IMO Shortlist
tilings
rhombus
Hall s marriage theorem
combinatorics

A holey triangle is an upward equilateral triangle of side length $n$ with $n$ upward unit triangular holes cut out. A diamond is a $60^\circ-120^\circ$ unit rhombus.
Prove that a holey triangle $T$ can be tiled with diamonds if and only if the following condition holds: Every upward equilateral triangle of side length $k$ in $T$ contains at most $k$ holes, for $1\leq k\leq n$.

[i]Proposed by Federico Ardila, Colombia [/i]

Every diamond contains an upward and a downward triangle with a common edge. Now in every upward triangle $ T$ with sidelength $ k$ we have $ \frac{k^{2}+k}2$ upward triangles and $ \frac{k^{2}-k}2$ downward triangles. The downward triangles are adjacent only with upward triangles from $ T$, so if more than $ k$ upward triangles are removed, there are not enough of them to match with the downward ones. This proves the easier part of the problem. Now let's prove the converse by induction on $ n$.
Call a sequence $ 2k+1$ of consecutive upward and downward triangles (starting and ending with upward triangles) from the same row a "block". The block above it of $ 2k-1$ triangles be called the "above block". We wish to cover all the triangles in the last row of the big triangle $ T$ with diamonds. There can be horizontal diamond, but there also can be vertical ones, and for them we will need to borrow upward triangles from the above row. When we finish, it is easy to conclude there will be $ n-1$ holes in the remaining triangle $ T'$ of side $ n-1$. We just need to pick up the diamonds in such a way that the induction hypothesis is respected. Call a triangle which contains a part of the base a "base" triangle. Call a triangle which contains as many holes as its sidelength "full", and one which contains more than its sidelength "deadly". We must not create "deadly" triangles after cutting the last row. It is clear a deadly triangle may appear only as a "base" triangle for $ T'$.
We claim that if a block from the base contained $ r$ holes, at most $ r$ holes are added to the block above it after removing the last row. We do the following procedure: the space between two consecutive holes in the last row can be filed in with diamonds except one downward triangle which has to be connected with a triangle above (clearly at least one of the downward triangles can be matched otherwise there would be too many holes breaking the condition). It is is easy to verify the claim after such a conclusion. Moreover it clear that any arrangement is of the above type otherwise we would create too many holes in $ T'$ (between any two consecutive holes we must crate a hole in $ T'$, and if we somewhere create more it would be over). Now we claim a "deadly" triangle arises from a full triangle. Indeed, if we have a base deadly triangle with sidelength $ j$ an $ j+l$ holes we can extend it one unit downward. This triangle has sidelength $ j+1$ and had originally also at least $ j+l$ holes according to our claim. So $ l=1$ and this triangle was full at the beginning. It only remains to manage the full base triangles, thus.

If two full triangles of sidelength $ k,l$ intersect in a triangle of sidelength $ m$, we deduce this small triangle is also full and the big traingle of length $ k+l-m$ is too. So we may do such unions unless $ k+l-m$ is not $ n$. When we finish, we have two possible cases:

a) we have some non-intersecting full triangles. Then every full base triangle is contained in one of them otherwise we could still unite it with one. Applying the induction hypothesis for them we can tile the bottom rows of them with diamonds without not making them full. Completing the tiling we achieve the goal.

b) We have intersecting triangles, but uniting them would form the big triangle $ T$ for which we can not apply the induction step. It is easy to see this is possible only when there are two such triangles of sidelength $ k,l$ which intersect in a traingle of sidelength $ m$ and $ k+l-m=n$. Then we have at most $ m$ holes in the common part. So at least $ k+l-m=n$ holes in the union of these two traingles.  It means the remaining parallelogram at the top contains no hole and then it can be tiled with diamonds. The two triangles can also be tiled. Moreover any tiling tiles also the common part of sidelength $ m$ completely. So we may tile one triangle, and tile the remaining part of the second. This is the desired tiling of $ T$.

QED

Let me post a little bit shorter solution :D.
Consider tiling with rhombus as matching upward and downward triangles and use Hall's theorem. All we have to do is check if Hall's condition is true. We know it's true for some special sets -  all downwards triangles contained in some equilateral upward triangles.
Suppose that Hall's condition is false and consider smallest set of downwards triangles and their neighbours which make that condition false. It has to be connected (connections by side). Consider smallest upward equilateral triangle containing this set (call its side's lenght k). Now we have to prove that in this set there are at most $k$ more upward triangles than downard but it can be easily proven by induction on $k$ that there are exactly $k$ more upward triangles :) (recall that we consider smallest equilateral triangle containing that set, so that set has to touch all sides of that triangle).

P.S. My 100th post :D!

We will prove the result with induction. The $n=1$ case is trivial, so proceed to the inductive step, with a holey triangle of size $n$. It is obvious that there is at least $1$ hole in the base, so consider all holes in the base, at positions $a_1,a_2,\ldots,a_k$, where position $1$ refers to the leftmost unit equilateral on the base. Clearly, we can tile the triangles between positions $1$ and $a_1$ with diamonds, as well as the positions $a_k$ to $n$. Now, consider the interval between $a_i$ and $a_{i+1}$, for all $i$. Within this range, there are an odd amount of unit equilaterals, so we will have to have at least one vertical diamond. Suppose we use exactly one, then note that it creates an effective hole in the upper equilateral triangle of size $n-1$, and the hole created by the diamond can be placed anywhere above the interval. Now, I claim I can place the hole such that the holey condition is still satisfied for the upper $n-1$-equilateral. Of course, we only need to consider sub-equilateral triangles which include the bottom row of the $n-1$-equilateral. Now, define a [i]maximal triangle[/i] to be a triangle of side length $k$ with base on the bottom row of the $n-1$-equilateral such that there are exactly $k$ holes inside of it. Note that as long as the hole isn't put inside a pre-existing maximal triangle, the new hole will preserve the holey-ness of the arrangement.

[b]Lemma: [/b] If the union of the bases of two maximal triangles is a line segment, then the triangle with base defined as that union is also maximal.

[b]Proof: [/b] Call the triangle in question the MEGA-triangle. Note that the amount of holes in the MEGA-triangle is equal to the sum of the side lengths of the two maximal triangles which form it, minus the amount of holes in their intersection. If the side length of the MEGA-triangle is $c$, note that the aforementioned quantity is at least $c$, however the amount of holes in the MEGA-triangle is at most $c$, due to the holey-ness of the arrangement. Therefore, there are $c$ holes in the MEGA-triangle, and the MEGA-triangle must be maximal as well.

Now, we will proceed sequentially through the ranges. Suppose that we have already placed the vertical diamonds in the ranges $a_i\rightarrow a_{i+1}$ for $1\le i < k$ (in positions $t_1$ through $t_{k-1}$ in the second to last row), and the holey-ness condition is still preserved for the top $n-1$-equilateral. I claim we can find a way to place the vertical diamond in the range $a_k$ to $a_{k+1}$. To do this, consider all currently existing maximal triangles (taking into account the newly formed holes.) Now, combine overlapping maximal triangles into MEGA-triangles, until we have a collection of disjoint maximal triangles which represent the union of the bases of all the maximal triangles. Now, if there exists a space within the range which isn't part of the base of a maximal triangle, we can just place our new hole there, and it will work perfectly.

Otherwise, we will have a maximal triangle which covers the entire range, which we will call the SUPER-triangle. Also, extend the SUPER-triangle down a row, to form a SUPERR-triangle. Note that if there are $a$ new holes in the SUPER-triangle, then in the original configuration, there were at least $a+1$ holes in the SUPERR-triangles at the beginning. So, before we added any holes, the SUPERR-triangle had to be maximal, or else there would've been a contradiction, which occurs when there are exactly $a+1$ holes in its base. In other words, if the position of the leftmost point in the SUPER-triangle is $l$ and the rightmost point of $r$, then we must have $r<a_{k+2}-1$, and $a_j < l \le t_j$, for some $j$. However, since we know that the SUPER-triangle isn't next to any other maximal triangle, location $l-1\ge a_j$ cannot be in a maximal triangle, so we can shift the hole at $t_j$ to $l-1$. Now, we repeat this process. As the sum of the positions of the holes strictly decreases after every adjustment, this algorithm eventually terminates, and after all the adjustments have been made, we will have a viable place to place the hole in our range. Now, repeat this for all intervals to get a viable way to place all our holes.

After this is done, note that we have filled the entire bottom row, and the upper $n-1$-triangle has exactly $n-1$ holes, and satisfies the holey condition, so we can apply inductive hypothesis to finish.

Ha, Hall’s Marriage Theorem is the shortcut that make it an ez problem.

Outline of solution from [i]Twitch Solves ISL[/i]:

The idea is to show that the condition in the problem is equivalent to the condition in Hall's Marriage Lemma, wherein we try to match down-triangles to non-hole up-triangles.
To this end, let's say two holes are [i]adjacent[/i] if they share a vertex, and a set of holes is [i]connected[/i] if they form a connected component with respect to the adjacency condition. Finally, the [i]bounding box[/i] of a set of holes is the smallest upwards equilateral triangle of some side length (by inclusion) that contains all the triangles.

[b][color=red]Claim:[/color][/b]  	Consider a connected set $D$ of down-triangles 	which has a bounding box of side length $k$. 	Then at least $\left\lvert D \right\rvert + k$ up-triangles are adjacent 	to at least one of the down-triangles in $D$.
[i]Proof.[/i]  	[Outline of proof] 	The idea is to go by induction on $|D|$, 	by deleting the uppermost triangle which touches 	the left boundary of the bounding box of $D$. 	(The catch is that doing this deletion may disconnect $D$, 	so this case needs to be handled as well.) $\blacksquare$

Suppose the problem condition holds; we prove Hall's condition. Consider a general set $\mathcal D$ of down-triangles. The $\mathcal D$ is the disjoint union $D_1 \sqcup \dots \sqcup D_k$ of $k$ connected components. The problem condition implies that now that there are at least $|D_i|$ up-holes neighboring some down-hole in $D_i$, and any up-hole is adjacent to down-triangles in at most one $D_i$; so Hall's condition holds for $\mathcal D$.

For the converse direction, if Hall's condition holds, taking $\mathcal D$ to be all the down-triangles in some equilateral triangle of side length $k$ implies the result. (This does not need the claim; it is the easy direction.)

Refer to unit upward and downward triangles as up-triangles and down-triangles respectively (upward/downward triangle still refer to triangles of any side length), and ignore the problem's definition of $T$. We will use Hall's Marriage Lemma to show that there exists a perfect matching between the down-triangles and the non-removed up-triangles. Call two unit triangles [i]adjacent[/i] if they share an edge, and a set of down-triangles [i]connected[/i] if the graph formed where shared vertices correspond to edges is also connected. Finally, let the [i]bounding triangle[/i] of a set of down-triangles be the smallest upward triangle that contains the set, and let $f(S)$ denote the side length of the bounding triangle of $S$. Note that two non-connected sets of down-triangles have non-overlapping bounding triangles, so no up-triangle can be adjacent to elements of both sets. The key claim is the following:

[b][color=#00f]Claim:[/color][/b] For any connected set $D$ of down-triangles, there are at least $|D|+f(D)$ (not necessarily non-removed) up-triangles adjacent to some element of $D$ (henceforth referred to as simply "adjacent to $D$").
[color=#00f]Proof:[/color] We use downwards induction on $|D|$. For the base case of $|D|=1$, note that the bounding triangle of $D$ has side length $2$, so at most $2$ of the $3$ adjacent triangles to the sole element of $D$ are removed, as desired. For the inductive step, we consider the following cases on the number of triangles in the lowest "row" of $D$:
[list]
[*] At least two triangles. Let the leftmost triangle on this row be $T$. Then the up-triangle $T'$ immediately left of $T$ (which evidently exists) is not adjacent to any element of $D$ except for $T$. Now remove $T$ from $D$ to yield $D'$. If $D'$ is connected, we have $f(D)=f(D')$, so there are at least $|D|+f(D)-1$ up-triangles adjacent to $D'$. Otherwise, $D'$ can be partitioned into two connected subsets $D'_1,D'_2$ (which are connected to the upper left and bottom corners of $T$). It is easy to see that we must have $f(D'_1)+f(D'_2)=f(D')$ (left and right edges should be shared, and the two bounding triangles should touch at the upper right corner of $T$, so by inductive hypothesis there are at least $|D'_1|+|D'_2|+f(D'_1)+f(D'_2)=|D|+f(D)-1$ up-triangles adjacent to $D'$. It then follows that there are at least $|D|+f(D)$ up-triangles adjacent to $D$, as desired.
[*] One triangle. Let this triangle be $T$. Then the up-triangles $T',T''$ immediately to the left and right of $T$ respectively are not adjacent to any other element of $D$. Remove $T$ from $D$ to yield $D'$. If $D'$ is disconnected, then it can be partitioned into $D'_1,D'_2$ which touch the top left and top right corners of $D'$ respectively. But since no other down-triangles lie in $T$'s row, both down-triangles above $T$ and sharing a vertex with $T$ are in $D'$, so $D'$ is connected. Thus we have $f(D)-1=f(D')$, so there are at least $|D|+f(D)-2$ up-triangles adjacent to $D'$ and then $|D|+f(D)$ triangles adjacent to $D$.
[/list]

Suppose the problem condition holds, and consider an arbitrary set $D$ of down-triangles. We partition it into disjoint connected subsets $D_1,\ldots,D_m$. Then the claim implies that there are at least $|D_1|+\cdots+|D_m|+f(D_1)+\cdots+f(D_m)$ up-triangles adjacent to $D$. Removing all holes, it follows that we have at least $|D_1|+\cdots+|D_m|=|D|$ non-hole up-triangles adjacent to $D$, so Hall's condition is satisfied and thus a perfect matching exists (this is the if direction).
For the only if direction, note that if a matching exists, we can take $D$ to be any complete set of down-triangles in an upwards equilateral triangle $T$ of side length $k$. Then we need at most $k$ holes in $T$, else there are fewer up-triangles than down-triangles in $T$ which contradicts the existence of a matching. $\blacksquare$

[quote=JuanOrtiz]I think  your shorter solution is incorrect. Proving that in the equilateral triangle there are at most $k$ more upward triangles than downward triangles does not complete the proof, because it doesn't tell you anything about your original set that violated Hall's condition. Also, you didn't prove that if the tiling exists then no triangle that violates the condition exists.

Here is my solution. First I prove that the condition implies the tiling exists.

