An equilateral triangle OBC has vertices at O, (1,0) and (1/2, 1/2 sqrt3), respectively. The point P has coordinates (x, y). The perpendicular distance from P to the line through C and O is h1; the perpendicular distance from P to the line through O and B is h2; and the perpendicular distance from P to the line through B and C is h3.
Show that h1 = 1/2 modulus(y-xsqrt3) and find expressions for h2 and h3.
Show that h1 + h2 + h3 = 1/2 sqrt3 if and only if P lies on or in the triangle OBC.

If P is inside the triangle, then h1, h2 and h3 are the altitudes of three triangles together making up the whole triangle, giving h1 + h2 + h3 = (sqrt 3)/2. Similarly for P on the triangle. If P is outside the triangle, the sum of the perpendicular distasnces can get larger and larger and so is not constant. Not sure about this and am still thinking about the converse. Any ideas?

I'm thinking "Simson line" but have got no further with the converse.


greatest common divisor
geometry
geometric transformation
reflection

Hey is ther a formula or an easy way of finding the solutions to these two problems?

1)  Consider the three integers 2718, 3875, and 4486.  Find the largest integer [i]n[/i] which yields the same remainder when each of these three integers is divided by n.  As an example, the largest integer [i]n[/i] for the three integers 12, 17, and 37 is 5.

2) Connect the points (5, 4) and (7, 1) by three line segments which meet the x-axis and y-axis as in the picture.  What is the length of the shortest such path joining these points?
[img]http://loprecruiter.com/images/connect.gif[/img]

SnowStorm wrote:1)  Consider the three integers 2718, 3875, and 4486.  Find the largest integer n which yields the same remainder when each of these three integers is divided by n.  As an example, the largest integer n for the three integers 12, 17, and 37 is 5.

Spoiler:

[hide]Since 2718, 3875, and 4486 are all the same number more than some multiple of n, 4486  - 3875 = 611 and 3875 - 2718 = 1157 must both be multiples of n.   The GCF of 611 = 13*47 and 1157 = 13*89 is 13, so this is the greatest possible value of n.  (Note that we didn't use the fact that 4486 - 2718 is a multiple of n, since this is already implied by the fact that 611 and 1157 both are.)[/hide]

For #2, try a simpler problem first - what if you only have to hit the y-axis?

Well if you still are using 3 line segments and only hitting the y-axis thats not very practical, but if you only have to hit the y axis using 2 segments it seems to me the shortest path would be drawing a diagonal from (5, 4) to (0, 1) and then a straight line from that point to (7, 1) since that's the longest straight line you can draw and you would want the most possible straight lines since the shortest distance between two points is a straight line.

Hint for #2: [hide]generally the shortest distance between two points is a straight line.  Can you transform your diagram in some way so that the restrictions (touch both x- and y-axes) go away and you're simply left with minimizing the distance between some two points?[/hide]

Reflect (7,1) across the y axis and again across the x axis.  Draw a line from (5,4) to the new (7,1) in the third quadrant.  This is the shortest distance.  To see this, reflect this line across the y axis and again across the x axis so that it meets backs at (7,1).  This distance is 13.  We can make a formula that the shortest disance between 2 points in the first quadrant, (a,b) and (c,d), which touches the x and y axis, is the distance between (a,b) and (-c,-d).


trigonometry
inequalities
inequalities proposed

Let $ ABC$ be a triangle for which $ a^2\plus{}b^2\plus{}c^2\equal{}1$.Prove that :
$ a \cdot sin^2A\plus{}b \cdot  sin^2B\plus{}c \cdot sin^2C \le \frac{9}{4}$.

[quote="soruz"]Let $ ABC$ be a triangle for which $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.Prove that :
$ a \cdot sin^2A \plus{} b \cdot sin^2B \plus{} c \cdot sin^2C \le \frac {9}{4}$.[/quote]
Can you check your problem? Because it's too easy :wink:
$ a \cdot sin^2A \plus{} b \cdot sin^2B \plus{} c \cdot sin^2C <sin^2A\plus{}sin^2B\plus{}sin^2C \le \frac {9}{4}$  :D

By Cauchy-Schwarz inequality we have :

$ a\sin^2A \plus{} b\sin^2B \plus{} c\sin^2C\le\sqrt {(a^2 \plus{} b^2 \plus{} c^2)(\sin^4A \plus{} \sin^4B \plus{} \sin^4C)} \\< \sin^2A \plus{} \sin^2B \plus{} \sin^2C < \frac {9}{4}$

The equality doesn't occur.


:sqrt: (1*2*3*4+1)=5
 :sqrt: (2*3*4*5+1)=11
 :sqrt: (3*4*5*6+1)=19
 :sqrt: (4*5*6*7+1)=29
 :sqrt: (50*51*52*53+1)=?

This was a National Team Round, but it was before calculators were used in Mathcounts, so you can't use one.

[quote="Gyan"]This problem has also appeared in the form:

prove for a natural  number number  n>0,   n(n+1)(n+2)(n+3)  is never a perfect square.[/quote]

Well, expand it, add 1, and try factoring that expression. If a number 1 greater than a certain number is square, then the number is never a square other than 0, and this expression can't be 0.

[quote]$\sqrt{1*2*3*4+1} = 5$

$\sqrt{2*3*4*5+1} = 11$

$\sqrt{3*4*5*6*+1} = 19$

$\sqrt{4*5*6*7+1} = 29$

$\sqrt{50*51*52*53+1} = \;?$[/quote][hide]
This is the solution based on the proof suggested by Gyan.

Consider the product of four consecutive integers: $n(n+1)(n+2)(n+3)$

Re-order: $[n(n+3)]\;[(n+1)(n+2)]$

Multiply: $[n^2 + 3n]\;[n^2+3n + 2]$

Modify: $[(n^2 + 3n + 1) - 1]\;[(n^2 + 3n + 1) + 1]$ ... we have the form $(a-b)(a+b)$

Multiply: $(n^2 + 3n + 1)^2 - 1$

Add 1: $(n^2 + 3n + 1)^2$

Hence: one more than the product of four consecutive integers is a square.

$\text{The }n^{th}\text{ term of the sequence is: }a_n = n^2 + 3n + 1$

$\text{And }a_{50} = 50^2 + 3\cdot50 + 1 = 2651$[/hide]

this was like an AIME question: sqrt(31*30*29*28 +1)
what we could do is reorder the terms, and write it as diff. of squares, like soroban did.

There was a paper in the American Mathematical Monthly that discussed generalizations of this problem:

  A van der Poorten
  Squares from products of consecutive integers.
  Amer. Math. Monthly 109 (2002), no. 5, 459--462.


[hide="answer"]um.. prime numbers are divisible by anything except for 1 and themselves so the answer is 1(just the 7)[/hide]

well, 1, right? 7 is the only prime which is divisible by 7.

[quote="aznmonkey1992"]yep!
yay! 236factorial didnt post this time :P[/quote]

Whoops, I pressed "submit".

The answer is 1, just 7 works (because anything other multiple isn't prime)  :P

man, 236 factorial, why do you have to spoil everything? i was pretty happy there :wallbash:  :wallbash_red:  :censored:

lol, the answer is 1 becuase prime numbers are divisible by themselves and 1.

[quote="aznmonkey1992"]i saw this in a sat book:
how many prime numbers are before 1000 and are divisible by 7[/quote]

1, since ony the prime 7 is divisible by 7.  :)

[hide]
Prime numbers are only divisible by 1 and themselves, so the only prime is 7, so the answer is 1. [/hide]

[hide="solution"]obviously, if the number is a prime and can b divisible by the prime # 7, it can only b itself.[b]7[/b]SO, THERS ONLY 1 SOLUTION. :rotfl: [/hide]


In programa de a 12-a de olimpiada se cere Grupul permutărilor, cicluri, descompunerea în produs de cicluri disjuncte. La ce se refera descompunerea in cicluri disjuncte?? In ce consta?

Incepi cu un element $k_1$ si calculezi $k_2 = \sigma (k_1)$, $k_3 = \sigma (k_2)$, ... pana dai din nou de $k_1$. Un ciclu ar fi $(k_1,k_2,\ldots,k_t)$.

Faci asta cu ce-ti mai ramane si gasesti descompunerea.


[quote="mofidy"]Let $f: [a,b] \to \mathbb{R}.$ Suppose $f'$,$f''$ exist on $[a,b]$ and $f'(a)=f'(b)=0.$
Prove for some $c\in (a,b)$
\[|f''(c)| \geqslant 4\frac{{|f(b)-f(a)|}}{{(b-a)^{2}}}. \]
[/quote]
try to use the fact: $f''(x)=f(0)+f'(0)x+\frac{f''(c)x^{2}}{2}$ where $c \in (0,x)$ or $(x,0)$.

[quote="Mathsniper"][quote="mofidy"]Let $f: [a,b] \to \mathbb{R}.$ Suppose $f'$,$f''$ exist on $[a,b]$ and $f'(a)=f'(b)=0.$
Prove for some $c\in (a,b)$
\[|f''(c)| \geqslant 4\frac{{|f(b)-f(a)|}}{{(b-a)^{2}}}. \]
[/quote]
try to use the fact: $f''(x)=f(0)+f'(0)x+\frac{f''(c)x^{2}}{2}$ where $c \in (a,b)$.[/quote]

The problem with Taylor's theorem is that it [b]depends[/b] on $x$ meaning the constant $c$ depends on what point $x$ you chose in $(a,b)$.

Two previous versions:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123255]Physics reference.[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123244]Math reference.[/url]

For example, I quote the content
[code]
A particle moves in a straight line. At time 0 it is at rest at position 0. By time T it has arrived at position L and is at rest there. There is a positive number A such that at some point during the trip, the magnitude of the acceleration (or deceleration) must exceed A. Find the largest possible value of A in terms of L and T.
[/code]
Use the result form the proved inequality. $L=f(T)$, $f(T)-f(0)=L$, $T-0=T$. Thus $|A| \geq \frac{4L}{T^{2}}$ where $A=\frac{d^{2}L}{dT^{2}}|_{T=T'}$.


graph theory
combinatorics proposed
combinatorics

A set od $1990$ persons is divided into non_intersecting subsets in such a way that
1-No one in a subset knows all the others in the subset,
2-Among any three persons in a subset,there are always at least teo who dont know each other ,and
3-for any three persons in a subset who dont know each other ,there is exactly one person in the same subset
knowing both of them .
(I)Prove that within each subset ,every person has the same number of acquaintances.
(II)Determine the maximum possible number of subsets.

(I):

Pick 2 arbitrary vertexs $A$ and $B$, let $d_A$ be $A$ friends that doest know $B$, define similar $d_B$, and let $d_{AB}$ be their common friends, clearly $|d_{AB}|<3$ and the sets $d_A$ and $d_B$ are independent. First let split things in 2 cases: $|d_A|>1$, pick any 2 guys from $d_A$ and $B$,, they must be joined to a guy in $d_B U d_{AB}$, so if we make a bipartite graph with partitions $d_B U d_{AB}$ and $d_A U d_{AB}$. we get both sides of the partition has same number of elements (just prove iff $|d_B|=k$, then $k(k-1)$ edges come out from the side $d_A U d_{AB}$ of the partition. So it is clear from here $A$ and $B$ have the same number of friends in their group.

In the secound case, we get that $A$ is a friend of at most 3 persons, also it means every person is a friend of at most every other 3 in the grioup (or we could take that person and name him $A$) I guess this case is not that hard, however I could only make lots of cases to deal with it...

Also for the first problem:

Let $x,y$ be two vertices which are connected. They have no common neighbours, and each vertex which is not connected to either of them must be connected to a neighbour of $x$ and to a neighbour of $b$. On the other hand, given a neighbour $n_x$ of $x$ and a neighbour $n_y$ of $y$, they must have exactly one common neighbour, which is not connected to any of $x,y$. All of this means that, for each pair $n_x,n_y$, there is precisely one vertex not connected to either of $x,y$, which is connected to both $n_x,n_y$, and all vertices not connected to $x,y$ are of this type.

If $d_x,d_y$ are the degrees of $x,y$ respectively, the total number of vertices must then be $2+d_x-1+d_y-1+(d_x-1)(d_y-1)=1+d_xd_y$, which is a constant (say $n$, the number of vertices). In particular, this means that any two vertices not connected to each other have the same degree (if $y,y'$ are not connected, then they have some common neighbour $x$, and $n=1+d_xd_y=1+d_xd_{y'}$).

It's easy to see that, since no vertex is connected to all the others, no vertex has degree $1$. It's easy to see that, in these conditions, each edge $uv$ is part of a $5$-cycle, and, using the fact that each two not connected vertices have the same degree, we can easily deduce, by reasoning with the vertices of this $5$-cycle, that $u,v$ have the same degree, so each two connected vertices also have the same degree, and this is what we need.

I hope it works.

As for the second question, we can simply divide the $1990$ people in groups of $5$ people, organizing each group as a $5$-cycle, can't we? Maybe I'm missing something? :?


In the game of PIG, a player tosses a pair of standard dice and loses all of her points if two 1's are rolled. If the dice are rolled 144 times, how many times would you expect to get a pair of 1's?

Well there is 1/6 * 1/6 = 1/36 probability that you'd get a pair of 1's.  144 divided by 36 is 4 so our answer is 4.

The first bit of info is irrelevant.

$ \frac {144}{6^2} \equal{} 4$.

Edited. My bad.

The Diety, I believe you meant 144/6^2= 4

[quote="TheDeity"]The first bit of info is irrelevant.
[/quote]

You mean the first bit of info here?

[quote="Maybach"]Well there is 1/6 * 1/6 = 1/36 probability that you'd get a pair of 1's.[/quote]

Because that's not irrelevant - it's the reason why you do $ 144 \times \frac{1}{36} \equal{} \frac{144}{36} \equal{} 4$. I could be misunderstanding you...

I think he was referring to the "In the game of PIG" part.

[quote="isabella2296"][quote="TheDeity"]The first bit of info is irrelevant.
[/quote]

You mean the first bit of info here?

