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AMC
USA(J)MO
USAMO

1. Find all positive integers a, b, and c, such that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 2001$

2. Find the number of integers x with $ |x| \le\ 1997$ such
that $ 1997$ divides $ x^{2} \plus{} (x \plus{} 1)^{2}$

3. For any real number $ t$, $ [t]$ is the largest integer that is less than or equal to $ t$
List all positive integers $ k$, such that $ k\leq 20$ and the area of the graph $ [x^2] \plus{} [y^2] \equal{} k^2$ is $ 4$

4. Find all ordered pairs $ (a,b)$ where $ a$ and $ b$ are both integers, while $ 3^{a} \plus{} 7^{b}$ is a perfect square.

5. An acute-angled triangle $ ABC$ is given in the plane. The circle with diameter $ AB$ intersects altitude $ CC'$ and its extension at points $ M$ and $ N$, and the circle with diameter $ AC$ intersects altitude $ BB'$ and its extensions at $ P$ and $ Q$. Prove that the points $ P,M,N,Q$ are concyclic.

6. Prove that the product of four consecutive natural numbers cannot be a perfect cube.

7. Find the number of zeroes at the end of $ 11^{(5!)!} \minus{} 1$.

[hide="1, from my blog"] Consider $ \pmod{9}$. All cubes are congruent to $ 0$ or $ \pm{1}\pmod{9}$, so $ a^3\equiv{b^3}\equiv{c^3}\pmod{9}$. Thus, all of $ a, b, c$ are congruent to one of $ 1, 4,$ or $ 7\pmod{9}$. Now note that $ 13^3 \equal{} 2197 > 2001$, so the possible choices for each of $ a, b, c$ are $ 1, 4, 7, 10$.

Now, consider $ \pmod{7}$. All cubes are congruent to $ 0$ or $ \pm{1}\pmod{7}$. From this, we find that $ a^3 \plus{} b^3 \plus{} c^3\equiv{6}\pmod{7}$ so $ a \plus{} b \plus{} c$ is either a rearrangement of $ 0 \plus{} 0 \plus{} 6$ or $ 1 \plus{} 6 \plus{} 6$ in modulo $ 7$. If it is the first, then two of $ a, b, c$ must be $ 7$ and we can check that does not work. Thus, it must be the latter and so the possible choices are $ 1, 4, 8, 11$ for the $ 1\pmod{7}$ and $ 2, 3, 5, 6, 9, 10, 12$ for the $ \minus{} 1\pmod{7}$.

The only ones that are in both are $ 1, 4, 10$, so we must have two of $ a, b, c$ equal to $ 10$ and we find the other is $ 1$. Thus, all solutions are $ (10, 10, 1), (10, 1, 10),$ and $ (1, 10, 10)$[/hide]

That was JBMO 2001

[hide="4"] Let $ 3^a \plus{} 7^b \equal{} x^2$. Examining $ \pmod{4}$, we find $ a$ and $ b$ are of opposite parities.

First, suppose $ a$ is odd and $ b$ is even. Let $ b \equal{} 2y$.

Then $ 3^a \equal{} (x \minus{} 7^{y})(x \plus{} 7^{y})$. Both terms in parentheses must be powers of threes Suppose that both are powers of $ 3$ greater than $ 1$. However, that means both terms are congruent to $ 0\pmod{3}$, which is clearly impossible. Thus, $ x \equal{} 7^y \plus{} 1$ and $ 3^a \equal{} 2\cdot{7^{y}} \plus{} 1$.

Suppose $ y > 0$. Consider $ \pmod{7}$. We must have $ 3^a\equiv{1}\pmod{7}$. $ 3^{a}\pmod{7}$ has period 6, with a cycle of 3, 2, 6, 4, 5, 1, so we must have $ 6|a$. However, we had $ a$ is odd, contradiction. Thus, $ y \equal{} 0$ so $ (a, b) \equal{} (1, 0)$ is the only solution here.

Thus, $ a$ is even and $ b$ is odd. Let $ a \equal{} 2k$.

Then $ 7^b \equal{} (x \minus{} 3^{k})(x \plus{} 3^{k})$. Both terms in parentheses must be powers of $ 7$. If both are divisible by $ 7$, their difference must be as well, a contradiction. Thus, $ x \minus{} 3^k \equal{} 1$ so $ x \equal{} 3^k \plus{} 1$. Thus, we have $ 7^b \equal{} 2\cdot{3^k} \plus{} 1$. Consider $ \pmod{9}$. If $ k\ge{2}$, we have $ 3|b$. The period of $ 7^b\pmod{19}$ is $ 3$, and it cycles as $ 7, 11, 1$. As $ 3|b$, $ 7^b\equiv{1}\pmod{19}$. On the other hand, $ 2\cdot{3^a}$ is never divisible by $ 19$ and so $ 2\cdot{3^a} \plus{} 1$ is never congruent to $ 1\pmod{19}$, so $ k < 2$.

Checking, we find $ k \equal{} 1$, so $ (a, b) \equal{} (2, 1)$ is the only solution.

Thus, $ (a, b) \equal{} (1, 0), (2,1)$ are the only solutions.[/hide]

Ugh it took me forever to think of (mod 19)...

6 is most likely well known; I know Erdos and Selfridge proved that the product of $ k > 1$ consecutive integers is never a perfect power. I'm too tired to go through this problem itself right now, but it probably can be done by expanding and factoring.

[quote="xpmath"][hide="1, from my blog"] Consider $ \pmod{9}$. All cubes are congruent to $ 0$ or $ \pm{1}\pmod{9}$, so $ a^3\equiv{b^3}\equiv{c^3}\pmod{9}$. Thus, all of $ a, b, c$ are congruent to one of $ 1, 4,$ or $ 7\pmod{9}$. Now note that $ 13^3 \equal{} 2197 > 2001$, so the possible choices for each of $ a, b, c$ are $ 1, 4, 7, 10$.

Now, consider $ \pmod{7}$. All cubes are congruent to $ 0$ or $ \pm{1}\pmod{7}$. From this, we find that $ a^3 \plus{} b^3 \plus{} c^3\equiv{6}\pmod{7}$ so $ a \plus{} b \plus{} c$ is either a rearrangement of $ 0 \plus{} 0 \plus{} 6$ or $ 1 \plus{} 6 \plus{} 6$ in modulo $ 7$. If it is the first, then two of $ a, b, c$ must be $ 7$ and we can check that does not work. Thus, it must be the latter and so the possible choices are $ 1, 4, 8, 11$ for the $ 1\pmod{7}$ and $ 2, 3, 5, 6, 9, 10, 12$ for the $ \minus{} 1\pmod{7}$.

The only ones that are in both are $ 1, 4, 10$, so we must have two of $ a, b, c$ equal to $ 10$ and we find the other is $ 1$. Thus, all solutions are $ (10, 10, 1), (10, 1, 10),$ and $ (1, 10, 10)$[/hide][/quote]

How did you know a,b, and c were congruent to 1,4, or 7?

You can easily verify this... :wink:

$ 1^3\equal{}1\equiv 1\pmod9$
$ 4^3\equal{}64\equiv 1\pmod9$
$ 7^3\equal{}343\equiv 1\pmod9$

I'm not sure if this is a legit explanation or not... :|

Okay, so $ a^3\plus{}b^3\plus{}c^3\equiv{3}\pmod{9}$. You can check that this means all three must be congruent to $ 1\pmod{9}$ as all cubes are $ 0, \pm{1}\pmod{9}$.

Now, go through and find for which possible $ a\pmod{9}$ is $ a^3\equiv{1}\pmod{9}$. You find that $ 1, 4, 7\pmod{9}$ work.

Then to find the possible values for $ a, b, c$, just find the numbers less than $ 12$ that are congruent to one of $ 1, 4,$ or $ 7\pmod{9}$.

Oh ok, thanks! I just wasn't sure if it was just guess and check or if you were using some other method.

I know I've seen #2 before somewhere...

Also, seeing as #1 was a JBMO problem (although this is not as difficult as USAMO, it is still beyond Intermediate in my opinion) and #4 was Canada #4 from this year, this hardly belongs in Intermediate.

Seven is simple though.  There's a pattern when dealing with 11 to the power of anything.

The last two digits of the following are:

11^1 : 11
11^2 : 21
11^3 : 31
...
11^10 : 01
11^11 : 11

Following this pattern we know the last two digits of 11^120 are 01.  The number of zeroes at the end of 11^120 is two.

Having the last three digits 001 only happens when you have 11^100 I think.

Well yes but .. the problem is asking for $ 11^{120!} \minus{} 1$  :P

[hide="5"]
Let $ H$ be the orthocenter, and $ O$ be the circle about $ AC$; then $ A' \in O$, and power of a point gives $ PH \cdot HQ \equal{} AH \cdot HA' \equal{} MH \cdot HN$ as desired.
[/hide]

[hide="7"]
Hint: applying binomial theorem gives $ (10\plus{}1)^{120!} \minus{}1 \equal{} \cdots \plus{} {120! \choose 1} 10$, and legendre's shows that last term leaves $ 29$ zeroes. Show the rest is indeed divisible by $ 10^{29}$ now.
[/hide]

[hide="2"]$ x^2 \plus{} (x \plus{} 1)^2 \equal{} 2x^2 \plus{} 2x \plus{} 1 \equal{} 2\left(x \plus{} \frac12\right)^2 \plus{} \frac12\equiv0\pmod{1997}$

$ \implies4\left(x \plus{} \frac12\right)^2 \plus{} 1\equiv(2x \plus{} 1)^2 \plus{} 1\equiv0\pmod{1997}$.

Note that $ 1997$ is a prime of the form $ 4k \plus{} 1$. Recall from Wilson's Theorem that $ (p \minus{} 1)!\equiv \minus{} 1\pmod p$, so if $ p \equal{} 4k \plus{} 1$ then

$ (4k)!\equiv\prod_{n \equal{} 1}^{2k}n(4k \plus{} 1 \minus{} n)\equiv\prod_{n \equal{} 1}^{2k} \minus{} n^2$

$ \equiv( \minus{} 1)^{2k}((2k)!)^2\equiv((2k)!)^2\equiv \minus{} 1\pmod{4k \plus{} 1}$.

Let $ i\equiv(2k)!$ so that $ i^2\equiv \minus{} 1$. Note that there can only be two solutions to $ j^2\equiv i^2\equiv \minus{} 1\pmod p\implies(j \minus{} i)(j \plus{} i)\equiv0\pmod p$, because $ p$ is a prime and thus $ p|j \minus{} i$ or $ p|j \plus{} i$, and the two solutions are hence $ j\equiv\pm i\equiv\pm(2k)!$.

Putting this back in the original problem, $ 2x \plus{} 1\equiv\pm998!\equiv\pm412\pmod{1997}$

$ \implies x\equiv\boxed{ \minus{} 1205, \minus{} 793,792,1204}\pmod{1997}$.[/hide]

I don't know a good way to find $ 998!\pmod{1997}$ without a calculator or computer...

[b]Edit:[/b] Lol I didn't realize it asked number of solutions. Thanks [b]Yongyi781[/b] for that and the correction (forgot $ 2x \plus{} 1\equiv\pm412$, not $ 2x \plus{} 1\equiv412$ somehow). :P

[b]Edit:[/b][quote="xpmath"]Thus, $ a$ is even and $ b$ is odd. Let $ a \equal{} 2k$.

Then $ 7^b \equal{} (x \minus{} 3^{k})(x \plus{} 3^{k})$. Both terms in parentheses must be powers of $ 7$. If both are divisible by $ 7$, their difference must be as well, a contradiction. Thus, $ x \minus{} 3^k \equal{} 1$ so $ x \equal{} 3^k \plus{} 1$. Thus, we have $ 7^b \equal{} 2\cdot{3^k} \plus{} 1$. Consider $ \pmod{9}$. If $ k\ge{2}$, we have $ 3|b$. The period of $ 7^b\pmod{19}$ is $ 3$, and it cycles as $ 7, 11, 1$. As $ 3|b$, $ 7^b\equiv{1}\pmod{19}$. On the other hand, $ 2\cdot{3^a}$ is never divisible by $ 19$ and so $ 2\cdot{3^a} \plus{} 1$ is never congruent to $ 1\pmod{19}$, so $ k < 2$.[/quote]

After you get $ 3|b$ let $ b \equal{} 3x$ (where $ x$ is odd and thus at least $ 1$ since $ b$ is odd), and it is probably more intuitive to do $ 2\cdot3^k \equal{} (7^x \minus{} 1)(7^{2x} \plus{} 7^x \plus{} 1)$. Because $ 7^{2x} \plus{} 7^x \plus{} 1$ is odd, $ 7^x \minus{} 1 \equal{} 2\cdot3^m\implies 7^x \equal{} 2\cdot3^m \plus{} 1$ for some $ m < k$. Now we're back where we started, except $ x \equal{} \frac b3$ (so it's smaller, and still odd). Obviously we can't continue like this forever, so we're done.

[quote="math154"]
Putting this back in the original problem, $ 2x \plus{} 1\equiv\pm998!\equiv412\pmod{1997}\implies x\equiv\boxed{1204}\pmod{1997}$. (OK, so $ \boxed{ \minus{} 1205}$ maybe also works.)[/quote]

If $ 1204$ works, then so does $ 1204 \minus{} 1997 \equal{} \minus{} 793$ and $ \minus{} 1205 \plus{} 1997 \equal{} 792$. :P

[hide="Another (bashy) way to do #2"]
Since 1997 is prime, just use the quadratic formula:
$ 2x^2 \plus{} 2x \plus{} 1 \equal{} 0\implies x \equal{} \frac { \minus{} 2\pm \sqrt { \minus{} 4}}4 \equal{} \minus{} \frac 12\left(1\pm \sqrt { \minus{} 1}\right)$. But $ \minus{} \frac 12\equiv 998\pmod{1997}$, so $ x \equal{} 998(1\pm \sqrt { \minus{} 1})$. Now, we will try to find $ k$ such that $ k^2\equiv \minus{} 1\pmod{1997}$... and we see that this gives us $ k \equal{} 412$. So, $ 998(1 \plus{} 412)\equiv 792\pmod{1997}\equiv \minus{} 1205\pmod{1997}$ and $ 998(1 \minus{} 412)\equiv 1204\pmod{1997}\equiv \minus{} 793\pmod{1997}$, and so the solutions are $ \{\minus{}1205,\minus{}793,792,1204\}$ and we're done.

