[hide](1+x^2)(1-x^3)
1-x^3+x^2-x^5
-x^5-x^3+x^2+1
D.[/hide]
You forgot to put the answer in descending order.  :P

[quote="Fish_Factorial"]You forgot to put the answer in descending order.  :P[/quote]

But he did put it in ascending order  :D

silverfalcon said:

[quote]Source: USA AMC 12 1987, Problem 1 [/quote]

[hide]i got d[/hide]

[hide]Answer D by foil[/hide]
I always keep thinking there's some kind of trick on these polynomial questions, but the subject topic sort of gave it away. :)

[hide]D[/hide]

This is totoally random but doesn't $\text\ FOIL$ stand for something?
I'm pretty sure it does but I can't remember what it stands for.
If not then why do they call it $\text\ FOIL$?

[b]F[/b]irst
[b]O[/b]uter
[b]I[/b]nner
[b]L[/b]ast

[quote="hoshattack"]this is kind of random, but why do 10 ppl have to put up the same answer to a problem?!  if it has been solved by a few ppl u dont have to solve it to!! :|[/quote]I guess people have a knack for 'showing' to others that they can solve the problem too, and by 'showing' I mean it literally  :roll:

but still, once it has been solved anybody could copy the same thing.  And this forum is not about showing off, but LEARNING

oh im sorry. i didnt read the posts right above me.  :blush:

Hey buzzer and Jose, you guys should post a solution at least even if there are many others already.

First:
$1\times1$
Outer:
$1\times-x^3$
Inner:
$x^2\times1$
Last:
$x^2\times-x^3$
then add the four together.
$1+x^2-x^3-x^5$
The answer is $\boxed{D}$

Just to clarify for everyone what foil is and how to do this problem :)

The answer is D due to FOIL.

First
Inner
Outer
Last


I'm asking this for my brother because he never posts on this forum:

Would the Intro books be good for preparing for the AIME if you know the general ideas but not the details?

Thanks in advance.

The Intro books are probably good for the first 3 to 5 problems on the AIME.

[quote="xpmath"]I don't know if the Intro books are great for AIME;if you just want to  make AIME, the challenge problems are good.[/quote]
His goal is to make the USAMO

Yeah, I seem to notice maybe 1 or 2 AIME problems at the end of the Intro books, but they seem as if they might be # 2 or 3 or something.

To make USAMO, the Intermediate C+P and Vol. 2 would be the best. And old problems.

[quote="gauss1181"]I'm not exactly sure about those ,but I do know that the 21st Century Contests CD can be a great help. It contains AIME competitions all the way back from 2001.[/quote]

which is what XPmath said about old problems.  :wink:


Let $\{a_k\}_{k\geq 0}$ be a sequence given by $a_0 = 0$, $a_{k+1}=3\cdot a_k+1$ for $k\in \mathbb{N}$.

Prove that $11 \mid a_{155}$

What is $a_0$? 0? 11? 121?

Using $b_n =a_n + \frac 1 2$, we deduce that $b_{n+1} = 3b_n$ so that $b_n = 3^nb_0$.
Therefore $a_n = 3^n(a_0 + \frac 1 2 ) - \frac 1 2$

It follows that $a_{155} = 3^{155}a_0 + \frac {3^{155} - 1} 2$

But, it is easy to see that $3^5 = 1$ mod[11]. Since $3^{155} - 1$ is even, we deduce that $a_{155} = a_0$ mod[11].

Thus, we are back to my initial question : what is $a_0$?

Pierre.

0. weird, apparently it vanished.

anyway, I think I proved it for a higher power of 11... any try? :P

That was computable. Let's try making it a bit generalized in various cases, one of these cases being below.
Prove that for any integer $k$, $a_{110k+54}$ is divisible by $11$

Divide by $ 3^{k+1}$ and put $ b_n= \frac {a_n} {3^n} $ to get $ b_{n+1} = b_n + \frac 1 {3^{n+1}} $
This directly gives $a_n= \frac  {3^n - 1}{2}$
Now we can generalize anyway we want.
As $Ord_{11} 3 = 5 $ we have $11|a_n$ for all $n$ divisible by 5
@above I think you meant $110k+55$


Everytime I log on, I check the "remember me" option, but when I come back, it still says that I haven't logged in.

Even though it says "View posts since last visit (0)", when I click on it, it tells me "Search found 22 matches". Before the update, only the new/edited posts are displayed, but now, it seems rather random.

About the remember me option, you must delete your cookies (if you can locate the mathlinks cookie, just that) and do the operation again. It will work this time.

About the unread flags, yes, I know, it's a bug that I coudn't fix up to know, but I am working on it.

Keep the reports coming in.


function
limit
trigonometry
algebra
domain
real analysis
topology

If $\forall \epsilon>0 \exists \delta(\epsilon): \text{ } \forall x,y \in (0,+\infty) : |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$ ( i dont know how it calls, equally continuous???) then what can we state about f(+0), f(+oo)?

[quote]i dont know how it calls[/quote]
i'm not sure if this is what you're asking for, but here it is

Just say i have a function $f$, and i think the limit of the function at point $a$ is $L$.

Epsilon delta basically makes a box.  the two vertical lines are at $a + \delta$ and $a - \delta$.  the horizontal lines are at $L + \epsilon$ and $L - \epsilon$.  If the limit is really the limit, then no matter how small epsilon and delta are, the limit will be in the center of the box.

For every value of epsilon greater than zero, a box can be created that the limit is inside of

for all $\epsilon$ $|L - f(x)| < \epsilon$ there exists $0 < |x - a| < \delta$

this [url=http://www2.scc-fl.edu/lvosbury/CalculusI_Folder/EpsilonDelta.htm]link[/url] animates that.

f has been defined such that a limit exists for every value between 0 and positive infinity, which means continuity

That's uniform continuity; the function is uniformly continuous.

At zero, the limit exists and is finite. At $\infty$, anything can happen.

i find that problem in the book Sadovnichiy Podkolzin "USSR students olympiad"
there was a solution like that:
since function is uniformly continuous we applying the Cauchy criteria get that the limit at +0 exist
can u tell me why it is so?
for the +oo its said that sinx is uniformly continuous and doesn't have a limit at +oo

Choose $\delta_n$ decreasing so that $|f(x)-f(y)|<\frac1n$ when $|x-y|<\delta_n$. Then $|f(c)-f(\delta_n)|<\frac1n$ whenever $0<c<\delta_n$. Then the sequence $a_n=f(\delta_n)$ is Cauchy ($|a_n-a_m|<\max\{\frac1n,\frac1m\}$), so it has a limit $A$. Since $|A-\delta_n|<\frac1n$, $|f(c)-A|<\frac2n$ when $c<\delta_n$, and $\lim_{x\to0}f(x)=A$.

A nice theorem on uniform continuity:
Suppose $X$ is a metric space and $Y$ is a subset of $X$. If $f$ is a uniformly continuous function from $Y$ to $\mathbb{R}$ (or any other complete metric space), $f$ can be uniquely extended to a uniformly continuous function from the closure $\overline{Y}$ to $\mathbb{R}$.

A quick paraphrase of what jmerry just said:

A continuous function maps all convergent sequences to convergent sequences.

A uniformly continuous functions maps all Cauchy sequences to Cauchy sequences.

To understand the distinction, consider $f(x)=\cos\left(\frac{\pi}x\right)$ on $(0,1].$

This function is continuous (on its domain) but not uniformly continuous.

Now consider the sequence $x_n=\frac1n.$ This is certainly a Cauchy sequence. However, restricting our field of vision to the domain of $f,$ namely, the interval $(0,1],$ we must say that this is not a convergent sequence. We say that because this sequence does not have a limit in that domain.

$f,$ being continuous, maps convergent sequences to convergent sequences. But since this $x_n$ is not a convergent sequence, $f(x_n)$ is free to behave badly. Indeed, $f(x_n)=(-1)^n$ and diverges.

But $x_n$ was a Cauchy sequence and $f(x_n)$ is not a Cauchy sequence. If we didn't already know that $f$ was not uniformly convergent, that would prove that it is not uniformly convergent.


AMC
AIME
probability
algebra
binomial theorem

For $\{1, 2, 3, \dots, n\}$ and each of its nonempty subsets a unique [b]alternating sum[/b] is defined as follows: Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers.  (For example, the alternating sum for $\{1, 2, 4, 6, 9\}$ is $9 - 6 + 4 - 2 + 1 = 6$ and for $\{5\}$ it is simply 5.)  Find the sum of all such alternating sums for $n = 7$.

[hide]Let $S$ be a non-empty subset of $\{1,2,3,4,5,6\}$. Then the "alternating sum" of $S$ plus the alternating sum of $S$ with 7 added is 7. (All elements of $S$ are moved over one, so each is added and subtracted.) There are 63 possible sets $S$. This sum adds up all possible subsets of $\{1,2,3,4,5,6,7\}$ except $\{7\}$, which has alternating sum 7. Then the final sum is $63*7+7=64*7=448$.[/hide]

(If only all AIMEs were as easy to me as the 1983 one...)

[hide="Scorpius 119s Modified Solution"]Let $S$ be a non-empty subset of $\{1,2,3,4,5,6\}$.

Then the alternating sum of $S$ plus the alternating sum of $S$ with 7 included is 7. In mathematical terms, $S+ (S\cup 7)=7$. This is true because when we take an alternating sum, each term of $S$ has the opposite sign of each corresponding term of $S\cup 7$.

Because there are $63$ of these pairs, the sum of all possible subsets of our given set is $63*7$. However, we forgot to include the subset that only contains $7$, so our answer is $64\cdot 7=448$.[/hide]

[hide="Well..."]All you really have to do is notice that the sums inclusive of $7$ cancel out those not inclusive of $7$, then count the number of sets with $7$ in them, which, by Binomial Theorem, is $2^6$, and $7\cdot 2^6=\boxed{448}$.[/hide]

without using the binomial theorem you can come up with the number of subsets in the power set:
each element (all seven) is either there or isn't there; thus each has two options associated with it. $2^7 = 128$.
The way I quickly worked the problem was:
Of all terms, $1/2$ have a seven. We add all these sevens (which all are positive) for $64*7$. From now on, we quickly see each sum is $0$. This is because (conceptually) each term is either positive or negative, with an equal probability of $1/2$. Thus they all cancel out.
(If these concept ideas don't work for you, you can work out for the first couple terms why this is so. For instance, $5$ is the greatest term in $1/4$ of the subsets, it is second greatest in $1/2$ subsets, and 3rd greatest in $1/4$ subsets. Since it is negative when it is second, the negative probability equals the postive probability ($1/4 + 1/4 = 1/2$).)

[quote="scorpius119"][hide]Let $ S$ be a non-empty subset of $ \{1,2,3,4,5,6\}$. Then the "alternating sum" of $ S$ plus the alternating sum of $ S$ with 7 added is 7. (All elements of $ S$ are moved over one, so each is added and subtracted.) There are 63 possible sets $ S$. This sum adds up all possible subsets of $ \{1,2,3,4,5,6,7\}$ except $ \{7\}$, which has alternating sum 7. Then the final sum is $ 63*7 \plus{} 7 \equal{} 64*7 \equal{} 448$.[/hide]

(If only all AIMEs were as easy to me as the 1983 one...)[/quote]

What do you mean by "alternating sum" of $ S$ plus the alternating sum of $ S$ with 7 added is 7?

Every subset with the element $ 7$ corresponds to a subset that does not contain the element $ 7$ (simply remove the $ 7$), and vice versa (simply append $ 7$). Let's take a look at an example of such a pair of sets:

$ \{1, 3, 5 , 7 \}$ and $ \{ 1, 3, 5 \}$.

The alternating sum of the first subset is $ 7\minus{}5\plus{}3\minus{}1$, and the alternating sum of the second subset is $ 5\minus{}3\plus{}1$.

If you add them, the 1's, 3's, and 5's cancel out, and you get 7.

So, we can pair subsets without $ 7$ with their corresponding subsets with $ 7$, and the sum of their alternating sums will be $ 7$.

There are $ 2^7\equal{}128$ subsets of $ \{ 1 \ldots 7 \}$, thus there are $ 64$ subset-pairs, and the sum is $ 64 \cdot 7 \equal{} 448$.

In general, for the set $\{ 1,2, \cdots , n \}$, the sum of all alternating sums is $n \cdot 2^{n-1}$.

Sorry for the revive, but I don't understand the second solution on the AoPS Wiki for this problem. It states:

[quote = AoPS Wiki] Consider a given subset $T$ of $S$ that contains 7; then there is a subset $T'$ which contains all the elements of $T$ except for 7, and only those. Since each element of $T'$ has one element fewer preceding it than it does in $T$, their signs are opposite; so the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 * 2^6 = \boxed{448}$. [/quote]

Why is it that the alternating sums of $T$ and $T'$ add to $7$? Consider $T = {5,6,7}$. Then, $T' = {5,6}$. The alternating sum of $T$ is $5-6+7 = 6$ and the alternating sum of $T'$ is $5-6 = -1$, but $6 + (-1) \ne 7$. Did I understand the solution wrong?

The alternating sum is [i]decreasing[/i] so it would be $7-6+5$ and $6-5$ which do cancel.

Two-liner.
Let $S$ be a set that does not contain 7 and let $U$ be the set $S$ but with 7 added as an element. We note that the sum of the alternating sums of these two sets is just 7. There are $64$ such pairs of sets like these so our answer is $64\cdot 7=448$.