[quote="Maybach"]Well there is 1/6 * 1/6 = 1/36 probability that you'd get a pair of 1's.[/quote]

Because that's not irrelevant - it's the reason why you do $ 144 \times \frac {1}{36} \equal{} \frac {144}{36} \equal{} 4$. I could be misunderstanding you...[/quote]

Sorry for the misunderstanding, I was addressing the question. The bit of info here

[quote="Snowpaw"] In the game of PIG, a player tosses a pair of standard dice and loses all of her points if two 1's are rolled. [/quote]


calculus
integration
floor function
ceiling function

$ s\equal{}\frac{1}{\sqrt{1}\plus{}\sqrt{2}}\plus{}\frac{1}{\sqrt{3}\plus{}\sqrt{4}}\plus{}\cdots\plus{}\frac{1}{\sqrt{99}\plus{}\sqrt{100}}$

Find the value of $ \lfloor s\rfloor$ .
:lol:

This is very easy  :lol:  :rotfl: $ \sqrt{100}\minus{}1\equal{}9$

[quote="Pirkuliyev Rovsen"]This is very easy  :lol:  :rotfl: $ \sqrt {100} \minus{} 1 \equal{} 9$[/quote]

Sorry , I can't understand.Would you offer the details,please.

$ f(n) \equal{} \frac {1}{\sqrt {2n \minus{} 1} \plus{} \sqrt {2n}} \equal{} \sqrt {2n} \minus{} \sqrt {2n \minus{} 1}$.
So,
$ \int_1^{50}f(x)dx \plus{} f(50) \le \sum_{n \equal{} 1}^{50} f(n) \le f(1) \plus{} \int_1^{50}f(x)dx$
$ \iff \frac {1031}{3} \minus{} \frac {2\sqrt {2}}{3} \minus{} 102\sqrt {11} \le s \le \frac {998}{3} \plus{} \frac {\sqrt {2}}{3} \minus{} 99\sqrt {11}$
$ \Rightarrow 4.43 < s < 4.79$
$ \Rightarrow \lfloor s \rfloor \equal{} \boxed{4}$

[hide="without calc"]

$ s \equal{} \sum_{k \equal{} 1}^{50} \frac {1}{\sqrt {2k \minus{} 1} \plus{} \sqrt {2k}} \ge \sum_{k \equal{} 1}^{50} \frac {1}{2\sqrt {2k}}$

$ \ge \frac {1}{2\sqrt {2}}\sum_{k \equal{} 1}^{50}\frac {1}{\lceil{\sqrt {k}\rceil}}$

$ \equal{} \frac {1}{2\sqrt {2}}\left(1 \plus{} \frac {3}{2} \plus{} \frac {5}{3} \plus{} \frac {7}{4} \plus{} ... \plus{} \frac {13}{7} \plus{} \frac {1}{8}\right)$

$ \equal{} \frac {3229}{560\sqrt {2}} \qquad(1)$

similarly

$ s \le \sum_{k \equal{} 1}^{50}\frac {1}{2\lfloor{\sqrt {2k \minus{} 1}\rfloor}} \equal{} \frac {1}{2}\cdot \frac {1063}{120} \qquad(2)$

From (1) and (2) it follows that $ \lfloor{s\rfloor} \equal{} 4$[/hide]

Denote $ t\equal{}\frac{1}{\sqrt{2}\plus{}\sqrt{3}}\plus{}\frac{1}{\sqrt{4}\plus{}\sqrt{5}}\plus{}\cdots\plus{}\frac{1}{\sqrt{98}\plus{}\sqrt{99}}$.
We have $ \frac{1}{\sqrt{1}\plus{}\sqrt{2}}>\frac{1}{\sqrt{2}\plus{}\sqrt{3}}>\cdots>\frac{1}{\sqrt{98}\plus{}\sqrt{99}}>\frac{1}{\sqrt{99}\plus{}\sqrt{100}}>\frac{1}{\sqrt{100}\plus{}\sqrt{101}}$.
Using the telescoping indicated, we have $ s\plus{}t \equal{} \sqrt{100} \minus{} \sqrt{1} \equal{} 9$. On the other hand we have $ 0< \sqrt{101} \minus{} \sqrt{100} < s \minus{} t < \sqrt{2} \minus{} \sqrt{1} < 1$.
Therefore, by adding the relations above, $ 9 < 2s < 10$, hence $ \lfloor s \rfloor \equal{} 4$.


perpendicular lines
convex quadrilateral
geometry
diagonals
IMO Shortlist

A convex quadrilateral has equal diagonals. An equilateral triangle is constructed on the outside of each side of the quadrilateral. The centers of the triangles on opposite sides are joined. Show that the two joining lines are perpendicular.

[i]Alternative formulation.[/i] Given a convex quadrilateral $ ABCD$ with congruent diagonals $ AC \equal{} BD.$ Four regular triangles are errected externally on its sides. Prove that the segments joining the centroids of the triangles on the opposite sides are perpendicular to each other.

[i]Original formulation:[/i] Let $ ABCD$ be a convex quadrilateral such that $ AC \equal{} BD.$ Equilateral triangles are constructed on the sides of the quadrilateral. Let $ O_1,O_2,O_3,O_4$ be the centers of the triangles constructed on $ AB,BC,CD,DA$ respectively. Show that $ O_1O_3$ is perpendicular to $ O_2O_4.$

For this one complex numbers should work just fine. Let E be the center of the equilateral triangle constructed on AB. F, G, H are the same for sides BC, CD, DA respectively. Let small letters (like a,b,c etc) be the complex numbers that represent the points with the correspondinc\g s\capital letters (A, B, C etc.) Let z=cos 120+i*sin 120. Then e=(a-bz)/(1-z), f=(b-cz)/(1-z), g=(c-dz)/(1-z), h=(d-az)/(1-z). If we calculate A=(e-g)/(h-f) and its conjugate and we take into account the fact that |a-c|=|b-d| we get A {conjugate} = -A, which is equivalent to A=(e-g)/(h-f) = k*i with k a real number and i=(-1)^1/2 (*). But (*) is the condition for the lines EG and FH to be perpendicular.

This is the usual method by which these prbs involving equilateral triangles can be solved (I think there are other methods, but for this specific problem this the best way I could find).

[Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=17861 ]

Let $ ABCD$ be a convex quadrilateral. We erect four equilateral triangles $ WAB$, $ XBC$, $ YCD$, $ ZDA$ on its sides $ AB$, $ BC$, $ CD$, $ DA$ exterior to the quadrilateral $ ABCD$, and denote the centroids of these triangles by $ S_{1}$, $ S_{2}$, $ S_{3}$, $ S_{4}$, respectively. Prove that: $ S_{1}S_{3}$ is perpendicular to $ S_{2}S_{4}$ if and only if $ AC \equal{} BD$.

Solution by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=154]Myth[/url]:

It is well known that $ S_1S_3\perp S_2S_4$ iff $ S_1S_2^2 \plus{} S_3S_4^2 \equal{} S_2S_3^2 \plus{} S_4S_1^2$.

Apply cosine theorem to $ AS_1S_4$ and $ CS_2S_3$. We obtain:
\[ S_4S_3^1 \plus{} S_2S_3^2 \equal{} \frac {1}{3}(AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} DA^2) \minus{} \frac {2}{3}AB\cdot AD\cdot\cos(A \plus{} \pi/3) \minus{} \frac {2}{3}CB\cdot CD\cdot\cos(C \plus{} \pi/3).
\]
We can write
\[ BD^2 \equal{} AB^2 \plus{} AD^2 \minus{} 2AB\cdot AD\cdot\cos(A) \equal{} CB^2 \plus{} CD^2 \minus{} 2CB\cdot CD\cdot\cos(C)
\]
and
\[ AB\cdot AD\cdot\sin(A) \plus{} CB\cdot CD\cdot\sin(C) \equal{} 2S_{ABCD}.
\]
So after some easy transformation we conclude that
\[ S_4S_3^1 \plus{} S_2S_3^2 \equal{} \frac {1}{6}(AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} DA^2) \plus{} \frac {1}{3}BD^2 \plus{} \frac {\sqrt {3}}{6}S_{ABCD}.
\]
Analogously,
\[ S_4S_1^2 \plus{} S_2S_3^2 \equal{} \frac {1}{6}(AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} DA^2) \plus{} \frac {1}{3}AC^2 \plus{} \frac {\sqrt {3}}{6}S_{ABCD}.
\]

[url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=652]Omid Hatami[/url] notes: Also this can easily proved with vector geometry.

Let $ABCD$ be a quadrilateral with equilateral triangles $ABE,BFC,CGD,DHA$ constructed externally.Then the coordinates of $E,F,G,H$ are

$e=b+(a-b)\epsilon$

$f=c+(b-c)\epsilon$

$g=d+(c-d)\epsilon$

$h=a+(d-a)\epsilon$

Thus coordinates of $O_1,O_2,O_3,O_4$ are

$o_1=\frac{a+2b+(a-b)\epsilon}{3}$

$o_2=\frac{b+2c+(b-c)\epsilon}{3}$

$o_3=\frac{c+2d+(c-d)\epsilon}{3}$

$o_4=\frac{d+2a+(d-a)\epsilon}{3}$

where $\epsilon=cos60^{\circ}+sin60^{\circ}$

After a bit of manipulation using the relations we get that $|o_1-o_2|^2=|o_2-o_3|^2=|o_3-o_4|^2=|o_4-o_1|^2$ and $o_1-o_3|^2=|o_2-o_4|^2 \Leftrightarrow |a-c|^2=|b-d|^2$(It won't be difficult at all if we use the realtion $|a-b|^2=(a-b)(\overline{a}-\overline{b})$).So we are through....

Another related problem:

On the sides of $AB,BC,CD,DA$ of quadrilateral $ABCD$ and exterior to the quadrilateral,we construct squares of centers $O_1,O_2,O_3,O_4$ respectively.Show that $O_1O_3 \perp O_2O_4$ and $O_1O_3=O_2O_4$

Haha this problem is crying for complex bash...

Let $a,b,c,d,o_1,o_2,o_3,o_4$ be the affixes of $A,B,C,D,O_1,O_2,O_3,O_4$, respectively.
Note that $a$ is the result of rotating $b$ by $\tfrac{\pi}{3}$ about $o_1$. We have the equation
\[ a-o_1=\varepsilon(b-o_1) \Rightarrow o_1=\dfrac{a-b\varepsilon}{1-\varepsilon} \]
where $\varepsilon=e^{\tfrac{2\pi i}{3}}$.
Similarly, $o_2=\tfrac{b-c\varepsilon}{1-\varepsilon}, o_3=\tfrac{c-d\varepsilon}{1-\varepsilon}, o_4=\tfrac{d-a\varepsilon}{1-\varepsilon}$.

We want to show $O_1O_3 \perp O_2O_4$, which from Orthogonality criterion is equivalent to
\[ \dfrac{\dfrac{a-b\varepsilon}{1-\varepsilon}-\dfrac{c-d\varepsilon}{1-\varepsilon}}{\overline{\dfrac{a-b\varepsilon}{1-\varepsilon}}-\overline{\dfrac{c-d\varepsilon}{1-\varepsilon}}} = -\dfrac{\dfrac{b-c\varepsilon}{1-\varepsilon}-\dfrac{d-a\varepsilon}{1-\varepsilon}}{\overline{\dfrac{b-c\varepsilon}{1-\varepsilon}}-\overline{\dfrac{d-a\varepsilon}{1-\varepsilon}}} \]
\[ \iff \dfrac{a-c-(b-d)\varepsilon}{\overline{a-c}-\overline{(b-d)\varepsilon}} = - \dfrac{b-d-(c-a)\varepsilon}{\overline{b-d}-\overline{(c-a)\varepsilon}}\]
\[ \iff \left[ a-c-(b-d)\varepsilon \right] \left[ \overline{b-d}-\overline{(c-a)\varepsilon} \right] = -\left[ b-d-(c-a)\varepsilon \right] \left[ \overline{a-c}-\overline{(b-d)\varepsilon} \right] \]
\[ \iff (a-c)\overline{b-d}-(a-c)\overline{(c-a)}\overline{\varepsilon}-(b-d)\overline{b-d}\varepsilon+(b-d)\overline{c-a}\varepsilon\overline{\varepsilon}=-(b-d)\overline{a-c}+(b-d)\overline{(b-d)}\overline{\varepsilon}+(c-a)\overline{a-c}\varepsilon-(c-a)\overline{b-d}\varepsilon\overline{\varepsilon} \]
Let $x=a-c$, $y=b-d$. (Should've done this earlier but whatever...)
\[ \iff x\overline{y}+x\overline{x}\dfrac{1}{\varepsilon}-y\overline{y}\varepsilon-y\overline{x} = -y\overline{x}+y\overline{y}\dfrac{1}{\varepsilon}-x\overline{x}\varepsilon+x\overline{y}\]
\[\iff x\overline{x}(\varepsilon+\dfrac{1}{\varepsilon})-y\overline{y}(\varepsilon+\dfrac{1}{\varepsilon})=0\]
\[\iff x\overline{x}=y\overline{y} \]
\[\iff AC=BD \]
which is given.

Really if is only if, like said the post #3 and this true for internal equilaterals too.

This is consequence of the area $\Delta '$ in the Napoleon triangle of $\triangle ABC$
$\Delta' = \frac{[ABC]}{2}+\frac{\sqrt{3}(a^2+b^2+c^2)}{24}$ (outer)
$\Delta'=\frac{a^2+b^2+c^2-4\sqrt{3}[ABC]}{6}$ (internal, this implies Weitzenböck's inequality)
Is sufficient prove that $O_1O_2^2+O_3O_4^2=O_2O_3^2+O_1O_4^2$ and this trivial, with this formule because area of a equilateral triangle is $\frac{l^2\sqrt{3}}{4}$

Let $\omega$ be a third root of unity. Then we can compute the centers are $\frac{a\omega - b}{\omega-1}$ and cyclic permutations, so it suffices to show that$$\frac{o_1-o_3}{o_2-o_4} = \frac{\omega(a-c)-(b-d)}{(b-d)\omega + (a-c)}$$is pure imaginary. The conjugate equals
\begin{align*}
\frac{\overline \omega \overline{a-c} - \overline{b-d}}{\overline{b-d}
\overline \omega + \overline{a-c}} &= \frac{\overline{a-c} - \overline{b-d} \omega}{\overline{b-d}-\overline{a-c} \omega} \\
&= \frac{\frac{\overline{b-d}(b-d)}{a-c} - \frac{\overline{a-c}(a-c)}{b-d} \omega}{\frac{\overline{a-c}(a-c)}{b-d} + \frac{\overline{b-d}(b-d)}{a-c}\omega} \\
&= \frac{(b-d) - (a-c) \omega}{(b-d) \omega + a-c},
\end{align*}which is the negative of the original, as needed.


