There are many possible meanings for $ |A|$. Which norm do you want?

Sorry, that means the determinant of $ A$.

\begin{align*}
\det A
&=\begin{vmatrix}
3 & 2 & - 1 & - 1 \\
6 & 3 & 4 & 1 \\
5 & 1 & 5 & 1 \\
1 & - 1 & 3 & 1
\end{vmatrix} \\
&=\begin{vmatrix}
3 & 2 \\
6 & 3
\end{vmatrix}
\times
\begin{vmatrix}
\begin{pmatrix}
5 & 1 \\
3 & 1
\end{pmatrix}
-
\begin{pmatrix}
5 & 1 \\
1 & -1
\end{pmatrix}
\begin{pmatrix}
3 & 2 \\
6 & 3
\end{pmatrix}^{-1}
\begin{pmatrix}
-1 & -1 \\
4 & 1
\end{pmatrix}
\end{vmatrix} \\
&=\begin{vmatrix}
3 & 2 \\
6 & 3
\end{vmatrix}
\times
\begin{vmatrix}
\frac{-22}{3} & \frac{-13}{3} \\
\frac{-20}{3} & \frac{-11}{3}
\end{vmatrix} \\
&=\begin{vmatrix}
3 & 2 \\
6 & 3
\end{vmatrix}
\times \frac{1}{9}
\begin{vmatrix}
-22 & -13 \\
-20 & -11
\end{vmatrix} \\
&= -3 \times \frac{1}{9} \times (-18) \\
&= \boxed{6}
\end{align*}

\begin{align*}
\det B
&= \begin{vmatrix}
1 & - 1 & - 1 & 1 \\
2 & - 1 & - 1 & 3 \\
- 1 & 2 & 3 & 3 \\
- 1 & 2 & 1 & - 2
\end{vmatrix} \\
&=\begin{vmatrix}
1 & -1 \\
2 & -1
\end{vmatrix}
\times
\begin{vmatrix}
\begin{pmatrix}
3 & 3 \\
1 & -2
\end{pmatrix}
-
\begin{pmatrix}
-1 & 2 \\
-1 & 2
\end{pmatrix}
\begin{pmatrix}
1 & -1 \\
2 & -1
\end{pmatrix}^{-1}
\begin{pmatrix}
-1 & 1 \\
-1 & 3
\end{pmatrix}
\end{vmatrix} \\
&=\begin{vmatrix}
1 & -1 \\
2 & -1
\end{vmatrix}
\times
\begin{vmatrix}
1 & 3 \\
-1 & -2
\end{vmatrix} \\
&= 1 \times 1 \\
&= 1
\end{align*}

So,
\begin{align*}
B^{-1}
&= \frac{1}{\det B}
\begin{pmatrix}
+\det \begin{pmatrix}
-1 & -1 & 3 \\
2 & 3 & 3 \\
2 & 1 & -2
\end{pmatrix}
&
-\det \begin{pmatrix}
-1 & -1 & 1 \\
2 & 3 & 3 \\
2 & 1 & -2
\end{pmatrix}
&
+\det \begin{pmatrix}
-1 & -1 & 1 \\
-1 & -1 & 3 \\
2 & 1 & -2
\end{pmatrix}
&
-\det \begin{pmatrix}
-1 & -1 & 1 \\
-1 & -1 & 3 \\
2 & 3 & 3
\end{pmatrix} \\
-\det \begin{pmatrix}
2 & -1 & 3 \\
-1 & 3 & 3 \\
-1 & 1 & -2
\end{pmatrix}
&
+\det \begin{pmatrix}
1 & -1 & 1 \\
-1 & 3 & 3 \\
-1 & 1 & -2
\end{pmatrix}
&
-\det \begin{pmatrix}
1 & -1 & 1 \\
2 & -1 & 3 \\
-1 & 1 & -2
\end{pmatrix}
&
+\det \begin{pmatrix}
1 & -1 & 1 \\
2 & -1 & 3 \\
-1 & 3 & 3
\end{pmatrix} \\
+\det \begin{pmatrix}
2 & -1 & 3 \\
-1 & 2 & 3 \\
-1 & 2 & -2
\end{pmatrix}
&
-\det \begin{pmatrix}
1 & -1 & 1 \\
-1 & 2 & 3 \\
-1 & 2 & -2
\end{pmatrix}
&
+\det \begin{pmatrix}
1 & -1 & 1 \\
2 & -1 & 3 \\
-1 & 2 & -2
\end{pmatrix}
&
-\det \begin{pmatrix}
1 & -1 & 1 \\
2 & -1 & 3 \\
-1 & 2 & 3
\end{pmatrix} \\
-\det \begin{pmatrix}
2 & -1 & -1 \\
-1 & 2 & 3 \\
-1 & 2 & 1
\end{pmatrix}
&
+\det \begin{pmatrix}
1 & -1 & -1 \\
-1 & 2 & 3 \\
-1 & 2 & 1
\end{pmatrix}
&
-\det \begin{pmatrix}
1 & -1 & -1 \\
2 & -1 & -1 \\
-1 & 2 & 1
\end{pmatrix}
&
+\det \begin{pmatrix}
1 & -1 & -1 \\
2 & -1 & -1 \\
-1 & 2 & 3
\end{pmatrix}
\end{pmatrix} \\
&= \frac{1}{1}
\times
\begin{pmatrix}
-13 & 5 & -2 & -2 \\
7 & -2 & 1 & 2 \\
-15 & 5 & -2 & -3 \\
6 & -2 & 1 & 1
\end{pmatrix} \\
&= \boxed{
\begin{pmatrix}
-13 & 5 & -2 & -2 \\
7 & -2 & 1 & 2 \\
-15 & 5 & -2 & -3 \\
6 & -2 & 1 & 1
\end{pmatrix}}
\end{align*}

For finding $ B^{-1}$, the method of applying elementary row operations on

\[ \begin{pmatrix}{1}&-1&-1&1&1&0&0&0\\2&-1 &-1 & 3&0&1&0&0\\-1&2&3&3&0&0&1&0\\-1&2&1&-2&0&0&0&{1}\end{pmatrix}\]

is in my opinion much quicker and efficient; it only takes 6 steps (I call a "step" the clearing (making 0) of all entries below the diagonal entry or the clearing of all entries above the diagonal entry in a column of the original vector through elementary row operations while applying the same row operation to the 4x4 matrix attached to the right).

Row Reduction is also an efficient way to compute determinants for large matrices. The operations of elementary row reduction preserve determinants.

Kouichi Nakagawa, do you have a nice proof of the 4x4 determinant relation you were applying? Namely, that $ \det \begin{pmatrix} A & B\\ C & D\end{pmatrix} \equal{} |A| \times |D \minus{} CA^{\minus{}1}B|$, where $ A,B,C,D$ are 2x2 matrices.

[quote="gauss202"]Kouichi Nakagawa, do you have a nice proof of the 4x4 determinant relation you were applying? Namely, that $ \det \begin{pmatrix} A & B \\
C & D\end{pmatrix} \equal{} |A| \times |D \minus{} CA^{ \minus{} 1}B|$, where $ A,B,C,D$ are 2x2 matrices.[/quote]
\[ \det (XY) \equal{} (\det X) \times (\det Y)\]
\[ \begin{pmatrix} A & B \\ C & D \end{pmatrix} \equal{} \begin{pmatrix} A & O \\ C & I \end{pmatrix} \begin{pmatrix} I & A^{\minus{}1}B \\ O & D\minus{}CA^{\minus{}1}B \end{pmatrix}\]
So trivial :)

Nice proof Kouichi Nakagawa. I'll try to remember that identity and see if it comes in handy. On obvious corollary which may be worth noting for the record is that if $ AC = CA$ or $ BD = DB$, then $ \begin{vmatrix}{A & B \\
C & D}\end{vmatrix} = |A\parallel{}D| - |B\parallel{}C|$.

EDIT: Corrected Latex mistake.

That's right. I often use the equality for 4*4 matrix.

In the related this: Here is  an excerpt from a book of Linear Algebra.

1. Let $ A,\ D,\ B$ be $ r*r,\ m*m,\ r*m$ matrices respectively.

Prove that det $ \left( \begin{array}{cc} A & B \\
0 & D \end{array} \right) \equal{} det A\cdot detD$.

2. Let $ C$ be an $ m*r$ matrix. Prove that det $ \left(\begin{array}{cc} A & 0 \\
C & D \end{array} \right) \equal{} det A\cdot det D$.

3. Let $ Y$ be an $ r*m$ matrix.

Prove that $ \left( \begin{array}{cc} A & B \\
C & D \end{array} \right)\left( \begin{array}{cc} 1 & Y \\
0 & 1 \end{array} \right) \equal{} \left( \begin{array}{cc} A & AY \plus{} B \\
C & CY \plus{} D \end{array} \right)$

4. Let $ A$ be a regular matrix.

Prove that

$ \boxed{det \left( \begin{array}{cc} A & B \\
C & D \end{array} \right) \equal{} det A\cdot det (D \minus{} CA^{ \minus{} 1}B)}$

5. If $ r \equal{} m$, then prove that

$ \left( \begin{array}{cc} 1 & \minus{} 1 \\
0 & 1 \end{array} \right)\left( \begin{array}{cc} A & B \\
B & A \end{array} \right)\left( \begin{array}{cc} \minus{} 1 & 1 \\
0 & 1 \end{array} \right) \equal{} \left( \begin{array}{cc} B \minus{} A & 0 \\
\minus{} B & B \plus{} A \end{array}\right)$

6. Prove that det $ \left(\begin{array}{cc} A & B \\
B & A \end{array} \right) \equal{} det (A \plus{} B) det (A \minus{} B)$.

7. Let $ A,\ B,\ C,\ D,\ X,\ Y,\ Z,\ U$ be $ m*m$ matrices.

Prove that $ \left(\begin{array}{cc} A & B \\
C & D \end{array} \right)\left(\begin{array}{cc} X & Y \\
Z & U \end{array} \right) \equal{} \left(\begin{array}{cc} AX \plus{} BZ & AY \plus{} BU \\
CX \plus{} DZ & CY \plus{} DU \end{array} \right)$

[quote="kunny"]1. Let $ A,\ D,\ B$ be $ r*r,\ m*m,\ r*m$ matrices respectively.

Prove that det $ \left( \begin{array}{cc} A & B \\
0 & D \end{array} \right) \equal{}$det$ A\cdot$det$ B$.
[/quote]
I think $ \begin{vmatrix} A & B \\ O & D \end{vmatrix} \equal{} \det A \times \det D$

Let $ (C_i),(L_i)$ denote the columns, lines of $ A$.
$ C_1\minus{}C_2\minus{}C_3\plus{}C_4\rightarrow{C}_1$ gives $ det(A)\equal{}\begin{vmatrix}3&4&1\\1&5&1\\\minus{}1&3&1\end{vmatrix}$.
$ L_2\minus{}L_1\rightarrow{L}_2,L_3\minus{}L_1\rightarrow{L}_3$ give $ det(A)\equal{}\begin{vmatrix}\minus{}2&1\\\minus{}4&\minus{}1\end{vmatrix}$.

About $ B^{\minus{}1}$ Duelist is right. Nakagawa shows what we should not write.

[quote="Kouichi Nakagawa"]

\[ \det (XY) \equal{} (\det X) \times (\det Y)\]

\[ \begin{pmatrix} A & B \\
C & D \end{pmatrix} \equal{} \begin{pmatrix} A & O \\
C & I \end{pmatrix} \begin{pmatrix} I & A^{ \minus{} 1}B \\
O & D \minus{} CA^{ \minus{} 1}B \end{pmatrix}\]
So trivial :)[/quote]

Isn't all math trivial - once it's understood? :)


[hide]We wish to show that
\[ \frac{bc}{b \plus{} c} \plus{} \frac{ca}{c \plus{} a} \plus{} \frac{ab}{a \plus{} b} \plus{} \frac{bc}{2a} \plus{} \frac{ca}{2b} \plus{} \frac{ab}{2c} \minus{} a \minus{} b \minus{} c \geq 0.\]
Putting this over a common denominator, it suffices to show that
$ \frac{a^4b^3\plus{} a^4c^3 \plus{} a^3b^4 \plus{} b^4c^3 \plus{} a^3c^4 \plus{} b^3c^4 \minus{} (a^4b^2c \plus{} a^4bc^2 \plus{} a^2b^4c \plus{} ab^4c^2 \plus{} ab^2c^4 \plus{} a^2bc^4)}{2abc(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq 0$
Multiplying both sides by $ 2abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$ and rearranging, it remains to show that
\[ a^4b^3\plus{} a^4c^3 \plus{} a^3b^4 \plus{} b^4c^3 \plus{} a^3c^4 \plus{} b^3c^4 \geq a^4b^2c \plus{} a^4bc^2 \plus{} a^2b^4c \plus{} ab^4c^2 \plus{} ab^2c^4 \plus{} a^2bc^4\]
But this is true by Muirhead's inquality. [/hide]

[quote="bboypa"]Let $ a,b,c$ be positive reals; show that $ \displaystyle a \plus{} b \plus{} c \leq \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} \plus{} \frac {ab}{a \plus{} b} \plus{} \frac {1}{2}\left(\frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\right)$
[i](Darij Grinberg)[/i][/quote]
Let $ a\equal{}yz,b\equal{}zx$ and $ c\equal{}xy,$ where $ x,y,z$ are some positive real numbers. The inequality becomes
$ \frac{1}{2}(x^2\plus{}y^2\plus{}z^2) \plus{}xyz \left( \frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x}\plus{}\frac{1}{x\plus{}y}\right) \ge xy\plus{}yz\plus{}zx.$
By the Cauchy-Schwarz Inequality, we have
$ \frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x}\plus{}\frac{1}{x\plus{}y} \ge \frac{9}{2(x\plus{}y\plus{}z)},$
and hence, it suffices to prove that
$ x^2\plus{}y^2\plus{}z^2\plus{}\frac{9xyz}{x\plus{}y\plus{}z} \ge 2(xy\plus{}yz\plus{}zx),$
which is Schur's Inequality.

