The equation $ \frac{2x^2}{x\minus{}1}\minus{}\frac{2x\plus{}7}{3}\plus{}\frac{4\minus{}6x}{x\minus{}1}\plus{}1\equal{}0$ can be transformed by eliminating fractions to the equation $ x^2\minus{}5x\plus{}4\equal{}0$.  The roots of the latter equation are $ 4$ and $ 1$.  Then the roots of the first equation are:

$ \textbf{(A)}\ 4 \text{ and }1 \qquad
\textbf{(B)}\ \text{only }1 \qquad
\textbf{(C)}\ \text{only }4 \qquad
\textbf{(D)}\ \text{neither 4 nor 1} \qquad
\textbf{(E)}\  \text{4 and some other root}$

[hide="Click for solution"]
Since $ x\equal{}1$ is not a solution of the original equation, the solution set of the first equation is the number $ 4$ only, or $ \boxed{\textbf{(C)}}$.
[/hide]

Please give other people a chance to solve the problems.  :wink:
You solved it ten minutes after posting. :mad:

hello, your equation is equivalent to
$ \frac {4(x \minus{} 4)(x \minus{} 1)}{x \minus{} 1} \equal{} 0$ so $ x \equal{} 4$ is the only solution.
So the answer is $ \text{C}$.
Sonnhard.


Hi,

I need a hint on a pre-calculus problem from a well-known textbook:

"Using f(x) = (x-k)q(x) + r, derive a cubic polynomial that will pass through (2,5) and rises to the right, and passes through (-3, 1)  and falls to the right".

I can create two cubics, one for each point:

f(x) = (x)(x-2)(x+3) + 5
f(x) = (x)(x-2)(x+3) + 1

for example, but I'm not clear about how to combine them. I think it has to do with:

f(x) = an(x-c1)(x-c2)(x-c3)...(x-cn), and finding the correct lead coefficient, an(meaning, "a sub n"). cn means "c sub n".

One other thing that's a bit confusing, is why or how, a cubic polynomial can rise and fall to the right, as it says in the problem, which is not, of course, my understanding of a graph of a typical cubic polynomial. Meaning, they either rise to right, fall to the left or rise to the left, fall to the right.

But please, only a hint!!!!

Thanks,
EPB

I'm not sure what [i]"Using f(x) = (x-k)q(x) + r"[/i] means, but here's what they're talking about:

Imagine a cubic that's increasing for very negative x, turns around and decreases for a while, and then increases everafter. Imagine the point (-3,1) being on the decreasing portion, and the point (2,5) being on the second increasing portion.

The solution set is huge, as you may imagine. [hide="Here is the general solution"]
$ \left( \frac {4}{5} x \plus{} \frac {17}{5} \right)\plus{}\alpha (x\minus{}\beta)(x\plus{}3)(x\minus{}2)$[/hide]

[u]Some questions to think about:[/u]
1. Where did I get this general solution from? What is it [i]doing[/i]?
[i]Hint: What two points of interest does that first linear factor pass through?[/i]
2. What are the bounds on $ \beta$?
3. What are the bounds on $ \alpha$? This will likely require calculus.

Hi,

Thank for replying, I don't have time at the moment to follow up on your questions, but I appreciate your post.

Ed Beaugard


Let f(z) = u(x,y) + iv(x,y) be an analytic function in a connected open set A. If au(x,y) +bv(x,y) = c in A, where a,b,c are real constants not all 0, prove that f(z) is constant in A.

If a,b,c are complex constants, is the result still valid?

(I assume you mean $ u,v$ are real functions)  Without loss of generality assume $ a\neq 0$.  Differentiate $ au \plus{} bv \equal{} c$ with respect to each variable and apply C-R equations to get
\[ \begin{cases} au_x \minus{} bu_y \equal{} 0 \\
au_y \plus{} bu_x \equal{} 0 \end{cases}
\]

\[ \Rightarrow \left(a \plus{} \frac {b^2}{a}\right)u_x \equal{} \left(a \plus{} \frac {b^2}{a}\right)u_y \equal{} 0
\]
So $ u$ is constant, and hence so is $ v$.  Nothing was assumed about the constants, so it holds just as well for complex $ a,b,c$.

so the coefficients vanish iff $ a^2 \plus{} b^2$ vanishes. This is certainly true in the real case but not immediately so in the complex case, so it looks like there is some assumption about the constants in your argument. However, if $ a^2 \plus{} b^2 \equal{} 0$, we may suppose $ a \equal{} 1, b \equal{} \pm i$. If we have the plus sign, then yes, $ f$ is certainly constant by assumption. If $ b \equal{} \minus{}i$, we may equate real and imaginary parts in the original expression to again obtain the result.


This has been bugging me for a while... I can't figure out what this problem is asking.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=848051#848051[/url]

I can't figure out what is meant by $ \text{Im}(f)$, with f being a function. The only thing I can think of that Im might mean is "imaginary part of a complex number", but I don't see how I could a have an imaginary part of a function of sets...

That's "Image", also known as the range of a function.


inequalities
calculus
integration
inequalities unsolved

for $ n\ge 2$ prove the inequality
\[ \sum_{k \equal{} 1}^{n \minus{} 1}\sqrt {1 \minus{} \frac {k^2}{n^2}} > \frac {n\pi}{4} \minus{} \frac {1}{\sqrt {8n}}\minus{}\frac {1}{2}
\]

Maybe integral sum can help?

Let take integral of $ \sqrt{1\minus{}x^2}$ in [0, 1].

[quote="Namdung"]Maybe integral sum can help?

Let take integral of $ \sqrt {1 \minus{} x^2}$ in [0, 1].[/quote]

It would work if we need upper bounduary, not lower  :P

- the integral easy gives $ \frac{n\pi}{4}>LHS$

Can someone say if [url=http://groups.google.com/group/sci.math/browse_thread/thread/3ae50b113d1c2993?hl=fi#]mr. Ullrich's[/url] approach is valid?


geometry
circumcircle
trigonometry
ratio
trapezoid
geometry unsolved

Through the vertices B, C of the triangle ABC parallels are drawn to the tangent to the circumcircle (O) at A, meeting AC, AB in A', A''. The line A'A'' meets BC in U. Show that AU and its analogues BV, CW are concurrent.

I will use my notations: we draw a tangent to $C$ and the parallels from $A, B$ meet $BC$ and $CA$ respectively at $S$ and $T$. Point $W$ is defined exactly the same.
Let: $\angle A = \alpha, \angle B = \beta, \angle C = \gamma$. Simple angle chasing shows that $\angle SAB = \alpha-\beta$ and $\angle SAC = \beta$. So from the Sine Theorem for the triangles $ABS$ and $ACS$ we have:
$\frac{|BS|}{\sin{(\alpha-\beta)}}=\frac{|AS|}{\sin{\beta}}$
and
$\frac{|CS|}{\sin{\beta}}=\frac{|AS|}{\sin{\gamma}}$
so from these to relations we get:
$\frac{|BS|}{|CS|}=\frac{\sin{(\alpha-\beta)}sin{\gamma}}{\sin^{2}{\beta}}$
Analogously we prove that:
$\frac{|CT|}{|AT|}=\frac{\sin^{2}{\alpha}}{\sin{(\alpha-\beta)}\sin{\gamma}}$
So:
$\frac{|CT|}{|AT|}\frac{|BS|}{|CS|}=\frac{\sin^{2}{\alpha}}{\sin^{2}{\beta}}$
But from the Menelaos Theorem we know that:
$\frac{|CT|}{|AT|}\frac{|BS|}{|CS|}\frac{|AW|}{|BW|}=1$
so: $\frac{|AW|}{|BW|}=\frac{\sin^{2}{\beta}}{\sin^{2}{\alpha}}$
In the same way we can prove that:
$\frac{|BU|}{|CU|}=\frac{\sin^{2}{\gamma}}{\sin^{2}{\beta}}$
and
$\frac{CV}{AV}=\frac{\sin^{2}{\alpha}}{\sin^{2}{\gamma}}$
so $\frac{|AW|}{|BW|}\frac{|BU|}{|CU|}\frac{|CV|}{|AV|}=1$ and because of the Ceva Theorem we are done.

A synthetic way, is to show that the segment lines $AU,$ $BV,$ $CW,$ are the symmedians of the triangle $\bigtriangleup ABC.$

I will post here later, an elementary proof for this result.

Kostas Vittas.

I think that's not hard to restate my proof in a pure synthetic way, since the ratio $\frac{\sin^{2}{\alpha}}{\sin^{2}{\beta}}$ is the same as $\frac{|BC|^{2}}{|CA|^{2}}$ so probably one can use triangle similarity instead of the Law of Sines...

We denote as $D,$ the antidiametric point $($ =antipode $)$ of $A,$ with respect to the circumcircle $(O)$ of $\bigtriangleup ABC$ and as $U',$ $U'',$ the orthogonal projections of $U,$ on the sidelines $AB,$ $AC,$ respectively. It is clear that the lines $BA',$ $CA'',$ are perpendicular to $AD.$

From the similar right triangles $\bigtriangleup U'BU$ $($ $\angle BU'U = 90^{o}$ $)$ and $\bigtriangleup CAD$ $($ $\angle ACD = 90^{o}$ $),$ $\Longrightarrow$ $\frac{UU'}{AC}= \frac{UB}{AD}$ $,(1)$

From the similar right triangles $\bigtriangleup U''CU$ $($ $\angle CU''U = 90^{o}$ $)$ and $\bigtriangleup BAD$ $($ $\angle ABD = 90^{o}$ $),$ $\Longrightarrow$ $\frac{UU''}{AB}= \frac{UC}{AD}$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $\frac{UU'}{UU''}= \frac{AC}{AB}\cdot \frac{UB}{UC}$ $,(3)$

Apply the Menelaus’s theorem for the transversal $A'A''$ of $\bigtriangleup ABC,$ we have that $\frac{UB}{UC}\cdot \frac{A'C}{AA'}\cdot \frac{A''A}{A''B}= 1$ $\Longrightarrow$ $\frac{UB}{UC}= \frac{AA'}{A'C}\cdot \frac{A''B}{A''A}$ $,(4)$

From $BA'\parallel CA''$ $\Longrightarrow$ $\frac{A''B}{A''A}= \frac{A'C}{AC}$ $,(5)$

From $(4),$ $(5)$ $\Longrightarrow$ $\frac{UB}{UC}= \frac{AA'}{A'C}\cdot \frac{A'C}{AC}= \frac{AA'}{AC}$ $,(6)$

From $(3),$ $(6)$ $\Longrightarrow$ $\frac{UU'}{UU''}= \frac{AC}{AB}\cdot \frac{AA'}{AC}= \frac{AA'}{AB}$ $,(7)$

Because of $BA'\perp AD$ and $DC\perp AC,$ we have that $\angle AA'B = \angle ADC$ $\Longrightarrow$ $\angle AA'B = \angle ABC$ $,(8)$

From $(8),$ we conclude that the sideline $AB$ of $\bigtriangleup ABC,$ is as tangent line of the circumcircle of $\bigtriangleup BCA',$ at it’s vertex $B$ and so, we have that $(AB)^{2}= (AC)\cdot (AA')$ $\Longrightarrow$ $\frac{AA'}{AB}= \frac{AB}{AC}$ $,(9)$

From $(7),$ $(9)$ $\Longrightarrow$ $\frac{UU'}{UU''}= \frac{AB}{AC}$ $,(10)$

From $(10),$ we conclude that the segment line $AU,$ is the symmedian of $\bigtriangleup ABC$ through vertex $A.$

Similarly for the segment lines $BV,$ $CW$ and the proof is completed.

Kostas Vittas.

[quote="vittasko"]A synthetic way, is to show that the segment lines $ AU,$ $ BV,$ $ CW,$ are the symmedians of the triangle $ \bigtriangleup ABC.$[/quote]
As a friend of mine [b][size=100]Silouanos Brasitikos[/size][/b] said, we can have the nicest ( I think ) proof of this result, as a neat applicatiobn of the [b][size=100]Antiparallel lines[/size][/b] theory.

So, because of the trapezium $ BA'CA'',$ we have that the line segment $ AU,$ bisects the segments $ CA'',\ BA'$ as well.

Because of now, $ \angle ACA'' \equal{} \angle 90^{o} \minus{} \angle OAC \equal{} \frac {\angle AOC}{2} \equal{} \angle B,$ we conclude that the line segments $ BC,\ CA'',$ are anrtipallel each other, with respect to the angle $ \angle A.$

Hence, the line segment $ AU,$ as the $ A$-median of the triangle $ \bigtriangleup ACA'',$ is the $ A$-symmedian of the triangle $ \bigtriangleup ABC$ and we are done.

Many thanks to [b][size=100]Silouanos[/size][/b], Kostas Vittas.

PS. It is used the same schema relative to the complicated solution I presented above, because it has been forgotten.

Denote the midpoint $ M$ of the segment $ [BA']$ . Observe that $ M\in AU$ and $ BA'$ is an antiparallel to the sideline $ BC$.
Since the geometrical locus of the midpoint of any antiparallel to $ BC$ is the $ A$-symmedian, then $ AU$ is $ A$-symmedian a.s.o.


this games okay, i don't understand it that much

[quote="k2c901_1"]http://www.godtower.com/

This is a particularly nice site, especially for those enjoying problem solving. They are not easy though... I am currently stuck on level 45, and would appreciate it if anyone discusses this game with me! (I can also provide hints for those levels lower than 45.)[/quote]

can you help me im stuck on level 15.

