Asymptote geometry circumcircle function

I know how to construct perpendicular bisectors now with straightedge and compass, and was hoping to try something similar with Asymptote (I know, it has something for the circumcenter), but how to you make a dot on the intersection points of two circles, lines, etc.  And also, how do you connect them?

Asymptote introduced a new function that gives an array of intersection points in version 1.21 where you can have intersectionpoint[color=red]s[/color](path p, path q). Note the s on the end of intersectionpoint[color=red]s[/color].
So make sure you have a recent version (the latest is 1.28) and then
[code]size(8cm,0);
import math;
import graph;
real r,s;
pair a,b, common;
path circ1, circ2;
r=1; s=1;
a=(0,0);
b=(1,0);
circ1=circle(a,r);
circ2=circle(b,s);
draw(circ1,linewidth(1bp));
draw(circ2,1bp+green);
pair [] x=intersectionpoints(circ1, circ2);
dot(x[0],3bp+blue);
dot(x[1],3bp+blue);
draw(x[0] -- x[1],1bp+red);[/code]will give
[asy]size(8cm,0); import math; import graph; real r,s; pair a,b, common; path circ1, circ2; r=1; s=1; a=(0,0); b=(1,0); circ1=circle(a,r); circ2=circle(b,s); draw(circ1,linewidth(1bp)); draw(circ2,1bp+green); pair [] x=intersectionpoints(circ1, circ2); dot(x[0],3bp+blue); dot(x[1],3bp+blue); draw(x[0] -- x[1],1bp+red);[/asy]


Does there exist DISTINCT reals $a,b,c,d$, such that $a+b=c+d$ and $ab=cd$?

Alternatively (and more simply),
[hide]Consider the two polynomials (x - a)(x - b) and (x - c)(x - d).[/hide]

Or, instead, [hide]set $c = a+r, d = b-r$ for some real number $r$.[/hide]

So, that's 3 distinct ways; do we have any more?

[hide]$a+b=c+d,ab=cd,a=c+d-b,b(c+d-b)=cd,bc+bd-b^{2}=cd,b(c-b)=d(c-b),b=d$
hence no solutions.[/hide]
I tried to generalise but I couldn't....is it possible?

[quote="srulikbd"]I tried to generalise but I couldn't....is it possible?[/quote]

It's not possible with just the sum and the product.  But, as my first solution suggests, it is possible provided you give more equations, such as the symmetric sums of the roots taken $n$ at a time.  There are many examples of failure for 3 numbers with just the sum and the product.  {0, r, -r} has the same sum and product for any r, for example.

[hide]substitute a, we get a=cd/b

we know that:
 a+b=c+d
cd+b²=cb+db
b²-db=cb-cd
b(b-d)=c(b-d)
b=c=impossible[/hide]

[quote="t0rajir0u"][hide="Solution"] $ab = cd \implies \log a+\log b = \log c+\log d$.  But by Jensen's this implies $\log$ is linear, contradiction.  Hence we cannot have such reals. [/hide][/quote]

Could you elaborate? I don't follow.

Ah, I should probably not have used that argument.  It's extremely unrigorous (and Jensen's wasn't what I really meant), but uses the concavity of $\log$:  the idea is to assume WLOG that $(a, b)$ majorizes $(c, d)$, which implies that the LHS must be strictly less than the RHS.

[hide]
WLOG, let $a<d, d<b$. Let $c=a+p$ and $b=d+q$.
Sub this into $ab=cd$ and we get $ap=dq$, and sub again into $a+b=c+d$ and we get $p=q$. This implies $a=d$, contradiction.
[/hide]


function
integration
trigonometry
calculus
real analysis
real analysis unsolved

Let f: R -> R be periodic with period 2pi, and Riemann-integrable on [-pi,pi]. Suppose that, for some real number alpha > 0, the function f is identically equal to zero in the interval (-alpha,alpha). Prove that for each beta in (0,alpha), the Fourier series of f converges uniformly on [-beta,beta].

$f_{n}(x)=(f*D_{n})(x) = \frac{1}{2 \pi}\int_{a<|t|<\pi}f(t) \frac{\sin((n+\frac{1}{2})(x-t))}{\sin(\frac{1}{2}(x-t))}dt$ is the partial sum of the Fourier expansion of $f$. Suppose for a moment that $f$ is $C^{\infty}$ periodic, and take $-a<-b<x<b<a$. Then
$f_{n}(x)=\frac{1}{2 \pi}\int_{a<|t|<\pi}[\frac{f(t) }{\sin(\frac{1}{2}(x-t))}] [\sin((n+\frac{1}{2})(x-t))] dt$
Now, integration by parts, and the fact that $|x-t|>a-b$ should conclude this particular case. For the general problem, remember that smooth functions are dense in $L^{1}$.


geometry
quadratics
perimeter
ratio
trigonometry
algebra
quadratic formula

Organized List of Geometry Solutions:
Problems are located at http://155.225.48.46/statemat/stuf2004/Geoprobs.htm

[u]Solutions[/u]
[hide="Number 2 by Joe"]
First, we use the fact that the altitude of a right right triangle is the geometric mean of the two segments that it divides the hypotenuse into: \[ \frac p{\frac 14(p+q)}=\frac{\frac 14(p+q)}q\Rightarrow p^2-14pq+q^2=0.  \]  Now, divide through by $q^2$: \[ \left(\frac pq\right)^2-14\left(\frac pq\right)+1=0.  \]  Using the quadratic formula, \[ \frac pq=\frac{14\pm\sqrt{196-4(1)(1)}}2=7\pm4\sqrt{3}.  \]  Since the problem tells you it is in the form $a-b\sqrt{3}$ you could be done right now.  But assume you didn't know that.  How would you decide between $7+4\sqrt{3}$ and $7-\sqrt{3}.$  Simple.  We were given that $p<q\Rightarrow \frac pq<1$ ($p$, $q$ must be positive).  If we look at $7\pm4\sqrt{3}$ as $\sqrt{49}\pm\sqrt{48}$ it makes it much easier to see that we must take the negative sign.  Thus, our answer is $7-4\sqrt{3}.$  The problem asked for the ordered pair, $(a,b)$, so $(7,4)$ is the answer to the problem.[/hide]
[hide="Number 3 by Joe"]

First, it is [i]really[/i] helpful to know the angle bisector theorem for this problem...so [url=http://www.cut-the-knot.org/Curriculum/Geometry/AngleBisectorRatio.shtml]here[/url]'s a link.

Now, by the angle bisector theorem, $7\cdot CD=5\cdot BD$.  But, $BD+CD=BC=3.$  So we have a pair of simultaneous equations for $BD$ and $CD$.  Solving, we get $BD=\frac 74$ and $CD=\frac 54.$

We do the same thing for $AE$ and $CE$ (use angle bisector theorem, get a pair of simultaneous equations) and solve to find that $AE=\frac 72$ and $CE=\frac 32.$

So then using the angle bisector theorem on $\triangle ABE$ we get $\frac xy=\frac 7{(7/2)}=2.$

Actually, I just realized that finding $BD$ and $CD$ was not necessary...anyway, I'm tired and I already wrote it, so it's staying.   [/hide]
[hide="Number 4 by Shobhit"]
Knowing tangents from each point are equal, we have $AB=50, BC=27+x, AC=23+x$, knowing this, we can use $rs=A=\sqrt{s(s-a)(s-b)(s-c)} \Longrightarrow 21(50+x)=\sqrt{50+x(x)(23)(27)} \Longrightarrow 21^2(50+x)=621x \Longrightarrow 22050=180x \Longrightarrow x=122.5$. From here, we see that the perimeter of the triangle is $\boxed{345}$.
[/hide]
[hide="Number 5 by Shobhit"]
Let $x=PS$, and draw in $OS$ and draw in $RQ$ such that $PR$ is the the diameter of Circle O. We see that triangle $POS$ is similar to $PRQ$ and are in a $1: 2$ side ratio, thus $\frac{x}{12-x}=\frac{1}{2} \longrightarrow x=\boxed{4}$[/hide]
[hide="Number 6 by Shobhit"]
Let $Z$ be the midpoint of $CB$ and so $ZA$ is the altitude of the triangle. So then $\triangle{AZD}$ is a right triangle with leg of $4$ and hypotenuse of $16$ so the altitude of the triangle is $\sqrt{240}$, so then $\triangle {AZC}$ is also a right triangle with leg of $x+4$ and $\sqrt{240}$ and hypotenuse of $17$ so $x=3$ and $CB=2x+8=\boxed{14}$[/hide]
[hide="Number 10 by Joe"]10. Triangle ABC is inscribed in a circle and BP bisects angle ABC. If AB = 6, BC = 8, and AC = 7, compute BP.

Using the angle bisector theorem we have:

\[ 6(CP)=8(AP)\Rightarrow AP=\frac 34 CP.  \]

We also have that $AP+CP=AC=7$.  Thus $\frac 34 CP+CP=\frac 74 CP=7$ meaning that $CP=4$.  Also, this makes $AP=3$.

Now, by Stewart's Theorem we have:

\[ 6^2(4)+8^2(3)=(7)BP^2+(3)(4)(7)\Rightarrow BP=6[/hide: d46efe42eb] [hide: d46efe42eb=number14byJoe]In the picture, the cevian from $: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebB$: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb is an altitude denoted by $: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebh_b$: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb (a 'cevian' is just a segment from a vertex of a triangle to the opposite side), the cevian from $: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebA$: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb is an angle bisector and the cevian from $: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebC$: 85b27eabbe: 5fbdbf67a8: 5f1ea05522: 4a74f95561: eeb94b27be: 765c56d723: 02ef952252: c79579760b: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb is a median. Everything else in the picture should be self-explanatory. [img: d46efe42eb]http: //img.photobucket.com/albums/v213/joml88/geom14ceva.png[/img: d46efe42eb] First we use the angle bisector theorem and get \[12m=15n\Rightarrow m=\frac 54 n.  \]

Putting that aside for a second, we know from Ceva's theorem that

\[ 4m(15-y)=6ny\Rightarrow m(15-y)=ny.  \]

Now we sub $\frac 54 n$ in for $m$

\begin{eqnarray*} \frac 54 n (15-y) &=& ny\\ \Rightarrow 75-5y &=& 4y\\ \Rightarrow y &=& \frac{25}3\\ \Rightarrow 15-y &=& \frac{20}3 \end{eqnarray*}

So now we know $y$ and $15-y$ in the picture.  Also, from the Pythagorean theorem we have

\begin{eqnarray} \left(\frac{20}3\right)^2+h_b^2 &=& 12^2\\ \left(\frac{25}3\right)^2+h_b^2 &=& (m+n)^2 \end{eqnarray}

Thus $(1)-(2)=\frac{400}9 - \frac{625}9 = 144-(m+n)^2 \longrightarrow \boxed{m+n=13.}$[/hide]
[hide="Number 12 by Shobhit"]Knowing tangents from each point are equal, let:

$a+d=18$
$d+c=21$
$b+c=17$
$b+a=?$

By adding the first three equations together we get $2(d+c)+a+b=56 \Longrightarrow \boxed{a+b=14}$[/hide]
[hide="Number 13 by Shobhit"]
We can see that $\frac{x-y}{2}=30 \Longrightarrow x-y=60$
We can also see that arc $ABD=230$ and arc $BAD=250$, and so $490-x-y=360 \Longrightarrow x+y=130$, so now we have a system of equations and we see that $x+y=190 \Longrightarrow x=\boxed{95}$[/hide]
[hide="Number 14 by Joe"]
In the picture, the cevian from $B$ is an altitude denoted by $h_b$ (a 'cevian' is just a segment from a vertex of a triangle to the opposite side), the cevian from $A$ is an angle bisector and the cevian from $C$ is a median.  Everything else in the picture should be self-explanatory.

