algorithm search function quadratics geometry MATHCOUNTS percent

Sure; as long as you can explain the code, I am pretty fluent with TI Basic.

[quote="12markkram34"]I'm sure most of you have heard about coin problems - "In how many ways can X cents of change be made using pennies, nickles, dimes, and quarters?"

A few months back I posted an algorithm to find the answer rather quickly. Now that science fair is past, I have time for other things. I am fluent in the C++ programming language - I could quite easily make a program to solve this type of problem extremely quickly.

However, I am not fluent in the TI programming language. I am wondering if anyone here would be able to convert C++ code I post here into a TI program, for everyone to download and use.

What are your thoughts? I am reluctant to start programming until I can find someone who can convert the code (I have some free time but I don't want to spend any until I can be guaranteed success in creating a TI program).[/quote]

Wait, do you remember which thread? :)

The all-mighty search function reveals [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=581360964&t=236178]this.[/url]

[quote="12markkram34"]
I have found a way to solve the problem easily. I was in chemistry class today and started listing out the combinations for some amounts. I noticed a pattern in the amounts of combinations and formed the following conclusion-method.

How to solve problems of said type:

First find the greatest multiple of 5 that is less than or equal to the amount you are trying to make.

Now consider this:

All of the possible coin combinations fit into exactly one of these two categories:

a. Includes quarters

b. Does not include quarters

Let's say you are trying to make 80 cents. We can find the amount of combinations in each of the two categories, and then add them together to find the answer. If you start listing the combinations in category b, this is what you get (D represents the number of dimes, N represents the number of nickels, and P represents the number of pennies):

| D | N | P |
| 8 | 0 | 0 |
| 7 | 2 | 0 |
| 7 | 1 | 5 |
| 7 | 0 | 10 |
| 6 | 4 | 0 |
| 6 | 3 | 5 |
| 6 | 2 | 10 |
| 6 | 1 | 15 |
| 6 | 0 | 20 |

As you can see, there is 1 possible combination using no quarters and 8 dimes, 3 possible combinations using no quarters and 7 dimes, and 5 possible combinations using no quarters and 6 dimes. This pattern continues all the way to 17 possible combinations using no quarters and no dimes. Hmm...this really looks like a list of consecutive odd integers. Therefore, there must be 9^2 = 81 combinations in category b since 17 is the 9th positive odd integer!

We can expand on this idea. A goal of 90 cents would therefore have 100 combinations in category b, following the same logic. Therefore, to find the amount of combinations in category b for any amount ending in a zero, truncate the ones digit, add one to the remaining number, and then square the result.

But what if the goal ends in a 5? Listing the combinations for 85 shows this:

| D | N | P |
| 8 | 1 | 0 |
| 8 | 0 | 5 |
| 7 | 3 | 0 |
| 7 | 2 | 5 |
| 7 | 1 | 10 |
| 7 | 0 | 15 |
| 6 | 5 | 0 |
| 6 | 4 | 5 |
| 6 | 3 | 10 |
| 6 | 2 | 15 |
| 6 | 1 | 20 |
| 6 | 0 | 25 |

As you can see, there are 2 possible combinations using no quarters and 8 dimes, 4 possible combinations using no quarters and 7 dimes, and 6 possible combinations using no quarters and 6 dimes. This pattern continues all the way to 18 possible combinations using no quarters and no dimes. Hmm...this really looks like a list of consecutive even integers. Therefore, there must be 9^2 + 9 = 90 combinations in category b since 18 is the 9th positive even integer!

We can expand on this idea. A goal of 95 cents would therefore have 110 combinations in category b, following the same logic. Therefore, to find the amount of combinations in category b for any amount ending in a zero, truncate the ones digit, add one to the remaining number, square the result, and then add on the truncated number resulting from the second operation.

If this doesn't make sense, please post and I will try to explain my thinking better.

Now that we can find the number of combinations in category b for any goal amount, we need to find out how to find the amount of combinations in category a for any goal amount. Listing for 80 shows this (Q represents the number of quarters, D represents the number of dimes, N represents the number of nickels, and P represents the number of pennies):

| Q | D | N | P |
| 1 | 5 | 1 | 0 |
| 1 | 5 | 0 | 5 |
| 1 | 4 | 3 | 0 |
| 1 | 4 | 2 | 5 |
| 1 | 4 | 1 | 10 |
| 1 | 4 | 0 | 15 |

But wait! This appears to be listing the amount of combinations total for the number 55, which is 25 less than 80. We can easily find the number of combinations in category b for the number 55, but how will be find out for a? But wait...the amount of combinations in category a for 55 is the same as the amount of combinations in all for 30! WOW! That means that the amount of combinations in category a for 80 is equal to the amount of combinations in category b for 55, plus the number of combinations total for the number 30 (can be found by listing or using the formula)!

We can expand this idea into an algorithm! If the number was 85 and we wanted to find the number of combinations in category a, we would find the number of combinations in category b for 60 (85 - 20) and the number of combinations total for 35 (60 - 25) using the formula, and then sum the two values!

Using my method, the number of combinations total for any amount can be calculated! In fact, I could write a C++ or TI-84 program to do it instantly!

So what do you think of my method that I devised today in chemistry class? Think it is useful? Need me to explain it better? Just post here!
[/quote]

Here's a simple program for TI-84 I made a while back to simulate this problem. It's sort of messy, but whatever.

[code]
DelVar KClrHome
Prompt X
int(X/5-->A
For(Q,0,int(A/5
5A-25Q-->B
For(D,0,int(B/10
B-10D-->C
For(N,0,int(C/5
K+1-->K
End
End
End
Output(8,1,K
[/code]

As for listing, in addition to the techniques quoted above, one can use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1396323#1396323]the lemma mentioned in this proof[/url] of a formula I found for $ 10\le n\le49$ (so if one has $ n \equal{} 84$, reducing it to $ n \equal{} 80$ can help in finding patterns and other things), or if it is restricted to pennies, nickels, and dimes, one can use the formula including quarters, and then subtract the number of possibilities with a quarter (which would involve finding the number of ways without quarters for $ n \minus{} 25$), remembering that $ 0\le n\le4$ have $ 1$ way, and $ 5\le n\le9$ have $ 2$ ways (they don't satisfy the formula in the link above!).

[b]Edit 1:[/b] Fixed something.
[b]Edit 2:[/b] Optimized something.
[b]Edit 3:[/b] Realized I forgot a line. Sorry for totally failing this.

@math154;

Are you sure that program works?  I put it on my TI-84 Plus but when I enter a value for $ x$ such as $ 26$, it gives me $ 196$.  The number of possible ways to get $ \$0.26$ is $ 13$...not $ 196$.

Sure it works?  For $ \$0.25$, it outputs $ 12$, whereas it should be $ 13$.

It works now. :D

Thanks for the program!  :D

Nice math154!!

I wish my brain worked the same way. Make some programs, "download" them into my brain and pwn everyone.  Sadly, my brain doesn't work at all.

EDIT: Does this program find the possibilities for $ x$ using pennies, nickels, dimes, and quarters?  What about half-dollars, will it tell you that?

[quote="007math"]EDIT: Does this program find the possibilities for $ x$ using pennies, nickels, dimes, and quarters?  What about half-dollars, will it tell you that?[/quote]

It doesn't show with half-dollars, but with the basic code math154 provided, it should be quite simple.

OK guys, I'm just going to put all of these programs as well as some other random programs here because of boredom, and for future reference.
[hide="PND"]
[code]
DelVar KClrHome
Prompt X
int(X/5→A
For(D,0,int(A/2
5A-10D→B
For(N,0,int(B/5
K+1→K
End
End
Output(8,1,K
[/code]
[/hide]
[hide="PNDQ"]
[code]
DelVar KClrHome
Prompt X
int(X/5→A
For(Q,0,int(A/5
5A-25Q→B
For(D,0,int(B/10
B-10D→C
For(N,0,int(C/5
K+1→K
End
End
End
Output(8,1,K
[/code]
[/hide]
[hide="PNDQH"]
[code]
DelVar KClrHome
Prompt X
int(X/5→A
For(H,0,int(A/10
5A-50H→B
For(Q,0,int(B/5
5B-25Q→C
For(D,0,int(C/10
C-10D→E
For(N,0,int(E/5
K+1→K
End
End
End
End
Output(8,1,K
[/code]
[/hide]
[hide="Base Converter"]
Have fun typing!

[code]
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-->Str1 (if you don't have lowercase, learn a new alphabet and replace all the lowercase letters  :P , or just don't type them; you'll have less bases, though)
ClrHome
DelVar NInput "Number?",Str2
Input "Base?",M
length(Str2)-1→P
For(I,1,length(Str2
inString(Str1,sub(Str2,I,1→A
If A>M or not(A
Then
Disp "Base Input Error
Return
End
(A-1)M^P+N→N
P-1→P
End
Input "New Base?",B
DelVar P0→dim(L_1 (that's a zero behind the P)
Repeat (B-1)B^P≥N
P+1→P
End
Repeat P<0
int(N/B^P→D
N-DB^P→N
D→L_1(1+dim(L_1
P-1→P
End
"0→Str2
For(I,1,dim(L_1
Str2+sub(Str1,L_1(I)+1,1→Str2
End
Repeat sub(Str2,1,1)≠"0
sub(Str2,2,length(Str2)-1→Str2
End
ClrHome
DelVar L_1DelVar Str1Str2
DelVar Str2Output(1,1,Ans
Output(1,1,"
[/code]
[/hide]
[hide="Equation Solver (not that great, but pretty much works)"]
[code]
ClrHome
Prompt G (input some value for a "guess" at an answer--the calculator will find the closest solution to the guess; if you don't know, just put like 0)
Input "",Str1 (LHS)
Input "=",Str2 (RHS)
Str1→Y_1
Str2→Y_2
solve(Y_1-Y_2,X,G→X
DelVar Y_1DelVar Y_2Output(8,1,X
[/code]
[/hide]
[hide="Prime Factorizer"]
[code]
ClrHome
Prompt N
For(A,1,E3 (E is "2ND"+"," and it will be for the rest of the E's)
A→K
If fPart(N/2
E3→A
If not(fPart(N/2
N/2→N
End
If K-1
Disp {2,K-1
For(A,3,√(N),2
If A>√(N
Goto X
For(B,1,E3
B→K
If fPart(N/A
E3→B
If not(fPart(N/A
N/A→N
End
If K-1
Disp {A,K-1
End
Lbl X
If N-1
Disp {N,1
Pause
ClrHome
Output(1,1,"
[/code]
Basically it will output in pairs $ \{p,k\}$, which corresponds to $ p^k$ as the largest prime power for each $ p$.
[/hide]
[hide="Shoelace"]
[code]
DelVar SClrHome
Prompt N
0→dim(L_1
0→dim(L_2
For(A,1,N
Prompt X,Y
X→L_1(1+dim(L_1
Y→L_2(1+dim(L_2
End
For(A,2,N
S+L_1(A-1)L_2(A)-L_1(A)L_2(A-1→S
End
.5abs(S+L_1(N)L_2(1)-L_2(N)L_1(1 (it's the command "abs(")
Disp Ans,Ans[Frac] (in place of the [Frac], put the thing from "Math:1")
Output(1,1,"
[/code]
[/hide]

[b]Edit:[/b] Fixed an error.

Prime factorizer: I created a program that does the same thing except it outputs in the format: {p,q,r...} for all prime factors.

So 34 -> {2,17}
and 1024 -> {2,2,2,2,2,2,2,2,2,2}
which is a pain in the neck to count out the number of twos.
However it is a bit more concise in numbers with lots of distinct prime factors like 510510.

well we can just make it an argument of the program

[quote]coins(x)
Func
Local 1,b,c,q,d,n,y
0 -> y
0 -> a
0 -> b
0 -> c
iPart(x/5) -> a
For q,0,iPart(a/5)
5a-25q -> b
For d,0,iPart(b/10)
b-10d -> c
For n,0,iPart(c/5)
y+1 -> y
EndFor
EndFor
EndFor
y
EndFunc
[/quote]
The -> is the STO store
Theres probably some unnecessary stuff in there though...

My pathetic factoring program:
[code]
<<FOR THE TI-83/84. CODE BEGINS NOW>>

Input "NUMBER:",A
{1}->L1
While fPart(A/2)=0 and A/2=/=1 <- =/= is not equals sign
A/2->A
augment(L1,{2})->L1
End
1->t <- Here and henceforth t = theta, ok?
While t+2<=sqrt(A) <- <= is less than or equal to sign
2+t->t
If fPart(A/t)=0
Then
While fPart(A/t)=0 and A/t=/=1
A/t->A
augment(L1,{t})->L1
End
End
End
If L1(1)=1
A->L1(1
SortA(L1
L1->L1
[/code]

Why L1 -> L1? Type the program and you'll see.

Try this shoelace:

[code]DelVar S
DelVar T
DelVar A
DelVar N
DelVar C
DelVar X
DelVar Y
DelVar L1
DelVar L2
ClrHome
Disp "N=NO. OF POINTS
Disp "
Disp "ENTER POINTS
Disp "IN ORDER
Disp "
Prompt N
ClrHome
For(C,1,N
Disp "POINT NUMBER:
Disp C
Prompt X
X->L1(C
Prompt Y
Y->L2(C
ClrHome
End
L1(N)L2(1)->S
L1(1)L2(N)->T
For(C,2,N
S+L1(C-1)L2(C)->S
T+L1(C)L2(C-1)->T
End
abs(S-T)/2->A
Disp "AREA=
Disp A
DelVar L1
DelVar L2[/code]

You can remove some of the extraneous stuff, but this works pretty well.

Is it just me, or did you forget a line of code, james?

Cuz in my L1, I got [b]a lot[/b] of twos.

Also, here is my (very basic!) FACTOR program.  I'm gonna tell you up front though, it sucks:

[code]Input A
0->B
Lbl 1
B+1->B
If B>sqrtA
Then
Stop
Else
If A/B=int(A/B)
Then
Disp B,A/B
Pause
End
Goto 1[/code]

Oh, for your factoring program, the Lbl/Goto loop can be replaced with a "While 1" or "Repeat 0" loop to improve speed.

For the program, I really don't see anything wrong...

can you tell me output for: A=1,2,3,4,5,6,7,8,9,10?