It was a bump from 2017. Also, if you are posting quality solutions, its okay. :)

[quote=HamstPan38825]Please don't bump threads from 15 years ago.[/quote]

Tell this to Evan Chen.
[quote=v_Enhance]3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.[/quote]
I don't see how I did anything wrong.

Ah, yes, @below, I see posting solutions are annoying. So that's how AoPS is now isn't it?

ah, its not against the rules but some of us find it kind of annoying. Sorry for being too authoritative.

[quote=OlympusHero]@2above, please stop being mean. Let's not have a flamewar on HSM. Goodness gracious.[/quote]

Why do you always say I'm being mean?
I'm just saying that @2above thinks posting solutions are annoying because that's what @2above literally said.
And by the way, you're the one that always says [quote]Let's not have a flamewar on [insert forum here][/quote] when nobody was going to have one anyways.

Alright, you can do whatever you want here. I know I'm being too controlling, but a lot of people consider posting solutions like this postfarming, unless you really have a new idea or approach. Well - at least it's better than pointless bumping. Sorry, though.

[quote=OlympusHero]@2above, please stop being mean. Let's not have a flamewar on HSM. Goodness gracious.[/quote]

Stop the hypocrisy :(

Compute the alternating sums to be $1$ for $n = 1$, $4$ for $n = 2$, $12$ for $n = 3$, and $32$ for $n = 4$. If we look at the sums with the subtraction and the addition, we notice that for $n = k$ all terms except the $k$'s cancel out. For $n = 1$, there are $1$ ones, for $n = 2$ there are $2$ twos, for $n = 3$ there are four threes, and so on. Thus we can see that $n = k$ gives a sum of $k \cdot 2^{k-1}$, for an answer of $7 \cdot 2^6 = \boxed{448}$.

why did you come to this thread and say something, then come back again a year later to post a solution lol

I wanted to explain my intuition for the (pretty much majority) solution to this problem. For problems like these where you need to take subsets, it is often a good idea to try and solve the problem for smaller $n$, like $1, 2, 3, 4$, and then see if you can use a recurrence relation for bigger $n$ (and then bash or look for a pattern). For this problem, we end up seeing that if $a_k$ is the number of solutions for $n=k$, then $a_{k+1}=a_k+(k+1)\cdot 2^k - a_k=(k+1)\cdot 2^k$

[b]Solution:[/b] Trying to look for a pattern, we notice that our answer is just $n \cdot 2^{n-1}$. So, our final answer is $448$.

[u]Motivation:[/u] It's a 1983 AIME. Just bash. GGs.

[i]Comments:[/i] Ok but seriously, this stems from the fact that "DURR WE WANT STUFF TO CANCEL." I mean, the alternative signs are just beggin' to do some canceling. Notice that when we take a subset (even an empty one) from ${1, 2, \dots, 6}$ and take its unique sum and add that to the unique sum of that same subset with $7$, we are just left with $7$. This means our answer is $7$ times the number of subsets (even the empty one) of ${1, 2, \dots, 6}$, which gives us our answer.

I do not have a rigirous solution. But we can find a pattern for smaller values of $n$ and make a claim. So we claim that this is just $n \cdot 2^{n-1}$, if we plug in $n=7$, we have this is $7 \cdot 64=\boxed{448}.$


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Is it true that every monic irreducible polynomial over $ \mathbb{Q}$ with all real roots is the characteristic polynomial of a symmetric matrix over $ \mathbb{Q}$?

The answer is no.
This is question 11321 from the American Mathematical Monthly, Vol. 114, No. 9, Nov 2007.
You should see answers in the AMM in a couple of months I think.


Find all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y) \quad \forall x, y \in \mathbb{R} \].

I think it's from some Belarus contest.

i think i have a start

[hide]

first you check for when $f(x) = c$ for some constant $c$, in which case you have:

$c + c^2 = c + c + c \implies c= 0$ or $2$

checking both $f(x) = 0$ and $f(x) = 2$ work

now we quickly see that $f(x) = x$ works

but from there im not sure how to prove that there are anymore/there aren't any more....

[/hide]

Thanks for replying. I also have some start...
[hide]

If we put $x=y=0$ we get that $f(0)(f(0)-2)=0$, and if $f(0)=2$ then (by putting $y=0$) $f(x)=2$ for all $x$ (first solution). So, $f(0)=0$.

$x=y=2$ yields $f(2)(f(2)-2)=0$, and I hoped to get that, if $f(2)=0$, then $f(x)=0$ for all $x$ (second solution), but I didn't manage to get that...

Also, if we put that $y=\frac{x}{x-1}$, we get that

$f(x)f(\frac{x}{x-1})=f(x)+f(\frac{x}{x-1})$ what is almost equivalent to $f(\frac{x}{x-1})=\frac{f(x)}{f(x)-1}$. That could help if we get that $f(x)=x$ for integers $x$.

[/hide]

Well, I solved it for all integers.
Subst $x=1, y=2$ to get $f(2)=2f(1)$. Also, $f(2)=0$ or $f(2)=2$ and because if $f(2)=0, f(x)=0$ the only case left is $f(2)=2$. Then $f(1)=1$.
Now subst $x=a, y=1$ to get:
$f(x+1)=f(x)+1$ which gets the solution $f(x)=x$ for all integers.

[quote="rem"]Well, I solved it for all integers.
Subst $x=1, y=2$ to get $f(2)=2f(1)$. [/quote]

Can you explain this, please? Don't we get $f(3)$ on the left side if we use that?
I feel like an idiot..

[quote="rem"]Also, $f(2)=0$ or $f(2)=2$ and because if $f(2)=0, f(x)=0$ [/quote]

This part too.  :( How do you get that if $f(2)=0$, then $f(x)=0$?

[quote="idioteque"][quote="rem"]Well, I solved it for all integers.
Subst $x=1, y=2$ to get $f(2)=2f(1)$. [/quote]

Can you explain this, please? Don't we get $f(3)$ on the left side if we use that?

This part too.  :( How do you get that if $f(2)=0$, then $f(x)=0$?[/quote]

I am so dumb :wallbash_red:
ignore my solution.
i'll get it hopefully

The following gives all possibilities for $f$ on the integers:[hide]Previous posters have established that $f(0) = 0$ or $f(0) = 2$, and that if $f(0) = 2$ then $f \equiv 2$.  Also, we've got that $f(2) = 0$ or $f(2) = 2$.  We just need one other thing: if we take $y\mapsto 1$ we get $f(x + 1) + f(x)f(1) = 2f(x) + f(1)$ and in particular if $(x, y) \mapsto (1, 1)$ we get $f(2) + f(1)^2 = 3f(1)$.  Therefore:

a) If $f(2) = 0$, we have $f(1)^2 = 3f(1)$ so either $f(1) = 0$ or $f(1) = 3$.
(i) If $f(1) = 0$ we have $f(x + 1) = 2f(x)$ and since $f(1) = 0$ we have $f(n) = 0$ for all integers $n$.
(ii) If $f(1) = 3$ we have $f(x + 1) = 3 - f(x)$ and since $f(1) = 3$ this gives $f(2n) = 0$ and $f(2n + 1)= 3$ for all integers $n$.

b) If $f(2) = 2$, we have $2 + f(1)^2 = 3f(1)$ so either $f(1) = 1$ or $f(1) = 2$.
(i) If $f(1) = 1$ then $f(x + 1) = f(x) + 1$ so $f(n) = n$ for all integers $n$.
(ii) If $f(1) = 2$ then $f(x + 1) = 2$ which is a contradiction since $f(0) = 0$, so no solution here.

Thus, there are exactly 3 possibilities for the values of $f$ on the integers.
[/hide]

very good! :)
How do you prove that these functions also work for [b]all[/b] reals? :?

Well, it's not clear at all how to "naturally" extend one of those three functions to all the reals.  And as for the other two, I haven't been able to figure out a nice way to pick up more than a few select values (like idioteque did).

I've got most of it.

\[ f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)\quad (0)  \]

Putting $x=y=0$ in (0), we get $f(0)(f(0)-2)=0$.
Case 1: $f(0)=2$, put $y=0$ in original to get $f(x)=2 \quad \forall x\in \mathbb{R}$.
Checking this we see that it is a solution.
Case 2: $f(0)=0$ Put $x=y=2$ in (0) to get $f(2)(f(2)-2)=0$, and $x=y=1$ to get that \[ f(2)=3f(1)-f(1)^2 \quad(1)  \].
Case 2.1: $f(2)=0$, now from (1) $f(1)(3-f(1))=0$.
Case 2.1.1 $f(1)=3$. Substitute $y=1$ in (0) to get that $f(x+1)=-f(x)+3$. Use that to prove by induction that $f(2n)=0$ and \[ f(2n+1)=3 \quad (2)  \] for all integers $n$. In (0) put $y=\frac{x}{x-1}$ to get \[ f(\frac{x}{x-1})=\frac{f(x)}{f(x)-1}\quad (3)  \]. Now we get that $f(\frac{1}{2})=f(\frac{3}{2})=f(\frac{3}{4})=\frac{3}{2}$, (put $x=-1, 3, -3$ in (3) and use (2)). But now put $x=\frac{1}{2}$ and $y=\frac{3}{2}$ in (0) to get contradiction. No solutions here.
Case 2.1.2 $f(1)=0$ Only case I haven't solved. I got only $f(x)=0 \forall x\in\mathbb{Z}$.

Case 2.2: $f(2)=2$, and from (1) $3f(1)-f(1)^2=2$.
Case 2.2.1: $f(1)=1$ Put $y=1$ in (0) to get
$f(x+1)=f(x)+1 \quad \forall x\in \mathbb{R}$. From this by
induction we get $f(x)=x$ for all integers $x$, and \[ f(x+n)=f(x)+n \quad (4)  \] for all reals $x$ and all integers $n$. Now in (0) put
$x=\frac{m}{n}$ and $y=n$ (where $m$ and $n$ are integers):
\[ f(\frac{m}{n}+n)+f(\frac{m}{n})f(n)=f(m)+f(\frac{m}{n})+f(n)  \]
\[ \Longleftrightarrow f(\frac{m}{n})+n+nf(\frac{m}{n})=m+f(\frac{m}{n})+n  \]
\[ \Longleftrightarrow f(\frac{m}{n})=\frac{m}{n}  \]
\[ \Longleftrightarrow f(x)=x \quad \forall x\in \mathbb{Q}  \].

Now I'll prove that $f(x+q)=f(x)+q \quad \forall x\in\mathbb{R}, q\in\mathbb{Q}$.

Subst $x=xn$ and $y=\frac{1}{n}$ in (0) to get
\[ f(xn+\frac{1}{n})=f(x)+f(xn)+\frac{1}{n}-\frac{1}{n}f(xn). \quad (5)  \]
Now substitute $x=xn+m$ and $y=\frac{1}{n}$ to get

\[ f(xn+m+\frac{1}{n})+f(xn+m)f(\frac{1}{n})=f(x+\frac{m}{n})+f(xn+m)+f(\frac{1}{n})  \]

Use (4) and (5) on this to get

\[ (x+\frac{m}{n})=f(x)+\frac{m}{n}  \] what was what I wanted, $f(x+q)=f(x)+q \quad \forall x\in\mathbb{R}, q\in\mathbb{Q}$.

Now put $y=q$ in (0) to get that $f(qx)=qf(x)$ if $q$ is rational. From this we get $f(-x)=-f(x)$, so the function is odd (is this the expression?). Put $y=-x$ and apply that function is odd to get $f(x^2)=f(x)^2$, meaning $f(x)\geq 0$ if $x\geq 0$ and $f(x)\leq 0$ if $x\leq 0$.

Now assume that for some $x$, $f(x)<x$. Take rational $q$ such that $f(x)<q<x$ (between any two real numbers there is one rational). Now $q>f(x)=f(x-q)+q\geq q$ (since $x-q>0$), which is an obvious contradition. Similiarly, if there is $x$ such that $f(x)>x$. Again choose rational $q$ such that $f(x)>q>x$. Now $r<f(x)=f(x-r)+r\leq r$ (since $x-r<0$), contradiction.

Hence $f(x)=x$ for all $x\in \mathbb{R}$. We check that this is also a solution.

Case 2.2.2: $f(1)=2$. Put $y=1$ in (0) to get $f(x+1)=2$ which contradicts $f(0)=0$.

If someone could help with case 2.1.2 ($f(0)=f(1)=f(2)=0$), I'd be grateful.

EDIT: You don't have to read the whole solution, but please try to see if you can get sth in case $f(0)=f(1)=f(2)=0$.


limit logarithms algebra function domain LaTeX calculus

How to prove that:
\[ \lim_{x \to 0} x^x =1  \]

How do you know that:
$\lim_{x\to 0} x\ln(x) = 0?$

For $x$ near 0, we may as well look at $\frac{-\ln(y)}{y}$ and let $y \to \infty$, and then we can use L'Hopital's Rule, which gives us $\frac{-1/y}{1}$ which clearly goes to 0 as $y$ grows.

You could also use the "fundamental" limit $\lim_{y \to \infty} \frac{\ln y}{y} = 0$.

[quote="t0rajir0u"]The limit as written is undefined.

