general case. I won't post it here because Myth already posted a much nicer one.vSo what abt the general case with n boys and n girls?

I think my estimation is asymptotically valid.


Maybe this is pretty simple for you ppl, and maybe i need no more than a little help, but:

$\vec c=s\cdot \vec a+t\cdot \vec b$

Where:

$\vec a=(1;2)$ $\vec b=(3;4)$ $\vec c=(-1;5)$

System of equations maybe? Using the x and the y components.

I am probably wrong though, I don't like vectors.

[quote="hydro"]I think you have just to solve
$-1=s+3t$
$5=2s+4t$[/quote]

$2=-2s-6t$
$5= 2s+4t$
$7=-2t$
$t=-\frac{7}{2}$

[quote="Sporally"]System of equations?

I tried solving this and ended up with $-s-t=-6$. But it doesn't work, right.... This is because we have two unknown values. You can easily solve this problem if we were working with one dimension only, but now that we have two it isn't that simple. I am wrong?[/quote]

What do you mean? you learned to solve system of equations in algebra 1!

Dow. My bad, could be a combination of the time of the day and the fact that i haven't done any work with what you call 'system of equations' lately. I made the two constants even instead of even out one of the $x$s and $y$s. Guess i got a little confused by the fact that they were called $s$ and $t$ :blush: Thx for your help on this one though.


induction
number theory unsolved
number theory

Prove that $ \forall n \in \mathbb{N}$
$ 0 < \sum_{k\equal{}1}^{n} \frac{g(k)}{k} \minus{} \frac{2n}{3} < \frac{2}{3}$ where $ g(k)$ denotes the greatest odd divisior of $ k$

[hide="hint"]use induction and the fact that $ \frac{g(2k)}{2k}\equal{}\frac{g(k)}{2k}$.[/hide]
see also [url=http://www.mathlinks.ro/viewtopic.php?t=150430]here[/url].


Mosalase ghaemolzavieye $ABC$ mafrooz ast.($\angle A=90$)
ertefa $AF$ ra mikeshim ta dayere mohitie $ABC$ ra dar $D$ ghat konad.az $A$ bar $DC$ amood karde ta an ra dar $H$ ghat konad.vasate $AH$ ra $M$ minamim.parekhate $CM$ ra emtedad midahim ta dayere mohiti $ABC$ ra dar $Z$ ghat konad.az $A$ bar dayere mohiti $ABC$ momasi rasm karde ta $BC$ ra dar $E$ ghat konad.sabet konid dayere mohit $AZE$ dar $E$ bar $BC$ momas ast. :D

fekr konam in masalaro baham hall kardim.

montaha to ba angle chasing hallesh kardi va man ba $INVERSION$.

in masale male short list e $98$ ast. ;)

badan link e morede nazar ro ezafeh mikonam.

[quote="mohamadhosein"]manam ba enekas halesh kardam. :P[/quote]

barikallah, vaghean khoshhalam kardi... :D


[hide][quote="Interval"]A room is 24 feet long, 18 feet wide, and 9 feet high. How many square yards of wallpaper are needed to paper the four walls of the room?[/quote]

8 yards by 6 yards by 3 yard
so then find 8*3+8*3+6*3+6*3=24+24+18+18=42+42=84[/hide]

I did not realize the question is so easy.

Thanks!


We have positive integer and its prime divisor p.Both p and m are sums of two perfect squares.Prove,that m/p is the sum of two perfect squeares too.

A number is expressable as the sum of two squares if and only if it is of the form Na where N is a square and a is the product of numbers of the form 4k+1and possibly 2. Since m is expressable as the sum of two squares, it must be of the form Na where N is a square and a is the product of numbers of the form 4k+1and 2. Since p is prime and expressable as the sum of two squares it is of the form 4k+1. Dividing we get Nb where b=a/p is the product of numbers of the form 4k+1and possibly 2. Then Nb is expressable as the sum of two squares as well.

More general manner: http://www.mathlinks.ro/Forum/viewtopic.php?t=66248
But see http://www.mathlinks.ro/Forum/viewtopic.php?t=87376 ...


algebra
polynomial
superior algebra
superior algebra unsolved

In trying to find all irreducible factors of a polynomial, say $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}w^{2}\in\mathbb{C}\left[x,y,z,w\right]$, is there a fairly quick method?  Or does it usually just come down to solving

\[ \left(ax\plus{}by\plus{}cz\plus{}dw\right)\left(ex\plus{}fy\plus{}gz\plus{}hw\right)\equal{}x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}w^{2}\]
for $ a,b,c,d,e,f,g,h\in\mathbb{C}$?

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2$ has nontrivial factors iff the variety $ x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2 \equal{} 0$ can be written as the union of two varieties.  Can you describe the graph of $ x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2 \equal{} 0$?

Another way is to use Eisenstein for the prime element $ y^2\plus{}z^2\plus{}w^2$ (it itself is prime by Eisenstein for $ z^2\plus{}w^2$).


pigeonhole principle
combinatorics unsolved
combinatorics

This has undoubtably been asked before (or something similar), but I can't find it...

A man has 42 cucumbers. Each day he eats at least 1. On the 28th day, he finishes them all. Show that there is a set of consecutive days such that the total number eaten is exactly 15.

let $ a_i$ be the number of cucumbers eaten on the $ ith$ day.

consider the sums

$ a_1$
$ a_1 \plus{} a_2$
$ a_1 \plus{} a_2 \plus{} a_3$
.
.
.

Suppose none of them is 15.

These sums are all distinct elements from {1,2,3...42}/{15}

Consider the following partition of this set:

{1,16,31},{2,17,32}.....{12,27,42},{13,28},{14,29},{30}

Now we have to pick 28 elements from here. If 2 of them are both in {13,28}, or both in {14,29} then we have the result established. If 3 of them are in the same class, the result is also established. Apply pigeonhole principle to show one of these cases must occur.

let $ a_k$ equal the number of cucumbers the man eats in the first $ k$ days. then
\[ 1\leq a_1 < a_2 < \cdots < a_{28}\leq 42\]

\[ 16\leq a_1 \plus{} 15 < a_2 \plus{} 15 < \cdots < a_{28} \plus{} 15\leq 57\]
if any of the $ a_k$ is equal to 15, we're done. so assume no $ a_k$ equals 15. then each of the 56 numbers $ a_1, ..., a_{28}, a_1 \plus{} 15, ..., a_{28} \plus{} 15$ has one of the 56 values in the set $ \{1,...,57\} \minus{} \{15\}$; call this set $ S$. if any two of the numbers are equal, we have $ a_i \equal{} a_j \plus{} 15$ for some $ i, j$, and we're done.

otherwise, every one of the values in $ S$ occurs exactly once. in that case, we must have $ a_k \equal{} k$ for $ k \equal{} 1, 2, ..., 14$. then $ a_k \plus{} 15 \equal{} k \plus{} 15$ for $ k \equal{} 1, ..., 14$. so far, 1, ..., 14, 16, ..., 29 are covered, the first 14 #'s by $ a_i$'s, the latter 14 #'s NOT by $ a_i$'s. thus we must have $ a_{15}\geq 30$. but then, since $ a_i > a_j$ for $ i > j$, it follows that $ a_{28}\geq 30 \plus{} 13 \equal{} 43$, a contradiction. this completes the proof.


