Homologation of Terminal Alkynes.

Devise methods for converting a terminal alkyne to its next higher homolog,

$RC \equiv CH \longrightarrow RCH_{2}C \equiv CH$,

and for converting an alcohol to its next lower homolog,

$RCH_{2}C \equiv CH \longrightarrow RC \equiv CH$.

[hide="Hint"]Use alkene as intermediate.[/hide]

lower homolog
1.ozonolysis
2.reduction to alcohol
3.elimnation
4.$ x_{2}$
5.alc.KOH
6.$ NaNH_{2}$

That's right, but why not "... 4. $ X_{2}$ then 5. 3 eq. $ NaNH_{2}$ and finally 6. $ NH_{4}Cl$"?

How about part 1?

why 3 eq of $ NaNH_{2}$...won't 2 suffice

3 eqs. of sodium amide are necessary because one eq. is consumed in an acid-base reaction between this base and the soon formed terminal alkyne (which are relatively acidic, and sodium amide is a very strong base), and 2 eqs. are necessary for the two elimination steps. That's why in the end we must treat the reaction mixture with acid, in order to convert the acetylide anion into the desired alkyne. On the other hand, when the desired product is an internal alkyne, exactly 2 eqs of amide should be used, because otherwise the internal alkyne product would isomerize to the terminal one.


An ordered pair of digits $ (a, b)$ is such that $ 4a5,b32$ is a multiple of 66. Find $ a\plus{}b$.

It must be divisible by 11. So, $ 4\plus{}5\plus{}3\minus{}2\minus{}a\minus{}b  \equal{} 0\pmod{11}$, so $ a\plus{}b\equal{}10 \pmod{11} \equal{} \{10,21,32,...\}$
But $ 0 \leq a,b \leq 9$, so that $ 0 \leq a\plus{}b \leq 18$
So, a+b=10.

This is the way I did it. If a number abc is divisible by 11, b=a+c, so if you plug that in a=4+5=9 AND b=3-2=1. 9+1=10  :play_ball:


Mary walked 4 laps around the park track. It took Mary 4 minutes to walk the first lap. The second lap took 25% longer than the first lap. The third lap took 20% longer than the second lap. The final lap took 50% longer than the third. What was Mary’s total walking time for the 4 laps?

[hide] The second lap took $ 4(1.25)\equal{}5$ minutes. Therefore, the third lap took $ 5(1.2)\equal{}6$ minutes. Finally, the last lap took $ 6(1.5)\equal{}9$ minutes. Thus, $ 4\plus{}5\plus{}6\plus{}9\equal{}\boxed{24}$.[/hide]

[hide] It took Mary 4 minutes to walk the first lap. The second lap took 25% longer than the first lap. The third lap took 20% longer than the second lap. The final lap took 50% longer than the third.

1st lap= 4min
2nd lap= 5 min
3rd lap=6min
4th lap= 9min.

total time = [b]24 minutes.[/b][/hide]


If a force of constant magnitude acts on a particle, always perpendicular to  velocity, prove that the particle performs circular motion. (The direction of $ \vec F \times \vec v$ is constant.)

Come on..
It's actually only a 1 to 2 line proof. :wink:

the particle gets uniformly accelerated,and its position vector keeps getting tilted  towards the direction of force.correspondingly the direction of the force is changed to keep it perpendicular to the velocity.this keeps happening continuosly and we get a circle

The force is always perpendicular to the direction of force.
Therefore, it does not have any effect on magnitude of velocity.
Thus, magnitude of velocity remains a constant.

Consider any moment. Assume that the body follows a curve, whose
[url=http://en.wikipedia.org/wiki/Osculating_circle]radius of curvature[/url] at that point is $ \rho$.
We can consider the motion to be circular motion for an infinitesimally small time period.
Therefore,
$ F\equal{}\frac{mv^2}{\rho}$.

Magnitude of F remains constant, m is constant. We have proved that magnitude of velocity remains constant. Therefore, $ \rho$ remains constant, no matter whichever instant of motion we consider. Thus, the particle always travels in a curve with constant radius of curvature everywhere. Thus, the curve is a circle.


Given a piece of sqaure cardboard whose sides are 10 cm in length. It is allowed to extract three circular discs from this cardboard. What is the maximum total area of the discs that can be taken out?

I think hat is 80.85 cm^2 for one inscribed disc and two smalls in the corners


$S_6$ - area of convex hexagon. Prove that there exists a diagonal which cuts off a triangle ABC, where $S_{ABC} \leqslant \frac{1}{6}S$

At least one of the colored quads has area$\le \frac 13S$. Suppose that's $ABCE$. At least one of the triangles $ABE,CBE$ has area $\le \frac 16S$. Suppose that's $CBE$. But one of the triangles $ABC,BCD$ has area $\le $ area$CBE$.


geometry
incenter
cyclic quadrilateral
angle bisector
geometry proposed

Let $ABCD$ be cyclic quadrilateral .$AC$ cuts $BD$ at $M$.$MA' \perp AB;MB' \perp BC;MC' \perp CD;MD' \perp DA(A' \in AB;B' \in BC;C' \in CD;D' \in DA)$.Prove that:$A'B'+C'D'=A'D'+B'C'$

[quote="chien than"][color=darkred]Let $ABCD$ be cyclic quadrilateral, and let $M=AC\cap BD.$ Define: $A',\,B',\,C',\,D'$ be the orthogonal projections of $M$ on $AB,\,BC,\,CD,\,DA,$ respectively. Prove that: $A'B'+C'D'=A'D'+B'C'$[/color][/quote]
[color=darkblue][b]Proof.[/b] It's equivalent to $A'B'C'D'$ has an incircle.
We have: $\widehat{A'B'M}=\widehat{A'BM}=\widehat{C'CM}=\widehat{C'B'M},$ thus $M$ lies on angle bisector of angle $A'B'C'$
Similarly, we also have: $M$ lies on angle bisectors of angles $B'C'D',\,C'D'A'$
Hence $M$ is incenter of the circle is inscribed in quadrilateral $A'B'C'D'$
This completes our proof.[/color]


Hi everyone.  Ive been trying to figure out this 3 equations for the longest.  Can anyone help explain how to do these problems.  Thanks. :)

Heres the link
[img]http://img187.imageshack.us/img187/184/11676890dh9.png[/img]
[/img]

Heres what I thought :
for 48.), I thought i 1st find the lcd, but i got a little confused on the multiplication

Er, these don't quite qualify as Olympiad level. :)   Try posting in High School Basics or Classroom Math.  And I hope they're not homework problems.

Most Compound Interesting Question can be answered using a simple formula. I am not gonna give you a answer but rather the method. Use it as many times as you need.

Interest=Principle*(1.00+Rate)^time.

so for a on the interest question, it would be 800(1.00+.10)?

No that is only part of the answer.
1. Find 800(1.0875). This is the sum of the other two
2. Find x(1.05). This would be the interest in 1 account.
3. Find (800-x)(1.10). This would be the interest in the other account.
4. Write the equation: x(1.05) + (800-x)(1-10) = 800(1.0875)

That should give you your answer. However I feel as if I am missing something.

Im kind of confused.  I was looking for the equation for a  at 5% and 10% and b , not the answer.


How many ways are there to write seven-element data lists of inegers in numerical order to fit the following criteria: the mean is 20; the median is 21; and the mode is 22 and unique; and the range is 10?

I get more answers than in this previous thread: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1601423807&t=29103[/url]

[hide]
I called the lowest number x, and the next lowest y.  Then the highest number is x+10.
Since the total is $20\cdot7=140$, the number just below 21 is $65-2x-y$.
$x,y,65-2x-y,21,22,22,x+10$
Since y must be less than or equal to $65-2x-y$, we have $65-2x-y\ge y$, or $2x+2y\le 65$.
Since x and y are integers, we can divide by 2 and round a bit to get $x+y\le 32$
Also, $65-2x-y\le 21$, $2x+y\ge 44$.  Substitute 32 for x+y, and $x\ge 12$.

Bruteforcing from there, I got:
12 20 21 21 22 22 22
13 19 20 21 22 22 23
14 17 20 21 22 22 24
14 18 19 21 22 22 24
15 16 19 21 22 22 25
15 17 18 21 22 22 25
as the only possibilities.

Hope that's right!

[/hide]

Yes, those are the numbers I got.  I solved it almost exactly the same way.  Thanks!

[quote="rcv"][quote="erdogankerem123"]is it [hide]6?[/hide][/quote]
Answers without solutions are not helpful to others.  Please start posting solutions along with your answers.  Thank you.

-- moderator rcv[/quote]
sry, i was running short on time, and was going to edit it on later, but others had already posted solutions...
Anyways, I did it the same as others


Does there exists a real number x such that for all real k we have:
  1/|sinx|+1/(2^k*|sin2x|)+1/(3^k*|sin3x|)+...=infinity?
   Harazi


For what real value of x does $64^{x-1}$ divided by $4^{x-1}$ equal $256^{2x}$?


For $ a,b,c\in [0,1]$,prove that
\[ 2(a^3\plus{}b^3\plus{}c^3)\minus{}(a^2b\plus{}b^2c\plus{}c^2a)\leq 3\]

It is from the following

\[ \sum_{cyc}(1\minus{}a^3)(1\minus{}b^3) \ge 0\]

[quote="great math"]It is from the following
\[ \sum_{cyc}(1 \minus{} a^3)(1 \minus{} b^3) \ge 0\]
[/quote]
Well done.Mine is $ a^3\plus{}b^3\minus{}a^2b\leq 1$


http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2001 - click on the numbers for threads on the problems.

It would help if you knew what number it was.

I put the entire question there. What would you search specifically?

[quote="aufha"]16. The mean of three numbers is 10 more than the least of the numbers and less
than the greatest. The median of the three numbers is 5. What is their sum?
(A) 5 (B) 20 (C) 25 (D) 30 (E) 36[/quote]
That seems to be almost the whole problem. At
http://courseweb.stthomas.edu/gps/amc/amc_10_2001,pdf
I found all the questions, including question 16 which stated:
" .... is 10 more than the least of the numbers and [b]15[/b] less
than the greatest. ...."

Ah, thank you. The version that I had found didn't say 15 so I was baffled when there was a second equation with 15 in it mentioned in the solution. Thanks, KMST.

Okay, that's annoying. I did the same problem yesterday, from the 2001 AMC 12.
[hide]Well, you could see it's obviously either 30 or 36. But, if you want to do the real way, let the mean be x. Then, if the numbers are a,b, and 5, we have a+b+5=3x, x=a+10=b-15. Set up a few equations and get 30.[/hide]
Woah, this AMC 10 must be pretty trivial, if you can brute force #16 really easily.

xpmath, why is it obviously either 30 or 36 before we start the problem?
It didn't say positive numbers.

Am I missing something obvious?

Let $ a$ be the smallest, $ b$ be the median, and let $ c$ be the greatest. Thus,  $ b \equal{} 5$. Since $ \frac{a \plus{} b \plus{} c}{3} \equal{} 10 \plus{} a$, $ a \plus{} b \plus{} c \equal{} 30 \plus{} 3a \geq 30$. Thus, $ a \plus{} b \plus{} c \equal{} 30$ or $ 36$. We see that if $ a \plus{} b \plus{} c \equal{} 36$, then $ a \equal{} 2$. Solving for $ c$, we get $ c \equal{} 29$. But since the mean is supposed to be less than $ c$, this is a contradiction, so $ a \neq 2$. Thus, $ a \equal{} 0$, so the sum is 30. (If you substitute $ a \equal{} 0$ in, you'll see that it meets the requirements).

[quote="xpmath"]
Woah, this AMC 10 must be pretty trivial, if you can brute force #16 really easily.[/quote]

You can brute force a lot of AMC 10 #16's easily... they don't get *hard* until 20s. e.g. [[2003 AMC 10A Problems/Problem 16]] [[2005 AMC 10A Problems/Problem 16]] and particularly [[2006 AMC 10B Problems/Problem 16]], with the only step of any difficulty or trickiness being taking the modulo, which is easily done.

Also, [[2006 AMC 10A Problems/Problem 16]] is pretty trivial geometry that has a analytical, brute force solution.


Let S be the set of polynomials anxn + an-1xn-1 + ... + a0 with non-negative real coefficients such that a0 = an ≤ a1

= an-1 ≤ a2 = an-2 ≤ ... . For example, x3 + 2.1 x2 + 2.1 x + 1 or 0.1 x2 + 15 x + 0.1. Show that the product of any

two members of S belongs to S.

Please use LaTex.

Let the first generalized member of $S$ be $f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_{n-1} x + a_n, a_k \le a_{k-1}, k = 1, 2, 3... n$, and let the second generalized member of $S$ be $g(x) = b_m x^m + b_{m-1} x^{m-1} + ... + b_{m-1} x + b_m, b_k \le b_{k-1}, k = 1, 2, 3... m$.

We now consider the coefficients of $f(x) g(x) = c_{n+m} x^{n+m} + c_{n+m-1} x^{n+m-1} + ...$.  The coefficient of the first term is $a_n b_m$.  The coefficient of the second term is given by all terms of $f(x)$ and $g(x)$ that multiply out to give an exponent of ${n+m-}$; there are two of these term pairs, $a_{n-1} x^{n-1} * b_{m} x^{m}$ and $a_n x^n * b_{m-1} x^{m-1}$, which add up to $a_{n-1} b_m + b_{m-1} a_n$.  Because $a_{n-1} \ge a_n$, it follows that $a_{n-1} b_m \ge a_n b_m$.  Applying the same argument to $b_{m-1} a_n$ gives $a_{n-1} b_m + b_{m-1} a_n \ge 2 a_n b_m$.

Repeating this analysis for the coefficient of the $(k+1)^{th}$ term from the beginning up to the center term of $f(x) g(x)$ gives $c_{n+m-k} x^{n+m-k} = \displaystyle\sum_{i=0}^{k} a_{n-i} b_{m-k+i} x^{n+m-k}$.  Each subsequent term of $c_{n+m-k}$ contains one more $a_{n-p} b_{m-k+p}$ than the previous; additionally, every term in the addition except the last satisfies this inequality: $a_{n-i} b_{m-(k+1)+i} \ge a_{n-i} b_{m-k+i}$, and since there is even an extra term, $c_{n+m-k} > c_{n+m-(k+1)}$ is guaranteed, thus satisfying one requirement of belonging to S.

