geometry
geometric transformation
homothety

There are two circles (as shown below) such that the smaller circle is tangent to the larger circle at point $A$. $m$ is the common tangent line for both circles. $B$ and $C$ are the points on the larger circle such that $BC$ is tangent to the smaller circle at point $D$. Prove that $AD$ bisects $\angle BAC$.

Are you sure you typed it correctly. I've drawn it and it don't need to be true.
Maybe radius of a small circle is related to the radius of a big circle?

Yeah, I think I typed everything correctly.
Actually the original problem just had a diagram and a brief question (so I added more explanations).

Did you see the diagram?

Anyone wants to try? :)

[hide="Hint"]
Let $E$ be the intersection of $AB$ and the smaller circle.
Let $F$ be the intersection of $AC$ and the smaller circle.

Play around with the angles...

[hide="Another hint"]
You need to show $EF\parallel BC$.[/hide][/hide]

[hide]
$(P,PA) :$ The big circle
$(R,RA):$ The small circle

The ray $AD$ intersects again the circle $(P,PA)$ at the point $E$.

$\angle CAE = \angle BAE \Leftrightarrow E$ is the midpoint of the arc $BC \Leftrightarrow PE \perp BC$

So, it's enough to prove that $PE \perp BC$

If $BC \parallel m$ then it is obvious.

So, suppose that $BC$ is not parallel to the line $m$.

[img]http://i8.tinypic.com/24vsms1.gif[/img]
[color=red]EDIT for the lost image[/color]

The triangles $\triangle ARD, \triangle APE$ are similar, because they are isosceles and they have a common base angle $\angle PAE$

So the other base angles are equal (or the angles on top are equal and we get the same result)

$\angle PEA = \angle RDA \Rightarrow EP \parallel DR$

$DR \perp BC \Rightarrow EP \perp BC$[/hide]

[quote="Paulina"]Does anyone know spanish??... i can only give my solution in spanish  :blush:[/quote]

At least, try to post in in English, I'm pretty sure you can manage to do so ;)

[quote="pontios"][hide]
$(P,PA) :$ The big circle
$(R,RA):$ The small circle

The ray $AD$ intersects again the circle $(P,PA)$ at the point $E$.

$\angle CAE = \angle BAE \Leftrightarrow E$ is the midpoint of the arc $BC \Leftrightarrow PE \perp BC$

So, it's enough to prove that $PE \perp BC$

If $BC \parallel m$ then it is obvious.

So, suppose that $BC$ is not parallel to the line $m$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=362[/img]

The triangles $\triangle ARD, \triangle APE$ are similar, because they are isosceles and they have a common base angle $\angle PAE$

So the other base angles are equal (or the angles on top are equal and we get the same result)

$\angle PEA = \angle RDA \Rightarrow EP \parallel DR$

$DR \perp BC \Rightarrow EP \perp BC$[/hide][/quote]
Nice proof :)

Does anyone have other way(s) to prove?
(If not, I'll post my proof)

[quote="frt"]
[color=blue]Hint[/color]
Let $E$ be the intersection of $AB$ and the smaller circle.
Let $F$ be the intersection of $AC$ and the smaller circle.

Play around with the angles...

[color=blue]Another hint[/color]
You need to show $EF\parallel BC$.[/quote]

Well, we need to prove that $\frac{BD}{CD} = \frac{BA}{CA}$

From power of point $B$ and $C$ w.r.t. the small circle we get:

$BD^2 = BE \cdot BA$
$CD^2 = CF \cdot CA$

We divide these two equations $\Rightarrow$

$\frac{BD^2}{CD^2} = \frac{BE}{CF} \cdot \frac{BA}{CA}$

So, what we need is to prove that $\boxed{\frac{BE}{CF} = \frac{BA}{CA}}$

Let $O,Q$ the center of the large and the small circle respectively

The triangles $OAB, QAE$ are similar, because they are isosceles $(OA=OB,QA=QE)$ and they have a common angle at $A$.

So we have $\frac{AE}{AB} = \frac{QA}{OA}$

Similarly, $\triangle OAC \sim \triangle QAF \Rightarrow  \frac{AF}{AC} = \frac{QA}{OA}$

hence

$\frac{AE}{AB} = \frac{AF}{AC}\ \ ($this means that $EF \parallel BC) \Rightarrow$

$\frac{BE}{AB} = \frac{CF}{AC} \Rightarrow  \boxed{\frac{BE}{CF} = \frac{AB}{AC}}$

[b][color=blue]frt[/color][/b], this is your proof ?

That's not what I have, but your proof is nice!

Here's my proof:
[hide]
$\angle DAB=\angle BDE$ ($\because BC$ is tangent to the smaller circle at $D$)

Similarly, we have $\angle CBE=\angle GEA$ (by looking at the smaller circle) and $\angle GBE=\angle CBA$ (by looking at the bigger circle).

Hence $\angle CEA=\angle CBA$, which implies $CB\parallel GE$.
Therefore, $\angle DEG=\angle BDE$ ($=\angle DAB$).

Finally, since $\angle DEG=\angle DAG$, we have $\angle CAD=\angle DEG=\angle BDE=\angle DAB$, which concludes the proof.[/hide]

I've got a proof using homothety.
Let the radius of a small and big circle be respectively $r_1$ and $r_2$.
Homothety $J_A^{\frac{r_2}{r_1}}$ maps small circle into big one. Furthermore:

$J_A^{\frac{r_2}{r_1}}(D)=D'$, so tangent to the small circle line $k$ is mapped into line $k'$, which is tangent to big circle.

$O'D'\bot k'$
$k\| k' \Rightarrow CB \bot O'D'$

Trangle $\triangle CD'B$ is isosceles, because line $O'D'$ halves the segment $CB$ and is perpendicular to it.
Hence $CD'$=$D'B$ and finally $\angle CAD'=\angle D'AB$

Well , as pontios wrote the problem is obvious if $m // BC$. (Please see Pontios's fingure)

   Let line  $BC$  meets line  $m$ at point $M$. Then

   $MA = MD$ , so $\angle MDA = \angle MAD$           or

   $\angle DAB + \angle B = \angle MAC + \angle CAD$ .     But

     $\angle MAC = \angle B$ ,     so

             $\angle DAB = \angle DAC$

  [u]Babis[/u]

[b] NOTE[/b]

 Pontios takes gold metal for his first solution. I mean it .Congatulations !

 I hope that my solution deserves a bronze metal - Joking !!!


12     9     13      9     20

What's the next number in the sequence that makes the most sense.

There really is too less numbers to work with.

I could propose
12 9 13 9 20 9, being all even sequence 9, odd sequence +1 +7 +13/+49
or
you can say the difference of two previous numbers have something to do with the third number, and find a quadratic line of best fit since there are only 3 sets (x,y)


linear algebra
matrix
vector
complex numbers

Let $ A$ be the complex matrix representing a transformation of the vector space $ C^{2}$ of $ 2$-tuples over the complex numbers into itself, relative to the natural ordered basis $ \{(1,0),(0,1)\}$.  Let $ A_{R}$ be a [i]real[/i] $ 4 \times 4$ matrix representing the [i]same[/i] transformation on the vector space of $ 2$-tuples of complex numbers over the field of [i]real[/i] numbers.

(i)  What is $ A_{R}$ relative to the ordered basis:  $ \{(1,0),(0,1),(i,0),(0,i)\}$ in terms of the entries of $ A$?

(ii)  If $ A$ is diagonal, what is $ A_{R}$ relative to the ordered basis:  $ \{(1,0),(i,0),(0,1),(0,i)\}$ in terms of the entries of $ A$?

(iii)  How can the trace and determinant of $ A_{R}$ be computed from the trace and determinant of $ A$?

(iv)  If $ A$ and $ B$ are similar $ 2 \times 2$ complex matrices, are $ A_{R}$ and $ B_{R}$, defined as above relative to the same basis, necessarily similar real matrices?  If $ A_{R}$ and $ B_{R}$ are similar $ 4 \times 4$ real matrices, are $ A$ and $ B$ then necessarily similar complex matrices?

The starting point here is that the field of complex numbers $ a+bi\in \mathbb C$ is isomorphic to the set $ \left\{\begin{pmatrix}a &-b \\b & a\end{pmatrix}\colon a,b\in\mathbb R\right\}$ with matrix multiplication and addition. You can turn an $ n$ by $ n$ matrix with complex entries into $ 2n$ by $ 2n$ matrix with real entries by replacing each complex number with a $ 2$ by $ 2$ block of the above form.

It has been a long time since I looked at this problem, but from the notes I left myself, I was having an issue with part (iv).  But since that is dependent on the other parts, I will share what I have:

Write $ A=\left[ \begin{array}{cc}a&b\\\noalign{\medskip}c&d\end{array}\right]$ with $ a=a_{1}+a_{2}i$, $ b=b_{1}+b_{2}i$, $ c=c_{1}+c_{2}i$, and $ d=d_{1}+d_{2}i$.

Then I can associate  $ \left[ \begin{array}{c}1\\\noalign{\medskip}0\end{array}\right] \rightarrow \left[ \begin{array}{c}1\\\noalign{\medskip}0\\\noalign{\medskip}0\\\noalign{\medskip}0\end{array}\right]$, $ \left[ \begin{array}{c}0\\\noalign{\medskip}1\end{array}\right] \rightarrow \left[ \begin{array}{c}0\\\noalign{\medskip}1\\\noalign{\medskip}0\\\noalign{\medskip}0\end{array}\right]$, $ \left[ \begin{array}{c}i\\\noalign{\medskip}0\end{array}\right] \rightarrow \left[ \begin{array}{c}0\\\noalign{\medskip}0\\\noalign{\medskip}1\\\noalign{\medskip}0\end{array}\right]$, and $ \left[ \begin{array}{c}0\\\noalign{\medskip}i\end{array}\right] \rightarrow \left[ \begin{array}{c}0\\\noalign{\medskip}0\\\noalign{\medskip}0\\\noalign{\medskip}1\end{array}\right]$.

So for (i) I get:

$ A_{R}=\left[ \begin{array}{cccc}{\it a_{1}}&{\it b_{1}}&-{\it a_{2}}&-{\it b_{2}}\\\noalign{\medskip}{\it c_{1}}&{\it d_{1}}&-{\it c_{2}}&-{\it d_{2}}\\\noalign{\medskip}{\it a_{2}}&{\it b_{2}}&{\it a_{1}}&{\it b_{1}}\\\noalign{\medskip}{\it c_{2}}&{\it d_{2}}&{\it c_{1}}&{\it d_{1}}\end{array}\right]$

For (ii), taking $ A=\left[ \begin{array}{cc}a&0\\\noalign{\medskip}0&d\end{array}\right]$, I get with respect to the new basis:

$ \left[ \begin{array}{cccc}{\it a_{1}}&-{\it a_{2}}&0&0\\\noalign{\medskip}0&0&{\it d_{1}}&-{\it d_{2}}\\\noalign{\medskip}{\it a_{2}}&{\it a_{1}}&0&0\\\noalign{\medskip}0&0&{\it d_{2}}&{\it d_{1}}\end{array}\right]$

For (iii), $ \text{trace}(A_{R})=2(a_{1}+d_{1})=2 \Re (a+d)=2 \Re \{ \text{trace}(A) \}$

For (iv), I defer to you.  Maybe the way I approached this problem makes this question less obvious, or maybe I am just missing something fundamental.

The answer to the first part is yes: if $ P^{-1}AP=B$, then $ P_{R}^{-1}A_{R}P_{R}=B_{R}$ because matrix multiplication plays well with blocks.
I suppose the answer to the second part is no, but don't have an example yet.


1) 6x+3(2x+3(x-2))=30x-6(2x+30)

2)4x|x|+36X^2-9x^2+16x^2+256=81x^2+16x(x+|x|)

3)There are 12 ordered pairs of integers(x,y) that satisfy x^2+y^2=25 what is the greatest possible sum?

For all of them show all work.

I think you're supposed to figure out for which x (or possibly multiple values of x) the statements hold.

I think what her question meant was, "Around which x are the parentheses supposed to be absolute values?"

Ars: to make the absolute value sign, press shift-backslash (which should be over on the right side of the keyboard around the [ and the ].

Number 1:[hide]

x=-54

6x+3(2x+3(x-2))=30x-6(2x+30)

6x+3(2x+3x-6)=30x-12x-180

6x+6x+9x-18=30x-12x-180

21x-18=18x-180

3x=-162

x=-54[/hide]

Number 2's parentheses are confusing

Number 3:[hide]

7

Hmm... greatest value for either variable is 5, least is -5. can't use negative intergers because we want greatest sum.

Since the only positive sums that will work are 3^2+4^2 or 5^2+0^2, and 4+3>5+0, the answer is 4+3=7.

[/hide]

Topper got #1 and #3 right. Also it is the parantheses around the x only.(x)

Ars wrote:1) 6x+3(2x+3(x-2))=30x-6(2x+30)

2)4x(x)+36X^2-9x^2+16x^2+256=81x^2+16x(x+(x))
Hint the parentheses around the x are supposed to absolute value.

3)There are 12 ordered pairs of integers(x,y) that satisfy x^2+y^2=25 what is the greatest possible sum?

For all of them show all work.

For #2, that "hint" isn't really a hint because without it, the problem has a completely different interpretation.  I believe the problem should read

2) 4x*|x| + 36x^2 - 9x^2 + 16x^2 + 256= 81x^2 + 16x(x+|x|)

Much of this simplication is really easy - it's just X:^2:[81-(36-9+16)] or something to that effect...

