This problem features the transposition of the carbonyl group of an ester (with conversion into a ketone): show how the following could be made,

$ C_6H_5\minus{}CH_2CH_2CO_2CH_2CH_3 \longrightarrow C_6H_5\minus{}CH_2COCH_3$.

The processes are
1. add conc. H+ to convert COOCH2CH3 into COOH
2. Perform Hell Volhard Zelinsky reaction with P/Br2 to get alpha bromaination.
3. Use P/HI to convert COOH to CH3
4. Use aq.NaOH to convert Br into OH
5. Use Cu/573K to oxidise CHOH to CO
 :P  :)

[quote="Madness"]1. add conc. H+ to convert COOCH2CH3 into COOH[/quote]

If you are going to do that then you will for sure set up an intramolecular Friedel-Crafts acylation. You have to propose another synthesis, because that will not work. Some more remarks:

[quote="Madness"]2. Perform - Volhard Zelinsky reaction with P/Br2 to get alpha bromaination.[/quote]

The reaction name is Hell-Volhard-Zelinski, or HVZ reaction.

[quote="Madness"]3. Use P/HI to convert COOH to CH3[/quote]

I am not aware of that functional group transformation.

[quote="Madness"]4. Use aq.NaOH to convert Br into OH[/quote]

Elimination would be a very competitive reaction, since there is a benzilic proton available.

" Elimination would be a very competitive reaction, since there is a benzilic proton available. "  :huh:

But I'm using aq. NaOH and not alc. NaOH so OH- prefers to act as a nucleophile
  :maybe:

I thought since you are using $ NaOH$ in water (polar protic solvent) and your alky halide is secondary (also benzylic) then you will have competitive $ S_N1$ and $ E1$ reactions occurring. Is that way Carcul meant?

Quick question for all of you: do you get the problems and work the solutions from knowledge of just Carey? or is there another more advanced orgo book you've all been studying?

These particular problems about transposition were adapted from an article of 1982 (and they don't appear in any Organic Chemistry book I know, since this technique is very specialized and so advanced), and so they are supposed to involve quite advanced knowledge (in order for the transformations to be made in very few steps). However, they are also very good for a student of Organic Chemistry to practice the knowledge he gets from introdutory books like Carey, Solomons, Morrison, and so on (naturally, in this case such a student is expected to provide a synthesis that will involve considerably more steps compared to the "advanced stuff").

[quote="OVB"]or is there another more advanced orgo book you've all been studying[/quote]

There are several books on Advanced Org. Chem., but they should not be studied before the material covered in the books I mentioned above is mastered - anyway, the only persons that need to study advanced material are those how are taking a chemistry course in university and want to specialize in organic chemistry.

Sorry Its not P/HI but its rather P/I2 which is used for converting COOH to CH3  :blush:

[quote="Madness"] but its rather P/I2 which is used for converting COOH to CH3[/quote]

I still have never heard that conversion. Can you provide a reference and a mechanism?

red P/I2 --> PI3
PI3 + RCOOH--> RCOI
RCOI ON REDUCTION WILL GIVE YOU ALKANE AS IT IS VERY UNSTABLE AND MOST REACTIVE WITH RED PHOSPHOROUS :)

[quote="Chemrock"]PI3 + RCOOH--> RCOI[/quote]

The reaction of carboxylic acids with phosphorus and molecular chlorine or bromine (or with the phosphorus trihalide) is known as the HVZ reaction, and the product is an alpha-haloacid. However, this reaction doesn't work with iodine (or fluorine). In the mechanism of this reaction there is in fact the formation of an acyl halide as an intermediate, but since there is no reaction with iodine, RCOI doesn't form.

[quote="Chemrock"]RCOI ON REDUCTION WILL GIVE YOU ALKANE[/quote]

In the first place, RCOI would be to unstable to exist. Second, I don't see what reagent can reduce an acyl halide RCOX to the alkane RCH3.

use rosenmund's reduction(H2/BaSO4) followed by clemmenson or wolff kishner red

That's not the point!!!! We were discussing about how RCOOH ---> RCH3 works using $ \ce{P}$/$ \ce{I2}$
Am sure we all know Clemensen and Wolff Kishner
Also Carcul disagrees about formation of $ \ce{RCOI}$

well that's not using $ \ce{I2}$ but using $ \ce{HI}$ Red Phosphorous is used to drive away $ \ce{I2}$ produced

YEP. YOU'RE RITE P/HI IS USED FOR REDUCING RCOOH TO RCH3 NOT P/I2. NOT SURE ABT THE MECH THO

[quote="Madness"]Am sure we all know Clemensen and Wolff Kishner[/quote]

But that doesn't work with carboxilic acids.

[quote="VARUNARASU"]YOU'RE RITE P/HI IS USED FOR REDUCING RCOOH TO RCH3[/quote]

Again I have never heard that reaction.

But Varun has converted the acid into aldehyde and only then used it  :huh:

yes he used Rosemund's on acyl halide then used Clemmensen

THERE IS SOMETHING WRONG WITH THE MECH OF P/HI PROPOSED BY CHEMROCK AS IT CAN REDUCE ALDEHYDES AND KETONES.

No it doesn't .
If it does wat do u think is its mechanism?

IF I KNEW DAT I WOULDNT BE ASKING U GUYS WUD I? BUT IT DOES REDUCE ALDEHYDES AND KETONES. AT LEAST DAT IS WHAT I SAW IN ONE OF THOSE LOUSY IIT CHEMISTRY BOOKS WHICH I KNOW ARE NOT VERY RELIABLE

yes i know they are not even i saw one where that's given

Reduction of aldehydes and ketones with P/HI? Reduction to what product and with what mechanism? I don't think so.

One thing is for sure: that reaction, if it really works, is certainly not a method used in synthesis.

[quote="VARUNARASU"]reduction to alkanes. dont know the mechanism[/quote]

Wait a minute, am I reading this right?

If we want to reduce Aldehydes/Ketones to alkanes, Wolff-Kishner Reduction should do the trick.

i know dat. we were discussing abt how p/hi reduces aldehydes, ketones and acids to alkanes
Carcul said
[quote]
One thing is for sure: that reaction, if it really works, is certainly not a method used in synthesis
[/quote]
how can you for sure dat dis method is not used during synthesis. is there any other method for reducing acids to alkanes in a single step?

[quote="VARUNARASU"]how can you for sure dat dis method is not used during synthesis.[/quote]

Because I am aware of the common synthetic methods used.

[quote="VARUNARASU"]is there any other method for reducing acids to alkanes in a single step?[/quote]

I have never seen a simple and direct method described for the reduction of acids to alkanes. There however, several indirect methods (which may involve or not a decarboxilation). As a matter of fact, the indirect methods would be much more usefull to you (from an educational point of view) than that reaction that you (or your book) claim to exist.


What is the least positive integer divisible by the four smallest odd, positive integers?

least integer=product of the four numbers=1*3*5*7=105


Prove that the product of the 99 numbers $ \frac{k^3\minus{}1}{k^3\plus{}1},k\equal{}2,3,\ldots,100$ is greater than $ 2/3$.


[quote="thanhnam2902"]Solve this equation with $ x \in R$:

$ \sqrt {3x \plus{} 1} \equal{} \minus{} 4x^2 \plus{} 13x \minus{} 5$[/quote]

Squaring gives the equation $ 16x^4 \minus{} 104x^3 \plus{} 209x^2 \minus{} 133x \plus{} 24 \equal{} 0$

Which may be written $ (4x^2 \minus{} 15x \plus{} 8)(4x^2 \minus{} 11x \plus{} 3) \equal{} 0$ and gives 4 roots : $ \frac {15 \minus{} \sqrt {97}}8,\frac {15 \plus{} \sqrt {97}}8,$ $ \frac {11 \minus{} \sqrt {73}}8,\frac {11 \plus{} \sqrt {73}}8$

from which only two give $ \minus{} 4x^2 \plus{} 13x \minus{} 5\ge 0$ and so are solutions of the original equation : $ \boxed{\frac {15 \minus{} \sqrt {97}}8\text{ and }\frac {11 \plus{} \sqrt {73}}8}$


function
real analysis
real analysis unsolved

$f: \mathbb{R}\to \mathbb{R}$ a continuous function.
$a$ is a local maximum of $f$ if there exists some $e>0$ satisfying:
if $x\in ]a-e;a+e[$ then $f(x) \leq f(a)$.

Let $E$ be the set of local maxima. Show that $m(f(E))=0$ where $m$ is the Lesbegue measure.

Strange.. I'm pretty sure $f(E)$ is countable.

For every $a\in E$, there is an open interval $I_{f(a)}$ with rational endpoints such that $f(a)$ is the maximal value of $f$ on $I_a$. Clearly, $I_{f(a)}\ne I_{f(b)}$ if $a,b\in E,\ f(a)\ne f(b)$, so there is an injection from $f(E)$ to the set of intervals with rational endpoints.

Anything wrong with that? :?

P.S.

As far as I know, Sard's Lemma is something completely different. It states that given a function $f: \mathbb R^n\to\mathbb R^m$ which is nice enough (smooth, for example; I don't know what the conditions are, exactly, but I'm sure it works for smooth, i.e. $C^\infty$ functions), and we call $x\in\mathbb R^n$ a critical point of $f$ if the rank of $Df(x)$ is $<m$, then $\mu(f(C))=0$, where $C$ is the set of critical points of $f$ and $\mu$ is the Lebesgue measure in $\mathbb R^m$.


Write the product $ \frac{35}{14} \bullet \frac{3}{125}$ in scientific notation.


calculus
integration
number theory unsolved
number theory

Solve this equation:
$ \ p^x\minus{}y^p \equal{}1$
where $ \ p$ $ \epsilon P$ and $ \ a,b$ $ \epsilon N$

[hide]
Well, http://en.wikipedia.org/wiki/Mihăilescu's_theorem trivializes it... without using that, we can:

We use the lemma at http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf

Two cases:

1. p is not 2. Then, we have $ p^x \equal{} y^p\plus{}1.$ Since p is odd, y+1 divides the right side, so y+1 divides $ p^x.$ y+1 therefore must be a power of p. We note that $ p^x \equal{} y^p\plus{}1 \equal{} y^p \minus{} (\minus{}1)^p.$ We can now use our lemma to get that if $ p^k$ is the greatest power of p that divides $ y\plus{}1 \equal{} y\minus{}(\minus{}1)$, then the greatest power of p that divides $ y^p \plus{} 1 \equal{} y^p \minus{} (\minus{}1)^p$ is $ p^{k\plus{}1},$ which implies that k+1=x and k = x-1. Therefore, y+1 is a multiple of $ p^x\minus{}1.$ Since p is a prime, it is greater than or equal to 2, so $ p^x \equal{} y^p\plus{}1 > y^2 \ge (p^{x\minus{}1}\minus{}1)^2.$ Also, $ p^{x\minus{}1}\minus{}1 \ge p^{x\minus{}2},$ so $ p^x > p^{2(x\minus{}2)},$ so x is strictly greater than 2(x-2), which implies that x is 1,2, or 3. If its 1, then we have p = $ y^p\plus{}1.$ However, if y does not equal 1, then $ y^p$ is greater than p, since p is at least 2. If y = 1, then we have that p = 2, a contradiction. If x = 2, then $ p^2 \equal{} y^p\plus{}1.$ Once again, since p is at least 3, the right side is greater than or equal to the left side unless y = 2 or y = 1. y can't equal 1, or else we have $ p^2 \equal{} 2,$ impossible, so  y = 2, and we get $ p^2 \equal{} 2^p\plus{}1,$ where the right side is greater unless p = 3. So x = 2, p = 3, y = 2 is a solution. If x = 3, then we have $ p^3 \equal{} y^p \plus{} 1.$ Once again, we get that  y = 1 or 2, and y = 1 is impossible, so  y = 2. Therefore, $ p^3 \equal{} 2^p \plus{} 1.$ Note that mod 7, $ p^3$ is either 1 or -1 or 0. However, $ 2^p$ can only be 2,4, or 1 mod 7, so $ 2^p\plus{}1$ is 3, 5, or 2 mod 7, a contradiction which implies that there are no solutions in this case.

2. p = 2. Then, $ 2^x \minus{} y^2 \equal{} 1.$ If x is at least 2, we can take mod 4 to get a contradiction. Therefore, x = 1, and y = 1. p=2,x=1,y=1 is another solution.[/hide]


A waiter served $ \$650.00$ worth of dinners in a busy restaurant.
She received tips in the amount of $ \$75.00$. How much less did the waiter receive in tips than she would have if she received her
expected rate of $ 15\%$ of the cost of the dinners?


Let $ a, b, c$ be positive integers. Show that $ \frac{1}{2a} \plus{} \frac{1}{2b} \plus{} \frac{1}{2c} \geq \frac{1}{a \plus{} b} \plus{} \frac{1}{a \plus{} c} \plus{} \frac{1}{b \plus{} c}$

[hide="Solution"]By AM-HM:
$ \frac{a\plus{}b}{2} \geq \frac{2}{\frac{1}{a}\plus{}\frac{1}{b}} \implies a\plus{}b \geq \frac{4}{\frac{1}{a}\plus{}\frac{1}{b}} \equal{} \frac{1}{\frac{1}{4a}\plus{}\frac{1}{4b}} \implies \frac{1}{4a}\plus{}\frac{1}{4b} \geq \frac{1}{a\plus{}b}$

Hence,
$ \frac{1}{4a}\plus{}\frac{1}{4b} \minus{} \frac{1}{a\plus{}b} \geq 0  \implies \sum_{cyc} \frac{1}{4a}\plus{}\frac{1}{4b} \minus{} \frac{1}{a\plus{}b} \geq 0$

Which expands to the original inequality.[/hide]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83115&search_id=821052012]Take a look here![/url]

Of course, the inequality is true in the positive reals.

