graph theory
combinatorics
Clique number
IMO Shortlist

Define a $ k$-[i]clique[/i] to be a set of $ k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="DusT"]This IS almost like the oficial solution. I mean, it is I think the only solution.[/quote]

Well, I don't see the problem if grobber comes up with the same idea as in the official solution. Dust, there are still lots of other unsolved ISL problems on the forums. Go and solve them if you want to have a gold medal at the next IMO in Mexico.  :D

Suppose for the sake of contradiction that the result is false. First, I will show that no two triangles can share an edge.

Suppose we have distinct triangles $ABC$ and $BCD$ sharing edge $BC$. If we remove $B$ and $C$, there must remain another triangle. This triangle clearly cannot have all of its vertices among $A,B,C,D$, so we can find some point $E$ distinct from $A,B,C,D$ belonging to a triangle. This 3-clique must not be disjoint with either $ABC$ or $BCD$, yet it cannot contain $B$ or $C$, so it must contain $A$ and $D$. Hence, $AD$, $DE$, and $EA$ are connected.

Suppose that upon removing $A$ and $D$, there still exists a 3-clique. It cannot be disjoint with triangle $ADB$, $ADC$, or $ADE$, yet it cannot contain either $A$ or $D$, so it must contain $B$, $C$, and $E$. It follows that $B$, $C$, and $E$ are connected, so $ABCDE$ is a 5-clique, which is a contradiction. Therefore, no two triangles can share an edge.

If we have one 3-clique, the result is trivial. If we have two 3-cliques, they may share only one vertex, say, $P$; in the first 3-clique, let the other two vertices be $Q$ and $R$, and in the second, let the other two vertices be $S$ and $T$. When $P$ and $R$ are removed, there exists a 3-clique among the remaining vertices. It cannot be among $Q$, $T$, and $S$, or else triangles $PQT$ and $QST$ will share edge $QT$, so there exists another vertex $U$ belonging in this 3-clique. This 3-clique cannot be disjoint with $PQR$, and it cannot contain $P$ and $R$, so it must contain $Q$. It is not disjoint with $PST$, so it must contain one of $S$ and $T$; without loss of generality, suppose it contains $S$. But then we have triangles $PQS$ and $QUS$ sharing edge $QS$, so we have reached a contradiction.

Here is my solution which may be equivalent to the ones above (I haven't read the ones above yet.)

Case 1: There exists a 4-klique
Let the 4 klique be ABCD. Then clearly, every 3-klique must intersect ABCD in at least 2 points. Now suppose that there exist kliques WLOG EAB and FCD. If E and F are the same point, then EABCD is a 5-klique. if they are not the same point, then EAB and FCD are disjoint, thus this is impossible. Therefore, suppose point E exists such that EAB is a 3 klique. Then ever other 3 klique MUST contain either A or B. Thus deleting A and B suffices.

Case 2: THere is no 4-klique but there exists ABCD such that B and C are not connected, but all other pairs are (so two triangles sharing an edge)
In this case, let us consider any 3-klique not containing A or D. Then, it clearly must contain B and C in order for the given constraint to hold. However B is not connected to C, thus all 3-klique's contain either A or D. Thus, deleting the two of them suffice.

Case 3: All 3-klique's intersect at one point exactly.
Suppose we have kliques ABC and ADE. Now, in order for any other klique to intersect both of these kliques but not go through A, it must contain WLOG B and D. However B and D are not connected by the given above, thus we have a contradiction. Hence all 3-kliques go through A which means deleting A suffices.

Suppose otherwise; let $f(X,Y)$ denote the assertion that there must exist a triangle not containing either of $X,Y$ and $g(UVW)$ for a triangle $UVW$ denote the assertion that each triangle in $G$ must contain at least one of $U,V,W$.

Start with a triangle $ABC$, so $f(B,C),g(ABC)$ force the existence of triangle $ADE$ with $\{D,E\}\cap\{B,C\}=\emptyset$. Then from $f(A,C),g(ACB),g(ADE)$, WLOG $BD$ is an edge, and from $f(A,B),g(ABC),g(ABD)$, $CD$ must be an edge. Finally, $f(A,D),g(ADC),g(ADE),g(ADB)$ together imply that $CEB$ must be a triangle, so $ABCDE$ form a $K_5$, which is impossible.

[hide=" Solution"]
Suppose there exists an arrangement contradicting the problem statement. The result is trivial if there is no more than one 3-cliques, so assume there are at least two 2-cliques.

First, suppose there exist persons $\{p_1, p_2, p_3, p_4\}$ such that $\{p_1, p_2, p_3\}$ and $\{p_4, p_2, p_3\}$ are both 3-cliques. Remove persons $p_2$ and $p_3.$ Then, there exists a person $p_5$ such that $\{p_1, p_4, p_5\}$ forms a 3-clique (if there are no 3-cliques remaining, we are done; furthermore, this 3-clique must share some person with the 3-cliques $\{p_1, p_2, p_3\}$ and $\{p_4, p_2, p_3\}$, so $\{p_1, p_4, p_5\}$ is the only possibility). Now, remove persons $p_1$ and $p_4$ (add $p_2$ and $p_3$ again). A similar argument shows that this forces $\{p_2, p_3, p_5\}$ to be a 3-clique. Hence, $\{p_1, p_2, \cdots , p_5\}$ forms a 5-clique, contradiction. We hereby assume any two 3-cliques share exactly one person.

Consider five persons $p_1, p_2, \cdots , p_5$ such that $\{p_1, p_2, p_3\}$ and $\{p_1, p_4, p_5\}$ both form 3-cliques. Removing $p_1$ and $p_2,$ we easily see that there exists a person $p_6$ such that $\{p_1, p_i, p_6\}$ and $\{p_2, p_i, p_6\}$ are both 3-cliques, where $i \in \{4, 5\}.$ This contradicts our first paragraph, so we're done. $\blacksquare$ [/hide]

If two triangles share exactly one vertex call that vertex a joint and if two triangles share an edge call the edge a hinge.

Take a triangle $abc$. Since all triangles have at least one vertex in common, every triangle forms a hinge or a joint with $abc$. If at most two of $a,b,c$ are joints, we can remove them (and possibly another one or two of $a,b,c$ and no triangle remains).

Suppose that all three of them are joints. Let $auv$, $bwx$, $cyz$ be the triangles that are jointed to $abc$. We will split into cases depending on how $auv$, $bwx$, $cyz$ are connected.

Case 1: All three are connected by hinges.

Note that since the three triangles don't share any edges with $abc$, the three triangles must all have a common hinge of $uv$, $wx$, $yz$. This forms a $K_5$ with $abcuv$, a contradiction.

We can't have exactly two of the pairs among $auv,bwx,cyz$ connected by a hinge because the hinge would have to be common.

Case 2: Two triangles are connected by hinges, and each are connected to the other triangle by a joint.

WLOG suppose that $u=w$, $v=x$. We can't have $\{y,z\}=\{u,v\}$ or they would all have a common hinge, so WLOG $y=u$ and $z\neq v$. We now have the triangles $cuz$ and $avb$ that are disjoint, a contradiction.

Case 3: All three are connected by different joints.

This means that we must have something isomorphic to $u=z$, $v=w$, $x=y$. But now $auv$ and $bxc$ are disjoint, a contradiction.

Case 4: All three are connected by a common joint.

Suppose that $u=w=y$, and $v\neq x\neq z$. Now, by considering the triangles $auv$, $bux$, $cuz$, we have that any triangle that does not have $u$ as a vertex must be of the form $pqr$, where $p\in\{a,v\}$, $q\in\{b,x\}$, $r\in\{c,z\}$. We can't have $vxz$ since that would be disjoint from $abc$. Now we have two distinct cases to take care of since everything is isomorphic to one of these. If it's $axz$ then it's disjoint from $buc$, a contradiction. If it's $abz$ then $abcuz$ is a $K_5$, a contradiction. Hence, there are no other triangles that don't have $u$ as a vertex and we can just remove $a$ and $u$.

First there is triangle ABC.Suppose claim is false.All triangles have at least one of points A,B, or C then if we remove some two of them there must exist another triangle containing third point.So there exist 3 triangles (number of triangles bigger than or equal to 4) such that they contain only one of A,B,C.Now analyze 2 cases how these triangles are connected (takes no more that 5 min) and we are done .

[quote=grobber]I'd love to see another solution..

It's clearly impossible to have more than two triangles s.t. each two have a common edge, so count this case out. Take two triangles $xyz,xuv$ s.t. $\{x,y\}\cap\{u,v\}=\emptyset$. Assume first that all of our triangles which don't have $x$ as a vertex have their vertices among $y,z,u,v$. There must be such a triangle which doesn't have $y$ as a vertex, and the same holds for $z,u,v$, so $xyzuv$ is a $K_5$, which can't be true. This means that among the triangles which don't have $x$ as a vertex, we can find one $tvz$ (for example) s.t. $t\notin\{x,y,z,u,v\}$.

Assume now that the conclusion doesn't hold. This means that we can find triangles which don't have $(x,v),(v,x)$ or $(z,x)$ as vertices. After noticing that $yu,ut,ty$ can't be edges, we can see that the only way this can happen is to have another vertex $p$ and triangles $pxt,pyv,puz$. This, however, means we have the $K_5$ $xyzvp$, a contradiction.[/quote]
why cannot  have more than two triangles s.t. each two have a common edge,?? for example {abc ,abd,abe}
i also wonder if in this graph there are no triangles at all then every persons departure leaves no triangle, is it a contradiction?

[quote=qq1075608078][quote=grobber]I'd love to see another solution..

It's clearly impossible to have more than two triangles s.t. each two have a common edge, so count this case out. Take two triangles $xyz,xuv$ s.t. $\{x,y\}\cap\{u,v\}=\emptyset$. Assume first that all of our triangles which don't have $x$ as a vertex have their vertices among $y,z,u,v$. There must be such a triangle which doesn't have $y$ as a vertex, and the same holds for $z,u,v$, so $xyzuv$ is a $K_5$, which can't be true. This means that among the triangles which don't have $x$ as a vertex, we can find one $tvz$ (for example) s.t. $t\notin\{x,y,z,u,v\}$.

Assume now that the conclusion doesn't hold. This means that we can find triangles which don't have $(x,v),(v,x)$ or $(z,x)$ as vertices. After noticing that $yu,ut,ty$ can't be edges, we can see that the only way this can happen is to have another vertex $p$ and triangles $pxt,pyv,puz$. This, however, means we have the $K_5$ $xyzvp$, a contradiction.[/quote]

i cant understand the meaning of this problem  totally confused... if ther exists 2 triangles like{a,b,c}and {a,d,e}, then only deleting a can leave no triangles if exist 2 triangles like {abc}{abd} then only a,b satisfy the reqirement ...but what is the use of the 5 clique?[/quote]

Consider the case where there are 5 people who all know each other. Clearly all 3-cliques intersect. But if two people leave, there is still a 3-clique. So it is necessary to assume no 5 cliques  ;)

Here is my solution.

Suppose $G$ is a graph with no $K_5$ such that every two triangles in $G$ share a vertex, but it requires at least three vertices to delete to get rid of all triangles. We will now derive a contradiction.

If every two triangles shared an edge, we could delete the two vertices of that edge to get rid of all triangles, so it must be the case that there is some triangle $abc$ and some triangle $axy$ (here all lettered vertices are assumed to be distinct). Since $abcxy$ can't be a $K_5$, one of $bx,by,cx,cy$ is not an edge. WLOG, assume $bx$ is not an edge.

[u]Claim:[/u] There exists a unique vertex $z$ such that $zby$ and $zcx$ are triangles.

[u]Proof of Claim:[/u] Consider a triangle $\Delta$ that doesn't contain $a$. It must have one of $b$ and $c$, and one of $x$ and $y$ (it has to share a vertex with $abc$ and $axy$). Therefore, one of $cy,by,cx$ is an edge of $\Delta$.

If there were no triangles that had $by$ as an edge, then we could simply delete $c$ and $a$, and any such $\Delta$ would get destroyed, and any triangle that did happen to have $a$ would also get destroyed. Therefore, there is a triangle with $by$ as an edge. Similarly we see that there is a triangle with $cx$ as an edge. Now those two triangles must intersect at one vertex, call it $z$.

We now show uniqueness of $z$. Suppose $z'$ was a vertex such that $z'by$ and $z'cx$ were triangles. But then $z'by$ and $zcx$ would have no vertices in common, so $z'$ can't exist, as desired. Thus the claim is proven. $\blacksquare$

We now have triangles $zby,zcx,abc,axy,aby,acx$. To prevent $abycz$ from becoming $K_5$, we must have that $cy$ is not an edge. We claim that every other triangle must have either $z$ or $a$. Suppose $\Delta$ didn't have $z$ or $a$. Therefore, it must have one from each of the pairs $(b,c),(x,y),(b,y),(c,x)$. Therefore, it must contain either $bx$ or $cy$, a contradiction. Therefore, all triangles contain either $z$ or $a$, so deleting $z$ and $a$ kills all triangles. But we had assumed that it takes at least three vertices deleted to accomplish such a feat, so we have arrived at the desired overall contradiction.

