analytic geometry
geometry
graphing lines
slope
quadratics
algebra

The first problem is about coordinates. Should you "Let y= 0, and then let X=0" and then solve it?

The line y=3-2x contains points in how many quadrants of the Cartesian coordinate plane?

The second problem is also about coordinates, when I draw the shape, I don't see how you could find the area of the quadrilateral.

How many square units are in the area of the convex quadrilateral with vertices (0,0), (3,0), (2,2) and (0,3)?

The third problem I have no idea what do do, but all I know is that if a square's sides are 1, then the diagonal is the root of two, but I'm not quite sure it will help...

Each point in the rectangular grid below is one units from its nearest horizontal and vertical neighbors. Which point is exactly the root of 10 units from point P?

And the last one I don't know how to do at all...

Two cylindrical cans have the same volume. The height of one can is triple the height of the  other. If the radius of the narrower can is 12 units, how many units are in the length of the radius of the wider can? Express your answer in simplest radical form.

I know these problems probably seem easy, but for some reason I can't solve them...   :what?:  :wacko:

The line y=3-2x contains points in how many quadrants of the Cartesian coordinate plane?

y=3-2x is in the form y=mx+b, where m=slope and b=y intercept.

Therefore, the line is going down 2 y units for every x unit and is above the orgin 3 points.

You could also graph this equation:

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-9,9,Ticks(f, 18, 0.5));
yaxis(-9,9,Ticks(f, 18, 0.5));
real f(real x)
{
return 3-2x;
}
draw(graph(f,-3,6));[/asy]

Therefore, it goes through the I, II and IV quadrants or $ \boxed{3}$

How many square units are in the area of the convex quadrilateral with vertices (0,0), (3,0), (2,2) and (0,3)?

Let's graph this question:
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-4,4,Ticks(f, 8, 0.5));
yaxis(-4,4,Ticks(f, 8, 0.5));
draw((0,0)--(3,0)--(2,2)--(0,3)--cycle);[/asy]

We can divide this into two triangles and a square:

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-4,4,Ticks(f, 8, 0.5));
yaxis(-4,4,Ticks(f, 8, 0.5));
draw((0,0)--(3,0)--(2,2)--(0,3)--cycle);
draw((0,2)--(2,2)--(2,0));[/asy]

The area of each of the triangles is $ \dfrac{1\cdot 2}{2} = 1$, so a total of 2.
Then the square is $ 2^2 = 4$

The total area is $ 4 + 2 = \boxed{6}$

Each point in the rectangular grid below is one units from its nearest horizontal and vertical neighbors. Which point is exactly the root of 10 units from point P?

[asy]dot((0,0));
dot((0,1));
dot((0,2));
dot((1,0));
dot((1,1));
dot((1,2));
dot((2,0));
dot((2,1));
dot((2,2));
dot((3,0));
dot((3,1));
dot((3,2));
dot((4,0));
dot((4,1));
dot((4,2));
dot((5,0));
dot((5,1));
dot((5,2));
label("$P$",SW,(0,0));[/asy]

To solve this, we must find the right triangle with hypotenuse $ \sqrt {10}$ so we can find the point that is that far from P.

$ a^2 + b^2 = 10$

We know that the sum of the squares in for the sides is 10. The only way that there are integer values for the lengths are length 1 and 3, making squares 1 and 9, which sum to 10.
The only way we can draw this triangle is:

[asy]dot((0,0));
dot((0,1));
dot((0,2));
dot((1,0));
dot((1,1));
dot((1,2));
dot((2,0));
dot((2,1));
dot((2,2));
dot((3,0));
dot((3,1));
dot((3,2));
dot((4,0));
dot((4,1));
dot((4,2));
dot((5,0));
dot((5,1));
dot((5,2));
label("$P$",SW,(0,0));
draw((0,0)--(3,0)--(3,1)--cycle);[/asy]

Which is point $ \boxed{H}$

Two cylindrical cans have the same volume. The height of one can is triple the height of the other. If the radius of the narrower can is 12 units, how many units are in the length of the radius of the wider can? Express your answer in simplest radical form.

The narrower can must have the larger height for the volumes to be equal.

First can:
Radius=$ 12$
Height=$ 3h$

Second can
Radius=$ r$
Height=$ h$

Now we set the volume equations equal to one another.

\begin{eqnarray*}
144\pi\cdot 3h&=&r^2\pi \cdot h
\\144\cdot 3&=&r^2
\\432&=&r^2
\end{eqnarray*}

Now the only positive solution is $ 12\sqrt{3}$

Ah, that makes much more sense now, thanks! :D

Sorry for double-posting, but isn't there in error in
\begin{eqnarray*}
3h(r^2-24r+144)\pi&=&r^2\pi h
\\3h(r^2-24r+144)&=&r^2 h
\\3(r^2-24r+144)&=&r^2
\\3r^2-48r+432&=&r^2
\\2r^2-48r+432&=&0
\\r^2-24r+216&=&0
\end{eqnarray*}

Where 3 times 24 is 72 rather than 48? or is it that I'm looking at the problem wrong?

Oh yeah sorry...
I dont think that matters in the final solution though.
 :blush:

EDIT:

Hmmm... now that I look at it again, it seems that there are two possible radii.

\begin{eqnarray*} 3h(r^2 - 24r + 144)\pi & = & r^2\pi h \\
3h(r^2 - 24r + 144) & = & r^2 h \\
3(r^2 - 24r + 144) & = & r^2 \\
3r^2 - 72r + 432 & = & r^2 \\
2r^2 - 72r + 432 & = & 0 \\
r^2 - 36r + 216 & = & 0 \end{eqnarray*}

Now apply the quadratic equation:
$ \dfrac{36\pm \sqrt {36^2 - 4(216)}}{2} = \dfrac{36\pm\sqrt {432}}{2} = \dfrac{36\pm 12\sqrt {3}}{2} = 18\pm 6\sqrt {3}$

This is interesting as the differences in absolute values is 12... I assume that this means the narrower can is the original answer.

EDIT 2:
Hmm this is odd. Are you sure the question doesnt say the difference is 24?

The two radii could be $ 18+6\sqrt{3}$ and $ -6+6\sqrt{3}$, resulting in a difference of 24.
Thats funny though how 24 is 12 times 2.

I did the problem in a different way and turns out I got 12 times the root of 3...

And it says just the way I wrote it, and I think I see another error, or at least Im not sure why you typed it... why would you subtract 12? I mean from the radius...

Oh whoops, read the problem wrong.

EDIT: fixed post.

The narrower can must have the larger height for the volumes to be equal.

First can:
Radius=$ 12$
Height=$ 3h$

Second can
Radius=$ r$
Height=$ h$

Now we set the volume equations equal to one another.

\begin{eqnarray*}
144\pi\cdot 3h&=&r^2\pi \cdot h
\\144\cdot 3&=&r^2
\\432&=&r^2
\end{eqnarray*}

Now the only positive solution is $ 12\sqrt{3}$

How many square units are in the area of the convex quadrilateral with vertices (0,0), (3,0), (2,2) and (0,3)?

We can use the [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]Shoelace Theorem[/url] after checking that the vertices are in order:

$ \frac{1}{2} | (0 \times 3 \plus{} 3 \times 2 \plus{} 2 \times 3 \plus{} 0 \times 0)\minus{}(0 \times 3 \plus{} 0 \times 2 \plus{} 2 \times 0 \plus{} 3 \times 0)|\equal{}\frac{1}{2}(12)\equal{}\boxed{6}$.


(a)
Given any positive integer n, show that there exist distinct positive integers x and y such that, for any j such that 1 \leq j \leq n, (x + j)/(y + j) is an integer.
(b)
Suppose x and y are positive integers such that (x + j)/(y + j) is an integer for any positive integer j. Prove that x = y.

The (b) - part is quite easy (I think) but what about (a) ?

For (b) I did the following :

Write x = hy + i (i < y).
Then (x + j)/(y + j) = (hy + i + j)/(y + j).
Now if we take j > hy then obviously (hy + i + j)/(y + j) < 2.
(Since (hy + i + j)/(y + j) < 2 <=> hy + i + j < 2y + 2j <=> (h - 2)y + i < j.
But we chose j > hy so (h - 2)y + i < (h - 2)y + y = (h - 1)y < hy < j.)
So we must have x + j = y + j since (x + j)/(y + j) < 2 is an integer.
Hence x = y.

For (a), you have to come up with a clever construction but I don't see how to do it.
(I am very bad at constructing solutions, see for example the post about xx = y3 + z3.)

Oops. It's easier than I thought.

Take y = 1 and x = (n + 1)! + 1.

Now j + 1 divides (n + 1)! since j \leq n. Hence k(j + 1) = (n + 1)!.

So (x + j)/(y + j) = ((n + 1)! + j + 1)/(j + 1) = k + 1.

Problem solved. Sorry :?


probability calculus conics hyperbola geometry integration logarithms

My professor isn't very good so I'm not getting this class at all...

1. Let X and Y have the pdf f(x,y) = 1, 0<x<1 and 0<y<1; zero elsewhere. Find the cdf and pdf of the product Z = XY.

2. Let 13 cards be taken, at random and without replacement, from an ordinary deck of playing cards. If X is the number of spades in these 13 cards, find the pmf of X. If, in addition, Y is the number of hearts in these 13 cards, find the probability P(X=2, Y=5). What is the joint pmf of X and Y?

I'm pretty sure I've the answers for question 2 but I'm not sure if there's supposed to be some calculus-based way of doing question 2.

3. Let X and Y have the joint pmf p(x,y) = xy/36, x = 1, 2, 3 and y = 1, 2, 3; zero elsewhere. Find the joint pmf of Z = XY and W = Y; then find the marginal pmf of Z.

I'll assure you that you don't need (or want) calculus for #2, and I'll focus on #1.

There are faster ways to do things, but I find that students are less confused if we approach things step-by-step using the cumulative distribution function. The cdf is a probability, and with any computation of a probability, we always have this question: which is easier to deal with: the set in question or its complement. Hence, a significant fraction of the time we actually compute the "reverse" or complementary cdf.

Let $ U\equal{}XY.$ Note that $ U\in[0,1]$ with probability $ 1,$ so in looking at the cumulative probability $ P(U\le u)$ we only need consider $ u\in[0,1].$

Look at the unit square cut by the hyperbola $ xy\equal{}u.$ It looks easier to compute the area above that hyperbola, so we do that (the reverse cdf). Since we have a uniform-in-area distribution, probability is just area.

$ P(U>u)\equal{}\int_u^1\int_{\frac{u}x}^11\,dy\,dx\equal{} \int_u^1\left(1\minus{}\frac{u}x\right)\,dx$

$ \equal{}1\minus{}u\minus{}u\ln 1\plus{}u\ln u\equal{}1\minus{}u\plus{}u\ln u.$

That's the reverse cdf; we subtract it from $ 1$ to get the usual cdf:

$ F(u)\equal{}P(U\le u)\equal{}u\minus{}u\ln u$ for $ u\in(0,1).$

(We have $ F(u)\equal{}1$ for $ u\ge 1$ and $ F(u)\equal{}0$ for $ u\le 0.$)

Differentiate this to get the density:

$ f(u)\equal{}\begin{cases}\minus{}\ln u&0<u<1\\0&\text{otherwise}\end{cases}.$


Simplify this expression to a common fraction:
$ \frac{1}{\frac{1}{(\frac{1}{2})^{1}}\plus{}\frac{1}{(\frac{1}{2})^{2}}\plus{}\frac{1}{(\frac{1}{2})^{3}}}$


Maybe this is off topic and non math, but i just wanted to say that today i saw a great turkish movie called Distant (Uzak).
are there more good turkish movies?

Yes,there are a lot of good Turkish films.I think you mentioned "film festival in north america"(boston turk film festival)The best 10 Turkish films were there.You can search that films and enjoy from them.By the way do you know Turkey?Thank you for your interest.
                                                                             Alper

What do you mean by "do you know Turkey"?
 :?

Sorry for my English,I mean what do you know about Turkey?For example did you see Istanbul or other province of Turkey?

Well I've never benn to Turkey so i don't know much... but my brother is a musician and he is interested in middle east and arab music and he knows about turkish music.. he plays the oud and the cumbus (i think they are two turkish intruments...)


geometry
geometric transformation
reflection
parallelogram
circumcircle

[hide="hint"] reflect E over the midpoint of AD[/hide]

[quote="Elemennop"]I got 20 degrees by guess and check

I don't see how reflecting E helps[/quote]

it makes...[hide]a parallelogram! then A is the circumcenter of E'CB[/hide]

[quote="Elemennop"]Because it's the same distance from A, E', and C?

ok I see that

but I still have no clue where to go from there[/quote]

if we do some angle chasing, we get that $\angle CED=80=\angle E'AC$, then using $E'A=BA$, we have that $\angle E'BA=\angle BE'A=20$, then let $E'B\cap AC=E''$, then $\angle AE''E'=80$, so $AE'=E''E'$, this shows that $E'$ is the same as $D$ (use triangle congruency...), just extended the other way, so $\angle BDC=\angle BE'A=20$

I get everything except why <CED = <E'AC

reflecting $E$ over the midpoint of $AD$ implies that E'AED is a parallelogram, so use the parallel lines...


[quote="Iversonfan2005"]It might be a little too easy:

How many numbers between 1 and 500, are multiples of 2 and 3, but not 6. Many of y'all have probably done a problem like this[/quote]

Do you wonna say "multiples of 2 or 3" ?

