Prove that there do not exist natural numbers $ n\ge 10$ having all digits different from zero, and such that all numbers which are obtained by permutations of its digits are perfect squares.

I did this problem today after school!!! xD

Well, if that number exists, first, it doesnt have the digits 2,3,7,8 (because no square can finish in that numbers, and some permutations take numbers with final digit 2,3,7,8).

So that number has to has this digits: 0,1,4,5,6,9.

But if it has a 0, then we can take any digit (for example 9) and take a permutation finishing in 90. This number is divisible by 10, but no by 100 so it cant be a square.

Now the number has to has this digits: 1,4,5,6,9

But if it has a 5, we can take some digit (obviously not equal to 2, for example 7) and we take the permutation finished in 75. This number is divisible by 5, so if its a perfect square it must finish in 25, contradiction.

So the number has only the digits 1,4,6,9

Using the same idea as before, take a permutation finishing in 14. We know that the number squared to get that permutation has to be even, so its (10a+2b)^2 (a natural and $ 0\le b\le 4$) and $ (10a\plus{}2b)^2\equiv 40ab\plus{}4b^2$ its a number with even number in the tens... so 14 and 94 fails.

Then the number can only have the digits 6 or 4.

If the number has both numbers, then exists a permutation that finishes in $ 46\equiv 6(8)$ but 6 is a non-cuadratic residue modulo 8, contradiction.

So the number must have only 4's or only 6's

But a number only consisting of 444..444 cant be a perfect square because of modulo 16 ($ 444...44\equiv 4444\equiv 12(16)$) but 12 is a non-cuadratic residue modulo 16 so it must consist of only 6's.

But $ 666...666\equiv 2(8)$ and 2 is a noncuadratic residue modulo 8 so the initial number cant exist.

There are obvious flaws. All that is proved at some point is that $1$ and $4$ cannot appear both as digits, and $9$ and $4$ cannot appear both as digits; this does not rule out the digits $1$ and $9$.


geometry
angle bisector
geometry proposed

[color=darkblue]Let be $PQRS$ a square, $E$ any point which lies in $\overline{QR}$ and $F$ the intersection point of the angle bisector $\measuredangle\ EPS$ with side $\overline{RS}$. Prove that

\[\overline{FS}+\overline{QE}=\overline{EP}\][/color]
[hide="Restriction"]
Without using trigonometry.[/hide]

[hide]Let $x$ denotes the direction of the ray $SP$ (in other words, $Px$ is the opposite ray of the ray $PS$)

The external angle bisector of $\angle EPS$ intersects the line $QR$ at $H$. Then $\angle EPH = \angle HPx$.

From the parallel lines $SP,QR$ we have $\angle EHP = \angle HPx$

Hence $\angle EHP = \angle EPH \Rightarrow  \triangle EPH$ is isosceles $(EH=EP)$

Then $EP=QE+QH$, so it's enough to prove that $QH=SF$.

But the triangles $SFP,QHP$ are equal, since they are right, have $PS=PQ$ and $\angle SPF = \angle QPH$ (these angles have perpendicular sides)[/hide]


hello, we have
$ \frac{6!\cdot5!}{6!\plus{}5!\plus{}5!}\equal{}\frac{6!\cdot5!}{5!\cdot6\plus{}5!\plus{}5!}\equal{}\frac{6!}{6\plus{}1\plus{}1}\equal{}\frac{720}{8}\equal{}90$.
Sonnhard.


sure  i'm in..... if there is any space for me

:rotfl:
Wrong Button, hmm¿ Try replying instead of new threads.

Mods.: please delete it.

It's unkind to insult the inexperience of new members :mad:  :roll:  :rotfl:

Take a look at his other threads  :wink:

[quote="ZetaX"]Take a look at his other threads  :wink:[/quote] :rotfl:  :rotfl:  :rotfl:

[quote="ZetaX"]Take a look at his other threads  :wink:[/quote]

ouch. i think we have a dilinquint on our hands.

Your spelling is creative kyle, but weren't you like, 3rd in the state spelling bee or something like that? :rotfl:

Nope. Only my astoneshing speling skillz! Serioslie!


Let S(n) denote the sum of all the permutations of a positive integer n which does not contain a digit 0. Thus S(52) = 5+2+52+25=84 and S(135) = 1+3+5+13+31+35+53+15+51+135+153+315+351+513+531. Find all positive integers n, 99<n<1000, such that S(n) is divisible by 2007.

[b]Case 1:[/b] Three distinct digits.

Let the digits be a, b, and c. Each will show up as a units digit 5 times, as a tens digit 4 times, and as a hundreds digit 2 times.

Hence, $S(n)=(2(100)+4(10)+5(1))(a+b+c)=245(a+b+c)$.

$\gcd (245,2007)=1 \implies 2007 | (a+b+c)$.

This is obviously impossible, so there are no solutions.

If $a, b, c$ are distinct, yes.  But what if, say, $a = b$?

Thanks.

[b]Case 2:[/b] The digits are a, a, b.

a, b, ab, ba, bb, aba, bab, bba $\implies S(n) = 123a+245b$

The maximum value of this is when $b=9$, $a=8$, giving $3189$, so $S(n)=2007$

Mod 3, we see that $b$ must be divisible by $3$. $41a+245 \frac{b}{3}=669$.
If $b=3$, then $41a=669-245=424$, so $b \not = 3$
If $b=6$, then $41a=669-490=179$, so $b \not = 6$
If $b=9$, then $4a = 669-3(245)<0$, so $b \not = 9$.

This exhausts all possibilities here.

[b]Case 3:[/b] The digits are a, a, a.

a, aa, aaa $\implies S(n) = 123a$

$123(9)<2007$, so there are no solutions here.

Thus, there are no solutions.


Solution:
I didn't know what to do other than that:
$ (a^{3}+b^{3}+c^{3})(a+b+c)^{2}-9(a^5+b^5+c^5)\geq (a+b+c)^5$
which after multiplying we have
$ \sum_{cyclic}{a^5}+10\cdot\sum_{sym}{a^3b^2}+20\cdot\sum_{sym}{a^4b}+20\cdot\sum_{cyclic}{a^3bc}\geq \sum_{cyclic}{a5}+5\cdot\sum_{sym}{a^4b}+10\cdot\sum_{sym}{a^3b^2}+30\cdot\sum_{cyclic}{a^2b^2c}+20\cdot\sum_{cyclic}{a^3bc}$
   and then
$ 15\cdot\sum_{sym}{a^4b}+20\cdot\sum_{cyclic}{a^3bc}\geq 30\cdot\sum_{cyclic}{a^2b^2c}+20\cdot\sum_{cyclic}{a^3bc}$
This is true becouse of Muirhead Inequality.

[quote="behdad.math.math"]$ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1.$ prove that: $ 10( a^3 \plus{} b^3 \plus{} c^3 ) \minus{} 9( a^5 \plus{} b^5 \plus{} c^5 ) > \equal{} 1$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59834
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63555
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161399


Show off your LaTeX skills with pictures!

$\text{(}>\text{\")}>\dagger$

$\text{biffanddoc}$
$\dagger \dagger \dagger \dagger \dagger \dagger \dagger \dagger \dagger$
$\text{kyyuanmathcount}$


function
inequalities
triangle inequality
real analysis
real analysis unsolved

How do you find the minimum of $ f: R\minus{}>R  f(x)\equal{}\sqrt{4x^2\plus{}28x\plus{}85}\plus{} \sqrt{4x^2\minus{}28x\plus{}113}$

Write as $ \sqrt{(2x \plus{} 7)^2 \plus{} 6^2} \plus{} \sqrt{(2x \minus{} 7)^2 \plus{} 8^2}$.  This is the sum of the distances from $ P \equal{} (2x, 7)$ to $ A \equal{} (\minus{}7, 1)$ and $ B \equal{} (7, 15)$.  By the triangle inequality, the minimum occurs when the three points are collinear and is $ \overline{AB} \equal{} \sqrt{14^2 \plus{} 14^2} \equal{} 14 \sqrt{2}$.  See [url=http://en.wikipedia.org/wiki/Minkowski_inequality]Minkowski's inequality[/url].

[quote="t0rajir0u"]Write as $ \sqrt {(2x \plus{} 7)^2 \plus{} 6^2} \plus{} \sqrt {(2x \minus{} 7)^2 \plus{} 8^2}$.  This is the sum of the distances from $ P \equal{} (2x, 7)$ to $ A \equal{} ( \minus{} 7, 1)$ and $ B \equal{} (7, 15)$.  By the triangle inequality, the minimum occurs when the three points are collinear and is $ \overline{AB} \equal{} \sqrt {14^2 \plus{} 14^2} \equal{} 14 \sqrt {2}$.  See [url=http://en.wikipedia.org/wiki/Minkowski_inequality]Minkowski's inequality[/url].[/quote]

Oh, your argument is very nice. But I wonder if we should modify by "$ P$ must be at between $ A$ and $ B$ in the line $ AB$"!


[quote="robinhe"][hide]After evaluating them, it is $4^0+4^1+4^2+4^3...4^9$.  The formula for the sum of an infinite geometric series is $\frac{a}{1-r}$ where a is the first term and r is the common ratio.  We can make two infinite series like $4^9+4^8+4^7...-(4^{-1}+4^{-2}...$.  Thus, our answer is $\frac{4^9}{\frac{3}{4}}-\frac{\frac14}{\frac34}$=$\frac{4^10}{3}-\frac13$=$\frac{4^10-1}{3}$=$\frac{(4^5+1)(4^5-1)}3$=(i'm going to skip the rest of my work w\o calculator) $349525$.[/hide]

I think.  ;)[/quote]
Yeah that's correct
how do you do this without making infinite sequences? is there a formula for a finite geometric sequence..?

Probably, but I wouldn't know.  I would know if someone would be helpful and post it here... :D

Yeah. $(x^n+x^{n-1}+...+x^2+x+1)(x-1)=x^{n+1}-1$, since all other terms cancel out. Divide both sides by $x-1$, set $x=4$ and $n=9$, and there you go.

The formula for the first $n$ terms of a geometric series is:
$\frac{t_1(1-r^n)}{1-r}$
Where $r$ is the common ratio, $t_1$ is the first term, and $n$ is the number


Let $ a,b,c\in \mathbb{R}$ such that $ a^3b\plus{}b^3c\plus{}c^3a\equal{}0$. Show that:

\[ a^4\plus{}b^4\plus{}c^4\plus{}2abc(a\plus{}b\plus{}c)\ge 0\]

[quote="Pain rinnegan"]Let $ a,b,c\in \mathbb{R}$ such that $ a^3b \plus{} b^3c \plus{} c^3a \equal{} 0$. Show that:
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc(a \plus{} b \plus{} c)\ge 0\]
[/quote]

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc(a \plus{} b \plus{} c)\ge {k(a^3b \plus{} b^3c \plus{} c^3a)}$

holds $ \forall (a,b,c)\in{R^3}$ iff $ k_0\equal{}\minus{}0.766158436...\le{k}
\le{1}$

$ (7k_0^4\minus{}8k_0^3\minus{}12k_0^2\minus{}64k_0\minus{}48\equal{}0).$

Hold on, let consider the LHS

\[ LHS \ge (ab\plus{}bc\plus{}ca)^2 \ge 0\]

Hence, the problem is proved without using the given condition. :)

[quote="great math"]Hold on, let consider the LHS
\[ LHS \ge (ab \plus{} bc \plus{} ca)^2 \ge 0\]
Hence, the problem is proved without using the given condition. :)[/quote]

yeah...

$ a^4\plus{}b^4\plus{}c^4\plus{}2abc(a\plus{}b\plus{}c) \equal{} \frac{1}{2}((a^2\minus{}b^2)^2\plus{}(b^2\minus{}c^2)^2\plus{}(c^2\minus{}a^2)^2) \plus{} (ab\plus{}bc\plus{}ca)^2$

 :lol:


As just about everyone (at least I would think everyone  :roll: ) in the US knows, the hurricane season in 2004 was pretty bad, in and 2005, it was worse. [b]Much[/b] worse. With Katrina, Rita, and Wilma, three hurricanes that were all at some point category 5, things were looking pretty bad, and they just kept getting worse. In fact, that year was the first year they started using Greek letters for hurricane names, such as Hurricane $\alpha$, Hurricane $\beta$, etc.

Considering all of these facts together, why have we had almost no hurricanes this year?  :huh:

[quote="Ignite168"][quote="nebula42"][quote="Ignite168"]The US has one of the lighter hurricane seasons. Even 2005 compared to an average Pacific hurricane season or Indian Ocean hurricane season was mild.[/quote]
I [b]seriously[/b] doubt that an [i]average[/i] Pacific hurricane season/Indian Ocean hurricane season is worse than the US 2005 hurricane season.  :wink:
I mean, many, many people suffered terrible losses because of hurricanes last year, and you're just saying it was "[i]mild[/i]"?

Besides, you're from Connecticut. What do you know about hurricanes?  :P  :roll:[/quote]

Instead of seriously doubting it, why don't you just Wikipedia it, while we only have Hurricane season's for about half the year, there is no end to the other hurricane seasons, as soon as they end another one begins.

Also, the Hurricane of New England was one of the most costiliest in history and we're about due for another one. Hurricanes that hit New England on average, cause the most damage because they are so few and when there is one, it hits hard.[/quote]
Even though their hurricane season is longer, that doesn't necessarily it's more devastating. Although they may have a few strong storms now and then (as we do), most of them do not actually make landfall or make a very expected landfall (because they form in an extremely large ocean, the Pacific, and they have to travel extremely far to reach land).

