geometry
3D geometry
tetrahedron
incenter
trigonometry
inequalities
vector

[quote="nutz_for2.718281828"]This forum didn't have an all subject marathon, so I will start one.

[color=brown][size=200]problem 1[/size][/color]

Define:
$^a^=a^2-a-1$
$!a!=-a^3$

Find the only two integers that $^a^=!a!$

Please say if you have an answr or a solution and hide them.

Any methods are allowed[/quote]
[hide] Set them equal and solve.
$-a^3=a^2-a-1$.
$a^3+a^2-a-1=0$.
$a^2(a+1)-1(a+1)=0$.
$(a^2-1)(a+1)=0$.
$(a+1)^2(a-1)=0$.
Solutions are $a=\boxed{-1, 1}$. [/hide]

[quote="nutz_for2.718281828"]This forum didn't have an all subject marathon, so I will start one.

[color=brown][size=200]problem 1[/size][/color]

Define:
$^a^=a^2-a-1$
$!a!=-a^3$

Find the only two integers that $^a^=!a!$

Please say if you have an answr or a solution and hide them.

Any methods are allowed[/quote]

I thought this forum already had a marathon...
[hide]Just set them equal to each other and divide and get $1,-1$[/hide]
Edit: a minute late...

A bit easy for Intermediate. If I'm supposed to post a question, I don't have one.  :?

[quote="ch1n353ch3s54a1l"]
I thought this forum already had a marathon...[/quote]
It seems that [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33505]Problem Solving Marathon 2[/url] is "dead" right now. Should we revive that marathon, or continue this marathon?

Maybe we can continue this marathon and keep it a bit easier than Problem Solving Marathon 2 (so that the marathon doesn't get stuck this time).

Ok then.
I'll make an attempt to create a next problem..
Problem 2
Two tetrahedrons are connected by their bases. The incenter of one face of the top tetrahedron is connected with the opposite two incenters on the other tetrahedron faces and those two faces on the bottom tetrahedron are also connected. What are the lengths of the resulting triangle's sides?

Since no one has answered that question, I'll post a new one.
[quote="frt"]Maybe we can continue this marathon and keep it a bit easier than Problem Solving Marathon 2 (so that the marathon doesn't get stuck this time).[/quote]

Problem 3

Find a formula for the area of a regular n-gon with side length s

I did it when I was 10 years old, you can too!
P.S. I'm NOT a computer

[quote="nutz_for2.718281828"]Problem 3

Find a formula for the area of a regular n-gon with side length s

I did it when I was 10 years old, you can too!
P.S. I'm NOT a computer[/quote]
I did it when I was bored in Pre-Calc. :roll:
[hide="Answer"] $\frac{ns^2}{4}\cot\frac{\pi}{n}$. [hide="Outline"] Divide polygon into n isoceles triangles. Find the radius by using Law of Sines. Then you can find the area with $k=\frac{1}{2}ab\sin{C}$. [/hide] [/hide]

[hide="Full Solution"] Divide the n-gon into n isoceles triangles. Each vertex angle is $\frac{2\pi}{n}$. Each base angle is $90-\frac{\pi}{n}$. Let $r$ be the congruent sides of the isoceles triangle or the radius of the n-gon. Using the Law of Sines, $\frac{\sin{(90-\frac{\pi}{n})}}{r}=\frac{\sin{\frac{2\pi}{n}}}{s}$. $r=\frac{s\sin{(90-\frac{\pi}{n})}}{\sin{\frac{2\pi}{n}}}=\frac{s\cos{\frac{\pi}{n}}}{2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}}=\frac{s}{2\sin\frac{\pi}{n}}$. The area of the triangle is $\frac{1}{2}(\frac{s}{2\sin\frac{\pi}{n}})^2\sin\frac{2\pi}{n}=\frac{s^2}{8\sin^2\frac{\pi}{n}}*2\sin\frac{\pi}{n}\cos\frac{\pi}{n}=\frac{s^2\cos\frac{\pi}{n}}{4\sin\frac{\pi}{n}}=\frac{s^2}{4}\cot\frac{\pi}{n}$. Therefore, the area of the n-gon is $\boxed{\frac{ns^2}{4}\cot\frac{\pi}{n}}$. :D [/hide]

[quote="nutz_for2.718281828"]Incorrect. I tried a square with side length 4. Try Alg 2.[/quote]
That gives $\frac{4*4^2}{4}\cot\frac{\pi}{4}=16*1=16$.  :huh: ;)

Oh yeah, sorry. My calculator was in degree mode :P

Try using degrees and not using cotangent.

[hide="hint"]It's similer to the answer you got.[/hide]

Umm...why did all the hide things go away?

I don't know. Here is my [hide="answer."]$\frac{ns^2tan(90-\frac{180}{n})}{4}$[/hide]

I could just use the identity $\cot{\theta}=\tan{(90-\theta)}$ to avoid cotangent. :D And I still find radians more...mathematical than degrees. :D Like natural log and common log.

[quote="ch1n353ch3s54a1l"]
Problem 2
Two tetrahedrons are connected by their bases. The incenter of one face of the top tetrahedron is connected with the opposite two incenters on the other tetrahedron faces and those two faces on the bottom tetrahedron are also connected. What are the lengths of the resulting triangle's sides?[/quote]

or another solution to nutzfore's question.

[quote="nutz_for2.718281828"]PROBLEM 4

Ice cream takes 10 minutes to make, cookies takes 25 minutes to make, iced tea takes 7 minutes to make and s'mores takes 5 minutes to make. At my party, I am to serve all of them, and I start each respective one exactly one minute after the last. Which item will be done first?[/quote]

s'mores
PROBLEM 5
(easy one)
prove
$\sqrt{(x_1^2+x_2^2)}+\sqrt{y_1^2+y_2^2}\geq \sqrt{(x_1+y_1)^2+(x_2+y_2)^2}$ :spider:

[quote="ch1n353ch3s54a1l"][quote="xime"][quote="nutz_for2.718281828"]PROBLEM 4

Ice cream takes 10 minutes to make, cookies takes 25 minutes to make, iced tea takes 7 minutes to make and s'mores takes 5 minutes to make. At my party, I am to serve all of them, and I start each respective one exactly one minute after the last. Which item will be done first?[/quote]

s'mores
PROBLEM 5
(easy one)
prove
$\sqrt{(x_{1}^{2}+x_{2}^{2})}+\sqrt{y_{1}^{2}+y_{2}^{2}}\geq \sqrt{(x_{1}+y_{1})^{2}+(x_{2}+y_{2})^{2}}$ :spider:[/quote]

[hide]Square, simplify, and divide by two to get $x_{1}y_{1}+x_{2}y_{2}\le \sqrt{(x_{1}^{2}+x_{2}^{2})(y_{1}^{2}+y_{2}^{2})}$. Square again to get $x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+2x_{1}x_{2}y_{1}y_{2}\le x_{1}^{2}y_{1}^{2}+x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}$. Simplify again to get $x_{1}^{2}y_{2}^{2}-2x_{1}y_{1}x_{2}y_{2}+x_{2}^{2}y_{1}^{2}\ge 0$, and then $(x_{1}y_{2}-x_{2}y_{1})^{2}\ge 0$. [/hide][/quote]

Square both sides of the inequality in your first line of work and we have Cauchy-Schwarz Inequality.

[quote="jhcreinhardt"]To keep the marathon alive...

If $k$ is an integer, what is the smallest $k$ that satisfies

$2^{k}\geq1,000,000$[/quote]

[hide]$k\geq \log_{2}5^{6}2^{6}$

$k\geq 6+6\log_{2}5$

$k=20$ [/hide]

If two random two digit numbers are chosen a and b where a>b, what is the probability $a-b\ge 10$?

[quote="jhcreinhardt"]To keep the marathon alive...

If $k$ is an integer, what is the smallest $k$ that satisfies

$2^{k}\geq1,000,000$[/quote]
[hide] A well known power of 2 is $2^{10}=1024$. $(2^{10})^{2}=2^{20}=1024^{2}> 1000000$. Since $1024^{2}< 2000000$, $k=\boxed{20}$. [/hide]

[quote="ch1n353ch3s54a1l"][hide]$k\geq \log_{2}5^{6}2^{6}$

$k\geq 6+6\log_{2}5$

$k=20$ [/hide]
[/quote]

 :o You used a calculator.

Hehe.  :roll:  :whistling:

I guess my posting of another problem makes up for it... :)

[quote="ch1n353ch3s54a1l"]
If two random two digit numbers are chosen a and b where a>b, what is the probability $a-b\ge 10$?[/quote]

[hide]
There are ${90\choose 2}$ pairs $(a, b)$. There are 89 pairs of 2 digit numbers 1 apart. This decreases by 1 as the interval goes up by 1. We want from 10 onward. There are 80 pairs 10 apart. So we take the sum $ \frac{80+1}{2}\times 80$ . Desired probability is equal to $\frac{{71\choose 2}}{{90\choose 2}}=  \frac{72}{89}$
[/hide]

[quote="ch1n353ch3s54a1l"]
If two random two digit numbers are chosen a and b where a>b, what is the probability $a-b\ge 10$?[/quote]

I get something different, but it's probably wrong.  What did I do wrong?

[hide]
We have $\displaystyle{90\choose 2}$ pairs $(a, b)$.  We want the number of pairs that are 10 apart, or more.  So we list them systematically:

$(10, 20)$ thru $(10, 99)$ = 80 pairs
$(11, 21)$ thru $(11, 99)$ = 79 pairs
....
$(88, 98)$ thru $(88, 99)$ = 2 pairs
$(89, 99)$ = 1 pair

So adding this up, we have $1+2+3....+80=\frac{80 \cdot 81}{2}=3240$ as our numberator.  Our denominator is $\displaystyle{90\choose 2}=4005$.  So $\frac{3240}{4005}=\boxed{\frac{72}{89}}$.[/hide]

[quote="mysmartmouth"]What did I do wrong?[/quote]

Nothing -- Aduna subtracted ten from ninety and got seventy, is all.

Next problem!

There are five courses at my school.
243 take algebra.
323 take language arts.
143 take scocial studies.
241 take biology.
300 take history.
213 take algebra and language arts.
264 take algebra and social studies.
144 take algebra and biology.
121 take algebra and history.
111 take language arts and social studies.
90 take language atrts and biology.
80 take language arts and history.
60 take social studies and biology.
70 take social studies and history.
60 take biology and history.
50 take algebra, language arts and social studies.
50 take algebra, language arts and biology.
50 take algebra, language arts and history.
50 take algebra, social studies and biology.
50 take algebra, social studies and history.
50 take algebra, biology and history.
50 take language arts, social studies and biology.
50 take language arts, social studies and history.
50 take language arts, biology and history.
50 take social studies, biology and history.
20 take algebra, language arts, social studies and biology.
15 take algebra, language arts, social studies and history.
15 take algebra, language arts, biology and history.
10 take algebra, social studies, biology and history.
10 take language arts, social studies, biology and history.
5 take all three.
None take none.

How many people are in my school?

[hide="hint"]pie, mmmmm![/hide]

Wow that's pretty easy, but someone who has the patience to read through all that is amazing.

[hide]Let A be the subset of students who take algebra, L-languages,S-social s.,B-biology,H-history, M the set of all students.We have: \[\mid M\mid =\mid A\mid+\mid L\mid+\mid S\mid+\mid B\mid+\mid H\mid-\mid A\cap L\mid-\mid A\cap S\mid-\mid A\cap B\mid-\mid A\cap H\mid-\mid L\cap S\mid-\mid L\cap B \mid-\mid L\cap H\mid-\mid S\cap B\mid-\mid S\cap H\mid-\mid B\cap H\mid+\mid A\cap L\cap S\mid+\mid A\cap L\cap B\mid+\mid A\cap L\cap H\mid+\mid A\cap S\cap B\mid+\mid A\cap S\cap H\mid+\mid A\cap B\cap H\mid+\mid L\cap S\mid H\mid+\mid L\cap S \cap B\mid+\mid S\cap B\cap H\mid+\mid L \cap B\cap H\mid-\mid A\cap L\cap S\cap B\mid-\mid A\cap L\cap S\cap H\mid-\mid A\cap L\cap B\cap H\mid-\mid A\cap S\cap B\cap H\mid-\mid L\cap S\cap B\cap H\mid =243+323+143+241+300-213-264-144-121-111-90-80-60-70-60+50+50+50+50+50+50+50+50+50+50-20-15-15-10-10+5=472\] [/hide]

Prove that for natural $a,b,c$: \[a^{a}b^{b}c^{c}\geq(\frac{a+b+c}{3})^{a+b+c}\]

I personally have no experience with inequalities (meaning I've never solved one). And no one is responding, so I guess I'll post a problem.

A paperboy delivers newspapers to 10 houses along Main Street. Wishing to save effort, he doesn't always deliver to every house, but to avoid being fired he never misses three consecutive houses. Compute the number of ways the paperboy could deliver papers in this manner.

Source: Mock AIME, by Mildorf :D

I have seen a problem a whole lot like this from the Primary Math World Cup team test from a year I can't remember
[hide="solution"]
Call the houses (in order that they are on the street) 1, 2, 3,..., 10
After delivering to house $x$, the paperboy must next deliver to house $x+1$, $x+2$, or $x+3$.
Now all we need to do is count to 10 using 1's 2's and 3's.
Using all 1's -> 1 way
8 1's, one 2 -> 9 nCr 1 = 9 ways
6x1, 2x2 -> 8 nCr 2 = 28 ways
4x1, 3x2 -> 7 nCr 3 = 35 ways
2x1, 4x2 -> 6 nCr 4 = 15 ways
5x2 -> 1 way
7x1, 1x3 -> 8 nCr 1 = 8 ways
4x1, 2x3 -> 6 nCr 2 = 15 ways
1x1, 3x3 -> 4 nCr 3 = 4 ways
5x1, 1x2, 1x3 -> (7 nCr 2)x(2 nCr 1) = 42 ways
3x1, 2x2, 1x3 -> (6 nCr 3)x(3 nCr 1) = 60 ways
1x1, 3x2, 1x3 -> (5 nCr 4)x(4 nCr 3) = 20 ways
2x1, 1x2, 2x3 -> (5 nCr 3)x(3 nCr 2) = 30 ways
2x2, 2x3 -> 4 nCr 2 = 6 ways
1+9+28+35+15+1+8+15+4+42+60+20+30+6=$\fbox{274}$[/hide]
Actually I think I undercounted somewhere...