The second requirement, that $c_{n+m-k} = c_k$, requires repeating the analysis starting from the constant term, $c_0$.  From the expansion of $f(x) g(x)$, it is obvious that $c_0 = a_0 b_0 = a_n b_n$.  In order to find $c_1$, it is necessary to find all terms in $f(x)$ and $g(x)$ that contribute a term of $x$ to $f(x) g(x)$; this gives, similar to the previous analysis, a sum of $c_1 x^1 = a_1 b_0 x^1 + a_0 b_1 x^1 = (a_{n-1} b_m + a_n b_{m-1}) x$.

Repeating this analysis for the coefficient of the $(k+1)^{th}$ term from the end gives $c_k x^k = \displaystyle\sum_{i=0}^{k} a_{i} b_{k-i} x^k = \displaystyle\sum_{i=0}^{k} a_{n-i} b_{m-k+i} x^k$; it is immediately obvious that $c_k = c_{n+m-k}$, satisfying the second requirement of belonging to S.  Therefore:

$\forall f(x), g(x) \in S: f(x) g(x) \in S$

Q.E.D.

Moreover, the title of the post is particularly poor.  You should try to provide some hints as to the nature of the problem, e.g, symmetric polynomials closed under multiplication, rather than just posting as much of the first line of your post as will fit in the subject line.


Welcome to the AoPS forums.

I'm moving your post to Games & Fun Factory which is our social forum.

Welcome to AoPS/ML!  :lol:

Btw- there is a 'Introduce Yourself' thread in Fun and Games.

:welcome:  :welcomeani:  :bye:
have fun on AoPS!!

Thanks and sorry for not posting at the rigth place!
The forum is really HUGE! It seem that all subjects of science are covered here! That's great!
But I got a bit confused about the place where I should introduce myself...
Anyway...
...Awsome forum! Congratulations!

introduce yourself [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15747]here[/url]

Welcome to AoPS, I hope you will know what math is really about.  :)

[quote="kylekatarn"]I'm new around here!
just want to say hello to everyone![/quote]

Hi kylekatarn and all other new users.

The best place to greet everyone and introduce yourself is the "Introduce yourself" thread in the stickies. You can imagine the obvious difficulty of navigating this forum if there were several hundred of these "hi" threads. :) Thanks.


:)
How many ways can you buy 4 pies from the "3.14 store" if the store only sells pies in flavors of apple, cherry, and blueberry?

I know where you got the idea for this from---AMC 10. I remember it from a long time ago when I took it for practice. Let's see if I still remember how to do these...

[hide]
$\text { apples } + \text { cherries } + \text { blueberries } = 4$
Number of solutions is $\binom{6}{2}=15.$

Or...

$*...* | *...* | *...*$
In this diagram thingy there are 4 total fruits divided into three categories, separated by 2 line thingies. There are 6 total thingies and you need to choose where the line thingies go, so $\binom{6}{2}=15.$ :D :D
[/hide]

I'm going to do this the ugly way . . .

[hide]  Assign a number 0 through 4 to apple.  Now we need to divide the remaining pies up among cherry and blueberry.  The number of ways to do this:

A=4: 0,0 (1)

A=3: 1,0  0,1  (2)

A=2:  2,0  0,2  1,1  (3)

A=1:  3,0  2,1  1,2  0,3  (4)

A=0:  4,0  3,1  2,2  1,3  0,4  (5)

So the number of ways to pick four pies is 5+4+3+2+1=15.[/hide]

[hide]
We wish to place 4 pies into 3 flavors or rather 4 objects into three boxes.
This can be done by using X's to represent the divisions:
So we would have ppppXX
So we only need the number of permutations of this string which is
${6\choose2} = 15$
[/hide]


[url]http://economicschallenge.ncee.net/[/url]

Nationals is in New York next weekend.  Any AoPSers attending?


Ok, while going through some old  probability problems I found this one and i'm not exactly sure how to do it. I have no idea where it is from so.....

Bender (apparently futurama was around) orders two robot painters to paint a 1000 meter long circular fence. One robot is ordered to paint a randomly selected 720 meter stretch of the fence while the other is ordered to paint a randomly selected 750 meter stretch of the fence. Find the probability that when both robots are done, the entire fence is painted.

I assume that a robot can't stand in one spot and paint it 700 times...  :rotfl:

If one robot paints a continuous expanse of 750 meters, there are 250 unpainted meters. So we need to find the probability that a given length of 720 meters covers the whole blank length of 250 meters. The beginning of the blank area can allign with anywhere form the 0th painted meter to the (720-250)th=470th one. Therefore, the probability that the whole fence is painted is 470/1000 which simplifies to 47/100.

Hope this helps.

Thanks, I didn't realize you could solve it like this.... hehe......


The rectangular region bounded by the lines with equations $ x \equal{} 1.2$, $ x \equal{} 2.6$, $ y \equal{} \minus{}0.2$ and $ y \equal{} d$ has area 14 square units. What is the greatest possible value of $ d$?
Express your answer as a decimal to the nearest tenth.

The "length" of the rectangle in the $ x$-direction is 1.4, so the "width" in the $ y$-direction has to be $ 10$ so that $ lw\equal{}14$. So $ d$ and $ \minus{}.2$ have a positive difference of $ 10$. $ d$ can therefore be $ 9.8$ or $ \minus{}10.2$, the greater of which is $ \boxed{9.8}$.


let a,b,c is non-negative such that:$ a\plus{}b\plus{}c\equal{}3$,no two of which are zero.prove that:
   $ \frac{a}{a\plus{}2bc}\plus{}\frac{b}{b\plus{}2ac}\plus{}\frac{c}{c\plus{}2ab} \leq 1$

[quote="tuantam1lan"]let a,b,c is non-negative such that:$ a \plus{} b \plus{} c \equal{} 3$,no two of which are zero.prove that:
   $ \frac {a}{a \plus{} 2bc} \plus{} \frac {b}{b \plus{} 2ac} \plus{} \frac {c}{c \plus{} 2ab} \leq 1$[/quote]

 :maybe: Try $ a\equal{}0,b\equal{}1,c\equal{}2$

[quote="tuantam1lan"]let a,b,c is non-negative such that:$ a \plus{} b \plus{} c \equal{} 3$,no two of which are zero.prove that:
   $ \frac {a}{a \plus{} 2bc} \plus{} \frac {b}{b \plus{} 2ac} \plus{} \frac {c}{c \plus{} 2ab} \leq 1$[/quote]

Assume: $ a\ge b \ge c$

Because : $ (a^2 \plus{} b^2 \plus{} c^2) \equal{} \frac {1}{3}(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \ge 3abc$

Hence:

$ \frac {a}{a \plus{} 2bc} \plus{} \frac {b}{b \plus{} 2ac} \plus{} \frac {c}{c \plus{} 2ab} \minus{} 1$

$ \equal{} \frac {2(a \minus{} bc)}{a \plus{} 2bc} \plus{} \frac {2(b \minus{} ca)}{b \plus{} 2ac} \plus{} \frac {2(c \minus{} ab)}{c \plus{} 2ab}$

$ \equal{} \frac {2(a^2 \minus{} abc)}{a^2 \plus{} 2abc} \plus{} \frac {2(b^2 \minus{} abc)}{b^2 \plus{} 2abc} \plus{} \frac {2(c^2 \minus{} abc)}{c^2 \plus{} 2abc}$

$ \ge \frac {1}{3}.2(a^2 \plus{} b^2 \plus{} c^2 \minus{} 3abc)( \frac {1}{a^2 \plus{} 2abc} \plus{} \frac {1}{b^2 \plus{} 2abc} \plus{} \frac {1}{c^2 \plus{} 2abc})$ (Chebyshev ineq)

$ \ge 0$

Done!

@too bad, it's so long

[quote="tuantam1lan"]let a,b,c is non-negative such that:$ a \plus{} b \plus{} c \equal{} 3$,no two of which are zero.prove that:
   $ \frac {a}{a \plus{} 2bc} \plus{} \frac {b}{b \plus{} 2ac} \plus{} \frac {c}{c \plus{} 2ab} \leq 1$[/quote]

See http://www.mathlinks.ro/viewtopic.php?p=1271938#1271938


A manufacturer of small parts ships them in lots of 20 to his customers.
5% of all the parts produced are defective.

If a customer received ten lots (each with 20 parts), what is the expected number of lots with no defectives?

If the customer recieved 10 lots, each with 20 parts, he has 200 parts.  Of these, 5% or 10 are defective.  Thus 190 parts should have no defects.  Since there are 20 parts to a lot, then 9.5 lots should have no defects.


What is the least positive integer divisible by 2, 3, 4, and 5?

[quote="math92"][hide]
i think you would do:
$5 \cdot 4 \cdot 3 \cdot 2$
which is 120[/hide][/quote]

Isn't 2 a multiple of 4, math92?  ;)

[hide]
yeah sorry, i wasn't thinking
you would do 5*4*3 which is [b]60[/b][/hide]

It is clear that if a is divisible by 4 it is divisible by 2 $\rightarrow$ we should find the least number divisible by 5,4,3 and we have (3,4,5)=1 $\rightarrow$ the least number is  $5 \times 4 \times 3=60$.


Let $ x;y;z\ge 0$
Prove that
$ (x \plus{} y \plus{} z)^6\ge 64(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$

[quote="Math pro"]I think you mistaken! :wink:[/quote]
I have a similar  ineq (from vasc i remember$ ((x\plus{}y)(y\plus{}z)(z\plus{}x))^2\ge 4(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$
I have two proof for it
@1:the proof of Vasc and Siloluan that was used  AMGM
@2:meUse  pqr we have a very short proof(quite nice)
Is it usefull

[quote="tranvanluan"]Let $ x;y;z\ge 0$
Prove that
$ 64(x \plus{} y \plus{} z)^6\ge (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$[/quote]
I think ineq must be : $ \frac{8}{729}(x \plus{} y \plus{} z)^6\ge (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$
If not first ineq is very loose ; and equal occur $ a\equal{}b\equal{}c\equal{}0$ . Not nice

[quote="onlylove_math"]
I think ineq must be : $ \frac {8}{729}(x \plus{} y \plus{} z)^6\ge (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$
If not first ineq is very loose ; and equal occur $ a \equal{} b \equal{} c \equal{} 0$ . Not nice[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.$ :wink:

Sorry
Here is my post
Let $ x;y;z\ge 0$
Prove that
$ (x\plus{}y\plus{}z)^6\ge 64(x^2\plus{}yz)(y^2\plus{}xz)(z^2\plus{}xy)$

[quote="Allnames"][quote="Math pro"]I think you mistaken! :wink:[/quote]
I have a similar  ineq (from vasc i remember$ ((x \plus{} y)(y \plus{} z)(z \plus{} x))^2\ge 4(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$
I have two proof for it
@1:the proof of Vasc and Siloluan that was used  AMGM
@2:meUse  pqr we have a very short proof(quite nice)
Is it usefull[/quote]
With it we only have $ ((x\plus{}y\plus{}z))^6\ge \frac{729}{16}(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$
And it is not enough,sorry
But ,any way,your problemcan be proves by pqr.I am sure

No I think it is $ 64$not$ \frac{729}{16}$

[quote="tranvanluan"]No I think it is $ 64$not$ \frac {729}{16}$[/quote]
Yes,i didn't say that,With $ 64$,this is a very well-know ineq (sorry if i have memory mistakes) and we can easyly prove it by pqr (Notice $ (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy) \equal{} q^3 \plus{} 8r^2 \minus{} 6qr \plus{} r)$there let $ p \equal{} 1$
the last we only must prove$ \frac{1}{64}\geq q^3 \plus{} 8r^2 \minus{} 6qr \plus{} r$ i think it is quite easy

[quote="Allnames"][quote="tranvanluan"]No I think it is $ 64$not$ \frac {729}{16}$[/quote]
Yes,i didn't say that,With $ 64$,this is a very well-know ineq (sorry if i have memory mistakes) and we can easyly prove it by pqr (Notice $ (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy) \equal{} q^3 \plus{} 8r^2 \minus{} 6qr \plus{} r)$there let $ p \equal{} 1$
the last we only must prove$ \frac {1}{64}\geq q^3 \plus{} 8r^2 \minus{} 6qr \plus{} r$ i think it is quite easy[/quote]
I don't think same as you
Please, Can you post detail proof? :wink:

[quote="bunhiacovski"][quote="Allnames"][quote="tranvanluan"]No I think it is $ 64$not$ \frac {729}{16}$[/quote]
Yes,i didn't say that,With $ 64$,this is a very well-know ineq (sorry if i have memory mistakes) and we can easyly prove it by pqr (Notice $ (x^2 + yz)(y^2 + xz)(z^2 + xy) = q^3 + 8r^2 - 6qr + r)$there let $ p = 1$
the last we only must prove$ \frac {1}{64}\geq q^3 + 8r^2 - 6qr + r$ i think it is quite easy[/quote]
I don't think same as you
Please, Can you post detail proof? :wink:[/quote]
Sorry i am quite busy,so i cant post it,Howerver,perhasp,my proof is not true
Let$ f(r)=q^3 + 8r^2 - 6qr + r$,$ f'(r)=0<=>r=\frac{1-6q}{16}$
Check$ f(1\27),f(0),f(\frac{4q-1}{9}),f(\frac{1-6q}{16})$
I have a belief that this problem can be splved by pqr because here is a problem in pvthuan'book,and it was written in achapter,where all problem can be solved by pqr :D


quadratics
algebra
polynomial
ceiling function
function

Find the number of quadratic polynomials, $ ax^2\plus{}bx\plus{}c$, which satisfy the following conditions:
(a) $ a, b, c$ are distinct;
(b) $ a, b, c \in \{1, 2, 3, . . . 2008\}$ and
(c) $ x \plus{} 1$ divides $ ax^2\plus{}bx\plus{}c$.
 :D  :lol:  :blush:

[hide="Solution?"]$ ax^2\plus{}bx\plus{}c\equal{}(x\minus{}1)(ax\plus{}b\minus{}a)\minus{}(a\minus{}b\plus{}c)$, so $ a\minus{}b\plus{}c\equal{}0\iff a\plus{}c\equal{}b$.
Then for each value of $ b$, there are $ 2(\lceil \frac{b}{2} \rceil \minus{}1)$ solutions, adding up to $ 2\plus{}2\plus{}4\plus{}4\plus{}\cdots\plus{}
2006\plus{}2006\equal{}2\cdot1003\cdot1004\equal{}2014024$.[/hide]

I got something different...