Also, severe hurricanes in New England occur slightly more often than once every hundred years. The last one was in 1938, which is quite a long time ago.
(Besides, I was just joking about the "What do you know...?".  :wink: )
Also, what do you mean by "when there is one, it hits hard"? There are many weak hurricanes that may hit the New England area too. Also, if you take a look [url=http://www.wunderground.com/hurricane/damage.asp]here[/url], you'll notice that Katrina, Rita, and Wilma were all in the top 6 costliest hurricanes, while the New England hurricane is #17.

The topic of discussion in this thread is getting pretty far from what I intended...  :wink:

The 2005 Atlantic Hurricane season had 28 named storms and 15 hurricanes. The average Pacific Typhoon season has about 27.625 named storms (average 1990-2005) and 17.625 hurricanes (average 1990-2005).

It depends on what you consider "worse." Strength of the storm or its impact (cost, deaths, etc.) In terms of strength, of the top 23 strongest recorded storms, only one (Wilma) was in the Atlantic. Also, out of the top 23, one was in the South Pacific and one was in the South Indian, the rest were in the West Pacific.
Also, ACE (Accumulated Cyclone Energy) is another way to measure a hurricane season. The average West Pacific season has an ACE of 305, while 87.5 is the average Atlantic season.

As you have already said, the typhoons must travel longer and generally don't affect land as severely in the Pacific.
The number of fatalities in the Atlantic is highly variable, with 11 deaths in 1997 and 18907 deaths in 1998.  :o The average for 1994-2005 is 2207 deaths per year and the median is 147.
In the Pacific, the average for 1994-2005 is 825 and the median is 570, much more constant.

Going back to the original question, [quote="Wikipedia"] They noted that conditions have become less favorable than they were earlier in the year. The sea level pressure and trade wind strength in the tropical Atlantic are above normal, while sea surface temperature anomalies are on a decreasing trend....They attributed the reduction to less favorable environmental conditions, a decrease in La Niña conditions, and the lack of a "very persistent upper-level ridge pattern over the eastern U.S. and western Atlantic." [/quote]
Also, [quote="Wikipedia"] As of September 23 the cumulative ACE is slightly ahead of the long term mean for that date, and the current pace of the season is about one tropical storm ahead of what is average. [/quote] $\text{4}\ne\text{almost zero}$ :D

For more information about Atlantic and eastern Pacific hurricanes, the definitive source is [url=http://www.nhc.noaa.gov/]here[/url]. Other agencies track storms in the central and western Pacific.

The NHC just stopped tracking Hurricane Helene, which at one point reached category 3 intensity. There have been three substantial hurricanes in a row - Florence, Gordon, and Helene - that all wheeled around in the open Atlantic without ever hitting land (except that Florence did hit the island of Bermuda).

The scariest storm of the season so far was Ernesto, which started up in the southeastern Caribbean. The earliest forecast track for it pointed it towards the Yucutan Channel (the gap between Cuba and Mexico) and into the Gulf of Mexico. The Gulf of Mexico has the warmest waters of the Atlantic, and storms can intensify rapidly there. But Ernesto kept veering to the north of the track, and lost most of its intensity in the mountains of Haiti and eastern Cuba. (It did track accross Florida, reintensify over the Gulf Stream, then come ashore over the Carolinas as a minimal hurricane - and an intense rainstorm for much of the eastern U.S.)

So it appears the original forecast of an active season for 2006 was overdone, but it is approximately a normal season - it's just that the strongest storms have been far from land.

The west coast of Mexico has seen several strong Pacific hurricanes this year.

Of course, it doesn't matter whether its a strong season or not - all that matters is the one storm in the wrong place. The second most damaging storm in U.S. history was Hurricane Andrew in 1992 - in an otherwsie below-average Atlantic hurricane year. And the greatest number of deaths ever in a tropical cyclone happened in Bangladesh, from an Indian Ocean storm.

I personally found it hard to find information on the Western Pacific typhoon seasons and storms. I had to take info from Wikipedia from individual years and average them and stuff. :D

[quote]The number of fatalities in the Atlantic is highly variable, with 11 deaths in 1997 and 18907 deaths in 1998.[/quote]
Tropical cylcones have three main deadly effects: wind damage (the primary agent in Hurricane Andrew's devastation of a portion of South Florida), storm surge or salt water flooding (the biggest problem with Hurricane Katrina), and fresh water flooding resulting from excessive rainfall. Most of the worst disasters, including the Bangladesh catastrophe and the 1900 Galveston event, have come from storm surge (salt water flooding) in low-lying coastal areas.

The 1998 deaths from Atlantic hurricanes are almost all the result of freshwater flooding and mudslides in mountainous inland areas - Hurricane Georges in the Dominican Republic and Haiti, and especially Hurricane Mitch in Honduras in late October. Mitch did reach category 5 intensity at one point but was nowhere near that strong when it came ashore. However, it was extremely slow-moving, and dropped 30 inches or more of rain on some parts of Honduras.


Let $ s$ be the side of the square, then $ s\sqrt {2}$ is the length of the diagonal, so:

$ s\sqrt {2} \plus{} s\sqrt {2} \equal{} 20$

$ 2s\sqrt {2} \equal{} 20$

$ s \equal{} \frac {10}{\sqrt {2}} \equal{} 5\sqrt {2}$

Perimeter $ \equal{} 4s \equal{} \boxed{20\sqrt {2}}$.

EDIT: thanks jxl

[quote="modularmarc101"]
$ s \equal{} \frac {10}{\sqrt {2}} \equal{} \frac {5\sqrt {2}}{2}$
[/quote]
I think you made a little mistake here.  It should be:  \[ s\equal{}\frac {10}{\sqrt {2}}\equal{}\frac {10}{\sqrt {2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\equal{}\frac{10\sqrt{2}}{2}\equal{}5\sqrt{2}\]
Therefore, our perimeter should be:
\[ 4\cdot5\sqrt{2}\equal{}\boxed{20\sqrt{2}}
\]

[quote="GameBot"]The sum of the lengths of the diagonals of a square is 20 inches. Find the number of inches in the perimeter of the square.[/quote]

Another way to do this is the find the area and take the square root. Each diagonal is obviously 10, so the area is $ \frac{10^2}{2}$, and a side is $ 5\sqrt{2}$.

A square has $2$ diagonals.

Sum of diagonals is $20$ (ignoring units for convince).

$\frac{20}{2}=10$ (length of diagonal)

Side length of square $=\frac{10}{\sqrt{2}}=5\sqrt{2}$

Perimeter of square $=4\cdot 5\sqrt{2}=\boxed{20\sqrt{2}}$


Let $ P$ be the point inside an acute triangle $ ABD$ outside of a triangle $ ABC$. Construct a regular triangle with side length $ AB$ outside and a regular triangle $ BPP'$ with side length $ BP$ outside of a triangle $ BCP$. Answer the following questions.

(1) If $ PA\equal{}4,\ PB\equal{}2,\ PC\equal{}3$, find the length of the broken line $ CPP'D$.

(2) Take a point $ Q$ inside $ \triangle{ABC}$. Find the measure of $ \angle{BQC},\ \angle{AQB}$ such that $ QA\plus{}QB\plus{}QC$ is least.

For the second, see [url=http://en.wikipedia.org/wiki/Fermat_point]Fermat point[/url].


The sequence a_1,..,a_n satisfies the following conditions: a_1=0,|a_i|=|a_(i-1)+1| for i=2,3,..,n. Prove that

a_1+....+a_n \geq -n/2

a(i) = a(i-1)+2a(i-1)+1

so 2* \sum ai = a(n) - n >= -n

It can't be that easy ??

You mean  2*Σai=[a_(n+1)]^2-n
Qute  problem!


I am new to this forum.  How to solve the following?
Find the value of a^3+b^3+c^3, whne a+b = 20, b+c=10, c+a=16

Regards
Sridhar

a+b = 20...(1)
(b+c)=10 ... (2)
(c+a)=16...(3)
(a+b)+(b+c)+(c+a)=46
i.e 2(a+b+c)=46
or (a+b+c)=23

Now 23- (1) gives c.
Proceed in this way to obtain a,b,c.

And then put the values of a,b,c.

Is there any solution without solving a, b and c.
Thanks
Sridhar

Solve for b in terms of a in the first equation: b = 20 - a. Plug that in for b in the second equation and solve for c in terms of a: c = a - 10. Plug that in for c in the third equation which gives a = 13. Now you know a = 13, b = 7, and c = 3. So it is 13^3+7^3+3^3 = 2567.

Or you can do it Nobelium's way, which is probably easier and more elegant.

And to answer your question: no, you have to solve for each individual variable.

[quote="yshridhar"]Million thanks Kunny for your formula.  I have no idea about this formula.
Sridhar[/quote]

You're welcome. :lol:


I sent my handwritten solutions by scanning them and then e-mailing them. Though I scanned all of the pages, I forgot to attach one of them.

Would I be able to resend it sometime?  :roll:


I am planning to write the iit jee, can anybody suggest me some books and how to prepare.
I am now at class XII.

hav u joined some coaching ...
plz provide some details
u should just do ur syllabus thoroughly :D

yes i am also under pressure about iit.

syllabus is screwing me

hey i am also in the same category but  hav taken advice from skand and took badminton  :rotfl:  :rotfl:  :rotfl:

how is it possible

badminton was my choice :mad:  :mad:

ok you take badminton,i take computer gaming :rotfl:


inequalities
number theory unsolved
number theory

Prove that equation
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\plus{}\frac{1}{xyz}\equal{}\frac{1}{x\plus{}y\plus{}z}$
have infinitely roots $ x,y,z\in N$

[quote="Algadin"]Prove that equation
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} \frac {1}{xyz} \equal{} \frac {1}{x \plus{} y \plus{} z}$
have infinitely roots $ x,y,z\in N$[/quote]
wlog say $ x\ge y \ge z$

$ LHS>\frac 1x\plus{}\frac 1y\plus{}\frac 1z\ge \frac 1x\plus{}\frac 1x\plus{}\frac 1x\frac xz$ and so $ LHS>\frac 1x(\frac xz\plus{}2)$

$ x\plus{}y\plus{}z\ge 3z$ and so $ RHS\le \frac 1{3z}$

And so $ \frac 1{3z}\ge RHS\equal{}LHS>\frac 1x(\frac xz\plus{}2)$ which implies $ \frac 13\frac xz>\frac xz\plus{}2$ and so $ \frac xz<\minus{}3$, impossible.

So there are no roots to this equation and [u][b]this problem is wron[/b][/u]g  :mad:

Could you pay more attention when you copy a problem in this forum, please ?

[quote="Algadin"]Prove that equation
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} \frac {1}{xyz} \equal{} \frac {1}{x \plus{} y \plus{} z}$
have infinitely roots $ x,y,z\in N$[/quote]

This equation is equivalent to $ (x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \plus{} x \plus{} y \plus{} z \equal{} xyz$ and By Am-Gm inequality we have $ (x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \plus{} x \plus{} y \plus{} z \ge 3\sqrt [3]{xyz}.3\sqrt [3]{x^2y^2z^2} \plus{} x \plus{} y \plus{} z \equal{} 9xyz \plus{} x \plus{} y \plus{} z > xyz$

So no solution  :maybe:


I just got this in the mail today

1. Western Washington: 35 + 37.5 = 72.5
2. Sofia Math Circle: 40 + 32 = 72
3. Colorado ARML: 33 + 37.5 = 70.5
4. Thomas Jefferson High School: 35 + 33 = 68
5. Long Beach Math Circle 33 + 34.5 = 67.5 [color=red]Yay![/color]
6T. Naperville North High School: 31 + 35.5 = 66.5
6T. Stuyvesant High School: 31 + 35.5 = 66.5
6T. Lynbrook High School: 37 + 29.5 = 66.5
9. Philips Academy: 32 + 34 = 66
10. Wayzata High School: 31 + 34.5 = 65.5
11. Illinois Math and Science Academy: 36 + 28.5 = 64.5
12. Walter Payton High School: 29 + 32.5 = 61.5
13. Hon. Vincent Massey SS: 28 + 31.5 = 59.5
14. Choate Rosemary Hall: 27 + 32 = 59
15. New Mexico Math League: 32 + 26.5 = 58.5
16. Western MASS A: 32 + 26 = 58
17T. Albany Area Math Circle: 29 + 26 = 55
17T. Georgia ARML: 34 + 21 = 55
19. Great Neck High School: 25 + 25 = 50
20T. Durham Academy: 29 + 20 = 49
20T. Danderyds Gymnasium (Sweeden): 31 + 19 = 49

Phillips Exeter Academy, Montgomery Blair HS, and AAST did not participate in the second round after scoring 37, 34, and 33 in the first round. Several other first round participants also did not participate in the second round.

East Chapel Hill High School scored 34 in the second round after not participating in the first round.

ARML Las Vegas should see some good power rounds - see listings above for #1, #3, #5, the third #6, and #15.

Question: which "Thomas Jefferson High School" is that in #4?

Where can one see the problems from this year's power contests? The ARML website only has one past event as an example, but unlike the actual ARML contest the site does not post the most recent event.

The green ARML problem book (1995-2003) has the power contests from that period.

I was curious about whether or not the Vegas site had a disproportionately large number of good power questions last year, so I took a look at the results.  Here are the power question rankings:
1. North Carolina A (39)
1. Bay Area A (39)
3. Colorado (38)
3. Montgomery A (38)
5. Upstate NY A (36)
6. NYC A (35)
7. Texas Gold (34)
8. Minnesota Gold (33)
9. Utah (32)
9. Lehigh Fire (32)
11. Exeter Red (31)
12. TJ A (30)
12. Alabama (30)
12. Indiana Gold (30)
12. NYC Y (30)
12. Taiwan C (30)

Of the 16 teams that got 30 or better, four were at Las Vegas, three were at Iowa, and the other nine were at Penn State.  Vegas 1/4 of those teams, whereas only 1/5 of the overall teams and 5/32 Division A teams were there.