You want $\lim_{x \to 0^{+} } x^x$.[/quote]

Usually, when one writes $\lim_{x \to 0} f(x)$, one also thinks about the maximum domain of definition of $f(x)$ (unless the domain is stated specifically). So, if $0$ is an accumulation point for $D_\textrm{max}$, then one can think about the limit. (which may exist or not)

By the way, what's with the weird $\LaTeX$?

[quote="perfect_radio"]By the way, what's with the weird $\LaTeX$?[/quote]

It's been showing up all over the site.  It's very strange  :huh:

[quote="perfect_radio"]By the way, what's with the weird $\LaTeX$?[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88819

[quote="Xevarion"]
$\lim_{x\to 0} \ln(y) = \lim_{x\to 0} x\ln(x) = 0$
$\lim_{x\to 0} y = e^0 = 1$[/quote]
I'm a high school student, and I didn't know that there's the formula:
\[ \lim_{x \to c}\ln f(x)=C \implies \lim_{x \to c}f(x)=e^C\ ?  \]

I want to ask, if the following one is exist:
\[ \lim_{x \to c}\ln f(x) = \ln \lim_{x \to c}f(x).  \]

Yes, if the function behaves nicely enough (we don't want the limit of $f(x)$ to be 0...). This is just from the formal definition of the limit.

$\lim_{x\to c} f(x) = L$ if $\forall \epsilon > 0\ \exists \delta > 0$ such that $|x-c|<\delta \Rightarrow |f(x) - L| < \epsilon$.

If you look at the limit of $\ln(f(x))$, you will have $|\ln(f(x)) - \ln(L)|$, and if $f(x)$ gets close to $L$, then $\frac{f(x)}{L}$ will get close to 1, so $|\ln(f(x)) - \ln(L)|$ also gets small. So $\ln(L)$ is the limit we're looking for, and we see that the log of the limit equals the limit of the log.  :) I hope this is clear.

If $\lim_{x \to a}g(x) = b$ and $\lim_{x \to b} f(x)= \ell=f(b)$ then $\lim_{x \to b} f(g(x))= \ell$

which is a bit more general than assuming continuity.

 :)

Edit : I messed it up a bit ... now it is correct i think.... :D

If you don't assume continuity, you have to add an additional condition: $g(x) \neq b$ for some neighbourhood $V \setminus \{ a \}$.


function
invariant
complex numbers
combinatorics proposed
combinatorics

I really enjoyed this one :).

We are working with words made up of the letters $a$ and $b$. We can change the words either by inserting a sequence of the type $www$, where $w$ is a word, anywhere inside a given word, or by deleting such a sequence from a word.

Can we reach $ba$ if we start with $ab$?

wow, i have reached a nice idea :heli:  :trampoline: .
I dont know if it is true but here it is:

I think the answer is $NO$.
Let us consider $a=f(x)$ and $b=g(x)$,such that

$(f\circ f \circ f)(x)=x,(g \circ g \circ g)(x)=x$

Now we have $ab=(f\circ g)(x)$  and if we add $aaa,bbb$ somewhere  then we wont have any change cause $aaa=(f\circ f\circ f)(x)=x=(g\circ g\circ g)(x)=bbb$.

So

Now all that we should do is find two functions like $f,g$ such that

$f(g(x)) \neq(f(g(x))$.

Am i right  :)  :) ?

[color=red]Here are two functions satisfying my condition.[/color]

Let $\alpha_1,\alpha_2$ be two other 3th roots of unit in complex numbers beside of $1$

Let $f(x)=x^{\alpha_1}$ and $g(x)=\alpha_2 x$

So we have $(f \circ f \circ f)(x)=x^{\alpha_1^3}=x$

and we have $(g \circ g \circ g )(x)={\alpha_2}^ 3x$

And we know that

$f(g(x))={\alpha_2x}^{\alpha_1} \neq \alpha_1x^{\alpha_2}=g(f(x))$

I think there might be something we could do.
For any sequence $S=a_1 a_2\ldots a_n$ $f : S\rightarrow \mathbb{Z}_3$ so that $f(S)=[\sum_{i=1}^n a_i\cdot i] $ mod $3$, where you assign to $a$ and $b$ different values mod $3$. Now $f(S_1)\neq f(S_2)$ and I hope $f$ is invariant under that transformation  :?

Yes, I think that works. I think I had essentially the same invariant, but I worded it differently :).

lomos_lupin, now that I understand what you wanted to do, I don't think that's enough. The triplicates $aaa,bbb$ are not the only ones. You also have $ababab$, for example, and infinitely many others. So $(f\circ f\circ f)(x)=(g\circ g\circ g)(x)=x,\ \forall x$ does not suffice. You also want, for instance, $(f\circ g\circ f\circ g\circ f\circ g)(x)=x,\ \forall x$, and so on.

Basically, what you're trying to do is construct the Burnside group $B(2,3)$, i.e. a group with two generators in which every element (except for the identity) has order $3$. This group has $27$ elements, so it's not an easy task to construct its multiplication table.

[quote="grobber"]

Basically, what you're trying to do is construct the Burnside group $B(2,3)$, i.e. a group with two generators in which every element (except for the identity) has order $3$. This group has $27$ elements, so it's not an easy task to construct its multiplication table.[/quote]

well, this is precisely what i had in mind as soon as i saw the question!

[quote]
lomos_lupin, now that I understand what you wanted to do, I don't think that's enough. The triplicates $aaa,bbb$ are not the only ones. You also have $ababab$, for example, and infinitely many others. So $(f\circ f\circ f)(x)=(g\circ g\circ g)(x)=x,\ \forall x$ does not suffice. You also want, for instance, $(f\circ g\circ f\circ g\circ f\circ g)(x)=x,\ \forall x$, and so on.

Basically, what you're trying to do is construct the Burnside group $B(2,3)$, i.e. a group with two generators in which every element (except for the identity) has order $3$. This group has $27$ elements, so it's not an easy task to construct its multiplication table.[/quote]

Maybe i am missing something really easy but
Would you please ,explain a little ,grobber,?
In the problem we can insert $aaa,bbb$ so i made the condition $fff(x)=x=ggg(x)$
to be free to add $f \circ f \circ f$ or $g \circ g \circ g$ where ever we want.
But why ababa :?:  :?:  :?  :?

Hmm.what is Burnsid group :fool: ? :blush:

The problem says that you can insert or delete [b]any[/b] triplicate. $ababab$ is also a triplicate: it's $ab$ repeated three times. So is $abaabaaba$, and so is $aaabaaabaaab$, and so on. You need to acount for these as well with your functions. Taking $f^3(x)=g^3(x)=x$ does not suffice; you need extra conditions.

As for the other question, I suggest you google it. I bet you'll find much more accurate answers than what I could provide :).

Let $f(w)=(s_{ab}, s_{ba}, s_{aa}, s_{bb})$ where $s_{ij}$ is the number of adjacent letters $i,j$ in the word $w$ modulo 3.

So $f(ab)=(1,0,0,0)$ and $f(ba)=(0,1,0,0)$.

It can be easily verified that if word $w'$ is obtained from $w$ by inserting or deleting some triplicated word, then
either $f(w')=f(w)$ or $f(w')=f(w)\pm (2,2,1,1)$ (all operations are performed modulo 3).

Since $f(ba)-f(ab)=(0,1,0,0)-(1,0,0,0)=(2,1,0,0)$ is not a multiple of $(2,2,1,1)$, the word $ba$ [b]cannot[/b] be obtained from $ab$ with insertions/deletions of triplicated words.

Nice solution ,Opposite to my stupid nonsenses :( .
Welldone.


vector
advanced fields
advanced fields unsolved

N.B. I can't type the symbol for vector, so instead i will type vect.

So how do you solve :

 | R - (vect.R' - vect.a)/|vect.R - vect.a| |

the term (vect.R' - vect.a)/|vect.R - vect.a| is less than one due to physics.

I need help expanding the expression i posted.

How would you expand 1-x? It seem already a good expression.  :)


geometry
geometric transformation
reflection
power of a point

In the diagram, $ P$ is any point on a diameter $ AB$ of a circle; $ QPR$ is a chord such that $ /angle APQ \equal{} 45$, prove that $ AB^2 \equal{} 2PQ^2 \plus{} 2PR^2$

Very nice!! :D
[hide]Let O be the center of the circle. Reflect R over AB to get S, on the circle. Then QP^2+PS^2=QS^2. We have to show QOS=90, or QOPS is cyclic. But $ \angle OQP \equal{} \angle ORP \equal{} \angle OSP$![/hide]

How does angle QOS being 90 degrees have to do with QOPS being cyclic?

He used the fact that $ QOPS$ is cyclic (proven by the last statement) to deduce that $ m\angle QOS\equal{}90^\circ$. This is because $ m\angle QOS\equal{}m\angle QPS\equal{}90^\circ$ as shown by $ QP^2\plus{}PS^2\equal{}QS^2$.

Oh, I knew that QOS and QPS are both right angles, but I didn't realize that QS is actually the diameter for the circle that circumscribes the quadrilateral.  :wink:

Another way you can do it is to draw $ QB$ and $ QA$ so that $ <AQB\equal{}90^{\circ}$.  Drawing the altitude from $ Q$ and labeling it $ a$, labeling $ AB$ as $ D$, $ AP$ as $ b$ and $ QP$ and $ QR$ as $ c$ and $ d$, respectively, we can use Power of a Point and $ 45\minus{}45\minus{}90$ triangles in order to get an equation solving for $ QP^{2}\plus{}PR^{2}$.  We can accomplish this by using the fact that we can find $ QP\plus{}PR$ in terms of $ D,a$ and $ b$.  We can also find $ (QP)(PR)$ from Power of a point and, using this information, we can go on to find $ QP^{2}\plus{}PR^{2}$.  After a little bit of work, multiplying by $ 2$, we get the desired result.

[quote="Z-coli"]Another way you can do it is to draw $ QB$ and $ QA$ so that $ < AQB \equal{} 90^{\circ}$.  Drawing the altitude from $ Q$ and labeling it $ a$, labeling $ AB$ as $ D$, $ AP$ as $ b$ and $ QP$ and $ QR$ as $ c$ and $ d$, respectively, we can use Power of a Point and $ 45 \minus{} 45 \minus{} 90$ triangles in order to get an equation solving for $ QP^{2} \plus{} PR^{2}$.  We can accomplish this by using the fact that we can find $ QP \plus{} PR$ in terms of $ D,a$ and $ b$.  We can also find $ (QP)(PR)$ from Power of a point and, using this information, we can go on to find $ QP^{2} \plus{} PR^{2}$.  After a little bit of work, multiplying by $ 2$, we get the desired result.[/quote]

Yeh, this is pretty much what i did, but i got stuck. What do you mean by power of a point, could you simply explain.

Namely that $ (AP)(PB)\equal{}(QP)(PR)$.
Therefore, you can find $ 2(QP)(PR)$ in terms of the other variables, helping you solve for $ QP^{2}\plus{}PR^{2}$.

It takes a bit of algebra, though.  It's not nearly as beautiful as Green's result, but it has its own light.

[quote="Z-coli"]Namely that $ (AP)(PB) \equal{} (QP)(PR)$.
Therefore, you can find $ 2(QP)(PR)$ in terms of the other variables, helping you solve for $ QP^{2} \plus{} PR^{2}$.

It takes a bit of algebra, though.  It's not nearly as beautiful as Green's result, but it has its own light.[/quote]

In the Australian education system, we get taught that $ (AP)(PB) \equal{} (QP)(PR)$. its really disappointing that they cannot even show the elegance of this power of a point theorem, but merely use it as a tool to memorize and apply.

Im just wondering, is there any other special result of this theorem, like some thing 'special'. I know that the tangent squared is equal to the bla bla, you know, but anything else you can do?

Oh yes, there are many many useful applications.
Here's an interesting one:
http://mathworld.wolfram.com/CirclePower.html

I will post a full proof of Z-coli's method.
As he said, let the foot of the altitude from Q to AB be H, let QH=HP=a, AP=b, PR=d, AB=D. Since AQB is a right triangle, we have $ a^2 \equal{} (b \minus{} a)(D \minus{} b \plus{} a)$, or $ D \equal{} \frac {a^2}{b \minus{} a} \plus{} b \minus{} a$. By Power of a point, we have $ ad\sqrt {2} \equal{} b(D \minus{} b)$, or $ d^2 \equal{} \frac {b^2(D \minus{} b)^2}{2a^2}$. Now,
$ 2(PQ^2 \plus{} PR^2) \equal{} 2(2a^2 \plus{} d^2) \equal{} 2(2a^2 \plus{} \frac {b^2(D \minus{} b)^2}{2a^2}) \equal{} 4a^2 \plus{} \frac {b^2(\frac {a^2}{b \minus{} a} \minus{} a)^2}{a^2} \equal{} \frac {a^4}{(b \minus{} a)^2} \plus{} 3a^2 \plus{} b^2 \minus{} 2ab \equal{} D^2$  :D

...and what does $ QP^{2} \plus{} PR^{2}$ equal?
You can find it by dropping the altitude from $ R$ to $ PB$ and using Power of a Point again.
All you have to do is find $ PR$, since you already know $ PQ$.

You should get what you got for $ D^{2}$, as I did.


algebra
linear equation
system of equations
AMC

I need serious help with this question. how many ordered triples of integers (a,b,c), where a is greater than or equal to 0, b is greater than or equal to 1 and c is greater than or equal to 2, equal both log_a b=c^2005 and a+b+c=2005? I tried system of equations to figure it out by turning the first equation into a linear equation, but that doesn't make sense because in a system of equations all the variables would cancel out and 0=2005. please help!