The question: How many odd numbers with middle digit 5 and no digit repeated are there between 20000 and 69999?

Case 1: Last digit 3. Here there are only the possibilities 2, 4, 5, and 6 for the first digit
...blah blah

How can the last or first digit be 5 if the middle digit must be 5 and no digits can be repeated?

(this is on pg. 301 in AoPS 1)

Well here's one way to do it. The number we are looking for is
_ _ 5 _ _, with the last digit odd. SO, the last digit can be 1,3,7 or 9. Let the first digit be 2, and let the last digit be 1. There are 7 more digits we can use to fill in the 2 blank digits. There are 7 *6 total possibilities. Since when the first digit is 2 there can be 1,3,7, or 9 as the last digit you can multiply this by 4 to get 168 possibilities for 2 as the first digit. This remains the same for 2, 4, and 6 as the first digit, so there are 168 *3=504 numbers so far. If 3 is the first digit, only 1,7 or 9 can be the last digit. SO, instead of 7*6*4 we have 7*6*3 = 126. 126 + 504 = [b]630[/b] total numbers. Am I right?

Magnus and I figured this out in class; he has a first edition AoPS in which this problem is messed up; white_horse, you are correct.


I have no idea what that equation means, but you seem to be the first person on Hunter's die list, so you may just be doing this to annoy us or something.

What, post random equations to annoy people?

[quote="sharkman"]List the intercepts and test for symmetry.

y = (x^4 + 1)/2x^5[/quote]

Don't post your homework and expect people to do it for you.


hey Valentine, great site you got here, really, the only thing is: we got so few smileys :(:(:(

can you put some more on them here? You know, adding extra ones now will NOT remove the old ones and will NOT replace the old ones in the field next to the typing window :)

So what are you waiting for??

[quote="Peter VDD"]hey Valentine, great site you got here, really, the only thing is: we got so few smileys :(:(:(

So what are you waiting for??[/quote]
you shouting of course :D:D ... just kiddin. i've put some up ... check them out.

anyothers you want, get them into a .zip file and attach them here. when I have the time I will put them.

:lol: :cool: :wink: :redface: :maybe:

great, that's bout the basic kit we need :D
sorry didn't look lately, just remembered that to say from when I last looked.

Oh well, I can get you a thousand smilies :D but I think these are the ones we'll most commonly need.

Thanks!!

I have changed some of the smilaes definition, in order to make them more natural.  Check them out.

I see. My previous posts looks pretty rediculous like that

test:  :D  :cool:  ;)  :blush:  :maybe:  :P   ==> succeeded

nice job :P

I am glad you are satisfied. I can go now to sleep, the local olympiad starts tomorrow in Bucharest, and I have to go correct some of the papers ... :D

I think we can have smileys and math symbols seperated.

Is it the local phase of the olympiad?
New smileys,great [img]http://community.the-underdogs.org/smiley/happy/icon10-3.gif[/img]
Valentin,you're the [img]http://community.the-underdogs.org/smiley/happy/biggrinking.gif[/img]
Keep up the good work,you a deserve a [img]http://community.the-underdogs.org/smiley/happy/pint.gif[/img]
[img]http://community.the-underdogs.org/smiley/happy/WalkingSmiley.gif[/img]
Going to sleep now,see ya!
[img]http://community.the-underdogs.org/smiley/alcoholic.gif[/img]

something makes me expect that those are not board smileys :D

where did you get those? (they're quite funny)

[quote="Valentin Vornicu"]... the local olympiad starts tomorrow in Bucharest...[/quote]

Bring us the problems of the local olympiad Bucharest
thx

[quote="Peter VDD"]ok

bring us some of the problems :D[/quote]

wait, I just came home from correcting the papers. The problems were badly chosen, because only one person choose the problems from many proposals, on subjective criteria.

I will present the "modified version". I really belive this was the worse local olympiad in years :(

Tiny note:

Is it possible to set the "View more Emoticons" window to be bigger when it opens up, so that we don't have to expand the window to see the full range of smileys?

Thanks.

yes, that's possible, the window size should be somewhere in the source code (it's even right in the posting.php I think, or elseway in the posting_body.tpl)


Let $ a,b,c \ge 0, a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \frac{a\sqrt{b\plus{}c}}{b^2\plus{}c^2}\plus{}\frac{b\sqrt{c\plus{}a}}{c^2\plus{}a^2}\plus{}\frac{c\sqrt{a\plus{}b}}{a^2\plus{}b^2} \ge 2\sqrt{2}$

[quote="NguyenDungTN"]Let $ a,b,c \ge 0, a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {a\sqrt {b \plus{} c}}{b^2 \plus{} c^2} \plus{} \frac {b\sqrt {c \plus{} a}}{c^2 \plus{} a^2} \plus{} \frac {c\sqrt {a \plus{} b}}{a^2 \plus{} b^2} \ge 2\sqrt {2}$[/quote]
By Holder we have:
$ (\frac {a\sqrt {b \plus{} c}}{b^2 \plus{} c^2} \plus{} \frac {b\sqrt {c \plus{} a}}{c^2 \plus{} a^2} \plus{} \frac {c\sqrt {a \plus{} b}}{a^2 \plus{} b^2})^2(\sum_{cyc} \frac{a(b^2\plus{}c^2)^2}{b\plus{}c}) \geq (a\plus{}b\plus{}c)^3$
Now we need prove that
$ (a\plus{}b\plus{}c)^3\geq 8\sum_{cyc} \frac{a(b^2\plus{}c^2)^2}{b\plus{}c}$
 :oops:


let a and b be non-negative real numbers. prove that
$\sqrt[3]{ab^{2}}+\sqrt[3]{a^{2}b}\le a+b$.

or just transform it to $(\sqrt[3]a+\sqrt[3]b)(\sqrt[3]a-\sqrt[3]b)^{2}\geq0$ :wink:


calculus
derivative
calculus computations

Sorry if this is in the wrong place, but i'm having a little trouble with a few limit evaluations involving L'Hopital's Rule. The first one is...

lim (1-x+lnx)/(1+cos(pi*x)
x->1

and the second is...

lim     ((2e^x)+(x^2))^3/x
x->+infinity

thanks for the help...

As for the first...

Because you have the indeterminate form 0/0 you must differentiate.  The first derivative will also be 0/0 at x=1 so you need to differentiate once more.  You will now have (-x^(-2))/(-(pi)^(2)(cos ((pi)(x)))).  You can now plug in 1 for x to determine that the limit is -1/(pi^2).

I will try to complete the second shortly.

[quote="lund"]I assume that in your notation you are saying that the exponent is 3 and the quantity is divided by x.  If that is the case you can differentiate because you have the indeterminate form infinity/infinity.  You can then directly find the limit as x approaches infinity to be infinity.  If the exponent is 3/x you have a different case entirely.[/quote]
 My bad, the exponent is 3/x


geometry
linear algebra
matrix
vector
analytic geometry
geometric transformation
LaTeX

is there an analogue?

so are all collineations of affine space induced by semilinear transformations? of form

f(x) = A*x^theta +b

with A a square invertible matrix and theta an automorphism of the field?

i asked my professor , what about the affine plane over the complex numbers, if you map each point

(a+b*i,c+d*i)---->(a-b*i,c-d*i)

is that a collineation of the plane?

he said yes, but it linear not semilinear?

can anyone explain or comment a bit?

thx

What is your definition of "affine space"?

I can define algebraic operations geometrically, so any automorphism of the incidence structure is an automorphism of the algebraic stucture.
Clearly, such isomorphisms are defined by a change of basis and a field automorphism.

Your example is a field automorphism with no change of basis.

sorry for getting back so late

well uhm affine ndim space is the set of all cosets of subspaces, from 0 to n-1 dimensional ones

(while my definition of projective n dim space is the set of all subspaces of a n+1 dim space, except the full space and trivial subspace)

my definition of collineation is a bijection of those subspaces preserving inclusion

and i was hoping that while in projective geometry every collineation is of form

A*x^theta  , with A a nonsingular matrix and theta a field automorphism

there is in affine geometry a theorem stating that every collineation is of form

A*x^theta+b with b a vector and A nonsingular, theta a field automorphism

so do you have that analogy
PgammaL contains PGL
AgammaL contains AGL?

thx

fred

Here's how I'd do it:
Start with an affine space.
Step 1: Choose an origin and a basis.
Step 2: Using this basis, define the algebraic operations on the points of one of the axes.
Let $0$ be the origin, and $1$ be the basis vector. Let the line we are working with be $\ell$.
Addition: Let $a,b$ be points on $\ell$. Choose a point $x$ and let $\ell'$ be the parallel to $\ell$ through $x$. The line through $a$ parallel to $\overline{0x}$ intersects $\ell'$ at $y$. The line through $y$ parallel to $\overline{xb}$ intersects $\ell$ at $a+b$.
Negatives: Let $a$ be a point on $\ell$. Choose a point $x$ and let $\ell'$ be the parallel to $\ell$ through $x$. The line through $0$ parallel to $\overline{ax}$ intersects $\ell'$ at $y$. The line through $y$ parallel to $\overline{x0}$ intersects $\ell$ at $-a$.
Multiplication: Let $a,b$ be points on $\ell$. Choose a point $x$ and let $\ell'$ be the line $\overline{0x}$. The line through $b$ parallel to $\overline{1x}$ intersects $\ell'$ at $y$. The line through $y$ parallel to $\overline{xa}$ intersects $\ell$ at $a\cdot b$.
Inverses: Let $a$ be a point on $\ell$. Choose a point $x$ and let $\ell'$ be the line $\overline{0x}$. The line through $1$ parallel to $\overline{ax}$ intersects $\ell'$ at $y$. The line through $y$ parallel to $\overline{x1}$ intersects $\ell$ at $a^{-1}$.

Step 3. Create coordinates by mapping these labels onto the other axes (sending basis vectors to basis vectors, and keeping the lines between equal labels parallel). By using lines/planes/etc. parallel to those generated by the axes, we can assign unique coordinates to each point.

The fact that this was $F^n$ for some field $F$ originally gives us additional geometric knowledge that allows us to say that our operations are well-defined and satisfy the field axioms.
The field generated must of course be isomorphic to the original field; these geometric definitions are compatible with the original algebraic definitions.
Now the theorem is obvious; every automorphism of the incidence structure is the composition of a translation, a change of basis, and a field automorphism.
The projective case is easily inherited from the affine case, except that translation is not allowed.

By the way- please learn $\LaTeX$.


[list]What exactly should i study and know for problems in competitions that would involve conics? Also, if you know what I'm talking about, is the focus-directrix property of any use? Third, where could i learn this?[/list]

thx.

Well, knowing how to recognize the different conic sections and their equation forms would be helpful.  For instance, x^2/a^2 + y^2/b^2 = 1 for ellipses, x^2/a^2 - Y^2/B^2 for hyperbolas, y -k = a(x-h)^2 or x-h = a(y-k)^2 for parabolas, etc.

As for the focus-directrix things, that would be useful.

For instance, knowing that for parabolas, a = 1/4p where F(h,k+p) or (h+p,k) and directrix is y=k-p or x = h-p...

As for where you can learn this, I'm sure that you could find it on Eric Weisstein's World of Mathematics at [url]http://mathworld.wolfram.com/[/url] or some of those other ones.  But most Algebra II or PreCalc books usually have this sort of thing.

[quote]But most Algebra II or PreCalc books usually have this sort of thing[/quote]

Actually, I have found that the directrix is almost never in normal algebra/precalc textbooks. dunno why...

I seem to remember that AoPS 2 has a nice section on conic sections. (I looked up something in AoPS 2 on conic sections a few months ago after being rather silly on a math contest that asked which type of conic section something was, but I got it right anyway.) But they occur on competitions rather infrequently, so I wouldn't worry too much about it. The ones with xy terms in them are much more annoying than ones without, and they very rarely show up in normal algebra 2 textbooks.

Well, both my Precalc and my Algebra II/Trig books had a lot on conics, but maybe I just had a mutant math teacher  :)

There are occasionally problems that involve knowing what a conic actually is (i.e. the locus definition) rather than simply pounding equations.  Sam comes up with one of these every once in a while for 'Brot, and sometimes you'll get them on the Olympiad level.  By this, I mean knowing that an ellipse is the locus of points such that the sum of the distances from the point to 2 given points is a constant, and so on.

If you want to spend some time with a true genius contemplating conic sections, go grab yourself a copy of Newton's Principia.  (Yeah, as in Isaac Newton)


calculus
topology
derivative
real analysis
real analysis unsolved

In the proof of Theorem 5-2 on p.111 of Calculus on Manifolds by Spivak, how does the proof that (C) implies (M) use part 1 of condition (C)?

Theorem 5-2 says that (M) implies (C), not the other way around. Maybe it's because I have that book in Russian translation. Anyway, here is condition (C) in English->Russian->English translation. :D

For every point $ x\in M\subset \mathbb R^n$ there is an open set $ U\subset \mathbb R^n$ containing $ x$, another open set $ W\subset\mathbb R^k$, and a differentiable one-to-one map $ f\colon W\to \mathbb R^n$ such that
1. $ f(W)\equal{}M\cap U$
2. $ f'(y)$ has rank $ k$ for all $ y\in W$.

If you take away part 1, the map $ f$ has no relation to $ M$ anymore. Sure, there exists a differentiable one-to-one map $ f\colon W\to \mathbb R^n$ with derivative of rank $ k$ (i.e. an inclusion map $ \mathbb R^k\subset \mathbb R^n$). This says nothing about $ M$ and can't be used to prove anything about $ M$.


