trigonometry
vector
linear algebra
matrix
summer program
MathPath

Suppose that $a,b,c,d$ are real number such that \begin{eqnarray*}a^{2}+b^{2}&=& 1 \\ c^{2}+d^{2}&=& 1 \\ ac+bd &=& 0. \end{eqnarray*} Show that \begin{eqnarray*}a^{2}+c^{2}&=& 1 \\ b^{2}+d^{2}&=& 1 \\ ab+cd &=& 0. \end{eqnarray*} [hide="My Solution"]Note that we can make the substitution $a=\sin x,$ $b=\cos x,$ $c=\sin y,$ $d=\cos y.$ Substituting, the third equation becomes \[ac+bd=\sin x \sin y+\cos x \cos y = \cos(x-y) =0.\] Hence, $x=y+\frac{\pi}{2}(2k+1).$

We can now use this to prove the other three equations true.

The first equation becomes \begin{eqnarray*}a^{2}+c^{2}&=& \sin^{2}(y+\pi/2(2k+1))+\sin^{2}y \\ &=& \sin^{2}(y+\pi/2)+\sin^{2}y \\ &=& \cos^{2}y+\sin^{2}y=1. \end{eqnarray*} The second becomes \begin{eqnarray*}b^{2}+d^{2}&=& \cos^{2}(y+\pi/2(2k+1))+\cos^{2}y \\ &=& \cos^{2}(y+\pi/2)+\cos^{2}y \\ &=& \sin^{2}y+\cos^{2}y=1. \end{eqnarray*} The final equation becomes \begin{eqnarray*}ab+cd &=& \sin x \cos x+\sin y \cos y \\ &=& \frac{1}{2}(\sin 2x+\sin 2y) \\ &=& \sin (x+y) \cos (x-y) =0. \end{eqnarray*} I used the sum to product formula in the second to last line.
[/hide]

Does anyone have a purely algebraic solution to this?   :)

This problem wasn't really intended to be solved purely algebraically.

[hide="But here"] Let's consider $\vec{v}= \left< a, b \right>, \vec{u}= \left< c, d \right>$.  We are given they are unit vectors with dot product zero, so they are perpendicular.

This means $d = \pm a, c = \mp b$ and we are done. [/hide]

[hide="solution"]
$ac=-bd$
$a=-bd/c$

You can put this into the 1st equation:
So,
$\frac{b^{2}d^{2}}{c^{2}}+b^{2}=1$
$b^{2}(\frac{d^{2}+c^{2}}{c^{2}})=1$
$b^{2}=c^{2}$
So,
$a^{2}+c^{2}=1$
and
$b^{2}+d^{2}=1$

To prove the last statement:
$ab=-cd$
We have $d^{2}=a^{2}$ and $b^{2}=c^{2}$
Now we can multiply these together to get $ab=-cd$
[/hide]

[quote="riddler"]To prove the last statement:
$ab=-cd$
We have $d^{2}=a^{2}$ and $b^{2}=c^{2}$
Now we can multiply these together to get $ab=-cd$[/quote]

Euhm, you get $a^{2}b^{2}= c^{2}d^{2}$, not $ab =-cd$, right?

I proved the third equation in another way:
[hide]Suppose that $ab+cd \neq 0$. We have that $a^{2}= d^{2}$ and $ac+bd = 0$, so $0 \neq (ac+bd)-(ab+cd) = (a-d)(c-b)$ and $0 \neq (ac+bd)+(ab+cd) = (a+d)(b+c)$, i.e. $a\neq d$ and $a\neq-d$, contradicting $a^{2}= d^{2}$. Therefore, $ab+cd = 0$.[/hide]

[quote="neelnal"][quote="Kurt Gödel"][quote="riddler"]To prove the last statement:
$ab=-cd$
We have $d^{2}=a^{2}$ and $b^{2}=c^{2}$
Now we can multiply these together to get $ab=-cd$[/quote]

Euhm, you get $a^{2}b^{2}= c^{2}d^{2}$, not $ab =-cd$, right?
[/quote]

Maybe he took the sqaure root of both sides and made one side a positive root and the other a negative?[/quote]

LOL, I'm not stupid, you know  :D I hope you realized while typing that 'explanation' that you've got to prove that "one side is a positive root and the other a negative". But yeah, from $ac+bd = 0$ you could prove that one of the numbers has another sign than the other three numbers, so that would prove that $ab = cd$ is impossible.

Do, I have to prove that $ab\neq{cd}$?
 :maybe:

[quote="nsato"][hide]
Confirm the following identity: \begin{eqnarray*}&& (a^{2}+b^{2}-1)^{2}+(c^{2}+d^{2}-1)^{2}+2(ac+bd)^{2}\\ &&\quad = (a^{2}+c^{2}-1)^{2}+(b^{2}+d^{2}-1)^{2}+2(ab+cd)^{2}. \end{eqnarray*} [/hide][/quote]

Wow, fantastic. I think this is the super solution.

This problem is also Nagoya university entrance exam/1957.

Explainig by matrix, $A^{t}A=E.$ that is to say, $A$ is orthogonal matrix.

This was on the MathPath qualifying quiz, I believe I was the only person to get the trig solution. However, I had a minor hole. I assumed that x=y.[/i]


inequalities
logarithms
inequalities proposed

Prove that the inequality $ (x^{k}\minus{}y^{k})^{n}<(x^{n}\minus{}y^{n})^{k}$ holds forall reals $ x>y>0$ and positive integers $ n>k$.

define $ a_{n}\equal{}\frac{(1\minus{}\alpha^{k})^{n}}{(1\minus{}\alpha^{n})^{k}}$ $ 0 <\alpha < 1$ $ n,k$ being  positive integers.
clearly $ a_{n}$ is decreasing in $ n$ .
observe $ a_{k}\equal{} 1$
so for $ n > k$ $ a_{n}< 1$
putting $ \alpha \equal{}\frac{y}{x}$  $ y < x$
we get $ (x^{k}\minus{}y^{k})^{n}< (x^{n}\minus{}y^{n})^{k}$

Put $ a\equal{}\frac{x}{y}>1$ , we need prove $ (a^{k}\minus{}1)^{n}<(a^{n}\minus{}1)^{k}$ or $ \frac{\ln (a^{k}\minus{}1)}{k}<\frac{\ln (a^{n}\minus{}1)}{n}$. This is true because $ f(t)\equal{}\frac{\ln (a^{t}\minus{}1)}{t}$ is increasing on $ (0,\infty)(f'(t)>0\forall t>0)$.


calculus
integration
logarithms
limit
real analysis
real analysis solved

The sequence $\{c_n\}\ (n=1,2,\cdots)$ satisfies $1\leqq c_n<2$ and the following equation.

\[\int^2_{c_n} \ln x dx=\frac{1}{n}\int^2_1 \ln x dx\]
Suppose that there exist  $\lim_{n\to\infty}n(2-c_n)$ , evaluate the limit.

We have $\int_{c_n}^2\ln x dx=(2-c_n)\ln x_n$ for some $x_n\in (c_n,2)$. We thus get $\int_1^2\ln x dx=n(2-c_n)\ln x_n$. $c_n\to 2$, so when $n\to\infty$, the RHS goes to $\lim_{n\to\infty}n(2-c_n)\cdot \ln 2$, whilke the LHS remains unchanged. The limit we're loking for is thus $\frac 1{\ln 2}\cdot\int_1^2 \ln x dx$.


Can you prove Newton's Law of Universal Gravitation? How did he get that formula?

Hi Visionary don't double post i deleted ur other topic...
As ar as ur question goes it was experimentally observed  and not derived

Newtons formula can be derived as a special case approximation of the Einstein-field equations. The derivation can be found e.g in Landau-Lifsic, second book: The classical theory of fields.

Why don't you just show the derivations if you already know about it? Thanks for giving me the title of the book by the way.

Well the derivation is not that easy..
Are you familiar with tensor analysis, tensor fiels, abstract index notation, differential equations etc. etc. ?

If not then there is not point in me writing it down here, because you wont understand it anyway.. :wink:


projective geometry
geometry proposed
geometry

Let CK,DH is diameter and E is the intersection of AD,BC.  It's easy to see K,A,Y collinear. H,B,Y collinear. Applying Pascal's theorem=> E,Y,O collinear. Similarly X,O,E collinear. We are done
 Sorry if I'm wrong

i cant see the image that you give me :(

MJ GEO can you post your solution! And @livetolovemath030894 where is point X in the figure?

hi.now my nice solution
$ AD,BC$ intersect at $ Z$. $ ZM,ZN$ are perpendicular to$ AC,BD$ such that $ M,N$ lie on $ AC,BD$. now we must prove that $ \frac{MA}{AC}\equal{}\frac{NB}{BD}$. by sin theorem we have $ \frac{BD}{AC}\equal{}\frac{ZB}{ZA}$. and $ \frac{NB}{AM}\equal{}\frac{ZB}{ZA}$  :)

Similarly we'll have E,O,X collinear. You can paint it yourself :)  :)  :)

[quote="MJ GEO"]$ ABCD$ is cyclic.$ YA,YB$ are perpendicular to $ AC,BD$ and $ XD,XC$ are perpendicular to $ BD,AC$.prove that $ AD,BC,XY$ are concur at one point.(I have simple and nice solution)[/quote]

Here is mine
Denote $ YA,YB$ meet $ (O)$ at $ P,Q$
Using Pascal theorem in $ PABQCD$ with : $ PA$ meets $ QB$ at $ Y$, $ DA$ meets $ BC$ at $ M$ and $ DQ$ meets $ CP$ at $ O$ then $ M,X,O$ are collinear

Similarly,$ A,X,O$ are collinear  :)


1:
if f(x)=x^2-3x+4, then find the values of x satisfying the equation
f(x)=f(2x+1)

2:
if for non-zero x, a f(x)+bf(1/x)=1/x-5 where a is not equal to b, then find f(x)

Thanx

1)$x^2-3x+4=(2x+1)^2-3(2x+1)+4$.Solve this

2)a f(x)+bf(1/x)=1/x-5---(1)

Replace x by 1/x to get another equation (2)

Multiply (1) by a and (2) by b and solve to get

$f(x)=\frac{1}{a^2-b^2}(\frac{a}{x}-bx)-\frac{5}{a+b}$

The second one i've seen in some R.D book ..?

Both of the ques are froom R.D Sharama 12th
Actually I m in 11th  but as I hv to study Concept of Physics..I need to study 12th R.D Sharama for Calculus  and before tht functions..

Studying on my own with no guidance..
Thanks shadysaysurspammed
Let me solve it on my own andp post later!

I'm doing the same thing..plus no guidance as well. :D

HI Could not understan the second question..
Well why u didi what u did?(Sounds funny!)

Why was this method used?

Pls explain a bit with reason!

Well for the second one,basically treat $f(x)$ as a variable we want to find,so i created another equation by substituting $x=1/x$ and once i get 2 equations,by simple methods i can eliminate $f(1/x)$ and thus get $f(x)$.


The equation $4x^{6}-5x^{5}+1=0$ should have two real solutions. The first is obvious, 1. Is it possible to find the other one?


Help. I have got part of the way through this question and was hoping to finish with Fermat's little theorem or Euler's theorem but m is not prime. Let m = (4^p - 1)/3, where p is a prime number exceeding 3. Prove that 2^(m-1) has remainder 1 when divided by m. (BMO round 2 1993, question 2.) Are solutions to BMO available on the web?

Note that $ 2p$ divides $ m\minus{}1\equal{}\frac{4}{3}(4^{p\minus{}1}\minus{}1)$ (by Fermat Little Theorem), implying that $ 2^{2p}\minus{}1\equal{}4^p \minus{}1$ divides $ 2^{m\minus{}1} \minus{} 1$.
Finally, since $ m$ divides $ 4^p \minus{}1$, it also divides $ 2^{m\minus{}1} \minus{} 1$.


Dear friends! Who can post "Ukranian Math Olympiath questions" ?

   Thanks...


trigonometry
function
geometry
calculus
analytic geometry
Euler
algebra

[quote="nebula42"][quote="i_like_pie"][quote="Arvind_sn"]Prove this: $sin^{2}x+cos^{2}x=1$[/quote]
I don't think this belongs in Classroom Math. :wink:[/quote]
I agree; this should be in at least High School Basics, because it involves both trig and proofs.
Just a note to remember: even if a question only requires something very simple to prove (such as the Pythagorean Theorem, in this case), it still shouldn't be in Classroom Math if it involves proofs and especially if it involves trig.  :wink:
(This is mostly because many people using the Classroom Math forum have not yet learned what trig is.)[/quote]
They teach trig in my school in 7th grade pre-alg...

[quote="bpms"][/quote][quote="nebula42"][quote="i_like_pie"][quote="Arvind_sn"]Prove this: $sin^{2}x+cos^{2}x=1$[/quote]
I don't think this belongs in Classroom Math. :wink:[/quote]
I agree; this should be in at least High School Basics, because it involves both trig and proofs.
Just a note to remember: even if a question only requires something very simple to prove (such as the Pythagorean Theorem, in this case), it still shouldn't be in Classroom Math if it involves proofs and especially if it involves trig.  :wink:
(This is mostly because many people using the Classroom Math forum have not yet learned what trig is.)[/quote][quote="bpms"]
They teach trig in my school in 7th grade pre-alg...[/quote]

look at my school, they don't even touch upon sine or cosine until pre-calc or trig at the high school

Can we do it anyways?
[hide]We have $\triangle ABC$ with angles $A,B,$ and $C$ and side lengths $a$, $b$, and $c$ (hypotenuse $c$). Assume WLOG that $x=A$. Therefore, $\sin^{2}x=\left(\frac{a}{c}\right)^{2}=\frac{a^{2}}{b^{2}}$ and $\cos^{2}x= \frac{b^{2}}{c^{2}}$. Adding these together results in $\frac{a^{2}+b^{2}}{c^{2}}$. By Pythagorean Theorem, $a^{2}+b^{2}=c^{2}$. Hence, $\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1$ as desired.[/hide]

[quote="Arvind_sn"]Prove this: $sin^{2}x+cos^{2}x=1$[/quote]

[hide]

That comes from the Pythagorean theorem:

$a^{2}=b^{2}+c^{2}$ so, $1 = \left(\frac{b}{a}\right)^{2}+\left(\frac{c}{a}\right)^{2}$

$1 =\sin^{2}\alpha+\cos^{2}\alpha$

[/hide]

This is easily proven by the Pythagorean Theorem, and really basic Trig functions!

