geometry rhombus perimeter ratio probability counting distinguishability

Greeting,

I have revived(the way I put it) my SC from long time.  The rules are changed a lot. Questions are harder and it's 18 multiple choice questions with each 5 points and one proof question worth 10 points.

All answers must be submitted to me by p.m. in two days which is 8/9.
Calculators are able except you should not use root finder for this test.

Here are the questions: Good luck!

1. The one of diagonals of a rhombus have lengths 48 and the perimeter of the rhombus is 100.  Then the length of other diagonal is:

A. 13
B. 14
C. 15
D. 16
E. None of them

2. Find what a^3 + b^3 equals if:
a/2 + b = 7
a + b/2 = 8
a/5 = 7/b
Thus, a^3 + b^3 is:

A. 100
B. -100
C. 50
D. -50
E. None of them

3. The perimeter of a small Florida-shaped pizza is 24 cm and the perimeter of a big Florida-shaped pizza is 36 cm.  If they are two similar figures, find the ratio of area from small to big.

A. 4/9
B. sqrt2/5
C. 3/7
D. sqrt3/9
E. None of them

4. What kind of triangle has sides of 5,7, and 11?

A. obtuse triangle
B. acute triangle
C. right triangle
D. equiangular triangle
E. None of them

5. A dart is thrown at a board 20 m long and 3 m wide.  There are 20 balloons attached to the board, each with radius 25 cm.  Knowing each balloons lie entirely on the board, find the probability that a dart that hit the board will also hit the board.

A. pi/50
B. 3pi/50
C. pi/100
D. 3pi/100
E. None of them

6. Calculate for b if a^6 - b^6 = 12746944 and a > b > or equal to 10 with square root of a - b = sqrt2

A. 13
B. 14
C. 15
D. 16
E. None of them

7. In how many distinguishable ways can the letters in word acacia can be written?

A. 20
B. 30
C. 40
D. 50
E. None of them

8. How many numbers are there that satisfy the following three conditions?
A) No digit contains digit 2
B) Greater than 100, less than 1000
C) Such that in abc where a,b,and c each represent different number a > b > c.

A. 52
B. 56
C. 60
D. 64
E. None of them

9. Three points that are not on a line determine three points.  How many lines are determined by ten points, no three of which are on a line?

A. 42
B. 45
C. 70
D. 100
E. None of them

10. The product of three consecutive multiple of 2 is 26880.  Then whats the sum of the three numbers?

A. 84
B. 90
C. 92
D. 94
E. None of them

11. The triangle ABC is scalene in which the length of sides are 5,9, and k.  If altitude is 8, what is the sum of largest and smallest possible area? (round to whole number)

A. 26
B. 28
C. 30
D. 32
E. None of them

12. Find the distance between points (1,-2,7) and (6,5,-9).

A. 18
B. sqrt330
C. 19
D. sqrt411

13. There are eight bridges that connects city D,E,F, and G.  The bridges are paired like this: 1 from D to F, 2 from F to E, 2 from E to G, 2 from G to D, and 1 from D to E.  If you must begin and return to the same point, which vertex should you start on? (You must cross all bridges)

A. D
B. E
C. F
D. G
E. Impossible

14. (a + b)^7  s second term and third term is?

A. 7a^6b^1 + 21a^5b^2
B. 13a^6+b^2 + 21a^5b^1
C. 21a^6b^2 + 7a^5b^1
D. 13a^6b^1 + 7a^5b^2
E. None of them

15. The lengths of the sides of triangle RST are ST = 12, TR = 13, and RS = 14.  If J is the midpoint of TR, and K is the point where TR is cut by the bisector of angle B, find JK.

A. 0.5
B. 1
C. 1.5
D. sqrt2
E. None of them

16. The triangle ABC is a right triangle with angle B of a right triangle and AC as hypotenuse in length of 10.  An altitude from right angle B is met with AC at P, resulting CP = 3 and PA = 7.  If altitude BP has length of h and other two legs of right triangle ABC has length of a and b, calculate a*b*h.

A. 120
B. 150
C. 180
D. 210
E. None of them

17. F (x) = a^5 - a^4 - 8a^3 - 24a^2 - 153a - 135 s one of solutions is 3i.  Find the sum of other real solutions.

A. 1
B. 2
C. 3
D. 4
E. None of them

18. Find the area of a regular hexagon with apothem, a line thats perpendicularly distanced to the center, 12.

A. 264 sqrt3
B. 272 sqrt3
C. 280 sqrt3
D. 288 sqrt3
E. None of them

* Proof *
Prove that a^5 + a^4 is divisible by any 2a such a > 0.

I don't get #2- why are there three equations and two variables- when you solve for a and b they don't work ini all of the equations...

For #5, shouldn't it be "find the probability that if a dart hits the board, it will also hit a balloon"?

Shouldn't  #9 read: three non-collinear points determine 3 lines. How many do 10 non-collinear points determine?

For #15, shouldn't it be the bisector of S?

Rep123max wrote:Ya, that last thing is useless/doesn't work with what a and b are.

Well...the way the problem was designed (poorly) it did have a use.  Sorry to ruin the problem, but it's no good to begin with:

[hide]a+2b=14, b+2a=16, a+b=10, a^2+2ab+b^2=100

ab=35

a^2+b^2=30

(a+b)(a^2+b^2-ab)=a^3+b^3=(10)*(30-35)=-50.[/hide]

Too bad that doesn't really work with real values of a and b.

For #18, just to clarify, an apothem is the segment with endpoints on the center and a side so that it is perpendicualr to that side.

Ya, I saw the way you were "supposed" to do it, but then I checked and it didn't work.. oh well...


inequalities
calculus
calculus computations

Two metrics $ d$ and $ \rho$ on a set $ M$ are [i]equivalent[/i] if they generate the same convergent sequences; that is, $ d(x_{n},x)\rightarrow 0$ iff $ \rho(x_{n},x)\rightarrow 0$.

Show that the metrics induced by $ \left\|\cdot\right\|_{1}$, $ \left\|\cdot\right\|_{2}$, and $ \left\|\cdot\right\|_{\infty}$ on $ \mathbb{R}^{n}$ are equivalent.

I know that $ \left\|x\right\|_{\infty}\leq\left\|x\right\|_{2}\leq\left\|x\right\|_{1}$, and that $ \left\|x\right\|_{1}\leq n\left\|x\right\|_{\infty}$ and $ \left\|x\right\|_{1}\leq\sqrt{n}\left\|x\right\|_{2}$.  I've already proved these.  I'm just not seeing how to connect them to the current problem.

Rearrange your inequalities a bit into this slightly more illuminating form:
\[ \frac {1}{\sqrt {n}}\left \Vert x \right \Vert _1\leq \left \Vert x \right \Vert_2 \leq \left \Vert x \right \Vert _1 \\
 \\
\frac {1}{n}\left \Vert x \right \Vert _{2} \leq \left \Vert x \right \Vert _{\infty} \leq \left \Vert x \right \Vert _{2} \\
 \\
\left \Vert x \right \Vert _{\infty} \leq \left \Vert x \right \Vert _1 \leq n \left \Vert x \right \Vert _{\infty}
\]
Now if $ \{a_k\}$ is a sequence of real numbers that converges to $ 0$, what can you say about the convergence of the sequence $ \{b_k\}$ if there are constants $ m,M$ such that $ m a_k \leq b_k \leq M a_k$ for all $ k$?

Use this reasoning to show that the first inequality proves $ (\text{convergence in }p \equal{} 1)\Rightarrow (\text{convergence in }p \equal{} 2)$, the second proves $ (\text{convergence in }p \equal{} 2)\Rightarrow (\text{convergence in }p \equal{} \infty)$, and the third proves $ (\text{convergence in }p \equal{} \infty )\Rightarrow (\text{convergence in }p \equal{} 1)$, so all three are equivalent.

Let me see if I've got this: From what you wrote, since $ \{a_{k}\}\rightarrow 0$, then $ c\{a_{k}\}\rightarrow 0$ as well for some constant $ c$.  So if $ ca_{k}\leq b_{k}\leq Ca_{k}$, then $ b_{k}\rightarrow 0$ as well since it is bounded by two sequences that converge to $ 0$.

Then from the first inequality, $ \left\|x\right\|_{2}$ converges to whatever $ \left\|x\right\|_{1}$ converges to, so they are equal.  From the second, $ \left\|x\right\|_{\infty}$ converges to whatever $ \left\|x\right\|_{2}$ converges to so they are equal, and from the third, $ \left\|x\right\|_{1}$ converges to whatever $ \left\|x\right\|_{\infty}$ converges to, so they are equal.  Hence all three are equal.


I'm asked to find some images or films about reaction between Phosphorus and other substances, such as $ Ca$, $ O_{2}$, $ Cl_{2}$, $ KClO_{3}$, $ KNO_{3}$, $ K_{2}Cr_{2}O_{7}$, condensed $ HNO_{3}$, and also, the preparation of $ P$ from $ Ca_{3}(PO_{4})_{2}$.
Can anyone post them here? Or give me link to download them?
Thank you very much.

For a start, what are the reactions that might happen between phosphorus and those substances?

Regarding the images/films, have you tried a google search?

I tried to google them, but the quality is very bad, not good enough for my presentation in my class (I need them to put in my PowerPoint)
Here are the reactions
$ 2P\plus{}3Ca\longrightarrow^{t^{o}}Ca_{3}P^{2}$
$ 4P\plus{}5O_{2}\longrightarrow2P_{2}O_{5}$
$ 2P\plus{}5Cl_{2}\longrightarrow2PCl_{5}$
$ 6P\plus{}5KClO_{3}\longrightarrow^{t^{o}}3P_{2}O_{5}\plus{}5KCl$
and so on ...
$ Ca_{3}(PO_{4})_{2}\plus{}3SiO_{2}\plus{}5C\longrightarrow^{t^{o}}3CaSiO_{3}\plus{}2P\plus{}5CO$

Thanks again for any help.

The third and fifth reactions you have written are wrong. You have also to specify in the reactions if you are using (or not) white phosphorus or other allotrope: because if you are using the white allotrope then all your reactions are wrong.

Check also [url=http://boyles.sdsmt.edu/pwithcl/reaction_of_white_phosphorus_and.htm]this site[/url].

Huh? What's wrong with my reactions? You mean I have to right $ P_{4}$ instead of $ P$ only?
Btw, thank you for your link.  :)

Yes, normally white phosphorus is used, so $ P_{4}$ should be written.


this is my first challenge (even though it is easy)

a theater has infinity seats. in the first row there is one seat. in the second row there is 3 seats and on the third there are there are 5 seats. if the seating order coninues like this, what seat number is behind [b]seat[/b] #144

[quote="Programmer_Xia"]the anwser is seat #168[/quote]
Try to show your work when answering a problem and welcome to AoPS. :)

Try not to post problem solving problems like this in the Classroom Math forum.  If a problem takes some ingenuity beyond middle school curriculum, it belongs in a different forum.

it is really 168 and you [b]have[/b] to draw a picture. If you do you can notice the numbers on the left gining up all are "that row number"^2 so you know that 144 is 12^2 and the number on the right is 13^2=169. Then go back one and you get 168


geometry
circumcircle
parallelogram
symmetry
perpendicular bisector
geometry solved

Given an isosceles triangle $ABC$ (with $AB=BC$). A point $X$ is chosen on a side $AC$. Some circle passes through $X$, touches the side $AC$ and intersects the circumcircle of triangle $ABC$ in points $M$ and $N$ such that the segment $MN$ bisects $BX$ and intersects sides $AB$ and $BC$ in points $P$ and $Q$.  Prove that the circumcircle of triangle $PBQ$ passes through the circumcentre of triangle $ABC$.

[b]proposed by Sergej Berlov[/b]

Let's look at it somewhat backwards.

If $X$ is fixed, there is clearly only one circle satisfying the conditions, so let's show that if $BX$ bisects the chord $MN$ of the circumcircle of $ABC$, then $M,N$ are the points we are looking for.

$BNXM$ is a parallelogram, so the entire figure is symmetric wrt the midpoint of $BX$. Since the parallel through $B$ to $AC$ is tangent to $(ABC)$, it means that $AC$ is tangent to $(MXN)$, so $M,N$ are the points mentioned in the problem.

We have shown that, if $T$ is the midpt of $BX$, then $P,Q$ are the intersections of $BA,BC$ with the perpendicular through $T$ to $OT$ (call this property $(*)$). On the other hand, if we take $P'\in BA,Q'\in BC$ s.t. $XP'\|BC,XQ'\|BA$, then it's easy to see that they satisfy $(*)$ and $\angle P'OQ'=\pi-\angle ABC$, so $P=P',Q=Q'$ and $\angle POQ=\pi-\angle ABC$, which is what we wanted.

I have just solved a generalized version of this problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=20669 post #4. In fact, in that post, I proved the following:

[color=blue]Construct the circle passing through the vertices A and B of triangle ABC and touching the side BC at the point B, and construct the circle passing through the vertices B and C and touching the side AB at the point B. These two circles intersect at B; let their second point of intersection be called F. Then, the circumcircle of triangle PBQ always passes through the point F.[/color]

This actually holds for arbitrary triangles ABC. Now, for isosceles triangles ABC such that AB = BC, the point F can be characterized in a simpler way: It is the circumcenter of triangle ABC. In order to prove this, note that, at first, the point F was defined in a way symmetric with respect to the vertices C and A; thus, since the triangle ABC is isosceles with base CA, the point F must lie on the symmetry axis of this isosceles triangle, i. e. on the perpendicular bisector of the segment CA. Hence, FC = FA. Now, as I showed in the above-mentioned post, the triangles FAB and FBC are directly similar; hence, FA : FB = FB : FC, so that $FA\cdot FC=FB^2$. Since FC = FA, this becomes $FA\cdot FA=FB^2$, so that $FA^2=FB^2$ and thus FA = FB. Combining this with FC = FA, we obtain FA = FB = FC, and thus the point F is the circumcenter of triangle ABC. So it follows that the circumcircle of triangle PBQ always passes through the circumcenter of triangle ABC. And the problem is solved.

