calculus
integration
college
trigonometry
function
inequalities
calculus computations

Modified from a UC Berkeley MATH H90 question:

Which is larger, $ \int_0^{\pi}e^{\sin^2x}\,dx$ or $ \frac{27\pi}{16}\,?$

(No calculators.)

(The original question had $ \frac{3\pi}2$ instead of $ \frac{27\pi}{16}.$)

I have no time now, but I give rough sketch of the solution.

$ \int_0^{\pi} e^{\sin ^ 2 x}\ dx >\int_0^{\pi} \left(1\plus{}\sin ^ 2 x\plus{}\frac{1}{2}\sin^ 4 x\right) dx$.

On the specific integration, better to use a reduction formula.

For $ n\ge 2,$ $ \int sin^nx\,dx\equal{}\minus{}\frac1n\sin^{n\minus{}1}x\cos x\plus{}\frac{n\minus{}1}{n}\int \sin^{n\minus{}2}x\,dx.$

Hence, $ \int_0^{\pi}\sin^nx\,dx\equal{}\frac{n\minus{}1}n\int_0^{\pi}\sin^{n\minus{}2}x\,dx.$

Starting from the $ n\equal{}0$ case we get

$ \int_0^{\pi}1\,dx\equal{}\pi,\ \int_0^{\pi}\sin^2x\,dx\equal{}\frac{\pi}2,\ \int_0^{\pi}\sin^4x\,dx\equal{}\frac34\cdot\frac{\pi}2\equal{}\frac{3\pi}8.$

There are reduction formulas for arbitrary powers of any of the six basic trigonometric functions. Each can be proved in the same way. Start with an integration by parts, setting $ dv$ to be something whose integral you can find painlessly. Then employ a Pythagorean identity (in this case, we replace a $ \cos^2x$ by $ 1\minus{}\sin^2x.$) Finally, finding the original integral on both sides of the equality, solve for it.

---

The method used by kunny and akech was what I had in mind when I changed the number to $ \frac{27\pi}{16}.$ The original $ \frac{3\pi}2$ would correspond to one fewer term of the series, although Prof. Kahan seems to have had something else in mind.

---

Here's something a couple of people in my class thought of. The function $ x\mapsto e^x$ is convex, so we can try to use Jensen's inequality. By Jensen,

$ \frac1{\pi}\int_0^{\pi}e^{\sin^2x}\,dx\ge
\exp\left(\frac1{\pi}\int_0^{\pi}\sin^2x\,dx\right)\equal{}e^{1/2}.$

Hence, $ \int_0^{\pi}e^{\sin^2x}\,dx\ge\pi e^{1/2}.$

But how big is $ e^{1/2}?$ We still don't have a calculator.

$ e^{1/2}\equal{}1\plus{}\frac12\plus{}\frac18\plus{}$ (some additional positive terms).

$ 1\plus{}\frac12\plus{}\frac18\equal{}\frac{26}{16}.$ Not quite there yet. What about the additional positive terms? Let's use that for $ n\ge3,\ n!\ge 1\cdot 2\cdot 3^{n\minus{}2}.$ Those additional terms are

$ \sum_{n\equal{}3}^{\infty}\frac1{2^nn!}\le
\sum_{n\equal{}3}^{\infty}\frac1{2^n\cdot2\cdot 3^{n\minus{}2}}\equal{}
\frac{\frac1{48}}{1\minus{}\frac16}\equal{}\frac1{40}<\frac1{16}.$

This gives us $ \int_0^{\pi}e^{\sin^2x}\,dx>\pi e^{1/2}<\frac{27\pi}{16}.$

This unfortunate non-chain of inequalities fails to prove anything. But it was an idea.


What is the positive difference between $ 2^5$ and $ 5^2$?

hello, we get $ 2^5\minus{}5^2\equal{}7$, so the searched positive difference is $ 7$.
Sonnhard.


Seven points are marked on the circumference of a circle. How many different chords can be drawn by connecting two of these seven points?

they're just different ways of visualizing it

[quote=Watermelon99]I did 6+5+4+3+2+1 instead of 7 times 6 times one-half.[/quote]

that is what i did too, and I don't understand the explanation it gives me...

[hide =My Solution with Explanation ]There are $7$ points marked on the circumference of a circle. A chord is just the connection of $2$ points on a circles circumference. So for the first point to choose, there are $7$ options, any of them. For the last point there are $6$ options because you can choose any of the $7$ original points except the one you chose in the beginning, so that is $6$ options. So there are $7 \times 6$ options. But this is wrong. Lets say you choose point $A$ for your first point and point $B$ for your second point. Then for another chord you choose first point $B$ and the point $A$. These two chord are the same chord so the answer is $\dfrac{7 \times 6}{2} = 21$
[/hide]

[hide=Solution Using Combinations]This is essentially choosing 2 marbles out of a hat with 7 marbles. This is $\binom{7}{2}=\frac{7!}{2! \times 5!}=\frac{7 \times 6}{2}=7 \times 3=\boxed{21}$.[/hide]

[hide=Solution]
Call the first point $A$. There are $6$ points to draw to to make a chord. For the next point, $B$, there are $5$ points to draw to to make a chord, because there is already a line between $A$ and $B$. This continues until there's no line connections left to draw, thus, the number of chords that can be drawn is $6+5+4+3+2+1=\boxed{21}$.
[/hide]

[hide=My solution][color=#9a00ff]Each of the 7 points can connect to each of the 6 other points. Then we divide by 2 so we do not double count the chords, and get[hide=this][color=#9a00ff][b][size=200]21 :-D [/size][/b][/color][/hide][/color][/hide]


One evening you met the person of your dreams. You ask the person for their phone number and the respond with the following riddle:
"The number formed by the digits in the 4th and 5th positions is a perfect square. So is the number formed by the digits in the 5th and 6th positions, as well as the number formed by the digits in the 6th and 7th positions. The exchange (first three digits) is a perfect cube, and is equal to the product of the last four digits. Call me if you can." What is your dream date's phone number?

*This is my 100th post!*

Congrats!

If all the numbers doesn't to be different, then I think [hide]729-364-9999[/hide] would work, but there'll be probably tons of them...maybe the girl was just kidding

Ah, came in at the same time. I posted the answer above.

no, only 7 digits. Since when is 99 a square number?

P.S. There's only one answer

? I thought it was xxx-yyy-zzzz, since you said the first three was the "exchange"...

Anyways, [hide]36, 64, and 49 was the swuares in my answer, not the last four[/hide]

The answer's [hide]216-1649[/hide]

the first three digits of a telephone number are called the 'exchange' and [hide]it's not 36[/hide]

Again!? :)

Anyways, I fixed it. I think there's only 3 ways to form the last four digits...

Kind of sad I don't even know what an exchange code is when my dad's in telecomunication...yet he works in Japanese, so oh well.

yes, that is correct...[hide]don't feel bad, I didn't know that the first three digits of a phone number were the exchange until yesterday afternoon.[/hide]

[hide]216-1649, basically, just list the perfect squares from 0-81, and just look at the last digit for the first digit, then that last digit for the next first digit etc.[/hide]

Well, technically from [hide]16-81[/hide], but oh well.

Mmm...not sure if that's considered a perfect number, or even legitimate numbers...besides, there's no perfect square ending with 0, unless it's 3-digits.

Well...I guess. it'd be 000-xxxx, but I'm pretty sure there's no extention that is 000.

yeah, i guess, but it never had to be true

:?: I think the problem implies it has to be a legitimate phone number...

[quote="MathFiend"]What is your dream date's phone number?[/quote]
It says "phone number"...

Let's ask our friendly telecomunications expert, MathFiend

...I'll go ask my dad.

*sigh* behave yourself. you're a comod for pete's sake, and if you don't act like a comod you could get...never mind.

Oh, and since 0 is reserved as a "special number," no telephone number can start with 0 or start with 00, 000, etc.

[quote="Tare"]*sigh* behave yourself. you're a comod for pete's sake, and if you don't act like a comod you could get...[/quote]
...what?

*sigh* even I don't want to discuss about it...

who said I was the telecommunications expert? This is just an elongated version of a problem we recieved in school. Plus, Im not sure if any phone number begins with a zero or not. I'll have to ponder that.

[color=green]Phone numbers cannot begin with 0 or 1, since those two digits are reserved to do special things -- 1 to indicate a call out of zone, 0 to connect to the operator.[/color]

There are also a few other things combinations phone numbers cannot begin with, such as 911.

hehehe. it'll be funny if it could... everytime someone tries to call 911-XXXX the police comes.

[quote="MathFiend"]One evening you met the person of your dreams. You ask the person for their phone number and the respond with the following riddle:
"The number formed by the digits in the 4th and 5th positions is a perfect square. So is the number formed by the digits in the 5th and 6th positions, as well as the number formed by the digits in the 6th and 7th positions. The exchange (first three digits) is a perfect cube, and is equal to the product of the last four digits. Call me if you can." What is your dream date's phone number?[/quote]

[hide]So the phone numbr is $abc-def-ghij$.
$10d+e=q^2$
$10e+f=w^2$
$10f+g=p^2$
$100a+10b+c=u^2=g\times h\times i\times j$[/hide]
Have to go do something else now.
Billy

I got $216-1649$... and u cant use zeros probably (just common sense). ANyways if you did use zeros, you would get something like $000-0164$, but in a real Mathcounts competition, play it smart and go for the more realistic answer


ratio
geometry
geometric transformation
homothety
similar triangles
geometry proposed

lol:)) the figure gets worse....and u can't use well that C(EFC)$\perp$C(EKD)...but if u're rambo try

Inversion with circle that center is D, than the figure becomes simpler and easier to solve.

[quote="stergiu"][quote="MathTeX"]Let $O$ be the center of a circle with diameter $\overline{AB}$. A straight line perpendicular to $\overline{AB}$ lies from a point $E$ (with $E$ in $\overline{AB}$) intersecting the circle in a point $D$. Another circle is tangent to $\overline{DE}$ and $\overline{BE}$ in points $K$ and $C$ respectively, and tangent to $\stackrel{ \frown}{AB}$ in a point $F$. Prove that $\measuredangle{\text{ }}EDC = \measuredangle{\text{ }}BDC.$[/quote]

[b]Well , it is a lovely problem ![/b]

[color=red]a)[/color] Points $A, K , F$ are on the same line, because $IK \parallel EC$ and triangles $IKF , OAF$ are similar ($O,I$ are the centers of these circles )Note that the ratio of the sides is $\frac{r}{R}$ where $r,R$ are the radii of the small and the big circle and angles $FIK, FOA$ are equal! ).
[color=red]b) [/color]$EKFB$ is inscribed , so :
  $AD^{2}= AE \cdot AB = AK \cdot AF = AC^{2}$ .
Thus $AD = AC$ . Since $\angle ABD = \angle ADE$  , we have
    $\angle ADC = \angle ACD$ or $\angle ADE+\angle EDC = \angle ACD = \angle ABD+\angle CDB$ . Hence $\angle EDC = \angle CDB$.

[u]Babis[/u]

[color=red] I'd like to see a detailed  solution by inversion ! [/color][/quote]

 [b] I must write some supplementary notes on the above proof :[/b]

Triangles $IKF , OAF$ are similar because :

  [color=blue] a)[/color] $IK \parallel OA$ as perpendicular to $ED$ , hence $\angle KIF = \angle AOF = \phi$
  [color=blue] b)[/color] $\frac{IK}{OA}= \frac{r}{R}= \frac{IF}{OF}$ , since $IK=IF = r , OA = OF = R$.

From the above similar triangles we take $\angle IKF = \angle OAF = \frac{180-\phi}{2}$

Since $IK \parallel OA$ and , $\angle IKF = \angle OAF$ , we conclude that points $A,K,F$ are colinear.

   Sorry , because my first solution was a little ... short.
[u] Babis[/u]

 [color=red]Mmm ! See also this :[/color]

 The homothety with center $F$ which transfers the small circle to the big circle , carries $I$ to $O$ , $IK$ to $OA$ and the tangent $DE$ to the tangent of the initial circle at $A$.Hence , this homothety carries $K$ to $A$ , so $F,K,A$ are collinear !!!

[color=darkblue]This nice and old problem is well-known.

$1\blacktriangleright$ [b][u]A metrical proof.[/u][/b] Suppose w.l.o.g. $E\in (AO)$ and $C\in (EB)$.