[quote="Kent Merryfield"]
Question: which "Thomas Jefferson High School" is that in #4?[/quote]

the great virginian power, TJHSST (Thomas Jefferson High School of Science and Technology, I think)

[quote="not_trig"]the great virginian power, TJHSST (Thomas Jefferson High School of Science and Technology, I think)[/quote]
I asked because there is a high school in Seattle, although I suppose anyone who goes to that Seattle high school was likely part of "Western Washington," the Hesterberg-led group.

At Las Vegas last year, our A team had a 29, just missing Si's list above - not that I'm happy with a 29. We ought to be able to do better than that. And Washington A had a 27. What really drives Si's comment is that Colorado and Utah were both [b]much[/b] better at the power round than at any other round - and that same Colorado group shows up as #3 on this Power Contest.

A number of the teams not listed as ARML teams do strongly overlap with ARML teams. "Western Washington" is clearly a large fraction of the Washington ARML team - just not those who live far from Seattle. (I would assume that such AoPS regulars as Paladin8 and T0rarjir0 had something to to with it.) "Long Beach Math Circle" is most of the Southern California ARML team - pretty much everyone except the Santa Barbara people (and Macky in the fall because it conflicted with the Caltech Rubik Cube tournament.) "Naperville North HS" should be a subset of Chicago ARML, "Lynbrook HS" a subset of SFBA ARML, and "Wayzata HS" a subset of Minnesota ARML.

Other similar pairings are Stuyvesant High School (NYC) and Albany Area Math Circle (Upstate New York).

[quote="Kent Merryfield"]...Colorado and Utah were both [b]much[/b] better at the power round than at any other round - and that same Colorado group shows up as #3 on this Power Contest. ...[/quote]
Thank you for noticing Colorado ARML!  There are many fine Power Contest writers in the State, this year.  With few high schools participating in the Power Contest, that allowed a high concentration of good power contest writers to participate on the Statewide team.

BTW, I thought Colorado had an adequate Relay Round at UNLV last year.

-- rcv, parent and chaperone

[quote="Kent Merryfield"]"Western Washington" is clearly a large fraction of the Washington ARML team - just not those who live far from Seattle. (I would assume that such AoPS regulars as Paladin8 and T0rarjir0 had something to to with it.) [/quote]

 :)

Considering we did the second power round with about $8$ people, this is a pleasant surprise.  I think we were able to foist off most of the work on Adam...

wow naperville/imsa did well...maybe chicago arml will have a stronger power score this year  :lol: ...one oddity...i made some comment (to an imsa student) along the lines of: i wish that i went to imsa because you guys get more of these math opportunities...

my comment was quickly followed by a "wtf you are from new trier"

... :mad:

but obviously, we don't do many of these extra contests that i would enjoy, admittedly, we do the amc series and the stupid local contests, but i still feel that i would have more math opportunity at another school...(also i am disappointed with the willingness of teachers at my school to do math outside of the curriculum...my calc teacher gave me a flat no...my precalc teacher would be a better bet, but he has been asking me to explain stuff...so i ono about that)

p.s...could anybody post/give a link to/pm me the contest, i would still like to look at it

In defense of your pre-calc teacher (and knowing absolutely nothing about that person), a very reasonable teaching technique is to have the student explain the topic. You may think you understand something in your mind, but being able to explain is when real comprehension occurs.

That said, maybe your pre-calc teacher really doesn't have a clue what you are talking about. (That happens WAY too often.)


calculus
derivative
algebra
polynomial
function
integration
calculus computations

[quote="benorin"]So the Laguerre Differential Equations are for fixed $ n\in\mathbb{N}$ and $ \gamma\geq 0$
\[ y L_{n}^{\prime\prime}\plus{}(1\plus{}\gamma\minus{}y)L_{n}^{\prime}\plus{}nL_{n}\equal{}0\]
where the primes denote differentiation w.r.t. $ y$ and $ L_{n}(y)$ denotes the $ n^{th}$ Laguerre polynomial.

The Laguerre polynomials are one solution to this DE, Mathematica gives a Hypergeometric function equivalent to the Laguerre polynomials as the other solution, but this is useless to me. Can one of you help me?[/quote]

Use Liouville's formula $ W(y)\equal{}W(y_{0})\exp\biggl(\minus{}\int_{y_{0}}^{y}p(t)dt\biggr)$ for DE $ z''\plus{}p(y)z'\plus{}q(y)z\equal{}0$.
From this formula $ z_{2}(y)\equal{}z_{1}(y)\int_{y_{0}}^{y}\frac{1}{z_{1}^{2}(t)}\exp\biggl(\minus{}\int_{y_{0}}^{t}p(s)ds\biggr) dt.$

For Laguerre DE we have $ p(y)\equal{}\frac{1\plus{}\gamma\minus{}y}{y}$, $ q(y)\equal{}\frac{n}{y}$, $ z_{1}(y)\equal{}L_{n}^{(\gamma)}(y)$.


Let P(x) be a polynomial with integer coefficients. Suppose that P(1000)=1000 and P(2000)=2000. Explain why P(3000) cannot be equal to 1000.

[hide="Solution"] Suppose that $ P(3000) = 1000$.  Let $ Q(x) = P(x)-1000$.  Then

$ Q(x) = (x-1000)(x-3000) R(x)$

For some polynomial $ R(x)$ with integer coefficients.  But then

$ Q(2000) = 1000(-1000) R(2000) = 1000$

And $ R(2000)$ must be an integer; contradiction. [/hide]


Due to an accident, Ted has become radioactive. His body will slowly undergo radioactive decay until he is no longer a danger to others. The reduction of his radioactivity follows the curve y=85e^kt as shown in the graph below. Mr. Smith (a scientist)  says that he can resume contact with others when only 5 kg of his body remains radioactive. How many days will this take?

Firstly , substitute the value (2,67) into the original equation to find the value of constant $k$ . After you found $k$ , substitute (t,5) into the equation again to get $t$ ..... :)

what did u get for k i don't think i got it right

Substituting (2,67) into the equation, we get:
67 = 85e^(2k)
Dividing both sides by 85, we get
e^(2k) = 67/85
Taking the natural log (ln) of both sides to eliminate the e, we get:
2k = -0.238
k = [b]-0.119[/b]


Hey,

What are some programs that give college credit for mathematics or science courses that are available to middle schoolers or rising high school freshmen?

Well i am in elementary school (fifth grade) and i am doing trig at a local college if the college board lets you take the course you want to take for credit then it will apply (all of it might not if you go to something like stanford or MIT) when you go to college


let $ n \in \mathbb{R}$ $ s(n)$ is the sum decimals of $ n$
prove
 If $ m$ , $ n$ $ \in \mathbb{N}^*$ , $ s(m \plus{} n) \le s(m) \plus{} s(n)$ and
- If $ 10^k \minus{} 1/n$, then $ s(n) \ge 9k$

math is a precise science. be precise when you write it.


let x be irrational number, f(nx)=nx-[nx] ,where n is a naturel number ,
show tha {f(nx),n in N} is dense in [0,1]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1507911329&t=27475]Here[/url] is one place where this has been discussed. There are probably others.

I think we can grow some collection about proof of density
here is one :
[quote="grobber"]I was not suggesting using Weyl's Theorem to prove denseness :). I was just saying that it had been discussed before, and I'm sure the denseness has received solutions on the forum without involving that powerful theorem.

Anyway, here's a proof.

Let's work on the circle of length $ 1$, and let's denote $ \{n\alpha\}$ by the point $ A_n$. As we add another $ \alpha$ (i.e. we add another point on the circle), we go around the circle in a counter-clockwise direction. First of all, since $ \alpha$ is irrational, no two points $ A_i,A_j,\ i\ne j$ are the same, meaning that there are infinitely many points on our circle. It means that we can find some $ m < n$ s.t. the length of the arc (in a counter-clockwise direction) $ A_mA_n$ is $ < \varepsilon$. All arcs of the form $ A_{m \plus{} k(n \minus{} m)}A_{m \plus{} (k \plus{} 1)(n \minus{} m)}$ have the same length $ < \varepsilon$, so, by looking only at these points, we see that for each arc $ I$ with length $ > \varepsilon$, at least one of these points must "fall" in $ I$.

This proves what we wanted, which was the denseness of $ (A_n)_n$ on the circle.[/quote]

and hre is another:
let me to prove that :
for every irrational $ x$ there exist infinity numbers $ p$ and $ q$ such as $ |x\minus{}\frac{p}{q}|<\frac{1}{q^{2}}$
or
$ |qx\minus{}p|<\frac{1}{q}$
it's easky seen  that i  this case $ p$ will be $ [qx]$
ok let's prove this lemma (it's wellknown Dirichlet Theorem)
let's consider intervakl $ (0,1)$
let's fixed some $ Q$ integer and divide our interval in $ Q$ part
we have
$ [0,\frac{1}{Q})$ , $ [\frac{1}{Q}, \frac{2}{Q})$ ...  $ [\frac{Q\minus{}1}{Q}, 1 ]$
we have $ Q\minus{}1$ parts  :)
now lets consider sequance
$ \delta_{n}(x)\equal{}nx\minus{}[nx]$ where $ 1\leq n \leq Q$
all points of this sequance will be in interval $ (0,1)$
and we have $ Q$ point's and $ Q\minus{}1$ parts therefore in some parts will be two points or more than two .
hence distance between this two points will be less than $ \frac{1}{Q}$
let denote this points by
$ \delta_{n_{i}}(x)$ and $ \delta_{n_{j}}(x)$ we have
$ |\delta_{n_{i}}(x) \minus{} \delta_{n_{j}}(x)|\equal{}| n_{i}x\minus{}[n_{i}x]  \minus{}  n_{j}x\plus{}[n_{j}x]|\equal{}| x(n_{i} \minus{}n_{j} )  \plus{} ([n_{j}x]\minus{}[n_{i}x] ) |<\frac{1}{Q}$
now let $ n_{i} \minus{}n_{j} \equal{} q$
and
$ [n_{j}x]\minus{}[n_{i}x]\equal{}p$
note that $ 1\leq p \leq Q$
diference will be not equal to 0 because $ x$ is irrational.
 and in finnaly I have
$ |xq\minus{}p|<\frac{1}{q}$
where $ p$ will be $ [qx]$ after you cab finish yourself because it's easy now how to prove that our sequance is dense on $ (0,1)$

 :wink:

is there any other proof?


Here's an interesting problem (that's hopefully in the correct forum):

Two children, Mario and Maria, are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun.  The gun is mounted on a table so that the marble leaves the gun at the edge of the table.  The center of the target box is a horizontal distance $ D$ from the edge of the table.  Mario compresses the spring a distance $ x_{1}$, but the center of the marble falls a distance $ d$ short of the center of the box.  What distance $ x_{2}$ should Maria compress the spring to score a direct hit (the marble hits the center of the box)?  Assume that neither the spring nor the ball encounters friction in the gun.  Express the answer in terms of given and known constants.

I generalized this problem from a textbook-ish problem.

Solution

By Hooke's law, the force $ F$ exerted by a spring  on an attached mass is directly proportional to, and acts in the opposite direction to, the displacement $ x$ of the mass. As a mathematical statement:

$ F \equal{}\minus{}kx$

By Newton's second law of motion:

$ m\frac{d^{2}x}{dt^{2}}\equal{}\minus{}kx$

$ \Rightarrow\frac{d^{2}x}{dt^{2}}\equal{}\minus{}\frac{k}{m}x$

Using the "energy" transformation:

$ \frac{1}{2}\frac{d}{dx}\left(\left(\frac{dx}{dt}\right)^{2}\right) \equal{}\minus{}\frac{k}{m}x$

$ \Rightarrow\left(\frac{dx}{dt}\right)^{2}\equal{}\minus{}\frac{2k}{m}\int x\, dx \equal{}\minus{}\frac{k}{m}x^{2}\plus{}c$

When $ x \equal{}\minus{}x_{1}$, $ \frac{dx}{dt}\equal{} 0\Rightarrow c \equal{}\frac{k}{m}x_{1}^{2}$

$ \Rightarrow\left(\frac{dx}{dt}\right)^{2}\equal{}\minus{}\frac{k}{m}x^{2}\plus{}\frac{k}{m}x_{1}^{2}$

We want to know the speed $ v_{1}$ at which the marble leaves the gun with. This is at the point where $ x \equal{} 0$.

$ \Rightarrow v_{1}^{2}\equal{}\frac{k}{m}x_{1}^{2}$

$ \Rightarrow v_{1}\equal{}\sqrt{\frac{k}{m}}x_{1}$

Since both marbles are fired from the same height, they will take the same time to fall. Hence the  horizontal range of the marbles are directly proporional to the horizontal component of velocity only (which remains constant throughout). Since this velocity is directly proportional to the compression of the spring then the horizontal range is directly proportional to the compression of the spring. Thus:

$ \frac{x_{2}}{D}\equal{}\frac{x_{1}}{D\minus{}d}$

$ \Rightarrow x_{2}\equal{}\frac{D}{D\minus{}d}x_{1}$

Yup, that's the same answer I got.  Although, I used equations rather than descriptions to reason through the projectile motion part.  Thanks for reasoning it through like that!  Awesome job!   :thumbup:

Yeah, at first I did this by considering elements of work done in compressing through a small distance. And from this, the potential energy stored in a spring can be derived. I then used conservation of energy to get the velocity. But I thought that was very boring, so I came up with this.