Spoiler:

[hide] 4x*|x| + 43x:^2: + 256 = 81x:^2: + 16x:^2: + 16x*|x|

256 = 54x:^2: + 12x*|x|

256 = 54|x|:^2: + 12x*|x|

Case 1: x > 0.  Then x = |x|.

256 = 54|x|:^2: + 12|x|*|x|

256 = 66|x|:^2:

|x|:^2: = 128/33

|x| = +- 8*sqrt(66)/33

Since |x| > 0, |x| = 8*sqrt(66)/33

x = +- 8*sqrt(66)/33

Since x > 0, x = 8*sqrt(66)/33

Case 2: x < 0.  Then x = -|x|.

256 = 54|x|:^2: + 12(-|x|)*|x|

256 = 42|x|:^2:

|x|:^2: = 128/21.

|x| = +- 8*sqrt(42)/21

Since |x| > 0, |x| = 8*sqrt(42)/21

x = +- 8*sqrt(42)/21

Since x > 0, x = - 8*sqrt(42)/21

So the two solutions are 8*sqrt(66)/33 and - 8*sqrt(42)/21, which are not happy numbers.[/hide]

Ars -- go back to the top, press the "edit" button at the top of your message, and put in the | |  instead of parenthesis for absolute values in the problem if you want it to be really clear.

As for #1 and #3...

Ars wrote:1) 6x+3(2x+3(x-2))=30x-6(2x+30)

[hide]6x+3(2x+3x-6)=30x-12x-180
6x+3(5x-6)=18x-180
6x+15x-18=18x-180
3x=-162
x=-54[/hide]

Ars wrote:3)There are 12 ordered pairs of integers(x,y) that satisfy x^2+y^2=25 what is the greatest possible sum?

These twelve ordered pairs are of the form (x,y) where

1) x and y are 5 and 0, vice versa, positive or negative {(5,0), (-5,0), (0,5), (0,-5)}

or

2) x and y are 3 and 4, vice versa, positive or negative {(3,4), (4,3), (3,-4), (-4,3), (-3,4), (4,-3), (-3,-4), (-4,-3)}

The greatest possible sum is 3+4=7.

Mathfanatic notice that if x=|x|.Then before you simplify you should see that everything is a square number.

Interesting problem ARSENIJ     :ninja:
Where did you get it?  USAMTS?

I had to it was my homework in a certian Art of Problem solving class. :)  ;)  :lol:  :P  :ninja:  :no:


geometry trapezoid rectangle ratio blogs rhombus parallelogram

I couldn't figure it out forwards or backwards, and neither could my team.

A farmer has previously fenced a field in the shape of a trapezoid as shown. The farmer divides this field into exactly 4 congruent trapezoids. How many additional meters of fencing are needed? Express your answer to the nearest whole number.

[code]
 ____40 m______
|               \
40 m            \
|                 \
|                   \
|                     \
|_____80 m_________\

[/code]

[hide="answer"]108[/hide]

Please help me!!

Thank you
Billy

Thank you!! :)  :)  :)  :lol:  :lol:  :lol:  :lol:

Now lets see if it comes on the test later...

Billy

[quote="Gyan"]BTW a classic problem (Has been around for a VERY long time) is after asking this question: ask a harder question.. suppose you add the 'missing triangle'  (So now the  plot looks like orinary rectangle )  and  question becomes [quote] Can you divide the new figures in five congruant figures? [/quote] :D[/quote]

[hide="Hint"]He's just saying "ask a harder question.." to confuse you[/hide]

What basic shape do you see here?

Since all of these trapezoids are equal, and all make up the area of the larger trapezoid, they smaller trapezoids are all similar to the larger trapezoids.

Thus the ratio of large to small is 2:1

So now all we have to do is find the length of that funky diagonal on the end, and continue. 45-45-90, so that makes it a $40 \sqrt{2}$

So the smaller diagonal is half of that, $20 \sqrt{2}$.  Add all of the similar lengths on the inside, $20+20+20+20+20 \sqrt{2} \Rightarrow 80+20 \sqrt{2}$ which is 108.2842712... Since the problem wants it to the nearest whole number, it is 108.

If you need more help, go to my blog and ask.

If you want to see the actual test go to my site, http://yashar.biz/mathcounts/.

2002 State Team:  The State Answer Key has the answers.

what bout the the one with the rectangle.

First of all, the side lengths of the smaller rhombus have to be half of the original length... (with the sauaring and square roots stuff) and so u kno dimensions are 40, 20, 20 and 20 :sqrt: 2 Then, you have to look at placement, and if you think about it, you cannot put the right angles on the bottom left corner (or else you would have to put a parallelogram next to it, so you kno that you have the 45 degree sides matching up on the bottom corner. The rest should fit in from there, as the other two rhombuses (? or rhombi or w/e) as the other rhombus fits in adjacent to the bottom one, and the last one fits in by process of elimination. This is really simpler than it sounds, you just have to 'see' it when solving.

Then, when you see how to divy up the big rhombus, just find out what walls you need: 4 "20"s (3 smaller bases and a leg, or 1 big base, 1 small base, and a leg; just don't double count) and a 20 :sqrt:2 wall, making your sum 80 +  20:sqrt:2

Hope this helps... xxxx xx xxxx... Xxxxxx xxx xxx xxx xxxxxx xx xxxxxxxx!!!

o and i forgot... u have to turn that into a whole number... so you get 80 + 20*1.41... which is about 28.2 so the answer is 108.2 when rounded for the answer is 108...


[quote="ch1n353ch3s54a1l"][quote="Phelpedo"]Let $E(n)$ denote the largest integer $K$ such that $5^{k}$ is an integral divisor of the product $1^{1}2^{2}3^{3}...n^{n}$. Find a closed-form formula for $E(5^{m})$, $m$ a positive i nteger.[/quote]

I don't know if this gets the problem anywhere (it's not in closed form...), but...
[hide]
$\frac{5\lfloor\frac{n}{5}\rfloor(\lfloor\frac{n}{5}\rfloor+1)}{2}+\frac{5^{2}\lfloor\frac{n}{5^{2}}\rfloor(\lfloor\frac{n}{5^{2}}\rfloor+1)}{2}+......$  :ninja: [/hide][/quote]

Well the problem does conveniently allow you to substitute $n = 5^{m}$.


Prove that 2003 cant be expressed as the sum of two squares

[quote="nat mc"]Since 3 is the units digit, you can subtract the square number units digit.
3-0=3
3-1=2
3-4=9
3-5=8
3-6=7
3-9=4

The only ones that work are 9 and 4.

And when you go through, none work[/quote]

  Yeah that was my solution too, any others?

I don't know if this belongs in the MATHCOUNTS forum, but it's worth a try.

[hide]We know that all even squares are congruent to 0 and all odd squares are congruent to 1, mod 4. 2003 is odd, which means it must be the sum of an odd square and an even one if it could be expressed as such a sum. Unfortunately, the only things expressable as the sum of an odd square and an even square are congruent to 1, mod 4. But 2003 is congruent to 3, mod 4. Therefore 2003 cannot be expressed as the sum of two squares. [/hide]

wow, that's good!  :clap2:
And yes, it's suitable for MATHCOUNTS.

That's the correct use of mods, right there

But is the converse also always true?

That is to say, even if it works with mods, can you necessarily express it as a sum of two perfect squares anyway?

For example, can you express 2005 as a sum of two perfect squares?


geometry
rectangle
symmetry
geometric transformation
reflection
analytic geometry
graphing lines

Can the following figure fit in a $ 5\times 5$ square. Why?

Yes, it can. Because I printed it and managed to put it inside a 5x5 square (though I was forced to fold it several times).

But this is not a precise resoning, is it? Maybe, it can be put into this square, because it has area equal to $ 18 < 25$? But wait, take e.g. rectangle $ 1x18$. Hmmm. Oh, but why its area is $ 18$? The scale is not given! Maybe, it is $ 36 > 25$. Then it's not possible. But what if the scale is $ 1$? Maybe it cannot be done because $ 4\sqrt {2} > 5$? But then how I managed to do that?

Nobody can post a solution yet?

I don't think this is such a hard problem that no mathlinker could solve it...

Note that a square of side 5 is the locus of points which have taxicab distance at most 5 from a fixed point. I don't think this belongs in the olympiad forum at any rate; maybe HSB or intermediate.

Could you post it in more details?
I'm very sorry to ask you this, but because of my poor English and knowledge, I didn't quite understand what you really meant.

Can someone explain me at last what fwolth meant?
Thank you very much.

Taxicab distance is shortest path parallel to a set of two perpendicular axes. In this case, we can take the axes at 45 degree angles to the grid lines and it's easy to show that no point is more than 5 from the center of the figure.

I don't think this works. Shouldn't the point be at most 2.5 from the the center of the figure?

No; its 5. But I did make an error; the axes are not at 45 degrees to the grid lines, they are parallel to the grid lines.

EDIT: whoops it's not actually possible to fit it into a square. but anyways you use the same method; just take the center and one of the farthest points and consider any pair of two perpendicular segments linking them.

[quote="fwolth"]but anyways you use the same method; just take the center and one of the farthest points and consider any pair of two perpendicular segments linking them.[/quote]

I can't see how this is going to work. Maybe you finish it with your idea? :D

Oh, I'm an idiot, it's $ 5\sqrt{2}/2$, not $ 5$. Anyways, doesn't change anything.

Here's an outline: Assume it can be contained in such a square. Prove that we can take the center $ O$ of the figure as the center of the square since by symmetry if we can take $ P$ we can take the reflection of $ P$ $ P'$ over $ O$, and since a square is convex, we can take $ \frac{1}{2}P\plus{}\frac{1}{2}P'\equal{}O$. Then assign $ O$ (0,0) in the coordinate plane and consider X(1,3) and Y(3,1), both in the figure. Choose an arbitrary slope $ m$ for the coordinate axes of the taxicab scheme and calculate the taxicab distance from $ O$ to $ X$ and $ Y$ in terms of $ m$; show they cannot be both simultaneously less than $ 5\sqrt{2}/2$.


logarithms
number theory proposed
number theory

[quote="pmvald"]For any integer $ n \geq 100$, there is an integer between $ n$ and $ 2.1n$ inclusively with the sum of its prime factors divisible by $ 10$.
(can $ 2.1$ be replaced by something better?)[/quote]

Let $ n\ge 15$ and $ k\equal{}[\log_2\frac{n}{15}]$. We have  $ k\ge 0$ and $ n<2^{k\plus{}1}15\le 2n$

And since  the sum of  prime factors of $ 2^{k\plus{}1}15$ is $ 2\plus{}3\plus{}5\equal{}10$ divisible by $ 10$, we get that for any integer $ n \ge 15$, there is an integer between $ n$ and $ 2n$ inclusively with the sum of its prime factors divisible by $ 10$.

Nice solution! Actually the original problem was to come up with some constant $ c$ such that, there is an integer between $ n$ and $ n\plus{}c$ inclusively with its sum-of-prime factors divisible by $ 10$. My approach was the same, I wanted to prove that there is an integer with the form $ 2^k3^l5^m$ between $ n$ and $ n\plus{}c$, but due to Chinese reminder theorem it is not true. I found this problem too difficult (is it?). Anyway, based on the above proof one can replace $ 2.1$ with any constant $ > 1$:

For any $ \epsilon > 0$, prove that there is an integer $ N$ such that, for any $ n \geq N$, there is an integer between $ n$ and $ (1\plus{}\epsilon)n$ inclusively with its sum of prime factors divisible by $ 10$.


number theory
relatively prime
algebra open
algebra

Find a closed form expression for $\frac{(2^{ab} - 1)}{(2^a - 1)(2^b-1)}$, where a and b are relatively prime positive integers.

What is the definition of "closed form expression"?

I'd be willing to accept any expression in a or b, actually.


a,b,c, are  positive
ab+ac+bc+abc=4  proof:

a^1/2+b＾１／２＋c^1/2>=(abc+8)^1/2

the problem  is  from  an  old  problem  which is Iran TST 2 008

hello, do you mean :If $ a,b,c>0$ and $ ab\plus{}ac\plus{}bc\plus{}abc\equal{}4$ then prove that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\geq\sqrt{abc\plus{}8}$.
Sonnhard.