$ \displaystyle \sum_{cyc}\left(\frac{bc}{b\plus{}c}\plus{}\frac{bc}{2a}\minus{}\frac{b\plus{}c}{2}\right)\geq 0$ iff $ \displaystyle \sum_{cyc}\left(\frac{\left(c\minus{}a\right)b^2}{2a\left(b\plus{}c\right)}\minus{}\frac{\left(a\minus{}b\right)b^2}{2a\left(b\plus{}c\right)}\right)\geq0$ iff $ \displaystyle \sum_{cyc} \left(b\minus{}c\right)a^2\left(\frac{1}{c\left(a\plus{}b\right)}\minus{}\frac{1}{b\left(c\plus{}a\right)}\right)\geq0$, which is a sum with three nonnegative addends.


$ ABC$ is an equilateral triangle and $ (C)$ a circle with diameter [$ BC$]. Let $ E$ and $ F$ be two points in [$ BC$] such that $ BE \equal{} EF \equal{} FC$.  Let [$ AE)$ and [$ AF)$ intersect the semicircle $ (C)$, which is situated out of $ ABC$, in $ U$ and $ V$. Prove that $ U$ and $ V$ divide the semicircle $ (C)$, which situated out of $ ABC$, in three equal arcs.

[color=red]EDIT by moderator: Moved from Geometry / Olympiad section.[/color]

Circle $ (B)$ with radius $ BC$ passes through $ C, A$ and it cuts the line $ AB$ again at $ X$ $ \Longrightarrow$ $ XC \perp CA$ and $ B$ is the midpoint of the segment $ AX$ $ \Longrightarrow$ $ CB$ is the C-median of the right angle $ \triangle AXC.$ $ EB \equal{} 3CB$ $ \Longrightarrow$ $ E$ is the centroid of this triangle. Its A-median $ AE$ cuts the opposite side $ XC$ at its midpoint $ U,$ $ UB \parallel CA$ is its midline $ \Longrightarrow$ $ \angle CBU \equal{} \angle BCA \equal{} 60^\circ$ and $ UB \perp (XC \equiv UC),$ $ U$ is on a circle with diameter $ BC.$ Similarly, $ \angle BCV \equal{} 60^\circ$ and $ V$ is on a circle with diameter $ BC.$


inequalities
geometry
invariant
geometric transformation
analytic geometry
AMC

What is the area of the region defined by the inequality $ |3x\minus{}18|\plus{}|2y\plus{}7|\le 3$?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ \frac{7}{2} \qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ \frac{9}{2} \qquad
\textbf{(E)}\ 5$

Area is invariant under translation, so you may as well solve the problem

$ |3x|\plus{}|2y|\leq 3$

and then it's pretty easy to see the diamond around the origin. If you divide it up into four right triangles (one in each quadrant, all congruent to each other), you get that each one has legs of $ 3/2$ and $ 1$. Then the area is $ 3/4$, so the answer is $ 4 \cdot 3/4 \equal{} 3$ or $ \boxed{A}$.

While you're at it, you may as well change it to 1/6 the area of $ |x|\plus{}|y| \equal{} 3$, then draw a square around it with coordinates (+/- 3, +/- 3) and be like, look, it's half that square! And then you can be like, 6^2 / 6 / 2 = 3.

[quote="cdymdcool"]I don't understand how can you apply "area is invariant under translation" to this problem.[/quote]

The equation is that |3x|+|2y|<=3 shifted

The bounded area doesnt change when you shift a graph

But it's not a simply shift from the original problem to |3x|+|2y|<=3

It sure is:

$ |3x\minus{}18|\plus{}|2y\plus{}7|\le 3 \iff |3(x\minus{}6)|\plus{}|2(y\plus{}3.5)| \le 3$

This is the same as $ |3x|\plus{}|2y|\leq 3$, except shifted $ 6$ units in the positive $ x$ direction and $ 3.5$ units in the negative $ y$ direction.

we test points (or at least thats the way i am most familiar with)

$x = 0$ , $y = 3/2 or -3/2$

$y = 0$ , $x = 1 or -1$

and those are the vertices of the figure so we can easily solve by splitting them up into 2 triangles

[quote="fryeggs"]how do you graph $|3x| + |2y| \leq 3$?[/quote]
Assume x or y is zero, since that is their minimum and then find the possible values of the other variable.

Yup -- alternatively, write it as $3|x| + 2|y| \leq 3$, and note that the region must be symmetric over both the $x$ and $y$ axes. This means you can just graph it in the first quadrant (where it is equivalent to $3x+2y \leq 3$) and fill in the other quadrants symmetrically.

Sorry to brink back an old feed, but, I'm still not understanding the translation. Please Help!

It's ok to bring back threads from the competition page when you have questions.

$ |3x-18|+|2y+7|\le 3\iff |3(x-6)|+|2(y+3.5)|\le 3 $

Since you have $(x-6)$, you simply shift up $6$ units in the $x$ direction to  obtain $x$.

Then,$ (y+3.5) $ allows you to shift $y$ down by $3.5$ units, so it becomes the translated form stated above.

Think about having a square with vertices $(0, 0) , (0,1), (1,0),$ and $(1,1)$.  If you shift this square to anywhere on the coordinate plane, it will still retain its area.

So the coordinates might as well be $(x, y), (x, 1+y), (1+x, y), $ and $(1+x, 1+y)$ because it is still a square of side length $1$, just translated up $x$ units in the $x$-direction and $y$ units in the $y$-direction.

[hide=solution]Area is invariant under translation, so it remains to compute the area of the region defined by the inequality $|3x|+|2y|\leq 3$. We will only consider the first quadrant for now, and then multiply by $4$ to get the total area. In the first quadrant we see that we have a right triangle with legs of length $\cfrac{3}{2}$ and $1,$ whose area is $\cfrac{3}{4}$. Then, the answer is $\cfrac{3}{4}\cdot 4=\boxed{\textbf{(A)} ~3}$[/hide]


What is the units digit of the product of $ 3^{35} \times 7^{35}$?

Thats the same as 21^35 which will always end in 1.

An important property is:

$ a^c \times b^c \equal{} (ab)^c$

Note that $ 21^{35}$ will always end in 1 because we just need to pay attention to the units digit, and we're basically doing $ 1^{35}$ = $ 1$.

we multiply the bases and keep the exponent

21^35

we just look at the units and we see

1^35=1

so 1 is the answer

rate my post please thanks :)

For the answer I got 1.895288449 and we look at the units we see, So the answer would be 1. I really can't give a solution because it did the problem with a calculator.

[quote="dyang"]Why cant you just do

3^35's unit digit is 7
(it cycles 3,9,7,1)
7^35's unit digit is 3
(it cycles 7,9,3,1)
7x3=21
Ends in
1!!!!!!!!!!!!!!!!![/quote]

You certainly can do it this way, and if you wished to double check your answer using a different method, this would be logical. However, in FTW, the other way (using a law of exponents) is better since it is much faster.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hello, i have some problems with the following exercise..I hope someone can help me..

Show that if H and K are subgroups of a group G and H [b]U[/b] K is a subgroup of G, then H is a subset of K or K is a subset of H.
Thanks

[url=http://www.mathlinks.ro/viewtopic.php?p=1124006&search_id=620483434#1124006]Solution in the forum[/url].

Could someone please explain this in more depth?..i didn't understand exactly how this contradiction shows that the one subgroup must be subset of the other or inversely.
Thank you for your patience..

You might want to start with elements $ a\in A\setminus B$ and $ b\in B\setminus A$ to derive a contraction.

Consider $ h\in H\setminus K\subset H$  and $ k\in K\setminus H\subset K$ ,
then we know that $ h,k\in H\cup K$. Since $ H\cup K$ is a group, so $ hk\in H\cup K$.
Hence $ hk$ must be in $ H$ or $ K$. If $ hk \in H$ then since $ h^{ \minus{} 1}\in H$, so $ k \equal{} (h^{ \minus{} 1}h)k \equal{} h^{ \minus{} 1}(hk)\in H$

so $ K\subset H$

similarly you can show the other way if $ hk\in K$


6 vertices are in a space such that no three are collinear , no four are coplane. Join a line between any two vertices and we color each line either blue or red . Prove that , we can always find a triangle with its three side having the same color .

The space part is just nonsense...
This is the statement $R(3,3) \leq 6$, where $R$ denotes Ramsey numbers.
It was posted before.


It's on March 11th at Grand Rapids. There's 21 Chapters.

[quote="yhh3002"]I have one question...

Where's the State round for the Michigan Mathcounts?

How many chapters are there?

I'm from East Hills.......

I'm advancing to the states

I hope it's not hard... :wink:[/quote]

I thought it was March 24th.  Check http://www.mathcounts.org.

If you can win in Oakland Chapter, you'll do fine at states.

[quote="yhh3002"]I have one question...

Where's the State round for the Michigan Mathcounts?

How many chapters are there?

I'm from East Hills.......

I'm advancing to the states

I hope it's not hard... :wink:[/quote]

i HAVEN'T really ever made it to States so I wouldn't know, if it is hard.

And btw, Allen, he is from Wayne, I though.

And one guy from Marquette seems to be really good at Math.

Wait a second, the 11th isn't a Saturday...

Mines is on the 17th. I actually don't care that much about MathCounts, so I'm not going to study.

[quote="mathgeniuse^ln(x)"]i HAVEN'T really ever made it to States so I wouldn't know, if it is hard.[/quote]
 :rotfl: uh huh...

Okay, whatever.

2 years ago, only 2 weeks between Chapters and States
1 year ago, 4 weeks
This year, 6 weeks

[quote="Fanatic"]MARCH 24th?? LUCKY.

Mine's on March 3rd /spasm.[/quote]
MARCH 3rd?? LUCKY.
Mine's on Marth 29th. /anticipation

Mine's on, huh gasp, I didn't make it.

March 29th. Heh, that is like a long way from here.

Why is some mathcounts chapters a month away? I didn't know the chapter would last for so long.


geometry
circumcircle
incenter
perpendicular bisector
angle bisector

Can somebody please write all possible names (in english) of all the [b]parts[/b] of a [b]triangle[/b]? Like,
altitude=perpendicular from a vertex to opposite side
       ?      = lines that join the middle points of the sides
        ?      = perpendicular through the middle points of the sides
         ?      = intersection of altitudes

etc, etc, etc (as many as you can)

thanks
Edursho :(

perpendicular bisector=perpendicular segment of a side drawn through the midpoint of that side
Their intersection is the circumcenter

median=segment drawn from a vertex to the midpoint of the opposite side
Intersect two-thirds of the way from the vertex to the midpoint.

altitude=segment drawn from a vertex that is perpendicular to the opposite side

angle bisector=segment that bisects an angle
Their intersection is the incenter.

incenter=center of the inscribed circle

circumcenter=center of the circumscribed circle

midpoint=middle of a side

[b]Post a definition of a term if you want another name.[/b]

This seems more appropriate for another forum...

Thanks Diophantus.
How about
                       =segments that joint the midpoints of the sides ?

 Interestingly enough, the word "median" in english does not correspond to the word "mediana" in spanish!

The midlines of a triangle join the midpoints of the sides.  The three midlines together are the edges of the medial triangle.  The feet of the altitudes are the vertices of the orthic triangle.

This may be overkill but: http://www.xtec.es/~qcastell/ttw/ttweng/portada.html


From [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1163307#1163307]this[/url], is there another way to improve the quality of the image on MS Word? I printed out a lab report a while ago with a section on calculations and the LaTeX was very blurry.

Where's the clip feature?  :maybe:  I can't find it on mine.

When you're in Adobe, there's a little camera icon that says "Snapshot tool" or something like that.  Once you click it, you can highlight the area you want to clip.

[quote="JRav"]I usually just do the calculations using Equation Maker.

Or, you can do all of your equations in $ \text{\TeX}$, generate the PDF, and then use the clip feature on the PDF to copy the image, and then just paste it into your word document.[/quote]

I'm not affliated with them, but Aurora is what you need -- http://www.elevatorlady.ca


real analysis
calculus
calculus computations

How to prove that if $ R$ is the radius of convergence for $ \sum c_n x^n,$ then $ \sum nc_n x^{n\minus{}1}$ has the same radius?

Try using the root test, in its limsup version.

Note that the question is only about the radius of convergence, not convergence on the boundary. The latter can change.

I've never worked with the $ \limsup$ version of the limit since my professor never mentioned it.

How does that work?


geometry
symmetry
geometric transformation
rotation
analytic geometry
trigonometry
AoPSwiki

The solutions of the equation $ z^4 \plus{} 4z^3i \minus{} 6z^2 \minus{} 4zi \minus{} i \equal{} 0$ are the vertices of a convex polygon in the complex plane.  What is the area of the polygon?

$ \textbf{(A)}\ 2^{5/8} \qquad
\textbf{(B)}\ 2^{3/4} \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 2^{5/4} \qquad
\textbf{(E)}\ 2^{3/2}$

We can look at the symmetry of the coefficients $ 1,4,6,4,1$. This immediately reminds us of the binomial expansion of $ (x\plus{}1)^4$.

However, we have to adjust this for the $ i$'s and such. Notice that the complex units of the coefficients are $ 1,i,\minus{}1,\minus{}i,\minus{}i$. We notice that this is close to the pattern formed by $ 1,i,i^2,i^3,i^4$.

Based on these two observations of symmetry, we can deduce that $ (z\plus{}i)^4$ is appropriate here. More specifically, we have that \[ (z\plus{}i)^4\equal{}1\plus{}i\equal{}2^{\frac{1}{2}}\cdot \text{cis}\, \frac{\pi}{4}\] We can apply a translation of $ \minus{}i$ and a rotation to eliminate the complex coefficients (since these operations obviously preserve area).