[hide]If we use the remainder theorem, we can determine that a-b+c=0, or a+c=b.
If a=1, c can be 2006 values, and b will vary accordingly.
Similarly, if a=2, c can be 2005 new values, and b will vary.
This implies that the answer is a sum of the integers from 1 to 2006.

2006*2007/2=2013021 quadratic polynomials.[/hide]

@[b]azn6021023[/b]:You overlook two facts. First, $ a$ has to be less than or equal to $ 2008$. Secondly, $ (a,b,c)$ are ordered, which means $ a\equal{}3, c\equal{}4$ is different from $ a\equal{}4,c\equal{}3$.

[b]timwu[/b], I don't think azn6021023 is overlooking any of those things. I got the same answer.

Did you mean Brut3Forc3?

Sorry, I misread [b]azn6021023[/b]'s solution when I posted that. But still, I don't think it's correct. The solution overlooked that when $ a > 1004$, $ a > 2008 \minus{} a$, which breaks the sequence. Using his method, the sequence should be (as $ a$ increases): $ 2006, ..., 1003, 1003, ..., 1$, which sums to $ 2014024$.

Your list is right. Note there are $ 2007$ columns, where the first has $ 2006$ elements and the last has $ 1$ element. It can't possibly be $ 2006\plus{}...\plus{}1$.

Yeah, I see my error at when a= 1004.  That number repeats in solutions instead of decreasing again, so 1003 + 2013021=2014024.

Well...

Plug in $ \minus{} 1$ to $ (c)$ to get $ a \plus{} c \equal{} b$.

Consider the function $ f(b)$, which gives the total number of possibilities given an initial $ b$ ($ f(1) \equal{} 0$, $ f(2) \equal{} 1$, etc.)

Comparing this to a 2008-sided fair die, we find that the function is symmetric to the line $ x \equal{} 2009$ (think 6 sided die)

Note that for every value of $ a$ there is a corresponding value of $ c$ that gives $ b \equal{} 2009$; thus $ f(2009) \equal{} 2008$

Therefore, $ f(1) \plus{} f(2) \plus{} ... \plus{} f(2008)$

$ \equal{} \frac {(f(1) \plus{} f(2) \plus{} ... \plus{} f(4016)) \minus{} f(2009)}{2}$

$ \equal{} \frac {2008 \times 2007 \minus{} 2008}{2} \equal{} 2008 \times 1003$


geometry
3D geometry
tetrahedron
combinatorics proposed
combinatorics

Determine the greatest positive integer $ n$ such that in three-dimensional space, there exist n points $ P_{1},P_{2},\cdots,P_{n},$ among $ n$ points no three points are collinear, and for arbitary $ 1\leq i < j < k\leq n$, $ P_{i}P_{j}P_{k}$ isn't obtuse triangle.

This is not a solution.

It looks like $ 8$ is the maximum number of points. Some trial and error shows tetrahedron does not work for us. Construct a cube with side length of $ 1$ wlog. Clearly we cannot place a point outside the cube. Consider the plane containing $ (0,0,0), (0,0,1),(1,1,1)$ and the other perpendicular to it that contains the other body diagonal. A plane partitions the interior space of the cube. Any point placed in one of the partition makes obtuse angle with a pair of points on the body diagonal. Since we cannot place the point at the centre of the cube, we cannot place a point in the interior of the cube.

Tetrahedron and cube are the only two regular convex polyhedra with internal angle between two faces $ \leq \frac{\pi}{2}$. Other than the four vertices of the tetrahedron we cannot place a point ouside or inside or on the surface of the tetrahedron. Similarly we cannot place a point (other than vertices) inside or outside or on the surface of the cube.
Hence we can place at most $ 8$ points with the given restriction.

First, given a regular tetrahedron there most certainly are points that I can add to create no obtuse triangles: each regular tetrahedron contains 4 vertices of a cube, and we can just add the other 4.  Second, that argument is worthless: just because you can't add another point once you've already chosen 8 points that are vertices of a cube definitely does not imply that you can't choose 9 points.

[quote="JBL"]just because you can't add another point once you've already chosen 8 points that are vertices of a cube definitely does not imply that you can't choose 9 points.[/quote]

It is clear that we cannot add a point once the cube is constructed. If we did have a polyhedron with greater than $ 8$ vertices, then at least one of the internal angle exceeds $ \frac{\pi}{2}$.(A regular convex polyhedron with more than $ 8$ vertices has internal angle between any two touching planes $ \geq \frac{\pi}{2}$).

Fine ?

[hide="@Srikanth:"][quote="I hate rating"]It is clear that we cannot add a point once the cube is constructed. If we did have a polyhedron with greater than $ 8$ vertices, then at least one of the internal angle exceeds $ \frac {\pi}{2}$.

Fine ?[/quote]
I will use similar arguments to prove the $ 2d$ case wrongly.

It is clear that we cannot add a point once a equilateral triangle is constructed. If we have a polygon with greater than $ 3$ vertices then at least one of the internal angle exceeds $ \frac {\pi}{2}$.

This is wrong because there are rectangles. This is why JBL didn't like your idea.

[quote="I hate rating"]A regular convex polyhedron with more than $ 8$ vertices has internal angle between any two touching planes $ \geq \frac {\pi}{2}$[/quote]
True, but $ 9$ points might not form a regular convex polyhedron. It is true for non-regular, non-convex polygons, in that case please prove it.

However, if you proved that "We cannot add a point once [b]any[/b] $ 8$ points are constructed(Not necessarily on a cube)" then it would be fine. :wink: [/hide]

[quote="JBL"]each regular tetrahedron contains 4 vertices of a cube.[/quote]
Um, I think there's some typo or something here.

I meant, "For any regular tetrahedron there exists a cube such that the vertices of the tetrahedron are a subset of the vertices of the cube."  In particular, if you have four points at the vertices of a regular tetrahedron, it is possible to add at least four additional points without violating the given condition.

[quote="JBL"]"For any regular tetrahedron there exists a cube such that the vertices of the tetrahedron are a subset of the vertices of the cube."[/quote]
Wow. I never knew that. {(0,0,0),(0,1,1),(1,0,1),(1,1,0)} is a tetrahedron. Never thought about it. Thanks JBL. I'm sorry. :blush:

Well...this is actually a well-know problem. The $d$ dimensional case was conjuctured by Erdos around $1950$ and solved by Ludwig Danzer and Branko Grunbaum in 1962. You may check further infomation from $Proofs$ $from$ $the$ $book$!


geometry
circumcircle
analytic geometry
function
graphing lines
slope
trigonometry

Here's another one of those tough Iranian things:

Let M and N be poitns on the side BC of the triangle ABC s.t. BM=CN and M lies between B and N. P and Q are points on AN, AM respectively s.t. angle PMC=MAB and angle QNB=NAC. Show that angle QBC=PCB.

[i]Iranian olympiad/2'nd round/2002[/i]

This is my idea. It's a bit clumsy but it's better than nothing. I won't write all computations because I don't have time, but if u follow my line of reasoning you will see they work out immediately (no more than 30 minutes to write it up).

First notice that MP is tangent to the circumcircle of MAB and NQ is tangent to the circumcircle of ANC. Therefore, angle(PMA)=angle(ABC) (1)  and angle(ANQ)=angle(ACB). (2)

Set up a coordinate system with B(-u,0),M(-m,0),N(m,0),C(u,0). We denote by f_XY the function of the form ax+by+c associated to the line XY. Also let s_XY be the slope of the line through X and Y.

f_MA = xv - y(u+t) + vt
f_MC = y

Therefore f_MP = f_MA - k f_MC. We will determine k from the angle condition (this works for acute B and C and a separate analogous case must be considered for obtuse B; then by simmetry it's true for obtuse C)

(s_MA - s_MP)/(1 + s_MA s_MP) = s_BA (equivalent to (1))

We get k = [(u+t)^2 + v^2]/(m-t)

Then we write the coordinates of P, the intersection between MP and AN :

f_MP = xv - y[(u+t)(u+m)+v^2]/(m-t) + vt
f_NA = xv - y(u-m) + vt

x = ... , y = ....

We compute Q in a similar manner and we are done. Does anyone have an ellegant solution ?

I think that we just only use the sine theorum. I'll give you my solution later. Thanks. :)  :)  :D  :)  :D

My solution also involves the sine thm. It's not significantly shorter than what iandrei posted, so..

I  have  better solution for this one ! at first assume that angle   PMC =BAM =x , MAN=z ,NAC=y=QNM, I want to use inversion twice  with center Q and then with center P (both power of inversion are equal)so we begin with center Q ,suppose that the inversion of A,B,N are A',B',N'. it is easy to see that angle A'B'Q=x, angle QB'N'=y. now I want to evaluate A'B'/B'N'. you know that A'B'=AB*K/QA*QB  (K is power of inversion)B'N'=BN*K/QB*QN   . so A'B'/B'N'=AB*QN/BN*QA=AB*sin(z)/BN*sin(QNA)=AB*sin(z)/BN.sin(c)=2R*sin(z)/BN (because QNA=C)  now I do as well for center P with same power now assume that the inversion of A,M,C are A", M' ,C' , you can see that M'C'A"=x+y ( we N'B'A'=x+y before ) now you can evaluate like before C'A"/C'M'=2R*sin(z)/CM we can see that C'A"/C'M'=A'B'/B'N'  so this two traingle are equal (A'B'N')=(M'A"C') so we can say that angle A"M'C'=A'N'B'  so it leads to QBC=PCB                    is it nicer? :D

A solution without trigo: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=463337&p=2597706.

Best regards,
sunken rock

OK, let the circumcircle of $MPA$ intersect $MN$ at $P'$ and analogously $k(TQA) \cap MN = Q'$. Then $\angle BQ'Q =\angle MAN = \angle CP'P$ so we need to prove that $\triangle P'CP \sim \triangle Q'BQ$ which is equivalent to $\frac{PP'}{P'C}=\frac{QQ'}{Q'B}$.This is equivalent to $\frac{PP'}{P'A}\frac{P'A}{P'C}\frac{BQ'}{Q'A}\frac{Q'A}{Q'Q}=1$ which, after transforming it in trigonometric form, is equivalent to the condition that $BM=CN$.

actually there is another beautiful solution

let $W_{1},W_{2}$ be the circumcirle of $\triangle BQN,\triangle CPM$ repectively  and let $AM \cap W_{1}=k, AN \cap W_{2}=L$ then we want to prove that $KLMN$ is cyclic so we need to prove that $AM.AK=AN.AL$ and this is easy because $\triangle ABK  \sim \triangle ACL $ and $\frac{AM}{AN}=\frac{AC}{AB} $   :)


$ x^{5}+3x^{4}-23x^{3}-51x^{2}+94x+120 = 0$

Keep in mind that all of the solutions of the equation are integers.

Yep :D Well technically , the final solution is $ x = \{-5,-3,-1,2,4\}$.  Do you mind showing your steps?

This may sound like cheating, but the first thig id notice is $ 120=5!$ so that'd be the first thing id try.


geometry
parallelogram
geometric transformation
circumcircle

8. Let $ABCD$ be a parallelogram. Suppose there exists a point $P$ in the interior of $ABCD$ such that $\angle ABP = 2\angle ADP$ and $\angle DCP = 2\angle DAP$. Prove that $AB = BP = CP$.

I tried working assuming that $AB = BP = CP$ first and then manipulating the circles but that didn't get far. Any approaches?

im not sure but perhaps work with the isosceles tirangles and try to find some similar tiranlge or something

produce DP AP BP and CP to the sides of the parallelegoram

any body can answer this?

hard?

or too easy?

[quote=shobber]Always think of Translation when you meet a parallelogram problem.

Do translation $T(\overrightarrow{BC})$. Then $P \to P'$. Thus $AB = DC$, $BP=CP'$. Now $\angle{P'CD}=\angle{PBA}=2\angle{PDA}$, $\angle{PCD}=2\angle{PAD}=2\angle{PP'D}$, therefore $C$ is the circumcenter of $\triangle{PP'D}$, thus $CP'=CD=CP$, which gives $CP=AB=BP$.[/quote]

I'm not sure how knowing $\angle{P'CD}=2\angle{P'PD}$ and $\angle{PCD}=2\angle{PP'D}$ would lead to that $C$ is the circumcentre of triangle $PP'D$, as I think that to know that, a pair of sides from $CP',CD,CP$ must be proven to be equal$. Can someone help out?

λυσηαργοτερα

Another way is to take reflections of $B, C$ about $AP,DP$ respectively and to see that the 2 reflected points coincide ( 2 circles have max 2 common points, one being $P$ ).

Best regards,
sunken rock


$ \int \sqrt {tg(x)}dx$

I know that it was posted before
But I couldn't find it with the search function

[b]Could somebody give me a link or solve it??[/b]

Did ever your parents teach you how to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26438]read[/url]?

Look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=168301]here[/url].


geometry
inradius
circumcircle
inequalities
inequalities proposed

Given triangle $ABC$ with side $a,b,c$ and inradius $r$ circumradius $R$ prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{1}{\sqrt{3}}(\frac{1}{R}+\frac{1}{r}+\frac{2}{15}(\frac{2}{R}-\frac{1}{r}))$

Unfortunately, this inequality is [b]wrong[/b].

