geometry
parallelogram
circumcircle
algebra proposed
algebra

Construct a quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal.

Take a 30-60-90 triangle. On the side on which 60 and 90 angles lie, outwards draw an isosceles triangle such that that side is not its base, and the angles of that triangle are 120,30,30. Obviously in the quadrilateral we have 2 opposite angles of 30, now we have to prove that the base of the isosceles triangle is equal to the side opposite to 30 angle in the right-angled triangle.

That goes easily because of lots of 30-60-90 triangles

There goes the quadrilateral!

My opinion:
Consider the parallelogram ABED with
$\mathop {\rm A}\limits^ \wedge   = \mathop {\rm E}\limits^ \wedge   < 90^ \circ  and\;AB = DE = a.$
We consider the circumcircle L of the
$\mathop {BDE}\limits^\vartriangle  .$Let (D, a)
the circle which recuts  the circle L at point
$C \ne E.$
Then ABCD is the demanded parallelogram.

S.E.Louridas

@Bugi, can you please post the diagram, too?

it is there in the book " challenge and thrill of pre college mathematics -micellaneous problems section"
with no solution

but you see i heavily doubt that quadrilateral made out of 30 60 90 triangles is a parellelogram


I'm not sure whether this belongs here...

But, I recently got into Photography.  I really want some feedback so I can improve.

Check out my pictures:
http://rp92.deviantart.com/

Thanks!

Not bad, young one... :D

For the invisible skateboarder one, the shoes look deflated; try putting something in them to make it more realistic. July flower is blurry... For endless strings, you might want to focus it a bit closer so the front is more clearly visible while the back gets more blurry and really gets "endless". :wink:

Try shooting some sunsets, they always turn out good. :)

Btw, what kind of camera do you have? I'm not in to photography too much, but my dad is; he has a Nikon D80.


Nine cyclists $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8, A_9$ are cycling in that order. Suddenly, they want to change positions in a way that the leader changes and no one is behind the same person. In how many ways can they do it?

We find a formula for $n$ cyclists. Call the number of ways for $n$ cyclists $f(n)$. Now call $g(n)$ the amount of ways they can rearrange so that no one is behind the former person, but without the first condition. Then $f(n)=g(n)-f(n-1)$ (we count all cases and substract those where the 1st cyclist remains in his place; then to cyclists $2, ..., n$ we have to apply $f$ because cyclist 2 can't be in his original position).

\[f(n)=g(n)-f(n-1) \\
f(n)=g(n)-g(n-1)+f(n-2) \\
f(n)=g(n)-g(n-1)+g(n-2)-f(n-3)\]

And so on. We need a recurrence for $g$. Take $n!$ and substract the illegal cases (where someone is behind the same cyclist). You say there's a fault if there are two cyclists $i, i+1$ one behind the other. To count the permutations with $k$ faults, you choose $k$ from the $n-1$ gaps between cyclists and join the ones on either side of a chosen gap into one deformed cyclist. Then you get an arrangement of $n-k$ cyclists, and since there are no more faults they are rearranged satisfying the conditions for $g$. So for $k$, there are $(C_k^{n-1})g(n-k)$ $k$-faulted permutations. Then

\[g(n)=n! - \sum_{i=1}^{n-1} (C_i^{n-1})g(n-i)\]

Well, now it's a matter of using this ugly recurrence to calculate the values of $g$ and plugging it in the first recurrence. Maybe there are some slick cancelations that make it look nice in the end, but I'm not looking for them. ;)

I couldn't help wonder if there is an easier solution


is this upcoming AMC 12 problem series going to be the same as the one that was given during the summer?
Are the problems the same or are they different like the AIME A vs. the AIME B class?

It's essentially exactly the same as the summer course.


inequalities
IMO Shortlist
graph theory
combinatorics
Partial Orders

This ISL 2005 problem has not been used in any TST I know. A pity, since it is a nice problem, but in its shortlist formulation, it is absolutely incomprehensible. Here is a mathematical restatement of the problem:

Let $k$ be a nonnegative integer.

A forest consists of rooted (i. e. oriented) trees. Each vertex of the forest is either a leaf or has two successors. A vertex $v$ is called an [i]extended successor[/i] of a vertex $u$ if there is a chain of vertices $u_{0}=u$, $u_{1}$, $u_{2}$, ..., $u_{t-1}$, $u_{t}=v$ with $t>0$ such that the vertex $u_{i+1}$ is a successor of the vertex $u_{i}$ for every integer $i$ with $0\leq i\leq t-1$. A vertex is called [i]dynastic[/i] if it has two successors and each of these successors has at least $k$ extended successors.

Prove that if the forest has $n$ vertices, then there are at most $\frac{n}{k+2}$ dynastic vertices.

mm actually the problem did appear in the Australian TST, Day 2 as Problem 3 (there are 4 problems on each of the 2 days).
My favourite solution is the one where you perform operations on the shortlist which only make the inequality stronger, until you reduce it to a triviality and in fact you can prove something a little stronger ... i might write it out properly sometime later.

Darij, I don't think your wording is exactly equivalent to the original wording :  I think that it is also possible for a vertex to have only one successor.  However, it doesn't really affect the solution.

I will call dynastic vertices [i]blossoms[/i].  We will call a vertex a [i]bud[/i] iff. it is a direct successor of a blossom but is not a blossom itself.  Let $m$ be the number of blossoms.  We prove the stronger bound
\[n \ge (m+1)(k+1)+m. \]
This is sufficient because $(m+1)(k+1)+m = m(k+2)+k+1 > m(k+2)$.

We proceed by induction on $m$.  For the base case, we have one blossom which must have two buds, each of which must have at least $k$ indirect successors.  Now, if we have many blossoms, then there exists one blossom none of whose indirect successors are blossoms.  We remove this blossom, one of its buds, and all indirect successors of that bud, and we make the remaining bud a direct successor of any vertex of which our blossom was a direct sucessor.  All other blossoms remain blossoms; all other buds remain buds.  We have decreased $n$ to some number $n'$ which is at least $k+2$ less than $n$; we have decreased the number of blossoms by one, and we have not changed $k$.  The inductive hypothesis gives
\[n-(k+2) \ge n' \ge m(k+1)+m-1, \]
which is sufficient to prove what we wanted.

This bound is sharp, for we can have equality when we have a sequence $A_{1}, \ldots, A_{m}$ of blossoms such that $A_{i+1}$ is a successor of $A_{i}$, and each bud has exactly $k$ indirect successors.

[quote="Boy Soprano II"]Darij, I don't think your wording is exactly equivalent to the original wording :  I think that it is also possible for a vertex to have only one successor.[/quote]

I am not sure about this - there are different ways to read the problem. However, you are right in that in that it doesn't affect the solution. We can even allow vertices to have several (more than $2$) successors, as long as a "dynastic vertex" is redefined as a vertex which has at least two successors each of which has at least $k$ extended successors.

[quote="Boy Soprano II"]I will call dynastic vertices [i]blossoms[/i].  We will call a vertex a [i]bud[/i] iff. it is a direct successor of a blossom but is not a blossom itself.  Let $m$ be the number of blossoms.  We prove the stronger bound
\[n \ge (m+1)(k+1)+m. \] [/quote]

Note: This holds only if $m>0$. (This is a subtle point; the first of the two proposed solutions makes a minor mistake here.)

Anyway, why I am posting this? Because there is a very cheap way to make the problem stronger:

[color=blue][b]Generalized problem.[/b] Let $T$ be a forest. Let $k$ be a nonnegative integer. For every vertex $v$ of $T$, we denote by $s_v$ the number of successors of $v$ each of which has at least $k$ extended successors. (Here, the notion of an "extended successor" is defined as in the original problem; in particular, no vertex is considered its own extended successor.) A [i]blossom[/i] of $T$ means a vertex $v$ of $T$ such that $s_v > 0$. Assume that $T$ has at least one blossom. Then, $T$ has at least $\left(k+1\right)+\sum_{v\text{ blossom of }T}\left(1+\left(s_v-1\right)\left(k+1\right)\right)$ vertices.[/color]

The solution is the same as the one given by Boy Soprano II for the original problem, but this time we have to remove $s_v-1$ buds at the induction step.

We assume the graph is a tree.

Call a dynastic vertex a "HUGE" vertex. Call the successors of huge vertices "big" vertices. Call the set of extended successors of big vertices a "family". We will assign $k+2$ distinct vertices to each huge vertex, and then we will finish.

To do this, divide the tree into levels. Define $m$ so that level $m$ is the highest: it has only one vertex. Level $1$ is the lowest level, consisting only of leaves. Call level $i$, $L_i$. Call $H_i$ the set of huge vertices in level $1$, $2$, ..., $i$. I will demonstrate I can assign to a huge vertex $H \in H_j$ $k+2$ distinct vertices (that are extended successors of some big vertex that is a successor of $H$), for any $1 \le j \le m$, using induction. Applying this to $j=m$ we will finish.

The base case: for a huge vertex, assign one of his successors and his family and the vertex itself to that vertex. Clearly we assigned at least $k+2$ vertex (remove vertices if needed). We cannot overlap vertices because huge vertices in this level cannot have huge extended successors (it is the lowest possible level for huge vertices).

Now for the inductive step. Take the assignment for $H_{j-1}$. Consider a huge vertex $X$ in $L_j$. Consider the set of all his extended successors. If this set has no huge vertices, assign to $X$ himself, one of his successors and his family. If this set has huge vertices, take such a huge vertex $Y$ in the lowest level. To $Y$ we already assigned some vertices, and all of these vertices are extended successors of $Y$. So, to $X$ we assign himself, and the family of the other big vertex that is a successor of $Y$ (and this big vertex itself).

So we are done.

First we can see that we only need to prove this for $1$ tree because if we do then summing up the maximum of dynastic vertexes for all trees(with $a_{1},a_{2},...,a_{s}$ vertexes) we get that the maximum is $\sum_{1\le i\le n} \frac {a_{i}}{k+1}=\frac {n}{k+2}$.
So we only need to prove for one tree.
First define the level of the tree $T$ as the largest path from the root of $T$ and denote the level of $T$ by $L(T)$.
Denote by $N(T)$ the number of vertices in $T$.
We will use a induction on $L(T)$. In fact we will say that there are at most $\frac {N(T)-k-1}{k+2}$ dynastic vertexes for $N(T)>k$ and there are $0$ for $N(T)\le k$.
Base: $L(T)=0$ is trivial.
Now consider that it's true for $L(T)\le s$ consider $L(T)=s+1$ now let $u,v$ be the $2$ successors of  the root of $T$ denoted by $r$. Denote by $U$ the parts of the tree containing $u$ and it's successors. Analougosly define $V$. Note that $L(U),L(V)<L(T)$ so we may use induction on $U,V$
[b]Proof: [/b]
Note that if $N(T)\le k$ then clearly there are no dynastic vertices.
If $N(T)\ge k+1$ then note that if $N(U)\le k$ then $r$ is not dynastic and if $N(V)\le k$ then there are $0\le \frac {N(T)-k-1}{k+2}$dynastic vertices and if $N(V)>k$ then there are at most $\frac {N(V)-k-1}{k+2}<\frac {N(T)-k-1}{k+2}$. We prove the desired result similarly for $N(V)\le k$.
If $N(V),N(U)>k$ then there are at most (since $r$ is a dynastic vertex)
$1+\frac {N(V)-k-1}{k+2}+\frac {N(U)-k-1}{k+2}=\frac {N(T)-k-1}{k+2}$
dynastic vertices. So we are done.
[b]end[/b]

Assume the forest is connected. The problem can be strengthened to $\frac{n-k-1}{k+2}$. We do this by induction on $n$.

Suppose a graph had a dynastic vertex. Then by definition, the graph must have at least $1+2+2k=2k+3$ vertices. Moreover, for a graph with $2k+3$ vertices, there can be at most 1 dynastic vertex. Thus the problem statement is true for $n\le 2k+3$.

For the inductive step, assume our graph had a dynastic vertex. Take a dynastic vertex such that none of its extended successors are dynastic - such a vertex clearly exists, as a dynastic vertex has at least $2k+2$ extended successors, and an extended successor of a vertex has less extended successors than the original vertex. Let this vertex be $V$. Let its successors be $A$ and $B$. Then $A$ and $B$ each have at least $k$ extended successors. Remove exactly $k+2$ of these extended successors. Note that $V$ may not be dynastic anymore. On the other hand, $V$ still has at least $2k+2-(k+2)=k$ extended successors. Thus, if $X$ is a dynastic vertex such that $V$ is an extended successor to $X$, after the operation, $X$ is still dynastic. Any successor to $V$ was not dynastic either, so no dynastic vertices are lost there either. Thus we remove at most $1$ dynastic vertex from our graph while removing $k+2$ vertices, implying there are at most $1+\frac{n-2k-3}{k+2}=\frac{n-k-1}{k+2}$ dynastic vertices in our graph by our inductive hypothesis, as desired.

[quote=darij grinberg]
Let $k$ be a nonnegative integer.

A forest consists of rooted (i. e. oriented) trees. Each vertex of the forest is either a leaf or has two successors. A vertex $v$ is called an [i]extended successor[/i] of a vertex $u$ if there is a chain of vertices $u_{0}=u$, $u_{1}$, $u_{2}$, ..., $u_{t-1}$, $u_{t}=v$ with $t>0$ such that the vertex $u_{i+1}$ is a successor of the vertex $u_{i}$ for every integer $i$ with $0\leq i\leq t-1$. A vertex is called [i]dynastic[/i] if it has two successors and each of these successors has at least $k$ extended successors.