Hence the name [i]Pythagorean[/i] trigonometric identities.

[hide]
\begin{eqnarray*}\sin^{2}x+\cos^{2}x&=&\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)^{2}+\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) \\ &=&\frac{\text{opposite}^{2}+\text{adjacent}2}{\text{hypotenuse}^{2}}\ \text{then, because of the Pythagorean theorem,}\\ &=&\frac{\text{hypotenuse}^{2}}{\text{hypotenuse}^{2}}\\ &=&1 \end{eqnarray*}
[/hide]

I think you forgot the square on the part $\frac{adjacent}{hypotenuse}$ in the first line.

[quote="Arvind_sn"]Prove this: $sin^{2}x+cos^{2}x=1$[/quote]

uhh pyth theorem

a^2/c^2+b^2/c^2
a^2+b^2/c^2
c^2/c^2
1

[quote="moogra"]yeah but you have to skip alg 1[/quote]
nono it's just that our school sucks terribly.
In Ohio, half the kids in 7th grade took algebra 1 and they took geometry the next year

[quote="Phelpedo"]Actually some schools offer geometry in 8th grade, which involves both proofs and trig. Anyway, it seems a little insulting for a high school forum...[/quote]
This is the smallest forum in which trig can be used period...

I learned this identity in Pre-Calc, a normally 12th grade course.  :|

[quote="neelnal"]by using Taylor-Maclaurian Series I believe.[/quote] Is that calculus?? Why not just use Pythagorus. :D

It is kind of calculus, more like sums.

I was just offering an alternative proof to the Pythagoras proof.

There is also a proof involving polar coordinates if you people want.

[quote="t0rajir0u"]The concept of polar coordinates requires the Pythagorean theorem to be sound, so that would be circular logic.

I, on the other hand, [b]redefined[/b] $\sin$ and $\cos$ in terms of their Taylor series.  My proof, working from my definitions, is totally valid. ;)

Of course, if you want to define $\sin, \cos$ one of two conventional ways (either on the unit circle, or as functions of an angle in a triangle) there are many easy proofs.[/quote]

Umm...look at my first reply. I used the e^([i]i[/i]x) series too.

[quote="Farenhajt"][quote="t0rajir0u"]
I, on the other hand, [b]redefined[/b] $\sin$ and $\cos$ in terms of their Taylor series.[/quote]

It's always SOOO refreshing to see Taylor series in High School Basics forum. Does it make you feel good when you're bragging about being oh so smart? ;)

Hey, dude, it's about time you spill out some tensor calculus or possibly topological, or even differential geometry approach to this problem. Any of that in your books? No? Shame...  :([/quote]

Lol. Yes I saw that the way he said that sounded kind of like bragging. But we(him and I) offered an alternative solution to anyone interested. Just as a point of interest, not to belittle*

*I speak for myself but I hope t0rajir0u shares these thoughts :|

[quote="neelnal"]Lol. Yes I saw that the way he said that sounded kind of like bragging. But we(him and I) offered an alternative solution to anyone interested. Just as a point of interest, not to belittle*

*I speak for myself but I hope t0rajir0u shares these thoughts :|[/quote]

But that hardly can be the point: High-schoolers who could already be interested in Taylor series (or in any kind of series) are WAY WAY past proving that $\sin^{2}x+\cos^{2}x=1$ (I would accept such proofs if the problem were to prove the statement in as many ways as possible). Therefore, most of those who would find high-end proofs enlightening would hardly even look at this topic.

If you're REALLY into broadening the scopes of young mathematicians, open a subforum for advanced high school pupils.

[quote="t0rajir0u"]Bragging?   :o

I was just making the point that the "proof" depends on the way in which $\sin$ and $\cos$ are defined, so saying "Pythagorean Theorem" is meaningless because the statement is itself the Pythagorean Theorem if $\sin$ and $\cos$ are defined in terms of a triangle.[/quote]

Not so, dear. Draw the trigonometric circle and use good old simple Pythagorean Theorem even if $\sin x$ or $\cos x$ are negative. (Not that the sign of those would affect the identity in any way.)

You're just bragging.

It hardly qualifies as a proof, especially if you define $\sin, \cos$ in terms of the unit circle.  The point I'm trying to make is that with either of the geometric definitions the statement is exactly equivalent to the Pythagorean Theorem, so the Theorem itself doesn't constitute a proof.

Aha, I see, and you would take this all the way down to proving that $2+2=4$? Since you won't take Pythagorean Theorem as as proven fact? BUT Taylor series is just fine and natural, we don't have to prove a THING there? Such as Euler exponential formula, for a start?

Have you ever heard of ANY non-trigonometric proof of Pythagorean Theorem? Consult your high-end books - there are just about a few hundreds of such proofs... Oh, sorry, your high-end books are too high-end for a measly no-words pure-geometry proof of Pythagorean Theorem... Then try [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23921&start=20]Proofs Without Words Contest Problems, Top of Page 2[/url], maybe you'll find that post illuminating.

Boy, wouldn't you do just anything to dazzle 9-graders with such arcana as Taylor series, generating functions, polynomial moduli, Euler formula for complex numbers etc. etc. (I'm already losing count of your specialties.)

[quote="neelnal"]This may be very advanced, but here we go:

e^([i]i[/i]x) = cos(x) + [i]i[/i]sin(x)

This identity can be used very effectively to prove this concept.

Use the fact that (a - b) (a - b) = a^2 - b^2
                                            = (cos(x))^2 - (-(sin(x))^2)
                                            =(cos(x))^2 + (sin(x))^2)
                                            = (e^([i]i[/i]x))^2 = 1
by using Taylor-Maclaurian Series I believe.

I am fairly confident this is the proof using the
e^([i]i[/i]x) = cos(x) + [i]i[/i]sin(x)  concept.[/quote]

This is wrong.  You want to expand $\exp(-ix)\exp(ix)=1$.  Also you need to be very careful to avoid circular definitions here.  Parametrizing the circle by $(\cos t,\sin t)$ essentially assumes the answer, whereas defining $\exp(it)=\cos t+i\sin t$ does not.

[quote="blahblahblah"]Parametrizing the circle by $(\cos t,\sin t)$ essentially assumes the answer, whereas defining $\exp(it)=\cos t+i\sin t$ does not.[/quote]

This is bull - $\sin x$ and $\cos x$ (for real $x$) can always be DEFINED via the unit circle. (The very name "tangent" for $\tan x$ comes from its definition on the unit circle.) Therefore, please humor me, accept THOSE definitions and PROVE Taylor expansion for $e^{x}, \sin x, \cos x$, the validity of the $e^{x}$ expansion in the complex domain, and Euler's exponential formula - all of that, of course, WITHIN the scope of 9th-10th grade math. If you please.

I'm not saying that this is appropriate for 9th or 10th grade math, I am merely saying that it is difficult to do so without reasoning circularly.

If $(\sin t, \cos t)$ are points on the unit circle $x^{2}+y^{2}=1$, then you have assumed the answer.

There is NO circular reason here: sine and cosine are DEFINED as the projections of a unit radius on the coordinate axes. FROM THAT you derive that the sum of their squares is $1$, using PREVIOUSLY proven Pythagorean Theorem.

We can claim that it's not much of a result, since it follows immediately from the definition, but it's not EQUIVALENT to the definition.


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {a^2 \minus{} bc}{3a \plus{} b \plus{} c} \plus{} \frac {b^2 \minus{} ac}{3b \plus{} a \plus{} c} \plus{} \frac {c^2 \minus{} ab}{3c \plus{} a \plus{} b}\geq0
\]
It seems that $ \sum_{cyc}\frac{a^3\minus{}bcd}{7a\plus{}b\plus{}c\plus{}d}\geq0$ is true too for positive $ a,$ $ b,$ $ c$ and $ d.$

Yes,arqady! we have  more generic
\[ \sum_{cyc}{\frac {a^{2m} \minus{} b^mc^m}{ka^n \plus{} b^n \plus{} c^n}}\ge0(a,b,c,m,n > 0,{0}\le{k}\le{1}).
\]


Prime number $ p$ is called cubic if every integer could be represented as $ x^{3}\plus{}y^{3}\plus{}pz$, where $ x$,$ y$,$ z$ are integers. Find all cubic numbers.

This is the same problem as one I "found" some time ago ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=90520


when I try building my file, it says it cannot find the idx. file for it along with the actual file. whats wrong?

That can be ignored if you haven't got an index, because in that case no idx file is required and hence won't be found. Similarly if you haven't got any citations or bibliography then you can ignore messages about them not being found.

Of course if you have got an index and used \makeindex then it is a problem; in that case please provide a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] giving the problem.


The Puzzle:

A girl, a boy, and a dog start walking down a road.

They start at the same time, from the same point, in the same direction.

The boy walks at 5 km/h, the girl at 6 km/h.

The dog runs from boy to girl and back again with a constant speed of 10 km/h. The dog does not slow down on the turn.

How far does the dog travel in 1 hour?

[size=200][color=red]May God help you, God bless you[/color][/size]

How far meaning how far is he from the starting point, or what was his total distance?

[hide="solution?"]If the latter, then the boy and the girl do not matter, since he's always travelling at 10 km/h. Multiply this by 1 hr, and you get $ 10km$[/hide]


So I am currently enrolled in a intro proof class and I am having a little trouble with this proof.

If a_n <= a_(n+1), then prove a_i <= a_n

If anyone can help it would be great. Thanks

Maybe you want to wrote something else... :)
Not that if $ a_n\leq a_{n\plus{}1}$ then $ a_i\leq a_n$ ;)


calculus
derivative
function
calculus computations

Hi, this is a more theory-based problem from my textbook that i need some help with:

Prove the assertion that a derivative, defined on an interval must have the intermediate-value property.
(Hint: Assume that f' exists on [a,b] and f'(a) ≠ f'(b). If k lies between f'(a) and f'(b), show that the function g defined by g(x) = f(x) - kx must have either a maximum value or a minimum value on [a,b] occurring at an interior point c in (a,b). Deduce that f'(c) = k).

thanks!

It tells you what to do. Where on that path are you falling off?

For a different argument, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=961184#961184]here[/url]. Let $ f$ be differentiable on an interval $ I$, and define $ T\equal{}\{(x,y): x,y\in I\text{ and }x>y\}$. Consider the difference quotients $ g(x,y)\equal{}\frac{f(x)\minus{}f(y)}{x\minus{}y}$ on $ T$. Since $ g$ is continuous on a connected set, it takes values on an interval. By the Mean Value Theorem, all of those values $ g(x,y)$ appear as $ f'(z)$ for some $ z\in (x,y)$, so $ f'(I)\supseteq g(T)$. In the other direction, the values of the derivative are limits of those difference quotients, and $ f'(I)\subseteq \overline{g(T)}$ (the closure). Every set between an interval and its closure is an interval, and we're done.


algebra
polynomial
Ring Theory
IMC
college contests

Suppose $ p$ is a polynomial with integer coefficients and $ a_1 < a_2 < \ldots < a_k$ are integers.

(a) Prove that there exists a $ a\in\mathbb{Z}$ such that $ p(a_i)$ divides $ p(a)$ for all $ i$. ($ p(a_i)\neq 0$)
(b) Does there always exists a $ a\in\mathbb{Z}$ such that the product $ p(a_1)p(a_2)\ldots p(a_k)$ divides $ p(a)$?

Could some of you change the titles,to write something more meaningful instead of "IMC 2008 Day 1"?How can one see the problems number,if the title is the same for all problems.
And after that delete this post.
Thanks in advance.

[quote="Peter"]I assume $ p$ needs integer coefficients? Or is this over a general polynomial ring (eg $ p: Z[x,y]\to Z[y]$ and divisibility in $ Z[y]$)?

Is $ \deg(p) \equal{} 0$ allowed for (b)?[/quote]

Integer coefficients, yes.

I don't know if there was any restriction for (b). Possibly the problem statement said that $ p$ should be non-constant, but I'll have to check it.

Well, in any case $ a$ is immediate using $ f(x\plus{}a)\equal{}f(x)$ mod $ a$ and b) is obviously false, as $ x^2\plus{}1$ shows (this is never a multiple of $ 4$).

[quote="Kurt Gödel"]
I don't know if there was any restriction for (b). Possibly the problem statement said that $ p$ should be non-constant, but I'll have to check it.[/quote]

No, there was no restriction... One could write the example p=5 and get all the points...


linear algebra
matrix
algebra
polynomial

Compute the determinant by minors of the first row, and you arrive at the recursion $\det(D_{n})=\det(D_{n-1})-\det(D_{n-2})$ with initial conditions $\det(D_{2})=0$ and $\det(D_{3})=-1.$ By explicit computation, that sequence proves to be periodic of period 6. (You can also use standard techniques for solving linear constant coefficient difference equations; the characteristic polynomial is $r^{2}-r+1,$ whose roots are sixth roots of unity.)

