probability
AMC
AIME
HMMT
search
expected value

In his spare time, Richard Rusczyke shuffles a standard deck of $ 52$ playing cards.  He then turns the cards up one by one from the top of the deck until the third ace appears.  If the expected number of cards Richard will turn up is $ \frac{m}{n}$ in simplest form, find $ m\plus{}n$.

[hide]Since there is an equal probability of any card turning up, it follows that on average, the four aces are scattered evenly throughout the deck.

Then, the expected value is $ \frac{3}{4} \cdot \left(52\plus{}1\right)\equal{}\frac{53 \cdot 3}{4}\equal{}\frac{159}{4}$.
Thus, $ m\equal{}159$ and $ n\equal{}4$, so $ m\plus{}n\equal{}\boxed{163}$.

Am I wrong? This seems too easy for a mock AIME problem, and I'm pretty insecure on probability.[/hide]

[quote="123456789"][hide]Since there is an equal probability of any card turning up, it follows that on average, the four aces are scattered evenly throughout the deck.

Then, the expected value is $ \frac {3}{4} \cdot \left(52 \plus{} 1\right) \equal{} \frac {53 \cdot 3}{4} \equal{} \frac {159}{4}$.
Thus, $ m \equal{} 159$ and $ n \equal{} 4$, so $ m \plus{} n \equal{} \boxed{163}$.

Am I wrong? This seems too easy for a mock AIME problem, and I'm pretty insecure on probability.[/hide][/quote]

yeah thats what i got too, but the answer key says 164, without a solution.  :?:

[hide="Answer"]Expected value is $ 3/5(53) \equal{} 159/5$, using basic pre-calc techniques. The answer is then $ \boxed{164}$. [/hide]

Suppose that we have a permutation of 52 items and we have 4 aces in this permutation. Suppose that the third ace occurs at $ k$ where $ 3\le k\le 51$. The probability times the number of cards flipped is \[ \frac{\binom{k\minus{}1}{2}\cdot \binom{52\minus{}k}{1} }{\binom{52}{4}}\cdot k\] Taking this sum from $ 3$ to $ 51$, inclusive, gives $ \frac{159}{5}$.

I'm pretty sure that this method is more rigorous than what you guys are doing....

[quote="MellowMelon"]I don't understand how you can come up with $ 3/4 \cdot 53$. By this method, you get $ 53$ for drawing the fourth ace, which is a big "wtf", and $ 53/2$ for drawing the second ace. However, by pairing together a deck of cards and a deck with reversed order, you get that the expected value for drawing three aces plus the expected value for drawing two aces is $ 53$. $ 53/2 \plus{} 159/4$ is not $ 53$.

altheman's answer makes sense. If you generalize the method the answers that result actually make the above patterns work.[/quote]
Oh yes, I messed up in my reasoning there. :oops:

Suppose that we have a permutation of 52 items and we have 4 aces in this permutation. Suppose that the third ace occurs at $ k$ where $ 3\le k\le 51$. The probability times the number of cards flipped is
\[ \frac {\binom{k - 1}{2}\cdot \binom{52 - k}{1} }{\binom{52}{4}}\cdot k
\]

We want a good way to evaluate this sum, consider \[ \sum_{k=3}^{51}\frac {\binom{k - 1}{2}\cdot \binom{52 - k}{1} }{\binom{52}{4}}\cdot k=\frac{3}{\binom{52}{4} }
\sum_{k=3}^{51} \binom{k}{3}\cdot \binom{52 - k}{1} \]

To evaluate this, consider if we have $ 53$ cards, and there are $ 5$ aces. Suppose the $ 4$th ace occurs at position $ k+1$, then the probability of this occurring is \[ \frac{\binom{k}{3}\cdot \binom{53-k-1}{1} }{\binom{53}{5}}\] The sum of this from $ k+1=4$ to $ k+1=52$ is equal to $ 1$ since $ k+1$ must occur in one of those positions. So \[ \binom{53}{5}=\sum_{k=3}^{51}\binom{k}{3}\cdot \binom{52-k}{1}\]

Hence our answer is \[ \frac{3}{\binom{52}{4} }\cdot  \binom{53}{5}=\frac{159}{5}\]

[hide="Direct Combinatorial Argument"]
Let us choose five items from a set of 53 (52 cards and a marker) as follows:
- The fifth choice is the 4th ace
- The fourth choice is the 3rd ace
- The first second and third choices are the 1st ace, the 2nd ace, and a marker card.

For each possible layout of 52 cards, each possible position of the marker card, which must be before the 3rd ace, adds 1 to the counter, giving a count of the number of cards before the 3rd ace. In fact, the third ace is excluded, causing an undercount by one position, but the existence of the marker card itself adds back this position.

Now, the marker card may either be the 1st, 2nd or 3rd selection of the five chosen cards, so we must multiply by 3 to get the sum of the possible positions of the third ace. In other words, the sum over all cases of the position of the third ace is $ 3 \binom{53}{5}$

The number of possible card arrangements is equivalent to choosing 4 aces of the 52 cards, so there are $ \binom{52}{4}$ cases.

We divide the sum of the positions by the number of situations to get $ E \equal{} \frac {3 \binom{53}{5}}{\binom{52}{4}} \equal{} \frac {159}{5}$, so the answer is $ 159 \plus{} 5 \equal{} \boxed{164}$.

[b]Inspiration:[/b] [i]The idea of including a counting marker came from the first official solution to HMMT 2007 Combinatorics #9, which used a nice combinatorial argument to compute $ \sum_{i \plus{} j \plus{} k \equal{} 17} ijk$[/i][/hide]

There is a way to solve this problem similar to 123456789s reasoning.

[hide]On average, the 4 aces will divide the deck into 5 equal sized groups of cards (meaning the cards between the aces). Each group will have (52-4)/5 cards. Therefore the 3rd ace will, on average, be after 3 of these groups plus two of the previous aces. So its average position will be 3*(52-4)/5 + 3 = 159/5. This method is justified due to linearity of expected value.[/hide]

[quote="gauss202"]This method is justified due to linearity of expected value.[/quote]

Can you explain this or be more explicit with a justification.  :)

Linearity of expected value was recently discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2044573078&t=193851]here[/url].  In my opinion, the step that could use some justification is not the linearity step but rather the observation that the aces divide the deck into five equal piles.  This observation is probably best justified by an idea close to what TZF did, namely that you have five special locations (four aces and one marker position for where the deck ends) distributed randomly among 53 symmetric positions.


Hello! I'm from Bryant Academy. I'll be posting questions and will accept any help you can give me.

Thanks in advance

You should post these kind of things in Fun & Games!!
can someone move this there?
Welcome. :)

[quote="math92"]the moderator moves topics but the moems moderator isn't on often.[/quote]
yeah I've noticed...

^Hataki could probably beat him if felt like it.  :ninja:


inequalities
calculus
calculus computations

For a series $\sum_{k=1}^\infty u_k$ with complex terms $u_k$, show that $\sum_{k=1}^\infty u_k$ is convergent if and only if $\sum_{k=1}^\infty \Re(u_k)$ and $\sum_{k=1}^\infty \Im(u_k)$ are convergent.

This should really be trivial, but I can't think of any good argument showing that if the first is convergent then the latter two are convergent (possibly due to lack of sleep). It must stem from that both the real and imaginary parts of a sequence must converge in order for a complex sequence to converge, but I'm not able to show this.

Just focus on the sequence case; don't worry about series, as that will follow.

We have the sequence $(z_n)$ which may or may not be converging to $z_0.$

Here's the key inequality:

$\max(|\Re z_n-\Re z_0|,|\Im z_n-\Im z_0|)\le |z_n-z_0|\le |\Re z_n-\Re z_0| + |\Im z_n-\Im z_0|$

Try some $\epsilon$-$\delta$ work with that.

The one extra thing you want to prove about series is this: if a complex series converges absolutely, then it converges.


Let $ n \geq 3$ be an integer and $ t_1, t_2, \dots, t_n$ be positive real numbers such that
$ n^2 \plus{} 1 > (t_1 \plus{} \dots \plus{} t_n)\left(\frac {1}{t_1} \plus{} \dots \plus{} \frac {1}{t_n}\right)$
Show that $ t_i,t_j,t_k$ are the side lengths of a triangle for all $ i,j,k$ with $ 1 \leq i < j < k \leq n$.


A circle graph is to be constructed to represent all the people who participated in the 2006 Bridge Run and Walk. There will be 16 non-overlapping sectors in the circle graph representing the types of participants identified in the chart below. One such type is``male, adult runners who did not complete the race.'' What is the sum of the measures of the central angles of the sectors representing adult runners who completed the race? Express your answer to the nearest whole number.

\begin{tabular}{|r|c|c|c|l|}
\multicolumn{5}{c}{\textbf{2006 Bridge Run and Walk}}\\\hline
\textbf{Gender}&\textbf{Adult Runners}&\textbf{Non-adult Runners}&\textbf{Adult Walkers}&\textbf{Non-adult Walkers}\\\hline
\textbf{Male}&6700 (90\end{tabular}\%$ )*\&300 (80$\%$ )*\&3000 (95$\%$ )*\&10,000 (85$\%$ )\\\hline
\textbf{Female}\&4800 (92$\%$ )*\&200 (90$\%$ )*\&10,000 (96$\%$ )*\&10,000 (80$\%$ )\\\hline
\multicolumn{5}{c}{\small{*Total number of participants in category (percent of participants in category who completed the race)}}\\
\end{tabular}$

This post needs to be edited; either the poster has to post the correct problem or AoPS Administrators have to modify the GameBot.

Hm, looking at my copy of 2008 Chapter Team:

[quote="2008 Chapter Team"]A circle graph is to be constructed to represent all the people who participated in the $ 2006$ Bridge Run & Walk.  There will be $ 16$ non-overlapping sectors in the circle graph representing the types of participants identified in the chart below.  One such type is "male, adult runners who did not complete the race."  What is the sum of the measures of the central angles of the sectors representing adult runners who completed the race?  Express your answer to the nearest whole number.

[img]http://i40.tinypic.com/20rv5fr.jpg[/img][/quote]

Here is my solution:

Looking at the information given, there were $ 6700$ adult runners who are males.  Of these, $ 90 \%\equal{}6030$ completed the race.  There were $ 4800$ females, of which $ 92 \%\equal{}44164$ completed the race.  So $ 4416\plus{}6306\equal{}10446$ adults finished the race.  The total number of people who participated in the race is $ 6700\plus{}4800\plus{}300\plus{}200\plus{}3000\plus{}10000\plus{}10000\plus{}10000\equal{}45000$ people.  The fraction of people who participated who were adult runners who completed the race is $ \frac{10446}{45000}\equal{}0.2321\overline{3}$.  Multiply by $ 360$ to find the number of degrees int his sector: $ 0.2321 \overline{3} \times 360 \approx 83.568$, so the answer is $ \boxed{84}$.

I don't know how you are supposed to do that in 45 seconds in For The Win. Even with a calculator.


topology
Functional Analysis
advanced fields
advanced fields unsolved

Let $ X$ be a Banach space.Prove that the unit ball in $ X$ is weak* dense in the unit ball of $ X$**.

Let $ \varphi: X \to X^{**}$ denote the canonical embedding. Let $ B \subset X$ be the closed unit ball of $ X$ and $ B^{**}\subset X^{**}$ be the closed unit ball of $ X^{**}$. Because $ \varphi$ is an isometry, $ \varphi(B) \subset B^{**}$.
I denote by $ \tau$ the weak$ ^{*}$-topology of $ X^{**}$. Banach-Alaoglu tells us that $ B^{**}$ is $ \tau$-compact and hence $ \tau$-closed. That is why $ \textrm{cl}_{\tau}\,\varphi(B) \subset B^{**}$.

Assume that there is an $ x_{0}^{**}\in B^{**}$ such that $ x_{0}^{**}\not \in \textrm{cl}_{\tau}\,\varphi(B)$. Because $ B$ is convex and balanced and $ \varphi$ is linear we have that $ \textrm{cl}_{\tau}\,\varphi(B)$ is closed, convex and balanced. Because $ X^{**}$ is locally convex we thus find $ x_{0}^{*}\in X^{*}$ such that $ |\langle x_{0}^{*}, x^{**}\rangle| \le 1$ for all $ x^{**}\in \textrm{cl}_{\tau}\,\varphi(B)$ but $ \langle x_{0}^{*}, x_{0}^{**}\rangle > 1$ (Theorem 3.7 in Rudin's Functional analysis + the definition of $ \tau$). We now get a contradiction easily. The first condition says that $ |\langle x, x_{0}^{*}\rangle| = |\langle x_{0}^{*}, \varphi x\rangle| \le 1$ with all $ x \in B$; so $ x_{0}^{*}\in B^{*}$ (the closed unit ball of $ X^{*}$). The second condition says that $ 1 < \parallel x_{0}^{*}\parallel \,\parallel x_{0}^{**}\parallel \le \parallel x_{0}^{*}\parallel $, because $ x_{0}^{**}\in B^{**}$, and hence $ x_{0}^{*}\not \in B^{*}$, a contradiction. We have proved that $ \textrm{cl}_{\tau}\,\varphi(B) = B^{**}$.


geometry
3D geometry
percent
percent increase

if the length of the edge of a cube is increased by 50%, what is the percent increase in the volume of the cube? express your answer to the nearest whole number.

how would u do this?