@math154

[code]DelVar KClrHome
Prompt X
int(X/5→A
For(H,0,int(A/10
5A-50H→B
For(Q,0,int(B/5
5B-25Q→C
For(D,0,int(C/10
C-10D→E
For(N,0,int(E/5
K+1→K
End
End
End
End
Output(8,1,K [/code]

The one for PNDQH doesn't seem right.  I put in $ 49$ and got 2015 while putting in 49 in PNDQ gets 39.

Darn, epic fail. :( Anyway, here are the updated versions of the programs:

[hide="PND"]
[code]
DelVar KClrHome
Prompt X
int(X/5→A
For(D,0,int(A/2
5A-10D→B
K+1+int(B/5→K
End
Output(8,1,K
[/code]
[/hide]
[hide="PNDQ"]
[code]
DelVar KClrHome
Prompt X
int(X/5→A
For(Q,0,int(A/5
5A-25Q→B
For(D,0,int(B/10
B-10D→C
K+1+int(C/5→K
End
End
Output(8,1,K
[/code]
[/hide]
[hide="PNDQH"]
[code]
DelVar KClrHome
Prompt X
int(X/5→A
For(H,0,int(A/10
5A-50H→B
For(Q,0,int(B/5
B-25Q→C
For(D,0,int(C/10
C-10D→E
K+1+int(E/5→K
End
End
End
Output(8,1,K
[/code]
[/hide]
[hide="Base Converter"]
Have fun typing!

[code]
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz→Str1 (if you don't have lowercase, learn a new alphabet and replace all the lowercase letters  :P , or just don't type them; you'll have less bases, though)
ClrHome
DelVar NInput "Number?",Str2
Input "Base?",M
length(Str2)-1→P
For(I,1,length(Str2
inString(Str1,sub(Str2,I,1→A
If A>M or not(A
Then
Disp "Base Input Error
Return
End
(A-1)M^P+N→N
P-1→P
End
Input "New Base?",B
DelVar P0→dim(L_1 (that's a zero behind the P)
Repeat (B-1)B^P≥N
P+1→P
End
Repeat P<0
int(N/B^P→D
N-DB^P→N
D→L_1(1+dim(L_1
P-1→P
End
"0→Str2
For(I,1,dim(L_1
Str2+sub(Str1,L_1(I)+1,1→Str2
End
Repeat sub(Str2,1,1)≠"0
sub(Str2,2,length(Str2)-1→Str2
End
ClrHome
DelVar L_1DelVar Str1Str2
DelVar Str2Output(1,1,Ans
Output(1,1,"
[/code]
[/hide]
[hide="Equation Solver (not that great, but pretty much works)"]
[code]
ClrHome
Prompt G (input some value for a "guess" at an answer--the calculator will find the closest solution to the guess; if you don't know, just put like 0)
Input "",Str1 (LHS)
Input "=",Str2 (RHS)
Str1→Y_1
Str2→Y_2
solve(Y_1-Y_2,X,G→X
DelVar Y_1DelVar Y_2Output(8,1,X
[/code]
[/hide]
[hide="Prime Factorizer"]
[code]
ClrHome
Prompt N
For(A,1,E3 (E is "2ND"+"," and it will be for the rest of the E's)
A→K
If fPart(N/2
E3→A
If not(fPart(N/2
N/2→N
End
If K-1
Disp {2,K-1
For(A,3,√(N),2
If A>√(N
Goto X
For(B,1,E3
B→K
If fPart(N/A
E3→B
If not(fPart(N/A
N/A→N
End
If K-1
Disp {A,K-1
End
Lbl X
If N-1
Disp {N,1
Pause
ClrHome
Output(1,1,"
[/code]
Basically it will output in pairs $ \{p,k\}$, which corresponds to $ p^k$ as the largest prime power for each $ p$.
[/hide]
[hide="Shoelace"]
[code]
DelVar SClrHome
Prompt N
0→dim(L_1
0→dim(L_2
For(A,1,N
Prompt X,Y
X→L_1(1+dim(L_1
Y→L_2(1+dim(L_2
End
For(A,2,N
S+L_1(A-1)L_2(A)-L_1(A)L_2(A-1→S
End
.5abs(S+L_1(N)L_2(1)-L_2(N)L_1(1 (it's the command "abs(")
Disp Ans,Ans[Frac] (in place of the [Frac], put the thing from "Math:1")
Output(1,1,"
[/code]
[/hide]

Some more programs to add to the collection. I made these while I was doing really tedious repetitive problems.
[hide="programs"]
Simpson's Rule is a more accurate way to approximate areas.
[code]
PROGRAM:SIMPSON
:0->K
:Prompt A,B,N
:(B-A)/N->I
:For(J,A+I, B-2I,2I
:K+4Y1(J)+2Y1(J+I)->K
:End
:K+Y1(A)+4Y1(B-I)+Y1(B)->K
:Disp KI/3

Percent 100 means right riemann
Percent 50 means middle riemann
Percent 0 means left riemann

PROGRAM:RIEMANN
:Disp "1. START","2. END","3. NUMBER RECT","4. PERCENT
:Prompt A,B,N,P
:(B-A)/N->D
:A+PD/100->X
:0->S
:For(K,1,N
:Y1+S->S
:X+D->X
:End
:Disp SD

PROGRAM:TRAPRULE
:0-K
:Prompt A,B,N
:(B-A)/N->I
:For(X,A+I,B-I,I
:Y1+K->K
:End
:(Y1(B)+Y1(A))/2+K->K
:Disp KI[/code][/hide]

[quote="james4l"]can you tell me output for: A=1,2,3,4,5,6,7,8,9,10?[/quote]

For every one of them, there is a {1} and an infinite output of {2}'s...

[quote="james4l"]Oh, for your factoring program, the Lbl/Goto loop can be replaced with a "While 1" or "Repeat 0" loop to improve speed. [/quote]

I've never used that.  Help?

[quote="ernie"]Help?[/quote]

Yeah. Use mine. :P

please do not revive old forums, and if you do, post something productive.  :D  :D  :D doesn't qualify as productive.

You can make any programs to your convenience for MC. Of course you can only use calc for target and team round.

one question, srry to post again, whats str on ti-84 plus??

str = string

Since you can't use a calculator on the sprint round, you should learn my algorithm (it's quoted on the first page).


Find a vector of length 1 which points in the exact opposite direction of $ p \equal{} (4, \minus{}2, 5)$

Well exact opposite direction would just be the negative of each entry, so
$ \textbf{v}\equal{}(\minus{}4,2,\minus{}5)$.
Now we need to normalize it, so we have
$ \textbf{u}\equal{}\frac{\textbf{v}}{|\textbf{v}|}\equal{}\frac{(\minus{}4,2,\minus{}5)}{\sqrt{(\minus{}4)^{2}\plus{}(2)^{2}\plus{}(\minus{}5)^{2}}}\equal{}\frac{(\minus{}4,2,\minus{}5)}{\sqrt{45}}\equal{}\minus{}\frac{4}{\sqrt{45}}\textbf{i}\plus{}\frac{2}{\sqrt{45}}\textbf{j}\minus{}\frac{5}{\sqrt{45}}\textbf{k}$
And that is your answer.

What's $ i, j$ and $ k$, is it any real number?

Nope, standard unit vectors.  $ \textbf{i}\equal{}(1,0,0)$, $ \textbf{j}\equal{}(0,1,0)$, $ \textbf{k}\equal{}(0,0,1)$.


Let $a;b;c$ be real numbers satisfying $1 \leq a;b;c \leq 2$.Prove that:
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \leq 10$

If fix $b,c$ then $\text{LHS}$ is convex for $a$, so we need only prove with $a,b,c\in\{1,2\}$. This is easy!

Similary, find maximum of $(x_{1}+x_{2}+...+x_{n})(\frac{1}{x_{1}}+...+\frac{1}{x_{n}})$, where $x_{i}\in [p,q]$ and $0<p<q$ be given.


Rationalize the denominator of the fraction
       $ \frac{1}{\sqrt 2 \plus{} \sqrt [3]{3} \plus{} \sqrt [5]{5}}$

Do you have any particular reason to suspect that the answer is nice?  The generic answer is to multiply the numerator and denominator by

$ \prod_{i\equal{}0}^{1} \prod_{j\equal{}0}^{2} \prod_{k\equal{}0}^{4} \left( (\minus{}1)^i \sqrt{2} \plus{} \zeta_3^j \sqrt[3]{3} \plus{} \zeta_5^k \sqrt[5]{5} \right)$

where $ \zeta_n \equal{} e^{ \frac{2\pi i}{n} }$ and $ (i, j, k) \neq (0, 0, 0)$, and generically you won't be able to do any better.


function
combinatorics proposed
combinatorics

How many permutations of $ n$ elements, when written in cycle notation, contain exactly $ k$ $ 2$-cycles?  Exactly $ k$ $ r$-cycles?

How many permutations $ \pi$ of $ n$ elements have the property that $ \pi^r$ has exactly $ k$ fixed points?

In the case $ r \equal{} 1$ we obtain the definition of the [url=http://en.wikipedia.org/wiki/Rencontres_numbers]rencontres numbers[/url] in both problems.

The exponential generating function for the first problem is at [url=http://www.research.att.com/~njas/sequences/A114320]A114320[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  That sequence is the case $ r \equal{} 2$, i.e., the triangle enumerating permutations by number of two-cycles.  It gives the general generating function (exponential in both variables, I believe) $ F(x, y) \equal{} \frac{1}{1 \minus{} x} e^{\frac{(y \minus{} 1)x^r}{r}}$.  That's a simple enough form that one can probably justify it directly.

The second problem doesn't seem to have an entry in the OEIS, though it's related in a rather natural way to the first.  I haven't really tinkered with it, yet, but I would think it should have a nice generating function.  (E.g., in the case $ r \equal{} 2$ we have the involutions running down the diagonal, and that's promising.)

I'm reasonably certain I have the correct EGF in both cases; it's exponential in $ x$ but not in $ y$.  I'm interested in combinatorial approaches :)


trigonometry
function
calculus
derivative
integration
real analysis
limit

Let $f\left(x\right)=\cos^{\frac{7}{3}}x$.  Without calculating the Fourier coefficients of $f$, show that $f$ can be represented by a uniformly convergent Fourier series:

\[f\left(x\right)=\sum_{n\in\mathbb{Z}}a_{n}e^{2\pi inx}.\]

$f\in C^{2}(\mathbb{T}).$ (It's fractionally smoother than that, but we don't really need the fraction.)

This implies that its Fourier coeffiecens $a_{n}=\widehat{f}(n)=O(n^{-2})$ as $|n|\to\infty.$ (In fact, $f\in C^{2}\implies \widehat{f}(n)=o(n^{-2})$ but that's also overkill.) Hence the Fourier series converges absolutely and uniformly by the Weierstrass M-test.

A sharper challenge would proving uniform convergence for the Fourier series of $f(x)=\cos^{\frac43}x.$

Thank you for the reply.  I understand what $f\in C^{2}$ means, but what does $f\in C^{2}\left(\mathbb{T}\right)$ mean?  Also, how does that imply that $a_{n}\in\mathcal{O}\left(n^{_}{-2}\right)$?

I think prof. Kent Merryfield uses $\mathbb{T}$ for torus

http://en.wikipedia.org/wiki/Torus#The_n-dimensional_torus

Yes, that is what I was using the symbol $\mathbb{T}$ for. Another way of saying $f\in C^{2}(\mathbb{T})$ is to say that $f$ is a periodic function with two continuous derivatives. (And the derivatives are themselves automatically periodic.)

However, now that I take a closer look at amcavoy's initial post, I see that there is a serious problem with it. His summation is the Fourier series expansion of a function that is periodic of period $1.$ But $f(x)=\cos^{7/3}x$ is not periodic of period $1.$ Rather, it is periodic of period $2\pi.$ If we try to extract a portion of the function over an interval of length $1$ and then extend that to be periodic, then for most choices we obtain a discontinuous function, and the Fourier series of a discontinuous function cannot converge uniformly. After all, the uniform limit of continuous functions is continuous.

So we'll assume the statement needs to be altered to match the periods. We could resolve that in either direction; I'll go with period $1.$ Let $f(x)=\cos^{7/3}(2\pi x).$

Define $\widehat{f}(n)=\int_{0}^{1}f(x)e^{-2\pi inx}\,dx.$

Then for sufficiently regular $f$ (which this is) we can write

$f(x)=\sum_{n=-\infty}^{\infty}\widehat{f}(n)e^{2\pi inx}.$

We can show that for differentiable $f$ we have $\widehat{f'}(n)=2\pi in\widehat{f}(n).$ We can extend that recursively, obtaining $\widehat{f''}(n)=-4\pi^{2}n^{2}\widehat{f}(n)$ or

$\widehat{f}(n)=-\frac{\widehat{f''}(n)}{4\pi^{2}n^{2}}.$

If $f''$ is continuous, then at the very least $f''$ is integrable and the Riemann-Lebesgue Lemma applies to it. Hence $\lim_{|n|\to\infty}\widehat{f''}(n)=0.$ Following up on that with the above identity, we get that $\widehat{f}(n)=o(n^{-2})$ as $|n|\to\infty.$


Find all continuous functions $ f: \mathbb{R}\to\mathbb{R}$ for which
\[ f(xy)=f\left(\frac{x^{2}+y^{2}}{2}\right)-f\left(\frac{x^{2}-y^{2}}{2}\right), \quad \forall x,y \in \mathbb R.\]

[quote="cnyd"]$ f: \mathbb{R}\mapsto\mathbb{R}$ and $ f$ is continuous
$ f(xy) \equal{} f(\frac {x^{2} \plus{} y^{2}}{2}) \minus{} f(\frac {x^{2} \minus{} y^{2}}{2})$[/quote]

[hide="My solution"]
Let $ P(x,y)$ be the assertion $ f(xy) \equal{} f(\frac {x^2 \plus{} y^2}2) \minus{} f(\frac {x^2 \minus{} y^2}2)$

$ P(0,0)$ $ \implies$ $ f(0) \equal{} 0$
$ P(0,\sqrt {2|x|})$ $ \implies$ $ f(|x|) \equal{} f( \minus{} |x|)$ $ \implies$ $ f(x) \equal{} f( \minus{} x)$

For $ x\geq 0$, let us define $ g(x) \equal{} f(\sqrt x)$. $ g(x)$ is continuous. Then, for $ x,y\geq 0$, $ P(x,y)$ becomes : $ g(x^2y^2) \equal{} g(\frac {x^4 \plus{} y^4}4 \plus{} \frac {x^2y^2}2) \minus{} g(\frac {x^4 \plus{} y^4}4 \minus{} \frac {x^2y^2}2)$

