algebra
polynomial
inequalities
quadratics
inequalities unsolved

Hello,
is there perhaps a good way, to show that:
$ (z^2\minus{}1)^2\plus{}(z\minus{}\frac{29}{315})^2>\frac{1}{4}$ for all $ z \geq 0$

Thanks a lot.

[quote="Atan"]Hello,
is there perhaps a good way, to show that:
$ (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2 > \frac {1}{4}$ for all $ z \geq 0$

Thanks a lot.[/quote]
For all real $ z$ we obtain:
$ (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2 \minus{}\frac {1}{4}\equal{}\left(z^2\minus{}\frac{3}{5}\right)^2\plus{}\frac{1}{5}z^2\minus{}\frac{58}{315}z\plus{}\frac{3}{4}\minus{}\frac{9}{25}\plus{}\left(\frac{29}{315}\right)^2>0.$

First of all, thanks a lot. I understand, that for all $ z \geq 0$ it is $ (z^2\minus{}\frac{3}{5})^2\plus{}\frac{1}{5}*z^2\plus{}\frac{3}{4}\minus{}\frac{9}{25}\plus{}(\frac{29}{315})^2>0$ but what about $ \minus{}\frac{58}{315}$?

[quote="Atan"]First of all, thanks a lot. I understand, that for all $ z \geq 0$ it is $ (z^2 \minus{} \frac {3}{5})^2 \plus{} \frac {1}{5}*z^2 \plus{} \frac {3}{4} \minus{} \frac {9}{25} \plus{} (\frac {29}{315})^2 > 0$ but what about $ \minus{} \frac {58}{315}$?[/quote]
Because $ \left(z\minus{}\frac{29}{315}\right)^2\equal{}z^2\minus{}\frac{58}{315}z\plus{}\left(\frac{29}{315}\right)^2.$ :wink:

I assume, you missunderstood me a little bit. Perhaps my question was not specific enough.

You wrote a transformation into this form:
$ (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2 \minus{}\frac {1}{4}\equal{}\left(z^2\minus{}\frac{3}{5}\right)^2\plus{}\frac{1}{5}z^2\minus{}\frac{58}{315}z\plus{}\frac{3}{4}\minus{}\frac{9}{25}\plus{}\left(\frac{29}{315}\right)^2>0$
I can fully follow your ideas, but i don't see, why the right side should be bigger than 0 for all z>0. That was my question:
It is clear, that $ (z^2 \minus{} \frac {3}{5})^2 \plus{} \frac {1}{5}*z^2 \plus{} \frac {3}{4} \minus{} \frac {9}{25} \plus{} (\frac {29}{315})^2$ is bigger than 0, but how to cope with the missing term $ \minus{}\frac {29}{315}*z$? This term is obviously smaller than 0 for all z>0. So, how can you one state, that the whole equation is bigger than 0, even if one term is constantly negative?

That sounds like to show, that the discriminant of the quadratic equation is always negative, so that there a no zeros belonging to this equation in the real numbers. Is that right?

Okay, I think, it's all clear. Thanks a lot for your help.

[quote="Atan"]Hello,
is there perhaps a good way, to show that:
$ (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2 > \frac {1}{4}$ for all $ z \geq 0$

Thanks a lot.[/quote]
another solution:
$ (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2\equal{}(z\plus{}1)^2\times{(z\minus{}1)^2}\plus{}
(z\minus{}\frac{29}{315})^2\geq{(z\minus{}1)^2\plus{}(z\minus{}\frac{29}{315})^2}$
and also:
$ (z\minus{}a)^2\plus{}(z\minus{}b)^2\equal{}2(z\minus{}\frac{a\plus{}b}{2})^2\plus{}\frac{(a\minus{}b)^2}{2}\geq\frac{(a\minus{}b)^2}{2}$
therefore:$ (z\minus{}1)^2\plus{}(z\minus{}\frac{29}{315})^2\geq\frac{(1\minus{}\frac{29}{315})^2}{2}\equal{}0.412174....>\frac{1}{4}$

Oh, thats also a very nice idea, especially because it seems to be a little bit shorter or at least less complicated. Thanks a lot.

[quote="Atan"]Hello,
is there perhaps a good way, to show that:
$ (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2 > \frac {1}{4}$ for all $ z \geq 0$
Thanks a lot.[/quote]

We need to prove, that

$ f(z) \equal{} z^4\minus{}z^2\minus{}\frac{58}{315}z\plus{}\frac{301039}{396900} > 0$

Case1: $ z\leq0$.

Let us notice that in this case $ \minus{}\frac{58}{315}z>0$ and $ \frac{301039}{396900}>\frac{1}{4}$, so
\[ f(z) \equal{} z^4\minus{}z^2\minus{}\frac{58}{315}z\plus{}\frac{301039}{396900} > z^4\minus{}z^2\plus{}\frac{1}{4}\]

But for all $ z$ the following identity is true:
\[ z^4\minus{}z^2\plus{}\frac{1}{4} \equal{} (z^2\minus{}\frac{1}{2})^2\]

This proves that $ f(z)>0$ for $ z\leq0$.

Case2: $ z>0$. Let us notice that in this case $ \frac{58}{315}<\frac{1}{5}$ and $ \frac{301039}{396900}>\frac{3}{5}$, so

\[ f(z) \equal{} z^4\minus{}z^2\minus{}\frac{58}{315}z\plus{}\frac{301039}{396900} > z^4\minus{}z^2\minus{}\frac{1}{5}z\plus{}\frac{3}{5}\]

But for all $ z$ the following identity is true:

\[ {z^4\minus{}z^2\minus{}\frac{1}{5}z\plus{}\frac{3}{5} \equal{} {\frac {19}{55}}\,{z}^{4}\plus{}{\frac {1}{660}}\, {z}^{2}\left( 6\,z\minus{}11 \right) ^{
2}\plus{}\frac{1}{60}\, \left( 6\,{z}^{2}\plus{}z\minus{}6 \right) ^{2}}\]

This proves that $ f(z)>0$ for $ z>0$.

Remark. We can eliminate viewing 2 cases, but in this case coefficients are not very nice.

Proof 2: For all $ z$ polynomial $ f(z) \equal{} (z^2 \minus{} 1)^2 \plus{} (z \minus{} \frac {29}{315})^2 \minus{} \frac {1}{4}$ is equal to

\[ {{\frac {6153308243593}{23590937493}}\,{z}^{4}\plus{}{\frac {98}{2367264077318409}}\, \left( 383542110\,z\minus{}2621215277 \right) ^{2}{z}^{2}\plus{}{\frac {1}{2709351}}\, \left( 1028657\,{z}^{2}\plus{}109620\,z\minus{}903117 \right) ^{2}}\]

This clearly proves it is non-negative...

Sergey Markelov


a man has infinite amount of money. he distribute his total money in  n peoples,such that each person does not get more than 4n-1 rupees...the total money distributed is multiple of 4.in how many ways he can do that. sir pl answer my question.


inequalities
geometry
triangle inequality
area of a triangle
inequalities proposed

Let $x,y,z,a,b,c>0$ so that $x+y\leq \sqrt{a+b}$, $y+z\leq\sqrt{b+c}$, and $z+x\leq\sqrt{c+a}$, prove that $4xyz(x+y+z)\leq ab+bc+ca$.

I could only prove that:

\[ 4xyz(x+y+z)\leq 4(x+y+z)^4\leq 9(a+b+c)^2\leq 27(a^2+b^2+c^2) \]

and that:

\[ 9(a+b+c)^2\geq ab+bc+ca \]

Could anyone finish the prove? :(

[b]pvthuan[/b], could you post your solution?

After putting $p=x+y$, $q=y+z$, $r=z+x$, $s^2=a+b$, $t^2=b+c$, $u^2=c+a$, we have to prove, with the conditions that $p\leq s$, $q\leq t$, $r\leq u$, that
$-p^4-q^4-r^4+2p^2q^2+2q^2r^2+2r^2p^2\leq -s^4-t^4-u^4+2s^2t^2+2t^2u^2+2u^2s^2$.
Could anybody continue, please?

i don't think this ineq is true. Take p,q,r = 1, 3/4, 3/4, s,t,u = 1,2,3.

[quote="toumaf"]Well, there is a small typo in leepakhin's post : we have to prove that for all p,q,r and s,t,u verifying the triangle inequality, with $p\leq s$, $q \leq t$ and $r \leq u$ :

\[ -p^4-q^4-r^4+2p^2q^2+2q^2r^2+2r^2p^2\leq -s^4-t^4-u^4+2s^2t^2+2t^2u^2+2u^2s^2,  \]

that is the area of a triangle with sides $p$,$q$,$r$ is smaller than the one of a triangle with sides $s$,$t$,$u$.[/quote]
surprising...what a nice solution...

Sorry for that stupid mistake.  :oops: I have corrected it.


These inequalities are in fact two equivalent forms of one inequality, because $(2)$ can be obtained form $(1)$ by replacing $x\leftrightarrow z$.


When he was discovered dead, Mr. Thompson had a bullet wound in his calf and a knife wound in his back. The prime suspects are Mr. Barton, Mrs. Scott and the elevator operator.

The following statements were in your “murder mystery” packet.
1. If Mr. Barton had an intruder in his apartment, then Mr. Barton shot the intruder.
2. Either Mr. Thompson was a CPA who worked at an accounting firm, or he broke into Mr. Barton’s apartment at midnight.
3. Mrs. Scott and Mr. Thompson were both employed at the local high school, and they were good friends.
4. Mr. Barton’s gun was found in his apartment, and the bullet in Mr. Thompson’s calf matched those used by Mr. Barton’s gun.
5. If the person who shot Mr. Thompson also stabbed him, then Mr. Thompson’s body was found in the parking garage.
6. Mr. Thompson’s body was found at 1:20 a.m.
7. Mrs. Scott would not kill anyone with whom she works.
8. Either Mr. Thompson’s body was found in the park, or his body was found before 1:00 a.m.
9. A knife was found in the parking garage, or Mr. Barton’s gun was found in the elevator.
10. Mrs. Scott waited in the lobby of the apartment building from 12:15 a.m. to 12:45 a.m. She was either waiting for her husband to get off work, or she was the person who killed Mr. Thompson.
11. If Mrs. Scott was friends with Mr. Thompson, then her husband was jealous of the friendship.
12. If Mrs. Scott was waiting in the lobby for her husband to get off work, then her husband was the elevator operator.

This are the logical sentences:
A. Mr. Barton had an intruder in his apartment.2
B. Mr. Barton Shot the intruder.
C. Mr. Thompson was a CPA who worked at an accounting firm
D. Mr. Thompson broke into Mr. Barton's apartment at midnight.
E. Mr. Thompson was employed at the local high school.
F Mrs. Scott was employed at the local high school.
G. Mr. Thompson and Mrs. Scott were good friends.
H. Mr. Barton's gun was found in his apartment.
I. The bulled in Mr. Thompson's calf matches those used by Mr. Barton's gun.
J. The person who shot Mr. Thompson also stabbed him.
K. Mr. Thompson's body was found in the parking garage.
L. Mr. Thompson's body was found in the park.
M. Mr. Thompson's body was found before 1:00 a.m
N. A knife was found in the parking garage.
O. Mr. Barton's gun was found in the elevator.
P. Mrs. Scott waited in the lobby of the apartment building from 12:15 a.m to 12:45 a.m.
Q. Mrs. Scott was waiting for her husband to get off work.
R. Mrs. Scott was the person who killed Mr. Thompson.
S. Mr. Thompson was jealous of the friendship.
T. Mrs. Scott's husband was the elevator operator.

I constructed this...
1. If Mr. Barton had an intruder in his apartment, then Mr. Barton shot the intruder.
A→B
2 .Either Mr. Thompson was a CPA who worked at an accounting firm, or he broke into Mr. Barton’s apartment at midnight.
C v D
3. Mrs. Scott and Mr. Thompson were both employed at the local high school, and they were good friends.
(E ^ F)^G
4. Mr. Barton’s gun was found in his apartment, and the bullet in Mr. Thompson’s calf matched those used by Mr. Barton’s gun.
H^I
5. If the person who shot Mr. Thompson also stabbed him, then Mr. Thompson’s body was found in the parking garage.
J→K
6. Mr. Thompson’s body was found at 1:20 a.m.
~M
7. Mrs. Scott would not kill anyone with whom she works.
(E^F)→~R
8. Either Mr. Thompson’s body was found in the park, or his body was found before 1:00 a.m.
L^M
9. A knife was found in the parking garage, or Mr. Barton’s gun was found in the elevator.
N V O
10. Mrs. Scott waited in the lobby of the apartment building from 12:15 a.m. to 12:45 a.m. She was either waiting for her husband to get off work, or she was the person who killed Mr. Thompson.
P^(Q V R)
11. If Mrs. Scott was friends with Mr. Thompson, then her husband was jealous of the friendship.
G→S
12. If Mrs. Scott was waiting in the lobby for her husband to get off work, then her husband was the elevator operator.
Q→T

Help me to find the murder!!!!

wow...i would never be able to do this......unless i had 2 or so hours

this is rather simple, i think, though i like these murder puzzles

2,3-> mr. thompson does not work at an accounting firm, so he broke into mr.barton's apartment at midnight
1-> mr barton shot mr thompson

2,7-> mrs. scott did not kill mr thompson
10-> mrs. scott was waiting for her husband to get off work
12-> the elevator operator is mrs. scott's husband

6,8-> mr thompson's body was found in the park
5-> mr barton did not stab mr thompson

3,11-> the elevator operator was jealous of mr thompson

4,9-> a knife was found in the parking garage

now, we face a linguistiv sublety: does "to kill" include "to stab someone who is already dead"? if yes, then the elevator operator stabbed mr thompson. if not, we have a problem.

[quote="abacadaea"]oh it probably just looks confusing[/quote

It is confusing.


A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

hello
[hide="solution"]
let r be the radius of the cylinder .
then $ \frac {12 \minus{} 2r}{r}$=$ \frac {12}{5}$
r=$ \frac {30}{11}$

[/hide]
Thank u.


MATHCOUNTS
induction
algorithm
number theory
relatively prime
Euclidean algorithm

okayy. so there is this one problem im having trouble on
its for algebra.

