The lattice shown is continued for 100 rows. What will be the third number in the 100th row?

$ \begin{tabular}{r c c c c c c c}
Row 1: & 1 & 2 & 3 & 4 & 5 & 6 & 7\\
Row 2: & 8 & 9 & 10 & 11 & 12 & 13 & 14\\
Row 3: & 15 & 16 & 17 & 18 & 19 & 20 & 21\\
Row 4: & 22 & 23 & 24 & 25 & 26 & 27 & 28

\end{tabular}$

$ 99\times7 \plus{} 3 \equal{} \boxed{696}$.
or
$ 100\times7 \minus{} 4 \equal{} \boxed{696}$.

Basically, all integers in the third row are congruent to

$3\pmod 7$

This implies that they can be written in the form

$7m + 3$

The first row corresponds to $m=0$, the second row corresponds to $m=1$, and thus, row $100$ corresponds to $m=99$. Plugging this into the expression, we get,

$7(99) + 3 = 696$


whats the 'nicest' way to do this?

find the smallest x such that 19x = 141 + 180y, where x y are both positive integers

[hide]

141/19 leaves remainder 8

180/19 leaves remainder 9

8 + 9z will be divisible by 19 first when z =16, or when 8+9z is 152.

z=y, so x=159.[/hide]

I'm not sure if this is the same way as rlee, but

x = (141+180y)/19

x = 7+9y+(8+9y)/19

let z = (8+9y)/19; so we want z to be as small of a positive integer as possible

y = (19z-8)/9

y = 2z+(z-8)/9

y is smallest when z = 8 since (z-8)/9 = 0

y = 16

x = 159

[hide]Generally, you want to try to find all the solutions so here it is....

x=7+9y+(8+9y)/19

z=(8+9y)/19; so x=7+9y+z

19z=8+9y --> y=(19z-8)/9=2z+(z-8)/9

t=(z-8)/9 thus y=2z+t, so 9t=z-8...

The parametric solution (the general equation to make 9t=z-8 work).... Is t=n, z=9n+8

Substitute this back in all the way up the ladder to get y=19n+16 and x=159+180n

Ta Da![/hide]


Alex and Becky are racing in a peculiar manner. They both begin at the same spot on a 1 km circular track and move in opposite directions. Alex runs  at 4 m/s and Becky runs at 5 m/s. How many times would they have passed each other if the race stops after 1 hour?

Convert into same units:
Run 1000m for 3600s. A runs 4m/s. B runs 5m/s
At the end of the hour,
A runs 14.4 laps
B runs 18 laps
Every lap they meet the other once, so in total, they met 32 times?


Find primes $ a$ and $ b$ such that $ ab\plus{}1$ and $ ab\minus{}1$ are also primes.

Is it $ a\equal{}2$,$ b\equal{}3$ ?

I guess it and I think its right. But how I can do it without guessing?

WLOG $ a \le b$. If $ a\equal{}2$, then $ b\equal{}3$ or one of $ 2b\plus{}1,2b\minus{}1\equal{}3$, so $ a\equal{}2,b\equal{}3$ or $ a\equal{}b\equal{}2$ are the solutions.
If $ a \neq 2$, then both $ ab\plus{}1$ and $ ab\minus{}1$ are even, i.e. $ \equal{}2$, contradiction.

Sorry but I didn't understand how you solve it. Consider that you didn't see the solutions I gave how you will do it?

He shows that the smaller number must be two and that the larger number can in turn be two or three. He didn't need to know the solutions beforehand to use his method.

[quote="hasan4444"]Find primes $ a$ and $ b$ such that $ ab \plus{} 1$ and $ ab \minus{} 1$ are also primes.

Is it $ a \equal{} 2$,$ b \equal{} 3$ ?

I guess it and I think its right. But how I can do it without guessing?[/quote]

You have a similar problem posted, in which the posters (again) misread the problem as well as you.

The problem statement just asks you to find primes $ a$ and $ b$ such that $ ab\plus{}1$ and $ ab\minus{}1$ are also primes, not find [i]unique[/i] primes $ a$ and $ b$ such that $ ab\plus{}1$ and $ ab\minus{}1$ are also primes or [i]all[/i] primes $ a$ and $ b$ such that $ ab\plus{}1$ and $ ab\minus{}1$ are also primes (or whatever you imagine it says).

So you've found one solution $ (a,b)\equal{}(2,3)$.  Guessing works rigorously in this case.  There is no further proof needed to solve the problem unless you wish to do so (kind of like solving a problem, generalizing it, and then solving the generalized version).

[quote="hasan4444"]Find primes $ a$ and $ b$ such that $ ab \plus{} 1$ and $ ab \minus{} 1$ are also primes.

Is it $ a \equal{} 2$,$ b \equal{} 3$ ?

I guess it and I think its right. But how I can do it without guessing?[/quote]

We know that one of these numbers is divisible by 3 $ ab\minus{}1,ab,ab\plus{}1$. since $ ab\minus{}1,ab\plus{}1$ are two primes, we have some cases:  :D
1- $ ab\minus{}1\equal{}3 \Rightarrow a\equal{}b\equal{}2$ and that's true !!.
2- $ ab\plus{}1\equal{}3$ and that's wrong.
3- $ ab$ is divisible by $ 3$ which means $ a\equal{}3$ or $ b\equal{}3$.

So let us work on case 3. Assume that $ a\ge b$. If $ a\equal{}3$, then $ b\equal{}2,3$ and we find $ (3,2)$. If $ a\ge 4$, then $ b\equal{}3$. We have $ a,3a\minus{}1,3a\plus{}1$ are three primes and that's cannot be happening because $ a$ is odd and $ 3a\minus{}1$ is even greater that $ 2$.

So $ (a,b)\in \{(2,2),(2,3),(3,2)\}$.


Prove that there exist no positive integers $x$ and $y$ such that both $x^{2}+y+2$ and $y^{2}+4x$ are perfect squares.

Suppose that $y^{2}+4x$ is a perfect square. Clearly, $y^{2}+4x > y^{2}$. Also, remark that $y^{2}+4x = (y+1)^{2}$ is impossible modulo 2. Therefore, $y^{2}+4x \geq (y+2)^{2}$, i.e. $x\geq y+1$. But then $x^{2}< x^{2}+y+2 \leq x^{2}+x+1 < (x+1)^{2}$, so $x^{2}+y+2$ can't be a perfect square.


For the most part, I know the rules of "Forms of Answers" for Mathcounts. Yes, I actually even sometimes listen when the guy is reading them before the sprint round. However, I wasn't sure if this would be okay- it came from a 2001 National Sprint-

Answer is: (2/pi)+1

I put: (2+pi)/pi

Is my answer valid? Or is it not simplified enough?

Well, they're not equivalent, I assume you meant that the answer was (2/pi)+1, and then it would be correct.

Oh, whoops, sorry. Thanks for catching that.. I'll make the edit.

Did it say to express in form $ \frac{A}{B\pi} \plus{} C$?

[quote="pinkmuskrat"]No, it didn't...[/quote]

How did it say to express it as?

I see it.

No, your answer is not simplified, you have a $ \frac{\pi}{\pi}$, which MATHCOUNTS would be pesky about.


Prove that there is at least one prime number between 1000 and 2000.

1009 is prime, and 1000 < 1009 < 2000. :)

Are you looking for a proof of Bertrand's Postulate (which Erdos proved)?


give some examples of curvature where we find in real life...why we study these concepts,i need simple introduction..because i know curvature formula nothing more than that...please any one help me..

Which curvature are you talking about?Please clarify..There are plenty of examples in real life(e.g.-Projectile motion,etc.)

I believe [b]barathisid[/b] is refering to the mathematical definition of curvature. Given position vector $ \overrightarrow{r}(t)\equal{}f(t)\overrightarrow{i}\plus{}g(t)\overrightarrow{j}\plus{}h(t)\overrightarrow{k}$, if we let
$ \overrightarrow{T}$ be its unit velocity vector and $ s$ be its arc length, then the curvature $ \kappa\equal{}|\frac{d\overrightarrow{j}}{ds}|$. Loosely speaking, curvature is the rate at which the unit velocity vector rotates.

I'm sure there are plenty of applications in theoretical physics, like curvature of spacetime for example.

One classic example is a circle: note that the smaller the circle is, the faster one turns on it. This especially applies to driving and road construction. If I'm not mistaken, most turns have a point set off the road from which a radius is extended to build a circular arc. This is why most turns can be driven by tilting your wheel at a constant angle.

But more so we define curvature by the rate of change $ \frac{d\phi}{ds}$. This stated bluntly, comes out to be the rate of change in angle per change in distance traveled along a curve. Take this concept back to the circle and we see that we must complete the same revolution for all circles, but with smaller or larger radii. It happens that curvature on a circle is constant (makes sense eh?) and is given by $ 1/r$.

Gooey gave a much more analytical explanation, but it's really quite nice.  :thumbup:

One a more complicated level, one can observe the normals of a curve have a lot to say about curvature. The Locus of centers of curvature are the evolute to a given curve, which is equivalent to the locus of its normal intersections. That is to say one finds the intersection point of one normal line to a curve and its instantaneous neighbor, then the locus of such points.

All food for thought.

The acceleration of a particle moving along a curve can be resolved into a component parallel to the curve and a component normal to the curve. The component parallel to the curve is equal to the rate of change of the speed (speed being defined as the magnitude of velocity) and is what we often naively think of when we think of "acceleration." The component of acceleration normal to the curve equals the curvature times the square of the speed.

So, in designing a road, or a railroad track, or anything like that, we have to ask: what lateral acceleration (lateral force) can our vehicle sustain and stay on the road or track? That limits the curvature we can build in, and relates the curvature to the speed.


[quote="Leon"]If $ a$, $ b$, $ x$, $ y$ are integers greater than 1 such that $ a$ and $ b$ have no common factor except 1 and $ x^a \equal{} y^b$ show that $ x \equal{} n^b$, $ y \equal{} n^a$ for some integer $ n$ greater than 1.[/quote]

Let any prime $ p$. Then $ a\cdot v_p(x)\equal{}b\cdot v_p(y)$ and, since $ \gcd(a,b)\equal{}1$ : $ v_p(x)\equal{}k_pb$ and $ v_p(y)\equal{}k_pa$ and so $ x\equal{}\left(\prod_pp^{k_p}\right)^b$ and $ y\equal{}\left(\prod_pp^{k_p}\right)^a$

Q.E.D.

[quote="pco"]
Let any prime $ p$. Then $ a\cdot v_p(x) \equal{} b\cdot v_p(y)$ and, since $ \gcd(a,b) \equal{} 1$ : $ v_p(x) \equal{} k_pb$ and $ v_p(y) \equal{} k_pa$ and so $ x \equal{} \left(\prod_pp^{k_p}\right)^b$ and $ y \equal{} \left(\prod_pp^{k_p}\right)^a$

Q.E.D.[/quote]

what's $ v_p(x)$?  :help:

$ v_p(x)$ is the number of times p divides x.

i feel the above theorem to be obvious fact that n^ab=n^ab :stink:


Let a,b,c are the side lenghts of a triangle, $ a+b+c=3$.Prove that:
$ \frac{\sqrt{a^{2}+b^{2}c^{2}}}{b+c}+\frac{\sqrt{b^{2}+c^{2}a^{2}}}{c+a}+\frac{\sqrt{c^{2}+a^{2}b^{2}}}{a+b}\ge \frac{3\sqrt{2}}{2}$  :welcome:

we have a similar inequality follow:
[Vasile Cirtoaje]Let $ a,b,c$ be positive real numbers. Prove that:
\[ \sum \frac{\sqrt{a^{2}+bc}}{b+c}\ge \frac{3\sqrt{2}}{2}\]


algebra
polynomial
inequalities
algebra unsolved

Let $ P(x)$ be a polynomial such that the following inequalities are satisfied:
\[ P(0) > 0;\]\[ P(1) > P(0);\]\[ P(2) > 2P(1) \minus{} P(0);\]\[ P(3) > 3P(2) \minus{} 3P(1) \plus{} P(0);\]
and also for every natural number $ n,$ \[ P(n\plus{}4) > 4P(n\plus{}3) \minus{} 6P(n\plus{}2)\plus{}4P(n \plus{} 1) \minus{} P(n).\]
Prove that for every positive natural number $ n,$ $ P(n)$ is positive.

I'm not sure how valid this is; motivated by [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]MellowMelon[/url]

The question can be restated as

$ \Delta^k f(0) > 0 \quad k \equal{} 0,1,2,3,4$

The fact that $ \Delta^4 f(n) > 0$ shows that $ \Delta^3 f(n)$ is increasing*. Since it starts larger than 0 we know $ \Delta^2 f(n)$ is increasing. Following this argument through we conclude that $ f(n)$ is increasing and since $ f(0) >0$ it is positive for all integers.

* increasing is to be taken over the integers

Actually $P(x)$ may be any function $P:X\rightarrow Y$ such that $Z\setminus Z_-\subseteq X\wedge Y\in R$ and the problem remains true.


[hide="Hint"]
Find the numbers with one 2, two 2s an so on with the three's too.[/hide]

[quote="xxrxxhxx"]How many five digit numbers have at least one 2 or one 3 among their digits?[/quote]

The way to solve this is to find the numbers that have no $ 2$'s or $ 3$'s. There are $ 7\cdot8^4\equal{}28672$. Now we subtract that from $ 9\cdot10^4\minus{}28672\equal{}90000\minus{}28672\equal{}61328$. I think.....  :maybe:


If a beast randomly throws a dart at a square dartboard with side length 2, what is the probability that the dart hits closer to the center of the dartboard than any corner, but within a distance of 1 from a corner?

