For every $ n \in N$  prove that the identity values ?

$ \dfrac{1}{1^2*2^2} \plus{} \dfrac{2}{3^2*5^2} \plus{} ... \plus{} \dfrac{n}{(2n\minus{}1)^2*(2n\plus{}1)^2}\equal{} \dfrac{n(n\plus{}1)}{2(2n\plus{}1)^2}$

Your solution is very nice!!  :lol:
Here is my uglier solution by induction on $ n$:
[b]Base case: $ \boxed{n \equal{} 1}$[/b]
$ \frac {1}{1^2*3^2} \equal{} \frac {1(1 \plus{} 1)}{2(2 \plus{} 1)^2}\implies \frac {1}{9} \equal{} \frac {2}{18}$ which is perfectly true.
[b]Induction Hypothesis:[/b]
Let it be true for $ n \equal{} k$; ie:
$ \frac {1}{1^2*2^2} \plus{} \cdots\frac {k}{(2k \minus{} 1)^2(2k \plus{} 1)^2} \equal{} \frac {k(k \plus{} 1)}{2(2k \plus{} 1)^2}$............(*)
[b]Final proof[/b]
We need to prove it true for $ n \equal{} k \plus{} 1$.
From (*);
$ \frac {1}{1^2*2^2} \plus{} \cdots\frac {k}{(2k \minus{} 1)^2(2k \plus{} 1)^2} \equal{} \frac {k(k \plus{} 1)}{2(2k \plus{} 1)^2}$
$ \implies \frac {1}{1^2*2^2} \plus{} \cdots\frac {k}{(2k \minus{} 1)^2(2k \plus{} 1)^2} \equal{} \frac {k(k \plus{} 1)}{2(2k \plus{} 1)^2} \plus{} \frac {k \plus{} 1}{(2k \plus{} 1)^2(2k \minus{} 1)^2} \equal{} \frac {k(k \plus{} 1)}{2(2k \plus{} 1)^2} \plus{} \frac {k \plus{} 1}{(2k \plus{} 1)^2(2k \plus{} 3)^2} \equal{} \frac {k \plus{} 1}{(2k \plus{} 1)^2}\left(\frac {k}{2} \plus{} \frac {1}{(2k \plus{} 3)^2}\right) \equal{} \frac {(k \plus{} 1)(k \plus{} 2)}{2(2k \plus{} 3)^2}$

Thus the identity is true for $ n \equal{} k \plus{} 1$ once it is true for $ n \equal{}k$; also it is true for $ n \equal{} 1$. So, from the principle of mathematical induction, the statement is true $ \forall n\in\mathbf{N}$.


In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358764#358764]this solution[/url], at the beginning, the author assumed $ a\plus{}b\plus{}c\equal{}1.$ Why are we allowed to do this?

eh isn't it because the ineqaulities cyclic?

In a following post, though, a poster assumes $ a\plus{}b\plus{}c\equal{}3$.  :wink:

right. I know its called a homogeneous inequality, but google failed me.

Suppose $ (a,b,c)\equal{}(2x,2y,2z)$. Then, plugging in those values and canceling, you'd get that the inequality holds true for $ (a,b,c)\equal{}(x,y,z)$ as well. In fact, if $ (a,b,c)\equal{}(x,y,z)$ is a solution, then $ (a,b,c)\equal{}k(x,y,z)$ is a solution as well -- in other words, we can scale the solutions. So it suffices to prove the inequality for a representative subset of the solutions.

We know that $ (a,b,c)$ works iff. $ \left( \frac{a}{a\plus{}b\plus{}c} , \; \frac{b}{a\plus{}b\plus{}c} , \; \frac{c}{a\plus{}b\plus{}c} \right)$ works, and those all add up to $ 1$, so it suffices to use the latter

A shortcut to see if you can assume $ a\plus{}b\plus{}c\equal{}1$ or other analogous sum, is when the inequality is homogeneous, or when every single term has the same degree. (Each variable counts when considering degree.) In that problem, each fraction had a numerator with terms of only degree 2 (terms that are $ a^2, b^2, c^2, ab$, etc.), and each denominator also had terms of only degree 2. Thus, each fraction had degree 0. The constant on the RHS also has a degree of 0, so that inequality was homogenous.

It is important that when you are thinking about "degree", you remember the spirit in which the word is used.

$ \frac{(a\plus{}b\plus{}c\plus{}1)(ab\plus{}bc\plus{}ca)}{abc\plus{}1}$, for example, could not be used in such a situation, because even though the net degree of the numerator is 3, you can't substitute in $ k(a,b,c)$ and get back the same expression.


Let $y_{n}$ be a sequence such that $y_{1}= 3$ and $y_{n+1}= \frac{1}{2}(y_{n}+\frac{4}{y_{n}})$

for any natural number $n$. Then, what is $y_{n}$?

Or you can put $y_{n}={p_{n}\over q_{n}}$ and find the recurrence system in $p_{n},q_{n}$. The answer is $y_{n}=\frac{2(5^{2^{n-1}}+1)}{5^{2^{n-1}}-1}$


limit
real analysis
real analysis unsolved


We have a triangle ABC. D is the midpoint of AB. We draw the circle with radius AD and center D, and mark the intersection points of this circle with AC and BC as E and F. Prove that the circumcenter of CEF, D, E, and F all lie on a circle.

Oh! it's easy but nice. I will post it soon.

$ \bullet$ Let $ AF$ meet $ BE$ at $ H$, since $ AF \perp BC$ and $ BE \perp AC \Longrightarrow C, E, F, H$ lie on a circle.

$ \Longrightarrow H \in$ circumscribed of $ \triangle CEF$.

$ \bullet$ We have $ \angle HEF \equal{} \angle HFE \equal{}\frac{1}{2}\angle HIE\equal{}\frac{1}{2}\angle HIF$.

$ \Longrightarrow \angle EIF\equal{}\angle EIH\plus{}\angle FIH\equal{}2\angle HEF\plus{}2\angle HFE$

$ \equal{}\angle HEF\plus{}\angle HFE\plus{}\angle FAB\plus{}\angle EBA$, (1).

$ \bullet$ We have $ \angle EDF\equal{}\angle EAF\plus{}\angle EBF$, (2).

$ \bullet$ Since (1) and (2) we have $ \angle EIF\plus{}\angle EDF\equal{}\angle EAB\plus{}\angle FBA\plus{}\angle HEF\plus{}\angle HFE$

$ \equal{}360^0\minus{}(\angle AEB\plus{}\angle AFB)\equal{}360^0\minus{}(90^0\plus{}90^0)\equal{}180^0$.

Oke!!

[hide="Using thanhnam2902 s diagram,"]
Since an inscribed angle is half of the central angle subtending the same arc, $ \angle EIF = 2\angle ACB$ and\begin{align*}\angle EDF & = 180 - \angle EDA - \angle FDB = 180 - 2\angle ABE - 2\angle BAF \\
& = 180 - 2(90 - \angle ABC) - 2(90 - \angle BAC) \\
& = 2(\angle ABC + \angle BAC) - 180.\end{align*}Thus $ \angle EDF + \angle EIF = 2(\angle ACB + \angle ABC + \angle BAC) - 180 = 180^{\circ}$, and so $ EIFD$ is cyclic.
[/hide]


vector
parameterization
analytic geometry
graphing lines
slope
algebra
system of equations

Discuss the solutions to
                                    $ 3x\minus{}2y\equal{}2$
                                    $ 6x\minus{}2y\equal{}k$

where k belongs to real numbers and give a geometric interpretation.

[quote="shadysaysurspammed"]Discuss the solutions to
                                    $ 3x \minus{} 2y \equal{} 2$
                                    $ 6x \minus{} 2y \equal{} k$

where k belongs to real numbers and give a geometric interpretation.[/quote]

Well, you could go from
3x - 2y = 2
6x - 2y = k

to

3x = 2 + 2y
6x = k + 2y

Then that means that k + 2y = 2(2 + 2y)
So k - 4 = 2y

Substitute that in for 2y in the equations to get;
3x - (k-4) = 2
6x - (k-4) = k

So
3x - k = -2
6x - k = k - 4

So
3x - k = -2
6x = 2k - 4

So 3x = k - 2
and 6x = 2k - 4
and then...
aww man!!! that is unsolvable.  :furious:

[quote="123456789"]
\[ 3x \equal{} k \minus{} 2
\]
[/quote]

That doesn't help at all, please don't post unless you have something good to say.

One of two things:

1. Either he was about to type something up, and then accidentally pressed submit instead of cancel

2. He's suggesting that you go back to that step, realize that you've almost solved the problem, and then solve it.

You need to understand the difference between a variable and a parameter. Here, $ k$ is a parameter, while $ x$ and $ y$ are variables. When you solve a system of equations, you need to eliminate or sub-out variables, not parameters. You want to solve for your variables explicitly in terms of constants and parameters.

Wickedestjr, as a less experienced member of this community, it would be prudent for you to think about what you're going to post before you post it.  Don't assume that if you don't understand a post, it's because there's a mistake in it - that's implicitly making the assumption that you know better than the average poster, and in a community like this, that assumption is frequently wrong.

I am not saying that is wrong, in fact, I would probably never question any problems in High School Forum or above, but it is just my disability to solve problems this hard.

So would you just solve for x,y in terms of K and constants ... thats easy

A natural first step would be to subtract the equations... so we get \[ 3x \equal{} k \minus{} 2\] so we know that $ x \equal{} \frac {k \minus{} 2}3$.

What is $ y$? Substitute into the first equation, and we get \[ 3(\frac {k \minus{} 2}3) \minus{} 2y \equal{} 2 \\
k \minus{} 2 \equal{} 2 \plus{} 2y \\
k \minus{} 4 \equal{} 2y \\
\frac {k \minus{} 4}2 \equal{} y\] So we have the parametric equations \[ x \equal{} \frac {k \minus{} 2}3 \\
y \equal{} \frac {k \minus{} 4}2\] Those are the solutions... If we want a geometric representation, we can convert this to a linear equation by eliminating $ k$: \[ 3x \equal{} k \minus{} 2 \\
2y \equal{} k \minus{} 4 \\
2y \minus{} 3x \equal{} \minus{}2\]

In slope-intercept form: \[ 2y \equal{} 3x \minus{} 2 \implies y \equal{} \frac 32 x \minus{} 1\]

Oh, I get it, so you never do find the values of x, and y, you just go immediately to a geometric representation. Cool!!!

[hide="tell me if im wrong"]from the first equation, $ 6x\minus{}2y\equal{}4$

and from the second equation, $ 6x\minus{}2y\equal{}k$

So if $ k\equal{}4$, then lines $ 3x\minus{}2y\equal{}2$ and $ 6x\minus{}2y\equal{}4$ are coincident.

If we let $ x\equal{}p$ (where p is a real number), then $ y\equal{}3p\minus{}2$
So, there are infinite solutions for $ x\equal{}p$

If $ k\neq 4$, then $ 3x\minus{}2y\equal{}2$ and $ 6x\minus{}2y\equal{}4$ are parallel.[/hide]

[quote="shadysaysurspammed"][quote="Wickedestjr"][quote="shadysaysurspammed"][hide="tell me if im wrong"]from the first equation, $ 6x \minus{} 2y \equal{} 4$

and from the second equation, $ 6x \minus{} 2y \equal{} k$

So if $ k \equal{} 4$, then lines $ 3x \minus{} 2y \equal{} 2$ and $ 6x \minus{} 2y \equal{} 4$ are coincident.

If we let $ x \equal{} p$ (where p is a real number), then $ y \equal{} 3p \minus{} 2$
So, there are infinite solutions for $ x \equal{} p$

If $ k\neq 4$, then $ 3x \minus{} 2y \equal{} 2$ and $ 6x \minus{} 2y \equal{} 4$ are parallel.[/hide][/quote]

In the problem, you said that 3x - 2y = 4 not 6x[/quote]

i said 3x-2y=2, not 4[/quote]

Well, that is not true either, because if
6x - 2y = 4, then 3x - 2y doesn't equal 2 unless y is 0.

[quote="Yongyi781"]Your error is here:

In your original problem statement, you said $ 3x \minus{} 2y \equal{} 2$.

Then I assume you multiplied that by 2 and supposedly got $ 6x \minus{} 2y \equal{} 4$.