Hence we need the area formed by the quadrilateral that has the roots of \[ (z)^4\equal{}2^{\frac{1}{2}}\] It is clear that the roots are created by rotating another ninety degrees about the origin, so we have a square. We can divide the square into four triangles to compute the area: \[ 4\cdot \frac{1}{2}\cdot (2^{\frac{1}{8}})^2\equal{}2^{\frac{5}{4}}\] so the answer is $ \boxed{D}$.

In the AoPS books, in what chapter are concepts such as this (complex plane) discussed?

Could someone please explain this part of the solution to me:

[quote="Altheman"]\[ (z \plus{} i)^4 \equal{} 1 \plus{} i \equal{} 2^{\frac {1}{2}}\cdot \text{cis}\, \frac {\pi}{4}
\]
We can apply a translation of $ \minus{} i$ and a rotation to eliminate the complex coefficients (since these operations obviously preserve area).

Hence we need the area formed by the quadrilateral that has the roots of
\[ (z)^4 \equal{} 2^{\frac {1}{2}}
\]
It is clear that the roots are created by rotating another ninety degrees about the origin, so we have a square. We can divide the square into four triangles to compute the area:
\[ 4\cdot \frac {1}{2}\cdot (2^{\frac {1}{8}})^2 \equal{} 2^{\frac {5}{4}}
\][/quote]

I don't quite get how do perform a translation of $ \minus{}i$.  Also, how were the geometric references gotten?

Thanks!

If you plot a point such that its x coordinate is equal to the real part of some number, and its y coordinate is equal to the imaginary part of the number, then the new point can be used to represent the original complex number in the plane. The geometry comes about when you consider the complex numbers as points on the plane.

A translation of $ z$ by $ \minus{} i$ simply means replacing $ z$ with $ z \minus{} i$. Remember that the imaginary part is in the y-coordinate direction. So if we reduce $ z$ by the value $ i$, visually, it's the same as sliding $ z$ downwards by 1 unit in the y direction. If you slide a shape down one unit, its area doesn't change.

So now, we started from $ (z \plus{} i)^4 \equal{} \sqrt {2} \cdot \text{cis} \left( \frac {\pi}{4} \right)$

Now, we'll replace $ z$ with $ z \minus{} i$, and we get $ (z \minus{} i \plus{} i)^4 \equal{} z^4 \equal{} \sqrt {2} \cdot \text{cis} \left( \frac {\pi}{4} \right)$.

Now, we can also rotate by an angle of -45 degrees counterclockwise, or $ \minus{} \pi/4$ in radian measure, and the area still won't change. To rotate a complex number around the origin by an angle of $ \theta$, you multiply it by $ \cos \theta \plus{} i \sin \theta$, abbreviated $ \text{cis} \theta$.

So let's do that: $ z^4 \equal{} \sqrt {2} \cdot \text{cis} \frac {\pi}{4} \cdot \text{cis} \frac { \minus{} \pi}{4}$

So now, the two rotations cancel out, since they are in opposite directions, and you get that the original area is the same as the area of the quadrilateral whose vertices are roots of $ z^4 \equal{} \sqrt {2}$

Now, if we solve that equation, we get the numbers $ \sqrt [4]{2}, \; \minus{} \sqrt [4]{2}, \; i \sqrt [4]{2}, \; \minus{} i \sqrt [4]{2}$. Now, plot those numbers in the plane. You have a square, and you can easily find the area of this square.

Nice explanation Zuton Force, except it should say at the end:

[quote]Now, if we solve that equation, we get the numbers $ \sqrt [8]{2}, \; \minus{} \sqrt [8]{2}, \; i \sqrt [8]{2}, \; \minus{} i \sqrt [8]{2}$. Now, plot those numbers in the plane. You have a square, and you can easily find the area of this square.[/quote]

[quote="Altheman"]\[ (z \plus{} i)^4 \equal{} 1 \plus{} i \equal{} 2^{\frac {1}{2}}\cdot \text{cis}\, \frac {\pi}{4}
\]
 [/quote]

why is this? can someone explain, i'm bad at math

Try expanding out $ (z\plus{}i)^4$ manually with the binomial theorem. Then take a look at the given equation. You'll notice the only difference between the two is a constant $ 1\plus{}i$, so we get the first equality. The second is a coordinate transformation.

I do not understand how the equation used in #23 on the 2008 amc 12a,

z^4 + 4z^3i - 6z^2 - 4zi - i = 0,

was factored with (z+i)^4, and then set equal to 1+i. This doesn't seem right when I expanded it.

(Solution from AoPSWiki)

Well if you expand $(z+i)^4$, remembering that $i^2 = -1$, you get $z^4 + \binom{4}{1}z^3i + \binom{4}{2}z^2i^2 + \binom{4}{3}zi^3 + i^4$, which turns out to be $z^4 + 4z^3i  + 6z^2(-1) + 4z(-i) + 1$, or $z^4 + 4z^3i - 6z^2 - 4z + 1$. The expression in the given equation is not the same as the one we found from expanding, but it only differs by a constant. If we add $1+i$ to both sides of the given equation, we get $z^4 + 4z^3i - 6z^2 - 4zi - i = 1 + i$, and now the left side is the same as our expanded expression, so we factor it as $(z+i)^4 = 1+i$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=506457]Please do not repost old contest problems[/url]. Merging.

You don't need to look for the thread. Just visit [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44]the USA page[/url] for a complete list of existing AMC problems. Clicking the problem numbers in any given year atkes you to the corresponding thread.

$(z+i)^4=z^4+4z^3i-6z^2-4zi+1$ so $(z+i)^4=i+1$

$|i+1|=2^\frac12$ so $|z+i|=2^\frac18$

so it is a square centered at $-i$ with radius $2^\frac18$ so area is $2^\frac54$


There is no nedd for Calculus .
We have that $xy+yz+zx\leq\frac{1}{3}$
By Cauchy and again Cauchy we have
$LHS\ \geq\frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}\geq\frac{9}{\sqrt{3(3+xy+yz+zx)}}\geq\frac{9}{\sqrt{10}}$


geometry
geometric transformation
reflection
circumcircle
analytic geometry
geometry unsolved

If G is the centroid of a triangle ABC, then prove that the reflections of the Brocard axes of triangles BGC, CGA, AGB in the sidelines BC, CA, AB concur at one point.

This result is verified using computer algebra. Does anybody have a synthetic proof?

  Darij

Sorry my ignorance  :blush: , but what are a Brocard axes??

It's the line $OK$, where $O$ is the circumcenter and $K$ is the Lemoine point (the intersection of the symmedians).

Yes, the Brocard axis of a triangle is the line joining the circumcenter with the symmedian point of this triangle. What could be perhaps of use is the fact that a point P lies on the Brocard axis of a triangle ABC if and only if $ \left( \frac{1}{b^{2}}-\frac{1}{c^{2}}\right) \cdot PA^{2}+\left( \frac{1}{c^{2}}-\frac{1}{a^{2}}\right) \cdot PB^{2}+\left( \frac{1}{a^{2}}-\frac{1}{b^{2}}\right) \cdot PC^{2}=0$.

  Darij

Today, [url=http://groups.yahoo.com/group/Hyacinthos/message/10523]Peter J. Moses has verified the concurrence of the reflections of the Brocard axes of triangles BGC, CGA, AGB in the sides BC, CA, AB using computer algebra[/url]. Just for fun, the barycentric coordinates of the point of concurrence are $\left( f\left(a,b,c\right) : f\left(b,c,a\right) : f\left(c,a,b\right) \right)$, where

$f\left(a,b,c\right) = a^2 \left( 49 a^8 - 133 a^6 b^2 + 120 a^4 b^4 - 29 a^2 b^6 - 7 b^8 - 133 a^6 c^2 + 21 a^4 b^2 c^2\right.$
$\left. + 69 a^2 b^4 c^2 + 23 b^6 c^2 + 120 a^4 c^4 + 69 a^2 b^2 c^4 - 48 b^4 c^4 - 29 a^2 c^6 + 23 b^2 c^6 - 7 c^8 \right)$

How about a synthetic proof?  :?

  Darij


inequalities
function
algebra
polynomial
vector
inequalities proposed

Let $a,b,c$ be positive real numbers and $abc=1$. Prove (or disprove) that \[\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{1+a+b+c}\ge 1.\] (conjecture because I haven't enough time to check the case $a=b$. But It's only expading).

Out of thought, Arqady. Very nice,  :)

I think the following inequality hold \[\frac{1}{(1+3a)^{2}}+\frac{1}{(1+3b)^{2}}+\frac{1}{(1+3c)^{2}}+\frac{13}{13+a+b+c}\ge 1 (2).\] Moreover, for all positive real numbers $k,r,p$ I think \[\frac{1}{(1+ka_{1})^{2}}+\frac{1}{(1+ka_{2})^{2}}+...+\frac{1}{(1+ka_{n})^{2}}+\frac{r}{p+a_{1}+a_{2}+...+a_{n}}\ge \min \left(1,\frac{n}{(1+k)^{2}}+\frac{r}{p+n}\right) (3).\] Where $a_{1},a_{2},...,a_{n}>0$ and $a_{1}a_{2}...a_{n}=1$. Furthermore, the number $p$ may be negative, but $p>-n.$

For both (2) and (3), I've just prove that the minimum values of the left-hand expressions are only attaned for $a_{1}=a_{2}=...=a_{n-1}$. But I hope a elementary solution for $(2)$.  :lol:

Actually, the result of part (b) is not true. But we can also prove that the left hand-side expression in only attained the maximum value when $n-1$ numbers equal. For some special value of $r,p$ we can archive some very nice result. But first prove that "why the left expression attained the maximum value when $n-1$ numbers equal".  :blush: I want to ask VASC if EV-method can be used (but I used nSMV theorem).

[quote="hungkhtn"]Out of thought, Arqady. Very nice,  :)

I think the following inequality hold \[\frac{1}{(1+3a)^{2}}+\frac{1}{(1+3b)^{2}}+\frac{1}{(1+3c)^{2}}+\frac{13}{13+a+b+c}\ge 1 (2).\] [/quote]

how about this nice inequality?

Not sure about EV-method but all of these problem can be done by ABC Method  :wink: .

[quote="Anh Cuong"]Yes, also for part 3, "old" ABC not works but ABC "nowadays" works. What do you mean by not creative, i prove it all by elementary  :wink: .[/quote]

Yes, I know your solution.  :lol:  It's only based on "a continous function defined in a compact set can attain the maximum and minimum value". Am I wrong?  :P  But this main principal is from "superior mathematics", you got it from "superior mathematics", and for me everything of elementary inequalities based on "superior mathematics" is not acceptable.

Yes, i told that to you before, do you forget and now guess something  :D . In fact using this is quite well-known and prove it by elementary is also easy. For this result, S.M.V and U.M.V is just a quick consequence from mixing variables.  :wink:

Oke, stop discus this topic, just rely on our concepts. In my concept, maths is maths, nice idea is nicea idea, never seperate it. In fact I never learn superior mathematic before but i still decude it from fixing idea.  :wink:
ABC still not stop at what you know, in fact all problems i post when i returned after quit at end of 10 grade is ABC consequence.  :wink:

You have so much argument over these methods! Do you realize how little it means to us who do not know what are these ABC and SMV theorems? Well if these are your secrets, don't bother to talk about them! If they are not your secrets, why won't you ever actually post a solution which uses them?

Yes, if there's no matter, I will post SMV-theorem next week. It's very simple but elementary.

Opps, please accept my apologies for all nonsense argurment above.
In fact I have nothing to hide, I posted ABC befored long time ago but it is Vietnamese. Also I have an English version now. You can download it at http://www.comp.nus.edu.sg/~anhcuong . However, this is just an old version, nowadays i improved it a lot, but still nice enough.  :)
You will know about it more clearly at IMO 2007 in VN, welcome  :D .
Some problems solved by ABC still unsolved for now:
http://www.mathlinks.ro/Forum/viewtopic.php?t=68171
http://www.mathlinks.ro/Forum/viewtopic.php?t=68162

Yeash... ABC is improved but the main idea is borrowed from "superior mathematics". "Old" ABC is very nice, but "improved ABC" is not.  :lol:

Not really, you still know too little about improved ABC  :D .It is not only what i told you. Wait for it at IMO 2007. :)

To Hungkhtn:
Oke, let the reader comment, you can try two problems i posted above and think how they can be solved without ABC. One more ABC is not a particular case of EV-theorem  :wink: .
You can think about the theorem:
All symmetric polynomial three variables degree less than or equal to five occurs the max and min when two variables are equal or one is equal to zero without follow my way  :D .
Also to image how ABC now, try to solve some cyclic inequality I posted  :) .

Ha ha, don't try to ruin all my creative by your naive comment, indeed the idea of proof is nice, and also what I speech in the theorem is the most important. The only thing I can see is that the problem you posted can be solved by EV as you said. Say so so I can say that all technique just follow $x^{2}\geq 0$. The ideas here, i called it abstract concreteness, are important and you should value it  :wink:,  . Oke stop here, just very private comment, i think the reader will know how strong ABC is .  :)
In fact still not really sure about EV-Theorem now  :| .

[quote="hungkhtn"]Moreover, with me all ideas which are borrowed from "superior mathematics" are not acceptable. [/quote]
Always I hear these words! (Not just from you but others also.) Surely you know that these famous inequalities of Cauchy-Buniakovsky-Schwarz, of Holder, of Minkowski, of Young -- these mathematicians were not solving elementary algebraic inequalities! They prove these inequalities as theorems for abstract Banach spaces, and in this forum we apply them for one particular vector space, our friend the Euclidean space!

