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What with the inaugural ARML Local competition but a fortnight away, I figured it was about time to resurrect ye olde probleme marathone. Without further adieu:

[size=150]Problem 1[/size]
Find all ordered pairs of real numbers $ (x,y)$ for which $ x^y\equal{}2$ and $ (2x)^{y^2}\equal{}64$.

That's a good solution, buuut... it's not quite complete  :wink:

y ln x = ln 2
ln x = ln2 / y

Other equation:
y^2ln(2x) = ln 64

y^2(ln2 + lnx) = 6ln2

Plug in

y^2(ln 2 + ln2/y) = 6ln2
y^2*ln2 + y * ln2 - 6ln2 = 0

ln2(y+3)(y-2) = 0

y = -3, 2

Therefore answers are
( cube root of (1/2),-3)(plus or minus the square root of 2 ,2)

So to make clear the way the marathon generally works:

1. Posting the first solution to a problem entitles a poster to put up the next problem.
1. Solutions in $ \text{\TeX}$ are teh bomb.
2. Problems generally come from past ARML / NYSML exams, or can come from anywhere of the appropriate level of difficulty.
3. As a courtesy, please "hide" your solutions if you're posting a complete solution. That way, people reading a problem will be able to give the problem a try without accidentally seeing your clever answer :wink:
5. If someone solves a problem but doesn't post a new one within a day or so, feel free to pick up the torch and post one yourself.

[hide="Solution, now with LaTeX!!!"]
\begin{eqnarray*}
(2x)^{y^2} &=& 64 \\
(2^y)^y(x^y)^y &=& 2^6 \\
\left(2^{y^2}\right)(2)^y &=& 2^6 \\
2^{y^2+y} &=& 2^6 \\
y^2+y &=& 6 \\
y^2+y-6 &=& 0 \\
(y+3)(y-2) &=& 0 \\
y=-3 &\text{or}& y=2
\end{eqnarray*}

After plugging these values of $ y$ into $ x^y=2$, we get these solutions: $ \boxed{\left(\frac{\sqrt[3]{4}}{2},-3\right),\left(2,\pm\sqrt{2}\right)}$.[/hide]

[b]Problem 2[/b]
Source: NYSML 1992, paraphrased

Two circles of radius $ r$ are in a triangle whose side lengths are 3, 4, and 5, as shown.  The circles are externally tangent to each other and are tangent to the sides of the triangle.  Compute $ r$.

[asy]pair A,B,C,O1,O2;
A=(0,3); B=(4,0); C=(0,0); O1=(5/7,11/7); O2=(13/7,5/7);
draw(A--B--C--A);
draw(Circle(O1,5/7));
draw(Circle(O2,5/7));[/asy]

Let the triangle be $ \Delta XYZ$, with $ XY\equal{}3,YZ\equal{}4,ZX\equal{}5$. Let the center of the circle tangent to $ XY,XZ$ be $ A$, and let the other center be $ B$. Construct $ C$ such that $ AC\parallel{}XY$, $ CB\parallel{}YZ$, and note that $ \Delta ABC\sim\Delta XYZ$. Let the common radius be $ r$. Drop perpendiculars $ AA_{z},CC_{z},CC_{x},BB_{x}, BB_{y},AA_{y}$ to $ XY,XY,YZ,YZ,ZX,ZX$, respectively. Clearly, $ YC_{x}\equal{}AA_{z}\equal{}CC_{z}\equal{}r$. From the similarity, $ BC\equal{}A_{x}B_{x}\equal{}\dfrac{8}{5}r$ and $ AC\equal{}AA_{z}\equal{}CC_{z}\equal{}\dfrac{6}{5}r$. Now, $ XA_{y}\equal{}XA_{z}\equal{}XY\minus{}YA_{z}\equal{}3\minus{}\dfrac{11}{5}r$. Similarly, we find that $ ZB_{y}\equal{}4\minus{}\dfrac{13}{5}r$. Also, $ AB\equal{}A_{y}B_{y}\equal{}2r$. Finally, we have $ XZ\equal{}XA_{y}\plus{}A_{y}B_{y}\plus{}B_{y}Z\equal{}5$, and substituting our values for the three lengths, we find that $ 7\minus{}\dfrac{14}{5}r\equal{}5$, and $ r\equal{}\boxed{\dfrac{5}{7}}$.

I shouldn't have clicked on the "view code" of the picture before trying to solve the problem, heh.

And also, please "hide" your solution so others can get a chance to think about a problem without accidentally seeing the answer  :P

It can also be solved using tangent half angle formulas and getting 4 little pieces (r+r+r/tanblah1+r/tanblah2) that add up to the hypotenuse

Let's keep this moving.  Here's one I made up:

[b]Problem 3[/b]
The lengths of the sides of a triangle form a geometric progression.  The measures of the angles of that same triangle form an arithmetic progression.  If the area of the triangle is 42, compute the length of the longest side of the triangle.

Also, HIDE YOUR SOLUTIONS.  Don't spoil the fun for everyone else.

[hide]Without loss of generality, let the angles be $ a\le b \le c$.  By the law of sines, $ \sin a, \sin b, \sin c$ is an arithmetic progression.  Since $ a,b,c \in [0,\pi]$, and $ x \mapsto \sin x$ is concave on that interval, it follows from Jensen's Inequality that
\[ \sin b \equal{} \frac{\sin a \plus{} \sin c}{2} \le \sin \left( \frac{a\plus{}b}{2} \right), \]
with equality only when $ a\equal{}c$.  Therefore $ a\equal{}b\equal{}c$, i.e., the triangle is equilateral.  Since the area of an equilateral triangle with side $ s$ is $ s^2\sqrt{3}/4$, it follows that the triangle's longest side is $ 2\sqrt{42}/\sqrt[4]{3}$, or however you want to simplify that.  $ \blacksquare$

Out of curiosity, did you have a more elementary solution in mind?[/hide]

This is from one of my favorite local competitions.

[size=134][b]Problem 4[/b][/size]
Let $ ABC$ be a triangle without right angles such that $ \tan B \equal{} \tan^2 A$ and $ \tan C \equal{} \tan^3 A$.  What is the value of $ \tan C \minus{} \tan A$?

[quote="Boy Soprano II"]Out of curiosity, did you have a more elementary solution in mind?[/quote]

Well, here was my solution to problem 3:

[hide="Problem 3"]Let the sides be $ a,ar,ar^2$.  Thus the angles are $ x - k,x,x + k$ for some positive reals $ a,r\geq 1,x,k$.  $ (x - k) + x + (x + k) = 180^\circ\Rightarrow x = 60^\circ$.

The side of length $ ar$ is opposite the angle whose measure is $ x$ (because both are the second-largest side/angle).  By the Law of Cosines:

\begin{eqnarray*} a^2r^2 & = & a^2 + a^2r^4 - 2a^2r^2\cos 60^\circ \\
r^2 & = & 1 + r^4 - r^2 \\
r^4 - 2r^2 + 1 & = & 0 \\
r & = & \pm1 \\
r & = & 1 \end{eqnarray*}

Thus the triangle is equilateral, and the rest is the same.[/hide]

[hide="Solution to Problem 4"]\begin{eqnarray*} \tan A + \tan B + \tan C & = & \tan A\tan B\tan C \\
\tan A + \tan^2 A + \tan^3 A & = & \tan^6 A \\
\text{Let }x & = & \tan A \\
x + x^2 + x^3 & = & x^6 \\
1 + x + x^2 & = & x^5 \\
x + x^2 & = & x^5 - 1 \\
(x^5 - 1) + x^3 & = & x^6 \\
x^5 + x^3 & = & x^6 + 1 \\
x^3(x^2 + 1) & = & (x^2 + 1)(x^4 - x^2 + 1) \\
x^3 & = & x^4 - x^2 + 1 \\
x^3 - 1 & = & x^4 - x^2 \\
(x - 1)(x^2 + x + 1) & = & x^2(x + 1)(x - 1) \\
x^2 + x + 1 & = & x^3 + x^2 \\
x + 1 & = & x^3 \\
x^3 - x & = & 1 \\
\tan C - \tan A & = & \boxed{1} \end{eqnarray*}

I'm wondering if there's a prettier solution.[/hide]

Oh, whoops, I misread the problem; I thought the sides were in arithemetic progression as well as the angles.  My argument is still valid, though, because then by AM-GM, we have
\[ \sin b \le \frac{\sin a + \sin c}{2}, \]
with equality only when $ \sin a = \sin c$.

[quote="lingomaniac88"][hide="Solution to Problem 4"]\begin{eqnarray*} \tan A + \tan B + \tan C & = & \tan A\tan B\tan C \\
\tan A + \tan^2 A + \tan^3 A & = & \tan^6 A \\
\text{Let }x & = & \tan A \\
x + x^2 + x^3 & = & x^6 \\
1 + x + x^2 & = & x^5 \\
x + x^2 & = & x^5 - 1 \\
(x^5 - 1) + x^3 & = & x^6 \\
x^5 + x^3 & = & x^6 + 1 \\
x^3(x^2 + 1) & = & (x^2 + 1)(x^4 - x^2 + 1) \\
x^3 & = & x^4 - x^2 + 1 \\
x^3 - 1 & = & x^4 - x^2 \\
(x - 1)(x^2 + x + 1) & = & x^2(x + 1)(x - 1) \\
x^2 + x + 1 & = & x^3 + x^2 \\
x + 1 & = & x^3 \\
x^3 - x & = & 1 \\
\tan C - \tan A & = & \boxed{1} \end{eqnarray*}

I'm wondering if there's a prettier solution.[/hide][/quote]
Well, this is simpler, I think.
[hide]Let us abbreviate $ k = \tan A$.  From the identity
\[ \tan A \tan B \tan C = \tan A  + \tan B + \tan C, \]
we have
\[ k^6 = k^3 + k^2 + k . \]
We notice that if we subtract $ k^2$ from both sides, both sides are divisible by $ k^3 +k$.  Motivated thus, we have the factorization
\[ (k^3+k)(k^3-k-1) = 0 . \]
Since $ k$ is the tangent of the smallest angle in a triangle, $ k>0$; hence $ k^3+k > 0$.  It follows that $ k^3-k-1=0$, or
\[ k^3 - k = 1, \]
as desired.[/hide]

The problem was from this year's Kennesaw University Math Competition.

[hide="More style advice"]
Use align* instead of eqnarray*, which is outdated.  You use a slightly different syntax; instead of
[code]\begin{eqnarray*}
a &=& b \\
&=& c \end{eqnarray*}[/code]
you use this:
[code]\begin{align*}
a &= b \\
&= c \end{align*} , [/code]
and the output is nicer.  If you wonder why align is so much better, read [url=http://www.tug.org/pracjourn/2006-4/madsen/]this article[/url].

Also, a good solution is usually more than just a list of equations.  Usually, it should be more like a running commentary punctuated by equations.  I recommend [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]How to Write a Solution[/url].[/hide]

Here's an old ARML individual event problem:

[size=150]Problem 5[/size]

Four consecutive terms in the binomial expansion of $ (a\plus{}b)^n$ are $ 2916, 4860, 4320,$ and $ 2160$. If $ a, b,$ and $ n$ are integers (and $ n$ is positive), compute the next term in the expansion.

[hide]I assumed that this was similar to finding the coefficients of the expansion of $ (ax\plus{}by)^n$.

For some value of $ r$, we have $ \binom{n}{r}a^r b^{n\minus{}r}\equal{}2916\equal{}2^2\cdot 3^6$ and $ \binom{n}{r\plus{}1}a^{r\plus{}1} b^{n\minus{}r\minus{}1}\equal{}4860\equal{}2^2\cdot 3^5\cdot 5$.  The second one can be changed into:
$ \binom{n}{r}\left(\frac{n\minus{}r}{r\plus{}1}\right)a\cdot a^r\cdot\frac{b^{n\minus{}r}}{b}$

If we combine these equations, we get $ \frac{n\minus{}r}{r\plus{}1}\cdot\frac{a}{b}\equal{}\frac{5}{3}$, which will come in handy later. $ (*)$

We got a 5 somewhere in the second equation.  It couldn't have come from multiplying by $ a$ (otherwise the first number would be a multiple of 5).  Thus it came from the $ n\minus{}r$, hence $ 5|n\minus{}r$.

And now we take a contest-problem-style gamble:

The prime factorizations seem to suggest that $ b\equal{}3$.  Thus we can say $ n\minus{}r\equal{}5$.  Substitution back into $ (*)$ yields $ \frac{a}{r\plus{}1}\equal{}1\Rightarrow a\equal{}r\plus{}1$.

We have $ \binom{n}{r\plus{}1}a\cdot a^r\equal{}2^2\cdot 3\cdot 5$.

The first equation suggests that $ a$ is a power of 2.  $ a\equal{}4\Rightarrow r\equal{}0$, so this doesn't work.  $ a\equal{}2\Rightarrow r\equal{}1$, which works nicely.  Thus $ n\equal{}6$.

We are given the terms when $ r\equal{}1,2,3,4$.  Thus when $ r\equal{}5$, $ \binom{n}{r}a^r b^{n\minus{}r}\equal{}6\cdot32\cdot3\equal{}\boxed{576}$.[/hide]

Someone else can post the next problem.