(Yes, calculating the square root of -1 mod 1997 required a calculator... :blush: [unless some advanced mathematical method allows us to do that without a calculator])[/hide]

Of course, #2 only required us to find the NUMBER of integers that satisfy that... which is 4. You can do what math154 and I did, or just notice that a quadratic equation has two roots $ x_1,x_2$, but since $ x_1 \plus{} x_2\not\equiv 0\pmod{1997}$, we have another two solutions $ \minus{} x_1, \minus{} x_2$.

[quote="xpmath"]I'm too tired to go through this problem itself right now, but it probably can be done by expanding and factoring.[/quote]

Indeed, [hide="stuff works."]$ x^3 \equal{} a(a \plus{} 1)(a \plus{} 2)(a \plus{} 3) \equal{} (a^2 \plus{} 3a)(a^2 \plus{} 3a \plus{} 2) \equal{} (a^2 \plus{} 3a \plus{} 1)^2 \minus{} 1$

$ \implies(a^2 \plus{} 3a \plus{} 1)^2 \equal{} x^3 \plus{} 1 \equal{} (x \plus{} 1)(x^2 \minus{} x \plus{} 1)$. (1)

Now, $ \gcd(x^2 \minus{} x \plus{} 1,x \plus{} 1) \equal{} \gcd(3,x \plus{} 1)$ after some silly computations. However, if $ x\equiv \minus{} 1\pmod3$, then taking both sides of (1) $ \pmod3$, $ (a^2 \plus{} 1)^2\equiv0\pmod3$, which is impossible.

So $ \gcd(x^2 \minus{} x \plus{} 1,x \plus{} 1) \equal{} 1$, and both $ x \plus{} 1$ and $ x^2 \minus{} x \plus{} 1$ are perfect squares. Let $ x^2 \minus{} x \plus{} 1 \equal{} n^2\implies\left(x \minus{} \frac12\right)^2 \plus{} \frac34 \equal{} n^2\implies(2n \minus{} 2x \plus{} 1)(2n \plus{} 2x \minus{} 1) \equal{} 3$. Obviously, both $ n,x$ are positive integers (well just assume $ n$ is positive because we can), so $ 2n \plus{} 2x \minus{} 1$ is positive, and $ 2n \minus{} 2x \plus{} 1$ is as well. Regardless of how we assign $ 2n \plus{} 2x \minus{} 1$ and $ 2n \minus{} 2x \plus{} 1$ to $ 1$ and $ 3$ (since $ 3$ is prime), adding up the two equations will yield $ 4n \equal{} 4\implies n \equal{} 1\implies x^2 \minus{} x \plus{} 1 \equal{} 1\implies x(x \minus{} 1) \equal{} 0$, but $ x$ is clearly greater than $ 1$, so we have a contradiction.[/hide]

Hm, can someone clarify on what area means in #3?

[color=darkred]Problem 2: Find the number of integers [/color]$ x$ [color=darkred]such that [/color]$ |x|\le\ 1997$ [color=darkred]and[/color] $ 1997$ [color=darkred]divides[/color] $ x^2 \plus{} (x\plus{}1)^2$.

Solution:  Writing the given divisibility condition in modulo form, we have $ x^2 \plus{} x^2 \plus{} 2x\plus{} 1 \equal{} 0 (\mod 1997)$  which is equivalent to  $ (2x\plus{}1)^2 \equal{} \minus{}1 (\mod 1997)$.
Notice that $ 1997$ is a prime of the form $ 4k\plus{}1$ and thus notice that we have  two different integral solutions for $ a^2 \equal{} \minus{}1 (\mod 1997)$ and also we notice that the solutions are non zero in our case  since $ 0$ is not a solution to $ (ax\plus{}1)^2 \equal{} \minus{}1 (\mod 1997)$ and  we then observe that there are 4 distinct non zero solutions for $ x$, $ 2$ in each, for values $ \minus{}1997$ to $ \minus{}1$ and $ 1$ to $ 1997$.{$ 0$ is not included here since it cannot be a solution }.
Thus we have 4 distinct non zero integral solutions for the given problem.
Infact, notice here that  we exploited the problem which just asked for the number of solutions rather than finding solutions itself.....


In Rudin's "Principles of Mathematical Analysis", I read that any ordered set with the least upper bound property is isomorphic to $ \mathbb{R}$. However, the proof was not included (probably because it was too difficult).
I was wondering if someone could include an outline of the proof, if its not  too complicated.

It's an ordered [b]field[/b], not just any set.

It's a long and tedious argument, but you pretty much have to go through it once. It's also essentially the same as constructing the real numbers.
First, we note that $ 1,1 \plus{} 1,1 \plus{} 1 \plus{} 1,\dots$ are all positive, and the rationals embed with order. Completeness implies the Archimedean property; the greatest lower bound $ c$ of $ 1,\frac12,\frac14,\dots$ must be zero, since $ c \equal{} \frac12c$. With the Archimedean property, every number in our field $ K$ is characterized by the rational numbers less than it, and that gives us Dedekind cuts.


Prompted by a comment from Arnav about measurability as a condition for Minkowski's theorem to hold, I [url=http://www.springerlink.com/content/h15tw63022ph5445/]learned[/url] that every convex subset of $ \mathbb{R}^n$ is in fact measurable.  Lang's discussion, however, does not make it clear (at least to me) whether the following assertion is true:

Is every convex subset of $ \mathbb{R}^n$ the union of a Borel set and a set of zero measure?

Every Lebesgue measurable set is a union of an $ F_\sigma$-set and a set of $ 0$ Lebesgue measure.

:oops:  Thanks.  (Obviously I need to learn more measure theory!)


abstract algebra
number theory
relatively prime
superior algebra
superior algebra solved

Hi, do you guys know how to do this question?  It's one of my assignment questions, pls help.......

[color=green]Q: Show that if n is a positive integer, and a and b are integers relatively prime to n such that (ord[size=75]n[/size]a, ord[size=75]n[/size]b)=1, then ord[size=75]n[/size](ab)=ord[size=75]n[/size]a x ord[size=75]n[/size]b[/color]

Thanks alot!!!   :blush:

It's not hard by direct checking and was also posted already.
Just note that it is true in any abelian group.


trigonometry
function
trig identities
Law of Cosines
geometry
Law of Sines

How do you apply sine to a non-right triangle. I think its possible, but how would you know which side is considered the hypotenuse, or is there something I'm missing about sine?
Also, regarding the law cosines, in setting up the cosine of the angle opposite the side you are looking for, how do you know which side goes on the numerator and which on the denominator? Would it just be the longer side?

Thanks,

You can't apply sine directly to a non-right triangle. The trig functions are specially designed for right triangles. However, by drawing an altitude, you can split up the triangle into right triangles which you can then apply trig to.

You can, but tis more complicated than a direct ratio.

This would be the law of Sines. http://www.artofproblemsolving.com/Wiki/index.php/Law_of_sines
Use it for non-right triangles.

Here's more detail on the law of Cosines.
http://www.artofproblemsolving.com/Wiki/index.php/Law_of_cosines
I'm not sure what you are referring to. If you are looking for a side, and you want to use the law of cosines, you shoul already know the angle oppsite that side, as well as the two other side lengths.

Well, reffering to the law of cosines, as in how can you have a cosine in a non-right triangle? If you are using the law of cosines, how would you apply it,? How do you know what Cos C is?

You use Law of Cosine to find the third side if you are given 2 sides and the included angle opposite side $ c$. The formula is $ c^2\equal{}a^2\plus{}b^2\minus{}2ab\cos C$. If you want to find the value of $ \cos C$ you solve for it to get $ \cos C\equal{}\frac{a^2\plus{}b^2\minus{}c^2}{2ab}$.

About your earlier post, what do you mean what goes on numerator and denominator? If you are thinking about $ \cos C\equal{}\frac{\text{adjacent}}{\text{hypotenuse}}$, that's only for right triangles, which is the appropriate definition of cosine.

If you are inquiring about how on Earth did mathematicians derive the Law of Cosine, please read pinkmuskrat's link above.

Hang on, How do we find CosC using c^2 when we need Cos C to find C^2?

Depending on what information about the triangle you are given you use one formula or the other, not both.

Eg. if you are given three sides you will use $ \cos C \equal{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab}$

If you are given two sides and an included angle you will use: $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos C$

Yes, how do you get Cos C from two sides and an included angle in a non-right triangle?

[quote="eli140"]Yes, how do you get $ \cos C$ from two sides and an included angle in a non-right triangle?[/quote]
It means if you are given an angle which lets say is angle $ C$ and two sides $ a$,$ b$ you use $ c^2 \equal{} ...$You get $ \cos C$ by taking the $ \cos$ of the angle...You will use the $ \cos C$ of a different angle for a different question. For the cosine law to be applied the angle has to be between 2 sides or there have to be 3 side lengths given.

[quote="tornado.adv4"]You use the Law of Sines if you are given ASA, AAS, or SSA.
The Law of Sines needs an angle and its opposite side.

You use the Law of Cosines if you are given SSS or SAS.

If you use the Law of Cosines, you can usually use the Law of Sines after (since it's shorter than the Law of Cosines).
For example, if you are given SAS, you can use the Law of Cosines to solve for the third side, and then the Law of Sines to get a second angle.[/quote]An observation should be added for SSA when applying Law of Sines: there are two angles possible for this configuration, which are supplementary to each other.

But how do you use the law of sines in a non right triangle?

You take the formula given. The law of sines and law of cosines use angles A, B, and C, as well as their respective opposite sides, a, b, and c. Now, based upon what information you have, assign a, b, c, A, B, C (all that are necessary) to different sides and angles. The only stipulation is that a goes opposite from A, etc.

To use a particular law, you must only have one variable remaining once all is plugged in. Then use the appropriate operations to solve for the variable.

Examples are shown in the links. :)

http://www.artofproblemsolving.com/Wiki/index.php/Law_of_sines
http://www.artofproblemsolving.com/Wiki/index.php/Law_of_cosines

Given angles $ A,B,C$ and sides $ a,b,c$, each one opposite to one angle respectively.
We have
$ \frac{a}{\sin a} \equal{} \frac{b}{\sin b}\equal{} \frac{c}{\sin c}$

Here is a basic problem that uses the Law of Sines.

Solve the triangle for $ A\equal{}35^{\circ}$, $ a\equal{}19$, and $ B\equal{}88^{\circ}$. (lowercase letters are opposite the uppercase angle.)

[hide="Solution"]
First we know that $ C\equal{}180^{\circ}\minus{}A\minus{}B\equal{}180^{\circ}\minus{}35^{\circ}\minus{}88^{\circ}\equal{}57^{\circ}$.
Using the Law of Sines, we have $ \frac{c}{\sin C}\equal{}\frac{a}{\sin A}$.

Now we plug in values to get $ \frac{c}{\sin 57^{\circ}}\equal{}\frac{19}{\sin 35^{\circ}}$.
Use a calculator to evaluate the sines, and we get $ c\approx 27.7814$.

Solve for $ b$ the same way. $ \frac{b}{\sin 88^{\circ}} \equal{} \frac{19}{\sin 35^{\circ}}$.
Then $ b\approx 33.1053$.
[/hide]


I took AMC12A in my school, and did horribly, so i didn't qualify for AIME.

And of course i took AMC 10B to qualify for AIME (and i got 145.5 - checked my answers with online answer keys) in Chapman University.

but since i didn't take 10B in my school, the math teacher who manages math competitions doesn't have my score for 10B (he only has the result for 12A)

i know that AIME is coming on Tuesday, and i have few questions.

1. Do i need my result for AMC 10B to take AIME?

2. Where can i get the result? From the professor in Chapman University? or do i just wait until i receive it?

i really need help on this. i don't want to miss AIME :(

You need to contact the test supervisor at Chapman. (I don't personally know anyone there, so I don't know if a message from me would do anything.)

Mr. Merryfield is correct.  With your math teacher or school contest manager, you need to contact the contest manager at Chapman Univerdsity.  The contact is listed on our web page.   By having your school math teacher present during the phone call, you can arrange the confirmation of score and transfer of AIME immediately.

Steve Dunbar
Director, American Mathematics Competitions

So, I just contact the contest manager at Chapman University with my school math teacher, and tell the contest manager to give my school teacher the score?

Does that mean I have to call the contest manager in my school with my teacherm so that he can get my score directly from the contest manager at Chapman University?


Let $ f: \mathbb{R}\mapsto\mathbb{R}$ be a convex function-not necessarily differentiable and let $ a$ be a positive real number.

prove that the function $ g: \mathbb{R}\mapsto\mathbb{R}$, defined by $ g(x)\equal{}f(a\plus{}x)\minus{}f(x)$ is increasing.

Try to prove that if $ a<c<b<d$ then

$ \frac{f(b)\minus{}f(a)}{b\minus{}a} \leq \frac{f(d)\minus{}f(c)}{d\minus{}c}$

In order to prove the one above use the fact that $ p_{\alpha}(x)\equal{}\frac{f(x)\minus{}f(\alpha)}{x\minus{}\alpha}$ is increasing.And for this, use the next characterization of a convex function:

 If $ f$ is convex and $ x_1<x_2<x_3$ , then

$ \frac{f(x_2)\minus{}f(x_1)}{x_2\minus{}x_1} \leq  \frac{f(x_3)\minus{}f(x_1)}{x_3\minus{}x_1} \leq \frac{f(x_3)\minus{}f(x_2)}{x_3\minus{}x_2}$


A bag of grapes is to be distributed evenly to 5 kids in a class, and the grapes that are left over will be thrown out. If each student receives the greatest possible number of grapes, what is the greatest possible number of grapes that could be thrown out?

