rectangle
inequalities
matrix
counting
combinatorics
graph theory
IMO Shortlist

Consider a $m\times n$ rectangular board consisting of $mn$ unit squares. Two of its unit squares are called [i]adjacent[/i] if they have a common edge, and a [i]path[/i] is a sequence of unit squares in which any two consecutive squares are adjacent. Two parths are called [i]non-intersecting[/i] if they don't share any common squares.

Each unit square of the rectangular board can be colored black or white. We speak of a [i]coloring[/i] of the board if all its $mn$ unit squares are colored.

Let $N$ be the number of colorings of the board such that there exists at least one black path from the left edge of the board to its right edge. Let $M$ be the number of colorings of the board for which there exist at least two non-intersecting black paths from the left edge of the board to its right edge.

Prove that $N^{2}\geq M\cdot 2^{mn}$.

Some extensions:

[i]1. The inequality is strict.[/i]
[i]2. Let $ n,$ the length, be fixed and let $ m$ vary, and put $ p_m \equal{} \tfrac{N(m,n)}{2^{mn}}.$ Then $ p_1 < p_2 < p_3 < p_4 < \cdots$[/i]
[i]3.
\[ \frac {M}{2^{mn}} \leq p_{m} p_{m \minus{} 1} \minus{} \sum_{i < m} p_i (p_i \minus{} p_{i \minus{} 1}) < p_m^2
\]
[/i]

Define a good path to be a path of black squares from the left edge to the right edge of the $m\times n$ grid. Define the set $A$ such that

\[A = \{ (p,q) | p,q \quad \text{are colourings with at least one good path} \}\]

Define the set $B$ such that

\[B = \{ (p,q) | p \quad \text{is a colouring with at least two non-intersecting good paths and} \quad q \quad \text{is any colouring} \}\]

It is clear that $|A| = N^2$ and $|B| = M \cdot 2^{mn}$.

Now we construct an injection $f : B \rightarrow A$ as follows. Let $(r,s)$ be such that $(r,s) \in B$. Since $r$ has two non-intersecting good paths, we can construct two new grid colourings as follows:

[b]1.[/b] Partitions $r$ into two sub-grids $a$ and $b$ such that the bottom square in each column of $a$ is the corresponding square in the uppermost good path in $r$.
[b]2. [/b] Partitions $s$ into two sub-grids $c$ and $d$ such that the shape of $c$ is the same as the shape of $a$ and the shape of $d$ is the same as the shape of $b$.
[b]3.[/b] Switches the positions of $a$ and $c$ to form the two new grids $a \cup d$ and $c \cup b$.

Let $f(r,s)=(a \cup d, c \cup b)$. Now note that $f$ is injective since it can easily be verified that $f(f(r,s))=(r,s)$.

Since there exists an injection $f : B \rightarrow A$, it follows that $|A| \ge |B|$ and that $N^2 \ge M \cdot 2^{mn}$. $\blacksquare$

Can someone check my solution? It feels fine, but there might be holes in it.

Let $\nu$ denote colorings with contain at least one black path connecting the left and right edges of the board.
Note that $N^2\ge M\cdot2^{mn}$ rearranges to $\frac{N}{2^{mn}}\ge \frac{M}{N}$.
We may interpret the left side of the inequality to be the probability that $\nu$ occurs given that each square is equally likely to be black or white. Similarly, the right side is the probability that two non-intersecting paths occurs given a random $\nu$.

[b]Lemma:[/b] Define $n$ of a not necessarily continuous or rectangular board $B$ as the number of columns in the smallest rectangular board that contains $B$. If square(s) are added to $B$ given that $n$ does not change, the probability that $\nu$ occurs does not decrease.

This can be seen by noting that $\nu$ of $B$ are unaffected, but some colorings that contain a path may gain a path.

If we take remove the shortest path in each $\nu$ and splice the top and bottom sections together (i.e. translate the bottom section up a unit), we obtain a $(m-1)\times n$ board that may contain holes. $\nu$ in this new board $B$ corresponds to colorings on the original board that have two colorings with intersections of at most one continuous square and $P(\nu | B)\ge \frac{M}{N}$. According to the lemma, we may add squares to $B$, including a whole row, to obtain the $m \times n$ board to obtain $\frac{N}{2^{mn}}\ge P(\nu | B)\ge \frac{M}{N}$. $\blacksquare$

The following stronger problem is from the solution given in the book The IMO Compendium, which is much easier to solve :P

Consider two rectangular $m \times n $ boards , each of them consists of $mn$ squares.Color all squares of the boards such that  there are $k$ '' fixed squares''  in each board, placed in similar positions, which are always colored using same color and there exists a black path from left to right of both the boards .The two paths can not intersect in any of the fixed $k$ squares. Suppose that the two boards can be colored in total $f(k_p) $ ways. Show that $2^k f(k_p) \le N^2$

Solution: Induction on $k$. The case $k=0$ is trivial. Assume that it is true for a couple of board having $k-1$ fixed squares i.e we have $2^k f({k-1}_q) \le N^2$. We add another fixed point. We have to show that $2^k f(k_p) \le N^2$. It suffices to show that $f({k-1}_q) \le 2 f(k_p)$, which is true because we can always change the color of the newly added ''fixed'' point in (at least)  one of the boards.

Now if we take $k=mn$, we get $f(mn_p) = M $, so $N^2 \ge M \cdot 2^{mn}$

[quote="Boy Soprano II"]We call a path of black squares which connects both sides of the board [i]good[/i].

First, it is necessary to show that if there is a good path, then there exists such a [i]lowest good path[/i] $M'$ such that in each column of the board, the highest square of $M'$ in that column is at least as low as the highest square in any other such path in that column.  This is not hard.

Now, we consider two boards, one with two disjoint good paths, and one colored any way we like.  There are $M \cdot 2^{mn}$ such boards.  Now, we transpose the lowest good path on the first board, and all squares beneath it, with the corresponding squares on the second board.  We thus obtain two boards, each with a good path.  This process is injective, for we can reverse it by transposing the lowest good path on the second board, and all squares beneath it, with the corresponding squares on the first board.  Thus $M \cdot 2^{mn}$ is at most the number of pairs of boards such that each board has a good path, and this latter number is $N^{2}$.  $\blacksquare$[/quote]
Hello,

I think transposing only the lowest path is sufficient to construct an injection, isn't it?

No.  Consider the following 3 by 3 examples:  in the first pair, the first grid is all black except the middle square in the bottom row, while the second grid is all white except for the middle square in the bottom row.  In the second pair, the first grid has black 1st and 3rd rows, while the second grid has black 2nd row.  Both pairs get sent by your proposed map to the pair in which the first grid has first row black and rest white while the second grid has first row white and rest black.

[hide=" Badly phrased solution."]
A coloring is called $k$-good if there are at least $k$ non-intersecting black paths. Consider two $m \times n$ boards. The LHS is the number of ways to color these two boards such that both colorings are 1-good, and the RHS is the number of ways to color these boards in such a way that at least one of them is 2-good.

Consider a 1-good coloring of a $m \times n$ board. Define the [i]height[/i] of a cell to be the distance from the bottom edge to the cell. Call a path [i]better[/i] than another path if each cell of the first path has more height than all cells of the second path in the same column. Consider the path which is better than all other paths. For each cell in the path, consider a grid rectangle with width 1 whose bottommost cell is this cell and uppermost cell is the cell with the maximum height in the same column. If two rectangles overlap (which means there are two cells in the same column), delete the smaller of them. Let $U$ be the union of these rectangles (we get one such rectangle for each column in our path). Consider another 1-good coloring of a $m \times n$ grid $G$. WLOG assume the path which is better than all other paths in $G$ is not better than the path which is better than all other paths in the first grid. Place $U$ on $G$ (replace the cells which were in the positions of $U$). Let $J$ be the set of cells which were replaced by $U.$ Replace the cells in $U$ of the first grid by the cells of $J.$ This gives us two rectangles one having a 2-good coloring.

Since we get different configurations for different 1-good colorings (applying the same operation on this configuration, we get the initial coloring again), it follows that the LHS must always be $\geq$ the RHS. $\blacksquare$ [/hide]

Call a path [i]leeky[/i] if goes from the left side to the right side. Also, denote the sets $A,B,C$ to denote the sets of all colorings, those with at least 1 leeky path, and those with at least 2. We will construct an injection $f: (A,C)\to B^2$, which of course will imply the result.

First, we claim that for all elements of $B$, there exists a highest leeky path, $l_H$, such that the black squares of any other path are below those of $l_H$ in the same column. Supposing the contrary, then for any path $a$, we can construct a path $b$ where at least one black square is above. If $a$ and $b$ are disjoint, then we must have $b$ is strictly above $a$. Otherwise, they intersect, and we can take a path from their union which ends up being above $a$. Hence, we can always find a path above any path. However, we can't continue picking higher and higher paths, so we have a contradiction, and there must be a highest path.

Now, suppose we have a pair of colorings $(P,Q)$, where $P\in A$, $Q\in C$, we can $P$'s highest path and all squares above its lower bound, and swap it with the same region in $Q$. Of course, this mapping is well defined, and brings our pair to an element of $B^2$. Now, to show it's injective, we show that the image of our mapping, $f$, is reversible. Of course, this is not hard, as the highest path of $P$ is now the highest path of $Q'$, so we can just perform the exact same map, except take the highest path of $Q'$ to $P'$. As we have constructed a valid injection, the problem statement is true.

[color=#f00]Could someone please check if this sol is correct as it seems quite different from the others sol which basically all use inj?[/color]

We can rearrange the condition as $N\geq \frac{M}{N}\cdot 2^{mn}$. Notice that $\frac{M}{N}$ is equal to $\frac{P(M)}{P(N)}$, where $P(M)$ denotes that the probability of a coloring having two non-intersecting black paths and $P(N)$ denotes the probability of a coloring having at least one black path. Notice then that the equation rearranges to $\frac{N}{2^{mn}}\geq \frac{P(M)}{P(n)}$. The expression on the left is the same as $P(n)$. Hence the problem reduces to proving that $P(M)\leq P(N)^2$. This is obvious, because obviously the probability of just picking two random boards that satisfy the first requirement does not guarantee that it would satisfy the second(stronger) requirement.

Consider a $2m\times n$ rectangular board.

Let $A$ be the set of colouring such that there exists a path from the left edge of the left $m\times n$ board to the right edge of it, and a path from the right edge of it to the right edge of the right $m\times n$ board.

Let $B$ be the set of colouring such that there exists two non-intersecting black paths from the left edge of the left $m\times n$ board to its right edge.

Obviously $|A|=N^2$ and $|B|=M\cdot 2^{mn}$.

We now create an injection from $B$ to $A$. Indeed, for each colouring in $B$, there are two paths. Since they do not intersect, one of them must be at the top of the other. We reflect that path with respect to the vertical axis of symmetry of the $2m\times n$ board. That is, for each square in the path we exchange it with the corresponding square across the axis

Then the resulting path must be an element of $A$. It is easy to see that this is an injection so we are done.

Consider any pair of boards, one of which is any board and the other is a color with two non-intersecting black paths from the left to the right edge. Since the paths are non-intersecting, if we cut the first board along the top border of the lower path, the upper path is unaffected.
\\$~$\\
Now, make make the same cut on the other board, which can be any board, and switch the lower sections of the paths. This creates two boards, each with at least one black path. Note that following these rules, there is at most one way to revert two boards with at least one black path into a board with at least two black paths. Thus, we are done.

It suffices to show that there is an injection from every pair of colorings represented in $N^2$ to one represented in $M \cdot 2^{mn}$.

Pick two boards, one board one of the $M$ that contain two non-intersecting paths, and another board which is arbitrary. Consider the following transformation, where we take the path through the first board that begins at the highest $y$-coordinate and swap it with the coloring of the corresponding squares in the second grid. This creates two grids, each with at least one crossing. Furthermore, if two pairs of grids mapped to the same grid, then they must have had the same coloring of all the squares that are not part of the topmost path in the first square, and also contained the same topmost path in the first square, thereby rendering the first two squares identical. As the pattern replicated from the second square must also be identical, and the remainders of the second squares are also identical, the other two squares must also be identical; thus, the preimages are also identical as well. Thus, this transformation is an injection onto the $N^2$ boards that have at least one path, so we have $N^2 \geq M \cdot 2^{mn}$, as desired.

Neat problem.

We consider [list][*]the set $S$ of [I]superior[/I] pairs of $m \times n$ boards, in which each board contains at least one path, and [*]the set $I$ of [I]inferior[/I] pairs of $m \times n$ boards, where the first board contains two disjoint paths and the second board is given any coloring.[/list] Note that $|S|=N^2$ and $|I|=M \cdot 2^{mn}$. Thus, it suffices to find an injection that maps $I$ to $S$.