[hide]Say the side length of the original cube is $x$.  Then the side length of the new cube is $1.5x$.  The volumes of these cubes are ${x}^{3}$ and ${1.5x}^{3}=3.375{x}^{3}$, respectively.  So the volume of the new cube is 3.375 times the volume of the original cube.  Expressed as a percent to the nearest whole number, the volume of the cube increased by 238%.[/hide]

[hide]let the normal side length be x.  Then the percent increase would be $100[(\frac{3}{2}x)^{3}-x^{3}]\approx\boxed{238}$[/hide]

[hide]
If each edge of the first cube=x then each edge of the second cube=1.5x. The area of the first cube would be x^3, and the area of the second cube would be 1.5^3*x^3, or 3.375x^3, which is equal to the first cube increased by 237.5% or to the nearest whole number, 238%[/hide]

[hide]237.5 and you round up, right? so you get $\boxed{238}$[/hide]

[quote="lotrgreengrapes7926"]Extension:
If the edge length of a cube is increased by $x\%$, what is the percent increase in the volume and surface area of the cube?[/quote]
[hide="Surface Area"]$100[(\frac{x+100}{100})^{2}-1]$[/hide]
[hide="Volume"]$100[(\frac{x+100}{100})^{3}-1]$[/hide]


calculus
integration
trigonometry
calculus computations

Compute
\[ \int_{0}^{\pi} \sin^{2n}{\theta} d\theta\]

http://en.wikipedia.org/wiki/Lists_of_integrals

Use the reduction formula.

Here is an idea, not sure how good it will work

[hide]

$ \int_0^{\pi}\sin^{2n}(x)dx\equal{}\int_0^{\frac{\pi}{2}}\sin^{2n}(x)dx\plus{}\int_{\frac{\pi}{2}}^{\pi}\sin^{2n}(x)dx$

=$ \frac{1}{2}B\left(n\plus{}\frac{1}{2},0\right)\plus{}\int_{\frac{\pi}{2}}^{\pi}\sin^{2n}(x)dx$

For the second one, maybe Lebniz's rule?

$ J(\theta)\equal{}\int_{\frac{\pi}{2}}^{\pi}\sin^{2n}e^{\minus{}\cos(x)\theta}dx$?

[/hide]

Put $ I_n$ be the given integral and obtain the recurrence relation for $ I_n$ by integration by parts.

it's similar with the previous post
http://www.mathlinks.ro/viewtopic.php?t=214671

[quote="kunny"]Hint:

$ \int_0^{2\pi} e^{i n\theta} d\theta \ (n\in\mathbb{Z})$.[/quote]
Maybe I am missing something but shouldn't it be

${ Im\{\int_0^{2\pi} e^{i n\theta} d\theta }\}$

Or am I mixing something up?

[quote="kunny"]I just gave the problem separately. :)[/quote]
Haha ok  :D


[quote="Peter"]The number $21982145917308330487013369$ is the thirteenth power of a positive integer. Which positive integer?[/quote]
So, we have that $A=21982145917308330487013369=B^{13}$.

The first thing to notice is that this number has 26 digits=>

$10^{25}<A<10^{26}$=>$B<100$

Next, it is not hard to show that $80^{13}<10^{25}$=>

$80<B<100$.

One more thing is that the last digit of $B$ should be $9$(=>B=89 or B=99) and that the sume of digits of $A$ does not dividisible by 9 =>B=89 :wink: .

Sorry, but could you show me how to decide 80^13 < 10^25? I am a beginner, though I understand the steps you are going. Thanks.

[hide=Alternate Solution]Let $x$ be the required number , then obviously $x < 100$

Also by euler's totient theorem $x^{4}\equiv 1 \pmod{10} $, hence  $ x^{13} \equiv 9 \equiv -1  \pmod{10}\implies x \equiv -1 \pmod{10}$

Similarly $x^{6}\equiv 1 \pmod{9} $, hence  $ x^{13} \equiv 8 \equiv -1  \pmod{9}\implies x \equiv -1 \pmod{9}$

So $x \equiv -1 \pmod{90}\implies x = 89$ [/hide]

Let $a^{13}$ be the large number. Note that $a<100.$ By Fermat's Little Theorem, $$a^{13}\equiv a\equiv 11 \pmod{13}.$$ Also observe that $$a^{13}\equiv a\equiv 8\pmod{9}.$$ Since it has to be less than 100 and $$89\equiv 8\pmod{9},$$ we are done.


abstract algebra
geometry
3D geometry
integration
geometric transformation
reflection
group theory

We say that the subset $P$ of an abelian group $(G,+)$ is an [i]order[/i] if the following conditions are satisfied:

(i) $P+P\subset P$, i.e. $P$ is closed under addition
(ii) $P\cup(-P)=G$
(iii) $P\cap(-P)=\{0\}$

Notice that this concept is equivalent to what is usually called an order. Indeed, given $P$, we can just define $x\ge y$ iff $x-y\in P$. This relation turns $G$ into a totally ordered group. Conversely, if $G$ is a totally ordered abelian group, we can simply take $P$ to be the cone of non-negative elements.

If, in addition, the group is locally compact, we say that $P$ is a [i]measurable order[/i] if $P$ is measurable with respect to the Haar measure on $G$.

Prove:

An infinite compact abelian group does not have a measurable order.

Let $G$ be an infinite compact abelian group, and suppose it has a measurable order $P$. Let $\mu$ be the measure on $G$, normalized so that $\mu(G)=1$.
Since $G$ is infinite, $\mu(\{0\})=0$.

Inversion preserves measure, so $\mu(-P)=\mu(P)$ and $\mu(P)=\frac12$.

Consider $I=\int_{P}\mu(x+P) dx$ (with respect to $\mu$, of course). $I$ is the product measure of the set $\{(x,y): 0\le x\le y\}$, which is $\frac18$ since the union of this set and its reflection is $P\times P$; the intersection is a subset of the zero-measure diagonal.
On the other hand, $\mu(x+P)=\mu(P)=\frac12$ for all $x$ by translation-invariance of measure. That makes $I=\frac14$, a contradiction.


Hi everyone!

I'm Manisha from ARML 2003/2002, but I couldn't make it this year :blush:  2 . Could some attendees tell me how it was? How many teams did we have? Where did we practice? How were the T-shirts? And all the other fun things about ARML...

We had 1 1/2 teams, we practiced at Penn State Friday morning and afternoon, and you do not want to know about the T-shirts.  (Nobody submitted a design.)  ARML was fun as usual:  they moved where we were staying at Penn state to somewhere more convenient.  We found a cool used book store with a fairly decent math book section, got kicked out of the boys' lounge, got lost, and had various other weird things happen.  We placed 19th or something, but if we had to lose to NYC, we did it with style:  at least Aaron won individual!  Yay!  We had lots and lots of inside jokes as usual.

apparently I'm outside now as I don't remember very many jokes from arml ... oh the agony of aging and being seperated from those around you  ;)


For every $(n,m)\in\mathbb{N}^{2}$ find an n-tuple $(x_{1},x_{2},...,x_{n})\in\mathbb{N}^{n}$ with the $x_{i}$ unequal and $m>1$ that satisfies the following equation
\[\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=\frac{1}{m}\]

:blush: i forgot to mention that the $x_{i}$ aren't equal.

i have another solution. let $x_{i}=(m+i-1)(m+i)$ for $1\le i\le n-1$ and $x_{n}=m+n-1$ :P

RDeepMath91 i bilieve that you are false in your equality. :wink:

let's see what RDeepMath91 has to say about this. :huh:

Well, as you can see, I have fixed the error..
Thx a lot, anyway..  :) I'm too careless...


function
limit
integration
LaTeX
real analysis
real analysis unsolved

Assume that $\{f_{n}\}$converges pointwise to $f$ on $[a,b]$  and that
$l.i.m_{n\rightarrow\infty}f_{n}=g$ on $[a,b]$. Show that $f=g$ if  both $f$ and $g$ are continuous on $[a,b]$.
PS:$l.i.m_{n\rightarrow\infty}f_{n}=g \Leftrightarrow f_{n}$ converges in the mean to $g \Leftrightarrow \lim_{n\rightarrow\infty} \int^b_a  \mid f_{n}(x) - g(x) \mid ^2 dx=0$

$\LaTeX\$  roxxx

Assume WLOG that $g\equiv 0$ and that $\inf_{[a,b]}f\ge 1$ (this is possible because of the continuity of $f$: if $f(x)\ge 2t>0$ for some $x$, then there is an entire interval $I$ around $x$ on which $f$ takes values $\ge t$, and we can scale it in order to replace $t$ with $1$, while restricting our attention to the interval $I$).

For any $n$, let $A_n$ be the set consisting of those $x\in[a,b]$ for which $|f_n(x)|\ge\frac 12$. Since $\|f_n\|_2\to 0\Rightarrow m(A_n)\to 0$, we can find a subsequence $(k_n)_n$ of $1,2,\ldots$ s.t. $m(A_{k_n})\le\frac 1{n^2}$ ($m(\cdot)$ is the usual measure), which means that $\sum_n m(A_{k_n})<\infty$.

Now we can conclude that the measure of the set on which $(f_{k_n})_n$ converges pointwisely to $f$ is zero, by applying the following nice theorem (which has been proved before around here somewhere :)): given a sequence of subsets $(S_n)_n$ of a measure space $X$ s.t. $\sum_n m(S_n)<\infty$, almost all $x\in X$ belong only to finitely many  $S_n$'s. We apply this to $S_n=A_{k_n}$ to conclude that for almost all $x\in[a,b]$, only finitely many $f_{k_n}$'s take values $\ge\frac 12$ in absolute value. However, if $f_{k_n}(x)$ converges to $f(x)\ge 1$, then infinitely many $f_{k_n}(x)$ should be $>\frac 12$.

I hope it's ok.


advanced fields
advanced fields unsolved

Let C * is the locus of centers of oskulating circles is a nonplane curve with a constant nonzero curvature. To prove that the curvature of C* is constant and to calculate torsion of C *. (Curve C * is called evolute of C and C - evolvent of C *.)

If you could give me guidance on the task will be very grateful!


Looking at The Round Table, I thought of a problem, but I don't know the answer:

Let's say that in a particular forum there are 10 topics (or, more generally, n) in a certain order.  If we randomly post k times (on not necessarily distinct topics), what is the probability that the topics end up in the same order?  Other predetermined orders?  Note that topics move to the top when posted on.  Other properties arising from this?

Well, I'd just start with some basic counting.

With 0 posts, there's only 1 way.

With 1 post, there's only 1 way.

With 2 posts, there are two ways.

With 3 posts, we can do: 321; 211; 221; 121; 111 = 5 ways.

With 4 posts, we can do: 4321; 3321; 3221; 3211; 3121; 2221; 2211; 2121; 1221; 2111; 1211; 1121; 1111; 1321; 2321 = 15 ways.

Well, I admit that I don't see the pattern yet.  I'm going to do a couple more and see if I get there.  Actually, it might have a Fibonacci-like recursion.  That's an interesting thought.  Huh.

With 5 posts, we can do: 54321; 44321; 43321; 43221; 42321; 41321; 43211; 43121; 42221; 42211; 42121; 41221; 42111; 41211; 41121; 16 comprised of only 1's and 2's; 14321; 24321; 34321; 27 which look like abc21, a,b,c in {1, 2, 3}, but we've already counted 8 of them which makes 19 new ones; 32211; 31211; 32111; 13211; 23211; 33211; 32111.  I think that's it.  That makes 60.

Am doing a little reasearch ..

Huh ... Sloane doesn't have any sequences that start this way.  Perhaps I've counted wrong?

Sorry for reviving such an old thread, but as it turns out, I did miscount.  The sequence of diagonal values is the Bell sequence, [url]http://www.research.att.com/~njas/sequences/A000110[/url].  The entire triangle of values is the partial sums of the Stirling numbers, [url]http://www.research.att.com/~njas/sequences/A102661[/url].  (Why triangle?  Because if $ k < n$, we get the same answer as if we reduce $ n$ to $ k$, so we might as well ignore those values.)


geometry rhombus perimeter parallelogram

I'm having trouble with geometry especially rhombus's and their diagnols >_@ so here it is: If a diagnal of a rhombus is three- fourths of another and the area is 24 square units, then find its perimeter and its height. thnx  :D

[hide="hint"]The area of a rhombus is $ \frac {1}{2}ab$ where a and b are the diagonals.[/hide]

We see that the area of any rhombus is $ A=\dfrac{1}{2}\times d_1 \times d_2$, where $ d_1$ and $ d_2$ are the diagonals of the rhomus.

Let's set $ d_1$ as $ x$. Since the other diagonal is three-fourths of it, the other diagonal, $ d_2$, is $ \dfrac{3}{4}x$.

Now we plug $ d_1$ and $ d_2$ back in the formula:

$ A=\dfrac{1}{2}\times x\times \dfrac{3}{4}x=\dfrac{3}{8}x^2$

We also know that the Area ($ A$) is 24, so plugging that in, we see:

$ 24=\dfrac{3}{8}x^2$

We see that we have reduced our formula to a one variable equation.

\begin{align*}
24&=\dfrac{3}{8}x^2\\
64&=x^2\\
\pm8&=x
\end{align*}

$ x$ must be positive, so the two diagonals are 8 and the other is 6.