If you are saying 2 or 3, [hide]There are 166 multiples of 3 between 1 and 500, and 250 multiples of 2.  166+250=416.  But here multiples of 6 are counted twice.  There are 83 multiples of 6 in the range, so we multiply this times two to get 166, and subtract this from 416 to get 250.  [/hide] But if you are saying 2 and 3, [hide]0[/hide]

[hide]there are 250 multiples of 2 from 1-500 and 166 multiples of 3 from 1-500. However, every other multiple of 3 is a multiple of 6 so we divide 166 by 2 to get 83. For multiples of 2, every 3rd number is a multiple of 6 so we divide 250 by 3 to get 83 1/3 and round down. Therefore, 83+83=166, which is the desired answer. [/hide]

I like plokoon51 much better....

much more detailed...

extension :

try solving up to 1000

I'm not exactly sure on this so correct me if I'm wrong
[hide="answer"] $ \frac {500}2$ = 250 . $\frac {500}3$ = 166 . and $\frac {500}6$ = 83. So 250+166=416 which is all the multiples of 2 or 3 together and the multiples of 6 counted twice so every 3rd mult. of two is a multiple of 6 and every 2nd mult. of 3 is a mult. of 6 so $\frac {250}3$ = 83 1/3 but round to 83 and $\frac {166}2$ = 83 and 83+83=166 like plokloon51 was saying [/hide]

1 - 1000...

500? I just calculated how many numbers there are and added: 333 + 167

[hide]

Multiples of 2: $\frac{1000}{2} = 500$.
Multiples of 3: $\frac{1000}{3} \approx 333$.
Multiples of 6: $\frac{1000}{6} \approx 166$.
(Multiples of 2) + (Multiples of 3) - 2(Multiples of 6) = $500 + 333 - 2 \times 166 = 500$.
[/hide]

You know, the origonal version of the question with the 2 and 3 clause could have very well been the actual question. The answer would have been zero.

[quote="Iversonfan2005"]sorry, i meant 2 or 3 :blush:[/quote]

Besides, what would be the fun in solving a problem if it was [i]and[/i]? :)

[quote="Iversonfan2005"]It might be a little too easy:

How many numbers between 1 and 500, are multiples of 2 [s]and[/s] or 3, but not 6. Many of y'all have probably done a problem like this[/quote]

From 1 to 498, half of those numbers fit. 499 doesn't, but 500 does.

1 + 299 = 300


Jamal wants to store $ 30$ computer files on floppy disks, each of which has a capacity of $ 1.44$ megabytes (MB). Three of his files require $ 0.8$ MB of memory each, $ 12$ more require $ 0.7$ MB each, and the remaining $ 15$ require $ 0.4$ MB each. No file can be split between floppy disks. What is the minimal number of floppy disks that will hold all the files?

$ \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 16$

i think it would b C 14.

the first type of files you cannot fit 2 on each so they would require 3 disks. (.8+.8 exceeds 1.44)

the second type of files you can fit only 2 on each so it would require 6 disks (.7+.7 =1.4 which doesnt exceed 1.44)

the third type of files you can fit only 3 on each so it would require 5 disks

nd the total disks is 14.

hope i am right lolz

[quote="bobthesmartypants"][quote="Aequilipse"][hide="Question"]
AMC SOLUTION:
http://grab.by/oqU2

I don't understand why we calculate the unused space for only the disks that hold both 1 0.8 and 1 0.4 file. If we do this, why don't we calculate unused space for all the disks, including the 0.7 ones which have 0.74 mb unused?

In other words can somebody explain the solution in more depth please? [/hide][/quote]
Actually, they have 0.74-0.7=0.04 unused space (two per disk) so the actual change this makes is negligible.[/quote]

Sorry, I don't think you understood my question. My question is, why are we choosing the case where 0.8 and 0.4 are stored on the disk versus the case where there are 3 0.4mb files, or one 0.7 and one 0.4?


A three-digit number has the same hundreds, tens and units digit.  The sum of the prime factors of the number is 47. What is the three-digit number?

the number is obviously 111x, and 111=3*37.

and 3+37=40.

sum of the prime factors are 47, so its 47-40=7.

answer : 777


linear algebra
matrix
vector
superior algebra
superior algebra unsolved

Let $ \mathbb{F}$ be a field. Prove that the ring $ M_2(\mathbb{F})$ of 2 x 2 matrices with entries in $ \mathbb{F}$ has no non-trivial two-sided ideals.

My thinking on this was as follows: let I be an ideal, and $ m \in I$. Then $ xm \in I$ for any $ x \in M_2(\mathbb{F})$. Then the product of m and any elementary matrix is also in I. So, if m is invertible, then it is possible to multiply m by elementary matrices until the product is the identity matrix, so that the identity matrix is in I; if the identity matrix is in I then I is the entire ring. Now, suppose m is not invertible...

And that's where I'm stumped.

I'm not sure, but I think it is possibly done in this way:
$ I$ is an ideal (both-sided) of the ring $ R\equal{}M_2(\mathbb F)$, where the latter is a vector space of dimension 4 over $ \mathbb F$. So, $ I$ is a subspace of $ R$.

Suppose $ I\ne\{O\}$.

The ideal $ I$ must include one of the elements $ E_{i,j}$, (which with these subscripts denotes a matrix with i,j-th element =1 and others =0).

Observe that $ \begin{bmatrix}
1&1\\
1&1\\
\end{bmatrix}\cdot
\begin{bmatrix}
0&0\\
0&1\\
\end{bmatrix}\equal{}
\begin{bmatrix}
0&1\\
0&1\\
\end{bmatrix}\in I$

Hence if $ E_{2,2}\in I$ then $ E_{1,2}\equal{}\begin{bmatrix}
0&1\\
0&1\\
\end{bmatrix}\minus{}E_{2,2}\in I$

Similarly, $ E_{2,1}\in I$, by multiplying the same matrices interchanging order of multiplication, and this is permitted since the ideal is both-sided.

Again, if $ E_{1,2}\in I$, then $ \begin{bmatrix}
0&1\\
0&0\\
\end{bmatrix}\cdot
\begin{bmatrix}
1&1\\
1&1\\
\end{bmatrix}\equal{}\begin{bmatrix}
1&1\\
0&0\\
\end{bmatrix}\in I$ and thus $ E_{1,1}\equal{}\begin{bmatrix}
1&1\\
0&0\\
\end{bmatrix}\minus{}E_{1,2}\in I$.
So, all the $ E_{i,j}$ are in $ I$; so as a subspace $ I$ exhausts the whole of $ M_2(\mathbb F)$. So as an ideal $ I$ is either $ \{O\}$ or is the whole of $ R$, i.e., can not be non-trivial.

The ideal $ I$ must include one of the elements $ E_{i,j}$. Well you don't know this a priori.
From the fact that I is a subspace of the big vector space it doesn't follows that it should contain some $ E_{i,j}$.

This works for $ M_n(\mathbb F)$:
If $ I\neq 0$ then there exists an element $ A\in I$ and some indices $ i,j$ so that $ A_{ij}\neq 0$. Now since $ I$ is an ideal it is closed under matrix multiplication, from both sides. Hence you multiply this matrix $ A$ with $ E_{ii}$ on the left and $ E_{jj}$ on the right and you get $ A_{ij}E_{ij}$ and now multiplying with $ A_{ij}^{\minus{}1}I_n$ you get $ E_{ij}$ which should be in $ I$.
Now multiplying this $ E_{ij}$ with permutation matrices you will get all the $ E_{kl}$-s, $ l,s\equal{}1,2,...,n$. Hence $ I\equal{}M_n(\mathbb F)$.


probability
real analysis
real analysis unsolved

Prove that we can't find a probability measure on $\mathbb N$ such that the probability of a number being a multiple of $k$ is $\frac1{k}$.

This is easy if one knows the classical problem about the absolutely convergent series $\sum_{n\ge 1}a_{n}$ such that $\sum_{n\ge 1}a_{kn}=0$ for all $k\ge 1$. The conclusion of that problem is that all $a_{n}=0$. Why is it relevant here? Let $p_{n}$ be the probability that our random number equals $n$. Fix $m$ and put $a_{n}=p_{n}-mp_{mn}$. Then $a_{n}$ satisfies the condition of the problem, so we must have $a_{1}=0$, i.e., $p_{m}=\frac{1}{m}p_{1}$. But this must be true for all $m$ and the harmonic series diverges.

A direct approach: Suppose we have such a measure. The probability that our number is zero is less than $\frac1k$ for all $k$ (since $k|0$), so the probability of zero is zero. By inclusion-exclusion, the probability that none of $p_{1},p_{2},\cdots, p_{k}$ divide our number is $\prod_{j=1}^{k}\left(1-\frac1{p_{j}}\right)$ for distinct primes $p$. For any nonzero $m$, the probability that our number is $m$ is thus at most $\prod_{p\nmid m}\left(1-\frac1p\right)$. That product diverges to zero when taken over all primes, so it still diverges to zero when taken over all but finitely many primes, and the probability that our number is $m$ is zero.
By countable additivity, the probability that our number is anything is zero; this contradicts the assumption that we had a probability measure, and we are done.


How many triangles can be formed with vertices in the following grid?

                  .    .    .    .
                  .    .    .    .
                  .    .    .    .
                  .    .    .    .

[hide="Hint"] How many combinations of three points can be selected? How many combinations of three or more collinear points can you find? Remember, a triangle is merely 3 non-collinear points.[/hide]

[quote="rnwang2"][hide]
$\dbinom{16}{3}-9\cdot\dbinom{4}{3}-2=522$.[/hide][/quote]

[hide="I think you still overcounted..."]
$\dbinom{16}{3}-10 \cdot \dbinom{4}{3}-4 \cdot \dbinom{3}{3}= \boxed{516}$

There are four vertical, four horizontal, and two diagonal lines of four; then four diagonal lines of three.
[/hide]

[quote="cincodemayo5590"][hide="I think you still overcounted..."]
$\dbinom{16}{3}-10 \cdot \dbinom{4}{3}-4 \cdot \dbinom{3}{3}= \boxed{516}$

There are four vertical, four horizontal, and two diagonal lines of four; then four diagonal lines of three.
[/hide][/quote]
 :o Wow, what was I doing??

Yeah I think your answer is right.


Can anyone find the sum of the series (3/2)+(3/4)+(3/6)...+(3/2n)? I had a false lead and came up with something that was off by [i]a factor of 2[/i], but it leads me to believe that it might be somewhere near 14.

Here are the first few sums.
1. (3/2)
2. (9/4)
3. (11/4)
4. (25/8)
5. (137/40)

The sum is +oo

(3/2)+(3/4)+(3/6)...+(3/2n) = 3/2 * H(n)

Where H(n) is the famous harmonic sum (see for instance http://mathworld.wolfram.com/HarmonicNumber.html).

H(n) = gamma + ln(n) + O(1/n)  (O = big "o" notation)
where gamma is the Euler number

It is easy (and very classical) to show that H(n) diverges :

H(2*n) - H(n) = 1/(n+1) + .... + 1/2*n > n/2*n = 1/2

So H(2^n) > n/2

Thanks. I was wayyyy off. :P

I'll need to teach myself calc to understand most of that resource, but it was interesting nonetheless.


induction
inequalities
combinatorics proposed
combinatorics

Let T  be a tree having vertex {1,2,...,n} and edge set denote by $E(T)$ . If $A_1,...,A_n\subset X$ ,prove that
$|A_1\cup...\cup A_n|\leq \sum_{i=1}^{n} |A_i| - \sum_{(i,j)\in E(T)}|A_i\cap A_j|$

Cute one :).

We can prove this by induction on $n$. Suppose WLOG that $1$ is a terminal vertex in the tree, and $2$ is its unique neighbour. Replace the LHS in the inequality with $|A_2\cup\ldots\cup A_n|$, and eliminate $|A_1|-|A_1\cap A_2|$ from the RHS. By the induction hypothesis, the inequality we end up with holds. Now, when we modify this expression back to the original one, we add $|A_1\setminus (A_2\cup\ldots\cup A_n)|$ to the LHS, and $|A_1\setminus A_2|$ to the RHS, i.e. the RHS increases at least as much as the LHS does, and we're done.


Gicen $ 0 < a \le b \le c \le 3 ;;2c \plus{} 3b \ge 3bc;;bc \plus{} 2ca \plus{} 3ab \ge 3abc$.Prove that:
$ a^5 \plus{} b^5 \plus{} c^5 \le 276$

[quote="tuandokim"][quote="quykhtn-qa1"]Gicen $ 0 < a \le b \le c \le 3 ;;2c \plus{} 3b \ge 3ab;;bc \plus{} 2ca \plus{} 3ab \ge 3abc$.Prove that:
$ a^5 \plus{} b^5 \plus{} c^5 \le 276$[/quote]
I think it's must be liked this:
$ 0 < a \le b \le c \le 3 ; 2c \plus{} 3b \ge 2bc;;bc \plus{} 2ca \plus{} 3ab \ge 3abc$.Prove that:
$ a^5 \plus{} b^5 \plus{} c^5 \le 276$[/quote]
Yes,you are right.I edited it.Thanks.Now,try it :)


inequalities
function
logarithms
geometry
3D geometry
sphere
inequalities proposed

Use $ \sqrt {1 \minus{} ab} \equal{} \sqrt {\frac {1}{\frac {1}{1 \minus{} ab}}}$ and use Jensen inequality for convex function $ \sqrt {\frac {1}{x}}$ you will get $ LHS\geq 3\sqrt {\frac {3}{\frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}}}$. It should be greater that $ \sqrt {6}$, but it is equivalent to show that $ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\leq \frac {9}{2}$ and this problem was already solved on mathlinks ;) :D, here: http://www.mathlinks.ro/Forum/viewtopic.php?t=36873

We can use logarithm and get equivalent:
$\ln(1-ab)+\ln(1-bc)+\ln(1-ca)\geq \ln\left(\frac{8}{27}\right)$
Now using Jensen inequality on $\ln{\frac{1}{x}}$ we get:
$\ln(1-ab)+\ln(1-bc)+\ln(1-ca)\geq 3\ln\left(\frac{3}{\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}}\right)\geq \ln\left(\frac{8}{27}\right)$
The last is one more time equivalent to the one once proved at mathlinks, I think, but I also doubt cause by the way of your post, Vasc, I think that it should not be solvable the same way as ineq in first post ;)
So wheres the miestake in my thoughts? :D

Of course, THis one is still true :

$\sqrt{3-abc}+\sqrt{3-bcd}+\sqrt{3-cda}+\sqrt{3-dab} \ge 4\sqrt{2}$

If $a^2+b^2+c^2+d^2=4$.