It's answer clearly ixxxxx
You must drink in GLASS only then you have that letter.
And the last one.What letter does the last place of him look like.

wow. ok i'll do this as soon as i finish notpron (on lvl 75 now)

I cant beat number 1, I dont understand this game at all.

haha im stuck on #1 too... i like notpron better cuz of the more informal, plain interface.

for number 1 i looked up the time zone and have tried c*c*s i*l**d and m*y**m*r, which are the two places google gave me, but neither works. does capitalization or spacing matter?

For level 1:[hide="spoilers"]look for a city, not a country[/hide]

For level 15:[hide="spoilers"]You are to look for five shapes that resemble letters. Example: for the last one, the "last place" of Christ was on a cross (t).[/hide]

Nothing I try works  !!!!!!!!!!!! Help me please...

Umm.. :blush: I haven't gotten past the first one...

[hide]Look for a city that is GMT +6.30. The map might give you a rough location of the city.[/hide]

er... is it just me, or are the "hints and rules" in another language?

yeah it is in another language...I like the "contact God" feature.

im on level 19
[hide]with a walkthrough lol[/hide]

Stuck on level 25 if anyone can help - anyone who needs hints on previous levels just ask.

Level 4 - The half of the sun on the second picture hints its something to do with halving something.  On the first picture there is a red sticker which if you turn your head to the right 90 degrees and split the 'writing' in half through the longest bit you get the answer.

Don't think that makes sense but hope you can make out the answer, else let me know and I  will try and re-explain it.

Level 25:
[hide]Listen to the number of times each sound appears in the recording. Type the numbers in its corresponding space, and you'll see that it looks like a brand name. [/hide]

wow this thing starts at a pretty hard level lol...notpron at least started with a "click here" level lol

Can anyone help me on level 24?
I got the zodiac signs, but have no idea what to do with them... A hint would be nice..

[quote="rawmat_05"]Help ME!!! I'm stuck on Level 7[/quote]

[hide]I don't know if this hint is too much, but try hitting shift 5 times.. Good luck![/hide]

I'm on 49l, and I'm stuck.  Are you still on 45?

I'm stuck at level 20 (one with the moon). Can anyone help?


MATHCOUNTS counting distinguishability AMC USA(J)MO USAMO

Ok guys, remember that SUPER MATHCOUNTS TOURNEY that I hosted way back in the winter of 2008? Well, yeah. As a heads-up, we shall recognize the fact that the top four scorers on the written round are eligible for participation in the MASTERS ROUND!!!!!!!!

As a heads-up, those people are:

1. [b]xpmath[/b] (Team PHAIL)
2. [b]pythag011[/b] (Team 1337)
3. [b]FantasyLover[/b] (Team 0)
4. [b]aleph0[/b] (Team 0)

(Bold indicates that the person is CERTAINLY participating--replied to my PM)

In order to see who will be the four participants, I am now sending out PMs to those people, saying whether or not they would participate. If some of them decline, I would take the next people in line based on their rankings in the tourney. I will update this thread when such things occur.

Wish everyone good luck! The questions will be posted right after we know who's doing it.

[quote="pythag011"]I will participate.

Make them absolutely crazy... everybody loves IMO-level master rounds...[/quote]

Sorry, but no. This is supposed to be a MATHCOUNTS level masters round, just like a National Masters Round... :wink:

ANd as a heads up, all four participants are entered! Yay!

The problems will be posted later this week, once I can get them up. Stay tuned, and good luck!

[quote="pythag011"][quote="gauss1181"]I'm back from my 2-day orchestra event, so now I'm planning to get the problems up and running tonight. Rules for participants would be posted at that same time.

Also, I would like to request two judges to assist me with the judging process. If you'd like to be a judge, please PM me. However, if you want to be a judge, please consider the following:

1. Of course, BE FAIR. No one is going to like it if you rate the person instead of his/her solution. Therefore, if you want to be a judge, please be one as long as you're not related to any of the participants in any special way (e.g. parent, schoolmate, teammate, close friend, close relative, teacher, etc.)

2. Use good common sense, and basically stick to the rubric guidelines that I will post along with the problems. Do not stress TOO much on certain criteria; treat everything equally. (e.g., do not mark someone down for the SOLE/MAIN purpose of spelling/grammar, formatting, etc.) We want to see more of the mathematical content, which is basically one of the core values of this Masters Round.

3. In addition to fairness, we would ask that if you're a judge, that you have NOT met any of the participants face-to-face in real life.

4. Absolutely no derogatory, sexist, religious, or racist comments would be allowed when judging.

5. Finally, be sincere and helpful in your judging. Don't just write brief comments like "Good" or "OK." Please be detailed in your feedback. That way, it can be a good learning experience for both those who are experienced with the Masters Round and those who aren't.

Thank you for reading these rules. If you're interested in being a judge, PM me.[/quote]

If I personally could, I would not be a contestant and be a judge, but I guess I have to be a contestant...[/quote]
well you've already said you're gonna participate, so you're just gonna be one.

EDIT: I guess I'd have more time to post the questions tomorrow...sorry for the delay.

Here are the problems! Remember that you get 30 minutes to prepare full, rigorous solutions to these. Please PM them to me when you're done. PLEASE PRESENT YOUR SOLUTIONS AS IF YOU WERE GOING TO PRESENT THEM IN FRONT OF A LARGE AUDIENCE! Remember that you're on your honor. We're watching you. Good luck! Note that you're being evaluated based on creativity, mathematical knowledge, and overall display/clarity of the solution.

Here is the problem:

Consider x balls to be placed in y boxes, where x>1 and y>1.

1. Start with the case where 2 balls are placed in y boxes. Find the number of ways the balls could be placed in the boxes when:
a. balls and boxes are distinguishable.
b. balls are distinguishable, but boxes aren't.
c. balls indistinguishable, boxes distinguishable
d. balls and boxes are indistinguishable.

2. Do the same process for 3, 4, and 5 balls to be placed in y boxes.

3. Find a formula for placing x balls in y boxes for each of the four distinguishability cases. If you can't use a specific formula, please EXPLAIN the general process/method of approaching each case.

[quote="gauss1181"][quote="xpmath"]I really hope the typing solutions part doesn't count as the 30 minutes.

Also, this is probably the WORST master's round topic I could ever get. Blah.[/quote]
The typing solutions part doesn't count as the 30 minutes. The 30 minutes was just the time allowed for PREPARING your solution. After time is up, you simply just have to submit whatever you have.

Received Solutions:

xpmath

Waiting:

pythag011
FantasyLover
aleph0

I have not set up a due date yet. Also, please do not expect that your scores and feedback will be returned to you immediately after submitting your solutions. It will have to take a couple of days for the judging process to fully be complete. We still are in need of two assistant judges who will help me with the grading. Remember, you can be a judge as long as you're fair AND you have not met any of the participants nor are related to them in any such way (e.g. friend, online friend, parent, teacher, relative, sibling, team mate, schoolmate, etc.)

@xpmath: I do realize that. It's the first and only masters round I've created on my own, so I tried to make it something that is solvable with MATHCOUNTS methods. (so it might not exactly be what you expect based on what you see in National Masters Rounds)[/quote]

Masters rounds are NOT problems that are just solvable with one application of a theorem.

Masters rounds are NOT well known theorems. The theorem in here is ridiculously well known, unfortunately.

Lawl the reason I said this was probably the worst is because I fail stuff like this.

Unfortunately, I do NOT know the theorem or formula or whatever. Thus, I failed. Seriously.

[quote="aleph0"]Sorry I can't participate after all...[/quote]
Why not? Are you busy? If so, I'll call the 5th place person to participate.

Ok, now I've set up a deadline.

DEADLINE: [size=200]February 28, 2009, at 10 PM CST!!!!![/size]

I've already finished making the judges' rubric so far. And xpmath is the only one who submitted his solution. pythag011 and FantasyLover still need to submit. (aleph0 withdrew)

Also, AIME15 and BOGTRO, those of you who requested to be judges, since I have a feeling that you have some connection to xpmath, FantasyLover or pythag011, sorry, but no. I will recruit some judges myself for now...and btw I need 2 judges to assist me. But Please feel free to PM me if you wanna be a judge, and remember: YOU CAN ONLY BE A JUDGE IF YOU ARE NOT CONNECTED WITH FANTASYLOVER, PYTHAG011, OR XPMATH IN ANY WAY, BUT NOT LIMITED TO: PARENT, FRIEND, RELATIVE, TEAMMATE, SCHOOLMATE, etc.

[quote="AIME15"]wait, so you're saying i have a connection to xpmath, or that i'm judging?  I don't even know xpmath...[/quote]
i have a feeling ur connected to pythag in some way...so i wanted to avoid any strange suspicions in the masters round scoring...sorry :(

EDIT: Twin Prime Conjecture is now a judge :)

Um, wouldn't it be easier for you to PM a judge the solutions WITHOUT saying who wrote them?  And I don't know pythag...I've talked to him, but that's about it.

Probably not, otherwise some may get confused between whose is whose, and such :maybe:

And many people have their own unique solution writing styles, so some may be easily noticeable :wink:

Blah, if it said "trivial" in it multiple times it's pythags :)

My writing style.... is basically leave out steps, outline a proof, and tell the grader to fill in details because I'm too lazy to and that part is trivial.

That got me a 2 instead of a 7 on USAMO 2008 #4 even though I had a correct solution... oh well.

FANTASYLOVER AND PYTHAG011 PLEASE SUBMIT ASAP!!!

A second reminder is that answers are due February 28, 2009 at 10 PM CST!!!! We are still in need for a second judge!

Submissions

xpmath

Waiting for

pythag011
FantasyLover

Judges

Twin Prime Conjecture
*we need a second judge!*
gauss1181 (Head Judge)

Thanks!

Updated info:

Submissions

xpmath

Waiting for

FantasyLover

Judges

Twin Prime Conjecture
shentang
gauss1181 (Head Judge)

EDIT: pythag011 has withdrawn. :(


geometry
trigonometry
incenter
angle bisector
congruent triangles

same as proving sin(180-x)/sin(2x-80) = 1 + sin(80-x)/sin100

and i guess this is not the right approach

[quote="Armo"]Let $ABC$ is a triangle with$ \measuredangle{ABC}=100$. Let $AK$ is an angle bisector, prove that
$AK+BK=AC.$[/quote]

If $\measuredangle{ABC} = 100^o$ and $AK$ is a bisector of the angle $\measuredangle{BAC}$, then $AK + BK = CA$ iff $\measuredangle{CAB} = \measuredangle{BCA} = 40^o$ (and not for other angles).

The problem should be stated either as

If $\measuredangle{ABC}=100^o$, $AK$ is a bisector of the angle $\measuredangle{CAB}$ and $AK + BK = CA$, find the angle $\measuredangle{CAB}$ (or $\measuredangle{BCA}$).

or as

If $\measuredangle{ABC}=100^o$, $\measuredangle{CAB} = 40^o$ (or $\measuredangle{BCA} = 40^o$) and $AK$ is a bisector of the angle $\measuredangle{CAB}$, show that $AK + BK = CA$.

Thanks for your reply [b]yetti [/b], excuse me all for wrong problem. :blush:
P.S. Actually I haven't proved this before posting it here, I just thought it easy. After this I'll post problems more carefuuly. :)

[quote="yetti"]If $\measuredangle{ABC}=100^o$, $\measuredangle{CAB} = 40^o$ (or $\measuredangle{BCA} = 40^o$) and $AK$ is a bisector of the angle $\measuredangle{CAB}$, show that $AK + BK = CA$.[/quote]

Assume that the angle $\measuredangle C = 40^o$ and we have to show $AK + BK = CA$. The angle $\measuredangle A = 180^o - (\measuredangle B + \measuredangle C) = 40^o$, i.e., the triangle $\triangle ABC$ is isosceles, $AB = BC$. Using sine theorem for the triangles $\triangle ABK, \triangle CBK$,

$\frac{BK}{AK} = \frac{\sin{\frac{\widehat A}{2}}}{\sin{\widehat B}} = \frac{\sin{20^o}}{\sin{100^o}} = \frac{\sin{20^o}}{\sin{80^o}}$

$\frac{AK + BK}{AK} = 1 + \frac{\sin{20^o}}{\sin{80^o}}$

$\frac{AK}{CA} = \frac{\sin{\widehat C}}{\sin{[180^o - (\widehat C + \frac{\widehat A}{2})]}} = \frac{\sin{40^o}}{\sin{120^o}} = \frac{\sin{40^o}}{\sin{60^o}}$

Multiplying these 2 equations,

$\frac{AK + BK}{CA} = \left(1 + \frac{\sin{20^o}}{\sin{80^o}}\right) \frac{\sin{40^o}}{\sin{60^o}}$

We have to show that this expression is equal to 1, i.e., that

$\sin{40^o}\ (\sin{80^o} + \sin{20^o}) = \sin{60^o} \sin{80^o}$

Imagine an isosceles triangle $\triangle XYZ$ with the angles $\measuredangle X = \measuredangle Y = 80^o$, $\measuredangle Z = 20^o$ and let $T$ be a point on the side $YZ$, such that the angle $\measuredangle YXT = 70^o$. Then the angles

$\measuredangle YTX = 180^o - (\measuredangle X + \measuredangle YXT) = 180^o - 150^o = 30^o$,
$\measuredangle ZXT = \measuredangle X - \measuredangle YXT = 80^o - 70^o = 10^o$,
$\measuredangle ZTX = 180^o - \measuredangle YTX = 180^o - 30^o = 150^o$.