Prove that if the forest has $n$ vertices, then there are at most $\frac{n}{k+2}$ dynastic vertices.[/quote]

Improvise the bound to $\tfrac{n-(k+1)}{k+2}$ and induct on $n \ge 1$; base cases $n \le 2k+3$ are trivial. We consider only the case when $G$ is a tree. For a dynastic vertex $w$; define the set of all its (extended) successors as its [i]dynasty[/i]. This [i]dynasty[/i] splits into two [i]clans[/i] determined by the immediate successors of $w$. Further we see that several oriented trees append to the clan-members. Call these the [i]relatives[/i] of $w$.

Pick a dynastic vertex $v$ with the largest dynasty. Let $A$ and $B$ be the two clans originating from $v$ and $C$ be the relatives of $v$. Let $G'=A\cup B \cup C \cup \{v\}$. Observe that $H=G/G'$  consists of oriented trees $P_1, \dots, P_{\ell}$ that can be identified by vertices $v_1, \dots, v_{\ell}$ with directed edge $\overrightarrow{v_iv} \in E(G)$ for all $1 \le i \le \ell$. Notice that for any $u \in G'$, no extended successor of $u$ lies in outside $G'$. Indeed we see that the unique (non-oriented) path from $u$ to a vertex outside $G'$ passes through $v$. However the orientations till $u$ to $v$ differ from that of $\overrightarrow{v_iv}$ for all $1 \le i \le \ell$.

Now no dynastic vertex in $H$ has an extended successor in $G'$ else the maximality of $v$'s dynasty is contradicted! Thus, $v_i$ is not a clan-member of any dynasty for all $1 \le i \le \ell$. Finally, we can apply IH to the sets $G'$, $P_1/\{v_1\}, \dots, P_{\ell}/\{v_{\ell}\}$ thus $G$ has no more dynastic vertices than $$\sum^{\ell}_{i=1} \frac{|P_i|-1-(k+1)}{k+2}+\frac{s-(k+1)}{k+2}=\frac{n-s-(k+2)\ell+s-(k+1)}{k+2}<\frac{n-(k+1)}{k+2}$$ unless $\ell=0$ (i.e. $G'=G$). If that is the case; delete $v$ and apply IH to the two trees so obtained. The conclusion still holds. $\blacksquare$

It suffices to prove the result if $G$ is a tree. We prove the stronger bound \[ \frac{n - (k+1)}{k+2} \] by induction on $n$ with the base cases $n \le k+1$ being clear.

Let $v$ be the dynastic vertex highest up in the tree, and let $T$ be the sub-tree rooted at $v$. Note that no vertex $w \notin T$ can be dynastic, otherwise any common ancestor of $v$ and $w$ would be dynastic too; thus discard vertices not in $T$. Now, the two children of $T$ induce two sub-trees $A$ and $B$. So the total number of dynastic vertices is at most \[ 	\frac{|A| - (k+1)}{k+2} 	+ \frac{|B| - (k+1)}{k+2} 	+ 1 	= \frac{(|A|+|B|-1) - (k+1)}{k+2} \] as desired.

Call a forest with all vertices having $0$ or $2$ children a [i]binary forest[/i].

[u]Claim:[/u] A binary forest with $k$ leaves has at most $2k-1$ vertices.

[u]Proof:[/u] WLOG, we may assume that we have a tree, since summing over the connected components gives the general result. We proceed by induction on $k$. If $k=1$, then our tree consists of a single vertex, which clearly satisfies the condition.

Suppose we're given a tree $\tau$ with $k$ leaves. Define the depth of a vertex to be its distance to the root. Consider a leaf $L$ with maximal depth, and let its parent be $P$. We see that $P$ has some other child $L'$, and since $L$ had maximal depth, $L'$ is also a leaf. Now, deleting $L$ and $L'$, we have a new tree $\tau'$ with $1$ fewer leaf (since $P$ becomes a leaf), and two fewer vertices, so the result follows by induction. $\blacksquare$

Given the forest $G$ on $n$ vertices, let $S$ be the set of vertices with at least $k$ descendants. We see that the induced subgraph for $S$ is a tree where each vertex has $0$, $1$, or $2$ children. We wish to show that the number of ``cherries'' in the induced subgraph of $S$ (things of the form $(P,\{A,B\})$ where $A$ and $B$ are the children of $P$) is at most $\frac{n}{k+2}$. To do this, create a graph $G'$ by taking the induced subgraph of $S$ and contracting all edges that are from a parent to an only child. The number of cherries in this graph is the same as the number of cherries in the induced subgraph of $S$, and furthermore, for each leaf of $G'$, we have at least $k$ new vertices of $G$ that are not in $S$ (in particular, this set of $k$ vertices is distinct for each leaf of $G'$). Letting $\ell$ be the number of leafs of $G'$, we then see that the total number of vertices in $G$ is at most $(2\ell-1)+k\ell$, so
\[(2\ell-1)+k\ell\le n\implies\ell\le\frac{n}{k+2}.\]
We have that the number of cherries in $G'$ is at most $(2\ell-1)-\ell\le\frac{n}{k+2}$ (each non-leaf contributes a cherry), so the number of cherries of the induced subgraph of $S$ is at most $\frac{n}{k+2}$, as desired.

[b]Remark[/b] Our proof shows that the estimate can be sharpened to
\[\frac{n+1}{k+2}-1.\]

Call a connected component a tree. I claim that if a tree has $x$ dynastic vertices, the tree itself must have at least
\[f(x) = (k+1)+x(k+2) \text{ nodes}\]
This will clearly suffice to prove the $\frac{n}{k+2}$ bound. We proceed with induction on $x$. For the base case, $x=1$, in order to have a dynastic vertice we need 1 node, two immediate children, and each of those children have $k$ children, for a total of
\[1+2+2\cdot k = 2k+1=f(1) \text{ nodes}\]
For the inductive step, take some tree with $m+1$ dynastic vertices. Call $P$ the node from which all others are descended.

Label it's children as $A$ and $B$.

Claim 1: If either $A$ or $B$ have legacy $<k$, then we can throw out the P and the bad child which would decrease total \# of nodes but not the number of dynastic.

Now, assume both $A$ and $B$ have $\geq k$ children. Assume that $A$'s family(including A) has $a$ dynastics, and $B$ has $m-a$ dynastics. For $a>0$, inductive hypothesis handles. However, when $a=0$, we use the earlier claim to bound below as $\geq k+1$ on A's side.

Thus, the total number of nodes is at least
\[1+f(a)+f(m-a) = 1 + 2\cdot (k+1) + (a+m-a)\cdot (k+2) = 2k+3 + m\cdot(k+2) = f(m+1)\]
And we are done! $\blacksquare$

WLOG the forest $G$ is a tree, since otherwise we just sum the result over all trees. Let the root be $v$, and for any other vertex in the tree, let its $\textit{rank}$ be the shortest distance from it to $v$. Let $v_0$ be the dynastic vertex with lowest rank; it follows that all other dynastic vertices are descendants of $v_0$, else if there is another dynastic vertex $u$ for which $u$ is not a descendant of $v$, then the earliest common ancestor of $v_0$ and $u$ must clearly be dynastic, and it has lower rank than $v_0$, contradicting the minimality assumption of $v_0$'s rank.

Let $T$ be the sub-tree of $G$ with root $v_0$. Since it is dynastic, it leads off to exactly two successors, $x$ and $y$, each of which are the roots of disjoint sub-trees $T_x$ and $T_y$ of $T$. At this point, it remains to induct, but straightforward induction does not work as the bound is not sharp enough.

Through some fudging, we see that the bound of at most $\tfrac{n - (k+1)}{k+2}$ dynastic vertices is sharp, and we proceed to strong induct on this value;\begin{align*}\text{dynastic}(G) = \text{dynastic}(T) &= 1 + \text{dynastic}(T_x) + \text{dynastic}(T_y)\\&\leq 1 + \frac{|T_x| - (k+1)}{k+2} + \frac{|T_y| - (k+1)}{k+2}\\&= \frac{|T_x| + |T_y| - k}{k+2}\\&\leq \frac{|T| - (k+1)}{k+2}\end{align*}completing the general inductive step. The base cases are easy; for $|G| \leq k+1$, we cannot have dynastic vertices, so the result generally follows from induction.

Hardest C2 ever?

Notice that proving the fact for a tree is sufficient.

Let the [i]family[/i] of a vertex be itself and all its descendants. Choose a dynastic vertex, cross it out, and color all the vertices in its family red. Then, color $k+2$ vertices of its family blue (so that a vertex that is colored red then colored blue is considered blue). Repeat this process until all dynastic vertices are chosen. We claim by induction on the number of crossed-out vertices that it is possible to color in such a way that no vertex is colored blue twice.

We prove these two statements simultaneously:
1. It is possible to color in such a way that no vertex is colored blue twice.
2. After each step, each vertex that is not in the family of a crossed-out vertex has at least $k+2$ family members that are not blue.
The case of one vertex being crossed out is trivial. Now, notice that statement 1 is routine by the inductive hypothesis. Choose a dynastic vertex that is not an ancestor of any dynastic vertex that is not crossed out. Now, look at its two children. Notice that one of these cannot be an ancestor of the vertex chosen in the previous step. By the inductive hypothesis, the chosen dynastic vertex still has $k+2$ family members that are not blue. Thus, our claim is proven.

Since each dynastic vertex corresponds to $k+2$ different blue vertices, there must be at most $\frac{n}{k+2}$ dynastic vertices.

Suffices to prove for tree. Strong induct on a stronger bound of $\frac{n-(k+1)}{k+2}$. By extremal, take the root to be dynastic. Applying inductive hypothesis on the two children of the root yields the result.

First, we'll show the same result for a tree instead of a forest. This is sufficient because a forest is just composed with trees. We claim that there are at most $\tfrac{n-(k+1)}{k+2}$ dynastic vertices. The result is obviously true for $n\le 2k+2$. Let $n$ be the smallest positive integer such that there is a tree with more than $\tfrac{n-(k+1)}{k+2}$ dynastic vertices. Then, take the root and remove it. We have removed one dynastic vertex. Let there be $n_1$ and $n_2$ vertices in the resulting trees, then we have remaining
\[\frac{n_1-(k+1)}{k+2}+\frac{n_2-(k+1)}{k+2}=\frac{n-2k-3}{k+2}\] and when we add back the one we removed, we are at $\tfrac{n-(k+1)}{k+2}$, absurd.

Very cool problem!
[hide = solution]
Obviously, we may replace the forest with just a single rooted tree, call it $T$. Color in red all the vertices of the tree with at least $k$ extended successors. Let the set of red vertices be $R$ and observe that if $u \in R$ then also $\text{parent}(u) \in R$. Observe that if $|R| \leq 2$ we are trivially done anyway, as there are $0$ red vertices in this case. It follows that the subgraph of $T$ induced by $R$ is another tree, say $T'$. We have two observations: [list][*] $T'$ is a binary tree rooted at the root of $T$. [*] The number of dynastic vertices is equal to the number of vertices of $T'$ with two children.[/list]
First, if we see a vertice with only one child, we truncate that edge by combining the two. Repeating this optimisation multiple times if necessary, each vertice of $T'$ has $0$ or $2$ children.

Suppose there are $m$ leaves in (the new) $T'$. The key is that since $T'$ is a binary tree, $|T'| \leq 2m-1$. Moreover, the extended successors of the $m$ leaves in $T$ are distinct. Accounting for their "distinct" existence along with the contracted tree $T'$ gives us that $2m-1+mk\leq n$ or that $m \leq \frac{n+1}{k+2}$. In particular, the number of dynastic vertices is \[|T'|-m \leq m-1 \leq \frac{n-(k+1)}{k+2} < \frac{n}{k+2}\] which finishes the problem.

[img]https://i.imgur.com/i5AchhM.png[/img]
[i]that's a lot of vertices[/i] -said no one ever
[/hide]


Hello,

I have a lot of random questions that I write down whenever I'm stuck on something.  I need someone to answer all of these questions.  If I took an independent study session, can I just ask these random questions, or am I only allowed to ask questions that are from the homework or my classes?

Of course, I'll still do my homework/classes and ask questions from there, but my main focus would be to finish asking my long line of random questions from all different parts of math.  Is this okay?

Please let me know, thanks.

during office hours, you can pretty much ask any question relating to math that you want.

Indeed, you can bring what questions you have, even if they do not come from the Independent Study.

The message board activity in the Independent Study is also pretty good.  There are many threads started by students each week about problems or topics that are discussed -- sometimes in great length.


[quote="Zed_darkx"]Find the sum of all the no.s formed from digits 1 3 5 7 9 (i)greater than 10000 , and (ii)smaller than 10000.

