inequalities
Euler
combinatorics unsolved
combinatorics

A connected graph has $1998$ points and each point has degree $3$. If $200$ points, no two of them joined by an edge, are deleted, show that the result is a connected graph.

I think the problem is "Prove that there exist such 200 points" , not for all independent 200 points.

terminology means: [i]"the technical or special terms used in a business, art, science, or special subject"[/i].

[quote="greentreeroad"][quote="BaBaK Ghalebi"]
[b]Lemma.[/b] let $ G$ be a trigraph such that the removal of any trivalent vertex disconnects $ G$. then $ G$ is planer. moreover $ G$ can be drawn in such a way that every vertex lies on the "outside" face; in other words, for any point $ P$ outside some bounded set, each vertex $ v$ of $ G$ can be joined to $ P$ by a curve which does not intersect any edges of $ G$ (except at $ v$).[/hide][/quote]

I don't quite understand the terminologies in this lemma. Could anyone explain to me "planar" and "outside face" and "bounded set"?[/quote]

Hi,planar here means [b]ichnography[/b]

Wow...A nice problem indeed,Thank you for the brilliant solution,BaBaK Ghalebi :lol:

[quote="Little Gauss"]Wow, Strange solution with planer graphs. Is it official solution?
I think my solution is basically same as yours, using $ 10K\le 1998$.
What is the word "terminology" mean?
There are many trigraph problems in russia. Is there are some theories about trigraph like above lemma?[/quote]
Hi friend,
Could you please present your nice solution? :lol:

[quote="BaBaK Ghalebi"]yeah you're right,the official problem is the following one:

[b]there are 1998 cities in Russia, each being connected (in both directions) by flights to three other cities. any city can be reached by any other city by a sequence of flights. the KGB plans to close off 200 cities, no two joined by a single flight. show that this can be done so that any open city can be reached from any other open city by a sequence of flights only passing through open cities.[/b]

[hide="Solution."]
we begin with some terminology. define a [b]trigraph[/b] to be a connected undirected graph in which every vertex has degree at most $ 3$. a [b]trivalent[/b] vertex of such a graph is a vertex of degree $ 3$. in this wording, the problem becomes: we have a trigraph $ G$ with $ 1998$ vertices, all of which are trivalent. we want to remove $ 200$ vertices, no two of which are adjacent, such that the remaining vertices stay connected.
we remove the vertices one at a time. suppose we have deleted $ k$ of the $ 1998$ vertices, no two of which are adjacent, such the trigraph $ G'$ induced by the remaining vertices is connected. we will show that if $ K < 200$, we can always delete a trivalent vertex of $ G'$ such that the graph remains connected. this vertex cannot be adjacent in $ G$ to any of the other $ k$ deleted vertices, because then its degree in $ G'$ would be less than $ 3$. hence repeating this $ 200$ times gives us the desired set of vertices.

[b]Lemma.[/b] let $ G$ be a trigraph such that the removal of any trivalent vertex disconnects $ G$. then $ G$ is planer. moreover $ G$ can be drawn in such a way that every vertex lies on the "outside" face; in other words, for any point $ P$ outside some bounded set, each vertex $ v$ of $ G$ can be joined to $ P$ by a curve which does not intersect any edges of $ G$ (except at $ v$).

[i]Proof.[/i] we induct on the number of trivalent vertices of $ G$. if $ G$ does not contain any trivalent vertices, then $ G$ must be a path or a cycle and the claim is obvious. so suppose $ G$ contains $ n\geq 1$ trivalent vertices and that every trigraph with fewer trivalent vertices can be drawn as described. if $ G$ is a tree the claim is obvious, so suppose $ G$ contains a cycle; let $ v_1,\ldots ,v_k$ $ (k\geq 3)$ be a minimal cycle. let $ S \equal{} \{v_1,\ldots ,v_k\}$, and let $ T \equal{} \{i\mid v_i \textrm{ is trivalent }\}$. ($ T$ cannot be empty, because then no $ v_i$ would be connected to a vertex of degree $ 3$.) for each $ i\in T$, let $ w_i$ denote the third vertex which is adjacent to $ v_i$ (other than $ v_{i \minus{} 1}$ and $ v_{i \plus{} 1}$), and let $ S_i$ be the set of vertices in $ G$ which can be reached from $ w_i$ without passing though $ S$. (for $ i\not\in T$, let $ S_i \equal{} \emptyset$.) we claim that the sets $ S,S_1,\ldots ,S_k$ partition the vertices of $ G$. first, note that if $ v$ is a vertex of $ G$ not in $ S$ then there is a shortest path joining $ v$ to a vertex $ v_i$ of $ S$; the penultimate vertex on this path must be $ w_i$, so $ v\in S_i$. now suppose that $ v\in S_i\cap S_j$ for some $ i\neq j$; then $ v_i$ and $ v_j$ are trivalent and there exist paths $ w_i\to v$, $ w_j\to v$ which do not pass though $ S$. we will show that there exists a path from every vertex of $ G \minus{} \{v_i\}$ to $ v$ which does not pass though $ v_i$. for $ k\neq i$ there is a path $ v_k\to v_j\to w_j\to v$; if $ w\in S_k$ for $ k\neq i$, then there is a path $ w\to w_k\to v_k\to v$; if $ w\in S_i$, there is a path $ w\to w_i\to v$. since $ S\cup S_1\cup\ldots\cup S_k \equal{} G$, we have shown that the graph obtained from $ G$ by deleting $ v_i$ is connected,a contradiction, as $ v_i$ is trivalent. therefore $ S_i\cap S_j \equal{} \emptyset$ for $ i\neq j$. obviously $ S_i\cap S$ is empty for all $ i$; hence $ S,S_1,\ldots ,S_k$ partition the vertices of $ G$. let $ G',G_1,\ldots ,G_k$ be the induced subgraphs of $ S,S_1,\ldots ,S_k$ in $ G$, respectively. by construction, the only edges in $ G$ which are not in one of the graphs $ G',G_1,\ldots ,G_k$ are the edges $ v_iw_i$ for $ i\in T$. now $ G_i$ is a trigraph with fewer than $ n$ trivalent vertices, since at least one of the $ n$ trivalent vertices in $ G$ is in $ S$. hence by the inductive hypothesis, we can draw each $ G_i$ in the plane in such a way that every vertex lies on the outside face. since $ v_1,\ldots ,v_k$ was a minimal cycle, there are no "extra" edges between these vertices, so the graph $ G'$ is a $ k$-cycle. now place the vertices of $ S$ at the vertices of a small regular $ k$-gon far from all the graphs $ G_i$; then we can draw a curve joining each pair $ v_i,w_i$. it is east to check that this gives us a drawing of $ G$ with the desired properties. $ \square$

now suppose we have removed $ k$ vertices from $ G$, no two of which are adjacent, such that the trigraph $ G'$ induced by the remaining vertices is connected, and suppose that removing any trivalent vertex of $ G'$ disconnects the graph; we must show $ k\geq 200$. by the lemma, $ G'$ is planner. we will call a dace other than the outside one a "proper face". let $ F$ be the number of proper faces of $ G'$; since $ G'$ has $ 1998 \minus{} k$ vertices and $ 2997 \minus{} 3k$ edges,

$ F\geq 1 \minus{} (1998 \minus{} k) \plus{} (2997 \minus{} 3k) \equal{} 1000 \minus{} 2k$.

we now show that no two proper faces can share a vertex. observe that each vertex belongs to at most as many faces as its degree; thus vertices of degree $ 1$ lie only on the outside face. no two proper faces can intersect in a vertex of degree $ 2$, or that vertex would not lie on the outside face, contracting the lemma. if two proper daces intersected in a trivalent vertex $ v$, each face would give a path between two of $ v$'s neighbors, so removing $ v$ would not disconnect the graph, by an argument similar to that of the lemma.
since each proper face contains at least $ 3$ vertices and no two share a vertex, we have $ 3F\leq 1998 \minus{} k$. combining this with the previous inequality gives

$ 3000 \minus{} 6k\leq 3F\leq 1998 \minus{} k$

so $ 1002\leq 5k$ and $ k\geq 200$, as desired.[/hide][/quote]

Another approach for the lemma:

Planars recall us the efficiency of Kuratowski theorem!!!That's what we need :D
$ d(v)\le 3$ so $ G$ contains no $ K_5$
$ G$ contains no $ K_{3,3}$ because $ G$ is a trigraph such that the removal of any trivalent vertex disconnects $ G$
Which completes the proof of lemma :D

by the way,Merry Christmas!!!

[quote="JBL"]Kuratowski's theorem says that every nonplanar graph contains $ K_{3, 3}$ or $ K_5$ as a [i]graph minor[/i], NOT necessarily as a subgraph.  Having all vertices of degree 3 or less does not rule out the possibility of a $ K_5$ minor.[/quote]

Very sorry,you're right :lol:
I missed the correct form of Kuratowski's theorem,anyway thanks for correction!!!
But I think there still exists an elementary proof for the lemma.I am just talking about the sketch of it:
pick out a cycle from the gragh,and any vertex having three adjacent points must have a segment which will split the gragh if being eliminated
From this fact,we are easy to get:if we regard every cycle as a special vertex(keep the vertices and segments if they don't belong to any cycle),we will get a tree :) (just to verify there can't be two or more cycles share a vertex or segment,and there can't be two or more ways connecting two cycles)
This surely ensures the gragh is a planar
Am I right now,my dear friend? :D

[quote="JBL"]Kuratowski's theorem says that every nonplanar graph contains $ K_{3, 3}$ or $ K_5$ as a [i]graph minor[/i], NOT necessarily as a subgraph.  Having all vertices of degree 3 or less does not rule out the possibility of a $ K_5$ minor.[/quote]

Happy new year :lol:
Firstly,sorry for my continuous posts in this topic,but I reall like this problem which is more than beautiful :lol:
I think Kuratowski theorem still solves this problem,um......Here we go:
Still use the fact [b]any vertex having three adjacent points must have a segment which will split the gragh if being eliminated [/b]
If the graph has a $ K_{3, 3}$ or $ K_5$ as a [i]graph minor[/i],we will find a vertex $ v$,$ d(v)\equal{}3$ and every segment of $ v$  is contained in a cycle,which contradicts the known fact...

[quote="Little Gauss"]Wow, Strange solution with planer graphs. Is it official solution?[/quote]
I don't think it's very strange. It's not hard to explicitly characterize the set $S$ of (connected) graphs $G$ disconnected by the removal of any $3$-vertex if you look at "small cases" (such graphs with few $3$-vertices). When you have none, $G$ is just a path or cycle, and otherwise, the condition on $3$-vertices $v$ is just that if $v_1,v_2,v_3$ are its neighbors and $S_i$ is the component containing $v_i$ when you remove $v$, we don't have $S_1=S_2=S_3$ (note that for distinct $i,j$, we either have $S_i=S_j$ or $S_i,S_j$ disjoint). So $S$ is just the set of graphs with vertex-disjoint minimal cycles that become trees when you contract all minimal cycles (you can inductively prove this either by looking at some $v$ with, WLOG, $S_1=S_2$, or some minimal cycle in $G$---the idea is the same in both cases). The Euler characteristic just simplifies the final calculation a bit.

Can someone explain what this lemma means:

Lemma. let $ G$ be a trigraph such that the removal of any trivalent vertex disconnects $ G$. then $ G$ is planer. moreover $ G$ can be drawn in such a way that every vertex lies on the "outside" face; in other words, for any point $ P$ outside some bounded set, each vertex $ v$ of $ G$ can be joined to $ P$ by a curve which does not intersect any edges of $ G$ (except at $ v$).

Thanks.

Here is a different solution.

The problem essentially boils down to the following.

[b] Lemma. [/b] Fix some nonnegative integer $0 \le x < 200.$ In any connected graph with $1998 - x$ vertices, each of degree at most $3$, where there are $\frac{1998 \cdot 3}{2} - 3 \cdot x$ edges, there exists a degree $3$ vertex which is not a cut-vertex. In other words, there is some vertex of degree $3$ that we can remove, which preserves the connectivity of the graph.

[b] Proof. [/b] Firstly, we will show that no two cycles share an edge. Indeed, suppose that $avb \dots$ and $cvd \cdots $ were two different cycles. Then, if there are exactly three distinct vertices among $\{a, b, c, d\}$ then we could simply remove $v$ while preserving connectivity of the graph. This is enough to imply that no two cycles share an edge, since if they did we could follow the edge until the point that the cycles "split."

In fact, now we know that no two cycles even share a vertex, since if they did, that vertex would either have degree $4$ or the two cycles would share an edge from that vertex. However, neither is possible, since all vertices have degree $\le 3.$

Let $c$ be the number of cycles in our graph. We therefore have now that $c \le \frac{1998-x}{3},$ since any cycle has at least $3$ vertices. Note that if we remove one edge from each of them, then our graph becomes a tree. This gives us that $\frac{1998 \cdot 3}{2} - 3x - c = 1998-x-1,$ so solving for $c$ gives:
$$c = 1000-2x \le \frac{1998-x}{3},$$
i.e. $3(1000-2x) \le 1998 - x \Rightarrow 1002 \le 5x,$ contradiction.