[size=150]Problem 6[/size]

For $ 0<x<2\pi$, find the difference between the solutions to
\[ \sec x \equal{} 1\plus{}\cos x\plus{}\cos^2 x\plus{}\cos^3 x \plus{} \cdots
\]

[hide=" Problem 6"]
Since $ | \cos x | \le 1$, the right side is a convergent geometric series.
So, $ \frac {1}{\cos x} \equal{} \frac {1}{1 \minus{} \cos x} \Rightarrow \cos x \equal{} 1 \minus{} \cos x \Rightarrow \cos x \equal{} \frac {1}{2}$
Thus, the solutions $ x \in (0,2\pi)$ are $ x \equal{} \frac {\pi}{3} , \frac {5\pi}{3}$ which have a difference of $ \frac {4\pi}{3}$.
[/hide]

[b]Problem #7[/b]
$ x^5\plus{}3x^4\minus{}3x^3\minus{}10x^2\plus{}4x\plus{}15 \equal{}$ $ a_0\plus{}a_1(x\plus{}1)\plus{}a_2(x\plus{}1)^2\plus{}a_3(x\plus{}1)^3\plus{}a_4(x\plus{}1)^4\plus{}a_5(x\plus{}1)^5$
is an identity for some constants $ a_0,a_1,a_2,a_3,a_4,a_5$. Find $ a_1\plus{}a_3\plus{}a_5$.

[hide="Problem 7"]Substituting $ x \equal{} 0$ yields $ 15 \equal{} a_0 \plus{} a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5$
Substituting $ x \equal{} \minus{} 2$ yields $ 7 \equal{} a_0 \minus{} a_1 \plus{} a_2 \minus{} a_3 \plus{} a_4 \minus{} a_5$
Subtracting these two gives $ 8 \equal{} 2(a_1 \plus{} a_3 \plus{} a_5)$, so our answer is $ \boxed{4}$[/hide]

[size=150]Problem 8[/size]
Find $ \sum_{k \equal{} 0}^{49}( \minus{} 1)^k\binom{99}{2k}$ where $ \binom{n}{j} \equal{} \frac {n!}{j!(n \minus{} j)!}$

By the way, I'm not totally sure of the source of the problem... I just found it on an old worksheet.

[hide]
$ \Re (1 \minus{} i)^{99} \equal{} {99\choose 0} \minus{} {99\choose 2} \plus{} ... \minus{} {99\choose 98}$
Also, we know that that is an eighth root of unity to the 99th power, multiplied by $ ( \sqrt {2})^{99}$.
It's the seventh eighth root of unity, and $ 99\equiv 3 \pmod{8}$, so it ends up at the 5th 8th root of unity.  Then we multiply this by $ 2^{49} \sqrt {2}$ and get $ \minus{} 2^{49} \minus{} 2^{49}i$.  The real part of this is $ \minus{} 2^{49}$.  So thats our answer.[/hide]
WHOOPS!

[quote="mgao"]Also, we know that that is an eighth root of unity to the 99th power, multiplied by $ ( \sqrt {2})^{99}$.[/quote]
[quote]Then we multiply this by $ 2^{98} \sqrt {2}$ and get $ \minus{} 2^{98} \minus{} 2^{98}i$.  The real part of this is $ \minus{} 2^{98}$.  So thats our answer. (is it right?)[/quote]

SO CLOSE!!!.  But $ ( \sqrt {2})^{99}\neq 2^{98} \sqrt {2}$.  I'll give you a chance to fix your answer...


What is the number of miles per hour in the average speed of a car that travels 210 miles in 3.5 hours?

The formula $ D\equal{}RT$ (D=distance, R=rate, and T=time.) is the key to this problem. So we have the distance and the time. So our equation would be $ 210\equal{}3.5R$. Dividing both sides by 3.5, we get $ 60$ mph.


[quote="NguyenDungTN"][quote="conann"]If a,b are real number and a>0 , b>0 , a+b=1
prove that ab + 1/ab >= 4 + 1/4[/quote]
Set $ ab=x$, we have $ x=ab \le \frac{(a+b)^{2}}{4}=\frac{1}{4}$
Inequality becomes:
$ x+\frac{1}{x}\ge \frac{17}{4}$
Or:
$ (4x-1)(x-4) \ge 0$
We are done![/quote]

pardon me, but i am not quite sure how this solution proves the original statement to its full extent... i would say that it is not a complete proof.
first of all, to answer your question, it is clearly not true that $ ab\geq 4$ since by the AM-GM inequality, $ \frac{(a+b)^{2}}{4}=\frac{1}{4}\geq ab$
Thus, the only case is $ ab\leq \frac{1}{4}$.
Anyways, this is what i would consider a complete solution for this problem:

we are given $ ab+\frac{1}{ab}\geq 4+\frac{1}{4}=\frac{17}{4}$
multiply $ 4ab$ on both sides and simplify:

$ 4a^{2}b^{2}-17ab+4=(4ab-1)(ab-4)\geq 0$

we have either $ ab\geq 4$ or $ ab\leq \frac{1}{4}$ since the first one is not true by AM-GM we just need to consider $ ab\leq \frac{1}{4}$.
Following this inequality, we have $ \frac{1}{ab}\geq 4$.

Now consider $ \frac{1}{ab}=4+x\geq 4$ for some small positive real $ x$. Then it follows that $ ab=\frac{1}{4+x}$

applying these substitutions to our original inequality, we have:

$ ab+\frac{1}{ab}=\frac{1}{4+x}+4+x\geq 4+\frac{1}{4}$

Getting rid of the $ 4$ on both sides we have:

$ \frac{1}{4+x}+x=\frac{x^{2}+4x+1}{4+x}\geq \frac{1}{4}$

Move the LHS to RHS and simplify, we have:
$ \frac{1}{4+x}+x=\frac{4x^{2}+16x+4-4-x}{16+4x}=\frac{4x^{2}+15x}{4x+16}\geq 0$

Since the denominator $ 4x+16$ and the numerator $ 4x^{2}+15x$ are both greater than $ 0$ since we supposed that $ x>0$, the final inequality is valid.
Therefore, our very first inequality must be true: $ ab+\frac{1}{ab}\geq 4+\frac{1}{4}$

[quote="conann"]Thanks for your response.
I understand that ab<= 1/4

Let x=ab
→  x+ 1/x >= 17/4
→ (4x-1)(x-4)>=0
→ x<=1/4 or x>=4
But I don't know that why is ab>=4???[/quote]
Because $ x\leq\frac{1}{4}$ therefore $ 4x-1\leq 0$ and $ x-4<0$ and we're done! :)

furthermore we have another solution , but not really nice
$ ab+\frac{1}{ab}=ab+\frac{1}{16ab}+\frac{15}{16ab}\geq \frac{1}{2}+\frac{15}{4}=\frac{17}{4}$


superior algebra
superior algebra solved

Prove that if $a$ is non-invertible and exist $b$ s.t. $ab=1$, then $a$ is zero divisor.

A very simple example: $a(ba-1)=0$. Since $a$ is not invertible, $ba-1$ is not zero.


How many methods you know to prove that $ \sum\limits_{k\equal{}0}^{\infty }{\frac{(\minus{}1)^{k}}{(2k\plus{}1)^{3}}}\equal{}\frac{\pi^3}{32}$ ?

I am aware of two methods: Fourier series, and power of $ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{(2n\plus{}1)}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223148]Cross-link to another version of the same problem.[/url]


[quote="vikas"]I have 3 marbles and 2 balls.

2 of the 5 objects is blue.

What is the probalbility that atleast 1 marble is blue?[/quote]
[hide]Complementary counting.

There are $ \binom{5}{2}=10$ possible outcomes. There is only one way in which there are no blue marbles, that is when there are 2 blue balls. There is only one way to do this. So we have:

$ 1-\frac{1}{10}= \boxed{\frac{9}{10}}$

This is clearly much easier than casework...[/hide]

[quote="vikas"]I have 3 marbles and 2 balls.

2 of the 5 objects is blue.

What is the probalbility that atleast 1 marble is blue?[/quote]

there are $ \binom{5}{2}(thats 10)$ ways that the marbles/balls can be blue. There is only 1 way where there can be no blue marbles. therefore, the answer is $ \frac{9}{10}$


trigonometry
analytic geometry
graphing lines
slope

I'm not sure where to post this topic.

Please see attachment. The issue is that of an approximation to the acceleration of rainfall down an inclined plane. Can it be approximated by a simple kinematical result?

As long as you are assuming no retarding forces (friction for example) or other unbalanced forces that may otherwuse impact the drop, then the acceleration is simply $ g\sin{\theta}$.

Thankyou for your reply. I imagine that once the surface is wet, subsequent rainfall on the surface will slide as though a lamina, down the incline.

Well once the incline is wet, there probably will be other forces contributing but it's hard to quantify them exactly.

Air resistence will play very noticeable role both on the slide and during the fall (anybody who ever drove a car could notice that at the speed 20mph or so, the drops travel up the windshield, not down, despite the slope is pretty large). Also, the same driving experience shows that the water drops sticks to the windshield pretty well until it gets completely wet. So there are some rather strong forces that keep them in place as long as they touch the surface, not other water molecules.

The travel of the water on the roof during the heavy rain is, most probably, like that of a very shallow mountain river. At any rate, the speed on the slope is rather the equilibrium speed that removes as much water from the roof as there gets there (with the thickness of the water layer proportional to the square root of the product of the intensity of the rain and the distance from the roof top) than the speed given by $ g\sin\theta$ acceleration. This agrees with the trivial observation that during a light rain the water drops from the edge almost vertically and during a heavy rain it shoots out for some noticeable distance. I forgot all fluid dynamics I was taught quite well  :blush: but I'm still pretty sure that what I described is much closer to the truth than the $ g\sin\theta$ model if the top is of any noticeable length. Any decent fluid dynamics textbook will give enough information to recover the details if my model is right.

The last remark is that if your task is to determine the point $ G$ in reality, forget it: even a relatively light wind will most likely move it much more than by the distance you will calculate using the complicated formulae for the free fall with air drag. So, unless you know the exact intensity of the rain and the wind speed and direction, you'll not be able to say anything certain.

As to the right place to post such things, it is Advanced Physics subforum, so I'm moving it there :).


Each row of the Misty Moon Amphitheater has 30 seats. Rows 12
through 22 are reserved for a special group. How many seats are
reserved for this group?

We know each row has $ 30$ seats. Thus, we just need to find how many rows are reserved for the special group. This is $ 22\minus{}12\plus{}1\equal{}11$. Hence our answer is $ 30\cdot11\equal{}\boxed{330}$.

ill answer this in a series of steps.

[hide="THIS IS MY EXPLANATION"]1.the group is from rows 12-22

2.there are 22-12+1=11

3.so there are 11 rows and each row there are 30 seats

4.so we get the equation 30*11=330

5.[hide="ANSWER"]330[/hide][/hide]

[hide="."][size=200][color=#804000][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!![/u][/i][/b][/color][/size][/hide]

:10: I HAVE 10/10 PERCENT CONFIDENCE

All we need to do is find the number of rows, there are 11 and so the answer is 30*11=330.


Could someone add me to the Florida community?  Thanks!  :D

You can request to join a community by going to the usergroups area (button at top of this posting area) and requesting to join in the Florida usergroup.


I really need some of the old state competitions. I really only need the sprint and target rounds but, team and countdown might help too. Please e-mail the links to me at nimishmittal07@yahoo.com along with posting them here. Thanks!


[b]Let a, b, c and x, y, z be nonnegative real numbers.[/b] Then, prove that:
$ x(b \plus{} c) \plus{} y(c \plus{} a) \plus{} z(a \plus{} b)$ $ \geq$ $ 2\sqrt{(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)}$

Rearrangement and then Cauchy-Schwarz:
\begin{align*}RHS & = \sqrt {(xy + yz + zx)(ab + bc + ca)} + \sqrt {(xy + yz + zx)(ab + bc + ca)} \\
& \le\sqrt {(x^2 + y^2 + z^2)(b^2 + c^2 + a^2)} + \sqrt {(x^2 + y^2 + z^2)(c^2 + a^2 + b^2)} \\
& \le(xb + yc + za) + (xc + ya + zb) \\
& = x(b + c) + y(c + a) + z(a + b) \\
& = LHS.\end{align*}
 :blush: (sorry, this solution is wrong)

Unfortunately, the sign from the second line to the third line is reversed.