:)

For your problem, Vasc. I think Expanding is the best way!  :)

Actually, we have

$(3-abc)(3-bcd)(3-cda)(3-dab) \geq 16$

and, better,

$(2-abc)(2-bcd)(2-cda)(2-dab) \geq 1$.

The last is nice.  ;)

Could anyone tell me if we could prove the following ineq with [color=blue]ONDROB[/color]'s method ?

$\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\leq \frac{9}{2}$

[quote="zhaobin"][quote="kunny"]Let $x=\sqrt{1-ab},y=\sqrt{1-bc},z=\sqrt{1-ca,}$  we have $x^2+y^2+z^2=3-(ab+bc+ca)\geq 3-(a^2+b^2+c^2)\geq 2.\left(x\geq \frac{1}{\sqrt{2}},y\geq\frac{1}{\sqrt{2}},z\geq\frac{1}{\sqrt{2}}\right)\cdots [*].$

For $(x.y,z)$ satisfying $[*],$ in $xyz$-plane, seeking the condition that the plane $x+y+z=k$ have the intersection points with $[*],$ we have

$\frac{|0+0+0-k|}{\sqrt{1^2+1^2+1^2}}\geq \sqrt{2}$ which is the radius of sphere $x^2+y^2+z^2=2$

$\Longleftrightarrow |-k|\geq \sqrt{6}.$ Since $k\geq 0,$ we obtain $x+y+z\geq \sqrt{6},$ which completes the proof.[/quote]
sorry,I don't think this "proof" is right.
if it is right then you can prove
$x+y+z\geq \sqrt{6},$  if  $x^2+y^2+z^2=2$.
but is certainly wrong :(
sorry,if I am wrong[/quote]

I should have shown when the equality holds.

The equality holds when the sphere $x^2+y^2+z^2=2$ contact with the plane $x+y+z=\sqrt{6}\Longleftrightarrow x=y=z=\frac{\sqrt{6}}{3}.$ I wonder my solution is wrong basically.Could someone point that?

kunny

[quote="Vasc"]Your method, Ondrob, works again. :lol:
But the last step follows, for the sharper inequality:
$(2-3ab)(2-3bc)(2-3ca) \geq 1$.
Who give a solution?[/quote]

It's easy but nice with AM-GM :

Let $t=\frac{xy+yz+zx}{3} \le 1$ we need to prove :
$(2-x)(2-y)(2-z) \geq 1$

Notice that : $(2-xy)(2-yz)(2-zx)+(2+xy)(2+yz)(2+zx) =16+12t$

By AM-GM : $(2+xy)(2+yz)(2+zx) \le (2+t)^3$

Because $t \le 1$, then $16+12t-(2+t)^3 \ge 1$. We have done.

Notice that $x=\sqrt{3}a,y=\sqrt{3}{b, z=\sqrt{3}c}$.

[quote="hungkhtn"]
Let $t=\frac{xy+yz+zx}{3} \le 1$ .
Notice that : $(2-xy)(2-yz)(2-zx)+(2+xy)(2+yz)(2+zx) =16+12t$[/quote]
There is a mistake in your calculation. ;)

[quote="Vasc"]Let $a,b,c$ be non-negative numbers such that $a^2+b^2+c^2=1$. Prove that

$\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-ca} \geq \sqrt{6}$.[/quote]
Vacs, I have a very simple solution to this inequality in my future book. The solution is due to a Mathematics Team in my country. It's involved in just AG and Cauchy.

And the harder problem is in my book, to.  :lol: It's problem of VASC.

Back to Vasc's original inequality:

Squaring both sides of the inequality gives
\[ 3-(ab+bc+ca)+2\sqrt{(1-ab)(1-bc)}+2\sqrt{(1-ca)(1-ab)}+2\sqrt{(1-bc)(1-ac)}\geq6. \]
Note that we have
\[ 1-ab=\frac{1+a^2+b^2+c^2}2-ab=\frac12+\frac{a^2+b^2}2-ab+\frac{c^2}2\geq\frac12+\frac{c^2}2. \]
By Cauchy,
\[ (1+u^2)(1+v^2)\geq(1+uv)^2. \]
Now, the desired inequality follows.

 :P

[quote="Vasc"]Indeed, it is right.
But what you say about the following:

$(1-ab)(1-bc)(1-ca) \geq \frac{8}{27}$ ?  :|[/quote]

Yes, it's right Vasc. It was solved by $p,q,r$ substitution \[p=a+b+c,\; q=ab+bc+ca,; r=abc.\]

[quote="Vasc"]
But the last step follows, for the sharper inequality:
$(2-3ab)(2-3bc)(2-3ca) \geq 1$.
Who give a solution?[/quote]
$\sum_{cyc}\frac{1}{2-3ab}\leq3$ is not true: $a=b=\frac{12}{17},$ $c=\frac{1}{17}.$
But $(2-3ab)(2-3ac)(2-3bc)\geq1\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a-b)^{2}(7a^{4}-10a^{3}b+15a^{2}b^{2}-10ab^{3}+7b^{4}+$
$+6a^{2}bc+6b^{2}ac-5c^{2}bc+c^{3}a+c^{3}b)\geq0$:D, which obviously true.

In "Algebraic Inequalities" there are other two nice solutions to the equivalent problem:
[i]If $a,b,c$ are non-negative numbers such that $a^{2}+b^{2}+c^{2}=3$, then
$(2-ab)(2-bc)(2-ca) \geq 1$.[/i]

[quote="zhaobin"][quote="arqady"]A.Gladkiсh promised to bring to me this book. :P  :)[/quote]
But what a pity that no one can bring it to me :([/quote]
it is too expensive for me... I can't afford it at all  :(

[quote="Vasc"]Actually, we have

$ (2 \minus{} abc)(2 \minus{} bcd)(2 \minus{} cda)(2 \minus{} dab) \geq 1$.
The last is nice.  ;)[/quote]

What is your solution, Vasc?


geometry
incenter
circumcircle
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, I the incenter of ABC, UVW the cevian triangle of I wrt ABC,
O the center of the circumcircle of ABC,
X the meetpoint of the parallel to AI with BC,
Y, Z symmetrically…
U' the symmetric of U wrt X,
V', Z' symmetrically…
A*B*C* the tangential triangle of ABC.
Prove that the lines A*U', B*V', C*W' are concurrent.
Sincerely
Jean-Louis

I can't understand your problem.
I'm intersted in but what is definition of X?
And $ AU$and $ AU'$ is isotomic line wrt $ angle BAC$ right?

Dear Yang,
yes I understand your observation. Sorry I missed to specifie that
"X the meetpoint of the parallel to AI pasing through O with BC".
Sorry again.
Sincerely
Jean-Louis

I will redifine points.
$ A^{*}, B^{*}, C^{*}$ to $ P, Q, R$
Then,
By cevian nest theorem,
If we want to show $ PU', QV', RW'$ are concurrent,
$ AP, BQ, CR$ are concurrent and $ AU', BV', CW'$ are concurrent.
But $ AP, BQ, CR$ are $ A \minus{} symmedian, B \minus{} symmedian, C \minus{} symmedian$ of triangle $ ABC$ respectively.
and with simple calculating, by ceva theorem, we can get $ AU', BV', CW'$ are concurrent.

So $ PU', QV', RW'$ are concurrent :lol:

if you want to know what is cevian nest theorem
Then see here

http://www.mathlinks.ro/viewtopic.php?t=6579&search_id=1788733684

It's sometimes very useful theorem.

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


topology
algebra
function
domain
real analysis
real analysis theorems

$ k \le n$

Let $ S \subset R^n$ be a smooth k-dimensional surface and let $ \varphi : I^k \toU$ be a local chart of S.

Does $ \partial \varphi (I^k ) \cap S$ have k-dimensional measure zero in $ R^n$*?

*In one sense, this means if $ f: I^k \to T$ is some (any) local chart of S, the set theoretic preimage $ f^{ - 1} (\partial \varphi (I^k ))$ has k-dimensional measure zero in $ R^k$.

I'm trying to investigate whether $ \partial \varphi (I^k )$ is 'negibligible' in $ R^n$ in a 'k-dimensional sense'.

If not, counterexample?

I don't think the question in your title is the same as the one in your post. The answer to the question in your title is yes- but that involves a measure-zero subset of the domain. Extending to the closure of your domain is not something that can be guaranteed.

Yes, the question in the title has nothing to do with what I'm asking here.  :blush:

What are the counterexamples then?


[quote="neuro"]So, I have to find all $ x,y,z \in \mathbb{N}$ s.t. $ \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y} \equal{} 3$. I've seen some similiar problems but I can't get it right. I tried to multiply by $ xyz$ and then to factor everything on the left side. Any hints?[/quote]

$ a^{2}\plus{}b^{2}\plus{}c^{2}\geq ab\plus{}ac\plus{}bc$ ----->

$ \frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{zx}{y}\equal{}3$  <-> $ 3xyz\equal{}(xy)^{2}\plus{}(yz)^{2}\plus{}(zx)^{2}\geq (xy)(yz)\plus{}(xy)(xz)\plus{}(yz)(zx)\equal{}(x\plus{}y\plus{}z)xyz$ ---->

$ 3xyz\geq (x\plus{}y\plus{}z)xyz$ ----> $ 3\geq x\plus{}y\plus{}z$ ---> $ x\equal{}y\equal{}z\equal{}1$

Thank you very much, can you point me to some similar problems?

I think you could use AM-GM at the beginning.
\[ \frac {xy}{z} \plus{} \frac {yz}{x} >\equal{}2y\]
\[ \equal{}>\frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}>\equal{}x\plus{}y\plus{}z\]
\[ \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y} \equal{} 3\]
\[ <\equal{}>3>\equal{}x\plus{}y\plus{}z\]
.....

Let $ x \leq y \leq z$

$ \frac {yz}{x} < 3 \implies z < \frac {3x}{y} \leq 3$

$ \therefore 1 \leq x \leq y \leq z \leq 2$

There are four possibilities:
$ (1,1,1) \qquad (1,1,2)\qquad (1,2,2) \qquad (2,2,2)$

The solution is $ (x,y,z) \equal{} (1,1,1)$

By AM-GM

$ \dfrac{xy}{z} \plus{} \dfrac{yz}{x} \plus{} \dfrac{xz}{y} \ge 3xyz$

$ \dfrac{1}{z^2} \plus{} \dfrac{1}{y^2} \plus{} \dfrac{1}{x^2} \ge 3$

Clearly the only solution to is $ x\equal{}y\equal{}z\equal{}1$

Shouldn't that be $ 3 \sqrt[3]{xyz}$?

Regardless, you'd still have $ 3\equal{}    \dfrac{xy}{z} \plus{} \dfrac{yz}{x} \plus{} \dfrac{xz}{y} \ge 3 \sqrt[3]{xyz} \implies xyz \leq 1$


Let polynomial f(x) in Z(x) ,deg f(x)=n  and
    f(2^i)= 1/2^i ( i=1,2,..n)
Find value f(0)

Let f(x) one is polinoms (may be find by Lagrange approximation), g(x)=(x-2)(x-4)...(x-2^n). Then for all whole m
f(x)+mg(x) satisfied your conditions. Therefore f(0) can not be find from your conditions, because g(0) is not equal 0.

If f(x) belongs Z(x), then f(x) is whole for all whole x. I think f(x) belongs Q[x], and f(2^i)=2^(-i) for all i=0,1,...,n. Then
f(x) can be find by Lagrange approximation.

Sorry for my topo
             f(x) must be in R(x)
Now help me

TRUE PROBLEM

[quote="bui_haiha2000"]Let polynomial f(x) in R(x) ,deg f(x)=n  and
    f(2^i)= 1/2^i ( i=1,2,..n)
Find value f(0)[/quote]

[quote="bui_haiha2000"]No one solve my problem???[/quote]
No one wait more than one day¿¿¿


The real numbers $s,t$ varies being satisfied with $s^2+t^2=1,s\geqq 0,t\geqq 0$.
Find the range of the value of the root for the following equation can be valued.
\[x^4-2(s+t)x^2+(s-t)^2=0\]

does that mean between some minimum and infinity, or...?

Sorry for my poor English, probability1.01.
I have just edited.Please rewrite this problem's sentense into plain English for my brushing my English up.

kunny

[hide]Using the quadratic equation:
$x^2=s+t\pm\sqrt{4st}$
Then substituting $s=sin(\theta)$ and $t=cos(\theta)$ for $0<\theta<\pi/2$ (I couldn't get less than or equal to or the greater to render correctly)

$x^2=sin(\theta)+cos(\theta)\pm\sqrt{2sin(2\theta)}$
The derivative of the right side is $cos( \theta)-sin( \theta)+\sqrt{2}\frac{cos(2 \theta)}{\sqrt{sin(2\theta)}}$ which = 0 at $x=\pi/4$ on the necessary interval.

So maximum $x^2=\frac{\sqrt(2)}{2}+\frac{\sqrt(2)}{2}+\sqrt{2}=2\sqrt{2}=2^{\frac{3}{2}}$
Therefore x max=$2^{\frac{3}{4}}$ and x min is negative that.[/hide][/hide]

Yes.you are right, Wumbate.

Here is my solution.

$s\geqq 0,t\geqq 0,s^2+t^2=1\Longleftrightarrow (s+t)^2=2(s^2+t^2)-(s-t)^2\leqq 2 \Longleftrightarrow 1\leqq s+t\leqq \sqrt{2}$,

equality occurs when $(s,t)=(0,1),(1,0),\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$.

Solving for quadratic equation to $x^2$, we have $x^2=s+t\pm\sqrt{4st}=s+t\pm\sqrt{(s+t)^2-(s-t)^2}$.

From $A.M.\geqq G.M.\  0\leqq s+t-\sqrt{4st},s+t+\sqrt{4st}\leqq 2(s+t)\leqq 2\sqrt{2}$.