If we draw a diagram of this rhombus, we would see:

[asy]size(250);
draw((0,0)--(-3,4)--(0,8)--(3,4)--cycle,linewidth(1));
draw((0,0)--(0,8),dashed);
draw((-3,4)--(3,4),dashed);
label("$d_1=8$",(0,6),E);
label("$d_2=6$",(1.5,4),S);
markscalefactor=0.05;
draw(rightanglemark((0,8),(0,4),(3,4)),red);[/asy]

We observe that each of the four smaller triangles is a right triangle with sides 3-4-5. The hypotenuse of all these triangles is 5, so the perimeter is 20.

To find the height of this rhombus, we simply turn it to the side, which reults in a parallelogram. The area of a parallelogram is $ A=bh$. We plug in $ A=24$ and $ b=5$. This results in:

\begin{align*}24&=5h\\\dfrac{24}{5}&=h\end{align*}

Therefore, our two answers are:

$ \boxed{P=20,h=24/5}$


geometry
trigonometry
parallelogram
angle bisector

Prove that a triangle with equal angle bisectors must be isosceles

That's also known as the Steiner-Lehmus Theorem (it appears with this name on "Geometry revisited" by Coxeter).
It also came out in some german TST, if I remember well...  :)

yeah, and I think that you can also prove it with Stewart's theorem

Hmmm, I saw this Problem in CPIG, before the chapter on Stewart's Theorem. I believe it is problem 6-8 or something. Is there a more elementary solution (one that does not use trig or anything derived from trig)?

You don't need trig for Stewart's theorem.

Of course I am aware that there are geometric proofs for this assertion; you should try and prove the contrapositive.

I think all these methods are unnecessarily complicated. The angle bisector proof is totally trivial and it takes like 5 mins to type up.

I have not found that the proof in Geometry Revisited presents techniques that are useful for other problems.

[quote="Altheman"]You don't need trig for Stewart's theorem.[/quote]

Hmmm, really? So how does one derive it without trig?

You can perform any derivation that uses trigonometry with an argument about similar triangles.  It just takes longer.

A nice trig proof of this theorem is available at the Forum Geometricorum Vol 8(2008) by Mowaffaq Hajja! Some proofs without trig are very cumbersome and should better be avoided!

[hide="Two simple proofs from the Crux Mathematicorum journal"]
[hide="A proof by Gardner"]
Let $ ABC$ be a triangle with equal internal angle bisectors $ BM$ and $ CN$, as in Figure 1. If the angles $ B$ and $ C$ are not equal, one must be less, say $ B<C$. Take $ L$ on $ BM$ so that $ \angle LCN\equal{}\frac12B$. Since this is equal to $ \angle LBN$, the four points $ L,N,B,C$ are concyclic. Since
\[ B<\frac12(B\plus{}C)<\frac12(A\plus{}B\plus{}C),
\] $ \angle CBN<\angle LCB<90^{\circ}$. Since the smaller chords of a circle subtend smaller acute angles, and $ BL<CN$,
\[ \angle LCB<\angle CBN.
\] We thus have a contradiction.[/hide]
[hide="A proof by a French engineer, M. Descube"]
Let $ ABC$ be a triangle with equal internal angle bisectors $ BM$ and $ CN$, as in Figure 2. Complete parallelogram $ NBMG$, so that the three angles denoted by $ b$ are equal (as well as the two angles denoted by $ c$). It will suffice to show that $ b\equal{}c$. If $ b\neq c$ then one is greater, say $ b>c$. Since $ \triangle$'s $ NBC$ and $ MBC$ have two corresponding sides equal and included angles $ b$ and $ c$, we have
\[ b>c\Rightarrow CM>BN.
\] In $ \triangle CGN$, we have $ CN\equal{}NG$, so that $ b\plus{}g\equal{}c\plus{}h$; hence
\[ b>c\Rightarrow g<h\Rightarrow CM<MG\Rightarrow CM<BN,
\] and we have a contradiction.[/hide][/hide]


[quote="Ignite168"]Find the radius of a circle inscribed in a 3-4-5 triangle.[/quote]

[hide]Area=$sr$=$\frac{1}{2}\cdot 3 \cdot 4=6$

where $s$ is semiperimeter $=\frac{12}{2}=6$ and $r$ is the inradius.

So $r=1$.

Or alternatively, use the well-known formula $\frac{a+b-c}{2}=\frac{3+4-5}{2}=1$.[/hide]


geometry
perimeter
circumcircle
inradius
trigonometry
geometry theorems

In any triangle,prove that:
a)$ r_a\equal{}p\cdot tan \frac{A}{2}$
b)$ S\equal{}(p\minus{}a)\cdot r_a$
c)$ S\equal{}\sqrt{ r \cdot r_a \cdot r_b \cdot r_c}$
d) $ {r_a}\cdot{r_b}\cdot {r_c}\equal{}p^2 \cdot r$
e) $ (r_a \plus{}r_b)(r_b \plus{} r_c)(r_c\plus{}r_a)\equal{}4p^2 \cdot R$
where p is half of perimeter,R is the circumradius,  r is the inradius and $ r_a,r_b,r_c$ is the exradius.

I'll just start it.The first one is obvious,because:
$ AC_1=p$,so $ \tan\alpha=\frac{r_a}{p}$
The second follows from there too,because
$ S=pr_a-r_a(BA_1+A_1C)=r_a(p-a)$.
The third follows from second because
$ S^3=(p-a)(p-b)(p-c)r_ar_br_c\Rightarrow$,since $ S^2=p(p-a)(p-b)(p-c)$,we got that
$ Sp=r_ar_br_c$,now using that $ S=pr$,finally we have:
$ S=\sqrt{rr_ar_br_c}$.
The fourth:
If we use that $ S=pr$ again and the third part,we have that:
$ p\sqrt{r}=\sqrt{r_ar_br_c}$.So after squaring we get the result.

[img]http://img62.imageshack.us/img62/6264/excirclenq5.png[/img]
P.S I'll post the proof to 5th part later,of course if nobody else do it before.

Speciall thanks for help .I want to know how you make the draw?What soft?Have nice day! :wink:

I used Geonext.It is very nice kind of soft,if i am not mistaken,geonext was created in Germany.


Janna's average score on five 100-point tests was 80. Her score on the sixth test raised her mean score two points. How many points did she score on the sixth test?

let us right the equation $ \frac{(80*5)\plus{}x}{6} \equal{} 82$

solving, x yields $ \boxed{92}$

To find averages, we add the numbers together and then divide by the number of numbers. For example, to find the average of $ a$, $ b$, and $ c$, we would do: $ \frac{a \plus{} b \plus{} c}{3}$. If you took the average, say $ d$, and multiplied it by $ 3$, you would find the sum of $ a$, $ b$, and $ c$. (Do you see why?)

So let's find the sum of her test scores. The average is $ 80$, and she took $ 5$ tests. $ 80 \times 5 \equal{} 400$.

We know that her score on the sixth test raised Janna's average by 2 points. That means her new average was $ 82$. The sum of all her scores, including the sixth test, is $ 82 \times 6 \equal{} 492$.

The difference between $ 400$ and $ 492$ is $ 92$. Hence, our answer is $ \boxed{92}$.

i use the method of just multipliying the number of tests(6) and the point change(+2) to get 12. then i add 12 to 80 and get 92. =P


Solve for x if 0<x<pi.  Give answers to the nearest thousandth.
sec(x)=1/5

(pi is like pi=3.141592654.....)

If you want to find the imaginary values use

$ \sec{z}\equal{}\frac{2}{e^{iz}\plus{}e^{\minus{}iz}}$

[quote="FightTheTide"]There are no solutions seeing as the range of the secant function is all reals whose absolute values are greater than or equal to 1.[/quote]
what about on a unit circle?

He's saying you can't solve $ \cos{x}\equal{}5$

cos and sine only exist for $ -1\pmx\pm1$. This can be seen on their respective graphs


vector
linear algebra
linear algebra unsolved

Let $ V$ be a vector space over $ Z_{p} \equal{} \{\overline{0},..., \overline{p \minus{} 1}\}$. Now, let $ W_1,...,W_n$ be subspaces of $ V$, where $ n < p \minus{} 2$. Suppose there exist an integer $ j \in \{1,...,n\}$ such that $ W_j$ is not contained in $ \bigcup_{i \neq j} W_i$. Also, suppose $ V \equal{} \bigcup_{i \equal{} 1}^{n}W_i$. Prove there exists $ k \in \{1,...,n\}$ such that $ V \equal{} W_k$.

  I need some help is this exercise, please!!!

I am sorry, $ V$ is a vector space over the field $ Z_p$.

If none of the $ W_i$ is equal to $ V$ then each of them contains at most $ \frac{1}{p}$ of the points of $ V$.  Then we certainly can't cover the whole space with $ n < p \minus{} 2$ subspaces.  Perhaps you meant something else?

If none of the $ W_i$ is equal to $ V$ then each of them contains at most $ \frac{1}{p}$ of the points of $ V$.

 Why????? Can you explain me!

How many points are there in an $ m$-dimensional vector space over $ \mathbb{Z}/p\mathbb{Z}$?

Some non-mathematical comments:
Really, one question mark would have done the trick ;)  (Also, there's a "quote" button that allows you to quote an earlier post in a way that the quoted text will not be confused with your own; though you don't really need to be copying my text at all, since who else could you possibly be responding to?)


Did any one use hash in C++? What is the header file for it? hash_set, hash_map,hash_set.h, hash_map.h none is supported by cygwin. Then what is the header file?

[code]#include<map>
using namespace std;

int main()
{

hash<int>H;

return 0;
}[/code]

This did not work


Prove that for all $ c > \sqrt {5}$ there are finitely many pair of $ (p,q) (p,q \in \mathbb{N})$ such that

$ \left|\frac {\sqrt {5} \minus{} 1}{2} \minus{} \frac {p}{q}\right| < \frac {1}{cq^2}$

Actually I know the proof of the problem above, but I'm not sure if the following is true:

For all irrational positive number $ x$ there are infinitely many pair of $ (p,q)$ such that

$ \left|x \minus{} \frac {p}{q}\right| < \frac {1}{\sqrt {5}q^2}$

If it's true, how to prove it? Thx~

I tried to find it in the forum but failed.

For $ \left|x \minus{} \frac {p}{q}\right| < \frac {1}{\sqrt {5}q^2}$, see http://www.esnips.com/doc/866ff532-c0c5-4dc4-b215-b3305188e8f8/Hardy,%20Wright%20-%20Introduction%20to%20Number%20Theory page 164.


geometry
3D geometry
real analysis
real analysis unsolved

Let $ f: \mathbb{R}\rightarrow \mathbb{R}$ be given.Assume that the square and cube of $ f$ are smooth. Is $ f$ smooth? That is , if $ f\cdot f\in C^{\infty}$ and $ f\cdot f\cdot f\in C^{\infty}$, does it follow that $ f\in C^{\infty}$?


geometry
circumcircle
Euler
geometry solved

given triangleABC inscribed (O).Line d through A parallel BC meet again
(O) at A'.similary with B',C'.AB' meet BA'at C" similary with B" A",Prove that A",B",C",O
collinear

Let's formulate a stronger assertion:

[color=blue][b]Theorem.[/b] Let ABC be a triangle. Let the parallel to the line BC through the point A intersect the circumcircle of triangle ABC at a point A' (apart from A), and similarly we define the points B' and C'. Let the lines AB' and BA' meet at C", and similarly we define the points A" and B". Then, prove that the points A", B", C" all lie on the Euler line of triangle ABC.[/color]

[i]Proof.[/i] Since AA' || BC, we have < C"BC = < BA'A. On the other hand, since the point A' lies on the circumcircle of triangle ABC, we have < BA'A = < BCA. Thus, < C"BC = < BCA. Similarly, < C"AC = < ACB, what rewrites as < CAC" = < BCA. Hence, < C"BC = < BCA = < CAC". Now, applying the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=25859 to the quadrilateral BCAC" (disregard the convexity condition - what is important is that the angles are properly directed), we see that the points O, H and C" are collinear, where O is the circumcenter and H is the orthocenter of triangle ABC. In other words, the point C" lies on the line OH, i. e. on the Euler line of triangle ABC. Similarly, the points A" and B" also lie on the Euler line of triangle ABC. Thus, the theorem is proved.

  Darij


A googol is the digit 1 followed by 100 zeroes. What is the least value of n such that n! is greater than a googol?
How do you solve this without a calculator?

The zeroes come [i]to the right[/i] of the 1...so 1000000...0000.

Factor googol. You'll need to have a product greater than that.

[quote="james4l"]Memorization: 70! is barely over a googol.

There exists, to my recall, absolutely no "non-bruteforce" method of calculating large factorials.[/quote]

You can use n! is somewhat close to (n/e)^n.

The trouble exists in multiple parts:

to get an approximation of (n/e)^n as in to get enough digits requires to know the logarithm of n/e, and Euler's number is hard to approximate, let alone its reciprocal.

Oops; didn't see the "without a calculator" part.


1. What does hybridization have to do with acidity? The context of what I mean is shown in the attached picture under the answer choices.

2. I am having trouble figuring out which of the four answer choices best explains why HCl is more acidic than water. [hide="My reasoning does give me a clear answer. "]
Someone told me that HCl is more acidic than water because of induction, but oxygen is more electronegative than chlorine so I find that particular reason hard to agree with. Oxygen and chlorine are not in the same row or columnn so periodicity cannot work and there is no resonance on HCl or H2O. That leaves us with induction or hybridization or both, but what do they have to do with acidity for these molecules?[/hide]

1. An sp hybridized compound is more acidic than an sp2 hybridized compound which is more acidic than an sp3 hybridized compound.  This has to do with the "s character" of the orbital.  Since electrons in s orbitals are closer to the nucleus than electrons in p orbitals, an orbital with more s character will have its electrons closer to the nucleus.  This increases the electronegativity of that atom which makes the overall compound more acidic.