Consider any pair $(I_1, I_2)$ in $I$. Note that $I_1$ contains a lowest path. Cut the part of the $I_1$ containing this lowest path such that no cell above the path is included, and call it $I_{1L}$. The remaining portion of $I_1$ can be called $I_{1U}$. Cut out subgrids from $I_2$ with exactly the same patterns, calling the upper portion $I_{2U}$ and the lower portion $I_{2L}$. Consider the map $f: I \to S$ defined by $f(I_1, I_2)=(I_{1U} \cup I_{2L}, I_{2U} \cup I_{1L})$, whence both elements of $f(I_1, I_2)$ are grids with at least one path. Since $f$ is an injection, we have $|I| \le |S|$, as desired.

injection? [tip=problem]2022 C5[/tip] says hi


function
floor function
combinatorics proposed
combinatorics

[u]Problem.[/u] Prove that for all $ n\in\mathbb{N}\cup\{0\}$,
\[ \sum_{k \equal{} 0}^n\frac {\binom{n}{k}\binom{n}{n \minus{} k}}{2^n} \equal{} \sum_{k \equal{} 0}^{\lfloor\frac {n}{2}\rfloor}\frac {\binom{n}{2k}\binom{2k}{k}}{4^k}.
\]
Enjoy! :)

this looks very familiar, i think it's been discussed here before

well here's my interpretation, cuz i see no one has given an explanation yet.

of course $ \sum_{k \equal{} 0}^n\binom{n}{k}\binom{n}{n \minus{} k} \equal{} \binom{2n}{n}$. also, an easy calculation shows that $ \binom{n}{2k}\binom{2k}{k} \equal{} \binom{n}{k}\binom{n \minus{} k}{k}$, and if we multiply throughout by $ 2^n$, the identity to be proven is

$ \binom{2n}{n} \equal{} \sum_{k \equal{} 0}^{\lfloor \frac {n}{2} \rfloor} 2^{n \minus{} 2k}\binom{n}{k}\binom{n \minus{} k}{k}$.

now consider a set of $ 2n$ objects partitioned into $ n$ pairs. obviously the LHS counts the number of ways to choose $ n$ of the $ 2n$ objects. the RHS counts the same thing, according to how many of the pairs contribute both their members.


Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its perimeter is equal to 1995?

I don't think so, because

$E:$ the area of the $\triangle ABC$
$2s:$ the perimeter of $\triangle ABC$

$E = \sqrt{s(s-a)(s-b)(s-c)}\Rightarrow$
$E^2 = s(s-a)(s-b)(s-c)$

on the other hand, $E = \frac{1}{2}ah_a \Rightarrow E^2 = \frac{1}{4}a^2h^2_a$

$s(s-a)(s-b)(s-c) = \frac{1}{4}a^2h^2_a \Rightarrow$

$4s(s-a)(s-b)(s-c) = a^2h^2_a$

$(s-b)(s-c)$ is not an integer, because it is $(\frac{1995}{2}-b)(\frac{1995}{2}-c)$

$2s=1995$ is an odd integer
$2(s-a)=1995-2a$ is an odd integer

So, the product $2s2(s-a)(s-b)(s-c)$ can't be an integer.

but $a^2h_a^2 \in \mathbb{Z}$


NC Mathcounts Team got all the team questions correct.

http://www.foxnews.com/video2/launchPage.html?051107/051107_live_math&Whiz%20Kids%21&Live_Desk&Eighth%20graders%20compete%20to%20be%20the%20%27National%20Math%20Champion%27&US&-1&Whiz%20Kids%21&Video%20Launch%20Page&News

Unfortunately, I can't watch that video for some reason.

Congratulations on eighth place! Any high individual placers?

wow - i just watched it and the only thing i have to say is that the anchorpeople are a bunch of  :cursing: IDIOTS.

i mean, come on.  wow.  i think i'll go home and cry and move to a country where people actually respect mathematicians.  like romania.  or china.  or SOMEPLACE LIKE THAT.

but yeah, to echo jb05 congrats guys.  gj to all.

No... thirty minutes too late to edit my old post. Why did they ever make that annoying rule?

Congratulations also for THREE individuals in the top 50!!! That's amazing!

27th-36th places had the score of 33. Jenny got a 33 (19 Sprint and 7 Target) but ranked down to 35th. She got the first problem correct (20) at first but changed to 40 later (she said she was thinking too hard).


geometry
perimeter
incenter
circumcircle
analytic geometry
geometry proposed

Let $\ell$ be the perimeter of an acute-angled triangle $ABC$ which is not an equilateral triangle. Let $P$ be a variable points inside the triangle $ABC$, and let $D,E,F$ be the projections of $P$ on the sides $BC,CA,AB$ respectively. Prove
that \[ 2(AF+BD+CE ) = \ell \] if and only if $P$ is collinear with the incenter and the circumcenter of the triangle $ABC$.

Here is a solution:

Any point $P$ on $OI$ can be represented as $\lambda O+(1-\lambda)I$ for some real $\lambda$ (we can regard this as a relation between the coordinates of $O,I,P$ on the axis $OI$). For any $P$ let $D(P)=D,E(P)=E,F(P)=F$. We then have $D(P)=\lambda D(O)+(1-\lambda) D(I)$ (this relation takes place on $BC$, where we have chosen origin $B$) and the like (i.e. analogous relations for $E(P),F(P)$; call these relations $(*)$), and since $AF(X)+BD(X)+CE(X)=\frac{\ell}2$ for $X\in\{O,I\}$, it means that it also works for any $P\in OI$ because of $(*)$.

We must also prove the converse.

For any $P$ satisfying the condition there's a $P'\in OI$ satisfying the condition, where $P'=PD(P)\cap OI$ (maybe $PD(P)$ doesn't intersect $OI$ inside $ABC$, but one of $PD(P),PE(P),PF(P)$ does, so we assume WLOG it's the first one). We have $CE(P)+AF(P)=CE(P')+AF(P')$ and $f(P)=f(P')$, where $f(X)=CE(X)^2-(b-CE(X))^2+AF(X)^2-(c-AF(X))^2$, and I guess these two equations give $CE(P)=CE(P'),AF(P)=AF(P')\Rightarrow P=P'$.

[color=red][Moderator edit: You can also find 99% of the solution of this problem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5694]another MathLinks thread[/url] (the rest is trivial).][/color]


conics
projective geometry
geometry proposed
geometry

Let $ ABC$ be a triangle. Let $ \omega$ be a circle passes through $ B$, $ C$, and intersect the segments $ AB$ and $ AC$ at $ D$ and $ E$, respectively. Let the tangent to $ \omega$ at $ E$ intersect the tangent to $ \omega$ at $ B$ at a point $ F$, and the tangent to $ \omega$ at $ D$ intersect the tangent to $ \omega$ at $ C$ at a point $ G$. Prove $ F$, $ G$ and $ A$ are collinear.

Polar maps kill it.  Ie,

$ F$ is the pole of $ BE$
$ G$ is the pole of $ CD$
$ A$ lies on the polar of the intersection of $ BE$ and $ CD$

Another (equivalent) way is to use Pascal's Theorem.  Ie, let $ H$ be the intersection of $ DE$ and $ BC$.  Then we can use Pascal's Theorem to show $ A, F,$ and $ H$ are collinear, and then to show $ A, G,$ and $ H$ are collinear.

But is there a non-projective way?

This problem have a nice generalization:

Take an 2n-agon $ P_1, \ P_2, \dots , \ P_{2n}$ such that $ \bigcap_{i\equal{}1}^{n} P_iP_{i\plus{}n}$ is a point, and such that $ \bigcup_{i\equal{}1}^{2n} P_i$ stay on a conic. Call the tangent line to the conic on $ P_i$, $ t_i$ for $ i\equal{}1,\ 2, \ \dots \ 2n$ and call $ Q_i \equal{} t_i \cap t_{i\plus{}1}$ for $ i\equal{}1,\ 2, \ \dots \ 2n$ (all mod 2n). Also call $ X_i : P_iP_{i\plus{}n} \cap P_{i\plus{}1}P_{i\plus{}1\plus{}n}$ for $ i\equal{}1,\ 2, \ \dots, \ n$ and $ Y_i : Q_iQ_{i\plus{}n} \cap Q_{i\plus{}1}Q_{i\plus{}1\plus{}n}$ for $ i\equal{}1,\ 2, \ \dots, \ n$.
Then prove that $ \bigcup_{i\equal{}1}^{n} (X_i, Y_i)$ are on the same line.

 :D


Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that

\[ \frac{a^2b}{4\minus{}bc}\plus{}\frac{b^2c}{4\minus{}ca}\plus{}\frac{c^2a}{4\minus{}ab} \le 1.\]

One very very nice and similar inequality can be found here

http://batdangthuc.net/forum/showthread.php?t=271

http://batdangthuc.net/forum/showthread.php?t=302

[quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that
\[ \frac {a^2b}{4 \minus{} bc} \plus{} \frac {b^2c}{4 \minus{} ca} \plus{} \frac {c^2a}{4 \minus{} ab} \le 1.
\]
One very very nice and similar inequality can be found here

http://batdangthuc.net/forum/showthread.php?t=271

http://batdangthuc.net/forum/showthread.php?t=302[/quote]
The inequality can be solved by rewritten the inequality as
\[ 4\minus{}\sum a^2b  \ge \sum \frac{a^2b^2c}{4\minus{}bc}
\]
Now using the know $ 4\minus{}\sum a^2b \ge abc$. It suffices to prove
\[ 1 \ge \sum \frac{ab}{4\minus{}bc}
\]
Or
\[ 64\minus{}32\sum ab \plus{}8\sum a^2bc \plus{}4\sum a^2b^2 \ge abc(\sum a^2b\plus{}abc)
\]
Using the above inequalit again, it suffice to prove
\[ 64\minus{}32\sum ab \plus{}8\sum a^2bc \plus{}4\sum a^2b^2 \ge 4abc
\]
or
\[ 16\minus{}8q\plus{}q^2\minus{} r \ge 0
\]
where $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$. Note that $ q^2 \ge 9r$ then it suffices to prove
\[ 16 \minus{}8q \plus{}q^2 \minus{}\frac{q^2}{9} \ge 0
\]
or
\[ (q\minus{}3)(q\minus{}6) \ge 0
\]
which is clearly true. Equality holds for $ a\equal{}b\equal{}c\equal{}1$ and $ a\equal{}2,b\equal{}1,c\equal{}0$ and their permutations. :)

How to prove the in equality :
    with a,b,c be nonnegative real numbers such that a+b+c =3
 We have : $ a^2b\plus{}b^2c\plus{}c^2a \leq 4$ :maybe:

Nice solution, can_hang2007.
This is also nice and easy:

$ \frac {ab}{4 \minus{} b} \plus{} \frac {bc}{4 \minus{} c} \plus{} \frac {ca}{4 \minus{} a} \le 1$,

for real nonnegative numbers $ a,b,c$ such that $ a \plus{} b \plus{} c \equal{} 3$.

[quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that
\[ \frac {a^2b}{4 \minus{} bc} \plus{} \frac {b^2c}{4 \minus{} ca} \plus{} \frac {c^2a}{4 \minus{} ab} \le 1.
\]
[/quote]
We can refine this inequality as follows:

Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Find the smallest value of $ p$ such that

\[ \frac {a^2b}{p \minus{} bc} \plus{} \frac {b^2c}{p \minus{} ca} \plus{} \frac {c^2a}{p \minus{} ab} \le \frac 3{p\minus{}1}.
\]
I think the smallest $ p$ is $ \frac 9{4}$. :D

[quote="Vasc"][quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that
\[ \frac {a^2b}{4 \minus{} bc} \plus{} \frac {b^2c}{4 \minus{} ca} \plus{} \frac {c^2a}{4 \minus{} ab} \le 1.
\]
[/quote]
We can refine this inequality as follows:

Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Find the smallest value of $ p$ such that
\[ \frac {a^2b}{p \minus{} bc} \plus{} \frac {b^2c}{p \minus{} ca} \plus{} \frac {c^2a}{p \minus{} ab} \le \frac 3{p \minus{} 1}.
\]
I think the smallest $ p$ is $ \frac 9{4}$. :D[/quote]
$ p\equal{}1^\plus{}$  :?:

$ \sum \frac{a^2b}{9\minus{}4bc} \equal{} \sum \frac{a^2b}{(a\plus{}b\plus{}c)^2\minus{}4bc} \le \sum \frac{a^2b}{a^2\plus{}2ab\plus{}2ac} \equal{} \sum \frac{ab}{a\plus{}2b\plus{}2c} \equal{}\sum \frac{ab}{6\minus{}a} \equal{}\sum (\frac{6b}{6\minus{}a}\minus{}b) \equal{} 6 \sum \frac{b}{a\plus{}2b\plus{}2c}\minus{}3 \le \frac{3}{5}$.

[quote="Vasc"]Yes. This is also my solution.

[quote="ehku"]
$ p \equal{} 1^ \plus{}$  :?: [/quote]
What means this ?[/quote]
I mean that if $ p\equal{}1$ then the inequality is true, but I am wrong :blush:   because... you are right  :lol:
We assume there exists $ p < \frac{9}{4}$ such that the inequality holds. Take $ b\equal{}c\equal{}0$ we get $ p >1$. We chose $ b\equal{}c<\sqrt{p} < \frac{3}{2}$ and take $ b \to \sqrt{p}^\minus{}$, then $ LHS \to \plus{}\infty$, this is a contradiction. Hence $ p \ge \frac{9}{4}$.

