algebra
system of equations
algebra unsolved

[color=darkblue]Hi, every One. This is a problem have posted at this forum, but I can find it. Can you give me a solution for this problem?

Let  $ a, b \in R$ and $  \{\begin {array}{c}a^4+8b=4(a^3-1)-16\sqrt{3}\\b^4+8a=4(b^3-1)+16\sqrt{3} \end {array}$.

Find $ S=a^4+b^4$[/color]

hello, i have computed $ a\equal{}1\plus{}\sqrt{3}$ and $ b\equal{}1\minus{}\sqrt{3}$ so we get
$ a^4\plus{}b^4\equal{}56$.
Sonnhard.

Please tell me, why you get $ a\equal{}1\plus{}\sqrt{3}$ and $ b\equal{}1\minus{}\sqrt{3}$ ?

hello, i have solved your equation system for $ a$ and $ b$.
Sonnhard.

is there any solution without solving $ a$ and $ b$ ?

Let x = a-1 and y = b-1. Then we have the system of equations:

$ x^4 \minus{} 6x^2 \minus{} 8x \plus{} 8y \equal{} \minus{} 16 \sqrt 3 \minus{} 9$ and

$ y^4 \minus{} 6y^2 \minus{} 8y \plus{} 8x \equal{} 16 \sqrt 3 \minus{} 9$

Adding we get

$ (x^2 \minus{} 3)^2 \plus{} (y^2 \minus{} 3)^2 \equal{} 0$

Subtracting we see that $ x \ne y$ so that $ x \plus{} y \equal{} 0$ and $ x^3 \plus{} y^3 \equal{} 0$

Since $ x^2 \equal{} y^2 \equal{} 3$, $ a^4 \plus{} b^4 \equal{} (x \plus{} 1)^4 \plus{} (y \plus{} 1)^4 \equal{} x^4 \plus{} y^4 \plus{} 6(x^2 \plus{} y^2) \plus{} 2$

$ \equal{} 18 \plus{} 36 \plus{} 2 \equal{} 56$

That is brilliant.

I knew there would be some transformation, but didn't think of that.

[quote="thanhnam2902"]Please tell me, why you get $ a \equal{} 1 \plus{} \sqrt {3}$ and $ b \equal{} 1 \minus{} \sqrt {3}$ ?[/quote]

Adding the given equations give $ a^4 \plus{} b^4 \plus{} 8(a \plus{} b) \equal{} 4(a^3 \plus{} b^3) \minus{} 8$

$ \Longleftrightarrow (a^4 \minus{} 4a^3 \plus{} 8a \plus{} 4) \plus{} (b^4 \minus{} 4b^3 \plus{} 8b \plus{} 4) \equal{} 0$

$ \Longleftrightarrow (a^2 \minus{} 2a \minus{} 2)^2 \plus{} (b^2 \minus{} 2b \minus{} 2)^2 \equal{} 0$

Since $ a^2 \minus{} 2a \minus{} 2,\ b^2 \minus{} 2b \minus{} 2$ are real numbers, we get $ a^2 \minus{} 2a \minus{} 2 \equal{} 0\ \cdots [1]\ and\ b^2 \minus{} 2b \minus{} 2 \equal{} 0\ \cdots [2]$

$ [1]\minus{}[2]: (a\minus{}b)(a\plus{}b\minus{}2)\equal{}0$, since $ a\neq b$, yielding $ a\plus{}b\equal{}2$

$ [1]\plus{}[2]: a^2\plus{}b^2\minus{}2(a\plus{}b)\minus{}4\equal{}0\Longleftrightarrow a^2\plus{}b^2\equal{}2(a\plus{}b)\plus{}4\equal{}8$

$ \therefore 2ab\equal{}(a\plus{}b)^2\minus{}(a^2\plus{}b^2)\equal{}\minus{}4\Longleftrightarrow ab\equal{}\minus{}2$.

$ \therefore S\equal{}a^4\plus{}b^4\equal{}(a^2\plus{}b^2)^2\minus{}2(ab)^2\equal{}\boxed{56}$

Dear [b]hsbhatt[/b] and [b]kunny[/b], your solution are very nice. Thank you very much.


[quote="#H34N1"]$ABCD$ is a square and $\triangle DCE$ is an equilateral triangle. Given $FE=1$ and $FE\parallel AD$, find $DC$. Express your answer in simplest radical form.

[img]7194[/img][/quote]
[hide]Label side BC as $2x$ and then BF as $x$.
Draw a line perpendicular to BC from E and the point of intersection is called point $G$.
Then, you have a rectangle ($BFEG$) and so the EG is $x$. Then you have a $30-60-90$ triangle so then CG is $x\sqrt{3}$. From the rectangle, BG is 1 and then CG is $2x-1$. Then set up a system.
$x\sqrt{3}=2x-1$
$3x^{2}=4x^{2}-4x+1$
$x^{2}-4x+1=0$
$x=\frac{4+\sqrt{16-4}}{2}$
$x=2+\sqrt{3}$.
Double to find the length of the side and you get $\boxed{4+2\sqrt{3}}$[/hide]

[quote="#H34N1"]$ABCD$ is a square and $\triangle DCE$ is an equilateral triangle. Given $FE=1$ and $FE\parallel AD$, find $DC$. Express your answer in simplest radical form.

[img]7194[/img][/quote]
[hide="hint"]30-60-90[/hide]

Another method:
[hide]Let $s$ be the side length.
If you divide the triangle into two 30-60-90's, you can see that the height is $\frac{s\sqrt{3}}{2}$ .  So, we can set up an equation:
$s-\frac{s\sqrt{3}}{2}=1$ .

$2s-s\sqrt{3}=2$
$s=\frac{2}{2-\sqrt{3}}$
in simplest radical form, this is
$s=4+2\sqrt{3}$ .[/hide]


Find all polynomials $f(x)$ with reel coefficients such that $f(x)f(x+1)=f(x^{2}+x+1)$ .

Bad title, but
[hide=" A hint"]http://www.mathlinks.ro/Forum/viewtopic.php?t=151076[/hide]


25. Right triangle $ \text{ABC}$ with right angle at vertex C has median $ \overline{\text{AD}}$ of length 5, and median $ \overline{\text{BE}}$ of length $ 2\sqrt{10}$. Find the length of the hypotenuse of triangle $ \text{ABC}$. Express your answer in simplest radical form.
[hide="Answer"]
$ 2\sqrt{13}$[/hide]

[hide]
Well, since it's a mathcounts question, you think of 3-4-5 for triangle ACD.
If you let AC be 4, you see that this also fits the other criterion: $ 2^2 \plus{} 6^2 \equal{} (2\sqrt {10})^2$.

So the triangle's hypotenuse has length $ \sqrt {4^2 \plus{} 6^2} \equal{} \sqrt {52} \equal{} \boxed{2\sqrt {13}}$.[/hide]

Oh, wow.

It's funny, because I used that exact same solution method on a similar problem today...

[quote="abacadaea"]$ AC^2 \plus{} CD^2 \equal{} 25$
$ CE^2 \plus{} BC^2 \equal{} 40$
$ 5/4 CD^2 \plus{} 5/4 BC^2 \equal{} 65$
divide by 5/4[/quote]

Sorry, I didn't quite understand where $ 5/4$ comes from. I can tell that you added two equations together...

whoops i screwed up my solution a bit. i meant CA instead of CD in the last equation but i will explain the 5/4

$ AC/2\equal{}BE$ so
$ AC^2\plus{}CE^2\equal{}5CE^2\equal{}5/4 AC^2$
similarly, we get $ BC^2\plus{}CD^2\equal{}5/4$ $ BC^2$
in my other solution, when we added the 2 equations together, we got
$ BC^2\plus{}CD^2\plus{}AC^2\plus{}CE^2\equal{}5/4(AC^2\plus{}BC^2)\equal{}65$
we know $ AC^2\plus{}BC^2\equal{}AB^2$ AB is the hypotenuese. we have
$ 5/4(AB^2)\equal{}65$
$ AB^2\equal{}52$
$ AB\equal{}2\sqrt{13}$

[quote="bfrock2000"]Is there a rule that AC/2 always equals BE? Does that go for the median of all Triangles?[/quote]
I believe he meant to type $ AC/2\equal{}CE$. But if the line segment is drawn from the right angle, which in this case is C, then you can say that the median going down to hypotenuse from the right angle is $ \frac12$ of the hypotenuse. You can prove this by drawing an circle, with the center on point C, that touches 3 vertexes of the right triangle.

@abacadaea: I see what you did. I couldn't see how it worked out. Thanks.

[quote="bfrock2000"]Is there a rule that AC/2 always equals BE? Does that go for the median of all Triangles?[/quote]
that IS the median of the triangle

There's no point just remembering a formula.

Here you have a 2-by-2 system using pythagorean theorem.

[quote="abacadaea"][quote="ajain"]I know this is off topic-a bit- but whats the formula to find the median and that other thingy having to do with a point to a point on a line?
 If the sides are ABC

BTW, will it work for ALL triangles[/quote]

im not sure i completely understand. do you mean the distance from a point to a line?

or do you mean the medians intersect at the centroid?[/quote]

The altitude and median.  Im weak on triangles (except for pyth... theoram and the law for 30-60-90 triangles and 45-45-90 triangles.)

ajain u seriously need to take honors geometry classes


We know that secant function is the reciprocal of cosine function. sec(angle)=1/cos(angle). I know how cosine graph looks like, but how can we base on cosine graph, to draw the secant graph. Suppose a point is (45,0.7) on the cosine graph, then is this point on the secant graph becoming the reciprocal (1/45,1/0.7)? I don't really think so. Hope you can help, thanks.

Just used the definition.
$ (x , f(x))$
$ (45^{\circ} , \cos 45 ^{\circ} ) \rightarrow \approx (45^{\circ} , 0.7 )$
$ (45^{\circ} , \sec 45^{\circ} ) \rightarrow \left( 45^{\circ} , \frac{1}{\cos 45^{\circ}} \right) \rightarrow \approx \left( 45^{\circ} , \frac{1}{.7} \right) \rightarrow \approx \left( 45^{\circ} , 1.4 \right)$.

The $ x$ value will remain the same, but the $ y$ value will be inverted.

i.e. if you have a point $ (x,\cos(x))$ on the cosine graph, the point $ (x,\sec(x))\equal{}(x,\frac1{\sec(x)})$ will be on the secant graph.


inequalities
induction
inequalities unsolved

n>2, show that
$ n!<(\frac{n\plus{}2}{\sqrt{6}})^n$ by AM-GM.

$ n!\leq(\frac{1\plus{}2\plus{}..\plus{}n}{n})^n<^{?}(\frac{n\plus{}2}{\sqrt{6}})^n$
$ ?\leftrightarrow n>\frac{\sqrt{6}\minus{}4}{2\minus{}\sqrt{6}}\leftrightarrow64>54$. Done

Well, the first inequality is true, but the second one...:
$ \frac{n\plus{}1}{2}<\frac{n\plus{}2}{\sqrt{6}}\iff n\sqrt{6}\plus{}\sqrt{6}<2n\plus{}4 \iff \\ n(\sqrt{6}\minus{}2)<4\minus{}\sqrt{4}\iff n<\frac{4\minus{}\sqrt{6}}{\sqrt{6}\minus{}2}$ and it does not give us our inequality
You can try simple induction here:
Assume:
$ n!(\sqrt{6})^{n}<(n\plus{}2)^{n}$
Proof:
$ (n\plus{}1)!*(\sqrt{6})^{n\plus{}1}\equal{}n!\sqrt{6}^{n}*(n\plus{}1)*\sqrt{6}<(n\plus{}2)^{n}*(n\plus{}1)*\sqrt{6}$
What is rest to prove is:
$ (n\plus{}2)^{n}*(n\plus{}1)*\sqrt{6}<(n\plus{}3)^{n\plus{}1}$ which is equivalent to:
$ (n\plus{}2)^{n}(n\plus{}1)(n\plus{}3)\sqrt{6}<(n\plus{}3)^{n\plus{}2}$
we know that:
$ (n\plus{}1)(n\plus{}3)<(n\plus{}2)^{2}$ so we wil prove a stronger inequality:
$ (n\plus{}2)^{n}(n\plus{}1)(n\plus{}3)\sqrt{6}<(n\plus{}2)^{n}(n\plus{}2)^{2}\sqrt{6}\equal{}(n\plus{}2)^{n\plus{}2}\sqrt{6}<(n\plus{}3)^{n\plus{}2}$
and the last inequality is equivalent to $ \sqrt{6}<(\frac{n\plus{}3}{n\plus{}2})^{n\plus{}2}\equal{}(1\plus{}\frac{1}{n\plus{}2})^{n\plus{}2}$ which is true for $ n\geq 5$, because:
for $ a_{n}\equal{}(1\plus{}\frac{1}{n\plus{}2})^{n\plus{}2}$ we have:
$ a_{5}>\sqrt{6}$
and it is well known that $ a_{n}$ is strictly monotonous and $ a_{n}\rightarrow e$, so our inequality is proved for $ n\geq 5$ (for less we can simply check the inequality) :wink:


geometry
parallelogram
perimeter
trigonometry
inradius
incenter
trig identities

Here's four geometry problems from the geometry written test of Mu Alpha Theta '93. I'm curious to see your solutions for these.

1. A circle is inscribed in a triangle with sides of lengths 3, 4, and 5. A second circle, interior to the triangle, is tangent to the first circle and to both sides of the  larger acute angle of the triangle. Find the radius of the second circle.

2. In triangle ACD, with sides AD (10), AC (16), and DC (14), a line is drawn from D to B on AC that creates a 135 degree angle at angle DBC and a 45 degree angle at angle ABD. Find the length of BC.

3. A convex octagon is inscribed in a circle and has four consecutive sides each of length 3, and four other sides each of length 2. Find the area.

4. PA and PE trisect one angle of parallelogram POEM. A is on OE so that OA = 4 and OP = 6. Find the perimeter of triangle APE.

[quote="surge"]Sorry to not provide any solutions, but where do you get Mu Alpha Theta problems from?[/quote]
Here are some of them:

http://www.mualphatheta.org/National_Convention/MAO_Past_Tests.htm

heres my idea for 3:

[hide] well I seem to recall that for cyclic polygons we can rearrange the sides and the area is preserved (if I got this wrong then my whole idea is flawed).  So rearange them 3-2-3-2-3-2-3-2 so its a truncated square.  Now its easy to just take 3+2sqrt(2) squared and subtract the 4 corners to get 13 + 12sqrt(2).[/hide]

I don't feel like doing number 2 but... [hide="Hint for 2"] Call $BC$ $x$ and call $DB$ $y$. Then use the Law of Cosines for triangles $BCD$ and $ABD$, and solve. [/hide]
Actually, I don't think this will work.  :?