Denote $AB=2R$ and $CE=r$. Prove easily that

$\{\begin{array}{c}CB=R-\sqrt{R^{2}-2Rr}\\\ EA=R-r+\sqrt{R^{2}-2Rr}\end{array}\ .$ But $(\frac{DE}{DB})^{2}=\frac{EA}{AB}\ .$

Thus, $\widehat{CDE}\equiv\widehat{CDB}$ $\Longleftrightarrow$ $\frac{DE}{DB}=\frac{CE}{CB}$ $\Longleftrightarrow$

$\frac{R-r+\sqrt{R^{2}-2Rr}}{2R}=(\frac{r}{R-\sqrt{R^{2}-2Rr}})^{2}$ $\Longleftrightarrow$

$\frac{R-r+\sqrt{R^{2}-2Rr}}{2R}=(\frac{R+\sqrt{R^{2}-2Rr}}{2R})^{2}$ $\Longleftrightarrow$

$2R(R-r+\sqrt{R^{2}-2Rr})=(R+\sqrt{R^{2}-2Rr})^{2}$, what is truly.

$2\blacktriangleright$ [b][u]A synthetical proof.[/u][/b] From an well-known propewrty obtain $K\in AF$.

Thus, the quadrilateral $BEKF$ is cyclically. Therefore,

$AC^{2}=AK\cdot AF=AE\cdot AB=AD^{2}$, i.e. $AC=AD$.

From the isosceles triangle $CAD$ obtain :

$\{\begin{array}{c}\widehat{ADC}\equiv\widehat{ACD}\\\ \widehat{ADE}\equiv\widehat{CBD}\end{array}\|\ \ominus\downarrow$ $\Longrightarrow$ $\widehat{CDE}\equiv\widehat{CDB}$.

[b]Remark.[/b] $m(\widehat{AFC})=45^{\circ}$ ![/color]


Nick's employer told him that he would receive a $ 0.5\%$ commission for all sales over $ \$5000$. Nick had $ \$40000$ in sales. How many dollars was Nick's commission?


If I do not understand a problem on the PROMYS application, can i post it on here so someone can explain the meaning of it?

Oy ... I would say probably not.  It's more probably okay to PM someone for an explanation of the problem, as you might ask a friend or teacher to help explain it to you.  However, you should in that case put on your solution that "I received help from so-and-so in understanding this problem, although the work on it is all my own."

Wait for someone else to back me up, of course.

why dont you email somebody at the program, I'm sure that is best as it will eliminate all plagiarism issues.

"Plagiarism issues..."

PROMYS could definitely reduce plagiarism issues if it didnt re-use 6/8 of its problems on applications...

Then again, second-year students like to see first-years squirm. :)


john is twice as old as  his son . in 42 years the ratio of their ages will be 4:3 what is the son's corrent age ?

We write 2 equations. $ J\equal{}2S, 3(J\plus{}42)\equal{}4(S\plus{}42)$. So $ 3S\equal{}4S\plus{}42$ $ 6S\equal{}4S\plus{}42$, so $ S\equal{}\boxed{21}$.


AMC
USA(J)MO
USAMO
MATHCOUNTS
AIME
AMC 10

i think i heard that an eighth grader this year made USAMO last year. I was just wondering, is he/she an AoPSer?

are any other Virginian AoPSers in middle school?

There are at least two people who went to Longfellow Middle School who qualified for USAMO as eighth graders last year. Those are In Young Cho (who I think may have made USAMO as a seventh grader- and who also got a perfect score on one or more of the non math sections in the SAT) and Andrew Kim. They both have accounts here but are not active. I wonder how in the world they do it...

I am not completely sure, but the user Sungmin 's younger brother Seung In, who goes to longfellow may have qualified as a seventh grader last year.

When people were trying out for the longfellow mathcounts team for districts, it was very funny to watch the three out-score everyone...

In Young is also here as one of the perfect scorers on the AMC 10:
http://www.artofproblemsolving.com/Wiki/index.php/AoPS_Community_Awards#Perfect_AMC_10_Scores

In Young Cho got a perfect on both reading and writing on the SAT...

im a middle schooler :D
8th grade
i think i can make usamo this year but i definetely will suck on allother sections of SAT
i just get an A in class and call it wuit ;)

I am also a 7th grader. I go to the same school as peregrine falcon.

I'm an eighth grader. I go to a rival school of peregrine and bobert ;)

Too bad my school isn't doing MathCounts this year. :(

I am a seventh grader. I qualified for the AIME (132 on the 10a and 102 on the 12b) but I'm not sure about USAMO...

I got 800 in the math section of the SAT.

I am a seventh grader going to Longfellow MS

LONGFELLOWIANS ROCK

im a seventh grader in longfellow also and w00t yeah LONGFELLOWIANS ROCK.

P.S. UNC is pwning Michigan right now

i'm in 7th-8th and in virginia and in longfellow's chapter (which makes getting to states very hard and so i'm the person who got 9th at chapter Mathcounts )

I seriously think that I am the only person going to Langston Hughes Middle School. :|

[quote="dolleee"]I go to Longfellow Middle School.[/quote]

Im a 5th grader, and I go to Nysmith School


On a given $ 2008 \times 2008$ chessboard, each unit square is colored in a different color. Every unit square is filled with one of the letters C, G, M, O. The resulting board is called [i]harmonic[/i] if every $ 2 \times 2$ subsquare contains all four different letters. How many harmonic boards are there?

Already posted last week: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221517

the answer is (2 power 2010) in 3 plus (-24)  :lol: !


Alec and Bianca play "Hackenbush". Alternately the two players delete an edge in the pic 331395 at: http://www.mathematik-olympiaden.de/Aufgaben/39/3/39133b.pdf . The player who has to make a "move" deletes an edge in the pic; in the case other edges loose their connection with the floor ("Boden") these edges get deleted as well. Then the next players makes his move. The winner is who deletes the final edge(s). Alec and Biance play "Hackenbush" with the "DEMO-2000-edge-building" displayed in the pic. Alec begins the game. Who of the players can force the victory ?

[i]source: German Mathematical Olympiad (DEMO) 2000, round 3, grade 11-13[/i]

This game is famous and can be solved using sprague grundy numbers wich are also famous in game theory.
 http://www.math.ucla.edu/~tom/Game_Theory/comb.pdf
Here the last problem is hakenbush, and the strategy for a general game is explained, maybe there is a simpler strategy for the game you proposed.


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Can anyone prove that $F_{n}$ divides all $F_{kn}$, $k \in \mathcal{Z}$? (or however that Z is supposed to look)

[hide="Proof 1"] Let $\phi, \varphi$ be the roots of $x^{2}= x+1$.  Then

$F_{n}= \frac{\phi^{n}-\varphi^{n}}{\phi-\varphi}$

So the conclusion follows from polynomial division. [/hide] [hide="Proof 2"] $F_{n+k}\equiv F_{k}\bmod F_{n}$ (very easy to prove.  Can you see why?) [/hide]

Nice inductive proof using following identity:
$F_{n+m}=F_{n-1}F_{m}+F_{n}F_{m+1}$

If $F_{n}|F_{kn}$ for some $k$ then $F_{n}|F_{n(k+1)}$

$F_{nk+n}=F_{nk-1}F_{n}+F_{nk}F_{n+1}\Rightarrow F_{n}|F_{n(k+1)}$

If you want to write Z just write:  \mathbb{Z}

The congruence: $F_{n+k}\equiv F_{k}\pmod{F_{n}}$ is wrong.
Take $n=4 \quad k=1$ and we obtain:
$F_{5}\equiv F_{1}\pmod{F_{4}}$
$5\equiv 1\pmod{3}$

Oops.  Haha.  [hide="I actually meant to write"] $F_{n+k}\equiv F_{n-1}F_{k}\bmod F_{n}$ [/hide]


Let $ x;y;z$ be poaitive real numbers satisfying $ xyz \equal{} 1$. Prove that$$ \frac {x}{y^3 \plus{} 2} \plus{} \frac{y}{z^3 \plus{} 2} \plus{} \frac {z}{x^3 \plus{} 2} \geq 1.$$
[url=http://www.artofproblemsolving.com/community/c6h166299p924788]here[/url]

[quote=arqady][quote=chien than]Let $ x;y;z$ be poaitive real numbers satisfying $ xyz \equal{} 1$
Prove that:
$ \frac {x}{y^4 \plus{} 2} \plus{} \frac{y}{z^4 \plus{} 2} \plus{} \frac {z}{x^4 \plus{} 2} \geq 1$[/quote]
I think the first step should be something like the following:
Let $x=\frac{b}{a}$, $y=\frac{c}{b}$ and $z=\frac{a}{c}$, where $a$, $b$ and $c$ are positives.
Hence, we need to prove that $\sum_{cyc}\frac{b^5}{a(c^4+2b^4)}\geq1$.
By C-S $\sum_{cyc}\frac{b^5}{a(c^4+2b^4)}\geq\frac{(a^3+b^3+c^3)^2}{\sum\limits_{cyc}(a^4bc+2a^5c)}$.
Thus, it remains to prove that $(a^3+b^3+c^3)^2\geq\sum\limits_{cyc}(a^4bc+2a^5c)$, which seems true, but I still have no a proof.[/quot

Thank you Arkady.

[quote=arqady][quote=chien than]Let $ x;y;z$ be poaitive real numbers satisfying $ xyz \equal{} 1$
Prove that:

Thus, it remains to prove that $(a^3+b^3+c^3)^2\geq\sum\limits_{cyc}(a^4bc+2a^5c)$, which seems true.[/quote]

You can try with DAC but it's still long and very calculate.

[quote=arqady][quote=Wangzu]

You can try with DAC[/quote]
What is it? :maybe:[/quote]

It's is Devide and Conquer. But I don't like this method much since it's very canculate

[quote=arqady]
Thus, it remains to prove that $(a^3+b^3+c^3)^2\geq\sum\limits_{cyc}(a^4bc+2a^5c)$, which seems true, but I still have no a proof.[/quote]

This appears to be stronger than Vasc's Inequality: $$(a^3+b^3+c^3)^2\geq\sum\limits_{cyc}(a^4bc+2a^5c)\ge \sum_{cyc}3a^{9/2}c^{3/2}$$

[quote=sqing]Let $ x;y;z$ be poaitive real numbers satisfying $ xyz \equal{} 1$. Prove that$$ \frac {x}{y^3 \plus{} 2} \plus{} \frac{y}{z^3 \plus{} 2} \plus{} \frac {z}{x^3 \plus{} 2} \geq 1.$$[/quote]
Let $x=\frac{b}{a}$, $y=\frac{c}{b}$ and $z=\frac{a}{c}$, where $a$, $b$ and $c$ are positives.
Hence, we need to prove that $\sum_{cyc}\frac{b^4}{a(c^3+2b^3)}\geq1$.
By C-S
$ \sum_{cyc}\frac{b^4}{a(c^3+2b^3)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3b+2ab^3)}$.
Thus, it remains to prove that $ (a^2+b^2+c^2)^2\geq \sum\limits_{cyc}(a^3b+2ab^3)$ ,  that's right .


function
real analysis
calculus
calculus computations

Prove that a uniformaly continous function defined on a set that is both bounded and closed cannot be unbounded.

I need help with this proof...my thoughts are that..if the set is bounded and closed..isnt that kinda make it that the function cannot be unbounded?..i could be wrong..

You're re-stating the problem as your proof.  Recall that a uniformly continuous function is necessarily continuous.  The extreme value theorem says that if $ [a,b]$ is an interval and $ f$ is continuous on that interval, then $ f$ is bounded and actually achieves it's max and min.

Did you need "closed" as a hypothesis in your problem? Would it still be true even if you left that out?

I don't think the problem is meant to be restricted to subsets of $ \mathbb{R}$, and the extreme value theorem only holds when closed and bounded implies compact (say, in $ \mathbb{R}^k$).

Well, if you're going to take that approach, "bounded" isn't useful by itself and "closed" is still not needed. They hypothesis you want is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245279]totally bounded.[/url]

[quote="t0rajir0u"]I don't think the problem is meant to be restricted to subsets of $ \mathbb{R}$, and the extreme value theorem only holds when closed and bounded implies compact (say, in $ \mathbb{R}^{k}$).[/quote]
Obviously, I was just solving it for one case.  Would "complete" probably be a more appropriate term?  Then we'd be getting into compactness.


Prove there is no prime in the series: 1001, 1001001, 1001001001, ...