From the first equation, you have $a^{(c^{2005})}=b$.  But, $b<2005$ and $a\ge 2$, so this means that $c^{2005}$ is either 0 or 1.  If $c^{2005}$ is 0, then $b=1$ and $a=2005-b-c=2004$.  If $c^{2005}=1$, then $c=1$ and so $a=b=\frac{2005-1}{2}=1002$.  Therefore there are exactly two integer solutions, $(a,b,c)=(2004,1,0)$ and $(a,b,c)=(1002,1002,1)$.


I'm not sure if I have already posted one of these questions.

1) The minimum number of edges of a regular polygon that is not constructible by straightedge and compass.

2) The number of 4-dimensional regular polytopes

3) The minimum number of incongruent squares that tile a (non-square) rectangle.

4) The chromatic number of the plane

5) The order of the symmetry group of the square

6) The number of partitions of {1,2,3,4} that contains 2 sets

7) The number of permutations of {1,2,3,4} that have no fixed point.

Thank you so much.

[quote="iamhe"]4) The chromatic number of the plane[/quote]
Depending on what is meant by this statement, this is an [url=http://maven.smith.edu/~orourke/TOPP/P57.html]open question[/url].  The answer is known to be between $ 4$ and $ 7$ inclusive.

I'm not quite sure what #6 is asking for, but I believe Stirling numbers of the 2nd kind may be the right topic.
for #7, look up derangements.

[quote="cheeseater"]1)7

2)6

3)9

5)8 maybe

6)6 i think

I have the exact set of problems. The math sudoku thing? Precal extra credit???[/quote]

yea...i solved it...PM me if u want the whole thing...

whats the rest of the answers
i have it to


Let $ a,b,c,d$ be positive real numbers such that
\[ \frac {1}{(a \plus{} 1)^2} \plus{} \frac {1}{(b \plus{} 1)^2} \plus{} \frac {1}{(c \plus{} 1)^2} \plus{} \frac {1}{(d \plus{} 1)^2} \equal{} \frac{16}{9}
\]
Prove that
\[ \sum_{cyc} a \minus{} 2\sum_{cyc} ab \minus{} ac \minus{} bd \plus{} 2\sum_{cyc} abc \minus{} 3abcd \ge 0.
\]
:)

[quote="can_hang2007"]Let $ a,b,c,d$ be positive real numbers such that
\[ \frac {1}{(a \plus{} 1)^2} \plus{} \frac {1}{(b \plus{} 1)^2} \plus{} \frac {1}{(c \plus{} 1)^2} \plus{} \frac {1}{(d \plus{} 1)^2} \equal{} \frac {16}{9}
\]
Prove that
\[ \sum_{cyc} a \minus{} 2\sum_{cyc} ab \minus{} ac \minus{} bd \plus{} 2\sum_{cyc} abc \minus{} 3abcd \ge 0.
\]
:)[/quote]
I am sorry. My problem is the following
Let $ a,b,c,d$ be positive real numbers such that
\[ \frac {1}{(a \plus{} 1)^2} \plus{} \frac {1}{(b \plus{} 1)^2} \plus{} \frac {1}{(c \plus{} 1)^2} \plus{} \frac {1}{(d \plus{} 1)^2} \equal{} \frac {16}{9}
\]
Prove that
\[ abc\plus{}bcd\plus{}cda\plus{}dab\plus{}ac\plus{}bd \ge 1
\]
I cannot edit my post... :(


USAMTS
probability
logarithms
graph theory
combinatorics proposed
combinatorics

prove that bipartite graph with $n$ vertices can be colored such that the color of each vertex is chosen from its list if the list of each vertex has more than $log_{2}(n)$ colors.

This looks a little to close to the USAMTS problem  :mad:

this the exercise of Alon-Spencer book .

Let $X,Y$ be the partitions of the graph. Let $C$ be the set of union of all lists, and for a vertex $v$ let $L(v)$ be its list. Construct a random subset $A$ of $C$ as follows: for every $c \in C$, include $c$ in $A$ with probability $1/2$. It is easy to see that $Pr[ L(v) \cap A = \emptyset] < (1/2)^{\log n}= 1/n$. Similarly $Pr[ L(v) \cap A^{c}= \emptyset ] < (1/2)^{\log n}= 1/n$. Therefore by union bound, \[Pr[ `\exists x \in X, L(x) \cap A = \emptyset' \vee `\exists y \in Y, L(y) \cap A^{c}=\emptyset'] < n \cdot \frac{1}{n}=1.\] Therefore there exists a set $A \subset C$ such that for every $x \in X, L(x) \cap A \neq \emptyset$ and for every $y \in Y, L(y) \cap A^{c}\neq \emptyset$. We can now get a proper coloring by coloring each vertex $x \in X$ with a color from $L(x) \cap A$ and each vertex $y \in Y$ with a color from $L(y) \cap A^{c}$.


number theory
greatest common divisor
number theory proposed

Determine all pairs $ (n,p)$ of positive integers, where $ p$ is prime, such that $ 3^p\minus{}np\equal{}n\plus{}p$.

If $ p \in \mathbb{P}\setminus \{2\}$ s.t. $ p \plus{} 1 \mid 3^p \plus{} 1$ then $ p \neq \text{ord}_{p \plus{} 1}(3) \mid \text{gcd}(\varphi(p \plus{} 1),2p) \le \varphi(p \plus{} 1) \le \frac {p \plus{} 1}{2}$, so $ \text{ord}_{p \plus{} 1}(3) \equal{} 2 \implies p \in \{3,7\}$ and only $ p \equal{} 3$ works. []


symmetry
geometry
3D geometry
number theory unsolved
number theory

[color=darkblue]Find $ x,y,z\in Z^\plus{}$, know that $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 2072$.[/color]

do it with the help of the formula
(x+y+z)^3=x^3+y^3+z^3+3(xy+yz+zx)

I really doubt that formula is true in general, rt_08. It isn't even homogeneous (four members are of degree $ 3$, three are of degree $ 2$). Any reason why it should be valid in this problem? Mind you, it isn't true for $ (12,7,1)$ which is one of the solutions to this problem.

[color=darkblue]Thank. But I think we only done it with $ z\in Z^\plus{}$[/color]

That makes it a whole lot easier - just limit it! The biggest one of them is not larger than $ 12$ and it is easy to show that at least one of them is divisible by $ 7$. After that, it's really not a problem.

hey guys chill i know i got it wrong!
 :)

[quote="thanhnam2902"][color=darkblue]Thank. But I think we only done it with $ z\in Z^ \plus{}$[/color][/quote]
Well. Now it is clearly easier...

See that $ 13^3 > 2072$ hence $ x,y,z \le 12$. Also note that $ 7 \mid 2072$. We cannot have $ 7 \mid x,y,z$ cause then would $ x \equal{} y \equal{} z \equal{} 7$... That means that one number is divisble by $ 7$, but the others are not. Say that $ x$ is that number.
Then we have $ 7^3 \plus{} y^3 \plus{} z^3 \equal{} 2072$ or $ y^3 \plus{} z^3 \equal{} 2072\minus{}7^3 \equal{} 2072\minus{}343 \equal{} 1729$. Because of symmetry we can assume that $ y^3 \equiv \minus{}z^3 \equiv 1 \bmod 7$.
Thus we have $ y \in \{1, 2, 4, 8, 9, 11\}$

Then we can just check:
$ y \equal{} 1 : z \equal{} (1729\minus{}1)^{\frac{1}{3}} \equal{} 12$
$ y \equal{} 2 : z \equal{} (1729\minus{}8)^{\frac{1}{3}} \equal{} Non\quad Integer$
$ y \equal{} 4 : z \equal{} (1729\minus{}64)^{\frac{1}{3}} \equal{} Non\quad Integer$
$ y \equal{} 8 : z \equal{} (1729\minus{}512)^{\frac{1}{3}} \equal{} Non\quad Integer$
$ y \equal{} 9 : z \equal{} (1729\minus{}729)^{\frac{1}{3}} \equal{} 10$
$ y \equal{} 11 : z \equal{} (1729\minus{}1331)^{\frac{1}{3}} \equal{} Non\quad Integer$

So the only solutions to $ (x,y,z)$ are permutations of $ (7,1,12)$ and $ (7,9,10)$

Yep as mornik says, because $ a^{3}\equiv 1,\minus{}1,0 mod7$ and $ 7|2072$. It's obviously that a,b,c<13 and there are 2 options :
1)$ x\equiv y \equiv z \equiv 0 (mod7)$ hence x=y=z=7, but $ 7^{3}\plus{}7^{3}\plus{}7^{3}\neq 2072$ so it doesn't work
2) $ x^{3}\equiv 1 mod7$. $ y^{3}\equiv \minus{}1 mod7$, $ z^{3}\equiv 0 mod7$
We know a,b,c<13 and $ 7|z^{3}$,$ 7|z , z\equal{}7$
$ x^{3}\plus{}y^{3}\equal{}1729$
Now just check what pass ! greetings

Yup, http://en.wikipedia.org/wiki/1729_(number) . It is, among other things, the smallest number expressible as the sum of two cubes in two different ways (Mathias_DK calculations have shown that $ 9^3\plus{}10^3\equal{}1729$ and $ 1^3\plus{}12^3\equal{}1729$).


geometry
inradius
IMO Shortlist
Triangle
incenter

The problem follows nicely by a "lemma" proposed by Virgil Nicula here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132338[/url]

Now, returning to the problem:
Let the incircles of $\triangle{ABX}$ and $\triangle{ACX}$ touch $BX$ at $D$ and $F$, respectively, and let them touch $AX$ at $E$ and $G$, respectively.

Clearly, $DE \| FG$.
If the line $PQ$ intersects $BX$ and $AX$ at $M$ and $N$, respectively, then $MD^{2}= MP\cdot MQ = MF^{2}$, i.e., $MD = MF$ and analogously $NE = NG$.

It follows that $PQ \| DE$ and $FG$ and equidistant from them. The midpoints of $AB, AC$, and $AX$ lie on the same line $m$, parallel to $BC$.

Applying the "lemma" to $\triangle{ABX}$, we conclude that $DE$ passes through the common point  $U$ of $m$ and the bisector of $\angle{ABX}$.
Analogously, $FG$ passes through the common point $V$ of $m$ and the bisector of $\angle{ACX}$. Therefore $PQ$ passes through the midpoint $W$ of the line segment $UV$ . Since $U, V$ do not depend on $X$, neither does $W$.

Let the incircle of $\triangle{ABC}$ meet $BC,CA,AB$ at $D_1,E_1,F_1$ and have inradius $r$. Then setting $u=AE_1=AF_1$ and so on, we have $r^2=uvw/(u+v+w)$. Similarly, let $\triangle{ABX}$'s incircle meet $BX,XA,AB$ at $D_2,E_2,F_2$ and let $\triangle{ACX}$'s incircle meet $CX,XA,AC$ at $D_3,E_3,F_3$. By simple length chasing, we find that $D_2D_3=E_2E_3=CD_1=CE_1=w$. Now define $M=BX\cap PQ$ and $N=AX\cap PQ$. Then $MD_2^2=MQ\cdot MP=MD_3^2$, and so $MD_2=MD_3=NE_2=NE_3=w/2$. Clearly we also have $D_3E_3 \| MN \| D_2E_2$, so
\[\angle{NMX}=\frac\pi2-\frac{\angle{AXB}}{2}=\frac{\angle{D_1BF_1}}{2}+\frac{\angle{E_2AF_2}}{2},\]and the tangent addition formula combined with some simple ratios and lengths gives us
\begin{align*}
\tan\angle{NMX}=\frac{\frac{r}{v}+\frac{s}{u+v-sv/r}}{1-\frac{r}{v}\frac{s}{u+v-sv/r}}=\frac{r(u+v)-sv+sv}{v(u+v-sv/r)-rs}
=\frac{r^2(u+v)}{v(r(u+v)-sv)-r^2s}
&=\frac{uvw(u+v)}{v(u+v+w)(r(u+v)-sv)-uvws}\\
&=\frac{uw(u+v)}{r(u+v)(u+v+w)-s(v+u)(v+w)}\\
&=\frac{uw}{r(u+v+w)-s(v+w)}.
\end{align*}Also,
\[BM=BD_2+MD_2=\frac{sv}{r}+\frac{w}{2}.\]Now consider the coordinate plane with $x$-axis $BM$ and origin $B=0$. The line $MPQN$ is
\[y=\frac{uw}{r(u+v+w)-s(v+w)}\left(x-\frac{sv}{r}-\frac{w}{2}\right).\]But it's easy to check that the point
\[\left(\frac{w}{2}+\frac{v(u+v+w)}{v+w},\frac{uvw}{r(v+w)}\right)\]always lies on this line, so we're done.