[img]http://img.photobucket.com/albums/v213/joml88/geom14ceva.png[/img]

First we use the angle bisector theorem and get

\[ 12m=15n\Rightarrow m=\frac 54 n.  \]

Putting that aside for a second, we know from Ceva's theorem that

\[ 4m(15-y)=6ny\Rightarrow m(15-y)=ny.  \]

Now we sub $\frac 54 n$ in for $m$

\begin{eqnarray*} \frac 54 n (15-y) &=& ny\\ \Rightarrow 75-5y &=& 4y\\ \Rightarrow y &=& \frac{25}3\\ \Rightarrow 15-y &=& \frac{20}3 \end{eqnarray*}

So now we know $y$ and $15-y$ in the picture.  Also, from the Pythagorean theorem we have

\begin{eqnarray} \left(\frac{20}3\right)^2+h_b^2 &=& 12^2\\ \left(\frac{25}3\right)^2+h_b^2 &=& (m+n)^2 \end{eqnarray}

Thus $(1)-(2)=\frac{400}9 - \frac{625}9 = 144-(m+n)^2 \longrightarrow \boxed{m+n=13.}$[/hide]
[hide="Number 17 by Shobhit"]

I drew in the chord who's length is $1$ and made a right triangle (the radius of the circle as the hypotenuse, $\frac{1}{2}$ and $1-x$ as the leg, and got $1-x=\frac{\sqrt{3}}{2}$ and then using trig, I got $tanx=\frac{\sqrt{3}}{3} \Longrightarrow x=30$ degrees and so the number of degrees before we hit the chord of length $1$ is $60$. So therefore if $k$ refers to the number of terms right after hitting $60$ degrees, we have $\boxed{11}$, since $\frac{10(10+1)}{2}=55$ degrees$< 60$ degrees. [/hide]
[hide="Number 18 by Shobhit"]
This is probably the easiest out of all the problems, no real trick involved, just do a little bit of angle chasing. We know that this is a regular heptagon and so each of the interior angles are $\frac{900}{7}$. Thus $\angle{EDC}=\frac{900}{7}$ and so $\angle{DEC}=\frac{180}{7}$ and so $\angle{CEF}=\frac{720}{7}$. We know that quadrilateral $FECB$ is cyclic and so $\angle{FBC}=\frac{540}{7}$ and so $\angle{PBA}=\frac{360}{7}$ and we know that $\angle{BAP}=\frac{180}{7}$ and so $\boxed{\angle{APB}=\frac{720}{7}}$[/hide]
[hide="Number 19 by Shobhit"]Using a hint that we need to use Ptolemy's Theorem to solve this problem, I first labeled the points of the pentagon $ABCDE$, where $DC=6$ and everything else is $16$. From here I decided to split the pentagon into triangles and realized that there were only two types of triangles the pentagon can be split up into (that we know atleast two sides of), triangles with two sides of $16$ and triangles with sides of $6$ and $16$, so therefore $AD=EB=AC$ and $CE=DB$. So then I called each of the first three segments $x$ and each of the second two segments $y$ and then I decided to split the pentagon into quadrilaterals so I can use Ptolemy's Theorem. First I had quadrilateral $ABDE$ and using Ptolemy's I had: $16^2+16y=x^2$ and then I then made the other possible quadrilateral which was $EBCD$ and used Ptolemy's to get $6x+16^2=y^2$ and the last possible quadrilateral $ABCD$ and got $yx=16x+96$
Okay so after simplifying for $x$ and getting the equation $x^3-512x-1536$, I used rational root theorem and using trial and error I found that only $x=24$ worked for this equation and so $y=20$ and so after using Heron's formula on the three triangles and adding their areas up, we have the area of the Pentagon is $\boxed{123\sqrt{7}}$[/hide]

[quote="Iversonfan2005"][quote="mysmartmouth"]Um, Shobhit, this isn't very organized.  You wanna put them in numerical order?[/quote]

well, you're a mod, i'm sure you can fix it :P[/quote]

Done.

[hide="Guess what?"]
You can put spaces in the hide tags![/hide]

I know.  I fixed that when I numerical-ized the post.  Also, a list of numbers we need solutions to is coming up...

$\text{We Need Solutions For}$

Number 1
Number 7
Number 8
Number 9
Number 11
Number 14
Number 15
Number 16

[quote="Iversonfan2005"][quote="joml88"]I thought we solved them all or at least most of them already? Hmmph.[/quote]

guess not...well...i'm on spring break so I'll finish the rest of em....

BTW..thanx for organizing them Sean..you're not as lazy as I thought :rotfl:[/quote]

Lol.  You're welcome.

[quote="joml88"]I thought we solved them all or at least most of them already? Hmmph.[/quote]

i thought we did too...


The Top 10 Countries in IMO 2005 Mexico:
1. China (235)  5 Golds 1 Silver
2. USA (213)    4 Golds 2 Silvers
3. Russia (212) 4 Golds 2 Silvers
4. Iran (201) 2 Golds 4 Silvers
5. Korea (200) 3 Golds 3 Silvers
6. Romania (191) 4 Golds 1 Silver 1 Bronze
7. Taiwan (190) 3 Golds 2 Silvers 1 Bronze
8. Japan (188) 3 Golds 1 Silver 2 Bronzes
9. Hungary (181) 2 Golds 3 Silvers 1 Bronze
10. Bulgaria (173) 2 Golds 3 Silvers 1 Bronze

Hey, as it says here

http://erdos.fciencias.unam.mx/results.htm

Ukraine has 181, so shouldn't it be 10th?

nope. equal 9th.

i made a ranking of the lot (although i thought we had 98 countries, not 91 - what happenned?):

rank, country, total score:

1, China, 235
2, United States of America, 213
3, Russian Federation, 212
4, Iran, 201
5, Korea, 200
6, Romania, 191
7, Taiwan, 190
8, Japan, 188
9, Hungary, 181
9, Ukraine, 181
11, Bulgaria, 173
12, Germany, 163
13, United Kingdom, 159
14, Singapore, 145
15, Vietnam, 143
16, Czech Republic, 139
17, Hong Kong, 138
18, Belarus, 136
19, Canada, 132
20, Slovakia, 131
21, Moldova, 130
21, Turkey, 130
23, Thailand, 128
24, Italy, 120
25, Australia, 117
26, Kazakhstan,112
27, Colombia, 105
27, Poland, 105
29, Peru, 104
30, Israel, 99
31, Mexico, 91
32, France, 83
33, Armenia, 82
33, Brazil, 82
33, Croatia, 82
36, India, 81
37, Georgia, 80
38, New Zealand, 77
39, Serbia and Montenegro, 75
40, Austria, 74
40, Belgium, 74
42, Indonesia, 70
42, Switzerland, 70
44, Denmark, 69
45, Estonia, 68
46, Argentina, 65
47, Latvia, 62
47, Netherlands, 62
49, Azerbaijan, 59
50, Greece, 58
51, Ireland, 55
52, Cuba, 54
53, Lithuania, 53
54, Macedonia, 50
55, Bosnia and Herzegovina, 49
55, Finland, 49
55, Slovenia, 49
58, Kyrgyzstan, 46
58, Spain, 46
60, Albania, 44
61, Sweden, 42
62, South Africa, 39
63, Macau, 38
63, Norway, 38
65, Costa Rica, 37
65, Uruguay, 37
67, Sri Lanka, 32
68, Philippines, 30
69, Portugal, 27
70, El Salvador, 25
71, Iceland, 23
72, Morocco, 18
72, Turkmenistan, 18
74, Ecuador, 17
75, Malaysia, 15
75, Venexuela, 15
77, Cyprus, 14
78, Trinidad and Tobago, 13
79, Paraguay, 12
80, Pakistan, 11
81, Tunisia, 9
82, Puerto Rico, 8
83, Guatemala, 6
84, Liechtenstein, 4
85, Bangladesh, 3
85, Kuwait, 3
85, Lexembourg, 3
85, Saudi Arabia, 3
85, Tajikistan, 3
90, Mozambique, 2
91, Bolivia, 0

[quote="tywebb"]

although i thought we had 98 countries, not 91 - what happenned?

[/quote]

I think Uzbekistan and Mongolia could not come because of visa/travel problems. Does anyone know any other teams which had similar problems? It's very bad that such things still happen, in this day and age where we have better communications. Arrangements should be made better in future years.

hey good job everyone. special congrats to the Canadian team and the Iranian team.

china did it again huh? what a great team. if I'm not mistaken the only lost 3 mark on average. :first:

fatehi, stop flooding all the topics and find it out in the oficial website.

[quote="nonIMO"]China:
2003 - 211
2004 - 220
2005 - 235
Looking forward for the perfect score (252) in 2006  :D[/quote]
Has there ever been a team getting 252 in history?

[quote="Idril"]Yes once there was a perfect score of 252

1994

IMO was held in Hong Kong.
69 countries were represented.
There were 30 gold medals, 63 silver medals, and 102 bronze medals awarded.
[b]For the first time ever in IMO history, all the members of a national team scored a perfect 42, earning them each a gold medal.[/b] This team was from the USA and consisted of Jeremy Bem, Aleksandr Khazanov, Jacob Lurie, Noam Shazeer, Stephen Wang, and Jonathan Weinstein. Their achievement was featured in TIME magazine.
Oleg Alexandrov earned a bronze medal and became the first Moldavian medalist.
Nasko Karamanov became the first Macedonian competitor to win a silver medal

(http://imo.wolfram.com/morefacts.html)[/quote]

Yes, I knew this. BTW, it looked like the world's combined team (pay attention at names of Aleksandr Khazanov, Jacob Lurie, Noam Shazeer, Stephen Wang). And I should rather name that IMO [i]the strangest ever[/i]. Because in [b]IMO 1993[/b] two of them (Stephen Wang and Jeremy Bem) scored 22 and 18 pts respectively with 151 pts totals for the USA. And in [b]IMO 1995[/b] another two of them (Aleksandr Khazanov and Jacob Lurie) scored 34 and 31 respectively  and USA became 11th with 178 pts. :huh:

Anyway, it would be nice to see one year the perfect China's performance.  :D

[quote="Elyot"][quote="fatehi"]by the way how many gold medals did the Canadian team get?[/quote]

Only 1... Yufei Zhao, with a score of 35.
We got also 2 silver and 2 bronze.
I'm amazed that we beat Moldova.[/quote]

It was Moldova that failed. I think, they just could not keep concentrated after such expectations.

[quote="siuhochung"]what valentin means, i believe, is that in the past, students cannot access training materials freely. Maybe for teams like China, Vietnam, the leaders can give them many many problems for practice. But for other participants, say for an example, a normal Hong Kong boy cannot access many materials. For me, I can't find much problems suitable for imo level before i found kalva and mathlinks. Team leaders don't give us problems enough to brush up certain type of skills or maybe in a general sense, to get a silver/gold medal. I think that's why some other countries are weaker although the students are not really worse than those in stronger countries.[/quote]

How true. Our team got A LOT of problems to do this year. And we made them write up solutions which we reviewed. We had the time to do this because we were mostly released from the army early this year and had nothing to do till college. So watch out, here we come! (Think if we aim for top 20 each year, it'll be a decent target, no?)

Of course, you can't say that students in one country are less capable then those in another... But certainly the education system plays a part. If a country has a strong educational system, then the training can start from a higher base.

In Singapore we do practically no geometry, so our geom training starts with... This is called a cyclic quadrilateral... It has these properties... Because *proof*...
Which is a little sad, but it works out in the end since our training starts when we're quite young, and we can teach a lot during that time. (We even have to teach things like induction because that isn't done in school till the last couple of years!)

[quote="Valentin Vornicu"]:o I suggest you think more before posting, as you are very close of not being able to visit this site again :ninja:
[/quote]

*gasp* Anger Yoda you must not!

[quote="Valentin Vornicu"]
There were no ways to buy olympiad books years ago either. You could not find them, or they were written in a language (different than english) that you would not understood. I had this problem myself at some point, and this was one of the reasons for founding MathLinks.[/quote]

I remember my team mates training with chinese books that I couldn't read (Failed Chinese). But there was still Crux, our own Math Medley, and the Kalva site...

[quote="Valentin Vornicu"][quote="khuong"]For Valentin Vornicu, I am sorry to type large font 'cause I think you are near-sighted..[/quote]:o I suggest you think more before posting, as you are very close of not being able to visit this site again :ninja:

[quote="khuong"]I never think that this country is smarter than the other! Anyway, you do not have to teach me about this..For the internet access, let me tell you that, as far as I know when I visit Vietnam, Internet in this country is just popular in great cities like Hochiminh and HANOI. Most VIETNAMSE students winning gold medals thorughout the past 25 years are very poor..they win by their own capabilites not by any technologies or computer..believe me![/quote]You do think Vietnam is much smarter than the other countries. As I have told you, the reasons for which some countries have done much better in the past are mostly because they have teachers that can feed their students materials and olympiad problems (such as Vietnam, which has some great teachers :) ) so their students were clearly more advantaged at the IMO. Now this advantage is decreasing.

[quote="shobber"][quote="siuhochung"]students cannot access training materials freely. [/quote]
You can buy a lot of books about math Olympiad.[/quote]There were no ways to buy olympiad books years ago either. You could not find them, or they were written in a language (different than english) that you would not understood. I had this problem myself at some point, and this was one of the reasons for founding MathLinks.[/quote]

Hello Valentin Vornicu,

The first thing I wanna say is again "I do not wanna have problems with you and any one else!". Its' your right to think in that wat but I was not regretful about my words..

My advice for you is to calm down..What I say in this forum is mainly based on my teaching experience in Computer Science and Maths in Southeast Asia, UK, Australia and US..I could not comment on your statement "VN has great teachers"..I think the world knows clearly about that...Again, IMO training materials, as I agree with some friends here, are popular and can be found almost everywhere :)) (even in Bukina Faso (Africa), as I visited there 7 years ago, you can find them in 3 book stores even though their team never participated!)..Concerning the training system, the students there just have a few days before going to IMO to review the materials and that's really can't be count on..I would think that even the committee members of IMO would disagree with the thought that training and materials are the key to success..Remember again, this is a competition on how you understand and solve the problems not on 'memorizing those meterials'! There's clearly a gap between 'knowing something' and 'solving something completely'.

You can do anything you want..everybody here, who loves the truth and maths, will know about it! I certainly know that you are admin here but the fact that 'you could stop me to express my viewpoint' is, I am sorry, meaningless to everyones! :))

I hope that you could travel to the above-mentioned countries to see what's the truth not just listening from somebody or seeing from somewhere..

I also believe that for a country to succeed, in addition to great passion for maths, tradition and natural-born abilities cannot be failed to mention. This also partly explained why we, human beings, up to the present, only have one Eistein (not thousands!)..I think Russia is a good example. They are not great Soviet Union any more but the spirit of Russian Mathematics is still there..and that's why they're still in the top 3 countries in so many years..