The $n$th term in the series is
\begin{eqnarray*} &\displaystyle\sum_{i=0}^{n-1}10^{3i}\\ =&\displaystyle\frac{10^{3n}-1}{10^3-1}\\ =&\displaystyle\frac{(10^n-1)(10^{2n}+10^n+1)}{999} \end{eqnarray*}
Clearly the first term is not a prime (it's a multiple of 11).  Suppose that term $n\geq 2$ is a prime p.  This implies that 999p is the product of two factors both bigger than 999, a contradiction.


What is the number of cubic yards of concrete needed to pour a patio 12 feet by 27 feet if the thickness is to be 6 inches?

Basically, we want to find the volume of a solid with dimensions 12' by 27' by 0.5'(notice I coverted everything to feet.) Multiplying this out, we get 162. But wait - we want it in cubic  yards. There are $ 3^3 \equal{} 27$ inches in one cubic yards, so we divide our original result by this number to get $ \boxed{6}$ cubic yards.

Or we can start out putting everything in yards. Its 4x9x(.5/3)=36x1/2x1/3=6.


induction
combinatorics unsolved
combinatorics

[quote="boxedexe"]I got the answer to be $F_{n+2}$.  Is that correct?[/quote]

Yes, you can prove it by induction:

Define $\binom{x}{y}=0$ for $x<y$ or $y<0$
$\sum_{i=0}^{n}{\binom{n-i}{i}}+\sum_{i=0}^{n-1}{\binom{n-i-1}{i}}$
$\sum_{i=0}^{n}{\binom{n-i}{i}}+\sum_{i=0}^{n}{\binom{n-(i-1)-1}{(i-1)}}$
$=\sum_{i=0}^{n}{\left(\binom{n-i}{i}+\binom{n-i}{i-1}\right)}$
$=\sum_{i=0}^{n}{\binom{n-i+1}{i}}$

Just check for $n=0,1$


function
geometry
cyclic quadrilateral
Pythagorean Theorem

One corner of a page of width $ a$ is folded over and just reaches the opposite side. Espress $ L$, the length of the crease, in terms of $ x$ and $ a$ where $ x$.

 [geogebra]fa9c3cd794dff0d180987f8f742241b2c9cc3011[/geogebra]

Sorry $ a$ isn't labeled on the diagram, but it is the width of the paper.

$ x$ is the length of the part of the paper [width] that is folded over.

Are you sure $ a$ is the width and $ x$ is the length? The diagram looks like it's the other way around.

No...

1990 Nats Sprint #30 is a question that would be on a chapter sprint now...

Find the sum of the value(s) of $ x$ that satisfy the given equation: $ 6\sqrt{(x\minus{}1)(x\minus{}2)(x\minus{}3)}\plus{}6\equal{}6$.

[quote="AIME15"]No...

1990 Nats Sprint #30 is a question that would be on a chapter sprint now...

Find the sum of the value(s) of $ x$ that satisfy the given equation: $ 6\sqrt {(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)} \plus{} 6 \equal{} 6$.[/quote]
=( I would have pwned 1990

Just to make it look tougher.

Anyways, does anyone know how to do this problem?  I was playing around with it for about 5 minutes but to no avail.

I would try bashing with Pythagorean Theorem...but that gives a lot of messy variables, etc. and there should be a better way...

[hide="One approach"] There's a cyclic quadrilateral here.  I think you can solve this problem analytically by exploiting the equation of the corresponding circle. [/hide]

[hide="Solution"]
Angle chasing and applying the Pythagorean Theorem once, we have the similar triangle relationship

$ \frac {\sqrt {b^2 \minus{} a^2}}b \equal{} \sqrt {1 \minus{} \frac {a^2}{b^2}} \equal{} \frac {a \minus{} x}x \equal{} \frac ax \minus{} 1\implies1 \minus{} \frac {a^2}{b^2} \equal{} \frac {a^2}{x^2} \minus{} \frac {2a}x \plus{} 1$

$ \implies b^2 \equal{} \frac a{\frac2x \minus{} \frac a{x^2}} \equal{} \frac {ax^2}{2x \minus{} a}$ and $ L \equal{} \sqrt {b^2 \plus{} x^2} \equal{} \sqrt {\frac {ax^2}{2x \minus{} a} \plus{} x^2} \equal{} \boxed{\sqrt {\frac {2x^3}{2x \minus{} a}}}$.
[/hide]

[quote="ernie"]Isn't this #30 on the 1991 Nats sprint?

Oops, I thought I saw it on 1990, I think it's 1991 though...[/quote]

Basically.

[quote="1990-1991 Nationals Sprint Round, Question 30"]
Corner $ A$ of a rectangular piece of paper of width $ 8$ inches is folded over so that it coincides with point $ C$ on the opposite side. If $ BC \equal{} 5$ inches, find the length in inches of fold $ l$.
[/quote]

I have to say, I am thoroughly impressed with what Geogebra can do. Hope this helps. The two circle's centers P,Q are movable.  [geogebra]ccc0515b63ca47234ec6b6431945e089c9d692f2[/geogebra]

 :) And thats my first time touching it. So easy to use. Professors would have KILLED for that 30 years ago hahaha.


Will they run an extermination through the Missouri forum again in 2010???

it's a ritual purge. however i dont think they know yet

as long as no idiot (cough max cough) decides to run around telling everyone MO is trying to outspam bay area, we should be fine

and by they i mean The Man

[quote="Maybach"][quote="Mewto55555"]
as long as no idiot (cough max cough) decides to run around telling everyone MO is trying to outspam bay area, we should be fine

[/quote]

But that's you![/quote]

o rly

lol
shorter threads with math-related topics would help.

[quote="Maybach"]mewto55555 = max schindler, who got 2nd written 3rd cd at nats.[/quote]

o rly?

wow erm maybach, maybe its for the best that you stop responding to mewto. hmm.  :maybe:  :wink:  :idea:

[quote="Maybach"]oops I mean mewto55555=max schindler[/quote]
[img]http://filer.case.edu/nrp6/O_RLY.jpg[/img]

also it would be nice if someonne correctly responded to this post

Was that "haha. haha." sarcastic?  I don't see what's funny.

I found a replacement for our "o rly" owl:

[img]http://www.hjo3.net/orly/gal1/orly_bush.jpg[/img]

[quote="007math"][quote="Maybach"]You're definitely a stalker[/quote]

Yup, Maybach from Munster, Indiana who goes to Wilbur Wright Middle School.

Note: The above information is probably right or wrong.[/quote]

How did you know which school I went to?

[quote]pretty clever[/quote]

[img]http://wiki.fifo.nl/images/1/17/O_RLY-Quite.jpg[/img]

meh...these owls are 10 seconds outdated.

Let's use tiny people!

[img]http://images.icanhascheezburger.com/completestore/2008/9/7/orly128652844336822948.jpg[/img]

meh...o rly just got outdated.

Time for the old-fashioned talk.

[quote]pretty clever[/quote]

Thank you.

[quote="007math"]I Googled Munster, Indiana (as thats what stalkers do) and found out there is only 1 middle school there. (http://en.wikipedia.org/wiki/Munster,_Indiana#Education)

Dang, I realized I am a stalker.[/quote]

Actually there are too other middle schools:  St. Thomas Moore and St. Paul.  But they're both private schools.

I go to Wilbur Wright.

[quote="007math"]BTW; you posted what school you go to right here: [url]http://www.mathlinks.ro/viewtopic.php?p=1725963#1725963[/url][/quote]

That's because someone asked me where Munster was!

at what point do we realize we dont care? i did a few pages ago

hm i sense the purge will come i change my vote

[quote="Maybach"]What do you mean by that?[/quote]

quackadoo [url]http://dictionary.reference.com/[/url]

[quote="kimt9"]So everybody here is from Missouri?[/quote]

[quote="007math"]Yup, Maybach from Munster, Indiana who goes to Wilbur Wright Middle School[/quote]

o_______________________o


Do you mean $a^{.66}$ instead of $a^{66}$?

[quote="calculuslover800"]This problem seems to have some issues with it.

I tried solving this problem in a few different ways and got different answers each time.

Then I realized that there is no unique $a$ that satisfies the equations given.  If we have $a^{.3} = 2$ then $.3\ln{a} = \ln{2} \Rightarrow a = 2^{1/.3} = 10.0794$ but, if $a^{.48}=3 \Rightarrow a = 9.8628$.  However, it's worth pointing out that in the range given the answer is always closest to A, 4.5.

[hide="1 method I used"]
First we need to find $x$ and $y$ such that $30x + 48y = 0 \mod 66$.  After solving, we have $30(1) + 48(9) = 66(7)$.  Now we can see that $a^{.3}*(a^{.48})^9 = (a^{.66})^7$.  It follows that $a^{.66} = 3*18^{1/7}$.  Since $18^{1/7}$ is irrational, greater than one, and less than two, it's apparent that $3 < a^{.66} < 6$, so given the choices it's closest to 4.5, or A.  It turns out the exact value is 4.53363.

However this isn't quite the same as what $e^{\pi*i} = -1$ got because there isn't actually a number that satisfies the requirements.

[/hide][/quote]

This is because jaden chose a number (10) to begin with. Then picked two other numbers (2 and 3) then solved:
$10^x=2$, $10^y=3$ to get
$x$ is about $0.30$, and $y$ is about $0.48$ so it's not exactly accurate.
e^ipi=-1 solved it algebraically. It's a good solution. :D

[quote="0714446459923"]Why bother with seventh powers?

Isn't the simplest solution 2(.48)-.3=.66
which yields 3(1.5) = 4.5[/quote]

Your right, you don't need 7th powers.  I'm just illustrating that there are multiple legitimate ways to the problem that results in different answers because of the roundoff in the original problem statement.


geometry inequalities trigonometry ratio AMC USA(J)MO USAMO

The 2008 COMC is coming up in under a week!!!  :wow:

How do you prepare for this contest?

Good luck to everybody writing it!!!!

**revives forum**  :D

hey

thanks for reviving the forum!

im doing mostly number theory and geometry review.  I am finishing Euclid's Elements, for example.  The contest has the odd question that doesn't really fit into a major category, too, so I did some general odds and ends preparation.  The book "Problem Solving Through Problems" is a good resource.

I hope I can at least make the repechage.  If the questions go my way, I think I have a chance.

Good luck to everyone writing it tomorrow!

LOL It is finally OVER!!! :lol:
Is this year's paper harder than the last year's?

I have no idea what last years paper was like... actually, no. Its on the waterloo website, I can't remember if it was hard or not though. Did you have any luck with # 4 of the long answer? I had no idea.

You can find the previous Sun Life math contests here :)
http://www.cemc.uwaterloo.ca/contests/past_contests.html#open

I think we should wait until Sunday to talk about the contest, because I don't know when ICAE is taking it (although, my best guess would be Saturday...). So, ICAE people, when are you taking it?
 :maybe:

Ye...
Don't talk about it at most until next week...

Still... I failed..:(

Ooh, very good idea. Never even thought about that. Whose ISCE? i think I got teh acrynym wrong... but you know what i'm referring to :)

Yeah, don't talk about it...
 :maybe:  ..I'm writing it on Friday (teacher in charge screwed up school's registration  :mad: )

ICAE is taking it on Saturday, but just to be safe, wait until at least next Monday. I can't take it on Saturday and I doubt I'll be able to take it on Sunday anyways, but whatever.

FAILED!  :rotfl:

Just a little fact I thought everyone should know. Good luck to late test-takers (ICAE)!

[quote="Nawahd Werdna"]ICAE is like this school in Michigan whose students do Canadian math contests... they usually get really good results.  :)[/quote]

Cool, never heard of it.  How come they get to do Canadian contests?

just curious...

so, are we all waiting till monday EST to discuss?

I think that it is now safe to discuss. So, let the discussion begin! Any objections?

Wait.. so, for the COMC,

I recognized that people with 72 are all 48th placed, and there are about 8 to 9 people with that score...

In that case, for the CMO cutoff, do the CMS sort the people according to their birth date?? (i.e. would the person with his/her birthday in december have a priority than the person with his/her birthday in march?)

Thx

I definitely do not think they would divide people who qualify to CMO based on their birth date... That would be really absurd. They probably compared the papers, but I am not sure.

Congratulations to all winners! :D

Do they actually invite exactly 50 students to write the CMO? If so I think that this would be kind of unfair, having to compare the papers again and distinguishing between people with the same scores.

ye..
I have a 72 and I'm definitely within the first 50 (tied with 8 other people)...