[hide="Overkill"] $ \frac {1}{x}$ is convex and $ (2a, 2b, 2c) \succ (a \plus{} b, b \plus{} c, c \plus{} a)$.  The conclusion follows by Karamata. [/hide]
[hide="As a sum of squares"] $ \frac {1}{2a} \plus{} \frac {1}{2b} \plus{} \frac {1}{2c} \ge \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \Leftrightarrow$
$ \frac { (a \minus{} b)^2}{4ab(a \plus{} b)} \plus{} \frac {(b \minus{} c)^2}{4bc(b \plus{} c)} \plus{} \frac {(c \minus{} a)^2}{4ca(c \plus{} a)} \ge 0$

This is based off of The Zuton Force's solution, which is straightforward. [/hide]

[quote="t0rajir0u"]
[hide="As a sum of squares"] $ \frac {1}{2a} \plus{} \frac {1}{2b} \plus{} \frac {1}{2c} \ge \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \Leftrightarrow$
$ \frac { (a \minus{} b)^2}{4ab(a \plus{} b)} \plus{} \frac {(b \minus{} c)^2}{4bc(b \plus{} c)} \plus{} \frac {(c \minus{} a)^2}{4ca(c \plus{} a)} \ge 0$

This is based off of The Zuton Force's solution, which is straightforward. [/hide][/quote]

I also solved as that, t0rajir0u. :wink:


so, i am wondering if the folowing inequality holds:

$\sum_{cyc}{x^{a_{1}}y^{a_{2}}z^{a_{3}}}\geq  \sum_{cyc}{x^{b_{1}}y^{b_{2}}z^{b_{3}}}$, where $(a_{1},a_{2},a_{3})\succ(b_{1},b_{2},b_{3})$ , of course if this is true my following questions is if it is true for the general case.

[quote="pohoatza"]it may have been already discussed[/quote]
Yes, it has been: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=103761[/url].


function
real analysis
real analysis solved

Is there a function f from [0,1] in the same set such that for all x the set $ x,f(x),f(f(x)),....$ is dense in [0,1]? What about a continuous one?

A continuous one clearly doesn't exist, since a continuous function from [$0,1$] to [$0,1$] has a fixed point. The other question seems to be very nice though.

Grobber, the second question was a joke meant to encourage the solver.  ;)

You succeeded in that: I feel very refreshed and encouraged :D:D.

This uses the Axiom of Choice, but I'm sure you won't mind :).

Partition [$0,1$] in classes $A_i,\ i\in I$ ($I$ is an uncountable set, of course) s.t. $x,y\ne 0$ are in the same class iff $\frac xy\in\mathbb Q$ (put $0$ in the same class as $1$). Each class is countable, so we can find an enumeration of the set $A_i$: $x_i(0),x_i(1),x_i(2),\ldots$. Now simply take $f(x_i(k))=x_i(k+1),\ \forall k\ge 0,\ \forall i\in I$.


A brick mantel over a fireplace consists of rectangles as shown. What is the total number of rectangles in the pattern?

[asy]for(int i=0; i<10; ++i) {
draw((i,0)--(i,4),linewidth(0.7));
}
draw((0,0)--(10,0)--(10,4)--(0,4)--cycle,linewidth(0.7));
draw((0,3)--(10,3),linewidth(0.7));[/asy]

Basically, pick two different points on the left side and pick two different points on the top. This effectively defines a rectangle.

So, there are 3c2*11c2=55*3=165.

if each small brick is an "a" and each large a is a "b" it would look like this:

aaaaaaaaaa
bbbbbbbbbb

there are ten all "a" rectangles with the far left "a" brick.
there are nine all "a" rectangles with the second from the left "a" brick.
              eight
               and so on
this makes 55 rectangles.
repeat this process with the all "b" bricks
and pairs of vertical a and b bricks:  a
                                                     b
this gives you 165.

[quote="AceOfDiamonds"][quote="generalmotors"]But some of them dont make rectangles, I'm kinda confused?[/quote]
call the 4 points $(0,y_1), (0,y_2), (x_1,a),(x_2,a)$ ($y_1\neq y_2, x_1\neq x_2$)
the rectangle these four points define has vertices $(x_1,y_1),(x_1,y_2),(x_2,y_1),(x_2,y_2)$[/quote]

You confused me even more.


The second and ninth terms of an arithmetic sequence are 2 and 30, respectively. What is the fifteth term?

After 7  numbers, the difference was 28, so the common difference is 4.
2+4(13) for the 15th term
$ 54$

There's a typo in the problem. "fifteth" could either be "fifteenth" or "fiftieth." It looks more like "fiftieth" ;)

$ 2\plus{}4(48)\equal{}2\plus{}192\equal{}\boxed{194}$ if it's the $ 50^{\text{th}}$ term.


The bound is very weak - my bound is 38. Split into two cases, squares with an even number of digits in them ($a_1^2$, $a_2^2$, ....), squares with an odd number of digits in them ($b_1^2$, $b_2^2$, ...). Let $l(x)$ be the number of digits in $x$.

Consider the numbers $a_i^2$, $a_{i+1}^2$, and let say the latter number has $2k$ digits more than former. Then, $a_{i+1}^2 > 10^{2k}a_i^2$, which implies that $a_{i+1}^2 \geq (10^{k}a_i + 1)^2 = 10^{2k}a_i^2 + 2a_i10^k + 1$ Now, the first term will given a "start" of $a_i^2$ iff the latter terms do not interfere. Since there are $2k$ zeroes after the first term, and the second term has $k + l(a_i)$ digits in all. Thus, $2k \geq k + l(a_i)$, which implies $k + l(a_i) \geq 2l(a_i)$, which implies $l(a_{i+1}) \geq 2l(a_i)$. Note that this argument also works for sequence $b$.

Thus, $l(a_i), l(b_i) \geq 2^{i-1}$. Since $2^{19} > 500000$ thus a square of a number with that many digits will have more than a million digits. Therefore, $i$ can only take values from $1$ to $19$, and so there are at most $38$ squares in all.

I am sure this can be sharpened still, but I feel it is more bruteforce from here.

[quote="Remike"]A person have written a non-zero digit (1, 2, 3, 4, 5, 6, 7, 8 or 9) on a blackboard. Then he wrote another non-zero digit at the right (for example, if the first digit was 7 and the second was 4 then it is written 74 on the blackboard). Then he continued such an operation until there was $10^6$ non-zero digits on the blackboard. Prove that squares of integers occured at most 100 times during the process.[/quote]
Sorry, I can't understand clearly your problem, in my understanding I think is wrong: I suppose you are saying that while u r writing some the chain $\overline{a_1a_2\cdots a_m}$ in your complete number is a square? for every $m$ in the number of digits of your oomplete number?  If is in this way what happend if I consider the number $55\cdots 5$ such times as you want.

Best Regards and sorry for my question.

RX

He said that the maximal number of squares occuring in $\overline{a_1a_2\cdots a_m}$ is $100$.

[quote="RobertuX"]Sorry, I can't understand clearly your problem, in my understanding I think is wrong: I suppose you are saying that while u r writing some the chain $\overline{a_1a_2\cdots a_m}$ in your complete number is a square? for every $m$ in the number of digits of your oomplete number?  If is in this way what happend if I consider the number $55\cdots 5$ such times as you want.

Best Regards and sorry for my question.

RX[/quote]
Sorry, but I can't understand clearly your question. What is "complete number"? What do you mean by the words "for every $m$ in the number of digits of your oomplete number?"? There is no predicate in your sentence, and I can't understand the sense of it. And finally, I suppose that the number $55\ldots 5$ can't be a square of integer, since it is divisible by 5, but it isn't divisible by 25. So, this number can't be considered as a counterexample to the claim of the problem.


Let $ a;b;c$ be positive real numbers satisfying $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}3$
Prove that:
$ \frac{ab}{c}\plus{}\frac{ac}{b}\plus{}\frac{bc}{a}\geq 3$

[quote="chien than"]Let $ a;b;c$ be positive real numbers satisfying $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$
Prove that:
$ \frac{ab}{c}\plus{}\frac{ac}{b}\plus{}\frac{bc}{a}\geq 3$[/quote]
We have
$ (\frac{ab}{c}\plus{}\frac{ac}{b}\plus{}\frac{bc}{a})^{2}\geq 3(a^{2}\plus{}b^{2}\plus{}c^{2})\equal{}9$
so
$ \frac{ab}{c}\plus{}\frac{ac}{b}\plus{}\frac{bc}{a}\ge 3$
 :blush:


a) Suppose that $A$ is a countable set of real numbers such that whenever $x<y$ are in $A$ there are $u<x<v<y<w$ all in $A$. Prove that one can find a strictly increasing bijection between $A$ and the set of rationals.
   b) Deduce that there exists a continuous bijective function from Q and Q (set of rationals) whose inverse is not continuous.

(a) See post #5 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Sierpinski&t=48347]this topic[/url].

(b) Let $(a_n)_{n\ge 1}$ be a strictly increasing sequence of rationals converging to an irational, and let $(b_n)_{n\ge 1}$ be a strictky increasing sequence of rationals converging to a rational $b$. Also, for each $i$, let $I_i$ be an open interval with irrational endpoints containing $a_i$ and no other $a_n,\ n\ne i$, and let $J_i$ be an open interval with irrational endpoints containing $b_i$ and no other $b_n,\ n\ne i$, and not containing $b$.

We can find an order isomorphism between $I_n$ and $J_n$ for every $n\ge 1$ (according to (a)), and another order isomorphism between $\mathbb Q\setminus\left(\bigcup_{n\ge 1} I_n\right)$ and $\mathbb Q\setminus\left(\bigcup_{n\ge 1} J_n\right)$. Together, these make up the desired bijection of $\mathbb Q$ onto $\mathbb Q$ (it’s easy to see that it’s continuous, but, on the other hand, it sends the non-convergent sequence $(a_n)_n$ onto the convergent sequence $(b_n)_n$).

I hope I didn't miss anything.


How many liters of water must be added to 40 liters of a $ 40\%$ salt solution in order to produce a $ 25\%$ salt solution?

If 40% of those 40 liters is salt, that means 16 liters are salt and 24 are water. Let the number of liters of water to be added be $ l$.

\[ \frac{24 \plus{} l}{40 \plus{} l} \equal{} 0.25(40 \plus{} l) \]

Solving yields $ \boxed{24}$.

Another solution: since the 40 liter solution is 40% salt, that means there is 16 liters of salt.  This must make up 25% of the new solution, so the new solution has 64 liters.  $ 64\minus{}40\equal{}\boxed{24}$.


Prove that
$ \frac{a}{b\plus{}c\plus{}1}\plus{}\frac{b}{c\plus{}a\plus{}1}\plus{}\frac{c}{a\plus{}b\plus{}1}\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c)\le 1$    for $ a,b,c\in [0,1]$

The inequality is convex in each of the variable, so it assumes the maximum for (x,y,z)=(0,0,0),(1,0,0),(1,1,0) or (1,1,1) according to some theorem(maybe it is called weistrass theorem)...


I am currently in Calc 1 and have been learning limits... my algebra is a little rusty so could someone help me find the limits of these two problems?

limit of (-2x-4)/(x^3+2x^2)  as x -> -2
and
limit of (5y^3+8y^2)/(3y^4-16y^2)   as y -> 0

I realize these are probably pretty easy for many of you, but I am stuck on how to manipulate the problems algebraically... any help would be much appreciated...  thanks!!


how many 'x's do they satisfy a^(a^x)=x (0<a<1)


geometry
geometric transformation
reflection

2 cities $ A$ and $ B$  are in the same side of a line. Where should be a bus  station $ C$ in this line such the sum $ AC\plus{}BC$ can be as small as posible.

[hide="Solution"]Reflect point $ B$ across the line, to the point $ B'$. The intersection of the line and $ AB'$ is where $ C$ should be.[/hide]

[quote="Elio (n)"]How can you prove your reponse of the problem?[/quote]He used the fact that in plane geometry, the shortest distance between two points is the line passing through them...

[quote="Elio (n)"]What about this :
$ M$ is the point in the midle of $ AB$. We draw the perpendicular  from the the line to $ M$.Where the perpendicular cut the line  there is $ C$.[/quote]
Um, what???

The Elio'solution is false!
Prove the 123456789'solution:
If D on the line and $ D\neq C$ then DA+DB > BB' = CA+CB.


find all the value natural of $ x$ and $ y$ such that
$ (2+3\sqrt{2008})^{x}=(5+2\sqrt{2008})^{y}$

[quote="Qruni"]WLOG x>y, we can write it in the form x=y+k, for a positive integer
k.
(2+3.sqrt(2008))/(5+2.sqrt(2008))^y=(5+2.sqrt(2008))^k
simplify the LHS[/quote]
can u plz show me how do u simplify the LHS and what do u get after doing it ,

No such x and y exist.
Let in contrary the equation $ (2+6\sqrt{502})^{x}=(5+4\sqrt{502})^{y}$
is satisfied by natural x and y.
Then the equation $ (2-6\sqrt{502})^{x}=(5-4\sqrt{502})^{y}$ is satisfied too.
Multiplication  these 2 equations gives $ (-18068)^{x}=(-8007)^{y}$
which is impossible because RHS has factor 3 whereas the LHS hasn't.


ARML AMC AIME HMMT Princeton college email

Tryouts are in Canton (though I hear it may move to Lexington), sometime in February (I think). You'll hear about it at a future MML meet probably, but if you don't you can just e-mail the coaches about it in like January (Martin Badoian from Canton runs it, badoianm AT hotmail DOT com).

Oh and the above was about Eastern Mass, I don't know anything about Western.

For Western Mass, the next practice is at Worcester Academy on November 15.  You can just show up, or contact Mr. Yanco.

how good do I have to be to make it? I dont want to waste my time...

[quote="cosinator"]If you can qualify for AIME you're probably good enough for A Team.  If not, there's still B and C Team.[/quote]  I'm not sure AIME qualification is the standard; last year Worcester Academy had fifteen AIME qualifiers, but I wouldn't have felt comfortable with a WA ARML team.  But the point to be made -- and I think you made it -- is that we try to accommodate any central or western Massachusetts student who is talented and interested in math.  AIME qualifiers probably have a good chance at our top team, while younger students who have ability but not as much knowledge get specific classes to remedy the lack of knowledge, plus a chance at the B or C team.

Since Mr. Dougherty stopped subsidizing a Rhode Island ARML team (I believe your last team was 2005), a small number of RI students have participated on the Western Massachusetts team.  Last year Karan Takhar from Wheeler School drove up.