HCl is more acidic than water because it gives H+ readily as compared to water as Cl- is highly stable which is not the case with OH-(remember it is not a good leaving group and protonates before leaving)

My reasoning of HCl vs H2O is bond strength of H and the other compound. The one with weaker bond breaks easily(and hetrolytically) and hence is more acidic.

Thanks.

I guess of the four answer choices, induction work is the only one left since the Cl in HCl and the O in H2O both have sp3 hybridization?

lol column/row periodicity because Cl is bigger than O because of that. induction/resonance/hybridization are def not the answer. probs just a badly worded choice


abstract algebra
superior algebra
superior algebra unsolved

Find Lie algebra of lie group $ SO(n)$ ($ SO(n)$ be orthogonal group)
Help me!

The Lie algebra of $ SO(n)$ is $ {%Error. "gothic" is a bad command.
so}_n \equiv \{A: A = -A^T\}$, the set of antisymmetric matrices.

Hint:
\[ \left.d (A A^T)\right|_{A=I} = \left.(dA\cdot A^T + A \cdot dA^T)\right|_{A=I} = dA + dA^T.\]


Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

1/3

2/5

1/2

3/5

2/3

Disclaimer: I'm bad at combinatorics in general.

[hide="Solution"]The total number of ways to select 3 people out of the 5 is $ {5\choose{3}} \equal{} 10$.
Out of these, there are the following cases:
- All men (this is one of the combinations because we have 3 men, so there's only one way to choose 3 of them)
- 1 man 2 women (there are three of these because we only have 2 women and the 1 man could be any of the 3)
- 2 men 1 woman (this is the remaining combinations after the above two are filtered out).

So the answer is $ \frac {10 \minus{} 1 \minus{} 3}{10} \equal{} \boxed{\frac {3}{5}}$[/hide]

What gm_trte did was complete the problem using
combinations, assuming order didn't matter
(which is correct in how he did it).  However,
you can also do the problem where order
DOES matter.  Let's assume that order
does matter then, so there are 5*4*3 ways
to arrange the 5 ppl into the 3 office spaces.
There are C(3,2) ways ( equals 3 ways) to choose
two spots for both of the men, and in these spots,
we can have 3*2 ways to order the men and
2 ways to order the women. THus, the probability is:
3*3*2*2/5*4*3=36/60=6/10=[u]3/5[/u]


In a game of darts, 2, 5, 8, or 11 points can be earned with each toss of one dart.   What is the fewest number of darts needed to earn exactly 100 points?

A.9	B.10	C.11    D.12    E. More than 12

[hide]The quickest way I can see is: 11,11,11,11,11,11,11,11 (88)+2 (90) +5,5 (100)

That's 11 throws, which is answer C.[/hide]

[hide="Another way"]Eight 11s, one 8, two 2s. Still, its $ \boxed{11}$, C.[/hide]


a square -6a + 9 - 36b square

h square - hk + hr - kr

I think their both irreducible, but need a second opinion.

You want us to factor

$ a^{2}\minus{}6a\plus{}9\minus{}36b^{2}$
$ h^{2}\minus{}hk\plus{}hr\minus{}kr$

[hide="1"] $ a^{2}\minus{}6a\plus{}9\minus{}36b^{2}\equal{} (a\minus{}3)^{2}\minus{}(6b)^{2}\equal{} (a\plus{}6b\minus{}3)(a\minus{}6b\minus{}3)$ [/hide]
[hide="2"] $ h^{2}\minus{}hk\plus{}hr\minus{}kr \equal{} (h\minus{}k)(h\plus{}r)$ [/hide]


I play an online game where you can purchase lottery tickets and win a jackpot. about 4,000 tickets are purchased by players in 1 week (7 days) and the winner wins the pot. Mathematically, are there certain times to purchase tickets to get a better shot at winning? Is there anyway I can beat the odds? :maybe:

"Mathematically," the winner is chosen at random from the set of lottery tickets bought in a particular time frame (here one week), so the only way to improve your odds is to buy more tickets.


A planet has orbital radius $ a$ $ \text{A.U.}$. Calculate the time for which it is in retrograde motion (It appears to move backwards in the sky). Assume planetary orbits are circular and coplanar.

Doesn't that also depend on the viewer's orbit around the sun and the velocity of the planet being observed?

Agree on the second part, about the velocity.  For the first part I assume he wants it to be Earth.

I was going to something about it but, I thought I may have been missing something.

Assume observer is on the Earth and $ a > 1$. Velocity can be calculated from $ a$;
$ v \equal{} \sqrt\frac {GM_{\odot}}{aa_{\oplus}}$; where, $ a_{\oplus}$ is orbital radius of Earth and $ M_{\odot}$ is mass of Sun.

shreyas_patankar , do you have the solution to the problem?  :)  Please post it.

No, I don't. :(
My approach involves lot of geometry and so on and the only concrete thing I have got is that at the start and end of retrograde period; components of Earth's and Planet's velocities along observers line of sight have to be equal.

The INAO had a problem which involved calculating the retrograde period using some experimental data. I feel that it should be possible to calculate it theoretically.


[quote="charlestreykang"]Can anyone explain to me what Riemann's Sum is and how it's related to sigma notations?[/quote] [url=http://en.wikipedia.org/wiki/Riemann_sum]Have you had a look here?[/url]  :ninja:


analytic geometry
combinatorics unsolved
combinatorics

For a pair $ A \equal{} (x_1, y_1)$ and $ B \equal{} (x_2, y_2)$ of points on the coordinate plane, let $ d(A,B) \equal{} |x_1 \minus{} x_2| \plus{} |y_1 \minus{} y_2|$. We call a pair $ (A,B)$ of (unordered) points [i]harmonic[/i] if $ 1 < d(A,B) \leq 2$. Determine the maximum number of harmonic pairs among 100 points in the plane.

I really like this!  Is this question original to the TST?  [hide]From our hundred points choose five, $ P_i \equal{} (x_i, y_i)$ for $ 1 \leq i \leq 5$ with $ x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5$.  By Erdős-Szekeres, the values $ y_1, y_2, y_3, y_4, y_5$ contain either a three-term increasing subsequence or a three-term decreasing subsequence, without loss of generality $ y_1 \leq y_2 \leq y_3$.  Then $ d(P_1, P_2) \plus{} d(P_2, P_3) \equal{} d(P_1, P_3)$, so not all three distances can be in the interval $ (1, 2]$.  Then the graph whose edges are harmonic pairs contains no $ K_5$ subgraph, so by Turán, there are at most $ \binom{4}{2} \cdot \left(\frac {100}{4}\right)^2 \equal{} 3750$ edges and thus at most this many harmonic pairs.  We can achieve this by taking twenty-five points within some $ \varepsilon$-neighborhood of each of the points $ (\pm3/4, 0)$ and $ (0, \pm 3/4)$.[/hide]

can anyone post solution not using grap?

Please check my solution :help: (this looks a bit different to that of #2)

Firstly note that we can [hide=change] We apply the well known trick to handle 2D Manhattan distance: rotate the plane by 45 degrees. Change $P = (x,y) \rightarrow P' = (x-y, x+y)$, so then $d(P,Q) = d'(P',Q')$  [/hide] the distance function to a new and easier to handle distance function $d'(P',Q') = \max \{ |P'_x - Q'_x|, |P'_y - Q'_y| \}$ without changing the structure of the problem. So now through the solution we work with the new distance function.

I claim the answer is $\frac{3*100^2}{4*2} = 3750$. To see that this bound is achieved first, put $25$ points each at some small neighborhood of the four points $(\pm \frac{1.0201082102011209}{2}, \pm \frac{1.0201082102011209}{2})$. Now we prove that this is the maximum we can achieve. Create a graph $G$ with $100$ vertices with two vertices joined iff the two points are mutually harmonic.

[b] Claim [/b] $G$ has no $K_5$
[b] Proof [/b]:  Firstly note this two facts, both of which are easy to verify by case-bashing:
[list=1][*] If a coloring of the edges of $K_5$ with two colors doesn't produces a monochromatic triangle, then it must have a monochromatic cycle with lenght $5$ [*] It's impossible to find three real numbers $A,B,C$ for which all the points $(A,0), (B,0), (C,0)$ are mutually harmonic.[/list]

Now, for each edge $\bar{PQ}$ already in $G$, color the edge $PQ$ red if $\max \{ |P_x - Q_x|, |P_y - Q_y| \} = |P_x - Q_x|$, or blue otherwise. Now suppose for the sake of contraction, there is a $K_5$ in $G$, with the points $A,B,C,D,E$. Then by #2, it has no monochromatic triangle, and so by #1, it has a monochromatic cycle of lenght $5$. WLOG the cycle is of red color, and let the cycle be $A \rightarrow B \cdots \rightarrow E$.

Note that now if $\max(A_y,B_y,C_y,D_y,E_y) - \min(A_y,B_y,C_y,D_y,E_y) > 2$, we have a contradiction (then the extreme points would not be harmonic), so $\max(A_y,B_y,C_y,D_y,E_y) - \min(A_y,B_y,C_y,D_y,E_y) \leq 2$. WLOG assume $\min(A_y,B_y,C_y,D_y,E_y) = A_y = 0$, so $\max(A_y,B_y,C_y,D_y,E_y) > 0$. Since $\max(A_y,B_y,C_y,D_y,E_y) \leq 2 + \min(A_y,B_y,C_y,D_y,E_y) = 2$, we have $A_y,B_y,C_y,D_y,E_y \in [0,2]$. So color the vertices with ordinate in the interval $[0,1]$ black and ordinate in the interval $(1,2]$ white.

Now we have reached a contradiction - firstly note that by transversing $A \rightarrow B \rightarrow \cdots E$, we change the color of the interval (because no two adjacent vertices in the cycle can't be in the same interval since $|P_y - Q_y| > 1$) each time, so that implies the odd cycle is bipartite, a contradiction ! $\square$

So by Turan's theorem on $G$, we now find that it's the strictest bound possible.

BTW, Let $P = (p_1,p_2,\cdots, p_n)$ and  $Q = (q_1,q_2,\cdots, q_n)$. Also define the  functions $d_1(P, Q) = |p_1 -q_1| +|p_2 -q_2|+ \cdots + |p_n -q_n|$ and $d_2(P, Q) = \max(|p_1 -q_1|, |p_2 -q_2|, \cdots, |p_n -q_n|) $. Also, let $T(m, n)$ denote the maximum number of edges in an $K_n$ free graph with $m$ vertices.

 Using the second function for defining harmonic pairs in $n$ dimensions, coupled with a [hide=generalization] Let $f(n)$ denote the minimum number for which no matter how the edges of a complete graph with $f(n)$ vertices are colored with $n$ colors there must exist an odd cycle. Then $f(n) = 2^n + 1$ [/hide] of a ISL 1983 problem found in PfTB Chapter 22, it's not hard to adopt the above proof to show that maximum number of harmonic pairs among $k$ points is $T(k, 2^n+1)$.

I have no idea what's the analogous answer when the first function is used for defining harmonic pairs, but I think it's $T(k, 2n+1)$.

diagram solves the problem.

[hide=motivational remarks]
graph theory interpretation is pretty natural, and so is the turan idea. so really the hard part is proving the $K_5$

the idea below is not too hard to find after you express algebraically? admittedly the property is quite magical, but i found that after putting the lengths together and just focusing on one point at a time the diagram below came pretty fast.
[/hide]

@above isn't that usamo 2008 3, not usa tst

We claim that the answer is $3750$. For the construction, put $25$ points near each of $(0.75,0),(0,0.75),(-0.75,0),(0,-0.75)$.

Connect two points if they are harmonic.

[b][color=#07f]Claim:[/color][/b] There cannot be a $5$-clique.
[i]Proof:[/i] Instead of doing normal stuff, we use the orz ERDOS SZEKERES theorem to show that there must be an increasing or decreasing subsequence of $3$ points. However, since taxicab distance is additive for monotonic subsequences, these $3$ points cannot all be harmonic.

By Turan’s theorem, we are done.

We claim that the answer is $3750$; this is achievable by taking $25$ points in a sufficiently small disk centered at each of $(0, \pm \frac{3}{4})$ and $(\pm \frac{3}{4}, 0)$. Construct graph $G$ on 100 points in the plane where we place an edge between two points if they form a harmonic pair.

[b]Claim:[/b] $G$ has no $5$-cliques.
[i]Proof:[/i] Assume to the contrary $G$ has a $5$-clique. By Erdos-Szekeres, we can find three vertices $P_1, P_2, P_3$ of this clique where $P_i=(x_i, y_i)$ for $1 \le i \le 3$ such that $x_1 \le x_2 \le x_3$, and $y_1, y_2, y_3$ are in either increasing or decreasing order. However $d(P_1, P_3)=d(P_1, P_2) + d(P_2, P_3)>2$ while $d(P_1, P_3) \le 2$, a contradiction. Thus there is no $5$-clique. $\square$

By Turan's theorem, the number of edges in $G$ is at most $t(100, 4)=100^2 \cdot \frac{4-1}{2 \cdot 4} = 3750$, and since we have a construction of $3750$ edges, the final answer is $3750$, as claimed.


How can we be sure if we are invited to ARML, how will they know our emails? How are teams selected?