[quote="Vasc"][quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that
\[ \frac {a^2b}{4 \minus{} bc} \plus{} \frac {b^2c}{4 \minus{} ca} \plus{} \frac {c^2a}{4 \minus{} ab} \le 1.
\]
[/quote]
We can refine this inequality as follows:

Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Find the smallest value of $ p$ such that
\[ \frac {a^2b}{p \minus{} bc} \plus{} \frac {b^2c}{p \minus{} ca} \plus{} \frac {c^2a}{p \minus{} ab} \le \frac 3{p \minus{} 1}.
\]
I think the smallest $ p$ is $ \frac 9{4}$. :D[/quote]
Yes, You're right, VasC. :) To prove the claim, we just prove the inequality
\[ \sum \frac{ab}{9\minus{}4bc} \le \frac{3}{5}
\]
or
\[ \sum\frac{9\minus{}4ab\minus{}4bc}{9\minus{}4bc} \ge \frac{3}{5}
\]
Note the $ 9\minus{}4ab\minus{}4bc \ge 0,9\minus{}4bc\minus{}4ca,9\minus{}4ca\minus{}4ab \ge 0$ then applying the Cauchy Schwarz, we have
\[ \sum \frac{9\minus{}4ab\minus{}4bc}{9\minus{}4bc} \ge \frac{(27\minus{}8\sum ab)^2}{\sum (9\minus{}4bc)(9\minus{}4ab\minus{}4bc)} \equal{}\frac{(27\minus{}8x)^2}{16x^2\minus{}108x\plus{}243\minus{}48abc} (x\equal{}\sum ab \le 3)
\]
By Schur inequality, $ abc \ge \frac{4x\minus{}9}{3}$, we have
\[ \frac{(27\minus{}8x)^2}{16x^2\minus{}108x\plus{}243\minus{}48abc}  \ge \frac{(27\minus{}8x)^2}{16x^2\minus{}108x\plus{}243\minus{}16(4x\minus{}9)}\equal{}\frac{3}{5}\plus{}\frac{(3\minus{}x)(207\minus{}68x)}{5(16x^2\minus{}172x\plus{}387)} \ge \frac{3}{5}
\]
:)

[quote="manlio"][quote="can_hang2007"]....To prove the claim, we just prove the inequality
\[ \sum \frac {ab}{9 \minus{} 4bc} \le \frac {3}{5}
\]
[/quote]

Why this and not

$ \sum \frac {a^2b}{9 \minus{} 4bc} \le \frac {3}{5}$[/quote]
well, because if we show $ \sum \frac{ab}{9\minus{}4bc} \le \frac{3}{5}$ then the inequality $ \sum \frac{a^2b}{9\minus{}4bc} \le \frac{3}{5}$ follows since it is equivalent to
\[ 4abc \sum \frac{ab}{9\minus{}4bc} \le \frac{27}{5} \minus{}\sum a^2b
\]
Using the known $ \sum a^2b \le 4\minus{}abc$ and the above inequality, it suffices to prove
\[ \frac{12}{5} abc \le \frac{7}{5} \plus{}abc
\]
or
\[ abc \le 1
\]
which is true. :)

Thank you very much. :)

Very nice proof, congratulations :wink:

[quote="Vasc"][quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that
\[ \frac {a^2b}{4 \minus{} bc} \plus{} \frac {b^2c}{4 \minus{} ca} \plus{} \frac {c^2a}{4 \minus{} ab} \le 1.
\]
[/quote]
We can refine this inequality as follows:

Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Find the smallest value of $ p$ such that
\[ \frac {a^2b}{p \minus{} bc} \plus{} \frac {b^2c}{p \minus{} ca} \plus{} \frac {c^2a}{p \minus{} ab} \le \frac 3{p \minus{} 1}.
\]
I think the smallest $ p$ is $ \frac 9{4}$. :D[/quote]

Can we use  case  $ p\equal{}\frac 9{4}$ to prove other cases？

[quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers with sum $ 3$. Prove that
\[ \frac {a^2b}{4 \minus{} bc} \plus{} \frac {b^2c}{4 \minus{} ca} \plus{} \frac {c^2a}{4 \minus{} ab} \le 1.\]
One very very nice and similar inequality can be found here

http://batdangthuc.net/forum/showthread.php?t=271

http://batdangthuc.net/forum/showthread.php?t=302[/quote]
See also here: http://canhang2007.wordpress.com/2009/11/02/inequality-37-p-k-hung/


My teacher recently showed me how to generate a function on a TI-83.  Does anybody know how to do this on a TI-89?

(Generating a Function Means:)

Given the points (x, y) (a, b) (c, d) (e, f) generate a function fitting this data.  (Topic came up on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=416194#p416194]this[/url] AMC problem.)

I don't know about calculators, but here's the theory behind it.

http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html


In my class we were a given a problem which ultimately came down to finding the roots of $12-2450x^{(-2)}=0$
When I multiplied through by $x^2$ my teacher said that I couldn't do that because I might be adding or removing a root. She said that in this case it works but it won't work in all cases. To defend her statement she used the equation
$12x^2=14x$
She said if you divide through by x you remove the root 0.

However I still believe I'm right in that you can multiply through by x because the only reason why you can't divide by x in the second equation is because you can't have x in the denominator since it might be 0. However what I did was multiply, not divide by x.

Anyone confirm me on this?

I think when you multiply you risk adding a root that doesn't satisfy the original equation. You can still do that, though. All you have to do is check each potential root in the original equation and make sure it works.

I don't like your teacher's explanation. When you divide through by x in the equation she gave, you must state that you are placing the condition that x cannot be 0. Therefore, you need to test the case in which x is 0, to see if 0 is a root as well. I believe that splitting the problem into two cases like this is mathematically and logically correct. Ask your teacher; you wouldn't want to do this on a test if your teacher disagrees.
I don't see anything wrong with multiplying through by x^2, because that's about the same as cross-multiplying if you put the two terms on different sides of the equal sign. If you were supposed to factor, you end up multiplying to solve for x anyway. Sometimes, you can add roots, by raising things to powers, like x=3 becoming x^2=9. I think you can add roots if you multiply by a constant when you're working in the complex plane. Just be careful; you should always check your "roots" to make sure that they satisfy the original equation. It wouldn't do if one of them resulted in division by 0 or indeterminate forms.


AMC
AIME
geometry
USA(J)MO
USAMO
number theory
AIME I

Practice AMC 12 Average: 118.25 exactly
Practice AIME Average (since november): 6 exactly

I hope to increase my AIME score by at least 4 points by the end of February. I've finished:

- AoPS volume 1
- AoPS volume 2 chapters that are useful for AIME (according to Mr. Rusczyk)

I have left to do:

- Challenging problems in geometry
- Challenging problems in algebra

Along with loads of practice AIME's, what would you recommend that I do in January and February? Keep in mind that I'm relatively good at algebra and number theory but I'm lacking in combinatorics and geometry. :D

[quote="chess64"]
Along with loads of practice AIME's, what would you recommend that I do in January and February? Keep in mind that I'm relatively good at algebra and number theory but I'm lacking in combinatorics and geometry. :D[/quote]

Whatever you feel like to do, anything on that list will help you.

Just keep working problems, even if they aren't AIME ones. As long as you keep working at the hard problems and keep your mind in that state, it's gonna help you.

chess64, you're only a freshman now. Even you fail this year (you won't :D ), you'll still have 3 more years. I personally don't worry about AIME much now because I think if you work hard enough to build good problem solving skills, you should be read for almost anything (except for USAMO because it's proof). That's what I'm doing now. Just working through problems, mostly later AHSMEs, and trust me, last year, I was lucky to get 1 on AIME and this year, on my first AIME, I got 5 and on the second one, I got 7. Of course, the second one was easy and I screw up pretty bad.  :D

But yeah, like Elemennop and many others said in the past (I can name their names that will fill this whole thread), only way you can get better is to keep doing problems. I suggest taking AMC-12's as well except without use of calculator because that will help you to get better at "stupid" calculation mistakes.

Good luck to all of AoPS users with AMC this year!  :D  :)


Given that in a triangle $ ABC$, $ AB \equal{} 3, BC \equal{} 4$ and the midpoints of the altitudes of the triangle are collinear, find all possible values of the length of $ AC$

This result will probably help:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84858[/url]


Its the first problem in 8.1 in ACoPs and I can't do it. I'm terrible at geometry. Here it is:

Let ABC be isosceles with vertex angle A and E and D be points on AC and AB respectively so that AE=ED=DC=CB. Find angle A.

Hints only!

[hide="Hint"]Let angle A be x. Draw out a picture, and do some angle chasing in terms of x. [/hide]

did you get an answer with that? I tried that for about half an hour and kept getting stuff like
180=180  :oops:

[geogebra]a034bf2047b9a645330e2690ed62c1d1d22bf506[/geogebra]
That's the image I end up with. So we have from the bottom triangle (or triangle ABC) $ 3x\plus{}3x\plus{}x\equal{}180 \implies 7x\equal{}180 \implies x\equal{}180/7.$

Wow. I just looked at all my tries and I got that solution but I regarded it as wrong because the result wasn't an integer and I assumed that the answer was probably an integer.


vector
linear algebra
matrix
circumcircle
combinatorial geometry
combinatorics proposed
combinatorics

Let $ P_0,P_1,P_2,....$ be a sequence of convex polygons such that for each $ k\geq 0$ the vertices of $ P_{k\plus{}1}$ are the midpoints of all sides of $ P_k$ .Prove that there exist   a unique point lying inside all these polygons .

[quote="pleurestique"]the unique point is the center of gravity of $ P_0$. (it's also the center of gravity of all $ P_k$.)[/quote]

It's true but what is your proof ?  :wink:

Let $ C_n$ be the circumscribed circle of $ P_n$ and let $ d_n$ denote one of diameters of $ C_n$. We choose $ d_n$'s which are on the same direction. Thus, we can consider $ d_n$'s as nested closed intervals. $ |d_n| \rightarrow 0$, so there exists a unique point lying inside all $ P_n$'s by  nested intervals theorem.

$ P_k$ need not have a circumscribed circle

Yes. I thought as if $ P_k$'s were regular :blush:

[quote="Albanian Eagle"]maybe you will find this topic enlightening :) ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=113248[/quote]
@Albanian Eagle:How can it be related to this problem :?:

I dont understand why cant we solve it in this easy and obvious way since it does not require the definition of that point(Center of gravity)(moreover it is a combinatorics problem):

Let $ A_k$ be the set of all points in the interior of $ P_k$. $ A_{k\plus{}1} \in A_k\ \ \forall\ k$ because of convexity. And $ A_k\not \equal{} \phi\ \ \forall\ k$. So there is a point $ \in A_{\infty}$. Thus that point $ \in A_k\ \ \forall\ k$.

I understand there could be some big flaw in my proof because none of you have tried like this. In that case I want to know what it is.

That there is [i]at least[/i] one point in the intersection of all these polygons is not difficult.  (That $ P_0 \equal{} \frac{1}{n} \sum_{i \equal{} 1}^n P_i$ is such a point is easy: it's the average of the vertices at each step, so it's in the convex hull at each step, so it's in the intersection of all the convex hulls.)  What we don't have is a proof that there is [i]only[/i] one such point.

Let $ A_n$ be the length of the longest diagonals (including sides) of $ P_n$. It can be shown that $ kA_n\ge A_{n \plus{} 1}$ for some $ k < 1$. The proof of this is straightforward but a bit complicated (that's why I'm lazy to type it here  :P )

You can extract the proof of the fact that each vertice converges to $ P_0$ from the topic [b]Albanian Eagle [/b] linked to.

choose the center of gravity as the origin and let $ v \equal{} (v_1, v_2, ..., v_n)$ be the $ n$-dimensional vector whose components are the vertices of $ P_0$. let $ A$ be the matrix with $ \frac {1}{2}$'s all along the main diagonal, $ \frac {1}{2}$'s along the diagonal above that, a $ \frac {1}{2}$ in the bottom left corner, and $ 0$'s everywhere else. then the vertices of $ P_k$ are given by $ A^kv^{\text{T}}$. so all we need to do is show that the entries of $ A^k$ go to $ 0$, and for this just compute $ A$'s eigenvalues...


[quote="yoyosir"]can it solve all ineq with 3-variables?[/quote]
What is S.O.S ? :blush:

I think it is wrong!!!
there are many problems can not solve by S.O.S method !!!


number theory
greatest common divisor
number theory proposed

There must be an error

with $ n \equal{} 3$ we get $ 7 \mid 728 \equal{} 3^6 \minus{} 1$ but $ 7 \nmid 26 \cdot 8 \equal{} (3^3\minus{}1)(3^2 \minus{} 1)$

[quote="PhilG"]There must be an error

with $ n \equal{} 3$ we get $ 7 \mid 728 \equal{} 3^6 \minus{} 1$ but $ 7 \nmid 26 \cdot 8 \equal{} (3^3 \minus{} 1)(3^2 \minus{} 1)$[/quote]

This means $ n\equal{}3$ is not a solution . You are required to find all $ n$ satisfying the conditions ,maybe you misread the question.

$ n^6\minus{}1 \equal{} (n\minus{}1)(n\plus{}1)(n^2\minus{}n\plus{}1)(n^2\plus{}n\plus{}1)$ while $ (n^3\minus{}1)(n^2\minus{}1) \equal{} (n\minus{}1)^2(n\plus{}1)(n^2\plus{}n\plus{}1)$.
We only need look at primes $ p \mid n^2 \minus{}n \plus{} 1$. But $ \gcd(n^2\minus{}n\plus{}1,n\minus{}1) \equal{} 1$, $ \gcd(n^2\minus{}n\plus{}1,n\plus{}1) \mid 3$, $ \gcd(n^2\minus{}n\plus{}1,n^2 \plus{} n \plus{}1) \mid 2$. Since $ n^2\minus{}n\plus{}1 \equal{} n(n\minus{}1) \plus{} 1$ is odd, the only possibility left is $ n^2\minus{}n\plus{}1 \equal{} 3^k$, with discriminant $ \Delta \equal{} 3(4\cdot 3^{k\minus{}1} \minus{}1)$, so $ k\equal{}1$ and $ n\equal{}2$. Indeed $ 2^6\minus{}1 \equal{} 3^2\cdot 7$ and $ (2^3\minus{}1)(2^2\minus{}1) \equal{} 3\cdot 7$.

Oops I should get more sleep  :blush:
Nice proof mavropnevma!