$\blacksquare$

By the lemma, we can employ a greedy algorithm. Namely, in any given moment we will simply take a degree $3$ vertex in the remaining subgraph which isn't a cut-vertex, and add it to the list of cities to be removed. By the lemma, we can repeat this $200$ times, and so after doing so we have $200$ cities which satisfy the conditions of the problem.

$\square$

@meagain: The wording is not as it has to be. Two cycles can share an edges, but in this case we can remove a vertex this edge is incident with and still remain with a connected graph. For example, in your picture, we can remove the top vertex (or any other). That's, if two cycles share an edge, we are done. So, it remains to see the other case. I suppose, this was the intended meaning.


If the area of the triangle shown is 40, what is $ r$?
[asy]import graph; size(200); defaultpen(linewidth(0.8));

real xmin=0;
real xmax=12;
real ymin=0;
real ymax=10;

xlimits(xmin, xmax);
ylimits(ymin, ymax);

xaxis("$x$",0,12,EndArrow);
yaxis("$y$",0,10,EndArrow);
dot(Label("$O$",align=SW),(0,0));
draw((0,8)--(10,0));
dot(Label("$(0,8)$",align=E),(0,8));
dot(Label("$(r,0)$",align=S),(10,0));[/asy]

We have:

\[\frac{8r}{2} = 40\implies 8r = 80\implies \boxed{r=10}\]


[u][b]The author of this posting is : hungkhtn[/b][/u]
____________________________________________________________________

$x_1$, $x_2$, ..., $x_n$ are positive reals such that:

$\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_n} = 1$.

Find the maximum:

$P = \frac{x_1}{4+x_1^2}+\frac{x_2}{1+x_2^2}+...+\frac{x_n}{4+x_n^2}$.

[color=red][Moderator edit: Isn't there a typo here in the second denominator?][/color]

I can't solve this problem. I only solve another problem:

Find the maximum:

$P = \frac{x_1}{n-1+x_1^2}+\frac{x_2}{n-1+x_2^2}+...+\frac{x_n}{n-1+x_n^2}$.

[u][b]The author of this posting is : siuhochung[/b][/u]
____________________________________________________________________

[quote="Myth"][quote="hungkhtn"]$x_1$, $x_2$, ..., $x_n$ are positive reals such that:

$\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_n} = 1$.

Find the maximum:

$P = \frac{x_1}{4+x_1^2}+\frac{x_2}{1+x_2^2}+...+\frac{x_n}{4+x_n^2}$.[/quote]

We have already discussed this problem twice. Try to search...[/quote]

his resembles CWMO 2003 no.7 so much. I think same method can be used.

[u][b]The author of this posting is : hungkhtn[/b][/u]
____________________________________________________________________

[quote="hungkhtn"]$x_1$, $x_2$, ..., $x_n$ are positive reals such that:

$\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_n} = 1$.

Find the maximum:

$P = \frac{x_1}{4+x_1^2}+\frac{x_2}{1+x_2^2}+...+\frac{x_n}{4+x_n^2}$.[/quote]

problem 2 I had solved. I want to ask the first problem!
I think it's difficult because when it's maximum $x_i$ not equa!


[quote="4everwise"]Three numbers, a, b, and c form an arithmetic progression. Adding 1 to a or c results in a geometric progression. Find the value of each of the variables.

A hint: [hide][hide]Set up systems of equations and work with them[/hide][/hide][/quote]

$\{a, a+k, a+2k\}=\{a, b, c \}$. $b=a+k$, $c=a+2k$.

$\{a+1, (a+1) \cdot r, (a+1) \cdot r^2\}=\{a+1, b, c \}$

I don't feel like doing it, nvm  :lol:

it's too hard for me, and i'm too lazy too do it.

It only involves setting up 3 equations and solving for one variable. Then the rest are simple.


Calculate: $2{}^{2}{}^{5}*5{}^{2}{}^{8}$
Express your answer in scientific notation

BTW, this is sprint so no calculators. I need a fast way to do this. Took me way too long.


[quote="Lovasz"]Let three nonnegative numbers. Prove the inequality:
\[{2a^{3}+abc \over (2a+b)(2a+c)}+{2b^{3}+abc \over (2b+c)(2b+a)}+{2c^{3}+abc \over (2c+a)(2c+b)}\geq{a+b+c \over 3}. \]
[/quote]
${2a^{3}+abc \over (2a+b)(2a+c)}+{2b^{3}+abc \over (2b+c)(2b+a)}+{2c^{3}+abc \over (2c+a)(2c+b)}\geq{a+b+c \over 3}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{2a^{3}+abc}{(2a+b)(2a+c)}-\frac{a}{3}\right)\geq0\Leftrightarrow\sum_{cyc}\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}\geq0.$
Let $a\geq b\geq c.$
Hence, $\frac{c(c-a)(c-b)}{(2c+a)(2c+b)}\geq0$ and $\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}+\frac{b(b-a)(b-c)}{(2b+a)(2b+c)}=$
$=(a-b)\left(\frac{a(a-c)}{(2a+b)(2a+c)}-\frac{b(b-c)}{(2b+a)(2b+c)}\right)=$
$=\frac{(a-b)^{2}(2a^{2}b+2ab^{2}+a^{2}c+b^{2}c-c^{2}a-c^{2}b+5abc)}{(2a+b)(2a+c)(2b+a)(2b+c)}\geq0.$ :)
It's equivalent to Tiks's problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66660


I thought there was a mathjam on friday for the last three USAMO problems. Is the transcript not finished, is there a problem with it, or am i halucinating about the whole thing? (This last part IS a possibility... I've had a hectic week.)

You may be hallucinating, but not about the Math Jam.  It ran as scheduled on Friday.  We had a little trouble with the transcripter.  It has been fixed now, and the transcript should be up on Monday.


Does anyone have any tips for countdown rounds?  I need to improve accuracy.

Slow down! Think through your answer quickly, and see if it makes sense.

[quote="mthiyaga"][quote="indianamath"]don't second-guess yourself. it will slow you down.[/quote]
I totally agree with indianamath. I came in 1st in CD both at chapters and states and that is what I did. Never second-guess yourself!![/quote]

never second guess yourself, but always make sure your doing the problem right.  i cant tell u how many times someone in the cd round did a problem wrong and was knocked off

[quote="mchoi815"]i have pretty good speed, but my accuracy drops because i answer too quickly.  i wanna know what can help me raise my accuracy.[/quote]

Hey Mario, I know you! (you won't respond to my pms  :( ) but I played you in chess at the MD state competition.

Ive never been to a countdown but heres a tip:

My friend alan got hyped up on caffiene before the countdown and he was one question away from first but instead he said something (i think the answer was 591 but he said 519). seriously. no caffiene before countdown...or mathcounts for that matter.

[quote="Fanatic"]At chapter I was hyped up on sierra mist and gummy bears and I DOMINATED... it's a person by person case.[/quote]

i thought u were on a sugar high during state CD too. no offense, u didn't exactly dominate (not that i'm saying i did, cuz i did the exact opposite of dominate if u know what i mean)...

No, I was sugar high too early at state. I got hyper, hit my sugar high, then deflated at state. Slowly but surely I lost my energy.

Who was there at my chapter... mathcrazed. He can tell you, that was domination [for a chapter, anyways... would be sucky at nats]

You may want to try reading differently.  Read a lot of books without saying the words in your head like most people read.  Sort of just see the words and make sure you understand them, and make your eyes go really fast.  I can get maybe 700-1000 words per minute like this.  It'll take your accuracy down a little at first, but then you can slow down in your super fast reading mode and still read faster than everyone else while still understanding more.

If you already do this, you shouldn't be worrying.  You'll knock out most peole anyways.  It can't hurt to practice a lot, too.

I'd also skip the sugar.  The round is too long for a sugar high to last all the way.  You'll go dead near the end, and trust me, that's not good.  I should know.

Also, try to skip over the unimportant words and look at the numbers. That's not saying skip all the words, but words such as modifiers, articles, etc. just slow you down.

I agree.  Of course, don't go too fast and mistake "radius" for diameter of 2 minutes per page for 2 pages per minute, as I did at state.  Then again I was so happy to get first that I really didn't care about countdown since it's not mandatory in Missouri.

Also, a risky move on long problems is to find the actual question like "How many pieces of candy does Bob have?" BEFORE reading the description part (i.e. everything else) It'll take you up a few seconds because you know what you're being asked, but the risk is it may take you a few seconds to actually FIND the question, so it's not that great.  On short questions it's useless though.[/b]

A little off topic, but personally I think it's easier to do the problems in the hot seat. When you're in the audience, you can't see the screen as well and therefore sorta have to listen to the moderator. I still got all but ONE question before the competitiors did at states.

jorian

lol... fanatic... u didn't dominate, u just buzzed in early and got lucky cause they were really easy, short questions... lol, the only person who had a chance against you died in the first round because they took 1-4, 5-8, 9-12, 13-16.  it sucked :D

lol get on gmail you crazy asian :p.

dude if i remember correctly,

I got question #1 [long question]
I got question #2 [long question]
I answered question #3 but missed it [long question], then tom answered and missed it with the same answer, then you got it
You got question #4 -- it was a freaking 1 sentence question. and you didn't give the answer in the right form so i was pissed :p.
I got question #5 [short question]

So really I got the long questions. #1 was the isosceles right triangle, which was about 5 sentences long. I forget #2, but i'm pretty sure it was pretty long. #3 was like 4 sentences long and involved the numbers 1500, 2000, and 400. #4 was like what is 2/5 of 40% or something really simple like that. #5 was how many ways can you rearrange 2992 so it was really simple.

I said it was domination for a [b] chapter [/b] [hence easy] contest and it would've sucked for nationals... check 4 posts ago.

mathcrazed: remember the kids that were asking for autographs? I was like o_O. And you were SOOOO pissed lol because you could've beaten me.

Edit: bleh, re-read this and it sounds like i'm attacking mathcrazed. the vocal tone that goes along with this post is joking/laughing.

in chapter, i got the practice one about oranges right when it flashed on the screen (and everyone's like "WOAH, THIS KID'S GONNA WIN"...and i did).

jorian


linear algebra
matrix
algebra
polynomial
number theory
greatest common divisor
superior algebra

Let A and B two nxn real matrices with real elements such that 2(AB)^2=BA+I, where I is the unity matrix. Prove that if n is odd then I-AB is singular.
   Harazi

AB=BA=-(1/2)*I [b](1)[/b]
2(AB) 2=(1/2)*I
BA+I=(1/2)*I
So 2(AB) 2=BA+I.

det(I-AB)=det(I+(1/2)*I)=(3/2) 2 <>0, so I-AB is not singular.

To prove that (1) is possible take A=I and B=-(1/2)*I.

Oh, soooo....ryyyyyyyyy! I was so...( Valentin doesn't let me say how).
 The correct statement is :
  If n is odd and 2AB=(BA)^2+I, where A and B are real nxn matrices, then I-AB is singular, but not lonely.

another little detalil, because I might be missin' something here: what is a lonely matrix? or was it just a game of words? :)

It's just a joke.
Singular but not lonely  :D

okay I have a solution. suppose BA-I is not singular, thus not lonely.
Then from the relation substracting 2I we obtain 2(AB-I)=(BA)2-I and taking this to det we get, as det AB-I = det BA - I <>0 that 2n = det BA + I.

let us denote by f the char. polynomial of BA (and AB in the same time since it's the same). We have that f(1) <> 0, also f(-1) <>0.
also from AB=1/2 (BA)2 + 1/2I we obtain that BA verifies also the polynomial f((x2+1)/2) which means that the minimal polynomial of BA which we will denote by m(x) will divide the GCD of f(x) and f((x2+1)/2), and this means that if x is a root of f, then also (x2+1)/2 is a root of f.

now as n is odd, we must have a real root of f (which has n as degree), say f(r)=0. If r>1 then the sequence xn = (xn-12+1)/2 with x1=r gives us an infinitely increasing sequence of roots of f contradiction.

further more, if r<-1 then (r2+1)/2 > 1 is also a root and we do it in the same way as above. as -1 and 1 cannot be roots, as supposed in the beginning, r can only lie in the interval (-1,1), but then the sequence xn defined as above is still increasing ( increasing not non-decreasing) and it can never reach 1, because (r2+1)/2 < 1 <=> -1 < r < 1 so again we have constructed an infinite number of roots of f, which is a contradiction.

therefore our loneliness assumption was wrong and we are done.
congrats harazi, very nice problem.

[quote="amfulger"]Then we notice that if x_0=a>1, then x_n is a decreasing sequence (I will not stress the fact that the monotony is strict because it seems you don't use this word  :) ) and lim x_n=1 and x_n>1 for all n, so x_n takes infinitely many values[/quote]

Seems I made a mistake here. As Valentin pointed out, x_n must be increasing.
Thanks to Valentin for removing the case a=-1. It was giving me some incredible headache.

I have another question: what if n is even? I cannot answer it, but I believe that for any even n we can find counterexamples.

it's a hard ball as still m(x) | f( (x^2+1)/2) - f(x) and also (x^2+1)/2-x | f( (x^2+1)/2) - f(x) which means that 1/2 (x-1)^2 | f( (x^2+1)/2) - f(x) and that means that for degree 2 the problem is still valid, as m(x) can only have 1 as a root, which means that 1 is a proper value.