We want to use Hall's theorem to prove the Downward and the Upward triangles (we will call them Dts and Uts from now on, for convenience) in the holey triangle can be "married". Since in the original big triangle there are $n$ more Uts than Dts, there is an equal number of Dts than Uts in the holey triangle.

Now, suppose they can't be married. Then by hall theorem there exists a set $S$ of Dts such that $N(S) < S$ (sorry, I want to put cardinality symbols but I don't know how to in this keyboard). ($N(S)$ is the set of Uts adjacant to a Dt in $S$).

Take the LARGEST such set $S$. Say $S=D_1,...,D_m$ (sorry, I don't know how to put curly brackets in this keyboard). We will call a "$p$-arrow" the set of Dts that are contained in an upward equilateral triangle of side $p$. For example, a $2$-arrow is just one Dt. If $A$ is an arrow, call $f(A)$ the "side" of $A$ (the such that $A$ is an $f(A)$-arrow), Take $S=A_1 \cup A_2 ... \cup A_x$ (where $A_i$ is an arrow) such that $x$ is minimal. (Such a thing is possible, since we could consider the union of $m$ $2$-arrows ($m$ Dts).)

Now, if the upward triangles corresponding to $A_1$, ..., $A_x$ were disjoint, then we could apply the condition about upward triangles to finish with a contradiction (Note: it is easy to see these triangles do not escape the boundary of the holey triangle). So assume we have $A_1 \cap A_2$ is not empty.

First, we have to establish two facts about $S$. First, if two consecutive Dts are both in $S$, then the Dt that is above these Dts must also be in $S$ (since it contributes only at most one new element to $N(S)$ and $S$ is maximal). Also, if two semiconsecutive Dts are both in $S$ (separated by one Dt), then this Dt must also be in $S$ by the same reasoning.

Using those propositions many times, we can conclude that $A'$ (the smallest arrow containing both $A_1$ and $A_2$, which is easy to see is contained in the holey triangle) satisfies $A'$ is a subset of $S$. Then we can make $x$ smaller by substituting $A_1$ and $A_2$ by $A'$. CONTRADICTION

Therefore we are done with the difficult part.

Now we do the easy part. This is easy. We also use Hall's Theorem, but backwards. The condition is equivalent to proving that $N(A)$ has more or equal elements than $A$, where $A$ is an arrow. This is true because of Hall's theorem.

So we are done![/quote]

How can you guarantee that when you replace $A_1$ and $A_2$ with $A'$ you don't get an upward triangle with extra Uts? Moreover, could you clarify why $S$ should contain $A'$? What do you mean by  "....using those propositions...? Do you increase the elements of $S$ or do you say the imposibility of $S$ having neither consecutive Dts nor semiconsecutive Dts because of the maximality of $|S|$? Could you help me to understand the solution? I want to understand the solution as it looks really nice because of the Marriage theorem. Thanks in advance.

Call upward and downward equilateral triangles of unit length u-Triangles and d-Triangles, respectively. Clearly, in an upward equilateral triangle of side length $k$, there are $k$ more u-Triangles than d-Triangles. In a tiling, each d-Triangle must be matched up with an adjacent u-Triangle. For each d-Triangle in this $k$-lengthed upward equilateral triangle, the only adjacent triangles are also inside the triangle, so if we remove more than $k$ u-Triangles then there are not enough u-Triangles to match up with all the d-Triangles. Thus, the condition is necessary. Now to prove that it is sufficient.

We proceed with Hall's. We will show that after the $n$ removals, we can match each d-Triangle with a distinct u-Triangle and thus completing the tiling. It suffices to prove that for any set of d-Triangles, there are as many u-Triangles that are adjacent to some d-Triangle in the set.

Suppose this set is not connected. That is, suppose that we can partition the set into two subsets such that no two d-Triangles in different subsets can be touching by a corner. In this way, the u-Triangles adjacent to the two subsets are also different, so it suffices to prove them separately.

We proceed with induction. We start with a single d-Triangle. For the induction, we take off the d-Triangle that that has the greatest impact on the number of u-Triangles adjacent to some d-Triangle. We switch to addition. We keep track of three variables as we add onto this single d-Triangle: $d$, then number of d-Triangles, $u$, the number of u-Triangles adjacent to some d-Triangle in the set, and $s$, the size of the smallest upward equilateral triangle that encompasses all of our current ones. At any point, the number of u-Triangles adjacent to some d-Triangle that wasn't removed is at least $u-s$.

We start with $d=1$, $u=3$, and $s=2$. If we add one to $d$, then we add zero, one or two to $u$. We cannot add three since that requires us to create a disconnected set of d-Triangles. We also add at most one to $s$.

If we add two to $u$, $u-s\ge d$ still must hold since $s$ increases at most $1$ and $d$ exactly $1$. If we add one to $u$ and add one to $s$, that's not possible since the added d-Triangle, and thus two of its adjacent u-Triangles, must be outside of the all-encompassing upward equilateral triangle. If we add zero to $u$, then we have filled in a "hole" that must be surrounded by other d-Triangles, contradiction to our selection criteria. Thus, we are done.


geometry
3D geometry
calculus
integration

Last time I checked, 2 wasn't a cube residue mod 9, so no, there are no integral values of N that satisfy the first equation. However, for your second equation, plugging in $ k=13$ provides $ N=5$.

[quote="mdk"]Is there an integer N such that N^3=9k+2, where k is an integer?

What about N^3=9k+8?[/quote]

[hide="Table (used in solutions)"]\begin{tabular}[t]{|c|c|}\multicolumn{8}{c}{Residue classes}\\\hline Number&\ Cube modulo 9\\\hline 0&0\\ 1&1\\ 2&8\\ 3&0\\ 4&1\\ 5&8\\ 6&0\\ 7&1\\ 8&8\\\hline \end{tabular}[/hide]

[hide="First one"]No. As Lazarus pointed out, 2 is not a cube residue mod 9.[/hide]

[hide="Second one"]Notice that 8 is a cube modulo 9, and it can be described as:
$ N = 3m-1 \textrm{, where }m \textrm{ is an integer such that }m\ge 1.$
You can cube this if you want, but I'm not going to.[/hide]


a) x=+or- y look at them they look lke the same thing if x=y and it doesn't mater if they are + or -
b) x=+or- I*y

Uh, if you want to say that, you'll still have to prove theres no more solutions (which there probably are)..

[quote="JBL"]Solve for x in terms of y:

a) y2(x4 + 1) - x2(y4 + 1) = 0

b) y2(x4 + 1) + x2(y4 + 1) = 0[/quote]

divide by x^2 or y^2 to get a quadratic with y^4
for e.g. y^4-(x^2+1/x^2)y^2+1=0 and you get (y^2-x^2)(y^2-1/x^2)
Y^2=x^2
i don't know why + or - 1/x doesn't work

[quote](y^2-x^2)(y^2-1/x^2)
Y^2=x^2
i don't know why + or - 1/x doesn't work[/quote]
Why doesn't  :pm: 1 work?

churchill: please provide full solutions in spoiler when you answer challenges.  Also, how sure are you that 1/y doesn't work?

juggle7, what do you mean?

a) y^2(x^4 + 1) - x^2(y^4 + 1) = 0

[hide]

y^2 = a

x^2=b

a(b2+1) - b(a2+1)=0

a(b^2)+a = (a^2)b+b.

b-a=ab^2-ba^2

b-a=b(ab)-a(ab)

b-a=(b-a)(ab)

ab must equal to one.

a and b must be reciprocals.

a and b can also be equal.

so x and y must be reciprocals.

[/hide]

b) y^2(x^4 + 1) + x^2(y^4 + 1) = 0

[hide]

y^2 = a

x^2=b

a(b2+1) + b(a2+1)=0

a(b^2)+a = -((a^2)b+b).

b-a=-(ab^2-ba^2)

b-a=-(b(ab)-a(ab))

b-a=-(b-a)(ab)

ab must equal to negative one.

a must be 1/k^2 and b must be -k^2 .

so x and y must be 1/k and -k.

[/hide]

Mostly good in the first one, but you haven't got all the solutions.  I only see one solution that you wrote down.

Hint: [hide]if a2 = 4, solve for all values of a.[/hide]

You have similar problems in the second one.  "Solve for x" in this case means "solve for x in terms of y."


email
geometry
geometric transformation
Support

According to the research, the average score of this exam in Chinese seniors in their high school is 109 points. Which means it just not hard to them.

But I think it would not easy to the high school seniors in other countries.

You can just have a look or you can do these questions.

basically, when Chinese student do this, they would not be allowed to get any helps or handouts or formula sheet. just let bring a calculator and must finish the exam in 120 minutes. But I think it would be too hard to students from other countries. So just relax. haha!!

If you finish this exam, you can email me if you really want to know what score you get or you can wait for several days until I give the answers

[img]http://fmn034.xnimg.cn/fmn034/20090523/1105/p_large_TZDQ_6120h206105.jpg[/img]

[img]http://fmn033.xnimg.cn/fmn033/20090523/1740/p_large_bpuo_96932i206099.jpg[/img]

[img]http://fmn027.xnimg.cn/fmn027/20090523/1105/p_large_E9hq_5970a206105.jpg[/img]

You might be correct as far as the [b]average[/b] high school senior goes, but the posters on this forum do not represent the average capability of a high school senior.  As it stands, the questions look both easier and more boring than those on the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]AMC[/url].  I have no doubt that many regulars of the forum would be able to score perfectly (up to translation errors; I am not 100% sure what Question 3 is asking.)

If you're trying to prove a point about the quality of education in other countries as compared to the quality of education in China, you should know that studies have also shown that students in many Asian countries are taught to memorize facts at the expense of understanding and that such students perform poorly at tasks not specifically designed to test exactly what they were taught.

[img]http://fmn027.xnimg.cn/fmn027/20090523/1105/p_large_YKxO_6237g206105.jpg[/img]

[img]http://fmn033.xnimg.cn/fmn033/20090523/1105/p_large_zl4O_6039o206105.jpg[/img]

hmmm, to #3 in part I, I think I has told clearly about that.

the question is a section intercept the ball and distance betweem between them is 1, calculate the volume of the ball.

--------------

PS: Just think some US students who like math are nice.

the test you says is harder ? are you sure?

I can't see the pictures, anyone knows why,or should I ask in Support?

it may needs the IE browser....I just use tht IE browser..

[quote="tonypr"]I tried it on Internet Explorer, Opera, and Safari and I can't see any of the pictures.[/quote]

just don't know what happhen to the appreance, maybe i can add an attachment.

just do it now  :)

5-7 ，  become harder ......alas

should be cautious to these...because it would be easy to get a wrong anwser.


abstract algebra
superior algebra
superior algebra unsolved

[quote="morova"]Suppose $ M$ is a projective module over a ring $ R$.   Prove that $ M$ is flat.  Is the converse true?
[/quote]

For the first part see any textbook in module theory! For the second part the answer is no. Think about the case where "free" is the same as "projective", e.g. the quotient field of a a PID.


The fractional part (after the decimal) of $ \sqrt {123456789}$ is closest to the fraction 1/n, where n is an integer.  What is the value of n?

I know how to take square roots by hand but I dont have the time to get the integer that fits in there first...like I dont know what integer^2 is the biggest perfect square less than sqrt(123456789)

[hide="Hint"]
Find $ (x^n \plus{} x^{n\minus{}1} \plus{} \cdots \plus{} x \plus{} 1 \plus{} x^{\minus{}1} \plus{} \cdots \plus{} x^{\minus{}(n\minus{}1)} \plus{} x^{\minus{}n})^2$[/hide]

If you actually know square roots by hand, you only double your calculation time by not being able to guess more than the first digit.

Incidentally, when I actually met this question in contest (ARML ... 2002?  2003?), I did attempt to calculate the square root by hand, but accidentally paired the terms incorrectly and so ended up calculating the fractional part of $ 1234567890$, a decidedly less-pretty quantity.


inequalities
Euler
inequalities proposed

Prove that

$ \frac{a^2}{r_a^2}\plus{}\frac{b^2}{r_b^2}\plus{}\frac{c^2}{r_c^2}\geq4$

[quote="Ligouras"]Prove that

$ \frac {a^2}{r_a^2} \plus{} \frac {b^2}{r_b^2} \plus{} \frac {c^2}{r_c^2}\geq4$[/quote]
It's equivalent to Schur.


Παρακαλώ μπορείτε να με βοηθήσετε μ΄αυτήν την άσκηση?

Δίνεται ο πίνακας νxν. Δείξτε ότι τα παρακάτω είναι ισοδύναμα"
1. Το ελάχιστο πολυώνυμο του Α είναι πρώτο με την παραγωγό του
2. Ο Α είναι διαγωνιοποιήσιμος
3. Κάθε αναλοίωτος υπόχωρος του Α έχει συμπληρωματικό αναλλοίωτο

Να σου δώσω μερικά hints:

1. Σκέψου πότε ένα πολυώνυμο έχει κοινές ρίζες με την παράγωγό του. Μετά σκέψου τι σημαίνει διπλή ρίζα για το χαρ/κό πολυώνυμο ενός πίνακα.

2. Σκέψου τι συμβαίνει στους υποχώρους που παράγονται από τα ιδιοδιανύσματα του πίνακα σου.

Cheerio,

Durandal 1707

Φίλε Παναγιώτη  από το 1 στο 2 νομίζω πως το απόδειξα,δηλαδη'
Εφόσον Α διαγωνιοποιήσιμος αυτό σημαίνει πως το ελάχιστο πολυώνυμο του, είναι είναι γινόμενο πρωτοβαθμίων παραγόντων.
Απ΄την άλλη αφού το ελάχιστό πολυώνυμο είναι πρώτο με την παραγωγό του θα αναλύεται σε γινόμενο πρωτοβαθμίων παραγόντων (γιατί ως γνωστό αν ένα πολυώνυμο είναι πρώτο με την παραγωγό του αναλύεται σε γινόμενο πρωτοβαθμίων παραγόντων, αυτό μου φαίνεται πως είναι ισοδυναμία)

Αλλά από το 2 στο 3 και αντίστροφα δεν το έχω καταλάβει ακόμα.
Σκέφτηκα αυτό που μου είπες, πότε ένα πολυώνυμο έχει κοινές ρίζες με την παραγωγό του
και σκέφτηκα πως πρέπει να αναλύεται σε γινόμενο πρώτων παραγόντων, γιατί τότε δεν μηδενίζεται η παράγωγος για καμία ρίζα που μηδενίζεται το πολυώνυμο.