Solve with solution please:

9. (Target) A point is selected inside a rectangle with sides 7 and 11 units. What is the probability that the point is more than 2 units from each vertex. Use Pi=3.14 . Express your answer as a decimal to the nearest hundredth.

Well, since it has to be more than 2 units away from a vertex, then the area of the place the point can't be is $ 2^2\pi$ Therefore, the probability of the point not being within 2 units of a vertex is $ \frac{77\minus{}4\pi}{77}$

thanks                                                                                .


Can anyone Please Help Me on this Problem !!

Q) In Triangle ABC ,angle A is equal to 120 degrees.Bisectore AP,BQ,CR meet at Point O. Prove that angle PRO =30 degrees

[b]** SORRY I DON'T HAVE LATEX[/b]
 :?:  :!:

The trick here is to notice that, in triangle $ APC$, $ AR$ is an external bisector, and $ CR$ is an internal bisector. Hence, $ R$ is an excenter of triangle $ APC$, and this implies that $ PR$ is also an external bisector.

$ PAC\equal{}60^{\circ}$, let $ PCA\equal{}\gamma$, then $ CPA\equal{}120^{\circ}\minus{}\gamma$, so $ BPA\equal{}60^{\circ}\plus{}\gamma$, and since $ PR$ is an external bisector $ RPA\equal{}RPO\equal{}30^{\circ}\plus{}\frac{\gamma}{2}$.

In triangle $ AOC$, $ OAC\equal{}60^{\circ}$, $ ACO\equal{}\frac{\gamma}{2}$, then $ AOC\equal{}ROP\equal{}120^{\circ}\minus{}\frac{\gamma}{2}$.

Hence in triangle $ POR$, $ PRO\equal{}180^{\circ}\minus{}(30^{\circ}\plus{}\frac{\gamma}{2})\minus{}(120^{\circ}\minus{}\frac{\gamma}{2})\equal{}30^{\circ}$ as wanted.


Suppose there's a group of people, with each person having at most 3 enemies within the group. Prove that it's possible to divide the group into two sub-groups such that each person has at most one enemy in his own sub-group.

[hide]Let A be a way of partitioning the group of people  into two sub-groups. Define d(person) to be the number of his enemies in his subgroups, and define f(A) to be the sum of d(B) as B ranges over all people. Consider the partition that minimizes f(A). (If there are multiple, choose any one. We can do this by the axiom of choice.) We claim this partition satisfies the desired property. Assume it doesn't. Then someone (call him person X) has d(X) > 1. Switch him to the other group. Then we get a partition Y with smaller f(Y) than A does. Contradiction.[/hide]

[quote="pythag011"](If there are multiple, choose any one. We can do this by the axiom of choice.)[/quote]  You don't need the axiom of choice to choose a single element from a finite set  :roll:

It's also easy enough to phrase this idea constructively: order the $ n$ people.  At step $ i$, add the $ i$th person to group A unless she has two enemies there, in which case add her to group B (where she can never have more than one enemy).

If the person X has at least 2 enemies in his current subgroup, and as he switches to the other group, isn't it also possible to increase f as there might be people in the group he's switching to that have him as an enemy?

"Having as an enemy" is presumably a symmetric relation.

So if we assume it's symmetric, (and considering the two-one scenario) by switching to the other subgroup he's increasing 1 more for his enemy (since the enemy will have one more enemy in the subgroup) while he decreases 1 for himself... meanwhile he's decreasing the 2 enemies in his old subgroup by 1 each thus decreasing overall by 2 (better off if he has 0 enemies in the other group)... thus f is strictly decreasing (and has to terminate at some point because this sequence is nonnegative)

however, what if the relation is not mutual?


trigonometry
real analysis
real analysis unsolved

For what $ k \in \mathbb{R}$ equation
 $ \minus{} u'' \minus{} \pi u \equal{} k u \plus{} \sum_{j\equal{}1}^n \cos \left( j \pi x \right), \quad x \in [0,1]$
has a solution such that $ u'(0) \equal{} u'(1) \equal{} 0$ ?  :huh:

We can rewrite equation : $ u'' \plus{} (\pi \plus{} k)u \equal{} \minus{} \sum_{j \equal{} 1}^{n}{\cos(j\pi x)}$
Now : $ u \equal{} u_1 \plus{} u_2$
Characteristic equation is : $ r^2 \plus{} \pi \plus{} k \equal{} 0$ and it's solutions are : $ r_1 \equal{} \minus{} i\sqrt {\pi \plus{} k}$ and $ r_1 \equal{} i\sqrt {\pi \plus{} k}$. So for $ u_1$ we can write :

$ u_1 \equal{} C_1 \cos (x\sqrt {\pi \plus{} k}) \plus{} C_2 \sin (x\sqrt {\pi \plus{} k})$

Let's find $ u_2$. Find it in case : $ u_2 \equal{} \sum_{j \equal{} 1}^{n}{B_{j}\cos(j\pi x)}$ for some constants $ B_j$.

Now : $ u_{2}' \equal{} \minus{} \sum_{j \equal{} 1}^{n}{B_{j}\cdot j \pi\cdot \sin(j\pi x)}$ and $ u_{2}'' \equal{} \minus{} \sum_{j \equal{} 1}^{n}{B_{j}\cdot j^2 \pi ^ 2\cdot \cos(j\pi x)}$. We get :

$ \minus{} \sum_{j \equal{} 1}^{n}{B_{j}\cdot (j^2 \pi ^ 2 \minus{} \pi \minus{} k)\cdot \cos(j\pi x)} \equal{} \minus{} \sum_{j \equal{} 1}^{n}{\cos(j\pi x)}$

And from this get : $ B_{j} \equal{} \frac {1}{j^2 \pi ^ 2 \minus{} \pi \minus{} k}$

Our solution is : $ u \equal{} C_1 \cos (x\sqrt {\pi \plus{} k}) \plus{} C_2 \sin (x\sqrt {\pi \plus{} k}) \plus{} \sum_{j \equal{} 1}^{n}{\frac {1}{j^2 \pi ^ 2 \minus{} \pi \minus{} k}\cos(j\pi x)}$.

$ u' \equal{} A_1 \sin (x\sqrt {\pi \plus{} k}) \plus{} A_2 \cos (x\sqrt {\pi \plus{} k}) \minus{} \sum_{j \equal{} 1}^{n}{\frac {j\pi}{j^2 \pi ^ 2 \minus{} \pi \minus{} k}\sin(j\pi x)}$ where $ A_1 \equal{} \minus{}C_1\sqrt {\pi \plus{} k}$ and $ A_2 \equal{} C_2\sqrt {\pi \plus{} k}$

From our conditions we have : $ u'(0) \equal{} A_2 \equal{} 0$ and $ u'(1) \equal{} A_1 \sin (\sqrt {\pi \plus{} k}) \minus{} \sum_{j \equal{} 1}^{n}{\frac {j\pi}{j^2 \pi ^ 2 \minus{} \pi \minus{} k}\sin(j\pi)} \equal{} 0$.
But $ \sin (j\pi) \equal{} 0$, so $ A_1 \equal{} 0$.

And our equation will have a solution : $ \boxed{u(x) \equal{} \sum_{j \equal{} 1}^{n}{\frac {1}{j^2 \pi ^ 2 \minus{} \pi \minus{} k}\cos(j\pi x)}}$

As we can see from this solution, our constant $ k$ shouldn't be $ j^2 \pi ^ 2 \minus{} \pi$ for $ j \equal{} 1..n$.

So, final answer will be : $ \boxed{k\ne j^2\pi^2 \minus{} \pi \; \; \forall j\in \left\{1,2,...,n\right\}}$


Let $ a,b,c\geq 0$ and $ a\plus{}b\plus{}c\equal{}3$
Prove $ \sum\frac{a^2}{1\plus{}b^2}\geq\frac{3}{2}$

[quote="Algadin"]Let $ a,b,c\geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$
Prove $ \sum\frac {a^2}{1 \plus{} b^2}\geq\frac {3}{2}$[/quote]
I don't think that it's new.
$ \sum_{cyc}\frac {a^2}{1 \plus{} b^2} \equal{} \sum_{cyc}\frac {a^4}{a^2 \plus{} a^2b^2}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2 \plus{} a^2b^2)}\geq\frac {3}{2}.$

What about the following inequality:
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {a^2}{1 \plus{} b^2} \plus{} \frac {b^2}{1 \plus{} c^2} \plus{} \frac {c^2}{1 \plus{} a^2}\geq\frac {3}{2}\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}
\]
:wink:

This is the original inequality. Not as nice a proof.

[hide="Solution"]

Let $ S \equal{} \sum_{cyc} \frac{a^{2}}{1\plus{}b^{2}}$

By Cauchy's Inequality:

$ S(3\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}) \geq (a^{2}\plus{}b^{2}\plus{}c^{2})^{2})$

Hence

$ S \geq \frac{(a^{2}\plus{}b^{2}\plus{}c^{2})^{2}}{(3\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}) }$

Since $ a\plus{}b\plus{}c \equal{} 3$ hence $ 3 \equal{} \frac{(a\plus{}b\plus{}c)^{2}}{3}$

Thus we have

$ S \geq \frac{3(a^{2}\plus{}b^{2}\plus{}c^{2})^{2}}{3(a^{2}\plus{}b^{2}\plus{}c^{2}) \plus{} (a\plus{}b\plus{}c)^{2}} \geq \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2}$

Now since $ \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{3} \geq (\frac{a\plus{}b\plus{}c}{3})^{2} \equal{} 1$

Hence $ S\geq \frac{3}{2}$.[/hide]

I will try Arqady's problem now.


Challenge Problem1: Find $ h^{\minus{}1}(\minus{}1)$ without finding $ h^{\minus{}1}(x)$ if $ h(x)\equal{}\frac {9^{2^{h(x)}}\minus{}2x} {5(5^{\minus{}1})\minus{}x}$ .Then, find $ h^{\minus{}1}(x)$ ,what can you conclude about h?

Please give some feedback on the difficulty(1-5),and enjoyment.Thanks!

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=297920[/url]


$ABC$ is an isosceles triangle $(AB=AC)$ and $\angle{A}=20^0$. $D$ is a point on $AC$ such that $AD=BC$.

Find $\angle{ABD}$.

Hm I got it down to:
[hide]
$tanx=\frac{sin20^2}{sin80-sin20cos20}$
[/hide]

Easy sinelaw:  (lets work in degrees)

BD / sinA = AD / sinABD

BD / sinC = BC / sinBDC = AD / sinBDC

Thus, sinA/sinABD = sinC/sinBDC

Put angle ABD = x.

Then we have sin20/sin80 = sinx/sin(x+20)

Put y = x+10.  The left side is $2 \cos 10$.

So we have $2\cos 10 \sin(y+10) = \sin (y-10)$

Expanding, we have $2\cos 10 \sin y \cos 10 + 2 \cos 10 \cos y \sin 10 = \sin y \cos 10 - \cos y \sin 10$

or $\tan y = \frac{ \sin 10 (1 + 2 \cos 10) }{ \cos 10 (1 - 2\cos 10) } = \tan 10 (\tan 5)^{-2}$

I used a calculator to find that y doesn't have a nice value, so we are done here.

***********

edit:  your right, mistake is above

singular i think you made a mistake, i plugged my expression into the calc and it gave me 10 as the angle

i still don't know how to simplify my expression into that tho...

I've seen this problem before. Actually, there's a nice way without using trig :)

Reflecting and copying the original triangle, with one more 20-80-80 triangle in between, we obtain the figure below.

$\triangle AGD$ is an equilateral triangle since $AG=AD$ and $\angle GAD=60^\circ$. Thus $GEBD$ is a rectangle.

$\angle DAB=20^\circ$ and $\angle ADB=90^\circ +60^\circ =150^\circ$, so $\angle ABD=\boxed{10^\circ}$

There are many other ways to do it.
[hide="solution 1"]Rotate $AB$ anticlockwise about B for $60^\circ$ to EB. Then $\triangle ADB\cong\triangle BCE \Rightarrow AB=BE$. Since $\angle ABE=60^\circ$, we know that $E, B, C$ all lie on the circle with centre A and radius AB. $\therefore \angle CEB=\frac{20^\circ}{2}=10^\circ \Rightarrow \angle ABD=\angle CEB=10^\circ $[/hide]
[hide="solution 2"]Construct an equilateral triangle BEC with point E lying inside $\triangle ABC$. We can prove that $\triangle ACE\cong \triangle BAD$. Therefore $\angle ABD=\angle CAE=\frac{20^\circ}{2}=10^\circ$.[/hide]


Find all the triples $(x,y,z)$   such that the number

$A=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$  is an integer

Do you assume that $x,y,z$ are positive integers ?

If yes, there are no solution for $A > 3$
If $A=3$ we must have $x=y=z=1$
If $A=2$ we must have wlog $x=1$, and then $1/y+1/z=1$ or $y=z=2$

If $A=1$, wlog $x \leq y \leq z$
Then $3/x \leq 1$ i.e $x=1,2$ or $3$
$x=1$ is not possible
If $x=2$ then $y \leq 4$ and manual check gives $2,3,6$ and $2,4,4$
If $x=3$ then $y \leq 3$ and manual check gives $3,3,3$


symmetry
graph theory
combinatorics unsolved
combinatorics

you can use double counting on that
in the way to solve or prepare the plane you can use graph theory
I mean you can use graph theory WITH double counting
 ;)  :D
is that what you want??
or you want to solve this problem only useing graph theory??

[quote="Barnacle"][quote="ehsan2004"]I know that this problem isn't hard, but I want to see the solution that using the graph theory . ;)

A difficult mathematical competition consisted of a part (a) and a part (b) with a combined total of $28$ problems. Each contestant solved $7$ problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who in part (a), solved either no problems or at least four problems.[/quote]

I think this is a ridiculous argument but someone tell me if this is true haha (its not a graph theory argument): for the desired statement to be false, it would need by symmetry on (a),(b) to be the case that each contestant solved 1,2, or 3 problems in each of part (a) and part (b). so that at most they could've solved 6 problems, contradicting the fact that you solved 7.[/quote]

can someone explain why that proof is wrong? coming back a few days later makes it seem more plausible.  am i severely misunderstanding the question? i didnt even use most of the facts they gave in the problem lol...