Edit:
Digits are not repeated.[/quote]
[hide]
(i)greater than 10000 ,
There are $ 5!$  numbers which are greater than $ 10000$ if digits are not repeatedly placed.
Their sum
$ \equal{}4![(1\plus{}3\plus{}5\plus{}7\plus{}9)\times 10^4 \plus{}(1\plus{}3\plus{}5\plus{}7\plus{}9)\times 10^3\plus{}(1\plus{}3\plus{}5\plus{}7\plus{}9)\times 10^2\plus{}(1\plus{}3\plus{}5\plus{}7\plus{}9)\times 10 \plus{}(1\plus{}3\plus{}5\plus{}7\plus{}9)]$

$ \equal{}4![250000\plus{}25000\plus{}2500\plus{}250\plus{}25]$
$ \equal{}24\times 277775$
$ \equal{}\boxed {6666600}$
[/hide]

[quote="vinskman"][quote="Zed_darkx"]^^ Above part i got.
The 2nd is killing me :(( and so is the case when the digits can be repeated::
1st part would be 5^5 instead of 4!...but what about less than 10000.[/quote]

[hide]
You may try to find
$ A \equal{}$ sum of all single digit numbers =$ 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9 \equal{} 25$
$ B \equal{}$sum of all double digit numbers $ \equal{} 4\times [10A \plus{} A]$
$ C \equal{}$ sum of all  triple numbers$ \equal{} 4\times3 \times[100A \plus{} 10A \plus{} A]$
$ D \equal{}$ sum of all  quadriple  numbers $ \equal{} 4\times3\times2\times[1000A \plus{} 100A \plus{} 10A \plus{} A]$

and then $ A \plus{} B \plus{} C \plus{} D$
[/hide][/quote]
And for repeatition being allowed : Multiplied by 5^4 right? Because we would have 1 digit fixed , and the rest can 4 places can take 5 values.


geometry
trigonometry
area of a triangle
Pythagorean Theorem

Prove Herons Theorem which is $\sqrt{s(s-a)(s-b)(s-c)}$ which enables you to find the area of a triangle given only three sides $a,b$ and $c$. Let $s=\frac{a+b+c}{2}$ and let $S$ equal the area of the triangle so the formula is read as so: $S=\sqrt{s(s-a)(s-b)(s-c)}$.

here is a pure geometry proof

[url]http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html[/url]

Which in my opinion is much more ingenius than the trig method that is taught in high schools.

[quote="K81o7"]You can do it with the Pythagorean Theorem alone if you want to.  Just draw the altitude in the triangle, write out the area in terms of the sides, and things start to unfold.  But it's not quite like the geometry proof on that page, where you can really see it.[/quote]

Yeah the thing is people back in Heron's time did not have good factoring tricks and people didn't know as much algebra as we do today...


Find the number of permutations $a_1,a_2,...,a_{10}$ of integers $1,2,...,10$ such that :\[ a_i > a_{2i} \qquad \textrm{and} \qquad a_i>a_{2i+1}  \]

For those who're wondering.. :
"et" means "and". ;)


geometry
3D geometry
pyramid
geometry unsolved

[color=blue][b]Let a pyramid $ S.ABCD$ with $ SC\equal{}x (0<x<\sqrt{3})$. All the remaining sides of it have the length of $ 1 cm$. Prove that the line $ SA$ is perpendicular to the line $ SC$ and find the volume of pyramid $ S.ABCD$ by $ x$. Specify values of $ x$ so that volume of pyramid $ S.ABCD$ is maximum.[/b][/color]

[color=blue][b]May this problem have no solution? Is it really difficult or easy?[/b][/color]  :blush:  :P


There is an ample supply of milk in a milk tank. Mr. Fat is given a 5-liter (unmarked) container and a 9-liter (unmarked) container. How can he measure out 2 liters of milk?

[hide]Fill the 5-liter and dump the milk in the 9-liter.  Fill the 5-liter again, and dump milk into the 9-liter until the 9-liter reaches its max capacity.  There will be one liter of milk left in the 5-liter.
Empty the 9-liter.  Dump the 5-liter milk in the 9-liter.  The 9-liter now has one liter.  Fill the 5-liter.  Dump all the contents of the 5-liter into the 9-liter.  The 9-liter now has six liters.
Now, fill the 5-liter. Dump the contents into the 9-liter, until the 9-liter reaches capacity. Now the 5-liter will have the desired 2 liters of milk in it.
  [/hide]
What is Bezout's Identity?

i remember it as the following:

given 2 functions f and g, f of degree m and g of degree n, the maximum number of times they intersect on a cartesian plane is mn.

there might be more but that's what i remember.

Bezout's identity is that for any integers $ m$ and $ n$, there is always a solution $ (x,y)$ to $ mx\plus{}ny\equal{}\gcd(m,n)$. This follows directly from the Euclidean Algorithm. This may not seem useful at first, but here's a technique involving Bezout's identity:

[color=blue]Sample Exercise:[/color]
Find all integer solutions to $ 7x\plus{}10y\equal{}u$, for any integer $ u$.

From Bezout's Identity, there exists a solution to $ 7x\plus{}10y\equal{}1$. We can easily see that the parametric form is $ x\equal{}10t\minus{}7$, $ y\equal{}7t\plus{}5$. The useful part comes here, for $ 7x\plus{}10y\equal{}u$ the solution $ x\equal{}10t\minus{}7u$, $ y\equal{}7t\plus{}5u$. This is a very quick way to solve otherwise annoying linear Diophantine equations, as opposed to the way that involves a lot of algebraic isolation and back-substitution and whatnot.

[quote="unimpossible"]given 2 functions f and g, f of degree m and g of degree n, the maximum number of times they intersect on a cartesian plane is mn.[/quote]

It is easy to get confused by the naming; this is known as Bezout's [b]Theorem.[/b]  It's actually possible for a curve of degree $ m$ and a curve of degree $ n$ to [i]always[/i] intersect $ mn$ times (with appropriate multiplicity), but it requires that

- we work over a plane with algebraically closed coordinates (for example, $ \mathbb{C}\times\mathbb{C}$), and
- we work in the [url=http://en.wikipedia.org/wiki/Projective_plane]projective plane[/url].

In fact, this stronger statement of Bezout's Theorem is one of the primary motivations for working in the projective plane.  :)


I'm pretty sure the problem writing is done independent of the NSA.

[quote="Pakman2012"]USAMTS is sponsored by the NSA.  Do they have any influence on the types of problems you all choose? Or even, any preference for that matter.

That'd actually be really cool if this was just a giant recruitment type thing for the NSA.[/quote]

Hi, I am the NSA influence on the problems in the USAMTS, along with my fellow mathematician David. After the Art of Problem Solving people write the first draft of a USAMTS problem set, they send it out to several mathematicians for their comments. That way, the AoPS can check for bad phrasing before the problems are posted. Two of us problem-checking mathematicians work for the NSA.

Also, many of the problems on the USAMTS have been invented and submitted by mathematicians across the world. Some of those mathematicians work for the NSA.

The NSA as a government agency has no influence on the USAMTS problems. If you check the history of the USAMTS (http://www.usamts.org/About/U_AbHist.php) you will see that the contest was founded independently of the NSA. The NSA became involved during year five (give or take a year, my memory is fuzzy), when George Berzsenyi needed more help grading all the entries he received. We mathematicians form a worldwide community, and one of the duties of the community is to encourage young mathematicians.

The NSA is not blind to the recruiting potential of the USAMTS. We like participants who are entering college soon to know about our scholarship, co-op, and summer hire programs (http://www.nsa.gov/careers/students.cfm). I worked with one young woman in the Stokes Educational Scholarship Program who had learned about it by paging around the NSA website back when the USAMTS problems were hosted on that site. And we need bright young math majors for the Director's Summer Program. We even have some work-study programs for high school students in the Baltimore-Washington area; alas, they focus more on clerical and computer-programming skills than on mathematical skills.

Erin Schram


Ring Theory
superior algebra
superior algebra solved

this is a nice one.Assume R is a commutative ring. show that the following are equivalent

1) R has exactly one prime ideal
2) every element of R is either nilpotent or a unit

$x$ is invertible $\iff$ $x$ is contained in no prime ideal
$x$ is nilpotent $\iff$ $x$ is contained in every prime ideal


calculus
integration
trigonometry
calculus computations

Find the integral of $ \int$$ \frac{1}{5\plus{}4\sin{x}} dx$

Let $ I\equal{}\int \frac{1}{5\plus{}4\sin x}\ dx,\ J\equal{}\int \frac{1}{5\minus{}4\sin x}\ dx$,

$ I\plus{}J\equal{}\int \frac{10}{9\plus{}16\cos ^ 2 x}\ dx$

$ J\minus{}I\equal{}\int \frac{8\sin x}{9\plus{}16\cos ^ 2 x}\ dx$

Or try the substitution: $ t \equal{} \tan \frac{x}{2}$

Notice that $ t \equal{} \tan \frac {x}{2} \implies dt \equal{} \frac {1}{2}\cdot \left(\tan^2 \frac {x}{2} \plus{} 1 \right)dx \implies \frac {2dt}{1 \plus{} t^2} \equal{} dx$

Next we want $ \sin x$ expressed in terms of t and you derive that as follows:

$ t \equal{} \tan \frac {x}{2}$---(1)

$ t^2 \equal{} \tan^2 \frac {x}{2} \implies t^2 \plus{} 1 \equal{} \tan^2 \frac {x}{2} \plus{} 1 \equal{} \frac {1}{\cos^2 \frac {x}{2}}$---(2)

$ \sin x \equal{} \sin \left(\frac {x}{2} \plus{} \frac {x}{2}\right) \equal{} 2 \sin \frac {x}{2} \cos \frac {x}{2}$ ---(3)

From (3) we get
$ \sin x \equal{} 2 \sin \frac {x}{2} \cos \frac {x}{2} \cdot \frac {\cos \frac {x}{2}}{\cos \frac {x}{2}}$ ---(4) multiplying by 1 does not change the result

$ \equal{} 2 \tan \frac {x}{2} \cdot \cos^2 \frac {x}{2}$ ---(5) We get this from rearranging (4)

$ \equal{} 2t \cdot \frac {1}{1 \plus{} t^2}$ We get this from substituting (1) and (2) into (5).

You are given the integral:

$ I \equal{} \int \frac {1}{5 \plus{} 4 \sin x} dx \equal{} \int \frac {1}{5 \plus{} 4\left(2t \cdot \frac {1}{1 \plus{} t^2}\right)} \cdot \frac {2dt}{1 \plus{} t^2}$

I hope from here you can go ahead and simplify and see if you will be able to finish the resultant problem.

Continuing akech's progress:
$ I \equal{} 2\int \frac{dt}{5t^2 \plus{} 8t \plus{} 5} \equal{} 10 \int \frac{dp}{(5p\plus{}4)^2 \plus{} 9}$
$ \equal{} \frac23 \int \frac{3 d(5p\plus{}4)}{(5p\plus{}4)^2 \plus{} 3^2} \equal{} \frac23 \arctan \left( \frac{5p\plus{}4}{3} \right) \plus{} c\equal{} \frac23 \arctan \left( \frac{5\tan \frac{x}{2} \plus{}4}{3} \right) \plus{}c$

[quote="kl2836"]any other ideas?[/quote]
$ \int\frac {1}{5 \plus{} 4\sin{x}}dx$
$ \implies \int\frac {1}{5(sin^{2}\frac {x}{2} \plus{} cos^{2}\frac {x}{2}) \plus{} 8sin\frac {x}{2}cos\frac {x}{2}} dx$
$ \implies \int\frac {sec^{2}\frac {x}{2}}{5 tan^{2}\frac {x}{2} \plus{} 8tan\frac {x}{2} \plus{} 5} dx$
$ \implies 2\int\frac {d (tan\frac {x}{2})}{5 tan^{2}\frac {x}{2} \plus{} 8tan\frac {x}{2} \plus{} 5}$
$ \implies \frac {2}{5}\int\frac {d (tan\frac {x}{2} \plus{} \frac {4}{5})}{(tan\frac {x}{2} \plus{} \frac {4}{5})^{2} \plus{} (\frac {3}{5})^2}$
$ \implies \frac {\frac {2}{5}}{\frac {3}{5}}\left(tan^{ \minus{} 1}\left(\frac {tan\frac {x}{2} \plus{} \frac {4}{5}}{\frac {3}{5}}\right)\right) \plus{} c$
$ \implies \frac {2}{3}\left( tan^{ \minus{} 1}\left (\frac {5 tan\frac {x}{2} \plus{} 4}{3}\right )\right) \plus{} c$


A print shop is going to print a book of $ 350$ pages printed on both sides of the sheets. $ 5000$ copies of the book have been ordered. The shop manager is trying to calculate how much paper is needed. She
knows that each ream contains $ 500$ sheets and each box holds $ 10$ reams. A pallet carries $ 20$ boxes. Find the minimum number of whole pallets she must order.

There is a total of $ 350 \times 5000 \equal{} 1750000$ sheets of paper, so there are $ 1750000 \div (500 \times 10 \times 20) \equal{} 17.5$ pallets needed. However, we need a whole number, so our answer is $ \boxed{18}$.

However, we are printing on both sides of the sheets therefore we are using exactly half the number of pages.  So the number of pallets needed is $\frac{17.5}{2}=8.75$; however, we need a whole number of pallets therefore the answer is $\boxed{9}$.


[quote="Ocean8"]Use numbers, written as words, to make the following sentence true: This sentence contains _____ e's, _____ t's and _____ s's.  What is the sum of the number of e's, t's, and s's in the previous sentence.

Is there a way to solve this question w/o trial and error?

[hide]Answer: 19[/hide][/quote]
Just count how many e's, t's, and s's there are in the sentence without the numbers and go from there.


limit
integration
real analysis
real analysis solved

1.Calculate n \lim \infty n((n)\sqrtn -1) and n \lim \infty   \sum (k=0,n)ln(n+k)/ \sqrt (n+k)
   ...and (n) \sqrt n is the root of order n from n  (radical de ordinul n din n)

n^1/n = e^(ln(n)/n) = 1 + ln(n)/n + o(1/n)
so n (n^1/n - 1) ~ ln(n)

ln(x)/ \sqrt (x) is decreasing (for x big enough)

 \sum  (k=0,n)ln(n+k)/ sqrt (n+k) >=  \int (x=n-1..2n-1) ln(x)/ \sqrt (x) -> +oo


limit
real analysis
real analysis unsolved

[I need to share with you this beauty :)]

Find the limit of the sequence: $x_n = \sum_{k=1}^n \frac{1}{6k-1} - \frac{1}{6}ln n, n \geq 1$

Probably there is a typo, since in this form the problem is absolutely trivial.

ohh, sorry for my rush, this is new edit...