Αλλά η διπλή ρίζα για το χαρακτηριστικό πολυώνυμο δεν ξέρω τι σημαίνει,
ίσως ότι ο πίνακας δεν διαγωνιοποιείται, αλλά είναι τριγωνίσιμος, άρα τα ιδιοδιανύσματα u (A*u=λu. λ πραγματικός ή μιγαδικός)  που προκύπτουν είναι λιγότερα από την τάξη του πίνακα και πρέπει να κάνουμε επέκταση, δηλαδή να βρούμε διανύσματα που δεν ανήκουν στους ιδιοχώρους V(λ) που παράγουν οι ιδιοτιμές!
ελπίζω να μην έγινα πολύ κουραστικός!
τότε αυτά τα διανύσματα επί τον πίνακα Α θα μας δίνουν γραμμικούς συνδιασμούς των ιδιοδιανυσμάτων u και των καινουργιων διανυσμάτων v που ορίσαμε, άρα αυτά τα διανυσματα δεν θα είναι αναλλοίωτα ως προς τον πίνακα Α, αφού Α*v=u+v
μήπως  όμως δεν έχουν σχέση αυτά που λέω
Κάθε ιδιόχωρος που παράγουν οι ιδιοτιμές δεν είναι αναλλοίωτος ως προς τον πίνακα Α?
ελπίζω να μην έγινα πολύ κουραστικός

[quote="spinos"]
Αλλά η διπλή ρίζα για το χαρακτηριστικό πολυώνυμο δεν ξέρω τι σημαίνει,
ίσως ότι ο πίνακας δεν διαγωνιοποιείται, αλλά είναι τριγωνίσιμος, άρα τα ιδιοδιανύσματα u (A*u=λu. λ πραγματικός ή μιγαδικός)  που προκύπτουν είναι λιγότερα από την τάξη του πίνακα και πρέπει να κάνουμε επέκταση, δηλαδή να βρούμε διανύσματα που δεν ανήκουν στους ιδιοχώρους V(λ) που παράγουν οι ιδιοτιμές![/quote]

Whoops, sorry, typo... δεν εννοούσα χαρ/κό πολυώνυμο στο post μου, αλλά ελάχιστο, sorry!

Με διπλή ρίζα στο χαρ/κό πολυώνυμο μπορείς μια χαρά να έχεις διαγώνιο πίνακα, πχ ο μοναδιαίος. Ωστόσο, η πολλαπλότητα μιας ρίζας του ελαχίστου πολυωνύμου είναι το μέγεθος του Jordan block που αντιστοιχεί στην ιδιοτιμή αυτή. Άρα...

Sorry για το μπέρδεμα! Cheerio,

Durandal 1707


topology
advanced fields
advanced fields unsolved

Here is some funny exercise I came across recently...

Determine which of the figures "$ 0 \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9$" (as subsets of $ \bf{R}^2$ where the lines are infinitely thin) are homeomorphic.

I managed to 'prove it intuitively' because I lack some Algebraic Topology. Of course, a rigorous proof is always very welcome. :)

You cannot prove this rigorously without exactly defining them. I'm pretty sure this is meant as a test of intuition^^
By the way, the answer:
$ 1,2,3,5,7$
$ 0,4,6,9$
$ 8$

You can only pull in tails if there's some thickness; a Y junction is topologically different from a segment, since cutting out a neighborhood leaves three pieces rather than two. In that "line" interpretation, they group as
$ 0$
$ 1,2,3,5,7$
$ 4$
$ 6,9$
$ 8$

I assume that they have some thickness...
Now you see what happens without definitions  :wink:


In triangle ABC, AB=8, BC=9, AC=6. Line segment DE is drawn parallel to side AC, with D on AB and E on BC, such that it divides the triangle into 2 equal areas. What is the length of DE?

What is the surface area in Sq. m of a right rectagnular pyramid of height 8 meters, base width 12, and base length 30?

[hide="1"]
The two triangles will be similar, and if the little triangle is to have half the area, its side lengths will be a factor of $\sqrt{2}$ less than the big triangles'. The side length of DE will be $\boxed{3\sqrt{2}}$
[/hide]

[hide="2"]
$\frac{(8)(12)(30)}{3}= \boxed{960}$.
[/hide]

# 2 is wrong. IT should be 864

with 17 times 12 equals two faces.

with 10 times 30 equals two faces.

and base with 360

so it equals 864.

surface area, right, thought it said volume.

oh ok. I figured out SA of pyramid after asking my friend .


1. Jenny wants to siwm from point a on one shore of a river to point b on the other shore. The shores are parallel straight lines, ab is perpendicular to both shores, and ab=300ft. Jenny swims at a constant rate of 4ft/sec, and the current is flowing at a constant rate of 3.2ft/sec. Find the fewest number of seconds for Jenny to swim from a to b.

2.  (1-cos(a)-tan^2(a/2))/(sin^2(a/2))=kcos(a)/(w+pcos(a)).  Find the value of k+w+p

[hide="(1)"]If Jenny swims in a straight line, her velocity against the current must be $3.2\ m/s$. Since her speed is $4\ m/s$, she will swim at a rate of $2.4\ m/s$ across the river by Pythagorean Theorem. The total time is $\frac{300}{2.4}\ s=125\ s$.[/hide]

[hide="(2)"]Let $x=\frac{a}{2}$.
\begin{eqnarray*} \frac{1-\cos a-\tan^2 x}{\sin^2 x}&=&\frac{1-(1-2\sin^2 x)-\tan^2 x}{\sin^2 x}\\ &=&2-\sec^2 x\\ &=&2-\frac{2}{1+\cos a}\\ &=&\frac{2\cos a}{1+\cos a} \end{eqnarray*}
making the sum 4.[/hide]

For (2), any sum could actually work. Here I assume k, w, p are positive integers such that no common factor other than 1 is shared by all.


The first round of the Power Contest has been scored, and the scores ran [i]very[/i] low. Here are the  top 20 or so teams on the first round. There were 40 points possible.

$ \begin{tabular}{ccc}
1T.&Long Beach Math Circle&25\\
1T.&Wayzata High School&25\\
3.&Lynbrook High School&24\\
4T.&AAST MuA&23\\
4T.&NYC Math Team&23\\
4T.&Colorado Math Circle&23\\
7T.&Belmont High School&21\\
7T.&Henry M. Gunn High School&21\\
7T.&Western Washington ARML&21\\
10.&State College Area High School (North)&19.5\\
11T.&Georgia ARML&19\\
11T.&Stuyvesant High School&19\\
13.&Paly Math Team&18\\
14T.&AAST Alpha&17\\
14T.&Thomas Jefferson High School&17\\
16.&Glenbrook North High School&16.5\\
17T.&Harvard\minus{}Westlake Advanced Seminar&16\\
17T.&Iowa City West High School&16\\
17T.&San Diego Math Circle&16\\
20T.&Albany Area Math Circle&15\\
20T.&MN ARML&15\\
20T.&Southwest High School&15\\
20T.&University of Chicago Lab Schools&15\end{tabular}$

Out of curiousity: is that all the teams that did the power round, or just the top 20?

Wow...

The power contest was really confusing... I couldn't figure out what strategy the player was playing with... so does he throw away a Yahtzee if he gets a Yahtzee? Etc...

There were about 60 teams. I only listed the first page of the results.

There was another sheet showing question-by-question breakdowns. It looked like this:

[quote]For each problem worth $ n$ points, and $ (n\plus{}1)$-tuple, $ (a_0,a_1,a_2,\dots a_n)$ is given, where $ a_k$ represents the number of teams that scored $ k$ points in the problem. Bracketed values indicate half points were earned. Data comes from the first 60 teams scored.

1a. (2, 58)
1b. (30, 30)
1c. (45, [1], 14)
1d. (54, [1], 5)
1e. (39, [1], 20)
1f. (38, [1], 31)
1g. (53, 7)
2. (33, 2, 25)
3. (46, 10, 4)
4. (10, 6, 44)
5. (13, 17, [4], 0, [1], 28)
6. (36, 22, 2, 0)
7. (27, 17, 1, 15)
8. (26, 18, 8, 8)
9. (14, 14, [1], 23, 8)
10. (32, 2, 0, 26, 0)
11. (36, 24, 0, 0, 0)
12. (56, 3, 1, 0, 0)[/quote]

Seems like everyone had trouble with it. Anyways, great job to everyone who participated!

are there results for foreign teams that took the test?

The listing doesn't give state or country. The only participating team that didn't appear to be from the U.S. or Canada was Indus Center for Academic Excellence at 14. There's also a listing for Olimpiadas Colombianas De Mathematicas but they didn't send in their paper. In previous years there's been a Sofia Math Circle, but I don't see them listed at all this year.

[quote="Math Geek"]Hmm, I guess GA ARML might have had more participants if the ARMLers and everyone else interested were informed about location and time a substantial amount of time beforehand.

Cuz I found out the date of the power contest the day before it took place and thus could not go due to scheduling conflicts. :([/quote]

Check your email more often.  According to the Rockdale website, we're doing Round 2 on February 8. :wink:

I have to say, this wasn't my favorite power question.  It seemed that the actual questions were not hard, but the rules of the game were so complicated and difficult to interpret that . . . well, you see the results.  Perhaps the question's authors thought that more people would already be familiar with the game.

I check my email daily.  It's just that the correct address for my email was never added to the ARML mailing list, so, like you said, my email's been bouncing all of the ARML team's messages. :|

Does anybody know whether Exter Academy does the Power Contes this year or not?I'm very surprised at the result above(TJ seems to have some problems this year, and Exter even doesn't appear on this list). :|

Exeter doesn't seem to be doing this in 2008-2009.

Final results for 2008-2009! And it was very, very, close.

Place, Team, Rnd 1, Rnd 2, Total
1T, Long Beach Math Circle, 25, 35, 60
1T, Wayzata High School, 27, 33, 60
3T, AAST $\mu$  A, 23, 36, 59
3T, Georgia ARML, 21, 38, 59
3T, NYC Math Team, 23, 36, 59
6, Colorado Math Circle, 23, 35.5, 58.5
7T, Thomas Jefferson HS, 17, 39, 56
7T, Western Washington ARML, 23, 33, 56
9, Palo Alto Math Team, 18, 35, 53
10T, Belmont Math Team, 21, 32, 53
12, Henry M. Gunn HS, 21, 28, 49
13, State College Area HS (N), 19.5, 29, 48.5
14T, Albany Area Math Circle,15, 33, 48
14T, Montgomery County, 16, 32, 48
16, Stuyvesant HS, 19, 28, 47
17, Oregon ARML, 12, 31, 43
18, Utah ARML, 14, 28, 42
19T, AAST $\alpha$, 17, 24, 41
19T, San Diego Math Circle, 16, 25, 41
21, Naperville North HS, 14, 24, 38
22, Glenbrook North HS, 16.5, 21, 37.5
23T, Central HS, 13, 23, 36
23T, IMSA, 14, 22, 36
25T, AAST $\mu$ B, 14, 21, 35
25T, MN ARML, 15, 20, 35

(And many more teams.)

Perhaps if they had told everyone what the Yahtzee player's strategy was (it was suboptimal) people might have scored higher.

Wayzata went from 25 points in round 1 to 27?

There was a re-grade of round 1 based on some objections to certain questions. We weren't in position to benefit from it.

May I ask which questions in round 1 were rescored?


geometry
parallelogram
geometry unsolved

[quote="kfrances"]Assume: triangle ABC  H is the orthocenter
                     angle A equals 60 degrees
                     M is the midpoint of AH an CD

         Show: BD = 2AH

Figure can be seen at the following link

http://gogeometry.blogspot.com/2008/06/elearn-geometry-problem-132.html

The proof online is incorrect any ideas about how to prove using only basic theormes and relationships?

Kfrances[/quote]

triangle ABC  H is the orthocenter ---> $ AH\equal{}2Rcos60^{o}\equal{}R$
                                                         $ BH\equal{}2Rcos\angle{B}$
M is the midpoint of AH an CD ----> ADHC parallelogram ---> DH//AC ---> $ \angle{}DHB\equal{}90^{o}$ ---->

$ BD^{2}\equal{}BH^{2}\plus{}AC^{2}\equal{}4R^{2}cos^{2}\angle{B}\plus{}b^{2}\equal{}4R^{2}cos^{2}\angle{B}\plus{}(2Rsin\angle{B})^{2}\equal{}4R^{2}(cos^{2}\angle{B}\plus{}sin^{2}\angle{B})\equal{}4R^{2}$

----> $ BD\equal{}2R$---->

$ BD\equal{}2AH\equal{}2R$

Thanks for the reply to my question however what does R stand for in the reply proof?

Kfrances


[quote]Mate, I wouldn't leave your e-mail address lying around here or it'll be spam galore before you know it. PM it to him.[/quote]

*deleted*

Why I enquire why your name is spelt differently here than on nRich?

Also, how do you know whether Chris Bryant has an account here?

He does: http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=14395

However, the point was that Chris and Dan were discussing on nRich (in Word Association Game 2) how to swap email adresses, and Tom suggested they posted here. Vicky deleted the conversation only a few minutes later.

Oh yes, I've just read that. Ooops. To be fair, NRICH should really have a private messaging system, even if it's moderated.


number theory
prime numbers
number theory proposed

Let $P_{1},P_{2},...$ be the sequance of prime numbers .
Let $m$ be a natural number.
Prove that there exist infinitely numbers like $i$ such that \[P_{i}-P_{i-1}>m \text{ , and ,}P_{i+1}-P_{i}>m\]

[hide="Is it known?"] That there are an infinite number of primes of the form $n! \pm 1$?  If so, I can give a solution. [/hide]

To t0rajir0u: No, this is not known. It is an open question in fact :(
Homever, you can prove that there is infinitely many composite $n!-1$ and infinitely many composite $n!+1$ (Apply Wilson's theorem in both).

[quote="pbornsztein"]For the original problem : Otherwise the density of primes would be at least $\frac{1}{m}$ which contradicts that it is $0$ (from the Prime Number Theorem for instance).
Pierre.[/quote]
I was too blind to see that , thanks Mr.bornsztein

Here is my solution :

Let $P_{1},P_{2},...,P_{2m}$ be different primes bigger than $m$
By chinese remainder the eqautions \[x =-m(mod, P_{1}P_{2})\]  \[x=-m+1(mod,P_{3}P_{4})\]  \[...\]  \[x=2m (mod ,P_{2m-1}P_{2m})\] has infinitely many answer which form an arithmetic sequance like $\{a+nb\}_{n=1}^{\infty}$ where $a=\prod_{i=1}^{2m}P_{i}$
Because $(a,b)=1$ by johann dirichlet theorem there exist infinitely many primes in this sequance .Those primes enjoy our problem condition .