[quote="swapnilshwetank"][b]Let a, b, c and x, y, z be nonnegative real numbers.[/b] Then, prove that:
$ x(b \plus{} c) \plus{} y(c \plus{} a) \plus{} z(a \plus{} b)$ $ \geq$ $ 2\sqrt {(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186636


[quote="PHONGVAN"]Let a,b,c >0 and $ a^5 \plus{} b^5 \plus{} c^5 \equal{} 3$ Prove that
 $ a^4 \plus{} b^4 \plus{} c^4\geq\ a^2 \plus{} b^2 \plus{} c^2$[/quote]
I think, it's easy.
$ a^4 \plus{} b^4 \plus{} c^4\geq\ a^2 \plus{} b^2 \plus{} c^2\Leftrightarrow\sum_{cyc}(a^4\minus{}a^2\minus{}\frac{2}{5}(a^5\minus{}1))\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}1)^2(2a^3\minus{}a^2\minus{}4a\minus{}2)\leq0.$
$ 2a^3\minus{}a^2\minus{}4a\minus{}2<0$ for all $ 0<a<1.86.$
Id est, if $ \max\{a,b,c\}<1.86$ your inequality is proved.
Let $ \max\{a,b,c\}\geq1.86$ and $ f(x)\equal{}x^4\minus{}x^2.$
Then $ f'(x)\equal{}4x^3\minus{}2x.$ Hence, $ f(x)\geq f\left(\frac{1}{\sqrt2}\right)$ for all $ x>0.$
Thus, $ \sum_{cyc}f(a)\geq f(1.86)\plus{}2f\left(\frac{1}{\sqrt2}\right)>0.$

I think, the following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^8 \plus{} b^8 \plus{} c^8 \equal{} 3.$ Prove that:
\[ a^5 \plus{} b^5 \plus{} c^5\geq a^4 \plus{} b^4 \plus{} c^4
\]
What about the following?
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^{2n} \plus{} b^{2n} \plus{} c^{2n} \equal{} 3,$
where $ n\in\mathbb N.$ Prove that:
\[ a^{n \plus{} 1} \plus{} b^{n \plus{} 1} \plus{} c^{n \plus{} 1}\geq a^n \plus{} b^n \plus{} c^n
\]


function
LaTeX
limit
logarithms
calculus
calculus computations

hi!

i have a problem in finding limit for this funciton f (x) = ((x/e)-x*((x/(x+1))^x)),where x --> infinity. i thought i will be able to use Lopital's rule, but cannot transformate funcition to uncertainity in order to use Lopital's rule. Do anyone have ideas? And I also think that there should be way to find limit without Lopital's rule, using formulas like e^x-1~x, but almost all of them are for x-->0. So maybe i must do y=1/x substitution.
But anyway, i have no ideas for the moment. Can anybody help?

thanks.

Please learn how to use LaTeX on this forum, your english combined with lack of LaTeX makes your post kind of hard to read.

You want to find

$ \lim\limits_{x\to\infty} \frac{x}{e} \minus{} x \left(\frac{x}{x\plus{}1}\right)^x \equal{} \lim\limits_{x\to\infty} x\left( \frac1e \minus{} \left(1 \minus{} \frac{1}{x\plus{}1}\right)^x\right)$ , correct?

[quote="hwstill"]So maybe i must do y=1/x substitution.  But anyway, i have no ideas for the moment. [/quote]  These two sentences seem to be in conflict ;)  Yes, the substitution $ y \equal{} \frac{1}{x}$ seems like a good idea.  Rather than using L'Hopital's rule at that point, I think that I would probably consider the second term and try to write down a few terms of its power series (for example, by taking a logarithm, writing down the power series of the result, then exponentiating; you only need to keep track of 2 or 3 terms).

[quote="Tyl"]Please learn how to use LaTeX on this forum, your english combined with lack of LaTeX makes your post kind of hard to read.

You want to find

$ \lim\limits_{x\to\infty} \frac {x}{e} \minus{} x \left(\frac {x}{x \plus{} 1}\right)^x \equal{} \lim\limits_{x\to\infty} x\left( \frac1e \minus{} \left(1 \minus{} \frac {1}{x \plus{} 1}\right)^x\right)$ , correct?[/quote]

i'm sorry, i'll try to get through that LaTeX.
But yes, that is the correct function.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I am having trouble in solving these two problems from Herstein. Any help would be appreciated.

1.Let $ G$ be a finite group and suppose that $ T$ is an automorphism of G such that $ T^3$ is the identity automorphism. Suppose further that $ T(x)\equal{}x$  implies $ x\equal{}e$. Prove that for every prime p which divides $ o(G)$, the p-sylow subgroup is normal in G.

2.Let $ P$ be a p-sylow subgroup of $ G$ an suppose $ a$, $ b$ are in the center of $ P$. Suppose further that $ a\equal{}xbx^{\minus{}1}$ for some $ x \in G$. Prove that there exists a $ y \in N(P)$ such that $ a\equal{}yby^{\minus{}1}$

I can't solve #1 either. However, I do have a solution for #2. It took me quite a while to come up with, but it is quite nice, so if you have the time I would encourage you to try to find it on your own as well. If not, however, here is a hint:

[hide="hint"]
Suppose that $ xPx^{\minus{}1} \equal{} Q$. Find an element that commutes with $ a$ but sends $ Q$ back to $ P$ through conjugation.
[/hide]

and another hint:

[hide="hint2"]
$ P$ and $ Q$ (as defined in the first hint) are both $ p$-Sylow subgroups of the centralizer of $ a$.
[/hide]

and here is my proof:

[hide]
We have $ a,b \in Z(P)$ and $ a \equal{} xbx^{ \minus{} 1}$. Observe that, since $ a \in Z(P)$, $ P \subset N(a)$, where $ N(a)$ is the centralizer of $ a$. Suppose that $ xPx^{ \minus{} 1} \equal{} Q$. Then, since the center is a characteristic subgroup, $ Z(Q) \equal{} Z(xPx^{ \minus{} 1}) \equal{} xZ(P)x^{ \minus{} 1}$. Thus since $ b \in Z(P)$, $ a \in Z(Q)$. Thus $ Q \subset N(a)$ as well.

Now, $ P$ and $ Q$ are both $ p$-Sylow subgroups of $ N(a)$. Thus they are both conjugate in $ N(a)$. So there exists $ y \in N(a)$ such that $ yQy^{ \minus{} 1} \equal{} P$. Thus $ yxPx^{ \minus{} 1}y^{ \minus{} 1} \equal{} P$, so $ yx \in N(P)$. However, since $ y \in N(a)$, $ yxbx^{ \minus{} 1}y^{ \minus{} 1} \equal{} yay^{ \minus{} 1} \equal{} a$. Thus $ yx$ is the desired element.
[/hide]

For #1: http://www.mathlinks.ro/Forum/viewtopic.php?t=196356 .

  darij


Alan and Peter are competing in a countdown round. Alan is beast, so he wins a problem with probability 1/2. Peter wins a problem with probability 1/3, and no one gets the problem with probability 1/6. They play a game to 4. find the probability that Alan wins.

I believe that this question is inaccurate. clearly alan wins with probability 1 and this kid peter (who sounds like he really sucks) wins with probability 0.

I am not sure; does playing a game to 4 mean the first one to 4 wins?

[hide="hints"]So we know, necessarily, that Alan wins the last problem. Further, Peter wins 0,1,2, or 3 games, while Alan wins 3 problems before he wins the last problem. Finally, we can just pretend that nobody-getting-it doesn't exist, and redo the probabilities of each one winning so that they add to 1.[/hide]


Suppose that $G$ is a finite simple group whose order is not multiple of 12 or a prime. Then the $p$-adic valuation of its smallest prime divisor is at least 3.

Hi Harazi.

Let $p^{k}$ be greatest power of p dividing |G|.
If k=1 we can conclude with [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=117205]this result[/url] that G ist not simple.

If k=2 and G is simple, then let S be a p-sylowgroup which is wlog isomorphic to $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ (if it wasn't it would be cyclic and again we can apply again the above result).
Then look at $N_{G}(S)/C_{G}(S)$ which is isomorphic to a subgroup of $Aut(S)\cong GL_{2}(\mathbb{F}_{p})$. Since S is abelian, $|N_{G}(S)/C_{G}(S)|$ is not divisible by p.
$|GL_{2}(\mathbb{Z}_{p})|=p(p-1)^{2}(p+1)$ and because $|N_{G}(S)/C_{G}(S)|$ is also a divisor of $|G|$ it must be a divisor of p+1.
If $|N_{G}(S)/C_{G}(S)|=1$, we can apply once again the above theorem to show that G is not simple.
If $|N_{G}(S)/C_{G}(S)|>1$, it must be equal to p+1 and prime, because p is the smallest prime dividing |G|. It it wasn't, p+1 and therefore $|N_{G}(S)/C_{G}(S)|$ would be divisible by a prime less than p, which is a contradiction.
Now we have that p is prime and p+1 is prime. It follows p=2 und hence $p^{2}(p+1)=12 | |G|$. This is also a contradiction.

k must therefore be greater than 2.


[color=green]Mod: Please post each of your problems in their own topics. (this is a split)[/color]

What about this here?
x,y,z -----naturals numbers.
h-greatest common divisor of x,y and z
To proof,that
a) $ h*x*y*z$
b)$ h*(y \minus{} x)$
are squares of natural numbers.

What if x, y, and z are primes? Then h=1 and $ h*x*y*z \equal{} x*y*z \equal{}$not a perfect square..

For b) you get also that its not true by finding a simple counter example. Such as when y=5 and x=3, then h=1. So you get $ 1*(5\minus{}3) \equal{} 2$ which is not a perfect square.

[quote="modularmarc101"]What if x, y, and z are primes? Then h=1 and $ h*x*y*z \equal{} x*y*z \equal{}$not a perfect square..

For b) you get also that its not true by finding a simple counter example. Such as when y=5 and x=3, then h=1. So you get $ 1*(5 \minus{} 3) \equal{} 2$ which is not a perfect square.[/quote]
I'am sorry.
x,y,z -----naturals numbers.
h-greatest common divisor of x,y and z
To proof,that
a)  $ h*x*y*z$
b) $ h*(y\minus{}x)$
if $ 1/x\minus{}1/y\equal{}1/z$
are squares of natural numbers.

It's really really late now and here's what I have for a) so far
$ gcd(x',y',z')\equal{}1\implies hxyz\equal{}h^4x'y'z'\equal{}h^4z'^2(x'\plus{}y')$ where $ x'y'\equal{}z'(x'\plus{}y')$
So $ x'\plus{}y'$ must be a perfect square.


analytic geometry
geometry
angle bisector
geometry proposed

Let $AD,BE,CF$ be the internal angle bisectors of $\triangle ABC$. Prove that there exists a point $P$ such that $PD,PE,PF$ are the bisectors of $\angle BPC,\angle CPA,\angle APB$, respectively.

well,since $PD$ is the angle bisector of $\angle BPC$ we get that $\frac{BD}{DC}=\frac{PB}{PC}$ so we get that $\frac{AB}{AC}=\frac{PB}{PC}$ so we get that $P$ most be on the [b]apollonius circle[/b] respect to $BC$...(the circle with diameter $DD'$ where $D'$ is the foot of the external bisector of angle $\angle BAC$...)

and in similar ways we get that $P$ must be on the other two [b]apollonius[/b] circles...

so now the only thing that remains is to show that the three apollonius circles of triangle $\triangle ABC$ are concurrent,which is a well known problem...

well i don't know much about Apollonius circles but here's my solution.

first i start with a problem i discovered: $B,C$ and $L$ are points such that $L\in\overline{BC}$ and $BL\neq CL$. Find the locus of the point $A$ such that $AL$ is the bisector of $\angle BAC$.

WLOG let $BL<CL$. it turns out that the locus is a circle with center on $BC$, passing through $L$ and of radius $\frac{BL\cdot CL}{CL-BL}$(maybe this is what you call Apollonius circle). I used coordinates to get this result. so each point $A$ on this circle satisfies $BA: AC=BL: LC$.

now for our original problem, we must find $P$ that satisfies $AB: AC=PB: PC,BC: BA=PC: PA$ and $CA: CB=PA: PB$. Consider the circle $\omega_{1}$ with center on $BC$ and passing through $A$ and $D$. define $\omega_{2}$ and $\omega_{3}$ in the same way for $B$ and $C$ respectively. the intersection $Q$ of $\omega_{1}$ and $\omega_{2}$ satisfies ${AB: AC=QB: QC}$ and $BC: BA=QC: QA$. multiplying these two equations we get $CA: CB=QA: QB$. this implies that $Q$ must lie on $\omega_{3}$ as well. so $P=Q$ and this completes the proof.$\square$

yeah...

[b]assume points $A,B$,then the locus of points like $P$ where $\frac{PA}{PB}=\frac{r}{s}$ is a circle with diameter $MN$...

(points $M,N$ are two points on segment $AB$ ($M$ inside $AB$ and $N$ outside $AB$) such that $\frac{MA}{MB}=\frac{NA}{NB}=\frac{r}{s}$)[/b]

this circle is called [b]apollonius circle[/b] now for a triangle $\triangle ABC$ the locus of points $P$ such that $\frac{PB}{PC}=\frac{AB}{AC}$ is a circle which is called [b]the apollonius circle respect to $BC$[/b] one can see that for these points we`ll
have $\angle APC-\angle APB=\angle B-\angle C$ to...

[color=blue]I propose to define the point $P$ for which $a\cdot PA=b\cdot PB=c\cdot PC$ as "[b]Nayel's point[/b]". [u]It is a joke, my dear Nayel ![/u]. [/color]


calculus
integration
inequalities
triangle inequality

A triangle is made of wood sticks of lengths 8, 15 and 17 inches
joined end-to-end. Pieces of the same integral length are cut from
each of the sticks so that the three remaining pieces can no longer
form a triangle. How many inches are in the length of the smallest
piece that can be cut from each of the three sticks to make this
happen?

The triangle inequality states that the longest side of a triangle must me shorter than the sum of the other two sides.

We have a triangle of $ 8$, $ 15$, and $ 17$ inches long. We sum the $ 8$ and $ 15$ to get $ 23$. Now, we subtract $ 2$ from $ 23$ every time we subtract $ 1$ from $ 17$ (because we joined the two sides). It becomes apparent that we must do this $ 23\minus{}7$ or $ 6$ times.


use the factor theorm to solve:

$12x^3+4x^2-3x-1$

u dont need to use factor theorem cuz it factors easily anyway by grouping.