[quote="srinath.r"]Find  all integers $ n > 1$ such that any prime divisor of $ n^{6} \minus{} 1$ is also a divisor of $ (n^{3} \minus{} 1)(n^{2} \minus{} 1)$[/quote]

Since $ (n\plus{}1)(n^3\minus{}1) \mid \text{gcd}\{n^6\minus{}1,(n^3\minus{}1)(n^2\minus{}1)\}$, it is enough that $ \text{rad}(\phi_3(\minus{}n)) \mid \text{rad}((n^3\minus{}1)(n^2\minus{}1))$ holds, where $ \phi_i(\cdot)$ is the i-th cyclotomc polyomial. Note that if $ n>1$ then $ \phi_3(\minus{}n)>1$. Now $ \text{gcd}(\phi_3(\minus{}n),n^3\minus{}1)\equal{}\text{gcd}(\phi_3(\minus{}n),(n\plus{}1)\phi_3(\minus{}n)\minus{}(n^3\minus{}1)) \mid 2$, but $ 2 \mid \phi_3(\minus{}n)\plus{}1$ for all $ n \in \mathbb{Z}$. On the other hand $ \text{gcd}(\phi_3(\minus{}n),n^2\minus{}1) \mid 3$ (with equality iff $ 3 \mid n\plus{}1$). So the only possible case is $ \omega(\phi_3(\minus{}n))\equal{}1$, but $ \phi_3(\minus{}n)\equal{}3^x$ for some $ x \in \mathbb{N}_0$ iff $ 3\left(\frac{2n\minus{}1}{3}\right)^2\plus{}1\equal{}4\cdot 3^{x\minus{}1}$ that is $ 1$ in $ \mathbb{F}_3$, so that the unique solution is $ n\equal{}2$.[]


What is the area, in square inches, of a right triangle with a 24-inch leg and a 25-inch hypotenuse?


What is the limit of $ (1\plus{}1/3n)^n)$?

$ \lim_{n \to \infty} \left( 1 \plus{} \frac{1}{3n} \right) ^n \equal{} \lim_{j\equal{}3n \to \infty} \left( 1 \plus{} \frac{1}{j} \right) ^{j \cdot \frac{1}{3}} \equal{} e^{\frac{1}{3}}$


we just have to prove

$\frac{1}{i\sqrt{i+1}}>\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}>\frac{1}{(i+1)\sqrt{i}}$

But the left two is just $\frac{1}{\sqrt{i}}>\frac{2}{\sqrt{i+1}+\sqrt{i}}$ and the right two is

$\frac{2}{\sqrt{i+1}+\sqrt{i}}>\frac{1}{\sqrt{i+1}}$

which is trivial

[quote="violetcraze"][quote="shyong"]we just have to prove

$\frac{1}{i\sqrt{i+1}}>\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}>\frac{1}{(i+1)\sqrt{i}}$

[/quote]

i don't get it... why is the "1" replaced by $\sqrt{i}$ ??      :([/quote]
Where is he doing that...?  His proof looks okay to me.

I think he's referring to the part where shyong took out the sigmas in the middle expression (so since Shyong came up with something that telescoped to the middle sum, he could take out the sigmas in that manner)

[quote="violetcraze"][quote="shyong"]we just have to prove

$\frac{1}{i\sqrt{i+1}}>\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}>\frac{1}{(i+1)\sqrt{i}}$

[/quote]

i don't get it... why is the "1" replaced by $\sqrt{i}$ ??      :([/quote]

If you can prove $\boxed{\frac{1}{i\sqrt{i+1}}>\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}>\frac{1}{(i+1)\sqrt{i}}}$

then you can write $\boxed{\sum^n_{i=1}\frac{1}{i\sqrt{i+1}}>\sum^n_{i=1}\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}>\sum^n_{i=1}\frac{1}{(i+1)\sqrt{i}}}$

And if you write out the center part for $i=1,2,3\cdots n$ and using telescoping method(or method of differences) , you will see that it is same as the original question  ;)

[quote="violetcraze"][quote="shyong"]we just have to prove

$\frac{1}{i\sqrt{i+1}}>\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}>\frac{1}{(i+1)\sqrt{i}}$

[/quote]

errr....one more question:

i can understand the working now. but if i'm tackling the question to solve it myself, how can i get the

 $2(1-\frac{1}{\sqrt{n+1}}) = \sum^n_{i=1}\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}$?

what method should we use to get to that? or is it just observation and trial and error?[/quote]

$\sum^n_{i=1}\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}=2\sum^n_{i=1}\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)$

Just write out a few term and see how they cancel each other  ;)

[quote="shyong"]

$\sum^n_{i=1}\frac{2(\sqrt{i+1}-\sqrt{i})}{\sqrt{i+1}\sqrt{i}}=2\sum^n_{i=1}\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)$

Just write out a few term and see how they cancel each other  ;)[/quote]

THanks  :)

It's good to be able to see $\frac{ f(i+1) - f(i) }{ f(i+1) f(i) } = \frac{1}{f(i)} - \frac{1}{f(i+1)}$ in series problems, especially when it is given that the problem involves telescoping.  :)


geometry
3D geometry
number theory unsolved
number theory

Hello mathematicians. I have problem with the following problem. I can't find solution and I don't have solution available to me now. Would you solve it.
Find the number of natural numbers a smaller than 2003 such that there exist a natural number n such that $ 3^{2003}|a\plus{}n^{3}$.

What are the possible values of cubes mod $ 3^{2003}$?

Of course, you should look at smaller powers of 3 first to understand the pattern. The role 2003 plays here is essentially that of a large number, so this question could be adjusted to any year.

[quote="jmerry"]
Of course, you should look at smaller powers of 3 first to understand the pattern.[/quote]
Can you explain please??

we can first get a+n^3 modulo a set of numbers co prime to each other then apply chinese remainder theorem

aadil- No, you don't. Everything is mod powers of 3, so nothing is relatively prime.

Cubes are either $ \pm 1\mod 9$ or divisible by $ 3$. If they're divisible by 3, ...

jimerry i was saying that we could get a+n^3 modulo some numbers which are co prime to themselves as well as 3


[quote="GameBot"]Two drivers leave city A at the same time for city B which is 450 miles away. The first driver travels at 60 mph and the second driver travels at 50 mph. How many minutes separated their arrival times in city B.[/quote]

B goes for 9 hours, A goes for 7.5, so 90 minutes.


Nice move, splitting physics into two separate forums! I believe this is the best solution to the 'homework' problem that has arisen over the last few days.

Now let's hope that people will post their links in right forums :)

Oh yeah....and I guess we need to thank both mysticterminator and the admins for that......and in any case, a slight bit of moderation ( which AoPS is more than capable of doing efficiently) will take care of ensuring that the right problem is at the right difficulty level.


inequalities
LaTeX
inequalities proposed

[u]1[/u] Prove that for positive reals a,b,c the following inequality holds good -
$ a^2(\frac {a \plus{} 2c}{3b}) \plus{} b^2(\frac {b \plus{} 2a}{3c}) \plus{} c^2(\frac {c \plus{} 2b}{3a}) \ge a^2 \plus{} b^2 \plus{} c^2$ - (own)

[u]2[/u] Prove that for positive reals a,b,c such that $ a \plus{} b \plus{} c \equal{} 1$ the following inequality holds -
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge 3(a^2 \plus{} b^2 \plus{} c^2)$
also mention when the equality occurs in both cases

The second inequality is equivalent to

$ LHS\minus{}RHS\equal{}\sum_{cyc}\left(\frac{a^2}{b}\minus{}2a\plus{}b\right)\plus{}a\plus{}b\plus{}c\minus{}3(a^2\plus{}b^2\plus{}c^2)\equal{}\sum_{cyc}\left(\frac{(a\minus{}b)^2}{b}\right)\plus{}(a\plus{}b\plus{}c)^2\minus{}3(a^2\plus{}b^2\plus{}c^2)\equal{}\sum_{cyc}\left(\frac{(a\minus{}b)^2}{b}\right)\minus{}\sum_{cyc}(a\minus{}b)^2\equal{}\sum_{cyc}\left(\frac {(a \minus{} b)^2(1 \minus{} b)}{b}\right) \ge 0$  :)

[quote="Spribo"]Both of the inequalities are simple applications of the Muirhead inequality. Just notice that (3,2,0) and (4,1,0) are majorized by (3,1,1). Expand and you'll get the answer.
For example, the second one:
$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\ge 3(a^{2} \plus{} b^{2} \plus{} c^{2}), (a \plus{} b \plus{} c)(\frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a})\ge 3(a^{2} \plus{} b^{2} \plus{} c^{2}), \sum_{cyc}(a^4c \plus{} a^3c^2)\ge2\sum_{cyc}a^3bc$, true by Muirhead...[/quote]
Muirhead only work for symmetric sums.
e.g. $ \sum_{cyc} a^3c^2 \ge \sum_{cyc}a^3bc$ is wrong . Try $ b\equal{}1,a\equal{}10,c\equal{}10^{10}$.

@Evariste-galois
very nice factorizations any hint on the inspiration ?
i dont know muirhead! the first easily solvable by AM-GM and the second with a little more than the AM-GM.

i have a new proof for 2 but the assumption that - $ a \ge b \ge c$  but i am wondering if i can make that assumption?

here are my proofs -
[hide="my solution to 1"]
rewrite the inequality as ,
$ \sum_{cyclic} (a \plus{} 2c) \frac {a^2}{b} \ge 3 (a^2 \plus{} b^2 \plus{} c^2)$,
$ \sum_{cyclic} (a \plus{} 2c) \frac {a^2}{b} \equal{} \sum_{cyclic}(\frac {a^3}{b} \plus{} \frac {a^2c}{b} \plus{} \frac {b^2a}{c}) \ge 3(a^2 \plus{} b^2 \plus{} c^2)$ and we are done!(the last one is Am-Gm for 3 positive reals) and the equality occurs for all positive real values such that $ a \equal{} b \equal{} c$[/hide]
[hide="my solution to 2"]
you could overkill the problem with SOS but that is not that nice compared to this,
assume WLOG: $ a \ge b \ge c$,
the inequality is equivalent to proving that-
$ \sum_{cyclic}(\frac {a^2}{b^2} \minus{} a^2) \ge 2(a^2 \plus{} b^2 \plus{} c^2)$
using $ a \plus{} b \plus{} c \equal{} 1$, we get,
$ \sum_{cyclic}(a^2)\cdot(\frac {a \plus{} c}{b}) \ge 2(a^2 \plus{} b^2 \plus{} c^2)$
observe that,
$ 2(a^2 \plus{} b^2 \plus{} c^2) \equal{} \frac {1}{3} \cdot 6 \cdot (a^2 \plus{} b^2 \plus{} c^2)$
and by AM-GM,
$ a^{2}(\frac{a\plus{}b}{c})\plus{}b^{2}(\frac{a\plus{}c}{b})\plus{}c^{2}(\frac{b\plus{}c}{a})\ge (\frac{1}{3})(a^{2}\plus{}b^{2}\plus{}c^{2})(\frac{a\plus{}b}{c}\plus{}\frac{a\plus{}c}{b}\plus{}\frac{b\plus{}c}{a})\ge 2(a^{2}\plus{}b^{2}\plus{}c^{2})$
the last inequality is true by chev' using the assumption.
thus we are left to prove that ,
$ a^2\cdot(\frac {a \plus{} c}{b}) \plus{} b^2\cdot(\frac {b \plus{} a}{c}) \plus{} c^2\cdot(\frac {c \plus{} b}{a}) \ge a^2 \cdot(\frac {a \plus{} b}{c}) \plus{} b^2\cdot(\frac {a \plus{} c}{b} \plus{} c^2\cdot(\frac {b \plus{} c}{a})$
or
$ a^2\cdot(\frac {a \plus{} c}{b}) \plus{} b^2\cdot(\frac {b \plus{} a}{c}) \ge a^2 \cdot(\frac {a \plus{} b}{c}) \plus{} b^2\cdot(\frac {a \plus{} c}{b})$
which is equivalent to after some modifications -$ a^2(c \minus{} b) \ge b^2(c \minus{} b)$ which is obvious from the assumption.
the equality holds when $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ and thus we are done  :)[/hide]
EDIT: Fixed Latex (thanks soarer!)

i am sorry but i cant at all understand what the mistake in latex in my post is! mods! help me!

Maybe it will be helpful if you post the $ \text{\LaTeX}$ code…

true! how dum of me to not think of it - CODE - a^2(\frac {a + b}{c}) + b^2(\frac {a + c}{b}) + c^2(\frac {b + c}{a}) \ge (\frac {1}{3})(a^2 + b^2 + c^2)(\frac {a + b){c} + \frac {a + c}{b} + \frac {b + c}{a}) \ge 2(a^2 + b^2 + c^2)
hopefully someone can fix the error

the frac in the second line has a wrong closing braces ), anyway
$ a^2(\frac {a \plus{} b}{c}) \plus{} b^2(\frac {a \plus{} c}{b}) \plus{} c^2(\frac {b \plus{} c}{a}) \ge (\frac {1}{3})(a^2 \plus{} b^2 \plus{} c^2)(\frac {a \plus{} b}{c} \plus{} \frac {a \plus{} c}{b} \plus{} \frac {b \plus{} c}{a}) \ge 2(a^2 \plus{} b^2 \plus{} c^2)$


I've decided to foray back into the area of Mathcounts (sadly, I'm done), and I thought I could write some mock tests since I'm not that busy over break.

This is my third(?) mock round and I'd be excited if I got around 15-20 people.

[b] Format: [/b] n progressively harder rounds of m questions each, from Chapter difficulty up to AMC 10 #25 level :) , 4 or 5 days each to submit (but won't take that long to complete, maybe an hour), total of all of the m*n problems.

Note: There will definitely be a bunch of problems, not an overwhelming amount, though. Say, about 8-10 rounds of 10?