For $ u,v\geq 0$, let $ x \equal{} \sqrt {\sqrt {u \plus{} v} \plus{} \sqrt v}$ and $ y \equal{} \sqrt {\sqrt {u \plus{} v} \minus{} \sqrt v}$ and insert these values in the above equation. We get : $ g(u) \equal{} g(u \plus{} v) \minus{} g(v)$

So, since $ g(x)$ is continuous,  $ g(x) \equal{} ax$ $ \forall x\geq 0$ and so $ f(x) \equal{} ax^2$ $ \forall x\geq 0$ and, since $ f(x) \equal{} f( \minus{} x)$ : $ \boxed{f(x) \equal{} ax^2}$ $ \forall x$ and it's immediate to check that this mandatory form indeed is a solution.
[/hide]


Prove $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\geq{abc(a \plus{} b \plus{} c) \plus{} abd(a \plus{} b \plus{} d) \plus{} acd(a \plus{} c \plus{} d) \plus{} bcd(b \plus{} c \plus{} d) \plus{} 3abcd}$

Sorry, this problem is wrong. I got it confused with another one I made up. :blush:

It should be:

Prove $ \frac{3}{2}(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4)\geq{a^2b^2\plus{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2\plus{}c^2d^2}$

It's too easy....isn't it? :lol:
Using AM-GM we have:a^4+b^4>=2a^2b^2
                               a^4+c^4>=2a^2c^2
                               a^4+d^4>=2a^2d^2
                               b^4+c^4>=2b^2c^2
                               b^4+d^4>=2b^2d^2
                               c^4+d^4>=2c^2d^2
Summing all of these inequalities we get: 3(a^4+b^4+c^4)>=2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2, which is equivalent to the original inequality(Q.E.D)
We can solve the generalized problem by the same method:
(n-1)/2(a^4+b^4+c^4+.....+n^4)>=sigma(cyclic)(a^2b^2) :wink:

[quote="Yongyi781"]$ \text{\LaTeX}$ifying:

[b]AM-GM Bashing:[/b]
\begin{align*} a^4 + b^4 & \geq 2a^2b^2 \\
a^4 + c^4 & \geq 2a^2c^2 \\
a^4 + d^4 & \geq 2a^2d^2 \\
b^4 + c^4 & \geq 2b^2c^2 \\
b^4 + d^4 & \geq 2b^2d^2 \\
c^4 + d^4 & \geq 2c^2d^2 \end{align*}
[b]Adding up:[/b]
\[ 3(a^4 + b^4 + c^4 + d^4) \geq 2(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2) \\
\frac 32 (a^4 + b^4 + c^4 + d^4) \geq a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2
\]
Which is what we wanted to prove.

We can solve the generalized problem by the same method.

$ \frac {n - 1}2\sum_{i = 0}^n a_i^4 \geq \sum_{\text{sym}} a_i^2a_j^2$[/quote]
Thank you very much for your help, Yongyi781! But the RHS of the generalized of this problem should be sigma cyclic, not symmetric :wink:

[quote="Yongyi781"]$ \sum_{\text{cyc}} a^2b^2 \equal{} a^2b^2 \plus{} b^2c^2 \plus{} c^2d^2 \plus{} d^2a^2$
$ \sum_{\text{sym}} a^2b^2 \equal{} a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2$

Which one do you want?[/quote]
I want the 2nd :oops: .I'm so dumb!

So solution to this problem would be:

$ a^4\plus{}b^4\equal{}\frac{2a^4\plus{}2b^4}{2}\ge\sqrt{2a^4\cdot2b^4}\equal{}2a^2b^2$

$ a^4\plus{}c^4\equal{}\frac{2a^4\plus{}2c^4}{2}\ge\sqrt{2a^4\cdot2c^4}\equal{}2a^2c^2$

...

$ c^4\plus{}d^4\equal{}\frac{2c^4\plus{}2d^4}{2}\ge\sqrt{2c^4\cdot2d^4}\equal{}2c^2d^2$

:)


Please solve the following:

Prove that: (sinx + acosx)(sinx +bcosx)  <=  1 + (a+b/2) 2


trigonometry
ratio
geometry
function
symmetry
LaTeX
power of a point

You have a circle with AT as a diameter of the circle, and with chord CD such that AT and CD are parallel. Let AC and TD intersect at point E, and $\angle AED = a$. Express the ratio of the area of $\triangle CDE$ to the area of $\triangle ATE$ in terms of a single trigonometric function of $a$.

No takers?
[hide="First step"]Drop a perpendicular from E through AT to the other side of the circle. Then use the power of a point theorem and work with the trig functions of a/2.[/hide]

obviously you have that the ratio is (DC/AB)^2 by similar triangles...so like easy angle chasing gives arc DC=180-2a...so you should be able to find a chord length in terms of the radius and the arc using trig....that gives 2Rsin 90-a...we have AB=2R...so we have cos^2 a

getting ready for naperville north going to totally own at state again haha

u guys are so going down in the junoir division though... :P

for the 2005 contest [taken as practice] we totally got like 8 questions more than first place that year

[quote="mathperson07"]is it (cos a)^2, its just two similar triangles,
and thers the fact that angle TDA is 90 degrees,
and that you can use 1/2*a*b*sinC for the area of a triangle...

I didnt use the Bictor717 hint of power of the point and a/2..but Im pretty sure that will work too.[/quote]

That's a method that didn't occur to me. Do you mind filling out the details? I can't see how that area formula leads to your answer, which, by the way, is the same as mine.

[hide="Solution"]The perpendicular hits the center of the circle, by a symmetry argument. Let it intersect the circle at $R$ and $S$, with $ER<ES$. Call the length of the radius of the circle $r$.
$ER=r\cot{(a/2)}-r$
$ES=r\cot{(a/2)}+r$
$\sin{(a/2)}=r/ET$
By PPT, $(ED)(ET)=(ER)(ES)$.
\begin{eqnarray*}(ED)(ET)&=&r^{2}(\cot{(a/2)}+1)(\cot{(a/2)}-1)\\ \frac{ED}{ET}&=&\frac{r^{2}}{ET^{2}}(\cot^{2}{(a/2)}-1)\\ &=&\sin^{2}{(a/2)}(\cot^{2}{(a/2)}-1)\\ &=&\cos^{2}{(a/2)}-\sin^{2}{(a/2)}\\ \frac{ED}{ET}&=&\cos a\end{eqnarray*}
The ratio of the areas of $\triangle CDE$ to $\triangle ATE$ is the square of the ratio of the lengths of corresponding sides $ED$ and $ET$. So the answer we want is $\cos^{2}a$.[/hide]

alright so the area of the two ratios is just
(DE)(EC)(sin180-a)(1/2)/(AE)(ET)(sin180-a)(1/2)
stuff cancel so (DE)(EC)/(AE)(ET)

since TDA and ACT (angles) are both right angles, we have that ED/AE=cos a and CE/TE=cos a

so DE=(AE)(cos a)
and EC=(ET)(cos a)

plug them back in and everything cancles and we are left with (cos a)^2

I have a question: how do you get those neatly typed things bolded?

Wow that was straightforward. :blush: I'm not sure why you had $\sin{(180-a)}$, when the diagram suggests the equivalent $\sin a$, though.

Use $\LaTeX$. There's a tab on the left that leads to a guide. Basically, you put your math in dollar sign tags. Please, please preview your work before submitting because it may turn out looking very unsightly, especially when you're unexperienced. To see the code I used, hover your mouse over the pretty math, or more easily, quote the post (but don't actually submit the quoted post).


Prove that in the set $ S\equal{}\{2008,2009,...,4200\}$ there are $ 125$ elements such that not any $ 3$ of them are in an arithmetic progression.

This problem is quite similar to IMO 1983/5. And here is the solution:
Denote $ t_i = a{}_1a{}_2...a_8_{(3)}$, while $ a_i$ takes only two values 0 or 1 and $ a_1 = 1$. And we choose numbers $ t_1 < t_2 < ...$.

 Clearly no three of which forms an arithmetic progresstion, and we also have
$ t_1 = 10000000_{(3)} = 2187 > 2008$
$ t_{128} = 11111111_{(3)} = 3280 < 4200$
So our proof is completed! Merry Christmas! (perhaps a bit late  :) )


number theory
greatest common divisor
number theory unsolved

Prove that for any $n\geq 2$ there exists a set $A_{n}$ of $n$ positive integers such that for any two distinct elements $i,j$, we have $|i-j|$ divides $i^{2}+j^{2}$.

[quote="probability1.01"]Do you think he meant them to have gcd 1?[/quote]

It doesn't say in the problem statement...

Maybe you can try that case.  :D


It's obvious that: $ 1 \minus{} c\frac {a \plus{} b}{c^2 \plus{} ab} \equal{} \frac {(a \minus{} c)(b \minus{} c)}{c^2 \plus{} ab}$.
Therefore, the inequality is equivalent to

$ \frac {a \plus{} b \plus{} c}{\sqrt [3]{a^2b^2c^2}} \minus{} \frac {ab \plus{} bc \plus{} ca}{abc} \plus{} \sum_{\textrm{cyc}}\frac {(a \minus{} c)(b \minus{} c)}{c^3 \plus{} abc}\ge 0$.

If we assume $ abc \equal{} 1$, then the inequality turns into
$ (a \plus{} b \plus{} c) \minus{} (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} \sum_{\textrm{cyc}}\frac {(a \minus{} c)(b \minus{} c)}{c^3 \plus{} 1}\ge 0$, which I solved with [i]Lagrange multipliers [/i].

However, something not so boring and long, that follow by the last is :

$ \sum_{\textrm{cyc}}\frac {(a \minus{} c)(b \minus{} c)}{c^3 \plus{} 1}\ge 0$ holds according to the Vornicu inequality.

Let $ a\le b\le c$. Then $ ab \equal{} \frac {1}{c}\le 1$ since $ c\ge 1$.
$ \Rightarrow a \plus{} b \plus{} c\ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$ $ \Leftrightarrow ab(a \plus{} b) \plus{} 1\ge a \plus{} b \plus{} (ab)^2 \Leftrightarrow (1 \minus{} a)(1 \minus{} b)\ge 0$.

So we conclude it's enough to see $ a\le 1\le b\le c$, which doesn't help on the whole :roll:

Nevertheless, the problem is really good and I will wait for more pleasant proof of it...  again :D
By the way, [b]nguoivn[/b], is this problem inspired by [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1396229#1396229[/url]  :roll:

[quote="nguoivn"]Given $ a, b, c > 0.$ Prove that: $ \frac {a \plus{} b \plus{} c}{\sqrt [3]{a^2b^2c^2}} \geq\ \frac {a \plus{} b}{ab \plus{} c^2} \plus{} \frac {b \plus{} c}{bc \plus{} a^2} \plus{} \frac {c \plus{} a}{ca \plus{} b^2}$
 :)
I found an ugly proof for it (by MIS)  :oops:[/quote]
It is posted here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=152419[/url] more than [b]one and a half year ago[/b] - and I do have a nice solution  :lol:

[quote="BG Yoda"]
By the way, [b]nguoivn[/b], is this problem inspired by [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1396229#1396229[/url]  :roll:[/quote]
No, as Duc said, it's his old problem (tranquoc- my friend isnt's the first people find it)  :)
[quote="BG Yoda"]
Nevertheless, the problem is really good and I will wait for more pleasant proof of it...  again :D
[/quote]
Yes, I agree with you :P
-------------------
@phamhuuduc: I think the hint of you is similar than my hint and can_hang's hint.
The last ineq we obtain is:
$ \frac {(a \plus{} b \plus{} c)^3}{(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {a \plus{} b}{ab \plus{} c^2} \plus{} \frac {b \plus{} c}{bc \plus{} a^2} \plus{} \frac {c \plus{} a}{ca \plus{} bc^2}$
Am I right?
But I have to prove it by MIS:
( $ \frac {\sum\ a^3 \plus{} \sum\ ab(a \plus{} b) \plus{} 18abc}{(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {a \plus{} b}{ab \plus{} c^2} \plus{} \frac {b \plus{} c}{bc \plus{} a^2} \plus{} \frac {c \plus{} a}{ca \plus{} bc^2}$  )
Of course, can_hang had a nicer proof  :)


$ \mathop{\lim}\limits_{x\rightarrow2^{ \minus{} }}\displaystyle{\frac {{x \minus{} 3}}{{x^2 \minus{} 4}}} \equal{} \mathop{\lim}\limits_{x\rightarrow2^{ \minus{} }}\displaystyle{\frac {{x \minus{} 3}}{{(x \plus{} 2)(x \minus{} 2)}}} \equal{} \mathop{\lim}\limits_{x\rightarrow2^{ \minus{} }}{\frac {x \minus{} 3}{x \plus{} 2}}\displaystyle\cdot\mathop{\lim}\limits_{x\rightarrow2^{ \minus{} }}{\frac {1}{{x \minus{} 2}}} \equal{}$
$ {\frac {2 \minus{} 3}{2 \plus{} 2}}\cdot( \minus{} \infty) \equal{} \plus{} \infty$.
$ \mathop{\lim}\limits_{x\rightarrow2^{ \plus{} }}\displaystyle{\frac {{x \minus{} 3}}{{x^2 \minus{} 4}}} \equal{} \mathop{\lim}\limits_{x\rightarrow2^{ \plus{} }}\displaystyle{\frac {{x \minus{} 3}}{{(x \plus{} 2)(x \minus{} 2)}}} \equal{} \mathop{\lim}\limits_{x\rightarrow2^{ \plus{} }}{\frac {x \minus{} 3}{x \plus{} 2}}\displaystyle\cdot\mathop{\lim}\limits_{x\rightarrow2^{ \plus{} }}{\frac {1}{{x \minus{} 2}}} \equal{}$
$ {\frac {2 \minus{} 3}{2 \plus{} 2}}\cdot( \plus{} \infty) \equal{} \minus{} \infty$.
So the $ \mathop{\lim}\limits_{x\rightarrow2}\displaystyle{\frac {{x \minus{} 3}}{{x^2 \minus{} 4}}}$ does not exists.
ADVICE: Beyond calculation you must grasp the idea of that calculation!
QUESTION: By $ \mathop{\lim}\limits_{x\nearrow2}$ you mean the left-side limit $ \mathop{\lim}\limits_{x\rightarrow2^{ \minus{} }}$ ? I never see it before. Which school (country) use that symbolism?


