my algebra II teacher gave us a challenge problem for bonus points going like this:
A prisoner has escaped from jail.  A guard and bloodhound "chase" after the prisoner.  The prisoner starts with a 10 mile head start, but the guard is 1 mph faster than the prisoner.  The blood hound runs at a speed at 10 mph and is trained to run to the prisoner, run back to the guard, back to the prisoner, etc. until the guard catches the prisoner.  How far does the bloodhound run?

I can do this problem with tedius fractions, but is there an easier way to do this?

Hahah, as i was going to sleep last night right after i posted I suddenly thought of the answer.

Here:[hide]the guy makes up 1 mile each time, so it takes 10 hours for the guard to cath up to the dog.  since the dog runs the whole time, he runs for 10 mph for 10 hours, so he runs 100 miles total.[/hide]

I didn't need help after all, but thanks anyways!

[quote="Wumbate"]Will the dog have been running that whole time?[/quote]

Are we assuming the dog makes turns in zero time?

[quote="SeXuaLSoYBeaN"][quote="Wumbate"]I'm not going to solve it since it's your homework, but here's a start:
Every hour the guard makes up 1 mile on the prisoner.  So you can find how many hours until the guard catches the prisoner.  Will the dog have been running that whole time?[/quote]

He said that he can solve it already, but with tedious fractions, and he wants to know if there's an easier way. I'll take a look at it after I finish my homework.[/quote]

He has already solved it elegantly (i.e. no tedious fractions).  I love this type of problem because it is simple but you have to think outside the box a little bit to get the answer easily.

I meant that they get pretty high considering you have to use fractions (since the prisoner is still running while the bloodhound is chasing after him) repeatedly until the guard catches up, in which the bloodhound can run many rotations.  And yes, turns are assumed to take 0 time

[quote="yif man12"][quote="SeXuaLSoYBeaN"][quote="Wumbate"]I'm not going to solve it since it's your homework, but here's a start:
Every hour the guard makes up 1 mile on the prisoner.  So you can find how many hours until the guard catches the prisoner.  Will the dog have been running that whole time?[/quote]

He said that he can solve it already, but with tedious fractions, and he wants to know if there's an easier way. I'll take a look at it after I finish my homework.[/quote]

Um, I really can't see any tedious fractions. Please tell me if there are any and I'll go upgrade my glasses.[/quote]

[quote="eryaman"]
I can do this problem with tedius fractions, but is there an easier way to do this?[/quote]

True, no tedious fractions, so no need to upgrade your glasses, but i do see tedius fractions, which is kinda close.  :P

Hi, I'm new to the forum, and I just wanted to see if I got the problem right.

If the guard makes up a mile on the prisoner every hour, and the bloodhound runs ten miles every hour, the hound will run 100 miles.

P.S. I dont see any fractions.

Hi you got the problem right, and the reason you didn't see any fractions was because you did it the easy way the first time. Good to see so many new members :D .

[quote="h_s_potter2002"]Are "tedius fractions" a typo, or is it actually a type of fraction?[/quote]

The "tedius" you see in the message is a typing mistake for "tedious."

[quote="h_s_potter2002"]Are "tedius fractions" a typo, or is it actually a type of fraction?[/quote]

I'm wondering the same thing.

I originally thought it was a typo, but it might be my lack of knowledge on math terminology.

If you would like a more wasteful solution, calculate the distance the dog runs from the guard to the criminal and back in terms of the velocity of the prisoner ( from here on). Turns out these distances form a geometric sequence which can be summed to cancel the velocity dependent terms.

  Say the  run of the dog is the  time he goes from guard to prisoner and back. Let the distance of that run be . For example,

.

There is a common ratio  of this sequence, , which can be intuited geometrically or proven:

.

Thus

which leads to the sum

(highlight to view final answer:) [hide]total distance=100[/hide]

  All this would be tedious if not for the fact that, for any  that satisfies , the sum of  to all nonegative integer powers is one divided by one minus r:

.

  You'll probably talk about series and sequences a little in Algebra II, but you don't need infinite sums for this problem, and you probably don't want to do the induction for the geometric progression.

There really is no way to solve the problem with "tedious fractions" without the use of limits, and, if you know how to use a limit, you must be in or past calculus.  If you are there, I would hope that you've seen this type of problem before and therefore know how to solve it without limits a.k.a. "tedious fractions"

-Added Later-
Whoops, I didn't see the second page, and I basically said what's already been said, but I'll leave it here for people to ponder my deep thoughts from my mind at 12:30 am.


Let x,y,z >0 satisfying xy+yz+zx =1. Prove that
    \sum ( \sqrt (x^2 -yz+1) + 1)^2  <= 4(x+y+z)^2

From Cauchy we have  \sum ( \sqrt x 2 -yz+1 +1) 2  \leq 2( \sum x 2 -2 \sum xy+6)=2 \sum x 2 +10
So we have:
2 \sum x 2 +10 \leq 4 \sum x 2 +8
2 \leq 2 \sum x 2
 \sum xy \leq  \sum x 2


Does anyone know when (or if) the UK National Mathematics Summer School is taking place this year? I haven't received any info about it yet, although I got full marks in the IMOK MacLaurin, so I hope I can do it.

It appears it is happening: but when?

https://nrich.maths.org/discus/messages/26/68918.html?1147385722

I have the dates now. I think it's 9th to 16th July or thereabouts.

Euclidean Geometry. It's the most overlooked of all the Olympiadesque topics.

However, from my Summer School experience, you'll have a lot of combinatorical stuff thrown at you. If I remember correctly, most of the topics are combinatorical themed - counting, colouring, logic, game theory.

Then you'll get one that seems totally out of the blue (like when I went, ruler and compass construction of a heptadecagon).

[quote="Android"]Im also going to be attending the summer school (scored 45 on the Maclaurin paper) but i havent really done much work outside the sylabus for afew years and was wondering if you could recommend anything which would be useful to help prepare for the summer school?[/quote]
What is the Maclaurin paper out of?

[quote]What is the Maclaurin paper out of?[/quote]

It is out of 60.


a,b and c addad haye mosbat hastand. hameye x,y va z ra biyabeed be tori ke:

\[ x+y+z=a+b+c \],
\[ 4xyz-[(a^2)x+(b^2)y+(c^2)z]=abc \]

behtare az $\LaTeX$ ham estefade konim ta zahmatahae Valentin doood nashe.
$x+y+z=a+b+c$
$4xyz-[(a^2)x+(b^2)y+(c^2)z]=abc$

kheili mamnoon

rastesh latex yad nadaram vali yad migeeram :blush:

sallam fatehi jan,nemikhad latex yad begiri.

faghat har harfi ro ke minvisi,beune $2$ ta alamate dollar gharar bedeh ;)

ha ha jeddi jeddi latex kar mikoneh

azat ye donya motshakkeram ashegh

injaa raahnamaaye latex hastesh:
[url]http://www.mathlinks.ro/Forum/files/latexhelp_140.pdf[/url]


:D  :)
Let $xy+yz+zx=1$, find the least value of:
$E=ax^2+by^2+cz^2$,
where $a,b,c$ are constants.

Considering that $a,b,c>0$ we have that
$(ax^2+by^2+cz^2)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq (|x|+|y|+|z|)^2\geq 3(|xy|+|yz|+|zx|)$$\geq 3|xy+yz+zx|=3$ so
$ax^2+by^2+cz^2\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$.
This might not be the best least value but it's a start.

The second:
Let $px^2+qy^2+rz^2=1$, find the greatest value of:
$E=axy+byz+czx$
$a,b,c,p,q,r$ are positive real constants.


If $ \mathcal{F}\equal{}\sigma(X)$, how do we interpret that $ \mathcal{F}$ is the information contained in $ X$?

for example, if $ X: \Omega \to \mathbb{R}$, you know the value of $ \phi(X)$ for every mesurable function $ \phi: \mathbb{R} \to \mathbb{R}$. This is like if you were reading a lot of information on $ X$. If $ X$ is a random variable, you can extract the distribution of $ X$, (hence all its moments, etc..) by taking $ \phi\equal{}1_{(\minus{}\infty;a)}$.


algebra proposed
algebra
Cauchy equation

Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(x^2+f(y))=(f(x))^2+y, \quad \forall x,y \in \mathbb R.\]

[quote="Kamyar1991"]f(x^2+f(y))=(f(x))^2+y sorry !!  :oops:[/quote]

Let $ P(x,y)$ be the property $ f(x^2+f(y))=f(x)^2+y$
Let $ a=f(0)$

$ P(0,x)$ implies $ f(f(x))=x+a^2$ and so $ f(x)$ is bijective.
Comparing $ P(x,y)$ and $ P(-x,y)$ we get $ f(x)^2=f(-x)^2$ and, since $ f(x)$ is injective, $ f(-x)=-f(x)$ $ \forall x\neq 0$
Since $ f(x)$ is surjective, it exists $ b$ such that $ f(b)=0$. Then, if $ b\neq 0$, we have $ f(b)=-f(-b)=0$, which is impossible ($ f(x)$ is injective). So $ b=0$ and $ f(0)=0$

So we have $ f(f(x))=x$.

$ P(x,0)$ gives $ f(x^2)=f(x)^2$
$ P(x,f(y))$ gives then $ f(x^2+y)=f(x)^2+f(y)=f(x^2)+f(y)$

So $ f(x+y)=f(x)+f(y)$ $ \forall x,y$ (If $ x\geq 0$, see line above; If $ x<0$, use $ f(-x)=-f(x)$ plus line above.

So we have a Cauchy equation with the supplementary characteristic that $ f(x)\geq 0$ $ \forall x\geq 0$, which implies $ f(x)$ increasing, so continuous, and so $ f(x)=cx$

Plugging back this necessary condition in the original equation, we get $ c=1$ and the unique solution is $ ^\boxed{f(x)=x}$


calculus
integration
real analysis
real analysis unsolved

hmm can't see to get this one can someone help? i have kind of given up

triple integral of (sin(x))/x)

triple integral of sinx/x   x: y*pi to 2*pi - z         z:  from 0 to Pi           y: from 0 to 1

+

triple integral of sinx/x  x: same as previous          z: from 0 to pi*(2-y)          y: from 1 to 2

says to compute the addition of both integrals

my guess is an inversion on the y and z plane and try to switch the order so that both end up  being the same integral, but that doesn`t work and even fi it did i would be stuck afterwards. help please[/code]

So are you the same person as "math beginner," or are you just cheating on the same work? ;)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257018


Well now that Round 2 is over, how would you draw the figure for problem 2 in Asymptote?

I used geogebra... I have geogebra installed, so I made a picture...

[img]http://img246.imageshack.us/img246/5127/uploadhw9.png[/img]

Why is it trivial that the diagram didn't look like this??? lol

Why didn't you use the letter "C" in your diagram? It's like AB_DEFGHI

because i thought i shouldnt use C since there already was one (circle C)  :lol:

[quote]@benzi455: The problem statement says that the fifth circle is externally tangent to the other four, which is impossible with your diagram.[/quote]

[quote]I think he was joking.[/quote]

I am not joking.  I know intuitively that there is no way to have a circle externally tangent to all four of those, but doesn't that require PROOF?

Example proof:

Let the centers of $ C_1, C_2, C_3$ and $ C_4$ be known as $ D_1, D_2, D_3$ and $ D_4$, respectively. Let the center of circle $ C$ be known as $ P$, and the radius of circle $ C$ be known as $ r$. We also know that $ D_1P \equal{} 1\plus{}r \equal{} D_2P$, $ D_3P \equal{} 3\plus{}r \equal{} D_4P$ and $ D_1D_4 \equal{} D_2D_3 \equal{} 4$ by tangency.

Lemma 1: $ D_1D_2D_3D_4$ is an isosceles trapezoid with sides $ D_1D_2$ and $ D_3D_4$ parallel.

Proof: Since $ D_1PD_4 \cong D_2PD_3$ by the side-side-side congruency of triangles, then $ \angle PD_1D_4 \cong \angle PD_2D_3$ as well as $ \angle PD_4D_1 \cong \angle PD_3D_2$ by the properties of congruent triangles. Since $ D_1P \equal{} D_2P$ and $ D_3P \equal{} D_4P$, then $ \angle D_2D_1P \cong \angle D_1D_2P$ as well as $ \angle D_4D_3P \cong \angle D_3D_4P$ by the properties of an isosceles triangle.

By the above, $ \angle D_1D_4D_3 \equal{} \angle D_1D_4P \plus{} \angle D_3D_4P \equal{} \angle D_2D_3P \plus{} \angle D_4D_3P \equal{} \angle D_2D_3D_4$ and $ \angle D_4D_1D_2 \equal{} \angle D_4D_1P \plus{} \angle D_2D_1P \equal{} \angle D_3D_2P \plus{} \angle D_1D_2P \equal{} \angle D_3D_2D_1$. This shows that $ D_1D_2D_3D_4$ is an isosceles trapezoid with sides $ D_1D_2$ and $ D_3D_4$ parallel.

[quote]By the above, $ \angle D_1D_4D_3 \equal{} \angle D_1D_4P \plus{} \angle D_3D_4P \equal{} \angle D_2D_3P \plus{} \angle D_4D_3P \equal{} \angle D_2D_3D_4$[/quote]

By using angle addition, you seem to be assuming that ray $ \overrightarrow{D_4P}$ is between rays $ \overrightarrow{D_4D_1}$ and $ \overrightarrow{D_4D_3}$, and similarly that ray $ \overrightarrow{D_3P}$ is between rays $ \overrightarrow{D_3D_2}$ and $ \overrightarrow{D_3D_4}$.  What is the justification for that?