My solutions

1.
[hide]
Let AB = 3, BC = 4, AC (hypotenuse) = 5.

[ABC] = 6, semiperimeter = 6, thus inradius = 1.

Let the incenter be Q.  AQ = $\sqrt{5}$.  Circle Q intersects AQ at P.  PQ = 1, so AP = $\sqrt{5} - 1$.

Draw $QH \bot AB$.  QH is the inradius of ABC and the radius of circle Q.  QH = 1.

We can draw JK parallel to BC so that the second circle, circle R, is the incircle of triangle AJK.  Triangle AJK is similar to triangle ABC.

Extend AQ to BC, so that it meets BC at X.  QX intersects circle Q at S.  AS = $\sqrt{5} + 1$.

Draw $RT \bot AB$.  RT is the radius of circle R, and the inradius of AJK.

Since the triangles are similar, $\frac{AS}{AP} = \frac{QH}{RT}$.  $\frac{\sqrt{5}+1}{\sqrt{5}-1} = \frac{1}{RT}$, so $RT = \frac{\sqrt{5}-1}{\sqrt{5}+1} = \frac{3 - \sqrt{5}}{2}$.
[/hide]

2.
[hide]
Use Law of Cosines to get $256 = 296 - 280\cos{ADC}, cos{ADC} = \frac{1}{7}$.  Thus $\sin{ADC} = \frac{4\sqrt{3}}{7}$, so [ADC] $= 40\sqrt{3}$.  Draw $DH \bot AC$, $DH = \frac{40\sqrt{3}}{16} = \frac{5\sqrt{3}}{2}$.

$\angle DHC = 90, \angle ABD = 45$, so $\angle HDB = 45$ and $DB = DH\sqrt{2} = \frac{5\sqrt{6}}{2}$

Note: [ADC] denotes the area of triangle ADC
[/hide]

4.
[hide]
Let $\angle OPA = \theta$.  Then $\angle POE = 180 - 3\theta$, so $\angle OAP = 2\theta$.  Thus, $\angle PAE = 180 - 2\theta$, and $\angle AEP = \theta$.  Now, we can say that $\angle PEM = 2\theta$, and $\angle PME = 180 - 3\theta$.  $\bigtriangleup PEM \sim \bigtriangleup PAO$, so $\frac{EM}{AO} = \frac{PM}{PO}$, so $PM = 9$ and $AE = 5$.  $AP = AE$, so $AP = 5$.  $\frac{PE}{PM} = \frac{PA}{PO}$, so $PE = \frac{15}{2}$.  Thus, the perimeter of $\bigtriangleup APE = 5 + 5 + \frac{15}{2} = \frac{35}{2}$.
[/hide]

OK, here's my solution for 3.

[hide]
Let AB=BC=CD=DE=3, and EF=FG=GH=HA=2.  Let Z be the center of the circumscribed circle.

Since ABCDEFGH is cyclic, $\frac{\angle AZB}{\angle AZH} = \frac{AB}{AH}$.  Applying this to the entire polygon, we can see that $\angle AZB = 54$ and $\angle AZH = 36$.  Let M be the midpoint of AB, and let N be the midpoint of AH.

AM = MB = 3/2, and AN = NH = 1.  $ZM \bot AB$, and $\angle MZA = \angle MZB = 27$.  $\angle ZAM = \angle ZBM = 63$.  Thus, the area of the hexagon ABCDEZ is $4*\left(\frac{3}{2}\right)^2*\tan{63} = 9\tan{63}$.

$\angle AZN = \angle HZN = 18$, so $\angle ZAN = \angle ZHN = 72$.  Thus the area of hexagon EFGHAZ is $4*(1)^2*\tan{72} = 4\tan{72}$.

Adding the areas and simplifying the tangents, we get the horrible-looking expression:

[ABCDEFGH] = $\frac{-9*\left(\left(\sqrt{5}-1\right)\sqrt{-2*\left(\sqrt{2*\left(\sqrt{5}+5\right)}-4\right)}-2*\sqrt{\left(\sqrt{5}+5\right)\left(\sqrt{2*\left(\sqrt{5}+5\right)}+4\right)}\right)}{2*\sqrt{\left(\sqrt{5}-1\right)*\sqrt{5-\sqrt{5}}*2\sqrt{2}-\left(\sqrt{5}-3\right)*\sqrt{2*\left(\sqrt{5}+5\right)}-\left(\sqrt{5}+5\right)*\sqrt{\sqrt{5}+5}*\sqrt{2}+32}}+\left(\sqrt{5}+1\right)*\sqrt{2*\left(\sqrt{5}+5\right)}$

which I don't think is right. :rotfl:

Dang.  That was hard to typeset. :mad:  :mad:  :mad:  :mad:  :mad:

The annoying thing is that nateharman1234 is probably right.  (Actually, he is - his idea is correct.  That property is true of cyclic polygons.)
[/hide]


function
calculus
integration
logarithms
advanced fields
advanced fields unsolved

A simple question on generating functions.

We know that: \[ \sum_{n\equal{}0}^{\infty} x^n \equal{} \frac{1}{1\minus{}x}.\]

With some work, we can show that:

\[ \sum_{n\equal{}0}^{\infty} nx^n \equal{} \frac{x}{(1\minus{}x)^2}\\
\sum_{n\equal{}0}^{\infty} n^2x^n \equal{} \frac{x(x\plus{}1)}{(1\minus{}x)^3}.\]

In general, what is the closed formula for $ \sum_{n\equal{}0}^{\infty} n^kx^n$?

Recently asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213847&search_id=137828664]here[/url].  See also [url=http://www.mathpages.com/home/kmath012/kmath012.htm]the link about Eulerian numbers[/url].

I have a problem with this also. But I want to know the closed form of the generating function when k is negative. You can help?

The [url=http://mathworld.wolfram.com/Polylogarithm.html]polylogarithm[/url] is not at all simple when (phrased in the variables of this problem) $ k$ is negative.

Thanks.

My original problem is similar to this, and for solve it in the same way I need the integral of a/(x-b) for a, b constants. This integrals have a closed form? I dont have calculus knowledge.

That would be $ a \ln (x \minus{} b)$ and is much easier than calculating the polylogarithm.


A tap, in the first minute, spills 50 liters of water and in each of the following minutes it will spill more 10% than it did in the minute before. Find the general term.

Could you help me in the exercise, please, and explain your thoughts?

think $compound %Error. "interest" is a bad command.
$.

 $50 \displaystyle\left(1 + \frac{10}{100}\right)^{n - 1}$
 is the amt spilled in the $n^{th}$ minute

[quote="madatmath"]think $compound %Error. "interest" is a bad command.
$.

 $50 \displaystyle\left(1 + \frac{10}{100}\right)^n$
 is the amt spilled in the $n^{th}$ minute[/quote]

Compound interest  :D

[quote]think compound \interest.

50 \displaystyle\left(1 + \frac{10}{100}\right)^{n - 1}
is the amt spilled in the n^{th} minute[/quote]

What do you mean by compound?

Compound means both "combine" and "intensify".

Think about it.

You have 50 liters of water. Now each period you add 10% more percent (which is 1.1). But you don't add 10% back on to the original 50 but rather the last number.

So... it's exponential growth (hence the $1.1^m$) for each minute added on.


geometry
geometric transformation
homothety

Here are two problems that I can't do (using mass points alone).  Help me!

1. If G is on BY , find x and BG : GY provided that
(a) 3B + 4Y = xG;
(b) 7B + xY = 9G.
2. In triangle ABC, D, E, and F are the trisection points of AB, BC, and CA nearer A,B,C, respectively. Let the intersection of BF and AE be J. Show that BJ : JF = 3 : 4 and AJ : JE = 6 : 1.

I don't understand how problem 1 has a solution... I thought $ x\equal{}7$ and $ BG : GY\equal{}\frac{4}{3}$ by (a), but (b) contradicts this.
As for problem 2, I can't think of a way to assign masses to A, B, and C...

I don't understand what you're asking for #1. B and G are points, not numbers, right?
For #2, Menelaus is all you need (I don't know much about mass points). We have
$ \frac {BE}{BC} \cdot \frac {CF}{AF} \cdot \frac {AJ}{JE} \equal{} 1 \implies \frac {AJ}{JE} \equal{} \frac {1}{6}$
Also,
$ \frac {AF}{AC} \cdot \frac {CE}{EB} \cdot \frac {BJ}{JF} \equal{} 1 \implies \frac {BJ}{JF} \equal{} \frac {3}{4}$

You'll have to know the basics of mass points at least in order to help me.  It's true that in #1, B and G are points, but the constants in front of them are their masses.  It's just basic mass point notation.

As for #2, I realize that Menelaus can be used.  However, that is the whole reason why I am using mass points: mass points is to ceva/menelaus as homethety is to similarity.  Just as homothety is a much quicker and powerful version of similarity, mass points is a much quicker, easier to understand and apply, and powerful version of ceva/menelaus.  That's why I'm asking for the mass points solution.

According to this site [url]http://mathcircle.berkeley.edu/mpgeo/mpgeo.html[/url], as listed under one of the problems in the basics section, a and b are listed as separate problems, thus not contradictory. So you are right. It says postulate 3 is equivalent to Menelaus, if that is of any help.

Problem 2:  Because $ E$ and $ F$ are trisection points, we have $ 3E \equal{} 2B \plus{} C$ and $ 3F \equal{} 2C \plus{} A$.  Let's eliminate $ C$ from these two equations.  We end up with $ A \plus{} 6E \equal{} 4B \plus{} 3F$.  Thus both sides must be a multiple of $ J$, namely $ 7J$.  Now you're home.


function
inequalities
search
calculus
derivative
inequalities proposed

Zeb, I found the general one!
Posted by vasc here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=178470
 :wink:

Weird - those two exactly follow from Vasc's :/

This one is stronger, though (same condition on $ f$):

$ f(20) \plus{} f(4) \plus{} f(3) \ge f(19) \plus{} f(8) \plus{} f(0) \ge f(15) \plus{} f(14) \plus{} f( \minus{} 2)$

(don't ask how I come up with these so quickly)

Yup it is stronger!
but the first one is wrong unfortunately
the second one follows by adding up:
$ f(19)\plus{}f(8)\plus{}f(9)\geq f(16)\plus{}f(15)\plus{}f(5)$ and $ f(16)\plus{}f(5)\plus{}f(0)\geq f(14)\plus{}f(9)\plus{}f(\minus{}2)$  :wink:

whooops!
I was wrong the first inequality holds too
and you can prove $ f(20x)\plus{}f(4x)\plus{}f(3x)\geq f(19x)\plus{}f(8x)\plus{}f(0)$
just notice that $ f'(x)$ is convex and use
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2080522680&t=104714
 :wink:

So I guess Zeb's result is smth like: $ f^{(m)}\geq 0$
$ \sum_{i\equal{}1}^n a_i^k\equal{}\sum_{i\equal{}1}^n b_i^k$ for $ k\equal{}1,2,\dots,m\minus{}1$ and
the $ a_i$ weighted by $ a_i^{m\minus{}1}$ respectively majorize the $ b_i$ weighted with $ b_i^{m\minus{}1}$.
(which when m=n equivalently could be replaced by:
$ \sum_{i\equal{}1}^n a_i^n\geq \sum_{i\equal{}1}^n b_i^n$)

will imply $ \sum_{i\equal{}1}^n f(a_ix)\geq \sum_{i\equal{}1}^n f(b_ix)$

[quote="zeb"]Prove that for any triply differentiable function $ f: [0,8]\to R$ with $ f'''\ge0$,

$ f(6) \plus{} f(2) \plus{} f(1) \ge f(5) \plus{} f(4) \plus{} f(0)$, and

$ f(8) \plus{} f(3) \plus{} f(1) \ge f(7) \plus{} f(5) \plus{} f(0)$.[/quote]
[quote="zeb"]
$ f(20) \plus{} f(4) \plus{} f(3) \ge f(19) \plus{} f(8) \plus{} f(0) \ge f(15) \plus{} f(14) \plus{} f( \minus{} 2)$.[/quote]
The following statement is known:

Let $ f$ be a three times differentiable function with  $ f'''\ge0$.
If $ x \plus{} y \plus{} z \equal{} a \plus{} b \plus{} c$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2 \plus{} b^2 \plus{} c^2$, then
$ E \equal{} f(x) \plus{} f(y) \plus{} f(z) \minus{} f(a) \minus{} f(b) \minus{} f(c)$
has the same sign as
 $ F \equal{} min\{x,y,z\} \minus{} min\{a,b,c\}$.

yes exactly! or the same sign with $ x^3\plus{}y^3\plus{}z^3\minus{}a^3\minus{}b^3\minus{}c^3$

[quote="Vasc"]
The following statement is known:

Let $ f$ be a three times differentiable function with  $ f'''\ge0$.
If $ x \plus{} y \plus{} z \equal{} a \plus{} b \plus{} c$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2 \plus{} b^2 \plus{} c^2$, then
$ E \equal{} f(x) \plus{} f(y) \plus{} f(z) \minus{} f(a) \minus{} f(b) \minus{} f(c)$
has the same sign as
 $ F \equal{} min\{x,y,z\} \minus{} min\{a,b,c\}$.[/quote]Shucks! Do you know where I can see the existing proof of that?

Here's a more interesting inequality, then (hopefully this one will defeat Gjergji's nefarious methods):

Let $ f$ be such that $ f''''\ge0$ (that's the fourth derivative!), prove that

$ f(0) \plus{} 6f(2) \plus{} f(4) \ge 4f(1) \plus{} 4f(3)$

Let $ g(x)\equal{}f(x\plus{}2)$ with the same condition ;)
now you can easily prove by differentiating twice and using karamata that: $ g(\minus{}2x)\plus{}6g(0)\plus{}g(2x)\geq 4(g(x)\plus{}g(\minus{}x))$
so we get your inequality for $ x\equal{}1$

[quote="Albanian Eagle"]yes exactly! or the same sign with $ x^3 \plus{} y^3 \plus{} z^3 \minus{} a^3 \minus{} b^3 \minus{} c^3$[/quote]
Let $ f$ be a three times differentiable function on $ I$.
If $ a,b,c, x,y,z \in I$ such that $ x \plus{} y \plus{} z \equal{} a \plus{} b \plus{} c$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2 \plus{} b^2 \plus{} c^2$,
then there is $ w\in I$ such that
$ f(x) \plus{} f(y) \plus{} f(z) \minus{} f(a) \minus{} f(b) \minus{} f(c)\equal{}\frac 1{2}(xyz\minus{}abc)f'''(w)$.