[quote="nttu"]Can you please explain to me why : "If $ X $ is fixed, there is clearly only one circle satisfying the conditions " ?[/quote]

Well, I don't understand why Grobber finds this so obvious, but it follows from one of my observations in http://www.mathlinks.ro/Forum/viewtopic.php?t=20669 post #4, actually from the following one:

[quote="darij grinberg"]First, it's time to involve the assumption that the radical axis of the circle k [b][note: this is our circle through the point X touching the side AC][/b] and the circumcircle of triangle ABC bisects the segment BX. In other words, if M is the midpoint of the segment BX, then this point M has equal powers with respect to the circle k and to the circumcircle of triangle ABC. Again, let's compute these powers: If the line BX meets the circle k at a point U (apart from X) and the circumcircle of triangle ABC at a point T (apart from B), then the power of the point M with respect to the circle k is $MX\cdot MU$, and the power of the point M with respect to the circumcircle of triangle ABC is $MB\cdot MT$. Since the two powers of M are equal, we thus have $MX\cdot MU=MB\cdot MT$. In other words, $MX\cdot MU=BM\cdot TM$. But since M is the midpoint of the segment BX, we have MX = BM, and thus MU = TM. Hence, BU = BM + MU = MX + TM = TX,[/quote]

In fact, this yields an easy construction of the circle k: Construct the point T as the point where the line BX meets the circumcircle of triangle ABC (apart from B), and construct the point U as the point on the line BX such that BU = TX (there is exactly one such point U, since we are working with directed segments). Then, the circle k must pass through the points X and U and touch the line AC at the point X; this is already enough for uniquely determining the circle k (you can construct its center as the point where the perpendicular to AC at X meets the perpendicular bisector of the segment XU). So the circle k is unique.

  Darij


What is the sum of the integers x that satisfy $-5\leq\frac{x}{\pi}\leq10$?

[hide]
Multiply everything by $\pi$. Since $x \geq -5\pi$, $x \geq -15$, since $5\pi < 16$ and $x$ is integral.

From the second part of the inequality, $x \leq  10\pi$, so $x \leq  31.4$, and since $x$ has to be integral, $x \leq  31$.

The numbers from $-1$ to $-15$ cancel the numbers from $1$ to $15$, so we are left with summing $16$ through $31$, which is $376$.[/hide]


If $ m(E) > 0$ then $ E \minus{} E \equal{} \{x \minus{} y: x,y\in E\}$ contains an interval centered around 0. The only other result that I know that could possibly be useful is that for any $ E$ such that $ m(E) > 0$ we have an interval $ I$ such that $ m(E\cap I) > \alpha m(I)$. I don't really see how this is going to be useful in proving the statement but here's what I have so far. If $ E \minus{} E$ doesn't contain an interval around 0 then for any neighborhood $ N$ of 0 there exists an $ a\in N$ such that $ a\not\in E \minus{} E$ which is equivalent to saying that $ E \plus{} a$ and $ E$ are disjoint. Since they are disjoint so are $ E\cap I \plus{} a$ and $ E\cap I$ where $ I$ is an interval such that $ m(E\cap I) > \alpha m(I)$. This means $ m(E\cap I \plus{} a\bigcup E\cap I) \equal{} m(E\cap I) \plus{} m(E\cap I) \equal{} 2m(E\cap I)$ which by assumption leads to $ 2m(E\cap I) > 2\alpha m(I) > m(I)$ if $ \alpha > \dfrac{3}{4}$. This is where I get stuck since I don't see where I can take it from here. Forgot to mention that $ \alpha<1$ otherwise the statement that $ m(E\cap I)>\alpha m(I)$ doesn't make sense.

Never mind. Apparently this is a theorem and the proof is at http://en.wikipedia.org/wiki/Steinhaus_theorem.

It has also been posted on this forum- see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113431]here[/url].


vector
inequalities
geometry proposed
geometry

Here are three brothers-problems (they are solved by the very same method):
1. Let $m_1,\cdots,m_n$ be positive reals and points $A_1,\cdots,A_n$
Let $M_0$ be a point s,t, $\sum_{i=1}^n m_i \overrightarrow{e_i}=\overrightarrow{0}$
where $\overrightarrow{e_i}=\frac{\overrightarrow{M_0A_i}}{M_0A_i}$
Show the inequality
$\sum_{i=1}^n m_iMA_i\geq \sum_{i=1}^n m_i M_0A_i$
where $M$ is an arbitrary point.
2. Given are positive reals $m_1,\cdots,m_n$ and points $A_1,\cdots,A_n$ with $\sum_{i=1}^{n-1} m_i\overrightarrow{e_i}|\leq m_n$
where $\overrightarrow{e_i}\frac{\overrightarrow{M_0A_i}}{M_0A_i}$
Show that $\sum_{i=1}^n m_i MA_i\geq \sum_{i=1}^{n-1} m_i A_nA_i$
where $M$ is an arbitrary point.
3. Given are positive reals $m_1,\cdots,m_n$ and convex polygon $B_1\cdots B_n$ inscribed in convex polygon $A_1\cdots A_n$ ($B_i \in [A_iA_{i+1}]$, $A_{n+1}=A_1$) and
$\overrightarrow{A_iA_{i+1}}(m_{i-1}\overrightarrow{e_{i-1}}-m_i\overrightarrow{e_i})=\overrightarrow{0}$
($A_{-1}=A_n$).
where $\overrightarrow{e_i}=\frac {\overrightarrow{B_iB_{i+1}}}{B_iB_{i+1}}$, $B_{n+1}=B_1$.
Show the inequality $\sum_{i=1}^n m_i C_iC_{i+1}\geq \sum_{i=1}^n m_i B_iB_{i+1}$ where
$C_1\cdots C_n$ is convex polygon inscribed in $A_1\cdots A_n$ the same way as $B_1\cdots B_n$
(i.e. $B_i \in [A_iA_{i+1}]$) and $C_{n+1}=C_1$.

Solution for the first problem:

$\displaystyle MA_i*M_0A_i\geq\overrightarrow{MA_i}*\overrightarrow{M_0A_i}$

$\displaystyle MA_i*M_0A_i\geq (\overrightarrow{MM_0}+\overrightarrow{M_0A_i})*\overrightarrow{M_0A_i}$

$\displaystyle MA_i-M_0A_i \geq\frac{\overrightarrow{MM_0}*\overrightarrow{M_0A_i}}{M_0A_i}$

Summing all terms multiplied $m_i$ times:
$\displaystyle \sum m_i(MA_i-M_0A_i)\geq 0$


Sin(x) = 1/7
Sin(y) = 7/8

What is sin (x+y)

Do I use one of the sum and difference formulas?

If so, which one?

[hide="hint"]
just use the formula $\sin(x+y)=\sin x\cos y+\sin y\cos x$

$\cos(x)$ and $\cos(y)$ can be evaluated using $sin^{2}\theta+\cos^{2}\theta=1$
[/hide]

Wouldn't there be four different values then?

yes, i suppose there would be, due to the fact that there are multiple values of x such that sin(x)=1/7, etc.


$ If a, b, c \ge 0such that abc \equal{} 1, then
(a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 7\ge 5(a \plus{} b \plus{} c).$


Hi everyone !
  I have a question. Do they have a stable number of participants in the USAMO every year ? Is that number around 250 or 160 ? I used to think it is 250, according to the USAMO website.

It is approximately 250. Selection process for this is in another thread.


email
search
real analysis
real analysis theorems

Kani or Kain, I don't remember your name. I'm so sorry. I deleted your email you sent to me. If you or every one who has this book "Problems in Mathematical Analysis I, II, III" of Kaczor and Nowak, please send to my mailbox: antueminh@yahoo.com  or gay_chay_so.
Thanks a lot. :)

You can search the member list (see the button above) for the name of your correspondent and contact them directly.
Whatever you do, [u]don't double-post[/u].


function
inequalities
LaTeX
complex numbers
complex analysis
complex analysis unsolved

Hey there, I realy need some guidance in the following questions:

1. Let f,g be entire functions and there exists a real constant M such as
 Re(f(z)) <= M * Re(g(z)) for every z in C.
Prove that there exist complex numbers a,b such as f(z) = a*g(z) +b for every z in C.

2. Let f(z) be analytic at the open unit circle and |f ' (z) | <= 1/ (1 - |z| )for every |z| < 1.
Prove that the coefficients in that taylor series f(z) = Sigma_ an*zn are : |an| < e for every n>=1.

About 1- If we'll define h(z) = f(z)/g(z) we'll get a constant function by Liouville's theorem. But I can't understand how to get the "b" in the expression...

About 2-I'm pretty sure we need to use Cauchy's Inequality but I can't figure out how...

I will be delighted to get some guidance around here...

Thanks a lot

In 1, why in the world do you think $ f/g$ is entire? You have an inequality on the real parts. Since $ Re f$ determines the size of $ e^f$, try exponentiating.

2. is a cute problem. Noting $ f^{(n)}(0) \equal{} (f')^{(n\minus{}1)}(0)$, apply Cauchy's estimates to $ f'$ on the circle of radius $ r$ centered at $ 0$, and obtain a simple estimate. Now choose the right $ r$ for each $ n$.

[quote="WannaBE"]

I will be delighted to get some guidance around here...

Thanks a lot[/quote]

We'll be delighted if you take the time to learn a little LaTex.

Hey there...
f/g isn't necessarily entire... I was thinking about taking f-Mg...
It's entire and we know that Re(f(z)) - M*Re(g(z)) <= 0  so by Liouville's theorem we know that there exists constant b such as: f(z)-Mg(z)=b. Hence, we can define M=a and then we'll get f(z)=ag(z)+b... Am I right in this one?

About the second one... I'll think about your guidance and if I'll have any further questions I'll ask here...

Hope you'll be able to verify my response...

Thanks a lot

Where can I learn some Latex?

And about the second one, if I'll take r= n / (n+1) it'll work, isn't it?

I'll be delighted if you'll be able to verify my two answers...I'm pretty sceptic about the first one because a,b are suppose to be complex numbers and not real ones...

Thanks a lot


algebra
function
domain
calculus
integration
inequalities
superior algebra

Let $ A$ be a normal domain, $ A \to B$ is an integral extension, $ B$ is torsion-free over $ A$. Let $ P$ be an ideal of $ B$ with $ p\equal{}P\cap A$. Show that the height(P)=height(p).

I can show using the going down theorem that height(p) $ \leq$ height(P). I don't know how to show the other inequality

I think only the condition integral extension is needed. If we have a chain of primes in $ B$, restrict each of them to $ A$. We are going to show that none of these restrictions coincide. This is because

Claim: If $ A \rightarrow B$ is integral, then for any $ p \subset A$ prime, its fiber (all the primes of $ B$ that restricts to $ p$) is of dimension 0. There is no containing relation.

Proof: $ A_p \rightarrow B_p$ is still integral. Moreover, $ pA_p$ is maximal, and we know that all the primes that restricts to $ A$ would be exactly the primes in $ B_p$ that restricts to $ pA_p$. By integrality, we know that only maximal ideals restrict to $ pA_p$, so there can't be any containing relation.


Ok, so I know everyone's going to laugh at this question, but I was just wondering...

Does anyone have any advice as to how to memorize the periodic table of the elements?

Er... don't?

The way I memorized the first few periods is by pronouncing the abbreviations out loud in chunks.  Here:

HHe
LiBeBCNOFNe
NaMgAlSiPSClAr
KCaScTiVCrMnFe CoNiCuZn GaGeAsSeBrKr
RbSrYZr NbMoCtRu RhPdAgCdIn SnSbTeIXe

I find that doing them in chunks helps. You can't expect to memorize it in one week but you can do 20 elements at a time. I did it two summers ago...but then, I've forgotten all of the elements past 50. =/

Why would you want to?

I've been told that Albert, when being teased for not knowing his own phone number said, why would I spend time learning something I can look up when I need it?


function
real analysis
real analysis unsolved

Prove that there doesn't exist any function $ f: (0,\infty)\rightarrow(0,\infty)$ such that $ f^2(x)\geq f(x\plus{}y)(f(x)\plus{}y)$.

Notes:
- no continuity of any sort is given
- obviously if it would exist, $ f$ should be decreasing

Pretty thanks and enjoy


Playing soccer with 3 goes as follows: 2 field players try to make a goal past the goalkeeper, the one who makes the goal stands goalman for next game, etc.

Arne, Bart and Cauchy played this game. Later, they tell their math teacher that A stood 12 times on the field, B 21 times on the field, C 8 times in the goal. Their teacher knows who made the 6th goal.

Who made it?

My attempt...

[hide]

Let's let A and C be represented by D

Clearly, B cannot stand in the goal two straight times. B also stood in the goal one time more than D. Therefore, B must have started in goal and ended in goal as such...