$ \textbf{(A)} \pi\qquad\qquad\qquad\textbf{(B)}
\dfrac{\pi\minus{}2}{4}\qquad\qquad\qquad\textbf{(C)}
\dfrac{\pi\minus{}4}{\pi}\qquad\qquad\textbf{(D)}
\dfrac{1}{4}\qquad\qquad\textbf{(E)}
\dfrac{1}{8}$

[quote="#H34N1"]If a beast randomly throws a dart at a square dartboard with side length 2, what is the probability that the dart hits closer to the center of the dartboard than any corner, but within a distance of 1 from a corner?

$ \textbf{(A)} \pi\qquad\qquad\qquad\textbf{(B)} \dfrac{\pi \minus{} 2}{4}\qquad\qquad\qquad\textbf{(C)} \dfrac{\pi \minus{} 4}{\pi}\qquad\qquad\textbf{(D)} \dfrac{1}{4}\qquad\qquad\textbf{(E)} \dfrac{1}{8}$[/quote]

The dart has to hit inside a $ \sqrt{2}$ by $ \sqrt{2}$ square formed by connecting the midpoints of the sides of the square.  Since it has to be withing 1 of a corner, we need to find the area of a quarter circle minus the isosceles triangle:

\[ \frac{\pi}{4}\minus{}\frac{1}{2}\equal{}\frac{\pi\minus{}2}{4}
\]

This area shows up 4 times, but we must divide by 4, so the answer is just $ \boxed{B}$


Let the function f : R--->R satisfy:
                       f(x) <= x
                       f(x+y) <= f(x)+f(y)
Prove that: f(x) = x.

Plugging $x=y=0$ in both conditions yields $f(0)=0$. Setting $y=-x$ in the second condition gives $f(x)+f(-x)\ge 0$, hence $x\ge f(x)\ge -f(-x)$, that is, $x\ge -f(-x)$ for all $x$. Replacing $x$ with $-x$ gives $-x\ge -f(x)$, or $x\le f(x)$, which, together with the first condition implies $f(x)=x$.


hi,
how many real (a,b) exits that both of below condition shold be trus for them(solve the equation)
1)1+(a^3)+3*a*b=(b^3)
and
2)1+(a^5)=(b^5)
 :lol:

I just give a hint. :D
[hide]$ 1\plus{} a^3 \minus{}b^3 \equal{} (1\plus{}a\minus{}b)(1\plus{}a^2\plus{}b^2\plus{}a \minus{}b\minus{}ab)\minus{}3ab$
and $ 1\plus{}a^2\plus{}b^2\plus{}a \minus{}b\minus{}ab \geq  0$[/hide]
([b]irantst[/b] try to use LaTeX  :wink: )


geometry
geometric transformation
rotation
modular arithmetic
algorithm
number theory
Euclidean algorithm

Find the sum of the first hundred numbers appearing in both the progressions 17,21,25 ... and 16,21,26

IN how many ways can 4 gentlemen and 7 ladies form a round so that no two gentlemen are consecutive?

For the first one, find the first few numbers that belong to each sequence and then note the pattern.  Once you observe this pattern, simply use the formula for sum of an arithmetic series.

Are the gentlemen and ladies distinct?

Place all the men together.  Note that we have already determined the places of at least four women.  Now count the places where the remaining three women must sit.

For the 1st sequence: $ a_1 \equal{} 17$ and $ d \equal{} 4$ so $ a_{100} \equal{} a_1 \plus{} 99d \equal{} 17 \plus{} 99*4 \equal{} 413$ and the sum is $ S_n \equal{} \frac {n(a_1 \plus{} a_n)}{2}$ so $ S_{100} \equal{} \frac{100(17 \plus{} 413)}{2} \equal{} 21500$.  For the second sequence, $ a_1 \equal{} 16$ and $ d \equal{} 5$ so $ a_{100} \equal{} 16 \plus{} 5*99 \equal{} 511$ and thus $ S_{100} \equal{} \frac {100(16 \plus{} 511)}{2} \equal{} 26350$  So the sum of the first hundred terms of both sequences is $ 21500 \plus{} 26350 \equal{} 47850$.

azn, what exactly do you mean?  I don't think that works... at least not very well.

by "round," I hope it is supposed to mean "circle."  There are $ 7!/7 *1/2 \equal{} 360$ ways to arrange the ladies in a circle by themselves (divide by 7 to account for rotations of the circle and divide by two because having the people arranged in the clockwise or counterclockwise directions is indistinguishable).  Then we have to choose $ 4$ of $ 7$ places to place the men, for $ 7!/3! \equal{} 840$ ways to place them.  Thus we have a total of $ 360*840 \equal{} 302400$ possible arrangements.

I think that the numbers must appear in both sequences.

$ 17\plus{}4a\equal{}16\plus{}5b \rightarrow 5b\minus{}4a\equal{}1$.  Applying Extended Euclidean Algorithm/Chinese Remainder Theorem.
$ n \equiv 1 \pmod {20}$.
$ \sum_{n\equal{}1}^{100}20n\plus{}1\equal{}20\frac{100\cdot101}{2}\plus{}100\cdot1$.
Or, $ 100\cdot\frac{21\plus{}21\plus{}99\cdot20}{2}$

This is $ 101100$.

[quote="azn6021023"]I think that the numbers must appear in both sequences.

$ 17 \plus{} 4a \equal{} 16 \plus{} 5b \rightarrow 5b \minus{} 4a \equal{} 1$.  Applying Extended Euclidean Algorithm/Chinese Remainder Theorem.
$ n \equiv 1 \pmod {20}$.
$ \sum_{n \equal{} 1}^{100}20n \plus{} 1 \equal{} 20\frac {100\cdot101}{2} \plus{} 100\cdot1$.
Or, $ 100\cdot\frac {21 \plus{} 21 \plus{} 99\cdot20}{2}$

This is $ 101100$.[/quote]

Can you plz explain this further.

I don't think all the euclidean algorithm is necessary.  First, by chinese remainder theorem, the new sequence formed has $ d\equal{}lcm(4,5)\equal{}20$.  Since the smallest term included in both sequences is $ 21$, the $ n$th term is $ a_n\equal{}21 \plus{} 20(n\minus{}1) \equal{} 20n\plus{}1$


Let x, y, z be positive real numbers such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that
\[ \frac {{x^2 \plus{} 1}}{{y^2 \plus{} 1}} \plus{} \frac {{y^2 \plus{} 1}}{{z^2 \plus{} 1}} \plus{} \frac {{z^2 \plus{} 1}}{{x^2 \plus{} 1}} \le \frac {7}{2}
\]

[quote="dvtvd"]Let a, b, c be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {{x^2 \plus{} 1}}{{y^2 \plus{} 1}} \plus{} \frac {{y^2 \plus{} 1}}{{z^2 \plus{} 1}} \plus{} \frac {{z^2 \plus{} 1}}{{x^2 \plus{} 1}} \le \frac {7}{2}
\]
[/quote]
What are $ x, y, z$? :?:

[quote="dvtvd"]Let a, b, c be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {{x^2 \plus{} 1}}{{y^2 \plus{} 1}} \plus{} \frac {{y^2 \plus{} 1}}{{z^2 \plus{} 1}} \plus{} \frac {{z^2 \plus{} 1}}{{x^2 \plus{} 1}} \le \frac {7}{2}
\]
[/quote]
quite a funny typing mistake anyway

[quote="nguoivn"]Yes, it's an old problem and very easy. But I haven't seen any nice proof on anywhere  :wink:
I'll post my proof later  :)[/quote]
Maybe it be easy to u coz u have done it previously however there are many who have not done it previously.for them I think it is another good inequality. :maybe:

Could you post your solution, please?  :blush:


Find all pairs $ (a, b) \in \mathbf{N}^2$ such that $ a$ is prime and $ a^b \plus{} b^a$ is prime.

EDIT: Prove that there are infinitely many such pairs.

Are you sure this can be solved? :maybe:

Here are solutions for $ a<10,b<100$ from Mathematica

$ (2,\{1,3,9,15,21,33\}),(3,\{2,56\}),(5,24),(7,54)$


Let $ a,b,c\in R^\plus{}$. Prove that

$ \frac{a\plus{}b}{b\plus{}c}\plus{}\frac{b\plus{}c}{c\plus{}a}\plus{}\frac{c\plus{}a}{a\plus{}b}\plus{}\frac{3(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2}\geq 4$

Is it a ineq in "Secrets in ineqs"-Pham Kim Hung-2006 ?

[quote="nguyentrang"]Let $ a,b,c\in R^ \plus{}$. Prove that

$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq 4$[/quote]
Assum $ c \equal{} max${a;b;c}

$ Ineq \leftrightarrow$
$ [\frac {1}{(a \plus{} c)(b \plus{} c)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} b)^2 \plus{} [\frac {1}{(a \plus{} c)(a \plus{} b)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} c)(b \minus{} c) \ge\ 0$

Which is obvious ! :wink:

We have done!  :)

Using one of my old results, we can prove the stronger:
$ \sum\ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {(6 \minus{} \sqrt []{3})(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \geq\ \frac {15 \minus{}\sqrt[]{3}}{3}$
 :)

[quote="onlylove_math"]
Assum $ c \equal{} max${a;b;c}
$ Ineq \leftrightarrow$
$ [\frac {1}{(a \plus{} c)(b \plus{} c)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} b)^2 \plus{} [\frac {1}{(a \plus{} c)(a \plus{} b)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} c)(b \minus{} c) \ge\ 0$
[/quote]
Can you show a few steps on how you got here?

[quote="dgreenb801"][quote="onlylove_math"]
Assum $ c \equal{} max${a;b;c}
$ Ineq \leftrightarrow$
$ [\frac {1}{(a \plus{} c)(b \plus{} c)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} b)^2 \plus{} [\frac {1}{(b \plus{} c)(a \plus{} b)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} c)(b \minus{} c) \ge\ 0$
[/quote]
Can you show a few steps on how you got here?[/quote]
Yes, I 'll show it . It 's quite simple

$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq 4$

$ \leftrightarrow \frac {b \minus{} a}{c \plus{} a} \plus{} \frac {a \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} b}{a \plus{} b} \plus{} \frac {(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \ge\ 0$

$ \leftrightarrow [\frac {b \minus{} a}{c \plus{} a} \plus{} \frac {a \minus{} b}{b \plus{} c}] \plus{} [\frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} b}{a \plus{} b}] \plus{} \frac {(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)}{(a \plus{} b \plus{} c)^2} \ge\ 0$

$ \leftrightarrow \frac {(a \minus{} b)^2}{(c \plus{} a)(b \plus{} c)} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{(b \plus{} c)(a \plus{} b)} \plus{} \frac {(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)}{(a \plus{} b \plus{} c)^2} \ge\ 0$

$ \leftrightarrow [\frac {1}{(a \plus{} c)(b \plus{} c)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} b)^2 \plus{} [\frac {1}{(b \plus{} c)(a \plus{} b)} \minus{} \frac {1}{(a \plus{} b \plus{} c)^2}](a \minus{} c)(b \minus{} c) \ge\ 0$

Besides, We have :

$ \frac{a\plus{}kc}{a\plus{}kb}\plus{}\frac{b\plus{}ka}{b\plus{}kc}\plus{}\frac{c\plus{}kb}{c\plus{}ka} \minus{} 3$

$ \equal{} \frac{k^2}{(ka\plus{}c)(kb\plus{}c)}(a\minus{}b)^2\plus{}\frac{k[(k\minus{}1)a\plus{}(k^2\minus{}k\plus{}1)b\plus{}kc]}{(a\plus{}kb)(b\plus{}kc)(c\plus{}kb)}(a\minus{}c)(b\minus{}c)$

P/s: It is SS method which was created by a Vietnamese :lol: ( I think so)


knd aveti subiectele de la sibiu....zicetimi si mie..:D....:P

[quote="Salajan Sabin"]ti le dau eu luni la scoala  :)[/quote]

Eu sunt din mures :D poate ma confunzi
daca le ai in format electronic te rog sa le postezi

pe http://www.tmmate.ro se gasesc subietele de la "Traian Lalescu".

[quote="AlinuS"]poate sa posteze cineva subiectele din anul acesta ?[/quote]
Ce clasa esti? Eu am subiectele de la a XI-a.


calculus
derivative
trigonometry
real analysis
real analysis theorems

Could someone explain to me the chain rule in English?

My calculus textbook has a lot of weird stuff that I can't understand.

see this link

[hide][url]http://en.wikipedia.org/wiki/Chain_rule[/url][/hide]

thanks..
I also asked someone else about it..

All I need to do now is to get myself cleared up with my trig stuff...

Thanks!!


Warning: This problem may seem [b]extremely[/b] easy, but do not let that deter you from solving it- esp. for those of you to who this problem may seem like a joke :D I am mainly interested in finiding more intricate solutions than the one i have found (i will post it after other ppl post their comments/solutions)

Anyway,

Coin A is flipped tree times and coin B is flipped 4 times. What is the probability that the number of heads obtained from flipping the two fair coins is the same.

I warned you it was easy...or so it seems :D

:? well I'd think there'll either be 0,1,2 or 3 coins. And a total of 2^7 = options

so 0 heads: 1
1 heads: 12
2 heads: 18
3 heads: 4

so for a total of 35/128

[quote="3X.lich"][quote="splintercelll990"] esp. for those of you to who[m] this problem may seem like a joke :D ... I warned you it was easy...or so it seems :D[/quote]
WOW, I can't believe this!!! What a joke! It is the easiest of the tough flipping coins problems  ;)  :D  :D[/quote]

i warned you 3x...anyway

[quote="Peter VDD]well I'd think there'll either be 0,1,2 or 3 coins. And a total of 2^7 = options

so 0 heads: 1
1 heads: 12
2 heads: 18
3 heads: 4

so for a total of 35/128 [/quote]

you got it right!, but what do you mean by 0,1,2 or 3 coins?, dont you mean 0,1,2,3 options to get heads?

whoops...the above positng was me (forgot to log in)

[quote="splintercelll990"]
you got it right!, but what do you mean by 0,1,2 or 3 coins?[/quote]

I meant 0,1,2 or 3 coins falling on heads in both cases (4 heads is not possible since one of them only has 3 coins)


The Riddle of Dracula is a series of riddles. All of these riddles deal with two types of inhabitants: humans and vampires. The inhabitants behave according to the following rules:

1. Humans always tell the truth, while vampires always lie.

2. Some humans, as well as some vampires, are sane, while others are insane. Sane creatures believe in what is true, while insane creatures believe in what is false. (So an insane human will never tell you the truth and an insane vampire will always tell you the truth)

An inhabitant says "I believe I am insane." What type is he?