However, I am sorry to say that it should've been $ 6x \minus{} 4y \equal{} 4$.[/quote]

Unless y is zero...   :rotfl:

Your first two equations are coplanar and have different slopes (one has slope 3, one has slope 3/2), so they necessarily intersect for any $ k \in \mathbb{R}$

This intersection can be written in terms of the parameter $ k$, as Yongyi did. This is really the only interesting result in this thread.

Yongyi continued by eliminating the parameter $ k$, resulting in a line. This line represents the set of all possible intersection points of the first two equations as $ k$ ranges through the reals. I [i]think[/i] (correct me if I'm wrong, Yongyi) that this is the point of his argument there.

However, if you notice, this line is exactly the same as the first of the two equations. Naturally, since all intersection points between equation (I), which is fixed, and equation (II), which is variable, must lie on the graph of equation (I).

oops yeah bad mistake .. yongyi's result looks right


Prove that we can find infinitely many disjoint translates of a ball of radius $ \frac{1}{4}$ inside the unit ball in a hilbert space.

You mean an infinite-dimensional Hilbert space.

yes infinite dimensional Hilbert Space, otherwise the question would make no sense :blush:


Prove that from any point inside an equilateral triangle, the sum of the measures of the distances to the sides of the triangle is constant.

[hide]
Let $ P$ be the point inside the triangle and $ d_{1},d_{2},d_{3}$ the distances to the sides.
Then $ S_{ABC}=S_{PAB}+S_{PBC}+S_{PAC}=\frac{ad_{1}}{2}+\frac{ad_{2}}{2}+\frac{ad_{3}}{2}=\frac{a(d_{1}+d_{2}+d_{3})}{2}=\frac{ah}{2}$, where $ a$ is the side of triangle, and $ h$ is the altitude of triangle.
So $ d_{1}+d_{2}+d_{3}=h$.
[/hide]


[quote="jelyman"]Whats the geq?[/quote]

\geq is the TeX code for "greater than or equal to" or ">=".

[quote=sqing][quote=georg]Prove that if x,y,z>0 so that xyz=1 then we have:
(x+y)(y+z)(z+y)\geq4(x+y+z-1).[/quote]
Prove that if $x,y,z>0$ so that $xyz=1$ then we have:$$
(x+y)(y+z)(z+y)\geq4(x+y+z-1).$$[/quote]
Let $a_1,a_2,\cdots,a_n>0$ and  $a_1a_2\cdots a_n=1$ $(n\geq 3)$ . [url=https://artofproblemsolving.com/community/c6h1486321p8791115]Prove that[/url] $$(a_1+a_2)(a_2+a_3)\cdots (a_{n-1}+a_n) (a_n+a_1)\geq\frac{2^n}{n-1}(a_1+a_2+\cdots +a_n-1).$$

[quote=sqing]Prove that if $a,b,c>0$ so that $abc=1$ then we have $$(a+b)(b+c)(c+a)\geq 4(a+b+c-1)$$[/quote]
[url]https://artofproblemsolving.com/community/c6h168544p936139[/url]


random links, random numbers, random dragonites, random calc books

for more infos:
http://www.senhorcarteiro.blogspot.com

[quote="randomdragoon"]Here we post links to flash games, youtube videos, news articles, and stuff.

First:
http://www.youtube.com/watch?v=49IDp76kjPw[/quote]

Wow.  XD

Curry and rice girl
[url=http://youtube.com/watch?v=jwVslAo8Cz8]http://youtube.com/watch?v=jwVslAo8Cz8[/url]

crazy DDR player

http://youtube.com/watch?v=HGrpGuDD0_Y

I don't like posting after myself but this site is awesome

http://www.fark.com


Thanks everyone. I learned lots from u guys. Good luck...

I second, third, and fourth that!

You guys, it was so great today.  I went to a state competition in Omaha that had all the good schools, including the guy who got 1st in Nebraska state (43) last year.  I was only 14th last year at state with a 30.  Anyhow, I had no idea how I might size up against this guy who totally intimidates me.

And I actually got first in the competition and 2nd in countdown.

This forum has helped me out a ton.  We can all really get busy in the next few weeks and see where it takes us.

Good luck.

The contest was set up like Mathcounts.  They had the exact type of tests, even called them Sprint, Target, Team, & Countdown.  A few things were different like: no calculators on target, only top 5 did countdown, and they arranged us randomly. I was in the first pair for the countdown round, so I thought I didn't place as high overall, especially because I knew that two of the guys last year were top in the state, I was just happy to be in the top 5 this year.  Then during awards, they announced the top 7th grader as me, then the top 8th grader (the guy who got first last year), then when it came time for the overall winner, I was certain it would be him, but I must have just barely beat him.  When they announced him winning the top 8th grade spot they also said, "With a perfect score on the Sprint, the winner was_________."  Anyhow, I am going on and on, but I am very happy.

This guy is so good that in the Fall competition at UNL's Math Day, they take the top 50 from the first test out of 1200 high school kids.  We both made that cut, (I was 48th or so of those 50.)  Then you take a 10 question test that is answered like an essay.  He placed 10th in that test which is very impressive for am 8th grader.  So for me to beat him at a competition is very big for me personally.

And yes, the sin/cos question was from that competition.

The competiton was held at Brownell Talbot in Omaha, NE.  They have a web site, with no info about the competition though.

wow, congrats!

i wish OK had pre-MC contests...

Good luck to all the new Mathcounts participants here, especially to the other teams that my team will meet at the City of Lakes South Chapter meet in Minnesota in three weeks.


How many odd numbers between 3000 and 7000 contain a 4 in the hundreds digit and a 2 in the tens digit?

[hide]There are 4 choices for the thousands digit: 3, 4, 5, and 6.  There is 1 choice for the hundreds digit: 4.  There is 1 choice for the tens digit: 2.  There are 4 choices for the units digit: 1, 3, 5, 7, and 9.  So the answer is $4\times1\times1\times4=16$.[/hide]

[quote="WindSlicer"]Eryaman almost had it :)

[hide]1,3,5,7,9 is 5 numbers therefore the answer is

$5\cdot4=\fbox{20}$[/hide]

Easy mistake to make[/quote]Dope.  I have no idea how i counted 4 instead of 5.  :oops:


Precede the dollar-sign with a backslash \\$ inside the TeX ... the backslash disappears on display...
Testing, eight dollars ... $ \$8$ right?

[quote="PowerOfPi"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=287094]Here[/url], on the first post, GameBot uses dollar signs, but they don't turn into LaTeX codes. The numbers are in LaTeX though, so LaTeX can't be disabled. How did it happen?[/quote]

\$[code]\$[/code]


9的-1次方,8的0次方,7的1次方,6的2次方,5的3次方. :P

还有其他的吗？

任意填啊, 填什么都可以构造通项.

所以答案是125

这种题目实在无聊，考查不到任何有价值的东西。


Square ABCD with area x squared lies within equilateral triangle JKL in a way that segment AB is coincident with side JK of the triangle, and one vertex of the square lies on each of the other two sides of the triangle, as shown in the diagram. If triangle JKL has perimeter 30, and the height of triangle JKL measures h, then find h/x.

it is just a triangle around a square picture, and it isn't drawn to scale.

yeah i got h equals 5 root 3, but i don't know the x.

coinicident- matching point for point i think or corresponding

uhh... would you please define coincident?

also, it says "shown in the diagram" but i see no diagram.

in any case. I managed to get h rather easily (pythagorean theorem) but i'm having trouble with x.

i'm really bad at geometry.

See the attached diagram. Notice that $\triangle LMD\sim\triangle LNJ$, hence $\frac{h-x}{x/2}=\frac{h}{a/2}\iff \frac{2(h-x)}{x}=\sqrt{3}\iff{h\over x}-1={\sqrt{3}\over 2}\iff{h\over x}=1+{\sqrt{3}\over 2}$

I had the same solution, but my computer started to screw up as I was posting it  :D


There is a total of 20 quarters stacked in four piles. The first pile has 3 fewer than the second pile. The second pile has 2 more than the third pile. The fourth pile has twice as many as the second pile. How many quarters are in the fourth pile?

Let the amount in the 2nd pile be x.

$ (x\minus{}3)\plus{}x\plus{}(x\minus{}2)\plus{}2x\equal{}5x\minus{}5\equal{}20$

Thus the amount in the 2nd pile is 5, so the amount in the 4th is $ 2\cdot5\equal{}10$


quadratics
inequalities
algebra
polynomial
sum of roots
triangle inequality

Suppose a, b, c are real numbers such that the quadratic equation
$ x^{2}$ − $ (a \plus{} b \plus{} c)x$ + $ (ab \plus{} bc \plus{} ca) \equal{} 0$
has roots of the form $ \alpha$ ± i$ \beta$, where  $ \alpha$ > 0 and  $ \beta$(non-zero) are real numbers.
Show that:
(a) The numbers a, b, c are all positive.
(b) The numbers √a,√b,√c are the sides of a triangle.

For (a),
[hide]
A is alpha,B is beta
Disc<0
therefore,
a^2 + b^2 + c^2 - 2ab - 2bc - 2ca < 0  -------- I

Sum of roots =(a+b+c )/2=2*A>0
therefore (a+b+c)>0     ----II

(a+b-c)^2=a^2 + b^2 + c^2 +2ab-2bc-2ac  >=0 -----III

III - I
4ab>0

Similarly taking (a -b +c) and (-a + b+c) instead of (a+b-c) in III,we get,
4bc>0
4ca>0

therefore,
ab>0
bc>0
ca>0
this will be true only if all are >0 or <0
If all are <0,then (a+b+c)<0   contradicts II
So,a,b,c are all positive

[/hide]

ii: To prove:√a,√b,√c are the sides of a triangle.

    solution: roots are of the form-(A+iB,A-iB)
                 therefore,
                 a+b+c>0......i
                 ab+bc+ca>0...........ii

                 frm the triangle-inequality-
                 √a+√b>√c
                => a+b+2√ab>c
                => 4ab>c^2+a^2+b^2+2ab-2ac-2bc
   therefore to prove :arrow:  2(ab+bc+ca)>a^2+b^2+c^2;

                 frm i,
                 b+c>-a...........iii

               frm ii,
            a(b+c)+bc>0

         frm iii&ii,
       -a^2+bc>0
   =>  bc>a^2.......iv
 since ii is a symmetric eqn-,so..
         ab>b^2........v
         ac>c^2.........vi

frm iv,v,vi...
 ab+ac+bc>a^2+b^2+c^2
=> 2(ab+ac+bc)>a^2+b^2+c^2
=> √a+√b>√c

similarly...
√a<√b+√c
√b<√a+√c

therefore...
√a,√b,√c are the sides of a triangle.   proved

[/quote]

[quote="abhinav1991"]
 frm i,
         b+c>-a...........iii

 frm ii,
          a(b+c)+bc>0

 frm iii&ii,
         -a^2+bc>0
 [/quote]

here you are mistaken, abhinav1991.

ii and iiii do not imply - $ a^{2}$ + bc  > 0

Gah. I thought this part was way easier than the first
But since arjun already posted a nice solution for the first i will do only the second part

If $ \sqrt a, \sqrt b, \sqrt c$ are sides of triangle then

$ \sqrt a\plus{} \sqrt b > \sqrt c$

And simplifying after squaring twice we have to prove that

$ 2ab\plus{}2bc\plus{}2ca > a^2 \plus{} b^2 \plus{} c^2$

But since the roots are imaginary the discriminant is negative so we have

$ (a\plus{}b\plus{}c)^2 \minus{}4(ab\plus{}bc\plus{}ca)$ is negative

That is the same as   $ 2ab\plus{}2bc\plus{}2ca > a^2 \plus{} b^2 \plus{} c^2$

Hence proved.


trigonometry
function
ratio
Pythagorean Theorem
geometry

I wasn't really sure how to do this:
Express as an algebraic expression: $ Sin(Tan(x))$

What's an "algebraic expression"?  There really isn't much simplification to do.  Is one of those supposed to be an arc-trig function?

Then no.  One of those should be an arc-trig function.

I'll assume that the problem was $ \sin\tan^{\minus{}1}(x)$.