$\det(D_{6m+2})=\det(D_{2})=0$
$\det(D_{6m+3})=\det(D_{3})=-1$
$\det(D_{6m+4})=\det(D_{4})=-1$
$\det(D_{6m+5})=\det(D_{5})=0$
$\det(D_{6m})=\det(D_{6})=1$
$\det(D_{6m+1})=\det(D_{7})=1$

So $\det(D_{3k})=(-1)^{k}.$

A further challenge:

The eigenvalues of $D_{2}$ are $0$ and $2.$

The eigenvalues of $D_{3}$ are $1-\sqrt{2},1,$ and $1+\sqrt{2}.$

The eigenvalues of $D_{5}$ are $1-\sqrt{3},0,1,2,$ and $1+\sqrt{3}.$

Can you find the general pattern here?


I need all the names of the properties... such as distributive and multiplicative identity and such... does anyone have them all? I've tried googling, but none of the results were relevant. Thanks in advance, dynamo729

These are all the basic arithmetic properties that I can recall of off the top of my head:
-Associative Property [a+(b+c)=(a+b)+c] [a*(b*c)=(a*b)*c]
-Commutative Property [a+b=b+a]  [a*b=b*a]
-Identity Property [a+0=a] [a*1=a]
(0 is also called the additive identity and 1 is called the multiplicative identity)
-Zero Property [a*0=0]
-Distributive property [a*(b+c)=ab+ac]

Hope that helps.

I did not know that there was a identity property :D

You missed a few, and I $ \text{\LaTeX}$ified them.

[b]Associative Property of Addition[/b]: $ a + (b + c) = (a + b) + c$

[b]Associative Property of Multiplication[/b]: $ a(bc) = (ab)c$

[b]Additive Identity[/b]: $ a + 0 = a$

[b]Multiplicative Identity[/b]: $ 1 \times a = a$

[b]Zero Property[/b]: $ a \times 0 = 0$

[b]Distributive Property[/b]: $ a(b + c) = ab + ac$

[b]Reflexive Property[/b]: $ a = a$

[b]Transitive Property of Equality[/b]: If $ a = b$, and $ b = c$, then $ a = c$.

[b]Symmetric Property of Equality[/b]: If $ a = b$ then $ b = a$.

[b]Commutative Property of Addition[/b]: $ a + b = b + a$

[b]Commutative Property of Multiplication[/b]: $ ab = ba$

EDIT: Oops.  :oops:  Fixed.

Izzy, it's commutative, not communitive.

And these are just the basic ones, that most people learn or review in Alg I. (These are probably the ones you asked for.) But there are also more complicated ones, like the trichotomy law, and laws applying to equations.


CHEF KIRBY!!

Lol. Cooking people is so awesome.

hah lol Pikachu Kirby's hand is censored.

aww haha samus kirby looks like spiderman

default this. lol

actually, get a sword kirby pic and default, since youre meta [b]knight[/b] kirby

hmm...pika-kirby is so enthusiastic  :lol:

awsome wow you really did get obsesseed with kirby how do you make a gallery anyway?????

wow who knew pikiby would be so.... happy. :rotfl:

ha just go to galleries on the top, click on the personal galleries thing
and then click make ur own or something like that

:D

lol nice album angie. so many kirbys!!!!!!!!!! :D

YAY its fun too watch as other charaters get cooked and seasoned! Link stew. hmmmm

too cute
wanna know more about the character.

why thank you kabi.
not that i made the picture...

I wanna  know about  KIRBY.
Anyways pics is really nice


calculus
integration
trigonometry
calculus computations

Evaluate using substitution $ \int \ {x^2}\sqrt{4\minus{}x^2} d\ x$

Let $ x \equal{} 2 \sin t$, $ dx \equal{} 2 \cos t dt$:

$ P \equal{} \int x^2 \sqrt{4\minus{}x^2}\,dx \equal{} \int 4 \sin^2 t \sqrt{4 \minus{} 4 \sin^2 t} \cdot 2 \cos t \,dt$

$ \equal{} 16 \int \sin^2t \cos^2 t\,dt \equal{} 4 \int \sin^2(2t)\,dt \equal{} 4 \int \frac12[1 \plus{} \cos(4t)]\,dt$

$ \equal{} 2t \plus{} \frac12 \sin(4t) \equal{} 2t \plus{} \sin(2t) \cos(2t) \equal{} 2t \plus{} 2 \sin t \sqrt{1 \minus{} \sin^2t} (1 \minus{} 2 \sin^2t) \plus{} C$

Now just substitute back.

hello, after a back substitution i have got
$ \int x^2\sqrt{4\minus{}x^2}dx\equal{}\minus{}\frac{1}{4}x(4\minus{}x^2)^{\frac{3}{2}}\plus{}\frac{1}{2}x\sqrt{4\minus{}x^2}\plus{}2\arcsin\left(\frac{1}{2}x\right)\plus{}C$.
Sonnhard.


function
MATHCOUNTS
search
number theory
Diophantine equation
algebra
binomial theorem

[quote="mathking123"]if the standard u.s. coins are 1cent, 5cent,10cent,and 25cent, how many ways are there to give change for a dollar?

is there any cool combinatoric or otherwise way to solve this or do i have  to solve a nasty 4 variable diophantine equation?[/quote]

uhhh

p+5n+10d+25q=100, look at it in mod 5

So p is a multiple of 5, makes it easier

5p+5n+10d+25q=100
p+n+2d+5q=20.

Now you have to split it up into a lot of cases (its computable but...) and then use balls and urns.

For example, first case

q=0, meaning p+n+2d=20, meaning p+n is even, meaing either both p and q are odd or both p and q are even,

so both even

2p+2n+2d=20, p+n+d=10, 12C2, typical balls and runs
2p+2n+2+2d=20 if p and n are odd, p+n+d=9, 11C2, typical balls and urns

now look at q=1 case, too lazy and tired to finish.

[quote="mathking123"]lolz thats exactly what im trying to aviod doing..,. :P[/quote]

well then you have to use generating functions, which is pretty pointless (its not required in mathcounts or even AMC/AIME for that matter)

so dont worry, this wont be a question on a mathcounts test (even nats)

Well, there are 293 ways to make change for a dollar, but I'm pretty sure that includes half-dollars.  Try finding the number of possibilities with a half dollar and subtract from 293?

Gosh, just google search "number of ways to make change for a dollar" (without the quotation marks).
And the first one that comes up, is
http://www.maa.org/features/mathchat/mathchat_4_19_01.html

I dunno, I hope my calculator can stand the generating function...

I really need to learn how to use the extended binomial theorem...

yay, my calculator survived

there are apparently 242 ways to get 1 dollar without half-dollars and only coins.  hmm, that's great news :roll:

Well, actually I memorized that.  Who knows?  It might come in handy!!

omg where are these random numbers coming from? 242? how did you derive that?

and no, dont give me a link, give me a formula i can use b/c in mathcounts i cant google search this during the competition....

There is no formula, because the coins aren't in nice increments.

What alan did (generating functions) is kind of advanced for mathcounts.

you really want to find the coefficient of $ x^{100}$ in the expansion of \[ (1\plus{}x^5\plus{}x^{10}\plus{}x^{15}\plus{}x^{20}\plus{}x^{25}\plus{}x^{30}\plus{}x^{35}\plus{}x^{40}\plus{}x^{45}\plus{}x^{50}\plus{}x^{55}\plus{}x^{60}\plus{}x^{65}\plus{}x^{70}\plus{}x^{75}\plus{}x^{80}\plus{}x^{85}\plus{}x^{90}\plus{}x^{95}\plus{}x^{100})^2\cdot (1\plus{}x^{10}\plus{}x^{20}\plus{}x^{30}\plus{}x^{40}\plus{}x^{50}\plus{}x^{60}\plus{}x^{70}\plus{}x^{80}\plus{}x^{90}\plus{}x^{100})\cdot (1\plus{}x^{25}\plus{}x^{50}\plus{}x^{75}\plus{}x^{100})\]

uh oh, that took three lines...

yeah, um, learn generating functions and the extended Binomial theorem (is it called that?) before trying to tackle this problem without casework and listing.

[quote="mathking123"]you think that the question is of a difficulty level higher than mathcounts nats? even team?

thanks for any help :)[/quote]

not difficulty level, its just not covered in mathcounts
(say find log(100), obviously thats super easy, easier than school questions, but mathcounts just doesnt cover them so you dont need to learn them)
this question mathcounts would not cover because its way too many cases (and using extended balls and urns really isnt a mathcounts technique) and way too huge of numbers, and obviously the other way, generating functions is 100% NOT covered in mathcounts. (infact ive never seen it on any math competition)

Uh... Alan, how do you do generating functions on a calculator?

Wow, this is the 3rd time i've seen a long discussion about this problem, or something related in my short AoPS life time.

Generating functions are not covered in Mathcounts, that's right. However, they can make a lot of questions much easier. The Mathcounts solution to this problem would be by using brute force. However, we AOPSers are trying to find a better, faster way to do it.

Actually, looking at this problem again, i think generating functions are not the easiest method with this problem; extended balls and urns does not require a calculator (or a long list of terms) and can be easily predicted by patterns (so just do the first case i described above and then the rest is predicable due to even/odd parity)

The answer is $ \binom{12}{2} \plus{} \binom{11}{2} \plus{} 2\binom{9}{2} \plus{} \binom{7}{2} \plus{} \binom{6}{2} \plus{} 2\binom{4}{2} \plus{} \binom{2}{2} \equal{} 242$

generating functions is more reserved for like $ x \plus{} 2y \plus{} 3z \equal{} 100$, which $ 1,2,3$ are relatively prime to each other.  But of course the latter question will NEVER be tested on mathcounts.  I doubt the one that is posted here will even be tested becuase mathcounts knows that if the best way to do it is above mathcounts level, they wont do it.


Is there a simple proof that for every $ a,m$ coprime, there exist [b]a[/b] prime in the arithmetic prog. $ mk \plus{} a$
which does not mention the fact that there are infinitely many (i.e. Dirichlet) ?

That theorem is no simpler than the "full" Dirichlet's theorem.
See http://www.mathlinks.ro/viewtopic.php?t=130206 .

I asked for the same thing [url=http://www.mathlinks.ro/viewtopic.php?t=201184]here[/url].

ZetaX has posted some useful links there.


Let a>0 , the sequence x_n s.t. : x_0=0;
x_(n+1)=(3*x_n - \sqrt (5*(x_n)  2  +4*a 2 ))/2
find a s.t.:x_n have a limit.
find a s.t.:when n in N , x_n in Z

Suppose x_n is convergent to L the
2L = 3L -  \sqrt (5L^2+4a^2)
L^2 = 5L^2 + 4a^2
0 < a^2 \leq L^2+a^2 = 0 contradiction
x_n is divergent sequence.

Thanks for ThanhBinh12T
Your answer is my mind.


topology
function
abstract algebra
superior algebra
superior algebra unsolved

Let $T$ be compact Hausdorff topological space and $C(T)$ unital comutative Banach algebra of continuous complex-valued functions (with standard operations and norm).
For closed subset $S \subset T$ we define
  $\mathcal I_{S}= \{f \in C(T) ; f(s)=0, \forall s \in S \}$.

Prove that with $S \mapsto \mathcal I_{S}$ is given bijective correspodence between set of all non-empty closed subsets $S \subset T$ and set of all closed ideals in algebra $C(T)$. Also prove that inverse bijection is given with $\mathcal I \mapsto T_{\mathcal I}$ where $T_{\mathcal I}= \{ t \in T ; f(t)=0,\forall f \in \mathcal I \}$.

I proved that definitons are good ($\mathcal I_{S}$ is closed ideal, and $T_{\mathcal I}$ is closed subset), and $S=T_{\mathcal I_{S}}$
I have problem with proving $\mathcal I = \mathcal I_{T_{\mathcal I}}$. Don't know how to show inclusion $\mathcal I_{T_{\mathcal I}}\subset \mathcal I$.

Pretty please help..

I've got a proof using methods of the theory of C*-algebras. an elementary proof can be found here:
https://digital.lib.washington.edu/dspace/bitstream/1773/2641/1/TTHT_Warner.pdf , page 2-22

let $ \mathcal{I}$ be a closed ideal of the commutative unital $ C^{*}$-algebra $ C(X)$. then $ C(X)/\mathcal{I}$ is a commutative unital $ C^{*}$-algebra, thus there's a compact space $ Y$ s.t. $ C(X) / \mathcal{I}\cong C(Y)$. namely $ Y$ is the character space of $ C(X) / \mathcal{I}$, whose elements corresponds to characters of $ C(X)$ which vanish on $ \mathcal{I}$, i.e. points of $ X$ whose evaluation map vanishes on $ \mathcal{I}$, i.e. points of $ \mathcal{T}_{\mathcal{I}}$. making the maps explicit, one checks that $ C(X) \to C(X)/I \cong C(\mathcal{T}_{\mathcal{I}})$ is just the restriction map, so that its kernel is $ \mathcal{I}= \mathcal{I}_{(\mathcal{T}_\mathcal{I})}$.


ratio
algebra
polynomial
limit
number theory unsolved
number theory

For a,b,t>0 and [tex]f(x)=x^2+ax+b[/tex].Moreover,2f(nt) are perfect square all  interger n.Show that:t is not ration

Assume that $t = \frac p q$ is rational, where $p>0.$
For $n=0$ we deduce that $2b = m^2$, where $m$ is an integer.

Replace $n$ by $nq$ to deduce that for all integers $n$, the number $2n^2p^2+2nap + m^2$ (1) is the square of an integer. Thus $a = \frac u v$ is rational, where $u > 0.$

Now, replace $n$ by $nv$ in (1) to deduce that, for all $n$, the number $2n^2p^2v^2+2nup + m^2$ is the square of an integer.
It follows that the polynomial $P(X) = 2X^2p^2v^2+2Xup + m^2$, with integer coefficient, is a square for all integers $X$, which is well-known to imply that $P$ is the square of a polynomial with integer coefficients.
Say $P(X) = (rX+s)^2$ where $r>0$ is an integer.
Identifying the coefficients leads to $r^2 = 2p^2v^2$ which is impossible by consideration of $2$-valuation. Hence result.

