In a triangle $ ABC$ with $ a>b>c$, the four vertices of a square lie on the three sides of $ ABC$, such as one vertex on side $ a$, one on side $ b$, and two on side $ c$ . Which of the 3 possible ways will give the maximum area?

Let $ PQRS$ be such a square with $ P,Q\in (AB), R\in (BC), S\in (CA)$ and denote $ PQ \equal{} x_c$. Since $ \triangle CSR\sim\triangle CAB$ we get $ RS \equal{} x_c \equal{} AB\cdot k_c \equal{} c\cdot k_c$, $ CS \equal{} CA\cdot k_c \equal{} b\cdot k_c$, where $ k_c$ is a real number in $ (0,1)$. Hence $ AS \equal{} b \minus{} b\cdot k_c \equal{} b(1 \minus{} k_c)$. Therefore $ x_c \equal{} PS \equal{} AS\sin A \equal{} b(1 \minus{} k_c)\sin A$. Thus we conclude that $ (1 \minus{} k_c)b\sin A \equal{} ck_c$ which implies $ \frac 12bc\sin A \equal{} \frac {c^2k_c}{2(1 \minus{} k_c)}$. This this is the area of $ \triangle ABC$ and is therefore constant. Thus applying similar arguments for $ a$ and $ b$ we conclude that
\[ \frac {a^2k_a}{1 \minus{} k_a} \equal{} \frac {b^2k_b}{1 \minus{} k_b} \equal{} \frac {c^2k_c}{1 \minus{} k_c} \equal{} 2[ABC] \equal{} k
\]
Therefore we get $ x_a \equal{} a\cdot k_a \equal{} \frac {ka}{k \plus{} a^2}$, $ x_b \equal{} b\cdot k_b \equal{} \frac {kb}{k \plus{} b^2}$, $ x_c \equal{} c\cdot k_c \equal{} \frac {kc}{k \plus{} c^2}$. Now note that $ \frac {kx}{k \plus{} x^2}\le\frac {ky}{k \plus{} y^2}\Leftrightarrow (x \minus{} y)(xy \minus{} k)\ge 0$. On the other hand, $ ab \equal{} 2\cdot\frac 12ab\ge 2\cdot\frac 12ab\sin C \equal{} k$ and similarly $ bc\ge k, ca\ge k$. Hence from $ a > b > c$ we conclude that $ \frac {ka}{k \plus{} a^2}\le\frac {kb}{k \plus{} b^2}\le\frac {kc}{k \plus{} c^2}$ and thus $ x_a\le x_b\le x_c$. In conclusion, the square with maximal area is the one which has two of its vertices on $ AB$.


trigonometry
geometry
inequalities proposed
inequalities

Let $ ABC$ be an acute triangle.Prove that $ cosA\plus{}cosB\plus{}cosC\geq1$

[quote="v235711"]Let $ ABC$ be an acute triangle.Prove that $ cosA \plus{} cosB \plus{} cosC\geq1$[/quote]

$ \cos A \plus{} \cos B \plus{} \cos C\equal{} 1\plus{} \frac{r}{R}$

[quote="Sergic Primazon"][quote="v235711"]Let $ ABC$ be an acute triangle.Prove that $ cosA \plus{} cosB \plus{} cosC\geq1$[/quote]

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} \frac {r}{R}$[/quote]

would you mind posting it´s proof?  :)

This formula is very nice.
We will prove that $ cosA,cosB,cosC$ is the roots of this equation:
$ x^3 - \frac {R + r}{R}x^2 + \frac {p^2 + r^2 - 4R^2}{4R^2}x + \frac {(2R + r)^2 - p^2}{4R^2} = 0$, $ (*)$
where, $ R,r,p$,respectively are the circum radius, incircle radius, and semipremeter.
[i]Proof[/i]:
Because $ a = 2RsinA = 2R\sqrt {1 - (cosA)^2}$ and $ p - a = rcot\frac {A}{2} = r\sqrt {\frac {1 + cosA}{1 - cosA}}$
Therefore, $ p = a + (p - a) = 2R\sqrt {(1 - cosA)(1 + cosA)} + r\sqrt {\frac {1 + cosA}{1 - cosA}}$.
Square both sides, we get:
$ p^2 = 4R^2(1 - cosA)(1 + cosA) + 4Rr(1 + cosA) + r^2\frac {1 + cosA}{1 - cosA}$
Time $ (1 - cosA)$ both sides, we get:
$ 4R^2(cosA)^3 - 4Rr(R + r)(cosA)^2 + (p^2 + r^2 - 4R^2)cosA + (2R + r)^2 - p^2 = 0$
Hence, we conclude that $ cosA$ is one root of the equation $ (*)$. Argue the same for $ cosB, cosC$.
Apply [b]Viete[/b] for $ (*)$, we shall get: $ cosA + cosB + cosC = \frac {R + r}{r}$, which implies the result you need.

$ \sum \cos A\equal{}1\plus{}\prod \sin \frac{A}{2}\equal{}1\plus{}\frac{r}{R}$

$ \cos A\plus{}\cos B\plus{}\cos C \geq \cos(\frac{\pi}{2})\plus{}\cos(\frac{\pi}{2})\plus{}\cos(0)\equal{}0\plus{}0\plus{}1\equal{}1$ by Karamata, since $ \cos(x)$ is concave for $ 0\leq x \leq \frac{\pi}{2}$ and $ (A,B,C)\prec (\frac{\pi}{2},\frac{\pi}{2},0)$.

because we can prove easily that cosA+cosB+cosC= 1+ 4.sinA/2.sinB/2.sinC/2 .A,B,C of triangle ABC


topology
real analysis
real analysis theorems

We can easily define uniformly continuous function's on metric spaces.
My book states that we cant define it in (general) topology spaces (it doesn't tell why). Now, my question is why not, and when do we encounter problems?  I would be glad if someone could provide an example.
What are the most general spaces where we [i]can[/i] define it (beside metric spaces )?[/i]

It's the "uniform" part- we need to be able to compare the "size" of neighborhoods of different points, and there's no way to do that in general.

The map $ x\mapsto x^{3}\plus{}x$ is a homeomorphism of the real line to itself in the standard topology. It does not preserve uniform properties; if $ f(x)$ is uniformly continuous, $ f(x^{3}\plus{}x)$ might not be.
Anything we define for general toplogical spaces has to be preserved by homeomorphisms. The only way to make uniform continuity work is to introduce extra structure. A metric is one way; there are some weaker notions that also work, but I'm not familiar with them.


If Don has 5 pairs of different colored socks and he loses 3 what is the probabilty that he will have only 2 complete pairs left?

[hide]The only way this can happen is if he loses 3 socks of different colors.
Lets see how many ways he can lose 3 of different colors.
The first sock he loses can be any color. 9 socks left.
Now the next must be any that is not the same color as the last one he lost. $\frac89$ chance of that. 8 socks left.
Third sock must be any of the three remaining pairs. $\frac68$ chance of that.

So our probability is $\frac89 \times \frac68 = \frac23$[/hide]

I think she meant 3 pairs of socks, not 3 socks ...

no it is asking if Robin losses 3 socks not 3 pairs

Oh. My bad.  :oops:  But where did Robin come from?

[hide]To lose three socks and only have 2 complete pairs, no combination of two socks of the three socks Don loses are matching.  There are a total of $\binom{10}{3}$ ways to lose three socks.  Let's make a pair of socks one sock, since they represent the same color. That makes $\binom53$ ways.  But we have to multiply by $2^3$ because each "sock" can be interchanged with the other sock sharing a color.

Thus, the answer is $\frac{2^3 \cdot \binom53}{\binom{10}3}$ which is $\frac23$.  [/hide]

I'm pretty sure robin came from your name >_>

I am a he and take offense to that but I ment Don

sorry man, i guess i failed to see your little gender symbol thing correctly

Hmm.. lol, I didn't notice those gender thingies before.


[quote="Mathermatics are beautifu"]Let $ a,b.c$ be positive real numbers.
Prove that $ \sum (a^4b^2 \plus{} a^3b^3 \plus{} 3a^2b^2c^2) \geq \sum (2a^3b^2c \plus{} 3a^4bc).$

Denote
$ \sum$ is symmetric sum.[/quote]
It's wrong. Try $ a\rightarrow\plus{}\infty$ and $ b\equal{}c.$ :wink:


function
algebra
polynomial
linear algebra
matrix
vector
probability and stats

It seems to me regression in 3d space doesn't work as good as in 2d? Also, there is very little literature on regression in 3d space. Even Matlab only has a 2d polyfit() function but not 3d.

Why does this happen?

i dont know if this is how matlab does it, but for my students when we ask to find the line (or polynomial ) of best fit, we do it with adjoints of matrices

say you are given a set of data points $ \{(t_j,y_j)\}_{j\equal{}1}^n$

then for a polynomial of degree k of best fit we create the matrix

$ A\equal{}[ a_k, a_2, ... a_1, 1 ]$ which is in $ M_{n\times k\plus{}1}(\mathbb{R})$
where each $ a_m$ is a column vector of the form $ [t_1^m,...,t_n^m]^T$
and $ 1\equal{}[1,...,1]^T$, defining $ y\equal{}[y_1,...,y_n]^T$

we then compute $ (A^*A)^{\minus{}1}Ay\equal{}c\equal{}[c_k,...c_1,c_0]^T$

then the desired polynomial is $ \sum c_jx^j$
since matlab is matrix based, i  imagine this is what goes on behind the scenes.. or maybe something even cooler.. but stretching this to another dimension seems hard-to-nearimpossible


If angle A = 20° and angle AFG = angle AGF, then angle B + angle D =

$80 \circ$ since angle afg and angle agf are the same. therefore, they are both $80 \circ$. On the straight line, pfg must equal $100\circ$. Since PDF is a triangle, the sum of the two must be $80 \circ$

I get your solution. good, except there was no angle P in the picture. It was a D.

if angle A is 20 and afg and agf are the same then 20+2x=180, x=80 so afg and agf are 80.
So angle bfd is 100 and angle b and d are equal then angle b is 40 and angle d is 40 so the some is $\boxed{80}$

[hide]
angle a is 20 degrees, so the other 2 angles
(since they are congruent)160/2 which equals 80.
Then, using complementary the angle next to it is angle is 100 degrees. The big triangle has one degree of 100, so the sum of B and D has to be $\framebox{80}$, since 180 - 100 = 80.[/hide]


Given four positive intergers a,b,c,d such that a>b>c>d>0 and ac+bd is divisible by a+b+c+d. Prove that $ a^mb^n\plus{}c^md^n$ is not a prime where m,n are odd numbers.

This problem was posted :
http://www.mathlinks.ro/viewtopic.php?p=1125575#1125575


...a copy of this years AMC 8?  Or any other AMC [8] resources...

Thank you so much.  AoPS is a VERY nice site.

You can order any AMC 8 from 1986-2003 from the AMC itself @
http://www.unl.edu/amc/d-publication/publication.html

Click on Order Form pdf, and mail/fax/call them.


Integer sequence
number theory
Calculate
IMO Shortlist

A sequence of integers $ a_{1},a_{2},a_{3},\ldots$ is defined as follows: $ a_{1} \equal{} 1$ and for $ n\geq 1$, $ a_{n \plus{} 1}$ is the smallest integer greater than $ a_{n}$ such that $ a_{i} \plus{} a_{j}\neq 3a_{k}$ for any $ i,j$ and $ k$ in $ \{1,2,3,\ldots ,n \plus{} 1\}$, not necessarily distinct. Determine $ a_{1998}$.

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

We prove that $a_{n}$ contains all positive integers of the form $3k+1$ or $9t+3$.
It can be esaily verified that $a_{1}=1$, $a_{2}=3$, $a_{3}=4$ and $a_{4}=7$. Also $a_{5}\neq8$, because $8+a_{1}=3a_{2}$.
Suppose that up till $n=9k$, the sequence $(a_{n})$ takes all values of the form $3t+1$ and $9t+3$.
Suppose $a_{m}=9k-2$.
Using trivial congruences mod $3$, we see that $9k+1$, $9k+4$ and $9k+7$ are certainly members of the sequence $(a_{n})$.
Now, because $9k+3=3(3k+1)$ and since $3$ and $3k+1$ are terms of $(a_{n})$, $a_{m+1}\neq9k$. Then $a_{m+1}=9k+1$.
Because $(9k+2)+1=3(3k+1)$, $a_{m+2}\neq9k+2$.
Because $9k+3+a_{l}\vdots 3$ iff $a_{l}\vdots3$, then $a_{l}=9r+3$, then $(9k+3)+a_{l}=3(3s+2)$, and $3s+2\notin (a_{n})$. Thus $a_{m+2}=9k+3$.
We have then $a_{m+3}=9k+4$. From $(9k+5)+7=3(3k+4)$ and $9k+6+9k+6=3(6k+4)$, we have $a_{m+4}=9k+7$ [since $7,3k+4,6k+4\in(a_{n})$].
Finally $9k+8+4=3(3k+4)$, so $9k+8\notin (a_{n})$.

Now, from $9k$ to $9k+8$ there are exactly 4 terms of the sequence and $a_{4t}=9t-2$. Then $a_{4t+2}=9t+3$, so $a_{1998}=4494$.

We will prove by induction that the sequence is made up of all numbers that are $1, 3, 4, 7\pmod{9}$. Assume that the sequence is like this until $a_n$ (inclusive).

Note that adding a number that's $1\pmod{3}$ is always possible without breaking the condition. So, all of the numbers that are $1\pmod{3}$ that are less than $a_{n+1}$ are in the sequence.