So the idea when you are doing these problems is to think of drawing a triangle and using that to simplify things. You have to be careful because the number that you plug into the $ \sin$, $ \cos$, or $ \tan$ is going to be the angle and the thing that you get out is going to be a ratio of sides of a triangle. (And if you use inverse functions, that switches).

So in this case lets call $ \theta\equal{}\tan^{\minus{}1}(x)$.

Now you should try and draw a triangle with an angle $ \theta$ and where $ \tan$ of that angle is $ x$ (the opposite side is $ x$ and the adjacent side is $ 1$, could I have chosen $ 2x$ and $ 2$? yes. why?). If this doesn't make sense, ask.

Now $ \sin$ is opposite over hypotenuse. The opposite side is $ x$. The hypotenuse can be calculated with the pythagorean theorem to get $ \sqrt{1\minus{}x^2}$. So we get $ \frac{x}{\sqrt{1\minus{}x^2}}$.

(what happens if $ x$ is negative? for your needs it probably doesn't matter but thats good to think about)

Altheman I think you meant the hypotenuse is $ \sqrt{1\plus{}x^2}$ to give $ \sin{\tan^{\minus{}1}(x)} \equal{} \frac{x}{\sqrt{1\plus{}x^2}}$


algebra
polynomial
number theory
prime numbers
number theory proposed

Hi,

Here you will find a primality test very simple.

S= n(n+1)(2n+1)/6 (sum of squares 1 to n)

S'=S-1

If n is not perfect square and if S' mod (n) = n - 1 then n is prime.

n=71
S'= 121835
121835 mod (71) = 70

You can check it for any prime number.

It always works.

Why? I do not know.

Try it please.

The condition you've written is equivalent to $ \frac {(n \plus{} 1)(2n \plus{} 1)}{6}$ being an integer, and it's trivial to find a non-prime such that this is true:  take, for example, $ n \equal{} 35$.

The true statement for $ n > 3$ is the [i]converse[/i]:  if $ n$ is an odd prime then $ (x \minus{} 1^2)(x \minus{} 2^2)...(x \minus{} (n \minus{} 1)^2) \equal{} \left( x^{\frac {n \minus{} 1}{2} } \minus{} 1 \right)^2$, and it's pretty obvious what the sum of the roots of this polynomial are $ \bmod n$.  Alternately, $ n \plus{} 1$ is even and $ n\plus{}1, 2n\plus{}1$ have different residues $ \bmod 3$.

Please check statements like these before posting them.

I'm sorry.
I checked more prime numbers than composite numbers.
But we can still use it to discard some composite numbers.
How many?
I do not know.

Thank you for your comment.

Look, this test doesn't do any better than checking for divisibility by $ 2$ and $ 3$: a necessary and sufficient condition is that $ n \equiv \pm 1 \bmod 6$.

Hi,

Instead of using the sum of squares I used the same formula above and the sum of power 4.
Same result : S' mod(p)=p-1

But for some values like 35 it did not work it give me 27 instead of 34.

So my question is :

What will happen if we use the sum of power 2^n with n bigger ?

I do not know how to sum i^(2^n) i varying for 1 to p.

Is there any formula to do it.

Example :

1^16 + 2^16 + 3^16 +.......+p^16 = ?

Thank you for any help.

Yes; it's called [url=http://mathworld.wolfram.com/FaulhabersFormula.html]Faulhaber's formula[/url], but it's not really necessary.  No test of this form will give you a primality test - again, it won't be any better than testing for divisibility.  You might be interested in examining the situations in which

$ 1^n \plus{} ... \plus{} (p\minus{}1)^n \equiv 0 \bmod p$

for number-theoretic interest, although again, [i]it still won't be a primality test.[/i]

Thanks a lot for the Faulhaber's Formula.
It will help to find something else I'm looking for.

10^10000000000000000  thank you!

[quote="t0rajir0u"]Yes; it's called [url=http://mathworld.wolfram.com/FaulhabersFormula.html]Faulhaber's formula[/url], but it's not really necessary.  No test of this form will give you a primality test - again, it won't be any better than testing for divisibility.  You might be interested in examining the situations in which

$ 1^n \plus{} ... \plus{} (p \minus{} 1)^n \equiv 0 \bmod p$

for number-theoretic interest, although again, [i]it still won't be a primality test.[/i][/quote]

It is not easy to find a composite number passing all the tests from power 2 to power 6.
I'm trying using Excel without result until now.
There is a way to check the primality of any number by incrementing the powers.
We test sum of pwer 2, then 3. then 4 and so on until it fails then we are sure that the number is composite.

Again:  [b]this test is just trial division in disguise, and is also much slower.[/b]  If I'm not mistaken, a necessary and sufficient condition for $ n$ to pass the test is that $ n$ isn't divisible by $ 2, 3, 5, 7$; in particular, try $ n \equal{} 121$.

If you like primality testing, are you familiar with [url=http://en.wikipedia.org/wiki/Solovay-Strassen_primality_test]the[/url] [url=http://en.wikipedia.org/wiki/Miller-Rabin_primality_test]standards[/url]?  And if you like primality tests with absurdly bad runtimes, try [url=http://en.wikipedia.org/wiki/Wilson%27s_theorem]Wilson's theorem[/url].

I tried for n=121 it works from 2 to 7 [u]but it did not work for power 8.[/u]

I want just to understand.
Until now I do not.

Do you understand either of the proofs I gave?  Try to adapt them and then try $ 13^2, 17^2, 19^2, 23^2, 29^2$ if you still think this is going to be efficient.  Sooner or later Excel will stop being able to handle the numbers you're using even though you'll still be in the range where [b]trial division works just fine.[/b]

I'm not looking for efficiency.
I'm just trying to understand why all the sum of any power minus 1 mod (p) are equal to p-1.
Thank you for all your comments.
I close this discussion even if I did not understand why it is working for all the prime numbers.


group theory
abstract algebra
superior algebra
superior algebra unsolved

What are all the subgroups of G, where G=the group of six permutations on {1,2,3}

G=the group of six permutations on {1,2,3} ==> one writes that down as $S_{3}$

ask urself the following question.
-what is the order of $S_{3}$
-if H is a subgroup of $S_{3}$: how are the order of H and $S_{3}$ related?
-How can I find a subset that satisfies the conditions to be a subgrp
-how many subgroups are there?
-do I have them all?
-why do I have them all?
-why did I forget the trivial subgroup?  :rotfl:

[quote="Jacobson"]
-why did I forget the trivial subgroup?  :rotfl:[/quote]

No, but you did forget the Improper subgroup  :rotfl:


geometry
circumcircle
geometric transformation
homothety
reflection
symmetry
geometry proposed

Given triangle $ ABC$, its centroid $ G$ and its symmedian point $ L$. Let $ A_1, B_1, C_1$ be the midpoints of $ BC, CA, AB$, respectively. $ (A_1,A_1G)\cap (B_1,B_1G)\equal{}\{G, A_2\}$. Similarly we define $ B_2, C_2$. Prove that the circumcenter of triangle $ A_2B_2C_2$ lies on $ GL$.

$ A',B',C'$ are the orthogonal projections of $ G$ on $ B_1C_1,$ $C_1A_1,$ $A_1B_1$ and $ L' \equiv X_{141}$ is the complement of $ L$ with respect to $ \triangle ABC,$ i.e.  the symmedian point of $ \triangle A_1B_1C_1,$ which lies on $ GL$ and satisfies $ LG \equal{} 2GL'.$ The six perpendicular feet from $ G,L'$ on $ B_1C_1,$ $C_1A_1,$ $A_1B_1$ lie on the pedal circle $ \mathcal{M}$ of the isogonal pair $ (G,L')$ WRT $ \triangle A_1B_1C_1$ and its center $ U$ is the midpoint of $ GL'.$ The homothety with center $ G$ and coefficient $ 2$ carries $ A',B',C'$ into the reflections $ A_2,B_2,C_2$ of $ G$ across the sidelines of  $ \triangle A_1B_1C_1,$ i.e.  $ \mathcal{M}$ is taken into $ \odot(A_2B_2C_2)$ and therefore its center $ U$ is taken into $ L'$ $ \Longrightarrow$ $ L'$ is the center of $ \odot(A_2B_2C_2).$

Nice problem dear Linh!

Actually $ A_2$ is reflection of $ G$ through $ B_1C_1$ similarly to $ B_2,C_2$, so $ AA_2,BB_2,CC_2$ are concurrent at $ B_G$-Begonia point of $ G$. We see an interesting result, $ B_G$ lies on $ GL$. This idea give us another problem as an extension

Given triangle $ ABC$ and $ A_1B_1C_1$ is cevian triangle of an arbitrary point $ P$. $ P^*$ is isogonal conjugate of $ P$. $ A_2,B_2,C_2$ are reflections of $ P$ through $ B_1C_1,C_1A_1,A_1B_1$, respectively. $ AA_2,BB_2,CC_2$ are concurrent at $ B_P$. Prove that circumcenter of triangle $ A_2B_2C_2$ lies on line $ P^*B_P$.

[quote="buratinogigle"]

Given triangle $ ABC$ and $ A_1B_1C_1$ is cevian triangle of an arbitrary point $ P$. $ P^*$ is isogonal conjugate of $ P$. $ A_2,B_2,C_2$ are reflections of $ P$ through $ B_1C_1,C_1A_1,A_1B_1$, respectively. $ AA_2,BB_2,CC_2$ are concurrent at $ B_P$. Prove that circumcenter of triangle $ A_2B_2C_2$ lies on line $ P^*B_P$.[/quote]
For convenience, I rewrite the problem as following :

Let $ \triangle A_1B_1C_1 $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ .
Let $ A_2, B_2, C_2 $ be the reflection of $ P $ in $ B_1C_1, C_1A_1, A_1B_1 $, respectively .
Let $ Q, R $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC, \triangle A_1B_1C_1 $, respectively .
Let $ \mathcal{B}_P $ be the Begonia point of $ P $ (ie. $ \mathcal{B}_P \equiv AA_2 \cap BB_2 \cap CC_2 $ ) .

Prove that $ \mathcal{B}_P, Q, R $ are collinear
____________________________________________________________
[b]Proof:[/b]

Let $ A_3, B_3, C_3 $ be the projection of $ P $ on $ BC, CA, AB $, respectively .
Let $ A_4, B_4, C_4 $ be the projection of $ P $ on $ B_1C_1, C_1A_1, A_1B_1 $, respectively .
Let $ O_3, O_4 $ be the center of $ \odot (A_3B_3C_3), \odot (A_4B_4C_4) $, respectively .

Easy to see $ O_3, O_4, A_4, B_4, C_4 $ is the midpoint of $ PQ, PR, PA_2, PB_2, PC_2 $, respectively .

Invert with center $ P $ and denote $ T' $ as the image of $ T $ .

Since $ P, A_1, A_3, B_4, C_4 \in \odot (PA_1) $ ,
so we get $ A_1', A_3', B_4', C_4' $ are collinear and $ PA_1' \perp B_4'C_4' $ .
Similarly, we can prove $ B_1', C_1' $ is the projection of $ C_4'A_4', A_4'B_4' $, respectively .
Since $ A_1(P, A_3; B_4, C_4 )=-1 $ ,
so $ PB_4A_3C_4 $ is a harmonic quadrilateral ,
hence we get $ A_3' $ is the midpoint of $ B_4'C_4' $ .
Similarly, we can prove $ B_3', C_3' $ is the midpoint of $ C_4'A_4', A_4'B_4' $, respectively .
Since $ P, B_3, C_3, A \in \odot (PA) $ ,
so $ A' $ is the projection of $ P $ on $ B_3'C_3' $ .
Similarly, we can prove $ B', C' $ is the projection of $ P $ on $ C_3'A_3', A_3'B_3' $, respectively .

Let $ A_5', B_5', C_5' $ be the reflection of $ P $ in $ B_3'C_3', C_3'A_3', A_3'B_3' $, respectively .