After substitution $ a \equal{} \frac {2x}{y \plus{} z},$ $ b \equal{} \frac {2y}{x \plus{} z}$ and $ c \equal{} \frac {2z}{x \plus{} y},$
where $ x,$ $ y$ and $ z$ are positive numbers, we need to prove that
$ \sum_{cyc}\sqrt {x(x \plus{} y)(x \plus{} z)}\geq2\sqrt {(x \plus{} y \plus{} z)(xy \plus{} xz \plus{} yz)},$ which follows from Holder:
$ \left(\sum_{cyc}\sqrt {x(x \plus{} y)(x \plus{} z)}\right)^2\sum_{cyc}\frac {x^2}{(x \plus{} y)(x \plus{} z)}\geq(x \plus{} y \plus{} z)^3.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221359

I know  the  way you do  the  problem(actually  we only need to  square it) .but I want to  use another way to  solve it  so I get the form like ab+ac+bc+abc=4
Is it ok to do the  inequality in this form? :blush:

[quote="stevenwang1992"]the problem  is  from  an  old  problem  which is Iran TST 2008, but I want to  use another way to  solve it[/quote]If $ a,b,c$ are  nonnegative numbers such that $ bc \plus{} ca \plus{} ab \plus{} abc \equal{} d$(constant), then $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\geq\sqrt {abc \plus{} 8}$ holds if and only if $ 4\leq d \leq 13 \plus{} 3\sqrt {33}.$

$ d \equal{} 4$ see also : http://www.mathlinks.ro/viewtopic.php?t=280805


function
inequalities
inequalities solved

For positive a,b,c
 \[  \frac{a}{7+b^3+c^3}+\frac{b}{7+a^3+c^3}+\frac{c}{7+a^3+b^3}\leq \frac{1}{3}\]
If someone interested in my solution here it is:
[hide]\[ \sum \frac{a}{7+b^3+c^3}\leq \sum \frac{a}{6+3bc}\leq \frac{1}{3}\][/hide]

Oooops I forgot this one  $0 \leq a,b,c \leq 1$

[hide]Since $ (1-b)(1-c) \geq0$ we have $1+bc \geq b+c$. Then $2+bc \geq 1+b+c \geq a+b+c$. So $2+bc \geq a+b+c.$
Assume that $a \geq b\geq c$ then $ \sum\frac{a}{2+bc}\leq \frac{a+b+c}{2+bc}\leq 1$ :D [/hide]

Here is my solution: (feedback would be appreciated)
[hide]
If you hold any two of a, b, c constant, the function is a convex function for the third. Therefore, when it reaches it's maximum, all of a, b, c are either 0 or 1. It is easy to check that the inequality holds when all of a, b, c are zero or one.
[/hide]

we have 1+a^3+b^3 >= 3ab >= 3abc then we can complete solution easily


[hide][quote="Interval"]A family spends 20 percent of its monthly income on food, 23 percent on rent, and 42 percent on other expenses and saves the balance. If the family saves 360 dollars per month, what is its monthly income?[/quote]

it saves 100-23-20-42=15%
.15x=360
.05x=120
x=2400[/hide]


calculus
integration
derivative
calculus computations

show that the integral from 0 to infinite of (x^2)*(e^(-x^2)) dx = integral from 0 to infinite 1/2*(e^(-x^2)) dx

 :blush: plz help me prove this


Solve the equation $y^2=1+x+x^2+x^3$ in integers

yeah the idea is good!  :D

the problem is not solved in the above link...


Find the minimum number of colors necessary to color the integers from 1 to 2007 such that if
distinct integers a, b, and c are the same color, then a does not divide b or b does not divide c.

1: if you have three integers, let a be the biggest and let c be the smallest. Thus you can have a monochromatic list.

The point was that the condition must hold true for all distinct a, b, and c, not just the ones you choose. For example, the ordered triplet (a,b,c)=(2,4,8) would have a|b and b|c, so it cannot be monochromatic.

[hide]
We need at least $ 6$ colors to partition the $ 11$ numbers $ \{2^0, 2^1, 2^2, \cdots, 2^{10}\}$ into sets, such that no set has more than two of those elements.

Consider a number $ n \equal{} \prod_{i \equal{} 1}^q p_i^{e_i}$, where $ p_i \in \{\text{primes}\}$, and let $ p(n) \equal{} \sum_{i \equal{} 1}^q e_i$, with $ p(1) \equal{} 0$ (in other words, $ p(n)$ is the number of prime factors of $ n$, including repeats). Suppose there are six colors, corresponding to the numbers $ 0,1,2,3,4,5$. We color $ n$ with the color corresponding to $ \left \lfloor \frac {p(n)}{2} \right \rfloor$.

Clearly $ 0 \le p(n) \le 10$ for $ n < 2^{11}$, so each number is colored. Also, we note that if $ a | b$ and $ a \neq b$, then $ p(a) < p(b)$. Thus if $ a|b$ and $ b|c$, then $ p(a) \le p(c) \minus{} 2$, so $ \left \lfloor \frac {p(a)}{2} \right \rfloor < \left \lfloor \frac {p(a) \plus{} 2}{2} \right \rfloor \equal{} \left \lfloor \frac {p(a)}{2} \right \rfloor \plus{} 1 \le \left \lfloor \frac {p(c)}{2} \right \rfloor$, and $ a,c$ will not have the same color. So $ 6$ colors are also sufficient.
[/hide]

[hide="Another Coloring"]For distinct $ a,b,c$, observe that $ a|b \implies b \geq 2a$ and $ b|c \implies c \geq 2b$. Thus, to prevent $ a|b \wedge b|c$ within a color set, it suffices to prevent instances of $ c \geq 4a$ within that set. This motivates the following coloring:

$ \{ 1 \} , \; \{ 2 \ldots 7 \} , \; \{ 8 \ldots 31 \} , \; \{ 32 \ldots 125 \}, \; \{126 \ldots 501 \} , \; \{ 502 \ldots 2007 \}$

Above, [b]azjps[/b] shows that we need at least this many sets, so we are done.[/hide]


I am looking for something to do this summer. Does anyone know of any math-related summer jobs or internships that are available to high-schoolers. Please post ideas here, thanks!

After poking around the NSA site while wishing the round 4 comments/short answers were up:

If you live near Washington/Baltimore or can find residence near there, the NSA's Gifted and Talented program sounds like a very sweet deal.

[url]http://www.nsa.gov/programs/employ/gat.cfm[/url]

Basically, you work 32 hours a week during the summer and get paid a GGD-01 salary ($15,125) for the first summer (after junior year) and a GGD-02 salary ($17,005) for the second summer.

However, I would check those figures.  Also, seeing that you live in Massachusetts, that might not be an option.  Still, it's a considerable amount of money for a high school student working only 32 hours per week, doing something that sounds pretty fun anyway.

[quote="NSA"]you will earn a wage consistent with a GGD-01($15,125) job classification during the first summer, and a GGD-02($17,005) the second summer. [/quote]
I believe these are yearly salaries at government grades 1 and 2.  So, for a summer, you should divide these numbers by approximately 4.

Hm...I take the part about "sweet deal" back.

mathfanatic:

Hmm, yes, that seems like quite an interesting program. Too bad the deadline was in November. :(

tokenadult:

Yes, tutoring is a very worthwhile experience. However, I was sorta looking for something more involved and large-scale, as opposed to tutoring, which I can do during the school year anyways.

Any other suggestions?

[quote="tokenadult"]Tutoring people who know less math than you is a worthwhile summer job.[/quote]

Wouldn't tutoring jobs be harder to find during the summer, seeing that most people won't be in school (and therefore won't be compelled to get a tutor)?


function
logarithms
real analysis
real analysis unsolved

Prove that there does not exist a twice differentiable function $f$, defined on $[0,\infty)$ such that \[f''(x)f(x)+1\leq 0,\forall x\geq 0.\]

[quote="Cezar Lupu"]Prove that there does not exist a twice differentiable function $f$, defined on $[0,\infty)$ such that
\[f''(x)f(x)+1\leq 0,\forall x\geq 0. \]
[/quote]

Let $y = f(x)$, and let $p = y'(x)$.

Then, $f''(x) = \frac{dp}{dx}= \frac{dp}{dy}* \frac{dy}{dx}= \frac{dp}{dy}p$.

Therefore, we have, $\frac{dp}{dy}p y+1 \leq  0$.

Thus, $-p dp \geq  \frac{dy}{y}$.

Then, $k-\frac{p^{2}}{2}\geq  \ln y$.

And then that should lead to a contradiction somehow...

Since $f(x)\ne 0$ for all $x\ge 0$, we may assume $f(x)>0$ without loss of generality. Then $f''(x)<0$, so $f$ is concave down. A concave function can stay positive only if it is increasing all the way. Then $f(x)\ge C>0$ for all $x\ge 1$, which implies $f''(x)\le-1/C$ for all such $x$. Integrating $f''$ over a sufficiently long interval, we find that $f'$ eventually becomes negative, a contradiction.

the original statement of the problem asserts that $f$ is a positive function.


do veterans have to re-register or do we just submit the first round?

Look no further than the [url=http://usamts.org/Forms/U_Forms.php]Forms[/url] page:
"Each participating student must submit a completed and signed a new Entry/Permission Form each year."


Find $ x \in R$ know that: $ 2x\plus{}\sqrt{x}\plus{}\sqrt{x\plus{}7}\plus{}2\sqrt{x^2\plus{}7x}<35$


4zelea ABCD  mohiti ast . nimsazhaye B va D ra rasm mikonim ta ham digar ra dar I ghat konand ..
BI ra emtedad dadeh tori ke zaviyeye MAB=90 va DI ra hamintor emtedad dadeh ta zaviyeye DAN = 90 Shavad . Sabet konid Khate AC bar MN amod ast .....

motmaenni soorate soalo doros neveshti? :maybe:

[quote="SoNYDvDPro"]va DI ra hamintor emtedad dadeh ta zaviyeye DNA = 90 Shavad .[/quote]
fek konam oon bayad bashe $ \angle DAN\equal{}90$ :maybe:

in soal ba ghazie karno kheyli rahat hal mishe :)
kafie neshoon bedim $ CM^{2}\minus{}CN^{2}\equal{}AM^{2}\minus{}AN^{2}$ .
hala az amoode $ MK$ ro bar $ BC$ rasm konin . kheyli asoon sabet mishe ke $ BK\equal{}BA$ va dar natije $ CK\equal{}|BC\minus{}BA|$ va alave bar oon $ MA\equal{}MK$ chon $ M$ rooye nimsaze $ B$ hast .
pas $ CM^{2}\minus{}MK^{2}$ = $ (BC\minus{}BA)^{2}$ .
baraye tarafe dige ham benevisin be rabeteye sadeye $ BC\plus{}AD\equal{}BA\plus{}CD$ miresin ke inam vase in doroste ke charzelyi mohitie :)


[quote="epatjn"]mathcount problem

what is the actual area of isosceles trapezoid 2 cm by 12 cm by 6 cm?  1 cm = 10 m
                    |
2|                 | 6
                    |
         above side
           is 12

again express answer in simple radical form.

answer is 800 times square root of 35....where and how did they get this answer?[/quote]

[hide]Each of the "|" things has a length of 2.  Use pythag theorem to get $\sqrt{12^{2}-2^{2}}=2\sqrt{35}$.  So that is the altitude.

The area is $\frac{1}{2}(2+6)(2\sqrt{35})=8\sqrt{35}$.

ummm..

i think i misunderstood the problem or something.  its kinda hard to understand ur diagram :maybe: [/hide]

Oh, I see...

Also, did I mention that the scale is 1 to 10, so that's why the answer is

800*square root of 35

thanks


For all (x1, y1) and (x2, y2) ∈R^2, define ρ((x1, y1), (x2, y2)) = |x2−x1|+|y2−y1|.
Show ρ is a metric on R^2.


geometry
number theory
modular arithmetic

not specifically, but just generally what do u discuss, my coach wants to know. :?:  :huh:

Maybe, you could have your coach make an account.

Yeah, and he could join the Coaches forum.

I'm not sure if this is what a coach would discuss during a meeting, but there are a few recurrent 'themes' in mathcounts. some of these are: modular arithmetic, geometry, combinatorics/probability, algebraic manipulations, etc. If i were a coach(and i am not), i would probably spend half a meeting teaching one of these themes, and use the other half presenting problems/examples to the students. so like if we meet every day, monday is combinatorics, tuesday is algebra, wednesday is.....

But keep in mind i am not a coach, so i wouldn't know....


[quote="climskywell"]I thought they doubled scores of (like if you get 6 right you score 12 points for your team)
each and everyone of our teammates...

okay then how did some people get a score of 9?[/quote]

1. It's just the number right.

2. The only thing I can think of is that you're looking at the columns in the ARML score sheet, which shows how many people on each team got question X right, not how many questions person Y got right.

Yeah, paladin8 is right, most likely.  If you're looking at the scores on ARML's official .xls file, then the columns are the number of people who got each question right on that team, which is the row.  Then you can get 9's.

Yeah, the format is very confusing. Confused me, and a few friends...


Find all positive integers $ n > 1$ such that for any two co prime factors
$ a, b$ of $ n$ the number $ a \plus{} b \minus{} 1$ is also a factor of $ n$.

[hide="my attempt"]
If any of $ a$ or $ b$ is $ 1$ then $ a \plus{} b \minus{} 1 \equal{} a$ or $ b$, so condition will be true.

If both $ a$ and $ b$ are $ > 1$ then

$ a*b *x \equal{} (a \plus{} b \minus{} 1)*y \equal{} n$

if  $ gcd(x,y) \equal{} d$

let,  $ h \equal{} x/d$ and $ k \equal{} y/d$
So, $ gcd(h,k) \equal{} 1$

Now,   $ a*b*h \equal{} (a \plus{} b \minus{} 1)*k$  -----------(1)

Since, $ (a \minus{} 1)(b \minus{} 1) > 0$
=>   $ a \plus{} b \minus{} 1 < ab$
=>   $ h < k$

From 1 we have,

   $ (ah \minus{} k)(bh \minus{} k) \equal{} k(k \minus{} h)$

Now how proceed from here????

Also if $ h \equal{} 1$, then we get ,

       $ (a \minus{} k)(b \minus{} k) \equal{} k(k \minus{} 1)$

Without loss of generality we can assume $ a > b$

So,   $ a \equal{} 2k$ and $ b \equal{} 2k \minus{} 1$

which seems to be the solution  but how can we get the values of n from here??

So please help.

[/hide]

[hide="attempt 2"]

Assume , a and b are prime numbers.