I simply can't agree what you have said. China is such a strong team for so many years. But then, think about their training systems. They have a CMO. They then have a training camp for such a long period. They keep on doing problems during that period. Then think about Hong Kong. We only has a once-a-week futile training sessions, which keep on repeating what has been taught in the previous years. In the past we can't even conveniently access past imo questions. So how can I get good results?
Of course, talent plays an important role in the results of imo. But if a country has a good training system, it can surely fully unleash all the potential of the students there.

Concerning some points you have mentioned,
"even in Bukina Faso (Africa), as I visited there 7 years ago, you can find them in 3 book stores even though their team never participated!)"
7 years ago I cannot find any good imo-level math book in Hong Kong bookstore. The case is not as good as you expect.

"Remember again, this is a competition on how you understand and solve the problems not on 'memorizing those meterials'! There's clearly a gap between 'knowing something' and 'solving something completely'. "

What if we don't even have the chance to "memorize" it? Of course we won't try to do so, but just imagine, how can we solve a problem that can be solved by inversion if we haven't even known what inversion is?? Take me as an example. Before I come to mathlinks, I don't know what muirhead is. I don't even know what Schur is. I haven't seen any real-time application of majorization. Then how can I master those skills well?

Of course I don't have the right to intervene your affinity towards Vietnam, but I only want to tell you that not every country can have so many imo training materials early on. If a country can access many materials in early years, the students of that country can of course be more able comparing to other students. However, concerning the natural flair of students, I don't think that varies much from place to place.

I also disagree with khuong.

[quote]Remember again, this is a competition on how you understand and solve the problems not on 'memorizing those meterials'! There's clearly a gap between 'knowing something' and 'solving something completely'. [/quote]
Even the most talented mathematicians have to train a lot to attain top level at IMO. And this training is not just to learn theorems, it is also very important to develop your 'feeling' with the problem, your ability to see in which direction you have to search for a solution. This skill is not just a talent you're born with, it has to be trained too. And the best way to do so is make a lot of exercises over a long period of time.

This fact is very striking in the VWO, the semi-national Olympiad in Flanders (the North part of Belgium). Almost no people really train for this Olympiad, so if you train yourself a bit, you can make a difference. I am an example of this. I really began to train a half year ago, and suddenly I won a first prize at the VWO and I was selected for the IMO, quite unexpected for most. Talent is certainly not everything, you need good training.

I can't really judge the availability training materials in Belgium in the past, but even nowadays (with the internet) there is very little material in my language, Dutch. The fact that you have only training materials in English, discourages people to train themselves.

The situation is certainly not ideal in Belgium. Hopefully it will improve with the website Arne and I and some others are making. It will provide Dutch training materials and will encourage young talents to train themselves. Because without any training, you may have as much talent as you want, but you won't succeed very much.

Lets just say there is not one key to succes, but you need two keys to unlock: both talent and good training.

[quote="khuong"]My advice for you is to calm down..What I say in this forum is mainly based on my teaching experience in Computer Science and Maths in Southeast Asia, UK, Australia and US..

You can do anything you want..everybody here, who loves the truth and maths, will know about it! I certainly know that you are admin here but the fact that 'you could stop me to express my viewpoint' is, I am sorry, meaningless to everyones! :))
[/quote]I fail to see how insulting me, or any other member of AoPS - MathLinks community is based on your teaching experience ... or maybe you haven't exactly taught math?  :?  :|

I will of course not stop you from expressing your viewpoint [b]on the subject at hand[/b], as long as it does not insult other people in the process. If you step over this line once again, you will suffer the consequences.

[quote="khuong"]I would think that even the committee members of IMO would disagree with the thought that training and materials are the key to success..Remember again, this is a competition on how you understand and solve the problems not on 'memorizing those meterials'! There's clearly a gap between 'knowing something' and 'solving something completely'.[/quote]I think that you make great confusions my friend. There is a [b]huge[/b] difference between working with a lot of materials and, to quote you again, "memorizing those meterials". If you fail to see that, then it's pointless to continue the discussion with you on this particular topic ;)

PS And I totally agree with Stijn's and siuhochung's opinions also. I couldn't have said it better myself guys  :roll:

i agree with valentin and i think everyone does here. i mean come on¡¡ training is essential in terms of success in the imo. i agree in what you say about the theorems i mean students can definitely discover these theorems in order to help solve the problems at hand but i think you have to think in the time factor, students having no training could definitly solve the problems but it would take them much more time to solve it as they would to discover some tools that with training would not have to be discovered , just applied¡¡¡ and in turn would not have time to solve problems. so again you have another reason to say that your opinion is wrong :ninja:  ;)  ;)

hmm i agreethat training materials are important (for the same reason marko said: time)
but... i mean.. if you love math, you have a sense of research, not only problem solving, but developing your own questions and developing your own method on how to attack it.. and i think SOMETIMES (in some cases) olympiads can destroy the sense of research and the sense of love to math. for example, many olympiad students have becaome so technical in solving problmes that this has corrupted their creativity and when working with a problem instead of thinking and solving the problem their own way looking for a beautiful work of art they just try 'dumbassing' or  remembering old problems that look like this kind of problem. and excesive OLYMPIAD training can affect this research sense (that is what math really is... when the human mind challanges logic through original questions).  so instead of looking and reading for OLYMPIAD training students justhave to look for MATH (articles, papers, results, books, inclusively olympiad books, problems) .. it doesn't matter if it is important for olympiads or not.. just look for math and produce math!
so i think that olympiads can be really harmful for the math world sometimes..
although i  see olympiads as a way of strengthening you math basis in general areas, and things that u must know.. and i like olympiads a lot!!!! and i like training for olympiads too!! i love to do problems, to see results, etc... but i don't do this to do well in olympiads.. it is because i love math. i do it because of curiosity. remember that you must LOVE math more than olympiads.. and hopefully this way of thinking will make me more efficent in olympiads and in my future research... because i will never get tired..never! i will solve my olympiad problems because of my great curiosity and love for math!! not because i want to win a medal.. and i think that cursiosity and love can lead u more tu succsess than the desire to win . :D and i am afraid to say that many people in this site love more olympiads because of winning medals than olympiads because of math.. and this will clearly affect their future math research (if they dare to be pure mathematicians  ;) )   :(

True manuel. It is indeed proven that training too much on a small area is sometimes devastating for the other areas. But I think most people are far away from 'too much'. Some may be there, but in general I think sites like this cause major improving to your math skills, also in reasearch perspectives. Afterall, you're busy with math and that also helps a lot in research.

I think however those "areas" you're talking about are defined by the field you're working in, and not by the fact that it's olympiad. When I solve analysis/matrix problems -far from olympiad stuff right?-, I encounter many 'easy/stupid/ugly' problems, which I dislike but which I can solve, and also harder problems, which I admire but cannot always solve. I mainly like the problems because they're hard, because they challenge me and - yes also - because I can solve them while most others cannot. :)
That is also partially what makes problems beautiful, their level. If you're challenged by $a^2+b^2\ge2ab$, I think you have a serious problem. ;)

If you mean by olympiad the IMO stuff, that what can be attained by just high school education, and just keep going further in that, then indeed you are right, you will reach a limit above which it doesn't help you any further. But if you mean problem solving in general I think you're wrong. Look in the college section, these are math problems that are much closer to nowadays mathematics, and who is great at them? The smart people with good education. Just like in the olympiads. If there would not have been olympiads, if there would not have been sites like this, if there would not have been IMO, etc... how much of them would have studied maths? Not me at least. Maths = boring conic intersections, that's the maths we see before we decide what to study! :!:

[quote="Peter VDD"]If you're challenged by $a^2+b^2\ge2ab$, I think you have a serious problem. ;)[/quote]

darn. how do you do this one? it looks like schur but I can't get it to work out right

you'r right peter. olympiads show people a glimpse of what real math is.. and friends that where into olympiads..
but i am a bit different because i started very interested in combinatorics and started researching in combinatorics without beggining to train in olympiad problems.. i was lucky to take some grad, courses on graph theory,, and i become like really strong in combinatorics.. but really weak in other areas like, geometry and algebra.. but .. now i've been this year traning hard and learning a lot of nice new stuff from other areas, and seeing things that are important for me to know..so i am working with every olympiad area right now to see if i can succeed in next year's imo..
 ;)  but yes

[quote="manuel"]and i become like really strong in combinatorics.. but really weak in other areas like, geometry[/quote]exactly the same here :)

But if you want to learn something from univ courses you must have luck being on a univ that is any good in it. ;) that's also an advantage of sites like this: you're not bound to the skills of the national training staff. I wish there was a site like this for university level. (and I mean more than the college section, I mean with effective papers/courses online etc...

[quote="Peter VDD"]I wish there was a site like this for university level. (and I mean more than the college section, I mean with effective papers/courses online etc...[/quote]

Same here!!!

[quote="Soarer"][quote="shobber"][quote="siuhochung"]students cannot access training materials freely. [/quote]
You can buy a lot of books about math Olympiad.[/quote]
I seldom go to shenzhen (the last time i went there was three years ago). In Hong Kong, the olympiad books are of very low quality and that they are mainly for local competitions which are far beyond imo level. Now, it's better, because some bookstores in Hong Kong sell books published in taiwan, especially "Jiu Zhang Chu Ban She". These books are of much higher quality. But in the past, these books are rare and expensive.[/quote]

Their local competitions is terribly easy but their IMO team is among the best in the world. (Hong Kong only has 7 million pop. While Canada has 38 million, but the performance of Hong Kong is much better than ours, except the 1 point)

I heard that Jiu Chang publisher, and the translated version can be bought in Calgary, Canada, but not in Halifax.


HCSSiM blogs AMC USA(J)MO USAMO AIME geometry

Yeah.
Massachusetts ftw.
Even though Texas will probably win

um wow ok that poll went away pretty fast
shall i start a new thread?

no way, West Virginia will beast it up at nats.  well use our hillbilly power to crush the competition.  itll be like wrestling cattle back home.  :rotfl:

My Prediction in order:
Texas
Massachusetts
Indiana
Michigan
Washington (moved up)
California (I dropped them after talking to Ubemaya...)
Virginia
Florida
Ohio
Illinois
Kentucky
Wisconsin
New York
Arkansas (moved up)
Pennsylvania
Colorado
Missouri
New Jersey
New Hampshire
Utah
Oklahoma
Delaware
Vermont
Maine
Montana
South Carolina
North Carolina
Georgia
Tennessee
Mississippi
Oregon
Connecticut
Nebraska
Idaho
West Virginia
Iowa
Arizona
New Mexico
Alabama
Nevada
Hawaii
Rhode Island
North Dakota
South Dakota
Wyoming

[quote="frost13"][quote="Aznwithabrain"]My Prediction in order:
California
Texas
Massachusetts
Indiana
Michigan
Virginia
Florida
Ohio
Illinois
Kentucky
Wisconsin
New York
Pennsylvania
Colorado
Missouri
New Jersey
New Hampshire
Utah
Oklahoma
Delaware
Vermont
Maine
Montana
South Carolina
North Carolina
Georgia
Arkansas
Tennessee
Mississippi
Washington
Oregon
Connecticut
Nebraska
Idaho
West Virginia
Iowa
Arizona
New Mexico
Alabama
Nevada
Hawaii
Rhode Island
North Dakota
South Dakota
Wyoming[/quote]

I'm offended.  I leave New Hampshire and now they have their 3rd best finish ever.  Was I bringing them down :(

I did have a team finish 15th a couple years ago at least.[/quote]

my prediction is probably wrong :rotfl:
i just named some random states based on my own info (which is not really good)
and
i just stereotyped some states (cause i didn't know anything about them) :rotfl:

[quote="frost13"][quote="Aznwithabrain"]My Prediction in order:
California
Texas
Massachusetts
Indiana
Michigan
Virginia
Florida
Ohio
Illinois
Kentucky
Wisconsin
New York
Pennsylvania
Colorado
Missouri
New Jersey
New Hampshire
Utah
Oklahoma
Delaware
Vermont
Maine
Montana
South Carolina
North Carolina
Georgia
Arkansas
Tennessee
Mississippi
Washington
Oregon
Connecticut
Nebraska
Idaho
West Virginia
Iowa
Arizona
New Mexico
Alabama
Nevada
Hawaii
Rhode Island
North Dakota
South Dakota
Wyoming[/quote]

I'm offended.  I leave New Hampshire and now they have their 3rd best finish ever.  Was I bringing them down :(

I did have a team finish 15th a couple years ago at least.[/quote]
Actually to the contrary, New Hampshire got the enthusiasm from your great coaching. We are hoping to make it better than what this list ranks us.

[quote="rd5493"]am i on one of those state teams (just look at my sig to see which state im from) that you stereotyped.  if so, im offended Aznwithabrain, ur from KY, which is almost as hillbilly country as WV.  besides, the hatfields were from KY[/quote]

sorry rd5493
i just don't know what WV has been ranked the last two years (if you can tell me. I'll think about editing the list)
hey
Kentucky's rank last two years:
13th (two years ago): Top ranker was 15th
and 16th (last year): Top ranker was 19th

[quote="vector2101"]I personally hope that Aznwithabrain's prediction is wrong, probably because I'm on the Washington team, and the entire team would, having come in second last year, rather not take 30th place this time around... but that's just me. Also, has anybody else noticed that only 45 states are listed, out of 57 competing states/territories/departments?[/quote]

again people
I KNOW THE LIST IS WRONG :rotfl:

and yeah theres only 45 because I was too lazy to list the rest of them and the departments/territories

Edit:
I moved Washington up ALOT. After hearing about what they got last year. Sorry guys.

