Everytime I turn on my computer, a window pops up telling me to register for the virtual printer and that my free trial will only last 14 days. What do I do about this? Am I supposed to register or ignore it?

Very odd - I've not ever had this happen, and I only installed mine a few weeks ago.  I hope they haven't changed their policy.  Anyone else having this problem?

I still don't get why one has to have virtual printer installed at all.

To create PDF images for inclusion in TeX documents (this is for folks without Acrobat Distiller).

Here's another program that produces PDF from PC applications:

[url=http://www.cutepdf.com/Products/CutePDF/writer.asp]CutePDF[/url]

It works for Windows98 and ME, but its images aren't quite as crisp as the Go2PDF application.  This one seems a bit more dedicated to remaining free than Go2, however.

Please do let us know if you lose the ability to use Go2PDF after the 14 days - we'll point people to something else if that turns out not to be free.


matrix
IMO
Ramsey Theory
combinatorics
Extremal combinatorics
IMO 2001
IMO Shortlist

[hide]
Set up a grid, with all $ 21^2$ boy-girl pairs marked along the horizontal axis and the problems marked along the $ y$ axis. Mark a $ 1$ in $ (i,j)$ if both members of pair $ i$ solved problem $ j$.
\[ \begin{bmatrix}
& \text{Pair 1} & \text{Pair 2} & \ldots & \text{Pair 441} \\
\text{Problem 1} & 1 & 0 & \ldots & 0 \\
\text{Problem 2} & 0 & 1 & \ldots & 1 \\
. & . & . & . & . \\
. & . & . & . & . \\
. & . & . & . & . \\
\end{bmatrix}
\]
What we want to do is show that there is a row with $ 1$'s from groups whose union contains three boys and three girls. Suppose to the contrary that the only $ 1$'s are in rows solved by at most two boys or two girls. Recall that all columns corresponding to a particular girl, say $ g_1$, can only have $ 1$'s in at most six different rows. In addition, there are at least $ 21$ $ 1$'s in columns corresponding to $ g_1$. Therefore, $ g_1$ contributes to at most $ 5$ rows solved by at most $ 2$ boys, for a total of at most $ 5(2) \equal{} 10$ $ 1$'s. Summed over all $ 21$ girls, we have at most $ 210$ $ 1$'s in rows solved by at most $ 2$ boys. Similarly, we have at most $ 210$ $ 1$'s in rows solved by at most $ 2$ girls. This totals to at most $ 420$ $ 1$'s which is obviously less than $ 21^2$. [/hide]

After reading Count Down, I decided to look up this problem. However, I've spent about 30 minutes reading this proof (for there is no other that I can find) and am still stumped.

[quote="Phelpedo"]Suppose to the contrary that the only $ 1$'s are in rows solved by at most two boys or two girls.[/quote]

This contradiction doesn't cover the case where there are 1's in rows solved by THREE boys, but at most two girls. Or does it? The bounds definitely don't work considering this.

Call a problem [i]bad[/i] if it's solved by at most $2$ boys, and [i]good[/i] if it's solved by at most $2$ girls. Assume every problem is either good or bad and argue by contradiction. Construct a $21$x$21$ grid where the entry at position $(i, j)$ gives a problem solved by both boy $i$ and girl $j$. By PHP at least $\lceil 441/2 \rceil = 221$ squares in this grid contain (WLOG) bad problems; PHP once again on the row these bad problems appear in gives a row with at least $\lceil 221 / 21 \rceil = 11$ squares with bad problems. Yet any bad problem is solved by at most $2$ boys, so any particular bad problem can appear in at most $2$ of the squares in this row. It follows that there are at least $\lceil 11 / 2 \rceil = 6$ distinct bad problems in this row. This is impossible; $6$ distinct bad problems are solved by at most $12$ boys, leaving $9$ boys who solved no problems in common with this particular girl, contradiction. It follows that there exists some problem solved both by at least $3$ boys and at least $3$ girls.

Assume the conclusion is false.

$(*)$If there exists a problem which was solved by only boys or only girls,we remove it.The configuration left still satisfies the hypothesis.
So now each problem has been solved by at least 1 boy and 1 girl.

From now on we work under the conditions of $(*)$, that is, every problem was solved by 1 of each gender. Say there are $m$ problems, and denote by $x_i$ and $y_i$ the number of boys respectively girls which solved problem $i$; due to $(*)$ we have $x_i>0,y_i>0$, and due to our negation of the conclusion we cannot simultaneously have $x_i \ge 3,y_i \ge 3$.Also, since every contestant solved at most 6 problems, there are exactly 21 of them, and $\sum x_i$ is the total number of problems solved by the boys,we get that $\sum x_i \le 21*6,\sum y_i \le 21*6(1)$.

Now consider the bipartite graphs $G_i$, $i=1,2,...,m$, which has 21 vertices on one side, name it the left side, and 21 on the other, name it the right side, and vertices $k$ on the left and $l$ on the right are connected if and only if boy $k$ and girl $l$ both solved problem $i$; we also consider the graph $G$, built by adjoining the $m$ $G_i$'s, and the corresponding edge sets $E_i$ and $E$.

It is obvious that $|E_i|$ is the number of boy-girl pairs that both solved problem $i$, equal to $x_iy_i$,while $|E|=\sum |E_i|=\sum x_iy_i(2)$; however,$|E|$ being the number of boy-girl pairs which solved a problem counted over all the m problems leads to $|E| \ge 21^2(3)$, due to every boy-girl pair having solved at least one problem in common, and there being $21^2$ of them. We also have the bound $x_iy_i \le 2x_i+2y_i-3(4)$, which is true due to being equivalent to $(x_i-2)(y_i-2) \le 1$, which can be easily proven using the conditions set on $x_i,y_i$.

Putting together $(1),(2),(3)$ and $(4)$ we obtain a bound on $m$:
$21^2 \le |E|=\sum x_iy_i \le \sum 2x_i+2y_i-3=(\sum 2x_i)+(\sum2y_i)-3m \le 21*6*4-3m \Rightarrow m \le 21$.

Now,of these $m$ problems,let's say $k$ have been solved by 2 or fewer boys;then the other $m-k$ have been solved by 2 or fewer girls.It is trivial to notice that $m \le 21$ forces $min(k,m-k) \le 10$;assume WLOG the minimum is $k$;then $k\le 10$.Now out of the 21 boys,there must be one,name him $B$, that didn't solve any of those problems, since there are at most $2*10=20$ boys who solved them; then he will have solved problems only from the $m-k$ remaining ones. However,now for each of the problems he solved, there exist at most 2 girls who also solved them,so there exist at most $6*2=12$ girls who could have solved a problem that $B$ also solved,leaving 9 girls who don't have any problem in common with $B$,contradiction.

This contradiction stems from our initial assumption,which was the negation of our conclusion,so our proof is finished.

A standard double count helps : My solution is basically same as the official one; though I named the problems differently, not [i]tough for boys[/i] or [i]tough for girls[/i] .  :D   :laugh:

Is there a way of solving this problem without finding a person that violates the first condition?

In other words can one show that there are no integer numbers $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ that satisfy ($a_i,b_i$ denote the number of boys and girls that solved problem $i$ respectively).

$0 \leq a_i,b_i \leq 21$ and $\sum a_i \leq 6\cdot 21$ and $\sum b_i \leq 6\cdot 21$ such that $\sum a_i b_i \geq 21\cdot 21$

With at least one of $a_i$ and $b_i$ being less than or equal to $2$ for each $i$. *

[hide="Solution"]
Assume that no problem is solved by at least three girls and three boys, and discard any problem that is solved by only boys and only girls (Clearly this can only help us). Let there be $n$ problems left, and for problem $i$, assign the ordered pair $(b_i, g_i)$ such that $b_i$ boys and $g_i$ girls solved the problem. Then, it's clear that $b_1g_1+\ldots +b_ng_n\ge 441$ and $b_1+\ldots +b_n\ge 126$ as well as $g_1+\cdots +g_n\ge 126$. Now, note that we must have $(b_i-2)(g_i-2)\le 1$ for all $i$, so $b_ig_1\le 2b_i+2g_i-3$, hence it follows that $441\le b_1g_1+\ldots +b_ng_n\le 504-3n$, so $n\le 21$.

Now, consider any boy. Since he solved $6$ problems, he must have solved a problem solved by at least $4$ girls. Label the boy with the number of this problem, and do the same with the girls. No three boys can have the same label, otherwise we have a contradiction. Then the boys have $11$ different problem labels among them, and same for the girls. But this is a contradiction since a boy and a girl can't share the same label, so we are done.

[/hide]

[quote=mcdonalds106_7]
Assume that no problem is solved by at least three girls and three boys, and discard any problem that is solved by only boys and only girls (Clearly this can only help us). Let there be $n$ problems left, and for problem $i$, assign the ordered pair $(b_i, g_i)$ such that $b_i$ boys and $g_i$ girls solved the problem. Then, it's clear that $b_1g_1+\ldots +b_ng_n\ge 441$ and $b_1+\ldots +b_n\ge 126$ as well as $g_1+\cdots +g_n\ge 126$. Now, note that we must have $(b_i-2)(g_i-2)\le 1$ for all $i$, so $b_ig_1\le 2b_i+2g_i-3$, hence it follows that $441\le b_1g_1+\ldots +b_ng_n\le 504-3n$, so $n\le 21$.

Now, consider any boy. Since he solved $6$ problems, he must have solved a problem solved by at least $4$ girls. Label the boy with the number of this problem, and do the same with the girls. No three boys can have the same label, otherwise we have a contradiction. Then the boys have $11$ different problem labels among them, and same for the girls. But this is a contradiction since a boy and a girl can't share the same label, so we are done.

e][/quote]

Wasn't it done from n =<. 21??
since it implies that bi,gi all equal 6 for all  i
and it's a contradiction. Maybe i'm misunderstanding something.

[quote=peregrinefalcon88]I have the same problem as eternica but i have an alternate solution. call a problem Legit if it is solved by atleast 3 boiz  and Legitimate if it is solved by atleast 3 gurlz. out of the 21 pairs formed with any gurl, there must be atleast 11 legit problems and with 21 gurlz there are 21*11 legit problems and by the same argument there are 21*11 legitimate problems making 22*21 problems but there are only 21*21 problems so some problems must be legit and legitimate.[/quote]

Is this solution good? It's a bit cloudy for me yet..

Or just give me a reason why are there 11 legit (and legitimate) problems

Nvm I figured it out, also he basically "marked" the problems in such way that we could count them more times, I don't get why there is only 21^2 problems with marking.

Not the prettiest solution, there is a bit of casework at the end that I didn't include. Basically I counted triples $(b_i, g_j, P_k)$ such that both $b_i, g_j$ solved $P_k$ and tried finding a contradiction that the # of triples is less than $21^2$.
[img]https://i.imgur.com/GaMPrF3.jpg/img]
[img]https://i.imgur.com/LjZ9gOc.jpg/img]

We will show the contrapositive. That is, assume that [list] 	[*]For each pair of a girl and a boy, 	there was at least one problem that was 	solved by both the girl and the boy. 	[*]Assume every problem is either solved 	mostly by girls (at most two boys) 	or mostly by boys (at most two girls). [/list] Then we will prove that then some contestant solved more than six problems.

Create a $21 \times 21$ grid with boys as columns and girls as rows, and in each cell write the name of a problem solved by the pair. Color the cell [b]green[/b] if at most two girls solved that problem, and color it [b]blue[/b] if at most two boys solved that problem. (G for girl, B for boy. It's possible both colors are used for some cell.)

WLOG, there are more green cells than blue, so (by pigeonhole) take a column (boy) with that property. That means the boy's column has at least $11$ green squares. By hypothesis, those corresponds to at least $6$ different problems solved. Now there are two cases: [list] 	[*]If there are any blue-only squares, 	then that square means a seventh distinct problems. 	[*]If the entire column is green, 	then among the $21$ green squares 	there are at least $11$ distinct problems solved 	in that column. [/list]

[hide="Remarks"]
[b][color=red]Remark:[/color][/b]  The number $21$ cannot be decreased. Here is an example of $20$ boys and $20$ girls who solve problems named $A$-$J$ and $0$-$9$, which motivates the solution to begin with.

\[
\begin{array}{c}
\text{\ttfamily{\color{green}0000000000}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}0000000000}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}1111111111}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}1111111111}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}2222222222}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}2222222222}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}3333333333}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}3333333333}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}4444444444}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{green}4444444444}{\color{blue}AABBCCDDEE}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}5555555555}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}5555555555}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}6666666666}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}6666666666}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}7777777777}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}7777777777}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}8888888888}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}8888888888}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}9999999999}} \\
\text{\ttfamily{\color{blue}FFGGHHIIJJ}{\color{green}9999999999}}
\end{array}
\]

[b][color=red]Remark:[/color][/b]  	This took me embarrassingly long, 	but part of the work of the problem seemed to be 	finding the right ``data structure'' to get a foothold. 	I think the grid is good because we want to fill each intersection, 	then we consider for each cell a problem to put.

I initially wanted to capture the full data by writing 	in each green cell the row index of the other girl who solved it, 	and similarly for the blue cells. 	(That led naturally to the colors, there were two types of cells.) 	This was actually helpful for finding the equality case above, 	but once I realized the equality case 	I also realized that I could discard the extra information 	and only remember the colors.
[/hide]

Let the boys and girls be $b_1, \dots , b_{21}, g_1, \dots , g_{21}$. Say a problem is [i]bad-for-girls[/i] if at most 2 girls solve it. For a pair $(b_i,g_j)$, we say that the pair is bad-for-girls if there was a bad-for-girls problem solved by both of them. Similarly define bad-for-boys problems/pairs.