Naturally we’ll speak in graphs. Ignore any vertex that doesn’t lie in any triangle. Start with any triangle and label its vertices $A,B,C$. Clearly any other triangle contains at least one of $A,B,C$.

We denote a triangle (distinct of $ABC$) as:
- an $A’-$triangle if it only intersects $ABC$ at $A$, and a $B’-$ and $C’-$triangle similarly.
- an $AB-$triangle if it intersects $ABC$ at both $A,B$, and similarly for $BC-$ and $AC-$triangles.

If there were (wlog) no $C’-$triangles, then all triangles contain $A$ and/or $B$, so removing them we’re done. Now assume there is at least one $A’-$, $B’-$ and $C’-$triangle; let them be $ADX$, $BEY$, $CFZ$.

We will prove the lemma: all triangles that intersect $ABC$ at only one vertex share a vertex $K$ themselves.
To prove this, we first prove any three of them concur. We consider two cases:

a) $D,E,F$ are not all the same vertex; wlog $D\neq E$. In order for $ADX$ and $BEY$ to intersect we must have $X=Y$. We notice that $ABX$ is a triangle. Now, assume neither of $F,Z$ is $X$; then since $CFZ$ is a $C’-$triangle, and since clearly $C\neq Z$, $CFZ$ does not intersect $ABX$, absurd. Then we set $K=X$.
b) $D=E=F$. Then we set $K=D$.

Notice that in both cases we have that $ABK$, $BCK$, $ACK$ are triangles. Assume wlog an $A’-$triangle doesn’t contain $K$. By definition it doesn’t contain $B$ or $C$ either, so it does not intersect triangle $BCK$, a contradiction. Similarly for $B’-$ and $C’-$triangles, we have proved the lemma.

We know focus on (wlog) $AB-$triangles. Assume such a triangle distinct from $ABK$ exists: let it be $ABN$. Take a $C’-$triangle $CFK$; in order for them to intersect, we must have $N=F$. Then $F$ is adyacent to $A,B,C,K$, meaning $ABCKF$ is a 5-clique, a contradiction.

Similarly for $BC-$ and $AC-$triangles, we conclude that all triangles other than $ABC$, $ABK$, $BCK$, $ACK$ intersect $ABC$ at exactly one vertex, hence they contain $K$. Thus, by removing $K$ the only triangle remaining is $ABC$, and removing any of its vertices, we are done.

[hide=Solution]
We shall consider the party as a graph, making each person a vertex and each acquaintance an edge.

First suppose a \(4\)-clique exists, say with vertices \(A,B,C,D\). At most one pair of vertices from \(\{A,B,C,D\}\) share a common neighbour. Suppose \(A,B\) is that pair, with the common neighbour \(X\). Then remove the vertices \(A\) and \(B\) (even if they don't share a common neighbour). We claim that now no triangles exist. Suppose a triangle still remains; then it must contain \(C\) or \(D\). Accordingly let \(CYZ\) be a triangle, where \(Y,Z\not = A,B\). Then the triangles \(YZC\) and \(XAB\) (before the removal) don't share any common vertex, contradiction.

Now suppose no \(4\)-clique exists. We further divide into two subcases:

Assume that there are two triangles which share a side, say \(ABC\) and \(DBC\). Remove \(B\) and \(C\). Now we show that no triangle exists. Suppose there was still a triangle; it must contain either \(A\) or \(D\) (not both, since then we would have a \(4\)-clique). So suppose \(XAY\) was a triangle. But then the triangles \(XAY\) and \(BCD\) didn't share a common vertex before the removal - contradiction.

The only case remaining now is when no two triangles share a common side. Consider any two triangles \(ABC\) and \(ADE\) which share the common vertex \(A\). Delete \(A\). Now clearly no triangles exist. Indeed, suppose a triangle existed. It obviously can't contain \(BC\) or \(DE\) as sides. But it must have one of \(BD,BE,CD,CE\) as sides. However, then \(ABC\) would share a side with one of \(ABD,ACD,AEB,AEC\), contradiction.

Hence, we are done.
[/hide]

Consider a graph with two people connected if they are acquainted. Suppose that it is not possible to delete two vertices and end with no triangles.

[color=#00f][b]Claim:[/color][/b]  There cannot be any 4-cliques.
[color=#00f][i]Proof:[/i][/color]  Suppose $abcd$ is a 4-clique for some vertices $a,b,c,d$. If there are no other vertices, just delete $a,b$. Else, suppose $x$ is a vertex, and suppose it creates a triangle, WLOG $\triangle bdx$. Then $acx$ cannot be a triangle, since otherwise $abcdx$ would be a 5-clique. Hence, any new triangle must have one vertex be $b$ or $d$, by considering $\triangle bdx$. So deleting $b,d$ leaves us with no triangles, contradiction.  $\blacksquare$

[color=#00f][b]Claim:[/color][/b]  There cannot be 2 triangles with a common edge.
[color=#00f][i]Proof:[/i][/color]  Suppose triangles $\triangle abx$ and $\triangle aby$ existed for some vertices $a,b,x,y$. Note that $xy$ is not an edge, else $abxy$ would be a 4-clique. Now, any other triangle must have one vertex be $a$ or $b$. So deleting $a,b$ leaves no triangles, contradiction.  $\blacksquare$

Now, suppose $\triangle ax_1x_2$ and $\triangle ay_1y_2$ are triangles and share vertex $a$. Consider the set $S=\{x_1y_1,x_1y_2,x_2y_1,x_2y_2\}$ of possible edges. Suppose one of these is an edge, WLOG $x_2y_2$. Suppose $x_2y_2$ creates a triangle; $\triangle x_2y_2z$, for some $z$. If $z\in \{x_1,y_1\}$, then we can delete $a,z$ to win. So $z\not\in \{x_1,y_1\}$. But then $\triangle ax_2y_2$ and $\triangle x_2y_2z$ share an edge, contradiction. Therefore, none of $S$ is an edge. We are almost done; finally notice that there cannot be a new triangle $\triangle ax_3y_3$, since any edge then would create two triangles sharing an edge. In conclusion, only $\triangle ax_1x_2$ and $\triangle ay_1y_2$ exist, and now we delete $a$ and win.

[hide=Remarks]The triangle condition is so strong that pushing it enough works. The claims follow naturally from experimentation. [/hide]

Done with [b]eisirrational[/b] and [b]Aryan-23[/b].

[hide=My writeup]

The main idea is we assume for the sake of contradiction that we cannot eliminate all $3$-cliques by deleting two points. We want to show that this forces a construction of a $5$-clique. We rephrase $3$-cliques as triangles for convenience.

Consider some initial triangle $\triangle ABC.$ We must have some triangle that does not have $B$ or $C$ as a vertex, since otherwise deleting $BC$ will still leave no triangle remaining. Also said triangle must have $A$ as a vertex since all triangles have a common vertex. So let this arbitrary triangle be $\triangle ADE$ - we want to show that $ABCDE$ forms a $5$-clique.

[asy]
size(4cm);
pair A=(0,5);
pair B=(-2,2);
pair C=(2,2);
pair D=(-1,1);
pair E=(1,1);
dot("$A$",A,N);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$D$",D,SW);
dot("$E$",E,SE);
draw (A--B--C--A);
draw (A--D--E--A,blue);
[/asy]

Note that we can still eliminate the entire diagram by deleting $A,$ so we need some triangle without $A$ as a vertex. It also must touch at least one of $\{B,C\}$ and $\{D,E\}$ because all triangles have a common vertex. Without loss of generality, say that it touches $\{C,E\}$ and we have a triangle $\triangle PCE$ for some random (and irrelevant) point $P.$ Since $CE$ is an edge, this means that $\triangle ACE$ is a triangle too.

[asy]
size(4cm);
pair A=(0,5);
pair B=(-2,2);
pair C=(2,2);
pair D=(-1,1);
pair E=(1,1);
dot("$A$",A,N);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$D$",D,SW);
dot("$E$",E,SE);
draw (A--B--C--A);
draw (A--D--E--A);
draw (C--E,blue);
[/asy]

But we can still break up all triangles by removing $A$ and $C,$ so we must construct some triangle that does not have $A$ or $C$ as a vertex. Thus, $BD$ is an edge since we have $\triangle QBD$. But note that said point $Q$ must be $E$ since $\triangle QBD$ has a common vertex with $\triangle ACE$ and we've already established that $A,C$ are not common vertices.

[asy]
size(4cm);
pair A=(0,5);
pair B=(-2,2);
pair C=(2,2);
pair D=(-1,1);
pair E=(1,1);
dot("$A$",A,N);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$D$",D,SW);
dot("$E$",E,SE);
draw (A--B--C--A);
draw (A--D--E--A);
draw (C--E);
draw (E--B--D,blue);
[/asy]

Finally, note that we can still eliminate all triangles by erasing $B,E.$ So we must have a triangle without those vertices. Also note that our final triangle must have $A$ as a vertex because of $\triangle ABE,$ it must have $C$ as a vertex because of $\triangle BCE,$ and it must have $D$ as a vertex because of $\triangle BDE.$ So $CD$ is a segment as well, and we are done.

[asy]
size(4cm);
pair A=(0,5);
pair B=(-2,2);
pair C=(2,2);
pair D=(-1,1);
pair E=(1,1);
dot("$A$",A,N);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$D$",D,SW);
dot("$E$",E,SE);
draw (A--B--C--A);
draw (A--D--E--A);
draw (C--E);
draw (E--B--D);
draw (C--D,blue);
[/asy]
[/hide]

[hide=eisirrational's writeup]
Assume for the sake of contradiction that every pair of $3$-cliques intersect, and that there does not exist two or fewer people whose departure leaves no $3$-clique remaining. We shall show that there exists a $5$-clique:

[list]
[*] It follows that for any $3$-clique $ABC$, there exists a triangle that does not contain $B$ or $C$, and so it contains $A$. Call this triangle $ADE$.
[*] Then, we similarly have that there exists a triangle through $C$ not containing $A$ or $B$; moreover, this triangle must intersect triangle $ADE$, so we can assume WLOG that $CE$ is connected.
[*] There exists a triangle through $B$ not containing $A$ or $C$ that intersects $ACE$, so $BE$ is connected.
[*] There exists a triangle through $D$ not containing $A$ or $E$ that intersects $ABE$, so $DB$ is connected.
[*] There exists a triangle through $D$ not containing $A$ or $E$ that intersects $ACE$, so $DC$ is connected.
[/list]
As such, $ABCDE$ is a $5$-clique, as desired. $\blacksquare$
[/hide]

[hide = Solution]
Assume not, so there are no two vertices we can delete such that the remaining graph is triangle-free.

Take a triangle $xu_1u_2$. Since deleting $u_1$ and $u_2$ doesn't work, there must exist a triangle not containing either; then the triangle must contain $x$. Let this triangle be $xv_1v_2$.

Now since deleting $x$ and $u_2$ doesn't work, there must be a triangle containing $u_1$ but not $x$ or $u_2$; then this triangle must contain either $v_1$ or $v_2$. So the graph contains either the edge $u_1v_1$ or $u_1v_2$.

[color=#0B5394][b]Claim:[/b][/color] If the graph contains the edge $u_1v_1$, then it must also contain $u_1v_2$.

[i]Proof.[/i] Deleting $x$ and $v_1$ doesn't work, so there must be a triangle not using either vertex. Then this triangle's overlap with the triangle $u_1xv_1$ must be $u_1$, and its overlap with $v_1xv_2$ must be $v_2$, so we must have the edge $u_1v_2$. $\blacksquare$

By repeatedly applying this argument, if we have one edge $u_iv_j$, then we have all of them. So then the graph contains a $5$-clique on $x$, $u_1$, $u_2$, $v_1$, and $v_2$, contradiction.
[/hide]

Assume towards a contradiction two or fewer people can't leave and leave no $3$-clique remaining. Represent the people as vertices on a graph that are connected if the two people are acquainted.

Then, there must exist a $3$-clique of vertices $A$, $B$, $C$. Note that for each pair, there must be a $3$-clique not including that pair. Applying this logic to $(A, B)$, $(B, C)$, and $(C, A)$ and because we find that each of $A$, $B$, and $C$ must be in a $3$-clique without the other two. Each pair of these $3$-cliques must share a vertex.

Assume towards a contradiction that they all share $2$ points. Then, the graph will have a $K_5$, a contradiction. So, there's a pair of these $3$ cliques, one connected to $B$ and one connected to $C$ (wlog) that only share $1$ vertex. Let $X$ be the shared vertex, and let $B'$ and $C'$ be the non-shared vertex besides $B$ or $C$ in the $3$-cliques connected to $B$ and $C$, respectively. So, the graph looks like a Sierpinski triangle with 4 small triangles with $A$, $B'$ and $C'$ being the main vertices.