Contact whoever is in charge of ARML in your region (which seems to be Texas in your case).

Everything about participating in ARML is local, and varies from region to regions. [url=http://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team]This[/url] contains information about a number of teams, although not (for now) Texas.


A set of positive numbers has the triangle property if it has three elements which are the side lengths of a non-degenerate triangle. Find the largest $ n$ such that every $ 10$-element subset of $ \{4,5,6,\ldots,n\}$ has the triangle property.

That link goes to a thread about AIME 2001 Problem 8 ...

[quote="mathcrazed"][hide="Solution?"]ermm...

Consider the subset $ (4,5,6,7,8,9,10,11,12,n)$

Largest $ n$ is then $ 22$[/hide][/quote]

[hide]4,5,6 make a non degenerate triangle, so no matter what n is that set has the triangle property.[/hide]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=714297#714297 anyway.

[hide="hint"]think of a fibonacci-like recursion[/hide]

[hide] Since we want the biggest that works, let's find the smallest that doesn't. This should happen when the two sides of triangle add up to the third, so we just add two consective numbers to find the next ones. The last one is 254, so it's 253.

I think.[/hide]


Find the number of square meters in the area of a regular hexagon inscribed in a circle of diameter 12 meters.

Since the hexagon is regular, it is comprised of 6 equilateral triangles.  If the diameter of the circle is 12, then the radius is 6, so each side of the hexagon has length 6.  Then the area is just 6 times the area of an equilateral triangle with side length 6.

Finishing, we have the area of an equilateral triangle of side length 6 to be $\frac{6\times3\sqrt{3}}{2}$ (we get the $3\sqrt{3}$ from a $30-60-90$ triangle by the height).  Then we multiply by $6$ to get $\boxed{54\sqrt{3}}$ square meters.


Hello.

Could someone PM me some previous MathCounts (state or nationals would be preferred) competition problems?

I really need some to practice because I am horrible at MathCounts... :blush:

Websites other than the mathcounts website and goodlymaths would also be helpful.

-Thanks.

Dude, you're really good at MathCounts.

Anyway, there's another thread that's really similar to this... that's still active...

[quote="snipefreak"]Have you tried Elias Saab's mathcounts practice site? It's really good, and it is pretty hard...sometimes. I'll give you the URL if you need it.[/quote]
Some of those problems are really hard! In one of them, I had to do $\left(\frac{2}{91}\right)^{3}$ and [i]it was a sprint round![/i]

I've tried that website...
It is too hard.
I need some other websites (not saab, not goodlymaths, not mathcounts)

There have been some websites mentioned in this forum, but I can't find it!

did you try [url]http://www.yashar.biz/mathcounts/[/url]


find all functions $f$,$Z\Rightarrow R$such that :$f(0)=2, f(1)=\frac{5}{2}$ and
$f(x)f(y)=f(x+y)+f(x-y)$

If we take  in the equation $x = 0$ then $f(y) = f(-y) \; \; \forall y \in Z$, and for this we can resolve the equation in number not negative.

The next step is take $y = 1$ in the original equation and the:

  $f(x+1) = \frac{2}{5}f(x)-f(x-1) \; \forall x \geq 1 \in Z_{+}$

and this is a classical recurrent equation and easy to resolve.

For $n\geqslant 0$ put $f(n)=f_{n}$. Then

${5\over 2}f_{n}=f_{n+1}+f_{n-1}$

The characteristic equation is $t^{2}-{5\over 2}t+1=0\iff t_{1}=2, t_{2}={1\over 2}$

Hence $f_{n}=A\cdot 2^{n}+B\cdot 2^{-n}$. Substituting $f_{0}, f_{1}$ we get

$A+B=2, 2A+{B\over 2}={5\over 2}\implies A=B=1$, hence

$f(n)=2^{n}+2^{-n}$

Plugging $f(x)=2^{x}+2^{-x}$, we see that the equation is satisfied for all integer $x,y$. Hence $f(x)=2^{x}+2^{-x}$ is the only solution.


trigonometry
function
ratio
geometry unsolved
geometry

we know that cos(sin(x)) is always greater than sin(cos(x)) .

Taking this concept to infinity, which of these functions is greater: C(x)=cos(sin(cos(......(sin(x))))...))), or S(x)=sin(cos(sin(......(cos(x))))...))) ?

How does the ratio C(x)/S(x) depend on x?

Let $C_{n}(x)$ be the n-th iterration of $\cos(\sin(x))$.
That is, $C_{1}(x) = \cos(\sin(x))$
and $C_{n+1}= \cos(\sin(C_{n}))$

Similarly define $S_{n}(x)$

There are two numbers $a,b$ such that
$\cos(a)=b$
$\sin(b)=a$

These numbers are
$a \approx 0.694819690730788$, and

$b \approx 0.768169156736796$

So, if $C_{n}(x)=b$ for some $n$, then we'll have $C_{n+k}(x)=b, \ \ \forall k>0$
Similarly for $S_{n}$

The interesting thing is that, [u]independently[/u] of $x$, these sequences tend very quickly to their limits
(after about 50 iterrations the difference between successive terms is less than $10^{-15}$)

How did you get these approximations ?

for a -> 0.6948... and,
for b -> 0.7681...

I didn't find them by hand.

I just put an initial value for $x$ in Microsoft Excel and then I calculated consecutively $\sin x, \cos(\sin x), \sin(\cos(\sin x)), ...$

Even if I changed $x$, I got the same result

Before I saw this I posted the problem: Find solutions to $\cos \sin x= \sin \cos x$ in another forum  :blush: Well I hope they don't look here! I came up with it while playing with [url=http://www.fooplot.com]FooPlot[/url].


function
vector
limit
complex numbers
superior algebra
superior algebra unsolved

and eum... when you call a real "of finite dimension"? Do you mean a vector of reals? (but in that case you'd have scalar+ vector) :?

I think that harazi meant that for every real $t\in\mathbb{R}$ the vectorial space generated by $g_t(x)=f(x+t)$ is of finite dimension

[quote="harazi"]Find all derivable functions $f$ from the reals to the complex numbers with the property that the vectorial space generated by  the  functions $ g_t(x)=f(t+x)$ where $t$ is arbitrary real is of finite dimension. I just love this problem![/quote]
I agree, it is quite nice... Maybe I will propose it for my students.

Unfortunatelly, I am slightly in hurry.

The answer is $f(x)=\sum_{k} P_k(x)e^{a_kx}$ [b](*)[/b]. You can easily see that $f(x)$ satisfies all conditions.

We will prove that in any case $f(x)$ satisfies to ODE with constant coefficients. Then it follows that [b](*)[/b] is correct.

let $\{f_k\}$ be a basis.

It is sufficies to prove that $f'(x)$ lies in the same vectorial space. Indeed, $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$. So we can write $f'(x)=\lim_{h\to 0} \sum a_{k,h}f_k(x)$. Let's prove that $a_{k,h}$ are bounded as $h\to 0$. Suppose the contrary, then there is $h_n$, s.t. $m_n=\max{|a_{k,h_n}|}\to\infty$. It means that $\sum \frac{a_{k,h_n}}{m_n}f_k(x)\to 0$ for all $x$. But coefficients $\frac{a_{k,h_n}}{m_n}$ are bounded and $\max |\frac{a_{k,h_n}}{m_n}|= 1$. So we can assume $a_{k,h_n}\to a_k$. It follows that $\sum a_kf_k=0$ for all $x$, $\max |a_k|=1$. Contradiction.
Since $a_{k,h}$ are bounded we may assume they converges for some sequence $h_n$, so $f'(x)$ lies in the same vectorial space.


calculus
integration
function
real analysis
real analysis unsolved

Solve  the equation : Integral(0 to x){f(t)^m.dt} = (Integral(0 to x){f(t).dt})^m
where  f:R*-->R*(non-negative real), f  continue,  m positive real
See you ...

A particular solution when f is not continuous:
f(t)=C_B(t) where C_B(t) is the caracteristic function
on B
C_B(t)= 1 if t \in B
C_B(t)=0 if t not in B

if m=1 we have identity! :D that is obious .. but it's the best I can do with an integral! :D

Hi Moubi,
What is B ? Is your solution an integrable fonction ?

I was reading a book in wich there was a problem little like your

Let f:R->R+ Lebesgue's integrable wich means \int_{R}f(x)dx<oo
and for all n,m positive integers we have
 \int _{R}(f(x))^m dx =  \int _{R}(f(x))^n dx
then f is equal a.e. (almost every where) to the caracteristic function
C_B(x) where B is a measurable set of finite measure.

Hmmm... It doesn't look to me like they're related...

Damned ! Nobody gives a solution ? Is it so hard ?

I think that if we derivate the expression we get f(x)/F(x)=k (a given constant),where f is the function from the pb and F(x) one of its primitives


induction
number theory proposed
number theory

hi,
prove that there are an infinite number of solutions $ (x,y,u,v) \in \mathbb{N}^{4}$ of
\[ x^{2} \minus{} y^{2} \equal{} u^{v},\]
where $ u,v$ are primes.

For every $ k\in \mathbb{N}$,
$ (x,y,u,v)\equal{}(2^{k}\plus{}2^{k\minus{}1},2^k\minus{}2^{k\minus{}1},2,2k\plus{}1)$ satisfies the equation and there are infinitely many $ k$ such that $ 2k\plus{}1$ is prime,so we are done.

If $ 0\in\mathbb{N}$ then take $ x\equal{}u\equal{}p,y\equal{}0,v\equal{}2$ where $ p\in\mathbb{P}$.

[quote="puuhikki"]If $ 0\in\mathbb{N}$ then take $ x \equal{} u \equal{} p,y \equal{} 0,v \equal{} 2$ where $ p\in\mathbb{P}$.[/quote]
actually i ment here that $ 0 \not\in \mathbb{N}$.
my solution is:
We construct a series $ (x_{n})_{n \geq 1}$ and $ (y_{n})_{n \geq 1}$ with $ x_{1} \equal{} 2$ and $ y_{1} \equal{} 1$ with
$ x_{n\plus{}1} \equal{} 2x_{n} \plus{} y_{n}$ and $ y_{n\plus{}1} \equal{} x_{n} \plus{} 2y_{n}$. With induction we have $ 3^{n} \equal{} x_{n}^{2} \minus{} y_{n}^{2}$. Now we take $ x \equal{} x_{p_{k}}$ and $ y \equal{} y_{p_{k}}$ and $ u\equal{}3$ and $ v\equal{}p_{k}$, where $ p_{k} \in \mathbb{P}$. Then the equality follows.

For $ K,L \in \mathbb{N}$ such that $ KL\equal{}u^v$ , $ K>L$ and $ K \equiv L\ (mod\ 2)$,
we can have

$ (x,y,u,v)\equal{}(\frac{K\plus{}L}{2},\frac{K\minus{}L}{2},u,v) \in \mathbb{N}^4$

[quote="arcsin1.01"][quote="Makoto Kohno"]For $ K,L \in \mathbb{N}$ such that $ KL \equal{} u^v$ , $ K > L$ and $ K \equiv L\ (mod\ 2)$,
we can have

$ (x,y,u,v) \equal{} (\frac {K \plus{} L}{2},\frac {K \minus{} L}{2},u,v) \in \mathbb{N}^4$[/quote]Did you notice that $ u,v$ must be primes?[/quote]

For every odd primes $ u,v$ we can get $ K,L\in \mathbb{N}$ as above.

So "there are an infinite number of solutions $ (x,y,u,v) \in \mathbb{N}^4$ " is shown.

[quote="Makoto Kohno"]For every odd primes $ u,v$ we can get $ K,L\in \mathbb{N}$ as above.[/quote]How do you prove that?

If  $ u$ and $ v\equal{}2r\plus{}1$ are odd primes ,

for every $ 0\le i \le r$ (which says $ i<v\minus{}i$) let $ L\equal{}u^i$ and $ K\equal{}u^{v\minus{}i}$

then $ K>L$, $ KL\equal{}u^v$ and then let $ x\equal{}\frac{K\plus{}L}{2},\ y\equal{}\frac{K\minus{}L}{2}$.

actually it is to find an infinite number of solution  :?:

the number of prime numbers is infinite.


[quote="arqady"][quote="M.A"]hello

i cant solve this problem. please help me.

 let a,b,c be positive numbers such that abc=1 prove that
(a/(b+3))^1/2 + (b/(c+3))^1/2 +(c/(a+3))^1/2>=3/2[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142431[/quote]

thank you very much arqady


Prove that

$ \sum_{cyc}\frac {x^3}{x^2 \plus{} xy \plus{} y^2}\ge \frac {x \plus{} y \plus{} z}{3}$

for $ x,y,z$ are all positive real numbers.