Since $ \mathcal{B}_P \equiv AA_2 \cap BB_2 \cap CC_2 $ ,
so $ \mathcal{B}_P' \equiv \odot (PA'A_2') \cap \odot (PB'B_2') \cap \odot (PC'C_2') $ ,
hence $ \mathcal{B}_P' $ is the midpoint $ P \mathcal{K}_P' $ where $ \mathcal{K}_P'=\odot (PA_4'A_5') \cap  \odot (PB_4'B_5') \cap \odot (PC_4'C_5') $ .
Since $ O_3' , O_4' $ is the image of $ P $ under the inversion WRT $ \odot (A_3'B_3'C_3'), \odot (A_4'B_4'C_4') $, respectively ,
so after homothety $ \mathcal{H}(P,2) \Longrightarrow $ we only have to prove $ O_3', O_4', P, \mathcal{K}_P' $ are concyclic .
(Notice that $ Q', R' $ is the midpoint of $ PO_3', PO_4' $, respectively )

Let $ H' $ be the orthocenter of $ \triangle A_4'B_4'C_4' $ .
Let $ A_6', B_6', C_6' $ be the projection of $ A_4', B_4', C_4' $ on $ B_4'C_4', C_4'A_4', A_4'B_4' $, respectively .

Since $ A_5' $ is the reflection of $ P $ in $ B_3'C_3' $ ,
so from symmetry we get $ A_6' \in \odot (PA_4'A_5') $ .
Similarly, we can prove $ B_6' \in \odot (PB_4'B_5') $ and $ C_6' \in \odot (PC_4'C_5') $ .

Consider the inversion $ \Phi $ with center $ H' $ which swap $ \odot (A_3'B_3'C_3') $ and $ \odot (A_4'B_4'C_4') $ .

Since $ A_4' \longleftrightarrow A_6' $ ,
so we get $ \odot (PA_4'A_5') $ is fixed under $ \Phi $ .
Similarly, we can prove $ \odot (PB_4'B_5'), \odot (PC_4'C_5') $ are fixed under $ \Phi $ ,
so we get $ P, \mathcal{K}_P' $ are the image of each other under $ \Phi $ and $ O_3', O_4', P, \mathcal{K}_P' $ are concyclic .

Q.E.D


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that for any polynomial of degree $n$ there exist $a_1,a_2,..a_n$ s.t. $Q(x)=\sum_{i=0}^Na_iX^{2^i}$ is divisible by $f$.

where is $f$ from? and $a_1,a_2,\ldots,a_n$ ?
thanks

I forgot that this is an algebra section.  :D There are real numbers, respectively , real polynomial.

isn't it $a_0+\displaystyle\sum_{k=1}^na_kX^{2^k}$? because if it is without $a_0$ and if I understood the problem correctly then I don't see the corresponding  polynomial for $X^2+X+1$...

I've edited it. The sum was from 0 , not from 1. Sorry! :blush:


I have done what seems like a million problems that require you to find the degree measure between the hour and minute hand is. Is there a formula or fast way to solve these kind of problems given the time (If so please post)? This applies to clocks with moving hour and minute hands.

Well, let the hour be x and the minutes be y.  At y minutes after the hour, the minute hand will be $\frac{y}{60}$  of the way around the clock and so the angle between 12:00 and the minute hand will be $6y$.  With the hour, using the same reasoning, we get that (if it didn't move) the hour hand would be $\frac{x}{12}$ of the way around the clock and so would have a degree measurement of $30x$.  Then we have
that every 60 minutes, the hour hand moves 5 minutes.  Thus the hour hand will have move 1 minute every twelve minutes, or the hour hand will move 6 degrees every 12 minutes (or 72 degrees). Then the degree is $(30x+\frac{y}{2})-6y$.  Then we must make some slight changes.  First, since the minute hand could be in front of the hour hand, we want $abs((30x+\frac{y}{2})-6y)$.  Finally, since the degree could be on either angle (on either side of the center of the clock), we know that the correct degree measurement (meaning the one that the problem is asking for) could be $abs((30x+\frac{y}{2})-6y)$ or $360-abs((30x+\frac{y}{2})-6y)$ (most problems will specify "the larger (or smaller) of the two angles."

I believe that works.  There may be a slight problem, so check it before you use it.  ;)

$\text{angle}=\text{time}\cdot 330^\circ$, where $\text{time}$ is in hours. After calculating this, you have to subtract $360^\circ$ as many times as possible. (Note: what this formula gives is the angle from the hour hand to the minute hand measured in clockwise direction.)

Example: $\text{time}=14^{20}=14{1 \over 3}={43 \over 3}$ hours. Then $\text{angle}={43 \over 3}\cdot 330^\circ=4730^\circ$. From this we subtract $360^\circ$ as many times as possible, and that's $13$, since $13\cdot 360^\circ=4680^\circ$. Hence, the answer is $50^\circ$.

Another example: $\text{time}=11^{50}=11{5 \over 6}={71 \over 6}$ hours. Then $\text{angle}={71 \over 6}\cdot 330^\circ=3905^\circ$. From this we subtract $360^\circ$ as many times as possible, and that's $10$, since $10\cdot 360^\circ=3600^\circ$. Hence, the answer is $305^\circ$.

I found some trouble with both methods: :huh:

1)samath's method seemed to work except when you plugged in a time like 1:55;
\[ \displaystyle abs((30(x)- \frac{y}{2})-6(y)) \Rightarrow \\\\ abs((30(1)- \frac{55}{2})-6(55) \Rightarrow \\\\ abs(\frac{115}{2} - 330) \Rightarrow \\\\ abs(\frac{-545}{2}) \Rightarrow\\\\ \frac{545}{2} = 272.5\\\\  \]
Since, this is no where near the estimation, so I subtracted it from 360
\[ \displaystyle 360-272.5=87.5  \]
But, that's too close to 90 sooooooooo, I am not sure......... :?

2)Farenhajt's method confuses me in that, does $14^{20}$ really mean  14^20? and how does that fit in? This turns out to be a huge number and doesn't seem to get anywhere. :wacko:

[quote="fireforce"]Farenhajt's method confuses me in that, does $14^{20}$ really mean  14^20? and how does that fit in? This turns out to be a huge number and doesn't seem to get anywhere. :wacko:[/quote]

Oh, sorry - from where i come, it's just usual notation for 14:20 (like 2:20pm) :)

First you figute our the distance in degrees between the clocks at XX:00 (where XX is a number, 01, 02, 03,....11,12) by 30(XX).

Then you find the amount of time the clock hand has travelled. And multiply if by 5.5. (Minute hand moves 6/min and hour hand moves 0.5/min so their diference is 5.5/min)

then you subtract that from what it was before.

I think that was pretty confusing so i'll do an example:

Find the time at 4:25

At 4:00, the time was $4\times30=120$ degrees

In 25 minutes, the min hand has moved towards/away from the hour hand $5.5\times25=137.5$ degrees

Therefore the distance between them is $\mid120-137.5\mid=17.5$


geometry
trigonometry
function
conics
Vieta
trig identities
Law of Sines

I have aops vol 1 and 2. I am currently in geometry class, but it turns out I am going to a tournament in April that is algebra 2 and under. Therefore, I will fail unless I learn algebra 2 until April 7.

What chapters would be helpful for competitive math in algebra 2 of both aops 1 and 2?

I also have an old algebra 2 textbook, but I don't know if that will help at all, as it is a school one.

I really need to prepare.

What the ... people, please.

Description for HSB says: [color=blue]For beginners to learn from more experienced problems solvers.[/color]

Description for Other PS Topics: [color=blue][b]Discuss resources[/b], teaching methods, current events in math, and other topics.[/color]

Post in the right forum.

I am a beginner, not an advanced problem solver.  :blush:

Well, I lost my copy of AoPS vol. I soon after completing it, and I've never taken a Algebra 2 course (skipped it and went straight to Precalc Honors), but I think Chapter 1 of vol. II should be good. It's about logs...you learn about those in Algebra 2, right?

[quote="Ignite168"]Chapter 1 of vol. 1 is about logs too ;)

Vol 2 covers everything vol 1 does but goes a lot more in depth, for example vol 1 gives you the definition of the basic trig functions ie. sin cos, etc. while vol 2 will give you identities and ways to manipulate and such. Of course vol 2 also covers a lot more things not touched on in vol 1 at all.[/quote]

so with alg ii and alg i

http://www.amazon.com/gp/reader/0764541358/ref=sib_dp_pt/104-4292984-4490300#reader-link
some random alg ii cliff note thing with it's table of contents
aops vol ii
http://www.artofproblemsolving.com/Books/Vol2/toc.pdf

basically tocs might help in your quest....
i wonder if bob miller has an algebra ii book?

EDIT BEFORE I LOG OFF!!!: http://www.xpmath.com/ebooks/math_ebooks.php which probably has been mentioned before but if you no longer go to the library like i do.

[quote="now a ranger"]what chapters in aops 2 are algebra 2? not precalculus[/quote]

I dunno, it seems like most of it is in Vol. I. The only part I can think of that's Algebra 2 is Chapter 1. You don't learn double-angle identities, the Law of Sines, Ptolemy's Theorem, conic sections, and Viete's formula in Algebra 2, do you? (That was stuff you learn in chapters 2, 3, 4, 5, 6)

[quote="Ignite168"]We did Law of Sines in geometry, Vieta's formulas in Algebra II, and possibly double-angle identities later this year? Conic Sections are definitely PreCalc though, no idea what Ptolemy's Theorem is.[/quote]
alg ii has almost no definition. i did conic sections.

Hmm, different classes may be different. For example, the rest of our Precalc class learned the Law of Sines for the first time last month.  :lol:


Ah, ok, I was trying to make it too complicated.


anyone ?

the message is too short, this is why it wont b e anymore

[quote="Lena"]why did you decide that those were the roots?[/quote]

I think it's because that-0.1<(3-x^2)/2x -1<0.1
=> 1-0.1< (3-x^2)/2x=f(x)<1+0.1


function
modular arithmetic
number theory
number theory proposed

Let $ m$ a positive integer and $ p$ a prime number, both fixed. Define $ S$ the set of all $ m$-uple of positive integers $ \vec{v} \equal{} (v_1,v_2,\ldots,v_m)$ such that $ 1 \le v_i \le p$ for all $ 1 \le i \le m$. Define also the function $ f(\cdot): \mathbb{N}^m \to \mathbb{N}$, that associates every $ m$-upla of non negative integers $ (a_1,a_2,\ldots,a_m)$ to the integer $ \displaystyle f(a_1,a_2,\ldots,a_m) \equal{} \sum_{\vec{v} \in S} \left(\prod_{1 \le i \le m}{v_i^{a_i}} \right)$.
Find all $ m$-uple of non negative integers $ (a_1,a_2,\ldots,a_m)$ such that $ p \mid f(a_1,a_2,\ldots,a_m)$.
[i](Pierfrancesco Carlucci)[/i]

[hide="Solution"]Let $ S_k \equal{} 1^{a_k} \plus{} 2^{a_k} \plus{} \ldots \plus{} p^{a_k}$. Observe that $ f(a_1, a_2, \ldots, a_m)$ factors as $ S_1 S_2 \ldots S_m$.

Thus, $ p | f(a_1, \ldots, a_m)$ if and only if $ p$ divides one of the terms in the sum. Clearly, if some $ a_k$ is 0, then $ p | S_k$. Thus, we may suppose each of $ a_1, a_2, \ldots, a_m$ are nonzero.

Lemma: If $ a_k > 0$, and $ p \minus{} 1 | a_k$, then $ p | S_k \plus{} 1$. Otherwise, if $ a_k > 0$ then $ p | S_k$
Proof: If $ p \minus{} 1 | a_k$, the result follows easily from Fermat's little theorem. Otherwise, let $ g$ be a generator mod $ p$. Observe that
$ S_k \equiv \sum{i\equal{}1}^{p\minus{}1} i^k \equiv \sum_{i\equal{}0}^{p\minus{}2} (g^i)^k \equiv \frac{(g^k)^{p\minus{}1} \minus{} 1}{g^k \minus{} 1} \pmod p$. Since $ g^k \not \equiv 1 \pmod p$ (since $ p \minus{} 1 \not | k$), and by Fermat's little theorem $ (g^k)^{p\minus{}1} \minus{} 1$ is a multiple of $ p$, $ S_k$ is a multiple of $ p$, as desired.

With the lemma, it follows that $ p | f(a_1, \ldots, a_m)$ if and only if some one of $ a_1, a_2, \ldots, a_m$ is 0 or not a multiple of $ p \minus{} 1$.