Support
trigonometry
function
calculus
derivative
calculus computations

Hello!

The question asks that:

$AX + BY = C$ and A,B,C are constants.

We are asked to transform the equation above into a polar equation in the form of $r = f(\theta)$

I would appreciate your help and support

[quote="kunny"]Let $X=r\cos \theta ,\ Y=r\sin \theta,$ plugg this into the given equation we have $r(A\cos \theta +B\sin \theta)=C,$yielding $r=\frac{C}{A\cos \theta +B\sin \theta}.$[/quote]
  But do I need to investigate all of the cases. For instance, what's happen when $A\cos \theta +B\sin \theta =0$ How can i show it?

I thought this question would come out. I didn't dare to say about this.

$X=r\cos \theta ,\ Y=r\sin \theta$ are to hold for arbiterary values of $\theta.$

If the equation of $A\cos \theta +B\sin \theta =0$ holds for arbiterary values of $\theta,$ then we have $A=B=0.$

however in this case the equation $Ax+By=C$ can't be the equation of a line.

By the way can you understand why $A\cos \theta +B\sin \theta =0$ holds for arbiterary values of $\theta,$ then we have $A=B=0,$ Gibbenergy ?

[quote="kunny"]
By the way can you understand why $A\cos \theta +B\sin \theta =0$ holds for arbiterary values of $\theta,$ then we have $A=B=0,$ Gibbenergy ?[/quote]

  Well, I don't know it. But my intuition idea is that I use $tan(\frac{\theta}{2})$ and write the equation above in term of $tan(\frac{\theta}{2})$ so after some calculations, it will turn into the quadacric form. Since this one is hold for abitrary $\theta$ so all of its coefficent must be zeroes , yeild that $A=B=0$ But I am not sure my approach is correct or not.

  I am looking forward to see your help, Kunny !

Since $A\cos \theta +B\sin \theta =0$ holds for arbiterary values of $\theta,$ so this holds especially for $\theta =0,\ \theta =\frac{\pi}{2}.$ Plugg these values of $\theta$ gives the equations $A=0$ and $B=0.$ Conversely when $A,B$ take value respectively, we can confirm that $A\cos \theta +B\sin \theta =0$ holds for arbiterary values of $\theta.$

Alternative answer: Rewrite $A\cos \theta +B\sin \theta =0$ as $\sqrt{A^2+B^2}\cos (\theta -\alpha)=0$ for some constant numbers $\alpha.$, yielding $A=B=0.$

Comment:

If the identy with respect to $x,y$, $ax+by=0$ holds for arbiterary real numbers $x,y$, we have $a=b=0,$ but

in case of $ax^2+bxy+cy^2=0,$ we can't conclude that $a=b=c=0$ imediately. Sounds strange!

[quote="kunny"]
in case of $ax^2+bxy+cy^2=0,$ we can't conclude that $a=b=c=0$ imediately. Sounds strange![/quote]

I inspired from your solutions above. Since it is hold for abitrary number $x,y$
Let $x=0$ $y=1$ get $c=0$
Let $x=1$ $y=0$ get $a=0$
Put $a=b=0$ back to its orginal equation get $bxy=0$ so $b=0$
Converserly, $a=b=c=0$ yield that equation is correct for abitrary number.
Is it correct?

In case $Acos(\theta)+Bsin(\theta)=0$ can be rewrite in $a_1t^2+a_2t+a_3=0$ (with $t=tan(\frac{\theta}{2})$) and it will yield $A=B=0$. Right?

[quote="Gibbenergy"][quote="kunny"]
in case of $ax^2+bxy+cy^2=0,$ we can't conclude that $a=b=c=0$ imediately. Sounds strange![/quote]

I inspired from your solutions above. Since it is hold for abitrary number $x,y$
Let $x=0$ $y=1$ get $c=0$
Let $x=1$ $y=0$ get $a=0$
Put $a=b=0$ back to its orginal equation get $bxy=0$ so $b=0$
Converserly, $a=b=c=0$ yield that equation is correct for abitrary number.
Is it correct?[/quote]

Your answer is almost correct . You should do like that.

Let $x=y=1$ get $a+b+c=0$ Solve the system equation of $a.b,c$, yielding $a=b=c=0.$
Converserly, $a=b=c=0$ yield that equation is correct for abitrary number.

Alternative solution:

We calssify two cases as follows.

Case 1: When $y=0, ax^2+bxy+cy^2=0\Longrightarrow ax^2=0,$ yielding $a=0.$

Case 2: When $y\neq 0,$ Divided by $y^2$ and let $t=\frac{x}{y},$ we can rewrite $ax^2+bxy+cy^2=0$ as $at^2+bt+c=0.$ finding the condition for which this equation holds for arbiterary numbers $t$ except $t=0,$ we obtain $a=b=c=0.$ Therefore the desired values are $a=b=c=0.$

Next, Sorry for that I didn't confirm your solution.

[quote="Gibbenergy"]

In case $t=tan(\frac{\theta}{2})$ can be rewrite in $a_1t^2+a_2t+a_3=0$ (with $t=tan(\frac{\theta}{2})$) and it will yield $A=B=0$. Right?[/quote]

Let $t=tan(\frac{\theta}{2}),$ we have $\cos \theta =\frac{1-t^2}{1+t^2},\ \sin \theta =\frac{2t}{1+t^2}.$

Plugg these into the equation $A\cos \theta +B\sin \theta =0,$ we have $At^2-2Bt-A=0$, yielding $A=B=0.$

Gibbenergy, Your idea is also good!

kunny

[quote="kunny"]
Next, Sorry for that I didn't confirm your solution.

[quote="Gibbenergy"]

In case $t=tan(\frac{\theta}{2})$ can be rewrite in $a_1t^2+a_2t+a_3=0$ (with $t=tan(\frac{\theta}{2})$) and it will yield $A=B=0$. Right?[/quote]

Let $t=tan(\frac{\theta}{2}),$ we have $\cos \theta =\frac{1-t^2}{1+t^2},\ \sin \theta =\frac{2t}{1+t^2}.$

Plugg these into the equation $A\cos \theta +B\sin \theta =0,$ we have $At^2-2Bt-1=0$, which isn't the identy of $t.$

kunny[/quote]

  I think you miscaculate in some where. Theoretically, it should yield the same conclusion as I guess. :o

Oops! :blush: you are right. i will edit.

kunny

[quote="blahblahblah"][quote="kunny"]By the way can you understand why $A\cos \theta +B\sin \theta =0$ holds for arbiterary values of $\theta,$ then we have $A=B=0,$ Gibbenergy ?[/quote]

What happens when you differentiate? :)

By the way, I agree that writing it as a single trig function is the best approach.[/quote]

The method by differentiation is good. The point of this appraoch is that we can differentiate the both sides of an identity many times and each identity which obtained whenever we differentiate is also identity.

Your idea reminds me of the next problems from  Japanese University entrance exam.

[color=blue]Problem 1: Prove that $\sin x$ can't be poyinomial. [/color]    1963 Nagoya University entarace exam.

[color=blue]Problem 2: Prove that $2^x$ can't be polynomial. [/color]      1980 Waseda University entrance exam/Politics and Economics

The idea of differentiating functions to establish their linear independence comes up elsewhere; e.g, if $a\exp(t_1x)+b\exp(t_2x)=0$ for all $x$ and $t_1\not=t_2$, then $a=b=0$.

Not sure I get it. May you tell me in more detail? For example, $A\cos \theta +B\sin \theta =0$ . We differnetiate it but it willl back to the original because it's a fuction of sin and cos. And...


logarithms quadratics algebra polynomial calculus integration

For some real numbers a and b, the equation:$8x^3+4ax^2+2bx+a=0$ has three distinct positive roots. If the sum of the base-2 logarithms of the roots is 5, what is the value of $a$?

[hide="answer"]Since the prodcut of the roots is $\frac{-a}{8}$, we have $\log_{2}{\frac{-a}{8}}=5$, so $a=-256$[/hide]

Old AMC problem I believe

How do you know that the product of the roots is -a/8 ? :-)

Rational root theorem. And AoPS volume 2 has a very nice proof of this (but I won't post it here unless someone really wants it).

[quote="Meghrabi"][quote="Pineappleperson"]It's some theorem in Algebra 2.

AX^n+BX^(n-1)+.......C=0

Product of roots are C/A.[/quote]

Thanks but, shouldn't abc be a/8 then ?[/quote]

I thought the formula for the sum ($-b/a$) and the product ($c/a$) were only for quadratic equations. Does it apply to all polynomials? Or is there a different polynomial for a degree 3? Anyone care to explain? thanks.

Ok, the reason that it was switched to negative was because it said all of the roots were positive, which means we have $(x-r_1)(x-r_2)(x-r_3)$ which gives the constant term $-r_1r_2r_3$. Now for any polynomial $a_1x^n+a_2x^{n-1}+\cdots+a_{n+1}$ the product of the roots is $\frac{a_{n+1}}{a_1}$ and the sum is $\frac{a_2}{a_1}$. Lastly, all rational roots can be expressed as $\frac{r}{s}$ where $r|a_{n+1}$ and $s|a_1$.
This clear things up a bit?

Look,
For any polynomial $P(x)$ with degree $n$, integral coefficients and a nonzero constant term:

$P(x)=a_nx^n+a_{n-1}x^{n-1}+... + a_0$

It is always true that if one of the roots of the equation ($P(x)=0$) is $x=\frac{p}{q}$, where $p$ and $q$ are nonzero integers with no common factor other than $1$, then $p$ must be a factor of $a_0$, and $q$ must be a factor of $a_n$

This is called The Rational Root Theorem.

Now, another important theorem for polynomial equations which I am not sure if there is a name for, is the folowing:

For the equation $a_nx^n+a_{n-1}x^{n-1}+... + a_0$

the sum of its roots is $\frac{a_{n-1}}{a_n}$

the prodcut of its root is $\frac{a_0}{a_n}$ if $n$ is even or it is $-\frac{a_0}{a_n}$ if $n$ is odd.

I don't remember the proofs for these thorems but I could look them up for you if that's what you want. Just let me know if you do.
I hope this helps you...


calculus
derivative
inequalities
function
calculus computations

Find the concavity of y=x/(x-1).  I used the second derivative test, y"=2/(x-1)^3  but there is no way for this second derivative to be equal to 0 since it would be undefined? Then should we find the value when the derivative fails to exist like the definition for the first derivative of the critical point?  Thanks for help.

The concavity of the graph $ y\equal{}f(x)$ is determined by the sign of $ f''$ (positive means concave up, negative means concave down). If $ \frac{2}{(x\minus{}1)^3}$ is positive, what can you say about $ (x\minus{}1)$?

[quote="mlok"]The concavity of the graph $ y \equal{} f(x)$ is determined by the sign of $ f''$ (positive means concave up, negative means concave down). If $ \frac {2}{(x \minus{} 1)^3}$ is positive, what can you say about $ (x \minus{} 1)$?[/quote]

okay, maybe I can just establish an inequality equation, like (x-1)^3>0 for the function to be positive and vice versa. Before I made it to be 0 to find the interval, then find out the concavity. Now in this case, your method should be used since it would be undefined when it's 0 denominator.

Or you can mark all the points where $ f''$ is either zero or undefined, then find its sign in each of the intervals determined by these points. Here $ f''$ is never zero, but is undefined at $ x\equal{}1$. So you test each of the intervals $ (\minus{}\infty,1)$ and $ (1,\infty)$.

[quote="mlok"]Or you can mark all the points where $ f''$ is either zero or undefined, then find its sign in each of the intervals determined by these points. Here $ f''$ is never zero, but is undefined at $ x \equal{} 1$. So you test each of the intervals $ ( \minus{} \infty,1)$ and $ (1,\infty)$.[/quote]
Yes, that's what I tried to ask that if I could find the values of f” when it's undefined for intervals. Got it! thanks.


function
calculus
derivative
real analysis
real analysis unsolved

A function $f$ has second derivative in $[a,b]$ and we have $f(a)=f(b).$ Prove that there exists $x_{1},x_{2}\in(a,b),$ such that $f'(x_{1})+f''(x_{2})\geq 0.$

This is false without additional conditions. Consider $f(x)=-x^{2}$ on $[-\frac12,\frac12]$.

You probably meant to ask a different question.


Two players are playing a game involving piles of coins. A legal move is to remove a coin or split a pile of coins into two non-empty piles. They are playing with $n_{1}, n_{2}, n_{3}..............$ piles of coins where $n_{k}=k$. Find a simple way to determine which player has the winning strategy.

under what conditions do you win the game?

I think the problem intends case works.

[hide]1. The one who removes the last coin wins.
2. The one who removes the last coin loses. [/hide]


[quote="Omid Hatami"]Suppose $H_1 $ is a convex n-gon. We construct $H_1,...,H_n$ recursively: vertices of $H_{k+1}$ are obtained from each vertex of $H_k$ from symmetry through it's $k$th neighbor (In trigonometric direction). Prove that $H_1$ and $H_n$ are similiar.[/quote]

Are you sure that you haven't forgotten a condition? I think that n must be a prime, else you can easily find a counterexample (just take n = 4). For prime n, the problem was solved by Myth on http://www.mathlinks.ro/Forum/viewtopic.php?t=2728 . Personally, I am far from understanding his proof, but the problem has almost nothing to do with geometry, that's clear. By the way, if that's a problem from your pre-TSTs, then I don't want to needlessly spend my time on trying your TSTs...

  darij

i think i have two more solutions to the problem, one using matrices, and the other using successions...
but i sincerely don't have time (and i don't want to) post them... too lazy, and not at all in the mood for it!


A bag contains 80 jellybeans, 20 of which are red, 20 are black, 20 are green, and 20 are yellow. The least number that a blindfolded person must eat to be certain of having eaten at least one of each color is:

a. 61 b. 23 c. 6 d. 5 e. NG

[hide] The worst case scenario is that he eats all the jellybeans of all the colors except for $ 1$. Thus, he must eat $ 20 \plus{} 20 \plus{} 20 \plus{} 1 \equal{} 61$ jellybeans that he must eat. Hence, the answer is A.[/hide]

[hide]I got the same solution as gaussintraining. SO you know that the worst scenario is he eats all the jellybeans of every color, except for one of them. Then he must eat 61 jelly beans. So the answer is A[/hide]


If $ 4^n\plus{}2^n\plus{}1$ is a prime, then prove that $ n$ is a power of $ 3$.