[quote="dduclam"]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc\ge\sum_{cyc}\frac {a^2(a^2 \plus{} bc)}{b \plus{} c} \plus{} \sum_{cyc} bc(b \plus{} c)
\]
When does the occours ? :)[/quote]

It's equivalent with
\[ \sum {\left( {a \minus{} \frac{{a^2 }}{{b \plus{} c}}} \right)\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}  \ge 0
\]
which is obviously true by Vornicu Schur. :)

Equality holds if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$. :)

[quote="dduclam"][quote="Honey_S"][quote="dduclam"]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc\ge\sum_{cyc}\frac {a^2(a^2 \plus{} bc)}{b \plus{} c} \plus{} \sum_{cyc} bc(b \plus{} c)
\]
When does the occours ? :)[/quote]

It's equivalent with
\[ \sum {\left( {a \minus{} \frac {{a^2 }}{{b \plus{} c}}} \right)\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \ge 0
\]
which is obviously true by Vornicu Schur. :)

Equality holds if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$. :)[/quote]

Are you sure, Loc ?  :wink:[/quote]

The equality holds again for $ a \equal{} b,c \equal{} a \plus{} b$.

[quote="Honey_S"][quote="dduclam"]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc\ge\sum_{cyc}\frac {a^2(a^2 \plus{} bc)}{b \plus{} c} \plus{} \sum_{cyc} bc(b \plus{} c)
\]
When does the occours ? :)[/quote]

It's equivalent with
\[ \sum {\left( {a \minus{} \frac {{a^2 }}{{b \plus{} c}}} \right)\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \ge 0
\]
which is obviously true by Vornicu Schur. :)

Equality holds if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$. :)[/quote]

Why is it true by Vornicu Schur ? Please show the detail, Honey_S. Thank you!  :maybe:


induction
number theory unsolved
number theory

For positive integers $ n$, $ k$ holds that $ n^2 \equal{} (k \plus{} 1)^3 \minus{} k^3$, prove that there exists $ a,\,b \in \mathbb{Z}$ such that $ a^2 \plus{} b^2 \equal{} n$.

Since $ ax^2 \plus{} by^2 \equal{} c$ has infinitive many solutions in integers if $ gcd(a,b)|c$, for $ a,b,c \in Z$

Since $ a$ and $ b$ are both 1, $ gcd(a,b) \equal{} 1$, therefore all we need to prove there are integer solutions for $ n \in N$ in equation $ n^2 \equal{} (k \plus{} 1)^3 \minus{} k^3$.

$ n^2 \equal{} (k \plus{} 1)^3 \minus{} k^3 \equal{} k^3 \plus{} 3k^2 \plus{} 3k \plus{} 1 \minus{} k^3 \equal{} 3k^2 \plus{} 3k \plus{} 1$

Since $ k \equal{} 7 \Rightarrow n \equal{} 13$ there is solution.

EDIT: NEVER MIND THIS IS WRONG

[quote="NapoleonXIV"]Since $ ax^2 \plus{} by^2 \equal{} c$ has infinitive many solutions in integers if $ gcd(a,b)|c$, for $ a,b,c \in Z$[/quote]
$ a\equal{}1, b\equal{}1, c\equal{}3$.

[quote]Since ax^2 + by^2 = c has infinitive many solutions in integers if gcd(a,b)|c, for a,b,c \in Z[/quote]
Oops. That works for $ ax \plus{} by \equal{} c$
 :blush:

But still $ 13\equal{}3^2\plus{}2^2$
            $ 104\equal{}10^2\plus{}2^2$
Those are only solutions for $ k\leq 1000$

This problem appeared on the Israel-Hungary competition, 1996.
My partial solution-
$ 3k^{2} \plus{} 3k \plus{} 1 < (2k)^{2} \minus{} > n^{2} \equal{} 3k^{2} \plus{} 3k \plus{} 1 \equal{} (2k \minus{} t)^2$
by solving for k, we get:
$ k \equal{} \frac {4t \plus{} 3\pm\sqrt {12(t \plus{} 1)^{2} \plus{} 1}}{2}$
$ n \equal{} 2k \minus{} t \equal{} 3t \plus{} 3 \plus{} \sqrt {12(t \plus{} 1)^{2} \plus{} 1}$
we can use pell equation's formula now - let's denote $ t \plus{} 1 \equal{} s$. then $ n \equal{} 3s \plus{} \sqrt {12s^{2} \plus{} 1}$
$ x^{2} \minus{} 12s^{2} \equal{} 1$
$ x_{1} \equal{} 7, s_{1} \equal{} 2$
$ x_{i \plus{} 1} \equal{} 7x_{i} \plus{} 24s_{i}$
$ s_{i \plus{} 1} \equal{} 2x_{i} \plus{} 7s_{i}$
and we need to prove that $ x_{i} \plus{} 3s_{i}$ is the sum of 2 squares. Should I use induction? It's morning, I'm tired now...

The official solution is much shorter and elegant, I will try to recall it.

I recalled it [sorry for double posting...]:

$ n^{2}\equal{}3k^{2}\plus{}3k\plus{}1 \equal{}> (2n\minus{}1)(2n\plus{}1)\equal{}3(2k\plus{}1)^{2}$
$ gcd(2n\minus{}1,2n\plus{}1)\equal{}1$, so we have 2 cases:
1.
$ 2n\minus{}1\equal{}3a^{2}$ for some a.
$ 2n\plus{}1\equal{}b^{2} \equal{}> b^{2}$ has remainder of 2 when divided by 3: a contradiction.
2.
$ 2n\minus{}1\equal{}a^{2}$ for some a. a is odd, therefore: $ 2n\equal{}1\plus{}(2t\minus{}1)^{2}\equal{}4t^{2}\minus{}4t\plus{}2, n\equal{}(t\minus{}1)^2\plus{}t^2 \equal{}>$ Q.E.D

[quote="bambaman"]I recalled it [sorry for double posting...]:

$ n^{2} \equal{} 3k^{2} \plus{} 3k \plus{} 1 \equal{} > (2n \minus{} 1)(2n \plus{} 1) \equal{} 3(2k \plus{} 1)^{2}$
$ gcd(2n \minus{} 1,2n \plus{} 1) \equal{} 1$, so we have 2 cases:
1.
$ 2n \minus{} 1 \equal{} 3a^{2}$ for some a.
$ 2n \plus{} 1 \equal{} b^{2} \equal{} > b^{2}$ has remainder of 2 when divided by 3: a contradiction.
2.
$ 2n \minus{} 1 \equal{} a^{2}$ for some a. a is odd, therefore: $ 2n \equal{} 1 \plus{} (2t \minus{} 1)^{2} \equal{} 4t^{2} \minus{} 4t \plus{} 2, n \equal{} (t \minus{} 1)^2 \plus{} t^2 \equal{} >$ Q.E.D[/quote]

Thanks!   :)


LaTeX
geometry
geometric transformation
limit
real analysis
real analysis unsolved

Let [sum]_n=0 to oo (a_n) be a convergent series of complex numbers.

Prove that lim_n->oo (1/(n+1))([sum]_j=0 to n (ja_j)) = 0.

$\LaTeX$ translation; quote to see the code.
[quote="Cezanne123"]Let $\sum_{n=0}^{\infty}a_{n}$ be a convergent series of complex numbers.

Prove that $\lim_{n\to\infty}\frac{1}{n+1}\sum_{j=0}^{n}ja_{j}= 0$.[/quote]

Let $S_{n}=\sum_{j=0}^{n}a_{j}$. The convergence of the series $\sum a_{n}$ implies that $\lim_{n\to\infty}S_{n}=L$ for some complex $L$.
By Abel's method of summation by parts,
$\sum_{j=0}^{n}ja_{j}=(n+1)S_{n}-\sum_{j=0}^{n}s_{j}$. Dividing by $n+1$, $\frac1{n+1}\sum_{j=0}^{n}ja_{j}=S_{n}-\frac1{n+1}\sum_{j=0}^{n}s_{j}$. By Cesaro's theorem, $\lim_{n\to\infty}\frac1{n+1}\sum_{j=0}^{n}s_{j}=L$, and thus $\lim_{n\to\infty}\frac1{n+1}\sum_{j=0}^{n}ja_{j}=0$.


If  $ a,b,c>0$  and $ abc\equal{}1$  real numbers

Prove that
 a)   $ \frac{ab}{a^5\plus{}b^5\plus{}ab}\plus{}\frac{bc}{b^5\plus{}c^5\plus{}bc}\plus{}\frac{ac}{a^5\plus{}c^5\plus{}ac}\leq1$
 b)   $ \frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}a\plus{}c}\plus{}\frac{1}{1\plus{}b\plus{}c}\leq1$

[quote="maksmaks24"]If  $ a,b,c > 0$  and $ abc \equal{} 1$  real numbers

Prove that
 a)   $ \frac {ab}{a^5 \plus{} b^5 \plus{} ab} \plus{} \frac {bc}{b^5 \plus{} c^5 \plus{} bc} \plus{} \frac {ac}{a^5 \plus{} c^5 \plus{} ac}\leq1$
 [/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113771
b) Let $ a\equal{}\frac{x^2}{yz},$ $ b\equal{}\frac{y^2}{xz}$ and $ c\equal{}\frac{z^2}{xy},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, $ \sum_{cyc}\frac{1}{1\plus{}a\plus{}b}\equal{}\sum_{cyc}\frac{xyz}{xyz\plus{}x^3\plus{}y^3}\leq\sum_{cyc}\frac{xyz}{xyz\plus{}x^2y\plus{}xy^2}\equal{}1.$


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AIME I

Yes.  You keep the test, and will get answers shortly thereafter.

No.  Low scores are expected and common.  Many people are very happy to get even one right.

[quote="kdt"]i took the amc and barely qualified this year. for the aime, do we get to keep the test, like in the amc? and answers will be posted a few days later?

and is there anything bad about getting a 0 or a low score on the aime, and will anything be recorded?[/quote]

yep, you keep the booklet
yep, answers will be posted
no external consequences about getting a 0
your score will, however, be listed as an anonymous statistic

do you tihnk i can go in, get the test and do the ones i can and absolutely skip the ones i have no chacne on and turn in what i have so i can go home early?

if you wish, i guess so. but i would always see the positive side: try to do ALL and try hard as you can!

First off congrats on making it to the AIME kdt. Second off, 0 is a perfectly respectable score.

Although I'm sure you could turn in your test early, given that your school allows it, I would highly recommend you just sit the full 3 hours. I know it seems like a long time, and it is. But getting one additional question right makes a lot of difference - not only for bragging rights, but also if you'd like to persue math as an extracurricular activity and for your college apps.

Most likely, you'll be able to keep your tests. And answers will USUALLY be posted several weeks afterwards. However, being the mathematically ambitious community that we are there will be members here who've programmed their computers to do the problems and "answers" will be posted as soon as the board reopens. The Math Jam on Thursday night is when the instructors here reveal their answers and solutions, which are generally always correct. So in theory, you should be able to find out your score on Thursday.

Good luck.

i think i'm one of your type student(who will take 1 or 2) however don't lose hope and try to be enthusiastic. i think if i will work hard and go to Canada/USA math camp i can even have chance to go to USAMO

[quote="kdt"]3 hours, and i know i wont get more than a 5 since i have never prepared for any of this junk before. i just plan on taking it next year and work hard for that test. aiming for a 10+ :)

for oyu guys is it like 3 hours after school? and where do you find out what the questions are on previous tests?[/quote]

My school has it during school so I get to miss class :). And you should really stay the full 3 hours; some people last year at my school left after 1 or 2, it was kind of weird (but reasonable, knowing how much people like math these days *cough*). You also want to get familiar with the format and if you are actually there for the full 3 hours I think it will help.

The forum will most likely be locked down during the AIME and for about 36 hours afterwards (for international students and whatnot). When it's reopened on Thursday, "answer" threads will go up, unless we've experienced a serious decrease in anxiousness since last year. You're welcome to help and/or post your own answers, of course. We should have all the answers by the end of the math jam Thursday night.

Maybe we should start just one answer thread. I've noticed that the administraters usually start a thread and everybody else starts like 100 new threads. Howabout we just use the administrater's thread? lot easier to find stuff

Before I take the AIME, I would like to thank AoPS(community and administraters) for all the resources and help they've  provided. Thank you!

ah, my post count is 198, I'm going to guess 198 on any question I can't get tommorow.

Stay for the full 3 hours definitely. You may get a sudden inspiration and come up with a great idea that may help you solve another question, and make you feel a lot better :)

Besides, at our school, we also miss class. I mean, you get to [i]miss[/i] class (I get to miss English thank god lol)

[quote="sponge008"]It's my first time too, so I know how you feel. After not advancing to the Matcounts nationals the day before yesterday, I decided to have a look at the AIME. I almost fainted when I saw it (not literally). Then, I decided to actually try it. Guess what? After a thorough analysis, the problems are broken down into simple algebra/geometry.

PS) Is there going to be an answer thread right after the AIME? I'm decent at programming, so I could probably help out. (don't look at my USACO results, there's [i]no way[/i] that could possibly be my score... :lol:)[/quote]
Ur lucky that you have states before AIMES... mine are the other way around so I have no preparation to my credit for the AIMEs... but to an eighth grader who doesn't have Algebra II, this test (esp. numbers 10-15) looks like a tough (I mean TOUGH) one... But I have no choice... since I go to the HS for math, I have to stay longer to take the AIMEs, and so because I had to pre-arrange transportation, I'm stuck in the HS for 4 classes. On a brighter note, I miss four classes at the middle school. O well I think it's better to stay for the 3 hours because:
a) You might have a miraculous discovery sitting through the third hour
b) It's annoying to see someone walk out of the room when you're testing, and it's disrespectful
c) You get to miss class (in my case 4/8!!) (doesn't apply to late test takers aka after school)
d) You can waste away at the third hour by figeting your thumbs, sleeping, thinking of how boring school is, or thinking of what to bubble in for your random guesses, and of course pray that you didn't get a zero.....