We want the largest remainder possible remainder after dividing by $ 5$, which is $ \boxed{4}$


number theory
relatively prime
number theory unsolved

Does anybody know whether this is true?

The decimal representation of m/n, where m and n are relatively prime positive integers and m < n, contains the string of digits a1 a2...ak.

For the smallest n that this is possible, is it true that there exists b/n (b is a positive integer, b < n) such that in the decimal representation of b/n, the string of digits a1 a2...ak occurs right in front (i.e. just after the decimal point)?

I hope I am not misunderstanding the question, but yes it is possible, suppose that m/n with m < n is .c1c2...cra1...ak..., then m/n*10r = c1c2...cr.a1...ak... then let d be the integer c1*10r-1 + c2*10r-2 + ... + cr

then m/n*10r - d = .a1...ak... which is (m*10r - n*d)/n, and 0 <.a1...ak... < 1, so (m*10r - n*d) is positive, less than n, and is an integer because m, 10r, n, and d are integers, so it is the b we are looking for.


I have some word problems that I am supposed to use Diophantine eqs for.

A farmer purchased 100 head of livestock for a total cost of $ 4000. Prices were:$120 for calves, $ 50 for lambs and$25 for piglets.  How many of each animal did he buy assuming that he must buy at least one of each.  I have the answer but all I did was plug numbers in to get it to work.  Is there a better approach to solving 3 variable diophantine eqs?

So after starting out with the two equations
\[ a\plus{}b\plus{}c\equal{}100,120a\plus{}50b\plus{}25c\equal{}4000\]
where $ a,b,c$ is number of calves, lambs, piglets, respectively, it helps to divide out as much as you can in the second equation. It first simplifies to
\[ 24a\plus{}10b\plus{}5c\equal{}800\]
Now this equation couldn't possibly have a solution unless $ a$ were divisible by 5, since the other terms are. So let $ a\equal{}5x$. Of course, the same reasoning says $ c$ is even, so let $ c\equal{}2z$. We now have
\[ 5x\plus{}b\plus{}2z\equal{}100,12x\plus{}b\plus{}z\equal{}80\]
At this point, we could just solve for $ b,z$ in terms of $ x$:
\[ b\plus{}z\equal{}80\minus{}12x,b\plus{}2z\equal{}100\minus{}5x\Rightarrow z\equal{}20\plus{}7x,b\equal{}60\minus{}19x\]
Hmm... there aren't too many possibilities where $ x$ and $ b$ are both positive. The solutions for $ (x,b,z)$ are
\[ (1,41,27),\ (2,22,34),\ (3,3,41)\]
In terms of our original variables, that means there's three possibilities: 5 calves, 41 lambs, 54 piglets; or 10 calves, 22 lambs, 68 piglets; or 15 calves, 3 lambs, 82 piglets.

I also have the problem: 100 bushels of grain are distributed among 100 persons in such a way that each man receives 3 bushels, each woman receives 2 bushels and each cild receives 1/2 bushel.  How many men, women and children are there??

I tried to do this problem the same way the last one was explained but I couldnt get it to work. Help please!

solve the system of equation in $ N$
$ a\plus{}b\plus{}c\equal{}100$ and $ 6a\plus{}4b\plus{}c\equal{}200$


A cirlce has 10 equi-distant points, how many regular hexagons can be obtained?

here goes, since there are 10 equidistant points, that already makes our hexagon regular right...wrong (i think) because you can skip a point, that would throw off your entire hexagon, from there i might count, but i dont think that is correct because you can choose one point to start, but that same point could be part of another hexagon thats exactly the same, (like if we had ABCDEF as a hexagon, we could start at D and that would be the same DEFABC) so if you can help, it would be appreciated  :D

I think the answer is either
[hide]10, 1, or 2[/hide]depending on how different hexagons are counted.

[quote="WindSlicer"]I'm thinking it's

[hide]0[/hide]

Just by drawing it and drawing some possible hexagons. *shrugs*[/quote]Forget my previous answer, I think you're right.  When I drew the diagram I drew pentagons instead of hexagons.  :blush:

[quote="Done"]is it zero?![/quote]

Yes, that's what I got at least...

I think the question was probably quoted wrongly.... :?

Do the vertices have to be on the points :P ?

[hide]Actually, I don't think it's zero. draw a circle, and draw a regular hexagon in it that has it's six vertices on the side of the circle. Put 4 dots randomly throughout the edge of the circle.. Voila! you got one regular hexagon by using points equidistant from a point[/hide][/hide]

[hide][color=red]I read the problem again.. I THINK IM WRONG >_< meh...[/hide]


Suppose $a_1>a_2>\dots a_n$ and $k=10^{a_1}+\dots+10^{a_n}$ is perfect square. Prove $n=1$

This has been posted before, and received no solution :(.

A propos, could someone explain me what a "perfect square"  is ? (that's stupid but I never found it out  :) )

[quote="Psycho"]a perfect square is the square of an integer... like 1,4,9,16,25,36,47,64,...  :)[/quote]
47 is not really perfect  :P

ok, that was obviously a typo  :blush:  :P


In a town election, candidates A and B were running for mayor. There
were 30,500 people eligible to vote, and 3/4 of them actually voted.
Candidate B received 1/3 of the votes cast. How many people voted for
candidate B? What percent of the votes cast, to the nearest tenth of a
percent, did candidate A receive?

[quote="Interval"]Is that the answer?[/quote]

He is saying that it "Cannot be determined from the given information"

Perhaps you might have a little bit more information to provide.

[quote="jli"]it depends on the number of people who are actually [b][color=red]Legible [/color][/b]to vote.[/quote]

Legible means able to be read... doesn't it?

If you meant eligible, there are $30500$, as Interval said, of which $\frac{3}{4}$ voted.


$Problem.$ Find all $x,y,z$ positive integers such that
\[3^x-2^y=$1987^z\]

$3^x-2^y=1987^z$ has the only solution $(x,y,z)=(1,1,0)$

Proof:

We have  $-(-1)^y=1[3]$ so $y$ is odd

- Case $y=1$:
$(-1)^x-2=(-1)^z[4]$ therefore we have two cases:
1) $x=2p$ and $z=2q+1$: we obtain an impossibility $modulo$ $8$
2) $x=2p+1$ and $z=2q$:
if $p=0$, we find the solution $(1,1,0)$
if $p>0$, we have, reasoning $modulo$ $9$: $q=2[3]$
and we find an impossibility $modulo$ $91$

- Case $y>1$:
We have $(-1)^x=(-1)^z[4]$ therefore we have two cases:
1) $x=2p$ and $z=2q$:
$2^y=(3^p-1987^q)(3^p+1987^q)$ consequently: $2^a=3^p-1987^q<3^p+1987^q=2^b$
so $2^a$ divide $3^p-1987^q+3^p+1987^q=2.3^p$
=> $a=0$ (impossible $modulo$ $3$) or $a=1$ (impossible:first case)
2) $x=2p+1$;$z=2q+1$ (and $y=2r+1$):
reasoning $modulo$ $7$, we obtain $p=0[3]$ and $r=2[3]$
and we find an impossibility $modulo$ $91$

________

blang (sorry for my french english!)


calculus
integration
limit
trigonometry
real analysis
real analysis unsolved

Find the limit: $ \lim_{n\to\infty}{\int_{0}^{\pi}{\cos (x^{n})\textbf dx}}$

Split this into $ \int_0^1\cos(x^n)\,dx\plus{}\int_1^{\pi}\cos(x^n)\,dx$

By the Dominated Convergence Theorem,

$ \lim_{n\to\infty}\int_0^1\cos(x^n)\,dx\equal{}\int_0^1\cos(0)\,dx\equal{}1.$

As for the second integral, try integrating it by parts:

$ \int_1^{\pi}\cos(x^n)\,dx\equal{}\left.\frac{\sin(x^n)}{nx^{n\minus{}1}}\right|_1^{\pi}\minus{} \int_1^{\pi}\frac{n\minus{}1}{n}\cdot\frac{\sin(x^n)}{x^n}\,dx$

But $ \left|\left.\frac{\sin(x^n)}{nx^{n\minus{}1}}\right|_1^{\pi}\right|\le\frac2{n}$ and thus tends to zero, and

$ \left|\int_1^{\pi}\frac{n\minus{}1}{n}cdot\frac{\sin(x^n)}{x^n}\,dx\right| \le\frac{n\minus{}1}{n}\int_1^{\infty}\frac1{x^n}\,dx\equal{}\frac1n$

so that also goes to zero.

$ \therefore\ \lim_{n\to\infty}\int_0^{\pi}\cos(x^n)\,dx\equal{}1.$


trigonometry
geometry
trig identities
Law of Sines

In a triangle ABC prove that $ a^{3}\cos (B\minus{}C)\plus{}b^{3}\cos (C\minus{}A)\plus{}c^{3}\cos (A\minus{}B)\equal{}3abc$

hello, let $ term_1\equal{}a^2\cos(B\minus{}C)$ then we get by applying the addition formulas
$ a^3(\cos(B)\cos(C)\plus{}\sin(B)\sin(C))$
by the law of cosine we have
$ \cos(B)\equal{}\frac{a^2\plus{}c^2\minus{}b^2}{2ac}$
$ \cos(C)\equal{}\frac{a^2\plus{}b^2\minus{}c^2}{2ab}$
and by the law of sines we get
$ \sin(B)\sin(C)\equal{}\frac{b}{c}(1\minus{}\cos(C)^2)$
plugging this in $ term_1$ and simplifying we get
$ term_1\equal{}\frac{a(a^2(b^2\plus{}c^2)\minus{}(b^2\minus{}c^2)^2)}{2bc}$
analogously we get $ term_2$ as
$ term_2\equal{}\frac{b(b^2(a^2\plus{}c^2)\minus{}(a^2\minus{}c^2)^2)}{2ac}$
$ term_3\equal{}\frac{c(c^2(b^2\plus{}a^2)\minus{}(b^2\minus{}a^2)^2)}{2ba}$
adding $ term_1,term_2$ and $ term_3$ we get after simplifying $ 3abc$.
Sonnhard.


Phil has 10 different marbles in a bag. What is the expected value of the minimum number of times for him to take the marbles out one at a time with replacement until each of the 10 marbles has been taken at least once?

[hide="Solution"]
Say that he has collected $ n$ marbles so far. If $ E_{n \plus{} 1}$ is the expected number of marbles he will need to get to get a new coupon, then $ E_{n \plus{} 1} \equal{} \left(1 \minus{} \frac n{10}\right) \plus{} \frac n{10}\cdot(1 \plus{} E_{n \plus{} 1}) \equal{} 1 \plus{} \frac n{10}\cdot E_{n \plus{} 1}\implies E_{n \plus{} 1} \equal{} \frac {10}{10 \minus{} n}$, and the expected value is thus $ \sum_{k \equal{} 0}^9E_{k \plus{} 1} \equal{} \sum_{k \equal{} 1}^{10}\frac {10}k \equal{} \boxed{\frac {7381}{252}\approx29.3}$.
[/hide]
This is the [url=http://en.wikipedia.org/wiki/Coupon_collector%27s_problem]coupon collector's problem[/url].


For each positive integer n, find the greatest common divisor of n!+1 and (n+1)!

$(n!+1,(n+1)!)=(n!+1, n+1)$.

If $n+1$ is prime, then both are divisible by $n+1$. [wilson's theorem]

If $n+1$ is not prime, then the gcd is $1$ because $n+1|n!$ when $n+1$ is composite.

Consider: $n+1=pq$, $1<p<q<n+1$. We can always make that choice except when $n=p^{2}$, $p$ a prime. However, the assertion is trivially true in that case.


A rook is given on a 8*8 chessboard. It can move to any neighbouring square (i.e. right, left, up, down) (so this is not the usual movement of the rook). After 64 moves, it is back on the square it was first placed on and all squares were visited exactly once. Prove that the number of vertical moves is different from the number oh horizontal moves.
This might look easy, but I think it’s hard.

Dear dzeta,

please be more specific about your source. Consider: http://www.mathlinks.ro/viewtopic.php?t=3518. I would appreciate to know to which year you refer, the main/advanced, junior or easy round of Tournament of Towns etc. Thank you.

Problem 6, 1999, spring session, main, advanced.

I remember it was solved some years ago on the forum sci.math :
http://groups.google.fr/groups?hl=fr&lr=&ie=UTF-8&threadm=38BDCE0B.6A6CF86C%40yesic.com&rnum=1&prev=/groups%3Fq%3Dtournament%2Bof%2Btowns%2Bproblems%2Bgroup:sci.math.*%26hl%3Dfr%26lr%3D%26ie%3DUTF-8%26group%3Dsci.math.*%26selm%3D38BDCE0B.6A6CF86C%2540yesic.com%26rnum%3D1

Look after the answer by Chas Brown.

Pierre.


geometry
parallelogram
rhombus
trigonometry
complex numbers
geometry proposed

[b][i]Suppose a parallelogram $ABCD$ then construct squares on the sides of the parallelogram. Prove that the centre of the squares make a new square.
[/i][/b]

Davron Latipov

Let P, Q, R, S be the square centers on the parallelogram sides AB, BC, CD, DA. The triangles $\triangle ASP, \triangle BQP, \triangle CQR, \triangle DSR$ are all congruent by SAS. For example, AS = BR, AQ = BQ and (WLOG assuming that $\angle DAB \le \angle ABC$)

$\angle SAP = \angle DAB + 2 \cdot 45^\circ = \angle DAB + 90^\circ$

$\angle QBP = 360^\circ - (\angle ABC + 2 \cdot 45^\circ) = 180^\circ - \angle ABC + 90^\circ = \angle DAB + 90^\circ$

Therefore, SP = PQ = QR = RS and the quadrilateral PQRS is at least a rhombus. But

$\angle SPQ = \angle AQB - \angle APS + \angle BQR = \angle AQB = 90^\circ$

so that the rhombus PQRS is a square.