[b]Counterexample :[/b] $\left\{\begin{array}{cccc}
a=1 \\ \\
b=3 \\ \\
c=3\end{array}\right\|\implies\left\{\begin{array}{ccc}
LHS\approx 1.66666 \\ \\
RHS\approx 1.66481\end{array}\right\|\implies\boxed{\ LHS\ \boxed{>}\ RHS\ }$


geometry
rectangle
trigonometry
geometry unsolved

N is positive integer non zero.

A rectangle (L, l) is formed by N rectangles $\left(L_i,l_i\right)$ for i=1,.......,N.

For any i=1,........N we have $L_i$ is positive integer or $l_i$ is positive integer.

Prove that L is positive integer or l is positive integer.

Suppose the contrary.

For each partition $P$, which doesn't satisfy that $L$ or $\ell$ is integer, let $N(P)$ be a sum of all integer sides of all rectangles from the partition $P$.
Consider partition $P$ with the minimal $N(P)$. Take its corner rectangle. This rectangle has an integer side $AA'$. Let's cut a vertical strip from the rectangle $ABCD$. We will obtain rectangle $A'BCD'$ and its proper partition $P'$ (i.e. all rectangles in this partition have at least 1 integer side). Moreover sides $A'B$ and $A'D'$ are not integer too, but it is easy to see that $N(P')<N(P)$. Contradiction.

I think it's worth reading [url=http://www.cs.toronto.edu/~mackay/rectangles.pdf]this[/url]

My proof was this:

I prooved that there exist $a_i$ for i=1 to n suchthat:

\[\displaystyle\sum^n_{i=1}{a_i\sin^2(L_i\pi)\sin^2(l_i\pi)}=\sin^2(L\pi)\sin^2(l\pi)\]

And the result follows.

Consider it like a new exercice :D

take \[f(x,y)=\sin(x\pi)\cos(x\pi)\sin(y\pi)\cos(y\pi)\]

And see the tittle.

Today, cretanman asked me to reveal some details of my solution, and I realized that it is completely wrong  :oops: Sorry for this nonsense. Unfortunately, noone pointed to my mistake.

<off-topic> It seems Myth's back! Isn't that great news? :D </off-topic>

How does one integrat $e^{2\pi i (x+y)}$ over the complex plane?


$ f: N\to R$, $ f(1) > 0$ and $ f(m^2 \plus{} n^2) \equal{} (f(n))^2 \plus{} (f(m))^2$

Use latex, is very pretty :D :lol:

I guess that is determined by the $ (f(3))^2\plus{}(f(1))^2\equal{}f(10)$ somehow

Yes, $ f(3)$ need $ f(10)$.

How can $ f(10)$ be determined?

Off course  by $ f(101) \equal{} (f(1))^2 \plus{} (f(10))^2$ but it is quite suspect because it gives us 2 roats :wink:

But we have $ f(n^2 \plus{} 1) \equal{} (f(1))^2 \plus{} (f(n))^2\Leftrightarrow f(1) \equal{} \pm \sqrt {(f(n^2 \plus{} 1) \minus{} (f(n))^2)}$

As $ f(1)$ is positive we get $ f(1) \equal{} \sqrt {f(n^2 \plus{} 1) \minus{} (f(n))^2}$ $ \forall n\in N$

But now we get the data that $ f(n^2 \plus{} 1) \minus{} (f(n))^2$ is constant :D

I would assume $ 0\not\in\mathbb{N}$.
If $ 0\in \mathbb{N}$,it is easy to see $ f(1) \equal{} 1$ and $ f(n) \equal{} n$ for most of $ n\in\mathbb{N}$.

I made one mistake translating to LaTex. $f$ may take natural values.


real analysis
trigonometry
calculus
integration
function
real analysis unsolved

Let

$ f(x) \equal{} |x|^{\minus{}3/2}|\sin x|$

for x in $ \mathbb{R}$.  Prove that $ f$ is Lebesgue-measurable. Is it integrable?

That's really "for all $ x\neq 0$, but Lebesgue integration doesn't care about values at one point.

It's piecewise continuous, so of course it's measurable. The question is whether it's absolutely integrable; for that, you need to examine the behavior near zero and near $ \infty$. Near zero, $ |\sin x|$ looks like $ |x|$. Near $ \infty$, it looks like 1.

A minor point of vocabulary: I'd rather reserve the words "piecewise continuous" for the narrow technical case of a function that has one-sided limits at each of finitely many exceptional points. This function, which tends to $ \infty$ at zero, doesn't meet that narrow technicality.

But of course it's measurable. $ \{x: f(x)>\alpha\}$ is, for each $ \alpha,$ not merely measurable but open.

Also, when jmerry said "Near $ \infty$, it looks like 1" concerning $ \sin x,$ that shouldn't be read too literally but rather in the sense that that is the useful comparison. (And yes, $ f\in L^1(\mathbb{R}).$)


inequalities
rearrangement inequality
inequalities proposed

Proof that $ xy\plus{}yz\plus{}zx\minus{}2xyz\le \frac{7}{27}$ if $ x\plus{}y\plus{}z\equal{}1$

[quote="Flakky"]Proof that $ xy \plus{} yz \plus{} zx \minus{} 2xyz\le \frac {7}{27}$ if $ x \plus{} y \plus{} z \equal{} 1$[/quote]
It 's here
http://www.mathlinks.ro/viewtopic.php?t=227598

$ (\sum_{cyc} xy)(\sum_{cyc} x)\minus{}2xyz \leq \frac{7}{27}(x\plus{}y\plus{}z)^3$
$ 6\sum_{sym} x^2y \leq 7\sum_{cyc}x^3\plus{}15xyz$
from schur inequality we have
$ 5\sum_{sym} x^2y \leq 5\sum_{cyc}x^3\plus{}15xyz$
and from rearrangement inequality we have
$ \sum_{sym} x^2y \leq 2\sum_{cyc} x^3$
so
$ 6\sum_{sym} x^2y \leq 7\sum_{cyc}x^3\plus{}15xyz$

easy
with the inequality of Schur
$ xy\plus{}yz\plus{}zx\leq\frac{1\plus{}9xyz}{4}  <\equal{}> xy\plus{}yz\plus{}zx\minus{}2xyz\leq\frac{1\plus{}xyz}{4}$
with AM-GM.
$ xyz\leq 1/27 <\equal{}> 1\plus{}xyz\leq 28/27 <\equal{}> \frac{1\plus{}xyz}{4}\leq\frac{7}{27}$
then:    $ xy\plus{}yz\plus{}zx\minus{}2xyz\leq\frac{7}{27}$


function
number theory
least common multiple
relatively prime

what is the remainder when
2^1990 divided by 1990?

could u plz explain the theorems involved in solving this question..

aviateurpilot has a minor arithmetic mistake, but that is easily rectified.

The Phi function $\phi(n)$ is the number of positive integers less than or equal to $n$ and relatively prime to it (sharing no common factors other than $1$).  For example, $\phi(12) = 4$ because $1, 5, 7, 11$ are the only four positive integers less than or equal to $12$ that are relatively prime to it.  Generally, we have the following results:

- $\phi(p^{k}) = (p-1)p^{k-1}$ for a prime $p$; this is because only the multiples of $p$ are not relatively prime to $p^{k}$.

- $gcd(a, b) = 1 \implies \phi(ab) = \phi(a) \phi(b)$

Hence

$\phi \left( 995 \right) = \phi(5 \cdot 199) = \phi(5) \phi(199) = 4 \cdot 198 = 792$

Now, a strong theorem known as [b]Euler's Totient Theorem[/b] states that for relatively prime integers $a, m$ we have

$a^{ \phi(m) }\equiv 1 \bmod m$

Hence

$2^{792}\equiv 1 \bmod 995$

Unfortunately, this is not good enough to solve the problem as is, but this is the reasoning behind aviateurpilot's solution as it stands.

[hide="A completion"] We have the stronger [b]Carmichael Theorem[/b] which states that

$a^{ \lambda(m)}\equiv 1 \bmod m$

Where $\lambda(m)$ is defined recursively:

$\lambda(1) = \lambda(2) = 1, \lambda(4) = 2$
${\lambda(2^{k+2}}) = 2^{k}, k \in \mathbb{N}$
$\lambda(p^{k}) = (p-1) p^{k-1}, p > 2$
$\lambda(a, b, c, d, ... ) = \text{lcm}\left( \lambda(a), \lambda(b), \lambda(c), \lambda(d) \right) ...$

Hence

$\lambda(995) = \text{lcm}(4, 198) = 396$

So we now know that

$2^{396}\equiv 1 \bmod 995$

And we proceed as aviateurpilot. [/hide]

When you say
\[\lambda(a, b, c, d, ... ) = \text{lcm}\left( \lambda(a), \lambda(b), \lambda(c), \lambda(d) \right) ... \]
You really mean something like: \[gcd(a,b,c,d, ... )=1\implies \lambda(a\cdot b\cdot c\cdot d \cdots ) = \text{lcm}\left( \lambda(a), \lambda(b), \lambda(c), \lambda(d) \right)\]right?

thanks t0rajir0u for the excellent explanation


Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}$.  Susan makes a list as follows: for each two-element subset of $\mathcal{S}$, she writes on her list the greater of the set's two elements.  Find the sum of the numbers on the list.

[quote="sen"][hide]
I am not sure, but is it simply?

Firstly arrange the digits in order: $1, 2, 3, 5, 8, 13, 21, 34$

$1$ will not be bigger than any. $2$ will only be bigger than $1$. $3$ will be bigger than $1$ and $2$...

$\implies 1\cdot 0 + 2\cdot 1 + 3\cdot 2 + 5\cdot 3 + 8\cdot 4 + 13\cdot 5 + 21\cdot 6 + 34\cdot 7 = 484$

[/hide][/quote]

It should be. You could also think of it as first forming all subsets with 34 in them, then removing 34 from the set. 21 becomes the largest number, so you can make all subsets with 21, etc.

[quote="Hokkage"]Does this seem really easy for an AIME question?[/quote]

err...it's only number three you know ;)

For $1$, there are $0$ ways. For $2$, there is $1$ way. For $3$, there are $2$ ways. For $5$, there are $3$ ways. So on, so the answer is $1 \cdot 0+2 \cdot 1+3 \cdot 2+5 \cdot 3+8 \cdot 4+13 \cdot 5+21 \cdot 6+34 \cdot 7 = \boxed{484}$.

$2$ repeats $1$ time. $3$ repeats $2$ times. $5$ repeats $3$ times. $8$ repeats $4$ times. $13$ repeats $5$ ways. $21$ repeats $6$ times. $34$ repeats $7$ times. Thus the sum of the elements in Susan's list becomes $$1\cdot 0 + 2\cdot 1 + 3\cdot 2 + 5\cdot 3 + 8\cdot 4 + 13\cdot 5 + 21\cdot 6 + 34\cdot 7 = \boxed{484}.$$

[quote=mahaler][hide = One-liner]$7(34) + 6(21) + 5(13) + 4(8) + 3(5) + 2(3) + 1(2) = \boxed{484}.$[/hide][/quote]

so basically the above sols but with no explanation


geometry
circumcircle
geometric transformation
homothety
geometry proposed

Let $ABCD$ be a cyclic quadrilateral. let $w$ be its circuncircle and $O$ its center. the bisector of $ABD$ meets $AD$ in $K$ and $w$ in $M$, the bisector of $CBD$ meets $CD$ in $L$ and $w$ in $N$. if $KL$ is parallel to $MN$, prove that the circuncircle of  triangle $MON$ pass trought the midpoint of $BD$

From the given condition we conclude that the circle $(BKL)$ is tangent to $(O)$ in $B$, and it's also tangent to $AD,CD$ in $K,L$ respectively. This means that we only need to prove the following conclusion, derived from the actual conclusion through a homothety of center $B$ which takes $M,N$ to $K,L$ respectively:

Let $O'$ be the circumcenter of $BKL$, and let $B'$ be the point where the $B$-symmedian of $BKL$ cuts the circle $(BKL)$. In this case, the circle $(O'KL)$ passes through the midpoint of $BB'$. I'll leave this as an exercise :). Just use the fact that the $B$-symmedian passes through the tangents in $K,L$ to $(BKL)$.


Find the sum of the terms in the arithmetic sequence $ \{3, 6, 9, ... ,90\}$.

Gaussian method, pair up and multiply.

93*15=1350+45=[b]1395[/b].

Alternative: 3 times 30th triangular number.

The equation for sum of all the terms in this arithmetic series is $\frac{3+90}{2}*30$ because the formula is ((first# + last#)/2) * (the amount of in the sequence. By substituting the numbers you will get $\frac{3+90}{2}*30$ which simplifies to $\frac{93}{2}*30 = \boxed{1395}$. This works since the average of the first number and the last number (in this case: $\frac{3+90}{2}$) is the average of all the terms in the sequence. Since its the average of all the terms in the sequence, you multiply the amount of terms (in this case it's $30$) by the average: $\frac{93}{2} * 30$. As you can see, this is the same equation we used before. Again we get the answer of $\boxed{1395}$. The other way you could do it is by adding it all up, but that isn't very efficient.


HMMT search Harvard college MIT algebra polynomial

I found this problem on the 2003 HMMT test

Find the sum of 2003C1+2003C4+...+2003C2002

the solution they gave was kinda weird so does anyone know a clear way to do this?

thanks

[quote="The Zuton Force"][hide="Quote in Solution"]
I leave the simplification of $\frac{2^{2003}+\omega^{2}(1+\omega)^{2003}+\omega (1+\omega^{2})^{2003}}{3}$ to you.
[/hide][/quote]

[hide="Hint to simplify"]
DeMoivre's Theorem.  :wink: [/hide]

I already know about the [hide="magic;"]roots of unity filter[/hide] but is there another (viable) way to tackle the problem?  :)

[quote="The Zuton Force"]I take it that it starts with 1 and increases by 3 each time?

Take a look at the method used in a very similar problem here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=352400967&t=119683

They're calling it a "roots of unity filter"

Basically, it involves expanding fairly simple binomials using the binomial theorem, which allows us to express a sum such as the one you have in closed form. In particular, look at [b]paladin8[/b]'s explanation.