[quote="pbornsztein"]For the original problem : Otherwise the density of primes would be at least $\frac{1}{m}$ which contradicts that it is $0$ (from the Prime Number Theorem for instance).

Pierre.[/quote]
Sorry, but I don't see how this is supposed to work.
You could have a twin prime, then a big gap for example.

Hmmm... I think I misread the 'and' in the original statement... :blush:

Pierre.


[img]http://img410.imageshack.us/img410/8633/pmwc200609teammt9.png[/img]

I don't understand this problem take from PMWC2006 (Po leung Kuk)

Can anyone explain me by some example. Please  :(

[hide]I get 26.
First, I made a table of the different ways to get a sum.
1:2
2:3
3:4
4:5
5:6
6:6
7:6
8:5
9:5
10:4
11:3
12:2
13:1
The worst case scenario is if you get the two 1 sums, the two 12 sums, and the  one 13 sum first. This is 5 numbers. Then, you might get two of each.
5+2+2+2+2+2+2+2+2+2+2=25
Still, you don't have a lucky set. The next card you draw will give you a lucky set, though.
25+1=26
[/hide]
Shouldn't this be moved to another forum, since it isn't from MOEMS?

Umm he said he doesn't understand the problem...

I'll reword it...
There are 52 cards numbered 1 to 52 in random order, stacked on a table.
How many cards do we need to pick up in order
to be sure we have 3 cards with the same digit sums?

@vamathletes: Possibly, but where would you move it to?
It's too easy for HSB (I should know) and it doesn't qualify as MC.

Thank you both of you.  :D
Really appreciate.  :coolspeak:


Let $ n>0$ be a natural number. Determine all the polynomials of degree $ 2n$ with real coefficients in the form
$ P(X)\equal{}X^{2n}\plus{}(2n\minus{}10)X^{2n\minus{}1}\plus{}a_2X^{2n\minus{}2}\plus{}...\plus{}a_{2n\minus{}2}X^2\plus{}(2n\minus{}10)X\plus{}1$,
if it is known that all the roots of them are positive reals.

[i]Proposer[/i]: [b]Baltag Valeriu[/b]

do all the odd-degree terms have coefficient $ 2n\minus{}10$, or just the $ x$ and $ x^{2n\minus{}1}$ terms?

Assuming it's only those two coefficients...

[hide="Solution"]By Vieta we have

$ \sum r_{i} \equal{} 10 \minus{} 2n$

$ \prod r_{i} \equal{} 1$

$ \sum 1/r_{i} \equal{} 10 \minus{} 2n$

By AM-GM we have $ \frac {10 \minus{} 2n}{2n}\ge 1$, so $ 5 \minus{} n\ge n$, and $ n\le 2$.  Now we just do some casework.

Case 1: $ n \equal{} 1$.  Then $ P(x) \equal{} x^{2} \minus{} 8x \plus{} 1$, which has positive real roots.

Case 2: $ n \equal{} 2$.  Then $ P(x) \equal{} x^{4} \minus{} 6x^{3} \plus{} (a \plus{} 2)x^{2} \minus{} 6x \plus{} 1$.  Consider the equation $ P(x) \equal{} 0$.  Divide both sides by $ x^{2}$ to get $ x^{2} \minus{} 6x \plus{} a \plus{} 2 \minus{} 6/x \plus{} 1/x^{2} \equal{} 0$.  We can rewrite this as $ (x \plus{} 1/x)^{2} \minus{} 6(x \plus{} 1/x) \plus{} a \equal{} 0$.  Let $ y \equal{} x \plus{} 1/x$, and note that $ y$ is positive iff $ x$ is positive. Therefore we want $ y^{2} \minus{} 6y \plus{} a \equal{} 0$ to have positive real roots, which is true as long as $ a$ is positive and $ 36 \minus{} 4a$ is non-negative.  Therefore, $ 0 < a\le 9$.  So our polynomials for this case are $ x^{4} \minus{} 6x^{3} \plus{} kx^{2} \minus{} 6x \plus{} 1$ for $ 2 < k\le 11$.[/hide]

EDIT: Good call rofler, my solution is wrong.  Look below for a correct solution.  :lol:

If the roots are $ b_1,b_2,...,b_{2n}$ we have

$ b_1b_2...b_{2n} \equal{} 1$ so no $ b_i \equal{} 0$

$ b_1 \plus{} b_2 \plus{} ... \plus{} b_{2n} \equal{} 10 \minus{} 2n$

$ b_1b_2...b_{2n}(1/b_1 \plus{} 1/b_2 \plus{} ... \plus{} 1/b_{2n}) \equal{} 10 \minus{} 2n$ so $ 1/b_1 \plus{} 1/b_2 \plus{} ... \plus{} 1/b_{2n} \equal{} 10 \minus{} 2n$

By AM-GM $ (b_1 \plus{} b_2 \plus{} ... \plus{} b_{2n})/2n \ge 1$

$ 10 \minus{} 2n \ge 2n$

$ 2.5 \ge n$

So n=1 or n=2.

n=2 gives the polynomial $ x^4 \minus{} 6x^3 \plus{} a_2x^2 \minus{} 6x \plus{} 1 \equal{} 0$

$ b_1b_2b_3b_4 \equal{} 1$

All of the roots can't be one (since their sum is 6), so WLOG, $ b_1 \ne 1$

Then it is easy to see that $ 1/b_1$ is also a root of this equation, WLOG, $ 1/b_1 \equal{} b_2$

Then $ b_3b_4 \equal{} 1$ so $ 1/b_3 \equal{} b_4$

Then we have $ (b_1 \plus{} 1/b_1) \plus{} (b_3 \plus{} 1/b_3) \equal{} 6$ as the only condition (yay).

And we need to find all possible values of $ a_2 \equal{} b_1b_2 \plus{} b_1b_3 \plus{} b_1b_4 \plus{} b_2b_3 \plus{} b_2b_4 \plus{} b_3b_4$

$ \equal{} 1 \plus{} b_1b_3 \plus{} b_1/b_3 \plus{} b_3/b_1 \plus{} 1/(b_1b_3) \plus{} 1 \equal{} 2 \plus{} (b_1 \plus{} 1/b_1)(b_3 \plus{} 1/b_3)$

Now, the only restrictions we have on the numbers $ b_1 \plus{} 1/b_1$ and $ b_3 \plus{} 1/b_3$ is that they are both $ \ge 2$ (since they are surjective on [2, \inf ) when $ b_1,b_3 \ge 0$ and $ b_i\plus{}1/b_i$ can't be <2 by AM-GM).

let $ b_1 \plus{} 1/b_1 \equal{} m$ and $ b_3 \plus{} 1/b_3 \equal{} k$

Then k and m are the roots of:

$ x^2 \minus{} 6x \plus{} a_2 \minus{} 2 \equal{} 0$

$ (x \minus{} 3)^2 \plus{} a_2 \minus{} 2 \equal{} 9$

$ (x \minus{} 3)^2 \equal{} 11 \minus{} a_2$

$ x \minus{} 3 \equal{} \pm\sqrt {11 \minus{} a_2}$

$ x \equal{} 3 \pm \sqrt {11 \minus{} a_2}$

So the restrictions on $ a_2$ are:

$ 3 \pm \sqrt {11 \minus{} a_2} \ge 2$

$ \pm \sqrt {11 \minus{} a_2} \ge \minus{} 1$

$ \minus{} \sqrt {11 \minus{} a_2} \ge \minus{} 1$

$ \sqrt {11 \minus{} a_2} \le 1$

$ 11 \minus{} a_2 \le 1$

$ 10 \le a_2$

But we also have to keep in mind that $ 11 \minus{} a_2 \ge 0$ so $ 11 \ge a_2$

Therefore, $ 10 \le a_2 \le 11$

Now, if n=1

$ x^2 \minus{} 8x \plus{} 1 \equal{} 0$

Which has 2 positive roots (check it yourself).

Therefore, the only solutions are $ x^4 \minus{} 6x^3 \plus{} a_2 \minus{} 6x \plus{} 1 \equal{} 0$ with $ 10 \le a_2 \le 11$

And $ x^2 \minus{} 8x \plus{} 1$

Q.E.D.

Your solution is a bit off matt276eagles, since we need $ x \plus{} 1/x \ge 2$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that if an infinite, noncommutative group $ G$ contains a proper normal subgroup with a commutative factor group, then $ G$ also contains an infinite proper normal subgroup.

[i]B. Csakany[/i]

We consider two cases:

A.) $ G'$ is infinite, then everything's clear, since it must be a proper subgroup by the first condition
B.) $ G'$ is finite. Then every conjugacy class $ \{gxg^{\minus{}1} | g\in G\}$ contains only finitely many elements (for it is contained in the left coset $ xG'$). So $ C_G(X)$ hase finite index and contains therefore a normal subgroup of finite index (which is therefore infinite). Since at least one of the conjugacy classes is nontrivial (G is non-abelian) one of these normal subgroups is proper.


I'm just curious as to what your wall paper is. Mine is a really cool intricate freehand drawing from someone I know.

Mine is a picture generated by [url=http://www.apophysis.org]Apophysis[/url]

mine is a deer in some woods. its animated though.

I have the blue default. if it were up to be it'd be something stereotypically girly. but it's my dad's computer. :(

None other than a 3 dimensional Mandelbrot fractal :D

Where's the other LoTR fans?

I've got a Gondor background currently. ^__^

[quote="DarkKnight"][url=http://img299.imageshack.us/img299/4488/desktop2dc.jpg]My Desktop[/url][/quote]wow, that's so cool.  where did you get it from?

[quote="DarkKnight"][url=http://img299.imageshack.us/img299/4488/desktop2dc.jpg]My Desktop[/url][/quote]Nice WinMX you have there :D

Anyway my desktop is ... attached.

[quote="Valentin Vornicu"][quote="DarkKnight"][url=http://img299.imageshack.us/img299/4488/desktop2dc.jpg]My Desktop[/url][/quote]Nice WinMX you have there :D
[/quote]

oh... I don't really use it anymore though.  :blush:

[quote="noneoftheabove"]doesnt that come with windows xp or something[/quote]No, the one that comes with WinXP is sort of a hill and looks worse. This one is from Windows Vista :D

I have things from http://www.digitalblasphemy.com

It's so cool!

I use Linux, so my computer has 6 different backgrounds, and they all look pretty cool.

Well, reading this thread I decided to change my background.

[url=http://img329.imageshack.us/img329/536/destinysending7pd.jpg]Wallpaper[/url]

wow, thats much cooler.  can you create these weird looking things with a program or something?

yeah a lot fo people create their own wallpaper which is pretty cool but i am horrible at those sort of things :(

a sand dune. it comes w/ win xp. yep. horray for lack of imagination.

[url=http://img394.imageshack.us/img394/4834/desktop4xp.jpg]Mine[/url]

Shrunken, I'll send full if you want.  :D

Is that english text? if so could you post that text here? Thanks.

Here's mine.  I made it in Apophysis yesterday.

Mine is Stonehenge right now, but I change it a lot.

a "Z" emblem on the front of a 300zx. (please joe, refrain from comments)

I have the default Mac OS X Tiger background, but I'll probably change it sometime.

I'm using a Homestarrunner background  :lol: ... but in the past ive used Windows ones, and ones I've created myself.

I have a hermione wallpaper similar to my blog background :) :)

I use a picture from http://www.worth1000.com


geometry
rectangle
function
real analysis
real analysis unsolved

Let $ Z$ be a set of measure 0.  Prove that $ Z^{2}\equal{}\{z^{2}\mid z\in Z\}$ also has measure 0.

One may proceed as follows:
Let $ \epsilon > 0$ and interval $ (a, b)$ be given. Since Z is a zero set, there is a covering of Z by open intervals $ I_{k}$ with $ \sum_{k}|I_{k}| < \frac{\epsilon}{(b \minus{} a)}$. Thus $ Z \times (a, b)$ is covered by rectangles $ I_{k} \times (a, b)$ whose total area is less than $ \epsilon$, so $ Z \times (a, d)$ is a zero set.

I'm not sure you interpreted the question correctly.  Here is the exact statement: Let $ Z$ be a subset of $ \mathbb{R}$ with measure zero.  Show that the set $ \{x^{2}\mid x\in Z\}$ also has measure zero.  So $ Z^{2}\neq Z\times Z$.  Unless I'm misinterpreting the question.

We can assume that $ Z$ is bounded, contained in some interval $ [\minus{}a,a]$. The function $ z \mapsto z^2$ is then Lipschitz continuous on $ Z$, and since $ Z$ can be covered by some arbitrarily small intervals, we can cover $ Z^2$ in the same way (by the images of the intervals under the map $ z \mapsto z^2$).

And if $ Z$ is not bounded, write $ Z\equal{}\bigcup_{n\equal{}1}^{\infty}Z\cap[\minus{}n,n],$ use Kalle's argument for each piece of that, and then say that the countable union of sets of measure zero has measure zero.


A circle is tangent to the circumcircle of triange ABC and to the rays AB and AC at points M and N, respectively. prove that the
excentre to side BC of triange ABC lies on the segent MN.


I didn't actually calculate the limit because I was too lazy  :D , but here's the way that I think works:

We can calculate l_2n and we find it to be ((3+5^1/2)/2)^n+((3+5^1/2)/2)^(-n) and 3=(3+5^1/2)/2+((3+5^1/2)/2)^(-1). We can actually calculate that sum S_n= (1/(l_0+l_2)+1/(l_2+l_2)+..+1/(l_2n+l_2)). We can generally calculate the sum E_n(a)= \sum {k=0 to n} 1/(a^k+a^(-k)+a+a^(-1)) like this: E_n(a)= \sum {k=0 to n} a^k/(a^(k+1)+1)(a^(k-1)+1) and this is a telescopic sum. We do the same for this problem because S_n=E_n((3+5^1/2)/2) and we know how to calculate that.

I had a hint for this problem: it was proposed at a Romaninan contest and I found it in a magazine. Exactly above this question (proposed for grades 11-12) was a question by the same author proposed for grades 9-10 which asked to calculate E_n(2), so I realized what the correct approach was.