[hide="Here is a proof with the complex numbers."][color=darkblue]I suggest you apply the complex numbers and I shall choose  the origin in the intersection $O$ of the diagonals $AC$ and $BD$. Thus, $A(a)$, $B(b)$, $C(-a)$, $D(-b)$, where $X(x)$ is the point $X$ with the affix $x$. Construct in the exterior of the parallelogram $ABCD$ the isosceles triangles $APB$, $BQC$, $CRD$,  $DSA$ in the vertices $P$, $Q$, $R$, $S$ respectively so that $\widehat {APB}\equiv \widehat {BQC}\equiv \widehat {CRD}\equiv \widehat {DSA}$. Denote the common value $\phi$ of these last angles and $\omega=\cos \phi +i\sin \phi$. If $P(p)$, $Q(q)$, $R(r)$, $S(s)$ then $p=\frac{b-\omega a}{1-\omega }$, $q=\frac{-a-\omega b}{1-\omega}$, $r=\frac{-b+\omega a}{1-\omega}$, $s=\frac{a+\omega b}{1-\omega}$. We observe that $p+r=q+s=0$, i.e. the quadrilateral $PQRS$ is a parallelogram with the same intersection $O\in PR\cap QS$ of the diagonals and $\frac{q-p}{s-p}=\frac{-(a+b)+(a-b)\omega }{(a-b)+(a+b)\omega }$.
If $\phi =\frac{\pi}{2}$, i.e. $\omega =i$ then $q-p=i(s-p)$, i.e. $PQ=PS$ and $PQ\perp PS$. Therefore the parallelogram $PQRS$ is a square with the centre in the intersection of the diagonals of the parallelogram $ABCD$.[/color][/hide]


Determine the number of digits in the value of $ 2^{12} \times 5^8$.


abstract algebra
superior algebra
superior algebra unsolved

Show that Z[i]/(1-i) is a field. How many elements does it have?

According to the 3rd(I think) isomorphism theorem we get:
$\displaystyle\frac{\mathbb Z[i]}{<a+i>}\simeq \frac{\mathbb Z[X]/<x^2+1>}{<a+x,x^2+1>/<x^2+1>}\simeq \frac{\mathbb Z[X]}{<a+x,x^2+1>}\simeq \\ \simeq \frac{\mathbb Z[X]/<a+x>}{<a+x,x^2+1>/<a+x>}\simeq\frac{\mathbb Z}{<0,a^2+1>}\simeq \mathbb Z_{a^2+1}$.

For $a=-1$ you get the desired result.


Must all of the intersection points of the graphs of $ f(x)$ and $ f^{\minus{}1}(x)$ be on the line $ y\equal{}x$?

Yes, because an inverse is the function reflected over that line. Therefore, any points of intersection have to be on y = x.

No, if the function graph is already symmetric with respect to line $ y\equal{}x$ that's not necessarily true.

Consider $ f(x) \equal{} 1\minus{}x$. Then $ f^{\minus{}1}(x) \equal{} f(x)$, and only one of the (infinite) points of intersection lies on $ y\equal{}x$.


Yeah I have the test but for some reason I can't seem to find the answers.  :o  Does anyone have a copy of the 89 Nationals Answers?  Thanks.

doesnt this belong on the stickie

ps. you can only get nats problems and answers from either your team coach or by buying them online, i think.  i dont think your allowed to trade online

What's wrong with giving out answer keys?

They got into a huge discussion about this on the mathcounts resources here stickie. You can look over there for some other opinions on this

Wait... are Mathcounts material copyrighted?

In that case, are coaches even allowed to reproduce them (for educational purposes)?


geometry
circumcircle
inradius
inequalities
function
area of a triangle
Heron\u0027s formula

[b]Prove that[/b]

$ \ \ 4(a \plus{} b \plus{} c)\ \le\ \boxed {\sum \frac {(b \plus{} c)^2}{b \plus{} c \minus{} a}\ \le 4(a \plus{} b \plus{} c) \plus{} \frac {2(4R \plus{} r)(R \minus{} 2r)}{r}\cdot\sqrt {\frac {4R \plus{} r}{16R \minus{} 5r}}}$

where $ R$ - circumradius and $ r$ - inradius ot $ \triangle ABC\ .$

The left inequality is easy ([u]with C.B.S.[/u])  [u]The right inequality is more  interesting[/u] (from the box).

[quote="Virgil Nicula"]Prove that $ \ \ 4(a \plus{} b \plus{} c)\ \le\ \sum \frac {(b \plus{} c)^2}{b \plus{} c \minus{} a}\ \le$ $ 4(a \plus{} b \plus{} c) \plus{} \frac {2(4R \plus{} r)(R \minus{} 2r)}{r}\cdot\sqrt {\frac {4R \plus{} r}{16R \minus{} 5r}}\ ,$

where $ R$ - circumradius and $ r$ - inradius ot $ \triangle ABC\ .$[/quote]

Well, the first one is obvious by CS, but the second one looks really tough...

Theoretically we can denote the whole inequality in the means of $ a,b,c$ (denoting $ R\equal{}\frac{abc}{4P}$ and $ r\equal{}\frac{2P}{a\plus{}b\plus{}c}$ and then using the Heron's formula), but the computations are unbearable :(

[quote="Mateescu Constantin"]The computations will not be unbearable if we make use of the identity : $\boxed{\ \sum\ \frac {(b+c)^2}{b+c-a}=8s+\frac {R-2r}r\cdot 2s\ }$ (its proof is left to the interest reader) :wink:
[/quote]
We just have to express the cyclic sum in function of $R$,$r$,$s$ using the following system :
$\begin{cases} a+b+c=2s \\ ab+bc+ca=s^2+4Rr+r^2 \\ abc=4pRr \end{cases}$


let a ,b be +ve integers such that gcd(1+ab) divides a^2 +b^2.show that the integer (a^2+b^2)/1+ab must be a perfect square

Please answer my question
I want the answer quickly
PLEASE

It's in PEN - [url]http://www.mathlinks.ro/viewtopic.php?t=150371[/url]


If $\nabla(a, b, c) = a + b + c$ and $N* = N^2$, find the sum of the values of $x$ such that $(\nabla(x, 4, 5))* = 144$.

Yep.  What would happen if 144 were replaced by any other positive number?

[hide="Comment"]Yep.  By your original method, the roots are $-9 \pm \text{something}$.  When you add them, the plus and minus terms cancel out, and so you just get two copies of $-9$.  That works even if the number you replace 144 by is not a perfect square.
[/hide]


search
function
trigonometry
real analysis
real analysis unsolved

Does anybody know the this example:
For differentible function $f(x)$ on $[0,1]$ such that $f'(x)$ is bounded,but not Riemann Integrable.

Let $K$ be a Cantor set of positive measure, and let $f(x)=d(x)^{2}\sin\frac{1}{d(x)}$, where $d(x)$ is the distance from $x$ to $K$.

Thank you mlok and perfect_radio.
But to mlok:I don't think you example works well,in your example,$f'(x)$ just discontinious on the boundary of the  Cantor set which has positive measure,but the boundary is still zero measure.

Cantor-type sets are totally disconnected; in particular, they have no interior points.

Oh,mlok,you give a nice example,Thank you.I am wrong with saying that the boundary is zero measure,sorry.


calculus
integration
analytic geometry
trigonometry
algebra open
algebra

what is baic difference between double and triple integral?what is difference cylindrical and spherical coordinates in integration?

[quote="sudarsh"]what is difference cylindrical and spherical coordinates in integration?[/quote]
In cylindrical coordinate, a point is described by $(r, \theta, z)$ where $x=r\cos\theta$, $y=r\sin\theta$, $z=z$. The integration is $\int\int\int_V f(r, \theta, z)r\,dr\,d\theta\,dz$.
In spherical coordinate, a point is described by $(\rho, \phi, \theta)$ where $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$. The integration is $\int\int\int_V f(\rho, \phi, \theta)\rho^2\,d\rho\,d\phi\,d\theta$.


This is the four tests of Brazilian Olympic Revenge.
All in Portuguese (No time to translations now, but I am trying to do it in a few weeks... Sorry :D).

Well, finally I finish the translations.
Please report any errors!

[quote="Rushil"]what's the program to read .dvi files. I haven't found any on the net!!![/quote]

I use Yap.

what's that???? Please give a net link if you have one

[quote="Rushil"]Please give a net link if you have one[/quote]

What about Google? ...

[url]http://math.ndsu.nodak.edu/resources/tex/bibliography/yap/yap.html[/url]

[quote="Rushil"]How do you open *.gz files. Any program??? Can you plz post thr .ps or the .pdf file separately? Thanks. Also generally, what's the program to read .dvi files. I haven't found any on the net!!![/quote]

Well, in Linux and generally Unix there exists txe XDVI. In Windows, the YAP in the MikTeX distibution is the best indicative.

And, for open *.gz files, use 7-zip(http://www.7-zip.org/) or the GNU tools (gzip, tar).


When are the MATHCOUNTS Chapter/State/National competitions for 2009?

Sorry if this has been asked.

~BOGTRO

The dates of chapter (February) and state (March) vary, but Nats is May 7-10.

The chapter and states competition is varied depending on where you live. The national competition is from May 7-10. You can check the mathcounts website for the date range for the chapter and states competition.


algebra
polynomial
number theory unsolved
number theory

Assume that $m$ and $n$ are positive integers, and $1\leq m\leq \phi (m)+n$. Prove that $m \mid \sum_{i=0}^{n} (-1)^i {n\choose i} i^m$.

Let $m=s\phi(m)+r$, where $r<\phi(m)$, then $n\geq m-\phi(m)\geq m-s\phi(m)=r$.

$\sum_{i=0}^{n}{(-1)^i {n \choose i} (i+k)^r}$ is the nth finite difference of the polynomial $x^r$ starting at k, with degree less than equal to n. It must be constant for all k. When k is -n/2, this sum is 0 by symmetry.

let $p^t$ be the highest power of prime $p$ that divids m.
So $\sum_{i=0}^{n}{(-1)^i {n \choose i} i^m} = \sum_{i=0}^{n}{(-1)^i {n \choose i} i^r}$ (Euler's theorem if gcd(i,p)=1, if not, $r\geq p^{t-1}\geq t$, so the reduction of power would still work)
$= \sum_{i=0}^{n}{(-1)^i {n \choose i} (i-n/2)^r} = 0$ (mod $p^t$)
Doing this for other prime divisors of m and we are done.
Peng Shi


What is the number of square centimeters in the area of the shaded region?

[asy]pair a,b,c,d,e; unitsize(.25cm);
a = (10,10); b = (0,-10); c = (10,-5); d = (10,5); e = (6,2);
fill(e-- rotate(90)*e -- rotate(180)*e -- rotate(270)*e -- cycle,grey);
for(int i = 0; i < 4; ++i){draw(rotate(i*90)*(a--b));
label("10 cm",rotate(i*90)*c,dir(i*90)); label("10 cm",rotate(i*90)*d,dir(i*90));
}
draw(a--(10,-10)--(-a)--(-10,10)--a);[/asy]

Let the area of the smallest triangle be $ a$. By triangle similarity, the area of the trapezoid region is $ 4a\minus{}a\equal{}3a$. This means that we want to find $ 400\minus{}16a$. Since one of the large right triangles ($ 10\minus{}20\minus{}\sqrt{50}$) has area $ 5a\equal{}\frac12\cdot10\cdot20\equal{}100\implies a\equal{}20$, our final answer is $ 400\minus{}320\equal{}\boxed{80}$.


calculus
integration
trigonometry
function
calculus computations

a question i asked for HW yesterday... :|

$ \int_{0}^{1}\;\;\sqrt{\;x\cdot\left(1\minus{}x\plus{}\sqrt{\;1\plus{}x\cdot(2\minus{}3x)\;}\right)\;}\;\;\textbf dx$

I could use a hint:

webMathematica (e.g. http://www.quickmath.com , http://www.integrals.com ) couldn't do it,

trig substitution is not pretty, and by not pretty I mean $ \frac{4}{9}\int_{0}^{\frac{2\pi}{3}}\sin (t)\sqrt{\sin\left(\frac{3t}{2}\right)\left(\sqrt{3}\cos\left(\frac{t}{2}\right)\plus{}\sin\left(\frac{t}{2}\right)\right)}dt$.

by parts maybe?

Beta functions?

Series expansion (yuck)?

Some clever substitution (my bet)?

Have you tried Euler's substitution?

I mean, $ \sqrt{1\plus{}2x\minus{}3x^{2}}\equal{} t(1\minus{}x)$. I think it works, but the computations are ugly :!:

Mathematica can't solve the integral as is, but after that substitution I get an integral \[ \int_{0}^{\infty}\frac{32 x^{2}\left( x^{2}\plus{}2\right)\left( x^{2}\plus{}1\right)}{\left( 3\plus{}\left( x^{2}\plus{}1\right)^{2}\right)^{3}}\, dx ,\] which is revealed to be
\[ \boxed{\frac{7\pi}{12\sqrt 6}}.\]
I'm not in the mood for contour integration, but that's the way you do it.

[quote="perfect_radio"]
... after that substitution I get an integral
\[ \int_{0}^{\infty}\frac {32 x^{2}\left( x^{2} \plus{} 2\right)\left( x^{2} \plus{} 1\right)}{\left( 3 \plus{} \left( x^{2} \plus{} 1\right)^{2}\right)^{3}}\, dx ,
\]
which is revealed to be
\[ \boxed{\frac {7\pi}{12\sqrt 6}}
\]
[/quote]

that is correct ! well done !  :coolspeak:


analytic geometry
graphing lines
slope
algebra
system of equations
absolute value

Positive integers a, b, and c are chosen so that a<b<c and the system of equations 2x+3y=2003 and
 y=|x-a|+|x-b|+|x-c| has exactly one solution. What is the minimum value of c?