[quote="amfulger"]Let X<>0 be an eigenvector of BA which corresponds to an eigenvalue a. Then (BA) 2 X=a 2 X and we get
(AB)X=((a 2 +1)/2)X, so (a 2 +1)/2 is an eigenvalue of AB.

It is known that AB and BA have the same eigenvalues, so (a 2 +1)/2 is an eigenvalue for BA.
We can repeat the process and obtain a sequence x_0=a, x_(n+1)=f(a), where f(x)=(x 2 +1)/2, of eigenvalues for BA.
Becasue BA has finitely many eigenvalues, x_n must have finitely many values.

Since the characteristic polynomial of AB has real coefficients and odd degree, it has a real root, so there exists a real eigenvalue of AB which is also eigenvalue for BA.

If we draw the grapf of f (i.e. y=f(x)), we notice that it is tangent to the first bisector (y=x) at x=1.

If x_0=a=1, then BA has 1 as eigenvalue, so det(BA-I)=0, so is singular.
It is known that det(AB-I)=det(BA-I), so AB-I is also singular, so BA-I is not lonely  :D  :D  :D
Suppose x_0<>1.
Then we notice that if x_0=a>1, then x_n is a decreasing sequence (I will not stress the fact that the monotony is strict because it seems you don't use this word  :) ) and lim x_n=1 and x_n>1 for all n, so x_n takes infinitely many values, so a<1. Similarly if a is in (-1,1) we get an increasing sequence x_n, convergent to 1 and different from 1, so we get a similar contradiction. (The contradictions too are not lonely  :D).
If a<-1, then f(a)>1 and goto the case a>1.
The only problem left is a=-1.
If a=-1 then AB has -1 as eigenvalue, so there exists an eigenvector X such that ABX=-X and X<>0. So (BA) 2X=-3X, so X is a eigenvector for (BA) 2 corresponding to the -3 eigenvalue.
So det((BA) 2 +3I)=0, so  i\sqrt 3 or -i \sqrt 3, but because A,B are real matrices we must have  \pm i \sqrt 3 are both eigenvalues for BA.
If we take g(x)= \sqrt (2x-1), we get same as for f that if y_k is an eigenvalue for AB then y_k+1={+ or -}g(y_k) is an eigenvalue for AB.
This should lead to a contradiction, but for now I'm not sure.[/quote]

Sorry but why AB and BA have the same eigenvalues?

because the have the same char. polynomial. Indeed, if one of A and B is invertible, say A, then
det ( AB+xI ) = det A(B+xA-1) = det A det (B+xA-1) = det (B+xA-1) det A = det (BA + xI) so they have the same char. polynomial.

if none of them is invertible, then we might have a problem. But don't worry, we'll pass the river :D

obivously there are an infinity of reals y such that det ( A+yI) <>0 thus we can apply the lemma above for B and A+yI, and obtain that

det (AB + By+xI) = det (BA+By+xI) for all real x, and for an infinity of reals y. But this equality is in fact an equality of polynomials in y, and because it's true for an infinity of variables, it is true for any real y, in particular y=0 thus det (AB+xI) = det (BA+xI) and we are done.

oh thx Vornicu. I thought about it today, and got if either one of A and B is invertible then AB and BA is the matrix representations of the same linear with respect to different ordered basis...


integration
real analysis
real analysis unsolved

Hello!:)

I have a problem with this:

Let $(E_n; n \in N)$ be a sequence of $\mu$- measurable sets and $m \in N$.
Let  $G$ be a set of all points, which are in at least $m$ sets of sequence $(E_n; n \in N)$
Proof that $G$ is measurable set and
$m* \mu(G) \leq \sum_{n=1}^{+\infty} \mu (E_n)$.

Thanks for any help. Regards!

For any finite subset $M$ of $N$, define $E_M = \bigcap_{n\in M} E_n$.
Then $G = \bigcup_{M} E_M$, where $M$ runs over all $m$-subsets of $N$. As a union of countably many measurable sets, $G$ is measurable.

lhs counts a point $m$-times if the point is in $G$, and does not count any point outside $G$. rhs counts a point as many times as the point is in $E_n$'s. so lhs $\leq$ rhs. More fancy way to say this is first say
\[ m\cdot\chi_G \leq \sum_n \chi_{E_n}  \]
and so
\[ lhs = m\int\chi_G \leq \int\sum_n \chi_{E_n} = rhs  \]


algebra
polynomial
floor function
geometry
parallelogram
linear algebra
matrix

Hello, we can discuss the admissions test now.

Also there was a typo on problem 10. The degree of the polynomial should have been 9. Nobody was penalized for not inferring the correct problem so there's no need to worry.

Anyway, I thought that there were a number of pretty cool problems...I cannot resist

[hide="5"] Let $ a \equal{} \{x\}$ and $ b \equal{} \lfloor x\rfloor$ then we have $ ab \equal{} \frac {4}{25}\cdot (a \plus{} b)^2$...[/hide]

[hide="6"] Make a parallelogram![/hide]

[hide="8"]Find the equality case and use that to do the appropriate AM-GM[/hide]

[hide="9"]Sum of squares... Comment: how can we come up with similar problems? Is there a geometric interpretation?[/hide]

[hide="10"]Define another polynomial such that $ Q(x)$ is some polynomial for $ x\equal{}1,2,...,9$. [/hide]

Also I'm not sure if an official solutions document is coming out but I will compile solutions that were written in this forum into a solutions document. So...don't write solutions like I did...haha

[hide="9"] The matrix of the linear transformation in question is orthonormal, hence preserves vector norm. [/hide]

[hide="6"]
You could compare areas after drawing in the cevian that goes through E.
[/hide]

For 10, I thought there was something wrong with the 2009...guess I was right :D[/hide]

[quote="ghjk"]#6 is killed easily by Laws of sines of two angles: AMB and AMC :)[/quote]

You don't need that. Its just one simple application of Menelaus.

[hide="crux move for 9"]Prove that the sum of the squares of the terms is invariant.[/hide]
I solved all but 6, 8, and 10. I can't do geometry that well, I fail at inequalities, and 10 was ugly.

I assure you that 10 has a nice solution :)

[quote="Zhero"]1-5 were largely trivial for me.

#6 was a pain for me. I first proved that angle AMC must be obtuse (that was the painful part.) After that, things worked out beautifully. I extended AM and took the point P such that AM = AP. Then, APC and AMB are congruent, yielding the desired result.

I wonder how others did #7. I took advantage of the fact that a number x is triangular if and only if 8x + 1 is a perfect square, and constructively showed that there are infinitely many such numbers (5050, 500500, 50005000, etc.)

For #8, the inequality, I ended up expanding and calc-bashing. I found the one extremum, showed it was a minimum, and it follows that that is necessarily the global extremum. I also showed that the function was positive at that value, so the desired inequality results.

#9... ah. All I noticed was that the sum remained the same, and it stopped there. That was clever, though. Is 20 the lowest integral upper bound, though? How would we compute the lowest possible upper bound (not necessarily integral?)

For #10, I tried doing what Altheman did, and failed. I ended up bashing it head-on; solved the system of equation. But then I got an unacceptable answer, so I gave up.

Could anyone post the full solution for 10?[/quote]

Heh, my solution wasn't too bad, basically just considering P(x)-x^3 and finding the remainder of x^2009 when divided by some polynomial using lagrange interpolation and then that happened to instantly solve the problem.

I'll post my solution when I have time.

BTW, I didn't actually apply to AMSP.

[quote="Zhero"]Hah, nice.

I figured it up to the remainder part. Then, I got stuck with the...what... Lagrange interpolation?

Can't wait for the full solution.[/quote]

Heh, you can find large remainders using Lagrange interpolation.

I think for #7 that I proved that 4999999.......95 worked for any number of nines. But I didn't know that 8x+1 was a square iff x was triangular...

For 7, what I did was show that as long as x was a power of 10, $ \frac{x(x\plus{}1)}{2}$ would be in the form 50...050...0 which has a digits sum of 10.

This is a really dumb question, but how do you do number 2?
$ 11111111111$ is divisible by $ 7$, but not $ 13$.
Can someone explain this problem please?

[quote="ghjk"]I don't see why x^2009 is wrong!! I meant from the given information, we still have 9 variables and 9 equations, which is clearly obvious to solve to get the coefficients(theoretical).Or you meant that if it has the term x^2009 , then it may have more than 9 roots?? But that still doesn't effect anything, does it??
@brightzu: On #2, you have to know the standard divisibility of 11. Then you need to have some clever brute-force that leads directly to the result. I don't remember what number I came up with though, but I believe  I used the right method.[/quote]
Maybe it's wrong because now they changed it to x^9?
If it had 2009 roots, you would need to account for them in the partial sums that are the coefficients.

[quote="serialk11r"]No no no, you DO NOT USE LAGRANGE WHATEVER ON NUMBER 10. It's supposed to be solvable by elementary methods, so we are immediately suspicious that the x^2009 may be a typo. It is now corrected apparently.
Solution:
Let x=1/n. Then we have for x=1, 1/2, 1/3...1/9, P(x)=x^3. Let Q(x)=P(x)-x^3. Then 1, 1/2, 1/3...1/9 are the roots of Q(x). Q(x)=2009(x-1)(x-1/2)...(x-1/9).
So for n=10, Q(1/10)+1/10^3=P(1/10)=
2009(1/10-1)(1/10-1/2)(1/10-1/3)...(1/10-1/9)+1/1000.[/quote]

Lagrange is elementary.

[quote="pythag011"]
Lagrange is elementary.[/quote]
Totally  :roll:

hey bobby shen, are you really coming again this year?

[quote="shelly32494"][quote="AIME15"]How many problems do you usually need to solve in order to be accepted into AMSP?

Oh, and I didn't really apply yet, I'm just wondering, considering the (small) amount of problems that I'm able to solve...[/quote]

It doesn't matter as much as you think it does..
i didn't solve very many problems and still got in.
as long as you have passion for math and it shows.
i think you'll be fine.
but i think i solved... 6-7 maybe lol. i dont really remember.[/quote]

I solved 6 (heh) and kinda messed up on one, and got in.
If I can get in, you def can.

Wow I didn't expect this...
I only did 6, because I was too lazy to write solutions until the last minute, so I only submitted 6 (solved all of them I think) since I was running out of time before the deadline :O

seriously how many questions you solve doesnt matter...

but you have to be able to solve at least SOME on the test to get in i suppose.
i mean there are people who solve all ten correctly and get in and etc,
but this camp's all about learning, and so should be the admission test.

there's three levels at the camp, so there will be people who solve only certain number of questions on the test.

keep in mind: SCORES DO NOT MAKE FRIENDS. PERSONALITIES DO.

i just hated how last year everyone kept posting their math competition scores and whatnot for introductions. that just annoyed the crap out of me.
why not get to know each other on personal level, rather than making prejudices about each other even before camp starts on each other's records and etc??

[quote="serialk11r"]Wait what are these levels you speak of :O
Does that mean I'm done for because I wasn't diligent enough on the test??[/quote]

Well, if you want to make advanced level all the problems except maybe 1/2 should be trivial for you... since most of the people in the advanced class last year solved all the problems really quickly...

I don't think that the admissions test makes a big difference in placing students.

Also

[b]Hello people, post your solutions. [/b]All it takes is posting your solutions files... I think that it would kind of stupid if I just post all the solutions that I wrote for these problems. Good solutions will be put in a 'official' solutions document.

[quote="pythag011"]
Well, if you want to make advanced level all the problems except maybe 1/2 should be trivial for you... since most of the people in the advanced class last year solved all the problems really quickly...[/quote]
I'm going to assume you got the wrong message. All the problems were trivial for me, but I was too lazy to write them. That's what "not diligent" meant.
Blargh so the solutions I wrote are here.

[quote="serialk11r"]Blargh so the solutions I wrote are here.[/quote]

Hmmm, now I have your phone number.

@Altheman: Can you please create threads for all the problems? Since I'm too lazy.

[quote="sooozyy"]
keep in mind: SCORES DO NOT MAKE FRIENDS. PERSONALITIES DO.
[/quote]
so true. ><

the "levels" are like classes by difficulty. there's "easy" "medium" "hard" classes.
but you can switch in or out of them at any time really. so it doesn't really matter what you got on your admissions test.


function
trigonometry
induction
algebra proposed
algebra

Let $f:R->R$ such that:
          a)$f(2x)=2(f(x))^2-1$  for all $x \in R$
          b)there exist a real number $x_0$ such that $f(2^nx_0) \leq 0$ for all number $n$
Determine the value of $f(x_0)$
I have the solution but i want to have the other solution.Can anyone help me? :)  :)

Since $f(x_0) \leq 0$ and $f(2x_0) \leq 0$ we deduce that $-\frac {\sqrt 2} 2 \leq f(x_0) \leq 0.$
Thus $f(x_0) = \cos (\theta)$ for some $\theta \in [\frac {\pi} 2 ; \frac {3 \pi } 4 ].$

Now, by induction and straightforward computations, we easily prove that for all $n$, we have :
$\frac {a_n\pi} {2^{2n-1} } \leq \theta \leq \frac {b_n \pi} {2^{2n}}$, where $a_1 = 1,b_1 = 3$ and $a_{n+1} = 4a_n+1,b_{n+1} = 4b_n-1.$

It is easy to deduce that $a_n = \frac {4^n - 1 } 3$ and $b_n = \frac {2^{2n+1} - 1 } 3.$

Thus, both $\frac {a_n} {2^{2n-1}}$ and $\frac {b_n} {2^{2n}}$ tend to $\frac 2 3$ as $n$ goes to $+ \infty$.
Thus $\theta = \frac {2 \pi } 3$ and $f(x_0) = - \frac 1 2.$

Pierre.