[quote="zeb"]  Do you know where I can see the existing proof of that?[/quote]
[quote="Albanian Eagle"]Vasc this is very nice, do you have a general form like the one I described above? ($ f^{m}\geq 0$)[/quote]
I attached a part of my article [b]A class of inequalities with second order convex functions[/b], which was published in Gazeta Matematica A, 3, 1989.

[quote="Albanian Eagle"]Let $ g(x) \equal{} f(x \plus{} 2)$ with the same condition ;)
now you can easily prove by differentiating twice and using karamata that: $ g( \minus{} 2x) \plus{} 6g(0) \plus{} g(2x)\geq 4(g(x) \plus{} g( \minus{} x))$
so we get your inequality for $ x \equal{} 1$[/quote]
Ok, this one appears resistant to using Karamata on shifts of the function ($ f'''\ge0$ again):

$ f(22) \plus{} 2f(20) \plus{} 4f(13) \plus{} 4f(12) \plus{} f(4) \plus{} 7f(3) \plus{} 4f(2) \ge 3f(18) \plus{} f(17) \plus{} 2f(16) \plus{} 2f(15) \plus{} f(9) \plus{} 2f(8) \plus{} f(7) \plus{} 2f(6) \plus{} 3f(5) \plus{} 3f(1) \plus{} 3f(0)$

BTW, the last inequality comes from this more general one:
$ {n\choose0}f(n) \plus{} {n\choose2}f(n\minus{}2) \plus{} ... \ge {n\choose1}f(n\minus{}1) \plus{} {n\choose3}f(n\minus{}3) \plus{} ...$,
if f has a positive nth derivative.


[quote="boxedexe"][i]You toss a fair coin.  If you get a head, you score 1 point.  If you get a tail, you score 2 points.  Find a recursive equation for a sequence modelling the probability of you getting exactly $n$ points after $n$ tosses.[/i]

Masoud Zargar[/quote]

perhaps you mean more than $n$, because you will always get at least $n$ after $n$ tosses because each roll contributes at least 1

as of now, the probability is just getting all heads, or $\frac{1}{2^{n}}$

or if you wish

$P_{n}$ is the probability that the sum is $n$, then we have

$P_{1}=\frac{1}{2}$ and $P_{n+1}=\frac{1}{2}*P_{n}$ for all natural numbers $n$


I find it hard to follow some of the solutions on AoPS Wiki, and the ones posted.  Would the official solutions be detailed enough so that I can follow them a lot more easily?

An extensive panel of professional mathematicians and educators write, review and edit the official solutions.  Problems and solutions go through 4 to 6 drafts before being finalized.  The solutions are meant to be understandable to a high school student who has studied the topic of the problem.

Note that for AIME problems, the cover of the AIME Solutions Pamphlet says "these solutions [are] examples of good mathematical exposition.  Routine calculations and obvious reasons for proceeding in a certain way are often omitted.  This gives greater emphasis to the essential ideas behind each solution."

Steve Dunbar
MAA Director, American Mathematics Competitions

Awesome.  I'll buy them then.

Thanks, Mr. Dunbar.

One more question; is there an expedited shipping option that would allow for delivery sooner?  Is the two to three weeks the average time or maximum?

Generally this time of year, shipments are sent within two days of receipt of the order.   At busy times of year, they are sent within a week.  Shipment for contest solutions is usually by U.S. Postal mail, which takes about a week for the continental U.S, longer for ALaska, Hawaii, and outside the U.S.

If you wish to arrange for expedited shipping, the easiest way to do that is to call the AMC office and place your order, asking for fastest shipping and getting a quote on the shipping expense.

Steve Dunbar
MAA Director, American Mathematics Competitions


Solve an equation in $ x,y \in \mathbb{N}_0$:
$ 47^x\plus{}17\equal{}2^y$

Pretty much the same as the other one ^^

The only solution for $ x < 2$ is $ (x,y) \equal{} (1,6)$
Since $ 47^x > 2^{4x}$ we have $ 2^{4x}|47^x \plus{} 17 \equal{} (3 \cdot 2^4 \minus{} 1)^x \plus{} 2^4 \plus{} 1$
But since $ (3 \cdot 2^4 \minus{} 1)^x \equiv (\minus{}1)^x \bmod 2^{4x}$ We have $ 2^{4x} | 17 \plus{} (\minus{}1)^x$ and then no solutions for $ x \ge 2$

[quote="Mathias_DK"]$ (3 \cdot 2^4 \minus{} 1)^x \equiv ( \minus{} 1)^x \bmod 2^{4x}$[/quote]hm, fail  :o

[hide]$ x\equal{}1,y\equal{}6$ is the only solution for $ x\le1$ or $ y\le6$. Assume now that $ x>1,y>6$.

$ 47^x\plus{}17\equal{}2^y$

$ 47^x\plus{}1\equal{}2^y\minus{}2^4$.

Since $ y>4$, 16|RHS, so $ x$ is odd. Let $ x\equal{}2c\plus{}1$.

$ 47^{2c\plus{}1}\plus{}1\equal{}2^4(2^{y\minus{}4}\minus{}1)$.

Since 3|LHS, then $ y\minus{}4$ is even, let $ y\minus{}4\equal{}2d$.

$ 47^{2c\plus{}1}\plus{}1\equal{}16(4^{2d}\minus{}1)$.

$ (47\plus{}1)(47^{2c}\minus{}47^{2c\minus{}1}\plus{}\ldots\plus{}1)\equal{}16(4\minus{}1)(4^{2d\minus{}1}\minus{}4^{2d\minus{}2}\plus{}\ldots\plus{}1)$

$ 47^{2c}\minus{}47^{2c\minus{}1}\plus{}\ldots\plus{}1\equal{}4^{2d\minus{}1}\minus{}4^{2d\minus{}2}\plus{}\ldots\plus{}1$.

LHS is even and RHS is odd. Hence there are no other solutions.[/hide]

[quote="Johan Gunardi"]$ (47 \plus{} 1)(47^{2c} \minus{} 47^{2c \minus{} 1} \plus{} \ldots \plus{} 1) \equal{} 16(4 \minus{} 1)(4^{2d \minus{} 1} \minus{} 4^{2d \minus{} 2} \plus{} \ldots \plus{} 1)$[/quote]$ (4 \minus{} 1)(4^{2d \minus{} 1} \plus{} 4^{2d \minus{} 2} \plus{} \ldots \plus{} 1)$
[quote="Johan Gunardi"]LHS is even[/quote]Unfortunately not  :blush:

And finally why in a one moment it appears: $ 4^{2d}$ instead of $ 2^{2d}$. If $ d \equiv 0 \ (mod \ 2)$ i got contraction mod 5 (that way it should be x odd and even at once). So: $ y \minus{} 4 \equal{} 4m \plus{} 2$. It entails $ x \equal{} 4k \plus{} 1$ (because $ RHS \equiv 3 \ (mod \ 5)$). Maybe that way?

Note that $ 47|2^y\minus{}1\implies 46|y$.

Let us write the equation as $ 47^x\plus{}1\equal{}2^4(2^{46y}\minus{}1)\equal{}A$. Clearly, $ x$ is odd. Hence, $ 1\equal{}(\minus{}47/2^{23}\minus{}1)\equal{}(2^{23}\minus{}1/47)$. But this implies that $ (\minus{}2/47)\equal{}1$ which is false.

[quote="Abel"]Note that $ 47|2^y \minus{} 1$[/quote]
Why?!!!
actually $ 47|2^{y\minus{}6}\minus{}1$

Okey, I was thinking today about this problem, and I solved it:
Let: $ x\minus{}1\equal{}a$, $ y\minus{}6\equal{}b$, (our equation is: $ 47(47^a\minus{}1)\equal{}2^6(2^b\minus{}1)$, by latest posts in this topic we can infer: $ 4|b, \ 4|a$, so: $ (47^4\minus{}1)|(47^a\minus{}1)$, but $ 47^4\minus{}1\equal{}2^6 \cdot 3 \cdot 5 \cdot 13 \cdot 17 \cdot 23$, so we get:
(1) $ RHS \equiv 0 \ (mod 13) \ \Rightarrow \ 12|b$
(2) $ RHS \equiv 0 \ (mod 17) \ \Rightarrow \ 8|b$

So: $ 24|b \ \Rightarrow (2^{24}\minus{}1)|(2^b\minus{}1)$, but: $ 2^{24}\minus{}1\equal{}3^2 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 241$, it implicates:
(3) $ LHS  \equiv 0 \ (mod 241) \ \Rightarrow \ 40|a \ \Rightarrow 8|a$, it causes:
$ (47^8\minus{}1)|(47^a\minus{}1)$, and we have also:

(4) $ 47^8\minus{}1 \equiv 0 \ (mod 2^7) \ \Rightarrow \ LHS \equiv 0 \ (mod 2^7)$, finally:
(5) $ RHS \equiv 64 \ (mod 2^7)$

(4) and (5) cause disagreement, uff  :D  :lol:


linear algebra
matrix
algorithm
induction
superior algebra
superior algebra theorems

Could anyone help me to solve the following problem:
    For any $n$, $SL_n(Z)$ is generated by only two elements.

Denote by for every $ 1 \leqslant i,j \leqslant n $ ,$ i \neq j $, by $ N_{ij} $
the matrix consisting of zeros and only in the place $ (i,j) $ we have $ 1 $, i.e
$ N_{ij} =
\begin{array}{ccccc}
   &  &  & j &  \\
  i & 0 & 0 & 1 & 0 \\
   & 0 & 0 & 0 & 0 \\
   & 0 & 0 & 0 & 0 \\
   & 0 & 0 & 0 & 0
\end{array}
$.
Then it is easy to see that for every $ i,j,k $ we have $ N_{ij} N_{jk} = N_{ik} $ and if
$ l \neq j $ then $ N_{ij} N_{lk} = 0 $, in particular $ N_{ij}^{2} = 0 $ for $ i \neq j $.
Denote $ A_{ij} := I + N_{ij} \in SL_{n}(\mathbb{Z}) $ , then $ A_{ij}^{-1} = 1 - N_{ij} $.
Now if we have two matrices $ M,N $ such that $ N^{2} = M^{2} = NM = 0 $ then we see that
$ (I-M)(I-N)(I+M)(I+N) = (I-M)(I-N+M-NM)(I+N) = (I-M)(I-N+M)(I+N) =
(I-N+M-M+MN-M^{2})(I+N) = (I-N+MN)(I+N) = I-N+MN+N-N^{2}+MN^{2} = I+MN $.
Therefore if we take $ i,j,k $ such that $ k \neq i $, then
$ N_{ij}^{2} = N_{jk}^{2} = N_{jk}N_{ij} = 0 $, so we get that
$ A_{ij}^{-1}A_{jk}^{-1}A_{ij}A_{jk} = A_{ik} $. Now define new matrix
$ S \in SL_{n}(\mathbb{Z}) $, which in standard basis $ e_{1},e_{2}, \ldots , e_{n} $
acts like that: $ e_{1} \rightarrow e_{2} ,  e_{2} \rightarrow e_{3} ,
\ldots , e_{n-1} \rightarrow e_{n} , e_{n} \rightarrow \pm e_{1} $, where we have
$ + $ sign when $ n $ is odd and $ - $ sign when $ n $ is even. Then it is easy to see
that $ S N_{ij} S^{-1} = \pm N_{ i+1 j+1} $, therefore
$ S A_{ij} S^{-1} = A_{i+1 j+1}^{ \pm 1 } $.
Let's now prove that $ S, A_{12} $ generate $ SL_{n} (\mathbb{Z}) $. Denote by $ G $
the group generated by $ S,A_{12} $. Then applying $ S A_{ij} S^{-1} = A_{i+1 j+1}^{ \pm 1 } $,
we get that each $ A_{i i+1}$ lies in $ G $. Now applying the identity
$ A_{ij}^{-1}A_{jk}^{-1}A_{ij}A_{jk} = A_{ik} $, we get that every $ A_{ij} \in G $.
Now observe that given matrix $ B $ ,multiplying by powers of certain $ A_{ij} $'s
to the left and to the right of $ B $ we get standard row,column operations over integers,
i.e adding some column multiplied by an integer to another column, or
adding some row multiplied by an integer to another row. Now it is enough to
show that for every matrix $ B \in SL_{n}( \mathbb{Z} ) $, applying those operations,
we can reach the identity $ I $. Indeed, if we have $ B = (b_{ij}), i,j =1, \ldots ,n $,
$ B \in SL_{n}( \mathbb{Z} ) $, then $ gcd(b_{11},b_{12}, \ldots ,b_{1n}) = 1 $, because
of $ det(B) = 1 $. Therefore applying Euclid algorithm to $ b_{11},b_{12}, \ldots ,b_{1n} $,
we can get $ \pm 1 $ at some place $ j $. Now, applying the column operations corresponding
to this algorithm, we will get new matrix $ B' = (b_{ij}'), i,j =1, \ldots ,n $, such that
$ b_{1j}' = \pm 1 $. Now, subtracting from the first column the column $ j $ with coefficient
$ \pm (b_{11}' - 1) $, we get that on the place $ (1,1) $ of the matrix we have $ 1 $.
Now subtracting from $ 2,3, \ldots n $ columns the first column, and then from
$ 2,3, \ldots n $ rows the first row, with appropriate coefficients, we can get that
the new matrix $ B'' = (b_{ij}''), i,j =1, \ldots ,n $ satisfies $ b_{11}'' = 1 $ and
$ b_{1j}'' = b_{j1}'' = 0 $ for $ j=2,3, \ldots , n $. Then, by induction, we can
make the $ (1,1) $ minor to be the identity $ I_{n-1} $ by some row, column operations,
so at the end we can get $ I_{n} $.

I hope it is correct.

Can somebody give a presentation of $SL_{n}(\mathbb Z)$ in terms of the two generators in Leva's proof?


Euler
search
number theory unsolved
number theory

Ehr... Fermat-Euler?! I think any factorization was intended to be carried out over the integers. :?  So trivially write $2^{24} + 1 = (2^8 + 1)(2^{16} - 2^8 + 1)$.

Sorry.. I wrote wrong...

Decompose in prime factors the number: $2^{32}+1$.