[hide="Hint"]The given line has slope -2/3, and the absolute value sum starts out at slope -3, then -1, then 1, then 3 (changing slopes at x=a, b, and c, respectively). Convince yourself that this means that the unique point of intersection must be at the bottom vertex -- namely, at x=b.[/hide]


Does anyone know any good tutorials on microprogramming? I'm having no luck with google at the moment.

micro programming as in micro controller programming?


The "board" on which the this 2-person game is played is a monicpolynomial
f(x) = x^(2n) + a_(2n-1)*x^(2n-1) + ... + a_1*x + 1 in which the 2n-1 coefficients a_(2n-1),...,a_1 are unspecified real numbers to begin with. The degree is set at any even integer \geq 4, and the players, starting with A, take turns specifying an undetermined coefficient until f(x) is completely defined. At the end, A wins if no root of f(x) = 0 is real, and B otherwise. Determine a winning strategy for B.

Note that at the beginning, there are n unidentified coefficients with odd subscripts, and n-1 unindentified coefficients with even subscripts. Since B plays after A, he is able to preserve this difference, until there are exactly 3 coefficients left to be determined : if A determine a_(2i) (resp. a_(2i+1) ) for some i, then B determine a_(2j+1) for some j (resp a_(2j) ), no matter what values are chosen at each step.

Thus, there are 3 terms left to be determined, one with even subscript and two with odd subscripts. When A makes its choice, it remains two, say a = a_p and b = a_q, for some p,q.

Let f(x) = P(x) + a*x p  + b*x q , where P is already determined.
There are two cases :

- If p is odd and q is even :
We have f(1) + f(-1) = P(1) + P(-1) + 2a, which is independant of b. Thus, B forces a win by choosing a = -(P(1)+P(-1))/2, since it leaves f(1) + f(-1) = 0 and by the intermediate value theorem, f must have a real root in [-1,1].

- If p and q are odd : With no loss of generality, we may assume that p < q. Then, we have :
f(2) + 2 p *f(-1) = P(2) + 2 p *P(-1) +a*(2 p - 2 q ).
Thus, B choose a so that f(2) + 2 p *f(-1) = 0, and forces a win as above.

Pierre.


floor function
number theory proposed
number theory

[quote="mohammad reza"]a,b,c,n,m are positive integers and gcd(m,n)=1
a^n+b^n=c^m[/quote]

You at least have the classical solution :

Let $ p,q$ such that $ pm\minus{}qn\equal{}1$, then $ (u(u^n\plus{}v^n)^q)^n\plus{}(v(u^n\plus{}v^n)^q)^n\equal{}((u^n\plus{}v^n)^p)^m$

at first thanks patrick for your solution
actually i want you show that this equation has infinite answers,as patrick showed
$ let(a,b,c)\equal{}((x^{(p\minus{}1)!\plus{}1}\plus{}xy^{(p\minus{}1)!})^\frac{(p\minus{}1)!}{p_1},(y^{(p\minus{}1)!\plus{}1}\plus{}yx^{(p\minus{}1)!})^\frac{(p\minus{}1)!}{p_2},(x^{(p\minus{}1)!}\plus{}y^{(p\minus{}1)!})^{1\plus{}2\left\lfloor \frac{(p\minus{}1)!\plus{}1}{2p}\right\rfloor})$
and$ x,y$are natural numbers
so it has infinite answers

[quote="pco"][quote="mohammad reza"]at first thanks patrick for your solution
actually i want you show that this equation has infinite answers,as patrick showed
$ let(a,b,c) \equal{} ((x^{(p \minus{} 1)! \plus{} 1} \plus{} xy^{(p \minus{} 1)!})^\frac {(p \minus{} 1)!}{p_1},(y^{(p \minus{} 1)! \plus{} 1} \plus{} yx^{(p \minus{} 1)!})^\frac {(p \minus{} 1)!}{p_2},(x^{(p \minus{} 1)!} \plus{} y^{(p \minus{} 1)!})^{1 \plus{} 2\left\lfloor \frac {(p \minus{} 1)! \plus{} 1}{2p}\right\rfloor})$
and$ x,y$are natural numbers
so it has infinite answers[/quote]

huh ?  Where are $ m$ and $ n$ ? What are $ p,p_1,p_2$ ? And since $ p_1$ and $ p_2$ do not appear in $ c$, The equality is surely wrong.[/quote]
we have this:$ (x^{n\plus{}1}\plus{}xy^n)^n\plus{}(y^{n\plus{}1}\plus{}yx^n)^n\equal{}(x^n\plus{}y^n)^{n\plus{}1}$as you had written before
let $ n\equal{}(p\minus{}1)!$ and $ p_1,p_2,p$ are primes such that $ p\neq p_1\neq p_2$ now with this lemma:
$ lemma: (p\minus{}1)!\plus{}1\equal{}p(1\plus{}2\left\lfloor \frac{(p\minus{}1)!\plus{}1}{2p}\right\rfloor)$we have:
$ a^{p_1}\plus{}b^{p_2}\equal{}c^p$
my solution looklikes yours,but i just want to use that beautiful lemma
sorry for my poor English :blush:

[quote="mohammad reza"]$ a^{p_1} \plus{} b^{p_2} \equal{} c^p$[/quote]

1) In your initial problem, powers of $ a$ and $ b$ were the same $ n$.

2) I thought $ m,n$ were given numbers and we had to find $ a,b,c$.

If we can choose $ m,n$ freely with the only constraint that $ gcd(m,n)\equal{}1$, just take $ n\equal{}1$ and any $ m$ and : $ u^1\plus{}(w^m\minus{}u)^1\equal{}w^m$

[quote="pco"][quote="mohammad reza"]$ a^{p_1} \plus{} b^{p_2} \equal{} c^p$[/quote]

1) In your initial problem, powers of $ a$ and $ b$ were the same $ n$.

2) I thought $ m,n$ were given numbers and we had to find $ a,b,c$.

If we can choose $ m,n$ freely with the only constraint that $ gcd(m,n) \equal{} 1$, just take $ n \equal{} 1$ and any $ m$ and : $ u^1 \plus{} (w^m \minus{} u)^1 \equal{} w^m$[/quote]
I'm sorry for answering to your questions late.
first of all i had to say:I made a big mistake,i read the source of problem and realize that the problem is:
$ a^{p_1}\plus{}b^{p_2}\equal{}c^p$and that question has another solution,as pco wrote
but about your first question:powers of $ a,b$ are the same $ n\equal{}(p\minus{}1)!$
$ a^{p_1}\equal{}(x^{(p\minus{}1)!\plus{}1}\plus{}xy^{(p\minus{}1)!})^{(p\minus{}1)!}$
$ b^{p_2}\equal{}(y^{(p\minus{}1)!\plus{}1}\plus{}yx^{(p\minus{}1)!})^{(p\minus{}1)!}$
so $ a^{p_1}\plus{}b^{p_2}\equal{}z^{(p\minus{}1)!}\plus{}w^{(p\minus{}1)!}\equal{}c^p\equal{}d^{(p\minus{}1)!\plus{}1}$


calculus
integration
trigonometry
calculus computations

For a positive integer $ m$, evaluate $ \int_0^{\frac {\pi}{2}} \cos ^ m x\ \sin 2mx\ dx$.

If $ m$ is odd:
$ I_m\equal{}\dfrac{1}{2^m}\sum_{k\equal{}0}^{m}\dfrac{(_k^m)}{3m\minus{}2k}$.


[quote="manos"]ασκηση 1)να δειχθει οτι $0.999....=1$  (με τη βοηθεια σειρων)

ασκηση 2)να εξετασετε ως προς τη συγκλιση τη σειρα
$\frac{1}{2}+\frac{1}{3}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{2^{n}}+\frac{1}{3^{n}}+...$ :!:[/quote]

Δεν υπάρχει λόγος να ξεφύγεις από την ανάλυση
υπάρχουν άτομα που ενδιαφέρονται για τέτοιες ασκήσεις ...  :roll:

και εγώ, ας ασχοληθούμε και με κάτι μοντέρνο! (και ενδιαφέρον κατά τη γνώμη μου όμως)

Μια πιό γενική μορφή της πρώτης είναι εδώ

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109795[/url] .

Για την δεύτερη αφού υπάρχουν οι $\sum_{k=1}^{ \infty}\frac{1}{2^{k}}= 1$ και $\sum_{k=1}^{ \infty}\frac{1}{3^{k}}= \frac{1}{2}$

έχουμε $\sum_{k=1}^{ \infty}\left( \frac{1}{2^{k}}+\frac{1}{3^{k}}\right) = 1+\frac{1}{2}=\frac{3}{2}$ .


Find a good value for $K >0$ such that for all $x_1,..,x_n$ positive reals we have

$\frac{x_1}{x_2+x_3}+\frac{x_2}{x_3+x_4}+..+\frac{x_n}{x_1+x_2} >Kn$

I have found for K a value , but I cannot if it can be improved. ;)

Let (x_0,y_0)   be the intersection point of the y-axe and the common tangent of  y=e^-x  and y=2/(e^x+e^(x/2)).it has been found that K=y_0/2   satisfies the condition for all n ,though
this is not the best bound. Also note that  y_0~0,989


calculus
conics
trigonometry
function
analytic geometry
AoPS Books

[quote="007math"]In my state, we have an math contest where you require knowledge of mathematics until calculus, but I am currently taking Algebra II.  What is the best thing to do to learn precalc? (preferably free, but other ideas are accepted (i know about AoPS books)).

I have AoPS V1 and V2, but I haven't gone through them.  If someone could tell me what chapters one would learn in precalc, that would also be great.

All replies are welcome, and thanks in advance![/quote]
I've found the chapters in AoPS V2 to be quite helpful for pre-calc. The chapters on limits, conics, matrices&vectors, and complex numbers.

You might also want to consider investing some time into learning trigonometry. Chapter 2 in volume two and certain sections of 103 Trig will do the job quite nicely.

Exactly what is the curriculum for Pre-Calc?

[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]This page[/url] should help with that question.

I wouldn't recommend following that chart. It leaves out all of the "Algebra 3" material like limits, conics, and some function stuff. It is also meant to have an instructor teaching it, which you do not have. If you did have a personal tutor, then by all means, give him that list and the algebra 3 list and you have yourself a class.

If you don't have a personal tutor (I am guessing this is you) do what ckck and mathemagician said to do with AoPS 2. Their combined list of topics + permutations/combinations/probability (if you haven't covered that already) covers what you need to know and more for almost all competitions.

Sorry for reviving, but I made the topic and just wanted some more info.

Sometimes I have a hard time understanding V1 and V2 of the AoPS books, i don't know why but it confuses me.  Is there any online site where I may be able to learn pre-calc?

Also, is trig considered pre-calc?  I know graphing trig equations and the unit circle is, but things like $ \sin{\alpha}\cos{\beta} \plus{} \sin{\beta}\cos{\alpha} \equal{} \sin{(\alpha \plus{} \beta)}$ and other common things like this included in it?  I mainly want to learn the trig stuff, and figured that pre-calc covered it so why not learn that.

Anyway, back to the topic.  Is there any online site or resource that is free which I can use to learn pre-calc?  Thanks in advance. (don't say aops, i kinda already know about it)

For the Pre cal class I took (not the AoPS one) Trig formulas and Polar form of coordinates were discussed.

For online sources of Pre Cal. Google is your best friend.


Let $ F$ be the midpoint of the side $ BC$ of a triangle $ ABC$. Isosceles right-angled triangles $ ABD$ and $ ACE$ are constructed externally on $ AB$ and $ AC$ with the right angles at $ D$ and $ E$. Prove that the triangle $ DEF$ is right-angled and isosceles.

Let $ D$ and $ E$ be the centers of the squares $ ABXY$ and $ ACPQ.$ Then $ \triangle ABQ$ and $ \triangle ACY$ are congruent by SAS criterion, due to $ AB \equal{} AY,$ $ AC \equal{} AQ$ and $ \angle BAQ \equal{} \angle YAC.$ Therefore, $ BQ \equal{} YC$ and $ \angle ACY \equal{} \angle AQB.$ If $ M \equiv{} YC \cap BQ,$ then quadrilateral $ AMCQ$ is cyclic $\Longrightarrow$ $\angle QMC$ is right. Lines $ FD$ and $ FE$ are respectively parallel to $ CY$ and $ BQ$ since $ D , E$ are midpoints of $ BY,CQ.$ Hence $FD \perp FE$ $\Longrightarrow$ $\triangle DEF$ is  isosceles right at $F.$

Let M and N be the midpoints of AB and AC. Then $ DM\equal{}NF\equal{}AB/2$ and $ EN\equal{}MF\equal{}AC/2$, and $ \angle DF\equal{}\angle DMB\plus{} \angle BFM\equal{}90\plus{}\angle A\equal{}\angle ENF$, so $ \triangle DMF$ is congruent to $ \triangle ENF$, so $ DF\equal{}FE$. Also we know $ \angle MFN\equal{}\angle A$ and $ \angle MFD\plus{} \angle NFE\equal{}\angle MFD\plus{} \angle MDF\equal{}180\minus{} \angle DMF\equal{}180\minus{}(90\plus{} \angle A)\equal{}90\minus{} \angle A$, so $ \angle DFE\equal{}90$.

We have $ \mathcal {R}^{\minus{}90^{\circ}}_{E}: C\mapsto A$, $ \mathcal {R}^{\minus{}90^{\circ}}_{D}: A\mapsto B$. Therefore $ \mathcal {R}^{\minus{}90^{\circ}}_{D}\circ \mathcal {R}^{\minus{}90^{\circ}}_{E}:$ $ C\mapsto B$. But $ \mathcal {R}^{\minus{}90^{\circ}}_{D}\circ \mathcal {R}^{\minus{}90^{\circ}}_{E}\equal{}\mathcal {R}^{\minus{}180^{\circ}}_{M}$. Therefore $ M\equiv F$ which is the midpoint of $ BC$. but $ \angle DFE\equal{}\angle DEF\equal{}\frac {90^{\circ}}{2}\equal{}45^{\circ}$. Hence $ \triangle DEF$ is right-angled and isosceles $ \square$.