Two boys, walking together, were 3/8 of the way across a railroad bridge when they saw a train approaching at 60 km/hr. They ran at the same speed towards opposite ends of the bridge, each arriving at an end of the bridge just as the train arrived at that end. How fast did the boys run, in km/hr?

[hide]Since the train arrives at the beginning of the bridge when the boys run 3/8 of it, if they keep running towards the end and get 3/4 through the bridge, then by that time the train would have reached the opposite end. So the time it takes for the boys to run 1/8 of the bridge is equal to the time it takes for the train to go across it. The boys have a speed of 60/8 = 7.5 km/hr.[/hide]

Well, once the first boy reaches the beginning of the bridge, the other one is $\frac{3}{4}$ of the way across, so he is four times slower than the train, which means that the boy's speed is 15 km/h.

Oh duh! I'm so stupid! I said that the boy has 1/8 left to go when its really 1/4... sorry! its 15.

[hide="Simple Answer"]
When the train had reached the bridge the boys had run $\frac{3}{8}b$ (b is the length of the bridge) and the last boy had $\frac{1}{4}b$ to go.  the train crosses the bridge in $\frac{b}{60}$ hours and the boy runs the remainder of the bridge in $\frac{b}{4r}$ (r is the boys rate of travel. Since they arrived at the same time $\frac{b}{4r}=\frac{b}{60}$ Therefore, the boys were running 15 km/h.
[/hide]

Hasn't an almost identical problem already been published here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30025


another ineq:
$ \frac{\sum a^3}{4abc}\plus{}\frac{1}{4}\geq {(\frac{\sum a^2}{\sum ab})}^2$

[quote="TRAN THAI HUNG"]Let a,b,c>0 Prove that:
$ \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{abc}} \plus{} 9\frac {{ab \plus{} bc \plus{} ac}}{{a^2 \plus{} b^2 \plus{} c^2 }} \le 12$
I know it's easy but It is nice. :wink:  :)[/quote]
It is instantly solved by SOS.. $ S_a \equal{} S_b \equal{} S_c \equal{} \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \minus{} 9abc}{2abc(a^2 \plus{} b^2 \plus{} c^2)} \ge 0$ (Btw. It should be $ \ge$)

[quote="TRAN THAI HUNG"]Let a,b,c>0 Prove that:
$ \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{abc}} \plus{} 9\frac {{ab \plus{} bc \plus{} ac}}{{a^2 \plus{} b^2 \plus{} c^2 }} \ge 12$
I know it's easy but It is nice. :wink:  :)[/quote]
Thank you, my dear friend for the nice ineq.
And by Am-Gm, we have the similar result:
With $ a, b, c > 0$ and $ k \geq\ 1$:
$ \sqrt[k]{\frac{a^3\plus{}b^3\plus{}c^3}{abc}} \plus{} 3. \sqrt[k]{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}} \geq\ 3 \plus{} \sqrt[k]{3}$
Because of your ineq and: $ (a^3\plus{}b^3\plus{}c^3)(ab\plus{}bc\plus{}ca) \geq\ 3abc(a^2\plus{}b^2\plus{}c^2)$
 :)

[quote="tuandokim"][quote="TRAN THAI HUNG"]Let a,b,c>0 Prove that:
$ \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{abc}} \plus{} 9\frac {{ab \plus{} bc \plus{} ac}}{{a^2 \plus{} b^2 \plus{} c^2 }} \le 12$
I know it's easy but It is nice. :wink:  :)[/quote]
It's must be like this:
$ \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{abc}} \plus{} 9\frac {{ab \plus{} bc \plus{} ac}}{{a^2 \plus{} b^2 \plus{} c^2 }} \geq 12$
$ LHS \minus{} RHS \equal{} (\sum a^2 \minus{} \sum ab)(\sum \frac {1}{ab} \minus{} \frac {9}{\sum ab})\geq 0$[/quote]
Yeah, you are right that's my mistake (typo):) .


algebra
polynomial
Rational Root Theorem

Hello
Resolve in IR:
$ \sqrt{x^2\minus{}4}\minus{}\sqrt{x\plus{}2}\equal{}x\minus{}2$

[quote="sami.d"]Hello
Resolve in IR:
$ \sqrt {x^2 \minus{} 4} \minus{} \sqrt {x \plus{} 2} \equal{} x \minus{} 2$[/quote]

[hide="solution"]We square:

$ x^2\plus{}x\minus{}2\minus{}2(x\plus{}2)\sqrt{x\minus{}2} \equal{} x\minus{}2$

$ x^2\equal{}2(x\plus{}2)\sqrt{x\minus{}2}$

We square again and move to one side:

$ x^4\minus{}4x^3\minus{}8x^2\plus{}16x\plus{}32\equal{}0$

From the [[Rational Root Theorem]], the possible rational roots are $ \pm 32$, $ \pm 16$, $ \pm 8$, $ \pm 4$, $ \pm 2$, $ \pm 1$. Those all are not roots from plugging them in.
[/hide]

The only way to solve that quartic that I know of is to bash with the formula for the roots: http://planetmath.org/encyclopedia/QuarticFormula.html

I leave finding the roots to you.

[quote="1=2"]$ x^2 \plus{} x \minus{} 2 \minus{} 2(x \plus{} 2)\sqrt {x \minus{} 2} \equal{} x \minus{} 2$[/quote]

You forgot to square the other side.  The resulting polynomial is actually cubic.  In any case, neither the cubic nor the quartic formula can be taken seriously as solutions.


geometry
geometric transformation
reflection

1) When an object absorbs or reflects ligth, it experiences a small force. Consider: a supernova occurs where the star becomes many times brighter than normal for a short period of time. ( a few weeks). Light from the star causes the deflection of a delicately balanced mirror. Does the action reaction law apply?

2) In a tug of war where the two teams exert forces equally in opposite directions, how can either team win?

1) Yes, because the light exerts a force on the mirror, so the mirror exerts and equal and opposite force on the beam of light.  In this case, it is easier to think of light as photons, since it is known that photons possess momentum, so since they reverse direction upon hitting the mirror, the mirror did indeed exert a force on the photons.