Yeah the California team, I believe is being underestimated. Although, many people who were expected to make it, didn't, I do believe that the team has some really good people like Pinsker. Plus, AIME and Mathcounts are totally different. There is a big difference so you can't really judge someone's Mathcounts skills by looking at their AIME score.

[quote="kstan013"][quote="bookaholic"][quote="yc"]
I bet North Carolina will do better than California. They have a USAMO qualifier even not making the team.  It's a super GOOD team.[/quote]
What gave you the impression that California only had one middle-school USAMO qualifier or something? We have eight, actually, if I counted right. Only one of which got onto the team. North Carolina appears to have just one middle-school qualifier, so if he didn't make it...[/quote]The NC guy didn't. He had made it in 6th grade, but I don't know what happened from there. Maybe he just stopped participating. btw his name is Bryce Taylor.[/quote]

Bryce I believe is an AoPSer, and if I have him connected properly, he's focusing on high school math competitions.

[quote="solafidefarms"][/quote][quote="kstan013"][quote="bookaholic"][quote="yc"]
I bet North Carolina will do better than California. They have a USAMO qualifier even not making the team.  It's a super GOOD team.[/quote]
What gave you the impression that California only had one middle-school USAMO qualifier or something? We have eight, actually, if I counted right. Only one of which got onto the team. North Carolina appears to have just one middle-school qualifier, so if he didn't make it...[/quote]The NC guy didn't. He had made it in 6th grade, but I don't know what happened from there. Maybe he just stopped participating. btw his name is Bryce Taylor.[/quote][quote="solafidefarms"]

Bryce I believe is an AoPSer, and if I have him connected properly, he's focusing on high school math competitions.[/quote]I suppose that makes sense. I DID see his name on the AMC 8 site though...peculiar. I wonder if he participates in MATHCOUNTS and is just indifferent on how he does (doesn't really prepare), or if he just avoids it all together. I guess it isn't really any of my business though :maybe:

He got 5th in NC according to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139815]this[/url]

[quote="yc"][quote="bookaholic"]I bet California will win. I know the captain, and I believe another of the team members is a 7th grader who qualified for the USAMO. It's a GOOD team.[/quote]

I bet North Carolina will do better than California. They have a USAMO qualifier even not making the team.  It's a super GOOD team.[/quote]

so does texas.
usamo qualifiers aren't neccisarily good as good at mathcounts.
on a different topic,  who is on the nc team?

[quote="LawOfSigns"]I think people are overrating florida because it has a high population.  really, 4th place got a 33- many states do better than that.  It's a good thing the AMC series of competitions is different from mathcounts- not just the same people dominating, but a chance to show what you're personally good at (I'm not good with the time pressure of mathcounts)[/quote]
Hmm, but this year Florida has a strong team, just watch.

[quote="LTSWANSON"]Yay Alabama is up there...Wow...Texas is good?Despite the school system here?[/quote]
It only takes 4 smart kids to win nats.
All 4 of the team members are quite good, and while we're counting, TX had 2 usamo qualifiers on the team and a usamo qualifier not making the team..

First of all, all Nats participants are smart! Well, all people in MathCounts pretty much are smart! (with a few exceptions, of course) And second, USAMO has nothing to do with MathCounts. They are [b]very[/b] different...

Even CONNECTICUT had the 10th place person get into USAMO!

jorian

[quote="fireninja0"][quote="Aznwithabrain"]I don't know rd5493

those ranks are pretty...trendy?

12
14
23
25

Looks like WV is going down each year
Must be the talent level :roll:
just kidding

Zmastr is right
we need to watch out for the defending champs[/quote]

OK, dude. I know you already apologized, but I'm on rd5493's team, and I am p-r-e-t-t-y sure that we'll get a pretty high rank this year. Our team capt. is a real beast, so watch out![/quote]

Woah, I just noticed prodigy called me a beast. Amazing.
I guess it's true.
I got 33 though I screwed up all of it with stupid mistakes.
Only one problem I didn't do.
Wish I had the motivational boost before competition not in high school.
I'm ashamed that I let down the team.

I'm pretty sure this is LAST year's MC?
Yeah.

A bit late for last year, a bit early for this year, you guys can make a new topic when the time comes.


3.0 amperes AC (effective value) is flowing in a 20-ohm heater. Find the maximum current and the power delivered.


function
calculus
integration
symmetry
probability
probability and stats

[img]http://i39.tinypic.com/rtnv9z.jpg[/img]

im having a bit of difficulty with part (ii) and (iii)

(ii) Do I just integrate the function from 0.5 to 0?

For ii), you integrate from $ \minus{}\infty$ to $ 0.5.$ Since the portion of the integral from $ \minus{}\infty$ to $ \minus{}1$ contributes nothing, this may in practice be written as an integral from $ \minus{}1$ to $ 0.5.$

For iii), you integrate from $ 0$ to $ 0.$ Of course, the result is zero - as would be the case for any continuous distribution (distribution with a density) at any single point.

This is even the case for functions with unbounded density. For instance, let $ Y$ have the density

$ g(y)\equal{}\begin{cases}\frac1{\pi\sqrt{1\minus{}y^2}}&\text{if }\minus{}1<y<1\\0&\text{otherwise}\end{cases}.$

Then it's even true that $ P(Y\equal{}1)\equal{}0,$ even though the density tends to infinity at that point.

yes i see it now

thanks for the reply prof.

Sorry, I misread ii). I thought it said $ P(X\le0.5)$ but it actually says $ P(|X|\le0.5),$ which is different.

$ P(|X|\le0.5)\equal{}P(\minus{}0.5\le X\le 0.5)\equal{}\int_{\minus{}0.5}^{0.5}\frac43(1\minus{}x^2)\,dx$

$ \equal{}2\int_0^{\frac12}\frac43(1\minus{}x^2)\,dx$ by symmetry -- $ f(x)$ is an even function.

As a check: this probability has to be greater than $ \frac12.$ It's half of the interval $ [\minus{}1,1],$ but is the portion of the interval on which $ f(x)$ is larger rather than the portion on which it is smaller. (Of course, being a probability, it also has to be less than $ 1.$)


$ \log_4(27\minus{}x)\equal{}\log_x 5$ . Find x .

Changing the bases, we get $ \frac {\log (27 \minus{} x)}{\log 4} \equal{} \frac {\log 5}{\log x}$, from which we can see that $ x \equal{} 2$ or $ x \equal{} 25$.


What is the ratio of the number of cubic inches in the volume of a cube to the number of the square inches in the surface area if one edge is $ s$ inches long?

Okay, so the surface area would be $ 6s^2$, and the volume would be $ s^3$.  So our answer is simply, $ \frac{s^3}{6s^2} \implies \boxed{\frac{s}{6}}$.

[quote=An_Absurd_Napkin]Thank you, but how do you know when they want it in the form s/6 and when in the form s:6[/quote]

It is almost always written as a fraction unless they tell you otherwise.


interesting...

$[(q-1)0]$

couldn't you just take this part out.

[quote="kennethkwek"]This only apply to value that has only the digit 9 in it(e.g 9999 or 99999999). This is the formula i discovered

{[(q-1)9]-8-[|(y-x)|9]-[(q-1)0]-1}
Where q is the number of digits of smaller no.
When both no. has the same value, take q as the number of digits of the number
y is the value of one no. while x is the other no.
Mutiply the number of digits, not the value.
-is to seperate sets of digits[/quote]

Formal proof available?

[quote="ch1n353ch3s54a1l"]The - separate the digits...
So 9(q-1)-8-0(q-1)-1 for 9999... q=the number of digits in the number.
So $99999^2$ is $9999800001$.  :P

They aren't supposed to be actual operations.[/quote]

this only works when you're squaring a #.

let $n=$# of 9's. then, $10^n-1$ represents the # you're squaring. $(10^n-1)^2=10^{2n}-2(10^n)+1$ is the square #. It's easy to see where the ones digit, the 1, comes from. it's easy to see that the # of digits is 2n.  in the last $n+1$ digits, there's an 8, one 1, and $n-1$ 0's. so you're left with $n-1$ digits which are 9's. i don't really know what you meant by the proof you were looking for but this is what i see.


analytic geometry
graphing lines
slope
vector
trigonometry
geometry

Find the distance between parallel lines [b][size=150]2y-3x+6=0 and 4y-6x-6=0[/size][/b].

The answer s [b][size=150]9/sqr(13)[/size][/b]

But i dont understand why this works. I tried to get a point from one of the lines. Then, I used the distance formula between a line and a point to get the distance between the 2 parallel lines. However, with different points, I get different answers and none of them equals the correct answer above. When should you use the distance formula? i'm confused.

so, i ended up getting the perpendicular line passing through them and i used the distance formula between the 2 points of intersection. Help please, thanks!:)

Let $ l_1 : 2y \minus{} 3x \plus{} 6 \equal{} 0\Longleftrightarrow 3x \minus{} 2y \minus{} 6 \equal{} 0$.

$ l_2 : 4y \minus{} 6x \minus{} 6 \equal{} 0\Longleftrightarrow 3x \minus{} 2y \plus{} 3 \equal{} 0$.

Arbiterary points $ (x,\ y) \equal{} (2t \plus{} 2,\ 3t)$ on the line $ l_1$, the distance in question is $ \frac {|3(2t \plus{} 2) \minus{} 2\cdot 3t \plus{} 3)|}{\sqrt {3^2 \plus{} 2^2}} \equal{} \boxed{\frac {9}{\sqrt {13}}}$.

Kunny's post above is a formulaic generalization of what you are trying to do

[b]Method 1:[/b] [i]Use a perpendicular line -- this is what you did[/i]

The idea is that the shortest distance between any two parallel lines is the [u]perpendicular[/u] distance -- that is, the segment along which the distance is taken should be perpendicular to both lines.

It is for this reason that we find any line that is perpendicular to both lines [i](the slope of these lines is 3/2, so we take any line of slope -2/3. y=-2/3*x makes for easy calculations)[/i]. If we find the intersections of this perpendicular line with these two lines, then we have an instance of the shortest distance between the two lines, and can use the distance formula.

The distance formula is always applicable. It is the definition of distance. However, unlike finding the distance between two points, if you are finding the distance between two lines, you need to first find a pair of points that you know will give you the right distance, and then use the formula.

[b]Method 2:[/b] [i]Minimize algebraically[/i]

This is another way to make sure that we have a pair of points with minimum distance.

Let's take some point on the first line -- for convenience, let's take the point $ (2,0)$. Now let's take any random point on the second line. Solving for $ y$, we get $ y \equal{} \frac {3}{2}x \plus{} \frac {3}{2}$, so we know that any point on the second line is of the form $ \left(x, \; \frac {3}{2}x \plus{} \frac {3}{2} \right)$

Now, let's use the distance formula: $ D^2 \equal{} (x \minus{} 2)^2 \plus{} \left(\frac {3}{2}x \plus{} \frac {3}{2} \right)^2 \equal{} \frac {13}{4} x^2 \plus{} \frac {1}{2} x \plus{} \frac {25}{4}$

We want to find the minimum value of $ D$, and therefore the minimum value of $ D^2$.

We note that we can complete the square to get $ D^2 \equal{} \frac {13}{4} \left( x \plus{} \frac {1}{13} \right)^2 \plus{} \frac {81}{13}$

The minimum value of that squared expression is $ 0$, which occurs when $ x \equal{} \minus{} \frac {1}{13}$. So, the minimum value of $ D^2$ is $ 0 \plus{} \frac {81}{13} \equal{} \frac {81}{13}$.

Taking the squareroot, the shortest distance between the two lines is $ \frac {9}{\sqrt {13}}$

Here's another way to do it using vectors.

Rearranging your two lines, we have

$ y_1 \equal{} \frac{3}{2}x \minus{} 3$ and $ y_2 \equal{} \frac{3}{2}x \plus{} \frac{3}{2}$

Two arbitrary points on $ y_1$ are $ A \equal{} (0, \minus{}3)$ and $ B \equal{} (2,0)$

An arbitrary point on $ y_2$ is $ C\equal{}(1,3)$

$ \overrightarrow{AB} \equal{} (2, 3)$

$ \overrightarrow{AC} \equal{} (1, 6)$

Letting the angle between these two vectors be $ \theta$ it follows that $ d \equal{} |\overrightarrow{AC}|\sin\theta$. Draw a picture and drop the perpendicular between both lines to see the right triangle I am looking at.

$ d \equal{} |\overrightarrow{AC}|\sin\theta \equal{} \frac{|\overrightarrow{AB}\parallel{}\overrightarrow{AC}|\sin\theta}{|\overrightarrow{AB}|}$

$ d \equal{} \frac{|\overrightarrow{AB} \times \overrightarrow{AC}|}{|\overrightarrow{AB}|}$

Computing the cross product (adding a 0 third term for sake of computation)
$ d \equal{} \frac{|(2,3,0) \times (1,6,0)|}{\sqrt{2^2 \plus{} 3^2}} \equal{} \frac{|(0,0,9)|}{\sqrt{13}} \equal{} \frac{9}{\sqrt{13}}$

You probably have not been exposed to vectors much, so you will likely not understand this (the cross product portion especially), but I think it's a more elegant solution. Vectors are a great bridge between algebraic coordinates and geometry (angles and lengths).