$ m , n , t$ are the roots of $ ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ prove that $ a^3(m^4 \plus{} n^4 \plus{} t^4) > b(10ad \minus{} bc)$

Am I overseeing something? We have that m,n,t are also roots of
$ \minus{}ax^3\minus{}bx^2\minus{}cx\minus{}d$,
but that would mean that inequalities
$ a^{3}(m^{4}\plus{}n^{4}\plus{}t^{4}) > b(10ad\minus{}bc)$
and
$ \minus{}a^{3}(m^{4}\plus{}n^{4}\plus{}t^{4}) > \minus{}b(10ad\minus{}bc)$
must hold at the same time => contradiction.

Moreover the inequality cannot be strict -- for a=d=1, b=c=3 equality occurs.

These inequality equavalent to $ a^3s_4 > a^3\sigma_1(10\sigma_3 \minus{} \sigma_1^2\sigma_2)$ and if $ a > 0$ equavalent to
$ s_4 > \sigma_1(10\sigma_3 \minus{} \sigma_1\sigma_2)$. I think he forget only about $ a_1 > 0$.
Because form not shange, when $ m,n,t\to km,km,kt$, we can consider only $ \sigma_1 \equal{} 1$, except case, when $ m \equal{} n \equal{} t\equal{}0$ and inequely is not strong..


Prove that all points on a right angled triangle are equidistant to the midpoint of the hypotenuse.

You mean all vertices right :D

This should go to getting started.

I would have put it there but I figured geometry is a better bet. I know the answer too by the way but I wan't to see if there is any other way of working it out.


function
inequalities
inequalities proposed

Let $a,b,c$ be sides of a triangle.
Prove that $2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\ge{\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+3}$
A solution not using functions would be more good.I have solved using this way. :)

Well, this is a super-nice problem. I remember that I solved this a long time ago. Anyway, here is a solution.

  [color=blue][b] Problem.[/b]

     Let $a, b, c$ be the side lenghts of a triangle. Prove the following inequality:

                              \[ 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{a}{c}+\frac{c}{b}+\frac{a}{c}+3.  \]

   [/color]

  [i] Solution. [/i]

     It is obvious that \[ \sum (a+b-c)(a-b)^{2}\geq 0\Leftrightarrow  \]
                              \[ \sum (a+b-c)(a^{2}-2ab+b^{2})\geq 0\Leftrightarrow  \]
                              \[ \sum (a^{3}+a^{2}c-a^{2}b-2a^{2}b-2abc+2ab^{2}+ab^{2}+b^{2}c-b^{3})\geq 0\Leftrightarrow  \]
                              \[ \sum 4a^{2}c-\sum 2a^{2}b-6abc\geq0\Leftrightarrow  \]
                              \[ 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{a}{c}+\frac{c}{b}+\frac{a}{c}+3.  \]

[quote="Cezar Lupu"]
                              \[ \sum (a+b-c)(a^{2}-2ab+b^{2})\geq 0\Leftrightarrow  \]
                              \[ \sum (a^{3}+a^{2}c-a^{2}b-2a^{2}b-2abc+2ab^{2}+ab^{2}+b^{2}c-b^{3})\geq 0\Leftrightarrow  \]
                             [/quote]
I think you are wrong here.(where is $-b^3$)
Btw:
1.the inequality:\[ \sum (a+b-c)(a-b)^{2}\geq 0  \] is true for all $a,b,c \ge 0$
2.The original inequality have been posted,please search.

[quote="tkhtn"]Use SOS (one solution of ANH CUONG in http://mathnfriend.net/index.php?showtopic=171    new web of math in VN).
Inequality is equivalen to $\sum(5a+5b-c)(a-b)^2\ge 0$.[/quote]

What is SOS?

http://www.mathlinks.ro/Forum/search.php

[b]darij grinberg[/b] explained clearly what it was, along with the Vornicu-Schur inequality.


sorry if this is a stupid question, but in (b) what is d? if d is an integer specific to m and n, then f(m)*f(n)|gcd(m,n) for all m,n but for every m we can choose n coprime to m ... so that is not what you mean.

Excuse me vinoth_90_2004.I think you can now understand the problem.

ive made some progress but i can't complete it  :(
in (b) put m=p^k,n=p^l, s.t. (k,l)=1 and k<l: f(p^(k+l))=p^k*f(p^l). putting k=1 into this we get f(p^m)=p*f(p^(m-1))=p^2*f(p^(m-2)=...=p^(m-1)*f(p). now in (b) it also follows iff (p,q)=1 then f(p)f(q)=f(pq). so it then follows that if
n=(p1)^(a1)*(p2)^(a2)...*(pk)^(ak) then f(n) = (p1)*(a1-1)...(pk)^(ak-1)*f(p1)*f(p2)...*f(pk) ... (#) . we further have from (#) that f(n)/n = (f(p1)/p1)*(f(p2)/p2)*...*(f(pk)/pk) where p1...pk are the prime divisors of n (##). from (a) we also have f(f(n))=f(n)^2 / n, so from (#) we have f(p1^(a1-1)*p2^(a2-1)...*pk^(ak-1)*f(p1)*f(p2)...*f(pk))=p1^(a1-2)*p2^(a2-2)...(pk)^(ak-2)*(f(p1)*f(p2)...*f(pk))^2 (###). Now suppose {f(p1),f(p2),f(p3)...f(pk)} have a set of prime divisors q1,q2...qs apart from p1,p2...pk. from (##) we now have that
f(q1)f(q2)...f(qk)=q1*q2...*qk. i cant get any further  :D


By how much does 3.5 exceed its reciprocal? Express your answer
as a common fraction.


Mến chúc thành viên sub_forum Viêt Nam nói riêng và toàn thể thành viên Mathlink nói chung một mùa giáng sinh hạnh phúc.
[img]http://upload.wikimedia.org/wikipedia/commons/f/ff/Juletr%C3%A6et.jpg[/img]

thanks link,
very happy when u said to members of Vietnam in merry chrismas "n" Happy new year 07.
[b] i fell warn my heart in cold winter, and i wish everyone have a merry chrismas full funny with ur family.... [/b]
          :P  :blush:

i'll go home at sartuday
  [b]i'm happy and calmer. i and members family are very happhy in Luner Tet's holiday.
    Tet's holiday of my total is 14 days, i'll visit my friend and uncle... all people whom i love you, and you ??? :?:  :oops: [/b]


Determine the relation between $ p,q \in \mathbb{C} $ s.t. the roots $ x_1,x_2,x_3 $ of $ x^3+px+q=0 $ satisfy $ x_3=\frac{1}{x_1}+\frac{1}{x_2} $. If $ p=q $ and $ p \in \mathbb{R} - \{ 0 \} $, then prove that the previous relation can't be satisfied.

Satisfying $ x_3=\frac{1}{x_1} + \frac{1}{x_2} $ for the polinomial $ x^3 + 0.x^2 + px + q=0 $ and using Viete's formula:

$ x_1 + x_2 + \frac{1}{x_1} + \frac{1}{x_2}=0 \implies \frac{1}{x_1} + \frac{1}{x_2}=-(x_1 + x_2) $ and:

$ x_1\cdot x_2 \cdot (\frac{1}{x_1} + \frac{1}{x_2})=-q \implies q=-(x_1 + x_2) \implies x_1 \cdot x_2=-1 $ then:

$ p=(x_1 + x_2).(\frac{1}{x_1} + \frac{1}{x_2}) + x_1\cdot x_2=-(x_1 + x_2)^2 - 1 $

If $ p=q $ we have:

$ -(x_1 + x_2)^2 - 1=-(x_1 + x_2) $, doing $ (x_1 + x_2)=k $:

$ -k^2 + k - 1=0 $ analising the discriminant of the equation:

$ \Delta={1^2-4.-1.-1}={-3} $,       it means that $ k=(x_1 + x_2) $ belongs to the complex numbers set,and since: $ p \in \mathbb{R} - \{0\} $ the condition to the third root do not satisfie when $ p=q $

[quote="Gabriel(fr)"]$ x_1\cdot x_2 \cdot (\frac{1}{x_1} + \frac{1}{x_2})=-q \implies q=-(x_1 + x_2) \implies x_1 \cdot x_2=-1 $[/quote]

how did you arrive at this relation?

[quote="perfect_radio"][quote="Gabriel(fr)"]$x_1\cdot x_2 \cdot (\dfrac{1}{x_1} + \dfrac{1}{x_2})=-q \implies q=-(x_1 + x_2) \implies x_1 \cdot x_2=-1$[/quote]

how did you arrive at this relation?[/quote]
$-q=x_1x_2(\dfrac{1}{x_1}+\dfrac{1}{x_2})=\dfrac{x_1x_2}{x_1}+\dfrac{x_1x_2}{x_2}=x_1+x_2$
Therefore, $q=-(x_1+x_2)$
Using $\dfrac{1}{x_1}+\dfrac{1}{x_2}=-(x_1+x_2)$ which has been proved one line above, we have $x_1x_2(\dfrac{1}{x_1}+\dfrac{1}{x_2})=-q=x_1+x_2=-(\dfrac{1}{x_1}+\dfrac{1}{x_2})\Rightarrow x_1x_2=-1$


$\left( {2x} \right)^{\log _b 2} - \left( {3x} \right)^{\log _b 3} = 0 \Leftrightarrow 2^{\log _b 2} \cdot x^{\log _b 2} = 3^{\log _b 3} \cdot x^{\log _b 3} \Leftrightarrow$

$\Leftrightarrow x^{\log _b 3 - \log _b 2} = \frac{{2^{\log _b 2} }}{{3^{\log _b 3} }} \Leftrightarrow {\log_b x} = \frac{1}{{\log _b 3 - \log _b 2}} \cdot \log _b \left( {\frac{{2^{\log _b 2} }}{{3^{\log _b 3} }}} \right) \Leftrightarrow$

$\Leftrightarrow {\log_b x} = - \frac{{\left( {\log _b 3} \right)^2 - \left( {\log _b 2} \right)^2 }}{{\log _b 3 - \log _b 2}} \Leftrightarrow {\log_b x} = - \log _b 6 \Leftrightarrow x=\frac 16$

Take log of both sides and use power rule to get $g(g+y) = h(h+y)$ where $g = \log_b 2, h = \log_b 3, y = \log_b x$.  Then $y = \frac{h^2 - g^2}{g - h} = - (h + g) = -\log_b 6$.  So $\log_b x = \log_b \frac16 \implies x = \frac 1 6$


function
algebra
polynomial
blogs
rational function
algebra unsolved

Let $ R(x) \equal{} \frac{P(x)}{Q(x)}$ be a rational function where $ P, Q \in \mathbb{Z}[x], \deg P < \deg Q$ and all of the Taylor coefficients of $ R$ are non-negative integers.  Prove or disprove that $ R$ is generated by the functions $ x$ and $ \frac{1}{1 \minus{} x}$ under addition, multiplication, and composition.  (Note that subtraction is not allowed.)

[hide="Remark"] Any rational function generated in such a manner describes a [url=http://en.wikipedia.org/wiki/Regular_language]regular language[/url].  This is more or less equivalent to the question I asked [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=270779]here[/url], but in a form that perhaps makes it more tractable.  I'm reasonably certain the answer is negative, by the way, so try to construct a counterexample. [/hide]

Some trivial bits:

- The composition should be restricted somewhat; we only take $ f\circ g$ if $ g(0)\equal{}0$. Without that, power series don't make sense.
- The degree condition is unimportant. We get polynomials as well, and mixed fractions.
- We absolutely need $ 1$ in the generating set. Without it, we miss things like $ \frac{1\minus{}2x}{1\minus{}3x\plus{}x^2}\equal{}1\plus{}\frac{1}{1\minus{}\frac{x}{(1\minus{}x)^2}}\cdot\frac{x}{1\minus{}x}$, and without the degree restriction, $ 1$ itself.

Other than that, I'm inclined to think it's possible.

[quote="jmerry"]We absolutely need $ 1$ in the generating set. Without it, we miss things like $ \frac {1 \minus{} 2x}{1 \minus{} 3x \plus{} x^2} \equal{} 1 \plus{} \frac {1}{1 \minus{} \frac {x}{(1 \minus{} x)^2}}\cdot\frac {x}{1 \minus{} x}$, and without the degree restriction, $ 1$ itself.[/quote]
Sorry; I'm thinking like a combinatorialist.  We automatically get $ 1$ as an empty product, but I guess I should make that clear.  By the way, that's an interesting example; I'd construct it as

$ \frac {1}{1 \minus{} \frac {x}{1 \minus{} \frac {x}{1 \minus{} x}} }$

for reasons that I explain in the blog post I just wrote.


geometry
AMC
AIME
MATHCOUNTS
Princeton
college
AMC 10

If you studied a lot and are consistently getting 144 or 150 on practice exams then yes it is feasible. Just don't make stupid mistakes!

The #1 factor isn't how much you studied, but how many stupid mistakes you make, so it's possible to understand the material, yes, but it will be difficult to get a 150 (138+ is a more realistic goal). Also, I wouldn't focus too much on getting a perfect score, and focus on doing well on AIME.

[quote="geomeister33"]Is it feasible for me to get a 150 on the AMC 10 if i got a 118.5 last year as a freshman without studying, and missing some of the geometry covered?[/quote]

I got a 118.5 on the AIME last year also as a freshman.

this year I'm aiming for a 130 on the AMC10's. I always fail on the later combinatorics problems, combinatorics is just SO HARD  :(

soltixnj, use Alcumus.  From not being able to solve a decent Mathcounts C&P problem, I can solve most AIME C&P problems.