[edited] :D

[quote="vinskman"]Prove that

$ \sum_{cyc}\frac {x^3}{x^2 \plus{} xy \plus{} y^2}\ge \frac {x \plus{} y \plus{} z}{3}$[/quote]

[hide="the first step"]$ \frac {x^3}{x^2 \plus{} xy \plus{} y^2}\equal{}\frac{x^3\plus{}x^2y\plus{}xy^2\minus{}x^2y\minus{}xy^2}{x^2\plus{}xy\plus{}y^2} \equal{} x\minus{} \frac{xy(x\plus{}y)}{x^2\plus{}xy\plus{}y^2}$[/hide]

[hide="Big hint"]
$ \sum_{cyclic} \frac {x^3}{x^2 \plus{} xy \plus{} y^2} \equal{} \sum_{cyclic} \frac {y^3}{x^2 \plus{} xy \plus{} y^2}$. You can check it's true. [/hide]

[quote="Potla"][quote="vinskman"]Prove that

$ \sum_{cyc}\frac {x^3}{x^2 \plus{} xy \plus{} y^2}\ge \frac {x \plus{} y \plus{} z}{3}$[/quote]

Are $ x, y, z$ positive reals or can they take arbitrary real/complex values?? :huh:
Please explain, vinksman. I think the Q is incomplete...[/quote]

Thanks for editing, vinskman! :D

[hide]
Using [b]Bugi's[/b] hint;

Proof of hint:
$ \sum_{cyc} \frac {x^3}{x^2 \plus{} xy \plus{} y^2} \minus{} \sum_{cyc} \frac{y^3}{x^2 \plus{} xy \plus{} y^2}\equal{}\sum\frac{(x\minus{}y)(x^2\plus{}xy\plus{}y^2)}{x^2 \plus{} xy \plus{} y^2} \equal{}\sum x\minus{}y \equal{}0 \qquad Q.E.D$

So the inequality is equavalent to
$ \sum \frac{x^3\plus{}y^3}{x^2 \plus{} xy \plus{} y^2} \ge \sum \frac{x\plus{}y}{3}$

Now consider
$ \frac{(x\plus{}y)(x^2 \minus{} xy \plus{} y^2)}{x^2 \plus{} xy \plus{} y^2} \ge \frac{x\plus{}y}{3}$

Rearrange to get
$ 3(x^2 \minus{} xy \plus{} y^2) \ge x^2 \plus{} xy \plus{} y^2 \Longleftrightarrow x^2\plus{}y^2 \ge 2xy$
Which is true by AM-GM

Apply the same method for each term and sum up to get the desired inequality
Equality occurs when $ x\equal{}y\equal{}z$[/hide]

[quote="PKMathew"][quote="HTA"][quote="vinskman"]Prove that

$ \sum_{cyc}\frac {x^3}{x^2 \plus{} xy \plus{} y^2}\ge \frac {x \plus{} y \plus{} z}{3}$[/quote]

[hide="the first step"]$ \frac {x^3}{x^2 \plus{} xy \plus{} y^2} \equal{} \frac {x^3 \plus{} x^2y \plus{} xy^2 \minus{} x^2y \minus{} xy^2}{x^2 \plus{} xy \plus{} y^2} \equal{} x \minus{} \frac {xy(x \plus{} y)}{x^2 \plus{} xy \plus{} y^2}$[/hide][/quote]

Can you post some more details?[/quote]
By HTA's suggestion and AM-GM, we have:
$ \sum_{cyc}\frac {x^3}{x^2 \plus{} xy \plus{} y^2}\equal{}\sum_{cyc}x \minus{} \frac {xy(x \plus{} y)}{x^2 \plus{} xy \plus{} y^2}\ge \sum_{cyc}x\minus{}\frac{xy(x\plus{}y)}{3xy}\equal{}\frac{x\plus{}y\plus{}z}{3}$


calculus
integration
inequalities
geometry
angle bisector
triangle inequality
geometry unsolved

Triangle $ ABC$ has integer-length sides and $ AC\equal{}2007$. The internal bisector of $ \angle BAC$ meets $ BC$ at $ D$. Given that $ AB\equal{}CD$, determine $ AB$ and $ BC$.

[hide=" My Solution"]
By the Angle Bisector theorem and by the given statement, we get
\[ \frac {AB}{2007} \equal{} \frac {AB}{AC} \equal{} \frac {BD}{DC} \equal{} \frac {BC}{AB} \minus{} 1\]

\[ \Longrightarrow AB^2 \plus{} 2007 AB \equal{} 2007\cdot BC\]
Since the triangle has integral sides,
Therefore, $ 2007|AB^2$ and thus $ 669|AB$

Let $ AB \equal{} 669k$, where $ k \in \mathbb{N}$
Therefore, the equation reduces to
$ 223 k^2 \plus{} AB \equal{} BC$
Using the triangle inequality $ AC > |BC \minus{} AB|$, we get
$ 2007 > 223 k^2 \Longrightarrow 3 > k$
Since $ k$ is a natural number, therefore $ k \equal{} 1$ or $ k \equal{} 2$,
$ k \equal{} 1$ violates the triangle inequality (makes $ AC$ greater than the sum of the other two sides)

Therefore
\[ \boxed{k \equal{} 1,\ AB \equal{} 1338, \ BC \equal{} 2230}\]
[/hide]

[quote="gokussj3"]small correction in last step k=2
otherwise nice soln  :wink:[/quote]
Thanks....But I think anyone who reads it carefully will get that typo... :)


Give an example of a number in decimal form which increases when it is rounded to two significant figures but decreases when rounded to the nearest integer and decreases when rounded to the nearest tenth.

4.449

The message is too small. Please make the message longer before submitting.

Please explain how you got this number, and especially what significant figures are. Thank you.

I do not think 4.449 is valid:

4.449 rounded to the nearest integer = 4
4.449 rounded to the nearest tenth = 4.4
[color=red]4.449 rounded to 2 sigfigs = 4.4[/color]

0.449 is, though

0.449 rounded to the nearest integer = 0
0.449 rounded to the nearest tenth = 0.4
0.449 rounded to 2 sigfigs = 0.45

since nobody actually said what significant digits are, i'll give an explanation

"Significant figures are the digits in any measurement that are known with certainty plus one digit that is uncertain" - taken from science worksheet =)

basically, it's a number that has only one uncertain digit
ex: you have a ruler and you just measured a line. the line was about 1/3 of the way between 12.3 and 12.4
the significant figure for this measurement is 12.33
since 12.3 is certain (due to markings on the ruler) and your final 3 is uncertain (due to estimating  the distance) you have your answer

now, this question doesn't talk about significant figures, it talks about significant digits

here are some rules to determine if a digit is significant:
1. digits from 1-9 are [b]always[/b] significant
2. zeros between two other significant digits are always significant
3. one or more additional zeros to the right of both the decimal place and another significant digit are significant
ex: 1.500, the zeros are significant because they are to the right of the decimal place and another significant digit (5)
4. zeros used solely for spacing the decimal point (place holders) are not significant
ex: 0.3456, the zero is not significant
however, in the number 504, the zero is significant because of rule 2

[quote="james4l"]0.449 is, though

0.449 rounded to the nearest integer = 0
0.449 rounded to the nearest tenth = 0.4
0.449 rounded to 2 sigfigs = 0.45[/quote]

I put that number down and had it marked wrong. (well, I only had .449 sans the zero)

Apparently, the correct answer is 559.21. But there are many correct answers...


Given triangle $ABC$ with an arbitrary point $P$ and three altitude $h_{a},h_{b},h_{c}$ prove that
$\frac{PA}{h_{a}}+\frac{PB}{h_{b}}+\frac{PC}{h_{c}}\ge 2$

i have a nice solution!

this is the same that prove that $aPA+bPB+cPC \geq 4[ABC]$.

let A', B' and C' the points, out of the triangle, such that APBC', BPCA' and APCB' are parallelograms.

we have: $[ACA'] = \frac{1}{2}AC.CA'.sin(ACA') \leq \frac{1}{2}AC.CA'$

$= \frac{1}{2}b.BP$

similary we have that $[ABA'] \leq \frac{1}{2}c.PC.$

adding, we get $[ABC]+[A'BC] \leq \frac{1}{2}(b.PB+c.PC)$. (*)

add (*) with your analogous, and the result comes..

Thank you it was wellknow inequality  :blush: but your solution is new and nice!


Let p be a given odd prime number. Prove that
$A=\sum_{k=1}^{p} \binom{p}{k}\binom{p+k}{k} - (2^p +1)$  is divisible by $p^2$
 ( $\binom{p}{k}$ is binomial coefficient  ).


ratio
trigonometry
geometry
circumcircle
geometry unsolved

[quote="Doctor A"](F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and $ \angle A \equal{} 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are such that $ \angle KAC \equal{} \angle LAB \equal{} 10^{\circ}$. Determine the ratio $ CK/LB$.[/quote]
See that $ |BC| \equal{} sin(50)*|AC|$, $ |BK| \equal{} sin(40)*|AC|$, $ |BL| \equal{} sin(10)*|AC|$.
So $ \frac{|CK|}{|BL|} \equal{} \frac{sin(50)\minus{}sin(40)}{sin(10)}$

i think you're wrong... we have that $ \frac{CK}{BL}\equal{}\frac{\frac{AC \sin 10}{\sin 80}}{\frac{AB \sin 10}{\sin 100}}\equal{}\frac{\sin 100}{\sin 50\sin 40}\equal{}2$...

I don't see anything's wrong with Mathias_DK's, but we can further simplify the expression into $ \frac1{\sqrt2\cos5^{\circ}}$.

I think campos's solutions is incorrect, because $ CK\equal{}\frac{AC\sin10^{\circ}}{\sin130^{\circ}}$ and $ BL\equal{}\frac{AB\sin10^{\circ}}{\sin80^{\circ}}$.

yes, it has a typo, however $ \frac{\frac{AC\sin 10}{\sin 50}}{\frac{AB\sin 10}{\sin 100}}\equal{}\frac{\sin100}{\sin 40\sin 50}\equal{}2$...

Sorry.. Must have been sleeping or something.. :S
$ |BL| \equal{} tan(10) * |AB|$,  $ |BK| \equal{} tan(40) * |AB|$, $ |BC| \equal{} tan(50) * |AB|$, $ |CK| \equal{} |BC| \minus{} |BK| \equal{} (tan(50) \minus{} tan(40))*|AB|$.
So $ \frac {|CK|}{|BL|} \equal{} \frac {tan(50) \minus{} tan(40)}{tan(10)} \equal{} 2$ :)

Seems to me that F. Nilov got to this problem working on Langley's 80-80-20 triangle.

   Here is PWW figure.

   M.T.

There is a very nice and short geometrical solution!
[hide]the distance from the circumcenter to the midpoint of a side in a triangle is half the distance from the orthocenter to the opposite triangle vertex[/hide]

1) See that ALC is an isosceles triangle with AL = LC ( 1 )
2) Take L' the symmetrical of L w.r.t. B. AL = AL' ( 2 ) The triangle AL'K is isosceles, with AL' = KL' ( 3 ).
From (1), (2) and (3) we get CL = KL', hence CK = LL', but LL' = 2 BL, so the ratio is 2.

Best regards,
sunken rock

I think that the configuration of the regular [b][size=100]9-gon[/size][/b], help us to solve this problem without words.

Kostas vittas.

This is my first post in this forum. I`m sorry because my english is not good.

Let N the midpoint of AC. Note that triangles KAC and BNL are similar(*) and the BD/AC=BL/KC=1/2 and hence CK/LB=2.

(* it`s not hard to prove this)[/code]

Draw KQ perpendicular to AK,KQ meet BC at Q;we have :△AKQ∽△ABL, Thus,BL/KQ=cos40;By angles,we also have :KQ=QC, Hence CK=2QK=2cos40*QK=2cos40*(BL/cos40)=2BL;obviously,CK/BL=2..   :lol:  :)

[quote=Doctor A](F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and $ \angle A \equal{} 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are such that $ \angle KAC \equal{} \angle LAB \equal{} 10^{\circ}$. Determine the ratio $ CK/LB$.[/quote]

$2sin(x)sin(90 - x) = 2sin(x)cos(x) = sin(2x)$ and $sin(x) = sin(180 - x)$.
We apply the $sinus$ $theorem$ in triangle $ALB$ and we get $\frac{LB}{sin(10)} = \frac{AB}{sin(80)}$.Now we apply $sinus$ $theorem$ in triangle $KCA$ and we get $\frac{KC}{sin(10)} = \frac{AC}{sin(50)}$ but $AC = \frac{AB}{sin(40)}$ $\implies$ $\frac{KC}{sin(10)} = \frac{AB}{sin(40)*sin(50)} = \frac{2AB}{sin(100)} = \frac{2AB}{sin(80)} = \frac{2LB}{sin(10)}$ and now we get that $KC = 2LB$ $\implies$ $\frac{KC}{LB} = 2$.

In ∆ ABC, sin ACB=AB/AC
AC/ AB=1 / sin 40°
In ∆AKC,
sin KAC/ CK= sin LKC/ AC
sin 10°/ CK= sin 130°/ AC
sin 10°= (CK sin 50° )/AC
In ∆ ABL,
sin BAL/ BL= sin ALB/ AC
sin 10°/ BL= sin 80°/ AB
sin 10°= (LB sin 80° )/AB
CK/ LB=( AC/ AB)* (sin 80°/ sin 50°)
= 2 sin 40° cos 40°/ sin 40° . sin 50°
= 2 cos 40°/ sin 50°
= 2
CK/ LB =2


AMC AIME USA(J)MO USAMO geometry limit function

...okay, I highly doubt that you can't apply [b]anything[/b] from AoPS I and II to the problems in question. And usually, by the time you're working on those later AIME questions and early USAMO, you wouldn't be doing AoPS II anymore, would you?

EDIT: And looking at the problems in question, I have no idea what you're talking about. You most definitely can use things you learned in AoPS books to solve at least most of them, if not all.

[quote="cdymdcool"]I just find out that AoPS book doesn't cover the AIME material, for example: #9-#15 on the first AIME in 2007, AoPS doesn't talk about those topics at all. :([/quote]

Look at the books more closely. . . Nearly all the fundamental tools needed for those problems are in Volume 1 or 2.  (Much of the geometry needed is actually in Volume 1, not Volume 2.)