This is my first year with the reins of the team (I coached from 2001-06 but haven't for two years).  I'm trying to move the focus of the team away from ARML specifically, towards a general goal of providing more opportunity to students in WOCOMAL (our coaches are all from WOCOMAL).  Opportunity for the best students means additional competitions:  HMMT February, PUMaC, ARML.  Opportunity for weaker students means the classes mentioned above, plus the same opportunity for additional competitions.

Where we just changed coaches, it'll be a practice or two before I figure everything out.  But the practice schedule is set:

November 15:  ARML Power Contest 1
December 13
January 10:  lecture at Holy Cross (this date is tentative)
February 14:  ARML Power Contest 2
March 7:  ARML Local
April 11
May 9:  NYSML (unofficially)

All practices run from 9 AM - 12:30 PM at Worcester Academy (except January 10).  The three competitions that our team will be attending are the Princeton University Math Competition on January 31, 2009; HMMT on February 21, 2009; and ARML on May 28-31, 2009.

If you want to be on our email list, send me a PM with your email address.  To answer your original question, no, there is no guarantee you'll make the team, but if you attend regularly and have ability, you have a shot.

Thanks.

Rick Yanco

What about the Eastern Mass team, do they have practices and lectures etc?


[quote="moldovan"]Let $ ABCDE$ be a regular pentagon of side length $ 1$. Let $ F$ be the midpoint of $ AB$ and let $ G$ and $ H$ be the points on sides $ CD$ and $ DE$ respectively $ \angle GFD \equal{} \angle HFD \equal{} 30^{\circ}$. Show that the triangle $ GFH$ is equilateral. A square of side $ a$ is inscribed in $ \triangle GFH$ with one side of the square along $ GH$. Prove that:

$ FG \equal{} t \equal{} \frac {2 \cos 18^{\circ} \cos^2 36^{\circ}}{\cos 6^{\circ}}$ and $ a \equal{} \frac {t \sqrt {3}}{2 \plus{} \sqrt {3}}$.[/quote]
hi all
here is my solution  :wink:
 [url=http://www.servimg.com/image_preview.php?i=318&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk19.jpg[/img][/url]
[url=http://www.servimg.com/image_preview.php?i=319&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk12.gif[/img][/url]
 :cleaning:  :cleaning:
mr mohamed : almonkez


Prove that $\prod_{n=2}^\infty \frac{n^3-1}{n^3+1} = \frac{2}{3}$. [i] This is also asked [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73970]here[/url], but it's somewhat buried there.[/i]

It is enough to notice that when we multiply the consequent terms $\frac{n^3-1}{n^3+1}\frac{(n+1)^3-1}{(n+1)^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\frac{(n+1-1)(n^2+3n+3)}{(n+2)(n^2+n+1)}$ the multipliers $n^2+n+1$ in the first numerator and the second denominator cancel out, so if we take the $N$'th partial product we must have
$P_{N}=\frac{1}{3}(\frac{1}{3} \frac{2}{4} \frac{3}{5} ... \frac{N-1}{N+1} \frac{N(N^2+3N+1)}{N+2})=\frac{2}{3}\frac{N^2+3N+1}{(N+1)(N+2)} \to \frac{2}{3}$ when $N \to \infty$


inequalities
calculus
derivative
parameterization
inequalities solved
4-variable inequality
asymmetric inequality

Prove that for any four nonnegative reals $ a$, $ b$, $ c$, $ d$, the following inequality holds:
\[\left(ab\right)^{\frac13}\plus{}\left(cd\right)^{\frac13}\leq\left[\left(a\plus{}c\plus{}b\right)\left(a\plus{}c\plus{}d\right)\right]^{\frac13}\]

I mean, it is so simple, that should be moved to beginners corner. It was proposed at 239 Lyceum traditional olympiad by myself.

I have spent a lot of hours on this problem and I can say that, for me, it is very hard to solve and not very easy.
I cannot understand if there is an original and simple trick to solve it or if it can be simply solved by means of usual inequalities (AM-GM, Cauchy,Holder,Minkwosky, Power Mean,.....).

Very interesting and very subtle inequality!!!!

It is true I am a bad solver of inequality but in my opinion I don't think it is a very easy inequality.
However the solution for this very nice inequality is very ingenious.

Thank you very much Fedor Petrov

\[ \frac {ab}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)} \equal{} \left(\frac {a}{a \plus{} c}\right)\left(\frac {a \plus{} c}{a \plus{} c \plus{} d}\right)\left(\frac {b}{a \plus{} c \plus{} b}\right)\]
And analogously,
\[\frac {cd}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)} \equal{} \left(\frac {c}{a \plus{} c}\right)\left(\frac {a \plus{} c}{a \plus{} c \plus{} b}\right)\left(\frac {d}{a \plus{} c \plus{} d}\right)\]
So by AM-GM,
\[3\left(\frac {ab}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)}\right)^{\frac {1}{3}}\leq \frac {a}{a \plus{} c} \plus{} \frac {a \plus{} c}{a \plus{} c \plus{} d} \plus{} \frac {b}{a \plus{} c \plus{} b}\]
And
\[3\left(\frac {cd}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)}\right)^{\frac {1}{3}}\leq \frac {c}{a \plus{} c} \plus{} \frac {a \plus{} c}{a \plus{} c \plus{} b} \plus{} \frac {d}{a \plus{} c \plus{} d}\]
Summing up the last two inequalities gives the desired result.

[quote]But the hardest mathematical competition for high school students is surely an Open Olympiad of 239 Saint-Petersburg school. It is usual that total 4 of 8 problems are solved by all the contestants together, and the maximum result is 3 (and it is also usual that a half a year after the contestants get 2-3 gold medals at imo, so they are not weak.)[/quote]

Fedor, that's what you said in some other thread, so I was wondering whether you really meant it when you said that this was really simple :D .  Anyway I got stuck on the inequality too.  Thanks for providing the proof.

Yes, it was not solved on olympiad, but it was proposed to 8-9 grades (the last grade is 11-th), because those older students could possible kill it by some derivatives (in what I, however, doubt).

I may also show, how this solution may be deduced (whatever it means). It concerns an important idea of quasiliearisation.

Let me start with example.

We know that 2ab<=a^2+b^2, of course. Introduce a parameter and get the following: 2ab<=t*a^2+b^2/t. Then read the last inequality from the other point: for any positive a, b there exist positive t such then 2ab=t*a^2+b^2/t (t=b/a). Or, we may write: 2ab=min(t*a^2+b^2/t).

How may this observation help? Put a=sum (x_i^2), b=sum (y_i)^2.
Then for appropriate t we have:
2ab=ta^2+b^2/t=sum (t*x_i^2+y_i^2/t)>=sum (2*x_i*y_i). So, we get CBS inequality. A lot of other inequalities also may be proved by this idea, and the problem of topic is just a little more subtle example.

We have 3(AB)^{1/3}<=Ax+By+1/(xy). And for any positive A and B there exist appropriate x and y, for which equality holds
(x=(AB)^{1/3}/A, y=(AB)^{1/3}/B).

Let A=(a+c+b), B=(a+c+d) in terms of initial problem. For some positive x, y we have
3(AB)^{1/3}=Ax+By+1/(xy)=(a+c+b)*x+(a+c+d)*y+1/(xy)=
(a+c+b)*x+(a+c+d)*y+1/(x*(x+y))+1/(y*(x+y))=
[a(x+y)+b*x+1/(x*(x+y))]+[c(x+y)+d*y+1/(y*(x+y))]>=
3(ab)^{1/3}+3*(cd)^{1/3} by AM-GM.

Very nice and interesting technique. To be sincerely, I also couldn't prove this inequality. Maybe it's not that simple. The solution is simple, but I doubt it can be found easily.

[quote="Fedor Petrov"]How may this observation help? Put $ a \equal{} \sum (x_i^2), b \equal{} \sum (y_i)^2$.
Then for appropriate $ t$ we have:
$ 2ab \equal{} ta^2 \plus{} \frac {b^2}{t} \equal{} \sum (tx_i^2 \plus{} \frac {y_i^2}{t})\geq \sum (2x_iy_i)$. So, we get CBS inequality.[/quote]
Wouldn't it be

[quote="Fedor Petrov(little edit by me)"]How may this observation help? Put $ a \equal{} \sum (x_i^2), b \equal{} \sum (y_i^2)$.
Then for appropriate $ t$ we have:
$ 2\sqrt {a}\sqrt {b} \equal{} ta \plus{} \frac {b}{t} \equal{} \sum (tx_i^2 \plus{} \frac {y_i^2}{t})\geq \sum (2x_iy_i)$. So, we get CBS inequality.[/quote]
 :?:
To understand the context of this beautiful idea, I latexified the posts of Fedor Petrov, so I'm posting it here for others.

[quote="Fedor Petrov"]Well,
$ \frac {ab}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)} \equal{}$
$ (\frac {a}{a \plus{} c})(\frac {a \plus{} c}{a \plus{} c \plus{} d})(\frac {b}{a \plus{} c \plus{} b})$
And, analogously,
$ \frac {cd}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)} \equal{}$
$ (\frac {c}{a \plus{} c})(\frac {a \plus{} c}{a \plus{} c \plus{} b})(\frac {d}{a \plus{} c \plus{} d})$
So by AM-GM,
$ 3(\frac {ab}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)})^{\frac {1}{3}}\leq \frac {a}{a \plus{} c} \plus{} \frac {a \plus{} c}{a \plus{} c \plus{} d} \plus{} \frac {b}{a \plus{} c \plus{} b}$
And
$ 3(\frac {cd}{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)})^{\frac {1}{3}}\leq \frac {c}{a \plus{} c} \plus{} \frac {a \plus{} c}{a \plus{} c \plus{} b} \plus{} \frac {d}{a \plus{} c \plus{} d}$
Sum up.

It was a joke, that it is very simple and for beginners.[/quote]

[quote="Fedor Petrov(little edit by me)"]Yes, it was not solved on olympiad, but it was proposed to 8-9 grades (the last grade is 11-th), because those older students could possible kill it by some derivatives (in what I, however, doubt).

I may also show, how this solution may be deduced (whatever it means). It concerns an important idea of quasiliearisation.

Let me start with example.

We know that $ 2ab\leq a^2 \plus{} b^2$, of course. Introduce a parameter and get the following: $ 2ab\leq ta^2 \plus{} \frac {b^2}{t}$. Then read the last inequality from the other point: for any positive $ a, b$ there exist positive $ t$ such then $ 2ab \equal{} ta^2 \plus{} \frac {b^2}{t}\ (t \equal{} \frac {b}{a})$. Or, we may write: $ 2ab \equal{} \min(ta^2 \plus{} \frac {b^2}{t})$.

How may this observation help? Put $ a \equal{} \sum (x_i^2), b \equal{} \sum (y_i^2)$.
Then for appropriate $ t$ we have:
$ 2\sqrt {a}\sqrt {b} \equal{} ta \plus{} \frac {b}{t} \equal{} \sum (tx_i^2 \plus{} \frac {y_i^2}{t})\geq \sum (2x_iy_i)$. So, we get CBS inequality. A lot of other inequalities also may be proved by this idea, and the problem of topic is just a little more subtle example.

We have $ 3(AB)^{\frac {1}{3}}\leq Ax \plus{} By \plus{} \frac {1}{xy}$. And for any positive $ A$ and $ B$ there exist appropriate $ x$ and $ y$, for which equality holds
$ x \equal{} \frac {(AB)^{\frac {1}{3}}}{A}, y \equal{} \frac {(AB)^{\frac {1}{3}}}{B}$

Let $ A \equal{} (a \plus{} c \plus{} b), B \equal{} (a \plus{} c \plus{} d)$ in terms of initial problem. For some positive $ x, y$ we have
$ 3(AB)^{\frac {1}{3}} \equal{} Ax \plus{} By \plus{} \frac {1}{xy} \equal{} (a \plus{} c \plus{} b)x \plus{} (a \plus{} c \plus{} d)y \plus{} \frac {1}{xy} \equal{}$
$ (a \plus{} c \plus{} b)x \plus{} (a \plus{} c \plus{} d)y \plus{} \frac {1}{x(x \plus{} y)} \plus{} \frac {1}{y(x \plus{} y)} \equal{}$
$ \big(a(x \plus{} y) \plus{} bx \plus{} \frac {1}{x(x \plus{} y)}\big) \plus{} \big(c(x \plus{} y) \plus{} dy \plus{} \frac {1}{y(x \plus{} y)}\big)\geq$
$ 3(ab)^{\frac {1}{3}} \plus{} 3(cd)^{\frac {1}{3}}$ by AM-GM.[/quote]

See also the scorpius119's proof here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186411

if u know the solution its easy,if u don't its extremly hard !

[quote="Fedor Petrov"]Prove that for any four nonnegative reals $ a$, $ b$, $ c$, $ d$, the following inequality holds:
$ \left(ab\right)^{1/3} \plus{} \left(cd\right)^{1/3}\leq\left(\left(a \plus{} c \plus{} b\right)\left(a \plus{} c \plus{} d\right)\right)^{1/3}$.[/quote]
My proof:
$ \iff 3\sqrt[3]{abcd}(\sqrt[3]{ab}\plus{}\sqrt[3]{cd}) \le a^2\plus{}2ac\plus{}c^2\plus{}bd\plus{}ad\plus{}bc$
$ \iff (2a^2 \plus{} bc \plus{} \frac{bd}{2} \minus{} 3\sqrt[3]{a^2b^2cd}) \plus{} (2c^2 \plus{} ad \plus{} \frac{bd}{2} \minus{} 3\sqrt[3]{abc^2d^2})  \ge (a\minus{}c)^2$
Now $ x\plus{}y\plus{}z\minus{}3\sqrt[3]{xyz} \ge (\sqrt{x}\minus{}\sqrt{y})^2 \iff 2\sqrt{xy}\plus{}z\minus{}3\sqrt[3]{xyz} \ge 0$ is true. Using the lemma we get:
$ 2a^2 \plus{} bc\plus{}\frac{bd}{2} \minus{} 3\sqrt[3]{a^2b^2cd} \ge 2\left ( a \minus{} \frac{\sqrt{bd}}{2} \right )^2$ and
$ 2c^2 \plus{} ad\plus{}\frac{bd}{2} \minus{} 3\sqrt[3]{abc^2d^2} \ge 2\left ( c \minus{} \frac{\sqrt{bd}}{2} \right )^2$.
So it is only necessary to prove:
$ 2\left ( a \minus{} \frac{\sqrt{bd}}{2} \right )^2 \plus{} 2\left ( c \minus{} \frac{\sqrt{bd}}{2} \right )^2 \ge (a\minus{}c)^2$.
Letting $ n \equal{} a \minus{} \frac{\sqrt{bd}}{2}, m \equal{} c \minus{} \frac{\sqrt{bd}}{2}$ it is equivalent to:
$ 2n^2\plus{}2m^2 \ge (n\minus{}m)^2 \iff (n\plus{}m)^2 \ge 0$ which is obvious.
Equality whenever $ b \equal{} \frac{a}{c}(a\plus{}c), d \equal{} \frac{c}{a}(a\plus{}c)$.

