USAMTS
induction
quadratics
geometric series
algebra
quadratic formula

I'm curious about how long people usually take to solve USAMTS problems

How do you quantify this?  Is it only the time you take sitting down and explicitly working on the problems, or do we also have to incorporate all of the other time when we might randomly think about the problem during the day or gain some insight into it?  For the first question, I'd say I spend about 5-8 hours, but I'll spend most of the month thinking about the problems on and off.

I guess just approximating the amount of time you clearly think about it.  Like sometimes I dream about solving USAMTS problems (seriously!!) but I won't count that as "thinking clearly"

Generally I spend 8 hours a day for 3 days solving the USAMTS problems, generally the 3 days before the deadline. Sometimes it's the first time I look at the problems.... But yes, I do sometimes solve USAMTS problems in my sleep. (Completely!)

I generally spend about three to four hours solving about two to three of the problems on the day that they come out.  Then, it takes about a day of brainstorming at school for each of the other problems (total maybe 1-2 hours per problem).

[quote="Erin J. Schram"][quote="calc rulz"]I'll have to say <1 hour due to the large number of easy USAMTS problems, especially those later in the year. However, in general they take me well over an hour to do, and I may think about it over a few days if need be (usually no more than 2-5 days per problem).

I also spend a day doing problem 5 because I procrastinate  :blush:  ;)[/quote]
As a problem evaluator, checking the first draft of the USAMTS problems to see whether any are too easy, too difficult, or too poorly phrased, it takes me on average half an hour per problem, for a total of two and a half hours per problem set. But I am extremely good at solving such problems. Anyone who manages less time than that is incredibly fast.

Problem 5/4/16 was a long one. It took me two hours.

Erin Schram[/quote]

Eh I thought it said on each problem.... :blush:

So I tend to solve the round over days. Sometimes there's an easy problem that takes me 5-20 minutes to solve (ie Round 3 Problem 2 or Round 4 Problems 1-3), but most times a problem takes me hours/days, and sometimes I don't even solve it.

In particular, I thought Round 4 Year 17 was one of the easiest rounds ever -
Problem 1 was trivial, that should be one of the earlier AMC Problems (just know about radii and circles and the solution should be obvious!). It took me 1-2 minutes to solve once I drew the diagram.
Problem 2 was also quite easy, once I found a formula for a centered hexagonal number (just an [i]exercise[/i] in induction) and algebraicly messed around with the sequences, it was easy to finish off the problem. That one took me about 5 minutes to do.
Problem 3 - well, first I misread the problem, I thought he doesn't take all the money out of the pot when a strike is obtained (only after 3 strikes), and solved it using infinite geometric series with quadratic numerators. Then I realized my mistake, so I looked at it again, and this problem was pretty easy, just use the idea of recursive EV used to solve Round 3 Problem 2.
Problem 4 - Meh, not too bad (though not a 5 minute problem like 1-3), I spent a World History class thinking about it and then took about 30 minutes to sovle it afterward. The crux move here was thinking to use the quadratic formula on the quartic, after that I just combined things and used some basic facts about rational numbers to arrive at the result.
Problem 5 - This, although it was nowhere as easy as the others, was relatively easy for a problem 5.

It often doesn't take long to see thepoint of the problem which leads to a solution.  I just take time making sure my argument is airtight, because there are often things I see which get me excited and make me think I have a solution, and I find the flaw and how to fix it when I rethink the solution.

I put down 15 + days.  I basically worked on the problems over the course of the whole round, between figuring them out, writing them up, and checking them over.  That being said, I worked on them only a little each day, whenever I got time.

JB

[quote="calc rulz"][quote="Erin J. Schram"][quote="calc rulz"]Problem 5 - This, although it was nowhere as easy as the others, was relatively easy for a problem 5.[/quote][/quote][/quote]
So problem 5 is usually significantly harder? (I noticed that for a few rounds, but I never knew that generally holds true.)

[quote="draperp"][/quote][quote="calc rulz"][quote="Erin J. Schram"][quote="calc rulz"]Problem 5 - This, although it was nowhere as easy as the others, was relatively easy for a problem 5.[/quote]
So problem 5 is usually significantly harder? (I noticed that for a few rounds, but I never knew that generally holds true.)[/quote][/quote][quote="draperp"][/quote]

Yeah, they tend to be in order of difficulty.

I spent about 3 hours on round 1 and another hour typing up the solutions. I'm pretty sure I'll be spending well over a week on the each of the other rounds though :wink:

[quote="draperp"]Most of the time I spend is retyping and rewording my theorems in an effort to make them sorter, simpler, and airtight.[/quote]

yea, because even though you've "solved" the problem, making the solution as nice as possible helps getting your solution commended...which most of us want :wink:

[quote="draperp"]Most of the time I spend is retyping and rewording my theorems in an effort to make them sorter, simpler, and airtight.[/quote]
That's the most fun part of the process - solving the problems is interesting, but solving and presenting them as elegantly as possible is the real challenge, and the most rewarding part once you have the finished product.

I haven't done USAMTS much but I'll say this much

Problems 1-3 shouldn't take me too long to do.  It'll usually take like 2 hours to solve the more basic problems and I'll focus on the harder aspects of it.  Then about another hour working on the more difficult part of the questions.

Problems 4 and 5 will take me about 45 minutes for each.  However, this is to solve, keep in mind.  I'll spend like 30 minutes writing up 1-3.  Fully justifying 4 and 5 is probably the more difficult thing to do in general, and I'll spend a solid hour or two making sure its fully airtight.

2+1+1.5+1+3.5.  About 8 hours.  Thats not that bad.

I mean this process does take place in a very varied time frame. I'll probably spend like two weeks from start to end.  Basically whenever I find time.

I probably spend about 45 minutes a day for 2 weeks.

While you can feel free to talk about past rounds in this discussion, do not talk about time spent on [b]this round[/b] until after this round has ended.

Hmm...I discover that as time progresses my time spent on USAMTS has been dramatically shorter...

In the beginning I spend bizillion hours and barely finish within the deadline, however, in the 3rd and 4th round last year I solve 4 problems in the first and second day, then spend about 2 weeks chewing on the last one...(usually number 4 or 5).


Find the number of solutions to the pair of equations:
\[ \begin{gathered}
  \sin \left( {\frac{{x \plus{} y}}
{2}} \right) \equal{} 0 \hfill \\
  |x| \plus{} |y| \equal{} 1 \hfill \\
\end{gathered}
\]

[hide]\begin{eqnarray}x + y & = & 2k\pi \\
|x| + |y| & = & 1\end{eqnarray}

Squaring and subtracting the equations, we get

$ xy - |xy| = 2k^2\pi^2 - {1\over 2}\quad(3)$

Hence $ |xy| = - xy$ (because $ |xy| = xy$ wouldn't satisfy the above equation), so $ x$ and $ y$ are of the opposite sign - and obviously neither is zero because the RHS of $ (3)$ is irrational - therefore $ (2)$ can be either $ x - y = 1$ or $ - x + y = 1$. Combined with $ (1)$, that yields

$ (x,y) = \left(k\pi + {1\over 2},k\pi - {1\over 2}\right)$ or $ (x,y) = \left(k\pi - {1\over 2},k\pi + {1\over 2}\right)$

Now with $ - 1\leqslant x\leqslant 1$ and $ - 1\leqslant y\leqslant 1$, we find that the only possible value for $ k$ is $ k = 0$. Therefore there are two solutions: $ \left({1\over 2}, - {1\over 2}\right)$ and $ \left( - {1\over 2},{1\over 2}\right)$[/hide]

\[ \begin{gathered}
  \sin \left( {\frac{{x \plus{} y}}
{2}} \right) \equal{} 0 \Leftrightarrow x \plus{} y \equal{} 2n\pi  \hfill \\
  |2n\pi | \equal{} |x \plus{} y| \leqslant |x| \plus{} |y| \equal{} 1 \Leftrightarrow |n| \leqslant \frac{1}
{{2\pi }} \Leftrightarrow n \equal{} 0 \hfill \\
  x \equal{}  \minus{} y \Leftrightarrow |x| \equal{} \frac{1}
{2} \Leftrightarrow x \equal{}  \pm \frac{1}
{2} \Leftrightarrow y \equal{}  \mp \frac{1}
{2} \hfill \\
\end{gathered}
\]
Hence 2 solutions :\[ \left( {\frac{1}
{2}, \minus{} \frac{1}
{2}} \right);\left( { \minus{} \frac{1}
{2},\frac{1}
{2}} \right)
\]
But the answer given in the book is 4...Are we missing something?


If (1/a)+(1/b)=(7/36) and a+b=21, what is the product of a and b?


calculus
integration
geometry
conics
parabola
calculus computations

Let $ a,\ b$ are positive constants. Determin the value of a positive number $ m$ such that the areas of four parts of the region bounded by two parabolas $ y\equal{}ax^2\minus{}b,\ y\equal{}\minus{}ax^2\plus{}b$ and the line $ y\equal{}mx$ have equal area.

[hide]Let $ A \equal{} (\sqrt {b/a},0)$, $ B \equal{} ( \minus{} \sqrt {b/a},0)$.  Let $ C \equal{} (k,mk)$, $ D \equal{} ( \minus{} k, \minus{} mk)$, where $ k > 0$ is a solution to $ mx \equal{} \minus{} ax^2 \plus{} b$.  Finally, let $ E \equal{} (h,mh)$, $ F \equal{} ( \minus{} h, \minus{} mh)$, where $ h > 0$ is a solution to $ mx \equal{} ax^2 \minus{} b$.  Then the four regions in question are

$ ACD$, $ BDC$, $ ACE$, $ BDF$.

The four regions bounded by the given curves occur in two congruent pairs, with $ |ACD| \equal{} |BDC| \equal{} \frac {1}{2}|ACBD|$ and $ |ACE| \equal{} |BDF|$ regardless of the values of $ a, b, m$.  Thus the four are equal if $ |BDC| \equal{} |ACE|$, or equivalently,

$ |ACD| \plus{} |BDC| \equal{} |ACD| \plus{} |ACE|$

or

$ |ACBD| \equal{} |AECD|$.

But

$ |ACBD| \equal{} 4 \int_{x \equal{} 0}^{\sqrt {b/a}} \minus{} ax^2 \plus{} b \, dx \equal{} 4 \left[ \minus{} \frac {a}{3} x^3 \plus{} bx \right]_{x \equal{} 0}^{\sqrt {b/a}} \equal{} \frac {8}{3}b \sqrt {b/a}$,

$ |AECD| \equal{} \int_{x \equal{} \minus{} k}^h (mx \minus{} ax^2 \plus{} b) \, dx \equal{} \frac {(4 ab \plus{} m^2)^{3/2}}{6a^2}$.

Equating the two and solving for $ m$ yields

$ m \equal{} 2 \sqrt {(2^{2/3} \minus{} 1) ab}$.[/hide]


How about starting a Getting Started marathon. :D
Answer the last question you see and write your own.

problem#1
How many numbers less than 1000 are not divisable by 2,3,or 10?

There's already one [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47851]here[/url]. :)

[hide="for your question though..."]
I'll just assume that by numbers you mean integers $n\geq 1$ There are a total of $999$ numbers for consideration. We can start by finding the number of positive integers that are divisible by either $2, 3$, or $10$ and subtract this from $999$. Every even number, $2\leq 2k\leq 998$ is part of this and this consists of $499$ integers. Every odd multiple of $3$, $(3,9,15...999)$ is also part of this and the total is $333$ integers. This adds up to $832$ integers. For divisibility by $10$, we subract the $99$ integers that we overcounted when we were counting multiples of $2$. This leaves $733$ natural numbers that are divisible by either $2,3 or 10$. Therefore, the answer is $999-733=266$. [/hide]


I heard of this site during IMO 2003 from Valentin... Could I join the contest? Also, I wanted to know more about Valentin's prospect about Mathlinks (eg database...)

[quote]This Account Has Been Suspended
Please contact the billing/support department as soon as possible[/quote]

....


Let $ABC$ be an acute-angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $\angle ACB$ and $\angle ABC$ with the line $DE$ and let $Z$ be the midpoint of $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $\angle A = 60^\circ$.

Let $I$ be the incenter. An angle chase shows that $B,I,X,D$ are concyclic, so, since $ID\perp AB$, it means that $BX\perp CI$. This means that $CXB$ is a right triangle, so $ZX=ZB=ZC$. In the exactly same way we show that $ZY=ZB$, so $XYZ$ is always isosceles.

Since $ZX=ZC$, we have $\angle ZXC=\angle ZCX=\angle ACX$, so $ZX\|AC$. We also have $ZY\|AB$.

From the above we conclude that $XYZ$ is equilateral iff $\angle XZY=60^{\circ}$, which is equivalent to $\angle A=60^{\circ}$.

Using my [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156195]lemma[/url] problem becomes obvious  :D

I don't understand why B,I,X,D are concyclic. Could someone explain to me?

because $\angle DXI =\angle DEC+\angle ECX=90+\frac A2+\frac C2$ so ,$\angle DXI+\angle DBI=180$

ahh.. from my drawing at first I thought X is outside the circle.
Bad drawing!

I see this problem is from Silk Road math olympiad 2004. I wonder that Balkan olympiad has problem from other olympiads !!!

What??? You're saying that the first problem is well known?

OMID HATAMI is true.1st problem was in silk road mathematical competition.(SRMC)
BUT somewhat another form.and i think difficult!!!

first we will show that BXC=90=BYC it is very easy because CXD=ABC/2 and BYE=BCA/2 in the same time XIY=180-BCA/2-ABC/2 so the ponits X,Y,D,E lie on the same line
now Z is the middle of BC so is the center of triangles XBC and BYC so XZ=ZY but XZC AND BZY are isosceles and BXYC is cyclic from here we obtain that ZXY=ZYX=BCA/2+ABC/2
and BAC=180-BCA-ABC now it is easy to observe that for ZXY=60 BAC=60 and viceverse

It follows from the lemma posted here by virgil nicula http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=132338&p=2154521#p2154521

From the lemma we conclude that if $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively then points $M,Y$ and $Z$ are collinear. Also points $N,X$ and $Z$ are collinear. So $\angle MZN= \angle YZX$. The conclusion follows.