[hide="Diagram"]
[asy]
import olympiad;
unitsize(2cm);
pair A=(3,4);
pair B=(0,0);
pair C=(4,0);
pair X=(10,0);
path C1=incircle(A,B,X);
path C2=incircle(A,C,X);
pair O1=incenter(A,B,X);
pair O2=incenter(A,C,X);
pair B1=foot(O1,B,X);
pair B2=foot(O1,A,X);
pair C_1=foot(O2,C,X);
pair C_2=foot(O2,A,X);
pair M1=(B1+C_1)/2;
pair M2=(B2+C_2)/2;
pair D=(A+B)/2;
pair E=(A+C)/2;
pair F=(A+X)/2;
pair k[]=intersectionpoints(D--F,M1--M2);
pair K=k[0];
pair p[]=intersectionpoints(C1,C2);
pair P=p[0];
pair Q=p[1];
draw(A--B--C--X--A--C);
draw(D--F);
draw(M1--M2);
draw(C1); draw(C2);
dot(A); dot(B); dot(C); dot(X); dot(P); dot(Q); dot(B1); dot(B2); dot(C_1); dot(C_2); dot(M1); dot(M2); dot(D); dot(E); dot(F); dot(K);
label("$A$",A,dir(90));
label("$B$",B,dir(-90));
label("$C$",C,dir(-90));
label("$X$",X,dir(-90));
label("$P$",P,dir(-135));
label("$Q$",Q,dir(110));
label("$B_1$",B1,dir(-90));
label("$B_2$",B2,dir(90));
label("$C_1$",C_1,dir(-90));
label("$C_2$",C_2,dir(90));
label("$M_1$",M1,dir(-45));
label("$M_2$",M2,dir(90));
label("$D$",D,dir(135));
label("$E$",E,dir(45));
label("$F$",F,dir(90));
label("$K$",K,dir(140));
[/asy]
[/hide]

Let the incircle of $ABX$ be tangent to $BX$ at $B_1$ and $AX$ and $B_2$.  Also, let the incircle of $ACX$ touch $CX$ at $C_1$ and $AX$ at $C_2$.  Let $M_1$ be the midpoint of $B_1$ and $C_1$, and $M_2$ be the midpoint of $B_2$ and $C_2$.   Since $PQ$ is a radical axis, it passes through $M_1$ and $M_2$.  Thus, it is sufficient to prove that, as $X$ varies, there is a constant point which $M_1M_2$ always passes through.

Let $D$, $E$, and $F$ be the midpoints of $AB$, $AC$, and $AX$ respectively.  Since $B$, $C$, and $X$ are collinear, the points $D$, $E$, and $F$ are collinear also.  I will prove that $K=DE\cap M_1M_2$ does not change as $X$ changes, this will show that all $M_1M_2$ pass through a common point.  It is sufficient to prove that $DK$ is a constant.

Since $B_1,B_2, C_1,$ and $C_2$ are points of tangency of incircles, $B_1X=B_2X=\frac{BX+AX-AB}{2}$ and $C_1X=C_2X=\frac{CX+AX-AC}{2}$. Thus, \[M_1X=M_2X=\frac{BX+CX+2AX-AC-AB}{4}\]Since $DF\|BX$, $M_2FK\sim M_2XM_1$.  Since $M_1X=M_2X$, $M_2F=KF$.  We can now compute the length of $DK$ (notice the signed lengths):
\begin{align*}
DK&=DF-KF\\
    &=\frac{BX}{2}-M_2F\\
    &=\frac{BX}{2}-M_2X+FX\\
    &=\frac{BX}{2}-\frac{BX+CX+2AX-AC-AB}{4}+\frac{AX}{2}\\
    &=\frac{BX-CX+AC+AB}{4}\\
    &=\frac{BC+AC+AB}{4}
\end{align*}
This is independent of $X$, so all $M_1M_2$ (and so $PQ$) intersect at a common point, as desired.

another solution :
Let , U is exenter of triangle ABC opposite to A , I is incenter of ACX , incircle of CXA is tangent to XB at point T , incircle of ABX is tangent to BX at point K , incircle of BCA is tangent to BC at point L
Easy to see that KT = LC
A - Excircle of ABC is tangent to XB at point H
Let line thru L' = midpont of LC and || to CI intersect line thru H' midpoint of BH and perpendicular to BU at point D , let line thru point D and perpendicular to XI intersect XB at point N
Easy to see that angle NDL' = IAC , angle L'DH' = CAU , angle DH'N = AUI , so H'L'ND ~ UCIA , UC/CI = H'L'/L'N = HC/CT = H'L'/CT , L'N = CT , so N is midpoint of KT , line ND = line PQ , D is fixed . done

Let $\omega_1$ and $\omega_2$ be the incircles of $\triangle ABX$ and $\triangle ACX$, respectively. Denote by $(P, \omega)$ the power of $P$ with respect to circle $\omega$. Define a function $f: \mathbb{R}^2 \to \mathbb{R}$ by\[f(P) = (P, \omega_1) - (P, \omega_2).\]This function is linear. We now use barycentrics with respect to $\triangle ABC$. Let $R = (x:y:z)$ be the constant point that lies on $PQ$. Since $R$ lies on the radical axis of $\omega_1$ and $\omega_2$, we have $f(R) = 0$. Let $BC = a$, $CA = b$, $AB = c$, $AX = p$, and $CX = q$. We claim that the point $R = (2a: a-b-c: a+b+c)$ works, which is independent of the position of $X$. We can compute
\begin{align*}
f(A) &= \left(\frac{a+c+p+q}{2}-a-q\right)^2 - \left(\frac{b+p+q}{2}-q\right)^2 \\&= \frac{1}{4}(-a+b+c+2p-2q)(-a-b+c), \\ f(B) &= \left(\frac{a+c+p+q}{2}-p\right)^2 - \left(\frac{b+p+q}{2}-p+a\right)^2 \\&= \frac{1}{4}(3a+b+c-2p+2q)(-a-b+c), \\ f(C) &= \left(\frac{a+c+p+q}{2}-c-q\right)^2 - \left(\frac{b+p+q}{2}-p\right)^2 \\&= \frac{1}{4}(a+b-c)(a-b-c+2p-2q).
\end{align*}
By linearity, we have $f(R) = xf(A)+yf(B)+zf(C)$. But now it's straightforward to check that indeed $2af(A)+(a-b-c)f(B)+(a+b+c)f(C) = 0$, implying that $R$ always lies on the radical axis of $\omega_1$ and $\omega_2$, as desired.

[hide="Remark"]
$R$ lies on the line passing through the midpoints of segments $AB$ and $AC$, which has been pointed out above; this is because $R$ lies on the line $x = y+z$.
[/hide]

Short Solution:

Take $B'$ and $C'$ fixed points that lie on $MN$ (line connecting midpoints of $AB$ and $AC$)  such that $AB'B = AC'C = 90$. Let $K$ be the midpoint of $B'C'$. I claim $K$ is the desired point. First I prove a lemma

LEMMA In triangle $XYZ$, let $X'$ and $Z'$ be the points of tangency of the incircle (centered at $I$) with $YX$ and $YZ$ respectively and let $W$ be the intersection of the bisector of angle $Z$ and $X'Z'$. Then, if $M$ and $N$ are the midpoints of $XY$ and $XZ$, we have $MNW$ are collinear and $XWZ=90$.
Proof: Angle chasing

Therefore, $B'$ lies on the line of the points if tangency of the incircle of $ABC$ with $AX$ and $BC$, and analogously $C'$. From this we see clearly $K$ lies on the radical axis, and we're done.

Let $\omega$ be the circle centered at the midpoint of the two incircles with radius the average of the other two. Let $Y$ be the point on $BC$ such that $\omega$ is the incircle of $AYX$. Note that
\[
AY+AX-YX=\frac{AX+AC-CX}{2}+\frac{AX+AB-BX}{2}
\]
so
\[
2AY+BY-CY=AB+AC
\]
There are finitely many points $Y$ on $BC$ satisfying this, and $Y$ as a function of $X$ is continuous, so $Y$ must be the same for all $X$. Note that $PQ$ meets $XY,XA$ at their tangency points with $\omega$, so it suffices to show that for all $X$ varying on ray $BY$ past $Y$, there exists a constant point on all $X$-touch chords of $AYX$. But by a well known lemma, the intersection of the bisector of $AYX$ and the circle with diameter $AY$ suffices.

Let the $A$-midline of $\triangle ABC$ intersect the internal bisector of $\angle B$ at $V$ and the external bisector of $\angle C$ at $W$. Let $L$ be the midpoint of $\overline{VW}.$ We will show that $L \in \overline{PQ}$ to prove the result.

Note that $\overline{PQ}$ is the mid-parallel of the $X$-touch chord of the incircles of $\triangle ABX$ and $\triangle ACX$. Point $V, W$ lie on the $X$ touch chord of $\triangle ABX$ and $\triangle ACX$, respectively (from the "right angles on the intouch chord" lemma). Evidently, $L$ being the midpoint of $\overline{VW}$ lies on $\overline{PQ}$ as desired.

[color=#00f]Lemma:In $\triangle ABC$ the $B$-bisector $A$-midline and the $C$-touch chord of the incircle are concurrent[/color]
[i]Proof:[/i]
Take the polar dual and it suffices to show that the line connecting $C$ and orthocenter of $\triangle CIB$ is perpendicular to $AI$ which is immediate.

Let the incircles of $\triangle ABX,\triangle $ touch AX,BC in $E,F,E_1,F_1$.Now let $EF,E_1F_1\cap \text{A-midline}=\{R,S\}$.By lemma $R$ lies on the B-angle bisector and
$S$ on C- outer angle bisector and hence they're fixed.$PQ||EF||E_1F_1$ and $PQ$ bisects $EE_1$ and hence it bisects $RS$ $\implies$ it passes thru the midpoint of $RS$ which is fixed.$\blacksquare$

Really nice problem!

Let the incircle of $\Delta ABX, \Delta ACX$ centered at $I_1, I_2$ and touches $BC$ at $E, F$ resp. Let $I_B$ be the center of $B$-excircle of $\Delta ABC$, which touches $BC$ at $T$. Let $K, M$ be the midpoint of $AT, EF$ and finally let $R$ be the point on $BC$ such that $AR\perp I_1I_2$.

Note that $M$ lies on radical axis of those incircles, which is $PQ$. Now we claim that $K$ also lies on $PQ$, which is the fixed point. It suffices to show that $MK\perp I_1I_2$, which is parallel to $AR$. Or equivalently, $M$ is the midpoint of $RT$.

To prove this, note by symmetry that $XR=XA$. So by side-chasing,
\begin{align*} ET &= CT - CF \\
&= \frac{AB + AC - BC}{2} - \frac{AC + CX - AX}{2} \\
&= \frac{AB + AX - BC - CX}{2} \\
&= \frac{AB + AX - BX}{2} \\
\end{align*}
and
\begin{align*}FR &= XR - EX \\
&= XA - \frac{XA + XB - AB}{2} \\
&= \frac{XA + AB - XB}{2} \\
\end{align*}
so $ET=FR$ which implies $M$ is midpoint of $RT$ and we are done.

[b]Note :[/b] The most effective way to claim the fixed point is to plug $X=C$ and $X={\infty}_{BC}$ and intersect the radical axii.

Nice algebra problem! Solved with [b]eisirrational[/b].

We use barycentric coordinates on $\triangle ABC$. Let $CX = d$ and $AX = x$. Moreover, let the incircles of $ABX$ and $ACX$ be tangent to line $AX$ at $M_1$, $N_1$, and to line $BX$ at $M_2$, $N_2$, respectively.

I claim that the point $R = \left( \frac{1}{2}, \frac{a-b-c}{4a}, \frac{a+b+c}{4a} \right)$ is the desired point. It suffices to show that $R$ lies on the radical axis of the two incircles, i.e. $R$, the midpoint of $\overline{M_1N_1}$, and the midpoint of $\overline{M_2N_2}$ are collinear. Call these midpoints $S_1$ and $S_2$.

We first compute some lengths. By equal tangents, we have that $N_1X = N_2X = \frac{x+d-b}{2}$ and $M_1X = M_2X = \frac{a+d+x-c}{2}$, so $S_1X = S_2X = \frac{2x+2d+a-b-c}{4}$. The coordinates of $S_2$ is readily calculated to be (with directed ratios)
$$(0:CS_2:BS_2) = (0:CX-XS_2:BX-XS_2) = \left(0,\frac{a-b-c+2x-2d}{4a},\frac{3a+b+c-2x+2d}{4a} \right).$$
The coordinates of $S_1$ are a bit harder, but still not too bad. Note the coordinates of $X$ are $(0:CX:BX) = \left( 0, -\frac{d}{a}, \frac{a+d}{a} \right)$, so
$$S_1 = XS_1 \cdot (1,0,0) + AS_1 \cdot \left( 0, -\frac{d}{a}, \frac{a+d}{a} \right) = \left( \frac{2x+2d+a-b-c}{4x}, \frac{-2xd+2d^2+ad-bd-cd}{4ax}, \frac{2ax+2dx-3ad-d^2-a^2+ab+bd+ac+cd}{4ax}\right).$$
To show that $X,S_1,S_2$ are collinear, we simply show that the displacement vectors $\overrightarrow{XS_1}$ and $\overrightarrow{XS_2}$ are proportional. In fact, because all of these coordinates are homogenized, we only to verify the proportion for the first two coordinates. Looking at the first coordinates, we receive a ratio of
$$\frac{\frac{1}{2} - \frac{2x+2d+a-b-c}{4x}}{\frac{1}{2}} = \frac{-2d-a+b+c}{2x}.$$
But for the second coordinates we also have
$$\frac{\frac{a-b-c}{4a} - \frac{2xd+2d^2+ad-bd-cd}{4ax}}{\frac{a-b-c}{4a} - \frac{a-b-c+2x-2d}{4a}} = \frac{(d-x)(2d+a-b-c)/4ax}{(2x-2d)/4a} = \frac{-2d-a+b+c}{2x}.$$
We thus have the desired.