It's obviously that way. Consider Apollonius' circles ([url=http://en.wikipedia.org/wiki/Circles_of_Apollonius]Linky[/url]). There is only 1 circle that is externally tangent to all 3 of the given circles. If we pick any 3 of the 4 circles $ C_1, C_2, C_3,$ or $ C_4$ then $ C$ (centered at $ P$) must be externally tangent to all 3 (as given by the problem) and then only circle that fulfills this condition lies in the triangle defined by the centers of the three circles we have previously selected.

Finally, an answer that satisfies me.  The solution to Apollonius' Problem plus a bit of extra commentary shows that there is only one possible circle C, which is obviously the one that we want.  Thanks, McDutchy.


I have no idea where to put this.  If a mod moves this, someone tell me where it is moved to.

I want to make an app for TI-84 silver plus OR TI-[i]N[/i]spire (not CAS) that can solve equations with one variable, such as:

$ 8x\minus{}1\equal{}3x\plus{}9$

Of course, there maybe some more difficult problems.  And I might need to add terms with $ x^3$ and $ x^2$ and etc.  Do any of you guys know how to program this, if it is possible?

And is it also possible for a program to solve a system of equations?

There is already a Polynomial Application that finds the roots of polynomials. All you have to do is input the highest power, and each coefficient. You might be able to take a look at the source code, but it's already an app.

And can a mod move this to Other Problem Solving Topics? Thanks.

On the TI-84 keypad or the N-spire keypad?

And where can I find it?

[quote="7h3.D3m0n.117"]I believe it is in every standard TI-84+ Silver Edition. That's where I got it though, I'm currently using it on my TI-83+.[/quote]

Oh yeah.  Thank god.  I bought both the TI-84 plus silver and the TI-N spire, to see which one is better.  So I just burned the whole apps CD on a blank one, cuz im returning my TI-84 (which i got in blue for 10 bucks cheaper at staples!!!  :D )

Thanks!  :)


inequalities
quadratics
Cauchy Inequality
triangle inequality
inequalities unsolved

This one was  posted before  but  there  was  no reply . So  i  remind  it to  you....

ABC  is  triangle , x,y,z are reals. Prove  that :
[tex] (\frac{b}{c}+\frac{c}{b}-1)x^2 + (\frac{c}{a}+\frac{a}{c}-1)y^2 + (\frac{a}{b}+\frac{b}{a}-1)z^2   \geq   (3-\frac{b+c}{a})yz +
(3-\frac{c+a}{b})zx + (3-\frac{a+b}{c})xy  \geq  0  [/tex]

well if x,y,z are all greater or equal to 0 or all smaller tahn 0 all is ok since (a/b+b/a-1)>=1 (easy cauchy inequality) and 3-(a+b)/c <=2 (triangle inequality) (if 3- (a+b)/c<0 we can majorize it by 0
 so the inequality becomes ....>=x^2+y^2+z^2 >=2*xy+2*yz+2*xz>=... which is pretty clear. when one or two of x,y,z are smaller than 0 the situation is something more complicated we can assume without loss of generality that one of x,y,z is smaller than 0 if all of the 3-(a+b)/c are greater than 0 then we can majorize the negative summands from the left by 0 and procede as higher. as you see there are so many possibilities and few would bother to deal with them. unless there is one solution for all cases (which i don't see to be straightforward) all in all there is a theory on quadratic forms that states that every quadratic form can be expressed in a canonical form so if you put all on the right you could transform it to    p*d^2+q*f^2+r*g^2 where p,q,r depend on a,b,c and d ,f,g are linear combinations of x,y,z) it's a sure way. i guess that p,q,r would eventually turn to be all >=0 but it is so laborous that i doubt whether you would like to try it (i won't)


Find all integers $ k$ for which there does not exist a function $ f: \mathbb N \to \mathbb Z$ which satisfies
(a) $ f(1995)\equal{}1996$, and
(b) For all integers $x$ and $y$,
\[ f(xy)\equal{}f(x)\plus{}f(y)\plus{}kf(\gcd(x,y)).\]

[quote="Xaenir"]For which integers $ k$, does ther exist a function $ f: N \to Z$ which satisfies:
$ (a)$ $ f(1995) \equal{} 1996$
$ (b)$ $ f(xy) \equal{} f(x) \plus{} f(y) \plus{} kf(gcd(x,y))$ ; $ \forall x,y \in Z$[/quote]

I suppose we must read $ \forall x,y \in N$ and not $ \forall x,y \in Z$. If so, let $ x\in\mathbb N$ :

$ f(x^2)\equal{}f(x)\plus{}f(x)\plus{}kf(\gcd(x,x))\equal{}$ $ (k\plus{}2)f(x)$
$ f(x^3)\equal{}f(x^2)\plus{}f(x)\plus{}kf(\gcd(x,x^3))\equal{}$ $ (k\plus{}2)f(x)\plus{}f(x)\plus{}kf(x)\equal{}$ $ (2k\plus{}3)f(x)$
$ f(x^4)\equal{}f((x^2)^2)\equal{}(k\plus{}2)f(x^2)\equal{}(k\plus{}2)^2f(x)$
$ f(x^4)\equal{}f(x^3\cdot x)\equal{}f(x^3)\plus{}f(x)\plus{}f(\gcd(x^3,x))\equal{}$ $ (2k\plus{}3)f(x)\plus{}f(x)\plus{}kf(x)\equal{}$ $ (3k\plus{}4)f(x)$

So $ (k\plus{}2)^2f(x)\equal{}(3k\plus{}4)f(x)$
So $ k(k\plus{}1)f(x)\equal{}0$ $ \forall x$

$ f(1995)\neq 0$ $ \implies$ $ k\equal{}0$ or $ k\equal{}\minus{}1$

It exists a function $ f(x)$ for $ k\equal{}0$ :
$ f(1)\equal{}0$
$ f(\prod p_k^{n_k})\equal{}\sum n_kf(p_k)$
$ f(3)\plus{}f(5)\plus{}f(7)\plus{}f(19)\equal{}1996$

It exists a function $ f(x)$ for $ k\equal{}\minus{}1$ :
$ f(1)\equal{}0$
$ f(\prod p_k^{n_k})\equal{}\sum f(p_k)$
$ f(3)\plus{}f(5)\plus{}f(7)\plus{}f(19)\equal{}1996$

So the answer is $ k\in\{\minus{}1,0\}$


geometry
3D geometry
algebra unsolved
algebra

the surface area of a cube with x+1 for length of each edge

I am stumped i have no clue how to do this. can some one help.

SO i have a cube. the length of each edge is x+1. what is the surface area

This is not really the appropriate section for a problem of this difficulty.  You probably want the Basics forum or thereabouts.

What is the surface area of a cube?  :)


find all $x,y,z \in \mathbb{N}$ such that
$\sqrt{\frac{2005}{x+y}}+\sqrt{\frac{2005}{y+z}}+\sqrt{\frac{2005}{x+z}}=n , n\in \mathbb{N}$
 :?:


analytic geometry
algebra solved
algebra

We have a chess table of size 1998*1998, and we put pions on its squares
a)Prove that we can put pions such that on every horizontal line and on every colons are exactly 42 pions
b)For every such amplasation the number of pions which are on black squares is even.

a) I don't know how to explain ... look at the picture below
[img]http://zortech.nm.ru/fig1.gif[/img]

b) Let A be the set of all cells that are chosen. Let x,y be the coordinates
of some cell (x=1,...,1998;y=1,...,1998). Then (x,y) is black if x+y is odd and white otherwise. Consider the following sum:
S = \sum (x,y) \in A (x+y) = \sum (x,y) \in A x+ \sum (x,y) \in A y. Each of these sums is exactly 42. (1+..+1998), because each of 1,...,1998 appears in each sum exactly 42 times. So  S is even. The number of black chosen cells equals to the number of (x,y) \in A, such that x+y is odd. But the number of such pairs is even, otherwise S is odd.


calculus
integration
logarithms
calculus computations

Find

$\int^{e^{n}}_{e}{\frac{1}{x\ln{x}}}\,dx$ where $n\in\Re$

This is just a case of \[\int_{a}^{b}\frac{f'\left(x\right)}{f\left(x\right)}\,dx=\ln f\left(x\right)\Big|_{a}^{b}=\ln\frac{f\left(b\right)}{f\left(a\right)}.\] In your case, $f\left(x\right)=\ln x$ so you have \[\int_{e}^{e^{n}}\frac{1}{x\ln x}\,dx=\boxed{\ln n}.\] Also, I think $n\in\mathbb{R}^{+}$.


calculus
integration
geometry
rectangle
area of a triangle

[quote="mathnerd314"]How would you solve a problem like this:

An object accelerates at 1 meter per second, and its initial velocity is 0 meters per second. How far will it have gone in 5 seconds?

Would one need calculus?

My knee-jerk response is that it goes 1 meter in the first second, 2 in the second, 3 in the third, 4 in the fourth, and 5 in the fifth, for a total of 15 meters, child's play. But somehow I don't think that's right.[/quote]

[hide]
Calculus helps in understanding
a(x) = 1; v(x) = x + c. c = 0. d(x) = x^2 + c. initial position is 0. so it goes 25 meters in 5 seconds.
Essentially, velocity is change in distance over time, acceleration is change in velocity over time. So you just integrate acceleration twice to get to distance. [/hide]

Agolsme:  The integral of $x$ is $\frac{x^2}2+C$, not $x^2+C$. :(

--

In Physics, you'll learn this equation:

$\Delta x=v_it+\frac12at^2$

$\Delta x$ = Displacement
$v_i$ = Initial velocity
$t$ = Time
$a$ = Acceleration

We have...

$v_i=0 m/s$
$a=1 m/s^2$
$t=5 s$

So...

$\Delta x=0+(\frac12)(1)(25)$
$\Delta x=\frac{25}2$

The answer is 12.5 meters.

Actually you can play around and get a lot of useful forms.  For example:

$v_f^2-v_i^2=2as$

[quote="fourierseries"]You don't need calculus for that proof.

Just use the fact that $v=\frac{\Delta v}{\Delta t}$ and the formula for acceleration.

For example in this case:

We have $\frac{v_i+v_f}{t}=a$.

Therefore, $v_f=v_i+at$

$s=v_{avg}t = \frac{v_i+v_f}{2}t$

Substituting our expression for $v_f$, we get the famous:

$s=v_0t+\frac{1}{2}at^2$[/quote]

OK, I have a few questions.

First, what are $v_i$ and $v_t$?

Second, how is $v=\frac{\Delta v}{\Delta t}$? The change in v over the change in t equals v?

Third, if $\frac{v_i+v_f}{t}=a$, wouldn't you get $v_f=-v_i+at$ instead of $v_f=v_i+at$?

Thanks so much!

[quote="mathnerd314"]Whoaaaaaaa. Calculus?  :huh: Why does that equation work? (proof)[/quote]The formula is obtainable using calculus, yes.  Even so, it is one of the simplest calculus problems to solve; so much so that it is used to demonstrate to beginners that calculus does indeed work the way it is supposed to -- you can solve using calculus, but it can be done with out it.

In this case your acceleration is constant, $ a = 1 \frac{\text{ m}}{\text{sec}^2}$.

That tells you that velocity is given by a line (y = mx +b): In this case, $v = at$. (The inital velocity, the y-intercept if you will, is zero.)

We know from grade school pre-algebra that $d = v t$.

So if you construct a rectangle that is $t$ long x $v$ high, you'll get an area $= v t$.

One might be tempted to say then that the distance travelled then is equal to the area of this rectangle, but recall that we started with zero velocity and ended at $ v = at$. In order to account for the change in velocity over time, we have to draw a diagonal across the rectangle. The distance travelled then is equal to the area of the lower triangle once the diagonal is drawn. Recalling from geometry that the area of a triangle is $\frac12 \text{ base x height}$. In this case, $\text{ base } = t\; \&\, \text{ height } = v = at$; therefore, the distance traveled, $ d = \frac12 at^2$

Finally, $d = \frac12 (1 \frac{\text{ m}}{\text{sec}^2})(5 \text{ sec })^2 = 12.5 \text{ meters }$.


Find all real solutions to:
\[ 4x^2 \plus{} 2xy \plus{} y^2 \minus{} y \plus{} 4 \equal{} 4\sqrt {3x^2 \minus{} y}\]

[hide]
Clever substitution... :D

Let $ z = \sqrt {3x^2 - y}$.
Then we have

\begin{align*} 4x^2 + 2xy + y^2 - y + 4 & = 4\sqrt {3x^2 - y} \\
(3x^2 - y) + (x^2 + 2xy + y^2) + 4 & = 4\sqrt {3x^2 - y} \\
z^2 + (x + y)^2 + 4 & = 4z \\
(x + y)^2 & = - z^2 + 4z - 4 \\
(x + y)^2 & = - (z - 2)^2 \\
x + y & = \sqrt { - (z - 2)^2} \end{align*}

Since $ (z - 2)^2$ is always positive [b]when z $ \neq$ 2[/b], $ - (z - 2)^2$ is always negative and the $ \sqrt { - (z - 2)^2}$ is a complex number. Therefore, [b]the only solution is when $ z = 2$[/b]. See solutions two posts below this.
[/hide]

Edit: Fixed :)

[hide="Nice Solution"]
Here's the real solution.
$ 4x^2 \plus{} 2xy \plus{} y^2 \minus{} y \plus{} 4 \equal{} 4\sqrt {3x^2 \minus{} y}$
$ 3x^{2} \minus{} y \minus{} 4\sqrt {3x^2 \minus{} y} \plus{} 4 \plus{} x^{2} \plus{} 2xy \plus{} y^{2} \equal{} 0$
$ (\sqrt {3x^2 \minus{} y} \minus{} 2)^{2} \plus{} (x \plus{} y)^{2} \equal{} 0$
Since we are looking for real solutions, we now have two equations and two variables
$ (\sqrt {3x^2 \minus{} y} \minus{} 2)^2 \equal{} 0$
$ (x \plus{} y)^{2} \equal{} 0$
$ \sqrt {3x^2 \minus{} y} \equal{} 2$
$ x \plus{} y \equal{} 0$
$ 3x^2 \minus{} y \equal{} 4$
$ 3x^2 \plus{} x \equal{} 4$
$ 3x^2 \plus{} x \minus{} 4 \equal{} 0$
Simple quadratic
$ \frac { \minus{} 1\pm \sqrt {1 \minus{} (4\cdot \minus{} 4\cdot3)}}{2\cdot3}$

$ \frac { \minus{} 1\pm \sqrt {1 \plus{} 48}}{6}$

$ \frac { \minus{} 1\pm 7}{6}$
Therefore, the two real solutions of $ (x,y)$ are
$ (1, \minus{} 1),( \minus{} \frac {4}{3},\frac {4}{3})$
However, the second solution causes $ \sqrt{3x^2\minus{}y}$ to become complex, so the only real solution is $ (1,\minus{}1)$
[/hide]
@Youngyi, your solution was correct, but you didn't think about $ z$ being $ 2$, causing $ (z \minus{} 2)^2$ to become $ 0$, which is neither positive or negative.