Pierre.

[quote="pbornsztein"]Please, do not post new problems directly into the solved section ;)

Pierre.[/quote]when i haven't solved it yet, i will post it into the unsolve section... :) >Here is my solution,I hope it's nicer than yours:
Given :[tex]u_n=\sqrt{2f((n+1)t)}-\sqrt{2f(nt)}
=\sqrt{2[(n+1)^2t^2+a(n+1)t+b]}-\sqrt{2(n^2t^2+ant+b)}
=\sqrt2 t[(n+1)\sqrt{1+\frac{a}{(n+1)t}+\frac{b}{(n+1)^2t^2}}-n\sqrt{1+\frac{a}{nt}+\frac{b}{n^2t^2}][/tex]
Put :[tex]a_n=\sqrt{1+\frac{a}{nt}+\frac{b}{n^2t^2}[/tex] and [tex]u_n=\sqrt2 t [(n+1)a_{n+1}-na_n][/tex].Since the hypothesis we obtain:[tex]u_{n+1}<u_n[/tex].We have:
[tex]\sqrt2 t a_{n+1}<u_n<\sqrt2 t a_n[/tex].because[tex] \lim_{n\to \infty}a_n=\lim_{n\to \infty}a_{n+1}=1[/tex][tex]-->\lim_{n\to \infty}u_n=\sqrt2 t [/tex].But [tex]u_n[/tex] is integer-->[tex]t[/tex] is not rational


inequalities
function
three variable inequality

Let a,b,c  be positive real numbers,a+b+c=3
 such that
look the a1.jpg

Prove the inequality

[color=#FF0000]Moderator edit: do not quote the whole post above.[/color]
My solution:
$ \sum\frac{1}{2+a^2+b^2}\le\frac{1}{2}\left(\sum\frac{1}{a+b}\right)$
So we have to prove that:
$ \left(\sum\frac{1}{a+b}\right)\le \frac{3}{2}$
Consider the function:
$ f(x)=\frac{1}{3-x}$,is concave.
So $ f(a)+f(b)+f(c) \le 3f\left(\frac{a+b+c}{3}\right)=3f(1)=\frac{3}{3-1}=\frac{3}{2}$

[quote="Dimitris X"]
So we have to prove that:
$ \left(\sum\frac {1}{a \plus{} b}\right)\le \frac {3}{2}$
[/quote]
I think the inverse inequality is true. The mistake  - in your last step. :wink:

[color=#FF0000]Moderator edit: do not quote the whole post above.[/color]
Yes arqady i saw my mistake after a while....
Neither the inequality is true nor the function is concave....
I'll try it later.

Should be locked.

The problem has been already posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1492633#1492633]here.[/url]

Dear Ji Chen, how do you manage to see all these inequalities sum of squares....which is...AMAZING! O.o


For $ 5^m\plus{}6^m \equiv 0 \pmod{11}$ where $ m$ is odd, can you just use the property: if $ a \equiv b \pmod$ then $ a^m \equiv b^m \mod{n}$ plus the addition property?

The addition property?  No, you're already done.  $ 5 \equiv \minus{}6 \bmod 11$ so $ 5^m \equiv (\minus{}6)^m \equiv \minus{}6^m \bmod 11$.


Determine the triples of integeres $(x, y, z)$ satisfying the equation $x^{3}+y^{3}+z^{3}= (x+y+z)^{3}$.

[hide]
$(x+y+z)^{3}-(x^{3}+y^{3}+z^{3})=3xy(x+y)+3yz(y+z)+3xz(x+z)+6xyz=3(x+y)(y+z)(x+z)$

So one or more of the following must be true:
$x+y=0 \implies y =-x$
$y+z=0 \implies z =-y$
$x+z=0 \implies z =-x$

[/hide]

How would someone who had never seen it before come up with that factorization?

you would have seen the problem before

this is a very standard factorization that you should know

else, if you consider it a polynomial in a, then f(-b)=f(-c)=0

The polynomial method is [b]very[/b] powerful.  What Altheman means is that if $f(-b) = f(-c) = 0$ when you consider the expression as a polynomial in $a$, then $a+b, a+c$ must be factors.

[hide="Alternately..."] $x^{3}+y^{3}= (x+y)(x^{2}-xy+y^{2}) = (x+y+z)^{3}-z^{3}= (x+y)( (x+y+z)^{2}+(x+y+z)z+z^{2})$

So $x+y$ is a factor.  Symmetrically, $y+z, z+x$ must also be factors.  Or you can divide out and continue:

$x^{2}-xy+y^{2}= x^{2}+2xy+y^{2}+2yz+z^{2}+2zx+zx+yz+2z^{2}\Leftrightarrow$
$3xy+3yz+3zx+3z^{2}= 0 \Leftrightarrow$
$3(y+z)(z+x) = 0$ [/hide]

I don't get the "polynomial" way to do it, could someone expand on that topic?

It will be easier to see if instead of using $x, y, z$ or $a, b, c$ we use $x, b, c$.

We want to find the roots of the polynomial

$(x+b+c)^{3}-x^{3}-b^{3}-c^{3}$

Clearly the $x^{3}$ terms cancel, so this polynomial is quadratic and has at most two roots.  Letting $x =-b$ we have

$(-b+b+c)^{3}-(-b)^{3}-b^{3}-c^{3}= c^{3}-c^{3}= 0$

And letting $x =-c$ we have

$(-c+b+c)^{3}-(-c)^{3}-b^{3}-c^{3}= b^{3}-b^{3}= 0$

So $-b,-c$ must be the two roots.  Then we must have

$(x+b+c)^{3}-x^{3}-b^{3}-c^{3}= k(x+b)(x+c)$

For some constant $k$.  It turns out that $k = 3(b+c)$, which is the coefficient of the quadratic term on the LHS, which is not hard to see.


Let O be the circumcentre of an acute triangle ABC and let A′, B′ and C′ be the
circumcentres of triangles BCO, CAO and ABO, respectively. Prove that the area of triangle ABC does not exceed the area of triangle A′B′C′.

(for a right-angle triangle the problem is obvios) $ [A'B'C'] \equal{} \frac{R^2}{4} \left ( \tan A \plus{} \tan B \plus{} \tan C \right ) \equal{} \frac{R^2}{4} \tan A \tan B \tan C$ and $ [ABC]\equal{} 2R^2 \sin A \sin B \sin C$.

Then we have to prove that $ \tan A \tan B \tan C \ge 8 \sin A \sin B \sin C$ that is equal to the well-know $ \cos A \cos C \cos C \le \frac{1}{8}$.

We have the equality iff $ A \equal{} B \equal{} C$.


search trigonometry geometry Duke college

I have an ACT coming up in February, I'm in 8th grade, and probably don't know some of the math concepts involved.  Just looking for some tips, hints, people who have taken it before and want to share experiences, or anything that could help in the least.  I'm pretty firm in the other subjects, I'm just not sure how hard they'll make the math section and want to be prepared.

[quote="raccoon"]I have an ACT coming up in February, I'm in 8th grade, and probably don't know some of the math concepts involved.[/quote]

I'll put here a link to ACT's official list of topics for the ACT math section.

http://www.act.org/aap/testprep/mathstrat.html

I think that list of topics goes beyond the math level assumed by the SAT I, but it's possibly math that would be reasonably easy for a well-prepared eighth grader to pick up. Of course you will be sure to take the practice test that you get when you register for the ACT. Are you testing for the Midwest Talent Search? I have some local friends who are taking the 24 January 2004 SAT I for that talent search. I only took the SAT I back when I was in high school, so I don't have personal been-there, done that advice on the ACT.

Good luck!

(After edit:) I have split off from this thread posts that pertained to the SAT Math Test (which now has its own thread), which makes this thread look a little disjointed. I would like to hear from other forum participants who have taken the ACT, which is the dominant college-entrance test in my region of the United States.

Yeah, I was picked for the midwest talent search, and testing at my own high school actually.  Thanks for the link, I'll check it out.

The biggest problem I had when I took it in 7th grade was that I knew no trig, and I remember  there being lots of trig on the test. So it would be really good to know all your trig stuff well (just the easy parts, which I'm sure you already know...) and the rest I remember was really easy algebra problems.

Yea, it goes thorugh Alg/II Trig with Geometry and Algebra I.  There's only 4 Trig question so don't worry about it too much (But do what MagnusRex said, learn your basic Sine, Cosine, Tagent, ect.).  The most important thing is to get the ones you know how to do right.  I took the SAT I in 7th grade for the TIP program offered by Duke and this year I looked back at the ones I missed (College Board will give you a copy of your answer sheet if you ask for it by phone) and realized that almost all of the mistakes were either I had missed a negative sign or added when I should have subtracted or some "dumb" mistake like that.  Other than that make sure you know the basic volume and area formulas (the SAT also provides these on the question document and I think the ACT does too) and also know basic Alg I since most of the question on there are Alg I and all of the question are worth the same anyway no matter how hard it is. The best way to get aquainted with the ACT is to take an old one or a sample one.  There are a lot of those out there.

Nope, we could only take one, not both, for the MTS.

Whoa, off subject, I'm only taking the ACT.


For all positive numbers $ a$ and $ b$, a function $ f$ satisfies the equation $ f(ab)=f(a)+f(b)$. If $ f(2)=x$ and $ f(5)=y$, what is the value of $ f(100)$, in terms of $ x$ and $ y$?

Easy, again. :lol:
[hide]We know that f(ab)=x+y, so f(10)=x+y.  Thus, f(10x10)=x+y+x+y=[size=150][u][i][b]2x+2y[/b][/i][/u][/size].[/hide]

theres about 1000 different solutions if we didnt need it in terms of x and yf(10)=x+y so f(100)=f(10*10) which is x+y+x+y which is 2x+2y


What is the value of $ \sqrt[3]{3^5 \plus{} 3^5 \plus{} 3^5}$?

[hide="Countdown Thought Process"]
This may be redundant but for MATHCOUNTS Countdown the best way to eyeball these problems is to see that there are [b]3[/b] $3^5$s so we must add $1$ to $5$ to get $6$ and the cube root means take $\frac13$ of that so $3^2=9$

Not sure but I think that is a bit faster to get to the answer! [/hide]


geometry
incenter
geometric transformation
reflection
geometry unsolved

The incircle of a triangle $ ABC$ touches $ BC$ at $ D$ and $ AC$ at $ E$. Let $ K$ be the point on $ CB$ with $ |CK|\equal{}|BD|$ and $ L$ be the point on $ CA$ with $ |AE|\equal{}|CL|$. Lines $ AK$ and $ BL$ meet at $ P$. If $ Q$ is the midpoint of $ BC$. $ I$ the incenter, and $ G$ the centeroid of $ \bigtriangleup ABC$, show that:
(a) $ IQ$ and $ AK$ are parallel.
(b) the triangles $ AIG$ and $ QPG$ have equal areas.

a,Denote $ I_a$ the A-excenter of triangle $ ABC, L$ the reflection of $ D$ wrt $ I$. We have $ \frac {IL}{I_aK} \equal{} \frac {IE}{I_aX} \equal{} \frac {AI}{AI_a}$
Therefore $ A,L,K$ are collinear. Since $ IQ$ is the midline of triangle $ LDK$ then $ IQ//AK$. Similarly, $ IN//BL$
b, Denote $ M$ the reflection of $ A$ wrt $ I, N$ the midpoint of $ AC$. We have $ BP//IN//MC$.
Let $ F$ be the tangency of $ (I)$ and $ AB, T$ be the reflection of $ F$ wrt the midpoint of $ AB$ then $ AK, BL, CT$ concur at $ P$- the Nagel point of triangle $ ABC$. Similarly we get $ CP//BM$.
Thus $ BPCM$ is a parallelogram. Hence $ Q$ is the midpoint of $ MP$. We obtain $ \frac {AP}{IQ} \equal{} 2 \equal{} \frac {AG}{GQ}$. Then $ I,G,P$ are collinear and $ GP \equal{} 2IG \Rightarrow S_{AIG} \equal{} S_{QPG}$


[quote="Utkirstudios"]let a,b,c positive integers then prove:
\[ a\sqrt{a^{2}+3bc}+b\sqrt{b^{2}+3ac}+c\sqrt{c^{2}+3ab}\ge 2(ab+bc+ac) \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33515


solve for c:

$abc+a^2-\frac{1}{b^2}+c^2$

you need an equation in order to solve for a variable.

well, i guess you can't.  Well, evaluate the expression if a=7, b=3, and c=-1/2.


cos :theta:  >=  :sqrt: 3/2

graphically, i can understand that  :theta: will be between 30 and -30, but i don't know how to algebraically get this.

it's probably something really obvious.  does it have something to do with cos:theta: being equal to cos-:theta:?

well, the function f(x)=cos x is strictly decreasing in the interval 0<=x<=90. So since cos 30 = :sqrt: 3 / 2, if x>30, cos x > :sqrt: 3 / 2, which isnt what we want. On the other hand, if x<30, cos x < :sqrt: 3 / 2, which is what we want. On the other hand, in the interval -90<=x<=90, f(x)=cos x is strictly increasing, and since cos (-30) = cos (30) = :sqrt: 3 / 2,  -30<=x<=0. So together -30<=x<=30 is the interval in which cos x>= :sqrt: 3 / 2 holds.


Yay! I'm done my exams now!

Bio, Eng, Geo, Frn....

Is anyone still taking their exams next week?

Oh man.

Math (nothing at all) on monday, Chem (not hard) on tuesday. Need 100 on math, at least 98 on chem.

SAT subjects Math 1 and 2, Physics (easy) on saturday. Need 800 on each of those buggers.

Busy, Busy, Busy days...

I am just wondering.  Why do you take SAT 1 and 2?  I've noticed from many people that many colleges do not pay attention to 1 if you take 2.