We claim $A$ is connected to $X$. There has to be a $3$-clique not including $B$ and $C$. Since it must share vertices with $ABC$ and $XBC$ and it doesn't have $B$ or $C$, it must have $A$ and $X$. By the same logic, $C'$ and $B$ are connected and $B'$ and $C$ are connected. Note $AC'$, $AB'$, and $B'C'$ cannot be edges since otherwise there would be a $K_5$.

Therefore, the $3$-clique without $B$ and $C$ (which includes $A$ and $X$) must include a vertex besides $A, B, C, B', X, C'$. Call that $Y$. By similar logic, the $3$-clique with $B'C$ and the one with $C'B$ must also include a vertex besides $A, B, C, B', X, C'$. Because the $3$-cliques with $BC'$ and $CB'$ must share a vertex with $AXY$, they must include $Y$. However, this creates multiple $K_5$'s, a contradiction.

Denote the people by $A_1,A_2,\dots,A_n.$ If there is only one clique the problem is trivial.

Suppose there are two $3$-cliques that share two people, say $A_1,A_2,A_3$ and $A_1,A_2,A_4.$ Now, removing $A_3,A_4$ would break every 3-clique unless there is 3-clique that contains neither $A_3$ nor $A_4.$ Then this clique must contain both $A_1$ and $A_2.$ Let this clique be $A_1,A_2,A_5.$ Now, we claim that removing $A_1,A_2$ does the job. If any 3-clique remains, it must contain $A_3,A_4$ and $A_5$ which creates a 5-clique, a contradiction.

If there are not two 3-cliques that share two people then let two $3$-cliques be $A_1,A_2,A_3$ and $A_1,A_4,A_5.$ We claim: removing $A_1$ does the job. Otherwise, if a 3-clique still remains it must have $A_2$ or $A_3$ and it must have $A_4$ or $A_5.$ WLOG, let it have $A_2$ and $A_4$ which means $A_1,A_2,A_4$ is a 3-clique. However, that creates two 3-cliques that share two people, so we're done by the previous part.

A pleasure to solve for certain!

The intuition behind this is that if we consider $K_5$ and remove two people, we are left behind with a $3$-clique. With this we may assume the contrary and derive a contradiction.

Assume $\exists$ two people when upon departing leaves behind a $3$-clique. Let the members of this initial $3$-clique be $A, B, C$. If people $D, E$ arrive once more, we have the following subsequent $3$-cliques: $ABD, ABE, ACD, ACE$, all with a common member $A$. Now we have that each person knows four other people. Hence we have a $5$-clique, which is a contradiction.

If a single person say $E$ departs, we have that there is a $4$-clique among the remaining $A, B, C, D$. Hence upon $E$ rekindling himself with the party, we have the new set of $3$-cliques $ABE, ACE, ADE$, which then form a $5$-clique. (In particular $4$-cliques aren't permitted). $\blacksquare$

Consider an arbitrary triangle(the result is obvious if there are no triangles) $ABC$ in the graph. Then every other triangle must share at least one vertex with $ABC$.

[b]Claim.[/b] If there exists three triangles $T_1$, $T_2$, $T_3$ such that $T_1\cap ABC=A$, $T_2\cap ABC=B$, $T_3\cap ABC=C$, then these three triangles share a common vertex.
[b]Proof.[/b] Assume FTSOC that there does exist $T_1=P_1P_2A$, $T_2=Q_1Q_2B$, $T_3=R_1R_2C$ that don't have a common vertex. Then consider $T_1\cap T_2$, $T_1\cap T_3$, and $T_2\cap T_3$.
[b]Case 1.[/b] All three of these have two points. WLOG $P_1=Q_1=R_1$ and $P_2=Q_2=R_2$, then $ABCP_1P_2$ is a $K_5$, contradiction.
[b]Case 2.[/b] Two of these have two points and the other has one. WLOG $Q_1=R_1$, $Q_2=R_2$, $P_1=Q_1$, $P_2=Q_2$. But then we necessarily have $P_1=R_1$, $P_2=R_2$, so $T_1\cap T_3$ also has two points, contradiction.
[b]Case 3.[/b] One of these has two points and the other two have one point. WLOG $Q_1=R_1$, $Q_2=R_2$, and $P_1=Q_1$. Then $P_1=R_1$ as well. But then $AP_1P_2$ and $BCR_2$ are two disjoint triangles, contradiction.
[b]Case 4.[/b] All of these have one point. Then WLOG $P_1=Q_1$.
[b]Subcase 1.[/b] $Q_1=R_1$. Then $P_1=R_1$ and we have a common vertex, contradiction.
[b]Subcase 2.[/b] WLOG $Q_2=R_1$. Then $R_2=P_2$. But then $AP_1P_2$ and $BCQ_2$ are two disjoint triangles, contradiction.

So if no three of these triangles exist, we can just remove two vertices of $ABC$ and finish. Otherwise...

[b]Claim.[/b] This common vertex is common to every triangle besides $ABC$ itself.
[b]Proof.[/b] Call the triangle $T$. If $T\cap ABC$ has one point, apply the previous claim to $T_1$, $T_2$, $T_C$, for example. Otherwise, $T\cap ABC$ has two points, say $AB$, so let $T=ABX$. $T$ must intersect $CR_1R_2$, so either $T=R_1$ or $T=R_2$. The former is what we want, so assume FTSOC that $T=R_2$. But then $ABCR_1R_2$ is a $K_5$, contradiction.

Thus we can remove the common vertex and $A$ to finish. QED.

[hide=Solution]
Assume the contrary. Also, assume that $G$ has at least one triangle, since otherwise the problem statement is obviously true by removing $2$ random vertices. We begin with a claim.
[b][color=#0ff]Claim 1:[/color][/b] Let $ABC$ be a triangle in $G$. Then, there exists triangles $T_1, T_2, T_3 \in G$ for which $ABC\cap T_1 = A$, $ABC\cap T_2 = B$, and $ABC \cap T_3 = C$.
[b][color=#0ff]Proof:[/color][/b] Assume for the sake of contradiction that there did not exist triangle $T_1$ for which $ABC\cap T_1 = A$. Then, for each triangle $T\neq ABC$ for which $A\in T$, we either have that $ABC\cap T = AB$ or $ABC\cap T = AC$. Then, we could remove vertices $B$ and $C$ and have no triangles, since if there existed $T \in G$ for which $B\notin T$ and $C\notin T$, then note that $ABC\cap T \neq \emptyset$, so we must have $ABC\cap T = A$, but we assumed the non-existence of such a $T$, so every triangle $T\in G$ intersects one of $\{B, C\}$, so we can remove vertices $B$ and $C$ and have no triangles. However, we are assuming that the problem statement is false, so this cannot occur, a contradiction. Thus, there exists a triangle $T_1\in G$ for which $ABC\cap T_1 = A$.  Similarly, we can show that there exists triangles $T_2, T_3 \in G$ for which $ABC\cap T_2 = B$ and $ABC\cap T_3 = C$. $\square$

Now, we proceed with another claim.
[b][color=#0ff]Claim 2:[/color][/b] $G$ has a $K_4$.
[b][color=#0ff]Proof:[/color][/b] Let $ABC$ be a triangle in $G$. Then, there exists a triangle $T_1\in G$ for which $ABC\cap T_1 = A$. Let $T_1 = ADE$ for some $D, E\in G$. Then, $B, C, D, E$ are all pairwise distinct. Now, note that there exists a triangle $T_3\in G$ for which $ABC\cap T_3 = C$. Then, we also must have $ADE\cap T_3 \neq \emptyset$, so $D$, $E$, or both is in $T_3$. If both $D$ and $E$ are in $T_3$, then it follows that $T_3 = CDE$, and $\{A, C, D, E\}$ is thus a $K_4$. Else, exactly one of $D, E$ is in $T_3$, WLOG suppose $E\in T_3$. Then, let $T_3 = CEF$ for some vertex $F$ not equal to any of $A, B, C, D, E$. Now, note that there exists a a triangle $T_2\in G$ for which $ABC\cap T_2 = B$. Then, since $ACE$ is a triangle, we must have that $ACE\cap T_2 \neq \emptyset$, so we must have that $E\in T_2$ since $A$ and $C$ are not in $T_2$. Thus, $T_2$ contains $B$ and $E$, so there is an edge from $B$ to $E$, and so $ABCE$ is a $K_4$. Thus, in all cases, we can find a $K_4$, as desired. $\square$

Now, we will prove that $G$ has a $K_5$, which will give our desired contradiction. Let $WXYZ$ be a $K_4\in G$. Then, by our first claim, there exists a triangle $T_1 \in G$ for which $WXZ\cap T_1 = W$. Then, note that since $XYZ$ is a triangle, $XYZ\cap T_1 \neq \emptyset$. Thus, since $X$ and $Z$ are not in $T_1$, we must have that $Y$ is in $T_1$. Thus, we can let $T_1 = WYH$ for some vertex $H$ not equal to $W, X, Y$, or $Z$. Now, note that by our first claim, there exists a triangle $T_2\in G$ for which $WYH \cap T_2 = H$. Then, since $WXY$ is a triangle, we have that $WXY\cap T_2 \neq \emptyset$. Thus, since $W$ and $Y$ are not in $T_2$, it follows that $X$ is in $T_2$. Also, since $WYZ$ ia a triangle, we have that $WYZ\cap T_2 \neq \emptyset$, so by the same logic $Z$ is in $T_2$. Thus, $T_2 = HXZ$, so $H$ is connected to $W, X, Y,$ and $Z$, so $HWXYZ$ is a $K_5$, which gives our desired contradiction.

Thus, $G$ has a $K_5$, as desired.

Does this sol work?
Claim: For any number of cliques at this party, the departure of two people results in no 3-cliques remaining.
We will solve this by induction.
Let the people at the party be vertices, and let there be an edge between two people if they are acquainted with each other.
Let $c$ denote the number of cliques.
Base Case: $c=1$
This is obviously true. The deletion of any vertex works.

Inductive step: Assume the property holds for $c=n$. We will prove that it works for $c=n+1.$
We can add one more clique and abide by the conditions of the problem by adding in one or two points.

Case 1: We add 1 vertex. Call this vertex $P$. Let $P$ be in a clique with vertices $R$ and $S$. Note that $deg(P)=2$. By the problem statement, every pair of 3-cliques has at least one person in common. This means that every 3 clique contains either $P,Q$ or $R.$ However, since $deg(P)=2,$ we can say that every 3 clique contains either $R$ or $S.$ so the deletion of those vertices result in no 3-cliques remaining.

Case 2: We add 2 vertices. Call them $A$ and $B.$ There exists a point that forms the 3-clique with $A$ and $B.$ Call it $C.$ Note that $deg(A)=deg(b)=2.$ By similar logic to case 1, every 3-clique contains $C$, so the deletion of vertex C results in no 3-cliques remaining.
Therefore we have just proved $c=n$ implies $c=n+1$ so our induction is complete.

Therefore for any number of cliques, the departure of 2 or fewer people results in no 3-cliques remaining.

[color=#00f][b]Case 1:[/b][/color] The graph contains a $K_4$.

Each additional triangle must share exactly two vertices with this $K_4$. If 3 or fewer vertices are used by the subsequent triangles, we're finished with Pigeonhole. Otherwise, all subsequent triangles have one common vertex in this $K_4$, from which we also conclude.

[color=#00f][b]Case 2:[/b][/color] The graph does not contain a $K_4$, but at least one pair of triangles share an edge.

Each subsequent triangle must have one of its vertices on this edge to avoid forming a $K_4$.

[color=#00f][b]Case 3:[/b][/color] The graph does not contain a $K_4$, and no two triangles share an edge.

There must be exactly one vertex which lies on every triangle to avoid a contradiction. $\blacksquare$

If there is only one triangle, we can remove any of its vertices and it will be gone.

Suppose there are two triangles $ABC$ and $ABD$. Notice that if we remove $C$ and $D$, any triangles remaining must contain $A$ and $B$, say it is $ABE$. Then, removing $A$ and $B$ would break any triangles since $CDE$ cannot exist (or else there is a $K_5$).

Suppose there are two triangles $ABC$ and $ADE$. Notice that if $A$ were to be removed, a triangle similar to $ABD$ would need to exist, whence we are done by the previous part. $\square$


Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=\frac{1}{2},\ (n-1)a_{n-1}=(n+1){a_{n}}\ (n\geq 2).$

Multipling by $n$ of both sides, $(n+1)na_{n}=n(n-1)a_{n-1}=\cdots = 2\cdot 1\cdot a_{1}=1,$ yielding $a_{n}=\frac{1}{n(n+1)}\ (n\geq 1).$


I faxed in Round 3 on the night of Jan. 4 to the number indicated on the USAMTS website.

However, I never received a Round 3 Rcvd message in my email - ever. If you did, around when did you get it?