Assume towards a contradiction that $a_{n+1}\equiv 2\pmod{3}$.

So, $a_{n+1}$ is of the form $9k+2, 9k+5$ or $9k+8$. In any of those cases, it is possible to add a previous term to get $9k+12=3(3k+4)$. However, $3k+4$ must be a previous term, a contradiction.

Assume towards a contradiction $a_{n+1}=9k+6$. Then, $a_{n+1}+a_{n+1}=18k+12=3(6k+4)$. However, $6k+4$ is a previous term in the sequence, a contradiction.

Assume towards a contradiction $a_{n+1}=9k$. Then, $a_{n+1}+a_2=9k+3=3(3k+1)$. However, $3k+1$ is a previous term in the sequence, a contradiction.

So, the next term must be $1, 3, 4, 7\pmod{9}$, all of which don't break the condition. So, the sequence is all of the numbers that are $1, 3, 4, 7\pmod{9}$. So, $a_{1998}=4494$.

We claim that the sequence contains all numbers $\equiv 1,3,4,7\pmod 9$ and no more. Note that this sequence works by considering $\pmod 9$. We write the first few terms of the sequence: $a_1=1,a_2=3,a_3=4,a_4=7,a_5=10,a_6=12,a_7=13,a_8=16,a_9=19.$
\\$~$\\
Suppose we know that sequence $a$ is $(1,3,4,7,\dots,9k-2,9k+1,\dots).$ If our next term is $9k+2$ then we have $(9k+2)+(9k+1)=3(6k+1)$ and $6k+1$ is a term of our sequence because it is congruent to $1\pmod 3.$ Thus, the next term is $9k+3$ and the next one is $9k+4$
\\$~$\\
Now, if the next term is $9k+5$ then $(9k+5)+(9k-2)=3(6k+1)$ which is a contradiction. If the next term is $9k+6$ then we have $(9k+6)+(9k+3)=3(6k+3)$, $(9k+6)+(9k-6)=3(6k)$, and $(9k+6)+(9k-15)=3(6k-3)$ and one of these must be a term of the sequence.
\\$~$\\
The next term is $9k+7$ and if the next term is $(9k+8)$ we have $(9k+8)+(9k+4)=3(6k+4)$ which is a contradiction. If the next term is $(9k+9)$ then $(9k+9)+(9k+3)=3(6k+4)$ which is another contradiction. Thus, the next term is $(9k+10)$ so we are done. $a_{1998}=4494$.

We claim that the sequence consists of all positive integers that are $1,3,4,7\pmod 9$. By considering $\pmod 9$ this sequence works. This would give $\boxed{a_{1998} = 4494}$. We prove by induction that this is the only solution.

First, notice that $a_1 = 1, a_2 = 3, a_3 = 4, a_4 = 7, a_5 = 10, a_6 = 12,$.

Suppose that $a_1, a_2, \ldots, a_{n-1}$ consists of positive integers only $1,3,4,7\pmod 9$ (so in fact, all positive integers $1,3,4,7\pmod 9$ at most $a_{n-1}$).  We will show that $a_n\equiv \{1,3,4,7\}\pmod 9$.

If $a_n = 9k$, then notice that $9k + (9(k-1) + 3) = 3(6k-2)$, contradiction.

If $a_n = 9k +2$, then we have $(9k+2) + (9k+1) = 3(6k+1)$, contradiction.

If $a_n = 9k + 5$, then we have $(9k+5) + (9k-2) = 3(6k+1)$, contradiction.

If $a_n = 9k + 6$, then $(9k+6) + (9k+6) = 3(6k + 4)$, contradiction.

If $a_n = 9k+8$, then $(9k+8) + (9k-5) = 3(6k +1)$, contradiction.

Thus, $a_n$ cannot be $0,2,5,6,$ or $8$ mod $9$, which proves $a_n \in \{1,3,4,7\}\pmod 9$, so the induction is complete.

We now conclude $a_i\in \{1,3,4,7\}\pmod 9$, which finishes.

The answer is $\boxed{4494}$. The sequence $a_i$ is the sorted sequence of $1$ mod $3$ or $3$ mod $9$ positive integers.

Manually check that the first seven terms are $1,3,4,7,10,12,13$. Now consider $m=a_{n+1}$ for $n\ge 7$ and use induction.

[b]Case 1.[/b] It is $2$ mod $6$. Then let $m=6k+2$. Since $3k+1$ is in the sequence, $(3k+1)+(6k+2)=3(3k+1)$, contradiction.
[b]Case 2.[/b] It is $5$ mod $6$. Then let $m=6k+5$. Since $3k-2$ is in the sequence, $(3k-2)+(6k+5)=3(3k+1)$, contradiction.
[b]Case 3.[/b] It is $1$ mod $3$. Since it is the new largest term, if there is a contradiction it must appear in the LHS. But the LHS cannot be a multiple of $3$ since the other term in the LHS cannot be $2$ mod $3$, so there is no contradiction and this always works.
[b]Case 4.[/b] It is $6$ mod $9$. Then let $m=9k+6$. Since $6k+4$ is in the sequence, $(9k+6)+(9k+6)=3(6k+4)$, contradiction.
[b]Case 5.[/b] It is $9$ mod $9$. Then let $m=9k$. Since $3$ and $3k+1$ are in the sequence, $(9k)+(3)=3(3k+1)$, contradiction.
[b]Case 6.[/b] It is $3$ mod $9$. Since it is the new largest term, if there is a contradiction it must appear in the LHS. But if the LHS is a multiple of $3$, the other term must also be $3$ mod $9$, so $a_k$ is $2$ mod $3$, which can't happen, so there is no contradiction and this always works.

It then remains to work out the computation. $\blacksquare$


calculus
geometry
integration
calculus computations

How can you use calculus to find the volume and surface area of a ball?  I know it involves integrating.  I've been trying to use the solid of revolution approach with a circle, but can't seem to set it up correctly.

Thanks for the help.

[quote="skimnc"][b]Volume[/b]

[hide]Assuming its a perfect ball we know that the equation of a semicircle is $y=\sqrt{r^{2}-x^{2}}~~\forall ~~y\geq 0$, where $r$ is the radius, and if we use the disc method, $V=\pi \int_{a}^{b}\left[ f(x) \right]^{2}~dx$, we get:

$V=\pi \int_{-r}^{r}\left[ \sqrt{r^{2}-x^{2}}\right]^{2}~dx$
$\iff V=2\pi \int_{0}^{r}(r^{2}-x^{2}) ~dx$
$\iff V=2\pi \left[r^{2}\cdot x-\frac{x^{3}}{3}\right]_{0}^{r}$
$\iff \boxed{V=\frac{4 \pi r^{3}}{3}}$[/hide]

[b]Surface Area[/b]

[hide]Again, we'll use $y=\sqrt{r^{2}-x^{2}}~~\forall ~~y\geq 0$ and the formula $SA = 2 \pi \int_{a}^{b}f(x) \cdot \sqrt{1+\left[ f'(x) \right]^{2}}~dx$

$SA = 2 \pi \int_{-r}^{r}\sqrt{r^{2}-x^{2}}\cdot \sqrt{1+\frac{x^{2}}{r^{2}-x^{2}}}~dx$
$\iff SA = 4 \pi \int_{0}^{r}\sqrt{r^{2}-x^{2}}\cdot \sqrt{\frac{r^{2}}{r^{2}-x^{2}}}~dx$; Because $0\leq x\leq r$ the square roots will both be entirely positive and we can combine them.  Note that if there were imaginary values we could not do this.
$\iff SA = 4 \pi \int_{0}^{r}r ~dx$
$\iff \boxed{SA = 4 \pi r^{2}}$[/hide][/quote]

ah so thats how you obtain those 2 formulas :o

thanks


search
geometry
geometric transformation
reflection
angle bisector
geometry proposed

I searched the site, but, with surprise, did not find this problem:

[color=blue]
In a convex quadrilateral ABCD. We have AB=AD, CB=CD, M is the intersection of AC and BD. Through M draw two lines. One line intersect AB and CD at P and Q. The other one intersect BC and AD at R and S.  PR meets BD at G and SQ meets BD at H.
If M is the midpoint of BD, prove: $GM=HM$.
[/color]

First of all, shobber, you have made a number of typos.
For this problem, I haven't tried, but I'm sure trigo works fine for it, and throughout the proof the trigo can be possibly "hidden" by applying Menelaus/Ceva/Transversal Theorems several times.

[quote="shobber"]I searched the site, but, with surprise, did not find this problem[/quote]
What keywords did you search?

[quote="mecrazywong"]First of all, shobber, you have made a number of typos.[/quote]
Thank you very much for pointing them out!  :)

[quote="Palytoxin"]is this problem somewhat related to your avatar, shobber? ;)[/quote]
Well, the proofs are nearly the same...... So try it!  :)

Maybe you guys think this problem is very easy.....So here are two more problems:

1 (1999 China) In quadrilateral ABCD, AC is the angle bisector of $\angle{BAD}$. F is on segment AC. BF meets CD at E and DF meets BC at G.
Prove $\angle{GAC}=\angle{EAC}$

2 In [b]any[/b] quadrilateral ABCD, AC bisect BD. M=AC $\cap$ BD. Through M draw two lines. One meets AB and CD at E and F, the other one meets AD and BC at G and H. GF meets BD at Q and EH meets BD at P.
Prove OP=OQ.

[i]Sir Shobber,[/i] I suggest you such a thing: looking at drawing,apparently I see a thetraedron ABCD. Can you [b]to raise in a space[/b] this problem ? (The point M be internal to the thetraedron ABCD and be in the bisector plane of the edge BD !!). Good chance !. I tried and I didn't managed. With friendship, [i]Levi[/i]

[quote="shobber"]Maybe you guys think this problem is very easy.....So here are two more problems:

1 (1999 China) In quadrilateral ABCD, AC is the angle bisector of $\angle{BAD}$. F is on segment AC. BF meets CD at E and DF meets BC at G.
Prove $\angle{GAC}=\angle{EAC}$
[/quote]

I've solved tons of such problems in the exact same way :) (actually, I'm getting pretty sick of this type of proof, so this is the last time I'm posting one :)):

As we move $F$ on $AC$, the map $AE\mapsto AG$ that this movement induces from the pencil of lines through $A$ to itself is a projective map. When $F=C,F=A$, and $F=AC\cap BD$, the lines $AE,AG$ are symmetric wrt $AC$, which means that our projective transformation coincides with the reflection in $AC$ (of the lines through $A$, remember) in three different positions. This actually means that the two transformations coincide, and we're done.

Your second problem has a very similar interpretation, leading to the same type of solution.

By chance, the your problem is named "the problem of the butterfly".
Look a equivalent expression of this problem:

"Let [b]ABC[/b] be a isosceles triangle ([b]AB=AC[/b]), the middlepoint [b]O[/b] of the side [b][BC][/b], the points $B_1\in ray(OB,\ C_1\in ray(OC$ such that $OB_1=OC_1$ and the line $d,\ A\in d,\ d\parallel BC.$
For any points $X,\ Y\in d$ I note the intersections: $\ M\in OX\cap AB,\ M_1\in OX\cap AB_1;\ N\in OY\cap AC,\ N_1\in OY\cap AC_1$.
Prove that the lines $MN_1,\ NM_1,\ AO$ are concurrently."

[b]Remark.[/b] The best and rapid solution is analytically. It is interesting the case
$\ \ XO\parallel AB\ (M\rightarrow\infty )\ \ \vee / \wedge\ \ YO\parallel AC\ (N\rightarrow \infty\ ).$ [b] See[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=47140

A another problem with the same name ("problem of butterfly") is following:

"Let $C(O,\ r)$ be a circle, a point $A$ which belongs to the interior of the circle $C$
and the points $A,\ B,\ M,\ N,\ P,\ R\in C$ such that $A\in BC\cap MN \cap PR$.
I note $X\in BC\cap PN,\ Y\in BC\cap MR$. Prove that $AX=AY\Longleftrightarrow OA\perp BC.\"$


Who will win the NL East? I vote for the Braves, they don't have a very big lead but they have the resources to beat out the Phillies.


what is the value of tan (pi/2).is it not define or is it also right to say + infinity.
if it is +infinity then what should be the value tan(3pi/2).is it +infinity or - infinity.
what i want to say is period of tan is pi .therefore if tan pi/2 is +infinity then it should be +infinity .but when i used tan(3pi/2)= sin(3pi/2)/cos(3pi/2) it comes out to be - infinity . where is the problem .graph gives both values +infinity and _infinity at 3pi/2
_________________

$ \tan\frac{\pi}{2}$ is undefined, you can't say it's $ \plus{}\infty.$ Same thing with $ \tan\frac{3\pi}{2},$ it's undefined.

Hmm...as far as I am concerned, $ \tan \frac{\pi}{2}$ is just undefined. $ \tan \frac{\pi}{2}$ is the same as $ \frac{\sin \pi/2}{\cos \pi/2}$ but the $ \cos \frac{\pi}{2}$ is 0. So basically, if you are thinking in terms of the graph, the value of $ \tan \frac{\pi}{2}$ is never reached by the tan function, and neither is $ \tan \minus{}\frac{\pi}{2}$ but the $ \tan$ function shoots higher and higher as the $ \cos$ function gets closer and closer to zero. So the range of the $ tan$ function is actually $ \minus{}\frac{\pi}{2} < \tan < \frac{\pi}{2}.$ So, I am not sure if $ \plus{}\infty$ is a good answer to that.