Does anyone know how to use the technique Mr. Fedor Petrov mentioned to prove elementary well known inequalities like chebyshev,re-arrangement etc..?

[quote="Fedor Petrov"]Prove that for any four nonnegative reals $ a$, $ b$, $ c$, $ d$, the following inequality holds:
$ \left(ab\right)^{1/3}\plus{}\left(cd\right)^{1/3}\leq\left(\left(a\plus{}c\plus{}b\right)\left(a\plus{}c\plus{}d\right)\right)^{1/3}$.[/quote]
by my knowledge, it was on Sank-Peterburg olympiad in 2004 or 2005 year

I think this solution is more intuitive:
Notice that $(a+c+b)(a+c+d) \geq (a+c)( \sqrt{b}+ \sqrt{d})^2$
and by Holder $(a+c)( \sqrt{b}+ \sqrt{d})( \sqrt{b}+ \sqrt{d}$ is not less than what we want.
At first I wanted $(a+c+b)(a+c+d) \geq 2(a+c)(b+ d)$ but it is obviously wrong.


geometry
trapezoid
trigonometry
inequalities
triangle inequality
geometry unsolved

this question was really two easy. why nobody solve it???????

[quote="ridgers"]prove that in every trapezoid the difference between the bases is bigger then the difference of the two legs[/quote]
We begin with the labelling like the diagram and Wlog we assume that, $ b>a,c>d$. (i.e. we set them according to their length)
Now, $ |a\minus{}b|\equal{}b\minus{}a\equal{}x\plus{}y\equal{}x\plus{}d\cos C$ and $ |c\minus{}d|\equal{}c\minus{}d$
and,$ x\plus{}d\cos C>c\minus{}d \iff x\plus{}d\cos C\plus{}d>c$
Here, left hand side minimizes when,$ d\equal{}h$ and we get,
$ x\plus{}h>c$ which follows from triangle inequality.

Let $ ABCD$ be the trapezoid with $ AD\parallel BC$. Then we sholud prove that $ |AD\minus{}BC|\equal{}|AB\minus{}CD|$. Suppose that $ BC>AD$. Then $ E \in BC$  such that $ ADCE$ is a parallelogram. Then $ AD\equal{}EC$ and therefore $ BE\equal{}BC\minus{}AD$. Also $ AE\equal{}CD$. Then by triangle inequality in $ ABE$ we have $ BE>|AB\minus{}AE|$ or $ |BC\minus{}AD|>|AB\minus{}CD|$, as desired.


[quote="mli7"]I know the remainder of 9/4 is 1.

But what is the remainder of 0/3,  1/2, 3/4.....

what is the formula for finding the remainder?

Any help is appreciated.  thank you.[/quote]

Isnt the remainder just what you get left over after you divide???

so the remainder of 1/2 is 1 and 3/4 is 3 and 0/3 would be no remainder. right?

i'm certain that the remainder of $\frac{0}{3}$ is 0.  I'm not to sure about the other ones.   :?

there is no formula for all fractions but for proper fractions the remainder is just the numerator because it goes into the denominator 0 times

I don't know of any formula like that.  :huh:

[quote="chess64"]Formula for a remainder??? :rotfl:[/quote]
It's a theorem called the "division algorithm". Basically you just perform the division and see what the remainder is. That's as close as you get to a formula :)

I think that the remainder can only be as big as the numerator itself, otherwise, it gets wierd.
So, the answers are 0, 1, and 3; right?

As they always say: Theres no such thing as a stupid question, but there is a such thing as a silly question. :P


The method of solving this problem I called Contradiction Method. For this problem, I remember to post this solution one time before.

Here is my solution:
Apply Shur's inequality:$a^4+b^4+c^4+abc(a+b+c) \geq \sum_{sym}a^3b$
                                 $\Longleftrightarrow 1-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c) \geq \sum ab(1-c^2)$
                                  $\Longleftrightarrow 1-2(ab+bc+ca)^2+5abc(a+b+c) \geq \sum ab-abc(\sum a)$
                                 $\Longleftrightarrow 1+6abc(a+b+c) \geq 2(ab+bc+ca)^2+\sum ab$(1)
Going back to the problem:
                             $(2-3ab)(2-3bc)(2-3ca) \geq 1$
                             $\Longleftrightarrow 8-12(\sum ab)+18abc(\sum a)-27a^2b^2c^2 \geq 1$
                             $\Longleftrightarrow 7-12(\sum ab)+18abc(\sum a)-27a^2b^2c^2 \geq 0$
From (1) $\Longrightarrow 7+18abc(a+b+c) \geq 4+6(\sum ab)^2+3\sum ab$
And $12(\sum ab)+27a^2b^2c^2 \leq 12(\sum ab)+(\sum ab)^3$
Put $t=ab+bc+ca$ ($t \in [0,1]$)
We have $t^3+12t-(4+6t^2+3t)=(t-4)(t-1)^2 \leq 0$
             $\Longrightarrow t^3+12t \leq 6t^2+3t+4$
Hence:$7+18abc(a+b+c) \geq 12(\sum ab)+27a^2b^2c^2$
So the problem is proved :D  :D

By Bernoulli, \[ \prod_{\mathrm{cyclic}}\Big(1+(1-3ab)\Big)\geq 1+\sum_{\mathrm{cyclic}}(1-3ab)=4-3(ab+bc+ca) \geq  \] \[ \geq4-3(a^2+b^2+c^2)=1. \]
 :)


number theory
prime numbers
number theory unsolved

[quote="Vegre nem butulok votabb"]Prove that there are 1000 consecutive positive integers among which there are exactly 10 primes.
---Hey friends---what if I don't feel confident enough in Math---[/quote]

Let $ f(n)$ the number of prime numbers in $ [n,n \plus{} 999]$. We have  :

$ |f(n \plus{} 1) \minus{} f(n)|\leq 1$
$ f(1) > 10$
$ f(1001!\plus{}2) \equal{} 0 < 10$

And so $ \exists n_0\in(1,1001!\plus{}2)$ such that $ f(n_0) \equal{} 10$


Fie a,b,c numere pozitive astfel incat abc=1.Sa se arate ca:
1/[a^3(b+c)] + 1/[b^3(c+a)] + 1/[c^3(a+b)] >= 3/2

Faimoasa problema, cu $n$şpe mii de soluţii.

De exemplu: [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=365576#365576]Link 1[/url] şi [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25778]Link 2[/url] (în cel din urmă e demonstrată o inegalitate puţin mai tare).

Habar n-am avut ca s-a mai dat odata pentru ca eu stau numai pe foruml Romanesc!!!!!!

hai ca cika am si eu o solutie relativ simpla la ea...

amplificam fiecare termen cu :

Primu cu b^2 ori c^2 si tot asa...

si avem inegalitatea:

suma din b^2 c^2 / a(b+c) >= 3/2

notam bc=x , ca=y, ab = z

<=> x ^2 / z +y + y^2 / x + z + z ^2 / x+y >= 3/2

Aplicam un amarat de cauchy in engel:

<=> ... >= (x+y+z)^2 / 2(x+y+z)

<=> ...>= x+y+z /2

mai ramane sa demonstram deci ca x+y+z >= 3....dar cum xyz = 1 ...rezultatu este clar din AM-GM

Cine poate gasi o generalizare a acestei inegalitati?

Fie $a, b, c$ trei numere reale strict pozitive astfel incat $abc=1$. Aratati ca: \[\frac{1}{a^{n}(b+c)}+\frac{1}{b^{n}(c+a)}+\frac{1}{c^{n}(a+b)}\geq\frac{3}{2}.\] p.s. Exista uan si mai generala ca asta. Daca e nevoie o postez. :wink:

Vezi ca le-au descoperit altii cu mult ianintea ta. Tu erai mic pe atunci. :wink:

Probabil ai dreptate.Pe atunci eram un mic proiect.


geometry
geometric transformation
reflection
Euler
analytic geometry
geometry unsolved

Let $ ABC$ be a scalene triangle, $ P$ is a point in the plane of triangle $ ABC,$ Let $ P_{1},P_{2},P_{3}$ be the reflections of $ P$ in the sides $ BC,CA,AB,$respectively. If $ AP_{1},BP_{2},CP_{3}$ concur at $ Q,$ then $ PQ$ has constant direction.

If the problem is true, I suppose that the line $ PQ$ is parallel to the Euler line of triangle $ ABC$.

You asked me to comment on this, but I can't contribute much with the time I have in these days.

By using barycentric coordinates, you can show that
[b](1)[/b] the locus of all points $ P$ such that the lines $ AP_1$, $ BP_2$, $ CP_3$ concur is the locus of all points $ P$ such that the line joining $ P$ to the isogonal conjugate of $ P$ with respect to triangle $ ABC$ is parallel to the Euler line of triangle $ ABC$.
Note that this locus is a cubic curve known as the [i]Neuberg cubic[/i] of triangle $ ABC$.

Now, applying the [url=http://www.mathlinks.ro/viewtopic.php?t=4337]Sondat theorem[/url] to the orthologic triangles $ ABC$ and $ P_1P_2P_3$, we see that if the lines $ AP_1$, $ BP_2$, $ CP_3$ concur, then their point of concurrence (which you call $ Q$) and the two orthologic centers of triangles $ ABC$ and $ P_1P_2P_3$ (which are $ P$ and the isogonal conjugate of $ P$ with respect to triangle $ ABC$) are collinear. Hence, from [b](1)[/b] we can conclude that the line $ PQ$ is parallel to the Euler line of triangle $ ABC$.

Of course, the above depends on two facts which I only know computational proofs for: the fact [b](1)[/b] and the Sondat theorem. I remember someone sending me a longer synthetic proof of the Sondat theorem some 2-3 years ago (I think it was a French Hyacinther) which I have never read, but probably he has already published it somewhere. Unfortunately my time for triangle geometry is very limited now.

  darij

The "French Hyacinther" who send the synthetic proof of Sondat's theorem to Darij Grinberg is Jean-Louis Ayme. I published this proof in Revistaoim n° 27
http://www.campus.oie.org/oim/revistaoim/
or Google : revistaoim
Sincerely
Jean-Louis


calculus
integration
logarithms
calculus computations

(i) $ \int_{0}^{1}\frac{x^{\pi}}{x^{\pi}\plus{}(1\minus{}x)^{\pi}}dx$.

(ii) $ \int[(xe^{x}\plus{}e^{x})\ln^{2}x\minus{}2\frac{e^{x}}{x}]dx$.

(iii) Ascertain the nature of the series
$ \sum_{n \equal{} 1}^{\plus{}\infty}\frac{\sqrt{\ln(n\plus{}1)}\plus{}\sqrt{\ln(n)}}{n^{x}}\int_{\sqrt{\ln(n)}}^{\sqrt{\ln(n\plus{}1)}}e^{t^{2}}dt$, $ x\in\mathbb{R^{\plus{}}}$.

(iv) $ F(x) \equal{}\int_{0}^{x}(4t^{2}\minus{}4t\minus{}1)e^{\minus{}t^{2}\plus{}t}dt$.
Find $ \int_{0}^{1}F(x)dx$.

for (i) sub in $ u\equal{}1\minus{}x$ and then use symmetry.

for (ii) i dont think there is a closed form because of $ \int\frac{e^{x}}{x}dx$

for (iv) my advice is sub in $ u\equal{}\minus{}t^{2}\plus{}t$ then u will have

$ F(x)\equal{}\int_{0}^{x}g(u)e^{u}du$ so $ \int_{0}^{1}F(x)dx\equal{}\int_{0}^{1}\int_{0}^{x}g(u)e^{u}dudx$

(i) Let $ I \equal{}\int_{0}^{1}\frac{x^{\pi}}{x^{\pi}\plus{}(1\minus{}x)^{\pi}}dx$. Then \[ \begin{aligned}1 & \equal{}\int_{0}^{1}dx \equal{}\int_{0}^{1}\frac{x^{\pi}}{x^{\pi}\plus{}(1\minus{}x)^{\pi}}dx\plus{}\int_{0}^{1}\frac{(1\minus{}x)^{\pi}}{x^{\pi}\plus{}(1\minus{}x)^{\pi}}dx\\ & \equal{} I\plus{}\int_{1}^{0}\frac{(1\minus{}x)^{\pi}}{x^{\pi}\plus{}(1\minus{}x)^{\pi}}d(1\minus{}x) \equal{} 2I\end{aligned}\]so $ I \equal{}\frac{1}{2}$.