[img]https://lh3.googleusercontent.com/-45dm7aNud6c/Xg7HbI2IrhI/AAAAAAAAFgY/AGXpRORVABEOpvje3wJgMpcR99dnnewOgCK8BGAsYHg/s0/2020-01-02.jpg[/img]
Let $I_1$ be the incenter of $ABX$ and let $I_2$ be the incenter of $ACX$. Let $D$ and $F$ be the contact points of the incircle of $ABX$ with $BX$ and $AX$ respectively, and let $E$ and $G$ be the contact of the incircle of $ACX$ with $CX$ and $AX$ respectively.

Let $P$ and $Q$ be the intersections of $DF$ and $EG$ with the $A$-midline respectively. The key claim is that $P$ and $Q$ are fixed as $X$ varies. This follows from the so called Iran lemma, as $Q$ is the intersection of the internal $C$-angle bisector with the midline and $Q$ is the intersection of the external $B$-angle bisector with the midline.

Note that the midpoints of $DE$ and $FG$ are on the radical axis of the two circles, so the intersection of the radical axis with $PQ$ is simply the midpoint of $PQ$, which is a fixed point, as desired.

Let the incircle of $\triangle ABX$ touch $XA$ and $XB$ at $S$ and $T$, and let the incircle of $\triangle ACX$ touch $XA$ and $XC$ at $U$ and $V$. Let the incenters of $\triangle ABX$ and $\triangle ACX$ be $I$ and $J$ respectively, and let $Y$ and $Z$ be the foots of perpendicular from $A$ to $\overline{BI}$ and $\overline{CJ}$. Let $M$ be the midpoint of segment $YZ$. We claim that $M$ is the desired fixed point. By the Iran lemma, it follows that $Y$ and $Z$ lie on $\overline{UV}$ and $\overline{ST}$ respectively. It is also easy to see that $UV\parallel ST$ since both of them are perpendicular to $\overline{IX}$. Moreover, the radical axis passes through the midpoints of segments $US$ and $VT$, so it is actually the midline of $\overline{UV}$ and $\overline{ST}$, and this passes through $M$ as desired.

First ISL G7.
Here's solution using famous [color=#f00][b]Iran Lemma[/b][/color]:
Let $I_{1}$ and $I_{2}$ be the incenters of $\triangle{ABX}$ and $\triangle{ACX}$, respectively. Denote by $T_{1}$, $T_{2}$ touchpoints of incircle of $\triangle{ABX}$ with $\overline{BX}$ and $\overline{AX}$, respectively. Similarly, let incircle of $\triangle{ACX}$ touches $\overline{CX}$ and $\overline{AX}$ at $T_{3}$ and $T_{4}$, respectively. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AX}$, respectively.
By Iran Lemma we have $T = T_{1}T_{2} \cup BI_{1} \cup MN$ and $S = T_{3}T_{4} \cup CI_{2} \cup MN$ are fixed, since $MN$, $BI_{1}$, and $CI_{2}$ are fixed.
Simple [hide=angle chasing]$\measuredangle{T_{2}T_{1}T_{3}} = \frac{\measuredangle{T_2XT_1}}{2}\ = \measuredangle{XT_4T_3} = \measuredangle{T_2T_4T_3}$ [/hide] gives us $T_2T_1T_3T_4$ is cyclic. From angle chasing we also get, that $T_1T_2 \parallel T_3T_4$. Hence, $TT_1T_3S$ is parallelogram. It's well-known, that $PQ$ bisects $\overline{T_1T_3}$; since $TT_1T_3S$ is parallelogram , it's also pass through the midpoint of $\overline{TS}$, which is fixed. $\blacksquare$

[hide=Solution(with geogebra)][quote=ISL 2004 G7]For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$  passes through a point independent of $ X$.[/quote]
Denote the incircle of $\triangle ABX$ as $\omega_1$ and that of $\triangle ACX$ as $\omega_2$.Say $\ell$ be the $A-\text{midline}$ of all the 3 triangles.Say $\omega_1$ meets $XA$ and $XB$ at $U_1$ and $V_1$ respectively and say $\omega_2$ touches $XA$ and $XB$ at $U_2$ and $V_2$ respectively.
[b]Claim :[/b]$U_1V_1 \cap \ell,u_2v_2 \cap \ell$ are fixed.
[b]Proof :[/b]As the point where the internal angle bisector of $\angle ABC$ and the external angle bisector of $\angle ACB$(or the internal angle bisector of $\angle ACX$) meet $\ell$ are fixed (as the midline can be determined by the midpoints of $AB$ and $AC$ irrespective of $X$).By the Well-known Iran-Incentre Lemma we are done $\blacksquare$
Now Say $U_1V_1 \cap \ell= L_1,U_2V_2 \cap \ell=L_2$ and say $L_3$ is the midpoint of $L_1L_2$,clearly $L_3$ is fixed.We claim that $P-Q-L_3$ or $L_3$ is the required point !
For the proof note that $PQ ||U_1V_1||U_2V_2$,say $PQ \cap XA=U_3,PQ \cap XB=V_3$.
Since $PQ$ is the radical axis of $\omega_1$ and $\omega_2$,$U_3$ and $V_3$ become the midpoints of $U_1U_2$ and $V_1V_2$.The conclusion follows trivially  $\blacksquare$[/hide]

Let incircles of $ABX,ACX$ meet $AX$ again at $D_1,D_2$ and $BX$ at $E_1,E_2$ respectively. Let also line homothetic to $BC$ wrt $A$ with coefficient $\frac{1}{2}$ intersect internal bisector of and $ABC$ and external bisector of $ACB$ at $X,Y$ respectively, so by Iran lemma $X\in D_1E_1,Y\in D_2E_2.$ But obviously $D_1XD_2Y$ is an isosceles trapezoid, therefore $PQ$ passes through midpoint of $XY,$ which is fixed.

Let $M,N$ be midpoints of $AB,AC$. Incircle of triangle $ABX$, which has center $I$ touches $XA,XB$ at $K,L$. Let incircle of triangle $ACX$ touches $AX,CX$ at $J,G$. Let $D=PQ\cap MN, E=PQ\cap BC, F=BI\cap MN$ and $R=PQ\cap AX$. From Iran Lemma $F\in KL$. Since $RK=RJ$ and $El=EG$, we get $KL||RE \implies FDEL$ is parallelogram
$\implies FD=EL=\frac{LG}{2}=\frac{XL-XG}{2}=\frac{(XA+XB-AB)-(XC+XA-AC)}{4}=\frac{BC+AC-AB}{4}$.
 Also $MF=MB=\frac{AB}{2} \implies MD=MF+FD=\frac{(BC+AC-AB)+2AB}{4}=\frac{AB+BC+CA}{4}$, which is constant when $ABC$ is fixed. So $D$ lies on fixed line $MN$ and length of $MD$ is constant, which implies $D$ is fixed point as $X$ varies. So we are done.

Let the incircle of $ABX$ intersect $BX$ at $D$ and $AX$ at $E$, and let $M$, $N$ be the midpoints of $AB$ and $AX$ respectively. Let $I$ be the incenter, then we claim that $BI$, $MN$, $DE$ are concurrent.
[center][img width=45]https://media.discordapp.net/attachments/1091890356796280852/1095004634659172352/Screenshot_2023-04-10_at_11.09.13.png?width=1692&height=1112[/img][/center]
Let $S$ be the intersection point of $DE$ and $MN$. Let $F$ be on $DE$ such that $AF\parallel BX$. Let $G$ be $AS$ intersection with $BX$. Let $R$ be the contact point on $AB$. Clearly, $AFGD$ is a parallelogram with center $S$. We have \[\angle AEF = \angle DEX = \angle EDX = \angle AFE\] so $AE=AF$. Thus, $DG=AF=AE=AR$. Also, $BD=BR$ so $AB=AG$. Since $AS=SG$, $\triangle ABS\cong \triangle GBS$ so $S$ is on $AI$ as desired.
[center][img width=45]https://media.discordapp.net/attachments/1091890356796280852/1095004634894057572/Screenshot_2023-04-10_at_11.12.41.png?width=1898&height=1048[/img][/center]
Now, similarly, if $J$, $K$ are the points of contact of the incircle of $ACX$ on $CX$ and $AX$, respectively then $JK$ passes through $T$, a fixed point on the midline and the $C$-angle bisector. Let $U$ be the midpoint of $ST$, also a fixed point with respect to $ABC$.

Since $XI\perp DE$ and $XI\perp JK$, $DE\parallel JK$. Now, since $PQ$ is the radical axes of the two circles, it bisects $EK$ and $DJ$. Thus, $PQ$ is the line right in the middle of $DE$ and $JK$, so it passes through $U$, as desired.

nice problem! glad i solved it in ~45 minutes? this is the last problem im doing on configgeo im done

Define the points as in the diagram, with all the lines given by Iran Lemma; note that $YJ\perp CD\perp VW\implies YJ\parallel VW\parallel PQ$. On the other hand, since PQ is radax which bisects the common tangent RK, PQ is the midline of JYWV, which passes through the midpoint of YV. We conclude. :surf:

also note that my diagram is extremely overkill i didnt need ALL of those info but its helpful to list out when you dont know what to do

Consider an arbitrary point $X$ on ray $BC$. Let $I$ be the incenter of $\triangle ABC$ with $\triangle DEF$ being the intouch triangle of $\triangle ABC$. Further, let $M_B$ and $M_C$ be the midpoints of $AC$ and $AB$ respectively. Let $M_{CD}$ and $M_{CE}$ be the midpoints of $CD$ and $CE$. We claim that all lines $PQ$ pass through $Z=\overline{M_{CD}M_{CE}} \cap \overline{M_CB_C}$.\\

Let $\omega$ be the incircle of $\triangle ABC$ and $\omega_B$ and $\omega_C$ the incircles of $\triangle ABX$ and $\triangle ACX$ respectively. Let $O_B$ and $O_C$ be the centers of $\omega_B$ and $\omega_C$ respectively. Now, let $G$ and $H$ be the tangency points of the incircle of $\triangle ABX$ with sides $AX$ and $BX$. Let $M$ be the midpoint of $AX$.

Clearly, $B-I-O_B$ and $C-O_C-O_B$ due to the common tangents. Further, $M_C-M_B-M$. By Iran Lemma on $\triangle ABC$, we see that $\overline{BI},\overline{M_BM_C}$ and $\overline{DE}$ concur  (say at $N$). Now, clearly $N$ also lies on $BI$ and $M_CM_B$ due to the above collinearities.Further notice that by Iran Lemma on $\triangle ABX$, we must have that lines $\overline{GH},\overline{M_CM}$ and $\overline{BO_B}$ concur. But clearly the latter two of these lines intersect at $N$. Thus, $GH$ also must pass through $N$.\\

Now, simply notice that $\overline{ED}\parallel\overline{M_{CD}M_{CE}}$ by Midpoint Theorem and $\overline{GH} \parallel \overline{PQ}$ since both these lines are perpendicular to $XO_C$. Thus, the intersection of $EF$ and $GH$ and the intersection of $M_{CD}M_{CE}$ and $PQ$ must form two triangles which are similar. Now, we prove the following. Let $D',E'$ be the tangency points of $\omega_C$ with $AX$ and $BX$. Then,

[b][color=#f00]Claim :   $GD'=CD$.[/color][/b]
[i]
Proof : [/i]    Note that $XD'$ and $XG$ are tangents to $\omega_B$ and $\omega_C$ respectively. Then, let $s_1$ and $s_2$ denote that semiperimeters of $\triangle ABX$ and $\triangle ACX$ and $s$ denote the semiperimeter of $\triangle ABC$. We have,
    \begin{align*}
        GD' &= XG-XD'\\
        &= s_1-AB - (s_2-AC)\\
        &= s_1-s_2 + AC - AB\\
        &= \frac{AX+XB + AB - AX - CX - AC}{2} + AC - AB\\
        &= \frac{BC + AB - AC}{2} + AC - AB\\
        &= \frac{AB+AC+BC}{2}-AB\\
        &= s-AB\\
        &= CD
    \end{align*}
    Thus, indeed $GD'=CD$ as claimed.

Now, let $R = \overline{PQ} \cap \overline{BX}$. It is well known that the radical axis bisects the common tangent. Thus, $RD'=\frac{GD'}{2}=\frac{CD}{2}=DM_{CD}$
Thus, the previously described similar triangles must infact also be congurent. This means, that the intersections of $EF$ and $GH$ and of $M_{CD}M_{CE}$ and $PQ$ must lie on a line parallel to $BC$. But clearly the former intersection point is $N$ which lies on $\overline{M_CM_B}$ and thus $Z'=  M_{CD}M_{CE} \cap PQ $ must also lie on $\overline{M_CM_B}$. This means, $Z'=Z$.\\

Thus, all lines $PQ$ must pass through a common point, this point being $Z=\overline{M_{CD}M_{CE}} \cap \overline{M_CB_C}$.