[color=green]problem:
during a special football game, a team scores 7 points for a touchdown & 3 points for a feild goal. There is no limit on the number of points a team can score. Using any combinations of touchdowns&&feild goals what is the most number of points a team [b]CANNOT [/b]score. Explain why you think your answer is correct solution to the problem. [/color]

 :huh:

so yeppss:]]

Using the Chicken McNugget formula, we find that the largest amount unable to be obtained with only 7 and 3 points are $ (3)(7)\minus{}3\minus{}7$ or $ 11$ points.

Or, if you're not allowed to use the Chicken McNuggets formula, you can make a list and cross out the ones that can be made.
11 can't be made.
12 can be made with 4 field goals.
13 can be made with 1 touchdown and 2 field goals.
14 can be made with 2 touchdowns.

Now, every score greater than 14 can be made if we add one or more field goals to 12, 13, or 14 points. For example, you can get 28 points by adding 5 field goals to 13 points.
So the answer is 11, like ZhangPeijin said.

Wait, whats the Chicken McNugget Formula?
Will it help in Mathcounts b/c I am taking that in two days.

http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem

This is quite useful.

Haha... i really like the name of that formula...

Just sec... how do you know that there won't be a score you can't score way higher than 11 like 3298472938472938429834792384 points?

above number if probably a possible point...

Read the page  :wink:
http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem

If two numbers $ a$ and $ b$ are [url=http://en.wikipedia.org/wiki/Coprime]Relatively Prime[/url], then the smallest number that cannot be made in the form $ an\plus{}bm$ where $ n$ and $ m$ are numbers, is $ ab\minus{}a\minus{}b$.

more correctly, $ n$ and $ m$ are whole numbers

[quote="fcms_brit"][color=darkblue][b]ehhh hi

well cood u guys explain tat formula cuz im still kinda confused o.0

thankss:]][/b][/color][/quote]

The general problem is called Frobenius' coin problem. The question is what is the largest amount of money that you can't make change for if you have only coins of denominations $ \left\{a_1,\, a_2,\, a_3,\,\ldots , a_n\right\}$?

Chicken-McNuggets is a example of Frobenius' coin problem where instead of coin denominations you have the number of McNuggets in a box. It makes a lot more sense to most people because hardly any one lives in a country where all the coins are relatively prime (have no common divisor other than one.)

A mathematican named Sylvester proved that when you only had two denominations of coins (two sizes of Chicken McNugget orders), $ \left\{a_1,\,a_2\right\}$, the largest number of Chicken McNuggets that you couldn't get from the order sizes available is: $ k\equal{} \left(a_1 \minus{} 1\right)\cdot\left(a_2 \minus{}1\right) \minus{} 1$. You can prove Sylvester's theorem by induction but I won't because it's complicated.

There exists no similar formula for $ n\geq 3$.

[quote="Dr. No"]A mathematican named Sylvester proved that when you only had two denominations of coins (two sizes of Chicken McNugget orders), $ \left\{a_1,\,a_2\right\}$, the largest number of Chicken McNuggets that you couldn't get from the order sizes available is: $ k \equal{} \left(a_1 \minus{} 1\right)\cdot\left(a_2 \minus{} 1\right) \minus{} 1$. You can prove Sylvester's theorem by induction but I won't because it's complicated.[/quote]

this seems like a bad idea.  what are you inducting on?

anyways this is a sketch of the proof of a slightly stronger result.

1)  prove that if $ k\geq (a_1 \minus{} 1)(a_2 \minus{} 1)$ then you can order $ k$ chicken mcnuggets.
2)  show that if $ m,n$ are nonnegative integers with $ m \plus{} n \equal{} (a \minus{} 1)(b \minus{} 1) \minus{} 1$, then it is possible to order either $ m$ chicken mcnuggets or $ n$ chicken mcnuggets, but not both.
3)  you can order 0 chicken mcnuggets
4)  you can't order $ (a_1 \minus{} 1)(a_2 \minus{} 1)\minus{}1$ chicken mcnuggets
5)  in fact, there are exactly $ (a_1 \minus{} 1)(a_2 \minus{} 1)/2$ order sizes of chicken mcnuggets which are invalid.

edit: to fix a typo in 4)

[quote="blahblahblah"][quote="Dr. No"]A mathematican named Sylvester proved that when you only had two denominations of coins (two sizes of Chicken McNugget orders), $ \left\{a_1,\,a_2\right\}$, the largest number of Chicken McNuggets that you couldn't get from the order sizes available is: $ k \equal{} \left(a_1 \minus{} 1\right)\cdot\left(a_2 \minus{} 1\right) \minus{} 1$. You can prove Sylvester's theorem by induction but I won't because it's complicated.[/quote]

this seems like a bad idea.  what are you inducting on?

...[/quote]

It isn't necessarily a bad idea. It can be done. (I found the proof while searching for Frobenius' coin problem.) It is just deplorably messy, and wholly inappropriate for Middle School math. It involves solving a system of Diophantine equations in $ \mathbb{L}^{2}$, and showing that if you can get $ k \plus{} 1$ then you can also get $ k \plus{} 2$.

[quote="Dr. No"]showing that if you can get $ k \plus{} 1$ then you can also get $ k \plus{} 2$.[/quote]

for the most part the combinations you use to get $ k\plus{}1$ and $ k\plus{}2$ can't be written explicitly (you get them from the euclidean algorithm), and they aren't really related.

Besides, that part (step 1) in the 'program' i listed above) is one of the easier steps:

if $ a_1$ and $ a_2$ are relatively prime, there exist integers $ m,n$ such that $ ma_1\plus{}na_2\equal{}k$, and we may choose $ m$ so that $ 0\leq m\leq a_2\minus{}1$.

Then

$ (a_1\minus{}1)(a_2\minus{}1) \leq ma_1\plus{}na_2 \leq a_1(a_2\minus{}1)\plus{}nb_2$, or $ 1\leq b_2(n\minus{}1)$.  This means that $ n>1$.  In particular, as $ m$ and $ n$ are nonnegative, this shows existence.


Let $ABCD$ be a convex quadrilateral. Is it possible to find a point $P$ such that the segments drawn between $P$ and the midpoints of the sides of $ABCD$ divide the quadrilateral into four sections of equal area? If $P$ exists, is it unique?

First of all, P lies on the newton's line of ABCD. Now it is easy to see P exists.
P is not unique if and only if ABCD is a parallelogram.

The Newton line of a quadrilateral is just the line connecting the midpoints of the diagonals, by the way (I think a lot of people don't know the name).


Can India please get a seperate forum with "Problem Solving" just like other countries have such as China.

It can post Indian Olympiad questions, IIT math problems, etc. on it.

Yes sure ..
If you are intereste in imo.you are most welcome.i am here in your section to help each other about all the things related to imo.

There is already an [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=513][b]India Problem Discussion[/b] Forum[/url]. That's because the latest poster of this thread has revived a 6-year old thread: not necessary.


[quote="zaya_yc"]$ a,b,c > 0$  prove:
$ (1 \plus{} abc)(\frac {1}{a(b \plus{} 1)} \plus{} \frac {1}{b(c \plus{} 1)} \plus{} \frac {1}{c(a \plus{} 1)})\geq 3$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161059


geometry
similar triangles
Pythagorean Theorem

Let $ ABC$ me a right triangle in $ A$. Of a point in the hypothenuse, we have $ \overline{DE}\perp\overline{AB}$ and $ \overline{DF}\perp\overline{AC}$ . Show that $ DB\cdot DC\equal{}EA\cdot EB\plus{}FA\cdot FC$

[quote="triplebig"]Let $ ABC$ me a right triangle in $ A$. Of a point in the hypothenuse, we have $ \overline{DE}\perp\overline{AB}$ and $ \overline{DF}\perp\overline{AC}$ . Show that $ DB\cdot DC \equal{} EA\cdot EB \plus{} FA\cdot FC$[/quote]

I assume you mean right angle at $ A$, $ D$ on $ BC$, $ E$ on $ AB$, $ F$ on $ AC$.

[hide]
First note that $ EDFA$ is a rectangle.

$ ED \parallel{} AC$, $ BA \parallel{} DF$; hence $ \triangle BED \sim \triangle DFC\Rightarrow DB\cdot DC \equal{} BD^2\cdot FC/ED$; but $ BD^2 \equal{} BE^2 \plus{} ED^2 \equal{} BE^2 \plus{} AF^2$, and $ ED \equal{} EF$; also $ BE\cdot FC/AF \equal{} EA$ by the similar triangles mentioned, and so the result follows.
[/hide]

I really, really, don't think this belongs in preolympiad, personally, especially since this can be solved with *solely* pythagorean theorem - I just added the similar triangles because they're fun.


inequalities
logarithms
calculus
integration
function
inequalities unsolved

plz solve this for me

Prove that for every integer n>=3, the following inequality is true:

n^(n+1) > (n+1)^n.

plz reply .................. :)

[hide="A hint"]See at $ f(x)\equal{}\frac{\ln x}{x}$ over $ [e,\infty)$.[/hide]

still not able to get

plz post ur soln

i think more generalised one is also true

(1 + 1/n)^n < 3

for all integral n

but i want to see ur soln to my problem

i.e

n^(n+1)>(n+1)^n

plz reply

hi, Jaswinder

(1 + 1/n)^n  is increasing in $ n$

and (1 + 1/n)^n -->  $ e$   as  n--> $ \infty$

this requires the knowledge of some calculus,
so i think you don't understand if you are a high school student

n^(n+1) > (n+1)^n  for $ n\ge 3$

$ \iff (n \plus{} 1)\log(n) > (n)\log(n \plus{} 1)$
$ \iff \frac {\log(n)}{n} > \frac {\log(n \plus{} 1)}{n \plus{} 1}$

the function $ f(x) \equal{} \frac {\log(x)}{x}$ is decreasing for $ x\in (e, \infty)$


I was reading through the AoPS volume one book (for the 4th time) and ran upon polar coordinates.   There was a question that I never asked about one of the examples in which the second item in the relation is not a degree.  It is a fraction with pi in it.  I also noticed that there wasnt anything written in the section explaining how to graph it (unless I missed it 100 times...)

The problem was as follows:

[i]EXERCISE 16-1[/i] ...and the polar points (2, 3$\pi$/4)...:?

How do you go about graphing that?

Transfer the radians to degrees! $2\pi=360^{\circ}$ My biggest complaint about AOPS V. 1 (don't get me wrong, its a great book) is that it I don't remember it giving an explanation of radians.

It had a brief explanation in Chapter 10.  But, yeah, it should have had a reminder in Chapter 16.

Oh yeah... Thanks a lot.

I thought it had something to do with radians... But wasnt sure.  Thanks a lot!

 :D


I'm trying to understand a little something about what I'm seeing about graphs.

Take for instance the game of Chess.  There's 8 rows of A-H and 1-8.  At the cross sections there are 8 possible cross sections, A1, B2, C3 and so forth.

How would I express something like this mathmatically when it comes to say a chess matchup.  There's 2 teams with 8 players on it.  Each player can play each other once, but only one and not each other.

This one I can do on a sheet of paper but what happens when it's 3 teams or even 6 teams?

I don't know if I'm making any since here or not but I'm just trying to figure out the mathmatical expression on to to have it so players wouldn't play themselves or anyone on their team.

Does no one know how to mathmatically do this?  There's got to be a formula out there or an array that expresses is.

Does no one know how to mathmatically do this?  There's got to be a formula out there or an array that expresses it.

I think what you are trying to figure out is how to set up a tournament bracket. That has nothing to do with chess notation (A1, B2, C3)

When it's just two teams, it should be easy to figure out, but 3 or 6 teams--  :o No idea. Maybe someone else knows, or there's a website/software that does it for you... that would be the easiest way

No it isn't a tournament bracket thing as you say.  I'm interested in finding out a quotiant that would basically express (A,B,C,D,E,F,G,H)*(1,2,3,4,5,6,7,8) or something along those lines and give each of the answers.  I see how it could be seen as a tournament but I'm only using chess as a metaphor here.  I'm looking for a forumula.

I guess I'm out of luck since most of you are convienced this is something for a sports related thing.  I was trying to find a quotiant or something mathmatical because I'm trying to do something that's about 1,000 x 1000 in deminsion and wanted to see where the common point (like center 500,500) exist for a design, not just the cross section \


Euler
limit
modular arithmetic
function
Gauss
number theory

Go Oiler!

I think your grandeur!  :lol:

Happy birthday...
i forgot ur cake though...  :mad:

Never give him a triangular cake unless you cut it along the Euler line :D

$e= \lim_{x \to \infty}\left(1+\frac{1}{n}\right)^{n}$

Happy 300th birthday Leonhard Euler!

MY GOD, HOW DID I NOT SEE THIS TOPIC!!!!!

Happy belated Birthday....

$a^{\phi{p}}\equiv a \pmod{p}$

Where p is a prime.

[quote="1=2"]

$a^{\phi{p}}\equiv a \pmod{p}$

Where p is a prime.[/quote]
$a^{\varphi (n)}\equiv 1\pmod{n}$, where $\gcd (a,n)=1$. :D

[quote="1=2"]It's the PHI function...[/quote]
p doesnt have to be prime but a and p have to be coprime which is what nttuan was saying
and no duh its the phi function you just forgot parenthesis
i suggest you check with mathworld or wikipedia when someone corrects your citation of a theorem...
(if p was prime, phi p=p-1, and the result is known as fermat's little theorem, not euler's)

Happy Birthday, O great mathematician. Leonhard 'Oiler'.

thanks everybody, happy bday for you too!

ohh, in fact, I'm not Euler, I'm Gauss

I'm no Oil, I'm Gas    :lol:


trigonometry
trig identities
Law of Cosines
geometry unsolved
geometry

(using  compas an ruler) Construct a triangle $ ABC$, given $ AB\plus{}AC$, $ AB\plus{}BC$ and $ \angle CAB$

Let $ AB\plus{}AC\equal{}x$, $ AB\plus{}BC\equal{}y$ and $ cos BAC\equal{}z$
$ \implies AC\equal{}x\minus{}AB$, and $ BC\equal{}y\minus{}AB$
By the law of cosines:
$ BC^2\equal{}AC^2\plus{}AB^2\minus{}2.AB.AC.cosBAC$
So with an algebraical work:
$ AB^2(2z\plus{}1)\minus{}AB.2(xz\plus{}x\minus{}y)\plus{}x^2\minus{}y^2\equal{}0$
Then we have the length of $ AB$, and the other part of the problem is easy.