[quote="xime"]OK, if it is too hard :?:
[hide]We have
$a \frac{1}{a},b \frac{1}{b}, c \frac{1}{c}$
from AM-GM we have
$[\frac{(\frac{1}{a}+\frac{1}{a}+...+\frac{1}{a})+(\frac{1}{b}+\frac{1}{b}+...+\frac{1}{b})+(\frac{1}{c}+\frac{1}{c}+...+\frac{1}{c})}{a+b+c}]^{a+b+c}>(\frac{1}{a})^{a}+(\frac{1}{b})^{b}+(\frac{1}{c})^{c}$
$(\frac{3}{a+b+c})^{a+b+c}>\frac{1}{a^{a}b^{b}c^{c}}$
$a^{a}b^{b}c^{c}>(\frac{a+b+c}{3})^{a+b+c}$[/hide] And the [b]NEXT PROBLEM[/b]
(easy one)
$\frac{1}{2}\frac{3}{4}\frac{5}{6}.... \frac{2n-1}{2n}< \frac{1}{\sqrt{3n+1}}$  :wink:[/quote]

Since it is an easy one I'll do it...
[hide]
When you see integers...INDUCTION!!! (btw you made this problem easier, because I think originally they have $\frac{1}{\sqrt{3n}}$ instead of what you have now).

So it suffices to check that if $\frac{1}{\sqrt{3n+1}}*\frac{2n+1}{2n+2}\le \frac{1}{\sqrt{3n+4}}$ for all positive $n$.  Simplifying get $0 \le n$, which is always true.
[/hide]

As for the other one:
[hide]
When we see something complicated like this I like recursion.  Suppose we note $p_{n}$, then for $p_{n+1}$ we just attach another house at the end of $n$ houses.  If the boy gives the paper to this last house, then any of the permutations before is still valid.  If not, then the last permutation doesn't work when the last two houses are not delivered.  If that's the case, then the 3rd house from the end must be delivered, and all the houses before can be delivered in any matter.  Therefore, our formula is:

$p_{n+1}= 2p_{n}-p_{n-3}$

going from 1 and up we obtain $504$.
[/hide]
I didn't get your answer....

[quote="pkerichang"]As for the other one:
[hide]
When we see something complicated like this I like recursion.  Suppose we note $p_{n}$, then for $p_{n+1}$ we just attach another house at the end of $n$ houses.  If the boy gives the paper to this last house, then any of the permutations before is still valid.  If not, then the last permutation doesn't work when the last two houses are not delivered.  If that's the case, then the 3rd house from the end must be delivered, and all the houses before can be delivered in any matter.  Therefore, our formula is:

$p_{n+1}= 2p_{n}-p_{n-3}$

going from 1 and up we obtain $504$.
[/hide]
I didn't get your answer....[/quote]
That is correct.

[size=200][color=green] Next Problem: [/color][/size]
[quote="ch1n353ch3s54a1l"]Two tetrahedrons are connected by their bases. The incenter of one face of the top tetrahedron is connected with the opposite two incenters on the other tetrahedron faces and those two faces on the bottom tetrahedron are also connected. What are the lengths of the resulting triangle's sides?[/quote]

I got weird numbers...
[hide]
$\frac{s\sqrt{33}}{9}$, $\frac{s\sqrt{33}}{9}$, $\frac{s}{3}$.

I assume the two are identical regular tetrahedron, and $s$ is the side of those two.
[/hide]


Let  A={1,2...n}. How many solutions $(X,Y,Z)$ with $2\leq|X|<|Y|<|Z|$ does the equation $X\cup Y\cup Z=A$ have?

Is there humane solution for this problem?

The one who proposed in the shortlist surely knows the humane solution of this.


[quote="bjh7790"]For $ a, b, c \in \mathbb {R}^{ \plus{} }$ $ a^3 \plus{} b^3 \equal{} c^3.$

Prove that $ a^2 \plus{} b^2 \minus{} c^2 > 6(c \minus{} a)(c \minus{} b)$

Thanks[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252171


search
parameterization
number theory unsolved
number theory

Prove that $n!$ is never a square for $n>1$.

Probably it's well known but I can't find it since most search engines filter the '!'.

There will always be a prime that occurs exactly once in $n! = 1 \cdot 2 \cdot ... \cdot n$ by Bertrands Postulate (there is always a prime between $n$ and $2n$).

It isteresting equation:
$x!+a=y^2, \ x,a,y\in Z_+.$
Case parametr a=0,a>1 is easy.


Prove that there is no $x$ rational number such that $\{x^2\}\+{x\}\=1$ where $\{x\}$ is the fractional part f $x$. ;)

[quote="pbornsztein"]You did not use $\{x \}$ in the statement... ;)

Pierre.[/quote]

 Actually it is $\{x^2 \}+\{x \}=1$  ;)

Simple.

Suppose x exists. Clearly x cannot be an integer. Let $x = p/q$ where $p,q$ are integers, $q>1$, and $gcd(p,q) = 1$.

Then $x^2 + x = p(p+q)/q^2$ and this has to be an integer. But it's impossible since p and q are relatively prime.


How many non-negative integers solutions to the equation:

$2x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10}=3$

[hide="Hint"] $x_{1}$ can only be $0$ or $1$[/hide]

[hide]
[b]Case 1[/b]
$x_{1}= 1$ suggests that only one of $x_{2}, x_{3}, \ldots, x_{10}$ can be $1$ while the rest are $0$. There are $9$ such possibilities.

[b]Case 2[/b]
$x_{1}= 0$ means that our equation is:

$x_{2}+x_{3}+\cdots+x_{10}= 3$

There are $\binom{3+9-1}{9-1}= \binom{11}{8}$ solutions to this.

So in total, there are $\binom{11}{8}+9$ solutions.
[/hide]


I have a question about writing proof for USAMTS.

I know how to use Latex, but I don't know how to draw diagrams....

What programs do u use? Do u guys use Mathematica?

there's an entire thread on geometry sketching programs

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49988]www.artofproblemsolving.com/Forum/viewtopic.php?t=49988[/url]

[quote="DPatrick"]We use [url=http://www.tug.org/metapost.html]MetaPost[/url].[/quote]
Sorry, I didnt really get it... So is Metapost a package for Tex??


Has anybody seen the following problem in any book or site? if so (or if he has a solution) please tell it to me:

prove that p!=-1 (mod p), where p is prime.

Thank you!

This is [url=http://www.utm.edu/research/primes/notes/proofs/Wilsons.html]Wilson's theorem[/url], very well known.

Moreover the statement is wrong because $p! = 0 \mod [p]$  :P
The good one is $(p-1)! = -1 \mod [p]$.

Pierre.

Thanks guys, I am just a beginner ij number theory...


integration
calculus
derivative
inequalities
real analysis
real analysis unsolved

Show that if $ f$ is of bounded variation on $ [a, b]$ then

$ \int^b_a |f'| dm \equal{} T_a^b(f)$ if and only if $ f$ is absolutely continuous.

(where $ T$ denotes the total variation)

I know how to do the backward direction. However, I do not see how to do the forward direction right now. Any hints would be very nice. Thanks in advance.

Let $ V_f (a,t)$ denote the total variation of $ f$ over the interval $ [a,t]$. Since $ f$ is of bounded variation, we have $ |f'(x)| \equal{} V_{f}^{'} (a,x)$ for a.e. $ x \in [a,b]$ (where $ V_{f}^{'} (a,x)$ denotes the derivative of $ x \mapsto V_{f} (a,x)$ w.r.t $ x$ at the point $ x$). Thus $ \int_{a}^{b} V_{f}^{'} (a,x) dm(x) \equal{} V_f (a,b) \equal{} V_f (a,b) \minus{} V_f (a,a)$, which is to say that $ V_f (a,x)$ is absolutely continuous on $ [a,b]$. Now the simple inequality $ |f(y) \minus{} f(x)| \le V_f (a,y) \minus{} V_f (a,x)$ ($ x \le y$) implies that $ f$ is absolutely continuous on $ [a,b]$ as well.


geometry
perimeter
combinatorics proposed
combinatorics

Given is a collection of squares with total area 1. Is it always pssible to pack these squares, without overlapping, into another square of area 2?

Another on same topic :-)
Sorry for posting this here as I did not want to start a fresh thread.

100 segments are drawn from centre point P inside a unit square to the perimeter, such that the square is divided into 100 parts each of perimeter p. Show that 1.4 < p < 1.5

For the original problem, have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?p=21783#21783

Pierre.

And here too :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=squares+area&t=6226

Pierre.

For the related problem, have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=square+perimeter&t=15797

Pierre.


A 2.00 -diameter vat of liquid is 2.90  deep. The pressure at the bottom of the vat is 1.40 . What is the mass of the liquid in the vat?

We are doing this problem for mastering physics and I submitted my answer and got it wrong and don't know why. Here is the working.

gauge pressure 1.4 atm-1.0atm=0.4atm
volume cylinder=pi(1)^2(2.9m)=12.25^3
p=pgh=0.4(101325)=40530Pa
40530Pa=P(9.8)(2.9)
P=1426Pa
p=m/V

m=pv=1426Pa(12.25m^3)=17470kg=17500kg

What do you mean by the last step?? :maybe:
[quote="victornumber"]m=pv=1426Pa(12.25m^3)[color=red]=17470kg=17500kg[/color][/quote]

By the way, I think some books use by convention atmospheric pressure to be $ 101336.3$ Pa. You might check the default value used by your book...otherwise I think there is no mistake in your procedure.

i think sunkern is right it might just be the value of default................. :roll:


For which integers $ n > 1$ do there exist natural numbers $ b_{1},\cdots, b_{n}$, not all equal, such that the number $ (b_{1} \plus{} k)(b_{2} \plus{} k)\cdots (b_{n} \plus{} k)$ is a power of an integer for each natural number $ k$? (The exponents may depend on $ k$, but must be greater than $ 1$.)


This is a real tough one... for me at least  :)

[b]Let A be the n*n matrix aI[size=59]n[/size] + B where B is STRICTLY upper triangular. Show that A is diagonalizable exactly when B = 0.[/b]

rahul1089,

The only eigenvalue for $A$ is $a$.

If $A$ is diagonalizable then it can be written $A=P(aI_n)P^{-1}=aI_n$.

JC


geometry
parallelogram
trigonometry
Pythagorean Theorem

Mods, please don't move this to the books forum. I am not asking about CPIG specifically, I'm just using it as an example.

I have CPIG. I got the book not very long ago and I started doing a couple of problems. I am on the 13th problem and I still haven't solved one without looking at the solution. I did intro to geo twice I am fairly sure that I know the material. How long should I work on each problem (I work on them for about 1/2 hour then I quit)? Should I be doing something different than what I'm doing now?

Maybe it's your strategy for attacking these problems. How do you normally approach a general geometry problem (especially proof)?

So say, it has something to do with a right triangle. Then I try different theorems that would apply. If this doesn't work, since there are triangles I would try to get something with congruences. After this, I would look at the hints. Then, I would try some more along the lines of the hints applying different theorems. And then, I would look at the solution.

Maybe that's where the problem is. You can't just restrict yourself to see where you can apply theorems; you also have to play around with the diagram itself to create easier figures to work with. For example, it would would be impossible to prove the Pythagorean Theorem if you just tried apply theorems on it. The most simple proof requires drawing an altitude.

So try to construct simpler figures to work with from the figure your given if you can't readily apply any theorems. It is common to construct parallelograms, right triangles, isosceles triangles, etc. Sometimes you just need to draw a line in a way such that it will help. If you don't know what line is appropriate, draw one line at a time and ask yourself how it will help. Usually. if you see two lines bisecting each other, you can construct a parallelogram and if you see those same lines perpendicular to each other, you can construct a rhombus. So be on the lookout for features that allow you to construct these simple quadrilaterals.

For example, for the following problem, drawing diagonal $ \overline{BC}$ makings it fairly obvious what you do next.

Let quadrilateral $ ABCD$ have lengths $ AB \equal{} a$, $ BC \equal{} b$, $ CD \equal{} c$, and $ DA \equal{} d$. Find the expression for $ \cos{A}$.


[b][size=134]Assume that $x-2$ is a factor of the polynomial $f(x)=x^3+ax^2+bx+2$ and that $f(x)$ gives a remainder of $-3$ when it is divided by $x+1$. Then what are a and b?[/size][/b]

By Factor Theorem

$f(2) = 0$
since $x-2$ divides $f(x)$
this gives us $8+ 4a+ 2b+ 2$ = $0$
or just $2a + b = -5$

By Remainder Theorem ,we have

$f(-1) = -3$
So,
$-1+ a- b+ 2 = -3$
or $a- b = -4$

Solving, $a= -3$ and $b= 1$
The polynomial is
 $f(x) = x^3 - 3x^2 + x + 2$


Find all triples of natural numbers $(a,b,c)$ such that $a$, $b$ and $c$ are in geometric progression and $a + b + c = 111$.

How do you know that the common ratio $k$ is an integer?