Sorry about that!  :oops:

Search at one of the big database things..

It just Fermat number...

Bomb

I found that numbers of the form: $2^{2^{n}}+1$ have prime divizors only of the form: $p=2^{n+2}k+1$.

But how about $2^{24}+1$? Is the a theorem for it? Or do I apply some tricks using modulo?

thank you!


Rank the following alcohols in order of increasing acidity: ethanol, fluoromethanol, tert-butanol, and cyclohexanol.

From my knowledge, we can rank alcohol acidity by looking at the stability of their alkoxide ions. This is dependent upon the types of groups attached to carbon of the $ C \minus{} O$ bond.

Am I right? If so, I will continue.

Edit: On second thought is it

[hide="Answer"] Acidity: tert-butanol < cyclohexanol < ethanol < fluromethanol[/hide]


One time I saw someone who could use parametric equations to graph a 5 pointed star, but I can't remember how they did it. Anyone know?

I just did it on my TI-89

x(t)=8sin(t)
y(t)=8cos(t)

tmin=0
tmax=4pi
tstep=4pi/5

Experiment around with it.  You can make all sorts of different stars and shapes by messing with the tmax and tstep variables.  Have fun :)

(Note:  The coefficient at the beginning is just so it looks bigger.  It's quicker than editing the x,y min,max values since the default is usually -10,10,-10,10)


I would appreciate if someone could show me in detail how to integrate the formulas in the picture. Thanks

Ben

This problem definitely doesn't belong in a High School forum.


It depends on what you mean by align. Do you mean maths or text?
Maths is aligned centrally with \ [... \] or the align environment. Text can be aligned by \begin{center}...\end{center}, \begin{flushleft}...\end{flushleft} or \begin{flushright}...\end{flushright}.
There are lots of other alignment possibilities so if you are more specific about what you need then we can give more specific help.

For example, how could I make $ \frac{4}{5}$ centered?

\[ \frac{4}{5}\]

To align equations, use \begin{align*} and \end{align*}.  So:
[code]\begin{align*}3x+5&=17\\3x&=12\\x&=4\end{align*}[/code]
\begin{align*}3x+5&=17\\3x&=12\\x&=4\end{align*}

[quote="ZhangPeijin"]Either that or double dollar signs.
[code]$$LaTeX Code Here$$[/code][/quote]
Please don't use double dollar signs as they are obsolete; see page 6 of [url=http://tug.ctan.org/tex-archive/info/l2tabu/english/l2tabuen.pdf]l2tabu[/url].

[quote="AIME15"]
[code]\[
\begin{align*}3x+5&=17\\3x&=12\\x&=4\end{align*}
\][/code][/quote]
In a document [i]align[/i] doesn't need \ [ ... \ ] but it does require \usepackage{amsmath}.


How do i put spaces in LaTeX code? For example, I typed [code]$I have two books$[/code] but there was no spaces  :?

Several possible commands, different spacing:

$a\ b$
$a\,b$
$a\;b$
$a\quad b$
$a\qquad b$

check out the code to see how this is done, the first one is the one used for regular spacing.

Please [b][u]DO NOT[/u][/b] put text in $\LaTeX$ formulaes on this site :) Thanks!

use the tilde (~) in math mode to insert a generic space between two "chunks" of text, or you can use as separators

\;    for a thick space
\:    for a medium space
\,    for a thin space

or you can specify exactly the amount of space you need by using

\hspace{1in}

where you can change

in (inches)

to either

cm (centimeters)

or

pt (points)

etc.  Google "latex spacing in math mode"

I repeat, this [b]does not mean that you can use text in[/b] $\LaTeX$ on this site. Please use normal text, and only put math variables in $\LaTeX$ code. This is both to avoid errors and also to speed up the loading of pages on the server.

[quote="Valentin Vornicu"][quote="amirhtlusa"]:huh:[/quote]What?[/quote]
nothing, i just thought, ther is something wrong with putting text in formulas...
but anyways.
for the text you can use the code: \text{}
then you will have spaces


What's the healthiest way to forget something or fall asleep?

so like sleeping pills or drugs or suicide are out of the question. seriously.

As soon as I read the title, I lost the game...

[quote="1=2"]Do lots and lots of math for forgetting, read lots and lots of books for falling asleep. Do not use for the game, though. I lost.[/quote]

Thanks for making me lose, I'd been going on for [b]2 years![/b]

i don't get what the game has to do with anything...

GEEZ read my blog!!!

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=222809

and i lost

omg no on read this thread cus I just lost too. Idiots...why du have to bring it up.

to forget something: fall asleep
to fall asleep: go to your bed and make some 1st grade math. That's boring :sleeping:


limit
logarithms
floor function
number theory
prime numbers
number theory theorems

$ \Pr {\text{ove that the }}\mathop {\lim }\limits_{x \to \infty } \frac{{\pi \left( x \right)}}
{{x/\log \left( x \right)}} \equal{} 1.$

Hint: Let $ p_n$ be the largest prime less or equal to $ n.$ Then $ p_n \sim n$ as $ n \rightarrow \infty.$

[quote="dubito"]$ \Pr {\text{ove that the }}\mathop {\lim }\limits_{x \to \infty } \frac {{\pi \left( x \right)}} {{x/\log \left( x \right)}} \equal{} 1.$[/quote]
What is $ \pi \left( x \right)$ ..?

$ \pi(x)$=the number of prime numbers less than or equal to $ x$. no straight formula has been found for it yet, but this that you see above (PNT:Prime Number Theorem) is the amazing work that Erdos and Selberg had done in 1948 and theres a famous storty of this named Erdos and Selberg Dispute!

I attached the paper! its really a nice one to study which also contains the nice proof of this nice theorem! :)

I Love this formula! :lol:

Actually there is a combinatorial formula for $ \pi(n)$ that can be also generalized to each real $ x$
sorry i dont have much time i'll post some here, but i'll promise to explain more next time :)

$ \textbf{Theorem: }\text{(Sieve Formula)}$ If  $ A_{1},A_{2},\cdots,A_{m}$ are the subsets of the finite set $ X$ then:

$ \left|A^{\prime}_{1}\cap A^{\prime}_{2}\cap\cdots A^{\prime}_{m}\right| = \left|X\right| - S_{1} + S_{2} - \cdots + ( - 1)^{m}S_{m}$

Where $ S_{k} = \sum_{\left\{i_{1},i_{2},\cdots,i_{k}\right\}\text{ is a subset of }\left\{1,2,\cdots,m\right\}}\left|A_{i_{1}}\cap A_{i_{2}}\cap\cdots\cap A_{i_{k}}\right|,\;\;\;\; k = 1,2,\cdots,m$

(Do you want to discuss about this or I can post the proof?!)

$ \textbf{Eratosthenes's Method: }$ The integer $ k\geq2$ is composite if and only if it is divisible by a prime number $ p$; $ p\leq\sqrt {k}$. Now let $ X = \left\{2,3,\cdots,n\right\}$, and $ r$ be the number of primes that are not bigger than $ \sqrt {n}$,in other words:

$ 2 = p_{1} < p_{2}{\cdots < p_{r}\leq\sqrt {n} < p_{r + 1}\;\;\;\;\;\text{(i)}}$

Now if $ A_{i}$'s ($ i = 1,2,\cdots,r$) are the subsets of $ X$ which include the multiples of $ p_{i}$, then $ \cup_{i}A_{i}$ is the set of all composite numbers of $ X$ and the first $ r$ prime numbers. Now by the theorem above lets evaluate $ \left|\cup_{i}A_{i}\right|$, here we have:

$ S_{1} = \left\lfloor\frac {n}{p_{1}}\right\rfloor + \left\lfloor\frac {n}{p_{2}}\right\rfloor + \cdots + \left\lfloor\frac {n}{p_{r}}\right\rfloor = \sum^{r}_{i = 1}\left\lfloor\frac {n}{p_{i}}\right\rfloor$

And in general:

$ S_{j} = \sum_{1\leq i_{1} < i_{2} < \cdots < i_{j}\leq r}\left\lfloor\frac {n}{p_{i_{1}}p_{i_{2}}\cdots p_{i_{j}}}\right\rfloor\;\;\;\;,\;\;\;\; j = 1,2,\cdots,r\;\;\;\;\;\text{(ii)}$

As a consequence:

$ \left|\cup_{i}A_{i}\right| = S_{1} - S_{2} + \cdots + ( - 1)^{r - 1}S_{r}$

Or:

$ \pi(n) = n - 1 + r - S_{1} + S_{2} - \cdots + ( - 1)^{r}S_{r}$

We have $ r$ (take a look at $ \text{(i)}$) and also $ S_{j}$'s so Done! :)

note: we can also take $ r = \pi(\sqrt {n})$!

[quote="mealmv"](How) Can we use this theorem to prove the following property about prime concentration?

Given any a > 0, there exists integer N (dependent on a) such that there is a prime p between n and n(1+a) for all real n > N.[/quote]
Suppose this is not true. Then the prime number theorem implies that $ n$th prime is asymptotically $ n\operatorname{log }n$. But $ a$ if fixed so $ n$th prime is asymptotically $ O(n)$. But
\[ \lim_{n\to \infty}\frac{n\operatorname{log}n}{n}\ne 1.\]
A contradiction with the prime number theorem.

It' not that easy. You need to show that gaps are asymptotically "small":
Assume there is some sequence $ n_i$ tending to $ \infty$ such that there is no prime in $ [n_i,(1\plus{}a) n_i ]$, then $ \pi((1\plus{}a) n_i)\minus{}\pi(n_i)\equal{}0$. Now this gives that $ 1\equal{} \lim_{i \to \infty} \frac{\pi((1\plus{}a) n_i)}{\pi(n_i )} \equal{} \lim_{i \to \infty} \frac{(1\plus{}a)n_i \log(n_i)}{n_i (\log(1\plus{}a)\plus{}\log(n_i))} \equal{} 1\plus{}a$, contradiction.


[quote="xpmath"]I don't think you can discuss on another thread on AoPS.[/quote]
perfect628 said I could post this.

Actually, I said you couldn't. Could a mod please lock this, it's for something in G&FF.


advanced fields
advanced fields unsolved

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122342]A recent discussion[/url] got me thinking: how does one prove that a given set $A\subset \mathbb N$ [b]does not[/b] contain arbitrarily long arithmetic progressions (APs)? To be precise, the goal is to prove that there is a number $M$ such that $A$ contains no APs with $M$ elements. The denser the set $A$ is, the more impressive the achievement. The ultimate goal is $\sum_{n\in A}n^{-1}=\infty$. :lol:

To begin with, consider $A_{m}=\{m^{k}\colon k\ge 1\}$ with $m\ge 2$. If $A_{m}$ contained an AP of length $3$, then the equation $m^{a}+m^{b}=2m^{c}$ would be solvable in integers $a<c<b$. Canceling $m^{a}$, get a contradiction.

Next step: $A_{m}=\{k^{m}\colon k\ge 1\}$ with $m\ge 2$. Anyone?

A blitheringly stupid example, which turns out not to invalidate anything:

Let $A=\bigcup_{k=1}^{\infty}{2^{k},2^{k}+1,\dots,2^{k}+k-1\}}$

Then $A$ does contain arbitrarily long arithmetical sequences but

$\sum_{n\in A}n^{-1}=\sum_{k=1}^{\infty}\sum_{j=0}^{k-1}\frac1{2^{k}+j}\le\sum_{k=1}^{\infty}\frac{k}{2^{k}}=2.$

But then, that's not the question you're asking.


[quote="Timothy"]The sum of two positive integers is 9.  What is the least possible sum of their reciprocals?  Express your answer as a decimal to the nearest hundredth.[/quote]

[hide]Possible values for the two numbers:

1,8
2,7
3,6
4,5

In order to make the sum of their recipricals the least, you must have both numbers the greatest, in which 4,5 works.

$\frac{1}{4}+\frac{1}{5}=\frac{5}{20}+\frac{4}{20}=\frac{9}{20}=0.45$[/hide]

[hide=" answer"]bigger the denominator smaller the fraction

4+5=9

1/4+1/5= 9/20

pianoforte you made a simple error (1/5+1/4 isnt[b] 2[/b]/20![/hide]

[quote="Drunken_Math"][quote="nonie"][hide=" answer"]bigger the denominator smaller the fraction

4+5=9

1/4+1/5= 9/20

Drunken_Math you made a simple error (1/5+1/4 isnt[b] 2[/b]/20![/hide][/quote]

That was pianoforte, not me.  ;)[/quote]
whoops Sorry...about that I'll edit it...

[quote="Drunken_Math"][quote="nonie"][hide=" answer"]bigger the denominator smaller the fraction

4+5=9

1/4+1/5= 9/20

Drunken_Math you made a simple error (1/5+1/4 isnt[b] 2[/b]/20![/hide][/quote]

That was pianoforte, not me.  ;)[/quote]

oops!  :P

[quote="risinglegend"]so wat's the answer? :?:[/quote]

scroll up a bit and you'll  see that the answer is : [hide]9/20[/hide]


[hide]I also got 90.

there are two kinds of seams:
between hexagon and hexagon
between hexagon and pentagon

since the average hexagon has 1.5 seans for "itself", and there are 20 hexagons, then there are 30 hexagon-hexagon seams. Since each pentagon has 5 seams, and there are 12 pentagons, there are 60 hexagon-pentagon seams. 60+30=90 [/hide]

[hide]There are 12*5=60 edges for the pentagons.
There are 20*6=120 edges for the hexagons.
Total is 60+120=180 edges. But each seam includes two edges. So you counted each edge twice. So there are 180/2=90 seams.[/hide]


I was just curious about the gender makeup of the round table (not AoPS because not everybody uses the round table).

I was really tempted not to make a N/A or both place to vote.

[quote="MagnusRex"]My prediction: It will be about 70-85% male.[/quote]
You're currently correct, 81%.

Mmm I recommend adding "both", "neither", and add "originally" to the question :P

Tare, I'm not adding both or neither unless you're saying that you're going to vote for one of them.

Whatever happened to confidentiality? hmm?

i was going to put both and N/A but I thought people like *coff coff* Tare *coff coff* would abuse it (Think about it.  it ends up 60% both and 40% niehter)

[quote="raccoon"]There are waay more guys than girls.  Oh well.  To bad for me.[/quote]

I guess it depends what age you are whether you think having your sex in the minority is an advantage or a disadvantage. On my part, having grown up with two younger sisters fairly close in age, and always feeling comfortable talking with girl classmates, I have regretted the lack of girls in my son's neighborhood social network and math program classes. At least now my son has a sister, but she is a good bit younger than he is, so I will always be on the lookout for girl companions for him just so that he grows up with perspective on the lives of girls, the better to understand women when he is grown up.