Indeed to me the beauty of inequalities is not the separation from "superior mathematics" but the connection! I am a student in university now. I don't care for solving IMO problems any longer. I study these "superior mathematics." And always it is amazing to me when I see more ways in which various distant parts of math are joined.
:lol: :D
We all have our own opinions on this matter and it is a futile quest to try to change others to agree with our own!

[quote="spanferkel"]I entirely agree!  :play_ball:  :clap2:  :icecream:[/quote]

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=113248

it's SMV theorem. I dont want to emerge it, but I like it because it's entirely belong to elementary mathematics.  :)

[quote="pvthuan"]Thank you, Xevarion. You have cleared up some selfish  thinking.  :thumbup:[/quote]

Not exactly sefish, thuan, you are wrong. You don't know why we must discuss so much about some thing not really nice as that. This matter's related to two tendancy of elementary mathematics. It's also related to all problems in the "world of elementary mathematics" nowaday. I'm faithful with the school that, elementary is only elementary, non-adding.

One reason for that kind of endless argument is your English.


I've got a really very clumsy solution to this problem :

Let (a, b, c, d, e, f, g, h, i) be a permutation of (1, 2, 3, 4, 5, 6, 7, 8, 9).
Find the smallest possible value of abc + def + ghi.

(It's from the French selection test for IMO 2003.)

Can anyone post an elegant solution ?

abc+def+ghi>=3*(abcdefghi)^(1/3)= 3*(9!)^(1/3) > 213
during the contest we were not allowed to have calculator ;-)

we guess that minimum is achieved when abc,def,ghi are near of
71 or 72. (because 72*3=216)

so we find:
9*8*1=72
6*4*3=72
7*5*2=70

then the minimum is
9*8*1+6*4*3+7*5*2=214>213

easy, isn't it ?

OK, I used the same strategy, but my solution was longer anyway.


I was just wondering, what are some of the activities that you engage in on weekends when you do not have school (except math and problem solving)?

Like do you kind of entertainment do you engage in?

i play [url=http://www.xgenstudios.com/play/stickarena/]this game[/url]

How about others? I just want to know what are interesting activities to engage in?

I work on my art, do online puzzles, some music, and do everything that isn't homework.

EDIT - Just tried stick arena and am now addicted. Add that to the list.

i go to work. and then come home and take a shower.  ;)

[quote="hello"]Its just that it always appears to me that people always go to all sorts of places. I dont really know where, but somewhere outside their home. Is that so, and if it is, where? It appears that not too many people are on AIM during the day.[/quote]

errands, maybe? we always have to go to the grocery store and a lot of times, that involves going "into town" because someone wants lunch "in town" and if that's the case, we end up going shopping at several other places because we're "in town"- like best buy, or the nice antique store, or bed bath and beyond- and then of course, someone wants to go to the library, so we go to the library, and someone else wants coffee so we go to a local coffee shop if we're lucky (or the evil empire, if there's not one around)... and by that time, we've spent the whole day.

so that's where. i guess.

Sleep, practice violin, read, and do other things which i have every weekend, ie violin lesson, karate, working at the library etc.
I used to play some games but i don't game much anymore...

I regard homework as a total waste of time; time which could otherwise be spent playing games, practicing music, and playing sports. Too bad my teachers check it every other day  :roll:

Planning what I am to do next weekend of course!


function
inequalities
calculus
derivative
real analysis
real analysis unsolved

Here is a nice problem that the authors gave it to me some time ago.Although it's quite easy, here it goes:

  Let $f:\mathbb{R}\to\mathbb{R}$ a function such that $f'''(x)\geq 0$ ,for any $x\geq 0$.Prove that for any $x\in\mathbb{R}$ the following inequality holds:

      $ f(a+4x)+2f(a+x)\geq 2f(a+3x)+f(a)$ for any $x\geq 0$ and $a\in\mathbb{R}$ ;)

Obviously, one can take $ a=0$ and next consider $LHS-RHS$ as a function $h$ of $x$. Its derivative is nonnegative since $f'$ is convex and thus $h$ is increasing and thus positive on the positive axis. By the way, what's the difference between beautiful and nice?  :P

Yes,indeed. I don't know how you do consider this problem. After all, eveybody with his taste. :D

No, it's nice but I was wondering about the difference between a beautiful problem and a nice one.


[quote="Vasc"]Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \ge \frac {27}{(p \plus{} 2)(a \plus{} b \plus{} c)^2}$.[/quote]

For the inequality
\[ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \le \frac {27}{(p \plus{} 2)(a \plus{} b \plus{} c)^2}.\]

The largest range of $ p $ is: $ \left[\frac{1}{7}  ,   4 \right] $.

Thus,we have two nice inequalities:

 \[ \frac 1{a^2 \plus{} 7b^2 \plus{} 7c^2} \plus{} \frac 1{b^2 \plus{} 7c^2 \plus{} 7a^2} \plus{} \frac 1{c^2 \plus{} 7a^2 \plus{} 7b^2} \le \frac {9}{5(a \plus{} b \plus{} c)^2}.\]
Equality holds if and only if $ a=b=c $ or $ a=4b=4c $ and cyclic permutations.

\[ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2} \le \frac {9}{2(a \plus{} b \plus{} c)^2}.\]

[quote="Vasc"]Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \ge \frac {27}{(p \plus{} 2)(a \plus{} b \plus{} c)^2}$.[/quote]
The largest of $ p $ is :$ \left[\frac{11+3\sqrt{13}}{2} , \infty \right).$

For $ p= \frac{11+3\sqrt{13}}{2} $,equality holds if and only if $ a=b=c>0 $ or $ a>0;b=c=0 $ and cyclic permutations.


Find all polynomials $P(x)$ with real coefficients satisfying the equation \[(x+1)^{3}P(x-1)-(x-1)^{3}P(x+1)=4(x^{2}-1) P(x)\] for all real numbers $x$.

For $x=1$ we get $8P(0)=0$, hence $x\mid P(x)$, so put $P(x)=xQ(x)$. Then

$(x+1)^{3}(x-1)Q(x-1)-(x-1)^{3}(x+1)Q(x+1)=4(x^{2}-1)xQ(x)$

or

$(x+1)^{2}Q(x-1)-(x-1)^{2}Q(x+1)=4xQ(x)$

Rewrite this as:

$(x^{2}+1)(Q(x-1)-Q(x+1))=2x(2Q(x)-Q(x-1)-Q(x+1))$

If $\deg Q(x)=n\geqslant 2$, then $\deg (Q(x-1)-Q(x+1))=n-1$ and $\deg (2Q(x)-Q(x-1)-Q(x+1))=n-2$, hence the LHS is of the degree $n+1$ and the RHS is of the degree $n-1$. Contradiction. If $Q(x)=ax+c$, then after rearranging we get $-2a(x^{2}+1)=0\implies a=0$. Therefore $Q(x)=c, c\in\mathbb{R}$ and $P(x)=cx$

But logically , we get the contradiction $\forall n \in N $

As shown above it remains to solve the eqution

$(x^{2}+1)(Q(x-1)-Q(x+1))=2x(2Q(x)-Q(x-1)-Q(x+1))$

Now look at the free coeeficent of LHS.
It should be 0 in order for this equation to make sense but the free coeeficent of LHS is -1*deg(Q) or 1-deg(Q) (the reader should work out the details)  so either the polinom is of degree 0 or 1


analytic geometry
graphing lines
slope
geometry
angle bisector

In Cartesian coordinate, if $\triangle ABC$ is isosceles triangle $(AB=AC)$. Let the equation of line $AB$ is $2x-y+5=0$ and $AC$ is $3x+6y-1=0$. We know that $I(2,-1)$ is a points in line $BC$. Find the equation of line $BC$.

as the triangle is isosceles the angle bisector will be perpendicular to the third side.  the slopes of the two given sides are 2 and -1/2, ergo the angle bisector has a slope of -3.  This means that the third side has a slope of 1/3, and given that it passes through the point (2,-1), the equation of the line is x-3y=5.

[hide]the equation of AB is $y=2x+5$, and the equation of BC is $y=\frac{-x}{2}+\frac{1}{6}$.  so they are perpindicular.  and if we set the values for y equal, we get
$2x+5=\frac{-x}{2}+\frac{1}{6}$
$12x+30=-3x+1$
$15x=-29$
$x=-\frac{29}{15}$

and $y=\frac{17}{15}$.  so the point A is $(\frac{29}{15},\frac{17}{15})$

now if you graph the stuff we have so far, you see that point I is sort of "under" AB and AC.  and it seems like we didn't need to find point A after all...

anyway, since the slope of AB is 2, you would go (from B to A) right some distance we will call $a$, and up a distance of $2a$.  from A to C you go right $2a$ and down $a$ (the slope is $-\frac{1}{2}$).  so from B to C, it's a distance of $3a$ to the right and $a$ up, so the slope of BC is $\frac{1}{3}$

and since we know the point (2, -1) is on BC, the equation of BC is $y=\frac{x}{3}-\frac{5}{3}$[/hide]


topology
trigonometry
function
arithmetic sequence

Consider the topology on the set $ \mathbb{Z}$ of integers, whose open sets are precisely the unions of (infinite-in-both-directions) arithmetic progressions.

i)  Prove that this topology is metrizable.

ii)  Construct a compatible metric.

fedja, can you explain the motivation behind that?  I was tinkering around with $ p$-adic norms for awhile but didn't really get anywhere.  And could we generalize to cosets of principal ideals in a commutative ring?

Well, we need 2 things:
a) if the distance $ d(a,b)$ is small, then $ b$ is contained in a given arithmetic progression containing $ a$.
b) if $ b$ is contained in a progression with properly chosen difference, then the distance is small.

You just start with progressions with [b]one[b] fixed difference $ k$ (they also determine a topology, though it is not Hausdorff). Here any distance function $ d_k$ between the cosets $ \mathbb Z/\mathbb Z_k$ gives you what you want (I chose $ \left|\sin\frac{\pi(a\minus{}b)}{k}\right|$ just to make the formula funnier; since you want the discrete topology on cosets, the more natural choice would be $ d_k(a,b)\equal{}\begin{cases}0,& k|a\minus{}b\\ 1, & k\!\not|\,\,a\minus{}b\end{cases}$). The idea to add all quasidistances up with small coefficients is quite a standard one.

You can do pretty much the same for other commutative rings $ R$ as long as the number of the ideals you want to use is countable and for every $ x\in R$, there exists an ideal in your family not containing $ x$.


Point P is x inches from line r.  If there are exactly 3 points that are 3 inches from line r and also 5 inches from P, what is the value of x?

[hide="solution"] The locus of points that are 3 in from a line are two lines parrallel to the given line and 3 in away from it.  The locus of points 5 in from $P$ is a circle centered at $P$.  Since there are exactly three intersection points of these loci, we can deduce that one of the lines parrallel to $r$ creates a chord in circle $P$ (creating two intersection points) and the other line is tangent to circle $P$ (creating one intersection point).  Since circle $P$ has a radius of 5 in, a perpendicular drawn from this tangent line to $P$ is 5 in long.  This perpendicular is also perpendicular to line $r$ (since the two lines are parrallel).  Now we know that $P$ is on the other side of line $r$ and 5 - 3 = 2 in away from line $r$.

Does this make enough sense?
[/hide]

[color=red]Your explanation is somewhat clear.  Try not using math jargon when providing students with notes.  [/color]

[size=150]This is what I did:

1) Drew point P x inches away from line r

2) I drew 3 points from line r

3) I then said:

3 + x = 5 inches

I subtracted 3 from 5 and x = 2.

Correct steps?[/size]

I think I can speak for most of us here at AOPS. We believe that "Math Jargon", while complicated at first to learn, actually simplifies most problems. Using words like "locus" and "perpendicular" makes perfect sense in this sort of context. If you are looking for a straightforward, no jargon answer, then ask for one. Otherwise, ask for an explanation of the "jargon", which is really just standard vocabulary. Dumbing down is not really our motus operandi here. Not to be on your case personally, I realize that many people ask for an answer, get one, and then proceed to say that the answer is wrong, or makes no sense, when in reality they simply don't understand it.

Summary: we're glad to help, just ask for an explanantion instead of asking for dumbed-down lingo.

DiscreetFourierTransform:

I never ask for the answer but for STEPS in number sequence:

Steps:

1.

2.

3., etc

This is what I seek for ALL my questions.

Guide me through the questions and if you desire to provide the answer, please do so WITH STEPS.

Certainly. That's a perfectly reasonable request. Sorry for the flame. I've posted a solution to a different problem, with the steps, in the intermediate forum.


topology
real analysis
real analysis unsolved

Prove that a totally bounded and complete metric space is compact.

Note: The definition of compactness I am using is if a metric space is the union of a collection of open covers, and it is also the union of a finite subset of this collection, then E is compact.

There is many proof, here is my:

it's particular case of this fact.
Fact : metric space is compact if any suqeance contains convergent subsequance. and if any suquance in metric space  contain convergent subseqquance then metric space is compact.
Fact:
In bounded metric space any sequance contains convergent subsequance.
really, if  at any point of sequance  constract $ \epsilon$ ball then one of them will be contains infinity points of our sequance.