[quote="ZhangPeijin"]@Youngyi, your solution was correct, but you didn't think about $ z$ being $ 2$, causing $ (z - 2)^2$ to become $ 0$, which is neither positive or negative.[/quote]

Oops... :blush: ... well continuing...

[hide]
\begin{align*} z & = 2 \\
3x^2 - y & = 4 \\
x + y & = 0 \\
3x^2 + x & = 4 \\
3x^2 + x - 4 & = 0 \\
(x - 1)(3x + 4) & = 0 \\
\end{align*}

Therefore, $ x = 1$ or $ - \frac {4}{3}$

So the two solutions are $ (1, - 1)$ and $ ( - \frac {4}{3}, \frac {4}{3})$.
[/hide]

@Peijin, the second solution doesn't lead to a complex number... I don't know where you got that... :huh:

[hide]
$ 4\sqrt {3\left ( - \frac {4}{3}\right )^2 - \left (\frac {4}{3} \right)} = 4\sqrt {\frac {16}{3} - \frac {4}{3}} = 4\sqrt {4} = 8$, not a complex number, because $ \left ( - \frac {4}{3}\right )^2$ is positive, not negative.
[/hide]

And my name is Yongyi, not Youngyi! :mad: No offense :lol:

[quote="CatalystOfNostalgia"]just as a comment, 8 is a complex number. Not that it matters in your solution.[/quote]

Yes, but 8 is a [b]simple[/b] complex number :wink:

Or I should just write [i]not not a real number...[/i]

[quote="Ihatepie"][quote="ZhangPeijin"]Find all real solutions to:
\[ 4x^2 \plus{} 2xy \plus{} y^2 \minus{} y \plus{} 4 \equal{} 4\sqrt {3x^2 \minus{} y}
\]
[/quote]

are you guys sure that this is mc level?[/quote]

Well, the solution doesn't involve anything beyond basic algebra and clever substitution...

This could very well be a nationals team round problem :wink:

[quote="alanchou"][quote="ZhangPeijin"][hide="Nice Solution"]
Here's the real solution.
$ 4x^2 \plus{} 2xy \plus{} y^2 \minus{} y \plus{} 4 \equal{} 4\sqrt {3x^2 \minus{} y}$
$ 3x^{2} \minus{} y \minus{} 4\sqrt {3x^2 \minus{} y} \plus{} 4 \plus{} x^{2} \plus{} 2xy \plus{} y^{2} \equal{} 0$
$ (\sqrt {3x^2 \minus{} y} \minus{} 2)^{2} \plus{} (x \plus{} y)^{2} \equal{} 0$
Since we are looking for real solutions, we now have two equations and two variables
$ (\sqrt {3x^2 \minus{} y} \minus{} 2)^2 \equal{} 0$
$ (x \plus{} y)^{2} \equal{} 0$
$ \sqrt {3x^2 \minus{} y} \equal{} 2$
$ x \plus{} y \equal{} 0$
$ 3x^2 \minus{} y \equal{} 4$
$ 3x^2 \plus{} x \equal{} 4$
$ 3x^2 \plus{} x \minus{} 4 \equal{} 0$
Simple quadratic
$ \frac { \minus{} 1\pm \sqrt {1 \minus{} (4\cdot \minus{} 4\cdot3)}}{2\cdot3}$

$ \frac { \minus{} 1\pm \sqrt {1 \plus{} 48}}{6}$

$ \frac { \minus{} 1\pm 7}{6}$
Therefore, the two real solutions of $ (x,y)$ are
$ (1, \minus{} 1),( \minus{} \frac {4}{3},\frac {4}{3})$
However, the second solution causes $ \sqrt {3x^2 \minus{} y}$ to become complex, so the only real solution is $ (1, \minus{} 1)$
[/hide]
@Youngyi, your solution was correct, but you didn't think about $ z$ being $ 2$, causing $ (z \minus{} 2)^2$ to become $ 0$, which is neither positive or negative.[/quote]

Just as a comment, there is rarely ever a "real solution" to any problem.  more correct solutions=better[/quote]

2 definitions of real:

Real = not involving $ i$ :D
Real = not fake

I think Peijin meant the first definition (not involving $ i$).


Outline a synthesis of 4,4'-difluorobiphenyl from benzene.

EDIT: TOtal rubbish sorry  :blush:  will write once more :)

[hide="My synthesis"]
1.a. benzene to nitrobenzene by nitration
1.b. nitrobenzene to aniline by reduction with $ \ce{Sn}/\ce{HCl}, \ce{OH \minus{} }$
1.c. Protect the $ \ce{NH2}$ group by acylation

2.a. benzene to nitrobenzene by nitration
2.b. nitrobenzene to aniline by reduction with $ \ce{Sn}/\ce{HCl}, \ce{OH \minus{} }$
2.c. Protect the $ \ce{NH2}$ group by acylation
2.d. Acylate to get para product.
2.e. Oxidise to get $ \ce{COOH}$ in the para position
2.f. Treat with $ \ce{SOCl2}$ to get $ \ce{COCl}$ in the para position.

3.a. react 1.c. and 2.f. to get para product.
3.b. Deprotect the two $ \ce{NH2}$ groups
3.c. Diazotise both the $ \ce{NH2}$ groups and use Balz Schiemann rn to get the reqd product.
:)
[/hide]

Is this okay now :) Sorry was confusing myself :D

Steps 2.d. and 2.e. are not certainly feasible.

Is it again bcos in step 2.e., the $ \ce{NRR'}$ group also gets oxidised?  :maybe:  And in that case, isnt there a good protecting agent to avoid that?  :huh:

Tha's also a reason. But tell me: how do you acylate acetanilide?

an easier method wud probably be hydrazobenzene + H+-> benzidine
benzidine + balz schiemann reaction to give the req compound. :huh:

Finnaly someone hited at the Benzidine Rearrangement!

:oops:  Sorry, i thought i was missing something. But cant another protecting group for the $ \ce{NH2}$ group take care of all these things? like any of [url=http://en.wikipedia.org/wiki/Protecting_groups]these[/url]?  :maybe:

Yes benzidine rearrangement  :roll:

Be more specific about that protecting group. Also, don't forget that protecting groups always have a drawback: we must use at least two additional steps. If there is an easier way, let's use it.

yes thats true :) ofc i didnt think of benzidine  :roll:


I need help a little bit  to solve those problems:

1.$ \int_{0}^{1} \frac{dx}{(x^{2}\plus{}1)(x^{2}\plus{}2)... (x^{2}\plus{}n)}$

2.$ \int_{0}^{\frac{\pi}{2}}\frac{x sin^{n}x}{sin^{n}x \plus{} cos^{n}x} dx$ with n from N

Try re-writing the first one in terms of partial fractions.

I have tried but I didn't get anything... could you give me more details

hello, i will show you an example. For the term $ \frac{1}{(x^2\plus{}1)(x^2\plus{}2)(x^2\plus{}3)}$ you must make the ansatz
$ \frac{1}{(x^2\plus{}1)(x^2\plus{}2)(x^2\plus{}3)}\equal{}\frac{Ax\plus{}B}{x^2\plus{}1}\plus{}\frac{Cx\plus{}D}{x^2\plus{}2}\plus{}\frac{Ex\plus{}F}{x^2\plus{}3}$ where are $ A,B,C,D,E,F$ are real numbers.
Sonnhard.

thanks a lot .those problems are for my analysis exam. I study at Tehnical University from Cluj-Napoca, Romania.

And for second one? Do you have some ideas?

I don't have time now, but I would try a few things. To me this integral screams with the possibility of symmetry. So try letting $ x\equal{}\frac{\pi}{2}\minus{}z$...if that doesnt work try letting $ x\equal{}2\arctan(z)$


'bad-math-days'

I was wondering if people encounter this phenomenon too, or am I the only one?

While studying (not obligated), I sometimes have long periods in which I don't make any progress on my work; everything goes wrong. My mind seems to be totally out of shape, no matter what I try to do (sleep more, relax, take a break for some days). I'm able to finish stuff in 1 week, what could have been done in 2 days. Then out of nowhere everything goes back well again, I seem to be more creative, more focussed, more driven,... Are there any tricks to be able to work at full capacity when wanted?

Or does somebody want to share my opinion that a person's mind is "not always open [i]to do [/i]mathematics"? Or are my bad-math-days just 'burn outs' of studying (too enthusiastic) and hence overloading my brain... :)

(p.s. my point of view is that of a student who learns primarily with a passion for science. So, I'm not talking about too much schoolwork, stress from deadlines, etc...)

One of my most unfortunate "bad-math days" occured during the Maryland MathCounts State Competition (where I failed to add four terms correctly).

The lack of product in your math might be the result of a poor environment. Unsharpen pencils, poor diagrams, unrealistic goals, etc. Try simplifying your workspace and making everything as neat as possible.

Also, try something else, besides sleeping or eating. What helps me is a quick jog outside no matter the temperature (if it's too cold or hot then limit your jogging to 3-5 min).

My math productivity is a function of several variables: amount of sleep the night before, teaching load during the day, level of physical activity, amount of caffeine consumed, etc. I think it's not monotone with respect to any of the variables, except possibly for teaching.


express the fraction $ \frac{1}{5}$ into the form $ (\frac{1}{a} \plus{} \frac{1}{b}) \times (\frac{1}{c} \plus{} \frac{1}{d})$

with condition

$ 25 > d > b > c > a > 1$
and $ a, b, c, d$ are whole numbers

I think this is correct...

$ (2,10,4,12)$


If I posted this in the wrong section, I apologise.

Could someone prove (or link to a proof of the AM_GM inequality?

thanks in advance,
Intrepidmath

Here is a proof in two variables:  [url]http://www.artofproblemsolving.com/Forum/topic-28069.html[/url]

:rotfl: I am laughing at how stupid I am, and how obvious the proof is. Thanks a lot Jimmy. You rule and you deserve a cookie.

Proof of General AM-GM (for any number of variables):

[b]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40857&postorder=asc&start=20[/b]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40857&postorder=asc&start=26

is the link :)


geometry
circumcircle
trapezoid
cyclic quadrilateral
geometry proposed

ABCD is a cyclic quadrilateral. The diagonals AC and BD meet at X. The circumcircles of ABX and CDX meet again at Y. Z is taken so that the triangles BZC and AYD are similar. Show that if BZCY is convex, then it has an inscribed circle.

Denote $a = AB, b = BC, c = CD, d = DA$ sides and $p = AC, q = BD$ diagonals of cyclic quadrilateral $ABCD$ with circumcircle $(O, R)$.
Parallels to $AC, BD$ through $D, A$ cut $(O)$ again at $D', A'$, respectively. $A'B = d = CD'$ $\Longrightarrow$ $A'BCD'$ is isosceles trapezoid $\Longrightarrow$ $A'C = BD' = r$.
By Ptolemy for $ABCD', A'BCD$ $\Longrightarrow$ $da + bc = pr$ and $ab + cd = qr$ $\Longrightarrow$ $\frac{da+bc}{ab+cd} = \frac{p}{q}$.

By problem condition, $\triangle BCZ \sim \triangle ADY$ are directly similar, with similarity coefficient $\frac{BC}{AD}= \frac{b}{d}$.
In addition, $\triangle ACY \sim \triangle BDY$ are spirally similar by ASA, with similarity coefficient $\frac{AC}{BD}= \frac{p}{q}$ $\Longrightarrow$
$\triangle CDY \sim \triangle ABY$ are spirally similar by SAS, with similarity coefficient $\frac{CD}{AB}= \frac{c}{a}$.
Calculating sides of quadrilateral $BZCY$ using these similarity coefficients,

$BZ = \frac{b}{d}\ AY,\ \ CZ = \frac{b}{d}\ DY = \frac{b}{d}\cdot \frac{q}{p}\ CY = \frac{b}{d}\cdot \frac{q}{p} \cdot \frac{c}{a}\ AY$

$BY = \frac{q}{p}\ AY,\ \ CY = \frac{c}{a}\ AY$

$\frac{BY+CZ}{BZ+CY}= \frac{\frac{q}{p}\left(1 +\frac{b}{d}\cdot \frac{c}{a}\right)}{\frac{b}{d}+\frac{c}{a}}= \frac{q}{p}\cdot \frac{da+bc}{ab+cd}= 1$

which means that quadrilateral $BZCY$ is tangential with incircle.


[quote="mr. math"]how would you do this problem without converting the numbers back to base 10?

73( base 8)- 27 (base 8)?[/quote]

[hide]Just do it the same way as if you were in base 10.