Can any limit of the form $\lim_n \sum_{k=1}^n \frac{1}{mk-1} - \frac{1}{m}ln n, n \geq 1$ be calculated?
 It have been posted a problem which ask to caculate $\sum \frac 1 {(2n-1)(3n-1)}$ which can be solved by calculatind two limits of this type. but i have noo ideea about how to find this limit.

do you mean this post ;)?

[url=http://www.artofproblemsolving.com/Forum/topic-74655.html]www.artofproblemsolving.com/Forum/topic-74655.html[/url]

you'll find some answers there :)


trigonometry
quadratics
algebra
polynomial
difference of squares
special factorizations

1) Find all possible ordered pairs $ (x,y)$
$ x \plus{} y \plus{} \sqrt {xy} \equal{} 28$
$ x^2 \plus{} y^2 \plus{} xy \equal{} 336$

2) If $ \displaystyle 2\cos^3 \frac {\pi}{7} \minus{} \cos^2 \frac {\pi}{7} \minus{} \cos \frac {\pi}{7}$ is a rational number in simplest form: $ \displaystyle \frac {a}{b}$, where a and b are integers, compute the integer $ \displaystyle a \plus{} b$.

3) Find the sum $ a \plus{} b \plus{} c \plus{} d \plus{} e$ if $ \sqrt {7 \plus{} \sqrt {40}} \equal{} a \plus{} b\sqrt {2} \plus{} c\sqrt {5} \plus{} d\sqrt {7} \plus{} e\sqrt {10}$, where a, b, c, d, and e are all integers.

4) For how many nonnegative integers n does $ x^3 \plus{} (n \minus{} 1) x^2 \plus{} (n \minus{} n^2) x \minus{} n^3$ have all integer roots?

Although I'm not a "Carolinan member", I still want to present my solutions for all of these problems:
[hide][hide]Problem 1:[/hide]
From the 2nd equation we will have: (x+y)= + or - square root(336+xy)
Now taking square both sides of the 1st equation we will have:
xy+square root(xy)*(x+y)=224
Replace the result we just calculated above of (x+y), we get the equation:
a^2-448a+50176=a(336+a) then calculate a=64
Substitute it into (1) we have: x+y=20
                                          x^2+y^2=272
Now simply substitute x=20-y we will have 2 results:(x,y)=(16,4) or (4,16)[/hide]
[hide][hide]
Problem 3:[/hide]
We have: square root(7+square root(40))=square root(5)+square root(2)
Now if we match all terms from both sides, we can get: a+b+c+d+e=1+1+0+0+0=2
However, I believe this problem is wrong!Why? Because we don't have restriction of a,b,c,d,e we can make-up many results as we can to make different sums(like if we give e=square root(2), then we find a=-2, and just give b=1,c=d=0 we get another result of a+b+c+d+e).Therefore, I think we lack of the restriction:"a,b,c,d,e is integer". At that time, my solution is not good enough to cover all of the sum of a+b+c+d+e(although the same methodand idea can be used if you consider that the equation: c.square root(5)+d.square root(7)=k.square root(2) or (5) or(10)(you should try to solve it :P ). And that's all for this problem.

Problem 4:
We can factorize this sum into:(x-n)(x^2+2nx+n^2-x)=0
Now, f(x) has all integer solutions when x^2+2nx+n^2-x=0 has all integer solutions. Knowing that it's a quadratic equation and the coefficient before x^2 is 1, we just need to have:
Delta=(2n-1)^2-4n^2=k^2(k, n are integers).
Unluckily, this equation has infinite solution(we can state that because the LHS is in the form: -4n+1, which is sufficient to lead a conclusion since if we let n is negative integer it becomes:4n+1)
Therefore, this problem also lacks of something(If I were you, i just give n is positive integer, then we will have only one solution) :wink:

 [/hide]
[hide]Problem 2:
We easily prove that this polynomial is equivalent to: -2.cos(pi/7).sin(pi/14).cos(2pi/7)=-sin(pi/7).cos(pi/7).cos(2pi/7)/cos(pi/14)=-sin(4pi/7)/4cos(pi/14)=-1/4
Therefore, a+b=3[/hide]

I think I have slighty more direct ways for the first two (although I'm not from South Carolina either :P ).
[hide="Problem 1"]We notice the difference of squares factorization
\[ x^2 + y^2 + xy = (x + y + \sqrt {xy})(x + y - \sqrt {xy}).
\]
Then
\begin{align*} x + y - \sqrt {xy} & = 12, \\
x + y + \sqrt {xy} & = 28,
\end{align*}
so $ x + y = 20$ and $ \sqrt {xy} = 8$.  It follows that $ x$ and $ y$ are the roots of the quadratic $ k^2 - 20k + 64 = 0$, so $ (x,y)$ is either (16,4) or (4,16).[/hide]

[hide="Problem 2"]Let $ \zeta$ denote $ e^{i\pi/7}$.  Then $ \cos \pi/7 = \text{Re}( \zeta)$, and $ \zeta^7 + 1 = 0$.  Evidently, $ \zeta \neq - 1$, so
\begin{align*} 0 & = 1 - \zeta + \zeta^2 - \zeta^3 + \zeta^4 - \zeta^5 + \zeta^6 \\
& = 1 - \zeta + \zeta^2 - \zeta^3 - \zeta^{11} + \zeta^{12} - \zeta^{13} \\
& = 1 - \zeta + \zeta^2 - \zeta^3 - \overline{\zeta^{3}} + \overline{\zeta^2} - \overline{\zeta} \\
& = 1 - 2\text{Re}(\zeta) + 2 \text{Re}(\zeta^2) - 2 \text{Re}(\zeta^3) . \end{align*}
Now, $ \text{Re}(\zeta^2) = \cos 2\pi/7 = 2 \cos^2 \pi/7 - 1$, and $ \text{Re}(\zeta^3) = \cos 3\pi/7 = 4\cos^3 \pi/7 - \cos \pi/7$.  Therefore
\begin{align*} 0 & = 1 - 2\cos \pi/7 + 2 (2\cos^2 \pi/7 - 1) - 2(4\cos^3 \pi/7 - 3 \cos \pi/7) \\
& = - 8 \cos^3 \pi/7 + 4 \cos \pi/7 + 4 \cos \pi/7 - 1 . \end{align*}
It follows that
\[ 2\cos^3 \pi/7 - \cos^2 \pi/7 - \cos \pi/7 = - 1/4,
\]
which solves the problem.[/hide]

P.S.  Thanks, Sean. :)

@Miles: That was pretty slick for #1... The way I did it required a lot more substitution.

@ghjk: I think you're right, #3 should state that a, b, c, d, and e are all integers.  I'll edit it.


trigonometry
analytic geometry
graphing lines
slope
function
calculus
calculus computations

At what point does the curve $x = 1-2\cos^{2}t$, $y = x\tan t$ cross itself? Find the equations of both tangents at that point.

So $(\tan t_{1})x = (\tan t_{2})x \Rightarrow \tan t_{1}= \tan t_{2}$

I got $t = 0$ and $t = \pi$ as the different values of $t$ that five the same x-y values. But when I tried using the point slope form, the slope ended up being infinity. What am I doing wrong?

Thanks

Since both $x$ and $y$ are periodic functions of $t$ with period $\pi$, the curve goes through the same points infinitely many times unless the values of $t$ are somehow restricted. However, this is not what is usually meant by saying that the curve crosses itself. The values of $t$ that are different by $\pi$ should not count as a self-crossing.

Note that $(\tan t_{1})x = (\tan t_{2})x$ implies that either $\tan t_{1}= \tan t_{2}$ or $x=0$. The curve crosses itself at $(0,0)$.


function
Functional Analysis
real analysis
real analysis unsolved

Show that in every Banach space $ X$ there is a subspace $ Y$ with $ \textrm{codim} Y < \infty$ that isn't closed.

This is false as stated, because finite-dimensional normed spaces are Banach. We need to specify that $ X$ is infinite-dimensional.

The existence of a finite-codimension non-closed subspace is equivalent to the existence of a non-closed subspace of codimension 1, which is equivalent to the existence of an unbounded linear functional. We construct the latter as follows: Choose a countably infinite linearly independent set $ x_1,x_2,\cdots$, normalized so that $ |x_i|\equal{}1$. Extend this set to a Hamel basis of $ X$. Define $ f(x_i)\equal{}i$, $ f(x)\equal{}0$ on each other element of the basis, and extended by linearity to everything else.

Why does closed nullspace imply continuous?

Let $ X$ be the space, $ f$ the functional, and $ Y$ the null space of $ f$. Since $ Y$ is closed, $ X/Y$ is a Banach space of dimension $ 1$, and $ f$ is defined on $ X/Y$. This makes $ f$ continuous on $ X/Y$; by the standard properties of functions on a topological quotient, $ f$ is continuous on $ X$.


probability
inequalities
inequalities unsolved

Prove that: |P(A intersects B)-P(A).P(B)|<=1/4.

Let $u$ denote $\displaystyle  u = \min_{k=1,..,n}(P(A_k))$. So I can write : $P(\displaystyle \bigcap_{k=1}^nA_k)-\prod_{k=1}^nP(A_k) \leq u - u^n \leq \frac{n-1}n.( \frac 1n)^{\frac 1{n-1}}$
 :cool:


The 4x4 square is a multiplicative magic square, the product of the numbers in each row, column, and diagonal is the same.
If all the entries are different positive integers,
what is the minimum value of the biggest number?

Generalization: How about an nxn square?

Here comes one example:

1	120	10	12
20	6	8	15
24	5	60	2
30	4	3	40

but i m sure is not minimal...

I ve got one with 36, is it the minimum?


I found that many people don't want to join this math chat because it's kind of too easy for them.. But I can't do anything to it since I can't handle any further things beyond algebra II...
So, here's my question:

Do you more likely to have math chat in specific area or like in competition?  Because some questions in competitions are pretty challenging while others are just simple computating.

I like having the math chats on a wide variety of topics( mostly ones from competitions) and then having people explain them because that is were you potentially attain the most knowledge.  But I would also like to have some chats on individual topics.  For example, I would really like to attend ones on geometry, modular aritmetic, and trig because those are my weak spots but I would pretty much attend anything if I were able to schedule wise.

[quote="Silverfalcon"]
Do you more likely to have math chat in specific area or like in competition?  Because some questions in competitions are pretty challenging while others are just simple computating.[/quote]

It seemed to me that the nature of the chat was that people shared basically whichever problems they found interesting and slightly challenging.  It's probably better to keep it mixed-topic and competition-question-ish.


Given a graph with $n$ vertices and $q$ edges numbered $1,...,q$, show that there exists a chain of $m$ edges, $m \geq \frac{2q}{n}$, each two consecutive edges having a common vertex, arranged monotically with respect to the numbering.

I believe this is a pretty simple graph theoretical problem if I understood the arrangement of the graph, however, you will probably not get a desirable reply in a high school forum.  I believe I have a solution using the
[hide]Handshaking Lemma[/hide]
Masoud Zargar


LaTeX
inequalities proposed
inequalities

[quote="nickolas"]ABC  triangle , x,y,z are reals. Prove that
(b/c+c/b-1)x 2 + (c/a+a/c-1)y 2 + (a/b+b/a-1)z 2  \geq  [3-(b+c)/a]yz + [3-(a+c)/b]xz + [3-(a+b)/c]xy \geq 0[/quote]

 :)

ABC  is a  triangle and  $x,y,z$ are  reals. Prove that :

$(\dfrac{b}{c}+\dfrac{c}{b}-1)x^2+(\dfrac{c}{a}+\dfrac{a}{c}-1)y^2+(\dfrac{a}{b}+\dfrac{b}{a}-1)z^2\geq(3-\dfrac{b+c}{a})yz+ (3-\dfrac{a+c}{b})xz+(3-\dfrac{a+b}{c})xy\geq 0$


geometry
circumcircle
IMO Shortlist
reflection
vector
parallelogram
moving points

Let $ ABC$ be a triangle, and $ H$ its orthocenter. Let $ P$ be a point on the circumcircle of triangle $ ABC$ (distinct from the vertices $ A$, $ B$, $ C$), and let $ E$ be the foot of the altitude of triangle $ ABC$ from the vertex $ B$. Let the parallel to the line $ BP$ through the point $ A$ meet the parallel to the line $ AP$ through the point $ B$ at a point $ Q$. Let the parallel to the line $ CP$ through the point $ A$ meet the parallel to the line $ AP$ through the point $ C$ at a point $ R$. The lines $ HR$ and $ AQ$ intersect at some point $ X$. Prove that the lines $ EX$ and $ AP$ are parallel.

Here is my solution.

[color=blue][b]Problem.[/b] Let ABC be a triangle and H its orthocenter. Let P be a point on the circumcircle of triangle ABC (distinct from the vertices A, B, C), and let E be the foot of the B-altitude of triangle ABC. Let the parallel to the line BP through the point A meet the parallel to the line AP through the point B at a point Q. Let the parallel to the line CP through the point A meet the parallel to the line AP through the point C at a point R. The lines HR and AQ intersect at some point X. Prove that EX || AP.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°.