[quote="Zenith Xan"][quote="citistarz"][hide="answer"]
$12x^3+4x^2-3x-1=4x^2(3x+1)-(3x+1)$=$(3x+1)(2x+1)(2x-1)$
$(3x+1)(2x+1)(2x-1)=0$
x=$-\frac{1}{3}$, $-\frac{1}{2}$, and $\frac{1}{2}$.
[/hide][/quote]

Correct, but you can't solve because it's not set to 0 (or anything for that matter).[/quote]

Yeah, I think he meant factorise in the original question :P

It was an assumption that it was equal to 0 because it says "solve using factor theorem".  :P


I was wondering if some one remembers a problem something like this:  $ 3\times3 \plus{} 3(3 \plus{} 3) \plus{} 3^3$? Im not sure weather the problem said to evaluate or to find the smallest possible value by adding parenthesis, but the answer was $ \minus{}114$.  I got 0 and so did everyone else in my game.  Any help would be apreciated. Thanks

[quote="zolojetto"]I was wondering if some one remembers a problem something like this:  $ 3\times3 \plus{} 3(3 \plus{} 3) \plus{} 3^3$? Im not sure weather the problem said to evaluate or to find the smallest possible value by adding parenthesis, but the answer was $ \minus{} 114$.  I got 0 and so did everyone else in my game.  Any help would be apreciated. Thanks[/quote]

The problem was actually $ 3^2\plus{}3(3\plus{}3)\minus{}3^3$  This simplifies to $ 9\plus{}18\minus{}27\equal{}27\minus{}27\equal{}0$

[quote="alkjash"]Hmmm...
I saw #H34N1 do the problem and get it right. And he took 17 secs so I'm pretty sure he wasn't meming...
Maybe there's some part of the problem I didn't read.
Can anyone post the complete problem?[/quote]

when I mem (an incorrect answer) I also take 17 seconds because I know that since everyone else is going to get it wrong, it doesnt matter how long I take.  Ill still get in first place since everyone else will put the wrong answer (since theres a misprint).

Also by doing 17 seconds, it makes it appear that it wasnt mem so people wont complain.

thats really strange and messed up, because the answer would appear to be 0.
 :huh:


Hello, I was looking through some books of mine, and I discovered this really difficult problem.

I have twenty statements below. Each one is either true or false. Try to figure it out.

1. The answers to statements 6 and 7 are the same.
2. Statement 1 is false.
3. The answers to statements 4 and 20 are different.
4. The answers to statements 3 and 20 are different.
5. The answer to this statement is different from the answer to statement 19.
6. Statement number 2 is true.
7. Statement 15 is true.
8. The answers to statements 11 and 19 are the same.
9. Statement 10 is true.
10. Statement 13 is false.
11. I'm not allergic to strawberries.
12. Statement number 16 is true.
13. Statement number 12 is true.
14. The answer to this statement is the same as the answer to statement 11.
15. At least half the problems contained in this problem are false.
16. At least half the problems contained in this problem are true.
17. The answers to statements 9 and 4 are the same.
18. Statement number 7 is true.
19. My middle name is Alan.
20. The answers to statements 3 and 4 are different.

I have about 8 clues that will help, and I will give you one every couple days, but only if I see you guys working hard on it, and making good discussion about it. gl

With 5 and 19: if 19 is true, then 5 can't be true, because then it would be false, and it couldn't be false, because then it would be true. Therefore, 19 is false, and 5 can still be true or false.

With 11 and 14: If 11 is false, then if 14 is false, then it's true, and if it's true, then it's false. Therefore, 11 is true.

With 11, 19, and 8: Since 11 and 19 are different, 8 is false.
So far:
1.?
2.?
3.?
4.?
5.?
6.?
7.?
8.F
9.?
10.?
11.T
12.?
13.?
14.?
15.?
16.?
17.?
18.?
19.F
20.?

Interesting logic. I am not going to tell you whether you are right or wrong right now though.  :)

There are 3 possibillities with 3,4 and 20:

3T,4T,20F
3T,4F,20T
3F,4T,20T

Good work finding that group. That should help you later on. Man, i wish that I were solving it too, because it is such a  good problem.  :lol:

Also, just to point out, 15 and 16 either must both be true, or only one be true. But that's probably going to be saved for last.

[quote="AIME15"]19 and 11 must follow some other ones, b/c we don't know them.[/quote]

erabel found that 11 - True, 19 - False

Look at 2nd post on this topic and you'll see  :wink:

Sorry, being stupid as usual...

Wickedestjr's middle name is NOT alan, apparnetly.

I have a few questions:

1) Are you allergic to anything? If yes, please state what it is.
2) What is your middle name?

[quote="tenniskidperson3"]I know where this is from...  :P The Sideways Arithmetic from Wayside School book.  I have the same exact book!

Anyways, here's some [hide="hints."]
Look at 1, 2, 6.  Try to deduce what 7 must be.  What does that tell you about 15 and 16?
After you figure that out, look at 3, 4, 20.  Which solution set works with 15 and 16?
[/hide]
The total is on erabel's post; nothing worth putting up has been deduced yet.[/quote]

Are you serious!!! I didn't think anybody would have that book. I can't believe it. please don't give them too many hints. Give them one like once a week.

Then 18 is false...

1.?
2.?
3.?
4.?
5.?
6.?
7.F
8.F
9.F
10.F
11.T
12.T
13.T
14.?
15.F
16.T
17.?
18.F
19.F
20.?

This is everything we know:

2 and 6 are the same thing, 1 is the opposite
two statements out of 3, 4 and 20 are true. 17 is true if 4 is false, and false otherwise.
5 can be anything at all.
14 can be anything at all.

Now, we have that 9 are guaranteed falses (the 7 we already have, plus 1, plus one of (3, 4, 20)).
As we can only have 9 false statements (according to 15/16), then

1.F
2.T
3.T
4.F
5.T
6.T
7.F
8.F
9.F
10.F
11.T
12.T
13.T
14.T
15.F
16.T
17.T
18.F
19.F
20.T


[quote="tonyteo"][quote="quykhtn-qa1"][quote="tonyteo"]Let {a_n}(n>=3)  : a_i >= 0 ( i = 1,2,...,n), (a_1)^2 + (a_2)^2 + ... + (a_n)^2 = 1.Prove that :
a_1 + a_2 + ... + a_n >= (a_1.a_2 + a_2.a_3+...+a_n.a_1).sqrt 3 :rotfl:[/quote]
This inequality in Mathematics and youth Magazine.You shouldn't post it .
Please moderator lock this topic[/quote]
What ??  "Mathematics and youth Magazine" ???? I don't know it, this inequality in a new ebook of my friend[/quote]
it's "Toán học và tuổi trẻ",you don't know huh?? :lol:


calculus
conics
parabola
derivative
analytic geometry
parameterization
calculus computations

Find the equations of both lines that pass through the point (2, -3) and are tangent to the parabola y = x^2 + x.

The answer is 11x - y - 25 = 0 and x + y + 1 = 0

I don't know exactly what i am supposed to do...

Suppose  equations of  lines is $y=k(x-2)-3$, we have system
\[k(x-2)-3=x^{2}+x,\]
\[k=2x+1.\]
has root.

Christine, can you use the differentiation?

now the parabola $y$ $=$ $x^{2}+x$
can be rewritten

 $y+\frac{1}{4}$ $=$ $(x+\frac{1}{2})^{2}$

now shift the coordinate system to $(\frac{-1}{2},\frac{-1}{4})$

and use the formulae for the equation to tangent to the general parabola
$x^{2}$ $=$ $4ay$

which is $x$ $=$ $my$ $+$ $\frac{a}{m^{2}}$

The method I would use:
Choose a generic point on the parabola, and find the tangent line through that point. This tangent line should depend on one parameter, which specifies the point of tangency.
We want this line to pass through $(2,-3)$; make that into an equation, and solve for the parameter. Once you have the solutions, give the line corresponding to each solution.


algebra
polynomial
Vieta
product of roots

[quote="surge"]Roots of unity just refer to the $n^{th}$ roots of one right? Not any other number...

The sum of the roots of unity is zero and the product is -1 if we have even roots and 1 if we have an odd number of roots. Can anyone of this be generalized for the $n^{th}$ roots of any number? (I don't think so.)[/quote]

Well, the $n^{\textrm{th}}$ roots of unity are the solutions of $z^{n}= 1$. Rewrite it as $z^{n}-1 = 0$ and then use Vieta to prove the things you said about their sum and their product.

So, the $n^{\textrm{th}}$ roots of, say, $u$ are just the solutions of $z^{n}-u = 0$, so their sum is zero ($n > 1$), and their product is $(-1)^{n+1}u$.

Meet [url=http://mathworld.wolfram.com/VietasFormulas.html]Vieta[/url].

[quote="surge"]Roots of unity just refer to the $n^{th}$ roots of one right? Not any other number...

The sum of the roots of unity is zero and the product is -1 if we have even roots and 1 if we have an odd number of roots. Can anyone of this be generalized for the $n^{th}$ roots of any number? (I don't think so.)[/quote]
[hide] We're solving the polynomial $x^{n}-k$ to find the $n^{th}$ roots of $k$. So, we can just use the formulas for finding sum and product of roots [/hide]
EDIT: Sorry, didn't see your post Kurt Godel. :D


Are you guys ready for nationals?
Only a few more days....let the countdown begin!

heh... wow... i feel like im gonna die at nationals... -_-... kevin chen and bobby shen better get sick :D

Denote x to be my ranking last year.
If my rank is less than or equal to $\varphi x$ then I'll be happy :D


what is the harder?
apmo(geometry) or imo shortlist(geometry)
I think imo-sl(geometry) is harder...
what do you think?
please answer this poor question ,
thank you... :juggle:

Imo-sl: there are problems which are usually proposed for 1st and 4th, and there are some proposed for 3rd and 6th, so various difficulties.


number theory
prime numbers
algebra unsolved
algebra

Find all triples $(p, q, r)$ of prime numbers such that their pairwise differences
$|p-q|, |q-r|, |r-p|$
are also prime numbers.

This is a bit easy for Algebra Unsolved, I think.

WLOG assume $p<q<r$. If all the numbers are odd, then $q-p, r-q, r-p$ are all even and $q-p\geqslant 2, r-q\geqslant 2, r-p\geqslant 4$, hence they can't be all prime. Therefore $p=2$, which gives $q-2\in\mathbb{P}, r-2\in\mathbb{P}$, and because of $q,r$ are odd primes, we have $2|r-q\land r-q\in\mathbb{P}\iff r-q=2$. Hence $q-2,q, q+2$ are all primes, and it's well-known and easily provable that only such triplet is $3,5,7$ (just consider primes of the form $6k\pm 1$). Therefore the only solutions are $(2,5,7)$ and permutations.


I coulda sworn #12 was on the AIME one year...Oh well, very doable with ample time even if you don't see the magic.

A lot of these are from the AIME/last problems of AMC12. Are you sure these are in the getting started scope?

Since the contest is over, can people  post solutions to the problems here (of course, it doesn't count)?

Last few problems were quite challenging.
I will post my solution to #14 since many beginners might not be familiar with inequalities.

Let [tex]k=\frac {3x}{2y}+\frac {4y}{z}+\frac{12z}{7x}[/tex]

multiply by 14xyz on both side

[tex]14xyzk= 21x^2z+56xy^2+24z^2y [/tex]
By Arithmetic Mean/ Geometric mean Inequality
[tex]14xyzk \geq 3 {((21)(56)(24){(xyz)}^3)}^\frac{1}{3}[/tex]
[tex]14xyzk \geq 12xyz ({441}^\frac{1}{3})[/tex]
  [tex]k \geq \frac{6}{7}({441}^\frac{1}{3})[/tex]
Therefore, the minimum value of k is [tex]\frac{6}{7}({441}^\frac{1}{3})[/tex]

You accidentally put +'s instead of x's.  But anyways, [tex]\frac{6}{7}\sqrt[3]{441} = 3*\sqrt[3]{\frac{72}{7}}[/tex]

I think our results are the same

[tex]\frac{6}{7} {441}^\frac{1}{3}=\sqrt [3]{\frac{72}{7}}, [/tex]but I don't think your result is simplified.

I don't understand  why I got 101/14 on number 14. I multiplied each term by 1 so that they were all in a common denominator and then got 101/14. Please tell me what I did wrong.

[quote="vinny919"]I multiplied each term by 1 so that they were all in a common denominator[/quote]

How did you do this part?

[quote="Rep123max"]13. If [tex]a<b<c<d<e[/tex] are consecutive positive integers such that [tex]b+c+d[/tex] is a perfect square and [tex]a+b+c+d+e[/tex] is a perfect cube, what is the smallest possible value of c?[/quote]

[tex]a+b+c+d+e = 5c,~b+c+d = 3c[/tex]

[tex]3m^3 = 5n^2[/tex]

Can't get the rest to [tex]\TeX[/tex] right, but it's not too hard from here.


Hi guys!

here's the [i]second[/i] round of the [b]"Oliforum contest[/b]"! You can find rules and problems of first round [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=227127]here[/url] :lol: .

i'll try to sum up again all details about it, [i]please be carefully[/i].

[b]1-[/b] The contest is made of some rounds(I still think 3 in total).