Let's recall a related, well-known problem:

or: let us consider the incenter as being the unit circle from the complex plane. every point can be calculated by depending only by the tangency points ( view cnum geometry.pdf from mathlinks)

Let $I$ be the incenter of $\triangle{ABC}.$
$\angle{BYX}=\angle{ADE}-\frac{\angle{B}}{2}=\frac{\angle{C}}{2}$
$\implies B,X,Y,C$ are concyclic.
$\angle{BIX}=\pi-\angle{BIC}=\angle{ADE}$
$\implies B,D,I,X$ are concyclic.
$\implies \angle {BXC}=\frac{\pi}{2}$
So $Z$ is the center of the circle passing through $B,X,Y,Z.$
$\implies ZX=ZY$
So $\triangle{XYZ}$ is equilateral $\iff \angle{XZY}=\frac{\pi}{3}$
$\iff \angle{A}=2\angle{XBY}=\frac{\pi}{3}.$

my solution =
lemma = in triangle $ABC$ let $D,E$ be point of tangency of incircle with $AB,AC$ and let $CI$ intersect $DE$ at $X$ than $\angle CXB = 90$

proof = $\angle BIX= 180-\angle BIC = 90-A/2$
$\angle EDB =\angle XDB =180 - (180-A)/2 = 90+A/2$
thus $BDXI$ is cyclic quad.
thus $\angle BXC = \angle BXI = \angle IDB = 90$
and we are done.

main solution = using above lemma we get $\angle CYB = \angle CXB = 90$
thus $BXYC$ is cyclic.

now $Z$ is centre of circle of $CBCY$ as $ZC=ZB$
thus $YZ=ZX$

now angle chasing we get $\angle CXY+\angle BYX = 60$ also
$\angle ZXC+\angle ZYB = B/2+C/2$
thus $\angle YXZ +\angle ZYX = B+X = 120$
and thus $\angle YZX = 60$ using this with $YZ=ZX$ implies that triangle $XYZ$ is equilateral.

and we are done :D

We begin by the following lemma,

[color=#00f]Lemma: Let $M$ and $N$ be touching points of incircle with sides $AB$ and $AC$ of triangle $ABC$, and $P$ intersection point of line $MN$ and angle bisector of $\angle ABC$, then $\angle BP C = 90$[/color]

[i]Proof[/i]: Let $MP\cap CX=D$, where $X$ is the incenter of $\Delta ABC$. Let $\omega_1$ and $\omega_2$ denote the circles $(MDX)$ and $(DPC)$ respectively.

[b]Claim: $\omega_1$ and $\omega_2$ intersect at $B$ other than $D$[/b]

[i]Proof[/i]: In $\Delta CND$,
\begin{align*}
\angle CND+\angle \frac{C}{2}+\angle CDN=180\\
\implies \angle CDN+90+\angle \frac{A}{2}+\angle \frac{C}{2}=180\\
\implies \angle CDN=\angle \frac{B}{2}
\end{align*}
Therefore $\angle MDX=180- \angle\frac{B}{2}$, which implies $B\in \omega_1$. Since $\angle PBC=B/2=\angle CDP$, it implies $B\in \omega_2$.$\blacksquare$
Therefore we have a spiral similarity with spiral center $B$ such that $MX\mapsto CP$. Hence we have $\angle BPC=\angle BMX=90$.$\blacksquare$

Therefore in our problem, we have $\angle BYC=\angle CXB=90$. Hence we can say that, $BXYC$ is cyclic with diameter $BC$, which implies that $XZ=YZ$.
Assume that $\angle A=60$, then we need to prove that $\angle XYZ=60$. We can see that $\Delta BZY$ is isoceles, then
\begin{align*}
\angle BYZ=\frac{\angle B}{2}
\end{align*}
We also have,
\begin{align*}
\angle XCB=\angle XYB=\frac{\angle C}{2}
\end{align*}
which implies that
\begin{align*}
\angle XYZ=\angle BYZ+\angle XYB=\frac{\angle B+\angle C}{2}=\frac{120}{2}=60^{\circ}
\end{align*}
[asy]
import graph; size(8.cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=4.5,xmax=11.5,ymin=-2.3,ymax=6.486217761047523;
pen zzttqq=rgb(0.6,0.2,0.);
pair A=(6.,6.), B=(5.,1.), C=(11.,1.), D=(5.394973141793125,2.9748657089656243), X=(5.878679755733847,3.1213203024309317), Y=(8.588348429744874,3.9417420470188533), Z=(8.,1.);
filldraw(A--B--C--cycle,invisible,linewidth(1.2)+zzttqq); filldraw(B--X--Y--C--cycle,invisible,linewidth(1.2)+blue); filldraw(X--Y--Z--cycle,invisible,linewidth(1.2));
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dot(A,linewidth(4.pt)+ds); label("$A$",(6.071617213404682,6.134821814771412),NE*lsf); dot(B,linewidth(4.pt)+ds); label("$B$",(4.7034546399703165,0.7273221197689224),NE*lsf); dot(C,linewidth(4.pt)+ds); label("$C$",(11.18593921409981,0.8413356675551195),NE*lsf); dot(D,linewidth(4.pt)+ds); label("$D$",(5.110645882063878,3.0401683748603485),NE*lsf); dot((8.181455294172352,3.818544705827648),linewidth(4.pt)+ds); label("$E$",(7.879546328300093,3.528797865372622),NE*lsf); dot(X,linewidth(4.pt)+ds); label("$X$",(5.811014818464803,3.2844831201164855),NE*lsf); dot(Y,linewidth(4.pt)+ds); label("$Y$",(8.612490564068503,4.164016203038577),NE*lsf); dot(Z,linewidth(4.pt)+ds); label("$Z$",(7.960984576718805,0.6621715210339526),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]


Hi. My final exams are coming up and I kind of forgot how to do a problem. It goes like this:

Sophie can paint a room in 5 hours, Frederick can paint a room in 3 hours, and Alexander can paint a room in 1.5 hours. How long would it take them to paint the room together?

Help is appreciated. Please show full solution on solving the problem. Thank you

[hide]Ok. Sophie's progress in one hour is 1/5 of the room.  Frederick's progress in one hour is 1/3 of the room.  Alexander's progress in one hour is 2/3 of the room.

In any kind of work problem we can set up the equation Rate times Time = Amount done. In this problem we want the amount done to be 1 (everything). Since all three people are working together, we can add their rates up. The number of hours worked can be defined in x, which is what we're solving for here.

We set up the equation $x\displaystyle\left(\frac{1}{5}+\frac{1}{3}+\frac{2}{3}\right)=1$
solving we get $x=\frac{5}{6}$
converting to minutes--[b]50 minutes[/b][/hide]


vector
linear algebra
linear algebra unsolved

1. If $ V$ is a vector space of dimention $ n$, if {$ v_1,v_2,v_3,\cdots,v_n$} is a subset of $ V$ which is linearry independent, prove that {$ v_1,v_2,v_3,\cdots,v_n$} is a basis of $ V$.
2. If $ A$ and $ B$ are subspaces of $ V$, prove that $ dim(A\plus{}B)\plus{}dim(A \cap B)\equal{}dim(V)$
Thanks in advance.

in the second part i think it is AUB instead of A+B :wink:  :wink:

The second one is not true, it should be $ \mathrm{dim}\left(A\plus{}B\right)\plus{} \mathrm{dim}\left(A\cap B\right)\equal{} \mathrm{dim}\left(A\right) \plus{} \mathrm{dim}\left(B\right)$.

OK, thanks for the information.
By the way, I need the help for problem number one. ANyone can help me please ? Thanks in advance.

Let ${ S = \{ v_1,\dots,v_k\ }$ be a linearly independent subset of $ V$, and $ \text{dim}(V) = n$. It follows from the definition of the dimension of a vector space that $ k\leq n$. If $ k = n$, then $ S$ is a basis for $ V$, because:
(1) $ S$ is linearly independent.
(2) $ S$ is a generator of $ V$. Else, choose $ v_{n + 1}\in V\setminus \langle S\rangle\neq \varnothing$. Clearly, $ S\cup v_{n + 1}$ is a $ n + 1$-element, linearly independent subset of $ V$, contradiction.


Good luck everyone in Nats even I can't go there but I want to at least watch it..

Does anyone know the times and stuffs in EST?

they're not doing it live. are they? last year they taped it and showed it on may 29th. i'm pretty sure they'll give the info later.

yeah you can bet it won't be until a lot later.

According to the MathCounts web site, the Nationals will be broadcast on ESPN2 on Thursday, June 3 at noon Eastern time.  See
[url]http://mathcounts.org/New/new.html[/url].


trigonometry
geometry
trig identities
Law of Sines

let $ a\equal{}\frac{\pi}{7}$ find  $ \frac{1}{\sin 2a}\plus{}\frac{1}{\sin 3a}$

For this type of problem, you just need to know how to convert radians to degrees.

degrees=radiansX$ \dfrac{180}{\pi}$

Therefore, the expression simplifies to $ \dfrac{1}{\sin 51.428571428571428571428571428571} \plus{} \dfrac{1}{\sin 77.142857142857142857142857142857}$  You could use a calculator from here I suppose?  Unless there is a more elegant way to solve this.

I'm not sure but I think it can be solved without a calculator  :roll:

But I think also that this type of problems that appear in competitions are useless I mean you can take a calculator and do it easily  :D

You can't find the actual value, but you can find a surprising identity.

[hide]

First, put them under a common denominator.

$ \frac{1}{\sin 2\frac{\pi}{7}}\plus{}\frac{1}{\sin 3\frac{\pi}{7}}  \equal{} \frac{(\sin 3\frac{\pi}{7}) \plus{} (\sin 2\frac{\pi}{7})} {(\sin 2\frac{\pi}{7})(\sin 3\frac{\pi}{7})}$

Then, apply the sum to product identity for two sines, which you can find here:

http://en.wikipedia.org/wiki/Trigonometric_identity#Product-to-sum_and_sum-to-product_identities

This gives $ \frac{2(\sin 5\frac{\pi}{14})(\cos \frac{\pi}{14})} {(\sin 2\frac{\pi}{7})(\sin 3\frac{\pi}{7})}$

Now use the fact the complimentary angle identity to change the cosine to a sine....

$ \frac{2(\sin 5\frac{\pi}{14})(\sin 3\frac{\pi}{7})} {(\sin 2\frac{\pi}{7})(\sin 3\frac{\pi}{7})} \equal{} \frac{2(\sin 5\frac{\pi}{14})} {(\sin 2\frac{\pi}{7})}$

Now you just have to note that $ \sin 2\frac{\pi}{7} \equal{} \sin 5\frac{\pi}{7}$ due to the angles being supplementary, and apply the double angle identity once more.

$ \frac{2(\sin 5\frac{\pi}{14})} {(\sin 2\frac{\pi}{7})} \equal{} \frac{2(\sin 5\frac{\pi}{14})} {(\sin 5\frac{\pi}{7})} \equal{} \frac{2(\sin 5\frac{\pi}{14})} {2(\sin 5\frac{\pi}{14})(\cos 5\frac{\pi}{14})} \equal{} \frac {1} {(\cos 5 \frac{\pi} {14})} \equal{} \frac {1} {sin \frac {\pi} { 7 } }$

The last step was just applying the complimentary angle identity again.

[/hide]

Well, I know a few identities that may be able to be used?

I know that $ \sin 2a \equal{} 2 \sin a \cos a$, and I know that $ \sin (a \plus{} b) \equal{} \sin a \cos b \plus{} \sin b  \cos a$.

(Sorry, I fail at trigonometry.  I only learned the basics, like sin is opposite by hypotenuse, etc etc.)

[quote="AdithyaGanesh"]$ \sin(a\plus{}b)\equal{}\sin a\cos b\plus{}\sin b\plus{}\cos a$[/quote]
Actually, $ \sin(a\plus{}b)\equal{}\sin a\cos b\plus{}\cos a\sin b$.  :)

[quote="gaussintraining"][quote="AdithyaGanesh"]$ \sin(a \plus{} b) \equal{} \sin a\cos b \plus{} \sin b \plus{} \cos a$[/quote]
Actually, $ \sin(a \plus{} b) \equal{} \sin a\cos b \plus{} \cos a\sin b$.  :)[/quote]

Yes, you're right.  That was a typo.

I edited my above post.  Thanks.   :)

Another method is to use the Law of Sines and Ptolemy's Theorem on the following quadrilateral:[asy]int i;
draw(unitcircle);
for(i=0;i<7;++i){
dot(expi(pi/2+2*pi/7*i));
}
draw(expi(pi/2+2*pi/7)--expi(pi/2+2*pi/7*3)--expi(pi/2+2*pi/7*4)--expi(pi/2+2*pi/7*5)--cycle);
draw(expi(pi/2+2*pi/7)--expi(pi/2+2*pi/7*4));
draw(expi(pi/2+2*pi/7*3)--expi(pi/2+2*pi/7*5));[/asy]

hello, see here
http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
Sonnhard.

Whenever you see a seven in the denominator (with trig), try to apply what Dr. Graubner showed.


Putnam trigonometry algebra polynomial ratio induction Vieta

Let $n$ be a positive odd integer and let $\theta$ be a real number such that $\theta/\pi$ is irrational. Set $a_{k}=\tan(\theta+k\pi/n),\ k=1,2\dots,n.$ Prove that
\[\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1}a_{2}\cdots a_{n}}\]
is an integer, and determine its value.

My solution, hidden for those who want to look at the problems before solutions.

[hide="Solution"]We solve this by finding an $n$th degree polynomial with the $a_{i}$ as roots. Let $z=e^{i\theta},iw=i\tan\theta=\frac{z-z^{-1}}{z+z^{-1}}$. Solving for $z$,
$iw=1-\frac{2z^{-1}}{z-z^{-1}}=1-\frac{2}{z^{2}+1}$
$z^{2}+1=\frac{2}{1-iw}$
$z=\sqrt{\frac{1+iw}{1-iw}}$. We can't tell which sign to take in the square root just from the tangent, of course; $\tan\theta=\tan(\theta+\pi)$.

Raise the above expression to the $2n$ power: $z^{2n}=e^{2in\theta}=\frac{(1+iw)^{n}}{(1-iw)^{n}}$
$(1+iw)^{n}=e^{2in\theta}(1-iw)^{n}$
$(1-e^{2in\theta})+(i+ie^{2in\theta})nw+$
$+\cdots+i^{n-1}(1-(-1)^{n-1}ne^{2in\theta})w^{n-1}+i^{n}(1-(-1)^{n}e^{2in\theta})w^{n}=0$
Using the fact that $n$ is odd and dividing by the nonzero leading coefficient $i^{n}(1+e^{2in\theta})$, this becomes
$\left(i^{-n}\frac{1-e^{2in\theta}}{1+e^{2in\theta}}+i^{-n+1}nw+\cdots+i^{-1}\frac{1-e^{2in\theta}}{1+e^{2in\theta}}nw^{n-1}+w^{n}\right)=p(w)=0$
This $p$ is an $n$-th degree polynomial, so it has at most $n$ roots. By our construction, $\tan\theta$ is one of these roots. Since $e^{2in(\theta+k\pi/n)}=e^{2in\theta}$ for each $k$, $a_{k}$ is also a root for each $k$. It is easily verified that the $a_{k}$ are distinct real numbers, and thus $p(w)=w^{n}+c_{n-1}w^{n-1}+\cdots+c_{1}w+c_{0}= (w-a_{1})(w-a_{2})\cdots (w-a_{n})$.