$ A,B$ are fixed points. Variable line $ l$ passes through the fixed point $ C$. There are two circles passing through $ A,B$ and tangent to $ l$ at $ M,N$. Prove that circumcircle of $ AMN$ passes through a fixed point.

These circles, one pass through A, other pass through B ? It seems to be not clear.

COULD YOU DRAW A FIGURE? I THINK M=N  :?:


function
combinatorics unsolved
combinatorics

For each of the following sets of binary strings, give an expression that generates each string in the set exactly once.

(a) Binary strings in which every block of 0s has even length.
(b) Binary strings that contain no occurrence of the substring 111.
(c) Binary strings that contain no occurrence of the substring 010.

For a) i got $ \frac{x\minus{}2}{1\minus{}2x^2\minus{}x\plus{}x^3}$

b) and c) have the same generating function; $ \frac{2x^4\minus{}x^3\plus{}1}{1\minus{}2x}$

Are these correct? Help please :)

Nevermind, I did them horribly wrong. Got the answers no =)


if x is a real number and x !=-3, find the minimum value of (x^2+2x-1)/(x^2+6x+9)

What I did was make it [(x^2+6x+9)-4x-10]/(x^2+6x+9) = 1-(4x+10)/(x^2+6x+9).  So we want to maximize (4x+10)/(x^2+6x+9), but how do i do that w/out calculus.  Or if not, i'd also like to see a completely different solution too.

Hint: You can write your final expression as 2[1 - ((x+2)/(x+3))2].

Thanks.  I don't see where that comes from, though.  How did you see that?

I started out by trying to express the numerator 4x + 10 as a function of x + 3.  In other words, 4x + 10 = 4(x + 3) - 2.  Then I played around with some algebraic manipulations.


geometry
calculus
integration
real analysis
real analysis unsolved

I'm having some trouble finding a few volumes by rotating an area bounded by two curves, or a curve and a line (which is where most of my problems are coming from).

Find the volume of the solid obtained by rotating the region bounded by
x=9 y^2, \ y = 1, \ x = 0,
about the y-axis.

With this one, I thought you would use cylindrical shells to calculate the volume, but I'm really not sure how to go about doing that.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line x=7

x = y^{4}, \ x = 1;

my professor didn't go over going around a line, and i'm just not really sure how to go about setting the radius up for these kinds of problems.[/code]

If you think about it, the line $x=1$ is just the $y$-axis, translated horizontally by 1. If we moved both the figure and the axis by the same amount, nothing would change- it's the distance to the axis that matters, not any absolute position.

so with the second problem,

would you just set up the integral with y^4 as the area?  And then the bounds would be from 1 to 7..?  I'm still kind of lost.


What are some good national or regional (florida) writing contests... essay contests, anything of the sort. I know of Ayn Rand's contest, Scholastic Art & Writing Awards, another scholastic contest sponsored by CitiBank, and another contest by Penguin Publishing contest. Do any of you compete in any of these or other writing contests?

i know this is a math forum. math kids like writing too....sometimes.

http://www.campbellssoup.com has one that sounds pretty good; hefty scholarship money up for grabs, so it's worth a try.

hmm for high school and junior high theres the Patriots Pen but the deadline just passed.  Theres several others I'll try to find the names.

Or rather, does anyone know a site that has a ton of links to writing contests?

thanks to those who replied. The campbells soup one sounds yummy.


Έφτιαξα δύο ανισότητες σήμερα (ελπίζω να σας αρέσουν)
και τις βάζω .Η πρώτη είναι εύκολη για χαλάρωμα όπως γράφω αλλά η δεύτερη είναι καλή.....
Περιμένω τις λύσεις σας.
Για τους θετικούς πραγματικούς $a,b,c>0$ να δειχθεί ότι
1)$\frac{2}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{1}{2a+b}+\frac{1}{2b+a}+\frac{1}{2c+a}+\frac{1}{2a+c}+\frac{1}{2b+c}+\frac{1}{2c+b}$

2)
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c}\geq\frac{3}{2a+b}+\frac{3}{a+2b}+\frac{3}{2b+c}+\frac{3}{b+2c}+\frac{3}{2c+a}+\frac{3}{c+2a}\]

Η λύση είναι "στο πόδι" οπότε αν μου 'φυγε λάθος κάποιο σύμβολο συγχωρείστε με..

Για την 1:

$\frac{1}{a}+\frac{1}{b}\geq\frac{3}{2a+b}+\frac{3}{2b+a}\implies\frac{a+b}{ab}\geq\frac{9(a+b)}{2a^{2}+2b^{2}+5ab}\implies\frac{2a^{2}+2b^{2}+5ab-9ab}{ab(2a^{2}+2b^{2}+5ab)}\geq0\implies\frac{2(a-b)^{2}}{ab(2a^{2}+2b^{2}+5ab)}\geq0$

Το ίδιο κάνουμε κυκλικά και για τις υπόλοιπες και στη συνέχεια προσθέτουμε κατά μέλη..

Η 2 θέλει ψάξιμο

Ωραία mostel ... :)
Απλά η λύση μου εκφράζει αυτό με τον εξής τρόπο
Είναι  από Cauchy (Andreescu)
$\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\geq\frac{9}{2a+b}$

...... :wink:


Find the sum of all values of k that satisfy:
$2^{2k}-2^{k+4}+48=0$

Are you sure this is an intermediate question?

[hide]

2^2k - 16(2^k) + 48 = 0

(2^k - 12)(2^k - 4) = 0

2^k = 12

k = (log12)/(log2)

2^k = 4

k = 2

[/hide]


[quote="kheg_k"][hide="hint"]first I would subtract 4 from both sides if the equation.[/hide][/quote]

could i get one more hint?? pls!

[quote="linzk3"]Solve for x in terms of a and b.

I just need some help getting started.

4+x/4-x=a/b[/quote]
Cross multiply. Then group the terms with $x$ on one side. Factor out $x$, and divide. :D

[quote="linzk3"]Solve for x in terms of a and b.

I just need some help getting started.

4+x/4-x=a/b[/quote]

does this mean:

$4+\frac{x}{4}-x = \frac{a}{b}$

or:

$\frac{4+x}{4-x}=\frac{a}{b}$?

oh and either way, I wouldnt subtract 4 from both sides...

Prove this and use it  $\frac{a}{b}= \frac{c}{d}\Leftrightarrow \frac{a}{b+a}= \frac{c}{d+c}$

 :)

[quote="inom"]$4+\frac{x}{4}-x = \frac{a}{b}$

$16+\frac{x}{4}-x^{2}-\frac{4a}{b}=0$[/quote]

What?? Are these 2 consecutive steps?

Anyway, it's a linear function.

[hide="solution"]

We start with the equation

$4+\frac{x}{4}-x=\frac{a}{b}=4-\frac{3x}{4}$

And solve.

$\frac{3}{4}\cdot x = 4-\frac{a}{b}$
$x = \frac{16}{3}-\frac{4a}{3b}$[/hide]

[quote="anirudh"][quote="linzk3"]Solve for x in terms of a and b.

$\frac{4+x}{4-x}=\frac{a}{b}$[/quote]

[hide="Small hint"]Both are fractions...[/hide]
[hide="Large Hint"]Simplify[/hide][/quote]

Wow, both are fractions and simplifying? I would've never guessed. Thanks for your help!

If you meant this...[quote="linzk3"]$\frac{4+x}{4-x}=\frac{a}{b}$[/quote]
[hide] $\frac{4+x}{4-x}=\frac{a}{b}$
Cross-multiply: $(4+x)b=(4-x)a$
$4b+xb=4a-xa$
$xb+xa=4a-4b$
$x(a+b)=4a-4b$
$x=\boxed{\frac{4a-4b}{a+b}}$ [/hide]

[quote="anirudh"][quote="linzk3"]Solve for x in terms of a and b.

$\frac{4+x}{4-x}=\frac{a}{b}$[/quote]

[hide="Small hint"]Both are fractions...[/hide]
[hide="Bigger Hint"]Cross multiply[/hide]
[hide="Large Hint"]Simplify[/hide][/quote]
Seriously, you should stop editing your posts when you say something stupid. It isn't good.

[quote="neelnal"]"Stupid" is a relative term. Stupidity is in contrast to something else. For example, I bet I can consider most people stupid relative to Einstein. But if I were to make this term relative to a rock, we would all be supergeniuses.

It is also an offensive term.

anirudh is not stupid; he is very intelligent, otherwise, he would not be on this forum, let alone on any math website. No one is stupid; they are HUMAN. And humans are smarter than 99.99999% of all living organisms.[/quote]
Let me rephrase that.
When you post something that you later regret, don't edit it.

He didn't edit it.  It was part of his original post.

Also, you shouldn't discuss anirudh's posting behavior here.  That too is considered spamming.


In the Baeyer villiger oxidation of aldehydes using percarboxylic acids.what is the order of migratory aptitude of groups - Is it
(1) H > phenyl > methyl
(2) phenyl > methyl > H
(3) phenyl > H > methyl
(4) methyl > H > phenyl
{Different books have given different orders}
Does the order change for different reactions?

Revived an old unanswered topic  :D

I think the answer is either 1 or 3 depending on the substitution in the benzene ring  :D

That's right: in "normal" reactions, H has the greatest migratory aptitude (that's way carboxilic acids are almost always obtained); when the benzene ring is activated, then aryl migration is preferred (the Dakin reaction); when the benzene ring is deactivated, alkyl migration can be preferred.


linear algebra
matrix
LaTeX
linear algebra unsolved

Let $ A,$ $ B$ and $ \Psi$ be positive definite matrices, $ D$ by $ D,$ where $ D$ is finite. $ \Psi$ can be assumed to be diagonal (all entries on the diagonal are positive numbers, the rest is zero). Denote by $ \Pi_n$ the $ D$ by $ D$ matrix obtained from $ \Psi$ by keeping only the first $ n$ entries from the diagonal of $ \Psi$ and completing the rest of the diagonal with zero (that is, $ \Pi_D\equal{}\Psi$).

Prove that

For any $ 1\leq n \leq D$, there is a rank-$ n$ orthogonal projector $ P_n\equal{}P_n^\dagger\equal{}P_n^2$ such that

$ \text{Trace}(P_n B \Psi A^2 \Psi B P_n) \leq \text{Trace}(B \Pi_n A^2 \Pi_n B)$

Thanks,

Vlad

Note: you can get $ \text{\LaTeX}$ to render on this site just by putting your code between dollar signs. I took the liberty of editing your post; you should check it yourself to see that it's OK.


i need help with this one:

find the sum:

$S=1^{2}*r+2^{2}*r^{2}+3^{2}*r^{3}+...+n^{2}*r^{n}$

a full solution would be greatly appreciated.

[hide="Hint"]
How can we reduce the power of $n$ on each of the terms?

[hide="Further Hint"]

Multiply $S$ by something, then subtract that from $S$. Rinse. Repeat until $S$ is easily summable.  :wink: [/hide] [/hide]


algebra
polynomial
function
algebra proposed

Suppose that $n$ be a positive integer and $P_n$ be a set of all polynomial with integer coefficient in form $a_0+a_1x+...+a_nx^n$ where $\mid a_i\mid \leq 2 , i=0,1,...,n .$ For positive integer $K$, find the number of elements of the set $A_n(K)=\{f(k)\mid f\in P_n \}.$

I think ,it's very interesting problem  :maybe:

If $ k\ge 5$, then it is obviously $ 5^{k}$ since we should have unique representation in like that base. The other four cases are not resolved as easily due to 'carrying'.

I agree that it is a nice problem.

You mean $ 5^{n}$, right?

[hide="Idea for the other values of k"]For $ k\leq 5$, consider the two functions in $ P_{n}$ with all coefficients -2 and all coefficients 2. Evaluate both at $ k$, and you should get $ A_{n}(k)$ to equal the closed interval between the two resulting values.[/hide]


geometry
circumcircle
incenter
geometry unsolved

Dear Mathlinkers,
let ABC a triangle, 0 the circumcircle, O the center of 0, I the incenter, A' the second intersection of AI with 0, A" the point of intersection of the parallel to AI through O with BC. Cyclically we have B', B", C', C".
Prove that A'A", B'B" and C'C" are concurrent.
Sincerely
Jean-Louis

Dear Mathlinkers,
two synthetic proofs of this known result originated from Clark Kimberling has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


geometry
3D geometry
inequalities proposed
inequalities


geometry
geometric transformation
reflection
articles

I am a math major but I also take the Moderen Chinese Language. I am studying the Chinese characters, and I have an assignment that we need to write a theme about Chinese characters. I want to write about the advantages and the disvantages of Chinese characters when compared with the words of English or other alphabetic languages. Could you please tell me some ideas?

In my opinion, Chinese, cuneiform, and hieroglyphics represent writing systems that reinforced the supremacy of the elite class.  In the corresponding societies, and under those systems, the literate elite could transmit and store information, a hugely useful bureaucratic tool, using an unwieldy written language not readily learned by the rest of the population.  Cuneiform and hieroglyphics died out when the societies using them were conquered or gradually lost power in other ways, to be replaced by societies using written languages more readily accessible to the public.

Chinese system is an interesting exception historically.  Beginning with the standardization of written Chinese by the Qin Dynasty, the Chinese writing system became an enshrined and fundamental aspect of Chinese culture; to the extent that Chinese characters have aesthetic beauty, writing became a form of visual as well as literary art.

That doesn't change my basic opinion, which is that the Chinese system is terrible compared to an alphabetic one.  The only reason the switch to an alphabetic system of writing Chinese would be awkward (to say nothing of its cultural implications) is the large number of homophones.

One possible disavantage of Chinese characters when comparared to an alphabetic language is that it is much more harder to learn to write in Chinese than in English for example. On the other hand, and as far as I know, the grammar structure of Chinese is quite simple. It would be also worth mentioning the "four-character idioms" (or chéngyu), which are, in my opinion, an interesting aspect of Chinese language that reflects an important cultural heritage and is not present, at least in the same extension, in most other languages.