Just [b]post here [/b]if you want to join - I'll edit this thread.

[b] First round starts Dec. 26. [/b]

Thanks, and have a good break!

Signed up:
athunder
FantasyLover
007math

-someperson

Signed up:
athunder
FantasyLover
007math
SuperMathBoy
Chopstiks
ernie
TPC

--hm - what's with the edit restrictions?

I'm new to the forum but I'll join :D and this things not that hard? ishh like hardest is AMC10 q25 right?

Hm... say maybe the #10s could be AIME-type, but only early AIME questions.

But otherwise, yup, not too hard.

I will join.

Will it have any non-mathcounts problems?

Most probably.

Signed up:
athunder
FantasyLover
007math
SuperMathBoy
Chopstiks
ernie
TPC
shentang
high_flyer10
Discrete_math
randomtext

oh my goddd misread #5

thought it was area of the hexagon / total area...

Scores will be posted below:

xpmath: 10 (good job)
arb95: 8.5
FantasyLover: 7
AIME15: 3/4 (Partial)

Update - gah those edit restrictions....

xpmath: 10 (good job)
shentang: 9
arb95: 8.5
andyliao: 8
FantasyLover: 7
AIME15: 3 (Partial)

Is it too late to join for later rounds?

Nope, that's the beauty of the round system.

xpmath: 10 (good job)
shentang: 9
arb95: 8.5
andyliao: 8
Discrete_math: 8
FantasyLover: 7
AIME15: 3 (Partial)

EXTENDING DEADLINE TO SUNDAY, 12 NOON

xpmath: 10 (good job)
shentang: 9
arb95: 8.5
andyliao: 8
Discrete_math: 8
FantasyLover: 7
randomtext: 5
AIME15: 3 (Partial)

lol i got the lowest score (besides AIME15 who didn't finish and everybody else who didn't submit and got 0's)

arb how'd you get an 8.5?

[quote="randomtext"]
arb how'd you get an 8.5?[/quote]
On problem 9, I lost partial credit for including two lattice points that are not in the first quadrant.
(The other points I provided were correct.) :oops:

lol which one did you miss? i epic failed by missing 4, 5, 7, 9, and 10.

OK, I think I'll post the next round tomorrow... Anyone know an easy way of mass pm?

There's no way to mass PM.  You have to send individually.

Why? Because if there were such an option, one could mass pm a heavily attached pm and crash the website? (I'm thinking *tinytim scandal* here xD.)

when will the answers to round 1 be revealed? at the very end?  :o


$ (X,M,\mu)$ is a measure space, if for each $ E\in M$ with $ \mu(E)\equal{}\infty$ there exists $ F\in M$ with
$ F\subset E$ and $ 0<\mu(F)<\infty$. Then if $ \mu(E)\equal{}\infty$, for any $ C>0$ there exists $ F\subset E,$ with
$ C<\mu(F)<\infty$

Some hints:
Suppose false.  Let $ a$ be the supremum of all finite $ \mu(F)$ with $ F \subseteq E$.
Show there is $ F_0 \subseteq E$ that achieves $ \mu(F_0) \equal{} a$.
Consider the complementary set $ E \setminus F_0$


i'm sorry.i don't know how to use Latex.
Rust's solution is nice and easy.but how to think of the solution when you meet the problem?


Find all positive integers $n$ for which the equation \[a+b+c+d=n \sqrt{abcd}\] has a solution in positive integers.

For $ n\equal{}1$: $ 4\plus{}4\plus{}4\plus{}4\equal{}\sqrt{4^{4}}$.
For $ n\equal{}2$: $ 2\plus{}2\plus{}2\plus{}2\equal{}2\sqrt{16}$.
For $ n\equal{}3$: $ 2\plus{}2\plus{}1\plus{}1\equal{}3\sqrt{4}$.
For $ n\equal{}4$: $ 1\plus{}1\plus{}1\plus{}1\equal{}4\sqrt{1}$.
Now, assume there was a solution $ a,b,c,d$ for $ n>4$, and take $ (a_{0},b_{0},c_{0},d_{0})$ the one where $ a\plus{}b\plus{}c\plus{}d$ is minimal, WLOG $ a_{0}\ge b_{0}\ge c_{0}\ge d_{0}$. Squaring both sides, it is natural to consider $ P(a)\equal{}a^{2}\minus{}a(n^{2}b_{0}c_{0}d_{0}\plus{}2(b_{0}\plus{}c_{0}\plus{}d_{0}))\plus{}(b_{0}\plus{}c_{0}\plus{}d_{0})^{2}$, the roots of which are integers $ a$ that are a solution of the problem. Since $ (b_{0}\plus{}c_{0}\plus{}d_{0})^{2}>0$, both roots $ a_{0}$ and $ a_{1}$ have the same sign, so both are solutions to the equation, and thus $ a_{1}\ge a_{0}\ge b_{0}\ge c_{0}\ge d_{0}$. So, $ 0\le P(b_{0})\leq (1\plus{}6\plus{}9)b_{0}^{2}\minus{}n^{2}b_{0}^{2}c_{0}d_{0}$, so $ n^{2}\le 16$, contradiction with $ n>4$.

The idea is correct but the writing is incorrect.
Consider P(a) = x^2-n*sqrt(bcd)*x+(b+c+d).
P(sqrt(b))>=0 leads to 4b>=n*b*sqrt(cd)>=n*b. Then n<=4.


Find $ Min(Max|\sqrt {x} \plus{} ax \plus{} b|)$, where $ a,b$ are 2 real numbers and $ x \in [0;16]$

Min and max over which variables in what order?

Sorry, i'll edit my problem:
Find $ a,b$ such that $ Max|\sqrt {x} \plus{} ax \plus{} b|$ is smallest possible, where $ a,b$ are 2 real numbers and $ x \in [0;16]$

Take $ x\equal{}y^2$, so $ y \in [0;4]$
Then $ ay^2\plus{}y\plus{}b$ has max value $ b$, $ 16a\plus{}4\plus{}b$, or $ \frac{1\minus{}4ab}{4}$
It seems like all three expressions could be made arbitrarily small (large negative) and therefore minmax is $ \minus{}\infty$

No, [b]me@home[/b]! The expression has a absolute value ( || ). Do you see it! :D It's not very easy!

It's not hard! kaka, please try to solve it!

[color=green]uhm, M&Y issue 1. Plz don't making a reply.  :) [/color]


HI, i was wondering if anyone could tell me why my avatar picture doesnt show up during class and how to fix this?

(My avatar picture is saved to my profile so thats not the problem.)

Next to the chat box in the bottom left corner, click that and you can change your avvy.

And note that there are restrictions on the avatar. (i.e. size, memory space)


algebra
polynomial
superior algebra
superior algebra unsolved

I think this is a classic problem, but searching for a polynomial seems impossible.

Why is $ x^p \minus{} x \plus{} 1$ irreducible in $ \frac{\mathbb{Z}}{p\mathbb{Z}}[x]$ for every prime?


algebra
polynomial
number theory unsolved
number theory

Give $ p$ integers $ x_1,x_2...x_p$ (not necessarily distinct).
Does that exist a polynomial $ Q \in Z[x]$ satisfied:
$ Q(i) \equiv x_i (mod p)$.
What is the minimal degree of $ Q$ (if it is existed).

we take $ P_i(X)=\prod_{i=1,i\neq j}^{p}(j-i)^{p-2}\prod_{i=1,i\neq j}^{p}(X-i)$
then $ \forall i\neq j: \ P(i)=0,P(j)=\prod_{i=1,i\neq j}^{p}(j-i)^{p-1}\equiv 1[p]$
so $ Q(X)=\sum_{i=1}^px_iP_i(X)$is solution.

did the problem say that $ p$ was prime?


limit
trigonometry
function
calculus
derivative
logarithms
calculus computations

1. Find $ \lim_{r \to 0} \dfrac{e^{rx} \minus{} 1}{r}.$

2. Find $ \lim_{x\rightarrow0}\frac {e^{x^2} \minus{} e^{3x}}{\sin(\frac {x^2}{2}) \minus{} \sin x}$

Expand those functions by using Taylor Series. :lol:

The first one is the definition of a derivative ... using Taylor series is excessive ;)

Thanks i managed to get both of them using l'hopitals theorem.

what about this one?

$ lim_{x \rightarrow \infty}(\frac{x}{2^x})$

Firstly, using l'hopital's rule, the limit is equal to 2^(-x) as x grows without bounds, which is 0.

Secondly, d/dx (x) =1 and d/dx 2^x=ln2*2^x, as such, even when x>0, the denominator grows faster than the numerator, meaning the limit as x approaches infinity is 0.

I've given #1 as a test question - and perhaps half of the students who set out to use L'Hôpital get it wrong! What happens is that people are so used to differentiating with respect to $ x$ that they fail to realize that they need to be differentiating that numerator with respect to $ r.$

JBL: Yes, I know it's a derivative. But that walks into the same human problem that I mentioned for L'Hôpital, of keeping straight what the variable is. People who expand out the series (and I do insist that my students have the power series for the exponential memorized) do tend to get the problem right.

Kono: that's another thing I insist that my students know. Call it the "Hierarchy of Size." Exponentials grow faster than polynomials. $ \frac{x}{2^x}$ may be an $ \frac{\infty}{\infty}$ case, but that exponential in the denominator grows faster than the mere power of $ x$ in the numerator. We [i]know[/i] that the limit has to be zero.

[quote="Kono"]Thanks i managed to get both of them using l'hopitals theorem.

what about this one?

$ lim_{x \rightarrow \infty}(\frac {x}{2^x})$[/quote]

Hi :) !!!

$ \frac{x}{2^x} \equal{} \frac{x}{e^{xln(2)}}\equal{} \frac{1}{ln(2)}\frac{xln(2)}{e^{xln(2)}}\equal{} \frac{1}{ln(2)} \frac{y}{e^y}$

$ \lim_{x \to \plus{} \infty} \frac{x}{2^x} \equal{} 0$

....
________________________________________
lahoucine

Hey,
[quote="mathema*"]
$ \frac {x}{2^x} \equal{} \frac {x}{e^{xln(2)}} \equal{} \frac {1}{ln(2)}\frac {xln(2)}{e^{xln(2)}} \equal{} \frac {1}{ln(2)} \frac {y}{e^y}$

$ \lim_{x \to \plus{} \infty} \frac {x}{2^x} \equal{} 0$
[/quote]
If we consider the limit,
$ {{\lim_{{x}\to{\infty}}}{\,}{\frac {x}{{a}^{x}}}} \equal{} {L}{\,}{\,}{,}{\,}{\,}{a > 0}$

Why was it needed to expand the numerator and denominator in to the form of,
$ {{\lim_{{x}\to{\infty}}}{\,}{\frac {x}{{a}^{x}}}} \equal{} {{\lim_{{x}\to{\infty}}}{\,}{{\frac {1}{C}}{\frac {Cx}{{b}^{Cx}}}}}{\,}{\,}{,}{\,}{\,}{b > 0}$

where, $ {{C} \equal{} {\log_{b}{a}}}$

It seemed redundant since if you just let $ {{C} \equal{} {1}}$ you just get back the original limit except with a different (constant) base in the denominator. It appears that the constants $ {a, b}$ don't change the limit $ {L}$ (even when one is bigger than the other) so long as both are greater than zero.

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]


A regular polygon has 27 diagonals. What is the number of degrees in the measure of an interior angle of the polygon?

We know that $ \frac{s(s\minus{}3)}2\equal{}27\implies s^2\minus{}3s\minus{}54\equal{}0\implies(s\minus{}9)(s\plus{}6)\equal{}0\implies s\equal{}9$.
Thus, our answer is $ 180^\circ\minus{}\frac{360^\circ}9\equal{}\boxed{140^\circ}$.

The sum of all exterior angles are 360, so 180-360/s=interior angle

@andyzhou,
If they already answered you don't need to reply again.  Mewto55555 was just making a helpful point


calculus
logarithms
trigonometry
probability
geometric series
calculus computations

This is already in several places. I'm making a new topic just so I know where to find it. Every calculus student should know these seven power series and be able to recall them instantly and to recognize them in a variety of settings. In each case, we have a series centered at zero (a Maclaurin series).

[b]Series related to the geometric series:[/b]

1. $ \frac1{1 \minus{} x} \equal{} 1 \plus{} x \plus{} x^{2} \plus{} x^{3} \plus{} \cdots \equal{} \sum_{n \equal{} 0}^{\infty}x^{n}$

* The geometric series itself; converges for $ x\in( \minus{} 1,1).$

2. $ \ln(1 \plus{} x) \equal{} x \minus{} \frac {x^{2}}2 \plus{} \frac {x^{3}}3 \minus{} \frac {x^{4}}4 \plus{} \cdots \equal{} \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \minus{} 1}x^{n}}n$

* The logarithm series; converges for $ x\in( \minus{} 1,1].$

3. $ \arctan(x) \equal{} x \minus{} \frac {x^{3}}3 \plus{} \frac {x^{5}}5 \minus{} \frac {x^{7}}7 \plus{} \cdots \equal{} \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^{k}x^{2k \plus{} 1}}{2k \plus{} 1}$

* The artangent series; converges for $ x\in[ \minus{} 1,1].$

[b]Series related to the exponential:[/b]

4. $ e^{x} \equal{} 1 \plus{} x \plus{} \frac {x^{2}}2 \plus{} \frac {x^{3}}6 \plus{} \frac {x^{4}}{4!} \plus{} \cdots \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x^{n}}{n!}$

* The exponential series; converges for all $ x.$

5. $ \cos x \equal{} 1 \minus{} \frac {x^{2}}2 \plus{} \frac {x^{4}}{4!} \minus{} \frac {x^{6}}{6!} \plus{} \cdots \equal{} \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^{k}x^{2k}}{(2k)!}$

* The cosine series; converges for all $ x.$

6. $ \sin x \equal{} x \minus{} \frac {x^{3}}6 \plus{} \frac {x^{5}}{5!} \minus{} \frac {x^{7}}{7!} \plus{} \cdots \equal{} \sum_{k \equal{} 0}^{\infty}\frac {(\minus{}1)^{k}x^{2k \plus{} 1}}{(2k \plus{} 1)!}$

* The sine series; converges for all $ x.$

[b]The binomial series:[/b] Let $ \alpha$ be an arbitrary real number.