Using the sine theorem for the triangles $\triangle XYZ, \triangle XYT, \triangle XTZ$,

$\frac{YT}{XY} = \frac{\sin{70^o}}{\sin{30^o}} = 2 \sin{70^o}$

$\frac{XY}{ZX} = \frac{\sin{20^o}}{\sin{80^o}}$

and multiplying these two,

$\frac{YT}{ZX} = 2 \sin{70^o}\ \frac{\sin{20^o}}{\sin{80^o}}$

$\frac{ZT}{ZX} = \frac{\sin{10^o}}{\sin{150^o}} = 2 \sin{10^o}$

$\frac{YT + ZT}{ZX} = 2 \sin{70^o}\ \frac{\sin{20^o}}{\sin{80^o}} + 2 \sin{10^o} = 1$

$2 \sin{70^o} \sin{20^o} + 2 \sin{10^o} \sin{80^o} = \sin{80^o}$

Subsituting $\sin{70^o} = \cos{20^o}$, $\sin{80^o} = \cos{10^o}$,

$2 \cos{20^o} \sin{20^o} + 2 \sin{10^o} \cos{10^o} = \sin{80^o}$
$\sin{40^o} + \sin{20^o} = \sin{80^o}$

Trisecting the angle $60^o$,

$\sin{60^o} = 3 \sin{20^o} - 4 \sin^3{20^o} = 4 \sin{40^o} \cos^2{20^o} - \sin{20^o}$
$\cos{60^o} = 4 \cos^3{20^o} - 3\cos{20^o}$

Multiplying $\sin{60^o}$ by $\sin{80^o}$,

$\sin{60^o} \sin{80^o} = (4 \sin{20^o} \cos^2{20^o} - \sin{20^o})(\sin{20^o} +  \sin{40^o}) =$

$= 4 \sin^2{20^o} \cos^2{20^o} - 4 \sin{20^o} \sin{40^o} \cos^2{20^o} - \sin^2{20^o} + \sin{20^o} \sin{40^o}) =$

$= \sin^2{40^o} + \sin^2{20^o}\ (8 \cos^3{20^o} - 2 \cos{20^o} - 1)$

Substituting $8 \cos^3{20^o} = 2 \cos{60^o} + 6 \cos{20^o} = 1 + 6 \cos{20^o}$,

$\sin{60^o} \sin{80^o} = \sin^2{40^o} + 4\sin^2{20^o} \cos{20^o} =$

$= \sin{40^o} (\sin{40^o} + 2 \sin{20^o}) = \sin{40^o}\ (\sin{80^o} + \sin{20^o})$

which is the equation we were supposed to prove.

[quote="yetti"]If $\measuredangle{ABC}=100^o$, $\measuredangle{CAB} = 40^o$ (or $\measuredangle{BCA} = 40^o$) and $AK$ is a bisector of the angle $\measuredangle{CAB}$, show that $AK + BK = CA$.[/quote]

Assume that $AK + BK = CA$ and we have to find the angles $\measuredangle A$ or $\measuredangle C$. Using sine theorem for the triangles $\triangle ABK, \triangle CBK$ as before,

$\frac{BK}{AK} = \frac{\sin{\frac{\widehat A}{2}}}{\sin{\widehat B}} = \frac{\sin{\frac{\widehat A}{2}}}{\sin{100^o}} = \frac{\sin{\frac{\widehat A}{2}}}{\sin{80^o}}$

$\frac{AK + BK}{AK} = 1 + \frac{\sin{\frac{\widehat A}{2}}}{\sin{80^o}}$

$\frac{AK}{CA} = \frac{\sin{\widehat C}}{\sin{[180^o - (\widehat C + \frac{\widehat A}{2})]}} = \frac{\sin{(\widehat A + \widehat B)}}{\sin{(\widehat B + \frac{\widehat A}{2})}} = \frac{\sin{(\widehat A + 100^o)}}{\sin{(100^o + \frac{\widehat A}{2})}} = \frac{\sin{(80^o - \widehat A)}}{\sin{(80^o - \frac{\widehat A}{2})}}$

Multiplying these 2 equations,

$\frac{AK + BK}{CA} = \left(1 + \frac{\sin{\frac{\widehat A}{2}}}{\sin{80^o}}\right) \frac{\sin{(80^o - \widehat A)}}{\sin{(80^o - \frac{\widehat A}{2})}} = 1$

We have to solve this equation for the angle $\measuredangle A$, i.e., the equation

$\sin{(80^o - \widehat A)} \left(\sin{80^o} + \sin{\frac{\widehat A}{2}}\right) = \sin{\left(80^o - \frac{\widehat A}{2}\right)} \sin{80^o}$

The last equation has period $720^o$ and the following 4 roots between $[0^o, 720^o)$:

$\measuredangle A = 0^o$ or $40^o$ or $240^o$ or $520^o$

The root $0^o$ is trivial, we already know the root $40^o$ from solving the opposite problem and the remaining 2 roots can be calculated by approximation methods. Only the root $\measuredangle A = 40^o$ can be an internal angle of a triangle.

I found a very sweet way to do this problem. (I just couldn't bear the thought that nobody posted a synthetic solution here :D)

Let $L$ be a point on $AK$ such that $AL = AC$ (where $A$, $K$ and $L$ are collinear in this order). Let $M$ be the intersection of $AB$ and $CL$. It is easy to see that $\angle{ACK} = \angle{MCK} = 40^{\circ}$, and hence $CK$ is the angle bisector of $\angle{ACM}$. Since $AK$ bisects $\angle{MAC}$, $K$ must be the incentre of $ACM$. Also, it is easy to see that $\angle{BKL} = 120^{\circ}$ and $\angle{AMC} = 60^{\circ}$. Hence $MBKL$ is a cyclic quadrilateral. Since $MK$ is the bisector of $\angle{AMC}$ (as $K$ is the incentre) we get $\angle{BLK} = \angle{BMK} = \frac{1}{2}\angle{AMC} = 30^{\circ}$ and similarly $\angle{KBL} = 30^{\circ}$. Hence $BKL$ is an isosceles triangle, thus $AC = AL = AK + KL = AK + BK$ and we are done.

[hide="elementary proof which involves congruent triangles only"]Extend $AK$ to $P$ such that $BK=PK$. Let also $Q$ be a point on $AC$ such that $AQ=AB$. Then $ \triangle ABK \cong \triangle AQK (S.A.S.)$. So $ \angle BKA=\angle QKA=60^\circ$. Then $\angle QKC=\angle PKC=60^\circ$. So $\triangle QKC\cong\triangle PKC (S.A.S.) \Rightarrow \angle KCQ=\angle KCP=40^\circ \Rightarrow \angle KPC=80^\circ$ and $\angle QCP=\angle KCQ+\angle KCP=80^\circ$. Therefore, $\angle KPC=\angle QCP \Rightarrow AC=AP=AK+KP=AK+BK$[/hide]

We had this problem at our training camp, but asked the other way round, (If this and that make this, what is this angle etc....)

Most gave the congruent triangle solution.


Since a similar topic was extremely popular, I decided to have a tally of the results

O and please post your replies... and I would be nice to disclose How many advanced to the states... For example Here in Greater Hartford, Connecticut, we have 300-400 peeps competing, and 50 individuals along with anyone else on a top 8 team advance...

in southwest Connecticut about 200 people competed and the top 5 teams and top 15 individuals not on a top team advanced to states

I think only the top three teams in my small chapter competition (only around 25) advanced to the state competition.

288...Maybe more..Thats the minimum though...

[quote="baseballfreak2"]163.[/quote]

Hmm.  That's exactly how many people were in my chapter.  Are you from Michigan also?

about 40 something people went, and the top 4 team and top 4 individual goes to state

Ours is a small chapter. (Seacoast, NH) just about 50-75 (11 schools)

um well i estimated that about 12 people came from each team and there were about 6 that i remember or imagined so im guessing 60. seriosly with my horrible memory, im not sure. :lol:  :D

WHAT!!!!!!!
I'm in a 150 person chapter, and we only send the top 3 individuals and top 2 teams!!!!!

What State are you from? It depends on that and the amount of people in your chapter compared to the entire state.


this is probably going to seem very trivial, but i just need a little explaining.. i'm working from aops vol1 that i started.
question 102..
given $9^6 = 531,441$, how would you represent $531,440$ in base $9$?
i mean it's obvious you need to get to $9^6 - 1 = 531,440$, and that $9^6 = 1000000$ base $9$, but where do you make out with the 1?

[hide]
$1000000_{10}-1_{10}=999999_{10}$
Likewise, $1000000_{9}-1_{9}=888888_{9}$

Here's an explanation, (not the best)
When you take 1 away from the first 2 digits, you are looking for the next digit down. That would usually be 9, but since we're in base 9, we can't have nine's and it'd be 8. That goes for all the digits.
[/hide]

Another way to think about it:
When you add 8+1 in the first column, you get $10_9$. You have to carry the 1 to the next column, and you add 8+1 again. So $888888_9+1=1000000_9$.

[hide]We have to consider base 9 in this problem.  We can show that $8_9+1_9=10_9$, very easily.  We can just continue this way to show that $888888_9+1_9=1000000_9$.

So this would mean that $888888_9=531440_{10}$[/hide]


A lattice point is a point with integer co-ordinates.
The circle x^2 + y^2 = 1.2^2 has 5 interior lattice points.
The circle x^2 + (y-0.5)^2 = 1.2^2 has 6 interior lattice points.
Prove or dsprove: For every positive integer n, there exists a circle with n interior lattice points.

Take a circle centered at $ (\sqrt{2},\sqrt{3})$. $ (a\minus{}\sqrt{2})^2 \plus{} (b\minus{}\sqrt{3})^2 \equal{} (c\minus{}\sqrt{2})^2 \plus{} (d\minus{}\sqrt{3})^2$ having a solution in integers $ a,b,c,d$ requires $ (a\minus{}c)\sqrt{2} \equal{} (b\minus{}d)\sqrt{3} \plus{} k$ for some rational $ k$. After squaring both sides this is seen to be impossible.

Thus every lattice point has a different distance from $ (\sqrt{2},\sqrt{3})$, so by gradually expanding this circle from radius 0 we add only one lattice point at a time.


trigonometry LaTeX calculus integration geometry trapezoid

Could someone give me a solution for this problem.  I am taking Physics B so please no calculus.  My teacher is new and isn't the smartest teacher you could have  ;)  so she didn't know how to do it.  Thanks:

A boy on a hill aims his slingshot up at an angle towards another boy in a tree a distance of d meters away.  The other boy is hanging onto a branch in the tree.  Show that if this boy were to let go at the same time as the water balloon is fired from the slingshot, he would get hit by it.

(assume no air resistance)

I had to paraphrase that because the problem in a book is an extension of a previous exercise and I didn't want to put the whole thing in this post.  So if some things are confusing ask and I will try to clarify them.

so it's just you shoot an object at him in such a way that while he falls from the tree and your object is attracted by gravity that it smoothes out so you hit him?

if so, plz give the height etc some names too, simplifies the stuff :)

I'll tell you how to do it, rather than doing it. If this isn't enough, let me know, and I'll type out something in detail whenever I get the chance and have internet access again (it might not be for a while... I'm moving up to college)

Decompose the velocity of the water balloon into a horizontal component and a vertical component using trig.

Use $D=d=v_ht$ to solve for $t$, the amount of time it takes the water balloon to get to where the boy is horizontally.

Use $D=\frac{a_vt^2}2+v_vt$ to solve for how high it is vertically at time $t$.

Use $D=\frac{a_vt^2}2+v_vt$ to solve for how high the boy is at time $t$.

If they're all the same, then the boy gets hit.

(Your physics book may call $D$ "$S$" instead. The difference is trivial)

Ok my book's formulas are slightly different than yours but are the same.  I will use the formulas that I have learned.

This is what I have so far:

We call the angle $\theta$ and the velocity that it is shot at $v$ .  So we can find that the horizontal compenent is $v\cos{\theta}$.  The vertical compenent is $v\sin{\theta}$.

Thus $\displaystyle t=\frac{x_1}{v_{x0}\cos{\theta}}$.

Then, let $y_b$ be the height of the boy and $y_s$ be the height of the balloon(that the slingshot shot).  Thus

$\displaystyle y_b=\frac{a_yt^2}{2}+y_0$     ($y_0$ is the initial height of the boy)

and

$\displaystyle y_s=\frac{a_yt^2}{2}+v_yt=\frac{a_yt^2}{2}+\frac{v_yx_1}{v_{x0}\cos{\theta}}$

I am stuck now... have I made a mistake?  Thanks for the help confuted.  I appreciate it.

PeterVDD:

The problem didn't have many specifics but I ll give you as much as I can.  Let the boy be d meters from the tree.  He is also on a hill but below where the hanging boy is.   The angle that he shoots the balloon is $\theta$.  There isn't too much more.

I see a few mistakes.

$v_{x0}=v_0\cos{\theta}$ as you said. But then
$\displaystyle t=\frac{x_1}{v_{x0}}$ (that may very likely have been a typo.)

Your equation for $y_b$ is correct. Substitute in $g=-9.8 m/(s^2)$ for $a_y$.

The equation for $y_s$ is okay (assuming that the slingshot is at height $y=0$), but you missed some substitutions. Also, I'll assume that you made the same typo here as you made before, with $v_{x0}$ meaning $v_0$. First, you can substitute in $\displaystyle  t^2=\frac{x_1^2}{v_{x0}^2}$, and you can solve for $v_{y0}$ with trig, just like you solved for $v_{x0}$. Also, $a_y=g=-9.8 m/(s^2)$. So then you'll have (it'd be best if you tried it before reading on):

\begin{eqnarray*}
\\
\displaystyle v_{x0}&=&v_0\cos{\theta}\\
\displaystyle v_{y0}&=&v_0\sin{\theta}\\
\\
\\
\displaystyle t&=&\frac{x_1}{v_{x0}}\\
\displaystyle t&=&\frac{x_1}{v_0\cos{\theta}}\\
\displaystyle t^2&=&\frac{x_1^2}{\left(v_0\cos{\theta}\right)^2}\\
\\
\\
\displaystyle y_b&=&\frac{a_yt^2}{2}+y_0\\
\displaystyle y_b&=&\frac{g\left(x_1^2\right)}{2\left(v_0\cos{\theta}\right)^2}+y_0\\
\\
\\
\displaystyle y_s&=&\frac{a_yt^2}{2}+v_yt\\
\displaystyle y_s&=&\frac{g\left(x_1^2\right)}{2\left(v_0\cos{\theta}\right)^2}+\frac{x_1v_0\sin{\theta}}{v_0\cos{\theta}}
\end{eqnarray*}
The $\text{\LaTeX}$ renderer seems to be acting up a bit again; I see a "to" on the first line of that last image, and I've very clearly not typed one. Ignore it.