[quote="tokenadult"]The book [url=http://www.amazon.com/Worlds-Writing-Systems-Peter-Daniels/dp/0195079930/]The World's Writing Systems[/url] is a good reference book with an article that gets into these issues. Other books on the same Amazon page may be helpful too, although I haven't personally looked at those. The first two replies mention some important issues.[/quote]
I have read the contents and I know it is a good reference for me, but I can't find the book in my university's library. On the other hand, I need writing the article in Chinese, and my english is not so good, so if I can find it, I need a lot of time to read it. Actually, I just want to know how people who speak alphabetic languages and know Chinese think of the Chinese charactors.

Well, interesting part of using of the Chinese characters is use in japanese - 漢字　（kanji), and korean - 한자 (hancha). Japanese&korean are two languages that are so much different that chinese, despite it chinese ideograms were (and still are, but in korean seldom nowadays) very good tool to write:)

Circa half of japanese words are sino-japanese;)

Well, I'm learning japanese, not chinese, but I see chinese ideograms as something necessary, because:
1. homophones
2. knowing it helps in making new words and remembering existing, e.g. 運転者, where 運 means 'transport, carry', 転 means 'change, turn around and '老' is one of prefixes meaning someone who is doing something (by the way, in japanese is read 'untensha', what's chinese reading? any similarity? XD),
3. learning text written by kana and nothing itself is pain in the ass:)

And, last but not least, they're beautiful:) It's good that they're more similar to traditional chinese than to simplified, cuz I find some simplified chinese characters somewhat ugly, e.g. 見　-> 见 or 風 -> 风.

By the way ifai, 您好 (nin2 hao3 in mandarin?) is something like 你好? :)

[quote="liu"]By the way ifai, 您好 (nin2 hao3 in mandarin?) is something like 你好? :)[/quote]
"您" is a respectful form of  "你 (ni3)". For instance, we say "您好" to your parents or your boss but "你好" to your friends. And we pronounce 你好 ni2hao3 instead of ni3hao3 since "3 3" is not natural.

it takes too much time a write a character. Words like 龍,鳳,費,聽,牽 are pretty ridiculous (not for simplified chinese of coz)
And the characters themselves don't tell us much about their pronunciations, which is so not convenient
especially for Cantonese, which even teachers manage to pronounce terribly

but then the characters could be written quite beatifully. You know, people in China used to learn calligraphy at school (and they valued it).  But I wouldn't expect to see the beautiful chinese handwriting 10 years later, at least not as much, as people from our generation don't really care about that.

PS I speak chinese, so you probably won't need my opinion

Well, how many times have you written text about dragons? ;)
I know what do you mean when you talk about calligraphy. Reading handwritten notes taken by e.g. younger japanese people is such a pain...  Especially in cases of characters consisting of many radicals, e.g. 繁.
But it's problem of all systems of writing, I can't stand seeing latin alphabet handwriting of many friends/teachers etc;) My own isn't very good, but at least it's readable;P And I know how to write using ,,school script" teached in elementary school, which it's rare nowadays. And with chinese characters I'm trying very hard to write them in way, that I would like to read. So there are people who care for that:)


A circle with center [i]O[/i] is tangent to the sides [i]AB, BC, CA[/i] and [i]DA [/i]of a convex quadrilateral [i]ABCD[/i] at [i]K, L, M,[/i] and [i]N [/i]respectively. The lines [i]KL [/i]and [i]MN[/i] intersect at point [i]S[/i]. Prove that [i]BD[/i] is perpendicular to [i]OS[/i].

did you mean side CA, or did you mean side CD?


let a,b,c,d be nonzero integers such that a is different from c, iff:

a/c=(a 2 +b 2 )/(c 2 +d 2 )

prove that a 2 +b 2 +c 2  is not a prime!

Where exactly is this from? Do you know what olympiad and/or grade it was proposed for? Maybe someone can find it (and a solution :)).

I found a solution:

from the hipotesis: (a-c)(b 2 -ac)=0, then b 2 =ac because a is different from c.

a 2 +b 2 +c 2 =
a 2 +(2ac-b 2 )+c 2 =
(a+c) 2 -b 2 =(a+c-b)(a+c+b). now, because
b 2 =ac we have a+c> =2:sqrt: ac=2b so a+c-b>1 and the number is not a prime

sorry but i don't understand one thing: when you posted the problem, did you mean $a/c=\frac{a^2+b^2}{b^2+c^2}$ instead of $.../(c^2+d^2)$ ???

oooopss

You are right, there is no d in the problem, sorry!


calculus
integration
algebra
polynomial
calculus computations

hi i need help on this certain problem

evaluate the definite integral
integral 2 to 9 ((10x^2)+4)/(sqrt(x)) dx

i came to integral 2 to 9  ((10x^3)/(sqrt(x))+(4/sqrt(x)) can someone point me in the right direction on how to take the antiderivative of this.

thanks for your help

Do you know how to find the antiderivative of $ax^b?$ Both terms are of this form.

yes but the sqrt(x)'s are throwing me off

You can use for $p\neq -1$,

$\int x^p dx=\frac{1}{p+1}x^{p+1}+Const.$

So we have $\int \sqrt{x} dx=\int x^{\frac{1}{2}}dx=\frac{2}{3}x^{\frac{3}{2}}+Const.$

But, since $\frac{d}{dx} \sqrt{x}=\frac{1}{2\sqrt{x}}$, we get $\int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+Const.$

This is worthwhile to memorize, because this integral appeares frequently.

for the first one you can do this:
$\int_{2}^{9}\frac {10x^2+4}{\sqrt{x}}dx=\int_{2}^{9} 10\sqrt{x^3}+\frac {4}{\sqrt{x}} dx=4\sqrt{x^5}+8\sqrt{x}$ from $2$ to $9$.
the final answer for the first one is: $996-24\sqrt{2}$

remember,just a piece of advice, you can solve many of these kind of integrals by division.  all you need to do is devide the numinator by the denominator and making a new polynomial, then integrate.


[quote="No Reason"]Let p be odd prime number and r be a positive integer.Prove that:
$ \binom{p^r}{p} \equiv p^{r \minus{} 1} (mod p^r)$[/quote]

We have:
$ \prod_{i\equal{}1}^{p\minus{}1}{(p^r\minus{}i)} \equiv \prod_{i\equal{}1}^{p\minus{}1}{(i)} (mod p^r)$
then
$ \binom{p^r}{p}\minus{}p^{r\minus{}1}\equal{}\frac{\prod_{i\equal{}0}^{p\minus{}1}{(p^r\minus{}i)}}{\prod_{i\equal{}1}^{p}{(i)}}\minus{}p^{r\minus{}1}\equal{}\frac{\prod_{i\equal{}0}^{p\minus{}1}{(p^r\minus{}i)}\minus{}p^{r\minus{}1}(\prod_{i\equal{}1}^{p}{(i)})}{\prod_{i\equal{}1}^{p}{(i)}}\equal{}\frac{p^{r\minus{}1}(\prod_{i\equal{}1}^{p\minus{}1}{(p^r\minus{}i)}\minus{}\prod_{i\equal{}1}^{p\minus{}1}{(i)})}{\prod_{i\equal{}1}^{p\minus{}1}{(i)}} \equiv 0 (mod p^r)$(becouse$ \prod_{i\equal{}1}^{p\minus{}1}{(p^r\minus{}i)} \equiv \prod_{i\equal{}1}^{p\minus{}1}{(i)} (mod p^r)$)

 therefore:
$ \binom{p^r}{p} \equiv p^{r \minus{} 1} (mod p^r)$

p.s.:we also have $ \binom{p^r}{p} \equiv p^{r \minus{} 1} (mod p^{2r\minus{}1})$

[quote="No Reason"]Let p be odd prime number and r be a positive integer.Prove that:
$ \binom{p^r}{p} \equiv p^{r \minus{} 1} (mod p^r)$[/quote]
First note that $ \binom{p^r}{p} \equal{} p^{r\minus{}1}\binom{p^r\minus{}1}{p\minus{}1}$. Then it comes down to proving $ \binom{p^r\minus{}1}{p\minus{}1} \equiv 1 \bmod p$. Which is obvious since $ \binom{p^r\minus{}1}{p\minus{}1} \equal{} \frac{(p^r\minus{}p\plus{}1)(p^r\plus{}2) \cdots (p^r\minus{}p\plus{}(p\minus{}1))}{1 \cdot 2 \cdots (p\minus{}1)} \equiv \frac{(p\minus{}1)!}{(p\minus{}1)!} \equal{} 1 \bmod p$. QED


A reference book lists a set of annual calendars. For any given year, there is a
calendar in the set that corresponds to it. How many annual calendars must be
included in the set in order to have a corresponding calendar for every possible year?

Is this question flawed??? It doesn't give any information at all, such as the number of years.....

Oh thanks.  i didn't see that different years can use teh same calendar.  then its 14?

Yes it's 14. How is $ 7\plus{}7\equal{}77$??

[quote="mewto55555"]Yes it's 14. How is $ 7 \plus{} 7 \equal{} 77$??[/quote]

quote me

[quote="GeoffreyS"]kind of unfair to people that don't know about leap year[/quote]

... WOW.

Tell me someone that doesn't

:D

[quote="mz94"]7 calendars must be used for years that start with each corresponding day of the week. another 7 must be used for leap years that start with each day of the week. so 7+7=[b]77[/b]

[size=9][color=white]Quote me and you'll find that the answer is 14[/color][/size][/quote]

BAM


Let $ S \equal{} \{1,3,4,9,10,12,13...\}$, where all the elements of $ S$ are powers of $ 3$ or sums of powers of $ 3$ (each power used once, if at all).  If all the elements are arranged from least to greatest, find the $ 100$th term.

Source: AIME

[hide]Consider the numbers when written in base 3. They form the set of positive integers when expressed in base 2. (In fact, all positive integers expressed in base 2 can be translated as if in base 3 back into base 10 and exist as an element of the given set.)

The binary representation of 100 is 1100100. Converting this from base 3 to base 10 yields $ 981$.[/hide]

out of curoiusity, what number/year aime is this?

[url=http://www.artofproblemsolving.com/Wiki/index.php/1986_AIME_Problems/Problem_7]AIME 1986 Problem 7[/url]. :)

I did the problem using combinatorics, kind of.
For all $ 3^{n}$, I proved that $ 3^{0} \plus{} 3^{1} \plus{} 3^{2} \plus{} .. \plus{} 3^{n \minus{} 1}$ is less than $ 3^{n}$.  I also was able to prove easily that between the powers $ 3^{n}$ and $ 3^{n \minus{} 1}$, there are $ 2^{n \minus{} 1} \minus{} 1$ different elements in $ S$.  So, I just listed out the powers and the number of elements between the powers, found what range the $ 100th$ term is, and from there, I just looked to see what I had to add to the lower range.  I too got $ \boxed{981}$.


geometry
logarithms
integration
calculus
calculus computations

Show that the curve whose polar equation is $ r \equal{} a \plus{} b \sec \theta$, where $ a > b > 0$, has a loop. Prove that the area enclosed by the loop is
\[ a^2 \arccos \frac {b}{a} \plus{} b\sqrt{a^2 \minus{} b^2} \minus{} 2ab \ln \frac {a \plus{} \sqrt {a^2 \minus{} b^2}}{b}.\]

Hello, any one wants to give this problem a try?

hello, you must use the formula
$ A=\frac{1}{2}\int_{\phi_1^}^{\phi_2}r^2\,d\phi$
Sonnhard.


Let $ [x]$ be the greatest integer that is less than $ x$ (I.E. $ [3.3]\equal{}3$). Let x and y be real numbers such that:

$ y\equal{}2[x] \plus{}3$
$ y\equal{}3[x\minus{}2] \plus{}5$

If $ x$ is a non integer number, then x+y  is:

(a) integer    (b)between -4 and 4       (c) Between 4 and 5

Find all solutions (if exsist)to the equation:

$ 3 ^{2x\plus{}2} \minus{}3^{x\plus{}3} \minus{}3^x \plus{}3 \equal{}0$

If $ a,b,c$ are real non-zero numbers such that

$ \frac{a\plus{}b\minus{}c}{c} \equal{}
\frac{a\minus{}b\plus{}c}{b}\equal{}\frac{\minus{}a\plus{}b\plus{}c}{a}$

,and if   $ x\equal{}\frac{(a\plus{}b)(b\plus{}c)(a\plus{}c)}{abc}$ , $ x<0$

Then the value of x is:

(a) -1     (b) -2   (c) -6

For the first problem:

You know that a) is untrue, because $ x$ is not an integer.

Simply solve the equations:

$ y\equal{}2z\plus{}3$
$ y\equal{}3z\minus{}1$

Subtract the first from the second.

$ 0\equal{}z\minus{}4$
$ x\equal{}z$

$ z\equal{}[x]$

$ 4\le x \le 5$

But $ z$ is not the same value fot both equations. In the first one $ z\equal{}x$, but in the second one $ z\equal{}x\minus{}2$

If $ n$ is an integer, $ \lfloor a \plus{} n\rfloor \equal{} \lfloor a \rfloor \plus{} n$ ;)

fishythefish does have a few typos that need fixing, though.