2) The tension in the rope is equal to the force with which it is being pulled.  The rope then exerts that same force on each team, pulling both inwards.  Then I guess the team that exerts less of a frictional force against the ground gets pulled in and across the line, so they lose.


please help me!   a,b  and c are integer numbers and   aX^2+bX+c  have 2 different  roots in (0,1)   prove that a>5 [/code][/list][/url][/b]

I see you are new. Welcome!

This forum is for questions about the website. The place to post mathematical questions is in one of the [url=http://www.artofproblemsolving.com/Forum/index.php?f=577]Mathematics Fora[/url], based on the level of the problem. Also, try $ \text{\LaTeX}$. For a start, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=700733#700733]here[/url].


Does anyone know of any good books on control theory, optimization, and calculus of variations?  I know close to nothing about it and it doesn't seem to be a course offered at the undergrad level.

Get the program "FlashGet" free on the internet.
Download the addon FlashGot for Mozilla

Go on google and type "Calculus of variations filetype: pdf" and put to show 100 results.

Open the first 5 pages and download everything.

after all you will have more than 500 books. so delete those who have less than 300 kb. and see the others one-by-one.

Good Luck.


ARML AMC USA(J)MO USAMO Princeton college calculus

I always thought it was really cool when you figured out where some problems came from especially when they were really from higher math.  I didn't realize other people might be interested too until I heard that a grader from MOSP was collecting such things.  Two examples come to mind which were both ARML power rounds: last years ARML power round was straight up p-adic stuff, and a while back there was one on finite affine and projective geometry.
I was wondering if people could post random things like this they have noticed.  This year's USAMO 5 also seemed a bit p-adic'y.

the IMO problem we did in class (n^p - p) was Chebotarev density theorem

the mock IMO #6 was (obviously) group theory, I guess...

I wanted to put a basic ergodic theory question (just about equidistribution of the fractional part of $\{n\alpha\}$, $\alpha$ an irrational) on the Princeton Math Competition calculus test, but my fellow test-writers wouldn't let me...

I would have wanted that on there.  That's too bad.  Arnav: does anybody actually know what that stupid theorem is?  I am wondering if USAMO 3 this year could have been solved by high powered techniques.  I mean anything with infinite sequences and number theory sounds like it could be solved by high powered stuff.


IMO Shortlist
number theory unsolved
number theory

Prove or disprove:
There exist infinitely many pairs of positive integers $ x,y,z,w$ $ (x,y,z,w)\equal{}1$
and $ x^4\plus{}y^4\plus{}z^4\equal{}w^4$
(I am wondering if it is a famous problem.I created it when I was solving an IMO shortlist problem.)

Just for reference,

"...Euler conjectured a fourth power cannot be written as the sum of 3 smaller fourth powers, but 200 years later this was disproven with:

$ 95800^4 \plus{} 217519^4 \plus{} 414560^4 \equal{} 422481^4$..."
from wikipedia [url]http://en.wikipedia.org/wiki/Fourth_power[/url]

Oh thank you!
Can this equation be solved:$ x,y,z,w$are positive integers $ (x,y,z,w)\equal{}1$ and $ x^4\plus{}4y^4\plus{}4z^4\equal{}w^4$
(I knew $ x^4\plus{}y^4\plus{}4z^4\equal{}w^4$ has infinitely many solutions,which has been solved in PEN)

[quote="aadil"]rewrite x^4 as a^2.y^4 as b^2  z^4 as c^2 and w^4 as d^2.now unless a^2 + b^2=k^2 and k^2 + c^2=d^2 the equation is not true[/quote]
Sorry I didn't really catch you.Could you explain more?


[quote="Erken"]Let $ \omega_{1}(O_{1},R_{1})$ and $ \omega_{2}(O_{2},R_{2})$ be to circles,such that $ R_{1}\equal{} R_{2}\equal{} 1,\omega_{1}\cap\omega_{2}\equal{} X,Y$.Suppose $ XY \equal{} 1$.Let $ C\in\omega_{1}$ and $ A,B\in\omega_{2}$,such that $ CA,CB$ are tangent lines to $ \omega_{2}$.If $ CB\cap\omega_{2}\equal{} A'$ then find $ AA'$?[/quote]

Did you mean $ CB\cap\omega_{1}\equal{} A'$?

Beacuse if CB is a tangent to $ \omega_{2}$ and $ B\in\omega_{2}$, then $ CB\cap\omega_{2}\equal{} B$.

Yes,you are right. :blush: But i can't edit my problem anymore.$ CB\cap\omega_{1}\equal{}A'.$ :wink:


Function f(x) is not defined for 0. For all non-zero real numbers x, $f(x) + 2(f(1/x))=3x$
The equation f(x)=-f(x) is sanisfied by what numbers?

[quote="PenguinIntegral"]Function f(x) is not defined for 0. For all non-zero real numbers x, $f(x) + 2(f(1/2))=3x$
The equation f(x)=-f(x) is sanisfied by what numbers?[/quote]

Okay I think this works:

If we plug in $x = 1/2$, then we have $3f(1/2) = 3/2 \Longleftrightarrow f(1/2) = 1/2$

Therefore, $f(x) + 1 = 3x \Longleftrightarrow f(x) = 3x - 1$

So by saying $f(x) = -f(x)$, we see that $f(x) = 0$ is the only way to satisfy the equation.  So $3x - 1 = 0 \Longleftrightarrow x = \frac{1}{3}.$

$f(x)=3x-2f(\frac1{2})$
$-f(x)=2f(\frac1{2})-3x$

So,

$\begin{eqnarray*}f(x)=-f(x)&\Rightarrow &3x-2f(\frac1{2})=2f(\frac1{2})-3x\\ &\Rightarrow &6x=4f(\frac1{2})\\ &\Rightarrow &x=\frac{2}{3}\cdot f(\frac1{2})$

$f(\frac1{2})+2f(\frac1{2})=\frac3{2}\Rightarrow f(\frac1{2})=\frac1{2}$

And the result is $\boxed{x=\frac1{3}}$

$f(\frac{1}{2})+2f(\frac{1}{2})=3$, so $f(\frac{1}{2})=\frac{1}{2}$ and $f(x)=3x-1$, so why is it not defined for $0$?  :?