And you meant 118.5 on AMC not AIME right? :)

Heh, I went from 109 on 10A and 76.5 (or so) on the 10B in sixth grade to 129 last year. As mentioned, 150 is difficult to get because one stupid mistake can kill you. So, that's why trying to get 150 can sometimes be a difficult thing to do. However, it's certainly possible to aim for at least a 132.

well, as working on not making stupid mistakes is important, learning how to solve EVERY type of AMC 10 problem is very crucial as well.
say there is one problem that you dont know how to solve: you're screwed out of a 144.
The best way to be able to solve almost any problem at the AMC 10 level:
-do every AMC 10 test. when you don't get a problem, work on it after for 15 min or so. if you still don't get it, read the solution. and i mean READ it. know it by heart. ive heard a lot of experienced users say:
learn the solution as if you were going to explain it to someone TOMORROW.
-do AMC 12's. the AMC 12 has a much, much bigger problem source, and many of the AMC 10 problems are taken from the past AHSME's, just changed a bit.

150 should be considered as just a number. Getting 150 makes you "feel" good that you know a lot more than other kids do. Maybe you do. Maybe you don't.

Colleges will not necessarily go, "Wow, you got 150 on AMC 10." But they will be (I believe Princeton University asks for AMC and AIME score? I'm not sure if it's both or either) if you make to USAMO.

I find that as much as shooting for 150 is definitely an ideal goal, one should focus more on AIME and attempt to score well on AIME than just receiving perfect on AMC. Of course, most students who do that well on AMC do well on AIME too.
Most do. But not all.

[quote="geomeister33"]Is it feasible for me to get a 150 on the AMC 10 if i got a 118.5 last year as a freshman without studying, and missing some of the geometry covered?[/quote]

Honestly....no, getting a 150 is largely dependent on luck (heck, even O Dorney didn't get a 150 last year)...I got a 145.5 last year and am aiming for a 138 this year, imo people who score above 132 can be anywhere between 132 and 150, depending on how many stupid mistakes are made...

DataBox, didn't evan get 138 on amc 12 last year?  Anyway, he's already aced AMC 10 before...

Anyway, the links in my signature can help.  Use AoPS volume 1 to help shore up on some weaker topics, then go to volume 2.

[quote] DataBox, didn't evan get 138 on amc 12 last year? Anyway, he's already aced AMC 10 before... [/quote]

....you realize that you're just proving my point right?

Here's my AMC 10 track:

6th grade: 80 (phail)
7th grade: 114 (ok, better, but missed AIME qualification by 3 points)
8th grade: 123 (half-pwnage)
9th grade (this is what i'm in right now) 144/150, depending on whether i mess up question 1 or not

For some reason I usually get most of 21-25 and I miss a few problems from 14-20, so I either barely pass 120 or barely miss it.
It's usually due to certain types of problems though. So I'm trying to work on those specific areas.

Maybe that will work for you too :)

I tend to the same thing... I usually can do most if not all the problems on the AMC 10, but miss the problems that are in the beginning. For example on the AMC 10 2002 A I missed #1... And like 6 and 8. I think it is better to check answers in the last 10 minutes instead of trying #25.

If you were confident that you had the first 24 problems right on an AMC10, and had no idea what to do for the last one, would you leave it blank (for a tiny bit higher chance of getting to USAMO) or would you answer it? (for a chance you'd get a perfect score)

Why does it matter that you didnt study last year and still got a 118.5...its about what you did this year.  I think if you read over AoPS 1+2 (for the AMC 10 sections), and did a lot of old contest problems, I think you will have a great shot at 150.

My friend went from a 120ish to a 150 last year (7th to 8th)

And I would always guess, even if I did not even eliminate all the answer and thus only have a 1/5 chance (so a disadvantage).  You would feel bad if you got 145.5 (sorry DataBox).

Besides, the only way it would matter is if the cutoff for 10th graders and below is like between 204 and 205.5 (144-145.5+6 on AIME), which I guess is possible, but Im willing to bet that most people that got the first 24 right on AMC 10 can easily do more than just 6 right on AIME (and thus get 210+ index, which is basically autoadmit unless the AIME is extermely easy)

what are the amc 10 sections?
not v2 geom (inversions homothety etc)
Combinatorics - probably not
Counting- Yes
Probability - Yes
V1 Geom and Counting

The AMC 10 has everything the 12 does minus the advanced algebra, analytic geometry, and trigonometry, essentially.

I mean I guess it has NT but that is pretty basic

[quote="#H34N1"][quote="brightzhu"]You should always be really sure the first 24 problems are correct. If you still have time, try the last problem. If you don't know how to do the problem at all, you should probably just leave it blank. If you know enough to eliminate a few answers, you might want to just guess.[/quote]

The last problem may at times be like really easy though comparatively? Check like 2004 10B #25 for example. :wink:[/quote]

I thought that the 2005 10B #25 was pretty easy too. And the #25 on the 2008 10B was trivialized with graph paper.


function
algebra
polynomial
geometry
complex analysis
algebra proposed

Suppose $a\in \mathbb{Z}\backslash\{0\}$ Prove that $p(x)=x^n+ax^{n-1}+\dots+ax-1$ is irreducible. I proved this using Rooshe (not sure the spelling) theorem.

I'll be satisfied if anyone send more simple solution.

Well isn’t it trivial? (I am not sure but I can’t find any mistake in my argument)

Say the function is irreducible then:
$p(x)=(x-s)(x^{n-1}+\cdots+t)$

So -st=1

As s and t both are integer then s = 1 or -1.

But p(1)=1+a(n-1)-1=a(n-1)=0. So either a=0 or n=1.

Again
p(-1)
= -2 if n=odd
= -a if n=even

So if p(x) is reducible then a=0 or n=1.

Is there any problem in my solution? I remember I used this method for edition 4 round 7 problem 3. but I don’t know whther that was correct or not. So if you can find any problem with my solution please tell me.

How did you said $p(x)=(x-s)(x^{n-1}+...+t)$?

My solution:
It's enough to prove for $a>0$ because: $x^np(\frac1x)=-(x^n-ax^{n-}-...-ax-1)$
$Q(x)=(x-1)p(x)=x^{n+1}+(a-1)x^n-(a+1)x^{n-1}+1$
Suppose $\epsilon$ is small number that: $a\geq 0$. Now if $f(x)=-(a+1+\epsilon)x$ and $g(x)=x^{n+1}+(a-1)x^n+1$ then for every $z$ on unit circle $f(z)>g(z)$ so number of roots of $f$ and $f+g$ in the unit circle are the same. So by Rouche theorem $f+g=x^{n+1}+(a-1)x^n-(a+1+\epsilon)x^{n-1}+1$ has one root in unit circle.
If we move $\epsilon$ near to $0$ then $(x-1)p(x)$ has $n$ roots on unit circle or out of it and because the only root of $(x-1)p(x)$ on the unit circle is $1$ then $p(x)$ has $n-1$ roots  like $z$ that $|z|>1$ now if $p(x)=q(x)r(x)$ then $q(0)=1$ and $r(x)=-1$ and WLOG all of roots of $q$ are out of unit circle so if $z_1,...,z_k$ are the roots of $q$ then $1=|q(0)|=|z_1\dots z_k|>1$. Contradiction proves the problem.

Edit: I added some details to my solutions

if p() is reducible then it must have a root say s. so p() has a factor (x-s). Am not i right?

*I used contradiction method.

Can't we use Eisenstein's critherion ,taking a prime factor of $a$,to state that the polynomial is irreducible?I think you are considering irreducibility over $Z[X]$,don't you?
I hope you are not talking about something completely different.

But it's not an Eisenstein polynomial. For one thing, that prime factor should also divide the free term, which is $-1$.

I believe it's a Romanian problem and it's been discussed before. If I am not mistaken, the idea was looking at the roots of the polynomial, and it didn't look neater than what Omid did.

What an idiot can I be sometimes! :blush:

[quote="ZetaX"]Why it must have a root when reducible¿  :?  :huh:[/quote]

it might not obviously, so I'm afraid mahbub solution is wrong.

Omid, please can you show your solution using Rouche's theorem  ?

Ähm... he has already given his solution...  ;)  :roll:  :?

:|  :huh:  :o  :rotfl:  I overlooked it, thanks ZetaX  :lol:

[quote="Omid Hatami"]How did you said $p(x)=(x-s)(x^{n-1}+...+t)$?

My solution:
$Q(x)=(x-1)p(x)=x^{n+1}+(a-1)x^n-(a+1)x^{n-1}+1$
[/quote]

I think it's $Q(x)=(x-1)p(x)=x^{n+1}+(a-1)x^n-(a+1)x+1$..

Can somebody please write Rouche's theorem you're talking about..

[quote="mahbub"]so what is the defenition of: reducible?

can it be like this (x^2+...)(x^(n-2)+...) or i mean (x^i+...)(x^(n-i)+...)?

i thought reducible in Z means that has at least one linear factor which has integer root. :blush:[/quote]

Polynom P(x) is reducible in Z if it can be represented as product of two (non-constant) polynoms with integer coefficients.. For example.. $P(x)=x^4+4x^2+3$ is reducible in Z because $P(x)=(x^2+1)(x^2+3)$. Notice that $P(x)$ doesn't have integer root.

Thanks that's what i guessed. But my idea was wrong at the first! :blush:

NO MAN IS PERFECT :lol:

Rouche's theorem:
Let be $\Gamma$ any closed curve in the complex plane that bounds an simply connected area (or what is enough for most cases: let $\Gamma$ be a circle in the complex plane)

If you have additionally two holomorphic functions $f,g$ (e.g. polynomials) defined on $\Gamma$ itself together with it's interior and the additional condition that $|f(x)|>|g(x)|$ for all $x\in \Gamma$, then $f$ and $f+g$ have the same number of zero's in the interior of $\Gamma$.


The figure shown is composed of congruent squares and has a total area of 216 square inches. When the shape is folded to form a cube, how cubic inches are in its volume?
[asy]size(100);
defaultpen(0.7);
for(int i = 0; i < 4; ++i){
 draw(shift(i,0)*unitsquare);
}
draw(shift(0,-1)*unitsquare);
draw(shift(2,1)*unitsquare);[/asy]

Notice that $ 216\equal{}6^3$. Because a cube has $ 6$ sides, area is squared, and volume is cubed, the surface area and volume would be equal specially for side length equal to six, since what we really have is the equation $ 6s^2\equal{}s^3$. Thus, the answer is $ \boxed{216}$ cubic inches.

[quote=Kirotori]We can make a simple observation here: 216 = 6^3. There are 6 faces on a cube and each one of them is equal, so 216/6 = 36. Also, since 36 = 6^2, we can find the edge of a cube simply doing sqrt36 = 6. The volume of a cube is s^3, and therefore, our answer is 6^3 = [b]216 cubic inches.[/b][/quote]

With $\LaTeX$:

We can make a simple observation here: $216 = 6^3$. There are $6$ faces on a cube and each one of them is equal, so $216/6 = 36$. Also, since $36 = 6^2$, we can find the edge of a cube simply doing $\sqrt{36}$$ = 6$. The volume of a cube is $s^3$, and therefore, our answer is $6^3 = $[b]$216$ cubic inches[/b].


floor function
logarithms
number theory unsolved
number theory

You are given three lists $A$, $B$, and $C$.  List $A$ contains the numbers of the form $10^k$ in base $10$, with $k$ any integer greater than or equal to $1$.  Lists $B$ and $C$ contain the same numbers translated into base $2$ and $5$ respectively:
$$\begin{array}{lll} A & B & C \\ 10 & 1010 & 20 \\ 100 & 1100100 & 400 \\ 1000 & 1111101000 & 13000 \\ \vdots & \vdots & \vdots \end{array}$$
Prove that for every integer $n > 1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.

Suppose that the i-th number in B has $a_i$ digits,and the i-th number in C has $b_i$ digits.
So $2^{a_i-1}<10^i<2^{a_i},5^{b_i-1}<10^i<5^{b_i}$,and it's obvious that both of $\{a_i\},\{b_i\}$ is strictly increasing.
If there exist $i,j\in\mathbb{N}$,$a_i=b_j$,
Since $2^{a_i-1}<10^i<2^{a_i},5^{b_j-1}<10^j<5^{b_j}$
so $10^{a_i-1}<10^{i+j}<10^{a_i},a_i-1<i+j<a_i$
A contradiction.($a_i-1,i+j,a_i$ are integers).
So there doesn't exist same number in $\{a_i\},\{b_i\}$
If there exists $k\in\mathbb{N},k>1$,$k\not\in\{a_n\}\cup\{b_n\}$.
Since $b_1=2,b_2=3,a_1=4$,so $k\geq5$,there must exist $i,j\in\mathbb{N}$ such that
$a_i<k<a_{i+1},b_j<k<b_{j+1}$.
So $2^{a_i-1}<10^i<2^{a_i}\leq2^{k-1}<2^k\leq2^{a_{i+1}-1}<10^{i+1}<2^{a_{i+1}}$
$5^{b_j-1}<10^j<5^{b_j}\leq5^{k-1}<5^k\leq5^{b_{j+1}-1}<10^{j+1}<5^{b_{j+1}}$
So $10^{i+j}<10^{k-1}<10^k<10^{i+j+2}$
$i+j<k-1<k<i+j+2$ A contradiction.
So for every integer $k$ , there is exactly one number in exactly one of the $\{a_i\},\{b_i\}$.
So for every integer $k>1$, there is exactly one number in exactly one of the lists B or C that has exactly $k$ digits.

Notice that the number of digits can be defined as:

$B = \lfloor k\log_2(10)\rfloor$, and $C = \lfloor k\log_5(10) \rfloor$.