Feel free to post your own riddles, as long as they are of this type.

[quote="G-UNIT"]Bob, I think you should credit the source in which you obtained the problem.  In this case I believe it's The Lady and The Tiger by Raymond Smullyan.[/quote]

I think it's from [i]What is the Name of this Book?[/i], but I'm not sure. So I'll just put Raymond Smullyan

How do we know that the vampire is insane?  A sane vampire would know that he were sane, and then lie about it.

1. AND 2. are INCONSISTENT as presented.

For example, "Humans always tell the truth" means, WITHOUT exception, you will never find any humans that do not tell the truth.

Combined with...

But then you typed "Some humans...are insane."  "...an insane
human will never tell you the truth..."

If these are part of Raymond Smullyan's own words taken out of
the originally intended order and context, then he was/is wrong.

Amend that to "Humans tell what they believe to be true".


$x+y+z=1$ are three real positive numbers. Prove that $\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}\le \sqrt{10}$

It's clear that $xy + yz + zx \leq \frac{(x+y+z)^2}{3}=\frac{1}{3}$

Jensen Inequality yields:

$\sum \sqrt{1+xy} \leq 3\sqrt{\frac{1}{3}(3 + xy + yz + zx)} = \sqrt{9 + 3(xy + yz + zx)} \leq \sqrt{10}$

QED.

[quote="TomciO"]It's clear that $xy + yz + zx \leq \frac{(x+y+z)^2}{3}=\frac{1}{3}$[/quote]

$(x+y+z)^2\geq 3(xy+yz+zx)$
$\Rightarrow  x^2+y^2+z^2+2xy+2yz+2zx\geq 3xy+3yz+3zx$
$\Rightarrow  x^2+y^2+z^2\geq xy+yz+zx$, which is well known.

However, is there a general way to prove it?

Here is an alternative solution with out Jensen. here it goes:

 [color=blue][b] Problem.[/b]

   Let $x, y, z$ three positive real numbers such that $x+y+z=1$. Prove that

                          \[ \sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}\leq\sqrt{10}.  \]

  [/color]
  [i] Solution. [/i]

    We will use this well-known inequality first: \[ (x+y+z)^{2}\geq 3(xy+yz+zx).  \]
    We get \[ xy+yz+zx\leq\frac{1}{3}.  \]
    On the other hand, by Cauchy-Schwarz inequality we have \[ a^{2}+b^{2}+c^{2}\geq\frac{(a+b+c)^{2}}{3}.  \]
     By putting $a=\sqrt{\alpha}, b=\sqrt{\beta}$ and $c=\sqrt{\psi}$ we obtain the following well-known inequality

                                                  \[ \sqrt{\alpha}+\sqrt{\beta}+\sqrt{\psi}\leq\sqrt{3(\alpha+\beta+\psi)}.  \]

    Now by making the substitutions $\alpha=1+xy, \beta= 1+yz$ and $\psi=1+zx$
    and by using the first inequality we get the conclusion easily.

[quote="Ezbakhe Yassin"]I think it could be extended to $x+y+z\leq 1$[/quote]

This follows trivially - for any $x + y + z = 1$ we can make a surjection to $x_1 + y_1 + z_1 < 1$ by reducing the value of one of them, which clearly reduces the value of the sum...


group theory
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ K$ be a field and $ A$ a valuation ring of $ K$ with value group $ \Gamma$.
If $ p$ is a prime ideal of $ A$ what are the value groups of $ A_p$ and $ A/p$ ?

[quote="Third Edition"]Assume $ \Gamma \equal{} K^*/A^*$.[/quote]
Well... that's the definition I use for value group.
For localisation I got $ \Gamma/\Delta_p$ where $ \Delta_p\equal{}<v(A\minus{}p)>$ (where $ v$ is the valuation and $ <X>$ means the group generated by $ X$) but for quotient I got something that doesn't look nice.

What is the order on $ \Gamma/\Delta_p$? And let's see what the nasty thing for the quotient looks like. ;)

The order on $ \Gamma/\Delta_p$ is $ \overline{x}\ge 0$ iff $ x\ge\Delta_p$. Here it is important that $ \Delta_p$ is isolated, i.e. $ 0\le u\le v$ and $ v\in\Delta_p$ implies $ u\in\Delta_p$.

For the quotient:
the invertible elements in $ A/p$ are $ A/p\minus{}m/p$ where $ m$ is the maximal ideal of $ A$ so I got $ K(A/p)^*/(A/p\minus{}m/p)$ and I don't know how to relate this with $ \Gamma$.

[quote="Kuba"]The order on $ \Gamma/\Delta_p$ is $ \overline{x}\ge 0$ iff $ x\ge\Delta_p$. Here it is important that $ \Delta_p$ is isolated, i.e. $ 0\le u\le v$ and $ v\in\Delta_p$ implies $ u\in\Delta_p$.[/quote]How did you derive this definition?
[quote]For the quotient:
the invertible elements in $ A/p$ are $ A/p \minus{} m/p$ where $ m$ is the maximal ideal of $ A$ so I got $ K(A/p)^*/(A/p \minus{} m/p)$ and I don't know how to relate this with $ \Gamma$.[/quote]Does $ K(A/p)\equal{}A_p/pA_p$ help you?

[quote="Third Edition"][quote="Kuba"]The order on $ \Gamma/\Delta_p$ is $ \overline{x}\ge 0$ iff $ x\ge\Delta_p$. Here it is important that $ \Delta_p$ is isolated, i.e. $ 0\le u\le v$ and $ v\in\Delta_p$ implies $ u\in\Delta_p$.[/quote]How did you derive this definition?[/quote]
Well, I think this is the natural order on a quotient (by an isolated subgroup), with a minor correction:
$ \overline{x}\ge 0$ if $ \overline{x}\equal{}0$ or if $ x>\Delta_p$. I think it's equivalent with: the positive elements in the quotient are exactly the images of the positive elements.


function
topology
real analysis
real analysis unsolved

The information missing is that the metric space must be compact,but i am failing to prove it.

It is not at all clear what you are asking.

Let me throw out an example, so you can tell me whether it has anything to do with what you're talking about.

Let $ X\equal{}\{a,b\},$ as set with two elements. Define $ d(a,b)\equal{}d(b,a)\equal{}1.$ Of course, we also have $ d(a,a)\equal{}d(b,b)\equal{}0.$ This is certainly a metric space, and a compact metric space at that. The ball centered at $ a$ of radius $ 2$ is all of $ X,$ and it is not connected.

I think the question is this:
Let $ X$ be a compact metric space. (btw what about locally compact or complete case?) If the closure of the open ball $ B(x,\epsilon) \equal{}\{y : d(y,x) <\epsilon\}$ in $ X$ is the closed ball $ \{y : d(y,x)\leq\epsilon\}$ for all $ x$ and $ \epsilon > 0$, then $ B(x,\epsilon) \equal{}\{y : d(y,x) <\epsilon\}$ is always connected.

This is exactly what i meant.
I have tried '' un raisonnement par l'absurde" supposing that htere is an open ball which is not connected and considered it as union of two disjoint nonempty  open substs , and considered the following continuous funtion:
d from the open ball to real number d(x) = d(a,x) where a is the center of the ball;and got stucked here.

You really should have put this in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165042]other thread[/url] rather than starting a new one.

A language note: In English, we use either "proof by contradiction" or the Latin form "reductio ad absurdum".

The classical proof of connectedness of $ [0,1]$ is to induce a contradiction of maximality of a value that quantifies a moment of crossing from one piece of $ [0,1]$ to the other piece of $ [0,1]$.

Let $ X$ be a metric space where all closed balls are compact and closure of any open ball is its corresponding closed ball. (example : $ \mathbb{R}^{n}$).

Suppose a closed ball $ \overline{B(x,R)}$ in $ X$ is not connected. Partition $ \overline{B(x,R)}$ into two nonempty disjoint compact sets $ K$ and $ L$. Assume $ x\in K$. Let $ r \equal{}\sup\{ e :\overline{B(x,e)}\subset K\}$. (The value of $ r$ is about the moment of crossing from $ K$ to $ L$.) Then $ 0 < r < R$ and the open ball $ B(x,r)$ is in $ K$. Therefore its closure $ \overline{B(x,r)}$ is also in $ K$.

The distance $ d \equal{} d(\overline{B(x,r)}, L)$ is positive because the two involved sets are compact.
Let $ r' \equal{}\min\{r\plus{}d/2, R\}$ then we have $ \overline{B(x,r')}\subset K$ contradicting to the maximality of $ r$.

We proved all closed balls in $ X$ are connected. An open ball is a union of closed balls with the same center, hence connected.


algebra
polynomial
geometry
geometric transformation
group theory
abstract algebra
Ring Theory

Suppose that p(x) in C[x], the polynomial ring in the complex field, is an irreducible polynomial, and let w be a root of p(x) in some field extension.  Let E be the splitting field of g(x)=(x^2 + 1)*(x), where the term (x) is the minimal polynomial of w over R, the reals.  Show that E contains the splitting field of p(x) over C, and that 2 divides |Gal(E/R)|.

Ok, so I know if we start with an irreducible polynomial, then Galois action is transitive.  I believe we WTS that f(w)=0 implies w is in E which implies w is in F since E generates F.
Also, let x in Gal(E/C).  Then 0 = x(f(w)) = f(x(w)).

Where do I go from here? Actually, where do I start, knowing all this already?

Already April 1st in some forsaken part of this planet?

  darij

Approximate translation: "Is this a joke?"

From your first two sentences, we have that $ p(x)$ is an irreducible factor of $ x^2\plus{}1$ in $ C[x]$, which makes it either $ x\plus{}i$ or $ x\minus{}i$. There's nothing here, because you're asking the wrong question.

This still looks extremely pointless. "An irreducible polynomial in $ \mathbb{C}\left[x\right]$" is synonymous to "a linear polynomial", and $ w$ must lie in $ \mathbb{C}$ by the fundamental theorem of algebra. Talking about the splitting field of a polynomial over $ \mathbb{C}$ is equally nonsense - it will just be $ \mathbb{C}$. Of course, $ \mathbb{C}$ contains $ \mathbb{C}$, and $ 2$ divides $ \left|\mathrm{Gal}\left(\mathbb{C}/\mathbb{R}\right)\right|$ (in fact, $ 2$ [i]is[/i] $ \left|\mathrm{Gal}\left(\mathbb{C}/\mathbb{R}\right)\right|$). But are you sure you understood the problem right?

  darij

We haven't learned the fundamental theorem of algebra yet.  I think we are supposed to get to the theorem based on this part of the question.  How can I do it without knowing the theorem?

Are you just asking for help with homework?  That's really not a great idea.

If you aren't, then you're a little too stuck on this idea of only using what you've seen already.  It's true that you need to build your mathematics from the ground up, but I've seen you reject legitimate solutions in other threads using this excuse.  A mathematics discussion should be a dialog, when you throw your hands up and make us do all the work it's not going to help you very much.

That said, your notion of "splitting field" is very imprecise.  To say that a splitting field over $ \mathbb{R}$ contains some splitting field over $ \mathbb{C}$ doesn't really work--you'd need to say that one can be isomorphically embedded in the other. (splitting fields don't really "exist somewhere"--you have to construct them)

But do you see what's going on?  Certainly you can arrange for $ E$ to contain the complex numbers, as it contains the reals and the roots of $ x^2\plus{}1$, and if you look closely I think you'll find that your inclusion of splitting fields is not very hard, fundamental theorem or no. (hint: what is the only property a field must have to contain a splitting field?)

The only other part is showing that $ 2$ divides $ |Gal(E/R)|$. (this is a separate problem, by the way--it's my taste to post separate problems in separate threads, or just solve one and post the other)  You can do that by showing that $ Gal(E/\mathbb{R})$ has a subgroup of index $ 2$.  Honestly, this is a trivial exercise in Galois theory, and I would reread the sections of your book on the basic theory and the Galois correspondence if you're having trouble with it.


Here's a problem I proposed some time ago to some of my students:

-Given an aqueous solution of a single solute (with molar mass M g/mol), whose composition is P % (m/m), and with density $ \rho$ g/mL, find a general expression for its molar concentration (mol/L).

[hide="is the ans "]$ \frac{10.P. \rho}{M}$  :maybe:  :maybe: [/hide]


[quote="verneer"]I would like to insert numberlines (showing inequalities) into Latex.[/quote]

I don't understand what you want.  Do you want somthing like this:

  .............
  .............
  .............
  .............
5.............
  .............
  .............
  .............
  .............
10.............

And about inequalities, I just have no idea what you want...

sorry

ledurt

I think he's talking about something like this:
<--2--3--4==5==6==>
(x>4)

As far as I know, you can't do this with LaTeX. You need to create a separate picture (but I don't know what would be used to make a number line...).

You can do this using some of the drawing packages, I prefer pgf.

Would something like this meet your needs?

<\cdots 2 \cdots 3 \cdots 4 --- 5 --- 6 \to

Geoff

that would look like
$<\cdots 2 \cdots 3 \cdots 4 --- 5 --- 6 \to
$

[quote="verneer"]I hope I'm posting this in the right place.  I would like to insert numberlines (showing inequalities) into Latex.  Is there a way to do it with Latex, or if not, what software is available which will allow me to do it?