[b]2-[/b] Each round is made of only 3 problems, and i will give points 0 to 7: [u]6 for all correct solutions, 7 for [b]only[/b] elegants and correct solutions[/u] (similar to ML contest).

[b]3-[/b] It is tought for high-school student, and also for undergraduate or higher student, but [i]you cannot use non-elementary-theory[/i], otherwise the solution is invalid (e.g. in the first round someone uses calculus or link at other solutions, please don't do that!)

[b]4- [/b]All begins from italian-oliforum and the classification is unique (it will be made only of contestant with score $ > 7$ points ).

[b]5-[/b] The second round starts on 12:00 of 4 october and it ends on 24:00 of 5 october, so saturday and sunday.
[b]NB. [/b]All times (date and hours) are according to [i]ROME MERIDIAN (+2GTM), please attention! [/i]

[b]6- [/b]Enrollment is not necessary, it sufficies to send me solution.

[b]7- [/b]I’ll take in consideration time of arriving solutions [i]only[/i] for the first place [i]if [/i]there are more full score 21/21 points (as i did in first round).

[b]8-[/b] [i]Way of sending solutions. [/i]
- You need to send solutions to both email address: pao_leo88@hotmail.it and leonettipaolo@gmail.com in a unique pdf file (so, with attachment) of a right size;
- The pdf file must be written with Latex (or a way that I can understand);
- You rename the pdf file as your nickname (the same of yours in ML);
- You need to write nothing in the email;
- Try to be [i]clear[/i] in solutions! I’ll estimate it!

{If [i](and only if[/i]) you are not able to create a pdf in latex model, you can send me a email in .doc format with solutions, as latex or in a way that i (and all correctors) can understand}.

I'll put below this topic the list of problems on the morning of saturday 4 october, at 12:00 (+2GTM).
Later, I'll make a list still below this topic where i'll write all names of who sent me solutions in right way.

For all doubts and questions, ask me in mp please, for clearity of this topic

Thanks for the attentions
And good luck

Paolo Leonetti
[url=http://olimpiadi.ing.unipi.it/oliForum/viewtopic.php?t=11877]Italian version[/url]

[b]OLIFORUM CONTEST- ROUND 2[/b]

[b]Problem 1[/b]
Let $ p>3$ be a prime. If $ p$ divides $ x$, prove that the equation $ x^2\minus{}1\equal{}y^p$ does not have positive integer solutions.

[b]Problem 2[/b]
Let $ \{a_n\}_{n \in \mathbb{N}_0}$ be a sequence defined as follows: $ a_1\equal{}0$, $ \displaystyle a_n\equal{}a_{[\frac{n}{2}]}\plus{}(\minus{}1)^{n(n\plus{}1)/2}$, where $ [x]$ denotes the floor function. For every $ k \ge 0$, find the number $ n(k)$ of positive integers $ n$ such that $ 2^k \le n < 2^{k\plus{}1}$ and $ a_n\equal{}0$.

[b]Problem 3[/b]
Let $ C_1,C_2$ and $ C_3$ be three pairwise disjoint circles. For each pair of disjoint circles, we define their internal tangent lines as the two common tangents which intersect in a point between the two centres. For each $ i,j$, we define $ (r_{ij},s_{ij})$ as the two internal tangent lines of $ (C_i,C_j)$. Let $ r_{12},r_{23},r_{13},s_{12},s_{13},s_{23}$ be the sides of $ ABCA^1B^1C^1$.
Prove that $ AA^1,BB^1$ and $ CC^1$ are concurrent.

(You can see the figure at this [url=http://img407.imageshack.us/my.php?image=aaaaafz3.png]link[/url])

[i]For any doubts contact me in mp e/o email
Good work [/i] :wink:

Here the list of contestant who sent me solutions begins:

Athinaios

Broniran

pak-man

Akashil

dima_ukraine

kaymar 1991

mod_2

giove

julio14

Rigel

elendil

sumita

[..]

You can also post your solutions here, waiting the gradation..

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229784]Problem 1[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229782]Problem 2[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229783]Problem 3[/url]

Here it is the gradation (with scores $ \ge 6$ points, so at least one problem correct):

[b][size=150]I . Giove (10pp)

II. sumita (9pp)[/size]

III. ex-aequo: Akashnil ,  mod_2 (7pp)

V. julio14 (6pp)[/b]

Thanks for partecipating also to:

[i]Rigel

pak-man

Kaymar1991

broniran

dima_ukraine

elendil

athinaios[/i]

* [u]Honorable mention to sumita[/u]: he's the only who solved the first problem
* [u]Honorable mention to giove [/u]: he's the only who gets at least one point on third problem (and getting nice observations about the first ptoblem)

(in those days we'll send email with personal score to all contestants)

Sum of partial scores (Iround+IIround with scores > 13 points)

[b]I: giove (31)

II: sumita (29)

III: Akashnil (28 )

IV: julio14  (26)

V: Kaymar1991 (23)[/b]

[b]VI, ex-aequo[/b]: silouan, tomci0 (21)

[b]VIII:[/b] Rigel (18 )

[b]IX:[/b] A Cave (16)

[b]X: [/b]mod_2 (14)

Hi. Sorry for not participating to the 2nd round... I had no internet for 9 days because i moved to another city (where the University is)

Does anyone know when the last round is?

hi Nicknafplio! dont worry..
however next round will be on next month, now i'm too busy with exams, i'll contact for more details..


inequalities
geometry
inradius
geometry proposed
100

hello, after the AM-GM inequality we have
$ \sqrt{\frac{ab(p\minus{}c)}{p}}\plus{}\sqrt{\frac{ca(p\minus{}b)}{p}}\plus{}\sqrt{\frac{bc(p\minus{}a)}{p}}\geq3\sqrt[3]{\sqrt{\frac{a^2b^2c^2(p\minus{}a)(p\minus{}b)(p\minus{}c)p}{p^4}}}$
since $ \sqrt{p(p\minus{}a)(p\minus{}b)(p\minus{}c)}\equal{}pr$ and $ abc\equal{}4AR\equal{}4prR$ we get that our formula above is greater or equal to
$ 3\sqrt[3]{4r^2R}$ and this greater or equal to $ 6r$ iff $ R\geq2r$.
Sonnhard.

[quote=April]Let $ a$, $ b$, $ c$ be the lengths of some triangle's sides; $ p$, $ r$ be the semiperimeter and the inradius of triangle. Prove an inequality \[ \sqrt{\frac{ab(p \minus{} c)}{p}}\plus{}\sqrt{\frac{ca(p \minus{} b)}{p}}\plus{}\sqrt{\frac{bc(p \minus{} a)}{p}}\ge 6r\]

[i]Author: D.Shvetsov[/i][/quote]
https://artofproblemsolving.com/community/c6h1738187p11292085

[quote=April]Let $ a$, $ b$, $ c$ be the lengths of some triangle's sides; $ p$, $ r$ be the semiperimeter and the inradius of triangle. Prove an inequality \[ \sqrt{\frac{ab(p \minus{} c)}{p}}\plus{}\sqrt{\frac{ca(p \minus{} b)}{p}}\plus{}\sqrt{\frac{bc(p \minus{} a)}{p}}\ge 6r\]

[i]Author: D.Shvetsov[/i][/quote]

Because:

\[\sum{\sqrt{\frac{ab(p \minus{} c)}{p}}\cos{\frac{B-C}{2}}}=\left( {\frac {b+c}{a}}+{\frac {c+a}{b}}+{\frac {a+b}{c}} \right) r\]


someone please help me with this one! :]
find the area of a ring formed by two circles. one circle is inscribed and one is circumscribed in a square with sides= 12.

[hide]
Since the side of the square is $ 12$, the radius of the small circle is $ 6$.
The area of the small circle is $ 36\pi$

The radius of the big circle is simply half of the diagonal of the square.
The diagonal of the square (d) is:

$ 12^2 \plus{} 12^2 \equal{} d^2$
$ 144 \plus{} 144 \equal{} d^2$
$ 288 \equal{} d^2$
$ d\approx17$

Thus the radius of the big circle is $ 8.5$
and it's area is $ 72.25\pi$

Therefore the area of the ring formed by the 2 circles is $ 72.25\pi \minus{} 36\pi \equal{} \boxed{36.25\pi}$[/hide]
hope that helps  :)

if could elaborate any further on the last solution, it would be the fact that

the radius of the bigger circle is $ 6\sqrt{2}$

then the rest should be simple

just the area of the big circle - area of the small circle


1) Prove that (a, b, c) = ((a, b), c).

2) Let b and g > 0 be given integers. Prove that the equations
(x, y) = g and xy = b can be solved simultaneously if and only if
g^2\b.

1) Consider three sets $ A, B, C$ which contain all positive divisors of $ a, b$ and $ c$ respectively. Then $ A\cap B\cap C\equal{}(A\cap B)\cap C$ and therefore the maximum element in $ A\cap B\cap C$ must equal the maximum element in $ (A\cap B)\cap C$.

2) Since $ (x,y)\equal{}g$, we may write $ x\equal{}gX, y\equal{}gY$, where $ (X,Y)\equal{}1$. Then $ xy\equal{}g^2XY$ which implies $ g^2$ must divide $ b\equal{}xy$. Similarly if $ g^2\mid b$ then we can find $ X,Y$ such that $ XY\equal{}b/g^2$ and $ (X,Y)\equal{}1$.

cool thanks a lot nayel.....

anyways anyone having solutions for niven or any students manual for the same book?


(-3/x+3y)+(4/x-2y)=9   find the point(s) of intersection.

i keep coming across problems like these... i know you are supposed to make m=1/(x+3y) and n=1/(x-2y) but i dont know where to go from there...pls and thanks =]

Points of intersection of what?  It's a single graph.

(3/x+3y)+(4/x-2y)=9  and  (5/x+3y)+(4/x-2y)=18  find the point(s) of intersection.

...sorry

just draw a graph and make the lines due to the accumulation of the pints.  It's a lot easier that way.


[code]$\varnothing$[/code] is supposed to be the symbol for empty set on $ \text{\LaTeX}$, but it doesn't seem to be working on TeXnic.  I know I have the right code, because it works on the forum.

[hide="For proof of that statement, click here"]
$ \varnothing$
[/hide]

Is there a special package I need? because TeXnic isn't recognizing it.

Thanks in advance![/code]

Include the following before \begin{document}:

[code]\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}[/code]
amssymb is the actual package that allows for the use of \varnothing, but since you're going to be doing math anyway, you might as well just include the whole shebang. :D

In fact amssymb loads amsfonts automatically so you only need
[code]\usepackage{amsmath}
\usepackage{amssymb} [/code]


linear algebra
matrix
vector
trigonometry

Let $D_n=\det(A_n)$, where
\[ A_n=\left[\begin{array}{rrrrrrr} 2 & -1 & 0 & 0 & \cdots & 0 & 0 \\ -1 & 2 & -1 & 0 & \cdots & 0 & 0 \\ 0 & -1 & 2 & -1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & 0 & \cdots & 2 & -1 \\ 0 & 0 & 0 & 0 & \cdots & -1 & 2 \end{array} \right]_{n \times n} \]
Find $D_n$ in terms of $n$ $(n \in \mathbb{N}, n\geq1)$.

Can someone move this to the appropriate forum?

First, "guess" the answer by trying some small values of $n$. To prove your guess, add the 2nd, 3rd, ..., $n$th columns to the first column. What happens? :)

[quote="Arne"]Can someone move this to the appropriate forum?[/quote]

Moved to Matrix Algebra.

$\det D_n=n+1.$

If you follow Arne's hint (well, I used rows instead of columns, but it's the same idea) ...

Add the sum of the other rows to the top row, and you get a top row with 1's in the first and last entries, and zeros elsewhere. Now do the determinant by minors of that top row. The minor associated with the upper left entry is simply $\det D_{n-1}.$ The minor associated with the upper right entry is the determinant of an upper triangular $(n-1)\times(n-1)$ matrix all of whose diagonal entries are $(-1)$. Thus,

$\det D_n=\det D_{n-1}+(-1)^{n-1}(-1)^{n-1}=\det D_{n-1}+1.$

The recursion, along with routine calculation for small $n$, proves the theoerem.

More fun facts about this matrix. (I'll fix $n$ and just call the matrix $D.$)

Let $v_m$ be the column vector whose $k$th entry is $\sin\left(\frac{\pi km}{n+1}\right).$

Then $Dv_m=\left(2-2\cos\left(\frac{\pi m}{n+1}\right)\right)v_m$

Hence we have found all of the eigenvectors and eigenvalues of $D.$

Since the determinant of the matrix is the product of the eigenvalues, we have the following product identity:

$n+1=\prod_{m=1}^n\left(2-2\cos\left(\frac{\pi m}{n+1}\right)\right) =4^n\prod_{m=1}^n\sin^2\left(\frac{\pi m}{2(n+1)}\right)$


trigonometry
inequalities
geometry proposed
geometry

[quote="Allnames"][quote="Ligouras"]Prove or disprove that
$ \sin(\frac {\pi}{7}) \plus{} \sin(\frac {3\pi}{14}) \plus{} \sin(\frac {5\pi}{21}) > \sin(\frac {25\pi}{42})$ .