Therefore $0\leqq x^2\leqq 2\sqrt{2}$, the desired  answer is $-2^{\frac{3}{4}}\leqq x\leqq 2^{\frac{3}{4}}$.


trigonometry
inequalities
inequalities proposed

Let real numbers $ a,b,c\in (0;1)$ satisfying $ (a\plus{}b\plus{}c\minus{}2)^2\equal{}2abc$. Prove that $ \sqrt{\frac{a}{2\minus{}a}}\plus{}\sqrt{\frac{b}{2\minus{}b}}\plus{}\sqrt{\frac{c}{2\minus{}c}}\ge \sqrt{3}$

[quote="thaithuan_GC"]Let real numbers $ a,b,c\in (0;1)$ satisfying $ (a \plus{} b \plus{} c \minus{} 2)^2 \equal{} 2abc$. Prove that $ \sqrt {\frac {a}{2 \minus{} a}} \plus{} \sqrt {\frac {b}{2 \minus{} b}} \plus{} \sqrt {\frac {c}{2 \minus{} c}}\ge \sqrt {3}$[/quote]
If $ a\plus{}b\plus{}c \ge 2$, then we have
\[ \sqrt{\frac{a}{2\minus{}a}} \plus{}\sqrt{\frac{b}{2\minus{}b}} \plus{}\sqrt{\frac{c}{2\minus{}c}} \ge \sqrt{\frac{a}{b\plus{}c}} \plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}} \ge 2>\sqrt{3}\]
If $ a\plus{}b\plus{}c \le 2$, the given condition can be rewriten as: $ a\plus{}b\plus{}c\plus{}\sqrt{2abc} \equal{}2$. This identity suggests us to put $ a\equal{}2\cos^2 A,b\equal{}2\cos^2B,b\equal{}2\cos^2C$ where $ A,B,C \in (0,\frac{\pi}{2})$, then we have $ \cos^2A\plus{}\cos^2B\plus{}\cos^2 C\plus{}2\cos A\cos B \cos C\equal{}1$, it shows that $ A,B,C$ are the angles of a triangle. Moreover, the above inequality becomes
\[ \cot A \plus{}\cot B \plus{}\cot C \ge \sqrt{3}\]
which is just a known inequality in triangle.
:)

How you get this?
[quote="can_hang2007"]
 then we have $ \cos^2A \plus{} \cos^2B \plus{} \cos^2 C \plus{} 2\cos A\cos B \cos C \equal{} 1$, it shows that $ A,B,C$ are the angles of a triangle. :)[/quote]

[quote="befuddlers"]Nice can hang.

By the way whatz ur real name?[/quote]

My real name is Vo Quoc Ba Can. You can call me Can. Thanks.  :blush:

Can_Hang, could u  please explain ur solution further ?  Especially the first few lines. PLease.


calculus
geometry
geometric transformation
reflection
vector

I didnt know where to post this so excuse me if its the wrong forum.
Anyway, I have not heard anything about ap statistics, on this website. Is anybody taking the class or exam, what book did you use, are you really preapred for the exam?
Did anybody study it by them selves? How long do you think it would take to study all of it?
If you used a review study guide book which did you use?

I think some old threads that have sunk out of view on the first page described experiences taking the AP statistics exam. The consensus here seems to be that the math on that exam is easy and that the concepts are common-sense enough that this is a good exam to self-study for.

For what it's worth I (somewhat) disagree with the general sentiment. I think some of the nuances of statistics can be hard to pick up by studying on your own, especially from a prep book alone. While some people might be able to do well on the exam by studying, it doesn't mean they got a full understanding of the subject. If someone does self-study for the exam I definitely advise them to get a good textbook. A lot of the more commonly used ones can be found here: http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/51924.html

But in general, self-studying for any AP exam is not for everyone, and people who give that advice are sometimes doing a disservice by assuming that what worked for them will work for everyone else. Just for comparison:

AP Calculus AB (2006)
Grade - Number - % At
   5       43,925    22.3
   4       40,353    20.5
   3       36,627    18.6
   2       30,486    15.5
   1       45,790    23.2

3 or Higher = 61.3
Mean Score = 3.03
Stand. Dev. = 1.48

AP Statistics (2006)
Grade - Number - % At
   5      11,154     12.6
   4      19,568     22.2
   3      22,357     25.3
   2      16,119     18.3
   1      19,039     21.6

3 or Higher = 60.2
Mean Score = 2.86
Stand. Dev. = 1.32

So there are a lot of people who found the AP Stats exam pretty hard, and there are a lot of people who really do need a teacher to explain this stuff to them. You might assume that none of those people are here on AoPS, but that may or may not be true.

I think each individual person has to give a self-assessment of whether self-studying for an AP Exam is right for them. Maybe it is, and maybe it isn't.

[quote="gauss202"]While some people might be able to do well on the exam by studying, it doesn't mean they got a full understanding of the subject. If someone does self-study for the exam I definitely advise them to get a good textbook.[/quote]

I agree with this. I see conceptual errors in thinking about statistics almost every day from the keyboards of students who claim to have taken an AP statistics class. The human brain has built-in cognitive illusions that tend to result in misunderstanding fairly straightforward concepts of statistics, so it takes practice and careful guidance to learn to think statistically.

The inferential reasoning used in statistics doesn't come naturally to all highly skilled in mathematics and deductive reasoning.  I wonder how many of those who self-studied "5's" last year could explain the topic to interested students this year.

When thinking about taking AP courses and/or AP exams, look carefully at the corresponding college courses.  For BC Calculus, the courses typically form the base for substantial further study in mathematics.

College courses that you might gain exemptions from by taking the AP Statistics exam are designed for majors in the social sciences, business, nursing, etc.  At most schools, these majors typically take just one course in statistics.

The AP statistics course is algebra-based.  Statistics courses designed for majors in statistics or mathematics will be calculus-based.  The credits/exemptions that you might earn taking AP Statistics might be of little value if you need a calculus-based Statistics course in college.

Similarly, introductory college science courses such as Physics are sometimes typically taught in two flavors, calculus-based and algebra-based.  If you major in the physical sciences (or math) you will probably start with the calculus-based courses.  The algebra-based courses are for students who are knocking out a science requirement and plan no further study of the topic.  AP credits for the algebra-based Physics courses won't mean much for such students.

Edited: As Kent Merryfield points out below, General Chemistry is typically taught with an algebra prereq, not calculus.  The AP Chemistry exam is useful for most folks.  The above post has been edited to reflect this.

College level General Chemistry is not a calculus-based course anywhere that I know of. The AP Chemistry exam does widely "count," especially for biological science majors - in fact, it's the AP course I'd suggest has the most value for a prospective biological science major.

AP Physics does come in the two "flavors" mentioned here; AP Physics B won't "count" for a physical science or engineering major.

I am self-studying Statistics this year, although in a different way than most people on these forums. During my vector calculus class I can choose to either stay in class, listen to lecture, or go outside and work on Stats on the computer. It appears on my high school transcript that I took the class "Early Bird", which means before school, and it is an AP credit and all of that.

However, the coordinator of the online course (from APEX) didn't get me the textbook which was explicitly stated as necessary, and I only realized this recently. I had been studying the online lectures and something felt like it was missing (they say in places "refer to resources section" which is a blank page). They also quiz over materail not lectured about.

For this reason, I got the school coordinator to order me a textbook and its due to come in sometime next week.

All in all, I will essentially be learning the whole course from next week to April or so.

For a very busy person, is this enough time to master the course? (I know that people here get 5's and do it in 2 weeks, so,....)

Check this really good [url=http://statland.org/MAAFIXED.PDF]review of statistics textbooks for mathematicians[/url] for the rationale of the AP statistics syllabus and some issues to look out for when studying statistics.

[quote="me@home"]
For a very busy person, is this enough time to master the course? (I know that people here get 5's and do it in 2 weeks, so,....)[/quote]

I'm going to go ahead and say no. Even though people sometimes rave about Statistics being easy, mastering the course definitely takes a lot more time input. The people you're talking about did simply that, "get 5's" and I'm pretty sure they don't quite remember much statistics now :D

There is definitely a difference between "master the course," and "gain a 5 on the AP test." You can see that my first answer in this thread related mostly to passing the test, and my second answer mostly to gaining an understanding of statistics. The important thing about any AP course is not to think that getting a 5 on the test means that you are done learning the subject.

Does anybody have or no where there might be good
programs for the graphing calculator either the ti-83 or 89, that could be use full for Ap statistics

The TI 83/84 calculators come with lots of useful standard features for AP Stats.  Normal, binomial, Poisson, etc. distributions are pre-programmed, eliminating the need for tables (2nd VARS - Distr.)  Univariate statistics and some rudimentary analysis of variance and regression output are all available.  There might be a package or two that might help with data entry or that expand the typical output, but I think virtually everything that you really need for AP Stats and much more is already on your calculator.

Yes, the latest versions of the 89 also have built-in stats software much like that that has long been on the 83/84 series.

I assure you that AP Statistics was many times easier than AP Calculus. It does not require much understanding as Calculus and most of the material consists of definitions and basic understandings. For example, the Free Response parts ask to do a statistical test. Instead of just finding the $t$ value, you have to state the null hypothesis, conditions, and a conclusion to get full points.

I think many of the AoPS-type people will find calculus easier to learn than stats.  Maybe I'm just biased because I found calculus to be more fun although more work.  However, I'm comfortable calling things easier than others even if the thing that I consider easier seems that way only because I like it more...

[quote="probability1.01"]AP stats is generally speaking not a hard AP, especially since the cutoff for 5 is low. However, I found it incredibly lame because you don't really get enough theory to actually understand 90% of what you learn. Also, it's not connected with most other math, so you will very likely forget most of the material if you're going into math or hard sciences. That's why I don't really consider stats to be actual math. If you want to do well on the AP, just memorize all the tests and their stipulations, understanding the reasons behind them when possible, and you should be fine.[/quote]

Yeah that's why I don't particularly like AP stats.

Statistics ISN'T math. See

http://statland.org/MAAFIXED.PDF

to get a better idea of what it is. Physics isn't math either, but that doesn't mean physics is a bad subject.

As a senior in high school i took AP statistics concurrently with Precalculus. Right now i am a college freshman in what is equivalent to AP Calculus BC. So far i find Calculus much easier than statistics, but maybe i am the oddball like that. Statistics isn't mathematically rigorous. It is all just simple computations that your calculator does for you anyways. The concepts in Statistics may seem easy at the start, but they are hard and it can be very difficult remembering all the procedures for all the inference tests and stuff. The AP test isn't too hard. I had "Senioritis" so i didn't study at all for the test and i got a 4.

It's unfortunate that good math students get the impression from AP Statistics that the field of statistics is not "mathematically rigorous."  Take a look at some of the leading journals in the field:  The Annals of Statistics, The Journal of the American Statistical Association, Biometrika, Technometrics, The Journal of the Royal Statistical Society, etc.  You will find that new ideas in the field are typically well-defined and "proven."

Before AP Statistics, most good math/stat students were introduced to statistics through courses in "Mathematical Statistics."  In these classes, not much effort was spent on the various statistical models you see in AP Stats.  However, basic concepts and properties of estimators were derived in a "rigorous" fashion.    After these classes, you will appreciate the difference between "least squares," "maximum likelihood" and other types of estimators.  You should be able to engage in a conversation about the properties of an innovative estimation technique or method.

Mathematical Statistics courses still exist at most universities.  I'd recommend taking a couple of these before giving up on the field of statistics.

I'm beginning to think that it's a real mistake to introduce top math students to the statistics field with the "AP Stats" course.  Most math/physical science/stat students will never be required to take the AP Stats equivalent in college.  It's a course designed for majors in business, life and social sciences.

Some statisticians are highly mathematical and spend their careers developing methodology or statistical theory (i.e. theorems and proofs.)  Many of these folks are in math or statistics departments at universities and a few work in government agencies.

Other statisticians are more applied and get their hands dirty designing experiments or surveys and analyzing real data.  Many applications point toward a need for new theory and techniques.  Those working on more demanding applications will develop the theory and methods required and publish in both theoretical and applied journals.  You'll find many of these statisticians in medical research and the pharmaceutical industry.

I still think that there is a good place in the high school curriculum for a course that emphasizes--as an AP statistics course should--that statistics is all about data, and mathematical analysis of data has to be connected to the real-world context in which the data were gathered.

In response to gt59

I apologize if i was unclear. I didn't mean that statistics as a whole was not mathematically rigorous, i just meant that the AP class specifically doesn't require much math beyond intermediate algebra, so the math isn't "hard" so to speak. I am aware that more advanced studies in statistics use much more "advanced" methods of math such as calculus and whatnot.


It has come to a policeman's ears that 5 gangsters (all of different height) are meeting, one of them is the clan leader, he's the tallest of the 5. He knows the members will leave the building one by one, with a 10-minute break between them, and too bad for him Belgium has not enough policemen to follow all gangsters, so he's on his own to spot the clanleader, and he can only follow one member.

So he decides to let go the first 2 people, and then follow the first one that is taller than those two. What's the chance he actually catches the clan leader like this?

Let ABCDE be the gangsters with A the tallest.

The ways that the first 2 gangstars can come out are AB AC AD AE BC BD BE CD CE DE  and the reverse, all with equal probability.  Then A is "marked" 40% of the time and caught 0%; B is marked 30%, C 20%, and D 10%.