[quote="confuted"]The $\LaTeX$ renderer seems to be acting up a bit again; I see a "to" on the first line of that last image, and I've very clearly not typed one. Ignore it.[/quote]

Well I would break too if I'd have to make such big images ;)

Really, it's a lot easier if you make it with a little less code ;)

I don't know anything about physics but I'll tell you my common sense.  If the boy aims the baloon directly at the kid in the tree and fires the baloon at the same moment that the other kid lets go,  the baloon will fall off its trajectory at the same rate the other kid falls from the tree.  Because the problem gave no specifics like angle measures and distances,  they don't matter; they can't matter or the problem would be unsolvable.  The upward angle could be 35 degrees or 0.1 degrees. so I used 0.1 in which the balloon falls off trajectory at 9.8 m/s/s or something like that.

Oh gosh! I'm doing physics right now, and our teacher gave that as an "extra credit" problem on one of our assignments.

I have just another question. I did calculus last year, and now we're studying centripetal force in physics. Is there any relation between the integral (finding the area under the curve by using the trapezoid method) and finding  :theta: of the angle of the velocity that the object is traveling in?

The boy will get hit, because the acceleration from gravity affects both the boy and the water balloon at the same rates. I guess I would just say that since both objects sink at the same rate and the water balloon was originally aimed for the boy, you would be doing the same thing to both the boy and the balloon, making for the same result.

This problem leaves a lot to be desired... and it assumes that hill and boy hanging in the tree are infinitely high (aka, the boy won't hit the ground before the balloon has time to reach him). I think it should be stated.

I think you should label some heights and get rid of your thetas using trig. Then just solve with the method above.


On a gel, bands are 2 mm wide. Red dye (shaded in picture) travels at 1 mm/minute and yellow dye (dotted in picture) travels at 1.5 mm/minute. At the start, the red and yellow bands are in exactly the same place, making orange (shaded and dotted, left figure). How many minutes will it be until the red and yellow bands no longer overlap, as seen in the right figure?

[asy]import graph;
real textsize = 10pt;
size(150);
picture p;
picture q;
filldraw((0,0)--(0,-2)--(6,-2)--(6,0)--cycle,gray(.6),black);
void drawdot(picture p,pair center)
{fill(p,Circle(center,.3),black);}
drawdot(p,(.33,-1));
drawdot(p,(1.2,-.6));
drawdot(p,(1.3,-1.4));
add(q,p);
add(q,rotate(10,(2.6,1))*shift(1.7*right)*p);
add(q,rotate(-3,(4.9,1))*shift(4*right)*p);
add(q);
filldraw(shift(8*right)*shift(5*down)*((0,0)--(0,-2)--(6,-2)--(6,0)--cycle),gray(.6),black);
draw((8,-9)--(14,-9));
add(shift(8*right)*shift(7*down)*q);
draw((0,0)--(0,-15)--(6,-15)--(6,0)--cycle);
draw(shift(8*right)*((0,0)--(0,-15)--(6,-15)--(6,0)--cycle));
label("Start",(3,1.5),fontsize(textsize));
label("End",(11,1.5),fontsize(textsize));
draw((-1.5,0)--(-.5,0)^^(-1,0)--(-1,-2)^^(-1.5,-2)--(-.5,-2));
label("2 mm",(-1,-1),W,fontsize(textsize));
draw((14.5,-5)--(15.5,-5)^^(15,-5)--(15,-9)^^(14.5,-7)--(15.5,-7)^^(14.5,-9)--(15.5,-9));
label("2 mm",(15,-6),E,fontsize(textsize));
label("2 mm",(15,-8),E,fontsize(textsize));[/asy]

We need there "distance" to be $ 2$mm. They move at a relative rate of $ 1.5\minus{}1\equal{}.5$ mm/minute to each other. Thus, using $ d\equal{}rt$, we have $ 2\equal{}.5t \implies t\equal{}\boxed{4}$.


geometry
3D geometry
tetrahedron
AoPS Books

Can someone give me some formulas and theorems regarding tetrahedrons?

i learned this 2 weeks ago while memorizing my formula sheet. while i am studying aops books and doing problems since then.

I think you already know this  :lol: I also saw this on past vestavia practice yesterday.

s cubed * root 2/ 12 is the volume ?

and surface area as well : root 3 * a squared?

side i believe ive used it be4

im taking geometry and practice and i think thats how its supposed to be maybe im wrong

[quote="now a ranger"]

s cubed * root 2/ 12 is the volume ?

and surface area as well : root 3 * a squared?[/quote]

So close...

$V=\frac{1}{12}a^{3}\sqrt{2}$

$A=a^{2}\sqrt{3}$

oh and A is a side. Yay . well i got that right on one of my tests but i like never remembered how i got the answer sorta.

"a" is side? which side?

would it not be more like the centroid to the corners of a regular tetrahedron?

By "side", it's meant the length of the edge of one of the faces.  In the diagram below, it would be the length of the dotted line.  Or any other line on here.
[img]http://www.jaist.ac.jp/~g-kampis/Lecture_One/tetrahedron.jpg[/img]
Keep in mind that a tetrahedron is made of four congruent equilateral triangles, so all sides are the same length.


Let $ a;b;c \ge\ 0 ; k \ge\ 3 and  a \plus{} b \plus{} c \equal{} k$. Prove that :
$ \frac {x \plus{} y^2}{\sqrt [k]{(x^2 \plus{} k)(y^2 \plus{} k)}} \plus{} \frac {y \plus{} z^2}{\sqrt [k]{(y^2 \plus{} k)(z^2 \plus{} k)}} \plus{} \frac {z \plus{} x^2}{\sqrt [k]{(z^2 \plus{} k)(x^2 \plus{} k)}} \ge\ \frac {(k \plus{} \sqrt {k})^2}{( \sqrt {k \plus{} k^2} \plus{} 2\sqrt {k})^{\frac {2}{k}}.3^{\frac {k \minus{} 1}{k}}}$

No one ? :( How about is it ? Solve it please. :)

[quote="nguoivn"]I think you should rewrite your ineq in case $ k \equal{} 3$, it' ll be nicer.
With me, I don't like the expression: $ \frac {(k \plus{} \sqrt {k})^2}{( \sqrt {k \plus{} k^2} \plus{} 2\sqrt {k})^{\frac {2}{k}}.3^{\frac {k \minus{} 1}{k}}}$[/quote]
Thanks! I think i have some mistake I 'll rewrite :

Let $ a;b;c \ge\ 0 ; k \ge\ 2 and a \plus{} b \plus{} c \equal{} k$. Prove that :
$ \frac {x \plus{} y^2}{\sqrt [2k]{(x^2 \plus{} k)(y^2 \plus{} k)}} \plus{} \frac {y \plus{} z^2}{\sqrt [2k]{(y^2 \plus{} k)(z^2 \plus{} k)}} \plus{} \frac {z \plus{} x^2}{\sqrt [2k]{(z^2 \plus{} k)(x^2 \plus{} k)}} \ge\ \frac {(k \plus{} \sqrt {k})^2}{( \sqrt {k \plus{} k^2} \plus{} 2\sqrt {k})^{\frac {2}{k}}.3^{\frac {k \minus{} 2}{k}}}$

With case k=2 :
$ \frac {x \plus{} y^2}{\sqrt [4]{(x^2 \plus{} 2)(y^2 \plus{} 2)}} \plus{} \frac {y \plus{} z^2}{\sqrt [4]{(y^2 \plus{} 2)(z^2 \plus{} 2)}} \plus{} \frac {z \plus{} x^2}{\sqrt [4]{(z^2 \plus{} 2)(x^2 \plus{} 2)}} \ge\ \frac {(2 \plus{} \sqrt {2})^2}{ \sqrt {6} \plus{} 2\sqrt {2}}$

[quote="onlylove_math"]
With case k=2 :
$ \frac {x \plus{} y^2}{\sqrt [4]{(x^2 \plus{} 2)(y^2 \plus{} 2)}} \plus{} \frac {y \plus{} z^2}{\sqrt [4]{(y^2 \plus{} 2)(z^2 \plus{} 2)}} \plus{} \frac {z \plus{} x^2}{\sqrt [4]{(z^2 \plus{} 2)(x^2 \plus{} 2)}} \ge\ \frac {(2 \plus{} \sqrt {2})^2}{\sqrt {6} \plus{} 2\sqrt {2}}$[/quote]
No one solve it  :(


Welcome to my riddle tourney. due to my dumbness  :rotfl: i will have "helpers" to make riddles
to become a helper pm me a riddle and an answer the 5 riddles i like the best will be my "helpers" and all riddles will be posted in first round!!!

pm me your answers for first round of elimination and after that i will choose a helper to post a riddle. the first to post the correct ans wins a point (added to the points of the first round) at the end of 10 rounds whoever has the most points wins!!!!!!! :first:

i no this is confusing if u have any qs just post and ill answer them. FOR THOSE OF U WHO WANT TO POST RIDDLES YOUR PM IS EXCPECTED THIS FRIDAY AT MIDNIGHT FORUM TIME!!!!!! JOIN NOW!!!!!!!!  :D


Why do we have 3 sigfigss for #2b in http://portal.acs.org/portal/PublicWebSite/e?

2. (11%) The ionization constant for water is 1.14 × 10–15 at 0 °C and 9.6 × 10–14 at 60 °C.
a. Write the equation for the ionization of water and determine the pH of water at 60 °C.
b. Calculate each value.
i. ΔHionization over this temperature range
ii. ΔG at 60 °C
iii. ΔS at 60 °C
c. State the significance of the sign of the sign of ΔS obtained in part 2b.iii, and explain how the process indicated
in 2a could lead to this sign.

We have a 9.6 × 10–14... (-14 is supposed to be an exponent but you know that)
Why do we have 3 sigfigss for #2b?

Logarithms work a little bit differently in terms of sig figs.  The way to figure out how many sig figs you use when taking the natural logarithm of something is to write it out in scientific notation.  The number of exponent in the scientific notation is used in determining the whole number of the answer.  For example, say the hydrogen ion concentration is 9.6x10^-14.  The 14 tells you the answer is going to be 13.something.  The number of sigfigs in that something is determined by the number of digits AFTER the decimal place in the original number.  In our example 9.6x10^-14, there is one digit after the decimal, a 6, meaning there is one digit in the 13.something so our answer is 13.0.  Some more examples are
log(5.45x10^13)=13.74
log(6x10^-15)=-14
If there's something wrong with these rules feel free to correct me, I'm not totally solid on them myself.


[color=darkblue]Let $z=\cos\frac{2\pi}{p}+i\cdot\sin\frac{2\pi}{p}$ be a complex number, where $p\in N\ ,\ p\ge 2$ is fixed.

If $S_{n}=z^{n}+\frac{1}{z^{n}}$ , where $n\in Z$, how many values are possible for $S_{n}$ ?
[hide="Answer."]$\boxed{\ \left[\frac{p}{2}\right]+1\ }\ .$[/hide]

[b]Remark.[/b] See a particular case to http://www.mathlinks.ro/Forum/viewtopic.php?t=110759[/color]

[hide="Solution"]$S_{n}=z^{n}+\bar z^{n}=2\cos{2n\pi\over p}$

What we're dealing with here are the vertices of a regular polygon which is symmetric around the $x$ axis. Hence we count only the vertices which are above or on $x$ axis. If $p=2t$, then there are $t+1$ such vertices, and if $p=2t-1$, then there are $t$ such vertices. Both cases are covered by $N=\lfloor{p\over 2}\rfloor+1$[/hide]


I didn't make USAMO but i have an attempt at number 4. Im almost positive theres something wrong with my logic but I can't find it. I was just wondering, am i allowed to post it here yet?

Yes, post it in the thread dedicated to #4 and we'll tell you if it looks correct or not.


Find $\sqrt{53-8\sqrt{15}}$

I used the form  $a+b\sqrt{15}$ to solve this problem, but I'm getting the wrong asnwer.  Can any one tell me what I did wrong?

squaring I get  $53-8\sqrt{15}= a^{2}+15b^{2}+ab\sqrt{15}$ for some number a and some number b.

so $2ab\sqrt{15}=-8\sqrt{15}$
                   $ab=-4$      $a=-4/b$
                   $a^{2}+15b^{2}=53$
           substituting in a I get:
    $16/b^{2}+15b^{2}=53$    mutiple everything by $b^{2}$
     $15b^{4}-53b^{2}+16=0$
     b^2=p  for some number p
     $15p^{2}-53p+16+0$
      $(5p^{2}-16)(3p-1)$
      so the roots are $16/5, 1/3$   $16/5=b^{2}$  so $b=4\sqrt{5}$    $a=-4/4\sqrt{5}=-\sqrt{5}$

so my answer  is
$12-\sqrt{5}$.   but  the book's answer is $4\sqrt{3}-\sqrt{5}$.  They used the form $a\sqrt{3}+b\sqrt{5}$ to solve the problem, but can't you solve this problem using my way?   :?

