There are several functions which fit the criteria given.

For example, any function of the form $ f(x) \equal{} kx$ fits, since then we have:

$ f(x \plus{} y) \equal{} k(x \plus{} y) \equal{} kx \plus{} ky \equal{} f(x) \plus{} f(y)$.

So I don't think there is enough information to solve the problem.

You are inducing that there exist $ {k}$ such that $ {f(x)}\equal{}kx$.
Now assume that there exist $ {m}$ such that $ {f(x)}\equal{}m^x$.
Try again

The second assumption is nonsense, and the first assumption can't be justified without some additional regularity hypothesis.

@21 7 15

My point was that there are several possible values for $ f(3)$ and $ f(4)$, and that the problem cnnot be solved without some additional information.

Let's assume there is such $ f(x)\equal{}m^x$.
Then $ f(2x)\equal{}m^{2x}\equal{}2m^x \implies m^x\equal{}2$ thus $ f(x)\equal{}2$. But $ f(x\plus{}y)\equal{}f(x)\plus{}f(y) \implies 2\equal{}2\plus{}2$ which is contradiction, therefore $ f(x)$ can't be of form $ m^x$.

If $ {f(1)}\equal{}2f(1)$,is it safe to imply that $ {f(1)}\equal{}0$?

[hide="Not sure if this is what you wanted..."]We prove by induction that $ f(x) \equal{} x \cdot f(1)$ for all positive integers.

First check $ x \equal{} 1$ which is obviously true since we have that $ f(1) \equal{} 1 \cdot f(1)$.

Now assume the result proven for $ n$ and prove it for $ n \plus{} 1$.

We have that $ f(n \plus{} 1) \equal{} f(n) \plus{} f(1) \equal{} n \cdot f(1) \plus{} f(1) \equal{} (n \plus{} 1) \cdot f(1)$.

So by the principle of mathematical induction, $ f(x) \equal{} x \cdot f(1)$ for positive integers $ x$.

So $ f(3) \equal{} 3 \cdot f(1)$, and $ f(4) \equal{} 4 \cdot f(1)$.[/hide]

[quote="21 7 15"]Functions
Inductive ability[/quote]That's like a [i]fineprint[/i] at the top of his original post. Next time, [u]please specify the conditions more clearly[/u].

KrazyFK solves the Cauchy equation for positive integers.


$\triangle ABC$ is inscribed in circle O, the radius of O is $12$ and $m\angle ABC=30^\circ$. A circle with center $B$ is drawn tangent to the line containing $\overline{AC}$. Let $R$ be the region which is within $\triangle ABC$, but outside circle $B$. Compute the maximum area of $R$.

I'm not sure if this is correct, but I don't have time to post right now. [hide="This is what I got:"]
\[R \leq \frac{108}{\pi}\]
[/hide]
I'll post my solution later.  :)


Additive Number Theory
Sum of Squares
number theory
Perfect Square
IMO Shortlist

Show that the set of positive integers that cannot be represented as a sum of distinct perfect squares is finite.

It is not true. Consider $ N\equal{}(k\plus{}1)^2\plus{}...(k\plus{}n)^2\equal{}\frac{n[2n^2\plus{}3n(2k\plus{}1)\plus{}6k^2\plus{}6k\plus{}1]}{6}$.
If $ N\equal{}p>2$ is prime must be $ n|6$. $ n\equal{}1$ give $ N\equal{}(k\plus{}1)^2\not \equal{}p$.
$ n\equal{}2$ give $ N\equal{}2k^2\plus{}6k\plus{}5\equal{}f_2(k)$,
$ n\equal{}3$ give $ N\equal{}3k^2\plus{}12k\plus{}14\equal{}f_3(k)$,
$ n\equal{}6$ give $ N\equal{}f_3(k)\plus{}f_3(k\plus{}3)\equal{}f_6(k)\equal{}6k^2\plus{}42k\plus{}91$.
there are infinitely many primes p, suth that $ p\not \equal{}f_2(k_1),f_3(k_2),f_6(k_3)$, because $ \pi(N)>3\sqrt N$ for big N.

Who said that these perfect squares have to be consecutive?

By a theorem due to Lagrange, any positive integer is expressible as a sum of 4 squares (possibly 0).
There are 4 possible cases for having 2 non-distinct squares:
(1) $ 4(a^2)$
(2) $ 2(a^2 \plus{} b^2), a \neq b$
(3) $ 3a^2 \plus{} b^2, a \neq b$
(4) $ 2a^2 \plus{} c^2 \plus{} d^2$, $ a,c,d$ are distinct.

Incomplete progress:
[hide]
The 1st case is OK - $ 4(a^2) \equal{} (2a)^2$
The 2nd case is OK when $ a, b > 0$, because of the identity $ 2(a^2 \plus{} b^2) \equal{} (a \minus{} b)^2 \plus{} (a \plus{} b)^2$.
If WLOG $ a \equal{} 0$, it is enough to express $ 2a^2$ as a sum of 2 distinct squares.
The number of representations by 2 squares is known: $ \prod_{p \equiv 1 \pmod 4} (1 \plus{} v_p(n))$. In our case, $ n \equal{} 2a^2$. When this is greater than 1, we're done. But this is always greater than 1, unless $ v_p(2a^2) \equal{} 0$ for all primes $ p \equiv 1 \pmod 4$, which implies that $ a$ is a product of primes $ p \equiv 3 \pmod 4$ and 2's.
The 3rd case is either solved or reduced to the 4th case if we replace $ 2a^2$ with 2 distinct squares like we did in the 2nd case.
The 4th case is solved if we replace $ 2a^2$ with the same of 2 squares, non of which equal to $ c^2$ or $ d^2$. As before, this is possible when $ \prod_{p \equiv 1 \pmod 4} (1 \plus{} v_p(n)) > 3$. It holds whenever $ a$ has at least 2 prime divisor of the form $ 1 \pmod 4$, or one prime divisor of such form such that $ v_p(a) > 1$.
[/hide]

Also, the number of representation of $ n$ as a sum of 4 squares is at least $ 8 \sum_{d|n} d$. But each solution is counted at most $ 4! \cdot 16$ times, so when $ \frac {8 \sum_{d|n} d}{24 \cdot 16} \equal{} \frac {\sum_{d|n} d}{48}$ is big enough (the least amount of distinct representations) we can probably avoid having equal squares.

[quote="bambaman"]...The number of representations by 2 squares is known: $ \prod_{p \equiv 1 \pmod 4} (1 \plus{} v_p(n))$...[/quote]
where can I find this forluma?

[quote="bambaman"]Also, the number of representation of $ n$ as a sum of 4 squares is at least $ 8 \sum_{d|n} d$.[/quote]
No  :wink:
The exact number is $ 8 \sum_{4\nmid d|n} d$, which is smaller in general (but shows that you just need to adapt for $ n$ having large $ 2$-adic valuation).

[quote="Flakky"]where can I find this forluma?[/quote]
In the formulary.

[quote="test20"][quote="hxy09"]I think it is a really nice problem.Can anybody post a solution different from the official (I privately think that the official solution is too hard to come up with.[/quote]

This actually is a well-known result.
Check out Sloane for some references:
http://www.research.att.com/~njas/sequences/A001422[/quote]
Thank you,but how can I find the proof of it? :(
And bambaman's approach is similar to mine,but I can't continue with complicated cases,I think we couldn't just use "representations by 2 squares "

[quote="hxy09"]I think it is a really nice problem.Can anybody post a solution different from the official (I privately think that the official solution is too hard to come up with :maybe: ) :)[/quote]
[quote="hxy09"]Thank you,but how can I find the proof of it?  :( [/quote]
I found this solution which I think is not too hard to come up with if finite (but long) time (or a computer) is given.

Statement 1:
1) All numbers in the interval $ [u,v)$ can be represented as sum of squares of distinct positive integers $ \le k$

Lemma 1: If statement 1 is true for some positive integers $ k,u,v$ such that $ v\ge u \plus{} (k \plus{} 1)^2$ then it is also true for $ k \plus{} 1$ in place of $ k$, $ v \plus{} (k \plus{} 1)^2$ in place of $ v$ (and u unchanged).

[hide="Proof"]Assume statement 1 is true for some $ k,u,v$
Add $ (k \plus{} 1)^2$ to each positive integer in the interval $ [u,v)$. The numbers will become in the interval $ [u \plus{} (k \plus{} 1)^2,v \plus{} (k \plus{} 1)^2)$. These numbers can be represented as sum of squares of distinct positive integers $ \le k \plus{} 1$ (we use their previous representation and just add $ \plus{} (k \plus{} 1)^2$ at the end of it.)
Also the original numbers in the interval $ [u,v)$ also can be represented as sum of squares of distinct positive integers $ \le k \plus{} 1$.
The union of these 2 intervals : $ [u,v) \cup [u \plus{} (k \plus{} 1)^2,v \plus{} (k \plus{} 1)^2) \equal{} [u,v \plus{} (k \plus{} 1)^2)$ because $ v\ge u \plus{} (k \plus{} 1)^2$[/hide]

Lemma 2: There is such a $ k$ satisfying statement 1.

[hide]Using C program I found that the least such $ k$ is $ 10$. All numbers $ > 128$ and $ \le 256$ can be represented as sum of squares of distinct positive integers $ \le 10$. This interval is greater than $ 121$ by a whisker.
I think it is possible without computer. The calculation is not as vast as it looks.
[/hide]

Now we have a base case for statement 1, lemma 1 implies $ k$ can get bigger and bigger and the interval $ [u,v)$ will get bigger and bigger with no end.
But numbers which cannot be represented as sum of distinct perfect squares cannot lie in that interval. So it must be less than $ u$ fir the least $ k$ (this $ u$ is actually $ 129$).

[quote="hxy09"][quote="test20"]This actually is a well-known result.
Check out Sloane for some references:
http://www.research.att.com/~njas/sequences/A001422[/quote]
Thank you,but how can I find the proof of it? :([/quote]

So did you take a look at  http://www.research.att.com/~njas/sequences/A001422 ???
This page provides references to the following five sources:

R. E. Dressler and T. Parker, "12,758", Math. Comp., 28 (1974), 313-314.

S. Lin, Computer experiments on sequences which form integral bases, pp. 365-370
of J. Leech, editor, Computational Problems in Abstract Algebra. Pergamon, Oxford, 1970.

Harry L. Nelson, The Partition Problem, J. Rec. Math., 20 (1988), 315-316.

J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 222.

R. Sprague, Ueber Zerlegungen in ungleiche Quadratzahlen, Math. Z. 51, (1948), 289-290.

The paper by Dressler and Parker has the following abstract:
Sprague [Math. Z. 51 (1948), 289--290; MR0027285 (10,283d)] proved that every positive
integer greater than 128 can be expressed as a sum of distinct perfect squares. He also
proved [ibid. 51 (1948), 466--468; MR0028892 (10,514h)] that, for every integer $ n\geq 2$,
there exists a largest positive integer $ r_n$ that is not expressible as a sum of distinct
$ n$th powers (of positive integers). In the light of the first reference above, we have
$ r_2\equal{}128$. Unfortunately, the technique used in the second reference above is existential
and does not give any idea of the size of $ r_n$ for $ n>2$. Our purpose here is twofold:
(1) we use a computer to find explicitly the value of $ r_3$ (it turns out to be 12,758), and
(2) in the process we give a new quantitative proof of Sprague's theorem for $ n\equal{}3$.

To Akashnil:
Thank you for such a nice proof I like it very much!
To test20:
Thank you for the precious information

This problem is a obvious consequence of the theorem of Bohman-Froberg-Riesel.

What is the theorem of  Bohman-Froberg-Riesel, please help me!

I am unsure if we can avoid bashing out the small cases (of course, elementarily speaking).

We begin by checking small cases and we observe that all numbers from $129$ to $324$ can be written as sums of distinct squares. (hooray!)

Now, let's induct. Assume that all numbers from $129$ to $k$ are good. Furthermore, assume that $i^2\leq k+1<(i+1)^2.$ Then, observe that:

$128=18^2-14^4\leq i^2-(i-4)^2<k+1-(i-4)^2\leq k$ so $k+1-(i-4)^2$ can be written as the sum of distinct squares.