To promote her school's annual Kite Olympics, Genevieve
makes a small kite and a large kite for a bulletin board display.
The kites look like the one in the diagram.

[asy]for ( int x = 0; x <= 6; ++x )
{
    for ( int y = 0; y <= 7; ++y )
    {
        dot((x,y));
    }
}
draw((0,5)--(3,7)--(6,5)--(3,0)--cycle);[/asy]

For her small kite, Genevieve draws the kite on a one-inch grid.  For the large kite
she triples both the height and width of the entire grid.
\\ \\
Genevieve puts bracing on her large kite in the
form of a cross connecting opposite corners of the kite.  How many
inches of bracing material does she need?

How did you get 117? As far as I know you just need to add: $21+18=\boxed{39}$

[quote]Since it's a $ 1$ inch grid, the braces in the diagram are $ 21$ and $ 18$. We must multiply by $ 3$ to get $ 117$.[/quote]

Actually, you already multiplied 7 and 6 by 3 to get 21 and 18. All you have to do is add them together to get 39. You don't have to multiply by 3 again.


If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then $b$, when written in base 10, equals
\[ \text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16 \]

[quote="dangvh"][quote="AntonioMainenti"][hide="hint"]\[5b^2+5b+4=(2b+4)^2\][/hide][/quote]
[hide] Umm i used Antonio's hint and solved for b... you get b=12 adn b=-1... i dont
think that -1 is possible, so is 12 (choice C) the correct answer?[/hide][/quote]

[hide="Dangvh"]
I think that Antonio's statement is correct in principle. However, when you solve for b, you need to be careful that you are not doing it in base 10. (I think that's the problem...)[/hide]

[hide="Hint"]
For this problem, I used $554 \equiv 24^{2} \equiv 4 \pmod {b}$. And of the answer choices, only b=12 is reasonable. [/hide]

[hide="Another Approach"]
Also, since this is a MC question (this isn't an elegant technique, but it does work), we know that b cannot be less than 10 because $24^2=576$. And a quick estimation shows that an answer of $b=12$ seems reasonable. If you're so inclined, you can substitue 12 for b and demonstrate that it works. But this quick estimation can save you some time (hopefully) if you need it. [/hide]

[size=59]I edited my LaTeX.[/size]

if the base b representation of the number is 24 then the base ten is obviously 4+2b.
then 554 is 5b2+5b+4 in base ten, then all you need to do is solve for b in this equation:
(b^2)-11b-12.
the number you get will be in base 10.
b=12

Obviously it has to be higher than base 10. So there's only two choices - test 'em both.

[hide]$24_b=2b+4$
$(2b+4)^2=4b^2+16b+16=5b^2+5b+4$
$b^2-11b-12=(b-12)(b+1)$
$b=12$

So, the answer is $C.$[/hide]

[b]Note[/b] - all numbers are in base 10 unless specified otherwise, including $b.$

[quote="AntonioMainenti"][hide="hint"]\[5b^2+5b+4=(2b+4)^2\][/hide][/quote]

that pretty much gave it away.

Oh, that was the hint? I didn't even look. :P


WHAT IS STRUCTURE OF     XeO2F2

WHY ARE F on the axial bonds

why not O

My guess:
$ Xe\minus{}O_2$ bond, being a double bond has smaller bond length than a $ Xe\minus{}Fe$ bond. We know that axial bonds are longer than equitorial bonds.Thus, it follws.

Q:   Why are axial bonds longer than equitorial bonds???

Q:   what is bents rule???

Q:   what is dragos rule???

Q:   if Xe-O bond is axial then the bond length will be shorter and so the molecule will be more stable?????????????
so why should axial bond be with F????

plz someone help yaar plz........ :o  :maybe:   :|

I think this is the wrong forum for Chemistry...


greatest common divisor
number theory
least common multiple

Hehe, got it :)
[hide]The other number is 18[/hide]

[hide]i got 18 by checking the factors of 36 :P  :lol:[/hide]

[hide]
36 doesn't work because then the largest common factor would be 12, so the answer is 18[/hide]

[hide][quote="math92"]I am thinking of two numbers. Their greatest common factor is 6. Their least common multiple is 36. One of the numbers is 12. What is the other number?[/quote]

Guess and check:

12 & 12: GCF: 12, (Nooo........)

12 and 18: GCF = 6 (Yay!!!), LCM = 36(WoooHoo!!!!)

18 is the correct answer.[/hide]


function
inequalities
real analysis
real analysis solved

[quote="TeeNKiT"] How to prove this: if f is continuous on(a,b) and f( (x+y)/2 )<=1/2 f(x) +1/2 f(y) for all x,y in(a,b) then f is convex?[/quote]
[edit] why does my post come out wrong? it shouldn't show the [quote=....etc.
[color=green]Because you checked the "Disable BBCode" option when posting, effectively prohibiting all markup ;) Fixed by a moderator :P[/color]

[quote="blahblahblah"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30871&sid=7313b52afb1e139290bebc61551cc7d5]See here[/url][/quote]
Isn't this jensens inequality?


quadratics
vector
geometry
geometric transformation
reflection
superior algebra
superior algebra unsolved

$ V$ is a finite dimentional vector space, and there is a quadratic form $ q$ defined on $ V$.
define $ q(v,w)$ by $ 2q(v,w): \equal{}q(v\plus{}w)\minus{}q(v)\minus{}q(w)$.

for a fixed $ v \in V$ ($ q(v) \not\equal{}0$), there is a linear mapping $ f_v$ of $ V \to V$:

$ f_v(w)\equal{}w\minus{}\frac {2q(v,w)} {q(v)} v$

1.how to show $ f_v$ is surjective and injective?

2.how to show $ f_v$ preserves $ q$, that is , $ q(f_v(w))\equal{}q(w)$ for all $ w \in V$?

I got it, it is just a reflection across the hyperplane perpendicular to $ v$.


Vestavia Hills High School Mathematics Tournament 2005, Geometery division
8.[i]TRON[/i] is a regular tetrahedron with sides equal to 7. Point [i]L[/i] is the midpoint of linesegment [i]TR[/i], and point [i]I[/i] is the midpoint of linesegment [i]ON[/i]. What is the length of linesegment [i]LI[/i]

A.$\frac{49\sqrt{2}}{2}$
B.9
C.7
D.$\frac{7\sqrt{2}}{2}$
E. None of the Above(NOTA)

since every triangel is equal , from $OTN$ we have $OI=IN$ so $TI=RI$ and at $TIR$ we have $TL=LR$ so we conclude that $LI$ perpendicular to $TR$. By using phytagorean identity we get $LI = \frac{7 \sqrt{2}}{2}$

I have a question from that Test its number 19. You need to download it.
Please include the solution I have the answer below:
 :ninja:

[hide]The Answer is B[/hide]

i HAVE the test lol i took it =)oooo thats the liney 1 i got it right!!!!!!(1 of the few) i'll give u a hint i [hide]The # of ways to get to a vertice is the sum of the number of ways to get to the vertices that can travel to it.tell me if you don't understand[/hide]

:huh: :huh:Soooooooooooooooooooooooooooooooooooooooooo, How do you find the sum of the ways you get to the verticies??

[hide]all the veritices right of Point A are $1$, All the vertices under point A are $1$, so then the vertice 1 right and $1$ down of point A is $2$, 1right 2 down is $3$,1 right 3 down is $4$, 1right 4down is $5$ and so on so forth if you keep following that pattern you'll get the right answer  :lol:[/hide]

The attached picture would help you visualize what NaoNao meant ;)

yes thats what i mean.. i just a newb at this and can't draw  :mad:

[hide]Thanks, I missed this problem at the competition, even though i had done exactly what you said but i counted one vertex when i should not have. In my picture, I forgot to include the vertex when i did my picture, but you can see it on the real test it is in the big 2x2 square, so I counted 64 ways instead of 63, but now i realize that vertex would not have made a difference because you can only move right or down. Thanks anyway!!!!!!! [/hide][/hide]

lol i need help w/ number 6

6. Three sides of a Quadrilateral circumscribed about a circle are 9,15,and 8. What is the smallest possible value for the fourth side.
A.1
B.2
C.4
D.14
E. NOTA

for a circumscribed quad, we have the sums of the opposite sides are equal, so the smallest is $9+8-15=2$ so $\boxed{B}$


[quote="archimedes1"]How many 12-digit bit strings exist such that no two zeroes are consecutive?[/quote]

[hide]Case 1: No zeros
$ 1$ way.

Case 2: 1 zero.
$ 12$ ways, depending on where you put the zero.

Case 3: 2 zeros
This is a balls and urns problem. The number is going to look like this:
some 1s|0| some 1s (at least 1)|0|some 1s.

Mathematically,
y|0|1+x|0|z,

where $ x+y+z = 9$. So there are $ \binom{9+2}{2}= \binom{11}{2}$ solutions.

Case 3: 3 zeros
Analogously to the last case,
w|0|1+y|0|1+x|0|z,

where $ x+y+z+w = 7$. So there are $ \binom{7+3}{3}= \binom{10}{3}$ solutions.

We start noticing the pattern, so there are:
$ \binom{13}{0}+\binom{12}{1}+\binom{11}{2}+\binom{10}{3}+\cdots+\binom{7}{6}=\boxed{377\textrm{ bits}}$[/hide]

EDIT: Thanks for pointing that out, Lazarus
EDIT2: Argh, I can't add...

There can't possibly be more than six zeros in a bit string that satisfies the conditions given.

[quote="da ban man"][quote="vishalarul"]
We start noticing the pattern, so there are:
$ \binom{12}{0}+\binom{11}{1}+\binom{10}{2}+\binom{9}{3}+\cdots+\binom{6}{6}=\boxed{233\textrm{ bits}}$[/quote]
vishalarul, Im  pretty sure tahts still wrong..
as your pattern is wrong..it starts at13C0 not 12C0 (although they give same answer your rest of the pattern will be off)

so its $ 377$..

(check out your 1 zero case, thats 12 not 11 and threw you off)[/quote]

Oops. I thought that the first number cannot be $ 0$ (I was thinking 12 digit number.)

There is a much nicer, more rigorous solution.

[hide=hint]Recursion[/hint]
[hide=bigger hint]Fibonacci[/hint]


[b]Ελκυστικό πρόβλημα[/b]

[color=red][b]ΑΣΚΗΣΗ 100[/b][/color]

Έστω  $ a,b,c$  θετικοί αριθμοί με $ abc \equal{} 1$. Να αποδείξετε ότι :

$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 7\geq 5(a \plus{} b \plus{} c)$

 Μπάμπης

Αρκετά ζόρικη και σφιχτή .Να η λύση μου

Καταρχήν $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1$

Επομένως θνδο $ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} 6\geq 5(a \plus{} b \plus{} c)$ ή

$ ab \plus{} bc \plus{} ca \plus{} \frac {6}{a \plus{} b \plus{} c}\geq 5$ . Θέτουμε $ a \equal{} x^2$ κτλ και έστω ότι ο x είναι ο μεγαλύτερος δλδ $ x\geq 1$

Tote

$ f(x,y,z) \minus{} f(x,\sqrt {yz},\sqrt {yz}) \equal{} (y \minus{} z)^2(x^2 \minus{} \frac {6}{(x^2 \plus{} y^2 \plus{} z^2)(x^2 \plus{} 2yz)}\geq \\  (y \minus{} z)^2(x^2 \minus{} \frac {2}{x^2 \plus{} \frac {1}{x}})$

to οποίο ισούται με $ (y \minus{} z)^2\frac {x^4 \plus{} 2x \minus{} 2}{x^3 \plus{} 2}\geq 0$

Επομένως αρκεί νδο $ f(x,\sqrt {yz},\sqrt {yz}) \equal{} f(x,\sqrt {\frac {1}{x}},\sqrt {\frac {1}{x}})\geq 5$

to οποίο είναι ισοδύναμο με
$ \frac {(x \minus{} 1)^2(2x^4 \minus{} x^3 \minus{} 4x^2 \plus{} 4x \plus{} 2)}{x^3}\geq 0$  (1)

Όμως $ \frac {(x \minus{} 1)^2(2x^4 \minus{} x^3 \minus{} 4x^2 \plus{} 4x \plus{} 2)}{x^3}\geq\frac {(x \minus{} 1)^2(2x^4 \minus{} x^3 \minus{} 4x^2 \plus{} 3x)}{x^3} \equal{} \frac {(x \minus{} 1)^4(2x \plus{} 3)}{x^2}\geq 0$

so qed  :D

Όντως δύσκολη. :(

Εγώ δοκίμασα εδώ να θέσω $ a\plus{}b\equal{}x$ κτλ. οπότε μετά από πράξεις ανάγεται στην:

Αν σε τρίγωνο $ xyz$ ισχύει $ t\equal{}1/r^2$ όπου $ t$ η ημιπερίμετρος και $ r$ η ακτίνα του εγγεγραμμένου τότε:

$ 4Rr\plus{}7r^2>\equal{}5$

Έχω δοκιμάσει όλες τις τριγωνομετρικές σχέσεις και όλες το μικραίνουν υπερβολικά. Μπορεί κανείς να βοηθήσει??? :maybe:


limit
trigonometry
real analysis
real analysis unsolved

Find the limit:
$ {\lim_{n \rightarrow {\infty}}}  \sum_{k=1}^{n} sin {\frac{k \pi}{n^{2}}}
$

Since
\[ \dfrac{\sin\dfrac{ k\pi}{n^2}}{\dfrac{k\pi}{n^2}}\xrightarrow[n\to+\infty]{}1\]

uniformly with respect to k. Then\[\lim_{n\to+\infty}\sum_{k=1}^n\sin\frac{k\pi}{n^2}= \lim_{n\to+\infty}\sum_{k=1}^n\frac{k\pi}{n^2}=\frac{\pi}2 \]

Theorem:
Let $a_{k,n},\;b_{k,n},\;n\in\mathbb N,\;k=1,2,...,n$  two sequences of real numbers with $b_{k,n}\ne0$. Suppose that $\dfrac{a_{k,n}}{b_{k,n}}\xrightarrow[n\to+\infty]{}1$ uniformely with respect to k. Then if $\lim_{n\to+\infty}\sum_{k=1}^nb_{k,n}$ exists :
\[
(*)\qquad\lim_{n\to+\infty}\sum_{k=1}^na_{k,n}=\lim_{n\to+\infty}\sum_{k=1}^nb_{k,n}
\]
Proof
By assumptions, $\dfrac{a_{k,n}}{b_{k,n}}=1+\epsilon_{k,n}$, where $\epsilon_{k,n}\to 0$, uniformly with respect to k. Thus
\[
\sum_{k=1}^na_{k,n}=\sum_{k=1}^nb_{k,n}+\sum_{k=1}^n\epsilon_{k,n}b_{k,n}
\]
Since $ \displaystyle\lim_{n\to+\infty}\sum_{k=1}^nb_{k,n}$ exists, there is an M > 0 such that $\left| \sum_{k=1}^nb_{k,n}\right|\le\txt{M},\;n\in\mathbb N$.
Morever, for any $\epsilon >0,\;|\epsilon_{k,n}|<\frac {\epsilon}{\text{M}}, $ for k=1,2,...,n, provided n is suffciently large. Hence $ \left|\sum_{k=1}^n\epsilon_{k,n}b_{k,n}\right|<\epsilon$, which means that  $\lim_{n\to+\infty}\sum_{k=1}^n\epsilon_{k,n}b_{k,n}=0$. Thus, by (*)
\[
\lim_{n\to+\infty}\sum_{k=1}^na_{k,n}=\lim_{n\to+\infty}\sum_{k=1}^nb_{k,n}
\]


given 4 non parallel lines in a plane prove that there is one and only one point from which the feet of the perpendiculars to the lines are collinear.