BDBDBDBDBDBDBDB....etc.... B

If B started in goal, D scored the first goal, B scored the second, etc... meaning D (A or C) scored all the odd numbered goals and B scored all the even numbered goals...

Therefore, B must've scored the 6th goal

[/hide]

eum.. yes.. but how do you proof there's nowhere ...BDDB... in the list? strings ..BCAB.. and ..BACB.. are not contradicting..

So you're still missing something...

Hey lucky you were almost there, why don't you continue? You've found the hardest part yet, but you miss the easy link to make it consistent :)

I didn't know I am that good at soccer :D

We know that A and C were in the goal 21 times. C was there 8 times, so A was goalie 13 times. Similar reasoning, B was goalie 4 times. So in total there were 25 rounds played, and C was on the field 17 times.

The only way for A to be the goalie 13 times in 25 games is if he starts out being goalie, and then is goalie as often as possible. This would mean he starts out being the goalie, then is it the third round, the fifth, the seventh.... The player who was goalie the seventh round was the one who made the sixth goal. So A made the sixth goal.


Hi, I'm new here people . There's a problem that i'd like to solve but i can't.

There are 27 numbers. You can use three times number 1, three times number 2, three times number 3, ... , three times number 8 and three times number 9.
When you use number 1 you must "jump" one space and use number one again. Ex: 1 (other number) 1 (other number) 1
When you use number 2 you must "jump" two spaces and use number two again. Ex: 2 (other number) (other number) 2 (other number) (other number) 2
With number 3 three spaces, number 4 four spaces ... number 8 eight spaces and number 9 nine spaces.
Now try to organize the 27 numbers :D, there's more than one solution.

There are 27 numbers _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .
You can use the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9, three times each one.
When you use number 1, you must skip one space and then put number one again and skip other space and put number 1 again. Exemple 1 _ 1 _ 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .
With number 2 you must skip to spaces then put another number 2 skip more two spaces and put another number two. Exemple 2 _ _ 2 _ _ 2 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .
With number 3 you must skip three spaces put aother number 3, skip  3 spaces and put another number 3.
Exemple : 3 _ _ _ 3 _ _ _ 3 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
...
With Number 9 you must skip nine spaces the put another number 9, skip nine spaces and put another number 9.
Exemple : 9 _ _ _ _ _ _ _ _ _ 9 _ _ _ _ _ _ _ _ _ 9 _ _ _ _ _ _ .

I think he meant the following:

Find all permutations $(a_1, a_2, ..., a_{27})$ of $(1, 1, 1, 2, 2, 2, ..., 9, 9, 9)$ such that $a_i = a_{i - a_i - 1}$ or $a_i = a_{i + a_i + 1}$ (if existent) for all $1 \leq i \leq 27$.


[hide="answer"]Not really sure how to explain how I got this...

$ 2x^{4}\plus{}3x^{3}\plus{}x^{2}\plus{}x\plus{}4\equal{}(2x^{2}\plus{}5x\plus{}4)(x^{2}\minus{}x\plus{}1)$

I just realized that it had to be of the form $ (2x^{2}\plus{}a_{1}x\plus{}b_{1}x)(x^{2}\plus{}a_{2}x\plus{}b_{2}x)$, and $ b_{1}\in\{1,2,4\}$, $ b_{2}\equal{}\frac{4}{b_{1}}$and you can very easily limit the possibilities even further.[/hide]


How many positive odd integers with middle digit 5 and no digit repeated are there between 10,000 and 69,999?

For certain integers $ n$, the expression $ n^2 \minus{} 3n \minus{} 126$ is a perfect square. What is the sum of all distinct possible values of n?

SINCE ALL NUMBERS ARE 5 DIGIT,
AND CENTER DIGIT 5 IS CONSTANT,
LET US ASSUME 4 CASES:HAVING LAST DIGIT 1,3,7,9
 [b] CASE1[/b]
FOR 1 AS LAST DIGIT:
EXCLUDING 5, 0, 1 THERE ARE 7 COMBINATIONS FOR FIRST DIGIT
FOR SECOND ALSO THERE ARE 7 DIGITS AS 0 CAN BE USED
FOR FOURTH, 6
  THEREFORE. COMBINATION=7*7*6
COMBINING ALL 4 CASE WE GET,
4(7*7*6)
=24*49
=1176
  EXCLUDING 69,999 THE ANS. IS 1175 :wink:

[quote="Poincare"]STOP SHOUTING  :P
the last digit has to be 1,3,5,7,9 because the number has to be odd.
Case 1(last digit 1 because the number has to be odd):
you have 10 choices for the first digit
you have 5 choices for the middle digit
you have 8 choices for the fouth digit
and only one choice for the last digit
so the answer for this case is $ 10*5*8$
all the next cases get the same answer so the final answer would be $ 10*5*8*5$ which is $ 2000$.[/quote]
?

The middle digit has to be 5 and the first digit can only be 1,2,3,4,6

thanks chemalete for pointing out

SINCE ALL NUMBERS ARE 5 DIGIT,
AND CENTER DIGIT 5 IS CONSTANT,
LET US ASSUME 4 CASES:HAVING LAST DIGIT 1,3,7,9
CASE1
FOR 1 AS LAST DIGIT:
EXCLUDING 5, 0, 1 and 7, 8, 9 THERE ARE 4 COMBINATIONS FOR FIRST DIGIT
FOR SECOND THERE ARE 7 DIGITS AS 0 CAN BE USED
FOR FOURTH, 6
THEREFORE. COMBINATION=4*7*6
 same as for 3
but for 7 and 9,
there are 5 choices for first digit.
COMBINING ALL 4 CASE WE GET,
4*7*6+4*7*6+5*7*6+5*7*6
=42*18
=756
EXCLUDING 69,999 THE ANS. IS 755

Why do you exclude 69,999 when you didn't count it? I think the answer is just 756.

[hide="thorough solution"]The middle digit is $ 5$. The last digit can be either $ 0, 2, 4, 6, 8$ and the first digit can be any number $ 1\minus{}6$, excluding 5.
2 Cases-
1) last digit $ (1,3)$
There are 2 ways to choose the last digit, 4 ways for the first digit, and (7)(6) ways to choose the middle two $ \implies (2)(4)(42) \implies 336$ ways.
2) last digit $ (7,9)$
2 ways for the last digit, 5 for the first, and 42 for the middle two digit for a total of $ (10)(42) \equal{} 420$ ways to the answer is $ 420 \plus{} 336$ = [b]756 ways[/b][/hide]


As in direct and inverse proportionality?

[hide="Direct"]
[hide="Problem"]
If $ a$ is directly proportional to $ b$ and $ a\equal{}7$ when $ b\equal{}13,$ find $ a$ when $ b\equal{}27.$ [/hide]

[hide="Solution"]
Since they are directly proportional, their quotient is constant. In other words, \[ \frac{a}{b}\equal{}\frac{7}{13}\] is the constant quotient between $ a$ and $ b.$ Now we find the value of $ a$ such that \[ \frac{a}{27}\equal{}\frac{7}{13}\Rightarrow a\equal{}\boxed{\frac{189}{13}}.\][/hide][/hide]

[hide="Inverse"]
[hide="Problem"]
If $ a$ is inversely proportional to $ b$ and $ a\equal{}17$ when $ b\equal{}24,$ what is $ a$ when $ b\equal{}15?$[/hide]

[hide="Solution"]
Since they are inversely proportional, their product is constant. In other words, \[ ab\equal{}17\cdot24\equal{}408\] is constant. Now we have to find $ a$ such that \[ 15a\equal{}408\Rightarrow a\equal{}\boxed{\frac{136}{5}}.\][/hide][/hide]

These are some simpler proportions. Some more complicated proportions involve more variables.

[quote="A.C"]Can some body give me an example from of direct variation and solve it so i can understand?[/quote]

If one quantity increases then other quantity also increases (or) If one quantity decreases then other quantity also decreases. It is called direct variation. May be seems little confusing.

Lets see this example:

Cost of 5 pens = $ \$$10

Would the cost of 10 pens will be more than the cost of 5 pens?

Yes, It will be more.

What would be the cost of 2 pens? Would it be less than cost of 5 pens?

As number of pens increases, the cost also increases, if the number of pens decreases, then the cost also decreases.

I hope you understand this.

This link may be useful to work with word problems.

[url]http://www.mathocean.com/[/url]


Ok so here I go.

I'm no math genius, so bare with me.

My problem is this.
I have a triangle, or at least part of it.
I know one of the the angles is 30deg, and the length of A to B, but I have no other measurements available.

Now, what I need to figure out is the length of B to C.

Can anyone help me with this?
There must be a way.

Thanks in advance.

A math genius would also be unable to compute a triangle given only one sidelength and one angle.

You need another measurement, else the "length of B to C" is not uniquely determined.

  darij

Ok darij, Yes I realize that. I have it figured out anyways.
Being that all angles added up will equal 180,  30deg + 60 deg + 90deg=180.

I forgot to mention that C to B, will always be 90 deg, leaving me with 60deg.

Thanks for the response anyways.

Off to do some calcs.

Which angle is $90^\circ$?
if $\angle C = 90^\circ$ , then $BC$ = 5cm
if $\angle B = 90^\circ$ , then $BC$ = $\frac{10\sqrt{3}}{3}$cm


A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 23-22.  Show that if the incident angle theta is small, the emerging ray is displaced a distance d = t (theta) (n-1)/n from the incident ray, where t is the thickness of the glass and theta is in radians.
(note: sin (theta) is approximately tan (theta) which is approximately theta for small thetas).

I don't know how to draw the figure, but I'll just describe it.

The figure consists of a rectangular piece of glass with the index n.
The surrounding medium is air, with index approximately 1.

The light ray enters the glass at angle.
The light ray is then refracted inside the glass.
When the light ray leaves the glass, it is refracted again.
When the light ray leaves the glass, the refraction angle is equal to the incident angle( the angle at which the light ray approaches the glass)

Refraction (Snell) Law:
\[ \sin\theta = n .  \sin \theta^'\]
With small angles you can assume : $ \cos \theta = \cos \theta^' = 1$,

so  $ \sin(\theta - \theta^') = \sin \theta - \sin \theta^'$, and at the end,

$ \sin \theta = \theta.$

Okay, I don't see how that proves d=t(theta)(n-1)/n where theta is in radians.

I don't about you guys, but I got $ d \equal{} \frac {t\theta}{n}$

is the original equation faulty? or am I missing something.

I find that d = tn/(n-1)

I have the angle of refraction, theta R, is equal to the angle of incidence divided by n (snell's law and sin theta = theta for small thetas)

In my drawing I have 2 triangles.
One with base d and height x and an angle theta R.

The other with base (t-d) and height x with angle theta I.

so then tan (theta R) = x/d
theta R = x/d
theta I / n = x/d
x=theta I * d / n

The other triangle gives me
tan (theta I) = x /( t-d)
theta I = x / (t-d)
x = theta I * ( t-d)

theta I * d / n = theta I * (t-d)
d/n = t-d
d=tn-dn
dn-d = tn
d(n-1) = t n
d = t n /(n-1)
Obviously my equation is wrong, but I'm not sure what I I did wrong.

I wish there was some way to draw my figure.

May be this will help:

[url=http://img195.imageshack.us/i/mathlinksdifraction.jpg/][img]http://img195.imageshack.us/img195/402/mathlinksdifraction.th.jpg[/img][/url]

From the rectangular triangle NCA , $ |AC| = {d \over {\cos \theta^'}}$.

From the rectangular triangle ABC , $ t = |AC|.\sin (\theta - \theta^')$

Sorry permuting t and d...

The main mistake that we both made was what our assumption of what the displacement.  As Immanuel has it, displacement is the perpendicular distance between the refracted light right at the point of refraction.  I think we thought it was the distance between the imaginary light and actual refracted light striking the other side of the glass.


Given that m+n people are waiting in a line at a box office and: n people have 5-dollar bills, other m people don't have less than 10-dollar bill. Now the ticket costs 5 dollars each.When the box office opens there is no money in the till.If each customer just buys one ticket,what is the probability that none of them have to wait for change?
PS: I clearly don't know how we arrange the line in case there are 6 people having 5-dollar bills and 4 people having 10-dollar bills.If we arrange the line in oder to have at least as many people with 10-dollar bills as people with 5-dollar bills ahead of the point, we don't get the result, do we? Because if we have at least 1 people with 10-dollar bill among others, that person will force the person right behind him to be waited. But if we arrange all of 6 people with 5-dollar bills in a row and then arrange 4 people in the rest of 4 positions, it won't work. So I don't think we can have any arrangements in this case?? I read the solution from the book but it's vague to me, not because of its solution but because of my argument I use to break their solutions :huh:.Am I wrong anywhere in the argument above?

By any chance, is the answer
[hide="?"]$ \frac{n\minus{}m\plus{}1}{n\plus{}1}$[/hide]

I'm not sure that I follow your logic; I think that 6 people paying with $ 5$ dollar bills followed by four paying with $ 10$ dollar bills will work.

Yes, your answer is right, azips(in the case m<n)! But how can you arrange 10 people as I show in the post above to make sure it works?

I'm not completely sure, but I believe you are mis-interpreting the question.

Suppose six people each pay $ 5$ dollars to buy tickets. Now, at the office, there are 6 $ 5$ dollar bills that are readily available to be exchanged. When any $ m < 6$ people with $ 10$ dollar bills approach the box office, each successive $ 10$ dollar bill can be exchanged for a $ 5$ dollar bills, thus receiving the correct amount of money and giving the correct amount of change for each of the $ m$ people.