Just finished my exams!!! Eng (was Wed), Chem (Thurs), Accounting + Bio (today)  :lol:

Gosh, am I the only one that's far from finished?

I don't know whether or not I need to take 1 or 2, but oh well, I'll just have get 800 on both.

Meh, 800 on 1 means no mistakes at all, and I'm sort of worried. :(


geometry
3D geometry
modular arithmetic
search
real analysis
quadratics

I suck at NT, but here's a start...

[hide="If x is odd..."] We first note that all perfect squares can be written as $ 4n \plus{} 1$ or $ 4n$

If $ x^2 \equal{} 4n \plus{} 1$, $ y^3 \equal{} 4n \plus{} 6$, which is never true, since an even perfect cube will always be a multiple of 4.[/hide]

[hide="Solution"]
Prove that the equation $ x^2\plus{}5\equal{}y^3$ has no solutions in $ Z$:

First note that $ x^2\equiv 0\pmod{4}$ or $ x^2\equiv 1\pmod{4}$.

Also note that $ x^2\plus{}5\equiv 1\pmod{4}$ or $ x^2\plus{}5\equiv 2\pmod{4}$.

If an integer $ w$ is a perfect cube, then $ w\equiv 0\pmod{4}$ or $ w\equiv 3\pmod{4}$, unless $ w\equal{}1$ in which case we note that the solution to $ x^2\plus{}5\equal{}1$ is not an integer.

Therefore, there are no integer solutions to the equation.

[/hide]

Wait, but doesn't $ 5^3 \equal{} 125\equiv1\mod 4$

Carbon57, you'r right I overlooked that. I'll try to fix it latter. Sorry about that.

It's a Lesbeguae equation(not sure if I spell the name correctly)! It's still unsolved for finding all positive integer k that makes the equation have at least one positive integer: x^2+k=y^3. The case when k=5 is already solved. In addition, the case when k<=100 is solved! Search on wiki or Mathworld to get more information if you want. :wink:

Lebesgue equation?  I know equations of this type by the name [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url].

[quote="goldengoal"]$ \text{Prove that the equation}$ $ x^2 \plus{} 5 \equal{} y^3$ $ \text{has no integer solutions.}$[/quote]
If $ y$ is even, then we have $ x^2 \equiv 3 \pmod4$ which is impossible.

If $ y \equiv \minus{}1 \pmod4$, we have $ x^2 \equiv2 \pmod4$ which is also impossible.

Therefore, we must have $ y \equiv 1 \pmod4$, and hence $ y^2\plus{}y\plus{}1 \equiv 3 \pmod4$. So $ y^2\plus{}y\plus{}1$ must have a prime factor $ p$ of the form $ 4n\plus{}3$.

Since $ x^2 \equal{} y^3 \minus{}5 \equal{} (y\minus{}1) (y^2\plus{}y\plus{}1) \minus{}2^2$, it follows that $ x^2 \equiv \minus{}2^2 \pmod p$ or \[ \left( \left( \dfrac{p\plus{}1}{2} \right) x \right)^2 \equiv \minus{}1 \pmod p.\] This is impossible because $ \minus{}1$ is not a quadratic residue modulo any prime of the form $ 4n\plus{}3$. So there is no integer solution to $ x^2 \plus{} 5 \equal{} y^3$.

I don't know why the solution by JZambrano201 is wrong...

JZambrano201's solution was wrong because he said that $ w \equal{} 1$ was the only possible value such that $ w\equiv1\pmod4$, when really, it is true for $ w \equal{} 1,125,729,\ldots$.


AMC
AIME
quadratics
algebra
polynomial
sum of roots

Source: 1997 AIME #7

A car travels due east at 2/3 mile per minute on a long, straight road.  At the same time, a storm in the shape of a circle, whose radius is 51 miles, moves southeast at :rt2:/2 miles per minute.  At time t=0, the center of the storm is 110 miles due north of the car.  At time t = t_1, the car enters the storm circle, and at time t = t_2 it exits the storm circle.  Find (t_1 + t_2)/2.

I just wanna say, that is a cool beans problem.

When it says south east, does it mean 45 degrees south east?

I beckon so. 7/4 :pi: radian, I presume.

mathfanatic wrote:A car travels due east at 2/3 mile per minute on a long, straight road.  At the same time, a storm in the shape of a circle, whose radius is 51 miles, moves southeast at :rt2:/2 miles per minute.  At time t=0, the center of the storm is 110 miles due north of the car.  At time t = t_1, the car enters the storm circle, and at time t = t_2 it exits the storm circle.  Find (t_1 + t_2)/2..

[hide]have a plane, where the origin is the initial position of the car. we write both equations in terms of t (time in minutes). the car will be at point (2t/3, 0), while the center of the storm will be at point (110-t/2, t/2). Since the radius of the storm is 51 mi, the equation of the storm (it's a circle) will be (x-(110-t/2)):^2:+(y-(t/2)):^2:=51:^2:. Now, at times t_1 and t_2, the car is on this circle. So we plug in the point (2t/3, 0). We get (2t/3-110+t/2):^2:+(2t/3-t/2):^2:=51:^2:=>(7t/6-110):^2:+(t/6):^2:=51:^2:

This yields a quadratic in t, but we just care about the first and second coefficients since we're going to find the sum of the roots. So the quadratic is 49t:^2:/36+t:^2:/36-770t/3+<constant>=0 => sum of roots will be (770/3)/(25/18)=770*18/25*3=924/5, which is not an integer :cry: ok so I probably made an arithmetical mistake somewhere but I can't find it, so this is the general method...[/hide]


ARML ratio inequalities function logarithms calculus integration

I am!
I'm missing an AP World History exam for this.

Dang, that's pretty intense...lol

I'm missing a day of French...whoo hoo. Anyone not from Northview's team?

I'm not.
So why was this thread started ,
arvind?

Ummm, who r u???

I started it to, u know, provide somewhere to talk about it...um...yeah.

i think it's cosmo, b/c i interrogated him yesterday concerning his sn, and he told me something like cbk.

Well then he is going. Ha. Btw I know what cbk stands for XD

ME. I heard the test is made this year by ARML members... this should be interesting  :)

Also, most of the harder questions were edited out by teachers. The test was written by "former students" aka the people Howard mentioned.

BTW: The tournament is on the 20th, right? Mr. Hedrick failed to answer that 1 question.
BTW2: Karthik, nice location: 01000111011001010110111101110010011001110110100101100001.

karth's location messes up my window size =(

Yea, I changed it back to USA... I wanted to see how long until someone would complain  :P

[quote="Xantos C. Guin"]BTW2: Karthik, nice location: 01000111011001010110111101110010011001110110100101100001.[/quote]

But does it actually mean anything? Or is it just a bunch of 1's and 0's. Now if it was binary encoded ASCII, THAT would be good! :-)

For instance, USA = 010101010101001101000001

It [i]is[/i] ASCII! (kudos to Karthik) (after Mr. Fulton mentioned that possibility, I fired up python and figured out that it was ascii :))

So it is. Very nice Karthik! I just got lazy and didn't figure it out myself. (And being the old guy that I am, I just read the ascii chart without the electronic help.)

What does Karthik's former location mean?  Yeah, I'm technologically challenged, so don't make fun of me... :blush:

Why, it's where he lives of course!  :D

Read it as groups of 8 and use that and this chart:

[url]http://www.asciitable.com/[/url]

[quote="captcha000"]oh yeah....varsity was me, eric, and santhosh...i think jv was written by ruby, henry, and sam. hope you guys have fun ;)[/quote]

...crap, i'm dead...i wish i could be a slacker and do jv but no...sigh

Hmm, I liked the test, because even though I couldn't finish, I looked it over and after you get past its original "omigosh"ness, it's actually pretty elegant...

Great test! I really commend Santosh, Howard, and Eric who spent much of their time making this test. I love how practically everyone got only 10 or so questions right on the written test and placed  :D

The ciphering was pretty tricky for me... I knew how to do many of them, but slipped up on some because of going to fast (for example, with the one about the infinite series, I forgot the 'positive' part and put 0 and the negative golden ratio for answers). Great test overall! I feel kinda sad that I won't be able to attend the tournament next year though  :(

[quote="Karth"]Great test! I really commend Santosh, Howard, and Eric who spent much of their time making this test. I love how practically everyone got only 10 or so questions right on the written test and placed  :D [/quote]

Seriously? Cause that's not what we were shooting for. Just out of curiosity, what were the top scores like?

Well, I got 15 out of the 18 I answered and got 2nd place. Harrison answered 22 or so, I think, and missed a few.

I don't feel I did too well, even though I got 10th...

I got a 25 on ciphering. A 25. Yes. Count 'em. 25...AARGH!!!

My written score is sad, like this smiley.  :(  Or possibly this one. :stink:

Should have read AoPS 2 better...lolz

Hoping to do better at GA Tech.

Yeah, haha, it's easy to write a test that's way too hard if you're not careful.  You have to have a really light touch or you'll become a mass murderer.

I'm glad you guys liked it. I figured it was kind of too hard; I hope everyone there was able to at least solve a few problems. So what were everyone's favorite problems?

[quote="Karth"]By the way, for the question on the last page which involved something about maximizing the square root of some algebraic expression, there's a pretty cool way to do it with Cauchy's Inequality... let's see if I can find the question sheet  :maybe:[/quote]

I believe the question reads as follows:

If $ 3x \plus{} 7y \equal{} 42$, find the minimum value of $ \sqrt {x^2 \plus{} y^2 \plus{} 6x \plus{} 8y \plus{} 25}$?

BTW: Where there any errors on the test?

No errors on the test, apart from perhaps a small typo towards the end... when you said $ A_i$ denoted the vertices, I interpreted it (fortunately) as the angles.

This was my solution to the above problem:

We have that

\[ x^2 \plus{} y^2 \plus{} 6x \plus{} 8y \plus{} 25 \equal{} (x\plus{}3)^2 \plus{} (y\plus{}4)^2
\]

Now, by Cauchy's Inequality,

\[ (3^2 \plus{} 7^2)(x^2 \plus{} y^2) \ge (3x \plus{} 7y)^2 \iff (3^2 \plus{} 7^2)[(x\plus{}3)^2 \plus{} (y\plus{}4)^2] \ge [3(x\plus{}3) \plus{} 7(y\plus{}4)]^2 \iff (x\plus{}3)^2 \plus{} (y\plus{}4)^2 \ge \frac{79^2}{58}
\]

Therefore, we have that $ \sqrt{x^2 \plus{} y^2 \plus{} 6x \plus{} 8y \plus{} 25} \ge \frac{79}{\sqrt{58}}. \blacksquare$

SHMOINER SQUARE!!! (Sitan's and my way of saying Halmos) :rotfl:  :rotfl:  :rotfl:

ok, my fav problem was probably either the one with the cubic roots of , or the hot dog and hamburger one. Either that, or number one (I believe it was the one with the function f(xy)=f(x)+f(y)). I like quick, fun things like that that are easier than they seem, even if they seem easy. Then they are just easier than easy, unless they appear so, in which case they are a bit easier than easier than easy.

I agree, that's why I liked it, I'm good at that stuffz.

But yea, I did it rigorously cuz i'm a nubcake.

[quote="Karth"]Hmm... one thing I must say about the test though was that it was unusually focused on functional analysis. About 3 or 4 problems (I think) were on functional analysis out of the 25... that's more than 12 or so percent. However, for the one about $ f(xy) \equal{} f(x) \plus{} f(y)$ a really easy/cheap way to do it is to let $ f(x) \equal{} \ln{x}$ and proceed. Or you could do it the rigorous way and factor 2008 and stuff...  :roll:[/quote]

I wrote most if not all of those problems. When writing problems, I wrote them in chunks on the same topic. To choose problems, I categorized everything, set approximate goals for each topic, then we voted on which problems we liked and came to a consensus. After that, the teachers requested changes which we made.

That first question was one that I didn't like a lot, but I was willing to put it in. There were some other cool combinatoric/advanced questions that were taken out. One involved a ratio of two integrals that were both relating to the volume of some object. The two integrals themselves were hard to evaluate, but the ratio was easy if you knew that both related to calculating the volume of a certain object.

[quote="Arvind_sn"]the hot dog and hamburger one.[/quote]Interesting - that's one of my least favorite questions that I wrote. It seemed too straightforward and wordy.

I liked the cube root one. I actually already memorized the formula, so it was just plugging in numbers. :lol:  I somehow messed up one the hotdog and hamburger one. That was weird.
Anyway, I'm happy with 4th right now.


geometry
incenter
trigonometry
similar triangles

[img]http://img47.imageshack.us/img47/3602/picturekt7.jpg[/img]

We know that $ IE \bot BD, IF \bot CD$ where $ I$ is the midle of incircle and $ D, E, F$ are like you may see above.
What's more $ |IE| \plus{} |IF| \equal{} \frac {|AD|}{2}$

$ \angle BAC \equal{} ?$

It would be nice to have one, but unfortunately I don't have any... :|

I haven't solved the problem, but here's a hint:

[hide="in here"]The incenter is the intersection of three special segments in a triangle. What are they?[/hide]

ok @1=2, but is this 'hint' really leading to the final solution?

To tell you the truth I don't know. When you don't know what to do, try random stuff(i.e. extending some lines that look special, finding similar triangles, angle chasing, etc.)

I also see quite a few cyclic quads, like IFDE and ABCD. If you make radii from I to AC and AB and call them X and Y respectively, we have two more: IFCX and IEBY. I think they would be useful in this problem somehow.

And if O is the center of that bigger circle, then $ OD\bot BC$.