$ (ah \minus{} k)(bh \minus{} k) \equal{} k(k \minus{} h)$

Let $ a > b$, then

$ a \equal{} 2k/h$  and  $ b \equal{} 2k/h \minus{} 1$           {Doubtful Step}
$ \equal{} > h \equal{} 1 or 2$

if $ h \equal{} 1,$  So, $ a \equal{} 2k$
$ \equal{} > a \equal{} 2$ and $ k \equal{} 1$
which is not possible since   h<k

So,  $ h \equal{} 2$
$ \equal{} > a \equal{} k$ and $ b \equal{} k \minus{} 1$

the only two consecutive primes are 3 ,2

So,   $ a \equal{} 3$ and $ b \equal{} 2$

$ \equal{} > n \equal{} 6x \equal{} 4y$

$ \equal{} > n \equal{} 12,24,36.......$

But 12 is the only number satifying the condition.......... :(

[/hide]

i request the moderator to move this post to the olympiad section


function
real analysis
real analysis unsolved

Find the cardinality of the set of monotone functions defined on the closed unit interval.

Let $ f$ be an increasing function. Add vertical segments to the graph of $ f$ at the points of discontinuity to obtain a continuous curve. Rotating this curve by $ 45$ degrees clockwise, we get the graph of some Lipschitz function. And the cardinality of the set of continuous functions is known to be continuum - they are determined by their values on $ \mathbb Q$.

Thanks for this, I had trouble with it at first since I didn't know what Lipschitz meant but Wikipedia cleared that up. As a side note, do you know of any more general results, e.g the size of the set of functions with at most countably many discontinuities?

Still continuum. First, you choose your points of discontinuity. There are continuum ways to do it. Secondly, you fix the values at the points of discontinuity and at all rational points. That is enough to recover the function and you still have continuum choices for this step. The union of continuum number of sets of cardinality continuum is still of cardinality continuum :).

I had a professor that used to comment on how the set of discontinuous functions was $ 2^{c}$, and thus is much larger than the set of functions with countably many discontinuities (including no discontinuities). Although this isn't surprising (easy to construct discontinuous functions), but it is kind of weird that we have all this theory about "nice" functions (only countably many discontinuities which can seem not nice at times) and still haven't said anything about most of the functions out there.

Fedja, your method was my first thought- this is how I solved the problem about monotone functions in fact. However, I then began to doubt myself, what about functions  discontinuities at all rational points in any neighbourhood of some irrationals; what about functions discontinuous at all rationals etc.
(I'm not trying to contradict you, I'm just confused. Sorry if this was a stupid query  :blush:)

All that doesn't matter :) If you have a point $ x$ of continuity, the value at that point can be recovered from the values at any sequence of points $ x_{k}$ converging to $ x$ [b]regardless[/b] of whether the function is continuous at $ x_{k}$ or not. So, even if $ f$ is discontinuous at all rationals, you still can use the values at rational points to recover $ f$ at any point of its continuity.

Awesome, thanks a lot. I would give you tons of those blue hexagons if I had the requisite authority  :D.


Find the area given by $ |x\plus{}y|<5$ and $ |x\plus{}y|>3$.

If you mean bounded by the axes, then the answer is $ 5^2\minus{}3^2\equal{}\boxed{16}$.

[quote="math154"]If you mean bounded by the axes, then the answer is $ 5^2 \minus{} 3^2 \equal{} \boxed{16}$.[/quote]

Where did you get that?
Is it really that simple?

$ |x \plus{} y| < 5\equiv \minus{} 5 < x \plus{} y < 5$ and $ |x \plus{} y| > 3\equiv x\plus{}y<\minus{}3\text{ or }x \plus{} y > 3$,

so the question now is if it is bounded by the axes.

[b]@ SmartGuy:[/b]

$ |x \plus{} y| < 5$, bounded by the axes, looks like this:

[code]
     y
     |\
     |  \
_____|____\ x
\    |
  \  |
    \|[/code]
Two right triangles with leg-length 5. What happens when you slide the lower one diagonally up so that the two hypotenuses coincide? How do you compute the area of the resulting figure?

Now, we want to subtract the area of $ |x \plus{} y| \leq 3$ because it does not satisfy the condition $ |x \plus{} y| > 3$. What is the graph of this region, if bounded by the axes? How can we compute it's area?

[b]@math154/OP:[/b] Perhaps the original question was, find the area of the region satisfying $ 3 \leq |x|\plus{}|y| \leq 5$?

[quote="math154"]If you mean bounded by the axes, then the answer is $ 5^2 \minus{} 3^2 \equal{} \boxed{16}$.[/quote]

Yes. That is how I got my answer.


One of the frequently asked questions is to evaluate the integral $\int_{-\infty}^{\infty}e^{-x^{2}}\,dx$.
It would be cool if newbies can search for "e^{-x^{2}}" and get to the related posts.

It would be cool, but right now it is not possible. I am thinking of different way to implement some sort of search but until then, I can't help that much with this issue.


logarithms
inequalities open
inequalities

Prove that :
$ SinA^{SinB}+SinB^{SinC}+SinC^{SinA}\geq 1+\frac{1}{\sqrt[e]{e}}$
$ A+B+C=\pi$


logarithms
integration
real analysis
real analysis solved

Guys I need help on this question,

$\sum_{n=2}^\infty \frac{1}{n\ln^2 n}$ approximate within 0.01.

one more

$\sum_{0}^{\infty}\frac{(-1)^n}{n!}$

thanks for the help

[i]Edited by Myth[/i]

For the second question we know the exact value. It is $e^{-1}$.

For the first problem we can write \[\sum_{n=2}^\infty \frac{1}{n\ln^2 n}=\sum_{n=2}^\infty \left(\frac{1}{n\ln^2 n}-\int_{n}^{n+1}\frac{dx}{x\ln^2 x}\right)+\sum_{n=2}^\infty\int_{n}^{n+1}\frac{dx}{x\ln^2 x}=\] \[=\sum_{n=2}^\infty\left(\frac{1}{n\ln^2 n}-\frac{1}{\ln n}+\frac{1}{\ln (n+1)}\right)+\frac{1}{\ln 2}.\]
The series $\sum_{n=2}^\infty\left(\frac{1}{n\ln^2 n}-\frac{1}{\ln n}+\frac{1}{\ln (n+1)}\right)$ converges much more faster and we can easily compute it with 0.01 precision. Answer is $\approx 2.1097$.


linear algebra
matrix
linear algebra unsolved

Let $A,B \in \mathcal{M}_{n}(\mathbb{C})$ s.t. $AB=e^{\frac{i2\pi}{m}}BA$, with $m\in \mathbb{N}^{*}$.
Show that $(A+B)^{m}=A^{m}+B^{m}$.


algebra
difference of squares
special factorizations

how do you factor:

$ 2b^2\minus{}5b\minus{}28\sqrt{3}$

I guess the best you can do is complete the square, then use difference of squares, but that turns out very ugly.  It's hard to factor because the coefficients are irrationa.

[quote]2b^2 - 5b - 97/2 = 0[/quote]

Too bad that, even though they're quite close, $ 28\sqrt{3} \approx 48.4974226 \neq 97/2 \equal{} 48.5$.

To factor $ 2b^2 \minus{} 5b \minus{} 28\sqrt{3}$ solve for $ b$:

$ b \equal{} \frac{\plus{}5 \pm \sqrt{25\plus{}4\cdot 2\cdot 28\sqrt{3}}}{4} \equal{} \frac{\plus{}5 \pm \sqrt{25\plus{}224\sqrt{3}}}{4}$

Therefore $ 2b^2 \minus{} 5b \minus{} 28\sqrt{3} \equal{} 2(b \minus{} \frac{\plus{}5 \plus{} \sqrt{25\plus{}224\sqrt{3}}}{4})(b \minus{} \frac{\plus{}5 \minus{} \sqrt{25\plus{}224\sqrt{3}}}{4})$. Ugly as h$ e$ll, but at least factored.


I know this is a stupid question, but it's frustrating me.

In LaTex, I want to say that some number n is greater than some integer.  What is the code/function/whatever for greater than?  All I can find is greater than or equal to, etc.

Thanks in advance.

I just use > , no special code or anything else.

E.g.

$10>n>0$

P.S.  This should go in the $\LaTeX$ forum.


The imaginary number $ i$ is defined such that $ i^{2} \equal{} \minus{}1$. What does $ i \plus{} i^{2} \plus{} i^{3} \plus{} ... \plus{} i^{23}$ equal?

$ S\equal{}i\plus{}i^2\plus{}\cdots \plus{}i^{23}$,
$ iS\equal{}\plus{}i^2\plus{}i^3\plus{}\cdots \plus{}i^{23}\plus{}i^{24}$

$ \therefore (1\minus{}i)S\equal{}i\minus{}i^{24}\equal{}i\minus{}1$, yielding $ S\equal{}\minus{}1.$

or remark $ i\plus{}i^2\plus{}i^3\plus{}i^4\equal{}0.$

[quote="kunny"]$ S \equal{} i \plus{} i^2 \plus{} \cdots \plus{} i^{23}$,
$ iS \equal{} \plus{} i^2 \plus{} i^3 \plus{} \cdots \plus{} i^{23} \plus{} i^{24}$

$ \therefore (1 \minus{} i)S \equal{} i \minus{} i^{24} \equal{} i \minus{} 1$, yielding $ S \equal{} \minus{} 1.$

or remark $ i \plus{} i^2 \plus{} i^3 \plus{} i^4 \equal{} 0.$[/quote]

Could you explain that more in-depth as if I've never seen a problem like that before?
What is $ S$, and why is it used?  Why did you add $ i^{24}$?

[quote="ephiny79"][quote="kunny"]$ S \equal{} i \plus{} i^2 \plus{} \cdots \plus{} i^{23}$,
$ iS \equal{} \plus{} i^2 \plus{} i^3 \plus{} \cdots \plus{} i^{23} \plus{} i^{24}$

$ \therefore (1 \minus{} i)S \equal{} i \minus{} i^{24} \equal{} i \minus{} 1$, yielding $ S \equal{} \minus{} 1.$[/quote][/quote]

Can you understand $ \boxed{1 \plus{} r \plus{} r^2 \plus{} \cdots \plus{} r^{n \minus{} 1} \equal{} \frac {1 \minus{} r^n}{1 \minus{} r}\ (r\neq 1)}?$

Expand:

$ (1\minus{}r)(1\plus{}r)$

$ (1\minus{}r)(1\plus{}r\plus{}r^2)$

$ (1\minus{}r)(1\plus{}r\plus{}r^2\plus{}\cdots \plus{}r^6)$

Now what dou you think about the result?

[quote="varunrocks"]cycles of four give you 0. Look:
i+i^2+i^3+i^4=
i-1+i(i^2)+1=
i-1-i+1=0
i^5 is the same as writing i so there is a cycle every four terms.
So all we need to calculate is the sum of the first three numbers of the sequence. which is:
i-1-i=-1 which is our answer.[/quote]

That's right. :)


show that (3^2n+2)*(4^2n+3)-12*(2^2n)*(6^2(n+1)) is a perfect square for all integers n.


inequalities
geometry
inequalities proposed

This is the inequality of my teacher.It's really ugly and I have my ugly solution for it. :D
Let a,b,c be the length of the sides of triangle ABC.S is the area of ABC.and 0<a<b<c .Prove that:
$ \sum (\frac{ab(a\minus{}b)}{(b\minus{}c)(c\minus{}a)})^2 \geq 4S$

Hint:
use this inq:
$ a^2x\plus{}b^2y\plus{}c^2z \geq 4\sqrt{xy\plus{}yz\plus{}zx} S$

This is my solution:
use Cauchy,and this inequality:$ abc \geq \frac{4}{3}*\sqrt{a^2\plus{}b^2\plus{}c^2}*S$
And I will prove this inequality:
$ \sqrt[3]{\frac{27abc}{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}}*\frac{4}{3}*\sqrt{a^2\plus{}b^2\plus{}c^2} \geq 1$
Which is obvious true.


How does one show that, for every prime p, one of 2, 3 or 6 is a square modulo?

[I know 2 is a square modulo p iff p is congruent to 1 or -1 modulo 8.]

Thanks

By the multiplicativity of the Legendre Symbol, $ (\frac{2}{p})*(\frac{3}{p})\equal{}(\frac{6}{p})$, so not all of these can be -1.


an experienced plumber made 600 dollars  for working on a certain job .his apprentice , who makes 3 dollars per hour less, also made 600 dollars . however ,the apprentice worked 10 hours more than the plumber. how much does the plumber mahe per hour

Let the number of dollars the plumber makes an hour be $ x$. The apprentice thus makes $ x \minus{} 3$ dollars per hour. The plumber thus worked for $ \frac{600}{x}$ hours. Thus, $ (\frac{600}{x} \plus{} 10) \cdot (x \minus{} 3) \equal{} 600$. Thus, $ x \equal{} \boxed{15}$.


AMC
AIME
USAMTS
pigeonhole principle
trigonometry
percent
AIME I

What is better, 123 AMC and 6 AIME or somewhere between 80-85 on the USAMTS. These are my scores, and for some reason I just do not do well on the USAMTS, no matter how much time I put into it. I feel like a failure.

You shouldn't feel like a failure - those are fine scores.  Remember when you're comparing yourself to people on this site both that there is a severe selection bias in what people are saying, and that some people are not as careful as they should be when they discuss their very high scores in derogatory terms.

You're doing fine.  Keep working at it like this and you'll see what I mean when you get to college and all those people who didn't spend time on things like this feel like a failure while you're cruising to A's.

If you are acing everything in sight, then you should feel like a failure because you aren't seeking big enough challenges.