The geometric progression ratio is clearly a rational number. So let it be $ \frac {p}{q}$, with $ (p,q) = 1$.
So we have $ b = \frac {pa}{q}$ and $ c = \frac {p^2a}{q^2}$, and since $ (p,q) = 1$, $ q^2|a$. So let $ a = q^2r$.
Then we have $ b = pqr$ and $ c = p^2r$, so we have $ q^2r + pqr + p^2r = 111$, then $ r|111$, so $ r\in\{1,3,37,111\}$.
If $ r = 1$, we have $ p^2 + pq + q^2 = 111$. It's easy to see the only possibilities are $ (p,q) = (1,10)$ and $ (p,q) = (10,1)$.
For the first one we have $ (a,b,c) = (100,10,1)$. For the second one we have $ (a,b,c) = (1,10,100)$.
If $ r = 3$, we have $ p^2 + pq + q^2 = 37$. It's easy to see the only possibilities are $ (p,q) = (3,4)$ and $ (p,q) = (4,3)$.
For the first one we have $ (a,b,c) = (48,36,27)$. For the second one we have $ (a,b,c) = (27,36,48)$.
If $ r = 37$ we have $ p^2 + pq + q^2 = 3$. It's easy to see the only possibility is $ (p,q) = (1,1)$, so we have $ (a,b,c) = (37,37,37)$.
If $ r = 111$ we have $ p^2 + pq + q^2 = 1$. It's easy to see this is impossible.
The solutions are:
$ (a,b,c) = (1,10,100)$
$ (a,b,c) = (100,10,1)$
$ (a,b,c) = (27,36,48)$
$ (a,b,c) = (48,36,27)$
$ (a,b,c) = (37,37,37)$

@ Dr. Ravi: You don't need to know that $ k$ is an integer. All you have to do is use the fact that a is an integer to get that $ 1\plus{}k\plus{}k^2$ is an integer as well, then find positive rational roots to $ 1\plus{}k\plus{}k^2\in \{ 1, 3, 37, 111\}$.

[quote="1=2"]All you have to do is use the fact that a is an integer to get that $ 1 \plus{} k \plus{} k^2$ is an integer as well[/quote]

$ \mathbb{Z}$ is a ring, not a field ;)

Joao Pedro Santos, you are assuming that $ a,b,c$ are [b]consecutive[/b] terms in the geometric progression.

The problem says "in geometric progression", not "in a geometric progression".  Based on that wording, I would say that $ a$, $ b$, $ c$ is a geometric progression; in other words, they are consecutive terms.  I agree that the wording could be clearer.


quadratics
number theory
greatest common divisor

Prove that if $ p$ and $ q$ are primes with at least 2 digits, then $ 240| p^4 \minus{} q^4$. Further prove that $ 240$ is the greatest common divisor of all such numbers that arise in this way.

[hide="Solution"]Writing the expression as $ (p^2 \plus{} q^2)(p \plus{} q)(p \minus{} q)$ we can extract some information from it.

First note that both the prime are of the form either $ 4k \plus{} 1$ or $ 4k \plus{} 3$. If they are both the same, then their sum is even and their difference is a multiple of 4. If they are both different then their sum is a multiple of 4 and their difference is even. Either way, this means that:

$ 8 \mid (p \plus{} q)(p \minus{} q)$.

Also note that $ p^2$ and $ q^2$ are both odd so their sum is even. This means that:

$ 16 \mid (p^2 \plus{} q^2)(p \plus{} q)(p \minus{} q) \equal{} p^4 \minus{} q^4$.

So it remains for us to prove divisibility by 15.

Both the primes are either of the form $ 3k \plus{} 1$ or $ 3k \plus{} 2$. This means that either their sum or their difference is divisible by 3. Now we have:

$ 3 \mid (p^2 \plus{} q^2)(p \plus{} q)(p \minus{} q)$.

And since we already know that it is divisible by 16 we have that $ 3 \cdot 16$ also divides it: that is

$ 48 \mid (p^2 \plus{} q^2)(p \plus{} q)(p \minus{} q) \equal{} p^4 \minus{} q^4$.

Now we need to prove divisibility by 5. Looking at quadratic residues mod 5, we see that the square of each prime is congruent to either 1 or 4 mod 5. If they are different, then their sum $ (p^2\plus{}q^2)$ is divisible by 5 and we are done. If they are both the same, then their difference $ (p^2 \minus{} q^2)$ is divisible by 5. Either way we now have:

$ 5 \mid (p^2 \plus{} q^2)(p^2 \minus{} q^2) \equal{} p^4 \minus{} q^4$

Since we already know it is divisible by 48, it is divisible by $ 5 \cdot 48 \equal{} 240$.

$ 240 \mid p^4 \minus{} q^4$.

In order to show that this is the gcd of all numbers formed this way, we just need to find a single pair of numbers of this form who share no divisor greater than 240. Such a pair is $ 13^4 \minus{} 11^4$ and $ 17^4 \minus{} 11^4$.[/hide]


Is there a multiple of $3^{2005}$ that
(a) ends in $2005$
(b) contains only $0$ and $1$ ?

Consider $3^{2005}$ modulo 10000.  The residue must be coprime to 10000, otherwise it has a factor of 2 or 5.  But powers of 3 always end in odd digits.  To show it doesn't end in a 5, we look at powers of 3 mod 10:  they are 3, 9, 7, 1, (repeat).

Thus, multiples of this residue must cycle throughout all possible residues.  So the answer to (a)  is _yes_.

--------

Take x = 3^2005.  We claim there is a k digit long string, X, of 1's that is divisible by 3^2005.

We have $9X+1 = 10^k$.  So it implies $10^k = 1 $  mod $3^{2007}$.  Now, $10$ is coprime to $3^{2007}$, so by Fermat's little theorem, $10^{3^{2007} - 1} = 1$  mod $3^{2007}$.  Then $k = 3^{2007} -1$ which completes the proof.

[quote="indybar"][quote="Singular"]

Thus, multiples of this residue must cycle throughout all possible residues.  So the answer to (a)  is _yes_.
[/quote]

Why? I can't see. Can you show?[/quote]
Suppose that among the first $10000$ multiples of $3^{2005}$, there are two the same modulo $10000$.
$a \cdot 3^{2005} \equiv b \cdot 3^{2005}$ mod $10000$ with $0 \leq a < b < 10000$.
So $(b-a)(3^{2005}) \equiv 0$ mod $10000$, but since $3^{2005}$ is coprime with $10000$, $(b-a)$ must be $0$ mod $10000$, so $a = b$. Contradiction.

So the first $10000$ multiples of $3^{2005}$ contain all $10000$ possible residues mod $10000$, hence $2005$ must be one of them.

Very nice! May I add
(c) contains only $7$
and
(d) contains all digits from $0$ to $9$

[quote="Peter VDD"][quote="warut_suk"]Very nice! May I add
(c) contains only $7$ [/quote]How do you mean? We had a 1-string... so that one x7? :) Or did you mean something different?[/quote]

Oops... I didn't see that. Sorry  :blush:

For d, can't you just use the all 1's thing (call it X again) and append 2X, 3X, 4X, etc. to it?

for b   if $m=3^{2005} $ or whatever :)
consider   $1,11,111,1111,111111,\cdots $ all mod  $m$
since each value is between $0$  and $m-1$ ..  two of them should be same mod $m$, now difference of these two number is obviously divisible by $m$ and is of the form $1111111\dots 0000$
for d, if nothing else, as some one has suggested, just write down above number nine times one after other and replace $1$ with $2,3,4 $ etc...


here's a problem I'm trying to help my 7th grade daughter with and I
just cant remember how to solve it...

"You are given 5 different weights, marked 1lb, 2lb, 5lb, 6lb, and
10lb. How many different weights can you obtain using these weights?"

My first thought was to do something with 5!, but that doesn't address
the need to get unique combinations.

I've tried searching in a number of different ways to find similar
problems, but can't find anything. Seems like it should be a simple
answer.

Can anyone help?

[quote="lunaclaire"]
"You are given 5 different weights, marked 1lb, 2lb, 5lb, 6lb, and
10lb. How many different weights can you obtain using these weights?"
[/quote]

First, what i would do is look to see if any combination of weights is equivalent to one other weight. the only one i see is 1+5=6. then, see if any combinations of weights are equivalent to two weights put together. we see 2+5=1+6, 1+10=5+6, 10+2=6+5+1. so all these equivalents are:

1+5=6

2+5=1+6

1+10=5+6

10+2=1+5+6

use the combinatoral identity:

\[ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\cdots\plus{}\binom{n}{n}\equal{}2^n\]

what we're doing (if we exclude 0) is:

\[ \binom{5}{1}\plus{}\binom{5}{2}\plus{}\cdots\plus{}\binom{5}{5}\equal{}31\]

subtract the 4 equivalents, since we counted them twice, so you get 31-4=27.

uh-oh. i got a different answer from mewto...that probably means i'm wrong.

thanks for the replies

I'm looking for an approach different from the brute force approach, 'cause I think she needs to show some logic in her solution.

So... the last reply above might be the way to go, though that looks like something more advanced that her advanced 7th grade algebra class would have gotten to yet... so, if there are any other ideas, I'd appreciate continuing our conversation a little longer

that said, can anyone give me a little more to look up re: the last reply's "combinatorial identity" as a way pf approaching this?

Generating functions make a more "organized" way to list out possibilities.

Just multiply $ (1 \plus{} x)(1 \plus{} x^2)(1 \plus{} x^5)(1 \plus{} x^6)(1 \plus{} x^{10})$--the coefficients of the terms just mean the number of ways to get that weight. (Do you see why this works? It may seem advanced at first, but it is pretty intuitive.)

Since we just want to see what possible coefficients there are, we can multiply this in an easier way--I will use the notation $ (a,b,c,\cdots)$ to represent which terms exist ($ (14)$ would mean that there is an $ x^{14}$ term):

$ (0,1)(0,2)\equiv(0,1,2,3)$

$ (0,1,2,3)(0,5) \equiv (0,1,2,3,5,6,7,8)$

$ (0,1,2,3,5,6,7,8)(0,6) \equiv (0,1,2,3,5,6,7,8,9,11,12,13,14)$

$ (0,1,2,3,5,6,7,8,9,11,12,13,14)(0,10)$

$ \equiv (0,1,2,3,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,23,24)$

Thus, those are the weights that we can make, and since the only ones that don't work (i.e. don't exist or are invalid) are $ 0,4$, giving us $ 24 \minus{} 2 \equal{} \boxed{22}$.

qwerty's solution is good as long as there are no repeats (i.e. coefficients greater than two), which is not the case here.

Have you thought it your self math154?  :) It looks geniusly  :w00t:

You can obtain 1lb with just the 1lb weight.
By allowing the 2lb weight, you can extend your reach to 3.
However you can't obtain 4lb. By using the 5/6lb you can get 10lb.
You can get 11lb with 10,1 and 12lb with 10,2.
From 13~24lb, basically we try to remove weights from the full set of 24lb.
However we can't remove 4lb, so 20lb is impossible.

Therefore, there are 24-2=22 ways.

That's essentially smart casework. :wink:

oh ok :blush:
but i still don't get why it doesn't work. could explain that to me?

You probably missed some repeat cases. I'm not sure though.


Just use the formula for the no. of divisors....I think you can answer this question on your own...

If $ n\equal{}p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ where the $ p_i$'s are prime.
Then the no. of divisors of $ n$ are
$ (\alpha_1 \minus{}1)(\alpha_2 \minus{} 1)\cdots (\alpha_n \minus{} 1)$
Using this formula, get the answer...
I'm feeling too lazy to do this myself

I don't understand very well, I tried, but I didn't find the answer.

$ 15\equal{}3\cdot 5$

Either the number is $ p^2q^4$ or $ p^{14}$ (where $ p,q$ are primes, obviously).

In the former case, it's clear that $ q\leq 3$. Checking $ (p,q)\equal{}(2,3)$ is the only solution.

In the latter case, $ 2^{5}$ is already $ 512$, so for any prime $ p$, $ p^{14}$ will be greater than $ 500$.

So $ 324$ is the only solution.

[quote="luiseduardo"]How many positive integer numbers less than 500 have exact 15 positive integer dividers ?

A) 0
B) 1
C) 2
D) 3
E) 4[/quote]
What does it mean for a number to have an odd number of positive divisors?

[quote="grn_trtle"]$ 15 \equal{} 3\cdot 5$

Either the number is $ p^2q^4$ or $ p^{14}$ (where $ p,q$ are primes, obviously).

In the former case, it's clear that $ q\leq 3$. Checking $ (p,q) \equal{} (2,3)$ is the only solution.

In the latter case, $ 2^{5}$ is already $ 512$, so for any prime $ p$, $ p^{14}$ will be greater than $ 500$.

So $ 324$ is the only solution.[/quote]

You forgot $ (3, 2)$.  $ 3^2\cdot2^4\equal{}144$ clearly works as well.  Therefore, there have to be $ \fbox{2}$ solutions.

[quote="grn_trtle"]$ 15 \equal{} 3\cdot 5$

Either the number is $ p^2q^4$ or $ p^{14}$ (where $ p,q$ are primes, obviously).

In the former case, it's clear that $ q\leq 3$. Checking $ (p,q) \equal{} (2,3)$ is the only solution.

In the latter case, $ 2^{5}$ is already $ 512$, so for any prime $ p$, $ p^{14}$ will be greater than $ 500$.

So $ 324$ is the only solution.[/quote]

Like the person above me, you missed 144, as well as 400.

In the latter, there are no solutions, but I think you meant $ 2^9\equal{}512$


The last number in each row is $ 1\plus{}2\plus{}\cdots\plus{}n$, and there are $ n$ numbers in each row.  So in the $ n$th row, the sum of the numbers is $ (1\plus{}\cdots\plus{}n)\plus{}(1\plus{}\ldots\plus{}n\minus{}1)\plus{}\ldots\plus{}(1\plus{}\cdots\plus{}n\minus{}n\plus{}2)\plus{}(1\plus{}\ldots\plus{}n\minus{}n\plus{}1)$.