For each $i$, let $B_i$ be the set of problems solved by $b_i$. By the given conditions we must have that some problem in $B_i$ is solved at least $\lceil \frac{21}{6} \rceil > 3$ times by girls; thus at most 5 problems in $B_i$ are bad-for-girls. By definition, this means that for any fixed $i$, at most 10 of the 21 pairs $(b_i, g_1), (b_i, g_2), \dots , (b_i, g_{21})$ are bad-for-girls. Summing across all $i$ we get that there are in total at most $10 \cdot 21 = 210$ bad-for-girl pairs.

Similarly, at most $210$ of the $441$ pairs of students are bad-for-boys. Since $210+210<441$ there must be a pair of students which is neither bad-for-girls nor bad-for-boys; then this pair must have solved a problem that was neither bad-for-girls nor bad-for-boys, as desired.

Consider a complete bipartite graph with the $21$ boys $b_1,\ldots,b_{21}$ on the left and the $21$ girls $g_1,\ldots,g_{21}$ on the right. Label the edge $b_ig_j$ by a common problem solved by the two (pick arbitrarily if many). Call a problem [i]good[/i] if at least $3$ girls solved it, and call an edge good if it is labeled by a good problem. We have the following key claim.

[b]Claim:[/b]  There are at least $21\cdot 11$ good edges.

[i]Proof:[/i]  Consider a boy $b_i$, and suppose he solved $m$ problems (note that $m\le 6$). Let the outdegrees from $b_i$ be $x_1,\ldots,x_m$ for each of the $m$ labels. We see that if $x_i\ge 3$, then those $x_i$ edges are good. Thus, there are at least \[\sum_{i:x_i\ge 3}x_i\] good edges coming out of $b_i$. It is easy to verify that this is always at least $21-5\cdot 2=11$, so we're done. $\blacksquare$

If we swap boys and girls, then the same claim implies that there at least $21\cdot 11$ "boy-good" edges, so some edge is boy-good and girl good. Then, that problem has at least three solves from both genders, so we're done.

Draw the graph $a_{1}, a_{2}, \ldots a_{21}, b_{1}, \ldots b_{21}$, where $a_{i}$ represents the ith boy, and $b_{j}$ represents the $j$th girl. Draw directed edge between $a_{i}$ and $b_{j}$ if there exists a problem that $a_{i}$ and $b_{j}$ both solved, and that problem has more than $3$ solutions from girls. Similarly, draw the directed edge in the other direction with boys and girls reversed.

I claim there are at least $21\cdot 11$ edges going from $a_{i}$ to $b_{j}$, for some $i$ and $j$. Observe that for any boy $a_{i}$, there will be some positive number of problems that are solved by at least $3$ girls and $a_{i}$. This means there are at least $21 - 5\cdot 2$ girls who solved a problem that $a_{i}$ did, in which the problem has more than $3$ solves.

Similarly, there are at least $21\cdot 11$ edges going from $b_{j}$ to $a_{i}$. Since $21\cdot 11 + 21\cdot 11 > 21^2$, there must be some edge that is bidirectional, so some problem is solved by at least $3$ boys and $3$ girls.

[quote=orl]Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.[/quote]

Fix a girl $G$ and consider all pairings with the boys.By PHP,at least 3 boys and the girl $G$ have attempted the same question.Do the same thing for the boys and for each boy let $$D_{b_x}=\{g_x,g_{x+1},g_{x+2} \}$$ be the set of the three girls who have attempted the same question as $b_x$;do the same thing for girls as well.
By PHP again there are atleast $\frac{21 \cdot 3}{21}=3$ boys and girls having solved the same question.

We count the number of boy-girl pairs that solved a problem that at most two people of the same gender have solved. For any person, at most five of the six problems they solved were solved by at most two people of the opposite gender. Thus the maximum number of these pairs is $21\cdot2\cdot5=210$. But there are $21^2>2\cdot210$ boy-girl pairs with one common problem solved. This proves the statement.

[hide=Generalization]Let $m,n \geq 2$ be integers and $k=2(m-1)(n-1)+2$. Then if $k$ boys and $k$ girls participate and nobody solved more than $m$ problems, then some problem was solved by at least $n$ boys and $n$ girls.[/hide]

Suppose that each problem is solved either by at most two girls or at most two boys. Construct a grid $21 \times 21$ the obvious way (the rows represent girls and the columns - boys; in each cell write the number of the problem solved by both the girl and the boy) and color a cell blue if it is a problem solved by at most two girls and red otherwise. By PHP at least $\lceil \frac {21^2}{2} \rceil =221$ cells are red (WLOG). Thus there is a row (a girl) having at least $\lceil \frac{221}{21} \rceil =11$ red cells (problems solved by at most two boys). Hence there are at least $6$ red numbers on this row (every red number can appear at most twice). But everyone has solved at most $6$ problems, so there are exactly $6$ different numbers on this row and we actually have $21$ red cells on this row. But using again the fact that each red number appears at most twice, we get that there are at least $11$ different numbers on that row, contradiction.

c8?????

say that a problem is [i]type 1 hard[/i] if it is solved by at most 2 boys, and say that a problem is [i]type 2 hard[/i] if it is solved by at most 2 girls.

Suppose FTSOC that each problem is either type 1 hard or type 2 hard (possibly both). Then, make a 21x21 grid, with the rows corresponding to the boys and the columns corresponding to the girls, and in each cell, put a problem that was solved by both of the.

Note that if every problem is either type 1 hard or type 2 hard, then there are at least 221 cells that contain type 1 hard problems. Consequently, there must be some column that contains at least 11 type 1 hard problems. However, note that each type 1 hard problem is solved by at most 2 boys, so there must be at least 6 different type 1 hard problems that appear in this column. This means that the chosen girl solved 6 type 1 hard problems, so they could not have solved any other problems. However, if this girl solved only 6 type 1 hard problems, then they can only share a problem with at most 12 boys, contradiction so we are done.

Call a problem buggy if it was solved by at most two boys, and glitchy if it was solved by at most two girls. Assume for the sake of contradiction that every problem was buggy or glitchy. Let a buggy solve be an ordered pair $(B,p)$ of a boy and a problem such that $B$ solved $p$ and $p$ was buggy. Let a glitchy solve be an ordered pair $(G,p)$ of a girl and a problem such that $G$ solved $p$ and $p$ was glitchy.

Now, consider each of the $441$ boy-girl pairs, and there exists a problem that they both solved. Thus, for each boy $B$ and girl $G$, there exists a problem $p$ so that $(B,p)$ was buggy or $(G,p)$ was glitchy. Without loss of generality, assume that of the $441$ such boy-girl pairs, at least $221$ of them has $(B,p)$ as a buggy pair. If it has a buggy pair, call the boy-girl pair bugged. Otherwise, call it glitched.

Now, there are $21$ girls, so by pigeonhole principle there exists a girl $G_0$ such that $(B,G_0)$ is buggy $11$ times. Each buggy problem can only be solved by two boys, so there are six buggy problems. These buggy problems clearly must be the only ones solved by $G_0$. However, these $6$ buggy problems can only be solved by $12$ boys, so for $9$ boys $B$, $(B,G_0)$ is not buggy. Thus, $G_0$ must've solved a problem that is none of the six buggy problems before, contradiction.

We reinterpret the problem as follows: for each pair of boy and girl, we assign exactly one [i]shared trait[/i] such that each person only has at most $6$ distinct traits. Since personality is a bit weird, the same trait might not express itself with everyone else, so it could be the case that a boy and a girl have the same trait, but it might not be the trait shared between them. Suppose that there are no traits held by at least three girls and three boys. Split the traits into two disjoint categories such that all [i]boyish[/i] traits are held by at most $2$ girls and all [i]girlish[/i] traits are held by at most $2$ boys.

Number the boys $1$ through $21$ and let $b_i$ denote the number of girlish traits the $i$-th boy has, so he has at most $6-b_i$ boyish traits. If $b_i=0$ for some $i$, then this boy only has boyish traits and thus can only share a trait with at most $6\cdot 2=12$ girls, which is bad (such are the consequences of toxic masculinity), so $b_i \geq 1$ always. Then the number of girls who he shares a girlish trait with is at least $21-2(6-b_i)=9+2b_i \geq 11$. Summing across all boys, we find that at least $11\cdot 21$ boy-girl pairs share a girlish trait. But the analogous argument for girls implies that at least $11\cdot 21$ boy-girl pairs share a boyish trait. Since each boy-girl pair only shares $1$ trait and boyish and girlish traits are disjoint, this easily yields a contradiction. $\blacksquare$


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]IMO 2007 HKTST 1[/url]
Problem 4

(a) Given five points on a plane such that no three of the points are collinear. Show that among the triangles which are drwan using any three of these five points as vertices, at least three of the triangles formed are not acute-angled triangles. (An acute-angled triangle is one in which all the three interior angles are acute angles.)

(b) Given any 100 points on a plane such that no three of the points are collinear. SHow that among the triangles which are drawn using any three of these 100 points as vertices, at least 30% of the trinagles are not acute-angled triangles.

(a) Consider the convex hull.  Show that if one of the points is inside a triangle whose vertices are some of the points, then we have at least 2 triangles there.  Show that if we have a convex quadrilateral whose vertices are some of the points, we have 1 at least triangle from those four.  Then consider the four cases: 1) triangle 2) quad 3) pentagon with two adjacent non-acute angles 4) pentagon without two adjacent non-acute angles.

(b) This follows from (a).  We have C(100,5) sets of five distinct points.  But each triangle occurs in C(97,2) of these.  Thus we have at least 3*C(100,5)/C(97,2) non-acute triangles.  We can check this value is exactly 30% of the total number of triangles.


calculus
integration
logarithms
trigonometry
calculus computations

@[b]didilica[/b]: After I finally managed to prove the formula you used (via a pretty long argument), I still have one question:
How do you justify the integral-sum interchange?

(I've always had problems with those ever-changing signs...)

[quote="misan"]

$ \int_{0}^{\frac{\pi}{2}}\;\left( \frac{x^{2}}{x^{2}+\ln^{2}{(2 \cos x)}}\right) \;dx\;=\;\boxed{\frac{\pi}{8}\left( \ln{(2\pi e)}-\gamma \right)}$

$ (where\;\;\;\gamma \;\;\;is \;\;\;Euler's \;\;\;constant)$

[/quote]

this one is hard !   :no: :wacko:

[quote="misan"][quote="misan"]

$ \int_{0}^{\frac {\pi}{2}}\;\left( \frac {x^{2}}{x^{2} \plus{} \ln^{2}{(2 \cos x)}}\right) \;dx\; \equal{} \;\boxed{\frac {\pi}{8}\left( \ln{(2\pi e)} \minus{} \gamma \right)}$

$ (where\;\;\;\gamma \;\;\;is \;\;\;Euler's \;\;\;constant)$

[/quote]

this one is hard !   :no: :wacko:[/quote]

not to mention, lengthy......  :(   :maybe:


There is a set of 1000 switches, each of which has four positions, called $A, B, C,$ and $D.$  When the position of any switch changes, it is only from $A$ to $B,$ from $B$ to $C,$ from $C$ to $D,$ or from $D$ to $A.$  Initially each switch is in position $A.$  The switches are labeled with the 1000 different integers $2^x3^y5^z,$ where $x, y,$ and $z$ take on the values $0, 1, \ldots, 9.$  At step $i$ of a 1000-step process, the $i$th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$th switch.  After step 1000 has been completed, how many switches will be in position $A$?

[hide]
For a switch label $2^{x_a}3^{y_a}5^{z_a}$ to divide another switch label $2^{x_b}3^{y_b}5^{z_b}$, $x_b \ge x_a, y_b \ge y_a, z_b \ge z_a$
after 1000 iterations, only switches that have the following property will be at A:
$4|(10-x)(10-y)(10-z)$
because the number of times that the switch is activated has to be a multiple of 4
(the number is 10 because you are counting the number $(x_k,y_k,z_k)$ itself and all other combinations with $x \ge x_k, y \ge y_k, z \ge z_k$)

So... a number is either odd, divisible by only 2, or divisible by a greater power of 2
We list the number of combinations of the first two cases.
For $(10-x)(10-y)(10-z)$ to be odd, all $x,y,z$ must be odd: there are $5^3$ cases.
(there are 5 odd integers from 0~9)
For $(10-x)(10-y)(10-z)$ to be divisible only by 2 and not 4: $(2)(5)(5)(3)$ cases.
(one of the variables is either 8 or 4; there are 5 odd integers for the other variables; there are 3 possibilities for choosing which variable to be even)

So $1000-125-150 = \boxed{725}$

[/hide]

[quote="minsoens"][hide]
For a switch label $ 2^{x_a}3^{y_a}5^{z_a}$ to divide another switch label $ 2^{x_b}3^{y_b}5^{z_b}$, $ x_b \ge x_a, y_b \ge y_a, z_b \ge z_a$
after 1000 iterations, only switches that have the following property will be at A:
$ 4|(10 \minus{} x)(10 \minus{} y)(10 \minus{} z)$
because the number of times that the switch is activated has to be a multiple of 4
(the number is 10 because you are counting the number $ (x_k,y_k,z_k)$ itself and all other combinations with $ x \ge x_k, y \ge y_k, z \ge z_k$)

So... a number is either odd, divisible by only 2, or divisible by a greater power of 2
We list the number of combinations of the first two cases.
For $ (10 \minus{} x)(10 \minus{} y)(10 \minus{} z)$ to be odd, all $ x,y,z$ must be odd: there are $ 5^3$ cases.
(there are 5 odd integers from 0~9)
For $ (10 \minus{} x)(10 \minus{} y)(10 \minus{} z)$ to be divisible only by 2 and not 4: $ (2)(5)(5)(3)$ cases.
(one of the variables is either 8 or 4; there are 5 odd integers for the other variables; there are 3 possibilities for choosing which variable to be even)

So $ 1000 \minus{} 125 \minus{} 150 \equal{} \boxed{725}$

[/hide][/quote]

Why is this solution wrong?

for a switch to return to position A, the number of steps it undergoes must be congruent to 0 mod 4. or, the number of labels its label divides must be congruent to 1 mod 4 or of the form 4n + 1. the number of labels a label $ 2^p 3^q 5^r$ divides is $ (9\minus{}p\plus{}1)(9\minus{}q\plus{}1)(9\minus{}r\plus{}1)$ or $ (10\minus{}p)(10\minus{}q)(10\minus{}r)$ so we have to find the number of $ (p, q, r)$ such that

$ (10\minus{}p)(10\minus{}q)(10\minus{}r)$ = $ 4n\plus{}1$.