This a problem I have seen a while ago.I think it was from an Argentinian Olympiad, and it was supposed to be for 8,9 grades.
Despite that I think it is a great problem with and amazing and nice solution.

Alice writes down in the blackboard a $ 37$ digit number, with all digits $ >0$. Then Bob has to erase some (but not all) numbers of the blackboard and the number left has to be a multiple of $ 271$.Can Alice write a number so that Bob can´t achieve his goal?

Hope you enjoy it... I really did it :D

Daniel

[hide="Cute."]
There are two big hints in this problem - the number $ 271$ and the number $ 37$. You can start with either one.

If you ask yourself "why $ 37$?", you may notice that Alice uses non-zero digits of which there are $ 9$, so $ 37$ is exactly the number of digits you need in order to guarantee that $ 5$ of them will be identical. Now, how can Bob possibly use $ 5$ identical digits? At this point you notice that $ 271$ divides $ 11,111$ and you're done.

If you ask yourself "why $ 271$?", you may consider it important to see which numbers are multiples of $ 271$, and it's not a stretch to think about such multiples that have interesting patterns in base $ 10$. If you have a calculator, one good thing to do is look at the decimal expansion of $ 1/271$; even a relatively simple calculator will quickly reveal that this fraction has a surprisingly short period, which tells you that $ 271$ divides some short number of the form $ 99\ldots 9$ and again, you're almost there. Alternatively, if you're patient or have good memory for divisors of repunits, you may discover directly that $ 41 \times 271 \equal{} 11,111$.
[/hide]


Let $ f: R\rightarrow R$ and f is cont. satysfying
$ f(x)f(f(x)) \equal{} 1$ and $ f(1000)\equal{}999.$
$ f(500)\equal{}?$

[quote="Przemo"]Let $ f: R\rightarrow R$ and f is cont. satysfying
$ f(x)f(f(x)) \equal{} 1$ and $ f(1000) \equal{} 999.$
$ f(500) \equal{} ?$[/quote]
$ f(500)\equal{}\frac{1}{500}$

I think:
$ f(f(x))\equal{}\frac{1}{f(x)}$
Because $ f: R\to R\Rightarrow f(x)\equal{}\frac{1}{x}$

It does not work. How can you explain that $ f(1000)\equal{}999$.

Let $ D\equal{}\{y| y\equal{}f(x) \ for \ some \ x\in R \}$
We get if $ y\in D$, then $ f(y)\equal{}f(f(x))\equal{}\frac{1}{f(x)}\equal{}\frac 1y.$ Therefore $ 999\in D,1000\not \in D, 0\not \in D$. From $ y\in D$ we get $ \frac 1y \in D$
Because f(x) continiosly exist $ 999\le a\le 1000$, suth that $ D\equal{}[\frac 1a,a]$ or $ D\equal{}(\frac 1a, a)$. For both case $ 500\in D$. Therefore $ f(500)\equal{}\frac{1}{500}$.

[quote="Rust"]Let $ D \equal{} \{y| y \equal{} f(x) \ for \ some \ x\in R \}$
We get if $ y\in D$, then $ f(y) \equal{} f(f(x)) \equal{} \frac {1}{f(x)} \equal{} \frac 1y.$ Therefore $ 999\in D,1000\not \in D, 0\not \in D$. From $ y\in D$ we get $ \frac 1y \in D$
Because f(x) continiosly exist $ 999\le a\le 1000$, suth that $ D \equal{} [\frac 1a,a]$ or $ D \equal{} (\frac 1a, a)$. For both case $ 500\in D$. Therefore $ f(500) \equal{} \frac {1}{500}$.[/quote]
Thank Rust,yesterday,I was too busy

I have other probelm
Let $ f$ be a continuous function on $ R$ such that $ f(2010)\equal{}2009$ and $ f(x).f(f(f(f(x))))\equal{}1$
Determine the value of $ f(2008)$


quadratics
LaTeX
algebra
quadratic formula

These sound like homework problems.  Please do not post homework.

1) Just take square roots.

2) Multiply by $x^2-3x$

3) Multiply by $x^2-4$

Then use whatever quadratic method you want.

You could probably ask your teacher to help you with these problems. Some first hand help would be good for learning new concepts.

1. $x^2+4x+4=17$
this is given so

all we have to do is make right side or the left side of the equation equal zero and use quadratic formula to solve

2.
you can solve it algebraically
$(10-12)/x = -1$
so we get
$x=2$

3.
multiply by $x^2-4$

(can't explain much farther because of the time limit I have)

If you are looking for assistance in homework, check out this wesite:

http://www.sosmath.com/

Actualy, could someone go through number three quickly? I can't find the elegant way, just the ugly way.

Are you sure number 2 was latexed correctly?  :?  fourierseries hint didn't really follow the problem as written by the mod.

[quote="frt"]I think #3 should be:

$\frac{x}{x-2}-\frac{1}{x+2}=\frac{8}{x^2-4}$

This would make the problem much easier.[/quote]

Following frt's correction:

$\frac{x}{x-2}-\frac{1}{x+2}=\frac{8}{x^2-4}$

$\Rightarrow \frac{x(x+2) - (x-2)}{x^2 -4} = \frac{8}{x^2-4}$

$\Rightarrow x^2 + x - 6 = 0$ __________________ $ (x\neq \pm 2)$

$\Rightarrow (x+3)(x-2) = 0$

So $x = -3$ because $x=2$ is not permissible.


A ten-letter word being split up into five words? How does that work...?

If there hadn't been those restrictions, I would have thought of what the Mississippi acorn said when it grew up:

Gee, I'm a tree! :roll:

[quote="me@home"]Find a word that can be split up into 5 words which form a sentence, in order.
[hide="hint 1"]
The sentence is about geometry
[/hide]
[hide="hint 2"]
The word is 10 letters long
[/hide]
[hide="hint 3"]
The sentence has a one-letter word "adjective" which wouldn't appear in normal conversation (although not uncommon in math)
[/hide]
[hide="hint 4"]
It starts with a d and ends with a t
[/hide]
[hide="hint 5"]
In the sentence formed, there is a certain word which if I tell you (and its position) would basically give away the sentence and the word (arguably)
[/hide]

Like I said, I think this is way too vague but good luck to anyone who tries it...[/quote]
is it an abbreviation? the 1 letter word?

also, could u try the ciphertext code?

[quote="kstan013"][hide]determinant?[/hide][/quote]
Look at hint #2. :wink:

Okay, someone has gotten the brainteaser correct already! (Possibly via PM)
Not to spoil it for other people, I will keep the teaser going.
[hide="hint 6"]
hint 3 refers the to usage of the word "P" as an adjective
[/hide]

edit: 1500 posts!!

Because there is a duplicate letter, I chose to use a couple of hints together this way for my offer:

[hide]"d is a p point." OR "d is a p-point."[/hide]

Okay, I guess with two solutions now the problem is officially broken :D

Good job guys


[quote="maksmaks24"]If $ m,n \in N$  then
Prove that
$ \mid\frac {m}{n} \minus{} \sqrt {2}\mid\geq\frac {1}{(\sqrt {3} \plus{} \sqrt {2})n^2}$[/quote]
See here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=256579[/url]


For a element of N and a is the square of the product of four different prime numbers. Let k and n be divisors of a such that k divides n.
How many pairs of (k,n) are there with given conditions.

Davron

it's just the number of ways of choosing two quadruples $(a, b, c, d)$ and $(a', b', c', d')$, where all components are from {0, 1, 2}, such that $a \leq a'$, $b \leq b'$, etc. ...


does ne1 know a good source of taxicab geometry stuff that i could look at? i have a book but i dun think it's that useful....


Consider an isosceles triangle $ ABC$ $ (AB\equal{}AC)$ with angle $ \angle{A}\equal{}\frac{\pi}{6}$. Let $ D$ a point on side $ AC$ sush that $ \frac{BC}{AD}\equal{}\sqrt{2}$. Find the angle $ \angle{DBC}$.

Let $ BC\equal{}2a$, then by the law of cosine, we can find that $ AB\equal{}AC\equal{}(\sqrt2 \plus{} \sqrt6)a$, which means that $ AD\equal{}\sqrt6a$. Hencs in $ \triangle{ADB}$, by the Law of Cosine, we can find that $ BD\equal{}(1\plus{}\sqrt3)a$. Finally, in $ \triangle{CBD}$, apply the Law of Sine, we find that $ \sin{\angle{DBC}}\equal{}\frac{\sqrt3}{2}$. So $ \angle{CBD}\equal{}60^o$.

Take $P$ inside the triangle so that $\triangle BPC$ is rightangled and isosceles at $P$; obviously, $BP=AD, \widehat{ABP}=\widehat{BAC}$, hence $ABPD$ is an isosceles trapezoid. Since $AP$ is bisector of angle $\angle BAC$, it follows $BD$ bisector of $\angle ABC$, so $\angle DBC=\angle DBP+\angle PBC=15^\circ+45^\circ=60^\circ$.

Best regards,
sunken rock


Is given a trapezium with bases's length respectively 12cm and 6cm and with perpendicular diagonals. Find his area.

This problem is really easy, for the 6-7th grade in Romania

If we denote by $ A$ the area of the trapezium then:

$ A\equal{}\frac {9 \cdot(12\plus{}6)} {2}\equal{}81cm^2$

[quote="moldovan"]This problem is really easy, for the 6-7th grade in Romania

If we denote by $ A$ the area of the trapezium then:

$ A \equal{} \frac {9 \cdot(12 \plus{} 6)} {2} \equal{} 81cm^2$[/quote]
But,could you please explain why the height is $ 9$?

Both the problem as well as the solution given by moldovan are wrong, obviously, the formula used by moldovan gives the maximum value of the area, and that should be the requirement of the problem, to be found the maximum value of the area.
The diagonals being perpendicular, their common point lies on the semicircle diameter 12, and the triangle made by diagonal and this diameter has maximum height 6, consequently, the maximum height of this trapezoid cannot exceed 9!.

Best regards,
sunken rock

If the problem will be so easy ,I wouldn't post it here. The trapezium culd not be isoscele.

Sorry for my mistake, I was in hurry when I posted the solution.

If we denote by $ ABCD$ the trapezium, by $ O\equal{}AC \cap DB$ and by $ M,N$ the feet of the altitudes from $ O$ in the triangles $ AOB$ and $ DOC$ and, finally, $ OC\equal{}a$ and $ OD\equal{}b$ then we obtain that $ OA\equal{}2a$ and $ OB\equal{}2b$, $ ON\equal{}\frac {ab} {6}$ and $ OM\equal{}\frac {ab} {3}$ $ \Rightarrow A_{ABCD}\equal{}\frac {9ab} {2}$, where $ a^2\plus{}b^2\equal{}36$. The trapezium can be isosceles if $ a\equal{}b\equal{}3\sqrt{2}$

The area of trapezium will be 81cm2-the maximum and the min of area goes to 0 cm2!
May be the length of bases are respectively 12cm and 6cm and the diagonals are perpindicular,but depends also how are the angels that diagonals have(formed) with the bases of trapezium.If the difference of the angels are near to 90',then the area of trapezium goes to 0cm2.{}


geometry
3D geometry
analytic geometry
integration
induction
calculus
calculus computations

Can someone help me with this problem?:

We will define a cone in n-dimensions as a figure with a cross - section along its height [tex]X_n [/tex] that has a constant shape, but each of its dimensions is shrunk linearly to 0.

a)let D be a cone in [tex] R^n [/tex] with height h [tex] (ie. [/tex]   [tex] X_n    [/tex]    [tex] \epsilon [/tex] [tex]   [0, h]) [/tex] and let the volume of its cross-section at h=0 be [tex] V_o [/tex].  Find the volume of D in terms of [tex]V_o [/tex].

b)Find the volume of the region defined by [tex]|x_1| +...+ |x_n| \le r    [/tex] in [tex] R^n [/tex], using a)

For a) pick a coordinate system in which the apex is on the plane $x=0$ and the base lies on the plane $x=h.$ If we slice this at height $h,$ we get an object the same shape as the base but with all linear dimensions scaled by $\frac{x}h.$ Hence the $(n-1)$-measure of this slice is $\left(\frac xh\right)^{n-1}V_0.$

So the $n$-measure of the cone is

$\int_0^h\left(\frac xh\right)^{n-1}V_0\,dx=\frac1nV_0h.$

For b), consider only the portion of the figure in which all coordinates are positive.

Let $V_n(r)$ be the $n$-measure of this object.  Using a base along one of the coordinate hyperplanes (which is the $n-1$-dimensional version of the same object) we get that

$V_n(r)=\frac rnV_{n-1}(r).$

Hence by induction, $V_n(r)=\frac{r^n}{n!}.$

The original object is symmetric across all of the coordinate hyperplanes; taking that into account gives us an answer of $\frac{2^nr^n}{n!}.$


[quote="Charlie"]During IMO 2000 I heard of a 14-yr-old named David from Norway who had participated in IMO since 1997. I don't remember if I see him in 2001, but he isn't in Norwaian Team since 2002. Do you know where is he now?[/quote]

I was also in Tokyo and I'm sure he was in the Norvegian team. He also participated in Glasgow in 2002. In 2002 he got a gold medal and now he got a silver one.