[quote="raccoon"]There are waay more guys than girls.  Oh well.  To bad for me.[/quote]

What would be so bad about that?  If I had the opportunity to spend lots of time with many good-looking girls, I'd jump at the chance.  And everyone knows we're all extremely good looking guys.  (Well, [i]almost[/i] all.)

Is it just me, or does it seem like more guys like math than girls? The class I took during the MMSS program was all guys except for me and I also noticed that most of the people who took the math courses were guys and the girls mostly stuck to the science courses.

Well...it may not be [b]because[/b] they're a girl, more of a social influence sort of a thing.

Wow, 3-1 boy, tough competition here :lol:

Really? Not a lot of here are in fact "both" nor "neither"...

Well, it not only depends on age, but the personality.  I actually don't mind that much, I don't enjoy it or hate it, I just act like I would around other girls because I'm used to it and I don't have that girly-girl personality where you shriek every two seconds for whatever reason you can find.  A lot of my honors track stuff is male-oriented, so I've gotten used to it.  Being in 8th grade, I've had crushes on guys, but I act the same around them as everyone else.  Big deal.  :D

[/quote]

What would be so bad about that?  If I had the opportunity to spend lots of time with many good-looking girls, I'd jump at the chance.  And everyone knows we're all extremely good looking guys.  (Well, [i]almost[/i] all.)[/quote]

Questionable.

Well, I guess it depends on who you ask if I'm good looking, I don't think I'm ugly or pretty, average I guess.  Kinda boring.


Este vorba despre urmatoarea problema aparuta in GM in urma cu vreo 10 ani:

Fie ABCD un tetraedru neortocentric şi A’, B’ şi C’ picioarele înălţimilor tetraedrului
duse din vrfurile A, B şi respectiv C. Notăm cu A”, B” şi cu C”ortocentrele triunghiurilor
A’BC, B’AC şi respectiv C’AB. Să se arate că punctele A”, B” şi  C” sunt coliniare.
                                                                                (Mihai Miculiţa,  ORADEA).

In attachment gasiti o frumoasa solutie a problemei (in 10 randuri!), insotita de un
scurt BREVIAR TEORETIC, in care sunt explicate notiunile si teoremele care au stat
la baza solutiei.

N-am gresit prea mult, problema se gaseste in GM., nr.3/1998 la pag. 135.
Este vorba despre problema C:2024.


Let $a,b,c$ be positive real numbers.Prove that the following inequality occurs: \[\left(\frac{1}{a+\sqrt{a^{2}+bc}}\right)^{2}+\left(\frac{1}{b+\sqrt{b^{2}+ac}}\right)^{2}+\left(\frac{1}{c+\sqrt{c^{2}+ab}}\right)^{2}>\frac{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}{(a+b+c)(a+b)(b+c)(c+a)}\]

We simply need :
$\left(\frac{1}{a+\sqrt{a^{2}+bc}}\right)^{2}\geq \frac{b+c}{(a+b+c)(a+b)(a+c)}$
which upon expanding becomes:
$(a+b+c)(a+b)(a+c)\geq (b+c)(2a^{2}+bc+2a\sqrt{a^{2}+bc})$
$a^{3}+2a^{2}(b+c)+a(bc+(b+c)^{2})+bc(b+c)\geq 2a^{2}(b+c)+bc(b+c)+2a(b+c)\sqrt{a^{2}+bc}$
$a^{2}+bc+(b+c)^{2}\geq 2(b+c)\sqrt{a^{2}+bc}$ which is trivial by AM-GM.
equality happens only when two of a,b,c are equal and the other is 0.

[quote="Albanian Eagle"]We simply need :
$\left(\frac{1}{a+\sqrt{a^{2}+bc}}\right)^{2}\geq \frac{b+c}{(a+b+c)(a+b)(a+c)}$
which upon expanding becomes..
[/quote]
You don`t need to expand :) Think Cauchy :wink:

Also,judging by the equality cases,wee need to have \[a^{2}=b^{2}+c^{2}+bc\]  \[b^{2}=c^{2}+a^{2}+ac\]  \[c^{2}=a^{2}+b^{2}+ab\] By summing these up we get that $a^{2}+b^{2}+c^{2}+ab+bc+ac=0$,which clearily means that we need to have $a=b=c=0$,which is a contradiction to the existence of the fractions,so the ineq is strict..

Are you sure about the equality case ?
Read my post again :wink:

I see..Here is the Cauchy solution(Somewhat messy :) ): \[ \left(a+(b+c)\right)\cdot\left(a(b+c)+a^{2}+bc\right)\geq(b+c)\left(a+\sqrt{a^{2}+bc}\right)^{2}\] ,which means \[ \left(\frac{1}{a+\sqrt{a^{2}+bc}}\right)^{2}\geq\frac{b+c}{(a+b+c)(a+b)(a+c)}\]


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[quote="ifai"]Did you watch the opening ceremony of Beijing Olympics on TV (or elsewhere)? What did you think about it?[/quote]
It was very impressive  :D more than wonderful ... :)
Hope Beijing Omlimpic will be successful as it's the opening ceremony was :blush:

[quote="ifai"]Did you watch the opening ceremony of Beijing Olympics on TV (or elsewhere)? What did you think about it?[/quote]
I just watched it in on NBC, and it was... AMAZING. I don't think I can put it in words. Btw, I think China already won a gold medal (weight lifting, I believe, or it might be shooting).

Yeah, I also just watched it a few hours ago and it was AWESOME...

Well, I didn't really expect anything less, since it is China's first Olympic games...

It was really fantastic. I have most loved the foots walking over the city and the lightning of the torch.

Wonderful ceremony.  :lol: It was very impressive. Big scale!

Definitely better than 1936! Actually, the ceremony was the most beautiful one I can recall.

I loved the opening ceremony, I'm really proud of China!  :D   China's already won 2 gold medals, leading in golds, second to USA in overall medals, who got 3 from women's saber fencing

The opening ceremony was phenomenal. Their ability to coordinate 2008 drummers or tai-chi masters so precisely shows a lot about how they view their role as the host nation. That blew me away. The screens, the fireworks, etc were also very impressive, though I am personally less fascinated by technology (but extremely interested in the artistic thought behind it all).

A show like this makes me proud of China, even though I'm not from China. I am impressed that a country can (and is willing to) come together and pull that off.

My one gripe (which has nothing to do with the spectacular display put on by the Chinese) was with the NBC commentators -- the just can't f-ing shut up! My God, I don't care to hear their hypotheses on symbolisms -- "what does the transition from the children to the soldiers symbolize? hmm blah blah blah blah". They did the same thing during the Wimbledon, too -- "that was a beautiful forehand by Nadal" no kidding, everyone just saw it. NBC commentators need to learn how to SHUT UP and let people watch.

I thought watching the opening ceremony would be a waste of time, and so I planned on watching like 10 minutes and leaving- I stayed for 3 hours! :D

[quote="TZF"]My one gripe (which has nothing to do with the spectacular display put on by the Chinese) was with the NBC commentators -- the just can't f-ing shut up! My God, I don't care to hear their hypotheses on symbolisms -- "what does the transition from the children to the soldiers symbolize? hmm blah blah blah blah". They did the same thing during the Wimbledon, too -- "that was a beautiful forehand by Nadal" no kidding, everyone just saw it. NBC commentators need to learn how to SHUT UP and let people watch.[/quote]
It seems that all the commentators in the world had the same behaviour. The CCTV commentators were very annoying too......

Hahaha, yeah, I think NBC's one of the worst for that.  I was watching the heats for one of the women's swimming races yesterday.  The American swimmer there, I don't remember her name, they kept focusing on her the entire time.

"Look at her go...she's trained so hard...blah blah blah."  Then suddenly "Oh, China won".  Of course, they did so many closeups of the American that you didn't even notice the Chinese swimmer was in the race for top 3 =D.

Opening ceremonies was amazing, the lighting of the torch was fantastic!

The opening ceremony was very nice.

The NBC commentators always seemed to have something to say about each nation that went through the parade.  For one othe nations for which the flag bearer was selected by a nationwide text message vote, the commentators just talked about "how can they asssure the authenticity of that?"  For Vietnam they made a reference to their succes at the IMO and said somthing like "I don't know about the Olympics but when it comes to the Pythagorean Theorem, they are all over that."

Was anyone else suprised when the last torchbearer started walking on the walls?

oh yeah (the thing about the pythagorean theorem). I was joking to my dad after i heard that, "How did they make the IMO if the only thing they're good at is the Pythagorean Theorem?"  :D

As for the opening ceremony, I watched it live since I live in the Northwest, and CBC also broadcasts their Olympic programs a little over the border. Even after watching it live, I was completely amazed at their attention to detail when I watched it again on NBC that night. My dad put the quality of the ceremony very well:

"There are only two people who are unhappy with the performance: The directors of the Vancouver 2010 Olympics and the London 2012 Olympics, beacuse thay have to try and top that performance."

:lol:

And mr. affable, that was really surprising, because even after everyone was sure that Li Ning (the last torchbearer) was hanging from a wire, it was so breathtaking to see that China did that much. Also, apparently to keep it a secret, he practiced the running at 1 A.M. each morning. Also, now that he's grown too old for athletics, he runs a clothes company. Imagine having to keep that a secret from your own employees!

There is an age limit because kids have the right to be kids and not be pushed around by crazy trainers. But clearly this doesn't stop some countries. The amount of training it would require for a kid to be an olympic medalist is child labor amount.

Did anyone see 4*100 m today? Again, Bolt + world record = true. He is incredible! No question about who the the king of this olympic is (no, it's not Phelps).

[quote="daboa"](no, it's not Phelps)[/quote]
Speaking of, I've found it really annoying how Phelps is being called "the greatest Olympic athlete" etc. He has won the most gold medals in a single Olympics by one person, but winning 8 is unrealistic in almost any sport. He can be called the greatest Olympic [i]swimmer[/i] of all time, but not [i]athlete[/i].

[quote="Sunny"][quote="daboa"]There is an age limit because kids have the right to be kids and not be pushed around by crazy trainers. But clearly this doesn't stop some countries. The amount of training it would require for a kid to be an olympic medalist is child labor amount.
[/quote]

Ironically, I've met tons of kids who were forced to study "for their own good" throughout their childhoods.  Some would say that's different, but I don't think so.[/quote]
Actually this came to my mind when typing (like Terence Tao who went to uni. as a kid). I get your point and agree to some extent. However, kids will be pushed harder if they are allowed to enter. To win you must sometimes do stupid and risky things. I believe they should be old enough to have their own opinions and being able to object to doing these things.  When you are 12 years old thing should be fun, not deadly serious. Also, one difference between doing "professional" math and professional gymnastics is that gymnastics includesjumps, which are required for that gold medal, are very risky if you are not trained enough. A professional gymnastics broke his neck one or two years ago.

Maturity is needed when it comes to professionalism. Or else why not remove the age limit for driving? If a youngster is a good at driving, why not allow him? It does not matter that he is 10. That argument clearly is not valid. There is more to it than just being a good driver or a good athlete.

daboa, where did you get the idea that these kids are getting pushed to do what they do?
They do what they do because they are good and they like it. The Americans, Russians, Romanians, etc. all start at young ages.

[quote="daboa"]

Maturity is needed when it comes to professionalism. Or else why not remove the age limit for driving? If a youngster is a good at driving, why not allow him? It does not matter that he is 10. That argument clearly is not valid. There is more to it than just being a good driver or a good athlete.[/quote]

hmm I find this argument pretty valid. what's the problem ?

[quote="MysticTerminator"]"Asians have different figures than people from the West, so that's what caused their suspicion," Huang said, referring to media inquiries. "They shouldn't be suspicious."

wtf?? (regarding gymnasts' ages)

in other news, 100 - 83

 :lol:[/quote]
Isn't the youngest competitor, like, 15???? (and they're pretty mature)

[quote="daboa"]serialk11r: I did not mean they are pushed like that. They pretty much do (well not all) what they love. However, do you honestly think the 12 year old little chinese chick would say "no, it's to dangerous" when her trainer says "do that jump"? These kids are not mature enough to represent themselves, even less a country.  I don't think it is fair to put so much pressure on a kid who is still wetting his bed.

MysticTerminator: So you would honestly not mind if the pilots on your next trip to e.g. France are 10 and 12 years old?[/quote]
They're not 12, and 12 isn't bed wetting. Also, the trainers are responsible for the kids if they get injured.
Honestly, if the pilot won some kind of award for best pilot and he was only 15, I'd feel fine.

[quote="perfect628"]Speaking of, I've found it really annoying how Phelps is being called "the greatest Olympic athlete" etc. He has won the most gold medals in a single Olympics by one person, but winning 8 is unrealistic in almost any sport. He can be called the greatest Olympic [i]swimmer[/i] of all time, but not [i]athlete[/i].[/quote]

Swimmers are the greatest athletes.

Also, an activity where it's advantageous to be as young and tiny a girl as possible should not be an Olympic sport.

After the last olympics a chinese gymnast admitted she was 13 at the time of the games. 3 years below the age limit. That is why there is so much suspicion aimed at the chinese gymnasts.

The issue is not just how the gymnasts look. The issue is that (as I read in the newspaper) last year they were 13 according to official documents and this year they turn 16.

[quote]Earlier this month, the AP found registration lists previously posted on the Web site of the General Administration of Sport of China that showed both He and Yang were too young to compete. He was born Jan. 1, 1994, according to the 2005, 2006 and 2007 registration lists. Yang was born Aug. 26, 1993, according to the 2004, 2005 and 2006 registration lists. In the 2007 registration list, however, her birthday has changed to Aug. 26, 1992.[/quote][url]http://www.nbcolympics.com/gymnastics/news/newsid=246060.html[/url]

[quote="durt"]
Also, an activity where it's advantageous to be as young and tiny a girl as possible should not be an Olympic sport.[/quote]

All sports give unfair advantages to people certain physical traits.

It's not only the Olympic Games where children compete. Does somebody watch MotoGP? In category of 125cc two 15-year olds were first and third on one race...

As the IOC director said, the age limits are fully determined solely by each sport's government. For example, diving's limit is 14, while gymnastics is 16. He also mentioned how the gymnastics government started a more detailed look into the age problem and that if there was a good amount of proof that any gymnast (or athlete for that matter) was breaking the age limit rule while competing, they would be automatically disqualified, no questions asked. Does that help any of your confusion? In addition, the director also stated how these age limits are to protect young children from over-practicing, and things like that, especially since sports such as diving and gymnastics can be mastered at such a young age due to flexibility.