[code]  73
 -27
-----
  44[/code]

Since $3-7$ is negative, borrow a 1 from the 7 and that leaves $13-7$ in base 8 which is $4$ (from 11-7 after conversion, but you can do this in your head).
Since 7 goes to 6, $6-2=4$ so your answer is $44$.[/hide]

[quote="mr. math"]how would you do this problem without converting the numbers back to base 10?

73( base 8)- 27 (base 8)?[/quote]
[hide="a suggestion in general for base work"]When stuck, think about what you do in base 10 and replace all 10s with 8s. Actually convince yourself you have 8 fingers, it really works...[/hide]

[hide="Click me!"]This problem doesn't seem too hard. Let $n_{8}$ be the number such that:
$73_{8}-27_{8}= n_{8}$

We seek to find $n_{8}$.

Well, let's just blindly do the subtraction just as we would do in base $10$:

$73_{8}-27_{8}= (46?)_{8}$

Indeed, $46_{8}$ is the answer.[/hide]
[hide="Similar problems"]
As an exercise, calculate the following, making sure that you don't get a $9$ as a digit.  :)

$73_{8}-24_{8}= ?$
$666_{8}-577_{8}= ?$[/hide]

vishalarul, see above. 46 is not the correct answer (try it in base 10 to convince yourself)

Oops. I forgot to "carry".  :blush:
Thus, the correct answer would be $44$.


algebra
polynomial
number theory
superior algebra
superior algebra unsolved

Factor or prove they are irreducible

1) X^2+X+1 in Z_2[X]
I think irreducible but how can I show that

2) X^3+X+1 in Z_3[X]
I factored it (X+1)(X^2+X-1)

3) X^4+1 in Z_5[X]
irreducible but could not prove. I tried using Eisenstein criterion but no results

4) X^p-X in Z_p[X]
I think it is reducible but I can't factor it in this general form

5) X^6+30X^5-15X^3+6X-120 in Z[X]
irreducible but nor proof

6) X^4+(5-2i)X^2+(3-7i)X+(12+i) in (Z[i])[X]
Irreducible but no proof.

Thank you

[hide="1"] $ x \equal{} 1 \implies x^2 \plus{} x \plus{} 1 \equal{} 1$ and $ x \equal{} 0 \implies x^2 \plus{} x \plus{} 1 \equal{} 1$, hence $ x^2 \plus{} x \plus{} 1$ has no roots in $ \mathbb{Z}_2$. [/hide]
[hide="4"] $ x^p \minus{} x \equal{} x(x \minus{} 1)(x \minus{} 2)...(x \minus{} (p\minus{}1))$.  This is a consequence of Fermat's Little Theorem and Legendre's Theorem. [/hide]

1,2) Use that for polynomials with degree $ \leq 3$ we have: irreducible $ \iff$ no roots.

3) Eisenstein's criterion is useless in a field. Either use some theory of caclotomic field extensions or use brute force (i.e. try $ x^4 \plus{} 1 \equal{} (x^2 \plus{} ax \plus{} b)(x^2 \plus{} cx \plus{} d)$ and show that this cannot happen).

4) see t0rajir0u.

5) Eisenstein's criterion.

6) Use Eisenstein's criterion for the prime $ 5 \minus{} 2i$.

[quote="ZetaX"]1,2) Use that for polynomials with degree $ \leq 3$ we have: irreducible $ \iff$ no roots.[/quote]

That should read "degree $ \leq 3$ and $ \geq 2$"  :P  owk

[quote="ZetaX"][quote="kunny"][quote="t0rajir0u"]
$ x^p \minus{} x \equal{} x(x \minus{} 1)(x \minus{} 2)...(x \minus{} (p \minus{} 1))$.  [/quote] :?:  :?:[/quote]

What's not clear about this¿[/quote]

What does this mean?

[quote="kunny"][/quote][quote="ZetaX"][quote="kunny"][quote="t0rajir0u"]
$ x^p - x = x(x - 1)(x - 2)...(x - (p - 1))$.  [/quote] :?:  :?:[/quote]

What's not clear about this¿[/quote][quote="kunny"]

What does this mean?[/quote]

The polynomial $ x^p-x$ factors into linear factors $ x \cdot (x-1) \cdot ... \cdot (x-(p-1))$ IN $ \mathbb Z / p \mathbb Z$.

Isn't this correct
$ x^p\minus{}x\equal{}x(x^{p\minus{}1}\minus{}1)\equal{}(x\minus{}1)\frac{x^{p\minus{}1}\minus{}1}{x\minus{}1}$

$ \equal{}(x\minus{}1)(1\plus{}x\plus{}x^2\plus{}\cdots x^{p\minus{}2})$ for $ x\neq 1$?

You lost a factor $ x$. After fixing this, it's correct.
But the factoring t0rajir0u gave is a complete factoring (thus $ x^{p \minus{} 2} \plus{} ... \plus{} x^2 \plus{} x \plus{} 1$ can be factored further in $ \mathbb Z$, and even more further in $ \mathbb Z / p \mathbb Z$).

[quote="ZetaX"]You lost a factor $ x$. After fixing this, it's correct.[/quote]

You are right.

But does this really holds $ x(x \minus{} 1)(x \minus{} 2)\cdots (x \minus{} (p \minus{} 1)) \equal{} x^p \minus{} x$?

Can someone explain in more detail?

I have tired with the argument. :wink:

$ k\equal{}0,1,2,...,p\minus{}1$ are different roots of $ x^p\minus{}x$, thus $ x\minus{}k$ is a factor of $ x^p\minus{}x$ and all these factors are different.


Suppose point C and D lie on the same side of segment AB, and let E and F be the midpoint of AC and BC, G and H be the midpoints of BD and AD respectively. Show that:
\[ \displaystyle S_{EFGH}\equal{}\frac12 \left| \triangle{ABC} \minus{} \triangle{ABD} \right|\]

[quote="shobber"]Suppose point C and D lie on the same side of segment AB, and let E and F be the midpoint of AC and BC, G and H be the midpoints of BD and AD respectively. Show that:
\[ \displaystyle S_{EFGH} \equal{} \frac12 \left| \triangle{ABC} \minus{} \triangle{ABD} \right|
\]
[/quote]
[hide="Solution"]
Let the distance from $ D$ to $ AB$ be $ h_1$ and that from $ C$ to $ AB$ be $ h_2$.  Thus, $ [ADB]\equal{}\frac{h_1AB}{2}$ and $ [ACB]\equal{}\frac{h_2AB}{2}\implies [ABC]\minus{}[ADB]\equal{}\frac{(AB)(h_2\minus{}h_1)}{2}$.  Furthermore, notice that $ EF$ and $ GH$ are the midlines of $ \triangle ACB$ and $ \triangle ADB$ respectively.  This means that $ EF\parallel AB\parallel GH$ and $ EF\equal{}\frac{AB}{2}\equal{}GH$, so $ EFHG$ is a parallelogram.  Hence, the distance from $ E$ to $ AB$ is $ \frac{h_2}{2}$ and the distance from $ G$ to $ AB$ is $ \frac{h_1}{2}$.  Therefore, the distance from $ E$ to $ GH$ is the distance from $ G$ to $ AB$ subtracted from the distance from $ E$ to $ AB$, which is $ \frac{h_2}{2}\minus{}\frac{h_1}{2}\equal{}\frac{h_1\minus{}h_2}{2}$.  Also, $ GH\equal{}\frac{AB}{2}$, so $ [EFGH]$ is the distance from $ EF$ to $ GH$ times $ GH$, which is $ \frac{AB}{2}\cdot \frac{(h_2\minus{}h_1)}{2}\equal{}\frac{\frac{AB(h_2\minus{}h_2)}{2}}{2}\equal{}\frac{[ACB]\minus{}[ADB]}{2}$.  [/hide]

Thank you! It's a nice and simple solution. But I still wonder if we can find a proof by translation. I have been thinking of it for one and a half year but still have no result.

Hi shobber, long time friend. If translation, why not try translating the smaller triangle by shifting it up until $ C'D\parallel AB$? This shift actually moves a unit equal to $ |h_1\minus{}h_2|$. The rest follows easily.

Hi! Long time no see! But can you explain your proof? I still don't see what happens after the translation...


logarithms
inequalities
blogs
inequalities proposed

Let $ a,b,c$ be reals s.t. $ a\geq b\geq c\geq 1$.Prove that:

$ log_clog_cb\plus{}log_b log_ba\plus{}log_alog_ac\geq 0$.

[quote="Inequalities Master"]Let $ a,b,c$ be reals s.t. $ a\geq b\geq c> 1$.Prove that:

$ log_clog_cb \plus{} log_blog_ba \plus{} log_alog_ac\geq 0$.[/quote]
$ log_b log_ba \equal{} log_blog_ca\minus{}log_blog_cb\equal{}\frac{log_clog_ca}{log_cb}\minus{}\frac{log_clog_cb}{log_cb}$;
$ log_a log_ac \equal{} \minus{}log_alog_ca\equal{}\minus{}\frac{log_clog_ca}{log_ca}$;
$ LHS \equal{} (1 \minus{} \frac{1}{log_cb})log_clog_cb \plus{} (\frac{1}{log_cb}\minus{}\frac{1}{log_ca})log_clog_ca \geq 0$          $ (log_ca \geq log_cb \geq 1)$; :)


Does anybody know about when the AMC office will post whose forms they have received? It's past the deadline and I still don't see it on the website.

Thanks.

I will ask the AMC Director when I see him tomorrow (we're both at the National Council of Teachers of Mathematics conference).  But I probably won't be able to post the reply until late Friday (they want $ \$99$, I kid you not, for wireless access at the convention center during this conference, so I will not be logging on anytime during the day tomorrow).


geometry
similar triangles
geometry unsolved

Over each side of a triangle $ ABC$ are constructed towards the exterior of the triangle, three similar triangles equally oriented; $ AC'B, BA'C$, y $ CB'A$. ¿Is it possible to reconstruct $ ABC$ from $ A'B'C'$?...


algorithm
calculus
complex analysis
complex numbers
combinatorics proposed
combinatorics

Place polls S, T and U in a circle with S at 12 o'clock. If n is even, place U at 4 o'clock and U at 8 o'clock. If n is odd, place T at 4 o'clock and U at 8 o'clock. The n disks start at S. Disk  1 is the smallest.

Repeat the following two steps until all n disks have been moved to pole T:

Step1: If disk 1 was not the last disk moved, then move disk 1 one position clockwise.

Step2: If disk 1 was the last disk moved, then make the only possible legal move that doesn't involve disk 1.

Prove the algorithm solves the Towers of Hanoi problem when n is even and when n is odd.

I'd appreciate any help, thanks!

Complex analysis is the study of calculus over the complex numbers, which has nothing to do with this question.  Hopefully, a moderator will move it soon.


If the angles of a trapezoid form an arithmetic sequence and the
smallest angle measures $ 60^{\circ}$, what is the measure of the largest
angle, in degrees?

The rule of thumb to apply is that the smallest and largest angles sum is 180. so 180-60= [b]120[/b]


If $S = i^{n}+i{}^{-}{}^{n}$ where n is an interger, how many distinct values are possible for S?

[quote="mustafa"]I'm sure that there is a better way to do this, but oh well...

[hide]If n=0, s=2. If n=1, s=$i+1/i$. If n=2, s=-2. If n=3, s=$-i-1/i$

$i^{x}=i^{(}x+4)$, so the only possible answers are $2,-2, i+1/i, and-i-1/i.$[/hide][/quote]

And now just try to simplify what you get for $n=1$ and $n=3$.

grrrr. I feel really stupid now. I'll finish up my work.

[hide]we have $i+1/i$, but we have a radical in the denominator of a fraction, so we'll multiply everything by $i$. When we do that, we get -1+1=0, and then we have $-i-1/i$. We will multiply that by $i$ also, so we get 1-1=0 also.[/hide]

OK everybody, what did I do wrong this time? :P

edit:lol, ok. Final answer, there are 3 different solutions to this problem. 0, 2, -2. [b] THE ANSWER IS 3[/b]

Apparently nothing, just state your final answer :)

[quote="mustafa"]I'm sure that there is a better way to do this, but oh well...[/quote] Checking the 4 values for $n$ is a pretty good method for this problem. :D

Also, mustafa, for your second post, I hope you used the fact that $xi=0\Longrightarrow x=0$, when you multiplied "everything" by $i$. you could also multiply $\frac{1}{i}$ by $\frac{i}{i}=1$ to get $\frac{i}{-1}=-i$.

Well, there ARE three values for $S$ - $0, 2,-2$...

[quote="beta9"]Very strange.  That's what I thought, but the answer is listed as 3.

http://gauss.math.oakland.edu/mmpc/previousexams.html

Check problem 35 from the 2004 exam.[/quote]
$3$ is the answer we got. Values for $S$ are $-2,0,2$.

EDIT: Sorry, didn't see Farenhajt's post.

While we're on the subject of $i$, I have a question. How would you find things like $5^{i}$, $3^{i}$ and even $i^{i}$? I don't think that this is a high school basics question, should I make a post like this for Intermediate High School?

Yep, that's an intermediate (or even a little beyond!) question.  Try asking in pre-Olympiad.

Whoops, sorry. I already posted in Intermediate. Maybe a mod could move it for me? :blush:

omg im so stupid I thought $i$ was a variable and I kept telling my self
"it must be infinity.  wait... how are there only four values of $n$?"  :rotfl:
[hide]plugging in the values of $n$ gives 2,0,-2,0 so $\boxed{3}$[/hide]

I had a problem like that on a different problem, but $i$ was the variable and I thought that it was $i$. :fool:

I never use $i$ as a variable. I get confused.

If you want to be really tricky, you can use $\imath$ to refer to $\sqrt{-1}$ rather than $i$.  ;)


For the first time ever, Firefox has overtook Internet Explorer as the most used browser on the site.