Alternatively, note that in going from the $ (n\minus{}1)^{th}$ to the $ n^{th}$ row, two times the average of the numbers increases by $ 2n\minus{}1$ (starting with $ 1\plus{}1$ in the first row), and we thus have $ S_n\equal{}\frac n2(2\plus{}3\plus{}5\plus{}\cdots\plus{}(2n\minus{}1)\equal{}\frac n2(1\plus{}(1\plus{}3\plus{}\cdots\plus{}(2n\minus{}1)))\equal{}\boxed{\frac{n(n^2\plus{}1)}2}$.


Determine the smallest real constant k which satisfies:

[i]Any sequence of 2004 reals has some consecutive terms,
the sum of which does not differ more than k from an integer.[/i]

Please DO NOT post anything here until I say you can!!!


Hi, i need help on this:

Four points are chosen on or inside a circle of radius 1 unit. Show that at least 2 of the points are within sqrt 3 units of one another.

I know that you divide the circle into 3 parts to use pigeon hole principle. now what nexT?

thanks and i appreciate workings....:D

Prove that the distance between any 2 points in each part is less than or equal $ \sqrt{3}$

uhh..that's what i was supposed to ask, i'm stuck at that part. how could you prove that it's less than sqrt3?
does it involve sectors and the like?

At least, 3 points are different from the centre O. Choose 2 points (say, $A_1,A_2$) with minimal angle $\angle A_1OA_2$. $\angle A_1OA_2$ can be at most $360^{\circ}/3=120^{\circ}$. Indeed, if $\angle A_1OA_2$ be greater, then the sum of all such possible angles will be greater than $360^{\circ}$, a contradiction. We have $\left |OA_1\right |\leq1$, $\left |OA_2\right |\leq1$. From cosine rule in triangle $A_1OA_2$, $\left |A_1A_2\right |\leq\sqrt{3}$.


Αν $ k,n$ θετικοί ακέραιοι και $ k\geq 2$ να δείξετε ότι:

$ \prod_{i \equal{} 1}^{n} (1 \plus{} \frac {1} {k^i}) < \frac {2k \plus{} 1} {k}$

Ένα hint για όσους θέλουν να την προσπαθήσουν:

[hide]Αποδείξτε την ισχυρότερη ανισότητα $ \prod_{i \equal{} 1}^{n} (1 \plus{} \frac {1} {k^i}) \leq \frac {2k \plus{} 1} {k}\minus{}\frac{1}{n}$ και μετά επαγωγή πάνω στο $ n$. [/hide]

Αλέξανδρος


trigonometry
calculus
integration
calculus computations

I've tried a lot of ways of approaching this question but I can't get it out:

Evaluate $I$ when \[I=\int^{\theta}_{0}{\frac{1}{1-a\cos{x}}}\,dx\] for $0<\theta <\pi$ and $-1<a<1$

[quote="perfect_radio"]Have you tried the substitution $y = \tan \left( \frac{x}2 \right)$? I think it works out...
(If not, sorry for the useless post.)[/quote]
Yes! perfect_radio is right. It's standart problems from trig integrals. In much times there works substitution $y = \tan \left( \frac{x}2 \right)$.
We consider two case :
When $-1<a<0$ and when $0<a<1$ in first case we take
$\int\frac{dx}{1-a\cos{x}}=\frac{2}{\sqrt{1-a^{2}}}arctan(\sqrt{\frac{1+a}{1-a}}\tan{\frac{x}{2}})$
seccond case I Think you can calculate :wink:


algebra
polynomial
LaTeX
superior algebra
superior algebra unsolved

Hi,
Our tutor set us this problem as an extra exercise (it's not worth any marks but it's interesting.)

"A real number, s, is algebraic if there exists a polynomial

p(x) = ax^k + bx^(k-1) + ... + yx + z

where a,b,...,y,z are integers such that p(s) = 0. (note, there isn't necessarily 26 of them I just can't get LATEX to install on my computer so can't do the constants as all a's with indexes)

Show that if s is algebraic then so is the square root of s.

I came up with a solution but would like to see if other people get the same as me, also it's a really nice proof so it's quite interesting to have a go at. Thanks.

hi,
Set $\alpha = \surd{s}$,Its easy to see that $\alpha$ satisfies the polynomial $p(X^2)$.
If you are new to the arena of algebraic numbers ,then i am sure u have'nt yet come across the most exciting things.
Cheers
Rohith

You need not to install $\LaTeX$ to use it here ;)
See http://www.mathlinks.ro/Forum/viewtopic.php?t=5261.

Hi.

In Fact, this is very simple:
Look at the extension $L_1=\mathbb{Q}(s)$ over $\mathbb{Q}$. Its degree is the degree of the minimum polynomial of s. Now take another extension $L_2=L_1(s')$ with $s'^2=s$. The degree of $L_2$ over $L_1$ is at most 2 and so the degree over $\mathbb{Q}$ is finite (at most $|L_1: \mathbb{Q}|\cdot 2$ ) and so it is an algebraic extension. Especially s' is an algebraic element.

Gockel.

Edit: I'm just too slow... :roll:

I am new to albegraic numbers, though I some about transcedental numbers, look forward to finding out more. My speciality is statistics really.
I tried using p(x^2) to start with but since there is no reason to assume that the same integers a,b,...,y,z satisfy p(x^2) being an algebraic polynomial I don't think that works.

p.s Thanks, will go look at the LATEX tutorial now!

hmmmm,
there is a reason why $p(X^2)$ works because $P(\alpha ^ 2) = P(s)$.
 :D

True, I'm just being really really dim! I did what Gockel did but that works too. Oops  :blush:


I feel that this one is known, but until a link is found let's discuss it here. The required maximum is $6$. This comes down to proving the inequality

$6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}$
$\geq\left(a+b\right)^{4}+\left(a+c\right)^{4}+\left(a+d\right)^{4}+\left(b+c\right)^{4}+\left(b+d\right)^{4}+\left(c+d\right)^{4}$

for any four reals $a$, $b$, $c$, $d$.

Using the identity $2\left(x^{4}+y^{4}+6x^{2}y^{2}\right)=\left(x+y\right)^{4}+\left(x-y\right)^{4}$, this becomes rather easy:

$6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}$
$=6\left(a^{4}+b^{4}+c^{4}+d^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2a^{2}d^{2}+2b^{2}c^{2}+2b^{2}d^{2}+2c^{2}d^{2}\right)$
$=2\left(a^{4}+b^{4}+6a^{2}b^{2}\right)+2\left(a^{4}+c^{4}+6a^{2}c^{2}\right)+2\left(a^{4}+d^{4}+6a^{2}d^{2}\right)$
$+2\left(b^{4}+c^{4}+6b^{2}c^{2}\right)+2\left(b^{4}+d^{4}+6b^{2}d^{2}\right)+2\left(c^{4}+d^{4}+6c^{2}d^{2}\right)$
$=\left(\left(a+b\right)^{4}+\left(a-b\right)^{4}\right)+\left(\left(a+c\right)^{4}+\left(a-c\right)^{4}\right)+\left(\left(a+d\right)^{4}+\left(a-d\right)^{4}\right)$
$+\left(\left(b+c\right)^{4}+\left(b-c\right)^{4}\right)+\left(\left(b+d\right)^{4}+\left(b-d\right)^{4}\right)+\left(\left(c+d\right)^{4}+\left(c-d\right)^{4}\right)$
$\geq\left(a+b\right)^{4}+\left(a+c\right)^{4}+\left(a+d\right)^{4}+\left(b+c\right)^{4}+\left(b+d\right)^{4}+\left(c+d\right)^{4}$

(because fourth powers are nonnegative).

  Darij


AMC
AIME
USA(J)MO
USAMO
AMC 12
MATHCOUNTS
AMC 10

As some of you may know, many consider this year's AIME very easy.

What do you predict the cutoff will be for sophomores and below?

I recommend you edit to give options for "10" and "11 and higher". It seems that a 10 cutoff may the case...

Im guessing a nine simply I so can be dissapointed now and then happy later if im wrong, and not vica versa. Right now im feeling frusterated both at myself and at the aime probelm writers. Seeing how it had alwasy been at most an 8, and how last time I made tons of stupid mistakes, I figgured id get 8 right, and then for the most part keep checking those 8, and thus I never spent a whole lot of work on any of the other seven.(except at the end, when I tried to do 13 but messed up there). Looking over the problems, I could have at least solved 11 and possibly 6 as well had I chosen to spend more times on the other problems, but as things stood, I played it safe and looks like its gonna kill me(ironically, the 10th grade and below qualifying bar is ment to help people like my make USAMO, not hurt them. But with my AMC score, had I been in 11th I wouldnt have played it safe, might have solved 2 more, and very likely wouldve scored high enough for USAMO  :P )

People always underperform what they "know they can do," so I predict a floor of 8.

[quote="guangweicao"]As of right now, it looks like an 8 (hopefully)[/quote]

What we say it "should" be is probably not going to affect what it will be. :-P But an 8 would be nice. It means we've overestimated the scores that are coming up.

Ahhhhhh. I'll cry if it's a nine. Not only did I get an eight, but I solved three of them afterwards. It's always more dissappointing when you solve more afterwards...

[quote="NightFlarer"]You guys are silly.  How could the AIME cutoff possibly be a 10?  Just because you and your friends on AOPS or wherever scored 10's and above doesn't mean the cutoff will be a 10 or higher!

I guess most people are assuming that because a lot of people scored highly the curve will be high, but this isn't true.  What about all those people who qualified for AIME by getting a 101 on the AMC 12?

So basically my guess is the curve will be an 8, if not, a 9.[/quote]

I'm not going by what me and my friends scored; I'm going by how hard I thought the test was, how it compared to previous AIMEs, and the cutoffs then.  And in my opinion, because the test was much easier than years with cutoffs of 8, it will be higher than an 8.

As for people with 101 on AMC12, they don't affect things.  We're only talking about 250 spots; the fact that there are many people with low AIME scores doesn't change the cutoff.

I have a feeling it will be a 9.  I think a 10 is too high for a cutoff on this test though.

It will be a 9. Easiest AIME ever (not including the 1980's AIME's), but still, it's not THAT easy.

Well I won't say anything about easiest ever.. but the first few were pretty cinchy... I can't believe that I made a silly error on the marching band problem!!! O well I'm in 8th grade so I have four more tries... so I think 8 will be the floor.. 9 is just insane...

Hopefully, for me at least, it will be a three, but that would never happen.

With a difficult AMC 12 this year, I tend to think that the floor value will be high (e.g, people who scored high on the AMC will likely do similarly well on the AIME).  I just skimmed over Albert's solutions, but I also did a few of the questions two nights ago with a friend, and there's no way that the floor is even an 8, I think.  I did question 12 in about 2 minutes - that shouldn't be happening.  Also, no really tricky counting questions with lots of cases (save for 13) showed up so there should be less marks lost to that than normal.

So, does alternate AIME change your predictions? I now think it might be an 8. Which makes me very happy.

I think AIME floor will be 8 will a 226 index. :)

Last year, the USAMO index was 210+ and the floor value was a 7.  That means that someone must have gotten a 140+ on the AMC 12, and only a 7 on the AIME.  This year, the top 160 AMC 12 scores were all above 138.  Even with the easy AIME, not all of those people have gotten 10s or higher.  Thus, I think the floor value will be a 9.  (and 138+90=228, so I think the USAMO index will be a 228+).

I agree with blahblahblah. It will be a 9. Last year, w/ a hard AIME, this USAMO index was 210+ and the floor value was 7, so one person ust have gotten a 140+ on the AMC 12 and a 7 on the AIME. This year, even with an easy AIME, one of the top 160 AMC 12 kids (who all got 138+) will end up with a 9 on the AIME.

[Moderator Edit: due to double post]

I'm a twelfth grader and I got a 237 index... How do you think? I'm also young for my grade, if that matters at all (two years skipped)

With a 237 you are very likely to make USAMO. Only your grade, not age, really matter for making USAMO.

Well, for us non-geniuses I think skipping 2 grades is a feat in itself. A 237 is an admirable score. :-)

I think you'll make USAMO with flying colors...lol

And Keone or Dimitar, would one of you guys take care of that double-post a few posts ago?

Am I the only one that's thinking a 245 index and 10 AIME?

Okay, I am going to say the index is 255 and the AIME floor is 11! Look, I am special! Nah, the index will be 227 and the cutoff will be a 9. As for the grade, not age, mattering for USAMO: Does this mean that a 20 year old home-schooler could pose as a freshman to make MOP easily? I am not a 20 year old home-schooler, but it is an interesting situation to think about.

[quote="JGeneson"]As for the grade, not age, mattering for USAMO: Does this mean that a 20 year old home-schooler could pose as a freshman to make MOP easily? I am not a 20 year old home-schooler, but it is an interesting situation to think about.[/quote]The age limit is 19.5.

And also, if someone had the mental capacity to make USAMO, they would most likely graduate on time. The only counterexample I personally know was a student who recently moved to the United States and chose to take an extra year of high school for English education purposes.

[quote="peterhuang"]Hi,
I just signed up.
I'm in 10th grade now.
I got 138 on the AMC 10, and 131 on the AMC 12, 10 on the AIME.
Do you think I can make the USAMO?[/quote]

A 10 on the AIME is almost sure to get you into the USAMO.