Okay I managed to solve it:

lim=(1+2sqrt(5))/10.

cheers! :D

You found it in the Gazeta Matematica too, didn't you?

No, cause .. as silly as it may sound, I only have the 1/2003 - 3/2003 and I am looking for other numbers!
:D

cheers!

You mean you only have these 3 issues? I only have 1 and 2 from 2003, but I also have all the issues from 2001 and 2002. This problem was in 11/2002.

I'll be going to sleep pretty soon.. But not too soon!  :D


Let $\mathbb{R}^{*}$ denote the set of nonzero real numbers. Find all functions $f: \mathbb{R}^{*}\to\mathbb{R}^{*}$ such that \[f(x^{2}+y)=f^{2}(x)+\frac{f(xy)}{f(x)}\; \forall x,y\in\mathbb{R}^{*},x^{2}+y\in\mathbb{R}^{*}.\]

[quote="N.T.TUAN"]Let $\mathbb{R}^{*}$ denote the set of nonzero real numbers. Find all functions $f: \mathbb{R}^{*}\to\mathbb{R}^{*}$ such that
\[f(x^{2}+y)=f^{2}(x)+\frac{f(xy)}{f(x)}\; \forall x,y\in\mathbb{R}^{*},x^{2}+y\in\mathbb{R}^{*}. \]
[/quote]

P1 : $f(x^{2}+y)=f^{2}(x)+\frac{f(xy)}{f(x)}\; \forall x,y\in\mathbb{R}^{*},x^{2}+y\in\mathbb{R}^{*}$

1) P2 : $f(x^{2}+1)=f^{2}(x)+1\; \forall x,\in\mathbb{R}^{*}$
Easy with P1 and $y=1$

2) $f(-x)=-f(x)$ $\forall x\neq 0$
$-x$ in P2 $\Rightarrow$ $f((-x)^{2}+1)=f^{2}(-x)+1=f(x^{2}+1)=f^{2}(x)+1$ $\Rightarrow$ $f(-x)=\pm f(x)$
Let then $\phi=\frac{1+\sqrt{5}}{2}$, we have $\phi^{2}=\phi+1$
If $f(-\phi)=f(\phi)$, then take $x=\phi$ and $y=-1$ in P1 and $f(\phi^{2}-1)=f^{2}(\phi)+1$ ,$\Rightarrow$ $f(\phi)=f^{2}(\phi)+1$ which have no real solution $\Rightarrow$ $f(-\phi)=-f(\phi)$
Then P1 gives : $f(\phi^{2}+\frac{x}{\phi})=f^{2}(\phi)+\frac{f(x)}{f(\phi)}$ but also $f((-\phi)^{2}+\frac{x}{\phi})=f^{2}(-\phi)+\frac{f(-x)}{f(-\phi)}$ and so $\frac{f(x)}{f(\phi)}=\frac{f(-x)}{f(-\phi)}$ and so $f(-x)=-f(x)$. Q.E.D.

3) $f(1)=1$
Let $a=f(1)$, $b=f(2)$ and $c=f(3)$
Then $x=1$ and $y=1$ in P1 gives $b=a^{2}+1$
Then $x=1$ and $y=2$ in P1 gives $c=a^{2}+\frac{b}{a}$
Then $x=2$ and $y=-1$ in P1 gives $c=b^{2}-1$
This system is easy to solve ang gives $f(1)=1$

4) $f(x+1)=f(x)+1$ $\forall$ $x\neq 0$ and $x\neq-1$
Just use P1 with $1$ and $x$

5) $f(x^{2})=f^{2}(x)$ $\forall$ $x\neq 0$ (and, as a consequence, $f(x)\geq 0$ $\forall$ $x>0$)
Immediate from points 1) and 4) above.

6) $f(a+b)=f(a)+f(b)$ for $a\neq 0$, $b\neq 0$, $a+b\neq 0$
P1 gives $f(x^{2}+y)=f^{2}(x)+\frac{f(xy)}{f(x)}$
Using $y+1$ instead of $y$, P1 gives $f(x^{2}+y+1)=f^{2}(x)+\frac{f(xy+x)}{f(x)}$
And, since (from 4) above), $f(x^{2}+y+1)=f(x^{2}+y)+1$ for $x^{2}+y\neq 0,-1$ :
$f^{2}(x)+\frac{f(xy+x)}{f(x)}=f^{2}(x)+\frac{f(xy)}{f(x)}+1$ and $f(xy+x)=f(xy)+f(x)$
Taking $x=b$ and $y=\frac{a}{b}$ gives the result.
The problem of "$x^{2}+y\neq 0,-1$", if  $b^{2}+\frac{a}{b}=0,-1$ may be solved (I've not checked) by using $y=b$ and $x=\frac{a}{b}$ or $y-1$ and not $y+1$ in P1.

7) $f(x)=x$
$f(a+b)=f(a)+f(b)$  gives easily $f(x)=x$ $\forall x\in\mathbb{Q}^{*}$
And since $f(b)\geq 0$ for $b>0$ (see 5)), then $f(x)$ is monotonous increasing and so $f(x)=x$ $\forall x\in\mathbb{R}^{*}$

And it is immediate to verify that this necessary condition works.


Find all pairs (n, k) of positive integers, for which the equation ... (seealso equation.gif)

Let $ A\equal{}2\frac{10^{k}\minus{}1}{9}\equal{}B*C$, were $ (B,C)\equal{}1$, therefore exist $ q,r$, suth that $ qB\equal{}rC\plus{}1$. It give solution
$ n\equal{}rC,n\plus{}1\equal{}qB,m\equal{}qr$and all solution can be represented in these form.

[quote="Rust"]Let $ A \equal{} 2\frac{10^{k}\minus{}1}{9}\equal{} B*C$, were $ (B,C) \equal{} 1$, therefore exist $ q,r$, suth that $ qB \equal{} rC\plus{}1$. It give solution
$ n \equal{} rC,n\plus{}1 \equal{} qB,m \equal{} qr$and all solution can be represented in these form.[/quote]

Hm... is that so? Please, find the values of [i]m[/i] and [i]n[/i], if [i]k[/i]=4.
Thank you in anticipation.
Nosy Parker

for $ k\equal{}4$, $ m\equal{}6$ and $ n\equal{}36$

[quote="mtphaovibul"]for $ k \equal{} 4$, $ m \equal{} 6$ and $ n \equal{} 36$[/quote]
m=6,n=36 work for k=3.

oops, for k=4 there is none, since 1111 can be factor to 11 and 101, so either n or n+1 must be divisible by 101. you can check that there is none.


Given a $ \odot$ and a line $ \overline{XY}$ through a point $ A$ , moveable along $ \overline{XY}$ , two tangents are traced. Show that the line that goes connects the tangent's contact points cuts the diameter perpendicular to $ \overline{XY}$ at a fixed point.

Inversion about $ \odot$ yields that $ G$, the point of intersection of the perpendicular bisector of $ XY$ and the two tangent points, is the polar inverse of $ C$, the point of intersection of $ XY$ and the perpendicular bisector of $ XY$, and so clearly only depends on $ AC$ (and the radius of $ \odot$).

Thank you for replying Yongyi781
Could you show a proof without using inversion?

[hide="Without inversion"]
[asy]pair O = (0, 0);
pair B = (0, 4);
pair C = (0, 1.5);
pair H = (-sqrt(13.75), 1.5);
pair I = (sqrt(13.75), 1.5);
pair A = (10, 1.5);
pair E = (1.02, 3.87);
pair F = (2.11, -3.4);
pair G = (0, 10.67);
pair D = (1.56, 0.23);
dot("O", O, SW);
dot("B", B, SW);
dot("C", C, SW);
dot("H", H, SW);
dot("I", I, SW);
dot("A", A, SW);
dot("E", E, SW);
dot("F", F, SW);
dot("G", G, SW);
dot("D", D, SW);
draw(H--A--E);
draw(A--F);
draw(F--G);
draw(G--O--A);
draw(O--E);
draw(O--F);
draw(Circle(O, 4));[/asy]

Let the radius of circle $ O$ be $ r$.
Since $ AE$ is tangent to circle $ O$, it follows that $ \triangle AEO$ is a right triangle with right angle at $ E$.

[b]Lemma 1[/b]: $ OA \perp EF$.
[i]Proof.[/i] $ \triangle OEA \cong \triangle OFA$ by congruent sides. Therefore, $ \angle EAO \cong \angle FAO$ and by SAS, $ \triangle EDA \cong \triangle FDA$. Therefore, $ \angle EDA \cong \angle FDA$ and since they are supplementary, they are right.

Now, $ \triangle GDO \sim \triangle ACO$ since they both have a right angle and share common angle $ \angle COD$. Thus,
\[ \frac {AO}{CO} \equal{} \frac {GO}{DO},
\]
or $ AO \cdot DO \equal{} CO \cdot GO$. But since $ \triangle AEO$ is a right triangle, $ AO \cdot DO \equal{} EO^2 \equal{} r^2$ Therefore, $ CO \cdot GO \equal{} r^2$ and $ GO \equal{} \frac {r^2}{CO}$, which does not depend on $ A$. $ \blacksquare$
[/hide]


In the number 34,5$ t$8, $ t$ is the tens digit. What value of $ t$ makes the number a multiple of 11?

Using divisibility rules for $ 11$, we have $ 8\plus{}5\plus{}3\equal{}4\plus{}t \implies t\equal{}12 \implies t\equal{}\boxed{1}$.

[quote="gaussintraining"]Using divisibility rules for $ 11$, we have $ 8 \plus{} 5 \plus{} 3 \equal{} 4 \plus{} t \implies t \equal{} 12 \implies t \equal{} \boxed{1}$.[/quote]
lol how can t=12 and t=1
maybe $ 8\plus{}5\plus{}3\equiv 4\plus{}t \mod11$ you meant :)


algebra
polynomial
quadratics
special factorizations
product of roots
sum of roots
quadratic formula

$1)$  iIf $x^2+ax+bc=0$ and $x^2+bx+ca=0$ have a common root,then show that the other roots will satisfy the equation $x^2+cx+ab=0$

$2)$  The real solution to the system:
            $x+y+z=0$
            $2xy-z^2=0$ is given by?

$3)$   The least value of the polynomial $2x^2-2xy+5y^2+5$ is?

$4)$    The greatest value of the ploynomial $-5x^2-5y^2+8x-6y-1$ is?

I myself do not have the answers to these qustions,sop i wanted to check my answers and please give your solutions as well :D

[hide="2"]
since $x+y=-z$ squaring both side $\implies x^2+y^2+2xy=z^2$ $\implies x^2+y^2=0$ so $x=y=0$ and hence $z=0$
[/hide]

[hide="3"]
completing the square to get
$2(x-\frac{y}{2})^2+\frac{9y^2}{2}+5\geq 5$
[/hide]

[hide="4"]
again , completing the square
$-5(x-\frac{4}{5})^2-5(y+\frac{3}{5})^2+4\leq 4$
[/hide]

completing square is like this

$x^2+2x+10=(x+1)^2+9$ . As for number 3 , it should be $5$ since the minimum occur when $x=y=0$ . What does your book gives ?  :ninja:

[quote="shadysaysurspammed"]Shyong according to my book,the answer to the third one is wrong,and what's "completing the squares?"[/quote]
Completeing the squares is how the quadratic formula can be proven.  When you want to complete a square, you take the constant term to the other side of the equal sign.  You then take the middle term, divide it by two, raise it to the power of two, and then add it to both sides.  You can then factor the left hand side with the highest power in a fashion such as $(a+b)^2$ or $(a-b)^2$.  You then take the square root of both sides and solve for x.

EDIT: Completing teh squares can be used in many different ways.  An example would be: [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=357749#p357749]My Proof[/url]

Masoud Zargar

[quote="scorpius119"]Here's a way to do #1:

[hide]If x is a root of both, subtract the equations to get (a-b)x+c(b-a)=0, so x=c or a=b. Assuming the polynomials are distinct, a is not equal to b and x=c. From product of roots,
other root of 1st=[b]b[/b], other root of 2nd=[b]a[/b]
(x-a)(x-b)=x^2-(a+b)x+ab. From sum of roots in 1st, -a=b+c so c=-(a+b). Substituting,
[b](x-a)(x-b)=x^2+cx+ab[/b], giving the result.[/hide][/quote]

Thats very nice.

[quote="shadysaysurspammed"]My book says put x=y=-1/4 :D[/quote]

That gives you the minimum = $5\frac{5}{16}$ but $x=y=0$ gives you minimum = $5$ . So guess who is smaller  :roll:


inequalities
number theory
inequalities solved

Let $a,b,c,d,$ be real and $ad-bc = 1$. Show that $Q = a^2 + b^2 + c^2 + d^2 + ac +bd$ $\not= 0, 1, -1$

[quote="Rushil"]Let $a,b,c,d, $ be real and $ ad-bc = 1$. Show that $Q = a^2 + b^2 + c^2 + d^2 + ac +bd $ $ \not= 0, 1, -1 $[/quote]

In fact, this is not a number theory problem but an inequality problem, and actually a very easy one: We always have $Q>1$, since

$Q-1=Q-\left(ad-bc\right)=\left(a^2+b^2+c^2+d^2+ac+bd\right)-\left(ad-bc\right)$
$=\frac12\left(\left(a+c\right)^2+\left(b+d\right)^2+\left(b+c\right)^2+\left(a-d\right)^2\right)>0$

(since squares are always $\geq 0$, and not all of the four expressions a + c, b + d, b + c and a - d can simultaneously become 0, since this would yield a = b = c = d = 0, contradicting ad - bc = 1).

  Darij


Let P be a cubic monic polinomyal with real coefficients and with 3 real roots.
let Q(x)=x 2 +x+2004. it is Known that P(Q(x)) doesnt have real roots. Prove that P(2004)>1/64

This is unexpectedly easy, isn't it?

We just use the fact that $\Delta<0$ for each of the equations $Q(x)-x_i=0$, where $x_i$ are the roots of $P$. We get $2004-x_i>\frac 14$, and we multiply these together.

exactly. I tought the same, it seemed hard because of the source and because of the conditions, however it was very very easy!


We have 2 right triangles. $\triangle{PQR}$ has sides $QR$ on the y-axis and another side $PR$ on the x-axis; its hypotenuse, $PQ$, is 12 units long. $R$ is the intersection point of sides $PR$ and $QR$; $\angle R=90^{\circ}$. We then have another $\triangle{PRS}$ that $\angle S=90^{\circ}$.  The hypotenuse of $\triangle PRS$ is $PR$, which is stuck to $\triangle PQR$.  What are the lengths of $PQRS$?