@ Zuton Force. hey i like solution number 2, it's creative... but i still don't understand. what i did was trying to get a point on l1 which is, say (0,-3). Now, i used the distance formula between (0,-3) and the line L2 but the answer is not 9/sqr13. why could it be so?

@chickendude- i know vectors VERY MUCH but i'm not supposed to use that in this part. just limit yourself to analytical geometry. thanks

for everybody: thanks for answering my posts, i got the formula wrong that's why i didnt end up with the right answer. sorry!:) thanks for the help and great solutions though.

btw, @zuton force-how old are you by the way? i wanna know because i think you're participating in ARML. what site are you in? in UNLV, penn state or iowa?

This formula is usually derived using the dot product (I believe that this is the most general solution).

The formula is \[ \frac{|ax_0\plus{}by_0\plus{}c|}{\sqrt{a^2\plus{}b^2}}\] for a line $ ax\plus{}by\plus{}c\equal{}0$ and a point $ (x_0,y_0)$.

[quote="Jakey"]thanks kunny. but what if you have say a specific point on l1, say (0,-3). the answer would be different right?[/quote]

No, it wouldn't. In such case, we would get the same answer as follows.

$ 4y \minus{} 6x \minus{} 6 \equal{} 0\Longleftrightarrow 6x \minus{} 4y \plus{} 6 \equal{} 0$, the distance is $ \frac {|6\cdot 0 \minus{} 4( \minus{} 3) \plus{} 6|}{\sqrt {6^2 \plus{} ( \minus{} 4)^2}} \equal{} \frac {9}{\sqrt {13}}$.

[quote="pi^2/6"][quote="kunny"]
Arbiterary points $ (x,\ y) \equal{} (2t \plus{} 2,\ 3t)$ on the line $ l_1$, the distance in question is $ \frac {|3(2t \plus{} 2) \minus{} 2\cdot 3t \plus{} 6)|}{\sqrt {3^2 \plus{} 2^2}} \equal{} \boxed{\frac {9}{\sqrt {13}}}$.[/quote]

I don't understand the numerator- did you plug in 2t+2 into the equation for l2? (is the last 6 in the numerator is supposed to be 3?)[/quote]

You are right. Thank you for pointing out the typo. I have just edited. :)


I need some book of Hojoo Lee
Can you help me?
thanks, very thanhks!

You can go to the download section at http://www.mathlinks.ro. :lol: There is a lot of hojoo lee file,such as PEN,etc.

http://ultrametric.googlepages.com/

This is the direct link to HoJoo Lee's website.
You can easily find PEN and TIN.


given two primes p and q of the type 2n+1, that satisfies: p is congruent to 7 mod 8, q is congruent to 2 modp and q is congruent to 3 mod4, calculate (p/q)

[hide]Parrot, there's infinitely many solutions.
For example,
if p=7, then q=23, 51, 79, or in general, q=28n-5.[/hide]


[b]A plane convex quadrilateral has area 32, and the sum of two opposite sides and a diagonal is 16. Determine all possible lengths for the other diagonal. [/b]

That is problem 1 of IMO 1976. :lol:

http://www.mathlinks.ro/Forum/viewtopic.php?p=367433#p367433

I think I've seen a a solution like below before.
[b]Solution.[/b] In a quadrilateral $ABCD$, let $16=BC+AD+AC$.  Then the area $A$ of quadrilateral $ABCD$ satisfies $A\leq AC\cdot\frac{AD+BC}{2}\leq \frac{16^{2}}{8}=32$.  Equality holds iff all of $BC, AD,$ and $AC$ are perpendicular to one another and $AC=BC+AD\implies AC=8=BC+AD$.  It then follows that $BD=8\sqrt{2}$.$\Box$


induction
inequalities
algorithm
linear algebra
matrix
combinatorics proposed
combinatorics

We are given $ 3^{2k}$ apparently identical coins,one of which is fake,being lighter than the others. We also dispose of three apparently identical balances without weights, one of which is broken (and yields outcomes unrelated to the actual situations). How can we find the fake coin in $ 3k\plus{}1$ weighings?

After I use the broken balance, can I identify it? or does the narrator take the balances and shifts them before any weighting?

The broken scale gives results totally random. You can think that there is an another player that controls the broken scale and doesn't want you to identify the fake coin.

Very nice problem! :) Though I wasn't able to prove it for $ 9$ coins, and $ 4$ weighings and this is all I need for the solution to be complete. Maybe someone can help with this case.
Clearly, if a balance works fine, then using it, by $ 1$ weighing we can reduce our attention to one third of the coins.

We show that if we have $ 3^{2k}$ coins, then we need at most $ 3k \plus{} 1$ weighings and if we have $ 3^{2k \plus{} 1}$ coins, $ k > 0$, we need at most $ 3k \plus{} 3$ weighings. Proceed by induction on $ n\ge2$, where $ n$ is the exponent of $ 3$ in the number of coins. As I said, I cannot prove the base case, $ n \equal{} 2$. But here's the induction step:

Let $ n > 2$.

[i]Case 1. $ n$ is odd. [/i]
Let $ n \equal{} 2k \plus{} 1$. Divide all coins in three groups: $ A$ , $ B$ and $ C$. Put $ A$ and $ B$ on balance $ B_1$ and then on balance $ B_2$. If we obtain the same result, since one of the balances works fine, this results is correct, hence we deduce in what heap - $ A$, $ B$ or $ C$ is the fake coin. According to the induction, we cand find the fake coin in $ 3k \plus{} 1$ weighings now, giving a total of $ 2 \plus{} 3k \plus{} 1 \equal{} 3k \plus{} 3$ weighings.
If $ B_1$ and $ B_2$ give different results, then one of them is broken, implying $ B_3$ works fine. Using only $ B_3$, we deduce in $ 2k \plus{} 1$ steps the fake coin, making $ 2k \plus{} 3 < 3k \plus{} 3$ weighings.

[i]Case 1. $ n$ is even[/i].
Let $ n \equal{} 2k$, $ k\ge2$. Take an arbitray balance, say $ B_1$. Let's presume it works just fine. Then using $ k$ weighings with it we obtain a heap $ H$ of $ 3^k$ coins, among which is the fake coin (we suppose so). Before the step we obtained $ H$, we had $ 3^{k \plus{} 1}$ coins, and we divided it into 3 heaps, say $ H,I,J$. We compared two of them, say $ A$ and $ B$, where $ A,B\in\{H,I,J\}$ and concluded that $ H$ was easier.
Now compare again $ I$ and $ H$ on another balance, say $ B_2$. If we do NOT obtain $ I > H$, then it means that either $ B_1$ or $ B_2$ is broken; or, that the third balance $ B_3$ works ok. Using $ B_3$ and all original $ 3^{2k}$ coins we can establish in $ 2k$ more weighings the fake coin, having done exactly $ 3k \plus{} 1$ weighings. If we do obtain $ I > H$ then it means that in $ H$ is indeed the fake coin! That is because either $ B_1$ or $ B_2$ is correct, so either all first $ k$ weighings have been correct (and then the fake coin is indeed in $ H$), or the $ k \plus{} 1$th weighing is correct, implying $ I > H$, thus $ H$ indeed contains the fake coin! At this moment we apply the induction hypothesis either for the $ 3^k$ coins of $ H$. If $ k$ is odd we will use $ \displaystyle\frac {3(k \plus{} 1)}2$ more weighings, giving a total of $ \displaystyle w \equal{} k \plus{} 1 \plus{} \frac {3(k \plus{} 1)}2$ weighings. Using the fact that $ k$ is odd and $ \ge3$ one can easily establish the inequality $ w\le 3k \plus{} 1$. If $ k$ is even then we use $ \displaystyle\frac {3k}2 \plus{} 1$ weighings, giving a total of $ w \equal{} k \plus{} 1 \plus{} \frac {3k}2 \plus{} 1\le 3k \plus{} 1$ weighings, what we needed.

Note that even if for $ 9$ coins we need at least $ 5$ weighings, then [b]the proof still works[/b] for $ k\ge4$ (that is for $ n > \equal{} 8$), [hide="because"] assuming we need at least $ 5$ weighings to find the fake coin among $ 9$ coins (this is certainly possible), the above proof still works for larger $ k$, namely $ k\ge 4$ (or $ n\ge8$), because the involved inequalities are strict for larger values of $ k$).
For example, if we denote by $ w_i$, $ i\ge2$ the minimal number of weighings necessary to find a fake coin among $ 3^i$ then if we take $ w_2 \equal{} 5$, the above algorithm gives $ w_3\le 2 \plus{} w_2 \equal{} 7$, $ w_4\le2 \plus{} 1 \plus{} w_2 \equal{} 8$, $ w_5\le 2 \plus{} w_4 \equal{} 10$, $ w_6\le 3 \plus{} 1 \plus{} w_3\le 11$, $ w_7\le 2 \plus{} w_6\le13$, $ \boxed{w_8\le4 \plus{} 1 \plus{} w_4 \equal{} 13 \equal{} 3\cdot4 \plus{} 1}$, $ w_9\le 2 \plus{} w_8 \equal{} 15$, $ \boxed{w_{10}\le5 \plus{} 1 \plus{} 10 \equal{} 16 \equal{} 3\cdot5 \plus{} 1}$, $ w_{11}\le 2 \plus{} w_{10} \equal{} 18$, $ \boxed{w_{12}\le\text{max}\{3\cdot 6 \plus{} 1,6 \plus{} 1 \plus{} w_6\} \equal{} 19}$, $ w_{13}\le 2 \plus{} w_12 \equal{} 21$, and so on. [/hide]

[quote]Very nice problem!  Though I wasn't able to prove it for $ 4$ coins, and $ 9$ weighings and this is all I need for the solution to be complete. Maybe someone can help with this case.
[/quote]
Me too :blush:

Base case $ 9$ coins and $ 4$ weighings.

Pick two coins $ C_1$ and $ C_2$ and with two weighings on $ B_1$ and $ B_2$ identify one good balance $ B_g$.
(same result then $ B_g\equal{}B_1$, different result then $ B_g\equal{}B_3$)

Put three coins on each plate of $ B_g$ then pick the lighter group of $ 3$ coins
Of the lighter group put one coin on each plate of $ B_g$ then pick the lighter coin.

Did I misunderstood the problem?

[quote="mszew"]Base case $ 9$ coins and $ 4$ weighings.

Pick two coins $ C_1$ and $ C_2$ and with two weighings on $ B_1$ and $ B_2$ identify one good balance $ B_g$.
[/quote]

How do you identify the good balance? If the results on the two balances are different, then you identify $ B_3$ as a good balance. But if results on $ B_1$ and $ B_2$ are the same then you cannot know for sure which balance is good. The broken balance need not give wrong results.

... but it was fun to decipher what was written in Russian!

I'll just solve the case $ k\equal{}1$, since it follows almost directly by induction after that (as remarked before).

Arrange the $ 9$ coins in a $ 3\times 3$ table:
\[ \begin{matrix}
1&2&3\\
4&5&6\\
7&8&9\\
\end{matrix}\]

The first weighting is to compare rows $ 1$ and $ 2$ on balance $ 1$ (that is, $ \{1,2,3\}\times \{4,5,6\}$). The second one is to compare columns $ 1$ and $ 2$ on balance $ 2$ (that is, $ \{1,4,7\}\times \{2,5,8\}$). The third weighting depends on the results of the previous two weightings.

Suppose, for instance, that balance $ 1$ indicates $ \{1,2,3\}< \{4,5,6\}$ and balance $ 2$ indicates $ \{1,4,7\} \equal{} \{2,5,8\}$. Let's see what information we can get.

If the first balance is broken, the only information we get is from balance $ 2$, that is, the lightest coin is in column $ 3$. If the second one is broken, then the lighest coin is in row $ 1$. If balance $ 3$ is broken, the lightest coin is $ 3$.

Unfortunately, we don't know which balance is broken, so the only thing we know is that the candidates for lightest coin are $ 1,2,3,6,9$ (row $ 1$ and column $ 3$). Then the third weighting is $ \{1,2\}\times \{6,9\}$ on balance $ 3$. Let's see what the third balance indicates in each of the three cases: in the first case, either $ \{1,2\} \equal{} \{6,9\}$ (in this case, the lightest coin is $ 3$) or $ \{1,2\} > \{6,9\}$; in the first case, either $ \{1,2\} \equal{} \{6,9\}$ (again, $ 3$ is the lighest coin) or $ \{1,2\} < \{6,9\}$; in the third case, ot can go either way, but we already know that $ 3$ is the lightest coin.

Now let's face it backwards: if balance $ 3$ indicates $ \{1,2\} > \{6,9\}$ either balance $ 1$ or $ 3$ is broken, anyway, balance $ 2$ is fine, and, additionally, the lightest coin is in column $ 3$. If balance $ 3$ indicates $ \{1,2\} < \{6,9\}$ then balance $ 1$ is fine and the lightest coin is in row $ 1$. If balance $ 3$ indicates $ \{1,2\} \equal{} \{6,9\}$ then the lightest coin is $ 3$.

Notice that now either we already know which coin is the lightest or we know a fine balance and a set of three coins, one of which is the lightest one, which can be determined by one more weighting.

The other cases are handled similarly.