[quote="minsoens"]For $ (10 \minus{} x)(10 \minus{} y)(10 \minus{} z)$ to be divisible only by 2 and not 4: $ (2)(5)(5)(3)$ cases.
(one of the variables is either 8 or 4; there are 5 odd integers for the other variables; there are 3 possibilities for choosing which variable to be even)[/quote]
Nah, the only mistake is that one of the variables can also be $ 0$. So there should be $ 3*5*5*3 \equal{} 225$ cases.
(yes, it sucks to have the whole answer wrong and don't get any credit just because you have a very very minor and careless error when you could have potentially gotten it :( )

By the way, either this problem is ill-defined, or my english sucks, but I think the problem SHOULD have specified that the $ i$th switch should be the $ i$th smallest label number, or else we wouldn't know which $ i$th switch it should be...

All the switches are the i-th switch at one point so the label order doesn't matter. However, this did confuse me at first.

[quote=rzsolt][quote="minsoens"][hide]
For a switch label $2^{x_a}3^{y_a}5^{z_a}$ to divide another switch label $2^{x_b}3^{y_b}5^{z_b}$, $x_b \ge x_a, y_b \ge y_a, z_b \ge z_a$
after 1000 iterations, only switches that have the following property will be at A:
$4|(10-x)(10-y)(10-z)$
because the number of times that the switch is activated has to be a multiple of 4
(the number is 10 because you are counting the number $(x_k,y_k,z_k)$ itself and all other combinations with $x \ge x_k, y \ge y_k, z \ge z_k$)

So... a number is either odd, divisible by only 2, or divisible by a greater power of 2
We list the number of combinations of the first two cases.
For $(10-x)(10-y)(10-z)$ to be odd, all $x,y,z$ must be odd: there are $5^3$ cases.
(there are 5 odd integers from 0~9)
For $(10-x)(10-y)(10-z)$ to be divisible only by 2 and not 4: $(2)(5)(5)(3)$ cases.
(one of the variables is either 8 or 4; there are 5 odd integers for the other variables; there are 3 possibilities for choosing which variable to be even)

So $1000-125-150 = /boxed{725}$

[/hide][/quote]

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/1999-aa/99aimans.htm , says '650' is the answer. Not '725'.

the way I got '650':  you notice a switch is hit once for each number that divides it, so by the fundemental therom of arithmetic, you want the number of solutions to the equation $4k = (x + 1)(y + 1)(z + 1)$, where $x, y, z$ are between $0$ and $9$ inclusive. to find the number of soloutions, take the compliment, better known as 'the number of ways to make $(x + 1)(y + 1)(z + 1)$ NOT a multiple of four. So there is ither one two, or zero two's among the three multiples. for there to be zero twos, there are $5^3$ ways to choose $x, y z$. to have a single two, there are ${3\choose 1}3*5^2$ ways. Summing, and subtracting from $1000$ we find the answer is $/boxed{650}$[/quote]

In this solution, I am confused as where the equation $4k=(x+1)(y+1)(z+1)$ is coming from? Specifically where does the number $4$ come into it? The problem is a little confusing, can anyone clear this up?


advanced fields
advanced fields unsolved

Show that $\{0\}\to [0,1]$ is a cofibration.


algebra
polynomial
system of equations
algebra unsolved

Solve in $ \mathbb{R}$ a following system of equations:

$ x_0 ^n \plus{}x_1 ^{n\minus{}1}\plus{}...\plus{}x_{n\minus{}1}\plus{}1\equal{}0$
$ x_1 ^n \plus{}x_2 ^{n\minus{}1}\plus{}...\plus{}x_{n}\plus{}1\equal{}0$
...
$ x_n ^n\plus{}x_0 ^{n\minus{}1}\plus{}...\plus{}x_{n\minus{}2}\plus{}1\equal{}0$

Comment: Also one could ask for a solution in $ \mathbb{C}$

I- If $ n$ is odd then an obvious solution is $ x_i\equal{}\minus{}1$.
II-Assume $ n\equal{}2$. Then $ x_2$ is a root of this polynomial:
$ (x^2\plus{}x\plus{}1)(x^6\minus{}x^5\plus{}4x^4\minus{}3x^3\plus{}7x^2\minus{}4x\plus{}5)$; this polynomial hasn't any real root.
III-Assume $ n\equal{}3$. Then $ x_3$ is a root of a polynomial as follows:
$ p_{72}(x)(x\plus{}1)(x^2\minus{}x\plus{}1)(x^2\plus{}1)^2$ where $ p_{72}(x)$ is a polynomial of degree 72 without any real root (by Sturm's method). Thus
1) there exists only one real value for $ x_3$: -1. Then $ x_0\equal{}x_1\equal{}x_2\equal{}\minus{}1$.
2) In the complex case there exist many solutions.
For $ n\equal{}4$ Maple is out of order.


function
algebra
polynomial
real analysis
real analysis unsolved

Let $f : \mathbb R \to \mathbb R$, $f \neq 0$, be a continuous, periodic function, $F$ be a primitive of $f$ and $g$ be a polynomial of degree at least $2$. Can $F \circ g$ be a periodic function?

[i]Problem 120 of the Romanian MO Shortlist 2005, 12th grade[/i]

Assume that $f$ have period $A$. Assume that $F\circ g(x+kT)-F\circ g(x)=0,\forall x,k$ for some $T$. Differentiating we get that: $f\circ g(x+kT)g'(x+kT)-f\circ g(x) g'(x)=0$. Since $g'(x+kT)\to \pm \infty$ when $k\to\infty$ we get that $f(g(x+kT))\to 0,\forall x$.

   Fix an $\epsilon$ arbitrarly. Denote by $F_n=\{x: f(g(x+kT))\le \epsilon,\forall k>n\}$. From Baire th we get that $F_N$ contains an interval $[a,b]$. And, since $F_n\subset F_{n+1}$, $[a,b]\in F_n,\forall n>N$. For sufficiently big $n$ we have that $g(b+kT)-g(a+kT)>A$. Hence, for any $y\in[g(a+kT),g(b+kT)]$ we have that $f(y)\le \epsilon$ and since  $g(b+kT)-g(a+kT)>A$, we get that $f(x)<\epsilon, \forall x\in \mathbb{R}$. Since $\epsilon$ was chosen arbitrary, we get that $f=0$. Contradiction.


well, thi sprobably shouldn't be in CLASSROOM MATH. Have a mod move it to NC local forum

EDIT FOR RANTING:so? NC don't only do it

No, Duke TIP is everywhere.  It's just like how College Board is based in Princeton, or whatever.

I did TIP when I was in 7th grade.  I got a 780 on math, and I don't remember what on English.  It sucked, because 1 more problem right and I would have got a $\$$4000 scholarship to a summer program.

o well I for one didnt do too great only a measly 28. looks like im one of the few who took the ACT

nah ACT math 28 isn't too bad. You would beat most regular juniors :rotfl:  I got a 29 (ACT) in 7th grade, but I didn't take it this year...

I took both tests in 7th grade. I got a 670 math on the SAT, and a 31 math on the ACT.

The ACT score was a 12 point improvement from 5th grade, and the SAT score was a 120 point improvement from 6th grade.

To qualify (I was in Canada for 7th grade) in 8th grade, I took the PSAT. I got 76/80 in Math, 79/80 in Crit. Reading, and 72/80 in Writing.

Yes...I did better in reading than in Math  :|

[quote="tridecalogism_lover"]yah, i remember taking that in 7th grade.....don't you dare comment on my scores..i took the SAT...

Math- 400
Critical Reading/Verbal- 450

Total Score: 850/1600

alright already! you can say that I sucked, but this was WAY before I got better at math, ok? :D[/quote]

did you get better in verbal/reading?

i haven't taken the SAT again, but when I take it, I hope to do better on both Math and Reading..:)

[quote="King Paul"][quote="bpms"]I did it last year. I got an 800 math, 640 verbal, 600 writing.[/quote]

u little beastly grasshopper  :rotfl:

I got 730 math, 510 verbal, and 590 writing

I was a dumb freak who could do nothing, and still can't

what a mumpy me[/quote]
I bombed it.
Besides, I speak Hindi not English so screw that.

[quote="Walk Around The River"]i got...

710 math
410 reading
310 writing.

shut up, i speak hebrew, not english.  screw this language stuff.[/quote]
comes from the person who's always correcting me on my english grammar

if you have desi parents you should have a fair guess on why my parents wont lemme do this duke tip thing

why r we continuing something that should be moved?  :P

[quote="Arvind_sn"]700 math
570 critical reading
650 writing
total 1910[/quote]

you tied me on math and reading [1270], i had +10 in math, -10 in reading, but you beat me by 140 on writing  :rotfl:

[quote="King Paul"][quote="bpms"]I did it last year. I got an 800 math, 640 verbal, 600 writing.[/quote]

u little beastly grasshopper  :rotfl:

I got 730 math, 510 verbal, and 590 writing

I was a dumb freak who could do nothing, and still can't

what a mumpy me[/quote]

i feel stupid.
800? is that even possible?

i got 690 on math, 590 verbal, and a dissapointing 550 writing...

not that bad... i guess?

[quote="now a ranger"]writing's okay. I just messed up on the 25 minute thingy. 25 minutes isn't a lot of time. Therefore, I think I made a 3 on the essay lol. I didn't do badly on the rest.

Verbal is my concern, and I'm constantly working to improve right now.  :P

and for people who are already almost done with geometry or above, 800 on SAT math isn't out of the question, especially competitive math people.[/quote]

yes... 25 minutes. impossible.

oh. an algebra 1 noob like me... cant say anything..

being only in algebra really jeopardized my chances of getting a higher SAT math score, possibly 800.

[quote="Walk Around The River"]it shouldnt be moved.  duke tip is not a north carolina thing.  its for all of the southeast.[/quote]
i don't remember ks being in the south east  :P

yeah...the only reason i didn't get more than 700 on math is because math was like the last few sections for me, and by then i looked something like this: :P  except i wasnt smiling.

780 math, 730 verbal, 670 writing.
You can get an 800 on writing without getting a 12 on the essay. One of my [high school] friends got an 800, with an 8 on the essay. Her grammar was really good though.
A lot of 7th graders manage to get perfect scores on the math or verbal! There were 20 in 2006..."Bevan Scholars", with a scholarship to one of the summer programs.
And I thought the math section wasn't too hard; I was in Algebra I, but there were only a few problems I guessed on. They weren't really tricky, just plain math.

I missed the test date, and the make up test date! My dad called, and they are allowing me to take it, but my score will be unofficial. Well, I'll still see what I'll get.

Geez I completely forgot to take it again this year in 8th grade. i might have even perfected the math section but o well. Besides, i'm still in Algebra 1 now like you guys were doing a year before me.  :(  :mad: Man... Our state is so behind....

I being dumb got 300-300-300 respectively

i got 650 Critical Reading - 660 Math - 580 Writing

No studying...  my essay sucked-9/12

in 7th grade, or this year.

[quote="lifesapain"]so who all took the duke tip and if so how high did u score on the math section? sry if this is random and irrelevant[/quote]
I didn't take SAT for Duke TIP I  took ACT, My Math was a 30, English 24, reading 23, science 21. In the 7th grade which i am in now

i hate your power of luck, i did that on Science and look what i got

[quote="lifesapain"]what are you talking about now a ranger you can take the duke tip many times over as long as you are in middle school. Ha I had a 34 in reading comprehension somehow for some reason even when randomly guessing for the last half of it. that's messed up.[/quote]

Haha. I know who you are.

No. The Duke talent search TIP program is for 7th graders only. 8th graders can take the SAT, but aren't eligible for the Duke talent search.

Well, I took the SAT for John Hopkins CTY.

I got 800 math, 740 reading, 630 writing. (Hehe, my essay got a 6)

The correct forum would be Other Contests and Programs, or else Round Table.

For those who are saying TIP is only for seventh grade - the Talent Identification Program is an umbrella title that covers, among other things,
 - Talent Search (4th/5th grade and 7th grade)
 - Summer Studies (7th through 10th grade)
 - Precollege (11th grade)
 - Field Studies (9th through 12th grade)
 - Exchange programs (9th through 12th grade)
 - Scholar weekends (8th through 12th grade)
 - Distance Learning (all levels)
 - Publications

The Duke Talent Search region covers Alabama, Arkansas, Florida, Georgia, Iowa, Kansas, Kentucky, Louisiana, Mississippi, Missouri, Nebraska, North Carolina, Oklahoma, South Carolina, Tennessee, and Texas.  However, many programs are open to anyone in the world who meets the qualifications.