In other words, at any given point in the line, the number of people who are paying with $ 5$ dollar bills must be greater than or equal to the number of people who are paying with $ 10$ dollar bills.

WOW! Thank you for your explanation azips! I think I understand the problem now! Just ask you a question to make sure:" So, the first 2 people will be with 5-dollar bill in every possible arrangements satisfying the restriction"??

[quote="ghjk"]So, the first 2 people will be with 5-dollar bill in every possible arrangements satisfying the restriction?[/quote]

Close: The first person must have a 5-dollar bill, because if the first person had a 10 dollar bill and there are no 5 dollar bills to exchange for, he would have to wait for change. However, after the first person buys a ticket, there is one 5 dollar bill stored, so a second person who is paying with a 10 dollar bill will be able to immediately receive change.

When $ m\equal{}n$ this becomes the [[Catalan number]]s.


Mesejele se pot edita :D. Este un mic butonel [i]edit[/i]

[quote="Cristi Sav"]Fie $\bigtriangleup ABC$ echilateral cu centrul in $O$ si $M$ in interiorul sau. Notam cu $P,Q,R$ proiectiile lui $M$ pe laturi. Daca Demonstrati ca \[\overrightarrow{MP}+\overrightarrow{MQ}+\overrightarrow{MR}=\frac{3}{2}\overrightarrow{MO}.\][/quote]

 Problema este draguta dar simpla.Tin minte c-am facut-o prin clasa a-9-a.Ideea de rezolvare a acestei probleme este foarte simpla. Se duc paralele la laturile triunghiului $ABC$ dupa care se formeaza trei drepte care au un punct comun.De aici problema devine banala aplicandu-se regulile binecunoscute ale operatiilor cu vectori.Daca cineva e interesat in a posta o solutie completa,am sa o fac,dar momentan mi-e lene. :P

Problema imi este cunoscuta apare undeva prin careva revista ruseasca "Kvant" sau in "Matematica v skole"!


a1=2,a2=8

a2n+1=(a2n+a2n-1)/2, a2n+2=a2n.a2n-1/2a2n+1 so prove that an convergent and   an-->4
Note : 2n,2n-1,2n+1,2n+2 they are subscript from a.
Thanks and best regards!

I'm going to latex this for clarity, and come back if I solve it:

$ a_1\equal{}2, a_2\equal{}8$
$ a_n$ is defined by $ a_{2n\plus{}1}\equal{}\frac{a_{2n}\plus{}a_{2n\minus{}1}}{2}, a_{2n\plus{}2}\equal{}\frac{a_{2n}a_{2n\minus{}1}}{2a_{2n\plus{}1}}$.

Prove that $ a_n$ is convergent and converges to 4.

Thank you very much :)I dont know how can i will use latex  :oops:

[hide="Are you sure this sequence converges to four?"]
Empirically, Excel has shown that this sequence does not approach 4.

This can also be seen since taking the product of any odd term and the term after it yields:

\[ a_{2n\plus{}1}a_{2n\plus{}2}\equal{}\frac{a_{2n}\plus{}a_{2n\minus{}1}}{2}\cdot \frac{a_{2n}a_{2n\minus{}1}}{2a_{2n\plus{}1}}\]
\[ a_{2n\plus{}1}a_{2n\plus{}2}\equal{}\frac{a_{2n}\plus{}a_{2n\minus{}1}}{2}\cdot \frac{a_{2n}a_{2n\minus{}1}}{a_{2n}\plus{}a_{2n\minus{}1}}\]
\[ a_{2n\plus{}1}a_{2n\plus{}2}\equal{}\frac{a_{2n}a_{2n\minus{}1}}{2}\]

That is, the product of an odd term and the term after it is half of the product of the previous odd term and the term after it. If the sequence did converge, the product of the terms would tend towards some constant product as well, which is not the case.
[/hide]


Only moderators may delete posts.  Other questions about the forum should be posted here:
[url]http://www.artofproblemsolving.com/Forum/index.php?f=144[/url]


wait, does nobody have apmo 2006 results...?

http://www.amt.canberra.edu.au/apmo2006.html
The overall results can be found by scrolling down the link.  This page is primarily for Australian results, though.

the link does not have the full results though(ie who got which award)
The results are not up on the official site. :(

[quote="rem"]
The results are not up on the official site. :([/quote]

They are now

http://www.kms.or.kr/competitions/apmo/data/APMO_Results_2006.pdf


[quote="shoki"]Let $ a_1<a_2<...<a_n$ be $ n$ positive integers satisfying this condition:
For all $ 1\le i<j\le n$ we have      $ a_j \minus{} a_i | a_i$ and $ a_j \minus{} a_i | a_j$
Prove that for all $ 1\le i<j\le n$ we have
  $ i\cdot(a_j) \le j\cdot(a_i)$
for example if $ n\equal{}2$ then we have $ a_2 \minus{} a_1 | a_1$ it means that $ a_2 \minus{} a_1\le a_1   \Rightarrow   1\cdot (a_2)\le 2\cdot(a_1)$.[/quote]

[quote="shoki"]Let a_1<a_2<...<a_n be n positive integers satisfying this condition:
For all 1=<i<j<=n we have      a_j - a_i | a_i and a_j - a_i | a_j
Prove that for all 1=<i<j<=n we have
  i*(a_j) <= j*(a_i)
for example if n=2 then we have a_2 - a_1 | a_1 it means that a_2 - a_1<= a_1   =>   1* (a_2)<= 2*(a_1).[/quote]
It seems not very hard problem . From the condition we have :
$ 1<\frac{a_j}{a_j\minus{}a_1}<\frac{a_j}{a_j\minus{}a_2}<...<\frac{a_j}{a_j\minus{}a_{j\minus{}1}}$ .
Notice that all numbers are integers and $ \frac{a_j}{a_j\minus{}1}>1$ , it is easy to prove that $ \frac{a_j}{a_j\minus{}a_i}\geq i\plus{}1$ or $ (i\plus{}1)a_i\geq ia_j$ ,because $ i\plus{}1\leq j$ so $ ja_i\geq ia_j$ for all $ j>i$ . That is what we must prove .
Added problem  :Show that for any n ,exist  a sequence that satisfies condition :$ a_j\minus{}a_i|a_j$ for all $ i,j\in \{1,..,n\}$


In how many ways can a convex n-gon be divided into triangles by nonintersecting
diagonals?

any help would be appreciated.
thank you so much!!

Is it gonna be n-2? I am not sure my answer,though.
Could somebody help me to take a look?
Thanks.

No, it's not $n - 2$.

Hint

[hide]Catalan numbers[/hide]


geometry
geometric transformation
homothety

The pairs of parallel lines make the triangles homothetic, and the lines AF, etc, are through corresponding points, so all 3 lines go through the center of homothecy, so they are concurrent

[quote="indybar"]Let $ABC$ and $DEF$ be triangles such that $DEF$ is inside $ABC$. It is known that $AB\parallel DE$, $BC\parallel EF$, $CA\parallel FD$. Show that $AD,BE,CF$ are concurrent.[/quote]

I'm not sure, but is this problem from the book "Geometry Revisited"???


calculus
integration
logarithms
calculus computations

[quote="Kent Merryfield"]Do you have any reason for thinking that there's anything interesting about this? You realize that if you just slap random functions together, the most likely result is a function whose antiderivative is not an elementary function.[/quote]

As Kent said in [url=http://www.mathlinks.ro/viewtopic.php?t=200744]here[/url], what's the point of this kind of "problem-making"?


Folks,

Check this out:
http://www.highiqsociety.org

Interesting part for people here is World Smartest Persons Puzzle Contest 2006.
Deadline for submission of the answers is 31.12.2006.

I can't find World Smartest Persons Puzzle Contest 2006.

me neither :blush: we must be awefully idiotic not to find that contest :o or it doesn't exist

http://www.highiqsociety.org/wsp_challenge.php

with the actual questions [url=http://www.highiqsociety.org/wsp_challenge.php]here[/url]

cule that kid lied to us about the deadline :mad:

[quote="Eros88"]Didn't lied on purpose.
 :mad:
I misread it in a hurry and thought the deadline was December 31st (not October 31st).Sorry,I don't have a time machine to fix that.
Cool puzzles I think anyway.[/quote]
oh i forgive you i have the worst problems with my eyes i miss problems on the IOWA test because my eyes are so bad :)

New IHIQS contest starts September $ 1^{st}$,2007.

First prize 1000 bucks  :)


Let $ O$ be the center of a quarter of circle. the radius that delimet this region are $ OA$ and $ OB$. A circle with center $ O'$ is inscribed in this region, Being tangent to $ OA,OB$ and the arc $ AB$ . The chord $ AB$ cuts the circle $ (O')$ in $ M,N$ . Find the value of $ cos\angle MO'N$

Prove also that this angle is close to $ 135^0$

[quote="Tiger100"]delimet[/quote]

What does this word mean?

Yes, sorry i wanted to mean [b]Delimit[/b]

Any ideas to solve this problem? it is nice in my opinion

I'm working on it.
Of course there is coordinate bash, I could ask pythag011 to do that... :P

EDIT: I have a solution:

Since OAB is a quarter sector, the midpoint of MN is the intersection of the perpendicular bisectors of OA and OB. WLOG let the radius of circle $ O'$ be 1. The radius of $ OB$ is therefore $ 1\plus{}1\sqrt{2}\equal{}\sqrt{2}\plus{}1$. The midpoint of $ MN$ (let's call it $ P$) is therefore $ \frac{\sqrt{2}\plus{}1}{2}$. Hence $ OP\equal{}\frac{2\plus{}\sqrt{2}}{2}$. Since $ OO'\equal{}\sqrt{2}$, we have $ O'P\equal{}\frac{2\minus{}\sqrt{2}}{2}$. Since $ MO'\equal{}1$, we have $ \cos{MO'P}\equal{}\frac{2\minus{}\sqrt{2}}{2}$, and $ \angle MO'N\equal{}2\angle MO'P$. We can now find $ \cos{MO'N}$, and it should be roughly $ \minus{}\frac{\sqrt{2}}{2}$, since $ \cos{135^{\circ}}\equal{}\minus{}\frac{\sqrt{2}}{2}$.


function
calculus
derivative
absolute value
calculus computations

SO here's the problem... seems simple, but I just can't get the jist of it because of the absolute value:

Find the local max/min, as well as the absolute max/min values of the function:

f(x) = |x^2 - 4x + 3|

in the interval (1,4].

Any help would be greatly appreciated!

Either $|x^2-4x+3|=x^2-4x+3$, or $|x^2-4x+3|=-(x^2-4x+3)$. You have critical points where the derivative is zero, and also where you switch formulas.

alright... I think that I've just been overcomplicating this.... is the derivative simply f'(x) = |2x-4| ? And if so, how do I prove where I switch formulas? I'm aware that it is at x=1 and x=3, so do I just solve for x-intercepts of the original function to prove this?

No, that's not the derivative. It has the same zeros, but its sign is only right some of the time.

The derivative is $2x-4$ where $x^2-4x+3$ is positive, and $4-2x$ where $x^2-4x+3$ is negative.
Never try to do calculus with a formula that still has an absolute value sign in it- you're always better off looking at it as the piecewise defined function it is.

If you write $|x|$ as $\sqrt{x^2}$, you get
\[ \frac{d}{dx}|x|=\frac{x}{|x|}  \]
for $x\not=0$, which is a reasonably nice way of looking at things.


The sum of the number of sides of two convex polygons is 10,

and the sum of the numbers of their diagonals is 11.

What type of polygons are described?

thank you~

If s is the number of sides, the number of diagonals is $\frac{n(n-3)}{2}$.

Therefore, $x+y=10$, and $x^{2}-3x+y^{2}-3y=22$.

This can be solved easily using substitution;

\begin{eqnarray*}22&=&(10-y)^{2}-30+3y+y^{2}-3y\\ &=&100-20y+2y^{2}-30\\ 0&=&2y^{2}-20y+48\\ &=&y^{2}-10y+24\\ 0&=&(y-6)(y-4) \end{eqnarray*}

The two polygons have 6 and 4 sides, respectively, so you have a square and a circle.

[quote="360_Fan"]Square and circle?   :P[/quote]

I wonder if there was a joke or an accident.


probability
function
real analysis
search
probability and stats

Find the distribution of $ \sum_{n\equal{}1}^{\infty}\frac{X_n}{3^n}$
$ X_n$ are iid with $ P(X_n\equal{}0)\equal{}P(X_n\equal{}2)\equal{}\frac{1}{2}$

This produces a "random" element of the usual Cantor set. Its cumulative distribution function $ F(x)$ is the Cantor-Lebesgue "devil's staircase" function. This distribution is continuous but not absolutely continuous. That is, it does not have a density function.

The distribution of $ \sum_{n\equal{}1}^{\infty}\frac{Y_n}{2^n}$ where $ Y_n$ takes on the values $ 0$ and $ 1$ each with probability $ \frac12$ is simply the uniform distribution on $ [0,1].$

and what if $ X_n$ are bernoulli random variable $ P(X_n\equal{}1)\equal{}1\minus{}P(X_n\equal{}0)\equal{}p \neq \frac{1}{2}$ ?