Given $ a, b, c > 0$. Prove that: $ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \plus{} \sqrt{\frac{3abc}{a^3\plus{}b^3\plus{}c^3}} \geq\ 4$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \sqrt {\frac {3abc}{a^3 \plus{} b^3 \plus{} c^3}} \geq\ 4$
 :)[/quote]
$ t\equal{}\sqrt[4]{\frac{a^3\plus{}b^3\plus{}c^3}{3abc}} \geq 1$
We have : $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \geq 3t$ ( [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=261534[/url] )
By AM-GM , we have :
$ LHS \geq 3t\plus{}\frac{1}{t^2}\equal{}t\plus{}(t\plus{}t\plus{}\frac{1}{t^2}) \geq 1\plus{}3 \equal{}4$
We have done :D

[quote="nguoivn"]I think it's not easy if don't use my lenma  :)[/quote]
Probably you are right :) But we still have another proof.
$ \sum \frac {a}{b} \ge \frac {10\sum a^2 \minus{} \sum ab}{(a \plus{} b \plus{} c)^2}$ which is equivalent to
$ \sum c(a \minus{} c)^2(a \minus{} 2b)^2 \ge 0$ obviously true.I remember this ineq was created by by Canhang
We add this to original.
All we need to proof is $ (\frac {7}{(a \plus{} b \plus{} c)^2} \minus{} \frac {a \plus{} b \plus{} c}{(\sqrt {\frac {3abc}{a^3 \plus{} b^3 \plus{} c^3}} \plus{} 1)(a^3 \plus{} b^3 \plus{} c^3)}) \sum (a \minus{} b)^2 \ge 0$
equivalent to $ 7(\sum a^3)(\sqrt {\frac {3abc}{\sum a^3}} \plus{} 1) \ge (\sum a)^3$
equivalent to $ (6 \sum a^3 \minus{} 3\sum ab(a \plus{} b)) \plus{} ( 2\sqrt {3abc(\sum a^3)} \minus{} 6abc) \plus{} 5\sqrt {3abc(\sum a^3)} \ge 0$ QED

p.s. I really appreciate your lemma.

[quote="nguoivn"]Yes, you are right. And the best result is:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {3\sqrt {3}}{2}. \sqrt {\frac {abc}{a^3 \plus{} b^3 \plus{} c^3}} \geq\ \frac {9}{2}$
We can still use my lenma above to solve it. I think it's not easy if don't use my lenma  :)[/quote]
I think for the following inequality your lemma still helps, but the proof by your lemma is very ugly:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=350792


A holiday fruit box contains a dozen each apples, oranges, and grapefruits. What is the least number of pieces of fruit must you pick
to guarantee that you have three of the same kind?

[hide="Solution"]

If you pick 4, you know that you have 2 of one of the colors. If you pick 7, you know that you have 3 of one of the colors.

Therefore, the answer is $ \boxed{7\text{ fruits}}$.

[/hide]

So worst case you have to pick 2 apples, 2 oranges, 2 gf, then 3 of something so 2+2+2=6 and 6+1=7


[quote="wangsacl"]For $ a,b\in (0,\frac {\pi}{2})$, prove that:
$ \tan a \plus{} \tan b\geq2\tan\sqrt {ab}$[/quote]

$ x\in (0,\frac {\pi}{2})$ ---> $ tanx$ concave function
$ \frac {1}{2}tana \plus{} \frac {1}{2}tanb\geq tan(\frac {1}{2}a \plus{} \frac {1}{2}b)\geq tan\sqrt {ab}$

[$ x\in (0,\frac {\pi}{2})$---> incriese function $ tanx$
$ \frac {1}{2}a \plus{} \frac {1}{2}b\geq\sqrt {ab}$----> $ tan(\frac {1}{2}a \plus{} \frac {1}{2}b)\geq tan\sqrt {ab}$]

[quote="arqady"][quote="zaya_yc"]

$ a,b\in (0,\frac {\pi}{2})$ ---> $ tanx$ concave function
[/quote]
Are you sure?[/quote]

sorry . $ x\in (0,\frac {\pi}{2})$ ---> $ tanx$ concave function

[quote="wangsacl"]Excuse me, but if I don't get it wrong, you're using Jensen's inequality?
Thanks for the responses.[/quote]

yes.  Jensen's inequality

[quote="zaya_yc"]

sorry . $ x\in (0,\frac {\pi}{2})$ ---> $ tanx$ concave function[/quote]
But $ \left(\tan x\right)''\equal{}\left(\frac{1}{\cos^2x}\right)'\equal{}\frac{2\sin x}{\cos^3x}>0.$ :wink:


[hide]I get it to ten. Probably wrong though.[/hide]

[quote="#H34N1"]How many rectangles of any size are in the figure below?

[img]7229[/img][/quote]
i counted 23
10 on top, 10 on bottom, 1 on left, 1 on right, and 1 in all

[hide="Solution"]The key to this problem is: [b]organization[/b]!

We start by counting the $8$ obvious ones. We then use the rectangles that are formed by 2 rectangles. There are $6$ of these in the top and bottom row. Then, we count the ones formed by three rectangles. THere are $4$ of these. Then, we count the $2$ horizontal ones (the top row and bottom row), and $2$ more (the two columns). Finally, we have the whole figure which is $1$ more. Adding them all up, we get $\boxed{23}$ rectangles.[/hide]


I figured out how to do enumeration but I want to know how you do it with letters.
Ex.
a.
b.
c.

Assuming you want to change just the top level of the list then use
\renewcommand{\labelenumi}{\alph{enumi}.}
You'll find more detail at [[LaTeX:Layout#Lists]]


[quote="Sashsiam_2"]I love mathematics and physics, although I don't know if I love them a lot. My dream is to work either on pure mathematics or on theoretical physics, maybe mathematical physics, but it isn't so simple. I feel fear if I could be a nule physicist, that don't contribute, that can't work with so advanced stuff. If this is the case, I can take a confortable carrer on engineering and leave the service to the true talented men. But I don't know myself enough to make this judgement. Do I have the habilities and competences to properly do this work? I don't want to be a genius that will revolutionize science/mathematics, I just want to contribute.
You've said in this forum that olympiad math isn't an indicator of research talent. How can I discover if I have or not that skills? Is there any type of test or something else that could say "go on, you can be a good mathematician if you dedicate yourself and work a lot" or "wait, you aren't born for this"? What do I have to can do to be a competent scientist? What type of reasoning should I have to be capable of doing?

Grateful for any reply. Sorry for bad English.[/quote]

work a lot, there's no such thing as being born with it. If you go through the list of best to "worst" mathematicians or musician or whatever in any school you'll likely find that the best are the ones who work the most, and the worst the ones who work the least. It's not just working with books that teachers give you ... you have to do it ... it's also how much you can do on your own. In my opininion you have to touch a lot of of subjects. Studying maths isn't only about studying maths, it's also about studying physics, neuroscience and various other subjects. Anybody can be great at what they as long as they cover the right material, get creative and just love what you do. Einstein wasn't the best of all students and you could make the case for several other great scientists.

a lot of people will tell you, hey look at Mozart, he composed his first symphony at the age of 7. Well it's true, but it just so happened his dad was one of the truly great music teacher of his time. Mozart learned all the rudiments, theories at a very young age. Anybody can learn how to write a symphony given the proper information. With that said, Mozart isn't remembered for his 1st symphony, he's remembered for his 40th and his requiem, which he wrote last. So nothing happens overnight. Einstein had to take more than a decade to work on his theory.

my 2 cents

I strongly agree with Nick -- the biggest key to your success will be your passion.  The second largest is probably luck.  Innate ability is probably at best third most important.  It sounds like you really want to do it, so you have much of the battle won (for now).  As you continue your studies, try to find opportunities to work in research environments as an intern (or even just fetching coffee).  That will give you some indication as to whether or not you really want to do it.  And even if you find out "no", that's valuable information.  I spent 2 summers in research labs, one in academia and one in industry.  That was enough for me to know that research is not where my future would be.


$x,y,z>0$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{27}{8}$.
Prove that $\sqrt{xyz}\leq \frac{x+y+z}{3}$

[quote="Ramanujan"]$\frac{x+y+z}{3} \geq \sqrt{\frac{xy+yz+xz}{3}} =\\ \\ \sqrt{\frac{xyz (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}{3}} \geq \sqrt{\frac{27}{8} \frac{xyz }{3}} = \sqrt{\frac{9 }{ 8}xyz} = !?!?!$

:what?: :wacko:[/quote]
:lol:  Don't worry this is also my solution ;) .
I have said"too easy ineq.".


I have this one computer program that teaches me geometry lessons (which isn't pretty good) but has pretty good tests. I did third one today and I totally screwed up. It has this level increasing as you get more questions and I got most of my questions screwed up as I got to third level which is advanced (the highest). I have many many questions so I'll post ones I can post now.

1. A square pyramid has a lateral area of 84 $cm^2$, and the area of the base is 49 $cm^2$. What is the slant height of the pyramid?

A) 10 cm
B) 6 cm
C) 5 cm
D) None of the above

2. Each of the circles A, B, and C are tangent to other two like this:

(I can't use geometry sketchpad so I'll just map it)

(A)((B))
**(((C)))

---> () means smallest, (()) means medium, ((())) means largest. ** are used to take the spaces of empty space. The place is important so.. i wish i can just use geometry sketch pad but...

AB = 7, AC = 9, and BC = 10. What are the radii of the circles A,B, and C?

A) 7, 9, 10
B) 4, 6, 8
C) 3, 4, 6
D) 3, 5, 7

What is AB, if AE = 5, CE = 4, and CD = 14?

A. 11.2
B. 17.5
C. 8
D. None of the above

Hint: Use power of a point for the last one.... (in case you don't know, try using some similar triangles.

1) at=84cm e2
ab=49cm e2
ab=s e2
s=7cm
ast = 84cm e2-49cm e2
=35cm e2
as=ast/4
=35cm e2/4
=8 3/4cm e2
as=1/2bh
8 3/4cm e2=1/2*7*h
17 1/2cm e2 = 7h
2.5cm=h

which doesnt make since because 2h>7

can someone tell me where i went wrong

2)
ab=7
ac=9
bc=10

ra+rb=7
ra+rc=9
rb+rc=10

ra+rb+2=ra+rc
=rb+2=rc

ra+rb+3=rb+rc
=ra+3=rc
so
ra+1=rb

just to list them:
ra+1=rb
rb+2=rc
ra+3=rc

ra+rb=7
=ra+ra+1=7
=2ra=6
=ra=3

ra+rc=9
=3+rc=9
=rc=6

rb+rc=10
=rb+6=10
=rb=4

ra=3
rb=4
rc=6

so the answer is c

[quote="Silverfalcon"]More are coming

If the circle is inscribed in the regular polygon, and the area of the polygon is 200 :sqrt: 3, what is the area of the circle?

A) 60 :pi:
B) 100 :pi:
C) 120 :pi:
D) None of the above[/quote]

area of a regular polygon=
1/2n*s*h
where n is the number of sides
200 :sqrt: 3=1/2n*s*h
400 :sqrt: 3=n*s*h
h=(400 :sqrt: 3)/ns
h=r of inscribed circle
area c=:pi:r e2
=:pi:h e2
=:pi:(160000*3)/(n e2)(s e2)
=(480000):pi:/(n e2)(s e2)
=:sqrt:(480000/(n e2)(s e2)):pi:

now i'm stuck since i can't find the side length or the number of sides

[quote]1.B quite simple...4(7h/2) = 84 simply use that formula h being slant height and radical 49 =7 the base of the triangle. Solve the equation[/quote]

Hmm.. Please state the formula...

demji, I don't get what you're saying about number 2 neither, what are you talking about? I lost you when you went talking about r. Is r = radius? Please explain..

For number 4, let see...

I just read power of point theorem again because i can't remember and here it goes:

AE*EB = CE*ED
5*x = 4*(14-4)
5x = 4*10
5x = 40
x = 8

Oh, I see now...

The question's asking for AB, which is AE+EB.

Since we know AE = 5 and EB = 8, AB = 13 and since no choices has that answer, it's D) None of the Choice.

Thanks, Ragingg

Well demji really has some weird notations that I find his solutions hard to read.  :?


Which is more acidic and why: acetic acid in water or acetic acid in methanol?

again i suppose acetic acid in water is more acidic, even tho methanol is slightly more basic than water, as water is more polar, and hydrates the ions formed better.

Yes, that's the correct answer. And the difference is almost 5 pKa units. Interesting, isn't it?

yes it is :yup: , i expected a minor diff.


geometry
incenter
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry unsolved

L et ABC  be a triangle with s(BAC)= 40 and s(ABC)= 60. Let D and E be points on AC and AB respectively such that s(CBD)=40 and s(BCE)=70.

BD and CE intersects at point F and AF intersects BC at point G.  Find the measure of the angle <GFC i.e. s(GFC)=?

I'm changing the enunciation according to my solution:

[b]Proposition.[/b] In the triangle $ \triangle ABC$ with $ \angle BAC \equal{} 40^{\circ}$ and $ \angle ACB \equal{} 60^{\circ},$ let $ P$ be an interior point such that $ \angle PBC \equal{} 70^{\circ}$ and $ \angle PCB \equal{} 40^{\circ},$ $ AP$ meets $ BC$ at $ Q.$ Then $ \angle BPQ \equal{} 20^{\circ}$ or  $ AP \perp BC.$

Take $ D$ on $ AC$ such that $ \triangle BCD$ is equilateral. Then $ \angle PBD \equal{} 70^{\circ} \minus{} 60^{\circ} \equal{} 10^{\circ}$ $ \Longrightarrow$ $ BP$ is bisector of $ \angle ABD.$ Further, $ BCDE$ is cyclic, $ \angle DEC \equal{} \angle DBC \equal{} 60^{\circ}$ and $ \angle EDB \equal{} \angle ECB \equal{} 40^{\circ}.$ But since $ BP$ and $ EP$ are bisectors of $ \angle EBD$ and $ \angle BED,$ it follows that $ P$ is incenter of $ \triangle BED$ $ \Longrightarrow$ $ \angle EDP \equal{} 20^{\circ}.$

For the same reason that $ BCDE$ is cyclic, we have that $ \angle AED \equal{} \angle ACB \equal{} 60^{\circ}$ and $ \angle ADE \equal{} \angle ABC \equal{} 80^{\circ}.$ These angles equal half of the supplementary angles of $ \angle DEP$ and $ \angle EDP,$ in other words, $ BE$ and $ DC$ are the external bisectors of $ \angle DEP$ and $ \angle EDP$ meeting at the P-excenter $ A$ of $ \triangle PED$ $ \Longrightarrow$  $ PA$ is bisector of $ \angle EPD.$ Therefore, $ \angle APD \equal{} \frac {_1}{^2}\angle DEP \equal{} 50^{\circ}$ $ \Longrightarrow$ $ \angle PAD \equal{} 30^{\circ},$ which yields $ AP \perp BC$ and $ \angle BPQ \equal{} 20^{\circ}.$

Other solution: Take $B'$ the reflection of $B$ in perpendicular bisector of $AC$. It can easily be seen that $BC = BB' = AB'$, then take $F'$ on the perpendicular bisector of $AC$ so that $\triangle BB'F'$ is equilateral; we get $BF' = BC$ and $\angle CBF'=40^\circ$, which makes $\angle ACF'=10^\circ$, hence $F'=F$; now we know that $F$ is on the perpendicular bisector of $AC$, hence $\angle GFC=20^\circ$.