Now we apply Viete's relations: $-i\frac{1-e^{2in\theta}}{1+e^{2in\theta}}n=c_{n-1}=-(a_{1}+a_{2}+\cdots+a_{n})$ and $(-i)^{n}\frac{1-e^{2in\theta}}{1+e^{2in\theta}}=c_{0}= (-1)^{n}a_{1}a_{2}\cdots a_{n}$. The desired ratio $\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1}a_{2}\cdots a_{n}}$ is therefore

$\frac{in\frac{1-e^{2in\theta}}{1+e^{2in\theta}}}{(-1)^{n}(-i)^{n}\frac{1-e^{2in\theta}}{1+e^{2in\theta}}}=i^{1-n}n=(-1)^{\frac{n-1}2}n$.[/hide]

Instead he set up a polynomial with roots $1/a_{i}$.

That works, and it's almost identical to my solution. Reciprocals don't change it very much.

A similar idea: define $f(t)=\prod_{k=1}^{n}\cos \left(t+\frac{k\pi}{n}\right)$ and $g(t)=\prod_{k=1}^{n}\sin \left(t+\frac{k\pi}{n}\right)$ and observe that the ratio in the problem equals $-f'(\theta)/g(\theta)$. By straightforward calculations, $f(t)=(-1)^{\frac{n-1}{2}}\frac{\cos nt}{2^{n-1}}$ and $g(t)=i^{n-1}(-1)^{\frac{n-1}{2}}\frac{\sin nt}{2^{n-1}}$ and from this point it is obvious.

Let $\alpha_k=\theta+\frac{k\pi}n$, then it's clear that $a_k=\tan(\alpha_k)=i\frac{1-e^{2i\alpha_k}}{1+e^{2i\alpha_k}}$.

Consider \[f(x)=\prod_{k=1}^n(x-e^{2i\alpha_k})=e^{2ni\theta}\prod_{k=1}^n\Bigg(\frac{x}{e^{2i\theta}}-e^{2ik\pi/n}\Bigg)=x^n-e^{2ni\theta}.\]

Now, \[\frac{f'(x)}{f(x)}=\frac{nx^{n-1}}{x^n-e^{2ni\theta}}=\sum_{k=1}^n\frac{1}{x-e^{2i\alpha_k}}.\]
Plugging in $x=-1$ we get (since $n$ is odd) \[\frac{n}{1+e^{2ni\theta}}=\sum_{k=1}^n\frac{1}{1+e^{2i\alpha_k}}.\]

Finally, \[\sum_{k=1}^n a_k=i\sum_{k=1}^n\Bigg(\frac{2}{1+e^{2i\alpha_k}}-1\Bigg)=\frac{ni(1-e^{2ni\theta})}{1+e^{2ni\theta}},\] \[\prod_{k=1}^na_k=i^n\frac{f(1)}{(-1)^nf(-1)}=\frac{-1}{i^n}\cdot \frac{1-e^{2ni\theta}}{1+e^{2ni\theta}},\] \[\frac{\sum_{k=1}^n a_k}{\prod_{k=1}^na_k}=(-1)^{\frac{n-1}{2}}\cdot n.\]

[hide="Solution"]
Lemma:
\[\tan(n\theta)=\frac{\binom{n}{1}\tan\theta-\binom{n}{3}\tan^3\theta+\binom{n}{5}\tan^5\theta-\dots}{\binom n0-\binom n2\tan^2\theta+\binom n4\tan^4\theta-\dots}\]
Proof: Straightforward induction on $n$.

Let $\tan(n\theta)=c$. If we let $\phi_k=\theta+k\pi/n$, then $\tan(n\phi_k)=c$ for all $k$. Thus, by the above Lemma, $a_k=\tan(\phi_k)$, $k=1,\dots,n$, are solutions of
\[\frac{\binom n1x-\binom n3 x^3+\binom n5 x^5-\dots}{\binom n0-\binom n2 x^2+\binom n4 x^4-\dots}=c\]
This rearranges to
\[\left[\binom n1x-\binom n3 x^3+\dots+(-1)^{\frac{n-1}2}x^n\right]-c\left[\binom n0-\binom n2 x^2+\dots+(-1)^{\frac{n-1}2}\binom n{n-1}x^{n-1}\right]=0\]
. Since this is an $n$-degree polynomial, its roots are precisely $a_1,\dots,a_n$. By Vieta's Formulas, this has sum of roots $cn$ and product of roots $(-1)^{\frac{n-1}2}c$. Therefore $\frac{a_1+\dots+a_n}{a_1\cdots a_n}=(-1)^{\frac{n-1}2}n$.
[/hide]

Slightly different proof,
\[a_k=\tan(\theta_k)=i\frac{1-e^{2i\theta_k}}{1+e^{2i\theta_k}}\]
Letting $x_k=e^{2i\theta_k}$ we see $x_k^n=e^{2ni\theta}=c$ for all $1\leq k\leq n$. And so, all of $x_k$ are just root of the equation $f(x)=x^n-c$ and all what we want is,
\[\frac{i\sum\frac{1-x_k}{1+x_k}}{i^n\prod  \frac{1-x_k}{1+x_k}}\]
And one way to get this considering $f(\frac{1-x}{1+x})$ wee need to look at the equation $(1-x)^n-c(1+x)^n=0$ and hence,
\[\frac{i\sum\frac{1-x_k}{1+x_k}}{i^n\prod  \frac{1-x_k}{1+x_k}}=i^{1-n}n=(-1)^{\frac{n-1}{2}}n\]

Set $\theta_k := \theta + k\pi/n$ and $z_k := e^{i\theta_k}$ for each $1\leq k\leq n$ for notational simplicity.  Denote by $P(z)$ the polynomial with roots $z_1,\ldots, z_n$.  Note that
\[
a_k =\tan\theta_k = \frac{\sin\theta_k}{\cos\theta_k} = \frac{(z_k - \overline{z_k})/2i}{(z_k + \overline{z_k})/2} = \frac{-i(z_k^2 - 1)}{z_k^2 + 1}\\
= -i - \frac{1}{i-z_k} + \frac{1}{-i-z_k}.
\]
Thus
\[\sum_{k=1}^na_k = -ni - \sum_{k=1}^n\frac{1}{i-z_k} + \sum_{k=1}^n\frac{1}{-i-z_k} = -ni - \frac{P'(i)}{P(i)} + \frac{P'(-i)}{P(-i)}\]
and
\[
\prod_{k=1}^na_k = (-i)^n\prod_{k=1}^n\frac{(1-z_k)(-1-z_k)}{(i-z_k)(-i-z_k)} = (-i)^n\frac{P(1)P(-1)}{P(i)P(-i)}.
\]
Dividing the second ratio by the first ratio implies that
\[
\frac{a_1 + \cdots + a_n}{a_1\cdots a_n} = \frac{-niP(i)P(-i) - P'(i)P(-i) + P'(-i)P(i)}{(-i)^nP(1)P(-1)}.
\]
It suffices to show that this is an integer.

The key observation now is that although $P(z)$ itself is a bit difficult to find, $P(z)P(-z)$ is not.  In particular, the roots of $P(z)P(-z)$ are precisely [i]all[/i] of the $2n$ roots of unity modulated by some $\theta_0\in\mathbb C$, so \[P(z)P(-z) = (\theta_0z)^{2n} - 1\qquad\text{for all }z\in\mathbb C.\] Now in particular $P(1)P(-1) = \theta_0^{2n} - 1$ and $P(i)P(-i) = -\theta_0^{2n} - 1$ (since $n$ is odd).  Furthermore, differentiating yields \[P'(z)P(-z) - P(z)P'(-z) = (2n)\theta_0^{2n}z^{2n-1},\] and so $- P'(i)P(-i) + P'(-i)P(i) = -2in\theta_0^{2n}$. Thus the expression reduces to
\[
\frac{ni(\theta_0^{2n}+1) -2in\theta_0^{2n}}{(-i)^n(\theta_0^{2n}-1)} = \frac{ni(-\theta_0^{2n} + 1)}{(-i)^n(\theta_0^{2n} - 1)} = (-1)^{\frac{n-1}2}n
\]
which is indeed an integer when $n$ is odd. $\blacksquare$

[i]Remark.[/i] The fact that we only use $P(z)P(-z) = z^{2n} - 1$ shows that the result holds whenever $\pi/n,2\pi/n,\ldots, \pi$ are replaced by any $n$ element subset $S$ of $\{\pi/n,\ldots, 2\pi\}$ such that for each $1\leq k\leq n$ exactly one of $k\pi/n$ or $(k+n)\pi/n$ is in $S$.

Let $b_k=\tan(k\pi/n)$ and $\tan(\theta)=\alpha$. By the tangent sum rule,
\[\tan(\theta + k\pi/n) = \frac{\tan(\theta)+ \tan(k\pi/n)}{1-\tan(\theta)\tan(k\pi/n)} = \frac{\alpha + b_k}{1-\alpha b_k}.\]
Then we find
\[\frac{\sum a_k}{\prod a_k} = \frac{\sum  \frac{\alpha + b_k}{1-\alpha b_k}}{\prod \frac{\alpha + b_k}{1-\alpha b_k}} = \frac{\sum  (\alpha + b_j) \prod_{j\ne k} (1-\alpha b_k)}{\prod (\alpha + b_k)}.\]

The numerator and denominator are degree $n$ polynomials in $\alpha$, denote them as $P(\alpha),Q(\alpha)$ respectively. We claim $Q$ divides $P$. Clearly the roots of $Q$ are $-b_k$ for $1\le k \le n$, and these are distinct. We claim that these are also roots of $P$, which will give the result.

We calculate
\[P(-b_i) = \sum_{k=1}^n (b_k-b_i) \prod_{j\ne k} (1+b_i b_j).\] Notice that \[b_{k-i} = \frac{b_k-b_i}{1+b_ib_k}\] by the tangent sum identity, so $b_k-b_i = b_{k-i}(1+b_ib_k)$. Thus \[P(-b_i) = \sum_{k=1}^n b_{k-i} \prod_{j} (1+b_i b_j).\] But $\sum_{k=1}^n b_{k-i}$ is $0$ is $n$ is odd since $b_k = -b_{n-k}$. Thus $Q$ divides $P$.

The polynomials have the same degree, so the ratio is a constant, which we may determine from the ratios of the leading coefficients.The leading coefficient of $Q$ is $1$, while the leading coefficient of $P$ is
\[\sum_{k=1}^n \prod_{j\ne k} -b_j = \prod_{j\ne n} -b_j = \prod_{j=1}^{n-1} \tan(j\pi / n),\] since $b_n=0$. This product is well known and is equal to $(-1)^{(n-1)/2}n$, so this is the desired ratio.


geometry
trapezoid
parallelogram
geometry proposed

Let the equilateral triangles ABC and AB'C' of centres O and O' have the same orientation. The perpendicular bisectors of BB' and CC' intersect at P.
Prove that A, P, O and O' form an isosceles trapezoid.

Best regards,
sunken rock

1.It is easy to prove B,H,E,B’  and  C,C’,H  are collinear ,respectively
Where H is intersection of ⊙O and ⊙O’
2.∠EPF=60  => P,F,E,H are concyclic
3.AEF is an equilateral triangle  => A,H,E,F are concyclic  => A,H,E,F,P lie on the same circle
4.denote the circle is ⊙O1, HP is the diameter of ⊙O1   OO’∥AP (both ⊥ AH)
 It is easy to show △AO1O’∽△AFC’  △AOO’∽△ACC’  =>  OO1=O1O’
5.OPO’H is a parallelogram    =>  OH=OA=O’P
DONE!


[hide=" really big hint"]pair up 1 and 1001
5 and 997 etc...[/hide]

[quote="Drunken_Math"]Find this sum:

$1+5+9+13+17+21+\cdots+997+1001$[/quote]

[hide]First find out the number of terms in the sequence, which is equal to 251,

Now find the sum of the first term and the last term, which is equal to 1002,

This means that the sum of all the terms is $1002*251/2$ which equals $125751$.[/hide]


Prove that in every triangle with sides $a,b,c$ the following inequality holds:
$a^3+b^3+3abc>c^3$

 :)

It's even simpler than that: \[a^3 + b^3 + 3abc = (a + b)(a^2 - ab + b^2) + 3abc > c(a^2 - ab + b^2) + 3abc = c(a + b)^2 > c^3.\]

[Arne wrote /quote] It's even simpler than that:$ \displaystyle a^3 + b^3 + 3abc = (a + b)(a^2 - ab + b^2) + 3abc > c(a^2 - ab + b^2) + 3abc = c(a + b)^2 > c^3$.[/quote]

This is also my solution!! :)  :lol:

[quote="Rushil"]By very simple manipulation, we get,

$ a^3 + b^3 + (-c)^3 - 3 ab(-c) = ( a + b -c) ( a^2 + b^2 + c^2 -ac +bc +ac) $
Now $ ( a^2 + b^2 + c^2 -ac +bc +ac) > 0 $ for all a,b,c. You can easily prove this using theory of quadratics.

Also by triangle inequality, $ a + b > c $. Thus it follows that above exp is positive and the required reult follows.

Shouldn't this come in Getting started of Middle school?!!!! :rotfl:[/quote]
   Is it too easy for you???

well the problem is indeed not Pre-Olympiad.

Not many middle schoolers have learned to solve inequalities so I would probably post this in the intermediate forum.

We tend to forget, as we learn more, the appropriate level in which a problem should be posted. Take me for example, I learned the triangle inequality in detail about 6 months ago (and i am a rising Junior), so this is DEFINITELY not for Getting Started. I would say this is in the right place, although some, like DPopov, might argue that it could be in intermediate forum. But again, I am not the one to decide.


We have k,n naturals, not 0.We know that there exist x1,x2,...,xn numbers
form the interval [k-1;k+1] such that :

x1 + x2 +...+ xn= n[2(k^2) -1)]/2k ;

(x1)^2 + (x2)^2 +...+(xn)^2 =n(k^2) .

Prove that n is a multiple of 4k !

(It seems tome that xi must be naturals, but the problem doesnt say that)

If xi are naturals then assume there are a of k-1, b of k and c of k+1 in the sequence xi.
We have a+b+c=n.
(k-1)a+kb+(k+1)c=n[2k 2 -1]/2k
(k-1) 2a +k 2b+(k+1) 2 c=nk 2
After some computations I obtained
b=0 and c=n(2k-1)/4k. c is obviously an integer and gcd(2k-1,4k)=1, so 4k|n.

Hope xi are integers.