The values of the sum would still be confined to the Cantor set. The cumulative distribution function $ F(x)$ would have the following properties. (I'll use the notation $ q \equal{} 1 \minus{} p.$)

$ F(x) \equal{} q$ for $ \frac13\le x\le\frac23.$

$ F(x) \equal{} q^2$ for $ \frac19\le x\le\frac29.$

$ F(x) \equal{} q \plus{} qp \equal{} 1 \minus{} p^2$ for $ \frac79\le x\le\frac89.$

And so on. We will still have $ F$ as the uniform limit of a sequence of continuous functions, and hence continuous. The distribution will still be continuous but not absolutely continuous.

If we do that with my alternative of $ Y_n,$ the result will be one of my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=101950603&t=118219]favorite examples.[/url]


geometry
circumcircle
modular arithmetic
geometric transformation
homothety
ratio
rotation

[u][b]Problem:[/b][/u]The vertices of the triangle ABC are lattice points and there is no smaller triangle similar to ABC with its vertices at lattice points. Show that the circumcenter of ABC is not a lattice point.
---------------------------------------------------------------

Suppose the contrary.

There are some cases:
1. $R^2=0\pmod{4}$. Make homotety with ratio $1/2$.
2. $R^2=2\pmod{4}$. Make homotety with ratio $1/2$.
3. $R^2=1\pmod{4}$. Make homotety with ratio $1/\sqrt{2}$ and rotate the triangle by $\pi/4$.


Why is our country going at length to ensure the rights of unborn fetuses? Now a woman can't even have an abortion without having law complications. It is kind of sad - just today Senate passed H.R. 1997 - The Unborn Victims of Violence Act. Isn't it enough that fetuses are constitutionally given the right to life no matter what the pregnant mom thinks? Now, if the mom is hurt during the pregnancy the "living" fetus can sue too. WHAT??!! That was exactly my reaction. The bill doesn't even specify the requirements for life, but it is assumed that fetuses are living. Why go to lengths to protect the unborn when those that are born are suffering even more. We look too far intot he future and obliviously ignore what is happening now sometimes.

I would like to pre-emptively remind people that, although this can be a contentious issue and can raise a lot of passions, you should still be polite and respectful in your discussions with your fellow forum-members.  Thank you.

I am religious, I don't believe in abortions. However, it is perfectly legal to abort if its danger the mother's health (which happened in my relatives). Recently in Utah a woman who was a looser (drug taker, child abuser, divorced, etc.). was convicted of murdering a stillborn child.

wow churchill, what a contrast coming from you :).

I believe that women should be allowed to choose whether or not they want an abortion without having to face legal consequences.

However, if a woman is going to have an abortion, she should have it done as soon as possible.  (Say, during the first six weeks.)  The longer she waits, the more taxing the baby will be on her body, the more she will become attached to the baby as a result of hormones, and the more developed the baby will be.

Some may argue that abortion should never, or almost never, be allowed.  But here are some objections to that viewpoint:

First, what if being pregnant puts the potential mother at risk of injury or death?  For instance, pregnancy in women with diabetes can result in hard-to-control blood sugar levels that require constant monitoring.  Even with constant attention, this can result in the loss of an eye, kidney, leg, etc.  Would you have a mother risk her life to deliver a baby?  What if this were someone you [i]knew[/i], like your sister or your friend?

In the absence of special, life-threatening circumstances, some argue that a woman should not be allowed to have an abortion because ``she was irresponsible".  But if this is the case, then forcing her to deliver the baby won't remedy this problem of her irresponsibility; it'll just result in a mom who can't care for her baby.  So you're really punishing the baby by placing it with a mother who cannot care for it.

And the suggestion that the ``irresponsible'' mother deliver the baby, then put her new baby up for adoption is no better.  First of all, no mother will give up her baby easily, perhaps because hormones she produces during the pregnancy make her feel attached to the baby.  Second of all, there are only a limited number of families that want to adopt a baby, so adding another baby to that pool of adoptees will displace some other baby.  Third, each extra baby will contribute to the overpopulation of the world.

Third, many abortions are given to teenagers.  Suppose a teenage girl is pregnant and wants to have an abortion but cannot, either because the law mandates that she cannot or because her friends and family cry out "you would kill your own baby?"  Then, she's a high-school age girl who has to worry about her own school life, her kid, and her source of income.  Chances are, if she was in high school, she'll start falling severely behind in school.  She'll probably drop out to go to work at Wal-Mart so that she can have a source of income to support her child.  This girl may have had a bright future ahead of her - maybe she was gonna go to college, get her degree, go to work somewhere, get married, raise a child together with her husband, etc.  But because she had to deliver this baby, her future went ka-put.

Finally, sometimes accidents happen (contraceptive measures are breached, so to speak) and a baby results unintentionally.  If a mother doesn't feel she's ready to have a baby yet (or ever) - maybe she want to focus on her job, maybe she hates kids, maybe she wants to devote 8 hours a day to math - then she shouldn't be forced to.

P.S. For those of you who are wondering, yes, I am religious.  Roman Catholic.  Does that change my view of abortion?  No.

Actually, white-horse's characterization of what the Bill says is incorrect. The Bill, as it stands right now, doesn't say anything explicitly about abortion. It says that if a pregnant woman is killed then the perpetrator can be charged with two crimes - one for the murder of the woman and one for the murder of the child.

Of course, it naturally brings up questions of the implications it will have for abortion; but that's not what [i]this[/i] Bill is addressing. This Bill is talking about the killing of an unborn child [i]without[/i] the woman's consent.

The Bill is very complicated, but it doesn't in itself say anything about abortion. It just opens the door to questions about what rights the fetus has. If the fetus is said to have stringent rights to life; just like any other person has - then pretty much all of mathfanatic's arguments (except for maybe the first one) become void. Because those stringent rights to life would superceed any "utilitarian" arguments - just like you can't kill any regular person, simply to make your life better.

What's always intersting to me though, is why do we value the lives of infants and children more so than adults? When we hear that a young child is killed, it seems to carry so much more weight than if we hear that an adult is killed. Yet, just a few months before they are born - we consider those [i]same[/i] lives worth so much less than the value of an adult's.

[quote="gauss202"]What's always intersting to me though, is why do we value the lives of infants and children more so than adults?[/quote]

Because children still have the prospect to do something impressive whereas adults have already blown it?
Because children are cute?
Because children are defenseless whereas adults theoretically have a chance to defend themselves or not enter themselves into dangerous situations in the first place?

As for the later question, there are certainly groups of people who value fetuses just as strongly.  That's why we have an anti-abortion movement -- it's not moderate religious Christians who maybe think abortion is unfortunate, it's the hardcore evangelicals who really do believe that the fetus is a person.  Most other people don't share that belief, so an abortion isn't such a tradgedy.

Again, another political issue I'm not entirely familiar with-but me being the gender that this most directly applies to I think I'll share what I believe.  First off, I'd like to say that I am a little bit offended at mathfanatic's remarks about a girl delivering a baby and her future going "kaput".  Yes, it's true that a girl who has a baby in her teenage years may not have all the advantages as a girl who doesn't-but it's not final. A successful future is relative, just because a girl doesn't win all the math contests after she's delivered a baby doesn't mean she's failed. Her own idea of a successful future might be just to be at peace with herself, and there's nothing wrong with that.  I am pro-choice because I believe a woman has the right to do what she wants with her own body. I don't understand this "unborn child rights".  Apparently it has to do maybe with the fact that biologically they're considered alive. Well, practically every month something biologically alive is killed in a woman's body, what about their rights?  Just what I think, I'm definitely no expert.

[quote="Tare"]I haven't read any of the above so I apologize if there are any duplicate opinions: Abortions are ok with me, after all I don't see any laws protecting cows and pigs from getting slaughtered and their carcass eaten, so why not babies as well? (even though very few cultures eat babies now-a-days)[/quote]
'tis [url=http://www.english.upenn.edu/~jlynch/Courses/95c/Texts/modest.html]A Modest Proposal[/url], to say the least.

A Modest Proposal is one of the funniest pieces of political writing I've ever seen.  I highly recommend you all go read it, although it's helpful to know a little bit of British-Irish history first.

I was just pointing out that laws against abortion are already, in my opinion, high strung. There are complications, but mothers shouldn't be completely deprived of the right to abortion. I was just saying as an allusion to those developing laws how I felt about giving fetuses the "power" to sue. I predict that this bill will be abused. Mothers may intentionally become hurt and the charge will be against the mother are the unborn fetus. Who takes the money for the damages? This is completely incredulous.

My madness isn't totally unjustified - babies are now given that right mentioned in the bill because they are considered to be living humans. Babies don't even have motor functions at that stage - how can they """""sue"""""? Another thing is, I'm concerned that the major reason this bill was proposed is because of not only sympathyzers but also people greedy for a quick buck. Also, money seems to be the best compensation for anything.

The argument for the bill seems based on the fact that a fetus is being deprived. Deprived of what? It isn't even really alive yet - no brain functions, motor functions, or any functions.

I'm concerned - what if people deliberately seek money through this method. People already overabuse sueing powers - people sue for everything.

I would like someone to weigh out the benefits vs. the negative potential effects and give a proper justification of the bill.

white_horse: one more mis-conception.  This law has nothing to do with sueing.  To sue is a matter for civil court.  In many jurisdictions, you can't do it until you are 18, although you might be able to get someone to sue on your behalf.  (An example of this is the case currently at the Supreme Court about the Pledge of Allegiance -- the man who brought the case brought it on behalf of his daughter, since she isn't old enough to sue and he doesn't have grounds for sueing.)  Anyhow, the law applies only to criminal cases.  I, personally, am worried very much about the math -- 1 mother + 1 fetus = 2 crimes; 1 mother = 1 crime.  Hopefully, none of our politicians will learn to subtract any time soon.


[quote="ragnarok23"]Let \[A=1, \ B=2, \ C=3, \ D=4, \ldots X=24, \ Y=25, \ Z=26\] Let the value of a word be the product of it's corresponding variables (i.e ALL=(1)(12)(12)=144). Take your name. Find the value of it. Multiply it by 200. Add 50 to it. Add 15 to it. Add 35 to it. Divide it by 2, subtract 27 from it and then divide it by 3. Subtract half the value of your name from it, then subtract 32(value of your name), and finally, subtract 5/6 of the value of your name from it. What are you left with? Express your answer AS A MIXED NUMBER.

Sorry for the uncombined sentences, but it takes all the fun outta it.  :D[/quote]
[hide]Lets just say that the value of my name is x. $\frac{\frac{200x+50+15+35}{2}-27}{3}-.5x-32x-\frac{5x}{6}=\frac{100x+23-1.5x-96x-2.5x}{3}=\frac{23}{3}=7\frac{2}{3}$[/hide]


What is the distance from (2,3) to (3,6)?

[hide]the distance is 10^0.5. the difference in x-coordinates is 1 and 3 in y-coordinates. therefore, applying the pyhagorean theorem, we get 1^2+3^2=10, so, the answer is 10^0.5[/hide]

That would be correct. $\sqrt{10}$ is the answer.

[hide]using the distance forumla:
i get $\sqrt{10}$ as my answer[/hide]


geometry
3D geometry
number theory proposed
number theory

a)Prove that there are infinitely many possitive integers α so φ(α) be perfect square.
b)Are there infinitely many possitive integers  α so φ(α) be perfect square and gcd(α,(φ(α))=1?
c)Are there infinitely many primes π so φ(π) is rerfect cube?
Thanks.

Oh, I thought it's $\phi (n)$... As I've read in many books...

a) Let $a=5^{2m-1}$ for any positive integer $m$. $\phi(a) = 5^{2m-2}(5-1) = (2\cdot 5^{m-1})^{2}$

c) Since $a$ is prime, then $\phi(a)=a-1$
If $a=2$, $\phi(a)=1^{3}$, If $a>2$, assume $\phi(a)=m^{3}$ for positive integer $m$, then $a=m^{3}+1>2 \Rightarrow m>1$.
$a=m^{3}+1 = (m+1)(m^{2}-m+1)$
Since $1<m+1<m^{3}+1$
$a$ is not prime. Contradiction.

c) equavalent to unsolved problem: there is infinetely many primes $p=x^{2}+1$.
b) can be solved used: for any k, there is x<k, suth that xk+1 is prime.

For a) generally for all α=(ν^2+1)^(2κ+1). :)


Prove that there are infinitely many primes $ p\equiv 5 \mod 6$.

This belongs to the class of special cases of Dirichlet's theorem with elementary proofs.

[hide="The elementary proof"] Suppose there are finitely many, $ p_1, p_2, ... p_k$.  Then $ P \equal{} 6p_1 p_2 ... p_k \minus{} 1$ is congruent to $ 5 \bmod 6$, hence (as can be easily verified) must have a prime factor congruent to $ 5 \bmod 6$; contradiction. [/hide]

Generally, a similar proof is possible for primes congruent to $ a \bmod m$ if and only if $ a^2 \equiv 1 \bmod m$.

There are infinitely many primes in the set {a,a+b,a+2b,a+3b,a+4b,...} where gcd(a,b)=1.

Sorry, but I would really like very much to know the proof of this....
Can anyone please provide with links for proofs? :oops:

The classical proof of Dirichlet's theorem is quite involved.  You can read, for example, the comments [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=205115]here[/url].  And the "elementary" proofs actually require more difficult arguments.


geometry
perimeter
algebra
polynomial
ratio
quadratics
number theory

Triangle ABC has perimeter 152, and angle BAC is a right angle. A circle with radius 19 and center O on AB is drawn so that it is tangent to AC and BC. Given that OP = m/n, where m and n are relatively prime positive integers, find m + n.