1.
Firefox
	54,207 	42.90%
2.
Internet Explorer
	53,553 	42.38%

(This includes all versions, all OSes).

I'm a Mac OS user, and I usually have both Firefox and Safari open at the same time. At some point, I divided most of the sites I visit into those that I thought worked better with Firefox and those that I thought worked better with Safari, with some in each category. I always had AoPS on the Firefox side of that split. Of course that was several versions ago for each browser; my reasons might not mean anything any more.


Let $ a,b,c$ be positive numbers such that: $ a\plus{}b\plus{}c\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$.Prove that:
$ \sqrt{2}(a\plus{}b\plus{}c) \ge \sqrt{a^2\plus{}1}\plus{}\sqrt{b^2\plus{}1}\plus{}\sqrt{c^2\plus{}1}$

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$.Prove that:
$ \sqrt {2}(a \plus{} b \plus{} c) \ge \sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2 \plus{} 1} \plus{} \sqrt {c^2 \plus{} 1}$[/quote]

$ \sqrt {2}(a \plus{} b \plus{} c)\equal{}\sqrt{2(a\plus{}b\plus{}c)^{2}}\equal{}\sqrt{2(a\plus{}b\plus{}c)(a\plus{}b\plus{}c)}\equal{}$
$ \sqrt{(a\plus{}b\plus{}c)(a\plus{}\frac{1}{a}\plus{}b\plus{}\frac{1}{b}\plus{}c\plus{}\frac{1}{c})}\equal{}\sqrt{(a\plus{}b\plus{}c)(\frac{a^{2}\plus{}1}{a}\plus{}\frac{b^{2}\plus{}1}{b}\plus{}\frac{c^{2}\plus{}1}{c})}\geq\sqrt{a*\frac{a^{2}\plus{}1}{a}}\plus{}\sqrt{b*\frac{b^{2}\plus{}1}{b}}\plus{}\sqrt{c*\frac{c^{2}\plus{}1}{c}}\equal{}\sqrt{a^{2}\plus{}1}\plus{}\sqrt{b^{2}\plus{}1}\plus{}\sqrt{c^{2}\plus{}1}$--->$ \sqrt {2}(a \plus{} b \plus{} c) \ge \sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2 \plus{} 1} \plus{} \sqrt {c^2 \plus{} 1}$

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


hi guys

why is [color=orange]calculator[/color] always wrote in [color=orange]orange[/color]?

because you made it [color=orange]orange[/color]?

no, use the search funtion and search for [quote]calculator.[/quote]

this is what shows up
[color=orange]
calculator[/color]

T!89

that's because it highlights the word you searched, just look at the URL

lol, its part of the search function. :rotfl:

sorry. :)

Wonder why AoPS decided to make search-function highlighting [color=orange]orange[/color]?

[quote="akalra1"]y not [color=violet]green[/color]?

jk

thnx guys[/quote]
Ummm...that's pink that you typed the word "green" in...

[quote="mtms5467"][quote="akalra1"]y not [color=violet]green[/color]?

[b]jk[/b]

thnx guys[/quote]
Ummm...that's pink that you typed the word "green" in...[/quote]

jk=just kidding

*sigh*

i was just kidding around


Gareth collects winged cats for his menagerie of strange Pymmian creatures. One day, a friend asked him, "How many winged cats do you have now, Gareth?"
Gareth answered, "I have two-thirds of their number plus two-thirds of a winged cat." How  many winged cats does Gareth have ?

[i]from "Fantastic book of Math Puzzles" by Margaret C. Edmiston[/i]

Please hide Your solution . Write inside Hide ... /Hide
:weightlift:  :P

[hide="solution"]
That would have to mean that the two-thirds of a winged cat makes up the remaining third of their number.
Set up a ratio: $ \frac {2}{3} \equal{} \frac {1}{3}x$
and $ x \equal{} \boxed{2}$[/hide]

Edited: Sorry, solution is now hidden.


function
trigonometry
algebra proposed
algebra

Find all function $ f: R \rightarrow ]\minus{}1,1[$ such that :
$ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)f(y)$

\[ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)f(y)\]
Put $ x\equal{}y\equal{}0$
\[ 2f(0)\equal{}2f^2(0)\]
$ \rightarrow f(0)\equal{}0$ because it can't be $ f(0)\equal{}1$.
Put $ y\equal{}0$
\[ 2f(x)\equal{}2f(x)f(0)\equal{}0\]
$ \rightarrow f(x)\equal{}0$.

why $ f(0)$ can't be $ 1$?

Also, there is another solution: $ f(x) \equal{} \cos kx$, when $ k$ is a constant.

[color=green]A bit of hard:

Find all continuous functionals $ f,g: \mathbb{R}\longrightarrow \mathbb{R}$ such that

\[ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)g(x)
\][/color]

@limsk1: the given function maps the reals to $ ]\minus{}1;1[$. This interval does not include $ \{1\}$, so $ f(0)\equal{}0$.

I saw $ ][$ these marks first. I was confused it with $ []$

[quote="Lepuslapis"]@limsk1: the given function maps the reals to $ ] \minus{} 1;1[$. This interval does not include $ \{1\}$, so $ f(0) \equal{} 0$.[/quote]

Sorry, I did typo. It should be :$ R \rightarrow [ \minus{} 1,1]$

[quote="quangpbc"][color=green]A bit of hard:

Find all continuous functionals $ f,g: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x)g(x)
\]
[/color][/quote]

[color=green]No body like it ?[/color]


Please send us  problems Silk Road Math Olympiad 2008?

Algebrageometry!
SMO 2008 hasn't hold yet.

Hi,Erken.
Could you post your solution for third problem.(about graph)
Thanks.

[quote="Erken"]I could post the problems,if someone wants :maybe: ...[/quote]Please do in English.

1st problem:
Let $ a,b,c,d$ be a natural numbers,such that $ d|a^{2b}\plus{}c$,and $ d\geq a\plus{}c$,prove that $ d\geq a\plus{}\sqrt[2b]{a}$.
2nd problem:
Let $ ABC$ be a triangle and $ A_0,B_0,C_0$ be the midpoints of $ BC,CA,AB$,respectively.And $ A_1,B_1,C_1$ be the midpoints of broken lines $ BAC$,$ ABC$,$ ACB$,respectively.Prove that $ A_0A_1,B_0B_1,C_0C_1$ are concurrent.
3rd problem:
Let $ G$ be a graph with $ 2n$ vertexes and $ 2n(n\minus{}1)$ edges.If we color some edge to red,then vertexes,which are connected by this edge,must be colored to red too.But not necessary that all edges from the red vertex are red.
Prove that it is possible to color some vertexes and edges in $ G$,such that all red vertexes has exactly $ n$ red edges.
4th problem:
Find all polynomials $ P\in\mathbb{R}[x]$,such that for all $ r\in\mathbb{Q}$,there exist $ d\in\mathbb{Q}$,such that $ P(d)\equal{}r$.

I'm waiting for your solutions :) ,especially for the fourth problem.And aslo if somebody knows the results of this olympiad in their own country,please post them.


Consider the following solitaire puzzle. The board is a circular ring of holes which may contain pegs. A move consists of removing a peg from some hole and then changing the state of its adjacent holes (that is, for each one, if it contained a peg take it away, if it was empty, add a peg). The starting position has just a single peg in one of the holes and the goal is to reach the position where there are no pegs in the holes. Is this possible if the board consists of 12 holes? What about 10 holes? In general?

What do you mean by adjacent holes. Like if you labeled the holes A B ..... L, and moved A to B, would you change L and C?

The holes are arranged in a circle. Two holes are adjacent if there's no other hole between them.

well if 3|n then number the spots 0,1,2......,n-1 and consider the number of pegs in the holes with a number wich is not divisible by 3, in each step exactly 2 of this holes change status so the parity of pegsa in this holes is invariant. i guess that this sum can initially be 1 but cant become 0


geometry
3D geometry
MATHCOUNTS
sphere
pyramid
prism

[quote="mr. math"]this seems like a simple problem, but i was wondering if there is any catch in it (need the answer and an explanation):

How many 1 inch cubes can be packed into a rectangular box with dimensions 8'' by 6'' by 6,5''?

 :huh:  :ninja:  :rotfl:[/quote]

[hide="Solution :yoda:"]It's simply the volume of the box:

$V_{\text{box}}= 8 \times 6 \times 6.5 = \boxed{312}$ 1 inch cubes.[/hide]

[quote="SplashD"][quote="lotrgreengrapes7926"][hide] I think it's $8\cdot6\cdot6=288$  :maybe:  [/hide][/quote]

Yeah because I don't think you're allowed to cut cubes in half.[/quote]

Of course, you are absolutly right. Sorry my stupid mistake. I am :sleeping:.

This is an old MathCounts handbook problem.  I remember how this stumped me and my teacher for about 5 minutes, before we realized how stupid we were being.  :blush:

[quote="leela"]Voume of box = 312 cc
Voume of 1 cube = 1cc

So 312 cubes can be packed into the box...simple as that![/quote]

Try to imagine this as a picture.  We can't fit a cube into that .5 space.  It will have to be wasted, because we can't cut a cube in half.  So the answer will be 288.

[hide]The answer is 8*6*6=288, because you can't put a cube into half a cm.
$288$[/hide]

What if we have the same problem with spheres of radius $\frac{1}{2}$?
Is that too hard?  :maybe:

the answer is [hide]$288$ because, if I'm not mistaken, they act as $1\times1\times\ 1$ cubes[/hide]

I'm not exactly sure about that, because with spheres you can have

.OOOOOOO
OOOOOOOO

where the spheres can be placed into the gaps left by already-packed spheres if you know what I mean

If so, [hide="the answer should be"] $\fbox{318}$
the bottom layer has (call the base the 8*6) 48 spheres.
on top of them can be packed 42 spheres so that each of those spheres goes into a "divot" created by 2 spheres on the base. stack 48 on top of them, 42 on those, 48 on them, 42 on those and 48 more on those. The current height of the stack is $1+3\sqrt{3}$ which is less than 6.5"
We now have 48+42+48+42+48+42+48=318 spheres[/hide]

Wouldn't the next layer be $7\cdot5=35$? :maybe:

I'm stacking them differently...

OOOOOO
OOOOOO
OOOOOO
OOOOOO
OOOOOO
OOOOOO
OOOOOO
OOOOOO

Imagine this as a pyramid to begin with...

.......O
......OO
.....OOO
....OOOO
...OOOOO
..OOOOOO
.OOOOOOO
OOOOOOOO

the height of this pyramid is $1+3\sqrt{3}$, about 6.1.
take 6 of these pyramids and place them next to each other so that they could be contained by a triangular prism

that's just a start.
here

OOOOOOOO
.OOOOOOO
OOOOOOOO
.OOOOOOO
OOOOOOOO
.OOOOOOO
OOOOOOOO

say you take 6 of the figure above and place them next to each other
it fits in the box
it has width 8, length 6, and height about 6.1

there's 318 spheres

[hide]Doing what lotrgreengrapes said, though, you can add an extra row, because the second set of spheres is in the divot made by 4 spheres, which is deeper.  Using a sqaure pyramid, you can figure out that each new row only adds $\frac{\sqrt{2}}{2}$ to the height rather than $\frac{\sqrt{3}}{2}$, showing that you can have that extra row ($1+\frac{7\sqrt{2}}{2}$ is less than 6.5).  So, there can be $4 \times 35+4 \times 48$ spheres, giving a total of $\boxed{332}$[/hide]

I had a feeling that the 48-35 would allow for 8 rows of spheres - I just wasn't sure how to calculate the height of that stack...

okay, [hide="I have now found a way to cram in"]
346 spheres

42
48
42
48
35
48
35
48

total height of stack = $1+2\sqrt{2}+\frac{3\sqrt{3}}{2}$ is between 6.4 and 6.5 if I'm correct

heigt of layer 1 is $1$

add layer 2 (35) to get $1+\frac{\sqrt{2}}{2}$

add layer 3 (48) to get $1+\sqrt{2}$

add layer 4 (35) to get $1+\frac{3\sqrt{2}}{2}$

add layer 5 (48) to get $1+2\sqrt{2}$

add layer 6 (42) to get $1+2\sqrt{2}+\frac{\sqrt{3}}{2}$

add layer 7 (48) to get $1+2\sqrt{2}+\sqrt{3}$

add layer 8 (42) to get $1+2\sqrt{2}+\frac{3\sqrt{3}}{2}$, approximately 6.4

$48+35+48+35+48+42+48+42=\fbox{346}$

and that should be the maximum
[/hide]

woohoo! 100th post!


logarithms
calculus
calculus computations

Let $a$ and $b$ be real numbers.  Show that $\log (1+a/b) = \sum_{\ell=0}^\infty \frac{2}{2\ell+1}\left( \frac{a}{2b+a}\right)^{2\ell+1}$. Thanks in advance for your help.

Start with this: for $|x|<1,$

$\ln(1+x)=x-\frac{x^{2}}2+\frac{x^{3}}3-\frac{x^{4}}4+\frac{x^{5}}5+\cdots$
$\ln(1-x)=-x-\frac{x^{2}}2-\frac{x^{3}}3-\frac{x^{4}}4-\frac{x^{5}}5+\cdots$

Subtract to get

$\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}3+\frac{x^{5}}5+\cdots\right)$

Now to continue: solve $\frac{1+x}{1-x}=1+\frac{a}{b}$ for $x.$

Where does that that get you?

(Shouldn't the original problem have some condition on $a$ and $b$ - say, $\frac{a}{b}>-1?$)

Yes, you are right.  Also, thank you that makes sense.


calculus
real analysis
real analysis theorems

erm, this probably doesn't have to really do with calculus, but here goes. in my calculus class, we have to write a song about calculus/math/etc. any ideas? thx! :)

Well, you asked for it :P.