[quote="Kuba"][quote]
In bounded metric space any sequance contains convergent subsequance. [/quote]
really?what if you take the space to be the interval $ (0,1)$ with the usual metric and the sequence $ \frac 1n$.
U need completeness![/quote]
yes sure need  completeness. thanks

For each $ \varepsilon_n\equal{}\frac1n$ cover the space with $ M_n$ balls of radius $ \varepsilon_n$. You can do this because the space is totally bounded.
Take a sequence $ (x_m)_m$ in your space. We want to prove that it has a convergent subsequence.
For $ n\equal{}1$ you will have in some ball (denote it $ B_1$) infinitely many members of this sequence, take one an denote it $ x_{k_1}$.
For $ n\equal{}2$ your $ M_2$ balls of radius $ \frac 12$ cover the space and hence your $ B_1$. Since $ B_1$ contains infinitely many members of the sequence $ (x_m)_m$ choose a ball of radius $ \frac 12$ (denote it $ B_2$) so that in $ B_1\cap B_2$ lie infinitely many members of the sequence and choose one element $ x_{k_2}$ so that $ k_2>k_1$. This is obviously possible.
For $ n\equal{}3$ the $ M_3$ balls of radius $ \frac13$ cover the space, hence your ball $ B_2$. Since $ B_2$ has infinitely many elements of the sequence $ (x_m)_m$ it follows that there is a ball $ B_3$ so that inside $ B_3\cap B_2$ lie infinitely many members of your sequence; now you choose one of them $ x_{k_3}$ so that $ k_3>k_2$.
....
Observe that $ d(x_{k_1},x_{k_i})<2, \forall i>0$, $ d(x_{k_2},x_{k_i})<2\cdot\frac12, \forall i>1$ and in general
$ d(x_{k_n},x_{k_{i}})<2\cdot\frac1n, \forall i>n\minus{}1$.
It follows that this subsequence $ (x_{k_n})_n$ is a Cauchy sequence and since the space is complete it follows that it converges. q.e.d.

In general, in a topological space it is not enough to prove the compactness with the convergence of some sequence, you need to work with generalized sequences.
But in our case (and more generally, when you have a countable fundamental system of neighborhoods for every point of the space) it is enough to work with countable sequences.


The 120 permutations of the AHSME are arranged in dictionary order as if each were an ordinary five-letter word. The last letter of the 85th word in this list is:

$ \textbf{(A)}\  \text{A}   \qquad
\textbf{(B)}\   \text{H}  \qquad
\textbf{(C)}\    \text{S} \qquad
\textbf{(D)}\  \text{M}   \qquad
\textbf{(E)}\     \text{E} $

Hint
[hide]
It starts with the letter $\fbox{M}$
[/hide]

[hide]If the first letter is A, then the word is anywhere from the 1st to 24th word, and so on.  Since the words that start with M are in the 73-96 range, and 84 is between 1 and 100, we know that the first letter is M.  We can keep doing this for the rest of the letters, and get MESHA, making the last letter A.[/hide]

Strange, I got a different answer:

[hide]You have the 73rd being MAEHS. So, the 78th is MASHE, so the 79th is MEAHS. So, the 84th is MESHA. But, the 85th is MHAES. So, I'm inclined to say S.[/hide]

I believe this is correct

[hide]
We can quickly see that $24$ words can be formed with of the letters as the first letter in the word. Hence, we conclude that the 85th word must start with the letter M since $72<84<96$. From here we see that

MA $3 \cdot 2 \cdot 1=6$
ME $3 \cdot 2 \cdot 1=6$

Hence, you can form $6$ words starting with the string MA and $6$ starting with the string ME. Adding this to our previous total of $72$ we obtain $84$. Thus the 85th word must start with MH. From here it is easy to see that the word must be MHAES since it is the first word starting with MH and it must be in alphabetical order. Thus, the last letter of the 85th word is $\fbox{S}$

[/hide]

There is an algorithm for this problem.
It was invented by a high school student in the 1960's.
I have the detailed buried somewhere in my files,
but I remembered enough of it to apply it.[hide]

I agree with white_horse_king88.

Writing the 5-letter "words" in alphabetical order, the 85th is  M-H-A-E-S.[/hide]

Wrong! Wrong! Wrong! Wrong! Wrong!

All of you're [b]WRONG!!![/b]

 ;) Guys, think again and post the answer again. This was only #10 on AHSME 1986, it shouldn't be that difficult!

 :D

OK
Since there are 24 different permutations for each first letter, M is in the 73-96 range. Breaking it down further, we find that a second letter of H is in the 79-85 range. Since 85 is last, the next three letters should be in reverse alphabetical order, making the word MHSEA, with a final letter of $\fbox{A}$

List.

73rd: MAEHS
74th: MAESH
75th: MAHES
76th: MAHSE
77th: MASEH
78th: MASHE
79th: MEAHS
80th: MEASH
81th: MEHAS
82th: MEHSA
83th: MESAH
84th: MESHA
85th: MHAES

Well, my chart must be wrong.

I think this is right.

According to [url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/ahsme1986sol.pdf]here[/url], E is the right answer.

And I did the 84th word, not 85th  :blush:

The answer in the packet is incorrect. They say that the 85th word is the 86th - 72nd word beginning with MH, which is incorrect. It is either the 85th - 72nd word, or the 86th - 73rd word. They overcounted by 1, and thus came up with a different answer. If you don't believe me, I listed all of them from 73 to 85.

I agree. I get S too.
Another case of never trusting model answers..

Hmm..

Wait,

1-24th are the first 24 words with A.

25-48th are the second 24 words with E.
(48-25 = 23+1 = 24)

49-72th are the third 24 words with H.
(72-49 = 23+1 = 24)

73-96th are the 24 words with M.

85th is between them so, it's M_ _ _ _.

73 - MAEHS
74 - MAESH
75 - MAHES
76 - MAHSE
77 - MASEH
78 - MASHE
79 - MEAHS
80 - MEASH
81 - MEHAS
82 - MEHSA
83 - MESAH
84 - MESHA
85 - MHAES

Yes, the answer packet must be wrong. :)

[quote="Soroban"]There is an algorithm for this problem.
It was invented by a high school student in the 1960's.
I have the detailed buried somewhere in my files,
but I remembered enough of it to apply it.[/quote]

I made an algorithm, too.  It works like a weird number-base system.
In a normal base-10 system, you have the places going like this:
$10^3;10^2;10^1;10^0;10^-1;...$

In this system, it looks like this:
$4!;3!;2!;1!$
Now, write the number minus 1 in that base (the 1st word=0, 2nd word=1, etc):
3200
And arrange the letters in alphabetical order:
AEHMS

For the first number slot (from the left), whatever digit you have, you take that letter+1.  So a 0 corresponds to A, 1 to E, 2 to H, 3 to M, 4 to S.

Then the second digit is 2.  Apply the same thing, only this time cross out the M.  So it goes AEHS, 2->H

Repeat for the third digit.  0->A
Repeat for the last digit.  0->E
Now the final letter is whatever is left: S

IMPORTANT:
When you use this, you have to remember that our "human" way of counting, starting from 1, is not how it works.  It actually starts from 0.  That is the reason why you have all these "plus 1" or "minus 1" in there.  It would be much more "elegant" if the things were arranged with the number 0 corresponding to the 1st element (like it does in some computer programming languages).

It appears that you are using some variation of the [url=http://www.mathpages.com/home/kmath165.htm]factorial base[/url], which is pretty cool although it doesn't strike me as incredibly useful, except perhaps in situations like this.  (If anyone happens to know of it being useful, I'd love to hear about it.)

Hmm -- it appears that factorial base is useful for larger versions of the program as well, from a computing standpoint:
http://www.roard.com/docs/cookbook/cbsu2.html

[quote="Silverfalcon"]One typical type of problems you see is the this problem:

I saw this kind of problem about 6 to 8 times in DIFFERENT math competitions.

The 120 permutations of the AHSME are arranged in dictionary order as if each were an ordinary five-letter word. The last letter of the 85th word in this list is:

A) A
B) H
C) S
D) M
E) E[/quote]

[hide=""The Answer""]
In alphabetical order, the letters are A, E, H, M, and S.

1 - 24 = A
25 - 48 = E
49 - 72 = H
[b]73 - 96 = M[/b]
97 - 120 = S

73 - 78 = A
79 - 84 = E
[b]85 - 90 = H[/b]
91 - 96 = S

[b]85 - 86 = A[/b]
87 - 88 = E
89 - 90 = S

[b]85 = E, S[/b]
86 - S, E

S
[/hide]

I keep getting S as the answer as well. However, on the Answer Key on AoPS it says the correct answer is E. The link where I found the problem is here.

https://www.artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_10

[quote=ishankhare]Why the downvotes?[/quote]

Check the date, this is an 11 year old thread...


algebra
polynomial
number theory proposed
number theory

Let $f(x)$ be a nonconstant integer polynomial. Show that the set of $n \in \mathbb{N}$ for which $f(n)$ has no prime factor smaller than $n$ has density $0$.

And conclude, by a standard Sieve argument and without using any analytic methods, that $\sum_{p \mid f(n)}\frac{1}{p}= \infty$, where the sum is over all primes dividing $f(n)$ for some $n$.

Happy Christmas!


geometry
ratio
Triangle
area
IMO Shortlist
pentagon

For any set $S$ of five points in the plane, no three of which are collinear, let $M(S)$ and $m(S)$ denote the greatest and smallest areas, respectively, of triangles determined by three points from $S$.  What is the minimum possible value of $M(S)/m(S)$ ?

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

I think it's achieved for the regular pentagon, so it's $\phi=\frac{1+\sqrt 5}2$.

First of all, let's notice that if it's not a convex pentagon, then we have a point inside a triangle, so that point is inside a triangle formed by the centroid and two vertices of the triangle, so the ratio would be at least $3>\phi$. This means that the pentagons which reach the minimal value are convex.

Furthermore, it's easy to see that, since the pentagon is convex, the triangle with the minimal area is formed by three consecutive vertices, while the one with the largest area is formed by three vertices which are not consecutive.

Assume our pentagon is $ABCDE$, and the triangle with minimal area is $BAE$. Since we can apply an affine transformation which maps $A,B,E$ to three consecutive vertices of a regular pentagon and this transformation preserves the ratios of areas, we may assume there is a regular pentagon $ABC'D'E$. Take $X$ s.t. $C'X\|BD',\ D'X\|EC'$. Since $S(C'AE)\le S(CAE),\ S(D'AB)\le S(DAB)$, and given the convexity of $ABCDE$, at least one of $C,D$ lies inside $C'XD'$ (just draw a picture to see why). Assume it's $D$. We can move $C,D$ along lines parallel to $BD$ s.t. the area of $BCD$ increases, and this time $C,D$ lie on the union of the two segments [$XC'$], [$XD'$]. It's easy now to see that we can enlarge the area of $BDC$ even more by moving $C,D$ s.t. they reach positions $C',D'$, but this means that the initial area of $BCD$ was smaller that that of $BAE$, and this is a contradiction, so the minimal ratio is reached for regular pentagons.

I know the argument is a bit murky, but it's not that hard to see why it works if you draw a picture.

[quote="grobber"]Furthermore, it's easy to see that, since the pentagon is convex, the triangle with the minimal area is formed by three consecutive vertices, while the one with the largest area is formed by three vertices which are not consecutive.[/quote]
While this is not necessarily true, the case in which the triangle with largest area consists of three consecutive vertices easily gives a ratio of $\ge 2$.

BTW, another way to deal with this convex case is to affine transform the largest triangle into the [url=http://mathworld.wolfram.com/GoldenTriangle.html]golden triangle[/url].

Indeed, if $ABC$ is the maximal triangle, then if $B'$ is selected so that $ABCB'$ is a parallelogram, then $D,E$ must lie within $AB'C$, which means that $[ADC]+[DEC]\le[ABC]$.

No suppose $ABD$ is the maximal triangle.  Take any line $g$ which passes through $B$ and intersects the interior of segment $AD$. Since $ABD$ is maximal, $E$ must lie within $ADB'$, where $ABDB'$ is a parallelogram.  So the line through $E$ parallel to $g$ intersects the interior of $AD$ as well.  Let $x,y$ be the distances from $B,E$ to $AD$ along an axis parallel to $g$, and let $a,b, y+z$ be the distances from $A,D,E$ to $BC$ along the same axis (positive distances).  For the purposes of this problem, we can assume WLOG, $a \le b$.  So $x,z \ge a$.

If $\theta$ is the maximal area ratio, then $y/x = [AED]/[ABD] \ge \frac{1}{\theta}$.  Also $(z+y)/a =[EBC]/[ABC] \le \theta$, which means that  $1+ \frac{1}{\theta}  \le \theta$, and we get the golden ratio solution, with equality in the case of the regular pentagon.

While not really any shorter than the other solutions, I thought I'd post this solution because it does give a sense why the golden ratio is the answer.