Whoops. I wasn't thinking; OP should be OB.

I remember I tried this problem before in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45994]this post[/url], but I couldn't complete it...

The equation I got was:

$ x+19+\sqrt{x^2-19^2}+2[(x+19)19/\sqrt{x^2-19^2}]=152$

kinda busy right now, I'll try to solve it later

Expanding out would give a third degree polynomial, so I guess try to guess a rational root?

i meant 31 and 2/3

95/3

gr at my bad latex.

I have a solution, problem sucked me in away from some work, hehe.

[hide]
First, we're going to shrink everything by a ratio of 19, to simplify the problem.

Let z be the desired length. Let D be the tangent on BC.
BD = $\sqrt{z^2-1^2}$

Also, DC=CA.

Since BDO and BAC are both right and share angle B, they are similar.

Therefore, 1/BD=AC/(z+1)

AC = $(z+1)/\sqrt{z^2-1^2}$

So, the whole perimenter is BO+OA+AC+CD+DB=$z+1+(z+1)/\sqrt{z^2-1^2}+(z+1)/\sqrt{z^2-1^2}+\sqrt{z^2-1^2}$

That's equal to 8 (152/19).

Simplifying:

$z+1+2(z+1)/\sqrt{z^2-1^2}+\sqrt{z^2-1^2}$=8

We get $9z^3-21z^2-5z+25=0$

I used the rational root theorem to get that 5/3 is a root. Dividing (3z-5) out, we find that 5/3 is a repeat root, and the final root is -1.

So, z=5/3, and the desired length is 19*5/3= $95/3$.
[/hide]

Krusty, please combine your posts by editing - you just did a double posts and one triple post.

 Another solution:

[hide]Extend O to the point where circle O is tangent to BC - this point is P.

Use standard abbreviations for the side lengths.

We have the following two ratios using the similarity $\bigtriangleup BPO \sim \bigtriangleup BAC$.

$\frac{c-19}{19} = \frac{a}{b}$ (1)

Since OC is the angle bisector of angle BCA, $\bigtriangleup OPC \cong \bigtriangleup OAC$. So, $PC = AC = b$.

$\frac{a-b}{19} = \frac{c}{b}$ (2)

You get a quadratic for $b$ after linear combination of (1) and (2) after substitution of $a = 152 - b - c$.

$b^2 - 76b + 1444 = 0 \Rightarrow b = 38$

So, substitute back into the original equations to obtain $c = \frac{152}{3}$. Subtract 19 to obtain OB. $OB = \frac{95}{3}$[/hide]


An organic compound A ($ C_{8}H_{10}$) gives the following chain of reactions:

$ A\plus{}N\minus{}bromosuccinimide/CCl_{4}\longrightarrow B\, (C_{8}H_{9}Br)$

$ B\plus{}(CH_{3})_{3}COK\longrightarrow C\, (C_{8}H_{8})$

$ C\plus{}O_{3}/then\,\, Zn/H_{3}O^{\plus{}}\longrightarrow D\, (C_{7}H_{6}O)$

$ D\plus{}CH_{3}MgBr/then\,\, H_{3}O^{\plus{}}\longrightarrow E$

$ E\plus{}PCC/CH_{2}Cl_{2}\longrightarrow F\, (C_{8}H_{8}O)$

$ F\plus{}N_{2}H_{4}/NaHO\longrightarrow A$.

Identify each of the compounds A – F.

Sorry to revive these old topics but I'm really low on confidence and morale these days... :(

A=ethyl benzene
B=(alpha brominate A)
C=styrene
D=benzaldehyde
E=1-phenyl ethanol
F=methyl phenyl ketone

Very good, your answers are correct.

You are welcome to revive these old topics. :)

This might sound like a silly question, Carcul, but I was just wondering if there is a reason that you write $ \ce{NaHO}$ instead of $ \ce{NaOH}$?

Finally someone noticed that. About two or three years ago, the Portuguese Chemical Society recomended (as a result of an IUPAC recomendation, I suppose) that the correct way to write the formula of hydroxide ion is HO- and not OH-: since this does, in fact, make sense (because the atom bearing the charge is oxygen, not hydrogen), since then I started to write the formula as such.


algebra
polynomial
Vieta
vector
inequalities
algebra proposed

hi,
consder the polynom $ P(x) \equal{} x^{n} \plus{} a_{1}x^{n\minus{}1} \plus{} ... \plus{} a_{n}$ with all roots positive. If there exists $ m,p \in \{1,2,...,n\}$ $ m \not\equal{} p$ s.t.
\[ a_{m} \equal{} (\minus{}1)^{m}\binom{n}{m}, a_{p} \equal{} (\minus{}1)^{p}\binom{n}{p},\]
prove $ P(x) \equal{} (x\minus{}1)^{n}$.

Set $ \Omega\equal{}\{1,\ldots,n\}$ and define for $ k\in\Omega$ the averaged symmetric polynomial via $ E_k(\bar{z})\equal{}\frac{1}{\left({n\atop k}\right)}\sum_{I\subseteq\Omega\atop |I|\equal{}k}\prod_{i\in I}z_i$. For example $ E_1(\bar{z})\equal{}\frac{1}{n}\sum_{i\equal{}1}^nz_i\,,\,E_n(\bar{z})\equal{}\prod_{i\equal{}1}^nz_i$ and for the special vectors $ \bar{z}\equal{}u\cdot (1,\ldots,1)$ we compute $ E_k(\bar{z})\equal{}u^k$.

Recollect [b]Newton-Maclaurin[/b]'s result for positive vectors: $ E_1(\bar{z})\geq E_2(\bar{z})^{\frac{1}{2}}\geq\ldots\geq E_n(\bar{z})^{\frac{1}{n}}$ with equality iff $ \bar{z}\in\mathbb{R}^\plus{}\cdot (1,\ldots,1)$

For its application to this post's problem let $ \bar{z}$ describe the positive roots of $ P$, then by assumption we have $ E_p(\bar{z})\equal{}1\equal{}E_m(\bar{z})$ for $ p\not\equal{}m\in\Omega$. This implies $ \bar{z}\equal{}u\cdot (1,\ldots,1)$ for some $ u>0$ and from $ u^p\equal{}u^m\equal{}1$ we conclude $ u\equal{}1$ and $ P(x)\equal{}(x\minus{}1)^n$.

P.S. Two nice articles on inequalities of symmetric polynomials:

[url]http://www.math.helsinki.fi/EMIS/journals/JIPAM/v4n2/116_02_www.pdf[/url]
[url]http://www.emis.de/journals/JIPAM/images/182_05_JIPAM/182_05.pdf[/url]


So I found a stack of old Descartes contests lying around my school and decided to open them up and start solving some problems, I stumbled across these two that I thought at first were going to be really straightforward but for some reason I can't seem to solve them. I've posted them here for your solving pleasure  :):

6.
Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be linearly dependent vectors in $R^{3}$ and let:
$\vec{u}= 3\vec{a}+2\vec{b}-\vec{c}$
$\vec{v}=-2\vec{a}+4\vec{c}$
$\vec{w}=-\vec{a}+3\vec{b}+k\vec{c}$
Determine $k$ so that $\vec{u}$, $\vec{v}$, and $\vec{w}$ are coplanar.

10 b).
In triangle $ABC$, sides $AB$ and $AC$ have fixed lengths $c$ and $b$ respectively, with $c>b$, and $BC$ has variable length $a$. Also, $\theta$ is the measure of $\angle CAB$ and $h$ is the length of the altitude from $A$ to $BC$.
Prove that:
$\frac{da}{d\theta}= h$

What grade level is this?  It seems like it is 13th Grade.

Yeah, Descartes was the OAC contest back when OAC existed.

10 b). is actually pretty simple
[hide]
$b^{2}+c^{2}-2bccos\theta=a^{2}$
Then take the derivative of $\theta$ on both sides
And it's obvious that $bcsin\theta=ah$
[/hide]

fOR 6, I  guess you just need to find the right combination.
[hide]$k=-1*\frac{3}{2}+4*\frac{11}{4}=9.5$[/hide]

10 b)... touche, probably should have persevered more.

6. Sorry but could you elaborate?? I have no idea what you did :P

well, vectors u v w are coplanar if and only if they are linearly dependent which means w would be a linear combination of finitely many u and v.

!!! I should definitely review my linear algerba basics (or... everything, it's not like I know much more than the basics  :P) I was totally only thinking of the definition
$\vec{u}\times \vec{v}\bullet \vec{w}$ $= 0$


function
calculus
integration
real analysis
real analysis unsolved

Distance(x)(cm)
0,1,5,6,8
Temp t(x) (celsius)
100, 93, 70, 62, 55

Metal wire is 8 cm.  and heated at one end.  distance x is how far from the heated end you are.  The function ti s decreasing and twice differentiable.
a) estimate t'(7)
b) write an integral expression of T(x) for avg temp of wire. estimate avg temp of wire using trapezoidal sum with 4 subintervals indicated by data
c) Find int up limit 8 lower limit 0. explain the meaning of this in the context of the problem
d)  Is that data consistent with teh assertion that T''(x)>0 for every x from 0 to 8. explain

That's just obnoxious. There's no reason to expect a response within a few hours- this forum moves more slowly than that.

sorry, that third bump was a double post by accident.  i am just bumping every 2 hours or so, not a big deal.

Between your first post and now, there has been only one other post in this subforum; even one bump was excessive.


algebra
polynomial
induction
algebra unsolved

Let $f(X)$ be polynomial with rational coefficients. Prove that to have $f$ taking integer values for integer arguments, it is necessary and enough to be able to write $f$ as:

$f(X) = \sum _{j=0}^{N} a_j C_{X}^j$

with integer $a_1,...,a_N$

Define $\Delta f (x) := f(x+1) - f(x)$ and observe that when $f(x) = \sum_{i=0}^n a_i \left( x \atop i \right)$ you get $\Delta f(x) = \sum_{i=0}^{n-1} a_{i+1} \left( x \atop i \right)$. So you can do it by induction on $n$.


find all function with reel numbers and wich stisfy
$f(x+y+z)=f(x)+f(y+z)+(y+z)(x+y+z)$. :lol:

$z=0\Rightarrow f(x+y)-f(x)-f(y)=y(x+y)$
LHS is symmetric in $x,y$, RHS isn't.
So there is no such function.


inequalities
trigonometry
geometry
function
inequalities unsolved

Let [tex]\alpha, \beta, \gamma[/tex] be angles in the interval [tex](0, \pi/2)[/tex] whose sum is [tex]{\pi}/2[/tex]. Then, prove that

[tex]{\sin \alpha \sin \beta \sin \gamma}(\sec^2 \alpha + \sec^2 \beta + \sec^2 \gamma}) \geq {\cos (2 \alpha) + \cos (2 \beta) + \cos (2 \gamma) - 1[/tex]

[color=blue][b]Problem.[/b] Let $ \alpha$, $ \beta$, $ \gamma$ be three angles in the interval $ \left]0,\frac{\pi}2\right[$ such that $ \alpha\plus{}\beta\plus{}\gamma\equal{}\frac{\pi}2$. Then, show the inequality

$ \sin \alpha \sin \beta \sin \gamma\left(\sec^2 \alpha \plus{} \sec^2 \beta \plus{} \sec^2 \gamma\right) \geq \cos \left(2 \alpha\right) \plus{} \cos \left(2 \beta\right) \plus{} \cos \left(2 \gamma\right) \minus{} 1$.[/color]

After three years of triangle geometry, one doesn't find such problems tough, but rather one finds them straightforward. At first, we can easily show

$ \cos \left( 2\alpha \right) \plus{} \cos \left( 2\beta \right) \plus{} \cos \left( 2\gamma \right) \minus{} 1 \equal{} 4\sin \alpha \sin \beta \sin \gamma$,

since

$ \cos \left( 2\alpha \right) \plus{} \cos \left( 2\beta \right) \plus{} \cos \left( 2\gamma \right) \minus{} 1$
$ \equal{} \cos \left( 2\alpha \right) \plus{} \cos \left( 2\beta \right) \minus{} 2\sin ^{2}\gamma \equal{} 2\cos \left( \alpha \plus{} \beta \right) \cos \left( \alpha \minus{} \beta \right) \minus{} 2\sin ^{2}\gamma$
$ \equal{} 2\sin \gamma \cos \left( \alpha \minus{} \beta \right) \minus{} 2\sin ^{2}\gamma \text{\ \ \ \ \ \ \ \ \ \ (since }\alpha \plus{} \beta \plus{} \gamma \equal{} \frac {\pi }{2}\text{)}$
$ \equal{} 2\sin \gamma \left( \cos \left( \alpha \minus{} \beta \right) \minus{} \sin \gamma \right)$
$ \equal{} 2\sin \gamma \left( \cos \left( \alpha \minus{} \beta \right) \minus{} \cos \left( \alpha \plus{} \beta \right) \right) \text{\ \ \ \ \ \ \ \ \ \ (since }\alpha \plus{} \beta \plus{} \gamma \equal{} \frac {\pi }{2}\text{) }$
$ \equal{} 2\sin \gamma \cdot 2\sin \alpha \sin \beta \equal{} 4\sin \alpha \sin \beta \sin \gamma$.

Therefore, your inequality simplifies to

$ \sec ^{2}\alpha \plus{} \sec ^{2}\beta \plus{} \sec ^{2}\gamma \geq 4$.