Then when B is marked, A is caught 100%.  When C is marked, either of A or B could come first with equal chances, so 50% prob.  When D is marked, A comes first 33% of the time.  So in total .40(0) + .30(1) + .20(.5) + .10(1/3) = 43.33... %


[quote="sanbittu"]problem 1
\[ {\rm{Solve using quadratic formula}},{\rm{ for }}x{\rm{ }}: {\rm{ }}a\left( {x{\rm{ }} \plus{} {\rm{ }}1} \right){\rm{ }} \equal{} {\rm{ }}x\left( {a{\rm{ }} \plus{} {\rm{ }}1} \right)
\]
problem2
\[ {\rm{Solve for }}x: {\rm{ }}\left( {x \minus{} 3} \right)\left( {x \plus{} 9} \right)\left( {x \minus{} 7} \right)\left( {x \plus{} 5} \right){\rm{ }} \equal{} 1680
\]
members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com[/quote]
1. $ a(x\plus{}1)\equal{}x(a\plus{}1) \Leftrightarrow x\equal{}a$
2. $ (x\minus{}3)(x\plus{}9)(x\minus{}7)(x\plus{}5)\equal{}1680 \Leftrightarrow [(x\minus{}3)(x\plus{}5)][(x\plus{}9)(x\minus{}7)]\equal{}1680$
$ (x^2\plus{}2x\minus{}15)(x^2\plus{}2x\minus{}63)\equal{}1680$
Set $ t\equal{}x^2\plus{}2x\minus{}39 \ge \minus{}40$
The equation becomes:
$ (t\minus{}24)(t\plus{}24)\equal{}1680 \Leftrightarrow t^2\equal{}2256 \Leftrightarrow t\equal{}4\sqrt{141}$ ( because $ t \ge \minus{}40$)
$ \Leftrightarrow (x\plus{}1)^2\equal{}4(10\plus{}\sqrt{141})$
$ \Leftrightarrow x\equal{}\minus{}1\minus{}2\sqrt{10\plus{}\sqrt{141}}$ or $ x\equal{}\minus{}1\plus{}2\sqrt{10\plus{}\sqrt{141}}$


geometry
advanced fields
advanced fields unsolved

I have to submit them by Monday and they are crazy....errr!!!!

1.   One in seven mathematicians is a philosopher, while one in nine philosophers
       is a mathematician. Are there more philosophers than mathematicians?

  2.   A study shows that the percentage of left-handed people among mathematicians
       is greater than the percentage of left-handed people among
       non-mathematicians. Prove that it follows that the percentage of
       mathematicians among left-handed people is greater than the percentage of
       mathematicians among right-handed people.

  3.   Three sailors were shipwrecked on an island. To make sure that they would
       have enough to eat, they spent the day gathering bananas. They put the
       bananas in a large pile and decided to divide the bananas equally among
       themselves the next morning. However, each sailor distrusted the other
       two, so after they were asleep, one of the sailors awoke and divided the
       pile of bananas into three equal shares. When he did so, he found that he
       had one banana left over; he fed it to a monkey. He then hid his share and
       went back to sleep. A second sailor awoke and divided the remaining bananas
       into three equal shares. He too found that he had one banana left over and
       fed it to the monkey. He then hid his share and went back to sleep. The
       third sailor awoke and divided the remaining bananas into three equal shares.
       Again, one banana was left over, so he fed it to the monkey. He then hid his
       share and went back to sleep. When the sailors got up the next morning, the
       pile was noticeably smaller, but since they all felt guilty, none of them
       said anything and they all agreed to divide the bananas. When they did so,
       one banana was left over and they again gave it to the monkey.
       What was the minimal possible number of bananas at the beginning?

  4.   Alice, Betty, and Carol took the same series of examinations. For each
       examination there was one mark of x, one mark of y, and one mark of z, where
       x, y, z are distinct positive integers. After all the examinations, Alice
       had a total score of 20, Betty a total score 10, and Carol a total score
       of 9.
       If Brabara was placed first in Algebra, who was placed second in Geometry?
       (Terence Tao, "Solving mathematical problems: a personal perspective",
        pages 86-89,       books.google.com)

  5.   A party was held at the house of the Schlobodkins. There were four other
       couples present (besides Mr. and Mrs. Schlobodkins), and many, but not all,
       pairs of people shook hands. Nobody shook hands with anyone twice, and
       nobody shook hands with his/her spouse. Both the host and the hostess shook
       some hands. At the end of the party, Mr. Schlobodkins polls every person
       present to see how many hands each person (other than himself) shook. Each
       person gives a different answer.
       Determine how many hands Mrs. Schlobodkins must have shaken.
       (Paul R. Halmos, "The Thrills of Abstraction",
        The Two-Year College Mathematics Journal", Vol. 13, No. 4 (1982)   JSTOR)

  6.   In general, a magic square is an arrangenment of the integers from 1 to n*n,
       in cells of an n-by-n square such that the numbers in each row, column, and
       diagonal give the same sum, the magic sum.
        (a) What is the magic sum for a 3-by-3 magic square?
        (b) What is the magic sum for an n-by-n magic square?
        (c) Place the numbers 1 through 9 into the nine cells of a 3-by-3 square
            to make it a magic square.

I have been busy with meetings and all and didn't get time to think about it....help!!!!!!!!!

- Problem 3 is similar to this one:

http://www.mathlinks.ro/viewtopic.php?t=185766

The minimal number of bananas is 79, according to my count.

- Problem 6 is the easiest one.

Solve part (b) first. The rows of a $ n \times n$ magic square partition the set $ \{1, 2, ..., n(n \minus{} 1), n^{2}\}$ into $ n$ disjoint subsets whose elements sum to the same number, say $ N$. Since $ 1 \plus{} 2 \plus{} ... \plus{} n^{2} \equal{} \frac {n^{2}(n^{2} \plus{} 1)}{2}$, it follows that $ N \equal{} \frac {n(n^{2} \plus{} 1)}{2}$.


A few questions on the format of the quiz answers:

In the qualifying quiz it says to "show all steps in your reasoning and in your computation". If, for example, you did a massive long division, do they want to see it, or does it assume use of calculators?

Also, do they accept typed quizzes?

Thanks.

I don't really know, but since you can use any resource except another person (textbooks, calculators, etc.) I don't think you need to show really long long division.


If someone could help me on these questions and explain the answers, I would be very appreciative...  :)

1. In rectangle $ ABCD$, points $ F$ and $ G$ lie on segment $ AB$ so that $ AF \equal{} FG \equal{} GB$ and $ E$ is the midpoint of segment $ DC$. Also, segment $ AC$ intersects segment $ EF$ at $ H$ and segment $ EG$ at $ J$. The area of rectangle $ ABCD$ is 70. Find the area of triangle $ AHF$.
[img]http://usera.ImageCave.com/jnma/untitled3.bmp.jpg[/img]

2. $ ABCD$ is a square. Parallel lines $ m$, $ n$, and $ p$ pass through vertices $ A$, $ B$, and $ C$, respectively. The distance between $ m$ and $ n$ is 7 units, and the distance between $ n$ and $ p$ is 9 units. Find the number of square units in the area of square $ ABCD$.

[hide="1"] DC is parallel to AB, so triangles EHC and FHA are similar. The ratio of their bases is $ \frac {x}{2}$ to $ \frac {x}{3}$, or 3 to 2 (let the base of the rectangle be x and you see that).

That means the ratio of their heights is also 3 to 2. Since the sum of their heights is the height of the rectangle, triangle FHA has a height that is 2/5 of the rectangle's and a base that is 1/3 of the rectangles. The ratio of it's area to the rectangle's is thus $ \frac {1}{15}$, so our answer is $ \frac {14}{3}$.[/hide]

[hide="2"] Utilize symmetry to construct more of these lines:
 [geogebra]afe69729a2fb205b74c8bd2c539ea39a5d66a0d3[/geogebra]

Now label a bunch of things with 7s and 9s, and you get four right triangles with legs 7 and 9, so the area from all four of those is 126. You also get a square with side (9-7)=2, which has an area of 4.

The total area is 130. Sorry for my bad drawing, I'm in a hurry.[/hide]

Wow xpmath! That was a terrific explanation. :D

Thanks again!


prove that among the numbers of the arithmetic progressions 3,7,11,15,19,23,... and 5,11,17,23,29,35,... there are an infinite nuber of primes.

It means that there exist infinite prime of the form $ 4k \minus{} 1$
You can induction.
Call $ p_1,..,p_n$ is the fisrt n prime of the form $ 4k \minus{} 1$
Consider
$ t \equal{} 4\prod_{i \equal{} 1}^np_i \minus{} 1$
because $ t\equiv 3(\mod 4)$ so it has a prime divisor in the form $ 4k \minus{} 1$ .Call it is m
But from $ \gcd(t,p_i) \equal{} 1$ so $ m\not\in\{p_i\}$
This problem claim.
To prove there exist infinite prime in the form consider
$ x^2 \plus{} 1$


I know you can use pascal's triangle to solve this, but you wouldn't want to generate down to the 12th row in a countdown round.  Is there any other trick?

It's the same as $ 12\choose{6}\equal{}\frac{12!}{6!6!}\equal{}924$ Erm, what's wrong with my latex??

$ 12\choose{6}$$ \equal{}\frac{12!}{6!6!}\equal{}924$


Try "\backslash".

This gives $\backslash$ ;)

[quote="jmadsen"]I want to display "R \ A" in LaTeX, but I cannot use a backslash as this has a different function. What do I use for this?
Also, when I use \mathbb{R} as in the Guide, the compiler says this is an undefined control sequence. Do I need to include a package?[/quote]

for displaying mathblackboard, you shoud include the amslatex package. Just to make sure that it will working just include everything here:

\usepackage{amsmath,amssymb,amsfonts}


Solve this Maths Problems within 60 Secs with Tricks

Plz provide your solution alongwith

[b][color=#FF0000]
1) 68% of 595 - 43% of 372 = ?

2) (96)² + (63)² = (?)² - (111)² - 8350

3) [(135)² ÷ 15 * 32] ÷ ? = 45 * 24

4) (739% of 383) ÷ 628 = ?

5) 814296 * 36 = ? * 96324

6) 628.306 + 6.1325 * 44.0268 = ?

7) (935.82)² = ?

8 ) 92.5% of 550 = ?

[/color][/b]

[color=#0000FF][b]Solve above problems within 60 secs..Without use of calculator (calc)

Use Paper, Pen & Brain only

Plz provide ur solution alongwith time u consumed[/b] [/color]

Not sure why you've posted this.

I did the first three, having skipped trying to spot the square root of $ 33856$ as question three was easy and cancelled nicely. Those three took me longer than $ 60$ seconds, so having accepted the challenge and failed, I look forward to seeing the tricks.


Let $ p$ а prime such that $ p\equal{}4k\plus{}1$ for some $ k\in\mathbb{N}$. Prove that $ k^{2k}\equiv1\left(\mod p\right)$.

You can make this problem better (and is an well-known exercise):
http://www.mathlinks.ro/viewtopic.php?t=251936


What is the sum of the real values of $ x$ for which the following expression is undefined? Express your answer as a common fraction.
\[ \frac{8}{\frac5x\minus{}\frac23}\]

x could be 0, and if it's not, then $ 5/x\equal{}2/3$ so x can be $ 15/2$
Sum: $ 15/2$


How many of the natural numbers from 1 to 800, inclusive,
contain the digit 6 at least twice? (The numbers 266 and 663
are two natural numbers that contain the digit 6 at least twice,
but 430 or 16 are not.)

There are $ 3$ choices for the non-six digit and $ 10$ choices for that digit, for a total of $ 27$.

However, we have missed $ 666$ and have erroneously counted $ 866$ and $ 966$, so we need to subtract $ 1$.

Our answer is $ \boxed{26}$.

Edit: Sorry, misread.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1056895860&t=217824]here[/url], post number 4. The answer is $ \boxed{26}$.

Why wouldn't the answer be 25? There are only 7 numbers with a 6 in the tens and units place. 866 is not allowed.

[hide="Casework"]There is 1 2-digit number that works: 66
We divide the 3 digit cases into four parts:
Case A) If the number is in the form of $66A$, where the digit A is not 6.
There are 9 values of A.
Case B) If the number is in the form of $A66$, where digit A is not 6.
There are 6 values of A.
Case C) If the number is in the form of $6A6$, where digit A is not 6.
There are 9 values of A.
Case C) All digits are 6- there is only one- $666$.
Therefore we have: $1+9+6+9+1=\boxed{26}$[/hide]

Unless you were being serious, you shouldn't revive to say a troll answer.

I DIDN'T KNOW WE CAN COUNT 2 DIDGIT NUMBERS

The solution seems wrong because it should be 66_ has 10 numbers. 6_6 has 10. _66 has 8 numbers, but then there should also be 66, so it should be 27. no?

@BudgieBricks I think you are counting $666$ twice, once in the 66_ case and once in the 6_6 case. I think there are only nine possibilities for 6_6 after doing the 66_ case.

[quote=roderickhuang]Here is what I did.

[hide="My Solution"]
2 digit: 66 That's 1.
3 digit:
166 606 661
266 616 662
366 626 663
466 636 664
566 646 665
666 656 666
766 676 667
866 686 668
966 696 669
There are 28 numbers here. But the numbers have to between 1 and 800 so 866 and 966 are eliminated making it $\boxed{26}$.
[/hide][/quote]

No offense, but that's a really sad way of doing it. lol

[hide="PIE"]
If the first two digits are 6: $10$ cases
If the first and last are 6: $10$ cases
If the last two digits are 6: $8$ cases
If all digits are 6: $1$ case

This results in a total of $10 + 10 + 8 - 1 - 1 - 1 + 1 = \boxed{26}$ cases by PIE.


quadratics
modular arithmetic
USAMTS
ceiling function
summer program
PROMYS
number theory

[quote="DanK"]Ok, so I understand what quadratic residues are now, but can someone please explain to me how you would go about finding the quadratic residues of a given $\pmod{n}$?[/quote]
I can give you an example, and that will make it very simple:
Lets say we are asked to find the quadratic residues in mod 7. We will look at the squares of the numbers from 0 to 6 in mod 7. So the distinct ones would be known as the [b]quadratic residues[/b] in mod 7. We have
$0^{2}=0,1,4,9\equiv 2, 16\equiv 2, 25\equiv 4, 36\equiv 1$. Thus, the quadratic residues are 0, 1, 4 and 2.
hope that helps.