Solution with coordinate
[hide="solution"]
If there exists a point $P$ inside a triangle formed by three other points $\triangle ABC$, then $[ABP] + [BCP] + [CAP] = [ABC]\implies \dfrac{M(S)}{m(S)} \ge 3$.
Otherwise, the convex hull of the 5 points must be a pentagon (because if not, then we arrive at the previous scenario again).
Since affine transformation preserves area ratios, WLOG let the triangle with maximal area have coordinates $A(1,0), B(-1,0), C(0,1)$.
Let $D(d_x, d_y), E(e_x, e_y)$. By maximality of area $D, E$ must be contained in the region $y\le 1, y \ge x-1, y\ge -x-1$. $\triangle ABC$ divides this region into 4 triangles congruent to it.
If $D, E$ are in the same region out of the three not contained in $\triangle ABC$, WLOG by affine transformation $D, E$ are in the region bounded by $y\le 0, y\ge x-1, y\ge -x-1$. I claim $\dfrac{M(S)}{m(s)}\ge 2$ in this case. Indeed, in order for $[DAB] \ge \dfrac{1}{2}[ABC]$, then $d_y \le -\dfrac{1}{2}$, and similarly $e_y \le -\dfrac{1}{2}$. But then $[ADE]\le \dfrac{1}{2}[ABC]$, proved.
Otherwise, $D, E$ are in different regions. WLOG let $D$ be in the region bounded by $y\le 1, y\ge x-1, y\ge 1-x$ and $E$ in the region bounded by $y\le 1, y\ge -1-x, y\ge 1+x$. I claim $\text{min}\{[DAB], [EAB], [CDE]\}\le \dfrac{\sqrt{5}-1}{2}$.
Proof:
Clearly we may increase $d_x$ and decrease $e_x$ as that increases the areas of all three triangles; thus, we may assume $d_y = d_x-1$ and $e_y=-e_x-1$.
Clearly $[DAB] = d_y=d_x-1$ and $[EAB] = e_y = -e_x-1$. By Shoelace we can find $[CDE] = \dfrac{1}{2}|d_xe_y-e_xd_y-d_x+e_x| = |e_xd_x+d_x-e_x|=|(e_x+1)(d_x-1)+1|$. Let $p=d_x-1, q=-e_x-1$; then $p, q\in [0, 1]$ we wish to prove $$\text{min}\{p, q, 1-pq\} \le \dfrac{\sqrt{5}-1}{2}$$ WLOG $p\ge q$; if $q\ge 1-pq\iff q\ge \dfrac{1}{p+1}$ then we can shift $p$ down until $q\in \left\{p, \dfrac{1}{p+1}\right\}$ and increase the minimum. Similarly, if $q\le 1-pq \iff q\le \dfrac{1}{p+1}$, we can shift $q$ up until $q\in \left\{p, \dfrac{1}{p+1}\right\}$ and increase the minimum. Thus, we just need to check those two cases.
Case 1: $p=q$. Then $\text{min}\{p, 1-p^2\}$ is maximized when $p=1-p^2\implies p=\dfrac{\sqrt{5}-1}{2}$.
Case 2: $q=\dfrac{1}{p+1}$. Then $\text{min}\left\{p, \dfrac{1}{p+1}, 1-\dfrac{p}{p+1}\right\} = \text{min}\left\{p, \dfrac{1}{p+1}\right\}$ is maximized when $p=\dfrac{1}{p+1}\implies p=\dfrac{\sqrt{5}-1}{2}$.
Either way, we see $\text{min}\{[DAB], [EAB], [CDE]\}\le \dfrac{\sqrt{5}-1}{2}$, so $\dfrac{M(S)}{m(S)} \ge \dfrac{1}{\frac{\sqrt{5}-1}{2}} = \boxed{\dfrac{1+\sqrt{5}}{2}}$
Equality case is obtained when $ABCDE$ can be affine transformed to a regular pentagon.
[/hide]

The answer is $\phi=\frac{1+\sqrt{5}}{2}$, achieved for example at a regular pentagon. Suppose for sake of contradiction that we had a set $S$ such that for any two triangle areas $\alpha$ and $\beta$, we have \[\alpha/\beta \in (1/\phi,\phi).\] We have the following claim.

[b]Claim:[/b]  The convex hull of $S$ has exactly five points.

[i]Proof:[/i]  First, suppose that it had only three points, say $ABC$, and $DE$ were inside triangle $ABC$. Then, one of $[DAB],[DBC],[DCA]$ is at most $\frac{1}{3}[ABC]$, so we can get a triangle area ratio of at least $3$, which is a contradiction.

Now, suppose that it had four points, say $ABCD$ in that order, with $E$ inside the quadrilateral. One of $[EAB],[EBC],[ECD],[EDA]$ is at most $\frac{1}{4}[ABCD]$, and one of $[ABC]$ and $[CDA]$ is at least $\frac{1}{2}[ABCD]$, so we have a triangle area ratio of at least $2$, which is a contradiction. Thus, the convex hull must have at exactly five points. $\blacksquare$

Suppose the points form a convex pentagon $ABCDE=A_1A_2A_3A_4A_5$ in that order.

[b]Claim:[/b]  The minimal area triangle is $\triangle A_iA_{i+1}A_{i+2}$ for some $i$ (indices mod $5$).

[i]Proof:[/i]  Suppose that WLOG $\triangle ABD$ was instead of minimal area, so in particular, $[ABD]\le [ABC],[ABE]$. Let $AB$ be the $x$-axis, with $A$ to the right of $B$. The condition is then saying that the $y$-coordinate of $D$ is at most that of $C$ and $E$.
[hide=Figure 1][img]https://lh3.googleusercontent.com/-xu6JL2HSp98/X5EhyxylVwI/AAAAAAAAF2w/77_yByEASEkqRy0HTPTCliTmMoRChJNawCK8BGAsYHg/s0/2020-10-21.jpg[/img][/hide]
Note that $D$ has to be in the region bounded by the half plane with line $BC$ on the same side as $A$, the half plane with line $AE$ on the same side as $C$, and the half plane with line $CE$ on the opposite side as $A,B$. This is a convex region, and by cutting with the line $y=N$ for large enough $N$ so that it is above $D$, we get a convex polygon that $D$ must lie in. Thus, the point in this polygon with smallest $y$-coordinate must be one of its vertices, and not on $y=N$ (or potentially $BC\cap AE$, as that is further along ray $BC$ than $C$), so it must be $C$ or $E$. Thus, one of $C$ or $E$ has a lower $y$-coordinate than $D$, which is the desired contradiction. $\blacksquare$

WLOG, let $\triangle ABC$ be of minimal area. Affine transform the picture so that $AB=BC$ and $\angle ABC=3\pi/5$. Note that affine transformations preserve ratios of signed areas, so it preserves that $ABCDE$ is a convex pentagon in that order, that $\triangle ABC$ is of minimal area, and all the unsigned triangle area ratios still lie in $(1/\phi,\phi)$.

So we have a convex pentagon $ABCDE$ in that order such that $AB=BC$ and $\angle ABC=3\pi/5$,$[ABC]$ is the minimal area triangle, and that all triangle area ratios are in $(1/\phi,\phi)$. We will derive a contradiction.
[hide=Figure 2][img]https://lh3.googleusercontent.com/-xOmgElUMNpc/X5Eh0JTk2gI/AAAAAAAAF20/giU-pN5a77M7FUlXjNahwUFQz6vI4SSPACK8BGAsYHg/s0/2020-10-21.jpg[/img][/hide]
Indeed, complete $ABC$ to a regular pentagon $ABCD'E'$. We see that $[ABD],[ABE]\ge [ABC]$, so $D$ and $E$ lie on the opposite side of $CE'$ as $A,B$, and similarly $D,E$ lie on the opposite side of $AD'$ as $B,C$. Furthermore, $[ABD],[ABE]\le[ABD']$ (else we have an area ratio of $>\phi$), so they lie on the same side of $\ell_1$ as $A,B$, where $\ell_1$ is the line through $D'$ parallel to $AB$. Similarly, $D,E$ lie on the same side of $\ell_2$ as $A,B$ where $\ell_2$ is the line through $E'$ parallel to $BC$. Let $X=AD'\cap CE'$ and $Y=\ell_1\cap\ell_2$.

All of this shows that $D$ and $E$ are in the region $XD'YE'$. For now, imagine that $D$ and $E$ vary over this region. We show that $[ADE]$ is maximized at when $\{D,E\}=\{D',E'\}$. Imagine fixing $D$, and consider a varying line parallel to $AD$. We see that $E$ must be on a line as far from $AD$ as possible, and since the direction of $AD$ is between that of $AD'$ and $AE'$, this implies that $E$ must be one of $D'$ or $E'$ (the direction is to eliminate the possibility of $X$ or $Y$). Now, holding $E$ fixed at that location, varying $D$ shows it must be at the other one of $D'$ or $E'$. Now, going back to the original locations of $D$ and $E$, this shows that \[[ADE]\le[AD'E'] = [ABC],\] so equality must hold (as $\triangle ABC$ is of minimal area), so $\{D,E\}=\{D',E'\}$, so $ABCDE$ is a convex pentagon. The ratio value is now $\phi\not\in(1/\phi,\phi)$, so we have our desired contradiction (finally!).

The source of this problem is the following article:

MR0286678
Paul Erdős, Amram Meir,  Vera T. Sós, Pál Turán:
On some applications of graph theory. II.
Studies in Pure Mathematics (Presented to Richard Rado) pp. 89–99
Academic Press, London, 1971.


group theory
abstract algebra
algebra
function
domain
Ring Theory
superior algebra

Prove that any finite local ring has prime-power cardinality.

If $ R$ is the ring and $ I$ is the maximal ideal, $ R/I$ is a field with characteristic $ p$. Since integers not divisible by $ p$ are invertible in $ R/I$, they are invertible in $ R$. If $ q|\#(R)$ for some prime $ q$, the additive group of $ R$ has some subgroup of order $ q$, and thus $ q$ is a zero divisor. By the preceding, $ q$ must be $ p$, and only one prime divides the size of $ R$.

This follows immediately from first properties of Artinian (DCC) rings.
First, every prime $ P$ is max, since $ R/P$ is a DCC domain so a field.
Since R by  hypothesis has only one max ideal it must be  $ (p) \equal{} char R$.
Finally, R is Noetherian, so every ideal contains a product of prime ideals.
In particular, $ (0)$ contains a product of the unique prime $ (p)$, i.e. $ p^n \equal{} 0$.


[quote=BOGTRO][quote="GameBot"]In an isosceles trapezoid, the lengths of the bases are 8cm and 16cm. The length of a diagonal is 13cm. How many square centimeters are in the area of the trapezoid?[/quote]

The height is clearly 5. Thus, $ 5 \cdot \frac{24}{2}\equal{}60$.[/quote]

The height being 5 is something we do not yet know (even though MATHCOUNTS® does generally provide you with integer answers unless otherwise specified). To get this, you can draw a picture. The top base is 8 and the bottom base is 16. Since the trapezoid is isosceles, we will drop down two altitudes on both sides of the top base. This divides the bottom base into segments of 4, 8, and 4. We were given a diagonal was 13, so we now have a triangle with sides of length $h$, 12, and 13 in the trapezoid where $h$ is the height. We now have it being 5. The area of the trapezoid is then $\boxed{60}$


Sometimes when i post and click "home" it takes me to MathLinks and sometimes to AoPS forums.

Anybody know why?


function
number theory
prime numbers
number theory proposed

Are ther  a function $f: N^*\to Q$ satisfinge  this 3 conditions
1)$f(x)\in{N}$ <=>$x\in{P^*}$
2)If $x\in{P^*}$ then $2(f(x))^2=y^2+z^2+2$ for some $y;z\in{N}$.
3)$f(x)=f(y)$<=>$x=y$

Remark: $P^*$ is the set of all prime numbers of $4k+1$ form.
$N^*$ is the set of all odd positive integer numbers.

[quote="jin"]I think we only need to discuss on $P^*$,And that's easy,isn't it?[/quote]
I can't understand what do you mean $jin$,please explane ;) .

[quote="jin"]I think we only need to discuss on $P^*$,And that's easy,isn't it?[/quote]
$jin$ you havn't answer to my question!,what did you mean?
Guys please try to solve this problem,I really want to know are there another solution,different from mine. :)

[quote="jin"]Because the two condition is only about $x\in P^*$,so if $x\not\in P^*$
$f(x)$ can be any rational but not integer.[/quote]
$jin$ if you have solved this problem,write solution with all detals,please.

[quote="jin"]For example,$f(x)=3,x\in P^*$,otherwise $f(x)=\frac{1}{3}$.Is it right?[/quote]
oops... :oops: ,sorry I have missed a condition in my problem,I have edited my post.

[quote="jin"]Then we just need to show there are infinite $m$
such that $2m^2-2$ can be expressed as the sum of two square.Isn't it?[/quote]
I am not sure that it is finishing solution :? .
Anyway let that there are infinite $m$
$2m^2-2$ can be expressed as the sum of two square,what then?

[quote="{x}"]Is $f$ supposed to be a bijection $\mbox{Odd} \longleftrightarrow \mathbb{Q}$, or just $\mbox{Odd} \longleftrightarrow \mbox{Im} f$ ?
In the first case $f$ cannot exist, because $f$ is not surjective since there is no $x \in P$ such that $f(x)=2$.
In the second case $f$ [i]should[/i] exist.[/quote]
So is there a function,satisfing to all 3 conditions? :?:

[quote="{x}"]Answer my question please. I [i]believe[/i] it should exist, in the case of a bijection between $\mbox{Odd}$ and $\mbox{Im } f$.[/quote]
No,that doesn't need to be a bijection between $odd$ and $Q$. ;)


I've been accepted by UC Irvine, UC Santa Barbara, UC Davis, and RPI. At the moment I'm leaning toward Davis or RPI, and have visited both. I'm going in as a pure math major, but I may switch to phyics, CS, engineering, or some combination thereof. Can anyone here tell me about the math departments at these schools? How are the professors? Are there any specific areas in which they are particularly good or bad? Are the students engaged in what they're learning outside of the classroom (this is particularly important to me)? I'm still undecided, and am trying to get all the input I can.

One of these things is not like the other...

Irvine, Davis, and Santa Barbara all have respectable mathematics research going on there.  In particular, I think of Irvine as being a department on the rise.  I know nothing about RPI.  If I were you, the quality of the math department wouldn't be the deciding factor for me.  (In particular, the overall reputation in science and engineering is probably important to you.)


For a real constant number $a,$ let $f(a)$ be the minimum value of the quadratic function $y=x^2-2ax-2x+3a^2+7a+3$ at $-1\leq x\leq 1.$

(1) How many solutions are there satisfying the equation with respect to $a,\ f(a)=6?$
(2) Find the minimum value of $a$ for which satisfies $f(a)=6.$

[quote="kunny"]For a real constant number $a,$ let $f(a)$ be the minimum value of the quadratic function $y=x^2-2ax-2x+3a^2+7a+3$ at $-1\leq x\leq 1.$

(1) How many solutions are there satisfying the equation with respect to $a,\ f(a)=6?$
(2) Find the minimum value of $a$ for which satisfies the condition $(2).$[/quote]

Hi kunny, nice problem but...
(2) is self-reference... can you explain a bit more where is condition (2)?  :?