This is the best post I've seen on this board in some time now, and perhaps on this board period.  I'm glad you took the time to say it.  This ought to be something that every student on the board has an opportunity to read.  I don't know if it actually IS posted elsewhere, but I haven't seen it if it is.

Discussions like one I recently waded through regarding the recent AIME - a  nauseating discussion on how "easy" it was - discourage fresh, undeveloped talent from becoming a part of this community, as well as students with less time to devote to math (and posting on this board), but still have an interest in improving that could be cultivated over time.  And its these exact people on the fringe that would drive the majority of the revenue from your classes, books, etc.  Needless to say, we all have an interest in expanding this community.

While you obviously can't keep discussions like that AIME one from occurring, you CAN make it clear that the elitist mentality of that thread (and many others) doesn't define the AoPS community itself.  A post exactly like what you wrote in response to this student that's put up front and center on this board would go a long way toward shaping a community mentality.  Hopefully you will consider putting it, or a similar statement, in a prominent place on the board.

Hear, hear!  I don't think AHSIMC's post needs any reiteration, but I think it's a brilliant analysis of the situation in a lot of "higher" high school math circles.  I've had several friends, all smarter than myself, who could run circles around me in math competitions (CAMLs and the AIME, for example), and who had nothing to comment on other than that the problems were "so easy."  I always felt intimidated and somewhat belittled by these comments, although they weren't intended at all to put me down, and they definitely were an impediment to my getting comfortable with math way beyond the normal high school level.  And it doesn't help our own reputation, either - would we rather be seen as an open group of bright, motivated math enthusiasts willing to help others with less experience, or as a group of elitists who look down their noses at those less skilled?  I think it would be great if we could all make a conscious effort when commenting on mathematics to say something more constructive than "that was so simple/easy."  Because being truly successful in mathematics isn't being the best at solving a problem (in my opinion) - it's being able to teach somebody else how you were able to solve it, and having an open and understanding atmosphere is a huge boon toward achieving that goal of success.

So which is better? AIME 6 or USAMTS 85? I'm wondering since I may not mention a low USAMTS score on apps and stuff if I can instead highlight my AIME score...

I really don't think you can compare AIME scores with USAMTS scores (unless its something obvious, like 98 on USAMTS and 1 on AIME). 6 on the AIME and 85 on the USAMTS are both very respectable scores. I think they are about the same; a 6 on the AIME may be a little better -- but again, you can't really compare them. In any case, 85 is definitely not a "low" score on the USAMTS. Besides, in your college application, just doing the competition in the first place will give you an adge over people who didn't.

85 is definently better than what I'm going to get... screwed up last round. Is this considered talking about the round? If it is just delete this portion. Anyways, 85 is really a good score.

(Note: There's a reason I'm at AoPS instead of Art of Essay Crafting or something.  I think I've used "we" here to mean "Those AoPSers who do not find the AIME easy, and who are slightly uncomfortable when other people do," or something to that effect. )

I can definitely understand how the discussion of, say, the 2005 AIME I ("ridiculously easy" - suuure...) would intimidate some people.  (Actually, I have to admit I was pretty intimidated by that thread too!)

But is it these students' comments, or merely their existence, that intimidates us?

Because many of us are the most talented math students we know in real life (come on, admit it) it can be pretty easy to fall into the trap of basing a sense of self-worth on this fact.  Fortunately, the Internet has made it harder for most of us to keep our eyes happily closed like this!  AoPS is the first experience some of us have interacting with students who are better at math than we are, and it can come as a bit of a shock when you're constantly being told how amazing your own accomplishments are.  But this is a good thing!  As Mr. Rusczyk is fond of pointing out, if you're the best student in the class you're in the wrong class.

I used to like math pretty much because I was good at it.  I mean, I saw that it was the easiest subject for me, and it was fun, and it wasn't an especial interest for either of my sisters so there wasn't the issue of competition.  This year, I've realized that a) I love math and b) I know next to nothing about it!  It turned out that "Because I'm good at it" wasn't a very good reason to like math, especially since this is *very* vulnerable to the realization that it's not true.  :)

I've started to love math for its beauty (call me superficial) and practicality - the ability to model and describe nature on pure logic is just - well, beautiful.  The "pure" branches of math are beautiful, too, but I'm no good at making a case for this - read Hardy's "A Mathematician's Apology."

There was a point to this rambling, I swear.  Mainly, that it's not anyone else's fault for "intimidating" us - it's our own fault for believing that someone else's relative strengths make us inferior.  (Of course, as there are many students who won't really believe this and might get turned off of math once they realize they're not the best at it, some sensitivity is required here.  "Deal with it, you're not that great even at your chosen interest" isn't too helpful.)

But we do have to consider the benefits of a "safe" spot to express opinions such as disappointment at the difficulty of the AIME which would generally be considered inconsiderate.  Why do we look down on an honest opinion?  Should we really be expecting some of the most profoundly mathematically gifted eighth-graders to consider, before every sentence, whether they're going to intimidate someone, when we don't expect this of any other students (or adults)?

I think many of us can relate to the boredom of classes at school.  They go slowly - they're too easy - whatever.  At school, naturally, we're not going to mention this, whether just to avoid getting beaten up or to facilitate friendships.  Here, it's okay not to "hide."

But wouldn't it stink to get *here* and not be allowed to speak freely, just because you:AoPS :: a typical AoPSer:regular classroom?  Given our own backgrounds, shouldn't we be able to empathize, if not agree that #15 was a gimme?

I think your overall assessment is fairly unrealistic.  Most high school students lack the necessary perspective to be unthreatened by the superior abilities of others, largely due to poor parenting, poor choice of friends, or any number of other things.

No one is saying to censor anyone, either, for any math-related opinion.  Though I would be interested to see what might happen if someone posted a giant "I HATE THE AIME" thread in the AMC folder.

This all boils down to a marketing issue for AoPS.  Its not my decision to make, though I've given my opinion on it.  I simply feel that, lately, the substance of posts around here has swung too far in a certain direction.  And I have felt for some time that a post containing the sentiments of rrusczyk's above post could make a fantastic opener to greet newcomers when they first arrive here.

Regardless, my two cents.  I will not be commenting further on this.

I don't think that this year's AIME was a blowoff test or anything like that.  However, I certainly agree that it was easier than other years' tests.  It also did seem to have math team style tricks, where either one knows or does not know the formula.  For someone who went into the AIME expecting a harder test, I don't think it's unreasonable for them to be a little disappointed in the product they received.

[quote="Magnara"]For someone who went into the AIME expecting a harder test, I don't think it's unreasonable for them to be a little disappointed in the product they received.[/quote]
Especially the people who consistently score 10+ anyway (that would not include me, just FYI).

Yes, this AIME could have been harder.  But I thought the first 10 were right about there.  I personally think that there should be at least one or two very easy (About 13-20 on AMC) AIME problems, such that [b]anyone[/b] who passes AMC 12/10 will almost be assured of at least a 1 (provided they don't make stupid mistakes).  This test, I felt, came very close to achieving this (actually with clearer wording on number 3 it would have).

This AIME was definitely very easy (comparatively).  This is coming from someone who barely made the AIME, (101, because I was too scared to answer those extra two questions, and just triple-checked the first eleven), and was hoping for just a nonzero score.

Part of the situation is not just that many of the kids on AoPS are smarter than us, or better at math, but that they've been exposed to different math, in classes specifically geared to make the AIME easy!  The pigeonhole principle, AM-GM, power of a point, etc. are referenced as if they're as well-known as Pythagoras' theorem. (Heck, until I taught it to my friends, I was the only one at my school who knew Hero's/Heron's formula.)
While the AIME only [i]requires[/i] basic high school math, through trigonometry, most of the elegant solutions found here use much more esoteric tools.

Are all these IMO-hopefuls a little intimidating? You bet.  "Oh no, I didn't get #25 on AMC. I only got a 144.  And to make matters worse, I only got a 14 on the AIME--that's only a 93.333.....333 percent! AAaaah.  I'm horrible at math!"
Yeah, that can scare people, but even worse, it can diminish the pride we get from awesome but not-as-good-as-IMO-team-member-number-two's scores.  After taking this AIME I felt great!  I had gotten eight of the problems.  Then after I realized some (five) stupid mistakes I had made.  I come on to AoPS, and see people crying about 12's.  That's a bit of a blow to the ego.  I ended up with a 2, which appears to be in the bottom ten percent.  Factor in who is using this forum, then factor in who is brave enough to actually post (notice at least 60% of the members haven't actually posted anything), then who passed the AMC 12 in the first place. [I'm making up numbers here:] About 200,000 kids took the AMC's. Approximately ten thousand kids took AIME.  Approximately 5000 got above a 2.  That puts you in the top 2.5% of AMC test takers.  Wait, who takes the AMC?  Do the kids in 12th grade algebra 1? Assume everyone who takes the AMC is already in the top five percent.  Then you're really in the [b]top .125%[/b].   Just making AIME would put you in the top .25%.  Yes, those were all made up numbers, but it should help put it into perspective.

If that doesn't work, just pretend they're lying.  That they have such low self esteem that they have to lie to make themselves feel smarter.  You're actually smarter than [name someone who says that an 11 is a low AIME score].  He's just seeing double.

I agree, you cant really compare AIME and USAMTS. AIME is short answers in 3 hours. It is very easy to make a little mistake that causes you to miss a problem. The key in te AIME is knowing how to solve the problems right away and then solving them. USAMTS, on the other hand, is proof based and a month long for 5 problems. Thus you have lots of time to try and solve them problems, the key in USAMTS is simply you general mathematical knowelge and your willingness to commit a lof of time to it. Thus the two tests, out side of highschool math as a key requirement  :D , take totally differnt skills and cannot be compared  :)


limit
integration
calculus
calculus computations

Not Formal, But

${\lim_{x\to \infty }\, f(x)=\infty }$
${\lim_{x\to \infty }\, g(x)=\infty }$

${\lim_{x\to \infty }\, f'(x)=\infty }$
${\lim_{x\to \infty }\, g'(x)=\infty }$

${\lim_{x\to \infty }\, f''(x)\neq \pm \infty }$
${\lim_{x\to \infty }\, g''(x)\neq \pm \infty }$

${\lim_{x\to \infty }\, \frac{f(x)}{g(x)}=\lim_{x\to \infty }\, \frac{f'(x)}{g'(x)}=\lim_{x\to \infty }\, \frac{f''(x)}{g''(x)}}$

But, Can i Assume the inverse!?

${\lim_{x\to \infty }\, \frac{f(x)}{g(x)}=\lim_{x\to \infty }\, \frac{\int f(x) \, dx}{\int g(x) \, dx}}$

Assuming

${f'(x)=g(x)}$

Yes, But, it is all true!?

$\lim_{x\to 0}\frac{\sin(x)}{x}=1$
but
$\lim_{x\to 0}\frac{\int\sin(x)dx}{\int xdx}\neq 1$

i had said the case of   x -> infinity   f[x] -> infinity


abc=1 a, b, c are positive real numbers.
prove 1+3/(a+b+c)>=6/(ab+bc+ca)

[quote="mathematical intelligence"]$ abc\equal{}1$  $ a, b, c$ are positive real numbers.
prove $ 1\plus{}3/(a\plus{}b\plus{}c)\geq 6/(ab\plus{}bc\plus{}ca)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80314

Just use $ (ab\plus{}bc\plus{}ac)^{2}\geq 3abc(a\plus{}b\plus{}c)$,and let $ x\equal{}ab\plus{}bc\plus{}ac$,the inequality is the same with
$ (x\minus{}3)^{2}\geq 0$. :)


We are given the following function:

(x - 1)
______   =  f(x)

(x + 1)

express the said function as a sum of an even and an odd function.

What? As in, $ x \minus{} 1 < f(x) < x \plus{} 1$ $ \forall x$?...

[hide="Rough outline"]Let $ f(x) = e(x) + o(x)$. We know that $ e( - x) + o( - x) = e(x) - o(x) = ( - x) = \frac {x + 1}{1 - x}$, so we have the system

$ \begin{eqnarray[*]} e(x) & - & o(x) & = & \frac {x + 1}{1 - x} \\
e(x) & + & o(x) & = & \frac {x - 1}{x + 1} \end{eqnarray[*]}$.

Solving, $ e(x) = \frac {x + 1}{2 - 2x} + \frac {x - 1}{2x + 2}$ and $ o(x) = \frac {x - 1}{x + 1} - \frac {x + 1}{2 - 2x} + \frac {x - 1}{2x + 2}$. However, $ e(x)$ isn't actually even and so there are no solutions(?) I might have made a mistake somewhere.[/hide]

J.Mayer: I believe that's a crude text representation of a fraction.

EDIT: Ah, I see my mistake. $ f( - x) = \frac {x + 1}{x - 1}$, not $ \frac {x + 1}{1 - x}$. Then
$ \begin{eqnarray[*]} e(x) & - & o(x) & = & \frac {x + 1}{x - 1} \\
e(x) & + & o(x) & = & \frac {x - 1}{x + 1} \end{eqnarray[*]}$.
so that $ e(x) = \frac {x + 1}{2x - 2} + \frac {x - 1}{2x + 2}$ and $ o(x) = \frac {x - 1}{x + 1} - \frac {x + 1}{2x - 2} - \frac {x - 1}{2x + 2}$.

We can simplify these to $ e(x)=\frac{x^2+1}{x^2-1}$ and $ o(x)=\frac{2x}{x^2-1}$, and these are indeed even and odd respectively.

hello, we have $ f(\minus{}x)\equal{}\frac{\minus{}x\minus{}1}{\minus{}x\plus{}1}\equal{}\frac{x\plus{}1}{x\minus{}1}$
Sonnhard.