OR maybe they decided to invite just 48 people directly :S :huh:

Yeah, I think they invite top 50 or less - 47 people in this case, or else they would be inviting 56 people if they include those who have 72.

last year, people with 72 were placed 50th and ppl with 71 placed 68th...

I believe non of them got invited ...
or, am i wrong?? :huh:

According to the CMO report, "the 67 top COMC scorers were invited outright; all but 7 accepted."  :)

I got 62, does that get any awards? Even a certificate?

62 is definitely enough for a certificate (in 2006 a mark of 62 would make Group 5...)

[quote="Jenkins"]If anyone still checks this, did anyone get an invitation (GR.8-10) to the summer camp yet. or did I just miss out on this?[/quote]
I think they send the invitations to summer camps much later than January (I think at least March, maybe even later).

I was soundly defeated (as far as CMO participation is concerned) but I am my college's champion, with 50/80. Remember, I live in Quebec, so college students can still compete.

Please, don't go out telling that Edouard Montpetit College has a bad record in the COMC!

Jenkins wrote:
If anyone still checks this, did anyone get an invitation (GR.8-10) to the summer camp yet. or did I just miss out on this?

I think they send the invitations to summer camps much later than January (I think at least March, maybe even later).

Damn, its May.

[quote="Jenkins"]Jenkins wrote:
If anyone still checks this, did anyone get an invitation (GR.8-10) to the summer camp yet. or did I just miss out on this?

I think they send the invitations to summer camps much later than January (I think at least March, maybe even later).

(I'll assume you were trying to say "damn" here), its May.[/quote]

I got mine in March, but I only got it in at the end of April, so I might not get accepted. And you have to hand in the forms by April 30th, otherwise they can't consider you.


combinatorial geometry
combinatorics
counting
circles
arrangement
IMO Shortlist
IMO Longlist

What is the maximal number of regions a circle can be divided in by segments joining $n$ points on the boundary of the circle ?

[i]Posted already on the board I think...[/i]

Let $R_n$ be the maximal number of regions for $n$ points. Clearly, this occur when any three chords are not concurrent.
Let's consider the configuration with $n+1$ points, say $A_1, \ldots, A_{n+1}$ in that order around the circle.
For each $i = 1, \ldots, n$ the chord $A_{n+1}A_i$ intersects each of the chords which have endpoints on the opposite sides of it, and none of the others. Since there are $i-1$ points in one side and $n-i$ on the opposite side, there are $(i-1)(n-i)$ intersections. The number of regions which have to be add with this chord is the number of intersected chord plus 1, thus :
$\displaystyle R_{n+1} = R_n + \sum_{i=1}^n [(i-1)(n-i)+1] = R_n + \frac {n^3 - 3n^2 + 8n} 6$.
Using that $R_1 = 1$ it follows that :
$\displaystyle R_{n+1}  = 1 + \sum_{i=1}^n  \frac {i^3 - 3i^2 + 8i} 6$, from which we easily deduce that :
$\displaystyle R_n = 1 + \frac {n(n-1)(n^2-5n+18)} {24}$.

Pierre.

[b]Claim.[/b] The maximal number of regions occurs when no three of the chords are concurrent.
[b]Proof.[/b] Let $l_1$, $l_2$, and $l_3$ be three chords that concur at a single point. Then, shift $l_3$ by an arbitrarily small margin such that $l_1,l_2,l_3$ no longer concur, and the intersections and regions $l_3$ creates with all chords other than $l_1$ and $l_2$ are not affected. In this manner, the number of regions by having these chords not concur increases (by $1$), since through this shifting, a new region determined by $l_1$, $l_2$, and $l_3$ is created. Applying this logic to every set of $3$ or more chords yields the desired result. $\Box$

Now, treat the optimal configuration as a graph. We first compute the number of vertices. Observe that there is a bijection between chord intersection points inside the circle and sets of $4$ points on the circumference, because each such intersection point is the intersection of the diagonals of a quadrilateral with its vertices on the circumference, and vice versa. As a result, there are $\tbinom{n}{4}$ intersection points, and combining this with the $n$ points on the circumference gives $\tbinom{n}{4}+n$ vertices in the graph.

We then compute the number of edges. There are $\tbinom{n}{2}$ chords connecting the $n$ points on the circumference, and each of the $\tbinom{n}{4}$ interior intersection points creates $2$ additional edges, because it splits each of the $2$ edges it lies on into $2$ smaller edges. As a result, there are $2\tbinom{n}{4}+\tbinom{n}{2}$ edges inside the circle, and combining this with the $n$ arcs forming the circumference yields $2\tbinom{n}{4}+\tbinom{n}{2}+n$ edges in the graph. Hence, by Euler's characteristic,
$$F=E-V+2=\left(2\binom{n}{4}+\binom{n}{2}+n\right)-\left(\binom{n}{4}+n\right)+2=\binom{n}{4}+\binom{n}{2}+2,$$
and subtracting the $1$ region outside of the circle gives $\boxed{\tbinom{n}{4}+\tbinom{n}{2}+1}$ regions inside the circle, as requested. $\blacksquare$


logarithms
limit
real analysis
real analysis unsolved

Let $ S_n = \ln \prod_{k=1}^{n} \left( 1 - \frac{2}{k+2}\right)^{\frac{k+2}{n}}$. Then

\begin{eqnarray*}
S_n & = & \frac{1}{n} \sum_{k=1}^{n} (k+2) (\ln k - \ln (k+2))\\
& = & \frac{1}{n} ( 2 \ln (n!) + \ln 4 - (n+1) \ln (n+1) - (n+2)\ln(n+2) )
\end{eqnarray*}

Using $ \ln (n!) \sim \ln \sqrt{2\pi n} + n \ln n - n$, we find $ \lim_{n\to\infty} S_n = -2$. Since the original limit is $ \lim_{n\to\infty} e^{S_n}$, it must be equal to $ e^{-2}$.

[quote="sos440"]Let $ S_n = \ln \prod_{k = 1}^{n} \left( 1 - \frac {2}{k + 2}\right)^{\frac {k + 2}{n}}$. Then

\begin{eqnarray*} S_n & = & \frac {1}{n} \sum_{k = 1}^{n} (k + 2) (\ln k - \ln (k + 2)) \\
& = & \frac {1}{n} ( 2 \ln (n!) + \ln 4 - (n + 1) \ln (n + 1) - (n + 2)\ln(n + 2) ) \end{eqnarray*}

Using $ \ln (n!) \sim \ln \sqrt {2\pi n} + n \ln n - n$, we find $ \lim_{n\to\infty} S_n = - 2$. Since the original limit is $ \lim_{n\to\infty} e^{S_n}$, it must be equal to $ e^{ - 2}$.[/quote]

How did you go from the first line to the second line after S_n (I mean, how did you get (1/n)(2ln(n!) + ln4 - (n+1) ln(n+1) - (n+2) ln(n+2) from the sum in the first line)?


On the table there are two bowls. One bowl contains p and the other one q coins (p,q \in N). The two forum members Bugzpodder and Siuhochung select alternately.  Bugzpodder begins.

One move is defined as:

(1) he takes one coin of one of the bowls away or
(2) takes from both bowls one coin each away or
(3) moves one coin from the one bowl into the other bowl.

The person who takes the last coin will win. Explain which conditions must be satisfied so Bugzpodder can force the victory. The same for Siuhochung !

I think the conditions are these: if both p and q are even the second player can force a win. If one of p and q is odd the first player can force a win. Is this correct? I think some kind of nasty double-variable induction works here. I only have half of the soln though. Could you at least tell me if I'm right? (I mean orl). Bye :D

Bugzpodder versus Siuhochung

why use our names...

how about grobber and orl? :D

Hi Siuhochung,

I thought it would make the questions more interesting and make the mentioned guys reply to it. Most of these questions rely on some invariant. Note that I have several other game questions !!!   :D  Anybody having the desire (not) to be included in the future should post me a message. If I don't get any I will choose the forum members by chance but probably the most active ones. It's easier to remember these guys than scanning the memberlist !!!  :D

Hi grobber,

you are right !!! If you like I could post a solution ?

Well, I basicly have a soln, it's, like I said, induction with 2 variables. It's not tht tricky, but post it if you want to (I'm writing from school and I don't really have tine to post solns.. :(

Bye! :D


This one is not too difficult.
Just notice that:
[tex] (a+b)^4-8(a^3+ab^3) =(a-b)^4 \geq 0 [/tex]
[tex] \Leftrightarrow\ 8(a^3b+ab^3) \leq (a+b)^4  [/tex]
So:
[tex] 4\sum a^3b \leq \frac12\ \sum (a+b)^4 [/tex]
That means:
[tex] \sum (a+b)^4 [/tex]
[tex]=3(\sum a^4+2\sum a^2b^2)+4\sum a^3b [/tex]
[tex] \leq 3+\frac12\ \sum (a+b)^4 [/tex]
[tex] \Leftrightarrow  \sum(a+b)^4 \leq 6  [/tex] :)


Let a,b,c,x,y,z>= 0  Prove that
$ \sum  x  \sqrt{\frac{2a}{a\plus{}b}} \leq xyz(\sum\frac{1}{x^2})$

I think this problem is:
Let a,b,c,x,y,z>0 such that  :

1/If  $ a \geq b \geq c$ then$ x \leq y \leq z$

2/If  $ a \leq b \leq c$ then $ x \geq y \geq z$
Prove that $ %Error. "sumx" is a bad command.
\sqrt {\frac {2a}{a + b}} \leq \sqrt {3(x^2 + y^2 + z^2)}$

By Cauchy_Schwarz and Chebyshev

      $ \sum\sqrt {\frac {2a(a + c)x^2}{(a + b)(b + c)}} \leq \sqrt {4(a + b + c)\frac {\sum a(b + c)x^2}{(a + b)(b + c)(c + a)}} \leq \sqrt {4(a + b + c)\frac {\frac {1}{3}[2(ab + bc + ca)](x^2 + y^2 + z^2)}{(a + b)(b + c)(c + a)}} \leq \sqrt {3(x^2 + y^2 + z^2)}$

Sorry for my bad English(Can you help me?) :rotfl:


inequalities
AMC
USA(J)MO
USAMO
inequalities unsolved

I recall seeing the solution before, but either didn't fully understand or just lost it. Now I cannot reconstruct the answer, I am familiar with all the AM, GM, HM inequality proofs, but have trouble with this one...please help!

Given:
   $ a, b, c \geq 0 ;    a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$, prove

          $ 0 \leq  ab \plus{} bc \plus{} ca \minus{} abc \leq 2$

Perhaps someone can show me the link to a solution and explanation. Thanks!


probability
Gauss
function
combinatorics unsolved
combinatorics

I have a question and I can't work it out. Please help me.

There are N(N>=2) vertices in plane to make a set A.
For any vertex P belongs to A, find the vertices closest to P, (maybe only one vertex, maybe many,  as long as they have the same shortest distance) assume we have X vertices closest to P, we call that X/(N-1) is the probability of being closest to P. For every vertex, we calculate such probability and sum up(known as f(N)). The question is what is the maximum of f(N)?

e.g. when N=3, we can find when the 3 vertices are on the three vertices of a equilateral triangle, the f(3)=1, this is the maximum of f(3).

I have a guess of f(N).
First define g(M)=[m/6](m-3[m/6]-3)+m+2+[(m+1)/6]
[x] is Gauss function
Then find the only g(m) that satisfy g(M)<=N<g(M+1)
then max(f(N))=(6n-2m-10)/(n(n-1)).

Thanks for reading and tolerance of my poor English.

Sorry, I made some mistake.
After count up the X/(N-1), we divide the sum by N, then it will be right.

An example for N=4, we put the 4 vertices to form a square, then for each vertex, the probability is 2/3, so the sum is 8/3, then divided by 4, get 2/3.
This is one condition of f(4), but not the maximum of f(4).

The problem could be nicer formulated, somehow like a more classical problem,(a believe u know it): n gunners stand on a field. At the same time they all shoot the closest enemies. What is the maximum number of bullets wasted?
       Well every gunner got shot at most 6 times (the hexagon reason), so at most $6n$ bullets, but we must subtract those gunners who couldn't receive that many bullets because they are on the "extremes" of the set of points. Well, i believe that your calculations were made by equilateral triangulation and u could calculate the maximum on the extremes by starting with a hexagon and scrolling around. But i think u already got that far since your guess has a really pretty form.