I'm thinking 9 and 230.  It was definitely easier than 2003.  Well, it's not like it matters for me.  I think I got about 5 on the AIME.  Stupid mistakes like whoa.

yes ednerd, you're the only one thinking that.


function
geometry
3D geometry
sphere
topology
real analysis
real analysis theorems

I am trying to figure out if I can say that a meromorphic function in a compact set on the Riemann sphere will only have finitely many poles. If there are an infinite number of poles I can by Bolzano-Weierstrass say that there is a sequence of distinct poles converging to some point, but I am not sure that this point will actually be in the compact set.

Is there some characterization of compact sets on $S_{2}$ like we can say in $R^{n}$ that a set is compact iff it is closd and bounded?

Thanks for all help.

Since the sphere is compact (and Hausdorff) a set on the sphere is compact if and only if it is closed.


function
real analysis
real analysis solved

hey
plz help me out with this one (and also the previous one i posted :D )
Find all twice differentiable functions on (0,infinity)    such that f'(x)>0  and
f(f'(x))=-f(x)
thanxss

Here's what I did, but there are some parts where I'm not sure it's correct:

Derive the equality (denote g(x)=f'(x)) and we get g(g(x))g'(x)=-g(x) (*). Because g(x):(0, infinity)->(0, infinity) g'(x)<0 for any x>0, (so g(x) is strictly decreasing). We have g(x)-> to l as x-> infinity. If we assume l>0, then we get (note that g'(x)->0 as x->infinity, this easy to prove for any function g which is strictly decreasing and bounded from below, and in our case g(x)>0 for any x>0) g(g(x))g'(x)->g(l)*0=0, so -g(x)->0, so l=0, contradiction, so the assumption l>0 is wrong, therefore g(x) decreases to 0 as x tends to infinity. The image of g is (0,a), so g:(0,infinity)-> (0, a) is bijective.

Then in (*) we may replace x with (g)^(-1)(x), which will be called h(x) (the inverse function of g). So, for any x in (0, a) we have g(x)g'(h(x))=-x, so -g(x)/x=1/g'(h(x))=h'(x). You can check that last part of the equality. We just use the fact that h(g(x))=x and we derive it. So -g(x)=xh'(x) for any x in (0, a). By adding h(x) we get h(x)-g(x)= (xh(x))' for all x in (0, a). Now we can replace x back with g(x). We have x-g(g(x))=(xg(x))'=g(x)+xg'(x) (**), and this works for any x>0. By combining (*) and (**) we get (1-g'(x))*(x-g(g(x))=0. But as g'<0, we can't have g'(x)=1 for any x so we must have x=g(g(x)), so in fact h(x)=g(x). Remember that we obtained h(x)-g(x)=(xh(x))'. By replacing h(x) with g(x) we get 0=(xg(x))' so xg(x) is a constant. The solutions are g(x)=k/x, and f must be a primitive of g. I think f has to be k*ln(x)+C, but probably C=0. I didn't check it out. Do you think it's correct?

Sorry! There's something wrong with the proof: if g is a decreasing function and it's bounded from below, it doesn't necessarily follow that g'(x)->0 as x-> infinity, but I think the proof of the problem can still be saved. We assume that the lower margin of g is l>0, as we did before, and take the bijective function g:(0, infinity)-> (l, a) instead of (0, a). I think that's it. I have no idea why the fact that g'->0 as x->infinity (which, as I said, I now think is false) seemed so obvious to me :?

Oh, and sorry about all those sad faces in the first post. I could swear I pressed the smiling green face... :D

Me again! :D  Sorry for bugging you. I just wanted to say that, even if it's not important, we can still prove that g->0 as x->infinity. We do the same as in the first post, but instead of using g'(x)->0 as x->infinity (which is not true) we use the fact that there is a sequence a_n->infinity such that g'(a_n)->0 as n tends to infinity, and the fact that g has a limit for x->infinity because it's strictly decreasing and bounded from below.

I also figured out the sad faces. When I presses : and ( to write g 'defined on (0, infinity)' that was transformed in the sad face. Stupid me! :D

Hey.  Something I've had trouble with before (when doing math over yahoo messenger) is that these programs have a tendency to interpret :( as the frowning smileyface, even in such contexts as f:(0, infinity)->(l, a).  Disabling graphical smileys will take care of that problem if you don't mind losing the cute little yellow faces.

math on yahoo messenger ? :) that sound spooky :D


algebra
polynomial
calculus
integration
linear algebra
algebra proposed

Let $ f(x)$ be a polynomial with degree at most 4. We have the following equation
\[ f(x\plus{}1) \minus{} f(x)  \equal{} x^3, \qquad x \equal{} 1,2,3,4
\]
and also $ f(0)\equal{}0.$ Determine $ f(x)$.

The most natural way of solving this problem is to let
\[ f(x) \equal{} a_4x^4 \plus{} \cdots \plus{} a_0,
\]
and from $ f(0)\equal{}0$ to deduce $ a_0\equal{}0.$
Then use the equations
\[ f(x\plus{}1) \minus{} f(x)  \equal{} x^3, \qquad x \equal{} 1,2,3,4
\]
along with Gaussian elimination to determine $ a_1, \ldots, a_4.$

I'm wondering if there are simpler and more elegant solutions. Please let me know.

[quote="lenny"]Let $ f(x)$ be a polynomial with degree at most 4. We have the following equation
\[ f(x \plus{} 1) \minus{} f(x) \equal{} x^3, \qquad x \equal{} 1,2,3,4
\]
and also $ f(0) \equal{} 0.$ Determine $ f(x)$.

The most natural way of solving this problem is to let
\[ f(x) \equal{} a_4x^4 \plus{} \cdots \plus{} a_0,
\]
and from $ f(0) \equal{} 0$ to deduce $ a_0 \equal{} 0.$
Then use the equations
\[ f(x \plus{} 1) \minus{} f(x) \equal{} x^3, \qquad x \equal{} 1,2,3,4
\]
along with Gaussian elimination to determine $ a_1, \ldots, a_4.$

I'm wondering if there are simpler and more elegant solutions. Please let me know.[/quote]
First,Because $ f(x)$ is  a polynomial with degree at most 4 and
\[ f(x \plus{} 1) \minus{} f(x) \equal{} x^3
\]
At four points so
\[ f(x \plus{} 1) \minus{} f(x) \equal{} x^3,\forall x
\]
Now ,I find $ f(x)$ when $ x$ is a positive integer.It is easy to see that
\[ f(x \plus{} 1) \equal{} x^3 \plus{} (x \minus{} 1)^3 \plus{} ... \plus{} 1 ,\forall x\in \mathbb Z^*
\]
so $ f(x) \equal{} (x \minus{} 1)^3 \plus{} (x \minus{} 2)^3 \plus{} .. \plus{} 1 \equal{} \frac {(x \minus{} 1)^2x^2}{4},\forall x\in \mathbb Z^*$
Because $ f(x)$ is  a polynomial with degree at most 4 we also get $ f(x) \equal{} \frac {(x \minus{} 1)^2x^2}{4}$ is the unique polynomial
 satisfies your condition.
Sorry for my bad English,and if my solution is wrong

Let $ g(x)\equal{}\frac{(x\minus{}1)^2x^2}{4}$
Then:
$ f(1) \minus{} g(1) \equal{} f(2) \minus{} g(2) \equal{} f(3) \minus{} g(3) \equal{} f(4) \minus{} g(4) \equal{} f(5) \minus{} g(5) \equal{} k$($ k \in R$)
Because $ deg f \leq 4,deg g \equal{} 4$ and the polynomial $ f(x) \minus{} g(x) \minus{} k$ has five different root in R then:$ f(x) \minus{} g(x) \equal{} k$ ($ \forall x \in R$)
But $ f(0) \minus{} g(0) \equal{} 0 \minus{} 0 \equal{} 0$,we deduce that k=0 and $ f(x) \equal{} \frac {(x \minus{} 1)^2x^2}{4}$
Q.E.D

There is a clever way to simplify the linear algebra that makes it very easy essentially because we pick a basis where everything becomes triangular; you won't even be able to tell that I'm doing any linear algebra.  See the method [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]here[/url] and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url], as well as [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]here[/url].  Demonstration:  the difference table for $ x^3$ is

\begin{tabular}{ | l | c | c | c | r | } x & P(x) & \Delta & \Delta^2 & \Delta^3 \\
\hline 0 & 0 & 1 & 6 & 6 \\
1 & 1 & 7 & 12 & \\
2 & 8 & 19 & & \\
3 & 27 & & & \\
\hline \end{tabular}

which gives $ x^3 = 6 {x \choose 3} + 6 {x \choose 2} + {x \choose 1}$.  The lemma in the first post then gives

$ f(x) = 6 {x \choose 4} + 6 {x \choose 3} + {x \choose 2}$

which simplifies to the usual answer.  You should think of this as "discrete integration," which has a very nice form in the Newton basis.  (The disadvantage of using the Newton basis is that while it is very easy to take sums and finite differences, it is very hard to take products.)

The other typical way to look at this problem is to generalize it to the problem of computing the sums $ \sum_{k = 1}^{n} k^r$, which is a polynomial of degree $ r + 1$ in $ n$; the answer is given by a rather [url=http://en.wikipedia.org/wiki/Faulhaber%27s_formula]curious formula[/url] involving a rather deep sequence.


It's not exactly HSB material.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=459558#p459558


advanced fields
advanced fields theorems

Hello,

I am looking for sources and general information about Chaitin's constant. In particular, I would like a proof of it's normality and transcendence.

Cheers,

Justin.

informations and references are given in the wikipedia article.


geometry
circumcircle
incenter
angle bisector
geometry proposed

Given an acute angle  triangle $ ABC$. $ \omega$ is a circle such that $ \omega$ touches $ AB$ , $ BC$ and circumcircle of $ \triangle ABC$ at $ P,Q,R$ respectively. Prove that angle bisector of $ \angle ARC$ intersects $ PQ$ at it's midpoint.

According, for example, to [url]http://www.mathlinks.ro/viewtopic.php?t=31739[/url], the midpoint of $ PQ$ is the incenter $ I$ of triangle $ ABC$.
See here for the rest: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=22925[/url].


How many positive four-digit even numbers can be written without using any of the digits 0,1,2,3 or 4?

Call the number $ abcd$. If we may not use 0,1,2,3,4, then we may only use 5,6,7,8,9. We have 5 choices for the first number, 5 for the second, and 5 for the third. However, for the last one, we only have 2 choices because 6,8 because the number must be even. Thus, we have: $ 5 \cdot 5 \cdot 5 \cdot 2 \equal{} \boxed{250}$.


Δινεται η ευθεια (ε) x=y=z/2 και το σημειο p(x1,x2,x3) του R^3.Εαν
Τ:R^3-->R^3 η γραμμικη απεικονιση που απεικονιζει το τυχαιο σημειο p του R^3 στην ορθη του προβολη πανω στην ευθεια (ε),εξεταστε εαν η Τ ειναι γραμμικος ισομορφισμος.

exume
 Kerf={v pu annikun ston R^3 tetia oste f(v)=0}
omos iparxun apira simia ton opion i or8i provoli pano stin ey8ia x=y=z/2  na ine to simio (0,0,0) tu R^3  [sigekrimena ine ena epipedo..]
diladi $dimKerf=2$,epomenos i f dn ine monomorfismos kai ara den ine isomorfismos.. :wink:

 :D


Given $ a, b, c \geq\ 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that: $ \frac {a^2 \plus{} bc \plus{} 4ab}{a \plus{} 8b} \plus{} \frac {b^2 \plus{} ca \plus{} 4bc}{b \plus{} 8c} \plus{} \frac {c^2 \plus{} ab \plus{} 4ca}{c \plus{} 8a} \geq\ 2$

@babyloverain and maiquothang: [hide]Chuc' mung` em moi' len chuc' CTV ^^!. Bi gio` box bdt co' Materazzi va` em la` anh yen tam rui`. Mong 2 em co' gang' hoan` thanh` tot' nhiem vu cua? minh`  :) [/hide]
PS: Best wishes for you, 2 younger brothers of me!

[quote="gb2124"]It really takes a lot of time to prove it.By the way,I've prove it by using Sos.[/quote]
SOS is a power method. But I really surprise with the way which you'll use SOS to solve this problem. Can you post your solution?
Thank you , gb2124 :)

I have something to do with these days so I will post the solution tomorrow.  :P  Wait for my coming back

All right.I'm coming back.Here is my solution by the method of Sos.

Wow, I really admire you because of your patience.  :huh:
But what a pity, your proof was wrong. In your proof, you proved:
$ \sum\ \frac {a^2 \plus{} bc \plus{} 4ab}{a \plus{} 8b} \geq\ \frac {2(a \plus{} b \plus{} c)}{3}$
But this ineq is not true. Try $ (a,b,c) \equal{} (2,1,0)$  :wink:

:blush:  :blush:  :blush:
There is a mistake in my solution.I feel that this leads to death to my method.
I feel terribly sorry for spamming.

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that: $ \frac {a^2 \plus{} bc \plus{} 4ab}{a \plus{} 8b} \plus{} \frac {b^2 \plus{} ca \plus{} 4bc}{b \plus{} 8c} \plus{} \frac {c^2 \plus{} ab \plus{} 4ca}{c \plus{} 8a} \geq\ 2$
[/quote]

@babyloverain: The hint: You should use the lenma: $ \sum\ \frac{a}{a^2\plus{}2bc} \leq\ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$
 :)


Well, I don't understand what's the real problem since there's no explanation about $n$..
Should we find the possible values of $n$ ?

Yes, find the great n. I forgot to post the request :blush:
So I will rewrite it!

Find the great n satisfying:
$n^{2}\leq 1000^{1000}+1001^{1001}+...+1997^{1997}\leq (n+1)^{2}$


whenever i made some posts, they didnt add to my post count, and i think there's a problem there. Also, i once went on and found that my post count got deducted. I dont know why, and i'd really appreciate it if you could help me here. thanks

Where are you posting? And the only way of losing post counts is if someone deletes a topic (I think).