[hide]Let $ \angle{BAD} \equal{} \alpha$.Then
$ \dfrac{1}{4}ID.AD \equal{} \dfrac{1}{2}ID.(IE \plus{} IF)$

$ \equal{} \dfrac{1}{2}BD.IE \plus{} \dfrac{1}{2}CD.IF \equal{} [BID] \plus{} [DIC]$

$ \equal{} \dfrac{ID}{AD}([BAD] \plus{} [DAC]) \equal{} \dfrac{1}{2}ID.(AB \plus{} AC).\sin \alpha$

whence $ \dfrac{AD}{AB \plus{} AC} \equal{} 2\sin \alpha.$

     Let $ X$ be the point on ray $ \vec{AB}$ different from $ A$ such that $ DX \equal{} DA$.
Because $ \angle{XBD} \equal{} \angle{DCA}$ and $ \angle{DXB} \equal{} \angle{XAD} \equal{} \angle{DAC}$, we have $ \triangle{XBD}$ and $ \triangle{ACD}$ are similar and $ BX \equal{} AC$. Then

$ 2\sin \alpha \equal{} \dfrac{AD}{AB \plus{} AC} \equal{} \dfrac{AD}{AB \plus{} BX} \equal{} \dfrac{AD}{AX} \equal{} \dfrac{1}{2\cos \alpha}$

so that $ 2\sin \alpha \cos \alpha \equal{} \frac {1}{2}$, and $ \angle{BAC} \equal{} 2 \alpha \equal{} 30$ or $ 150$[/hide]


Given the two sets of complex numbers, $A = \{z : z^{18} = 1\}$, and $B = \{z : z^{48}=1\}$, how many distinct elements are there in $\{zw : z \in A, w \in B\}$?

You have to type \{ and \} to get $\{ \}$ in $\LaTeX$.

[hide]
This is just a fancy way to ask how many possible products there are when you multiply an 18th root of unity and a 48th root of unity. The 18th roots of unity are in the form $e^{\pi i j/9}$ for $0\le j \le 17$, and the 48th roots of unity are in the form $e^{\pi i k/24}$ for $0\le k \le 47$.

The product of these is then in the form $e^{\pi i(j/9+k/24)}=e^{\pi i (8j+3k)/72}$. Thus we need to find all possible values that $8j+3k$ can attain that are not greater than 144. By the chicken mcnugget thm, the highest value that it cannot attain is 24-11=13. A quick check tells us that it can attain 12, 11, 9, 8, 6, 3, and 0 but not 10, 7, 5, 4, 2, or 1. However... it [i]can[/i] attain 10+144=154, 7+144=151, 5+144=149, 4+144=148, 2+144=146, and 1+144=145. This means that the answer is $\boxed {144}$.
[/hide]

Don't know if that's right though...

You are correct.
This was #10 on 1990 AIME by the way.

I am kind of confused though:
You say
"Thus we need to find all possible values that 8j+3k can attain that are not greater than 144."

But go on to say
"However... it can attain 10+144=154, 7+144=151, 5+144=149, 4+144=148, 2+144=146, and 1+144=145."

Aren't these greater than 144?

Or is this because they repeat after $2\pi$ ?  I've never really worked with complex numbers in this sense so I am somewhat confused.

Right, we had $e^{\pi i (8j+3k)/72}=e^{\pi i (8j+3k)/72+2\pi i}=e^{(8j+3k+144)/72\pi i}$, so even though we can't get 10, 7, 5, 4, 2, or 1, we can get the numbers that are 144 greater than them.

(I think...)

[hide="another solution"]set A has elements that have angles of $20n$, and set B has elements $7.5m$, since we add angles in complex multiplication, we can thing of the problem as how many different numbers $p$ can be made by $20n+7.5m=p$ with $p\in[0,360)$ or $8n+3m=q$ $q\in[0,144)$
by CRT, since 8 and 3 are coprime, we can find integer solutions for all integers $q$, so all $q$ in that range are solutions, $\boxed{144}$[/hide]

[hide]Rewrite the elements in terms of $e^{\frac{2 \pi i}{144}}$, since the set of 144th roots of 1 contains all 18th roots of 1 and all 48th roots of 1. Observing how the exponents multiply with eachother, the number of distinct products is just $lcm(18,48)=144$[/hide]


[quote="8th Army"]Thanks, is China Olympiad harder than American Olympiad?[/quote]
Definitely, and China TST should be the hardest around the world.

前苏联的几个加盟共和国的队都廷强的

[quote="8th Army"]前苏联的几个加盟共和国的队都廷强的[/quote]
恩,百足之虫,死而不僵.
苏联的光辉不散啊!

俄罗斯太穷了,没钱给队员进行强化训练

[quote="8th Army"]俄罗斯太穷了,没钱给队员进行强化训练[/quote]
啊!?

Of course, Russia's economy is not very good now, but the people study hard though

[quote="shobber"][quote="8th Army"]Of course, Russia is pretty poor now, but the people study hard though[/quote]
It seems that Myth is a teacher in Russia. His students get pretty high scores. He seems to give his students lots of exercise...don' t what the rest of teachers do...[/quote]
Myth is actually a postgraduate now, and, I believe that Russia's economy is at least better than China's now.

[quote="shobber"]I thought he was a teacher...He mentioned: "My students get..." when he was talking about All Russia MO...
So he is still pretty young!

As for the Chineses economy: It is not very good now. But the Chinese people pay a lot on IMO......You can figure it out if you go into any mailand bookstore...there are mountains of books about math contest![/quote]
Of course he is very young...about 22-25 years old, I cannot recall :blush:
I have been buying MO books from China. However, for each time I have to spend much time choosing due to the varied qualities. I don't think quantity is any important, but quality. In other words, 少而精 ;)
Anyway, it's true that Chinese people really pay a lot on MO...but this is what I believe to be too extreme: they should spend some time on "non-MO math". Actually a more precise saying will be they just focus on what they
[i]have to[/i] learn. This can be seen by the fact that there are rarely(or even none!) any recreational math books in China. In fact, some famous Chinese MO trainers [i]do[/i] write books about it, but because of the unfavourable market in China, they are all published by "ChiuChiang" in Taiwan.

[quote="shobber"]我前几天看了一篇文章,是中国队的一名教练写的(好象是熊斌?). 他在文章中指出,中国的奥林匹克数学教育并不会对孩子有太多的启发.相反,这是不利于许多孩子在数学方面的发展的. 文章说,过早的让孩子接触那些高深的数学知识会让他们对数学失去兴趣,这对他们今后的发展是极为不利的. 但由于中国目前高考中考形势严峻,家长们又迫使孩子学习这些他们本没有兴趣的东西. 这就造成了书目泛滥,但真正能出成绩的却不多。最后竹篮打水,只有书商占了便宜.[/quote]
以后真正学数学的竞赛尖子确实很少...不过你也不能否认数学竞赛对个人在其他学科的帮助，如物理、计算机等。

我不否认, 可问题是, 他们真正对这些东西感兴趣么? 如果对学习没有兴趣, 那有帮助也没什么用.

[quote="shobber"][quote="Enya"]:?  :ninja:  :huh: 真牛逼,居然连集训队的试题都弄到了,不简单!
该不会是集训队的成员吧,那么我就更加佩服不已拉 :D[/quote]
他好象是老师吧? Wpolly 是CMO冠军,当然是集训队成员啦!
还有, 牛逼的 逼 字应该是"人+肉"吧? 谁知道怎么打?[/quote]

红楼梦中有不少这类的词, 就看你读得细心不细心了.

[quote="shobber"]我前几天看了一篇文章,是中国队的一名教练写的(好象是熊斌?). 他在文章中指出,中国的奥林匹克数学教育并不会对孩子有太多的启发.相反,这是不利于许多孩子在数学方面的发展的. 文章说,过早的让孩子接触那些高深的数学知识会让他们对数学失去兴趣,这对他们今后的发展是极为不利的. 但由于中国目前高考中考形势严峻,家长们又迫使孩子学习这些他们本没有兴趣的东西. 这就造成了书目泛滥,但真正能出成绩的却不多。最后竹篮打水,只有书商占了便宜.[/quote]

应该是丘成恫吧？到处都是他的文章。


Definitely doing problems. The problem is, I don't have desk for my laptop ( :( ), so its stacked up on a few moving boxes and I don't have enough space to work on either side :(

But, I come up with a new weird solution every time :)

Recapitulating myself with Cornell Notes :)

b[hide="oo"]oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo[/hide]k

[color=green]Mod edit: No need for so many o's.[/color]


Let a,b,c be non-negative numbers, no two of which are zero.then:

\[\sum_{cyc}{\sqrt{\frac{a}{{4a+5b}}}\le 1}\]

[hide]
First let $x=\sqrt{\frac{a}{4a+5b}}$ etc. so that we are given
\[(1-4x^{2})(1-4y^{2})(1-4z^{2})=125x^{2}y^{2}z^{2}\]
and we need to prove $x+y+z\leq 1$. Suppose for the sake of contradiction that $x+y+z>1$. Then
\[(1-4x^{2})(1-4y^{2})(1-4z^{2})<\]
\[(3x+y+z)(3y+z+x)(3z+x+y)(x+y-z)(x+z-y)(y+z-x)\]
(This follows from homogenizing, factoring, and noting that we can multiply the inequalities because $x,y,z\leq \frac{1}{2}$.) So a contradiction will follow from
\[(3x+y+z)(3y+z+x)(3z+x+y)(x+y-z)(x+z-y)(y+z-x)\leq 125x^{2}y^{2}z^{2}\]
Let $y+z-x=2p$ etc. so $q+r=x$. Now at most one of $p,q,r$ negative (since the sum of any two is). But if exactly one is negative, above inequality trivial. Otherwise, $p,q,r$ all nonnegative, and the above inequality reduces to
\[125(p+q)^{2}(q+r)^{2}(r+p)^{2}\geq 64pqr(p+2q+2r)(q+2r+2p)(r+2p+2q)\]
To verify this, first note that
\[(p+q)(q+r)(r+p)=(p+q+r)(pq+qr+rp)-pqr\geq\]
\[\sqrt{3}(p+q+r)^{3/2}\sqrt{pqr}-\frac{(p+q+r)^{3/2}}{3\sqrt{3}}\sqrt{pqr}=\frac{8}{3\sqrt{3}}(p+q+r)^{3/2}\sqrt{pqr}\]
by $p+q+r\geq 3\sqrt[3]{pqr}$ and $(pq+qr+rq)^{2}\geq 3pqr(p+q+r)$. The remaining inequality,
\[125(p+q+r)^{3}\geq 27(p+2q+2r)(q+2p+2r)(r+2p+2q)\]
is just AM-GM on $p+2q+2r,q+2p+2r,r+2p+2q$. Done.
[/hide]

Very nice proof, scorpius119. :lol:
It's old Vasc's inequality. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549&start=20

[color=#FF0000]it's wrong![/color]

[hide]my proof to the more general 4 varibles:

[quote="red3"]my proof:

[hide]it suffice to prove :

$ \prod (\frac{a+b+c}{3})^{2}\ge abcd(\frac{a+b+c+d}{4}) ^4$

assume $a+b+c+d=4 $

it suffice to prove that :          $(4-a)(4-b)(4-c)(4-d) \ge 81\sqrt{abcd}$

or :              $16(ab+bc+cd+da+ac+bd)-4(abc+bcd+cda+dac) +abcd\ge 81\sqrt{abcd}$

by$ AM-GM$          $ \frac{27}{2}(ab+bc+cd+da+ac+bd)\ge 81\sqrt{abcd}$

it suffice to show that :

$5(ab+bc+cd+da+ac+bd)-8(abc+bcd+cda+dac) +2abcd\ge 0$

let $f(a,b,c,d)=5(ab+bc+cd+da+ac+bd)-8(abc+bcd+cda+dac) +2abcd\ge 0$

$LEMMA$   if$ c+d\le 2 $,l$et c+d=2t $,$then f(a,b,c,d) \ge f(a,b,t,t)$

WLOG,assume $c=t+x,d=t-x$ $0<x\le t$

then from $f'(x)=16(a+b)x-4abx -10x >0 $,we have proved the lemma

now,assume $a\ge b\ge c\ge d$,

$f(a,b,c,d) \ge f(a,b,\frac{c+d}{2},\frac{c+d}{2}) \ge  f(a,\frac{2b+c+d}{4},\frac{2b+c+d}{4},\frac{c+d}{2}) \ge ...$

take the limmit ,we should only prove $f(a,\frac{4-a}{3},\frac{4-a}{3},\frac{4-a}{3},)$

or$(1-a)^2(10-a)\ge 0 $,done![/hide]

i haven't checked it  yet[/quote][/hide]


algorithm
function
quadratics
real analysis
number theory
Euclidean algorithm
totient function

What book should I use to study number theory?
I'm very weak in number theory, that i only know euclidean algorithm, simple congruence, FlT, totient function theorem and some quadratic residues. (I know something about dirichlet convolution too, like mobius inversion formula, but i don't how to apply them at all)
I don't know where should I start.....
(I really want to learn number theory because I like it very much, while I find it difficult for me to discuss in the number theory forum because the level of people there are far beyond me.. :() Thanks for your kind attention..

I would first advise you to stop using chat speak (lvl, ppl, etc.) Secondly NT is learnt by solving lots and lots of problems (if you already have all the basic knowledge about theorems, Euler's function, Fermats' theorems, Pell's equation and so on).

Lastly this is hardly an International Mathematical Olympiad topic, so this goes into it's proper place ;)

The book they use for the number theory class at my college is by Niven, Montgomery, and Zuckerman. I haven't read it, but it's generally considered to be a very good book.

Valentin, I have fixed my post, and sorry for posting in a wrong forum..

ComplexZeta, I think I can borrow this book from my school library after easter holiday. (It's introduction to the theory of numbers right?)

I don't remember the title, but [i]An Introduction to the Theory of Numbers[/i] is by Hardy and Wright (which is a fantastic book, but alas there are no problems). The Niven book may have the same title.

Yes, I used the NZM book for my first number theory course, and I really liked it. For me, it was somewhat easier to understand than Hardy and Wright's book.