Furthermore, $k+1-(i-4)^2<(i+1)^2-(i-4)^2<(i-4)^2$ so those squares are different from $(i-4)^2.$ Therefore, $k+1$ can be written as the sum of distinct squares.

Hence, all $N>128$ have the desired property.

Here's a different proof without bashing small cases:
[rule]
Let $S$ be set of all representable numbers. We want to show $| \mathbb Z_{>0} \setminus S|$ is finite.

[b][color=#f00]Claim 1: [/color][/b] There is a constant $C$ such that for any $n \in \mathbb Z_{>0}$, there is a element $n' \in S$ with $0 \le n- n' \le C$.

[i]Proof: [/i] For each $n \in \mathbb Z_{>0}$, define $f(n)$ as the least non-negative integer of set $n - S$. To prove $f(1),f(2),\ldots$ is bounded. Basically, if $n$ is large, then $$f(n) \le f \left(n - \lfloor n \rfloor^2 \right)$$
where $f(0) = 0$ by convention. More particularly, there is a constant $D$ such that for all $n > D$, we have $f(n) \le f(n_0)$ for some $n_0 < n$. It follows
$$ C = \max(f(1),f(2),\ldots,f(D) ) $$
just works. $\square$

[color=#f00][b]Claim 2: [/b][/color] There is a constant $k$ such that the set
$$ S \oplus (\{0,4\} \oplus \{0,8\} \oplus \cdots \oplus \{0,4k \}) $$
contains all sufficiently large integers.

[i]Proof: [/i] Pick a sufficiently large $n$. Define
$$ \lambda(n) = \begin{cases} n \qquad &\text{if } n \equiv 0 \pmod{4} \\ n - 1^2 \qquad & \text{if } n  \equiv 1 \pmod{4} \\ n - 1^2 - 3^2 \qquad & \text{if } n \equiv 2 \pmod{4} \\ n - 1^2 - 3^2 - 5^2 \qquad &\text{if } n \equiv 3 \pmod{4} \end{cases} $$
By definition $\lambda(n)$ is sufficiently large and divisible by $4$. By using only numbers divisible by $4$ in representation of $\lambda(n)$, we obtain it suffices to have the set
$$ S \oplus (\{0,1\} \oplus \{0,2\} \oplus \cdots \oplus \{0,k\}) $$
to contain all sufficiently large numbers $\lambda(n)$. But this just follows from the facts that
\begin{align*}
f(\lambda(n)) &\le C \\
\{0,1\} \oplus \{0,2\} \oplus \cdots \oplus \{0,k\} &= \left\{ 0,1,2,\ldots,\frac{k(k+1)}{2} \right\}
\end{align*}
This proves our claim. $\square$

We return to our main problem now. Pick a large $M$. We will in fact show the set
$$ T = 4S \oplus ( \{0,3^2,5^2\} \oplus \{ 0,7^2,9^2\} \oplus \cdots \oplus \{0,(4M-1)^2,(4M+1)^2 \} ) $$
contains all sufficiently large numbers $n$ (above set is clearly a subset of $S$). Pick $N \in [M-3,M]$ with $n \equiv N \pmod{4}$. Let,
$$ n - (3^2 + 7^2 + \cdots + (4N-1)^2) = 4\alpha $$
As $\alpha$ is sufficiently large (as $M$ was fixed and $n$ was large), so by [b]Claim 2[/b], we obtain
$$ \alpha \in S \oplus (\{0,4\} \oplus \{0,8\} \oplus \cdots \oplus \{0,4N\} ) $$
which then implies that
$$ n \in  4S \oplus ( \{3^2,5^2\} \oplus \{ 7^2,9^2\} \oplus \cdots \oplus \{(4N-1)^2,(4N+1)^2 \} ) $$
completing the proof. $\blacksquare$

Can you explain $$f(n) \le f \left(n - \lfloor n \rfloor^2 \right)$$  ?

Say $k^2 \le n < (k+1)^2$. So if you write $n = k^2 + (n-k^2)$, where $n - k^2 < k^2$ since $n$ large, so one directly gets desired conclusion. Basically, all numbers while writing a number $\le n- k^2$ will be $< k^2$, so we don't need to worry about repetition. After that, $f(n) \le f(n-k^2)$ pretty much follows by definition.

Quite a tricky N6[hide=Solution(from IMO compendium)]Suppose a natural number $N= a_1^2+a_2^2 \dots a_k^2$ and $2N= b_1^2+b_2^2 \dots b_l^2$ such that none of $\frac{a_i}{a_j}, \frac {a_i}{b_j},\frac {b_i}{a_j} \frac {b_i}{b_j}$ are powers of 2.
[color=#f00]Claim 1[/color]: Every $n >  $$\sum_{i=0}^{4N-2} (2iN+1)^2$ work
[hide=Proof]Suppose $n = 4Nq+r$ $0 $$\leq$$    r  $$\leq$$  4N-1$ then
[b][size=150]$n = 4Ns +$$\sum_{i=0}^{r-1} (2iN+1)^2$[/size][/b] (that is why we took 4N-2)
it's enough to show $4Ns$ is a sum of distinct even squares (that is why we took N like that)
let $s = \sum_{c=1}^{C} 2^{2u_c} + \sum_{d=1}^{D} 2^{2v_d+1}$
then
$4Ns =  \sum_{c=1}^{C}\sum_{m=1}^{k} (2^{u_c+1}a_i)^2+$$\sum_{d=1}^{D}\sum_{j=1}^{l} (2^{u_d+1}b_j)^2$
Hence the claim, Now we may pick N=29 [/hide] [/hide]

[quote=ngv]It says that, there are only $31$ numbers that cannot be expressed as the sum of distinct squares: $2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128$.[/quote]

Where do I read more about this theorem?, I can't find a reliable source on searching


probability
expected value
combinatorics unsolved
combinatorics

You have two decks of 26 cards. Each card in each of the two decks has a different letter of the alphabet on it. You pick at random one card from each of the two decks. A vowel is worth 3 points and a consonant is worth 0 points. Let X be the sum of the values of the two cards picked. Find E(X), V(X), and the standard deviation of X.

Note:
E(X) stands for expected value
V(X) stands for variance

This is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197787]double post[/url] -- please don't post your questions in more than one section of the forum.


This question was in the AoPS vol.2, ch.19: Probability.
I read the solution for 310 and I didn't get :( . I would really appreciate it if someone could explain it to me.

Could you please post the problem; many people do not have the AoPS books.  ;)

Problem 310. A circular coin of radius 1cm is dropped near the center of a chessboard(8 squares on a side) comprised of squares with sides of 2cm. What is the probability that the coin lands so that it is entirely within one square? (That is, it may touch but not cross any of the lines.) A line is considered to have zero width.

Click below to reveal the answer:
[hide]0[/hide]

Please can someone explain to me as to how this happens to be?

P.S. Masoud I really think you should get the AoPs books, they are really really great, but get them with the solutions manual because answers are not included at the back.

The places in which it can land make up a continuous space (very very very small distances on the board) and there are only a finite number of positions that satisfy the condition (64, like jrshoch said) so you get 0. It's like getting one particular value in a continuous distribution model.

Let's just examine one square. Because the coin has radius $1$, it has diameter $2$, the same length as a side of the square. Hence, the only way to have to coin fit is if its center were perfectly on top of the square's center. This is 1 [i]point[/i] in the entire square.

The square contanes an infinite number of points, so the probability is $\frac1\infty$, which is theoretically $0$.

Try the problem where the radius is $\frac12$. It won't be 0 this time. :D

I like your solution 4everwise  :lol:  thanks

If you have a set of countably infinite points that satisfy the given conditions, would the probabilty still be 0?

never think of these questions as number of points, always as an area to area ratio

[quote="4everwise"][quote="jrshoch"]It just seems like there are an infinite number of ways it could land but only 64 different ways that this condition would be satisfied... so $\frac{64}{\infty}=0$?[/quote]Wouldn't the probability still be $\frac1\infty$, since the number of squares doesn't matter?  :huh:[/quote]You're both right. You were just counting different spaces, which you both counted correctly. Note $\frac{64}{\infty} = \frac{1}{\infty}$.

i suppose that this problem can be thought of intuitively

if you drop a coin, it has to land [i]exactly[/i] in the center of one of the squares, but obviously, the chances of an exact landing are 0

[quote="4everwise"]
Try the problem where the radius is $\frac12$. It won't be 0 this time. :D[/quote]

How do you do this?? Wait, would the answer be $\frac{1}{4}$? Looking at one square, The center of the coin must land in the center fourth of the square. :D

Yes, I think it would be $\frac14$.  I get that from answering the questions, "Where can the center land?" and "What is the probability it will land there?"

if the radius was less than 1 then you could trace out a smaller square with the coin and then the area of that square divided by the area of 4 for the 2x2 would be the probability.  In this case there is only the point in the center where it could land.  And since a point has no area $\frac{0}{4}=0$


I am having trouble figuring out what to do here.  Here is the problem:

I am to cut an M&M in half and create the curve of best fit for the chocolate and candy shell.  Than I am to use Calculus to calculate the ratio of the candy shell to the chocolate.  Once I have that completed, I am to check the volume with another type of measurement, i.e. displacement or other type of mathematical calculation (or explain the difference if there is one).

Can anyone get me started in the right direction?

Well, I would. But I don't know. P.S. Unless you have 15 years to solve this, I don't think you need help anymore.


Let $a^2+b^2 =c^2+d^2=5$
 Find the maximum and minimum value of
 $S =\sqrt{5-a-2b}+\sqrt{5-c-2d}+\sqrt{5-ac-bd}$

$(a+2b)^2=(a+\frac{b}{2}+\frac{b}{2}+\frac{b}{2}+\frac{b}{2})^2\leq 5\left(a^2+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}\right)=5(a^2+b^2)=25$
So $a+2b\leq 5$.
Analogically:
$(c+2d)\leq 25\Rightarrow c+2d\leq 5$
We also have: $(ac+bd)^2\leq (a^2+b^2)(c^2+d^2)=25 \rightarrow ac+bd\leq 5$
We have equality in all cases when $a=c=1,b=d=1$.

Using it we have: $S =\sqrt{5-a-2b}+\sqrt{5-c-2d}+\sqrt{5-ac-bd} \geq \sqrt{5-5}+\sqrt{5-5}+\sqrt{5-5}=0$.


AMC AIME USA(J)MO USAMO Harvard college Yale

I would like to know, having not done many AIMES before, how much is a decent score on the AIME? How much points do USAMO qualifiers generally get? I want something to work up towards  :lol:

Thank you for your help.

The difficulty varies quite a bit between years.

If you're using recent tests, one number to look at is the "floor" score, which has been a minimum qualifying standard for 10th graders and younger.

USAMO qualification minimums in recent years:

12th and younger, taking AMC 12:
2008: 204 index
2007: 197.5 index
2006: 217 index
2005: 233 index for AIME 1, 220.5 for AIME 2
2004: 210 index

10th and younger, taking either test:
2008: 202.5 index (and 6 AIME)
2007: 6 AIME and 181.0 index
2006: 8 AIME
2005: 9 AIME
2004: 7 AIME
2003: 8 AIME
2002: 6 AIME

I don't know the senior standards for 2002 and 2003, and the system was different before 2002. In all cases, "index" is AMC score plus 10*AIME score.

The USAMO dramatically expanded between 2005 and 2006, nearly doubling the number of participants. This didn't actually make a huge difference in the qualifying levels; I estimate the difference as a little more than one AIME question.

my goal is a 7 personally - that should get me to usamo with a good ten score. Textangle, how are u preparing for aime?

a 9 (possibly an 8) will almost certainly get you in.

I think the question is ill-posed. The answer depends very much on your training, experience and certain other environmental factors. For some students 2-5 problems might be great, others are aiming at 13+ problems. And jmerry interpreted the question in the way that a decent score implies you are qualifying for the next higher stage USAMO which requires a certain floor score for junior or seniors in the qualification formula he presents.