Yeah, I saw that you could figure out the individual square in mod $n$ like that, but I didn't know when to stop.  So I guess I understand how to do it now, but I don't understand one thing about why it works: how do we know all of the possible quadratic residues will occur within the squares of numbers less than $n$?

[quote="DanK"][quote="Elemennop"]$(kn+r)^{2}\equiv r^{2}\pmod{n}$[/quote] :wallbash_red:  :wallbash: Sometimes I really wonder how I can be so stupid...[/quote]
I doubt you're stupid, after all you helped me. Sometimes it takes time for the idea to reach your head, it does for me.

Yeah, I just get annoyed with myself sometimes...

At least I understand it now.  They kept talking about in in one of the solutons to a USAMTS problem from last year, and it was driving me crazy...

[quote="lotrgreengrapes7926"]You only have to check the first $\lceil\frac{n}{2}\rceil$ squares. The residues repeat, but backwards. For example, $0,1,4,2$ and $2,4,1$. This is because in $\mod 7$, numbers can be written as $0,1,2,3,-3,-2,-1$. $(-k)^{2}=k^{2}$, giving the same residue.[/quote]

Nice.  I always forget to check negative results from mods...

Well, if $n=p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{k}^{a_{k}}$, then \[x^{2}\equiv r\mod{n}\iff x^{2}\equiv r\mod{p_{i}^{a_{i}}}\] for $1 \le i \le k$.  Then, we use a simple divisibility argument to show that there are exactly zero or two solutions to that congruence mod $p_{i}$, and then, using an inductive method, prove there are exactly zero or two solutions mod $p_{i}^{a_{i}}$.  Then, given a bunch of congruences in each $p_{i}^{a_{i}}$, there are two possible residues in each, so it seems there are either zero or $2^{k}$ solutions mod $n$ for each possible $r$, so then we have $\frac{\phi(n)}{2^{k}}$ quadratic residues...

I don't know if this is making any sense if you haven't done some number theory, but it's an interesting chain of reasoning (woot for PROMYS!)...start off with proving $x^{2}\equiv r\mod{p}$ has either zero or two solutions, and if two, they are additive inverses of each other mod p, and go from there.

How do you use quadratic residues in problem solving?

For example last post from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102730[/url].

Or: there exists infinitely many primes $\equiv 1 \mod 3$:
Assume there are only finitely many, call their product $P$. Then consider the number $P^{2}+3$ and let $q$ be a prime divisor of it. Then $P^{2}\equiv-3 \mod q$, thus $-3$ is a quadratic residue $\mod q$. By quadratic reciprocity we get that $q$ is a quadratic residue $\mod 3$. Thus $q \equiv 1 \mod 3$ ($q \neq 3$ should be clear). But $q$ divides $P$ by definition, which gives $q|3$, acontradiction.


A baby goes to sleep at 8 p.m., wakes up after two hours, stays awake for 15 minutes, and then goes back to sleep. If the baby repeats this sleep pattern until she is awakened at 6 a.m., what percent of the time from 8 p.m. to 6 a.m. was the baby asleep?

In the first $ 9$ hours from $ 8$ p.m. to $ 5$ a.m., there are $ 8$ hours of sleep, as four complete cycles have been completed. Then, there is $ 1$ more hour of sleep until she is awakened, so the answer is $ \frac{8\plus{}1}{9\plus{}1}\equal{}\boxed{90\%}$.


Find all $ x,y,z \in N$ that $ 3^x\plus{}4^y\equal{}5^z$

don't you think the topic should be more specific ?

it was posted some time ago

the sketch of solution:

1. $ 2|x$

2. $ 2|z$ (cong. mod 3)

3. then $ 4^y\equal{}5^z\minus{}3^x\equal{}5^{2Z}\minus{}3^{2X}\equal{}(5^Z\minus{}3^X)(5^Z\plus{}3^X)$ so both of them are powers of 2.
The rest isn't difficult.

answer: $ 2,2,2$ (only)


the centers of fence posts are 30 inches apart.  they enclose a triangular region 20 feet by 20 feet by 10 feet.  How many posts will be needed?

i think that equals [b]20[/b].  [hide]Wouldn't the corners of the fence be shared by two sides?  I think it would then equal [b]17[/b][/hide][/hide]

[quote="Xantos C. Guin"][hide]22
The sides are 240in, 240in, 120in
Divide each by 30 and you get:
8,8,4
8+8+4= [b]22 ANS[/b][/hide][/quote]
[quote="4everwise"][hide]22
The sides are 240in, 240in, 120in
Divide each by 30 and you get:
8,8,4
8+8+4= 22 ANS [/hide][/quote]

WOW!
those 2 posts are exactly identical
exept 4everwise didn't put the [hide]22 ANS[/hide] in bold.
I guess great minds do think alike.

8+8+4=$20$
2 people made a careless error!
[b]2 people!!!!!!!!!!!!!!!!!!!!!!!!![/b]
and it was the SAME error!

Notes to penguin:
Stop making careless errors.

Yeah, if the answer was 22, then there will be two posts stacked on top of another.

oh well what if you have 60 yards of fence and a built in wall that expands 2 miles.  What is the max area?  Make a rectangle using the wall as one side and using the fence for 3 sides.


trigonometry
function
symmetry
inequalities
quadratics
algebra
system of equations

Please help with these problems:

1.

By choosing different real values for $ a$ in the system of equations:\[ x^{2}\minus{}y^{2}\equal{}0\]\[ (x\minus{}a)^{2}\plus{}y^{2}\equal{}1\]the number of real-valued solutions can be:

a) 0, 1, 2, 3, 4, or 5     b) 0, 1, 2, or 4     c) 0, 2, 3, or 4 m    d) 0, 2, or 4     e) 2 or 4

2.

The graph of the equation $ y\plus{}\mid y\mid \equal{} x\plus{}\mid x\mid$, restricted to $ x < 0$, is:

a) the empty set     b) a line     c) shaped like a V     d) quadrant of a plane     e) two quadrants of a plane

3.

How many real number solutions does $ \left(\frac{x\minus{}21}{2x}\right)^{x^{2}\minus{}x\minus{}12}\equal{}1$ have?

a) 0     b) 1     c) 2     d) 3     e) 4

I seem to get 3 for this problem but the answers say 4 can someone explain

4.

If $ \sin\theta\plus{}\cos\theta\equal{}\frac{1}{5}$ and $ 0\le\theta<\pi$, then $ \tan\theta$ is:

a) $ \minus{}\frac{4}{3}$     b) $ \minus{}\frac{3}{4}$     c) $ \frac{3}{4}$     d) $ \frac{4}{3}$     e) not completely determined by the given information

5.

What is the set of values for $ k$ for which the function $ f(x) \equal{}\cos^{4}x\minus{}k^{2}\cos^{2}(2x)\plus{}sin^{4}x$ is constant?

a) $ {\frac{1}{\sqrt{2}},\minus{}\frac{1}{\sqrt{2}}}$     b) $ \{\frac{1}{2},\minus{}\frac{1}{2}\}$     c) $ \{\frac{1}{2}\}$     d) $ \{\frac{1}{\sqrt{2}}\}$     e) cannot be determined

[hide="(1)"]
The answer is (c).

Graphically, the first equation is the pair of lines $ x \equal{} y$, $ x \equal{}\minus{}y$. The second equation is a circle with center $ (a,0)$ and radius 1. When we place a circle of radius one along the x-axis, we find that when the center is near the origin we can intersect at four points, when the center is at $ (1,0)$ it intersects at 3 points, when the center is at $ (\sqrt{2},0)$ its tangent to the two lines and so there it intersects at 2 points. If we drag the circle away from the lines there are 0 intersections. However, we cannot get 1 point because of symmetry across the x-axis.

Not sure if it is possible algebraically. (also there's an "m" next to choice (c), though I'm assuming that's a typo)[/hide]

[hide="(2)"]
The answer is (d).

Let's break up the absolute value signs:

$ y\plus{}|y|\minus{}x \equal{} |x|$

When $ x < 0$, we have

$ y\plus{}|y|\minus{}x \equal{}\minus{}x$
$ y\plus{}|y| \equal{} 0$

This will be true when $ y$ is negative; so we have the inequalities $ x < 0, y < 0$ which is the third quadrant of the plane.
[/hide]

[hide="(3)"]
The solution you are probably missing is $ x \equal{} 7$.

We find that $ x^{y}\equal{} 1$ is true when $ x \equal{}\boxed{\pm}1$ (under real numbers; the negative is true when $ y$ is even) or when $ y \equal{} 0$.

Case 1: $ x \equal{}\pm 1$

Check the parity of $ x^{2}\minus{}x\minus{}12$: $ x$ being even causes $ x^{2}\minus{}x\minus{}12$ to be even, and $ x$ being odd also causes that to be even. I don't think this is rigorous under real numbers ... oh well.

Then $ \frac{x\minus{}21}{2x}\equal{}\pm 1\Longrightarrow x\minus{}21 \equal{}\pm 2x\Longrightarrow x \equal{} 7,\minus{}21$.

Case 2: $ x^{2}\minus{}x\minus{}12 \equal{} 0$

Factors into $ (x\minus{}4)(x\plus{}3) \equal{} 0\Longrightarrow x \equal{} 4,\minus{}3$.

So solutions: $ \boxed{\minus{}21,\minus{}3,4,7}$
[/hide]

[hide="(4)"]
Just by looking at the solutions (which imply that $ \sin\theta \equal{}\frac{4}{5}$ and $ \cos\theta \equal{}\minus{}\frac{3}{5}$ is a possibility), guess and check quickly tells us that the answer is either (a) or (e). On a contest I would just guess (a) and continue, but here I'll be rigorous.

Notice that $ \theta$ must be in the second quadrant; in the first, both $ \sin\theta$ and $ \cos\theta$ are positive and at least one is $ \ge\frac{1}{\sqrt{2}}$. Hence $ \tan\theta < 0$.

Square the equation and we get $ \sin^{2}\theta\plus{}\cos^{2}\theta\plus{}2\sin\theta\cos\theta \equal{}\frac{1}{25}\Longrightarrow\sin 2\theta \equal{}\frac{\minus{}24}{25}$. Hence $ \theta \equal{}\frac{\sin^{\minus{}1}\left(\frac{\minus{}24}{25}\right)}{2}$.

Use the tangent half-angle formula:

$ \tan\left(\frac{\theta}{2}\right) \equal{}\pm\sqrt{\frac{1\minus{}\cos (\sin^{\minus{}1}\frac{\minus{}24}{25})}{1\plus{}\cos (\sin^{\minus{}1}\frac{\minus{}24}{25})}}$

$ \cos (\sin^{\minus{}1}\frac{\minus{}24}{25}) \equal{}\frac{7}{25}$ (just draw it out, you get a 7-24-25 triangle in the second quadrant), so we get

$ \tan\left(\frac{\theta}{2}\right) \equal{}\pm\sqrt{\frac{1\minus{}\frac{7}{25}}{1\plus{}\frac{7}{25}}}\equal{}\pm{\frac{32}{18}}\equal{}\pm\frac{4}{3}$. It's in the second quadrant, take the negative root, and our answer is (a).

[/hide]

[hide="(5)"]
The answer is (a).

$ f(x) \equal{} (\cos^{2}x\plus{}\sin^{2}x)^{2}\minus{}2\sin^{2}x\cos^{2}x\minus{}k^{2}\cos^{2}2x$
$ f(x) \equal{} 1\minus{}\frac{1}{2}\sin^{2}2x\minus{}k^{2}\cos^{2}2x$

Since we want it to be constant, we want the last two terms to be constant; we will use the identity $ \sin^{2}\theta\plus{}\cos^{2}\theta \equal{} 1$. Then $ k^{2}\equal{}\frac{1}{2}\Longrightarrow k \equal{}\pm\frac{1}{\sqrt{2}}$

[/hide]

[quote="azjps"]Not sure if it is possible algebraically. (also there's an "m" next to choice (c), though I'm assuming that's a typo)[/quote]

[hide="Algebra"] The first equation gives $ y^{2}\equal{} x^{2}$, so a substitution into the second produces $ (x\minus{}a)^{2}\plus{}x^{2}\equal{} 1$.  A quadratic can have $ 0, 1, 2$ solutions (varying $ a$ appropriately).

Generally, each value of $ x$ corresponds to two values of $ y$; however, when $ y \equal{} 0$ this does not occur, so there can be $ 0, 2, 3, 4$ solutions.  [/hide]


Let $x, y$ be natural numbers such that

$2x^{2}-1=y^{15}$

Show that if $x>1$, then $5\mid x$.

Thanks a lot... I was wondering where it had come from...


can you show that :D

$ \sum_{n\equal{}1}^\infty \; \frac{1}{(1\plus{}\pi^2\,n^2)^3}\;\equal{}\;\boxed{\frac{11\minus{}31e^2\plus{}41e^4\minus{}5e^6}{16\;(e^2\minus{}1)^3}}$

strange thing is we have $ \pi$ on the thing that we are trying to sum, but the sum itself actually involves $ e$  :|

it's simple exercises for complx analysis and it's not important there will be $ \pi^{2}$ or not . we can find
$ \sum \frac {1}{(1 \plus{} d^{2}n^{2k})^{m}}$

OK, let me make it easier, you [b]cannot[/b] use complex analysis.

[quote="kenn4000"]this question [u]is[/u] a complex analysis question[/quote]
Yes, but I think there is also some way by real analysis.
For example when Ramanujan proved many interesting results, He did not know  complex analysis.