It is unfortunate that my final answer was incorrect for this problem... oh well. [/hide]


I wonder why are you giving a problem in the "unsolved section" naming it "really easy"...

some one told me it's really easy and i didn't solve it yet mahbub would U send your solution

well after taking sqrt(y) to rhs u get:
x=1376-2sqrt(1376)y+y

note x and y are natural number. but sqrt(1376) is not a full square number. hence u get a contradition.

Errr... you made a mistake.  The equation should be

\[x=1376-2\sqrt{1376}\sqrt{y}+y\]

not

\[x=1376-2\sqrt{1376}y+y\]

so there are solutions (one such solution is $(111,999)$).

I'm pretty sure the only positive integer solutions are $(111,999)$, $(444,444)$, and $(999,111)$, but I'm not sure how to prove it on the spot.

OOPS
 :oops_sign:

I will try it again! Just wait... :P

Nice one. Really nice one!

Not so easy as you were saying. And to dts, no those are not the only solutions.

Here is the solution:
$\sqrt{x}+\sqrt{y}=\sqrt{1376}$
$x=1376-2\sqrt{1376y}+y$
$x=1376-2\sqrt{2^5 43y}+y$

hence $y=2*43*k^2=86k^2$

Putting back,
$x=1376-688k+86k^2$

hence $x=86p$

Putting back,
$p=16-8k+k^2=(k-4)^2$

So
$x=86(k-4)^2$ and $y=86k^2$

The real game will start now. If you are simply going to put those values and solve for k then you will not be able to get the solutions!!!

What we have to do is this:

We will consider three cases:
1) $k<0$
2) $0 \leq k \leq 4$
3) $k>4$

1) Putting the values in the expression: -k+4-k=4 => k=0. Contradiction.
3) k+k-4=4 =>k=4 Contradiction.
2) k+4-k=4 => 4=4 true for all k.

Hence our solution is:
$x=86(k-4)^2$ and $y=86k^2$

For all $k$ such that $0 \leq k \leq 4$

Bingo!

:oops_sign: myself... That's what I get for trying to factor 1376 in my head... :fool:

I think it's as easy as Mathx said.

let $(x,y)=d,x=x'd,y=y'd$ therefore $\sqrt{x'}+\sqrt{y'}=\sqrt{\frac{1376}{d}}------>x'+y'+2\sqrt{x'y'}
=\frac{1376}{d}$ so x'y' has to be square of a number in$\mathbb{Q}$and(x',y')=1 so $\frac{1376}{d}$
is perfect square.

There is a similar AIME problem: Find all possible integers $n$ such that$\sqrt{n} +\sqrt{n+60} =\sqrt{m}$ for some non-square integer $m.$

err for the AIME problem i got:

[hide]
$n+60 = m - 2\sqrt{mn} + n$
$60 -  m = -2\sqrt{mn}$
$\displaystyle\frac{(m-60)^2}{4m} = n$
$\displaystyle\frac{m}{4} - 30 + \frac{900}{m} = n$

so from there i listed out all the factors of 900, eliminated the ones that were squares and less than 60, and the ones that werent divisible by 4

so only possible values of m are $60,180,$ and $300$

finding n for all of those, you see that when $m=60, n = 0$
similarly, $m=180, n=20$, $m=300, n=48$

so possible values of $n=0,20, 48$

[/hide]

very nice :surf: i like your solution muchhhhh more than the kalva one

[quote="Mathx"]find all answer of $\sqrt{x}+\sqrt{y}=\sqrt{1376}$ in $\mathbb{N}$[/quote]
16th IRMO (iran mathematica olympiad),  level one.
 it has 2 solutions: $(344,344);(86,744)$ look
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%5C%5Csqrt%7B1376%7D&t=35780]here[/url]


Find 2 four digit numbers whose product is $ 4^{8}\plus{}6^{8}\plus{}9^{8}$.


[quote="Evariste-Galois"]I just edited the inequality  :lol:

$ \text{Let}$ $ a,b,c$ $ \text{be positive real numbers Prove that: }$
$ \frac {a(a \plus{} b)}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b(b \plus{} c)}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c(c \plus{} a)}{(2c \plus{} a \plus{} b)(a \plus{} b)}\geq\frac {3}{4}$[/quote]
We have:
\[ \sum (\frac {a(a \plus{} b)}{(b \plus{} c)(2a \plus{} b \plus{} c)} \minus{} \frac {a \plus{} b}{4(b \plus{} c)})\]

\[ \equal{} \sum \frac {a \plus{} b}{b \plus{} c}(\frac {a}{2a \plus{} b \plus{} c} \minus{} \frac {1}{4})\]

\[ \equal{} \sum \frac {a \plus{} b}{b \plus{} c}.\frac {a \minus{} b \plus{} a \minus{} c}{4(2a \plus{} b \plus{} c)}\]
We 'll prove this ineq:
\[ \sum \frac {a \plus{} b}{b \plus{} c}.\frac {a \minus{} b \plus{} a \minus{} c}{2a \plus{} b \plus{} c} \geq 0\]

\[ < \equal{} > \sum (a \minus{} b)[\frac {a \plus{} b}{(b \plus{} c)(2a \plus{} b \plus{} c)} \minus{} \frac {b \plus{} c}{(a \plus{} c)(2b \plus{} a \plus{} c)}] \geq 0\]

\[ < \equal{} > \sum (a \minus{} b)[\frac {(a^2 \plus{} ab \plus{} bc \plus{} ca)(2b \plus{} a \plus{} c) \minus{} (b^2 \plus{} ab \plus{} bc \plus{} ca)(2a \plus{} b \plus{} c)}{(b \plus{} c)(a \plus{} c)(2a \plus{} b \plus{} c)(2b \plus{} a \plus{} c)}]\]

\[ < \equal{} > \sum (a \minus{} b)^2\frac {ab \plus{} (a \plus{} b)(a \plus{} b \plus{} c) \minus{} ab \minus{} ac \minus{} bc}{(b \plus{} c)(a \plus{} c)(2a \plus{} b \plus{} c)(2b \plus{} a \plus{} c)} \geq 0\]
It is true :D

[quote="Evariste-Galois"]I just edited the inequality  :lol:

$ \text{Let}$ $ a,b,c$ $ \text{be positive real numbers Prove that: }$
$ \frac {a(a \plus{} b)}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b(b \plus{} c)}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c(c \plus{} a)}{(2c \plus{} a \plus{} b)(a \plus{} b)}\geq\frac {3}{4}$[/quote]

Let $ a \plus{} b \plus{} c \equal{} 1$

$ f(x) \equal{} \frac {x}{1 \minus{} x^2}$ - convex and increase $ (0,1)$

$ \Leftrightarrow (1 \minus{} c)f(a) \plus{} (1 \minus{} a)f(b) \plus{} (1 \minus{} b)f(c) \ge 2f(\frac {1 \minus{} ab \minus{} bc \minus{} ca}{2}) \ge 2f(\frac {1}{3}) \equal{} \frac {3}{4}$


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Here are some useful formulas on top of my head.

1+2+3+....n=n(n+1)/2
Sum of the first n odd numbers= n^2
1+3+5+.....n=2n-1= n^2

Got any more useful formulas?

There are TONS of formulas?  What kind do you want?

An example, the sum of the factors of a number $p_1^{i_1}\cdot p_2^{i_2}$...

is $(p_1^0+p_1^1+...+p_1^{i_1})(p_2^0+p_2^1+...+p_2^{i_2})$...

If you multiply it out, you will see that every number is created in the prime factorization of the number, then they are added.

Useful formulas in MathCounts that will almost certainly be on a test.

Your second formula doesn't make sense. What it should be is $1+3+5+...+(2n-1)=n^2$.

Other ones that might show up are:

$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$
$1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2$
If $n=p_1^{e_1}p_2^{e_2}...p_n^{e_n}$, then the number of positive divisors of $n$ is $(e_1+1)(e_2+1)...(e_n+1)$.
The number of times which a number $n$ (generally a prime) appears in a factorial is found by dividing the factorial by $n$, then dividing the quotient by $n$, and repeating until the quotient is too small to divide again, then adding up each quotient. For example, how many $5$'s are found in $236!$? $\frac{236}{5}\Rightarrow 47$, $\frac{47}{5}\Rightarrow 9$, $\frac{9}{5}\Rightarrow 1$, $47+9+1=\boxed{57}$ $5$'s appear in $236!$.

Number of rectangles you can make in a x by y grid
(1+2+3+...x)(1+2+3+...y)

Greatest number you can't get by adding two relatively prime numbers m and n
mn-m-n

There are a lot here:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28130&postorder=asc&start=180]www.artofproblemsolving.com/Forum/viewtopic.php?t=28130&postorder=asc&start=180[/url]

Don't ask me how I found out about that.  :lol:

Well, make sure you understand these!!!

Often, the MATHCOUNTS people are evil and twist up the problems so you really need to know the derivation.

Try this:  Prove that any orthodiagonal quadrilateral has an area of $\frac{d_1*d_2}{2}$.  Also: Prove that for all rhombi (often easier in countdown).

yes, yes it is really important to know the derivation of the formula and why they work.
so you can manipulate them if necessary and it REALLY helps, trust me.

If a question asks how many total handshakes will be in room if everybody shakes once and there is 10 people you'll need to know this formula. n(n-1)/2. Therefore there are 10(10-1)/2 handshakes which equals 45.

knowing how to calculate degrees in an arc might help... :|

To find the area on a geoboard.

Picks Formula= [color=red]B+I/2-1[/color]
B=Border Dots(exactly on line)
I=Interior Dots(exactly inside)

[quote="b-flat"]On how many rectangles are in an $x$ by $y$ grid...

There are ${x\choose2}\cdot{y\choose2}$ total rectangles.

This is assuming there are $x$ vertical lines and $y$ horizontal lines, or vise-versa.[/quote]

I dont think this formula works. Biffanddoc's formula (1+2+3+...X)(1+2+3+...Y) is correct.

[quote="jhredsox"]huh, what do you mean
-jorian[/quote]

Huh? what do you mean? :maybe:

[quote="13375P34K43V312"]1001=7*11*13, and multiplying a 3-digit number abc by that results in abcabc. I agree with bpms, everyone should know why that works.[/quote]
I was actually referring to the formula for the sum of divisors, but whatever.

To williamshakespeare:

If you still don't get the sum of factors formula, download this. Credit frost13(Mr. Frost) for making it. :D

[quote="jhredsox"][quote="jhredsox"][quote="13375P34K43V312"]Here's an example without the solution right underneath:

Find the sum of the factors of 6006.[/quote]

2,3003
3,1001
11,91
7,13
13,1

$3*4*12*8*14=144*112=16128$

-jorian[/quote]

wait. so was this correct?

-jorian hoover[/quote]

The prime factorization of 6006 is $2*3*7*11*13$. Using the formula, it would be $(2^{0}+2^{1})(3^{0}+3^{1})(7^{0}+7^{1})(11^{0}+11^{1})(13^{0}+13^{1})=16128$ so yes you are correct.

[quote="mysmartmouth"]Hm, most likely if you need a median formula for MathCounts, it is this one:

In a right traingle, the median to the hypotenuse is half the length of the hypotenuse.

This is a more complex formula, with a proof beyond MathCounts.  You won't need it for MathCounts, but it can help you in a tight spot.  I wouldn't reccommend relying on this, however, because all you can do is memorize it:

Median to side $c$ of triangle $ABC$:

$m_{c}= \sqrt{\frac{a^{2}}{2}+\frac{b^{2}}{2}-\frac{c^{2}}{4}}$

EDIT: I'm a bit confused about the quarter-circle formula.  Are we saying that R is the radius of the sector, and r is the radius of the inscribed circle, and $r = R\sqrt{2}-R$?  Because that doesn't make sense...r would be bigger than R...I must be understanding wrong.[/quote]
Did you get this from stewart's theorem by letting m=n and a=2m?

It's a theorem involving the length of a cevian given the sides of the triangle and the ratio in which the cevian cuts the base, but it doesn't appear on MathCounts. It's more of a trig theorem, even though there's no trig in the actual formula.