(ii) Let $ U \equal{}\int e^{x}\ln xdx$. Integration by parts yields $ \int (xe^{x}\plus{}e^{x})\ln^{2}xdx \equal{}\int\ln^{2}xd(xe^{x}) \equal{} xe^{x}\ln^{2}x\minus{}2U$ and $ \int\frac{e^{x}}{x}dx \equal{}\int e^{x}d(\ln x) \equal{} e^{x}\ln x\minus{}U$. Therefore, \[ \begin{aligned}\int\left((xe^{x}\plus{}e^{x})\ln^{2}x\minus{}2\frac{e^{x}}{x}\right)dx & \equal{} xe^{x}\ln^{2}x\minus{}2U\minus{}2(e^{x}\ln x\minus{}U)\\ & \equal{} e^{x}\ln x(x\ln x\minus{}2)\end{aligned}\]

[quote="Spasty"]
(iv) $ F(x) \equal{}\int_{0}^{x}(4t^{2}\minus{}4t\minus{}1)e^{\minus{}t^{2}\plus{}t}dt$.
Find $ \int_{0}^{1}F(x)dx$.[/quote]
[hide="Only an idea"]
$ \int_{0}^{1}F(x) dx$
Integration by parts,
$ xF(x)\big|_{0}^{1}\minus{}\int_{0}^{1}xF'(x) dx$
Fundamental theorem,
$ F(1)\minus{}\int_{0}^{1}x(\minus{}2x\plus{}1)^{2}e^{\minus{}x^{2}\plus{}x}dx$
But, I cant get passed that.
[/hide]

[quote="Spasty"]
(iv) $ F(x) \equal{}\int_{0}^{x}(4t^{2}\minus{}4t\minus{}1)e^{\minus{}t^{2}\plus{}t}dt$.
Find $ \int_{0}^{1}F(x)dx$.[/quote]

i get $ F(x)\equal{}e^{\minus{}x^{2}\plus{}x}\cdot(\minus{}2x\plus{}1)\minus{}1$
and so $ \int_{0}^{1}F(x)dx\equal{}e^{\minus{}x^{2}\plus{}x}\minus{}x\bigg|_{0}^{1}\equal{}\minus{}1$

Perfect!!
A solution of iv) is here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=123895

iii) The series converges for $ x>1$ and diverges otherwise.

Using the Mean Value Theorem we get that

$ \left(\sqrt{\ln(n\plus{}1)}\minus{}\sqrt{\ln(n)}\right)n<\int_{\sqrt{\ln(n)}}^{\sqrt{\ln(n\plus{}1)}}e^{t^{2}}dt<$

$ <\left(\sqrt{\ln(n\plus{}1)}\minus{}\sqrt{\ln(n)}\right)(n\plus{}1)$.

Thus, if $ S$ is the value of your series we get that

$ \sum_{n\equal{}1}^{\infty}\ln\left(\frac{n\plus{}1}{n}\right)^{n}\frac{1}{n^{x}}<S<\sum_{n\equal{}1}^{\infty}\ln\left(\frac{n\plus{}1}{n}\right)^{n}\frac{n\plus{}1}{n}\frac{1}{n^{x}}$,

and hence

your sum behaves like $ \sum\frac{1}{n^{x}}$ which converges for $ x>1$.


geometry
3D geometry
pyramid
calculus
integration

Prove that the volume of a circular cone is 1/3*pi*r^2*h

I believe proofs are above Classroom Math level.

Yeah, I think this is a very rigorous proof.  I don't know how to prove it, however.

Imagine that a cone is a pyramid whose base is a regular polygon with million (or more) vertixes. Pyramid volumes formula - $ 1/3 \cdot S \cdot H \equal{} 1/3 \cdot \pi r^2 \cdot H$

Too prove pyramid's volume's formula is n times harder...

What happened to no calculus in this forum?
Look up surface of revolution on google.
Volumes, surface areas, and arc lengths all use calculus

Easy Integration but whateverhttp://www.mph.net/coelsner/JSP_applets/cone_ex.htm

The Calculus way: Dividing the cone into a large number of thin disc's and aproximating the volume of the cone by the sum of the volumes of the disc's.

Volume disc at distance x from the top of cone with thickness $ \Delta x$: $ \frac{r}{h} \equal{} \frac{r_x}{x}$ gives $ r_x \equal{} \frac{r}{h}x$ and $ V_x \equal{} \pi r_x^2\Delta x$

Total volume $ V \equal{} \sum_{n\equal{}0}^{N}V_{x_n} \equal{} \sum_{n\equal{}0}^{N}\pi\left(\frac{r}{h}\right)^2 x_n^2 \Delta x$ when we let $ N\to\infty$ ($ \Delta x \to 0$) we get the integral

$ V \equal{} \int_{0}^{h} \pi\left(\frac{r}{h}\right)^2 x^2 dx \equal{} \frac{\pi r^2 h}{3}$

lol even calculus problems don't belong HERE (for the record, it was moved from classroom math to intermediate, where it is still too hard)


Do you mean $ \lim_{x \to 0} \frac{(1\plus{}x)^2(log(1\plus{}x))^2\minus{}x^2\minus{}x^2log(1\plus{}x)}{(x(1\plus{}x)log(1\plus{}x))^2}$     ?

the limit exist.
Ans is 5/12. Online calculator calculates the same ans as given in the book.
 :D  :rotfl:

I have solved it, but it's too much lengthy.
I want shorter method.

Method of Taylor series.  Series for $ \log(1\plus{}x)$ is known.  Take enough terms to work out Taylor series in numerator and denominator including at least one term that doesn't cancel.  Then divide.

log(1+x) = x - x^2/2 + x^3/3 + O(x^4)
(log(1+x))^2 = x^2 - x^3 + 11 x^4/12 + O(x^5)
numerator = 5x^4/12 + O(x^5)
denominator = x^4 + O(x^5)
quotient = 5/12 + O(1), as desired.

The first time I took only 2 terms of the log and everything cancelled, so I went back and used 3 terms.


Yea, I just created another account, and I didnt do anything...and now its deleted, this is the third account I created today.  Can you bring my other account back

Alright, lets get something straight.  I have a dynamic IP address so I cant possibly have the exact same address as someone else.  If you go check my accounts, I have never spammed anyone, maybe lets say change the score a bit, but that isnt spamming.  I was trying to help jason out.  Furthermore, I can still access this website (how?), if you banned this IP address.  My IP address goes from the 67.1 to 76.4.  And my internet is the high speed one.  I know atleast like 4 people that go here that has it, so its not me.

Email me your original username from the account to which that username is registered.  I understand the dynamic IP address issue - it's just that the spamming is always coming from an IP address that is used by one of two regular users, no matter which dynamic IP address is allocated.

And if you do know several people that are coming here the same way are, then you probably know the guilty party.


is this pre-olympiad level?Seems little easy

he is probably hinting at a slightly clever solution...but in my opinion, it is faster to multiply it out and solve the linear system than to think about it

$a+b+c=0$
$(-5a)+(-4b)+(-3c)=1$
$6a+3b+2c=5$

and this is not that bad to solve by whatever technique you choose...

[quote="PenguinIntegral"][b] PenguinIntegral's Effort at being a unique and friendly poster Post #2[/b]

Find a, b, and c so that:

$\frac{x+5}{(x-1)(x-2)(x-3)}=\frac{a}{x-1}+\frac{b}{x-2}+\frac{c}{x-3}$[/quote]

Standard method:
$x+5 \equiv a(x-2)(x-3) + b(x-1)(x-3) + c(x-1)(x-2)$
Sub. $x = 1$, $x = 2$ and $x = 3$ in turns.


How many non-congruent rectangles with integer sides have not more than 25 square units in their areas?

How do you know when you've found them all?

Good idea, trying to arrange them by size.

But... [hide]are you sure you have them going in the right order?[/hide]

dinoman wrote:[hide]10

u can have 5x4, 5x3, 5x2, 5x1, 4x3, 4x2, 4x1, 3x2, 3x1, 2x1
[/hide]

How about [hide] 6x4, 6x3, 6x2, 6x1, 4x4, 3x3, 2x2 and 1x1? [/hide]

In my opinion, the easiest way is to [hide]take 2 integers 1-25 n and m. Start m at 25. Make sure that n is always less than or equal to m, so you don't repeat pairs. just count down for each m. You'll find m=13-25 there is only 1 possible n. For m=9-12, there's 2 n. For 7 or 8, there's 3 n. For 6, there's 4 n and so on, until you get to m=2.[/hide]

Well I happen to love guess and check :P .

But I see how your way works...


geometry
angle bisector
similar triangles

Let $ABCD$ be a parallelogram. The bisector of the angle $B$ intersects $AC$ in $B^\prime$ and the parallel through $B^\prime$ to $AB$ intersects $AD$ and $BD$ in $P$ and $Q$.

Prove that $\triangle PAQ$ is isosceles.

[i]Ion Safta[/i]

[quote="Andreas"][hide]By Angle Bisector Theorem $\frac{AB}{BC}= \frac{AB'}{B'C}= \frac{AB}{AD}$. Since parallelism $\frac{AB}{AD}= \frac{PQ}{PD}$, and $\frac{AP}{AB'}= \frac{AD}{AC}$. Hence $PQ = \frac{AB}{AD}\cdot PD$ and $AP = \frac{AB}{AC}\cdot B'C$. By Thales $\frac{PD}{AD}= \frac{B'C}{AC}$, and so $AP = PQ$.[/hide][/quote]

What's Thales?  It looks like you were just using similar triangles in that step...

Yeah, you're right. Do problems before going to bed it's not a good thing  :o


[quote="pvthuan"]Prove that if $ a,b,c,d\geq0$, then
\[ (a \plus{} b \plus{} c \plus{} d)^{\frac32}\geq (a \plus{} b)\sqrt c \plus{} (b \plus{} c)\sqrt d \plus{} (c \plus{} d)\sqrt a \plus{} (d \plus{} a)\sqrt b.
\]
[/quote]
Nice joke.
Assume that $ a\plus{}b\plus{}c\plus{}d\equal{}4$, then we have to prove
\[ (a \plus{} b)\sqrt c \plus{} (b \plus{} c)\sqrt d \plus{} (c \plus{} d)\sqrt a \plus{} (d \plus{} a)\sqrt b \le 8\]
But this is true since
\[ 2LHS \le (a\plus{}b)(c\plus{}1)\plus{}(b\plus{}c)(d\plus{}1)\plus{}(c\plus{}d)(a\plus{}1)\plus{}(d\plus{}a)(b\plus{}1) \equal{} 8\plus{}2ac \plus{}2bd \plus{}(a\plus{}c)(b\plus{}d) \le 8 \plus{}\frac{(a\plus{}c)^2}{2}\plus{}\frac{(b\plus{}d)^2}{2}\plus{}(a\plus{}c)(b\plus{}d) \equal{}8\plus{}\frac{1}{2}(a\plus{}b\plus{}c\plus{}d)^2 \equal{}16\]
Therefore $ LHS \le 8$.
Done!

[quote="pvthuan"]Yes, that is it. A more difficult (symmetric) one is
\[ (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^3\geq \sum_{\rm cyclic} (a^3 \plus{} bcd)(a \plus{} b)(a \plus{} c)(a \plus{} d).
\]
[/quote]
Try $ a\equal{}b\equal{}0,c\equal{}\frac{1}{3},b\equal{}\frac{4}{3}$ ;)


Let $m$ be a natural number such that there exist integers $x,y$, with $(x,y)=1$, such that \[m = ax^{2}+2bxy+cy^{2}\] (where $a,b,c \in \mathbb{Z}$).
Prove that $b^{2}-ac$ is a perfect square (mod $m$).

$b^2-ac$ is a perfect square $\mod m$ iff $az^2+2bz+c \equiv 0 \mod m$ has a solution. But clearly $z = \frac{x}{y}$ works.


geometry
incenter
circumcircle
geometry unsolved

Let $ I$ be the incenter of the triangle  $ ABC$ . A circle $ K$  passes through $ I$ and touches both side $ BC$ and the

circumcircle of $ \triangle ABC$.

Prove that $ AI$ is a tangent to the circle $ K$.

[u]Babis[/u]

there is a lemma which solves directly the problem:
if $ (K)$ is a circle tangent to $ AD$ at point $ Q$ where $ D \in BC$ and also tangent to $ BC$ at point $ F$ and finally tangent to the circle $ (O)$ where $ (O)$ is the circumcircle of $ \Delta ABC$ then $ QF$ passes through $ I$.
note that if we define similarly $ (K')$ on the other side of $ AD$ then the same property will appear and also we'll have $ I \in KK'$

[quote]Let $ I$ be the incenter of the triangle  $ ABC$ . A circle $ K$  passes through $ I$ and touches both side $ BC$ and the circumcircle of $ \triangle ABC$. Prove that $ AI$ is a tangent to the circle $ K$.[/quote]

Just want to introduce an inversive proof.

Let $ A_1,B_1,C_1$ respectively be the tangency points of $ (I)$ with respect to $ BC,CA,AB$ and $ r$ be $ (I)$'s radii. Denote $ K_1,K_2$ respectively by the tangency points s of $ (K)$ with respect to $ BC$ and $ (ABC)$. Conisder the inversion through pole $ I$, power $ k \equal{} r^2$, we have $ \mathcal {I}(I,k):$ $ K_1\mapsto K'_1$, and $ K_2\mapsto K'_2$. Also, $ \mathcal {I}(I,k):$ $ BC\mapsto ([IA_1])$ and $ (ABC)\mapsto (E)$, which is the $ 9 \minus{}$ point circle wrt $ \triangle A_1B_1C_1$. Since $ (K)$ tangents to both $ BC$ and $ (ABC)$, thus, $ K'_1K'_2$ also tangents to both $ ([IA_1])$ and $ (E)$, moreover, $ K'_1\in ([IA_1])$ and $ K'_2\in (E)$ $ (*)$.

Let $ M_b,M_c$ respectively be the midpoints of $ A_1C_1$ and $ A_1B_1$, it is followed immediately that $ M_b,M_c\in ([IA_1])$ and $ M_b,M_c\in (E)$, which implies that $ M_bM_c$ is the radical axes of $ (E)$ and $ ([IA_1])$. But it is easy to prove that the radii of $ (E)$ and $ ([IA_1])$ are equal. Combine with $ (*)$, we obtain, $ K'_1K'_2\perp M_bM_c$. But $ M_bM_c\| B_1C_1$ and $ B_1C_1\perp IA_1$. Therefore, $ IA_1\|K'_1K'_2$. Note that $ IA_1$ is mapped into itself by the inversion through pole $ I$, power $ k$. As the result, we have $ IA_1$ is a tangent to $ (K)$. Our proof is completed then. $ \square$

Let $K$ touch $BC,\odot ABC$ in $E,F$ respectively and let $AI\cap \odot ABC=\{Q\}$
Consider the homothety $\mathcal{H}_{F}$ taking $K\mapsto \odot ABC$.It takes $E$ to the midpoint of the arc $BC$ and hence $F,E,Q$ are colinear.On the other hand $Q$ is the center of inversion interchanging $BC$ and $\odot ABC$ $\implies$ $QE\cdot QF=QI^2$ which iplies the result.