Best regards,
sunken rock

[quote="sunken rock"]The above proof is related (vertices names) to the original problem.

Other method, related to [b]Luis[/b] naming:
Take $B'$ the reflection of $B$ in perpendicular bisector of $AC$, see that $F$ is the reflection of $B$ in $CB'$ and $AB\cap CB'$ is the circumcenter of $\triangle BFB'$, this will give $F$ - the incenter of $\triangle AB'C$, next is very easy.

Best regards,
sunken rock[/quote]

it is ok thenks...


See
http://www.mathlinks.ro/Forum/viewtopic.php?t=28587
http://www.mathlinks.ro/Forum/viewtopic.php?t=4132
http://www.mathlinks.ro/Forum/viewtopic.php?t=36055

  dg

[quote="indybar"]Let $a,b,c$ be sides of an acute triangle.
Prove that $(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3)>4(a^6+b^6+c^6)$[/quote]

          I don't know if  it is well how I think, but we can take c=max{a,b,c}.Then the problem resumes at to prove that 27c^6>4c^6, which is real, because 27>4.Could you give me a message and tell me if it is the right way? :) .Or , because a,b,c are the sides of a triangle, we have a<b+c, and because it is a acute triangle it is very easy to prove that a^2<b^2+c^2 and a^3<b^3+c^3.Then we must prove that 8a^6>4a^6.-real.I wish I knew which is the surse of the problem.If you have a better solution, will you give me it? :D

[quote="hendrata01"]it's not true.... if $c \geq a,b$ then $RHS \leq 4c^6$ but then $LHS \leq 27c^6$ and it doesn't give you anything.
Proving that $27c^6 > 4c^6$ is in the wrong direction that required.[/quote]

But the second solution?Is it false?It's very easy to prove that a^2<b^2+c^2 and a^3<b^3+c^3.

doesn't work still.... because then you want the RHS to be $12a^6$ and not $4a^6$.


geometry
LaTeX
area of a triangle
Heron\u0027s formula

Quote:2. The equation  has real roots , , and . Find the area of the triangle with sides , , and .

[hide]

By Viete's formula,

r+s+t = 4

rs + rt + st = 5

rst = 1.9

By Heron's formula,

.

.

[/hide]

[quote="white_horse_king88"]That sure is in spoiler Mathfanatic!!! HEHE.[/quote]

Indeed.  I like LaTeX far too much.


$ x \equal{} 1$ & $ y \equal{} 1$ is trivial .

[quote="vijaymenon"]$ x = 1 \& y = 1$ is trivial .[/quote]

Yes, it's a trivial solution, but I want to find ALL solutions.

but how come $ x\equal{}y\equal{}\frac{1}{2}$ SATISFIES THE GIVEN EQUATION ?

[quote="vijaymenon"]but how come $ x \equal{} y \equal{} \frac {1}{2}$ SATISFIES THE GIVEN EQUATION ?[/quote]

Sorry, I made a mistake. But an other trivial solution: $ x \equal{} \frac {2}{3}$ and $ y \equal{} \frac{4}{3}$.

[quote="MTristan"][quote="vijaymenon"]but how come $ x \equal{} y \equal{} \frac {1}{2}$ SATISFIES THE GIVEN EQUATION ?[/quote]

Sorry, I made a mistake. But an other trivial solution: $ x \equal{} \frac {2}{3}$ and $ y \equal{} \frac {4}{3}$.[/quote]

Could somebody help me?

x and y are different from 0, so the given equation can be simplified in (x^2)*(y-1)+x*(y^2)-y=0 . Solving this equation to the variable"x" (using Bhaskara formula) we find that
x=[(-y^2)+-sqrt((y^4)+4(y-1)y)]/(2(y-1)). So, all the pairs (x,y) satisfies the original equation. Obviousily, y should be different from 1.

I hope I've helped you.


function
logarithms
inequalities unsolved
inequalities

Let us first prove that for any $a,b>1$ we have:

$(\frac{a+1}{a-1}) (\frac{b+1}{b-1}) \geq (\frac{\sqrt{ab}+1}{\sqrt{ab}-1})^{2}$.

Since $a+b \geq 2\sqrt{ab}$ thus $ab-2\sqrt{ab}+1 \geq ab-(a+b)+1$. Therefore

$\frac{1}{ab-(a+b)+1}\geq \frac{1}{ab-2\sqrt{ab}+1}$ which then should be clear that

$\frac{ab+(a+b)+1}{ab-(a+b)+1}\geq \frac{ab+2\sqrt{ab}+1}{ab-2\sqrt{ab}+1}$.

The above result shows that the function $\ln (\frac{e^{x}+1}{e^{x}-1})$ is convex on $(0,+\infty)$.

Now just apply all that to the original expression.

[quote="BanishedTraitor"]Now just apply all that to the original expression.[/quote]

How?

  dg


limit
trigonometry
calculus
calculus computations

prove that :
$ \lim_{n\rightarrow\infty}nsin(n!{\pi}{e^{-1}})=\pi$


$ \{x,y\}\subset \mathcal R^*_ + \ ,\ x^2 + y^2 = 1\Longrightarrow \{\begin{array}{c} x + y\le \sqrt 2\ \ ,\ \ \frac {1}{xy}\ge 2\ . \\
 \\
\boxed {x + y + \frac {1}{xy}\ge 2 + \sqrt 2}\ .\end{array}$

From inequality $ \frac {x + y}{2}\leq\sqrt {\frac {x^{2} + y^{2}}{2}}$ it follows that: $ x + y\leq\sqrt {2}$

Denote $ xy = t$. Then $ t\leq\frac {x^{2} + y^{2}}{2} = \frac {1}{2}$

We have $ (x + y)^{2} = 1 + 2xy = 1 + 2t$ so $ x + y = \sqrt {1 + 2t}$

Our inequality becomes equivalent to:

$ \sqrt {1 + 2t} + \frac {1}{t}\geq 2 + \sqrt {2}$  which is equivalent to:

$ t\sqrt {1 + 2t}\geq(2 + \sqrt {2})t - 1.$          (*)

Now if $ t\leq\frac {1}{2 + \sqrt {2}} = \frac {2 - \sqrt {2}}{2}$ then inequality is obviously true becouse $ LHS\geq 0$ and $ RHS\leq 0$.

Now assume that ${ t\geq\frac {2 - \sqrt {2}}{2}}$. Then both sides of (*) are positive so it can be squared in order to get its equivalent inequality:

$ 2t^{3} - 5t^{2} - 4\sqrt {2}t^{2} + 4t + 2\sqrt {2}t - 1\geq 0$

And this one is equivalent to:

$ (t - \frac {1}{2})(2t^{2} - t(4 + 4\sqrt {2}) + 2)\geq 0$

Last inequality is true since $ t\leq\frac {1}{2}$ and it can be easily shown that second bracket is strictly  negative  (using $ t^{2}\leq\frac {1}{4}$ and $ - t\leq - \frac {2 - \sqrt {2}}{2}$).

So both brackets are nonpositive so their product is nonnegative.

Equality holds when $ t = \frac {1}{2}$ which means $ x = y = \frac {1}{\sqrt {2}}$

[quote="Virgil Nicula"]$ \{x,y\}\subset \mathcal R^*_ \plus{} \ ,\ x^2 \plus{} y^2 \equal{} 1\Longrightarrow
\boxed {x \plus{} y \plus{} \frac {1}{xy}\ge 2 \plus{} \sqrt 2}$[/quote]
With same conditions the following stronger inequality true too:
\[ 4\sqrt2(x\plus{}y)\plus{}\frac{1}{xy}\geq10\]
which equivalent to the following problem:
let $ a,$ $ b$ and $ c$ are positive numbers such that $ a^2\plus{}b^2\equal{}2.$ Prove that
\[ 2(a\plus{}b)\plus{}\frac{1}{ab}\geq5\]

Question to[b] Virgul Nicula[/b]

You wrote that this is your own inequality. Please can you just tell us when, for first time, you have published it?

The your [u]proof[/u], [b]Delegat[/b] and the your [u]extension[/u], [b]Arqady[/b] are very nicely !

B.t.w. the last reply, I wrote "Own ([b][u]?![/u][/b])"and it wasn't published. This question is of small account ! The upper inequality is as a simple joke for me. Sorry ! If you want, I"ll  modify "own" in "classic". O.K. ? Is much too great this feel of the property for ... a trifle. When I wrote "own", I mean "I found alone" ...

See http://www.mathlinks.ro/viewtopic.php?t=170431

Let $ x\equal{}\sin a$, $ y\equal{}\cos a$. Because $ x, y>0$, so we only have to consider $ a\in\left(0, \frac{\pi}{2}\right)$
The given inequality is equivalent to \[ \sin a\plus{}\cos a\plus{}\frac{1}{\sin a\cos a}\ge 2\plus{}\sqrt 2\] \[ \iff \sqrt 2\sin\left(a\plus{}\frac{\pi}{4}\right)\plus{}\frac{1}{\sin a\cos a}\ge 2\plus{}\sqrt 2\] \[ \iff \sqrt 2\left(\sin\left(a\plus{}\frac{\pi}{4}\right)\minus{}1\right)\plus{}\frac{\left(\sin a\minus{}\cos a\right)^2}{\sin a\cos a}\ge 0\] \[ \iff \frac{2\cos^2\left(a\plus{}\frac{\pi}{4}\right)}{\sin a\cos a}\minus{}\frac{\sqrt 2\cos^2\left(a\plus{}\frac{\pi}{4}\right)}{\sin\left(a\plus{}\frac{\pi}{4}\right)\plus{}1}\ge 0\]
what is truly, because $ \sin a\cos a\le\frac{\sin^2 a\plus{}\cos^2 a}{2}\equal{}\frac{1}{2}<\sqrt 2\plus{}\sqrt{2}\sin\left(a\plus{}\frac{\pi}{4}\right)$. Equality occurs for $ a\equal{}\frac{\pi}{4}$, i.e. $ x\equal{}y\equal{}\frac{1}{\sqrt 2}$


Let $ p,q,r$ be distinct real numbers that satisfy: $ q\equal{}p(4\minus{}p), \, r\equal{}q(4\minus{}q), \, p\equal{}r(4\minus{}r).$ Find all possible values of $ p\plus{}q\plus{}r$.

Let $ a,b,c \equal{} p \minus{} 2,q \minus{} 2,r \minus{} 2$ respectively, and the system becomes $ b \equal{} f(a), c \equal{} f(b), a \equal{} f(c)$ where $ f(x) \equal{} 2 \minus{} x^2$.

Then $ a \equal{} f(f(f(a))) \Longrightarrow f(f(f(a))) \minus{} a \equal{} 2 \minus{} (2 \minus{} (2 \minus{} a^2)^2)^2 \minus{} a \equal{} 0 \quad (*)$. The $ a^7$ term has a coefficient of zero. Thus the eight roots of $ f(f(f(a))) \equal{} a$, two of which are trivially found from $ f(a) \equal{} a \Longrightarrow a \equal{} 1, \minus{} 2$, sum to zero. The remaining six roots (hopefully which are distinct/real) can be partitioned into two triples $ \{a_1, f(a_1), f(f(a_1))\}$, $ \{a_2, f(a_2), f(f(a_2))\}$ and sum to $ 1$.

(disclaimer: I [u]blatantly[/u] cheated and used a computer to actually factor these. Is there an easier way - on an actual contest, finding cubic factors is not quite as feasible?) The polynomial factors as $ (*) \equal{} (a \minus{} 1)(a \plus{} 2)(a^3 \minus{} 3a \minus{} 1)(a^3 \minus{} a^2 \minus{} 2a \plus{} 1) \equal{} 0$. And because it factors so nicely, we suspect that the roots of each of the cubics will give the desired triples. It suffices to show that if given $ a^3 \minus{} 3a \minus{} 1 \equal{} 0$, then $ f(a)^3 \minus{} 3f(a) \minus{} 1 \equal{} a^6 \minus{} 6a^4 \plus{} 9a^2 \minus{} 1 \equal{} 0$, which after long division turns out to be true, and similarly for the other cubic factor. Then the two triples sum to $ 0$ and $ 1$, respectively, and $ p \plus{} q \plus{} r \equal{} a \plus{} b \plus{} c \plus{} 6 \equal{} \boxed{6,7}$.


conics ellipse algebra polynomial quadratics AMC

1. The graph of $2x^2+xy+3y^2-11x-20y+40=0$ is an ellipse in the first quadrant of the $xy$-plane. Let $a$ and $b$ be the maximum and minimum values of $\frac{y}{x}$ over all points $(x,y)$ on the ellipse. What is the value of $a+b$?