Anyway, I sent Mr. Kelly an email about it on Jan. 12, and he never wrote back. After another week without a response (Jan. 19) I faxed another copy of my solutions to that number, with a letter on top of the cover page explaining my situation.

As of now, I still have not yet received either a response from Mr. Kelly, a "Round 3 Rcvd" email, or my scores (has anyone gotten their Round 3 scores yet?).

Is this normal or something I should be concerned about?

Also, they said the Round 4 problems would be released last week of Jan or first week of Feb., and it hasn't happened.

[i]EDIT - They said late Jan or early Feb. Well, there are technically only four more days which you can call "early" Feb. :D[/i]

The USAMTS is horribly organized, probably due to the increased participation this year. Like this is the first year I participated--I didn't participate previous years because of the poor typesetting, and this year they LaTeX'd it (FINALLY). They used to be able to keep an active connection between contest management and the participants, but I guess not anymore... but it is free... I guess the incompetence is excusable?

The questions are supposed to be posted at least 30 days before the deadline.  Since there are only 28 days in February and the deadline is March 14th, they are supposed to be posted by February 12th, i.e. thursday.  I've checked the website at least 3 times each day, and still no cigar...

i cant remember if i got round 3 scores yet...

But i know when i faxed in my stuff i never got an email saying it was received, but I did get my score.  So dont worry about it.

Don't worry. I didn't get a message saying that they received mine either, but I got my scores back today. (I mailed mine though).

I have not received the confirmation email either.....

[quote="mathfanatic"]Oops, forgot about the leap year.  Bah.

But as for the confirmation e-mail, I got one for round 3, but I haven't recieved my scores yet.[/quote]

same here.

Now that Im thinking about it, I never did recieve a confirmation letter. And i havent recieved my scores either.

i havent received the email or my scores. does anyone know when the aime qualifying cutoff will be announced?

I received my scores in the mail today and noticed that there was  [b]no score for Problem #3[/b], even though I faxed it in.

What can I do about this?

received my solutions today, yay! hmm my score has decreased.

Got my scores today.  To quote someone*, m00t.

[hide]*Neal[/hide]


AHHHH! geez! thank you so much stevem! my texnic center hadn't been working for about 3 months now! thanks again!  :D

I had a similar problem, but I can't figure out how to fix it.  How did you do it?! Help!

you must open an acrobat-reader application, because by some reason the texnic-center isn't able to execute it itself, but if the acrobat reader is already running then everything works fine (at least with my computer).

<edit>

stevem is right and has the better solution. the above strategy works but isn't very fine work. So how to fix the problem:
1. open texniccenter
2. go to Build>Define Output Profiles>Viewer
3. select the Latex=>PDF - profile
4. Enter the Path of your acrobat reader under Path of executable

Now it should work.
I missed the moment when my system broke and didn't really care about it, because I found a (not so pretty) solution). The thing was I installed a newer version of the acrobat reader, so the executable path wasn't correct anymore, so my above post is selfevident.

[quote="pkothari13"]i have a problem with my texnic center.... everytime i try to compile my file, it always says:

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
Cannot Execute the Command.

i would really like to get this fixed, i tried uninstalling mixtex and texnic center and i reinstalled them, but still the same problem!! can anyone help?????  :([/quote]

Check your acrobat

No wonder I get non-existent messages. Now it's fixed.  :)

Aha!. I know now.. first let me know if you are running any kind of firewall.

In the second post above it says:
[quote="stevem"]In TexnicCenter check the paths in the Profiles (press Alt+F7) are correct.

Click on LaTeX=>PDF and on LaTeX=>PS=>PDF then on Viewer tab and make sure that Path of executable is pointing to the correct path for Adobe Reader.[/quote]Have you done this?

Yes, I checked it.
In Viewer I have: C:\PROGRAM\ACROBAT 7.0.9\READER\ACRORD32.EXE

Could the mistake be connected to the Postprocessor?

For prosper as I understand one needs to run LaTeX=>PS=>PDF???

I also see error message: latex.exe: The remote package repository is corrupted. What can one do?

In TexnicCenter check the paths in the Profiles (press Alt+F7) are correct.

Click on LaTeX=>PDF and on LaTeX=>PS=>PDF then on Viewer tab and make sure that Path of executable is pointing to the correct path for Adobe Reader.

[quote="tigu"]Yes, I checked it.
In Viewer I have: C:\PROGRAM\ACROBAT 7.0.9\READER\ACRORD32.EXE[/quote]Is acrord32.exe actually in C:\PROGRAM\ACROBAT 7.0.9\READER? The only reason I ask is that I have version 7.0.9 and my path is C:\PROGRAM FILES\ADOBE\ACROBAT 7.0\READER\ACRORD32.EXE
[quote="tigu"]Could the mistake be connected to the Postprocessor?[/quote]You can check by compiling [code]\documentclass[12pt]{article}
\begin{document}
Just testing
\end{document}[/code]using LaTeX=>PDF and see if you have any errors.
[quote="tigu"]I also see error message: latex.exe: The remote package repository is corrupted. What can one do?[/quote]It sounds like MiKTeX is trying to install a package. Compare the packages you are loading with \usepackage with MiKTeX's Browse Packages. If something needs installing, try doing it by using Browse Packages. Right-click on the package and click on Install.

Hi,
thanks a lot for your reply but...it is still not working

In Viewer I have: C:\PROGRAM\ACROBAT 7.0.9\READER\ACRORD32.EXE
Is acrord32.exe actually in C:\PROGRAM\ACROBAT 7.0.9\READER? The only reason I ask is that I have version 7.0.9 and my path is C:\PROGRAM FILES\ADOBE\ACROBAT 7.0\READER\ACRORD32.EXE

Yes, I have checked that.

tigu wrote:
I also see error message: latex.exe: The remote package repository is corrupted. What can one do?
It sounds like MiKTeX is trying to install a package. Compare the packages you are loading with \usepackage with MiKTeX's Browse Packages. If something needs installing, try doing it by using Browse Packages. Right-click on the package and click on Install.

How can I do it in Texniccenter?

P.S. Earlier have I used WibEdt, but now have to use TexnicCenter and ...

[quote="tigu"]In Viewer I have: C:\PROGRAM\ACROBAT 7.0.9\READER\ACRORD32.EXE
Is acrord32.exe actually in C:\PROGRAM\ACROBAT 7.0.9\READER? The only reason I ask is that I have version 7.0.9 and my path is C:\PROGRAM FILES\ADOBE\ACROBAT 7.0\READER\ACRORD32.EXE

Yes, I have checked that.[/quote]I find it odd that Adobe is installed in c:\program\... and not c:\program files\.... In TexnicCenter, Tools, Define Output Profiles, Latex=>PDF click on Viewer then on the top button labelled ... and navigate to where acrord32.exe is and press Open. Repeat for Latex=>PS=>PDF, then OK and try compiling again.

Unfortunately, this doesn't help:((

I haven't installed Adobe this is our Administrator work. But it would be odd if TexnicCenter cannot work just because of that.
I have never had any problem with WinEdt when running Prosper.

[quote="tigu"]Unfortunately, this doesn't help:(([/quote]Was it c:\program files rather than c:\program?
[quote="tigu"]I haven't installed Adobe this is our Administrator work. But it would be odd if TexnicCenter cannot work just because of that.
I have never had any problem with WinEdt when running Prosper.[/quote]TexnicCenter will work provided that it knows where Adobe Reader is and that DDE is working. Is a pdf file being produced and the only problem is that it isn't being opened?

Adobe is located in c:\program in my case.

"TexnicCenter will work provided that it knows where Adobe Reader is and that DDE is working.

Is a pdf file being produced and "

Exactly. It cannot open it.

If you are sure the path is correct then in Build, Define Output Profiles, LaTeX=>PDF, Viewer, View project's output, try changing DDE to command line argument and click on the arrow to the right of the box and choose main file's full path.

still getting message:

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
Cannot Execute the Command.

And in the log-file it is written (among others) that dvi-file cannot be opened...

[quote="tigu"]still getting message:

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
Cannot Execute the Command.[/quote]Then you can't have made the change from DDE correctly since such a command wouldn't exist any more. Are you using LaTeX=>PDF or LaTeX=>PS=>PDF? Have you made the change for the relevant one?

[quote="tigu"]And in the log-file it is written (among others) that dvi-file cannot be opened...[/quote]You need to give a lot more information for this to be analysed.

I have a weird problem here, when I compile with LaTeX => PS => PDF, the pdf file that i got, will not show my header. as in the paper size is smaller. In the log, it says "unknown" paper size even i have A4 set in MikTeX. Another weird thing is I had this profile functioned before but it didn't work out as it should after a re-installation.

I found a solution at Rug BER MikTeX page: http://tex.aanhet.net/miktex/download.html

His LaTeX => PS => PDF VBscript works on my box. Finally, after 3 hours of trials and errors.

Hope this help.

this actually just happened to me.  Make sure that you have the     \end{document}    tag at the end of your tex file.

Since this comes up first on google on this subject, I'll add this very delayed piece of information anyway. If you get the "Cannot execute the Command" window when trying to view a output pdf in TeXnicCenter, check that the name of the folder in which the file resides, does not contain any spaces. Windows (I've only tested in XP) allows these folder names, but I'm guessing that the reader doesn't allow spaces in the path of the file, and would rather prefer 8.3 style names or simply no spaces.

[quote="awyarbr"][quote="awyarbr"]I'm receiving this error, but it turns out that my .pdf file isn't being generated!  But the file compiles with no errors.

Interestingly, the pdf [i]is[/i] produced using LaTeX => PDF instead of LaTeX => PS => PDF, except that none of my .eps files are recognized.[/quote]

<sigh> Okay, I found a working installation and found the problem.  Somehow, all my post-processing commands (on the PostProcessor tab of the Profiles dialog) were deleted!  These should include, at a minimum, dvips.exe in the miktek installation and gswin32c.exe in the ghostscript installation.  There are a bunch of arguments that should be included, which I will post if anybody is interested.

Since this generated the same error, I thought I'd bring it to your attention here.[/quote]

I would be very interested in all the arguments. I believe I have the same problem, thanks.

[quote="tigu"]Hi,
When I try to run prosper-document in TexnicCenter, I get
[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
Cannot Execute the Command.
What could be wrong?[/quote]
I had exactly the same problem. Before it everything used to work properly. This forum helped me, but the way I used was wizard in "define output profiles". Unfortunately, I began to look for the solving of
this problem in the internet after a big pause in working.


Give an equilateral triangle ABC and a point M moves in the plane (ABC). Let A',B',C' be respectively the symmetric of A,B,C wrt M.
(1) Prove that the perp bisector of AB',BC',CA' all pass through a point P.
(2) Denote M be the midpoint of A, and N be the intersection of DM with AP. Prove that the circle (MNP) always passes through a fixed point when M moves (M doesn't concide with D).


Onomazoume affine metasximatismous tou epipedou sinartiseis tis morfis g(x)=Ax+B, opou A,B einai migadikoi arithmoi (kai i g einai mia sinartisi apo to C sto C). Gia kathe tetoia sinartisi grafo to A os d(g). Paratiroume oti i g exei ena stathero simeio x sto C (i.e simeio x oste g(x)=x) an kai mono an eite B=0 eite d(g)<>1. Se autin tin periptosi groufoume to stathero simeio x os fix(g) (kai einai monadiko, giati ?) Episeis ean f,g einai affine metasximatismoi, tote fg simbolizei tin sinthesi f(g(x)) ton f kai g.

Problima :

Esto f,g,h affine metasximatismoi etsi oste d(fg), d(gh), d(hf) kai d(fgh) den einai 1, kai esto j=d(fgh)
Tote to (a) kai (b) einai isodinama :
(a) $ f^3g^3h^3 \equal{} 1$, diladi $ f^3g^3h^3(x) \equal{} x$ gia kathe migadiko x.
(b) $ j^3 \equal{} 1$ kai $ a \plus{} j*b \plus{} j^2*c \equal{} 0$ opou a=fix(fg), b=fix(gh), c=fix(hf).

Endiaferon erotima :

Poia gnosti protasi geometrias einai isodinami me to parapano problima ? !

To deutero erotima exei poli endiaferon giati parexei mia isodinamia metaksi enos geometrikou problimatos kai tin afirimeni algebra. Argotera sto panepistimio tha deite oti to arxiko problima isxuei opos leme se kathe algebriko soma (i apla division algebra) kai sinepos anadiknuei tin diikotita metaksi geometrias kai algebras, simantikes kai bathies idees pou odigoun se poli omorfa mathimatika opos p.x. tin algebriki geometria. ( Kai opou o monos tropos na ta mathete einai [b]mono[/b] meso kapoiou tmimatos mathimatikon  :wink:  )


geometry
rectangle
search
combinatorics unsolved
combinatorics

Prove that one can't put $ 25$ pawns on a chessboard such that no four of them forms a rectangle with sides parallel to the boundary of the board.