_____________
???  :([/quote]
It is true by computer :lol:[/quote]

----------------
The computer is not a mean disposition for the elementary mathematics... I wait for an elegant resolution without computer. Good 2009

 :wink:

\begin{eqnarray*}
\sin \left( \frac{\pi}{7}\right) + \sin \left( \frac{3\pi}{14}\right) + \sin \left( \frac{5\pi}{21}\right)
& > & 3\sin\left( \frac{\pi}{8} \right) = \frac{3}{2}\sqrt{2-\sqrt{2}} \\
& > & \frac{3}{2} \sqrt{0.49} = 1.05 \\
& > & \color{red}{1} \\
& > & \sin \left( \frac{25\pi}{42} \right)
\end{eqnarray*}

$ \frac {1}{3}\left(\frac {1}{7} \plus{} \frac {3}{14} \plus{} \frac {5}{21}\right) \equal{} \frac {5}{12}$

Apply Jensen to $ f(x) \equal{}\sin x$

[quote="kunny"]$ \frac {1}{3}\left(\frac {1}{7} \plus{} \frac {3}{14} \plus{} \frac {5}{21}\right) \equal{} \frac {5}{12}$

Apply Jensen to $ f(x) \equal{} \sin x$[/quote]

[b]Jensen[/b]
\[ \sin \left( \frac {\pi}{7}\right) \plus{} \sin \left( \frac {3\pi}{14}\right) \plus{} \sin \left( \frac {5\pi}{21}\right) < 3\sin\left(\frac {1}{3}\cdot\frac {25\pi}{42}\right)
\]
But, This problem is
\[ \sin \left( \frac {\pi}{7}\right) \plus{} \sin \left( \frac {3\pi}{14}\right) \plus{} \sin \left( \frac {5\pi}{21}\right) > \sin\left(\frac {25\pi}{42}\right)
\]
How do you use Jensen?
And, what is the meaning of $ \frac {5}{12}$?

[quote="kunny"]Sorry, my idea was meaningless.[/quote]

I thought about the same thing first, too.
$ " \ \frac{\pi}{7} \plus{} \frac{3\pi}{14} \plus{} \frac{5\pi}{21} \equal{} \frac{25\pi}{42} \ "$ may have some relation.


superior algebra
superior algebra unsolved

Let $A$ be a ring such that for every $x\in A$ there exist $n$ such that $x^n \in Z(A)$. Prove that $A$ is abelian.

$Z(A)=\{x: xa=ax,\forall a\in A\}$, the center of the ring $A$.


MATHCOUNTS probability calculus factorial

there is a 3x3 grid on a geoboard.  What is the probability that four randomly selected points on the geoboard shown below will be vertices of a square?  Express your answer as a common fraction

erg solution please and I don't think that's right...

incorrect, but show your solution anyways

alright...someone just say how many square there are...and we'll work on the denominator later...

close...the sides of the square don't need to be parallel to the sides of the geoboard...

I've seen the same problem as 7th grade extra credit problem...actually, it was even more difficult since it only said "quadrilateral"

Hint:[hide]tilt the square[/hide]

I believe that is correct. Now, the denominator will be the number of ways to choose 4 vertices from 9 possible vertices. So, what is it?

Hint[hide] think combinations[/hide]

MathFiend wrote:[hide]6[/hide]

yup, that's right. the rest is easy. you're going straight to the answer!

um, it would actually be 9!/5!4!, or 9*8*7*6/4*3*2*1, which would give you a slightly different answer, but that's ok, you've got the general idea.

Could you generalize this to a 4*4 square (i think i saw this version on a nat)?

Could you generalize this to a n*n square?

sorry,

okay, so there are 6 different square. so thats your numerator, then for the denominator, it is 9C4 which is 126. so then 6/126=1/21

if you dont understand the denominator part, just ask

? technically, I think 4*4 isn't generalizing, but oh well, who cares.

and I thought [hide]1/504[/hide] was the answer (same as MathFiend, but he mis-divided)

I mean, can you explain why it's 9*8*7*6/4*3*2*1?

First of all, I'm not a 'he,' I'm a she. that might explain why I'm so stupid. Also, I get the denominator part, I was just thinking the wrong thing.

humbug, i hate it when people say that theyre stupid

Wow! Sexism alert! :)
I mean. there's lots of female math geeks...I mean, [b]mathematicians[/b] :D For example, Grace Murray Hopper. You should know her if you're a programmer.

Well you get my point.

and can anyone explain to me why the denominator is the way it is?
[size=67]who's stupider than you? me! :twisted: </beingajerk>[/size] :)

the denominator would be the number of different ways of choosing 4 dots (to form a square) from 9 dots (the whole geoboard). so then its 9C4 (which is 9 choose 4)=126

do you know that formula thingie?

Nope, I don't even know C, though I know C as in the computer language :)

C denotes choose for the combination thingie. so 3C2=3 choose 2

the formulae thingie is n!/(r!(n-r)!), where n would be 3 in the above, r would be 2, and n-r being the difference 1. so then it would be 3!/(2!(3-2)!)=6/(2 x 1)=3

need any more clarification?

try this one: 4C2

3!

Anyway, aside from that confusing remark, just know you'll learn that in Precalc/Alg II (cant remember)

anyways, would someone like to try generalizing?

Wait, is there another formula? I mean, how could 4!/2!2! be 3! without calculating first?

and I skipped this stuff...I know, I know, I shouldn't have, but if I wanted to study calculus in the US I had to skip at least two grades if not three.

how did you skip if you didnt know the stuff?

I know, I ask that question too :)

The thing is, factorials were on the graduation test but I still got it right because it was really easy, but I haven't been taught about the specifics. I only knew the definition. Heck, I didn't even [b]heard[/b] of factorials until a month before the test :)
Yup, that's the American educational system at it's best :twisted:
I mean, our school is really good in all the rankings, but the inside... :twisted: *evil snicker* :)


topology
vector
real analysis
advanced fields
advanced fields unsolved

Are there any two homeomorphic but non-isometric normed vector spaces ?

Sure. $\mathbb{R}^2$ with the Euclidean norm is not isometric to $\mathbb{R}^2$ with the $\infty$-norm.

Sorry. I meant "homeomorphic but not isomorphic" ...  :blush:

The answer is [i]no[/i] in finite dimensions, since homeomorphic => same dimension => isomorphic.
The answer is [i]yes[/i] for infinite-dimensional spaces. In fact, any two infinite-dimensional separable Banach spaces are homeomorphic [Kadec]. For an explicit example, consider the sequence spaces $\ell_{p}$ and $\ell_{q}$ with $1\le p<q<\infty$, and define $f\colon\ell_{p}\to\ell_{q}$ by \[f(\{x_{n}\})=\{|x_{n}|^{p/q-1}x_{n}\}\] The problem becomes more interesting if one requires the homeomorphism to be uniformly continuous both ways. Still, uniformly homeomorphic spaces may be non-isomorphic [Ribe].


Let be $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$ . Show that :

$ \frac{1\plus{}a\plus{}b}{2\plus{}c}\plus{}\frac{1\plus{}b\plus{}c}{2\plus{}a}\plus{}\frac{1\plus{}c\plus{}a}{2\plus{}b}\ge \frac{15}{7}$

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$ . Show that :

$ \frac {1 \plus{} a \plus{} b}{2 \plus{} c} \plus{} \frac {1 \plus{} b \plus{} c}{2 \plus{} a} \plus{} \frac {1 \plus{} c \plus{} a}{2 \plus{} b}\ge \frac {15}{7}$[/quote]
Thanks, Alex, it's nice but easy  :)
This ineq is equivalent to:
$ \frac {1 \plus{} a \plus{} b}{2 \plus{} c} \plus{} 1 \plus{} \frac {1 \plus{} b \plus{} c}{2 \plus{} a} \plus{} 1 \plus{} \frac {1 \plus{} c \plus{} a}{2 \plus{} b} \plus{} 1\ge \frac {36}{7}$
We have:
$ LHS \equal{} (3 \plus{} a \plus{} b \plus{} c)\left(\frac {1}{2 \plus{} c} \plus{} \frac {1}{2 \plus{} a} \plus{} \frac {1}{2 \plus{} b}\right)$
$ \ge (3 \plus{} a \plus{} b \plus{} c)*\frac {9}{6 \plus{} a \plus{} b \plus{} c} \text{ (Apllying CS Ineq)}$
$ \equal{} \frac {36}{7} \equal{} RHS$
So, my proof is complete  :)


Prove or disprove:

If $x_{i}, i=1,2,...,n, n\geq 2$ are positive real number and it
stands
\[\sum x_{i}^{2}=1\] then \[\sum x_{i}\geq \sqrt{2}\]
Bye

[quote="SpongeBob"]Prove or disprove:

If $x_{i}, i=1,2,...,n, n\geq 2$ are positive real number and it
stands
\[\sum x_{i}^{2}=1 \]
then
\[\sum x_{i}\geq \sqrt{2}\]
Bye[/quote]
FALSE!

Ex: $x_{1},x_{2},...,x_{n-1}\to 0$ and $x_{n}\to 1$ :D

So N.T. Tuan's proof is invalid. That said, I am not sure how to prove/disprove this.

[quote="agolsme"]0 is not a positive real number.[/quote]
$x_{1}=x_{2}=...=x_{n-1}=\frac{1}{k}$ and $x_{n}=\frac{\sqrt{k^{2}-n+1}}{k}$. Now, let $k\to\infty$ !  :wink:

P/S: $k$ indepent for $n$

It is clear by that if $\sum x_{i}^{2}=1$ for positive $x_{i}$, then $x_{i}\leqslant 1$, so $\sum x_{i}\geqslant \sum x_{i}^{2}=1$ and this lower bound is reached for one of $x_{i}$ equal to $1$ and all other equal to $0$. As far as the upper bound is concerned, we have $\sum x_{i}\leqslant n\cdot \sqrt{\frac{\sum x_{i}^{2}}{n}}= \sqrt{n}$ and there is an equality for $x_{1}=\ldots=x_{n}= \frac{\sqrt{n}}{n}$.

Of course, maybe I wrote the sign incorectly, maybe it should stand $\sum x_{i}\leq \sqrt{2}$, I don't remember right now, never mind. Thanks!


I know there is the very well known site, kalva demon.

But as it says, for copyright reasons, they changed some of the wordings which often makes confusion...

Presumably due to my weak comprehension, 7 out of 10 times I find myself solving the wrong thing...  Are there any other compilation of AIME problems that has some decent wording?

Buy them. $1 a piece, original, with a detailed solution packet. Help out AMC.

[quote]we are already paying a lot of cash per contestant...[/quote]

Really? Is that due to the AMC or your school? I pay about $3 per test.

is it $3 per contestant for AMC?  that sounds reasonable.  i thought it was higher, but the school pays for it.

The fee is not a lot. But even if it was, that $1 per test packet is still a good price.

ya.. but the trouble is, I dun live in the US rite now...

I'll be going there soon, but I was wondering whether if there were any online resources...

Try http://www.gotmath.com/aime.pdf

It clearly doesn't have all of the AIMEs (or any solutions) but hope that helps!

[quote="Magnara"]The shipping on those is damned expensive though.[/quote]
Well, they do ship via UPS, which is generally much more reliable then the regular mail, because you basically have real-time tracking, so the orders don't get lost. Also, there is the mounting cost of fuel surcharges and other expenses which they have to pay, seeing that they are a national organization taking orders from all over the country. I think that its worth paying for because they try to offset the shipping costs by keeping the contests at a low price. In fact, I recently ordered some AIME's as well as some AMC 12's myself, which I think is a valuable resource, because you get detailed solutions as well (as opposed to Kalva, or anyother online place where you just get answers, or just sketches of solutions)

[quote="SirErnest"]Where can you get old ASHMEs?[/quote]

Harold Reiter's Homepage: [url]http://www.math.uncc.edu/~hbreiter/[/url]


When DNA replicates, one of the strands is replicated "backwards." This strand is synthesized in Okazaki fragments. Does each fragment need its own RNA primer?

yes.
infact the lagging sequence needs more than 1 primer :D


calculus
derivative
parameterization
analytic geometry
function
calculus computations

Prove that $ \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \equal{} \frac{dy}{dx}$ where both $ x$ and $ y$ can be related through some variable $ t$ (eg/ $ y \equal{} t^{2}, x \equal{} 3t$)

I'm asking this because I have trouble grasping why such an operation is allowed in the following article, and can't conjure up a proof for it (so far I've tried to re-arrange the chain rule but doubt whether the steps I took are valid).

http://en.wikipedia.org/wiki/Polar_coordinates#Differential_calculus

$ \frac{dy}{dt}\equal{}\frac{d}{dt} y(x(t))\equal{} \frac{dy(x(t))}{dx}\frac{dx}{dt}$

If $ \frac {dx}{dt}\neq 0$ then the statement is equivalent to $ \frac {dy}{dt} \equal{} \frac {dy}{dx}\frac {dx}{dt}$, which as ken4000 noted above is merely the statement of the chain rule if we have $ y \equal{} y(x(t))$.  The only potential difficulty is proving that $ y$ can indeed be seen as a function of $ x$ in some neighborhood of any point for which $ \frac {dx}{dt}\neq 0$, but this can be resolved using the [url=http://en.wikipedia.org/wiki/Implicit_function_theorem]implicit[/url] or [url=http://en.wikipedia.org/wiki/Inverse_function_theorem]inverse function theorem[/url].


modular arithmetic
combinatorics unsolved
combinatorics

x and y are real  numbers  with 0<x,y<1 , such that  the $2^{n}$-th digit of x in its decimal representation (after the comma) is the same as the $n$-th digit of y  in its decimal representation(after the comma).  Prove that  if x  is rational then  y  is rational

This is immediate from the fact that the sequence $a_{n}= 2^{n}\pmod k$ is eventually cyclic for any $k$, so the decimal representation of $y$ eventually repeats.