I think jmerry was also using the USAMO qualifying score as a possibly useful proxy for thinking about the meaning of lower scores. Unfortunately, it isn't that useful a proxy: for instance, the floor/index scores for 2008 weren't particularly high (although some of that was the relative difficulty of the AMC), but it was clearly easier in 2008 to score in the 3-5 range on the AIME than it has been in most years.

Well, I've never even gotten to AIME, seeing that my AMC 10 scores are rock bottom... So i decided to do the 2008 AIME I for a bit of a practice. I got an 8 (so I was like cool, I made USAMO when I saw this  :lol: ).

I suppose then i should study the more advanced problems/techniques to prepare? Or is there anything special?

Actually, you wouldn't have made it in 2008 without also having at least a 122.5 AMC 10. 2008 combined a relatively easy AIME with a hard AMC 12, which led to index scores as the only factor in qualification.
Also, many people do better in practice than on the real test. The adrenaline makes a big difference, and there's no way to know how you respond to that except experience.

I really had two reasons to post that:
- getting all of the information in one place so I can refer to it again.
- answering the question in the second sentence of the OP.

imo, you should aim for a perfect score. study as if you're wanna get one. and when you don't, whatever, you most likely did well.

i'm aiming for a 15. realistically, who cares?

[quote="jmerry"]I don't know the senior standards for 2002 and 2003, and the system was different before 2002. In all cases, "index" is AMC score plus 10*AIME score. [/quote]
I don't know the senior standards either, but from some local records I have, it's clear that the senior index standard for 2002 was at least 201, and the senior index standard for 2003 was higher than 221 (but not higher than 238.5).

I scored a 7 on the AIME... would this be impressive to a college like Harvard or Yale? Would it be a significant factor to the admissions committee?

It would certainly be impressive and could help you get in if coupled with more things that show mathematical passion...

It really depends on what your goal is.
If it's to make USAMO, you probably aim (haha, get it) for 1/2 right.
Personally, I just try my best and hope :)


Find all real $x$, satisfying the equation
[list]$\sqrt{n}^{3x-1}=\sqrt[n]{x}^{2+3x}$[/list]

hello, and for what stands $n$ here?
Sonnhard.


calculus topology search geometry function derivative integration

For this question, Im mainly looking for either people who have already been throuhg both books or at least are knowledgeable about them.

I am going to be learning calculus this summer, and before I do that, Id like to get a book that I feel is best for me. This is what I am looking for:

1. A book that teaches calculus indepth and explains the why's, i.e. a book that is very proof based and doesnt just say something, but explains it.
2. A book that is challenging and is more than just standard textbook problems, i.e. one that involves proofs and has problems that gain in difficulty in each chapter.
3. A book that doesnt just dive right into the retardedly hard, but also at the same time, doesnt take baby steps.

From these conditions, Ive heard (aka complexzeta) that Calculus (not manifolds) by Spivak is the best choice. I found Stewart for an extremely low cost, but was wondering what it is like compared to Spivak and other books.

If you name a book, please list its pros and cons (your opinion please).
Thanx in advance to those who post.

Proof by eminent reference: Paul Zeitz told me that Spivak is the book to use.

For what you seem to be looking for I can also vouche for Spivak.

As a (far distant) second choice I would recommend [u]One-Variable Calculus with an Introduction to Linear Algebra[/u] by Tom Apostol.

I own the Spivak textbook. I have looked at the Stewart calculus textbook from time to time and own Stewart's precalculus textbook. I own about nine other calculus textbooks, and my dad owns five more from his college days. I have also downloaded three or four freebies from the Web. (The best of the freebies is by Neal Koblitz, quite a good free download, in [url=http://www.math.washington.edu/~duchamp/124notes/k124-main.pdf]two[/url] [url=http://www.math.washington.edu/~duchamp/125notes/k125-main.pdf]parts[/url].) For what you want to do, you want the Spivak textbook. Google up "Michael Spivak Calculus" and you will see a SCARY degree of unanimity among mathematicians that the Spivak textbook is the best available (in English).

Hope this helps!

Hey, token, what would be the best online complex analysis textbook?  You seem to be incredibly organized, and I figure if anyone knows its you.

[quote="Magnara"]Hey, token, what would be the best online complex analysis textbook?[/quote]

I'm far from proficient enough in mathematics to judge for sure, but [url=http://www.math.lsu.edu/~neubrand/notes.pdf]Neubrander's lecture notes[/url] look all right, as does [url=http://www.math.gatech.edu/~cain/winter99/complex.html]George Cain's online textbook[/url]. I found these by following links from a Google search "'complex analysis' 'lecture notes'" and will note here that usually "lecture notes" is an excellent phrase to add to a Google search on any broad mathematical topic.

Hope this helps!

P.S. It's scary to think that anyone would ask moi for advice on how to study complex analysis, but I'll take this as an invitation to share a bibliographic finding.

[quote] As a (far distant) second choice I would recommend One-Variable Calculus with an Introduction to Linear Algebra by Tom Apostol. [/quote]

This is what I will be using next year in calculus in college.

I am also taking Calc I & II at URI this summer.  This is the text we will use: "Hughes-Hallet, et. al., CALCULUS - Single Variable   (THIRD Edition)" Anyone know anything about it?[/quote]

[quote="JS1527"]This is the text we will use: "Hughes-Hallet, et. al., CALCULUS - Single Variable   (THIRD Edition)" Anyone know anything about it?[/quote]

The textbook by Hughes-Hallet is quite notorious as an example of a "reform calculus" textbook that makes things easy for non-math-majors by taking a lot of the math out of a calculus course. It wouldn't hurt you to take a summer course based on it, but you'll learn a lot more during your school year course with the Apostol textbook. The same eminent reference cited by ComplexZeta praises Hughes-Hallet for her guidance of his career, but we note that Paul Zeitz recommends Spivak's textbook rather than hers when he chooses which calculus textbook to recommend.

Hmmm....  This is not good news.  These two classes are supposed to replace a year of HS calculus.  Having "calculus experience" is required to take the class I want to take in college, and since I couldent take calc in HS am taking this class instead.  Do you have any suggestions?  Should I buy the apostol calc text this summer and study it?  Or do you think this class will be sufficent to teach me the basics and I can learn the rest in college?  Here are the descriptions of the courses:

Calc I w/analytic geom:  Topics in analytic geometry, functions and their graphs, limits, the derivative, applications to finding rates of change and extrema and to graphing, the integral, and applications.

Calc II w/analytic geom: Continues the study of calculus for the elementary algebraic and transcendental functions of one variable. Topics include the technique of integration, improper integrals, indeterminate forms, and calculus using polar coordinates.

That looks more or less like the basic BC Calc sylabus.  But it doesn't really matter how much breadth you get, rather, how much depth.  And I'm not in a position to tell you how much you'll get from that class.

I'm getting ready for Calc 3 or multivariable. The one class I'm going to take is freaky. Like 2-3 weeks in a row every day. Its going to be very exciting, and intensive. The University here uses Varberg's Calculus 8th Edition. "Great Well Written Concise Book for all Calc 1-3"

[quote="JS1527"]Hmmm....  This is not good news.  These two classes are supposed to replace a year of HS calculus.[/quote]

If it's any consolation, there are plenty of high schools for which the same textbook would indeed be used for a year of HS calculus. The Apostol calculus textbook, which is second ONLY to Spivak's in the esteem of many mathematicians, starts at the beginning, so what you mostly need in your summer course is a first look at calculus, which you will surely get, for orientation.

An old [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8699]Round Table thread on Textbooks[/url] includes my tentative thoughts on preparing for a university calculus course, which I hope are helpful to you.

Going from Hughes-Hallet to Apostol is a good route to go for learning calculus. Hughes-Hallet (which is used by Harvard in it's introductory Calculus classes) will give you a good intuitive understanding of calculus; and Apostol will give it a more broad and abstract treatment (volume II of this book, which I think is excellent, is used for the Honors Calculus and Linear Algebra class at Princeton).

I always liked to get a good intuitive feel for something before getting into it's more abstract treatment, so the Hughes-Hallet to Apostol approach seems good to me. Of course this all depends on whether you learn better from the book, or from the teacher. And if it's the latter - how good your teacher is for the course.

Thanks for all the replies.  This makes me feel a lot better, and I feel that I will be prepared. Thanks again!

Ditto for me. I dont usually post, just lurk, but thanx. You have persuaded me to spend an extra 60 bucks to get spivnak + sols instead of stewart.

Just be warned that while Spivak is a very good book, some people find it hard to use as an [i]introduction[/i] to Calculus - like for people who are seeing Calculus for the first time. You said you were looking for a book that provided proofs of things, which Spivak does, but some people find that difficult if they don't have an intuitive understanding of things first.

It also doesn't have as many problems of "direct" application to sciences and humanities as say Stewart does. And it doesn't have as many "computational" exercises.

But from your description of what you were looking for, Spivak fits best. Just be aware that you may want to supplement it with a more traditional "high school AP textbook" like Stewart or Larson or Anton or Finney for intuition or experience with computation.

I picked up the only two books on advanced math in the bookstore, they are
Mathematics, Its content, Methods, and meaning by Aleksandrov, Kolmogorov and Lavrent'ev
and Advanced Calc by Widder
are they any good?

[quote="bugzpodder"]I picked up . . . Mathematics, Its content, Methods, and meaning by Aleksandrov, Kolmogorov and Lavrent'ev[/quote]

That's a very good book, one of the best of the Dover reprints. Kolmogorov should need no introduction to anyone studying advanced mathematics. Everyone should read some math books by Russian authors--the Russian national tradition of excellence in mathematics education appears to go all the way back to the time when Euler taught at the St. Petersburg Academy.

There's a reason why my avatar on this site is a Russian.


When filled with an cold water using a particular cold water tap, a tank will be full in 14 minutes. In 21 minutes, the full tank could be emptied by opening a hole on the base of the tank. If the cold water tap and the hot water tap are opened simultaneously (allowing hot and cold water fill the tank), and the hole on the base of the tank is opened, the tank will be full in 12.6 minutes. Determine the number of minutes needed to fill the tank with hot water until the tank is full, assuming at first the tank is empty and the hole is closed.


induction
combinatorics unsolved
combinatorics

In the cells of the $ n\times n$ board are written $ n^2$ different positive integers. In each column of the chessboard the cell with the greatest number is colored red. A set $ S$ of $ n$ cells is called admissible if no two cells from $ S$ lie in the same column or row. Prove that the admissible set of cells with the greatest sum of numbers contains at least one red cell.

Suppose, for a contradiction, that the admissible set with the greatest sum has no red cells.  We claim that there exists some set of columns such that if we take the rows which the cells of our admissible set in those columns are in, then we have in each of these rows, exactly one red cell.  If this is the case, then we can switch those red cells in for the other cells and increase the sum, causing a contradiction.  So it suffices to prove that such a set of columns exists.

Order the columns in the following way.  Begin with the first one.  Now go to the column such that the cell we have selected in this second column is in the same row as the red cell in the first column.  Then third, go to the column such that the cell we have selected in that column is in the same row as the red cell in the previous column.  And so on.  Eventually we either have that this process cycles (because we have a finite number of columns), and the columns in that cycle satisfy the desired property.  This completes the proof.

Here is a somewhat random induction argument. Suppose the result holds for $ n\minus{}1$, and suppose for sake of contradiction we have an admissible set $ S$ for the $ n$ case that contains no red cell. Then take any subset $ T$ of $ n\minus{}1$ cells in $ S$, and suppose that the remaining cell in $ S$ is in row $ i$. Then the elements of $ T$ determine a $ n\minus{}1$ by $ n\minus{}1$ sub-array (the cells in $ T$ lie on $ n\minus{}1$ distinct rows and columns), within which at least one element $ s$ is the largest in its column, say $ j$, (restricted to the sub-array) by the inductive hypothesis. Since this cell $ s$ is not red, the red square in column $ j$ lies in row $ i$. Since the choice of $ i$ was arbitrary, it follows that the set of red squares is itself an admissible set and clearly has maximal sum.