Can it be done with analyse of some picture (classic construction work)?


I think you need cflculate \[ S\equal{}\sum_{k\equal{}1}^{n}\frac{k^{2}}{2^{k}}.\]
Consider $ S_{2}(x)\equal{}\sum_{k\equal{}1}^{n}k(k\minus{}1)x^{k}$ and $ S_{1}(x)\equal{}\sum_{k\equal{}1}^{n}kx^{k}$. And you get $ S\equal{}S_{2}(\frac{1}{2})\plus{}S_{1}(\frac{1}{2})$.

But you doesn't ans my question.......... :maybe:


conics
parabola
ratio
geometry
geometric transformation
dilation
rotation

Let $ \mathcal P$ be a parabola, and let $ V_1$ and $ F_1$ be its vertex and focus, respectively. Let $ A$ and $ B$ be points on $ \mathcal P$ so that $ \angle AV_1 B \equal{} 90^\circ$. Let $ \mathcal Q$ be the locus of the midpoint of $ AB$. It turns out that $ \mathcal Q$ is also a parabola, and let $ V_2$ and $ F_2$ denote its vertex and focus, respectively. Determine the ratio $ F_1F_2/V_1V_2$.

[hide="hint"]
Try the trivial case of $ y\equal{}x^2$ as P.

it was the best i could come up with  :rotfl:
[/hide]

Rather than "try the trivial case," you could have said "without loss of generality." Do you see why this is a representative example? :)
[hide="Hint to above"]You can turn any parabola into another parabola by at most a translation, a dilation, and a rotation about the vertex[/hide]
If you don't see how, you can still make a legitimate solution by using the only slightly "uglier" and more intuitively representative form $ y \equal{} kx^2$

[hide="Solution"]Let $ A(a,a^2)$ and $ B(b,b^2)$ be on $ y \equal{} x^2$. Rotate $ (b,b^2)$ 90 degrees clockwise to get $ (b^2, \minus{} b)$. This must be collinear with $ (0,0)$ and $ (a,a^2)$, thus $ \frac {a^2}{a} \equal{} \frac { \minus{} b}{b^2}$. Multiply by $ ab^2 \neq 0$ to get $ a^2 b^2 \equal{} \minus{} ab$.

Thus, the midpoint of $ (a,a^2)$ and $ (b,b^2)$ is $ \left( \frac {a \plus{} b}{2} , \; \frac {a^2 \plus{} b^2}{2} \right) \equal{} \left( \frac {a \plus{} b}{2} , \; 2 \left( \frac {a \plus{} b}{2} \right)^2 \plus{} 1 \right)$

To find the locus of points of this form, we treat $ t \equal{} \frac {a \plus{} b}{2}$ as a parameter, so our points are of the form $ (t,2t^2 \plus{} 1)$, and the curve is $ y \equal{} 2x^2 \plus{} 1$

Finally, we can just calculate $ V_1 V_2 \equal{} 1 \minus{} 0 \equal{} 1$ and $ F_1 F_2 \equal{} (1 \plus{} 1/8) \minus{} 1/4 \equal{} 7/8$, so the answer is $ \frac {7}{8}$[/hide]


Let $ a,\ b,\ c$ be positive real numbes such that $ (a^2 \plus{} 1)(b^3 \plus{} 2)(c^4 \plus{} 3) \equal{} 24$.
Prove that $ abc\leq 1$.

Usin AM-GM:
\[ a^2 \plus{} 1\ge2a \\
b^3 \plus{} 2 \equal{} b^3 \plus{} 1 \plus{} 1\ge 3\sqrt [3]{b^3\cdot 1 \cdot 1} \equal{} 3b \\
c^4 \plus{} 3 \equal{} c^4 \plus{} 1 \plus{} 1 \plus{} 1\ge\sqrt [4]{c^4\cdot 1 \cdot 1\cdot 1} \equal{} 4c
\]
so:
\[ 24 \equal{} (a^2 \plus{} 1)(b^3 \plus{} 2)(c^4 \plus{} 3) \ge 2a\cdot 3b\cdot 4c \equal{} 24 abc\Rightarrow abc\le 1
\]

[quote="kunny"]That's right. :)

Generalization:

[color=blue]Let $ a_i > 0\ (i \equal{} 1,\ 2,\ \cdots \, n)$ for a positive integer $ n$ such that $ (a_1^2 \plus{} 1)(a_2^3 \plus{} 2)\cdots (a_n^{n \plus{} 1} \plus{} n) \equal{} (n \plus{} 1)!$.

Prove that $ a_1a_2\cdots a_n\leq 1$.[/color][/quote]
 it's  easy and  killed  by AM-GM  ineq
 $ LHS  \geq (n\plus{}1)! a_1a_2..a_n$  hence  we done completely.


What is the size of the angle opposite the side of length 7 in a 3-7-8 triangle?

Use the law of cosines since you already have the three side lengths.

$ 49 \equal{} 9 \plus{} 64 \minus{} 2(3)(8)\cos{\theta}$
$ 24 \equal{}  48\cos\theta$
$ \cos\theta \equal{}  \frac {1}{2}$
$ \theta \equal{} 60$

[quote="brightzhu"]$ 49 \equal{} 9 \plus{} 64 \minus{} 2(3)(8)\cos{\theta}$
$ 24 \equal{} \minus{} 48\cos\theta$
$ \cos\theta \equal{} \minus{} \frac {1}{2}$
$ \theta \equal{} 120$[/quote]

Doesn't sound right... quick mental check says that the angle opposite the 8 should have the largest measure, so $ \theta < 90$. You slipped in an unnecessary minus sign; the last step should read $ \theta \equal{} 60$.

Yes. (I was initially surprised that the answer is 60 degrees, which is why I posted the question.)

yeah, I messed up :oops_sign:
It's fixed now.


[quote="ch1n353ch3s54a1l"][quote="mr. math"]Mrs. Smale's class filled out a survey. Here are the results for her 30 students.

14 like hot dogs
18 like cheeseburgers
16 like tacos
8 like both hot dogs and cheeseburgers
7 like tacos and cheeseburgers
6 like tacos and hot dogs
1 likes none of these

How many stundents like all three?

If there is a good method or a formula to do this please give it to me. :)[/quote]

[hide]Since for each person that likes two things is counted twice in the people that like individual things, just subtract those out of the sum of the individual things. $14+18+16-8-7-6+1=28$ so $30-28-2$ people[/hide][/quote]

or you could draw a venn diagram

[quote="mr. math"]Mrs. Smale's class filled out a survey. Here are the results for her 30 students.

14 like hot dogs
18 like cheeseburgers
16 like tacos
8 like both hot dogs and cheeseburgers
7 like tacos and cheeseburgers
6 like tacos and hot dogs
1 likes none of these

How many stundents like all three?

If there is a good method or a formula to do this please give it to me. :)[/quote]

[hide]29=14+18+16-8-7-6+x

x= the number of people who like all three

x=2[/hide]

Use either Principle of Inclusion-Exclusion, like ch1n353ch3s54a1l, or draw a Venn-Diagram suggested by SplashD.


rotation
geometry
geometric transformation
LaTeX
I always get this wrong
Hard problem
this is so hard

How many ways can the four unit squares of a $ 2\times2$ board be colored with any number of 5 colors, if two colorings are considered the same if they can be rotated to each other?
[asy]draw(unitsquare);
draw((.5,0)--(.5,1));
draw((0,.5)--(1,.5));[/asy]

We split this into 5 cases: 1 color, 2 colors, 3 colors, 4 colors, 5 colors

1 color: obviously 5 ways
2 colors: first, we have 5C2=10 ways to choose the two colors.  Then we have $ 2*2*2*2 \equal{} 16$ ways for each of the four corners, divide by $ 4$ for rotations to get 4
3 colors: first we have 5C3=10 ways to choose 3 colors.  then we have $ 3*3*3*3\equal{}81$ ways for each of the corners, divide by 4 to get 81/4

WAIT!

what am i doing wrong?  you can't have 81/4 ways.

some of those cases only have 1 color, other's only have 2. same mistake for the 2 color case.

unfortunately I don't have time to post my own solution right now

The answer is supposed to be 165
This is what I did, but I think I forgot some arrangements...:
[hide="when all four squares are 1 color"] There are 5 choices...intuitively [/hide]
[hide="when there are two colors present"]
There are two possibilities here: a [hide="checkerboard style"]
A B
B A    there are (5*4)/2 color combos here[/hide] and a [hide="column style"]
A B
A B    there are also (5*4)/2 combos here[/hide]. [/hide]
[hide="when there are three colors present"]
There are three possibilities for arrangements here: [hide="checkerboard"]
A C    <- arrangement where a, b, and c are different colors. there are (5*4*3)/6 color combos here.
B A
[/hide], [hide="column1"]
A B    <- Arrangement where a, b, and c are differnt colors. there are also (5*4*3)/6 color combos here.
A C
[/hide] and [hide="column2"]
A C    <- arrangement where a, b, and c are different colors. there are (5*4*3)/6 color combos here.
A B
[/hide]. [/hide]
[hide="when there are 4 colors present"]
There are 6 possibilities: [hide="arrangement1"]
A B        (5*4*3*2)/24 color combos
C D
[/hide], [hide="arrangement2"]
A C        (5*4*3*2)/24 color combos
B D
[/hide], [hide="arrangement3"]
A B        (5*4*3*2)/24 color combos
D C
[/hide], [hide="arrangement4"]
A D       same # of combos as above
B C
[/hide], [hide="arrangement5"]
A C       same # as above
D B
[/hide], and [hide="arrangement6"]
A D       same # as above
C B
[/hide]
[/hide]
[hide="the sum of all that..."]
is
5+(5*4)/2+(5*4)/2+(5*4*3)/6+(5*4*3)/6+(5*4*3)/6+(5*4*3*2)/24+(5*4*3*2)/24+(5*4*3*2)/24+(5*4*3*2)/24+(5*4*3*2)/24+(5*4*3*2)/24=5+10+10+10+10+10+5+5+5+5+5+5=85
[/hide]

We proceed by casework on types of permutations with different numbers of rotations.

Permutations with only one rotation: all four colors must the the same, so $ 5$ permutations.
Permutations with two rotations: this happens if there are two sets of colors that are diagonally opposite, so $ \binom52\cdot2 \equal{} 20$ permutations.
Permutations with four rotations: all other cases, so there are $ 5^4 \minus{} 5 \minus{} 20 \equal{} 600$ permutations.

However, for the first case, we have to divide by $ 1$, for the second, $ 2$, and for the third, $ 4$ to account for rotations, giving an answer of $ \frac51 \plus{} \frac {20}2 \plus{} \frac {600}4 \equal{} 5 \plus{} 10 \plus{} 150 \equal{} \boxed{165}$.

[b]Comment:[/b] For smaller numbers of colors, something like AugmentedSixth's solution works just fine, but this method is much easier for larger amounts.

thanks math154. Do you know where I missed some permutations?
hmm...I should figure out how to do LaTeX

augmentedsixth you forgot about the:

A A
A B

A B
B B

How is the motivation of the Alcumus solution found? I didn't think about it that way at all...

[quote=reaganchoi]Plis dont revive :([/quote]

This is the [b]official[/b] thread in the Alcumus question. It is okay to revive for a REASON. It is NOT ok to revive NEEDLESSLY.

I think they should be more specific that you can use the same color more than once. I got the problem wrong by thinking that I had to use distinct colors.

[quote=williamzhouawsome]I'm just wondering, but couldn't it be possible that each square has a different color?[/quote]

Yes, that is possible. I think it's counted in the 600/4 case.

[quote=nan101]12 year bummppp lol[/quote]

it was only a couple of months bump actually

@above, is that burnside's lemma? What do the variables mean and how do you apply it?

[hide=burnside go brrnside]
rotation 0 = 625
rotation 90 = 5
rotation 180 = 25
rotation 270 = 5
total is 660/4 = 165
BURNSIDE GO BRRNSIDE
[/hide]

[quote=ilovepizza2020]Burnside's lemma looks crazy complicated but it's actually quite simple...[/quote]

I still haven't learned it in these past 6 months lol

Me neither, too lazy to read brilliant's articles.

[quote=ilovepizza2020][quote=Owlly81]Could post #1 and 4 plez hide their solutions[/quote]

Don't spam the forums and maybe read the stickied threads before posting

There was no hide answers rule back then.[/quote]

yeah, and it was 12 years ago, so neither person has been active in 5+ years.

[quote=centslordm]Burnside Lemma was useful for this! :)[/quote]

stop being cringe and post a solution instead of remarking

follow my example

rotation 0 = 625
rotation 90 = 5
rotation 180 = 25
rotation 270 = 5
total is 660/4 = 165
BURNSIDE GO BRRNSIDE $\color{white} source: tenebrine$

lol I got this problem 3 times in a week, but I forgot the answers  :rotfl: Got it right again tho

it was simple to just count and then correct with division

[quote=DramaticFlossKid]I'm confused, how do we know that in all the other 600 cases, we will always quadruple count? Do we test all of them?[/quote]

There are 4 rotations for the 600 cases so we quadruple count them(so you have to divide by 4).


geometry
inradius
perimeter
circumcircle
area of a triangle
Heron\u0027s formula

Given a triangle $ ABC$ with side length $ a\equal{}3,\ b\equal{}6,\ c\equal{}7$. Find the area of the triangle, exradius $ R_1$ and inradius $ R_2$.

[hide="Area of Triangle"]
By heron's formula we get $ \sqrt{8\cdot5\cdot2\cdot1}\equal{}4\sqrt{5}$
[/hide]

[hide="Inradius"]
The area of triangle is $ sr$ were s is the semi perimeter and r is the inradius. From the previous part, $ r\equal{}\frac{4\sqrt{5}}{8}\equal{}\frac{1}{2}$
[/hide]

don't you have to find the exradius also? I forgot the formula to that.