I just found another purely synthetic way to prove the formula for the length of a median.
[hide="an outline of the proof"]I won't post the algebra at each step since it's a little messy.
Let the triangle have sides a, b, and c. WLOG assume the median is hitting side a. Use heron's to find the area (heron's can be proven with only synthtic geometry). Find the length of the altitude. Find the distance from where the altitude hits side a to the nearest vertex. Now subtract that from half the length of a to find the length of the little segment connecting the point where the median hits the side to where the altitude hits the side. Then use pythagorean with the legs being that little segment and the altitude and the hypotenuse being the median.

Like I said, it's messy :D.[/hide]

[quote="pianoforte"][quote="Phelpedo"]The best way to find medians is to rotate the triangle 180 degrees to form a parallelogram.[/quote]
[hide]WLOG, let the median, m, be to side c. Draw lines through A and B, parallel to a and b respectively, and let them intersect at D. This makes a parallelogram, ACBD. 2m and c are diagonals of ACBD. For a parallelogram, the sum of the squares of the sides=the sum of the squares of the diagonals (i think), so $2(a^{2}+b^{2})=(2m)^{2}+c^{2}\implies m=\sqrt{\frac{a^{2}}{2}+\frac{b^{2}}{2}-\frac{c^{2}}{4}}$.[/hide][/quote]
don't you need to prove that the sum of the squares of the sides=the sum of the squares of the diagonals and that the diagonals bisect each other? I think they take methods really complicated, but maybe not. Can you outline their proofs, please? Anyways, my proof is 10x as messy so...

k. First, you don't need to prove anything in mathcounts. But if this were on the USAMO or something, no you do not need to prove that the diagonals bisect eachother in a parallelogram. As for the sum of the squares of the sides etc., it's a well known fact, but if you want to prove it use law of cosines and it pretty much just falls out.


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algebra unsolved

For every positive integer $x$ we define $S(x)$ in this form:

$S(x)={\{a_n\mid a_0=x , a_{n+1}=a_n!+(n+2)!  , n\geq 0}\}$ .Prove that there exist

a sequence $(x_k)_{k\geq 0}$ s.t. $\bigcup_{k \geq 0}S(x_k)=\mathbb{N}$and for

every $k$ is not equal to $j$  , $S(x_k)\cap S(x_j)=\emptyset$

I know it's a little bit off topic, but could you tell us what olympiad that is, ehsan2004? I have never heard of "Bamayari".

After Fishagoures and Batlamious my best guess is Bay Area MO ;)

According to your statement, I assume that $\mathbb {N}$ does not contain $0$ am I wrong?

First note that for a given positive integer $x$ the corresponding sequence $(a_n)$ is strictly increasing, and that $a_{n+1} \geq 1 + (n+2) = n+3.$ (*)

Now, define the sequence $(x_i)$ as follows :
$x_0 = 1$ and for each $i$, $x_{i+1}$ is the least positive integer (if any) which does not belong to $S(x_0) \cup \cdots \cup S(x_i).$

The sequence $(x_i)$ is therefore strictly increasing and starts with $x_0=1,x_1=2,x_2=5, \cdots$
Moreover, the construction implies obviously that any positive integer will appear at least once in the union of the $S(x_i)$.

Now assume that there exist distinct $x_i$ and $x_j$ such that $S(x_i)$ and $S(x_j)$ are not disjoint. For convenience, we will denote by $(a_n)$ and $(b_n)$ the corresponding sequences of $x_i$ and $x_j$ respectively.
Thus, there exists $n,p$ such that $a_n = b_p.$ Wlog, we may assume that $n+p$ is minimal.
Thus, assumming that $n,p \geq 1$, it follows that $a_{n-1} \neq b_{p-1}$ and $a_{n-1}! + (n+1)! = b_{p-1}! + (p+1)!.$ (**)
Wlog, we may assume that $n \leq p.$ The above equality leads to $a_{n-1} > b_{p-1}.$ And then to $n < p.$
If $b_{p-1} \geq n+2$ then $\frac {b_{p-1}!} {(n+1)!} + \frac {(p+1)!} {(n+1)!} -\frac {a_{n-1}!} {(n+1)!} = 1 $ where each of the three first terms is a integer divisible by $n+2$. A contradiction.
Thus $b_{p-1} \leq n+1.$
Thus, using (*), we deduce that if $p -1 \geq 1$ we have $a_{n-1} > b_{p-1} \geq p+1$ so that $a_{n-1}! + (n+1)! > (p+1)! + b_{p-1}!$, which contradicts (**).

Therefore $p = 1$ and $n=0$, but if $i>j$ then $a_0 = x_i$ cannot be equal to any integer already reached by the $S(x_k)$ by our definition of the sequence $(x_n)$, and in particular $a_0 \neq b_1.$
And if $i < j$ then $a_0 = x_i < x_j < b_1$.
Thus in any case, we have a contradiction, and the result is proved.

Pierre.

Hmm.. I'm wondering if the question is also asking for a proof that the sequence $(x_k)$ is infinite, but this last result holds clearly by 'density' argument ($a_n > n!$).

Anyway, Enescu, I thought that Bay area olympiad started in 1999  ;)

Pierre.

thanks pbornsztein , your solution is correct and nice  ;)

[quote="pbornsztein"]
Anyway, Enescu, I thought that Bay area olympiad started in 1999  ;)
[/quote]Right! I found the origin of the problem. It is an international competition held in the Romanian town Baia Mare in september 1997. It was the second (and last) edition of a contest named "Unesco" competition, initiated by the Romanian Unesco Committee in 1996, to celebrate 50 years of existence.
I attended the first competition in 1996. Here's a nice problem proposed then by Mihai Baluna:

Let $p\ge 2$ be an integer and $f$ be a polynomial with integer coefficients such that $\sqrt[p]{f(0)},\sqrt[p]{f(1)},\sqrt[p]{f(2)},\ldots$ are rational numbers. Prove that there exists a polynomial $g$ with integer coefficients such that $f=g^p$.

[quote="enescu"]
Let $p\ge 2$ be an integer and $f$ be a polynomial with integer coefficients such that $\sqrt[p]{f(0)},\sqrt[p]{f(1)},\sqrt[p]{f(2)},\ldots$ are rational numbers. Prove that there exists a polynomial $g$ with integer coefficients such that $f=g^p$.[/quote]

Classical Harazi's type problem  :D

Pierre.

That was exactly my thought when I saw it :).

Anyway, let $f=g_1^{a_1}g_2^{a_2}\ldots g_k^{a_k}$ be a decomposition of $f$ into irreducible integral polynomials of degree $\ge 1$. We may assume WLOG that $0<a_i<p$ for all $i$. The hypothesis says that $f(n)$ is a $p$'th power of an integer for all integral $n$ (well, it says less than that, but this can easily be deduced).

Suppose we prove that there are infinitely many primes $q$ for which there is some $n$ s.t. $q|g_1(n),\ q^2\not|g_1(n)$ (written as $q\|g_1(n)$ from now on). Then we can prove that such a prime $q$ exists with the additional property that $q\not|g_j(n),\ \forall j\ge 2$, because for every $j\ge 2$ we can find integers $\alpha_j,\beta_j,\gamma_j$ s.t. $\alpha_jg_1+\beta_jg_j=\gamma_j$, and all we need to do is choose a prime $q\|g_1(n)$ among the infinitely many which do not divide any of the $\gamma_j$'s. Now, having found such a prime, we conclude that, since its exponent in $f(n)$ is at least $p$, we must have $a_1\ge p$, so we get a contradiction.

All that's left now is to prove this:

$(*)$ Given an irreducible integral polynomial $g$, there are infinitely many primes $q$ for which there exists some $n$ (depending on $q$) s.t. $q\|g(n)$.

I'll assume it well-known that there are infinitely many primes among the prime factors of the numbers $g(n)$. Since $g$ is irreducible, it's coprime to its derivative, so we can find integers $\alpha,\beta,\gamma$ s.t. $\alpha g+\beta g'=\gamma$. Now pick an $n$ s.t. there is a prime factor $q$ of $g(n)$ which does not divide $\gamma$. Then $q\not|g'(n)$. For any integral $k$, we have the following relation modulo $q$: $\frac{g(n+kq)}q=\frac{g(n)}q+kg'(n)$, and since $q\not|g'(n)$, we can choose $k$ s.t. $q\not|\frac{g(n+kq)}q$, meaning that $q\|g(n+kq)$, as desired. Since this can be done for infinitely many primes $q,\ (*)$ is proven.


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I am looking for a solution to this.
Let D be a division ring and let a belong to D. If a has finitely many conjugates show that it is in the center of D.

Let $G$ be the subset of $D$ of elements that commute with $a$. It is immediate that $G$ is a division ring. Also, $G$ contains the center $k$ of $D$.
$D^{*}$ acts trasitively by conjugation on the finite set of conjugates of $a$. The stabiliser of $a$ by this action is $G^{*}$. We deduce that $G^{*}$ is a subgroup of finite index in $D^{*}$. Let $(g_{i})_{i=\overline{1,n}}$ represent $D^{*}/G^{*}$ i.e. $D^{*}=\sqcup_{i=1}^{n}g_{i}G^{*}$ (disjoint union).
Let $G_{i}=g_{i}G$ for all $i$. Since $G$ is a division ring and a vector space over $k$, the $G_{i}$ are all subvector spaces of $D$ seen as a vector space over its center (multiplication by $g_{i}$ is a vector space endomorphism of $D$). Moreover the intersection of any two distinct $G_{i}$ is $0$.

Assume that $a$ is not central. Then $G\neq D$. Using the translation $x\to g_{i}x$ (or the disjoint union above), we see that this is equivalent to $G_{i}\neq D$ for any $i$.

So what we assume is that $D$, as a vector space over $k$, is the union of finitely many ($n$) proper sub-vector spaces any two of which have $0$ intersection. Clearly this implies $n>1$.

Assume $k$ is infinite. Let $x$ and $y$ be nonzero elements of $G_{1}$ and of $G_{2}$. Because of the mutual zero intersection, $x$ and $y$ are linearly independent. For infinitely many $\alpha\in k$ we have $x+\alpha\cdot y$ belongs to $G_{l}$ for some $l\in\overline{1,n}$. Take two such different and non-zero elements of $k$: $\alpha$ and $\alpha'$. Then $(x+\alpha y)-(x+\alpha' y)\in G_{l}$ because $G_{l}$ is a vector space. Since $\alpha\neq\alpha'$ we get $y\in G_{l}$. Because $G_{2}\cap G_{l}=0$ if $l\neq 2$, and because $y\neq 0$, we get $l=2$. But just as well we could have interchanged $x$ with $y$ and $\alpha,\ \alpha'$ by their multiplicative inverses and got $l=1$. Or look at $x+\alpha y\in G_{2}$ for infinetely many (actually at least two or at least one nonzero) $\alpha$ and at $y\in G_{2}$ to see that $x\in G_{2}$ which contradicts $G_{1}\cap G_{2}=0$.
We have our contradiction in case $k$ is infinite.

If $k$ is finite the previous algorithm does not work just as easy, but we can smooth thigs up.
First, $D$ is infinite and infinite dimensional over $k$ otherwise $D=G=G_{i}$ because $D=k$ by Wedderburn's "every finite division ring is a field".
Since the $G_{i}$ are in bijection to each other by multiplicative left translations, we see that all $G_{i}$ are infinite dimensional over $k$.
Let $y_{1},\ldots, y_{m}$ be linearly independent elements of $G_{2}$ for some $m>\max\{|k|,n\}$ and take $x\in G_{1}$, $x\neq 0$. Then for at least two distinct elements $\alpha=(\alpha_{1},\ldots,\alpha_{m})$ of $k^{m}$ we have that $x+\sum_{j=1}^{m}\alpha_{j}y_{j}\in G_{l}$ for some $l\in\overline{1,n}$. Taking their difference we find that $l=2$. Since all $y_{j}$ belong to $G_{2}$ and $x+\sum_{j=1}^{m}\alpha_{j}y_{j}\in G_{2}$, we get that $x\in G_{2}$, so $G_{1}$ and $G_{2}$ have nonempty intersection. Again a contradiction.

I hope everything is ok with this proof.