Is the problem that you're expecting the books to teach methods that make the problem obvious? No book will directly teach the solutions to difficult contest problems, such as late AIME ones. You'll always need some degree of creativity to solve them.

As Rusczyk said, AOPS will give you the fundamental tools, and even give you a few tips on what to do with them. Ultimately, you're the one that has to put them to use.

The AoPS books are not Algebra II textbooks, and the AIME is not the SAT. For example, in an Algebra II textbook, you may see 50 problems in a row that look like the following:

Find the minimum value of $ 2x^{2}\plus{}3x\plus{}7$.

Then on the SAT you find the following question:

Find the minimum value of $ 3x^{3}\minus{}5x\plus{}9$.

The point of the AIME is not to test how well you have read the AoPS books and how many of the problems you've memorized the methods for; it's to test your creative thinking and problem solving.

that could not have been better said 1337. :D

If math was just as easy as memorizing a few books, we'd have a world full of math geniuses.  Math is about thinking.  The fact is, unless you want to be an engineer or physicist, you'll never need to use advanced math.  [b]However[/b] you will always be able to use the deeper thinking capabilities you learn from doing math.

Have you done all the exercise problems in the AoPS books? As I was doing some of them, I could have sworn the topics weren't covered in the chapter either ;)

Also, if you can get 8 on the 2007 AIME (#1-8) you'd be in pretty good shape anyway.

Well, if you're good at math, figure out how to do them without reading a book to tell you.

I am not sure I understand the premise of the whole discussion.

Why should be there any corelation between AIME problems and AoPS books?

People that invent problems for AIME have, most likely, never read any AoPS book and do not feel obliged to cover any "topic" for the exam. They just invent good, but elementary, problems and propose them.

On the other hand, why is an AoPS book that does not help you with AIME useless? If you enjoyed reading it and learned something from it, it served its purpose.

Because there is a correlation between AOPS books and pretty much every high school mathematics competition that doesn't suck. If the math competition is good, there's not a specific formula that helps you get better. You get better at one, you usually get better for almost all of the others. AOPS happens to be one of the best resources for preparing for high school competitions. Replace AIME with almost any other competition and this thread still holds true.

I agree with shunka somewhat. I don't particularly feel that AoPS is some sort of AIME-preparation book. It is a good book to learn the basics of various topics covered by the AIME, and quite a few that are not covered. However, I think the best way to practice for the AIME is to try past tests.

[quote="13375P34K43V312"]The AoPS books are not Algebra II textbooks, and the AIME is not the SAT. For example, in an Algebra II textbook, you may see 50 problems in a row that look like the following:

Find the minimum value of $ 2x^{2} \plus{} 3x \plus{} 7$.

Then on the SAT you find the following question:

Find the minimum value of $ 3x^{3} \minus{} 5x \plus{} 9$.

The point of the AIME is not to test how well you have read the AoPS books and how many of the problems you've memorized the methods for; it's to test your creative thinking and problem solving.[/quote]

What Algebra II book are you using?

I felt that the aops books taught me a way of thinking [or at least guided me towards one]. It depends on the person. For me, doing the AoPS books helped a lot for AIME problems. cdmycool, it seems that you are not reading the books in a way that will help you develop problem solving ability. Perhaps, you are not ready for the books/ you capable of learning them on your own just yet.

[quote="MysticTerminator"][quote="MellowMelon"]Yeah, I won't disagree with that. Provided you're at the right level of knowledge, the best way to prepare for any competition is probably to do past tests. But there is more than one way to look at this. If doing well on a test is your primary goal, do the past tests then. If competitions are simply a means to an end, a way for you to be exposed to more mathematics and to get better at it (rather than you getting better at math for the competition), AOPS and similar books will probably suit you more.

Last year someone told me that he spent most of the year studying some topics in college-level mathematics, and was then pleasantly surprised by how easy high school contest problems seemed afterwards. He was a USAMO winner.[/quote]

wait was this me? probably not. I am confused.[/quote]

This is probably Krishanu. By the way, the Aops books are useful. QED

[quote="13375P34K43V312"]The AoPS books are not Algebra II textbooks, and the AIME is not the SAT. For example, in an Algebra II textbook, you may see 50 problems in a row that look like the following:

Find the minimum value of $ 2x^{2} \plus{} 3x \plus{} 7$.

Then on the SAT you find the following question:

Find the minimum value of $ 3x^{3} \minus{} 5x \plus{} 9$.

The point of the AIME is not to test how well you have read the AoPS books and how many of the problems you've memorized the methods for; it's to test your creative thinking and problem solving.[/quote]
im going to risk going on a rather off-topic tangent here, but how do you find the minimum value of $ 3x^{3} \minus{} 5x \plus{} 9$? I thought it would be $ \minus{}\infty$  :maybe:

I'm pretty sure that was a hypothetical example that wasn't actually supposed to be a real problem. You are correct.

As for the person I've mentioned, the last I'll say about that is that none of you are right. Probably should have just not said that at all.

Yea, that's right. Let $ \lim x \rightarrow \minus{}\infty.$

[quote="erdogankerem123"]
im going to risk going on a rather off-topic tangent here, but how do you find the minimum value of $ 3x^{3} \minus{} 5x \plus{} 9$? I thought it would be $ \minus{} \infty$  :maybe:[/quote]

Technically, that would result in $ 3( \minus{} \infty)^{3} \minus{} 5( \minus{} \infty) \plus{} 9 \equal{} \minus{} \infty \plus{} \infty \plus{} 9 \equal{} \boxed{9}$, which is no lower than the same expression for zero. Like Karth said.

Technically (again), the answer should be that there is no minimum.

[quote="Temperal"][quote="erdogankerem123"]
im going to risk going on a rather off-topic tangent here, but how do you find the minimum value of $ 3x^{3} \minus{} 5x \plus{} 9$? I thought it would be $ \minus{} \infty$  :maybe:[/quote]

Technically, that would result in $ 3( \minus{} \infty)^{3} \minus{} 5( \minus{} \infty) \plus{} 9 \equal{} \minus{} \infty \plus{} \infty \plus{} 9 \equal{} \boxed{9}$, which is no lower than the same expression for zero. [/quote]

Um...no. That is highly illegal.

[quote="Karth"]Just so you know, -infinity + infinity = infinity.[/quote]

Um...no? I'm not sure whether or not you are trying to make a general statement, but it doesn't work in the above problem either...

[quote="Karth"]Make very sure you understand what a limit is... don't try to apply formulae. Graph out the function and see for yourself. Just so you know, -infinity + infinity = infinity.[/quote]

Because we all know that as $ x$ approaches infinity, $ \minus{}x\plus{}x$ approaches infinity as well.

Well, okay my statement wasn't perfectly correct, as $ \minus{}\infty \plus{} \infty \equal{} \pm \infty,$ where it depends. My statement should have been:

If $ f(x)$ and $ g(x)$ are two functions and $ x \rightarrow \infty,$ $ f(x) \minus{} g(x) \rightarrow \infty$ if $ f(x)$ increases faster than $ g(x).$ Now, since $ 3x^3 \rightarrow \minus{}\infty$ faster than $ \minus{}5x \rightarrow \infty$ when $ x \rightarrow \minus{}\infty,$ the entire expression collapses to $ \minus{}\infty.$ In general, go and understand why limits work how they do. Technically, you can't even plug in $ \infty$ as $ x,$ because $ \infty$ doesn't denote an actual number. It denotes a number which may be arbitrarily large. So if you really wanted to, plug in a ridiculously large number and see what happens.

[quote="Lazarus"][quote="Karth"]Well, okay my statement wasn't perfectly correct, as $ \minus{} \infty \plus{} \infty \equal{} \pm \infty,$ where it depends.[/quote]

What the ****? Did you even read what I wrote?[/quote]

Yes, and what you wrote is incorrect.  :wink: You keep confusing yourself as to what $ \infty$ is. It's not a number. However, you do bring up a good point that $ \minus{}\infty \plus{} \infty$ could equal anything, and thus it is undefined. In general, to evaluate such limits, you have to take a look at the graph if you're horribly confused, or resort to algebraic methods, etc.

Now, if we let $ f(x) \equal{} \minus{}x \plus{} x,$ according to your statement, $ \lim_{x \rightarrow \infty} f(x) \equal{} \lim_{x \rightarrow \infty} \minus{}x \plus{} x \equal{} \lim_{x\rightarrow\infty} \equal{} \lim_{x\rightarrow\infty}0 \equal{} \infty,$ which is terribly incorrect. So I don't see what you're trying to say.

Is it just me or does anyone else think that Daesun was being sarcastic?

I don't see any treething posts... is Lazarus daesun?

Oh, geeze, it was just a hypothetical example, possibly even a typo! Everyone is getting way off topic here.

I wouldn't know what standard AIME is, but the books contain lots of problems which I would expect to see in the AMC and gives good generalised approaches. The Solutions Manual is invaluable though, it would be good if it were just stuck on the end of the main text. Without it you keep repeating bad habits, with it you learn subtle tricks and the most efficient method.

That would be tantamount to declaring my allegiance to the Order of the Phoenix...

Seriously, though, I think that's a bad idea. In fact, I think the Solutions Manual is already too successful. I haven't heard of anyone who had the willpower to go through the problems the "right" way, that being by spending at least 30 minutes before looking at the solution.

[quote="Phelpedo"]
Seriously, though, I think that's a bad idea. In fact, I think the Solutions Manual is already too successful. I haven't heard of anyone who had the willpower to go through the problems the "right" way, that being by spending at least 30 minutes before looking at the solution.[/quote]

Does it count if i go over the problem in 5 minutes and check to see if my answer's wrong?

It counts if you find that it's wrong and then proceed to read the solution.

@Karth and Arnav: Actually I think he was referring to me.

I always spend at least a half hour on a problem before going to the solution, unless it's obvious that there's no way I'll solve it.

[quote="Karth"][quote="Lazarus"][quote="Karth"]Well, okay my statement wasn't perfectly correct, as $ \minus{} \infty \plus{} \infty \equal{} \pm \infty,$ where it depends.[/quote]

What the ****? Did you even read what I wrote?[/quote]

Yes, and what you wrote is incorrect.  :wink: You keep confusing yourself as to what $ \infty$ is. It's not a number. However, you do bring up a good point that $ \minus{} \infty \plus{} \infty$ could equal anything, and thus it is undefined. In general, to evaluate such limits, you have to take a look at the graph if you're horribly confused, or resort to algebraic methods, etc.

Now, if we let $ f(x) \equal{} \minus{} x \plus{} x,$ according to your statement, $ \lim_{x \rightarrow \infty} f(x) \equal{} \lim_{x \rightarrow \infty} \minus{} x \plus{} x \equal{} \lim_{x\rightarrow\infty} \equal{} \lim_{x\rightarrow\infty}0 \equal{} \infty,$ which is terribly incorrect. So I don't see what you're trying to say.[/quote]

HES MAKING FUN OF YOU HOW DO YOU NOT SEE THIS


[quote="cdymdcool"]So basically i just finished some important chapters of the AoPS volumn2 and all of the AMC problems, but i still barely score a 5 on most of the AIMEs.  Should i start doing Problem-Solving Strategies or Art and Craft of Problem Solving while i'm working on the AIMEs?[/quote]

Hmm...Art and Craft of Problem Solving is good but you should review AoPS volume 2.

ya im only halfway with aops V2 and i can get about a 7-8 usually..

dude contests are pretty much all experience

you don't really have to know like 1029471209571 facts; you just have to know when to use the facts you know

ok, so what math contests are equivalent to AIME and worth to do for preparation?

ARML problems are good, although the time limit is different. I also like using Mandelbrot problems and perhaps HMMT?

Should i do the AMC12 before 1980s too?  Or should i jump into the ARMLs?  BTW, where can i buy the past ARML and Mandelbrot problems?


Let a,b,c,d be 4 real numbers such that
$ (133a\plus{}29b\plus{}7c\plus{}2d\minus{}7)(91a\plus{}25b\plus{}7c\plus{}2d\minus{}7)<0$ prove that exist 2 real numbers x,y such that
$ x\plus{}y\equal{}7$ and $ a(x^3\plus{}y^3)\plus{}b(x^2\plus{}y^2)\plus{}c(x\plus{}y)\plus{}2d\minus{}7\equal{}0$

If $ g(x,y) \equal{} a(x^3\plus{}y^3) \plus{} b(x^2\plus{}y^2) \plus{} c(x\plus{}y) \plus{} 2d\minus{}7$, then $ g$ gives a continuous function out of the set $ S \in \{(x,y) | x\plus{}y \equal{} 7\}$, and
$ g(2,5)g(3,4) < 0$, so $ g(2,5)$ and $ g(3,4)$ have opposite signs. By continuity there is some $ (x,y)$ belonging to $ S$ such that $ g(x,y) \equal{} 0$.


Hi

\documentclass{book}
\usepackage[pdftex]{graphicx}
\begin{document}
\includegraphics[scale=0.6]{myimage.png}
\end{document}

This is the code to generate the image in the PDF

Below is the image in PDF

But the real image is

[img]http://img380.imageshack.us/img380/908/myimageyk2.png[/img]

If you realize, the image in the PDF is not the same that the original

How can I fix the problem?

[quote="MathTeX"]If you realize, the image in the PDF is not the same that the original[/quote]What's not the same?