[img]http://www.boxedexe.tamranda.com/images/image.JPG[/img]

Masoud Zargar

[hide]Alright so if we draw a diagram, or do it algebraically we obtain that since the side PR of the right trangle PQR, we get PS^{2}+SR^{2}+RQ^{2}=PQ^{2}.  From here we get that, we recognize this analagous to the equation of euclidian distance in three dimensional space.  So PS=PQcos{\alpha}, SR=PQcos{\beta}, RQ=PQcos{\gamma}, and PQ=\sqrt{PS^{2}+SR^{2}+RQ^{2}}.  This is as far as I have done so far.  I wonder if someone could finish this off?  I have to go, but I might come back with a solution.[/hide]

I agree with Iversonfan2005, it seems more information is need for the problem to be solved.

I will attach a diagram for brevity's sake.

EDIT: Diagram attached.:)

[quote="Iversonfan2005"][quote="DGS"]I agree with him, it seems more information is need for the problem to be solved.[/quote]

Yes, because wouldn't you need an additional side? I could put in any numbers which satisfy pythagorus with hyp of $12$[/quote]

Actually you can't, because if you place a different value than it's supposed to be, the value for $\angle S$ will change.

I don't see how you can get exact answers. for example, I could have $QR = 6$ and $PR = 6\sqrt{3}$ and then $SR=6$ and $PS=6\sqrt{2}$.  OR, I can say $QR = 3$ and $PR = 3\sqrt{15}$ and then $SR=3\sqrt{6}$ and $PS=9$

I can easily assign arbitrary values that will satisfy the diagram.  Are you sure you accidentally forgot to include some piece of information?

Yes I agree with you all, there are no integer values found and if they are arbitary then you can fit any in. :?

This isn't homework, but my teacher just gave it to me.  I believe it's missing some info too.  The only thing I can do is measure the angle iff the diagram is exact, which may be on the paper.  I measured $\angle QPR$ and got $30^{\circ}$.  I'll just say it is missing some info, however, if the diagram is exact, we can figure out angles and solve it with trigonometry.  Now, let's solve if $\angle QPR$ is $30^{\circ}$...the rest should be easy.

You need one more side. It can be any side, but you will need one more in order to solve the problem. :)

Let's say I have solved for $QR$ and $PR$.  How would I go about solving for $PS$ and $RS$?
EDIT: mvm, it's solved.


calculus
derivative
induction
algebra
binomial theorem

[quote="Meatman"]Prove ${ (fg)^{(n)} = {{C_n}^0}f^{(n)}g^{(0)} + {{C_n}^1}f^{(n - 1)}g^{(1)} + ... + {{C_n}^{n - 1}}f^{(1)}g^{(n - 1)}} + {{C_n}^{n}}f^{(0)}g^{(n)}$[/quote]

Counter example: $ n=1, f=1, g=\pi$.

I think you mean

$ (f+g)^n=\binom{n}{0}f^ng^0+\binom{n}{1}f^{n-1}g^1+\cdots+\binom{n}{n-1}f^1g^{n-1}+\binom{n}{n}f^0g^n$.

You are basically saying "Prove the [[Binomial Theorem]]."

[hide="hint"]We know that there are terms in the form $ constant \cdot f^j g^{n-j}$, we just need to find $ constant$.[/hide]

This should be moved to the Intermediate forum.

[quote="Farenhajt"]You didn't get it, 1=2. Exponents in parentheses denote derivatives, so what needs to be proven is Leibnitz's formula for the derivative of the product of functions.[/quote]

 :wallbash_red:

I'll shut up.

[quote="Meatman"][quote="SimonM"][quote="Meatman"]Prove ${ (fg)^{(n)} = {{C_n}^0}f^{(n)}g^{(0)} + {{C_n}^1}f^{(n - 1)}g^{(1)} + ... + {{C_n}^{n - 1}}f^{(1)}g^{(n - 1)}} + {{C_n}^{n}}f^{(0)}g^{(n)}$[/quote]

I'm not going to prove it (I have before, so don't see the point in doing it again), but the general idea is induction in a similar way to binomial[/quote]

If you've allready proved it at this forum, could you give me a link?[/quote]

Nope, it was a homework assignment


[quote="missloverly"]i need the solution to 16x16x2-8x+1=0
or
16x multiplied 16x squared minus 8x plus 1 = 0[/quote]
$16x(16x)^{2}-8x+1=0$
$256x^{2}-8x+1=0$
Now, we use the quadratic formula to find $x=\frac{1\pm i\sqrt{15}}{64}$ which are complex roots. I'm curious to know where you got this problem from because it is not classroom math. :)
Edited

[quote="missloverly"]i need the solution to 16x16x2-8x+1=0

or

16x multiplied 16x squared minus 8x plus 1 = 0[/quote]
$16\left(16x^{2}\right)-8x+1=0$

$256x^{2}-8x+1=0$

Use the quadratic formula: $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$x=\frac{8\pm\sqrt{64-1024}}{512}=\frac{1\pm i\sqrt{15}}{64}$

sorry...im just wondering...does i represent the negative square root?? just want to make sure, as im in 6th grade,and we have not learned this yet...:blush:

i is imaginary number.
without using i, i dont remmeber how they presented the answer

You can simply leave an answer as something such as $\sqrt{-15}$, if you do not want to use $\imath$.

Also, all the previous solutions appear to be incorrect on my understanding of the problem  :maybe:

$16x(16x^{2})\neq256x^{2}$
$16x(16x)^{2}\neq256x^{2}$

Just my thought, but I could still be interpreting the problem incorrectly.

So, to the original poster, do you mean
$16x(16x^{2})-8x+1=0$

or

$16x(16x)^{2}-8x+1=0$?

$i$ is defined mathematically as $\sqrt{-1}$.  We factor out the square root of -1, and thus are left with a positive radicand.

[quote="b-flat"]
So, to the original poster, do you mean
$16x(16x^{2})-8x+1=0$

or

$16x(16x^{2})-8x+1=0$?[/quote]

The way it's written up there, that's what it would mean.  I'm not sure quite how to go about solving $256x^{3}-8x+1=0$ though, unless you just graph it on a calculator...

I thought maybe some type of factorization thing, since we have $2^{8}$ and $2^{3}$ as coefficients, but I haven't come up with anything yet...

I just thought I'd let you know that tt's possible to find the answer (or atleast an aproximate answers) using Newton's Method, but that would require some calculus. I'd just use a calculater.

Yeah, I was meaning without using calculus.  It should be fairly easy with Newton's method...

[quote="b-flat"]So, to the original poster, do you mean
$16x(16x^{2})-8x+1=0$

or

$16x(16x^{2})-8x+1=0$?[/quote]
Unless I'm going blind, those are the same... :maybe:

Look above that - it was a small typo on my part  :blush:

$16x(16x)^{2}-8x+1=0$

or

$16x(16x^{2})-8x+1=0$?


i want to  find the shortest route for a mailboy to take.
Let's say he starts at the origin (0,0) but has to go to the houses (x_1,y_1) , (x_2,y_2), ... (x_n, y_n)
and then he returns back to (0,0) he takes straight line paths between the points.

My issue with this problem is testing all possible cases. Can Anyone help??

If you're using C++ and you want an easy implementation, look up the next_permutation function.

Otherwise, there are multiple ways to generate the permutations by yourself. One of the simpler ways would be to do it recursively: at each number try each index that has not already been used, and continue for the next number. For example this might be how it would work for the permutations of [1, 2, 3]:

First, we try to place a 1 at the first location:
1 _ _
Then we move on to the second location. We've already used a 1 so we can't use it again, so we then place a 2 there.
1 2 _
Then we do the same for the third, for which we can only place 3.
1 2 3
This is the first permutation. Next we backtrack to 2 numbers filled in.
1 2 _
We then try the next possible value for the second number, namely 3.
1 3 _
Then we can only place 2 in the third position.
1 3 2
This is the second permutation. Our recursion then backtracks twice, taking us back to
1 _ _
Then we use a 2 for the first position.
2 _ _
We continue in this manner until we've found all six permutations.

This gives you an O(n * n!) solution, where n is the number of houses. In addition, using dynamic programming it's possible to obtain a solution to this problem that is O(n^2 * 2^n) in complexity. In fact, there may even be a better solution involving the fact that the points are coordinates.

Not built in, no. Though its possible to write one yourself (I've got a version I wrote which I use every so often).

I suggest you use the second method above, as its probably what you are meant to do. Just keep a boolean array saying which houses you've visited, and call a recursive function which loops through each possible unvisited house, sets visited to true, recursively calls itself, sets visited back to false and tries to next house, etc.

I'm pretty certain that in $ \mathbb R^2$ you can do much better than full search. Unfortunately, the only relevant result I know of is that of Barvinok and it concerns maximizing the length of the route rather than minimizing it (you can find it on [url=http://www.math.lsa.umich.edu/~barvinok/papers.html]his webpage[/url], look for the article with the words "travelling salesman" in it).

guess what I found a really easy way to do it :

I know that there are n! different permutations  of a string with n different objects and each time I find one permutation I increment my counting variable.

I find my permutations by using  generator.nextInt(n); and using  a couple tests
There is a small probability that my algorithm won't work or will take an extremely long time since I am using the Random class but it seems to always work


[quote="marin.bancos"]For $ a,b$ positive real numbers, prove that
$ \left.\left(\frac {a}{b} \plus{} 1\right)^2 \plus{} \left(\frac {b}{a} \plus{} 1\right)^2\geq \frac {(3a^2 \plus{} b^2)(3b^2 \plus{} a^2)}{(a \plus{} b)^4} \minus{} 4\frac {(a \plus{} b)^2}{a^2 \plus{} b^2}\right.$[/quote]
I think your inequality must be
$ \left(\frac {a}{b} \plus{} 1\right)^2 \plus{} \left(\frac {b}{a} \plus{} 1\right)^2 \ge \frac {\color{red}{16}\color{black}(3a^2 \plus{} b^2)(3b^2 \plus{} a^2)}{(a \plus{} b)^4} \minus{} \frac {4(a \plus{} b)^2}{a^2 \plus{} b^2}.$

Sorry for my typo! You are really speedy!

[quote="marin.bancos"]For $ a,b$ positive real numbers, prove that
$ \left.\left(\frac {a}{b} \plus{} 1\right)^2 \plus{} \left(\frac {b}{a} \plus{} 1\right)^2\geq 16\cdot\frac {(3a^2 \plus{} b^2)(3b^2 \plus{} a^2)}{(a \plus{} b)^4} \minus{} 4\cdot\frac {(a \plus{} b)^2}{a^2 \plus{} b^2}\right.$[/quote]
I think this proof is not nice and your proof is nicer (I am sure of that), but I like it because in the proof, we only use one tool - the AM-GM Inequality.

The inequality is equivalent to
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{4}{a^2\plus{}b^2} \ge \frac{16(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{(a\plus{}b)^6}.$
By the AM-GM Inequality, we have
$ (a\plus{}b)^4\equal{}[(a^2\plus{}b^2)\plus{}2ab]^2 \ge 8ab(a^2\plus{}b^2).\;\;\;\; (*)$
Therefore,
$ \frac{16(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{(a\plus{}b)^6} \le \frac{2(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{ab(a^2\plus{}b^2)(a\plus{}b)^2}.$
It suffices to prove that
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{4}{a^2\plus{}b^2} \ge \frac{2(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{ab(a^2\plus{}b^2)(a\plus{}b)^2},$
or
$ \frac{(a^2\plus{}b^2)^2}{a^2b^2}\plus{}4 \ge \frac{2(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{ab(a\plus{}b)^2}.$
Now, we see that
$ \begin{aligned} (3a^2\plus{}b^2)(3b^2\plus{}a^2)&\equal{}3(a^2\plus{}b^2)^2\plus{}4a^2b^2 \equal{}3(a^2\plus{}b^2)^2\plus{}2ab(a\plus{}b)^2\minus{}2ab(a^2\plus{}b^2)\\ &\equal{}(a^2\plus{}b^2)(3a^2\plus{}3b^2\minus{}2ab)\plus{}2ab(a\plus{}b)^2,\end{aligned}$
hence
$ \frac{2(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{ab(a\plus{}b)^2}\equal{}\frac{2(a^2\plus{}b^2)(3a^2\plus{}3b^2\minus{}2ab)}{ab(a\plus{}b)^2}\plus{}4.$
Thus, the above inequality can be written as
$ \frac{(a^2\plus{}b^2)^2}{a^2b^2}\plus{}4 \ge \frac{2(a^2\plus{}b^2)(3a^2\plus{}3b^2\minus{}2ab)}{ab(a\plus{}b)^2}\plus{}4,$
or equivalently,
$ \frac{a^2\plus{}b^2}{a^2b^2} \ge \frac{2(3a^2\plus{}3b^2\minus{}2ab)}{ab(a\plus{}b)^2}.$
Using the inequality $ (*)$ again, we get
$ \begin{aligned} \frac{2(3a^2\plus{}3b^2\minus{}2ab)}{ab(a\plus{}b)^2} &\equal{}\frac{2(a\plus{}b)^2(3a^2\plus{}3b^2\minus{}2ab)}{ab(a\plus{}b)^4}\\ & \le \frac{2(a\plus{}b)^2(3a^2\plus{}3b^2\minus{}2ab)}{8a^2b^2(a^2\plus{}b^2)}. \end{aligned}$
Therefore, it is enough to check that
$ \frac{a^2\plus{}b^2}{a^2b^2} \ge \frac{2(a\plus{}b)^2(3a^2\plus{}3b^2\minus{}2ab)}{8a^2b^2(a^2\plus{}b^2)},$
or
$ 4(a^2\plus{}b^2)^2 \ge (a\plus{}b)^2(3a^2\plus{}3b^2\minus{}2ab),$
which is true because by the AM-GM Inequality, we have
$ (a\plus{}b)^2(3a^2\plus{}3b^2\minus{}2ab) \le \frac{1}{4}[(a\plus{}b)^2\plus{}(3a^2\plus{}3b^2\minus{}2ab)]^2 \equal{}4(a^2\plus{}b^2)^2.$

[quote="marin.bancos"]For $ a,b$ positive real numbers, prove that
$ \left.\left(\frac {a}{b} \plus{} 1\right)^2 \plus{} \left(\frac {b}{a} \plus{} 1\right)^2\geq 16\cdot\frac {(3a^2 \plus{} b^2)(3b^2 \plus{} a^2)}{(a \plus{} b)^4} \minus{} 4\cdot\frac {(a \plus{} b)^2}{a^2 \plus{} b^2}\right.$[/quote]
A simpler proof:

Like the above proof, we have to prove that
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{4}{a^2\plus{}b^2} \ge \frac{16(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{(a\plus{}b)^6}.$
By the AM-GM Inequality, we have
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{4}{a^2\plus{}b^2} \ge 2\sqrt{\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\right)\frac{4}{a^2\plus{}b^2}}\equal{}\frac{4}{ab}.$
Therefore, it suffices to prove that
$ \frac{1}{ab} \ge \frac{4(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{(a\plus{}b)^6},$
or
$ (a\plus{}b)^6 \ge 4ab(3a^2\plus{}b^2)(3b^2\plus{}a^2).$
Using the AM-GM Inequality again, we have
$ 4ab(3a^2\plus{}b^2)(3b^2\plus{}a^2) \le [b(3a^2\plus{}b^2)\plus{}a(3b^2\plus{}a^2)]^2\equal{}(a\plus{}b)^6.$
The proof is completed.