The value of $ \tan \frac{3\pi}{2}$ is the same story. It is exactly between $ \pi$ and $ 2\pi$ so it is never reached by any of the graphs. Thus, it is also undefined, as $ \cos \frac{3\pi}{2} \equal{} 0$ and division by zero is undefined. Again, I am not sure about $ \plus{}\infty$ or $ \minus{}\infty,$ I think the best is just that it is undefined.

What is the value of $ \tan(\pi/2)$?  The most elementary thing to say is "not defined".  Do not say $ \plus{}\infty$, since values of $ \tan(x)$ are negative when $ x>\pi/2$.

A more advanced thing to say, when you study complex analysis, is:  $ \tan(z)$ has a pole at $ z\equal{}\pi/2$ with residue $ \minus{}1$.  But $ \tan(z)$ also has a pole at $ z\equal{}3\pi/2$ with the same residue $ \minus{}1$ ... as you note the period is $ \pi$ so these two points are [b]the same[/b] for $ \tan(z)$.


inequalities
calculus
derivative
inequalities unsolved

If $ a,b,c$ are reals show that:
$ \frac{(b\plus{}c\minus{}a)^2}{(b\plus{}c)^2\plus{}a^2}\plus{}\frac{(c\plus{}a\minus{}b)^2}{(c\plus{}a)^2\plus{}b^2}\plus{}\frac{(a\plus{}b\minus{}c)^2}{(a\plus{}b)^2\plus{}c^2}\geq\frac{3}{5}$

[quote="Dimitris X"]If $ a,b,c$ are reals show that:
$ \frac {(b \plus{} c \minus{} a)^2}{(b \plus{} c)^2 \plus{} a^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{(c \plus{} a)^2 \plus{} b^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{(a \plus{} b)^2 \plus{} c^2}\geq\frac {3}{5}$[/quote]
By Cauchy we obtain:
$ \sum_{cyc}\frac {(b \plus{} c \minus{} a)^2}{(b \plus{} c)^2 \plus{} a^2}\equal{}\sum_{cyc}\frac {(b\plus{}c)^2(b \plus{} c \minus{} a)^2}{(b \plus{} c)^4 \plus{} a^2(b\plus{}c)^2}\geq$
$ \geq\frac{\left(\sum(b^2\plus{}c^2\plus{}2bc\minus{}ac\minus{}bc)\right)^2}{\sum(2a^4\plus{}4a^3b\plus{}4a^3c\plus{}8a^2b^2\plus{}2a^2bc)}\equal{}$
$ \equal{}\frac{2(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^4\plus{}2a^3b\plus{}2a^3c\plus{}4a^2b^2\plus{}a^2bc)}\geq\frac{3}{5},$ where the last inequality is obviously true after expanding.

Alternative solution to the problem:

Since the inequality is homogeneous in the three variables a,b,c, we may assume that $ a + b + c = 1$.

The inequality is equivalent to

$ \sum_{cyc} \frac {(1 - 2a)^2}{(1 - a)^2 + a^2} = \sum_{cyc} \frac {1 - 4a + 4a^2}{1 - 2a + 2a^2} \# = \sum_{cyc} ( 2 - \frac {1}{1 - 2a + 2a^2} ) \geq \frac {3}{5}$
This is equivalent to $ \frac {27}{5} \geq \sum_{cyc} \frac {1}{1 - 2a + 2a^2}$

[color=red][b]Lemma.[/b] [/color] [color=blue]For all positive x, the following inequality holds:[/color]
$ \frac {1}{1 - 2x + 2x^2} \leq \frac {54}{25}x + \frac {27}{25}$
Proof is trivial - just send all terms to the LHS and factor them.

By the lemma, $ \sum_{cyc} \frac {1}{1 - 2a + 2a^2} \leq \sum_{cyc} ( \frac {54}{25}a + \frac {27}{25}) = \frac {27}{5}$

[quote="qwerty414"]
[color=red][b]Lemma.[/b] [/color] [color=blue]For all positive x, the following inequality holds:[/color]
$ \frac {1}{1 \minus{} 2x \plus{} 2x^2} \leq \frac {54}{25}x \plus{} \frac {27}{25}$
Proof is trivial - just send all terms to the LHS and factor them.

[/quote]

Could you show this in your way, please?

[quote="epsilonist"]
Could you show this in your way, please?[/quote]

$ \frac {1}{1 \minus{} 2x \plus{} 2x^2} \leq \frac {54}{25}x \plus{} \frac {27}{25}$
$ \Leftrightarrow (3x\minus{}1)^2(6x\plus{}1)\ge0$, It is obvious true.

[quote="NextPeace"][quote="epsilonist"]
Could you show this in your way, please?[/quote]

$ \frac {1}{1 \minus{} 2x \plus{} 2x^2} \leq \frac {54}{25}x \plus{} \frac {27}{25}$
$ \Leftrightarrow (3x \minus{} 1)^2(6x \plus{} 1)\ge0$, It is obvious true.[/quote]
But it's wrong for real $ x.$  :wink:

[quote="arqady"]

But it's wrong for real $ x.$  :wink:[/quote]

Oh, Sorry. I don't see the condition :blush:

But, if one or two or three (^^) of the numbers are negative, then the LHS increases and so we can assume that $ a,b,c$ are positive or at least non-negative real numbers.   :P
And then your proof is valid and indeed very nice.

[quote="arqady"]But it's wrong for real $ x.$  :wink:[/quote]

Yes, but $ 1\minus{}2a\plus{}2a^2 \geq 1\minus{}2|a|\plus{}2|a|^2$

so $ \frac{1}{1\minus{}2a\plus{}2a^2} \leq \frac{1}{1\minus{}2|a|\plus{}2a^2}$

[quote="hungnsl"][quote="arqady"]But it's wrong for real $ x.$  :wink:[/quote]

Yes, but $ 1 \minus{} 2a \plus{} 2a^2 \geq 1 \minus{} 2|a| \plus{} 2|a|^2$

so $ \frac {1}{1 \minus{} 2a \plus{} 2a^2} \leq \frac {1}{1 \minus{} 2|a| \plus{} 2a^2}$[/quote]
What we doing with $ a\plus{}b\plus{}c\equal{}1$ $ ?$ :wink:

sorry arqady but i do not understand what you mean if $ a\plus{}b\plus{}c\equal{}1$ ???

[quote="hungnsl"]sorry arqady but i do not understand what you mean if $ a \plus{} b \plus{} c \equal{} 1$ ???[/quote]
Why after your estimation $ \frac {1}{1 \minus{} 2a \plus{} 2a^2} \leq \frac {1}{1 \minus{} 2|a| \plus{} 2a^2}$ we having $ |a|\plus{}|b|\plus{}|c|\equal{}1$ $ ?$ :wink:
It's better: what your full proof? Thank you!

yeah , i see my mistake  :blush: really sorry

this is not very easy as i thought

Let a+b+c=d. Then the inequality becomes Σ{[(d-2a)^2)]/[(d-a)^2 +a^2]}>=3/5.
Let f(x)=[(d-2x)^2)]/[(d-x)^2 +x^2]. Then f''(x)>0, for every real x, so it is convex and we can apply the Jensen inequality:
RHS>=3*[(d -2d/3)^2)]/[(d-d/3)^2 +(d/3)^2]=3/5.
Equality holds if and only if a=b=c.

[quote="Spribo"]
Let f(x)=[(d-2x)^2)]/[(d-x)^2 +x^2]. Then f''(x)>0, for every real x,...[/quote]
Are you sure?

simply apply cauchy... the rhs $ \ge$ 3\5 cross multiply u get a condition which holds true by am-gm?

i hope i'm nt making any mistakes.

plz let me know if i'm going wrong

[quote="sfsreporter"]simply apply cauchy... the rhs $ \ge$ 3\5 cross multiply u get a condition which holds true by am-gm?
[/quote]
Explain it in more detail. Thank you!
[quote="Spribo"]Yes, I am totally sure: we get positive derivative.[/quote]
Write the second derivative, which you  obtained. Thank you!

i have done it like this..chances of error are there..so let me know if i'm right or wrong and plz also mention why. :maybe:  :maybe:

$ \sum \frac{{(a\plus{}b\minus{}c)}^2}{a^2\plus{}{(b\plus{}c)}^2} \ge \frac{{(a\plus{}b\plus{}c)}^2}{a^2\plus{}b^2\plus{}c^2\plus{}{(a\plus{}b)}^2\plus{}{(b\plus{}c)}^2\plus{}{(c\plus{}a)}^2}$

$ \ge \frac{3}{5}$

therefore

$ 5{(a\plus{}b\plus{}c)}^2 \ge 9(a^2\plus{}b^2\plus{}c^2)\plus{}2(ab\plus{}bc\plus{}ca)$

on simplification

$ {(a\plus{}b\plus{}c)}^2 \ge 3(ab\plus{}bc\plus{}ca)$

which holds true by am- gm  ???

i hope i have avoided mistakes.... plz let me know :blush:
thank you

Well, in mathematics and science logic estimations can often be very useful. Here, if the derivative was negative, then the inequality wouldn't hold  :!:   But the exercise itself states that it should hold. I simply follow the path of books that find, behind the long calculations of derivatives, the solution, and so I didn't actually calculate it. Yes, Jensen wouldn't be very useful in a competition, as it may take time, but it does remain a solution.
sfsreporter, in the second step you forget to multiply 2(ab+bc+ca) with 3...

[quote="sfsreporter"]
therefore

$ 5{(a \plus{} b \plus{} c)}^2 \ge 9(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2(ab \plus{} bc \plus{} ca)$

[/quote]
It's wrong. Try $ a\rightarrow\infty.$ :wink:

the inequality, $ 5(a\plus{}b\plus{}c)^2>9(a^2\plus{}b^2\plus{}c^2)\plus{}2(ab\plus{}bc\plus{}ca)$ is true only if $ a,b,c$ are the lengths of the sides of a triangle.

[quote="Spribo"]... and so I didn't actually calculate it.[/quote]
I have obtained $ f''(x)\equal{}\minus{}\frac{4d^2(6x^2\minus{}6dx\plus{}d^2)}{(2x^2\minus{}2dx\plus{}d^2)^3},$ which is can be negative. :wink:

Yes, I see it.  :blush:  I shouldn't be so absolute. But why didn't Jensen work? I thought it always worked!

[i]Straight From the Book[/i] by Titu Andreescu and Gabriel Dospinescu contains a nice proof using substitution on p. 15:

Let $x=\frac{b+c}{a}, \quad y=\frac{a+c}{b}, \quad z=\frac{a+b}{c}.$  Then the problem becomes one of proving that for any positive numbers $x,y,z$ such that $xyz=x+y+z+2$ we have
\[ \frac{(x-1)^2}{x^2+1}+\frac{(y-1)^2}{y^2+1}+\frac{(z-1)^2}{z^2+1} \ge \frac{3}{5}. \]
Applying the Cauchy-Schwarz inequality (in the "Titu's lemma" form), we see that
\[ \frac{(x-1)^2}{x^2+1}+\frac{(y-1)^2}{y^2+1}+\frac{(z-1)^2}{z^2+1} \ge \frac{(x+y+z-3)^2}{x^2 +y^2 +z^2+3}. \]
Now we have to prove that the last quantity is at least $\frac{3}{5}.$  Let $S=x+y+z$ and $P=xy+yz+zx,$ so that the inequality becomes
\[ \frac{(S-3)^2}{S^2-2P+3} \ge \frac{3}{5}.\]
Notice that since $\frac{a}{b} +\frac{b}{a} \ge 2$ and cyclic permutations of this, we have $S \ge 6.$  Also, we can show that $P \ge 2S$ [Problem 7 on p. 9 of the book mentioned above]. Therefore
\[ \frac{(S-3)^2}{S^2-2P+3} \ge \frac{(S-3)^2}{S^2-4S+3} = \frac{S-3}{S-1}=1- \frac{2}{S-1} \ge \frac{3}{5}.\]

[quote="PolyaPal"]
Notice that since $\frac{a}{b} +\frac{b}{a} \ge 2$ [/quote]
It's wrong of course! Try $a=1$ and $b=-1$. :wink:

Note that the original problem for the 1997 Japanese Mathematical Olympiad stated that $a,b,$ and $c$ are positive real numbers.  This is what Andreescu's solution assumes and this is why $\frac{a}{b} + \frac{b}{a} \ge 2$ is valid.

[quote="PolyaPal"]Note that the original problem for the 1997 Japanese Mathematical Olympiad stated that $a,b,$ and $c$ are positive real numbers.  [/quote]
But the starting inequality from another Olympiad. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1408011#p1408011
 :wink:

It may be that the problem from "another Olympiad" is misstated.

[url=http://www.artofproblemsolving.com/community/c6h146p537]Japan MO 1997, problem #2[/url]


analytic geometry
linear algebra
matrix
vector
conics
hyperbola
calculus

(1) In the coordinate plane, for a given constant $ m$, let $ H$ be the foot of the perpendicular drawn from the point $ P$ to the line $ y \equal{} mx$. Find the matrix $ T_m$ of the linear transformation $ f$ which maps the point $ P$ to the point $ H$.

(2) For the linear transformation by $ T_{\sqrt {2}}$, find the image of the curve $ x^2 \minus{} 3y^2 \equal{} 1$.