Number Theory
by George E Andrews

It might be a bit basic for you, but it's really well written


Mine are Donald Duck, Dr. Zachery Smith from "Lost in Space", Fred Flinstone, etc. The list goes on.....

...I like billy, and mandy, and the grim reaper and anyone from that show...

I like Donald Duck too, and Sheen (sp?) from Jimmy Neutron.  :D

James Roday (Shawn Spencer, Psych), Hugh Laurie (Doctor House, House) and William Petersen (Gil Grissom, CSI)


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

$D$ is a diagonal matrix of size $n \times n$ and all of its diagonal elements are positive. $A$ is a $n\times n$ matrix s.t. $A^T = -A$. Show that $\det(A+D) > 0$.

Let $D=diag[d_1,d_2,\ldots,d_n]$ with $d_k>0$ for $k=\overline{1,n}$.

Let $f(x)=\det(A\cdot x+D)$, for $x\in\mathbb{R}$.
$f(x)=\det(A\cdot x+D)=\det(A\cdot x+D)^{T}=\det(A^T\cdot x+D^T)=\det(-A\cdot x+D)=f(-x)$.

If we expand the determinant in f, we see that f is a polynomial of degree least or equal to n.

We see that $f(0)=\det(D)=\prod_{k=1}^nd_k>0$. We have to prove that $\det(A+D)=f(1)>0$.

Assume $f(1)\leq 0$.

Because f(0)>0, by Darboux, there exists $t\in\mathbb{R}$ such as f(t)=0.
This means $\det(A\cdot t+D)=0$.
Let $T=A\cdot t$. We have $T^T=-T$ and $\det(T+D)=0$. So there exists $u\in\mathbb{R}^n^*$ such as $(T+D)\cdot u=0$. By transposing, we get $u^T\cdot (-T+D)=0$.

$u^T(T+D)u=u^T((T+D)u)=u^T\cdot 0=0$.
$u^T(-T+D)u=(u^T(-T+D))u=0\cdot u=0$.

These two imply $u^TTu=u^TDu=0$.
Let $(u{}_i){}_{i=\overline{1,n}}$ be the components of u.
We have $0=u^TDu=0\Rightarrow \sum_{k=1}^n u_k^2\cdot d_k=0$.
Because the $d_k$'s are all positive, the last means $u_k=0\ \forall k=\overline{1,n}$. This contradicts the choice of u in $\mathbb{R}^n^*$. So f(1)>0.

[quote="liyi"]Your proof is quite good. Thanks, amfulger[/quote]

Thanks  :blush: . I live for this kind of appreciation.


prove that there are an infinite number of composites

Doesn't this follow directly from there are infinitely primes?

I agree.

In another thread you posted, someone already proved that there are infinite primes. If there are infinite primes, there are also infinite composites because composites are basically created by using primes (whether it's adding or multiplying. Pick your favorite)

it just enough to cnsider a prime number like $2$, and so $2*2,2*3,2*4,.....,2*n,...$ so we have infinte composite ;)

It is not enough just to say that there are infiintie primes. You prove it almost the same way as infinite primes. Say there are a finite number of composites, $c_1,c_2,c_3,...,c_k$. Multiply them together but don't add $1$. $c_1\cdot{c}_2\cdot{c}_3\cdots{c}_k$. This is not equal to one of the finite primes $c_n$, but is divisible by all $c_n$, so it is composite

take a prime number say 2, and multiply it buy itself. 2*2=4, so 4 is composite.  So if you take a prime number, n, and square it $n^2$ will always be composite.


Putnam
inequalities
pigeonhole principle
college contests

Prove that, for every set $X=\{x_{1},x_{2},\dots,x_{n}\}$ of $n$ real numbers, there exists a non-empty subset $S$ of $X$ and an integer $m$ such that
\[\left|m+\sum_{s\in S}s\right|\le\frac1{n+1}\]

My approach could probably be trimmed this up a bit, but here is
my solution:

$\forall S\subset X, S\neq\varnothing$, we have $\sum_{s\in S}s \in \mathbb{R}$ since $s\in X$ and $\forall x_{i}\in X$, we are given $x_{i}\in\mathbb{R}$ .  So, $\exists t\in\mathbb{Z}$ s.t:

$\lvert\sum_{s\in S}s\rvert \leq t$.  Therefore, we know

$-t\leq \sum_{s\in S}s \leq t$
$\Longrightarrow$ $-t+\frac{1}{n+1}\leq \sum_{s\in S}s+\frac{1}{n+1}\leq t+\frac{1}{n+1}$ for $n\in\mathbb{N}$

If we consider the right hand side of the inequality, we see that:
$\sum_{s\in S}s \leq t+\frac{1}{n+1}$ since $\frac{1}{n+1}>0$.

So,

$-t+\sum_{s\in S}s \leq \frac{1}{n+1}$

Thus, for some $m=-t \in\mathbb{Z}$, we have

$\lvert m+\sum_{s\in S}\rvert\leq \frac{1}{n+1}$.  QED.

Let's divide the interval $[0, 1]$ into $n+1$ closed subintervals of length $\frac{1}{n+1}$.  Let's think of those subintervals as "pigeonholes".

Let's consider the $n+1$ subsets $\emptyset$, $\{x_{1}\}$, $\{x_{1}, x_{2}\}$, ..., $\{x_{1}, x_{2}, \ldots, x_{n}\}$.  To each such set, let's map it into its sum modulo 1.  As a special case, let's map the empty set into two numbers (0 and 1).  So we have $n+2$ numbers in $n+1$ subintervals.  By the pigeonhole principle, two of those numbers are in the same subinterval.  By subtracting, we get the desired nonempty subset.

I don't think I said it very well, but hopefully you get the idea.

@FlatSphere:
Nice thought, but it doesn't work: when you get to $-t+\sum_{s\in S}s \leq \frac{1}{n+1}$, you're still good, but you can't get to the next step: you also need $-\frac1{n+1}\geq-t+\sum_{s\in S}s \leq \frac{1}{n+1}$, but you've given that up by dropping the $\frac1{n+1}$.

This one falls to the (somewhat classic) trick of considering the sets $S_{1}= \{x_{1}\}$, $S_{2}= \{x_{1}, x_{2}\}$, . . ., $S_{n}= \{x_{1}, \ldots, x_{n}\}= X$ and arguing either than one of them is of the appropriate size or (by pigeonhole) the difference between two of them is.

wow, i just realized my solution to this problem is almost identical to the official solution :)

I used a version of the dirichlet approximation theorem: for all a in R, for all N in natural numbers, there exist n <=N and there exist p in integers such that

n*abs(a-(p/n))<(1/N)

And i thought the problem is exactly the same thing as dirichlet approx thm.  then how come I didn't get any point for this problem.  something's wrong!  but how?[/code]

[quote="jmerry"]You didn't get anything because that theorem doesn't apply in this situation. We have many different numbers, not many copies of the same one.[/quote]

i dont get it.  not many copies of the same number?  what do you mean?

It suffices to find a $S$ such that the sum has fractional part in $\left[0, \frac{1}{n+1} \right)$ or $\left[ \frac{n}{n+1}, 1 \right).$ Let $S_{i}=\{x_{1}, \ldots, x_{i} \}$ for $i=1,2,\ldots,n$ and consider the fractional parts of their sums. If any of them is in one of the two intervals above; we are done. Otherwise, we find $S_i \subset S_j$ such that the sum in both cases has fractional part in $\left[\frac{k}{n+1}, \frac{k+1}{n+1}\right)$ and then $S= S_{j}\setminus S_i$ satisfies the condition.

[hide=Definitions]Denote $f(x)$ to be the fractional part of $x \in \mathbb{R}$. I.e.,
$$f(x) = x - \lfloor x \rfloor ,$$
where $\lfloor x \rfloor$ denotes the greatest integer not larger than $x$. Denote $\Sigma i = \sum _{k=1} ^i x_k$ for $1 \leq i \leq n$.[/hide]

[hide=Lemma 1]Lemma 1: Let $x,a$ be real with $0 \leq a \leq \tfrac{1}{2}$. If $f(x) \leq a$ or $f(x) \geq 1-a$, then there exists an integer $m$ such that
$$| m + x | \leq a.$$

Proof: If $f(x) \leq a$, we can write $x=n+r$, where $n \in \mathbb{Z}$ and $0 \leq r \leq a$. Then we can choose $m=-n$ to obtain
$$|m+x| = |-n + n + r| = |r| = r \leq a.$$
If $f(x) \geq 1-a$, we can write $x=n+r$, where $n \in \mathbb{Z}$ and $1-a \leq r < 1$. Then we can choose $m=-n-1$ to obtain
$$|m+x| = |-n -1 + n + r| =^1 |r-1| =^2 1-r \leq a$$
where we used $r<1 \Rightarrow r-1 < 0$ for $^1$ and $1-a \leq r \Rightarrow 1-r \leq a$ for $^2$.[/hide]

[hide=Lemma 2]Lemma 2: Let $x,y,a$ be real with $0 \leq a \leq \tfrac{1}{2}$. If $|f(x) - f(y)| \leq a$, then
$$f(|x-y|) \leq a \text{ or } f(|x-y|) \geq 1 - a.$$

Proof: Let $x = n_x + r_x$ and $y = n_y + r_y$, where $n_x, n_y \in \mathbb{Z}$ and WLOG $0 \leq r_y \leq r_x < 1$. Thus,
$$|f(x) - f(y)| \leq a \Rightarrow |r_x - r_y| \leq a \Rightarrow 0 \leq r_x - r_y \leq a.$$
If $x \geq y$, we have
$$f(|x-y|) = f(x-y) = f(n_x + r_x - n_y - r_y) = f(r_x + r_y) \leq a.$$
If $x < y$, we have
$$f(|x-y|) = f(y-x) = f(n_y + r_y - n_x - r_x)$$
$$= f(1 + r_y - r_x) \geq 1 - a \text{ for } r_y \neq r_x, \text{ } = f(0) \leq a \text{ for } r_y = r_x$$
where we used the fact that $0 \leq r_x - r_y \leq a \Rightarrow 1 \geq 1 + r_y - r_x \geq 1 -a.$[/hide]

[hide=Solution]If $f(\Sigma i) \leq \tfrac{1}{n+1}$ or $f(\Sigma i) \geq \tfrac{n}{n+1}$ for some $i$, $1 \leq i \leq n$, then by lemma 1 we could find an integer $m$ such that
$$|m+\Sigma i| \leq \frac{1}{n+1}$$
$$\Rightarrow \left| m+ \sum _{s \in S} s \right| \leq \frac{1}{n+1}$$
where $S = \{ x_1, \dots, x_i \}$. Note that for $n=1$, $x_1$ always satisfies $f(x_1) \leq \tfrac{1}{2}$ or $f(x_1) \geq 1 - \tfrac{1}{2} = \tfrac{1}{2}$, so choosing $S = \{ x_1 \}$ satisfies. Thus, assume $n \geq 2$ and $\tfrac{1}{n+1} < f(\Sigma i ) < \tfrac{n}{n+1}$ for all $1 \leq i \leq n$. Place each of $\Sigma 1, \Sigma 2, \dots, \Sigma n$ into $n-1$ containers corresponding to $\tfrac{k}{n+1} < f(\Sigma i ) \leq \tfrac{k+1}{n+1}$ for all $1 \leq i \leq n$. By the pigeonhole principle, there must exist some container with at least two $f(\Sigma i)$. I.e., for some $1 \leq i < j \leq n$ and $1 \leq k \leq n-1$, we have
$$\frac{k}{n+1} < f(\Sigma i), f(\Sigma j) \leq \frac{k+1}{n+1}$$
$$\Rightarrow \left| f(\Sigma j) - f(\Sigma i) \right| \leq \frac{1}{n+1}.$$
Thus, by lemma 2 we have
$$f(\Sigma j - \Sigma i) = f \left( \sum _{k=i+1} ^j x_k \right) \leq \frac{1}{n+1}$$
or
$$f(\Sigma j - \Sigma i) = f \left( \sum _{k=i+1} ^j x_k \right) \geq \frac{n}{n+1}.$$
And by lemma 1, there must exist an integer $m$ such that
$$\left| m+\sum _{k=i+1} ^j x_k \right| \leq \frac{1}{n+1}$$
$$\Rightarrow \left| m+ \sum _{s \in S} s \right| \leq \frac{1}{n+1}$$
where $S = \{ x_{i+1}, \dots, x_j \}$.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let the group G have order 110. How many elements of order 29 are there in G? What are the possible number of elements of order 406 having a different number of elements of order 2?

- There can be no elements of order 29 in a group of size 110. That's an immediate consequence of Lagrange's theorem.

Sry i made a mistake the group is of order 406 not 110

[b]The question actually is:[/b]

Let the group G have order 406. How many elements of order 29 are there in G? What are the possible number of elements of order 7 in G. Give examples of different groups of order 406 having a different number of elements of order 2?

[b]G[/b] has only one [b]29[/b]-Sylow subgroup. Therefore, there are [b]28[/b] elements of order [b]29[/b] in [b]G[/b]. On the other hand, [b]G[/b] has either one or twenty-nine [b]7[/b]-Sylow subgroups. Hence, [b]G[/b] has either [b]6[/b] elements of order seven or [b](6)(29) = 174[/b] of them.

Thanks... And what about the last part of the question please?


Q.Find the probability that in any randoly assembled group of n people atleast two people have the Birth on the same day of a year (the year may be different but the month and the date have to be the same).Assume all to be born on a non leap year

[hide="My approach"]
let $ P_{n}$ be the required probabilty
 $ p_{(n,m)}$ be the proabbility that [b]exactly[/b] m people in a gathering of $ n$ people have the birthdays same
then
Assume
$ P_{n} \equal{} \sum_{m \equal{} 2}^{n}p_{(n,m)}$
probability that eactly m people out of n have the Birthday same. is sum of the probability taht each group of m people have the birthdays on the same day.hence $ \frac { \binom{n}{m}}{365^{m}}$
so $ P_{n} \equal{} \sum_{m \equal{} 2}^{n} \frac {\binom{n}{m} }{365^{m}}$ tis gives $ P_{n} \equal{} (1 \plus{} \frac {1}{365})^{n} \minus{} (1 \plus{} \frac {n}{365})$
[/hide]

The number $ \frac{\binom{n}{m}}{365^m}$ is not the probability you think it is. In general, what you called $ p_{(n,m)}$ is painfully difficult to calculate. You must worry about overlaps, and inclusion-exclusion, and that all the other birthdays are different.