Thanks in advance!

     -- Pawel[/quote]

This is exactly the task I faced a few days ago.  The easiest way to do this, IMO, is to use the Eukleides package, which relies on PSTricks (get it at http://perso.wanadoo.fr/obrecht/).

Here is how it works:  You write your special Eukleides code in a file, say file.euk.  You then compile it using the binary eukleides (or eukleides.exe if you're using Windows) and redirect the output to some file, say output.  You then copy and paste the output into your LaTeX document.

You can also invoke the euktoeps shell script (or euktoeps.exe on Windows) to directly convert your .euk source file into an EPS file, which you can later include in your LaTeX document.

For example, the following Eukleides source code

---------------------------------------
% ineq1.euk
ninfty = -16
infty = 22
frame(ninfty,-2,infty,16,0.3)

l1 = 15
l2 = 12
l3 = 9
l4 = 6
l5 = 3
l6 = 0

draw(segment(point(-11,l1),point(15,l1)), full)
draw("$-11$", point(-11,l1), .4, 270:)
draw("$15$", point(15,l1), .4, 270:)

draw(segment(point(-6,l2),point(ninfty,l2)), full, arrow)
draw(segment(point(-4,l2),point(infty,l2)), full, arrow)
draw("$-6$", point(-6,l2), .4, 270:)
draw("$-4$", point(-4,l2), .4, 270:)

draw(segment(point(-2,l3),point(ninfty,l3)), full, arrow)
draw(segment(point(0,l3),point(infty,l3)), full, arrow)
draw("$-2$", point(-2,l3), .4, 270:)
draw("$0$", point(0,l3), .4, 270:)

draw(segment(point(2,l4),point(ninfty,l4)), full, arrow)
draw(segment(point(6,l4),point(infty,l4)), full, arrow)
draw("$2$", point(2,l4), .4, 270:)
draw("$6$", point(6,l4), .4, 270:)

draw(segment(point(7,l5),point(ninfty,l5)), full, arrow)
draw(segment(point(9,l5),point(infty,l5)), full, arrow)
draw("$7$", point(7,l5), .4, 270:)
draw("$9$", point(9,l5), .4, 270:)
---------------------------------------

produces the output in the attached ineq1.gif, and the following source code

---------------------------------------
% ineq2.euk
ninfty = -16
infty = 22
frame(ninfty,-2,infty,4,0.3)

l1 = 3
l2 = 0

draw(segment(point(-11,l1),point(-6,l1)), full, doublearrow)
draw("$-11$", point(-11,l1), .4, 270:)
draw("$-6$", point(-6,l1), .4, 270:)

draw(segment(point(-4,l1),point(-2,l1)), full, doublearrow)
draw("$-4$", point(-4,l1), .4, 270:)
draw("$-2$", point(-2,l1), .4, 270:)

draw(segment(point(0,l1),point(2,l1)), full, doublearrow)
draw("$0$", point(0,l1), .4, 270:)
draw("$2$", point(2,l1), .4, 270:)

draw(segment(point(6,l1),point(7,l1)), full, doublearrow)
draw("$6$", point(6,l1), .4, 270:)
draw("$7$", point(7,l1), .4, 270:)

draw(segment(point(9,l1),point(15,l1)), full, doublearrow)
draw("$9$", point(9,l1), .4, 270:)
draw("$15$", point(15,l1), .4, 270:)

draw(point(0,l2))
thickness(3)
xaxis = segment(point(ninfty,l2),point(infty,l2))
draw(xaxis, full, doublearrow)
draw("$0$", point(0,l2), .4, 270:)
---------------------------------------

produces the output in the attached ineq2.gif.
IMPORTANT:  Eukleides offers only a limited number of the great selection of PSTricks arrowheads.  I achieved the empty circles (open interval) by adding {o-o} after \psline in the output, or changing {->} to {o->}.  For more information and tweaking, such as creating full circles to denote closed intervals, refer to the PSTricks documentation that explains how to use arrowheads.

Good luck.

A similar solution is to use MetaPost.  It is not that easy to learn to use, but once you get used to it, it produces very high quality diagrams.

Learn about MetaPost [url=http://www.tug.org/metapost.html]here[/url].

I don't know how to put real photos in LaTex-Is it possible anyway?

I prefer excellent program GCLC made by professor Predrag Janicic. Visit this web page:

http://www.matf.bg.ac.yu/~janicic/gclc/


algebra
polynomial
calculus
integration
algebra proposed

Prove that any polynomial $\in \mathbb{R} [X]$ can be written as a difference of two strictly increasing polynomials.

The problem is not very hard, but it states an interesting fact.

:) Cute.

Let $f$ be our polynomial, which we assume has degree $n$, and let $m$ be an even number larger than $n-1$. The polynomial $x^m-f'$has limit $\infty$ at both $\pm\infty$, so there is a positive constant $c$ such that $g(x)=x^m+c-f'(x)>0,\ \forall x\in\mathbb R$. This means that we have succeeded in writing $f'(x)$ as the difference between two polynomials $x^m+c$ and $g(x)$ which take only positive values. Now $f$ will be the difference between a primitive of $x^m+c$, which is strictly increasing, and a primitive of $g$ which is strictly increasing as well.

Congratulations, grobber. One of the most elegant proofs I have ever seen.

This shows me that I'm far from using analysis at its (full) potential.

Other method: write $f'$ as the half difference between the integrals of $(f')^2+f'+1$ and $(f')^2-f'+1$.

Is there a way to prove this without the use of calculus? My knowledge of it is extremely limited.

[quote="perfect_radio"]Prove that any polynomial $\in \mathbb{R} [X]$ can be written as a difference of two strictly increasing polynomials.[/quote]
$P(x)=\left(x+P(0)+\int_0^x\left(\frac{P'(t)+1}2\right)^2dt\right)$ $-\left(x+\int_0^x\left(\frac{P'(t)-1}2\right)^2dt\right)$


From the picture, at which acceleration should the $ 5$kg man climb up/down the rope so that the pulley $ P$, assumed to be light,

(a) doesn't move

(b) move downwards at $ 2$ m/s

Have you already tried to solve this? If not, start by finding all the forces applied to each body.

[b]@Carcul[/b] well should we(everyone) make it mandatory that before posting any 'homework-type' question which is not an challenge(which will be posted in the Marathon)one needs to show some work.That will benefit the the one who posted also

OK, let the tension in the rope hanging the $ 4$kg box be $ T$. Then the tension in the rope hanging the $ 1$kg box must be $ T$ as well.

We have $ 40\minus{}T\equal{}4a, T\minus{}10\equal{}a$, which gives $ a\equal{}6, T\equal{}16$.

Since the pulley is assumed to be light, we have that the tension in the rope above is $ (2)(16)\equal{}32$.

So the man is pulled up with $ T\equal{}32$.

So looking at the man, $ mg\minus{}32\equal{}ma$... $ 50\minus{}32\equal{}5a$, so $ a\equal{}3.2$.

So the man moves downward at $ 3.2 \frac{m}{s^2}$, in both cases?


Hi everyone,

I'm in need for help. I want to add a color in the background of a minipage. Here is my code

[code]
\newcommand{exemple}[1]
   {
      \vspace{5pt}
      \noindent
      \textbf{Exemple}
      \begin{center}
         \begin{minipage}{0.9 textwidth} {#1}
         \end{minipage}
      \end{center}
      \vspace{5pt}
   }
[/code]

The purpose of adding the color in the background is to emphasize the "exemple" in the document. I've already written the

[code]
\definecolor{Light}{gray}{0.90}
[/code]

in the header of my document.

I thank everyone in advance for their help!

Peter

You can use the \colorbox command (in the color or xcolor package):

[code]
\newcommand{exemple}[1]
   {
      \vspace{5pt}
      \noindent
      \textbf{Exemple}
      \begin{center}
         \begin{minipage}{0.9 textwidth} {\colorbox{Light}{#1}}
         \end{minipage}
      \end{center}
      \vspace{5pt}
   }
[/code]

ok it works, but doesn't do what I want.  :mad:  and the box can't display math formula!

The result I'm searching is to have a light gray rectangle of 90% of the width of the page, with the text of the exemple written in it! Not a gray background only behind the text!

Peter

OK, try adding a parbox:

[code]\newcommand{\exemple}[1]
   {
      \vspace{5pt}
      \noindent
      \textbf{Exemple}
      \begin{center}
         \begin{minipage}{0.9\textwidth} {\colorbox{Light}{\parbox{\textwidth}{#1}}}
         \end{minipage}
      \end{center}
      \vspace{5pt}
   }
[/code]

OMG !!

That's exactly what I wanted !!!!

I'm releaved now!!

A big thank to you DPatrick :lol:

Peter


geometry
incenter
perimeter
angle bisector
geometry proposed

$ABCD$ is inscribed and circumscribed. Line passing through $O$ (the incenter) and parallel to $BC$ intersects $AB$ and $CD$ in $K$ and $L$ respectively. Prove that $KL$=$\frac P4$ ($P$-perimeter of $ABCD$).

Consider the quadrilateral $AKLD$. It is clear $AKLD$ is cyclic. Apply now IMO19985 pr1. We see that $KL=AK+LD$. On the otherhand, the triangles $KBO$ and $LCO$ are isosceles ($OK=KB$ and $OL=LC$). Therefore $2KL=AB+CD=\frac{1}{2}P$.  :D

P.S. It is a quick sketch only. Try to complete all things that I wrote here. It is a good training :)

[Edit: I am very sorry. I made very stupid typos here. Corrected :blush: ]

I have another solution. I would even call it simpler than Myth's one, since it doesn't use any auxiliary results ;) . What follows is a kind of do-it-yourself form of the solution; straightforward steps are banned into footnotes.

[color=blue][b]Problem.[/b] Let ABCD be a bicentric quadrilateral (i. e. a quadrilateral which is both inscribed and circumscribed). Through the incenter O of this quadrilateral, we draw a line parallel to the side BC; this line intersects the sides AB and CD at the points K and L, respectively. Prove that $KL=\frac{AB+BC+CD+DA}{4}$.[/color]

[i]Solution.[/i] In order to make our configuration more symmetric, we consider not only the points K and L where the parallel to the line BC through the point O meets the lines AB and CD, but also the points M and N where parallel to the line DA through the point O meets the lines AB and CD.

We have to prove that $KL=\frac{AB+BC+CD+DA}{4}$. In order to prove this, it will be enough to show that KL = NM and that $KL+NM=\frac{AB+BC+CD+DA}{2}$.

An easy angle chase [footnote 1] proves that the two triangles OKM and ONL are similar. Moreover, their altitudes issuing from the vertex O are equal [footnote 2]. Hence, these two triangles are congruent. It follows that OK = ON, OM = OL and KM = NL. We thus have

  KL = OK + OL = ON + OM = NM.

Thus, KL = NM is proven. It remains to show that $KL+NM=\frac{AB+BC+CD+DA}{2}$. We can easily see [footnote 3] that $\frac{AB+BC+CD+DA}{2}=AB+CD$; thus, in order to prove the equation $KL+NM=\frac{AB+BC+CD+DA}{2}$, it will be enough to show that KL + NM = AB + CD.

It is easily seen [footnote 4] that the triangle BKO is isosceles, with OK = BK. Similarly, OM = AM, OL = CL and ON = DN. Thus,

  KL + NM = (OK + OL) + (ON + OM) = (BK + CL) + (DN + AM) = (BK + AM) + (CL + DN).

Now, let's WLOG assume that the points A, K, M and B lie on the line AB in this order. Then, automatically, the points D, N, L and C lie on the line CD in this order. As a consequence, BK + AM = AB + KM and CL + DN = CD - NL. Thus,

  KL + NM = (BK + AM) + (CL + DN) = (AB + KM) + (CD - NL) = AB + CD + KM - NL.

Since KM = NL, this becomes KL + NM = AB + CD, and we are done.

[b]Explanations of the footnotes:[/b]

[hide="footnote 1"]
Why are the triangles OKM and ONL similar? Well, since OK || BC, we have < OKM = 180° - < ABC. Since the quadrilateral ABCD is inscribed, < ADC = 180° - < ABC. Thus, < OKM = < ADC. Since ON || DA, we have < ADC = < ONL. Hence, < OKM = < ONL. Similarly, < OMK = < OLN. Thus, the triangles OKM and ONL are similar.[/hide]
[hide="footnote 2"]
Why are the altitudes of triangles OKM and ONL issuing from the vertex O equal? Well, these altitudes are simply the distances from the point O to the lines AB and CD; and these distances are equal, since the point O is the incenter of the quadrilateral ABCD and thus is equidistant from all sidelines of this quadrilateral.[/hide]
[hide="footnote 3"]
Why do we have $\frac{AB+BC+CD+DA}{2}=AB+CD$ ? Well, since the quadrilateral ABCD is circumscribed, we have AB + CD = BC + DA, and thus

$\frac{AB+BC+CD+DA}{2}=\frac{\left(AB+CD\right)+\left(BC+DA\right)}{2}$
$=\frac{\left(AB+CD\right)+\left(AB+CD\right)}{2}=AB+CD$.[/hide]
[hide="footnote 4"]
Why is the triangle BKO isosceles? Since OK || BC, we have < KOB = < OBC, while, since the point O is the incenter of the quadrilateral ABCD and therefore lies on the angle bisector of the angle ABC, we have < OBC = < OBK. Thus, < KOB = < OBK. Hence, the triangle BKO is isosceles, with OK = BK.[/hide]

  Darij


[quote="Elio (n)"]A bird fly 10 meters for $ 2.7sec$ to the north and 5 meters for $ 1sec$ to the same direction. What is the medium velocity of the bird flying?[/quote]

Is that the velocity at the median displacement or time?