Noting that

$ \sec ^{2}\frac {\alpha \plus{} \beta \plus{} \gamma }{3} \equal{} \sec ^{2}\frac {\pi }{6} \equal{} \left( \frac {2}{\sqrt {3}}\right) ^{2} \equal{} \frac {4}{3}$,

we see that the above inequality follows from the convexity of the function $ f: x\rightarrow \sec ^{2}x$ in the interval $ \left( 0;\;\frac {\pi }{2}\right)$. In fact, here is, for the sake of completeness, a proof that this function is convex in this interval: If x and y lie in the interval $ \left( 0;\;\frac {\pi }{2}\right)$, then we must show

$ \sec ^{2}x\plus{}\sec ^{2}y\geq 2\sec ^{2}\frac{x\plus{}y}{2}$.

By multiplying with $ \cos ^{2}x\cos ^{2}y$, we "simplify" this to

$ \cos ^{2}y\plus{}\cos ^{2}x\geq \frac{2\cos ^{2}x\cos ^{2}y}{\cos ^{2}\frac{x\plus{}y}{2}}$.

But remembering

$ 2\cos \frac{x\plus{}y}{2}\cos \frac{x\minus{}y}{2}\equal{}\cos x\plus{}\cos y$,

our inequality becomes
\[ \cos ^{2}y \plus{} \cos ^{2}x\geq \frac {2\cos ^{2}x\cos ^{2}y}{\left( \cos x \plus{} \cos y\right) ^{2}/\left( 4\cos ^{2}\frac {x \minus{} y}{2}\right) },
\]
or
\[ \left( \cos ^{2}y \plus{} \cos ^{2}x\right) \left( \cos x \plus{} \cos y\right) ^{2}\geq 8\cos ^{2}x\cos ^{2}y\cos ^{2}\frac {x \minus{} y}{2}.
\]
Denoting $ X \equal{} \cos x$ and $ Y \equal{} \cos y$, with $ 0\leq X\leq 1$ and $ 0\leq Y\leq 1$, and $ T \equal{} \cos ^{2}\frac {x \minus{} y}{2}$, with $ 0\leq T\leq 1$, the inequality we must prove becomes

$ \left( X^{2} \plus{} Y^{2}\right) \left( X \plus{} Y\right) ^{2}\geq 8X^{2}Y^{2}T$.

Now, we know $ X^{2} \plus{} Y^{2}\geq 2XY$ and $ \left( X \plus{} Y\right) ^{2}\geq 4XY$; thus, $ \left( X^{2} \plus{} Y^{2}\right) \left( X \plus{} Y\right) ^{2}\geq 8X^{2}Y^{2}\geq 8X^{2}Y^{2}T$. And we are done.

Of course, rather than proving the convexity of $ f: x\rightarrow \sec ^{2}x$ as I did, one could also bash it by calculus.

  Darij


combinatorial geometry
geometry
combinatorics
length
polygon
IMO
IMO 1982

Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.

There is sth wrong ith my idea:

The length of $X,Y>4*99-1$ is quite trivial in one direction, but if they ean the shortest distance, we just take points very close to the boundary and with angles close to $180$ degrees except in the corners, what is the mistake in those ideas?

See John Scholes' solution at [url]http://mks.mff.cuni.cz/kalva/[/url].

The condition should be $A_0\neq A_n$. That's the statement appearing on the official IMO website.

The solution is difficult to understand because it involves compact sets. Can anyone explain the properties of compact sets


function
quadratics
algebra
domain
induction
linear algebra
matrix

Let
$f(x)=\frac{ax+b}{cx+d}$
$F_n(x)=f(f(f...f(x)...))$ (with $n\ f's$)

Suppose that $f(0) \not =0$, $f(f(0)) \not = 0$, and for some $n$ we have $F_n(0)=0$,
show that $F_n(x)=x$ (for any valid x).

Consider the equation $f(x) = x$, which gives the fixed points of $f$:
\begin{eqnarray*} f(x) = \frac{ax + b}{cx + d} &=& x \\ \Leftrightarrow \ ax + b &=& x(cx + d) = cx^2 + dx. \end{eqnarray*}

If $c = 0$, then $f(x) = (ax + b)/d$, a linear function, so every iterate $F_n$ is also linear, and the above equation becomes
\[ ax + b = dx.  \]
Suppose $a = d$.  Then $f(x) = x + b/a$.  If $b = 0$, then $f(0) = 0$, contradiction, so $b \neq 0$.  But then $F_n(x) = x + nb/a$, so $F_n(0) = nb/a$ can never be zero for $n \ge 2$, contradiction, so $a \neq d$.

That means the equation $ax + b = dx$ has a solution; in other words, $f$ has a fixed point, which means that $F_n$ has a fixed point for all $n$.  We know this fixed point is not 0, since $f(0) \neq 0$.  But we also know that $F_n(0) = 0$ for some $n$.  For this $n$, $F_n$ has at least two fixed points, and is linear, which means that $F_n(x) = x$ for all $x$.

Now we can handle the case $c \neq 0$.  In this case, the above quadratic
\[ ax + b = cx^2 + dx  \]
has two roots (we can extend the domain of $f$ to the complex numbers), so $f$ has at least two fixed points.  We know that 0 is not one of these fixed points, since $f(0) \neq 0$.

Again as before, we know that $F_n(0) = 0$ for some $n$.  For this $n$, $F_n$ has at least three fixed points, and by a straight-forward induction argument,
\[ F_n(x) = \frac{a_n x + b_n}{c_n x + d_n}  \]
for some constants $a_n$, $b_n$, $c_n$, and $d_n$.  Note that
\[ F_n(x) = x \ \Leftrightarrow \ a_n x + b_n = x(c_n x + d_n).  \]
But the quadratic can only be satisfied by three different values if and only if it is identically 0, i.e. $c_n = b_n = 0$ and $a_n = d_n$.  Hence, $F_n(x) = x$ for all $x$.

Note that the condition $f(f(0)) \neq 0$ is not used anywhere.  This doesn't seem useful anyway, since you can check that $f(f(0)) = 0$ if and only if $b(a + d) = 0$, and $a + d = 0$ implies that $F_2(x) = x$ for all $x$.

Nice problem and solution  :) .
Only one question :How do we know that $f(f(f(...f(x)...))$ is defined for any $x \neq -\frac{d}{c}$ ?
Maybe its a stupid question cant see it  :?

Alternatively, one could use the following result:
\[ f^{n}\left( x\right) =\frac{a_{n}x+b_{n}}{c_{n}x+d_{n}},  \]
where $\left( \begin{array} [c]{cc} a_{n} & b_{n}\\ c_{n} & d_{n} \end{array} \right) =A^{n},$ the matrix being $A=\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) .$ This is easy to prove by induction.

Now, if $f^{n}\left( 0\right) =0$ for some $n,$ then $A^{n}$ has the form $\left( \begin{array} [c]{cc} \alpha & 0\\ \beta & \gamma \end{array} \right) .$ Since $A^{n}\cdot A=A\cdot A^{n},$ we obtain $a\alpha +b\beta=a\alpha$ and $b\gamma=b\alpha.$ Since $f\left( 0\right) =b\neq0,$ it follows that $\beta=0$ and $\alpha=\gamma,$ so $f^{n}\left( x\right) =\frac{\alpha x}{\alpha}=x$ for all $x.$

Well, in the day of this exam, the professor Edmilson Motta discovered that the condition $f(f(0)) \not = 0$ is superfluous. The demo is along the last we post here...

Nicely done, enescu!  I tried the matrix approach, but I couldn't get it to work myself.


I'm having some trouble writing my summer science program essay, when they ask "Tell us something unique about yourself to help us understand you better as a person."  what would be considered unique?  Should I tell an anecdote about myself or state facts?  would being bilingual consider unique, not anymore I guess.  I like to solve problems and figure how things work, but I'll bet everyone whoes applying to SSP and on AoSP likes to do that. being the only speech club member on math team? They also ask me why math is important to me, the first answer that comes to mind is that I'm happy when I do math and  that I like figuring how to solve interesting problems with twists.  Its also important to me because I want to become an enginneer, not to mention math and my other favorite subject, physics, are intertwined. But my answer seems so bland and cliche.

any advice?
thanks

While your answer may seem bland and cliche to you, it's all in how you word it.  I read all of the friendly letters submitted with HCSSiM applications, and most of them have the same basic themes---but they're all expressed in different ways.  Your personal passion should come through (just let it out) and that's what I think the purpose of these questions is.

Anyone else applying to SSP?
When do the results come?
According to the CC guys it looks like a week or so.... :huh:


geometry
inequalities
inequalities unsolved

Let $ABCD$ be an inscribable quadrilateral of sides $a,b,c,d$ , diagonals $p,q$, semiperimeter $s$ and area $F$. Prove the two inequalities

$ \frac{1}{p^2} + \frac{1}{q^2} \leq (\frac{s}{2F})^2  $

$ 2pq > (\frac{2F}{s})^2 $

[quote="manlio"]Let $ABCD$ be an inscribable quadrilateral of sides $a,b,c,d$ , diagonals $p,q$, semiperimeter $s$ and area $F$. Prove the two inequalities

$\frac{1}{p^2} + \frac{1}{q^2} \leq (\frac{s}{2F})^2$

$2pq > (\frac{2F}{s})^2$[/quote]

 :D  :D


I know the basic method of induction is to show that it's true for $n=1$, and then showing that if by assuming it's true for $n=k$, then it's true for $n=k+1$, thus showing it's true for all $n$.

However, I have a question on a stronger form of induction: Is it still true  you can still show it is true for $n=1$ and $n=2$, then assuming it's true for $n=k$ and $n=k+1$ you prove it's true for $n=k+2$? I believe it's a valid form of induction, but I'm unsure of whether it's actually true or not.

Yes!  In fact, this stronger form of induction is called Strong Induction :D  You can go even further, of course.  You can assume it is true for $n=k, n=k+1,\ldots, n=k+(j-1)$ and prove it is true for $n=k+j$.  You have to show that the first $j$ base cases work as well.

If you understand why regular induction works, this should seem like a logical extension.  There are even more ways you can use induction as well such as reverse induction and so forth (well, reverse is all I can think of right now...but I'm sure there are other well-known ways to induct).

[quote="joml88"]Yes!  In fact, this stronger form of induction is called Strong Induction.[/quote]

Not quite true.  Suppose we want to prove the statements P(1), P(2), P(3), ... .  In strong induction, we assume that [b]all[/b] statements P(1), P(2), ..., P(k) are true to establish P(k + 1).

[quote="Elemennop"]However, I have a question on a stronger form of induction: Is it still true  you can still show it is true for $n=1$ and $n=2$, then assuming it's true for $n=k$ and $n=k+1$ you prove it's true for $n=k+2$? I believe it's a valid form of induction, but I'm unsure of whether it's actually true or not.[/quote]

This is valid, because this argument shows that P(1) and P(2) imply P(3), and P(2) and P(3) imply P(4), and so on, and it is then "clear" that P(n) is true for all positive integers n.  This is a fairly simple variation of the standard induction argument, so in a proof I would just call it induction.


limit
algebra
binomial theorem
calculus
calculus computations

For (a) I'll use the binomial theorem:

$\sqrt{4n^{2}+1}=2n\left(1+\frac1{4n^{2}}\right)^{\frac12}= 2n\left(1+\frac12\cdot\frac1{4n^{2}}+O(n^{-2})\right)$

$=2n+\frac14+O(n^{-1})$

$\sqrt{n^{2}+3n+1}=n\left(1+\frac3n+O(n^{-2})\right)^{\frac12}$

$=n\left(1+\frac12\cdot\frac3n+O(n^{-2})\right) =n+\frac32+O(n^{-1})$

So, $\sqrt{4n^{2}+1}+\sqrt{n^{2}+3n+1}-3n$

$=2n+n+\frac14+frac32-3n+O(n^{-1})=\frac74+O(n^{-1}).$

The limit is $\frac74.$

The same methods will get us that the limit in (b) is $\frac32.$

[quote="Kent Merryfield"]For (a) I'll use the binomial theorem:

$\sqrt{4n^{2}+1}=2n\left(1+\frac1{4n^{2}}\right)^{\frac12}= 2n\left(1+\frac12\cdot\frac1{4n^{2}}+O(n^{-2})\right)$

$=2n+\frac14+O(n^{-1})$

$\sqrt{n^{2}+3n+1}=n\left(1+\frac3n+O(n^{-2})\right)^{\frac12}$

$=n\left(1+\frac12\cdot\frac3n+O(n^{-2})\right) =n+\frac32+O(n^{-1})$

So, $\sqrt{4n^{2}+1}+\sqrt{n^{2}+3n+1}-3n$

$=2n+n+\frac14+frac32-3n+O(n^{-1})=\frac74+O(n^{-1}).$

The limit is $\frac74.$
[/quote]
I maybe wrong here but I think the answer to [b](a)[/b] is $\frac32.$

First, I think it is possible to justify that

$\lim_{n\rightarrow \infty}(\sqrt{4n^{2}+1}+\sqrt{n^{2}+3n+1}-3n) = \lim_{x\rightarrow \infty}(\sqrt{4x^{2}+1}+\sqrt{x^{2}+3x+1}-3x)$

Now, we make the substitution $y = \frac1{x}$, where $y > 0.$

So, $\lim_{x\rightarrow \infty}(\sqrt{4x^{2}+1}+\sqrt{x^{2}+3x+1}-3x) = \lim_{y \to 0}\frac{\sqrt{4+y^{2}}+\sqrt{1+3y+y^{2}}-3}{y}$

$= \lim_{y \to 0}\frac{\frac{y}{\sqrt{4+y^{2}}}+\frac{3+2y}{2\sqrt{1+3y+y^{2}}}}{1}$ ... (L'Hospital's rule)

$= \frac32$

For part [b](b)[/b] and [b](c)[/b], I would again use the same technique.