Personally, the summer studies programs are very highly recommended, of course.


function
algebra
polynomial
induction
number theory
prime numbers
number theory unsolved

[quote="romano"]Let f: Z+ --> Z+ satisfying two conditons :
  $ f(p)= 1 \forall $ prime numbers p
  $ f(ab) = af(b) + bf(a) $
 Prove that :
 There exists only funtion f . Find n such that : f(n) = n[/quote]
Well clearly $f$ is hardly identified, I mean if $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ then
\[
f(n)=n(\alpha_1p_1^{-1}+\cdots+\alpha_kp_k^{-1}).
\]
Then, if $f(n)=n$ implies that $\alpha_1p_1^{-1}+\cdots+\alpha_kp_k^{-1})=1$ an $n$ must be $3^3$.

Best Regards.

RX TCM

Assume $n=\prod_{i=1}^k p_i$ where $p_i$ are (not necessary distinct) prime numbers.
Lemma: $f(n)=s_{k-1}(p_1,...,p_k)=n \cdot \sum_{i=1}^k \frac{1}{p_k}$ where $s_i$ are the elementary symmetrical polynomials ( $s_k(x_1,...,x_n) = \sum_{1 \leq a_1 < ... < a_k \leq n} x_{a_1} \cdot ... \cdot x_{a_k} $ )
Prove (induction on $k$):
start $k=1$: given by $f(p)=1$
step: $f(p_{k+1} \cdot \prod_{i=1}^k p_i)= p_{k+1} \cdot f(\prod_{i=1}^k p_i) + \prod_{i=1}^k p_i \cdot f(p_{k+1}) = p_{k+1} \cdot s_{k-1}(p_1,...,p_k) + \prod_{i=1}^k p_i = s_k(p_1,...,p_{k+1})$

Now let be $n=\prod_{i=1}^k p_i^{v_i}$ where the $p_i$ are distinct this trimes.
$f(n)=n \iff n=n \cdot \sum_{i=1}^k \frac{v_i}{p_k} \iff  \sum_{i=1}^k \frac{v_i}{p_k} = 1 $. Since the primes are distinct you get $p_i|v_i$ and thus only $n=p^p$ are solutions.

[quote="ZetaX"]RobertuX: why only $3^3$ ¿  :?[/quote]
The question is to find an $n$ no every $n$, so I have given an example, that's all. :D  :blush:

I assume that $\mathbb {Z}^+$ means $\{ 0 ,1, 2, \cdots, \}$ otherwise you would have a problem to evaluate $f(1)$...

Now, it is easy to see that $f(0)=f(1)=0.$
A straightforward induction (on $k$) leads to $f(p^k)=k^p{k-1}$ for all prime $p$ and non-negative integer $k$.
Another straightforward induction (on $k$) leads to :
If $n = p_1^{\alpha _1} \cdots p_k^{\alpha_k}$ is the prime decomposition of $n$ then $f(n)= n( \frac {\alpha_1} {p_1} + \cdots + \frac {\alpha_k} {p_k} ).$

This proves the unicity.
Conversely, the function defined as above clearly satisfies the problem.

Clearly, $n=0$ is a solution of $f(n)=n$ but $n=1$ is not.
Now assume that $n>1$ and $n = p_1^{\alpha _1} \cdots p_k^{\alpha_k}$ so that $f(n)= n( \frac {\alpha_1} {p_1} + \cdots + \frac {\alpha_k} {p_k} ).$
Let $T = p_1 \cdots p_k$ and $T_i = \frac T {p_i}.$
If $f(n)=n$ then $\frac {\alpha_1} {p_1} + \cdots + \frac {\alpha_k} {p_k} = 1.$ (1)
That is $T = \sum_{i=1}^k \alpha_iT_i$.
It follows that for all $i$, we have $\alpha_i = 0 \mod [p_i]$.
Since $\alpha_i > 0$ we ten have $\alpha_i \geq p_i$ so that $\frac {\alpha_i } {p_i} \geq 1$.
From (1), we deduce that $k=1$ and $\alpha_1 = p_1$.
Conversely, it is clear that this gives a solution of the problem.

Thus, the solutions of $f(n)=n$ are $n=0$ and $n=p^p$ for any prime $p$.

Pierre.

Why didn't I see the other posts before posting mine??? ;)  ;)  ;)

Pierre.

[quote="pbornsztein"]Why didn't I see the other post before posting??? ;)  ;)  ;)

Pierre.[/quote]
Well, you did slowwwwwwly!!!   :D  :D  :D  Because we was discusing for a long time before you put this post.


The vertices of a triangle are the points of intersection of the lines y=-x-1, x=2, and y=x/5+13/5. Find an equation of the circle passing through all three vertices.

[hide]
Solving the 3 pairs of 2 equations, we find that the points of intersection are $(2,-3)$, $(2, 3)$, and $(-3, 2)$. Noticing that these are all have $\pm 3$ and $\pm 2$ for both coordinates, we find that the equation of the circle is $x^{2}+y^{2}= 13$.[/hide]

There has to be an easier way to do it but...
[hide]
Rewrite the equations of the lines as $x+y=-1, x=2, and-x+5y=13$ (this step is personal preference) Then, find the vertices of the triangle. They are $(2,-3),(2,3), and (-3,2)$. Two perpendicular bisectors are $x-y=0$ and $y=0$ Therefore the circumcenter is the origin. The distance from the origin to anyone of these points is $\sqrt{13}$ The final equation is $x^{2}+y^{2}= 13$
[/hide]

I don't think $x-y = 0$ is a perpendicular bisector... however, $x+y = 0$ is![/hide]

Is it really? Why? Maybe i'm just slow... :blush:


linear algebra
matrix
search
linear algebra unsolved

We denote the set M, which has all $3\times 3$ matrixes with elements from $\{-1,+1\}$.

(a) If $A\in M$ then $6\leq \det(A)\leq 6$

(b) Find $P,Q\in M$ such that $rank(Q)=1,rank(P)=2$.

(c) If $A\in M$ then $\det(A)$ divides $4$.

(d) Show if $A\in M$ then $A\in \{-4,0,+4\}$

(e) Show if we have $B\in M$ a inversable matrix then $B^{-1}$ is not in $M$.

(f) Show if $A\in M$, then $A^{2006}$ has every element different by $0$

(g) card(M)=?

thanks! I didn't search it properly  :blush:


Prove that for each graph $G$: \[\chi_{\ell}'(G)\leq\frac{n+\omega(G)}{2}\]($\omega(G)$ is size of biggest subgraph of  $G$ isomorph to $K_{m}$, and $\chi'_{\ell}$ is [url=http://en.wikipedia.org/wiki/List_edge-coloring]list edge chromatic number[/url] of $G$.)

Are you sure this is correct? It doesnt seem to be...

If G is a star, i.e. one point connected with n-1 other points, then
omega(G)=2. You didnt specify what is n, but if it is a number of
vertices, then in this case we cannot color n-1 edges with (n+2)/2 or
less colors, because any two edges are neighbours to each other.
Do I miss something?

You're right.
Consider $\chi_{\ell}(G)$ instead of $\chi_{\ell}' (G)$


function
trigonometry
limit
calculus
integration
real analysis
real analysis unsolved

Let $ f$ be the function defined by $ f(x)\equal{}x$ if $ x$ is irrational and $ f(x)\equal{}p \sin (1/q)$ if $ x\equal{}p/q$ in lowest terms. At what points is $ f$ continuous?

The following is what I have so far:

$ f$ is discontinuous at each nonzero rational:
Let $ r$ be a nonzero rational. Let $ \varepsilon \equal{} |r\minus{}f(r)|$. If $ r>0$, given $ \delta>0$, choose an irrational $ y$ in $ (r,r\plus{}\delta)$. Then $ |f(y)\minus{}f(r)|\equal{}y\minus{}f(r)>r\minus{}f(r) \equal{} \varepsilon$. If $ r<0$, given $ \delta>0$, choose an irrational $ y'$ in $ (r\minus{}\delta,r)$. Then $ |f(y')\minus{}f(r)|\equal{}f(r)\minus{}y'>f(r)\minus{}r \equal{} \varepsilon$.

$ f$ is continuous at 0:
Given $ \varepsilon>0$, let $ \delta\equal{}\varepsilon$. If $ |x|<\delta$, then $ |f(x)| \leq |x|<\delta\equal{}\varepsilon$.

I need help with the irrationals.

Suppose $ x$ is an irrational number. Note that $ f(x)\equal{}x.$

Now let $ r_n$ be a sequence of rational numbers such that $ r_n\to x.$

Let $ r_n\equal{}\frac{p_n}{q_n}.$ Note that $ q_n\to\infty,$ and that $ |p_n|\le C q_n.$

$ f(r_n)\minus{}r_n\equal{}p_n\left(\sin\left(\frac1{q_n}\right)\minus{}\frac1{q_n}\right).$

But we know that $ |\sin x\minus{}x|\le\frac{|x|^3}6.$ Hence,

$ |f(r_n)\minus{}r_n|\le|p_n|\cdot\frac16\cdot\frac1{q_n^3}\le \frac{C q_n}{q_n^3}$

This goes to zero as $ r_n\to x,$ so

$ \lim f(r_n)\equal{}\lim (f(r_n)\minus{}r_n)\plus{}\lim r_n\equal{}0\plus{}x\equal{}x\equal{}f(x).$

Since we also have $ \lim f(x_n)\equal{}x\equal{}f(x)$ for any sequence of irrational $ x_n$ that tend to $ x,$ we have that $ f$ is continuous at every irrational.

$ f$ is Riemann integrable over any bounded interval, and its integral is the same as the integral of $ x$ over that interval.

[quote]Note that $ q_{n} \rightarrow \infty$[/quote]

I need some help with this part. I can't quite formalize the argument here.

Suppose for some $ M$ we can for every $ n_0$ find $ n \geq n_0$ such that $ q_n < M$.

Consider the rational numbers $ p/q$ with $ q < M$. One of these numbers has minimal distance $ d$ to $ x$ ($ |p| < (x \plus{} 1)M$, so there are only finitely many numbers to consider). Hence we cannot have $ r_n \to x$


inequalities
calculus
geometry
topology
linear algebra
IMO

I think we should not stop at posting problems from IMO as the main topic
We should open topic for the other competitions such as in Poland, Austrian, Baltic-Way Holand ...
What do you think ?

hello to you too :)

if you haven't been paying attention on the forum: in the advanced section there are a lot of problems from these competition and many other competitions.

in this forum, we discuss only IMO problems (they were posted exactly when the IMO took place, and hopefully MathLinks will re-do the same things again this year ;))

to post any other problem just post it in the right forum (ie algebra, inequalities, etc.) or if it involves college questions (linear algebra, field algebra, analisys, calculus, algebraic geometry, topology) try the College Playground.


Let $ ABCD$ be a convex quadrilateral such that the sides $ BC$ and $ CD$ have equal lengths and $ 2\widehat{A} \plus{} \widehat{C} \equal{} 180^0$. Let $ M$ be the midpoint of the line segment $ BD$. Prove that $ \widehat{MAD} \equal{} \widehat{BAC}$.

Let $ OBCD$ be a cyclic kite, $ \angle BOD \plus{} \angle BCD \equal{} 180^\circ$ and $ \angle OBC \equal{} \angle ODC \equal{} 90^\circ.$
$ A$ is on a circle $ (O)$ with radius $ OB \equal{} OD$ tangent to $ BC$, $ DC$ at $ B, D.$ $ AM$ is the A-median and $ AC$ the A-symmedian of the $ \triangle ABD.$


let a , b odd numbers
show that a^2 +b^2 will not be complete square
i.e  a^2+b^2 not equal c^2 for any c

If $ a$ and $ b$ are odd numbers ,so is $ a^{2}$ and $ b^{2}$ .
Ok suppose that $ a^{2} \plus{} b^{2} \equal{} c^{2}$
Since $ a$ and $ b$ is odd . We have $ a^{2} \equiv 1\bmod{4}$ and $ b^{2} \equiv 1\bmod{4}$
$ \implies a^{2} \plus{} b^{2} \equal{} c^{2}\equiv 2\bmod{4}$ which is not possible as a perfect square is either congruent to $ 0,1$ mod $ 4$
Also if $ a,b,c$ form a pythagorean triple such that $ a^{2} \plus{} b^{2} \equal{} c^{2}$ we have $ 12\mid ab$ which implies both $ a$ and $ b$ cannot be odd .
Also the general solution of pythagorean triples is $ a \equal{} 2mn$ ,$ b \equal{} m^{2} \minus{} n^{2}$ and $ c \equal{} m^{2} \plus{} n^{2}$ . So we have $ a$ to be even .

[quote="srinath.r"]
Also the general solution of pythagorean triples is $ a \equal{} 2mn$ ,$ b \equal{} m^{2} \minus{} n^{2}$ and $ c \equal{} m^{2} \plus{} n^{2}$ . [/quote]

I thought the formula only gives primitive triples i.e. $ (a,b,c) \equal{} 1$.

But all triples are given by multiplying a primitive by some constant... that is ka,kb,kc. But still, one of them is even, so srinath.r's solution is correct.


Let be $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(x)\plus{}2f(\frac{1}{x})\equal{}x^2\plus{}\frac{2}{x^2}\plus{}3 , (\forall)x\in \mathbb{R}$ . Calculate :

$ \sum_{k\equal{}2}^{2007}\frac{1}{f(k)\minus{}2}$

We put $ x\equal{}a$ and we get
$ f(a) \plus{} 2f\left(\frac{1}{a}\right) \equal{} a^2 \plus{} \frac{2}{a^2} \plus{} 3$     (1)

We put $ x\equal{}\frac{1}{a}$ and we get
$ f\left(\frac{1}{a}\right) \plus{} 2f(a) \equal{} \frac{1}{a^2} \plus{} 2a^2 \plus{} 3$    (2)

Adding up (1) and (2) and dividing by $ 3$ we have
$ f(a) \plus{} f\left(\frac{1}{a}\right) \equal{} \frac{1}{a^2} \plus{} a^2 \plus{} 2$    (3)

We substract (3) from (2) and finally
$ f(a) \equal{} a^2 \plus{} 1$.