[quote="Elio (n)"]A bird fly 10 meters for $ 2.7sec$ to the north and 5 meters for $ 1sec$ to the same direction. What is the [b]medium velocity[/b] of the bird flying?[/quote]Can you define what 'medium velocity' is? Perhaps it is 'average velocity'?  :maybe:

Medium velocity is           $ v_{mes} \equal{} \frac {l_{t}}{t_{t}}$
$ t_{t}$-total time
$ l_{t}$-total way

medium velocity =  avarage velocity

AM I CLEAR :)

I do not understand why no one  answered to the question maybe you do not understand the concept of the medium velocity or this problem is to easy to answer?  :huh:

medium velocity is the total distance covered by total time taken
so ${ v_{mean}=\frac{15}3.7} ms^{-1}$


geometry
circumcircle
geometric transformation
reflection
ratio
angle bisector
geometry proposed

Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$. Let $M$ be a point on the side $BC$ such that $\angle BAM = \angle DAC$. Further, let $L$ be the second intersection point of the circumcircle of the triangle $CAM$ with the side $AB$, and let $K$ be the second intersection point of the circumcircle of the triangle $BAM$ with the side $AC$. Prove that $KL \parallel BC$.

I "solved" the problem in the following way.

Let D a point on BC, K be the (second) common point of AC and c(ABD) and L be the (second) common point of AB and c(ACD).
Let J be the common point of c(ALK), c(BLD) and c(CDK). As <LAK=<CDK=<CJK, then J lies on CL. If, by ipothesys, <LAJ=<CAD then, since <ALC=<ADC, it results that  <ACD=<AJL=<AKL. This means that LK and BC are parallel. From here is immediate that the midpoint of BC lies on AJ.

PS
The problem I solved shoul, at first insigth, be equivalent to the problem given.

Comment on that are wellcome.

We multiply the following relations:

$\frac{MB}{MC}=\left(\frac{AB}{AC}\right)^2$ ([u]Steiner's theorem[/u]),

$LB\cdot AB=MB\cdot BC,MC\cdot BC=KC\cdot AC\Longrightarrow\frac{LB}{KC}=\frac{AB}{AC}\Longrightarrow LK\parallel BC$.

[u]Remark.[/u] Generally, $\frac{BD\cdot BL }{CD\cdot CK}=\frac{AB}{AC}\Longleftrightarrow \sigma [BLD]=\sigma [CKD].$

My solution:

Since < BAM = < DAC, the line AM is the reflection of the line AD in the angle bisector of the angle CAB. But the line AD is a median of triangle ABC; hence, the line AM is a symmedian of triangle ABC. Since a symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the two adjacent sides, we have $\frac{BM}{CM} = \frac{BA^2}{CA^2}$. On the other hand, since the points A, C, L and M lie on one circle, the Intersecting Chords Theorem yields $BM\cdot BC = BL\cdot BA$, and since the points A, B, K and M lie on one circle, the Intersecting Chords Theorem yields $CM\cdot CB = CK\cdot CA$. Hence,

$\frac{BA^2}{CA^2} = \frac{BM}{CM} = \frac{BM\cdot BC}{CM\cdot CB} = \frac{BL\cdot BA}{CK\cdot CA}$.

Consequently,

$\frac{BA}{CA} = \frac{BL}{CK}$.

This yields KL || BC after Thales.

[b]PS.[/b] I was asked about how to prove the following fact which I used:

[quote="darij grinberg"]Since a symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the two adjacent sides, we have $\frac{BM}{CM} = \frac{BA^2}{CA^2}$.[/quote]

Now I posted a proof of this fact at http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 (post #1 Theorem 5). To make it short, it is a particular case of the Steiner theorem, which tells that if M and N are two points on the line BC such that the lines AM and AN are isogonal cevians with respect to triangle ABC, then $\frac{BM}{CM}\cdot\frac{BN}{CN}=\frac{BA^2}{CA^2}$.

  Darij

LAK=LMB.   KMC=LAk.  So LMK=180-2A.   We most prove that  LM=MK.   LM/AC=BM/BA.    And MK/AB=MC/AC if LM=MK sow we must prove that  BM/MC=AB/AC*AB/AC.       BM/MC=(sin BAM*AB)/(sinMAC*AC).           sinBAM/sinMAC=sinDAC/sinBAD=(DC*AB)/(DB*AC).      BD=DC.    So it proved

Bonus Question:
Prove my corollary below:

$Corollary:$
Let $E$ be the point of intersection of line segments $BK$ and $CL$. Then points $A$, $E$ and $D$ are collinear.

It just so happens that $BK$ and $CL$ intersect at $E$ such that $A$ and $D$ all happen to line up with $E$. This means that $E$, $A$, and $D$ all lie on the same line, which follows that $A$, $E$, and $D$ are collinear by definition.

[quote=Ambroseyang]It just so happens that $BK$ and $CL$ intersect at $E$ such that $A$ and $D$ all happen to line up with $E$.[/quote]

Proof?

Let $P$ be $A$ - Humpty point of $\triangle ABC$
We have: $\dfrac{BM}{CM} = \dfrac{AB^2}{AC^2} = \dfrac{BP^2}{CP^2}$
Then: $PM$ is $P$ - symmedian of $\triangle BCP$
So: $\angle{BPM} = \angle{CPD} = \angle{ACB} = \angle{BKM}$ or $B$, $K$, $P$, $M$ lie on a circle
Similarly: $L$, $C$, $M$, $P$ lie on a circle
Hence: $\angle{BPL} = \angle{BPM} + \angle{LPM} = \angle{ACB} + 180^o - \angle{ACB} = 180^o$ or $B$, $P$, $L$ are collinear
Similarly: $C$, $P$, $K$ are collinear
So: $KL$ $\parallel$ $BC$

Let $CL$ meet $AD$ at $S$.
Claim: $B,S,K$ are collinear.
Proof : $\angle SCD = \angle BAM = \angle CAD$ so $DC^2 = DS.DA = DB^2$ so $\angle DBK = \angle MAK = \angle BAD = \angle DBS \implies B,S,K$ are collinear.

From Ceva Theorem we have $\frac{AK}{KC} . \frac{CD}{DB} . \frac{BL}{LA} = 1 \implies \frac{AK}{KC} = \frac{AL}{LB}$ so $KL || BC$.


The numbers $ 1$ through $ 10$ are written on the board. By deleting some of them, the remaining numbers can be divided into two groups such that the product of the numbers in each group is the same. What is the least number of numbers that need to be deleted?

[hide]Factor $ 10!$ into $ 2^8\cdot3^4\cdot5^2\cdot7$

We notice that this isn't a perfect square, so the answer can't be 0.

We notice that we can make two groups, each with product of 720, plus 7 as the leftover, so the answer is 1.[/hide]

Wow....that's pretty short. An example of the groups (we must show they exist!) is $ {1,2,3,4,5,6}$ and $ {8,9,10}$.


$\sum \frac{x^2-2}{y}=\sum \frac{(x-y)^2}{y}+\frac{x^2+y^2+z^2}{\sqrt{xyz(x+y+z)}}\geq \sqrt{3}$
was not difficult but nice!

[quote="I"]Let $x,y,z>1$ satisfy $x+y+z=xyz$, then find the least value of:
$\frac{x^2-2}{y}+\frac{y^2-2}{z}+\frac{z^2-2}{x}$[/quote]

Sorry, I got a typo here, but no matter :) .

[quote]Let $x,y,z>1$ satisfy $x+y+z=xyz$, then find the least value of:
$\frac{x-2}{y^2}+\frac{y-2}{z^2}+\frac{z-2}{x^2}$[/quote]


calculus
integration
calculus computations

Define $ I(\alpha)\equal{}\int_0^\infty \exp\left(\minus{}x^2\minus{}\frac{\alpha^2}{x^2}\right) dx$

then $ I'(\alpha)\equal{}\int_0^\infty \exp\left(\minus{}x^2\minus{}\frac{\alpha^2}{x^2}\right) \frac{\minus{}2\alpha \, dx}{x^2}$ ,

which is after substitution $ x\to\frac{\alpha}{x} \qquad \minus{}2 I(\alpha)$ .

The differential equation $ I'(\alpha)\equal{}\minus{}2I(\alpha)$ has the

solution $ I(\alpha)\equal{}C\, e^{\minus{}2\alpha}$, whereat $ C\equal{}I(0)\equal{}\frac{\sqrt{\pi}}{2}$.

So $ I(\alpha)\equal{}\frac{\sqrt{\pi}}{2}\, e^{\minus{}2\alpha}$.


Be $ABCD$ convex quadrilateral, the line $MN$ (where $M,N$ are the midpoints of $AC,BD$) pass trough $P,Q$ on $AB,CD$. Now: $AP\cdot DQ = PB\cdot CQ$. Any reference?

Menelaus' theorem for the quadrilateral $ABDC$ (in that order) implies this, as 2 of the 4 ratios are 1.


AMC
AIME
logarithms
AMC 12
AMC 12 B
USA(J)MO
USAMO

So (since the test window is now officially closed) what did you think of this year's AIME? Did it seem hard compared to last year? How many do you think you got? And (most importantly), what do you think the cutoff will be for 9th and 10th graders?

gah. I got #4 wrong because i multiplied s by 1000 instead of 100...
sob........ I did 9 problems but I made at least 3 reading mistakes.

many people in my school believe that this year's AIME is harder. So the floor will probably be a 6.....

i think it was pretty easy in general, but i made 3 stupid mistakes... 3!!! blah... i prbly only got 5 or 6 but couldve got 7 or 8... i think #4 is 86, at least thats what most of the ppl at my school got... anyways, any1 have an answer for #9? i hope its 35...

Well, I didn't get 54 or 55 on #4--I had 86. I either got 6 or 7 right, depending on whether the answer to #6 is 882--can anyone confirm this?

Great, I wasn't even close! I'm thrilled. well, not thrilled, but at least I don't have to kick myself for making a careless mistake...

I also got 35 for #9...

for #6 i got 798, but it's probably wrong, i ALWAYS miss those kind of problems...

i got 588 on #7... is that right? there was a pattern, right?

[quote="applepuppy"]I did 10, I thought it was easier than last years....

I also missed #4, i multiplied wrong and got 55 instead of 54 (is that the answer?? that's what I keep getting with a calculator)

oh well. *cry sob* there's always next year... anyone have the answers? My teacher couldn't find her copy of the solutions...[/quote]

I don't believe teachers are given answer pamphlets.

Re: Mithsapprentice

okay, thanks... i thought i read somewhere that they get answers, but I guess not...

(sorry, I'm not going to attempt the quote thing just yet...)[/quote]

I got 586 for #7... everyone else I've talked to that's taken the test got 588. Grr.

well, I got the same answers on 1, 2, 7 and 9... 4 problems, weee, that's one less than last year... I guess we'll just have to wait til they post the answers on the AMC website.

[quote="dareonion"]Can anyone post a list of correct answers?
My answers were
1. 217
2. 201
3. 241
4. 086
5. 849
6. 882
7. 588
8. 484
9. 035
10. 268
11. 512
12. 014
13. 641
14. 813
15. 298

I'm not sure about problem 14, and problem 15 was a complete guess.  Anyone want to compare answers?[/quote]

your number eight is wrong.  it is most definitely 199.  I'm projecting an 8.  how about the rest of you?

If Wyra, Jack Po, darkquantum, or anyone else going with the NYC team to NYSML this year reads this -- if you could bring me a copy of the AIME and this years AMC's, I would be eternally greatful :)

On #12, the logarithm one, did anyone get 139? (I think the fraction was 40/99, so 40+99).

I've got a question about the AIME floor....this may have been mentioned before, but if it was I can't find the thread. The question is, if the USAMO index cutoff for 11th/12th graders is, say, 206.5, but there was no one who had at least that index with an AIME score less than, say, 9, would the AIME floor for 9th/10th graders be a 9 or a 6?

According to the rules on the AMC website, the AIME floor should be 9. But I don't believe that it would be. I also don't believe that that will ever happen.

Why would the AIME floor this year be 9?

I was responding to JSRosen3's ``hypothetical situation.''

#12 is most definitely 5/9. Just use the formula (a/1-r) and you'll get it easily.

For #3, I got 247 instead of 241. There are 19 space-lines from each vertex, and 26 vertices. 19*26 = 494, and when you divide by 2 (since each will be counted twice) you get 247.

Indeed so.  There have been years when Wyoming has no USAMO qualifiers, because nobody in Wyoming even made AIME.

[quote="rlee"]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2004-12a/04amc12statstats/WY.HTML

I do not believe that. Does that mean that USAMO qualification in Wyoming is almost free?[/quote]

It's not quite that simple. Here's how I did it:

There are 26 vertices, so there are (26 choose 2)=325 lines connecting two vertices. There are 60 edges, so we remove those to get 265. Then we remove the diagonals of each quadrilateral (2 on each of 12) to to get 241.

I thought about using 26 choose 2 as well, so I knew the max was 325. I then tried subtracting but went for the other way.

What is the flaw in my solution?

I think what his idea is is that if each vertex belonged to exactly one quadrilateral and one triangle (which shared an edge), there would be four other vertices with which it shared a face, and so the lines reaching out to the 26 - 4 - 1 other vertices would all make space diagonals.  Unfortunately that doesn't work, because the thing doesn't close up that way.

[quote="ComplexZeta"]How do you know there are 19 from each vertex?[/quote]

[quote="tokenadult"][quote="rlee"]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2004-12a/04amc12statstats/WY.HTML

I do not believe that. Does that mean that USAMO qualification in Wyoming is almost free?[/quote]

I'm sure the statistics are correct. Participation in USAMO is done PARTLY on [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]the basis of state representation[/url] (rule 5), which means that a few low-population states, and a few higher-population states with wimpy performance on the AMC 10/12-AIME tests, have a state winner invited to take the USAMO with MUCH lower index scores than most other USAMO participants. Doubtless the theory is that anyone who can be conspicuous in math in a place where math performance is not nurtured by a surrounding "math culture" ought to be given the chance to show what he or she can do on the USAMO.