Is there someone who thinks this problem is not solved?
I mean: is there someone who thinks that this problem can be solved if xi are reals?
I ask because I want to move the problem to solved.


Prove that the number of ways to tile a 2 by n board with non-overlapping dominos (2 by 1 tiles) is equal to the nth Fibonacci number.

Sorry, didn't know that the problem had been posted before.


inequalities
function
logarithms
integration
number theory proposed
number theory

Let $ x,k\in \mathbb{N}$. Prove that
\[ \frac{k}{\varphi(k)}\sum_{d\le x.(d,k)\equal{}1}\frac{\mu^2(d)}{\varphi(d)}\geq \log x\]

Proof when $ k \equal{} 1$:

note that $ \mu^2(d)/\phi(d) \equal{} \sum_{rad(n) \equal{} d} 1/n$, where $ rad(n)$ is the product of the primes dividing $ n$. Thus,

$ \sum_{d \le x} \mu^2(d)/\phi(d) \equal{} \sum_{rad(n) \le x} 1/n \ge \sum_{n \le x} 1/n \ge \log x$

[edited] Yup! it's right.

now what about other cases? :)

$ \displaystyle \frac {k}{\varphi(k)}\sum_{d\le x,(d,k) \equal{} 1}{\frac {\mu^2(d)}{\varphi(d)}}\equal{}$ $ \displaystyle \sum_{d\le x,(d,k) \equal{} 1}{\left(\frac {\mu^2(d)}{d}\cdot \frac{d}{\varphi(d)} \cdot \frac{k}{\varphi(k)}\right)}$ $ \equal{}\displaystyle \sum_{d\le x,(d,k) \equal{} 1}{\left(\frac {\mu^2(d)}{d}\cdot \prod_{p \mid d}\left(1\minus{}\frac{1}{p}\right)^{\minus{}1} \cdot \prod_{p \mid k}\left(1\minus{}\frac{1}{p}\right)^{\minus{}1} \right)}$ $ \equal{}\displaystyle \sum_{d\le x,(d,k) \equal{} 1}{\left(\frac {\mu^2(d)}{d}\cdot \sum_{p \mid d}\left(\sum_{i \in \mathbb{N}}{\frac{1}{p^i}}\right) \cdot \sum_{p \mid k}\left(\sum_{i \in \mathbb{N}}{\frac{1}{p^i}}\right)\right)}$ $ \equal{}\displaystyle \sum_{d\le x,(d,k) \equal{} 1}{\left(\frac {\mu^2(d)}{d}\cdot \left(\sum_{j \ge 1, p \mid j \implies p \mid dk}{\frac{1}{j}} \right)\right)}$ $ \displaystyle \ge \sum_{d \le x}{\frac{1}{d}} \ge \int_1^x{\frac{\delta x}{x}} \equal{} \ln(x)$


conics Harvard college Yale Stanford geometry Support

As we know from PhD comics, see image below, further education, in particular graduate school gets very attractive during economic downturns. Enrollment in science and engineering graduate schools is quite linearly related to the unemployment rate. [url=http://www.guardian.co.uk/education/2008/nov/11/postgraduate-masters-courses-retraining]Students are enrolling in master programs to avoid the job turmoil.[/url] But also some private schools do well in raising funds at these times: [url=http://www.nytimes.com/2008/11/29/nyregion/29private.html?_r=1]Private Schools Say They’re Thriving in Downturn.[/url]

In the past some universities as Harvard and Yale did well in stock market and others as the University of Washington did less well. But now the endowment of a considerable number of US universities that put heavy focus on the stock market  melts away: [url=http://www.nytimes.com/2008/11/26/business/26endowment.html]Beyond the Ivied Halls, Endowments Suffer.[/url] [url=http://www.mysanantonio.com/news/education/Academias_ivory_tower_not_above_fiscal_storm.html]Here[/url] and [url=http://discussionleader.hbsp.com/now-new-next/2008/11/academia-and-the-downturn-bold.html]there[/url] are two more articles.

The richest university is Harvard which got an endowment of $ \$37$ billion. So far the Harvard Management Company (HMC) had a mean annual growth of 13.8% for 1998 to 2008 by trading stocks in the US market and commodities in Europe and Asia. As most US universities Harvard uses 30% of the endowment for its annual budget.

The rating agency Moody's Investor Services anticipates a decline of 30% in endowment for US universities. Dartmouth College was down $ \$220$ million, Brown University stops academic recruiting, Cornell also stops recruiting but also pauses on any constructions until at least March 2009, Stanford reduces its budget by $ \$45$ million and Harvard stops the building of a science park south of Charles rivers that its includes a science component, museum and parks. State University University of Wisconsin had an endowment of $ \$1.8$ billion in 2007 and is now down 18%.

I assume alumni that made a fortune, e.g. in the finance sector, are less likely to donate to their universities just now. And many parents who put savings for their children's university education in stocks probably might urge their children to look for cheaper universities. To balance out their losses Universities are likely to increase their tuition fees: [url=http://www.nytimes.com/2008/11/29/business/yourmoney/29money.html]Colleges Are Not Going Hungry, but Are in Need.[/url]

Universities in California got several cuts in state funding and are about to unable to fund their operations. Thus they decided to artifically increase their admission standards to admit fewer students to cope with the lower budget. I assume Prof. Merryfield might be able to say more about it on how in particular it effects his institution.

What will be the expected impact on financial aid for international students at the undergraduate and graduate level?

[quote]I assume Prof. Merryfield might be able to say more about it on how in particular it effects his institution. [/quote]
What can I say, orl? The short answer is that we don't know how bad it's going to get. The state legislature has not yet faced reality, and we in the CSU are even more directly dependent on state financing than the UC system.

As for admissions: we have very blunt tools for controlling enrollment. Our undergraduate admissions aren't a competitive pool where we compare students and make offers to the best $ n$ for some pre-selected $ n.$ Instead, we have a simple rule and formula based process that sorts students into just two piles - those that meet a certain standard, and those that don't. We distinguish between a local service area (just Long Beach and a very small group of adjacent cities - much, much smaller than the whole metropolitan area) and everyone else. The standard for local admission is quite low, and we can't adjust it. We can adjust the standard for non-local admission and have been setting it quite a bit higher than for locals. We still got surprised last year by an out-of-trend yield rate: a greater fraction of those we admitted chose to come than we expected. I don't know where we'll set our cutoff this year. The difference between locals and non-locals is likely to be the equivalent of 1000 SAT points or 1.25 points of high school GPA, which is a lot.

As for international admissions and out-of-state admissions: we don't have all that many of either, and we don't offer financial aid to international students at all. The standards will look like those for non-local Californians.

Of course, orl's post is talking mostly about graduate admissions. We have a master's program only, not a Ph.D. program. We support (as TA's and GA's) only a minority of the students we accept. We have the capacity to add more graduate students if that's what happens - the effect there will be fairly small.

For those who don't understand California governmental finance - well, no one really understands it - here's a quick rundown.

Though the system of initiative referenda, which are propositions placed on the state ballot by citizen petitions, the voters of California have spoken with a loud, clear, self-contradictory voice. The voters of California want:
- Schools and colleges, for which they're willing to support infrastructure bond issues
- Roads, and the principle that gasoline taxes should only pay for roads and nothing else
- All sorts of other infrastructure projects, such as water projects (always with "conservation" in the title.) Just passed: high speed trains.
- Criminals should be locked away for a very long time.
- In principle, K-12 schools should have a high budget priority.
- Taxes should be low.
- It should be very, very difficult to ever raise taxes.
- It should be very, very difficult even to agree on a budget. (There's a requirement for a 2/3 vote of the legislature to pass a budget. As a result, budgets are months late and often wholly fictional.)
- And did I mention that someone [i]else[/i] should pay taxes for schools, prisons, and roads? Not me!

By the standards of the other states in the United States, what does California's tax structure look like?

1. Compared to other states, property taxes are quite low, and also grossly unfair and discriminatory. The discrimination is in favor of those who are already property owners and against newcomers. Also, over time, the property tax burden has shifted onto residential property rather than commercial property, since commercial property changes hands less often. Because property taxes are low, schools cannot be financed from local taxes and must depend on the state general fund.

2. Compared to other states, sales taxes are high but narrow. The sales tax percentage is high compared to most other states, but it applies to fewer goods than in most states.

3. Income taxes are high and somewhat progressive.

That means that the income tax is the most important source of revenue for state government. The income tax is also highly volatile and exquisitely sensitive to general economic conditions. When times are flush and income tax collections are high, it is politically impossible to run a surplus and hold it back for harder times to come - either the money must be spent, or some other tax must be lowered. A number of years ago during a time of high tax collections, a governor (by executive action) lowered the portion of the vehicle license fee that depends on the car's resale value. That money went to local governments, so the state had to divert some of its income tax revenue stream to replace that money in support of local governments. A few year later, facing a state deficit, another governor (Gray Davis) reversed that executive action and restored the license fee (the "car tax") to its former levels. The resulting wave of popular anger swept Davis from office in a recall election (which is how Schwarzenegger became governor.)

Now: the state depends on income taxes, and income taxes collections are falling off a cliff. A special session of the state legislature just concluded indecisively - since 1/3 of the body can block any tax increase, and since most actions that would actually reduce state expenditures (like releasing half of the state's prison inmates) would be wildly unpopular, no resolution of the budget crisis is in sight.

I'm sure that those of you from countries with parliamentary-style systems must be really scratching your heads - do I really mean that a majority of the state legislature can't even adopt a budget? A parliamentary ruling coalition that couldn't pass a budget wouldn't be the ruling coalition any more. Yes, I really mean that.

Thanks for your interesting discussion on the topic, Kent!

[quote="orl"]
The richest university is Harvard which got an endowment of $ \$37$ billion. So far the Harvard Management Company (HMC) had a mean annual growth of 13.8% for 1998 to 2008 by trading stocks in the US market and commodities in Europe and Asia. As most US universities Harvard uses 30% of the endowment for its annual budget.
[/quote]

[i]"As has been reported, the value of the University’s endowment was $ \$$36.9 billion on June 30, 2008, the end of the last fiscal year. Since then, the severe turmoil in the world’s financial markets has affected all major asset classes in which the endowment is invested... Harvard Management Company, using standard industry practices for valuing assets, has calculated investment losses of approximately 22 percent from July 1 through October 31... With those considerations in mind, and in view of the uncertain outlook ahead, we continue to plan for a scenario in which our endowment is down 30 percent in value for the year."[/i] [url=http://www.president.harvard.edu/speeches/faust/081202_economy.html]Harvad: Financial Update[/url]

On Tuesday, 2nd December Drew Galpin Faust, president of Harvard, reports to the council of deans that Harvard was down 22% from $ \$$ 36.9 billion which is about $ \$$ 8.1 billion in 4 month. The anticipated scenario of 30% would mean a loss of $ \$$ 11.1 billion.

This week it is even covered in the Economist: [url=http://www.economist.com/finance/displaystory.cfm?story_id=12778077]Ivory-towering infernos.[/url]

Not sure why nobody cares to comment but this situation may have an impact on your financial aid, scholarships, recruitment for permanent academic jobs, teaching quality etc. There was yet another article in the NY Times [url=http://www.nytimes.com/2008/12/17/business/17yale.html?_r=1&scp=2&sq=yale&st=cse]Yale’s Endowment Drops 13.4%.[/url] This loss is still way better than the anticipated loss of possibly more than 22% for Harvard. I do not know whether anybody has read any of the books or articles about/by David Swensen who leads the endowment management of Yale. But he is quite a well-known institutional investmentor.

[quote="orl"]Not sure why nobody cares to comment...[/quote]
Well, there is not much to say really. The budgets get pretty tight with all the usual implications. The most obvious ones are freezes in hiring and abandoning postdoctoral programs. They aren't talking about salary reductions yet but clearly there will be no noticeable raises in the next couple of years. Also we may expect downsizing the faculty and increasing teaching loads. An interesting side effect is that even the second rate universities now have a decent chance to get top recent PhD's, so if there are some aggressive administrators over there who want to raise their department rankings, it is a good time for the "faculty hunt" (all you need is to have some free money and to look a bit farther than your nose :P). As to the "increased number of graduate school applications", we haven't noticed this yet in Wisconsin: our graduate applicant pool this year is fairly poor, perhaps even smaller than last year :(.


Let $ f: I \rightarrow R$,$ \forall k,m,n \in I$;$ f(km) \plus{} f(kn) \minus{} f(k)f(mn) \geq 1$
find all function.

[quote="aviateurpilot"][quote="apollo"]Let $ f: I \rightarrow R$,$ \forall k,m,n \in I$;$ f(km) \plus{} f(kn) \minus{} f(k)f(mn) \geq 1$
find all solution.[/quote]
$ I\subset \mathbb{N}\ or \ \mathbb{R}??$[/quote]
$ I\equal{}Z$

Put $ m \equal{} n \equal{} k \equal{} 1$ and $ m \equal{} n \equal{} k \equal{} 0$. In the first case we get
$ 0 \geq (f(1) \minus{} 1)^2$ and in the second we get $ 0 \geq (f(0) \minus{} 1)^2$ so $ f(0) \equal{} f(1) \equal{} 1$.

Now put $ m \equal{} n \equal{} 0$. We get $ 2 \minus{} f(k) \geq 1$, or $ 1 \geq f(k)$, where $ k$ is any integer.
Put $ m \equal{} 0, k \equal{} 1$. We get $ f(n) \geq 1$ where $ n$ is any integer.
So $ f(n) \equal{} 1$ for all integers $ n$.


geometry
analytic geometry
graphing lines
slope
angle bisector
geometry proposed

A, B, and C are the points (1,2), (0,0), and (-1, 3) respectively. What is the slope of the bisector of angle ABC?

I tried solving it using the half angle formula with the tan(22.5) and got the answer 2^(1/2) - 1, but I'm not sure that is correct.

Let $BE$ be the internal angle bisector of $\triangle ABC$.

Let $C'$ be the symmetric point of $C$ w.r.t. $B$. So $C'=(1,-3)$.