Now,

$ \sum^{2007}_{k\equal{}2} \frac{1}{f(k)\minus{}2} \equal{} \sum^{2007}_{k\equal{}2} \frac{1}{k^2\minus{}1} \equal{} \sum^{2007}_{k\equal{}2} \frac{1}{k\minus{}1} \minus{} \frac{1}{k} \equal{}$

$ \equal{} 1 \minus{} \frac{1}{2} \plus{} \frac{1}{2} \minus{} \frac{1}{3} \plus{} \frac{1}{3} \minus{} ... \minus{} \frac{1}{2006} \plus{} \frac{1}{2006} \minus{} \frac{1}{2007} \equal{} 1 \minus{} \frac{1}{2007} \equal{} \boxed{\frac{2006}{2007}}$.

[quote="uglysolutions"]Oops... I suck at calculations... =P[/quote]
Me too :P I forgot the $ \frac{1}{2}$.. So the answer is:
$ \frac{1}{2}\left ( 1\plus{} \frac{1}{2} \minus{} \frac{1}{2007} \minus{} \frac{1}{2008} \right )$


Mary had a coin purse with fifty coins (pennies, nickels, dimes, and quarters) that total exactly 1.00 dollar .
She has at least one of each coin. Unfortunately, while counting her change, she dropped one coin.
What is the probability that the dropped coin was a penny?

a. 0.5   b. 0.6   c. 0.7   d. 0.8   e. 0.9

[hide=" click to reveal content"]

By considering $ p\plus{}n\plus{}d\plus{}q\equal{}46$ and $ p\plus{}5n\plus{}10d\plus{}25q\equal{}59$, since she has one of each coin as well, we get $ p\equal{}44, n\equal{}3, d\equal{}0, q\equal{}0$.

So she has $ 45$ pennies, $ 4$ nickels, $ 1$ dime and $ 1$ quarter, giving a probability of $ \frac{45}{50} \equal{} 0.9$

[/hide]

Hi;

She can have:

45 pennies 2 nickels  2 dimes 1 quarter, making the answer .9

Hi;

Sorry, but I thought that's the reason you hide an answer, so other people can answer it too.

@AndrewTom:

So how did you go about solving those equations? Is it just guess/check or is there a method using number theory?

Well, you have two equations.

$ p\plus{}n\plus{}d\plus{}q\equal{}50$.

$ p\plus{}5n\plus{}10d\plus{}25q\equal{}100$.

Subtracting the first equation from the second yields $ 4n\plus{}9d\plus{}24q\equal{}50$. Clearly if $ q\equal{}2$ (or anything greater than two), then $ n$ and $ d$ are not integers, which cannot happen. since each of $ p$, $ d$, $ n$, and $ q$ have to be a positive integer at least $ 1$, we have $ q\equal{}1$. Therefore,

$ 4n\plus{}9d\equal{}26$.

The only solution in positive integers for this equation is $ d\equal{}2$ and $ n\equal{}2$. Plugging this back into either of our equations, we see that $ p\equal{}45$. The rest follows.

@mathemagiacian1729: Sorry, I've just spotted this now as I'm looking for questions to try.

I did much the same as gaussintraining did, except that I took one of each coin out of the equation (no pun intended) first to get the first two equations in my post.

Hope that's OK. I can write out a solution if you like, but, looking at gaussintraing's post, I guess it's not necessary.

Hi Andrew Tom:

I did it very simply. I exploited a weakness in the question. I started from the .9 because that means 45 pennies and only 5 coins left to think about. Less possibilities. Now I only had to make 55 cents out of 5 coins. Easy to just try a few, especially since you need one of each.

With one of each it has to be one of the combinations of 2,2,1. Only 3 to try.

2 dimes 2 nickels 1 quarter.

Hi bobbym. I must say that I didn't think of doing it that way. I supppose if you're given possible answers, you can consider them.

Hello Bobbym and AndrewTom
[quote="bobbym"]Hi Andrew Tom:

I did it very simply. I exploited a weakness in the question. I started from the .9 because that means 45 pennies and only 5 coins left to think about.[/quote]
Since there are 45 pennies and 50 coins in total, 45/50=0.9
 :idea:
Rooooobiks


Evaluate: $ \sqrt \frac{6!\cdot4!\cdot2!\cdot0!}{5 \cdot 3}$

This is $ \sqrt {\frac {720 \cdot 24 \cdot 2 \cdot 1}{15}}$.  Compute $ 720 \div 15$.  This is $ 48$.  Thus, the answer should be $ \sqrt {48 \cdot 24 \cdot 2 \cdot 1}$, which can be simplified to $ \sqrt {48 \cdot 48}$, which can be further simplified to $ \boxed{48}$.

Note that $ 0! \equal{} 1$.

i thought that 0! was 0 but it turns out i was wrong :)


[u][i] Problem[/i][/u]
 [i]Let the sequence $ (x_{n})$ defined by
 $ x_{0} \equal{} 2007 ,  x_{1} \equal{} 2008$
  $ x_{n\plus{}1} \equal{} \frac{ (n\minus{}1)x_{n} \plus{} (n\plus{}1)x_{n\minus{}1}}{2n}$  with $ n\equal{} 1,2,....$
  Find $ \lim_{n\rightarrow\infty}x_{n}$
 [/i]

[i]
  Love An Forever[/i]

[i]Noone can hlep me? :(

  I think the problem is very nice :maybe: [/i]

The recurrence implies $ x_{n\plus{}1}\minus{}x_n\equal{}\minus{}\frac{n\plus{}1}{2n}(x_n\minus{}x_{n\minus{}1})$ giving
\[ x_{n\plus{}1}\minus{}x_n\equal{}(\minus{}\frac{1}{2})^n(n\plus{}1)(x_1\minus{}x_0)\equal{}(\minus{}\frac{1}{2})^n(n\plus{}1)\]
and so
\[ x_n\equal{}2007\plus{}\sum_{k\equal{}0}^{n\minus{}1} \frac{(\minus{}1)^k(k\plus{}1)}{2^k}\]
As a result, $ x_n$ converges and its limit is $ 2007\plus{}\sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k(k\plus{}1)}{2^k}\equal{}\boxed{2007\plus{}\frac{4}{9}}$.


limit
inequalities
function
algebra proposed
algebra

Find:

\[ \lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}  \]

Well, the limit is equal with $e.$ By Cesaro-Stolz lemma we have:

                          \[ lim_{n\to\infty} \frac{n}{\sqrt [n] {n!}}=\lim_{n\to\infty} {\frac{1}{{\sqrt [n+1] {(n+1)!}-{\sqrt [n] {n!}}}}=e.  }\]

 The last one follows from Lalescu's sequence. ;)

 P.S. Another solution is using D'alembert criterion, but I think you know that solution. :)

Well, it was almost an accident that I came across this limit.  I found it because of the inequality \[ (n-1)! \le n^ne^{1-n} \le n!  \]
which I proved using calculus.

(To, prove that, I found the upper and lower sums of a function (I forget what it was) and compared that to the integral.  Then, I used the fact that \[ a \le b \le c  \] implies \[ e^a \le e^b \le e^c  \].)


Let  $a,b,c$  real numbers . Prove that

$a(b+c-a)^3+b(a+c-b)^3+c(a+b-c)^3\leq 4abc(a+b+c)$

Here it is  :cool:

[hide]$4abc(a+b+c)\geq a(b+c-a)^3+b(a+c-b)^3+c(a+b-c)^3 \\ \\ \Leftrightarrow (a^2+b^2+c^2-2ab-2bc-2ca)^{2}\geq 0$[/hide]


Can anyone help me with this one:

Let f(x)= x^4 + ax^3 + bx^2 + cx + d. If the graph of y= f(x) intersects the graph of y= 2x -1 at x=1, x=2, and x=3, then what is the value of f(0) + f(4) ?

[hide="Solution"]
We are given 3 solutions to $ g(x) \equal{} f(x) \minus{} 2x \plus{} 1 \equal{} 0$. Because $ \mathrm{deg}(f(x) \minus{} 2x \plus{} 1) \equal{} 4$, we know that there are a total of four solutions by the Fundamental Theorem of Algebra. Let the fourth solution be $ r$. We write $ g(x)$ in factored form like so:

$ g(x) \equal{} f(x) \minus{} 2x \plus{} 1 \equal{} (x \minus{} r)(x \minus{} 1)(x \minus{} 2)(x \minus{} 3) \implies f(x) \equal{} (x \minus{} r)(x \minus{} 1)(x \minus{} 2)(x \minus{} 3) \plus{} 2x \minus{} 1.$

We have $ f(0) \equal{} ( \minus{} r)( \minus{} 1)( \minus{} 2)( \minus{} 3) \plus{} 0 \minus{} 1 \equal{} 6r \minus{} 1$ and $ f(4) \equal{} (4 \minus{} r)(3)(2)(1) \plus{} 8 \minus{} 1 \equal{} 24 \minus{} 6r \plus{} 7 \equal{} 31 \minus{} 6r.$ Therefore, $ f(0) \plus{} f(4) \equal{} 6r \minus{} 1 \plus{} 31 \minus{} 6r \equal{} \boxed{30}.$
[/hide]


trigonometry
Pythagorean Theorem
geometry

A plane keeps an airspeed of 550 miles per hour in a southwesternly direction. The velocity of the plane stream is a constant 80 miles per hour from the west. Find the actual speed and direction of the plane.

If the plane is flying directly SW, then using the pythagorean theorem, the x and y components of the velocity are the same, both $ \frac{550}{\sqrt{2}}$.  The wind is of length 80 directly to the right, and so the length of the horizontal velocity we're interested in is $ \frac{550}{\sqrt{2}-80=\frac{550-80\sqrt{2}}{\sqrt{2}}$.  Then from Pythagoras, the hypotenuse (total velocity) is $ 590.34$.  So that is the velocity.  Now, use trig to get the angle and you're done.


For an even-sided bicentric polygon, the diagonals are concurrent at the limiting point of the two circles.
For an odd-sided bicentric polygon, the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point.
Who can tell me how to prove it?

[url]http://www.mathlinks.ro/viewtopic.php?t=295083[/url]
Do not double post.


superior algebra
superior algebra theorems

hello,

where can I read a proof of the Multinomial theorem ?

See here //http://planetmath.org/encyclopedia/MultinomialTheoremProof.html .


algebra
polynomial
induction
algebra proposed

Let $ c\in\mathbb C$ and $ A_c \equal{} \{p\in \mathbb C[z]|p(z^2 \plus{} c) \equal{} p(z)^2 \plus{} c\}$.

a) Prove that for each $ c\in C$, $ A_c$ is infinite.

b) Prove that if $ p\in A_1$, and $ p(z_0) \equal{} 0$, then $ |z_0| < 1.7$.

c) Prove that each element of $ A_c$ is odd or even.

Let $ f_c \equal{} z^2 \plus{} c\in \mathbb C[z]$. We see easily that $ B_c: \equal{} \{z,f_c(z),f_c(f_c(z)),\dots\}$ is a subset of $ A_c$. Prove that in the following cases $ A_c \equal{} B_c$.

d) $ |c| > 2$.

e) $ c\in \mathbb Q\backslash\mathbb Z$.

f) $ c$ is a non-algebraic number

g) $ c$ is a real number and $ c\not\in [ \minus{} 2,\frac14]$.

To simplify the notation, we use $ f(x)$ instead of $ f_c(x)$, $ f^{(n)}(x)$ as $ n$ th composition of $ f$.
The idea is similar to http://www.mathlinks.ro/viewtopic.php?t=256478.
a)c) are trivial.
b) For $ c \equal{} 1$, we can easily show $ A_1 \equal{} B_1$. Suppose $ |z_0| > 1.7$. Obviously $ p(z) \equal{} z$ will not hold. Let $ p(z) \equal{} f^{(n)}(z)$.
We can show by induction in $ n$ that $ |p(z_0)|\ge 1.7$. Contradiction!

For d)e)f)g), it suffices to show the sequence $ 0,f(c),\cdots,f^{(n)}(c)$ has infinitely many distinct terms.
Suppose it is not true, then there exists $ n\ne m$ such that $ f^{(n)}(c) \equal{} f^{(m)}(c)$.
If we define $ g_1(x) \equal{} x^2 \plus{} x,g_n(x)\equal{}g_{n\minus{}1}(x)^2\plus{}x$. It is equivalent to say $ G(x) \equal{} g_n(x) \minus{} g_m(x) \equal{} 0$ has root $ c$.
Now f) is straightforward.
e) is also trivial because $ G(x)$ is a monic polynomial, then rational root has to be integer.

For d), we define $ z_0 \equal{} c^2 \plus{} c,z_n \equal{} f(z_{n \minus{} 1})$.
We can prove by induction that $ |z_n| \equal{} |z_{n \minus{} 1}^2 \plus{} c| > |z_{n \minus{} 1}|$ as follows:
Suppose by induction assumption $ |z_{n \minus{} 1}| > |c|$, then
$ |z_n|\ge |z_{n \minus{} 1}|^2 \minus{} |c| > |c|^2 \minus{} |c| > |c|$.
Then $ |z_n| \minus{} |z_{n \minus{} 1}|\ge |z_{n \minus{} 1}|^2 \minus{} |z_{n \minus{} 1}| \minus{} |c| > |c|^2 \minus{} 2|c| > 0$.