Let ABC be a triangle. Prove that:
bc- a 2 /4  \leq  ma 2   \leq  bc+ a 2 /4

Hi Maverick,

do you know the formula 4*ma^2 = 2*(b^2+c^2) - a^2 and its cyclic permutations?
If not, derive it from Stewart's theorem about the length of Cevians (I can explain it if you like).

With this formula the left-hand side is equivalent to (b-c)^2 >= 0, and the right-hand side to a >= b - c which is indeed a triangle inequality.

Cheers!
-Eckard

Yeah, it works perfectly here:
For the first ineq we use (b^2+c^2)/2 >= bc and for the second one we use (b-c)^2<a^2, so I guess the second one is strict.


is there anyone that actually uses this forum in the first place?
and if there are, are any of you from central ohio?
i am from Dublin, OH, even though my location says Columbus. :|

I'm from Pickerington.  We're in the same conference as you in high school sports.

It's nice to know that there at least some Ohioans on AoPS.

From Sycamore meself, down in the South (Cincinnati). Our school wins at trivial contests and sucks at everything else.

well, as long as we're extending from Central Ohio down to Sycamore, I suppose I will mention that I am also from the Cincinnati area and go to Lakota West


[quote="Interval"]( x + 5 ) ( 7 + x ) =
( x + 5 )•( 7 )+( x + 5 )•( x )
is an example of which property?[/quote]

[hide]Naturally, the distributive.

You could even carry it out further using FOIL.[/hide]

This is exactly what I suspected.

Thanks!


According to [url=http://www.wolframalpha.com]wolframalpha[/url], it doesn't factor with integer/rational number coefficients.


function
trigonometry
calculus
derivative
calculus computations

Let $f(x) = {(2x - \pi)^3} + 2x - {\cos x}$ . Then the value of ${\frac d{dx}} {f^{-1}(x)}$ at $x=\pi$ is

$(A){3\pi^2 + 2}$                               ${(B) -2}$
$(C)\frac 1{3\pi^2 + 2}$                     ${(D) \frac 1{3}}$

The correct answer given was $(D)$.

In the days to come I would be putting up other problems of the same test paper.

Let $y = f^{-1} (x)$
Then, $f(y) = x \rightarrow \frac{d f(y)}{dy} \cdot \frac{dy}{dx} = 1$
$\frac{d f(y)}{dy} = 3(2y - \pi)^2 (2) + 2 + \sin(y)$

At $x = \pi \rightarrow \pi = (2y - \pi)^3 +2y - cos(y)$
By inspection one solution is $y = \frac{\pi}{2}$

The derivative at $y = \frac{\pi}{2}$ is $6 (0) + 2 + 1 = 3$

Therefore, $\frac{dy}{dx} = \frac{1}{3} \rightarrow \frac{d}{dx} f^{-1}x = \frac{1}{3}$

Unfortunately, I don't know a good way to find $y$ for $f(y) = \pi$, so all I could use were my eyes.

In the solution provided to us. The question was solved by the same way as you have solved.

Thanks. :)


algebra
polynomial
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let $K\subseteq L$ be a finite extension of fields and let $M$ be an algebraically closed field containing $K$. Prove that the extension $K\subseteq L$ is separable if and only if the tensor product of $L$ and $M$ over $K$ has no nilpotent elements except $0$.

One implication:

Suppose $K\subset L$ is a finite separable extension, and let $K\subset M$ be [i]any[/i] field extension (we do not need algebraic closure). Then, $A=M\otimes_{K}L$ is a product of fields.

Being finite and separable, the extension $K\subset L$ is simple, i.e. of the form $L=K(\theta)$. Let $f$ be the minimal polynomial of $\theta$ over $K$, and let $f=f_{1}\cdot\ldots\cdot f_{t}$ be its decomposition into irreducible factors in $M[X]$. Also let $d$ be the degree of $f$. Note that $f_{i}$ are distinct, since $\theta$ is separable over $K$. Now, $A$ is an $M$-algebra with basis $1,\theta,\ldots,\theta^{d-1}$ (identifying $\theta$ with $1\otimes\theta\in A$), with $f$ being the minimal polynomial of $\theta$. This identifies $A$ with $M[X]/(f)$, which is the direct product of the fields $M[X]/(f_{i}),\ i=\overline{1,t}$.

The other implication:

Let $K\subset L$ be a finite inseparable field extension, and let $M$ be an algebraically closed field containing $K$. Then, $A=M\otimes_{K}L$ is not reduced (i.e. it has non-trivial nilpotent elements).

Assume the contrary. Being a finite-dimensional $M$-algebra, $A$ is Artinian. Since we assumed its Jacobson radical is trivial (its nilradical actually, but the two notions coincide in commutative Artinian rings), it must be a product of field extensions of $M$. The only such field extension is $M$ itself (because $M$ is algebraically closed), so $A$ must be of the form $A=M^{t}$ for some positive integer $t$.

Because $K\subset L$ is inseparable, the trace map $\mbox{tr}_{L/K}: L\to K$ is identically zero. This carries over to the trace map from $A$ to $M$ (here trace is defined as usual: pick a basis $e_{i}$ of $A$ over $M$, and define the trace of $a\in A$ to be the trace of the $M$-linear map $x\mapsto ax$). However, the element $(1,0,\ldots,0)\in M^{t}$ has trace $1$. We have the desired contradiction.


hi all. i'm from india, currently pursuing BSc Mathematics(somewhat like the BS u have in US). mine is a  3 yr course( i'm in the 3rd yr) and the amount of math that we do, we are way behind what ppl in the US and Canada do. almost all the math that i know, ive read on my own since we dont have very good professors here. now my aim is to get in to a decent grad school in the US or Canada, to pursue MS and then PhD both in math. my friend says that i must have 16 yrs of education before entering grad school, whereas when i finish my BSc i'll have only 15. so do universities strictly require ppl to have 16 yrs of education or can i get in? also should i give the GRE subject test? i'm writing the general GRE test on aug 21. i've already written TOEFL and my score is 277/300.

also could any of you give me some info regarding universities elsewhere like Australia,germany etc?

I have never heard of any US university requiring a certain number of years of education before entering graduate studies.  As long as you have a B.S. (or equivalent), you should be considered for admission.

If you are applying to a graduate program that has a corresponding GRE Subject exam, it will almost certainly be required.  In particular, almost any math M.S. or Ph.D. program will require the GRE Mathematics Subject test.

Some Canadian students have run into this problem - not all U.S. graduate programs will accept a 3 year degree as equivalent to a 4 year US degree.  One Canadian student in particular had completed grade 13 in high school then a 3 year degree from a respectable Canadian university.  She had to complete a 2nd undergraduate degree from my US university before being considered for US-based masters/PhD programs in business.

Typically, foreign transcripts will be evaluated with a translation service (even those in English) before being accepted by most US schools.  If the translation service will not accept the degree as equivalent to a 4 year US degree - and many use time on task as a determining factor - then many schools will not accept the degree.

Following up on what DPatrick and gt59 said, yes, the key issue is not how many years you took to get an undergraduate degree (because learners sometimes skip grades in school) but how strong the degree program itself is. Here in the United States, some accredited degree programs are NOT adequate preparation for graduate school, but some are, and that is also true in other countries. So it is correct that your undergraduate degree will be evaluated by the universities to which you apply, to decide whether or not you can be admitted to a graduate program.

Just one more point: it is often not necessary and not expected to get a master's degree before getting a Ph.D. Many Ph.D. programs begin right after the student has completed a (strong) undergraduate degree, and many master's degree programs are terminal degrees--they don't lead on to a Ph.D. So as you look for information about universities (which you can do by searching on university Web sites), look for details like that.

as you say , many MS programmes might not lead to a PhD but thing is that i'm sure that i wont be "strong enough " in math ,to go straight to a phd. so i most def want to do an MS first and then PhD. or do u think even if i'm a little behind i can catch up in the first couple of years say 2?


Let $ a,b,c>0$ satisfying $ a\plus{}b\plus{}c\equal{}1$.Prove that
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}2abc$≥$ \frac{247}{54}$

[quote="SUPERMAN2"]Let $ a,b,c > 0$ satisfying $ a \plus{} b \plus{} c \equal{} 1$.Prove that
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} 2abc$≥$ \frac {247}{54}$[/quote]
The following stronger inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}1.$ Prove that:
\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} 13.5abc\geq5\]

[quote="SUPERMAN2"]Let $ a,b,c > 0$ satisfying $ a \plus{} b \plus{} c \equal{} 1$.Prove that
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} 2abc$≥$ \frac {247}{54}$[/quote]

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\equal{}\sum \frac{c}{a\plus{}b} \plus{} 3$
Prove $ \frac{1\minus{}2q\plus{}3r}{q\minus{}r}\plus{}2r \geq 85/54$

have $ r \geq (4q\minus{}1)/9$
$ 2r^2 \leq \frac{2}{27}r$ =>>


algebra
system of equations
algebra proposed

Find all real solutions $ (a,b,c)$ to the system of equations

$ (a\plus{}b\plus{}c\plus{}2)^2\plus{}2abc\equal{}abc(a\plus{}2)(b\plus{}2)(c\plus{}2)$
$ (a\plus{}1)(b\plus{}1)(c\plus{}1)\equal{}8.$

let $ a\plus{}1\equal{}x,b\plus{}1\equal{}y$ and $ c\plus{}1\equal{}z$ ,from the first equation we get $ x^2y^2\plus{}y^2z^2\plus{}z^2x^2\equal{}x^2y^2z^2\minus{}2xyz\equal{}48$
and we have $ x^2y^2z^2\equal{}64$. and by AM-GM for $ x^2y^2,y^2z^2$ and $ z^2x^2$ we have :
$ 48\equal{}x^2y^2\plus{}y^2z^2 \plus{} z^2x^2 \ge 3(x^4y^4z^4)^{\frac{1}{3}}\equal{} 48$ so $ x^2y^2\equal{}y^2z^2\equal{}z^2x^2\equal{}16$ so $ x^2\equal{}y^2\equal{}z^2\equal{}64/16\equal{}4$  so $ (x\equal{}2 ,y\equal{}\minus{}2,z\equal{}\minus{}2)$ or $ (x\equal{}2,y\equal{}2,z\equal{}2)$ or $ (x\equal{}\minus{}2,y\equal{}2,z\equal{}\minus{}2)$ or $ (x\equal{}\minus{}2,y\equal{}\minus{}2,z\equal{}2).$...


The perimeter of a rectangle is 42 meters. The length of the rectangle is four less than four times its width. What is the number of square meters in the area of the rectangle?

Let the dimensions be $ l$ and $ w.$ We have \[ 2(l\plus{}w)\equal{}42\implies l\plus{}w\equal{}21\] and \[ l\equal{}4w\minus{}4.\] Substitute the value for $ l$ into the first equation to get \[ 4w\minus{}4\plus{}w\equal{}21\implies 5w\equal{}25\implies w\equal{}5.\] Substituting this into the other equation gives \[ l\equal{}4\times5\minus{}4\equal{}20\minus{}4\equal{}16.\] The area is thus \[ 5\times16\equal{}\boxed{80}.\]

The width is x, so the length is 4x-4

combining it we get 5x-4 =  half the perimeter = 21

x = 5

So the width is 5 and the length is 16. Therefore the area of the rectangle is 80.


inequalities
quadratics
function
geometry
inequalities proposed

I found this inequality 20 years ago.
It holds for any $ a_{i}$ and $ b_{i}$ real numbers such that $ a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} a_{3}a_{1} > 0$.
We may write it as $ D \ge 0$, where $ D$ is the discriminant of the quadratic function

$ f(x) \equal{} (a_{1}x \minus{} b_{1})(a_{2}x \minus{} b_{2}) \plus{} (a_{2}x \minus{} b_{2})(a_{3}x \minus{} b_{3}) \plus{} (a_{3}x \minus{} b_{3})(a_{1}x \minus{} b_{1})$.