[quote="Temperal"]
We can simplify these to $ e(x) \equal{} \frac {x^2 \plus{} 1}{x^2 \minus{} 1}$ and $ o(x) \equal{} \frac {2x}{x^2 \minus{} 1}$, and these are indeed even and odd respectively.[/quote]
These add up to $ \frac {x \plus{} 1}{x \minus{} 1}$, not $ \displaystyle \frac {x \minus{} 1}{x \plus{} 1}$. So $ o(x)$ is actually just the negative of what you put, $ \minus{} \frac {2x}{x^2 \minus{} 1}$


Alright I am stuck on the last 3 problems of the Inequalities chapter in the text "Mathematical Circles". Unfortunately there is not solution for these in the book so hopefully you guys can help me out.

83. Suppose a, b, c, d are positive. Prove that at least one of the following 3 inequalities are false:
a) a+b < c+d
b) (a+b)cd < ab(c+d)
c) (a+b)(c+d) < ab + cd

84. Assume a,b,c,x,y,z are all positive. Show at least one of the following 3 inequalities is false:

a) a*y / (a+y) < b*x / (b+x)
b) b*z / (b+z) < c*y / (c+y)
c) c*x / (x+c) < a*z / (a+z)

85. Prove that if x + y + z >= xyz then x^2 + y^2 + z^2 >= xyz

Thanks for your help

[hide="83"]$ (a\plus{}b)^2>ab\plus{}cd$ (a and c)
$ (a\plus{}b)cd<ab\left(\frac{ab\plus{}cd}{a\plus{}b}\right)$ (b and c)
$ \Leftrightarrow (a\plus{}b)^2<\frac{ab(ab\plus{}cd)}{cd}$
$ \Leftrightarrow cd<ab$ (*)
$ \Leftrightarrow (a\plus{}b)(c\plus{}d)<2ab$ (1 and *)
Obviously $ cd<ab\plus{}cd$ and therefore we get that $ c\plus{}d<a\plus{}b$. Contradiction.[/hide]

[quote="Spacecow"][quote="Temperal"][hide="83"]$ (a \plus{} b)^2 > ab \plus{} cd$ (a and c)
$ (a \plus{} b)cd < ab\left(\frac {ab \plus{} cd}{a \plus{} b}\right)$ (b and c)
$ \Leftrightarrow (a \plus{} b)^2 < \frac {ab(ab \plus{} cd)}{cd}$
$ \Leftrightarrow cd < ab$ (*)
$ \Leftrightarrow (a \plus{} b)(c \plus{} d) < 2ab$ (1 and *)
Obviously $ cd < ab \plus{} cd$ and therefore we get that $ c \plus{} d < a \plus{} b$. Contradiction.[/hide][/quote]

I believe (a) and (c) give the opposite inequality  :huh:[/quote]

Erm, huh? Could you explain?

Here's is x,y,z are positive
[hide="85"]
The last equation says $ x(x \minus{} yz) \plus{} y^2 \plus{} z^2\ge 0$. This is true if $ x\ge yz\ge 0$. So we can assume $ x\le yz$ if they are all positive. Since we're assuming $ x\le yz$, $ y\le xz$, $ z\le xy$, we have $ xyz\le x \plus{} y \plus{} z\le xy \plus{} yz \plus{} xz\le x^2 \plus{} y^2 \plus{} z^2$, where the last is by AM-GM. [/hide]

[quote="Temperal"][quote="Spacecow"][quote="Temperal"]$ (a \plus{} b)^2 > ab \plus{} cd$ (a and c)
[/quote]

I believe (a) and (c) give the opposite inequality  :huh:[/quote]

Erm, huh? Could you explain?[/quote]

a+b < c+d and (a+b)(c+d) < ab + cd

=> (a+b)^2 < ab + cd hence I don't think you can conclude cd < ab later on.

[hide="83 (correct solution)"]
Assume there exists a solution.
Notice that as long as we keep ab constant, we can decrease a+b and still have the inequalities hold.

$ a\plus{}b\geq2\sqrt{ab}$ with equality holding if and only if a=b. Therefore, a+b is minimal when a=b.

Therefore, we can let a=b.

$ a<\frac{c\plus{}d}{2}$ (1)
$ cd<a\frac{c\plus{}d}{2}$ (2)
$ 2a(c\plus{}d)<a^2\plus{}cd$ (3)

(2) applied to R.H.S. of (3) gives

$ 2a(c\plus{}d)<a^2\plus{}a\frac{c\plus{}d}{2}$
$ 2(c\plus{}d)<a\plus{}\frac{c\plus{}d}{2}$
Apply (1) now to the R.H.S.
$ 2(c\plus{}d)<c\plus{}d$
$ c\plus{}d<0$
Contradiction!
[/hide]

[quote]
84. Assume $ a,b,c,x,y,z$ are all positive. Show at least one of the following 3 inequalities is false:

a) $ \frac{ay}{a+y}<\frac{bx}{b+x}$
b) $ \frac{bz}{b+z}<\frac{cy}{c+y}$
c) $ \frac{cx}{c+x}<\frac{az}{a+z}$
[/quote]

[hide]Since all variables are positive, write the three given inequalities as \begin{eqnarray*}
\frac{b+x}{bx}&<&\frac{a+y}{ay}\\
\frac{c+y}{cy}&<&\frac{b+z}{bz}\\
\frac{a+z}{az}&<&\frac{c+x}{cx}\end{eqnarray*} By splitting the numerator, this is equivalent to \begin{eqnarray*}
\frac1b+\frac1x&<&\frac1a+\frac1y\\
\frac1c+\frac1y&<&\frac1b+\frac1z\\
\frac1a+\frac1z&<&\frac1c+\frac1x\end{eqnarray*} If all three of these inequalities were true, then the sum of the inequalities would be true. However, the sum is
\[ \frac1a+\frac1b+\frac1c+\frac1x+\frac1y+\frac1z<\frac1a+\frac1b+\frac1c+\frac1x+\frac1y+\frac1z,\] which is clearly false. Thus, at least one inequality is false.[/hide]

Actually, I think my proof applies to any x,y,z now that I look again. If all aren't positive, then exactly 2 are negative since the second given says xyz is positive. If x(x-yz) is negative and x and y are negative then x-yz is positive and hence $ x\le yz$. Also, the inequality $ xy \plus{} yz \plus{} xz \le x^2 \plus{} y^2 \plus{} z^2$ applies to any numbers, because if it works when they are positive, if you sub in -x, -y the right would stay the same but the left would decrease.

[quote="Spacecow"][/quote][quote="Temperal"][quote="Spacecow"][quote="Temperal"]$ (a + b)^2 > ab + cd$ (a and c)
[/quote]

I believe (a) and (c) give the opposite inequality  :huh:[/quote]

Erm, huh? Could you explain?[/quote][quote="Spacecow"]

a+b < c+d and (a+b)(c+d) < ab + cd

=> (a+b)^2 < ab + cd hence I don't think you can conclude cd < ab later on.[/quote]
agh, sorry about that. it was a misreading of what i had written myself on my paper.

Whoops, I can't delete my last post now, but it is wrong about my proof applying to negatives.


a)find all the solutions of the equation $(m,n)$ are positive integer:
      $7^m=3^n+4$
b) solve the equation:
      $\frac {x}{y^2z^4}+\frac {y}{z^2x^4}+\frac {z}{x^2y^4}=t$   $(x,y,z,t \in Z^{+})$
It isn't hard but really nice .I'm so satisfactory while i have the solution with two problems.But it's long .So i want the other solution :D  :D  :D  :D

Hmm..

I'm way below this level :( but I'll give it a shot for the first one.

$7^m = 3^n + 4$ to be true on how many $(m,n)$ pairs?

Okay. I have a good idea for finding the solution but I'm not sure how to work on this. If anyone can gives me hint, I'll try to do my best on it.

I think if I can find number of $n$ that makes $3^n \equiv 3 \mod 7$, then I'm done.

The reason I think this way is since $4$ is $4 \mod 7$ (obviously), to be divisible by 7, other $3^n$ must be 3 in mod 7.

By calculator, I found $n = 1,7$ work. But this might go infinitely long time unless I find the answer to my question.

I think the question "may" be simple if I can get the answer to that question.

 :) Help!

Here's an interesting factorization:
We can show that $m$ and $n$ are both odd. If $m=2r+1$ and $n=2s+1$, \[4\cdot\left(7^r-1\right)\cdot\left(7^r+1\right)= 3\cdot\left(3^s-7^r\right)\cdot\left(3^s+7^r\right)\]
I'm not sure what to do from here, but this feels useful.

I think if we use it we will have a lot of cases and difficult to find the solution to each case.For problem a) I use modular and i have the solution ($mod 27$ and $mod 19$)

[quote="blahblahblah"]Silverfalcon, that congruency is necessary but not sufficient.  Furthermore, we have

$3^n\equiv 3\bmod 7$ for all multiples of $7$ by Fermat's Theorem.[/quote]

From Mathworld..

Fermat's Theorem states that: If $p$ is a prime number and $a$ is a natural number, then:

$a^p \equiv a \mod p$

So, $p = 7$ works because of this. :D

And I see why just this information is not sufficient because I don't know how to go from here. :?

I just fixed a typo in my factorization. On the other hand, maybe I should have used $7(7^{2r}-1)=3(3^{2s}-1)$ instead.


I think we should have a quick reply at the bottom of every topic so you can easily post things without having to wait for the post screen to load.  My computer is very slow and sometimes i become impatient of the post screen to load.  Its just a suggestion. :)

That would be great. A forum I regularly go to has it and it is much faster. I also do not like it says "Your message has been posted".
I would prefer if it just went back to the thread. You can go back to prev forum from there and it takes the same amount of steps.

The quick reply will not be added very soon. The site is CSSised, and once you've visited it once, whenever you post you are actually receiving a very, very small quantity of data (so it doesn't matter how slow your connection is).

The confirmation message has other purposes also, and will remain there.

I have a suggestion. Is it possible to change the 'who is online section to a kind of a java applet. an example would be yahoo chess, you immidiately know when someone is offline or online. And, then maybe... you could spread it to the rest of the site (know when a new topic is posted). It sounds very hard, but as math92 said, it's just a suggestion. :D

[quote="Inspired By Nature"]It sounds very hard, but as math92 said, it's just a suggestion. :D[/quote]
i know, but that would be cool. :)


induction vector trigonometry ratio geometry parallelogram logarithms

Hello, can we discuss the CMOR yet? Also, despite all of this talk presenting 8 as the hardest, I thought personally that 5 was the hardest. Was there an inductive solution to 5 that did not involve finding the leg lengths in terms of the kth and other Fibonacci numbers

We can probably discuss. The deadline to turn in the solution was more than 15 hours ago EST.

Yeah there was an induction solution (athunder used one iirc), but thats not the way I did it. I proved a relationship of how an odd fibonacci number always equals the sum of two squares, and then if you square the odd fibonacci number, you can do some basic manipulating to get a sum of two squares, as desired.

#8 was pretty hard imo, but I used vectors to produce a very nice solution. Personally, I though #7 was a joke. Number 6 was probably my favorite question, but still very easy. Number 4 could have gone either way. If you decide to just list the factors, it would be pretty stupid, but the way I did it was really nice. My number 3 was ugly but got the job done, and #2 and 1 were trivial.

Also, when do you generally get the results?

nice solution for #8! That basically makes the whole "parallel" part extraneous information, right. In other words it would hold for any way we drew the three quadrilaterals?

You dont quite need law of cosines, you can use ratios and AM-GM to finish the whole question.

Well, #1 -> #7 were pretty easy.

For #5, I showed that: f(2k+1) = f(k)^2 + f(k+1)^2

You give a tiling proof for this:

f(2k+1) = # ways to tile 1*2k board with 1*1 and 1*2 tiles.

Regard kth tile of a 1*2k board. If the board is breakable at that point (i.e. you can split it into two boards of lengths k each), then there is f(k+1)^2 ways to tile the board (f(k+1) for first k tile, and f(k+1) for last k tiles). If it isn't breakable at the kth tile (i.e. if there's a 1*2 tile laying over tile k and tile k+1), then the # ways to tile the board is f(k)^2 ((f(k) for first k-1 tile, and f(k) for all the tiles after the 1*2 tile). Therefore   f(2k+1) = f(k)^2 + f(k+1)^2.

From here it is merely manipulation:
f(2k+1)^2 =  (f(k)^2 + f(k+1)^2)^2
              =   (2*f(k)^2 + f(k-1)^2 + 2*f(k)*f(k-1))^2
=(2*f(k)^2 + f(k-1)^2 + 2*f(k)*f(k-1)) * (2*f(k)^2 + f(k-1)^2 - 2*f(k)*f(k-1)+ 4*f(k)*f(k-1))
=4*f(k)^2 + f(k-1)^2 + 8*f(k)^3*f(k-1) + 4*f(k)*f(k-1)^3 + 8*f(k)^2*f(k-1)^2
=(2*f(k)*f(k+1))^2 + (f(k-1)*f(k+2))^2

Therefore f(2k+1)^2 =(2*f(k)*f(k+1))^2 + (f(k-1)*f(k+2))^2 for all k >= 2.

#8 was a good question. I set the sides of P1 to X, and Y. And P3's sides to rX and sY (where rX is parallel to X and sY is parallel to Y). Then if we add the areas of the triangles we get when drawing P1, P2 and P3 and P3's area, that equals the area of P1 (draw diagram to notice what I am talking about). After A lot of hard work (:rotfl:), you get that r + s >= 1. In other words, r >= 1/2 or s >= 1/2. That means either rX >= 1/2*X or sY >= 1/2*Y. QED.