Note that

\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{(1 + \pi^2 k^2)^3}
& = & \sum_{k=1}^{\infty} \sum_{n=0}^{\infty}  \frac{(n+1)(n+2)}{2} \frac{(-1)^n}{(\pi k)^{2(n+3)}}\\
& = & \sum_{n=0}^{\infty}  \frac{(n+1)(n+2)}{2} (-1)^n \frac{\zeta(2n+6)}{\pi^{2(n+3)}}\\
& = & \sum_{n=1}^{\infty}  \frac{(n-1)(n-2)}{2} (-1)^{n-1} \frac{\zeta(2n)}{\pi^{2n}} \quad \quad \cdots (1)
\end{eqnarray*}

Using the identity $ (-1)^{n-1} \frac{\zeta(2n)}{\pi^{2n}} = \frac{B_{2n}}{(2n)!} 2^{2n-1}$ and $ \frac{x}{e^x - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n,$ we have

$ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\zeta(2n)}{\pi^{2n}} x^{2n-1} = \frac{1}{2} \left( \frac{e^{2x} + 1}{e^{2x} - 1} - \frac{1}{x} \right). \quad \quad \cdots (2)$

Differentiate $ (2)$ once and twice, put $ x = 1$ to obtain

$ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\zeta(2n)}{\pi^{2n}} = \frac{1}{e^2 - 1}$

$ \sum_{n=1}^{\infty} (2n-1) (-1)^{n-1} \frac{\zeta(2n)}{\pi^{2n}} = \frac{1}{2} - \frac{2e^2}{(e^2 - 1)^2}$

$ \sum_{n=1}^{\infty} (2n-1)(2n-2) (-1)^{n-1} \frac{\zeta(2n)}{\pi^{2n}} = \frac{1+e^2+7 e^4-e^6}{(e^2 - 1)^3}$

Finally, choose a suitable linear combination of these three series to obtain $ (1)$. Then we have the result.

But I think it is not a direct way to obtain the value of the given series. There may be an easier, direct way.

[quote="sos440"]
... Using the identity $ ( \minus{} 1)^{n \minus{} 1} \frac {\zeta(2n)}{\pi^{2n}} \equal{} \frac {B_{2n}}{(2n)!} 2^{2n \minus{} 1}$ ...[/quote]
I only note that this identity   :lol: can be proved without using complex analysis.

touche. although bernoulli numbers and zeta functions might be hard to find in a noncomplex textbook =P

There is a book called "Proofs from the Book" that gives famous theorems and elegant proofs.  In the chapter "Cotangent and the Herglotz Trick" they derive that zeta formula using only real analysis.  Anyone who wants an interesting read on mathematics and beautiful proofs should pick up a copy.

here is elementary proof by using  [b]Papadimitriou's[/b] idea for $ \zeta(2)$.

Except Papadimitriou's I any more did not hear about modern mathematicians of Greece  :(


The product of any two of the positive integers 30, 72, and $N$ is divisible by the third.  What is the smallest possible value of $N$?

[hide]The least is $\displaystyle 60$.

This was done using prime factorization.

$\displaystyle 30=2\cdot 3\cdot 5$
$\displaystyle 72=2^3 \cdot 3^2$
Therefore $\displaystyle N$ must be a factor of $\displaystyle 2^4 \cdot 3^3 \cdot 5$.

$\displaystyle 2^3 \cdot 3^2 | 2 \cdot 3 \cdot 5 \cdot N$
So $\displaystyle N$ is divisible by $\displaystyle 2^2 \cdot 3$

$\displaystyle 2 \cdot 3 \cdot 5| 2^3 \cdot 3 ^2 \cdot 5 \cdot N$
So $\displaystyle N$ is divisible by $\displaystyle 5$

$\displaystyle 2^2 \cdot 3 \cdot 5=60$
[/hide]


geometry parallelogram circumcircle trigonometry function

Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA \equal{} \angle QBA$.

Let $ (X) \equiv \odot(AA_1P)$ cut $ DA$ again at $ A_2$ and let $ (Y) \equiv \odot (CC_1P)$ cut $ CD$ again at $ C_2.$

$ \angle CQP \equal{} \angle PC_1B \equal{} \angle PAA_2 \equal{} \pi \minus{} \angle A_2QP$

$ \Longrightarrow$ $ C, Q, A_2$ are collinear and similarly, $ A, Q, C_2$ are collinear. From $ (X)$, $ A_1Q$ is antiparallel of $ AA_2 \parallel BC$ WRT angle $ \angle (AB, A_2C)$ $ \Longrightarrow$ $ A_1BCQ$ is cyclic, and $ A_2P$ is antiparallel of $ AA_1 \parallel DC$ WRT angle $ \angle (DA, CA_1)$ $ \Longrightarrow$ $ CDA_2P$ is also cyclic. From $ \odot (A_1BCQ), \odot(CDA_2P),$

$ \angle QBA \equal{} \angle QBA_1 \equal{} \angle QCA_1 \equal{} \angle QCP \equal{} \angle A_2CP \equal{} \angle A_2DP \equal{} \angle ADP.$

[color=darkblue]From  [url=http://www.mathlinks.ro/viewtopic.php?p=1486021#1486021][b]Virgil Nicula[/b]'s extension[/url] we have

$ \frac {\sin \angle ABQ}{\sin \angle CBQ} \equal{} \frac {AA_1}{CC_1}$

Denote by $ E$ the intersection of the lines $ AD$ and $ CP$. Then

$ \dfrac{\sin \angle ADP}{\sin \angle CDP} \equal{} \dfrac{EP}{PC} \cdot \dfrac{CD}{ED}$

But

$ \dfrac{EP}{PC} \equal{} \dfrac{EA}{CC_1}$

and

$ \dfrac{AD}{ED} \equal{} 1 \minus{} \dfrac{EA}{ED} \equal{} 1 \minus{} \dfrac{AA_1}{CD} \equal{} \dfrac{CD \minus{} AA_1}{CD}$

$ ED \equal{} \dfrac{CD \cdot AD}{CD \minus{} AA_1} \equal{} \dfrac{CD \cdot AD}{AB \minus{} AA_1} \equal{} \dfrac{CD \cdot AD}{BA_1}$

So

$ \dfrac{\sin \angle ADP}{\sin \angle CDP} \equal{} \dfrac{EA}{CC_1} \cdot \dfrac{CD}{\dfrac{CD \cdot AD}{BA_1}} \equal{} \dfrac{EA}{BC} \cdot \dfrac{BA_1}{CC_1} \equal{} \dfrac{AA_1}{BA_1} \cdot \dfrac{BA_1}{CC_1} \equal{} \frac {AA_1}{CC_1} \equal{} \frac {\sin \angle ABQ}{\sin \angle CBQ}$

As function $ f(x) \equal{} \dfrac{\sin (\alpha \minus{} x)}{\sin x}$ is strictly decreasing on $ (0,\pi)$ we get that $ \angle PDA \equal{} \angle QBA$.[/color]

Sorry to revive this old thread, But I have an alternative solution to this beautifull problem.
[hide=" My solution"]
From Miquel's theorem, we have $Q\in \odot(A_1BC)$. Then $\angle{QCB}=\angle{AA_1Q}=\angle{APQ}$. But also $\angle{PAQ}=\angle{PA_1Q}=\angle{CBQ}$ it follow that $\triangle{PAQ}\sim \triangle{CBQ}$ therefore $\frac{AP}{PQ}=\frac{BC}{QC}=\frac{AD}{QC}$, then $\triangle{PAD}\sim \triangle{PQC}$ because $\angle{PQC}=\angle{PAD}$ it follow that $\angle{PDA}=\angle{PCQ}=\angle{ABQ}$ Which end the proof.
[/hide]

Use complex numbers.  Let $Q=0$.  Note that in angle measure we have $QAP=QA_1P=QA_1C=QBC$, $QPA=QA_1A=QCC_1=QCB$ since $Q$ takes $AC_1$ to $A_1C$.  So $Q$ takes $AP$ to $BC$ and $pb=ac$.  So $p(a+c-d)=ac$, $-pd=-ap-cp+ac$, and $p(p-d)=(p-a)(p-c)$.  Thus, $P$ takes $AD$ to $QC$.  So now $PDA=PCA=A_1CQ=A_1BQ=QBA$, so we're done.

Let $APQ$ interrsect $AD$ at $X$ and let $CPQ$ intersect $CD$ at $Y$.

Observe that $\angle XAP = \angle PC_1B = \angle PQC$. Thus $X, Q, C$ are collinear. Similarly, $A, Q, Y$ are collinear.

This mean that $DAPY$ is cyclic and $BA_1QC$ is cyclic (properties of the Miquel Point). This gives us

$\angle PDA = \angle PYA = \angle PYQ = \angle PCQ = \angle A_1CQ = \angle ABQ$ as required.

Pick $A_2 \in \overline{AD}, C_2 \in \overline{CD}$ such that $AA_1A_2PQ$, $CC_1C_2PQ$ are concylic. Observe that $Q$ is the Miquel point of $PA_1BC_1$.

$\angle PQA_2 + \angle PQC = 180 - \angle PAA_2 + 180 - \angle PC_1C = 180$, and hence $A_2, Q, C$ are colinear. Similarly $C_2, A, Q$ are colinear. Thus $P$ is the Miquel point of $QC_2DA_2$.

Now we're done since $\angle QBA = \angle QC_1A = \angle PC_1Q = \angle PC_2Q = \angle AC_2Q = \angle ADP$

Let circle $AA_1P$ and $CC_1P$ meet $AD$ and $CD$ at $S,K$.
Claim $: BA_1QC$ is cyclic.
Proof $:$ Note that $\angle CQA_1 = \angle CQP + \angle PQA_1 = \angle BC_1A + \angle PAB = \angle 180 - \angle A_1BC$.
Claim $: A,Q,K$ and $C,Q,S$ are collinear.
Proof $:$ Note that $\angle PQA = \angle PA_1B = \angle PCK = \angle 180 - \angle PQK$. we prove the other one with same approach.
Claim $: CPSD$ is cyclic.
Proof $:$ Note that $\angle PSA = \angle PQA = \angle PCK = \angle PCD$.
Now Note that $\angle QBA = \angle QCA_1 = \angle QCP = \angle SCP = \angle SDP = \angle PDA$.
we're Done.

Denote by $R$ reflection of $P$ wrt center of parallelogram. In $A_1BC_1P$ Miquel point $Q$ is the isogonal conjugate of $\infty_{BR},$ since $BR$ is homothetic to it's Gauss line with center $P$ and coefficient $2,$ therefore $$\angle PDA\stackrel{\text{symmetry}}{=}\angle RBC=\angle QBA.$$

[hide = solution]
Remark that $Q$ is the centre of the spiral similarity sending $AA_1 \mapsto CC_1$. Making use of the complete quadrilateral configuration, we deduce that $Q$ also maps $AP \mapsto BC$. Therefore $QB/BC = BC/AP$. Observe that $DA = BC$ and thus we may rewrite the condition earlier as $QB/AD = QA/AP$. Moreover, observe that $\angle BQA = \angle BC_1A = \angle C_1AD = \angle DAP$ since $DA \parallel BC$. This along with the previous length condition implies that $\triangle BAQ \simeq \triangle DPA$. In particular, $\angle PDA = \angle QBA$, as desired.
[/hide]


limit
probability
expected value
probability and stats

hi kunny could you explain this differently.. im not sure i understand what X is..

if you take more than one card out shouldnt you get multiple values for X?

[quote="kunny"]There are $ n$ cards labbeled 1 to $ n$ in a box. You take one card from the box and take it back $ k$ times repeatedly.
Denote by $ X$ the number written in the card took out.

(1) For $ n \equal{} 4,\ k \equal{} 3$, find $ P(X \equal{} 2)$ and the expected value of $ X$.

(2) For $ k \equal{} 3$, express the expected value of $ X$ in terms of $ n$.

(3) Two persons $ A,\ B$ make the procedure as follows.

$ A$ conducts the procedure for $ k \equal{} 1$, $ B$ conducts the procedure $ k \equal{} a > 1$. Find $ \lim_{n\to\infty} P(X_A\leq X_B)$.[/quote]

To Moderators:

Could you correct the context of thar part ''Denote by $ X$ the number written in '' into ''Denote by $ X$ the [color=red]maximum[/color] number written [color=red]on[/color]?

Thank you for your troubles.

kunny

let $ Y_j$ be the number drawn on the jth try
$ P(Y_j\equal{}m)\equal{}\frac{1}{n}$ for each $ 1\le m \le n$

and $ Z_k\equal{}\max\{Y_1,...,Y_k\}$ so
$ p_k(m)\equal{}P(Z_k\equal{}m)\equal{}P(Z_{k\minus{}1}\equal{}m \cap Y_k\le m)\plus{}P(Z_{k\minus{}1}<m \cap Y_k\equal{}m)$
$ \equal{}p_{k\minus{}1}(m)\frac{m}{n}\plus{}\sum_{j\equal{}1}^{m\minus{}1}p_{k\minus{}1}(j)\frac{1}{n}$

with $ p_1(m)\equal{}\frac{1}{n}$ for each n, so

$ p_1(m)\equal{}\frac{1}{n}$
$ p_2(m)\equal{}\frac{m}{n^2}\plus{}\sum_{j\equal{}1}^{m\minus{}1}\frac{1}{n^2}\equal{}\frac{2m\minus{}1}{n^2}$
$ p_3(m)\equal{}\frac{m(2m\minus{}1)}{n^3}\plus{}\sum_{j\equal{}1}^{m\minus{}1} \frac{2j\minus{}1}{n^3}\equal{}\frac{3m^2\minus{}3m\plus{}1}{n^3}$

so $ p_3(2)\equal{}\frac{7}{64}$ when n=4
$ E[X]\equal{}\sum_{j\equal{}1}^4 j\cdotp_3(j)\equal{}\frac{1}{64}\sum_{j\equal{}1}^4 3j^3\minus{}3j^2\plus{}j\equal{}\frac{220}{64}\equal{}\frac{55}{16}$

for general n,
$ E[X]\equal{}\sum_{j\equal{}1}^n j\cdot p_3(j)\equal{} \frac{1}{n^3} \sum_{j\equal{}1}^n 3j^3\minus{}3j^2\plus{}j$
$ \equal{}\frac{1}{n^3}\left[ 3\frac{n^2(n\plus{}1)^2}{4}\minus{}3\frac{(2n\plus{}1)(n\plus{}1)n}{6}\plus{}\frac{n(n\plus{}1)}{2}\right]$
$ \equal{}\frac{1}{12n^3}\left[ 9(n^4\plus{}2n^3\plus{}n^2)\minus{}6(2n^3\plus{}3n^2\plus{}n)\plus{}6(n^2\plus{}n)\right]$
$ \equal{}\frac{9n^4\plus{}6n^3\minus{}3n^2}{12n^3}\equal{}\frac{1}{4n}(3n^2\plus{}2n\minus{}1)$

for part (3)

$ P(X_A \le X_B)\equal{}\sum_{m\equal{}1}^n P(X_A\equal{}m)P(m \le X_B)$
$ \equal{}\frac{1}{n}\sum_{m\equal{}1}^n P(m \le X_B)\equal{}\frac{1}{n} \sum_{m\equal{}1}^n (1\minus{}P(m>X_B))\equal{}1\minus{}\frac{1}{n}\sum_{m\equal{}1}^n \left(\frac{m\minus{}1}{n}\right)^a$
$ \equal{}1\minus{}\frac{1}{n^{a\plus{}1}}\sum_{m\equal{}0}^{n\minus{}1} m^a\to 1\minus{}\frac{1}{a\plus{}1}$

That's correct answer. I have nothing to add more than this, but

Let me say only one thing, most students those who can solve use usually use the fact:

$ \boxed{P(X\equal{}k)\equal{}P(X\leq k)\minus{}P(X\leq k\minus{}1) \ (k\geq 2)}$


Fie x,y,z numere reale strict pozitive astefel incat x+y+z=xyz. Sa se demonstreze ca: 4(xy+xz+yz) <= 9+x^2*y^2*z^2 (inegalitatea "mai mic sau egal").