By the way, it also says in the booklet that it is possible to determine the lightest among $ 3^{n(n\plus{}1)}$ coins with at most $ (n\plus{}1)^2$ weightings (there are still three balances, one of which is broken). Anyone?

russian balance problems will be the death of me

It suffices to show that given nine coins, we can either find the lightest in $3$ moves or find the lightest in $4$ moves [i]and[/i] identify a non-broken scale. Then we are done by induction (by breaking the $3^{2k}$ coins into nine "mega-coins"), since once a non-broken scale is identified we can clearly find the lightest coin among $3$ in $1$ move using it.

Number the coins $1$ through $9$. First use two different scales to weigh $1,2,3$ against $4,5,6$, and then weigh $1,4,7$ against $2,5,8$. For symmetry reasons WLOG let both scales say that the group containing the coin $1$ is lighter. Then use the third scale to weigh $2,3$ against $4,7$. Finally,
[list]
[*] If the result is that $2,3$ and $4,7$ are balanced then the $1$ coin must be the lightest, since any pair of correct results alone implies that $1$ is the lightest coin, so we are done.
[*] Otherwise WLOG the result is that $2,3$ is lighter (so one scale provides an incorrect result). Then the only possible candidates for lightest coin are $1,2,3$, since anything else implies that at least two scales lied. It therefore cannot be the case that the first scale (used to weigh $1,2,3$ against $4,5,6$) provided an incorrect result, implying that it is guaranteed to be non-broken. Then use it to weigh $2$ against $3$; the (correct) result will indicate the lightest coin among $1,2,3$.
[/list]
This finishes the problem. $\blacksquare$


vector
linear algebra
matrix
algebra
polynomial
calculus
integration

Define a cyclic diffrence operator $\Delta $ on an integer $n-$tuple $a=(a_1,a_2,\dots,a_n), \Delta (a)=(a_2-a_1,\dots,a_n-a_{n-1},a_1-a_n)$.Determine all solutions $\Delta ^k(a)=a$.

Let's consider $a$ to be a column vector, and $\Delta$ to be a matrix acting on such vectors. It's easy to find the expression of $\Delta$ and to see that it's characteristic polynomial $\det(xI_n-\Delta)$ equals $(x+1)^n+(-1)^n$.

In order for the equation $\Delta^kx=x$ to have a non-trivial solution (I mean $x$ different from the null vector), $\Delta^k$ must have $1$ as one of its eigenvalues, so $\Delta$ must have an eigenvalue $x_0$ of absolute value $1$. We have $|x_0|=|1+x_0|=1$, so $x_0\in\{\varepsilon,\bar\varepsilon\}$, the two cubic roots different from $1$. This means that $1+x_0\in\{\omega,\bar\omega\}$, where $\omega=e^{\frac{\pi i}3}$.
The equation becomes $\omega^n+(-1)^n=0$, and we can easily check that this does not hold for any $n$, which means that there are no non-trivial solutions for our equation.

I'm afraid I'm not entirely sure about this. I didn't use the fact that the vectors have to have integral entries, for example :?.

grobber, you're overlooking the possiblities in the $n=6$ case.

Let $x=(1,1,0,-1,-1,0).$ Then $\Delta^3x=x.$

Yes, I've found my mistake: the characteristic polynomial is $(x+1)^n-1$, so some cases did slip out. Thanks for pointing that out.

When $n$ is even, we do have a chance whenever it's also a multiple of $6$ (because we must have $\omega^n-1=0$). In such cases, $\varepsilon$ is, indeed, an eigenvalue of $\Delta$, and in order to have $1$ as an eigenvalue of $\Delta^k$, the necessary and sufficient condition is to have $3|k$.

I guess it's not solved, then, it's just reduced to the cases $6|n,3|k$ :(. I knew it couldn't have been that easy.. :)

I'm looking at it from a discrete Fourier transform point of view. I haven't settled the details, but I'm starting to think that if you want integer-valued solutions, the example I posted in #3 is essentially the only example. (Oh, you can rotate it around, and multiply by integer scalars but that's it.)

Actually, the work is almost entirely done :). Let's assume, as said above, that $6|n,3|k$.

$\Delta$ is diagonalizable, since its characteristic polynomial clearly has simple roots. This means that $\Delta^k$ is also diagonalizable, so the algebraic multiplicities of its eigenvalues are equal to the geometric multiplicities. The only $k$'th roots of unity which are also eigenvalues of $\Delta$ are $\varepsilon,\bar\varepsilon$, and these have multiplicity $1$. This means that $1$, as an eigenvalue of $\Delta^k$, has multiplicity $2$, which means that its eigenspace has dimension $1$. Since we can find two linearly independent eigenvectors, $(1,1,0,-1-1,0,\ldots)$ and $(0,-1,-1,0,1,1,\ldots)$, it means that the complex vectors fixed by $\Delta^k$ are precisely the linear combinations of these two.

Since we want vectors with integral entires, it means that the coefficients of the linear combinations must be integers, and we have found our integral solutions.

I hope it's Ok this time.

Let $T$ be the operator of translation $a_i \to a_{i+1}$ on the space of cyclic n-tuples.
Then the problem is (literally) asking for simultaneous solution of

$T^n = 1$
$(T-1)^k = 1$

Eigenvalues of $T$ and $(T-1)$ are roots of unity that differ by 1,
so the only possibilty is $n$ a multiple of $6$, $k$ a multiple of 3, and $T^2 - T + 1 = 0$.
This gives the same solutions as above.

[b]Question[/b]: classify the solutions in characteristic $p$ (or, equivalently, find the ideal in $\mathbb{Z}[T]$ generated by $T^n-1$ and $(T-1)^k-1$).


[quote="orl"]Find all plane triangles whose sides have integer length and whose incircles have unit radius.[/quote]
 use  this : $S=\sqrt {p(p-a)(p-b)(p-c){}}$ and $S=r.p$


[hide]The answer here is 2. You know the numbers are different, so it cannot be one. The integers values are insignificant, as many combinations will give you the correct answer.

For example:

100/50 = 2
102/51 = 2
104/52 = 2

ect...[/hide]

[hide]well, since it's looking for the least number, it can't be one cuz one is 3 digit and the other is 2 digit..... so it much be 2[/hide]

[quote="solafidefarms"]A positive 3-digit integer is divided by a positive 2-digit number, yielding a integer quotent and zero remainder. What is the smallest possible positive integer quotient? (95 State Sprint#7)[/quote]

There was another problem from a school or chapter level that said :

A 3-digit integer is divided by a  2-digit number, yielding a integer quotent and zero remainder. What is the smallest possible integer quotient?

If memory serves the given answer ended up being incorrect.  Answers?


geometry
geometric transformation
national olympiad

It'd be great if you translate these problems  :)

Shobber, why? Avoiding translation make nothing but distingushs Chinese-speaking people from not-Chinese-speaking, is it a goal?

Read this and you know why.  :)
http://www.mathlinks.ro/Forum/viewtopic.php?p=476480#476480

Is there a problem-only version? Thanks.

[quote="shobber"]If you still remember what Mr. Zuming Feng said last year, then do not translate the problems until July.[/quote]
Now, can you post all those problems?

can someone translate this problems now?


How many 2 digit numbers are there where the tens digit is greater than the ones digit?

how would you do this problem?

[hide]If the one's digit is 0, there are 9 possible ten's digits. If the one's digit is 1, there are 8 possible. etc.

9+8+7...+2+1=45
[/hide]

wow i am impressed.

I would just list them starting with 10 up that work lol, never thought of it that way.

In general, the number of strictly increasing $n$ digit numbers is $\dbinom{9}{n}$ and the number of strictly decreasing $n$ digit numbers is $\dbinom{10}{n}$, with the exception that there are 9 strictly decreasing 1 digit numbers. I sort of derived this, so it's not any formal rule that I know of, but I'm pretty sure it's true.

[quote="t0rajir0u"][hide="Alternately"] The problem is also easily solved with symmetry.

$\text{Number of 2-digit numbers}= \text{Number of 2-digit numbers where the tens digit is greater than the ones digit}$ $+\text{Number of 2-digit numbers where the tens digit is less than the ones digit}$ $+\text{Number of 2-digit numbers where the tens digit is equal to the ones digit}$

(Where we're looking at the numbers $00, 01, 02, ... 99$ as having "two digits." )

Note that if we reverse a 2-digit number from the first category we get a 2-digit number in the second category, so they are of equal size.  Let that size be $x$.  Then

$100 = 2x+10$
$x = 45$

We know we have not counted any numbers with $0$ as their first digit because they are all in the second or third groups.

This kind of reasoning is very instructive for more difficult probability problems where a straightforward summation would be difficult but where a symmetry is easy to take advantage of.
[/hide][/quote]

Ooo, very nice solution t0rajir0u!  :clap2:

[hide]Well for every choice of two digits, there is exactly one working order. So it is just $\binom{n}{2}$[/hide]


Let $a,\,b,\,c$ are positive numbers satisfying the condition $abc=1,$ prove that: \[(a+b)(b+c)(c+a)\ge 2(1+a+b+c)\]

[hide="identity usage"]
Notice \[(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc = (a+b+c)(ab+bc+ca)-1 \]
Now, \[(a+b+c)(ab+bc+ca)-1\ge 2+2(a+b+c)\iff (a+b+c)(ab+bc+ca-2) \ge 3 \]
But we have that $\sum a \ge 3$ and $\sum ab \ge 3$ (by Am-Gm) thus the inequality \[(a+b+c)(ab+bc+ca-2) \ge (3)\cdot(1)=3 \]
follows and is equivalent to the problem.
[/hide]

[quote="SplashD"]$2abc+\sum_{cyc}(a^{2}b+a^{2}c)\ge 2+2\sum_{cyc}a^{\frac{4}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}$[/quote]

How does that work? I don't think $a^{2}b+a^{2}c \ge 2a^{\frac{4}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}$ is true. For example, try $a = \frac{1}{10}$, $b = 10$, $c = 1$.

On the other hand, we can use

$\frac{5}{12}a^{2}b+\frac{5}{12}a^{2}c+\frac{1}{12}b^{2}c+\frac{1}{12}c^{2}b \ge \sqrt[12]{a^{20}b^{8}c^{8}}= a$,

but that might be unnecessarily complicated if I'm just misunderstanding you. It follows from weighted AM-GM.

Well SplashD was just using muirhead... to say that $\sum_{sym}a^{2}b\geq \sum_{sym}a^{\frac43}b^{\frac13}c^{\frac13}$ is just to say that $[2,1,0]\succ[4/3,1/3,1/3]$ which is clear.
I guess you went ahead and solved for the specific Am-Gm...
but SplashD wasn't implying any specific inequality by the way he grouped the terms... at least i don't think

Yeah, that makes more sense. But the way he wrote it suggested otherwise, beside the fact that Muirhead only applies for symmetric sums.

here is a related (weaker) result is $(a+b)(b+c)(a+c) \geq 8$.

You can prove this result by expanding to get the sum ${ 2+\sum_{\sigma}\frac{a+b}{c}=2+\sum_{\sigma}\frac{a}{c}+\frac{b}{c}= 2+\sum_{\sigma}\frac{a}{b}+\frac{b}{a}\geq 2+3(2)=8}$.

This is weaker since $1+a+b+c \geq 4$ whenever $abc=1$.

Now here's a way to prove the original result directly. Expand to get ${ 2+\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}}$. Without loss of generality, suppose $a \geq b \geq c$. Then $b+c,c+a,a+b$ is a nondecreasing sequence, as is $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Thus by Chebyshev,

$2+\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 2+\frac{1}{3}(2(a+b+c))(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})={1}{3}(2(a+b+c))(ab+bc+ca)$. Now applying AM-GM to $ab+bc+ca$ yields that the original expresion is greater than or equal to $2+2(a+b+c)$, which is what we wanted to show.

[quote="Arne"][quote="drunner2007"]
This is stronger since $1+a+b+c \leq 4$ whenever $abc=1$.
[/quote]

No, $1+a+b+c \geq 4$ whenever $abc = 1$. Your result isn't stronger.[/quote]

haha yeah i realized that about an hour later. i should have said weaker result.


For a finite sequence $ A \equal{} (a_1, a_2, \ldots, a_n)$ of numbers, the [i]Cesaro sum[/i] of $ A$ is defined to be
\[ \frac {S_1 \plus{} S_2 \plus{} \cdots \plus{} S_n}{n},
\]
where $ S_k \equal{} a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k$.  If the Cesaro sum of the $ 99$-term sequence $ (a_1,a_2,\ldots ,a_{99})$ is $ 1000$, what is the Cesaro sum of the $ 100$-term sequence $ (1,a_1,a_2,\ldots , a_{99})$?

[quote="IW@IT"]If I'm not misunderstanding the problem
The answer is $ 991$[/quote]

Could you please explain what you did?  Thanks!

EDIT: This is not correct, I have misunderstood your problem....

Since

1+99=100
2+98=100
3+97=100
4+96=100
5+95=100
6+94=100
7+93=100
8+92=100
.
.
.
49+51=100

and only left number you have is 50 which is unpaired...

Sk=4950

you can do it this way too:

Sk=[n*(n+1)]/2

using this:

Sk=[99*(99+1)]/2

Sk=99*100/2

Sk=4950

EDIT: I have misunderstood your problem... I will maybe later psot the correct answer...

@[b]miyomiyo[/b]
How did you know $ S_1 \equal{} 1$

EDIT: sorry I misread the question

Because 1 is the first term of the sequence, so $ S'_1 \equal{} 1$.
I suppose you asked for $ S'_1$, because we don't know $ S_1$.