Probabmy something like that
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=389965051&t=14949

Pierre.


function
real analysis
real analysis solved

Prob:Let f be a real surjective function satisfied :for any sequences (x _n) such that f(xn) converges then (x_n) converge.Prove that f is a continuos function.

could it be you posted this twice?

mods plz delete the other :)

The proof holds for all separable spaces A,B .  The function is f:A-->B
Let y be in B, so there is t\inA such that y=f(t), because of  f surjective.
Let z(n) be a sequence in B such that z(n) converges, and z(n)---> y.
For each z(n) there is x(n)\in A such that z(n)=f(x(n)), because of  f surjective.
So the sequence x(n) converges, and let L be its limit (unic of course, very important)
Let u(n) be a sequence such that u(2*n)=t, u(2*n+1)=x(n)
Manifestly f(u(n)) converges, f(u(n))---> y=f(t) , so u(n) converges, so t=L, and t is unic.
Therefore f is injective, so bijective , and f<sup>-1</sup>(x) is continous, so f(x) is continous.

I study this proof carefully, it does not show that f is injective.
It shows that there exist t in A, there exist L in A,
with the sequence u(n) you have t=L and you deduce f(L)=f(t).

It does not prove that  f is injective
For injectivity  for any a in A, and any b in A such that f(a)=f(b) then a=b

Umm...Is injective on f required? My solution does not require f to be injective in order for f to be continuous on A.

This is the first problem from the final round of the RMO in 1995, 11'th grade:

f:R->R is a surjective function with the property that if the sequence (xn)n>=1 is [b]divergent[/b] then so is the sequence (f(xn))n>=1. Show that f is continuous.

Maybe dreammath was referring to this problem in the first place, because they just look too similar... Anyway, try this one (even if it's a totally different problem).

Oops! Sorry! I just realized that the 2 problems are the same (they're equivalent, since the negation of "convergent" is "divergent"):D.

Anyway, here's how we can prove the injectivity: Let's assume there exist a=/=b s.t. f(a)=f(b). We take the sequence (xn)n>=1 to be a, b, a, b, a, b etc. Then (xn)n>=1 is divergent, but (f(xn))n>=1 isn't, so we have a contradiction.

Woahou great grobber your idea is like light  :)  at midnight time
thank you very much man

Let's assume that f(x) doesn't tend to f(t) as x->t for some real t. Then there is a sequence xn->t s.t. f(xn)->c=/=f(t). There exists a number a s.t. f(a)=c. Then the sequence x1, a, x2, a, x3, a etc. is divergent, but f(the terms of the sequence) tends to c, so it's convergent, so we have a contradiction.

This works for c\in R; I haven't worked out the case c=00.

Never mind that.

We know f is injective, so it's bijective. Denote f-1 by g. Then, we know that for xn convergent g(xn) is also convergent. The conclusion that f(x)->f(t) as x->t follows easily from here.


geometry
geometric transformation
reflection
AMC
AIME
AMC 10
AMC 10 A

you might want to edit the title of the post to better reflect the content

my school offers the A to everyone in Honors math, and then the B to some people who barely didn't make it. however, there is a close by program called EPGY that offers both tests to anyone wishing to take it

Our school will probably just do one.

If they do it at all.

just A except when we do just B instead for strange reasons

Actually my school did the AMC 10 A and the AMC 12 B last year . . . the AMC A date was the day of our first semester finals, so we had a relatively small number of students take it. But, at least I got to take both tests.

[quote="breez"]Does your school do both forms of the tests (A and B) or just one out of the two?[/quote]

For us, it really depended on who wanted to take both and if there are enough people (i think amcs come in packs of 12?), then we'd do both.

Our school does AMC 12A to all people who qualified for part II in MMPC. (MIchigan)

Then our school does 10B and 12B, to anyone.

Our school does AMC 10/12 A/B.  I don't think we do AIME II though.

Our school takes the 10/12 A. I should probably try to get them to offer the B tests as well  :lol:

I don't know what my school does...

(Since it's a new school)

I'd guess both. I'd HOPE both.

this year my school was going to do A but it was mysteriously cancelled the day of the test, so we took the B instead.

[quote="akalra1"]r skool takes AMC but i dunno weter it is A or B[/quote]
No leet speak please. It makes the forum seem more unprofessional

Our school only wrote AMC B because we had exams on the AMC A date.

my school did A and B just round 1 test A and round 2 test B

[quote="amirhtlusa"]over here, every year my schooled took the B test (only one test per year both 12 and 10).
but this year was the first time we took the A and i got 150, so i recomend A instead of B. :)[/quote]

relative difficulty varies

i think my school does 2, but it is optional...

My school does both, but B is over our vacation this year so I'm expecting turnout to be lower than normal this year...

However, offering both tests helps the underclassmen do well on the AMC 10 while giving them training for the AMC 12.

[quote="Phelpedo"]Some say that the B is usually harder.[/quote]

That's definitely not true. "Harder" and "Easier" on the AMCs and AIMEs almost categorically depend on what type of score you're aiming for. Last year (2005), for instance, AMC 12 A had one perfect score (omg go zabel!) but B had...well...quite a few.

I said it was a rumor, not a fact. In any case, number of perfect scorers is a very poor way of judging difficulty.

[quote="Phelpedo"]In any case, number of perfect scorers is a very poor way of judging difficulty.[/quote]

I'm not sure that there exists a very good way to judge difficulty. Tests like the AMC have takers of such a wide spectrum that "hard" for one group is very easily "easy" for another (relatively speaking in both examples, of course). You could use the mean scores, median score, etc. but then that fails to identify "difficult" as it pertains to any particular person at all. I mentioned the number of perfect scores because at the top of the scale, that's a  simple and often discussed statistic.


Let $ ABCDEF$ be a hexagon inscribed in a circle $ O$, with $ BD$ perpendicular to $ CF$. Let $ G\equal{}BD\cap CF, H\equal{}AC\cap BE, I\equal{}AD\cap CE$. Assume that $ AI\equal{}CI$. Prove the relation

$ CH\equal{}AH\plus{}DE\leftrightarrow GH\cdot BD\equal{}BC\cdot DE$

Can you tell where? Or show the solution?


vector
geometry
parallelogram
analytic geometry
trigonometry
trig identities
Law of Cosines

[quote="236factorial"]One force of 20 pounds and one force of 15 pounds act on a body at the same point so that the resultant force is 19 pounds. Find, to the nearest degree, the angle between the two original forces.[/quote]

[hide]Do you just draw a triangle with lenghts 20, 15, and 19 and find the angle opposite the 19 side?

64$^\circ$[/hide]

No ch1n353ch3s54a1l, that's the subtraction of two vectors. The resultant is the sum of the vectors.

[hide="Hint#1"]Parallelogram law[/hide]

[hide="Hint#2"]Choose one as the horizontal and find the coordinates of the other vector.[/hide]

[hide="Hint#3"]Dot product[/hide]

[hide="Lastly a very slick trick you can do here..."]Angles are supplement because of parallelogram law :D[/hide]

[hide]Make parallelogram $ABCD$ so that $AB=20$, $BC=15$, and $\angle B$ is acute.  We know that $AC=19$.  We have two 15-19-20 triangles.  Now, we just use the law of cosines to find $m\angle DAB$, which is $m\angle DAC+m\angle CAB$.  Let $X=\angle DAC$ and $Y=\angle CAB$.
$20^{2}=15^{2}+19^{2}-2(15)(19)\cos X \Leftrightarrow 400=586-570\cos X \Leftrightarrow 186=570\cos X \Leftrightarrow \cos X=\frac{31}{95}\Leftrightarrow X=\arccos\frac{31}{95}\approx 70.955^{\circ}$
$15^{2}=19^{2}+20^{2}-2(19)(20)\cos Y \Leftrightarrow 225=761-760\cos Y \Leftrightarrow 536=760\cos Y \Leftrightarrow \cos Y=\frac{67}{95}\Leftrightarrow Y=\arccos\frac{67}{95}\approx 45.149^{\circ}$
Add them up to get approximately $\boxed{116^{\circ}}$.[/hide]

That's an interesting way of solving it lingomaniac88.

[hide="My solution"]
Place point A at the origin and draw parallelogram ABCD with $|AB|=20$ and $|AD|=15$ and put AD on the x-axis. Make $|AC|=19$. The angle made by vector AB is the angle between the two original forces.

In coordinate vectorial notation, $AD=(15,0)$. Let $AB=(x,y)$. Then, $x^{2}+y^{2}=20^{2}$

Since $AB+AD=AC$ by parallelogram law, the addition of vectors is the sum of the individual components, and so $AC=(15+x,y)$. So, $(15+x)^{2}+y^{2}=19^{2}$

Solve and you get $x=-\frac{44}5$

By the dot product, $\cos\angle BAD=\frac{AB\cdot AD}{|AB\parallel AD|}$

So, $\cos\angle BAD=\frac{-\frac{44}5\cdot15+y\cdot0}{20\cdot15}=-\frac{11}{25}$ and $\angle BAD=\arccos\left(-\frac{11}{25}\right)\approx116^\circ$
[/hide]

[hide="pythagorean theorem way?"]
let the vector with magnitude 15 lie on the x-axis, and let $\theta$ be the angle separating the two vectors.
The net horizontal displacement is equal to:
$20\cos\theta+15\cos{0}$
The net vertical displacement is equal to:
$20\sin\theta+15\sin{0}$
The net displacement vector with magnitude 19 is represented by: $<20\cos\theta+15\cos{0},20\sin\theta+15\sin{0}>$,
so we can use the pythagorean theorem:
$(20\cos\theta+15\cos{0})^{2}+(20\sin\theta+15\sin{0})^{2}=19^{2}$
$400\cos^{2}\theta+600\cos\theta+225+400\sin^{2}\theta=361$
$400+600\cos\theta+225=361$
$\theta=116$

but my method probably isn't much different from the ones above..
[/hide]


geometry
rectangle
calculus
integration
algebra unsolved
algebra

A $ 6\times 6$ square is dissected into $ 9$ rectangular regions by lines parallel to its sides in such a way that all these rectangles have only integral sidelengths. Prove that, out of these nine rectangles, at least two have to be congruent.

the sums of the areas must be greater than 1+2+3+4+5+6+7+4+6=28+10=38>=36, because 4 and 6 can be writen in 2 products and the others have just a way to be writen

Suppose the nine rectangles are different. But the smallest nine different rectangles have dimensions $1\times 1, 1\times 2, 1\times 3, 1\times 4, 2\times 2, 1\times 5, 1\times 6, 2\times 3$ and $2\times 4$. Their areas add up to more than 36, and this proves the claim.


number theory
relatively prime
number theory unsolved

There is a pair of positive integers $(a,b)$ such that both:
$a^2+b^2$ and $a^2+9b^2$
are perfect squares?

First what we can see, that WLOG we can assume that $(a, b)=1$.
Suppose that $(a,b)$ is a solution with minimum sum $a^2+b^2$ and
put $a^2+b^2=c^2$ and $a^2+9b^2=d^2$, then $3\not | d$. Really, if
$3|d$ then $3|a$ so $a=a_1\cdot 3$ and then $d^2=a_1^2+b^2$ and
$c^2=9a_1^2+b^2$ so we get another solution with smaller sum
$a_1^2+b^2$ and that is not possible.

So from that we have $c^2=a^2+b^2$, $d^2=a^2+9b^2$, $(a,b)=1$ and
$3\not | d,a$. Then $8b^2=d^2-c^2=(d-c)(d+c)$ and
$8a^2=(3c-d)(3c+d)$. Now it can only be $d\equiv_4 c$ or $d\equiv_4 -c$ because $d$ and $c$ are odd. Suppose that $d\equiv_4 c$.

$8b^2=(d-c)(d+c)\Rightarrow (2b)^2=(d-c)\frac{d+c}2$. $(d-c, \frac{d+c}2)=1$ because if we suppose that $(d-c,\frac{d+c}2)=e$
then $e|2\cdot \frac{d+c}2 - (d-c)=2c$ so $e|c$ because
$\frac{d+c}2$ is odd (remember $d\equiv_4 c$). In other hand
$e|4b^2$ so $e|b^2$ and $e|c^2$ so $e|a^2$. As you can se, numbers
$d-c$ and $\frac{d+c}2$ are relatively prime and their product is
perfect square, so they are both perfect squares. Similarly, from
$8a^2=(3c-d)(3c+d)$ and $3\not |d, a$ we have that numbers
$\frac{3c-d}2$ and $3c+d$ are relatively prime with a square as
product so they are perfect squares to. So, for some integers $k, l, m, n$ it stands $d-c=k^2, \frac{d+c}2=l^2, \frac{3c-d}2=m^2, 3c+d=n^2$ where $k$ and $n$ are even and $l$ and $m$ are odd. So
$-1\equiv_4 n^2-l^2=\frac{5c+d}2\equiv_4 \frac{c+d}2 = l^2$
 contradiction!!