What books and resources do you find useful in learning solving proof-type of questions?

Well, it depends. No book taught me how to "write" a proof - I learned mostly be experience (participation in USAMTS, various other contests.) However, the AoPS series contains a chapter on proofwriting. An excellent resource is the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]"How to Write a Solution"[/url] article.

The books I've found most helpful for learning methods for more difficult problems (those that would be found on a "proof/essay" contest) are of course, AoPS, and Engel's "Problem Solving Strategies." For geometry, I liked "Challenging Problems in Geometry" by Posamentier.


AMC AIME USA(J)MO USAMO MIT college geometry

In terms of reasoning/thinking/natural ability, how do the AMC/AIME/USAMO/IMO compare to the SAT Math?

On a scale of 1-10, if the SAT math is a '5', where would the aforementioned tests lie on the scale? (the scale can go past 10 if need be...)

Thanks, all!
-Arachnotron

>.< The SAT is a joke compared to the american math competitions.
It doesn't even COME CLOSE in difficulty in comparison. And plus, you're trying to compare different thigns. The SAT throws a bunch of routine problems that you can solve with your methods right out of the book. You don't have to think at all. You just plug and chug numbers into the right formula and fill in the right answer.

The AMC tests your problem solving ability. You can't just apply your cookie cutter methods and end. You have to actually think, which is something that no SAT problem has even required.

For example, last year I got a 118.5 on my AMC 10, a 3 on my AIME, and an 800 on my SAT for math.

[quote="AIME15"]If SAT math is 5, then AMC 10 is 8-10, AMC 12 is about 10, AIME is about 9-13, USAMO and IMO about 15-20?

Disregard this if someone else turns up and says I'm totally wrong.[/quote]

You can't really compare these.

Though if I could, I would probably say that AMC 10 is 10, AMC 12 is 15, AIME is 30, and USAMO/IMO are 100.

But then I'm probably wrong?

[quote="Arachnotron"]So the AMC/AIME/USAMO/IMO problems are all "creative thinking" and require math ability whereas the SAT math is, in comparison, straightforward and routine?

Any others care to use the scale to compare them? :)[/quote]

OK, you can't really compare them with a scale.

[quote="soltixnj"]
For example, last year I got a 118.5 on my AMC 10, a 3 on my AIME, and an 800 on my SAT for math.[/quote]

I think if I tried for three weeks, I couldn't come up with a better explanation for the shortcomings of the SAT at the top end than this.  Thank you, soltixnj.

(To be fair, that's not what the SAT is for, which is why MIT asks for AMC scores...  But still...)

In my opinion, there's usually one problem on the math SAT that requires more than about ten seconds of thinking and that's usually the last multiple choice problem. From my experience, it is usually a geometry problem. But AMC is much harder than SAT. I'd say SAT is about 3, AMC 10 about 7, AMC 12 (from the practice tests I've done) 8 and AIME around 12 or 13.

So the AMC is for sure a much better of mathematical ability/creativity than SAT math? Do MIT and other top techie schools place more emphasis on AMC/AIME scores than SAT Math?

Oh, and is AIME much harder than AMC? And is USAMO much harder than AIME?

[quote="Arachnotron"]So the AMC is for sure a much better of mathematical ability/creativity than SAT math?[/quote]
Yes, because as soltixnj already said, SAT math is just applying formulas and "plug and chug". In the AMC, you actually have to think about what to do.

[quote="Arachnotron"]Do MIT and other top techie schools place more emphasis on AMC/AIME scores than SAT Math? [/quote]
I'm not an expert on this, but I would hazard a guess to say yes.

[quote="Arachnotron"]Oh, and is AIME much harder than AMC? And is USAMO much harder than AIME?[/quote]
The first few questions of the AIME aren't that much harder than the AMC12; if you can qualify for the AIME, you could probably automatically get the first 3-4 problems. Near the end, however, the problems are fairly difficult compared to the AMC, and need a lot more thinking.
The USAMO is something else entirely (it is much harder than AIME), because the questions themselves are hard, and you need to rigorously prove your answer is right to get full credit. It's basically the most difficult high school math competition in the US (the difficulty is pretty much equivalent to the IMO).
Example: Last year, I got 118.5 AMC 12 and 11 AIME, but only 3 on the USAMO.

andersonw, 3/42? Wow, that test must be hard!

So basically, the reason AMC/AIME's average scores are so low isn't because people haven't learned the material, but because they're just darned hard? :)

For most people on AoPS, I think SAT math would present no more than a minor irritation. As many people have said already, I don't think they're comparable.

I'd say if SAT math is a 2, AMC 10 is around 8-10, AMC 12 is 10-12, AIME is like 11-20, and USAMO is like 50.

As for the reason of difficulty, it's not so much that AMC/AIME math is so "hard," its just that you need to be clever. It really is a good test of your intellect (or of how much you practiced) SAT math is like the final you take in your math class, or like standardized testing. IMHO, it's tedious and pretty much useless to measure anything.

I would say AMC 8 was about equal w/ SAT math...the problems are all easy, you just need to not make any careless errors.

ACT math is harder though...requires learning precalculus (I think).

I think you can only compare the SAT to the AMC competitions, because SAT is more about speed than thinking while, since in AIME and especially USAMO type competitions you have much more time per problem, its more actually thinking than how fast you can solve the problems

Here's an example of an SAT problem that seems AMC level...tell me if I'm wrong:

[img]http://img366.imageshack.us/img366/466/harderamclx8.jpg[/img]

Don't tell me that this doesn't require at least some reasoning skill?

this problem is still very easy compared to the AMC
Please, posting SAT problems here is getting annoying  :!:

Just look at the AMC problems!
It's quite obvious that AMC cant be compared to the SAT!!!

Oh wow, Arachnotron, bringing back the memories...

I remember that problem on a REAL TIMED SAT I did! Lol...I also remember number 20 on that one there were answer choices I, II, and III but I forgot which combination was right...there was also one about sphere inscribed inside cube on that SAT (on another section in same SAT)

It must have been when I was like in 8th grade when I took this SAT (i remember because this was first/only SAT I took (two PSAT), and I think I guessed 9, which rechecking seems to be correct lol)...

Too bad I got like 740 for math on that SAT lol

and btw it might be AMC 10, but only like a number 10 or so which most people on this forum can do in 20 seconds.

You might see a problem like that on AMC 8/10.

Anyway, a solution:  the quadrilateral's angle's degrees sum to 360, and x+y=80, so the other two sum two 360-80=280.  Then they each are 140 degrees.  So there are 9 sides to the polygon, or B.

Yep! This is January 2006. One of the only SAT's where -1 = 800.

Haha, sorry if I'm annoying you guys with the problems; believe me, I do not want to annoy you guys because I think this place is going to be my internet "home". I want to become a better problem solver and I think I can learn a lot from you guys.

But yeah, what I did for that problem:

360-80 = 280

280/2 = 140

Ext Angle = 180-140 = 40

Sum of ext/angle measure = sides; 360/40 = 9.

Yeah.  For example, the 2008 AIME I had ridiculously easy #1 and 2.  I thought #2 belonged at the end of an AMC 8.

I got that one wrong  :huh:  Then again, I only had a 700 on the Sat Math...it was in sixth grade though.

[quote="soltixnj"][quote="the future"]actually last couple of years AIME number 1s and 2s are like the SAT number 20s.

but thats only because they wanted to give people a chance to atleast solve 1 or 2 problems[/quote]

Last year #1 on the AIME was
Find 100^2 + 99^2 -98^2 - 97^2 + 96^2 + 95^2... -2^2 - 1^2, where + and - signs alternate in pairs.

This is much harder than 20 on the SAT.[/quote]

i was talking about the dinner party, something about boys and girls, the #1 on the AIME I from 2008...im confused where that question came from...was that AIME II?

and #2 on that AIME I was probably an SAT like problem 16...it really was only like an AMC 8 problem

Yes the dinner party one and the extremely easy geometry one seemed like they came from a hard amc 8 or something.

sorry for reviving this topic.

I am a more extreme example of having a high amc score than SAT, last year (7th grade) I was the school(middle+high) AMC 10 winner, and qualified for the AIME (i got 3-4 right). I took the SAT the same year, and got 720 on the math, probably from all of my careless mistakes (i agree, there are no problems on the sat that really make you think.)

Consider this: According to one source, there were 289 perfect (2400/2400) SAT scores last year.  Compare that to the ~30 perfect AMC 10 or 12 perfect scores (and the AMC is a 25-question, multiple choice test!) or the [b]1[/b] perfect AIME score.


I asked that question a while back [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=piecewise&t=15365]here.[/url]

[quote="Monkey"]I'm having trouble figuring out how to LaTex piecewise functions.

Does anyone know?[/quote]

You can see the link given by joml88 above. Another method is by using the amsmath package (also included amssymb,amsfonts if it is not working)

Using the following code:
$\$$f(x)=\begin{cases}
x&\text{if } x\geq 0\\
-x&\text{if } x<0\end{cases}$\$$

to obtain
$f(x)=\begin{cases}
x&\text{if } x\geq 0\\
-x&\text{if } x<0\end{cases}$

Personally, for systems of equations etc, or piecewise functions, I use

\begin{aligned}
\left\{
xy + yz + zx & = 12 \\
xyz & = x + y + z + 2
\right.
\end{aligned}


Let (x_n) be a bounded sequence so that:
x_(n+2)>=x_(n+1)-x_n/n! for any n>=1. Prove that the sequence (xn)n>=1 converges.

I'll just outline it here, so it won't be very rigorous:

There is an M s.t. |x_k| <= M for all k.

First of all, we can see that y_n=x_1/1!+..+x_n/n! is bounded. Let's assume we have 2 subsequences x_i(n) and x_j(n) which have limits l1 < l2, respectively. For sufficiently large n, all terms of the sequence x_j(n) are 'really close' to l2 and all terms of x_i(n) are 'really close' to l1, but for all such sufficiently large n's there are numbers m s.t. i(m)>j(n) => x_i(m) >= x_(i(m)-1)-x_(i(m)-2)/(i(m)-2)! >= .. >= x_j(n) - x_(i(m)-2)/(i(m)-2)! - .. - x_(j(n)-1)/(j(n)-1)! But we can choose the vicinities of l1 and l2 in which we want our terms to be in such a way that all sufficiently large n's become as large as we want, and we make these n's large enough that M/(i(m)-2)!+..+M/(j(n)-1)!+... (>= x_(i(m)-2)/(i(m)-2)! + .. + x_(j(n)-1)/(j(n)-1)!) is so small that, starting with x_j(n) and decreasing it with x_(i(m)-2)/(i(m)-2)! + .. + x_(j(n)-1)/(j(n)-1)! we can't reach the vicinity of l1, where x_i(m) should be.

This means that there aren't 2 subsequences which have different limits, and this means the sequence x_n has a limit, and since it's bounded, it must be convergent.


calculus
integration
limit
logarithms
trigonometry
calculus computations

1) When exists $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$ then also exists $\lim_{n\to\infty}\sqrt[n]{a_{n}}$ and

$\lim_{n\to\infty}\sqrt[n]{a_{n}}= \lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$.

So:

$\lim_{n\to\infty}\frac{n}{n!^{\frac{1}{n}}}= \lim_{n\to\infty}\sqrt[n]{\frac{n^{n}}{n!}}$
$= \lim_{n\to\infty}\frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^{n}}{n!}}= \lim_{n\to\infty}(\frac{n+1}{n})^{n}= e$.

1. Let's consider $\lim_{n \to \infty}\ln \frac{n}{\sqrt[n]{n!}}$

\begin{eqnarray*}\lim_{n \to \infty}\ln \frac{n}{\sqrt[n]{n!}}& = & \lim_{n \to \infty}\frac{1}{n}\ln \frac{n^{n}}{n!}\\ & = & \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\ln \frac{n}{k}\\ & = & \lim_{n \to \infty}-\sum_{k=1}^{n}\frac{1}{n}\ln \frac{k}{n}\\ & = &-\int_{0}^{1}\ln x \, dx\\ & = & \int_{0}^{\infty}t e^{-t}\, dt\\ & = & 1 \end{eqnarray*}

where $x = e^{-t}$. So the original sequence $\left(\frac{n}{\sqrt[n]{n!}}\right)$ converges to $e$.