Since H is the orthocenter of triangle ABC, the lines AH, BH, CH are the altitudes of this triangle; hence, the point E, being the foot of the B-altitude, must lie on the line BH, and we have $AH\perp BC$, $BH\perp CA$ and $CH\perp AB$, so that < (AH; BC) = 90°, < (BH; CA) = 90° and < (CH; AB) = 90°. Thus,

 < CHA = < (CH; AH) = < (CH; AB) + < (AB; BC) - < (AH; BC)
       = 90° + < (AB; BC) - 90° = < (AB; BC) = < ABC.

But since the point P lies on the circumcircle of triangle ABC, we have < ABC = < APC, and since CR || AP and AR || CP, we have < (AP; CP) = < (CR; AR), or, equivalently, < APC = < CRA. Thus, < CHA = < ABC = < APC = < CRA, and it follows that the points C, A, H and R lie on one circle. Thus, < HRC = < HAC. In other words, < (HR; CR) = < (AH; CA). Since CR || AP, we have < (HR; CR) = < (HR; AP), and this becomes < (HR; AP) = < (AH; CA). But

 < (AH; CA) = < (AH; BC) + < (BC; CA) = 90° + < (BC; CA) = 90° + < BCA,

and since the point P lies on the circumcircle of triangle ABC, we have < BCA = < BPA. Thus, < (AH; CA) = 90° + < BPA, so that < (AH; CA) - < BPA = 90°. Hence,

 < (HR; BP) = < (HR; AP) - < (BP; AP) = < (AH; CA) - < BPA = 90°.

Thus, $HR\perp BP$. Since AQ || BP, this becomes $HR\perp AQ$. Thus, < HXA = 90°. On the other hand, since E is the foot of the B-altitude of triangle ABC, we have < HEA = 90°. Thus, the points X and E lie on the circle with diameter HA. This yields < EXA = < EHA, or, equivalently, < (EX; AQ) = < (BH; AH). But

 < (BH; AH) = < (BH; CA) + < (CA; BC) - < (AH; BC)
            = 90° + < (CA; BC) - 90° = < (CA; BC) = < ACB,

and since the point P lies on the circumcircle of triangle ABC, we get < ACB = < APB. Since AQ || BP, we have < (AP; BP) = < (AP; AQ). Thus,

 < (EX; AQ) = < (BH; AH) = < ACB = < APB = < (AP; BP) = < (AP; AQ),

so that EX || AP, and the problem is solved.

  Darij

[i]Darij,[/i] your enunciation is complicated. The points $R$ and $Q$ are the reflections of the point $P$ w.r.t. the midpoints of the sides $AC$, $AB$ repectively ! I think I make no mistakes. I wish you a happy New Year - 2000 !

thank you for solved problem

Happy new year :roll:

First, we use vectors to show that $BP\perp RH$.
Put the origin at the circumcenter of ABC, so that $\vec{H}=\vec{A}+\vec{B}+\vec{C}$ and $|\vec{A}|=|\vec{B}|=|\vec{C}|=|\vec{P}|$. Because of parallelogram APCR, $\vec{R}=\vec{A}+\vec{C}-\vec{P}$, and now we calculate $\vec{BP}\cdot\vec{RH}=(\vec{P}-\vec{B})\cdot(\vec{P}+\vec{B})=0$, i.e. $BP\perp HR$.

This means that <AXH = 90 (since AX || BP), and so <AXH = <AEH, i.e. AEXH is cyclic. From here, a simple angle chase (directed, modulo 180 degrees) gives

<AXE = <AHE =  <BCA = <BPA = <XAP,

and so EX || AP, as desired.

I remember I had solved this a long way back.Sorry for posting late.Its pure euclidean geometry.

Let $D,E,F$ be the foot of altitudes from $A,B,C$ respectively and wlog let $P$ belongs to minor arc $BC$ (the proof is exactly same in all the cases).Now $APBQ$ is a parallelogram so $\angle{BQP}=C$.So points $A,H,B,Q$ are concyclic $\Rightarrow \angle{HQA}=\angle{HBA}=90-A$.Similarly points $H,A,R,C$ are concyclic.So $\angle{HRC}=\angle{HAC}=90-C$.Also $\angle{QHX}=\angle{QHA}+\angle{AHX}=\angle{QBA}+\angle{ACR}=\angle{BAP}+\angle{CAP}=A$ so from $\triangle{QHX}$ we get that $\angle{HXA}=90^{\circ}$.Hence points $A,X,E,H$ are concyclic $\Rightarrow \angle{HXE}=\angle{HAE}=90-C \Rightarrow XE \parallel RC \parallel AP$.

Let the reflection of $\bigtriangleup ABC$ across $AH$ be $\bigtriangleup AB'C'$ and  the reflection of $E$ across $AH$ be $D$.
$1)\ EX \| AP \iff AEHD$ is cyclic $\iff \angle HXA = 90^{\circ}$(simple angle chasing) so problem now reduces in proving $RX \perp XA$
$2)\ \angle {B}'DA = 90^{\circ}$, so, $RX \perp XA \iff \angle XAD = \angle RHB'$ (simple transformation geometry tricks)
$3)\ AHCR{B}'$ and $AH{C}'QB$ are cyclic(simple angle chasing)
$4)\ \angle XAD = \angle QA{C}' = \angle RA{B}' = \angle RH{B}' ($from $(3)$ and some simple angle chasing)
Now, $(2)$ and $(4)$ solves the problem.

Dear Mathlinkers,
by applying a spacial case of the Reim's theorem

http://jl.ayme.pagesperso-orange.fr/Docs/6%27.pdf

we avoid an angle chasing.

Sincerely
Jean-Louis

$RH\cap PB={K}, RC\cap BP={T}, AH\cap BC={L}$
$\angle{AHC}=\angle{ARC}=180-\angle{C}$, therefore $(ARHC)$ cyclic.
$\rightarrow \angle{KRT}=\angle{HAC}=90- \angle{C}$, also $\angle{RTK}=\angle{APB}= \angle{C}$
So $HR \perp QA$ and $HR \perp BP$. This gives a new pair of cyclic quadrilaterals, namely $(XAEH),(BHKL).$
$\rightarrow \angle{EAP}=\angle{CAP}=\angle{CBP}=\angle{LBK}=\angle{LHK}=\angle{XHA}=\angle{XEA} \leftrightarrow AP \parallel XE \blacksquare$

WLOG assume that $P$ lies on arc $AC$ not containing $B$.

[b]Claim 1[/b]- $AHRC$ is cyclic.
Proof- $\angle AHC=180-\angle CAH-\angle ACH=180-(90-\angle C)-(90-\angle A)=\angle A+\angle C=\angle APC=\angle ARC$.

[b]Claim 2[/b]-$APHX$ is cyclic.
Proof- As $AH$ is diameter, we need $\angle AXH=90$. This can be achieved-
$$\angle AXH=180-\angle XAH- \angle XHA=180-(\angle XAB+\angle BAH)-(\angle ACR)$$
$$=180-(\angle ABP+(90-\angle B))-(\angle PAC)=180-(90-\angle CBP)-(\angle CBP)=90^{\circ}$$

Moving to the problem,
We have $$\angle AXE=\angle AHE=90-\angle HAC=\angle C=\angle APB=\angle AQB$$.
Hence, $EX \parallel QB \parallel AQ$, as desired. $\blacksquare$

Assume $P$ lies on arc $BC$ not containing $A$. Then, $\angle AQB=\angle C=180^{\circ}-\angle AHB$ $\implies$ $AQBH$ & similarly $AHCR$ are cyclic. $BQ$ $=$ $AP$ $=$ $CR$ $\implies$ $AH$ $\perp$ $QR$ and then note that,
$$\begin{cases} \angle QHR=\angle QBA+\angle RCA=\angle BAP+\angle CAP=\angle BAC \\ \angle QAR =360^{\circ} - (\angle BAC+180^{\circ})=180^{\circ} -\angle BAC \end{cases} \implies A \text{ is the orthocenter of } \Delta QHR$$
Hence, $AHEX$ is cyclic and By Reim's Theorem $\implies$ $XE||QB||AP$

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf  p. 25...

Sincerely
Jean-Louis

Here's another solution: Animate $P$ on $\odot (ABC)$. Then taking homothety from the midpoint of $AB$ with ratio $-1$, we get that $P \mapsto Q$ is a projective map. Similarly $P \mapsto R$ is also projective. Also, taking perspectivity from $A$ to the line at infinity, we see that $P \mapsto \infty_{AP}$ is a projective map. Let $$X_1=AQ \cap \odot (AH), X_2=RH \cap \odot (AH), X_3=E \infty_{AP} \cap \odot (AH) $$
As $A, E, H$ are fixed, so we get that the points $X_1, X_2, X_3$ are all projective related to $P$. So it suffices to show $X_1=X_2=X_3$ for three positions of $P$. One can easily see this to be true for $P=A, C$ and when $P$ is the antipode of $B$ in $\odot (ABC) $ (It requires some work, but I am really not in the mood of writing all that :D). Hence,  done. $\blacksquare$

[asy]
size(10cm);
defaultpen(fontsize(10pt));

pair A = dir(125), B = dir(210), C = dir(330), P = dir(280), Q = A+B-P, R = A+C-P, H = orthocenter(A,B,C), E = foot(B,C,A), X = extension(A,Q,R,H);

dot("$A$", A, dir(90));
dot("$B$", B, dir(260));
dot("$C$", C, dir(330));
dot("$P$", P, dir(280));
dot("$Q$", Q, dir(90));
dot("$R$", R, dir(90));
dot("$X$", X, dir(0));
dot("$H$", H, dir(270));
dot("$E$", E, dir(45));

draw(unitcircle);
draw(A--B--C--A, linewidth(1.2));
draw(R--H--B--P--C--H--A--X--E--H^^C--R--A--Q--B^^A--P);
draw(circumcircle(A,B,H)^^circumcircle(A,C,H)^^circumcircle(A,E,H), dotted);
[/asy]

Let $\measuredangle$ denote a directed angle modulo $180^\circ$.

[b]Claim 1:[/b] $AHBQ$ and $AHCR$ are cyclic.

[i]Proof.[/i] We have $$\measuredangle AQB = \measuredangle BPA = \measuredangle BCA = \measuredangle AHB,$$ so $AHBQ$ is cyclic. The other case is analogous. $\square$

[b]Claim 2:[/b] $\overline{QA} \perp \overline{RH}$.

[i]Proof.[/i] Since $\overline{QB} \parallel \overline{AP} \parallel \overline{RC}$, we have $$\angle QXR = \angle XQB + \angle XRC = \angle APB + \angle CAH = \angle ACB + \angle CAH = 90^\circ$$ as desired. $\square$

It follows from claim 2 that $AXEH$ is cyclic. Therefore $$\angle HRC = \angle HAC = \angle HXE,$$ and we are done. $\blacksquare$

Solved with [b]Smileyklaws[/b] and pretty big hints from [b]tree_3[/b] and [b]mathgirl199[/b]

Note that $AHRC$ is cyclic since $$\measuredangle AHC = \measuredangle CBA = \measuredangle CPA = \measuredangle ARC$$Now, since $QA \parallel BP$ and $AR \parallel PC$, we have $$\measuredangle XAR =\measuredangle QAR = \measuredangle BPC = \measuredangle BAC$$But remark that $$\measuredangle ARX  = \measuredangle ARH = \measuredangle ACH = 90^{\circ}-\measuredangle BAC$$so it follows that $RX \bot AQ$. Then $AXHE$ is cyclic, so $$\measuredangle AXE = \measuredangle AHE = -\measuredangle ACB = \measuredangle BPA = \measuredangle AQB$$which means $XE \parallel QB \parallel AP$, as needed. $\blacksquare$

Notice that $PBC$ and $AQR$ are congruent and corresponding sides are parallel. Let the orthocenter of $PBC$ be $H'$; it's well known that $PH'=AH$ so that a translation bringing $PBC$ to $AQR$ brings $H'$ to $H$ so that $H$ is the orthocenter of $AQR$. Thus $AQ\perp HR$ and $A,X,H,E$ are concyclic. So $\measuredangle EAP=\measuredangle PBC=\measuredangle XHA=\measuredangle XEA$ as desired.

pls point out any issues with my directed angles, thanks!

I'll just outline my solution.

[hide=Solution Outline]
Prove that $AQR$ and $PBC$ are collinear. Prove that $AHCR$ are concyclic. Let $F$ be the foot of the perpendicular of $QR$ from $A$. Use the previous results to prove that $XQFH$ are concyclic. This implies that $\angle AXR = \frac{\pi}{2}$ so $AXHE$ is concyclic. Then $\angle EXA = \angle EHA = \angle ACB = \pi - \angle APB = \angle XAP$ which implies the result.
[/hide]


In ∆ ABC, m<A = 33, a = 12, and b = 15. What is m<B to the nearest degree?

(1) 41      (3) 44
(2) 43      (4) 48

I had no idea this question involved trigonometry.


What is the positive value of the expression $ \sqrt{x^3 \minus{} 2^y}$ when $ x \equal{} 5$ and $ y \equal{} 2$?

Simply plug the values for $ x$ and $ y$ in and the answer chugs out.

$ \sqrt{x^3\minus{}2^y}\equal{}\sqrt{5^3\minus{}2^2}\equal{}\sqrt{125\minus{}4}\equal{}\sqrt{121}\equal{}\boxed{11}$


If $U$, $S$, $A$ are three positive real numbers, then prove that

$\frac{U}{\left(1+U\right)\left(1+S\right)}+\frac{S}{\left(1+S\right)\left(1+A\right)}+\frac{A}{\left(1+A\right)\left(1+U\right)}\geq\frac{3USA}{\left(1+USA\right)^{2}}$.