SO the formula is:
[ABC]=abc/4R
so we know that the are is 4sqrt5
so 4sqrt5=126/4R
R=126/16sqrt5.
R=53/8sqrt5
R=53sqrt5/40

That's the formula for the circumradius.

$ p\equal{}\frac{3\plus{}6\plus{}7}{2}\equal{}8$, Heron gives $ \triangle{ABC}\equal{}\sqrt{8\cdot(8\minus{}3)\cdot(8\minus{}6)\cdot(8\minus{}7)}\equal{}\boxed{4\sqrt{5}}$

$ 4SR_1\equal{}abc\Longrightarrow R_1\equal{}\boxed{\frac{63}{8\sqrt{5}}}$

$ \frac{a\plus{}b\plus{}c}{2}R_2\equal{}\triangle{ABC}\Longrightarrow R_2\equal{}\boxed{\frac{\sqrt{5}}{2}}$

Wait. That doesn't make sense for the exradius. There should be 3 answers not 1.

Because the formula I know for the exradius is $ \frac{[ABC]}{s\minus{}a}$ where $ a$ is a corresponding side.

Also, I made a typo up there. It should be $ \frac{\sqrt{5}}{2}$ not $ \frac{1}{2}$

Sorry, I meant it as the radius of circumcircle. :blush:
How dou you call the radius of circumcircle in English?


Compute $\cot{(\alpha+\beta)}$ in the following figure:


prove that a divides bc only and only if a/gcd(a,b) divides c

We have to prove that $ a|bc$ iff $ \frac{a}{gcd(a,b)}|c$. Let $ gcd(a,b)\equal{}d$ and let $ a_{1}d\equal{}a$ and $ b_{1}d\equal{}b$.Very well-know fact is that $ gcd(a_{1},b_{1})\equal{}1(*)$. We have $ a_{1}d|b_{1}dc \Leftrightarrow a_{1}|b_{1}c$ and using $ (*)$ we conclude that $ a_{1}|c$


[hide]$ 2a^2\plus{}bc\equal{}2(b\plus{}c)^2\plus{}bc\equal{}2b^2\plus{}5bc\plus{}2c^2\equal{}(2b\plus{}c)(b\plus{}2c)\equal{}(b\minus{}a)(c\minus{}a)$

Similarly for the remaining denominators, so the expression becomes

$ {a^2\over (a\minus{}b)(a\minus{}c)}\minus{}{b^2\over(a\minus{}b)(b\minus{}c)}\plus{}{c^2\over(a\minus{}c)(b\minus{}c)}$

and then its really easy to show that the expression is equal to $ 1$ (provided $ a\neq b\neq c\neq a$, but otherwise the initial expression isn't defined in the first place).[/hide]


[quote="leonariso"]Problem 4
Prove that number 6 divides the product of every three consecutive integer numbers and then show that 240 divides the product
$(5\nu+1)(5\nu+3)(5\nu+5)$, where $\nu$ is any odd number.[/quote]
First part: obviously there are at least one even number and at lest one multiple of $3$ in any three consecutive untegers.
Second part: at first we notice that $240= 16\cdot 3\cdot 5$. Obviously $5\mid (5\nu+1)(5\nu+3)(5\nu+5)$. Also $3\mid (5\nu+1)(5\nu+3)(5\nu+5)$, because all three factors give different remainders modulo $3$. Since $\nu$ is odd, $2\mid 5\nu+1, 5\nu+3, 5\nu+5$. If we prove that at least one of numbers $5\nu+1, 5\nu+3, 5\nu+5$ is divisible by $4$, we will be done. Let's assume that $5\nu+1, 5\nu+3, 5\nu+5\not\equiv 0\pmod{4}$. It is equivalent to $\nu+1,\nu+3\not\equiv 0\pmod{4}$. But since $\nu$ is odd, also $\nu,\nu+2\not\equiv 0\pmod{4}$. Thus $\nu,\nu+1,\nu+2,\nu+3\not\equiv 0\pmod{4}$ - contradiction and we are done.

A bit simpler: since $\nu$ is odd, put $\nu=2\mu+1$. Then the product becomes $8(5\mu+3)(5\mu+4)(5\mu+5)$. By (1), the product of parentheses is divisible by $6$, and $5\mu+5$ is divisible by $5$, hence the whole thing is divisible by $8\cdot 6\cdot 5=240$


ratio
search
complex numbers
number theory unsolved
number theory

Let d be a positive integer, and let S be the set of all positive integers of the form $ x^2\plus{}dy^2$, where x and y are non-negative integers.
(a)Prove that if a belongs to S and b belongs to S, then ab belongs to S
(b)Prove that if a belongs to S and p belongs to S, such that p is a pime and p|a, then a/p belongs to S
(c)Assume that the equation $ x^2\plus{}dy^2\equal{}p$ has a solution in non-negative integers x and y, where p is a given prime. Show that if d>1, then the solution is unique, and if d=1, then there are exactly two solutions

[quote="greentreeroad"]Let d be a positive integer, and let S be the set of all positive integers of the form $ x^2 \plus{} dy^2$, where x and y are non-negative integers.
(a)Prove that if a belongs to S and b belongs to S, then ab belongs to S
(b)Prove that if a belongs to S and p belongs to S, such that p is a pime and p|a, then a/p belongs to S
(c)Assume that the equation $ x^2 \plus{} dy^2 \equal{} p$ has a solution in non-negative integers x and y, where p is a given prime. Show that if d>1, then the solution is unique, and if d=1, then there are exactly two solutions[/quote]

[hide="(a)"]
$ (a^2 \plus{} d b^2)(x^2 \plus{} d y^2) \equal{} (ax \minus{} dby)^2 \plus{} d(ya \plus{} bx)^2$
Motivation: In the complex numbers, the magnitude of a product equals the product of the magnitudes. Now consider $ a \plus{} b\sqrt { \minus{} d}$ and $ x \plus{} y\sqrt { \minus{} d}$, and square the equation of the previous sentance.
[/hide]
[hide="start for (b)"]
It is sufficient to show that if you represent $ a$ and $ p$ as the squares of magnitudes of complex numbers of the form $ x \plus{} y\sqrt { \minus{} d}$, then the ratio of those two complex numbers is also of the form $ x \plus{} y\sqrt { \minus{} d}$. Then since the square of the magnitude of that number is $ a/p$, you'd be done.
[/hide]

I suck at complex number, can you avoid using it? :oops:

The underlying motive here is the ring of [url=http://en.wikipedia.org/wiki/Algebraic_integer]algebraic integers[/url] in the [url=http://en.wikipedia.org/wiki/Algebraic_number_field]algebraic number field[/url] $ \mathbb{Q}[ \sqrt { \minus{} d} ]$.

So anyone gives a complete solution for (b) and (c)?

Completely posted before. Take a search next time.
http://www.mathlinks.ro/viewtopic.php?t=104473

OK, sorry for not searching. Also thanks ZetaX.
I have another question, which I'm pretty sure that it was posted before. But I don't know which key word I need to type in to search, so I just ask the question here. Please help me.
(a) If p is a prime and p is of form 8k+3 or 8k+1, then there exist positive integer x and y such that $ p\equal{}x^2\plus{}2y^2$
(b) If p is a prime and p is of form 6k+1, then there exist positive integers x and y such that $ p\equal{}x^2\plus{}3y^2$


Define $a_n$ as follows: $a_{n+2}=\frac{2}{a_n+a_{n+1}}$, $a_1,a_2>0$.
Show that ${\lim_{n\to\infty}a_n=1}$ for all $a_1,a_2$.

[i]Edited by Myth[/i]

What do you mean? Noone say it is easy... for the time being...

Here is the solution:
Choose $\alpha >1$ such that $a_0,a_1\in (\frac{1}{\alpha},\alpha)\Rightarrow a_n\in (\frac{1}{\alpha},\alpha)\forall n\geq 0$
Put :
$M_n=\sup_{i\geq n}a_i;n_n=\inf_{i\geq n}a_i$,
\[\Rightarrow M_1\geq M_2\geq \cdots \rightarrow M,m_1\leq m_2 \leq \cdots \rightarrow m\]
There exist $\{a_{n_i}\}\to M$ when $i\to +\infty$.We have:
$a_{n_i}=\frac{2}{a_{n_i-1}+a_{n_i-2}}\leq \frac{1}{m_{n_i-2}} \to \frac{1}{m}$ when $i\to +\infty,\Rightarrow m\cdot M\leq 1(1)$
Similarly,$\exists \{a_{n_j}\}\to m$ then we have $m\cdot M\geq 1(2)$
From $ (1)(2)$ we deduce that $m\cdot M=1$.Thus: $a_{n_i}\to M=\frac{1}{m},\Rightarrow a_{n_i-1}+a_{n_i-2}\to 2m,\Rightarrow a_{n_i-2}\to m(3),a_{n_i-1}\to m$
On the other hand,$a_{n_i-1}\to m,\Rightarrow a_{n_i-2}+a_{n_i-3}\to 2M,\Rightarrow a_{n_i-2}\to M(4)$
From $(3)(4)$ we deduce that: $m=M=1$.So we can conclude that $\lim_{n\to \infty}a_n =1$

I think my solution same as yours pigfly,
anyway because of spending time on solving and typing it:
First of all we notice that no three consecutive are greater or less that 1.
Suppose $a_{n+3} \geq 1$ than we will prove $a_{n+3} \leq max(a_{n+1},a_{n},a_{n-1})$

We have $ \frac{2}{a_{n+3}} = (a_{n+2}+a_{n+1}) = \frac{2}{a_{n+1}+a_{n}}+ \frac {2}{a_{n}+a_{n-1}}\geq \frac {2}{max(a_{n+1},a_{n},a_{n-1})} $

Hence $1 \leq a_{n+3} \leq max(a_{n+1},a_{n},a_{n-1})\leq . . . \leq max(a_{1},a_{2},a_{3})$
It remains to prove that equality doesnt hold.
From upper thoughts there are two cases.
    1)$a_{n+3} =  a_{n+1}= a_{n} = t $
so  $a_{n+2} = 1/t , a_{n+3} = 2/(t+1/t) < 1$ contradiction.
    2) $a_{n+1} = 1/t, a_{n+2} = 2/(t+1/t), a_{n+3} = \frac {2}{2/(t+1/t)+1/t)}= t $
Or $ t= 1$ hence $ a_{n+1}= a_{n} = 1 $ and all others equal to 1.
So it decreses and is bounded and therefor $ lim a_{n} = 1$ q.e.d.


PLZZ help on level 40 the TERRANOVA THINGY..

PLZZ help us figure out the TERRANOVA THINGY..

We do not know what you are talking about. Could you give a description of the game, or a link, or something.

google is your friend.
is [url=http://www.terranovatrading.com/]this[/url] what you were talking about


Let $ ABCD$ be a trapezoid with $ AB$ and $ CD$ parallel, $ \angle D \equal{} 2 \angle B, AD \equal{} 5,$ and $ CD \equal{} 2.$ Then $ AB$ equals

A. 7
B. 8
C. 13/2
D. 27/4
E. $ 5 \plus{} \frac{3 \sqrt{2}}{2}$

Check if I did this right. I'm not quite sure :roll:
[hide="solution"]
Because $ AB$ and $ CD$ are parallel, $ ABCD$ is an isosceles trapezoid. Opposite angles in this trapezoid add up to $ 180$ degrees, so $ m<D \equal{} m<C \equal{} 120$ and $ m<A \equal{} m<B \equal{} 60$. We then draw the perpendicular from $ C$ to $ \overline{AB}$, and we call the point of intersection $ E$. Triangle $ BCE$ is a 30-60-90 triangle, and its hypotenuse it $ 5$. Therefore, $ BE \equal{} 5/2$. We can do likewise on the other side of the trapezoid, and therefore $ AB \equal{} 2\plus{}5/2\plus{}5/2 \equal{} 7$, and the answer would be $ (A)$
[/hide]

[quote="xscapezaer"]Check if I did this right. I'm not quite sure :roll:
[hide="solution"]
Because $ AB$ and $ CD$ are parallel, $ ABCD$ is an isosceles trapezoid. Opposite angles in this trapezoid add up to $ 180$ degrees, so $ m < D \equal{} m < C \equal{} 120$ and $ m < A \equal{} m < B \equal{} 60$. We then draw the perpendicular from $ C$ to $ \overline{AB}$, and we call the point of intersection $ E$. Triangle $ BCE$ is a 30-60-90 triangle, and its hypotenuse it $ 5$. Therefore, $ BE \equal{} 5/2$. We can do likewise on the other side of the trapezoid, and therefore $ AB \equal{} 2 \plus{} 5/2 \plus{} 5/2 \equal{} 7$, and the answer would be $ (A)$
[/hide][/quote]
Hmmm...where does it say it's isoceles?

(sigh)... that's a good question. Apparently not, then. Oh well.

Let  DE bisector of angle ADC ( E point of AB ) . Tha angle AED=EDC=ADE ,  then DEBC is parallelogram  and EB=DC=2 .
 The triangle ADE is isosceles , then AE=AD=5 . And AB=5+2=7.