Define line $\ell$ as the $A$-midline and the touch points of the incircles of $\triangle ABX$ and $\triangle ACX$ to $AX$, $BX$ to be $J_1$, $J_2$ and $K_1$, $K_2$. If we suppose $J = J_1J_2 \cap \ell$ and $K = K_1K_2 \cap \ell$, we know
[list]
[*] $J$ and $K$ are fixed as they lie on the $B$-angle bisector and $C$-external angle bisector, respectively, through Iran Lemma.
[*] $PQ$ bisects both $J_1K_1$ and $J_2K_2$ by power of a point, and $J_1J_2 \parallel PQ \parallel K_1K_2$.
[/list]

Thus $PQ$ passes through the midpoint of $JK$, which is fixed. $\blacksquare$

I'm surprised no one found this clean linpop solution!

Let $\omega_1$ and $\omega_2$ denote the incircles of $\triangle ABX$ and $\triangle ACX$ respectively. Define $f(\bullet) = \text{Pow}(\bullet, \omega_1) - \text{Pow}(\bullet, \omega_2)$. We claim that the intersection of the $A$-midline and the radical axis of $\omega_1$ and $\omega_2$ is fixed. In order to prove this, it suffices to show that the powers of $f((A+B)/2)$ and $f((A+C)/2)$ are independent of $X$, because it follows that the point on the $A$-midline that $f$ vanishes on is independent of $X$. This can be done by bashing.

Here's an example of what the bash would look like; consider the example of proving $f((A+B)/2)=(f(A)+f(B))/2$ constant (proving $f((A+C)/2)$ is analogous). Let all lengths be signed. Then using the intouch points, we calculate \[ f(A) = \frac{1}{4} \left[ (AB+AX-BX)^2 - (AC+AX-CX)^2\right] \] and \[ f(B) = \frac{1}{4} \left[ (AB+BX-AX)^2 - (2BC+AC+CX-AX)^2\right]. \] Now we can bash out $f(A)+f(B)$, and utilizing the fact that $BC+CX=BX$, we compute this quantity to be independent of $X$, as desired.

This problem is so cute!

I claim that the fixed point $K$ lies on the $A$-midline $\overline{MN}$. Let $E$ and $F$ are the tangency points of the incircle of $ABX$ to $\overline{BX}$ and $\overline{AX}$, respectively, and define $G$ and $H$ similarly. Let $I_1$ and $I_2$ be the incenters of triangles $ABX$ and $ACX$, and recall that $R = \overline{BI_1} \cap \overline{MN}$ lies on the tangent chord $\overline{EF}$. Similarly, $S = \overline{CI_1} \cap \overline{MN}$ lies on the tangent chord $\overline{GH}$. Note that $R$ and $S$ are fixed points.

Now, by radical axis, $\overline{PQ}$ is the midline of trapezoid $EGHF$, i.e. $K = \overline{PQ} \cap \overline{MN}$ is the midpoint of $\overline{RS}$. It follows that $K$ is the desired fixed point.


(a,b)=1. a,b E Z , and a,b is not 0
a^2+6ab+b^2 is square. find a,b

there is no sollution in non-zero numbers to:
$a^{2}+6ab+b^{2}=n^{2}$ because if we supposed so then
$(a^{2}+6ab+b^{2})(a-b)^{2}=(a+b)^{4}-(4ab)^{4}=m^{2}$ fot some $m\in \mathbb{Z}$ which is impossible unless $a=b$ or $ab=0$
in the case $a=b$ we get $n^{2}=8a^{2}$ and $\sqrt{2}\in \mathbb{Q}$ :D


if $ a,b,c > 0$ and $ a\plus{}b\plus{}c\equal{}1$ prove that $ \frac{1}{a\plus{}b^2\plus{}c^3}\plus{}\frac{1}{b\plus{}c^2\plus{}a^3}\plus{}\frac{1}{c\plus{}a^2\plus{}b^3} \leq 4\plus{}\sqrt{\frac{1}{a^3}\plus{}\frac{1}{b^3}\plus{}\frac{1}{c^3}}$


calculus
integration
probability
probability and stats

The continuous random variables X and Y have joint pdf f(x,y) where
f(x,y)=k(x^2 +2xy) for 0 < x < 1 and 0 < y < 1,

i)Find k.

Do i integrate first with respect to x and then w.r.t y? Then do i put the answer equal to 1 and find k? When i do this i get k=6/5.

ii)Find the marginal pdf of X and of Y.

is the marginal pdf of X the integral from 0 to 1 of f(x,y) wrt y?
then the marginal pdf of Y is calculated similarly to this - integral from 0 to 1 of f(x,y) wrt x?

iii)find prob[Y<0.5|X<0.5].
iv)find prob[Y<0.5|X=0.5].
v)find prob[X<Y].
I have a good idea how to do these three questions but its part (vi) that im not sure about

vi)Find E[Y|X=0.5]?

Any help is appreciated! Thanks.!

the joint density distribution is
$ f(x,y) = k (x^2 + 2 xy)$
where as you said $ \int_{[0;1] \times [0;1]} f(x,y) dx dy = 1$

Now, I think that you got that if ${ E \subset [0;1] \times [0;1]}$ is an "arbitrary" subset, then
$ P[(X,Y) \in E ] = \int_{E} f(x,y) dx dy$

Now suppose that you know the value $ x_0$  of $ X$ ($ x_0 = \frac {1}{2}$ in your example). Then, if $ \int_{y \in [0;1]} f(x_0,y) dy > 0$, then
$ d_{x_0}(y) = \frac {f(x_0,y)}{\int_{y \in [0;1]} f(x_0,y) dy}$
can be seen as the probability distribution of $ Y$ conditionally to the fact that $ X = x_0$.
Hence: $ E[Y | X = x_0] = \int_{y \in [0;1]} y d_{x_0}(y) dy$

Another way to do that, if you know the notion of conditional expectation, is to write
$ E[Y | X] = f(X)$
and then $ E[Y | X = x_0] = f(x_0)$


probability
search
combinatorics unsolved
combinatorics

Let A  and B be two vertices of a tree with 2006 edges. We move along the edges starting from A  and would like to get to N without turning back. At any vertex we choose the next edge among all possible edges(i.e. excluding all the one wich we arrived) with equal probability.Over all possible trees and choices of vertices A and B. Find the minimum probability of getting from A to B.

What is the source of this problem?

[hide="All the setup"]Well, the nice property of trees is that there is a unique such path between any $A$ and $B$.  So, for a given tree $T$, suppose this path is $A = a_{0}, a_{1}, \ldots, a_{n}= B$ and that the degree $d(a_{i}) = d_{i}$.  Then our probability of winning is $(d_{0}\cdot (d_{1}-1) \cdot (d_{2}-1) \cdots (d_{n-1}-1))^{-1}$.  We want to choose a tree to minimize this probability.  Note that if there exists an edge joining two vertices, neither of which are an $a_{i}$, we can decrease the success probability by breaking off this portion of the tree and reattaching it to one of the $a_{i}$.  Repeating this process as much as possible shows that the minimum occurs for some tree in which every vertex either is one of the $a_{i}$ or is connected only to one of the $a_{i}$.  Thus, what we are really looking for is an integer $n$ and integers $d_{0}, d_{1}, \ldots, d_{n}$ such that $1 \leq d_{0}, d_{n}$, $2 \leq d_{2}, \ldots, d_{n-1}$ such that $(d_{0}-1)+(d_{1}-2)+(d_{2}-2)+\ldots+(d_{n-1}-2)+(d_{n}-1) = 2006-n$ and $(d_{0}\cdot (d_{1}-1) \cdot (d_{2}-1) \cdots (d_{n-1}-1))$ is maximized.

It's clear that we want to set $d_{n}= 1$.  Then, taking $c_{n}= d_{0}$ and $c_{i}= d_{i}-1$, we can re-phrase this as a search for an integer $n$ and $n$ integers $c_{1}, \ldots, c_{n}$, such that $c_{1}+\ldots+c_{n}= 2006$ and $c_{1}\cdot c_{2}\cdots c_{n}$ is maximized.[/hide]


probability
geometry
3D geometry
geometric transformation
rotation

20. Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

(A) $ \frac{1}{4}$   (B) $ \frac{5}{16}$   (C) $ \frac{3}{8}$   (D) $ \frac{7}{16}$   (E) $ \frac{1}{2}$

[b]HINTS ONLY[/b]

Thanks.

Well since you want a hint, how many ways can you paint the cube? How many satisfy the given conditions?

$ 2^6$ ways to paint the cube.

I'm not sure how many ways satisfy the given conditions.

[quote="mewto55555"]Well for it to satisfy it, how many ways can you choose the 4 faces?[/quote]

how do i find that? (i need to brush up on my counting skills)

I'll go through my entire solution, and have hide tags between steps.

[hide="Step 1"]
We want to find all ways to get 4 sides. It is clear that the other 2 sides are opposite each other
[hide="Next"]
We have 3 sets of opposite sides. Therefore the number of ways to choose the 2 sets that we make the same color is
[hide="Next"]
3c2 or 3. We also have 2 ways to choose the color for this. Thus we have

6 ways to choose and color the sides. For the other 2 sides

2x2=4 ways so there are a 24 ways
$ \frac {24}{64} = \frac {3}{8}$[/hide][/hide][/hide]

Is this correct?

You force me to look at the solution now.

The answer is: [hide]B[/hide].

Your step 3 is right and the other steps following it are wrong.

[hide="Hints"]
There are $ 2^6 \equal{} 64$ choices total--these include rotations of all cubes, which is important.
[hide="What cubes have this property?"]
The only ways to have this possible is to have at least four of one color, so
i. RRRRRR or BBBBBB
ii. RRRRRB or BBBBBR
iii. RRRRBB or BBBBRR
Now calculate the number of possibilities (including rotations!) of these figures.
[hide="Case i"]
Obviously, there are exactly $ 2$ cubes for this (only $ 1$ rotation for each RRRRRR and BBBBBB).
[/hide]
[hide="Case ii"]
In RRRRRB, there are $ 6$ places for the blue cube, and then the reds are fixed.

The same holds for BBBBBR, so there are $ 12$ possibilities in all for this case.
[/hide]
[hide="Case iii"]
In RRRRBB, the blue faces must be opposite each other, and the reds are forced.

There are $ 3$ ways for this (why aren't there $ 6$ ways?).

This is the same for BBBBRR, so there are $ 6$ total possibilities for this case.
[/hide]
[hide="Answer"]
$ \frac {2 \plus{} 12 \plus{} 6}{64} \equal{} \frac5{16}\implies\boxed B$.
[/hide]
[/hide]
[/hide]

You're counting each of RRRRRR and BBBBBBB three times, i.e. for each of these cubes, there are three choices of the pair of sides, then the rest is fixed.

Oh ok so since I counted those 3 times each instead of 1 time each you subtract 4 and get 20 possibilities.


Ten distinct points are identified on the circumference of a
circle. How many different convex quadrilaterals can be
formed if each vertex must be one of these 10 points?

For each set of 4 points, there is one possible convex quadrilateral that can be formed. Therefore, there are $ \binom{10}{4} \equal{} \boxed{210}$.


[quote="captcha000"]Can someone help me out? I downloaded the MikTex stuff, but then I go through the setup, and it wants me to choose a folder to run the setup from, but it doesn't seem to work no matter what folder i choose. It always gives me something like "this is not a valid directory"[/quote]You should choose the folder where your download has been stored. Or maybe you've not downloaded the whole thing, just the webinstalled. In that case you won't find any local repositories. You can install it online, using the internet package repository (if you have a cable connection of course).

The Mik$\text{\TeX}$ download is a bit strange. You have to download the installer first and then the program. So it's a two-pass installation.


I need some information about air resistance...specially in projectile motion...
Any ideas?

i read an article on Projectile motion with air resistance last night, which involves lots of Calc and stuff (it is from an old College math journal)
i can post some stuff from that here, ifyou want,
but specifically what are youlooking for??


geometry
3D geometry
sphere
tetrahedron
perimeter

This problem is similar to an old MOEMS problem, but I added some more to it:

Given a circle, square, equilateral triangle, and regular pentagon that all lie in the same plane, what is the maximum number of intersection points that can be formed?

And here's an even better one:

EDIT: Given a sphere, a cube, and a tetrahedron, what is the maximum number of intersection points that can be formed?

EDIT: An intersection point is defined as a point which is common to the perimeters of at least two geometrical objects

The circle can intersect each polygon twice on each side: 6 points for the triangle, 8 for the square, 10 for the pentagon. The triangle can have 6 points with the square and 6 for the pentagon. The square can have 8 points with the pentagon. These numbers give a total of $ 6 \plus{} 8 \plus{} 10 \plus{} 6 \plus{} 6 \plus{} 8 \equal{} \boxed{44}\text{intersection points}$.

As to your challenge problem with the 3-D objects, how can one have an intersection between a sphere and a cube? If they have intersection points, then either the sphere is tangent to the cube at one point, or the sphere intersects the cube at an infinite number of points on its surface.

Thus, their is no limit to how many intersection points they share.

Also, your definition for intersection points does not apply to a circle, whose perimeter contains no line segment. I think the definition should be "Any point that is common to the perimeters of two or more figures."

Oh....when I drew a sphere and a cube I got that they could intersect at a maximum of $ 24$ points ($ 3$ at each vertex of the cube)

Is this correct?

Oh, I see what you're saying.....If you count intersection points as a point where the edge of a polyhedron meets the face of another 3-D figure, then it sorta works to be 24. Thanks.