Sorry for that, I have just edied. :(

kunny

[hide]$y'=2(x-a-1)=0$, $x = a+1$
Substituting x=a+2, $f(a)=2a^2+7a+2=6$, $2a^2+7a-4=0$, $a=-3$ or $a=\frac{1}{2}$

So I guess that makes
1. two solutions and
2. makes $a=-4$ the minimum a value?
[/hide]

EDIT: 0+1 = 2... I cant add :( Answer is -3

shouldnt it be $x=a+1$? but im still not sure i follow your solution..

Anyway, three cases must be discussed: if $y$ is increasing on $[-1,1]$, then $f(a)=y(-1)$; if $y$ is decreasing on $[-1,1]$, then $f(a)=y(1)$; and otherwise $f(a)=y(x_0)$ where $x_0$ is such that $y'(x_0)=0.$

[hide]

letting $g(x)=y$ ,then $g'(x)=2x-2a-2$ . Hence $g'(x)=0$ gives $x=a+1$

(i) When $-1\leq a+1\leq 1$ or $-2\leq a\leq 0$ , the min is when $x=a+1$ , hence

$f(a)=g(a+1)=2a^2+5a+2$

(ii) When $a+1<-1$ or $a<-2$ , the min is at $x=-1$ so

$f(a)=g(-1)=3a^2+9a+6$

(iii)When $a+1>1$ or $a>0$ the min is at $x=1$ so

$f(a)=g(1)=3a^2+5a+2$

(1) for $f(a)=6$ ,
(i) case gives $2a^2+5a-4=0$ $\implies a=\frac{-5\pm\sqrt{57}}{4}\not\in [-2,0]$
(ii) case gives $3a^2+9a=0$ $\implies a=-3$  because $0\not\in (-\infty,-2]$
(iii) case gives $3a^2+5a-4=0$ $\implies a=\frac{-5+\sqrt{73}}{6}$ because $\frac{-5-\sqrt{73}}{6}\not\in [0,\infty)$

So there are only 2 solutions

(2) from above cases , the minimum is $a=-3$
[/hide]

[quote="K81o7"]Isn't #2 trivial because you can plug in $a$ for $x$, yielding a quadratic?

As is #1...[/quote]

No, because $f(a)\neq y(a)$


Sa se demonstreze ca pt orice $n \in \mathbb{N}$ numarul:

                                 $x=20^{n}+1$ nu este patrat perfect

Daca $n$ este par atunci $20^{n}+1=M_{3}+2$, evident, nu e patrat perfect.
Daca $n$ este impar treaba e cu ceva mai complicata.
$20^{n}+1=M_{19}+2$
Dupa o verificare stresanta iese ca un patrat perfect nu poate da restul $2$ la impartirea cu $19$

Vezi [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=perfect&t=91539]aici[/url] o problema mai tare.

Apropo, $\left( \frac2{p}\right) = (-1)^{\left( p^{2}-1 \right)/8}$, unde $\left( \frac{t}{p}\right) = \begin{cases}1, \, \, \, \textrm{daca}\, \, t \, \, \textrm{este patrat modulo}\, \, p \\-1, \, \, \, \textrm{daca}\, \, t \, \, \textrm{nu este patrat modulo}\, \, p \\ 0, \, \, \, \textrm{daca}\,\, p \mid t \end{cases}$ este simbolul lui Legendre.
Rezultă că $\left( \frac2{19}\right) = (-1)^{45}=-1$, deci $2$ nu este rest pătratic modulo $19$.


.
Also how to deduce, that $ R$ not homeomorphic $ R^{n}$ at $ n \geq 2$?

Answer to the second question: If you remove a point from $ \mathbb R$ it becomes non-connected. But that won't happen for $ \mathbb R^{n}$

yes you are right. But you must say that :
if $ X$ \ {$ a$} is not Homeormophic to $ Y$ \ {$ b$} for all $ a\in X$  and for all $ b \in Y$
then $ X$ is not homeormophic to $ Y$.
 :blush:


geometry
3D geometry
pyramid
geometric transformation
rotation
reflection
symmetry

http://img5.imageshack.us/img5/7826/74271509.jpg
Find Area of DEF If
ar(ADE) = 5 sq units
ar(EBF) = 4 sq units
ar(DFC) = 3 sq units
Find ar(DEF)

...............................Changing units, assume wlog a unit mass for the ball,
let $g,f(v), h,t_0,t_1, v_0,v_1$ be the gravity, air resistance, height, up and down times,
initial and terminal speed;
we use the Integral Chebyshev inequality:
"for any continuous monotone $g,h$ with the same monotonicity, $\int_a^bgh\le\frac 1{b-a} \int_a^bg.\int_a^bh, \; a<b$,
with reversed inequalities for different monotonies".

[b]1)[/b] First assume $v$ vanishes just before the ball is caught ($h$ is maximum):
define $(\varphi,\psi)=(\frac  1{g-f},\frac 1{g+f})$;
$\bullet$  for the up flight, the Newton's principle yields $\begin{cases}\; dv=-\big(g+f(v)\big)dt\Rightarrow~t_0=\int_0^{v_0}\psi\\\;
vdv=-\big(g+f(v)\big)dx\Rightarrow~h=\int_0^{v_0}v\psi(v)dv\end{cases}$
$\bullet$  for the down flight, we get changing the sign of $g:\;\begin{cases}\; t_1=\int_0^{v_1}\varphi\\\; h=\int_0^{v_1}v\varphi(v)dv\end{cases}$
now $f$ increasing as $\varphi$ and $-\psi$ yields by the lemma :
$v_0t_0=\frac{v_0^2}{v_0}\int_0^{v_0}\psi\le 2\int_0^{v_0}v\psi(v)dv=2\int_0^{v_1}v\varphi(v)dv\le \frac {v_1^2}{v_1}\int_0^{v_1}\varphi=v_1t_1$,
hence the equalities with $\varphi>\psi$ forces $v_1<v_0$ and $v_0t_0<v_1t_1\Rightarrow~t_0<t_1$;
the down flight is the longest.

[b]2)[/b] Now for the general case, the value of $t_0$ is lower by $\int_V^{v_0}\psi<\int_0^{v_0}\psi=t_0, \; V>0$
 and $t_1$ is still the same, hence the down flight is still the longest.


algebra
polynomial
vector
linear algebra
matrix

Let $ T$ be a linear operator on the finite dimensional vector space $ V$. Suppose $ T$ has a cyclic vector. Prove that if $ U$ is any linear operator which commutes with $ T$, then $ U$ is a polynomial in $ T$.

More generally, a matrix $ A$ is nonderogatory if and only if every matrix that commutes with $ A$ is a polynomial in $ A$. See [i]Matrix Analysis[/i], by Horn, Johnson $ 1985$.
[url=http://books.google.com/books?id=9wTacOjHE6IC&pg=PA135&lpg=PA135&dq=nonderogatory+if+and+only+if+polynomial&source=web&ots=1yHmkV_JAY&sig=4gkFvrsWTVKHYXoBNN-rdqHI3po&hl=en&sa=X&oi=book_result&resnum=8&ct=result#PPA135,M1]Available in google books[/url]

Hi Captain,
"More generally" is not exact. In fact it's an equivalence by the  "cyclic vector theorem".

Cyclic vector theorem is nothing but the argument that I have stated in my previous post. By more generally, I wanted to say that somehow the converse to the statement in the original post is also valid. But as you said it's not really "generalization". Anyway thanks for pointing that out, loup blanc. I've always appreciated your posts.


$ \frac {6!}{4!} \equal{} \frac{6\times5\times\cancel{4\times3\times2\times1}}{\cancel{4\times3\times2\times1}}\equal{} 6\times5 \equal{} \boxed{30}$.


on 15 October 2008, tomorrow, application for cambridge will be closed, do you think tomorrow will be possibile to apply? o


Anyone play, and what's your USCF rating if you do?

My USCF is 1613.

I'm in the San Francisco Bay Area, a hotbed of chess activity :)

I play, USCF 1477 but outdated about 3 years or something.  I think I've improved since then but probably not by that much.  I don't play real openings :D.

I don't play real openings. I play the stupidest openings and get away with them because of the strength of my opposition... lol.

[quote="MithsApprentice"]Haha. You missed the big chess tournament last summer. :-P

Ah well. There's a lot of very talented chess players here. ^_^[/quote]

There's big chess tournaments every other week... at least in the adult chess world, there are.

I play in both kids and adult tournaments.  :D

Starwars and DDR are just too cool.

:starwars:  :starwars:  :starwars:  :ddr:  :ddr:  :ddr:

we dont have any chess tournaments around where i live that i know of... only one we had was for a club at school, but we only met once...=/ (well i'm not that good at chess anyways lol)

Someone at my school (David Chock) has USCF 1913

Peter, I remember playing you two years ago at some tournament I can't remember which one. I played your brother Andrew as well. In any case, I'm well aware of Saratoga's chess team and how they've got the Garg brother/sister combo and David Chock. I also seem to remember at BAMM that they're pretty good at bughouse.

<-- 1590 but my membership expired and i havnt played any tournements this year. I dont think i have a rating anymore. aznphatso, I want to play you sometime. Do you play Yahoo Chess?

our chess club stinks  :furious:  :furious:  :furious:  :furious:  :help:  :help:  :help:

Well, noneoftheabove, you're still doing better than I am--my school doesn't have a chess club  :(   So instead, I play my friends, none of whom are very good at all (I've lost exactly once and had exactly one stalemate out of, oh, the last 30 games I've played against them)  I don't have an 'official' rating because I'm too lazy to find out how to get one.
Every now and then, I get a newsletter in the mail about local chess tournaments, where I generally get 2nd or 3rd.

i used to play.... 4 years ago?
my rating was 1275 when i quit. i wish i hadnt though.

My Yahoo username is [i][b]seans41k[/b][/i] and I have a rating in the 1500s.  If someone wants to play me, pm me.  How about every list your Yahoo chess usernames?  ;)

My yahoo is Ameem91. Its in the 1500's too but it used to be like 1700, untill i started to play more Blitz  :D

A google search possible could have helped (e.g. "go game").  It's an Asian board game (at least it's most popular there) that is based on surrounding territory using stones (no special pieces).  You can find more details with a google search.

I used to be around 1000 in 5th grade, but I stopped playing.

I like math more than chess--plus, math is more important anyhow.

it seems people who are good at math are usually good at chess i wonder why

i don't have a uscf rating but my friend got a 1450 recently (his first tournament) 4.5 wins, 7th grader playing in a 7-12 grade competition and beat everybody else by at least one point. Good for a seventh grader who just started playing chess last year don't you think?! (yeah he is one of the best in our school)


Hi. I run on a Mac. I'm currently typing a mathematics assignment with TeXShop. How do you tpye in pictures? Like a triangle?

If you look around this site you will find:
1. [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]Pictures[/url] where you draw the picture outside of LaTeX then include it in your document
2. [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#xy]xypic[/url] which interprets LaTeX code to produce drawings
3. Look for graphics packages recommended by others in this forum and in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url]. Some of them are visual and generate LaTex code.

Many of the suggestions will work on the Mac.


[hide="hint"]$ 1 \plus{} \tan^2\theta \equal{} \sec^2\theta$ and $ 1 \plus{} \cot^2\theta \equal{} \csc^2\theta$.
Just a matter of plugging in now.[/hide]

-Don't title your topics LIKE THIs.
-This belongs in High School Basics.

[hide="Solution"]
So we plug in from xpmath's formula.

$ \frac{\tan ^2 \theta }{\sec ^2 \theta } \plus{} \frac{\cot^2 \theta }{\csc^2 \theta } \equal{} 1$

which is obviously true, since $ \cot A\equal{}\frac{1}{\tan A}$ and $ \sin^2 A\plus{}\cos^2 A\equal{}1$[/hide]


[quote="john lai"]10*9*8*7*6*5*4*3*2*1 = 10!
Can this be true?! Why or why not?[/quote]
It is absolutely true.
I think you must not send like questions to this forum.

[quote="john lai"]10*9*8*7*6*5*4*3*2*1 = 10!
Can this be true?! Why or why not?[/quote]
That is the definition of 10!. What you said is similar to saying $3^{4}=3*3*3*3$ or $3+3+3+3=3*4$. It is the definition of a factorial.

[quote="john lai"]10*9*8*7*6*5*4*3*2*1 = 10!
Can this be true?! Why or why not?[/quote]
It [i]is[/i] true. This is because what you wrote is a definition.
In case you're confused, the '!' is the sign that means factorial, and factorial means multiply that number times every whole number less than it until you get to one. Factorials are very useful in counting problems, so look some up using the search function at the top of your screen

[quote="b-flat"][quote="moogra"][quote="john lai"]10*9*8*7*6*5*4*3*2*1 = 10!
Can this be true?! Why or why not?[/quote]

well of course it can be true. 10! is defined as that (assuming that is is base 10)[/quote]

It doesn't matter the base (so long as it is $\geq10$ so that there would be the digit "9").[/quote]
No, that's not right. For example, if that was written in base 11 and then we converted it back to base 10 we would get

11*9*8*7*6*5*4*3*2*1=11!

Oops, forgot about that.

It would work in any base if it was $9!$  :blush:

It would work in bases $k$ and $l$ if it was $(\min(k,l)-1)!$ :D

I think the original author was intending to have two different interpretations:

10*9*8*7*6*5*4*3*2*1 = 10 (exclamation mark)
and
10*9*8*7*6*5*4*3*2*1 = 10! (factorial)

So depending on how you interpret it, the equation can go both ways.

But the question "can..."... that's sort of dumb  :roll:

I think he might have seen somewhere where it said

10*9*8*7*6*5*4*3*2*1=10!

not realizing it was factorial, instead thinking that the LHS was simply 10. This might have been the reason for "can..."


A man plants twelve rows of beans, each row twenty feet long. The plants in each row are spaced four inches apart. If the first and last plants in each row are four inches from the ends of the row, then the
total number of plants is:

a. 732 b. 720 c. 718 d. 708 e. NG

First, how many beans are in each row?
Since each bean takes up 4 inches of space, and there is an extra 4 inches, then the number of beans in each row is $ \frac{20\cdot12\minus{}4}{4}\equal{}59$ and of course the total is $ 59\cdot12\equal{}\boxed{d. 708}$

Pascal's way is the same way I did it. Btw, sri um is this your homework, because you have been spamming this whole forum.