Sorry, I didn't explain my notation.  I'm calling $ S'_k$ the $ k^{th}$ sum of the new sequence, while $ S_k$ is the $ k^{th}$ sum of the old sequence.

[quote="miyomiyo"][hide="Solution"]
$ S'_1 \equal{} 1$

$ S'_2 \equal{} 1 \plus{} a_1 \equal{} 1 \plus{} S_1$

$ S'_3 \equal{} 1 \plus{} a_1 \plus{} a_2 \equal{} 1 \plus{} S_2$

$ \vdots$

$ S'_{100} \equal{} 1 \plus{} a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{99} \equal{} 1 \plus{} S_{99}$

$ \Downarrow$

$ S'_1 \plus{} S'_2 \plus{} \cdots \plus{} S'_{100} \equal{} S_1 \plus{} S_2 \plus{} \cdots \plus{} S_{99} \plus{} 100$

$ \frac {S'_1 \plus{} S'_2 \plus{} \cdots \plus{} S'_{100}}{100} \equal{} \frac {1000 \times 99 \plus{} 100}{100}$

$ \equal{} \boxed{1090}$
[/hide][/quote]
My approach is exactly the same, but it seems to me that $ \frac {1000 \times 99 \plus{} 100}{100} \equal{} (99000 \plus{} 100)/100 \equal{} 991$.

Haha, I didn't divide the $ 100$.  Your answer is correct.


Well they are choosing the U.S. team...slowly.  Theres at least 2 maybe 3 Texans on the team this year. A Gymnast from Grand Prarie and a runner from Texas Tech I think.  Anyone know if they have chosen the equestrian team yet?  If so whose on it.  For the U.S. that is.

For all you peoples from other coutries Do you know whose representing your country in the games?

Well the only Canadian that I know on our team is one gymnast who was in my class in 7th grade (she was in 8, but we had a split level class).  However, she moved away before highschool since her hometown got a proper gymnastics center. The rest I only know by name/reputation...

[quote=mathleticguyyy]Well winter Olympiad happened half a year ago![/quote]

[quote=mikimoto12][quote=mathleticguyyy]Well winter Olympiad happened half a year ago![/quote]

PLEASE STOP[/quote]

THIS THREAD IS 14 YEARS OLD, wait I'm actually older than it wow

It's kinda sad: this thread os older than me.....


The given expression is equal to $ \frac {2^{0}(2^{2010} \minus{} 1)}{2 \minus{} 1}$ the last digit of $ 2^{2009}$ is 2 so the answer will be 3 that is $ B$
EDITED

[hide="Solution"]
This is a geometric series with first term 1 and common ratio 2. Using the formula for a geometric series, \begin{align*}2^0+2^1+2^2+\dots+2^{2009}&=\frac{2^0\left(2^{2009+1}-1\right)}{2-1}\\&=2^{2010}-1\end{align*} is the value of the sum. The last digit of powers of two cycle in fours, so the last digit of $ 2^{2010}$ is the same as $ 2^2=4.$ Finally, subtract 1 from this to get the last digit of $ 3,$ which is $ \boxed{\textbf{(B)}}.$[/hide]


\[ {\rm{Using the quadratic formula}},{\rm{ solve the equation}}: a^2 b^2 x^2 \minus{}\minus{}{\rm{ }}\left( {4b^4 {\rm{ }}\minus{}\minus{}{\rm{ }}3a^4 } \right)x{\rm{ }}\minus{}\minus{}{\rm{ }}12a^2 b^2 {\rm{ }} \equal{} {\rm{ }}0
\]

[quote="sanbittu"]
\[ {\rm{Using the quadratic formula}},{\rm{ solve the equation}}: a^2 b^2 x^2 \minus{} \minus{} {\rm{ }}\left( {4b^4 {\rm{ }} \minus{} \minus{} {\rm{ }}3a^4 } \right)x{\rm{ }} \minus{} \minus{} {\rm{ }}12a^2 b^2 {\rm{ }} \equal{} {\rm{ }}0
\]
[/quote]

Not quite sure what is going on due to the $ \minus{} \minus{}$ which is completely meaningless.

Still...

Let

$ 0 \equal{} (a^2b^2)x^2 \plus{} (4b^4 \plus{} 3a^4)x \plus{} (12a^2b^2)$
Using the quadratic formula, taking $ a \equal{} (a^2b^2), b \equal{} (4b^4 \plus{} 3a^4), c \equal{} (12a^2b^2)$

$ x \equal{} \frac { \minus{} (4b^4 \plus{} 3a^4)\pm \sqrt {(4b^4 \plus{} 3a^4)^2 \minus{} 4(a^2b^2)(12a^2b^2)} }{2(a^2b^2)}$
$ x \equal{} \frac { \minus{} (4b^4 \plus{} 3a^4)\pm (4b^4 \minus{} 3a^4)}{2(a^2b^2)}$
$ x \equal{} \frac { \minus{} 3a^2}{b^2} \mbox{ or } x \equal{} \frac { \minus{} 4b^4}{a^2}$


geometry
geometric transformation
reflection

We have a $2 \times 8$ billiards table, with a single ball at is center. The ball is sent in a straight line, and after 29 meters it stops at a corner of the table. How many times did the ball bounce on the borders of the table?

Note: The ball bounces obeying the law of reflection, that is, the angle of incidence between the path of the ball and the border of the table is equal to the angle of reflection.

[hide="Solution"]
Let the number of times the ball hits a short edge (i.e. it travels $8$ feet) be $a$, and the number of times it hits a long edge (i.e. it travels $2$ feet) be $b$.

Then, by reflections, we are solving the equation
\[29^{2}= 16 \cdot (2a+1)^{2}+(2b+1)^{2}\] in non-negative integers. We test the few possible values of $a$, and find the only solution to be $(a,b)=(2,10)$. So, not counting the final pocketing of the ball, we have a total of $\boxed{12 \text{ bounces off of a wall}}$.[/hide]


function
algebra unsolved
algebra
functional equation

I have this problem i suspect i know how to solve it but i cant because i dont have a nice series of identities concerning cubes.
find all functions from the nonegative integers to theirselves such that:

$f(a^3+b^3+c^3)=f(a)^3+f(b)^3+f(c)^3$ for all $a,b,c$ nonnegative integers.

I seem to remember Pierre solving this one on the board, but i'm not sure at all.

We used it for our correspondance program. You may find a solution (un french, but it doesn't matter here) at :

http://www.animath.fr/tutorat/dossier_03044sol.pdf

Pierre.

Well...in my opinion, it won't...but who knows...

Pierre.

I have solved this problem before.
The main lemma is : for each integer number  n>=4 we have  x,y,z,t,w  integer numbers s.t their [b]absolute[/b] values less than  n   and:
       n^3=x^3+y^3+z^3+t^3+w^3

To prove this ,we may use this identity:
 (2n+1)^3=(2n-1)^3  +  (n+4)^3  -  (n-4)^3  -  5^3  -  1^3

Really?Actually,i didn't read pierre's solution. :)
But the problem i solved was to find     f(x):   Z   ->    Z  , not  f(x):     Z+  ->    Z+
For your one i think we should use the following identity:
    [b]a^3 + (5a)^3 + (2n+a)^3 =(2n-a)^3 + (n+4a)^3 + (4a-n)^3[/b]
 there [b]a,n[/b] are nonnegative integers and [b]a<n<4a[/b]
The rest work can be done by induction.


Euler
inequalities
number theory solved
number theory

Let sigma(n) denote the sum of the positive divisors of the positive integer n and phi(n) the Euler phi-function. Prove that sigma(n) and phi(n) always average at least n, that is sigma(n) + phi(n) \geq 2n.

trivial. use induction.
first let n= p^k .
then sigma +phi >2p^k
let n be composite with at least 2 prime divisors and let n=ab where gcd(a,b)=1, a>b
then sigma(n)+phi(n)=sigma(a)sigma(b) +phi(a)phi(b)
>=1/2( sigma(a)+phi(a)) (sigma(b)+phi(b) >=2ab=2n by induction.

Here's what I did here (it might look A LOT more complicated, but once you write it down you'll see it isn't):

First of all, it's easy to see that the sum of all divisors of
n= \pi {1<=i<=k}pi^ai is (1+p1+..+p1^a1)*(1+p2+..+p2^a2)*..*(1+pk+..+pk^ak)= \pi {1<=i<=k}(pi^(ai+1)-1)/(pi-1). The inequality is, after subtracting phi(n) from both sides and multiplying both sides with
 \pi {1<=i<=k} (pi-1),

 \pi {1<=i<=k} (pi^(ai+1)-1) >=  \pi {1<=i<=k} (pi^(ai+1)-pi^(ai-1)), but this is obviously true because pi^(ai+1)-1 >= pi^(ai+1)-pi^(ai-1) for all i (because pi^(ai-1)<=1 for all i).


function
integration
calculus
calculus computations

Find all infinitely differentiable functions $ f: \mathbb R\to\mathbb R$ such that
\[ f(x)\plus{}f'(x)\equal{}2x\]
for all $ x\in\mathbb R$.

[hide]Mutliplication by $ e^x$ gives finally $ f(x)\equal{}2(x\minus{}1)\plus{}\frac{c}{e^x}$[/hide]

$ \{e^xf(x)\}'\equal{}2xe^x\Longrightarrow e^xf(x)\equal{}2(x\minus{}1)e^x\plus{}C$. $ \therefore f(x)\equal{}2(x\minus{}1)\plus{}Ce^{\minus{}x}$.

hello, solving the homogeneous part of the equation we get
$ \frac{dy}{dx}=-y$
$ \frac{dy}{y}=-dx$
$ y_h=Ce^{-x}$
For the inhomogenius part we make the ansatz $ y=a\cdot x +b$, solving this we get the solution
$ y=C\cdot e^{-x}-2+2x$
Sonnhard.

[hide="solution"]
as this is a first order linear differential equation .
$ I.F. \equal{} e^{x}$
$ ye^{x} \equal{} \int2xe^{x}$
$ y \equal{} 2(x \minus{} 1) \plus{} Ce^{ \minus{} x}$[/hide]
thanks

[hide="Solution:"]$ \{f(x) \plus{} f'(x)\}e^x \equal{} 2xe^x \\
 \\
\frac {d}{dx}\{e^xf(x)\} \equal{} 2xe^x \\
 \\
e^xf(x) \equal{} \int 2xe^xdx \\
 \\
e^xf(x) \equal{} 2xe^x \minus{} 2e^x \plus{} 2c \\
 \\
\;\boxed{f(x) \equal{} 2\left(x \minus{} 1 \plus{} \frac{c}{e^x}\right)}$[/hide]

y+y'=2x
e^xy'+e^xy=2xe^x
d(e^xy)/dx=2xe^x

What so intersting in this simple question of yours?
 :oops:


quadratics
number theory unsolved
number theory

The digits of a number $n$ are 0,1,2,3,4,5,6,8,9 in some order. Prove that $n$ is not a square.

$38 \equiv 2 \bmod 9$ is not a quadratic residue.


We consider the polynomial $P(z)=z^2-mz+m$, where $m \in \mathbb Z$. We have $\Delta = \sqrt{m(m-4)}$. Take any $m \equiv 0 \left( \bmod ~ 8 \right)$, $m > 0$ $\displaystyle \left( \Delta = 4 \sqrt{2k(2k-1)} \, \, \textrm{for some } \, \, k \right)$. Then the roots of $P$ are the numbers we are looking for.

Right?

[quote="JohnConstantine"]$xy-x-y+1\in \mathbb{N}$, $(x-1)(y-1)\in \mathbb{N}$, $x=a+\sqrt{b}+1$, $y=a-\sqrt{b}+1$, $a,b \in \mathbb{N}$ :D[/quote]

Did you forget about the condition $xy = x + y$?

[quote="Valentin Vornicu"]Prove that there exists an infinity of irrational numbers $x,y$ such that the number $x+y=xy$ is a nonnegative integer.[/quote]

 $\large{x+y\,\in\,N\,\,\Longleftrightarrow{\,\,x\,=\,a_1+\sqrt{b}\,\,\,\textrm{and}\,\,\,y\,=\,a_2-\sqrt{b}\,\,\,|\,a_i\,\in{\mathbb{Z}}\,\,\textrm{and there is no}\,\,\exists{c}}\,\in{\mathbb{Z}\,\,|\,b\,=\,c^2}}$

And :

$\large{(a_1+\sqrt{b})(a_2-\sqrt{b})\,=\,a_1a_2-b+\sqrt{b}(a_2-a_1)\,\in\,\,\mathbb{N}\,\,\Longleftrightarrow{\,a_1\,=\,a_2}}$

We have :

$\large{x\,=\,a+\sqrt{b}\,,\,y\,=\,a-\sqrt{b}}$

Then :

$\large{2a\,=\,a^2-b\,\Longrightarrow{b\,=\,a^2+2a}}$

We FINALY  have :

$\large{x\,=\,a+\sqrt{a^2+2a}\,=\,a+\sqrt{(a+1)^2-1}\,\,,\,\,y\,=\,a-\sqrt{(a+1)^2-1}}$

PS.: See that $\large{(a+1)^1-1}$ CANNOT be a square perfect for $\large{a>0}$

$\large{\mathbb{C.Q.D.}}$


pl help...........

1)2^1001 mod 1001

2)Another one.............................