Show that equality holds if and only if $U=S=A=1$.


Find the sum of all the coefficients in the following expansion:
\[ (1+x^2)+(1+x^2)^2+(1+x^2)^3+ \cdots +(1+x^2)^{20} \]

Hm. [hide]Now the sum of the coefficients of (1+a)^n=2^n. If we let x^2=a, we have for our sum of coefficients 2^1+2^2+2^3...2^20, which is equal to 2^21-2. I cannot presently evaluate this, but it is simple. [/hide]

@solafidefarms

[hide]
it follows from Newton's bynomial. In fact, if we consider $(1+x^2)^n$, the sum of coefficients is $\sum_{i=0}^n \binom ni=2^n$, so the sum of coefficient of the expression is $2^1+2^2+...+2^{20}=\sum_{i=1}^{20}2^i=\frac{2^{21}-1}{2-1}-1=2^{21}-2$[/hide]


a function a(x,y) i sdefined by the following rules:

a(0,n)=n+1
a(m,0)=a(m-1,1)
a(m,n)=a(m-1,a(m,n-1))

for m and n natural numbers.
a(2,3)=k, where k is a whole number. find the value of k.

is there any "pretty" way or a shortcut, or do you just have to brute force this. even though the numbers are small, the problem gets very messy. what shortcuts work here?

you find just a few helpful patterns im pretty sure but you have to brute force it for the most part.
incase youre not sure how to brute force it, you have to find the value of function a for other values of m and n
since this is team you should have at least 2 people working on this at the same time: 1 person finding values and one person checking
if you need a full solution please say

OK THANKS...
SO FROM WHAT I UNDERSTAND IS THAT UR SAYING THAT I SHOULD BRUTE FORCE IT??

PS. HOW DO I JOIN THE CLUB??

stop typing in caps lol
[hide="Solution, with some extra stuff"]$ a(0,0)\equal{}a(\minus{}1,1)\equal{}1$
$ a(1,0)\equal{}a(0,1)\equal{}2$
$ a(1,1)\equal{}a(0,a(1,0))\equal{}a(0,2)\equal{}3$
$ a(1,2)\equal{}a(0,a(1,1))\equal{}a(0,3)\equal{}4$
notice a pattern??
In general $ a(1,k)\equal{}2\plus{}k$ (not hard to prove, use induction).

$ a(2,1)\equal{}a(1,a(2,0))\equal{}a(1,a(1,1))\equal{}a(1,3)\equal{}5$
$ a(2,2)\equal{}a(1,a(2,1))\equal{}a(1,5)\equal{}7$
In general, $ a(2,k)\equal{}a(1,a(2,k\minus{}1))\equal{}a(1,2k\plus{}1)\equal{}2k\plus{}3$ (induction again).

$ a(2,3)\equal{}9$[/hide]

yes, sometimes you can find shortcuts like mathcrazed did to check your work quickly
MATHCOUNTS will rarely give you any thing that you have to brute force *especially on sprint* with more than ~20 terms
if it seems like you do, there is probably a shortcut and thats where you post here

P.S. paste my sig onto yours...


(Okay, Simon, I'm obliging.)

I don't know if this has anything to do with a math contest, but in class we learned the formula(?) to find the projection of u onto v.  (I imagine it could be useful to find distances given endpoints or something.)  This is:

If u and v are nonzero vectors, the projection of u onto v is
proj_v u = [(u . v)/|v|^2]v.

(I'm not familiar with symbols, so the . is the dot product.  For anyone who doesn't know |v| is magnitude and not absolute value.)  Can anyone help me find the proof of this?  (I hate memorizing things without understanding where they come from.)  Thanx.

I will continue to use your notation that u . v is the dot product. Also, |x| represents the length of vector x. Well, draw a triangle with sides of vectors u and v. The included angle between the two sides we many name :theta:, as the angle between them. Now, we know that u . v=|u||v|cos:theta:, so we drop a perpendicular from the tip of vector u onto vector v, and see that |u|cos:theta:=|projv u|. Thus, u . v =|v||projv u|. Now, when we divide by |v|, we get the length that we want, but we don't have the direction, so we multiply by a unit vector in the direction of v, or v/|v|. So, ((u . v)/|v|:^2:)*v=|projv u|*v/|v|=the vector in direction v with length |projv u|=projv u.


Determine the smallest $ a\in R$ such that for any real numbers $ x,y,z\geq a$ satisfying $ x\plus{}y\plus{}z\equal{}3$, it holds that $ x^3\plus{}y^3\plus{}z^3\geq 3$

Another problem from this year's Polish Mathematical Olympiad (Problem 9)
This time I'll simply give you a link.
http://www.om.edu.pl/zadania/om/om59_1r.pdf
:wink:


Line segment AB is 3 units in length. Find the probability that three randomly selected points on AB will not be within 1 unit of each other. (for example, if the endpoints of AB were (0,0) and (3,0), then a valid arrangement of points would be: (0,0), (1.5,0), and (3,0) )

you mean:

Line segment AB is 3 units in length. Given three random points on this line segment, find the probability that each point will be at least 1 unit away from the other points.


Can someone help me with this, It does look quite easy but im struggling.

1/2 <= 4(2-3x)/5  <= 2/3

Im new to these boards so im not sure if this in the right place or anything, but any advice and help is much appreciated.  :)

Well I think the important thing to know is that whatever you do on one side of the inequality, you must do on the other... The rest is just simple algebra!

No, because remember that the middle expression is already over 5, so it wouldn't be 20(2-3x), but rather 4(2-3x), then the rest comes!

Thats what I thought....  :)

So, it would be.

5/2  4(2-3x)  10/3

Take the left side first.
5/2 <= 8-12x
12x<= 10.5
[b]x    <= 10.5/12  which is the same as err...7/8[/b]

Then the right side.
8-12x <= 10/3
11.33 <= 12x
[b]11.33/12 <= x[/b]

[b]11.33/12 <= x <= 10.5/12[/b] Is this right then....or are the numbers the other way around?

[quote="LynnelleYe"]You occasionally add when you ought to subtract, and if you do something to two of the expressions you should do it to the third as well. For example if you subtract 5/2 from the left side you have to subtract it from the right side too. That should be easily correctible though (is that a word?) so you'll get the right answer soon enough...[/quote]

Im quite unsure what you mean, sorry its me being very stupid.

so for example,

5/2 <=8-12x
12x<= 10.5

Or is it,
5/2<=8-12x
12x >= 6.5x
[b]x >= 6.5/12[/b]

And with the right side.

8-12x <=10/3
4.66  >= 12x
[b]4.66/12  >= x [/b]

Leaving,      [b]   6.5/12 <= x <=  4.66/12 [/b] ?

Im very unsure about the 6.5/12.

[quote="LynnelleYe"]You have to work on all three parts of the inequality at once. For example, what you have right now is
$\frac{5}{2}\le 8-12x\le \frac{10}{3}$
If you plan to add 12x, you have to do it to all three sides, not just left and center. I don't think that's a very good idea anyways, because then you have a variable term in two of the parts instead of one. It would be better to get the constant out of the center by subtracting 8 from [i]all three parts[/i] and then dividing by -12. (Oh yeah and don't forget to reverse the signs when multiplying/dividing by a negative number. I used to have serious trouble with that.)
Did I explain that clearly enough?[/quote]

Thanks for your help so far.

Would it be something like this.

$\frac{-11}{2}\le -12x\le \frac{-14}{3}$

and then, divide by 12: $\frac{-45}{100}\le -x\le \frac{-38}{100}$

reverse the signs.

[b]0 .45  >= x >= 0.38[/b] ??

I could leave it as the original fractions though.

[quote="LynnelleYe"]Almost...you did something weird when you divided by 12, because -11/2 divided by 12 should be -11/24 and -14/3 divided by 12 should be -14/36=-7/18, but you've got the concept down.[/quote]

Yea ive been using my calculator to help me!

Thanks for the help and that, im not usually this rusty at maths. Im very glad of your help  :)

[quote="yeoviltown_fc"]Oopps I need to check just one more problem.

16-9x/2 > 1-6x/3

times by 2:  16-9x > 2-12x/3

times by 3: 48 - 27x > 2-12 x

gives me: 46 > 15x
[b]
46/15 > x[/b] ??[/quote]

The first time, when you multiplied by 2, you forgot to multiply the 16 by 2 as well.  And it would probably be easier to multiply by 6 instead of 3 and 2...it gives you twice as many places to make arithmetic mistakes otherwise.

So try that problem over again, remembering to multiply everything.


Ten people are sitting around a round table.  Three of them are chosen at random to give a presentation.  What is the probability that the three chosen people were sitting in consecutive seats?

$ \dbinom{10}{3}$ is a combination. This stands for $ \frac{10!}{3!(10\minus{}3)!}$.

Similarily, in general, $ \dbinom{n}{k}\equal{}\frac{n!}{k!(n\minus{}k)!}$, and represents the number of ways to choose $ k$ objects from $ n$ objects, where order doesn't matter (so ABC=BAC=CBA=...).

There are many cool things you can do with combinations. For one, $ \dbinom{n}{k}\equal{}\dbinom{n}{n\minus{}k}$. You can prove this by just expanding them both out.

Some other identities are $ \binom{n\minus{}1}{k}\plus{}\dbinom{n\minus{}1}{k\minus{}1}\equal{}\dbinom{n}{k}$, which is known as Pascal's identity.

$ \dbinom{n}{0}\plus{}\dbinom{n}{1}\plus{}\hdots\plus{}\dbinom{n}{n}\equal{}2^n$

For more information see Introduction to Counting & Probability by David Patrick.

[quote="louiethechewy123"]I wonder why you write it like that.[/quote]

This question can be expanded for everything. For example, $ 5\plus{}3$. Why do we write it like that?

Since no one gave a solution yet, here it is...

[hide="Solution"]We can think of all the three people, A, B, and C, as one person, ABC, to be picked, and then just find the number of ways to arrange the the 8 people (7 people and ABC) over the number of ways to arrange 10 people.

[hide="7 people and ABC"]In a circle with 8 people, there are $7!$ ways to arrange them.  However, ABC can be CBA, ACB, etc., so there are $3!=6$ arrangements for ABC, thus a total of $7!*6$ ways to arrange the 8.[/hide]

[hide="10 people"]In a circle with 10 people, there are $9!$ ways to arrange them.[/hide]

[hide="Final Answer"]Therefore, the answer is $\frac{7!*6}{9!}=\frac{6}{8*9}=\boxed{\frac{1}{12}}$
[/hide]
This is the correct answer, but is anyone else surprised the probability is (relatively) so high?[/hide]

[hide=sol]Let the first person be fixed. Now there are 9 choose 2=36 ways to choose 2 more people, only 3 (left and right of 1st person, both left or both right of 1st person) will have 3 consecutive people seating, so the probability is $\frac{3}{36}=\frac{1}{12}$[/hide]

Sorry for the revive but that third post was really helpful :blush:


for the first line i think you mean

$\sqrt{5+\sqrt{2}}=a+b\sqrt{2}$ then it follows that..
$5+\sqrt{2}=a^{2}+b^{2}+2ab\sqrt{2}$

[quote="mathgeniuse^ln(x)"][hide]And $\frac{5+2\sqrt{6}}{2}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}=\frac{sqrt{6}}{2}+1$[/hide][/quote]
uh...how is that true?

i think he meant

$\left(\sqrt{3}+\sqrt{2}\right)^{2}$

I thank you all for your help and suggestions.

[quote="Quevvy"][quote="bpms"][quote="Interval"]What is the easiest way or method to solve questions involving square roots within square roots?

Sample:

Find sqrt[5 + sqrt(2)] without using a calculator.[/quote]
[hide]
$5+\sqrt{2}=a+b\sqrt{2}$
$5+\sqrt{2}=a^{2}+2b^{2}+2ab\sqrt{2}$
$a^{2}+2b^{2}=5$
$ab=\frac{1}{2}$
$a^{2}+\frac{1}{4a^{2}}=5$
$a^{2}=x$
$4x^{2}+1=20x$
$4x^{2}-20x+1=0$
$x=\frac{20\pm\sqrt{384}}{8}$
This clearly isn't simpler, so this one can't be simplified.[/hide]
I think most of your questions belong in Classroom Math.[/quote]
Actually, when you get to this: $x=\frac{20\pm\sqrt{384}}{8}$, 384's prime factorization is $2^{7}\cdot3$ so then $\sqrt{384}=8\sqrt{6}$
Then: $x=\frac{20\pm\sqrt{384}}{8}=\frac{20\pm8\sqrt{6}}{8}=\frac{5\pm2\sqrt{6}}{2}$[/quote]
Yes, but $x=a^{2}$ so you'd have to square root that one as well.

All the great replies and help is greatly appreciated.

All answers all wrong or right?

Is there is more than one right answer to this question?