Thanks

But

$(12-\sqrt5)^{2}$ does not equal $53-8\sqrt{15}$

 :)
So the answer can't be $12-\sqrt{5}$
What your book might have done is equate the rational and irrational parts as -
Let $\sqrt{53-8\sqrt{5}}= (a-b)^{2}$
$53-8\sqrt{5}= a^{2}+b^{2}-2ab$

Hence,
$a^{2}+b^{2}= 53$
Also, $-2ab =-8\sqrt{15}$
$\Rightarrow ab = 4\sqrt{15}$

By trial we see that $a = 4\sqrt{3}$ and $b = \sqrt{5}$
On checking we see that-
$(4\sqrt{3}-\sqrt{5})^{2}= 53-8\sqrt{15}$

Heh yeah, you gotta remember to try all factors of the number under the radical sign.

Yes, composite numbers are tricky since you might need to use other factors of the numbers.  But $4\sqrt{3}-\sqrt{5}$, is correct.

wait... so
$\sqrt{53-8\sqrt{15}}= (a-b)^{2}$   I thought the sign should've been negative, but in the book as an example they had

$\sqrt{34-24\sqrt{2}}=(a+b\sqrt{2})^{2}$  (aops book1 pg. 82)
 should the sign be negative or positive for (a+b)^2
 and do you need to have the $\sqrt{15}$.
$(a-\sqrt{15}b)^{2}$

Thanks!

[quote="myc"]$\sqrt{34-24\sqrt{2}}=(a+b\sqrt{2})^{2}$  (aops book1 pg. 82)
 should the sign be negative or positive for (a+b)^2
 and do you need to have the $\sqrt{15}$.
$(a-\sqrt{15}b)^{2}$[/quote] It can be either positive or negative. If it is positive, b will be negative.
The $\sqrt{15}$ will not guarantee an answer. If it doesn't work, you could try $a\sqrt3+b\sqrt5$.


Hello!

Let $ P$ be a point inside of a triangle $ ABC$, $ \Gamma$ is the circumcircle of $ \Delta ABC$ and $ d$ is a line through the point $ P$. If
$ \left\{A,A'\right\}\equal{}AP\cap\Gamma$, $ \left\{B,B'\right\}\equal{}BP\cap\Gamma$, $ \left\{C,C'\right\}\equal{}CP\cap\Gamma$, $ \left\{D\right\}\equal{}d\cap BC$, $ \left\{E\right\}\equal{}d\cap CA$ and $ \left\{F\right\}\equal{}d\cap AB$
then prove that the lines $ A'D$, $ B'E$ and $ C'F$ are concurrent and the point of concurrency lies on the circle $ \Gamma$.

Best regards, Petrisor Neagoe :)

Let $ U$ be the second intersection of $ A'D$ with $ \Gamma.$ By Pascal theorem for the non-convex hexagon $ UA'ABCC'$ we get that $ D \equiv A'U \cap BC,$ $ P \equiv AA' \cap CC'$ and $ F' \equiv AB \cap C'U$ are collinear $ \Longrightarrow$ $ F \equiv F'$ $ \Longrightarrow$ $ U \equiv A'D \cap C'F.$ Again by Pascal theorem for the non-convex hexagon $ UA'ACBB',$ the intersections $ D \equiv A'U \cap BC,$ $ P \equiv AA' \cap BB'$ and $ E' \equiv AC \cap B'U$ are collinear $ \Longrightarrow$ $ E' \equiv E$ $ \Longrightarrow$ $ U \equiv A'D\cap B'E.$

Dear Mathlinkers,
this result for a circle is well-known...

http://jl.ayme.pagesperso-orange.fr/Docs/La%20P-transversale%20de%20Q.pdf

Sincerely
Jean-Louis


geometry
rectangle
power of a point
radical axis

[quote="Quevvy"]What do you mean "circumferences intersect at M and N"? circumference is the distance of a closed curve, so distances cannot intersect...[/quote]

 ok, sorry , it is actually circles.But this shouldn´t have gotten in the way, since by your explanation the question would make no sense at all. You could have figured by yourself that it is actually circles.

[quote="v235711"][quote="PKMathew"][quote="v235711"]circunferences $ S_1$ and $ S_2$ intersect at  M and N.a rectangle ABCD has vertices A and C at $ S_1$ and B and D at $ S_2$. prove that the intersection of the diagonals lies on MN.[/quote]

Would you post a diagram of your question here.
 :blush:[/quote]

that´s part of the problem![/quote]

believe me ,it is right.

[quote="v235711"]Subject very ,very easy one[/quote]
[quote="v235711"]nobody solved it yet?!...[/quote]

Subjective descriptions don't belong in these subject lines.
And posts of impatience and desperation don't belong either.

Here could be a correct restating of the problem, with the
attentiveness given to forming proper sentences, which there
could have been the first time:

"Circles $ S_1$ and $ S_2$ intersect at M and N.  A rectangle ABCD has
vertices A and C on $ S_1$, and B and D on $ S_2$.  Prove that the
intersection of the diagonals lies on MN."

[hide]Let the diagonals intersect at $ T$. Since $ ABCD$ is a rectangle and hence cylclic, $ TA*TC \equal{} TB * TD$. But this means that $ T$ has equal power to both circles, so it lies on the radical axis, which is simply $ MN$.[/hide]


hi ... where i can find MOSP problems???plz help me.. :?:

I have some in them . What is your email?

http://www.unl.edu/amc/a-activities/a6-mosp/archivemosp.shtml   :wink:


function
geometry
parallelogram
Functional Analysis
advanced fields
advanced fields unsolved

Let $ \{ A_{i}\}$ be a nested decreasing sequence of nonempty closed bounded convex sets in the Hilbert space $ H$. Show that the intersection of all the $ A_{i}$'s is nonempty.

Give a counterexample that it is not true in a Banach space.

In any reflexive Banach space, closed convex and bounded sets are weakly compact, so the result follows if we pass to the weak topology.

For a counterexample, consider the space $ X=C([0,1])$ of continuous real-valued functions on the unit interval. For each $ \alpha\in[0,1)$ let $ A_\alpha$ be the set of elements of $ X$ which are $ 0$ on $ [0,\alpha]$ and $ 1$ in $ 1$. As $ \alpha$ increases to $ 1$, the (obviously closed, convex, and bounded) sets $ A_\alpha$ decrease, and their intersection in the space of all real-valued functions on the unit interval is the characteristic function of $ \{1\}$, which is not continuous.

Here is another proof (of course grobber's idea is more general).

It is well known that in every closed and convex subset of a Hilbert space there exists an element with minimal norm. Let $ x_{n}\in A_{n}$, $ \|x_{n}\| \equal{}\mathrm{dist(0,A_{n})},$ be a sequence of such elements ($ x_{n}$ is even unique).

$ (\|x_{n}\|)_{n}$ is real, bounded, non-increasing sequence, hence it is Cauchy.

Then by the parallelogram identity we have $ (m < n)$
\[ \|x_{m}\minus{}x_{n}\|^{2}\equal{}2\|x_{m}\|^{2}\plus{}2\|x_{n}\|^{2}\minus{}4\left\|\frac{x_{m}\plus{}x_{n}}{2}\right\|^{2}.\]
Since $ x_{m}, x_{n}\in A_{n}$ and $ A_{n}$ is convex, $ \left\|\frac{x_{m}\plus{}x_{n}}{2}\right\|\geq\|x_{n}\|$, thus
\[ \|x_{m}\minus{}x_{n}\|^{2}\leq{}2(\|x_{m}\|^{2}\minus{}\|x_{n}\|^{2})\leq M(\|x_{m}\|\minus{}\|x_{n}\|),\]
which goes to $ 0$ as $ m, n\to\infty$, by Cauchy condition.

This proves that $ (x_{n})$ is Cauchy sequence in our Hilbert space.


calculus Putnam search AMC AIME USA(J)MO USAMO

I'm a senior in high school this year (in a couple of months) and am going to be taking Calculus III. I am a very avid mathematician (have trained for olympiads, etc.) and wish to do well on the Putnam (top 50 would be nice). Our class will be using the book "Calculus: One & Several Variables" by Salas, Hille, and Etgen. Has anyone used this book before? Also, can someone recommend books (or a series of books) that I could get regarding linear algebra which would help me on the Putnam level given the following:

- I know absolutely no linear algebra
- I know no real analysis
- I know no advanced math past AP Calculus BC, apart from AP Statistics, which isn't even advanced at all (but if this would narrow my search, okay  :) )

My problem solving level is as such:

- I can consistently score double digits on an AIME
- I can consistently get at least one question correct on a USAMO

Thank you very much,

Karth

[quote="Karth"]I'm a senior in high school this year (in a couple of months) and am going to be taking Calculus III. I am a very avid mathematician (have trained for olympiads, etc.) and wish to do well on the Putnam (top 50 would be nice). Our class will be using the book "Calculus: One & Several Variables" by Salas, Hille, and Etgen. Has anyone used this book before? Also, can someone recommend books (or a series of books) that I could get regarding linear algebra which would help me on the Putnam level given the following:[/quote]

Im happy for you that you are good in math. Let me tell you something that should make you very comfortable. The next several years of math are going to be the easiest stuff you have ever taken. Seriously compared to problem solving all undergradute math is super easy.

I happen to know this, because I used to be a freshman in the begging of the year before I got kicked out  :blush:  . I was taking advanced courses, it was baby stuff, I used to help out my classmates, thats how easy it was.

[quote]
- I know absolutely no linear algebra
[/quote]
You aint missing a lot. The only fun stuff in a basic course in linear algebra is "Vector Spaces" and basically the only thing you need in pure math. The rest is boring computation of matrices and determinants.
[quote]
- I know no real analysis
[/quote]
Most colleges call it advanced calculus. Real analysis is a study of integration and is a graduate course.

Take that after Calculus III as fast as you can. I fell in love with analysis (or advanced calculus). It is amazing. You will love it.
[quote]
I know no advanced math past AP Calculus BC, apart from AP Statistics, which isn't even advanced at all (but if this would narrow my search, okay  :) )
[/quote]
I think you are much above an average person your age in mathematics. You shouldnt be concerned about that. I think good things are waiting for you.

I can weigh in with my thoughts on Linear Algebra.

 - Try going through the video lectures on MIT Open Courseware with Professor Strang; this will cover the basics and you can proceed at your own pace.

e.g. [url]http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/index.htm[/url] (there is more than one though)

 - If you get comfortable with the material, a good source of challenging problems can be found at:  [url]http://math.berkeley.edu/graduate_prelim.html[/url] and solutions are available in book form:  "Berkeley Problems in Mathematics" by Paulo Ney de Souza and Jorge-Nuno Silva.

 - Another good book that fills in some of the gaps of a first course and captures all of the interesting odds and ends is "Linear Algebra" by Peter Lax.

   Once you get through the elementary manipulations and properties of matrices, you will come to realize the power of the subject.  It is fundamental to much of numerical analysis and it is an excellent launching off point for infinite dimensional vector spaces which are at the heart of functional analysis and PDEs.

Enjoy!


[b]Find the period of $ \sin{\frac{ax}{b}}\plus{}\cos{\frac{cx}{d}}$ where $ a,b,c$ and $ d$ are integers and $ \frac{a}{b}$ and $ \frac{c}{d}$ are in lowest terms[/b]


algebra
polynomial
calculus
derivative
advanced fields
advanced fields unsolved

Let $ f \in k[x,y,z]$ be a homogeneous polynomial, let $ Y \equal{} Z(f) \subset \mathbb{P}^2$ be the algebraic set defined by $ f$, and suppose that for every $ P \in Y$, at least one of $ (\partial f/\partial x)(P)$, $ (\partial f/\partial y)(P)$, $ (\partial f/\partial z)(P)$ is nonzero.  Show that $ f$ is irreducible, and hence that $ Y$ is a nonsingular variety.

We can write $ Y \equal{} Y_1 \cup \cdots \cup Y_n$, where each $ Y_i$ is closed, irreducible and no one of the $ Y_i$ contains any other.  Can we conclude anything about the $ Y_i$?  What polynomial(s) define them?  I want to use the fact that any two curves have nonempty intersection.  Also, I don't really see how the fact that there are no singular points on $ Y$ helps me.  Any hints/help would be appreciated.

For the first part, assume that $ f \equal{} ab$ factors. If both are nonconstant, you can find a point $ P$ of intersection of the curves to $ a$ and $ b$. Then check that all partial derivatives of $ f$ at $ P$ are $ 0$.


How many positive 3-digit integers less than 400 satisfy each of the following conditions:

A)  Each of teh three digits is prime.

B)  The sum of the three digits is prime.

C)  The number is prime.

[quote="tinytim"]0. :wink:

a three digit number consisting of three prime digits either has a digit sum thats even or a digit sum of 15.[/quote]
337 :P

Prime numbers under $ 10$: $ 2,3,5,7$
Possible Hundred Digits: $ 2,3$
Possible Tens digits: $ 2,3,5,7$
Possible Units digits: $ 2,3,7$
Prime numbers under $ 3 \plus{} 7 \plus{} 7$: $ 2,3,5,7,11,13,17$

Total Numbers to consider: $ 2*4*4$
If hundred digit is $ 2$, then there has to be another $ 2$, and the last digit must be $ 7$. So that yields $ 2$ possibilities to look at: $ 232$ and $ 223$. $ 232$ is not prime. $ 223$[b] is prime[/b]. That is $ 1$ possible value so far.
If hundred digit is $ 3$, then there must be a $ 2$, and another prime, or 3 primes. The last digit cannot be prime, because then it wouldn't be prime. So possible values $ 323$, $ 325$, $ 327$. $ 325$ is obviously not prime. $ 327$ is divisible by $ 3$ because sum of digits is $ 12$. $ 323$ [b]is prime[/b].
Now if value has $ 3$ primes as digits, more possibilities!
$ 333$, $ 337$, ($ 335$ has no point of being looked at), $ 353$, $ 357$, $ 373$, $ 377$. $ 333$ is obviously divisible by $ 111$. $ 337$ [b]is prime[/b]. $ 353$ [b]is prime.[/b] $ 373$ [b]is prime[/b]. This gives $ 3$ more primes

$ 1 \plus{} 1 \plus{} 3 \equal{} 5$

oops. :blush: Thought the three digits had to be different.


geometry calculus trigonometry vector integration search algebra

How many of u are going to write the INMO (next) year :D
especially 12th of which  only 6 will be selected at most.