If you assume that $d\equiv_4 -c$ all is the same. Put
$(2b)^2=(\frac{d-c}2)(d+c)$ and $(2a)^2=(3c-d)(\frac{3c+d}2)$. Then
again they are relatively prime so $k^2=\frac{d-c}2, l^2=d+c, m^2=3c-d, \frac{3c+d}2=n^2$, and now $k,n$ are odd and $l,m$ are
even. Now $-1\equiv_4 l^2-n^2=\frac{d-c}2 = k^2$ so again
contradiction!!


trigonometry
Asymptote
real analysis
real analysis unsolved

Let $ A\equal{}\{n_{1},n_{2},...\}$ be a set of positive integers such that the sequence $ (\cos n_{k})_{k\geq 1}$ is convergent. Prove that $ A$ has zero density.

Converting $ \cos n_{k\plus{}1}\minus{}\cos n_{k}$ into a product, you should be able to conclude that $ n_{k\plus{}1}\minus{}n_{k}\to\infty$.

Erm, no. There is a possibility that $ n_{k\plus{}1}\minus{}n_{k}\equal{}1$ infinitely often: ;)
[asy]import cse5; size(150); D(CR(D((0,0)),1)); MP("\approx n_{k+1}",D(dir(29)),NE); MP("\approx n_k",D(dir(-29)),SE);[/asy]

Fair enough.

Converting $ \cos n_{k\plus{}1}\minus{}\cos n_{k}$ into a product, we conclude that
1) either $ n_{k\plus{}1}\minus{}n_{k}$ is large, or $ n_{k\plus{}1}\plus{}n_{k}$ is close to a multiple of $ 2\pi$.
For the same reason,
2) either $ n_{k\plus{}2}\minus{}n_{k\plus{}1}$ is large, or $ n_{k\plus{}2}\plus{}n_{k\plus{}1}$ is close to a multiple of $ 2\pi$.

From 1) and 2) it follows that $ n_{k\plus{}2}\minus{}n_{k}\to\infty$. Therefore, both subsequences $ n_{2j}$ and $ n_{2j\plus{}1}$ have density zero, and so does the whole sequence.


Amy, Betty, Cathy, and Daisy have $ 36$ apples that they plan to split among themselves in a ratio of $ 2: 3: 3: 4$, respectively. How many apples will Betty and Cathy have together?

The question asks for how much Betty and Cathy have together. We can model this in respect to what fractional part they have of the whole without even solving for any x.

$ \left(\dfrac{3x\plus{}3x}{2x\plus{}3x\plus{}3x\plus{}4x}\right)(36)\equal{}\left(\dfrac{1}{2}\right)(36)\equal{}\boxed{18}$

$(2+3+3+4)k = 36$, because ratios can be simplified by any number: we call this number $k$ for constant.
Then just solve. $12k = 36$, thus making the original numbers of apple be 3 times of the ratio.
Our "new" ratio is this: $6 : 9 : 9: 4$. Read the problem again; it states that you want Betty's and Cathy's apples, which is $3 + 3$, making the answer $\boxed{6}$ $\textbf{Ans.}$

[hide="Solution"]Looking at the ratios, we see that Betty and Cathy get $6$ parts, and A and D get $6$ parts.  Thus, B and C  get half the apples, or $36 \div 2$ \[\boxed{18}\][/hide]


[quote="shobber"]In a convex pentagon $ABCDE$, $\triangle{ABC}=\triangle{BCD}=\triangle{CDE}=\triangle{DEA}=\triangle{EAB}=1$, find the area of $ABCDE$.[/quote]

[hide]I got $ABCDE\approx9.105929973$[/hide]


Find the largest value of $ y$ which is a solution of $ y^3 \minus{} 12y^2 \plus{} 11y \equal{} 0.$

We factor this as $ y^3\minus{}12y^2\plus{}11y\equal{}y(y\minus{}1)(y\minus{}11)\equal{}0\implies y\equal{}\{0,1,\boxed{11}\}$.

hello, because $ y^3\minus{}12y^2\plus{}11y\equal{}(y\minus{}11)(y\minus{}1)y$ the searched value is $ \boxed{y\equal{}11}$.
Sonnhard.


function
linear algebra
matrix
algebra
linear equation

Question:
Let A and B be invertible n×n matrices and b be an n×1 vector.

Write a MATLAB function with inputs (A,B, b) to solve the equation x=B^−1*(2A^−1 + 1)b

Make use of functions "LU Facotrization, Forward Substituion and Backwards Substitution" and DO NOT calculat any matrix inverse.

-------

Apart from trying to write the code, Im kind of lost in understanding the math - which is much more important.

I want to LU-factorize A and B to get and upper and lower matrix for each of A and B.

With these matracies, how would I go abouts get it's inverse?

Forword substitution solves for x in Lx = b
and
Backword substitutions solves for x in Ux=b

I do not know where to go from here..

I have no idea how to arrive at x=B^−1*(2A^−1 + 1)b

Could somebody help me out please?

[quote="jmerry"]If we're using $LU$ factorization, we don't find inverses. The $LU$ factorization is designed to make it easy to solve multiple linear equations with the same matrix without actually finding inverses.

With that in mind, we need to rewrite the equation without inverses.

$x=B^{-1}(2A^{-1}+I)b$
$Bx=2A^{-1}b+b$
$ABx=2b+Ab$

Calculate $2b+Ab$; that's cheap. I'm not sure whether you want to compute and factorize $AB$ then solve the linear equation, or factorize $A$ and $B$ separately, then solve the chain $Ay=2b+Ab,y=Bx$.[/quote]

Ohh, I see what you are saying..that is so awesome, I didn't think of it like that.

We have:

$ABx=2b+Ab$

We know what $2b+Ab$ is.  Let $2b+Ab = C$

Therefore;

$ABx=C$

Let $AB = D$

Therefore,

$Dx=C$

Then use LU factorization on D and behold,

$LUx=C$

Then use forward and backward substitution to solve for x!

Now that I have the math correct...Onto the matlab function

Thank you so much for your help jmerry, I appreciate that!


calculus
derivative
calculus computations

An marathon athelet spent 2.2 hours finishing the New York Marathon which is about 26.2 miles. Please prove that the athelet at least two times to run with the speed of 11 miles per hour.

I think there may be some thing wrong in the question but I am not sure.Could you answer it?

The key to this problem: the marathoner starts and ends at speed zero. His average speed is greater than 11, so somewhere along the track his speed is greater than 11. Derivatives have the intermediate value property.

Yes,there is a idear of reason in your answer.
But I think there is an instance which has not appeared in the practice:may be the marathoner speed up all the 2.2 hours.And his speed is not at zero in the end.
What about your opinion??

True enough. The marathoner might not drop below 2.2 mph until after crossing the finish line.

On the other hand, I've never seen a marathon runner who didn't stop immediately after the finish.

:) :lol: This is an exercise in <Thomas' calculus > .
I quite agree with what you say.
Sorry,my English is so poor!I'm going to improve it.


If the product of $ n$ numbers is equal to $ \equal{}n$ and the sum of this number is equal to $ \equal{}0$ prove that $ n$ is divisible by $ 4$

If $ n$ is odd then all the factors have to be odd, too, and as there is an odd amount of them they will add up to an odd number, i.e. not to 0.

If $ n$ is divisible by 2 then exactly one of the factors is even, and the remaining $ n\minus{}1$ factors add up to an odd number, i.e. not 0.

So $ n$ has to be divisible by 4. :)


calculus integration trigonometry logarithms LaTeX

If I am not wrong, that antiderivative also doesn't seem to be expressable as a finite sum of elementary functions.
However, it is interesting to note that the same doesn't happen with

\[\int \sqrt{\tan x}dx\]

Carcul

[quote="melissimon"][url=http://imageshack.us][img]http://img227.imageshack.us/img227/9586/zazaavm0.gif[/img][/url][/quote]

This problem does not belong in High School Basics.

It goes in the calculus forum.

And yeah the sine is complicated, the tan is complicated but nice.

$\frac{1}{2\sqrt{2}}(-2\arctan(1-\sqrt{2}\sqrt{\tan{x}})+2\arctan(\sqrt{2}\sqrt{\tan{x}}+1)+\ln(-\tan{x}+\sqrt{2}\sqrt{\tan{x}}-1)-\ln(\tan{x}+\sqrt{2}\sqrt{\tan{x}}+1)$

Just as a pointer, I would recommend that you just use LaTeX to post math.  You don't have to upload any pictures or anything, because that takes longer.


Find all real polynomials p such that
$(x-16)p(2x)=16(x-1)p(x)$

[hide]
I got $p(x)=0$ as the only solution... is it right?
[/hide]

No.  :wink:

$x=16 \Rightarrow p(16)=0$
$x=1 \Rightarrow p(2)=0$

So $p(x)=(x-2)(x-16)q(x)$ for a real polynomial $q(x)$.

Substituting this: \[16(x-1)(x-2)(x-16)q(x) = (x-16)(2x-2)(2x-16)q(2x)\]

$x=8 \Rightarrow q(8)=0$
$x=2 \Rightarrow q(4)=0$

So $q(x)=(x-8)(x-4)r(x)$ for some real polynomial $r(x)$.

Substituting this: \[(x-1)(x-2)(x-4)(x-8)(x-16)r(x)=(x-1)(x-2)(x-4)(x-8)(x-16)r(2x)\]

Now we can see that there are values $z$, different from $1,2,4,8,16$ so that $r(z)=r(2z)=r(4z)=r(8z)=...$
Since $r(x)$ is a polynomial, it has to be constant.

So finally, $p(x)= a(x-2)(x-4)(x-8)(x-16)$, with $a \in \mathbb{R}$.
Checking shows us that all these functions satisfy the conditions.


geometry
circumcircle
ratio
geometry proposed

Let ABC be the acute triangle. M is its medians' intersection, H is its orthocenter. The lines perpendicular to MA, MB, MC by A, B, C intersects each other at A1, B1, C1. G is A1B1C1's medians' intersection. Prove that H, G, M are collinear. :blush:

I solved it by complex number but too long :wallbash:

We basically have to prove that the symmedian point is the centroid of its pedal triangle. Thus, let us prove the following lemma

[b]Lemma:[/b] Let $P$ be a point in the plane of $\triangle ABC$ and $\triangle XYZ$ its pedal triangle. $X \in BC,$ $Y \in CA,$ $Z \in AB.$ Then $P$ is the centroid of $\triangle XYZ,$  if and only if $P$ coincides with the symmedian point of $\triangle ABC.$

$\frac{[\triangle PXY]}{[\triangle  ABC]}=\frac{PX \cdot PY}{BC \cdot CA} \ \ , \ \ \frac{[\triangle PXZ]}{[\triangle  ABC]}=\frac{PX \cdot PZ}{BC \cdot AB}$

Thus,  $[\triangle PXY]=[\triangle PXZ]$ $\Longleftrightarrow$ $\frac{_{PY}}{^{PZ}}=\frac{_{CA}}{^{BA}}$ $\Longleftrightarrow$ $P$ moves of the A-symmedian of $\triangle ABC.$ By similar reasoning, loci of points $P$ such that $[\triangle PYZ]=[\triangle PYX]$ and $[\triangle PZY]=[\triangle PZX]$ are the B- and C- symmedian of $\triangle ABC.$ Consequently, the symmedian point $K$ of $\triangle ABC$ is the unique point that satisfies

$[\triangle KXY ]=[\triangle KYZ]=[\triangle KZX]$ $\Longrightarrow$ $K$ is the centroid of $\triangle XYZ.$

Consider $ \triangle A_1B_1C_1$ and point $ M$, we have $ ABC$ is the pedal triangle of $ M$ wrt $ \triangle A_1B_1C_1$ and $ M$ is the centroid of $ \triangle ABC$, i.e., $ M$ is the Lemoine's point (or the sym-median point) of $ \triangle A_1B_1C_1$. Thus, if $ O$ is the circumcenter of $ \triangle ABC$, we deduce, by the property of Lemoine's point, that $ O$ is the midpoint of $ MG$.

With this, the problem is solved.

P/S: In order to prove the properties of the Lemoine's point of $ \triangle ABC$, you can also apply its bary-centric ratios, but there is of course a synthetic proof for them. :lol:

with http://www.mathlinks.ro/viewtopic.php?t=228911
we conclude G=O,where O is the circumcenter of triangle ABC , thus we are done

PS:
Hi,Mashimaru:
Are you sure about that?
I think O=G,but not O is the midpoint of MG

No,Mashimaru was right.Because Lemoine point and centroid of $ A_1B_1C_1$ are isogonal conjugate wrt $ A_1B_1C_1$ then the midpoint O' of them pass through vertices of their pedal triangles.Obviously $ O' \equiv O$


can you tell me which are are most important conceptual books with problems

for a student entering in iit jee .there should be in depth concepts(particularly physics)

Go for Concept of Physics by H.C Verma
Try Irodov for prblems....

[quote="Abhishek"]The most recommended conceptual book for physics is "Fundamentals of Physics" by Halliday, Resnick & Walker[/quote]

Exactly what I would say. That has been my favourite book throughout. I never got round to learning Surface Tension and Viscosity as H, R, & W excluded them.