Anyway, the answer has to be $\frac{1}{3}$ because the only way $f(x)=-f(x)$, is if both of them equal $0$, but I don't know why it's not defined for $0$ if it's linear ;)

answer to frt's
[hide]
$f(x) + 2f(\frac 1 x) = 3x$
Plug in $\frac1x$ for $x$
$f(\frac1x) + 2f(x) = \frac3x$
Solving the system, $f(x) = \frac2x - x$
Setting equal to zero, we get $x = \pm \sqrt 2$
[/hide]

[quote="frt"]If this was meant to be the exact same problem as 1981 AHSME #17, then he made a typo. The correct equation is $f(x)+2 f\left(\frac{1}{x}\right) =3x$.[/quote]

Ahah, now I get it...
[hide]
Allright, we can substitute $\frac{1}{x}$ for $x$, so we have $f(\frac{1}{x})+2f(x)=\frac{3}{x}$, so after we solve for $f(x)$ from both the two equations, we have $f(x)=\frac{2-x^2}{x}$, so $f(x)=-f(x)=\frac{x^2-2}{-x}=\frac{2-x^2}{x}$....hmmm....I'm getting all real numbers for this.... :? so I'm thinking the problem should read $f(x)=f(-x)$ ;) ,
From this, we have $\frac{2-x^2}{x}=\frac{2+x^2}{-x}$, and $x=+-\sqrt{2}$...[/hide]

P.S.: PenguinIntegral, you have to change two things about this problem ;) (f(x)=f(-x)) and f(1/x)


Edited.
Pictures of others without permission are not allowed.

I have permission! :spam:  :spam:  :spam:


Hey all, mega Latex newbie here,

I'm writing a summary of some research I did this summer in school, and I need to do what is done in this paper:

http://arxiv.org/PS_cache/astro-ph/pdf/0201/0201534v1.pdf

especially page 3 - how do you get a nicely-centered table or figures (or multiple figures) on a page when the page is formatted in two columns?

Thanks,

Adam

Just use the starred version of  the table or figure environments:
\begin{table*} ... \end{table*}
or
\begin{figure*} ... \end{figure*}

Indeed, that at least centers the table, but it doesn't put it on the page I want it on.

Also, the last page of my document has both columns of the page filled partially with text, instead of the left column being filled up completely before moving to the right column. How do I fix this?

The float can only appear at the top of a page (as in the document you quoted) so you just need to arrange that accordingly.
If you want to balance the columns at the end of the document you will either have to do that manually with \pagebreak or use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=multicol]multicol package[/url].

Thanks, I got it where I want it.

Now how would I create that little "Note" under the table, and how would I put the title at the top? the \caption command always puts the title at the bottom!

EDIT: err, never mind, I'm just not paying attention. I found another tutorial online that deals with this stuff, so I'll take a look there before posting more questions.

Thanks for your help.

Okay, so the tutorial wasn't very helpful. I still don't know how to put a caption on the table like it's done above.

I'm not sure what the problem is since you can put the caption and note wherever you please: note
[code]\begin{table*}
\caption{This is a table}
\begin{tabular}{cc}
...
\end{tabular}\\
This is a note
\end{table*}[/code]


let$A$ and $B$ be two matrixs  same rank
 let$B^{t}$ be transposed matrix  of $B$   and such that : $B^{t}A =0$
 prove that : $r(A)+r(B) =r(A+B)$ .
  where $r(A)$  is rank of matrix $A$

Suppose $u$ is in the range of $A$ and $v$ is in the range of $B.$ Then $u=Ax$ and $v=By.$ Now compute the inner product:

$\langle u,v\rangle=v^{T}u=(By)^{T}(Ax)=y^{T}B^{T}Ax=0$ since $B^{T}A=0.$

Hence the range of $A$ and the range of $B$ are orthogonal, which in particular means that their intersection is $\{0\}.$ Thus

${{\dim(\text{range}(A+B))=\dim(\text{range}(A)+\text{range}(B}))= \dim(\text{range}(A))+\dim(\text{range}(B}))$

$r(A+B)=r(A)+r(B).$

Could u explain more clearly [b][size=150]Im(A+B)=Im(A)+Im(B)[/size][/b]
[quote]Sorry for my poor knowledge on linear algebra[/quote]


AMC
AIME
ARML
USA(J)MO
USAMO
MATHCOUNTS
AMC 10

Well, if you're studying for math competitions in general (like MC, AMC, AIME, Arml, etc...), aren't you preparing for state mathematics league competitions, as well as any math competition, at the same time?

I don't think you should study [b]for[/b] a competition, but gain some insight and tips/tricks along the way. Ususally a competition is to determine how good you are now, not how good you could be if you study a lot.

My mom views that a competition is where you learn stuff, and that's I think thats right.
And then... well maybe for math league... you dont' need to study... but for MC, you have to like practice countdown stuff... and that kinda stuff...

you dont need to worry bout mathleague martin. mathleague prep comes with the other prep.

youre right bout the mathcounts part though

nah, youll still do well

but if you want to roxor...

[quote="Chinaboy"]Hmmm... true... .but what is the purpose of the math league anyways??? Is it just so that there's a competition for determining who's the best in state???
It doesn't lead to anywhere unlike AMC's  to AIME to USAMO and then above...[/quote]

The purpose of doing math competitions isn't to go on to another competition, or for prizes, etc.  It's for the joy of doing math!

For that matter, IMO doesn't lead to anything (other than, in theory, to college and your career), but people participate in it.

[quote="Tare"]Then I am toast...I'm concentrating on my AoPS 1 rather than MC stuff (which I don't have anyways)[/quote]
Actually, Mass is ridiculously competitive. Don't count on making nats, because three of the MA team members were 7th graders, and I know at least 3 people who were on my teeam who are capable of making nats this year

OK...did the three 7th graders skip grades?

[quote="Chinaboy"]Do you think the State Mathematics Leagues are worth the time studying for?[/quote]

For some people, the state mathematics leagues (we have at least two here in Minnesota at the pre-Math Counts level) are the first experience in "competition culture" in math. For a learner who starts early enough, the state competition is fun, not too stressful, and it's a good background to take into more serious competitions (AMC, Mathcounts, etc.) a few years later.

Anything worth doing is worth doing well, and in my opinion (as an adult of a generation that missed out on "competition culture" in my part of the United States), math competitions are worth doing. All of them, any of them.

i dont see why not to count on making nats. if you practice/learn hard enough, you will catch up to those who are better than you now. also, does it matter if the 7th graders skipped grades? you can be in any math class at your school if you practice at home. also, i dont think you guys should put absolutely everything you have into mathcounts. there are a lot of contests afterwards too.

[quote="hello"]I was just wondering whether anyone here participates in the atlantic pacific math league?[/quote]

I took them. Unfortunately, my school sent the final round results in late, so our results didn't count.

i think you should actually work hard for math counts though, since it'll give you a solid foundation for high school contests.  If I did MC, then it would have saved me a year of getting adjusted to math contests.

Mmm, perhaps so. For me even though AMC 10 was my first math competition, I didn't panic or anything...I guess it depends on the definition of "adjustment" since I don't quite know what that refers to...it might mean stuff like "how to manage time and score the most; Strategies, strategies: Get a lot of easy ones or go for the challenge?" or "how to stay organized so you can double-check faster"

Oh, and if former MathletesSM can post their strategies for MATHCOUNTS in the MATHCOUNTS forum, that'll really be helpful.