This, of course, suggests a strategy for people who are really determined to make USAMO: try to persuade mom and dad to move to the northern Great Plains or to the Rocky Mountains. One still has to make AIME first, of course, and won't have a very vital local math club, but it is one possible strategy. As for me, I'll keep living in Minnesota, which always has several people who get into USAMO the hard way.[/quote]

Or you could just move to Arkansas(Andrew Spann graduated, so it is no longer necessary to kill to usamo qual) or really down south.

Anywhere but Florida though.  Trust me, honest to g** that area is loaded with people.


how many questions are there total, because i can only access 1072.

Can you tell me what error message you get?

there really is no reply, every time i try to answer a question, its one of my old questions that i didn't get right.

Weird. When I try getting a problem through your user id I get new problems without any issues.

this is freaky- i haven't done many problems at all!!!  :maybe:

Are you guys still receiving problems you've already completed? Whenever I try under your account, I'm getting new problems as you are suppose to.


In pentagon $ABCDE$, $AB=BC=BD=BE=9$ and $CD=DE=EA=6$.

Find $AC$.

Am I visualizing this right? Is it a pentagon divided into three congruent isosceles triangles with sides 9, 9, and 6?

Yes, $\frac{46}{3}$ is correct. So riddler, please post your solution :)

[hide="solution"]
Lets say the intersection of $AC$ and $BE$ be $P$,
We will find out that $\triangle{APE}~\triangle{BED}$
So,
$AE=AP$
and $EP=4$
So, $BP=5$ because,
$BE=9$
Lets say the intersection of $AC$ and $BD$ be $Q$,
We will find out $AC\parallel ED$
So,
$\triangle{BPQ}~\triangle{BED}$
we need to find $PQ$,
which is,
$\frac{PQ}{6}=\frac{BP}{BE}$
So, $PQ=\frac{10}{3}$
we know that $AP=CQ=6$ ($\triangle{APE}~\triangle{ABE}~\triangle{CDQ}$)
So, $AC=6+6+\frac{10}{3}$
$AC=\frac{46}{3}$
[/hide]

*head hurts* could u explain it any clearer?

[hide="tip"]when seeing two triangles right next to each other ($\triangle{ABC}~\triangle{XYZ}$) he actually means those two are similar. Hope that makes it more clear. [/hide]

in latex how do you show similarity sign?


What is the greatest common factor of $ 357$ and $ 561$?

Using divisibility rules, it should be relatively easy to see that both are divisible by $ 3$. Quickly factoring, we also see that $ 17$ is a factor of both numbers. So the GCF of the two numbers is $ 3\times{17}\equal{}\boxed{51}$. Or just recognize instantly that they are both factors of $ 51$  :D .


algebra
polynomial
integration
function
calculus
derivative
linear algebra

Find $ p\in P_3(\mathbb{R})$ such that $ p(0) \equal{} 0$, $ p'(0) \equal{} 0$, and
$ \int^1_0 |2 \plus{} 3x \minus{} p(x)|^2 \,dx$

is as small as possible.

Yea I have no idea how to start this.

Every $ p$-candidate has the form $ p(x)=ax^2+bx^3$ with $ b\not=0$. We will consider the function $ F(a,b)=\int_0^1|2+3x-ax^2-bx^3|^2\,dx=\int_0^1(2+3x-ax^2-bx^3)^2\,dx$

Direct calculation yields $ \frac{\partial F}{\partial a}=-\frac{4}{3}-\frac{3}{2}+\frac{2a}{5}+\frac{b}{3}$ and $ \frac{\partial F}{\partial b}=-1-\frac{6}{5}+\frac{a}{3}+\frac{2b}{7}$ and the Hessian is given by $ \left(\begin{array}{cc}\frac{2}{5} & \frac{1}{3}\\[0.1cm]\frac{1}{3} & \frac{2}{7}\end{array}\right)$ which is positive definite. By setting the partial derivatives to zero, we determine a global minimum for $ a=24\,,\,b=-\frac{7\cdot 29}{10}$.

Where did this come from? It was also recently posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174825]here[/url].

Kent, these problems are from my Linear Algebra Done Right 2nd Edition by Sheldon Axler.


It is hard, because B is not multiplicative. For example $1^{3}+2^{3}=9\in B, \ 1^{3}+3^{3}=28\in B$, but $9*9=81\not \in B, \ 9*28=252\not \in B$.

[quote="conceive"]Let

$A=\{a; a\in N,a=c^{2}+d^{2}$ for some $c,d\in\ N \}$
$B=\{b; b\in N,b=x^{3}+y^{3}$ for some $x,y\in\ N \}$

Determine all $x\in N$ such that $x\in A$ and $x\in B$.[/quote]

We have at least all members of $C=\{u; u\in N,u=p^{6}+q^{6}$ for some $p,q\in\ N \}$  :)

--
Patrick


Prove that $ \alpha, \beta, \gamma \in \mathbb{C}$ are the vertices of an equilateral triangle if and only if $ \alpha^2 \plus{} \beta^2 \plus{} \gamma^2 \equal{} \alpha \beta \plus{} \beta \gamma \plus{} \gamma \alpha$.

That's just Muirhead or AM-GM or $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\equal{}0$ or whatever, you must have $ \alpha\equal{}\beta\equal{}\gamma$, so all the angles are equal, so it's equilateral.

None of those are enough if $ \alpha, \beta, \gamma$ aren't real.

[hide="Method 1"] Write as $ (\alpha \plus{} \omega \beta \plus{} \omega^2 \gamma)(\alpha \plus{} \omega^2 \beta \plus{} \omega \gamma) \equal{} 0$. [/hide]
[hide="Method 2"] The given equation has both a scaling symmetry and a translational symmetry.  Why can you reduce the problem to $ \alpha \equal{} 0, \beta \equal{} 1$? [/hide]

[hide="Another method"]Write it as $ (\alpha\minus{}\beta)^2\plus{}(\beta\minus{}\gamma)^2\plus{}(\gamma\minus{}\alpha)^2\equal{}0$

Put $ u\equal{}\alpha\minus{}\beta, v\equal{}\beta\minus{}\gamma$. Then obviously $ \gamma\minus{}\alpha\equal{}\minus{}u\minus{}v$, hence

$ u^2\plus{}v^2\plus{}(u\plus{}v)^2\equal{}0\iff u^2\plus{}uv\plus{}v^2\equal{}0\iff {v\over u}\in\{\omega,\bar\omega\}$. Hence $ BC$ is obtained by rotating $ AB$ around $ B$ for $ \pm 60^\circ$ and the conclusion follows.[/hide]


Suppose that $ D,\ E$ are points on $ AB,\ AC$ of $ \triangle{ABC}$ with area $ 1$ respectively and $ P$ is $ BE\cap CD$. When $ D,\ E$ move on $ AB,\ AC$ with satisfying the condition $ [BCED] \equal{} 2\triangle{PBC}$, find the maximum area of $ \triangle{PDE}$.

Let $ PD\equal{}b$, $ PC\equal{}a$, $ EP\equal{}d$ and $ PB\equal{}c$, $ [PDE]\equal{}L$
$ 2[PBC]\equal{}[BCED]\equal{}[PBC]\plus{}[PDB]\plus{}[PCE]\plus{}L$
$ [PBC]\equal{}[PDB]\plus{}[PCE]\plus{}[PDE]$

$ 1\equal{}\frac{b}{a}\plus{}\frac{d}{c}\plus{}\frac{db}{ca}$

$ 2\equal{}(\frac{b}{a}\plus{}1)(\frac{d}{c}\plus{}1)$   $ (1)$

It is easy to see that:

$ L\equal{}\frac{b}{a\plus{}b}[CED]\equal{}\frac{b}{a\plus{}b}\frac{CE}{AC\plus{}CE}[ACD]$

$ \equal{}\frac{b}{a\plus{}b}\frac{CE}{AE\plus{}CE}\frac{AD}{AD\plus{}DB}[ABC]$

And in the same way:
$ L\equal{}\frac{d}{d\plus{}c}\frac{BD}{BD\plus{}DA}\frac{AE}{AE\plus{}CE}[ABC]$

Then:
$ \frac{L*(AE\plus{}CE)(AD\plus{}DB)}{CE*AD}\equal{}\frac{b}{a\plus{}b}$

So $ \frac{CE*AD}{CE*AD\minus{}L*(AE\plus{}CE)(AD\plus{}DB)}\equal{}\frac{b}{a}\plus{}1$

And in the same way: $ \frac{BD*AE}{BD*AE\minus{}L*(BD\plus{}DA)(AE\plus{}CE)}\equal{}\frac{d}{c}\plus{}1$

And multiplying this equalities we get:

$ \frac{CE*AD*BD*AE}{(CE*AD\minus{}L*(AE\plus{}CE)(AD\plus{}DB))(BD*AE\minus{}L*(BD\plus{}DA)(AE\plus{}CE))}$

$ \equal{}(\frac{b}{a}\plus{}1)(\frac{d}{c}\plus{}1)\equal{}2$

Then $ L\equal{}\frac{CE*AD\plus{}BD*AE\pm \sqrt{CE^2AD^2\plus{}BD^2AE^2}}{2(AE\plus{}CE)(AD\plus{}DB)}$

But $ L\equal{}\frac{CE*AD\plus{}BD*AE\plus{} \sqrt{CE^2AD^2\plus{}BD^2AE^2}}{2(AE\plus{}CE)(AD\plus{}DB)}$ is impossible because:

$ \frac{CE*AD\plus{}BD*AE\plus{} \sqrt{CE^2AD^2\plus{}BD^2AE^2}}{2(AE\plus{}CE)(AD\plus{}DB)}$

$ \equal{}L\equal{}\frac{d}{d\plus{}c}\frac{BD}{BD\plus{}DA}\frac{AE}{AE\plus{}CE}$

Then  $ \frac{CE*AD\plus{}BD*AE\plus{} \sqrt{CE^2AD^2\plus{}BD^2AE^2}}{2*AE*BD}\equal{}\frac{d}{d\plus{}c}<1$, which is cleary impossible because $ CE*AD>0$

So $ L\equal{}\frac{CE*AD\plus{}BD*AE\minus{} \sqrt{CE^2AD^2\plus{}BD^2AE^2}}{2(AE\plus{}CE)(AD\plus{}DB)}$

If $ \frac{CE}{AE}\equal{}x$ and $ \frac{BD}{AD}\equal{}y$:

$ L\equal{}\frac{x\plus{}y\minus{} \sqrt{x^2\plus{}y^2}}{2(x\plus{}1)(y\plus{}1)}$

By Menelaus we get $ \frac{b}{a}\equal{}\frac{y}{xy\plus{}x}$,and $ \frac{d}{c}\equal{}\frac{x}{xy\plus{}y}$
So using $ (1)$:

$ x^2\plus{}y^2\equal{}x^2y^2$

But it is not difficult to prove that $ \frac{x\plus{}y\minus{} \sqrt{x^2\plus{}y^2}}{2(x\plus{}1)(y\plus{}1)}\le 10\sqrt{2}\minus{}14$
And the equality case is when $ x\equal{}y\equal{}\sqrt{2}$

Then the answer is $ 10\sqrt{2}\minus{}14$, when $ DE\parallel{}BC$

[quote="kunny"][color=darkred]Let $ ABC$ be a triangle with $ [ABC] \equal{} 1$ . Let $ D\in (AB)$ , $ E\in (AC)$ be two mobile

points for which  $ P\in BE\cap CD$ and $ [BCED] \equal{} 2\cdot [PBC]$ . [b]Find[/b] $ \max\ [DPE]$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ [BPD] \equal{} x$ , $ [CPE] \equal{} y$ and $ [BPC] \equal{} z$ . Observe that $ [DPE] \equal{} z \minus{} (x \plus{} y)$ and $ [ADE] \equal{} 1 \minus{} 2z$ .

$ \left\|\begin{array}{ccc} [DEP]\cdot [BPC] \equal{} [BPD]\cdot [CPE] & \Longleftrightarrow & z^2 \equal{} z(x \plus{} y) \plus{} xy \\
\ [ACD]\cdot [ABE] \equal{} [ADE]\cdot [ABC] & \Longleftrightarrow & (1 \minus{} z \minus{} x)(1 \minus{} z \minus{} y) \equal{} 1 \minus{} 2z \end{array}\right\|$ $ \implies\ z \equal{} \frac 12 \minus{} \frac {xy}{x \plus{} y}\ \Longleftrightarrow$

$ [DPE] \equal{} \frac 12 \minus{} (x \plus{} y) \minus{} \frac {xy}{x \plus{} y}$ and $ (x \plus{} y \minus{} 2xy)^2 \equal{} 2(x \plus{} y)^3$ a.s.o. I denoted $ [XYZ]$ - the area of the triangle $ XYZ$ .[/color]


A person sits down and begins flipping a coin repeatedly.  What is the probability that he will get a string of at least four heads at some point before he first gets a string of at least two tails?

Thanks

OK, could you please show your solution?

I'm kind of busy with other math, but all I did was separate the problem into cases that start with a tail followed by a successful collection of throws, the desired throws "head,...,head" four times, and the throws "head", "head, head",  and "head, head, head" followed by a successful set of 4 head beginning with a tail.  Then I constructed a recursion in terms of the probability of a successful series of throws beginning with a head and likewise a tail.  Solved the system of probabilities and added them to get $\frac{1}{6}$.  The solution seems long, but it is in fact short if you try it.


Ρε Σιλουανέ είναι ειδική περίπτωση της γενίκευσης που λέγαμε...