Posts in G&FF,Polls, and Test Forum don't count. (The only ones that I know)


pigeonhole principle
geometry
3D geometry
number theory proposed
number theory

Prove that there exist a partition of the set $A=\{1^{3},2^{3},...,2000^{3}\}$ into $19$ nonempty subsets such that the sum of elements of each subset is divisible by $2001^{2}$

Does this use some kind of pidgeonhole with cubic residues? How can you deal with those messy cubes?...

My answer is likely to be wrong; however, I'm not capable of finding the mistake.

First of all, given a set $a_{i}\in N$ with $1\leq a_{i}<2001$ such that $\forall i,j \; a_{i}+a_{j}\neq 2001$, the sum of distinct elements of the set  $\left\{ 1^{3}, 2^{3}, 3^{3}\ldots 2000^{3}\right\}$, $\sum_{i=1}^{n}(a_{i}^{3}+(2001-a_{i})^{3}) = \sum_{i=1}^{n}(2001^{3}-3 \times 2001^{2}a_{i}+3\times 2001a_{i}^{2}\equiv 3 \times2001\sum_{i=1}^{n}a_{i}^{2}\;\;(mod \; 2001^{2})$, which is  $\equiv 0$ if   $\sum_{i=1}^{n}a_{i}^{2}\equiv 0 \;\;(mod \; 667)$.
Consider an arbitrary prime $p$ and one of the primitive roots $\zeta$ $(mod\; p)$ and take $k>2$, $k%Error. "divides" is a bad command.
(p-1)$. Then, for any $0\leq r<\frac{p-1}{k}$,   $\sum_{i=0}^{k-1}(\zeta^{ \frac{p-1}{k}i+r})^{2}\equiv \zeta^{2r}\sum_{i=0}^{k-1}\zeta^{\frac{2(p-1)}{k}i }\equiv$

$\equiv \zeta^{2r}\frac{\zeta^{2(p-1)}-1}{ \zeta^{\frac{2(p-1)}{k}}-1}\equiv 0 \;\;(mod \;p)$, since $\frac{2(p-1)}{k}<(p-1)$  and so ${\zeta^{\frac{2(p-1)}{k}}}-1 \neq 0$. This means that the sums of the squares of the elements of each of the $\frac{p-1}{k}+1$ subsets of the set of all residues modulo $p$, of wich they are a partition, $\left\{ 0\right\}=S_{p,k,0}$ and $S_{p,k,r}=\left\{ \zeta^{\frac{p-1}{k}i+r}| 0\leq i<p-1 \right\}$ for each $1\leq r\leq \frac{p-1}{k}$ is $\equiv 0$.
For all $0\leq i<23$ $0\leq j<30$ call  $1\leq r(i,j)<2001$ the unique number in that interval such that $r(i,j)\equiv i \;\;(mod\; 23)$, $r(i,j)\equiv j \;\;(mod \;29)$ $r(i,j)\equiv 1 \;\;(mod 3)$   [$667 =23\times29$]. For each ¡,j,k,l,  $r(i,j)+r(k,l) \equiv 2 \;\; (mod \;3)$, so $r(i,j)+r(k,l) \neq 2001$; notice also $r(i,j)\neq 2001$. Now consider the sets  $X(a,b)=\left \{ r(i,j) | i \in S_{23,11,a}\wedge j \in S_{29,4,b}\right\}$ where $0\leq a\leq 2$ and $0\leq b\leq 7$, there are 24 such sets, which are disjoint. Since  $\sum_{x \in X(a,b)}x^{2}\equiv 29\times 0 \equiv 0 \;\;(mod \;23)$ and $\sum_{x \in X(a,b)}x^{2}\equiv 23\times 0 \equiv 0 \;\;(mod \;29)$, $\sum_{x \in X(a,b)}x^{2}\equiv 0 \;\;(mod \;667)$, so
$\sum_{x \in X(a,b) \vee (2001-x) \in X(a,b)}x^{3}\equiv 0 \;\;(mod \;2001^{2})$.    On the other hand
$\sum_{i=1}^{2000}i^{3}=(\sum_{i=1}^{2000}i)^{2}= (\frac{2000 \times 2001}{2})^{2}\equiv0 \;\;(mod \;2001^{2})$, so the sum of the cubes of all the elements not belonging to one of the $X(a,b)$ nor summing 2001 with one of the elements of one of the $X(a,b)$ is also $\equiv 0 \;\;(mod \;2001^{2})$ and in this way we have found a partition of the given set in 26 subsets (possibly impovable by considering more in detail multiples of 3) which satisfies the property. But there must be something wrong in the whole thing.
I tried also to find a solution involving the last algebric identity, but I only managed to obtain a partition into 7 different subsets.

Editing: still trying to fix this piece of *****. Sorry again.


topology
function
geometry
3D geometry
sphere
analytic geometry
advanced fields

Let M be a closed ball in R^n viewed as a manifold with boundary. Show that the inclusion map of M to R^n is smooth.

I am not sure...I thought that a map from an arbitrary set A to R^n is smooth when it is smooth on some open set U containing A.

So, can we answer this question by just saying that since an inclusion map of a slightly bigger open ball to R^n is smooth, then it's restriction to M is smooth?

The key here is to look at what the definition of smoothness is for a manifold with boundary. In the interior, it's like an ordinary manifold; we have a family of charts to open subsets of $ \mathbb{R}^n$, which have smooth transition maps. A function to $ \mathbb{R}^m$ is smooth if its composition with the inverse of all appropriate charts is smooth on $ \mathbb{R}^n$. For the closed ball, we have a natural chart for the interior: the identity map. Inclusion is obviously smooth with respect to this chart.
On the boundaries, we have charts to relatively open pieces of a closed half-space $ H$, with the base point going to a point on the boundary of $ H$. These maps are not so obvious for the ball, as we can't simply use the identity. We need to find something that makes sense for "the closed ball", and then verify that the identity is smooth with respect to it.

Thanks.
So, we need to come up with some smooth structure on a (closed) disk $ {\mathbb B}_n$, so that each chart is homeomorphic to an open set of $ {\mathbb H}_n$ (closed upper half plane).

...is there any standard structure on a closed ball? (like the one we have for a sphere)

I am thinking something similar to the following would do:

If $ {\mathbb B}_2$ is a closed disk in $ {\mathbb R}^2$ then we can have 4 charts, where, for example, one chart $ U_1 \equal{} \{ (x, y) s.t. x^2 \plus{} y^2 \leq 1 and y > 0 \}$. The coordinate map to an open subset of  $ {\mathbb H}_2$ will be $ (x, y) \minus{} > (x, \sqrt {1 \minus{} x^2} \minus{} y)$ (this will basically map $ U_1$ to the set $ \{ (x, y) s.t. x^2 \plus{} y^2 < 1$ and $ y \geq 0 \}$ which is an open set in $ {\mathbb H}_2$.)

Yes, that's one way to do it.

The right way to think of the standard structure is that the sphere embeds to the boundary. Our boundary charts should extend charts of the sphere to an additional dimension.


geometry
circumcircle
parallelogram
geometry proposed

Let $ H$ the orthocenter of a triangle $ ABC$.If $ K,L,M$ are the circumcenters of triangles $ HAB,HBC,HCA$ , prove that the circle $ (K,L,M)$ which passes through points $ K,L,M$ is equal to the circle $ (A,B,C)$ which passes through points $ A,B,C$.
 [u]
 Babis[/u]

This problem is rather obvious,because it is easy to prove that $ CL \equal{} AK$,$ CL\|AK$,so $ KL \equal{} AC$.Thus $ \triangle KLM \equal{} \triangle ABC$,
here is an image:
[img]http://img341.imageshack.us/img341/8664/stergiu2zb8.png[/img]
P.S:I'm sure that there is much easier solution,for example using coordinates... :roll:
Best regards,Erken.

Just want to explain my solution in details,
It is obvious that $ \angle BKL \equal{} \angle BAH \equal{} \angle HCB \equal{} \angle KLB$,so $ KB \equal{} BL$.Thus $ KA \equal{} KB \equal{} BL \equal{} CL$.With counting angels we can obtain that $ KA\|CL$.Hence $ KACL$ is a parallelogram and $ KL \equal{} AC$.Similarly $ KM \equal{} BC$ and $ LM \equal{} AB$,thus $ \triangle KLM \equal{} \triangle ABC\Rightarrow R_{KLM} \equal{} R_{ABC}$. :)


Find all funtion $ f: R\to R$ and $ g: R\to R$ satisfying
$ f(x^3\plus{}2y)\plus{}f(x\plus{}y)\equal{}g(x\plus{}2y)$ $ \forall x,y\in R$

[quote="Algadin"]Find all funtion $ f: R\to R$ and $ g: R\to R$ satisfying
$ f(x^3 \plus{} 2y) \plus{} f(x \plus{} y) \equal{} g(x \plus{} 2y)$ $ \forall x,y\in R$[/quote]
[hide="The result without proof"]$ f(x) \equal{} \frac {a}{2},g(x) \equal{} a$where $ a\in R$
I will post the proof later or [b]Rust[/b] help me agian  :lol: ,I am too busy now [/hide]


Does anybody know in what extent physics is included in GRE subject test in mathematics?
What themes from physics can occured there?

i don't recall having to know any physics at all (which is good, because i really don't know any).

blahblahblah is correct: it's not there at all.

Ok I believe you are right, but there is obviously some misunderstanding here.
What do we mean by 'physics'? Some problems about forces, work etc. DO occur
and I think it is considered as physics, more precisely mechanics which is a part of
physics. Am I not right?


If $ x\in Q^\plus{}$, compute \[ [x] \plus{} [2x] \plus{} [3x] \plus{} ... \plus{} [nx].\]

( $ [x]$ is the greatest integer less than or equal $ x$ )

if tex m, n are positive integers  we have x=m/n so then
\[ [x] \plus{} [2x] \plus{} [3x] \plus{} ... \plus{} [(n\minus{}1)x] \equal{} (mn \minus{} m \minus{} n \plus{} NWD(m,n))/ 2 .
\]


function
trigonometry
calculus
derivative
real analysis
real analysis unsolved

I was looking for a proof that a differentiable function (say on a real interval, with real values) is locally lipschitz. I had a look at http://www.etsu.edu/math/gardner/5510/cspace.ps and that guy shows a proof but I do not understand why it works, i.e, the so-called Lipschitz constant he picks depends on the point $x$, since there's no reason for $f$ to be uniformly differentiable ... What do I miss ?

Counterexample: the function $f(x)=x^{2}\sin\frac{1}{x^{2}}$ is differentiable at every point of $\mathbb R$ including zero ($f'(0)=0$). If $f$ were Lipschitz in some neighborhood of $0$, its derivative would be bounded in that neighborhood. However, for $x\ne 0$ we have $f'(x)=2x\sin\frac{1}{x^{2}}-\frac{2}{x}\cos\frac{1}{x^{2}}$.

I personally find differentiable functions neither interesting nor useful. In contrast to continuously differentiable, weakly differentiable, differentiable a.e., Lipschitz, etc.

Right, that's what I thought ... a funny mistake coming from a college professor (especially when it is included in a paper as a theorem)...


As you'll notice, the contests are no longer stickies.  I've un-stickied them for the following two reasons:

1) There are way too many of them atop the forum.  The active ones will stay atop the forum through use, and that's the way it should be.

2) Some of them are partially way too hard or entirely way too hard for this forum.  I've moved one or two of them to Intermediate.  We can't have way too hard stuff stuck to the top of the forum.

In general, this forum has drifted far too difficult once again.  I encourage the mods to be more aggressive about moving posts to Intermediate, and I will be doing so frequently myself as well.  Those of you posting in this Forum should be careful not post items that are too difficult.  Read the guidelines, and if you're still not sure, post in Intermediate instead.

Just in case we post a difficult problem,Could you please give us a warning before banning us?

Indeed - I'll generally PM people whose posts I have to frequently move.

I apology for making my contest going to way difficult than Getting Started. This 5th COP is the last one I'm going to do till I finish taking AMC-12 this year so no more worries on that one.  :D As for other stickied threads, I wasn't sure what to do with them because contest seemed to be keep going on. Also, some contests didn't have definite endings so that was one problem.

I would be more appreciated if people who host their contest to put the first round in one thread and second in another and so forth because that way, I don't have to sticky the one thread for long time.

Thanks.  :)

[quote="rrusczyk"]In general, this forum has drifted far too difficult once again.[/quote] Hmm, I thought that the difficulty level of the posts was right on track. Guess I'll spend more time looking out for overly difficult problems from this day forward.

[quote="rrusczyk"][quote="4everwise"][quote="rrusczyk"]In general, this forum has drifted far too difficult once again.[/quote] Hmm, I thought that the difficulty level of the posts was right on track. [/quote]

I think that's probably because you're getting a lot better at math.[/quote]

That's what I thought in the Mathcounts forum until a moderator gave me a warning.  :lol:

It's what happens, like rrusczyk said, when you improve  :)

Also, I'm going to move most of the proof problems to Intermediate from now. The word "proof" is not a getting started level material, and this decision was already made in Moderator's Lounge forum few weeks ago.

So, if you post any proof problems that are not VERY VERY basic, expect it to be moved.  :)

I have had a few users notify me that the difficulty of Getting Started is getting harder again.
Can we try to keep the difficulty level on track? Thanks.  :)


Find the smallest 14-digit natural number $ n$, such that $ 2^{14}|n$ and $ n$ does not contain any digit other than 8 and 9.

[hide="tedious solution"]Note that if $ 2^k|n$ then $ 2^k$ divides the last $ k$ digits of $ n$.
After finding the last $ k\minus{}1$ digit of $ n$, we can use the above property to test whether $ 8$ or $ 9$ is the $ k$th digit from the right.
Also, since $ 2^k\nmid 10^{k\minus{}1}$, only one of $ 8$ or $ 9$ can be the $ k$th digit from the right.