Arne

I just checked the IMO 2000 and 2001 contestant lists that he's there. So he's the one who has participated the most IMOs? a 7-timer already, and will turn 8 next year (He's only 17) (Though the IMO 2002 and 2003 webiste really sucked, I don't even get to see the names of contestants.)

Another nice thing to look is that I see Yamanoto again. He's a contestant for Japan in 2000 and a guide for SCG this year.

I met David in Tokyo, it was the seventh and last time for him.. Though he already got many gold medals, he just got a silver one this year.

Now I wouldn't say he obtained "many" golds, just one or two I think ... but that's still brilliant of course, I'd never be able to obtain gold.

Btw, ool, what's your real name ? Were you in the French team in Tokyo ?
If so, we probably met each other there ... (I was in the Belgian team)

This is post number 3000 :D

:D 3000 posts wow - anyway Arne don't get your hops up because it not that hard to score gold - you can score gold by solving only the 4 easy / medium problems which are not so hard, and getting one extra point on questions no.3 and no.6  ... many contestants have done that in the past two years - so has David.

Even that will be too difficult for me ...

Ok, maybe it's not impossible to solve the four "easy" (?) and "medium" problems if you are well - trained but in Flanders we have almost no training ... don't forget about that ! Being well - trained is in my eyes still very important to do well at the IMO. You know, I live in Flanders, and the French speaking part of Belgium is Wallonia. The Wallonians have some training (at least 7 weekends each year + correspondence) but we've only two weekends ! So it's no surprise that the Wallonians did much better at the IMO (Wallonia got 47, Flanders got 23) but there will always be 3 Flemish guys and three Wallonian guys in the team ! You see how important training is ? I'm trying to do some things on my own (i absolutely wanna have a medal next year :D I was VERY close this year, stupid mistake in boring calculs, I told this before I think ; so I wanna have one next year. It's my second and last chance ...)

<daydreaming> oh, a gold medal !

 :)

you train by solving lots and lots of problems - which is what you do right now :D

I don't remember the order, but David got  1 nothing, 1 bronze, 3 silvers and 2 golds. OK, the "many" was a little optimistic, but it's however impressing. :D

Don't be so bitter about the lack of preparation and money.
It is not an isolated case, if this helps. For IMO 2002, we had 5 days of preparation, and 50 preparation problems, no theory lectures at all, which I think is essential for solving higher-level problems.I know you gain profficency by solving problems on your own, but you need at least some basic theory knowledge before that, and you can't learn everything by yourself.
( Not to mention the 0 days of preparation  for the BMO).
We had to buy books on our own, to find problems on the net, to study on our own... And, we had to pay our plane tickets to UK ( or to find sponsors). It happens year after year...
The explanation is always the same : insufficient funds.
That's why I gave up math :-( and started studying business.
Hope  there are ones who don't quit as easy as I do.
Best luck to you all!
Irina


trigonometry
induction
algebra unsolved
algebra

2 sequences of real numbers $ x_{1},x_{2},...$ and $ y_{1},y_{2},...$ are defined as follows :
$ x_{1}\equal{}y_{1}\equal{}\sqrt{3}$
$ x_{n\plus{}1}\equal{}x_{n}\plus{}\sqrt{1\plus{}x^{2}_{n}}$
$ y_{n\plus{}1}\equal{}\frac{y_{n}}{1\plus{}\sqrt{1\plus{}y^{2}_{n}}}$.
for all $ n\geq 1$.Prove that $ 2 <x_{n}y_{n} <3$,for all $ n>1$.

Very easy just make substitution and the problem will become obvious :wink:
$ x \equal{} \tan{60},y \equal{} \cot{30}$.

You also prove by induction  :wink:

Same problem :

Two sequences $ a_0,a_1,...$ and $ b_1,b_2,...$ difined by  $ a_0 = \frac {1}{\sqrt {2}} ,a_{n + 1} = \sqrt {\frac {1 - \sqrt {1 - a_n^2 }}{\sqrt {2}}}$ and $ b_0 = 1,b_{n + 1} = \frac {\sqrt {1 + b_n^2 - 1}}{b_n}$  .

Prove that:

$ 2^{n + 2} a_n < n < 2^{n + 2} b_n$

[hide="hint"] prove $ a_n = \sin\frac {\pi}{2^{n + 2}},b_n = \tan\frac {\pi}{2^{n + 2}}$[/hide]

[quote="quangpbc"]You also prove by intruction  :wink:
[/quote]
You mean induction :wink:
Best regards,Erken.

Exactly $ x_n \equal{} ctg\frac {\pi}{3*2^n}, y_n \equal{} tg\frac {\pi}{3*2^{n \minus{} 1}}.$

I forget about \[ x_ny_n\equal{}\frac{2}{1\minus{}tg^2\frac{\pi}{3*2^n}}\to 2.\]


we have :
  Caychy's  ineq :
 $ (a \plus{} b \plus{} c)(a^3 \plus{} b^3\plus{}c^3) \geq (a^2 \plus{} b^2 \plus{} c^2)^2 \geq \frac {(a \plus{} b \plus{} c)^4}{9}$
  hence : $ 27(a^3 \plus{} b^3 \plus{} c^3)^2 \geq \frac {(a \plus{} b \plus{} c)^2}{3} .(a \plus{} b \plus{} c)^4 \geq (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^4$


suppose that f(x) is iriducible polynomial in z[x].
asume that $ \alpha$ is a root of f(x) with norm greather than $ \frac{3}{2}$.
prove that $ \alpha^3 \plus{} 1$ cant be a root of f(x).

farz konim$ \alpha$  va $ \alpha^3 \plus{} 1$ rishe haye f bashand sbet mikonam bar har $ \beta$ ke $ f(\beta)$=0
angah $ f(\beta^3 \plus{} 1) \equal{} 0$

midanim $ (x^3 \minus{} \alpha^3)|f(x^3 \plus{} 1)$
pas darim $ (x \minus{} \alpha)|f(x^3 \plus{} 1)$

vali chon $ f(x)$ chand jomei minimal $ \alpha$ ast pac $ f(x)|f(x^3 \plus{} 1)$
pas agar $ x \equal{} \beta$ darim  $ f(\beta^3 \plus{} 1) \equal{} 0$

hala farz konid $ |\beta| > \frac {3}{2}$
hala $ |\beta^3 \plus{} 1| > |\beta^3| \minus{} 1 > |\beta|$ (chon max rishe  $ x^3 \minus{} x \minus{} 1$ az$ \frac{3}{2}$ kochektar ast)
 pas be donbale namotenahi az risheha miresim ke tanaghoz ast


inequalities
function
algebra proposed
algebra

solve this equation:
 with x,y,z>0
   $x^{3}=27+18y(2y-3)$
   $8y^{3}=27+81z(z-1)$
   $3z^{3}=3+x(x-3)$
my E is very very bad so everybody can help me for do some exercises i post thanks :D b-(

it is the proof of my teacher and it is regular i want another proof use inequality (mine) :D :blush:
P/S i will post more exercices about it i think it is pure

You can post other exercises, but not in THAT section! :mad:

Pierre.

oh but i think it semilar :D sorry because i spam :D and can u solve my problem by inequality

here is my proof,easy to demonstrate that x,y,z>0
$(x-3)(x^{2}+9+3x)=18y(2y-3)$
$(2y-3)(4y^{2}+9+6y)=81z(z-1)$
$3(z-1)(z^{2}+z+1)=x(x-3)$
 $(\frac{x^{2}+9+3x}{18y})(\frac{4y^{2}+9+6y}{81z})(\frac{3(z^{2}+z+1)}{x}=1$
by AM-GM inequality we have
$x^{2}+9+3x\geq\ 9x$
$4y^{2}+9+6y\geq\ 18y$
$z^{2}+z+1\geq\ 3z$we have"="when$y=\frac{3}{2},$z=1,x=3
i will post more but i have final exam :D Rust i hope u will help me :D i post this topic in my country 4rum but may be they are not interested in it :D the last thanks Rust


Find all n such that n^1999+n+5 is divisible by product of the digits of n as well as the sum of the digits of n.


here's a difficult problem from spoj...any help is more than welcome:)

A set S of positive integers is called strongly triple-free if, for any integer x, the sets {x, 2x} and {x, 3x} are not subsets of S. Let's define F(n) as a number of strongly triple-free subsets of {1, 2, ..., n}, where n is a natural number.

You need to write a program which being given a number n calculates the number F(n) modulo 1 000 000 001.

Input
The first line of input contains integer T (1 ≤ T ≤ 500) - the number of testcases. Then descriptions of T testcases follow.

The description of the testcase consists of one line. The line contains an integer number n (1 ≤ n ≤ 100 000).

Output
For each testcase in the input your program should output one line. This line should contain one integer number which is the number F(n) modulo 1 000 000 001.

Example
Input:
5
3
1
10
20
39

Output:
5
2
198
43776
971827200

now i am able  to calculate number of subsets starting from a particular number (using matrices) but i am recalculating too much...can any come up with a neat and succinct solution
best regards

I've thought about this problem for a looooong time , and I dont think there is anything mathematical about this problem, But i found a paper related to this problem in Wolfram's mathworld site .
U can just google it out . they paper also gives the algo i guess :)

[quote="galois"]

now i am able  to calculate number of subsets starting from a particular number (using matrices) but i am recalculating too much...can any come up with a neat and succinct solution
best regards[/quote]
Use dynamic programing to optimize the recursion.


modular arithmetic
Euler
number theory
algebra proposed
algebra

[size=200][size=150]it's a prime number or not  :?: [/size][/size]

yeah if they're allowed to be zero, then choose $ab=cd=0$ then we can find solutions.  If none of them is zero, consider $a^2+b^2+c^2+d^2$ mod 4.  If it is 2 mod 4, then the prime is 2, in which case we must have them be zero.  Else, it must be 1 or 3 mod 4.  that means that there is either 1 even and 3 odd, or 3 odd and 1 even, so then $ab \neq cd$ based on parity argument

examples $(a,b,c,d)=(1,0,1,0), (2,0,1,0)$

I don't think you solved the case for $a^2+b^2+c^2+d^2 \equiv 1\pmod{4}$.

woops you're right, sorry.  thats silly.

edit: i am not sure that way will go anywhere.  write $a/d=c/b=p/q$ where $(p,q)=1$.  let $(a,d)=g_1$, $(c,b)=g_2$.  then we have $a=g_1p$, $d=g_1q, c=g_2p, b=g_2q$.  then we claim $g_1^2(p^2+q^2)+g_2^2(p^2+q^2)$ is prime, but it factors as $(g_1^2+g_2^2)(p^2+q^2)$.  As before, assume none of $a,b,c,d$ are zero, then both factors are at least 2, so its not prime.  if $a,b,c,d$ are zero, we said there's solutions.

1) Please post these kind of problems [b]Fermat-Euler[/b] in the number theory section (!) next time

2) There are no such primes $p$ since $(a+b)^2+(c-d)^2=(a-b)^2+(c+d)^2=p$ (if $c-d=a-b$ and $c+d=a+b$ then pairs $(a,b)$ and $(c,d)$ are the same - what's impossible) and
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=47169]www.mathlinks.ro/Forum/viewtopic.php?t=47169[/url]


i searched before posting but i didn't find.so this is the problem
Let $ n > 1$ be an integer such that $ n$|$ 2^{\phi(n)} \plus{} 3^{\phi(n)} \plus{} ... \plus{} n^{\phi(n)}$
Let $ p_1$, $ p_2$ ,...,$ p_{k}$ be all distinct prime divisors of $ n$.
Prove that $ \frac{1}{p_1} \plus{} \frac{1}{p_2} \plus{} ... \plus{}\frac{1}{p_k} \plus{} \frac{1}{p_1...p_{k}}$ is an integer.

It is here :
http://www.mathlinks.ro/viewtopic.php?p=1090526#1090526


arithmetic sequence
number theory
relatively prime

Determine all pairs $ (p,q,r,s)$ of primes such that

$ pq\minus{}2r\equal{}pr\plus{}2q\equal{}s$.