Find the sum of the x-coordinates of the distinct points of intersection of the plane
curves given by x^2 = x + y + 4 and y^2 = y- 15x + 36.

hello, by substituting $ y\equal{}x^2\minus{}x\minus{}4$ we get the equation $ (x^2\minus{}x\minus{}4)^2\equal{}x^2\minus{}x\minus{}4\minus{}15x\plus{}36$. By factorization we have $ (x^2\plus{}2x\minus{}4)(x\minus{}2)^2\equal{}0$.
Sonnhard.


$ \log 125$ equals:

$ \textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25$
$ \textbf{(D)}\ 3 \minus{} 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)$

[hide]
Look at the answer choices individually.
(A) is equivalent to $ log 1.25^{100}$.
(B) is equivalent to $ log 3^5$ or $ log 243$
(C) is equivalent to $ log 25^3$ or $ log 15625$
(D) is equivalent to $ 3 log 10 \minus{} 3 log 2$ $ \Longrightarrow$ $ log 10^3 \minus{} log 8$ $ \Longrightarrow$ $ log \dfrac{1000}{8}$ $ \Longrightarrow$ $ log 125$

Therefore the answer is $ \boxed{D}$ $ 3 \minus{} 3 log 2$
[/hide]


Here is a tricky geometry problem:

Triangle ABC has vertex A at the origin, vertex B at the point (2,0), and vertex C on the circle with center (3,2) and radius 1. What is the maximum possible area for such a triangle?

we want the height to be the furthest from C perpendicular to AB, so point C should be at the top of the circle, point (3,3). our area is then 2(3)/2=3

The third vertex would have to be (3,3).

and using the shoelace method, we  get the area as being 3 units squared.

Ha. Wow that was fast. :)

Here is another hard one:

Sue and Barry are trying to find the area of triangle ABC. Sue mistakenly uses AB and the height from A (instead the height from C), and Barry mistakenly uses BC and the height from C (instead of the height from A).

Sue finds an area of 120 and Barry finds an area of 150. What is the area of triangle ABC?


geometry
rectangle
induction
graph theory
linear algebra
combinatorics proposed
combinatorics

Every cell of m*n square is white.Each cell can be white or black.Each time we can choose a square and and change its color and its neighbors.(each cell has at most 4 neigbors).Prove that we can make all cells black

You said "m*n square". What exactly does it mean? Is it supposed to be "m*n rectangle" or "n*n square"?

I thought it was a rectangle, so I was pretty amazed by the fact that this worked for any rectangle, but I didn't manage to find a solution for 3*7. The procedure is equivalent to putting 1 or 0 in each cell s.t. the sum of the numbers in a cell and it's neighbor cells is odd for every cell. It's easier to look for solutions if you restate the problem like this. However, I tried really hard to find such an arrangement of 0's and 1's for a 3*7 rectangle, but failed, so I started asking myself if it wasn't "n*n square".

So how about it Omid?

The arrangement for the $3 \times 7$ rectangle is unique, and it's the following:

$\begin{array}{ccccccc}
0 & 0 & 1 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 0 & 1 & 0 & 0
\end{array}$

IMHO, the statement is true for every rectangle.

Oh excuse me, I wanted to say ''m*n rectangle''.
It is a vey beautiful problem.Well if you want me I will send my solution.

Thanks for clarifying it. I think I'll work on it a bit longer :)..

Please send your solution!
However, I can't find an example for 5*6

We construct a bipartite graph with $m\times n$ vertices at each side. Suppose $t=mn$. Number the cells arbitrarily with $1,2,3,\dots,t$. We construct a graph with $S_{1},S_{2},S_{3},\dots,S_{t}$ at one side and $L_{1},L_{2},L_{3},\dots,L_{t}$ at other side. For cell with number i put one vertex $S_{i}$ at one side and $L_{i}$ at other side. Between $S_{i}$ and $L_{j}$ is edge if and only if $i$ and $j$ are neighbors. So we see Between $S_{i}$ and $L_{i}$ is edge, and between $S_{i}$ and $L_{j}$ is edge if and only if betwee $S_{j}$ and $L_{i}$ is an edge.
Now because of a famous graph problem you can select some $S_{i_{1}},S_{i_{2}},\dots,S_{i_{k}}$ of $S_{1},S_{2},\dots,S_{t}$ that each $L_{i}$ is connected two odd number of these selected vertices.
(You can solve this problem with linear algebra or with induction)
Now Select cells with nubers $i_{1},i_{2},,\dots,i_{k}$ and all cells become black.  :P  :P


N(1<=N<=100) friends gathered in order to play chess, according to the following rules. In the first game, two of the N friends will play. In the second game, the winner of the first game will play against another friend (maybe even the same friend who lost the first game). In the third game, the winner of the second game will play against someone else and so on.. No game will end as a draw (tie). Given the number of games each of the N friends played, find a schedule for the games, so that the above rules are obeyed. The number of games will always be less than 10000.

At each step, let the player with the maximum number of games to go beat the player with the minimum (>0) number of games to go.

Is it possible that this doesn't work? If so, we must end up with one player who still has games to go, while everybody else has 0. But in that case, if you work backwards, that player must have played and won every single game, which means at the start he must have more games to go than everybody else combined, which is impossible.


geometry
inequalities
angle bisector
triangle inequality

1.  Given triangle ABC with angle bisector AA' (A' on BC).  The incircles of triangles ABA' and ACA' are drawn and are found to be tangent to each other.  Prove that AB = AC.

2. Given triangle ABC with angle A = 150 :^o:.  Let d(X, YZ) represent the distance between the point X and the line YZ.  A' is chosen on BC and points P and Q are chosen so that the following are true:
(a) d(P, AB) = d(P, BC) = d(P, AA')
(b) d(Q, AC) = d(Q, BC) = d(Q, AA')
(c) d(Q, AA') + d(P, AA') = PQ (the distance between P and Q)
(d) d(P, AA') = :rt3: * d(Q, AA')

Compute the measure of angle PAQ.

kinda tough with Paint... sorry.

[hide]

To understand the labelling, x,y,w,z are all lengths.  Length x goes from vertex B to the point of tangency, length y goes from vertex A to the points of tangency, etc.

I got the diagram by using the fact that tangents are equal in length to the point of tangency.  Then using the angle bisector theorem, i got (y+x)/(x+z)=(y+w)/(z+w).  Simplifying, it comes out to w(y-z)=x(y-z).  Then I used the triangle inequality on triangle ABC.  So y+x+y+w>w+z+z+x, which simplifies to y>z.  So we can cancel out y-z, and we get w=x.  Then AB = x+y, and AC=w+y, and since w=x, the two are equal.[/hide]

Ah, nicely done -- quite different from my method, actually.

The 2 circles have equal radii (cal the radius r), so let the foot of the bisector be D. Then r/z=tg(<ADC/2)=tg(90-<ADC/2)=1/tg(<ADC/2), so tg(<ADC/2)=1, so <ADC/2=45 deg, so <ADC=90 deg, so the bisector is also an altitude, so AB=AC.

z is the same as in cats' notation.


geometry
circumcircle
geometric transformation
homothety
projective geometry
similar triangles
geometry proposed

Let  $ \triangle A'B'C'$ be the pedal triangle of the $ \triangle ABC$ WRT a point Q, with pedals $ A' \in BC,\ B' \in CA,\ C' \in AB.$ Any $ \triangle DEF$ with $ D \in BC,\ E \in CA,\ F \in AB$ directly similar to the pedal $ \triangle A'B'C'$ is obtained by a spiral similarity with center Q. If some other $ \triangle DEF \sim \triangle A'B'C'$ existed, which could not be obtained in this way, one could pick the $ \triangle D'E'F'$ obtained form the pedal $ \triangle A'B'C'$ by the spiral similarity with center Q and equally inclined as $ \triangle DEF.$ The $ \triangle DEF \sim \triangle D'E'F$ would have parallel sides, they would have to be centrally similar with some similarity center, but this similarity center would have to be simultaneously identical with the vertices A, B, C of the original $ \triangle ABC.$ If A', B', C' are midpoints of BC, CA, AB, the medial triangle $ \triangle A'B'C'$ of the $ \triangle ABC$ is directly similar to it and it is the pedal triangle WRT the circumcenter O. Since $ \frac{A'B}{A'C} \cdot \frac{B'C}{B'A} \cdot \frac{C'A}{C'B} \equal{} 1,$ $ AA', BB', CC'$ concur (at the centroid P). Any $ \triangle DEF \sim \triangle ABC \sim \triangle A'B'C'$ with $ D \in BC,\ E \in CA,\ F \in AB$ is obtained by a spiral similarity with center O. But then either $ BD > BA',\ CE > CB',\ AF > AC'$ and $ CD < CA',\ AE < AB',\ BF < BC'$ or the other way around, $ \frac{DB}{DC} \cdot \frac{EC}{EA} \cdot \frac{FA}{FB} > 1$ or $ < 1,$ and AD, BE, CF do not concur.

[size=150]This my solution:
We see $ \triangle DEF \sim \triangle ABC$ and they are same directly
-> existed a transform propitious:
$ D \mapsto A$
$ E \mapsto B$
$ F \mapsto C$
and the transform  propitious only has one immovable: O
after that we'll prove that: $ O\equiv P$

 :lol:[/size]

[quote="Erken"]Does anybody know the name and the proof of the following theorem:
Let $ \triangle ABC$ and $ \triangle A'B'C'$ be two similar triangles,such that $ P\in AA'\cap BB'\cap CC'$.Prove that $ P$ is homothety center.[/quote]

It is called "Wrong Theorem".

Take two circles k and k' intersecting at two points P and Q, and three points A, B, C on k. Let the lines AP, BP, CP intersect k' again at A', B', C'. The triangles ABC and A'B'C' are directly similar, and P lies on the lines AA', BB', CC', but the triangles ABC and A'B'C' are only homothetic if the circles k and k' are tangent to each other.

  darij

Uuups   :blush: ,maybe i misunderstood something because one guy solved it using this theorem,and he score max points for this problem....

[quote="darij grinberg"]
It is called "Wrong Theorem".
  darij[/quote]

 :)  :)  I was just ready to ask , if such a theorem is valid , but your post is much better !!!

 Babis

[quote="Erken"]Let $ P$ be an interior point of an acute angled triangle $ ABC$. The lines $ AP,BP,CP$ meet $ BC,CA,AB$ at points $ D,E,F$ respectively. Given that triangle $ \triangle DEF$ and $ \triangle ABC$ are similar, prove that $ P$ is the centroid of $ \triangle ABC$.[/quote]

Just assume that $FE$ and $BC$ aren't parallel. Let $FE$ and $BC$ intersects in $A_1$, simillary denote $B_1$ and $C_1$. WLOG assume that $A_1$ lies on extension $CB$ behind $B$. Then by simple angle chasing we get that $B_1$ lies on extension $AC$ behind $C$ and $C_1$ lies on extension $BA$ behind $A$. But by Desargues theorem, we know that $A_1$, $B_1$ and $C_1$ are collinear which is clearly impossible...

[quote="Erken"]Does anybody know the name and the proof of the following theorem:
Let $ \triangle ABC$ and $ \triangle A'B'C'$ be two similar triangles,such that $ P\in AA'\cap BB'\cap CC'$.Prove that $ P$ is homothety center.[/quote]
if we add one condition that P is an interior point of $ \triangle ABC$, I guess the claim must be true. if anyone would post a proof for this? thanks ~

Label the angles $\angle CDE = \angle 1$, $\angle FDB = \angle 2$, and so on in a clockwise order.

Assume for the sake of contradiction that $\angle 1 \neq \angle B$, so without loss of generality let $\angle 1 > \angle B$. Check that
\begin{align*}
\angle 1 &> \angle 4 \\
\angle 3 &> \angle 6 \\
\angle 5 &> \angle 2
\end{align*}
by the given similarity. Then as $\angle 1 + \angle 4 = 2 \angle B < \pi$, we have $\sin \angle 1 > \sin \angle 4$, and similarly $\sin \angle 3 > \sin \angle 6$ and $\sin \angle 5 > \sin \angle 2$. Thus, $$\frac{CE \cdot BD \cdot AF}{CD \cdot BF \cdot AE} = \frac{\sin \angle 1 \cdot \sin \angle 3 \cdot \sin \angle 5}{\sin \angle 2 \cdot \sin \angle 4 \cdot \sin \angle 6} > 1,$$ which contradicts Ceva.

Thus $\angle 1 = \angle B$, so $\overline{DE} \parallel \overline{AB}$. Ceva now implies $F$ is the midpoint of $\overline{AB}$, so it follows that $P$ must be the centroid.


[quote="mcaddict"]I'm new to AoPS. I'm sure there are a lot of experienced people out there in AoPS. What is the best thing on AoPS for MATHCOUNTS?     :D[/quote]

There are many resources for MC on AoPS. Where you have just posted, the MathCounts forum has many problems and questions posted, which are completely free to look at and solve/look at solutions.

Also, AoPS has its very own MC classes, a basics and mastering. Unfortunately, they just finished, and the won't start up again until next year.

I would also recommend AoPS Volume 1 to learn the basics, which should give you enough understanding the master mathcounts. AoPS Volume 2 has some interesting concepts, but many of them will not help for MC.

Also, if you have the time and the money, read the introduction series. They go more into detail, and helped me a lot.

There are many topics in the MC forum, many with solutions already. You can look through them.

http://www.mnspe.org/mathcounts/coach.htm

This website has quite a feq tests you can do for practice. If there is anything you don't understand, you can post as you just did, and ask other AoPSers for help.

Welcome to AoPS! I joined a few years ago and the resource has been indispensible.

I hope it will be the same way for you.

~BOGTRO~

[quote="john alabbad"]Sorry to spam, but I think that jonathanchou711, aznkid711, and mcaddict are all the same person.