Let $ u$ be a unit vector in $ \mathbb{R}^n.$ The linear orthogonal projection of $ x$ onto the (line, plane, hyperplane) perpencdicular to $ u$ is given by $ Px$ where $ P \equal{} I \minus{} uu^T.$ Note that $ P^2 \equal{} (I \minus{} uu^T)^2 \equal{} I \minus{} 2uu^T \plus{} uu^Tuu^T \equal{} I \minus{} uu^T \equal{} P$ since $ u^Tu \equal{} 1.$

A vector perpendicular to the line $ y \equal{} mx$ is $ \begin{bmatrix} \minus{} m \\
1\end{bmatrix}.$ That's not a unit vector, so we normalize it: $ u \equal{} \frac1{\sqrt {m^2 \plus{} 1}}\begin{bmatrix} \minus{} m \\
1\end{bmatrix}.$

Then $ uu^T \equal{} \frac1{m^2 \plus{} 1}\begin{bmatrix}m^2 & \minus{} m \\
\minus{} m & 1\end{bmatrix}.$

$ P \equal{} T_m \equal{} I \minus{} uu^T \equal{} \frac1{m^2 \plus{} 1}\begin{bmatrix}1 & m \\
m & m^2\end{bmatrix}.$

That makes $ T_{\sqrt {2}} \equal{} \frac13\begin{bmatrix}1 & \sqrt {2} \\
\sqrt {2} & 2\end{bmatrix}.$

One can brute force the second question by geometry/calculus methods. I'm trying to find a way to cast it strictly as linear algebra, but am not happy about what I have just yet.

Edit: the image of that curve (a hyperbola) under $ T_{\sqrt{2}}$ is the points $ (x,\sqrt{2}\,x)$ such that $ x\in\left(\minus{}\infty,\minus{}\frac1{\sqrt{3}}\right]\cup\left[\frac1{\sqrt{3}},\infty\right).$

Wasn't pretty. Set $ y\equal{}\pm\sqrt{\frac{x^2\minus{}1}3}.$ Ran that through $ T.$ Decided that I had to minimize (for $ x>0$) the expression $ x\pm\sqrt{\frac{2(x^2\minus{}1}3}.$ Used calculus on that.


Euler
induction
number theory
relatively prime

Prove that the set of integers of the form $2^{k}-3$ ($k=2,3,\cdots$) contains an infinite subset in which every two members are relatively prime.

Nice!

Let $2^{a_i} - 3$, with $0\leq i \leq n$, numbers with every two relative primes.

to find our '$a_{n+1}$'  the idea is see that iff $2^k - 1$ is multiple of all of this numbers, $2^k - 3$ is relative prime with all of this!

to find this k, is very easy  :grin:

[quote=Moonmathpi496][quote="Peter"]Prove that the set of integers of the form $ 2^{k} - 3$ ($ k = 2,3,\cdots$) contains an infinite subset in which every two members are relatively prime.[/quote]
We assume to the contrary that all such subsets are finite. Now we take a subset of maximal size. Let it be, $ A = \{ 2^{a_1} - 3,2^{a_2} - 3,\cdots 2^{a_m} - 3 \}$
Now, we shall prove that, there exists a $ m > a_m$ such that $ (2^m - 3,2^{a_l} - 3) = 1$; $ 1\leq l\leq m$.
We take $ q = \prod ( 2^{a_l} - 3 )$. We have that among, $ 2^0 - 3,\cdots, 2^{q} - 3$, two of the integers will give same residue, when divided by $ q$.
That is $ q  | 2^i - 2^j$; where $ 0\leq j\leq i\leq q$.
i.e. $ q  | 2^{(i - j)} - 1$. As $ 2^{(i - j)} - 1$ is odd, $ 2^{(i - j)} - 3$ is relatively prime to $ 2^{(i - j)} - 1$, and apparently so to the elements of $ A$. Also, $ 2^{a_l} - 3 < q\leq 2^{(i - j)} - 1$.
(That is we may take $ m = (i - j)$)
Which contradicts our assumption, so $ A$ is infinite.[/quote]

Can we not just choose $m=t\phi(q)$ where $t$ is sufficiently large such that $m>q$?


Find all functions $ f: \mathbb{N} \rightarrow \mathbb{N} \minus{} \{1\}$  such that for all $n \in \mathbb{N}$, \[ f(n) \plus{} f(n\plus{}1) \equal{} f(n\plus{}2)f(n\plus{}3) \minus{} 168.\]

[quote="rajelaz"]Find all function $ f: \mathbb{N} \rightarrow \mathbb{N} \minus{} \{1\}$  then $ \forall n \in \mathbb{N}$ satisfy $ f(n) \plus{} f(n \plus{} 1) \equal{} f(n \plus{} 2)f(n \plus{} 3) \minus{} 168$[/quote]

Let $ P(n)$ be the assertion $ f(n)\plus{}f(n\plus{}1)\equal{}f(n\plus{}2)f(n\plus{}3)\minus{}168$

Subtracting $ P(n)$ from $ P(n\plus{}1)$ implies $ f(n\plus{}2)\minus{}f(n)\equal{}f(n\plus{}3)(f(n\plus{}4)\minus{}f(n\plus{}2))$ and, since $ f(n\plus{}3)\geq 2$, $ |f(n\plus{}4)\minus{}f(n\plus{}2)|\leq|\frac{f(n\plus{}2)\minus{}f(n)}2|$

So, for $ n$ great enough, $ f(n\plus{}2)\equal{}f(n)$ and the sequence is $ a,b,a,b,a,b,...$ with $ a\plus{}b\equal{}ab\minus{}168$ $ \iff$ $ (a\minus{}1)(b\minus{}1)\equal{}169\equal{}13^2$

So the sequence ends with either $ 14,14,14,14, ...$, either with $ 2,170,2,170,2,170, ...$

And, since $ f(n)\equal{}f(n\plus{}2)f(n\plus{}3)\minus{}f(n\plus{}1)\minus{}168$, we have only three solutions :

$ 2,170,2,170,2,170, ....$
$ 170,2,170,2,170,2, ....$
$ 14,14,14,14,14,14,...$


AMC
USA(J)MO
USAMO
number theory unsolved
number theory

We see that y=6k+3. So 2^y=1 mod7
But x and y must both be odd.
So 3^x-19^z=1 mod 7.
So the remainders must be 6 and 5 mod 7.
So x=6k+3.
Hence, 3^(x/3) -2^(y/3) =19^t @
Since 3^9-2^9 does not divide 19^z, we cannot have x and y divisible by 3.
So @ has no solution.
Hence the only solution is x,y,z=3,3,1.

By the way this isnt USAMO 1998, I looked it up.


[quote="GameBot"]A non-square rectangle has integer dimensions. The number
of square units in its area is numerically equal to the number
of units in its perimeter. What is the number of units in the
perimeter of this rectangle?[/quote]

$ 2(a\plus{}b)\equal{}ab$
$ ab\minus{}2a\minus{}2b\plus{}4\equal{}4$
$ (a\minus{}2)(b\minus{}2)\equal{}4$
$ a\minus{}2\equal{}1, b\minus{}2\equal{}4$
$ a\minus{}2\equal{}2, b\minus{}2\equal{}2$
$ a\minus{}2\equal{}4, b\minus{}2\equal{}1$.

The second case is impossible, the first and third case are analogous and give a perimeter of $ 18$.


Let ABC be a triangle.
Prove that:

3/4* \sum (a 2 -b 2 ) 2  +36S 2  \geq(a*ma+b*mb+c*mc) 2

I don't know where you are in the stage of proving this inequality but I think the following inequality should be useful :

a*ma+b*mb+c*mc<=2p(R+r)

so you must prove that:
3/4* sum (a^2 -b^2 )^2 +36S^2 >= [2p(R+r)]^2>= (a*ma+b*mb+c*mc)^2

hope it works! cheers! :D :D


calculus
integration
trigonometry
logarithms
function
absolute value
algebra

Original problem: $ \int \sqrt{5x^2+4}\,dx$

My Solution:

$ \int \sqrt{5x^2+4}\,dx=2\int \sqrt{\left(\dfrac{\sqrt{5}} {2}x\right)^2+1}\,dx$
Doing a u-substitution where $ u=\dfrac {\sqrt {5}} {2}x$ and $ du=\dfrac{\sqrt{5}} {2}dx \rightarrow dx=\dfrac{2\sqrt{5}} {5}du$ we get $ \dfrac {2\sqrt{5}} {5}\int \sqrt{u^2+1}\,du$
Doing a trig-substitution where $ u=\tan\left({v}\right)$ and $ du=\sec^2 \left({v}\right)dv$ we get $ \dfrac{2\sqrt{5}} {5}\int \sqrt{\tan^2\left({v}\right)+1}\sec^2\left({v}\right)\,dv$
$ =\dfrac{2\sqrt{5}} {5}\int \sec^3\left({v}\right)\,dv=\dfrac{2\sqrt{5}} {5}\int \dfrac{1} {\cos^3\left({v}\right)}\,dv$$ =\dfrac{2\sqrt{5}} {5}\int \dfrac{1} {\cos\left({v}\right)}\cdot\dfrac{1} {1-\sin^2\left({v}\right)}\,dv$
Doing another u-substition where $ m=\sin\left({v}\right)$ and $ dm=\cos\left({v}\right)dv \rightarrow dv=\dfrac{1} {cos\left({v}\right)}dm$ we get $ \dfrac{2\sqrt{5}} {5}\int \dfrac{1} {\cos^2\left({v}\right)}\cdot\dfrac{1} {1-m^2}\,dm$
$ =\dfrac{2\sqrt{5}} {5}\int \left(\dfrac{1} {1-m^2}\right)^2\,dm=\dfrac{2\sqrt{5}} {5}\int \left(\dfrac{1} {2}\left(\dfrac{1} {1-m}+\dfrac{1} {1+m}\right)\right)^2\,dm$ ${ =\dfrac{\sqrt{5}} {10}}\int \left(\dfrac{1} {\left(1-m\right)^2}+\dfrac{1} {\left(1+m\right)^2}+\dfrac{2} {1-m^2}\right)\,dm$ $ =\dfrac{\sqrt{5}} {10}\left(\dfrac{1} {1-m}-\dfrac{1} {1+m}+\int\left(\dfrac{1} {1-m}+\dfrac{1} {1+m}\right)\,dm\right)$ $ =\dfrac{\sqrt{5}} {10}\left(\dfrac{1} {1-m}-\dfrac{1} {1+m}-\ln \left|1-m\right|+\ln \left|1+m\right|\right)$.
Then if you wanted to get the indefinite integral in terms of x, then we would just work backwards substituting variables back in.

Flaw I found in my solution was in the second and third lines.  It's supposed to look like this because of the square root:

$ \dfrac{2\sqrt{5}} {5}\int \sqrt{\tan^2\left({v}\right)+1}\sec^2\left({v}\right)\,dv=\dfrac{2\sqrt{5}} {5}\int \left|\sec^3 \left({v}\right)\right|\,dv=\dfrac{2\sqrt{5}} {5}\int \dfrac{1} {\left|\cos^3\left({v}\right)\right|}\,dv$

I've tried evaluating the integral with the absolute value but I don't know how to do it.  Maybe the correct solution will look something like my solution I have here.  There also might be a much easier way to do it that I'm just not seeing.  I also tried plugging the integral into WolframAlpha but in the intermediate steps it used a reduction formula: $ \int \sec^m\left({u}\right)\,du=\dfrac{\sec^{m-1}{\left(u\right)}\sin{\left(u\right)}} {m-1}+\dfrac{m-2} {m-1}\int \sec^{-2+m}{\left(u\right)}\,du$.  I don't want a solution that uses formulas, but is this an important formula I should remember in the future?  If it is, can you show me how/why it works?  I don't like just memorizing things without knowing how/why they work.  Can someone help me find the correct solution to this indefinite integral?

In the integral $ \int\sqrt{u^2\plus{}1}\,du,$ you used the trigonometric substitution $ u\equal{}\tan v.$ That's written "backwards"; written forward, this substitution would be $ v\equal{}\arctan u.$ When you look at it that way, that implies that $ \minus{}\frac{\pi}2<v<\frac{\pi}2.$ And that means that $ \cos v>0.$ There's no need for the absolute value that is bothering you.

As for the integral of $ \sec^3x,$ or that formula you found for the integral of $ \sec^mx:$ that's a so-called reduction formula. Here is the plan for deriving reduction formulas.

1. Do an integration by parts. Choose something to integrate that you can integrate easily; differentiate the rest. (In the case of the secant powers, the thing you can integrate easily is $ \sec^2x\,dx.$)

2. Use a Pythagorean identity inside the integral that results, production two terms: one with the original power of the trig function in question, one with the power reduced by 2. (The appropriate Pythagorean identity for secants is $ \sec^2x\equal{}1\plus{}\tan^2x.$)

3. You have an equation in which the unknown integral appears on both sides. Solve this (as a linear equation) for the unknown integral.

This works for powers of all six trigonometric functions.

hello, my solution is
$ \int\sqrt{5x^2\plus{}4}\,dx\equal{}\frac{1}{2}x\sqrt{5x^2\plus{}4}\plus{}\frac{2}{5}\sqrt{5}arsinh(\frac{1}{2}\sqrt{5}x)\plus{}C$
Sonnhard.

Hyperbolic substitution $ x\equal{}\sinh t$ gives $ \int\sqrt{x^2\plus{}1}dx\equal{}\frac{1}{2}(x\sqrt{x^2\plus{}1}\plus{}\ln{x\plus{}\sqrt{x^2\plus{}1}})\plus{}C$.

hello, i think it must be
$ \int\sqrt{x^2\plus{}1}\,dx\equal{}\frac{1}{2}(x\sqrt{x^2\plus{}1}\plus{}\ln(x\plus{}\sqrt{x^2\plus{}1}))\plus{}C$.
Sonnhard.


calculus
derivative
function
search
real analysis
real analysis theorems

Does anybody know anything about a jarnik theorem, which has to do with derivative functions and Darboux' property?
It seems to be one of the things needed to know for the romanian national olympiad, 11-th class, but i think they made a mistake about the name, since I couldn't find it anywhere on the web.