[quote="Chalao"]
Find all solutions (if exsist)to the equation:

$ 3 ^{2x \plus{} 2} \minus{} 3^{x \plus{} 3} \minus{} 3^x \plus{} 3 \equal{} 0$

[/quote]

This will simplify to

$ 3^2 \cdot 3^{2x} \minus{} 3^x \cdot 3^3 \minus{} 3^x \plus{} 3 \equal{} 0$

Let $ u \equal{} 3^{x}$, then

$ 9u^2 \minus{} 27u \minus{} u \plus{} 3 \equal{} 0$

$ 9u(u \minus{} 3) \minus{}(u \minus{} 3) \equal{} 0$

$ (u \minus{} 3)(9u \minus{} 1) \equal{} 0$

$ \therefore u \equal{} 3, 3^{\minus{}2}$

ie

$ 3^x \equal{} 3^{1} \implies x \equal{} 1$

and

$ 3^x \equal{} 3^{\minus{}2} \implies x \equal{} \minus{}2$

Hence we have the solutions,

$ \boxed{x \equal{} 1, \minus{}2}$

[quote="Chalao"]

If $ a,b,c$ are real non-zero numbers such that

$ \frac {a \plus{} b \minus{} c}{c} \equal{} \frac {a \minus{} b \plus{} c}{b} \equal{} \frac { \minus{} a \plus{} b \plus{} c}{a}$

,and if   $ x \equal{} \frac {(a \plus{} b)(b \plus{} c)(a \plus{} c)}{abc}$ , $ x < 0$

Then the value of x is:

(a) -1     (b) -2   (c) -6[/quote]

Begin by cross-multiplying the equations:

$ ab\plus{}b^2\equal{}ac\plus{}c^2$
$ ab\plus{}a^2\equal{}bc\plus{}c^2$

Thus, $ ab\plus{}b^2\minus{}ac\equal{}ab\plus{}a^2\minus{}bc \implies bc\minus{}ac\equal{}a^2\minus{}b^2$

$ c(b\minus{}a)\equal{}(a\minus{}b)(a\plus{}b)$

$ \minus{}c(a\minus{}b)\equal{}(a\minus{}b)(a\plus{}b)$

$ a\plus{}b\equal{}\minus{}c$ OR $ a\equal{}b$.

Case 1: $ a\plus{}b\equal{}\minus{}c$.

$ \frac{a\plus{}b\minus{}c}{c}\equal{}\minus{}2$. $ \frac{a\minus{}b\plus{}c}{b}\equal{}\minus{}2 \implies a\plus{}c\equal{}\minus{}b$

$ \frac{\minus{}a\plus{}b\plus{}c}{a}\equal{}\minus{}2 \implies b\plus{}c\equal{}\minus{}a$.

$ (a\plus{}b)(b\plus{}c)(a\plus{}c)\equal{}\minus{}abc$.

$ \frac{\minus{}abc}{abc}\equal{}\minus{}1$.

Case 2: a=b.

By cross multiplying,
$ ab\plus{}a^2\equal{}bc\plus{}c^2$
$ ac\plus{}a^2\equal{}bc\plus{}b^2$.

$ c^2\minus{}ab\equal{}b^2\minus{}ac$

$ a(b\minus{}c)\equal{}(c\minus{}b)(c\plus{}b)$

$ \minus{}a(c\minus{}b)\equal{}(c\minus{}b)(c\plus{}b)$

If $ c\minus{}b\equal{}0$, $ a\equal{}b\equal{}c$, and $ \frac{(a\plus{}b)(b\plus{}c)(a\plus{}c)}{abc}\equal{}\frac{8a^3}{a^3}\equal{}8$, which isn't negative.

$ c\plus{}b\equal{}\minus{}a$, which is the same as the previous case.

~BOGTRO

I have a different solution than BOGTRO (but I get the same answer of course)

$ \frac{a\plus{}b\minus{}c}{c}\equal{}\frac{a\minus{}b\plus{}c}{b}\equal{}\frac{\minus{}a\plus{}b\plus{}c}{a}$ can be rewritten as

$ \frac{a\plus{}b}{c}\minus{}1\equal{}\frac{a\plus{}c}{b}\minus{}1\equal{}\frac{b\plus{}c}{a}\minus{}1$ which gives us

$ \frac{a\plus{}b}{c}\equal{}\frac{a\plus{}c}{b}\equal{}\frac{b\plus{}c}{a}\equal{}m$

Lets add all three fractions.

$ \frac{a\plus{}b}{c}\plus{}\frac{a\plus{}c}{b}\plus{}\frac{b\plus{}c}{a}\equal{}\frac{a^2b\plus{}ab^2\plus{}a^2c\plus{}ac^2\plus{}b^2c\plus{}bc^2}{abc}$

$ \equal{}\frac{(a\plus{}b)(a\plus{}c)(b\plus{}c)\minus{}2abc}{abc}\equal{}\frac{(a\plus{}b)(a\plus{}c)(b\plus{}c)}{abc}\minus{}2\equal{}m^3\minus{}2$

Since the addition of all three fractions yields $ 3m$, we have $ m^3\minus{}2\equal{}3m$ or $ m^3\minus{}3m\minus{}2\equal{}(m\plus{}1)^2(m\minus{}2)\equal{}0$
$ m\equal{}\minus{}1$ or $ 2$ but $ 0>x\equal{}m^3$ so only $ m\equal{}\minus{}1$ would give $ x$ a negative value of $ x\equal{}\minus{}1$.

P.S. How do you box answers in latex?

Hmm...I probably do have a few typos, but the general idea is the same.

I have a quick solution to #3, but it involves the fact that if $ \frac{x_1}{y_1} \equal{} \frac{x_2}{y_2} \equal{} \ldots \equal{} \frac{x_n}{y_n}$ for $ y_i \not\equal{} 0$ and $ \sum_{i \equal{} 1}^{n}{y_i}\not\equal{} 0$, then $ \frac{x_1}{y_1} \equal{} \frac{x_2}{y_2} \equal{} \ldots \equal{} \frac{x_n}{y_n} \equal{} \frac{x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n}{y_1 \plus{} y_2 \plus{} \ldots \plus{} y_n}.$

Using this, we have:

$ \frac{a \plus{} b \minus{} c}{c} \equal{} \frac{a \minus{} b \plus{} c}{b} \equal{} \frac{\minus{}a \plus{} b \plus{} c}{a} \equal{} \frac{a \plus{} b \plus{} c}{a \plus{} b \plus{} c} \equal{} 1.$

Thus,

$ \frac{a \plus{} b \minus{} c}{c} \equal{} 1\implies a \plus{} b \equal{} 2c\\
\frac{a \minus{} b \plus{} c}{b} \equal{} 1\implies a \plus{} c \equal{} 2b\\
\frac{\minus{} a \plus{} b \plus{} c}{a} \equal{} 1\implies b \plus{} c \equal{} 2a.$

Substituting this, we obtain: $ x \equal{} \frac{(a \plus{} b)(b \plus{} c)(a \plus{} c)}{abc} \equal{} \frac{8abc}{abc} \equal{} \boxed{8}$.

Elefont: it's [code]\boxed{answer}[/code]. In the future, you can click any LaTeX to see the code used to generate it.

TachyonPulse: $ x < 0$ though, yet I can't see a flaw in your answer.

Whoops, I forgot about that. I assumed that $ a \plus{} b \plus{} c \not\equal{} 0$; that is probably what is wrong with my solution.


Can someone help to show this?

$ (A \times B)^C=B^C \times A^C$

With C I mean the complement

Thanks guys

No, because it's false.  Any example in which neither $ A$ nor $ B$ is the universe nor the empty set suffices.

ow sorry stupid of me
I need to prove that there is a bijection between the two
they of course don't need to be equal

Do you know a one-to-one function which satisfies this?

Thanks for your help :blush:

I don't know what your universe is, but in general those sets do not necessarily have the same size.  For example, suppose our universe for $ A, B$ is $ \{1, 2, 3\}$ and $ A \equal{} B \equal{} \{1\}$.  Then $ (A \times B)^C$ has size 8 while $ B^C \times A^C$ has size 4.

Is it possible if C is a set?
(they use C also for complement, so i must have read it wrong)

If it's a set, you don't have to work very hard. There is a natural bijection between functions from $ C$ to $ A\times B$ and ordered pairs $ (f,g)$ of functions from $ C$ to $ A$ and $ C$ to $ B$ respectively.


Could you please rephrase that? Do you mean 1\6 of his remaining money or his original?

I think he means remaining because $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ already.

[hide]$6.25*2=12.50$.

$12.50=\left(1-\frac{1}{3}\right)x$

x is the money he had before eating lunch, so he spent $\$6.25$ on lunch.[/hide]

Actually all of my fractions were of what was remaining.

[hide]$6.25 i had to "work backwards" when i realised that i hadn't answered the question instead i had the starting amount of money.[/hide] :oops_sign:

[quote="nutz_for2.718281828"][hide="solution"]$*.5 .5*\frac{5}{6}*\frac{2}{3}*n=\frac{5n}{36}$
$\frac{5n}{36}=6.25$
$\boxed{n=45}$[/hide][/quote]
We are solving for the lunch price, not the original money he had.

[hide]
$6.25\cdot2=12.50$
$12.50\cdot\frac{3}{2}=19.50$
12.50 is what he had after he bought lunch, so:
$19.50-12.50=7$
[/hide]
I didn't get what all of you guys got...

[quote="Arvind_sn"]Billy Bob went to the Renaissance Festival. When he was there, he spent one half of his original money buying a black staff. Then, he spent one sixth of his money on a glass sculpture. After that, Billy Bob took a lunch break. He ordered medieval coke, some cheesy mushroom pizza, and cheese fries (Boy, doesn't this make you hungry?) for one third of his money. Then, he spent one half of that on a hat. He had six dollars and twenty five cents left. How much did his lunch cost? :ninja:[/quote]

$\$6.25\times\frac{2}{1}=\$12.50$ (hat)

$\$12.50\times\frac{3}{2}=\$19.50$ (lunch)

$\$19.50-\$12.50=\boxed{\$7.00}$

[quote="akalra1"]$\$12.50\times\frac{3}{2}=\$19.50$ [/quote]
That's incorrect.  :wink:


Ring Theory
superior algebra
superior algebra unsolved

Prove that a Dedekind with a finite number of prime ideals is PID.

By the approximation theorem, you can find an $x$ having the same valuations as a given ideal at all of the finitely many primes. Now $(x)$ has the same prime ideal decomposition.


probability
inequalities
search
triangle inequality

[quote="thegreatindiankhali"][size=150]Three real numbers are chosen at random between 0 and 1. What is the probability that they
form the side lengths of a triangle? Prove your answer.
[/size]
PLZ dont answer that:
0<=Probability<=1.
A rigorous and concrete answer is the need.[/quote]

probability = 1/2.
After picking a biggest number (N1), problem is simplified to finding probability for sum of 2 random numbers (between 0 and 1) is greater then 1. Which is obviously 1/2. (to satisfy triangle inequality)

That doesn't make sense. If you're biggest number (N1) is 1/2, then the two smaller numbers can't possibly add up to one, but they can still form a triangle.

You can divide all three numbers by the largest number so that that 0 < a <= b <= 1, where a and b are random numbers.
By the triangle inequality a + b > 1, so a > 1 - b;
Substituting in the first inequality gives 0 <= 1 - b < b <= 1, which will be true if b > 1/2.
This has a probability of 1/2.

This has been asked many times. You should try searching. (I've asked the same question too, though  :P )
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1845987662&t=280826
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=568364215&t=271275
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1188310357&t=279618

Similar problems:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1845987662&t=288922
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=568364215&t=26040


For $a,b,c$ positive reals such that $abc=1$ prove that

$\frac{1}{2+a} +\frac{1}{2+b}+\frac{1}{2+c} \leq 1$

Dear Manlio, I think the general case of this inequality was already discussed at least 5 times. Don't you think it's time for this "old" inequality to rest a little bit?

This is the special case for n = 2 of the inequality http://mathlinks.ro/Forum/viewtopic.php?t=19230 ...

  Darij

Sorry very much, Harzi. You are right that this is "old".

I posted it to see if somebody has a direct proof.

But I realize this is a stupid question :blush:  :blush:, sorry.

Okay. Calculations show that any three positive numbers a, b, c with abc = 1 satisfy

$1-\left(\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\right) = \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3}{\left(2+a\right)\left(2+b\right)\left(2+c\right)}$.

Since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3$ (from AM-GM, since abc = 1), the right hand side is $\geq 0$ and the inequality is proven.

Alas, this proof is not generalizable, but at least it's a proof...

  Darij

Thank you. I am so stupid and stupid and stupid that I didn't see this simple proof.

I am really ashamed for that  :blush: !!!!!

My brain is 0.

Stupid? I needed a computer to find this identity...

  Darij

Dear Darij, it is a simple algebraic computation. Why do you say you need a computer ?


geometry
Ross Mathematics Program
circumcircle
geometry theorems

The following problem was posed in IMO 2008 :

"Let H be the orthocenter of an acute-angled triangle ABC. The circle centered at the midpoint of BC and passing through H intersects the sideline BC at points A1 and A2. Similarly, define the points B1, B2,  C1 and C2

Prove that the six points  A1, A2, B1, B2, C1 and C2 are concyclic."

The solution to the problem was already there in Ross Honsberger's Book "Episodes in 19th & 20th Century Euclidean Geometry'. Moreover the following fantastic dual statement is also proved there:

'Let O be the circumcenter of an acute-angled triangle ABC. The circle centered at the foot of the height from A to BC and passing through O intersects the sideline BC at points A1 and A2. Similarly, define the points B1, B2,  C1 and C2.

Prove that the six points  A1, A2, B1, B2, C1 and C2 are concyclic.'

Remarkably, these two circles are centered at H and O respe ctively and have same radius ! :!:  :|

Nice. I posted to say IMO presents old problems also.
Moreover Honsberger addresses both problems simultaneously.

Mathlinkers have discussed that too. See here: http://www.mathlinks.ro/viewtopic.php?t=216051


Here is two problems from the Vietnamese qualifying test in 2003 and 2004:

1/ Let a,b,c be 3 positive real numbers such that a+b+c<=3/2. Prove that
sqrt(a^2+1/a^2)+sqrt(b^2+1/b^2)+sqrt(c^2+1/c^2)>=sqrt(81)

2/ Determine 3 angles of the acute triangle ABC such that
cos2A+3sqrt(3)(cosB+cosC)=3

Uh, Peter:

I may be wrong. But, when you say f is at least 17/4, I think you mean f is at least sqrt(17/4). Thus, 3f < sqrt(81), so this does not solve the proof? Perhaps I misunderstand something in the proof though?

-JGeneson

ah no i see :)

will correct it immediately.

Eh, I think the problem itself might need to be corrected. For some reason, I am getting a contradiction for the (1/2, 1/2, 1/2) case.

Hmm wait a sec... no way of correcting it! :?

$(a,b,c) = (\frac12,\frac12,\frac12)$ fulfills the requirement.