Σωστά  :wink:
Αλλά αφού η γενίκευση δεν έχει λυθεί (και πολλοί μπερδεύονται με τα $\sum$  :P  ) είπα να βάλλω την πρώτη υποπερίπτωσή της ώστε να δούμε καμια λύση .....

A Nick για τη γενίκευσή μας βρήκα και μια άλλη λύση ....

[quote="silouan"]Λοιπόν βρήκα μια πιο ισχυρή έκδοση της εν λόγω ανισότητας .
Αν $a,b>0$ είναι πραγματικοί αριθμοί (στη χειρότερη περίπτωση κάντε τους και ακεραίους αν δε σας βολεύει ) να αποδειχθεί ότι

$(\frac{a^{2}+b^{2}}{a+b})^{a+b}\geq a^{a}\cdot b^{b}$

Γιατί είναι πιο ισχυρή ? :wink:[/quote]

 [color=red]Επειδή πολλοί μαθητές δεν γνωρίζουν την ανισότητα των βαρών , θα μου επιτρέψετε να δώσω τη λύση στην άσκηση αυτή , ώστε να δουν και ένα τρόπο εφαρμογής της.[/color]

[b]ΑΝΙΣΟΤΗΤΑ ΒΑΡΩΝ.[/b]

Αν $x , y >0$ και $x+y = 1$,τότε  $xa+yb \geq a^{x}b^{y}$
 για θετικούς $a , b$.

 [hide]Σύμφωνα με αυτή την ανισότητα , αν θέσουμε $x=\frac{a}{a+b}, y =\frac{b}{a+b}$ , τότε $x+y = \frac{a}{a+b}+\frac{b}{a+b}= 1$ , και έτσι παίρνουμε  :

 $\frac{a^{2}+b^{2}}{a+b}= \frac{a}{a+b}a+\frac{b}{a+b}b \geq a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}$

 Υψώνουμε τώρα και τα δύο μέλη στην $a+b$ και παίρνουμε τη ζητούμενη.Η ισότητητα ισχύει όταν $a = b$.[/hide]

[u] ΜΠΑΜΠΗΣ[/u]


Prove that the system of equation $x+y=z^{2}$ and $xy=\frac{z^{2}-z}{3}$ is not solveable in nonzero integers

from the first equation we get that $x+y$ is positive and from the second one we have that xy is positive because $z^{2}-z\geq 0$ for integers so both $x,y$ are positive...we also know $z^{2}-z$ cannot be zero so we rule out $z=1$.
if $z$ were positive we'd have $x+y>3xy\geq 3 max(x,y)>2 max(x,y)\geq x+y$ so this gives no solution.now we know that
$x+y=z^{2}$ $3xy=z^{2}+z$ where all are natural numbers.
$3xy=x+y+\sqrt{x+y}$ if $y=1$ easily obtain there is no solution. if $x\geq y\geq 2$ then
$(x-1)(y-1)\geq 1$ $\to $ $xy\geq x+y$ so $3xy\leq xy+\sqrt{xy}$ or $xy\leq \frac{1}{4}$


Show that $ f(x_1,x_2) \equal{} \frac {x_1 \cdot x_2}{\sqrt {x_1^2 \plus{} x_2^2}}$ if $ (x_1,x_2) \not{ \equal{} } (0,0)$ and $ 0$ if $ (x_1,x_2) \equal{} (0,0)$ is continuous at $ (0,0)$.

[hide="What I have so far"]
$ |f(x_1,x_2) \minus{} f(0,0)| \equal{} \left| \frac {x_1 \cdot x_2}{\sqrt {x_1^2 \plus{} x_2^2}} \minus{} 0 \right| \equal{} \frac {|x_1| \cdot |x_2|}{\sqrt {x_1^2 \plus{} x_2^2}}$ and we WTS this is $ < \epsilon$ for some $ \delta$ such that $ \parallel{}(x_1,x_2) \minus{} (0,0)\parallel{} < \delta$[/hide]

$ (x_1 \minus{} x_2)^2 \geq 0 \Rightarrow x_1 \cdot x_2 \leq (x_1^2 \plus{} x_2^2)/2$.  Then, $ f \leq \sqrt{x_1^2 \plus{} x_2^2}/2$ which can be made arbitrarily small as $ (x_1,x_2) \rightarrow 0$.


What is the 13th number in the arithmetic sequence 7, 10,
13, ...?

[hide]
[b]Formula for the $n$th term of an arithmetic sequence where $d$ is the common difference and $s$ the starting term: $d(n-1)+s$[/b]

The starting term is obviously $[b]7[/b]$, $ n = 13$,  and the common difference is $10 - 7 = [b]3[/b]$
Plugging in the values, the $13$th term is:
 $3(13-1)+7 = 3(12)+7 = 36+7 = \boxed{43}$
[/hide]


Show that if $2+2\sqrt{12n^{2}+1}$ is an integer, then it is square.

(I know this is really well known - I let the setters know how I felt about this afterwards)

Go into the Intermediate forum, and you'll see this same problem already discussed.


let $ a$ , $ b$ ,$ c$ lenghts of a triangle

is this inequality true ?

$ a^2\plus{}bc \geq ab\plus{}ca$


Here's a problem from Mathcounts that I need help with...as I can't even come up with a possible solution. :huh:

One day I found a receipt from my grandfather’s store. The bill was faded so that I couldn’t read the price of the gadgets at all, and I could read only 3 of the 5 digits in the total. The bill showed the following:
	72 gadgets at _._ _  each = _74.5_
How many dollars did one gadget cost? Express your answer as a decimal to two places.

The underscores represent the places of missing digits. I couldn't add dollar symbols w/o invoking LaTeX. Sorry. Plz help. Thanks.

Because 72 gadgets cost $ a74.5b$, the number $ a74.5b$ is divisible by 72
If a number is divisible by 72, it is divisible by both 8 and 9

If a number is divisible by 9, the sum of its digits is divisible by 9
Therefore, $ a \plus{} b \plus{} 16 \equal{} 18$ or $ 27$
This gives that $ a \plus{} b \equal{} 2$ or $ 11$

If a number is divisible by 8, then its last three digits are divisible by 8
Therefore, $ 45b$ is divisible by 8, quick division shows that $ b \equal{} 6$

The only value of $ a$ that corresponds with this value of $ b$ is $ a \equal{} 5$
Therefore, the cost of 72 gadgets is $ 574.56$
Dividing by 72 gives that the cost of one gadget is $ 7.98$ F.S.

Oh.....my............. gosh. DUH!!! Thanks a bunch. That makes the problem soooo much easier. :P

lol state target 4 means that it cant be thaat hard! :D

Weeeeeeeeeeeeelll, yes, it is, for the mentally challenged people like me... :P I only got 10 on last year's state target round. 32 in total.

Last years target round was moderate.....i got 14  :(  missed #7


AMC USA(J)MO USAMO Putnam AIME geometry email

I am trying to decide which book to read after AoPS 2: the art and craft of problem solving by zeitz or mathematical olympiad challenges by andreescu and gelca. I know the ACoPS is often referred to as "AoPS 3", but I am not sure whether its price of $70 coupled with its lack of a solutions manual is worth it. I am also not sure whether Mathematical Olympiad Challenges is too difficult for someone just finishing AoPS 2, nor whether its contents are thorough enough for making the jump to olympiad-type problems. Could someone who has read one or both of these books tell me more? Thanks.

I've heard ACoPS is pretty good FyTY...you may want to do that one before the USAMO book...

i havent read it...but thats what i recommend

If you want to win the USAMO, and you think you're ready to work toward that goal, get [i]Mathematical Olympiad Challenges[/i]. Otherwise, get [i]The Art and Craft of Problem Solving[/i].

Although I [i]have[/i] heard good things about it too, I don't think I should comment on it as I haven't read it (I wouldn't want to give a false opinion on it).  Sorry, Fierytycoon (and BTW, why does everyone call you FyTy?  Do you like that nickname?)

[quote]Who calls me ComplexZ? Is it really worth it to save 3 characters?[/quote]

Well, I call you Simon, so that saves quite a few...

[quote]If you want to win the USAMO, and you think you're ready to work toward that goal, get Mathematical Olympiad Challenges. Otherwise, get The Art and Craft of Problem Solving.[/quote]

What do you mean by this? What would working on ACoPS be working towards?

[color=cyan]A more nebulous goal of "mathematical and problem-solving knowledge."[/color]

[quote="MithsApprentice"]nah...its juss me that calls him FyTy...im too lazy to write is all out...juss like how Syntax Error is juss Syntax and ComplexZeta is juss ComplexZ and how MysticTerminator is juss Mystic and how Rippledance is juss Ripply or Ripples...and etc...[/quote]
Mmm, what about me? :)
I like Fiery better :D or Fieries :P </jk>

By the way, [url=http://www.usfca.edu/math/Faculty/]here's Paul Zeitz's homepage[/url] (@ University of San Francisco), but you probably knew that already :)

Tare, ur name is too short for me to try to abbreviate it...

Yeeeeeah,[b] that's[/b] the funny part, haha? no? :)

I mean we could just call you "The T" or something odd like that...but that's also 4 letters long...

Well ...

Comments on "Olympiad Challenges" :
http://www.kalva.demon.co.uk/books/andree2.html

Comments on TAACOPS :
http://www.kalva.demon.co.uk/books/zeitz1.html

The following book is probably the best problem - solving book :

http://www.kalva.demon.co.uk/books/engel1.html

The whole list :

http://www.kalva.demon.co.uk/book1.html

these are typically pretty advanced problems though arent they?

Fine, but the more mathematicians there are, the higher level the competitions are.

whatever just don't diss indians cause somewhere in delhi there's enough intelligence to blow up the world so there

jus look at IIT IIT rocks even tho i won't go there probably

im not dissing...

there are many talented indians that work in a variety of fields...

we're just dicussing the difficulty of the national olmpiads...

yes im sure the indian olympiad is hard tare....
and yes mystic IIT rocks...are you indian?
but traditionally many indians become doctors or engineers....and what does this have to do with national olympiads?

*nods somp*...its completely irrevelent...totally disregards topicality...

Tare, you don't do debate do you?

india would win like each time if they had a good way to select people for their team other than a, "hey you kid, lets go to the imo!"

Man, all of the country olympiads are hard...well...except for the ones for Canada and the UK  :D

its not the population of a country that makes its imo team any good, its really the country's mathematical culture. look at bulgaria, it won this year. its population compared to china and india is extremely small. i took a look at china's national olympiads. they are no doubt hard, but not much more so than some of the usamos. just different kinds of problems.

*nods*...the USAMO is full of problems that we're used to attacking after MC training and stuff...but like IMO problems...too many types...

[quote="Arne"]I've got a pdf document which is called "The art and craft of problem - solving, Instructor's resource manual". It contains chapter - to - chapter hints and solutions for the book. Certainly not all problems are discussed, only the more difficult ones I think.[/quote]

Ok guys, since I get tons of PM's asking me to send the document by email:

I don't have this document anymore, I guess I somehow deleted it (or it might be lost in a computer crash or something - it wasn't of any use to me anyway). But I remember sending it to quite a lot of people back then (that's like two years ago :huh:) so I guess some people here should still have it. Anyone?

Hi I would like a copy of the solutions to Art and Craft of Problem Solving by Zeitz. If you have it, please send it to me. My email  is fdasdfasdf AT gmail.com

[quote="Tare"]I'm not dissing Indians, I'm saying they're smart so they probably have better, harder math competitions then we do.

...uh oh, the DoHS has arrived... :) , logging out...[b]now[/b][/quote]

What is DoHS?

Also, what are the "Chinese Olympiad books" referenced in the kalva comments???

Geek sense tingling...


In Duck Creek, a bicycle license plate consists of one letter
followed by one digit; for example, $ Q7$ or $ J1$. How many different
license plates are possible?

Why did you post this, then? (or if you didn't, why did you comment?)

The answer is $ 26\cdot10\equal{}\boxed{260}$.


I don't get it, but it seems like my post number becomes smaller every few days.  Like yesterday...I posted like 6 posts, which my cummulative total 242 or something, but today when I logged on, it said that I have 235 posts.  Is this happening to anyone else?  Any explanations?

-interesting_move

Yeah..since I post in the Etc... board, and and Etc... board has posts deleted every once in a while.

[quote="interesting_move"]I don't get it, but it seems like my post number becomes smaller every few days.  Like yesterday...I posted like 6 posts, which my cummulative total 242 or something, but today when I logged on, it said that I have 235 posts.  Is this happening to anyone else?  Any explanations?[/quote]

Conversations in Etc. are auto-deleted if they are idle for 7 days.

[quote="ComplexZeta"]Do they actually get deleted, or do they just get hidden?[/quote]

Richard set that up, so I'm not 100% positive, but I think they are getting deleted now, but when we switch to our own server I doubt we'll delete anything, though I could be wrong.  It may depend on how large this site becomes.

[quote="Tare"][quote="ComplexZeta"]Do they actually get deleted, or do they just get hidden?[/quote]
What's the point of hiding posts?[/quote]

1. So that they are not taking up active server space.
2. To keep the bulk of the discussion math related.

So if our posts get hidden/deleted, will our post count go down?

No, he already answered that somewhere...(oh yeah, [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=966]here[/url]:) )

[quote="MCrawford"]
Conversations in Etc. are auto-deleted if they are idle for 7 days.[/quote]

I've noticed a couple of posts haven't been deleted even though they've been idle for more than 7 days.  I believe that Mr. Rusczyk specially set them aside because they were relevant and could likely come up again.

Not sure why interesting's total goes down - maybe you've posted on threads that the moderator has manually deleted.  This would have that effect.

As for the threads that are still around seemingly way too long, those are a result of people still voting in the polls.


A mathematics teacher wants her two intelligent students P and S to derive the exact value of a 2-digit natural number by revealing the number of positive divisors of n to S and sum of digits of n to P. A brief conversation follows between S and P as follows:
P: I cannot determine n.
S: I can't either but I know whether n is even or not.
P: Now I know what n is.
S : So do I know.
Assuming both the students are honest and that their statements are backed by perfect logical reasoning, determine n and justify.