7. $ (1 \plus{} x)^{\alpha} \equal{} 1 \plus{} \alpha x \plus{} \frac {\alpha(\alpha \minus{} 1)}2x^{2} \plus{} \frac {\alpha(\alpha \minus{} 1)(\alpha \minus{} 2)}{3!} \plus{} \cdots$

$ \equal{} \sum_{n \equal{} 0}^{\infty}\binom{\alpha}{n}x^{n}$ where $ \binom{\alpha}{n} \equal{} \frac {\alpha(\alpha \minus{} 1)\cdots(\alpha \minus{} n \plus{} 1)}{n!}.$

* Newton's favorite series. The radius of convergence is $ 1$ (unless $ \alpha$ is a nonnegative integer, in which case it is a polynomial). Convergence or lack of convergence at the endpoints depends on the value of $ \alpha.$

There are, of course, other series that aren't too far behind. The hyperbolic cosine and hyperbolic sine could be placed in the exponential grouping - their series are quite easy to remember, anyway, if you know the trigonometric sine and cosine. The inverse hyperbolic tangent, which can also be written as $\frac12\ln\left(\frac{1+x}{1-x}\right),$ could be added to the "related to the geometric series" grouping. You could differentiate geometric series (and further manipulate) to get things like $\sum_{n=1}^{\infty}nx^{n}.$ You could integrate a binomial series to get a series for the arcsine. Some would leave the binomial series off of this list - for my own sake, I find it far too useful to ignore.

I hand out this list every time I teach Calculus II - and many times in following courses (probability, differential equations, and so on) when I think students need to be reminded of the series they're supposed to know.

Hey what about the logarithmic series for the right endpoint?
Dont you like it because Abel's theorem is advanced for a Calculus class?
I would still mention it.

I did mention it - I said that series converges for $x\in(-1,1].$ I was explicitly including the right endpoint, as well as both endpoints for the arctangent series, which is the same issue.


The sequence of [i]Ulam numbers[/i] $ u_1, u_2, u_3,\ldots$ is defined by $ u_1\equal{}1$, $ u_2\equal{}2$, and for $ n>2$, $ u_n$ is the smallest number greater than $ u_{n\minus{}1}$ that can be uniquely expressed as the sum of two distinct Ulam numbers. Prove that there are infinitely many Ulam numbers.

[hide="Solution"] The algorithm that generates Ulam numbers fails if the set of numbers greater than $ u_{k}$ that can be uniquely expressed as the sum of two distinct Ulam numbers in $ \{ u_1, u_2, .. u_k \}$ is empty.  We claim that this set, which we will abbreviate $ U_k$, is nonempty for all $ k$.

Clearly $ 3 \in U_2$.  Our inductive hypothesis is that $ U_2, U_3, ... U_k$ were all non-empty.  Then we have at least $ k$ Ulam numbers $ u_1, u_2, ... u_k$ and we wish to show that $ U_{k\plus{}1}$ is non-empty.

Note that the definition of Ulam numbers requires that the sequence be monotonically increasing.  Hence $ u_{k\minus{}1} \plus{} u_{k} \in U_{k\plus{}1}$.  The conclusion follows. [/hide]


The sum of a positive number and its square is 156. What is the number?

We have $ x^2\plus{}x\equal{}156 \implies x^2\equal{}x\minus{}156\equal{}0 \implies (x\plus{}13)(x\minus{}12)\equal{}0$. Since $ x$ is positive $ x\equal{}\boxed{12}$.


The wheels of a car have a 24-in. diameter.  When the car is being driven so that the wheels make 10 revolutions per second, how far will the car travel in one minute?

Please show all work in detail.

Thank you.

[hide="answer"]The formula for the circumference of a circle is $\pi d$, so the circumference of a wheel is $24\pi$ inches, or $2\pi$ feet, and the car goes $10\times 2\pi=20\pi$ feet in 1 second.  Therefore, the car goes $60\times 20\pi=1200\pi$ feet in 1 minute.[/hide]

First we need to find out how many revolutions are made in one minute.
If 10 revolutions are made in a second, that means that 600 are made in a minute.
So we simply multiply the circumference of the wheel which is $24\pi$ times the number of revolutions made in a minute which is $600$, and the answer is $14400\pi$ $in.$or about $45238.934$ $in$.

dank you should have multiplied by $20\pi$ and gotten $1200\pi$  feet

Yeah, I just noticed that when I read your first post and changed it...

So, we are in agreement that Jhonny is correct, yes?

Thank you guys for your efforts.  :)

[quote="K81o7"]Should this even be in Intermediate?
...Seems suspiciously like homework...[/quote]

Thank you for your concern.  This is not homework.  It is for a detailed PowerPoint presentation I am giving to be presented to investment professionals, where I make mathematical comparisons to investment models.  I do not want to be incorrect for the presentation.

I have several more questions, but if this great forum would rather me not post, I will respect the decision.

Thank you.

Yeah that's right that could have been posted in the Getting started forum.

I don't see anything wrong with you posting the problems, although this probably should have gone in Getting Started...

It should probably go to a different forum though...
Still...for a professional, this problem should be simple...you could just check your answers with a calculator...


(groan)
come on!
This is the 3rd thread on movies currently active, not to mention that ruddy sticky!
What are the moderators doing I say?


geometry circumcircle perimeter ratio trigonometry

In an acute triangle $ ABC$, let $ D$ be a point on $ \left[AC\right]$  and $ E$ be a point on $ \left[AB\right]$ such that $ \angle ADB\equal{}\angle AEC\equal{}90{}^\circ$. If perimeter of triangle $ AED$ is 9, circumradius of  $ AED$ is $ \frac{9}{5}$ and perimeter of triangle $ ABC$ is 15, then $ \left|BC\right|$ is

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\  \frac{24}{5} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\  \frac{48}{5}$

[hide="Solution"]
Because $ BCDE$ is cyclic, $ \angle ABC = \angle ADE$ and $ \angle ACB = \angle AED$. So $ ADE \sim ABC$. The similarity ratio is $ \frac {9} {15}$.
$ \dfrac {AD} {AB} = \dfrac {9} {15} = \dfrac {3} {5}$. So $ \sin  \angle A = \frac {4} {5}$.
The ratio of circumradius is also included in similarity. So the circumradius of $ ABC$ is $ \frac { \frac {9} {5}} {R} = \frac {9} {15} \Rightarrow R = 3$.
In the triangle $ ABC$, $ \frac {BC} {\sin\angle A} = 2R \Rightarrow R \equal{} \frac {24} {5}$
$ \boxed B$.
[/hide]


that's correct ... this problem was used in the JBMO Rom TST in 2001, at Targu Mures.
Very Easy problem.


geometry
circumcircle
geometric transformation
geometry proposed

Let A,B,C be the points on sides $B_1C_1,C_1A_1,A_1B_1$ of $\triangle A_1B_1C_1$ such that $\triangle ABC\sim\triangle A_1B_1C_1$. Suppose O and H be the orthocenter and circumcenter of $\triangle ABC$, H' the orthocenter of $\triangle A_1B_1C_1$. Prove that OH=OH'.

Assuming it's well-known that $H$ is actually the circumcenter of $A_1B_1C_1$ :):

Fix the triangle $A_1B_1C_1$ and move $A,B,C$ s.t. the conditions are satisfied. The locus of $O$ is the image of the locus of $A$ through the spiral similarity centered at $H$ which takes $A$ to $O$, and, since the locus of $A$ is the line $B_1C_1$, it means that the locus of $O$ must be a line which makes with $B_1C_1$ the same angle that $OH$ makes with $HA$. Since the figures $ABCHO$ and $A_1B_1C_1H'H$ are similar, this angle is equal to the angle that $H'H$ makes with $H'A_1$, and, because $H'A_1\perp B_1C_1$, we find the locus of $O$ to be orthogonal to $H'H$.

When $A$ is the midpoint of $B_1C_1$, the triangle $ABC$ is the medial triangle of $A_1B_1C_1$, and $O$ is the midpoint of $HH'$. This means that the locus of $O$ is a line passing through the midpoint of $HH'$ and perpendicular to $HH'$, and this shows that $OH=OH'$.

Nice proof, grobber :)
My proof is more or less the same as yours, but only considering $\triangle A_1B_1C_1$ instead.

[quote="suweijie"]I am interested in the so called [color=red]geometry problems collection[/color],where did you find it,net or books?[/quote]
I found it from nowhere, but the file in my computer-it was created by me :)


Why is there no Heinsenberg uncertainty in calculating the path of a bullet?

If by "bath" you mean "path", then it's because a bullet is pretty huge...


geometry
geometric transformation
rotation

Source: Massachusetts Mathematics League December 2005

$A, B,$ and $C$ are consecutive vertices of a regular polygon with less than 25 sides. A rotation centered at $B$ maps $A$ onto $C$ and $C$ onto $P$. If $A, B$, and $P$ are not collinear find the difference between the maximum and minimum possible values of $m\angle {ABP}$.

[quote="Altheman"][quote="cincodemayo5590"][hide]Drawing a diagram, we can see that $m\angle {ABP} = 360 - 2\theta$, where $\theta$ is the interior angle of the polygon. Call $n$ the number of sides of the polygon. Then,

$m\angle {ABP}=360-2(\frac{180(n-2)}{n}) = 360(\frac{2}{n})$.
So the difference is $360(\frac{2}{3})-360(\frac{2}{25}) = \boxed {\frac{1056}{5}}$[/hide]

(edit: forgot to hide)[/quote]

1. i believe the angle should be $0<\theta<180$, i suppose the problem is ambigious, but that is still the convension when we describe angles
2. there cannot be 25 sides[/quote]

[hide]ok so the difference is $120-360(\frac{2}{25}) =\boxed{\frac{456}{5}}$.  What do you mean by there can't be 25 sides?[/hide]

edit-darn hiding....

[quote="Altheman"][quote="cincodemayo5590"]

[hide]ok so the difference is $120-360(\frac{2}{25}) =\boxed{\frac{456}{5}}$.  What do you mean by there can't be 25 sides?[/hide]

edit-darn hiding....[/quote]

[quote="weatherbottle"] a regular polygon with [b]less than[/b] 25 sides.[/quote]

also, there is still a larger angle...lol[/quote]
oh.. meh

fine... $120-\frac{360(2)}{24}=\boxed{90}$

(grr...)


Find the sum of all $ 3$-digit natural numbers which contain at least one odd digit and at least one even digit . This problem carries 15 marks

This is elementary. Required sum= sum of all 3 digit natural numbers - sum of numbers with all digits even - sum of numbers with all digits odd.
To find that sum, take mathematics teacher and take subjunior paper and you will see the solution

@goutham:
are you attending rajesh sirs classes?


Is it feasible for me to take an intro abstract algebra course and calc 3 (stewart). I am pretty good at writing proofs. What do you guys think?

Thanks

There are some introductory abstract algebra courses that are aimed at that level, so if you are willing to do the work that could be possible. See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]FAQ on Preparing for College Math Study[/url] for suggestions about how to get ready.

tokenadult, isnt abstract algebra completely independent of calculus? Also I am studying Courants Differential and Integral Calculus, and Spivaks calculus. Then moving on to Rudin. So I guess abstract algebra shouldnt be that difficult? I have had linear algebra (Hoffman/Kunze).

They are independent, but abstract algebra assumes a level of mathematical maturity which is well beyond the standard calculus sequence. You have the content prerequisite (linear algebra), and it sounds like you can probably handle the course.

Here is the book being used. From what you can see, is this manageable?

[url=http://www.amazon.com/First-Course-Abstract-Algebra-Groups/dp/1584885157/]A First Course in Abstract Algebra[/url]

[quote="ragingbullrocky"]Here is the book being used. From what you can see, is this manageable?
[/quote]

Yes.  Anderson/Feil is totally manageable (and I'm glad to see that it's back in print).

Thanks for the reply. Yeah, one of them is a professor at my school.


[b]Find the amplitude of $ f(x)\equal{}3\sin{x}\plus{}\cos{x}$[/b]

please show your complete though process including any identities you are using?

Isn't there a formula that the amplitude of $ y\equal{}a sin x\plus{}b cos x\equal{}\sqrt{a^{2}\plus{}b^{2}}$.
But if you don't wanna use that, I guess you can compare it to $ sin(x\plus{}y)\equal{}sin x cos y\plus{}cos x sin y$? and look for values? But I find that a little time-consuming...

it's true there is such a formula ( if u can call it that), but that's exactly what im trying not to use  :blush: because i didn't understand the book's derivation of that rule, so i wanted to see how you guys would explain it.