I've changed my preferences

This is the code

\documentclass{book}
\usepackage[pdftex]{graphicx}
\begin{document}
\begin{figure}[h]
\includegraphics[width=5.6cm,keepaspectratio]{mov.png}
\end{figure}
\end{document}

This is the image that I wanna put in PDF (it's a test image)

[img]http://i121.photobucket.com/albums/o212/janyo_bucket/mov.png[/img]

This is the image that appears in the PDF

[img]http://i121.photobucket.com/albums/o212/janyo_bucket/1-1.png[/img]

As you can see, the image put in the PDF it's a little blurred

PNGs are bitmaps so don't scale well. Try to make the image the same size as you want in the document so it isn't resized. You could also try saving the png to pdf format first and including that instead in the tex file.

For the best images you could use LaTeX to draw the image itself, though of course that will need time to learn. Or use vector images which scale better such as the ones [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_About]Asymptote[/url] produces.

[quote="stevem"]PNGs are bitmaps so don't scale well. Try to make the image the same size as you want in the document so it isn't resized. You could also try saving the png to pdf format first and including that instead in the tex file.[/quote]

Thanks, but how can I do it?

Do what? You don't say how you drew the image but assuming it's a decent graphics program just resize it in that program.

To convert to pdf there are hundreds of programs out there or use one that prints to pdf. Some are listed at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=go2pdf&t=101681]word=>pdf[/url]. There are even online places that will convert images. The first one I came across in Google is https://www.pdfonline.com/convert_pdf.asp , there will be many more.

Forget all, I'm new in this.

Please help me with this

Suppose that I wanna put this image in the document

[img]http://img251.imageshack.us/img251/7373/movam4.png[/img]

Tell me the steps to generate this image in the PDF

Thanks

As before you are not making yourself clear about what you want to do. Do you want to
1) make a pdf from the diagram just drawn which you then include in your latex code?
or
2) do you want to write latex code to draw the image?

1) convert the image to pdf which I have already said how to do in a previous post above then use \includegraphics to put it into your document.

2) You will either have to learn all about the picture environment or else download [url=http://jpicedt.sourceforge.net/site/index.php]JPicEdit[/url]. JPicEdit is a free drawing program where you can draw your picture but it also generates the LaTeX code to insert into your document.


induction
inequalities
number theory proposed
number theory

Let $f:\mathbb{N^*}\to\mathbb{N^*}$ be bejective function.
[b](1)[/b]$f(m\cdot n)=f(m)\cdot f(n),\forall m,n\in\mathbb{N^*}$
[b](2)[/b]$m|n\iff f(m)|f(n)$
Prove that:$(1)\iff (2)$
I had poor solution  :( ,and I need simpler solution.Can you help me?

Isnt it quite straightforward? But, maybe is that your 'poor' solution... ;)

Pierre.

What? An answer or a nice answer?

Pierre.

- If $f(mn) = f(m)f(n)$ for all $m,n$.
Then, if $m$ divides $n$ we have $n=km$ so that $f(n)=f(m)f(k)$ and $f(m)$ divides $f(n)$.
Conversely, if $f(m)$ divides $f(n)$ then using that $f$ is injective, we may find $k$ such that $f(n)=f(k)f(m)=f(km)$. Using injectivity again, we deduce that $m=kn$, so that $m$ divides $n$.
Thus $m|n$ iff $f(m)|f(n)$.

- If $m|n$ iff $f(m)|f(n)$ :
Clearly $f(1)$ divides all $f(n)$ and since $f$ is surjective, we deduce that $f(1)$ divides $1$, so that $f(1)=1$. Note that it follows $f(n) > 1$ for all $n>1$ from the injectivity.
Now, let $p$ be a prime number. Let $q=f(p)$. Suppose that $q$ is not a prime, then it has a prime divisor $q'=f(p')$ where $q' < q$ so that $1 < p' \neq p$. But $p'$ has to divide $p$, a contradiction. Thus $q$ is a prime.
Using induction and the same reasoning, we easily deduce that for all $n \geq 1$ we have $f(p^n) = q^{\alpha_n}$ where $\alpha_1 = 1 < \alpha_2 < \alpha_3 < \cdots $ (we have strict inequalities using injectivity).
In another hand, it is clear that using the same lines as above, if $f(m) = q^a$ for some positive integer $a$ then $m$ is a power of $p$.
Thus, the $\alpha_i$'s form a permutation of $\mathbb {N}^*$ and the increasing property leads to $\alpha_i = i$ for all $i$.

Now, if $\gcd(a,b) = 1$. Assume that $\gcd(f(a),f(b)) =d>1$. Then $d=f(c)$ with $c>1$, and $c$ divides both $a$ and $b$, which is absurd. Therefore $\gcd (f(a),f(b)) = 1.$
It follows that, for $\gcd (a,b) = 1$ we have $f(a)f(b)$ divides $f(ab)$.

I think that from now, and using the prime decomposition, and the same lines as above, the rest is easy, am I wrong?

Pierre.

Uff... Thank you, Pierre!
Indeed, it is straightforward.

Why didn't you just trust me  :D  ?

Pierre.


My teacher has given this example for homework, but I have no idea while there are different degrees of roots

$\lim_{x\to \infty}(\sqrt[3]{x^3+1}-\sqrt{x^2+x+1})$

p.s. I hope this a appropriate board

$f(x)=\sqrt[3]{x^3+1}-\sqrt{x^2+x+1}=x\sqrt[3]{1+\frac{1}{x^3}}-x\sqrt{1+\frac{x+1}{x^2}}$
now you can use the fact that for $\epsilon \rightarrow 0$, $(1+\epsilon)^\alpha\sim1+\alpha\epsilon$

so, since $\frac{1}{x^3},\frac{x+1}{x^2} \rightarrow 0$ for $x\rightarrow +\infty$

$\displaystyle f(x)\sim x\left(1+\frac{1}{x^3}*\frac{1}{3}\right)-x\left(1+\frac{x+1}{x^2}*\frac{1}{2}\right)=x+\frac{1}{3x^2}-x-\frac{x+1}{2x}=\frac{1}{3x^2}-\frac{1}{2}-\frac{1}{2x}\rightarrow -\frac{1}{2}$

p.s. how do you type symbol ~ with LaTeX?

[quote="hydro"]
p.s. how do you type symbol ~ with LaTeX?[/quote]

\sim, http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php


If you got into MIT, and you wil probably go to MIT post your name and where are u from !

I am Manuel Rivera, from PR. MIT is my first choice, i'll just send UMichigan, Harvard and Princeton apps but don't know why because i'll probably end in MIT

A lot of universities will pay for plane fare to visit them, so I recommend that you apply to other schools - how else are you going to get a 60 dollar airplane ticket? :)

Fred Sala, from Troy, Michigan

Appwise, I got into UMich and Caltech. I don't know if I'll apply elsewhere.

Hi Manuel!

blablablabla, awsome idea!!!  :D
Hi Fred, congrats

My name is Inigo Montoya.  You killed my father.  Prepare to die.

Uhhh, I'm David Chen, from Maryland... see you guys at CPW?

Geehoon Hong from New York
Class of '09 :D

[quote="manuel"]If you got into MIT[/quote]
but you did not specify which year...

yeah there is no year i wanna meet people from mit and aops :D

[quote]MIT is my first choice, i'll just send UMichigan, Harvard and Princeton apps but don't know why because i'll probably end in MIT  [/quote]

XD I'm only a freshman (in highschool) but that's the exact list of colleges I'm planning on applying to.

[quote="probability1.01"][quote="mel"][quote="Albanian Eagle"]i have applied to mit and now im waiting.
does mit give great importance to olympiads?[/quote]

This has been discussed many times before. Run a search for more in-depth discussion.

While anything short of an international olympiad will not give you the magic key into the door, participation in olympiads certainly would not hurt your chances.[/quote]

Even an intenational olympiad doesn't guarantee anything. I heard that one person who won gold medals in like 3235435 different olympiads got rejected or waitlisted from one college (of course he was accepted everywhere else).[/quote]

Yeah, I should have noted that there are no guarantees in admissions. Typically, international olympaid participants that are U.S. citizens have the upper hand for admissions into U.S. colleges and universities. International admits are more or less a lottery.

Sungyoon Kim from Korea. Class of 09. The spring term is gonna start tomorrow!


linear algebra
matrix
induction
graph theory
combinatorics theorems
combinatorics

Let G be a pseudo-digraph with n vertices- multiple edges and loops are allowed. The adjacency matrix A(G) of G is the nxn matrix in which (a)ij is the number of edges joining the vertices i and j(so the entries can be greater than 1). Then (a) ij gives us the number of arcs from i to j.

And here is a theorem about the number of walks of length k :

THM : Suppose we label the vertices of D and let A be the adjacency matrix according to this labelling. Then the number of walks of length m from vertex i to vertex j equals the (a) ij th entry of the matrix A^(m). I guess this theorem can be proved by induction, but is there another good way to show this is true?

Okay, let me change the question and replace "pseudo-graph" by "digraph".Hence multiple edges and loops are not allowed. In this case, how can we prove the theorem? Your help is appreciated because i have some difficulty proving these kind of graph theory theorems related to (adjacency) matrices.

The result isn't any more obvious for simple graphs than for graphs with multiple edges or loops.  I don't really know what to say about this except that if you examine the expression that results from matrix multiplication, each entry of the resulting matrix is exactly what you would calculate if you summed the number of ways to make the trip over all possible paths.  The 0 terms correspond exactly to times when you've taken an edge that doesn't exist; the 1s correspond to times when there is exactly one path between the two relevant vertices, and so on.


Prove that:$\frac{1}{x^{2}+y}+\frac{1}{y^{2}+z}+\frac{1}{z^{2}+x}\geq \frac{27}{4}$
where $x;y;z$ are positive real numbers satisfying $x+y+z=1$

We use CBS for $(\frac{1}{\sqrt{x^{2}+y}},...)$and for $(\sqrt{x^{2}+y},...)$.
So we get
 $S=\sum x^{2}+y \sum\frac{1}{x^{2}+y}\ge 9$
 But $\sum x^{2}+y =\sum x^{2}+3$
  and   $\sqrt \sum \frac{x^{2}}{3}\ge \frac{x+y+z}{3}=frac{1}{3}$
  so $\sum\frac{x^{2}}{3}\ge \frac{1}{3}$
So we get the conclusion.


function
limit
integration
real analysis
real analysis unsolved

For a given function $x(t)$ , what does it mean to find the $l^{p}$ norm for $p\rightarrow 0$?

What if the function is discrete?

regards,

bhupala

What is "the $l^{p}$ norm" for $0<p<1$?
The limit $\lim_{p\to 0}\int |x(t)|^{p}dt$ is the measure of the set on which $x$ is nonzero (under some mild assumptions regarding $x$). $\lim_{p\to 0}\left(\int |x(t)|^{p}dt\right)^{1/p}$ is probably either $0$ or $\infty$.


Consider the reaction $N_{2}O_{4(g)}\to 2NO_{2(g)}$ at constant temperature with a starting pressure of $N_{2}O_{4(g)}$ of 0.100 atm.  After some time the pressure increases from 0.100 atm to 0.148 atm.  The partial pressure of $N_{2}O_{4(g)}$ is....

a) 0.052 atm
b) 0.048 atm
c) 0.128 atm
d) 0.036 atm
e) 0.096 atm

Correcto....here's another one...

A 44.8 liter vessel at 273 K contains 2.00 mol of a gas X.  The pressure is measured to be 0.800 atm.  These data imply that...

A) A small fraction of X has dissociated
B) Effects due to attractive forces are more important than those due to excluded volume
C) Effects due to attractive forces and excluded volume are equally important
D) The gas is ideal
E) Effects due to attractive forces are less important than those due to excluded volume

Oh and I have another question: Is the average velocities of $O_{2(g)}$ and $H_{2(g)}$ the same at a given temperature? or are they different?  I think they're the same because of the Kinetic Molecular Theory (average kenetic energy same at given temperature)..am I wrong? I definitely know their speeds are not the same but I'm not sure about their average velocity.

Thanks!

[hide]I'm not totally sure, but if it were an ideal gas, the pressure should be around 1 atm. Since it is 0.800, I suppose that the attractive forces are a large factor, and the answer is B[/hide]

Also, the average kinetic energy is the same, but the average velocity is not.
RMS velocity = root(3RT/M)
Since Hydrogen gas has the smaller molar mass, it will be moving with a higher average velocity.

ohh ok i get it now for the hydrogen and oxygen question..thanks

and your answer was correct...also if you noticed, if you divide the liters by mol, you get 22.4 L/mol which should be at STP...but the pressure is not 1 atm...so the pressure is lower, b/c like you said of the attractive forces..

here's the last one...I can't figure this one out:

For a certain gas at atmospheric pressure and $-50^{o}C$, (PV/nT) = 0.5 R.  This deviation from ideal behavior results from the breakdown of which of the following statements?

A) Molecules are in constant, random motion
B) All collisions are elastic
C) Average kinetic energy of a molecule is proportional to the absolute temperature
D) Vol. occupied by molecules is negligible compared with the vol. of container
E) Molecules do not attract one another

I was thinking...D, not too sure because I think it could be E, but I can't prove it...

I think D because (PV/nT) = (.5)(0.08206) which would be a small number and I assume we have 1 mol of gas and we have 223 K...so (PV/223), PV would need to be small to get the 0.04103 for R.  Hence I thought volume of molecules would be small enough to be not negligible...

NVM I got the answer already...anyone can still try if they want to...