I too noticed the AM-GM method. However, I failed to work out the technicalities. Besides, I find this way better  :lol:.

I believe that the cutoff for CMO will be something like 37, or 38 out of 40 (where each question is out of 5).

I did same thing as AJ for $ 5$. Also, my solution for #8 works I think. It was synthetic. I will post it later (gotta go).

Thanks athunder. So, unless they take marks off for the writing up of the solution for #8, I think I should be ok (and so should a lot of people here based on their posts). Did anyone else use a similar method to mine for #8? I mean, it is true that the sum of the ratios of two sides of P3 to their corresponding sides of P1 is => 1 (and athunder's proof seconds this notion).

Hey guys I have a couple of questions:
1) What do you guys think the cutoff score for CMO will be? (looks like you will need a perfect :( )
2) When do results come out?

dude judging by everyone else here it appears as if the cutoff will be 40 $ \pm$ 2..yeah thats right PLUS minus 2.
 :(

haha @Thunder365

Well, I believe that a more realistic cutoff is 37 - 40. Also, you should be getting a mail telling you your score (well, the interval in which your score fell in) and another one letting you know whether you get invited or not to CMO near the end of this month (well, I received my CMOR results on 22nd and the CMO invitation on January 30th last year).

@athunder - No, I don't believe Binet's formula helps for #5...I mean, I haven't worked it out yet, but if it works, it'll be ugly.

math154 showed me a vector method for #8...almost no insight but it ends up working without too much computation  :lol:

[quote="A.J"]haha @Thunder365

Well, I believe that a more realistic cutoff is 37 - 40. Also, you should be getting a mail telling you your score (well, the interval in which your score fell in) and another one letting you know whether you get invited or not to CMO near the end of this month (well, I received my CMOR results on 22nd and the CMO invitation on January 30th last year).
[/quote]

One thing I would like to know is how hard the graders are. What do they look for in a solution? I asked Bacteria and he said that they look for a correct solution mostly, but since he has never taken the CMOR, it would be better if someone with previous experience answered this.

[quote="A.J"]No, I don't believe Binet's formula helps for #5...I mean, I haven't worked it out yet, but if it works, it'll be ugly.
[/quote]
Binets actually helped, and surprisingly it worked without much computation. But your solution using tile colorings is MUCH more elegant than using Binets.

[quote="Bacteria"]Can someone post the link to the problems for us non-CMORs? Thanks[/quote]
http://cemc.uwaterloo.ca/cmor_files/2010/2010CMOQR.pdf

Hm.. you could try calling (613-733-2662) or send an email to: chair-cmoc@cms.math.ca
Your go-to guy is Kalle Karu (Chair of Canadian Mathematical Olympiad Committee)

Did anyone actually get an email, or only the letter that your math teacher got?

Wow...I actually might not make it...that's VERY depressing....

I mean, I should have at least gotten an email of rejection if I didn't make it...I didn't get anything at all. So there still might be a chance (eventhough people already got the emails two days ago ...)

Update...

Apparently, the person I know of who got in was not informed by email. So, I still don't know anyone who's received a final acceptance or rejection email after getting 30-40. We can continue to hope!

So, I asked my math teacher if anything turned up, and he says that he has nothing...

I just hope they send the mails soon (and that I make it lol)

Thnk you Uno!! Thanks to you I contacted the people and have registered now. :)

[quote="A.J"]Wait, what? They already handed everything out?

I mean, I didn't even get an email of rejection yet...and I thought I solved everything perfectly...

This IS depressing...[/quote]

same here... :(

they dont send rejection emails, only if you made it.

Actually I know for a fact that they do send rejection emails. I know a couple of people who got it last year and this year. Also, on the email that tells you the range of your score on CMOR, it states that they'll email you whether you get invited or not....

But I thought I should make it...I mean, I solved everything ....damn, I just guess I fail @ math  :(  :(

[quote="Email From CEMC People"]
Thank you for your submission to the CMO Qualification Repêchage.  We were very pleased with the number and quality of the submissions.

Each of the eight problems was marked out of a possible 5 marks.  Your total score was between 30 and 40. We are not releasing exact scores.  We encourage you to try the problems that you did not solve initially -- talk to your friends or teachers too.  We will not be posting solutions.

We are unable to tell you at this time whether or not you will be invited to write the 2010 Canadian Mathematical Olympiad.  You will hear directly from the Canadian Mathematical Society if (and only if) you are invited to the CMO.[/quote]

Specifically the last paragraph

Yes, that should be correct although last year they accidently sent out rejection emails spewing a wave of confusion here on AOPS...

thanks for asserting the fact that I am bad at math...I appreciate it...  :(

I didn't get anything....so I have failed.

I finally got my invitation, 3 weeks after the letter was sent. The reason? School beaurocracy. Between the principal (who's very nice) and some of the math teachers (Who are also very nice), it took 3 weeks (of pain/anxiety) to finally get my invitation - with 3 days to spare.

I don't know if other schools take this so lightly - but I guess my advice would be to call CMO if you even have an inkling of belief that you made it. Or maybe it's just my school.

Well, I know the reason I didn't make it. To quote what Mr. Ian VanderBurgh sent me:

"
Hi AJ,

Thanks for the message.  My recollection is that your score was quite high, but the decision was made that only people officially invited to the CMOQR would be considered for the CMO.

Ian.

"
To top things up, on the AMC 12 A contest, I forgot to bubble in my answers into the scantron... :(

Also, I finished 1->7 in about 3 hours...and I eventually finished #8...and I thought they would like my solution to #5...

apparently, I wasn't even considered this time round..


Given that (BC)^r = (B/C)^p, express C in terms of B.

1) What does the phrase "express C in terms of B" really means in simple words?

2) What is the answer?

1) so you have c=blahblah..and the blah is just b's..

Come again???

Do you know how to play with this question?

i think thats what it saying..hmm some other person had a different approach but i think it leads to the same result..


Suppose today is tuesday,  what day of the week will it be 100 days from now?

[hide] 100/ 7= 14 r 2    wednesday, [b] thursday [/b] [/hide]

[hide]100 divided by 7 gives us 14 r 2, so 2 days will be added to Tuesday, giving us Thursday. [/hide]

[hide="Solution"]Thursday

Remaider of $\frac{100}{7}$ is $2$. Tuesday + 2 days = Thursday[/hide]


On April 26, discussion may begin.  Until then, even stuff like "____ was really hard/easy/surprising/etc." is not really appropriate.

I know I'm not a moderator or anything, but let's wait for the ACS deadline, OK?  Thanks.

Could we begin discussion on the 25th, since by then, all the students will have take the exam already?  :?

Hm... as I read it, the ACS will accept an exam delivered on the 26th, which may allow testing on the 25th.  But you could always call Ms. Hernandez and ask her tomorrow.  It's not like I work for ACS; I just know what's online.

I would say just wait until the 26th then.  It's not like anything will change in that span of one day, and it's worth putting forth a little extra effort in order to maintain exam integrity.  That's not to say that I'm not eager to discuss the exam, but I think it should just be kept in perspecitve.


I just took this test for practice, and is it just me, or does it seem like this AMC 10 is far easier than all the others? I worked through most problems with minimal effort, yet still had about 20 minutes to check over it. Does this test represent the average difficulty of an AMC 10?

insanely easy...not at ALL representative of regular AMC 10 difficulty level


I was pondering about this curious question.

Let us have a two point bodies on a plane. One of the points say $ p_1$ is fixed at a certain point and an inverse square law holds (ie., For bodies $ A$ and $ B$, $ A (alt\, B)$ attracts the other with a force proportional to $ \frac{1}{r^2}$ in the direction of the ray originating at $ B (alt\, A)$ and pointing at $ A(alt\, B)$  where $ r$ is the distance between $ A$ and $ B$ at that instant of time.). Since $ p_1$ is fixed, the other body $ p_2$ executes a motion if it possessed a initial velocity component in the direction perpendicular to the line joining the two bodies, whose locus is a [url=http://en.wikipedia.org/wiki/Orbit_equation]conic section[/url].

Suppose at $ t\equal{}0$, $ p_2$ has a velocity $ v$ in the direction perpendicular to line joining the two bodies separated by a distance $ r_i$. Is it true that $ \forall \, \parallel{}v\parallel{}<k'$ for some positive real number $ k'$, $ p_2$ excecutes an ellipse?

Yes, $ k'$ is the escape velocity and is easily calculated by requiring that the total kinetic energy plus potential energy is zero.

http://en.wikipedia.org/wiki/Escape_velocity


projective geometry
geometry unsolved
geometry

Let $ABC$ be a right triangle at $A$.Draw the circle $(B)$ with center at $B$ and radius $BA$ and draw a diameter $AD$.Choose two points $E;F$ on the line $BC$ such that $B$ is their midpoint.$DE$ and $DF$ meeyd with $(B)$ at $M$ and $N$ respectively.Prove that $M;N$ and $C$ are colinear.

$MN$ intersects $AC$ at $C'$, $AE$ intersects $(B)$ at $K$.$A$ symmetric with $D$, $E$ symmetric with $F$ to $B$, so $K$ symmetric with $N$ to $B$. Apply Pascal's theorem with the cyclic hexagon $AAKNMD$, we have $AA \cap MN = \{ C' \}, AK \cap MD = \{ E \}, KN \cap DA = \{ B \}$, so $C', E, B$ are collinear. It means that $C \equiv C'$, so $C,M,N$ are collinear.


[quote="chickendude"]Let $S = \{1,2,..... 3n+1\}$ and $T ={1,2,..... 3n}$
Let $x$ and $y$ be the number of ways 3 distinct numbers can be chosen fron the sets $S$ and $T$ respectively such that the sum of the three numbers is divisible by $3$. The order in which the numbers are drawn is irrelevant.

given $x-y = 330$ compute $n$

I have been trying to solve this for a while and keep getting non-integer solutions.[/quote]

[hide]$\binom{n-1}{3}+2\binom{n-2}{3}+(n-1)(n-2)^{2}-3\binom{n-2}{3}-(n-2)^{3}=330$

If you factor and cancel and stuff, you get (I think) $3n^{2}-13n-646=0$ and $\boxed{n=17}$?[/hide]

yep,

here's what I did (it didn't work, I can't figure out why)
[hide]
since $x-y = 330$ then, set $S$ has 330 more of these groups of 3. All these extra groups must be attributed to the new member, $3n+1$

All the new groups of 3 have $3n+1$ as one of their members
This number is $1 \mod{3}$

the other two members of these new sets of 3 numbers must be
both $1 \mod{3}$
or
$2 \mod{3}$ and $0 \mod{3}$

the former has $\binom{n}{2}$ possibilities
the latter has $n^{2}$ possibilities (since the sets are exclusive)

$\frac{n(n-1)}{2}+n^{2}= 330$
$3n^{2}-n-660 = 0$
$(3n+44)(n-15)$
$n = 15$
I'm not going to lie: I have no idea what I did differently just now, but it worked this time around.
[/hide]

[quote="The Zuton Force"]Massachusetts State Meet?[/quote]

I got this one correct at the meet!  :lol:

Nice.

I was on Shrewsbury; the only team round questions we got right were ones I did  :(  this wasn't one of them, but I tried it out after the meet.


[hide="Touch here and see a nice inequality"]
$\triangle ABC,x>0,y>0\Longrightarrow$$\max\{xa+yh_a,xb+yh_b,xc+yh_c\}\le \frac{4x^2+y^2}{8x} (a+b+c).$
[/hide]

Nice problem!We will prove that $\displaystyle xa+yh_a\leq \frac{4x^2+y^2}{8x}(a+b+c) \leftrightarrow 4x^2(b+c-a)+y^2(a+b+c)\geq \frac{8xyS}{a}$,where S is the area.From $4x^2(b+c-a)+y^2(a+b+c)\geq 4xy\sqrt{(a+b+c)(a+b-c)}$ we just need to prove that $a\geq\sqrt{(p-b)(p-c)}$,which is trivial where p is the semiperimeter,using heron formula for area. :)


Two distinct points are selected at random form the six-point set $ \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)\}$. what is the probability that the distance between them on a Cartesian grid is a rational number? Express your answer as a decimal.

There are $ \binom{6}{2} \equal{} 15$ possible ways to select the two points. We now look for rational values with the distance formula. We see that the probability is $ \frac {3 \plus{} 2 \plus{} 1 \plus{} 2 \plus{} 1}{15} \equal{} \frac {3}{5} \equal{} \boxed{.6}$

Is there another way to do this other than trial and error?


Vieta
AoPSwiki
algebra
polynomial
ratio
complex numbers
binomial theorem

Hi everyone. Alright, so I have about 5 hours of time, before I go to sleep, so 5 hours of math, basically. I have a tournament tomorrow, but I really need to learn Vieta's in a few hours time, if possible, or try to, at least, instead of giving up. How can I do so? I've previously checked out other websites like wolfram's and aopswiki, but it just hasn't made sense to me. I tried to get people explain it to me online, but that failed as well.

Does anyone know how to explain it in a simple way, or know a place that has been proven to work? I'm taking algebra 2 at school this year. I think examples would also really help, as in a problem, a simple one that requires it to be solved.

Assume we have a polynomial of degree n with roots $ r_1$, $ r_2$, ..., $ r_n$.  Then this polynomial must be:
$ P(x)\equal{}c(x\minus{}r_{1})(x\minus{}r_{2})\cdots (x\minus{}r_{n})$

Multiply this out and see what happens.