Since $ 2^1$ divides the last $ 1$ digit of $ n$, the last $ 1$ digit of $ n$ must be $ 8$.
Since $ 2^2$ divides the last $ 2$ digits of $ n$, the last $ 2$ digits of $ n$ must be $ 88$.
Since $ 2^3$ divides the last $ 3$ digits of $ n$, the last $ 3$ digits of $ n$ must be $ 888$.
Since $ 2^4$ divides the last $ 4$ digits of $ n$, the last $ 4$ digits of $ n$ must be $ 9888$.

Continuing this process, we find that $ n\equal{}98898989989888$.[/hide]


Let $ a,b,c\ge0$.Prove that
 $ (\frac a{2a \plus{} b \plus{} c})^2 \plus{} (\frac b{2b \plus{} a \plus{} c})^2 \plus{} (\frac c{2c \plus{} b \plus{} a})^2 \ge \frac3{16}$

 $ (\frac a{0.5a \plus{} b \plus{} c})^2 \plus{} (\frac b{0.5b \plus{} a \plus{} c})^2 \plus{} (\frac c{0.5c \plus{} b \plus{} a})^2 \ge \frac {12}{25}$

[quote=dduclam]Let $ a,b,c\ge0$.Prove that
 $ (\frac a{2a \plus{} b \plus{} c})^2 \plus{} (\frac b{2b \plus{} a \plus{} c})^2 \plus{} (\frac c{2c \plus{} b \plus{} a})^2 \ge \frac3{16}$
[/quote]
https://artofproblemsolving.com/community/c6h1805527p12019955
[quote=anhduy98][quote=cxyh]$Assume$   $a,b,c>0$   $abc=1$
$Prove:$$\sum_{cyc} (\frac{1}{2+a+ab})^2\geq \frac{3}{16}$[/quote]

$$\bullet abc=1\rightarrow \frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}=1$$
$$\bullet LHS=\sum \frac{1}{\left ( 2+a+ab \right )^2}=\sum \frac{(a+ab)(a+ab-2)^2}{32(1+a+ab)(2+a+ab)^2}+\sum \left [ \frac{9}{32(1+a+ab)}-\frac{1}{32} \right ]$$
$$\rightarrow LHS=\sum \frac{(a+ab)(a+ab-2)^2}{32(1+a+ab)(2+a+ab)^2}+ \frac{3}{16} \geq \frac{3}{16}=RHS$$[/quote]


[color=blue]Estoy interesado en crear un foro para mi grupo universitario (ya ven por qué no voy más a la I.M.O.).

Me encanta la opción usar $ \text{\LaTeX}$ en este foro. Pero no sé cómo crear un foro que admita este código (y que no tenga costo alguno).

Pregunté hace muchas semanas a Yoda Vornicu (despachado a la antigua yo), pero parece que está muy ocupado desde entonces.

¿Alguien puede darme consejos, por favor?[/color]


Hey. . .

I think finding the product and sum of the roots of a polynomial come up sometimes on the AIME, do you think someone could post the formulas (not just for quadratics, but for higher degree polynomials, is there a formula?)

This was on 1983/3

Thanks.

Of course there is, but I'm going to let you figure it out.

what is the form of

(x-r_1)(x-r_2)(x-r_3)....(x-r_n), the polynomial with n roots r_i (1<=i<=n). Multiply it out for small cases and see what you get.

Have fun.

Yay! Whatever you do, don't learn the formula, do what Nukular said.

Also keep in mind that you can find the sum of the kth powers of the root, where k is a positive integer less than or equal to n, where n is the degree of your polynomial. I'll leave you to figure out why


Hi,

I am helping my children with a computer game based on Logic and Reasoning. The player has to solve different puzzles as the game progresses. I am struggling with a matching puzzle where the player has to form a string of the game characters based on their traits. Please follow this link to see a screenshot of the puzzle. [url]http://img246.imagevenue.com/img.php?image=12040_Zoombinis_122_220lo.jpg[/url]

[color=blue] I am not sure how I should approach such a problem. I also wish to know the branch of Mathematics that studies these kinds of problems.[/color]

[b]Brief background[/b]: [i]The characters in the game are called Zoombinis, with each of them having four special characteristics : hair, eyes, nose, and foot. Based on 4/5 variations of the each of the charactristics, the game randomly produces zoombinis. The player has to guide around 16 zoombinis at a time through various puzzles.[/i]

At the above mentioned matching puzzle, the game produces a string of holes (one zoombini can be placed over one hole) . Between any two holes a picture of one of the 4 characteristics is shown. The two zoombinis on the either side of a picture have to have the same trait. For example, if there is an eye picture between two holes, then the zoombinis on the either side of that picture have to have similar types of eyes; both of them can have sleepy eyes, or wide-open eyes, or eyes covered with glasses.

Thanks,

MG.

I remember playing an earlier version of that game, years ago.  :)

Now I can't really tell what some of the symbols are. I can tell the eyes, but what are the ones that look like a twig (1 branch from left, 2 exit on right), the mop shape thing, and the rock (never touches left or top?)

(It's probably just my bad screen resolution that's doing this.)

Oh man! I loved the Zoombinis.  I think I got past that puzzle by making a chart of intersections and seeing how many intersections should be spread over how many similarities.   It took me a very long time but in the end it was much easier than my original method ( guessing  :lol: ).  I think this would be similiar to studies of the branch of mathematics called "graph theory" but I am not sure.  I think that puzzle shares similarities to "Instant Insanity". -http://en.wikipedia.org/wiki/Instant_Insanity


[quote="TripleM"]Do you think it would be a good idea, as bugzpodder suggested, to have SOLVED/UNSOLVED in the subject of each msg, which can be edited after the problem is solved?[/quote]

Yes, that would be helpful.  :)

[quote="Fierytycoon"]What would be even better is if there could be a search option that allowed users to search for only solved/unsolved problems (so that people can work on the unsolved ones). However, this idea is probably difficult/bothersome to implement.[/quote]

Well, if we did this then you could just search for the word "unsolved" in the subject area. So you could really do that.

[quote]Well, if we did this then you could just search for the word "unsolved" in the subject area. So you could really do that.[/quote]

I guess that's true.


linear algebra
matrix
vector
induction
algebra
polynomial

prove that the determinant of an antisymetric matrix is a square.(in any field)

I have an argument. It's not really mine, but I'm passing it along. I don't have time to write it all up now, but will by tomorrow.

Some starting points:

Every element in a field of characteristic 2 is a square, so we may assume WLOG that the field in this problem is not of characteristic 2.

Since zero is a square, we're done if $\det A=0.$ Hence WLOG, we may assume that $A$ is invertible.

$\det A = \det A^T = \det(-A)=(-1)^n\det A$ where $A$ is an $n\times n$ matrix.  Given that we've assumed that the field is not of characteristic 2 and that $\det A\ne0,$ we have that $n$ is even.

Call $B$ [u]congruent[/u] to $A$ if there exists an invertible $P$ such that $B=P^TAP.$ Although $A$ and $B$ may not have the same determinant, they share the property of the determinant being a square, since $\det B=(\det P)^2\det A.$ Furthermore, if $A^T=-A$ then $B^T=-B.$

The objective of the proof will be to show that $A$ is congruent to a matrix with the block structure $\begin{bmatrix} S & 0 \\ 0 & A' \end{bmatrix}$ where $S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$

Here's the rest of the argument.

Since the transpose of a $1\times 1$ matrix is itself, $u^TAu=u^TA^Tu=-u^TAu.$ Hence, $u^TAu=0$ for all $u.$

Take some nonzero $u$. Then $Au$ is also nonzero. (Remember, we are assuming $A$ invertible.) Since the column vector $Au$ is of rank 1, there must be $v$ such that $v^TAu\ne0.$ (In fact, the set of such $v$ is of dimension $n-1$.) Multiply by the appropriate scalar so that $v^TAu=1.$

Note that $v$ is clearly not a scalar multiple of $u$ so $\{u,v\}$ is independent. Furthermore, since $A$ is invertible, $\{Au,Av\}$ is also independent.

Let $W$ be the subspace of vectors orthogonal to ${Au,Av}.$ (I'm not assuming we have an inner product space - this is just a matter of rank.) That is, $W$ consists of the vectors $w$ such that $u^TAw=0$ and $v^TAw=0.$ Let $\{w_1,w_2,\dots w_{n-2}\}$ be a basis for $W.$

Neither $u$ nor $v$ lie in $W,$ because $v^TAu=1$ and $u^TAv=-1.$ Hence, the set $\{u,v,w_1\dots w_{n-2}\}$ is an independent set. Let $P$ be the matrix with those columns. Then $AP$ is the matrix whose columns are $\{Au,Av,Aw_1,\dots,Aw_{n-2}\}.$ Now compute $P^TAP.$ The $2\times2$ upper left corner of that is $S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$ The remaining elements in the top two rows are of the form $u^TAw$ or $v^TAw.$ But we have chosen $W$ so that all of those are zero.

The transpose of  $P^TAP$ is $P^TA^TP=-P^TAP$ so $P^TAP$ is antisymmetric. Hence, now that we've found the top two rows, the block structure must be

$P^TAP=\begin{bmatrix} S & 0 \\ 0 & A' \end{bmatrix}$ where $A'$ is antisymmetric.

$\det A = \frac1{(\det P)^2}\,\det(P^TAP)=\frac1{(\det P)^2}\cdot 1 \cdot \det(A').$

The theorem then follows by induction on the size of $A.$

Hey, Alekk, I think in one of the latest issues of Review des Mathematiques speciales there is an article about the characteristic polynomial of an antisymmetric matrix. You can find there many details.

Yes Kent, that is also the proof I known. Moreover my teacher said that there were some applications of this result, especially if the field has not a lot of squre as $Q$, $Q(X)$,... Nevertheless he didn't give any example. Do you know how to use this result ?


For a positive real number $\alpha$, define \[S(\alpha)=\{ \lfloor n\alpha\rfloor \; \vert \; n=1,2,3,\cdots \}.\] Prove that $\mathbb{N}$ cannot be expressed as the disjoint union of three sets $S(\alpha)$, $S(\beta)$, and $S(\gamma)$.

Solution by jmerry: http://www.mathlinks.ro/viewtopic.php?t=127810

we assume that $\alpha<\beta<\gamma$ and if $\alpha<1 \implies S(\alpha)=N$

now assume that three number exist such as that $1<\alpha<\beta<\gamma$ and they satisfy the problem, so for every natural number $n$  if $n$ isn't divisible by $a,b,c$ , the number of members of  $S(\alpha)$ that are less than $n$ is equal to $\lfloor\alpha^{-1}n\rfloor$ and respectively we get $\lfloor\beta^{-1}n\rfloor$ for $S(\beta)$ and $\lfloor\gamma^{-1}n\rfloor$ for $S(\gamma)$.

from now on suppose that $n$ isn't divisible by $a,b,c$ and because they are infinitive natural numbers with this property this wouldn't be a problem.

because the three sets are disjoint and all numbers $1$ to $n-1$ belong exactly to one of these sets then we must have $\forall n>1 \in N :\lfloor\alpha^{-1}n\rfloor+\lfloor\beta^{-1}n\rfloor+\lfloor\gamma^{-1}n\rfloor=n-1$ .

I didn't make any progross from here!


Except for the first two terms, each term of the sequence $1000, x, 1000-x,\ldots$ is obtained by subtracting the preceding term from the one before that.  The last term of the sequence is the first negative term encounted.  What positive integer $x$ produces a sequence of maximum length?

[quote="joml88"]What positive integer $x$ produces a sequence of maximum length?[/quote]
What do you mean ?

By the way we have $u_n = u_{n+12}$  :roll:

[hide]$x=618$
By carrying out the sequence in terms of $x$, you get \[ 1000, x, 1000-x, 2x-1000, 2000-3x, 5x-3000, 5000-8x, 13x-8000, 13000-21x, 34x-21000 \] and if you set each greater than 0, you get in succession, \[ 0<x<1000 \Rightarrow 500<x<1000 \Rightarrow 500<x<667 \Rightarrow 600<x<667 \Rightarrow 600<x<625 \Rightarrow 616<x<625 \Rightarrow 616<x<619 \Rightarrow 618<x<619 \] and by carrying out both $618$ and $619$, you find that $618$ carries out to $14$ terms including the first negative, while $619$ carries out only $11$.[/hide]

You know, its so cool how you always find Fibonacci hidden in AIME problems. I swear, I've seen on average 1.5ish problems on each AIME that have Fibonacci hidden in there somewhere. I always get excited when I see them. OK, I'll stop writing now..... but it is cool.

An easy way to solve the problem:

Figuring out the terms we have:

Let $a_n = 1000,x,-x+1000,2x-1000,-3x+2000,5x-3000,...$

Observe that the $n$th term, for $n \geq 3$ is $F_{n-2}*1000*(-1)^{n+1} + F_{n-1}*x*(-1)^n$, where $F_n$ is the $n$th term of the Fibinochi sequence.

Setting this term to be greater than zero, it follows that:

$F_{n-2}*1000*(-1)^{n+1} + F_{n-1}*x*(-1)^n > 0$
$F_{n-1}*x*(-1)^n > F_{n-2}*1000*(-1)^{n}$
$x*(-1)^n > \frac{F_{n-2}*1000*(-1)^{n}}{F_{n-1}}$

At this point we see that

$x > 1000*\frac{F_{n-2}}{F_{n-1}}$ if $n$ is even, or
$x < 1000*\frac{F_{n-2}}{F_{n-1}}$ if $n$ is odd.