The inequality yields

(1) $ xa^{2} \plus{} yb^{2} \plus{} zc^{2}\ge 4S\sqrt {xy \plus{} yz \plus{} zx}$

for $ xy \plus{} yz \plus{} zx\ge 0$ and $ a,b,c$ side-lengths of a triangle with area $ S$;

(2) $ (b \plus{} c)^{2}x \plus{} (c \plus{} a)^{2}y \plus{} (a \plus{} b)^{2}z \ge 4\sqrt {abc(a \plus{} b \plus{} c)(xy \plus{} yz \plus{} zx)}$

for $ a,b,c$ and $ x,y,z$ nonnegative real numbers.

[Moderator edit: See
http://www.mathlinks.ro/viewtopic.php?t=90999 for (1),
http://www.mathlinks.ro/viewtopic.php?t=122316 for (2).]

To prove this nice ineq I tried to find two values $m,n$ such that $f(m) \geq 0$ and $f(n) \leq 0$ so that $D$ must be positive, but I cannot find them. A help please.

I would like to see a solution, please and how to deduce the two Vasc ineq from this very nice inequality.

 :blush: Thank you very much.

[quote="manlio"]Any help, please.  :(
Thank you very much.[/quote]
Manlio,
try $f(x_{0})$,
where $(a_{1}x_{0}-b_{1})+(a_{3}x_{0}-b_{3})=0$.

Thank you very much, I will try just now :)

[quote="Vasc"]
The inequality yields

(1) $ xa^{2} \plus{} yb^{2} \plus{} zc^{2}\ge 4S\sqrt {xy \plus{} yz \plus{} zx}$

for $ xy \plus{} yz \plus{} zx\ge 0$ and $ a,b,c$ side-lengths of a triangle with area $ S$;

(2) $ (b \plus{} c)^{2}x \plus{} (c \plus{} a)^{2}y \plus{} (a \plus{} b)^{2}z \ge 4\sqrt {abc(a \plus{} b \plus{} c)(xy \plus{} yz \plus{} zx)}$

for $ a,b,c$ and $ x,y,z$ nonnegative real numbers.[/quote]

I have spent a lot of hours to deduce $ (2)$ but I failed so I would like to have another hint, please.

If I put

$ a_{1} \equal{} bc, a_{2} \equal{} ca , a_{3} \equal{} ab$ and
$ b_{1} \equal{} x, b_{2} \equal{} y, b_{3} \equal{} z$ the inequality yields

((2) $ (b \plus{} c)ax \plus{} (c \plus{} a)by \plus{} (a \plus{} b)cz \ge 2\sqrt {abc(a \plus{} b \plus{} c)(xy \plus{} yz \plus{} zx)}$

for $ a,b,c$ and $ x,y,z$ nonnegative real numbers, but now I cannot go on.

Every help is very appreciated as I am really in difficult. Thank you very much. :(  :(

To prove (2) replace the side-lengths $ a,b,c$ by $ q \plus{} r,r \plus{} p,p \plus{} q$, respectively (Ravi's substitution). Finally, replace $ p,q,r$ by $ a,b,c$.

Thank you very much, Vasc. I will try.  :)

P.S. But in (2) a,b,c are not the sides of a triangle :|

Sorry, Vasc, but my problem is how to prove (1) and what to choose for

$a_{1}, a_{2}, a_{3}$ as I suppose that
$b_{1}=x, b_{2}=y, b_{3}=z$

Thank you very much.  :oops:

[quote="manlio"]Sorry, Vasc, but my problem is what to choose for
$a_{1}, a_{2}, a_{3}$ as I suppose that
$b_{1}=x, b_{2}=y, b_{3}=z$
Thank you very much.  :oops:[/quote]
Choose
 $a_{1}=-a^{2}+b^{2}+c^{2}$, $a_{2}=a^{2}-b^{2}+c^{2}$, $a_{3}=a^{2}+b^{2}-c^{2}$.

Thank you very much.

Now I will enjoy myself to deduce your supernice inequality.

Many thanks again.  :)


wah... Level 4: 74... Level 6: FAILED (肥佬佐) ... :oops:  :P
didn't do level 8...

4级考了个68，六级不及格……八级与他人合作才得到50。 :blush:


Let $ n$ be a positive integer. Find the number of $ 2n$-digit positive integers $ a_1a_2\ldots a_{2n}$ such that
(i) none of the digits $ a_i$ is equal to $ 0$, and
(ii) the sum $ a_1a_2 \plus{} a_3a_4 \plus{} \ldots \plus{} a_{2n \minus{} 1}a_{2n}$ is even.

Followup: What if we mandate that $ a_1a_2 \plus{} a_2a_3 \plus{} a_3a_4 \plus{} \ldots \plus{} a_{2n \minus{} 1}a_{2n}$ be even instead?

[hide="my progress"]

Disclaimer: the numbers may be inaccurate

I got that the solution to the first version was
\[ \sum_{k\equal{}0}^{n} \binom{2n}{2k} 25^k 56^{n\minus{}k} \equal{} \frac{1}{2} \left( \left(5\plus{}\sqrt{56}\right)^{2n} \plus{} \left(5\minus{}\sqrt{56}\right)^{2n} \right)
\]

The second version seems harder. I was able to obtain some recursive relations but I couldn't solve them. [/hide]

[quote="Phelpedo"]Let $ n$ be a positive integer. Find the number of $ 2n$-digit positive integers $ a_1a_2\ldots a_{2n}$ such that
(i) none of the digits $ a_i$ is equal to $ 0$, and
(ii) the sum $ a_1a_2 \plus{} a_3a_4 \plus{} \ldots \plus{} a_{2n \minus{} 1}a_{2n}$ is even.

Followup: What if we mandate that $ a_1a_2 \plus{} a_2a_3 \plus{} a_3a_4 \plus{} \ldots \plus{} a_{2n \minus{} 1}a_{2n}$ be even instead?

[hide="my progress"]

Disclaimer: the numbers may be inaccurate

I got that the solution to the first version was
\[ \sum_{k \equal{} 0}^{n} \binom{2n}{2k} 25^k 56^{n \minus{} k} \equal{} \frac {1}{2} \left( \left(5 \plus{} \sqrt {56}\right)^{2n} \plus{} \left(5 \minus{} \sqrt {56}\right)^{2n} \right)
\]
The second version seems harder. I was able to obtain some recursive relations but I couldn't solve them. [/hide][/quote]

I think your result is wrong. For example, it gives $ 81$ for $ n\equal{}1$, which is wrong, since we have exactly $ 56$ numbers $ \overline{a_1a_2}$ such that $ a_1a_2$ is even ($ 36$ numbers where $ a_1$ is even and $ 20$ numbers where $ a_1$ is odd).

For a number $ x\equal{}\overline{a_1a_2\ldots a_{2n}}$, let us call $ p(x)\equal{}a_1a_2\plus{}a_3a_4\plus{}\ldots\plus{}a_{2n\minus{}1}a_{2n}$.

If we call $ S_n$ the required number (count of numbers $ x$ without 0, with length $ 2n$ and such that $ p(x)$ is even), we can say that $ S_{2(n\plus{}1)}$ is :
The count of numbers with length $ 2n$ and $ p(x)$ even ($ S_n$) multiplied by $ 56$ (the number of possibilities for $ a_{2n\plus{}1}a_{2n\plus{}2}$ even).
Plus the count of numbers with length $ 2n$ and $ p(x)$ odd ($ 9^{2n}\minus{}S_n$) multiplied by $ 25$ (the number of possibilities for $ a_{2n\plus{}1}a_{2n\plus{}2}$ odd).

And so $ S_{n\plus{}1}\equal{}56S_n\plus{}25(81^n\minus{}S_n)$ and $ S_1\equal{}56$

This formula is quite easy to solve (compute first $ T_n\equal{}\frac{S_n}{81^n}$) and gives :

$ \boxed{S_n\equal{}\frac{31^n\plus{}81^n}{2}}$


In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:
$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad\textbf{(C)}\ \frac{\sqrt{5}\minus{}1}{2} \qquad\textbf{(D)}\ \frac{1\minus{}\sqrt{5}}{2} \qquad\textbf{(E)}\ \frac{2}{\sqrt{5}}$

Let $ a_n\equal{}a_1q^{n\minus{}1} (a_1>0, q>0)$,

$ a_n\equal{}a_{n\plus{}1}\plus{}a_{n\plus{}2}\Rightarrow q^2\plus{}g\minus{}1\equal{}0\Rightarrow q\equal{}\frac{\sqrt 5\minus{}1}{2}$.


Given a circle, its centre, and a point outside the circle, construct the tangents from this point to the circle using only a straightedge and a compass.

I feel stupid, because I definitely know how to do this (or otherwise why it's impossible) - but I just can't remember tonight.

Hmm.. how about this:

[hide]

If P is the point and O is the centre of the circle, and T is that point of tangency that you want, then OTP is 90 degrees, so you can construct midpoint of OP and draw a circle with diameter OP..

[/hide]

OF COURSE! Thank you so much. Now I feel even more stupid... :?

In the same context, only MUCH harder (I completely forgot how to solve this one, but it's a classical problem):

You have a circle. Find its center using only the compasses. Notice that there is no straightedge...

Here's one I like.  You know what the operation circle inversion is, right?  So, construct an inversion for any point outside (or inside -- it doesn't really matter, it's the same thing in reverse, roughly) a given circle.  (Both straighedge and compass may be used here.)

[hide]Make two chords, and construct their perpendicular bisectors.  The point of intersection is the center of the circle.[/hide]

Bookworm - you have no straight edge. Ie you can't draw straight lines, which include chords and perpendicular bisectors.

Yup, no straight edge. You just draw lots of circles and circle arcs, but I forgot the solution, as mentioned before  :(

Do a Google search for Mascheroni construction. That's what compass-only constructions are called. Straightedge-only constructions are called Steiner constructions.


Let $ P\equal{}(1\minus{}\frac{1}{2})(1\minus{}\frac{1}{3})(1\minus{}\frac{1}{4})...(1\minus{}\frac{1}{n})$. What is the value of $ P$ if $ n\equal{}2007$? Express your answer as a common fraction.

This becomes \[ \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2006}{2007} \] Notice that most of the terms are eliminated, leaving us with $ \boxed{\frac{1}{2007}}$.


[b]a)[/b] Let $n_1,...,n_{39}$ be the consecutive integers in that order.
Among the first 20, there are two such that their unit digit is 0. They differ by 10 and at least one, say $n_k$, has its second digit from the right which is not 9 (thus it is at most 8). Then $n_k \leq n_{20}$.

Let $s(x)$ be the sum of the digits of $x$. Then $s(n_{k+i}) = s(n_k) + i$ for $i=0,1,...,9$.
moreover $n_{k+10}$ has unit digit 0, and the digit before the last is one more than the one of $n_k$, thus $s(n_{k+10}) = s(n_k)+1$, and $n_{k+10} \ leq n_{30}$.

It follows that $s(n_{k+10+i}) = s(n_k) + i + 1$ for $i=0,...,9$.
It follows that $s(n_j)$ takes each integer value from $s(n_k)$ to $s(n_k)+10$ which are 11 consecutive integers. One of them is divisible by 11.

Could somebody explain the last line of Pierre's proof to me
Thanks

I used the fact that among any 11 consecutive integers, there is one which is divisible by 11, since they give all the residues $0,1,2,...,10$ modulo 11, in some order.

Pierre.

[quote="Iris Aliaj"]a) Prove that it is possible to choose one number out of any 39 consecutive positive integers, having the sum of its digits divisible by 11;

b) Find the first 38 consecutive positive integers none of which have the sum of its digits divisible by 11.[/quote]

Solution to b)999981,999982,…,1000018.


What is the probability that if 6 dice are rolled, the numbers 1, 2, 3, 4, 5, and 6 will appear? (In any order)

you have 6 spots. Each spot has 6 different numbers that can appear. so $ 6^6$ total possibilities. The number of successes are the different arrangements of a 1, 2, 3, 4, 5, and 6. This is $ 6!$. So success over total = $ \frac{6!}{6^6}\equal{}\frac{5}{324}$


I know this question has been posted previously on this site but I want to rerun it as much more people joined the forum.