Their emails are jonathanchou711@gmail.com, aznkid711@gmail.com, and mcaddict711@gmail.com respectively, and when posting, they always type something into the subject, and the way they talk are similar.[/quote]

I don't know who is jonathanchou711 or aznkid711. i just chose mcaddic711@gmail.com cause it was just a... random choice? I really don't know. You know there are a lot of 911 's and like 411's and stuff like that. Sorry. I really don't know. Are you allowed to post your name in this forum? Cause if I could, i would tell you my real name.

The way you talk, your usernames, and your join dates are definitely enough information to conclude that you are aznkid711 and jonathanchou711.

Let's move this discussion to FF or just stop it completely.

I agree with AIME15, this discussion is against the ToS; having multiple accounts is not allowed, giving out real names, etc.

Even then, why in the world would we be talking about multi accounts here?

EDIT: That means this discussion shouldn't continue maybe? :wink:

Well some parts of it obviously [i]have[/i] to be applicable for the whole site.


geometry
3D geometry
tetrahedron
geometry unsolved

Can we use n sticks to make n triangles which have each side is one of those sticks?

yes, in space.
  I have another small problem. I hope we can use the idea of this problem to solve the first one:  6 sticks -> 4 equil triangles (it's rather easy).

Yes, you can use them as the edges of a regular tetrahedron, that's why I asked you about space.

Pierre.

why do you have a tiny problem like this : 6 sticks --> 4 equil triangles? (It's too easy for thinking about a tetrahedron in space)
We are discussing about n sticks --> n equil triangle, aren't we?

I think we should fit many regular tetrahedrons to have the figure satisfies the problem. (those tetrahedrons have common face)

So that the regular tetrahedron i have the common face with the reg tetrahedron i-1.
             We infer : n sticks -->n-2 triangles (but not for all n)
             Then I hope there is k so that the reg tetrahedron k have the common face with the first reg tetrahedron. ( it means we reduce the numbers of the sticks).
             Can you solve and improve it?


[quote="GameBot"]At what time is the sum of the digits which represent the hours
and minutes on a 12-hour digital watch the greatest?[/quote]

Obviously it's either 12:59 or 9:59. From there, you can see it's the latter option.


[color=indigo]My favorite author is Tamora Pierce.  I highly reccomend any of her books, especially any in the Song of the Lioness series.  Also, If you like her books, go to http://www.tamora-pierce.com or http://www.sheroesfans.com or http://www.sheroescentral.com.

Her fans call her Tammy.[/color]

I've heard of her, but never really been interested...ever read Sabriel? or Lirael?  or Abhorsen?  (by Garth Nix) Those had a she-ro in them too (actually two), and were pretty good.

garth nix rocks. have you read shade's children, which he also wrote? i just finished re-reading it. he also wrote the seventh tower series, which is pretty cool too. all of his books are good. and tammy pierce is sweetness too. there was something of hers that i read that REALLY stayed with me, but for the life of me i cant remember what its called now. hm.

Yep, Shade's children is a good one.  Kinda morbid, but cool.  I read the Sabriel-Lirael-Abhorsen series, too.  Sabriel was good.  Lirael was fabulous.  Abhorsen was boring.  What did you think of the series?  I liked Lirael because it had lots of surprises, whereas I hated Abhorsen because it didn't - you knew what was going to happen when you started reading the book!  Let's see... hm. Lirael, Sam, and the others rescue Nick, but not before Orannis is unearthed.  Everyone fights against Abhorsen and eventually they win.  Classic fantasy plot.  Ah well.  Lirael WAS really good, though.

its morbid, yeah, but interesting. i like how nix is able to create alternate... realities?... and make them pretty believable. i dont remember sabriel much its been a while. wasnt there something about a wall that divided the "safe" city from the "unsafe" wilderness?


[b]IF[/b] real numbers $ a$ , $ b$ , $ c$ are positive, [b]THEN[/b]

$ \boxed{\ \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} b^2}{c \plus{} b} \plus{} \frac {a^2 \plus{} b^2}{a \plus{} b}\ \ge\ \sqrt {3\left(a^2 \plus{} b^2 \plus{} c^2\right)} \plus{} \frac {1}{2(a \plus{} b \plus{} c)^2}\cdot \left[a(b \minus{} c)^2 \plus{} b(c \minus{} a)^2 \plus{} c(a \minus{} b)^2\right]\ }\ .$

See and [url=http://www.mathlinks.ro/viewtopic.php?p=1390033#1390033][color=red][b]here[/b][/color][/url]

[quote="Virgil Nicula"][b]IF[/b] real numbers $ a$ , $ b$ , $ c$ are positive, [b]THEN[/b]

$ \boxed{\ \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} b^2}{c \plus{} b} \plus{} \frac {a^2 \plus{} b^2}{a \plus{} b}\ \ge\ \sqrt {3\left(a^2 \plus{} b^2 \plus{} c^2\right)} \plus{} \frac {1}{2(a \plus{} b \plus{} c)^2}\cdot \left[a(b \minus{} c)^2 \plus{} b(c \minus{} a)^2 \plus{} c(a \minus{} b)^2\right]\ }\ .$
[/quote]
$ \sum_{cyc}\frac{a^2\plus{}b^2}{a\plus{}b}\minus{}\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\minus{}\frac{1}{2(a\plus{}b\plus{}c)^2}\cdot\sum_{cyc}c(a\minus{}b)^2\equal{}$
$ \equal{}\sum_{cyc}\left(\frac{a^2\plus{}b^2}{a\plus{}b}\minus{}\frac{a\plus{}b}{2}\right)\minus{}\left(\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\minus{}a\minus{}b\minus{}c\right)\minus{}$
$ \minus{}\frac{1}{2(a\plus{}b\plus{}c)^2}\cdot\sum_{cyc}c(a\minus{}b)^2\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)^2\left(\frac{1}{2(a\plus{}b)}\minus{}\frac{1}{a\plus{}b\plus{}c\plus{}\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}\minus{}\frac{c}{2(a\plus{}b\plus{}c)^2}\right)\geq$
$ \geq\sum_{cyc}(a\minus{}b)^2\left(\frac{1}{2(a\plus{}b)}\minus{}\frac{1}{2(a\plus{}b\plus{}c)}\minus{}\frac{c}{2(a\plus{}b\plus{}c)^2}\right)\equal{}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)^2c^2}{2(a\plus{}b)(a\plus{}b\plus{}c)^2}\geq0.$


algebra
binomial theorem
combinatorics unsolved
combinatorics

First off, I do not know if this problem has already been solved, but I am not sure in which forum to put it...
Also, this is my first time using Latex. and i havent yet found out how to put arguments in the sum and prod signs.

Anyway, my question is related to the binomial theorem:

$(x+y)^n = \prod(x+y) = \sum \left(\begin{array}{c}n\\k\end{array}\right) x^k y^{n-k}$

The prod is from i= 1  to i = n, but because there is no i in the argument it simplifies to the leftmost argument
In the sum k is k from 0 to n and:

$\left(\begin{array}{c}n\\k\end{array}\right) = \frac{n!}{k!(n-k)!}$

My teacher has asked me to derive a similar formula for:

$\prod (x+1)$

This one also n times, with x=0, 2, 4, 6, 8 etc

Example for n = 4:
1*3*5*7

Example for n=5:
1*3*5*7*9

What is the fast way to do this? Im completely stuck on this problem and the more I think about it the less I understand it.

So you want the product of the first $n$ odd positive integers¿
Hint: You know $(2n)!$ as the product of the first $2n$ integers, so you have just to kill the even ones.

Ah, I understand now,

$\frac{(2n)!}{n! 2^n} = 1*3*5*7*...*n$

Still now im curious, is there any general case for this problem if x is not the even numbers but some random sequence and you already know the product of all x taken together?


The least common multiple of positive integers a, b, c and d is equal to (a + b + c + d). Prove that abcd is divisible by at least one of 3 and 5.

This is an odd question, because it is not so very hard to fully categorize all solutions.  So, I give a solution and some new questions:
[hide="Solution"]WLOG say $a \leq b \leq c \leq d$.  Note that $d, d, d, d$ is never a solution, so we have $a + b + c < 3d$.  However, $d | a + b + c + d$, so $a + b + c = d$ or $a + b + c = 2d$.  In the first case, $a, b, c | 2d$ and $a, b, c < d$.  If $d$ is not divisible by either 3 or 5, then $a, b, c \leq \frac27d$ which is a contradiction (since $a + b + c = d$).  In the second case, $a + b + c + d = 3d$ so the lcm of the four numbers is divisible by 3 so one of the numbers must be divisible by 3 so the product is divisible by 3.[/hide]

My questions:
1) Characterize all pairs $(a, b)$ such that $lcm(a, b) = a + b$.
2) Characterize all triples $(a, b, c)$ such that $lcm(a, b, c) = a + b + c$.
3) Characterize all 4-tuples $(a, b, c, d)$ such that $lcm(a, b, c, d) = a + b + c + d$.
[hide="Answer (not solution) to 3)"]Every quadruple of integers which works is an integer multiple of one of the following:
{1, 1, 4, 6}
{1, 2, 2, 5}
{1, 2, 6, 9}
{1, 3, 4, 4}
{1, 3, 8, 12}
{1, 4, 5, 10}
{1, 6, 14, 21}
{2, 3, 3, 4}
{2, 3, 10, 15}[/hide]

No one likes my questions?  (Bump.)
[quote="JBL"]My questions:
1) Characterize all pairs $(a, b)$ such that $lcm(a, b) = a + b$.
2) Characterize all triples $(a, b, c)$ such that $lcm(a, b, c) = a + b + c$.
3) Characterize all 4-tuples $(a, b, c, d)$ such that $lcm(a, b, c, d) = a + b + c + d$.[/quote]

The first case I cannot find any solutions to.
[hide]
WLOG assume $a \leq b$. Then since $b,b$ cannot be a solution, $a<b$, strictly. However, since $b|a+b$, $a=b$ which contradicts. Thus there cannot be any solutions.[/hide]

I will see if i can solve for the other questions.


Can Someone introduce a good book on combinatorics in the olympiad level which introduces the main ideas and their applications?

[color=#FF0000][Mod: Illegal link deleted.][/color]

hi all

can you  show me please combinatotics  book? but  in french because im not  good in english
thank's for your help

enigmation :)


What interger 'n' has the property that 5^96 is greater than n^72 and 5^96 is less than (n+1)^72?

Thank you!

I have another solution. Note that $ \gcd(72,96) \equal{} 24$ so we can take the $ 24$th root.

$ n^{72} < 5^{96} < (n \plus{} 1)^{72}\implies n^{3} < 625 < (n \plus{} 1)^{3}.$

With simple trial and error using bounds, we find that $ n \equal{} 8$ satisfies this.

Okay, I epicly screwed up.  That doesn't make any sense.


Find the number of quadratic polynomials $ax^2 + bx +c$ which satisfy the following:

(a) $a,b,c$ are distinct;

(b) $a,b,c \in \{ 1,2,3,\cdots 1999 \}$;

(c) $x+1$ divides $ax^2 + bx+c$.

[hide]
The question is equivalent in finding the pair of solution for $a+c=b$ where $a,b,c$ are distinct integer from 1 to 1999 .

When $b$ is 1999 , $a$ has $1998$ choice . when $b$ is 1998 , $a$ has $1997$ choice and so on until when $b$ is 3 , $a$ has 2 choice . But then we notice that for even $b$ , there will exist one pair of solution such that $a=c$ . Hence the total way is

$2+3+4+...+1998 - 998=1996002$
[/hide]

[quote="seamusoboyle"]What about b=2?[/quote]

then $a=c=1$ absurd  :)

Oops, should read the question next time... :blush:

I tried it the other way by fixing the value of a and finding no of solutions for b, and c..It came 1997+1996+...+2+1-998
Where am I wrong?


function
calculus
derivative
calculus computations

We define 2 new functions $cosh x = \frac{e^{x}-e^{-}x}{2}$ and $sinh x=\frac{e^{x}-e^{-}x}{2}$
a) Show that each of these fns is the derivative of the other

----------------
Differentiate, either by expressing as a power of e, or by taking logs of boths sides and using implicit differentiation
y=$x^{logx}$

acordin to your definition $sinh = cosh = Untrue$

http://en.wikipedia.org/wiki/Hyperbolic_function
http://mathworld.wolfram.com/HyperbolicFunctions.html

read the definition

$\cosh x$  is actually $\frac{e^{x}+e^{-x}}{2}$.  These are easy derivatives, especially if you split the fractions up.
[hide]$\frac{d}{dx}\left(\frac{e^{x}}{2}+\frac{e^{-x}}{2}\right) = \frac{e^{x}}{2}-\frac{e^{-x}}{2}= \sinh x$.  Use similar steps for $\frac{d}{dx}\sinh x$.[/hide]


number theory
greatest common divisor
algorithm
HMMT

A sequence of positive integers is defined by $a_{0}=1$ and $a_{n+1}=a_{n}^{2}+1$ for each $n\geq 0$. Find $gcd(a_{999},a_{2004})$.

[hide="Solution"] We will calculate the sequence $\bmod a_{999}$.  This gives us

$a_{999}\equiv 0$
$a_{1000}\equiv 1$
...
$a_{1000+k}\equiv a_{k}$

So $a_{2004}\equiv a_{4}\bmod a_{999}$.  Now,

$a_{1}= 2$
$a_{2}= 5$
$a_{3}= 26$
$a_{4}= 677$

Hence $gcd(a_{999}, a_{2004}) = gcd(a_{999}, a_{4})$ by Euclidean Algorithm.  Now we use our technique again and evaluate the sequence $\bmod a_{4}$.

${a_{5+k}\equiv a_{k}}$

Giving us $a_{999}\equiv a_{4}\equiv 0 \bmod a_{4}$.  Hence the gcd is $\boxed{ 677 }$. [/hide]

Great use of the Euclidean Algorithm, BTW this problem was from HMMT 2004 Algebra topic test.


induction
modular arithmetic
function
inequalities
combinatorics proposed
combinatorics

The sequence $\{a_n\}_{n \geq 1}$ of $0$s and $1$s is defined recursively by $a_1=0,a_2=1$, and the following rules for $n \in \mathbb{N}$:

$a_{4n} = 1 -a_{n+1}$;
$a_{4n+2}=a_{n+2}$;
$a_{2n+1} = a_n$.