If I am not wrong, it has to be one of the following numbers: $11, 59, 89.$ If this is correct, then I will work on my solution further, else no point working too much on the problem!

Yes you are right I too got the three numbers, and I can't rule out any two of them, so I don't know if the problem is worded wrongly or something like that.


Each lateral edge of a regular hexagonal pyramid is 13 inches long. The altitude is 12 inches long. How many inches are in the perimeter of the hexagonal base? Express your answer is simplest radical form.

I believe the correct [hide=" answer is"]10sqrt3[/hide]

[quote=kz190][quote="Sephiroth"]We see a 5-12-13 triangle, so the radius of the base is 5. Since a hexagon is composed of 6 equilateral triangles, the side length is 5, so the perimeter is 30.[/quote]

But it says that it needs to be in simplest radical form though.[/quote]
30 is simplest radical form

[quote=EpicSkills32]I believe the correct answer is 10sqrt3[/quote]
[I][b]Actually[/I][/b]
[hide = Steps]$13^2-12^2=x^2$
$169-144=x^2$
$25=x^2$
$\sqrt{25}=x$
$x=5$
$6\cdot 5$[/hide]
This means each side length of the regular hexagon is [hide = ' the final answer of steps']$6\cdot 5$[/hide].
Therefor the [hide = ' answer actually is: ']
$\boxed{30}$
[/hide]

oh I think I misread the problem lol I can't picture hexagonal pyramids.


i wanna join in to the pool waddyamacallit, but when do the games start counting?
also if there is a set date when this starts, i wanna be told (plz)

sry about that i meant it in the other subject :blush:  :blush:  :blush:


analytic geometry
geometry
perpendicular bisector
Pythagorean Theorem

A girls' camp is located $ 300$ rods from a straight road. On this road, a boys' camp is located $ 500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is:
$ \textbf{(A)}\ 400\text{ rods} \qquad\textbf{(B)}\ 250\text{ rods} \qquad\textbf{(C)}\ 87.5\text{ rods} \qquad\textbf{(D)}\ 200\text{ rods}\\
\textbf{(E)}\ \text{none of these}$

[hide="Sounds Wrong"]
I imagined the girls' camp at $ (300,0)$ on the coordinate plane and the straight line to be $ y\equal{}0$. The boys camp would be $ 500$ rods away and on the line $ y\equal{}0$, so draw a diagonal which makes a right triangle that is a Pythagorean triple, $ 300,400,500$ with $ 400$ being the distance from the origin.

Now, we have a point located at $ (x,0)$ with distance $ x$ where the two distances equal. That gives $ 400\minus{}x\equal{}300^2\plus{}x^2 \implies 0\equal{}x^2\plus{}x\plus{}89600$ where we need to build the canteen. $ x$ is irrational though...
[/hide]

The mistake is here :
[quote]
$ 400\minus{}x\equal{}300^2\plus{}x^2 \implies 0\equal{}x^2\plus{}x\plus{}89600$
[/quote]
 It should be
$ (400\minus{}x)^2 \equal{} 300^2 \plus{} x^2 \implies 160000\minus{}800x\plus{}x^2 \equal{} 90000\plus{}x^2\implies$
$ x\equal{}87.5 \implies \boxed C$.

I seem to be getting $ 312.5$. If I understand the problem correctly, the road is taken to be a straight line.  Let $ G$ be the girl camp and $ B$ for the boy camp. Let $ C$ be the canteen. Let $ P$ be the projection of $ G$ on line $ BC$. Since $ BC \equal{} CG$, $ C$ is the intersection of the perpendicular bisector of $ BG$ and $ BC$. Let the midpoint of $ BG$ be $ M$. Then from similar triangle, $ CB \equal{} BM * BG/BP\equal{} 250*500/400$
$ \equal{} 125*5/2 \equal{} 125*(2 \plus{} 0.5) \equal{} 250 \plus{} 62.5 \equal{} 312.5$.

You're right, what I found as $ x$ earlier  $ (87.5)$ was not the distance from the camp to the canteen, the correct distance should be $ 400\minus{}x \equal{} 312.5$.

Here is how I did it:

Lets place camps in coordinate system such that girls' camp is placed on $ A (300,0)$ and boys' on $ B (0,400)$ (we got $ 400$ from Pythagorean theorem). Let canteen be on $ C (0, 400 \minus{} d)$ (where $ d$ is desired distance). Now we have right triangle $ ACO$ with sides of length $ 400 \minus{} d, 300$ and $ d$. From Pythagorean theorem we get $ d \equal{} 312.5$

The answer must be $ \textbf{(E)}$.

[size=67](this post was edited due to mz94's complaint)[/size]

[quote="Flame"]I got $ 312.5$ too. The answer must be $ \textbf{(E)}$.[/quote]

...tahts not a very useful post you know...unless you did it a diferent way, refrain from posting. we dont need to know everyone who got (E) as their answer

[quote="Flame"]Oh I'm sorry. I thought it might be useful for someone to confirm the result.  :oops_sign:[/quote]

its okay. next time, maybe say your confirming the answer. that might be more useful than just saying the answer without us knowing the real reasoning behind it


So on the fifth MML meet, Lexington was originally penalized six points because one student took out his cell phone to turn it off, because he kept getting an audible low battery signal. This was completely understandable, as in the rules (at mathmeets.com), it's clearly stated that cell phones are not allowed inside the testing rooms. However, the other 24 points we accumulated in that round were later penalized from us as well, giving us a 0/30 and putting us down 15 points overall this year, instead of up 9.

I'm not complaining, but I'm wondering what exactly in the rules calls for this penalization. I couldn't find anything, I was hoping someone could enlighten me. If indeed the rules state that if a cell phone is discovered, the team is disqualified for that round, fine; the rules are the rules. But unless my memory deceives me, phones have gone off in rounds before, and I don't recall anything like this ever happening.

Carl

Carl, it sounds stupid. They can't do that to you guys because it doesn't make sense to penalize the other 24 points which were unrelated. If the student who took out his cell phone was penalized, it can be understandable.

Also, note that there is some bias against Lexington since you always come out in first place. I hope that penalization didn't affect your school's score that much.

Thanks for the support, but I'm not really looking for debate on the issue; obviously all of us at Lexington aren't happy that we were penalized, but we're willing to accept any decision that the MML makes. I was just wondering what part(s) of the rules influenced/dictated the decision, because Mr. Roos told me that it was in the rules to disqualify the whole team, stupid rule or not.

http://www.mathmeets.com/mml/public_MML_RULES_2008_2009_01.php

Under Contest Procedure #4: NO CELL PHONES are allowed in the testing area.

And then: The following are also NOT allowed inside the testing room:

    * Cell Phones
    * Pagers and Beepers
    * Powerbooks and handheld computers
    * Pocket Organizers
    * Electronic Writing Pads or any such devices
    * Devices with QWERTY keyboards
    * ipods

However, it is a little ambiguous as to whether the student should be penalized or the whole team should be penalized.

Those are strict rules. Do they have that in the GBML?

[quote="essman"]From the MML statement of Proctoring Responsibilities:  NO cell phones are allowed in the testing area. If a student is caught will a cell phone, that school will forfeit the round.[/quote]

Oh, I was looking in the general rules. I found it now, thanks.

Those are some strict rules... Is that why Lexington isn't in first place in the MML?

[quote="calc rulz"]I could see taking off just his points, but taking off the entire team's is totally not right.[/quote]

I agree, I am glad the GBML doesn't have these rules.

those rulz are a bit harsh.

but it'll serve a nice reminder to other ppl to turn their cell phones off or not bring it into the room.

I guess the judges did it to show other teams what will happen if one member is caught w/ a cell phone.

No, the judges did it because it is a rule. They followed the rules, and next time we will too. Let's not turn this into a debate on how good/not good the rule is, please...

Actually it's pretty common in sports to disqualify teams of individuals who commit sufficiently significant offenses.

[quote="CatalystOfNostalgia"]instead of up 9.[/quote]

Technically, had the phone not gone off we would be up by 15.


topology
calculus
integration
real analysis
analytic geometry
advanced fields
advanced fields unsolved

Why is there no analogue of Fubini's theorem for Hausdorff measures?

what do you mean, xevarion?

something like $H_{a_{1}}\times H_{a_{2}}= H_{s}$, where $s = a_{1}+a_{2}$, $H_{a_{i}}$ is a measure on $(X_{i}, d_{i})$, and $H_{s}$ is a measure on $(X_{1}\times X_{2},\sqrt{d_{1}^{2}+d_{2}^{2}})$?

Well you want to be able to integrate wrt one of the measures, and then wrt to the other measure, and get the integration wrt to the product measure. But I have read that this doesn't work with Hausdorff measures. Has it something to do with Hausdorff measures are not $\sigma$-finite? As far as I can tell this presents some difficulties with developing the integration theory, but I really don't understand well enough yet :(:(

Fubini's theorem is proved for general $\sigma$-finite measures in Folland's "Real Analysis". So yes, the issue is the lack of $\sigma$-finiteness.

Folland also gives this counterexample: Consider the product of two copies of $[0,1]$, one with the Lebesgue measure $\lambda$, and the other with counting measure $\nu$. Let $f(x,y)=1$ if $x=y$ and $f(x,y)=0$ otherwise. Then the three integrals $\int\left(\int f(x,y)\,d\lambda(x)\right)d\nu(y)$, $\int\left(\int f(x,y)\,d\nu(y)\right)d\lambda(x)$, and $\iint f(x,y)d(\lambda\times\nu)$ are all different.

And what is the product of Hausdorff measures anyway? Let $X$ be the interval $[0,1]$ equipped with measure $\mathcal H^{s}$ for some $0<s<1$. Then $X\times X$ is equipped with the product measure $\mu=\mathcal H^{s}\times\mathcal H^{s}$. The measure $\mu$ assigns the value $\mathcal H^{s}(A)\mathcal H^{s}(B)$ to any product set $A\times B$. Let's consider the diagonal $D=\{(x,x)\colon x\in [0,1]\}$. I claim that $\mu(D)=\infty$.

First, observe that the projection of $D$ onto either coordinate axes is a bi-Lipschitz map with the constant $\sqrt{2}$. Therefore, for any product $A\times B$ we have $\mathcal H^{s}(D\cap (A\times B))\le 2^{s/2}\min\{\mathcal H^{s}(A),\mathcal H^{s}(B)\}$. Since $D$ is not $\sigma$-finite for $\mathcal H^{s}$, it cannot be covered by countably many sets $A_{i}\times B_{i}$ with $\mathcal H^{s}(A_{i})\mathcal H^{s}(B_{i})<\infty$. Thus, $\mu(D)=\infty$.

The above implies $\mu\ne\mathcal H^{2s}$ at least when $s\ge \frac12$. Replacing $D$ with a Cantor subset of the diagonal, one can show that $\mathcal H^{s}\times \mathcal H^{s}\ne \mathcal H^{2s}$ for any $0<s<1$. Conclusion: the product of Hausdorff measures is something strange and probably useless, so we should not worry about not having a Fubini theorem for it.  :)


In a group of cows and chickens, the number of legs was $ 14$ more than twice the number of heads. The number of cows was:

$ \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$

[hide="simple"]

$ Let:$

$ x$- be the number of heads of all the cows.
$ y$- be the number of heads of all the chickens.
$ 4x$- be the number of legs of all the cows.
$ 2y$- be the number of legs of all the chickens.

[quote="Brut3Forc3"]the number of legs was $ 14$ more than twice the number of heads. [/quote]
That is: $ 4x \plus{} 2y \equal{} 2(x \plus{} y) \plus{} 14$.

[quote="Brut3Forc3"]The number of cows was
[/quote]
$ \fbox{(B) 7}$

[/hide]


Να υπολογιστεί το όριο  lim[ln(x-e).(lnx-1)]  με το χ να τείνει στο e. Με κάποιους πρώτους υπολογισμούς το όριο δεν προσδιορίζεται. Μήπως πρέπει να παραγωγίσουμε;

[quote="ypsonas"]Να υπολογιστεί το όριο  $I= \lim_{x\to e}{[\ln (x-e) \cdot (\ln x-1)]}$. Με κάποιους πρώτους υπολογισμούς το όριο δεν προσδιορίζεται. Μήπως πρέπει να παραγωγίσουμε;[/quote]

Με de L'Hospital . Είναι :

 $I = \lim_{x\to e}{\frac{ln(x-e)}{\frac{1}{lnx-1}}}= ... =-e \lim_{x\to e}{\frac{(lnx-1)^{2}}{x-e}}= ... = 0$

 Ρίξε μια προσεκτική ματιά στις πράξεις γιατί τις έκανα πολύ γρήγορα. Αν κάτι ξέφυγε, ... εδώ είμαστε !

 Μπάμπης

[quote="ypsonas"]Φίλε  και εγώ κάτι παρόμοιο έκανα αλλά με αγχώνει αυτό το μηδενικό στον παρονομαστή..Το όριο του x-e παρονομαστή μας δίνει e-e=0.[/quote]

 Τελικά κατάλαβες τη λύση ή να γράψω κάτι ακόμα ; Εκτός και αν έχω κάνει λάθος. Στο τελευταίο όριο να πάρεις ξανά  de L' Hospital  για να βρεις το τελικό αποτέλεσμα.
 Το μήνυμα αυτό στο γράφω , διότι λες ότι σε  '' αγχώνει το $0$  του παρονομαστή '' . Εμένα δε με αγχώνει , διότι δε βλέπω να υπάρχει πρόβλημα. Αν χρειάζεσαι κάτι ακόμα ,  ευχαρίστως να βοηθήσω.

  Φιλικά - Μπάμπης