[hide]We can write this as $ \frac{2^{3n}\minus{}1}{2^n\minus{}1}$.
Lemma: $ \frac{(x^{3n}\minus{}1)(x^k\minus{}1)}{(x^n\minus{}1)(x^{3k}\minus{}1)}$ is a polynomial with integer coefficients, where $ k\equal{}3^{v_p(n)}$.
Proof: By the division algorithm, it suffices to show that this is a polynomial.  Consider a root  of the denominator of this fraction.  If it is a root of either factor, it is clearly a root of $ x^{3n}\minus{}1$.  If it is a root of both factors, it is a root of $ x^k\minus{}1$, since $ k\equal{}gcd(n,3k)$.  But it is also a root of $ x^{3n}\minus{}1$, so it also has multiplicity 2 in the numerator.  Thus the Lemma is proven.

This means that our original prime is divisible by $ \frac{2^{3k}\minus{}1}{2^k\minus{}1}$.  Thus $ \frac{2^{3k}\minus{}1}{2^k\minus{}1}\equal{}\frac{2^{3n}\minus{}1}{2^n\minus{}1}$, so $ k\equal{}n$.  But k is a power of 3, so we are done.[/hide]

This was already posted [url]http://www.mathlinks.ro/viewtopic.php?t=312354[/url]


function
real analysis
real analysis solved

Let $f$ and $f_n$, $n=1,2,...$ be functions from $\mathbb{R}$ to $\mathbb{R}$. Assume that $f_n(x_n) \to f(x)$ as $n\to\infty$ whenever $x_n \to x$. Show that $f$ is continious.

Suppose there are $a$ and $a_n\to a$ s.t. $f(a_n)>f(a)+\varepsilon$. For each $a_n$ there is $m_n$ s.t. $|f_k(a_n)-f(a_n)|<\epsilon/2$ for $k>m_n$, so $|f_k(a_n)-f(a)|>\epsilon/2$. We can assume $m_n$ is increasing. Then construct sequence $\{b_l\}$ s.t. $b_l=a_n$ for all $l\in \{m_n+1,...,m_{n+1}\}$. We see that $f_l(b_l)\not\to f(a)$.

I hope it is correct.

[Edit: I forgot to write $k>m_n$. Corrected]

If $f_n(x_n)\to f(x)$, signifying"converging continously", then $ \{f_n\} $converges pointwise to the same limit function f(x). Then f is continous even the $f_n$ are not themselves continous.

We can show that the condition  "$f_n(x_n) \to f(x)$ as $n\to\infty$ whenever $x_n \to x$" is equivalent to the condition "$f$ is continuous and the $f_n$ converge uniformly to $f$ on bounded sets".

[quote="blahblahblah"]$\vert f_n(x_n)-f(x)\vert\leq \vert f_n(x_n)-f_n(x)\vert+\vert f_n(x)-f(x)\vert$
The first one is small from continuity, the second one is small from (pointwise) convergence.[/quote]
"The first one is small from continuity" ?

[quote="Myth"]Suppose there are $a$ and $a_n\to a$ s.t. $f(a_n)>f(a)+\varepsilon$. For each $a_n$ there is $m_n$ s.t. $|f_k(a_n)-f(a_n)|<\epsilon/2$, [/quote]
Where is $m_n$?

[quote="blahblahblah"]$\vert f_n(x_n)-f(x)\vert\leq \vert f_n(x_n)-f_n(x)\vert+\vert f_n(x)-f(x)\vert$

The first one is small from continuity, the second one is small from (pointwise) convergence.[/quote]

Sorry, I misread the problem.


what is the maximum number of acute angles that convex n-gon can have?

$3.$

Assume that a convex $n$-gon has $p>0$ acute angles, say $a_{1}, \cdots , a_{p}$ and $n-p$ obtuse or right-angles, say $b_{p+1}, \cdots, b_{n}.$

Note that $\frac{a_{i}}{\pi}< \frac{1}{2}$ and $\frac{b_{j}}{\pi}< 1.$

The sum of all the angles of the $n$-gon is $(n-2)\pi$, so that
$n-2 = \sum_{i}\frac{a_{i}}{\pi}+\sum_{j}\frac{b_{j}}{\pi}< \frac{p}{2}+(n-p).$
This leads to $p<4$.

Just consider an equilateral triangle to see that $p=3$ is possible.

Pierre.


[b]Let a, b, c, d be integers with a > b > c > d > 0.[/b]

Suppose that

ac + bd = (b + d + a - c)(b + d - a + c).

Prove that ab + cd is not prime.

This must look like a Algebra problem, but it has a wonderful Geometry solution... Try to find it!

The given condition is equivalent to
$ b^{2}\plus{}bd\plus{}d^{2}\equal{}a^{2}\minus{}ac\plus{}c^{2}$ $ \Longleftrightarrow$
$ b^{2}\plus{}d^{2}\minus{}2bd \cos 120^\circ\equal{}a^{2}\plus{}c^{2}\minus{}2ac \cos 60^\circ$.

So, if we assume a quadrilateral MNPQ, with MN=$ a$, NP=$ c$, QP=$ b$, QM=$ d$ and with $ \angle$ MNP=60$ ^\circ$ and $ \angle$ PQM=120$ ^\circ$, then its diagonals are MP, $ MP^{2}\equal{}a^{2}\minus{}ac\plus{}c^{2}$ and NQ, $ NQ^{2}\equal{}\frac{(ac\plus{}bd)(ab\plus{}cd)}{bc\plus{}ad}$.

Now we apply Ptolomey's theorem on MNPQ and we obtain that $ (ab\plus{}cd)(bc\plus{}ad)\equal{}(ac\plus{}bd)(a^{2}\minus{}ac\plus{}c^{2})$.

The rest follows easily, but not using geometry.

However, I don't find it so wonderful as you says [b]LucasDGF[/b]... Do you have some other geometry solution?


ratio
calculus
integration
trigonometry
real analysis
real analysis unsolved

Here is my [color=blue]1000[/color]th post!

Let $a,b$ are positive integers such that $a>b$.

\[x=e^{at}\cos bt-e^{bt}\sin at\]

\[y=e^{at}\sin bt+e^{bt}\cos at\]

and let \[\vec{v}=\frac{d}{dt}(x,y),\ \vec{V}=\frac{d}{dt}(x+y,\ x-y)\]

Evaluate

\[\frac{\int_0^{2\pi} |\vec{v}|^2 dt}{\int_0^{2\pi} |\vec{V}|^2 dt}\]

Use an algebraic identity and hope the denominator is nonzero?


(i) Find all primes $p,q$ such that $p^2 + q^2 + 7pq$ is a perfect square.
(ii) Prove that no number in the sequence $<11,111,1111,...>$ is a perfect square.
(iii) find all integers $n$ such that $\frac{n^3 + 3}{n^2 + 7}$ is an integer.

(III)

$\frac{n^3+3}{n^2+7}=n-\frac{7n-3}{n^2+7}$, therefore $\frac{7n-3}{n^2+7}$ must be integer.

a) Solving the inequation $\frac{7n-3}{n^2+7}\geqslant 1$ gives $2\leqslant n\leqslant 5$, of which only $n=2$ and $n=5$ work.

b) Solving the inequation $\frac{7n-3}{n^2+7}\leqslant -1$ gives $-6\leqslant n\leqslant -1$, of which neither work.

c) The equation $\frac{7n-3}{n^2+7}=0$ has no integer solution.

Therefore, the only solutions are $n=2$ and $n=5.$

(ii) Prove that no number in the sequence $<11,111,1111,...>$ is a perfect square.
They are congruent to 3 mod 4, so they can`t be perfect squares.

1.: please don't double-post (former thread number 80151) or at least mark one of them for deletion.
2.: please only one problem per thread.

[quote="Hoang_manh_hung_1988"]1.p=q                                                           :D[/quote]
Proof please instead of just giving an answer that is easy to guess!

$p^2+q^2+7pq=n^2$ which is equivalent to $5pq=n^2-(p+q)^2=(n+p+q)(n-p-q)$

So $n+p+q=5p, pq, 5pq$

Case 1:

$n+p+q=5p$ which implies that $q=n-p-q$ hence $5p-q=2p+2q$ so $p=q$

Case 2:

$n+p+q=pq$ implies that $n-p-q=5$ hence $pq-5=2p+2q$ so $(p-2)(q-2)=9$
This gives us the solutions $(5,5)$ and $(11,3)$

Case 3:

$n+p+q=5pq$ implies that $n-p-q=1$ hence $5pq-1=2p+2q$ so $(p-2)(q-2) + 4pq = 5$
This case doesn't have solutions in primes $p, q$.


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

Prove using elementary methods (i.e., without Cauchy's theorem or Sylow):

Let $ G$ be a finite Abelian group of order $ n$ and let $ d$ be a positive divisor of $ n$. Then $ G$ has a subgroup of order $ d$.

Let $ G$ be a group and let $ g$ be an element of finite order $ m$. Then $ g^{k}\equal{}1$ if and only if $ k | m$.

[quote="dDI"]Prove using elementary methods (i.e., without Cauchy's theorem or Sylow):

Let $ G$ be a finite Abelian group of order $ n$ and let $ d$ be a positive divisor of $ n$. Then $ G$ has a subgroup of order $ d$.

Let $ G$ be a group and let $ g$ be an element of finite order $ m$. Then $ g^{k}\equal{} 1$ if and only if $ k | m$.[/quote]

For the second question if $ ord(g)\equal{}m$ then $ g^{k}$ cannot be equal to the identity when $ 0<k<m$. it should be $ g^{k}\equal{} 1$ if and only if $ m | k$

The easiest way to prove 1. is propably the structure theorem for finite abelian group (which can be proven without using the theorems of sylow and cauchy!).
An even more elementary way is the following induction: If $ |G|$ is prime or if $ d \equal{} 1,|G|$, the result is trivial. If $ |G|$ is not a prime and d is a nontrivial divisor of $ |G|$ then choose any element $ g\in G$ of order p where p is prime. Let $ U: \equal{}\langle g\rangle$.

Case 1: $ p | d$
The factor group $ G/U$ is abelian and its order if less than $ |G|$, so we can apply the induction hypothesis. So choose a subgroup $ V/U\leq G/U$ with $ |V/U| \equal{}\frac{d}{p}$. The corresponding subgroup $ V\leq G$ has d elements.

Case 2: $ p\nmid d$
Then consider the homomorphism $ G\to G: x\mapsto\ x^{p}$. The image $ G^{p}$ is a proper subgroup of G because U lies in the kernel. The kernel must be a subgroup of order $ p^{k}$ for some k, so $ |G^{p}| \equal{}\frac{|G|}{p^{k}}$. d is still a divisor of $ |G^{p}|$ so we apply the induction hypothesis once again to obtain a $ V\leq G^{p}\leq G$ with $ |V| \equal{} d$.

The second assertion is really basic knowledge about groups and a proof should be in every single book on elementary group theory! Anyway:
$ m|k\Rightarrow g^{k}\equal{}1$ is trivial and should be clear. For the other direction let $ k\equal{}qm\plus{}r$ with $ 0\leq r<m$. We get $ 1\equal{}g^{k}\equal{}g^{qm\plus{}r}\equal{}g^{r}$. Because m is the least positive number with $ g^{m}\equal{}1$ r must be zero, so $ k\equal{}qm$ which means nothing else than $ m | k$.

How do we know that we can choose a subgroup $ V/U\leq G/U$ with $ V/U\equal{}\frac{d}{p}$?

I wrote that in the sentence before: "..., so we can apply the induction hypothesis". What's unclear about that?


calculus
integration
trigonometry
geometry
trig identities
Law of Sines

Find the lengths of the sides of the smallest triangle with integral side lengths for which the measure of one angle is exactly 3 times the measure of another angle.

I think I have a solution with a lot of trigonometry (thus very ugly  :( ) : (8,10,3)

I would be more than interested by a "geometric solution".

Law of sines : a/sinA = b/sin3A = c/sin(:pi: - 4A) = c/sin4A

Now sin3x = 4sin(x)cos(x)-sin(x), sin4x = 8sin(x)cos(x)^3-4sin(x)cos(x)

Some algebra (easy but tedious to write) gives :
 c = (b-a) :sqrt:((b+a)/a)

Now a,b,c must be integer > 0, some tries give (8,10,3) as first solution. I checked this is correct.

Others small solutions are (16,20,6), (16,20,6), (24,30,9), (32,40,12), (40,50,15), (48,60,18) ...

Indeed you are right!! Although I think showing more of the steps would be good for other people that want to learn from how you did it.

I solved it pretty much the same way that you did. I can't think of a purely "geometic" solution; and it may not even be possible, just because of the nature of the problem.


It asks to simplify i dont know if thats right?

That's as simplified as it gets. So I'd say it's pretty right.


Hi~
I wrote here first,, I think this site is very good!

Can somone help me with this problem;;
http://www.oi.edu.pl/php/show.php?ac=e171100&module=show&file=zadania/oi12/akc
I think this problem is so difficult to code it;
Plz help me~

Well, can u plz make me understand the problem description? I am not understanding it clearly.

What I can understand given some soldier landing location and their survival radius.

But where is the bomb locations? And.. well I can’t understand.. so plz help to understand the problem!

Essentially, the problem reduces to finding the smallest number of commandos that can jump such that for every point on the plane (bomb location), at least one of the commandos is located sufficiently far away from that point to survive.

well, can u plz describe the sample input output? :oops:

There are 5 commandos. There are places commando 1~3 will die, but there isn't places commando 1~4 will die.(at least one will survive.) So sample output should be 4.


Let be given two reals $a,b$ such that $a-2b+2=0$. Prove that:
\[ \sqrt{(a-3)^2+(b-5)^2}+\sqrt{(a-5)^2+(b-7)^2} \ge 6.  \]

I don't think this problem is ugly. I think if we construct an a-b coordinate plane, then the expression with become the sum of distance from a line to two points and it is easy from here!

Note that if we change $\ge$ to $=$, we have an ellipse on the $ab$ coordinate plane.  If the line $a-2b+2=0$ intersects it at only one point, then it must be tangent.  If so, all other points on $a-2b+2=0$ are outside the ellipse and consequently the LHS would be greater than 6.