Now, we evaluate $\dfrac{1}{\log_2(10)} + \dfrac{1}{\log_5(10)}$.

$\Rightarrow \dfrac{\log_2(10) + \log_5(10)}{\log_2(10) \cdot \log_5(10)}$.

Let $\log_2(10) = p, \log_5(10) = q \implies 2^{pq} = 10^{q}, 5^{pq} = 10^{p}$.

Thus, $10^{pq} = 10^{p+q} \implies p + q = pq$

$\implies \dfrac{1}{\log_2(10)} + \dfrac{1}{\log_5(10)} = \dfrac{\log_2(10) + \log_5(10)}{\log_2(10) \cdot \log_5(10)} = 1$.

Hence, by Beatty's theorem on that sequence, it follows every number length appears in that sequence, and the conclusion follows $\blacksquare$.

Somewhat minor, but your computation could be simplified greatly:

\[\frac{1}{\log_2{10}} + \frac{1}{\log_5{10}} = \log_{10}{2} + \log_{10}{5} =  \log_{10}{10} = 1\]

I did it in a much simpler manner: I defined the number of digits in $y$ (base $x$) as $\lfloor\log_x(y)\rfloor+1.$ Using that, it's easy to prove that the number of digits in $2^x$ plus the number of digits in $5^x$ is $x+1.$ Knowing that, I said that when we increment $x$ to be $x+1,$ exactly one of our two numbers is going to have one more digit than before. If it's $5^{x+1},$ then $5^x<10^a<5^{x+1},$ and otherwise, $2^x<10^b<2^{x+1},$ those being the other definition of number of digits (base $5$ or $2$), so either $10^a$ has $x+1$ digits base $5,$ or $10^b$ has $x+1$ digits base $2$, and one of them is always true, and only one can ever be true, so we have our proof.


calculus
logarithms
integration
calculus computations

1.\[ (\frac{\minus{}1}{y}\plus{}\frac{x^2}{(x\minus{}y)^2})dy \plus{} (\frac{1}{x}\minus{}\frac{y^2}{(x\minus{}y)^2})dx\equal{}0\]

2.\[ ysin2xdx\equal{}dy(1\minus{}k\sqrt{y}cosx)\]   ,where k is a known fixed constant

Surely this should be in College Playground: Calculus Computations and Tutorials.

[quote="Dr Sonnhard Graubner"]hello to 1), i don't think that there is a solution which can be expressed by the known elementary functions.
Sonnhard.[/quote]

Really?
My answer is $ \frac{x y}{x\minus{}y} \plus{} \log \frac{x}{y} \equal{} C$
by　exact differential equation

[url=http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html]Exact First-Order Ordinary Differential Equation[/url] - MathWorld
[url=http://en.wikipedia.org/wiki/Exact_differential_equation]Exact differential equation[/url] - Wikipedia

... which can't be solved for $ y$ by elementary functions.  I.e., what you wrote is consistent with* what Sonhard wrote.

* and also more helpful than

Kouichi Sir,can u please show the process for question 1.Showing the working will surely help.Or give some detailed hint for it.

Also,please everyone solve question 2 also.

Thanks everyone....

\[ \left\{\begin{array}{l}
f(x, y)dx\plus{}g(x, y)dy\equal{}0 \\
\text{and} \\
\frac{\partial f}{\partial y}\equal{}\frac{\partial g}{\partial x}
\end{array}\right.
\Rightarrow
\begin{array}{l}\left\{\begin{array}{l}
\int f(x, y) dx \plus{} \int g(x, y) dy \minus{} \int \left( \frac{\partial}{\partial y} \int f(x, y) dx \right) dy \equal{} C \\
\text{or} \\
\int f(x, y) dx \plus{} \int g(x, y) dy \minus{} \int \left( \frac{\partial}{\partial x} \int g(x, y) dy \right) dx \equal{} C
\end{array}\right. \\
\hfill  (C \in \text{Const.})
\end{array}\]

Thanks Sir,

What about second question???Anyone??

[b][color=blue]1. I get C=x-y.[/color][/b]


Supposed that each pair of eight tennis players either played exactly one game last week or did not play at all. Each player participated in all but 12 games. How many games were played among the eight players?

(a) 10
(b) 12
(c) 14
(d) 16
(e) 18

[quote="JesusFreak197"]I don't understand the question, especially the last part. :huh:[/quote]

THat would be because I mean "eight" instead of either  :blush:

I've also added choices  :D

Ahh, now I get it. :P

[hide="Answer"]We draw out an 8x8 grid to indicate who plays whom. If we mark in each box whether one players plays another, then we have $8n$ games played, where $n$ is the number that each player plays, but we have overcounted the number of games by two times, so we have $4n$ games, and $4n-n=12\Rightarrow n=4$, so there were $16$ games played. $\boxed{D}$[/hide]


limit
probability
integration
real analysis
real analysis unsolved

Suppose $n$ is a positive integer. Let
$S_{n}= \sum_{k = 0}^{n}2^{2k}\binom{n}{k}^{2}$
$f(n) = \frac{3^{2n}}{\sqrt{n}}$*
Show that as $n \to \infty$, $S_{n}= O(f(n))$.

Can we find
$\lim_{n\to \infty}\frac{S_{n}}{f(n)}$
 :?:

* I found this myself so it might actually be wrong. But I checked with Maple and it seems reasonable for $n$ about as large as my computer can handle (which isn't that large except in a computational sense). If $S_{n}$ is too big, maybe the $\sqrt{n}$ isn't supposed to be there.

Let $p_{n}(x)=(1+2x)^{n}(2+x)^{n}$; $S_{n}=\sum_{k=0}^{n}2^{k}\binom{n}{k}\cdot 2^{k}\binom{n}{n-k}$ is the coefficient of $x^{n}$.

Now consider the probability distribution $X_{n}$, where $P(X_{n}=r)$ is $3^{-2n}$ times the $x^{r}$ coefficient of $p_{n}(x)$. $X_{n}$ is the sum of $n$ independent copies each of two fixed distributions (probability $\frac13$ of zero and $\frac23$ of 1, or vice versa).

The Central Limit Theorem says that $X_{n}$ is approximately normal, with mean $n$ and variance $\frac{4n}{9}$.
$X_{n}$ takes only integer values, so $\frac{S_{n}}{3^{2n}}=P(X_{n}=n)\approx \int_{-\frac12}^{\frac12}\frac1{\sqrt{2\pi}\cdot\frac{2\sqrt{n}}{3}}\exp(-\frac9{8n}x^{2})\,dx\approx \frac{3}{\sqrt{8\pi n}}$

The convergence I need is stronger than that provided by the Central Limit Theorem (the intervals I'm looking at shrink relative to the standard deviation). The above is not actually a proof of this formula, but I will say that $\lim_{n\to\infty}\frac{S_{n}}{3^{2n+1}/\sqrt{8\pi n}}=1$.

I wasn't thinking about probability theory when I tried to solve this but I think my estimate was effectively the same as yours anyway. I used Stirling's approximation first to estimate the terms. Then I estimated how many of the central terms contribute significantly to the sum. Thence my result. But I lost some constants along the way so I was unable to determine the limit.

Also, I think since my method is basically the same as yours in a different form, it's no more solid as proof. Certainly we can say the sum is $O(3^{2n}\sqrt{n})$ by replacing every term with the largest one, but that's a bit crude. ;)


How many ways can you arrange 6 different charm on a bracelet without a clasp?

i thought it was a clasic combination problem, the answer was easy (6-1)!
=120 ways

But my classmate said the correct answer should be 60 ways.

Pls help me!

Um, your classmate's right, because you can not only rotate the bracelet, but you can also it flip it over/reflect it.  That takes away half of the results, so the answer is $ \frac {120} {2}\equal{}\boxed {60}$


How many different isosceles triangles have integer side lengths and perimeter of $ 81$ units?

Let the sides be $ x,y,y$. Then $ 2y>x$ is the only nontrivial statement from the Triangle Inequality. Since $ x\plus{}2y\equal{}81$, we have $ x\equal{}81\minus{}2y$, and substituting, we get $ 2y>81\minus{}2y\implies4y>81\implies y\geq21$. Since $ x>0$, $ 21\leq y\leq40$, giving us $ 40\minus{}21\plus{}1\equal{}\boxed{20}$ as our answer.


You know that a polynomial $ f(x)$ has 4 integer varablies for which $ f(x)\equal{}7$, prove that there is no an integer solution for $ f(x)\equal{}14$.

I have something but I do not know if it is a solution :)

[hide="Solution"] Let $ g(x) \equal{} f(x)\minus{}7$.  Then for distinct integers $ a, b, c, d$ we can write

$ g(x) \equal{} (x\minus{}a)(x\minus{}b)(x\minus{}c)(x\minus{}d)h(x)$

We want to know whether there exists $ e$ such that $ g(e) \equal{} 7$.  But $ g(e)$ is the product of at least four distinct integers $ e\minus{}a, e\minus{}b, e\minus{}c, e\minus{}d$, and $ 7$ can only be written as the product of at most three distinct integers (for example, $ 1,\minus{}1,\minus{}7$).  Hence $ e$ does not exist. [/hide]

That is what I had, but how do you proof that $ h(x)$ must be a polynomial with integer coefficients ?

The problem is false unless $ f(x)$ is a polynomial with integer coefficients, in which case the fact that $ h(x)$ is a polynomial with integer coefficients follows from the division algorithm / the factor theorem.

Can you give me a link about that theorem :)

It will be faster to prove it.

[b]Theorem:[/b]  Let $ A(x)$ be a polynomial with integer coefficients.  Then for any [i]monic[/i] polynomial $ B(x)$ with integer coefficients there exist polynomials $ Q(x), R(x)$ with integer coefficients such that

$ A(x) \equal{} B(x) Q(x)\plus{}R(x)$

and either $ R(x) \equal{} 0$ or $ \text{deg}R <\text{deg}B$.

[hide="Proof"] When $ \text{deg}A <\text{deg}B$ we let $ Q(x) \equal{} 0, R(x) \equal{} A(x)$.

When $ \text{deg}A\ge\text{deg}B$, we induct on $ d \equal{}\text{deg}A\minus{}\text{deg}B$.

[b]Base case:[/b]  $ d \equal{} 0$.  Suppose $ A(x)$ has leading term $ a$.  Then we let $ Q(x) \equal{} a, R(x) \equal{} A(x)\minus{}a B(x)$, which has degree strictly less than $ B(x)$.

[b]Induction:[/b]  Suppose the division algorithm holds for $ d \equal{} 0, 1, 2, ... k$.  Let $ A(x), B(x)$ be polynomial such that $ d \equal{} k\plus{}1$.  Again suppose that $ A(x)$ has leading term $ a$.  Then

$ A(x) \equal{} ax^{k\plus{}1}B(x)\plus{}R_{1}(x)$

where $ R_{1}(x)$ has degree strictly less than $ A(x)$.  We can then apply the division algorithm for $ d\le k$ to $ R_{1}(x)$, giving us

$ A(x) \equal{} ax^{k\plus{}1}B(x)\plus{}B(x) Q_{1}(x)\plus{}R_{2}(x)$

Letting $ Q_{2}(x) \equal{} ax^{k\plus{}1}\plus{}Q_{1}(x)$, we have

$ A(x) \equal{} B(x) Q_{2}(x)\plus{}R_{2}(x)$

as desired.  QED. [/hide]


function
algebra
functional equation
real analysis
real analysis solved

I had yesterday a disscution with one of my teachers and he said that the functional equation below is simple. Anyway, after one hour of hard work I still couldn't solve it. So, I need help:
   Given k>0,find all bijective functions from (0, infinity) into itself such that f(f(x))=f(x)+kx.

mate
are u sure the function isnt continuous?in case it is then the following recurrence will help .define x_(n+1)=f(x_n) and then proceed to solve the recursion

What is also what I asked , but the answer was no. Even with the condition of continuity, the problem is hard, since the recursion doesn't give much information.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72217[/url] for the solution.


Consider a convex polygon with 2009 vertices.A scissor is used to cut along all of its diagonals,thereby dividing the original polygon into smaller convex polygons.How many vertices are there of a resulting polygon with the greatest number of edges?

I'm sorry, but this is a problem in Mathematics and Youth contest, and the deadline is 31st July. PLEASE MOVE THIS TOPIC IMMEDIATELY./.


Group N1 (top 15 but not top 5):

Adam Hesterberg, Princeton
William Johnson, University of Washington

Group N2 (top 25 but not top 5)

Nathan Pflueger, Stanford

Honorable Mention:

Jeffrey Wang, Stanford

Group I (in the top 100 but not HM):

Group II (top 200 but not top 100):

Qiaochu Yuan, MIT

Top 500 but not top 200:

Nate Bottman, U. Washington
Keyun Tong, U. Washington

Anyone else I should look for?

Nathan Pflueger, Stanford (Group N2)

I'd say Washington had a pretty good showing this year...

BTW, Will Johnson is listed as group N1 in the Putnam results thread; I believe that is probably more correct, as he said he solved 10 problems or so.

Oops, I knew that about Will Johnson. I've edited, and added in Nathan Pflueger as well.


Prove that if in a quadrilateral ABCD

     AB*BC*CA + AC*CD*DA  =  BC*CD*DB + BD*DA*AB

     then it is a cyclic one.  Using this identity prove concyclicity
     of the four incenters of non-overlapping triangles made by sides and
     diagonals  of a circumscribed (with incircle) quadrilateral.