If one is given a balance, and n weights of 1,3,3^2,...,3^(n-1)!
lb. respectively, show that it is possible, by placing some
weights in one pan and some in the other, to weigh out any
weight of N lb., with n an integer >= 1 and <= 1/2 (3^n- 1)

hello, to your first question, we have $ 2^{10} \equiv 23 \mod 1001$, from here we get $ 2^{100}\equiv23^{10}\equiv562\mod1001$ and further $ 2^{1000}\equiv562^{10}\equiv 562\mod 1001$ so we get as the result $ 2^{1001}\equiv123\mod 1001$.
Sonnhard.

i wont post the complete solutions...
will just give u hints...

1.
2^6 == 1 (mod 7)
2^12 == 1 (mod 13)
2^10 == 1 (mod 11)

( == stands for.. is congruent to)

2.
we see that for 1, 3
we can form 1, 2, 3, 4;
i.e. it is true for n = 2; (and n = 1)

rest is simple induction
assuming it true for n
=> we can form >= 1 and <= 1/2 (3^n- 1)

now we have a new weight 3^n
we could already form >= 1 and <= (1/2) (3^n- 1) [without using 3^n]

we can form [3^n - (1/2) (3^n- 1)] upto 3^n
by keeping 3^n on one side, and putting k on the other one
k => 0 and <=  (1/2) (3^n- 1)

and [3^n + 1] upto (1/2)(3^(n+1) -1 )
by keeping 3^n on one side, and putting k on the same side
k => 1 and <=  (1/2) (3^n- 1)

keeping k on one side... means that we know the arrangement required to weigh k
it does not mean that all smaller weights be kept on the same side

sorry for not using latex
(i dont kno how to)

and Sonnhard..
we are not allowed calculators in competitions
so it would not be so nice to calculate 562^10
:)

hello, but you can write $ 562^{10}\equal{}(562^2)^5\equiv529^5\equiv562\mod 1001$.
Sonnhard.

[quote="Arnav...!!!"]how can the first ques be solved by using crt-chinese remainder theorem......???[/quote]

I Hope this Helps !!


Putnam
logarithms
integration
probability
analytic geometry
college contests

For each positive integer $ n,$ let $ f(n)$ be the number of ways to make $ n!$ cents using an unordered collection of coins, each worth $ k!$ cents for some $ k,\ 1\le k\le n.$ Prove that for some constant $ C,$ independent of $ n,$
\[ n^{n^2/2\minus{}Cn}e^{\minus{}n^2/4}\le f(n)\le n^{n^2/2\plus{}Cn}e^{\minus{}n^2/4}.\]

[hide]
First, we find an approximation for $ \prod_{i\equal{}1}^n i^i$ which will figure in both parts of the inequality.
Let $ g(x) \equal{} \frac{x^2}2 \ln x \minus{} \frac{x^2}4.$ Then $ g'(x) \equal{} \frac{x}2 \plus{} x \ln x  \minus{} \frac{x}2 \equal{} x \ln x.$ Because $ x \ln x$ increases on $ [1,\infty),$
\[ \sum_{i\equal{}1}^n i \ln i \le \sum_{i\equal{}1}^n \int_i^{i\plus{}1} x\ln x dx \equal{}\int_1^{n\plus{}1} x\ln x dx \equal{} g(n \plus{} 1) \plus{} \frac14
\]
and
\[ \sum_{i\equal{}1}^n i \ln i \ge \sum_{i\equal{}2}^n \int_{i\minus{}1}^i x\ln x dx \ge  \int_1^n x\ln x dx\ge g(n).
\]
This implies
\[ e^{g(n)} \le \prod_{i \equal{} 1}^n i^i \le e^{g(n\plus{}1) \plus{} \frac14}.
\]

Observe that the quantity $ f(n)$ we need to bound is the number of sequences $ (k_i)_{i\equal{}1}^n$ of nonnegative integers satisfying $ \sum_{i\equal{}1}^n k_i i! \equal{} n!.$

We know instantly that $ k_i \le \frac{n!}{i!}.$ Therefore, $ k_i$ can take at most $ \frac{n!}{i!} \plus{} 1 \le 2\frac{n!}{i!}$ different values. Thus,
\[ f(n) \le \prod_{i\equal{}1}^n 2\frac{n!}{i!} \equal{} 2^n \prod_{i\equal{}1}^n i^{i \minus{} 1} \le 2^n \prod_{i \equal{} 1}^n i^i \le 2^n e^{g(n\plus{}1)\plus{} \frac14}.
\]
We have
\[ e^{g(n\plus{}1) \minus{} g(n)} \le \frac{(n\plus{}1)^{\frac{(n\plus{}1)^2}2}}{n^{\frac{n^2}2}} \equal{} \left(1 \plus{} \frac1n\right)^{\frac{n^2}2} (n\plus{}1)^{n \plus{} \frac12} \le e^{\frac{n}2} (2n)^{2n} \le n^{C_1n}
\]
for some constant $ C_1.$ Therefore, $ f(n) \le e^{g(n)} n^{C n} \equal{} n^{\frac{n^2}2 \plus{} C n} e^{\minus{}\frac{n^2}4}$ for some constant $ C.$

Let us set $ k_n \equal{} 0$ and pick $ k_i \in \left[1, \frac{(n\minus{}1)!}{i!}\right]$ for $ i \equal{} 2,3,\ldots, n\minus{}1.$ Then
\[ \sum_{i\equal{}2}^n k_i i! \le (n \minus{} 1)(n \minus{} 1)! < n!.
\]
Therefore, for every such choice we can always find $ k_1$ so that $ \sum_{i\equal{}1}^n k_i i! \equal{} n!.$ This yields
\[ f(n) \ge \prod_{i \equal{} 2}^{n\minus{}1} \frac{(n\minus{}1)!}{i!} \equal{} \prod_{i \equal{} 3}^{n \minus{} 1} i^{i \minus{}2} \equal{} (n!)^{\minus{}2} n^{\minus{}n} \prod_{i \equal{} 1}^n i^i \ge e^{g(n)} n^{\minus{}3n} \equal{} n^{\frac{n^2}2 \minus{} 3n} e^{\minus{}\frac{n^2}4}.
\]
[/hide]

Since we have a solution to the standards of the problem already up, here's a much tighter estimate:

For reasonably large $ n$, $ f(n)\approx \frac1{(n\minus{}2)!}\frac{(n!)^{n\minus{}2}}{2!3!\cdots (n\minus{}1)!}$.
In terms of exponentials and powers of $ n$, this is asymptotically close to $ A\cdot n^{\frac{(n\minus{}1)(n\minus{}2)}{2}\plus{}\frac1{12}}e^{\minus{}\frac{n^2}{4}\plus{}2n}$ for some constant $ A$ as $ n\to\infty$.

There are $ n\minus{}1$ ways to make $ n!$ using only one kind of coin (other than $ 1$ cent). We subtract those, and ignore them due to insignificant size.
For $ k\equal{}2,3,\dots,n\minus{}1$, we choose between $ 0$ and $ \frac{n!}{k!}\minus{}1$ coins of denomination $ k!$ uniformly at random. If the sum of the values of the coins chosen this way is at most $ n!$, there is exactly one way to choose 1-cent coins to complete our pile to $ n!$; otherwise, there is no way. The total number of ways is thus $ \frac{(n!)^{n\minus{}2}}{2!3!\cdots (n\minus{}1)!}\cdot p\plus{}n\minus{}1$, where $ p$ is the probability that the sum is at most $ n!$.

To estimate $ p$, we approximate the discrete uniform distributions of coin number by continuous uniform distributions; $ p$ is approximately the probability that a sum of $ n\minus{}2$ independent uniform distributions on $ [0,1]$ is at most $ 1$. This probability is the volume in $ \mathbb{R}^{n\minus{}2}$ of the simplex bounded by the coordinate hyperplanes and the hyperplane $ x_1\plus{}x_2\plus{}\dots\plus{}x_{n\minus{}2}$, which is $ \frac1{(n\plus{}2)!}$

lol, this question isn't even true for $ n\equal{}1$

You're right- powers of 1 don't give a lot of room to maneuver.
Note that my estimate only makes sense for $ n\ge 2$.

I have a question. In the solutions next to B6 it says (suggested by Oleg Goldberg). Does this mean that he suggested the problem or suggested the particular solution? I only ask because all the other problems have "solution by" next to them or something of the sort while this one doesn't.

Seeing that Oleg Golberg was a contestant, it was probably the solution that he proposed.

Note that this problem is very similar to [url=http://www.mathlinks.ro/viewtopic.php?p=356713&sid=6eca5a8d6067b2cf1e0fde92894916ff#356713]IMO 1997.6[/url]. The only difference is that there the coins are size $ 2^n$ instead of $ n!$. I believe the same proofs (with some modifications) still work though.


[quote="dragon96"]Is there a prebuilt funtion that allows you to put ticks in without manually drawing the ticks? Thanks.[/quote]
For graphs you can take a look to [url]http://piprim.tuxfamily.org/asymptote/graph/index.html[/url].
If you want markers on paths only, [url=http://piprim.tuxfamily.org/asymptote/odoc/index.html#fig0126]here an example[/url] (figure 0126).


Doe's anyone know what formula represents the total number of gifts received in the song " The 12 days of Christmas" I know the total number of gifts is 364 but I would like to know how to derive it from a formula. :?

Thankyou so much for your help. We were able to derive part of the formula but not all of it.


Sorry, I don't know how to type some TEX. You can see my solution in the attached file.


modular arithmetic
number theory unsolved
number theory

Find all solutions if any so that

$ q,r,p$ odd primes

$ qr \equiv 2 \mod p^2$
$ (q\minus{}1)(r\minus{}1) \equiv 2 \mod p$

holds simultaneously

There are way too many solutions.
For example, this is just one infinite series of solutions:
\[ \begin{cases}
p\equal{}11\\
q\equiv 106\pmod{121}\\
r\equiv 16\pmod{121}
\end{cases}\]

Such series exist for every prime $ p$ such that $ p\equiv  1,\;9,\;11,\;15,\;23,$ or $ 25\pmod{28}$.


function
search
algebra unsolved
algebra

Let $ D$ a line in the complex plane . Find all functions $ f: \mathbb{C} \to \mathbb{C}$ with the propriety that $ \forall \text{ distinct } z_1,z_2 \in \mathbb{C}$ , the point $ z \equal{} \frac{f(z_1)\minus{}f(z_2)}{z_1\minus{}z_2} \in \mathbb{D}$ .

$ \mathbb{D}\equal{}a\mathbb{R}\plus{}b$ for some $ a,b\in\mathbb{C},a\not\equal{}0$, and we find $ g: \mathbb{C}\to \mathbb{R}$ with
$ {f(x)\minus{}f(0)\over x}\equal{}ag(x)\plus{}b$ for all $ x\in\mathbb{C}\setminus\{0\}$.
Then for $ c: \equal{}f(0)$ we have $ f(x)\equal{}axg(x)\plus{}bx\plus{}c$ for all $ x\in\mathbb{C}$,
which gives $ {f(x)\minus{}f(y)\over x\minus{}y}\equal{}a{xg(x)\minus{}yg(y)\over x\minus{}y}\plus{}b\in \mathbb{D}\equal{}a\mathbb{R}\plus{}b$
and therefore $ {xg(x)\minus{}yg(y)\over x\minus{}y}\in\mathbb{R}$ for all $ x,y\in\mathbb{C},x\not\equal{}y$.

But then for $ x\in\mathbb{C}\setminus\mathbb{R}$ and $ y\in\mathbb{R}\setminus\{0\}$ we have
$ {xg(x)\minus{}yg(y)\over x\minus{}y}\equal{}g(x)\plus{}(g(x)\minus{}g(y)){y\over x\minus{}y}\in\mathbb{R}$
with $ g(x)\in\mathbb{R}$ and $ g(x)\minus{}g(y)\in\mathbb{R}$ and $ {y\over x\minus{}y}\not\in\mathbb{R}$.
So $ g(x)\equal{}g(y)$ for all $ x\in\mathbb{C}\setminus\mathbb{R}$ and $ y\in\mathbb{R}\setminus\{0\}$,
which gives easily $ g(x)\equiv g(1)$ constant for all $ x\in\mathbb{C}\setminus\{0\}$
and then $ f(x)\equal{}(ag(1)\plus{}b)x\plus{}c$ for all $ x\in\mathbb{C}$.

So $ f$ is linear and $ {f(x)\minus{}f(y)\over x\minus{}y}$ is constant.

Thank you for your solution [b] olorin [/b] . Please , can you tell me how is this representation (of a line in a complex plane) called , so I can search more about that ? ( I didn't know this reprezentation  :blush:  ) .

[quote="olorin"]$ \mathbb{D} \equal{} a\mathbb{R} \plus{} b$ for some $ a,b\in\mathbb{C},a\not \equal{} 0$ [/quote]

Edit : Oh silly me ... :blush:    I didn't notice that it derives from the line equation :ewpu:

How else would you define a line?
It's the formal definition [url=http://en.wikipedia.org/wiki/Line]here[/url].
I simply identify $ \mathbb{C}\equal{}\mathbb{R}^2$.


Has anyone tried to solve the sample problems posted on the AHSIMC website (http://www.ahsimc.com)? How does everyone feel about discussing them and sharing solutions?

I've tried to solve them, and I'd welcome a discussion