[quote="Ignite168"]Square roots can never be negative, so almost all your answers were wrong.[/quote]
Then what is your answer?

by the way, the easiest way to do this problem is with a calculator...  :P

[quote="SM4RT"]k bpms' solution has a tiny error and its not complete we all just stopped at $x$ when we want to find what $\sqrt{5+\sqrt{2}}=$
[hide]
$\sqrt{5+\sqrt{2}}=a+b\sqrt{2}$
$5+\sqrt{2}=a^{2}+2b^{2}+2ab\sqrt{2}$
$a^{2}+2b^{2}=5$
$ab=\frac{1}{2}$
$a^{2}+\frac{1}{2a^{2}}=5$
$a^{2}=x$
$2x^{2}+1=10x$
$2x^{2}-10x+1=0$
$x=\frac{10\pm\sqrt{92}}{4}=\frac{5\pm\sqrt{23}}{2}$
$a=\sqrt{\frac{5\pm\sqrt{23}}{2}}$
$b=\frac{1}{2a}=\frac{1}{2\sqrt{\frac{5\pm\sqrt{23}}{2}}}$

so $\sqrt{5+\sqrt{2}}=\sqrt{\frac{5\pm\sqrt{23}}{2}}+\frac{\sqrt{2}}{2\sqrt{\frac{5\pm\sqrt{23}}{2}}}$[/hide]

You can simplify it if you wnat i'm not bothering..lol..[/quote]

I'm not really sure that is any simpler than what is given in the problem...

Also, why the $\pm$?  $\sqrt{5+\sqrt2}$ can only be one thing...

k ur right on the not simpler part..however the $\pm$ is actually correct..if you check it u see that both solutions give the same value..

C'mon man, learn to use $\LaTeX$, it won't kill your time.

Now, if we've got por example $\sqrt{a+\sqrt b}$, then:
\[\sqrt{a+\sqrt b}=\sqrt{\frac{a+m}{2}}+\sqrt{\frac{a-m}{2}}\]
Where $m=\sqrt{a^{2}-b}$

Hence:
\[\sqrt{5+\sqrt 2 }= \frac{{\sqrt{10+2\sqrt{23}}+\sqrt{10-2\sqrt{23}}}}{2}\]
And the deal is off...

Do you mean
\[\sqrt{a+\sqrt b}=\sqrt{\frac{a+\sqrt{m}}{2}}+\sqrt{\frac{a-\sqrt{m}}{2}}\]
Where $m=a^{2}-b$

I forgot something, must be $m=\sqrt{a^{2}-b}$ instead $m=a^{2}-b$


function
real analysis
real analysis unsolved

for every measurable function $f$ there exists borel function $g$ which differ from $f$ in just nullset.

Let $(q_{n})_{n}$ be an enumeration of the rationals.

Assume $f^{-1}((q_{n},\infty))$ is $B_{n}\cup T_{n}$, with $B_{n}$ Borel and $T_{n}$ of measure zero (every measurable set can be decomposed like this). Furthermore, we can assume that $T=\bigcup T_{n}$ is disjoint from $B=\bigcup B_{n}$. Now keep $f$ as it is on $B$, and make it map $T$ onto $0$. The modified function $g$ is what we are looking for.


function
trigonometry
algebra proposed
algebra

I have function -see attachment. Does anobody know what to do?

[/hide][/code]

Define sequence.you can see a problem from shostlist IMO 2003

"Behavior" is quite a broad term.  What do you want to know?

This is the hyperbolic cosine, also known as a catenary.

http://en.wikipedia.org/wiki/Hyperbolic_function
http://en.wikipedia.org/wiki/Catenary


Let $ a,b,c$ be sides of a triangle, and $ m_a,m_b,m_c$ the corresponding medians, and $ R$ the circumradius. Prove that

$ \frac{a^2\plus{}b^2}{m_c}\plus{}\frac{b^2\plus{}c^2}{m_a}\plus{}\frac{c^2\plus{}a^2}{m_b}\leq 12R$.


i'm curious to know how play videogames :
witch is your favorite game console?

X-Box.

Might be because it's the only one I have.  ;)

What about SNES, it is great! Or the original "Sega Console"!

In general there are many options missing, but may be it is because I used to play a lot of videogames 10 years ago!

PSP by far. I dont own one but my friend does and i've used it before. The graphics are AMAZING and you can watch movies on it and listen to music. Its definently the best.

i don't have a PSP, so i'll have to go with X-Box, which is the only system i have besides n64. Actually, n64 isn't that bad, i play it if i get bored with halo (extremely rare :D)

Computer because I can download emulators and there are PC games on their own. ;) Full access of the keyboard rocks. ANd my parentsare actually willing to buy it.

i think that we all forget the first console with 128Bits Dreamcat, it was a good console with nice games like Phatasy star online or the famous
Shenmou I & II.I don't know exactly whey it's finish :? but now i have PS2 its .......  :first:

ps2 rocks, so many games, so many new cool stuffs coming out, like the psp,and i heard the ps3 will be out soon.  but x-box got way better graphics(but the best graphics are still on PC)

I like the PS2 solely because of the FF series and GT series... umm i usually dont mind the console that much because i'd rather be playin a good game... like mario and zelda for nintendo... the legend of zelda looks sweet... cant wait :D

[quote="MathFiend"][quote="djimenez"][quote="MathFiend"]what about the good old super nintendo? i love that one. we also have an N64. that works too, but i love the super nintendo. it's awesome. super mario is great, as are donkey kong 1-3, and i like the zelda for that one also. :)[/quote]

Just in case, just before you I talked about the SNES, it stands for Super Nintendo Entretainment System!  :P[/quote]

oops :blush: ... i wondered what that meant but decided i wasn't going to look dumb and ask. look where that got me  :lol:[/quote]
It is never dumb to ask, you weren't born learned, and one of the most useful techniques to acquire knowledge is asking to the one who already has such knowledge  :D .

Em-yoo-lay-turs.

There are actually legal ones out there, so don't start screaming at me saying "OMG THEYRE ILLEGAL U DIE NAO".

I'd probably get a PSP after the price goes down. I'm pretty sure it has an mp3 feature, and I like listening to the mp3s I remix.

GameCube, mainly because that's the main system I own (I also have an N64 and a very outdated GBC).

I'm not used to the XBox's controller (and I rarely go anywhere that I can play it) and I have never tried any Playstation system.

The only game I've played I've played is Super Mario 64 DS unless you count the Metroid Prime Hunters Demo. (...I spent a LONG time replacing all of the high scores in the Morph Ball Training Mode - now they're all 72,250, which I believe is the maximum.) I might have put DS if there were more/better games for it.


Here goes an easy one...

The product of all odd integers less than $10000$ is...

(Factorial form...please)

[hide]
The product of all the integers 1-10000 is $10000!$.  We want to divide out the even integers 1-10000, which is $(2^{5000})5000!$, so the product of the odd integers 1-10000 is $\frac{10000!}{(2^{5000})5000!}$
[/hide]


Okay this one is also simple! :D :D

look at this: sum(2ab/(b^2+ca)) =sum(2/(b/a+c/b))<=sum(1/2(a/b+b/c))=a/b+b/c+c/a

cheers! :D :D


Solve for integers $a,b,c$
\[ (a+b+c)^3 + \frac{1}{2} (b+c)(c+a)(a+b) = 1 - abc  \]

I think it's reall well-known:the given equation is equivalent to:
$(b+c+2a)(c+a+2b)(a+b+2c)=2$.The rest is trivial!

I'm sorry that I'm not aware. Can you plz explain?!

See
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=438681&highlight=#438681[/url]

 :)

basic expansions before seeing the proof

$(x+a)(x+b)(x+c)=x^3 + (a+b+c)x^2 +(ab+bc+ca)x +abc$
$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$

let's assume that $a+b+c=x$

then the given expression can be converted as

$2x^3+ (x-a)(x-b)(x-c)=2-2abc$
$2x^3+x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=2-2abc$

now since $a+b+c=x$

therefore
$2x^3+x^3-x^3+(ab+bc+ca)x-abc=2-2abc$
$2x^3+(ab+bc+ca)x+abc=2$
$x^3+x^3+(ab+bc+ca)x+abc=2$
now one of $x^3$ can be written as $x^2(x)$
that is also equal to $x^2(a+b+c)$
therefore the expression becomes
$x^3+x^2(a+b+c)+(ab+bc+ca)x+abc=2$
and thus
$(x+a)(x+b)(x+c)=2$
$(2a+b+c)(a+2b+c)(a+b+2c)=2$
now the expression becomes easy to analyze


There is a 3*3 matrix of Pigeon Holes such that each Pigeon Hole can accomodate at the most one pigeon only.You are having 4pigeons,in how many ways can u place the 4 pigeons into this matrix such that there is atleast one pigeon in each of the rows and in each of the coloumns.

Edit: Assume that the pigeons are identical by nature

There are $ 3!$ ways to put three pigeons such that there is one in every row and column. The remaining pigeon can go in 6 different spots, for a total of $ \boxed{36}$ ways.

While taking 3! ways of putting pigeons such that there is atleast one in each row and in each coloumn,you have counted the cases in which one is placing a pigeon in any particular row and the rest of the two are placed in the other two rows but in different coloumns only.There will be some extra cases which wont be obtained using this method but still they will be feasible.The answer is more than 36

Can someone verify the correct answer please?

My point is that in any placement of 4 pigeons which satisfies the problem, exactly 3 of them are such that they satisfy the condition of one being in each row and column.

Varun Nakra is right: you missed some possibilities.  (In other words, the claim in your last post is false.)  We have 9 additional cases: take any square and color the four squares in the same row or column.

We can consider the row and column thing seperately when counting. If you have 4 numbers 1 to 3 inclusive, 1,2,3 covered, then there are 3 distinct ways of doing that. Now we need to match the rows and columns...how many ways are there to match 1,1,2,3 and A,A,B,C? Well lets fix the first set, then permute the other set. We can have AABC, AACB, ABAC, ABCA, ACAB, ACBA, BCAA which is $ 7$ ways [note 11-AB is the same as 11-BA, but 23-AB is not the same as 23-BA]. In total we have $ 3*3*7\equal{}\boxed{63}$ ways.

Altheman, I can't exactly figure out what you're doing, but your 7 is wrong.  We have to pair elements from $ \{1, 1, 2, 3\}$ with elements from $ \{A, A, B, C\}$, and there are only 5 ways to do this:
$ 1B, 1C, 2A, 3A$
$ 1A, 1B, 2A, 3C$
$ 1A, 1C, 2A, 3B$
$ 1A, 1B, 2C, 3A$
$ 1A, 1C, 2B, 3A$

The first case corresponds to the 9 possibilities ubemaya missed.

Edit: okay, I did figure out what you did wrong.  In the cases "AABC" and "AACB," you've double-selected the square 1A.


projective geometry
geometry proposed
geometry

Let $ABC$ a triangle and let $C_{1}$ a circle in its interior which has no common points with the sides of the triangle .
The tangents from $A$ to $C_{1}$ meet $BC$ at $D,\ E$ ($D$ is near to $B$).
Similarly we define $K,\ L$ on $AC$ ($K$ is near to$C$ ) and
$M,\ N$ on $AB$ ($M$ is near to $A$) .
The lines $KE$ ,$ML$ , $DN$ form a triangle $A_{1}B_{1}C_{1}$ .
Prove that $A_{1}B_{1}C_{1}$ and $ABC$ are perspective.

[quote="fotis"] [color=blue]A question to the Greek MLnkers [/color]
 Μια απορία για τους έλληνες εδώ....

Οταν χρησιμοποιούμε το θεώρημα Desargue λέε πχ ότι το το εφαρμόζουμε διότι τα τρίγωνα "τάδε και τάδε" είναι "perspective"....Σχηματικά ξέρω πότε δύο τρίγωνα είναι perspective για να χρησιμοποιήσω το θεώρημα αλλά δυστυχώς δεν ξέρω ούτε τον ελληνικό όρο ούτε τον ακριβή ορισμό.....Θα μπορούσε κάποιος να μου τα πει;Ευχαριστώ! :D[/quote]

 [b]The greek word for '' perspective triangles  '' is '' προοπτικά τρίγωνα ''.[/b]

It is a good opportunity for a solution without metrical relations.

By the tangents of the circle $(O_{1},$ through vertices $A,$ $B,$ $C$ of $\bigtriangleup ABC,$ it is formed a hexagon circumscribed to $(O_{1}),$ considering instead of $(C_{1}).$

We denote the vertices of that hexagon, as follows:

$A'\equiv BL\cap CM$ and $A''\equiv BK\cap CN.$

$B'\equiv AD\cap CN$ and $B''\equiv AE\cap CM.$

$C'\equiv AE\cap BK$ and $C''\equiv AD\cap BL.$

$\bullet$  As well known, by Brianchon’s theorem, we have that the diagonals $A'A'',$ $B'B'',$ $C'C'',$ of the above hexagon $A'B''C'A''B'C'',$ are concurrent at one point, so be it $P.$

We conclude now, by Desarques’s theorem, that the triangles $\bigtriangleup A'B'C''$ and $\bigtriangleup A''B''C',$ are perspective and so, the points $A\equiv B'C''\cap B''C',$ $B\equiv A'C''\cap A''C'$ and $S\equiv A'B'\cap A''B'',$ are collinear.

That is the point $S,$ as the intersection point of $A'B',$ $A''B'',$ lies on the sideline $AB$ of $\bigtriangleup ABC.$

Similarly, we can prove that the point $Q\equiv B'C'\cap B''C'',$ lies on the sideline $BC$ and the point $R\equiv A'C'\cap A''C'',$ lies on the sideline $AC.$

Because of $A'A''\cap B'B''\cap C'C''\equiv P,$ we conclude also that the triangles $\bigtriangleup A'B'C',$ and $\bigtriangleup A''B''C'',$ are perspective and so, we have that the points $Q,$ $R,$ $S,$ are collinear.