In honor to the truth, I had never listened of such a theorem before. Nevertheless, I have not found it directly on the web, but at least this reference.

Sondow, J. "Irrationality Measures, Irrationality Bases, and a Theorem of Jarnik." Submitted to the proceedings of Journées Arithmétiques, Graz 2003 in the Journal du Theorie des Nombres Bordeaux.

Hey, afterwards (this is, I am editing the post) I found this one
http://www.math.jyu.fi/research/pspdf/303.pdf
Theorem 2.3, Just in case, I don't know what they mean by SEDN, but the theorem says that

For a tipical $f\in C[0,1]$ we have $SEDN(f,x)=\overline{\mathbb{R}}$ for almost every $x\in(0,1)$.

I hope it works.

Best,

thanks for your search djimenez, but I am afraid this is not the wanted theorem. It surely is a much more simple fact, and  it surely has somethin tot do about $\frac {f'}{g'}$.
As a matter of fact I am almost convinced that it is the simple fact that $\frac {f'}{g'}$ has PD, for $g'(x)\neq x. \forall x$ :)

I think you mean the following generalization of [i]Darboux[/i] theorem.Given $f,g:[a,b]{\rightarrow}\mathbb{R}$.Then for every value $C$ between $\frac{f'(a)}{g'(a)}$ and $\frac{f'(b)}{g'(b)}$ there exists $c\in(a,b)$ s.t. $\frac{f'(c)}{g'(c)}=C$.
Proof:$g'(x){\neq}0$ thus $g'(x)>0$ or $g'(x)<0$ for all $x\in[a,b]$.Lets take $g'(x)>0$.Moreover let's
assume that $\frac{f'(a)}{g'(a)}<C<\frac{f'(b)}{g'(b)}$.Now let's look at the function $h(x)=f(x)-Cg(x)$.$h'(x)=f'(x)-Cg'(x)$ with $h'(a)<0$ and $h'(b)>0$ so using Darboux theorem we conclude that there exists $c\in(a,b)$ s.t. $h'(c)=0$ or $\frac{f'(c)}{g'(c)}=C$.
We are done.    :)


Amy was born on a Tuesday. What is the probability that exactly two of her three best friends were also born on Tuesday? Express your answer as a common fraction.

Anyways, back on topic:
TTx = 6 ways
TxT = 6 ways
xTT = 6 ways
Total: 18 ways/343 possible, so 18/343 is our answer.

I think the problem should have included a statement to the effect of "Given all days are equally likely...".  However, we assume this, and get

\[\binom{3}{2}\cdot \dfrac{1}{7} \cdot \dfrac{1}{7}\cdot \dfrac{6}{7}\]

This is choosing the two friends, and the probabilities of each individual.  Thus, our answer is $\boxed{\dfrac{18}{343}}$.

[quote=sophia]This is a very nice problem, but I thought I'd point out that there's an implicit assumption in the problem that birth is equally likely on any day of the week.

It's interesting to note that this is not, in fact, empirically true.  Tuesday is actually the most popular day of the week for giving birth.

According to a recent National Center for Health Statistics report, [url]http://www.cdc.gov/nchs/data/nvsr/nvsr56/nvsr56_06.pdf[/url], births are significantly more likely to happen on a weekday than on a weekend day.  Moreover, Tuesday is the most popular day for births, with a rate 16% above average.  (Saturday was 25% below the average and Sunday was 35% below the average.)

This is very unlikely to be random chance, given the discrepancies involved.

A large proportion of births these days are scheduled in advance, or at least the timing is somewhat subject to discretionary manipulation by the physician.  Over 30% of births are Caesarian sections and 22% are  induced labors and doctors usually avoid scheduling births on the weekends.  (This is not just for the obstetrician's own convenience, by the way.  It's also the case that the hospital is not as fully staffed on weekends, so there are fewer--and less senior--nurses, anesthetists, neonatal specialists etc. around to support the birth and aftercare for the mother and child.  It's in the mother and baby's best interest to avoid a weekend delivery.)

The Wall Street Journal's "Numbers Guy" did a story on this not that long ago.

The Journal of Political Economy also did a somewhat related article a few years back on the fact that a large number of births have now been accelerated from early January back to late December due to the passage of new and more generous tax breaks for dependent children.[/quote]

But technically (because this is a math forum  :) ), I think that we have to consider equally?

[quote=monks][quote=sophia]This is a very nice problem, but I thought I'd point out that there's an implicit assumption in the problem that birth is equally likely on any day of the week.

It's interesting to note that this is not, in fact, empirically true.  Tuesday is actually the most popular day of the week for giving birth.

According to a recent National Center for Health Statistics report, [url]http://www.cdc.gov/nchs/data/nvsr/nvsr56/nvsr56_06.pdf[/url], births are significantly more likely to happen on a weekday than on a weekend day.  Moreover, Tuesday is the most popular day for births, with a rate 16% above average.  (Saturday was 25% below the average and Sunday was 35% below the average.)

This is very unlikely to be random chance, given the discrepancies involved.

A large proportion of births these days are scheduled in advance, or at least the timing is somewhat subject to discretionary manipulation by the physician.  Over 30% of births are Caesarian sections and 22% are  induced labors and doctors usually avoid scheduling births on the weekends.  (This is not just for the obstetrician's own convenience, by the way.  It's also the case that the hospital is not as fully staffed on weekends, so there are fewer--and less senior--nurses, anesthetists, neonatal specialists etc. around to support the birth and aftercare for the mother and child.  It's in the mother and baby's best interest to avoid a weekend delivery.)

The Wall Street Journal's "Numbers Guy" did a story on this not that long ago.

The Journal of Political Economy also did a somewhat related article a few years back on the fact that a large number of births have now been accelerated from early January back to late December due to the passage of new and more generous tax breaks for dependent children.[/quote]

Interesting thought process! I wonder what the probability would be if the question accounted for these discrepancies.[/quote]

It would be a big bash then.

[quote=sophia]This is a very nice problem, but I thought I'd point out that there's an implicit assumption in the problem that birth is equally likely on any day of the week.

It's interesting to note that this is not, in fact, empirically true.  Tuesday is actually the most popular day of the week for giving birth.

According to a recent National Center for Health Statistics report, [url]http://www.cdc.gov/nchs/data/nvsr/nvsr56/nvsr56_06.pdf[/url], births are significantly more likely to happen on a weekday than on a weekend day.  Moreover, Tuesday is the most popular day for births, with a rate 16% above average.  (Saturday was 25% below the average and Sunday was 35% below the average.)

This is very unlikely to be random chance, given the discrepancies involved.

A large proportion of births these days are scheduled in advance, or at least the timing is somewhat subject to discretionary manipulation by the physician.  Over 30% of births are Caesarian sections and 22% are  induced labors and doctors usually avoid scheduling births on the weekends.  (This is not just for the obstetrician's own convenience, by the way.  It's also the case that the hospital is not as fully staffed on weekends, so there are fewer--and less senior--nurses, anesthetists, neonatal specialists etc. around to support the birth and aftercare for the mother and child.  It's in the mother and baby's best interest to avoid a weekend delivery.)

The Wall Street Journal's "Numbers Guy" did a story on this not that long ago.

The Journal of Political Economy also did a somewhat related article a few years back on the fact that a large number of births have now been accelerated from early January back to late December due to the passage of new and more generous tax breaks for dependent children.[/quote]

They should add:
assuming births on every day of the week are equal

[hide=900 posts, thanks so much olympushero for letting me do this]
Assuming births on each day are equally likely, we have $\dfrac{1}{7} \cdot \dfrac{1}{7} \cdot \dfrac{6}{7} = \dfrac{6}{343}.$ There are $3$ ways to choose one of her best friends to not be born on Tuesday, so we have $3 \cdot \dfrac{6}{343} = \boxed{\dfrac{18}{343}}.$
[/hide]

[quote=Mewto55555]Alternatively, its just $$\binom{3}{2}(\frac{1}{7})^2(\frac{6}{7})=\frac{18}{343}$$[/quote]

ftfy

I need some help. Sorry for the bump. Why does the solution say [quote=solution]Since her friends' birthdays are completely independent of her own, each of them has a $\frac{1}{7}$ probability of being born on a Tuesday, and a $1 - \frac{1}{7} = \frac{6}{7}$ probability of not being born on a Tuesday. If exactly 2 friends were born on a Tuesday, the probability of this occurring is ${3 \choose 2} \cdot \frac{1}{7} \cdot \frac{1}{7} \cdot \frac{6}{7} = 3 \cdot \frac{6}{343} = \boxed{\frac{18}{343}}$. [/quote=solution]
can someone explain clearly


Prove that the sum of three consecutive squares cannot be another perfect square.

[quote="Andreas"][hide]Let $a, a+1, a+2$, be three consecutive numbers. Then one of them is divisible by $3$. So $a^{2}+(a+1)^{2}+(a+2)^{2}\equiv 0+1+1 \pmod 3$, with all possible permutation of $(0, 1, 1)$ for the residues. Hence the sum is $\equiv 2 \pmod 3$, but $2$ is not a quadratic residue $\bmod 3$.[/hide][/quote]

Nice :) But using $a-1,a,a+1$ we get $(a-1)^{2}+a^{2}+(a+1)^{2}=a^{2}-2a+1+a^{2}+a^{2}+2a+1=3a^{2}+2$. Done :)

Nice. I was trying to use mod 4, but it didn't work.


inequalities
trigonometry
function
inequalities unsolved

Prove $2\sin x+3\tan x\geq 3x$ for $0\leq x<\frac{\pi}{2}$

Consider the function
$f: \left[0,\frac{\pi}{2}\right)$ ,$f(x)=2sinx+3tgx-3x$
we have $f'(x)=2cosx+\frac{3}{cos^{2}x}-3\geq 0$. Denoting $t=cosx$ $\rightarrow$ $t\in (0,1)$  we have that
$f'(x)\geq 0$ $\Leftrightarrow$ $2t^{3}-3t^{2}+3\geq 0$ $\Leftrightarrow$ $(t-1)(2t^{2}-t-1)+2\geq 0$ obviously.
Therefore $f$ is increasing.Because $f(0)\geq 0$ we are done.


inequalities
calculus
integration
function
real analysis
real analysis unsolved

By Jensen's inequality, if $ f$ is integrable on $ [0,1]$, then $ \int \exp(f(t)) \; dt \geq \exp(\int f(t) \; dt)$. When does equality hold?

Since exp is strictly convex - that is, it touches its tangent line at only one point - an examination of the proof shows that inequality will occur only when $ f$ takes on only one value. That is, we have equality only when $ f(t)\equal{}a$ almost everywhere for some constant $ a.$

I was looking at a proof of Jensen's inequality for convex $ \varphi$ and integrable $ f$. In the proof, $ \alpha$ was defined to be $ \int f$ and the supporting line $ y\equal{}m(x\minus{}\alpha)\plus{}\varphi(\alpha)$ was used. So $ \varphi(f(t)) \geq m(f(t)\minus{}\alpha)\plus{}\varphi(\alpha)$ and then both sides were integrated to get the desired inequality.

Now when $ \varphi\equal{}\exp$, I guess $ m$ would be $ \exp(\alpha)$. How do I continue from here to see that equality holds when $ f$ is constant a.e.?

There is the completely general theorem that if $ g\ge 0$ and $ \int g\equal{}0,$ then $ g\equal{}0$ a.e. Find a way to use that.

No, go into the proof itself. Let $ \alpha\equal{}\int_0^1f.$ Then there is a linear (affine) function $ L(x)$ such that $ e^{\alpha}\equal{}L(\alpha)$ and $ e^x>L(x)$ for $ x\ne\alpha.$

$ \int_0^1(e^f\minus{}L(f))\equal{}\int_0^1e^f\minus{}L\left(\int_0^1f\right)\equal{}\int_0^1e^f\minus{}L(\alpha).$

Since $ e^f\minus{}L(f)\ge0,$ the right hand side above is nonnegative. But looking a little closer, since $ e^f\minus{}L(f)\ge0,$ the only way for the integral to be zero is for $ e^f\minus{}L(f)\equal{}0$ a.e. By the essential properties of $ L,$ that can only happen if $ f\equal{}\alpha$ a.e.

I see. Thanks a lot.
By the way, what was the mistake that I made?

You went from $ \int_0^1e^f\minus{}e^{\alpha}\equal{}0$ to concluding that $ e^f\equal{}e^{\alpha}$ a.e.

But $ e^f\minus{}e^{\alpha}$ is not nonnegative; it is sometimes positive and somtimes negative. It is quite possible for such a nonzero function to integrate to zero.

A correct proof [b]must[/b] involve the function $ L(x)\equal{}e^{\alpha}(x\minus{}\alpha)\plus{}e^{\alpha}.$ (That's known as the supporting line.)


Hi,
Could anyone help me with this inequality :

(a+2b)/(a+2c)+(b+2c)/(b+2a)+(c+2a)/(c+2b)>=3, for a,b,c>0

Thx, gelo

We have to show that for positive reals $ a$, $ b$ and $ c$ the following inequality holds true:
\[ \sum_{cyc} \frac {a \plus{} 2b}{a \plus{} 2c} \ge 3.\]
Cauchy implies $ \left( \sum \frac {a \plus{} 2b}{a \plus{} 2c}\right) \left( \sum (a \plus{} 2b)(a \plus{} 2c)\right) \ge 9(a \plus{} b \plus{} c)^2$. Thus, it's left to prove that $ 3(a \plus{} b \plus{} c)^2 \ge \sum (a \plus{} 2b)(a \plus{} 2c)$ or euqivalently $ \sum (a \minus{} b)^2 \ge 0$, which is true.


geometry proposed
geometry
U sviti matematyky

Squares $ ABCD$ and $ AXYZ$ are located inside the circle $ \omega$ in such a way that quadrilateral $ CDXY$ is inscribed into the circle $ \omega.$ Prove that $ AB\equal{}AX$ or $ AC\perp XY$.