If you still don't get it, consider:

[hide="this"]
Consider $\omega$ such that $\omega^{2}+\omega+1=0$. That is, $\omega$ is a third root of unity (but not 1). Then, $\omega^{3}=1$. Also, $(\omega^{2})^{3}=1$.

Now consider the expansions of:
$(1+1)^{2003}= 1^{0}\binom{2003}{0}+1^{1}\binom{2003}{1}+1^{2}\binom{2003}{2}+\ldots$
$\omega^{2}(1+\omega)^{2003}=\omega^{2}( \omega^{0}\binom{2003}{0}+\omega^{1}\binom{2003}{1}+\omega^{2}\binom{2003}{2}+\ldots )$
$\omega (1+\omega^{2})^{2003}=\omega ( (\omega^{2})^{0}\binom{2003}{0}+(\omega^{2})^{1}\binom{2003}{1}+(\omega^{2})^{2}\binom{2003}{2}+\ldots)$

Using the fact that $\omega^{3}= (\omega^{2})^{3}= 1$ and multiplying out, this simplifies to:

$(2)^{2003}= 1 \binom{2003}{0}+1 \binom{2003}{1}+1 \binom{2003}{2}+\ldots$
$\omega^{2}(1+\omega)^{2003}=\omega^{2}\binom{2003}{0}+1 \binom{2003}{1}+\omega \binom{2003}{2}+\ldots$
$\omega (1+\omega^{2})^{2003}=\omega \binom{2003}{0}+1 \binom{2003}{1}+\omega^{2}\binom{2003}{2}+\ldots)$

And now if you add these, you get:

$2^{2003}+\omega^{2}(1+\omega)^{2003}+\omega (1+\omega^{2})^{2003}$ $= (\omega^{2}+\omega+1) \binom{2003}{0}+(1+1+1) \binom{2003}{1}+(\omega^{2}+\omega+1) \binom{2003}{2}+\ldots$ $=0\binom{2003}{0}+3\binom{2003}{1}+0\binom{2003}{2}+\ldots$

By $\ldots$, I mean that the pattern will continue 0,3,0,0,3,0,0,3,0,...
So now if you divide by $3$,

$\frac{2^{2003}+\omega^{2}(1+\omega)^{2003}+\omega (1+\omega^{2})^{2003}}{3}=\binom{2003}{1}+\binom{2003}{4}+\ldots+\binom{2003}{2002}$

Although $\omega$ and $\omega^{2}$ both have complex parts, they will cancel out and the sum will be real.

I leave the simplification of $\frac{2^{2003}+\omega^{2}(1+\omega)^{2003}+\omega (1+\omega^{2})^{2003}}{3}$ to you.

[/hide][/quote]
Nice solution.

But thats actually the same soultion on the HMMT website (except you explanined it alot better so thanks)

But yea I was also wondering is there a better way to do this because I really dont understand that way.

[quote="cincodemayo5590"]I already know about the [hide="magic;"]roots of unity filter[/hide] but is there another (viable) way to tackle the problem?  :)[/quote]

Not a way that wouldn't be equivalent to the roots of unity filter.  If there were a purely combinatorial way, it would probably be harder.

[quote="NeverOddOrEven"]can someone show me how
$(1+1)^{2003}= \omega^{2}(1+\omega)^{2003}= \omega(1+\omega^{2})^{2003}$[/quote]

Er... they're not.

[quote="da ban man"]But yea I was also wondering is there a better way to do this because I really dont understand that way.[/quote]

I believe AoPS II has a chapter (chapters?) on complex numbers and roots of unity, so that would be a good resource for you.  This is not, in fact, an easy problem, and as I already stated I don't think there's an easier way to do it.

This is an easy problem if you have seen it already. This is one of those problems that you just kind of remember. For some reason, I feel like I have seen this a million times on AoPS. I guess some people feel proud that they can pull out such a random solution. This has limited use in contests. The case for $n=2$ might show up...

The Harvard-MIT Mathematics Tournament !
http://web.mit.edu/hmmt/www/

That'll do it, Vishalarul.

And yes, this general problem is showing up all over the place (a bit too often, really).

But although it is a bit limited in application, it's very useful where it does apply, and its a good thing for people to know.

The general idea is also pretty helpful - using roots of unity to filter out terms in a polynomial sum. And it trivializes certain kinds of sums, such as $\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!}$, which can be helpful.

Ideally, you're supposed to see a bunch of binomial coefficients and be like OMG its a binomial expansion! and then think of polynomials and roots of unity filtering.

In reality, you just have to know that in some cases, combinatorics problems can be converted into algebra through the Binomial Theorem, and then if you see a combinatorics problem that could potentially use it, it'll be there in your head.

[quote="worthawholebean"]So what would cause someone to think of this? I mean, when I look at combinatorics problems my thoughts generally don't go towards roots of unity.[/quote]

Well, I think it would start as the roots-of-unity result that, for example, third roots of unity have $1^{n}+\omega^{n}+(\omega^{2})^{n}\neq 0$ iff $3 | n$ for integer $n$. Then you could see how to have this occur when expanding three different binomials in the form $(1+x)^{n}$, and that's where the binomial coefficients come in. I don't think the original creator(s) of these types of problems had, from the beginning, binomial coefficients in mind; especially since the original result and others similar to it can be taken in many different directions, such as the sum that [b]The Zuton Force[/b] gave.

oh whoah i definitely thought there were equal signs on those parts of the solution nevermind  :blush:


Euler articles trigonometry function calculus algebra polynomial

http://www.artofproblemsolving.com/Wiki/index.php/Mathematical_problem_solving
This article states that Euler used the fact that zeros of sines are $0$, $\pi$, $2\pi$, etc. I am confused as to what if he assumed the zeros on the degree scale rather than the radian scale. Wouldn't the resulting solution be completely different? Some one please explain why assuming zeros based on radians yields this mathematical truth.

I am not sure, but I am guessing that the (I think it is called Taylor series) sinx equation assumes that the sine function is calculated in radians.  If it wasn't calculated in radians, the equation might not be valid.

I just remember seeing that equation in a calculus book (I don't know calculus, I just have a book)

[quote="Hikaru79"]On a related note, it would be nice to see a proof of Euler's "leap" as quoted in the Wiki. Just because two polynomials have the same zeroes does not neccesarily make them equal. Can anyone rigorously show the identity?[/quote]

The rigorizing process is decidely non-trivial -- it uses quite a bit of complex analysis, unfortunately.


Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Here's what I did.

[hide="Solution"]Let the two arithmetic sequences be $a,$ $a+m,$ $a+2m,...$ and $b,$ $b+n,$ $b+2n,\ldots.$

Note that we seek the value of
\[(a+7m)(b+7n)=ab+7an+7bm+49mn,\] so we need to find the values of $ab,$ $an+bm,$ and $mn.$

Multiplying respective terms,
\begin{eqnarray}ab&=& 1440 \\ (a+m)(b+n)&=&ab+an+bm+mn=1716 \\ (a+2m)(b+2n)&=&ab+2an+2bm+4mn = 1848. \end{eqnarray}
Substitute $ab$ into equations $(2)$ and $(3)$ to get a system of linear equations. (The variables are $an+bm$ and $mn$.)

Solving, $an+bm=348$ and $mn=-72.$ Hence, the answer is $1440+7(348)-49(72)=348.$
[/hide]

[quote="The Zuton Force"][hide]
Notice that all are divisible by 12. Dividing out 12, we get:
120,143,154

$120 = (2^{3})(3)(5)\qquad143=(11)(13)\qquad154=(2)(7)(11)$

We can easily get the arithmetic sequence $15,13,11$ out of those factors, and checking, the quotients, $8,11,14$, also form an arithmetic sequence.

The 8th term of the first sequence is $15-7(2)=1$ and the 8th term of the second sequence is $8+7(3)=29$.

The 8th term of the original sequence is 12 times the product of these (since we took the 12 out to start with):
$a_{8}=12(1)(29)=\boxed{384}$

[/hide][/quote]

typo lol..

$12*1*29=348$

amctrivial grind

We find that the two sequences are

$45,39,33,27,21,15,9,3$

and $32,44,56,68,80,92,104, 116$

So the answer is $116\cdot3=\boxed{348}$.

p12: Let the sequences be $a, a+d, \ldots$ and $b, b+e, \ldots$ so $ab = 1440, (a+d)(b+e) = 1716, (a+2d)(b+2e) = 1848$, and we want $(a+7d)(b+7e) = 1440 + 7(ea+db) + 49(de)$. Expanding the second equation gives $(ea+db) + de = 276, 2(ea+db) + 4(de) = 408$, so $ea+db = 348, de = -72$. The answer is $1440+7 \cdot 348 - 49 \cdot 72 = \boxed{348}$.


No one is interested in this ineq?This ineq is very nice I think.

[quote="SUPERMAN2"][quote="tuandokim"][quote="SUPERMAN2"]Prove that:
$ \frac {16}{9}(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})$≥$ \frac {1}{r(4R \plus{} r)}$[/quote]
I think your ineq is this:
$ \frac {4}{9}(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})$≥$ \frac {1}{r(4R \plus{} r)}$[/quote]
I think we can check this when ABC is equilateral.[/quote]
The inequality of tuandokim is true :)


in a trapezium $AB$ is parallel to $CD$ and $AB>CD$
$K$ and $L$  are points on $AB$ and $CD$ such that :
$\frac{AK}{BK}$ $=$ $\frac{DL}{CL}$

On the line $KL$ there are points $P$ and $Q$ such that
$<APB = <DCB$
$<DQC = <ABC$

Prove that
$P,Q,B,C$ are concyclic points.

$ \frac {AK}{BK}$ $ \equal{}$ $ \frac {DL}{CL}$ meens that KL passes through the intersection $ X \equiv AD \cap BC.$

$ \angle APB \equal{} \angle DCB$ meens that P is on a circle $ \mathcal C_B$ with radius $ r_B$ passing through A and tangent to BC at B.
$ \angle DQC \equal{} \angle ABC$ meens that Q is on a circle $ \mathcal C_C$ with radius $ r_C$ passing through D and tangent to BC at C.
Since $ \angle APB \equal{} \angle DCB \equal{} 180^\circ \minus{} \angle ABC \equal{} 180^\circ \minus{} \angle DQC,$ X is the external similarity center of the circles $ \mathcal C_B,\ \mathcal C_C.$ Let $ \mathcal C_B$ meet KL again at R and let $ \mathcal C_C$ meet KL again at S.

$ XB^2 \cdot XC^2 \equal{} (XP \cdot XR) \cdot (XQ \cdot XS) \equal{}$
$ \equal{} \left(XP \cdot XQ\ \frac {r_B}{r_C}\right) \cdot \left(XQ \cdot XP\ \frac {r_C}{r_B}\right) \equal{} XP^2 \cdot XQ^2$

$ XB \cdot XC \equal{} XP \cdot XQ$

which means that BCPQ is cyclic.

[quote="pardesi"]in a trapezium $ AB$ is parallel to $ CD$ and $ AB > CD$
$ K$ and $ L$  are points on $ AB$ and $ CD$ such that :
$ \frac {AK}{BK}$ $ \equal{}$ $ \frac {DL}{CL}$

On the line $ KL$ there are points $ P$ and $ Q$ such that
$ < APB \equal{} < DCB$
$ < DQC \equal{} < ABC$

Prove that
$ P,Q,B,C$ are concyclic points.[/quote]
This problem has been posted before: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=875014#875014

Very nice solution, [b]yetti[/b]!

My solution is the same as [b]Tomici0's[/b] that is shown in the link, so unfortunately, I cannot post an alternate solution  :(


geometry
3D geometry
tetrahedron
geometry unsolved

Let ABCD be a tetrahedron satisfying the condition:
$AB^2+CD^2= BC^2+DA^2=CA^2+BD^2$
Then prove that there's at least one of four sides of the tetrahedron has three acute angles.

No, it's an old problem. I don't remember solution, but in it is used the theorem of cosines.

Yes, thanks Pavel.
I know, I tried but..., can anybody help me?


In the figure, it is given that angle $ C \equal{} 90^{\circ}, \overline{AD} \equal{} \overline{DB}, DE \perp AB, \overline{AB} \equal{} 20$, and $ \overline{AC} \equal{} 12$. The area of quadrilateral $ ADEC$ is:
[asy]unitsize(7);
defaultpen(linewidth(.8pt)+fontsize(10pt));
pair A,B,C,D,E;
A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10);
draw(A--B--C--cycle); draw(D--E);
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE);
draw(rightanglemark(B,D,E,30));[/asy]$ \textbf{(A)}\ 75 \qquad\textbf{(B)}\ 58\frac {1}{2} \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 37\frac {1}{2} \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
Since $ AB\equal{}20$ and $ AC\equal{}12$, $ BC\equal{}16$.  Because $ \angle B \cong \angle B$ and $ \angle D \cong \angle C$, we have $ \triangle BDE \sim \triangle BCA \implies \frac{[BDE]}{[BCA]}\equal{}\frac{10^2}{16^2}$.  We have $ [BCA]\equal{}\frac{1}{2} \cdot 12 \cdot 16\equal{}96 \implies [BDE]\equal{}37 \frac{1}{2}$, so $ [ADEC]\equal{}96\minus{}37 \frac{1}{2}\equal{}58 \frac{1}{2}$, or $ \boxed{\textbf{(B)}}$.
[/hide]


Set $A=1.3.5...2005+2.4.6...2006$. Prove that $A$ is divisible by $2007^{4}$.
I think this problem is wrong  :huh: I think $A$ is only divisible by $2007$


On this 5 by 5 grid of dots, one square is shown in the diagram. Including this square, how many different sizes of squares can be formed using four dots of this array as vertices?