Retell [url=http://en.wikipedia.org/wiki/History_of_calculus]the history of calculus[/url] (or something similar) in verse. Just make sure you have (at least some trace of) rhime and metre: the content is already there, so you can easily start like

In ancient Egypt men could find
the volumes of solids in space
but it is Greece that comes to the mind
as calculus' real birthplace.

and finish like

From Democritus' first attempt
to Newton and Leibnitz designs
from things confusing and unkempt
to modern impeccable lines

of textbooks written in every tongue
the story was complex and long
going far beyond the things we sung
of in this calculus song!

I can, probably, write more, but my purpose was just to demonstrate that what I suggested is doable in principle and I guess you've got the idea by now. :P

P.S. Needless to say, it would be much more useful if your teacher had made you learn some math. instead of poetry... :roll:

P.P.S. Moving it to "Other questions".


1-Let $ a,b,c\ge 0$ such that $ abc\le 1$.Prove that $ a\plus{}b\plus{}c\ge ab\plus{}bc\plus{}ca$
2-Let $ a,b,c\ge 0$ such that $ abc\equal{}1$.Prove that $ (a\plus{}b\plus{}c)^3\ge \frac{8}{3}(a^2\plus{}b^2\plus{}c^2)^2\plus{}a\plus{}b\plus{}c$

[quote="every"]1-Let $ a,b,c\ge 0$ such that $ abc\le 1$.Prove that $ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ca$
2-Let $ a,b,c\ge 0$ such that $ abc \equal{} 1$.Prove that $ (a \plus{} b \plus{} c)^3\ge \frac {8}{3}(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} a \plus{} b \plus{} c$[/quote]
1) Try $ a \equal{} 2,b \equal{} 5,c \equal{} \frac {1}{10}$
2) $ c\equal{}\frac{1}{a}$, $ a\equal{}1294359359640680458346^{18291992412842525}$ :lol:

[quote="dima ukraine"][quote="every"]1-Let $ a,b,c\ge 0$ such that $ abc\le 1$.Prove that $ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ca$
2-Let $ a,b,c\ge 0$ such that $ abc \equal{} 1$.Prove that $ (a \plus{} b \plus{} c)^3\ge \frac {8}{3}(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} a \plus{} b \plus{} c$[/quote]
2) $ c \equal{} \frac {1}{a}$, $ a \equal{} 1294359359640680458346^{18291992412842525}$ :lol:[/quote]
$ b\equal{}???$

[quote="every"][quote="dima ukraine"][quote="every"]1-Let $ a,b,c\ge 0$ such that $ abc\le 1$.Prove that $ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ca$
2-Let $ a,b,c\ge 0$ such that $ abc \equal{} 1$.Prove that $ (a \plus{} b \plus{} c)^3\ge \frac {8}{3}(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} a \plus{} b \plus{} c$[/quote]
2) $ c \equal{} \frac {1}{a}$, $ a \equal{} 1294359359640680458346^{18291992412842525}$ :lol:[/quote]
$ b \equal{} ???$[/quote]$ b\equal{}1$


algebra
polynomial
function
trigonometry
absolute value

Find the absolute value of $(x-e^{i\frac{\pi}{n}}) (x-e^{i\frac{2\pi}{n}}) (x-e^{i\frac{3\pi}{n}})...(x-e^{i\frac{(n-1)\pi}{n}})$

Huh, how do you find the absolute value of a polynomial?

$x$ real or complex?

Edit:  Actually, it might not matter.  I think it's possible to find as a function of $|x|$.

Edit:  Ah, I figured out what was confusing me.  I'll use the assumption $\bar{x}= x$.  [hide="Solution"] Let our given expression be $P(x)$.  Then

$P(x) \overline{P(x)}(x^{2}-1) = x^{2n}-1$

Since $z \cdot \bar{z}= |z|^{2}$ we have

$|P(x)| = \sqrt{1+x^{2}+...+x^{2n-2}}$ [/hide]

I made it a little harder than intended. Find it for $x=1$.

EDIT: t0r, can you elaborate a little on this line?

[quote]$P(x) \overline{P(x)}(x^{2}-1) = x^{2n}-1$ [/quote]

[quote="Phelpedo"]EDIT: t0r, can you elaborate a little on this line?

[quote]$P(x) \overline{P(x)}(x^{2}-1) = x^{2n}-1$ [/quote][/quote]

$P(x)$ has $n-1$ of the $2n^{th}$ roots of unity, the ones in the upper half plane.  Then $\overline{ P(x) }$ contains $n-1$ more roots in the lower half plane.  We're only missing the roots $1,-1$ which I supply using $x^{2}-1$ and then the LHS has all of the $2n^{th}$ roots of unity as its roots.

Quite a nice solution.

In the context of one of my problems, the answer (for $x=1$) should be $2^{n-1}\sin{\frac{\pi}{n}}\sin{\frac{2\pi}{n}}...\sin{\frac{(n-1)\pi}{n}}$
These don't look equal, but maybe they are?

What's $\theta$?  Should that be $n$?

It looks like that expression might be equal to the square of mine.  This is only from checking $n = 3$, though :P

[hide]
To match Phelpedo's answer...

$|1-e^{i\frac{k \pi}{n}}|^{2}= |(1-\cos{\frac{k \pi}{n}})+i \sin{\frac{k \pi}{n}}|^{2}= 2(1-\cos{\frac{k \pi}{n}})$.

So...

$\prod |1-e^{i\frac{k \pi}{n}}|^{2}= 2^{n-1}\prod (1-\cos{\frac{k \pi}{n}})$,

where the summations are from $1$ to $n-1$. But notice that

$\begin{eqnarray*}(1-\cos{\frac{k \pi}{n}})(1-\cos{\frac{(n-k) \pi}{n}}) &=& (1-\cos{\frac{k \pi}{n}})(1+\cos{\frac{k \pi}{n}}) \\ &=& 1-\cos^{2}{\frac{k \pi}{n}}\\ &=& \sin^{2}{\frac{k \pi}{n}}\\ &=& \sin{\frac{k \pi}{n}}\sin{\frac{(n-k) \pi}{n}}$

so

$2^{n-1}\prod (1-\cos{\frac{k \pi}{n}}) = 2^{n-1}\prod \sin{\frac{k \pi}{n}}$.

Note: This is actually the square, so if we can prove that this equals $n$, we match the two answers above. [/hide]

I'll provide that last part then:

We set $z=\cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$ We know that $x^{n-1}+x^{n-2}...+x+z=(x-z)(x-x^{2})...(x-z^{n-1})$. If we input $1$ for $x$, we get the desired result.

EDIT: I left out a factor of 2 in the exponents in the original question. It should be: show that

$(1-e^{i\frac{2\pi}{n}})(1-e^{i\frac{4\pi}{n}})...(1-e^{i\frac{2(n-1)\pi}{n}})$

The only thing is, this doesn't appear to be the square of the problem I gave, so why did we get the square root as the answer?

EDIT2: [quote="paladin8"]$|1-e^{i\frac{k \pi}{n}}|^{2}= |(1-\cos{\frac{k \pi}{n}})+i \sin{\frac{k \pi}{n}}|^{2}= 2(1-\cos{\frac{k \pi}{n}})$. [/quote]

$|a+bi|=a^{2}+b^{2}$, so $|(1-\cos{\frac{k \pi}{n}})+i \sin{\frac{k \pi}{n}}|$ should equal $(1-\cos{\frac{k \pi}{n}})^{2}+(\sin{\frac{k \pi}{n}})^{2}$- in other words, why do you have another exponent 2?

Actually... $|a+bi| = \sqrt{a^{2}+b^{2}}$... http://mathworld.wolfram.com/ComplexModulus.html


Find all continuous functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $x\in\mathbb{R}$ we have $f(x)=f(x^{2}+c)$, where $c\in \mathbb R$ is a constant number!

For $x\in R$ define set $C(x)=\{x_{0}=x,x_{n+1}=x_{n}^{2}+c\}$
For any x,y and $\epsilon$ exist $z,t, |z|\le |c|+1,|t|\le |c|+1$, suth that $x\in C_{z}, y\in C_{t}, |z-t|<\epsilon$.
Therefore only constant functions work.

[quote="Rust"] Therefore only constant functions work.[/quote]

Surely not.
If $c>\frac{1}{4}$, for example, there are infinety many nonconstant solutions :

Let g(x) be any continuous function defined on $[0,c]$ and such that $g(c)=g(0)$
Then let $a_{i}$ be :
$a_{0}=0$
$a_{n+1}=a_{n}^{2}+c$ $\forall n\geq 0$

Since $c>\frac{1}{4}$, $a_{n}$ is strictly increasing and $\lim_{n\rightarrow+\infty}a_{n}=+\infty$

Then we can define $f(x)$ :
For $x\in[a_{0},a_{1})$ $f(x)=g(x)$
For $x\in[a_{n},a_{n+1})$ $f(x)=f(\sqrt{x-c})$ $\forall n>0$
For $x<0$, $f(x)=f(-x)$


I was walking down Mulberry Lane,
I met a man doing the same.
He tipped his hat and drew his cane,
And in this rhyme I said his name.

What was the man's name?

[hide="hint"]
you have to look between words for the name, like this:
say you are looking for the word panda...it could be like this:
there was a [b]pan da[/b]ta.
(that makes no sense)
but you get the idea.[/hide]


induction
number theory
relatively prime
Additive Number Theory

Let $a_{1}, a_{2}, \cdots, a_{k}$ be relatively prime positive integers. Determine the largest integer which cannot be expressed in the form \[x_{1}a_{2}a_{3}\cdots a_{k}+x_{2}a_{1}a_{3}\cdots a_{k}+\cdots+x_{k}a_{1}a_{2}\cdots a_{k-1}\] for some nonnegative integers $x_{1}, x_{2}, \cdots, x_{k}$.

Prove by result:
$ \sum_{i \equal{} 1}^na_ix_i \equal{} f(n)$
If $ f(n)\geq (k \minus{} 1)a_1a_2...a_k \plus{} 1$ then the equation has positive solution.
I think to induction this problem is quite easy. :)
We prove that $ f(n) \equal{} (k \minus{} 1)a_1..a_k$ then the equation has no positive solution.
obivious because $ a_i|x_i,\forall i \equal{} 1,..,k$
Because $ x_i>0$ so $ x_i\geq a_i$
So $ \sum_{i \equal{} 1}^na_ix_i\geq ka_1..a_k>(k\minus{}1)\prod_{i\equal{}1}^ka_i$ (tradition)
So $ f(n) \equal{} ka_1...a_k$

Can you please explain yourself? What is $ n$? Why are there (no) positive solutions in both cases?

I'm sorry, I just don't get a word from this proof?  :maybe:


search
number theory proposed
number theory

Let $n \in \mathbb{Z_{+}}$. If $4^{n}+2^{n}+1$ is a prime, prove that $n$ is a power of three.

hint
[hide]
if $n$ is even the expresion is divisible by $3$ and if $n$ has an odd prime divisor different from 3 then
put $n=p*k$ and prove that $(4^{k}+2^{k}+1)|(4^{n}+2^{n}+1)$.
[/hide]
solution
[hide]
put $a=2^{k}$
we know that $\frac{a^{3}-1}{a^{2}+a+1}=a-1$ so $a^{3}\equiv 1$ mod $(a^{2}+a+1)$
if $p=3l+1$ then $a^{2p}+a^{p}+1\equiv (a^{3})^{2l}a^{2}+(a^{3})^{l}a+1\equiv a^{2}+a+1\equiv 0$ mod $a^{2}+a+1$
if $p=3l+2$ then $a^{2p}+a^{p}+1\equiv (a^{3})^{2l+1}a+(a^{3})^{l}a^{2}+1\equiv a+a^{2}+1\equiv 0$ mod $a^{2}+a+1$
[/hide]

[quote="paqui_2590"]2^3=8=1 mod7
=> 4^3=1 mod7 => 2^3k=4^3k=1 mod7 => if n=3k+i with 1<=i=<2 then 2^(3k+i) +4^(3k+i) +1= 2^i + 4^i + 1 mod 7,
if i=1 or 2, 2^i + 4^i + 1=0 mod 7
then, if i isn't 0 7divides 2^n + 4^n + 1. So, if it is prime it must be a power of 3. In particular if n=3^0 =1 fullfills it.[/quote]
No, this only proves $n=1$ or $n$ is a multiple of 3. It doesn't weed out numbers like 6. :wink:

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5467 or search for even more threads with this.


A  particle   moves along a straight line such that its displacement s is given by

$S=t^3-6t^2+3t+4$

question is what is the velocity of the particle when its acceleration is zero.

a)-9m/s
b)9m/s
c)-12m/s
d)3m/s

[quote="kunny"]The velocity is $v=\frac{dS}{dt},$ and its accelaration is $\frac{dv}{dt}.$[/quote]

cant we solve this problem in other method other than calculus
can you solve the problem

[quote="kunny"]There is no method other than  calculus, differentiation.[/quote]

i am not able to do it.please send me the answer

i have done the following .please verify.
$s=t^3-6t^2+3t+4$

$v=ds/dt=3(t^2)-12t+3$

$a=dv/dt=6t-12=0$

6t=12;t=2

when we substitute in $v=3t^2-12t+3$
we get v=-9m/s

[quote="kunny"]Sorry for I didn't reply. Your answer is correct.

kunny[/quote]

it's ok thanks for the interaction


The sum of nine consecutive integers is 171. What is their mean value?

So find the sum, divide the total by the number of terms. Thus, the mean value is $ \frac{171}{9}\equal{}\boxed{19}$.