Prove that $ 3^{5a} \plus{} 4^{5b\plus{}2} \plus{} 5^{5c\plus{}1}$ is divisible by $ 11$ for any $ a, b, c \in \mathbb{N}$.

[hide]Writing out $ 3^n$, $ 4^n$ and $ 5^n$ mod 11 we see they are periodic with period 5.

$ 3^{5k} \equiv 1 \bmod{11}$
$ 4^{5k\plus{}2} \equiv 5 \bmod{11}$
$ 5^{5k\plus{}1} \equiv 5 \bmod{11}$

Adding these gives the desired result.[/hide]


If 6abc - 2ab + 3ac + 6bc - a - 2b + 3c + 2008 = 0, where a, b and c are integers, find the minimum value of a + b + c.

Hint:
[hide]
$ (a \plus{} 1)(2b \plus{} 1)(3c \minus{} 1) \equal{} \minus{} 2009 \equal{} \minus{} 7^2\cdot41$...
[/hide]
and from hint to solution:
[hide]
Then take $ c \equal{} 0$, $ b \equal{} \minus{} 1$, $ a \equal{} \minus{} 2010$, and $ a \plus{} b \plus{} c \equal{} \minus{} 2011$.  Assume that this is not the minimum.  So at least one of $ a,b,c$ is smaller than $ \minus{} \frac {2011}{3}$, and $ \min(a,b,c)\leq \minus{} 671$.  If $ c\leq \minus{} 671$, then $ 3c \minus{} 1\leq \minus{} 2013$ absurd, while if $ b\leq \minus{} 671$, then $ 2b \plus{} 1\leq \minus{} 1341$, or $ 2b \plus{} 1 \equal{} \minus{} 2009$, yielding $ (a \plus{} 1)(3c \minus{} 1) \equal{} 1$, for unique solution $ a \equal{} c \equal{} 0$, obviously larger.  Now, $ a\leq \minus{} 671$ yields $ a \plus{} 1\leq \minus{} 670 < \minus{} \frac {2009}{3}$, and $ a \plus{} 1 \equal{} \minus{} 2009$, for the solution that we claimed to be minimum, done!
[/hide]

Note: I solve for the minimum of $ a \plus{} b \plus{} c$, as stated.  How about the minimum of $ |a \plus{} b \plus{} c|$?

I can't find a way to solve this without using trial and error. After finding twelve solutions, I found out that the solution a = -8,
b = 3,  c = 14 gives the minimum |a + b + c| which is 9.


geometry
perimeter
geometric transformation
rotation
videos
Support
ARML

i'm wondering about other people's predictions about who will go all the way in the NBA. personally, i'm going with the Houston Rockets  to go all the way cause of T-Mac, Yao Ming, and a great bunch of supporting players :D !!!

I live in the Bay Area, but its not going to be GS.  :blush:  They have Richardson and Fisher, but thats all. Kings(Go Kings), Suns , or the Spurs will go all the way. If some underdog wins it all, it'll be the Sixers with Iverson scoring 60 Pts.

Rockets all the way. T-mac, Yao and Sura.

I like the Miami Heat, mainly cause of Shaq.  I think the Spurs also have a very good chance though.

Tough call really. I'd like to say Rockets since here in h-town,  it's generally good to have rockets pride or w/e...but I dont think it's happening. We'll see...

Not the Raptors

The Spurs are lookin good right about now.

Here comes my long ranting....
Course i'm rooting for the jazz, but the early injury to kirilenko and the getting familiar with the system for boozer and okur hurt them after a great start...
I think the west and east is still wide open.. like the T-Wolves could get back into contention and the Clippers could make the playoffs with a "real" team of role players and not whole team of ballhogs. Suns have a very, very good chance, except for the fact that only the main 6 players get like 90% of the time and will be dead tired by playoffs, so i'm counting them out. Spurs look good (even though i think ginobolli was a wasted pick for all-star game), i don't know if they can beat out a better team with a great perimeter game, like the Sonics, cuz they got specialists in all areas except for inside scoring. As for the Rockets, don't think T-Mac can step it up, unless a Mr. Sura feels that he will. Grizzlies ain't much of a shot since teams now know how to handle the deep rotation now. I feel that Mavs and Kings could be sleeper picks, especially with the recent pickup of Mobley. But in the end I think it'll be the Sonics that'll dominate and go to the finals.

The east is kinda of wide open, but i think the standings will stay that way to the end, except for the Pacers getting in. Heat look very good, and if Mourning joins aboard, they will be the best no doubt. Pistons look kinda sluggish this time around with big ben not doing as much on D as he did before, but if they trade Darko something before the deadline, they're sure to be better. The only other true competitor for the Heat are the Cavs because of how they have such a great team overall. First there's Lebron, and if he can't perform, Big Z is there, then Gooden is next.

I'm gonna say that the Cavs and Sonics end up in the finals and 4-2, the Cavs win, with Ray Allen getting injured in game 4 and Fortson setting a new record for techs in a series. Plus Lebron and Big Z will be Co-MVP's if that's possible.

Finally! A basketball thread.

I don't have much to say about the West - the Spurs are typically amazing, and the Sun, Sonics are just awesome.

However, being a Detroiter, I have strong opinions on the East.

[quote="theone853"]Pistons look kinda sluggish this time around with big ben not doing as much on D as he did before[/quote]

*Slaps theone853 in the face*

How can you say that? I'm going to admit that the Pistons have faced certain distractions, but we're back on track. 7 wins last 8 games. Last three - runaway wins, 25 point average margin. And tonight we're going to take care of the league's hottest player, AI.

Seriously though, the Pistons have the 2nd best record in the East, and we're pulling far ahead of the Cavs. Yeah, LeBron is good. Their center whose name I prefer not to spell is too, but they really don't match up to a motivated pistons team.

I don't know about the Darko thing - he's been really trying lately. Career highs in scoring last two games.

Playoff Predictions: Pistons vs. Heat in conference finals. Ben Wallace fouls out early vs Shaq. McDyess gets tired early. Darko dunks on Shaq :)

Houston's last 30 or so games involves:
2 at apurs
2 w/ kings
1 w/ dallas
3 w/ suns
4 w/ sonics
1 w/ memphis
and 1 w/ wolves, heat, cavs, lakers, bulls, wizards and celtics

even though rockets beat spurs twice already, they still dont stand much of a chance against duncan at san antonio (spurs 23-1 at homecourt).  Kings are doing poorly, so rockets might pick them off.  Dallas is tough and houston needs a good day to win.  of course, I dont need to say much about suns, rockets probably will take 1 out of the 3.  sonics and rockets havent played this season so i dont know how the match up is gonna be.  and memphis game is not going to be easy, nor does playing the other 7 teams i've mentioned above (but i gotta say, 30 point win over wizards yesterday is pretty nice).  rockets is just too unstable  :S
and houston needs to get rid of taylor and strickland... they are useless

i think the rockets can at least get past the first round, although if they stay in 6th place, they're sure to be in tons of first round trouble. They haven't played the Sonics but like playing the Suns or Dallas, it could be trouble because of the uptempo game of the other teams.
In the east, I like Miami with Shaq and Wade. Zo might be able to help. Detroit will be a team contending for the finals, as they are getting back on track, Cleveland is fairly good with King James, Z, and Gooden but they don't seem to be playing at the same level as the other teams.

Kings will win in 6 games, with Webber winning the MVP.

From boston but we all know that the celtics have no chance in winning the championship...  I like the heat with shaq and wade but Spurs are doing awesome...  Heat vs Spurs  its going to game 7 but i don't know who.

elvenchamp777

never say never!!! the celtics r doin pretty good this year

Right now, it looks as if the Spurs or the Suns will win the NBA finals. The Eastern conference teams are just not as good.... No disrespect to teams such as the Hear, Pistons, etc. but they just don't compare to Western Conference teams. The Suns have the best offense in the league and they're fun to watch. Steve Nash is easily the MVP for helping the Suns to first place. But lack of experience will hurt them in the playoffs. The Spurs have proven to be winners in the playoffs. Plus, they have this guy named Tim Duncan..... So now my final pick is the Spurs to win it all! (Even though I hate them because I'm a Mavericks fan....Go DIRK!)

We've heard the stories of how weak the East is compared to the West before... specifically during last year's playoffs. They no longer really hold weight. And if you're basing yourself on record, the Pacers had last year's best record but couldnt get to the finals.

[quote="shobber"][quote="juicybooty911"]it's game 7[/quote]
when?[/quote]

8 ET

Heat is down by 2 with 11 sec left in the game
wowwww

Heat just lost 82-88, Pistons is going for 2 in a row.  :flex:

Finally...Heat lost with 3-4.
So Spurs vs Pistons. Won't be too much scores but will be very fascinating!

[quote="noneoftheabove"]miami all the way although i want spurs to beat suns
this is my ranking of the teams i like
1.heat
2.pistons
[/quote]

[quote="noneoftheabove"]
heat will win if shaq is in the game
[/quote]

[quote="mcalderbank"]Miami just beat the Wizards handily, I think they will be Eastern Conference Champs.[/quote]

[quote="juicybooty911"]it looks like Heat is unbeatable, won't be suprised if they 4-0 pacers or pistons[/quote]

This is the part where we Detroiters point and laugh.

Detroit was obviously gonna beat the Heat, but Detroit will lose to the Spurs quickly.

Reminds me of last year....

(of course Detroit was going to beat the Pacers. They're too young, without enough experience. But they have no chance whatsoever versus an experienced Lakers team with four hall of famers... )

pistons vs. spurs: things to watch for

1) bowen annoying hamilton to death
2) FT shooting: in a series with defense as no.1 priority, clutch FT shooting will be key. S.A. is the team in question, as detroit showed last night they are clutch
3) duncan is the best player for S.A., but parker and ginobili must be able to penetrate to open up 3-point shooters.

The spurs had a long time to rest, while the pistons will have tues, and wed off before thurs. game. i don't think rest will be that much of a factor, as detroit showed the same thing vs. miami

SPURS over piston in 7. It will be close. Low scoring games don't leave much room for blowouts

Rest won't be a factor, and I hope rustyness contributes to a Game 1 win for the Pistons.

Predictions: Bowen will chase Rip for 48 minutes and promptly find himself getting rehydrated at the local hospital.

Tim Duncan will go 1-17 at the three-throw line to lose the game. :)

Seriously, Detroit in 6 (a chance of 7.) We're hungry for another, so we don't have to spend all of our time defending ourselves of the "the Lakers were collapsing" attacks. Detroit is a better defensive team, and considering with the defensive stats Phoenix put up, the Spurs aren't conditioned to this kind of play.

[quote="stupidkid"][quote="GoBraves"]I think Manu's going to be the lead scorer, not Duncan. Detroit plays good defence against big men, but Spurs in 5.[/quote]
Wow, very good prediction.[/quote]

Manu may have had a good game, but come on, did you see Duncan. 25 points, 17 rebounds. The rebounds killed Detroit, SA had so many second chance points.


geometry
rectangle
ratio
AMC 10
similar triangles
area of a triangle
Pythagorean Theorem

practicing for AMC 10

[img]http://www.hpphoto.com/servlet/LinkPhoto?GUID=6c6139a9-5904-6673-61bf-541314877976&size=lg[/img]

Incase the questions are too small for you guys to read....

19)Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

20) In rectangle ABCD, AB =5 and BC =3. Points F and G are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of Triangle AEB

thx
bigger image of the one above>>> http://www.hpphoto.com/servlet/LinkPhoto?GUID=6c6139a9-5904-6673-61bf-541314877976&size=

This probably shouldn't be in getting started, but thats OK.

Some hints in spoiler - highlight to read:

[hide]

For the first one, if you draw all the semi circles in full, you get two little trianglar-like shapes but with two of the edges rounded. One you calculate the area of one of those you are fine.

To calculate that bits area, use the area of a sector formula a couple of times, and then subtract out the bit that you've counted twice.

[/hide]

For the second one:

[hide]

Use similar triangles to calculate the height of the triangle.

[/hide]

JBL wrote:For the first question, I'll give you another question that, if you can solve, will show you what you need to do:
[hide]Given two circles, A and B, of the same radius r.  If the distance between A and B is also r (that is, A lies on circle B and B lies on circle A), what is the area of the lens-shaped region bounded by the two circles?[/hide]

And for the second, just a hint: [hide]Look for similar triangles.[/hide]

can't get the one you gave you

i'm a ninth grader if your wondering y i'm so stupid

AEB :sim: FEG  and their ratio is 2:5

that's the farest i could get

TripleM[hide] i belive much easier is that its just to just say that we have 2 equaleteral triangles and 5 out of 6 sectors of a whole 1unit raidus circle. Maybe this is what you mean but i don't understand it.... So the answer would be 2pi-5/6pi- :rt3: /2[/hide]

[quote="JBL"]The relevant issue here is experience, not age -- you don't have much.  That's not stupid, it's un-taught, or something.  Anyhow:

Let me borrow some of Stephen's terminology:  If you look at the lens shape that I've suggested, it looks sort of like "two little trianglar-like shapes but with two of the edges rounded."  (The images I've attached for kicks should illustrate.)