The opposite inequality is correctfor all real numbers(that don't make any of the donominators 0, of course).
For a short proof, apply ab<=(a^2+b^2)/2 and then change x=a^2+b^2 and so on
The inequality will then be reduced to (x+y+z)(1/x+1/y+1/)>=9 for positive x,y and z and the last one holds according to Cauchy-Schwartz.


floor function
inequalities
induction
ratio
Divisibility Theory

Let $m$ and $n$ be arbitrary non-negative integers. Prove that \[\frac{(2m)!(2n)!}{m! n!(m+n)!}\] is an integer.

For all primes $p$ let $v_{p}(n)$ be the $p$-adic (normed, additive) valuation of $n$, thus the greatest integer $k$ such that $p^{k}|n$.

Lemma 1 (well known):
$v_{p}(n!) = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^{k}}\right\rfloor$.

Lemma 2 (trivial):
$m|n \iff v_{p}(m) \leq v_{p}(n)$ for all primes $p$.

Proof of the theorem itself:
By lemma 2, we just need to show that $v_{p}(m! n! (m+n)!) \leq v_{p}((2m)! (2n)!)$ (for all primes $p$).
By lemma 1, we have:
\[v_{p}(m! n! (m+n)!) = v_{p}(m!)+v_{p}(n!)+v_{p}((m+n)!) = \sum_{k=1}^\infty \left\lfloor \frac{m}{p^{k}}\right\rfloor+\left\lfloor \frac{n}{p^{k}}\right\rfloor+\left\lfloor \frac{m+n}{p^{k}}\right\rfloor\]
and
\[v_{p}((2m)! (2n)!) = v_{p}((2m)!)+v_{p}((2n)!)= \sum_{k=1}^\infty \left\lfloor \frac{2m}{p^{k}}\right\rfloor+\left\lfloor \frac{2n}{p^{k}}\right\rfloor ,\]
thus it suffices to show that
\[\left\lfloor \frac{m}{x}\right\rfloor+\left\lfloor \frac{n}{x}\right\rfloor+\left\lfloor \frac{m+n}{x}\right\rfloor \leq \left\lfloor \frac{2m}{x}\right\rfloor+\left\lfloor \frac{2n}{x}\right\rfloor\]
for all positive reals $x$ (here $m,n$ can be real, too).

If we write $m=ax+r$ and $n=bx+s$ with $0\leq r,s <x$, we get that the required inequality is equivalent to
\[a+b+a+b \left\lfloor \frac{r+s}{x}\right\rfloor \leq 2a+\left\lfloor \frac{2r}{x}\right\rfloor+2b+\left\lfloor \frac{2s}{x}\right\rfloor,\]
so we simply want to show that $\left\lfloor \frac{r+s}{x}\right\rfloor \leq \left\lfloor \frac{2r}{x}\right\rfloor+\left\lfloor \frac{2s}{x}\right\rfloor$.
But thats rather trivial:
We have $0 \leq r+s < 2x$, and if $r+s<x$, the statement is obvious. Otherwise we have $r+s \geq x$ implying that at least one of the numbers $r,s$ (w.l.o.g. $r$) is $\geq \frac{x}{2}$.
We then have $\left\lfloor \frac{r+s}{x}\right\rfloor =1 = \left\lfloor \frac{2r}{x}\right\rfloor \leq \left\lfloor \frac{2r}{x}\right\rfloor+\left\lfloor \frac{2s}{x}\right\rfloor$ as desired.

But we can always use combinatorics to prove this!

[quote="TTsphn"]Here is an short proof for the inequalities:
$ \lfloor 2x\rfloor \plus{} \lfloor 2y\rfloor \geq \lfloor x\rfloor \plus{} \lfloor y\rfloor \plus{} \lfloor x \plus{} y\rfloor ,\forall x,y > 0$
We has :
$ \lfloor 2x\rfloor \equal{} \lfloor x\rfloor \plus{} \lfloor x \plus{} \frac {1}{2}\rfloor$
Then the inequalities equivalent:
$ \lfloor x \plus{} \frac {1}{2}\rfloor \plus{} \lfloor y \plus{} \frac {1}{2}\rfloor \geq \lfloor x \plus{} y\rfloor$
$ \lfloor \{x\} \plus{} \frac {1}{2}\rfloor \plus{} \lfloor \{y\} \plus{} \frac {1}{2}\rfloor \geq \lfloor \{x\} \plus{} \{y\}\rfloor$
It is true.
In the same method we can prove that exist infinite $ (m,n)\in N$ for $ \frac {(2m)!(2n)!}{m!n!(m \plus{} n)!}$ is an even number.[/quote]
Yes ,TTsphn,we only need $ n > \sum _{\{k \equal{} 1\to \infty\}} [\frac {n}{2^k}]$

[b]t0rajir0u  wrote[/b]:[quote]Would you like to give a solution?[/quote]

Well I am giving you a hint try the rest yourself. See my post Related Problem:[quote]Let $ m$ and $ n$ be positive integers then prove that $ m!(n!)^{m}$ divides $ (mn)!$.[/quote]

There are obviously $ \frac{(mn)!}{m!(n!)^{m}}$ ways to split $ mn$ people into $ m$ groups implying the result.

Let $ f(m,n)\equal{}\frac{(2m)!(2n)!}{m!n!(m\plus{}n)!}$ then we have $ f(m,0)\equal{}\frac{(2m)!}{m!m!}$ which is an integer for all $ m$. Also we note that $ f(m,n)\equal{}4f(m,n\minus{}1)\minus{}f(m\plus{}1,n\minus{}1)$.
Now it is easy to prove the result by induction!

I dont know if this is correct. Can anyone check this?

Seems to be correct.
If I get it correctly you want to use induction on $ m\plus{}n$ and after that use the recursive relation(I haven't checked it yet) many times till one of the term in the sum
will be of the form $ f(m\plus{}n,0)$,what is whole number,am I right?

[quote="t0rajir0u"]I am aware of the combinatorial technique, but I have seen this problem posted several times and no one has ever offered a combinatorial solution, so I would be very interested in seeing that solution.[/quote]

Note that the expression in question is the same as $ N = \frac{(2m+2n)!}{m!n!(m+n)!} \div \frac{(2m+2n)!}{(2m)!(2n)!} = \binom{2m+2n}{m,n,m+1}  / \binom{2m+2n}{2m,2n}$

Imagine a group of $ 2m+2n$ people and a $ 2 \times 2$ grid of cells $ C{}_1{}_1$, $ C{}_1{}_2$, $ C{}_2{}_1$, $ C{}_2{}_2$ which can accomodate $ m$, $ n$, $ m$ and $ n$ people respectively.
Define a type A partition as consisting of $ C{}_1{}_1$, $ C{}_1{}_2$ and $ S = C{}_2{}_1 \cup C{}_2{}_2$ of sizes $ m$, $ n$ and $ m+n$ respectively.
Define a type B partition as consisting of $ D =C{}_1{}_1 \cup C{}_2{}_1$, $ E = C{}_1{}_2 \cup C{}_2{}_2$ of sizes $ 2m$, $ 2n$ respectively.

[ There are  $ \binom{2m+2n}{m,n,m+n}$ ways to slot the group of people into a type A partition and $ \binom{2m+2n}{2m,2n}$ ways to slot the group of people into a type B partition.]

Each type B partition can be associated with a type A partition in a fixed number of ways (and $ N$ = the ratio of the number of type A partitions to the number of type B partitions is an integer constant) as follows:-
  From column $ D$, choose $ m$ people to go to $ C{}_1{}_1$ and the rest go to $ C{}_2{}_1$.  [There are $ \binom{2m}{m,m}$ ways of doing this.]
  From column $ E$, choose $ n$ people to go to $ C{}_1{}_2$ and the rest go to $ C{}_2{}_2$.  [There are $ \binom{2n}{n,n}$ ways to do this.]
We then merge cells $ C{}_2{}_1$ and $ C{}_2{}_2$ to form $ S$, obtainig a type A partition.

[b]Note[/b]: $ N \not= \binom{2m}{m,m} \binom{2n}{n,n}$ because: analysing in reverse from type A partition, we note that the people in $ S$ could have come from
type B partition in $ \binom{m+n}{m,n}$ no. of ways, because there are that many ways to split the people in $ S$ back into cells $ C{}_2{}_1$ and $ C{}_2{}_2$ i.e.
the product $ \binom{2m}{m,m}\binom{2n}{n,n}$ would have overcounted by a constant factor of $ \binom{m+n}{m,n}$.  Thus $ N = \binom{2m}{m,m}\binom{2n}{n,n} / \binom{m+n}{m,n}$ which can also be
verified by expanding the latter expression in terms of factorials.

[quote="rkeele"]Solution attached.[/quote]
Hello,

Mr. Keele, you unfortunately misunderstood the problem. The problem you suggested is trivial.


Putnam
analytic geometry
function
inequalities
college contests

Putnam 1981 B2) Determine the minimum value of

\[(r-1)^{2}+\left(\frac{s}{r}-1 \right)^{2}+\left(\frac{t}{s}-1 \right)^{2}+\left(\frac{4}{t}-1 \right)^{2}\]

for all real numbers $r,s,t$ with $1\leq r\leq s\leq t\leq 4$.

Note: I'm looking to find a nice geometric solution akin to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=116695]this one[/url] since the expression to be minimized is the square of the distance between the points $\left( r, \frac{s}{r},\frac{t}{s},\frac{4}{t}\right)$ and $(1,1,1,1)$ or else the square of the distance between the same two points with any two corresponding coordinates interchanged. Ideas?

The ordering of $r,s,t$ and the fact that they lie in $[1,4]$, which 'appear' in the function to be minimized really suggests a substitution.  Make this substitution, and you need to minimize $\sum (a-1)^{2}$ subject to the condition $abcd=4$.  It's trivial to see that
\[(x-1)^{2}-[2\sqrt{2}x-1-2\sqrt{2}]\geq 0 \]
since the term in square parentheses is the tangent line to a convex, increasing function.  But AM-GM yields
\[\sum [2\sqrt{2}a-1-2\sqrt{2}] \geq 8\sqrt{2}\sqrt[4]{abcd}-4(1+2\sqrt{2})=16-8\sqrt{2}\]

[b]Solution 1)[/b]  Looking for a geometric solution is an excellent idea.

Let

\begin{eqnarray*}x & = & r\\\\ y & = &\frac{s}{r}\\\\ z & = &\frac{t}{s}\end{eqnarray*}

Then $ w=\frac{4}{t}=\frac{4}{\frac{t}{s}\frac{s}{r}r}=\frac{4}{xyz}$.

So the problem becomes finding the shortest distance between the point $ (1,1,1,1)$ and the surface $ (x,y,z,\frac{4}{xyz})$.  A good analogy to make to this would be finding the shortest distance between the point $ (1,1,1)$ and the surface $ (x,y,\frac{4}{xy})$ which happens when $ x=y=\frac{4}{xy}$.

[b]Solution 2)[/b]  On a different note, there is also a succinct solution using classic inequalities that is not in the book or online.

\begin{eqnarray*}\left((r-1)^{2}+\left(\frac{s}{r}-1\right)^{2}+\left(\frac{t}{s}-1\right)^{2}+\left(\frac{4}{t}-1\right)^{2}\right)\left(1+1+1+1\right) &\geq &\left(r-1+\frac{s}{r}-1+\frac{t}{s}-1+\frac{4}{t}-1\right)^{2}~~~~~~~~~~~~~(\mbox{Cauchy})\\ &\geq &\left( 4\sqrt[4]{r\frac{s}{r}\frac{t}{s}\frac{4}{t}}-4\right)^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\mbox{AM-GM})\\ & = &\left(4\sqrt{2}-4\right)^{2}\end{eqnarray*}

Both inequalities are sharp when $ r=\frac{s}{r}=\frac{t}{s}=\frac{4}{t}$ which gives the minimum.

[b]Solution 3)[/b]  Alternatively, we can use Jensen's inequality.  The original problem is equivalent to finding the minimum value of

$ (x_{1}-1)^{2}+(x_{2}-1)^{3}+(x_{3}-1)^{2}+(x_{4}-1)^{2}$ such that $ x_{1},x_{2},x_{3},x_{4}\geq 1$ and $ x_{1}x_{2}x_{3}x_{4}=4$

Let

$ w_{1}=\ln(x_{1})$

$ w_{2}=\ln(x_{2})$

$ w_{3}=\ln(x_{3})$

$ w_{4}=\ln(x_{4})$

So we wish to find the minimum value of

$ (e^{w_{1}}-1)^{2}+(e^{w_{2}}-1)^{3}+(e^{w_{3}}-1)^{2}+(e^{w_{4}}-1)^{2}$ such that $ w_{1},w_{2},w_{3},w_{4}\geq 0$ and $ w_{1}+w_{2}+w_{3}+w_{4}=\ln(4)$

The function $ (e^{x}-1)^{2}$ is convex on $ [0,\infty)$ so Jensen's inequality gives

\begin{eqnarray*}(e^{w_{1}}-1)^{2}+(e^{w_{2}}-1)^{2}+(e^{w_{3}}-1)^{2}+(e^{w_{4}}-1)^{2}&\geq & 4 (e^{\frac{w_{1}+w_{2}+w_{3}+w_{4}}{4}}-1)^{2}\\ & = & 4 (e^{\frac{\ln(4)}{4}}-1)^{2}\\ & = & 4(\sqrt{2}-1)^{2}\end{eqnarray*}

The inequality is sharp when $ w_{1}=w_{2}=w_{3}=w_{4}$ which gives the minimum.