Suppose that $ \frac{cosx\plus{}cosy\plus{}cosz}{cos(x\plus{}y\plus{}z)}\equal{}\frac{sinx\plus{}siny\plus{}sinz}{sin(x\plus{}y\plus{}z)}\equal{}a$
Prove that: $ cos(x\plus{}y)\plus{}cos(y\plus{}z)\plus{}cos(z\plus{}x)\equal{}a$


inequalities
search
inequalities proposed

Prove that if a,b,c>0,then:
$ \frac{a^2}{(a\plus{}b)^2}\plus{}\frac{b^2}{(b\plus{}c)^2}\plus{}\frac{c^2}{(c\plus{}a)^2}\plus{}\frac{2abc}{(b\plus{}c)(c\plus{}a)(a\plus{}b)}$≥1

hello, simplifying the left hand-side of your inequality we get
$ \frac{a^4b^2\plus{}b^4c^2\plus{}a^2c^4\minus{}3a^2b^2c^2}{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}\geq0$, this is clear because
$ \frac{a^4b^2\plus{}b^4c^2\plus{}a^2c^4}{3}\geq\sqrt[3]{a^6b^6c^6}\equal{}a^2b^2c^2$.
Sonnhard.

Here is my solution:
We need to prove the equivalent problem:
Let x=a/(a+b),y=b/(b+c),z=c/(c+a)
=> xyz=(1-x)(1-y)(1-y)
We need to prove that $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz$≥1
So I prove the equivalent problem:
If a,b,c≥0 satisfying $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz$=1,then xyz≤(1-x)(1-y)(1-y)
Because $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz$=1,so we can let x=cosA,y=cosB,z=cosC(A,B,C are angles of acute-angled triangle ABC)
xyz≤(1-x)(1-y)(1-y) becomes
cosAcosBcosC≤(1-cosA)(1-cosB)(1-cosC)(this ineq is easy to prove)

I have also posted this inequality almost two weeks ago. Here you can find other solutions.

http://www.mathlinks.ro/viewtopic.php?t=255869

By the way, I can't see why the inequalities are equivalente. Could you explain with details?

:)

This is called "contradicting method".For example,if we have x+y+z=1,then $ x^2\plus{}y^2\plus{}z^2$≥$ \frac{1}{3}$,the equivalent problem is "If $ x^2\plus{}y^2\plus{}z^2$=$ \frac{1}{3}$,then x+y+z≤1"
When we have $ x^2\plus{}y^2\plus{}z^2$=$ \frac{1}{3}$,we have decrease the value of x,y,z,so we need to prove x+y+z≤1.You can find this in some books.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77407


geometry
vector
calculus
probability
trigonometry
geometric transformation
rotation

please give feedback - especially on problem 3 and 5

It looks like we got all the same answers, but most of your solutions appear to be more complicated than mine. For example, my solution to #2 was half a page. Have a look.

Actually, your #4 answer doesn't include the case that n = 1... but other than that it's the same answer as mine, just written a different way.

[quote="JesusFreak197"]Can someone help me figure out why having $X$ be the incenter of the triangle in #5 doesn't work?[/quote]Well, just because there's a point there doesn't mean that it is the [i]incenter[/i], right?

For it to be the incenter, the point would have to be an equal distance from each of the sides, which we don't know is true.

Does that work?  :huh:

[quote="TheAmazingOne"]For example, my solution to #2 was half a page. Have a look.[/quote]I liked how you just jumped straight to the 6th degree conjugate, while everybody else used a cubic and a quadratic conjugate.  :D

[quote="zanttrang"]AmazingOne: you didn't need to include the $n=1$ case, because the problem specifically excludes it.  PkKerichang, your solution to #1 seemed far more complicated than it needed to be, but all of your solutions seem to be correct (they're the same as mine anyway).  Also, I don't think you need to show how to solve closed-form for the recursive equation; it's a pretty well-known technique.  I've attached my own solutions - feel free to criticize them as you wish[/quote]Your solutions are really nice. They make all the problems seem simple.  :)

I proved that it does work, as far as I can tell.

[quote="JesusFreak197"]Can someone help me figure out why having $X$ be the incenter of the triangle in #5 doesn't work?[/quote]

Because it's simply not true. $X$ is the orthocenter, not the incenter. If you somehow proved that $X$ is the incenter, there was a flaw in your proof.

Any comments on my solutions? I tried to make them pretty brief. Are they confusing?

4everwise, I got to the part I of Pg. 8 in your solutions before I had to fax it. :(  I had the exact same method before that. :)

[quote="solafidefarms"][quote="4everwise"]Feel free to critique mine.

[quote="solafidefarms"]Mine: check out how ugly 5 is![/quote]
No, you have to check out how ugly [i]my[/i] number 5 is!  :P
Anyways, the answer was 140 degrees right?[/quote]

OOOOOPSSS!!!!! :blush:  :blush:  :blush: I added 75 and 65 to get 130....[/quote]
I minor addition error rarely loses you points - however, you used approximations a lot during your proof, and I'm not sure whether that constitutes a "rigorous" proof (I found a similar method, but decided to try to find another method that didn't rely heavily on computation).

My solutions are...not too good :blush: You can ignore what i wrote for #3. :rotfl: For some reason I didn't feel like finding a equation to describe the numbers 4, 7, 16, ...  Oh and I also accidentally included 1.

http://www.artofproblemsolving.com/Forum/files/17_2_solutions_404.pdf

I might be wrong about this because no one else seems to have this, but I think that you guys got something wrong in problem 3. I wasn't able to solve the problem because of this, so I hope that you are right and I'm flawed, but here it is: I don't think that every point that is not around the boundaries is able to be circled at once. If any point is completely circled, ahy point that shares a touching triangle can't be circled because the triangle would need two arcs.

If I'm wrong, let me know how so I can kill myself for not figuring out why :P

And, TheAmazingOne, you double counted the corner points when you subtracted the boundaries, you need to add 3.

[quote="TheAmazingOne"]
But hey, maybe the reason you struggle with geometry is that you took the Geometry class so long ago, and when much younger than "recommended." So when faced with geometrical problems you end up turning to messy algebra because it's what you're more familiar with.[/quote]
I think I'd agree with that sentiment.  I took geometry in something like fifth or sixth grade, and when I look back through my old textbook, I recognize many theorems that I've just recently re-discovered; I feel like part of my weak grounding geometry is due to the fact that I was so young when I took it and I never really took the course seriously, i.e. as something to be remembered long-term (that was WAY before I was interested in contest math).

[quote="ZennyK"][quote="zanttrang"][quote="K81o7"]GAh...
I just recently learned the cross products.
The funny thing is, shortly after I solved #5, we started working in 3-D land in math class.  I could have used so many tools here...[/quote]
3-d vectors are nice, but I've found that relying on them too much can actually reduce the elegance of your solutions.  Vector analysis often ends up with large messy solutions that require a lot of simplification, and/or they're not very aesthetically pleasing; I've found it more satisfying, even if I have a vector solution, to go back and try to find a purely geometric solution that is more elegant (this is not to say, of course, that vector solutions are ALWAYS inelegant; sometimes the best proofs are vector-based, though I believe that's not the case for this particular problem).[/quote]
Vector solutions are usually inelegant. However, they're also very natural for me, since I learned vectors in an algebra 2/trig class in 8th grade, precalculus in 9th grade, calculus BC in 10th grade, multi-variable calculus in 11th grade, and now linear algebra in 12th grade. :lol: So I guess I'm pretty comfortable with vectors. :P

My problem with geometry problems is that I tend to turn beautiful geometry into ugly algebra. I need to take an AoPS geometry class or something. I dunno. Thanks for the feedback everyone. It is extremely helpful for me.[/quote]

WOW...you're speeding thru the academic curriculum... :rotfl: Unfortunately, when I just came from Taiwan to America, I dont know any english so i cant answer all the word problems on the math level deciding test, i got a bad grade and went to pre-algebra in 7th grade. My middle school and high school never let me skip any classes, so i'm stuck with pre-calc as a junior in high school.... Good thing is that i learn ahead of myself(that's why im doing USAMTS and reading buch of books) so i never have to study for math tests and i finish homework in 5-10 min instead of 30min-1hour  :rotfl:  I'm feeling lazy...(my pre-algebra teacher in seventh grade even give me separate homework (way ahead of the class) and never care if i pay attention in class(i just do homework in math class))

I guess we can all be in agreement in our discontent for the way the public school system deals with the smartest math students. Right now I'm having to take Calc BC even though I basically already know the whole course. I wanted to skip straight to Advanced Calculus (probably similar to Multivariable Calculus in curriculum) but BC was a pre-requisite. My teacher doesn't care if I don't do any math homework and she lets me work on stuff for other classes during her class... but it sucks that I'm sitting there wasting an hour a day going through the process just so I can "officially" know Calc BC. And if I could have been in a higher math course it would have been a big honor, too. It's not that often that you see a sophomore in a second-year college Calculus class =P I'm sure I would have a 99% or 100%.

[quote="TheAmazingOne"]But hey, maybe the reason you struggle with geometry is that you took the Geometry class so long ago, and when much younger than "recommended." So when faced with geometrical problems you end up turning to messy algebra because it's what you're more familiar with.[/quote]
That's probably it.
[quote="TheAmazingOne"]ZennyK, you're lucky that you even got a chance to take Algebra II in 8th grade;[/quote]
Yep -- it's called homeschooling! :lol:

Also, the Multivariable Calculus class I took last year was actually a "college in the classroom" class -- Advanced Calculus (like TheAmazingOne mentioned). It's combination of Calculus III and Multivariable Calculus.

pkerichang ~
I think Mr. Rusczyk would appreciate your situation in math. ;) You're way above your level in regular cirriculum, but you haven't fallen into "the calculus trap" like too many of us have. Good for you (despite the ridiculousness of making someone with your talent take easy classes).

[quote="TheAmazingOne"]Right now I'm having to take Calc BC even though I basically already know the whole course. I wanted to skip straight to Advanced Calculus (probably similar to Multivariable Calculus in curriculum) but BC was a pre-requisite. My teacher doesn't care if I don't do any math homework and she lets me work on stuff for other classes during her class... but it sucks that I'm sitting there wasting an hour a day going through the process just so I can "officially" know Calc BC. And if I could have been in a higher math course it would have been a big honor, too. It's not that often that you see a sophomore in a second-year college Calculus class =P I'm sure I would have a 99% or 100%.[/quote]
Yeah, schools are a real pain like that sometimes! I've had to do that same thing with non-math classes  -- taking something I've already learned, just so I could "officially" know it. Obviously it's not quite the same, but it's still seriously a major bummer.

[quote="pkerichang"][quote="ZennyK"]pkerichang ~
I think Mr. Rusczyk would appreciate your situation in math. ;) You're way above your level in regular cirriculum, but you haven't fallen into "the calculus trap" like too many of us have. Good for you (despite the ridiculousness of making someone with your talent take easy classes).[/quote]

Thank you :) . btw, since i have little experience with vector (learn about it in physics and some self-studying), what is the application of vector in a geometry problem??More specificly, what kind of geometry problem is it good for, and how do you use it???[/quote]
I think AoPS V 2 has a good section on vector-based geometry;  essentially, vectors allow you to more precisely define lines in a geometry problem, and many properties of vectors (for example the cross product) allow you to manipulate the vectors and find relationships between them, which can give you lengths and angles in the problem.  Maybe somebody else will be able to give a more helpful description.

[quote="ZennyK"]Vector solutions are usually inelegant. However, they're also very natural for me, since I learned vectors in an algebra 2/trig class in 8th grade, precalculus in 9th grade, calculus BC in 10th grade, multi-variable calculus in 11th grade, and now linear algebra in 12th grade. :lol: So I guess I'm pretty comfortable with vectors. :P

My problem with geometry problems is that I tend to turn beautiful geometry into ugly algebra. I need to take an AoPS geometry class or something. I dunno. Thanks for the feedback everyone. It is extremely helpful for me.[/quote]
Heh, sounds a lot like me. I have trouble with Geometry since I started it at the end of 6th grade and finished halfway through 7th, then I took Algebra II second semester of 7th, did Pre-Calc over the summer, supposedly started Calc AB in 8th but didn't really do anything since I was busy with MathCounts and just didn't really want to :whistling: , then tried BC in 9th for first semester and switched back to AB, and now as a sophomore and homeschooled, I'm using the Calculus book that CalTech and MIT use. :P I'm not sure I'm going to take any more math after this year, though.

Hmm, i've been held back quite a bit...

6th grade: Did algebra outside school
7th grade: Was made to do algebra again
8th grade: Geometry
9th grade: Supposedly an Alg2/Trig course, but had just about no trig in it, and most of the stuff I already knew.
Now, 10th grade: Precalc BCH with an amazing teacher who teaches all sorts of stuff outside the curriculum.  Or at least extends out.  Trig, parametrizations with cycloids etc, matrices, determinants, linear algebra (stuff in 3d with vectors and planes), and rotations so far, pretty much.  And I'm doing calc outside school, so I should be able to take the AP for that at the end of the year.

Have you seen the leaderboard?  Does this mean they are loading the scores??

JB

Apparently. Looks like they go bottom up, so your cumulative scores are displayed, and all the smarter AoPSers haven't got their score yet. :D

I find it very interesting to read all this. In my school district most kids go into alg1 in either 8th or 9th grade, and continue on from there. I did a special thing (usually 1-3 kids in the grade do it) where I took alg 1 in 7th grade, geo in 8th, and now alg2 in 9th grade.

[i]Regarding geometry[/i], I actually had some trouble with it compared to other areas of math. I had trouble memorizing all those theorems and recognizing them when trying to prove something on a test. And now, the year after I took that course, I don't remember many of the theorems, making it difficult for me to solve geometry problems on contests. I wonder if an AoPS geometry course would be useful for me. Or can anyone reccommend anything else that could help get me on par with geometry?