Try looking at those lens-shapes and finding their areas.  Stephen's other hint should help.  If you need more, come on back :)[/quote]

the radius is 1 right? so the area of the triangles should be 1/2?

so the lens is 2

[quote="JBL"][quote="Unanswered"]AEB :sim: FEG  and their ratio is 2:5

that's the farthest i could get[/quote]

Good.  Now, what do you need to find the area of AEB?  (Or, for that matter, any triangle?)[/quote]

length times width

but i got  :sqrt: of 10 for AF and got  2 times the  :sqrt:  of 10 over 3 for EF

and i got 25 * square root of 10 over 6

Right -- to find the area of a triangle, you should use 1/2 * b * h.  (one half base times height.)  So you need to find a height for your triangle.

theone853 wrote:unanswered, you are finding unecessary measures

area of triangle = [(side)(altitude drawn to that side)]/2

Tip:[hide]assign variable to altitude of triangle efg, which is drawn to base fg and use ratios[/hide]

so should i cross multiply like this.....

x over 2 is  :equiv:  to x+  :sqrt: of 10 over 5?

where as X is the variable for EF and  :sqrt: of 10 is FA

You're on the right track -- but why do you need EF?

[quote="JBL"][quote="Unanswered"]the radius is 1 right? so the area of the triangles should be 1/2?

so the lens is 2[/quote]

Okay, you've made three statements, but you didn't show any logic for getting from one to the other.

The standard method for the lens problem is this:
We know how to find the area of a sector of a circle, right?  Each half-lens is composed of 2 overlapping sectors, one from each circle, right?  Then, the overlap is composed of a triangle.  So, what you have to do is to find the area of the sectors and subtract the area of the triangle.

Then, you can use that method and apply it to the problem you started with.[/quote]

you have to find the area of the circle subtracted from the overlapping sector....but the problem is i don't know the area of the overlapping sector

Okay, I'll start from scratch with that problem, because I think we're having trouble communicating here.

Look at the 4 pictures below.  The first one shows the original diagram, where we want to find the area bounded by the darker curves.

To do this, we want to find the area inside the three small semicircles, and subtract it from the area of the larger semicircle.  So, we need to find the area inside the small semicircles.  Half of that area is shown in the second diagram.  Now, the area (in the second diagram) composed of the yellow and green regions is a quarter circle, while the area of the blue and green regions is a semicircle.  If we add these two regions together, we get the area we want, except that the green part is counted twice.  So we need to find out what the area of the green part is (third image).  Now, if we draw in radii in the third image, we divide the interior of that shape into three parts, which are light blue, blue and purple in the fourth image.  The light blue and blue regions together make up a sector of a circle.  You should be able to figure out what the area of this sector is.  Similarly, the blue and purple regions also make up a sector of the circle, whose area you should also be able to figure out.  If you add the two sectors together, you get the area of the shape in the third image, but with the blue triangle in the 4th image counted twice.  So you need to figure out the area of that triangle and work backwards from there.

Now, what [i]you[/i] need to do is do exactly what I've written, but use the numbers in the problem to figure out actual areas of things.  Then, once you've gotten down to the last step I wrote, you have to go back through the steps plugging in until you get the original area that you wanted.  Post that here (including every step, and the reasons for going from one step to another) just to make sure that we're on the same wavelength.  Okay?

[quote="JBL"] So you need to figure out the area of that triangle and work backwards from there.
[/quote]

ok

so the radius of the small semi circle is one which mean it's base and height are both 1

base * height /2  = area

1*1  divide by 2 = 1/2'

Area of the blue triangle is  1/2

[quote="JBL"]he light blue and blue regions together make up a sector of a circle. You should be able to figure out what the area of this sector is[/quote]

i don't know how to get the area of the light blue sector

[quote="Unanswered"]so the radius of the small semi circle is one which mean it's base[i] [b]and height[/b][/i] are both 1[/quote]

Are you sure about that?  What type of triangle are we dealing with here?

equalitarial triangle
if i'm not mistaken....the light blue and blue sector of the cirlce is
 1/3 of the entire semi circle...am i right?

[quote="Unanswered"]equalitarial triangle[/quote]
Correct.  Is the height of an equilateral triangle equal to its base?

[quote="Unanswered"]if i'm not mistaken....the light blue and blue sector of the cirlce is 1/3 of the entire semi circle...am i right?[/quote]

Yes!  Now, put in the correct area of the triangle and we'll be getting somewhere.

[quote="JBL"][quote="Unanswered"]equalitarial triangle[/quote]
Correct.  Is the height of an equilateral triangle equal to its base?

yes

[quote="Unanswered"]if i'm not mistaken....the light blue and blue sector of the cirlce is 1/3 of the entire semi circle...am i right?[/quote]

Yes!  Now, put in the correct area of the triangle and we'll be getting somewhere.[/quote]

isn't it 1/2?

btw is the area of the small semi circle's  7 :pi: /6?

[quote="JBL"]I promise you, from the great deepest depths of my heart, that the area of an equilateral triangle with side 1 is not 1/2.  Draw in an altitude and use the Pythagorean Theorem.

How do you find the area of a circle?[/quote]

 :sqrt:  of 2?  so the area is  :sqrt: of 2? :?

y not 1/2  it's equilateral all sides of the triangle is = to 1

area of a circle= radius x radius x  :pi:

[quote="Unanswered"]area of a circle= radius x radius x  :pi:[/quote]

Good.  So the area of the semicircle is what?

You are assuming, with no justification, that the sides of an equilateral triangle are perpendicular to each other.  If I give you a triangle with sides 3 and 7, that does not mean the area of the triangle is going to be 3*7/2, any more than for a triangle with sides of 1 and 1 and 1 the area would be 1*1/2.

Call the equilateral triangle ABC.  If you draw an altitude from A, it goes to the midpoint of BC.  Call that point M.  Now, use the Pythagorean theorem in triangle ABM.

1*1* :pi:  = full circle

 :pi: /2  = semi circle

ok

Ab= 1  x  0.5 (1/2 of CB)

itis 1/4

[url=http://mathworld.wolfram.com/PythagoreanTheorem.html]Pythagorean Theorem.[/url]
[url=http://www.cut-the-knot.org/pythagoras/index.shtml]More Pythagorean Theorem.[/url]

The second one is probably more worthwhile looking at.

Your area for the semicircle is correct.  Now, what is the area of that sector?

sorry, i did the area instead of the PythagoreanTheorem

.5sq +xsq = 1sq

.25 +x squared =1

x squared= .75

x =  :sqrt:  of .75

can you explain...how you got  :rt3: /2?

is this alright?

 :sqrt:  of .75 =  :sqrt: of .25  *  :sqrt:  of 3

=.5* :sqrt: 3 cuz this is wat i got

[quote="JBL"]Yes, which is nicer to write as :rt3: / 2.  Now, using that, we're pretty much all set.  You can then get the area of the triangle and proceed from there.[/quote]

1 (base)*  :rt3: /2 (height) = :rt3: /2 divide by 2 =  :rt3: /4

Good.  Now, use that to figure out the area of the thing in my 3rd and 4th picture.

i'd love to but i don't have a clue how to get the area of the light blue or purple part of the circle

Wait -- but you know how to get the [i]combined[/i] area of light blue + blue, don't you?  You already said it was 1/3 of the semicircle, and you already found the area of the semicircle.  And since the area you want, (dark blue + light blue + purple), is the same as ((dark blue + light blue) + (dark blue + purple) - dark blue), and you know all three quantities in this second sum, you can find the area of the whole thing.

i got

14 :pi: - 3 :rt3: over 12

which simplifies

to  7 :pi:  over 6  subtract  :rt3:  over 4

as the answer

i think i solved problem #20

2/5  = x/x+3

then i crossed multiplied....

Thus....2x+6=5x

solve for x and i got 2

is the answer choice (D) 25/2?

I am thinking that your #19 is right and ml2 is wrong. Congrats!

On the website with 43 proofs of the Pythagorean theorem, I'm appalled that they don't have the proof by Euclid, the "windmill" proof?  It's on the MathWorld one though.


Suppose we have 2 fair 6-sided dice numbered from 1 to 6.  If we roll them both, what is the probability that the two numbers shown sum to 7?

I have a simpler solution than the one offered by Alcumus:

Call x the number that we roll on the first die. In order to roll a 7, we must roll 7-x on the second die. This can only happen with probability 1/6 because a die has 6 faces.

What if you have 9, though?  You can't do that, because if you roll a 1 on the first die, you can't roll an 8 on the second.

Soltixnj provided a clever solution for this particular problem. In general, it would not work though.

I have done so many problems with two six sided dice, I have this memorized.

Oops. I realize I just bumped this post badly.  :oops_sign:


group theory
abstract algebra
modular arithmetic
superior algebra
superior algebra unsolved

hi

i'm stuck at this problem

suppose G is a group with 350 elements

show that there is one sylowfive group
show that there is one sylowseven group (by showing that every sylow seven group commutes with every
sylow five group)
determine all orders for which there are subgroups

the last one i cant do

i found them for all divisors except 70 and 10
i made use frequently of the theorem that if H is subgroup and N normal subgroup, HN is also a subgroup

if there is one for 70, there is one for 10 , and the other way around

any ideas?

Just the first two (so it's not really helpful to you fredbel, since you can solve these :():

The number of Sylow $5$ subgroups divides $\frac{350}{25}=2\cdot 7$ and must be $1\pmod 5$, so it must be $1$. Call this subgroup $N$ (because it's normal :)). For similar reasons, the number of Sylow $7$ subgroups is either $1$ or $50$. Let's assume it's $50$ and let $H$ be such a group.

Since $N$ is normal, we have $HN=NH$ ($HN$ is a group), and $|HN|=175=25\cdot 7$. The number of Sylow $7$ subgroups of the group $HN$ is $1\pmod 7$ and divides $25$, so it's $1$. This means that all elements of $N$ belong to the normalizer of $H$. On the other hand, the normalizer of $H$ in $G$ ($G$ is the initial group, the one with $350$ elements) has order $\frac{350}{50}=7$, so it consists only of the elements of $H$. We have thus found a contradiction, so the number of Sylow $7$ subgroups can't be $50$. This means it's $1$, as required.

Regarding the second part.

Every divisor except for $10,70$ is easy to find, as you ahve already mentioned. By using the fact that $N$ normal, $H$ subgroup imply $HN$ subgroup (these notations are not the same as above), and the fact that if a prime $p$ divides the order of a group, then that group has a subgroup of order $p$, we easily find that there are subgroups of order $1,2,5,7,2\cdot 7,5\cdot 7,5^2,2\cdot 5^2,5^2\cdot 7,2\cdot 5^2\cdot 7$. We are thus left with $2\cdot 5$ and $2\cdot 5\cdot 7$.

Let's focus on the group $T$ with $2\cdot 5^2$ elements, and see whether we can prove that it always contains a subgroup with $10$ elements.

It has a subgroup $H$ with $5^2$ elements, which must be normal, because it has index $2$. Let $Q$ be a subgroup with $2$ elements. In this case, $T$ is the semi-direct product of $Q$ and $H$. There are two cases we have to consider here: $H$ is cyclic or $H$ is isomorphic to $C_5\times C_5$, where $C_5$ is the cyclic group with $5$ elements.

Any semi-direct product is determined by a homomorphism $\theta: Q\to Aut(H)$. The composition law on the semi-direct prod of $Q$ and $H$ is then defined by $(q_1,h_1)(q_2,h_2)=(q_1q_2,h_1\cdot \theta(q_1)(h_2))$.

[b]a)[/b] Assume now that $H$ is cyclic. The homorphism $\theta$ maps the non-null element of $Q$ to an automorphism of order $2$, which must either be the identity (and we obtain the direct product, which has a subgroup of order $10$), or $x\mapsto x^{-1}$. In this latter case, it's easy to see that the group generated by the non-null element of $Q$ and an element of order $5$ from $H$ has order $10$ as well. It means that in this case, there are groups of order $10$ and thus of order $70$.

[b]b)[/b] We take $H\cong C_5\times C_5$. Again, we need a homomorphism $\theta:Q\to Aut(H)$ of order $2$. I'm tired right now, so we'll talk about it tomorrow. :)


Hi!

find all polynomials $P\in{\mathbb Q}[X]$ such that
\[P({\mathbb R}\setminus{\mathbb Q})\subset{\mathbb R}\setminus{\mathbb Q}.\]

If the solution to this problem is not readily apparent to the reader, then he will never be a first-class mathematician.

Here's a start:

$P(x) = ax+b$ (rational first-order polynomial) satisfies the given condition.

$P(x) = ax^{2}+bx+c$ (rational second-order polynomial) never does for $a \neq 0$.  We can show this as follows:

$P(x) = a\left(x+\frac{b}{2a}\right)^{2}+c-\frac{b^{2}}{4a}.$

For $x = \sqrt{k}-\frac{b}{2a},$ $P(x)$ is always rational, so we've found an irrational input such that the output is rational.  Therefore such a polynomial cannot satisfy the given condition.

Also, any $P$ satisfying the given condition cannot have irrational roots, because in that case such a root would give a rational output for irrational input.