[hide]Adding $ 2r \minus{} 2q$ to the original equation we get
$ (p \minus{} 2)q \equal{} (p \plus{} 2)r$, which implies $ \dfrac{q}{r} \equal{} \dfrac{p \plus{} 2}{p \minus{} 2}$.
$ \dfrac{q}{r}$ is already reduced so
$ p \plus{} 2 \equal{} kq$ [1]
for some natural number $ k$.
Also, $ p \minus{} 2 \equal{} kr$ [2].
Subtracting [2] from [1] we get
$ 4 \equal{} k(q \minus{} r)$.
$ q \minus{} r$ divides 4, so $ q \minus{} r$ can have one of three values: $ 1, 2$, and $ 4$.
If $ q \minus{} r \equal{} 1$, then $ q$ and $ r$ are of opposite parities. One of $ q$ and $ r$ must be even, and since it is even it must be $ 2$, the smallest possible prime, and be less than the other variable. $ r$ is the lesser of the two variables, so $ r \equal{} 2$. $ q \equal{} 3$, and substituting into the original equation we get
$ p(3) \minus{} 2(2) \equal{} p(2) \plus{} 2(3)$.
Solving for $ p$ we get
$ p \equal{} 10$, not a prime, so we have a contradiction.
If $ q \minus{} r \equal{} 2$ then substituting $ q \equal{} r \plus{} 2$ into the original equation we get
$ p(r \plus{} 2) \minus{} 2r \equal{} pr \plus{} 2(r \plus{} 2)$.
Solving for $ p$ we get
$ p \equal{} 2(r \plus{} 1)$. 2 divides $ p$, so $ p \equal{} 2$.
$ 2 \equal{} 2(r \plus{} 1)$, so $ r \equal{} 0$ which is not a prime. Contradiction.
By elimination $ q \minus{} r \equal{} 4$ must hold, if the original equation is to hold.
Substituting $ q \equal{} r \plus{} 4$ into the original equation we get
$ p(r \plus{} 4) \minus{} 2r \equal{} pr \plus{} 2(r \plus{} 4)$.
Solving for $ p$ we get
$ p \equal{} r \plus{} 2$.
$ (r, p, q) \equal{} (r, r \plus{} 2, r \plus{} 4)$ is an arithmetic sequence with common difference $ 2$. Since the three numbers cannot all be $ 2$, the only even prime, they cannot be even. They must be odd, and they must be, in some order, 1, 3, and 5 mod 6. $ 3$ is the only prime that is 3 mod 6, eliminating all but three possibilities for $ (r, p, q) \equal{} (r, r \plus{} 2, r \plus{} 4)$: $ ( \minus{} 1, 1, 3)$, $ (1, 3, 5)$ and $ (3, 5, 7)$. $ 1$ is not prime so
$ (r, p, q) \equal{} (3,5,7)$ is the only possibility.
Substituting into the original equation we get
$ s \equal{} 5 \cdot 3 \plus{} 2 \cdot 7 \equal{} 29$.
$ (p, q, r, s) \equal{} (5,7,3,29)$ is the only solution, and we are done.[/hide]

Note that
$ 2s^2 \equal{} (pq \minus{} 2r)^2 \plus{} (pr \plus{} 2q)^2 \equal{} (p^2 \plus{} 4)(q^2 \plus{} r^2)$
So $ q^2 \plus{} r^2 \equal{} 2s$ and $ p^2 \plus{} 4 \equal{} s$ (We easily see p=2 doesn't work).
So $ p^2 \plus{} 4 \equal{} pr \plus{} 2q \equal{} pq \minus{} 2r \equal{} \frac {q^2 \plus{} r^2}{2} \ge qr$
So $ pq \ge qr \plus{} 2r$, so $ p \ge r$
But $ r \equal{} \frac {q(p \minus{} 2)}{p \plus{} 2}$
$ p \plus{} 2$ and $ p \minus{} 2$ are relatively prime, so $ p \plus{} 2$ divides $ q$
But if $ p \plus{} 2 \le q$, then $ r \ge p$, contradiction
So $ p \plus{} 2 \equal{} q$, $ r \equal{} p \minus{} 2$
So the primes are $ r$,$ r \plus{} 2$,$ r \plus{} 4$
So $ r \equal{} 3$ (otherwise one is divisible by 3)
So the only solution is $ (3,5,7,29)$


This is a marathon using an identity or formula of some sort on which everyone contrives a problem using that identity.
Who wants to post the first problem, and decide the identity to use??

find a problem that uses

$\frac{a-b}{a+b}=\frac{\tan\frac{A-B}{2} }{\tan\frac{A+B}{2} }$

I'm stuck, any body have any suggestions??


I canot solve the following AMM problem:

Prove that if p is a prime, p = 4k+1, there exists a prime q <  \sqrt p for which q p-1 is not congruent to  1 (mod p 2)

(AMM E2956 by Barry Powell)

Anyone can help?

Namdung

(Thank Arne for correction)

Because p = 1 mod 4, there exist positive integers m and n,  so that
m 2 + n 2  = p.  Now we show that either
m (p-1) != 1 mod p 2   or n (p-1) != 1 mod p 2 , for if this is not the case, define k = (p-1)/2 (k is even), and consider n (2k) = (m 2 - p) k = m 2k - p*k*m 2(k-1) + p 2 *(some other terms), and then looking mod p 2 we see that  p*k*m 2(k-1) = 0 mod p 2 or k*m 2(k-1) = 0 mod p.  But, (m, p) = 1 so this forces k = 0 mod p, a contradiction.

So, WLOG m p-1 != 1 mod p, and m must have a prime divisor q so that q p-1 != 1 mod p

then q   \leq m =  \sqrt (p-n 2 ) <  \sqrt p

Very nice solution for very nice result. Thank you zscool! I spent much time for it.

Namdung

finally we some some of our fellow americans joining the battle with the problems :D welcome zscool!


For the first problem go read [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130162]here[/url]. For the 3rd [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111400]here[/url].

(*) $P(x,y)=2P(x-y,x+y)$
Then iterating we get
$P(x,y)=2P(x-y,x+y)=4P(-2y,2x)\\ \hspace*{0.65in}=8P(-2(x+y),2(x-y))=16P(-4x,-4y)$.
So if we write $P(x,y)=\sum_{i,j\geq 0}\alpha_{ij}x^{i}y^{j}$, then
$16P(-4x,-4y)=\sum_{i,j\geq 0}16\alpha_{ij}(-4)^{i+j}x^{i}y^{j}$, and then comparing coefficients we must have
$\alpha_{ij}=16\alpha_{ij}(-4)^{i+j}$ and $\alpha_{ij}=0$ for all $i,j\geq 0$. So $P\equiv 0$ is the only solution of (*).

Let's also solve
(**) $P(x-y,x+y)=2P(x,y)$
Then analog we get
$P(x,y)={1\over 2}P(x-y,x+y)={1\over 4}P(-2y,2x)\\ \hspace*{0.65in}={1\over 8}P(-2(x+y),2(x-y))={1\over 16}P(-4x,-4y)$
and $16\alpha_{ij}=\alpha_{ij}(-4)^{i+j}$ for all $i,j\geq 0$, where $P(x,y)=\sum_{i,j\geq 0}\alpha_{ij}x^{i}y^{j}$.
So $\alpha_{ij}=0$ for all $i+j\not =2$ and $P(x,y)=ax^{2}+bxy+cy^{2}$ for some $a,b,c\in\mathbb{C}$.
Then ${1\over 4}P(-2y,2x)=ay^{2}-byx+cx^{2}$, and again comparing coefficients
we get $a=c,b=0$ and $P(x,y)=a(x^{2}+y^{2})$, which is indeed a solution of (**) for all $a\in\mathbb{C}$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

I'll rewrite if for you, but I don't know how to solve this stuff..

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

I'll try my best:

Let $ u=\tan(x)$, then
$ \frac{du}{dx}=\sec^2(x)$
$ du=\sec^2(x)dx$
$ dx=\frac{1}{\sec^2(x)} du$

So,

$ \int{\sqrt{\tan(x)}}dx\&=\frac{1}{\sec^2(x)}\int u^{\frac{1}{2}}du\\
                                \&=\frac{1}{\sec^2(x)} * \frac{3}{2} u^{\frac{3}{2}}+K\\
                                \&= \frac{3}{2\sec^2(x)}\tan^{\frac{3}{2}}(x)+K\\
                                \&=\frac{3\tan^{\frac{3}{2}}(x)}{2\sec^2(x)}+K$

I think it should be this.

Substitutions don't work that way. The d_ is part of the integral; whatever you get out of it needs to stay inside, and pulling out a factor that depends on the variable being integrated makes no sense. You must account for that $ \sec^2$ inside, either by finding a factor of it or converting it into the new variable. The first is difficult here. The second is easiest to do by writing the substitution backwards, as $ x\equal{}\arctan u$.

Hi every body :) !!!

i think that you look for: $ \int \sqrt {tan(x)}dx$

by changing vraible $ (t \equal{} tan(x))$ and some other technique (IPP) we can find $ \int \frac {\sqrt {t}}{1 \plus{} t^2}dt \equal{} \frac {\sqrt {2}}{4}ln \left( \frac {\sqrt {2t} \minus{} (1 \plus{} t)}{\sqrt {2t} \plus{} (1 \plus{} t)} \right) \plus{} \frac {\sqrt {2}}{2} \left( arctan(1 \plus{} \sqrt {2t}) \minus{} arctan(1 \minus{} \sqrt {2t}) \right) \plus{} C$

and thx

$ \tan x \equal{}t \\ \\
\sec ^2 x dx \equal{} dt \\ \\
dx \equal{} \frac{dt}{1\plus{}t^2} \\ \\
\int \sqrt{\tan x}\cdot dx \equal{} \int\frac{\sqrt{t}\cdot dt}{1\plus{}t^2}$

[quote="jmerry"]The second is easiest to do by writing the substitution backwards, as $ x \equal{} \arctan u$.[/quote]  Or actually in this case it's even easier, since $ \sec^2 x \equal{} \tan^2 x \plus{} 1$.  (Rajiv and mathema* used this implicitly.)

Edit: typo corrected.

[quote="JBL"][quote="jmerry"]...Ravij and mathema* used this implicitly....[/quote][/quote]

You mean Rajiv, don't you?  :mad:

How can u get this one?
$ \int\frac{\sqrt{t}}{1\plus{}t^{2}}dt \equal{}\frac{\sqrt{2}}{4}ln\left(\frac{\sqrt{2t}\minus{}(1\plus{}t)}{\sqrt{2t}\plus{}(1\plus{}t)}\right)\plus{}\frac{\sqrt{2}}{2}\left( arctan(1\plus{}\sqrt{2t})\minus{}arctan(1\minus{}\sqrt{2t})\right)\plus{}C$


limit
function
trigonometry
calculus
calculus computations

Claim: suppose $ f$ is a locally invertible function and $ f(x)\equal{}x\plus{}ax^{3}\plus{}O(x^{5}).$ Then $ f^{\minus{}1}(x)\equal{}x\minus{}ax^{3}\plus{}O(x^{5}).$

We can get this just by solving for $ x$ in $ y\equal{}x\plus{}ax^{3}\plus{}O(x^{5})$ by assuming that $ x$ can be written as a power series in $ y.$

Hence:

$ \sin x\equal{}x\minus{}\frac{1}6x^{3}\plus{}O(x^{5})$ so $ \arcsin x\equal{}x\plus{}\frac{1}6x^{3}\plus{}O(x^{5}).$

$ \tan x\equal{}x\plus{}\frac{1}3x^{3}\plus{}O(x^{5})$ so $ \arctan x\equal{}x\minus{}\frac{1}3x^{3}\plus{}O(x^{5}).$

$ \arcsin x\minus{}\sin x\equal{}\frac{1}3x^{3}\plus{}O(x^{5})$

$ \arctan x\minus{}\tan x\equal{}\minus{}\frac{2}3x^{3}\plus{}O(x^{5})$

Take the quotient, let $ x\to 0.$ The limit is $ \minus{}\frac{1}2.$

many thanks sir but is there any high-school solution ?

Limits are calculus, and calculus is not "high-school level" by the standards I follow. High school contests don't generally have calculus problems, and calculus classes have very large enrollment at most colleges.

Is there a solution which comes earlier in the calculus sequence? Yes, there is. L'Hopital's rule is normally taught before power series, and it does work here. It's also a lot uglier than the power series method, and it's easier to make mistakes.


geometry
perimeter
area of a triangle
geometry solved

Let $ABC$ be an equilateral triangle (i. e., a triangle which satisfies $BC=CA=AB$). Let $M$ be a point on the side $BC$, let $N$ be a point on the side $CA$, and let $P$ be a point on the side $AB$, such that $p\left(ANP\right)=p\left(BPM\right)=p\left(CMN\right)$, where $p\left(XYZ\right)=YZ+ZX+XY$ denotes the area of a triangle $XYZ$.
Prove that $\triangle ANP\cong\triangle BPM\cong\triangle CMN$.

@Bui_haiha2000: I have changed the name of the topic so that it is halfways meaningful now. Next time, please do this on your own. Also, please specify the source of the problem (e. g. if you have read it in a book, write the name of the book, or if it is from a contest, the name of the contest).

The problem is harder than the two similar ones you posted before. Here is my solution.

[color=blue][b]Theorem.[/b] If ABC is an equilateral triangle and M, N, P are points on the segments BC, CA, AB such that the triangles ANP, BPM, CMN have equal perimeters, then these triangles ANP, BPM, CMN are congruent.[/color]

[i]Proof.[/i] We first show a useful lemma:

[color=blue][b]Lemma 1.[/b] If ABC and XYZ are two triangles with equal perimeters and satisfying < CAB = < ZXY, then $AB\leq XY$ holds if and only if $AC\geq XZ$.[/color]

[i]Proof of Lemma 1.[/i] Since the two triangles ABC and XYZ have equal perimeters, they must also have equal semiperimeters. Denote the common semiperimeter of triangles ABC and XYZ by $s=\frac{BC+CA+AB}{2}=\frac{YZ+ZX+XY}{2}$.

Let B' and C' be points on the rays AB and AC such that AB' = XY and AC' = XZ. Since we also have < C'AB' = < CAB = < ZXY, this yields that the triangles AB'C' and XYZ are congruent. Thus, the semiperimeter of triangle AB'C' equals to the semiperimeter of triangle XYZ, which, in turn, equals s.