$\sqrt{\frac{17}4}+\sqrt{\frac{17}4}+\sqrt{\frac{17}4} < \sqrt{81}$ :?

ANMT10122603, could you please correct the problem statement?

it's not true... (a=b=c=1/2 is a counterexample)

oh you already posted this while I was posting :)

sorry didn't see it.

Sorry to everyone. I've made a mistake. The correct condition for the first problem is a+b+c<=1 and for the second:
Determine 3 angles of the triangle ABC such that
cos2A+2sqrt2(cosB+cosC)=3

Ok then the final answer for number 1:

Convexity of the function $f(x) = \sqrt{x^2+\frac1{x^2}}$ means $f(a)+f(b)+f(c) \geq 3\cdot f\left(\frac{a+b+c}3\right)$

The said property means that $\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}} \geq 3\cdot f \left(x\leq \frac13 \right)$
And $3\cdot f \left(x\leq \frac13 \right)$ is at least $3\cdot\sqrt{\frac{82}9}=\sqrt{82}$ which is strictly greater than $\sqrt{81}$ :)

Could it be the $\sqrt{81}$ has to be replaced by $\sqrt{82}$? Or the $\geq$ replaced by $>$? ;)


calculus
integration
function
real analysis
calculus computations

Let function f(x) is integrareble on the segment [0;1] and $\int_{0}^{1} f(x)>0$
Prove that there exist a,b s.t. 0<a<b<1 and f(x) is positive on this segment.
to view all the Some problems from the Uzbekistan universities olympiads visit
[url]http://www.mathlinks.ro/Forum/topic-81099.html[/url]

What about $f(x)=\chi_{\mathbb{R}\setminus\mathbb{Q}}$?  Clearly $\int^1_0 f(t)dt=1$, but $f$ is strictly positive on no segment.  If we're working with Riemann integration, use the characterization that an integrable function is continuous a.e; the set of points where $f(x)>0$ has to have positive measure so that ${f^{-1}((0,\infty])\cap \{x: f\text{ is continuous at }x}\}\not=\emptyset$.  Take some point in that intersection; $f$ is continuous there and $f>0$ there, so we can find our segment.

Looking at the contrapositive, we need to prove that a Riemann-integrable function which is nonpositive on a dense set $S$ has nonpositive integral. This is almost trivial; given any partition, we can choose points in $S$ to evaluate at so that Riemann sum is nonpositive. Taking a limit, the integral is nonpositive.

If the integral is positive, the set on which $f$ is not positive is not dense and its complement contains an interval $(a,b)$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=37648]here[/url] for a proof that a Riemann-integrable function is continuous at a dense set without using measure theory.
I thought I would use this to stay with elementary methods, but it was redundant.


geometry
trigonometry
ratio
analytic geometry
geometry proposed

I think this is a nice, but tough problem...

A line through the centroid $G$ of $\Delta ABC$ intersects the sides of the triangle at points $X,Y,Z$. Using directed line segments, prove that \[\frac{1}{GX}+\frac{1}{GY}+\frac{1}{GZ}= 0\]

[quote="gemath"]I have a similar problem for you!
$\triangle ABC$ with excenter $J_{a}$ a line through $J_{a}$ cut three segments (not ray) $BC,CA,AB$ at $A',B',C'$ prove that
$\frac{a}{J_{a}A'}=\frac{b}{J_{a}B'}+\frac{c}{J_{a}C'}$
here we don't use directed segments it is geometric length! :D[/quote]

Thanks gemath  :)

I'm afraid I can only come up with a trig solution...

Draw the points $B_{1}$ and $C_{1}$ so that $B_{1}C_{1}$ is parallel to $BC$ and touches the circle. Obviously, $\Delta ABC$ and $\Delta AB_{1}C_{1}$ are similar.

Now, multiplying with the right ratio $k$, we can transform the problem into: \[\frac{B_{1}C_{1}}{J_{a}A_{1}'}= \frac{AC_{1}}{J_{a}B'}+\frac{AB_{1}}{J_{a}C'}\] Now call $\alpha = \angle J_{a}A_{1}'C_{1}$

equivalent statement: \[\frac{B_{1}C_{1}.\sin{\alpha}}{r}= \frac{AC_{1}.\sin{C+\alpha}}{r}+\frac{AB_{1}.\sin{\alpha-B}}{r}\] Or \[B_{1}C_{1}.\sin{\alpha}= AC_{1}(\sin{\alpha}\cos{C}+\sin{C}\cos{\alpha})+AB_{1}(\sin{\alpha}\cos{B}-\sin{B}\cos{\alpha})\] Or \[B_{1}C_{1}= AC_{1}\cos{C}+AB_{1}\cos{B}\] Which is obviously true...

Do you have a nice solution like your previous?

ok thank Jan for your good solution !  :)  :)

[color=darkred][b][u]Extension.[/u][/b] Let $ ABC$ be a triangle and let $ P$ be a point with the barycentrical coordinates

$ (x,y,z)$ w.r.t. $ \Delta\ ABC$. Prove that for any line which intersects the sidelines $ BC$, $ CA$, $ AB$

at the points $ X$, $ Y$, $ Z$ respectively exists the relation $ \frac {x}{\overline {GX}} \plus{} \frac {y}{\overline {GY}} \plus{} \frac {z}{\overline {GZ}} \equal{} 0$.

[b]Particular cases.[/b] $ P\in\{\mathrm {G,H,O,I,I_a,I_b,I_c,N,\Gamma}\ldots\}\ .$[/color]


function
calculus
derivative
algebra
domain
real analysis
real analysis unsolved

What are the solutions of the differential equation $f'(x) = f^{-1}(x)$?

..if there are any such functions at all..   :maybe:

It may help to recall that if $y=f(x),$ and if $g: =f^{-1}$ so that $x=g(y),$ then we have (by implicit differentiation) $1=g^{\prime}(y)y^{\prime}$ which gives $f^{\prime}(x)=\frac{1}{g^{\prime}\left( f(x)\right)}$ if that helps at all...

[quote="Ottem"]..if there are any such functions at all.. [/quote]
There are, as long as we restrict to one side of zero. Consider functions of the form $f(x)=ax^{b}$, $f^{-1}(y)=\left(\frac{y}{a}\right)^\frac1b$. We have $f'(x)=abx^{b-1}$, so we need $b-1=\frac1b$ and $ab=a^{\frac{-1}{b}}$.
This gives $b=\frac{1+\sqrt{5}}{2}$ and $a=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}$ (the other root doesn't work due to sign problems).

This is a solution for $x>0$.
There is a similar solution for $x<0$: $f(x)=-\left(\frac{-1+\sqrt{5}}{2}\right)^{\frac{1+\sqrt{5}}{2}}(-x)^{\frac{1-\sqrt{5}}{2}}$

There is no solution valid on all of $\mathbb{R}$; this would require that $f'(f(x))=x$ for all $x$, and therefore $f'$ takes on all real values. Then $f$ is not monotone, and $f'$ is not well-defined.

Specifying the domain is essential to the problem, because of that inverse. I wouldn't expect a general solution to be feasible.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=inverse&t=37916]here[/url] for $\mathbb R^{+}$.


Consider an interior point $ P$ of a given $ \triangle ABC$ . Denote the projections $ X$ , $ Y$ , $ Z$

of point $ P$ on the sides of given triangle. Prove that $ \boxed {\ \sum PA\ \ge\ 2\cdot\sum \frac {a}{\sqrt{bc}}\cdot PX\ }\ .$

Let $ h_a$ be A-altiude,

\[ PA\geq h_a \minus{} PX \equal{} \frac {1}{a}a(h_a \minus{} PX) \equal{} \frac {1}{a}(2S_{ABC} \minus{} 2S_{PBC}) \equal{} \frac {1}{a}(2S_{PCA} \plus{} 2S_{PAB}) \equal{} \frac {1}{a}(bPY \plus{} cPZ)\]

So

\[ \sum PA\geq\sum\frac {1}{a}(bPY \plus{} cPZ) \equal{} \sum aPX(\frac {1}{b} \plus{} \frac {1}{c})\ge\sum\frac {2a}{\sqrt {bc}}PX\]


[i]Βάζω ένα ωραίο πρόβλημα γεωμετρίας κι ελπίζω να μην είναι γνωστό[/i]

Εστω $ ABCD$ εγγράψιμο.  Θεωρούμε μεταβλητά σημεία $ E,F$ στις πλευρές $ AB,CD$ αντίστοιχα, τέτοια ώστε $ \frac{AE}{EB}\equal{}\frac{CF}{FD}$. Εστω επίσης $ P$ σημείο πάνω στην $ EF$ τέτοιο ώστε $ \frac{PE}{PF}\equal{}\frac{AB}{CD}$. Να αποδείξετε ότι ο λόγος των εμβαδών των τριγώνων $ APD$ και $ BPC$ είναι ανεξάρτητος απ' την επιλογή των $ E,F$.

Θα πρότειναι να αποφύγετε την τριγωνομετρία!!!

Κανένας? :maybe:

Πάντως είναι ωραία άσκηση και θα πρότεινα να την κοιτάξετε πριν να βάλω τη λύση μου. Αλλιώς θα βάλω τη λύση μου σε 1-2 μέρες.

Για να δώσω ένα hint:
[hide]Βρείτε τον γεωμετρικό τόπο του $ P$[/hide]

Kι ένα σημαντικότερο hint:
[hide]Προκύπτει $ \frac{APD}{BPC}\equal{}\frac{AD}{BC}$[/hide]

Είναι όντως πολύ ωραία άσκηση .
Έστω ότι οι ευθείες $ AD,BC$ τέμνονται στο σημείο $ K$ . Τότε έχουμε ότι $ \angle{KAE}\equal{}\angle{KCF}$
και $ \frac{AE}{CF}\equal{}\frac{AB}{CD}\equal{}\frac{KA}{KC}$ και επομένως τα τρίγωνα $ KAE,KFC$ είναι όμοια .
επομένως $ \angle{AKE}\equal{}\angle{CKF}$ . Έστω ότι η διχοτόμος της γωνίας $ \angle{EKF}$ τέμνει την $ EF$ στο $ P'$
Τότε $ \frac{P'E}{P'F}\equal{}\frac{KE}{KF}\equal{}\frac{KA}{KC}\equal{}\frac{AB}{CD}\equal{}\frac{PE}{PF}$

Επομένως $ P\equiv P'$ και επομένως το $ P$ βρίσκεται στη διχοτόμο της γωνίας $ EKF$ και άρα βρίσκεται και στη διχοοτόμο της γωνίας $ DKC$  . Επομένως οι αποστάσεις του $ P$ από τις $ AD,BC$ είναι ίδιες και άρα ο λόγος των εμβαδών των τριγώνων που ψάχωουμε είναι ίδιος με το λόγο των $ \frac{AD}{BC}$ το οποίο είναι σταθερό και QED

Μπράβο ρε Sil. Και μένα περίπου αυτή ήταν η λύση μου. Το κλειδί είναι η προέκταση και μετά η παρατήρηση ότι $ KE, KF$ είναι ισογώνιες κι άρα η διχοτόμος τους είναι και η διχοτόμος της γωνίας $ AKB$.


I have the following two doubts:-
1.As we know mass can be converted into energy and energy can be converted into mass.My question is,in stars like our sun,why doesn't the energy created from mass again form mass?enough energy is present in the center of stars..
2.Every second,tonnes of hydrogen is converted to helium in the stars leading to the conversion of mass into energy.Then the stars must be losing some mass.This effect can be significant if we consider the lifespan of a star like the sun.Then why isn't such a thing observed?The sun ought to be less massive than it was some million years ago.Is it true?
Correct me if i am wrong.

Mass to energy conversion holds true only for certain atoms in atomic fusion. Atoms lighter than FE can produce energy when they fuse.

For example we know that in AMU's the atomic weight sum of 2 protons, 2 neutrons and 2 electrons is 4.0322980. On the other hand, the atomic weight in AMU's of the Helium is 4.00260. This difference is what is called the bound energy of the atom, so if you try to fusion all these particles into one you get some energy in the proccess (recall that $ E \equal{} mc^2$). The case reverses for FE.

The second effect that you mention is also true. That's exactly why stars "die".

Mass to energy conversion takes place always when a system produce energy; the mass decreases $ \;\; {E \over {c^2}\;\;}$.

In classical processes the fraction of mass converted in energy is so small that one can't afford to measure it.

In nuclear processes like fusion or fission it's likely to be detected: for instance, in the 235 Uranium fission, about 0,15% of mass is converted in energy

I'm sorry for my poor english.

What i wanted to say was that in atomic fusion, not all atoms release energy. Only the ones that are lighter than FE realease it, massive atoms NEED energy to fuse. So a star that's fusing elements lighter than FE will always produce energy.

[quote="Aashay THE ROKR"]But if we concentrate large amount of energy which is greater than the binding energy of nuclei,can't the nuclei be formed out of energy?Such extreme conditions may exist in the center of stars. :maybe:[/quote]

That's what actually happens. The kinetic energy of the atoms, i.e. the temperature it's so high in the core of stars that they are able to fuse and form a greather atom. The sun itself it's fusing Helium-3 in the core (2 protons, 1 neutron on the nuclei) to Helium-4 (2 protons, 2 neutrons).

OK..Now got it QBoy..thanks for the explanation.


superior algebra
superior algebra solved

Let $A,B\in {M_n(\mathbb{C})}$ such that $AB-BA=B$. Prove that $B$ nilpotent

It's obvious that $Tr(B^k)=0,\ \forall k\in\mathbb N$, and it's well-known that this implies the fact that $B$ is nilpotent.


[hide]By converting the [...] form into numbers, it becomes $[1,1,1]$ which becomes $2$.[/hide]

[hide]
As [a,b,c] means a+b/c, we can take the solutions from the numbers in the sets to find the final answer. For all three of the sets, we see that the first two numbers, a and b, have a sum of the last number,c. Therefore, we have a final set [a,b,c] as [1,1,1]. 1+1/1=2/1=2, the final answer.[/hide]

[hide]Notice that in each of the three numbers given, the sum of the first two numbers equal the third. Thus we can reduce that to be $[1, 1, 1]$ which is $2/1=2$. [/hide]