How is the USAMO index calculated if you take the USAMTS? (what is the formula?)

Try the "search" tool... ;)

Anyways, your index if you qualified for AIME through USAMTS is 100 + 10*(AIME Score)

[quote="MCrawford"][quote="4everwise"]So if you were trying to qualify for the USAMO, it would put you at a disadvantage against other AIME qualifiers?[/quote]

Yes.  If your goal is to qualify for the USAMO, you're better off taking an AMC exam than relying purely on your USAMTS score.[/quote]

This depends on your grade level.  If you are 10th grade or below, your qualification will likely depend solely on your AIME score.  Therefore, the USAMTS is essentially the same as the AMC for these students.  Regardless, I would highly recommend doing both.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $\{S_i: i\in I\}$ be a family of supgroups of a group $G$, let $\{S_it_i: i\in I\}$ be a family of right cosets,and let $D=\cap S_i$.Prove that either $\cap S_it_i=\emptyset$ or $\cap S_it_i=Dg$ for some $g$.
[i]
Rotman[/i]

[quote="amfulger"]Assume $M=\cap S_it_i\neq\emptyset$
Let $x\in M$. There exists $g\in G$ such that $x\in Dg$. $D$ contains at least the neutral element $e\in G$, so $D$ is nonempty. $D$ is a subgroup, so the set of right cosets of $D$ covers $G$.
There exists $d\in D$ such that $x=dg$. Then $dg\in S_it_i\ \forall i$, so there exist $s_i\in S_i$ such that $dg=s_it_i$ for all $i$. Since $d\in D\subset S_i$, it is easy to see that $S_it_i=S_ig$, so $M=\cap S_it_i=\cap S_ig=(\cap S_i)g=Dg$.[/quote]

Thank you :)


How many ways can 9 people be grouped into 3's?

Is the answer (9C3X6C3)/3 or (9C3x6C3)/3!. I think it's the latter...i just wanted to check. Thanks!

(9C3 x 6C3 x 3C3) / 3!.
i think it is the latter as well.


Who else is taking the AMC 12 after a full day of school?  What tips do you guys have to deal with it?  I don't know about you, but I'm always exhausted at the end of a school day.  Maybe I should take some chocolate or something to ride the sugar high during the AMCs?

hmm... im not sure i would "ride the sugar high" because, for me, i would crash right in the middle of the test which woudnt be a good thing, but i dont know you.  The best thing that i can think of is just get lots of sleep the night before

I'm taking the 10A and 12B after school...specifically at 7:30 on both days after a 1 hour 20 min ride. I'll just try to get a really good sleep the night before and not to spend too much energy at school.

Sugar is definitely a bad idea, for the AMCs and especially for the longer tests.

You might be tempted by caffeine but I've seen mixed results for its usage. Sometimes the caffeine is too quick to wear out, sometimes its too slow to kick in.

I'd suggest eating a hearty breakfast and lunch (bring your own if your school's lunch is not good quality), and maybe something along the lines of potato chips / carrot sticks before the test. Don't overdo it though, no more than 1 serving.

Get a good night's sleep so you start off the day with a higher base level of energy.

Also, if your parents won't mind signing you out of school, more power to you. (j/k this piece of advice didn't come from me)

/takes AMC's first thing in the morning


Does anyone know what STL documentation is available in the IOI?

I think that a more or less exact copy of the official STL Programmer's Guide ( http://www.sgi.com/tech/stl/ ) should be available. And I think that's what we had in Poland too...


complex analysis
calculus
calculus computations

I'm not sure how to go about evaluating a sum like the following....

$ \sum_{n=1}^{\infty}\frac{ (-1)^{n+1}\cos(n \pi) \sin((2n+1)\frac{\pi}{2})}{n^{2}}$

I would just simplify it first.

$ \cos(n \pi)=(-1)^{n}$ and $ \sin((2n+1) \frac{\pi}{2})=(-1)^{n}$ so the sum just reduces to:  $ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}$ which is known to equal:  $ \frac{\pi^{2}}{12}$.

If you are asking how to calculate that last sum, I would use complex analysis to show $ \sum_{n=1}^{\infty}\frac{1}{n^{2}}= \frac{\pi^{2}}{6}$ by looking at the residues of $ \frac{\pi \cot (\pi z)}{z^{2}}$ over a suitable contour.  From this sum you can deduce that the sums of over the evens or the odds will be half as much which is enough information to deduce the value of your sum.


Can somebody please post the answers for the AMC 12A?

17. A
19. B
22. B
23. B

All after the test...

D
B
B
A
B
B
C
D
A
B
E
D
D
D
C
D
A
A
B
E
C
B
B
B
C

Just follow them in that order.

Does anyone have a scanned version of test I can have?

If so, please do not post it here (for copyright implications).


Suppose I'm solving a geometry problem. How do I do this?

[img]http://img518.imageshack.us/img518/1859/erreyrtyrk0.png[/img]

(Solution in words, of course.)

[quote="stevem"]Use minipages - something like
[quote]\begin{minipage}{3in}
\includegraphics{...}
\end{minipage}
\hfill
\begin{minipage}{3in}
solution goes here
\end{minipage}[/quote][/quote]

or even use "0.3\linewidth" and "0.6\linewidth" respectively to replace the "3in".


Prove that any three consecutive odd numbers are pairwise relatively prime.

I believe using the Euclidean Algorithm on the three possible pairs works.

[hide="Proof"]If the numbers are $ a$, $ a\plus{}2$, $ a\plus{}4$:

Pair 1: $ (a,a\plus{}2)\equal{}(a,2)$, which is equal to 1 because $ a$ is odd.
Pair 2: $ (a,a\plus{}4)\equal{}(a,4)$, which again must equal 1.
Pair 3: $ (a\plus{}2,a\plus{}4)\equal{}(a,2)\equal{}1$

And we are done.
[/hide]


Consider the dark square in an array of unit squares, part of which
is shown. The first ring of squares around this center square
contains 8 unit squares. The second ring contains 16 unit squares.
If we continue this process, the number of unit squares in the
$ 100^{th}$ ring is

[asy]fill((2,2)--(7,2)--(7,7)--(2,7)--cycle,gray(0.8));
fill((3,3)--(6,3)--(6,6)--(3,6)--cycle,gray(0.6));
fill((4,4)--(4,5)--(5,5)--(5,4)--cycle,black);
for (int i=0; i<10; ++i) {
draw((0,i)--(9,i));
draw((i,0)--(i,9));
}[/asy]

Each new ring has 8 more squares, so it's an arithmetic sequence. $ 99 \times 8 \plus{} 8 \equal{} \boxed{800}$


algebra
polynomial
calculus
integration
superior algebra
superior algebra solved

Prove that for any $n>1$ the polynomial $X^{n}+X+3$ is irreducible over $\mathbb{Q}$.

What about the coefficient of $X$ in the new polynomial. It doesn't seem to be divisible by $p$. Can somebody clear this thing out ?

[quote="yannick"][quote="sissa"]Apply the Eisenstein criterion to $(X+3)^{n}+(X+3)+3$: It has constant coefficient $3\cdot(3^{n-1}+2)$, which is not divisible by $3^{2}$.

This works for $X^{n}+X+p$ with $p$ prime $\neq 2$.[/quote]

I don't understand why checking that $X^{n}+X+3$ is irreducibel is equivalent to checking that  $(X+3)^{n}+(X+3)+3$ is irreducibel. Can someone explain this to me please.

Thanks :)[/quote]
Assume that $(x+3)^{n}+(x+3)+3$ is irreducible while $x^{n}+x+3$ is not. Then let us look at the factors a(x) and b(x)  of $x^{n}+x+3$ where a(x)*b(x) = $x^{n}+x+3$ and neither is constant in x. Clearly a(x+3)*b(x+3) = $(x+3)^{n}+(x+3)+3$, so one of the two must be a constant - but then it would be a constant in x, so the assumption is wrong, and if $(x+3)^{n}+(x+3)+3$ is irreducible, then $x^{n}+x+3$ is irreducible.

in general, $f \in R[X]$ is irreducible if and only if $f(aX+b) \in R[X]$ is irreducible, where $a \in R^{*}, b \in R$ are fixed. instead of dealing with factorizations to prove this, just remark that $X \to aX+b$ induces (e.g. by the universal property of $R[X]$) an automorphism of $R[X]$.

Finister is right; Sissa's proof is not a proof!

If one knows some bounds on the coefficients (like here because they are constant), one can try Rouche's theorem:

Assume that $x^{n}+x+3=a(x)b(x)$, w.l.o.g. $a(x)$, $b(x)$ integral and monic with $a(0)=\pm 1$, $b(0)= \pm 3$.
By Rouche's theorem, there are no roots of $x^{n}+x+3$ in or on the unit circle, thus the same holds for it's factor $a(x)$. But the product of all roots of a monic polynomial is the constant coefficient, and the product of numbers of (complex) norm $>1$ has again norm $>1$, thus $|a(0)|>1$, contradiction.

Smart Zetax; your proof works for x^n+x+p where p is a prime>=3.

I just wanted to say that we can replace Rouche's theorem by simply considering the modulus.


[quote="Nameless"]a,b,c>0
abc=1.
Prove that: $\sum \frac{a^{4}-1}{(a^{4}+1)a^{2}}\ge0$[/quote]
try $a-\to 0,b,c\to \infty$


geometry
parallelogram
geometry unsolved

The tangent lines from a point $ P$ meet a circle $ k$ at $ A$ and $ B$. Let $ X$ be an arbitrary point on the shorter arc $ AB$, and $ C$ and $ D$ be the orthogonal projections of $ P$ onto the lines $ AX$ and $ BX$, respectively. Prove that the line $ CD$ passes through a fixed point $ Y$ as $ X$ moves along the arc $ AB$.

Denote M be the midpoint of AB and N be the mipoint of PM. Since two cyclic quadrilaterals PCMA, PDMB, we have the parallellogram PCMD, then CD always passes through N.


algorithm
number theory
greatest common divisor
Euclidean algorithm

For every integer n prove that the fraction

21n + 4
---------
14n + 3

cannot be reduced any further.

[hide="Solution"]
Use the Euclidean Algorithm (I will use the notation $ (a,b)\equal{}\text{gcd}\ (a,b)$)

$ (21\plus{}4,14n\plus{}3)\equal{}(14n\plus{}3,7n\plus{}1)\equal{}(7n\plus{}2,7n\plus{}1)\equal{}1$, so $ (21n\plus{}4,14n\plus{}3)$ is $ 1$, which implies the desired result.
[/hide]

[quote="AIME15"][hide="Solution"]
Use the Euclidean Algorithm (I will use the notation $ (a,b) \equal{} \text{gcd}\ (a,b)$)

$ (21 \plus{} 4,14n \plus{} 3) \equal{} (14n \plus{} 3,7n \plus{} 1) \equal{} (7n \plus{} 2,7n \plus{} 1) \equal{} 1$, so $ (21n \plus{} 4,14n \plus{} 3)$ is $ 1$, which implies the desired result.
[/hide][/quote]

Thank you ! Could you explain to me how to use the Euclidean Algorithm in expressions that include variables ?

You have all the IMO-s and other competitions in Resources (ML) or Contests (AoPS).

Here is the problem:

http://www.mathlinks.ro/viewtopic.php?p=341470#341470

[quote="AIME15"][hide="Solution"]
Use the Euclidean Algorithm (I will use the notation $ (a,b) \equal{} \text{gcd}\ (a,b)$)

$ (21n \plus{} 4,14n \plus{} 3) \equal{} (14n \plus{} 3,7n \plus{} 1) \equal{} (7n \plus{} 2,7n \plus{} 1) \equal{} 1$, so $ (21n \plus{} 4,14n \plus{} 3)$ is $ 1$, which implies the desired result.
[/hide][/quote]

Once you get to $ (14n \plus{} 3, 7n\plus{}1)$, you get

$ 14n\plus{}3 \equal{} 2(7n\plus{}1) \plus{} 1$

So it should be

$ (7n\plus{}1,1) \equal{} 1$

Which proves it.

Two consecutive integers much be relatively prime to each other so you don't need that extra step.

We both end up with the same number of steps.  It doesn't really matter either way.


Αν $ a,b,c,d$ μη αρνητικοί πραγματικοί να δείξετε ότι $ \left(a^{2}\minus{}a\plus{}1\right)\left(b^{2}\minus{}b\plus{}1\right)\left(c^{2}\minus{}c\plus{}1\right)\left(d^{2}\minus{}d\plus{}1\right)\geq \left(\frac{1\plus{}abcd}{2}\right)^{2}$.