Now, if the A-excircle of triangle ABC touches its sideline AC at a point U, then we know that AU = s because s is the semiperimeter of triangle ABC. On the other hand, if the A-excircle of triangle AB'C' touches its sideline AC' at a point U', then AU' = s because s is the semiperimeter of triangle AB'C'. Thus, AU = AU', so the points U and U' coincide (because both of them lie on the ray AC = AC' and satisfy AU = AU'); in other words, the A-excircle of triangle ABC and the A-excircle of triangle AB'C' touch the line AC at one and the same point U = U'. Similarly, these two excircles touch the line AB at one and the same point. Thus, these two excircles must coincide (in fact, if they didn't coincide, then, since they both touch the line AC at the point U = U', they would be tangent to each other at this point U = U' on the line AC, but similarly we could show that they would be tangent to each other at a point on the line AB, so they would be tangent to each other at two distinct points, what is impossible).

Denote the common A-excircle of triangles ABC and AB'C' by k. Let this excircle k touch the side BC of triangle ABC at a point P, and the side B'C' of triangle AB'C' at a point P'. Since the point P lies inside the segment BC, the ray AP lies inside the angle CAB, so this ray AP must intersect the segment B'C'. Let the ray AP intersect the segment B'C' at a point Q. Since the points A and P lie on different sides of the line B'C' (in fact, the point P lies on the A-excircle of triangle AB'C' and thus it lies on the opposite side of the sideline B'C' as the point A), this point Q must lie inside the segment AP. Similarly, the ray AP' intersects the segment BC at a point Q' which lies inside the segment AP'. Now, since the points Q and Q' lie inside the sides AP and AP' of triangle APP', the segments PQ' and P'Q must intersect (in their interior). Since these segments PQ' and P'Q are parts of the segments BC and B'C', it follows that the segments BC and B'C' intersect. Hence, the point B lies inside the segment AB' if and only if the point C lies on the extension of the segment AC' beyound C'. In other words, $AB\leq AB^{\prime}$ holds if and only if $AC\geq AC^{\prime}$. Since AB' = XY and AC' = XZ, this rewrites as follows: $AB\leq XY$ holds if and only if $AC\geq XZ$. This proves Lemma 1.

Now to the proof of the actual Theorem:

Since triangle ABC is equilateral, BC = CA = AB and < CAB = < ABC = < BCA = 60°. Now, WLOG assume that $AN\leq BP$. Since the triangles ANP and BPM have equal perimeters and satisfy < PAN = < MBP (this is equivalent to < CAB = < ABC), according to Lemma 1, the relation $AN\leq BP$ yields $AP\geq BM$. Thus, $BP=AB-AP=BC-AP\leq BC-BM=CM$. So we saw that $AN\leq BP$ yields $BP\leq CM$. Similarly, $BP\leq CM$ yields $CM\leq AN$, so we get $AN\leq BP\leq CM\leq AN$, what is only possible if AN = BP = CM. Hence, AP = AB - BP = BC - CM = BM. From AN = BP, AP = BM and < PAN = < MBP, it follows that triangles ANP and BPM are congruent. Similarly, triangles BPM and CMN are congruent. Thus, all three triangles ANP, BPM and CMN are congruent, and the Theorem is proven.

  Darij

no,darij grinberg .  This isn't the contest , my teacher creat this problem


trigonometry
vector
function
geometry
search
complex numbers

Find $\min f$ and $\max f$ where $x$ and $y$ are real numbers and

$f(x,y)=2\sin(x)\cos(y)+3\sin(x)\sin(y)+6\cos(x)$

$f(x,y)=\sqrt{13}\sin(y+\arctan{2\over 3})\sin x+6\cos x$

Put $g(y)=\sqrt{13}\sin(y+\arctan{2\over 3})$. Then

$f(x,y)=g(y)\sin x+6\cos x=\sqrt{g^{2}(y)+36}\sin(x+\arctan{6\over g(y)})$

Then its range is obviously $[-\sqrt{g^{2}(y)+36}, \sqrt{g^{2}(y)+36}]$, and since $|g(y)|\leqslant\sqrt{13}$, the maximum and minimum are obtained when $g^{2}(y)=13$, in which case $f_{min}(x,y)=-7$ and $f_{max}(x,y)=7$. As for the points in which they are reached, first choose $y$ such that $\sin(y+\arctan{2\over 3})=\pm 1$, and then for such $y$ choose $x$ such that $\sin(x+\arctan{6\over g(y)})=\pm 1$

[quote="me@home"]Doesn't Cauchy work for this problem? Then it if you look at it as vectors, that gives the negative (min) also?[/quote]

what do u mean by this?

[quote="me@home"]Doesn't Cauchy work for this problem? Then it if you look at it as vectors, that gives the negative (min) also?[/quote]

You don't have to look at vectors, because Cauchy gives $\left(f(x,y)\right)^{2}\leq 49$, so $|f(x,y)| \leq 7$, so $-7 \leq f(x,y) \leq 7$.

thanks for the solutions.
i was wondering if it was possible to find $x$ and $y$ such that the function attains its minimum and maximum values?

We've applied Cauchy on $(\sin(x)\cos(y),\sin(x)\sin(y),\cos(x))$ and $(2,3,6)$. To have equality, we should have that \[\frac{\sin(x)\cos(y)}{2}\buildrel{(1)}\over{=}\frac{\sin(x)\sin(y)}{3}\buildrel{(2)}\over{=}\frac{\cos(x)}{6}.\] Now remark that if $\sin(x) = 0$, then $\cos(x) = \pm 1$, so we would have that $0 = 0 = \pm 6$. Therefore, we can assume that $\sin(x) \neq 0$. (1) becomes $3\cos(y) = 2\sin(y)$, i.e. $9\cos^{2}(y) = 4\sin^{2}(y)$, i.e. $9 = 13\sin^{2}(y)$, i.e. $\sin(y) = \pm\frac{3}{\sqrt{13}}$. So $y = \arcsin(\pm\frac{3}{\sqrt{13}})$. (Of course, the negative one will give the minimum, the positive one the maximum.)
Now (2) becomes $2\sin(x)\sin(y) = \cos(x)$, i.e. $\frac{9}{13}\cdot 4\sin^{2}(x)\sin^{2}(y) = \cos^{2}(x)$, i.e. $\frac{49}{13}\sin^{2}(x) = \cos^{2}(x)+\sin^{2}(x) = 1$, i.e. $\sin(x) = \pm\frac{\sqrt{13}}{7}$. So $x = \arcsin(\pm\frac{\sqrt{13}}{7})$.

[quote="me@home"]Oh right... so I guess Cauchy without vectors encapsulates everything you need without looking at those vectors...[/quote]

It's not bad to think of Cauchy as vectors. It's always about vectors, even if you don't explicitly have to say anything about it. Cauchy-Schwarz is true is much more general cases than you usually realize when solving these kinds of problems -- and it's all about vectors. :D

[quote="me@home"]I've been teaching myself a lot of vector stuff and a lot of it seems hard to completely absorb... are they really that powerful?[/quote]

Yes!  They certainly are.  It's certainly worth your time to learn everything you can about a technique as powerful as vector algebra.  :)

Especially once you start doing geometry using complex numbers ;)

I was under the impression that it was worth putting more effort into actually learning complex number algebra than vector algebra, because vectors are more limited. I'm a Y11 student though, so I've done neither properly :).

Complex numbers have their own very important uses, but they're restricted to two dimensions.  Vector algebra, on the other hand, you can do in as many dimensions as you want :D

Yay! n-dimensional giant squid to explain the existence of plot holes in Pirates 3!!!
do a web search of SIMUW

anyway if vectors have the advantage of n dimensions then I am definitely concentrating on that!


[quote="GameBot"]Forty-five percent of the area of circle $ O$ is contained in the shaded region. How many degrees are in the measure of the non-reflex angle $ \angle MON$?
[asy]size(150); defaultpen(linewidth(0.8)); import graph; import patterns;
draw(Circle((0,0),5));
path a = Arc((0,0),5,-90,72);
add("hb",hatch(3,NW));
filldraw(a--origin--cycle,pattern("hb"));
label("$O$",origin,W);
label("$N$", (0,-5),S);
label("$M$", 5*dir(72),N);[/asy][/quote]

$ \frac{45}{100}\equal{}\frac{9}{20}\equal{}\frac{162}{360}$, so 162.


Given a circle and a chord,  how to draw an intersecting chord so that the 4 circles
  inscribed in 4  'sectors' had concyclic centers?

   Thank you.

    M.T.

While you trying the other construction ( two triangles cut by a line),

   please do not forget this still unsolved construction problem.

   If you are not used to this much freedom for the cut point of the chords,

   then  fix your own  point on the chord given.

   After this construction is done, maybe ML will petition the mayor of Tokyo

   to hang this solution carved on wooden board in the garden next to City Hall.

   Thank you and a Happy New Year to all.
   M.T.

[quote="armpist"]Given a circle and a chord,  how to draw an intersecting chord so that the 4 circles
inscribed in 4  'sectors' had concyclic centers.[/quote]
I think that the midperpendicular of the given cord, separate the circle in to four sectors of which the centers of their inscribed circles, form an isosceles trapezium and so, they are concyclic.

Sorry if I misunderstood your problem dear [b][size=100]armpist[/size][/b] and also a happy new year to all.

Kostas Vittas.

dear Vittasko

   thank you for your reply.

   This problem was without any solution for too long, that is why
   just to get a feeling of some accomplishment in the holiday
   season I allowed an unrestricted choice of the cut point.
   Out of so many possible points on the chord you picked the
   right one. Very nice!

   Comes new year I hope you will look at it again and try to do
   something about a few other points on the chord.

   Happy New Year!

   M.T.


x^y = y^x

Express this equation in terms of y.

Hmm,  I meant to write prove that it is equal to y=x. Not sure why it came out so wrong when I typed it, I must be tired....

$x^y = y^x$
$y \ln x = x \ln y$
$\frac{\ln x}{x} = \frac{\ln y}{y}$
$x^{1/x} = y^{1/y}$

An obvious solution is then $x=y$.

If both are greater than $e$ then there is no other solution than that.
If either one is negative, I am scared, because I'm not sure what I should do.
If one is greater than $e$, and the other is positive, but less than $e$ then there is another solution.

Can anyone elaborate on this more, or am I erring?

One observation that $x=y$ is not the full solution: let $x=2$ and $y=4$; clearly equation is satisfied but $x=2\ne4=y$  ;)

[hide]going along with what Fmako wrote,

we have $\frac{\ln x}{x} =\frac{\ln y}{y}$

lets define $g(x)=\frac{\ln x}{x}$

so we want $(x,y)$ that satisfy  $g(x)=g(y)=k$

look at the graph of $g(x)$ and $y=k$

clearly we see that there is a maximum value (that occurs at $e$)

1) so if $k \ge e$, then the only way it can be true if $x=y$ because the graphs do not intersect, so there are not 2 inputs that produce $k$

2) but when $0< k< e$, it is not a one-to-one function. i.e. there are 2 values of $x$ that give the same output

3) if $x$ and $y$ are negative, we have a problem because those are imaginary, i have an idea, but i am not sure if this works

let $x=-m$ and $y=-n$, where $m,n >0$

$\frac{\ln -m}{-m} =\frac{\ln -n}{-n}$

$\frac{\ln m+\ln -1}{-m} =\frac{\ln -n+\ln -1}{-n}$

$\frac{\ln m}{-m} + \frac{\pi*i}{-m} =\frac{\ln n}{-n} + \frac{\pi*i}{-n}$

so equating the imaginary parts, we have that $m=n \implies x=y$

4) and finally, when $x =0$ or $y=0$, we get an undefined expression ($0^0$)[/hide]

I was more curious about the positive range less than e and greater than e, but I like your interesting approach to dealing with the negative values.  Congrats Altheman.


Find all$ t $such that $\forall x,y,z\in R^*:
2(x^3+y^3+z^3)+3(1+3t)xyz\geq (1+t)(x+y+z)(xy+yz+zx)$

Believe me, I've just opened a random file and found this problem... It's Crux Mathematicorum problem #2105, by Christopher J. Bradley, with solution in Crux Mathematicorum 1/23 (1997), pages 54-55. Freely downloadable from http://www.journals.cms.math.ca/CRUX .

  Darij

[quote="darij grinberg"]Manlio, just to tell you what file I meant: look at http://math4u.de/ , more exactly at "[b]Over 1111 problems involving inequalities and extremal problems[/b]". But please don't post [i]all of them[/i] at MathLinks  :)

  Darij[/quote]

Some time ago when the old links section was still valid (the new one is a lot shorter) I already posted a link. But he seemed to have updated and extended his file now.  :)

Manlio, please post [b]ALL[/b] the problems.  :D


SuMAC
pigeonhole principle
AMC
USA(J)MO
USAMO
number theory

So, who's in for 2006?

Acceptences are out.

(I am not quite sure if I am going for 2006 yet, but uh it's quite likely)

i was accepted and will be attending.  See you there!

I'm in too, and I'll probably go.  It would be a shame not to after putting so much time into the application.  The last part of the last problem took forever.

hmm really? do you remember what the problem was? I recall it not being very long(some application of the pigeon hole principle) but I might've misread the problem.