Let $BD$ be the internal angle bisector of $\triangle ABC'$

We have $BD \perp BE$, so $m_{BD}\cdot m_{BE}=-1$, where $m_{k}$ is the slope of the line $k$

The line $AC'$ has equation $x=1$. So let $D(1,y)$

$\frac{BC'}{AB}= \frac{DC'}{AD}\Rightarrow $

$\frac{\sqrt{10}}{\sqrt{5}}= \frac{y-(-3)}{2-y}\Rightarrow $

$y =\frac{2\sqrt{2}-3}{1+\sqrt{2}}= (2\sqrt{2}-3)(\sqrt{2}-1) = 7-5\sqrt{2}$

So the slope of $BD$ is $m_{BD}= \frac{y}{1}= 7-5\sqrt{2}$

and $m_{BE}=-\frac{1}{m_{BD}}= 7+5\sqrt{2}$


advanced fields
advanced fields theorems

We define the cardinal number of $X$ as the smallest ordinal number equinumerous to $X$, namely,
$|X| = \mbox{card}(X) = \inf \{\alpha\in \mbox{Ord} : \alpha \sim X \}$
(where $A\sim B$ means that there exists a bijection from $A$ to $B$)

Prove that $|A^B|=|A|^{|B|}$, where $A\neq \emptyset$.

cardinal arithmetic is defined in these terms. you can't use ordinal arithmetic here, because it doesn't coincide. so basically there can't be proved anything ;)


[quote="marc1217"]Of the students in a Math 17 class, 95% have desktop computer, 75% have both Desktop and Laptop, and
60% have both Desktop and PDA. At least what percent of students in the class have all three kinds of
devices?[/quote]

I guess, we need more information!

is the answer [b]5 %[/b] by any chance??

[i]even im bad at the venn diag type questions[/i]


calculus
integration
calculus computations

this has already been discussed here:
http://www.mathlinks.ro/Forum/post-515467.html#515467

unfortunately, you can only solve it numerically!  you can also express it in taylor series expansion..

This is simply a try. Please see whether I am right in the proceeding. Also I am not complete.

Let, $ y \equal{} x^{x}$ . To provide a closed form, take log on both sides we get $ log(y) \equal{} xlog(x)$. This changes as $ y \equal{} x^{x} \equal{} e^{xlog(x)}$. Using the expansion of $ e^{x}$, we get,

$ \int x^{x}.dx \equal{} \int e^{xlog(x)}.dx \equal{} \int 1 \plus{} \sum_{n \ge 1} \frac {{x^nlog(x)}^n}{n!}.dx$.

Someone help me to proceed after this.


geometry
circumcircle
incenter
ratio
inradius
trigonometry
perpendicular bisector

Could anyone find a nice solution to this problem :?:

Denote by $(O, R)$, $(I, R_{a})$ the circumcircle, and the excircle of angle $A$ of triangle $ABC$. Prove that

$IA.IB.IC = 4R.R^{2}_{a}$

Assume that nice means without trigonometry. Let X, Y, Z be tangency points of the A-excircle $(I_{a})$ with BC, CA, AB. Inversion in $(I_{a})$ with radius $r_{a}$ carries A, B, C into the midpoints A', B', C' of the excircle chords XY, YZ, ZX and the circumcircle (O) into the 9-point circle (O') of the $\triangle A'B'C'.$ Denote a = BC, b = CA, c = AB sides of the $\triangle ABC,$ a' = B'C', b' = C'A', c' = A'B' sides of the $\triangle A'B'C',$ and x = YZ, y = ZX, z = XY sides of the $\triangle XYZ.$ The $\triangle A'B'C'$ is the medial triangle of the $\triangle XYZ,$ hence x = 2a', y = 2b', z = 2c'. By the basic inversion properties,

$B'C' = BC \cdot \frac{r_{a}^{2}}{I_{a}B \cdot I_{a}C},$ cyclic exchange,

$\frac{xyz}{8}= a'b'c' = abc \cdot \frac{r_{a}^{6}}{(I_{a}A \cdot I_{a}B \cdot I_{a}C)^{2}}$

Substituting $abc = 4R |\triangle ABC|,$ $xyz = 4r_{a}|\triangle XYZ|,$ and rearranging,

$\frac{2R}{r_{a}}\cdot \frac{|\triangle XYZ|}{|\triangle ABC| }= \left(\frac{4Rr_{a}^{2}}{I_{a}A \cdot I_{a}B \cdot I_{a}C}\right)^{2}$

Let I be the triangle incenter and $I_{b}, I_{c}$ the remaining B-, C-excenters. The triangles $\triangle XYZ \sim \triangle II_{b}I_{c}$ are similar, having parallel sides, and the triangles $\triangle ICB \sim \triangle II_{b}I_{c},$ because the quadrilateral $BCI_{b}I_{c}$ is cyclic on account of the right angles $\angle I_{b}BI_{c}= \angle I_{b}CI_{c}= 90^\circ.$ Thus the triangles $\triangle ICB \sim \triangle XYZ$ are also similar, hence their areas are in the ratio of squares of their circumradii. (This seems simpler than a direct angle chase.) Let D be the midpoint of BC and let the perpendicular bisector of BC meet (O) at M on the arc opposite to A and at N on the arc containing A. Then M is the circumcenter of the $\triangle ICB$ and $\rho = MB = MC$ its circumradius, so that

$\frac{|\triangle ICB|}{|\triangle XYZ|}= \frac{\rho^{2}}{r_{a}^{2}}$

Let r be inradius of the $\triangle ABC,$ s its semiperimeter, and $h_{a}$ its A-altitude. Areas of the $\triangle ABC,\ \triangle ICB$ with the common base BC are in the ratio of their altitudes $h_{a},\ r.$

$\frac{|\triangle ABC|}{|\triangle XYZ|}= \frac{|\triangle ABC|}{|\triangle IBC|}\cdot \frac{|\triangle ICB|}{|\triangle XYZ|}= \frac{h_{a}}{r}\cdot \frac{\rho^{2}}{r_{a}^{2}}$

The right angle triangles $\triangle MNB \sim \triangle MBD$ with a common angle at M are similar,

$\frac{MB}{MN}= \frac{MD}{MB},\ \ \rho^{2}= MB^{2}= MN \cdot MD = 2R \cdot MD$

The right angle triangles $\triangle BMD \sim \triangle AYI_{a}$ are similar, having equal angles $\angle YAI_{a}\equiv \angle CAM = \angle CBM \equiv \angle DBM,$ hence

$\frac{r_{a}}{s}= \frac{I_{a}Y}{AY}= \frac{MD}{BD}= \frac{2 MD}{a}$

Substituting this to the formula for circumradius $\rho$ of the $\triangle ICB,$ we get

$\rho^{2}= \frac{a R r_{a}}{s}$

Substituting this into the area ratio of the $\triangle ABC,\ \triangle XYZ,$ reducing by $r_{a}$ and than by $sr = \frac{ah_{a}}{2}= |\triangle ABC|,$

$\frac{|\triangle ABC|}{|\triangle XYZ|}= \frac{h_{a}aRr_{a}}{ srr_{a}^{2}}= \frac{2R}{r_{a}}$

It follows that

$\left(\frac{4Rr_{a}^{2}}{I_{a}A \cdot I_{a}B \cdot I_{a}C}\right)^{2}= \frac{2R}{r_{a}}\cdot \frac{r_{a}}{2R}= 1$

$I_{a}A \cdot I_{a}B \cdot I_{a}C = 4Rr_{a}^{2}$

Yes, you got it  :D . With trigonometry it's easy but long, we can even find $IA.IB.IC$ by using the measures of the sides.

First, we prove that the power of I wrt (O, R) is $2RR_{a}$. Let IA meet (O, R) again at X. It's easily shown by angle chase that XB = XI, so it remains only to show that $XB \cdot IA = 2RR_{a}$. Indeed, $\frac{XB}{2R}= \sin \frac{A}{2}= \frac{R_{a}}{IA}$.

Let IB and IC meet (O, R) again at Y and Z, respectively. Then the equation we are trying to prove reduces upon multiplication by IY and IZ to

$IY \cdot IZ = IA \cdot R$.

This is readily apparent because a quick angle chase shows $AYI \sim AOZ$.

Here:
$IA.AP=AB.AC$ ($P$ is the incenter of $ABC$, well known theorem)
$IB.IC=R_{a}.IP$ (due to isogonal's theorem, where $IP$ is the diameter of circumcircle of $IBC$ )
Now:$IA.IB.IC.AP=AB.AC.R_{a}.IP=2R.h_{a}.R_{a}.IP$ ($h_{a}$ altitude from $A$ of $ABC$)
Then:$IA.IB.IC=\frac{2R.h_{a}.R_{a}.IP}{AP}$
But:$\frac{h_{a}.IP}{AP}=2R_{a}$
Hence:$IA.IB.IC=4R.R_{a}^{2}$


inequalities
calculus
2-variable inequality

Let a,b>0 Prove that
$ \frac{a}{\sqrt{a^2\plus{}3b^2}}\plus{}\frac{b}{\sqrt{b^2\plus{}3a^2}} \geq1$

[quote="tran300491"]Let a,b>0 Prove that
$ \frac {a}{\sqrt {a^2 \plus{} 3b^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}} \geq1$[/quote]
By Holder $ \left(\frac {a}{\sqrt {a^2 \plus{} 3b^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}}\right)^2(a(a^2\plus{}3b^2)\plus{}b(b^2\plus{}3a^2))\geq(a\plus{}b)^3\equal{}$
$ \equal{}a(a^2\plus{}3b^2)\plus{}b(b^2\plus{}3a^2).$ :)

Very nice !
Thank great math  and  arqady  very much
And this is new inequality(easy too)
$ \frac{a}{\sqrt[3]{a^3\plus{}7b^3}}\plus{}\frac{b}{\sqrt[3]{b^3\plus{}7a^3}} \geq 1$

We use my method above, how effectively it works! Now, we check it together.
We set $ x \equal{} \frac{a}{\sqrt[3]{a^3\plus{}7b^3}}$
and $ y \equal{} \frac{b}{\sqrt[3]{b^3\plus{}7a^3}}$
In addition, we get the condition $ 48(xy)^3\plus{}x^3\plus{}y^3\equal{}1$
and then, we assume that $ x\plus{}y < 1$ that leads to $ 48(xy)^3\plus{}x^3\plus{}y^3  > (x\plus{}y)^3$
Expanding, we have $ 16(xy)^2 > x\plus{}y$ but we have the fact that ( easily checking by Cauchy-Schwartz) $ 16(xy)^2 \leq (x\plus{}y)^4$ . It implies $ x\plus{}y >1$ which is wrong.
In conclusion, we have $ x\plus{}y \geq 1$
That's all.

[quote="tran300491"] [color=darkred]Let $ a > 0$ and $ b > 0$ .  Prove that $ \frac {a}{\sqrt {a^2 \plus{} 3b^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}} \geq1$ .[/color] [/quote]
[color=darkblue][b][u]Proof (a simple entertainment).[/u][/b] Denote $ \frac ab \equal{} t > 0$ . Then our inequality becomes $ f(t)\equiv \frac {t}{\sqrt {t^2 \plus{} 3}} \plus{} \frac {1}{\sqrt {3t^2 \plus{} 1}}\ge 1$ .

Observe that $ f(1) \equal{} 1$ and $ f\left(\frac 1t\right) \equal{} f(t)$ . Thus can suppose w.l.o.g. that $ t > 1$ .

Must prove only  $ \boxed {\ t > 1\ \implies\ \frac {t}{\sqrt {t^2 \plus{} 3}} \plus{} \frac {1}{\sqrt {3t^2 \plus{} 1}} > 1\ }\ (*)$ . Therefore,

$ \frac {t}{\sqrt {t^2 \plus{} 3}} \plus{} \frac {1}{\sqrt {3t^2 \plus{} 1}} > 1$ $ \Longleftrightarrow$

$ \frac {t}{\sqrt {t^2 \plus{} 3}} \minus{} \frac 12 > \frac 12 \minus{} \frac {1}{\sqrt {3t^2 \plus{} 1}}$ $ \Longleftrightarrow$

$ \frac {2t \minus{} \sqrt {t^2 \plus{} 3}}{\sqrt {t^2 \plus{} 3}} > \frac {\sqrt {3t^2 \plus{} 1} \minus{} 2}{\sqrt {3t^2 \plus{} 1}}$ $ \Longleftrightarrow$

$ \frac {4t^2 \minus{} \left(t^2 \plus{} 3\right)}{\sqrt {t^2 \plus{} 3}\left(2t \plus{} \sqrt {t^2 \plus{} 3}\right)} > \frac {\left(3t^2 \plus{} 1\right) \minus{} 4}{\sqrt {3t^2 \plus{} 1}\left(\sqrt {3t^2 \plus{} 1} \plus{} 2\right)}$ $ \Longleftrightarrow$

$ \frac {3\left(t^2 \minus{} 1\right)}{\sqrt {t^2 \plus{} 3}\left(2t \plus{} \sqrt {t^2 \plus{} 3}\right)} > \frac {3\left(t^2 \minus{} 1\right)}{\sqrt {3t^2 \plus{} 1}\left(\sqrt {3t^2 \plus{} 1} \plus{} 2\right)}$ $ \Longleftrightarrow$

$ \left(3t^2 \plus{} 1\right) \plus{} 2\sqrt {3t^2 \plus{} 1} > \left(t^2 \plus{} 3\right) \plus{} 2t\sqrt {t^2 \plus{} 3}$ $ \Longleftrightarrow$

$ t^2 \minus{} 1 > t\sqrt {t^2 \plus{} 3} \minus{} \sqrt {3t^2 \plus{} 1}$ $ \Longleftrightarrow$

$ t^2 \minus{} 1 > \frac {t^2\left(t^2 \plus{} 3\right) \minus{} \left(3t^2 \plus{} 1\right)}{t\sqrt {t^2 \plus{} 3} \plus{} \sqrt {3t^2 \plus{} 1}}$ $ \Longleftrightarrow$

$ t^2 \minus{} 1 > \frac {t^4 \minus{} 1}{t\sqrt {t^2 \plus{} 3} \plus{} \sqrt {3t^2 \plus{} 1}}$ $ \Longleftrightarrow$

$ t\sqrt {t^2 \plus{} 3} \plus{} \sqrt {3t^2 \plus{} 1} > t^2 \plus{} 1$ $ \Longleftrightarrow$

$ t\left(\sqrt {t^2 \plus{} 3} \minus{} t\right) > 1 \minus{} \sqrt {3t^2 \plus{} 1}$ $ \Longleftrightarrow$

$ \frac {t\left[\left(t^2 \plus{} 3\right) \minus{} t^2\right]}{\sqrt {t^2 \plus{} 3} \plus{} t} > \frac {1 \minus{} \left(3t^2 \plus{} 1\right)}{1 \plus{} \sqrt {3t^2 \plus{} 1}}$ $ \Longleftrightarrow$

$ \frac {1}{\sqrt {t^2 \plus{} 3} \plus{} t} > 0 > \minus{} \frac { t}{1 \plus{} \sqrt {3t^2 \plus{} 1}}$ .