[quote="bilot4"][i]Λύση αφιερωμένη στο Σιλ για το μάθημα του[/i] :wink:

Αρχικά θα αποδείξουμε ότι $ (a^2 \minus{} a \plus{} 1)^2\geq \frac {1 \plus{} a^4}{2}$ (παρατηρείστε ότι αυτό είναι ισοδύναμο με την ζητούμενη γιατην περίπτωση που a=b=c=d). Το παραπάνω είναι ισοδύναμο με $ (a \minus{} 1)^4\geq 0$ κι άρα ισχύει! Βάσει αυτού:
$ (LHS)^2\geq \frac {1 \plus{} a^4}{2}\cdot \frac {1 \plus{} b^4}{2}\cdot\frac {1 \plus{} c^4}{2}\cdot\frac {1 \plus{} d^4}{2}\geq (RHS)^2$
κι έτσι δείξαμε το ζητούμενο.[/quote]

Χρησιμοποίησες το κλειδί της άσκησης δηλαδή Holder. Μια άλλη εκδοχή είναι να αποδείξουμε ότι $ 2(a^{2}\minus{}a\plus{}1)(b^{2}\minus{}b\plus{}1)\geq 1\plus{}a^{2}b^{2}$ γινόμενο κυκλικά και χτυπάμε Holder ανα 2. Anyway, well done :D


What are the reagents under and above the arrow?

1. What is the source of this problem?
2. Are you sure about the structure of the product?

Yes, I suspected it was your book. But wat book is it? What is its title? I am asking this because, unless I am not seeing something obvious, this seems a quite advanced problem.

well the book is just an preparatory book for iit-jee no author name it's just a book from an institute even i doubt if there exists a simple way to describe it

Can you even say what is the context of the problem? That is, in what chapter is it "inserted" in?

surprisingly it is in the chemical properties of $ \text{Hydrocarbons}$
let me quote the book
[quote="Book"]
Iodine is a very good electrophille(?) for effective intramolecular nucleophillic addition reaction on alkenes .Reaction of Iodine with Carboxylic acids having Carbon-Carbon double bond placed to permit intramolecular reaction results in formation of $ \text{Iodolactones}$[/quote]
also for what is $ \text{NaHCO}_{3}$ used normally.

Ok, but it happens that in your picture above there is no lactone in the product. That's why I asked you if the product structure was correct.


induction
number theory solved
number theory

Let k be a positive integer.  The sequence a1, a2, a3 ... of positive integers satisfies \sum d|n ad = kn, for all n \geq 1.

Prove that n|an for all n \geq 1.

I would be rather pleased if somebody could find a solution without induction. :)

I think the Mobius inversion formula works here:

f(n)=k n. We know that  \sum d|n a_d=f(n), so by the inversion formula we have a_n= \sum d|n u(d)f(n/d) = \sum d|n u(d)*k  n/d , so what we have to do is show that n divides this last thing for all n.

Basicly, I know how to continue, but I haven't tried everything out. A good idea would be starting with n of the form p a with p prime to see what happens and then extending the proof for all n.

I have a solution,but it use induction.
For n=1, we have 1|u_1
If we have k|a_k for all k<n
Then,  \sum [size=75]d|n[/size] u_d=k^n
Let be p a prime divisor of n and p^a such that p^a|n but p^(a+1) don't divises n. If we prove that p^a|u_n so we have n|u_n(because it is true for all primes)
Then we have: \sum [size=75]d|n[/size] u_d=\sum [size=75]d|n/p[/size] u_d+\sum [size=75]d|n/p^a,d<n/p^a[/size] u_(d*p^a)+u_n=k^n
But we have \sum [size=75]d|n/p[/size]u_d=k^(n/p) and p^a|\sum [size=75]d|n/p^a,d<n/p^a[/size] u_(d*p^a)
So \sum [size=75]d|n/p^a,d<n/p^a[/size] u_(d*p^a)+u_n=k^n-k^(n/p)
But k^n-k^(n/p)=k^(n/p)(k^(n(p-1)/p)-1)
If p|k then p^a|k^(n/p) because n/p>=a(n/p>=p^a/p=p^(a-1)>=a)
So we have p^a|u_n.
If k and p are coprime, then phi(p^a)=p^(a-1)(p-1)|n(p-1)/p
So p^a divises k^(n(p-1)/p)-1 and p^a divises u_n.
So in each case, p^a|u_n and n|u_n

Here's a proof without induction:

Consider strings formed with the "digits" 1, 2, ..., k.  Clearly, there are k^n possible n-digit strings.  Let bd be the number of strings of length d and period d (i.e. it is not made by repeating smaller parts of it).  Then \sum d|n bd = k^n, for all n \geq 1.  Thus an = bn for all n \geq 1.

Now it's easy to see that n|bn because of how bn is defined.


Let $ f\equal{}\Sigma_{i\equal{}0}^{\infty}a_{i}x^i \in R[[x]]$we consider $ f^m$. Let $ t$ be the smallest integer $ k$ for which $ a_{0}^{n}$ does not accure in every summand used in cumputing the $ k^{th}$ coefficient of $ f^m$ (where $ m\geqslant n$)
Why the $ t^{st}$ coefficient  of $ f^m$ is  this form
$ ca_{0}^{n\minus{}1}a_{1}^{m\minus{}(n\minus{}1)} \plus{}$ other terms, each of whitch has $ a_{0}^{n}$ as a factor

Did you try the work over the quotient ring $ R/a_0^nR$ and use some of the results of your other postings? ;)


Replace each blank with the correct digit.

          4 3 _ 2
            4 5 _
          _ 1 2 7
         ---------
          8 8 9 3
 :rotfl:
 :?:

Salahcool, have you mistyped something here?

Maybe the 4 5 _ is supposed to be moved over one?

[hide="solution"]Well, you go through each column through guess+check.
Units column:
2+_+7= either 3,13, or 27. It's easy enough to find that the blank is 4. Continuing with this method, you get
4312
 454
4127
-------
8893[/hide]

Are you supposed to add these?
Cuz it doesn't say.

[hide="If it's adding..."]

It's impossible to solve since the ones digits add up to 9 not 3. [/hide]

Scamper, did you read my solution?

It could add up to 13, and the 10 would just go in the next column.
Also, it's most likely adding, since you don't subtract this way with 3 numbers, and the result is way too small for multiplication.

I'm sorry pascal12,

your solution correctly 100%

But unstructured

thanks
salah


calculus
integration
calculus computations

Evaluate $ \int_{\minus{}1}^1 \frac{xe^x\plus{}e^x\plus{}1}{x^2(e^x\plus{}1)^2}\ dx$.

Very easy:

$ \frac{xe^x\plus{}e^x\plus{}1}{x^2(e^x\plus{}1)^2}\equal{}\frac{d}{dx}\left(\minus{}\frac1{x(e^x\plus{}1)}\right)$
So the integral is :

$ \minus{}\frac1{x(e^x\plus{}1)}\Big|^1_{\minus{}1}\equal{} \minus{}\frac1{e^{\minus{}1}\plus{}1}\minus{} \frac1{e\plus{}1}$

[hide]$ f(u)\equal{}\minus{}u^{\minus{}1}\Longrightarrow (f(xe^x\plus{}x))^{\prime}\equal{}\frac{xe^x\plus{}e^x\plus{}1}{(xe^x\plus{}x)^2}$
$ \therefore f(x)\equal{}\minus{}\frac{1}{xe^x\plus{}x}$

And $ f(1)\minus{}f(\minus{}1)\equal{}\minus{}\frac{1}{e\plus{}1}\minus{}\frac{e}{e\plus{}1}\equal{}\minus{}1$[/hide]
 :)


An environmental agency needs to hire a number of new employees so that 85 of the new employees will be able to monitor water pollution, 73 of the new employees will be able to monitor air pollution and 27 of the new employees will be able to monitor both. What is the minimum number of employees that need to be hired?

This principle is called PIE, Or Principal of Inclusion/Exclusion.

The general formula for two sets, $ S_1$ and $ S_2$, where you want to find the size of the union, $ U$(like in this problem), given the size of the intersection is:

$ |S_1| \plus{} |S_2| \minus{} |S_1 \plus{} S_2| \equal{} |U|$


group theory
abstract algebra
linear algebra
matrix
complex numbers
superior algebra
superior algebra unsolved

Let $ U(n)$ be the set of unitary matrices, which have the property $ UU^{\dagger} \equal{} I$. Show that
\[ SU(n) \equal{} \{A \in U(n): \; \det(A) \equal{} 1\}\]is a normal subgroup of $ U(n)$ and that $ U(n)/SU(n)$ is isomorphic to $ S^1$, the group of the unit circle in $ \mathbb{C}$ under multiplication.

Show that $ SU(2)$ contains the quaternion group $ Q_8$ as a subgroup.

I've shown that it's a normal subgroup, I don't know how to do the last 2 parts. I played around with eigenvalues having modulus 1, but no go. Any clues?  :)

Note that $ \det: U(n)\to S^1,$ that $ \det$ is a homomorphism, and $ SU(n)$ is the kernel of that homomorphism. That establishes the isomorphism, and is also another way of establishing that $ SU(n)$ is normal in $ U(n).$

There appear to be several different ways of finding $ Q_8$ as a subgroup of $ SU(2).$ Here's one: take the matrices

$ \begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}i&0\\0&\minus{}i\end{bmatrix},\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix},\begin{bmatrix}0&\minus{}i\\\minus{}i&0\end{bmatrix}$

along with their negatives.

[quote="Kent Merryfield"]Note that $ \det: U(n)\to S^1,$ that $ \det$ is a homomorphism, and $ SU(n)$ is the kernel of that homomorphism.[/quote]

I thought $ \det: U(n) \to \{\pm 1\}$?  :maybe:

[quote="Kent Merryfield"]There appear to be several different ways of finding $ Q_8$ as a subgroup of $ SU(2)$.[/quote]

I just read [url=http://en.wikipedia.org/wiki/Quaternion_group#Matrix_representations]matrix representations[/url] of the quaternion group on Wiki, are these isomorphisms?

The one presented on Wiki is:
\[ 1 \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\
i \mapsto \begin{pmatrix} \sqrt{-1} & 0 \\ 0 & -\sqrt{-1} \end{pmatrix} \\
j \mapsto \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \\
k \mapsto \begin{pmatrix} 0 & \sqrt{-1} \\ \sqrt{-1} & 0 \end{pmatrix}\]but based on what you said, it's not unique?

Suppose $ A\in U(n).$ That means that $ A^*A\equal{}I.$ Note that $ \det(A^*)\equal{}\det(\overline{A}^T)\equal{}\det(\overline{A})\equal{}\overline{\det(A)}.$ So $ \overline{\det(A)}\det A\equal{}1,$ which means that $ |\det(A)|\equal{}1$ or $ \det(A)\in S^1,$ where $ S^1$ is identified with the complex numbers of modulus $ 1.$

And for any $ e^{i\theta}\in S^1,$ we can find $ A\in U(n)$ with $ \det(A)\equal{}e^{i\theta}.$ Just let $ A\equal{}\text{diag}(e^{i\theta},1,1,\dots,1).$

I think you were confusing this with $ \det: O(n)\to\{1,\minus{}1\}.$ And the quotient $ O(n)/SO(n)$ is a two-element group.

As for the representation of the quaternions: that seems to be the same set that I wrote, but with a different identification of $ i,j,k.$ But I would strongly doubt that there's any kind of uniqueness.

[quote]I would strongly doubt that there's any kind of uniqueness.[/quote]

There is some kind of uniqueness: every such embedding is conjugate in GL(2,C) and I think in U(2,C).

Obviously though the normalizer of the image is not the entirety of GL(2,C), and so there are many conjugates.

[quote="Kent Merryfield"]Suppose $ A\in U(n).$ That means that $ A^*A \equal{} I.$ Note that $ \det(A^*) \equal{} \det(\overline{A}^T) \equal{} \det(\overline{A}) \equal{} \overline{\det(A)}.$ So $ \overline{\det(A)}\det A \equal{} 1,$ which means that $ |\det(A)| \equal{} 1$ or $ \det(A)\in S^1,$ where $ S^1$ is identified with the complex numbers of modulus $ 1.$

And for any $ e^{i\theta}\in S^1,$ we can find $ A\in U(n)$ with $ \det(A) \equal{} e^{i\theta}.$ Just let $ A \equal{} \text{diag}(e^{i\theta},1,1,\dots,1).$

I think you were confusing this with $ \det: O(n)\to\{1, \minus{} 1\}.$ And the quotient $ O(n)/SO(n)$ is a two-element group.[/quote]
This explains it, I was indeed confusing $ SO(n)$ and $ SU(n)$  :blush:  How about, say, $ SO_3(\mathbb{C})$? That's again different from the unitary group?

Is $ U(n)/SU(n)$ an infinite group?

For matrix representations of the quaternion group, it sounds like there are an infinite number of representations o.0

[quote="aidan"]How about, say, $ SO_3(\mathbb{C})$? That's again different from the unitary group?[/quote]
Is that a notation that anyone would use? It doesn't make any sense to me.

[quote]Is $ U(n)/SU(n)$ an infinite group?[/quote]
Of course it's infinite, and we've identified it in this very topic. It's the circle group.

I first saw the notation $ SO_3(\mathbb{C})$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1717786#1717786]here[/url].

I assume it's the group of $ 3 \times 3$ orthogonal matrices $ Q$ with complex entries, but I don't know whether they should satisfy $ Q^TQ \equal{} I$ or $ Q^{\dagger}Q \equal{} I$ where $ T$ is the ordinary transpose, $ \dagger$ the conjugate transpose.

[url=http://mathworld.wolfram.com/OrthogonalGroup.html]Wolfram mathworld[/url] also uses the notation $ O(3, \mathbb{C})$ to denote the complex orthogonal group, which it says is different from the unitary group.