geometry
geometric transformation
homothety
2002 aime ii 14
ratio
analytic geometry
algebra

Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6),$ and the product of the radii is $68.$ The x-axis and the line $y=mx$, where $m>0,$ are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt{b}/c,$ where $a,$ $b,$ and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a+b+c.$

let $O_{1}$, $O_{2}$ be the centers of the circles, $O$ the origin, $T_{1}$, $T_{2}$ points of tangency to x-axis, let $\angle O_{1}OT_{1}=\theta$, let $I$ be an inversion about the circle centered at $O$, passing through (9,6), note $I(\mathcal{C}_{1})=\mathcal{C}_{2}$, $OO_{1}O_{2}$ is a line, since $I(T_{1})=T_{2}$, $R^{2}=OT_{1}*OT_{2}\implies 9^{2}+6^{2}=[r_{1}\cot\theta] [r_{2}\cot\theta]$, we want $\tan2\theta=\frac{2\cot\theta}{\cot\theta^{2}-1}=\frac{12*\sqrt{221}}{46}\implies a+b+c=282$

i also must say that this is a nice problem, although it is somewhat contrived, and there is only one nice solution...

[hide="homothety!"]
Consider a circle of radius 1 that is tangent to the lines $y=0$ and $y=mx$, and let $(a,1)$ be the center of the circle. The line $2x=3y$ intersects this circle at two points. If $O$ is the origin and $X,Y$ are the two points, then by Power of a Point,
\[OX\cdot OY=a^{2}\]
Now expand the circle by dilations of ratio $r_{1}$ and $r_{2}$ centered at $O$, such that the first sends $Y$ to $(9,6)$ and the second sends $X$ to $(9,6)$. From this, we get
\[OY\cdot r_{1}=OX\cdot r_{2}=3\sqrt{13}\]
Multiplying the two,
\[68a^{2}=117\Rightarrow a=\frac{3\sqrt{221}}{34}\]
Finally, using the double-angle formula for tangent,
\[m=\frac{2/a}{1-1/a^{2}}=\frac{2a}{a^{2}-1}=\frac{12\sqrt{221}}{49}\Rightarrow \boxed{282}\]
[/hide]

[quote="Altheman"]let $ O_{1}$, $ O_{2}$ be the centers of the circles, $ O$ the origin, $ T_{1}$, $ T_{2}$ points of tangency to x-axis, let $ \angle O_{1}OT_{1} \equal{} \theta$, let $ I$ be an inversion about the circle centered at $ O$, passing through (9,6), note $ I(\mathcal{C}_{1}) \equal{} \mathcal{C}_{2}$, $ OO_{1}O_{2}$ is a line, since $ I(T_{1}) \equal{} T_{2}$, $ R^{2} \equal{} OT_{1}*OT_{2}\implies 9^{2} \plus{} 6^{2} \equal{} [r_{1}\cot\theta] [r_{2}\cot\theta]$, we want $ \tan2\theta \equal{} \frac {2\cot\theta}{\cot\theta^{2} \minus{} 1} \equal{} \frac {12*\sqrt {221}}{46}\implies a \plus{} b \plus{} c \equal{} 282$[/quote]

What's an inversion of the circle?

Here's a fast overview: [url=http://en.wikipedia.org/wiki/Circle_inversion]here[/url]

Let the x-coordinates of the centres be $ p,q$ respectively, and let them lie on the line $ y\equal{}kx$.

The radii of the 2 circles are $ kp$ and $ kq$ respectively, so we have $ k^2pq\equal{}68$.

$ (p\minus{}9)^2\plus{}(kp\minus{}6)^2\equal{}(kp)^2$ gives $ p^2\minus{}(18\plus{}12k)p\plus{}117\equal{}0$

Similarly, $ (q\minus{}9)^2\plus{}(kq\minus{}6)^2\equal{}(kq)^2$ gives $ q^2\minus{}(18\plus{}12k)\plus{}117\equal{}0$

Vieta tells us that since $ p,q$ are distinct roots of $ t^2\minus{}(18\plus{}12k)t\plus{}117\equal{}0$, we have $ pq\equal{}117$

Hence, $ k\equal{}\sqrt{68}{117}$

Then use $ m\equal{}\frac{2k}{1\minus{}k^2}$ (tangent double angle formula).


Let $ \triangle ABC$ be an acute triangle.The bisector of $ \angle ACB$ intersects $ AB$ at point $ L$,The feet of the perpendiculars from $ L$ to $ AC$ and $ BC$ are denoted by $ M$ and $ N$,respectively.Let $ P$ be the intersecting point of $ AN$ and $ BM$.Prove that $ CP \perp AB$

Dear Anonymous 111 and mathlinkers,
this problem can be solved with the Blanchet' theorem.
Sincerely
Jean-Louis

call K the feet of the altitude form C, we have $ \frac{AM}{MC} \cdot \frac{CN}{NB} \cdot \frac{BK}{KA} \equal{} \frac{AL \cos A}{BL \cos B} \cdot \frac{a \cos B}{b \cos A} \equal{} 1$ so for ceva AN,BM,CK concur

Let H is the foot of altitude of C. Suppose that P is not on the CH. But if we show to P is on the CH, we solve this quesiton.
K,M,C,N and L points are on the circle. \angle NLC= \angle CLM as for \angle NHC= \angle CHM. it is well known and it is Blancket' theorem. therefore, P is on the CH.


linear algebra
matrix
invariant
superior algebra
superior algebra unsolved

1) Prove that the eigenvalues of a matric do not form a complete set of invariants under similarity.