$a=2b-2$, so substiution yields $\sqrt{(2b-5)^2+(b-5)^2}+\sqrt{(2b-7)^2+(b-7)^2}=6 \Rightarrow \sqrt{5b^2-30b+50}+\sqrt{5b^2-42b+98}=6$
Isolating the left radical and then squaring gives:
$\sqrt{5b^2-30b+50}^2=\left(6-\sqrt{(5b^2-42b+98}\right)^2$
$5b^2-30b+50=36-12\sqrt{(5b^2-42b+98}+5b^2-42b+98$
$0=-12b+84-12\sqrt{(5b^2-42b+98}$
Isolating the right radical and then squaring gives:
$\sqrt{5b^2-42b+98}^2=(-b+7)^2$
$5b^2-42b+98=b^2-14b+49$
$4b^2-28b+49=0$
The discriminant $(-28)^2-4(4)(49)=784-784=0$, so there is only one solution for $b$.  Since the line is not orthogonal, there is only one intersection, proving the desired inequality.


function
trigonometry
calculus
calculus computations

I dont have much to do these. But what about the standard $ (n\plus{}1/2)(\pm 1\pm i)$ contour approach for the function $ f(z) \equal{}\pi\cot\pi z\tan^{\minus{}1}\left(\frac{1}{z^{2}}\right)$.

So,
$ 2\sum_{n\equal{}1}^{\infty}\tan^{\minus{}1}\frac{1}{n^{2}}\equal{}\minus{}\mbox{res}(f,0)$

Now compute the residue. Which seems to be very very messy.

This [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=149843]link[/url] could be helpful.


What is the smallest integer value of $ n$ such that $ 3^{\minus{}n} < \frac{1}{250}$?

Since a negative exponent just indicates a reciprocal, we can manipulate the equation into $ 3^n>250$. Since $ 3^5\equal{}243$, just below $ 250$ by a bit, the smallest integer value of $ n$ that satisfies the inequality is $ 3^{\boxed{6}}\equal{}729$.


geometry
perimeter
trigonometry
ratio
angle bisector
similar triangles

The triangle $ABC$ have dimensions $AB=11$, $AC=18$ and $BC=20$. Find the perimeter of the triangle $AMN$ if $MN$ is parallel to $BC$, $OB$ is the bisector of the angle $ABC$ and $OC$ is the bisector of the angle $ACB$.
Thanks

I just made this picture to help understand the problem. (Mods: please move it, if you think that it not belongs to here). sorry for the bad english. Thanks for any help.

Have you tried extending $\overline{CO}$ and $\overline{BO}$ and then applying the angle bisector theorem?

Then maybe similar triangles?

I get 29 (three sides lengths are 319/49, 6380/539 and 5742/539)  If this is right, ill post a solution

start by lableing ange OBC as alpha and see what you get with angle chasing, and i didnt have to extend anything..

let $OB\cap BC=P$, using mass points, we can easily find $\frac{AO}{AP}$, the answer is $49*\left(\frac{AO}{AP}\right)^{2}$

b-flat,  i didn't tried that theorem (i was looking for some simple solution)
Art of Owna your answer is correct, can you show your solution?
Altheman i don't know what is mass points :oops: .
Thanks for the replies, i had no idea to attack this problem :D

[quote="math_music"]b-flat, i didn't tried that theorem (i was looking for some simple solution)
Art of Owna your answer is correct, can you show your solution?
Altheman i don't know what is mass points  .
Thanks for the replies, i had no idea to attack this problem [/quote]

ok first, glen, you must have done something wrong because the answer is not 33.68, and although you can use sines and cosines, those are not normal angles so you cannot possibly caculate them without a caculator..

ill give a hint first, if you cant use that, then ill post a solution

[hide="hint"]first start angle chasing and see what you get, you will get two isosclees triangels..lable AN as x and solve for x using similar triangles and substitution of side lengths for the two equal sides of the isoscles triangle

[/hide]

k maybe my answer is wrong, I'll check it out again I did it late last night..

however your statement "cannot do it possibly without a calculator.." you need a calculator to find the first angle in the first place.. its not an equilateral triangle..

edit::
K, I did it again and came up with 28.955 as the perimeter, which is safe to say 29.
[hide="how i went about it"]
First I find all the angles of the main triangle.
then as art of owna said, focus on triable BOC, and find side length BO.
then i focus on triangle BMO and find side length BM

subtract BM from 11 and you get your AM length, then I developed a ratio for the side lengths since they are similar triangles to find the side lengths AN and MN.[/hide]

If this can be done without a calculator, id like to know how :!:

Art of Owna 'almost' got the shorter solution (there is no need to find similar triangles or mass points). All the required information is given and Art of Owna's hint really gave away the main idea to solve this VERY EASY problem.

[hide="Additional hint"]Because of two isosceles triangles, deduce $MN=BM+NC$ and hence the perimeter of AMN is $AB+AC$[/hide]

holy schnikes 10000th user!

thats incredible.. I wish I'd seen that.. wow.

Very Nice 10000th User, that was very simple and elegant!


Sally has $ 624$ quarters in her piggy bank.  In dollars, how much money is this?

4 quarters makes a dollar so you do
$ \frac{624}{4}\equal{}156$


MATHCOUNTS
summer program
MathPath
Support
Mathcamp

So come out of hiding - I know a lot of people going must also use AoPS -- that's just a logical assumption, isn't it?  :?:   I just thought if a bunch of us knew others going we could say "hi" - maybe start a dialogue before the first day of camp? Anyone interested from any state?

Right underneath the post with all of the last year posts is the current thread of people going.... Just post there instead of here....

I went ahead and unlocked this (and deleted the discussion about why it was locked in the first place) -- karate is asking two different questions in his two threads (who from Nats is going to MathPath, and who will be at MathPath that participated in MATHCOUNTS in general, not just Nats).  I would guess that karate asks these questions separately because the information karate will get from the two is different: from the Nats question, karate finds out who he/she met at Nats will be at MathPath, while from the other question, karate learns of other AoPSers with a MATHCOUNTS background who will be at MathPath.

As for whether or not this is appropriate for the MATHCOUNTS forum, I think it's fine.  One of the great things about MATHCOUNTS and about MathPath is that they build communities.  That these communities overlap is one of the great strengths of both programs, and discussing these overlaps is very relevant to both programs' success.  (I will, of course, support deleting a thread that is about something irrelevant to MATHCOUNTS such as 'How many MATHCOUNTS participants play XBox?'  That sort of discussion doesn't do anything for MATHCOUNTS.)

Mr. Rusczyk, thank you for your kind reply and understanding. I appreciate your unlocking my thread and hope anyone who plans on attending eventually finds their way to this posting before camp begins.

              Karatemagic7

Please say if you were accepted and if you are planning to attend. If you were accepted, but not attending, why not? What are you doing instead? If you are attending, if you want to state your state, go ahead...

i may not be the opposite of possibly going[hide=":)"]that means im going[/hide]

Uh... I didn't do Mathcounts but I am going to MathPath. Yay.

[quote="Poincare"]Don't you have to pay like $ \$$ 200 dollars to go?[/quote]

umm...i think its around $ \$$3000

Yeah 200 would be a really really good deal. And how much are the scholarships? Like $ \$0$???

most mathcamps aren't known for their generosity.


A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one $2$ is tossed?

$\displaystyle \textbf{(A)} \ \frac{1}{6} \qquad \textbf{(B)} \ \frac{91}{216} \qquad \textbf{(C)} \ \frac{1}{2} \qquad \textbf{(D)} \ \frac{8}{15} \qquad \textbf{(E)} \ \frac{7}{12}$

[hide="Answer"]We have either a $2$ in the first two tosses or a $2$ as the third toss. If the latter, the only possibility is $(1,1,2)$. If the former, we have one of the numbers a $2$ and the other of the first two $1$ through $4$ so that the sum does not exceed $6$. We have $4$ combinations here, but $3$ of them can be doubled to account for permutations ($(2,2,4)$ cannot be doubled). The total number of possibilities is $15$, so our answer is $\frac{8}{15}\Rightarrow \boxed{D}$.[/hide]


the above term can be sımplıfıed.........

         TI n-1/n    n: ranges from 2 to 100...
      then u have,   1/2. 2/3. 3/4.................99/100=1/100

  sorry,ı dont ve LATEX on the computer whıle ı solved thıs prob

[quote="shaun ong"](1-1/2)(1-1/3)(1-1/4)(1-1/5)..................................(1-1/100)[/quote]
Latex: $ (1\minus{}\frac{1}{2})(1\minus{}\frac{1}{3})(1\minus{}\frac{1}{4})(1\minus{}\frac{1}{5})...(1\minus{}\frac{1}{100})$
[hide="Solution"]Simplifying each term in the parentheses you get: $ (\frac{1}{2})(\frac{2}{3})(\frac{3}{4})...(\frac{99}{100})$
The denominator of the first cancels out with the numerator with the next, and the pattern continues so it reduces down to $ \frac{1}{100}$ :D[/hide]

thanks....and i thought how hard could it be

It's something called telescoping series. One example is 1/1*2+1/2*3+..... You can solve this by observing that 1/1*2=1-1/2. 1/2*3=1/2-1/3, and so on. You can cancel all of the terms except the first and the last.

-rts2007

Well, here we have a telescoping [i]product[/i], but it's the same idea.


[quote="turs12"]Given x,y,z>0 what's the best way to go about showing

x+y+z<= 2*( x^2/(y+z) + y^2/(x+z) + z^2/(x+y) )[/quote]
If I  understand correct you mean
$ 2(\frac{x^{2}}{y\plus{}z}\plus{}\frac{y^{2}}{x\plus{}z}\plus{}\frac{z^{2}}{x\plus{}y})\geq x\plus{}y\plus{}z$
Just use Cauchy-Schwarz in Engel form
$ 2(\frac{x^{2}}{y\plus{}z}\plus{}\frac{y^{2}}{x\plus{}z}\plus{}\frac{z^{2}}{x\plus{}y})\geq \frac{2(x\plus{}y\plus{}z)^{2}}{2(x\plus{}y\plus{}z)}\equal{}x\plus{}y\plus{}z$   :)

$ LHS \minus{} RHS \equal{} (x\plus{}y\plus{}z)\cdot(\sum_{cyc} (x\minus{}y)^{2}\cdot(\frac{1}{(y\plus{}z)\cdot(x\plus{}z)})) \geq 0$  :P


counting
derangement
search
combinatorics unsolved
combinatorics

A disarranged number from the set 1~9 is an arrangement of these integers so that no number takes up its 'usualy' position. (ex. 1 is in any place other the the first position, 2 is in any place other than the second position....all the way to 9.)

For instance, 514327698 is an example of a disarranged number.

How many disarranged numbers are possible from the set 1~9?

its called the parishes (dont know the exact spelling) of number $ n$ denote as $ D(n)$...
we have:
$ D(n)=n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...+(-1)^{n}\frac{1}{n!})$

In American mathematical English, a permutation with no fixed points is called a [i]derangement[/i]; the sequence of numbers may sometimes be called the subfactorial or rencontres numbers.  (I'm sure that last word is suppose to have some accents on it*; in the case it's called subfactorial, it is often denoted $ !n$.)  The sequence is [url=http://www.research.att.com/~njas/sequences/A000166]A000166[/url] in the OEIS.  Probably the most straightforward way to do it (not already knowing the answer) is to prove the identity $ D(n) = (n-1) (D(n-1)+D(n-2))$, arguing based on the position of $ n$ when the permutation is written in cycle notation.  Perhaps the easiest way to calculate it is to note that it is exactly equal to the nearest integer to $ \frac{n!}{e}$.

* mathmanman informs me via PM that there is, in fact, no accent on the word rencontre.  Also, the derangement problem has appeared on the forum many times before -- do a search!


algebra
polynomial
AMC
AMC 12
system of equations

hello

how to solve question like that please help
thanx

a) Solve the recurrence relation
an = an−1 + 20an−2
with a0 = −1 and a1 = 13.

b) Solve the recurrence relation
an = 6an−1 − 12an−2 + 8an−3
with a0 = −5, a1 = 4, and a2 = 88.

[hide="1"] The characteristic equation is $x^{2}-x-20 = 0$ and hence the solution has the form

$a_{k}= \alpha \cdot 5^{k}+\beta \cdot (-4)^{k}$

Where we solve for $\alpha, \beta$ using the initial values:

$\alpha+\beta =-1$
$5\alpha-4 \beta = 13$
$9 \alpha = 9$
$\alpha = 1$
$\beta =-2$

So

$a_{k}= \boxed{ 5^{k}-2(-4)^{k}}$

The other proceeds similarly, but with a cubic characteristic equation (three roots).  [/hide]

If you want a more detailed explanation, I'll point you to my three posts that have to deal with this. Most have full solutions written out.

From most recent to least

[hide="Post1"]
Given $s_{0}= 3$, and $s_{1}= 4$, under the recurrence relation $s_{n}= 4s_{n-1}-3s_{n-2}$, find $s_{2008}$ .

$(\text{A}) \frac{12+2^{2005}}{2}$

$(\text{B}) \frac{3^{2008}+5}{2^{1004}}$

$(\text{C}) \frac{4^{2008}+7}{2^{2007}}$

$(\text{D}) \frac{3^{2008}+5}{2}$

$(\text{E}) \frac{2^{4016}-3^{2008}+5}{2}$

[hide="My Solution"]:
The characteristic polynomial of the recurrence relation is $s^{2}=4s-3$, or $s^{2}-4s+3 = 0$. This polynomial factors as $(s-3)(s-1)$, with roots $r_{1}= 3$ and $r_{2}= 1$.

By Binet's theorem, we can write the general form for the Nth term of a linear recurrence as $s_{n}=A(r_{1})^{n}+B(r_{2})^{n}$ with constants A and B.

Now we solve a system of equations using our starting values.
$3 = A(3)^{0}+B(1)^{0}\longrightarrow 3 = A+B$
$4 = A(3)^{1}+B(1)^{1}\longrightarrow 4 = 3A+B$
Obtaining values of A = 0.5, B = 2.5, we complete our closed form expression.
$s_{n}= 0.5(3)^{N}+2.5(1)^{N}\longrightarrow s_{n}= 0.5(3)^{N}+2.5$

Our value is then $\frac{3^{2008}+5}{2}$, or Choice D.
[/hide]

Any comments? I designed this problem to be ~23 level on the AMC12, or 4-5 level on the AIME.
[/hide][/hide]

[hide="Post 2"]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=polynomial+characteristic&t=133223]
Fibonnaci Recursion with Binet's Formula
[/url]
[/hide]

[hide="Post 3"]

[url=http://www.artofproblemsolving.com/Forum/posting.php?mode=quote&p=698257]
An example about a specific recursion
[/url]
[/hide]


Congratulations to the following qualifiers of USAMO 2005 (hope I didn't miss any names).