I used many times that $ t > 1$ . In conclusion, $ \frac {a}{\sqrt {a^2 \plus{} 3b^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}} \geq1$ with equality iff $ a \equal{} b$ .[/color]

Let $ a_i \geq0$ Prove that
$ \sum\frac {a_1}{\sqrt [n]{a_1^n \plus{} ((n\plus{}1)^n \minus{} 1)a_2a_3...a_{n \plus{} 1}}} \geq1$

[quote="tran300491"]Let $ a_i \geq0$ Prove that
$ \sum\frac {a_1}{\sqrt [n]{a_1^n \plus{} ((n \plus{} 1)^n \minus{} 1)a_2a_3...a_{n \plus{} 1}}} \geq1$[/quote]
This inequality  is true .We can use Holder  to prove .this is an extension of  IMO2001
And  with a,b>0 We have
$ \sqrt [3]{\frac {a^{11}}{2b^3(a^8 \plus{} 3b^8)}} \plus{} \sqrt [3]{\frac {b^{11}}{2a^3(b^8 \plus{} 3a^8)}}\geq1$
Good Lucky!

[quote="great math"]
that lead to  the condition: $ 8(xy)^2 \plus{} x^2 \plus{} y^2 \equal{} 1$

 then we have $ 4(xy)^2 \geq 1$     ( from the above condition, you can check it properly) [/quote]
I have checked it. Let $ x \equal{} \sqrt {\frac {8}{17}}$ and $ y \equal{} \frac {1}{3}.$
Then $ 8(xy)^2 \plus{} x^2 \plus{} y^2 \equal{} 1$ but $ 4(xy)^2 < 1$ and $ 4xy<1.$ :maybe:

[quote="tran300491"]Let $ a_i \geq0$ Prove that
$ \sum\frac {a_1}{\sqrt [n]{a_1^n \plus{} ((n \plus{} 1)^n \minus{} 1)a_2a_3...a_{n \plus{} 1}}} \geq1$[/quote]
[hide][img]http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=15522[/img][/hide]

Hey, Arqady, I have to justify clearly that I assume $ x \plus{} y \leq 1$ and then it leads to the contradiction.
Your example with 2 terms which its sum is greater than 1 ($ \sqrt{\frac{8}{17}} > \frac{2}{3}$)

[quote="tran300491"][quote]And  with a,b>0 We have
$ \sqrt [3]{\frac {a^{11}}{2b^3(a^8 \plus{} 3b^8)}} \plus{} \sqrt [3]{\frac {b^{11}}{2a^3(b^8 \plus{} 3a^8)}}\geq1$
Good Lucky![/quote]
I want you  prove this inequality(not  IMO 2001)

(An Inequlity  easy
If a,b>0 then
$ \frac {2a \plus{} 3b}{2a^2 \plus{} 3b^2} \plus{} \frac {2b \plus{} 3a}{2b^2 \plus{} 3a^2} \leq \frac {4}{a \plus{} b}$
--->I not SPAM :rotfl: )[/quote]
An inequality  with a,b>0
$ \sqrt{\frac{a\plus{}2b}{a^2\plus{}2b^2}}\plus{}\sqrt{\frac{b\plus{}2a}{b^2\plus{}2a^2}} \leq \sqrt{\frac{8}{a\plus{}b}}$

[quote="great math"]Well, well. The problem is rather good although it only contains 2 variables a,b. Interesting enough!
Here is my solution.
We set that $ x \equal{} \frac {a}{\sqrt {a^2 \plus{} 3b^2}}$   and   $ y \equal{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}}$
that lead to  the condition: $ 8(xy)^2 \plus{} x^2 \plus{} y^2 \equal{} 1$

Assume that $ 4xy \leq (x \plus{} y)^2 \leq 1$, then we have $ 4(xy)^2 \geq 1$     ( from the above condition, you can check it properly) .As a result, we have the contradiction.
 :)[/quote]
[quote="great math"]Hey, Arqady, I have to justify clearly that I assume $ x \plus{} y \leq 1$ and then it leads to the contradiction.
Your example with 2 terms which its sum is greater than 1 ($ \sqrt {\frac {8}{17}} > \frac {2}{3}$)[/quote]
OK. It can be " we have $ 4xy \geq 1.$ " :wink:

Let a,b>0 such that $ a^2\plus{}b^2\equal{}2$
Prove that
$ \frac{a^2}{b}\plus{}\frac{b^2}{a} \geq2$

hello, by inserting $ b\equal{}\sqrt{2\minus{}a^2}$ in your inequality we get after a simple calculation $ a^3 \geq\sqrt{2\minus{}a^2}(a^2\plus{}2a\minus{}2)$. For all $ a$ with $ 0<a \le\sqrt{3}\plus{}1$ is this true. After squaring both sides and factoring we get $ (a^4\plus{}4a^3\plus{}6a^2\minus{}4)(a\minus{}1)^2 \geq 0$, which is true for all $ a$ with $ a\geq \sqrt{3}\plus{}1$.
Sonnhard.

[quote="tran300491"]Let a,b>0 such that $ a^2 \plus{} b^2 \equal{} 2$
Prove that
$ \frac {a^2}{b} \plus{} \frac {b^2}{a} \geq2$[/quote]
$ LHS\equal{}\frac {a^4}{a^2b} \plus{} \frac {b^4}{ab^2}\ge\ \frac {(a^2\plus{}b^2)^2}{ab(a\plus{}b)} \ge\ 2$

[quote="tran300491"]Let a,b>0 such that $ a^2 \plus{} b^2 \equal{} 2$
Prove that
$ \frac {a^2}{b} \plus{} \frac {b^2}{a} \geq2$[/quote]
The following inequality is true too.
Let $ a$ and $ b$ are positive numbers such that $ a^9 \plus{} b^9 \equal{} 2.$
Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{a} \geq2\]

[quote="arqady"][quote="tran300491"]Let a,b>0 such that $ a^2 \plus{} b^2 \equal{} 2$
Prove that
$ \frac {a^2}{b} \plus{} \frac {b^2}{a} \geq2$[/quote]
The following inequality is true too.
Let $ a$ and $ b$ are positive numbers such that $ a^9 \plus{} b^9 \equal{} 2.$
Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{a} \geq2
\]
[/quote]
$ \frac {a^2}{b} \plus{} \frac {b^2}{a} \geq2$

$ \leftrightarrow a^3\plus{}b^3 \ge\ 8ab$

$ \leftrightarrow 2\plus{}3(a^6b^3\plus{}a^3b^6) \ge\ 8a^3b^3$

$ \leftrightarrow a^6b^3\plus{}a^3b^6 \plus{}1 \ge\ 3a^3b^3\rightarrow$ right

[quote="arqady"][quote="onlylove_math"]
$ \leftrightarrow 2 \plus{} 3(a^6b^3 \plus{} a^3b^6) \ge\ 8a^3b^3$

$ \leftrightarrow a^6b^3 \plus{} a^3b^6 \plus{} 1 \ge\ 3a^3b^3\rightarrow$ right[/quote]
Why $ 2 \plus{} 3(a^6b^3 \plus{} a^3b^6) \ge\ 8a^3b^3$ is true? Explain please.[/quote]
By AM_GM
$ 3(a^3\plus{}b^3)^4\plus{}16\geq 8(a^3\plus{}b^3)^3$
$ \minus{}>a^3\plus{}b^3\geq2ab$
And this is new inequality
If a,b>0 such that $ a^2\plus{}b^2\equal{}2$
Prove that
$ \sqrt[3]{\frac{a^7}{b^5}}\plus{}\sqrt[3]{\frac{b^7}{a^5}} \geq2$

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219238

[quote="Math pro"][quote="arqady"][quote="tran300491"]
$ 3(a^3 \plus{} b^3)^4 \plus{} 16\geq 8(a^3 \plus{} b^3)^3$
$ \minus{} > a^3 \plus{} b^3\geq2ab$
[/quote]
Why? Explain please.  :wink:
[quote="tran300491"]
And this is new inequality
If a,b>0 such that $ a^2 \plus{} b^2 \equal{} 2$
Prove that
$ \sqrt [3]{\frac {a^7}{b^5}} \plus{} \sqrt [3]{\frac {b^7}{a^5}} \geq2$[/quote]
The following inequality is true too.
Let $ a$ and $ b$ are positive numbers such that $ a^{22} \plus{} b^{22} \equal{} 2.$ Prove that:
\[ \sqrt [3]{\frac {a^7}{b^5}} \plus{} \sqrt [3]{\frac {b^7}{a^5}} \geq2
\]
[/quote]
dear,argady I think this is tran300491 's solution
$ \minus{} > a^3 \plus{} b^3\geq2\geq2ab$
but It is not true because $ a^3 \plus{} b^3\le2$->alway true[/quote]
$ 3(a^3 \plus{} b^3)^4 \plus{} 16\geq 8(a^3 \plus{} b^3)^3$
->$ 3(a^3\plus{}b^3)^4\plus{}8(a^9\plus{}b^9)\geq 8(a^3\plus{}b^3)^3$
->$ 3(a^3\plus{}b^3)^4 \geq 8[(a^3\plus{}b^3)^3\minus{}(a^9\plus{}b^9)]$
->$ 3(a^3\plus{}b^3)^4 \geq 24a^3b^3(a^3\plus{}b^3)$
->$ (a^3\plus{}b^3)^3 \geq 8a^3b^3$

[quote="arqady"][quote="arqady"][quote="tran300491"]
$ 3(a^3 \plus{} b^3)^4 \geq 8[(a^3 \plus{} b^3)^3 \minus{} (a^9 \plus{} b^9)]$
->$ 3(a^3 \plus{} b^3)^4 \geq 24a^3b^3(a^3 \plus{} b^3)$
[/quote]
Why? :wink:[/quote]
[quote="tran300491"]USE
$ (x \plus{} y)^3 \equal{} x^3 \plus{} y^3 \plus{} 3xy(x \plus{} y)$[/quote]
I don't understand you. :maybe: Explain please.[/quote]
We have:
$ (x\plus{}y)^3\equal{}(x\plus{}y)(x\plus{}y)(x\plus{}y)\equal{}x^3\plus{}3x^2.y\plus{}3x.y^2\plus{}y^3\equal{}x^3\plus{}y^3\plus{}3xy(x^2\plus{}y^2)$
Thus
$ (a^3\plus{}b^3)^3\equal{}a^9\plus{}3a^3b^3(a^3\plus{}b^3)\plus{}b^9$
 :wink: I don't think that it's really difficult ,argady  :huh:

[quote="tronghieu"][/quote][quote="arqady"][quote="arqady"][quote="tran300491"]
$ 3(a^3 + b^3)^4 \geq 8[(a^3 + b^3)^3 - (a^9 + b^9)]$
->$ 3(a^3 + b^3)^4 \geq 24a^3b^3(a^3 + b^3)$
[/quote]
Why? :wink:[/quote]
[quote="tran300491"]USE
$ (x + y)^3 = x^3 + y^3 + 3xy(x + y)$[/quote]
I don't understand you. :maybe: Explain please.[/quote][quote="tronghieu"]
We have:
$ (x + y)^3 = (x + y)(x + y)(x + y) = x^3 + 3x^2.y + 3x.y^2 + y^3 = x^3 + y^3 + 3xy(x^2 + y^2)$
Thus
$ (a^3 + b^3)^3 = a^9 + 3a^3b^3(a^3 + b^3) + b^9$
 :wink: I don't think that it's really difficult ,argady  :huh:[/quote]
OK. But why $ 3(a^3 + b^3)^4 \geq 24a^3b^3(a^3 + b^3)$ is true? Explain it.
I think the tran300491's solution is total wrong.

[quote="arqady]OK. But why $ 3(a^3 \plus{} b^3)^4 \geq 24a^3b^3(a^3 \plus{} b^3)$ is true? Explain it[/quote]

I agree with you,arqady,i think it is not true

[quote="Allnames"][quote="arqady"]OK. But why $ 3(a^3 \plus{} b^3)^4 \geq 24a^3b^3(a^3 \plus{} b^3)$ is true? Explain it[/quote]
I agree with you,arqady,i think it is not true[/quote]

No! $ 3(a^3 \plus{} b^3)^4 \geq 24a^3b^3(a^3 \plus{} b^3)$ is true for all positive $ a$ and $ b$ such that $ a^9 \plus{} b^9 \equal{} 2.$
But the tran300491's proof for it is totally wrong.

[quote="arqady"] But why $ 3(a^3 \plus{} b^3)^4 \geq 24a^3b^3(a^3 \plus{} b^3)$ is true? Explain it.[/quote]
\[ {\left( {a}^{3} \plus{} {b}^{3} \right) ^{4} \minus{} 8\,{a}^{3}{b}^{3} \left( {a}^{3} \plus{} {b}^{3} \right) \equal{} \left( a^3 \plus{} b^3 \right) \left( {a}^{3} \minus{} 2\,ab \plus{} {b}^{3} \right) \left( {a}^{6} \plus{} 2\,{a}^ {4}b \plus{} 2\,{a}^{3}{b}^{3} \plus{} 4\,{a}^{2}{b}^{2} \plus{} 2\,a{b}^{4} \plus{} {b}^{6} \right) }
\]
Now it's clear that the inequality $ 3(a^3 \plus{} b^3)^4 \geq 24a^3b^3(a^3 \plus{} b^3)$ is true if, and only if, $ {a}^{3} \plus{} {b}^{3} \geq 2ab$. Of course, this inequality does not hold in general, try $ a \equal{} b \equal{} \frac 1 2$. To guess, whether this inequality is true or false under the condition $ a^2 \plus{} b^2 \equal{} 2$ or $ a^{22} \plus{} b^{22} \equal{} 2$, let's plot all 3 graphics:

[img]http://i13.photobucket.com/albums/a287/Gaz_v_pol/inequality2.jpg[/img]

Now it's clear that $ {a}^{3} \plus{} {b}^{3} \geq 2ab$ holds if $ a^2 \plus{} b^2 \equal{} 2$, but does not hold under the condition $ a^{22} \plus{} b^{22} \equal{} 2$. Try $ a \equal{} 0.95$ and $ b \equal{} 1.024$. For these values $ a^{22} \plus{} b^{22} \minus{} 2 > 0$ and $ {a}^{3} \plus{} {b}^{3} \minus{} 2ab < 0$. However, this method does not answer the question, whether this inequality is true if $ a^9\plus{}b^9\equal{}2$ because the plots are near the same in this case.