2. The polynomial $x^3-2004x^2+mx+n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?

[hide="1."]Let y/x = k, so that y = kx.  Substituting this in the given equation and grouping by order of x yields
\[(3k^2+k+2)x^2 + (-11-20k)x + 40 = 0\]
From the quadratic equation,
\[x = \frac{11+20k \pm \sqrt{(11+20k)^2 - 4(3k^2+k+2)(40)}}{2(3k^2+k+2)}\]

In order for $x$ to be real, the discriminant ($(11+20k)^2 - 4(3k^2+k+2)(40)$) must be greater than 0:

\[(11+20k)^2 - 4(3k^2+k+2)(40) \ge 0\]
\[-80k^2 + 280k - 199 \ge 0\]

Since the equation \[-80k^2 - 280k - 199 = 0\] has negative leading coefficient, it points down, so the only points where it will be greater than 0 are between its two roots.  So the min. and max. values of $k$ are the two roots, and the sum is the sum of the roots, $-\frac{280}{-80} = \frac72$.  [/hide]

[hide="2."]
Let the roots be [tex] a,b,c [/tex] with [tex]a>b,c[/tex].
[tex] a+b+c = 2004 [/tex] so [tex] a=b+c=1002 [/tex].  But we also have that [tex] n = -(abc) = -1002(bc) [/tex], which means that [tex] bc [/tex] has to be an integer. Since [tex] b,c [/tex] are not integers, let [tex] b=\frac{p}{q} [/tex] and [tex] c = 1002-\frac{p}{q} [/tex], where [tex] gcd(p,q) = 1 [/tex].
Then [tex] \displaystyle bc = \left(\frac{p}{q}\right)\left(1002-\frac{p}{q}\right) = \frac{p(1002q-p)}{q^2} [/tex]. Since [tex] gcd(p,q) = 1 [/tex], [tex] q^2|(1002q-p) \Rightarrow q|(1002q-p) \Rightarrow q|p [/tex]. Contradiction. So there are no solutions.

So now we consider if [tex]b,c[/tex] are irrational. Let [tex]c=1002-b[/tex] this time. So [tex]m = ab+bc+ca = a(b+c)+bc = 1002^2+bc[/tex] so [tex]bc[/tex] is an integer. Then [tex]bc = b(1002-b)  = 1002b-b^2 = k[/tex] for some integer [tex]k[/tex]. We can solve the quadratic [tex]b^2-1002b+k = 0[/tex] to get [tex]b = 501+\sqrt{501^2-k}[/tex] (we only look don't look at [tex]\pm[/tex] because then you get [tex]b[/tex] and [tex]c[/tex]. However, for [tex]501[/tex] values of [tex]k[/tex], [tex]b[/tex] is an integer so we have to subtract those away. So [tex]n[/tex] can have [tex]501^2-501 = 501(500) = 25500[/tex] possible values.
[/hide]

[quote="paladin8"][hide="2."]
Let the roots be [tex] a,b,c [/tex] with [tex]a>b,c[/tex].
[tex] a+b+c = 2004 [/tex] so [tex] a=b+c=1002 [/tex].  But we also have that [tex] n = -(abc) = -1002(bc) [/tex], which means that [tex] bc [/tex] has to be an integer. Since [tex] b,c [/tex] are not integers, let [tex] b=\frac{p}{q} [/tex] and [tex] c = 1002-\frac{p}{q} [/tex], where [tex] gcd(p,q) = 1 [/tex].
Then [tex] \displaystyle bc = \left(\frac{p}{q}\right)\left(1002-\frac{p}{q}\right) = \frac{p(1002q-p)}{q^2} [/tex]. Since [tex] gcd(p,q) = 1 [/tex], [tex] q^2|(1002q-p) \Rightarrow q|(1002q-p) \Rightarrow q|p [/tex]. Contradiction. So there are no solutions.
[/hide][/quote]

b, c doesn't necessarily have to be rational, in fact, you have shown that it's impossible that they are rational numbers, this implies they have to be irrational.


If $ n$ is integer greater than 1 such that for some positive integer $ b$ number $ \frac {b^{n} \minus{} 1}{b \minus{} 1}$ is power of prime than prove that $ n$ is prime.


Guess what levans looks like in real life.

Hm I will google him and find a pic. I think 5'6, nerdy glasses, short brown hair.

apparently Larry Evans is a chess cahmpion.... :rotfl:

Marjorie L. Hoffer is even harder to find. And so is Richard Meyer's birthday.

Edit: Speaking of which, I should practice Compute This. :wink:

What's levans first name? He's not on the wiki. Does anyuone have a California phonebook?

hmm.. I wonder when levans will post in this thread... if he ever will....


How many subsets of {1,2,3,4,5,6,7,8} will sum to a multiple of 3?
null subset sum=0

[hide]Change all of the numbers in the set to mod 3.  Then we get {0,0,0,1,1,1,-1,-1,-1}.  The 0's we choose doesnt matter so we look at the 1's and -1's.

We can get a multiple of 3 by:

3 1's + 0 -1's(1 way)
3 1's + 3 -1's (1 way)
2 1's + 2 -1's (9 ways)
1 1's + 1 -1's (9 ways)
0 1's + 0 -1's (1 way)
0 1's + 3 -1's (1 way)

There's a total of 22 ways before we consider the 0's.  After considering the 0's we have $22\times 8=176$ sets.[/hide]

Actually there are only two numbers $\equiv 0\pmod3$


greatest common divisor
number theory solved
number theory

Let $p$ and $q$ be primes such that $\text{hcf}(p-1,q-1)=2$. Let $m=pq$. Show that there is some integer $a$ prime to $pq$ such that $\text{ord}_{pq}(a)=\frac{(p-1)(q-1)}{2}$.

Thanks for you help.
Mark.

Asumming $hcf$ is the greatest common divisor:
Take a primitive root $a \mod p$ and a primitive root $b \mod q$ and then by the chinese remainder theorem a $c$ equivalent to both $\mod$ their modules. Done. (do this for all prime divisors of some integer $n$ and you get the result I stated in the other thread concerning this)


algebra polynomial search AMC USA(J)MO USAMO factorial

If $ P(x)$ denotes a fifth degree polynomial such that $ P(k)\equal{}\frac{k}{k\plus{}1}$ for $ k\equal{}0,1,2,3,4,5$, determine $ P(6)$.

can you do it like this?

P(x)= ax^5+bx^4+cx^3+dx^2+ex+w
then plug in the values 0-6, solve the resulting system for the coefficients; and then plug in 6?

what is $ P(n\plus{}1)$?what does it mean ?

[quote="mathemonster"]can you do it like this?

P[i](x)= ax^5+bx^4+cx^3+dx^2+ex+w
then plug in the values 0-6, solve the resulting system for the coefficients; and then plug in 6?[/quote]
Yea, but wouldn't that be difficult?

I got an answer of 1 after a long solution.
We have P(0)=0, P(1)=1/2, P(2)=2/3, P(3)=3/4, P(4)=4/5, P(5)=5/6.
Note that the following polynomial works (find out why):

$ x(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)\cdot(1/144) \plus{} x(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)(x \minus{} 5)\cdot(1/48) \plus{} x(x \minus{} 1)(x \minus{} 3)(x \minus{} 4)(x \minus{} 5)\cdot( \minus{} 1/18) \plus{} x(x \minus{} 1)(x \minus{} 2)(x \minus{} 4)(x \minus{} 5)\cdot(1/16) \plus{} x(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 5)\cdot( \minus{} 1/30).$

Subbing in 6 gives 1 as the answer.
This is easier than solving a system of 5 equations but still pretty long. There is probably a combinatorial trick since factorials are written all over this problem.

[quote="mathemonster"]can you do it like this?

P[i](x)= ax^5+bx^4+cx^3+dx^2+ex+w
then plug in the values 0-6, solve the resulting system for the coefficients; and then plug in 6?[/quote]

Both this approach and dgreenb's solution (hi, David!  It's Qiaochu :) ) are unnecessarily long.  A much nicer approach is outlined in the link I gave.

[quote="dgreenb801"]Note that the following polynomial works (find out why)[/quote]

Lagrange Interpolation?

I would just put the equations into a matrices and solve it using a TI-84 :D

It's too bad that you can't use a TI-84 on the USAMO.

Thank you for posting that, t0rajir0u. My solution wouldn't cut it if it was a USAMO problem, that one is more insteresting. I wondered why it came out to exactly 1, and if there was such a pattern.

dbgreen what is this lagrange interpolation and why does it work here
plus it would only take me like 30 mins tops to solve the resulting system

[url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange interpolation[/url] is a standard method for interpolating a polynomial, but it works best if the data points given are arbitrary (and even then, using Newton polynomials is much faster).  The information given here is far from arbitrary.

And one last time:  the standard solution takes [b]under a minute.[/b]

[quote="t0rajir0u"]Practically speaking, using [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]Newton[/url] [url=http://en.wikipedia.org/wiki/Newton_polynomial]polynomials[/url] (note the two different links) would be faster and explains the factorials; see the second solution in the link.  There are also other techniques; see [url=http://en.wikipedia.org/wiki/Polynomial_interpolation#Non-Vandermonde_solutions]the Wikipedia article[/url].

Again, the technique of examining $ Q(x) \equal{} (x \plus{} 1) P(x) \minus{} x$ is both standard and results in a nicer solution.[/quote]

this means Q(x) = x(x-1)(x-2)(x-3)(x-4)(x-5)

i think?

[quote="unimpossible"][quote="t0rajir0u"]Again, the technique of examining $ Q(x) \equal{} (x \plus{} 1) P(x) \minus{} x$ is both standard and results in a nicer solution.[/quote]

this means Q(x) = x(x-1)(x-2)(x-3)(x-4)(x-5)

i think?[/quote]

Close.  We don't know the constant term of $ Q(x)$ (and in fact it must be chosen carefully).  Again, see the link in my first post.

[quote="Aravind Srinivas L"]Hello guys this is Aravind Srinivas from India.
I just saw ur posts and read through some problems here and there.
How old r u guys and from which grades?
I have just finished 8th grade.[/quote]

Most people who browse this forum are anywhere from the ninth to twelfth grade. There are probably some younger kids though. And some older ones who drop by. It's pretty active so feel free to browse around.

Having said that, you may want to check out the round table if you just wanna chat. Or even AoPS for the win might be a good. You can find it [url=http://www.artofproblemsolving.com/Edutainment/g1/index.php]here[/url]. There's also a link on the main menu.

[quote="t0rajir0u"]The generalization is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1045699011&t=116027]USAMO #3 1975[/url].  The technique is standard.[/quote]

That was a very nice solution (the first post). For that $ C$ value, can we just say: $ Q(x)\plus{}x\equal{}(x\plus{}1)P(x)$ only if $ Q(\minus{}1)\minus{}1\equal{}0$, instead of its fraction-polynomial argument?

BTW, is the premise of this problem true for all $ n$?

That argument is equivalent to the one Farenhajt gave by the Factor Theorem.  What do you mean by the premise of the problem?

Sorry for not being clear. I'm asking if for each $ n$ there does exist such a polynomial.

A polynomial of degree $ n$ is determined by $ n \plus{} 1$ points.

Hm, just in my opinion, it should be that a polynomial of degree [b]no greater than[/b] $ n$ is determined by $ n \plus{} 1$ points. For example, $ (1,1), (2,2), (3,3)$ determine a linear polynomial rather than a quadratic one. So I'm basically asking if for each $ n$, there is a $ n^{th}$ degree polynomial as stated in the problem.

Which statement do you disagree with, iniquitus? $ n\equal{}1$ yields $ P(x)\equal{}\frac{x}{2}$, which works perfectly fine.

Farenhajt gives an explicit form of the polynomial for all $ n$.  I don't see what the problem is.  And when I say "a polynomial of degree $ n$" I mean "a polynomial of the form $ P(x) \equal{} a_n x^n \plus{} ... \plus{} a_0$ where some of the leading coefficients may be zero."

@Tim

oh ok i see what you mean (I think)

Well since k starts from 0, for P(n) you actually have n+1 terms, which does satisfy the conditions for finding a polynomial.

@t0rajir0u:
[quote]A polynomial p(x) of degree n satisfies p(0) = 0, p(1) = 1/2, p(2) = 2/3, ... , p(n) = n/(n+1).
[/quote]
I was thinking about the possibility that these $ n \plus{} 1$ points actually gives a polynomial with a degree lower than $ n$, in which case $ Q(x)$ will have fewer than $ n \plus{} 1$ zeros. But I just realized they must be degree $ n$, since $ Q(x)$ does have $ n\plus{}1$ roots.... Yeah, there is no problem.  :)

@iniquitus:
That's just what I mean.

ok Tim

see you at ARML practice on Sunday!


trigonometry
geometry
rhombus
geometry unsolved

Let triangle $ABC$ . $AP$,$BQ$ are the bisector of angle $A,B$ such that $P,Q$ in the segment $BC,CA$ , respectively.
Suppose that $AB$+$BP$ = $AQ$ +$QB$ and angle $A$=$60$ degree.Calculate angle $B,C$ of triangle $ABC$

Let $I$ be a center of inscribed circle and angle $\angle ABI=\alpha$. Then, we have $\angle BAI=30$. We also have $\angle AQB=120-\alpha$ and $\angle APB=150-2\alpha$  We'll prove the thesis using sines: \[AQ=\frac{QB\sin\alpha}{\sin60}\]  \[AB=\frac{QB\sin(120-\alpha)}{\sin60}\]  \[BP=\frac{AB\sin30}{\sin 150-2\alpha}=\frac{QB\sin(120-\alpha)\sin30}{\sin60\sin(150-2\alpha)}\]  \[PA=\frac{AB\sin2\alpha}{\sin(150-2\alpha)}=\frac{QB\sin(120-\alpha)\sin2\alpha}{\sin60\sin(150-2\alpha)}\] So the thesis is now: \[\frac{QB\sin\alpha}{\sin60}+QB=\frac{QB\sin90\sin30}{\sin60\sin90}+\frac{QB\sin90\sin2\alpha}{\sin60\sin90}\] Now, we divide both sides by QB. We can see, that when we put $\alpha=30$ we'll get: \[\frac{\sin30}{\sin60}+1=\frac{\sin90\sin30}{\sin60\sin90}+\frac{\sin90\sin60}{\sin60\sin90}\] What is obciously true. So angles of this triangle are $60$, $60$, $60$.

$\triangle ABC$ definitely is not equilateral. You certainly can solve the problem with trigonometry, but you have to do it right. Observe that if $\angle B = 80^\circ$ and $\angle C = 40^\circ,$ then $\triangle QBC$ is isosceles with both base angles $40^\circ$ and a QB = QC, so that AQ + QB = AQ + QC = AC. On the other hand, if $C' \in AC,$ such that $\triangle ABC'$ is equilateral, then AB = AC', $BC' \perp AP.$ CPC'I is a rhombus, because it has perpendicular diagonals bisecting its internal angles. $\triangle PCC'$ is then also isosceles with C'P = C'C, because $PC' \parallel BQ,$ so that BP = C'P = C'C and AB + BP = AC' + C'C = AC.

OK, so I will try to improve my solution:)

big thanks for finding my mistakes:)