I think it was definelty discovered and not invented but have a hard time convincing someone else. He is trying to convince me for example (even though this has little to do with the question at hand) that 1+1 doesnt always equal 2, I dont think thats true. I mean in regular arithmetic, not boolean stuff, the concept cant change if we r on earth or different universe. sorry for combining the topics like that, they have only some relevance to each other.

I added the poll now so that I can prove by your votes that math was discovered, I know that is not a prove but nevertheless.

I think it's discovered because it has always existed. Nobody invented the concept of numbers. There were always animals and stuff, and because they existed in some quantity, there were numbers...  :roll: it's not easy to explain.

Mathematics was in use from the very first human being (or caveman, or something :)) It just advanced over the years. So, in a way, it was neither invented or discovered, just integrated into life automatically.

--- This text is written from an atheistic view. I'm not antireligious this is what I believe.

 :idea:  I guess it was invented. Mathematics is all about defining stuff. If some person would ask a mathematician why 1+1=2, most of the times the mathematician would either not know the answer or just start a monologue without giving an answer. But the answer is quite simple: because we defined 1+1=2.

 :!:  From the beginning of human there was mathematics. Even before there was language, the first mathematics already existed. It was not even human's first invention, the animals we came from already invented it. There is no part in mathematics which it's "true" or something like that.

 :arrow:  Although for us, the first counting and dividing and such things seem really logical, those animals/humans might have had some other way to express things and their quantity. Maybe we can't imagine, but we're to much indoctrinated with mathematics, we don't know any better.

I dont agree, because I think that u cant calculate stuff in any other manner, even though others like the guy I was talkin aobut thinks u can.

This is a great question, since that really is a good mystery...

Maybe an ancient genius thought of the concept of numbers, and contributed all his life to create the language of mathematics...though that might date back to about 5 million years ago.

what r u talking about? r u just spamming, what u r saying has no logic whatsover. Anyways, math was discovered and not invented period. I dont believe that this is subject to argument and all mathematicians will agree. At least they should.

Then why did you start this topic if you think there is no discussion possible about the right answer? Although, you said you didn't think there is another way to calculate stuff, but that's exactly what I said you'd answer. Of course we don't know. But there might be a manner. Since we have no proof of any God, this does not mean there can't be any or something. If you don't know any proof that something is true, this does not make something false at all. As long as we don't know, we can't be sure about it but we have to guess or do some nice arguments.

I think math just came to people. No one actually discovered it. This topic is sort of pointless, I don't know who's really spamming here  ;)

Math is more recent than you think. About 25,000 years ago, only the most intelligent humans could understand the concept of numbers. "one" is such a strange abstraction for someone not knowing it. How can one person be "one" and another also be "one" if they are different people? People learn how to count by using their fingers, but what happens after someone runs out of fingers? These were very difficult issues to settle for our ancestors.

But they still were solvable in the imaginaruy form, that makes it a discovery rather then an invention.

And they could think in quantities. Five bannanas is better than one bannana... they could grasp the concept of inequalities.

So u agree with me right? I mean how can math be invented, it isnt logical to say that right?

Anyone know where I can find the last time this thread existed, I am trying to prove it to this guy about math being discovered, can anyone help?

People, especially IMOers and USAMOers, can u please vote, the last time we did this was a long time ago.

Hmmm...I think we agree completely!  :huh:

I now just make the one extra claim: speaking of mathematics outside of this context is meaningless. Proofs that we consider perfectly correct can be wrong. If all of humans died out, and then a new intelligent species arose, their mathematics could be completely different.

Crap, OK, 14 hour day does not make for productive thinking right now. I will see if I still believe this in the morning. (I realized that I mixed to layers here: proofs can be wrong is a statement within a formal system, and the second is about the foundations of the formal system itself).

[quote="HilbertThm90"]
Proofs that we consider perfectly correct can be wrong. If all of humans died out, and then a new intelligent species arose, their mathematics could be completely different.
[/quote]
The first claim may be true about each particular proof but is very hard to believe that all or even the majority of them are wrong. We try to do our best to eliminate possible errors, so we have some (but, of course, not 100%) confidence in our verbal arguments.

The second claim deserves a separate thread and it is going to be a very long story which will range from the question of which mathematical statements can be set up as physical experiments and observations, so they could be viewed as an empirical truth, to the question whether you can design something deserving to be called mathematics without any concept of "formal deductive proof" in it at all, to the question whether our true-false logic with its modus ponens is the only or even the most natural way of argumentation to... To be honest, I haven't thought of it much, so I can only pose some relevant questions and try to convince you that everything is not as simple here as it might seem. Perhaps, there is some good book on the subject we can read before plunging into such a discussion but I am completely unaware of it.

*12 year bump noises*

As a Platonist, I generally think that all mathematics has a Platonic existence independent of anyone's minds, so I think that mathematics is discovered instead of invented.

[quote=Kalle]I don't know formal mathematics, so I can't really argue this. So let me just ask: AFAIK we don't know if arithmetic with real numbers is contradiction-free. Doesn't this interfer with the absolute truths of mathematics?[/quote]

@Kalle Maybe arithmetic with real numbers or whatever is contradictory. In this case it was never the "right" theory of mathematics.

Math was definitely discovered. A lot of animals can count as well...

I love how this was bumped after 12 years

[quote=hexafandra]I feel like it's a little bit of both. It was invented because I don't think most animals know what is 5-3. Rather I think that we invented the characters and discovered the actually math and counting part.[/quote]

I think animals know what 5-3 is, but when you show them the characters 5 - 3, they won't know. If hyenas can tell that they can't fight two lions but can fight one, they clearly can do math.

Yes that's what I am saying they know what 5-3 is, but they don't know the charecters and so they won't be able to do our style math. Thus we invented the characters and discovered math.

I think we invented math, because it's a non-physical thing. If we were to "discover" it, therefore it'd mean that other species would be able to [i]by themselves[/i] without assitance.

I think the concept was discovered, but the notation was invented.

I'd have to disagree, since just like English, Japanese, and Spanish, mathematics is a language. And like all languages, it was invented as a way to communicate ideas and way to do things without having describe it in a complex way

Yet animals can understand math as well. I think you're (@above) talking about the characters in math, not the math in general.

Discovered. Because what if one caveman (as an example LOL) was like,[i] on ston pluz on ston ekualz to stons[/i],
and figured out the first 1+1 = 2?

Half half in my opinion

Math was invented, for math is our interpretation of concepts, and understanding fundamental things in the world. For example, $+$ is the interpretation of an addition, or $\div$ is the interpretation of splitting things in groups of $x$. We also invented many systems, like the metric system

However, math was discovered, since Math is the language of the universe, and our interpretations are to interpret the mathematical discoveries we made, such as we discovered addition, we discovered what a regular hexagon looks like and we discovered the area formulas.

In conclusion, Math is sort of half discovered half invented.


Let $ ABC$ be a triangle, and let $ M$, $ N$ be points on $ \overline{AB}$ and $ \overline{AC}$, respectively, such that the tangents from $ B$ and $ C$ to the circumcircle of $ \triangle AMN$ have the same length. Prove that the midpoints of $ \overline{AB}$, $ \overline{AC}$, $ \overline{BN}$, and $ \overline{CM}$ are cyclic.

By the way, I'm not very good with latex. How do you express $ AB$ as a line segment (with the bar over it)?

After a few minutes of experimenting...

Try $ \overline{AB}$ ;)

[code]\overline{AB}[/code]

Hey, you're from Missouri too.

Thanks Amadeus. Maybe if I could actually get my signature to work you'd know who I am.

Oh, I know who you are :o

...from your email address :D

As for the original problem, would you happen to have a diagram?

[quote="ColbertCo"][color=darkred]Let $ ABC$ be a triangle, and let $ M\in (AB)$, $ N\in (AC)$ be points such that the tangents from the points $ B$ and $ C$ to the circumcircle $ w$ of $ \triangle AMN$ have the same length. Prove that the midpoints $ F\in (AB)$, $ E\in (AC)$, $ U\in (BN)$, $ V\in (CM)$ of the mentioned segments are cyclically. [/color][/quote]
[color=purple][size=150]Welcome on M.L.S., Missouri, Yakima Valley "te salutant" ![/size][/color]

[color=darkblue][b]Proof.[/b] Denote the midpoint $ D$ of the side $ [BC]$. Observe that $ U\in DF$ and $ V\in DE$. The tangents from the points $ B$ and $ C$ to the circle $ w$ have the same length $ \Longleftrightarrow$ $ p_{w}(B)=p_{w}(C)$, where $ p_{w}(X)$ is the power of the point $ X$ w.r.t. the circle $ w$. Thus, $ {BM\cdot BA=CN\cdot CA}\ \ (1)$. Therefore, the points $ F,E,U,V$ are cyclically $ \Longleftrightarrow$ $ DU\cdot DF=DV\cdot DE$ $ \Longleftrightarrow$ $ \frac{1}{2}\cdot CN\cdot \frac{1}{2}\cdot CA=\frac{1}{2}\cdot BM\cdot\frac{1}{2}\cdot BA$ $ \Longleftrightarrow$ $ CN\cdot CA=BM\cdot BA$, i.e. the true relation $ (1)$.[/color]


Triangle $ ABC$ obeys $ AB = 2AC$ and $ \angle{BAC} = 120^{\circ}.$ Points $ P$ and $ Q$ lie on segment $ BC$ such that
\begin{eqnarray*}
AB^2 + BC \cdot CP = BC^2 \\
3AC^2 + 2BC \cdot CQ = BC^2
\end{eqnarray*}
Find $ \angle{PAQ}$ in degrees.

[color=darkblue]
[b]Let $ AB\equal{}2a, AC\equal{}a$. By cosine theorem in $ ABC$ we have:
$ BC^2\equal{}a^2(1\plus{}4\minus{}2\cos{120})\equal{}7a^2$.
Then:
$ AB^2 \plus{} BC \cdot CP \equal{} BC^2 \Leftrightarrow 4a^2\plus{}\sqrt{7}a\cdot CP\equal{}7a^2 \Leftrightarrow \boxed{CP\equal{}\dfrac{3a}{\sqrt{7}}}$
$ 3AC^2 \plus{} 2BC \cdot CQ \equal{} BC^2 \Leftrightarrow 3a^2\plus{}2\sqrt{7}a\cdot CQ\equal{}7a^2 \Leftrightarrow \boxed{CQ\equal{}\dfrac{2a}{\sqrt{7}}}$.
So points lie in order $ C\minus{}Q\minus{}P\minus{}B$ and $ \boxed{CQ\equal{}\dfrac{2a}{\sqrt{7}}, PQ\equal{}\dfrac{a}{\sqrt{7}}, PB\equal{} \dfrac{4a}{\sqrt{7}}}$.
By Stewart for $ AQ$ in $ ABC$ we have:
$ AQ^2\cdot BC\equal{}AC^2\cdot QB\plus{}AB^2\cdot CQ\minus{}BC\cdot CQ\cdot BQ \Leftrightarrow$
$ \Leftrightarrow  AQ^2\cdot \sqrt{7}a\equal{}a^2\cdot \dfrac{5a}{\sqrt{7}}\plus{}4a^2\cdot \dfrac{2a}{\sqrt{7}}\minus{}\sqrt{7}a\cdot \dfrac{10a^2}{7} \Leftrightarrow \boxed{AQ\equal{}\sqrt{\dfrac{3a}{7}}a}$.
By Stewart for $ AP$ in $ ABC$ we have:
$ AP^2\cdot BC\equal{}AC^2\cdot PB\plus{}AB^2\cdot CP\minus{}BC\cdot CP\cdot BP \Leftrightarrow$
$ \Leftrightarrow  AP^2\cdot \sqrt{7}a\equal{}a^2\cdot \dfrac{4a}{\sqrt{7}}\plus{}4a^2\cdot \dfrac{3a}{\sqrt{7}}\minus{}\sqrt{7}a\cdot \dfrac{12a^2}{7} \Leftrightarrow \boxed{AP\equal{}\dfrac{2a}{\sqrt{7}}}$.
By cosine theorem in $ PAQ$ we have:
$ a^2(\dfrac{3}{7}\plus{}\dfrac{4}{7}\minus{}4\dfrac{\sqrt{3}}{7}\cos{\angle{PAQ}})\equal{}\dfrac{a^2}{7}$
$ 7\minus{}4\dfrac{\sqrt{3}}{7}\cos{\angle{PAQ}}\equal{}1$
$ \cos{\angle{PAQ}}\equal{}\dfrac{\sqrt{3}}{2} \Rightarrow \boxed{\angle{PAQ}\equal{}30^\circ.}$[/b]