P/s: I am sorry for the first terrible proof above.

This is the shortest solution to this inequality! It's also my solution. Thank you very much!
The inequality is strong, because it states that
$ \frac{(a\plus{}b)^2}{4ab}\geq \frac{(3a^2\plus{}b^2)(3b^2\plus{}a^2)}{(a\plus{}b)^4}\geq 1$


calculus
integration
logarithms
limit
function
calculus computations

Let $ n\geq 2$ be natural number. Answer the following questions.

(1) Evaluate the definite integral $ \int_1^n x\ln x\ dx.$

(2) Prove the following inequality.

$ \frac 12n^2\ln n \minus{} \frac 14(n^2 \minus{} 1) < \sum_{k \equal{} 1}^n k\ln k < \frac 12n^2\ln n \minus{} \frac 14 (n^2 \minus{} 1) \plus{} n\ln n.$

(3) Find $ \lim_{n\to\infty} (1^1\cdot 2^2\cdot 3^3\cdots\cdots n^n)^{\frac {1}{n^2 \ln n}}.$

[quote="kunny"]Let $ n\geq 2$ be natural number. Answer the following questions.

(1) Evaluate the definite integral $ \int_1^n x\ln x\ dx.$

(2) Prove the following inequality.

$ \frac 12n^2\ln n \minus{} \frac 14(n^2 \minus{} 1) < \sum_{k \equal{} 1}^n k\ln k < \frac 12n^2\ln n \minus{} \frac 14 (n^2 \minus{} 1) \plus{} n\ln n.$

(3) Find $ \lim_{n\to\infty} \{1^1\cdot 2^2\cdot 3^3\cdots\cdots n^n)^{\frac {1}{n^2 \ln n}}.$[/quote]

[color=blue]
My solution

[hide="(1)"]

Let $ I\equal{}\int_1^n x\ln x\ dx$

Applying Integration by parts by taking  $ \ \ln x$  first function and  $ \ x$  second function we get,

$ \boxed{\boxed{I\equal{}\frac {n^2}{2}\ln n \minus{}\frac {1}{4}(n^2\minus{}1)}}$
[/hide]

[hide="(2)"]

Let $ \ f(x) \equal{} x\ln x$

As we know that $ f(1)\plus{}\int_{1}^{n}f(x)dx<\sum_{k \equal{} 1}^{n}f(k)<\int_{1}^{n}f(x)dx\plus{}f(n)$

$ \implies 0\plus{}\int_1^n x\ln x\ dx <\sum_{k \equal{} 1}^n k\ln k<\int_1^n x\ln x\ dx \plus{}n\ln n$

$ \boxed{\boxed{\implies\frac {1}{2}n^2\ln n \minus{} \frac {1}{4}(n^2 \minus{} 1) < \sum_{k \equal{} 1}^n k\ln k < \frac {1}{2}n^2\ln n \minus{} \frac {1}{4} (n^2 \minus{} 1) \plus{} n\ln n}}$
[/hide]

[hide="(3)"]

Let $ l\equal{}\lim_{n\to\infty} \{1^1\cdot 2^2\cdot 3^3\cdots\cdots n^n)^{\frac {1}{n^2 \ln n}}\equal{}\lim_{n\to\infty}\left(\prod_{k\equal{}1}^{n}k^k\right)^{\frac {1}{n^2 \ln n}}$

$ \implies l \equal{} e^{\lim_{n\to\infty}\frac {1}{n^2 \ln n}\left(\sum_{k \equal{} 1}^{n}k\ln k\right)}$

$ \boxed {\implies e^{\lim_{n\to\infty}\frac {1}{n^2 \ln n}\left(\frac {1}{2}n^2\ln n \minus{} \frac {1}{4}(n^2 \minus{} 1) \right)}\le l\le e^{\lim_{n\to\infty}\frac {1}{n^2 \ln n}\left(\frac {1}{2}n^2\ln n \minus{} \frac {1}{4} (n^2 \minus{} 1) \plus{} n\ln n\right)}}$

$ \implies e^{\frac 12}\le l \le e^{\frac 12}$

$ \boxed{\boxed{\implies\lim_{n\to\infty} \{1^1\cdot 2^2\cdot 3^3\cdots\cdots n^n)^{\frac {1}{n^2 \ln n}}\equal{}e^{\frac 12}}}$
[/hide]
[/color]


linear algebra
matrix
linear algebra unsolved

Let $ A\in M_{n}\mathbb{(R)}$ such that $ a_{ii} > 1$ with $ \forall i\in$ [b]{1,2,...,n}[/b]
 and : $ \sum_{i \neq j}{a_{ij}^{2}} < 1$ .
Prove that : A is invertible .
[b] Love Math !!! Love Math !!! Love Math !!! Do math forever  :P [/b]

Write $ A\equal{}D(I\plus{}E)$ where $ D\equal{}diag(a_{11},\ldots,a_{nn})$ ($ \det D > 1$ by assumption), and  $ E\equal{}D^{\minus{}1}B$ where $ B\equal{}A\minus{}D$ is the matrix $ A$ without the diagonal ($ \parallel{}B\parallel{}<1$, and hence $ \parallel{}E\parallel{}<1$, by assumption.)
Then we may define $ A^{\minus{}1}: \equal{}(I\minus{}E\plus{}E^2\minus{}E^3\plus{}E^4\minus{}\ldots)D^{\minus{}1}$ which is clearly an inverse to $ A$.


algebra
polynomial
quadratics
algebra unsolved

A polynominal is given:  $ W(x)\equal{}x^{4}\minus{}3x^{3}\plus{}5x^{2}\minus{}9x$
Find all pairs $ a,b$ such that $ W(a)\equal{}W(b)$.
($ a$ and $ b$ are integers and $ a\neq b$)

$ W(a)\equal{}W(b)$ $ \Leftrightarrow$ $ (a\minus{}b)[(a\plus{}b)(a^2\plus{}b^2)\minus{}3(a^2\plus{}ab\plus{}b^2)\plus{}5(a\plus{}b)\minus{}9]\equal{}0$ $ \Leftrightarrow$
$ (a\plus{}b)(a^2\plus{}b^2\plus{}5)\equal{}3(a^2\plus{}ab\plus{}b^2)\plus{}9$ $ \Rightarrow$ $ \frac{3(a^2\plus{}ab\plus{}b^2)\plus{}9}{a^2\plus{}b^2\plus{}5} \in \mathbb{Z}$ $ \Leftrightarrow$ $ \frac{3(a^2\plus{}b^2\plus{}5)\minus{}15\plus{}3ab\plus{}9}{a^2\plus{}b^2\plus{}5}$ $ \Leftrightarrow$ $ a^2\plus{}b^2\plus{}5 | 3(ab\minus{}2)$ $ \Rightarrow$
$ a^2\plus{}b^2\plus{}5 | ab\minus{}2$ since $ a^2\plus{}b^2\plus{}5 >3$ .So $ a^2\plus{}b^2\plus{}5 \leq |ab\minus{}2| ,a,b \in \mathbb{Z}$ but this is impossible , therefore it's necessay that $ ab\minus{}2\equal{}0$ .and from here it's easy

Hmmm.... It seems to be correct but it is not.I have answers (but do not have complete proof)
Answers $ a, b$
$ (\minus{}1, 3), (3,\minus{}1), (1,2), (2,1)$

From your solution it is: $ (1,2), (2,1), (\minus{}1, \minus{}2), (\minus{}2,\minus{}1)$

We can´t say $ a^2 \plus{} b^2 \plus{} 5|3(ab \minus{} 2) \Rightarrow a^2 \plus{} b^2 \plus{} 5|ab \minus{} 2$ because $ a^2 \plus{} b^2 \plus{} 5 > 3$... We would need $ gcd(a^2 \plus{} b^2 \plus{} 5, 3) \equal{} 1$ to say it.

So, can anyone prove this correctly, please?

We have:
$ a^2 \plus{} b^2 \plus{} 5 | ab \minus{} 2$
So it can be $ ab\minus{}2\equal{}0$ or  $ a^2 \plus{} b^2 \plus{} 5 \equal{} ab \minus{} 2$ or $ 2(a^2 \plus{} b^2 \plus{} 5) \equal{} ab \minus{} 2$ or $ 3(a^2 \plus{} b^2 \plus{} 5) \equal{} ab \minus{} 2$......
But: it cannot be $ 2(a^2 \plus{} b^2 \plus{} 5) \equal{} ab \minus{} 2$ because $ a^2 \plus{} b^2 \plus{} 5 > ab \minus{} 2$.

So:
1. $ ab\minus{}2\equal{}0$
2. $ a^2 \plus{} b^2 \plus{} 5 \equal{} ab \minus{} 2$
From 1. it is only $ (1,2), (2,1)$
But how to prove 2.? How to solve $ a^2 \plus{} b^2 \plus{} 5 \equal{} ab \minus{} 2$  I know it is not hard, please can anybody tell me how to solve thos quadratic equation?

for part 2. when we have $ a^2\plus{}b^2\plus{}5\equal{}ab\minus{}2$ we add 2ab-4 to both sides and thus get =>
$ a^2\plus{}b^2\plus{}5\plus{}2ab\minus{}4\equal{}3ab\minus{}6\equal{}3(ab\minus{}2)$ =>
$ (a\plus{}b)^2\plus{}1\equal{}3(ab\minus{}2)$, but
$ (a\plus{}b)^2$ is congruent to 0 or 1 mod 3 hence the LHS expression is congruent to 1 or 2, but it needs to be congruent to 0 to suffice => no solution

Sorry -  $ a^2 \plus{} b^2 \plus{} 5 \equal{} ab \minus{} 2$ is not correct, my mistake

$ a^2 \plus{} b^2 \plus{} 5 \equal{} 3(ab \minus{} 2)$ - this is correct equation which I cannot solve. I really need help... Thanks.


The number [i]m[/i] is such that each pair of the numbers 24, 42, [i]m[/i] has the same highest common factor and each pair of the numbers 6, 15, [i]m[/i] has the same least common multiple. Find [i]m[/i].


geometry
circumcircle
geometric transformation
homothety
incenter
geometry unsolved

Dear Fang-jh, Mathlinker and Moderator,
I reformulate my problem because I omit one hypothesis:
let ABC be a triangle, 0 the circumcircle of ABC, 1 the A-mixtilinear incircle of ABC, P the point of contact of 1 with 0,
X, Y the incentres wrt triangles PAB, PAC,
D the meet point of AX with the perpendicular to AY at Y,
and 2 the circumcircle of the triangle PXY.
Prove: D is on 2.
Any reference, synthetic proof?
The moderator can delete my first email. Sorry
Sincerely
Jean-Louis

Similarly, let $ E$ be the intersection of a perpendicular to $ AX$ at $ X$ with $ AY.$

Internal bisectors $ AX, AY$ of angles $ \angle PAB, \angle PAC$ cut the circumcircle $ (O)$ again at $ U, V,$ internal bisectors $ PX, PY$ of angles $ \angle APB, \angle APC$ at $ K, L,$ and internal bisectors $ BX, CY$ of angles $ \angle ABP, \angle ACP$  at $ M, N.$

$ \angle KCN \equal{} \frac {\angle ACP \minus{} \angle ACB}{2} \equal{} \frac {\angle BPC}{2} \equal{} \angle UCP$

$ \Longrightarrow$ $ CU, CN$ are isogonals WRT the angle $ \angle PCK,$ $ PU \equal{} KN,$ $ PUNK$ is isosceles trapezoid. $ \triangle UXB$ is isosceles with $ UX \equal{} UB$ $ \Longrightarrow$

$ \angle UXP \equal{} \angle UPX \equal{} \angle UPK \equal{} \angle NKP$

$ \Longrightarrow$ $ KN \parallel XU \equiv AX$ and similarly, $ LM \parallel YV \equiv AY.$ $ MN$ is a diameter of $ (O)$ $ \Longrightarrow$ perpendicular to $ KN$ at $ K$ cuts $ LM$ at $ M \in (O)$ and perpendicular to $ LM$ at $ L$ cuts $ KN$ at $ N \in (O).$

The mixtilinear A-incircle $ (1)$ touches $ AB, AC$ at $ Q, R$ and $ AQ \equal{} AR.$ $ P, Q, K$ are collinear and $ P, R, L$ are collinear due to similarity of $ (1), (O)$ with external similarity center $ P$ and their paralel tangents at $ Q, K$ and at $ R, L.$ $ X, Y$ are feet of internal bisectors $ AX, AY$ in the $ \triangle APQ, \triangle APR.$

$ \frac {PQ}{PK} \equal{} \frac {PR}{PL}$ and $ \frac {PX}{XQ} \equal{} \frac {AP}{AQ} \equal{} \frac {AP}{AR} \equal{} \frac {PY}{YR}$  $ \Longrightarrow$ $ \frac {PX}{PK} \equal{} \frac {PY}{PL}$

As a result, the pentagons $ PDXYE \sim PNKLM$ are centrally similar with similarity center $ P.$ Since $ PNKLM$ is cyclic, inscribed in the circumcircle $ (O)$ of $ \triangle PKL,$ so is $ PDXYE,$ inscribed in the circumcircle $ (2)$ of $ \triangle PXY.$