The classic question in approaching any combinatorics or combinatorial probability question: which set is easier to describe? The set we want, or its complement? In this particular case, the complement is far, far easier to describe. $ 1\minus{}P_n$ is simply the probability that all $ n$ people have different birthdays. Work from that side.


[quote="monkeygirl12"]Do I Have to get parentral permission if I am an aops member?
When does the one month to finish the problem set start?
Can I use my AOPS username?[/quote]

yeah, to sign the entry form (assuming you're under 18)
whenever you feel like starting. the due date is set, but you can work whenever (it's not actually one month exact)
sure, if you want to.


function
linear algebra
linear algebra theorems

p=D(q)=60-3/4q            Demand

p=S(q)=3/4q                 Supply

My text book has these formulas under our supply and demand section, but i'm not completly sure if the -3/4 is part of the formula or the world problem. The world problem makes no referece to -3/4 so I think its the formula in itself.

What do you mean, "Is this the correct formula?" What you're looking at is something artificial and grossly oversimplified. It's just a textbook trying to say, "Just so you can get an idea about supply and demand, suppose that those functions were given by these simple, artificial formulas." Don't assign any more meaning to it than that.


geometry
parallelogram
rectangle
congruent triangles

This question is one of the questions I saw from the recent math book and derived into my wording and description so that I can put this on my 100 Math Problems which is online resource I'm planning to create. I don't have much problems yet but I'm trying to get good questions for 100 Math Problems but I'm trying my best to create some.

This is actually #1 question. You might say, "Hey, I saw something like this!" but I tried to make it little bit different.

I posted the question with the answer. Please check both of them.

Thanks

Prove that if ABCD is a parallelogram, then EFGH, which is a quadrilateral formed by plotting the midpoints of the ABCD, is going to be also parallelogram.

Answers

[hide]AE = EB = DG = GC

AH = HD = BF = FC

 :ang: A = :ang: C

 :ang: B = :ang: D

Thus, AE = CG, AH = CF, :ang: A = :ang: C, resulting  :tri: AHE = :tri: CGF

Also, DH = BF, DG = EB, :ang: B = :ang: D, resulting :tri: DGH, :tri: BEF.

Congruent triangles have congruent lengths so we know HE = GF and GH = EF.

Since those legs make up the quadrilteral and one of the thing we can do to prove the quadrilateral as parallelogram is proving the opposite sides to be congruent, we just proved that EFGH is a parallelogram.

[/hide]

Hope this is right!

That's a correct proof. However, to be a bit of a nit-pick, it should be triangle AHE congruent to triangle CFG.
Another way, which is a bit shorter than using congruent triangles, is to draw in the diagonals, and so by midsegment theorem, EH and FG parallel to BD, and EF and HG parallel to AC, so by transitivity of parallel lines, EH parallel FG and EF parallel HG. Therefore, EFGH is a parallelogram.

Hello, Silverfalcon!

Quote:Prove that if ABCD is a parallelogram, then EFGH, which is a quadrilateral
formed by plotting the midpoints of the ABCD, is going to be also parallelogram.[hide]

My proof is similar to what nolachrymose suggested . . .

Draw diagonal BD.

In triangle ABD, EH joins the midpoints of two sides.

Hence, it is parallel to and one-half the length of the third side, BD.

In triangle CBD, FG joins the midpoints of two sides.

Hence, it is parallel to and one-half the length of the third side, BD.

In quadrilateral EFGH, we have a pair of opposites sides (EH and FG)

which are both equal and parallel.

Therefore, quadrilateral EFGH is a parallelogram.[/hide]

Awesome.. :D

I haven't seen that part.

Hey, I'm going to remove the image. Instead, I'll try to post another geometry question (hopefully #2 on [i]100 Math Problems[/i] with answers so that people can try on it.

Good job and thanks to everyone!

Silverfalcon

Ok.. Here goes the second problem of 100 Math Problems

This is bit harder than the first one.

Such rectangle exists that F,G,H, and I each represent midpoint of the segments on the rectangle. Then prove that :tri: JEI  :cong: :tri: MEI.

Answer

[hide]First, this question has a lot of information. So, when you have a lot information, it's rather HARD to pick out the ones you need. So, after doing some eliminations, we get:

IC = BI, EI = EI, EB = CE(congruently bisected diagonal in rectangle)

:tri: BEI :cong: :tri: CEI.

Notice that FBIE and EICH are both rectangles with their two diagonals on each meet at J and M, respectively, which means that BJ = EJ = J = FI and EM = CM = HM = IM and transitive property (recall BE = CE tus BE must CE), we get JE = JI = MI = ME, and EI is reflexive side, resutling us to get this conclusion:

:tri: JEI :cong: :tri: MEI.

[/hide]


[quote="thetruthisfake"]Given that f(x) = cos x and g(x) = sin x.
Find the value of x if f[f[f[f(x)]]] = g[g[g[g(x)]]][/quote]

There is no solution, as far as http://www.quickmath.com
shows in solving equations, and the graphs do not
intersect when doing the plotting there insofar
as graphing them together for certain domains of x.

I don't know how to work with them algebraically/trigonometrically.


[quote="GlenthemanN"][hide="possible answer?"]
$\frac{i-1}{i}$

using the fact that
$i = \sqrt{-1}$
$i^{2}=-1$
$i^{3}= \sqrt{-1}$
$i^{4}= 1$

this pattern continues, so i found 2005, 2006 and 2007 in modulo 4 and came up with the following expression. So i think its right..
[/hide][/quote]
Um $i^{3}$ shouldn't be the same as $i$.

Keep in mind that some of the rules for working with radicals don't apply to negative or imaginary numbers (otherwise you can prove obviously false things, such as $1=-1$.

$i^{3}=i^{2}\times i=-1 \times i=-i$

Extension:
Find the value of $\sum_{n=1}^{2004}i^{2004-f(n)}$ where $f(n)$ represents the Fibonacci Sequence 1, 1, 2, 3, 5, 8...

I think thats too much to stay in HSB.. you should repost thread in pre olymp... fibonacci mod 4? Unless there is a stupid answer

It's a five minute ARML individual round question from today's SFBA meet.
It's kinda nice once you see the pattern. However, I will repost in Pre-Olymp. for all who want to do it there.


[quote="luimichael"]Question No. 6

Prove that for any positive integers a, b, c , d > 0

$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$.[/quote]

$(a+c)(\frac{1}{a+b}+\frac{1}{c+d})\geq \frac{4(a+c)}{a+b+c+d})$

 :D


Compute $ \displaystyle\sum_{k\equal{}1}^{n} k!(k^2\plus{}k\plus{}1)$

$ \sum_{k\equal{}1}^{n}k!(k^{2}\plus{}k\plus{}1)$

$ k^{2}\plus{}k\plus{}1$
$ \equal{} (k\plus{}1)^2 \minus{} k$
$ \equal{} (k\plus{}1)(k\plus{}1) \minus{} k$

So,
$ k!.((k\plus{}1)(k\plus{}1) \minus{} k)$
$ \equal{} (k\plus{}1)!.(k\plus{}1) \minus{} k.k!$

On expanding the sigma , you will observe that the first term of the $ i^{th}$ group and the second term of the $ (i\plus{}1)^{th}$ group get cancelled.

So, you will be left with two terms in the end:
one : the second term of the first group and
two : the first term of the last group,
 since they are the ones that remain uncancelled.

The answer is therefore : $ n!.((n\plus{}1)(n\plus{}1) \minus{} 1!.1)$

i.e, $ (n\plus{}1)!.(n\plus{}1) \minus{} 1$

Plz correct me if i am wrong


As Fedja (in the locked topic, so I answer here) and other people ask for 239 problems, I post a zip archive with Russian variants of last years (the very last variant may be found in the forum and it is very easy). If somebody is not lazy to translate them, I would appreciate. In English, I have only 2002, but in .pdf and with many other materials. Maybe, I get it in .tex and upload in few days.

From open olympiad of 239 lyceum in Saint-Petersburg.

My guess was right.  :)  Thank you for information!

[quote="Gil"]Could you please post a PDF version of the TEX files ? Thank you very much.[/quote]
I would appreciate that too since my computer does not allow to see Russian symbols in *.tex files, no wonder about compiling them.

But I can't open the file,that's strange. :(


a,b,c>0 prove that
\[ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge 3\sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}}}
\]

[quote="kimnimalar"]a,b,c>0 prove that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3\sqrt {\frac {{a^2 \plus{} b^2 \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}}}
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202347


Find all positive integers $ k$ for which equation $ n^m\minus{}m^n\equal{}k$ has solution in positive integers.

[hide="Answer"]The ordered pair $ (n,m) \equal{} (k \plus{} 1,1)$ satisfies the equation for all $ k \in \mathbb{Z}^{ > 0}$.[/hide]


4 cowboys get mad at each other. Each cowboy decides to shoot one of the other three cowboys. What is the probability that exactly two cowboys are shot?

there are 81 ways the cowboys can shot each other(3*3*3*3)
there are 6 ways to choose the two cowboys (4C2)
A cowboy can't shot himself therefore he must shot the other cowboy being shot
the two not being shot at can shoot either cowboy being shoot at therefore there are 2 possibilities for both cowboys not being shot at
thus 6*2*2 is the number of ways exactly two cowboys can be shot and the probability exactly two are shot is 24/81 or 8/27

[quote="JesusFreak197"][hide="Answer"]If two cowboys are shot, then we have the following posibilities:

1. Two cowboys are each shot twice; there are $6$ ways this can be done.
2. One cowboy is shot once and the other is shot three times; there are $12$ ways this can be done.

Therefore, there are $18$ total ways that two cowboys can be shot.

As for the total number of ways... only two, three, or four cowboys can be shot. There is one way to shoot all four, meaning 19 ways so far. Three cowboys can only be shot in one pattern: $(1,1,2)$ bullets, and there are $12$ ways this can be done, so the probability that exactly two are shot is $\frac{18}{31}$.[/hide]

That should be right, unless who shoots whom matters, as well...[/quote]

We're looking for the [b]probability[/b] Jesusfreak  ;)

[hide]Name the cowboys $A$,$B$,$C$, and $D$.  Since each cowboy can shoot one of three other cowboys, the number of possibilities is $3^4$.  Each possibility can be described by writing the letter of the cowboy that person $A$ shoots at, then the letter of the cowboy that person $B$ shoots at, and so on.  For exactly two cowboys to be killed, the possibility we choose must be of the form $XXYY$ or $XXXY$, however, we can change the order of the letters.  Now, we must add the number of possibilities of each form.

For the first form, $XXYY$, we must choose which two people are shot at $(\frac{4\times3}{2})$.  Now, we must choose which person shoots at which other person.  Since person $X$ must shoot person $Y$ and vice versa, we only need to consider the other two people.  There are two possibilities for the people the other two people shoot at, so the number of possibilities of the first form is $4\times3$.

For the second form, we also must choose which two people are shot at $(4\times3)$.  Person $X$ must be the only person who doesn't shoot at himself, so we already know which person shoots at which other person.

Our answer is \[\frac{4\times3+4\times3}{3^4}=\frac{8}{27}\][/hide]

[quote="JesusFreak197"][hide="Answer"]If two cowboys are shot, then we have the following posibilities:

1. Two cowboys are each shot twice; there are $6$ ways this can be done.
2. One cowboy is shot once and the other is shot three times; there are $12$ ways this can be done.

Therefore, there are $18$ total ways that two cowboys can be shot.

As for the total number of ways... only two, three, or four cowboys can be shot. There is one way to shoot all four, meaning 19 ways so far. Three cowboys can only be shot in one pattern: $(1,1,2)$ bullets, and there are $12$ ways this can be done, so the probability that exactly two are shot is $\frac{18}{31}$.[/hide]

That should be right, unless who shoots whom matters, as well...[/quote]

No :(  first of all, look at the total number of ways they can be shot. Each cowboy has $3$ options of shooting another one (only 1 bullet is used). so the total is $3^4$, now think about how many combinations they have of only two being chosen to be shot...hint hint

Okay, so that means that it does matter who shoots whom, as I said.

[b]EDIT:[/b] And since it's already been proven twice the other way, I'll just leave my proof as what (I think) is correct if who shoots whom does not matter. It's more difficult to do it that way, though, so I might be wrong.

[quote="JesusFreak197"]Okay, so that means that it does matter who shoots whom, as I said.[/quote]

Sorry, didn't see that :blush:

They do shoot at exactly the same time right?  And the bullets can't bounce off each other or anything like that, right?
Anyways, exactly two cowboys are shot if
Cowboy A is shot twice, Cowboy B is shot twice
Cowboy A is shot once (by Cowboy B) and Cowboy B is shot 3 times (by each of the others, probably kills him)
Cowboy A is shot 3 times, Cowboy B once.
A and B can be replaced by each other or C/D

The probability that the second/third scenario will occur is exactly 1/81.  Then there's replacement by C or D/each other.  The number of possibilities for this scenario is 12, and thus the probability is 12/81.
For the first scenario, the number of possibilities is 6, and the probability is 2/81, I think...  A must shoot B, B must shoot A, C can shoot either A or B and D must shoot the other one.  So add it up and you get 24/81=8/27.  That is, considering hat it matters who shoots whom.  But it does matter who shoots whom, because that determines the probability.