I know the 5 out of the6 people who r going to write from TN See
http://amtionline.com/rmoresult.htm

hey is ROhit.K u
and i beleive Sanjith the 6th ranker is Sathej's brother

Yes he's also now called the Alchemist now. But he's in 11th and in my school.
The rohit k thingy made me both mad  :mad:  and sad bcos all along i was praying that me(" rohit k ") should be selected  :rotfl:

I am anirudh from mumbai in 11th standard and i'm writing inmo next year. hopefully, w'll have some nice geometry problems and practically (nil) combinatorics problems....

@Anirudh
an perfect indian attitude toward olympiad problems  :D

what the hell i love combinatorics  :mad: and hate geometry

a rare show
i love geometry especially with the modern stuff and combi....

i love combinatorics , number theory and algebra and calculus and trig
Hate geometry 2d and 3d and vectors and stats and matrices (mechanical topic)

hey i am abhishek, i am in class 12 n will be riting the inmo from delhi region
best of luck to all of you!!

Pardesi u never told me you were selected  :mad:
 :wink:

sorry :blush:
i thought u understood it  :(

Dont take it soooooo serious da .
P.S. Look at the  :wink:  I posted in the previous post

Best of Luck to all who are going to appear for the INMO 2008 tomorrow.

[b][size=200][color=blue]
ALL THE BEST FOR INMO '08  :)
[/size]
[size=59]
and a special best of luck to you pardesi incase u didn't notice  :D
[/color][/size][/b]

[size=200][color=blue]I NOTICED THAT[/color][/size]
[size=59]Thanks[/size] :D

The junior camp - 29 Apr - 27 May

Senior : 06May  27 May

Yes SOUMYASHANT  thou shall find info abt chappli on the 1st of may dont u worry as i promised  :D  :D  :D

best of luck for inmo 2008 everybody
i thought this time to save myself from 5th time failure blushes so i didnt appear at rmo even

My guess for today's paper:

2 geometries
1 combinatorics
1 functional equation
1 stupid elementary number theory
1 polynomial kind question(algebra)

highest marks:75
cutoff : 36 for junior students
           48 for XIIthers

i would have loved an inequality this year :rotfl:

I would love the cutoffs to be what Soumyashant has predicted!

i would like it to be lesser for 12thers :D

Did soumyashant's prediction come true?  :D
Then pardesi u could've asked him earlier and not wasted time on inequalities  :D  :D

no he was wrong the special part(functional equation) :P
rest were obviuos :wink:

:D yes and we had an easy geometry instead of an easy number theory

but it has become a 'fashion' to habve a stupid number theory just for the sake of it each year :P

Since the exams are over i am locking this topic  :D  :lol:


trigonometry
complex numbers
irrational number
algebra proposed
algebra

Prove that there is n rational number  $r$ such that   $cosr\pi=\frac{3}{5}$

thank you. but Can anyone find an elementary solution for this very nice problem ?

A truely elementary solution seems impossible since it's very closely related to extensions of the rationals, I think.
But Ravi B's solution is very near to be totally elementary. What you don't like on that solution¿

You have right. His solution is very close to be elementary but he used the Chebychev's theorem which is not elementary. The other solutions posted at the above used complex numbers. I do not think that the proposer of an IMO TST from Serbia use in his official solution complex numbers .So I wait a more elementary solution .That's all. Thanks

[quote]but he used the Chebychev's theorem which is not elementary[/quote]

Quite on the contrary, Chebyshev's theorem is elementary. (in the sense that it does require advanced knowledges) Its proof is trivial, but I don't think it is learnt in school, so some people may not know about it.

As a side remark, there was a problem in the 2004 romanian tst whose solution was based on complex analysis. I'll never understand why. That's really a problem which needs a more elementary solution. Have a look over it...

[later edit] http://www.mathlinks.ro/Forum/viewtopic.php?t=5450

Perfect-Radio, what problem are you reffering to? :?  Barbu Berceanu's one.
If so, it was a really tough one. :)

Suppose $\cos{r\pi}=\frac35$ for some rational $r$. Then $\sin{r\pi}=\frac45$ and using de Moivre's formula, we deduce that $\left(\frac35 +i\frac45 \right)^n=1$ for some positive integer $n$.
Now, observe that $\frac35 +i\frac45 =\frac{2+i}{2-i}$, hence $(2+i)^n=(2-i)^n$. But
\[ (2+i)^n=(2-i+2i)^n=(2-i)^n+\binom{n}{1}(2-i)^{n-1}\cdot 2i+\ldots +(2i)^n. \]
We deduce that $(2i)^n=(2-i)(a+bi)$ for some integers $a,b.$ Taking the absolute values yields
\[ 2^{2n}=5(a^2+b^2), \]
a contradiction.

I have something like a solution but I don't know if it is true.

We have

$cos r\pi=1-\frac{r^2(\pi)^2}{2!}+\frac{r^4(\pi)^4}{4!}-\frac{r^6(\pi)^6}{6!}+....$

which is always an irrational number because  $(\pi)^2,(\pi)^4,....$  are irrationals.

$irrational=\frac{3}{5}$   which is absurb. Please tell me if it is right. Thanks

[quote="silouan"]I have something like a solution but I don't know if it is true.

We have

$cos r\pi=1-\frac{r^2(\pi)^2}{2!}+\frac{r^4(\pi)^4}{4!}-\frac{r^6(\pi)^6}{6!}$[/quote]
Note that the sum is not finite!

[quote]which is always an irrational number because  $(\pi)^2,(\pi)^4,....$  are irrationals.[/quote]
The argument doesn't work. The sum of several irrational numbers may  be rational as well.

[quote="enescu"] The sum of several irrational numbers may  be rational as well.
[/quote]

Yes, but this time we have to deal with the same irrational  the number  $\pi$ and

the sum of these can't be rational .(We cannot have equal terms in the sum with different sign.)I think it is under discussion

What do you mean ?The number $\pi$ doesnot ''leave'''

Silouan: as already said, your prove doesn't work. Set $r=1$, you would prove that $-1$ is irrational!

Yes.you are right .Could we improve my solution to be correct ?


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

There are 2n people, A1,A2,...,An, B1,B2,..., Bn sitting around a round table, such that Ai, Bi do not sit beside each other. Find the number of ways of sitting.

Are two dispositions which can be deduced using rotations (and symmetries?) considered as different?
In other words, are the seats numbered from $1$ to $2n$?

Pierre.


induction
combinatorics proposed
combinatorics

In a permutations $(x_{1}, x_{2}, ..., x_{n})$ of the set ${\{1, 2, ..., n}$ we call a pair $(x_{i}, x_{j})$ discordant if $i<j$ and $x_{i}>x_{j}$. Let $d(n, k)$ be the number of such permutation with exactly $k$ discordant pairs. Find $d(n, 2)$

What you are calling [i]discordants[/i] are usually called [url=http://mathworld.wolfram.com/PermutationInversion.html]inversions[/url].

[hide="Solution"]I claim that for $n \geq 2$ we have $d(n, 2) = \binom{n}{2}-1$.

Proof: It is clearly true for $n = 2$ and $n = 3$.  Suppose it is true for $2 \leq n \leq k$ for some $k \geq 3$.  Now, consider a permutation $(x_{1}, \ldots, x_{k+1})$ of $\{1, \ldots, k+1\}$ with two discordant pairs.  We classify these permutations as follows:
Class 1: those for which $x_{k+1}= k+1$.
Class 2: those for which $x_{k+1}= k$ and $x_{k}= k+1$.
Class 3: those for which $x_{k+1}= k-1$ and $x_{k}= k+1$ and $x_{k-1}= k$.
Class 4: those for which $x_{k+1}= k$ and $x_{k}= k-1$ and $x_{k-1}= k+1$.
Any permutation which does not fall into one of these classes must have at least 3 discordant pairs.  Let us count the number of elements in each class:
There is a bijection between Class 1 and permutations of $\{1, \ldots, k\}$ with two discordant pairs which we get by dropping $x_{k+1}$.  Thus, Class 1 has $d(k, 2) = \binom{k}{2}-1$ members.

There is a bijection between Class 2 permutations and permutations of $\{1, \ldots, k-1\}$ with one discordant pairs which we get by dropping $x_{k}$ and $x_{k+1}$.  Thus Class 2 has $d(k-1, 1) = k-2$ members (those in which we transpose two consecutive elements).

Classes 3 and 4 each contain a unique element.

Thus, $d(k+1, 2) = \binom{k}{2}-1+k-2+2 = \binom{k+1}{2}-1$ and by induction we're done.[/hide]

A similar approach could be used for successive values of $d(n, k)$.

It can also be done directly: [hide]starting from the order $(1, \ldots, n)$ we must either move a single element to the left twice ($n-2$ choices, since neither 1 nor 2 has two places to its left) or move two elements each one place to the left ($\binom{n-1}{2}$ choices, since neither one can be 1).  $n-2+\binom{n-1}{2}= \binom{n}{2}-1$[/hide]

Sorry for double-posting, but could not edit.

This problem is from $1984$ IMO Shortlist.


[hide]If you draw the cube (10x10), then the only cubes that are painted red on 3 sides are the vertices, so there are always 8.

Then, you notice that the only cubes that are painted red on 2 sides are the ones on the edges that are not vertices.  So, on each edge, there are 8.  $8 \times 12=96$.

Thus, there are 104 cubes.[/hide]

[quote="goldendomer"]I never know how to do these cube problems. Ex.

-1999 state
-A cube is painted red and then cut into 1000 congruent cubes. How many of these cubes are painted red on atleast two faces?

Can you tell me how to get the answer?[/quote]

[hide]The cube lets say that the big cube is cut into 1000 cubes, let them be unit cubes. The big cube is a 10x10x10. The sides that are red on at least 2 faces are the edges and corners of the big cube. You can then count. $\(104$ should be the answer....unless I added incorrectly.... 40 for each of the sides plus 8 times 8 for the other sides.[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Suppose that f is a homorphism from finite group G to G and kernel of f is nilpotent. If set of orders of elements of G is same as ones of f(G), then what conditions  do f satisfy ?

[quote="rulin"]If set of orders of elements of G is same as ones of f(G)[/quote]
Can you explain what does it mean?

Are you sure about the problem statement, e.g., if $ G$ is a nilpotent group itself, than any normal subgroup $ N$ with $ o(G/N)\equal{}o(G)$ will imply a homomorphism $ f$ as required, however, no generic additional structure on $ f$ or $ N,G$ can be concluded !?

[quote="rulin"]set of orders of elements of G is {o(g)|for all elements g in G}, where f(G) ={f(g)|all g in G}[/quote]
Just consider the abelian group $ \mathbb{Z_{2}}\times\mathbb{Z_{2}}\to\mathbb{Z_{2}}$,$ f$ needn't to be isomorphism


inequalities
algebra
polynomial
inequalities proposed

Let $a,b,c$ be positive reals such that $a+b+c=1$. Find $k$ maximum such that the following inequality holds
\[ 1-27abc\geq k\left(ab+bc+ca-9abc\right)  \]

here is an useful lemma in   Hojoo lee's article.
Let $P(u,v,w) \in R[u,v,w]$ be homogeneous symmetric polynomial with degree 3.Then the following two stataments are  quavilent:
(a)$P(1,1,1),P(1,1,0),P(1,0,0) \ge 0$
(b)$P(x,y,z) \ge 0$ for all $x,y,z \ge 0$

Zhaobin , can you post a proof for Hojoo Lee's article ? Where can I find it ?

I get it from [url=http://ultrametric.googlepages.com/tin2006new.pdf]Topics in Inequalities[/url]


quadratics
conics
parabola
calculus
algebra
polynomial
analytic geometry

If in applying the quadratic formula to a quadratic equation
\[ f(x) \equiv ax^2 + bx + c = 0, \]
it happens that $c = b^2/4a$, then the graph of $y = f(x)$ will certainly:

(A)	have a maximum
(B)	have a minimum
(C)	be tangent to the $x$-axis
(D)	be tangent to the $y$-axis
(E)	lie in one quadrant only

[hide="Answer"]The quadratic formula is $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If $c=\frac{b^2}{4a}$, then we have $\frac{-b\pm\sqrt{b^2-b^2}}{2a}=\frac{-b}{2a}$, meaning that the quadratic has only one distinct solution, so it is of the form $(x-n)^2$. This means that the parabola must have a minimum, so our answer is $\boxed{B}$.[/hide]

what if its $(x+n)^2$  :lol:
if both roots are equal
[hide="then"]
$\boxed{C}$[/hide]

[quote="JesusFreak197"]No, not necessarily. And $(x+n)^2$ is the same thing as $(x-n)^2$ as far as this problem is concerned; it's still only one root. What if you have $(x-2)^2=x^2-4x+4$? That won't be tangent to the x-axis. My point in proving that it must be $(x-n)^2$ is that you can't have a negative coefficient for the $x^2$ term, so it must have a minimum.[/quote]

sorry i meant

what about $-(x+n)^2$?   :D
$f(x)=(x-2)^2=x^2-4x+4$ is tangent to the x-axis in $(2,0)$ because $D(f)(2)=0$. But $-f(x)=-(x-2)^2=-x^2+4x-4$ has no minimum.