[hide="Very brief solution"]

Create $ E$ on ray $ DC$ such that $ DE \equal{} AB$ and $ EB \equal{} DA \equal{} 5$ (Yes, this is possible). This gives parallelogram so $ \angle EBC \equal{} x$ and by AIA, $ \angle ECB \equal{} x$ as well in respect to segment $ BC$. So, $ EC \equal{} 5$ as well and $ AB \equal{} DE \equal{} 2\plus{}5 \equal{} 7$. A. [/hide]


Given $ a,b,c$ be real numbers .Prove that:
\[ 5(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} 34(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \ge (a \plus{} b \plus{} c)^4 \plus{} (a \plus{} b)^4 \plus{} (b\plus{} c)^4 \plus{} (c \plus{} a)^4 \plus{} 2(ab \plus{} bc \plus{} ca)^2
\]
:)

[quote="forevercontinue"][quote="quykhtn-qa1"]Given $ a,b,c$ be real numbers .Prove that:
\[ 5(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} 34(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \ge (a \plus{} b \plus{} c)^4 \plus{} (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \plus{} 2(ab \plus{} bc \plus{} ca)^2
\]
:)[/quote]
The inequality is equivalent to :

$ A \equal{} \sum a^4 \plus{} 15\sum a^2b^2 \minus{} 4 \sum ab(a^2 \plus{} b^2) \minus{} 8abc\sum a \geq 0$
We have
$ (\sum (a \minus{} b)^2)A \equal{} (2\sum a^3 \plus{} 24abc \minus{} 5\sum ab(a \plus{} b))^2 \plus{} 75(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$
$ \Rightarrow A \geq 0$
Our proof is completed[/quote]
Nice proof,  :lol:
Here is my solution:
The inequality is equivalent to:
$ \sum{(a\plus{}b\minus{}4c)^2(a\minus{}b)^2} \ge 0$
Which is obvious true. :)
equality occurs if and only if :$ (a,b,c)\equal{}(t;t;t),(t;t;2t),(t;2t;t),(2t;t;t)$

[quote="nguoivn"][quote="quykhtn-qa1"]Given $ a,b,c$ be real numbers .Prove that:
\[ 5(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} 34(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \ge (a \plus{} b \plus{} c)^4 \plus{} (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \plus{} 2(ab \plus{} bc \plus{} ca)^2
\]
:)[/quote]
Sorry but I don't think it's nice  :maybe: . And this hint is very old.
Example, see problem 22 in here: http://www.maths.vn/forums/showthread.php?t=9244
I think it's nicer  :wink:[/quote]
Yes. your inequality nicer :wink:


Let $ a,b,c$ be positive numbers having product $ 1.$ Prove or disprove the following inequality
$ \frac{a^3}{(2a^4\plus{}1)^2}\plus{}\frac{b^3}{(2b^4\plus{}1)^2}\plus{}\frac{c^3}{(2c^4\plus{}1)^2} \le \frac{1}{3}.$

I think it is true but I dont know why I can't prove it. Hope someone will help me.

[quote="can_hang2007"]Let $ a,b,c$ be positive numbers having product $ 1.$ Prove or disprove the following inequality
$ \frac {a^3}{(2a^4 \plus{} 1)^2} \plus{} \frac {b^3}{(2b^4 \plus{} 1)^2} \plus{} \frac {c^3}{(2c^4 \plus{} 1)^2} \le \frac {1}{3}.$

I think it is true but I dont know why I can't prove it. Hope someone will help me.[/quote]

The inequality  holds!
   ( Let  $ a\equal{}\frac{y}{z},b\equal{}\frac{z}{x},c\equal{}\frac{x}{y},$ and
     $ x\equal{}p\plus{}q\plus{}r,y\equal{}p\plus{}q,z\equal{}p ,$ or
     $ {x\equal{}p\plus{}q\plus{}r,z\equal{}p\plus{}q,y\equal{}p },{p,q,r}\ge{0}$)

[quote="fjwxcsl"][quote="can_hang2007"]Let $ a,b,c$ be positive numbers having product $ 1.$ Prove or disprove the following inequality
$ \frac {a^3}{(2a^4 \plus{} 1)^2} \plus{} \frac {b^3}{(2b^4 \plus{} 1)^2} \plus{} \frac {c^3}{(2c^4 \plus{} 1)^2} \le \frac {1}{3}.$

I think it is true but I dont know why I can't prove it. Hope someone will help me.[/quote]

The inequality  holds!
   ( Let  $ a \equal{} \frac {y}{z},b \equal{} \frac {z}{x},c \equal{} \frac {x}{y},$ and
     $ x \equal{} p \plus{} q \plus{} r,y \equal{} p \plus{} q,z \equal{} p ,$ or
     $ {x \equal{} p \plus{} q \plus{} r,z \equal{} p \plus{} q,y \equal{} p },{p,q,r}\ge{0}$)[/quote]

Method of violence... :o


function
algebra
polynomial
LaTeX
calculus
derivative
calculus computations

The roots of the cubic f(x) = x^3 - 6x^2 + 9x - 2 = (x-2)(x^2 - 4x + 1) are x = 2, x = 2 + sqrt(3) and x = 2 - sqrt(3).

The mean average of the roots is (2 + (2+sqrt(3)) + (2 - sqrt(3)))/3 = 2.

Differentiating the function gives f'(x) = 3x^2 -12x + 9 = 3(x-1)(x-3) and this has roots x = 1 and x = 3.
The mean average of the roots of f'(x) = 0 is (1+3)/2 = 2.

Is it a coincidence that the two means are the same?

What about (i) other cubics, (ii) other polynomuials?

Take $ f(x) \equal{} x^3 \minus{} x^2 \minus{} x \plus{} 1$. Then $ f(x) \equal{} (x \minus{} 1)(x \plus{} 1)^2$. The mean average of these roots is $ \frac{1 \plus{} (\minus{}1) \plus{} (\minus{}1)}2 \equal{} \minus{}\frac12$.

$ f'(x) \equal{} 3 x^2 \minus{} 2x \minus{} 1 \equal{} (x \minus{} 1)(3x \plus{} 1)$. The mean average of these roots is $ \frac{1 \plus{} \left(\minus{}\frac13\right)}{2} \equal{} \frac13$.

Seems like coincidence. There will probably be some condition the coefficients of your polynomial have to satisfy in order for that to work out, but I don't know it.

AndrewTom, please use [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ) -- since you're already using many of the codes, you just need to enclose your equations in dollar signs.  Also, what is the source of all the problems you've been posting recently?  (There is a bar called "Subject description" under "Subject" in which you can type it when you make posts in the future.)

The sum of the roots of $ f(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$ is $ \minus{} a$, so the average is $ \frac { \minus{} a}{3}$.  The derivative is $ f'(x) \equal{} 3x^2 \plus{} 2ax \plus{} b$, and the sum of its roots is $ \frac { \minus{} 2a}{3}$, with average $ \frac { \minus{} a}{3}$.  Tyl, to take the average of three numbers, divide by 3, not 2  :wink:  It's easy to see that the same computation will work for any polynomial of degree 2 or larger.

Thanks JBL. Will try. They're from some sheets of problems someone gave me and I thought they would be of interest.


i think it's too dofficulte so can you post a solution


I think inequality is very difficult
In addition to memorize things nothing else.
How do you study?
What books do you read?
(Sorry .. my english is poor)

http://www.mathlinks.ro/viewtopic.php?t=325628

who can help me , thanks :lol:


First, we have:

$ (U_n ): \left\{\begin{array}{l} U_1 \in R \\
U_n \equal{} U_{n \minus{} 1}^2 \plus{} aU_{n \minus{} 1} \plus{} 1. \\
\end{array} \right.(n \equal{} 1,2,3,4,.......)$

Question: Find $ \mathop {\lim }\limits_{n \to \infty } \left( {\frac {{U_1 U_2 U_3 .....U_{n \minus{} 1} }}{{U_n }}} \right)$ with:

1. $ a \equal{} 1$

2. $ a \equal{} \minus{} 1$

[b]I edited (2) to (-1).[/b]

um, maybe you wanna post this in intermediate or even the calculus forum cause i think its too hard for HSB
EDIT: set builder notation is messed up...  :P  :P


If a * b = a + b - ab, then what is the value of (4 * 2) * (3 * 3) ?

First evaluate 4 * 2 by plugging in the numbers.

$ 4 \plus{} 2 \minus{} 8 \equal{} \minus{}2$

Now evaluate 3 * 3.

$ 3 \plus{} 3 \minus{} 9 \equal{} \minus{}3$

Finally, evaluate -2 * -3.

$ \minus{}2 \minus{} 3 \minus{} 6 \equal{} \minus{}11$

EDIT: Beaten.

As mewto55555 and izzy demonstrated, you have to work with the parentheses first.  But don't be intimidated by these problems.  Just simply plug in the corresponding values of $ a$ and $ b$ in the parentheses first, before you work into the bigger operation.

[quote="mewto55555"]On the last step, -2x-3=[b]+[/b]6[/quote]
Yes but its -ab so it would be -6.


[quote="quykhtn-qa1"]Given $ a , b , c > 0$ .Prove that:
$ \frac {a^4}{1 \plus{} a^2b} \plus{} \frac {b^4}{1 \plus{} b^2c} \plus{} \frac {c^4}{1 \plus{} c^2a}\geq \ \frac {abc(a \plus{} b \plus{} c)}{abc \plus{} 1}$.[/quote]
use Cauchy reduce it to
$ \frac{(\sum a^2)^2}{3\plus{}\sum a^2b} \ge \frac{abc\sum a}{abc\plus{}1}$
Then use $ (\sum a^2)^2\ge 3abc\sum a$ and $ (\sum a^2)^2 \ge \sum a\sum a^2b$
The last is valid for  $ (\sum a^2)^2 \ge 3\sum a^3b$ :)

[quote="hxy09"][quote="quykhtn-qa1"]Given $ a , b , c > 0$ .Prove that:
$ \frac {a^4}{1 \plus{} a^2b} \plus{} \frac {b^4}{1 \plus{} b^2c} \plus{} \frac {c^4}{1 \plus{} c^2a}\geq \ \frac {abc(a \plus{} b \plus{} c)}{abc \plus{} 1}$.[/quote]
use Cauchy reduce it to
$ \frac {(\sum a^2)^2}{3 \plus{} \sum a^2b} \ge \frac {abc\sum a}{abc \plus{} 1}$
Then use $ (\sum a^2)^2\ge 3abc\sum a$ and $ (\sum a^2)^2 \ge \sum a\sum a^2b$
The last is valid for  $ (\sum a^2)^2 \ge 3\sum a^3b$ :)[/quote]
Nice proof :lol: .But the last inequality we can only use AM-GM :)

[quote="quykhtn-qa1"]Nice proof :lol: .But the last inequality we can only use AM-GM :)[/quote]
It's only the similar type of sum of square.
[hide]@quykhtn-qa1: Am-Gm chi? la` cach viet khac cua? pt tong? bp thoi em a. Va` anh nghi~ ve` ban? chat, no' gan` nhu ko the? tim` ra neu em chua doc. n~ cach CM bang` pt tong? binh` phuong  :) [/hide]

[quote="quykhtn-qa1"]Given $ a , b , c > 0$ .Prove that:
$ \frac {a^4}{1 \plus{} a^2b} \plus{} \frac {b^4}{1 \plus{} b^2c} \plus{} \frac {c^4}{1 \plus{} c^2a}\geq \ \frac {abc(a \plus{} b \plus{} c)}{abc \plus{} 1}$.[/quote]
By the Cauchy-Schwarz Inequality, we have
$ \sum \frac{a^4}{1\plus{}a^2b} \ge \frac{\left( \sum a^2\sqrt{c}\right)^2}{\sum c(1\plus{}a^2b)} \equal{}\frac{\left( \sum a^2\sqrt{c}\right)^2}{(a\plus{}b\plus{}c)(abc\plus{}1)}.$
Therefore, it is enough to prove that
$ \left( \sum a^2\sqrt{c}\right)^2 \ge abc(a\plus{}b\plus{}c)^2.$
This is equivalent to
$ \sum a^2\sqrt{c} \ge \sqrt{abc}(a\plus{}b\plus{}c),$
or
$ \sqrt{\frac{a^3}{b}}\plus{}\sqrt{\frac{b^3}{c}}\plus{}\sqrt{\frac{c^3}{a}} \ge a\plus{}b\plus{}c.$
By the AM-GM Inequality, we have
$ \sqrt{\frac{a^3}{b}}\plus{}\sqrt{\frac{a^3}{b}}\plus{}b \ge 3a.$
Adding this and its analogous inequalities, the conclusion follows.


If anyone can go to the wiki, the solution for problem 20 given on the wiki looks like it requires trig. Is there a way to do it without trig?

I assume you mean this year?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1420460#1420460]this[/url].

I've replaced the solution in the wiki. In the future, you can go to the "Resources" or "Contests" section, which will be linked from the toolbar at the top of this page. Find the problem you're looking for and click on the problem number - post all discussion in that thread instead of starting a new one.


conics
parabola
integration
vector
analytic geometry
calculus

It seems like $ s(t)$ needs to be vector-valued to me... then again, I don't know too much about "Advanced Physics"  :oops:

[quote="perfect_radio"]I'm trying again to understand physics... Not really successfully!

Let's consider the parabola $ z \equal{} a x^2, y \equal{} 0$ and an object with initial speed $ v_0$ found in the origin (in the initial problem, the parabola was rotating around $ Oz$, but that's pretty much irrelevant to my question). I want to derive the differential equation that the motion of the object satisfies.

Is anything wrong with the following method?

Let $ s(t)$ be the (signed) distance from the origin to the current position (measured on the parabola). Next, I want to use Newton's Law: $ m \ddot{s} \equal{} F$. I take the weight, project it on the tangent to the parabola, stick a $ \minus{}$ in front, and I get an equation $ m \ddot{s} \equal{} F(x)$. Next, I write $ s(t) \equal{} f(x(t))$, where $ f(t) \equal{} \int_0^t \sqrt {1 \plus{} 4 a^2 u^2} \, du$, and I differentiate that twice, so that I get a second order differential equation in $ x$.

If I do it with Lagrangians, I get the same answer as in the book, but if I use the above "method", then one of the three terms involved has a different sign. I'm suspecting that it's wrong altogether (at a conceptual level) :blush:[/quote]

Your integral equation for $ s\left(t\right)$ is correct so far as I can tell. But why would you do the problem that way if you can write the Lagrangian for it?

Incidentally, I differentiated your expression for $ s\left(t\right)$ once, to obtain $ \dot{s}$; squaring that gives the proper expression for the Lagrangian. Therefore, I suspect the misstep is in your calculation of $ \ddot{s}$.