By the way you better ask Rushil (AIR 7) than me (AIR 1154!!!!!!!)

Note: The sign '!' does not always stand for factorials, and not at all in this post.

I thiunk Krane is better than walker,thats what all my friends say as well


Which is the problem? What would be equivalent?

P means that a problem can be solved in polynomial time.
NP means that given an input, we can check the output in polynomial time.

It's then obvious that P is a subset of NP, but a big problem is whether NP is P, which is open now.


I really like resolving problems and stuff like that but I also like multiplying, dividing, adding and subscracting by heart. I dont know the word as I'm not familiar with the english language (I'm from montreal) but basically whoever finishes the operations first wins. I'm pretty good at this... anyone up for a contest or something? Thoughts?

[quote="b-flat"]Could you give an example of some type of problem you are talking about?  Is it verbal, or do you see it on a piece of paper.  Sounds like someone would post a question after several people were viewing the forum, then they would solve them and post their reply as soon as possible.  You could turn something like that into a contest.  You would get 2 pts for answering first correctly and 1 for answering correctly second.  If you could get a date and tell me how hard of problems I should write, I could host the contest (i.e posting the problems and keeping track of the score.).[/quote]

just a suggestion...you could add a variety of ranges of math problems at random so they're not all the same type each time...

Nobody said anything, so I didn't post any problems.  Please post saying if you wish to participate.  Then, I will set a new date.

I dont exactly understand what you are getting at, but oh well. Here is the answer to the problem:

[hide]Let Dick be D, Jane be J, and Spot be S.

D+J=180
J+S=225
D+S=192

Adding them together gives 2D+2S+2J=597
D+S+J=298.5

Since we already know D+S=192, J=298.5-192=[size=150]106.50[/size][/hide]

Your answer is correct, but because nobody seems to understand what I am saying, I will just not do the speed contest...

[quote="b-flat"]Good idea, monoversketcher!  How about tomorrow, Saturday 16th at 11:00 a.m Eastern??  Everyone should be up by then.  After each part of the contest is over, I will post when the next time will be.  How does this sound as a sample problem:

Dick and Jane have a total of $\$180$.  Jane and Spot have a total of $\$225$.  Dick and Spot have a total of $\$192$.  How much money does Jane have?  Express your answer to the nearest cent.

I won't make the problems too hard, because then it wouldn't be based on speed.

[size=150][color=red][b]NO CALCULATORS!![/b][/color][/size][/quote]

[hide]So Dick=d, Jane=j, and Spot=s

D+J=180
J+S=225
D+S=192

Add them all up.
2(D+J+S)=597
Then divide by 2.
D+J+S=298.5

Then 180+S=298.5
[b][color=red]S=118.5[/color][/b]
D+118.5=192
[b][color=red]D=73.5[/color][/b]
83.5+J=180
[b][color=red]J=106.5[/color][/b]

Therefore Jane has $\$106.50$ (I did extra work) :rotfl:  :rotfl: [/hide]


Bob works at 6 sph (stuff per hour - whatever it is) after eating (having finished eating in the past 3 hours), and 3/4 of that rate before eating. Lisa works at 7 sph after eating, and 4/5 of that rate before eating. Bob works the shift before breakfast (5-6 AM), Lisa works the shift after breakfast (7 to 10 AM). Bob works the shift from 10 AM to 12 Noon. 12 Noon is lunch. They then switch off at 2 hour intervals until they finish working at 5 PM. How many 'stuffs' did they do?

[quote="math92"]this problem is complicated  :blush:
but i'll get it.[/quote]

lol, math92.

Yes, this problem is difficult. Even though the concept probably is Classroom level, its very wordy, making this prblem complex.

I'll figure it out later. :blush: :D


superior algebra
superior algebra unsolved

Let K be a division ring. Prove that K cannot be writen as a finite union of proper subdivision rings.

Assume $ K\equal{}K_1\cup \cdots\cup K_n$ with proper subdivision rings $ K_1,\ldots, K_n$ of $ K$ and $ n$ minimal.
First assume $ K$ is infinite. Then some $ K_i$ is infinite
and we find $ \emptyset \not\equal{}I\subseteq \{1,\ldots,n\}$ maximal with $ F\equal{}\bigcap_{i\in I}K_i$ infinite.
Then for all $ i\equal{}1,\ldots,n$ we have $ i\in I\Leftrightarrow F\subseteq K_i\Leftrightarrow F\cap K_i$ infinite.

Assume $ I\subsetneq \{1,\ldots,n\}$.
Then we find some $ x\in K\setminus \bigcup_{i\in I}K_i$ by minimal choice of $ n$.
But for $ i\equal{}1,\ldots,n$ with $ (x\plus{}F)\cap K_i\not\equal{}\emptyset$ we must have $ i\not\in I$,
and then for $ y\in (x\plus{}F)\cap K_i$ we get $ (x\plus{}F)\cap K_i\equal{}y\plus{}(F\cap K_i)$ finite.
So $ x\plus{}F\equal{}\bigcup_{i\equal{}1}^n(x\plus{}F)\cap K_i$ must be finite, and then $ F$ also. Contradiction!

This gives $ I\equal{} \{1,\ldots,n\}$ and $ F\equal{}\bigcap_{i\equal{}1}^nK_i$ infinite.
By minimal choice of $ n$ we find $ x\in K\setminus K_1$ and $ y\in K\setminus (K_2\cup \cdots\cup K_n)\subseteq K_1$.
Then $ S: \equal{}\{x\plus{}\alpha y: \alpha \in F\}$ is also infinite, as $ F\to S,\alpha\mapsto x\plus{}\alpha y$ is bijective. ($ y \not\equal{}0$ is invertible in $ K$)
So $ S\cap K_i$ infinite for some $ i\equal{}1,\ldots,n$ and $ x\plus{}\alpha y,x\plus{}\beta y\in K_i$ for some $ \alpha,\beta\in F,\alpha\not\equal{}\beta$.
But $ x\not\in K_1$ and $ \alpha,y\in K_1$ gives $ x\plus{}\alpha y\not\in K_1$ and $ i>1$.
Then $ (\alpha\minus{}\beta)y\equal{}(x\plus{}\alpha y)\minus{}(x\plus{}\beta y)\in K_i$ and $ 0\not\equal{}\alpha\minus{}\beta\in F\subseteq K_i$ is invertible in $ K_i$.
And this gives the contradiction $ y\in K_i$ with $ i>1$.

And now if $ K$ is finite, then by [url=http://en.wikipedia.org/wiki/Wedderburn_theorem]Wedderburn[/url] we get $ K$ a finite field. (short but tricky!)
Then $ K^\times\equal{}K\setminus\{0\}$ is cyclic, say $ K^\times \equal{}\langle\zeta\rangle$,
and if $ \zeta\in K_i$ for some $ i\equal{}1,\ldots,n$, then $ \langle\zeta\rangle\subseteq K_i$ and $ K\equal{}K_i$. Contradiction!


[quote="The Zuton Force"]In a particular game using a standard 52-card deck, a hand consists of 10 cards, and cards can be grouped into 3 of a kind, 4 of a kind, or any same-suit sequence of 3 or more consecutive cards. Ace is equivalent to 1, and sequences cannot wrap around.

Define the [i]advantage[/i] of a hand to be the number of remaining cards that, when added to the hand, will form a new group of cards (ie, they do not simply add to an existing group).

What is the maximum possible advantage of a hand, and how many hands share this value?[/quote]
[hide="an attempt"]
Let a b c and d be the suits, and the numbers be the numbers.
For example, take the hand a2 b2 c2 d2 a3 b3 c3 d3 d4 d5.
The grouping is a2 b2 c2; a3 b3 c3; d2 d3 d4 d5.

In this case, the advantage is $ 42$, because you can add all the remaining cards, and added together it forms 13 groups of 4 cards of the same number, or 4 groups of 13 consecutive cards.
It is also the maximum advantage because there are only 52 cards in the deck.

Since there are 240 possible ways ($ 10\times 6 \times 4 \equal{} 240$) to have the kind of hand I just described, I think at least $ 240$ hands have this advantage. All the other kinds of hands should also have this advantage since they form the same set of cards when added to the remaining cards (it's the same 52-card deck).
[/hide]

[hide="a different interpretation"]
If only one separate group can be added, and the previous cards can't be regrouped, then I think $ 13$ is the maximum possible advantage.

This is because the biggest possible group that can be created is $ 13$ cards long (The samesuit sequence of 13 consecutive cards). This can't be added on to the way described above (Because of the existence of this sequence in the remaining cards, it means that there is at least 1 suit that does not exist in the hand).

The ways to make a hand have this advantage, using this interpretation, is either to have the hand be a 10-card consecutive sequence of a different suit than the advantage ($ 4 \times 4 \equal{} 16$ ways to have such a hand), or to have a 7-card consecutive sequence and a 3-of-a-kind with the missing suit NOT being the suit of the 7 card sequence ($ 4 \times 7 \times 15 \times 3 \equal{} 1260$ ways to have such a hand), or to have a 4-card consecutive sequence and two 3-of-a-kinds with the missing suit NOT being the suit of the 4 card consecutive sequence ($ 4 \times 10 \times 36 \times 3^2 \equal{} 12960$ ways to have such a hand). No other hands are possible because having a 4-of-a-kind would not allow the remaining cards to have a 13-card sequence, and having more 3-of-a-kind's would not leave enough cards for the consecutive sequence in the original hand. (10 is not divisible by 3, so there has to be a consectuve sequence in the original hand)
In this case, the number of hands with this advantage is $ 14236$.

[/hide]


The lines $ y\equal{}ax\plus{}1$ and $ y\equal{}1x\plus{}a$, with $ a\neq1$, intersect in exactly one point.  What is the $ x$-coordinate of that point?

[hide="Solution"]
When they intersect, that means the $ y$ values are equal. So \[ ax\plus{}1\equal{}a\plus{}x.\] Regrouping and using SFFT, \[ (a\plus{}1)(x\plus{}1)\equal{}0\Rightarrow x\equal{}\boxed{\minus{}1}.\][/hide]

Isn't it suppose to be:
$ ax \plus{} 1 \equal{} a \plus{} x$
$ ax \minus{} a \minus{} x \plus{} 1 \equal{} 0$
$ a(x \minus{} 1) \minus{} 1(x \minus{} 1) \equal{} 0$
$ (a \minus{} 1)(x \minus{} 1) \equal{} 0$
$ x \equal{} \boxed{ 1 }$

hello, we have
$ ax\plus{}1\equal{}x\plus{}a$
$ ax\minus{}x\plus{}1\minus{}a\equal{}0$
$ x(a\minus{}1)\minus{}(a\minus{}1)\equal{}0$
$ (a\minus{}1)(x\minus{}1)\equal{}0$
and since $ a\ne1$ it must be $ x\equal{}1$.
Sonnhard.

For speed, you can assume that x will stay constant if the question is worded that way. Then just plug in a random value for a (not 1) like zero.

But obviously, do it right way if you're not in a hurry.

[hide="Hm"]
This has to be a better way:
$ax+1=1x+a \implies ax-a=x-1 \implies a(x-1)=(x-1)$. Since $a \neq 1$, we have $x-1=0 \implies x=1$ (otherwise, we could divide by $x-1$ to get a=1).[/hide]


geometry
circumcircle
incenter
geometry open

[quote="Johan Gunardi"]Let $ ABC$ be a triangle, $ I,O$ is its incenter and circumcenter. Extend $ AI$ to meet the circumcircle at $ K$, where $ K\ne A$. Prove that $ OK\perp BC$.[/quote]
It is obvious,because $ K$ is the midpoint of arc $ \overarc{BC}$.


calculus
function
algebra
domain
analytic geometry
inequalities
trigonometry

Dear math addicts,

I got this problem from class.

I need to integrate the function,

f(x,y)=[1-x^2-y^2] / [(1+x^2+y^2)^2]

over the domain

x^2+y^2 <= 2x  and  x >= 1/2

either in coordinates (x,y) or in polar coordinates.

Until now all my efforts lead me nowhere.

If someone can give some hints would be very helpful.

Thanks

"primitive" is the wrong word here. This is a definite integral.

The function is obviously nicest in polar coordinates; its value depends only on $r$. The region is reasonably nice in polar form as well: the two inequalities become $r^2\le 2r\cos\theta$ and $r\cos\theta>\frac12$.

You should restrict to $-\frac\pi2\le\theta\le\frac\pi2$, since the entire region is in the reghi half-plane. The restriction is actually tighter than that, and is determined by where the two parts of the boundary intersect.

Can you write down this integral in polar form, given what I've just said?

From my calculations it should be

Integral in $\theta$ from $\frac {-\pi} 3$ to $\frac {\pi} 3$

and integral in $r$ from
$\displaystyle \frac1{2\cos\theta}$
 to
$\displaystyle 2\cos\theta$

of
$\displaystyle \frac {1 - r^2} {(1 + r^2)^2} r$

an integration order dr d(theta)

Would this be correct?