We have
$ a\cos x\plus{}b\sin x$

Consider

$ \frac{a}{\sqrt{a^{2}\plus{}b^{2}}}\cos x\plus{}\frac{b}{\sqrt{a^{2}\plus{}{b^{2}}}}\sin x$

$ \left(\frac{a}{\sqrt{a^{2}\plus{}b^{2}}}\right)^{2}\plus{}\left(\frac{b}{\sqrt{a^{2}\plus{}b^{2}}}\right)^{2}\equal{} 1$ so there exists a $ \theta$ such that $ \sin\theta \equal{}\frac{a}{\sqrt{a^{2}\plus{}b^{2}}}$ and $ \cos\theta \equal{}\frac{b}{\sqrt{a^{2}\plus{}b^{2}}}$. Then, $ a\cos x\plus{}b\sin x \equal{}\sqrt{a^{2}\plus{}b^{2}}(\sin\theta\cos x\plus{}\cos\theta\sin x) \equal{}\sqrt{a^{2}\plus{}b^{2}}\sin(\theta\plus{}x)$, so the amplitude is $ \sqrt{a^{2}\plus{}b^{2}}$.

Another derivation using calculus:

Let $ f(x) \equal{} a\cos{x}\plus{}b\sin{x}$ hence we have $ f'(x) \equal{}\minus{}a\sin{x}\plus{}b\cos{x}$ and $ f''(x) \equal{}\minus{}f(x)$.

When $ f'(x) \equal{} 0\Rightarrow\frac{b}{a}\equal{}\tan{x}$. Therefore at this point we have $ f(x) \equal{}\frac{a^{2}}{\sqrt{a^{2}\plus{}b^{2}}}\plus{}\frac{b^{2}}{\sqrt{a^{2}\plus{}b^{2}}}\equal{}\sqrt{a^{2}\plus{}b^{2}}$.

Since this value of $ f(x) > 0$ and so $ f''(x) < 0$ at this point and therefore it is a maximum.

So the amplitude of $ f(x)$ is $ \sqrt{a^{2}\plus{}b^{2}}$.

[quote="mihail911"][b]Find the amplitude of $ f(x) \equal{} 3\sin{x}\plus{}\cos{x}$[/b]

please show your complete though process including any identities you are using?[/quote]

Or we can write the given equation in form $ A\sin(x\plus{}y) \equal{} A\sin{x}\cos{y}\plus{}A\cos{x}\sin{y}$ Now we get that $ A\cos{y}\equal{} 3$ and $ A\sin{y}\equal{} 1$ since $ \sin{\theta}^{2}\plus{}\cos{\theta}\equal{} 1$. So we get that $ {3}A^{2}\plus{}{1}A^{2}$ and A is our answer.

EDIT: yea thanks t0rajirou I guess what I said was confusing since it wasn't exactly accurate

[quote="sonny"][quote="mihail911"][b]Find the amplitude of $ f(x) \equal{} 3\sin{x}\plus{}\cos{x}$[/b]

please show your complete though process including any identities you are using?[/quote]

Or we can write the given equation in form $ \sin(x\plus{}y) \equal{}\sin{x}\cos{y}\plus{}\cos{x}\sin{y}$ Now we get that $ \cos{y}\equal{} 3$ and $ \sin{y}\equal{} 1$[/quote]

This is a somewhat contorted way to express your thoughts.  Ideally, you should instead write

$ f(x) \equal{} A\sin (x\plus{}y) \equal{} (A\cos y)\sin x\plus{}(A\sin y)\cos x$

Where $ A$ is the amplitude and $ A\cos y \equal{} 3, A\sin y \equal{} 1, A^{2}(\cos^{2}y\plus{}\sin^{2}y) \equal{} A^{2}\equal{} 10\implies A \equal{}\sqrt{10}$.  This is just a rewording of Hamster1800's solution.


abstract algebra
group theory
superior algebra
superior algebra unsolved

If G is an abelian group of order (p^t)m, and (p,m)=1, show that G(p) has order p^t

and G(p) = {a e G| |a|=p^m  where m is a natural number}

any suggestions

if $ G$ is a abelian torsion group, $ G$ is the direct sum of the $ G(p)$ where $ p$ runs through the primes (the directness is easy to see, and the sum decomposition comes from bezout's lemma applied to the prime powers in the order of an element of $ G$).

lol...kk...But G isn't necessarily a torsion group?

Well...Actually, I could use Cauchy's theorem for Abelian Groups...

"If G is a finite abelian group and p is a prime that divides |G|, prove that G contains an element of order p...."

we know that p divides |G|...and p2 divides the order of G...and so on and so on until we have pt divides the order of G...so all these elements exist in G(p)..then use the (p,m)=1 to show that each one of these elements is uniquely the product of m with some interger...I think that is it...

[quote="Frank1987"]But G isn't necessarily a torsion group?[/quote]Hmm, in a  [b]finite[/b] group $ G$, $ g^{|G|}\equal{}e$ for every $ g\in G$ (Lagrange). Hence $ G$ is a torsion group in your case.  :wink:

Nevermind..the cauchy way will NOT work..at all... That was bad thinking...

So, I think I need to do the G = G(p1) + G(p2)+..+ G(pk) where these are the distinct primes that divide the order of G....Now one of these is surely my G(p)....So G = G(p) + M, where M is the sum of all the other stuff... Even with this...how do I know that |M| = p^t*(m-1) ...or more formally that |G(p)| = p^t ....

a finite $ p$-group (i.e. the order of every element is a power of $ p$) has $ p^n$ elements for some $ n$. this follows from cauchy's theorem. apply this to the $ G(p)$.

Nice...So we know that G(p) has order of some power of p...but still doesn't say that it must be p^t....There is something fundamental that I a missing here...or you are making giant leaps to conclusions...

you could fill in the details if you made some effort ..

every $ G(p)$ has order $ p^{v_p}$ for some $ v_p$ and because $ G$ is the direct product of the $ G(p)$, the order of $ G$ is the product of the $ p^{v_p}$. thus the $ v_p$ are the exponents of $ p$ in $ |G|$.


group theory
abstract algebra
limit
topology
linear algebra
linear algebra unsolved

Let $H$ be a subgroup of ${\mathbb R}$. Show that $\forall x \in {\mathbb R}, \exists x_n \in H$ such that $\lim_{n \to \infty}x_n = x$.

 :ninja:

This is obviously false. What did you really mean?

Well, it is true if you suppose that the group $H$ isn't generated by a single element, that is, $H$ isn't of the form $a\mathbb{Z}$ with $a\in\mathbb{R}$.
That's to say : "Every subgroup of the real numbers is either dense in the euclidean topology or generated by a single element."


It's a kind of problem that most people struggle with...

Insert parentheses to make each statement true...

4 + 16 / 2 + 3 * 5 = 20

This is pretty interesting

[color=darkred](made it more legible)[/color]

[quote="shinwoo"]that's correct!

how long did it take you to solve the problem?

it actually took me for awhile  :blush:[/quote]

 umm.. it took me about...close to a minute.

An interesting extension to the problem: How many different values can you make by just adding parentheses?

I'm guessing that your guys used guess and check to find the answer.  If you are, then you will probably get different times for solving the problem, since one person might have a lucky guess, and another might not.

Is there any other way to solve this problem besides guess and check though?

I'd suggest putting that extension in MATHCOUNTS or even Getting Started!

I'll put it up on the Getting Started section

Insert parentheses to make each statement true...

4 + 16 / 2 + 3 * 5 = 20

This isn't exactly another way to do the problem, but it can narrow down the number amount of time spent on guess and check on the problem.

Parentheses only make a difference when placed around signs that should go later; therefore placing parentheses around 16 / 2 or 3 * 5 makes no sense because it will not change the problem. That cuts down the number of guesses needed. :ninja:

Err...the last post was 4 years ago (since 2009 is starting) and you are posting 4 years later..the problem was solved anyway...so don't revive topics unless you have a whole different way to solve it...


Why can't I type Latex today, even for simple expression like $a+b\geq\sqrt{ab}$? (I was typing [b]a+b\geq\sqrt{ab}[/b] )
After some previewing, I find that most commands beginning with [b]\[/b] cannot be shown. Besides, all the newly posted topics have this problem.
Is there any problem in Mathlinks today?

Indeed there is a problem. I will see to it ...


What is the reciprocal of $ \frac{1}{2}\plus{}\frac{2}{3}$?

$ \textbf{(A)}\ \frac{6}{7} \qquad
\textbf{(B)}\ \frac{7}{6} \qquad
\textbf{(C)}\ \frac{5}{3} \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ \frac{7}{2}$

[hide]
$ \frac{1}{2}\plus{}\frac{2}{3}\equal{}\frac{7}{6}$, so the reciprocal is $ \frac{6}{7}$.
[/hide]

Hey, I'm going to demonstrate what I'm talking about when I say that postfarming is on the rise. Please try to learn from what I say here.

[quote=sugar_rush][hide=solution]$\cfrac{1}{\cfrac{1}{2}+\cfrac{2}{3}}=\cfrac{1}{7/6}=\boxed{\textbf{(A)} ~\cfrac{6}{7}}$[/hide][/quote]

This solution doesn't have any explanation, just mindless computation. It's also already been done above. Therefore, this post is basically pointless. However, it's not [i]horrible[/i], since it at least provides a semblance of a good solution.

[quote=OTMS_MATHCLUB]was this bump really necessary?[/quote]

I understand this comment, but making it a standalone is completely unnecessary. Please try to include something relevant other than your comment when making posts like this.

[quote=pog][hide=Solution]
Note that $\left(1-\dfrac12\right)+\left(1-\dfrac13\right)=\dfrac12+\dfrac23$, so since $\dfrac1a+\dfrac1b=\dfrac{a+b}{ab}$, we get that $$\left(1-\dfrac12\right)+\left(1-\dfrac13\right)=2 - \left(\dfrac12 + \dfrac13\right)=\dfrac{12-5}{6},$$ which has reciprocal $\dfrac{6}{12-5}=\boxed{\textbf{(A) }\dfrac67}$.
[/hide][/quote]

This is an interesting new solution. It's a little more complicated than necessary but I see nothing wrong with it. Great post.

[quote=RohanQV][quote=OTMS_MATHCLUB]was this bump really necessary?[/quote]

 If you have nothing to contribute, don't comment. :thumbup:

[hide=Solution]First, we can find a common denominator by multiplying $\frac{1}{2}$ by $3$ and $\frac{2}{3}$ by $2$ getting $\frac{3}{6}+\frac{4}{6}=\frac{3+4}{6}=\frac{7}{6}.$ Now, we take the reciprocal, getting $\frac{1}{\frac{7}{6}}=\frac{1}{1}\cdot\frac{6}{7}=\frac{6}{7}$ so we get the answer of $\boxed{A}.$[/hide][/quote]

Fine post, though this solution was already basically stated above. For the record, I have to agree with the first part of your post.

[quote=RedFireTruck]$\frac{1}{\frac12+\frac23}=\boxed{\frac67}$[/quote]

This is quite plainly not a good post, though as with sugar_rush it's not [i]horrible[/i]. How did you get that expression and why is that what you want? There is no explanation and it's hard to understand what you're doing. You also haven't hidden your solution (though I forget to do so a lot, don't worry).

[quote=cugel2020]de answer is $a$[/quote]

This post has no proper punctuation, and are you saying the answer is the variable $a$ or the answer choice? Low quality post.

[quote=PEKKA]this thread is older than me[/quote]

Completely off-topic. Please state such observations only in your own head.

[quote=RedFireTruck][quote=PEKKA]this thread is older than me[/quote]

this thread is 8 1/2 months older than me :o[/quote]

Same as above.

[quote=PEKKA]2 1/2 for me[/quote]

Same as above.

[quote=Ladka13]Bruh legit this thread is older than me. But I'm catching up. The difference in our ages was once a few months and now it's .... still a few months :huh:?[/quote]

Same as above.

To cap this off, I have no intention of post farming, so here's an example of a solution I would deem to be fairly good, despite the fact that this is a fairly easy problem.

[hide=Solution]Firstly, we calculate the value of the expression we need to take the value of. This expression is just $\frac{1}{2}+\frac{2}{3}$. Writing over a common denominator, we get $\frac{3}{6}+\frac{4}{6}=\frac{7}{6}$. Taking the reciprocal gives $\frac{6}{7}$, or $\boxed{\text{(A)}}$.[/hide]

[quote=MWZ]just a quick question: why is problem 2 of an AMC12 so easy? (I haven't done AMC12s, but I thought they were supposed to be hard)

also, @above back with the trolling, so REPORTED[/quote]

Typically, the first few on AMC 12 are really trivial.

[quote=MWZ]just a quick question: why is problem 2 of an AMC12 so easy? (I haven't done AMC12s, but I thought they were supposed to be hard)

also, @above back with the trolling, so REPORTED[/quote]

this is a 2008 contest and usually older tests are easier

Oh ok. Well, I knew the older ones were easier, but I didn’t know 2008 would be this easy :/
Ty everyone though!

And these days, I’ve been seeing more trolls...

[quote=VroomVroom][quote=PEKKA]this thread is older than me[/quote]

yeah apparently it's 2 years older than me[/quote]
[i]
Completely off-topic. Please state such observations only in your own head.[/i]


induction
Gauss
vector
binomial coefficients
combinatorics unsolved
combinatorics

let $0 \leq k \leq n $and $ k , n \in  Z $ . $C(n,k )$ are numbers with following conditions such that:

1-$C(n,n)=C(n,0)=1$
2-$C(n+1,k)=2^kC(n,k)+C(n , k-1)$  , $n \geq k \geq 1$

Show that : $C(n , k)=C(n, n-k)$   ,   $n \geq k \geq 0$

easy problem (jane to :P )
for every natural $n$ , $A_n=\prod_{i=1}^n (2^i-1)$ we can prove with induction $C(n,k)=\frac{A_n}{A_k.A_{n-k}}$,  $(1)$

now from $(1)$  we have $C(n,k)=C(n,n-k)$.