If A1,A2,A3 are the angles of a triangle with sides a1,a2,a3 inradius r,
area K and semi-perimeter s .Prove the following inequalities :
(a) (A1a1)*(A2a2)*(A3a3) >=(2*pi/3)^2 *rK
(b) (pi-A1)a1*(pi-A2)a2*(pi-A3)*a3>=(4pi3/9)^3*sK

hello, do you mean
a) $\alpha a\cdot \beta b\cdot \gamma c\geq \left(\frac{2\pi}{3}\right)^2K\cdot r$
b) $(\pi-\alpha)a(\pi-\beta)b(\pi-\gamma)c\geq \frac{4\pi^3}{9}sK$?
Sonnhard.


[quote="Interval"]A standard deck of 52 cards is shuffled and one card is drawn.  What is the probability that the card is:

a.  a black card?

b.  a king?

c.  an eight of spades?

NOTE: Whoever replies, keep in mind that I know nothing about deck of cards.[/quote]

There are always 26 black and red cards in a standard deck.

[hide="a"]Black card is $\frac{1}{2}$.[/hide]

[hide="b"]There are always 4 kings in a standard deck(one for every suit). So its $\frac{1}{13}$.[/hide]

[hide="c"]Theres  only one eight of spades so its $\frac{1}{52}$.[/hide]

Standard deck of cards = 52 cards: 13 spades (black), 13 hearts (red), 13 clubs (black), 13 diamonds (red). Each suit has 2-10, J, Q, K, A.

[quote="moogra"][quote="Interval"]A standard deck of 52 cards is shuffled and one card is drawn.  What is the probability that the card is:

a.  a black card?

b.  a king?

c.  an eight of spades?

NOTE: Whoever replies, keep in mind that I know nothing about deck of cards.[/quote]

[hide="a"]
There are 26 black cards in a deck. So it's $26/52=1/2$
[/hide]
[hide="b"]
There are 4 kings in a deck. So it's $4/52=1/13$
[/hide]
[hide="c"]
There is only 1 eight of spades in a deck. So it's $1/52$[/hide]
Where do you get all these problems? You post like 10 threads a day[/quote]

hes studying to pass the math section of a teacher's exam or something...

I like every answer given and every method used.


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Solve this with x,y are the integer numbers:
$2x^{2}+9y^{2}-6xy-6x-12y+2004 = 1995$

Come on guys, this is not an ugly problem... You can easyley reduce it

[hide]EDIT: rubbish... as i expected  :lol: [/hide]

I'm in a weird situation. I don't seem to see any mistakes in the above solution, but I don't believe in it, it seems too 'cheap'  :lol:
Could someone verify please ?

[quote="i_like_pie"][quote="Phelpedo"]Eww, yucky!!!

I say do a suitable transformation to get rid of the $6xy$ and then you just have a simple equation for an ellipse, or hyperbola, or whatever it's called...[/quote]
I don't feel like doing that equation. :ninja:

I think you mean parabola.[/quote]
It's an hyperbola with rotated axes. I am however disinclined to solve it.

[quote="Jan"].

Get back to the first equation: \[(x-3y)^{2}+(x-3)^{2}= 12y\] Now, clearly $x-3y \equiv x-3 \equiv 0 \pmod 4$, because LHS must be divisible by $4$.
[/quote]

Jan,
$x-3y \equiv 2 \pmod 4$ and/or  $x-3 \equiv 2 \pmod 4$ is equally well possible, unless you ruled out this case some way or another.

Exactly! That's what I was looking for, thank you very much B23.

[quote="B23"]You can make it yourself a LOT easier if you write the original equation as:
$(x-3y+2)^{2}+(x-5)^{2}=20$
And there [b]is[/b] a solution (only one): x=9 and y=3  :wink:[/quote]

Thank you!  I was looking for precisely such a form, but didn't think to add a $2$ into my $(x-3y)^{2}$ term.   :oops:


probability
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probability and stats

A and B decide to meet at a cinema hall between 2 pm to 3 pm . it was decided by them that who so ever comes first will wait for the other for  exactly 15 mins before leaving.What is the probability that they meet?

Please solve in all possible ways(geometrical,non-geometrical) and can someone provide a link to such problems with theory :)

I'm not sure about links but this doesn't seem to be something requiring any theory. The implicit assumption seems to be that the arrival times are independent and uniformly distributed in the 1-hour time interval in question. Then the joint density function is just the characteristic function of the unit square and the question reduces to finding the area of the set $ \{(x,y): x,y\in[2,3],|x\minus{}y|<\tfrac 14\}$. If you want, you can set it as double integral but it is easier just to find the areas of 2 triangles you should drop:
[asy]size(70);
path P=(0,0)--(1,0)--(4,3)--(4,4)--(3,4)--(0,1)--cycle;
fill(P,yellow);
D((0,0)--(4,0)--(4,4)--(0,4)--cycle,linewidth(1));
D((0,0)--(4,4),dashed);[/asy]
As an extra exercise, try to find out what happens if the first person arrives at a randon time between $ 2$ and $ 2: 45$ and the second at the random time between $ 2: 15$ and $ 3$.

can u please explain those assumptions?
what do u mean by their arrival time is 'uniformly' distributed

Exactly what I said: [url=http://mathworld.wolfram.com/UniformDistribution.html]uniformly distributed[/url]. In other words, the probability that the arrival time is in a given interval (or, more generally, Borel set) $ E$ is proportional to the length of $ E$. So, if a random variable $ \xi$ is uniformly distributed on the interval $ [a,b]$ and $ E\equal{}[c,d]\subset [a,b]$, then $ P(\xi\in E)\equal{}\frac{d\minus{}c}{b\minus{}a}$. Are you familiar with this notion at all? And how about "independence"? Should I explain it too?

thank u ...u need not explain the independence  :)  i only meant the uniform distribution  :)

The graphical approach works well here because the probability that the two people arrive at any one time is equal throughout. If it were not so, then perhaps something like a weighted graph will have to be used.

And Pardesi, A [url=http://www.mathlinks.ro/viewtopic.php?p=1073742#1073742]bird in the hand[/url] is worth two in a bush. (But I shouldn't have given this away :()

One more question for you:

Consider the cartesian plane $ R^2$, let X denote the subset of points for which both co-ordinates are integers. A coin of diameter 1/2 is tossed randomly onto the plane. Find the probability that the coin covers a point of X.

[quote="VARUNARASU"]i dunno abt any links, pardesi. but i have seen the same problem and also the one shre posted in interactive algebra by goyal(arihant publications). many such problems are actually there in that book.[/quote]

No only 3-4. And one of them have been posted by Pardesi (above) and 2 by me.


trigonometry
quadratics
function
algebra
system of equations
quadratic formula

Solve in the set of real numbers the following system of equations:

\[ \begin{cases}
\sqrt{y^2\minus{}7\plus{}\sqrt{x\plus{}y^2\minus{}7}}\equal{}x\\
\sqrt{x^2\plus{}2\plus{}\sqrt{y\plus{}x^2\plus{}2}}\equal{}y
\end{cases}\]

Another problem:
Find all real solutions to the equation:

$ 2\cos x \equal{} 2^x \plus{} 2^{\minus{}x}$

[quote="tonypr"]Solve in the set of real numbers the following system of equations:
\[ \begin{cases} \sqrt {y^2 \minus{} 7 \plus{} \sqrt {x \plus{} y^2 \minus{} 7}} \equal{} x \\
\sqrt {x^2 \plus{} 2 \plus{} \sqrt {y \plus{} x^2 \plus{} 2}} \equal{} y \end{cases}
\]
Another problem:
Find all real solutions to the equation:

$ 2\cos x \equal{} 2^x \plus{} 2^{ \minus{} x}$[/quote]

[b]1)[/b] $ \begin{cases} \sqrt {y^2 \minus{} 7 \plus{} \sqrt {x \plus{} y^2 \minus{} 7}} \equal{} x \\
\sqrt {x^2 \plus{} 2 \plus{} \sqrt {y \plus{} x^2 \plus{} 2}} \equal{} y \end{cases} \Rightarrow \begin{cases} (\sqrt {x \plus{} y^2 \minus{} 7})^2  \plus{} \sqrt {x \plus{} y^2 \minus{} 7} \equal{} x^2\plus{}x \\ (\sqrt {y \plus{} x^2 \plus{}2})^2  \plus{} \sqrt {y \plus{} x^2 \plus{}2} \equal{} y^2\plus{}y \end{cases}$

Then by quadratic formula, $ \begin{cases} x\plus{}y^2\minus{}7\equal{}x^2...........(1) \\y\plus{}x^2\plus{}2\equal{}y^2.........(2) \end{cases}$ (bacause $ x,y$ are non-negative).

$ (1)\plus{}(2)$ we get $ x\plus{}y\equal{}5$, then from $ (1), x\minus{}7\equal{}(x\plus{}y)(x\minus{}y)\equal{}5(x\minus{}y) \Rightarrow 4x\minus{}5y\plus{}7\equal{}0 \Rightarrow (x,y)\equal{}(2,3)$ as the only solution after checking.

[b]2)[/b] $ 2 \cos x\equal{}2^x\plus{}2^{\minus{}x}\equal{}(2^{\frac{x}{2}}\minus{}2^{\minus{}\frac{x}{2}})^2\plus{}2 \ge 2 \ge 2 \cos x$

Equality holds $ \Rightarrow x\equal{}0$ is the only solution.

I have a question: in an olympiad how would you justify that $ x\equal{}\sqrt{u\plus{}\sqrt{u\plus{}x}}\implies x\equal{}\sqrt{u\plus{}x}$

[quote="mathwizarddude"]I have a question: in an olympiad how would you justify that $ x \equal{} \sqrt {u \plus{} \sqrt {u \plus{} x}}\implies x \equal{} \sqrt {u \plus{} x}$[/quote]

Square both sides of the equation and then add $ x$ to both sides, we will have

$ x^2\plus{}x\equal{}(\sqrt{u\plus{}x})^2\plus{}(\sqrt{u\plus{}x})$, now let $ \sqrt{u\plus{}x}\equal{}y$ be the variable and $ x$ fixed we get

$ y^2\plus{}y\minus{}(x^2\plus{}x)\equal{}0$ , using quadratic formula we get $ y\equal{}\frac{\minus{}1 \pm \sqrt{4x^2\plus{}4x\plus{}1}}{2}\equal{}x$ or $ \minus{}1\minus{}x$ , as both $ x$ and $ y$ are positive we reject $ \minus{}1\minus{}x$.

Yes, of course this can be easily shown for only 2 iterations, but what if we have say $ n$ of them, such that

$ x=\sqrt{u+\sqrt{u+\sqrt{u+...\sqrt{u+x}}}}$ where the iterations takes place $ n$ times?

I'm pretty sure this will reduce to $ x=\sqrt{u+x}$ but how would you prove it in an olympiad?

[quote="mathwizarddude"]Yes, of course this can be easily shown for only 2 iterations, but what if we have say $ n$ of them, such that

$ x = \sqrt {u + \sqrt {u + \sqrt {u + ...\sqrt {u + x}}}}$ where the iterations takes place $ n$ times?

I'm pretty sure this will reduce to $ x = \sqrt {u + x}$ but how would you prove it in an olympiad?[/quote]

$ x>\sqrt{u+x} \Rightarrow x = \sqrt {u + \sqrt {u + ... \sqrt {u +\sqrt {u +x}}}}$ $ <\sqrt {u + \sqrt {u + ... \sqrt {u+x}}}$ $ <....<\sqrt{u+x}$, contradiction.

$ x<\sqrt{u+x} \Rightarrow x = \sqrt {u + \sqrt {u + ... \sqrt {u +\sqrt {u +x}}}}$ $ > \sqrt {u + \sqrt {u + ... \sqrt {u+x}}}$ $ >....>\sqrt{u+x}$, contradiction.

So $ x=\sqrt{u+x}$ is the only solution.

[quote="mathwizarddude"]I have a question: in an olympiad how would you justify that $ x \equal{} \sqrt {u \plus{} \sqrt {u \plus{} x}}\implies x \equal{} \sqrt {u \plus{} x}$[/quote]

One way of seeing this is if $ f(x)$ is an increasing function then $ f(f(x)) \equal{} x$  and  $ f(x) \equal{} x$ are equivalent. So in this case setting $ f(x) \equal{} \sqrt {u \plus{} x}$, you get that $ f(f(x)) \equal{} \sqrt {u \plus{} \sqrt {u \plus{} x}} \equal{} x$ so from the definition, $ \sqrt {u \plus{} x} \equal{} x$.