For g), $ c < \minus{} 2$ is covered by d). Now assume $ c > \frac14$.
Still by induction, $ z_n \minus{} z_{n \minus{} 1} \equal{} z_{n \minus{} 1}^2 \minus{} z_{n \minus{} 1} \plus{} c > (z_{n \minus{} 1} \minus{} \frac12)^2 > 0$.

Q.E.D.


function
Functional Analysis
advanced fields
advanced fields unsolved

Let $ M(K)$ denote the set of Borel, regular, complex measures on a  compact set $ K$. Prove that $ \parallel{}v\parallel{} \equal{} |v|(K)$ - variation of the measure, is a norm on $ M(K)$ that makes it a Banach space.

It is immediately when we can use the Riesz representation theorem, which says that $ M(K)$ is the dual of $ C(K)$, space of continuous functions on $ K$, then $ M(K)$ being a dual of a Banach space is Banach itself.

Is there any easier way to prove that, without invoking the Riesz representation theorem.

You can reduce this to the completeness of $ L^1$. Indeed, if $ \sum \|\nu_k\|$ converges, then each $ \nu_k$ is absolutely continuous with respect to $ \nu: \equal{} \sum |\nu_k|$, and therefore can be identified with an element of $ L^1(\nu)$.


blogs function calculus derivative integration

[quote="ProtestanT"][quote="Maaske"]hey could anyone help me with theis riddle plz ?

What is the next number in this sequence?

52, 28, 33, 63, 73, 23, 56,...

Anyone knows plz ?[/quote]
it's 52

since it doesn't say what type of sequence it is, it could very well be a repeating sequence of 6 numbers.

52, 28, 33, 63, 73, 23, 56, 52, 28, 33, 63, 73, 23, 56, 52, 28, 33, 63, 73, 23, 56,.....[/quote]

then why can't it be a repeating sequence of a lot more numbers

in your theory

it could be any number

-jorian

yep, so thats why i said it coudl be a repeating sequence

Thanx guys , suppose it could be 24 ,

I have the answer in a week so :s lets hope its right

[quote="Fabiomr1"]I'd say it is 24

look:

52, 28, 33, 63, 73, 23, 56,...

try to sum the first with the second, the third with the fourth, the fiveth with the sixth:

$52+28 = 80$
$33+63 = 96$
$73+23 = 96$

if the sums of the middles repeat, the sums of the external parts should match too:

$56+x = 80$

so:

$x = 24$[/quote]

wait, how would you then determine the second number that you don't have?

jorian

hey Mr. Jihad. suppose that u got a sequence and wants to determine the n-nth term of it. well. if they don't provide you with previous or coming numbers, you shouldn't be able to solve it. thats why, in that case, we need the second term. if it was to evaluate the second one, some more previous numbers would come...

...I think

in fact, I think the next number in the sequence is $n$

maybe $x$

or, who knows, $y$?!

I like how thales and fabio's posts are always within minutes of eachother.

hey Mr. Kauan Account, it's because he's bullying me   :(

r u bullying me too?!!

and their blog
:LOL:

hi tall guy and mr. fabio

hi mister function $dx$

are you a gradient? or sony?

i am a[color=white][size=59]gradient[/size][/color]
what is a sony?

by the way, cool eggmaster, Mr.Function

but these one are cooler: http://senhorcarteiro.blogspot.com/2006/10/eggmaster.html

cool mr. fabio
 i can't read that language though

function

suppose

x=y

now you can read this?

tall guy,
well my eyes sort of hurt, but yes i can see that clear as a crystal
suppose
x=y
AND
x=-y
what is x?
what is y?

notice this is really easy so i leave the rest as an exercise for the reader

multiple these 2 equatiions

you get

$x.x =-y.y$
$x.x+y.y = 0.0$

so, these is the sides of a triangle

with the hipotenuse = 0

x^2+y^2=0
interesting,
well i was looking for a less imaginary answer,
(0,0) by probably graphing or addition

if you choose

a very large 0, suppose $0 \to  1$

so

we get a circle

draw a triangle in this circle and paint this blue

hey mr. integral function of many variables, u r one of the only people in here that got the right things to see the good math humour

in fact, only u and Mr.Buda got math sense of humour.

so I think that u already read a good calc book or read the info of http://www.senhorcarteiro.blogspot.com

salutes

actually mr. fabio i am in the process of reading a good calc book
a really good one

thank you

$\vec{\nabla}f$

hey mister function

you have already done the laplace transform? by the way, you are a jude, no!?

hmm....

too dangerous

so. backing

why fabio's have  1 post?!

she ever had 0!!!

is because 0!= 1!?

hm..


algorithm AMC AIME function algebra polynomial calculus

[quote="Altheman"][quote="PenguinIntegral"]Let x and y be positive integers with x < y. Find all possible
integer values of
$P=x^3-y/1+xy$[/quote]

do you mean:

$\frac{x^3-y}{1+xy}$

?

if so
[hide="hint"]euclidean algorithm[/hide][/quote]
Yes, that is what I mean. Do you have a complete solution?

Setting y=x+k and setting n as the integer, I get:
$-x^3+nx^2+(nk+1)x+(n+k)=0$

I think I made an error (it's late) but now we can use poly techniques. Though I think we should be solving for n.

[hide]
Let $\frac{x^3-y}{1+xy}=k$ for some $k \in \mathbb{Z}$. Solve the expression in function of $y$, and find that $y=\frac{x^3-k}{1+xk}$.

So we can substitute back into the first expression: $\frac{x^3-\frac{x^3-k}{1+xk}}{1+x \left(\frac{x^3-k}{1+xk} \right)}=k$ to get

$\frac{x^4k+x}{1+x^4}=k \implies x=k$. Re-subtitute and get $y=\frac{k(k-1)}{1+k^2}$. Now, $\gcd (k,1+k^2)=1$ and

$\gcd_{max}(k-1,1+k^2)=2$, so the only possible values  for  $k$ would be $1+k^2=2 \rightarrow k= \pm 1 \implies y=0$, that doesn't work.
So there are no solutions
[/hide]

I didn't include 0 cause the problem asked for positive integers, and 0 isn't positive, right? (I never remember if we have to include it or not...)

hydro you did mistake. you get after substituting $\frac{x^4k + x}_{1 + x^4}$ $= k$ which is false. check it you get $\frac{x^4k + k}_{1 + x^4}$ $= k$ but this expression is of no use so try some other mathod.

:blush:  yes I made an error doing calculus... I'm very sorry...  :blush:

Now try some other methods for getting the result.

[quote="Altheman"]

$\frac{x^3-y}{1+xy}$

?

if so
[hide="hint"]euclidean algorithm[/hide][/quote]

How would you use the euclidean algorithm here? (I know what it is, but I don't see a good way to apply it here)

[quote="chess64"][quote="PenguinIntegral"][quote="chess64"]I think I remember this from IMO or maybe IMO shortlist, I can't remember.[/quote]
Nah, the isn't [i]that[/i] hard. Think hard AIME level.[/quote]

Well where did you get it from?[/quote]Note the similarity to [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1994]IMO 1994 Pr 4[/url] although it is not exactly the same.

Let x=y-k
then we can get the numerator as y^3-3ky^2+(3k-1)y-k^3

1+xy = (y^2-ky+1)

On polynomial division we get the remainder as k^3-7k^2*y+4k+2y=0
On subsituting the integral values of y>1 we can draw grpahs of k^3 and
(7k^2*y-4k-2y) we can see that the graphs dont meet at one another for any real values..hence no values exist

[quote="10000th User"][/quote][quote="chess64"][quote="PenguinIntegral"][quote="chess64"]I think I remember this from IMO or maybe IMO shortlist, I can't remember.[/quote]
Nah, the isn't [i]that[/i] hard. Think hard AIME level.[/quote]

Well where did you get it from?[/quote][quote="10000th User"]Note the similarity to [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1994]IMO 1994 Pr 4[/url] although it is not exactly the same.[/quote]
Oh, that's probably what I was thinking of.

Sorry Varun Nakra but your solution is completely off from being correct, the expansion is wrong and the graphs part has no algebraic support ...

Seems a bit hard for Intermediate. Off it goes to PreOlympiad.

Hmm, still no solution?

@Chess
I got it from a MOP document. It was the red level.


Probably not.  There wasn't one last year.

[quote="knexpert"]Does anyone know if a list of people attending Mathcamp this year will be emailed to everyone?[/quote]

We do not send such a list before camp, since we respect the privacy of your personal data.  However, in case you want to start meeting people right now, this year I have collected a few returning alumni who volunteered to correspond with new campers before camp.  You should have already received an email about that.


$ f$ is a function defined for all integers $ m, n$.

(i) $ f(m)$ is an integer;

(ii) $ f(2)\equal{}2$;

(iii) $ f(mn)\equal{}f(m)f(n)$;

(iv) $ f(m) > f(n)$ for $ m>n$.

Find $ f(3)$.

[hide="Solution"]
By (iii) and (ii), $ f(4) \equal{} f(2)f(2) \equal{} 2 \times 2 \equal{} 4$
By (iv) $ 2 \equal{} f(2) < f(3) < f(4) \equal{} 4$
By (i) $ f(3) \equal{} 3$
[/hide]

This happens to be in a book I'm reading, which says that the Australian MC 1984 asks you to find f(1983).

Can someone prove that f(m)=m? Im pretty sure that is true, but I dont have a proof yet

I think I've proven that $ f(n)\equal{}n$.

[hide="Proof"]
A couple of base cases to start things off. First of all, $ f(m)\equal{}f(m)f(1)$ for all integers $ m$ by (iii), so $ f(1)\equal{}1$.
$ f(0)\equal{}f(0)^2$ by (iii), so $ f(0)\in\{0,1\}$. However, by (iv), $ f(1)>f(0)\implies f(1)\minus{}f(0)>0$ which means $ f(0)$ must be $ 0$.
We're given that $ f(2)\equal{}2$.

Now, we will show that $ f(n)\equal{}n$ for positive integers $ n$ by strong induction on $ n$. Our base cases have been established above, so we'll assume that $ f(n)\equal{}n$ for every integer $ n\in\mathbb{Z}^\plus{}$ such that $ n$ is less than or equal to an arbitrary positive integer $ k$. Let us further assume that $ n,k>2$.

Suppose $ k\plus{}1$ is composite. We may express $ k\plus{}1$ as $ \prod p_i^{\alpha_i}$ for primes $ p$ and positive integers $ \alpha$, where each $ p_i<k\plus{}1$. We previously assumed that for all integers $ n\leq k$ that $ f(n)\equal{}n$, and it follows from (ii) that $ f(n^a)\equal{}f(n)^a$, where $ a$ is a positive integer. From the previous assertions, we conclude that $ f(p_i^{\alpha_i})\equal{}p_i^{\alpha_i}$, and so $ f\left(\prod p_i^{\alpha_i}\right)\equal{}f(k\plus{}1)\equal{}k\plus{}1$.

This leaves only the case where $ k\plus{}1$ is prime. In this case, however, $ k\plus{}2$ must be composite and, furthermore, divisible by 2 (recall that we assumed $ k>2$). Therefore, $ \frac{k\plus{}2}{2}$ is an integer which is clearly less than $ k\plus{}1$, and we can express $ f(k\plus{}2)$ as $ f\left(\frac{k\plus{}2}{2}\right)f(2)$, which is $ 2f\left(\frac{k\plus{}2}{2}\right)$ by (ii). From our prior assumptions, $ f\left(\frac{k\plus{}2}{2}\right)\equal{}\frac{k\plus{}2}{2}$, so $ f(k\plus{}2)\equal{}k\plus{}2$. By (iv), $ f(k\plus{}2)>f(k\plus{}1)>f(k)\implies k\plus{}2>f(k\plus{}1)>k\implies f(k\plus{}1)\equal{}k\plus{}1$.

By mathematical induction, $ f(n)\equal{}n$ for all $ n\in\mathbb{Z}$ $ \Box$
[/hide]

You have it a little too complicated, like I was thinking before I read the [hide="solution."]
So yeah, $ f(1) \equal{} 1$.
Again using strong induction on n, proving that $ f(n) \equal{} n$ starting from n = 3:
we split into two cases.
Case 1. $ n$ is even. Let $ n \equal{} 2k$.
Then, since 2 < n and k < n we have $ f(n) \equal{} f(2) f(k) \equal{} 2k \equal{} n$.
Case 2. $ n$ is odd; let $ n \equal{} 2k \minus{} 1$.
Okay, by strong induction $ f(2k \minus{} 2) \equal{} 2k \minus{} 2$
and $ f(2k) \equal{} f(2) f(k) \equal{} 2k$,
so $ 2k \minus{} 2 < f(2k \minus{} 1) < 2k$
so $ f(2k \minus{} 1) \equal{} 2k \minus{} 1$.
[/hide]

Follow-up question: (ii) becomes "for some n > 1 we know f(n) = n".

I guess that's a bit simpler, though there's really not a huge difference between casework on parity and casework on primality


Prove that 93 cannot be written as the sum of two prime numbers.

[i]Note[/i]: If you find this problem easy, please do not write in a solution until some beginners have tried it.

Also, if you can't finish your proof, feel free to post your progress so far and I (or anyone else) will be glad to help.

[hide]all prime numbers except for 2 are odd. odd numbers can only be the sum of an even and an odd number. so the only possibility for this is 2+91=93, but 91=7*13, so 93 can't be the sum of two primes. is this enough?[/hide][/hide]

Well done! Anyone else want to give this a shot? (just don't look at jeffrey's solution)

Maybe I am being stupid, but I think that the only way to do it is[hide]write out all the prime numbers less than 93 and show all the possible sums, and show that none of them = 93 [/hide].  Would this be a valid way to prove something?