Because now, of the collinearity of $Q,$ $R,$ $S,$ we conclude that the triangles $\bigtriangleup ABC,$ and $\bigtriangleup A'B'C',$ are perspective and hence the lines $AA',$ $BB',$ $CC',$ are concurrent at one point, so be it, $T.$

$\bullet$  We denote as $D',$ $K',$ $M',$ the intersection points of $BC,$ $AC,$ $AB$ respectively, from the segment lines $AT,$ $BT,$ $CT.$

Also we denote the points $D''\equiv BC\cap ML,$ $K''\equiv AC\cap DN$ and $M''\equiv AB\cap EK.$

From the complete quadrilateral $AMA'LBC,$ we have that the point $D'',$ is harmonic conjugate of $D',$ with respect to $B,$ $C.$ Similarly for the points $K''$ $($ harmonic conjugate of $K',$ wrt $A,$ $C$ $)$ and $M''$ $($ harmonic conjugate of $M',$ wrt $A,$ $B$ $).$

Because of $AD'\cap BK'\cap CM'\equiv T,$ we conclude that the points $D'',$ $K'',$ $M''$ are collinear, as well known and easy to prove $($ by Ceva’s and Menelaus’s theorems $).$

Because of the collinearity of the points $D'',$ $K'',$ $M'',$ by Desaques’s theorem, we conclude that the triangles $\bigtriangleup ABC,$ $\bigtriangleup A_{1}B_{1}C_{1},$ are perspective and the proof is completed.

$($ We have been denoted as $A_{1}\equiv DN\cap EK,$ $B_{1}\equiv EK\cap LM$ and $C_{1}\equiv DN\cap LM$ $).$

Kostas Vittas.

Because of the segment lines $AA',$ $BB',$ $CC',$ are concurrent at one point, here the point $T,$ it is easy to prove that the segment lines $AA'',$ $BB'',$ $CC'',$ are also concurrent at one point, so be it $V$ and that the points $V,$ $P$ $($ concurrency point of $A'A'',$ $B'B'',$ $C'C''$ $),$ $T,$ are collinear.

In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=108652 , there is a reference about a theorem of ''concurrent'' segments, based on which, may be proved the above results.

Kostas Vittas.


ABCD tetraedru echifacial. Notam $A_{b}, A_{c}$ punctele de intersectie ale muchiei $[DA]$ cu bisectoarele unghiurilor $\angle DBA$ si respectiv $\angle DCA$, cu $B_{a}si B_{c}$ punctele de intersectie ale muchiei $[DB]$ cu bisectoarele unghiurilor $\angle DAB$ si respectiv $\angle DCB$, iar cu $C_{a}si C_{b}$ punctele de intersectie ale muchiei $[DC]$ cu bisectoarele unghiurilor $\angle DAC$, respectiv $\angle DBC$. Sa se arate ca $A_{b}B_{a}, B_{c}C_{b}, C_{a}A_{c}$ sunt paralele cu acelasi plan.

Cred ca am gasit rezolvarea.

   $ABCD$ tetraedru echifacial, deci sa notam pe $AB=CD=a, BC=AD=b, AC=BD=c$.
 Din teorema bisectoarei, avem: $\frac{DB_{a}}{B_{a}B}=\frac{b}{a}$ etc.
 Apoi obtinem: $\frac{DB_{a}}{c}=\frac{b}{a+b}, \frac{A_{b}D}{b}=\frac{c}{a+c}$ etc., si in final:  $DB_{a}=\frac{bc}{a+b},DA_{b}=\frac{bc}{a+c},DA_{c}=\frac{ab}{a+c},DC_{a}=\frac{ab}{b+c},DB_{c}=\frac{ac}{a+b},DC_{b}=\frac{ac}{b+c}$
   Fie $M, N, P$ puncte astfel incat $M\in [BD], N \in [CD], P \in [AD],$ iar $DM=x, DN=y, DP=z.$ Fixam $z (0<z<b)$ si luam $x, y$ astfel incat
$x=\frac{a+c}{a+b}z; y=\frac{a+c}{b+c}z.$
    Atunci $\frac{DB_{a}}{DM}=\frac{DA_{b}}{DP}\leftrightarrow \frac{bc}{(a+b)x}=\frac{bc}{(a+c)z}\leftrightarrow (a+b)x=(a+c)z$, ceea ce este adevarat, deci $\frac{DB_{a}}{DM}=\frac{DA_{b}}{DP}\rightarrow A_{b}B_{a}\parallel (MNP).$ Analog pentru celelalte. In final, dreptele sunt paralele cu $(MNP)$.


Hello, could you help me with some nice geometry problems with Simson line and their solutions? Thank you  :lol:

Let quadrilateral $ ABCD$ inscribes the circle $ (O). M$ lies inside $ (O). X,Y,Z,T,U,V$ are the projections of M on $ AB,BC,CD,DA,AC,BD. N,P,Q$ are the midpoint of $ UV,XZ,YT$. Prove that $ N,P,Q$ are collinear


inequalities
geometry
incenter
circumcircle
trigonometry
geometry proposed

Let I and O be the incenter and the circumcenter of a triangle ABC, respectively. Assume that A > 60°.

Prove $R\left(1+2\cos A\right) > 2\cdot IA\cdot\cos\frac{B-C}{2}$.

[color=red][Moderator edit: Rewrote the problem text.][/color]

I think there is a mistake: for example, $A=120^{\circ}\longrightarrow$ false.

Must you verify again your inequality !. I note the $D\in AI\cap BC$. Then,

$\frac{IA}{ID}=\frac{b+c}{a},\ \cos \frac{B-C}{2}=\frac{r}{ID}\Longrightarrow 2\cdot IA\cdot \cos \frac{B-C}{2}=\frac{2(b+c)r}{a}$.

$A\ge60$ $\Longleftrightarrow$ $\cos A\le \frac 12$ $\Longleftrightarrow$ $b^2+c^2\le a^2+bc$ $\Longleftrightarrow$ $4s(s-a)\le 3bc$.


The fact is, I've never had any real training for math olympiads of this sort. I can go on my intuition to some extent, but when we get into areas where there are certain theoroms and such that are particularly useful, I just don't know them.

I'm particularly bad on inequalities, so I think I'll concentrate on those first. Does anybody have any good suggestions for resources (websites, books, whatever)? I know there are plenty of books about this type of problem solving in general (Amazon.com has a number listed by, e.g., Titu Andreescu) I don't know if those will have enough explanation (more than just "we now use such-and-such inequality to simplify to..." when I haven't a clue what the inequality in question is) for me to learn anything from them.

Any help would be greatly appreciated.

Thanks,

Jonathan

first and foremost go and download kiran kedlaya's short text on inequalities and synthetic geometry from that usa olympiad website (AMC website) :D
and secondly if u can possible afford tutors of the calibre of Harazi or Namdung , u are in for something great :D  :D (jokin)

The AOPS II book should help a little, or maybe that's too simple. Just look at the solutions on past olympiad tests.

seriously stop reviving really old threads. you will get banned if you keep doing this.


Let $ a,b,c$ be positive real numbers such that $ a \le b \le c$ and $ abc\equal{}1$. Prove that
\[ a\plus{}b^2\plus{}c^3 \ge \frac{1}{a}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^3}\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \le b \le c$ and $ abc \equal{} 1$. Prove that
\[ a \plus{} b^2 \plus{} c^3 \ge \frac {1}{a} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^3}
\]
:)[/quote]
I have already posted this inequality but It still without a proof:

See here:

http://www.mathlinks.ro/viewtopic.php?t=201274

If $ a \equal{} b \equal{} c \equal{} 1$, we have equality. Otherwise we must have either $ a\le1\le b\le c$ or $ a\le b\le1\le c$.

[u]Case 1[/u]: $ a\le1\le b\le c$

Putting $ a \equal{} \frac {1}{bc}$ we have

$ \color{white}.\hspace{7mm}.$ $ a \plus{} b^2 \plus{} c^3\ \ge\ \frac {1}{a} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^3}$

$ \iff\ \frac {1}{bc} \plus{} b^2 \plus{} c^3\ \ge\ bc \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^3}$

$ \iff\ bc^2 \plus{} b^4c^3 \plus{} b^2c^6\ \ge\ b^3c^4 \plus{} c^3 \plus{} b^2$

$ \iff\ b(c^2 \minus{} b) \plus{} (b^4 \minus{} 1)c^3 \plus{} b^2c^4(c^2 \minus{} b)\ \ge\ 0$

[u]Case 2[/u]: $ a\le b\le1\le c$

Putting $ c \equal{} \frac {1}{ab}$ the inequality becomes $ 1 \minus{} a^6b^6\ \ge\ a^2b^3(1 \minus{} a^2) \plus{} a^3b(1 \minus{} b^4)$ – which is the problem I posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219697 – and it appears a solution has been found already. :lol:

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \le b \le c$ and $ abc \equal{} 1$. Prove that
\[ a \plus{} b^2 \plus{} c^3 \ge \frac {1}{a} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^3}
\]
:)[/quote]

It is a nice one,but easy. :)
AM-GM is worked here.

$ c\geq1$ so it suffices to prove $ a \plus{} b^2 \plus{} c^2 \geq \frac {1}{a} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$
or $ (b^2 \plus{} c^2) \minus{} (\frac {1}{b^2} \plus{} \frac {1}{c^2})\geq\frac {1}{a} \minus{} a$
 or $ (b^2 \plus{} c^2)(1 \minus{} \frac {1}{b^2c^2})\geq \frac {1 \minus{} a^2}{a} \equal{} (1 \minus{} a^2)bc \equal{} (1 \minus{} \frac {1}{b^2c^2})bc$
or $ b^2 \plus{} c^2\geq\ bc$ which is clear.

Can the author post his solution?

to [b]shfdfzhjj[/b]: How do you use AM-GM?

[quote=arqady]Let $0<a\leq b\leq c\leq d$ such that $abcd=1$. Prove that:
\[ a+b^2+c^3+d^4 \ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}\][/quote]

[img]http://s1.sinaimg.cn/middle/0018zOAxgy70jPyo24we0&690[/img]


logarithms
number theory proposed
number theory

Show that the number $ 3^{4^{5}} \plus{} 4^{5^{6}}$ is a product of 2 intengers, each of which is larger than $ 10^{2002}$

The number is divisible by 5 so it is a composite but has a factor $ \leq 10^{2002}$.

[quote="zweig"]But you have to prove that [b] each [/b] its factor is [b]bigger[/b] than $ 10^{2002}$[/quote]
I can't prove it since that is not true! I gave a counterexample.

Yeah sorry, I did a bad translation. I rewrited the exercise.


Prove that $f(x)=sin^{2}(x+a)+sin^{2}(x+b)-2cos(a-b)sin(x+a)sin(x+b)$ does not depend on x, without using the fact that $f '(x)=0$

Transform

\begin{eqnarray*}\sin^{2}(x+a)+\sin^{2}(x+b)={1-\cos 2(x+a)+1-\cos 2(x+b)\over 2}=1-\cos(2x+a+b)\cos(a-b)\end{eqnarray*}

and

$\sin(x+a)\sin(x+b)={1\over 2}(\cos(a-b)-\cos(2x+a+b))$.

The expression reduces to $\sin^{2}(a-b)$


A printer is used to print out the squares of the first 15 positive
integers. How many digits are printed by the printer?

yes 3 is for 1 digit perfect squares
24 is for 2 digit
18 is for 3 digit
sum 33


Aww, I feel the same way.  Would like to know what the answer to this is too.

I think the cat may remember your scent, I know it'll remember smells on inanimate objects you haven't washed

well lets put it this way- dogs/cats/other animals remember being beaten for the REST OF THEIR LIVES. like dogs that were beaten by men will be scared of men forever. so i would think this would also apply for someone who takes care of them and loves them. they will pretty much remember forever. animals arent stupid.

A goldfish's memory span is about 3 seconds.

A goldfish's memory span is 3 seconds? That's hard to believe. My goldfish seems to remember me...   :)            jk

As far as I've heard, that fact is probably the most commonly stated that is completely wrong. I've heard of people doing tests that have them remembering things from over a week earlier.

never heard the 3 seconds fact, but.. wont' surprise me

my goldfish says blub every 3 seconds and really no use to reply him "You just said that!" :maybe:

What if someone got turned into a goldfish? Would they be in a ton of trouble, because they wouldn't remember that they used to be a person and just think they're a goldfish?


If we count by $ 3$s starting with $ 1$, the following sequence is obtained: $ 1$, $ 4$, $ 7$, $ 10$, $ \dots$ . What is the $ 100$th number in the sequence?

The $ x$th term in the sequence is defined by $ 3x \minus{} 2$. So the $ 100$th term in the sequence is $ 3(100) \minus{} 2 \equal{} \boxed{298}$.


have to find certain topics with special genes. Like the longest topic ever.
The newest topic ever.
The oldest topic ever.
The shortest topic ever.
etc.
I would like a few people to sign up.

Just sort them in different ways and you'll get these. And what do you mean by newest topic ever?

Forget about the newest.
I'm making this a free-for-all

We are now finding the LARGEST TOPIC

Do I hear the never ending story talking?

the longest posted on in GGF is[color=green] Never Ending Story[/color] with 2,839 and most views is [color=green]notpron[/color] with a whopping 105,141 views! It beats the next topic by more than 50,000 views!

OOH OOH HERE'S A GOOD TOPIC

[hide="Click to reveal AWESOME topic idea"]3[/hide]


Give your real name or fragments of it. Even your pet name will do. You may add your age.

My name is Ritam.

My nickname is the notes I write is "Moon." lol.

I'm [b]"The"[/b] Nilesh Rao Pawar.

My nickname is Sunny

daesun yim

it's like an invitation for stalkers at a paranoid party

Hi, I'm Chuck Norris. Nice to meet you guys :)