Just suppose, for the sake of argument, that you have a large block of stone. While staring at this block of stone, it dawns on you that perhaps you should carve it into something. Luckily, you have a hammer and chisel handy.

But it's no ordinary hammer and chisel. Somehow, no matter how hard you hit the chisel, exactly one-half of one percent of the block chips off.

How many times do you have to hit the chisel before the block is less than half of its original size?

Can Someone plese help me with a solution, I just can't get it.

This is in the GS forum aswell I dont think you should have double posted.

Well Mcrawford said that the problem was to difficult and didn't belong in the GS forum

well suppose $V(0)$ is the volume of the block in the beginning and $V(n)$ is the volume after $n$ hittings.
then $100\%-0.5\% = \dfrac{4999}{5000}$. so after each hitting the volume $V(n)=\dfrac{4999}{5000} \cdot V(n-1)$

altogether then we get that for all natrual $n$: $V(n) = V(0) \cdot \left(\dfrac{4999}{5000}\right)^n$
now we want to know after how many times the volume is less than half of the starting volume: $\frac12 V(0) \ge V(n) = V(0) \cdot \left(\dfrac{4999}{5000}\right)^n$
this implies that $n\ge \dfrac{\log \frac12}{\log \left(\dfrac{4999}{5000}\right)}$, so just take the smallest number which is greater than that ...

ps: this sounds a whole lot like a school assignment ... and the purpose of this forum isnt quite to help on homework. it is a problem solving community ;)

every time you hit the block, you remove $\dfrac{1}{200}$ of the block, and that leaves $\dfrac{199}{200}$ of the block each time you hit it
so we have $1*\left(\dfrac{199}{200}\right)*\left(\dfrac{199}{200}\right)*...\left(\dfrac{199}{200}\right)* < \dfrac{1}{2}$ where $\left(\dfrac{199}{200}\right)$ is the amount after hitting the block once, and condensing this, we have

$\left(\dfrac{199}{200}\right)^n<\dfrac{1}{2}$

$\log{\left(\dfrac{199}{200}\right)^n}>\log{\dfrac{1}{2}}$

$n>\dfrac{\log1-\log2}{\log{199}-\log{200}}$

$n>138.28...$ since n must be an integer, we have $\boxed{n=139}$

Actually, I found it on Neopets math problem solving game and I couldn't find the solution.

Thanks for the help.

[quote="Altheman"]every time you hit the block, you remove $\dfrac{1}{200}$ of the block[/quote]

ahh, fractions ... i never was good at that ;) $\dfrac{1}{200}$ is right of course  :blush:


The value of $ [2 \minus{} 3(2 \minus{} 3)^{\minus{}1}]^{\minus{}1}$ is:

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ \minus{}5\qquad
\textbf{(C)}\ \frac{1}{5}\qquad
\textbf{(D)}\ \minus{}\frac{1}{5}\qquad
\textbf{(E)}\ \frac{5}{3}$

$ \frac{1}{2\minus{}3(\minus{}1)}\equal{}\boxed{\frac{1}{5}}$.

Answer: C


geometry
circumcircle
trigonometry
geometry proposed

Let BC be a fixed chord of (O) and A varies in the bigger arc BC.
  Let M be the midpoint of BC and D be the point in BC such that AD is the bisector of  BAC.
  The circumcircle of ADM intersects AB, AC and (O) at N, P and Q.
   Let  D_b and D_c be the projections of D in OB and OC.
      a) Prove that ND_b = PD_c.
      b) Denote I and J the intersections of DD_b and QB, DD_c and QC.
 Find A such that BI.CJ is maximum.

No point in leaving this unsolved… :)

[b]a)[/b] Let $S=DD_b\cap AB,T=DD_c\cap AC$. It’s easy to show (by sine law applied to $DBS$ and $DCT$ together with some angle chasing, for example) that $BS=CT$. If we show that $N,P$ are the midpoints of $BS,CT$ respectively, we’ll be done, because $ND_b=\frac{BS}2=\frac{CT}2=PD_c$.

It suffices to show that $N$ is the midpoint of $BS$, so let’s do that.

The triangles $ABC,DBS$ are similar, so what we want to show is that $\angle BAM=\angle BDN$. This is equivalent to $\angle BND=\angle AMB$, which follows from the fact that $ANDM$ is cyclic, so $\angle BND=\pi-\angle AND=\angle AMB$.

[b]b)[/b] Since $\frac{BI}{BD}=\frac{CJ}{CD}$, and they are constant (don’t depend on the position of $A$), all we need to do is find the position of $D$ for which $BD\cdot CD$ is maximized, and since $BD+CD$ is constant, this happens when $D$ is the midpoint of $CD$, i.e. $A$ is the midpoint of the arc $BC$ on which it moves.

P.S. I forgot to mention: I used the obvious fact that $Q$ is the midpoint of the arc $BC$ on which $A$ moves.


Here are some awesome goals:

[url]http://video.google.com/videoplay?docid=-5675272436036532702&q=marek+malik&hl=en[/url]

[url]http://video.google.com/videoplay?docid=2636503729289931176&q=goal&hl=en[/url]

[url]http://video.google.com/videoplay?docid=-442116740003613893&q=hockey+goal&hl=en[/url]

First one is pretty good, third one is great, but the second one is sure to knock you out!

the second one might have been pure luck.

but the third one was sick!!!!


function
algebra proposed
algebra
Functional Equations
Germany
TST
Team Selection Test

Find all functions $f: \Bbb{R}_{0}^{+}\rightarrow \Bbb{R}_{0}^{+}$ with the following properties:

(a) We have $f\left( xf\left( y\right) \right) \cdot f\left( y\right) =f\left( x+y\right)$ for all $x$ and $y$.
(b) We have $f\left(2\right) = 0$.
(c) For every $x$ with $0 < x < 2$, the value $f\left(x\right)$ doesn't equal $0$.

[b]NOTE.[/b] We denote by $\Bbb{R}_{0}^{+}$ the set of all non-negative real numbers.

Could you be more precise? What is the difference with IMO 1986?

Pierre.

Setting $ x = 2: f(x+2) = f(xf(2))f(2) = 0$ implying $f(x) = 0$ for all $x \geq 2$.  For $y < 2$, $f(y)f((2-y)f(y)) = f(2) = 0$.  Therefore, condition c) implies that $(2-y)f(y) \geq 2$ or $f(y) \geq \frac 2{2-y}$.
If $f(z) > \frac 2{2-z}$, we can find some $2 > w > z$ such that $f(z) = \frac 2{2-w}$.  This is possible because $f$ is increasing on $(0,2)$.  However, then $f(2-w+z)=f((2-w)f(z))=f((2-w) \frac 2{2-w}) = f(2) = 0$ implying $2 = 2-w+w > 2-w+z \geq 2$, a contradiction.  Therefore, $f(z)= \frac 2{2-z}$ for all $z<2$.
The function defined as above is the unique function which can satisfy the given equation.  We easily verify that it does.