Dear Mathlinkers,
do you think about the Miquel's point ?
Sincerely
Jean-Louis


given the polynomial $a_0 +a_1x +....a_n x^n$.
It is known that $a_0 +\frac {a_1} {2^2} +\frac {a_2}{3^2} + \cdots +\frac {a_n} {(n+1)^2}=0.$
Prove there exist root from $(0,1)$ of this polynomial.

Edited by pbornsztein to make it readable.

let's call $p$ such polynomial, and define $ q = \sum_{i=0}^{n} {a_i \over {(i+1)^2}}x^{i+1}$.
the given condition is $ q(1) = 0$, and obviously holds $q(0) = 0$.
then, for Rolle theorem, there exist a point $y \in (0,1)$ such that $ 0 =q'(y) = \sum_{i=0}^{n} {a_i \over {i+1}}x^{i}$.
now we have that $ p(x) = (xq'(x))'$, and $0q'(0) = yq'(y) = 0$, so there exist a point $\alpha \in (0,y) \subset (0,1)$ such that $p(\alpha) = 0$

Let $P(x) = a_0 + a_1x + \cdots + a_nx^n.$

Suppose for a contradiction, that $P(x) \neq 0$ for all $x \in (0;1).$
Then $P(x)$ has a constant sign on $(0;1)$, say $P(x) > 0$ for $x \in (0,1)$ (otherwise let's consider $-P$). It follows that $Q(x) = a_0x + \frac {a_1} 2 x^2 + \cdots + \frac {a_n} {n+1} x^{n+1}$ is strictly increasing on $[0;1]$.
Since $Q(x) = 0$, we deduce that $Q(x) > 0$ for all $x \in (0;1).$
Thus $\frac {Q(x) } x = a_0 + \frac {a_1} 2 x + \cdots + \frac {a_n} {n+1} x^n >0$ for all $x \in [0;1]$.
Therefore $R(x) = a_0x + \frac {a_1} {2^2} x^2 + \cdots + \frac {a_n} {(n+1)^2} x^{n+1}$ is strictly increasing on $[0;1].$
Since $R(0)=0$ it follows that $R(1) > 0$ contradicting the hypothesis.

Pierre.

10x a lot.
this is a problem from bulgarian test for students from the university(1,2 year) for the olympiad.


probability
Support
limit
expected value
probability and stats

1. The numbers from 1 to 100 are arranged in a 10×10 table so that no two adjacent numbers have sum less than S. Find the smallest value of S for which this is possible.

2. A box contains ‘p’ white balls and ‘q’ black balls. Alongside is a pile of black balls. Two balls are taken out at random from the box.If the balls are different, then the white ball is replaced and the black ball is removed from the box. If the balls are similar, then both balls are removed and replaced with a black ball from the pile. The process is continued till only one ball is left. Find the probability that the last ball left is white.

3. Let n, k, p be positive integers such that k ≥ 2 and k (p+1) ≤ n. Determine the number of ways to color n labeled points on a circle in blue or red, so that exactly k points are colored blue, and any arc whose endpoints are blue but contains no blue points in its Interior contains at least p red points.

4.  The positive integers from 1 to n2 are placed arbitrarily on squares of an n x n chessboard. Prove that there exist two adjacent squares (having a common vertex or a common side) such that the difference of the-numbers placed on them is not less than n + 1.

5. Does there exist a natural number N which is a power of 2 whose digits (in the decimal representation) can be permuted to form a different power of 2? Support your answer with a proof.

6. A box contains 4 white balls and 4 red balls, which we draw from the box in some order without replacement. Before each draw, we guess the color of the ball being drawn, always guessing the color more likely to occur (if one is more likely than the other). What is the expected number of correct guesses?

for number one i can only guess, but i think 109?

and for 2 we can get
 $ P[A_{n \minus{} 1} \equal{} (p \minus{} 2,q): A_n \equal{} (p,q)] \equal{} \frac {p(p\minus{}1)}{n(n\minus{}1)}$

extremely ugly tree plus brute force gives me $ \frac{383}{70}$ for problem six

To the OP:
1) Choose less bad titles for your threads.
2) It's better to put only one problem per thread (or perhaps a small number of problems, if they are closely related) -- doing a problem dump like this makes it much harder to discuss any given problem.

Thinking about a closed form for number 6: if we replace 4 with 1, 2, 3, we get $ \frac {3}{2}, \frac {17}{6}, \frac {41}{10}$, respectively.  This pattern means basically nothing to me.  But, here's a sketch of how to generate it: let $ e_n$ be the number we're after.  We have a total of $ \binom{2n}{n}$ sequences.  Of these, $ 2 C_{k \minus{} 1} \binom{2(n \minus{} k)}{n \minus{} k}$ have the property that the first time we've drawn an equal number of white and black marbles is on the $ 2k$th draw.  The expected value from such a sequence is $ \frac {1}{2} \plus{} k \plus{} e_{n \minus{} k}$.  So we get something like $ e_n \equal{} \sum_{k \equal{} 1}^n 2 C_{k \minus{} 1} \binom{2(n \minus{} k)}{n \minus{} k} (\frac {1}{2} \plus{} k \plus{} e_{n \minus{} k})$, with $ e_0 \equal{} 0$.  This suggests you've made an arithmetic (or typographical) error somewhere: it gives $ \frac {373}{70}$ as the answer to the given problem.

Follow-up question: what's the limit $ \lim_{n \to \infty} \frac{e_n}{n}$?  In other words, how does this strategy compare to random guessing for large $ n$?

Excuse me [b]msg_1989[/b], I was checking your problems and would like to show them to some of my friends, so could you tell me about the actual source of the problems?? (In particular I'd like to know that of problem 2) :) Thank you!

By the way [b]kenn4000[/b], I think your solution is unclear - what is $ n$? I realized that when you take out 2 black balls, 2 white balls, and 1 black ball and 1 white ball from the box, the number $ (p,q)$ of balls of each type changes to $ (p,q\minus{}1)$, $ (p\minus{}2, q\plus{}1)$ and $ (p,q\minus{}1)$ respectively. Hence the parity of $ p$ remains unchanged, which means that if $ p$ is even it will eventually decrease to 0, and if it's odd it will eventually become 1. So the required probability is 0 if $ p$ is even and 1 if $ p$ is odd. Isn't this approach correct? :)


Find all $ f: R\rightarrow R$ such that
$ i) f(1)\equal{}1$
$ ii) f(f(x)y\plus{}\frac{x}{y})\equal{}xyf(x^2\plus{}y^2)$ $ \forall x \in R, y \in R$ \ $ \{0\}$

you need to play around giving values to x and y. It is similar with the problem: find all function $ f(x)$ that satisfy $ f(a\plus{}b)\equal{}f(a)\plus{}f(b)$.
In you case if $ f(x)y\plus{}\frac{x}{y}\equal{}1$ this will imply that $ f(x)\equal{}\frac{1}{y}\minus{}\frac{x}{y^{2}}$, but also implies that $ xyf(x^2 \plus{} y^2)\equal{}1$ i.e. $ f(x^2 \plus{} y^2)\equal{}\frac{1}{xy}$. Use the formula above for $ f(x)$ and you get that
$ f(x^2 \plus{} y^2)\equal{}\frac{1}{y}\minus{}\frac{x^2\plus{} y^2}{y^{2}}\equal{}\frac{1}{y}\minus{}\frac{x^2}{y^2}\minus{}1\equal{}f(x^2)\minus{}1$. Now if you make $ x\equal{}1$ you get $ f(y^2 \plus{}1)\equal{}0$, and from here you can figure out...at least try to see if you can do it.
Please check my computations because i usually do mistakes...
And do not forget to look for the example i told you...it will give you an idea of what you need to do next.

mikiduta

It's hard to put this nicely: your post is riddled with errors.

First: $ f(x)y \plus{} \frac{x}{y} \equal{} 1$ may not have a solution. In fact, for $ f(x) \equal{} 1/x$ which satisfies the original equation it doesn't have a solution. Second: you try to turn $ x$ into $ x^2 \plus{} y^2$ in $ f(x)\equal{}\frac{1}{y}\minus{}\frac{x}{y^{2}}$, but that hinges on $ f(x)y \plus{} \frac{x}{y} \equal{} 1$; if you change $ x$ that may not hold anymore. Third: Similar mistake to the fisrt: you try to set $ x \equal{} 1$, but you don't have the freedom to do that because of the $ f(x)y \plus{} \frac{x}{y} \equal{} 1$ constraint.

On a more relevant note, this has been posted many, many times. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231718]here[/url] for example.


calculus
function
limit
integration
trigonometry
real analysis
real analysis unsolved

Let $ f$ be e function increasing, continuous, and $ f(0)=0$. We know that
\[ \lim_{x\rightarrow 0} \int^x_0f(t)\frac{t}{\sqrt{x^2-t^2^}}=1\]
Does  the limit $ \lim_{x\rightarrow 0}\frac{f(x)}{x}$ always exist?

This doesn't make any sense, since for any function $ f$ that is bounded in a neighborhood of $ 0$ we have
\[ \lim_{x \to 0} \int_0^x f(t) \frac{t}{\sqrt{x^2\minus{}t^2}} dt \equal{} 0,\]
which can be seen using the standard substitution $ t \equal{} x \sin s$. (I hope I didn't do any miscalculation.)
Maybe you made a mistake in the statement of your problem?

Sorry, I really had a mistake, the statement is:
\[ \lim_{x\rightarrow 0}\frac{1}{x^2}\int^x_0\frac {tf(t)}{\sqrt {x^2 - t^2^}} = 1\]

Making solyaris's substitution, I get that
\[ \lim_{x\to 0}\int_0^{\frac {\pi}2}\frac {f(x\sin s)}x\,\sin s\,ds \equal{} 1.\]
Starting with a partial converse: suppose $ f'(0) \equal{} \lim_{x\to0}\frac {f(x)}x$ does exist. Then, interchanging limit and integral (justified by uniform convergence), we would have
\[ \int_0^{\frac {\pi}2}f'(0)\sin^2s\,ds \equal{} 1\]
from which we see that we must have $ f'(0) \equal{} \frac {2}{\pi}.$

So if the limit exists, we know what it must be - but must that limit exist? I'm not there yet - it may require arguing in a different direction.

(Also, I'm moving this topic out of the "computations and tutorials" section; there's too much going on for it to belong there.)


function
real analysis
real analysis unsolved

Let $ f$ be funtion differentiable at $ [0,1]$ and satisfy conditions
\[ f(0)\equal{}0\ \text{and}\ |f'(x)|\leq K|f(x)|,\ K\in R^\plus{}\]
Prove that $ f(x)\equiv 0$.

A possible proof employs Lagrange. $ f(x) \minus{} f(0) \equal{} f'(c_1)x$ thus
$ \vert f(x)\vert \le x K \vert f(c_1)\vert,$ $ 0 < c_1 < x.$ Doing
the second step we obtain $ \vert f(x)\vert \le x \cdot c_1 K^2 \vert
f(c_2)\vert,$ $ 0 < c_2 < c_1 < x$ and interating  $ \vert f(x)\vert
\le (xK)^n \max_{0\le x \le 1}\vert f(x)\vert.$ The iteration goes
to zero provided $ 0 \le x < 1/K.$ To go beyond $ 1/K$ you start a
new iteration at $ x_0 < 1/K$ instead of $ x_0 \equal{} 0.$

[quote="suadnovic"]Let $ f$ be funtion differentiable at
$ [0,1]$.... [/quote]

And what if we define function
$ g(x) \equal{} e^{ \minus{} Kx}f(x)$?