If $ 0<a,b<1$ and $ p,q\geq 0$, $ p\plus{}q\equal{}1$ are real numbers, prove that

$ a^pb^q\plus{}(1\minus{}a)^p(1\minus{}b)^q\leq 1$.

note that $ 0 < a,b < 1$ so by AM-GM we have:

$ a^pb^q = \sqrt [p + q]{a^pb^q}\leq\frac {\overbrace{a + a + \ldots + a}^{p\textrm{ times}} + \overbrace{b + b + \ldots + b}^{q\textrm{ times}}}{\underbrace{p + q}_{ = 1}} = ap + bq$

and similary:

\begin{eqnarray*}(1 - a)^p(1 - b)^q\leq\frac {\overbrace{(1 - a) + (1 - a) + \ldots + (1 - a)}^{p\textrm{ times}} + \overbrace{(1 - b) + (1 - b) + \ldots + (1 - b)}^{q\textrm{ times}}}{\underbrace{p + q}_{ = 1}} = p(1 - a) + q(1 - b)\end{eqnarray*}

now summing up these two inequalities we get that:

$ a^pb^q + (1 - a)^p(1 - b)^q\leq pa + qb + p(1 - a) + q(1 - b) = p + q = 1$

QED

[quote="BaBaK Ghalebi"]

$ a^pb^q \equal{} \sqrt [p \plus{} q]{a^pb^q}\leq\frac {\overbrace{a \plus{} a \plus{} \ldots \plus{} a}^{p\textrm{ times}} \plus{} \overbrace{b \plus{} b \plus{} \ldots \plus{} b}^{q\textrm{ times}}}{\underbrace{p \plus{} q}_{ \equal{} 1}} \equal{} ap \plus{} bq$

[/quote]

Well, actually $ p,q$ are real numbers, but I guess weighted AM-GM works in that case anyway...


Does someone of u know how to solve Euler's cirlce. pls

   Help me Lagrangia :D  :D

Hi! :D

I am posting from school so I can only give you some reference! check these links:

http://mathworld.wolfram.com/Nine-PointCircle.html
http://mathforum.org/epigone/geometry-pre-college/all

cheers! :D :D :D

Dear Mathlinkers,

an original way to get the Euler's circle is to use a Morley's circle.
See : http://perso.orange.fr/jl.ayme vol. 2 Les cercle de Morley, Euler...

Sincerely
Jean-Louis


Find $ n$ min such that $ A \equal{} 2^{16} \plus{} 2^{9} \plus{} 2^n$ is a square
$ (n \in N)$

Posted before many times, that $ 2^x \plus{} 2^y \plus{} 2^z,x\ge y\ge z$ is square if and only if $ x,z \minus{} \ even \ y \equal{} 1 \plus{} \frac {x \plus{} z}{2}.$

[quote="Jure the frEEEk"]if $ n > 9$ than $ 2^{n} \plus{} 2^{16} \plus{} 2^{9} \equal{} 2^{9}(2^{n \minus{} 9} \plus{} 2^{16} \plus{} 1)$ which obviously can't be square

if˘$ n\leq9$ than $ 2^{16} \plus{} 2^{9} \plus{} 1 \equal{} (2^{8} \plus{} 1)^{2} < 2^{16} \plus{} 2^{9} \plus{} 2^{n}\leq2^{16} \plus{} 2^{10} < 2^{16} \plus{} 2^{10} \plus{} 4 \equal{} (2^{8} \plus{} 2)^{2}$

obivously no solutions for $ n$[/quote]
Oh yes use this we can solve general problem .


1. How many stereoisomers are possible for Farnesol (structure and information [url=http://en.wikipedia.org/wiki/Farnesol]here[/url])?

2. What product(s) might be formed when farnesol is treated with sulphuric acid?

[hide="answer"]
1.4 stereoisomers (2 geometrical centres).
2. allylic rearrangement to give conjugated diene:
3,7,11-trimethyl 1,3,6,10-dodecatetraene.
[/hide]

[hide="possible answer"]
1,2,2-trimethyl-1-(4'-methyl-3'-cyclohexenyl)-cyclopentan-3-ol
1-methyl-2-(1'-hydroxy-1'-methyl ethyl)-1-(4'-methyl 3'-cyclohexenyl)-cyclobutane[/hide]

[quote="mythili"][hide="possible answer"]
1,2,2-trimethyl-1-(3'-cyclohexenyl)-cyclopentane
1-methyl-2-isopropyl-1-(3'-cyclohexenyl)-cyclobutane[/hide][/quote]
what is the mechanism and are my eyes playing tricks on me or is there really one carbon atom less in the products.

sorry wrong naming :blush:  there will be a 4' methyl at 3'cyclohexenyl.
the mechanism is first the protonation of the $ \ce{OH}$ followed by ring closing steps...

i very much doubt if cyclization can take place as the hydrogens are not very acidic. but will have to wait for carcul to post his answer.  :)

well this is an inspired procedure from the the closing of nerol to give $ \alpha$ terepinol under similar conditions...well i have edted some parts of the answer see the edited answer ....i posted that in a hurry :blush:

Mythili's answer is also wrong. Just one ring will form predominantly.

is atleast a part of the answer right or is it a one big ring :(

What rings form more easily in cyclizations?

well i had both of them in one of the answers...well aren't my answers even partially right :rotfl:


Suppose X={x_1,x_2,...,x_n} and Y={y_1,y_2,...,y_n} are two sequences of natural number and satisfying:
  Multiply_{k=1}^{n}{x_k}=Multiply_{k=1}^{n}{y_k}=N,
where N is a constant.
  Suppose Var(X),Var(Y) are variances and Mean(X), Mean(Y) are means of two sequences.
  I guess that: if Var(X)<Var(Y),then we have Mean(X)<Mean(Y).
  For example, for X={2,3} and Y={1,6},Var(X)<Var(Y) and Mean(X)<Mean(Y).
  Then how to prove or disprove it? Thanks very much!

And thanks go to every guys who pay attention to this question.

Also, I am very sorry for a double post. It's because I'm a new comer and make a mistake. I promise will not do it again :blush: .

Thanks again :lol:


My code

\begin{bmatrix}
0& 0.60& 0& 0& 0& 0& 0& 0& 0& 0& 0
\end{bmatrix}

keeps outputting the last zero on the 2nd row. Is there any reason why it does that? The error message I keep getting is

X! Extra alignment tab has been changed to /cr.
<recently read> \endtemplate

The default maximum number of columns in bmatrix (and pmatrix etc) is 10 and you have 11 columns. All you need to do is, before the bmatrix environment, add the line
\setcounter{MaxMatrixCols}{11}


Quite easy
$ a,b,c\in Z$
$ \frac{\sqrt{3}a\plus{}b}{\sqrt{3}b\plus{}c}$ is a rational number
Prove that
$ a\plus{}b\plus{}c|a^{2}\plus{}b^{2}\plus{}c^{2}$
2)Find all prime $ p,q$
$ p^{2}\minus{}p\plus{}1 \equal{} q^{3}$
3)$ n > 1$
   $ a > 0$
p is a prime(p>2)
$ a^{p}\equal{} 1(\text{mod}p^{n})$
Prove that
$ a \equal{} 1(mod p^{n\minus{}1})$
Solve equation
$ x^{n}\equal{} p^{z}\plus{}1$(n>1)

[hide="1"]
$ \frac{\sqrt{3}a\plus{}b}{\sqrt{3}b\plus{}c}\equal{}\frac{(\sqrt{3}a\plus{}b)(\sqrt{3}b\minus{}c)}{3b^{2}\minus{}c^{2}}$
equating the radical part to $ 0$ we get
$ b^{2}\equal{} ac\rightarrow a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} (a\plus{}b\plus{}c)(a\minus{}b\plus{}c)$
[/hide]


function
limit
calculus
calculus computations

Let $f(x)$ be differentiable function such that $f'(0)=k.$
Define the sequence $\{a_{n}\}$ as follows.
\[a_{1}=1,\ \ a_{n}=\lim_{x\to\infty}x^{2}\left(f\left(\frac{a_{n-1}}{x}\right)-f(0)\right)^{2}\ (n=2,\ 3,\ \cdots). \]
Find the range of the value of $k$ for which the sequence $\{a_{n}\}$ converges.

Suppose $a_{n-1}\neq0$. Then

$a_{n}$

$=\lim_{x\to\infty}x^{2}\left(f\left(\frac{a_{n-1}}{x}\right)-f(0)\right)^{2}$

$=\lim_{h\to 0}(a_{n-1})^{2}\left( \frac{f(h)-f(0)}{h}\right)^{2}$ where $h=\frac{a_{n-1}}{x}$

$= \left( ka_{n-1}\right)^{2}$

If $a_{n-1}=0$, Then $a_{n}= 0$.

You can check that $a_{n}$ converges if and only if $|k|\leq1$


hey this might seem easy but i am lost:)
given a circle with N numbers written on it.
write a O(N) algorithm which finds the consecutive block with largest sum

well i can't figure out how to use dp for this

best regards
vasu

Hint: split it into 2 cases: one where the block of numbers does not 'wrap around' from N to 1, and one where it does.

[edit]Actually, theres another way to do it.. involving going around the circle twice, storing the maximum sum so far as well as the length of the sequence ending there, stopping when that length gets bigger than N.

well as u said earlier
i did this
leave the first element
find the max consecutive block using dp.call it "maxbest". and also find the total sum including the first number
also find the min consectuive block leaving the first number,call it minbest.
now ur answer is max(maxbest,sum-minbest)
right:)


Solve in reals:

1.$ \frac {x^2}{(x\plus{}1)^2}\plus{}x^2\equal{}3$

2.$ 2^{\frac {2\sin x}{\sin x\plus{} \cos x}} \equal{} 1 \plus{} 2^{\frac {\sin x\minus{} \cos x}{\sin x\plus{} \cos x}}$

[quote="hell_ever"]2.$ 2^{\frac {2\sin x}{\sin x \plus{} \cos x}} \equal{} 1 \plus{} 2^{\frac {\sin x \minus{} \cos x}{\sin x \plus{} \cos x}}$[/quote]
$ \Leftrightarrow 2.2^{\frac {\sin x \minus{} \cos x}{\sin x \plus{} \cos x}} \equal{} 1 \plus{} 2^{\frac {\sin x \minus{} \cos x}{\sin x \plus{} \cos x}}\Leftrightarrow 2^{\frac {\sin x \minus{} \cos x}{\sin x \plus{} \cos x}} \equal{} 1$
$ \Leftrightarrow \sin x \minus{} \cos x\equal{}0\Leftrightarrow \tan x\equal{}1\Leftrightarrow x\equal{}\frac{\pi}{4}\plus{}k\pi$

[quote="hell_ever"]Solve in reals:

1.$ \frac {x^2}{(x \plus{} 1)^2} \plus{} x^2 \equal{} 3$

[/quote]

Let A = x/x+1
B = x

$ AB \equal{} A \minus{} B \equal{} X^2/X \plus{} 1$

$ A^2 \plus{} B^2 \equal{} 3$

Squaring equation 1

$ 3 \minus{} 2AB \equal{} A^2B^2$

Solving AB = 1, -3.

Due to obvious reasons -3 is not possible.

$ (A \plus{} B)^2 \minus{} 2 \equal{} 3$

A + B = Root(5)

$ x^2 \minus{} Root(5)x \plus{} 1 \equal{} 0$

Solving x = $ \frac {Root(5) \plus{} 1}{2}$
$ \frac{Root(5) \minus{} 1}{2}$

I might have made some stupid calculation error. I least bother. Hopefully you got the method.

Nice one tdl!

[quote="hell_ever"]Solve in reals:

1.$ \frac {x^2}{(x \plus{} 1)^2} \plus{} x^2 \equal{} 3$

2.$ 2^{\frac {2\sin x}{\sin x \plus{} \cos x}} \equal{} 1 \plus{} 2^{\frac {\sin x \minus{} \cos x}{\sin x \plus{} \cos x}}$[/quote]

Another method for question 1:
[hide]
Completing the squares:
Add and subtract $ (2 \cdot x \cdot \frac {x}{x\plus{}1}) \implies (\frac {x}{x\plus{}1} \minus{} x)^2 \plus{} 2 \cdot x \cdot \frac{x}{x\plus{}1} \equal{} 3$
$ \implies (\frac {x^2}{x\plus{}1})^2 \plus{} \plus{} 2 \cdot x \cdot \frac{x}{x\plus{}1} \equal{} 3$
Solve from now on  :)
[/hide]


abstract algebra
superior algebra
superior algebra unsolved

Let $ S \equal{} \{1,2,4,..,2^n,..\}$ and consider the fraction ring $ Z_S$. Show that $ Z_S/Z$ is an Artinian non-Noetherian Z-module.

Every proper submodule is either $ 0$ or is generated by $ 2^{ \minus{} n}$.


topology
vector
geometry
3D geometry
sphere

call $ X$ the space of unit length vectors in the tangent space to the sphere $ S^2$ (with respect to the standard metric, but that's not important), and $ Y$ the union of $ X$ and the set of tangent vectors of length $ \le 1$ to a fixed point of $ S^2$ (say it's the north pole).

1. show that $ X$ is homeomorphic to $ \mathbb{RP}^3$.
2. compute $ \pi_1(Y)$.

Should $ X$ be unit length vectors in the tangent bundle, or are we fixing some point there as well?

no special point there, just the unit tangent bundle.

bonus question: what is the homotopy type of $ {\rm Fr}(S^n)$, the (total space of) the frame bundle of $ S^n$?
or, if you prefer, what is the total space of the space of orthormal oriented frames over $ S^n$?