The picture above was made by the following command in Maple 12:

[i]
plots[implicitplot]([a^3+b^3-2*a*b,a^2+b^2-2,a^22+b^22-2],a=0..2,b=0..2,numpoints=200000,color=[red,black,blue],legend=[`a^3 + b^3 = 2ab`,`a^2 + b^2 = 2`,`a^22 + b^22 = 2`],labels=[a,b]);
[/i]

Sergey Markelov

Sergey, $ a^{22} \plus{} b^{22} \equal{} 2$ ( $ a > 0,$ $ b > 0$ ) is a condition for the $ \sqrt [3]{\frac {a^7}{b^5}} \plus{} \sqrt [3]{\frac {b^7}{a^5}} \geq2.$ :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1214771#1214771

For the $ a^3 \plus{} b^3\geq2ab$ my condition was $ a^9 \plus{} b^9 \equal{} 2,$ like you said. :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1211543#1211543

Let $ t=a^3+b^3>0$ By AM_GM
 $ t^4+t^4+t^4+16  \geq 8t^3$

----->Why  not true?[quote="tran300491"][/quote][quote="Math pro"][quote="arqady"][quote="tran300491"]
$ 3(a^3 + b^3)^4 + 16\geq 8(a^3 + b^3)^3$
$ - > a^3 + b^3\geq2ab$
[/quote]
Why? Explain please.  :wink:
[quote="tran300491"]
And this is new inequality
If a,b>0 such that $ a^2 + b^2 = 2$
Prove that
$ \sqrt [3]{\frac {a^7}{b^5}} + \sqrt [3]{\frac {b^7}{a^5}} \geq2$[/quote]
The following inequality is true too.
Let $ a$ and $ b$ are positive numbers such that $ a^{22} + b^{22} = 2.$ Prove that:
\[ \sqrt [3]{\frac {a^7}{b^5}} + \sqrt [3]{\frac {b^7}{a^5}} \geq2
\]
[/quote]
dear,argady I think this is tran300491 's solution
$ - > a^3 + b^3\geq2\geq2ab$
but It is not true because $ a^3 + b^3\le2$->alway true[/quote][quote="tran300491"]
$ 3(a^3 + b^3)^4 + 16\geq 8(a^3 + b^3)^3$
->$ 3(a^3 + b^3)^4 + 8(a^9 + b^9)\geq 8(a^3 + b^3)^3$
->$ 3(a^3 + b^3)^4 \geq 8[(a^3 + b^3)^3 - (a^9 + b^9)]$
->$ 3(a^3 + b^3)^4 \geq 24a^3b^3(a^3 + b^3)$
->$ (a^3 + b^3)^3 \geq 8a^3b^3$[/quote]

[quote="tran300491"]Let $ t \equal{} a^3 \plus{} b^3 > 0$ By AM_GM
 $ t^4 \plus{} t^4 \plus{} t^4 \plus{} 16 \geq 8t^3$

----->Why  not true?[/quote]
OK. $ t^4 \plus{} t^4 \plus{} t^4 \plus{} 16 \geq 8t^3$ is true.
Why $ a^3\plus{}b^3\geq2ab$ is true? Explain it.

[quote="tran300491"][/quote][quote="Math pro"][quote="arqady"][quote="tran300491"]
$ 3(a^3 + b^3)^4 + 16\geq 8(a^3 + b^3)^3$
$ - > a^3 + b^3\geq2ab$
[/quote]
Why? Explain please.  :wink:
[quote="tran300491"]
And this is new inequality
If a,b>0 such that $ a^2 + b^2 = 2$
Prove that
$ \sqrt [3]{\frac {a^7}{b^5}} + \sqrt [3]{\frac {b^7}{a^5}} \geq2$[/quote]
The following inequality is true too.
Let $ a$ and $ b$ are positive numbers such that $ a^{22} + b^{22} = 2.$ Prove that:
\[ \sqrt [3]{\frac {a^7}{b^5}} + \sqrt [3]{\frac {b^7}{a^5}} \geq2
\]
[/quote]
dear,argady I think this is tran300491 's solution
$ - > a^3 + b^3\geq2\geq2ab$
but It is not true because $ a^3 + b^3\le2$->alway true[/quote][quote="tran300491"]
$ 3(a^3 + b^3)^4 + 16\geq 8(a^3 + b^3)^3$
->$ 3(a^3 + b^3)^4 + 8(a^9 + b^9)\geq 8(a^3 + b^3)^3$
->$ 3(a^3 + b^3)^4 \geq 8[(a^3 + b^3)^3 - (a^9 + b^9)]$
->$ 3(a^3 + b^3)^4 \geq 24a^3b^3(a^3 + b^3)$
->$ (a^3 + b^3)^3 \geq 8a^3b^3$[/quote]
-->$ a^3+b^3\geq \sqrt[3]{8a^3b^3}=2ab$

tran300491, I have understood your proof. It's true and nice. I am sorry. :(

[quote="tran300491"][/quote][quote="tran300491"][/quote][quote="Math pro"][quote="arqady"][quote="tran300491"]
$ 3(a^3 + b^3)^4 + 16\geq 8(a^3 + b^3)^3$
$ - > a^3 + b^3\geq2ab$
[/quote]
Why? Explain please.  :wink:
[quote="tran300491"]
And this is new inequality
If a,b>0 such that $ a^2 + b^2 = 2$
Prove that
$ \sqrt [3]{\frac {a^7}{b^5}} + \sqrt [3]{\frac {b^7}{a^5}} \geq2$[/quote]
The following inequality is true too.
Let $ a$ and $ b$ are positive numbers such that $ a^{22} + b^{22} = 2.$ Prove that:
\[ \sqrt [3]{\frac {a^7}{b^5}} + \sqrt [3]{\frac {b^7}{a^5}} \geq2
\]
[/quote]
dear,argady I think this is tran300491 's solution
$ - > a^3 + b^3\geq2\geq2ab$
but It is not true because $ a^3 + b^3\le2$->alway true[/quote][quote="tran300491"]
$ 3(a^3 + b^3)^4 + 16\geq 8(a^3 + b^3)^3$
->$ 3(a^3 + b^3)^4 + 8(a^9 + b^9)\geq 8(a^3 + b^3)^3$
->$ 3(a^3 + b^3)^4 \geq 8[(a^3 + b^3)^3 - (a^9 + b^9)]$
->$ 3(a^3 + b^3)^4 \geq 24a^3b^3(a^3 + b^3)$
->$ (a^3 + b^3)^3 \geq 8a^3b^3$[/quote][quote="tran300491"]
-->$ a^3 + b^3\geq \sqrt [3]{8a^3b^3} = 2ab$[/quote]
sorry tran,mycountryman,I didn't understand U,but now I see that your proof is very nice! :wink:
I have found a nice solution for general problem of argady,It is here:
the general problem:
$ X,Y$ are non-negative numbers such that X+Y=2,find min:
$ A = \sqrt [n]{X} + \sqrt [m]{Y}$
 with $ m \geq n$
we have $ X + Y = 2$ asume $ X\geqq$
set:$ X = 1 + t,Y = 1 - t$ with $ 1\geqq\geq0$
Then derivative A we easy have done,I'm sure that! :wink:
@allnames:Are you spaming? :rotfl:

[quote="tran300491"]Let $ a,b > 0$. Prove that

$ \frac {a}{\sqrt {a^2 \plus{} 3b^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}} \geq1.$[/quote][quote="great math"]Set $ x \equal{} \frac {a}{\sqrt {a^2 \plus{} 3b^2}}$ and $ y \equal{} \frac {b}{\sqrt {b^2 \plus{} 3a^2}}$, we have[/quote]$ (x \plus{} y \minus{} 1)(x \plus{} y \plus{} 1)(2xy \plus{} 1)\equal{}2xy(x \minus{} y)^2\geq0\Longrightarrow x \plus{} y\geq1$.

We can prove the original directly through Jensen! (this is becoming my favorite inequality).
Using Jensen on $ f(x)\equal{}\frac{1}{\sqrt{x}}$ say the given $ \ge \sqrt{\frac{(a\plus{}b)^3}{a^3\plus{}3a^2b\plus{}3ab^2\plus{}b^3}}\equal{}1$.


Hello AoPS users. I am beginning a new game. It is a game from Argentina. You probably have never heard of it. Basically twelve people sign up, I randomly select four of them. These four will compete in a challenge. The challenge will also be randomly selected. There are only four to choose from. The top 2 contestants at each challenge will move on to the next round, while the other two lose. This process continues until there are only two people left. Then these two people compete in a challenge. The winner of this challenge will compete to win money, and it will be a risk challenge, where they can either keep on going and risk losing it all, or keep what they have. Once a person has finished that challenge, then we start again.

Sign-ups start now. If there are any questions about the game, then please feel free to ask away.


Prove that any quadrilateral with a pair of [b][u]opposite obtuse angles[/u][/b] can be divided into at least [b][u]six[/u][/b] isosceles triangles.


Prove that the medians drawn to the legs of an isosceles triangle are congruent.

My answers
[hide]Call the triangle ABC and the R and S as the midpoint of the AB and AC, which is the two legs with the same length. Call the point O where two medians intersect. Then, we have:

\[AR = BR = CS = AS\]

And because of AB = AC, <ABC and <ACB are congruent as wel.

Okay. Now, construct a segment that connects the two midpoints, R and S. Let's call that RS. Then we have:

triangle ARS ~ triangle ABC because they share one angle (angle A) and the two of its sides are in proportional, 1:2. So, it provides us that <ASR = <ACB, providing us tha RS || BC. This let us to say that <RSB = <SBC by alternate interior angles. Same thing occur for <SRC and <RCB. And <ROS = <BOC by vertical angles, letting us to say:

triangle ROS ~ triangle BOC. Since those two sides are equal, it doesn't matter which way you call it.

Anyway, this tells us that BO/OS = CO/OR, and by segment addition postulate, we allowed to say BO+OS = CO+OR as well.

\[BO+OS =\]median\[_1\]
\[CO+OR =\]median\[_2\]
Thus, median\[_1\] = median\[_2\] which was the question asked to prove.
  [/hide]

This question might've been little tougher than other two proof question I posted.

Really, are all of these necessary?

Draw triangle $ABC$ with $M,N$ being the midpoints of $AB, AC$ respectively.

By the condition we have $AB=AC$ and $\angle ABC=\angle ACB$.

Then $BM=BN$, $\angle MBC= \angle NCB$, and $BC=BC$, so that $MBC\equiv NCB$ and we're done.

Ohhhh :blush: I knew my proof was too extreme in length.

Yeah, you're right. Then after what you post, BN = MC by CPCTC.

Thanks, blahblahblah


inequalities parameterization conics parabola algebra function domain

Solve the inequality, where $ a$ is a parameter:

$ x^2\minus{}(2\plus{}\sqrt{1\minus{}a})x\plus{}2\sqrt{1\minus{}a}<0$

First find the roots of the equivalent equality and you get

$ x_1 \equal{} 2$

$ x_2 \equal{} \sqrt {1 \minus{} a}$

For this just use the formula $ x_{1,2} \equal{} \frac { \minus{} B \pm \sqrt {B^2 \minus{} 4AC}}{2A}$ where $ A \equal{} 1$, $ B \equal{} \minus{} 2 \minus{} \sqrt {1 \minus{} a}$ and $ C \equal{} 2\sqrt {1 \minus{} a}$.
Now, since the parabola is always concave, you get:

$ x \in (2, \sqrt {1 \minus{} a}) , \, a \in ( \minus{} \infty,1]$

$ x$ should be satisfied : $ x^2 \minus{} (2 \plus{} \sqrt {1 \minus{} y})x \plus{} 2\sqrt {1 \minus{} y} < 0\ \wedge y \equal{} a$
$ \Longleftrightarrow (x \minus{} 2)(x \minus{} \sqrt {1 \minus{} y}) < 0\ \wedge y \equal{} a$ gives the domain as attached figure, not including the boarders.

Answer:

$ a < \minus{} 3\ \cdots 2 < x < \sqrt {1 \minus{} a}$

$ a \equal{} \minus{} 3\ \cdots \phi$

$ \minus{} 3 < a \leq 1\ \cdots \sqrt {1 \minus{} a} < x < 2$

$ a > 1\ \cdots \phi$


calculus
integration
real analysis
real analysis solved

Compute the integral:

$\displaystyle \int \frac{xe^{\arctan x}}{(1+x^{2})\sqrt{1+x^{2}}} dx,$ $x \in R.$
(You are not allowed to use the change of variables).

cheers! :D :D

\[\int \frac{x}{(1+x^2)^{\frac{3}{2}}} dx = -\frac{1}{\sqrt{1+x^2}}\]

Denote $I = \int \frac{xe^{\arctan x}}{(1+x^{2})\sqrt{1+x^{2}}} dx$

Integrate by part, we have
\[
I = \int e^{\arctan x} d\left(-\frac{1}{\sqrt{1+x^2}} \right) = -\frac{e^{\arctan x}}{\sqrt{1+x^2}} + \int \frac{e^{\arctan x}}{(1+x^2)\sqrt{1+x^2}}dx = -\frac{e^{\arctan x}}{\sqrt{1+x^2}} + J
\]

Notice that \[
\int \frac{1}{(1+x^2)^{\frac{3}{2}}} dx = \frac{x}{\sqrt{1+x^2}}
\]

Hence (integrate by part)
\[J = \int e^{\arctan x} d\left(\frac{x}{\sqrt{1+x^2}} \right) = \frac{x e^{\arctan x}}{\sqrt{1+x^2}} - I\]

Therefore

\[I = \frac{(x-1)e^{\arctan x}}{2\sqrt{1+x^2}}\]


Let $ \triangle ABC$ with incenter  $ I$ and $ \left|\ \begin{array}{c}
 E\in AC\ ,\ BE\perp AC\\\
 F\in AB\ ,\ CF\perp AB\end{array}\ \right|\ .$

Prove that  $ I\in EF\ \Longleftrightarrow\ \cos B\plus{}\cos C\equal{}\cos A\ .$

By theorem 1 of http://www.mathlinks.ro/viewtopic.php?p=570413#570413, we have $ I \in EF \Longleftrightarrow \cos{A}\equal{}\frac{b\plus{}c}{a\plus{}b\plus{}c}$.

Since $ \cos{B}\equal{}\frac{BF}{a}$, $ \cos{C}\equal{}\frac{CE}{a}$. So,

$ \cos{B}\plus{}\cos{C}\equal{}\frac{BF\plus{}CE}{a}\equal{}\frac{c\minus{}b\cos{A}\plus{}b\minus{}c\cos{A}}{a}\equal{}\frac{(1\minus{}\cos{A})(b\plus{}c)}{a}$.

So if $ \cos{A}\equal{}\frac{b\plus{}c}{a\plus{}b\plus{}c}$, then $ \cos{B}\plus{}\cos{C}\equal{}\frac{a}{a\plus{}b\plus{}c}\cdot\frac{b\plus{}c}{a}\equal{}\frac{b\plus{}c}{a\plus{}b\plus{}c}\equal{}\cos{A}$.

If $ \cos{B}\plus{}\cos{C}\equal{}\cos{A}$, we will have $ \cos{A}\equal{}\frac{b\plus{}c}{a\plus{}b\plus{}c}$.


Let $ p>0$ be a prime number . Solve the equation in $ \mathbb{Z} \times \mathbb{Z}$ :
\[ x^4 \plus{} x^2y^2 \plus{} y^4 \equal{} p\] .