I'm definitely going (got accepted). See you there kevivek!
Oh, and the last part of the last problem was about how many tickets he had to buy to be sure of winning a 5 bit lottery.

If there was a simple solution, I must have missed it when I was Writing up my unnecessarily long proof.  If you know a concise way to answer this one, please post it.  (Part D)

8. In the game of n-bit Lottery players try to correctly guess a randomly chosen sequence of n binary
digits. The digits 0 and 1 are binary digits, and we call a sequence of n binary digits an n-bit sequence.
For example, 01110 is a 5-bit sequence. A player in n-bit Lottery can purchase one or more lottery
tickets, each of which has an n-bit sequence of the player’s choice. Then an n bit sequence is selected at
random. A prize is awarded for each ticket that is either an exact match of the randomly chosen one,
or differs from the randomly chosen sequence in only one place. For example in Four-bit Lottery, if the
randomly chosen sequence is 0100 then the following tickets win prizes: 0100, 1100, 0000, 0110, 0101.

(a) In terms of n, what is the smallest number of tickets you need to purchase in n-bit Lottery in order
to guarantee that you match the randomly chosen sequence exactly.

(b) In terms of n, how many di_erent winning tickets are there in n-bit Lottery?

(c) What is the smallest number of tickets you need to purchase to guarantee that you win in 4-bit
Lottery? Justify your answer.

(d) What is the smallest number of tickets you need to purchase to guarantee that you win in 5-bit
Lottery? Justify your answer.

for any lottery ticket of length n, there are $n+1$ lottery tickets that differ in 0 or 1 place from this particular ticket. so by the pigeon hole principle, you always need to buy $2^n - (n+1) + 1$ tickets to guarentee a win. at least that's what i wrote.

Sorry, but that doesn't work.  According to that, you need to buy 27 tickets to guarentee that you win a 5-bit lottery.  It is possible to do so with as few as 7.  Proving that was the hard part.

sumac is great. youll have a blast there.

and uh, you guys probably shouldnt talk about the problems on the forum. (you can pm/im each other) but these application tests are reused (this years was the same as two years ago), so Rick wouldnt be too appreciative of this nonsense about pigeon holes and whatnot. plus leah, why'd you use pigeonhole?! its a covering space problem. (whereas the ABA island problem last year was packing spaces). remember?! (aren't covering space problems NP?)

Sorry.  I didn't realize that they reused questions.

hmm, I fell asleep during that part of the lecture :P doh.

Oh yeah, I think I also confused myself thinking that the tickets you choose were picked randomly rather than a specific strategy.  :blush:

and ironically, I am not coming back to SUMaC, for sure now; unless i die, and my ashes are interred in palo alto or something.

[quote="beta9"]I'm officially not going.  A similar free program near by was too good to turn down.[/quote]

I'm curious---what program is similar to SUMaC and also free?

yeah me too...whats the name of the program?

for anyone who knows...whats the daily schedule like at sumac? Is it class intensive or more heavy on open time to do problem sets?

If I remember the schedule correctly (it's in the emailed packet) it's class time from after breakfast to lunch. After that is problem set time/downtime. So it seems that class time isn't the majority.

Sorry, I haven't really been checking this.

The program is in Michigan, It's called Oakland University Summer Mathematics Institute (OUSMI).  It's not really like SUMAC because no one boards, everyone commutes.

For free about 40 kids (You have to apply, but of course it's not as selective as SUMAC) get to take two advanced college classes and have access to whatever they want on the Oakland University Campus (Whatever we wants is really just the cafeteria, computer lab, and basketball courts).  This year we are taking classes in number theory and linear algebra.

As far as the caliber of students go, I think they are probably just about the same as SUMAC (probably more than 10 or more USAMO qualifiers, but I haven't really asked).

Also, I know a few USAMOers who do stuff on these forums.  If you guys see this, don't bother asking who I am.  If people knew who I was, I might be afraid to ask stupid questions on these forums.


logarithms
Euler
real analysis
real analysis unsolved

http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant

http://en.wikipedia.org/wiki/Harmonic_number

[color=red]How  to  prove  the   following:[/color]

$ H_n \equal{} \ln n \plus{} \gamma \plus{} \frac {1}{2n} \minus{} \frac {1}{12n^2} \plus{} \frac {1}{120n^4} \minus{} ....$

Thanks!

The Euler-Maclaurin formula will probably give this.


group theory
abstract algebra
invariant
superior algebra
superior algebra theorems

How many up to isomoporhism abelian groups $A$ are there, which could be included in the following exact sequence: $0 \to \mathbb{Z}\to A \to \mathbb{Z}/6000\mathbb{Z}\to 0$ ?

After some brief analyzing and some well-known facts about short exact sequences we see that the problem essentially boils down to what groups have $\mathbb{Z}$ as a normal subgroup and have $\mathbb{Z}/6000\mathbb{Z}=\mathbb{Z}/A$ as the factor group.

This is an interesting problem and I would like to know what other people think, though.

I find $\mathbb Z$ and $\mathbb Z \oplus \mathbb Z / 6000 \mathbb Z$

Would the IUM exam require you to know such things as $\text{Ext}$ and $\text{Hom}$, or can the solution be stated without category theory? Although, I guess if the problem is dealing with exact sequences, some category theory isn't unlikely.

[quote="amfulger"]The answer is $|Ext^{1}(\mathbb Z/600, \mathbb Z)\simeq \mathbb Z/600|=600$.[/quote]

Unfortunately, the answer I gave above is not the answer to the question asked. It is the number of equivalence classes of extensions of $\mathbb Z/600$ through $\mathbb Z$.

The correct answer is the number of isomorphism classes of abelian groups of the form $\frac{\mathbb Z\oplus\mathbb Z}{(\beta,-600)\mathbb Z}$ with $\beta\in\mathbb Z$. By looking at the invariant factors we see that such a group is isomorphic to $\mathbb Z/m\oplus\mathbb Z$ for some $m|600$.

If you want to try a proof for these, this should be a good recipe:

- First let $m|600$. We take $A=\frac{\mathbb Z\oplus\mathbb Z}{(m,-600)\mathbb Z}$,
$i: \mathbb Z\to A$ defined by $i(x)=\overline{(x,0)}$ and $p: A\to \mathbb Z/600$ by $p(\overline{(x,y)})=\hat y$, where $\hat y$ is the class of $y$ through the natural application $\mathbb Z\to\mathbb Z/600: 1\to \hat 1$.
Prove that $0\to\mathbb Z\stackrel{i}{\to}A\stackrel{p}{\to}\mathbb Z/600\to 0$ is a short exact sequence.
Prove that $A\simeq Z\oplus\mathbb Z/m$.

- Now take $A$ an arbitrary extension: $0\to\mathbb Z\stackrel{i}{\to}A\stackrel{p}{\to}\mathbb Z/600\to 0$. Let $s\in A$ be an element such that $p(s)=\hat 1$. Since $p(600\cdot s)=\hat 0$, there exists $m\in\mathbb Z$ such that $i(m)=600\cdot s$.
Define $B=\frac{\mathbb Z\oplus\mathbb Z}{(m,-600)\mathbb Z}$.
Let $F: \mathbb Z\oplus\mathbb Z\to A$ be defined by $F(x,y)=i(x)+sy$.
Prove that $F$ factors through $B$ i.e. there exits $f: B\to A$ such that $f(\bar u)=F(u)$ for all $\bar u\in B$.
Use the Five Lemma to prove that $f$ is an isomorphism.
Use whatever to prove that $B$ is isomorphic to some $\mathbb Z/r\oplus\mathbb Z$.

$600$ has $24$ distinct divisors.


Prove or disprove that if $ a$ and $ b$ are two natural numbers such that

$ a^{2}\minus{}ab\plus{}b^{2}$ divides $ a^{2}b^{2}$ then $ a\equal{}b$.

Nope. Take $ a\equal{}6,b\equal{}3$. More generally, if $ g,m,n$ are such that $ m^2\minus{}mn\plus{}n^2$ divides $ g^2$, then $ a\equal{}gm,b\equal{}gn$ works.

If we add the assumption $ (a,b)\equal{}1$ then it becomes true.  And if course if we changed it to $ (a^2\minus{}ab\plus{}b^2)|ab$ it would be true as well. :P


Good evening. Hope everyone had a great week.

I just received an exam back and I answered these two questions incorrectly. I am still not sure as to why my solutions are incorrect.  :oops: If someone could please explain how to solve these I would really appreciate it. Thanks!

1. An ac generator has emf ε = εmsinωdt, with εm = 33.8 V and ωd = 519 rad/s. It is connected to a 9.64 H inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is -16.9 V and increasing in magnitude, what is the current?

2. An ac generator with εm = 275 V and operating at 596 Hz causes oscillations in a series RLC circuit having R = 309 Ω, L = 197 mH, and C = 32.2 μF. Find (a) the capacitive reactance XC, (b) the impedance Z, and (c) the current amplitude I.

[color=blue][b](1)[/b][/color] Instead of working with sines and cosines, it is easier to define complex emf

$\hat \epsilon = \epsilon_0 \exp (\text i \omega t)$

In this case, the real emf is

$\epsilon = \epsilon_0 \sin (\omega t) = \text{Im}(\hat \epsilon)$

The coil impedance is $\hat Z = \text i \omega L$ and the complex current is

$\hat I = \frac{\hat \epsilon}{\hat Z} =\frac{\epsilon_0 \exp (\text i \omega t)}{\text i \omega L} = \frac{- \text i \epsilon_0 \exp (\text i \omega t)}{\omega L} =$

$= \frac{\epsilon_0}{\omega L}\ \exp\left(-\frac{\text i \pi}{2}\right) \exp (\text i \omega t) = \frac{\epsilon_0}{\omega L}\ \exp \left[\text i \left(\omega t - \frac \pi 2 \right)\right]$

Since the real emf is $\epsilon = \text{Im}(\hat \epsilon)$, the real current is

$I = \text{Im}(\hat I) = \frac{\epsilon_0}{\omega L}\ \sin \left(\omega t - \frac \pi 2 \right) = I_0 \sin \left(\omega t - \frac \pi 2 \right) = -I_0 \cos (\omega t)$

where $I_0 = \frac{\epsilon_0}{\omega L}$ is the current amplitude (maximum current). Thus in a coil, AC current follows AC voltage. When the current is maximum, $\sin \left(\omega t - \frac \pi 2 \right) = 1,$ $\omega t - \frac \pi 2 = \frac \pi 2 + 2n \pi,$ $\omega t = (2n + 1)\pi$, where n is an integer. But the emf is then

$\epsilon = \epsilon_0 \sin(\omega t) = \epsilon_0 \sin [(2 n + 1) \pi] = 0$

When $\epsilon = \epsilon_0 \sin(\omega t) < 0$ and increasing, then $\frac{3 \pi}{2} < \omega t < 2 \pi,$ $-\frac \pi 2 < \omega t - 2 \pi < 0$ and arcsin defined in [-1, 1] with the range $\left[-\frac \pi 2,\ \frac\pi 2\right]$ can be used to calculate $\omega t - 2 \pi.$

$\omega t - 2 \pi = \arcsin \frac{\epsilon_1}{\epsilon_0},\ \omega t = 2 \pi + \arcsin \frac{\epsilon_1}{\epsilon_0}$

where $\epsilon_0 = 33.8\ \text V$ and $\epsilon_1 = -16.9\ \text V$, so that arcsin comes out negative. This $\omega t$ is then substituted into the above equation for AC current.

[color=blue][b](2)[/b][/color] The angular frequency is $\omega = 2 \pi f$, where $f = 596\ \text{Hz}$. The reactance is $\hat X_R = R,$ indutance $\hat X_L = \text i \omega L,$ and capacitance is $\hat X_C = \frac{1}{\text i \omega C}.$ When connected in series, these add for a total impedance

$\hat Z = \hat X_R + \hat X_L + \hat X_C = R + \text i \omega L + \frac{1}{\text i \omega C} = R + \text i\ \frac{\omega^2 LC - 1}{\omega C}$

$Z = |\hat Z| =\sqrt{R^2 + \frac{(\omega^2 LC - 1)^2}{\omega^2 C^2}} = \frac{\sqrt{(\omega R C)^2 + (\omega^2 L C - 1)^2}}{\omega C}$

Assume AC voltage $\epsilon = \epsilon_m \sin(\omega t) = \text{Im}(\hat \epsilon)$, where $\hat \epsilon = \epsilon_m \exp(\text i \omega t)$ and $\epsilon_m = |\hat \epsilon|$. Then AC current is $I = \text{Im}(\hat I)$, where

$\hat I = \frac{\hat \epsilon}{\hat Z}$

and its amplitude is

$I_m = |\hat I| = \frac{|\hat \epsilon|}{|\hat Z|} = \frac{\epsilon_m}{Z}$

where $Z = |\hat Z|$ is given above.

chrislikephysics, kindly keep in mind that in the future, before asking for help, you should explain what part you were unable to do or you couldnt understand. Atleast tell us what you did before expecting us to solve your problem fully. Thank you.

Rushil - wait a second.

If you think that I come here for a complete solution without doing any work than you are completely mistaken. If only you saw the pads of paper I have here with formulas written in attempts to solve a particular problem. I only request that someone help me with a couple of the problems that I am unaware of how to even begin or not sure really what to do.