Thanks.

What about

$\lim_{x\rightarrow \infty}(\sqrt{2x-\sqrt{x^{2}-1}}-\sqrt{x+\sqrt{x+1}})$


Let x_0=1 and let x_(n+1)=3x_n+[x_n* \sqrt 5]. What is x_2004?

[quote="harazi"]Let $ \ x_0\equal{}1$  and let $ \ x_{n\plus{}1}\equal{}3x_n\plus{}\left\lfloor (\sqrt {5}) x_n\right \rfloor.$ What is $ \ x_{2004}?$[/quote]

$ \ x_0\equal{}1\implies x_1\equal{}5,\ x_2\equal{}26,\ x_3\equal{}136,\ x_4\equal{}712,\ x_5\equal{}3728,$ etc.

It can be proved that $ \ x_{n}\equal{}6x_{n\minus{}1}\minus{}4x_{n\minus{}2}$

$ \implies\boxed{\boxed{x_n\equal{}\left(\frac {\sqrt{5}\plus{}2}{2\sqrt 5}\right)\left(3\plus{}\sqrt 5\right)^n\plus{}\left(\frac {\sqrt{5}\minus{}2}{2\sqrt 5}\right)\left(3\minus{}\sqrt 5\right)^n}}$ or $ \boxed{\boxed{x_n\equal{}\frac {1}{2}\left(\left(3\plus{}\sqrt 5\right)^n\plus{}\left(3\minus{}\sqrt 5\right)^n\right)\plus{}\frac {1}{\sqrt 5}\left(\left(3\plus{}\sqrt 5\right)^n\minus{}\left(3\minus{}\sqrt 5\right)^n\right)}}$


$ \triangle ABC$ has consecutive integral sides and the bisector of  $ \angle A$ ($ AB$ is the smallest and $ AC$ is the largest) meets $ BC$ in $ D$ while $ E$ & $ F$ are mid points of $ AC$ and $ BC$ respectively. If quadrilateral $ ADFE$ has an area that is the smallest non-zero, integer possible then find the sides of the triangle.

[quote="mathbel"]I found  the numbers 3 , 4 , 5  (with Heron's formula).[/quote]

If the sides of the triangle be $ AB\equal{}3, BC\equal{}4 and AC\equal{}5$ then one may easily verify the following: $ \triangle ABD \equal{} 2.25$, $ \triangle EFC \equal{} 1.5$ and $ QUAD ADFE \equal{} 2.25$, which is not an integer.

If I may give you a hint, assume the sides to be $ (x\minus{}1)$, $ x$ and $ (x\plus{}1)$ and determine area of $ \triangle ABC$. Now $ \triangle ABD$ and $ \triangle EFC$ can be determined in terms of $ x$ and thus one can write an equation for the area $ ADFE$ in terms of $ x$. Call this area $ n$. Now find an integral solution ($ n.x$) where $ n\ge 0$.

[$ n\equal{}0$ actually gives $ 1, 2, 3$  as the solution but this isn't a triangle].

I thought the same way to solve this problem , but I took F as AB midpoint and ... :( .
 I try again and the result is AB=13 , BC=14 and AC=15 . Is it OK now??


Let $a_1,\cdots a_n$ be positive integers such that $a_1 \leq \cdots \leq a_n$. Prove that
\[ \frac{1}{a_1}+ \cdots +\frac{1}{a_n} =1 \Rightarrow a_n<2^{2!} \]

I remember this is a famous problem written in Problem-Solving Strategies from Arthur Engel. You can buy see it on http://www.amazon.com/gp/product/0387982191/102-0259262-9615320?v=glance&n=283155&n=507846&s=books&v=glance

A highly recommended book, i must add. i have it at home, and learnt basically all of my knowledge of olympiad level maths single-handedly from it. Its worth whatever price it is.


A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?

$ \textbf{(A)}\ \frac{1}{8} \qquad
\textbf{(B)}\ \frac{5}{32} \qquad
\textbf{(C)}\ \frac{9}{32} \qquad
\textbf{(D)}\ \frac{3}{8} \qquad
\textbf{(E)}\ \frac{7}{16}$

[hide]Following the rule of replacing any bead with a red bead, we have the possible orders RGG, GRG, or GGR. The sum of the probabilities is
\[ \frac {1}{2} \cdot \frac {1}{2} \cdot \frac {1}{4} \plus{} \frac {1}{2} \cdot \frac {3}{4} \cdot \frac {1}{4} \plus{} \frac {1}{2} \cdot \frac {1}{4} \cdot 1 \equal{} \frac {2}{32} \plus{} \frac {3}{32} \plus{} \frac {4}{32} \equal{} \frac {9}{32} \rightarrow \boxed{\textbf{C}}.
\]
[/hide]

You're allowed room for one red bead to pull out. (In fact, you must pull out at least one red bead...)

Case RGG: $ \frac 12 \cdot \frac 12 \cdot \frac 14 \equal{} \frac 1{16}$.

Case GRG: $ \frac 12 \cdot \frac 34 \cdot \frac 14 \equal{} \frac 3{32}$.

Case GGR: $ \frac 12 \cdot \frac 14 \cdot 1 \equal{} \frac 18$.

Adding up, $ \frac 1{16} \plus{} \frac 3{32} \plus{} \frac 18 \equal{} \frac 9{32}$, which is choice $ \mathbf C$.

From this, I can tell that the 2003 AMC 10B was rather easy...


Is it true that every positive rational number can be written in the form (a^2 + b^3)/(c^5 + d^7) where a,b,c,d are positive integers ?

It seems really close to an ISL one.

As Myth says : 'Trust but check'.
So, come on Vess, give a proof.

Pierre.

Yes, thrust but check. But first, check the solved problems section, I am sure I have posted a solution for this problem.


An original permutable inequality of 3 positive variables

Consider 3 positive real variables $ x, y, z$. Prove that:
$ \sum\limits_{cyc} \frac {x(y \plus{} z)^2}{2x \plus{} y \plus{} z} \geq \sqrt {3(x \plus{} y \plus{} z)xyz}$

mods lock this topic this problem i suppose is from an ongoing competition and has been posted several times directly and indirectly  and all have been locked
[url=http://www.mathlinks.ro/viewtopic.php?t=168941]here[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167965]here[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167949]here[/url]
 :huh:

You're false. It's from Uzebekistan MO 1998! We can solve it!

it may be from whatever but it's now a question from an ongoing c :) ompetition and if u don't beleive me contact the mods of inequalities section.


topology
real analysis
real analysis unsolved

Show that if $ f: X\to \mathbb{R}$ is measurable, and $ g: \mathbb{R}\to\mathbb{R}$ is continuous, then $ g\circ f$ is measurable.

$ (g(f(\minus{}\infty, c)))^{\minus{}1} \equal{} f^{\minus{}1}(g^{\minus{}1}(\minus{}\infty,c))$. Note that $ g^{\minus{}1}(\minus{}\infty,c)$ is open in $ f(X) \subset \mathbb R$, then $ g^{\minus{}1}(\minus{}\infty,c) \equal{} f(X) \cap G$ for some open set $ G$ in $ \mathbb R$. Now $ f^{\minus{}1}(g^{\minus{}1}(\minus{}\infty,c)) \equal{} X \cap f^{\minus{}1}(G)$.


algebra
polynomial
superior algebra
superior algebra unsolved

Hello,

I don't know how to proof the following proposition:
"Let $ p$ be a prim number and $ f \in \mathbb F_p[X]$ an irreducible polynomial with $ deg(f)\equal{}m$. Then $ f$ divides the polynomial $ X^{p^m}\minus{}X \in \mathbb F_p[X]$.

My first idea was, that there must exist a polynomial $ g \in \mathbb F_p[X]$ and $ f \cdot g \equal{} X^{p^m}\minus{}X \in \mathbb F_p[X]$, but I think this is not the reasonable way to proof it. Then I thought, it is enough to show, that the roots of $ f$ are also the roots of $ X^{p^m}\minus{}X \in \mathbb F_p[X].$
But how to continue?

Bye,
Edgar

Are you the one I talked to earlier today¿  :D

Take a root $ a$ of $ f$ and consider the field $ \mathbb F_p[a]$. Especially, what can you say about powers of $ a$ in the multiplicative group of that field(bigger hint: Fermat)¿


inequalities
linear algebra
linear algebra unsolved

Let $E={\cal M}_n(\mathbb R)$. We define for $A,B\in E$ the scalar product: $(A|B)=\textrm{tr}(^tA.B)$.

Prove that : $\forall A, B\in E,~~\| AB\|\leq \|A\|\ \|B\|$.

For any pair of $n\times m$ matrices $A,B$ with entries $a_{ij},b_{ij}$ respectively, $\text{tr}(A^TB)=\sum_{i=1}^m\left(\sum_{j=1}^n a_{ji}b_{ji}\right)$; the $ij$ entry of $A^T$ is $a_{ji}$, and that inner sum is the $ii$ entry of the product.

If we identify $M_{m\times n}(\mathbb{R})$ with $\mathbb{R}^{mn}$ using matrices with only one nonzero entry as a basis, this inner product is the Euclidean inner product on $\mathbb{R}^{mn}$. You should know how to prove the inequality over there.

More specifically, http://www.mathlinks.ro/Forum/viewtopic.php?t=84339 .


Prove that: $ F_{3n}\equal{}F_{n\plus{}2}^3\minus{}F_{n\minus{}1}^3\minus{}3F_{n}F_{n\plus{}1}F_{n\plus{}2}$

All of these problems are computations with binet's formula. I would be more interested in how these problems were thought of.

Hi! Use this problem we can prove Fibonacci 8 problem!
[url]http://www.mathlinks.ro/viewtopic.php?t=182727[/url]


In a $ 7\times 7$ board $ 19$ cells are colored. We say that a row or column is colored if it contains at least four colored cells. How many colored rows and columns at most can there be in the board?

At most $ 4$ colored rows and at most $ 4$ colored columns gives $ 8$...

Right... what if $ 20$ cells are colored?


There are some pop ups that say there are new messages wating for you.
But sometimes mine say there are no new messages waiting for you!!  :huh:

It happens when the message pops up, you navigate to a new page, and then click the back button.

sometimes i get that; dont hit the back button if u navigate to a new page or it'll mess everything up!


What is the maximal number of intersection points that six circles can have?

A. 15
B. 24
C. 28
D. 30
E. 36

[hide]The maximum occurs when each circle intersects (at two points) with all the others.
2 x 6C2 = 30. D[/hide]

[quote="Done"]maximal number [/quote]

You mean maximum, right? :) [/i]

well..it's from an Australian Contest  :)

wouldnt it be infinity because if the circles were on top of eachother....

Well, by looking at the answer choices, you can assume they meant distinct circles.

would bezout's theorem not apply to this?

lol englighten us? bezout's theorem?

although i agree with sky9073's method... just test a few out, and you'll see it works...
actually for the intersection of n triangles, the maximum is twice the n'th triangular number... and the nth triangular number may be written as n(n+1)/2 or nC2
so your answer is really 2*n(n+1)/2 = n(n+1) ORRR as sky said, 2*nC2


calculus
integration
function
real analysis
real analysis unsolved

(X,M,m) is a msr space, 1<p<inf and {f_n} is in Lp(X,M,m). If sup{||f_n||p:n>=1}<inf and integral over E  f_{n}dm----->0
for every set E in M (m(E) finite) as n----->inf prove that integral (f_n)(g)dm------>0 for every g in Lq, p and q conjugate.
If (X,M,m) is a sigma finite msr space, prove the same for p=1 and q=inf.

$(X,M,m)$ is a measurable space, $1<p<\infty$ and $f_n \in L_p(X,M,m)$. If $\sup_{n \geq 1} |f_n|_p < \infty$ and $\int_E f_{n}(x) dm(x) \to  0$ with $n$ for every set $E \in M$, $m(E) < \infty$, prove that $\int_X f_n(x) g(x) dm(x) \to  0$ for every $g \in L_q$ where $\frac{1}{p}+\frac{1}{q}=1$.
If $(X,M,m)$ is a sigma finite measurable space, prove the same holds for $p=1$ and $q=\infty$.

The second assumption shows that $|f_n e|_1 = \int_X f_n(x)e(x) dm(x) \to  0$ for every simple function $e=1_{E_1}+1_{E_2}+...+1_{E_k}$ where the $E_k$ are measurable subsets of finite measure. And one knows that these simple functions are dense in $L_p(X,m)$. So, if one choose $g \in L_q$ and $\epsilon > 0$, one can find a simple function $e$ such that $|g-e|_q < \epsilon$. Hence $|f_n g|_1 \leq |f_n (g-e)|_1 + |f_n e|_1 \leq |f_n|_p |g-e|_q + |f_n e|_1$. This is why, because of the boundedness of $f_n$ in $L_p$, $|f_n|_p<C$, one can conlude that $\limsup |f_n g|_1 \leq C \epsilon$, and this for every $\epsilon > 0$. This is precisely the desired conlusion.