2. Mathematica yielded non-elemenatary antiderivative...

or by Stirling $n!$~$\sqrt{2\pi{n}}(\frac{n}{e})^{n}.$
we can more: find equivalent $\sum^{N}_{n=1}\frac{n}{n!^{\frac{1}{n}}}$
it will be $Ne+c''+o(1)$

2) $P = \int \frac{\sqrt{x}\sqrt{1-x^{2}}}{(1-x^{2})(1+x^{2})}dx = \int \frac{\sqrt{x}}{(1+x^{2})\sqrt{1-x^{2}}}dx$.

Now change variables according to $x = \tan t, x' = \sec^{2}t$, and we get

$P = \frac{1}{\sqrt{2}}\int \sqrt{\tan 2t}dt$.

Now change variables again according to $\tan 2t = 1/u^{2}, t' =-\frac{u}{1+u^{4}}$, and we get

$P =-\frac{1}{\sqrt{2}}\int \frac{1}{1+u^{4}}du$.

This last primitive is easily computed (I will leave that to you), and we get

$\int \frac{1}{1+u^{4}}du = \frac{1}{4\sqrt{2}}\ln\left(\frac{u^{2}+\sqrt{2}u+1}{u^{2}-\sqrt{2}u+1}\right)+\frac{1}{2\sqrt{2}}\arctan\left(\frac{\sqrt{2}u}{1-u^{2}}\right)+C$.

Now you just have to make the "reverse" substitutions.


inequalities
geometry
incenter
inequalities proposed

[quote="The soul of rock"]Prove that in all triangles $ABC$ we have
$\sum r_{a}^{2}.Sin\frac{A}{2}\le \frac{1}{6}\sum\ \left(r_{a}^{3}\right)\left(\sum\frac{1}{ r_{a}^{2}}\right)$[/quote]

Is $r_{a}$ the distance from incenter of a triangle to vertex $A$?

If yes, then take an equilateral triangle $ABC$ and inequality then will be:
$\frac{3r_{a}^{2}}{2}\le \frac{1}{6}(3r_{a}^{3})\left(\frac{3}{r_{a}^{2}}\right)$

It is equivalent to:
$r_{a}^{2}\le r_{a}$

This inequality is false with $r_{a}> 1$.

I think, given inequality should have a bit different form.

[quote="N.T.TUAN"]I thinks $r_{a}$ is radius of excricle respond to $A$.[/quote]

Thanks. But still, in equilateral triangle $r_{a}=r_{b}=r_{c}$ and inequality is wrong.

Uhm, Yes. I think problem must is

Prove that in all triangles $ABC$ we have
$\sum r_{a}^{}.Sin\frac{A}{2}\le \frac{1}{6}\sum\ \left(r_{a}^{3}\right)\left(\sum\frac{1}{ r_{a}^{2}}\right)$


I found a major problem with the test version:
The fact that you can't [b]GIVE UP[/b] !!

I was very discouraged with one problem but I couldn't give up and I hate guessing so I was very SAD  :(

So, you guys should try to fix that :) [/u]

This forum is specifically for the problems in FTW. Post this in the FTW Question & Support Forum.


geometry algebra polynomial AMC AIME USA(J)MO USAMO

So I take a look at a few past USAMO's and find that there is at least one geometry problem on every single one of them (as opposed to say, number theory or polynomial problems, which only appear about half the time). However, I then take a look at the actual geometry problems and find that they are quite difficult for me. I can handle AIME-level geometry, but do you guys have any suggestions on how I would take the next step to bridge the gap between geometry problems on the AIME and geometry problems on the USAMO?

I agree, USAMO geometry is tough.

A couple resources that might help you would be Geometry Revisited by Coxeter and Greitzer (a number of MOPpers swear by this book) and Kiran Kedlaya's geometry packet the AMC student corner: http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html  The latter also has a load of good problems for practice.

i have read kedlaya's packet, but a lot of the problems seem quite tough for me, even after reading the sections. are there any guides, or problems collections that are only moderately difficult?

Okay, so it isn't an olympiad geometry bible, but honestly I don't know of one.  However, it is an excellent book about geometry, and working through the proofs and exercises is good practice for olympiad problems.  (And some of the stuff in there is quite useful, even if some is just nifty geometry facts.)

For stuff easier than Kiran's notes, all I can think of is the hard geometry from AOPS 2, although some of that is closer to AIME stuff.

Also, if you are already good at AIME geometry, you probably have most of the machinery needed to tackle USAMO stuff.  What matters more is patience and ingenuity (and being able to brute-force it when all else fails.)  Just keep at it!

[quote="Fierytycoon"]Hmm, I have taken a look at "Geometry Revisited" before, but do you really think its THAT useful? It just seems to discuss a bunch of random theorems/constructions in geometry (ie pedal triangles, simson lines, napoleon triangles)...[/quote]

Well, thanks for the suggestions. I guess I'll just go over the book and the lecture notes, and keep on trying to do olympiad problems/reading the solutions and hopefully I will be able to do some after a while.

By the way, what did you mean by "being able to brute-force it when all else fails"?

Fiery, are you referring to my post in the other thread?

you could always take geo olympiad clats

A number of problems can just be done by the "dumb-ass method" -- either by using coordinates or complex numbers to reduce it to mere calculation or by using massive trig to compute all the lengths/angles involved in the problem instead of using the clever olympiad trick.  It's usually long and ugly and not worth it, though.

[quote="Fierytycoon"]
By the way, what did you mean by "being able to brute-force it when all else fails"?[/quote]

[quote]It's usually long and ugly and not worth it, though. [/quote]

Dunno...with the 4.5 hours on the USAMO, I think I'd be willing to go through some disgusting trig for a few side lengths :D

Heh.  I'm terrible at brute force because I know I'll make n careless errors.

[quote]
Dunno...with the 4.5 hours on the USAMO, I think I'd be willing to go through some disgusting trig for a few side lengths :D[/quote]

[quote]Heh. I'm terrible at brute force because I know I'll make n careless errors. [/quote]

Yeah, but since its a proof-type contest, you'd still get most of the credit.

[quote="Fierytycoon"]I can handle AIME-level geometry, but do you guys have any suggestions on how I would take the next step to bridge the gap between geometry problems on the AIME and geometry problems on the USAMO?[/quote]

I'm gonna have to figure out how to find time (maybe I need to write a grant proposal) to translate into English the textbooks used in the key schools in China. They have pages upon pages of excellent geometry problems at all levels, and of all natures, synthetic, analytic, trigonometic, you name it. It hasn't escaped my notice that young people whose parents came from the Chinese-speaking world make up a HUGE, disproportionate part of the United States IMO team most years, and that China is a perennial powerhouse country in IMO. I think having much better geometry preparation provides some of that advantage. That's why I have a shelf full of Chinese-language problem books at home.

[quote="Fierytycoon"]Yeah, but since its a proof-type contest, you'd still get most of the credit.[/quote]

I doubt it. Graders do not appreciate people who use really ugly methods and enjoy giving such people 0s.

tokenadult:

Yes, although I have heard that American students are particularly weak in the field of geometry compared to other countries.

ComplexZeta:

Hmm..but isn't that just..wrong? Ugly solutions are valid solution, after all.

[quote="Fierytycoon"]ComplexZeta:

Hmm..but isn't that just..wrong? Ugly solutions are valid solution, after all.[/quote]

Fiery, you were asking ComplexZeta, not me, that question, but I think what he said is completely plausible. Have you read Hardy's [url=http://www.amazon.com/exec/obidos/tg/detail/-/0521427061/learninfreed]A Mathematician's Apology[/url] yet? His famous quotation from that book is "there is no permanent place in the world for ugly mathematics." Mathematics-for-engineers is about getting answers, even by quick-and-dirty methods, but mathematics-for-mathematicians is all about what mathematicians usually call "elegance." There is no such thing as ugly elegance. The IMO, and hence the USAMO as a screening test for the IMO, is judged by a mathematician's standard of what a good solution is, not by an engineers.

To see more about elegant solutions, check out the Clay Institute's [url=http://www.claymath.org/Education/Olympiad_Scholar/]Olympiad Scholar[/url] prize for a solution more elegant than the "official" solution in each year's USAMO.

Ugly solutions are also a major pain to grade, so it's not surprising that the graders have a bias against them.  After all, they only have a finite amount of time to spend grading.  Besides, they won't be happy at all if with my n careless errors I end up proving something decidedly false.  (I once spent a while using complex numbers on problem only to find I'd shown a decidedly scalene triangle to be equilateral, or something of the sort...)

I bought Geometry Revisited a little while ago and enjoyed reading it-- until my mom realized that we didn't have a christmas present for my math teacher.  So, off went Geometry Revisited.  Sigh.  I ordered another one, though, and it should be here soon.  I can't wait!

[quote="tokenadult"]

Have you read Hardy's [url=http://www.amazon.com/exec/obidos/tg/detail/-/0521427061/learninfreed]A Mathematician's Apology[/url] yet? His famous quotation from that book is "there is no permanent place in the world for ugly mathematics." Mathematics-for-engineers is about getting answers, even by quick-and-dirty methods, but mathematics-for-mathematicians is all about what mathematicians usually call "elegance." There is no such thing as ugly elegance. The IMO, and hence the USAMO as a screening test for the IMO, is judged by a mathematician's standard of what a good solution is, not by an engineers.[/quote]

As much as I respect Hardy, I have to disagree with this statement. There is a permanent place for any mathematics that has a wide variety of uses - be it ugly or not. I personally find nothing inelegant or ugly about coordinate geometry at all. In fact I find it more universally appealing than synthetic methods most of the time because you can see how changes in the original problem affect each step all the way through. Making it more useful for generalizations.

But there are lots of examples of "ugly" mathematics that still persists and are useful to this day. I imagine the problem with very algebraic solutions is that mistakes are harder to pick up than in synthetic geometric solutions. But a clever and patient grader should realize when a method is leading towards a correct solution. I can't imagine that they would have a personal bias against coordinate solutions versus synthetic solutions.

Fiery - you're not the only one wondering what to do about geometry.  It has been easily the most requested subject since we started the school, which is why we are designing the Geometry Olympiad Problem Series which will start next month.  It will cover the tools, how to use them, and clues on when to use them, which are most useful for Olympiad geometry problems.

[quote="ComplexZeta"]What I was saying was that if you give a flawed solution using a really ugly method, you can count on the grader to be really stingy with partial credit, if you get any at all. Even a very slightly flawed solution using an ugly method can bring the score down to a 1 or even a 0. If you give a perfect solution using an ugly method, there's not much the grader can do besides giving you 7 points.[/quote]

I had a slightly flawed solution using coordinates on last years APMO and ended up with a 6.  considering I've labeled a bunch of angles and sides on the diagram without any assertion, getting a 6 on the question would probably be a very lucky thing for me.

AIME and USAMO geometry are not that different... USAMO just make uses of theorems that are a little bit tougher (rarely do you see Ceva on AIME for example, but last year's USAMO require Ceva).  and I agree, Geometry Revisited isnt as good as some ppl claims.  I read it, did half of the exercises, got bored, and tossed it.  Try Challenging problems in geometry for a better taste

Ugliness is subjective, relative, and sometimes contradictory.  There also happens to be a distinction between "ugly" and "messy".

Personally, I find the method of complex numbers in geometry to be extremely cool and beautiful.  That doesn't keep me from finding certain complex numbers proofs to be sort of messy, if not downright ugly, but even an ugly complex numbers proof is nicer than a trig proof of the same complexity.

Another example:  Lagrange Multipliers in the abstract are not an inherently ugly method for constrained optimization problems  -- some problems do require them.  However what is ugly is using them on problems that can be done in half a page of AM-GM.

Just my thoughts.

Okay, and honestly, I never finished Geometry Revisited either, at least I didn't work through all the exercises.  But I definitely did learn from what I did.  Any one serious about olympiad geometry should read that book, if only because you might see something similar to one of the problems in an olympiad.  (This has happened to me a couple times.)

Oh yes, and some problems just are not pretty in the first place -- like the APMO problem bugzpodder mentioned.  Not that it was ugly either, it just wasn't very fun.


What score do you need to make it?
Thanks.

Wow... I've never made Nats, but I think maybe 43+ or something.

Guys please don't say random scores if you have absolutely no clue how people do :\.

If you want to know, last year's cutoff was a 37, but it was 25 sprint, 12 target, so you'd probably need 38 or more.

The year before it was 36, but it was 20 sprint, 16 target, so if you got 36, you'd probably make it.