[asy]size(50);
for(int i = 0; i < 5; ++i)
{
for(int j = 0; j < 5; ++j)
{
dot((i,j));
}
}
draw((2,0)--(3,1)--(2,2)--(1,1)--(2,0)--cycle,linewidth(2));[/asy]

[hide="Solution"]The possible side lengths are $ 1,\sqrt2,2,\sqrt5,2\sqrt2,3,\sqrt{10},4$, giving us an answer of $ \boxed8$.[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide all solutions.[/url][/color]

[hide]where are the squares with sqrt5 side length?[/hide]

[hide="Solution"] The possible side lengths are: $sqrt2$, $sqrt5$, $sqrt10$, $2sqrt2$, $1$, $2$, $3$, and $4$. Therefore the answer is $\boxed{8}$. [/hide]

[quote=amazingxin777]can someone please draw out each diagram without mixing together all of it
[quote=Person][hide="Solution and Diagram"]There are the 4 squares that have sides that are parallel to (or are) the sides of the lattice grid.  Then we have the diagonal squares $(\sqrt2, \sqrt5,2\sqrt2, \sqrt{10})$, giving us an answer of 8.
[asy]
size(50);
for(int i = 0; i < 5; ++i)
{
for(int j = 0; j < 5; ++j)
{
dot((i,j));
}
}
draw((0,2)--(1,4)--(3,3)--(2,1)--cycle,blue+linewidth(2));
draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)--cycle,green+linewidth(2));
draw((2,2)--(2,4)--(4,4)--(4,2)--(2,2)--cycle,magenta+linewidth(2));
draw((1,1)--(1,3)--(3,3)--(3,1)--(1,1)--cycle,yellow+linewidth(2));
draw((0,0)--(0,1)--(1,1)--(1,0)--(0,0)--cycle,cyan+linewidth(2));
draw((2,0)--(0,2)--(2,4)--(4,2)--(2,0)--cycle,grey+linewidth(2));
draw((1,0)--(0,3)--(3,4)--(4,1)--(1,0)--cycle,black+linewidth(2));
draw((2,0)--(3,1)--(2,2)--(1,1)--(2,0)--cycle,red+linewidth(2));[/asy][/hide][/quote]
I like your diagram, but it's a little too colorful, can you (or somebody else) instead do seperate diagrams for each one.  ;) I keep on getting six as an answer.  :wallbash:
everyone: posted way before 2020
me: hi guys, um, is this thread that old? for one of the posts, I wasn't even born!
and could someone tell me the trick for these problems? everyone seems to be able to do it (unless they looked at the alcumus solution and posted basically the exact thing)
I am really confused about how to stay organized in this type of counting, there's tons of ways to miss out on a square!  :wacko:[/quote]

it might be helpful to go to problem 2.4 in your counting and prob book. You can find the obvious ones like 1X1 and 3X3 easily but in order to find the "kite" shaped squares you need to remember that because of the Pythagorean theorem the side lengths of the squared must be in the form $\sqrt{m^2+n^2}$. Because our squares can only  be lower then 5x5 we can then find all of the answers of the side lengths that are lower then 4.

I hope I explained that ok. Someone correct me if I am off please.


Euler
advanced fields
advanced fields theorems

Homology of S^2 by Identifying the north pole with the south using the Mayer-viectoris squence, i´ve used it and it gave me the caracteristic equal to 0 , and i don´t know if its right.

Any way if anyone could show me the resolution using the Mayer-viectoris squence i would be pleased.

Thank you all

I assume you are talking about the homology of (S^2 with poles identified).

Seems more natural to use excision rather than Mayer-Vietoris.  The Euler characteristic of the homology shouldn't be zero.  (Perhaps you're looking at the reduced homology?)


calculus
integration
logarithms
function
trigonometry
special factorizations
algebra

$ \int \frac {1}{t^4\plus{}1} dt$

I've got 2 solutions for these...

but I'm looking for some simpler or more creative solutions ^^  :lol:

So maybe you could tell us which ones (I mean those two methods you've used) do you have to prevent somebody from doing the same  :)

hmm.

1 is by using complex numbers and Euler's formula...

another is by adding terms of x...

[quote="zhen"]$ \int \frac {1}{t^4 \plus{} 1} dt$

I've got 2 solutions for these...

but I'm looking for some simpler or more creative solutions ^^  :lol:[/quote]

It's classic!
$ \int \frac {1}{t^4 \plus{} 1}dt \equal{} \frac {1}{2\sqrt {2}}\int{\frac {t \plus{} \sqrt {2}}{t^2 \plus{} \sqrt {t} \plus{} 1}dt} \minus{} \frac {1}{2\sqrt {2}}\int{\frac {t \minus{} \sqrt {2}}{t^2 \minus{} \sqrt {t} \plus{} 1}dt}$

$ \equal{} \frac {1}{4\sqrt {2}}\ln{\frac {t^2 \plus{} t\sqrt {2} \plus{} 1}{t^2 \minus{} t\sqrt {2} \plus{} 1}} \plus{} \frac {1}{2\sqrt {2}}\arctan{\frac {t\sqrt {2}}{1 \minus{} t^2}} \plus{} C$

[quote="hash_include"][quote="zhen"]another is by adding terms of x...[/quote]

were you thinking of this one?
adding and subtracting $ x^2$,  the integral reduces to

$ \frac 1 2\left( \int \frac {(1 \plus{} x^2)dx}{1 \plus{} x^4} \plus{} \int \frac {(1 \minus{} x^2)dx}{1 \plus{} x^4} \right)$

now, divide the numerator and denominator of both integrals by $ x^2$  and substitute:

$ p \equal{} x \minus{} \frac 1 x$ in the first integral, and $ q \equal{} x \plus{} \frac 1 x$ in the second integral, and solve  :)

It's quite painful, i know. but i've seen much worse.  :roll:[/quote]

ya you got it right...  :lol:

well, my first way to solve it is by complexifying,

i.e. $ t^4 \plus{} 1 \equal{} (t^2 \plus{} i)(t^2 \minus{} i)$

it doesn't appear to be any simpler, but at least it's another solution ^^

I didn't notice it has classical solution as NguyenDungTN's and Yuriy Solovyov's too  :oops:

The solution I've seen relies on completing the square in a clever way and then using some partial fractions.  I'm wondering, is it possible to extend such an argument to
$ \int\frac{dx}{1\plus{}x^{8}}$?

A formula for $ \int\frac {dx}{x^{2^{n}} \plus{} 1}$ is posted on Wikipedia's [url=http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions]List of Integrals of Rational Functions[/url] page, last equation on the page.

That is terribly ugly.  How did someone figure that out?

It's partial fractions based on the known roots $ \exp\left(\frac{2\pi ik}{m}\right)$ and some tricks with symmetric polynomials. If we match up conjugate roots, it becomes a real expression.
In general, $ \frac1{x^m\minus{}1}\equal{}\minus{}\frac1n\sum_{k\equal{}0}^{m\minus{}1}\frac{r_k}{x\minus{}r_k}$, where $ r_k\equal{}\exp\left(\frac{2\pi ik}{m}\right)$.

The only reason to restrict to powers of two is so that we can compute the trig functions with iterated square roots.


Let ABC be a acute triangle, EB and FC be altitutes. Let the intersection points of the tangents from the circumcircle of AEF through C be U and V. Prove that B, U and V lie on a line.

if i understood clearly, your problem is a well-know result from pole and polar.(UV is the polar and B belongs to it).


function
algebra
functional equation
algebra unsolved

find all function $f:R^+\rightarrow R^+$which that:
  $f(xf(y))=f(xy)+y$

Hmm... First of all I think the answer is not $f(x)=x+1$.Second: I think that this is a bad practice to write only the answer to problems related with functional equations.Because in many such problems it is easy to see what function satisfies to the given relation and probably the author already guessed the answer, so it is just useless to write down only the unswer without any proof(It doesn't give any information).It is only my opinion. Hope you agree with me.
 :)

Hmm...first of all I think the answer isn't exactly $f(x)=x+1$.

grodij, if you plug $x+1$ in the functional equation you get $xy+x+1$ on the left side and $xy+y+1$ on the right side. I think that's reason enough for the answer not to be $x+1$... :)

[quote="amir2"]find all function $f: R^+\rightarrow R^+$which that:
  $f(xf(y))=f(xy)+y$[/quote]
If $R^+$ means $]0,\infty[$ then a such function does'nt exist . :D

Let $a$ be such that $a=f(1)$ , so $\Longrightarrow f(x)=f(ax)-1$ . I take an integer $n > f(1)$, so :
$f(\frac 1{a^n}) = f(\frac 1{a^{n-1}})-1 = ... = f(1) - n < 0$ .. manifestly false ..

 :cool:


algebra
polynomial
trigonometry
function
system of equations

Find all real solutions to the following system of equations
x^3 - 3x = y
y^3 - 3y = z
z^3 - 3z = x

plz try to solve my problems

no one seems to be interested in my problems :(

plzzzzzzzzz solve them..............

Taking the value of $ y$ in the second equation and $ z\equal{}(x^3\minus{}3x)^3 \minus{} 3 (x^3\minus{}3x)$ in the third equation for $ x$, we get that      $ x(x\minus{}2)(x\plus{}2)(x^6\minus{}7x^4\plus{}14x^2\minus{}7)(x^3\minus{}x^2\minus{}2x\plus{}1)(x^3\plus{}x^2\minus{}2x\minus{}1)(x^6\plus{}x^5\minus{}5x^4\minus{}4x^3\plus{}6x^2\plus{}3x\minus{}1)(x^6\minus{}x^5\minus{}5x^4\plus{}4x^3\plus{}6x^2\minus{}3x\minus{}1)\equal{}0$. Solve for $ x$ and then for $ y$ and $ z$.

thanks aravind for getting the question readable no i will see the sum

Thats SOME answer Arvind!  :rotfl: If this were an objective MCQ, easiest shortcut would be to consider
symetry with no. of equations=no. of variables.
That way setting x=y=z=a
You get a(a+2)(a-2)=0 So 3 solns. will be -2, 0,+2 :p
But again thats only jugadh for MCQ

probably i think he has solved the problem

from the first three factors,we get $ x \equal{} 0,2, \minus{} 2$
for others just factorise the polynomials in parentheses and you'ld get answers

the problem i found is that the factorisation is not that easy for me  :)

I wasn't able to do this problem myself.

I'm not in mood to post down the whole official solution but here's the main idea.

First we find all solutions in the range $ [ \minus{} 2,2]$.

Replace  $ x \equal{} 2.cos\;u,\;y \equal{} 2.cos\;v,\;z \equal{} 2.cos\;w$.

Observe that the expressions turn up to be triple angle formulas for cosine function and simplify into the following:

$ cos\;3u \equal{} cos\;v$
$ cos\;3v \equal{} cos\;w$
$ cos\;3w \equal{} cos\;u$

One will find that these subs give $ 27$ solutions in total.

Now these are the only solutions as the maximum number of solutions is $ 3.3.3 \equal{} 27$. [Check the maximum power of each equation.]


Six less than a number is multiplied by 6. Eight less than the same number is multiplied by 8. The two results are equal. What is the number?

Let's translate words into algebra, to get the following equation:

$ 6(x\minus{}6)\equal{}8(x\minus{}8)$

$ 6x\minus{}36\equal{} 8x\minus{}64$

$ 28\equal{} 2x$

$ x\equal{}14$

Was it a coincidence that we just got 6+8 in the end? Perhaps not, since 6(8)=8(6), and we can quickly see x as 14 from that equation.


Euler
quadratics
modular arithmetic
number theory unsolved
number theory

Find all pairs (a, b) of positive integers such that  ( a^2+b^2)/(a^3+1)
is a positive integer

[color=red]mathematicien@naseej.com[/color]

do you have a source for this problem, Fermat-Euler?

it's some kind of a "think up" problem ;), suppose that $a^3+1$  has only prime divisors of the form $4k+1$ then such $b$ that $b^2 \equiv -a^2 (\mod a^3+1)$ obviously exists.

[quote="maxal"]Numerical experiments suggest that $a$ must be a square. Can anybody (dis)prove that?[/quote]
We can only conclude that $a$ is a quadratic residue $\mod a^3+1$.

Rafal has actually solved this problem already.

Note that for this problem, we only need to characterize the solutions for $a$, since if $b$ satisfies the conditions, then so does $k(a^3 +1) \pm b$, where $k$ can be any arbitrary integer. We can easily see that if $b$ is a solution satisfying $0 < b < a^3+1$ for a fixed $a$, then the only other solution for $b$ for the same fixed $a$ would be $a^3 + 1 - b$. This problem is actually equivalent to finding all $a$ such that $-a^2$ is a quadratic residue $\mod a^3 + 1$, and all the solutions for $b$ corresponding to this fixed $a$ will follow. And to characterize such $a$, it is indeed what Rafal has said, $a$ only needs to satisfy the condition that all prime divisors of $a^3+1$ are $\equiv 1 \pmod{4}$.

Ok, for completeness regarding my previous post, we also need to consider the case when $a^3+1$ is even. For odd $a^3+1$, which means $a$ is even, the only condition for $a$ is that all prime divisors of $a^3+1$ are $\equiv 1 \pmod{4}$. For even $a^3+1$, which means $a$ is odd, we just need the additional condition that $-1$ is a quadratic residue $2^{\alpha}$, where $2^{\alpha}$ is the highest power of $2$ dividing $a^3+1$, but $-1$ is a quadratic residue $\mod 2^{\alpha}$ iff $\alpha = 1$, so this condition is reduced to the case that $a$ must be $\equiv 1 \pmod{4}$. But this is already implied in the condition that all odd prime divisors of $a^3+1$ are $\equiv 1 \pmod{4}$, since $a^3+1 = (a+1)(a^2-a+1)$ and $a \equiv 3 \pmod{4} \Leftrightarrow a^2-a+1 \equiv 3 \pmod{4}$, implying that $\exists$ an odd prime divisor $\equiv 3 \pmod{4}$, contradicting our original assumption.

Thus for the sake of completeness, the condition for $a$ should be all [u]odd[/u] prime divisors of $a^3+1$ are $\equiv 1 \pmod{4}$.

There's just a little bit more to say.  For a number $n$ to be a sum of two squares $x^2+y^2$ it is necessary and sufficient that all primes dividing $n$ which are $\equiv 3 \bmod 4$ divide $n$ to an even power: and if so then these primes divide both $x$ and $y$.  Suppose that such a prime $q$ divides $a^3+1$.  Then $q \mid a^2 + b^2$ and this implies that $q \mid a$ and $q \mid b$: a contradiction.