Actually, you could also try [hide]indirect proof[/hide].

[hide]Assume temporarily that 93 can be expressed as the sum of two prime numbers, i.e. [tex]p_1+p_2=93[/tex].  Then [tex](p_1+p_2)^2 = 93^2[/tex] which expands to [tex]{p_1}^2 + 2p_1p_2 + {p_2}^2 = 93^2[/tex]. [/hide] Blah, stuck.

Ah shoot I worked on this for a while and didn't notice that someone else had already answered it. MY BAD!

EDIT: Okay NVM I will not look at Jeffery's solution.  Can anyone help me finish mine?

It would be a valid proof to try out all the possible sums, but there's an easier way. Hint: [hide]think about the parity (even or odd) of two numbers that add up to 93[/hide].

[hide]It takes one even and one odd number to add up to an odd number like 93. The only even prime is 2, so the other prime has to be 93-2=91. But 91 isn't prime. So no two primes add up to 93.[/hide]

I just wish I thought of this without the hint.


Hello

Where should I ask some questions I have? They are:

1. What is the difference between mathlinks and aops?

2. How do I decide whether to have my mother sign my little brother up for aops classes, send him to a pre Olympiad style math camp, force him to study at home, or make him do any combination of these? Are there any other ways good to get him to do math?

He picks up mathematical concepts pretty quickly, and I want to introduce him to the math enthusiast community early on and have him study hard. His personal choice is not my concern. I want to guide him because I think he'll play too much if left alone.

3. Which methods are the most time effective or preparing for contests?
Would trying many previous math competitions be more time effective than studying several math topics from the aops textbooks in depth?

[quote="aufha"]
1. What is the difference between mathlinks and aops?
[/quote]There is no major difference. They used to be two different sites way in the past, but now they are just two different ways of presenting information (so layout differences, not content - for most part at least).

[quote="aufha"]3. Which methods are the most time effective or preparing for contests?
Would trying many previous math competitions be more time effective than studying several math topics from the aops textbooks in depth?[/quote]
I found that doing both is best for me, but I can't say which is better. Reading textbooks is very informative, but you should always practice applying concepts that you learn in the textbooks. Also, since different competitions tend to have different types of questions, it always helps to study past competitions to know what you're likely to see and be ready for it.

If you're preparing for a hard competition (AIME and above), then I suggest textbooks + past competitions, but for most other competitions, sometimes past competitions are all you need. Try different ways of preparing (on not-so-important competitions), and see which works best for you.

[quote="aufha"]
2. How do I decide whether to have my mother sign my little brother up for aops classes, send him to a pre Olympiad style math camp, force him to study at home, or make him do any combination of these? Are there any other ways good to get him to do math?[/quote]

It depends how advanced your brother is. If he can handle a pre-olym. camp, try to get him to MOSP through the AMC tests. WOOT is also  good for that level.


I am not sure that this is the right place to post it, but, anyway:
The general form of Ceva's theorem ( the natural extension for a polygon with an odd number of vertices) is called the theorem of.......?
Does anybody know?
Thanks! ;)

[color=red][Moderator edit: See http://www.mathlinks.ro/Forum/viewtopic.php?t=45945 post #7 for one case of the general form of the Ceva theorem.][/color]

[url]http://mathworld.wolfram.com/CevasTheorem.html[/url] appears to give such a theorem but does not attribute it to anyone in particular. Apparently it was drawn from Grünbaum, B. and Shepard, G. C. "Ceva, Menelaus, and the Area Principle." Math. Mag. 68, 254-268, 1995.


A circular track has a diameter of 100 meters. Bert and Ernie start at the same position on the track and run in opposite directions around the track. Bert's speed is $ 20\%$ faster than Ernie's. When Bert returns to the starting position for the third time, how many meters will Ernie have run? Use 3.14 as an approximation for $ \pi$ and round your answer to the nearest whole number.

Bert travels 6/5 times faster (hence, as much distance) than ernie.

Bert travels 300 pi meters, so Ernie travels 250 pi meters. 250*3.14=[b]785[/b].


It says India 1995 in [i]The Art and Craft of Problem Solving[/i] but I could not find it there...

Anyway, it's an interesting problem.

Find all positive integer solutions $x,y,z,p$, with $p$ prime, of the equation $x^p+y^p=p^z$.

[hide]

We will use the notation $p^s||a$ to mean that $p^s$ divides $a$ but $p^{s+1}$ does not divide $a$.

First suppose that $p$ is odd.

$x^p+y^p=p^z$.

$(x+y)(x^{p-1}+\ldots{}+y^{p-1})=p^z$, so $x+y=p^a$ for some $a \in Z$

Then $y=p^a-x$

$x^p+(p^a-x)^p = p^z = x^p + p^{ap} - \binom p1 p^{a(p-1)}x + \ldots{} + \binom{p}{p-1}p^ax^{p-1} - x^p = p^{ap} - \binom p1 p^{a(p-1)}x + \ldots{} + \binom{p}{p-1}p^ax^{p-1}$

Note that $p^1 || \binom pi$ for $1 \le i \le p-1$, since this coefficient has one $p$ in the numerator and none in the denominator.

We now take mod $p^{2a+1}$. All terms disappear except $\binom{p}{p-1}p^ax^{p-1}$. So $p^a | x^{p-1}$ and thus $p|x$.

Now suppose $p^s||x$. Put $x=p^sx'$.

Then $y=p^a-p^sx'=p^s(p^{a-s}-x')$. So $p^s||y$. We divide through by $p^s$ to obtain

$x^m+y^m=p^n$ where $p$ divides neither $x$ nor $y$.

$x^m+(p^a-x)^m=p^n$. Take mod $x$. Then $p^{am} \equiv p^n$. So $x|p^{am}-p^n$. This is a contradiction unless $am=n$.

$x^m+(p^a-x)^m=p^{am}$.

So $(p^a-x)^m=p^{am}-x^m$.

$p^{am}-\binom p1 p^{a(m-1)}x + \ldots{} + \binom{p}{p-1}p^ax-x^m=p^am-x^m$.

Take mod $p^{a+2}$.

$\frac{p-1}{2}x - x^m \equiv -x^m$. So $p^{a+2}|\frac{p-1}{2}x$ and $p|x$. Contradiction.

However, we assumed there were at least three terms in the binomial expansion; this is not true if $m=1$.

(If $m=0$, we have a clear contradiction.)

Then we have $x+y=p^{z-p-1}$ as our solutions in this case.

Now suppose $p=2$.

$x^2+y^2=2^z$.

Say $2^s||x$.

Divide out by $2^{2s}$ so both $x'$ and $y'$ are odd.

Then $x'^2+y'^2 \equiv 2$ (mod 4). Then $z-2s=1$ and $z=2s+1$.

So $x=2^sx'$, $y=2^sy'$, $p=2$, and $z=2s+1$ is our other solution.

[/hide]


Find a closed form for the sequence

$\sum_{n = 1}^{\infty}{\frac{x^{n}}{x^{2n}-y}}$

do you mean a closed form for the sum?  or for the partial sums?

might be a tad easier if you limit $y \ge 0$, then you can write $y = z^{2}$ and factor as difference of squares.

actually, I don't know if that necessarily converges...never mind...it's sorta geometric so it should for sufficiently small $x$

partial fractions: $\frac{x^{n}}{x^{2n}-z^{2}}= \frac{1}{2}\left(\frac{1}{x^{n}-z}+\frac{1}{x^{n}+z}\right)$

those would be nasty to sum, however.


Suppose f (x) is a rational function such that 3f(1/x)=2f(x)/x=x^2 for x not equal to zero.

Find f (−2) .

I don't get it, if $ 2f(x)/x\equal{}x^2, then f(x)\equal{}x^3/2$ is a possible function but since $ 3f(1/x)\equal{}x^2$, $ f(x)\equal{}1/(3x^2)$ so it is only defined when $ 1/(3x^2)\equal{}x^3/2$, or $ x\equal{}\sqrt [5] {2/3}$ ?

I think the question is
$ 3f(1/x)\plus{}2f(x)/x\equal{}x^2$

This is standard, then.

[hide]Original equation: $ 3f(1/x)\plus{}2f(x)/x\equal{}x^2 \implies 6xf(1/x)\plus{}4f(x)\equal{}2x^3$
Plug in 1/x into original: $ 3f(x)\plus{}2xf(1/x)\equal{}1/x^2 \implies 9f(x)\plus{}6xf(1/x)\equal{}3/x^2$

Subtracting the first (rightmost form) from the second (rightmost form), we get $ 5f(x) \equal{} 3 / x^2 \minus{} 2 x^3 \implies f(x) \equal{} \frac{3 \minus{} 2 x^5}{5 x^2}$

Therefore, $ f(\minus{}2) \equal{} \frac{67}{20}$[/hide]


Let $ x_1,x_2,x_3,...$ be a sequence of a nonzero real numbers satisfying -
$ x_n \equal{} \frac{x_{n\minus{}2}x_{n\minus{}1}}{2x_{n\minus{}2} \minus{} x_{n\minus{}1}}$ for $ n\equal{}3,4,5...$
Establish the necessary and sufficient conditions on $ x_1$ and $ x_2$ for $ x_n$ to be an integer for infinitely many values of n

it must be an old problem
we must try to prove that $ x_n \equal{} \frac {x_1x_2}{(n \minus{} 1)x_1 \minus{} (n \minus{} 2)x_2}$ (by induction) and then deduce that $ x_1 \equal{} x_2 \in Z \minus{} \{0\}$
(if not for sufficienly large $ n$ , $ x_n,x_{n \plus{} 1},...$ won't be integers)

[quote="The Enigma"]Let $ x_1,x_2,x_3,...$ be a sequence of a nonzero real numbers satisfying -
$ x_n \equal{} \frac {x_{n \minus{} 2}x_{n \minus{} 1}}{2x_{n \minus{} 2} \minus{} x_{n \minus{} 1}}$ for $ n \equal{} 3,4,5...$
Establish the necessary and sufficient conditions on $ x_1$ and $ x_2$ for $ x_n$ to be an integer for infinitely many values of n[/quote]

We know thru problem statement that $ x_1x_2\ne 0$ and so that $ x_n\ne 0$ $ \forall n$

Let then $ v_n\equal{}\frac{x_{n\plus{}1}}{x_{n}}$. We have the relation $ v_{n\plus{}1}\equal{}\frac 1{2\minus{}v_n}$

and so $ v_n\equal{}\frac{(1\minus{}v_1)n\plus{}2v_1\minus{}1}{(1\minus{}v_1)n\plus{}v_1}$

In order to have $ x_n$ defined for any $ n$, we need $ v_n\ne 2$ $ \forall n$ and so $ v_1\ne 1\plus{}\frac 1k$ $ \forall k\ge 1$

So we have $ v_n\equal{}\frac{x_{n\plus{}1}}{x_n}\equal{}\frac{(1\minus{}v_1)n\plus{}2v_1\minus{}1}{(1\minus{}v_1)n\plus{}v_1}$ $ \equal{}\frac{(x_1\minus{}x_2)n\plus{}2x_2\minus{}x_1}{(x_1\minus{}x_2)n\plus{}x_2}$ $ \equal{}\frac{h(n)}{h(n\plus{}1)}$ where $ h(n)\equal{}(x_1\minus{}x_2)n\plus{}2x_2\minus{}x_1$

Hence $ x_n\equal{}x_1\frac{h(1)}{h(n)}$ $ \equal{}\frac {x_1x_2}{(x_1\minus{}x_2)n\plus{}2x_2\minus{}x_1}$

And obviously, if $ x_1\ne x_2$, $ |(x_1\minus{}x_2)n\plus{}2x_2\minus{}x_1|>|x_1x_2|$ for $ n$ great enough and $ x_n$ cant be an integer.

So we need $ x_1\equal{}x_2\ne 0$ and it's immediate to see that this necessary condition is sufficient.

[u][b]edit [/b][/u]: too late once again :)

[quote="The Enigma"]thanks for replying shoki and pco
pco i think u missed the very last conclusion ofcourse it is obvious - that all are [b]equal to 1[/b][/quote]

Surely not, $ x_1\equal{}x_2$ non zero integer is enough.

For example, $ x_1\equal{}x_2\equal{}2$ $ \implies x_n\equal{}2$ $ \forall n$

And no need for $ x_1\equal{}x_2\equal{}1$

ofcourse i am very sorry abt that i overlooked that the recursive equation is only true from $ n\equal{}3,4..$ sry pco :oops:


perhaps one could start with ozonolysis (ozone; sodium borohydride). Then protect one of the hydroxyl groups via TBSCl or some other protecting group. Reduce the other hydroxyl group to its next lower homologue (as Carcul posted in classic problem previously on the forum). Then convert to grignard and attack on the deprotected alcohol; finish by oxidation (TsCl; NaBr; TBAF; Mg; PCC).

[quote="MysticTerminator"]Then convert to grignard and attack on the deprotected alcohol; finish by oxidation[/quote]

I am not following this. What is the Grignard going to attack?

oops, sorry. I forgot to oxidize the alcohol back to the aldehyde oxidation state. The last steps should then be tosyl chloride; sodium bromide; TBAF; PCC; magnesium metal; acid quench; PCC.

When you treat the protected alcohol group, TBSi-O-R, with TBAF in order to deprotect it, the fluoride attacks silicon in SN2 departing an alcoxide ion, -O-R, which can attack intramolecularly the bromide on the other side. Or am I missing something?