The year before, I'm not sure, but I know it was 35 or 36.

2005 was known to be pretty scary, but I'm not sure.

Does it really matter? What if this year is easier than any previous year and you need a 44? If you though a 40 would get you in and scored just enough so you were confident of 40, you wouldn't have made it.

EDIT: Yeah I thought 2005 was 30 as well, but I dunno the breakdown and 30 seems quite low compared to other cutoffs...

how many people get in again?wasnt it like 15 or sometihng?

12 people get into Countdown. The top 4 written round winners get a bye, which means they automatically become quarterfinalists at the start of the round.

Um, have you seen national problems?  Getting 37 is no small feat.

Totally. I saw these Nationals once..... The heavens opened up!!! It was literally saying to me: Calculate 1999999^912384791384 + 1. I was like... Holy!

Are you saying it was hard or it was easy?

Dude... Nationals is particularly hard. Just a little.

[quote="james4l"]What??? A 37 was all that was necessary to get in!?

That is missing 9 sprint OR 4 target!

(I would assume people were smarter than that!)[/quote]
I would say the most pressing issue there is time. When I did it, I definitely could've gotten a 44 if I had 20 more minutes. But I didn't, I got like a 25.

i took a national sprint the other day...and if it wasnt just me, but they are very tediously irritating......and not particulary hard but a lot of concentration...........which i just so happen to lack

[quote]
Atleast you don't have carelessness resees.
I always get the harder questions right but the first 12 wrong........Something like that.[/quote]

HAHAHA same....i sometimes get the whole last page rite, and end of missing 2-3 from the first page...

Well the first and second page, I get in total about 90% right. But then when I get to the 3rd and 4th pages, I die....  :surrender:


algebra
polynomial
functional equation
algebra unsolved

Suppose $ f(x)\equal{}a_0x^n\plus{}a_1x^{n\minus{}1}\plus{}\ldots\plus{}a_{n\minus{}1}x\plus{}a_n$ ($ a_0\neq 0$) is a polynomial with real coefficients satisfying $ f(x)f(2x^2) \equal{} f(2x^3 \plus{} x)$ for all $ x \in\mathbb{R}$. Prove that $ f(x)$ has no real roots.

Let $ p(x) \equal{} 2x^3\plus{}x$. Note that if $ f(x) \equal{} 0$ then $ f(p(x)) \equal{} 0$ as well. Now $ p(x) > x$ if $ x>0$, and $ p(x) < x$ if $ x < 0$, so if $ x_0 \neq 0$ is a root of $ f(x)$ we get infinitely many roots $ x_0, p(x_0), p(p(x_0)), \ldots$ which is impossible. So if $ f$ has any real root it must be $ x_0\equal{}0$.

Suppose that $ f(0)\equal{}0$. Then $ f(x) \equal{} x^k g(x)$ for some polynomial $ g(x)$ with $ g(0) \neq 0$ and $ k \geq 1$. Substituting that into the functional equation we get

$ x^k g(x) (2x^2)^k g(2x^2) \equal{} (2x^3\plus{}x)^k g(2x^3\plus{}x)$

$ 2^k x^{3k} g(x)g(2x^2) \equal{} x^k(2x^2\plus{}1)^k g(2x^3\plus{}x)$

$ 2^k x^{2k} g(x) g(2x^2) \equal{} (2x^2\plus{}1)^k g(2x^3\plus{}x)$

but now at $ x\equal{}0$ we once again get $ 0 \equal{} g(0)$, contradicting our assumption that $ g(0) \neq 0$. QED

[quote="matex (L)(L)(L)"]I have one question:

in this step: 2^k x^{3k} g(x)g(2x^2) = x^k(2x^2+1)^k g(2x^3+x) You have divided both sides by x^k and thys x can't be 0
but in the next line You have substituted x=0 which contradicts with the previous one( x not equal to 0) and You get  g(0) = 0

Could You explain it?[/quote]
I should have made that clearer. Both sides of the equation are polynomials which are identical at infinitely many points (in fact, they have the same value at every $ x$ except possibly at $ x \equal{} 0$). Therefore they must also be identical at $ x\equal{}0$. In other words, before we divided by $ x^k$, the coefficient of $ x^k$ must have been the same on both sides, but it is obviously $ 0$ on the left (since we have a factor of $ x^{3k}$) and non-zero on the right (since we assumed that $ g(0) \neq 0$, so $ g(2x^3\plus{}x) \equal{} b_0 \plus{} b_1x \plus{} \ldots$ with $ b_0 \neq 0$).

more over we can find: $ f(x)\equal{}(x^2\plus{}1)^n$

Here is my solution for this problem
[b]Solution[/b]
For proving $f(x)$ has no non - zero real roots, I have the same idea as randomgraph
For proving $f(x) \ne 0$, we just need to prove that $a_n \ne 0$
Suppose $a_n = 0$. Let $k$ be the maximum index number which satisfies $a_k \ne 0$
We have: $p(x) = f(x)f(2x^2) = a^2_0 . 2^n x^{3n} + ... + a^2_k . 2^{n - k} x^{3(n - k)}$ and $q(x) = f(2x^3 + x) = a_0 . 2^n x^{3n} + ... + a^2_k x^{3(n - k)}$
Since: $n - k > 0$ then: $3(n - k) > n - k$
But: $p(x) = q(x)$ so: $a_k = 0$, conflict with $a_k \ne 0$
Hence: $a_n \ne 0$


In pascal, when we declare a pointer, it is immediately set to nil or we need to initialize it?
Thank you.

I'm not sure if this will help, but I know that in C++, it is important to initialize pointers before using them, but not immediately.  It's good to initialize them immediately, though!

Are you having problems or are you simply curious?

In C++, a pointer is treated like any other variable. When created, it contains whatever junk that was in the same memory address before. Thus, initializing the pointer to zero ensures that no errors occur.

Now, I do not know for sure (I have never used pascal), but I would assume that if variables in Pascal contain a junk value when first created, then a pascal pointer would too. This works vice versa.

You can do a simple test to find out some information regarding the initialization of variables (its in pseudocode, kinda). It won't be exactly accurate, because the junk value could be zero:

test #1

integer variable;

if variable = 0, display message "Initialized to zero."
else display message "Not initialized to zero."

integer * pointer;

if pointer = 0, display message "Initialized to zero."
else display message "Not initialized to zero."

EDIT: I must also mention that pascal may not support numerical use of pointers. I.e., you can't have a statement pointer++:


Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$.  Suppose further that
\[ |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|. \]
What is the smallest possible value of $n$?

[hide]We have that $19 + \left|x_{1} + x_{2} + \cdot\cdot\cdot + x_{n}\right| < n$.
Obviously $-1 < x_{i} < 1$. The smallest value of $n$ occurs when for example all the $x_{i}$ are equal to $0$.
Then $n = 20$.[/hide]

[hide="Solution"]Since $|x_{i}|<1$ for $i=1,2,\ldots,n,$ \[19+|x_{1}+x_{2}+\dots+x_{n}|=|x_{1}|+|x_{2}|+\dots+|x_{n}| < n.\] This means that $n\ge 20.$ Naturally, we want to test if the smallest possible value of $n$ works. Note that when $x_{i}$ alternates between $\frac{19}{20}$ and $-\frac{19}{20},$ \[19+\left|\frac{19}{20}-\frac{19}{20}+\cdots+\frac{19}{20}-\frac{19}{20}\right|=10\left|\frac{19}{20}\right|+10\left|-\frac{19}{20}\right|=19.\] Hence, $n=20.$[/hide]

[quote="Andreas"][hide]We have that $19+\left|x_{1}+x_{2}+\cdot\cdot\cdot+x_{n}\right| < n$.
Obviously $-1 < x_{i}< 1$. The smallest value of $n$ occurs when for example all the $x_{i}$ are equal to $0$.
Then $n = 20$.[/hide][/quote]

Err, it doesn't work when all the $x_{i}$ are zero... in case that was confusing to anyone.

Just to make sure on something,

If the 19 was a 20 instead, would the answer be 22?

this post was a long time ago, but no the answer would be 0.  In general if the 19 is an odd number (like it is in the problem) then the answer is n+1 and if the 19 is an even number, then the answer is n. (The number obviously can't be negative lol)

[quote="yankeesrule007"] the answer would be 0.  [/quote]

You mean 20  :wink:

p4: (p3 was locked) If the $x_i$'s could be equal to $1, -1$, then we could just use $1,-1,1 \ldots$ 19 times. But this is not possible, so the answer is $\boxed{20}$.

[hide=Solution]I claim the answer is $\boxed{20}$.

Note that if $n\leq19$, \[\lvert x_1\rvert+\lvert x_2\rvert+\cdots+\lvert x_n\rvert<n\leq19\leq19+\lvert x_1+x_2+\dots+x_n\rvert,\]
so equality cannot be reached.

A suitable construction for $n=20$ is $x_i = \frac{19}{20}\cdot(-1)^i$ (alternating negative and positive $\frac{19}{20}$s).[/hide]


function
real analysis
probability and stats

Let $ (F_j)_j$ be distribution functions which converge in distribution to a [i]continuous[/i] function $ F$.  Show that
\[ \sup_x |F_n(x)\minus{}F(x)| \to 0
\]

Let M so that for x>M F(x) is close to 1 and for x<-M F(x) is close to 0.  Since F is continuous F will be uniformly continuous on [-M, M].  Then for any epsilon we can partition [-M, M] into a finite number of intervals [xi, xi+1] such that F(xi+1) - F(xi) < epsilon.  We can now show that Fn converges uniformly to F on each of these intervals by using the hypothesis of weak convergences and the monotonicity of Fn and F.  Since there are only finitely many of these intervals Fn will converge uniformly to F on [-M, M].  A similar argument will work for |x|>M.


calculus
integration
trigonometry
LaTeX
logarithms
calculus computations

Evaluate the integral of sqrt(tan x) dx...

Perform the change of variables $ \tan x \equal{} u^2$:

$ \int \sqrt {\tan x}\,dx \equal{} 2 \int \frac {u^2}{1 \plus{} u^4}\,du$.

Can you proceed from here?

Alternatively, you can make $ \tan x \equal{} \frac{1}{u^2}$, for $ x \neq 0$, yielding $ \minus{}2 \int \frac{1}{1\plus{}u^4}\,du$.

thanks Mr. calcul...
but i actually had 3 substitutions...and finally arrive at this..

That look ok, except that you seemed to have messed up some signs and arctan coefficient. Also, that sum of arctan can be simplified. And a word of advice: try to learn who to type here in Latex instead of attaching pictures.

thanks again mr. calcu..now i get
$ \boxed{\int \sqrt{\tan x}\mathrm{d}x\equal{}\frac{\sqrt{2}}{2}\tan^{\minus{}1}\left ( \frac{\tan x\minus{}1}{\sqrt{2\tan x}} \right )\plus{}\frac{\sqrt{2}}{4}\ln \left | \frac{1\plus{}\tan x \minus{} \sqrt{2\tan x}}{1\plus{}\tan x \plus{} \sqrt{2\tan x}} \right |\plus{}C}$


algorithm
algebra
polynomial
geometry
3D geometry
function
absolute value

Does anyone here know what actually happens when a real number is raised to an irrational power?

For example, what does $2^e$ actually mean? I recognize that when we enter this expression into our calculators, our calculator approximates $e$ to the nearest rational number $p/q$, and then multiplies $p$ $2$'s and takes the $q$th root. But what does it mean, [i]theoretically[/i] even, to raise a number to an irrational power, even if it can't be done?

Umm... I don't think there's really a better way to explain it. That's just what it is.

[quote="K81o7"]That's something i have always wondered.  Well here's one way to think of it.
Take a series which tends to e.  Let the elements be S1, S2, S3, ...
Take the series 2^S1, 2^S2, ...
Find the limit of this series, whic should be 2^e.[/quote]

Wow, that actually makes sense. Thank you.

just a side note
$e=\sum_{n=0}^{\infty}\frac{1}{n!}$

That side note is wrong (the summation should start at 0).

k8 is exactly correct -- in fact, axiomatically, the irrational numbers are constructed essentially by taking limits of rational numbers, so operations on irrational numbers follow logically as limits of operations on rational numbers.

There are some issues of complication that arise when you talk about complex numbers.  Leaving that aside, though: use this idea to explain why the expression $(-2)^{\sqrt{2}}$ does not exist if we operate only in the real numbers.

thanks for correcting JBL, yes it had to start at zero

Transcendental numbers, i think theyre called.