[b]Boris Braverman[/b] 10 Sir Winston Churchill High School Calgray AB
[b]Farzin Barekat [/b]11 Sutherland Secondary School N Vancouver BC
[b]Daniel Ye[/b] 9 Fredericton High School Fredericton NB
[b]Francis Chung[/b] 12 London Central Secondary School London ON
[b]Lin Fei[/b] 10 Don Mills Collegiate Institute Don Mills ON
[b]Alan Guo [/b]10 O'Neill Collegiate Institute Oshawa ON
[b]Kent Huynh [/b]10 University Of Toronto Schools Toronto ON
[b]Oleg Ivrii [/b]12 Don Mills Collegiate Institute Don Mills ON
[b]Andy Kong [/b]10 Vincent Massey Secondary School Windsor ON
[b]Taotao Liu [/b]12 Vincent Massey Secondary School Windsor ON
[b]Peng Shi [/b]11 Sir John A Macdonald Collegiate Institute Toronto ON
[b]Geoffrey Siu[/b] 12 London Central Secondary School London ON
[b]Chen Sun[/b] 7 London Central Secondary School London ON
[b]Shaun White[/b] 12 Vincent Massey Secondary School Windsor ON
[b]Yufei Zhao[/b] 11 Don Mills Collegiate Institute Don Mills ON

Just a reminder that you might still be able to write USAMO [i]unofficially[/i] even if you didn't qualify. Traditionally, you would need to be either a Winter Camp participant, or a sufficiently high scorer on APMO and CMO.

also, watch out for that Grade 7, our IMO team is getting younger and younger in a lightning fast pace  :lol:

btw, i deleted the part I posted several hours ago, didn't read yufei's post carefully  :blush:

So for the unofficial contest, do you have any idea when the invitiation would arrive?

These are all people who wrote the American math contests (AMC and AIME) - that's the only way to qualify officially.  So the really good mathematicians who didn't qualify either didn't write the AMC and AIME, or made some mistakes on the contests.  I assume Yufei got these results from http://www.unl.edu/amc .
Oh - and Chen Sun is an amazing London mathematician.  He does Olymon, too.  He'll definitely be an IMO contender in the next couple years.

[quote="geoffreynham"]Oh - and what were your USAMO indexes, for anyone who made it?  WOW, Yufei - your score is impressive!!  14 on AIME - which one did you miss?[/quote]

Well I haven't gotten the official score yet  :blush: [edit: I do now]. I made a mistake on problem 7 (AIME I), where I flipped a sign wrong.

Heh... my school doesn't do any of those American contests... I still want to write unofficially, if possible, but how would I go about doing that?  I assume I'd be allowed cuz I went to winter camp.  Is there somebody I should email to discuss this or some kind of registration form I need?

You probably will be getting an email from Dr. Sands in the near future.

supposedly Richard Peng got USAMO email, but I didn't??

darn.

[quote="geoffreynham"]Does anyone know who Daniel Ye is, by the way?

Oh - and what were your USAMO indexes, for anyone who made it?  WOW, Yufei - your score is impressive!!  14 on AIME - which one did you miss?

:) I've finally reached P vs NP level[/quote]

he is a goddess man. he got 57 out of 60 on the maritime math competition. the next highest score person gets a score in the mid 40s


Let $ f(x)\equal{}x^2\minus{}c$
Does there exist positive real $ c$ so that $ \{x|f^{(n)}(x)\equal{}x,n\equal{}1,2,3,....\}$ is a finite set.($ f^{(n)}(x)\equal{}f(f^{(n\minus{}1)}(x)),f^{(1)}(x)\equal{}f(x)$)

Anyboby any ideas please :)
This problem is my conjecture.I guess the answer is that there is no $ c$ satisfies it  :maybe:


[hide="Hint:"]try the first few cases; something cool happens[/hide]

[hide="Solution"]
Following Altheman's hint,

We can find that $f_{n}(x) = f^{n}(x)$. Trying out the first few cases of the problem
\begin{eqnarray*}f_{9^{k}}(1125) &=& f^{9^{k}}(1125) \\ &=& f^{{9^{k}}-1}\left(-\frac{1}{1124}\right) \\ &=& f^{{9^{k}}-2}\left(\frac{1124}{1125}\right) \\ &=& f^{{9^{k}}-3}(1125)\\ \end{eqnarray*}
The values repeat every multiple of 3. So following this pattern, we can conclude that
$f_{9^{k}}(1125) = \boxed{1125}$.
[/hide]


[hide] By doing percentages, we see that the team made $ 19$ 1 point shots, $ 14$ 2 point shots, and $ 4$ 3 point shots. Thus, we have $ 19\plus{}14(2)\plus{}4(3)\equal{}59$. Hence, the answer is E.[/hide]


Given triangle $ PQR$ with $ \overline{RS}$ bisecting $ \angle R$, $ PQ$ extended to $ D$ and $ \angle n$ a right angle, then:
[asy]path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false)
{
    pair M,N;
    path mark;
    M=t*0.03*unit(A-B)+B;
    N=t*0.03*unit(C-B)+B;
    if(flip)
        mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));
    else
        mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));
    return mark;
}

unitsize(1.5cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

pair P=(0,0), R=(3,2), Q=(4,0);
pair S0=bisectorpoint(P,R,Q);
pair Sp=extension(P,Q,S0,R);
pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp);
pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R);

draw(P--R--Q);
draw(R--Sp);
draw(P--D--M);

draw(anglemark2(Sp,P,R,17));
label("$p$",P+(0.35,0.1));
draw(anglemark2(R,Q,P,11));
label("$q$",Q+(-0.17,0.1));
draw(anglemark2(R,Np,D,8,true));
label("$n$",Np+(+0.12,0.07));
draw(anglemark2(R,M,D,13,true));
label("$m$",M+(+0.25,0.03));
draw(anglemark2(M,D,P,29));
label("$d$",D+(-0.75,0.095));

pen f=fontsize(10pt);
label("$R$",R,N,f);
label("$P$",P,S,f);
label("$S$",Sp,S,f);
label("$Q$",Q,S,f);
label("$D$",D,S,f);[/asy]$ \textbf{(A)}\ \angle m \equal{} \frac {1}{2}(\angle p \minus{} \angle q) \qquad \textbf{(B)}\ \angle m \equal{} \frac {1}{2}(\angle p \plus{} \angle q)$
$ \textbf{(C)}\ \angle d \equal{} \frac {1}{2} (\angle q \plus{} \angle p) \qquad \textbf{(D)}\ \angle d \equal{} \frac {1}{2}\angle m \qquad \textbf{(E)}\ \text{none of these is correct}$

[hide="Click for solution"]
$ \angle m\equal{}\angle p \plus{} \angle d$, $ \angle d \equal{} \angle q \minus{} \angle m \implies \angle m \equal{} \angle p \plus{} \angle q \minus{} \angle m$, so $ \angle m \equal{}\frac{1}{2} (\angle p \plus{} \angle q)$ or $ \boxed{\textbf{(B)}}$.
[/hide]

In case AIME15's solution is not clear enough in the second step:
Call the point where $\angle m$ is A, and the point where $DA$ intersects $RQ$ as B.
$\angle DBQ = \angle RBA = \angle RAB = \angle m$
$\angle RQD = 180-\angle Q$
$\therefore \angle D = 180 - (\angle m + 180 - \angle q) = \angle Q - \angle M$

This part of the solution was the only vague part when I read it. Everything else is self-explanatory.

Thanks!


{(1+Sqrt(1+x)}TanA ={(1+Sqrt(1-x)}

Prove  x =Sin4A

$(?)\ \ \tan \alpha = \frac{1 + \sqrt{1 - \sin 4 \alpha}}{1 + \sqrt{1 + \sin 4 \alpha}}$

$\sqrt{1 + \sin 4 \alpha} = \sqrt{\cos^2 2\alpha + 2 \sin 2\alpha \cos 2\alpha + \sin^2 2\alpha} = \pm(\cos 2\alpha + \sin 2\alpha)$

$\sqrt{1 - \sin 4 \alpha} = \sqrt{\cos^2 2\alpha - 2 \sin 2\alpha \cos 2\alpha + \sin^2 2\alpha} = \pm(\cos 2\alpha - \sin 2\alpha)$

Here, we have to be careful: $\tan \alpha$ has the period $\pi$ and

$-\infty < \tan \alpha \le 0$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac \pi 2, 0\right)$, [color=white].[/color] $0 \le \tan \alpha \le +\infty$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(0, +\frac \pi 2\right)$

$\cos 2 \alpha + \sin 2 \alpha \le 0,\ \cos 2 \alpha - \sin 2 \alpha \le 0$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac{\pi}{2}, -\frac{3\pi}{8}\right)$

$\cos 2 \alpha + \sin 2 \alpha \le 0,\ \cos 2 \alpha - \sin 2 \alpha \ge 0$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac{3\pi}{8}, -\frac{\pi}{8}\right)$

$\cos 2 \alpha + \sin 2 \alpha \ge 0,\ \cos 2 \alpha - \sin 2 \alpha \ge 0$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac{\pi}{8}, +\frac{\pi}{8}\right)$

$\cos 2 \alpha + \sin 2 \alpha \ge 0,\ \cos 2 \alpha - \sin 2 \alpha \le 0$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(+\frac{\pi}{8}, +\frac{3\pi}{8}\right)$

$\cos 2 \alpha + \sin 2 \alpha \le 0,\ \cos 2 \alpha - \sin 2 \alpha \le 0$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(+\frac{3\pi}{8}, +\frac{\pi}{2}\right)$

Thus the following signs have to be used:

$(?)\ \ \tan \alpha = \frac{1 + \sqrt{1 - \sin 4 \alpha}}{1 - \sqrt{1 + \sin 4 \alpha}} = \frac{1 - (\cos 2 \alpha - \sin 2 \alpha)}{1 + (\cos 2 \alpha + \sin 2 \alpha)}$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac{\pi}{2}, -\frac{3\pi}{8}\right)$

$(?)\ \ \tan \alpha = \frac{1 - \sqrt{1 - \sin 4 \alpha}}{1 - \sqrt{1 + \sin 4 \alpha}} = \frac{1 - (\cos 2 \alpha - \sin 2 \alpha)}{1 + (\cos 2 \alpha + \sin 2 \alpha)}$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac{3\pi}{8}, -\frac{\pi}{8}\right)$

$(?)\ \ \tan \alpha = \frac{1 - \sqrt{1 - \sin 4 \alpha}}{1 + \sqrt{1 + \sin 4 \alpha}} = \frac{1 - (\cos 2 \alpha - \sin 2 \alpha)}{1 + (\cos 2 \alpha + \sin 2 \alpha)}$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(-\frac{\pi}{8}, +\frac{\pi}{8}\right)$

$(?)\ \ \tan \alpha = \frac{1 + \sqrt{1 - \sin 4 \alpha}}{1 + \sqrt{1 + \sin 4 \alpha}} = \frac{1 - (\cos 2 \alpha - \sin 2 \alpha)}{1 + (\cos 2 \alpha + \sin 2 \alpha)}$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(+\frac{\pi}{8}, +\frac{3\pi}{8}\right)$

$(?)\ \ \tan \alpha = \frac{1 + \sqrt{1 - \sin 4 \alpha}}{1 - \sqrt{1 + \sin 4 \alpha}} = \frac{1 - (\cos 2 \alpha - \sin 2 \alpha)}{1 + (\cos 2 \alpha + \sin 2 \alpha)}$ [color=white].[/color] for [color=white].[/color] $\alpha \in \left(+\frac{3\pi}{8}, +\frac{\pi}{2}\right)$

In all intervals, we have

$(?)\ \ \tan \alpha = \frac{1 - (\cos 2\alpha - \sin 2\alpha)}{1 + (\cos 2\alpha + \sin 2\alpha)} =  \frac{\cos^2 \alpha + \sin^2 \alpha - \cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha}{\cos^2 \alpha + \sin^2 \alpha + \cos^2 \alpha - \sin^2 \alpha + 2 \sin \alpha \cos \alpha} = $

$= \frac{2 \sin^2 \alpha + 2 \sin \alpha \cos \alpha}{2 \cos^2 \alpha + 2 \sin \alpha \cos \alpha} = \frac{\sin \alpha\ (\sin \alpha + \cos \alpha)}{\cos \alpha\ (\cos \alpha + \sin \alpha)} = \frac{\sin \alpha}{\cos \alpha}$

Obviously true, but the original relation $\tan \alpha = \frac{1 + \sqrt{1 - \sin 4 \alpha}}{1 + \sqrt{1 + \sin 4 \alpha}}$ with positive square roots holds only for $\alpha \in \left(+\frac{\pi}{8}, +\frac{3\pi}{8}\right)$.


inequalities
linear algebra
linear algebra unsolved

Let $A,B$ be two complex commutative operators. Prove that they have common eigenvector.

This and stronger problems have been solved here. For this one, try searching for "Schur", "Schur's Theorem", etc. (in this subforum).


a convex quadrilateral ABCD has AD=CD and DAB=ABC<90.the line through d and the midpoint of BC intersects line AB in point E.prove that BEC=dac.(note:the problem is valid without the assumption ABC<90.)

use $ latex$ next time.

name midpoint of $ BC$ point X.draw a parallel line to$ AD$ through $ X$ till intersect $ AB$ in $ Y$.
we have $ \frac {DC}{DE}\equal{}\frac {DA}{DE}\equal{}\frac{XY}{XE}\equal{}\frac{XB}{XE}\equal{}\frac{XC}{XE}$.use sin law we have
$ \angle DCB \plus{}2\angle BCE \equal{}180$ but $ \angle DCB \plus{}2\angle CAB$ so $ \angle BCE\equal{}\angle CAB$ .this means
 $ \angle BEC\equal{}\angle CAD$

What did you have in mind posting this as a problem on probability¿ Choose the subforum more carefully next time!