But evaluating the first integral would give a very complicated result in cos(theta), which I would then need to integrate in theta,  right? Is there any work-arround this?

PS - By the way how does the code Latex works?

[quote="Conqueror"]PS - By the way how does the code Latex works?[/quote]
I've just edited your post above. Click on the edit button for your post so you can see the details.

Among the things to note:

1. The "Code" tag has nothing to do with $\text{\LaTeX}.$ To activate $\text{\LaTeX},$ simply put dollar signs on both sides of the code in question. For example:

$\$\int_0^{\infty}\frac{e^{-x}dx}{1+e^{-x}}$\$ becomes

$\int_0^{\infty}\frac{e^{-x}dx}{1+e^{-x}}$

2. Use brackets {} to enclose elements, such as numerators, denominators, or exponents. Don't use parenthesis for this purpose - $\text{\LaTeX}$ only understands {} for those purposes.

3. \displaystyle is not necessary.

Oh - and this also belongs in the "Calculus Computations and Tutorials" section, which is where I'm sending it.

There's really no way around a little ugliness. For the inner integral, try the substitution $u=1+r^2$.

The alternative is to do everything in rectangular coordinates, which makes things algebraic- but you might have to use trig-substitution somewhere anyway.

It's a complicated function and a complicated region- you have no right to expect things to work out nicely.


Let $ N$ denote the set of natural numbers. Let $ \phi: N\rightarrow N$ be a bijective function and assume that there exists a finite limit
\[ \lim_{n\rightarrow\infty}\frac{\phi(n)}{n}\equal{}L.
\] What are the possible values of $ L$?

1. Assume there exists $N$ such that $\phi(n) > n$ for all $n\geq N$. Then $\phi(n) > N$ for all $n\geq N$, leaving not enough values to fill the set $\{1,2,\ldots,N\}$. Therefore for any $N$ there exists some $n\geq N$ such that $\phi(n)\leq n$, and so $\liminf_{n\to \infty} \frac {\phi(n)} {n} \leq 1$.

2. Assume there exists $N$ such that $\phi(n) < n$ for all $n\geq N$. Let $m = \max \{\phi(k) \mid 1\leq k\leq N\}$. Clearly $m \geq N$, but then $\max \{\phi(k) \mid 1\leq k\leq m+1\} = m$, absurd. Therefore for any $N$ there exists some $n\geq N$ such that $\phi(n)\geq n$, and so $\limsup_{n\to \infty} \frac {\phi(n)} {n} \geq 1$.

3. So, if $\lim_{n\to \infty} \frac {\phi(n)} {n}$ does exist, it can only be equal to $1$. A trivial example is for $\phi(n) = n$ for $n\geq 1$, but there are other examples, like $\phi(2n-1) = 2n$, $\phi(2n) = 2n-1$ for $n\geq 1$.


Find all ordered pairs of integers $(a,b)$ such that $a^3-b^3=65$.


Please give attention to this and reply to this question. I want to know when the TST's in each country take place. Please give this information. thank you

the brazilian TST's will occur between march and may, i think.  :)

the Brazilian selection process consists of three or four training lists and three selection tests.  :)

[quote="Megus"][quote="silouan"]Thank you. BTW in Greece we write only ONE TST  :rotfl: and this take place in March[/quote]

well, in Poland there's no TST at all  :P[/quote]

And how is the team selection made?? :P

Maybe from their olympiad but I am not sure.

Exactly, the first six in olympiad are going to IMO

Lithuanian TST's (usually it's only one set of problems) are usually held in the begining of April.


trigonometry
integration
function
calculus
topology
real analysis
real analysis theorems

Hi. I'm a little confused about something. We say rationals do not follow the completeness axiom. Wiki and alot of other references seem to say that since

$ \sup \, \{ x \in \mathbb{Q} : x^2 < 2 \} \equal{} \sqrt{2}\,$

thus rationals are incomplete. I seem to be unable to grasp how we are making this deduction from the above example. This particular example has both an upper bound and a supremum. So where is the problem? Where are we violating the least upper bound property?

If anyone could give some other proof of the incompleteness of rationals, that would be really awesome...

Thanks

The set $ \{x \in \mathbb{Q} : x^2 < 2 \}$ doesn't have a [i]rational[/i] supremum, despite being bounded above in the rationals, so the rationals are not complete.

Thank you for the reply....
But we do not demand from other subsets of real numbers that their supremum be a part of that subset as well. How come we are doing this over here? Or more to the point, the completeness axiom itself does not place any such restriction on us.

Well, when you think about the completeness of $ \mathbb{Q}$, forget about it being embedded in $ \mathbb{R}$ - as if you never heard of the real numbers, and you only knew the rationals (a little like Pythagoras did). Now, where and what is the supremum of the set $ \{ r \in \mathbb{Q} \ \mid \ r^2 < 2\}$ ?

Heh, I had a student asking about this just yesterday!

$ \bf R$ is the completion of $ \bf Q$.  If you start with any metric space $ \bf X$, which has completion $ {\bf X}^*$, every point that "should" be in $ \bf X$ [i]is actually a member of[/i] $ {\bf X}^*$.  But some points that we feel "should be" in $ \bf X$ aren't: we can find them [i]only[/i] in $ {\bf X}^*$.

When you look at the sorts of subsets of $ \bf R$ that you can write down without much work, it's very easy to tell whether they are complete or not, [i]because you already know what their completion is[/i].  For example, the open interval $ (0, 1)$ (of $ \bf R$) is very easily seen to not be complete: if it were complete, it "should" contain the points $ 0$ and $ 1$, but to actually find these points we have to look in its completion $ [0, 1]$.  Similarly, to find the limit points of $ \bf Q$, it's not enough to look in $ \bf Q$; we have to go look to its completion, $ \bf R$.  This is [i]what it means[/i] to not be complete.

Thanks alot guys.... all cleared up, I just want to confirm something now....

For reals we say: Every non-empty subset of reals which is bounded above has a supremum.

Now to say that some other set of numbers is complete we have to seperately define the property of least-upper bound [i]tailor made for that set.[/i] We cannot use this definition for finding completeness of some other set. Say for $ S \equal{} \{x \in \mathbb{R} : 0 < x < 1 \}$ the set would be called complete if and only if it satisfies the condition that,

For every non-empty subset of S which is bounded above has a supremum.

I'm sorry if this is trivial. I'm doing Apostol, not much about metric spaces or very long discussions about completeness. Just some basic theorems and defn's and then very tough problems....

Here's the essence of completeness: every real number that you think ought to be there, is there. Or to put the same thing in another way: limits exist even when you don't know a name for the number in question. That's a feature, not a bug: it allows you to talk about real numbers that you don't know explicit names for.

Some examples of real numbers that I don't know explicit names for (or at least, I don't know shorter or simpler names than what I'm about to give):

The unique real $ x$ such that $ x\equal{}\cos x.$ (Existence guaranteed by the Intermediate Value Theorem, which is a direct consequence of completeness.)

$ \sum_{n\equal{}1}^{\infty}\frac{\ln(n\plus{}1)}{\sqrt[3]{n^5\plus{}2n\plus{}7}}.$ (I know the series converges by some comparison test; comparison tests rely on the fact that monotone bounded sequences converge, which is fully equivalent to completeness.)

$ \int_0^1\sqrt[3]{x^5\plus{}2x\plus{}7}\,dx.$ (Continuous functions on closed bounded intervals are Riemann integrable. The Riemann integral is defined as a limit - but to show that there is a real number to be equal to that limit, we need completeness.)

[quote="maliksaim"]Now to say that some other set of numbers is complete we have to seperately define the property of least-upper bound [i]tailor made for that set.[/i] [/quote]  Ah, good question: in an arbitrary metric space, instead of looking at sups and infs we want to look at [i]Cauchy sequences[/i].  Informally, these are sequences in which eventually all the members of the sequence are close together.  I think it's intuitively reasonable that all convergent sequences are Cauchy (if eventually your entire sequence is near some fixed point, then all of its members must also be near each other), and this is true in every metric space; but the converse is not true.  Taking the completion adds in limit points of all of these Cauchy sequences that "should" converge.

Right.  If you want to use infs and sups to talk about completeness, you're not working in a metric space: you're working in an order topology.  (It's a nice property of $ \mathbb{R}$ that its order and metric topologies coincide.)


limit
logarithms
calculus
calculus computations

prove :maybe:   :blush:

$ \lim_{n\to\infty}\;\left[ \;\left(\sum_{j=0}^\infty\;\frac{\;\textbf H_{j}\cdot\left( \frac{n^{j}}{j!}\right)^{2}\;}{\;\left(\sum_{j=0}^\infty\;\left( \frac{n^{j}}{j!}\right)^{2}\right)\;}\right)\;-\;\ln n\;\right]\;=\;\boxed{\gamma}$

where $ \textbf H_{j}$ is the $ j^{th}$ harmonic number $ \left( =\sum_{k=1}^{j}\frac{1}{k}\right)$,
and $ \gamma$ is Euler's constant.

Is $ j$ in the denominator independent to the $ j$ of outer sum?


Just for the forum, perhaps a way to automatically log in every time you visit the site as is common in most forums? (if this is not possible already)

Done (I think).  Let me know if that works for you (you'll see a little box next time you log in).

Wow, that was quick! Seems to be working fine.

How about a whiteboard/chat area where we can just chat amongst ourselves? Not necessarily a Math Jam but more of a plain chat room...

-Isaac

ooohhhh...sounds nice....maybe java with seperate rooms too!!!

At this point, the issue is primarily a legal one.  The federal government has all sorts of rules and regulations (many of which haven't been clarified by the courts yet) regarding websites which cater to folks under age 18.  Some of these laws make it difficult for a site which many people under 18 use to support community tools like a chat room - in fact, when some of these laws were passed a few years ago, many sites shut down or stopped accepting minors as users.  Large companies like Disney can still work around (or with) the laws - they've got the legal resources to wade through the legislation.  We can't.

:evil:
I guess we'll have to stick with AIM right now. How about a AIM screenname link in the users' public profiles, i.e. when you click on their usename? Many other sites have them. Also, is there a way to private message people on this site?

-Isaac

Can you point me to a few sites that target folks under the age of 18 which have these features?  I'll look into it.

Thanks...

One of the sites that has the more detailed member pages is gamefaqs, at http://www.gamefaqs.com. Here's a link to a member profile there:
[url]http://s1.cgi.gamefaqs.com/boards/whois.asp?user=593740&board=20052&topic=8057718[/url]
The FAQ mentions private messages, and I quote,
[quote]I cannot send private messages!
There are three reasons for this; you are not registered and/or not logged on, the board administrator has disabled private messaging for the entire board, or the board administrator has prevented you individually from sending messages. If it is the latter case you should try asking the administrator why.
Back to top[/quote]
By the way, this is a great site. Good luck and thanks!

-Isaac


suppose that the set A is denumerable, the set B is denumerable, and the sets A and B are disjoint; show that the union A u B is denumerable.

Merci.

Besides the fact that another thread has been open, isn't the relation to the automata a bit ... tenuous ?  :)


function
calculus
derivative
integration
calculus computations

How do you prove that a C1 function is differentiable?
in particular, given that the partials of a function all exist and are continuous, how do you prove that the function is differentiable at each point in U:

f: U ->  R^m,  where U in R^n,

I have a proof of this, as it is a classic result, but I was wondering if there was a more compact or clever proof, as mine is quite ugly

The problem can be reduced to the following special case: all first-order partial derivatives of $f$ are continuous and equal to zero at the origin $0$. Furthermore, $f$ is a scalar function and $f(0)=0$. Our goal is to prove that for any $\epsilon>0$ there is a neighborhood of the origin in which $|f(x)|\le \epsilon \parallel x\parallel $.

Choose a ball centered at zero in which all partials of $f$ are less than $\epsilon/n$. Write $f(x_{1},\dots,x_{n})-f(0,0,\dots,0)$ as the sum of $n$ integrals like $\int_{0}^{x_{2}}\frac{\partial f}{\partial x_{2}}(x_{1},t,0,\dots,0)\,dt$. Each of the integrals does not exceed $\epsilon \parallel x\parallel /n$ in absolute value. QED.


The elements in a set S, are {0,1,2,3,4,5,6,7,8}. If you randomly select two distinct elements from S, what is the probability that their sum is less than their product?

I thing it shoulde be 22,2 %
sorry...i havent read the subject carefully..this was wrong :P

[hide]Well, the chance that the sum is less than the product is 1-P(sum is greater than product).  Those numbers only come with 0 and (1,2,3,4,5,6,7,8) for 8 numbers and then 1 and (2,3,4,5,6,7,8) for 7 numbers. So there are 15 total numbers that don't fit the pattern.  There are 9C2=36 total different combinations, so it is (36-15)/36=[b]7/12 chance[/b][/hide]
Edit: is this correct?