[hide="my method"]
1. Oxidative ozonolysis
2. Pimelic acid + $ \ce{ThO_2}$ --> cyclohexanone (Ruzcika reaction)
[/hide]


The Screamers are coached by Coach Yellsalot.  The Screamers have 12 players, but two of them, Bob and Yogi, refuse to play together.  How many starting lineups (of 5 players) can Coach Yellsalot make, if the starting lineup can't contain both Bob and Yogi?

We could also do complementary counting; there are $ \binom{12}{5} \equal{} 792$ ways to take 5 people to be in the starting lineup, and we have to subtract the number of ways Bob and Yogi can be in it together. If Bob and Yogi are already in the lineup, that leaves $ \binom{10}{3} \equal{} 120$ ways to choose the other three people, and thus 120 ways to have Bob and Yogi in the lineups together. Therefore, the answer is $ 792 \minus{} 120\equal{}\boxed{672}$.

[quote=Problemsolving17]Can't you use Casework in this problem. Oops 10 years bump[/quote]

lol  :rotfl:

[quote=EProfesser]Why is it 10 instead of eleven?[/quote]

because you are assuming 2 players, bob and yogi, are already on the team together

[quote=EProfesser]Why is it 10 instead of eleven?[/quote]

wait u talking about the 10 year bump or the actual problem? the bump is 10 years because at that time the guy posted it in 2018, not 2019.


algebra
polynomial
number theory
group theory

Let $ \sigma_k$ denote the sum of all combination of $ 1,2,3,....,p \minus{} 1$ taken $ k$ at a time .If $ p$ is a prime , prove that $ p$ divides $ \sigma_k$ for all $ k$ within  {$ 0,1,...,p \minus{} 2$}

The argument you're thinking of is trivial; I'd like to see a combinatorial argument using [url=http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind]Stirling numbers[/url].

[hide="Trivial argument"]$ k!\sigma_k$ includes every ordering of each combination once. Pair every $ a_1a_2\cdots a_{k-1}a_k$ term with the corresponding $ a_1a_2\cdots a_{k-1}\left(p-a_k\right)$; their sum $ a_1a_2\cdots a_{k-1}p$ is divisible by $ p$, so $ p|k!\sigma_k$, and since $ p$ doesn't divide $ k!$, $ p|\sigma_k$, as desired.[/hide]

[hide="The trivial argument I was thinking of"] Fermat's little theorem is equivalent to the factorization

$ x^p \minus{} x \equiv x(x \minus{} 1)(x \minus{} 2)...(x \minus{} (p\minus{}1)) \bmod p$. [/hide]

[quote="worthawholebean"][hide="Trivial argument"]$ k!\sigma_k$ includes every ordering of each combination once. Pair every $ a_1a_2\cdots a_{k - 1}a_k$ term with the corresponding $ a_1a_2\cdots a_{k - 1}\left(p - a_k\right)$; their sum $ a_1a_2\cdots a_{k - 1}p$ is divisible by $ p$, so $ p|k!\sigma_k$, and since $ p$ doesn't divide $ k!$, $ p|\sigma_k$, as desired.[/hide][/quote]
Wouldn't this trick also work if we consider $ \sigma_k$ instead of $ k!\sigma_k$ (take $ p = 5.k = 3$ for example - 123, 234, 124, 134)? But how do you prove such pair always exist? Also why does this only work for prime p?
Also t0rajir0u how do you prove that congruence? (I see the LHS is congruent to 0 but how do you prove the expression is congruent to the expression on the RHS?)

I'm not sure how worthawholebean's argument works anymore, since it requires that $ p \minus{} a_k$ isn't equal to $ a_1, a_2, ... a_{k \minus{} 1}$.

As for my proof, the key is that $ \mathbb{F}_p$ is a field, hence polynomials $ \bmod p$ have unique factorization: a polynomial of degree $ n$ has at most $ n$ roots.  The polynomial $ x^p \minus{} x$ has $ p$ roots by Fermat's little theorem and the rest follows.  (Here $ x$ is an indeterminate, not a number.)

Anyway:

[hide="The combinatorial argument I was thinking of"] $ \sigma_k$ counts the number of permutations of $ p$ elements with $ k \plus{} 1$ cycles.  There is an obvious action of the cyclic group $ C_p$ acting by renaming the $ p$ elements; if you prefer, take the $ p$ elements to be $ C_p$ itself and take the action to be addition.  This action has no fixed points, so every orbit has size $ p$. [/hide]

Oh, that's a good point. I didn't think of that. This starts to get a little fuzzy: we can separate all combinations with both $ a_k$ and $ p\minus{}a_k$ into a separate category, and then make the combination $ b_1$ through $ b_{k\minus{}2}$ and continue the process... It's late and the other argument is just better, so ignore mine.


Find all rational solutions of $\{x^{2}\}+\{x\}=0.99$

where $\{x\}$ is the fractional part of $x$.

i think there are an infinity of solutions
we guess a solution: 13/10 and then we try to generalize it: x= 10k+13/10 I think is a solution to the problem, where k integer.

$x$ may take infinitely many values but i think he was asking for $\{x\}$ which is unique.


Problem:

Given a real valued function f(x)=sqrt((x+k)/(w+2x)) where k and w are integers.  If there is only one real number which is not in the natural domain of f, find the ratio of k to w.

Since the radicand approaches 1/2 as abs(x) becomes large, I decided to set (x+k)/(w+2x) equal to 1/2 and see what happens.  This equation implies k/w = 1/2 -- the correct answer according to the solution key.

Does this constitute a valid solution?  I'm bothered by the fact that I arrived at the correct result by setting two quantities equal that will never actually be equal (since there are NO values of x that actually makes this happen).   Can this logic be justified or was this just an accident?

Well you can think of it this way...
Suppose x=-w/2.  That makes w+2x=0, so the function is undefined.  This is probably the value which isn't in the domain.
Now, in order for that to be the ONLY real value of x for which f(x) is undefined, the number (k+x)/(w+2x) has to NEVER be negative (or it would be undefined over reals).  Therefore, when k+x is positive, so is w+2x.  When k+x is negative, w+2x is negative as well.  In order for this to be the case, w+2x and k+x have to have a constant ratio with no numerical difference.  Since you have 2x on the bottom and x on top, w must equal 2k so that they have a constant ratio.
So then in a sense, your arrival at the correct answer was not pure luck.  You found the 1/2 in the correct way...you just didn't realize that that would solve the problem.

[quote="SJS"]Why does (numerator, denominator with same sign) necessarily imply they are in a constant ratio here?[/quote]
If the numerator and denominator are not always in a constant ratio, then their ratio will at some point be negative, thus creating another undefined value.  Suppose you graphed the function y=(k+x)/(w+2x).  Put in random numbers for w and k.  You find that the x-value at which k+x becomes 0 (-k) and that at which w+2x becomes 0 (-w/2) are not always the same.  Therefore there is an area of overlap where one is negative and the other is positive.  Thus the function has some part below the x-axis.  In order to never have the function drop below the x-axis, w and k have to be in the ratio 2:1 to make -k=-w/2 so that the ratio will always be a positive number (except when the denominator is 0).


Prove that for every integer x, there's an integer y such that

[y^2 - 2] / [x^4 + 1] is an integer.

Take an arbitrary odd prime factor of $x^{4}+1$.If $x$ is an even number,it is sufficient to show that $y^{2}\equiv 2(\mod p^{\alpha})$ has solutions in $\bf{Z}$
$\Leftrightarrow p\equiv-1,1(\mod 8)$.So we only need to check  that  $p\equiv-1,1(\mod 8)$.
Let $t$ be the order of $x$ modular $p$.Since $x^{4}\equiv-1(\mod p)$ and $x^{8}\equiv 1(\mod p)$,we certainly know that $t=8$.By Fermat Little Theorem,we have $t|p-1$.Thus we get $p\equiv 1(\mod 8)$.
If $x$ is an odd number,clearly the power of $2$ in $x^{4}+1$ is $1$,so just let $y$ be an even number.


inequalities
number theory proposed
number theory

[quote="tdl"]Prove that: $ |12^m \minus{} 5^n|\geq 7\ \forall m,n\in N^*$.[/quote]
this problem is very easy  :D
link: http://www.mathlinks.ro/viewtopic.php?p=849706#849706
(Problem H 63)

[quote="tdl"][quote="kunny"]$ 12^m \minus{} 5^n \equal{} (12 \minus{} 5)(12^{m \minus{} 1} \plus{} 12^{m \minus{} 2}\cdot 5 \plus{} 12^{m \minus{} 3}\cdot 5^2 \plus{} \cdots \plus{} 12\cdot 5^{n \minus{} 2} \plus{} 5^{n \minus{} 1})$[/quote]

Sorry, but it is wrong! May be you too hurry![/quote]

 :blush:  Thanks.


I am confused. On Friday, we did a lab where we added a drop of hydrochloric acid to different stuff to test the prescence of calcium carbonate. For example, one of the items was marble. We observed that the marble started fizzing quickly and bubbles formed. It asks me to write a balanced equation for the reaction. What does it mean?
Thanks

You are adding $\mbox{HCL}$ to various substances to test the for the presence of $\mbox{CaCO}_3$.  If there is $\mbox{CaCO}_3$ in the sample then the calcium is going to be replaced by hydrogen, forming carbonic acid.  This is a very unstable acid and almost immediately breaks up into carbon dioxide and water, hence the fizzing.

So the balanced reaction is
[hide]
$2\mbox{H}^{+} + \mbox{CO}_3^{2-} \rightarrow \mbox{H}_2\mbox{O} + \mbox{CO}_2$
(The $\mbox{Cl}^-$ and $\mbox{Ca}^{2+}$ are spectator ions.)
[/hide]

so... what would a general equation be like? something like

hydrochloride + calcium carbonate yields carbon dioxide + water ???

but then, where's the calcium in the product?

The calcium forms calcium chloride which is soluble, so essentially the calcium and chloride ions just remain floating around in solution and don't really do anything.

ok, so would it then be

hydrochloride + calcium carbonate yields carbon dioxide + water + calcium + chlorine ???

Oops the limestone is solid, ok so its
$\mbox{HCl} + \mbox{CaCO}_3 \rightarrow \mbox{H}_2\mbox{O} + \mbox{CO}_2 + \mbox{CaCl}$
And the net ionic would be
$\mbox{H}^+ + \mbox{CaCO}_3 \rightarrow \mbox{H}_2\mbox{O} + \mbox{CO}_2 + \mbox{Ca}^{2+}$

The balanced equation is:

2HCl(l) + CaCO3(s) = CaCl2(l) + CO2(g) + H2O(l)

The net ionic equation would be:

2H+(l) + CaCO3(s) = Ca+2(l) + CO2(g) + H2O(l)

This is a common test and the amount of solid dissolved to generate quite a visable bubbling won't visabilty damage the surface.  It also works with most acids, such as strong vinger.

and thats how rain damages structures  :(


Prove that in any triangle $ ABC$ exists the relation $ \sum \left(b^2\plus{}c^2\right)\cdot\left[2(b\minus{}c)^2\minus{}(a\minus{}b)(a\minus{}c)\right]\ge 0\ .$

equivalent to $ \sum_{cyc} a^2\left(4(a\minus{}b)(a\minus{}c)\plus{}(b\minus{}c)^2\right)\geq 0$


Give  a  integer $ n $ which is odd and $ n $ integers $ k_1<k_2<...<k_n $.
Let $ a=(a_1,...,a_n) $ is a permutation of $ (1,2,...,n) $ and $ S(a)= \sum k_i \cdot a_i. $
   Prove that there exists two permutation $ b $ and $ c( b\neq c) $ satisfying $ S(b)-S(c) $ is divided by $ n!$.

Isn't this from some IMO? It's been posted before.

Yeah, 2001. Theres n! permutations all different mod n!, so you can add them up that way, and also sum them up the usual way and they aren't equal.


geometry
circumcircle
incenter
trigonometry
geometry open

Let $D,E,F$ be points on the circumcircle of $\triangle ABC$ such that $ID,IE,IF$ are perpendicular to $BC,CA,AB$ respectively where $I$ is the incenter of $\triangle ABC$. Prove that $AD,BE,CF$ are concurrent if and only if $\triangle ABC$ is isoscele.

Using trig Ceva, we can prove that the diagnols a cyclic haxagon $ABCDEF$ are concurrent [b]if and only if[/b]

\[\frac{AB}{BC}\cdot \frac{CD}{DE} \cdot \frac{EF}{FA}=1\]
So the only if part becomes trivial.

I haven't completed my proof for the "if" part, but I think that assume $\angle{A} < \angle{B} < \angle{C}$ and then use the formula again to get contradiction will work.


Salam omidvaram hale hame khoob bashe.

 yek soal :chetory mishe ba (Microsoft office word 2007)to mathlink soal nevesht

(hagme latex baraye download ziade)

 ba tashakor

asan niazi be downloade chizi nadari,haminja mitooni harchio ke mikhai type koni be zabane $ \text{\LaTeX}$  :D
faghat kafie yad begiri chetori be zabane $ \text{\LaTeX}$ benevisi,ke oonam kheili asoone,kafiue dotarafe ebaratet alamate dolar bezari (shift+4) masalan age mihai adade 2+2=4 ro benevisi kafie do tarafe ebaratet dota alamate doalr bezari mishe:

$ 2 + 2 = 4$

hala baraye yad gereftane inke andiso tavano kasro radikalo in chizaro chetori benevisi ye file ke kafie donwloadesh koni,oontoo bet amoozesh dade:

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ye jaiam has ke mitooni toosh tamrin konio emtehan koni ke doors chap mishe ya na:

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age soaliam dari rajebe latex mitooni haminja beporsi ya be engelisi too inja:

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