The $2^k$ can be replaced by $q^k$ and the result is $q$-binomial coefficients used by Gauss and Euler.  They enumerate subspaces of dimension $k$ in an $n$-dimensional vector space over $\mathbb{F}_q$.   It is a bad problem for competitions, as it is trivial if you already have seen it, and difficult otherwise.


function
real analysis
real analysis solved

The problem I posted is eronated, because the limit is oo and it has no utility! :)

I think it should be:
Compute lim(n-->oo) n*((4an)-2^(en)).
case(oo*0)

what do you think? any idea to compute it?
cheers! :D :D

[quote="Lagrangia"]... (an)n>=2;
an=[2*(2*...*2^(1/n))^(1/3)]^1/2.
[/quote]

the expression of a(n) is not clear, could you precise a(2);a(3);a(4);a(5)

i think it is equivalent with this:
a_n=2^(1/2!+1/3!+1/4!+...+1/n!)

Kuba is 100% right, the sequence is:
a_n=2^(1/2!+1/3!+1/4!+...+1/n!)

I think that o(n) is Landau's notation, and f(n) is a function which tends to 0!

cheers!


vector
algebra
polynomial
linear algebra

Find an example of a vector space $ V$ and a proper subspace $ S$ of $ V$ for which $ V \approx S$ ( $ V$ isomorhic to $ S$)

My question is what is a proper subspace and how does it differ from a subspace/vector space? Also, do all proper subspaces have to be isomorphic to the vector space?

Any subspace of $ V$ is a proper subspace, except for $ V$ itself.
A proper subspace of a finite-dimensional space $ V$ cannot be isomorphic to $ V$, because it has smaller dimension. Therefore, the example you are looking for must be infinite-dimensional.

well then what are simple examples of a proper subspace?

and what do you mean by

"Any subspace of $ V$ is a proper subspace, except for $ V$ itself. "?

I'm not sure how mlok could have been any clearer with that statement than he was.

Here's an example. Let $ V$ be the vector space of all polynomials (in the variable $ x.$) Let $ S$ be the set of all polynomials in $ V$ whose constant terms are zero.

Define $ T: V\mapsto S$ by $ T(f)(x)\equal{}xf(x).$ $ T$ is an isomorphism from $ V$ onto $ S,$ so $ V$ and $ S$ are isomorphic.

sorry, it was really late at night when i read that statement, that is a REALLY clear definition of a proper subspace. whoops.

but thank you!


geometry
3D geometry
sphere
topology
advanced fields
advanced fields unsolved

show that $(L^1([0;1], \lambda),|.|_1)$ is not the dual space of any Banach space.

Can this work?

Suppose the contary: $L^1([0,1])$ is the dual of $X$. Then, by Banach-Alaoglu theorem we must have that the unit  sphere in $L^1([0,1])$ must be compact in weak* topology, which would imply by Krein-Milman that this unit sphere must have extreme points. It is only left to prove that the unit sphere of  $L^1([0,1])$ has no extreme points.

It is what I thought  :)  (the non existence of extreme points is straightforward ..)


n is an natural number. assume that f1, f2,...,fn are polynomials with integer coefficients .prove the existance of polynomial g(x) with integer coeficients such that g(x) is reducible but fi(x)+g(x) are irreducible.

[quote="sam-n"]$n$ is an natural number. Assume that $f_{1}, f_{2}, \ldots, f_{n}$ are polynomials with integer coefficients. Prove the existance of polynomial $g(x)$ with integer coeficients such that $g(x)$ is reducible but $f_{i}(x)+g(x)$ are irreducible.[/quote]

[hide="Hint"]Use Eisenstein's criterion.[/hide]

i remember myth even solved without eisenstain criteria, i dont find the thread.anyway with this wellknown criteria the solution is cuter :)

Here is the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5051]link[/url]. Problem can be solved with Einstein Criterion.


Show that the product of the two real roots of the equation $ x^4\plus{}x^3\minus{}1\equal{}0$ is a root of the equation $ x^6\plus{}x^4\plus{}x^3\minus{}x^2\minus{}1\equal{}0$

[quote="mkn19"]Show that the product of the two real roots of the equation $ x^4 \plus{} x^3 \minus{} 1 \equal{} 0$ is a root of the equation $ x^6 \plus{} x^4 \plus{} x^3 \minus{} x^2 \minus{} 1 \equal{} 0$[/quote]

Let $ x^4\plus{}x^3\minus{}1\equal{}(x^2\plus{}ax\plus{}b)(x^2\plus{}cx\plus{}d)$
We get :
$ a\plus{}c\equal{}1$
$ b\plus{}d\plus{}ac\equal{}0$
$ ad\plus{}bc\equal{}0$
$ bd\equal{}\minus{}1$

So :
First line implies $ c\equal{}1\minus{}a$ and last line implies $ d\equal{}\minus{}\frac 1b$. Setting this in third line, we get $ a(\minus{}\frac 1b)\plus{}b\minus{}ab\equal{}0$ and so $ a\equal{}\frac{b^2}{b^2\plus{}1}$ and $ c\equal{}\frac{1}{b^2\plus{}1}$

Setting these values in second line, we get $ b\minus{}\frac 1b\plus{}\frac{b^2}{(b^2\plus{}1)^2}\equal{}0$ and so

$ b^6\plus{}b^4\plus{}b^3\minus{}b^2\minus{}1\equal{}0$

Hence the result (which is true for the product of any two roots of the original equation, not only the real ones).


calculus
probability
vector
geometry
trigonometry
topology
number theory

What is your favorite subtopic of math?
Personally, I like algebra the most.

Please justify your answers below.  I just like to know what other people like :lol:.

I'd argue that "Proofs" isn't really a subtopic of math, it's more of a method to show that you solved something... I picked Number Theory though, it's IMO the most elegant subject and is the "queen of mathematics."

I dunno, I like all the branches of math :D  I went with Number Theory though.  Yeah, proofs isn't a separate topic.  Insteads proofs supply the rigor needed to build up [b]all[/b] branches of mathematics from the axioms.  Also, Combinatorics was a popular one that you left off as well as Calculus

I find probabilty and stats, vector calculus, and linear algebra to be appealing, along with most physics topics.

Right now, I like algebra but the fact that I'm pretty bad at analysis probably has a lot to do with that. Those are pretty much the only two branches of mathematics I've learned to any real extent so far.

I like geometry the most because I'm good at it :)
I don't like probability because I'm not good at it :P

I like all.
But what exactly is 'proofs'?

i like complex numbers and the ways you can connect that with trig (at the high school math level- i haven' t studied past calculus yet)

i kinda like geometry

finding a perpendicular bisector is fun! (with a compass and a straight edge)

I put proofs in there for people who don't really have another favorite subtopic, but they like proofing things.  I know its not really a subtopic, but I thought it might be a good thing to have in there.

For me, tied is trig and geometry.

I've noticed some people are really lopsided in their strength in certain math areas. Like Iversonfan (the Geometry master :) !). Others, like me, aren't particularly good in any subject.

don't really have a favorite. i guess algebra and trig are at the top of the list.

[quote="curl"]My current favorite is analysis, mostly because of how it relates well to physics. Proofs and constructions are often very elegant as well. My avatar shows off my favorite theorem from the subject  :)[/quote]
Real or complex?

I voted for number theory because it tends to have the most elegant solutions. I like all of those topics, though. ;)

I like Number Theory most (voted for it), and some closely related topics: abstract Algebra (is it part of the algebra mentioned in the poll¿) and complex analysis.

i like algebra over geometry because its [b]so[/b] much easier

[quote="daveed"]Haha yeah, geometry solutions sometime seem to come out of nowhere... ; :huh:[/quote]
Which is why geometry is so fun! :)

Number Theory Rocks ! (I love analytic num. theory and everything that is more than division theory ;))
Also abstract/linear algebra is nice.
Edit BTW: Why didn't you put analysis ? Topology ? etc.

[quote="PenguinIntegral"][quote="curl"]My current favorite is analysis, mostly because of how it relates well to physics. Proofs and constructions are often very elegant as well. My avatar shows off my favorite theorem from the subject  :)[/quote]
Same here![/quote]
What he said.

By geometry do you mean university kind of geometry, finite fields, and polarities and stuff, or euclidean geometry?

I like Number Theory best, it is the most basic kind of maths and all other kinds serve it.
Next I like algebra , then geometry (university kind)
probability i don't really like

It is my understanding at university that many people like algebra/number theory and all that stuff, but that nobody really tends to kick on calculus/complex analysis/functional analysis

Are the 'young' not interested in analysis (i am not by the way, analysis is one idea and a thousand of seemingly meaningless conditions on top of it)

[quote="fredbel6"]Are the 'young' not interested in analysis (i am not by the way, analysis is one idea and a thousand of seemingly meaningless conditions on top of it)[/quote] I think it is the way the subject is taught. Algebra is problem solving and open-book exam, statistics/dynamical systems/all other applied maths is open book and problem solving, but analysis is closed book and purely brainless reproduciton and memory torture. I hope it will change, but if this does not change I'm not going to continue in analysis either...

I got analysis from an assistant in Ghent doing his doctoral thesis in geometry/coding theory, simply because nobody was doing research in analysis  :o


Define $ X\equal{}\mathbb{Q}\cap [0,1]$. Is this true : $ X \cong X\times X$  ?

yes. it is true. both are countable sets,

we can rewrite $ X\equal{}\{r_1, r_2, ...\}\cong Z\cong Q\cong Z\times Z \cong X\times X$

I don't buy that. We're talking about homeomorphism. Your justification is that they are both countable. Not all countable sets are homeomorphic.

oops, i have a problem that i dont read all the parts of questions i see (my advisor can vouch for this)

In [url]http://at.yorku.ca/p/a/c/a/25.htm[/url] you can find a proof of a classic Sierpinksi theorem that all countable metrisable spaces without isolated points are homeomorphic.
You can apply this to your problem to conclude the spaces are indeed homeomorphic (and also homeomorphic to the rationals as a whole).

You must tell us ( specify) the kind of morphism you are interested in. Set morphism ( in whixh case the countability argument is ok) or continuity in which case with what topology should the morphisms respect .


If $ c\geq b\geq a$, then

$ \frac{ac}{a^2\plus{}b^2}\plus{}\frac{ba}{b^2\plus{}c^2}\plus{}\frac{cb}{c^2\plus{}a^2}\leq\frac{2}{3}(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2$

Sorry, but I think we can prove that for any positive real numbers $ a$,$ b$,$ c$ (not necessarily satisfying $ c\leqslant b\leqslant a$) the following holds:
\[ \frac{2}{3}\left(\sum\frac{a}{b}\right)^2\geqslant \sum\frac{4ab}{b^2\plus{}c^2}\]
 :maybe:

[quote="Martin N."]Sorry, but I think we can prove that for any positive real numbers $ a$,$ b$,$ c$ (not necessarily satisfying $ c\leqslant b\leqslant a$) the following holds:
\[ \frac {2}{3}\left(\sum\frac {a}{b}\right)^2\geqslant \sum\frac {4ab}{b^2 \plus{} c^2}\]
:maybe:[/quote]

Dear [color=red][b]Martin N.[/b][/color], The inequality proposed by you is longer. Do you have a resolution? who has written the inequality? thanks

[quote="Ligouras"]Dear [color=red][b]Martin N.[/b][/color], The inequality proposed by you is longer. Do you have a resolution? who has written the inequality? thanks[/quote]
"My" inequality from above is just what I got trying to solve filomen's original one...my opinion was (and is) that he just forgot the factor $ 4$ because his original one doesn't even have an equality case...
Furhermore, I don't know where he took his condition that the variables are sorted from  :huh:

However, my solution is quite simple:
Due to the trivial $ b^2\plus{}c^2\geqslant 2bc$, we have to prove
\[ \left(\sum\frac{a}{b}\right)^2\geqslant 3\cdot\sum\frac{a}{c}\equal{}3\cdot\sum\frac{a}{b}\cdot\frac{b}{c}\mbox{,}\]
which is clearly true because $ (x\plus{}y\plus{}z)^2\geqslant 3(xy\plus{}yz\plus{}zx)$.  :)


The equation $ \sqrt {x \plus{} 10} \minus{} \frac {6}{\sqrt {x \plus{} 10}} \equal{} 5$ has:
$ \textbf{(A)}\ \text{an extraneous root between } \minus{} 5\text{ and } \minus{} 1 \\
\textbf{(B)}\ \text{an extraneous root between } \minus{} 10\text{ and } \minus{} 6 \\
\textbf{(C)}\ \text{a true root between }20\text{ and }25 \qquad\textbf{(D)}\ \text{two true roots} \\
\textbf{(E)}\ \text{two extraneous roots}$

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Let $ y\equal{}\sqrt{x\plus{}10}$. So, we now have $ \frac{y^2\minus{}6}{y}\equal{}5$. Cross multiplying and moving all the terms to one side, we get $ y^2\minus{}5y\minus{}6\equal{}(y\minus{}6)(y\plus{}1)\equal{}0$. Testing $ y\equal{}6,\minus{}1$ shows that $ x\equal{}\minus{}9$ is extraneous, so we have $ \textbf{(B)}$

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topology
advanced fields
advanced fields theorems

For sets like R or R^2 what would the usual topology be?

Can a set like (-infinity,0] appear in R with the usual topology?

The "usual topology" of a set is the topology we most frequently use, i.e. it is a term of convention, not a formal definition.  The usual topology on $\bf R$ is the topology it inherits from its metric (so a basis is all sets of the form $(a, b)$ for reals $a < b$).  The usual topology on ${\bf R}^{2}$ is the product topology of $\bf R$ with itself, which is also the standard Euclidean topology on the plane.