Facem substituţia $a = \frac1{yz}, \, b = \frac1{zx}, \, c = \frac1{xy}$.

Condiţia devine $a+b+c=1$, iar inegalitatea de demonstrat este $4(bc+ca+ab) \leq 9abc+1$:
$ \, \Longleftrightarrow \, \, 4 \left( a+b+c \right) \sum_\textrm{cyc}bc \leq 9abc+\left( a+b+c \right)^{3}$
$ \, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( 4abc+4 b^{2}c+4 b c^{2}\right) \leq 9abc+\sum_\textrm{cyc}\left( a^{3}+3 a^{2}b+3 a b^{2}\right)+6abc$
$ \, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( b^{2}c+b c^{2}\right) \leq \sum_\textrm{cyc}a^{3}+3abc$
$ \, \Longleftrightarrow \, \, 0 \leq \sum_\textrm{cyc}a(a-b)(a-c)$, care-i Schur.

[color=darkblue][b]Putina geometrie ![/b] Pentru triunghiul $ABC$ notam : semiperimetrul $p$ , $N$- punctul lui Nagel , $C(I,r)$- cercul inscris , $C(O,R)$- cercul circumscris.

Atunci exista identitatile evidente : $\{\begin{array}{c}(p-a)+(p-b)+(p-c)=p\\\\ (p-a)(p-b)+(p-b)(p-c)+(p-a)(p-b)=r(4R+r)\\\\ (p-a)(p-b)(p-c)=pr^{2}\end{array}$

adica ecuatia $t^{3}-pt^{2}+r(4R+r)t-pr^{2}=0$ are radacinile $x=p-a$ , $y=p-b$ , $z=p-c\ .$

Se stie ca $NI^{2}=p^{2}+5r^{2}-16Rr\ .$ Asadar, $\boxed{\ p^{2}+5r^{2}\ge 16Rr\ }\ \ (*)\ .$ Acum vom transforma echivalent inegalitatea $(*)$ :

$(*)\Longleftrightarrow$ $p^{2}+9r^{2}\ge 4r(4R+r)$ $\Longleftrightarrow$ $p^{3}+9pr^{2}\ge 4p\cdot r(4R+r)$ $\Longleftrightarrow$ $\boxed{\ (x+y+z)^{3}+9xyz\ge 4(x+y+z)(xy+yz+zx)\ }\ \ (**)\ .$

Se stie ca inegalitatea $(**)$ este echivalenta cu [b][u]inegalitatea Schur[/u][/b] (asa cum a aratat si Perfect_radio) : $\boxed{\ \sum [x(x-y)(x-z)]\ge 0\ }\ \ (***)\ .$[hide]
$\{\begin{array}{c}(\sum x)^{3}=\sum x^{3}+3\prod (y+z)\\\\ \prod (y+z)=\sum [yz(y+z)]+2xyz\\\\ \sum [yz(y+z)]=\sum [x^{2}(y+z)]\end{array}$ $\Longrightarrow$ $\{\begin{array}{ccc}\boxed{\ \sum [x(x-y)(x-z)]\ }& = & \blacktriangleleft S\blacktriangleright\\\\ \sum x^{3}-\sum [x^{2}(y+z)]+3xyz & = & \blacktriangleleft C\blacktriangleright\\\\ (\sum x\right)^{3}-4\sum [yz(y+z)]-3xyz & = & \blacktriangleleft H\blacktriangleright\\\\ (\sum x)^{3}+9xyz-4\{\sum [yz(y+z)]+\sum (xyz)\}& = & \blacktriangleleft U\blacktriangleright\\\\ (\sum x)^{3}+9xyz-4\sum [yz(x+y+z)]] & = & \blacktriangleleft R\blacktriangleright\\\\ \boxed{\ (\sum x)^{3}+9xyz-4\cdot\sum x\cdot\sum (yz)\ge 0\ }& \mathrm{\ }& \mathrm{O.K.}\end{array}$

Asadar , $\boxed{\ x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)\ge 0\Longleftrightarrow (x+y+z)^{3}+9xyz\ge 4(x+y+z)(xy+yz+zx)\ }\ .$[/hide]
[b]Concluzie.[/b] Cele trei inegalitati (dintre care una geometrica) sunt echivalente. Cu aceasta ocazie am obtinut si [b][u]o demonstratie geometrica a inegalitatii Schur ![/u][/b]

[b]Observatie.[/b] Inegalitatea propusa este inegalitatea $(**)$ transformata prin conditia $x+y+z=1$ si substitutia $x: =\frac{1}{yz}$ etc. [/color]

[quote="Virgil Nicula"][color=darkblue][b]Putina geometrie ![/b] Pentru triunghiul $ABC$ notam : semiperimetrul $p$ , $N$- punctul lui Nagel , $C(I,r)$- cercul inscris , $C(O,R)$- cercul circumscris.

Atunci exista identitatile evidente : $\{\begin{array}{c}(p-a)+(p-b)+(p-c)=p\\\\ (p-a)(p-b)+(p-b)(p-c)+(p-a)(p-b)=r(4R+r)\\\\ (p-a)(p-b)(p-c)=pr^{2}\end{array}$

adica ecuatia $t^{3}-pt^{2}+r(4R+r)t-pr^{2}=0$ are radacinile $x=p-a$ , $y=p-b$ , $z=p-c\ .$

Se stie ca $NI^{2}=p^{2}+5r^{2}-16Rr\ .$ Asadar, $\boxed{\ p^{2}+5r^{2}\ge 16Rr\ }\ \ (*)$
[/quote]

superba demonstratia `geometrica` pt inegalitatea lui gerretsen  :blush:


geometry
trigonometry
trig identities
Law of Sines

Let $ a$, $ b$, and $ c$ be the side lengths of $ \triangle ABC$. Let $ R$ be the circum-radius of $ \triangle ABC$. Prove that

\[ [ABC]\equal{}\frac{abc}{4R}\]

where $ [ABC]$ denotes the area of $ \triangle ABC$.

[hide="hint"]Prove that the area is equal to the $ (ab \sin C)/2$, and then use the Extended Law of Sines to get rid of the angle.[/hide]

Since $ A\equal{}\dfrac{bh}{2}$, and $ h\equal{}b\sin C$, the area of the triangle is $ \dfrac{ab\sin C}{2}$. Also $ \dfrac{2\sin C}{c}\equal{}\dfrac{1}{4R}$.  Need some help there :dry:

[△ABC] = 1/2*sinA*bc = 1/2*sinB*ac = 1/2*sinC*ab    -------(1)

by 'sine rule'
2R= a/sinA = b/sinB = c/sinC

so you can change the (1) to [size=150][△ABC] = abc/4R [/size](because, [b]sinA is a/2R[/b])

[quote="#H34N1"]Also $ \dfrac{2\sin C}{c} \equal{} \dfrac{1}{4R}$.  Need some help there :dry:[/quote]That's not correct. It should be $ \dfrac{\sin C}{2c} \equal{} \dfrac{1}{4R}$.


geometry
perpendicular bisector
congruent triangles

Derive the equation of the locus of all points that are equidistant from points A and B if
a.  A= (-3,5) and B= (3,-5)
b.  A= (-4,-1) and B= (10,5)

HINT: [hide][How is the locus related to the line segmant AB?][/hide][/hide]

HINT Define a line that goes through $A$ and $B$ and find a perpendicular line such that the intersecting of the two lines is at the center of segment $\overline{AB}$.

[hide="Hint"]
$\cdot \text{It is a trace of a point in an equidistance from two fixed point A, B}\iff \text{A perpendicular bisector of AB}$
$\cdot z,\alpha,\beta\in\mathbb{C}\ s.t. \ |z-\alpha|=|z-\beta| \ (\alpha \not= \beta) \iff \text{Point z is a point on a perpendicular bisector of segment of a line}\ \alpha \beta$[/hide]

[hide="A solution"]
1.  Draw the line $AB$ (Construction)
2.  Draw a point $P$ in the locus (Construction)
3.  Drop a perpendicular from $P$ to $AB$ at $M$ (Construction)
4.  $AP = BP$ (Given)
5.  $MP = MP$ (Reflexive property)
6.  $\Delta AMP \cong \Delta BMP$ ($HL$ from steps 4 and 5)
7.  $AM = MB$ (Corresponding parts of congruent triangles are congruent)
8.  $P$ lies on the perpendicular bisector of $AB$ (Steps 3 and 7)
[/hide]

would the solution be 3/5x=y for a
and y=3/7x+5/7 for b
??


Which number equals the sum of all numbers [i]a [/i]between 0 and 2 :pi:   that satisfy :     tan [i]a [/i]- 2003tan [i]a [/i]+ 1 = 0

It's zero
[hide]
Consider for what tan(a) it holds and then consider graph of function tan(x)
[/hide]

[quote="Silverfalcon"]I know this is offtopic but where I can see the Flanders Math Olympiad problems? I don't think the site offers it.[/quote]Seach for Flanders on the forum, and you will find some of the problems proposed in the past years.


geometry
3D geometry
tetrahedron
counting
distinguishability
rotation
symmetry

Each face of a regular tetrahedron is painted either red, white or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 54 \qquad \textbf{(E)}\ 81$

We can do casework.
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If the tetrahedron is all one color, there are $3$ ways to do this.

If the tetrahedron has three faces of one color and one of another, there are $3*2=6$ ways to do this.

If the tetrahedron is two and two, there are $3$ ways to do this that are distinguishable.

If the tetrahedron is two, one, and one, there are $3$ ways to do this because many end up being reflections.  Thus there are only three distinguishable types, and you get to choose only what color is used for two faces.

The total is $15$, or $A$.[/hide]

[quote="Allan Z"]#red faces=r, #blue faces=b, #white faces=w

r+b+w=4,
there are $6C2=15$ different tetrahedrons  :D[/quote]

Wow, why didn't I think of that?   :D  Nice solution. Lol, I love how balls and urns can be applied to almost ANYTHING.   :)

(Oh, and you can use [code] \dbinom{6}{2} [/code] to get it to show up as $\dbinom{6}{2}$ if you want.)

[quote="Allan Z"]#red faces=r, #blue faces=b, #white faces=w

r+b+w=4,
there are $6C2=15$ different tetrahedrons  :D[/quote]

hah, great.

[quote="iamagenius"][quote="Allan Z"]#red faces=r, #blue faces=b, #white faces=w

r+b+w=4,
there are $6C2=15$ different tetrahedrons  :D[/quote]
are you sure that's valid?[/quote]

I think it is, because of all the symmetry and rotations. And on an AMC, nothing needs to be rigorous, as long as it works (quickly).  :)

[quote="iamagenius"][quote="Allan Z"]#red faces=r, #blue faces=b, #white faces=w

r+b+w=4,
there are $6C2=15$ different tetrahedrons  :D[/quote]
are you sure that's valid?[/quote]

Yes, I'm pretty sure the rotations make the individual triangle sides indistinguishable.

Not quite: the symmetry group of a tetrahedron up to rotations has 12 elements, while the symmetric group on four elements has 24 elements.  (For those of you who don't speak group theory, this just means that there are 12 different ways you can rotate the four faces of the tetrahedron, but 24 permutations of four objects.  This means that there are some permutations of the faces that aren't achieved only by rotations.  In fact, you can get these extra arrangments if you also allow reflection.)  However, it takes 4 colors of paint (one for each side) in order for you to tell a tetrahedron apart from its reflection.  With only three colors, each tetrahedron looks the same as its reflection and so this calculation is valid.  I wonder if this solution occured to the problem writers, and whether they thought about what I just said.

[quote="JBL"]With only three colors, each tetrahedron looks the same as its reflection and so this calculation is valid.[/quote]

Pity I only realized this immediately after I'd finished all my casework.   :oops:

yes, it worked because its just a tetrahedron, with 4 faces. So it's just basically asking how many nonnegative integer set (r,b,w) to make r+b+w=4.. If you have the same (r, b, w), you can rotate them and see that they are indistinguisable

[quote=iamagenius][quote="Allan Z"]#red faces=r, #blue faces=b, #white faces=w

r+b+w=4,
there are $6C2=15$ different tetrahedrons  :D[/quote]
are you sure that's valid?[/quote]

If it isn't, then the answer is at most $15$. Seeing that there are no other answer choices less than $15$, it's the answer.