Hmm, that's true... but is there a way to prove it without Calculus? The AMC/AIME/USAMO never require Calculus to solve.

well suppose it has two roots, that would mean it has two intersections with the x-axis. As these roots come closer to each other, the curve will become more tangent to the x-axis.

or, supposed that you know that in the minimum/maximum of a parabola the tangent to the curve is horizontal this follows immediately (or however you write that last word) .

Oh, right, only one x-value that gives a y-value of $0$, so only one intersection with the x-axis. :roll:

[hide]Plug the c value in and you get, after simplifying, $\frac{-b}{2a}$. THere is only one solution. I don't know about the discussion above, but I think it must have a minimum, since all the other choices are impossible (especially C, D, and E). [/hide]

Heh, I thought so too, but try actually reading the discussion above. ;)

After some posts above i'm thinking seriously about to burn my books of Calculus.  :huh:

[hide]If $f(x)=ax^2+bx+c$, then it is a polynomial function. So $f(x)$ can be writen like $f(x)=a(x-x_1)(x-x_2)$,

where $x_1, x_2$ are the roots. Therefore if $a>0$, $f(x)$ have a minimum, and if $a<0$, $f(x)$ have a maximum.

Since we don't know $a$, we just have $\triangle = b^2 - 4 \cdot a \cdot \frac{b^2}{4a} = 0$. Thus $x_1=x_2=-\frac{b}{2a}$ and the graph of $f(x)$ is tangent to the $x$-axis.

Now, using calculus, $f'(x) = 2ax + b \Rightarrow f'\left(-\frac{b}{2a}\right)=2a \cdot \left(-\frac{b}{2a}\right) +b = 0$ and we are donne.

I'm missing something?[/hide]

$f(x) \equiv ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}) = a(x+\frac{b}{2a})^2$

If I understood the problem correctly. I think it is phrased a bit weirdly. Anyway, this quadratic curve is not translated upwards or downwards, so it is tangent to the x-axis in the point $x = -\frac{b}{2a}$. It confuses me as to why everyone says this is wrong though.

It only has 1 solution so it touches the x-axis at only one point.  Because this point is the vertex of the parabola the line tangent at that point will have a slope of zero.  Therefore the graph is tangent to the x-axis.  It doesnt neccessarily have a minimum or maximum because the graph could be pointing either upwards or downwards (consider the graphs $(x)^2$ and $-(x)^2$).  Note that the example of $x^2$ follows the property that $c=(b^2)/(4a)$, lies in more than one quadrant, and is not tangent to the y-axis.  So the only choice that is true is $C$.  I hope this provides a clear explanation for anyone who may be confused.

allow me to add just this one thing: no polynomial at all lies in one quadrant only or touch the y-axis so D and E shouldn't even be considered

the answer is C, we know that there is one root, so there is a double root, so it is tangent to the x-axis, you cannot say that there is a minimum because we do not know that a>0


Could you please put the picture on? It could really help.  :maybe:

Use http://www.photobucket.com, and sign up there, or use the AoPS Galleries>geometry pictures or http://www.imageshack.us

[quote="ckck"]There is a triangle ABC. On side AB there is a point G that seperates AG and GB in a ratio of 2:3. There is also a point F on side AC that seperates AF and FC in a ratio of 1:3.
There is another point E on side BC. BF, AE and GC all meet at a point P. What is the ratio of CP to PG?

$\text{A. }5: 1\qquad\text{B. }10: 3\qquad\text{C. }10: 9\qquad\text{D. }3: 10$
I have no idea where to start with this problem.
(sorry, there was supposed to be a picture with this problem, but I have no idea how to put a picture in it.)[/quote]
[hide]Let $\triangle ABC$ be a right triangle with $A(0,0),\, B(0,10),\, C(24,0).$ In this triangle, $AB=10,\, BC=26,\, AC=24.$ We know $G(0,4)$ and $F(6,0).$

Taking points $C$ and $G,$ we get the equation $x+6y=36.$ Then, we take points $F$ and $B,$ and get the equation $5x+3y=30.$

Solving these two equations, we get $x=\frac{8}{3}$ and $y=\frac{50}{9}.$ This is the intersection of the two lines, so we know $P\left(\frac{8}{3},\frac{50}{9}\right).$

Using the distance formula, $CP=\sqrt{\left(0-\frac{50}{9}\right)^{2}+\left(24-\frac{8}{3}\right)^{2}}=\sqrt{\frac{2500}{81}+\frac{4096}{9}}=\frac{\sqrt{39364}}{9}=\frac{2\sqrt{9841}}{9}$ and $PG=\sqrt{\left(4-\frac{50}{9}\right)^{2}+\left(0-\frac{8}{3}\right)^{2}}=\sqrt{\frac{196}{81}+\frac{64}{9}}=\frac{\sqrt{772}}{9}=\frac{2\sqrt{193}}{9}.$

The ratio is $\sqrt{9841}: \sqrt{193}.$[/hide]

At the moment I have it narrowed down to either A or B.

I messed up somewhere and nobody else has solved it, so why doesn't ckck tell us the answer and solution?

[quote="ckck"]
I have no idea where to start with this problem.
[/quote]

I think he's implying he hasn't ready any solution :) .

-jorian

I don't think this problem is suitable for Mathcounts. Other than analytically, I don't think there's a way to do this without Stewart's theorem.

[quote="ckck"]There is triangle ABC. On side AB there is a point G that seperates AG and GB in a ratio of 2:3. There is also a point F on side that seperates AF and FC to a ratio of 1:3.
There is another point E on side BC. BF, AE and GC all meet at a point P. What is the ratio of CP to PG?

A.5:1   B.10:3    C.10:9    D.3:10
I have no idea where to start with this problem.
(sorry, there was supposed to be a picture with this problem, but I have no idea how to put a picture in it.)[/quote]

deleted my note, i misread it

[quote="jhredsox"]the picture is very distorted. i think that line fb is supposed to be more to the left

-jorian[/quote]

FIGURES ARE NOT NECCESSARILY DRAWN TO SCALE

i know that, im just saying that for clarity for others :lol:

-jorian

i do think it should move over to the left :P

ok, so I solved this with just basic area geometry and some logic, and some similar triangle thinking.  Not sure how well I can explain it here, but I'll give it a try.

If you don't draw this picture and follow along my work step by step I am sure this will be confusing.  If you do, it should make sense.  Frost

[hide]

Since AF:CF = 1:3, I called the Area of $\triangle{APF}= B$ and so the area of $\triangle{CPF}= 3B$.  I know this to be true since the two triangles have the same height and their bases are in ration of 1:3.

Similarly I called the are of $\triangle{APG}=2A$ and the area of  $\triangle{BPG}= 3A$ for the same reasons as above.

Then I said the ratio of  $\triangle{ABF}$ :  $\triangle{CBF}$ would be equal to 1:3 as well since they have the same height and bases in ratio of 1:3.  So based on the parts of the triangles whose areas are already labeled, the two parts of  $\triangle{BPC}$ must be equal to 15A(since they have to be three times as big as $\triangle{BPA}$ which is 5A.

Then I did the same with  $\triangle{CGA}$ and  $\triangle{CGB}$ which must be in ratio 2:3.  And that means that  $\triangle{CPA}$ and  $\triangle{CPB}$ must be in ratio 2:3 as well.  Since  $\triangle{CPA}=4B$,   $\triangle{CPB}=6B$.

Ok, now we know 15A = 6B.  So $A = \frac{2}{5}B$.  So in your picture, replace  $\triangle{APG}$ label of 2A with $\frac{4}{5}B$.

Now we will examine  $\triangle{ACG}$.  The area of the whole thing is  $4\frac{4}{5}$.  The area of $\triangle{APG}=\frac{4}{5}$.  Since those two triangles have an area ratio of $\frac{1}{6}$ once you simplify and the base is the same, the height must be in ratio 1:6 as well.  Now while CG and PG are not the heights of the triangles, they will have the same ratio.  So $\frac{CG}{PG}= \frac{6}{1}$ and therefore $\frac{CP}{PG}= \frac{5}{1}$.

Hurray the ratio is 5:1

[/hide]


inequalities
calculus
algebra solved
algebra

Looking through a gazette from Romania (Galati, nr 16, 2000) I found three problems, which I cannot solve! pls help me!

1)Show that the following inequality holds for any real numbers x,y,z:

[sqrt(x^2+y^2)+sqrt(y^2+z^2)+sqrt(z^2+x^2)]^2>=6(xy+yz+zx).

2) Solve in natural numbers the equation:
(4k+3)^x-2^x=y^2.

3) Let n belong to N (n>=2), x(1),x(2),..,x(n)  belong to R with the property that :
 |x(k+1)-x(k)|<=1, 1<=k<=n. Prove that:
    n
   \sum (|x(k+1)-x(k)|) <= 2*[n/2], with  x(n+1)=x(1).
   k=1

   x(k) is read "x index k"
   [n/2]=n/2-{n/2}.

Thank you very much!

here's my solution to the inequality.substitute
a=x+y  ,  b=y+z  ,  c =z+x
then the given inequality(after squaring) is equivalent to proving the following inequality
a^2+b^2+c^2 - 2xy- 2yz- 2zx + sqrt[{a^2 + (b-c)^2}{b^2 + (c-a)^2}]
+ two other similar terms under sq roots >= 6(xy+yz+zx)
transferring the xy,yz,zx terms on lhs to the rhs we get the following inequality
a^2+b^2+c^2 + (the same sq root terms)>= 2(2ab+2bc+2ca-a^2-b^2-c^2) which is equivalent to 3(a^2+b^2+c^2) - 4(ab+bc+ca) + sqrt{a^2 + (b-c)^2}{b^2 + (c-a)^2} + sqrt{b^2 + (c-a)^2}{c^2 + (a-b)^2} +
sqrt{c^2 + (a-b)^2}{a^2 + (b-c)^2} >= 0
now split the whole thing
3(a^2+b^2+c^2-ab-bc-ca) + [sqrt{a^2 + (b-c)^2}{b^2 + (c-a)^2} + sqrt{b^2 + (c-a)^2}{c^2 + (a-b)^2} + sqrt{c^2 + (a-b)^2}{a^2 + (b-c)^2} -ab-bc-ca] >=0
now it is well known that term in the first parentheses is greater than zero
and the other can be easily seen to be greater than zero becoz each term in the sqrt is greater than ab,bc,ca resp.hence our inequality is proved.
note that it suffices to prove the inequality for positive reals only coz if ane of the terms are negative then the rhs remains unchanged while lhs decreases.though my proof dint assume them to be positive reals.
i hope my proof  is correct.in case u have got a different proof yourselves post it
cheers!!

This is a solution to the diophantine equation. The solution is pretty nice, using congruences (what else?)

So, let's see that x=1 means the equation becomes 4k+1=y ^2 , so
x=1, y=n odd, k=(n ^2 -1)/4 satisfies. Let's show that if x>1 there are no solutions.

Let's consider the equation modulo 4. We have (-1)x+0=0, if y even or (-1)x+0=1, if y odd. The first is impossible and the latter means that x even. So x=2zm, where m is odd.

The equation thus becomes [(4k+3)x/2-2x/2][(4k+3)x/2+2x/2]=y ^2. But the two paranthesis above are coprime. Indeed if there were a prime d to divide both of them by substraction we would have d|2*2x/2, so d=2. But that can't be since the paranthesis are both odd.

So we have to coprime numbers whose product is a square. So both of them must be squares. In particular, (4k+3)x/2-2x/2 is a square. By using the same argument, we have that   (4k+3)x/4-2x/4 is a square, and so on, until finally we obtain (4k+3)m-2m is a square.

But as we already have shown, that cannot be unless m=1(because m is odd). So m=1 and x is a power of 2. Evidently we cannot have x=2, because then (4k+3) ^2 and y ^2 would be two squares whose difference is 4.Impossible.

Now suppose x>2, so we have
(4k+3)x/2-2x/2=a ^2    and
(4k+3)x/2+2x/2=b ^2 .

Let's take this mod 3.
We have kx/2-1=0 or 1   (mod 3) and
              kx/2+1=0 or 1     (mod 3)
That would mean that kx/2 must be 2(mod 3), but that can't be, since it is a square.
This completes the proof.

I think that I have another solution for your ineq
  From Cauchy ineq we have   \sqrt x*x+y*y>=(x+y)*( \sqrt 2)*1/2
  That means that  \sum  \sqrt x*x+y*y>= \sqrt 2*(x+y+z)
  So now we have to prove that 2(x+y+z)^2>=6(xy+yz+xz)
  And after some more calculs we will have to prove that
  (x-y)^2+(y-z)^2+(z-x)^2>=0 which is true.

   Hope it is corect!

For the 3rd we have that:

y_k=x_(k+1)-x_k satisfies : sum y_k=0 and |y_k|<=1. We have to prove now that sum |y_k| <=2[n/2]. Because we have simetry we can presume that: y_1,..,y_j<=0 and y_(j+1),.., y_n >=0 so -(sum_(p=1,j) y_p)=(sum_(p=j+1,n) y_p) which is that  (sum_(p=1,j) |y_p|) =(sum_(p=j+1,n) |y_p|). It's obiuos that j and n-j are smaller or equal than [n/2] so (sum_(p=1,n) y_n)=2(|y_1|+..+|y_j|)<=2[n/2].

that's all!!!
cheers! :D :D :D

Wow these problems are hard. Thx for solutions! I was working on them.

However, It took really long finding this through post all the way to 2003.