[color=red]کسی از درس پنجم امسال خبری نداره؟
سال قبل این درس جبرخطی و سال قبلش احتمالات و سال قبلش تحدب بوده امسال چی میتونه باشه؟ [/color]

چرا این قدر بی حالین؟ من که نمی فهم یعنی هیچکی اینجا خبری از درس پنجم نداره  حتی امید و ... سال قبل که یادمه همه باخبر بودن حتی رامتین!!!!!
 :rotfl:

[quote="JACOBY VDD"]darse pangom ehtemalan yeki az se darse zire
1.hesaban
2.jabre mojarad
3.analys[topolozy][/quote]
بعید میدونم مباحث اینقدر کلی باشه به هر حال ممنون             :D  :D

[quote="JACOBY VDD"]JABRE KHATY VA EHTEMAL HAM MABHAS HAYE KOLLI HASTAN !![/quote]
در هر صورت کلیت این مباحث رو ندارن در هر صورت بعید میدونم از معادلات دیفرانسیل استفاده کنن چون این یکی که فوق العاده کلی هستش.
بازم مر30 به خاطر کمکت.
میشه بگی منبعت در این مورد کیه؟ممنون :D  :P  :lol:

hich vaght nemian darse panjomo bezaran HESAABE DIFRANSIEL VA ... ina male olampiaz nist.


Let $ ABC$ be a triangle with centroid $ G$ and incircle $ C(I,r)$ . Denote the midpoint $ M$ of the side $ [BC]$

and the point $ P\in (AB)$ which has the distance  $ 2r$ to the line $ BC$ . The $ A$ - exincircle touches the side $ [BC]$

in the point $ D$ and the sideline $ AB$ in the point $ T$ . Prove that $ IG\perp AB\ \Longleftrightarrow\ DA\perp DT\ \Longleftrightarrow\ P\in MI$ .

could you post a proof please....it seems like I can't manage to solve it...is it something kind of Nagel's point?

[quote][color=darkred]Let $ ABC$ be a nonisosceles triangle with centroid $ G$ and incircle $ C(I,r)$ . Denote the midpoint $ M$ of the side $ [BC]$

and the point $ P\in (AB)$ which has the distance  $ 2r$ to the line $ BC$ . The $ A$ - exincircle touches the side $ [BC]$

in the point $ D$ and the sideline $ AB$ in the point $ T$ . Prove that $ IG\perp AB\ \Longleftrightarrow\ DA\perp DT\ \Longleftrightarrow\ P\in MI$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the diameter $ XY$ of the incircle, where $ X\in (BC)$ . The relations $ Y\in (AD)$ , $ MX \equal{} MD \equal{} \frac 12\cdot |b \minus{} c|$

and $ MI\parallel \overline {AYD}$ are well-known. Observe that $ \frac {PY}{BD} \equal{} \frac {h_a \minus{} 2r}{h_a} \equal{}$ $ \frac {ah_a \minus{} 2ar}{ah_a} \equal{}$ $ \frac {2pr \minus{} 2ar}{2pr}$ $ \implies$ $ \boxed {\ PY \equal{} \frac {(p \minus{} a)(p \minus{} c)}{p}\ }$ .

$ \blacktriangleright$ $ \boxed {\ IG\perp AB\ }$ $ \Longleftrightarrow$ $ IA^2 \minus{} IB^2 \equal{} GA^2 \minus{} GB^2$ $ \Longleftrightarrow$ $ \frac {bc(p \minus{} a)}{p} \minus{} \frac {ac(p \minus{} b)}{p} \equal{} \frac 49\cdot (m_a^2 \minus{} m_b^2)$ $ \Longleftrightarrow$

$ 9c(b \minus{} a) \equal{} \left[2\left(b^2 \plus{} c^2\right) \minus{} a^2\right] \minus{} \left[2\left(a^2 \plus{} c^2\right) \minus{} b^2\right]$ $ \Longleftrightarrow$ $ 9c(b \minus{} a) \equal{} 3\left(b^2 \minus{} a^2\right)$ $ \Longleftrightarrow$ $ \boxed {\ a \plus{} b \equal{} 3c\ }$ .

$ \blacktriangleright$ $ \boxed {\ P\in MI\ }$ $ \Longleftrightarrow$ $ PY \equal{} MD$ $ \Longleftrightarrow$ $ 2(p \minus{} a)(p \minus{} c) \equal{} p(b \minus{} c)$  $ \Longleftrightarrow$ $ \boxed {\ a \plus{} b \equal{} 3c\ }$. Othewise (without calculus $ PY$) .

$ \boxed {\ P\in MI\ }$ $ \Longleftrightarrow$ $ \frac {BP}{BA} \equal{} \frac {BM}{BD}$ $ \Longleftrightarrow$ $ \frac {2r}{h_a} \equal{} \frac {a}{2(p \minus{} c)}$ $ \Longleftrightarrow$ $ \frac ap \equal{} \frac {a}{2(p \minus{} c)}$ $ \Longleftrightarrow$ $ p \equal{} 2c$ $ \Longleftrightarrow$ $ \boxed {\ a \plus{} b \equal{} 3c\ }$ .

$ \blacktriangleright$ Since $ BD \equal{} BT$ obtain $ \boxed {\ DA\perp DT\ }$ $ \Longleftrightarrow$ $ BD \equal{} BA$ $ \Longleftrightarrow$ $ p \minus{} c \equal{} c$ $ \Longleftrightarrow$ $ \boxed {\ a \plus{} b \equal{} 3c\ }$ .

[u][b]Remark.[/u][/b] In the right trapezoid $ BPYX\ ,\ YX\perp BX$ the incircle with the diameter $ [XY]$ is tangent to the side $ [BP]$ . Thus,

$ IB\perp IP$ $ \Longleftrightarrow$ $ IC^2\equal{}CP\cdot CB$ $ \Longleftrightarrow$ $ IC^2\equal{}PY\cdot BX$ $ \Longleftrightarrow$ $ PY\equal{}\frac {r^2}{p\minus{}b}$ $ \Longleftrightarrow$ $ \boxed {\ PY\equal{}\frac {(p\minus{}a)(p\minus{}c)}{p}\ }$ .[/color]


Solve the following eqasion in natural numbers:

[size=150]$ a^{3}\minus{}b^{3}\equal{}c!$[/size]

I don't understand your problem ,  :( . $ c$ is constant number :?:

$ c! \equal{} c(c\minus{}1)(c\minus{}2)(c\minus{}3)...\cdot2\cdot1$

where $ a,b,c$ are positive integers

Oh my god, i have understood that $ a^{3}\minus{}b^{3}\equal{}c$ and ! to complete problem . :P

I could have poseted it more clearely:D unfortunatelly that $ !$ quite complicates things
I bassically don't know if it has been solved at all. I just came acros a problem, that as it was later announced had a typing mistake:D, that deduced to this problem here. I'm stil curious about if and how it's solvable however


function
trigonometry
floor function
inequalities
quadratics

The number of solutions for:

$ 8[x^2 \minus{} x] \plus{} 4[x] \equal{} 13 \plus{} 12 [\sin x]$ are?
$ [.]$denotes the Step Function.

hello
there r just 3 cases:$ [sinx]\equal{}\minus{}1,0,1$
form three dif. eq. with these values .
I hope u can do further.
Thank u

[quote="kabi"]3 cases:$ [sinx] \equal{} \minus{} 1,0,1$
form three dif. eq. with these values .
I hope u can do further.
Thank u[/quote]
$ \sin x \equal{} \minus{} 1 \; 1, \; or \; 0$.
Anyway, I got the answer. It has no solutions.

Um, $ \left\lfloor\sin \frac{\pi}{2}\right\rfloor\equal{}1$.

[hide][quote="Rajiv"]The number of solutions for:

$ 8[x^2 - x] + 4[x] = 13 + 12 [\sin x]$ are?
$ [.]$denotes the Step Function.[/quote]
Here's my partial solution(I'm too lazy to write everything out but I'm sure it works):
We have: [sinx]={-1,0,1}
Using the inequality: $ [x^2-x]>=(x^2-x-1)$ and $ 4[x]>=(4x-4)$, we have: $ LHS>=8x^2-4x-12$.
Therefore, we have the intervals of x in three cases: $ 8x^2-4x-12=<{25,1,13}$.Solving these inequalities and get the intervals of x, plug in each interval to solve the quadratic equation(some cases show that $ 13+12[\sinh]-4[x]$ is not divisible by 8, which can be eliminated immediately).
By the way, a nice problem Rajiv :wink: [/hide]

i have a simpler soln

if you notice closely
RHS can take only odd values
LHS can take only even values

so no solution


[quote="bos1234"]Use the work-energy theorem to solve this

At the base of a frictionless ice hill that rises at $ 25^o$ above the horizontal, a tobaggon has a speed of 12m/s toward the hill. How high vertically above the base will it fo before stopping?[/quote]
[hide="Solution"]
\[ \Delta PE \equal{} \minus{}\Delta KE \\\Rightarrow mg(y_f \minus{} y_i) \equal{} {1\over{2}}m(v_{y_i}^2 \minus{}v_{y_f}^2)\]

From the problem statement, $ y_i \equal{} 0 \quad v_{y_f} \equal{} 0$,

\[ \Rightarrow y_f \equal{} \frac{v_{y_i}^2}{2g}\]

$ v_{y_i} \equal{} 12\sin 25^{\circ}\quad \text{m/s}$

\[ \Rightarrow y_f \equal{}\frac{(12\sin 25^{\circ}\, \text{m/s})^2}{2\cdot 9.81\, \text{m}/\text{s}^2}\approx 1.31\quad \text{m}\]
[/hide]


quadratics
inequalities
algebra
algebra solved

A student concentrates on solving quadratic equations in $\mathbb{R}$. He starts with a first quadratic equation $x^2 + ax + b = 0$ where $a$ and $b$ are both different from 0. If this first equation has solutions $p$ and $q$ with $p  \leq  q$, he forms a second quadratic equation $x^2 + px + q = 0$. If this second equation has solutions, he forms a third quadratic equation in an identical way. He continues this process as long as possible. Prove that he will not obtain more than five equations.

Let's say that he starts with x^2+a1x+b1=0 and continues with x^2+a2x+b2=0 and so on... (for easier notation)

Solution:

We certainly must have a1^2+b1^2<>0 (otherwise we get that an=bn=0 for any n from N*). Let's consider the n-th (n>=3) equation with real roots x^2+anx+bn=0. Then we have that an^2>=4bn, an<=bn, anbn>=-(an+bn). We can see that an=0 <=> bn=0. we will prove that an*bn>=0. Lets presume that an*bn<0, an<0<bn and an^2*bn<=-an^2-an*bn<=-4bn-an*bn, from where we get that an^2<=-4-an, whish is false. Lets consider now the 5th equation x^2+a5*x+b5=0, and presume that it has real roots. Then a5*b5>=0. If a5>=0 and b>=0 then a4<0<b4, imposible. If a5<0,b5<0, then a4>=0, b4>=0 from which we get that a3<0<b3 again false. So we can only build less than 5 of these equations!!!

cheers! :D

A equation that has five steps is: x^2-26x-1925/4=0.

cheers! :D

How did you get the equation x - 26x - 1925/4 = 0 ?

Since something strange happened ...
Your solution is exactly the same as the solution which is given in the official booklet of the 2001 Flanders Mathematics Olympiad. But also the example you gave is exactly the same as in the booklet ! Is this a coincidence ?

hmmm..... strange...  :?

It's pure coincidence.. I didn't know about that..
The 1st time I saw this problem was on this forum.
The equation I gave was a simple equation I found for those who learn to solve 2nd degree ecuations.. computing delta and things.

The problem was like so.

The following equation is given:
 x - 26x - 1925/4 = 0.

You start with it. if this first equation has solutions a1 and b1 ( real numbers) then you form a second quadratic equation x + a1x + b1 = 0. If this second equation has real solutions, you form a third quadratic equation in an identical way. Find out how many times can you do this?

look what a simple trainee problem can do in a contest!! strange... :?

cheers! :D

I have a different solution:

Assume, for the sake of contradiction, that the student obtains at least six equations. Let $x^2+a_1x+b_1=0$ be the equation the student has started with; for every positive integer n, if the quadratic equation $x^2+a_nx+b_n=0$ has real roots, then we denote these roots by $a_{n+1}$ and $b_{n+1}$ in such a way that $a_{n+1}\leq b_{n+1}$. Then, according to the assumption, the student obtains at least six equations $x^2+a_1x+b_1=0$, $x^2+a_2x+b_2=0$, $x^2+a_3x+b_3=0$, $x^2+a_4x+b_4=0$, $x^2+a_5x+b_5=0$, $x^2+a_6x+b_6=0$, each of them having as coefficients the roots of the preceding one. From the recurrent definition of $a_i$ and $b_i$, it follows that $a_2\leq b_2$, $a_3\leq b_3$, $a_4\leq b_4$, $a_5\leq b_5$, $a_6\leq b_6$, and also, for every positive integer n, since $a_{n+1}$ and $b_{n+1}$ are the roots of the quadratic equation $x^2+a_nx+b_n=0$, the Viete theorem yields $a_n=-\left(a_{n+1}+b_{n+1}\right)$ and $b_n=a_{n+1}b_{n+1}$.

Particularly, $a_3=-\left(a_4+b_4\right)$, $b_3=a_4b_4$, $a_2=-\left(a_3+b_3\right)$, $b_2=a_3b_3$. Thus,

$a_2-b_2=-\left(a_3+b_3\right)-a_3b_3=-\left(-\left(a_4+b_4\right)+a_4b_4\right)-\left(-\left(a_4+b_4\right)\right)a_4b_4$
$=a_4\left(b_4-1\right)^2+b_4\left(a_4-1\right)^2+3a_4b_4$.

Now, if $a_4>0$, then, since $a_4\leq b_4$, we have $b_4>0$ as well, and thus, according to this formula, $a_2-b_2>0$, so that $a_2>b_2$, contradicting with $a_2\leq b_2$. Hence, we cannot have $a_4>0$. Thus, we have proven that $a_4\leq 0$. Similarly, $a_5\leq 0$ and $a_6\leq 0$. Since $a_4=-\left(a_5+b_5\right)$, the inequality $a_4\leq 0$ becomes $-\left(a_5+b_5\right)\leq 0$; thus, $a_5+b_5\geq 0$, what, in view of $a_5\leq 0$, yields $b_5\geq 0$. Similarly, $b_6\geq 0$. Now, as $b_5$ and $b_6$ are $\geq 0$, while $a_6\leq 0$, the equation $b_5=a_6b_6$ can hold only if $b_5=0$. Then, of course, $b_4=a_5b_5=0$, $b_3=a_4b_4=0$, $b_2=a_3b_3=0$, $b_1=a_2b_2=0$, but the latter equation contradicts a condition of the problem, namely that the quadratic equation the student starts with has nonzero coefficients. This contradiction shows that our assumption was wrong, and the student indeed cannot get more than five equations.

  Darij