[quote="pbornsztein"]Could you be more precise? What is the difference with IMO 1986?[/quote]

Well, the difference is minimal: in our problem, (c) was assumed to hold for every 0 < x < 2, while in the IMO 1986 problem, this was assumed for 0 <= x < 2. But this is, of course, not a real difference!

@ Michael Lipnowski: yes, I've solved the problem in the same way...

  Darij

$f(0)=1$
claim 1: $f(x)=0$ if $ x \geq 2$
claim 2: $ 0 <f(x) \leq 1$ if $ 0<x<2 $
claim 3: $ f(x)=1 iff x=0 $
so we have $ 0< f(x) < 1 $ with $ 0<x<2$
we try to prove f is injective on $ (0,2) $
assume that we have $0<a<b<2$
take $ y=a $ and $ y=b$ we have
$ f(x+a)= f(x+b) $
take $ x= 2 - b $ we have contradiction
so f is injective on $(0,2)$
let $ x= x/{f(y)} $ and f is injective on $(0,2)$  $(1)$
we have
 $ f(x).f(y) = f( x/{f(y)} + y) = f(y/{f(x)} +x)$  $(2)$
Notice that $f(z) > 0 $ iff $ x<2$  $(3)$

Let $P(x,y)$ denote the assertion $f(xf(y))f(y)=f(x+y)~\forall x,y \geq 0.$
$P(x,2)\implies f(x)=0 ~\forall x\geq 2,$ which works.
$P(0,0)\implies f(0)=1$ or $f(0)=0,$ the latter implies $f\equiv 0$, false.
For $x\in (0,2),$ we have $f(x)\geq 2/(2-x),$ by $P(2-x,x).$

Assume $f(x)>2/(2-x)$.
$\exists u,v\in (0,2): f(u)>2/(2-u)$ and $f(v)=2/(2-u).$
Case 1. $v<u\implies f(v)=2/(2-v)\implies u=v,$ contradiction.
Case 2. $v>u.$ But then, $P(2-v,u)\implies u\geq v,$ contradiction.
Thus the solution for $x\in [0,2)$ is $f(x)=2/(2-x),$ which works.


Given ten 0's and ten 1's. How many ways 0-1 binary sequences can be formed such that no three or more than three 0's are together ? (For example, 01001001010011101011 is such a sequence, but not 01001000101001110111)

We can think of a permissable sequence like

01001001010011101011

as instead

12212001100.

So we are looking for 11 digit base 3 numbers with the sum of digits adding to 10.  This simplifies things considerably.

A strategy of:

If there are 5 twos, then ....
If there are 4 twos, then...

will finish the count, but I am lazy.


prove that if n divides 2^(n)+1 then n=3 or n divides 9

It seems like you could do it like this:

Suppose toward a contradiction that  $n\neq 3$ and n does not divide 9 but n divides $2^{n}+1$.
Applying division algorithm to 9 and n, we know that there are eight cases to check for instance
$9 = nk+1$
$9 = nk+2$
.....
But some of the cases are very easy to check so you only have a few cases. I mean you can know what values n will take under each situation and then plug your result for n into the statement n divides $2^{n}+1$, this statement will not be statisfied unless n is 3 or a multiple of 3.


solve this system
a^2-1=c and b^3+1=c and a;b;c are integers

[quote="motaedzi"]Solve the system in $\mathbb{Z}$:
\[ a^2-1=c  \]
\[ b^3+1=c  \][/quote]

This should be moved to Number Theory section.

Does rewriting as $a^2+b^3=2c$ help at all?

It is equavalent to solve elliptic equation:
$a^2=b^3+2$
It had only these solutions $(\pm 1,-1)$ in integers.

[quote="motaedzi"]Solve the system in $\mathbb{Z}$:
\[ a^2-1=c  \]
\[ b^3+1=c  \][/quote]
Maybe $a^2+1=c,$ $b^3-1=c$ and $b^3=a^2+2$ ? ;)
If so, this is very nice, but very old. :)


[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ 2(a^5 \plus{} b^5 \plus{} c^5) \plus{} 9(x^3y^2 \plus{} y^3z^2 \plus{} z^3x^2)\geq11(x^2y^3 \plus{} y^2z^3 \plus{} z^2x^3)$[/quote]
I think, you are mean that we need to prove that
$ \sum_{cyc}(2a^5 \plus{} 9a^3b^2 \minus{} 11a^2b^3)\geq0.$
Let $ a \equal{} \min\{a,b,c\}$ and $ b \equal{} a \plus{} x,$ $ c \equal{} a \plus{} y.$
Then $ x\geq0,$ $ y\geq0$ and your inequality equivalent to the following:
$ 12(x^2 \minus{} xy \plus{} y^2)a^3 \plus{} 3(6x^3 \plus{} 7x^2y \minus{} 13xy^2 \plus{} 6y^3)a^2 \plus{}$
$ \plus{} 2(5x^4 \plus{} 9x^3y \minus{} 3x^2y^2 \minus{} 9xy^3 \plus{} 5y^4)a \plus{} 2x^5 \plus{} 9x^3y^2 \minus{} 11x^2y^3 \plus{} 2y^5\geq0,$
which true because $ 6x^3 \plus{} 7x^2y \minus{} 13xy^2 \plus{} 6y^3\geq0,$
 $ 5x^4 \plus{} 9x^3y \minus{} 3x^2y^2 \minus{} 9xy^3 \plus{} 5y^4\geq0$ and $ 2x^5 \plus{} 9x^3y^2 \minus{} 11x^2y^3 \plus{} 2y^5\geq0.$


search MATHCOUNTS USAMTS Duke college email articles

First time I learned about it was through a Google search (not particularly looking for a math forum) that I did a long time ago.  I decided to join sometime in the middle of April this year.

One of the other Mathletes going to Nationals from Maine told my coach, who told me about it. :D

I heard about the AoPS books from somebody, (probably during Mathcounts) and then while buying them, I saw the community thing, and I joined :D

I did duke tip in 2005, and the class' textbook was Aops 1&2.

I read the books partly because I was forced to. then I started to like it and I saw the ad of aops.com in the book and came here

actually my dad was searching inf for IMO and found this forum where some friends of mine already were and he told me that and then I came here andandandand..............................

I could be wrong (it's been 3.5 years) but I think I received an email from someone, possibly Valentin  Vornicu himself, when I was on a temporary IMO 2002 mailing list.