For the first question: Assume x,y,z are positive, as them being negative only decreases the RHS.

$$\left(\sum_{ }^{ }\sqrt{x^2+y^2}\right)^2\ge6\sum_{ }^{ }xy$$
After expanding:
$$2\sum_{ }^{ }x^2+2\sum_{ }^{ }\sqrt{\left(x^2+y^2\right)\left(y^2+z^2\right)}\ge2\sum_{ }^{ }x^2+2\sum_{ }^{ }\sqrt{4\left(\frac{x+y}{2}\right)^2\left(\frac{y+z}{2}\right)^2}=3\sum_{ }^{ }x^2+3\sum_{ }^{ }xy\ge6\sum_{ }^{ }xy$$

Which is clearly true


In how many ways can you arrange the letters of the word $CINEMA$ such that the order of the vowels does not change ??

[hide]
I actually multiplied $3!$ into each probability of the arrangement of the letters I, E, and A  and then added up the the whole thing but then that was pretty time consuming....I'm basically wondering if there's an easier way of doing the whole thing..[/hide]

[hide="One solution"]$\binom{6}{3}$ to choose spaces for the vowels and consonants.
Then $3!$ to choose the order of the consonants.
The answer is $\frac{6!}{3!3!}\times 3!=\frac{6!}{3!}=\boxed{120}$.[/hide]

[hide="Another solution"]There are $6$ places to put the C, then $5$ places to put the M, then $4$ places to put the N.  The places of the vowels will then be uniquely determined.  The answer is $6\times 5\times 4=\boxed{120}$.[/hide]

my way:

[hide]
just consider six blank spots, there are 6*5*4 = 120 ways to arrange the non-vowels (I dont know what they are called although I'm in AP English...) and then the the three left over will be for the vowels.  Since the vowels always stay in the same order, we're done. so $120$.
[/hide]

my way:

[hide]The letters $I$, $E$ and $A$ are fixed in this sequence. The other three are free and we have $3! = 6$ ways of arranging. But the three free spots can permut with the other 3 fixed spots, which give us $P_6^{(3,3)} = 20$ ways. So, the answer is $20*6 = 120$ ways.[/hide]

[hide][quote]But the three free spots can permut with the other 3 fixed spots, which give us $P_6^{(3,3)} = 20$ ways[/quote]

Well, what exactly does $P_6^{(3,3)}$ mean ?? [/hide]

[hide]It means permutation of 6 elements with 3 equals (the three free spots) and 3 equals (the three vowels spots).
The 'geral' formula is $P_n^{(\alpha, \beta, \gamma)} = \frac{n!}{\alpha!\beta!\gamma!}$.

Since I don't know how to explain it, I will give an example:
To calculate the number of anagrams of AABBBCCCC, we use the formula I used, and we'll have $P_9^{(2,3,4)} = \frac{9!}{2!3!4!} = 1260$ anagrams

In the exercise in question, I used to count all the possible ways of $\underline{3} I \underline{2} E \underline{1} A$, which could be $\underline{3}\underline{2}\underline{1}IEA$, $IEA\underline{3}\underline{2}\underline{1}$, $I\underline{3}EA\underline{2}\underline{1}$, and so on, notice that is the same case as the example above.

Well, as you saw I'm not good to make myself clear, but I hope you have understood it  :blush: [/hide]

[quote="Zellex"]consonance (sp?)[/quote]

Consonants ;)

I thought of it like this:

[hide]take the 3 consonants first. The vowels we will replace with x's

We are looking for how many ways to arrange xxxCNM

It is $6!/3!$ because the x's are the same.

The vowels are then placed in the x spots in the correct order.

We get $6 * 5 * 4 = \boxed{120}$[/hide]

Number of distinct permutations divided by number of permutations of the vowels.

6!/3!

[b]120[/b]


tui lap cai nay de cac than dan VN ai quan tam thi ghe vo !
             tinh tui chua thay gi  ca ? nhung nghe noi Ha Tinh co 1 giai nhat , 1 giai 2 va 5 giai 3 , 1 trat   .
 tui chi biet the thoi ! khong biet dung khong ?

Dinh chinh :HaTinh 1 giai 1 ; 1 giai 2; 2 giai kk
con thong tin moi de :
 Gia Lai :1 giai 2 ; 4 giai 3 ; 2 giai kk

hmm... :( , please: http://www.mathlinks.ro/Forum/viewtopic.php?t=72927


geometry
circumcircle
perpendicular bisector
geometry proposed

The circles $ S_1$ and $ S_2$ intersect at points $ M$ and $ N$. Let $ A$ and $ D$ be points on $ S_1$ and $ S_2$, respectively, such that lines $ AM$ and $ AN$ intersect $ S_2$ at $ B$ and $ C$, lines $ DM$ and $ DN$ intersect $ S_1$ at $ E$ and $ F$; and $ A,E,F$ lie on one side of line $ MN$, and $ D,B,C$ lie on the other side. Prove that there is a fixed point $ O$ such that for any points $ A$ and $ D$ that satisfy the condition $ AB\equal{}DE$, $ O$ is equidistant from $ A,F,C$ and $ D$.

[quote="nayel"]The circles $ S_1$ and $ S_2$ intersect at points $ M$ and $ N$. Let $ A$ and $ D$ be points on $ S_1$ and $ S_2$, respectively, such that lines $ AM$ and $ AN$ intersect $ S_2$ at $ B$ and $ C$, lines $ DM$ and $ DN$ intersect $ S_1$ at $ E$ and $ F$; and $ A,E,F$ lie on one side of line $ MN$, and $ D,B,C$ lie on the other side. Prove that there is a fixed point $ O$ such that for any points $ A$ and $ D$ that satisfy the condition $ AB \equal{} DE$, $ O$ is equidistant from $ A,F,C$ and $ D$.[/quote]
We have $ \Delta END\equal{}\Delta ANB$ then $ BN\equal{}DN$
We get $ \angle NBD\equal{}\angle NDB$ then $ \angle AMN\equal{}\angle DMN$
$ \Rightarrow \angle AFN\equal{}\angle DCN$ so AFCD is cyclic.
Draw a diameter NL of $ (O_1), NT$ of $ (O_2)$
It's easy to show that L,M,T are collinear. The perpendicular bisector of AC always passes through the midpoint of LT. It's point O and O is a fixed point.


ARML geometry AMC AIME USA(J)MO USAMO probability

Congratulations to North Carolina for winning ARML!!

Also, Montgomery ( :D ) came in 3 points behind for second. I don't know about the details, but we (Montgomery) had a near perfect relay round.

So people are going to be getting Mathematica as prizes :lol:.

Does anyone have more detailed information?

CONGRATULATIONS SAMUEL DITTMER OF INDIANA GOLD!!!
Perfect score! WoW! One of only two to get #8! Just a soph! Nice Job!

IN Gold has three in the top 10.

Texas Gold did well too with 2 of the top 3.

there was at least 2 more students who got #8 @ ARML.

One student from NYC and one student from Lehigh Valley Fire.

The Ohio A team was 4th in the B division, so we get to be in the A division next year! Yay!!!

Just from memory (help me out !)

Individual:
Name--------------------State
1. Samuel Dittmer--------Indiana Gold
2. Jeffrey Chen----------Texas Gold
3. Zachary Abel---------Texas Gold
4. Brian Lawrence--------Montgomery County A
5. Dan Schafer----------Thomas Jefferson A
6. Jennifer Iglesias-------Chicago A
7. Charles Tam----------Indiana Gold
8. Nathan Benjamin------Indiana Gold
9. Shaunak Kishore------(Pennsylvania somewhere)
10T. Jeffrey Wang-------Washington
10T. Jeff Manning-------Southern California A

Team (Division A)
1. North Carolina A
2. Montgomery A
3. Texas Gold
4. Lehigh Valley Fire
5. Chicago A
6. San Francisco Bay Area A
7. Phillips Exeter Red

Team (Division B)
1. Colorado
2. Western Massachusetts
3. Chicago Blue
4. Ohio A
5. Montgomery B

Montgomery B tied for 4th in division B.

6th Division A was SFBA A I think. And yes, I'm from Washington. JAM is from Southern California.

[quote="Ars"]Shaunak Kishore was 9th and Lehigh Valley Fire was tied for third exact same score. Montgomery got second in Division A[/quote]
Texas won the the tiebreaker with a higher Power

[quote="nat mc"]
10T. JAM----------------??
[/quote]

That would be me.

I'm Jeff Manning from SoCal A.

Filling in some of the gaps:

"dschafer" is me, Dan Schafer from TJ A.

Brian Lawrence was on Montgomery County A.

Exeter Red was 7th for Division A Overall.

Oh, and if you want my name: Jeffrey Wang.

[quote="4everwise"][quote="Sly Si"]Wow, only 4?  So we can start comparing it to the two really hard ones on the 1997 individual (8 and 6 right answers on them).

No one got it at Las Vegas.[/quote]I say it was too hot. ;)[/quote]

 :stretcher:

We had to take the test in a room with no air circulation and the air conditioning wasn't on.  I did fine for the first four individual questions and then just sort of died on #5 onward.  My brain melted.  :(

samath (AoPS sn, a member of my team) managed to get the answer to #8 about half a minute after time was called...  :P

Well, on a different note, I am more than 90% sure I'm related to Samuel Dittmer.  I think he is my second cousin once removed, or in other words, my mom's second cousin.  I am still not totally sure, but here is my reasoning.

First off, I know that I am related to an Andrew Dittmer, who made it to the 1993 IMO.  So then I knew their family lived in Indiana, so I guessed that we were related.  To top things off, I asked my grandpa today if their was a Samuel in their family, and he said yes, so there is only the slight possibility that there is another Samuel Dittmer in Indiana that would make this false.

Anyway, for now, I'll say I'm related to the perfect scorer!!!

[quote="Sly Si"]First, it's rather time-consuming, especially if you need time to think about how to set it up.  Second, it's just plain intimidating. [/quote]

Third, most people I know are terrible at combinatorics.
Fourth, Las Vegas was [b]hot[/b].
Fifth, ... :P

Add the last line, "The probability can be expressed as a fraction $\frac{m}{n}$ in lowest terms. What is $m+n?$" Then it would be an AIME problem, and not one of the simpler AIME problems. With an AIME time limit, you'd have a better chance.

Ouch.  Oh well, improvement is always good.

The main reason your scores on the relays went up from last year to this year is that it was easier this year - the easiest of the last four years. In this case, when relays are easier, the scores have a wider spread and they matter more. The fact that relays always been a relative weakness for us mattered this year - our A team got 20 this year, up from 8 each of the last two years, but below what one might expect for a team of our overall strength. (By the way, the scoreboard shows Washington Gold with 22, not 24.)

Here's a table of relay scores, using the same format as in post #67:

[code]Year    overall   top teams
2006     12.3      27.3
2005      4.1      12.3
2004      6.9      18.3
2003      5.2      18.4[/code]

[quote="Teki-Teki"]Going back to the subject of the relay round, here's something more embarrassing.

What was NC A's relay score in 2005?

Answer:
[hide]
Are you sure you want to know?
[hide]
Do you agree not to laugh at us?
[hide]
Ok, fine then, we'll tell you.

(2)

[/hide]
[/hide]
[/hide]

This year we got a 28, though.

One thing about #8 that made it particularly hard was the case with 4 wires.  If you didn't take out the 10 cases where two computers were linked and the three others were linked on a separate 'network', then you got 369/512 as your final answer, which was what at least 3 people on NC A got.  Those 10 cases probably caused many of the low scores.  The AIME limit wouldn't have helped much unless you caught that tricky bit.

Personally, I tried for some insane reason to use transition matrices till I realized you just might need an infinite sum, and then I went to combinatorics but didn't have enough time.[/quote]

NC A's relay got a 6 last year.  :?

My dad and I took 20 minutes to get #8. He usually gets these problems in 5 minutes or less...

[quote="Sly Si"]This year's individual #8 is an interesting case.  If you go by numbers alone, it probably ranks as the hardest ARML question ever given (3 correct answers nationally).

However, IMHO, it's not that difficult a problem.  After working through it, I can see why the writers felt it was an appropriate problem for that spot.  However, I think two things were working against it.  First, it's rather time-consuming, especially if you need time to think about how to set it up.  Second, it's just plain intimidating.  Nonetheless, if you ask me, WAY more people should have answered #8.

Take my opinion with a grain of salt, though; combinatorics is my strong suit, so I could be underestimating the difficulty of the problem.[/quote]

I would like to add that any question where the answer is Case Analysis with 6 questions should never be on a short contest. I would bet that many people saw this and then looked for a better solution that they couldn't find. As it took me 8 minutes to do the calculations alone, I would conjecture that spending even a minute looking for a better solution before giving up would put one in a bad position towards solving the problem. This isn't meant to demean any of the 3 people who solved it, I absolutely say congratulations to them, I merely think that a better question could have been picked.

[quote="xxreddevilzxx"][quote="h_s_potter2002"]Well, Ohio's relay was way worse than anyone else's here.  We got 0 points last year on the relay.  But we upped it to 18 this year.[/quote]

well look at this.

Washington last year 6 this year 24. Same improvement.[/quote]

No, not really.  You see, Ohio improved by an infinty percent, while Washington only improved by only 300%.

YAY FOR NC A

and putting random case-analysis questions on #8 is just annoying

OK SO WE CAN'T DO A BILLION CASES IN 10 MINUTES

Actually, there were a total of 8 cases, 6 of which were trivial.

[quote="futuremontrealian"]it's not fair that missouri has to be A division... we stink for the most part :(.[/quote]
You were 7th overall in 2004. That's a piece of your history that "we stink for the most part" doesn't exactly capture. Then you dropped to tied for 45th overall in 2005, then bounced back to tied for 23rd overall this year. In the same three years, our (Southern California) team has been 17th, 13th, and 14th. That may be more consistent than you, be we don't have a top-10 finish and you do. I think it's quite fair that you're in Division A.

So who did you have who graduated in 2004?

P.S. - as for your signature: I'm a violist.