Chao tat ca cac ban !
Xin loi vi van de minh sap viet len day khong lien quan den toan hoc cua cac ban .MInh rat muon gui thu cho admin nhung lai khong ranh` tieng anh nen danh viet vao day .hi vong co ai biet tieng anh co the viet ho minh len nhung trang chinh cua dien dan` .
  MInh co mot nguoi ban rat than ,cau ay co mot nguoi em dang mac phai 1 can benh hiem ngheo .va` bon minh rat can tien nhung chi la` cong nhan thoi nen khong co du tien de em cau ta chua benh.MInh nghe theo loi cua mot so nguoi nen da quyet dinh len mang de tim` su giup do~
Chi con 1 ngay nua thoi cau ay can phai co du tien de nop vien phi
Vang! Neu cac ban co the ung ho dc cho bon minh thi bon minh rat cam on.Chi? can mot chut' thoi hi vong se cuu song dc mot mang nguoi .neu cac ban khong the minh cung rat cam on vi da doc bai viet nay .chi xin cac ban dung cuoi minh.Vi minh it hoc nen khong con cach nao` ca?
  neu co the xin cac ban ung ho mnh qua tai` khoan ngan hang cong thuong ma` minh van dung` de lay luong:
           Vu Van Hai
tai khoan: 711A.05379786
ngan hang cong thuong Viet Nam
Cam on tat ca moi nguoi

Bạn nên đem sang các diễn đàn có đông người Việt tham gia mà đăng cái này, ttvnol.com, vietnamnet.vn,... Để ở đây sẽ không có nhiều người quan tâm đâu.

bên diễn đàn toán học thấy có bạn bảo là đã gửi cho anh 80K rồi đó  :D em là học sinh nghèo ko có tiền nên ko giúp gì đc   :huh:


limit
real analysis
real analysis unsolved

Let $( a_n )_{n = 1}^{\infty}$ be a strictly decreasing sequence such that $a_n \geqslant 1$ for all $n$ and $\lim_{n \rightarrow \infty} a_n = 1$. Does it follow that
\[ a_1^{a_2^{a_3^{a_4^{.^{.^.}}}}}  \]
converge ?


superior algebra
superior algebra solved

[quote="Kuba"]Let $K\subseteq \mathbb C$ be a field with the operations from $\mathbb C$ s.t.
i) K has exactly two endomorphisms, namely f and g
ii) if f(x)=g(x) then $x\in\mathbb Q$.
Prove that there exists an integer $d\neq 1$ free from squares so that $K=\mathbb Q(\sqrt d)$.[/quote]

[b]'ta corrongo!![/b] (Sorry, I don't really know how to translate very accurately the experession, but it means basically, that it is a nice problem!)

[hide="My Solution"]
Well, let's see. I hope I don't have to prove that $\mathbb Q$ has only one endomorphism. I skip that. Well, given that, it is obvius that $\mathbb Q\subset K$ is an strict inclussion. One of the endomorphism has to be the identity $id$ (obviusly), and the other let's call it $\phi$, then $\phi\circ\phi=id$.
Take $\alpha\in K\setminus\mathbb Q$, then if $\beta=\phi(\alpha)$, then $\phi(\beta)=\alpha$. Note that $(x-\alpha)(x-\beta)\in\mathbb Q[x]$, because $\alpha+\beta=\phi(\alpha+\beta)$ and $\alpha\beta=\phi(\alpha\beta)$. Then, $\alpha=\frac{-b\pm c\sqrt{d}}{2a}$ for some integers $a,b,c,d$, with $d$ square free (otherwise you just take the square out of the root and multiply it to c). But then, $K=\mathbb Q(\alpha)=\mathbb Q(-b+c\sqrt{d})=\mathbb Q(c\sqrt{d})=\mathbb Q(\sqrt{d})$, because $2a,-b,c$ are rationals!
[/hide]

Best,


function
trigonometry
calculus
calculus computations

could someone help me with the analysis of the following function:
$f(x)=1-\frac{4}{x}+\frac{2}{x}\frac{\tan{\frac{2\pi}{x}}}{\tan{\frac{\pi}{x}}}$


What is the reciprocal of $ .\overline{63}$? Express your answer as a common fraction.

We have $ .\overline{63}\equal{}\frac{63}{99}\equal{}\frac{7}{11}$. Thus, the reciprocal is $ \frac{1}{\frac{7}{11}}\equal{}\boxed{\frac{11}{7}}$.


[quote="Zmastr"]when 15 marbles are removed[/quote]
Umm... what type are the marbles that are removed when the first fifteen are cuz i think you need to know. :?

Set up a system of equations, if the ratio is 2:1, then you need to multiply the side with one by 2 to make it equal

p=number of purple marbles
g=number of green marbles

2g-30=p
g-15=5p-225

2g-30=p
2g-30=10p-450

p=10p-450
9p=450

50=p

reenter it in and 40=g

I think it is 40 green marbles

sorry, I am new to this, what do those stars next to your name mean?

sorry, at the beginning, 15 GREEN marbles

[quote="Zmastr"]There are purple and green marbles in a bag in an unknown ratio, but when 15 marbles are removed, the ratio becomes 2 purple marbles for each green marble.  Then, when 45 purple marbles are additionally removed, the ratio becomes 5 green marbles for each purple marble.  How many green marbles were in the bag before any were removed?

Help![/quote]

[hide="hint"]g-15/p=1/2
g-15/p-45=5
proportions lead to systems...
[/hide]
This should help...
[hide="answer"]
40[/hide]
EDIT: POST NUMBER 50


There are two families of convex polygons in the plane. Each family has a pair of disjoint polygons. Any polygon from one family intersects any polygon from the other family. Show that there is a line which intersects all the polygons.

I don't understand the last step, what does $ z_r and z_b$ represent?, what is extreme point. At last, why the diagnol of quadilateral formed by preimages works?

Can any master read the solution in above link and explain, thanks :D


geometry
circumcircle
geometric transformation
reflection
Euler
geometry proposed

Let $ ABC$ be a given triangle and let $ G$, $ K$ be its centroid, and symmedian (Lemoine) point, respectively. Denote by $ X$, $ Y$, $ Z$ the second intersections of $ AG$, $ BG$, $ CG$ with the circumcircle of $ ABC$, and let $ X'$, $ Y'$, $ Z'$ be their reflections in the sidelines $ BC$, $ CA$, and $ AB$, respectively. It is known that the points $ H$, $ X'$, $ Y'$, $ Z'$ are on a same circle $ \mathcal{H}_{G}$. Prove that the antipode of $ H$ with respect to $ \mathcal{H}_{G}$ lies on the line $ GK$.

For a more general version, see Hyacinthos message #16602.

Dear Pohoatza and Mathlinkers,
in my proof of the P-Hagge circle (P general point not on the circumcircle of ABC), I consider the anticomplement of the isogonal of P.... In preparation on my website.
Now I come again with my problem when P is the median point G.
In two of my last equivalent problems on Mathlinks, I have the same idea.
Let U be the antipode of H and V the orthogonal projection of U on AH.
How can we prove that the points V, G and X' are collinear.
I research a synthetic proof.
Any ideas?
Sincerely
Jean-Louis

This is solution of generallization by jayme.
Let $ G$ be the centroid of triangle $ ABC$ and let $ P'$ be the isogonal point of $ P$ wrt $ ABC$, $ Q$ be the anticomplement of $ P'$ wrt $ ABC$. We will show $ \angle HX'Q = \angle HY'Q = \angle HZ'Q = 90$. Let $ AP'$ meet circumcircle of triangle $ ABC$ at $ S$($ \neq$ $ A$). It is easy to see quadrilateral $ X'BSC$ is parallelogram. Let $ M$ be the midpoint of $ BC$ and let $ QM$ meet $ AS$ at $ N$.From $ AG: GM = QG: GP' = 2: 1$, $ G$ is centroid of triangle $ AQN$. So $ M$ is midpoint of $ QN$ and we also know $ M$ is midpoint of $ X'S$. Hence $ X'Q\parallel NS$, or equivalently $ X'Q\parallel AP'$.
Since $ X'H$ is line that connecting orthocenter of triangle $ ABC$ and reflection of $ X$ on side $ BC$, $ X$ is Anti-Steiner point of line $ X'H$. So Simson line of $ X$ wrt triangle $ ABC$ is parallel to $ X'H$ but Simson line of $ X$ wrt triangle $ ABC$ is perpendicular to the isogonal line of $ AX$ wrt $ \angle A$, i. e. line $ AP'$. So we conclude $ X'H\perp AP'$. Together with fact that $ X'Q\parallel AP^'$, we get $ X'H\perp X'Q$, or equivalently $ \angle HX'Q = 90$ and similarly $ \angle HY'Q = \angle HZ'Q = 90$. So we are done.

Dear Leonhard Euler and Mathlinkers,
thanks you for your interest... but I have always no synthetic solution for my question in my above message...
I continue to search.
Any ideas?
Sincerly
Jean-Louis

Dear Mathlinkers,
you can see for example
http://perso.orange.fr/jl.ayme/   vol. 5  Le P-cercle de Hagge p. 24.
Sincerely
Jean-Louis


superior algebra
superior algebra unsolved

$ R$ is a unital ring which has no non-trivial automorphism($ f(x)\equal{}x$ is the only automorphism of $ R$) .Prove that the set of nilpotent elements of $ R$ form an ideal.

What has the automorphism to do with that? The set of nilpotent elements is the intersection of all prime ideals of $ R$.

For $ x\in R$ nilpotent, say $ x^n\equal{}0$,
we have $ 1\minus{}x$ invertible with $ (1\minus{}x)^{\minus{}1}\equal{}1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{n\minus{}1}$.
Then $ R\to R, y\mapsto (1\minus{}x)y(1\minus{}x)^{\minus{}1}$ is an automorphism,
and by assumption $ (1\minus{}x)y(1\minus{}x)^{\minus{}1}\equal{}y$ for all $ y\in R$.
So $ 1\minus{}x\in Z(R)$ and then also $ x\in Z(R)$.

And now if $ r,x,y\in R$ with $ x,y$ nilpotent, say $ x^n\equal{}y^m\equal{}0$,
then $ (rx)^n\equal{}r^nx^n\equal{}0$ and $ (x\plus{}y)^{n\plus{}m}\equal{}\sum_{k\equal{}0}^{n\plus{}m}\binom{n\plus{}m}{k}x^ky^{n\plus{}m\minus{}k}\equal{}0$,
as $ x$ commutes with $ r,y$. So $ rx$ and $ x\plus{}y$ are again nilpotent.

So the set of nilpotent elements of $ R$ form an ideal.


college
Harvard
Princeton
University of Chicago
MIT
Stanford
Yale

I come from Thailand and I decide to learn Undergraduate in Maths Major at US University.
What College should I choose? My GPA is approximate 3.7-3.8 (not finish high school yet) from 4.00.
And I got all 4.0 in Maths except Statistics which I got 3.5  :(
Thanks for reply :lol: .

I found that UC Berkeley is good at Maths, but there admit only 2% of International Students :maybe: .

All of the best universities in USA will have low acceptance rates for international students, probably.

Schools like Harvard and Princeton will probably offer the best financial aid. The best schools for math are (in no particular order) something like UChicago, Princeton, MIT, Harvard, maybe CalTech.

If you can waive the application fee, you might as well apply to as many of these schools (and others) as you can. But if not, the application fee might be a deterrent, so you should decide how much you can afford and make a list of schools to apply to.

Of course, since I am a student at Princeton, I would highly recommend Princeton. ;) Some other good schools that I didn't mention already are Stanford, Yale, Brown, Dartmouth, Cornell, Rice, Duke, Harvey Mudd. I'm sure other people would post more. Maybe you should ask Dr. Merryfield about how his school treats international students; his school may not be really famous but that definitely does not make it bad. See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106786.

I am an international applicant (but I might have an advantage over international applicants from abroad on the SATs since I have lived in the US for a while, so I can get decent scores on the CR and Writing sections), and, according to personal experience, if you have a non-trivial financial need, some of the best chances you have are at Harvard, Princeton, and Yale (because those 3 schools are the only 3 that treat you on a completely equal basis as all other domestic students, meaning that your admission process is need-blind just as for the domestic students).

Most other schools that Xevarion mentioned have 20-50 spots for international students requiring financial aid, and you must remember that very few of those can be used for prospective math majors (there are many academic areas other than math in which students apply). Basically, if you don't have a very strong record of performance on national and/or international competitions, you'll probably end up wasting the money for the application fees (like I did...).

I don't know about UC Berkley because I was a CA resident, so that might have helped.
The feeling I get about UC's, however, is that housing is hard to find, classes are really hard to get into (there are limits to how many students can get in a class), and you're pretty much on your own for most things.

[b]Definitely consider CSU Long Beach[/b]. There is a thread devoted to that in this forum.

[quote="mlok"]I think that, unless your family is moving to the USA, you should seriously consider undergraduate math programs in your home country.[/quote]
How come? Do you refer to the fees?

[quote="shbmsc90"] Basically, if you don't have a very strong record of performance on national and/or international competitions, you'll probably end up wasting the money for the application fees (like I did...).
[/quote]

Is this the same for PHD applicants? I would like to go Cornell (Operation Research and Ing.) to obtain a Phd (but I haven't decided that finally  :wink: ),  but so far I havent been on competitons (except the one's that I have perticipated during high-school education) . In fact, my question realy is ; what is needed for a Phd candidate???

Thanks in advance.

[quote="Mr.Doe"]
Is this the same for PHD applicants? I would like to go Cornell (Operation Research and Ing.) to obtain a Phd (but I haven't decided that finally  :wink: ),  but so far I havent been on competitons (except the one's that I have perticipated during high-school education) . In fact, my question realy is ; what is needed for a Phd candidate???

Thanks in advance.[/quote]

Graduate schools mostly look at your undergraduate research and letters of recommendation. Most schools also ask you to take standardized tests (GRE, TOEFL). Universities will also look at your transcript and GPA (average grade) - it is better to take several advanced/graduate level courses.

You do not need to participate in math competitions in order to apply to the grad school.

[quote="Stewie"]How come? Do you refer to the fees?[/quote]
To the combination of higher cost and not necessarily higher quality. Of course, I don't know what's available to [b]mathpk[/b] in Thailand.

[quote="Mr.Doe"]Yury: Can you recommend me  some courses, but keep in mind that I am a Finance and Business Mathematic's major.
I plan to take (next year) : Topology,Functional Analysis and Probability Theory.[/quote]

Sounds good. As far as I understand, what really matters is that you take some advanced courses, and it is not that important what particular courses you take. By the way, my suggestions are just general suggestions for applying to a grad school - I do not know much about programs in Operations Research or Financial Mathematics.