And when I started, it was MathLinks, and I still call it MathLinks.

My mum saw a short article about AoPS in her Princton Alumni magazine and cut it out and gave it to me.  Some time later when I rediscovered it from under a heap of papers on my desk I checked it out and made an account.

Hmm, I seem to belong to the Nobodys ;)
I got to MathLinks because it was discussed between Darij and Annika, thus I asked what it is (that was at the time they were already merged at, but I knew nothing about AoPS at the beginning).

[quote="fredbel6"]Nobody discovered it through MathLinks?   :maybe:   Have we forgotten that part of history? :([/quote]

I also subscribed Math Links, and probably found it yeah, surfing the web. I don't remember well why I quitted, maybe there wasn't a lot of activity a few years ago. Then I rediscovered it, and I really mean it because I was actually going to start an account with my user name and it goes and tells me that that username is already taken. And then I remembered and was able to get my username back, good luck I hadn't forgotten my previous passwords!, although I would probably be able to recover them.
So that's why I have very few posts. The community is much larger now and active as rarely seen. Very nice site.

Daniel

some crazy kid i met in mathcounts collected everyone's e-mail address that was there and sent a strange letter 2 pages long about AoPS

i've heard about MATHLINKS in one math magazine i've read

I found a link on the MATHCOUNTS resources for students section, albeit 3 years after I'd been in MC.  The name sounded interesting, and potentially useful, and it's been one of the most influential things to ever happen to me because, besides my huge math improvement, I never knew so many people liked math.  It was always just me and maybe a kid in another grade who I never saw and some random kids in other schools.

Mathlinks has its own topic as well:

http://www.mathlinks.ro/Forum/viewtopic.php?p=715289
 :lol:

I googled it, and I found it a GREAT site for math help.

If not for this, I wouldn't have known about anything, and I wouldn't even know what AMC is!  :rotfl:

I went to HCSSiM, and people there mentioned it to learn TeX.  Also, they had AoPS volume 2 in the library and I read part of it.

[quote="nutz_for2.718281828"]My Mathcounts coach when I got onto the Scoring Team last year. From there, I heard of the Mathcounts website. Wierd, isn't it?

P.S. I improved a lot since last year...[/quote]

I IMPROVED A LOT since last year too.  :lol:

AoPS really makes me do more math and get more into it.

I didn't even know what Mathcounts is till I found AoPS, and I don't do mathcounts because our schools and stuff don't do it.

After the awards banquet for GA State Mathcounts was over, solafidefarms, one of my teammates going to nationals, told me about this website. I joined sometimes that weekend.

[quote="kate25"]I got a brochure in the mail advertising this website's classes.  :)[/quote]

Cool. That means that you aced something.

[quote="D3m0n Shad0w"]from my brother. actually, he didnt advise me, but i wanted to. hes 7he.D3m0n.117![/quote]

well, there are some similarities between his and your username  :P

One thig maybe your brother just used leet on some letters and you unleeted them


trigonometry
LaTeX
ratio
calculus
limit
integration
complex numbers

[quote="I Am Me"]The first one doesn't make any sense. Is that supposed to be $2n+1$ instead of $2n-1$? Assuming it is...

[hide="1"]
My $LaTeX$ was acting up. I suggest Ratio Test.[/hide]

[hide="2"]
Obviously,$\cos^{2}\frac{n\pi}{3}\leq 1$ for all $n$. Dividing both sides by the always-positive $2^{n}$,

$\frac{\cos^{2}\frac{n\pi}{3}}{2^{n}}\leq \frac{1}{2^{n}}$ for all $n$.

$\sum_{n=1}^{\infty}\frac{1}{2^{n}}$ obviously converges (geometric with ratio $\frac{1}{2}$), so by the Direct Comparison Test, so does $\sum_{n=1}^{\infty}\frac{\cos^{2}\frac{n\pi}{3}}{2^{n}}$. Since $a$ is constant, it is irrelevant to the convergence (notice that we could simply factor it out in front of the entire series). [/hide]

Don't know for 3. Looks like some sort of comparison?[/quote]
the first by delambert theorem. the seccond it's easy by $cos^{2}nx<1$ and th third by tayler I mean when n is big then we can write instead $arcsin\frac{1}{\sqrt{n}}$ his equivalent $\frac{1}{\sqrt{n}}$ wich is big than harmonic series

In the seccond I think there must be written instead $cos^{2}{n(\pi/3)}$ this $\cos{n(\pi/3)}$ in this case we can do next: lets count this sum $\sum_{0}^{n}(\frac{e^{\frac{i\pi}{3}}}{2})^{n}$it's can be easly seen that the real part of this sum is our sum $\sum\frac{\cos{n(\pi/3)}}{2^{n}}$. therefore we have geometric progress in this case

$\sum_{0}^{n}(\frac{e^{\frac{i\pi}{3}}}{2})^{n}$ wich is equal
$\frac{(\frac{e^{\frac{i\pi}{3}}}{2})^{n+1}-1}{\frac{e^{\frac{i\pi}{3}}}{2}-1}$ and the real part of this is equal $-\frac{(\cos{(n+1)(\pi/3)}-2^{n+1})\frac{3}{2}+\sin{(\pi/3)}\sin{(n+1)}((\pi/3)}{2^{n-1}/9}$ if I don't have any mistake in calculus this answer is right  :wink:

For the third one:

$\lim_{n\to\infty}\frac{arcsin\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}}= 1$

Besides this and the other methods already posted, it is also a good exercise to use the Integral Test.

[quote="Extremal"]In the seccond I think there must be written instead $cos^{2}{n(\pi/3)}$ this $\cos{n(\pi/3)}$[/quote]

$cos^{2}x = \frac{1}{2}(1+cos2x)$

Carcul


Many really productive people I know (including some of my busiest professors) have these [b]huge[/b] screens hooked up to their computers. Many have the 30" Apple Cinema display, and many others have displays that are 24"+. All the computers in the engineering library have at least 22" inch screens.

I never thought about screen size and productivity until my laptop screen fried out, and I had to use library computers. I noticed my productivity going up. At first, I thought it was because I was in the library, and not in my distraction-filled dorm. Then, I got an external display for my room, and I still see a dramatic improvement in productivity while on the computer.

[b]1. [/b]From your experience, or from that of people you know, how helpful are large displays?
[b]2. [/b]Also, are there any other suggestions on improving computer productivity?[i]
(I'm not looking for things like, don't go on facebook/check your email obsessively)[/i]

One way I can see productivity improving is if you have such a large viewing area you can have several important windows open at once, and not having to scroll between them. When typing up lab reports for example if I want to cite data from a spreadsheet it is somewhat difficult since it requires copious amounts of alt-tab or clicking. Situations like that, where you can have a window with resources and information for citing, and a spreadsheet or document that you are typing, I think a large display or multiple displays helps a lot.
However I'm not sure if going from a 12" laptop screen to a 15" monitor would help that much. I can imagine for very intensive work, a triple 22" display could make things a lot easier, but you have to be pretty hardcore to have all those monitors set up :O.