Demonstration the if $ f(x)$ is an increasing function then $ f(f(x)) \equal{} x$ and $ f(x) \equal{} x$ are equivalent:

If $ x_1$ is a solution to the equation $ f(x) \equal{} x$ then $ f(f(x_0)) \equal{} f(x_1) \equal{} x_1$, then $ x_1$ is also a solution to the second equation. Now let $ x_1$ be such that $ f(f(x_1)) \equal{} x_1$ Suppose $ f(x_1)\not \equal{} x_1$. If $ f(x_1) > x_1$, then given that $ f(x)$ is increasing, $ f(f(x_1))\ge f(x_1) > x_1$, which contradicts $ f(f(x_1)) \equal{} x_1$. The same method is used to show if $ f(x_1) < x_1$.

Corolary: If $ f(x)$ is an increasing function, then the equations
\[ f(x) \equal{} x \text{ and } \underbrace{f(f(\cdots f(}_{\text{k times}}x)\cdots )) \equal{} x
\]
Are equivalent. .. the proof is left as an excercise to the reader.

Edit: stephencheng beat me again, but this post shows for the general problem.


inequalities
geometry
articles
number theory

hi
i'm new here
i wanted to know where i can find olympiad level math text on the internet
basically i was looking for something on inequalities and number theory.
thank to all of u in advance.

udi

Go on google and type the following:

Inequalities filetype:pdf
Number Theory filetype:pdf
Euclidean Geometry filetype:pdf

Enjoy.

Have you tried the articles on this page?  http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php

Also, try using the Google Scholar service. Some of the stuff may be too high-level, but certainly there will be a few gems in there too :)


[color=blue]   Estoy hace un tiempo en el foro. Me agrada como se trabaja, especialmente en la cuestión del respeto hacia los demás. Ya identifiqué a algunos viejos conocidos entre ustedes. Aún así, quisiera saber quién es quién dentro de esta comunidad. ¿Pueden escribir a este topic con algunos de sus datos personales y una breve descripción propia? Estaré muy agradecido.

   [b]P.D.[/b]
1. Me inspiré en lo que hicieron los peruanos hace ya bastante tiempo para crear este topic. ¡Muchas gracias a ellos!
2. Perdónenme por no presentarme al instante. Estoy un poco apurado. Más tarde lo haré.[/color]

[color=blue]Soy Rafael Guillermo Arias Michel (de ahí viene [b][i]R.G.A.M.[/i][/b]…). Resido en la capital del Paraguay. He nacido el 1º de Junio de 1.989.

Participo de las olimpiadas nacionales de matemáticas desde el 2.002 (¡muchos años ya…!). Participé de la Olimpiada Matemática del Cono Sur del año pasado y de la I.M.O. de este año. Dentro de poco voy a Ecuador para la Olimpiada Iberoamericana.

Me dedico a rivalizar con los siths, aunque no significa que esté a favor de los jedis… De hecho, me considero una mezcla muy extraña de luz con oscuridad. Y lo digo sin temor, puesto que la oscuridad no tiene por qué considerarse maldad…

Como ya mencioné antes, fui invitado por Daniel Chen para participar de este foro. Y no me arrepiento, mas parece que mis propuestas no son muy interesantes. Me consta decir que la única vez que me respondió alguien fue para avisarme que faltaba el archivo que debía adjuntar… Así que ruego encarecidamente que me diga alguien qué debo hacer para captar el interés de ustedes.
Sé que soy desconocido para muchos, y que aún no pude tratar los problemas que escribieron. Cuando cuente con más tiempo podré participar más activamente.

El que siga que tome la posta.

[b]P.D.:[/b] ¿Alguien me puede proponer algunas frases para el "signature"? Me parece muy aburrido el mío, sabiendo que el nombre de mi contacto ya aparece a la izquierda de cada post que envío.[/color]

Buenos dias R.G.A.M. ,

Espero q la falta de respuesta no te desanime a seguir participando ..
te recomiendo que postees Preguntan de olimpiadas en la seccion: Olimpiadas de Matematica, alli la gente entra he intenta los problemas q vamos posteando ...

ademas, en la seccion: Discuciones generales, podes postear enlaces a  paginas q hayas visitado y q te hayan parecido muy interesantes, asi ,,, alguien q quisiera revisar tal o cual teoria se ahorra la busqueda en la web ...

Y en la 3ra seccion, puedes postear problemas de nivel basico, o sino postear examenes de seleccion o entrenamiento que sean interesantes y q quisieras compartir con la comunidad ...

Ademas, envia mensajes privados a los moderadores y miembros q veas q participan mas a menudo, y has las consultas q creas convenientes para iniciar alguna iniciativa q tengas para la comunidad ...

Y claro esta, tienes todo el forum internacional donde encontraras gente de todo el mundo con quienes puedes interactuar ... solo hace falta algo de ingles y muchas ganas ...

Carlos Bravo
Lima - PERU


Me in a Countdown round vs. Edward Hou from Appleton

A race for third place against Iris Xu from Madison

The winner of this countdown round wins first place...

[quote="i_like_pie"][quote="junggi"]wait does this mean you didn't make nats[/quote]
Notice "2006 State Competition," and the title, "Yay! I won!"[/quote]
notice the "countdown."
I'm sure they don't use countdown to choose who goes to nats.

Me with former Wisconsin champ Dan Mulder

yes, we do actually :wink: In 6th grade I worked my way up from 5th to 3rd to go to nats and in 7th I worked from 4th to 1st.

It was the only one on the WSPE site for MathCounts :wink:

what was ur written score kstan?

are u in 8th grade this year?

-jorian

I rather bombed the test and I will say that I could have had a 39 if it weren't for not going with my gut :mad: but yes I am in 8th grade :wink:

My name in tiny print on a huge screen .

Our team with Herb Kohl, founder of the Kohl Center, Wisconsin senator, and owner of the Milwaukee Bucks

Our team with the other Wisconsin senator and '08 VP candidate Russ Feingold

The four from Wisconsin that went to nats '06

Mr/Dr. Rusczyk with our coach on competition day

At the Iwo Jima memorial we met someone who fought in Iwo Jima

[quote="funcia"]wow i always picture you guys as your avatars too bad you aren't  :maybe:[/quote]I know. Aren't I one sexy beast :D

Wow, your coach makes Mr.Rusczyk look pretty short.  :rotfl:

Actually, uh... our coach isn't what you would call tall ;)

I noticed that in the classroom. That Tommy fellow kept calling you Kyle :wink:

you so know you want to get on your feet

[quote="funcia"]wow i always picture you guys as your avatars too bad you aren't  :maybe:[/quote]

That would be really funny if I looked like my avatar :rotfl: . Nobody can look like Albert Hitchcock, because Albert Hitchcock is just too savage to have any imitators. If you tried to look like Albert Hitchcock, his ghost would come from the grave and beat you down!

Cruelly defeating mustafa in 2005, crushing his dreams and spirit :( . Especially because I almost beat kstan, and I could have easily beaten the next guy, because he had a lame afro.

[quote="funcia"]wow i always picture you guys as your avatars too bad you aren't  :maybe:[/quote]
agree

[quote=SoxFan34]So youre Kyle Stankowski? That's a coincidence; my real name is also Kyle.[/quote]

wow... you have a Cphill avatar


[quote="arkhammedos"]Find all real functions wich satisfy:
$ (x \plus{} y)[f(x) \minus{} f(y)] \equal{} (x \plus{} y)f(x \plus{} y)$.
good luck[/quote]

Are you sure there is no typo?
$ (x \plus{} y)[f(x) \minus{} f(y)] \equal{} (x \plus{} y)f(x \plus{} y) \Rightarrow f(x \plus{} y) \equal{} f(x) \minus{} f(y)$.

oh sorry i have corrected now
many thanx.

It's from Georgian Team Selection Test 2001.

[quote="arkhammedos"]Find all real functions wich satisfy:
$ (x \plus{} y)[f(x) \minus{} f(y)] \equal{} (x \minus{} y)f(x \plus{} y)$.
good luck[/quote]

Let $ P(x,y)$ the property $ (x \plus{} y)[f(x) \minus{} f(y)] \equal{} (x \minus{} y)f(x \plus{} y)$.

$ P(x\plus{}1,\minus{}x)$ gives E1: $ f(x\plus{}1) \minus{} f(\minus{}x)] \equal{} (2x\plus{}1)f(1)$

$ P(\minus{}x,x\minus{}1)$ gives E2: $ \minus{}f(\minus{}x) \plus{} f(x\minus{}1) \equal{} (1\minus{}2x)f(\minus{}1)$

$ P(x\plus{}1,x\minus{}1)$ gives $ 2x[f(x\plus{}1) \minus{} f(x\minus{}1)] \equal{} 2f(2x)$ and so E3 : $ f(x\plus{}1) \minus{} f(x\minus{}1) \equal{} \frac{f(2x)}{x}$

Then E1-E2-E3 gives : $ 0\equal{}(2x\plus{}1)f(1)\minus{}(1\minus{}2x)f(\minus{}1)\minus{}\frac{f(2x)}{x}$

And so $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ for some $ a,b,c$

Plugging back this in the noriginal equation, we get $ c\equal{}0$

And so the general solution $ f(x)\equal{}ax^2\plus{}bx$


[quote="Euclid_great"]Find $ x$

in a war, 70% soldiers lost one eye, 80% one ear, 75% one hand, 85% one leg and x% lost all four limbs. Find minimum value of $ x$.[/quote]
Do you mean $ x$% lost at least the combo (one eye + one ear + one hand + one leg)? In that case I have a hint:[hide]The 30% that still has both eyes, cannot be unscathed because we know that only 15% came home with both legs. Assuming that 15% (2-legged soldiers) still have both eyes, that leaves 15% with eyes and legs, and 15% with 2 eyes and 1 leg. How many of those could still have an ear or a hand?[/hide]

By limbs, I assume you mean, people who lost an ear, an eye, an arm and a leg.

[hide="Answer"]Suppose there are 20 people. 17 lost a leg, 16 lost an ear, 15 lost a hand, and 14 lost an eye.

[b]Legs and Ears:[/b] Let's start with the leg and the ear. Suppose the three people that didn't lose a leg lost an ear, and the four that didn't lose an ear lost a leg. This means:

3 people lost one ear but not a leg, 4 people lost a leg but not an ear, and 13 people lost both.

[b]Hands:[/b] Now, let the 7 people who still have either a leg or an ear lose a hand. This leaves 8 of the 13 people that lost both to lose a hand as well. So so far, 8 people have lost an ear, a leg, and a hand.

[b]Eyes:[/b] Now, let the 12 people who did not lose all of the first 3 lose an eye. This leaves 2 of the people who lost a leg, an ear, and a hand to lose an eye as well.

[b]Answer:[/b] 2 out of 20 is 10%, thus x=10[/hide]
What a sadistic problem!

[hide="My approach"]Suppose there are 20 people.  6 did not lose an eye, 4 did not lose an ear, 5 did not lose a hand, and 3 did not lose a leg.  To minimize $ x$, assume that no person was left with more than 1 limb.  That means $ 6\plus{}4\plus{}5\plus{}3$ or $ 18$ people kept a limb and $ 2$ did not.  The answer is thus $ \frac{2}{20}$ or $ 10\%$.

I definitely think this is simpler, if it counts as a legitimate solution.[/hide]

Here is a more rigorous (and probably easier) arugument:
Let k% of the people lose both an eye and an ear. It is clear that $ k \ge 50$.
Let m% of the people lose both a hand and a leg. It is clear that $ m \ge 60$.
So we want to find the smallest k and m can overlap, clearly 10%.


An international firm has 250 employees, each of whom speaks several languages. For each pair of employees, $(A,B)$, there is a language spoken by $A$ and not $B$, and there is another language spoken by $B$ but not $A$. At least how many languages must be spoken at the firm?

For each $A$, let $L_A$ be the set of languages spoken by $A$. The statement of the problem says that the sets $L_A$ form an antichain. But, from Sperner's theorem, the maximal length of an antichain from a set $E$ of $n$ elements is $C_n^{[\frac n 2 ]}.$
Thus the desired minimum is the least $n$ such that $C_n^{[\frac n 2 ]} \geq 250.$

Direct checking leads to $n=10$.

Pierre.