Thanks for all the replies.  :)

What other irrational numbers which tend to them?

Is there one for every sqrt?

[quote="JesusFreak197"]I think it'll be imaginary.[/quote]

Well yea, but why?

Consider the following question: What would the graph of $y = x^x$ look like for negative values of x?

[quote="banblink182"][quote="JBL"]...the expression $(-2)^{\sqrt{2}}$ does not exist if we operate only in the real numbers.[/quote]

Is this just because there is no sequence which approaches ${\sqrt{2}}$?[/quote]

Of course not - it's easy to find a sequence which approaches $\sqrt{2}$ in rational numbers (consider a newton iteration with a rational seed).  The problem lies in the fact that we can make this sequence (in parity), in lowest terms, of the form

odd/odd

and then

even/odd

and it should be evident from here that such a sequence cannot converge to a limit.

By the way, the usual definition for $a^b$ is $\sup_{r\in\mathbb{Q},r\leq b} a^r$ where $\sup$ just denotes the least upper bound (indeed it is sometimes denoted l.u.b, though this is non-standard)

[quote="blahblahblah"]Okay, if you know Newton's method, it's not hard to show that using a rational seed, we can create a sequence of rationals which goes to $\sqrt{2}$

But anyways, consider a sequence of rationals $\{a_n\}$ which goes to $\sqrt{2}$ which is of the form

odd/odd and then even/odd

Then the sequence (excuse the abuse of notation) $(-2)^{\{a_n\}}$ is alternately even and odd and does not go to zero.  Thus it has no limit.  Thus $(-2)^{\sqrt 2}$ does not exist.[/quote]

I'm sincerely sorry about how I still do not understand your point, even after you have explained it twice, but I must ask again.

You have said that according to Newton's method (which I do not know), it is possible to create a sequence of rationals which goes to $\sqrt{2}$. Then later on you said that due to the fact that the sequence alternates between even and odd, this sequence has no limit. I do not understand either why the sequence MUST alternate between even and odds, or how there [i]is not[/i] a contradiction between what you first stated and your conclusion.  :(

Newton's method constructs a sequence of rationals which tend to $\sqrt{2}$.  This shows that the answer to your first question, "Is this just because there is no sequence which approaches ${\sqrt{2}}$?", is in the negative.  You may look it up on mathworld - it is also known as the bablyonian algorithm for calculating square roots.

Next, we want to show that $(-2)^{\sqrt 2}$ is not well-defined.  To do this, we consider a sequence of rationals with the desired parity properties that I've been alluding to.  I don't know an explicit way to show that such a sequence exists, though it's not hard if we consider the following.  Take our intervals to be successive newton iterations.  For large iterations, these intervals are strictly contained in $[1,2]$.  There exist large primes $p$ such that $n/p,(n+1)/p$ both lie inside any of these intervals for some $n$.  We may choose either $n/p$ or $(n+1)/p$ so that our rational, in parity, is either even/odd or odd/odd and so it alternates.  These are the numbers that form our sequence $\{a_n\}$.  They are bounded by successive newton iterations, so it converges to $\sqrt{2}$.

It is obvious that the sequence $\{(-2)^{a_n}\}$ is alternately even and odd (I showed this earlier) and does not converge to zero, because $a_n$ in absolute value is larger than $2^{\sqrt{2}-\epsilon}$ for large $n$.

Following up on some things which were said;
Transcendental numbers are those which cannot be expressed as solutions to any polynomial equation with rational coefficients.  For instance, we know that sqrt[2] is not transcendental because is solves $x^2=2$.
However, there are sequences which tend to transcendentals.  Take, for instance, the sequence which begins with 4, which is denoted by a(1).  For a(n), if n is even, then a(n)=a(n-1)*(n/n+1), and if n is odd, a(n)=a(n-1)*(n+1)/n.  This sequence tends to pi.

(-2)^sqrt[2] will probably be imaginary, because the sequence of rationals which tends to sqrt[2] probably has 2s in the denominators.  Or it might not...if you find a sequence for sqrt2, then you can find out.
However, (-2)^pi probably wouldn't be irrational, because the sequence doesn't have 2s in the denominator.  However, you could probably find a sequence with 2s in the denominator...

As far as imaginary powers go, you can work off Euler's equality...e^(pi*i)=-1

So, just to address some things:

What JesusFreak and K8 said about it being imaginary:
In the complex numbers, if I take a square root (of anything but 0), I get 2 values.  If I take a cube root, I get 3.  In general, if I take an nth root, I get n values.  This is the same as saying, "In the complex numbers, the expression $z^{\frac{1}{q}}$ is a set of $q$ distinct values when $z \neq 0$."  However, irrational numbers aren't rational -- they are the limits of sequences of rational numbers.  To make such a sequence, the denominator has to get arbitrarily large (that is, approach infinity).  Thus, the number of values we get when raising a number to successive terms of the sequence has to approach infinity.  Indeed, in the complex numbers, the function $f(z) = z^e$ is not a function at all in the usual sense, but in fact takes [i]infinitely many[/i] values in the complex numbers.

If we're operating only in the real numbers, the story is somewhat different.  Blahblahblah made his sequence much too confusing -- far more obviously, just use the decimal expansion.  $\sqrt{2} = 1.41421356...$.  We can express this as a limit of fractions easily: just take the sequence $\{1, \frac{14}{10},  \frac{141}{100}, \ldots \}$.  If we tinker slightly with this sequence -- adding or subtracting 1 to the numerator or denominator of the terms, for example -- we can leave the limit unchanged while putting it into the form $\{ \frac{a_n}{b_n} \}$ where:
1) $a_i$ and $b_i$ are odd for all $i$ OR
2) $a_i$ is even and $b_i$ is odd for all $i$ OR
3) $a_i$ is odd and $b_i$ is even for all $i$.

In case 1, every term of the sequence $\{s_n\} = \{2^{\frac{a_i}{b_i}}\}$ is negative, so the sequence approaches the limit $-\left(2^{\sqrt{2}}\right)$.  In case 2, every term of $\{s_n\}$ is positive, so it approaches $2^{\sqrt{2}}$.  In the third case, every single member of the sequence is undefined.  However, in order for the number $\left(-2\right)^{\sqrt 2}$ to be defined, it must have the same value no matter what sort of sequence we choose.  Thus, we have a contradiction -- $(-2)^{\sqrt 2}$ does not exist as a well-defined real number.

I like my sequence more, because it has certain immediate montonicity properties and it alternates nicely.  I agree that it's a bit complicated, but every time I think of rational sequences converging to $\sqrt{n}$, $n$ not a perfect square, I think of Newton iterations.  And, as they say, when the only tool you have is a hammer, everything starts to look like a nail  :)


modular arithmetic
function
number theory proposed
number theory

Let $ p \geq 3$ be a prime, and let $ p$ points $ A_{0}, \ldots, A_{p-1}$ lie on a circle in that order. Above the point $ A_{1+\cdots+k-1}$ we write the number $ k$ for $ k=1, \ldots, p$ (so $ 1$ is written above $ A_{0}$). How many points have at least one number written above them?

My understanding of this problem is that $ A_{i}=A_{i\pmod{p}}$.  If this is the case, then let $ f : \{1,2,\dots, p\}\to \{0,1,\dots, p-1\}$ be a function such that $ k\mapsto [0+1+\cdots+(k-1)]\pmod{p}$.  Then, if for some distinct $ a,b\in\overline{1,p}$ we have $ f(a)=f(b)$, then

$ f(a)=f(b)\Leftrightarrow \frac{1}{2}(a^{2}-a)\equiv\frac{1}{2}(b^{2}-b)\pmod{p}\Leftrightarrow a^{2}-a\equiv b^{2}-b\pmod{p}$
$ \Leftrightarrow (a-b)(a+b-1)\equiv 0\pmod{p}\Leftrightarrow a+b=p+1$.

Thus, we know that if $ S_{p}=\{f(x)|x\in\overline{1,p}\}=\{f(x)|1\le x\le \frac{p+1}{2}$ and $ x\in\mathbb{N}\}$, then the number of vertices with a number above them is $ |S_{p}|=\frac{p+1}{2}$.


[quote="daermon"]Can anybody explain to me what the principle of inclusion and exclusion is?  thanks  :D[/quote]

Here's an example for three sets: if we want to find $A\cup B\cup C$, then we  add $A$, $B$, and $C$, but we have counted elements that are in two of the sets twice, so we must subtract them to count them only once, which means we must then subtract $A\cap B$, $B\cap C$, and $C\cap A$. However, now we have added elements in all three of $A$, $B$, and $C$ three times and subtracted them three times, so we must add them back to make them counted once, meaning we must add back $A\cap B\cap C$. Thus our total equation is
$A\cup B\cup C$=$A$+$B$+$C$-$A\cap B$-$B\cap C$-$C\cap A$+$A\cap B\cap C$.

Here are some links you can look at:
http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html
http://en.wikipedia.org/wiki/Inclusion-exclusion_principle

Why not try our own AoPSWiki and I highly disrecommend MathWorld as a resouce. It makes even simple things very complex and is not for the beginner problem solver like me and probably way above classroom level.

http://www.artofproblemsolving.com/Wiki/index.php/PIE

If you go to chess64's website, there used to be a paper on PIE.  I will see if I can find it...

Umm why was this moved to Intermediate? PIE is not a really advanced topic at all, in fact, it is found in a lot of problems at the state level of MathCounts (not even Nationals!)

[quote="ShoeFactory"]Indeed, PIE is more like common sense. I used the principle without knowing its official name.[/quote]

Hmm, yes.  However, what if we had a 5-set example?  It takes more than common sense, unless you are very good with mental pictures (perhaps like many AoPSers?).  Take a look at the 5-set example on the AoPS Wiki:

http://www.artofproblemsolving.com/Wiki/index.php/PIE

[quote="quantum leap"]Why is there a 5-set example anyway?  Wouldn't the 2- and 3-set (maybe 4-set as well) examples be sufficient with a thorough explanation?  The thing is, that 5-set example ends up going off the page and doesn't look as good as other articles.[/quote]

I fixed that.


trigonometry
function
Pythagorean Theorem
geometry
algebra unsolved
algebra

Compute the arithmetic mean of the solutions to the equation:

arctan(3x)+arctan(5x)=arctan(7x)+arctan(2x)

I have no clue how to start....someone give me a hint?

The short version is that if $x$ is a solution, $-x$ must also be, so the average of the solutions must be zero.

In order to actually find the (positive) solutions of this equation, we look at it geometrically:
Let $\Gamma$ be a circle, and let $AB$ be a chord of length $72x$. Choose points $E$ and $F$ on the chord so that $AE=16x$ and $BF=45x$. The perpendiculars from $E$ and $F$ meet the circle at $C$ and $D$ respectively (on the same side of chord $AB$). We know that $\angle ACE+\angle ECB=\angle ADF+\angle FDB$. If $CE=8$ and $FD=9$, then $\angle ACE=\arctan 2x,$ $\angle ECB=\arctan 7x,$ $\angle ADF=\arctan 3x,$ and $\angle FDB=\arctan 5x.$

Now, assume we have such a configuration ($CE=8$, $FD=9$). Let $O$ be the center of $\Gamma$ and $y$ be the (signed) distance from $O$ to $AB$. We have $OC^2=OA^2=OD^2$, so by the Pythagorean Theorem
$(20x)^2+(8+y)^2=y^2+(36x)^2=(9x)^2+(9+y)^2$
$20^2x^2+8^2+16y=36^2x^2=9^2x^2+9^2+18y$
$50x^2+8+2y=162x^2$
$144x^2=9x^2+9+2y$
Adding these last two equations to eliminate $y$, we have
$(50+144-162-9)x^2=9-8$
$23x^2=1$

Therefore, the single positive root of this equation is $\frac1{\sqrt{23}}$. We also have a trivial root at zero and another root at $-\frac1{\sqrt{23}}$ by symmetry.

If $\theta=\arctan\frac2{\sqrt{23}}+\arctan\frac7{\sqrt{23}}= \arctan\frac3{\sqrt{23}}+\arctan\frac5{\sqrt{23}}$, $\cos\theta=\frac1{\sqrt{6}}$

[quote="jmerry"]The short version is that if $x$ is a solution, $-x$ must also be, so the average of the solutions must be zero.[/quote]

how do you know that???

EDIT: never mind. I didnt read the "arc". =(