 :!: Another problem for this triangle:
If $ M$ is midpoint of $ BC$ prove that $ AM\perp AC$
[/color]

[quote="worthawholebean"] [color=darkred]The triangle $ ABC$ obeys $ \begin{array}{c} AB = 2\cdot AC \\
\ m(\widehat {BAC}) = 120^{\circ}\end{array}$ . Points $ \{P,Q\}\subset [BC]$

such that $ \begin{array}{c} AB^2 + BC \cdot CP = BC^2 \\
\ 3AC^2 + 2BC \cdot CQ = BC^2 \end{array}$ . Find $ m(\widehat{PAQ})$ in degrees .[/color] [/quote]

[quote="Ahiles very nicely "][color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. $ \boxed {\ b = 1\ }$ . Thus $ \boxed {\ c = 2\ }$ . Apply the [b]generalized Pythagoras' theorem[/b] to $ [BC]$

in $ \triangle ABC$ : $ a^2 = b^2 + c^2 - 2bc\cos A$ $ \implies$ $ \boxed {\ a = \sqrt 7\ }$ . The metrical relations of hypothesis become :

$ \|\begin{array}{ccc} PC = \frac {a^2 - c^2}{a} & \Longleftrightarrow & \boxed{\ PC = \frac {3}{\sqrt 7}\ } \\
 \\
QC = \frac {a^2 - 3b^2}{2a} & \Longleftrightarrow & \boxed {\ QC = \frac {2}{\sqrt 7}\ }\end{array}$ . Observe that $ CP > CQ$ , i.e. $ Q\in (CP)$ and $ \begin{array}{c} \boxed {\ PQ = \frac {1}{\sqrt 7}\ } \\
 \\
\boxed {\ PB = \frac {4}{\sqrt 7}\ }\end{array}$ .

Apply the [b]Stewart's relation[/b] in $ \triangle ABC$ to the cevians :

$ \begin{array}{cccc} AP\ : & c^2\cdot PC + b^2\cdot PB = PA^2\cdot a + a\cdot PB\cdot PC & \implies & \boxed {\ PA = \frac {2}{\sqrt 7}\ } \\
 \\
AQ\ ; & c^2\cdot QC + b^2\cdot QB = QA^2\cdot a + a\cdot QB\cdot QC & \implies & \boxed {\ QA = \frac {\sqrt 3}{\sqrt 7}\ }\end{array}$

Apply the [b]generalized Pythagoras' theorem[/b] to the side $ [PQ]$ in the triangle $ ABC$ :

$ PQ^2 = AP^2 + AQ^2 - 2\cdot AP\cdot AQ\cdot\cos \widehat {PAQ}$ $ \implies$ $ \cos\widehat {PAQ} = \frac {\sqrt 3}{2}$ $ \implies$ $ \boxed {\ m(\widehat {PAQ}) = 30^{\circ}\ }$ .

[b]Remarks.[/b]

$ 1\blacktriangleright$ Observe that $ AC^2 = QA^2 + QC^2$ $ \implies$ $ QA\perp QC$ .

$ 2\blacktriangleright$ Denote the midpoint $ M$ of $ [BC]$ . Apply the [b]theorem of median[/b] to $ AM$ in $ \triangle ABC$ :

$ AM^2 = \frac {b^2 + c^2}{2} - \frac {a^2}{4}$ $ \implies$ $ AM = \frac {\sqrt 3}{2}$ . Since $ MC^2 = MA^2 + MC^2$ $ \implies$ $ AM\perp AC$ .[/color][/quote]

Remark:

First relation is equivalent to: $AB^2=BC\cdot BP$, i.e. $AB$ is tangent to circle $(ACP)$ and $\angle APC=60^\circ$, and after calculation we see that $AQ\bot BC\ (AB^2-AC^2=BQ^2-CQ^2)$, so easily $\angle PAQ=30^\circ$.

Best regards,
sunken rock

Small correction to previous solution:
2nd given equation is equivalent to: $3AC^2=BC^2-2BC\cdot CQ\iff AB^2-AC^2=BC(BC-2CQ)\iff$  $AB^2-AC^2=(BQ+CQ)(BQ-CQ)\iff AB^2-AC^2=BQ^2-CQ^2$, hence $AQ\bot BC$, thus avoiding annoying calculations.

Best regards,
sunken rock


Calculate:

$1^k+2^k+\dots+n^k=?$
$(\ k\in \mathbb{N} \ )$

In general, there isn't really any nice "closed formula" (well, except for some small values of $k$). You can write down some sort of recurrence relation though.

I am not sure about whether there is a nice general formula for this sum, but I know that there is a general method to find it. Here is the brief idea.
First let us begin with the following formulas:
\[1+1+1+...+1=n\]
\[1+2+3+...+n=\frac{1}{2}n(n+1)\]
\[1+3+6+...+\frac{1}{2}n(n+1)=\frac{1}{3!}n(n+1)(n+2)\]
\[1+4+10+...+\frac{1}{3!}n(n+1)(n+2)=\frac{1}{4!}n(n+1)(n+2)(n+3)\]
or the general formula is:
\[C_r^r+C_r^{r+1}+C_r^{r+2}+...+C_r^{r+n-1}=C_{r+1}^{r+n}\]  (this is an arithmetic sequence of r th order)
By some knowledge in difference (of which I don't remember the exact stuff), we can always express $p^k$ in terms of $C_1^p, C_2^{p+1}, C_3^{p+2}, ...$. When p takes value from 1 to n and the sum is taken, we get the formulas above, which together form the formula for $\sum_{p=1}^n p^k$.
(Sorry for my poor English  :blush:. Hope you can understand.)


function
inequalities
logarithms
real analysis
real analysis solved

Consider $f(a,b)=a\ln(1+a)+e^b-1-ab-b.$

Then $\frac{\partial f}{\partial b}=e^b-a.$

Assume $a>1$ and consider $f$ as a function of $b$. It achieves its minimum when $b=\ln a$ and that minimum value is $a\ln\left(1+\frac1a\right)+a-1-\ln(a).$ Since the first term is clearly positive  and since $\ln a\le a-1$ (the tangent line), this minimum value is positive, hence the function is always positive.

For $0< a \le1$, $\frac{\partial f}{\partial b}>0$ for all $b>0$ and the minimum occurs when $b=0.$ This minimum value is $a\ln(1+a)$ and is positive. Hence $f(a,b)>0$ for all $(a,b)$ in the first quadrant.

[quote="kunny"]Let $a,b$ be positive real numbers.Prove that following inequality.
\[a\ln (1+a)+e^b>1+ab+b\][/quote]
Whatis the problem? I don't know. ;)
We can write $(LHS-RHS)'_b=e^b-(a+1)$, so $b_{\min}=\ln(a+1)$. But then $LHS-RHS=a-\ln(a+1)\geq 0$.
That's all.

Thank you for your replies,Kent merryfield and Myth.

This problem can be solved in using Young's inequality +$\alpha$

kunny

I will post another problem which will be become hint for this problem.

kunny


I'm writing from some proxy because I have the following problem:

Whenever trying to open http://www.mathlinks.ro (or similar: AOPS) on the normal way, nothing loads and after some time the message of a timeout while connecting occurs.
This happens only on ML, all other sides I tested worked well. But when using proxy servers it works.
I will try to change my IP later on, but at least I find this curious... (Am I banned¿)

I hope this can be solved or at least explained to me.
Thanks.

Edit:
changing my standard IP worked now.

You are not banned :)
However I fail to see how this can happen from a certain IP (page not loading) and then loading from another IP. If you were banned a message stating so would be seen.


calculus
integration
inequalities
real analysis
real analysis unsolved

I add $ f$ is continuous

$ \int\int_{[0;1]^2}f(x)f(y)(x \minus{} y)(f(y) \minus{} f(x))dxdy \geq 0$

with Fubini you get it

My solution use Riman's sum.
Sorry Yuriy Solovyov, how you get
\[ \textbf{(i)}\int_{0}^{1}xf(x)^2dx\leq\int_{0}^{1}xf(x)dx \int_{0}^{1}f(x)dx\]


Is there any solution of
$x^{2}+y^{2}+z^{2}+u^{2}+v^{2}=xyzuv-65$
if $x,y,z,u,v$ are integeres greater than 1998?

I can't find the solution anywhere.... Can anyone help ??

we have $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55=120-65=5!-55$ and we saw that if $(x,y,z,u,v)$ is a solution, then $(yzuv-x,y,z,u,v)$ is also a solution. So, we can obtain $x,y,z,u,v$ greater than 1998.

(Excuse me if my english is bad, I am french...)

One such solution is $(1,2,3,4,5).$ Let us consider one solution $(x,y,z,u,v)$ with $x-min$ or one of the $min.$ Then since $x^{2}-x(yzuv)+y^{2}+z^{2}+u^{2}+v^{2}+65=0$ and $yzuv$ and $y^{2}+z^{2}+u^{2}+v^{2}+65$  are positive integers, then the equation has other solution $x'$ such that $x+x'=yzuv$ and therefore $x'>\max \(y,z,u,v \).$ Thus we have another solution $(y,z,u,v,x')$ with bigger maximal element. Thus proceeding we can have a solution with big enough larger component.


Consider a sequence $a_n$ with $a_2=1$,$a_3=3$, and the following recurrence holds:
$a_n = (\sum_{m=1}^{r} \binom{n}{m} \bullet \binom{n-2}{m-1} \bullet a_m \bullet a_s) + f(n)$, where $r$ is the largest integer less than or equal to $\frac{n-1}{2}$, $s=n-m$, $f(n) = 0$ when n is odd, but $f(n) = \frac{\binom{n}{\frac{n}{2}} \bullet \binom{n-2}{\frac{n-2}{2}} t^2}{2}$ when n is even, where $t = a_r$ where $r = \frac{n}{2}$. .
Explicitly, determine $a_n$.

c'mon people, this problem is modified from the 05 shortlist ...
that should be enough motivation for people to solve it !! ..


geometry
parallelogram
geometry proposed

[quote="Fermat -Euler"]$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square[/quote]

Erm this was the easiest INMO problem ever, I think.
We prove that the triangles QKB and QMD are congruent.
 which implies that MD = KB and so AB=CD
 also AD is parallel to CD as a pair of alternate interior angles are equal.From the above congruency.
  So,ABCD will become a parallelogram.
With this we can show that the angles QMD,QMC,QMA,QMB are all equal to $\pi/2$(this is a direct result of congruency) and so
  all the angles of the parallelogram will be $\pi/2$  and now the quadrileteral becomes a rectangle. As by congruency LK/LM=1,BC/CD=1, i.e BC=CD .
 Now ABCD becomes a square,and we are done.
    Ashwath :D


calculus
integration
inequalities
function
real analysis
real analysis unsolved

let f(x) be a continuous function in [1,2] satisfying $ \int_1^x f^2(t)dt \leq \frac {x^3 \minus{} 1}3$ for all $ x \in [1,2]$
prove that $ \int_1^2 f(x)dx \leq \frac32$

This is trickier than it seems. Some observations with no solution:

1. The simplest application of the Schwarz-Buniakovskii inequality shows that
\[ \int_1^2f(x)\,dx\le\sqrt{\frac73}.\]
But $ \sqrt{\frac73}$ is larger than $ \frac32,$ so this is no solution.

2. The condition can be written as $ \int_1^xf^2(t)\,dt\le\int_1^xt^2\,dt.$

Starting from there, it's not hard to see that if $ f(x)\equal{}x$ the equality becomes an inequality.