Alright, well I multiplied it out, up to x-r3, and I got:

$ cx^3 \minus{} [c(r3) \plus{} c(r2) \plus{} c(r1)] x^2 \plus{} [(c(r2)(r3) \plus{} c(r1)(r3) \plus{} c(r1)(r2)] x \minus{} C(r1)(r2)(r3)]$ I guess we just see how many roots there can be up to, and go to that power? I don't even know if I did this right =()

I don't understand why the constant is there.

edit: fixed the +c(r1)part

That is almost correct for the case n=3 (Look at the plus sign that is in red below).  Now just look at the ratios between the 2nd, 3rd, and 4th coefficients with the first one...

2nd coefficient/ 1st coefficient = $ \minus{}$($ r_{3} \plus{} r_{2}$ [color=red]+[/color] $ r_{1}$)
3rd coefficient/ 1st coefficient = $ r_{1}r_{2} \plus{} r_{1}r_{3} \plus{} r_{2}r_{3}$
4th coefficient/ 1st coefficient = $ \minus{}r_{1}r_{2}r_{3}$

Now, can we generalize this?

Can someone post something about how we use it here as well.

And for the record, Casa De Mathematica, you can put stuff into 'subscript' using $ \verb\plus{}a_n\plus{}$ $ a_n$

I have thought for 5 minutes, and don't know how this can be generalized?

I think I'm missing something.   :huh:

Look at the patterns of the signs and of the combinations of roots

[quote="krsattack"]

2nd coefficient/ 1st coefficient = $ \minus{}$($ r_{3} \plus{} r_{2}$ [color=red]+[/color] $ r_{1}$)
3rd coefficient/ 1st coefficient = $ r_{1}r_{2} \plus{} r_{1}r_{3} \plus{} r_{2}r_{3}$
4th coefficient/ 1st coefficient = $ \minus{} r_{1}r_{2}r_{3}$

Now, can we generalize this?[/quote]

Well the 4th/1st divided by the 2nd/1st equals the 3rd/1st... I don't know if that's it.

I made a chart with the coefficients for x^k-1 etc. for polynomials with degree 2, 3, and 4, but I still don't see the big picture. =/

If anyone can help, that would be awesome!

Consider a $ n^{th}$ degree polynomial $ P(X) = a_nX^n + a_{n - 1}X^{n - 1} + ... + a_1X + a_0$
Since $ a_n \neq 0$, let's factor $ a_n$ out; for the sake of convenience, we use the notation that $ \frac {a_i}{a_n} = a_i^'$, so
$ P(X) = a_n(X^n + a_{n - 1}^'X^{n - 1} + ... + a_1^'X + a_0^')$
On the other hand, among complex numbers, there are $ n$ roots for this $ n^{th}$ degree polynomial (this is the fundamental theorem of algebra), which means we can factor it out like
$ P(X) = a_n(X - r_1)(X - r_2)...(X - r_n)$
Ignore the coefficient $ a_n$ for a second and see what happens if we multiply out these binomials. Since every term is a product of $ X$'s and roots, we can certainly combine the like terms, which have the same number of $ X$'s, and so get back into the first polynomial.
Try $ X^n$. Its coefficient is surely $ 1$ (before multiplied by $ a_n$). Now see $ X^{n - 1}$. We need to choose $ (n - 1)$ $ X$'s and one root (with negative sign), so there are $ n$ combinations: $ - r_1, - r_2, ..., - r_n$. So the coefficient $ a_{n-1}=a_n\cdot a_{n - 1}^'=- a_n\cdot (r_1 + r_2 + ... + r_n)$.
You can try $ X^{n - 2}$ and so on and so forth. In fact, all of these coefficients are symmetric sums with $ (n - i)$ of the roots, where $ i$ is the power of $ X$. You can determine the number of terms by binomial theorem when regarding all the roots as same (or simply use combination, which is exactly the same thing)


Σε τριγωνο ΑΒΓ πανω στις ΑΓ,ΑΒ,ΒΓ παιρνω τα ισα μεταξυ τους τμηματα ΑΔ,ΒΕ,ΖΓ αντιστοιχα.Το τριγωνο ΔΕΖ ειναι ισοπλευρο.Να δειξετε οτι ΑΒΓ ισοπλευρο.[/img]

Η άσκηση είναι σίγουρα σωστή;

Μανιέ σωστη πρέπει να είναι

Λυνεται σε 2 λεπτα με χρηση διανυσματων σε συνδιασμο με το νομο των συνημιτονων.
 Πχ βγαζοντας οτι $ \vec{CD}\cdot\vec{ZD} \equal{} \vec{AE}\cdot\vec{DE}$ {Γ=C, Δ=D} αναλυοντας τα εσωτερικα γινομενα και επειτα απο το νομο των συνημιτονων εφαρμοσμενο καταλληλα, παιρνουμε ευκολα ΑΕ=ΔΓ και ομοια για ΑΕ=ΒΖ οποτε με ισοτητα τριγωνων κλπ....

 Λυνεται και με χρηση του νομου ημιτονων μονο, εφαρμοσμενου στις γωνιες του ΑΒΓ για τα τριγωνα, ΑΕΔ,ΔΖΓ,ΒΕΖ, πχ ημα=ημβ=ημγ συνεπαγεται α=β=γ για γωνιες α,β,γ στο [0,π/2].
 Εδω πρεπει να αποδειξουμε οτι το ΑΒΓ δεν μπορει να ειναι αμβλυγωνιο αλλα αυτο ειναι ευκολο.

Βγαίνει πράγματι έτσι, αλλά πώς γίνεται μόνο με ύλη της Α' λυκείου;


[quote="mathmasterman"]does a aime state winner get a plaque or something?[/quote]

I don't think so.

What plaques are given for the AMCs and AIMEs?

nevr heard of a plaque for state winners
or any prize for that matter...

i think there is a plaque for amc 12 team state winner

Check out the listing and display of intramural (within-school) awards and certificates at:
http://www.unl.edu/amc/d-publication/d4-awards/1012awards.html
It's in the AMC 12 section, under the link Intramural School Awards.

Check out the listing of National Awards in the Teachers Manual at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/html/06tm-04-text.html
It's too lengthy to repeat here.

Steve Dunbar
AMC Director


calculus
integration
trigonometry
LaTeX
calculus computations

Evaluate

$ I\equal{}\int_{0}^{\pi} \frac{x}{1\plus{}\sin x}\ dx$.

$ I\equal{}\int_{0}^{\pi} \frac{x}{1\plus{}sinx}dx
\equal{}\int_{0}^{\pi}\frac{\pi \minus{} x}{1\plus{}sinx}dx$
so $ 2I \equal{} \int_{0}^{\pi}\frac{\pi}{1\plus{}sinx}dx$
substituting  $ sinx \equal{} \frac{2\tan\frac{x}{2}}{1\plus{}\tan^2\frac{x}{2}}$
we should get the result. :maybe:

Thanks both of you.

joshisunny16, I like the $ \sec^{2}x -\tan x\sec x$ which I guess you got by multiplying above and below by $ 1-\sin x$. It seems to me that your $ \text{\LaTeX}$ only needs a dollar sign at the start and end of each of your lines.


Define a sequence of real numbers $ a_1$, $ a_2$, $ a_3$, $ \dots$ by $ a_1 = 1$ and $ a_{n + 1}^3 = 99a_n^3$ for all $ n \ge 1$. Then $ a_{100}$ equals

$ \textbf{(A)}\ 33^{33} \qquad
\textbf{(B)}\ 33^{99} \qquad
\textbf{(C)}\ 99^{33} \qquad
\textbf{(D)}\ 99^{99} \qquad
\textbf{(E)}\ \text{none of these}$

[hide="Click for more explained solution"]
Since $ a_{n\plus{}1}\equal{}\sqrt[3]{99} \cdot a_n$ for all $ n \ge 1$, it follows that $ a_1,a_2,a_3,\ldots$ is a geometric sequence whose first term is $ 1$ and whose common ratio is $ r\equal{}\sqrt[3]{99}$.  Thus:

$ a_{100}\equal{}a_1 \times r^{100\minus{}1}\equal{}(\sqrt[3]{99})^{99}\equal{}99^{33}$

which corresponds to answer choice $ \textbf{(C)}$.
[/hide]

[hide="My solution"]
since, $a_{n+1}^3=99a_3^3$ for every positive integer $n \geq 1.$ We have, $$ a_{n+1}^3=\frac{a_{n+1}^3}{a_1^3}= \prod_{i=1}^{n}\frac{a_{i+1}^3}{a_i^3}=99^n.$$
Therefore, $a_n= 99^{\frac{n-1}{3}}$ for every positive integer $n \geq 1.$ Thus, $a_{100}=99^\frac{100-1}{3}=99^{33}.$ Which corresponds to answer choice $ \textbf{(C)}$.
[/hide]


[quote="Rushil"]This question has a mix of theory and numerical. I've solved the numerical part but am not able to understand the phenomenon behind the problem. Please help!!! :help:

The numerical problem was: If a laser beam of diameter $D$ is incident at an angle $ \theta_{i} $ on the interface of air and glass with refractive index $ n_g$, find the diameter of the lase beam in glass?

It's quite easy to solve the problem using a bit of geometry and Snell's law. Even , geometrically, it seems the width should change!

However, I am not able to '[i][u]understand[/u][/i]' this. After all, if you consider the beam to be composed of infinite many rays and consider that if you rotate(change) each ray by the same angle, the relative seperation does not change.(Am I wrong here??)

Can somebody plz help me in understanding this!!![/quote]

Rushil,

Here's my take on the problem.

The laser beam will form an Airy pattern on a screen some fixed distance from the aperture of the laser.

The radius of the central bright spot in an Airy pattern is directly proportional to wavelength.

$r_0 \propto \lambda$

But we know that for light in media,

$\lambda \propto \frac{1}{n}$

$\therefore r_0 \propto \frac{1}{n}$

As you have said, it gets quite obvious when you draw a picture. The problem with rotating the rays is that you don't rotate them all at the same place. (See my illustration.) You could say that the light becomes "less dense".

By the way, did you know that there was a connection between the refractive index of the material and the speed at which light travels in it? In materials such as glass light travels slower than in vacuum, more precisely

speed of light in material g = (speed of light in vacuum) x (refractive index of material g).

And that's also why the laser beam becomes wider - same thing as with liquids and the continuity equation.

[quote="Rushil"]I didn't quit understand what you are trying to say and it's connection to my question???[/quote]

[quote="Rushil's question:"]However, I am not able to 'understand' this [how the diameter of the laser beam changes]. After all, if you consider the beam to be composed of infinite many rays and consider that if you rotate(change) each ray by the same angle, the relative seperation does not change.(Am I wrong here??)[/quote]

A laser beam forms an Airy pattern when it strikes the surface of a target.

The radius of the central bright spot in the Airy pattern is given by:

$r_0 = 1.22 \frac{z \lambda}{2a}$ , where $z$ is the distance between the aperture and the target, and $a$ is the dimension of the aperture.

The energy of a photon, $E = \frac{hc}{\lambda}$, is constant.

In a medium, $E=\frac{h}{n \lambda \sqrt{\epsilon_0 \mu_0}}$

$\therefore \lambda \propto \frac{1}{n}$

$r_0 \propto \lambda \: \& \: \lambda \propto \frac{1}{n} \Rightarrow r_0 \propto \frac{1}{n}$

Suppose $n_g = 1.5 ;\; n_{\text{air}} = 1.0$ then $\frac{r_0_{\text{glass}}}{r_0_{\text{air}}} = \frac{n_{air}}{n_g} = \frac{1.0}{1.5}$

$\therefore r_0_{\text{glass}} = \frac{2}{3} r_0_{\text{air}}$

[quote="Rushil"]I think bus has got my answer. Thanks!!!![/quote]

[quote="bus"]By the way, did you know that there was a connection between the refractive index of the material and the speed at which light travels in it? In materials such as glass light travels slower than in vacuum, more precisely

[b]speed of light in material g = (speed of light in vacuum) x (refractive index of material g).[/b][/quote]

Rush,

Bus got the [b]bold[/b] bit completely wrong. If light travels more slowly in a medium than in vacuum; and it does, then [i]speed of light in material g $\neq$ (speed of light in vacuum) x (refractive index of material g)[/i], since $n_g > 1$

Be careful whose advice you follow.

I'm sorry, my mistake. It should be:

speed of light in g = (speed of light in vacuum) x (refractive index of g)

And the analogy between light and liquids is not as good as I thought it would be, but's still an interesting thing to think about...

[quote="Rushil"]I was just refering to his diagram. That explained things. And yes , I agree, the other things were wrong!![/quote]

No problem, Rush; but next time you should state what parts you agree with and what parts you don't. Physics is not done by concensus.

[quote="bus"]I'm sorry, my mistake. It should be:

[b]speed of light in g = (speed of light in vacuum) x (refractive index of g)[/b]

And the analogy between light and liquids is not as good as I thought it would be, but's still an interesting thing to think about...[/quote]

Bus, that's precisely the same formula as you had before and it's still wrong.

The speed of light in a medium is inversely proportional to the index of refraction in that medium: $c_n=\frac{c_{vac}}{n}$.

$c_n=f\lambda_n=\frac{f\lambda}{n} \Rightarrow \lambda_n = \frac{\lambda}{n}$

And yes, the same thing holds for liquids too.

[quote="Rushil"]Thanks for pointing out - Dr. No! Really appreciate![/quote]

You're welcome. Good luck.