I finally got identical results! I messed up the Lagrangian computations: $ \frac{\partial L}{\partial \dot{x}}$ was correct (it involved both $ x$ and $ \dot{x}$), but when I applied $ \frac{d}{dt}$ to it, I differentiated only $ \dot{x}$ (I think I was tired from so much physics...) :blush:

[quote="Djole"]The object on the parabola is not a free particle; the fact that it is constrained to move on a curve affects the acceleration, since the direction of the velocity vector will change without regard to the change in its velocity. In your approach, you don't take this change into account -- and this change is important (e.g. it causes 'centripetal acceleration'). If you want to use Newtonian mechanics, you shouldn't switch away from cartesian coordinates (however, the calculation can get slightly messy), so Lagrangians are probably the way to go here.[/quote]
I've got me thinking here...  :!: But since I (eventually) obtained identical results, I guess the method works, because I care only how object moves on the parabola and what forces are applied to keep it going; in particular, I'm not interested how the velocity changes direction. Even with these said, I wouldn't bet my life savings (if I had any) on the validity of the approach.

[quote="t0rajir0u"]Alternate (valid) methods should give the same answer.  If they don't, it calls that method into question, and understanding why that method does not work deepens one's understanding of the subject.  :)[/quote]
Exactly!


(1)How many permutation of $ (a,a,a,b,b,b,c)$
(2)How many permutation of $ (a,a,a,a,a,b,b,b,b,b,c)$

1) There are $ \binom {7}{3}$ ways to choose the places of the $ a$'s.
For each of these coices, there are $ \binom {4}{3}$ ways to choose the places of the $ b$'s.
Then the $ c$ is placed.
Thus there are $ \binom {7}{3} \cdot \binom {4}{3}$ permutations.

2) Same reasoning.

Pierre.

\sum and \prod are "big" automatically -- \bigsum and \bigprod are not standard commands.  Are you aware that there's a preview button, so you can check your post for accuracy before you post it?

[quote="apollo"]How many permutation of  (a,a,a,b,b,b,c) around a circle[/quote]
if you mean that (abcde)=(bcdea)
then there are $ \frac{(3\plus{}3\plus{}1)!}{(3\plus{}3\plus{}1)\times 3!\times 3!\times 1!}\equal{}20$


Can someone plese explain the rearrangement inequality. I looked it up on AoPSWiki but didnt fully understand.

Thank You.

hello, i hope this will help you
http://www.math.auckland.ac.nz/~olympiad/Rear.pdf
Sonnhard.


HOW MANY WAYS are there to form a four letter sequence using {a ,b ,c ,d ,e, f }?
containing the letter a in which repitition is not allowed

please in details

thank you

[hide="Hint"]How many possibilities are there for the first letter?  The second letter?  The third letter?  The fourth letter?[/hide]

hmm.. i would do caseworks like this
no pairs(need 4 letters)
EXACTLY 1 pair(need 3 letters)
2 pairs(need 2 letters)
3 of a kind(need 2 letters)
4 of a kind(need 1 letter)

too lazy to work it out.. ROFL, it's like poker?!  :rotfl:

Ummm either I'm reading the question wrong, or it's really simple.

If you can reuse letters, it should just be $6^{4}=1296$.

If you can't, it would just be $\frac{6!}{2!}=360$.

Ya, this is way to easy. But what if I said order does not matter?

sorry the question should be like that

HOW MANY WAYS are there to form a four letter sequence using {a ,b ,c ,d ,e, f }?
containing the letter a in which repitition is not allowed

Ok, now taht uve changed the quetion.. there are 4 places where you can put a, the rest 3 places, there is 5*4*3 ways so 240

thanx to all

so what is the first position is a and what is the second....

please show how thsi works
thanx

[quote="1=2"]I hope this helps!

[hide="solution"]

We have the set {$a,\alpha,\beta,\gamma$} and there are 24 ways to arrange them.

[hide="Why?"]We have four spaces:

_ _ _ _

The first space can be filled with one of four letters. The second space can be filled with one of three ketters. The third, one of 2. The fourth, the remaining letter.

4*3*2*1=24[/hide]

Since $\alpha$, $\beta$ and $\gamma$ all have to be different letters(we can't use the a since we've used it), there are 5 ways to choose a letter for $\alpha$. There are 4 ways to choose for $\beta$, and 3 ways to choose for $\gamma$. 5*4*3=60. But we didn't count arrangements. So 24*60=14400
[/hide][/quote]

Actually, by rearranging them, you have counted multiple times.  There are actually 5 choose 3=10 ways to choose $\alpha$, $\beta$, and $\gamma$, because the order of these three doesn't matter since we are rearranging them at the end.  So there are $24\times 10=240$ sequences.


function MATHCOUNTS geometry trapezoid parallelogram

The following questions are pretty basic questions that I hope everyone in this forum should be able to do:

1. We have this sequence: 1, 1/2, 1/6, 1/24, ..... can you find the sixth term? And if you're advanced, can you find the equation of this sequence?

2. Given the function f (x) = x^2, find the relationship of this function with the originial function f (x) = (x - 3)^2 without the aid of the calculator.  State which way it has been moved.

3. Find the real roots of this equation: y = (x^2+3x+2)(x^2+4x+4)

4. Simplify (10!/8!)^2.

I'll post more soon.. So, please try to do here and for number one, there are two questions  :lol:

[hide]1. 1/n! works so far, so I'll say 1/720 ?

2. The new function is shifted to the left 3, as x can now be 3 less than it was before and give the same y value as when the original function is used on x. Sorry, that was a poor explanation.

3. y = 0 when x = 2, 2, -2 or -1?

4. Simplify (10!/8!)^2.

4. (10!/8!)^2 = (10*9)^2 = 90^2 = 8100[/hide]

1.[hide]The 6th term is 1/720.  The equation is the nth term = 1/(n!)[/hide]

2.[hide]It has been moved to the right[/hide]

3.[hide]0 = (x^2+3x+2)(x^2+4x+4)

                     0 = (x^2+3x+2) or 0 = (x^2+4x+4)

                     0 = (x+1)(x+2) or 0 = (x+2)(x+2)

                     x = -1 or x = -2

                     the roots are -1 and -2[/hide]

4.[hide](10!)/(8!) * (10!)/(8!) = 10*9*10*9 = 100*81 = 8100[/hide]

Would these be about MATHCOUNTS level problems?

[quote="Silverfalcon"]Very good.  And Rep, yes, you're right, it was my bad on that wrong word.
 :) .

Then three another problems:

1. 2/3*3/4*4/5*5/6*6/7*7/8 = ?

2. 1^5+2^5+3^5....35^5?

3. What is the sum of digits for 2^202*5^204? (If the number is 125 it's 1+2+5=8)

Good luck~[/quote]

1. 1\4

2. it is well known that $S_{5}=1^{5}+ \ldots+ n^{k}=\frac{n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}$

3. Observe that it is $10^{202} \cdot 5^{2}$

Silverfalcon wrote:Very good.  And Rep, yes, you're right, it was my bad on that wrong word.
  .

Then three another problems:

1. 2/3*3/4*4/5*5/6*6/7*7/8 = ?

2. 1^5+2^5+3^5....35^5?

3. What is the sum of digits for 2^202*5^204? (If the number is 125 it's 1+2+5=8)

Good luck~

1.[hide]7!/(8!-2) = 2/8 = 1/4[/hide]

I could probably do number 2, but I really don't feel like thinking right now.  3 just boggles my mind.  I really don't get it.[/hide]

mikewd wrote:[hide]2nd and 3rd have me beaten at first glance...[/hide]

Look very closely at number 3...

ok, if  2^202*5^204 = 10^202*5^2,
then the answer is 7.

2+5+ all of those zeroes at the end. (202 of them.)

2^202*5^204 = 10^202*5^2 ?

I was looking for some sort of manipulation but couldn't find one. How did you find this?

[quote="mlee06"]well, if you pair up every "5" and "2" together (202 of them) then you have 10^202[/quote]

Of course. I'm annoyed with myself for not seeing that.

Mikewd, Don't be so hard on yourself. I've seen some of your other posts, and you can be quite inventive. By the way, you live in South England... that's so much more interesting than Palo Alto, California for me.

Haha, thanks. I'm always a bit like that because that one the pairing up seems really obvious now, they all do once you've had the answers revealed to you. I guess the hint was add up the digits, so most of them would be zeros and so 10^a would be in the rearranged expression.

Anyway thanks, it's quite nice here but boring now that college is out. I'm going to the USA soon on holiday, who knows I might see you! (not very likely though). :cool:

It seems like all people have been doing really good job :)

So, let's get little bit higher~

1. There is a triangle ABC, with M as midpoint of AB and N as midpoint of CA. If AB is 7, CA is 7, and BC is 10, find the length of MN.  Now, if you get the correct answer, prove it!

2. There is a trapezoid ABCD.  If Z is a midpoint of AB and X is a midpoint of CD, find the length of ZX if length are: AB = 7, BC = 8, CD = 5, and DA = 11.

These are quite geometrical(right?) problems.

Silverfalcon wrote:It seems like all people have been doing really good job

So, let's get little bit higher~

1. There is a triangle ABC, with M as midpoint of AB and N as midpoint of CA. If AB is 7, CA is 7, and BC is 10, find the length of MN.  Now, if you get the correct answer, prove it!

2. There is a trapezoid ABCD.  If Z is a midpoint of AB and X is a midpoint of CD, find the length of ZX if length are: AB = 7, BC = 8, CD = 5, and DA = 11.

These are quite geometrical(right?) problems.

I'm only going to do number 1 right now.

[hide]Okay, the triangle is an isoceles triangle, since two of the sides are equal.

So basically, we have a similar triangle, with all of the sides cut by 1/2.  We know the triangle is similar, becuase MN goes from the midpoint of AB to and the midpoint of CA, and both CA and AB are 7.  Since the midpoint of AB is the same distance from BC as the midpoint of CA, we can say that MN is parallel to BC.

AM and CN are proportional to AB and CA, so the other side of the triangles also have to be proportional, NM has to be 1/2 of BC, since AM and CN are 1/2 of AB and CA.

Therfore, NM is equal to 1/2 of 10, or 5.[/hide]

I hope you guys understand it.  I'm not really good at explaining myself.

[hide]1. There is a triangle ABC, with M as midpoint of AB and N as midpoint of CA. If AB is 7, CA is 7, and BC is 10, find the length of MN.  Now, if you get the correct answer, prove it!

Call the midpoint of BC P. As ABC is an isoceles triangle, by pythagoras:

√

5 + AP = 7

AP = 24

AP = √24 = 2√6

Now the midpoint of AP is √6 from A (and obviously from P as well). Call this point Q. The triangle ANQ is also a right angle therefore, and by pythagoras,

QN + AQ = (7/2)

QN = (49/4) - 6

QN = 6.25

QN = 2.5

And MN = 2QN = 2*2.5 = 5[/hide]

Both are really good.  But anyone can think of way using the definition of parallelogram?

1. (with parallelogram)

[hide]Take triangle ABC and double it, flipping the second triangle and putting it next to it, thereby creating a parallelogram.  Now, extend line MN to a point P on the other side of the parallelogram.  MP (or NP, depending on which way you created your parallelogram) is double MN.  Since MP is halfway between the bases of the parallelogram, it is equal to one of the bases: MP = 10.  Using what we found earlier, MP = 2MN, MN is found to be 5.[/hide]

silverfalcon, for number two, you do not specify whether lengths ab and cd or bc and ad are parallel. I think it would help to know which way the trapezoid is orientated. Thank you...

edited: parallelogram to trapezoid

I haven't really thought of parallelogram on number two.

But for number one, here's one approach you might want to do:

[hide]Is there anyway you can have parallelogram INSIDE the triangle?[/hide]

Just wondering, what's the answer to number two?

I believe it's [hide]9.5[/hide]

(Base1+Base2)/2 --> Answer

Oh, I don't think it said the problem said AD and BC were parallel

Which configuration does your #2 refer to?

I think it's the second one, because AD and BC are parallel, otherwise the answer couldn't be [hide]9.5[/hide]

I'm new to this site and i read over these questions. They are mediocrely challenging. But this is beginner's right? You guys are really smart... :D I don't think anyone posted an answer to this question:

2. 1^5+2^5+3^5....35^5?

1*1*1*1*1*2*2*2*2*2...35*35*35*35*35 =
35!^5

There was this answer by Langria but i don't get it.
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/44c667977f78d72a843812cb7ef7b80b.gif[/img]

[quote="icare"]Hi, im new on this server...

Have a look at it:

What is the last number (the last digit) of:

(1/5)^2004

I don't know if you will understand my English, but, for exemple, the last number of 123 is 3  :D[/quote]

[color=white]5[/color]

icare wrote:Hi, im new on this server...

Have a look at it:

What is the last number (the last digit) of:

(1/5)^2004

I don't know if you will understand my English, but, for exemple, the last number of 123 is 3

Solution in spoiler:

[hide](1/5)^2004 = (2/10)^2004 = (2^2004)/(10^2004). Dividing by some power of 10 is equivalent to moving the decimal place of a number, which will not change its final digit. So the problem reduces to finding the last digit of 2^2004. Since the units digit of powers of two repeats every four numbers, the units digit of 2^2004 is the same as the units digit of 2^4, which we find readily to be 6.[/hide]

Instead of creating a 10 factor, you just could say; (1/5)=0,2

the two reapeats itselfeach time you multiplicate it 5 times :

2^5 has a final digit which is 2.

write 2004 like 5k+k' with  0=<k'<5 :

2^2004= 2^(5x400+4) =( (2^5)^400 )x(2^4) => 2^2000 ends with a 2

2^4=16

SO, (1/5)^2004 ends with a 2 (2x16=32)


calculus
real analysis
calculus computations

Dear All;

Can anyone offer assistance in the solving of below 2 question?

1) $ max(a,b)\equal{}\frac{a\plus{}b\plus{}|a\minus{}b|}{2}$

2) $ min(a,b)\equal{}\frac{a\plus{}b\minus{}|a\minus{}b|}{2}$

Thank You.

I think it's not calculus. :lol:

1)if $ a\geq b\rightarrow a\minus{}b\geq 0$ then $ \frac{a\plus{}b\plus{}|a\minus{}b|}{2}\equal{}\frac{a\plus{}b\plus{}a\minus{}b}{2}\equal{}a\equal{}max(a,b)$

if $ b>a\rightarrow b\minus{}a>0$ then $ \frac{a\plus{}b\plus{}|a\minus{}b|}{2}\equal{}\frac{a\plus{}b\plus{}b\minus{}a}{2}\equal{}b\equal{}max(a,b)$

Similar method can be applied for 2)

No, this is not calculus. Its real analysis. :-) Thanks TimeStopper_STG

That's is not calculus or real analysis, just algebra.