[quote="rgep"]For a number $n$ to be a sum of two squares $x^2+y^2$ it is necessary and sufficient that all primes dividing $n$ which are $\equiv 3 \bmod 4$ divide $n$ to an even power[/quote]
Not true... $-1$ is a quadratic non-residue $\forall$ primes $\equiv 3 \pmod{4}$. $n$ can never have any prime divisors $\equiv 3 \pmod{4}$.

[quote]$n$ can never have any prime divisors $\equiv 3 \pmod{4}$.[/quote]
\[245 = 7^2 + 14^2 = 5.7^2\]

Let me clarify myself. You said you suppose $\exists$ a prime $q \equiv 3 \pmod{4}$ dividing $a^3+1$, then you considered the case when $q$ divides $n = a^2 + b^2$. What I'm trying to say is that you do not even need to consider such a $q \equiv 3 \pmod{4}$, whether it divides $n$ or not, because $-1$ is a quadratic non-residue, which would result in a contradiction. So you don't even need to use the fact that all primes $\equiv 3 \pmod{4}$ dividing $n$ must divide to an even power.

The condition for $a$ that all odd prime divisors of $a^3 + 1$ are $\equiv 1 \pmod{4}$ is sufficient. $a^3+1$ cannot have any prime divisors $\equiv 3 \pmod{4}$, for if such a prime $q$ exist, then we have $b^2 \equiv -a^2 \pmod{q}$, which is a contradiction since $-1$ is a quadratic non-residue $\mod q$. We won't run into any problems since $\gcd (a^2, a^3+1) = 1 \Rightarrow \gcd (a, q) = 1$.


We are sorry for the inconvenience. We've tried to start serving foreign customers. So we've honored few orders but there were problems. The first one is the shipping. The romanian postal office is not secure, meaning that they don`t guarantee you that the package will be delivered. So we went to DHL. It is ok, their service is obviously professional but the problem is the cost. There was an order to Iran, and the cost only for the shipping was more then 150 US $. I say this is too much but we as other solutions are not available, we can't do anything. There is a program, SkyPack at the romanian postal office that started and we will try it. It has cheaper (much!) costs but yet again, we don`t know the quality. For orders, please write to sales@gil.ro. We have recieved alot of e-mails and every one of them will be honored. I hope this will clarify you.

By the way, another problem is the payment. Until now we have accepted only trough Western Union, as bank payment is not accessible. Banks also progressed in Romania so maby there is a chance for things to get better. Until then, we can give no good news.  :maybe:

Best regards,
GIL

P.S. New book launched !!!
Title: Balkan Mathematical Olimpiads 1984-2006
Authors: Mircea Becheanu, Bogdan Enescu
 :D

I hope to read nice books from Gil, too. So in every condition, I will try to buy. Bringing them to Vietnam this IMO is a good solution to me  :lol: because the shipping is very easy (no \$).


[hide]The square roots are ugly, so let's get rid of them. We know that $\sqrt{52+6\sqrt{43}}=x+y\sqrt{43}$ for some $x$ and $y$.

Squaring both sides yields $52+6\sqrt{43}=x^2+43y^2+2xy\sqrt{43}$, so $x^2+43y^2=52$ and $xy=3$. It is easy to see that $x=3$ and $y=1$, so $\sqrt{52+6\sqrt{43}}=3+\sqrt{43}$. Similarly, $\sqrt{52-6\sqrt{43}}=\sqrt{43}-3$.

Therefore, our answer is $(3+\sqrt{43})^3-(\sqrt{43}-3)^3=2(3^3+3(3)(43))=\boxed{828}$.[/hide]

You could have also simplified the radical by setting it equal to $ \sqrt{a} \plus{} \sqrt{b}$.  For $ \sqrt{52 \plus{} 6\sqrt{43}}$ I got $ 3 \plus{} \sqrt{43}$.  But for $ \sqrt{52 \minus{} 6\sqrt{43}}$ I got $ 3 \minus{} \sqrt{43}$.  By using $ 3 \minus{} \sqrt{43}$ I got a different answer.  I don't think I did anything wrong in the simplification.  Should I have used something other than $ \sqrt{a} \plus{} \sqrt{b}$ or could I easily have seen which one to pick, if so how?

$ 52 \minus{} 6 \sqrt{43}$ is positive.  Is $ 3 \minus{} \sqrt{43}$ positive?

[quote=4everwise]We know that $\sqrt{52+6\sqrt{43}}=x+y\sqrt{43}$ for some $x$ and $y$. [/quote]

I'm probably just overthinking this or something, but how do we know that $\sqrt{43}$ is part of the simplified form of $\sqrt{52+6\sqrt{43}}$?

[quote=Not_a_Username]Well, there will always be [i]some[/i] constant $x, y$ such that $\sqrt{52+6\sqrt{43}}=x+y\sqrt{43}$; we choose $\sqrt{43}$ in hope that $x, y$ will be integers.[/quote]

Moreover, if we "pull that out of the hat" (i.e. the form $x + y\sqrt{43}$) and it works, no further justification is needed. Of course if we need to for whatever reason prove this is the [i]only[/i] way to simplify to the form $x + y\sqrt{z}$, that's harder, but not what we're asked to do here.

[quote=Python54]Okay, thanks guys. :)

When simplifying any expression in the form $\sqrt{x+y\sqrt{5}}$ (I'm using $\sqrt{5}$ as an example), should I try to find the simplified value in the form $a+b\sqrt{5}$?[/quote]

Yes, because 5 is a prime, so it cannot be multiplied:
$(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}$, so $ab=5$

In general, try $\sqrt{a}+\sqrt{b}$. If $a=9$ or something then $\sqrt{a}$ will be an integer, but it still works.

[quote=Python54]Okay, thanks guys. :)

When simplifying any expression in the form $\sqrt{x+y\sqrt{5}}$ (I'm using $\sqrt{5}$ as an example), should I try to find the simplified value in the form $a+b\sqrt{5}$?[/quote]

Try $\sqrt{3+\sqrt{5}}$

[hide=Solution]We first find $\sqrt{52+6\sqrt{43}}$. Since $3^2+\sqrt{43}^2=52$ and $3 \cdot 2 = 6$, we have $\sqrt{52+6\sqrt{43}}=3+\sqrt{43}$. Similarly, $\sqrt{52-6\sqrt{43}}=\sqrt{43}-3$. We now want $(3+\sqrt{43})^3-(\sqrt{43}-3)^3 = \boxed{828}$ from Difference of Cubes.[/hide]

[hide=Solution]Note that we are able to write $\sqrt{52+6\sqrt{43}}$ in the form $a+b\sqrt{43}$. Solving for $a$ and $b$, we have $$(a+b\sqrt{43})^2=52+6\sqrt{43} \implies a^2+2ab\sqrt{43}+43b^2=52+6\sqrt{43}.$$ Undetermined coefficients shows that $$a^2+43b^2=52$$ and $$2ab=6 \implies ab=3.$$ Given these pieces of information, we are able to see that $a=3$ and $b=1$, meaning that we can also write the nasty square root in the form $3+\sqrt{43}$. Thus, our answer is $$(3+\sqrt{43})^3-(-3+\sqrt{43})^3=\boxed{828}.$$ (The last step follows from expansion or difference of cubes.)[/hide]


If $f(x)$ is a monic quartic polynomial such that $f(-1)=-1,$ $f(2)=-4,$ $f(-3)=-9,$ and $f(4)=-16,$ find $f(1).$


In triangle $ ABC$, $ P$ is the midpoint of side $ BC$. Let $ M\in(AB)$, $ N\in(AC)$ be such that $ MN\parallel BC$ and $ \{Q\}$ be the common point of $ MP$ and $ BN$. The perpendicular from $ Q$ on $ AC$ intersects $ AC$ in $ R$ and the parallel from $ B$ to $ AC$ in $ T$. Prove that:
(a) $ TP\parallel MR$;
(b) $ \angle MRQ\equal{}\angle PRQ$.

[i]Mircea Fianu[/i]

$ BT\parallel{}AC$ $ \implies$ $ \Delta QTB \sim \Delta QNR$, therefore $ \frac{RQ}{QT}\equal{}\frac{NQ}{QB}$.
The same way $ MN\parallel{}BC$ from where $ \frac{NQ}{QB}\equal{}\frac{MQ}{QP}$
Therefore $ \angle MQR\equal{}\angle TQP$ $ \implies$ $ \Delta MQR\sim \Delta TQP$ from where $ MR\parallel{}TP$

Note that $ S$ is the common point of the lines $ TP$ and $ AC$.
$ BP\equal{}PC$ and $ TB\parallel{}CS$ $ \implies$    $ TBSC$ is a parallelogram and therefore $ P$ is the middle of $ TS$ $ RP\equal{}\frac{1}{2}TS\equal{}TP$ from where $ \angle TRP\equal{}\angle RTP$. On the other side $ \angle RTP\equal{}\angle MRT$ .Therefore $ MRT\equal{}TRP$

Another solution for (b)
Assume that $MRQ=a$,$PRQ=b$ thus $RTP=a$,$PRC=90-b$
We know $BP/sin(BTP).sin(PRC)/PC=TP/sin(TBP).sin(RCP)/PR$
so we have $sin(90-b)/sin(90+a)=sina/sinb$
thus $cos2a=cos2b$ so $a=b$
QED

Yet another proof for b):
$RT$ is the projection of $BC$ onto the direction perpendicular to $AC$ and $M$, the midpoint of $BC$, is equally apart from the ends $R, T$ of the projection; with the result from a), done.

Best regards,
sunken rock


geometry
integration
calculus
calculus computations

1.find the area of the region bounded by $ y\equal{} x^{2}$ ,$ y\equal{}x\plus{}2$ and $ y\equal{}0$

2.find the area of the region bounded by the lines $ y\equal{}3x\plus{}2$ ,$ y\equal{}0$, $ x\equal{}\minus{}1$,$ x\equal{}1$

1.for 1 the [hide="answer"]$ \frac{9}{2}$[/hide] .i got the same thing as urs but the answer doesn't match.

 2.is there any region which is bounded by all these lines

[quote="Pardesi"]i got the same thing as urs but the answer doesn't match[/quote]

Then the solution in your book is probably wrong.

[quote="Pardesi"]is there any region which is bounded by all these lines[/quote]

No.

yes i think the answer in the book is wrong

so if there is no region then what is the meaning of finding the area bounded by these lines

Because there are two regions bonded by those four lines: one bonded by y = 3x+2, y = 0, and x = -1, and the other bonded by y = 3x+2, y = 0, and x = 1. So our area is made up of two regions. In this kind of problems, "find the area of the region bounded by the lines " doesn't necessarily mean the region is a single one bonded by all lines given: it can have two or more "subregions" that share a point (like your example) or even two or more "subregions" that don't share any point at all.


modular arithmetic
number theory unsolved
number theory

Find all $ x,y$ positive interger such that: $ x \plus{} y \equiv 0 \pmod 2$ and:
$ x^n \minus{} y^n \equal{} (\frac {x \plus{} y}{2})^n$, where $ n$ is odd. :D

[hide="Solution"]By Fermat's Last, $ n<3$
1) n=1

$ 2x\minus{}2y\equal{}x\plus{}y$
$ x\equal{}3y$
$ (x,y)\equal{}(3y,y),y\in\mathbb{R}$

2) n=2

$ 4x\minus{}4y\equal{}x\plus{}y$
$ 3x\equal{}5y$
$ (x,y)\equal{}(\frac53y,y),y\in\mathbb{R}$[/hide]

@Bugi - that's an overkill, but on the other hand - you get -all- the solutions without the restriction on $ n$.

Another solution - since  the equation is homogeneous, we can first find the primitive solutions - $ (x,y) \equal{} 1$.
Looking mod 2, we get: $ \frac {x \plus{} y}{2} \equiv 0 \pmod 2$, or: $ 4 | x \plus{} y$.
If $ p$ is an odd prime divisor of $ x \minus{} y$, then:
$ p|x \minus{} y \implies p| x^n \minus{} y^n \implies p | \left( \frac {x \plus{} y}{2} \right)^n \implies p | \frac {x \plus{} y}{2} \implies p| x \plus{} y \implies p | 2x, 2y \implies p | (x,y) \equal{} 1$
So it follows that $ x \minus{} y \equal{} 2^s$, $ s \in \mathbb{N}$. Since $ 4|x \plus{} y$ already and $ (x,y) \equal{} 1$, we must have $ s \equal{} 1$, for otherwise $ x$ and $ y$ are even.
$ x \minus{} y \equal{} 2 \implies (y \plus{} 2)^n \minus{} (y \plus{} 1)^n \minus{} y^n \equal{} 0$. Looking mod $ y \plus{} 1$, we get $ 0 \equiv 2 \pmod {y \plus{} 1} \implies y \equal{} 1$
This shows that $ x \equal{} 3$ and  that $ (3,1)$ is the only primitive solution and $ y(3,1)$ is the general one.


I was reading a question solved before in this forum and modified a little bit:

Do there exist a finite set $A\,(|A|\geq 2)$ of distinct points in the plane $\mathbb{R}^{2}$ such that for every $x,y\in A,x\ne y$ there are exactly two points $z,w\in A, z\ne w$ such that $z$ and $w$ are in the normal passing through the midpoint of the line segment $xy$?

Yes there really do. Draw four points making a square. Then add four points in the plane, in the exterior of a square, such that you can form four equilateral triangles.


Umm...what does this have to do with exponents?

I was going to make up a problem with exponents but then i changed it and forgot to change the title. :blush:

[quote="math92"]Find x if
3x+7=13[/quote]

[hide]We subtract seven from both sides to get

$3x=6$.