The first few terms are $0,1,0,0,1,0,0,1,0,0,1,1,\ldots$. Prove that this sequence is not periodic.

--Vesselin

Suppose the sequence is periodic. Then for some $k$, $a_{n+k}=a_n$ for all $n$. It is clear that $k \geq 2$. Write $k$ in the form $m.2^{r+1}+2^r+1$, where $m, r \in Z, \geq 0$. Now we have $a_{n.2^{r+1}+m.2^{r+1}+2^r+1}=a_{n.2^{r+1}}$. If $r \geq 3$, then
\[a_{n.2^r+m.2^r+2^{r-1}}=1-a_{n.2^{r-1}+1}\]
\[a_{n.2^{r-2}+m.2^{r-2}+2^{r-3}+1}=a_{n.2^{r-2}}\]
which has the effect of replacing $r$ by $r-3$. So we may assume that we have this property for $r \in \{0, 1, 2\}$. If $r=2$, then
\[a_{8n+8m+5}=a_{8n}\]
\[a_{4n+4m+2}=1-a_{2n+1}\]
\[a_{n+m+2}=a_n\].
If $r=1$, then $a_{4n+4m+3}=a_{4n}$, and $a_{n+m}=1-a_{n-1}$. Hence $a_{n+2m-1}=a_{n-1}$. If $r=0$, then either $a_{4n+4m}=a_{4n}$, in which case $a_{n+m+1}=a_{n+1}$, or $a_{4n+4m+2}=a_{4n}$, in which case $a_{n+2m+3}=a_{n+1}$. In all of these cases we can replace $k$ by a smaller number. So, since $k$ is clearly quite large, it follows by induction that there is no such $k$. So the sequence is aperiodic.

Hmm, I should have given a more complicated recurrent definition  :) (but nice solution, by the way). My solution (try this!) is algebraic: evaluate $\alpha := \sum_{k \geq 0} \frac{x_i}{2^i}$ explicitly and show that it converges too fast to be rational (the irrationality of $\alpha$ is equivalent to the aperiodicity of the sequence). I can actually show that every similarly defined non-trivial sequence is aperiodic; try the following more general problem:

[b]Theorem. [/b] Let $a_i \pmod{b_i}, 1 \leq i \leq k$ be a finite exact covering of the integers with residue classes. Consider linear functions $d_ix+c_i, 1 \leq i \leq k$, such that:

(i) $c_i,d_i \in \mathbb{N}_0$ with at least one $d_i$ strictly positive;
(ii) $\forall i \in \{1,\ldots,k\}$, $c_i \leq a_i$ and $d_i \leq b_i$, at least one of these two inequalities being strict.

For $g,h : \mathbb{N} \to \{0,1\}$, call $f$ a [i]binary[/i] function of $g$ if either $h \equiv g$ or $h \equiv 1-g$.

Assume that the sequence $\{x_n\}_{n \geq 1}$ of zeros and ones has the property that, for each $i \in \{1,\ldots,k\}$, $x_{b_in+a_i}$ is a binary function of $x_{d_in+c_i}$ (note that this means that the sequence posseses a simple recursive definition). Then, if $\{x_n\}$ is non-constant, it is aperiodic.

--Vesselin


Mikela drove 500 miles on her three-wheeler. She rotated a spare tire with the other tires so that all four tires got the same amount of wear. How many miles of wear did each tire accumulate?

Driving 500 miles on 3 wheels gives a total of 1500 miles of tire-wear.

Splititng this into 4 wheels gives 375 miles each.

We could also have done this by 3/4(500)=375.


1- Find all functions f : R --->R , verifying:
 For all x non-zero real, $f(x+1/x)=f(x)+f(1/x)$.
2- n integer and k real given , find all functions f: R --->R , verifying:
 For all x non-zero real,
[b]f2(xn+1/xn)=f(xn)+f(1/xn)+k[/b]

[u]NB: [/u] f 2 (x)= f(x).f(x)

Very simple.
I do the first part first:

For every a>0 define X(a) ={xi}i>=0 as follows: x(0)=a, x(n)= x(n-1) + 1/x(n-1) for all n>=1.
Now note that for every x>=2 there is a unique number 1<=a<2 such that x belongs to X(a).
This is easy to prove. For each x>0, let the respective a, be called g(x).
Define the function arbitrarily on (0,2). We will now show that f(x) is uniquely defined for all x>2.
Note that for any y>2, y belongs to the following sequence: X(g(y)), and so if we know f(g(a)) and f(1/g(a)) we can find f(y).
no problem is caused by defining f(x) arbitrarily for 2>=x>1. Note that we know f(g(a)) since g(a)<2, and also 1/g(y)<1 So we know f(1/g(a)).
Hence for every y>2 we can find f(y). But would we get two values of f(y) or just one value?
The answer is that we get one value. For we know that g(y) is uniquely defined.

The same can be said for all x<0 . You are also free to choose f(0).
Summary: All you have to do is to difine f(x) on (-2,2) The rest takes care of itself.
So the first part of the problem is solved.

part b) Note that agian as above we can define for each a, a sequence X(a) , and similarly for each x>=2 there is a number 1<=y<2
such that x belongs to X(y). This comes from the fact that if y>2, x^n + 1/x^n - y changes sign for x in (1,2)
Also note that the number of solutions in (1,2) is unique. This  is true since x^n +1/x^n - y is an increasing function for x>(y/2)^(1/n)
So we can define g(x) in the exact same manner as above.
Now agian if we define f(x) on (0,2) arbitaririly we can define all f(x) for all x>0 too. One has to only make sure that for each x
such that x^n and 1/x^n <2 we have that f(x^n) +f(1/x^n) +k>0. In this way we are done.
Note that for each x>2 we can have two values for f(x) , depending on taking the positive square root or negative square root.
Also if n be odd we can do the same for x<0.
If n be even. then f(x)=f(-x) for all x<-2 and so one has to take this in consideration.

Any ways the whole problem is just about making a general construction as i did here.
-Ali


inequalities
function
geometry proposed
geometry

prove that in each triangle we have:

ma(a-b)(a-c) + mb(b-a)(b-c) + mc(c-a)(c-b)>=0.

ma, mb, mc are the medians

obviously it looks like the schur inequality, but how can you proof that?

I think you can't use Schur directly - i.e. They aren't similarly sorted sequences. However , you [i]can[/i] use Valentin Vornicu's generalisation of the Schur Inequality. Check out http://www.artofproblemsolving.com/Wiki/index.php/Schur's_Inequality for this. :)

I reduced the first inequality to proving that \[\sum\left(\frac{1}{a}+\frac{1}{b}\right)(a-c)(b-c)\geq 0,\] where $a,b,c$ are positive reals. From here I have a nice solution.

from there it's just vornicu schur : $\sum{c(s-c)(c-a)(c-b)}\geq 0$, where $s=a+b+c$.  :)

but, we don't need, the transformation, andy, as we can see, $f(a) = m_{a}$ is decreasing an vornicu- schur holds for al monotonic functions, so, the inequality is trivial , by only verifying that f is (monotonic) decreasing: $a\leq b \Longleftrightarrow m_{a}\geq  m_{b}$, but it`s very interesting to notice the "creative spirit" of this problem's "author".

Yeah Vornicu Schur pwns it, but for the sake of compleetness, I solved it without just stating the theorem: just the standard arrangement and small computations(which are the base of the proof for Vornicu-Schur itself)


Show that there exists a continuos function $f: [0,1]\rightarrow [0,1]$ such that it has no periodic orbit of order $3$ but it has a periodic orbit of order $5$.

$x$ is a periodic point of order $n$ if $f^n(x)=x$ and for every $m<n$ we have $f^m(x)\neq x$.
And a periodic orbit of order $n$ is $n$ different points $x_1,x_2,\ldots,x_n$ such that $f(x_i)=x_{i+1}$ for $1\leq i\leq n-1$ and $f(x_n)=x_1$.


How many diagonals does a regular hexagon have?

The formula for the number of diagonals in a regular $ n$-gon is $ \frac{n(n\minus{}3)}{2}$. Plugging in 6 yields $ \boxed{9}$.

Or if you don't know the formula, you can do 6C2, but you have to subtract 6 because a side isnt a diagonal.

6C2-6=15-6=9

Um, yeah.  My friend posted this and he doesn't know that much about this stuff.  So a really really long explanation:

    [asy]pair A=(0,0),
    B=A+5*dir(0),
    C=B+5*dir(60),
    D=C+5*dir(120),
    E=D+5*dir(180),
    F=E+5*dir(240);
    label("A", E, N);
    label("B", F, W);
    label("C", A, S);
    label("D", B, S);
    label("E", C, NE);
    label("F", D, N);
    draw (A--B--C--D--E--F--cycle);[/asy]

This is our hexagon.  The problem asks for a diagonal, which would be a line like this:

    [asy]pair A=(0,0),
    B=A+4*dir(0),
    C=B+4*dir(60),
    D=C+4*dir(120),
    E=D+4*dir(180),
    F=E+4*dir(240);
    label("A", E, N);
    label("B", F, W);
    label("C", A, S);
    label("D", B, S);
    label("E", C, NE);
    label("F", D, N);
    draw (A--B--C--D--E--F--cycle);
    draw(F--D);[/asy]

Because our figure is so small, we can simply draw out all the diagonals:

    [asy]pair A=(0,0),
    B=A+4*dir(0),
    C=B+4*dir(60),
    D=C+4*dir(120),
    E=D+4*dir(180),
    F=E+4*dir(240);
    label("A", E, N);
    label("B", F, W);
    label("C", A, S);
    label("D", B, S);
    label("E", C, NE);
    label("F", D, N);
    draw (A--B--C--D--E--F--cycle);
    draw(F--D);
draw(A--C);
draw(A--B);
draw(A--D);
draw(A--E);
draw(A--F);
draw(B--C);
draw(B--D);
draw(B--E);
draw(B--F);
draw(C--D);
draw(C--E);
draw(C--F);
draw(D--E);
draw(E--F);[/asy]

Wow that looks pretty :D.

We count our diagonals, and get 9 as our answer.  But what if we have a larger polygon, maybe with 18 sides?  From our diagram it is clear that point A has 3 diagonals extending, or 6-3.  We can write this as 6-3 because A has a diagonal connecting to every point, except for itself and two adjacent points.  So for an n-gon, we would have $ n(n\minus{}3)$ diagonals, right?  Well, we have to divide by 2, because we've counted each diagonal twice:  starting from the first point, and starting from the second point!  So we divide by 2 to get $ \frac{n(n\minus{}3)}{2}$, or isabellas formula.  Hope i helped you nerdy97.


From the first 2005 natural numbers, $k$ of them are arbitrarily chosen. What is the minimum value of $k$ to ensure that there is at least one pair of numbers such that one of them is divisible by the other?

[hide="click"]
isn't it just $1004$?

in the worst case, you take all the numbers from $1003$ till $2005$, which are obviously not divisibly by one another...
But from the moment you take a smaller number with them, you get at least one number from $[1003,2005]$ which is a multiple of that number.

[/hide]

You have to justify the "worst case" part of that solution.  It might just be possible, for example, that there exists a set of 1004 numbers with none dividing any other, unless you can prove none such exists.

You can prove with 1003 of those numbers 2 must divide each other.


When I posted in the Mod Forum, I got this, but the post still went through.
[code]
SQL requests not achieved

DEBUG MODE

SQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 2580) GROUP BY ug.user_id' at line 7

SQL Request: SELECT MAX(a.auth_value) as allowed, ug.user_id FROM phpbb_auths a, phpbb_user_group ug WHERE a.obj_type = 'f' AND a.obj_id = 373 AND a.auth_name = 'auth_watch_forum' AND a.group_id = ug.group_id AND ug.user_id IN (, 2580) GROUP BY ug.user_id
Line : 1183
File : functions_post.php
[/code]

I think that's the bug we have/had already 3 threads about :D

Should be fixed now (although I did not see anything in the code that might provoke it  :maybe: ).


calculus
real analysis
algebra unsolved
algebra

1. Consider the sequence in modulo 11: 5,0,7,3,5,0,7,3,5,0,7,3,.... So the 2002nd term is equivalent to the 2nd term, which is divisible by 11. Q.E.D.

2. It is wrong. $ u_1\equal{}0$ is not a prime number.

@: Reply
[hide][quote="arcsin1.01"]1. Consider the sequence in modulo 11: 5,0,7,3,5,0,7,3,5,0,7,3,.... So the 2002nd term is equivalent to the 2nd term, which is divisible by 11. Q.E.D.

2. It is wrong. $ u_1 \equal{} 0$ is not a prime number.[/quote]

with $ n \ge 2$[/hide]

[quote="L_Euler"]with $ n \ge 2$[/quote]$ u_2\equal{}1$ is not a prime. $ u_4\equal{}1$ is not a prime either.

[quote="L_Euler"]1. Let sequence $ (u_n)$:

$ \left\{\begin{array}{l} u_1 \equal{} 5;u_2 \equal{} 11 \\
u_{n \plus{} 2} \equal{} 2u_{n \plus{} 1} \minus{} 3u_n \\
\end{array} \right.$
Prove that: $ u_{2002} \vdots 11$

2. Let sequence $ (u_n)$:

$ \left\{\begin{array}{l}u_1 \equal{} 0;u_2 \equal{} 1 \\
u_{n \plus{} 2} \equal{} 1999u_{n \plus{} 1} \minus{} u_n \\
\end{array} \right.$

Prove that $ u_n$ is prime number.

:( :( :o[/quote]
2) is not true.Notice that $ u_4\equal{}1992.1990$
By the way only when $ n\equal{}2$ then $ u_n$ is a prime.And $ 2$ is also the unique number satisfying that.
Hint:You can get
\[ u_{n\plus{}1}^2\minus{}2.1991u_{n\plus{}1}u_n\plus{}u_n^2\equal{}1
\]
Ofcourse,We can replace $ 1991$ by  $ p,p\in \mathbb P$
the problem which I mentioned will be true

[color=darkblue]This prblem can be done by use method of difference to find formula of this sequences. Hence easy we done for your question.

But this problem must post at Calculus - Real Analysis forum.

http://www.mathlinks.ro/index.php?f=67[/color]