$ (a\plus{}b)^2\minus{}(a\minus{}b)^2\equal{}4ab\Longleftrightarrow \left(\frac{a\plus{}b}{2}\right)^2\minus{}\left(\frac{a\minus{}b}{2}\right)^2\equal{}ab$

$ \Longleftrightarrow \left(\frac{a\plus{}b}{2}\plus{}\frac{|a\minus{}b|}{2}\right)\left(\frac{a\plus{}b}{2}\minus{}\frac{|a\minus{}b|}{2}\right)\equal{}ab$

$ \therefore \frac{a\plus{}b}{2}\plus{}\frac{|a\minus{}b|}{2}\equal{}Max\{a,\ b\}$

$ \frac{a\plus{}b}{2}\minus{}\frac{|a\minus{}b|}{2}\equal{}\min\{a,\ b\}$

Thanks Kunny. I extract this from a real analysis textbook. So this should be real analysis topic but it seems the way to solve it is using Algebra. :-)

Sorry I exaggerated that, maybe your equalities is needed to study $ \epsilon \minus{} \delta$ reasoning, isn't it? :lol:

Actually, this is problem 13 from Chapter 1 of Spivak's [i]Calculus  [/i]:)

Wow. It seem this is a famous question. I extract this question from an analysis book published in India only. I am surprise the same question appear in another book.


I need some help finding the inverse function of h(x)√(x-3) -5.
(-5 is not included in the square root)

I tried and got h^-1(x)=(x+5)^2+3
is this right?
if not can you please help me out on doing it correctly!!
pls and THANKS!! =]

Actually that appears right. You can check whether your answer is right because the composition of a function with its inverse is always x. (i.e. $f^{-1}(f(x))=x$)


What is the probability that six persons have their birthdays in any 2 months of the year.
You can assume all months have equal number of days.

[quote="fourierseries"]Okay the first and second birthday months we designate as THE two months.  Therefore our probability becomes:

$(\frac{1}{6})^4$[/quote]

Is it necessary that the first two birthdays are in different months? I want to say that they can be in the same month also.

Also, I want to say that [b]I don't know the answer[/b] and I have many confusions.

The probability I got is $\frac{\binom{12}{2}(2^6-10)}{12^6}$

Well, it wouldn't make a difference if it were the first and second, first and third, etc. that were in different months...

I think that fourierseries has a good method but it needs to be modified.

There are two possible cases:

Case 1.  All six people were born in the same month.  The probability of this is $(\frac{1}{12})^6\times 12=\frac{1}{248832}$, since there are 12 months.

Case 2.  All six people were born in two differents months.  The probability is $(\frac{1}{6})^6\times 66=\frac{11}{7776}$, since there are 66 different poossible sets of two months.

Than the total probability is $\frac{1}{248832}+\frac{11}{7776}=\frac{122}{248832}=\frac{61}{124416}$.  I don't know if this is right; I hope it is though.

So basically. there are a possible $\displaystyle{12 \choose 2}$ pairs of months.

We want to drop 6 people somewhere in these 2 months. Each person has $2$ choices of months. except all 6 can't choose one month so their choices are: $2^6 - 1$.

There should then be $\displaystyle{12 \choose 2}(2^6 - 1)$ ways for this to happen.

Oh, probability. Well then, I think there should be $12^6$ on the denominators.

I guess, basically, I agree with some earlier post.

Let's consider for smaller numbers of people. Say that $p_n=$probability for n people.
 Obviously $p_1 = p_2 = 1$ (2 ppl only occupy 2 months).

Now $\frac{1}{12}$ of the time the first two people occupy the same month, so the probability is one. The other $\frac{11}{12}$ of the time, there's a $\frac{1}{6}$ chance that the third will occupy one of the 2 months of the first two people. So we get \[ \frac{1}{12} + \frac{11}{12} \times \frac{1}{6} = \frac{6}{72} + \frac{11}{72} = \frac{17}{72}  \].
Notice that the probability that they all occupy the same month is $\frac{1}{12^2}$, or more generally $\frac{1}{12^{n-1}}$.

So we have the probability if they're all in the same month. Now we'll get the probability if we know they're in 2 months. So with 2 people, they don't take up more than 2 months. With 3 people, we multiply a factor of $\frac{1}{6}$. So adding another person? Just multiply again.  So we get $\frac{1}{6^{n-2}}$, but must also put in a factor to remember that this is assuming they're in at least 2 months. So the final result:\[ \frac{1}{6^{n-2}} \times \frac{12^{n-1} - 1}{12^{n-1}} + \frac{1}{12^{n-1}} = p_n  \]
\[ p_6 = \frac{1}{6^{4}} \times \frac{12^{5} - 1}{12^{5}} + \frac{1}{12^{5}} = \frac{250127}{322486272} \approx .00077562  \]
(you can evaluate that for yourself)

This solution was inspired by the MathJam solution of USAMTS year 17, round 1, problem 2.

My answer is 11/31104.
 I dont knoe if it is right.
My logic:-
12c2=66 combinations possible.
Let 1st be born in Jan
P=1/12
2nd can be born in any of 12 months
p=1/12
3rd 4th 5th 6th born in either of 2 months
total p=1/(12*12*6*6*6*6)
66 such conditions possible
there fore answer = 11/31104


algebra
polynomial
superior algebra
superior algebra unsolved

let $ M\equal{}\mathbb{R}[x]/(x\minus{}2)(x\plus{}1)\bigoplus\mathbb{R}[x]/(x\minus{}2)(x^{2}\plus{}3)$, and let $ T : M\to M$ denote the $ \mathbb{R}$-linear transformation "multiplication by x". Find the following for $ T$:

a) Minimal polynomial

b) characteristics polynomial

c) determinant

d) rational canonical form

thanks so much

Well, it's really four questions, and you don't give any indication how much you understand already, how much you're having trouble on, or why you're asking us for help in the first place.


trigonometry
HMMT
Harvard
college
MIT
trig identities
geometry

Let $ C_1$ and $ C_2$ be externally tangent circles with radius 2 and 3, respectively. Let $ C_3$ be a circle internally tangent to both $ C_1$ and $ C_2$ at points $ A$ and $ B$, respectively. The tangents to $ C_3$ at $ A$ and $ B$ meet at $ T$, and $ TA \equal{} 4$. Determine the radius of $ C_3$.

[hide]We know that $ C_{1}T = \sqrt{20}$ and $ C_{2}T = 5$ by Pythagorean.
Since $ C_{2}T = C_{1}C_{2}$, $ \triangle C_{1}C_{2}T$ is isosceles,
and $ \angle C_{2}C_{1}T = \angle C_{1}TC_{2}$ which we can call $ \alpha$.
Thus using the law of sines:
$ \frac{sin{\alpha}}{5} = \frac{sin{2\alpha}}{\sqrt{20}}$
and using $ sin{2\alpha} = 2sin{\alpha}cos{\alpha}$
we have $ cos{\alpha} = \frac{1}{\sqrt{5}}$
which with some manipulation leads us to $ sin{2\alpha} = \frac{4}{5}$
which actually might seem a little funny at first because ${ sin{\angle TC_{2}B} = \frac{4}{5}}$
but after some thought we realize that $ \angle TC_{2}B = 180 - 2\alpha$
And thus using law of sines on $ \triangle C_{1}C_{2}C_{3}$ we have

\[ \frac{sin{4\alpha}}{r-2} = \frac{sin( \angle C_{2}C_{1}T + \angle TC_{1}A)}{r-2} \]

using some common trig identities and values we have calculated earlier, our calculations boil down to:
\[ \frac{20}{25}r - \frac{40}{25} = \frac{24}{25}r - \frac{72}{25} \]
which with a little algebra gets us
\[ r = 28 \][/hide]
although it bugs me that there probably is a way with less computation...

[quote="Math4tots"][hide]We know that $ C_{1}T = \sqrt {20}$ and $ C_{2}T = 5$ by Pythagorean.
Since $ C_{2}T = C_{1}C_{2}$, $ \triangle C_{1}C_{2}T$ is isosceles,
and $ \angle C_{2}C_{1}T = \angle C_{1}TC_{2}$ which we can call $ \alpha$.
Thus using the law of sines:
$ \frac {sin{\alpha}}{5} = \frac {sin{2\alpha}}{\sqrt {20}}$
and using $ sin{2\alpha} = 2sin{\alpha}cos{\alpha}$
we have $ cos{\alpha} = \frac {1}{\sqrt {5}}$
which with some manipulation leads us to $ sin{2\alpha} = \frac {4}{5}$
which actually might seem a little funny at first because ${ sin{\angle TC_{2}B} = \frac {4}{5}}$
but after some thought we realize that $ \angle TC_{2}B = 180 - 2\alpha$
And thus using law of sines on $ \triangle C_{1}C_{2}C_{3}$ we have
\[ \frac {sin{4\alpha}}{r - 2} = \frac {sin( \angle C_{2}C_{1}T + \angle TC_{1}A)}{r - 2}
\]
using some common trig identities and values we have calculated earlier, our calculations boil down to:
\[ \frac {20}{25}r - \frac {40}{25} = \frac {24}{25}r - \frac {72}{25}
\]
which with a little algebra gets us
\[ r = 28
\]
[/hide]
although it bugs me that there probably is a way with less computation...[/quote]

[hide="Comment"]I do recall that the answer is just 8, not 28, so I think you might have messed up your computation somewhere.[/hide]

[quote="zolojetto"]uhhh....wow lol whats the HMMT???[/quote]

Harvard-MIT Mathematics Tournament

[quote="Math4tots"][hide]We know that $ C_{1}T = \sqrt {20}$ and $ C_{2}T = 5$ by Pythagorean.
Since $ C_{2}T = C_{1}C_{2}$, $ \triangle C_{1}C_{2}T$ is isosceles,
and $ \angle C_{2}C_{1}T = \angle C_{1}TC_{2}$ which we can call $ \alpha$.
Thus using the law of sines:
$ \frac {sin{\alpha}}{5} = \frac {sin{2\alpha}}{\sqrt {20}}$
and using $ sin{2\alpha} = 2sin{\alpha}cos{\alpha}$
we have $ cos{\alpha} = \frac {1}{\sqrt {5}}$
which with some manipulation leads us to $ sin{2\alpha} = \frac {4}{5}$
which actually might seem a little funny at first because ${ sin{\angle TC_{2}B} = \frac {4}{5}}$
but after some thought we realize that $ \angle TC_{2}B = 180 - 2\alpha$
And thus using law of sines on $ \triangle C_{1}C_{2}C_{3}$ we have
\[ \frac {sin{4\alpha}}{r - 2} = \frac {sin( \angle C_{2}C_{1}T + \angle TC_{1}A)}{r - 2}
\]
using some common trig identities and values we have calculated earlier, our calculations boil down to:
\[ \frac {20}{25}r - \frac {40}{25} = \frac {24}{25}r - \frac {72}{25}
\]
which with a little algebra gets us
\[ r = 28
\]
[/hide]
although it bugs me that there probably is a way with less computation...[/quote]
Because cos$ C_1TC_2$ =cos$ C_3TA$, we find that r=$ C_3A$=8, which don't need doing as you!

Let the center of $C_i$ be the center of $O_i$ for $i \in [1,3]$. Further, suppose that $\angle TO_1A = a$ and $\angle TO_2B = b.$ Since we know $\sin b = \frac{4}{5}$ and $\sin a = \frac{2}{\sqrt{5}}$, we see that:
$$\sin (2b) = 2 \sin b \cos b = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$$
and
$$\sin (2a) = 2 \sin a \cos a = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}.$$
We know that, by the Law of Sines:
$$\frac{O_3O_2}{O_3O_1} = \frac{\sin (180 - 2a)}{\sin (180 - 2b)} = \frac{\sin (2a)}{\sin (2b)} = \frac{5}{6}.$$
We also know that $O_3O_2 + 3 = O_3O_1 + 2$, so $O_3O_1 = 6$ and the radius of the circle $C_3$ is $\boxed{8}.$


A triangle has lengths $ 18$,$ 24$, and $ 30$.  Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle.

Source: $ HMMT$

This is a right triangle,
let A(0,0), B(0,18), C(24,0) be the 3 points of the triangle.
Then the centriod is at (8,6), the circumcenter is at (12,9) and the incenter is at (6,6)
Thus, the base is 2 units and height is 3 the area is 3


circumcircle
combinatorics unsolved
combinatorics

Let n be an integer at least 4. In a convex n-gon, there is NO four vertices lie on a same circle. A circle is called circumscribed if it passes through 3 vertices of the n-gon and contains all other vertices. A circumscribed circle is called boundary if it passes through 3 consecutive vertices, a circumscribed circle is called inner if it passes through 3 pairwise non-consecutive points. Prove the number of boundary circles is 2 more than the number of inner circles.

Call a triangle [i]big[/i] if its circumcircle covers the whole $n$-gon.I claim that all the [i]big[/i] triangles constitute a certain triangulation of the $n$-gon,note that this will lead to the solution immediately.

Start from a [i]big[/i] triangle say $\triangle ABC$(one surely exists by simply taking a side and selecting the smallest angle formed by some other vertex) and pick one of its side say $AB$.Note that all the the vertices on the same side of $AB$ as $C$ subtend a smaller angle with A,B as the endpoints,while all the vertices $X$ on the other side fulfill $\measuredangle AXB+\measuredangle ACB>\pi$.Taking point $X_1$ so that $\angle AXB$ is minimal and $X,C$ on the different sides of $AB$ will also create $\triangle AXB$ into a [i]big[/i] one,indeed :

$\bullet$ $\forall$ points  $D$ on the same side of $AB$ as $X$ ,$\angle AXB$ s the smallest
$\bullet$  among all of angles $\angle ADB$ so that $X,D$ are on different sides of $AB$,$\angle ACB$ is the smallest and hence $\angle ADB+\angle AXB>\pi$
And hence the desired.