Then we divide both sides by 3

To get

$x=2$[/hide]

[quote]Find x if 3x+7=13[/quote]

[hide]$3x+7=13$

Subtract 7 from both sides

$3x=6$

Divide by three on both sides

$x=2$[/hide]


superior algebra
superior algebra unsolved

$ R$ is a ring wich has no non-zero nilpotent element.and for each $ a,b \in R$ $ (ab)^2\equal{}(ba)^2$.Prove that $ R$ is commutative.

for all $ a,b,c$, we have $ (a(b \plus{} c))^2 \equal{} ((b \plus{} c)a)^2$, which expands to (using $ (ab)^2 \equal{} (ba)^2$ and $ (ca)^2 \equal{} (ac)^2$)

(1) $ abac \plus{} acab \equal{} baca \plus{} caba$.

setting $ c \equal{} 1$, we get

(2) $ aab \equal{} baa$,

i.e. squares lie in the center of $ R$. setting $ c \equal{} a$ in (1), we get $ abaa \plus{} aaab \equal{} baaa \plus{} aaba$, i.e. according to (2)

(3) $ 2(aaab \minus{} baaa) \equal{} 0$.

using these three equations and the assumption $ baba \equal{} abab$, we compute

$ (aba \minus{} baa)^2 \equal{} abaaba \minus{} ababaa \minus{} baaaba \plus{} baabaa$
$ \equal{} aaabba \minus{} abaaab \minus{} aa baba \plus{} bbaaaa$
$ \equal{} aaaabb \minus{} abaaab \minus{} aa abab \plus{} aaaabb$
$ \equal{} aaaabb \minus{} abaaab \minus{} abaaab \plus{} aaaabb$
$ \equal{} 2a(aaab \minus{} baaa)b$
$ \equal{} 0$

since $ R$ has no nilpotents besides $ 0$, this implies

(4) $ aba \equal{} baa$,

and then $ (ab \minus{} ba)^2 \equal{} abab \minus{} abba \minus{} baab \plus{} baba \equal{} 2(aba \minus{} baa)b \equal{} 0$. thus, $ ab \minus{} ba \equal{} 0$, and $ R$ is commutative.

I hope it's correct ..

[quote="-oo-"]for all $ a,b,c$, we have $ (a(b \plus{} c))^2 \equal{} ((b \plus{} c)a)^2$, which expands to (using $ (ab)^2 \equal{} (ba)^2$ and $ (ca)^2 \equal{} (ac)^2$)

(1) $ abac \plus{} acab \equal{} baca \plus{} caba$.

setting $ c \equal{} 1$, we get

(2) $ aab \equal{} baa$,

i.e. squares lie in the center of $ R$. setting $ c \equal{} a$ in (1), we get $ abaa \plus{} aaab \equal{} baaa \plus{} aaba$, i.e. according to (2)

..[/quote]
The proof is true only when $ R$ has 1!!!
How you prove that $ R$ has 1? :maybe:

[quote="-oo-"]we may assume that $ R$ is unital by considering the unitalization $ R \oplus \mathbb{Z}$. this ring has also no nilpotent elements $ \neq 0$.[/quote]
OK.
But why $ (ab)^2\equal{}(ba)^2$ holds for $ R \oplus \mathbb{Z}$?

yeah I forget that. my proof works only when $ R$ is unital. I assumed that your ring is unital, which is often the case when you talk about rings. but I won't try to fix my proof, it was difficult enough :-)


calculus
integration
trigonometry
calculus computations

Prove that

$ \int_{0}^{\pi} x f(\sin x)\ dx \equal{} \frac{\pi}{2} \int_{0}^{\pi} f(\sin x)\ dx$.

Evaluate $ \int_{0}^{a} \frac{x \sin x}{1\plus{} \cos^{2} x}\ dx$ when $ a\equal{}\pi$ and when $ a\equal{} 2\pi$.

hello, the integral
$ \int_{0}^{2\pi}\frac{x\sin(x)}{1\plus{}\cos(x)^2}\,dx\equal{}0$,
the other integral is also not elementary.
Sonnhard.

Thanks Sonnhard, perhaps I made a mistake: I got $ \frac{\minus{}\pi^{2}}{2}$ and for the other one I have $ \frac{\pi^{2}}{4}$. I'll check again later.

hello,probably i have made a mistake.
Sonnhard.

Hi Sonnhard, I think I've got it. Here goes:

Letting $ y \equal{} \pi \minus{} x$,

$ \int_{0}^{\pi} x f(\sin x)\ dx \equal{} \int_{\pi}^{0} (\pi \minus{} y)f(\sin y) ( \minus{} dy)$

$ \equal{} \int_{0}^{\pi}(\pi \minus{} y)f(\sin y)\ dy$

$ \equal{} \int_{0}^{\pi}(\pi \minus{} x)f(\sin x) \ dx$

$ \equal{} \pi\int_{0}^{\pi} f(\sin x)\ dx \minus{} \int_{0}^{\pi} xf(\sin x)\ dx$.

Therefore $ \int_{0}^{\pi} xf(\sin x) \ dx \equal{} \frac {\pi}{2}\int_{0}^{\pi} f(\sin x) \ dx$.

Now, since $ f(\sin x) \equal{} \frac {\sin x}{(1 \plus{} \cos^{2}x)}$,

$ \int_{0}^{\pi} \frac {x \sin x}{1 \plus{} \cos^{2}x}\ dx$

$ \equal{} \frac {\pi}{2} \int_{0}^{\pi} \frac {\sin x}{1 \plus{} \cos^{2}x}\ dx$

$ \equal{} \frac {\pi}{2}\int_{1}^{ \minus{} 1} \frac { \minus{} 1}{1 \plus{} z^{2}}\ dz$, where $ z \equal{} \cos x$

$ \equal{} \frac {\pi}{2}[\tan^{ \minus{} 1}z]_{ \minus{} 1}^{1}$

$ \equal{} \frac {\pi^{2}}{4}$.

$ \int_{\pi}^{2\pi}\frac {x\sin x}{1 \plus{} \cos^{2}x}\ dx \equal{} \int_{0}^{\pi}\frac {(y \plus{} \pi)( \minus{} \sin y)}{1 \plus{} \cos^{2}y}\ dy$

$ \equal{} \minus{} \int_{0}^{\pi} \frac {y\sin y}{1 \plus{} \cos^{2}y}\ dy \minus{} \pi \int_{0}^{\pi} \frac {\sin y}{1 \plus{} \cos^{2}y}\ dy$

$ \equal{} \minus{} \frac {\pi^{2}}{4} \minus{} \frac {\pi^{2}}{2}$.

Therefore, $ \int_{0}^{2\pi} \frac {x \sin x}{1 \plus{} \cos^{2}x}\ dx \equal{} \frac {\pi^{2}}{4} \minus{} \frac {\pi^{2}}{4} \minus{} \frac {\pi^{2}}{2} \equal{} \minus{} \frac {\pi^{2}}{2}$.


For given $u_0$,find the general form of $u_n$ if $u_n=a u_{n - 1}^2 + b u_{n - 1} + c$.

I have read ABOUT ab=2 and in other cases there was no example.


geometry
search
combinatorics unsolved
combinatorics

Given 18 points in the plane, no three collinear, so that they form 816 triangles. The sum of the area of the triangles is $ A$. 6 are colored red, 6 green and 6 blue. Show thaat the sum of the ares of the triangles whose vertices are the same color does not exceed $ A/4$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1657994991&t=3300

Pierre.


geometry circumcircle inradius parameterization calculus integration

Find necessary and sufficient conditions such that a triangle with integer side lengths $a, b, c$ has altitudes of rational length.

[hide] Rational area? Rational circumradius? Rational inradius? :wink:

Or maybe you wanted something in terms of $a, b, c$. [/hide]

No need to hide that question and have all of us think it's an answer.

An altitude has the property that it's length l satisfies sqrt(a^2-l^2)+sqrt(b^2-l^2)=c

Can someone do something with this?

If it's possible.  Otherwise, I think there is some [hide="terrible secret"] rational parameterization of $a, b, c$.  Maybe. [/hide]

Writing l=p/q we find

qc=sqrt(q^2a^2-p^2)+sqrt(q^2b^2-p^2)

so q^2a^2=k^2+p^2 and q^2b^2=l^2+p^2

and k+l=qc so -2p^2+2kl=(k+l)^2=0 (mod q^2) so p^2=kl (mod q^2/2)

Not much of a condition, since we don't know what $L$ is beforehand.

Altheman, that's exactly what I was looking for, thanks.

I was trying to get a condition. I couldn't get rid of the ps, qs, ks, and ls. However, note that Heronian triangles have integral area, not rational altitudes.

[quote="Bill The Maniac"]note that Heronian triangles have integral area, not rational altitudes.[/quote]

Rational area implies rational altitudes if the sides are rational because $A = \frac{1}{2}bh$.

I saw integral area because all triangles given were integral, which is a sufficient but not necessary condition.

So we now have it down to (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 is a square, or (a+b+c)(-a+b+c)(a-b+c)(a+b-c) is a square. This is still a pretty messy equation. Can anyone simplify it?

Probably not the best approach.  The one outlined in Altheman's link is good.  (And the reason it only gave triangles with integer areas is because it multiplied by the least common denominator to get the smallest integer representation of the "similarity class.")


ratio
real analysis
real analysis unsolved

Let $(a_n)_n$ be a sequence of positive irational numbers.
a) Prove that for every $n\in\mathbb N^*$, the binomial development $(1+a_n)^n$ admits a unique maximum term and determine its rank $r_n\in\{1,2,\ldots,n+1\}$.
b) We consider the sequences $x_n=a_n\sqrt n, n\in\mathbb N^*$ and $y_n=(1+a_n)^{r_n}, n\in\mathbb N^*$. Prove that $(x_n)_n$ is convergent if and only if the sequence $(y_n)_n$ is convergent.

[i]Eugen Paltanea[/i]

For a) we can use well known trick. Consider ration $\frac{a^{k+1}C_n^{k+1}}{a^kC_n^k}=\frac{a(n-k)}{(k+1)}$, so this ratio decreses and we are to determine the moment when it becomes less than 1, i.e. $k> n-\frac{n+1}{a+1}$. It means that $r_n\sim \frac{n+1}{a_n+1}$ :?
It is quite easy to derive b) from here...

Yeah Myth. Only that the official solution took like 3 pages, and they requested every single simple statement to be proved. That would be one of the reasons I got 2p on this one  :(


inequalities
induction
inequalities proposed

Let $ x_n=2^{2^n} +1 $, where $n\in\mathbb{N}$.Prove that the following inequality holds:

              $\frac{1}{x_1}+\frac{2}{x_2}+\frac{2^2}{x_3}+...+\frac{2^{n-1}}{x_n} < \frac{1}{3} $ ;)


[hide]
After clearing denominators, it's equivalent to
\[n!=\sum_{j=1}^{n}\left( j(-1)^{j+1}\binom{n}{j}\prod_{k\neq j}(x+k) \right)\]
If we can show that they are equal as polynomials, we'll be done, but this requires only showing that they are equal for $n$ different values of $x$, because the degree of the difference is clearly either zero or of degree at most $n-1$. Guess what, we can look at each $-j$ from $j=1$ to $j=n$. Here every term except for the $j$th in the summation disappears, so it remains to prove
\[j(-1)^{j+1}\binom{n}{j}\prod_{k\neq j}(k-j)=n!\]
But the product on the right can be simplified:
\[\prod_{k\neq j}(k-j)=\prod_{k>j}(k-j) \prod_{k<j}-(j-k)=(n-j)!\cdot (-1)^{j-1}(j-1)!\]
When multiplied by $j(-1)^{j+1}\cdot \frac{n!}{j!(n-j)!}$, it in fact equals $n!$.
[/hide]


How do you do fractions (using $\LaTeX$) such that they look like [img]http://i176.photobucket.com/albums/w169/stitch4eva/fraction.jpg[/img] instead of $\frac{1}{2}$?

[quote="Valentin Vornicu"]Hmm, you mean $1/2$? [code]$1/2$[/code][/quote]

No. I want the 1 to be smaller and above and the 2 to be the same size as the 1 but below.

Is there some other way that's not [code]$^1 / _2$[/code]

Well does $^{1}/_{2}$ look good enough? If so, you can make a command that does it for you.

In your document $\tfrac{1}{2}$ is the normal size so you don't need to do anything. However, if you are in displaymode or on the forum you can force it to display this rather than $\frac{1}{2}$ by using \tfrac instaed of \frac.


trigonometry
LaTeX
calculus
calculus computations

I am trying to prove something, but I don't see where it goed wrong
I enclosed the file

Thanks :oops:

No you are wrong
Use
$ \text{cos}x \minus{}\text{ cos}y \equal{} \minus{} 2\text{sin}(\frac {x \plus{} y}{2})\text{sin}(\frac {x \minus{} y}{2})$

[quote="Fys"]Can't I solve it using
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
and
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

Because I had to use this, but I don't know where I make the mistake

Thanks[/quote]
$ \text{cos}(a + b) = \text{cos}a \text{cos}b - \text{sin}a\tex{sin}{b}$
 $ \text{cos}(a - b) = \text{cos}a\text{cos}b + \text{sin}a \text{sin}b$

TT sphn thanks for the help, but can you also see where I made the mistake? :blush:

In the second line after "So I get" you lost the term $ - \cos nx$ from the previous line. In general, if you want to find a mistake the effect of which is that the final formula is false for $ x = 0$, plug $ x = 0$ into every formula in your derivation and look for the first wrong identity. Most likely, the problem will be there. By the way, you would get the answer sooner if you used $ \text{\LaTeX}$ in your post rather than attaching a Word file :)

Rather than using "\text{cos}" every time you want cosine, use \cos (and likewise for the sine): it looks better (the spacing around it is nicer) and is easier to write.

Oke I found the mistake,
thanks for help
And I really have to learn to work with latex

Thanks all


The instructions says

[quote]"In order to create a finished LaTeX document, we need to compile the source file. First, make sure that the bar near the top of the TeXnicCenter window reads "LaTeX => PDF"; if it doesn't, change it so that it does."[/quote]

What does the TeXnicCenter window look like and where is the bar?

I think you mean http://www.artofproblemsolving.com/Forum/viewtopic.php?p=768853#768853


Arnold must water his cacus plants every 9 days and he must feed them every 60 days. He watered and fed them last Tuesday. What day of the week will it be the next time he water and feeds them on the same day?

$ [9,60]\equal{}180$, and $ 180\equiv5\pmod7$, so it will be Sunday.