I find having a large monitor very useful, especially when coding. Being able to look at many source files at once is extremely nice. When I was at Google over the summer, almost all FTEs had two big monitors (maybe 22" or 24"). It definitely helps.

My dad bought a 40 inch wide screen. I'm not used to it yet, but sometimes the graphics are better but it looks weird cause the screen is kinda stretched out.  :ninja:

He's probably driving the screen at a resolution with the wrong aspect ratio. But wow, 40 inches, that would require a lot of neck turning, no? Good for movies though.

FTE probably means full-time employee.

Large monitors are extremely useful for writing LaTeX documents and for writing code (need one window for writing the code, another to view the results).  If you want to be able to read documents on your screen in the same size as they would be on paper, you also need at least a 24" monitor.

My home setup is a Dell 2407WFP (24") hooked up to my 15" Macbook Pro.  I keep things like mail.app, iTunes, bittorrent, and adium on the macbook display, and work with documents on the large display.  I originally bought the monitor so I could multitable online poker, but it's turned out to be useful in many different contexts.  I don't think I could go back to a smaller display, and in fact, I'm considering getting a 30" display soon, as I think the extra screen space would come in handy (for example, when I am writing LaTeX and writing code at the same time!)

There are no 'true' 40" LCD monitors on the market.  The best large LCD is probably the 30" Dell 3008WFP.  You can obviously hook up an LCD TV to a computer but the picture is just not as sharp -- most LCD TVs top out at around 1920x1080, which is the same resolution as my Dell monitor but over a much larger space.  The aforementioned 3008WFP has a native resolution of something like 2560x1600 and looks unbelievable.

I vaguely remember seeing a 3840X2560 display on the internet before or something like that. If you have a monsterly graphics setup, and that thing, darn...


analytic geometry
geometry
algebra
polynomial
IMO Shortlist

In a rectangular coordinate system we call a horizontal line parallel to the $x$ -axis triangular if it intersects the curve with equation \[y = x^4 + px^3 + qx^2 + rx + s\] in the points $A,B,C$ and $D$ (from left to right) such that the segments $AB, AC$ and $AD$ are the sides of a triangle. Prove that the lines parallel to the $x$ - axis intersecting the curve in four distinct points are all triangular or none of them is triangular.

Actually, it has nothing to do with geometry :?..

Let $r_i(t),\ i\in \overline{1,4}$ be the four real roots (arranged in increasing order from $i=1$ to $4$) of the polynomial $P_t(x)=P(x)-t,\ t\in \mathbb R$. We have to show that for all $t$ for which $P_t$ has four real roots, the expression $r_1(t)+r_4(t)-r_2(t)-r_3(t)$ has a constant sign. We can observe that $\displaystyle \sum_{1\le i\le 4} r_i(t),\ \sum_{1\le i<j\le 4}r_i(t)r_j(t),\ \sum_{1\le i<j<k\le 4}r_i(t)r_j(t)r_k(t)$ do not depend on $t$, as they depend only on the first $4$ coefficients of $P_t$, which are the same for all $t$. This means that $(r_1+r_2-r_3-r_4)(r_1+r_3-r_2-r_4)(r_1+r_4-r_2-r_3)$ is constant for all $t$ (I wrote $r_i$ instead of $r_i(t)$), and from the fact that $r_1+r_2-r_3-r_4<0,\ r_1+r_3-r_2-r_4<0$, we derive what we want: $r_1+r_4-r_2-r_3$ has a constant sign for all $t$ for which $P_t$ has four real roots.


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 12 A
AMC 12 B

I think Mildorf got a 150 AMC + 13 AIME last year for exactly 280.  Other than on this forum I don't him personally.

You can ask him, but I think he got an 11 on the AIME last year.

[quote="keta"]Does anyone know anyone who has reached over 280 on their AMC + AIME index from 2004? I'm just curious since the AMC webpage stopped posting the index scores and the AMC forum is locked thus far.[/quote]

Let's see, Brian Rice I think had it--but he's off to college a year early.  I know there are a couple others, but the trouble is that--by my guessing--anyone who got a perfect or even near-perfect AMC last year would've been expecting to get into the USAMO so that unless they were really looking for a perfect 300, they would've spent all their time studying up for USAMO and IMO qualification rather than sticking with AIME.  Just a hunch.  (Although none of those people could be expected to not put their best forward in any case.)

Also, not everybody's scores are posted, [i]ie[/i], if they opt not to, they do not get their scores published anyway.  This is more common than you might think.

I heard somewhere that Tiankai Liu got a perfect score on the AIME, a perfect score on the AMC12A, and a perfect score on the AMC12B his senior year, thus giving him a USAMO index of 300.

Uh, there were at least two perfect AIMEs that I knew of last year, so that would almost certainly be two such people.

[quote="keta"]C'mon, everyone knows that the two perfect AIME'rs were Martha Stewart and the Iron Chef.  :lol:[/quote]
Ah, so THAT is why Michiba San retired. :P

[quote="fedwinri"]Let's see, Brian Rice I think had it--but he's off to college a year early.[/quote]

Actually, not true for a couple of reasons.  First of all, I never had a really good index --  I had a 232.5 last year, for instance.  And second, I am at college, but it's not a year early.  I went through all four years of HS; I didn't qualify for USAMO my freshman year (if that's what you're going by).  Maybe you're confusing me with someone else?

[quote="goodyfresh741"]I'd be willing to bet that Zachary Abel has achieved over 280 before.  I don't know him except from his posts in the AOPS forums and from hearsay, but he is apparently absolutely amazing.  If you want, you might ask him what he's gotten in the past---his alias on this site is zabelman.[/quote]

Well, thanks for the praise! But Fred (=tetrahedr0n) is right. Last year I had a 144 AMC12 + 11 AIME = 254 index.

But don't let Index fool you. Last year my index was about 20 points higher than Brian's, but he can still run circles around me!

Should we expect a 280 or more from you this year, Zach?

i got a 130 amc
so i still hav a chance for 280  :D

[quote="zabelman"]but he can still run circles around me![/quote]

I wouldn't say so.  But thanks for the compliment! :)

[quote="zabelman"]

But don't let Index fool you. Last year my index was about 20 points higher than Brian's, but he can still run circles around me![/quote]

So now u think ur the center of the universe huh?  :P  ;)  :D
btw it would look sorta stupid running circles around somebody. no offense.