Let's assume $\text{deg}P\geq 2$
First replace $P(x)$ by $\pm (P(x+n)-P(n))$ for some $n\in\mathbb{N}$ large enough,
such that $P(0)=0$ and $P$ is strictly increasing on $[0,\infty)$.
Next replace $P(x)$ by $mP(\frac{x}{n})$ for some $m,n\in\mathbb{N}$ (this doesn't change the degree or the conditions above!)
and assume $P\in\mathbb{Z}[X]$ with leading coefficient $1$.
Now for every $n\in\mathbb{N}$ the equation $P(x)=n$ has a real solution $\lambda_{n}\in [0,\infty)$, with $0<\lambda_{1}<\lambda_{2}<\cdots$.
Then by assumption $\lambda_{n}\in\mathbb{Q}$ for all $n\in\mathbb{N}$.
But then $\lambda_{n}\in\mathbb{Z}$, as it is root of $P-n\in\mathbb{Z}[X]$ with leading coefficient $1$ for all $n\in\mathbb{N}$. $(\text{deg}P\geq 2)$
And now $0<\lambda_{1}<\lambda_{2}<\cdots$ gives $n\leq \lambda_{n}$ and $P(n)\leq P(\lambda_{n})=n$ for all $n\in\mathbb{N}$, which contradicts $\text{deg}P\geq 2$.


geometry
rectangle
trigonometry
induction
inequalities
modular arithmetic
function

Funny one:
A rectangle with side lengths  $a_0, b_0$ is dissected into smaller rectangulars with side lengths $a_k, b_k,\, 1\le k\le n$. The sides of smaller rectangulars are parallel to the corresponding sides of the big rectangular. Prove that $|\sin a_0\sin b_0|\le \sum_{k=1}^n|\sin a_k \sin b_k|.$

Are you sure of the statement of the problem? Take $a_0 = b_0 = \frac{\pi}{2}$ and $a_k = b_k = \frac{\pi}{6}$ for $1 \le k \le 3$. It is
$|1\cdot 1| \le 3|1/2 \cdot 1/2|$?

Should it be $\sum_{1 \le j, k \le n} |\sin(a_j)\sin(b_k)|$?

Then I think this is the approach: it is sufficient to prove it when we divide the original rectangle into squares. Then the statement simplifies to an easier expression; I think we can conclude by taking the (positive) square root of each side and considering Taylor series. I'll try to post full solution tonight (unless someone else does first, or if I realize I'm wrong :P)

yep, you have to sum over all the small rectangles. (in your example there are $9$ smaller rectangles)

All of them! Then I don't think my idea works, at least not as easily as I hoped... Still I am sure we should find some kind of induction from a simple case in order to win.

Actually, isn't it enough if we prove it for dividing a rectangle into two pieces? The inequalities just add up and are all the direction we want.

Then we just need to show that if $a_0 = a_1 + a_2$ then
$|\sin(a_0)\sin(b_0)| \le |\sin(a_1)\sin(b_0)| + |\sin(a_2)\sin(b_0)|$
right? Which is of course just $|\sin(a_0)| \le |\sin(a_1)| + |\sin(a_2)|$.
In case of $a_1 = a_2$, we have $|\sin(2a_1)| \le 2|\sin(a_1)|$ which is obvious from double angle identity for sine. But really we need to look at minimum of $|\sin(x)| + |\sin(a-x)|$, neh? Wait, by AM-GM an arithmetic mean is always minimized by equality, right? So we just want $|\sin(x)| = |\sin(a-x)|$, which means $2x-a \equiv 0 \pmod{\pi}$. Then we may as well take $a_0' = a_0 \pmod{\pi}$ because |sine| is determined uniquely up to that. Then we can be sure of minimizing either when $x = a_0'/2$ or $x=0$, and in both cases the inequality is obvious.

I hope there is no mistake....  :D  :?

[quote="Xevarion"]Actually, isn't it enough if we prove it for dividing a rectangle into two pieces? The inequalities just add up and are all the direction we want. [/quote]

What about this dissection?
:? :?$\begin{picture}(30, 30) \put(0, 0){\line(1, 0){30}}\put(0, 10){\line(1, 0){20}} \put(10, 20){\line(1, 0){20}}\put(0, 30){\line(1, 0){30}} \put(0,0){\line(0, 1){30}} \put(20, 0){\line(0, 1){20}} \put(10, 10){\line(0, 1){20}} \put(30, 0){\line(0, 1){30}} \end{picture}$ :? :? :?

:o

Well, that particular one can be rearranged into the kind that I like. However I am not sure how to prove that this works in general. (I am not even sure it's correct...)

 :blush:

edit: in fact, it surely isn't. Just make the bottom right rectangle narrower, and stretch the other two to fill the space, and then I can't rearrange it any more.  :(

Alekk, what was you idea for this problem ?

[quote="fedja"]Seems we are forgetting olympiad classics :([/quote]
Too bad my experience with olympiads is pretty much $O(f(x))$ ($x \to \infty$) where $f$ is in the Schwartz class...

[quote="fedja"]Seems we are forgetting olympiad classics :(: remember the problem about rectangles with one integer side and $e^{2\pi i(x+y)}$? ;)[/quote]

Ooops, really seems we(at least me) forgot this classic.

As Fedja said, if we integrate the function $e^{2\pi i(x+y)}$ over our big rectange(say it is $R$) the value of this integral is the same as the sum of integrals over little rectangles $R_{k}$, i.e

$\int_{R}e^{2\pi i(x+y)}dx dy= \sum \int_{R_{k}}e^{2\pi i(x+y)}dx dy$ (*), but

$\int_{a}^{b}e^{2 \pi i x}dx = \frac{1}{2 i \pi}e^{2\pi i a}(e^{2\pi i (b-a)}-1)$ and taking module in (*) we get that

$\mid (e^{2\pi i a_{0}}-1)(e^{2\pi i b_{0}}-1) \mid \leq \sum_{k=1}^{n}\mid (e^{2\pi i a_{k}}-1)(e^{2\pi i b_{k}}-1) \mid $ and now it is only left to notice that

$\mid (e^{2\pi i t}-1)\mid = \sqrt{( \cos{2\pi t }-1)^{2}+(\sin{2\pi t})^{2}}= \sqrt{ 2-2\cos{2 \pi t}}= 2 \mid \sin{\pi t}\mid $.


calculus
integration
trigonometry
calculus computations

Prove that the integration $\int_{0}^{+\infty}\frac{\mid sin(x) \mid}{x}$ is diverging.
I have proved that $\int_{0}^{+\infty}\frac{sin(x)}{x}$ is converging but this one is a little more difficult.

It's funny, because the convergence of $\int_{0}^{+\infty}\frac{sin(x)}{x}$ is more difficult than the divergence of  $\int_{0}^{+\infty}\frac{\mid sin(x) \mid}{x}$    :D

$ \int_{0}^{+\infty}\frac{|sin(x)|}{x}>\sum_{k=0}^{+\infty}\int_{k\pi+\frac{\pi}4}^{k\pi+\frac{3\pi}4}\frac{|sin(x)|}{x}>\sum_{k=0}^{+\infty}\frac{\pi}2\frac{\frac{\sqrt 2}2}{k\pi+\frac{\pi}4}$ .. and so on
 :cool:

Or $\int_{0}^{\infty}\frac{|\sin x|}{x}dx\ge \int_{1}^{\infty}\frac{\sin^{2}x}{x}dx=\frac12 \int_{1}^{\infty}\left(\frac{1}{x}+\frac{\cos 2x}{x}\right)dx$. The integral of $\frac{\cos 2x}{x}$ is convergent for the same reason as $\frac{\sin x}{x}$.

just a small correction- $sin^{2}(x)=\frac{1-cos(2x)}{2}$ . the proof is valid otherwise


An advertisement read, "Take an additional $ 10\%$ off any item which is already discounted $ 30\%$." A clerk with MATHCOUNTS training said, "those two combined discounts are the same as one $ n\%$ discount." What is the value of $ n$?


V = 4/3 x pi x r^3 or in other words V = 4/3 x pi x r to the third power

Solve for r... please provide a detailed explanation

I myself got r = the cubed root of v times 4/3 divided by pi or in other words
r =  the cubed root of v times 4/3 over pi.

It's very hard to understand what you're doing without parentheses.  In any case, the correct expression:

$ V \equal{} \frac{4}{3} \pi r^3 \Leftrightarrow$
$ \frac{3}{4 \pi} V \equal{} r^3 \Leftrightarrow$
$ r \equal{} \sqrt[3]{ \frac{3}{4 \pi} V }$

Uh, yeah that's my equation, but I didn't have the keys on my computer to type it. But um, how did you get 3/4 x pi = r^3 in the second equation??? What did you do on both sides??

[quote="Noldi"]Uh, yeah that's my equation, but I didn't have the keys on my computer to type it.[/quote]

Even without $ \text{\LaTeX}$ (which doesn't require any special keys, just dollar signs and some notation) you should be able to use parentheses.

[quote="Noldi"]But um, how did you get 3/4 x pi = r^3 in the second equation??? What did you do on both sides??[/quote]

Multiplied by $ \frac{3}{4 \pi}$.  Note that $ \frac{4}{3} \pi = \frac{4\pi}{3}$.

Oh right, you multiply by the reciprocal to cancel it out. I didn't know that 4/3TT is equal to 4TT/3, mind clearing that up a bit too?

Do you understand what a fraction is?

(Please don't take this question the wrong way.  Many otherwise intelligent people have been failed so badly by their math curriculae or their math teachers that they never properly learn their basics.  I am asking for your honest evaluation of how well you understand the concept of a fraction.)

The fact that $ \frac{a}{b} \cdot c \equal{} \frac{ac}{b}$ is a basic property of fractions.

Oh sorry, of course I know my basics, besides I have a 99 in math on my report card and I'm in the 8th grade. The fact that it was a fraction made me think a bit differently. But now that I look at it again, it's like 5 x 4 divided by 2. You could either do 4 divided by 2 and then multiply by 5, or multiply 5 by 4 and then divide by 2. Simple, really.

That math problem up there is actually from a 9th grade New York City test. My teacher made me study some of it.


Suppose that $ T$ is a stopping time such that for some $ N$ in $ \mathbb{N}$ and some $ \varepsilon > 0$, we have, for every $ n\in\mathbb{N}$:
$ P(T\leq n \plus{} N|\mathcal{F}_n) > \varepsilon$ a.s.
Then, show that $ E(T) < \infty$.

you could try to do something like:
\[ P[T>(k\plus{}1)N] \equal{} P[T>(k\plus{}1)N | T>kN ] P[ T>kN ] < (1\minus{}\epsilon) P[ T>kN ] \]
henceinduction shows that $ P[T>kN] < C (1\minus{}\epsilon)^k$ for a certain constant $ C$. You should be able to conclude from here, isn't it.


topology
algebra
function
domain
complex analysis
complex analysis unsolved

Here's a problem I thought of that follows the same type of ideas of problems that Arne have recently posted.

Suppose $ f$ is analytic in the open unit disc and extends continuously to the closed disc such that $ \overline{f(z)} \equal{} af(z)$ when $ |z| \equal{} 1$ for some constant $ a$. What is $ f$?

Note that if $ f$ extends analytically to the closed disk, then if $ f$ is not constant there are a lot of points on the unit circle where $ f'(z_0) \neq 0$, so at such points $ f$ would be both conformal and anti-conformal, which won't be possible.

The assumption $ \overline{f}\equal{}af$ is equivalent to $ \arg f \equal{} \theta$, where $ e^{\minus{}2i\theta}\equal{}a$. Simply put, $ f$ maps the unit circle into a line. This is impossible because the image of unit disk is bounded, and no subset of a line can be the boundary of a bounded domain.

That's neat. Not what I had in mind. How does one prove this intuitively obvious claim? Topology is hard.

Suppose $ \Omega\subset \mathbb C$ is a domain such that $ \partial\Omega\subset \mathbb R$. Pick a point $ z\in\Omega\setminus\mathbb R$. Since the half-line $ \{tz\colon t\ge 1\}$ does not intersect $ \partial \Omega$, it is contained in $ \Omega$. Therefore, $ \Omega$ is unbounded.


Find all function $ f: R \to R$ such that $ f(x\plus{}y)\equal{}x^2f(\frac{1}{x})\plus{}y^2f(\frac{1}{y})$ for all $ x,y \neq 0$.

Update: TTsphn wrote below that "this is a problem from Math and Young magazine and now we shouldn't discuss it ". So I removed my original post.

[quote="Yury"]I will assume that $ f$ is continuous; otherwise, it seems that the problem has "nasty" solutions (however, I haven't checked that).

First, note that $ f(2x) \equal{} f(x \plus{} x) \equal{} 2x^2 f(1/x)$. Hence
\[ f(1/x) \equal{} \frac {f(2x)}{2x^2}.
\]
Therefore,
\[ f(x \plus{} y) \equal{} x^2 f(1/x) \plus{} y^2 f(1/y) \equal{} \frac {f(2x) \plus{} f(2y)}{2}.
\]
Let $ a \equal{} 2x$, $ b \equal{} 2y$. Then
\[ f\left(\frac {a \plus{} b}{2}\right) \equal{} \frac {f(a) \plus{} f(b)}{2}.
\]
Since $ f$ is continuous, for every $ \lambda \in (0,1)$ we have
\[ f(\lambda x \plus{} (1 \minus{} \lambda)y) \equal{} \lambda f(x) \plus{} (1 \minus{} \lambda) f(y).
\]
We conclude that $ f(x) \equal{} c x \plus{} d$ where $ c$ and $ d$ are constants. From the condition, we get that $ d \equal{} d(x^2 \plus{} y^2)$. Hence $ d \equal{} 0$. That is, $ f(x) \equal{} cx$. It is easy to see that $ f(x) \equal{} cx$ is indeed a solution.[/quote]
Condition $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ is true if $ x,y,x\plus{}y$ is different from 0 .
The condition continuous doesn't neccessary /
And the last ,this is a problem from Math and Young magazine and now we shouldn't discuss it  :)