     Capt. Pestich

[quote="pestich"]Prove that if in a quadrilateral ABCD

     AB*BC*CA + AC*CD*DA  =  BC*CD*DB + BD*DA*AB

     then it is a cyclic one.[/quote]

This is not quite correct - the statement above is true only for [b]convex[/b] quadrilaterals ABCD:

[b]Theorem 1 (converse of 2nd Ptolemy theorem):[/b] If in a convex quadrilateral ABCD, we have

$AB\cdot BC\cdot CA+AC\cdot CD\cdot DA = BC\cdot CD\cdot DB+BD\cdot DA\cdot AB$,

then the quadrilateral ABCD is cyclic.

[i]Proof.[/i] Using the notations a = AB, b = BC, c = CD, d = DA, p = AC, q = BD, we can rewrite the above equation

$AB\cdot BC\cdot CA+AC\cdot CD\cdot DA = BC\cdot CD\cdot DB+BD\cdot DA\cdot AB$

in the form $abp+pcd=bcq+qda$, or, equivalently, $\left(ab+cd\right)p=\left(ad+bc\right)q$. This rewrites as

$\frac{p}{q}=\frac{ad+bc}{ab+cd}$.

Now, post #2 of the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=14058 shows that, if we denote by A, B, C, D the angles of the quadrilateral ABCD, then we have the following implications:

If A + C > 180°, then $\frac{p}{q}<\frac{ad+bc}{ab+cd}$.
If A + C < 180°, then $\frac{p}{q}>\frac{ad+bc}{ab+cd}$.

Thus, we have neither A + C > 180° nor A + C < 180°. Consequently, A + C = 180°, and this yields that the quadrilateral ABCD is cyclic. Thus, Theorem 1 is proven.

[quote="pestich"]  Using this identity prove concyclicity
     of the four incenters of non-overlapping triangles made by sides and
     diagonals  of a circumscribed (with incircle) quadrilateral.[/quote]

I am almost sure that such a proof, if possible at all, would be not better than all the proofs of the CBC we know. Also, since Theorem 1 demands the quadrilateral ABCD to be convex, such a proof will rely on convexity and thus on arrangement, what will have the unpleasant consequences that it will be possibly not extensible to the case of excenters rather than incenters, or to the case of a quadrilateral with an excircle rather than with an incircle.

  Darij


Let $ a,b,c,d>0$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}1$. Prove that

$ 19(a^3\plus{}b^3\plus{}c^3\plus{}d^3)\plus{}51(abc\plus{}bcd\plus{}cda\plus{}dab)\geq4$

[quote="Ligouras"]Let $ a,b,c,d > 0$ such that $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that

$ 19(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 51(abc \plus{} bcd \plus{} cda \plus{} dab)\geq4$[/quote]

$ LHS\equal{}f(a,b,c,d)$

Let $ a \le b \le c \le d$

1. $ f(a,b,c,d) \ge f(a,b,\frac{c\plus{}d}{2},\frac{c\plus{}d}{2})\Leftrightarrow 57(c\plus{}d)(d\minus{}c)^2 \ge 51(a\plus{}b)(d\minus{}c)^2$ $ \Rightarrow min(LHS)$ holdes for $ c\equal{}d$
$ a \le b \le c \equal{} d$

2. $ f(a,b,c,c) \ge f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2},c)\Leftrightarrow 57(b\plus{}c)(c\minus{}b)^2 \ge 51(a\plus{}c)(c\minus{}b)^2$ $ \Rightarrow min(LHS)$ holdes for $ b\equal{}c$
$ a \le b \equal{} c \equal{} d$

3. $ a\equal{}x$ , $ b\equal{}c\equal{}d\equal{}\frac{1\minus{}x}{3}$ , ($ 0 \le x \le \frac{1}{4}$)

$ LHS\equal{} g(x)\equal{}19x^3\plus{}4(1\minus{}x)^3\plus{}17x(1\minus{}x)^2\equal{}32x^3\minus{}22x^2\plus{}5x\plus{}4$

 $ g'(x) \equal{} (24x\minus{}5)(4x\minus{}1)$ , $ g(\frac{1}{4}) > g(0) \equal{} 4$

So $ min(LHS) \equal{} 4$ and holdes for $ a\equal{}0$ , $ b\equal{}c\equal{}d \equal{} \frac{1}{3}$


function logarithms algebra domain Gauss ARML geometry

Here's an interesting problem that appeared in this month's magazine.

For what real value(s) of $a$ do the graphs of $y = \log_a x$ and $y = a^x$ intersect at exactly one point? For different values of $a$ how many intersections will there be?

I like this question.  It should ask for positive real $a$ other than 1, although I wonder if you could do it for negative $a$ where those functions are simply restricted to the domains on which they are defined.

Gauss, will you be at ARML this year?

Yes I will. At the Penn State site. Maybe we'll get together for lunch or dinner sometime.

And I like this question because it has lots of different "layers" to it. You can look into this question as deeply as you want, and extend it to other interesting questions.

The graphs are symmetric about $y = x$, hence, they will intersect at one point iff either graph intersects $y = x$ exactly once; that is, iff

$x = a^x = \log_a x$

Has exactly one solution.  Now, we can write this as

$\frac{\ln x}{x} = \ln a$

Examine the graph of $f(x) = \frac{ \ln x}{x}$.  It has a maximum and falls off on both sides, approaching $0$ on the right; that is,

For $a < 1$ we must have only one solution because $\frac{\ln x}{x}$ only takes on a specific negative value at one point;

For $a_{max} > a > 1$ we have two solutions, one small and one large;

For $a = a_{max}$ we have one solution.

So what is $a_{max}$?

$f'(x) = - \frac{\ln x}{x^2} + \frac{1}{x^2} = \frac{1 - \ln x}{x^2}$,

So the maximum occurs at $x = e$, or

$\ln a = \frac{1}{e}$
$a = \boxed{ e^{ \frac{1}{e} } }$.

(I realize I have omitted some rigor.  The rigor can be filled in easily  :P )

[quote="gauss202"][quote="t0rajir0u"]For $a < 1$ we must have only one solution because $\frac{\ln x}{x}$ only takes on a specific negative value at one point[/quote]

Your answer to what happens when $a > 1$ is absolutely correct. But for the statement above, I think you're pre-supposing that there's only one solution when you derive the equation $\frac{\ln x}{x} = \ln a$. So you can't use it in turn to justify there being only one solution.

EDIT:
Actually, I think your problem originates further back with the "if and only if" statement here. Why is this true?

[quote="t0rajir0u"]The graphs are symmetric about $y = x$, hence, they will intersect at one point iff either graph intersects $y = x$ exactly once[/quote][/quote]

Suppose the two graphs intersect at a point that is not on the line $y = x$.  Then they will also intersect at the reflection of that point across $y = x$, and if those two points are distinct the graphs will intersect at two points.

(This is also justifiable using the fact that $a^x$ is convex.  If there is one intersection point, then

$\log_a x = a^x$

Has one solution, or

$x = a^{ a^x}$

Has one solution.  Using the fact that $a^x$ is convex, you can show that this impiles

$x = a^x$,

Which leads to my discussion of $f(x) = \frac{ \ln x}{x}$.)

Now, for $x < e$ the function is convex increasing, so only one unique solution to $f(x) = r$ ($r$ is a negative real, the case $a < 1$) exists.  This is evident from the graph (I told you the rigor was not there :D )

[quote="mathmanman"]Also discussed at : http://www.mathlinks.ro/Forum/viewtopic.php?t=86600 . :)[/quote]

That's a similar question, but not quite the same. The graphs here have a little bit different features so the answer is a little bit different.

I found it to be quite unexpected.

[quote="gauss202"][quote="t0rajir0u"]
Suppose the two graphs intersect at a point that is not on the line $y = x$.  Then they will also intersect at the reflection of that point across $y = x$, and if those two points are distinct the graphs will [b]intersect at two points[/b].[/quote]

Could they intersect at more than 2?

[quote]
If there is one intersection point, then...
$x = a^{ a^x}$ Has one solution.
Using the fact that $a^x$ is convex, you can show that this impiles $x = a^x$, which leads to my discussion of $f(x) = \frac{ \ln x}{x}$[/quote]

Here you're assuming that there's one intersection point from the outset. So you can't use anything derived from that assumption to conlude that there is only one point of intersection! That's circular reasoning.[/quote]

I should probably have been more clear about the casework I was implicitly using.

[b]Case:[/b]  The two graphs intersect nowhere.  This case is irrelevant to the problem.

[b]Case:[/b]  The two graphs intersect at one point on $y=x$ and nowhere else.  This leads to the previous discussion.

[b]Case:[/b]  The two graphs intersect at at least one point not on $y=x$.  Then they intersect at one more point and this case is irrelevant to the problem.


linear algebra
matrix
complex numbers
linear algebra unsolved

$A,B$ are two real matrices such that $ e^{tA}$ is bounded for nonnegative values of $t$. Prove that $ e^{tA+B}$ has the same property.

It works for complex matrices as well, I believe.

We may assume that $A$ is in Jordan form, since if $S$ is invertible and $(A_n)_n$ is a sequence of matrices, $(A_n)_n$ is bounded iff $(SA_nS^{-1})_n$ is. Let $N$ be the nilpotent part of $A$, and $D$ the diagonal part (s.t. $A=N+D,\ ND=DN$).

If $A=N$, it's easy to see that we must in fact have $A=O$.

If $A$ only has purely imaginary non-zero eigenvalues, then the problem is easily reduced to the previous case by using $e^{tA}=e^{tN+tD}=e^{tN}e^{tD}$.

If $A$ has some non-zero eigenvalue $\lambda$ with non-zero real part, then for any fixed matrix $X$ we can find $t\ge 0$ large enough so that $|e^{t\lambda}|>\|X\|$ (we can use the norm $\|X\|=\max_{i,j}|x_{ij}|$, for instance), and by putting $X=t_0A+B$ for arbitrary fixed $t_0\ge 0$ we see that $\sup_{t\ge 0}\|e^{tA}\|\ge\sup_{t\ge 0}|e^{t\lambda}|\ge\sup_{t\ge 0}\|e^{tA+B}\|$, and the conclusion follows.

I hope it's correct.


geometry
ratio
analytic geometry
symmetry
parallelogram

You can use coordinate geometry, which from what I know, gets pretty ugly.

You can use similarity and proportion:

Let $A_1, B_1, C_1$ be the midpoints of $BC,CA,AB$ respectively. If $G$ is the centroid of $\triangle ABC$, then it's also the centroid of $\triangle A_1B_1C_1$. Let $A_1B_1$ intersect $CC_1$ in $M$. If $CC_1=m$, then $CM=\frac{m}{2}$, and because of the similarity, $G$ must divide $CC_1$ and $C_1M$ in the same proportion. If $CG=x$, then we have

\begin{eqnarray*}\frac{CG}{GC_1}=\frac{GC_1}{GM} &\implies& \frac{x}{m-x}=\frac{m-x}{x-\frac{m}{2}}\\ &\implies& x^2-\frac{mx}{2}=m^2-2mx+x^2\\ &\implies& \frac{3mx}{2}=m^2\\ &\implies& x=\frac{2m}{3}\end{eqnarray*}

Hence $G$ divides $CC_1$ in the ratio $1: 2$

[hide="proof"]Let our triangle be $\triangle ABC$ with midpoints $M$, $N$, and $P$ opposite $A$, $B$, and $C$ respectively. THe centriod is $G$.
Using mass points, assign masses such that $w_A=w_B=w_C=1$. Then using the addition for mass points, $w_M=w_N=w_P=2$. Then $w_A*AG=w_M*GM\implies\frac{AG}{GM}=2$. By symmetry, we have the other ratios as well. [/hide]

[hide="comment"]This proof is essentially the same as frt's proof because mass points are just a simplification of triangle areas.[/hide]

[hide="Yet another approach"]
Let $M$ be the midpoint of $BC$ (in $\triangle ABC$).
Take point $D$ outside of $\triangle ABC$ so that $ABDC$ is a parallelogram.
Note that $D$ is on $\overleftrightarrow{AM}$ (because $M$ is the midpoint of $BC$ and we know that the two diagonals of a parallelogram bisect each other.)

Let $N$ be the midpoint of $AC$.
Since $AC=BD$, we have $BD=2\cdot AN$.
Then, since $AN\parallel BD$, it follows that $NG: GB=AN: BD=1: 2$ ($G$ is the centroid), and we are done.
[/hide]

[quote="bkunchanapalli"]how do we prove that a centroid divides the median in the ratio 2:1[/quote]

[hide] A very known proof starts by knowing that the medians of a triangle divides it in six smaller triangles with equal area. Then, you can continue with proportion of segments and areas[/hide]

José, just have a look at that :) :
[quote="frt"]If you use the fact that the 6 small triangles created by the three medians have equal areas, the proof is easy.

[hide="Proof"]
Given $\triangle ABC$, $D$, $E$ and $F$ are the midpoints of $AB$, $BC$ and $CA$, respectively, and let $G$ be the centroid.

Since $\triangle ADG=\triangle AGF=\triangle FGC$, we have $\triangle AGC=2\triangle ADG$.
Therefore, $CG: GD=2: 1$ ($\because$ $\triangle AGC$ and $\triangle ADG$ have the same height, so $CG: GD= \triangle AGC: \triangle ADG$)
Similarly, we have $AG: GE=2: 1$ and $BG: GF=2: 1$.
[/hide][/quote]

mass points would probably be the easiest way to prove this i think  :P

frt, why is triangle ADG = AFG? The heights aren't the same because the Centroid doesn't have the same distance from all three sides, which would be only for the incenter. There's only one side in common, which is AG, so you can't use any of the congruence postulates.
Thanks!!


Find three positive integer numbers $a,b,c$ that satisfy this system:
$a+b=[gcd(a,b)]^2$
$b+c=[gcd(b,c)]^2$
$c+a=[gcd(c,a)]^2$