It is known that $\displaystyle\lim_{n \rightarrow \infty}{\frac{F_{n-1}}{F_n}} = \phi - 1= \frac{1 + \sqrt{5}}2 - 1 \approx 0.61803$.  I do not know how to justify this, but there is a lot of information about the number $\phi$ available at http://goldennumber.net/.  It is worth memorizing 0.618, because it comes up a lot with fibinochi.

So, as $n$ tends to infinity in the equations for this problem, we see that $x$ will be largest at $1000*\phi \approx 618$.

In case you're interested, the exact value that will produce a series of infinite length is $1000*(\frac{1+\sqrt{5}}2-1) \approx 618.03398875$.  However, according to my calculator, this approximation gives only 35 terms.

p7: Computing a few more terms we get $1000, x, 1000-x, 2x-1000, 2000-3x, 5x-3000, 5000-8x, 8000-13x$, and so on. We can see that term $n$ is $1000(F_{n-2}) - xF_{n-1}$ if $n$ is odd, and $xF_{n-1} - 1000(F_{n-2})$ otherwise. We want $1000F_{n-2} \geq xF_{n-1}$, or $1000 \frac{F_{n-2}}{F_{n-1}} \geq x$. We know that $\frac{F_{n-2}}{F_{n-1}}$ approaches $\frac{2}{\sqrt5+1}$, so the answer is $\frac{2000}{\sqrt5+1}$ rounded down, or $\boxed{618}$.


Prove that, for all $x,y\in$ ${Z}$, the number $(x^{2}y^{2}-4x^{2}y)(x^{4}+x^{2}-2)$ is divisible by $216$


function
inequalities
real analysis
real analysis theorems

Let x ba a positive real number and n is a negative real number , prove that we have only one number y such that : y^n=x

No matter if $ n$ is negative or positive (as long as $ n \neq 0$), the function $ f(y) \equal{} y^n$ is strictly monotonous on $ (0, \infty)$, and its range is $ (0, \infty)$, therefore it meets value $ x>0$ exactly once.

Boy , They are similar clauses :) , You must show one in two clauses

Who is Boy? And which are the two clauses?

Oh I am sorry ^^ i think you are a high shool student  :( . I think your 1st comment and my problem are similar  :) .

This is a Proof :

Put A={z | $ z^{n} \leq x$}

(+) If we put $ z=\frac{x}{x+1}$ , we have : $ z^{n}\leq z\leq x$ then : $ A \neq \phi$
Put $ y=Sup(A)$ we 'll prove that : $ y^{n}=x$

Lemma : $ a,b$ are 2 positive real numbers such that : $ 0<a \leq b$ then we have $ b^{n}-a^{n} \leq nb^{n-1}(b-a)$

We can prove this lemma by cauchy's inequality  :) .

Come back our problem
We assumed  $ y^{n}\neq x$ , we have 2 case : ${ y^{n}<x}$ or $ y^{n}>x$

(+) If $ y^{n}<x$ :

Let h be a positive real number and $ h<1$ , by the lemma we have : $ (y+h)^{n}-y^{n}\leq n(y+h)^{n-1}h<n(y+1)^{n-1}h$
Put $ h=\frac{x-y^{n}}{n(y+1)^{n-1}}$ then we have : $ (y+h)^{n}<x$ . It is unresonable because : $ y=sup(A)$ :)

(+) If $ y^{n}>x$ :

Put :$ k=\frac{y^{n}-x}{n.y^{n-1}}$
If $ (y-k)^{n}>x$ then we have $ y^{n}>(y-k)^{n}>x>z^{n}$ with $ z \in A$ it is unresonable
because $ y=Sup(A)$ . Hence , $ (y-k)^{n}<x$ (1)

If $ t$ exists such that : $ x>t^{n}>(y-k)^{n}$ then by the lemma : $ y^{n}-t^{n}\leq y^{n}-(y-k)^{n}  \leq ny^{n-1}.k=y^{n}-x$ then $ t^{n}>x$ , it is unresonable (!) . In short , t does n't exist (2)

From (1) and (2) we have : $ y-k= Sup(A)$ It also is unresonable .  :)

Hence , we have : $ y^{n}=x$


Hi, all --

The Hampshire College Summer Studies in Mathematics webpages (at http://www.hcssim.org) have been updated.  One of the new pages is the first announcement of and information about Yellow Pig Math Days (count the letters), a conference we're hosting July 15 - 18, 2006 in celebration of the 34th Summer Studies.  The conference is intended for alumns and friends of HCSSiM as well as participants in HCSSiM, 2006.  If there's information you want that's not on the YPMD page ( http://www.hcssim.org/ypmd.html), then please ask---of course, the information might not be there because we don't have it yet *grin*.

Hi, all --

The YPMD webpage has been updated, and we're now accepting pre-registration.  If you're thinking of going, [i]please[/i] preregister so that we have some clue of how many people will show up.  (Plus, this will help us argue for how many rooms Summer Programs should allocate to us for the weekend.)

Should people who may be going to the '06 HCSSiM register?

No, potential HCSSiM '06 participants shouldn't pre-register.  Alumns from '05 (and, of course, all other years) should pre-register.

But I think you're really asking: what should alumns who have applied for readmission (but not gotten a response yet) do?  I'd say that you should wait until you hear back from us, as if we invite you to HCSSiM '06 you won't need to pre-register.


Came up with this while looking at another problem:

$ A_0, A_1, B_0, B_1, C_0, C_1$ are points in the plane such that $ A_0, B_0,$ and $ C_0$ are collinear, and $ A_1, B_1, C_1$ are collinear.  For any $ t \in \mathbb{R}$, define $ A_t, B_t,$ and $ C_t$ to be the points on lines $ A_0A_1, B_0B_1,$ and $ C_0C_1$, respectively, such that
\[ \frac{\vec{A_0A_t}}{\vec{A_0A_1}}\equal{}\frac{\vec{B_0B_t}}{\vec{B_0B_1}}\equal{}\frac{\vec{C_0C_t}}{\vec{C_0C_1}}\equal{}t\]
If there exists some $ t_0 \in \mathbb{R}$ other than $ 0$ or $ 1$ such that $ A_{t_0}, B_{t_0},$ and $ C_{t_0}$ are collinear, then show that for any real $ t$, the points $ A_t, B_t,$ and $ C_t$ are collinear.

Seems pretty generic, and so it's probably well-known.  But the solution is pretty nice, so...

[hide="Spam, Spam, Spam, Spam, Spam, Baked Beans, Spam, Spam, Spam and Spam"]
We have $ (C_0-B_0)=\alpha_0(B_0-A_0)$ and $ (C_1-B_1)=\alpha_1(B_1-A_1)$ so
\begin{align*}
(C-B)&=(tC_0+(1-t)C_1-tB_0-(1-t)B_1) \\
&= t(C_0-B_0)+(1-t)(C_1-B_1) \\
&=t\alpha_0(B_0-A_0)+(1-t)\alpha_1(B_1-A_1)\\
&=\alpha_0(t(B_0-A_0)+(1-t)(B_1-A_1))+(\alpha_1-\alpha_0)(1-t)(B_1-A_1)\\
&=\alpha_0(B-A)+(\alpha_1-\alpha_0)+(\alpha_1-\alpha_0)(1-t)(B_1-A_1)\\
&=(\alpha_0-\alpha_1)t(B_0-A_0)+\alpha_1(t(B_0-A_0)+(1-t)(B_1-A_1))\\
&=(\alpha_0-\alpha_1)t(B_0-A_0)+\alpha_1(B-A)
\end{align*}

This implies that if $ A,B,C$ are collinear for some $ t\neq 0,1$, either $ \alpha_1=\alpha_2$ or $ A_0B_0\parallel A_1B_1$ which implies that these vectors are always collinear...
[/hide]


Find all positive integer $ n$ such that $ 1! \plus{} 2! \plus{} 3! \plus{} \cdots \plus{} n!$ is a perfect square.

[hide="Hint"]
Consider modulo some numbers to find a finite number if them
[/hide]

[hide="Bigger hint"]
5
[/hide]

[hide] Consider the expression (mod 10). For all n>4, $ n!\equiv0\pmod{10}$.

1!+2!+3!+4!=1+2+6+24=33. Hence, for all n>3, $ 1!\plus{}2!\plus{}...\plus{}n!\equiv3\pmod{10}$, which is not a quadratic residue.  We only need to check for n=1,2, and 3, and we find it is a perfect square for n=1,3.[/hide]

[hide="solution"]The units digit of a square is either 0, 1, 4, 5, 6, or 9. Let's look at the units digit for some small values of n:

n=1: 1
n=2: 3
n=3: 9
n=4: 3
n=5: 3
n=6: 3
.
.
.

Since $ 10|n!$ for $ n>4$, the units digit won't change after $ n\equal{}4$, so for all $ n>3$, the given expression is not a square. The only $ n$ such that the expression is a square are $ n\equal{}\boxed{1,3}$.[/hide]


Also prove that AA' is perpendicular to B'C'.

hmm, i don't know about a geometric proof, but i know that the analytic proof would be pretty straight forward

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 22
Sincerely
Jean-Louis

[quote="fourierseries"]On the three sides of triangle ABC, construct squares facing outwards. Let A', B', C' be the centeres of the squares constructed on sides BC, CA, and AB, respectively.  Prove that:

dist(A,A')=dist(B',C')[/quote]
[b][u][color=#0000FF]Vector Proof:[/color][/u][/b]
Now let the squares are ABED,BCFG,ACHM,and also let T and S are midpoints of AB and AC.
We have:

$R^{90^\circ }(\vec{C'B'})= R^{90^\circ }(\vec{C'T}+\vec{TS}+\vec{SB'})=R^{90^\circ }(\vec{C'T})+R^{90^\circ }(\vec{TS})+R^{90^\circ }(\vec{SB'})=$

$=R^{90^\circ }(\frac{1}{2 }\vec{EB})+R^{90^\circ }(\frac{1}{2}\vec{BC})+R^{90^\circ }(\frac{1}{2 }\vec{AM})=\frac{1}{2}\vec{ED}+\frac{1}{2}\vec{FC}+\frac{1}{2}\vec{CA}=$

$=\frac{1}{2}(\vec{FA}+\vec{BA})=\vec{A'A}$

$ => A'A\bot C'B'$ and $A'A=C'B'$.


compute
lim sum ([(2k^2-3k-5)*3^k]/(k+2)!), k=1, n and lim when n -->oo.
really need help with this one!
bye bye!

I would really love to see a fully detailed solution of this one!!
PLEASE!


In the local frisbee league, teams have 7 members and each of the 4 teams takes turns hosting tournaments. At each tournament, each team selects two members of that team to be on the tournament committee, except the host team, which selects three members. How many possible 9 member tournament committees are there?

This is just $ \binom 7 3 \cdot \binom 7 2 \cdot \binom 7 2 \cdot \binom 7 2\equal{}35 \cdot 21 \cdot 21 \cdot 21\equal{}324125$.

Sorry for reviving.

Actually, I forgot to multiply by $4$ because there can be $4$ teams hosting the tournament, so there are $4$ choices for the team that gets $3$ members. Thus, the answer is $324125 \cdot 4 = 1296540$.

I think it is unclear that you have to take in the fact that each school can host it.

[hide=Solution]
In each tournament, the host team selects three members and the other three each select two, so we have
$$\binom{7}{3}\cdot\binom{7}{2}^3=324135.$$
However, since each of the four teams takes turns hosting tournaments, we multiply by four to get $\boxed{1296540}$ ways to choose tournament committees.
[/hide]

why did i use permutations instead of combinations  :censored: Im so mad


The game of chess is played on an eight by eight grid of squares.  In one move, the king may be moved to any of the squares which adjoin the square it currently occupies or share an edge with it.  If the king starts in a corner square, how many different squares could it occupy after exactly 4 moves?

[hide]I drew all possible sides, and I got a 5x5 square, so 25?[/hide]

Yeah, that's what I got too...  Is there be a better way to do the problem than drawing the paths?

It's simple, you just need to see that he could be in any of 25. I'm not sure how else to explain it, but it should be obvious.

the way the problem is worded it seems like you can't go diagonal. if that's the case, the answer is [hide]
9[/hide]

Kings can always go diagonal. The problem specifies either squares that adjoin the one which the king is currently on OR share an edge with it. The latter obviously means diagonally, although "edge" isn't the correct word there, or else there's no need to specify. And if you couldn't, the answer would be $15$, not $9$.

we start from A1
so the longest four moves are either
A5 or E1 or E5

five squares
and since it covers everything within the square, it's $5^2$ or just $25$

Bleh.  Sorry.  I was too lazy to type it all last time.  Here's the exact wording:

The game of chess is played on an eight by eight grid of squares. In one move, the
king may be moved to any of the squares which adjoin the square it currently
occupies, either along an edge or at a corner. If the king starts in a corner square,
how many different squares could it occupy after exactly four moves?

[hide]Say it starts on the far left corner. The most it can move up is 5 spaces and max to the right is also 5. Multiply 5 and 5 and that is 25.[/hide]

[hide]The places it could be is a 5x5 square. 5x5=25.[/hide]

No it's 25. I drew it all out. Not that that is the best way to do it


Let a,b,c be fixed natural numbers. Suppose that, for every positive integer, there is a triangle whose sides have lengths a^n,b^n,c^n respectively. Prove that these triangles are isosceles.

[hide]Consider three positive numbers $ a > b\geq c$.

Note that eventually,
$ \left(\frac{a}{b}\right)^{n}> 2\implies a^{n}> 2 b^{n}\geq b^{n}\plus{}c^{n}$.

In other words, if there is a unique largest side, $ \exists k :\forall n\geq k,\; a^{n}\not\leq b^{n}\plus{}c^{n}$; that is, the Triangle Inequality is not achieved.

Hence, there is not a unique largest side, and said triangle is isosceles.[/hide]