But, I already know alg2 anyway (and most of trig and calc) so my teacher will probably let me only come to class during a test or quiz so I'll get credit for the course but won't have to listen to material I already know. Well, even this was hard to get, we have to get it approved by the head of our math department and the principal and all.

Next year I'm supposed to take trig, but I may test out of it and go to BC Calc (in sophmore year). That's going to be a bit hard, not because the subject is hard, but just because the teacher is known for assigning insane amounts of homework :o  and being very very picky about details on tests. Though hopefully calc shouldn't be too hard, already learned it at home  :lol: (well, maybe I didn't learn every little detail of it, but I am reasonably familiar with limits, derivatives, integrals, etc).

I actually think it's a good idea to learn calculus early; I must say, calculus is a very interesting and elegant piece of mathematics. I just loved that fact that one can use it to solve problems that nobody would know how to do using the ideas presented in earlier math.

My school does offer a multivariate calculus "course", but it's not like a regular course, no credits, no grade, etc. If I take BC Calc next year, they have nothing in line for me in senior year, I wonder if I could take a course at a college.

Well, let's just say none of us are alone in trying to deal with school districts. I (and well, my parents) have met multiple people who had trouble with their school systems. My school, although one of the best public schools in it's state, is known for being particularly troublesome for advanced students (which I've found to be true in practice).[/i]

After this year, i'm pretty much stopping math at my school and doing stuff through EPGY.

[quote="asquared"]I found the EPGY program to be decent, but our schools dealings with the people were horrible. It may have to do with the fact that I'm at a private school, but still... It took almost three months for one of the cds to come in so in the meantime, I had to go to the regular class. Eventually, I just stopped doing them.[/quote]

I can't do them for credit at my school...
But right now, I'm doing calculus through EPGY, so I'll just not do it in school next year.

Meh I'm doing trig/precalculus this year(10th grade)  :blush: , I should've skipped algebra II last year because it was pointless. Although I don't think I've forgotten anything yet because of AMC/AIME/USAMO.

BTW calc rulz for geometry read AoPS books, they do a pretty good job covering geometry.


In 1988 AIME, there was a question about $ f(n)$ denoting the square of the sum of the digits of $ n$.  It stated that $ f^{2} (n)$ denotes $ f(f(n)), f^3 (n)$ denotes $ f(f(f(n)))$ and so on. So, here's the modification problem that uses the same function $ f(n)$.

Find the sum of all values of $ f^{1} (n)$ such that for $ f^{2} (n) \equal{} f^{3} (n) \equal{} \cdots \equal{} k$, a constant value called $ k$, if the values of $ n$ are in the set $ \{1,2,3,....99 \}$ (all the positive integers from 1 to 99).

Level: Easy AIME

Let $ y \equal{} f^2(n)$, then $ f^2(n) \equal{} f^3(n)\implies f(y) \equal{} y$, which has the only solution $ y \equal{} 1$.

Now we need $ f(k) \equal{} 1$, which is either $ 1$ or $ 10$.

So we check that $ f(1) \equal{} 1$, and $ f(13) \equal{} 10$, so those exist, the sum is 11.

How come f(13)=10?shouldn't it be 16?? :?:

I think you have misinterpreted it. $ f(n)$ is the sum of all digits of $ n$ (in base 10), not the sum of the digits of $ n^2$. So $ f(13)\equal{}1^2\plus{}3^2\equal{}10$, and $ f(13)\not\equal{}1\plus{}6\plus{}9\equal{}16$.

Ohhh damn okay i did: sum of the squares of the digits, when it is sum of the digits squared...no wonder i got only one output lol

Let $ y \equal{} f(f(n))$, then $ f(f(n)) \equal{} f(f(f(n)))\implies f(y) \equal{} y$. Anyway, we solve $ f(y) \equal{} y$...$ y$ is a perfect square less than $ 100$. We easily see that $ 1,81$ are the only solutions. Let $ k \equal{} f(n)$.

Case 1: $ f(k) \equal{} 1$ has $ 1,10$. But $ k$ is a perfect square so we only have $ f(n) \equal{} 1$ (which occurs since $ f(1) \equal{} 1$)

Case 2: $ f(k) \equal{} 81$ So the sum of the digits is $ 9$, so potentially

$ k \equal{} 9, 18, 27, 36, 45, 54, 63, 72, 81, 90$. But $ k$ is a perfect square so

$ f(n) \equal{} 9,36,81$

$ n$ exists for all of these since $ f(3) \equal{} 9$, $ f(6) \equal{} 36$ and $ f(9) \equal{} 81$.

The sum is then $ 1 \plus{} 9 \plus{} 36 \plus{} 81 \equal{} 127$.

It is a good AIME problem.

[quote="Brut3Forc3"]I think you have misinterpreted it. $ f(n)$ is the sum of all digits of $ n$ (in base 10), not the sum of the digits of $ n^2$. So $ f(13) \equal{} 1^2 \plus{} 3^2 \equal{} 10$, and $ f(13)\not \equal{} 1 \plus{} 6 \plus{} 9 \equal{} 16$.[/quote]

No, what Cronus translated is correct.

Thanks to Altheman for solving other problem though. Here is my solution (since there has been confusion about the wording of problem... hmm, I got the phrase "square of the sum of the digits" from Kalva though...).

[hide="Solution"]
Because the numbers range from 1 to 99, the sum ranges from 1 to 18. So, find $ f^{1} (n)$ by squaring 1,2,...18. Of all the resulted values, the problem holds true if and only if in two cases (this is also covered in Altheman's method too):

Case 1: $ f^{1} (n)$'s sum of its digits is 1. This can be achieved by either 1 or 100. So, Case 1 has two numbers 1 and 100 for $ f^{1} (n)$.

Case 2: $ f^{1} (n)$'s sum of its digits is 9. Why? Because if their sum is 9, the square is 81 and this'll repeat! So, the all $ f^{1} (n)$ [which is a perfect square by the way] that have sum of its digits to be 9 in the range is 9,36,81,144,225, and 324.

Their sum is $ \fbox{920}$. [/hide]

I'm not sure why the confusion occurred. Like... square of 7 is 49. Then square of sum of 3 and 4 is $ (3\plus{}4)^2 \equal{} 49$. So, using similar logic, it is obvious to see how the wording meant.  :maybe:

[quote="Altheman"]Anyway, we solve $ f(y) \equal{} y$...$ y$ is a perfect square less than $ 100$. We easily see that $ 1,81$ are the only solutions. Let $ k \equal{} f(n)$.
[/quote]

How come y is a perfect square? Why are 1 and 81 the only solutions?

[hide="My ‘solution’"]If $ n \in \{1, 2, 3, \dots , 99\}$, then $ f(n) \in \{1, 4, 9, 16, 25, 36, \dots , 324\}$. Now, $ f(f(n)) \equal{} f(f(f(n))) \equal{} k$, which implies that $ f(k) \equal{} k$, therefore $ k$ is a perfect square whose digits sum to its square root. Of course, we have only two such numbers in the interval $ [1, 99]$: $ 1$, and $ 81$ (we can also argue that they are the only ones with this characteristic in the set of the first 18 perfect squares). Thus, we have to find values of $ f(n)$ whose digit sum is either $ 1$ or $ 9$; as [b]Silverfalcon[/b] pointed out, these are $ 1, 9, 36, 81, 100, 144, 225, 324$.

Our answer is $ \boxed{920}$.[/hide]

[quote]How come y is a perfect square? Why are 1 and 81 the only solutions?
[/quote]Cause it's the square of the sum of the digits.And the sum of the digits is an integer. :D
And 1 and 81 are the only solutions 'cause in order to satisfy the cond.s of the question the digit sum is required to be 1 or 9.For detyails read Silverfalcon's solution.

Yea this is why I do badly on the AIME...I make stupid assumptions like if $ 1\le x\le 99$ $ f(x)$ is bounded the same.

Haha, you're not the only one: I kept relating the possible values of $ n$ with the domain of $ f$. Oftentimes, the wording of the problem determines the accuracy of my ‘solution.’

[quote="Silverfalcon"]
[hide="Solution"]
Case 1: $ f^{1} (n)$'s sum of its digits is 1. This can be achieved by either 1 or 100. So, Case 1 has two numbers 1 and 100 for $ f^{1} (n)$.[/hide]
[/quote]

Wait, if you plug in $ 10$ into the function f(n), you also get 1, right? I think I'm misinterpreting something...

Yes, you're right, but we're considering values for $ f(n)$; in the case you mentioned, the value we would consider would be $ 1$.

[quote="Silverfalcon"]

Case 1: $ f^{1} (n)$'s sum of its digits is 1. This can be achieved by either 1 or 100. So, Case 1 has two numbers 1 and 100 for $ f^{1} (n)$.

[/quote]

Wait, then why is 100 okay and ten not okay? Aren't we considering 1?

Again, we're considering the values of $ f(n)$, which are the first eighteen [positive] perfect squares. If $ n \equal{} 10$, $ f(n) \equal{} 1$. We consider $ 100$, because it is one of the first eighteen positive perfect squares — a requisite for being a value of $ f(n)$ — and the sum of its digits equals the square root of a number — in other words, $ f(100) \equal{} f(f(100)) \equal{} \dots \equal{} 1$.

Ten is not ‘okay’ because it is not a value of $ f(n)$ (namely, because it is not one of the first eighteen positive perfect squares).


geometry
MATHCOUNTS
3D geometry
Pythagorean Theorem

[hide]
The diagonal is $12\sqrt{2}$. Therefore the side is 12 and the area is 144.
[/hide]

[hide]

If the diagonal of a square is $12\sqrt{2}$, Then the side has to be 12 because $12\sqrt{2}^2 = 144; \sqrt{144} = 12$ So $12 * 12 = \framebox{144}$[/hide]

[quote="math92"]The diagonal of a square is $12\sqrt{2}$ centimeters. What is the area of the square in centimeters?[/quote]
[hide]
Since the diagonal is $12\sqrt{2}$, each side is $12$, so the area is $144$ cm. squared.[/hide]

[quote="math92"]The diagonal of a square is $12\sqrt{2}$ centimeters. What is the area of the square in centimeters?[/quote]

look at the diagram. half base times height will result in 72*2=144

You really don't need a diagram, especially for MathCounts. In MathCounts, you are expected to know that the diagonal is $s\sqrt2$, and if you don't, then you could even use pythagorean thereom

I'm not sure if I get his diagram...  :? I guess I like computations better.

[quote="mathnerd314"]Related question:

The space diagonal of a cube is $24\sqrt{3}$ centimeters. What is its surface area in square centimeters?[/quote]

[hide]Its side is 24. The surfact would be 6*24*24 = 3456 squared centimeters. [/hide]

Quick proofs for those who don't know:

Square with side $x$ has diagonal $x\sqrt{2}$ because of Pythagorean Theorem: $x^2+x^2=d^2 \Rightarrow d^2=2x^2 \Rightarrow d=x\sqrt{2}$

Cube with side $x$ has diagonal $x\sqrt{3}$ because of PT as well. Diagonal of one face is $x\sqrt{2}$, as already proven; then, $(x\sqrt{2})^2+x^2=d^2 \Rightarrow d^2=3x^2 \Rightarrow d=x\sqrt{3}$.

But you still need to prove that the face diagonal and the edge are perpendicular...

I'll do it for you.
PROOF
The edge is perpendicular to two lines in the plane containing the face; hence it is perpendicular to the plane (line-plane perpendicularity theorem) and perpendicular to every line in the plane. QED

Someone else can prove the perpendicularity theorem, that if a line is perpendicular to two lines in a plane, it is perpendicular to the whole plane. I saw a proof in my textbook and tried to read it without success. (The proof, not the book!)

If the diagonal is 12 sqrt. 2 then the side of the square is 12. Area will be 12^2=144  :starwars:  :surf:  :stretcher:

Formula: diagonal of a square is xroot2 where x is the side length...
find x, which is 12.
12^2 = 144


Find the max coefficient of the terms when expand $ P(x,y,z)\equal{}(2x\plus{}3y\plus{}4z)^{20}$

I believe there is a way to solve this kind of problem for $ (a_{1}x_{1}\plus{}...\plus{}a_{k}x_{k})^n$

[hide="tentative solution?"]
The coefficient of $ x^ay^bz^c$ is $ \frac{20!}{a!b!c!}*2^a3^b4^c$.
So we want to maximize $ \frac{2^a}{a!}\frac{3^b}{b!}\frac{4^c}{c!}$.
First, set $ a, b, c \equal{} 0$ and then find the largest factor you multiply by when you increment one of $ a, b, c$ by 1, and increment that one by 1. Repeat 20 times, and I found $ a\equal{}4, b\equal{}7, c\equal{}9$ which makes the desired coefficient $ \boxed{\frac{20!*256}{1225}}$?[/hide]

[quote="alkjash"][hide=tentative solution?]
The coefficient of $ x^ay^bz^c$ is $ \frac {20!}{a!b!c!}*2^a3^b4^c$.

Can you explain this?

That's a consequence of the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]multinomial theorem[/url], which is in turn a consequence of the following counting argument:  a term of the form $ x^a y^b z^c$ is essentially found by taking a string of three symbols such that there are $ a$ of the first symbol, $ b$ of the second symbol, and $ c$ of the third symbol.  How many ways are there to arrange the symbols?

Well, if we pretend that the symbols are distinct, then there are $ (a \plus{} b \plus{} c)!$ ways to do it.  But since $ a$ of them are the same and $ b$ of them are the same and $ c$ of them are the same, we have overcounted by a product of the number of ways to permute each of the three kinds of symbols, so our final answer is $ \frac{(a \plus{} b \plus{} c)!}{a! b! c!}$.