3. If you let $ f(x)\equal{}x\plus{}g(x),$ we can rephrase what we need as this:

If $ \int_1^x2tg(t)\plus{}g^2(t)\,dt\le0\,\forall x\in [1,2],$ show that $ \int_1^2g(x)\,dx\le 0.$

there is an easier problem:
 for f is continuous in [1,2] and $ \int_{x_1}^{x_2} f^2(t)dt \leq \frac{x_2^3\minus{}x_1^3}3$ for all $ x_1, x_2 \in [1,2]$.
prove that $ \int_1^2 f(t)dt \leq \frac32$
with solution: from the first we conclude $ |f(x)| \leq x$ for all $ x \in [1,2]$ and get the desire.
but i haven't found the solution of the first yet.

Here's one idea: write the function $ \mathbf{1}_{[1,2]}$ (the function which is identically one on $ [1,2]$ and zero elsewhere) as a positively weighted sum/integral of the functions $ h_x(t)\equal{}2t\cdot \mathbf{1}_{[1,x]}$ The appropriate weighting is $ \frac14\delta(x\minus{}2)\plus{}\frac1{2x^2}$, so $ \int_1^2 1\cdot g(t)\,dt\equal{}\int_1^2\frac1{2x^2}\cdot\int_1^x 2tg(t)\,dt\,dx\plus{}\frac14\int_1^2 2tg(t)\,dt$ for any $ g$.

Now, apply this to $ g$ as defined by Kent. Since $ g^2\ge 0$, we have that $ \int_1^x 2tg(t)\,dt\le 0$ for all $ x\in [1,2]$. That makes both terms on the left-hand side nonpositive, and hence $ \int_1^2 g(t)\,dt\le 0$ as well.

A generalization, based on the ideas of my previous post:

Suppose $ g$ is a nonnegative increasing function on $ [a,b]$, and $ \int_a^x f^2(t)\,dt\le \int_a^x g^2(t)\,dt$ for all $ x\in [a,b]$. Then $ \int_a^b f(t)\,dt\le \int_a^b g(t)\,dt$.

thank jmerry,
here is the complete solution of the problem I posted:
$ \int_1^x f^2(t)dt \leq \int_1^x t^2dt
\Rightarrow (\int_1^x tf(t)dt)^2 \leq (\int_1^x t^2 dt)(\int_1^x f^2(t)dt) \leq (\int_1^x t^2dt)^2$
$ \Rightarrow \int_1^x tf(t)dt \leq \int_1^x t^2dt$
$ \Rightarrow F(x)\equal{}\int_1^x t[t\minus{}f(t)]dt \geq 0$ for all $ x \in [1,2]$
$ \int_1^2 (x\minus{}f(x))dx\equal{} \int_1^2 \frac1x [x(x\minus{}f(x)]dx\equal{} \frac{F(x)}x|^2_1\plus{}\int_1^2 \frac{F(x)}{x^2}dx\equal{}\frac{F(2)}2\plus{}\int_1^2 \frac{F(x)}{x^2}dx \geq 0$
$ \Rightarrow \int_1^2f(x)dx \leq \int_1^2xdx\equal{}\frac32$
I think your problem you've posted can be solve in the way similar to that, but the condition is "f is continuous"

You don't need $ f$ to be continuous. Everything works for any measurable $ f$.


I can't believe Richard wanted my Patriots to lose...

I angry. He wants my favorite football team to lose.  :mad:

[quote="GameBot"]The Best Football Weekend Ever

We find the probability that the best football weekend ever occurs.[/quote]

What is the best football weekand?

Watch the videos.  Here's the link to all of them that are discussed in this forum:  http://www.artofproblemsolving.com/Resources/videos.php?type=counting


we say that a set of convex figures has the property $k$,if we can part them into $k$ disjoint sets s.t every set is intersectional(set $X$ is intersectional if there is a point that is belonged to every element of $X$.)

now suppose that we have a set $X$ of some  convex figures in $R^{n}$.is there a $m$ that satisfies the following statement?
if every $m$subset of $X$ has the property $k$,then the set $X$ has the property $k$?

note that $m$ should only related in $k$ and $n$.

note that if $k=1$ then $m=n+1$ works.this is indeed helly theorem.


[color=green]Grandma has a chandelier containing [b]n[/b] identical candles. She lights up the candles for [b]n[/b] consecutive Saturdays in the following manner:

the first Saturday, she lights up one candle for one hour....

the second Saturday, she lights up two candles, conveniently chosen, for one hour....

and continues in the same fashion, increasing the number of candles lighted each Saturday by one..

Until, in the [b]n[/b]th Saturday, she lights up all the [b]n[/b] candles for one hour.....

For what values of [b]n[/b] is it possible for all the [b]n[/b] candles to be of equal lengths right after the [b]n[/b]th Saturday?[/color]

  [color=white].....................[/color]:noo:  :fool:  :ewpu:  :wallbash_red:  :stretcher:

[hide="Hint"]
Total number of lightnings is n(n+1)/2, and if every candy is lightened same amount of times, it must be (n+1)/2. So you kicked out even numbers n. For odd, let your imagination do the trick...
[/hide]


In a sequence of five integers the third integer is the sum of the previous two, the fourth integer is the sum of the previous three and the fifth integer is the sum of the previous four.  If the sum of the five integers is $ 248$, what is the third integer in the sequence?

Let $ a$ be the first term and $ b$ be the second term.

We are given that \[ a\plus{}b\plus{}(a\plus{}b)\plus{}(2a\plus{}2b)\plus{}(4a\plus{}4b)\equal{}248 \Rightarrow 8a\plus{}8b\equal{}248\] The third term is $ a\plus{}b$, so dividing both sides of the equation by $ 8$ be get \[ a\plus{}b\equal{}\boxed{31}\]


Euler
geometry
geometric transformation
reflection
circumcircle
trigonometry
geometry proposed

Let $ ABC$ be a triangle. Given two points $ X$ and $ Y$ on the lines $ AB$ and $ CA$, respectively, such that the line $ XY$ is perpendicular to the Euler line of triangle $ ABC$.
Let $ P$ and $ Q$ be points on the lines $ AB$ and $ CA$, respectively, satisfying $ CB\equal{}CP$ and $ BC\equal{}BQ$. Prove that the points $ P$, $ Q$, $ X$, $ Y$ are concyclic.

[quote="Little Gauss"]Oh, Are you Korean? I guess you are Han-sol Shin.
Now, I have a nice solution of your nice problem using similar triangles.
It is enough to show that reflection of $ PQ$ to bisector of $ A$ is perpendicular to Euler line of $ ABC$.
Let $ H, O$ be a orthocenter and circumcenter of triangle $ ABC$ repectively,
and $ O'$ is reflection of $ O$ to $ AC$.
I will show that $ BPQ \sim O'OH$  :!:  :!:  (it is enough)
It is easy to show that $ \angle BCH=\angle ACO=\angle ACO' \Rightarrow \angle HCO'=\angle C$.
Because circumcircle of $ AHC$ is reflection of $ (ABC)$ to $ AC$, $ O'$ is circumcenter of $ AHC$.
$ \Rightarrow BO'=R=CO=CO'$.
$ \Rightarrow \frac{OO'}{BP}=\frac{2R \cos B}{2a \cos B}=\frac{R}{a}=\frac{BO'}{BC}=\frac{BO'}{BQ}$.
Now we must show that $ \angle PBQ=\angle BO'O$.
But $ \angle BO'O=\angle OO'C-\angle BO'C=\angle O'OC-(180-2\angle C)$
$ =\angle B-\angle CBQ =\angle PBQ$.
Therefore $ BPQ \sim O'OH$.   :lol:[/quote]

hi, I'm han-sol shin.
nice solution.But I don't understand why it is enough to show $ BPQ \sim O'OH$

My solution
Lema
$ U,V$ is on $ AB,AC$ such $ UV$ is perpendicular to $ AO$. then $ V,B,C,U$ is cyclic

$ M,N$ is a point on $ AB$ such $ OM,HN$ is perpendicular to $ AB$. $ AP=AM+MP=MB+MP=2MN$. $ S$ is a intersection with line which through O and  parallel to $ AB$ and circle which center is$ H$,radius is $ HO$. then $ OS=2MN=AP$. Similarly define $ T$. Then $ OST$ is congruent $ APQ$ and this two triangle is parallel. Let $ O_{1}$be a circumcenter of triangle $ APQ$. Since $ H$ is circumcenter of triangle $ OST$, $ AO_{1}$ is parallel to $ OH$.But $ OH$ is Euler line of triangle $ ABC$. So $ AO_{1}$ is parallel to Euler line of $ ABC$. then by Lema, this problem is proven


See my figure below, I would like to create a math font link this, but I dont know which package I should use. :( Please help me, thank you very much.

[quote="stevem"]Do you mean you want a math font like the picture? You will fnd samples of math fonts at [url=http://ctan.tug.org/tex-archive/info/Free_Math_Font_Survey/survey.html]A Survey of Free Math Fonts for TeX and LaTeX[/url]. Fonts like Utopia use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=mathdesign]mathdesign package[/url]. Have a look also at [url=http://www.tug.dk/FontCatalogue/]The LaTeX Font Catalogue[/url] and [url=http://www.ctan.org/tex-archive/info/fontsampler/sampler.pdf]Free typeface sampler[/url].[/quote]
Thank you very much, mr Stevem. Now I can do it. :)


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

[quote="Lusus Naturae"]Q: Compute $ \displaystyle \frac {\int^{\frac {\pi}{2}}_{0} \sin x \sqrt {1 \plus{} \cos ^2x} \ \mathrm{dx}}{\int^{\frac {\pi}{2}}_{0} \sqrt {1 \plus{} \cos ^2x} \ \mathrm{dx}}$

The numerator is easily evaluated by the substitution $ \cos x \equal{} \sinh u$ , however it seems that the denominator cannot be expressed in terms of elementary functions. Is there some way we can use the fact about the quotient and/or the simple limits to compute the integral? I cannot see any way...  :blush:

Any thoughts / sketches / solutions are welcome.  :)

(I am not sure how difficult this question is, so please move it if necessary.)[/quote]

$ \int_0^{\pi/2} \sqrt{1\plus{}\cos^2 x} dx \equal{} \int_0^{\pi/2} \sqrt{2 \minus{} \sin^2 x} dx \equal{} \sqrt{2} \int_0^{\pi/2} \sqrt{1 \minus{} \frac{1}{2} \sin^2 x} dx \equal{} \sqrt{2}E\left( \frac{\sqrt{2}}{2} \right)$.
Where $ E$ is the complete elliptic function of the first kind.


[quote="bpms"]Is there a closed form for the equation $\sum^{k}_{n=1}n!$?

Thanks,
bpms.[/quote]

No, but there is something for n!*n?  You figure it out.

You might want to refer to

http://mathworld.wolfram.com/FactorialSums.html

He's expecting you to understand the notation.

There isn't exactly a nice formula I am assuming right?


For some reason, when I pres ctrl+F5 to build and view output on TeXnicCenter, it is not doing anything. There is no error message or anything, it just does not respond. I've done this the previous two years, and it has worked then... I downloaded a new version of TeXnicCenter yesterday because I reset my computer, and now ctrl+F5 does not work.

I have included all the files, etc. in the directory...

EDIT: could it be because i did not complete the configuration wizard (distribution directory)? How do I complete it?

Go to Build-->Define Output Profiles, and then click wizard in the lower left corner.  Follow these instructions:

[quote="LaTeX Entry on Wiki"]When it askes you for the "full path of the directory where the executables" are located, type C:\Program Files\MiKTeX 2.7\miktex\bin (Note: if you changed the default location when installing MiKTeX, then you'll need to replace C:\Program Files\MiKTeX 2.7 with the directory to which you installed MiKTeX.)
If it asks you to pick a PostScript viewer, you may just leave everything blank and just click "Next". Similarly if it asks you to pick a DVI viewer, just leave everything blank and click "Next". (It may or may not ask you these things, depending on how your computer is configured.)[/quote]

hm, i do not have a "MiKTeX 2.7" folder in my Program Files.

You probably do not have MiKTeX installed - instructions are [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]here[/url].

I know nothing about any of this, but isn't it Ctrl+F7? F5 is for viewing on my thing.