Continuing this process we reach a triangulation of the $n$-gon and call in $S$.Also note that any triangle from the triangulation uniquely determines the rest.Suppose now there exists triangle $T$ not belonging to the previous triangulation but still being [i]big[/i].Go ahead and apply the same procedure as in the previous case and suppose we reach a triangulation $S'$.Fix a certain side of the $n$-gon say the $m$-side.Note that side $m$ is part of only one [i]big[/i] triangle be it form $S,S'$ and hence $S,S'$ share a triangle  however than we have $S\equiv S'$ and hence we're done.$\blacksquare$


Quadrilateral $ ABCD$ is inscribed in a circle $ \mathcal C$. Lines $ l_{b}$ and $ l_{c}$ are tangent to $ \mathcal C$. Let $ l_{b}\cap CD=M$ and $ l_{c}\cap AB=N$. Show that $ AD,BC$ and $ MN$ are concurrent.

Assuming $ \ell_{B}, \ell_{C}$ are tangent at $ B, C$ respectively, the statement is equivalent to Pascal on $ BBCCDA$.


Fina all naturals n such that for all primes of the form nk+1, there are a,b such that it is = to a^2+nb^2.

It is true for n=3, 4. Numerical evidence suggests all such n are powers of 2 multipled by powers of 3.

All help please.

Thanks. :D  2

Hi, do you mind posting it in the correct forum (number theory), and not flooding here?


At the risk of starting another controversial topic, I would like to ask these questions, what is your opinion?  Feel free to give arguments and describe the situation.

Here in belgium many women still want to be called by their original name

In my own opinion that is just not normal.

I find it unbelievable that in the USA, a nation that is trying a lot to be politically correct, women lose their name.  When they are Lisa Tyler, and they marry Bob Smith, divorce and then marry Zach Cameron, they will be Lisa Smith Cameron.

My mother strongly wishes to keep her original name, when she was in the USA and needed a doctor, she feared that might become an issue in the overwhelming bureaucracy she had to go through as a foreigner to get medical attention

So, what do you think?

This is also for practical reasons. I told my fiancée I would prefer her to have my name when I marry her. Why? I find it normal that something in her identity links her to the other members of the family. Of course, the maiden name still appears on the ID, so it doesn't really disappear. Also, I wouldn't mind taking my fiancée's name, but for tracking purpose one choice has to be made and people have to stick to it. In my case I 'll ask her to keep my name - and not the other way - because if we come back to France it's better for her name to sound French.

In Chinese tradition, a married woman keeps her name. I didn't vote though, as I don't see anything wrong with changing or keeping your name. I feel it's just a personal choice.

[quote="fredbel6"]I find it unbelievable that in the USA, a nation that is trying a lot to be politically correct, women lose their name.[/quote]
I don't think you can say that women [i]lose[/i] their name - they have every right to keep it if they choose. It's just that most of them choose not to.

I think she should be able to choose.

My mom retains her maiden name(while I have my dad's last name) -- I don't think that makes her any less related to me.

In the United States, a woman's name remains the same as it was before marriage, from a legal standpoint.  However, in all states (this is a state issue, not a federal issue) the woman may use her new husband's name by convention.  This convention has recently (past 40 years of so) been expanded to allow the husband to change his name (by convention) or use a combination of the two last names.  The name of the children is legally set by what is written on the birth certificate (or equivalent form specified by the state).

The problem of adding names is that they quickly get too long.  Folks that change their status (for death, their choice, or the choice of the one they chose) quickly have a burden on children.  In the past ten years, as the number of children who have parents changing their names increase (in number and percentage) this has less of a stigma for the children, but it seems to me that the parents deisre to avoid convention creates a burden that falls most heavily on their children.  It seems to me that it should be avoided.

The only problem I really have with the whole name-changing thing is with women who are doctors. For instance, let's say Susan Smith is a woman, who happens to be unmarried and a doctor. Now, all of Susan's patients know her, of course, as Dr. Smith. Well, after Susan's been practicing a few years, she meets a tall, dark, and handsome man, Bob Brown. Susan and Bob fall in love and decide to get married. Now we have the whole name change mess. If Susan changes her name to Susan Brown, does that mean she's now Dr. Brown? This doesn't really seem fair, since Susan SMITH, not Bob Brown, went through medical school and got the degree. At any rate, her many patients know her as Dr. Smith! Does she, perhaps, keep her professional name as Dr. Susan Smith and adopt the name at home of Mrs. Susan Brown? This would make it very difficult for a patient to contact her at home, because they would be looking under "Smith" in the phone book, while her number would actually appear under "Brown." So there are a few more options but they all seem so confusing.  :mad: So that's my opinion. Name changes basically suck.  :P

Going back to bleumoose's post, yes it does get confusing.  My mom kept her name, but she's learned to answer to my last name (my dad's), and she's constantly getting asked if her husband died, or if she's married or divorced.  And then people confuse my dad's last name with hers...hehehe.  Its amusing.

When did Western woman first lose their name after marriage?

No option for "A woman should be able to do whatever she wants with her name if and when she marries"?

My cousin, a Canadian, kept her last name when she married, but in Canada (or perhaps just Quebec, I'm not sure) you have to file official paperwork to change your name when you marry (unlike in the United States, where that would be unnecessary) and she just never went to the trouble.

The style in Europe of a woman changing her name is quite old -- I imagine it's probably as old as the younger of last names and conventional western marriages.  The reason I guess this is that the traditional western marriage is in fact a property transaction, with the woman being (one part of) the property being exchanged.


I'm thinking of organizing a chat for users who are currently enrolled in an AoPS class and have access to the online classroom. My question is,will this work? Is it possible to get a transcript of this "chat" after it's completed?
Even if transcripts are unavailable, the chat will still be organized. For the people who want to join: can you guys help me work out a time schedule for this chat if it works?

(don't worry, I plan to make this an awesome chat if it works, I will bring some awesome problems and some even more awesome solutions! :lol: )

I think there are different classrooms for each class...
Edit: Never mind, see below.

There is an Introduction classroom (for students taking Introduction-series classes), an Intermediate classroom (students taking intermediate classes) and an Advanced classroom (WOOT).

O.K., that means I can only organize the chat in an intro classroom because I'm enrolled in intro to geo...

So now people can start signing up...

I plan to first post a thread similar to this on the intro to geo class message board. Then I will find a suitable time and date and post it here.<---something like that...


prove that in convex quadrilateral, where a, b, c, d are sides and e, f diagonals:

aˇ2+bˇ2+cˇ2+dˇ2>=eˇ2+fˇ2

It is easy, based on the first median's theorem.

If $M,$ $N,$ are the midpoints of $AC,$ $BD$ respectively, we have that $a^{2}+b^{2}+c^{2}+d^{2}= 4(MN)^{2}+e^{2}+f^{2}.$

Kostas Vittas.


conics
ellipse
analytic geometry
graphing lines
slope
trigonometry
quadratics

Answer the following questions.

(1) Find the equation of the tangent lines with slope $ m$ for an ellipse $ C: \frac{x^{2}}{a^{2}}\plus{}\frac{y^{2}}{b^{2}}\equal{}1$.

(2) Show that 2 tangent lines drawn from the point on a circle $ x^{2}\plus{}y^{2}\equal{}a^{2}\plus{}b^{2}$ are orthogonal.

[size=150][b]Very nice problem, Kunny ! Thank you for it[/b].[/size]
[quote="kunny"][color=darkred]Show that [b]tangent lines[/b] to the [b]ellipse[/b] $ \mathcal E\ :\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$ drawn from the any point $ M$ of the [b]circle[/b] $ \mathcal C\ :\ x^{2}+y^{2}= a^{2}+b^{2}$ are [b]orthogonally[/b].[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] The equation of the [b]tangent line [/b]in the point $ P(x_{0},y_{0})\in\mathcal E$ to the [b]ellipse[/b] $ \mathcal C$ is $ \frac{x_{0}\cdot x}{a^{2}}+\frac{y_{0}\cdot y}{b^{2}}=1$ (it is well-known).

Consider [b]two different points[/b] $ \{\begin{array}{cccc}A & (\ am\ ,\ bn) & , & m^{2}+n^{2}= 1\\ \\ B & (\ ap\ ,\ br) & , & p^{2}+r^{2}= 1\end{array}$ of the [b]ellipse[/b] $ \mathcal E$.

The [b]tangent lines[/b] to the [b]ellipse[/b] $ \mathcal E$ in these points are $ \{\begin{array}{c}AA\ :\ bm\cdot x+an\cdot y = ab\\ \\ BB\ :\ bp\cdot x+ar\cdot y = ab\end{array}$. Suppose that $ AA\cap BB\ne\emptyset$,

i.e. $ mr\ne np$. Since $ m^{2}+n^{2}= p^{2}+r^{2}= 1$, from the [b][u]C.B.S.[/u]'s identity[/b] results $ \boxed{\ mp+nr\ne 1\ }$. Denote  $ M\in AA\cap BB$.

Then the its [b]coordinates[/b] are $ x_{M}=\frac{a(r-n)}{mr-np}$ and $ y_{M}=\frac{b(m-p)}{mr-np}$. Observe that $ \boxed{\ AA\perp BB\Longleftrightarrow a^{2}\cdot nr+b^{2}\cdot mp = 0\ }\ .$

Therefore, $ M\in\mathcal C$ $ \Longleftrightarrow$ $ x^{2}_{M}+y^{2}_{M}= a^{2}+b^{2}$ $ \Longleftrightarrow$  $ a^{2}\cdot (r-n)^{2}+b^{2}\cdot (m-p)^{2}= (mr-np)^{2}\cdot (a^{2}+b^{2})$ $ \Longleftrightarrow$

$ a^{2}\cdot (r^{2}+n^{2}-2rn-m^{2}r^{2}-n^{2}p^{2}+2mnpr)+$ $ b^{2}\cdot (m^{2}+p^{2}-2mp-m^{2}r^{2}-n^{2}p^{2}+2mnpr) = 0$ $ \Longleftrightarrow$

$ a^{2}\cdot[r^{2}(1-m^{2})+n^{2}(1-p^{2})-2rn+2mnpr]+$ $ b^{2}[m^{2}(1-r^{2})+p^{2}(1-n^{2})-2mp+2mnpr] = 0$ $ \Longleftrightarrow$

$ 2a^{2}\cdot nr(mp+nr-1)+2b^{2}\cdot mp(mp+nr-1) = 0$ $ \Longleftrightarrow$ $ (a^{2}\cdot nr+b^{2}\cdot mp)\cdot(mp+nr-1) = 0$ $ \Longleftrightarrow$

$ a^{2}\cdot nr+b^{2}\cdot mp = 0$. [b]In conclusion[/b], $ \boxed{\ M\in\mathcal C\Longleftrightarrow a^{2}\cdot nr+b^{2}\cdot mp = 0\Longleftrightarrow AA\perp BB\ }$, q.e.d.

[b][u]Remark.[/u][/b] You can use the [b]trigonometrical coordinates[/b] for a point $ P\in\mathcal E$, i.e. $ x_{P}= a\cdot\cos\phi$ and $ y_{P}= b\cdot\sin\phi$, where $ \phi\in[0,\pi)$.

We can say equivalently that [b]the circle $ \mathcal C$ is the geometric locus of the points $ M$ from which the tangent lines to the ellipse $ \mathcal E$ are orthogonally ![/b][/color]

Thank your for your admiration, Virgil Nicula. :)

I will post my solution later.

kunny

Here is my solution.

(1) Let $ l: y = mx+n$ be the tangent line in question, $ C$ touches with $ l\Longleftrightarrow b^{2}x^{2}+a^{2}(mx+n)^{2}= a^{2}b^{2}$, i.e. quadratic equation $ (a^{2}m^{2}+b^{2})x^{2}+2a^{2}mnx+a^{2}(n^{2}-b^{2}) = 0$ has double root. By $ D/4 = a^{2}m^{2}n^{2}-a^{2}(a^{2}m^{2}+b^{2})(n^{2}-b^{2}) = b^{2}(a^{2}m^{2}+b^{2}-n^{2}) = 0$,
yielding $ n =\pm\sqrt{a^{2}m^{2}+b^{2}}$. Ans. $ l: y = mx\pm\sqrt{a^{2}m^{2}+b^{2}}$.

(2) The tangent line drawn from the point $ (s,\ t)$ on the given circle can be written as $ y = m(x-s)+t\Longleftrightarrow y = mx-(ms-t)$.Comparing the result of (1), we have $ n =-(ms-t)$. By $ n^{2}= a^{2}m^{2}+b^{2}$, we have $ (ms-t)^{2}= a^{2}m^{2}+b^{2}\Longleftrightarrow (a^{2}-s^{2})m^{2}+2stm+b^{2}-t^{2}= 0$. Let $ m_{1},\ m_{2}$ be the roots of the equation, for case 1: $ a^{2}\neq s^{2}\Longleftrightarrow s\neq\pm a$, we have $ m_{1}m_{2}=\frac{b^{2}-t^{2}}{a^{2}-s^{2}}$ since the point $ (s,\ t)$ are on the given circle, we have $ s^{2}+t^{2}= a^{2}+b^{2}\Longleftrightarrow b^{2}-t^{2}=-(a^{2}-s^{2})$, yielding $ m_{1}m_{2}=-1$. Then for case 2: $ a^{2}= s^{2}\Longleftrightarrow s =\pm a$, the tangent lines are $ x =\pm a,\ y =\pm b$, which completes the proof.

equation of tangent to the given ellipse in terms of its slope,m is

y=mx+sqrt((a^2)(m^2)+b^2)

Let any point on the circle be (rcosp,rsinp) where r=sqrt(a^2 + b^2).

substituting in eqn. of tangent we get a quadratic in m.Product of roots of this quadratic comes

out to be -1 which means that the two tangents so formed are perpendicular


Let $ABC$ be a triangle with sides $a, b, c$. Show that
  $\sum_{cyclic}{a^{13}} + \sum_{cyclic}{a^8b^5} + \sum_{cyclic} {a^5b^5c^3 \geq 2 \sum_{cyclic}{a^9b^4}}$

No one has any interest ;)
 [hide="Hint"]
 The inequality is equivalent to
 $\prod(b^2+c^2-a^2).(a^2+b^2+c^2)(a^5+b^5+c^5) \leq a^3b^3c^3(ab+bc+ca)^2$
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