[hide="Solution"]
Apply an inversion with arbitrary radius around $ A$, mapping any point $ Z$ to $ Z'$.  Since the $ A$-mixtilinear is internally tangent to the circumcircle of $ \triangle ABC$, $ AB$, and $ AC$, its inverse is tangent to $ AB'$, $ AC'$, and $ B'C'$ externally, making it the $ A$-excircle of $ \triangle AB'C'$ (notice that this gives another proof for how the external center of homothety between the incircle and circumcircle and the Nagel point of a triangle are isogonal conjugates).  Now, $ X'$ is the excenter of $ \triangle AB'P'$ and $ Y'$ is the excenter of $ \triangle AC'P'$, where $ P'$ is the tangency point of the $ A$-excircle of $ \triangle AB'C'$ and $ B'C'$.  Let this circle meet $ AB'$ at $ M$, the $ A$-excenter and incenter of $ \triangle AB'C'$ be $ K$ and $ I$ respectively.  Now, since $ \angle AYD \equal{} 90$, we see that $ D'$ is the foot of the perpendicular from $ Y'$ to $ AX'$.  Since $ AY'$ and $ AX'$ are the respective bisectors of $ \angle P'AC'$ and $ P'AB'$, we see that $ \angle Y'AD' \equal{} \frac {\angle BAC}{2} \equal{} \angle KAM$ and $ \angle Y'D'A \equal{} 90 \equal{} \angle KMA$, meaning that $ \triangle Y'AD'\sim \triangle KAM$, so a spiral similarity maps $ \triangle D'AM\sim \triangle Y'AK$.  Notice that $ Y'$ lies on the exterior bisector of $ \angle AC'B'$, so $ K$, $ Y'$, and $ C'$ are collinear.  Thus,\[ \angle D'MA \equal{} \angle AKY' \equal{} \angle IKC' \equal{} \angle IB'C' \equal{} \angle P'MA\]so $ M$, $ D'$, and $ P'$ are collinear.  Now, since $ P'X'$ and $ P'Y'$ are exterior bisectors of $ \angle AP'B'$ and $ \angle AP'C'$ respectively, we have that $ \angle X'P'Y' \equal{} 90 \equal{} \angle X'D'Y'$, so $ X'D'P'Y'$ is cyclic, so $ XDPY$ is cyclic. [/hide]

<XDY = π/2 - <XAY = (π - α)/2.
<XPA = <XPB = γ/2 and <YPA = <YPC = β/2, so <XPY = (β+γ)/2 = (π-α)/2.
<XDY = <XPY => XYPD is cyclic.

Beautiful Problem!!! It combines the mixtilinear incircle configuration and the incircle configuration from IMO SL 1999 G8.

[i]Problem: Let $ ABC$ have circumcircle $ \omega$ and the $ A$-mixtilinear incircle of $ ABC$ touches $ \omega$ at $ P$. $ X$;$ Y$ are the incenters of $ ABP$;$ ACP$. $ D\in AX$ such that $ DY\perp  AY$. Then $ XYPD$ is cyclic.
[/i]

Notation: $ X_2;Y_2$ are midpoints of small arcs $ AB$;$ AC$. The $ A$-mixtilinear incircle touches $ AB$;$ AC$ at $ X_1$;$ Y_1$. $ B_1$; $ C_1$ are midpoints small arcs $ BP$;$ CP$. Let $ AX$ intersect the circumcircle of $ \triangle PXY$ again at $ D'$. It is clear that $ X\equal{}X_2P\cap AB_1$; $ Y\equal{}Y_2P\cap AC_1$.

Lemma 1: $ PX_2: PY_2\equal{}AX_2: AY_2$. Proof: well-known (consider $ \triangle AX_1X_2\sim\triangle PAX_2$ and $ X_1Y_1\parallel X_2Y_2$).

Lemma 2: If $ AI$ intersects the circumcircle of $ ABC$ at $ A'$, then $ BA'\equal{}CA'\equal{}IA'$. Proof: well-known.

Now $ X_2X: Y_2Y\equal{}X_2A: Y_2A\equal{}PX_2: PY_2$ by lemma 1 & lemma 2. From this relation, we get $ XY\parallel X_2Y_2$ (1). So $ XYP\sim X_2Y_2P$, so the homothety, $ \mathcal{H}$ at $ P$ that maps one triangle maps one circumcircle to another. Thus $ \omega$ and the circumcircle of $ \triangle XYP$ are tangent at $ P$ (2).

Let $ Q$ be on the circumcircle of $ \triangle ABC$ so $ QX_2\parallel XD$. Now $ \mathcal{H}(D')\equal{}Q$ since the image of $ D'$ is on the circumcircle of $ ABC$ and it lies on the parallel to $ XD$ through the image of $ X$, $ X_2$.

Claim: $ QY_2\perp AC_1\qquad (3)$. Proof: the angle between the lines $ AB_1$ and $ QY_2$ is $ \frac{1}{2}\cdot (\widehat{B_1Y_2}\plus{}\widehat{QA})$. [hide="easy angle chasing"]

$ \bullet$ First, $ \widehat{B_1Y_2}\equal{}\widehat{B_1C}\plus{}\widehat{CY_2}\equal{}\frac{1}{2} \widehat{CP}\plus{}\frac{1}{2}\widehat{AC}\equal{}\frac{1}{2}\widehat{ACP}$.

$ \bullet$ Second, $ \widehat{BB_1}\equal{}\widehat{B_1P}\equal{}2a$, $ \widehat{AX_2}\equal{}\widehat{BX_2}\equal{}2b$. $ \angle QAB\equal{}\angle QAB_1\minus{}\angle BAB_1\equal{}\angle X_2QA\minus{}\angle BAB_1\equal{}b\minus{}a$. Then $ \widehat{QX_2}\equal{}\widehat{X_2B}\minus{}\widehat{QB}\equal{}2b\minus{}2(b\minus{}a)\equal{}2a$. So $ \widehat{QA}\equal{}2a\plus{}2b\equal{}\frac{1}{2}(\widehat{BP}\plus{}\widehat{AB})\equal{}\frac{1}{2}(\widehat{ABP})$.

So the angle between $ AB_1$ and $ QY_2$ is $ \frac{1}{2}\cdot\left(\frac{1}{2}\widehat{ACP}\plus{}\frac{1}{2}\widehat{ABP}\right)\equal{}\frac{360}{4}\equal{}90$, as desried.[/hide]

Since a homothety maps lines to parallel lines, $ \mathcal{H}^{\minus{}1}(AB_1)\perp D'Y$ and $ \mathcal{H}^{\minus{}1}(AY) \parallel AY$. Line $ AY\equal{}AB_1$  so $ D'Y\perp AY$. It follows that $ D\equiv D'$ since they are on $ AX$ and $ DY\perp AY$. This finishes the claim!


You have a group of 90 students.

In year 1, you want to split them into 2 equal groups
Year 2, 3 groups (as equal as possible)
Year 3, 4 groups
Year 4, 5 groups
Year 5, 6 groups

****You want students to be with students they haven't been in a group with in the previous years (as much as possible)

What fromula could you use?

f(x) =  90/(x+1)

Gives you number of people in a group each year as a function of x, the year.

I applaude you for misunderstanding of the year! I think Marc means is how to keep students apart who have been in the same class some previous year as best as possible...

Yes.  Seamusobyle is correct.  how can you keep students apart who have been in the same class some previous year as best as possible?

I don't think there's any way to address the problem of separating 90 people "as much as possible" except to look at the problem step by step.  (divide into 2 groups, then try to divide into 3 groups, etc.)  Here's a problem I just made up that relates to your original question:

There are some number of people in a club.  The members of the club know how to play a game that can be played between any number of people $\ge 2$.  Next week they will play only games with 2 people, and the week after they will play only games with 3 people, and so on.  Each person will play exactly one game each week, and no two people will play the game with each other twice.  Derive a formula relating

the number of weeks $w$ they plan to play

with the number of people $p$ they need to play for that number of weeks.

I don't know if this is possible... but give it a try. :roll:

I think the original question is pigeonhole

Sorry I don't understand what a "pigeon holed"  question means.  I am starting to realize that this problem may be more difficult than I thought.  I realize that you can't avoid that some students will be in the same groups as other students they have been with previously.  The only way I knoe how to work on this problem is to create the lists and then move students around manually by eye, and try to separate them  from year to year as best as possible.  Some else suggested I look at:
Quasirandom sequences, Hammersly Point Sets, van der Corput Sequences and Uniform Distribution Theory.  I will start to research these and see if they help.
Another suggestion was to look into "Numerical Recipes in FORTRAN:  the art of scientific computing, 2nd ed.

Pigeon hole is just a maths principle :)

http://www.cut-the-knot.org/do_you_know/pigeon.shtml


trigonometry
geometry
inequalities
rearrangement inequality
inequalities unsolved

Prove that:

If $x,y,z > 0,\alpha ,\beta  \in \left( {0,2\pi } \right)$ and $\alpha  + \beta  > \pi $

then:
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{{\sqrt {y^2  + z^2  - 2yz\cos \left( {\alpha  + \beta } \right)} }}{{2yz\sin \left( {\alpha  + \beta } \right)}} \le \frac{{\sqrt {x^2  + y^2  - 2xy\cos \alpha } }}{{2xy\sin \alpha }} + \frac{{\sqrt {x^2  + z^2  - 2xz\cos \beta } }}{{2xz\sin \beta }}$

Actually geometric interpretation is quite obvious to consider (cosinus theorem to get rid of those squares, and e.g. 1/2 xy sin L interpret as area of the triangle) but I have no idea for this $\frac{1}{x}$. I'd be glad to see the solution and surely I don't think it is so easy  :)

Surely there is an elegant way of doing this...

Bbomb

well it is the first thing to think about the geometry but is it said that x, y, z are sides and $\alpha, \beta$ are angles. No! So how can u directly use geomtry? :?

Am i wrong?

Well tomorrow is my exam so I can’t think more on this problem. Well how did u bring that state using $\alpha, \beta < \pi$? But if that is the fact then I think th problem becomes easy now. I am giving the possible idea please some one solve it.

Have u saw the proof of Erdos-Mordell inequality? It is almost like that.

If ABC is a triangle in which P is a point da, db and dc are the perpendicular distance from P to BC, CA, AB respectively. Then PA+PB+PC >= 2(da+db+dc)

Say X, Y and Z are the foot of da, db and dc respectively.

Almost similar! So plz look at the steps and u will understand that these two problems are almost same. I think by changing some steps you can solve this problem. So here are the steps:

By corollary of Ptolemy’s theorem: PAsin<A = YZ

Use cosine law on YZ

Bring it to:
PAsinA >= dcsinB + dbsinC

Divide by sinA.

Add all the three inequalities.

Apply rearrangement inequality on x+1/x >= 2.

And get the Erdos-Mordell inequality.

So plz some1 solve it!

[quote="ondrob"]If this:
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq \frac{{\sqrt {y^2 + z^2 - 2yz\cos \left( {\alpha + \beta } \right)} }}{{2yz\sin \left( {\alpha + \beta } \right)}} + \frac{{\sqrt {x^2 + y^2 - 2xy\cos \alpha } }}{{2xy\sin \alpha }} + \frac{{\sqrt {x^2 + z^2 - 2xz\cos \beta } }}{{2xz\sin \beta }} $
what we are trying to prove (sign $\leq$ replaced) the problem can be remade to (with one more condition $\alpha,\beta < \pi$):

Let's have triangle $ABC$ with interior point $P$. Let distances from $P$ to vertices be $a,b,c$ and to sides $k,l,m$. Then prove:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq \frac{1}{2k}+\frac{1}{2l}+\frac{1}{2m}$[/quote]
yeah,i have seen the new problem you write before,but I haven't solve it,can you help me?

EDIT: After few minutes I realized that something stupid was written here  :blush:  :blush:


function
trigonometry
advanced fields
advanced fields unsolved

Show that $ \sum_{n\equal{}1}^{\infty}e^{\minus{}n}sin(nz)$ in analytic in the region A={z|-1<Im z<1}

Where is your difficulty? The series converges absolutely and uniformly on every compact subset of the strip and consists of analytic functions (that is, if you prefer analysis to algebra; if it is the other way around, just use Eiler's formulae for the sine and sum two resulting geometric progressions to get a closed form expression for the function in question).

Do we take the fact that since $ \minus{} 1 < Im z < 1$ then $ \left| \sin \left( {\it nz} \right) \right| \equal{} 1/2\, \left| {{\rm e}^{ {\it inz}}} \minus{} {{\rm e}^{ \minus{} {\it inz}}} \right| \leq 1$ and then $ \sum_{n \equal{} 1}^{\infty}e^{ \minus{} n}$ converges uniformly and is thus analytic for the region A? How is this shown exactly?


A vertical cylinder of radius $1$m is fixed on a horizontal, ideally smooth tabletop. There is a string of length $2$m attached to the bottom of the cylinder, while to the other end of the string a small disc, whose mass is $5$g and radius $2$cm, is fixed. The small disc is given a velocity $v_{0}$ which is perpendicular to the tightened string. The disc slides on the tabletop while the string is wound round the cylinder.

a) What is the velocity of the small disc when it hits the wall of the cylinder? (Give both the direction and the magnitude.)
b) What is the tension at this moment?

The magnitude of speed is $v_{0}$ the direction is just geometry,
the tension is $T=m\frac{v_{0}^{2}}{\sqrt{(R+r)^{2}-R^{2}}}$

Well, the only force exerted on the small disc is the tension in the rope. Because the direction of this force is perpendicular to the direction of the discs velocity it doesnt have work on the disc, so the magnitude of the velocity doesnt change. :D

Well, what I don't see is why the radius of circular motion is $\sqrt{(R+r)^{2}-R^{2}}$...

[quote="Thaakisfox"]the disc is touching the cylinder, so the length of the loose rope is that with pythagoras theorem... :D[/quote]

Which length do you mean?  :maybe:

You can reduce the process mathematically to $2$ circles.When the smaller circle is touching the bigger one,the circles are tangent.Let $P$ be the point from where the line is not touching the big circle,in contact moment,of course.Then you need to fiind $PO_{2}-r$ ,$O_{2}$ being the center of the smaller circle.The next thing you must do is put the circles in $xOy$ system for easy calculations and you put the condition that the ''free'' line(tangent to the big circle) passes through $O_{2}$.You fiind the ''wanted'' point on the big circle and after you fiind the distance you want.

I reached $l=\sqrt{r(R+2r)}-r$