$ x^4 \plus{} x^2y^2 \plus{} y^4 \equal{} (x^2 \plus{} y^2 \plus{} xy)(x^2 \plus{} y^2 \minus{}xy)$, they are both positive, so $ x^2 \plus{} y^2 \minus{} xy \equal{} 1$ and $ x^2 \plus{} y^2 \plus{} xy \equal{} p$, but $ x^2 \plus{} y^2 \minus{} xy \geq |xy|$ so $ 1\geq |xy|$, so $ x\equal{}y\equal{}1$.

More than .
Find all solution of the equation :"
$ x^4\plus{}x^2y^2\plus{}y^4\equal{}p^a$

It is easy. $ x^2\plus{}y^2\plus{}xy\equal{}p^r,x^2\plus{}y^2\minus{}xy\equal{}p^q$, if $ xy>0$ (we can consider only solutions $ x>0,y>0$, all solutions put by $ (\pm x, \pm y)$), then $ r>q$.  If $ r\equal{}q$ one of them 0, and $ x^4\equal{}p^a\ to x\equal{}p^k,a\equal{}4k.$
Let $ r>q$, then $ 2(x\minus{}y)^2\equal{}3p^q\minus{}p^r.$ Therefore $ p\le 3$. If $ p\equal{}3$, then $ x\equal{}y\to x^4\equal{}3^{a\minus{}1}\to x\equal{}3^k,a\equal{}4k\plus{}1$.
If $ p\equal{}2$, then $ r\equal{}q\plus{}1,a\equal{}2q\plus{}1,q\ge 1, q\equal{}k\plus{}1$ and $ x^2\plus{}y^2\equal{}3*2^k,xy\equal{}2^k\to x\equal{}2^t,y\equal{}2^s, t\not \equal{}s.$
Let $ t>s$, then $ 2^{2m}\plus{}1\equal{}3*2^{k\minus{}2s}$ - contradition ($ 3\not |2^{2m}\plus{}1.$)


How many positive multiples of 7 that are less than 1000 end with the digit 3?

If the multiple is $ 7k$ for some positive integer $ k$, then $ k\equiv9\pmod{10}$ (or, more generally, that $ k$'s last digit is $ 9$). Therefore, $ k$ can be any integer from $ 0$ to $ 13$ ($ 7\times{9} \equal{} 63$ and $ 7\times{139} \equal{} 973$). This is a total of $ \boxed{14}$ numbers.


linear algebra
matrix
limit
function
calculus
calculus computations

Use algebra to evaluate the limits if they exist:

${ f(x) = \left\{\begin{matrix}
x^2 -4, & 0 < x < 6\\
4, & x = 6\\
8x + 2, & x>6
\end{matrix}\right.}$

$ \lim_{x\rightarrow 6^+}{f(x) =}$
$ \lim_{x\rightarrow 6^-}{f(x) =}$
$ \lim_{x\rightarrow 6}{f(x) =}$

The first limit can be evaluated by considering the third function case.  That's just 50.  The second limit is handled by the first case, so the limit is 32.  The third limit can't be evaluated because the left and right-hand limits are not the same.

In other words, f(x) have a discontinuity of the first kind (or jump discontinuity) at x = 6.


Due to the copyrights, it's not available in here.

[quote="theone853"][quote="Silverfalcon"][quote]2. How many combinations are for word SILVERFALCON?
[/quote]

There are twelve letters so it's 12!. But since there are two L's, it's 12!/2! which is 239500800.
[/quote]

Actually I believe that it ought to be divide 2, not 2!, even though they yield the same result, you have to understand the concept for different problems or else you will make mistakes on those.[/quote]

You divide by 2! because there are 2 "L" 's. The concept would be that if there were 3 "L" 's, you would divide by 3!, and if there were 2 "L" 's and 3"A" 's you would divide by 2!3!.

[quote="RC-7th"][quote="Silverfalcon"][quote]1. What is the maximum number of intersection of a circle, triangle, and square given that they lie on the same plane?
[/quote]

First, find the maximum intersection of circle and triangle, triangle and square, square and circle. The total three figure's intersection is basically union of those numbers, which is 10.
[/quote]

Union of circle and triangle,6.

Union of triangle and square,6.

Union of square and circle, 8.

Although my drawing was kinda bad...[/quote]
Yes, that's what I got too.

[quote="LynnelleYe"][/quote][quote="RC-7th"][quote="Silverfalcon"][quote]1. What is the maximum number of intersection of a circle, triangle, and square given that they lie on the same plane?
[/quote]

First, find the maximum intersection of circle and triangle, triangle and square, square and circle. The total three figure's intersection is basically union of those numbers, which is 10.
[/quote]

Union of circle and triangle,6.

Union of triangle and square,6.

Union of square and circle, 8.

Although my drawing was kinda bad...[/quote][quote="LynnelleYe"]
Yes, that's what I got too.[/quote]

yep, same here.  its probably a typo then...

Silverfalcon, for number seven, aren't they asking for the actual vector, not its magnitude?

Fixed for number one.

For seven, it was asking for vector, not just a magnitude. According to what I know, magnitude simply means the coordinate.

<24,3> is the vector. The magnitude is a scalar value. However, I didn't see that you had given the actual vector, since you called it the magnitude. Anyways, you seem to have alternated your subtration, as you got <-24,3>.

[quote="h_s_potter2002"]Are vectors really getting started problems?  And would they be on the AMC's?[/quote]

Vectors are not getting started and you shouldn't see a problem referring to vectors on one of the AMC's.  You should, however, know how to find the distance between two points which is the magnitude of a vector. Also, even though you don't absolutely need to know vectors you still might be able to use them on a geometry problem.  You will never absolutely need them but supposedly they can help in a few situations.  I wouldn't worry about them for the AMC.

[quote="Silverfalcon"]Yes, that wasn't quite appropriate question for the COP.

I was going to give credit to ones who would put the coordinate but I think no one did and impressively, most people got that right.
Only possible mistake was during the Pythagorean Theorem. :D[/quote]
Or not simplifying because I didn't realize 585 was divisible by 9  :blush:

[quote="alan"][/quote][quote="LynnelleYe"][/quote][quote="RC-7th"][quote="Silverfalcon"][quote]1. What is the maximum number of intersection of a circle, triangle, and square given that they lie on the same plane?
[/quote]

First, find the maximum intersection of circle and triangle, triangle and square, square and circle. The total three figure's intersection is basically union of those numbers, which is 10.
[/quote]

Union of circle and triangle,6.

Union of triangle and square,6.

Union of square and circle, 8.

Although my drawing was kinda bad...[/quote][quote="LynnelleYe"]
Yes, that's what I got too.[/quote][quote="alan"]

yep, same here.  its probably a typo then...[/quote]

Silverfalcon didn't count my answer wrong.

Huh?

I thought I graded them right. Hmm.. There's no way I can check them now since I nearly emptied my pm box for Mock AMC G.

Hmm... What was your answer? ;)

Mine was the one I posted. Why would I be posting a wrong answer?

So, you had the right answer (20) when you sent me right?

 ;)


inequalities
function
rearrangement inequality
triangle inequality

It appears the inequality sheet that I posted was lost.  I still have it though, and I'm posting it again.  Enjoy!

Feel free to download, print, and distribute it as you see fit.  If there is something that I left out that would be suitable for someone who is still relatively new to serious inequality solving, please post it.

[u][b]The author of this posting is : blahblahblah[/b][/u]
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You should add:

For convex/concave functions:
Majorization (Karamata)
Popoviciu

For homogeneous inequalities:
Muirhead/Bunching

Oh, and Holder generalizes to $i$ weights which sum to 1, not just two.

[u][b]The author of this posting is : dts[/b][/u]
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[quote="blahblahblah"]You should add:

For convex/concave functions:
Majorization (Karamata)
Popoviciu

For homogeneous inequalities:
Muirhead/Bunching[/quote]

Where can I find those? I've never heard of them.

[quote="blahblahblah"]Oh, and Holder generalizes to $i$ weights which sum to 1, not just two.[/quote]

I said that. It was part of the text under the inequality.

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

You can google majorization and muirhead, or use the forum search.

Popoviciu is as follows. Say that $f$ is convex and let $m_{abc}$ be the arithmetic mean of $f\left(a\right), f\left(b\right), f\left(c\right)$.

Then

$f\left(x\right)+f\left(y\right)+f\left(z\right) + m_{xyz} \geq \frac43 \left(m_{xy}+m_{yz}+m_{xz}\right)$.

[u][b]The author of this posting is : billzhao[/b][/u]
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[quote="blahblahblah"]Popoviciu is as follows. Say that $f$ is convex and let $m_{abc}$ be the arithmetic mean of $f\left(a\right), f\left(b\right), f\left(c\right)$.

Then

$f\left(x\right)+f\left(y\right)+f\left(z\right) + m_{xyz} \geq \frac43 \left(m_{xy}+m_{yz}+m_{xz}\right)$.[/quote]

Not quite. If $f$ is convex in $I$, and $x, y, z \in I$, then

$f\left(x\right)+f\left(y\right)+f\left(z\right) + 3f\left(\frac{x+y+z}{3}\right) \geq 2\left( f\left(\frac{x+y}{2}\right) + f\left(\frac{y+z}{2}\right) + f\left(\frac{z+x}{2}\right) \right)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19097&highlight=popoviciu]Here[/url]'s a generalization of popoviciu.

[u][b]The author of this posting is : blahblahblah[/b][/u]
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Unsurprisingly, you're correct. I wasn't thinking and posted one of the inequalities out of kedlaya's packet. Popoviciu actually trivializes the inequality I posted.

[u][b]The author of this posting is : ThAzN1[/b][/u]
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You forgot rearrangement, and $k>0$ is sufficient for Schur. You can also add Bernoulli's inequality, though it isn't very widely used.

I think it'd also be helpful to start listing commonly used strategies for proving inequalities.

[u][b]The author of this posting is : billzhao[/b][/u]
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[quote="ThAzN1"]You forgot rearrangement, and $k>0$ is sufficient for Schur.[/quote]

Actually, $k \in \mathbb{R}$ is sufficient for Schur, as shown by Pierre Bornsztein in his inequality booklet.

[u][b]The author of this posting is : dts[/b][/u]
____________________________________________________________________

Well, I looked into the inequalities that blahblahblah suggested. Majorization and Popoviciu didn't seem like they would be useful very often (click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]here[/url] if you want the definition of Majorization). As for Muirhead, I found it, and it looks rather complex-I don't intend this sheet to include every inequality that might ever be useful, but rather to be a simple reference sheet of basic, useful ones.
Thanks for the suggestions, though.

As for

[quote="ThAzN1"]You forgot rearrangement....[/quote]

Yes, I did. I should have included that one. I will be revising it shortly. Thanks!

[u][b]The author of this posting is : billzhao[/b][/u]
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[quote="dts"]Majorization and Popoviciu didn't seem like they would be useful very often[/quote]

You're gonna hear a lot of oppositions to that one.

[u][b]The author of this posting is : dts[/b][/u]
____________________________________________________________________

[quote="billzhao"][quote="dts"]Majorization and Popoviciu didn't seem like they would be useful very often[/quote]

You're gonna hear a lot of oppositions to that one.[/quote]

I probably will, but I wanted to keep this list (relatively) simple, for those (like myself) who haven't had years of experience with inequalities.

Anyway, here is the revised sheet. I added the Rearrangement Inequality and a variant of the Triangle Inequality, and made a slight change to the wording of Chebyshev.

[color=red][Moderator edit: This revised sheet is the one which is currently attached to the [b]first message[/b] of this thread.][/color]

[u][b]The author of this posting is : blahblahblah[/b][/u]
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[quote="billzhao"][quote="dts"]Majorization and Popoviciu didn't seem like they would be useful very often[/quote]

You're gonna hear a lot of oppositions to that one.[/quote]

Yes. I mean, you can do most inequalities with AM-GM and ordering if you're creative enough, but really, who has the time for that? The big guns like majorization are just a lot more useful.

[u][b]The author of this posting is : dts[/b][/u]
____________________________________________________________________

I'm don't mean that the only useful inequalities are the ones on the sheet (sorry if you thought I was implying so). But my goal was to make a useful sheet for someone who was just starting to solve Olympiad level inequalities, rather than for someone who has spent a long time doing them. By "not very useful", I meant for a beginner, someone who still hasn't fully gotten the hang of such things as Cauchy-Schwartz (like me). That's why I posted this in the Pre-Olympiad forum. A list that would be appropriate for someone with more inequalities experience (as you seem to have) would be out of place in Pre-Olympiad; a better place would be the Inequalities forum.

Someday (not now, though... I have two difficult AoPS online class Final Challenges with their due dates approaching), perhaps I will compile a more complete list with the experienced inequality solver in mind.

[u][b]The author of this posting is : TripleM[/b][/u]
____________________________________________________________________

[quote="dts"]Someday (not now, though... I have two difficult AoPS online class Final Challenges with their due dates approaching), perhaps I will compile a more complete list with the experienced inequality solver in mind.[/quote]

Have a look in the inequalities section in advanced topics - in the useful theorems + formulae bit, theres about 9 pages full of inequalities.


Problem 1 :
http://www.mathlinks.ro/viewtopic.php?t=198297
Problem 2 :
 Find all $ m\in N$ such that there exist some polynomials like $ P,Q ,R$with real coefficients satisfying for all a,b such that $ a^m = b^2$ then we always have:
$ P(R(a,b)) = a,Q(R(a,b)) = b$
Problem 3 :
 For each n natural number we denote T(n) the set of n first natural numbers (positive integers). A subset S of $ T(n)$ is called "bad" if there is a number c not exeeding $ \frac {n}{2}$ such that for all$ s_1, s_2$ two arbitrary elements of S we have $ |s_1 - s_2|$ is different from $ c$. How many elements a such "bad" subset S can have at most?
Problem 4 :
 Let given 2 positive integers $ (m.n)$ Prove that
$ 6m|(2m + 3)^n\Leftrightarrow 4m|3^n + 1$
Problem 6 :
M is the set of the first 2008 positive integers.Each element of M to M is coloured in  one of 3 colours (green , yellow, or red). Plus all the colours are used. Consider a subset ${ A = {(x,y,z}}$ of M (x,y,z) all are coloured in only one colour $ x + y + z$ is divisible by 2008. and $ B = ..$ another subset of M whose (x,y,z) are colours in  difference colours and $ x + y + z$ is divisible by 2008
Prove that the $ |A| < \frac {|B|}{2}$
Problem 5:
http://www.mathlinks.ro/viewtopic.php?t=197652

Vietnam TST 2008 problems can be found in the resources section [url=http://www.mathlinks.ro/resources.php?c=186&cid=41&year=2008]here.[/url]