2) Show that $ \tau \in \mathcal{L} (V)$ is invertible $ \iff$ $ 0$ is NOT and eigenvalue of $ \tau$

Thanks a lot.

1) what do you mean by complete set of invariants?
2) suppose $ 0$ is an eigenvalue of $ \tau$, then $ \tau v \equal{} 0$ for some $ v$. This means $ v\in ker(\tau)$ which means it is not of maximal rank hence not invertible.

[quote="LkNsngth"]1) what do you mean by complete set of invariants? [/quote]

I am not sure what it means.


There are Odd consecutive numbers from 9 to 111, 5 numbers in this series are added together making distinct sums. How many different sums can you make?

[hide="Solution"]The minimum possible sum is $ 9\plus{}11\plus{}13\plus{}15\plus{}17\equal{}65$. The maximum possible sum is $ 103\plus{}105\plus{}107\plus{}109\plus{}111\equal{}535$. There are $ (535\minus{}65)/2\plus{}1 \equal{} 236$ different odd integers between the given minimum and maximum, so the answer is $ \boxed{236}$.[/hide]

can you explain why you divide by 2 and add 1?  :huh:

The division by $ 2$ is to compensate for the fact that not all integers are included; only the odd integers are included. I add one to compensate for the fact that the integers are undercounted in the subtraction; as a simple analogy, consider the number of odd integers between $ 1$ and $ 3$. Clearly, there are $ 2$ odd integers, but $ (3\minus{}1)/2\equal{}1$.

Or in other words, since you can only get odd numbers, you might want to prove this, and that all the odd numbers between 65 and 535 are possible sums.

For the first, you must choose 5 odd numbers, so the sum is odd(odd)=odd.

For the second, when you take 9+11+13+15+17, just add 2 to the 9. Then add 2 again. And again. And again. ... Soon you get: 111+11+13+15+17. Now you add 2 to 11. And again. And again. ... And if you continue this, you will soon cover all the odd numbers in between.


modular arithmetic
number theory unsolved
number theory

If 2009^n-1 is divisible by n,
show that there exists a prime p satisfying the following.
"2009^(np)-1 is divisible by np."

Take any prime dividing $ 2009^n \minus{} 1$.
Proof - Let $ x \equal{} 2009^n$. You're looking for a $ p$ such that $ p | \frac {x^p \minus{} 1}{x \minus{} 1}$. If $ x \equiv 1 \pmod p$, then $ \frac {x^p \minus{} 1}{x \minus{} 1} \equiv 1 \plus{} x \plus{} \cdots \plus{} x^{p \minus{} 1} \equiv 1 \plus{} 1^1 \plus{} \cdots \plus{} 1^{p \minus{} 1} \equal{} p \equiv 0 \pmod p$.

Similar to IMO 2000\5.

thanks a lot bambaman!!
you are a life saver


This Question needs to be done in 30 seconds or less:

$\text{Each number represents a non-zero digit. What is the value of }t?\\ \begin{eqnarray*}c+o&=& u\\ u+n&=& t\\ t+c&=& s\\ o+n+s&=&12 \end{eqnarray*}$

[hide]Substitute systematically and start with the first equation.
$(((c+o)+n)+c)+o+n=12$
$2(c+o+n)=12$
$c+o+n=6$
$u+n=6$
$t=6$[/hide]

[quote="indianamath"]ADD:

O= U-C
N=T-U
S=T+c
---------
O+N+S=2T

12=2T
[b]T=6[/b][/quote]

Nice original solution. Mine took me 2 mins trying to sub with the vars that they gave us.


pirates have guns

see what guns did to feudal japan?

[i]Who would win in the ultimate battle between Ninjas and Pirates? [/i]
If it's on sea, pirates would win (ninjas aren't as experienced on ships or operating ships, pirates might even be able to sink the ninja ship before contact).
If it's on land, it is more balanced, but Ninjas would probably win because Ninjas are used to land, while pirates are more used to ships.

[i]Who would win in most cases? [/i]
In most cases, pirates would win just because they are more leet. (Better planning, better psychological warfare / terror, and easier to understand tactics)

[i]Which one is cooler[/i]
Ninjas are cooler in appearance, but in reality, i think I'm the coolest :D

[i]Which one is manlier[/i]
They both are made of mostly men from what i know, but Ninjas have probably a higher porportion of man:woman. Pirates could somehow be manlier...they are slightly more massive and strong because Ninjas have to be skinny to be fast and versatile.

[i]What if both teams had super powers. (Pirates of Carribean, Naruto Powers, One Piece... You get the picture) Then who would win? [/i]
It depends on the powers. If pirates get the win-no-matter-what (aka invincibility) superpowers, then they'd win. If ninjas get the win-no-matter-what (aka invincibility) superpowers, then they'd win.

Haha there was an application on Facebook and I joined the ninjas  :D . So much cooler.

Ninjas. That has the same number of letters as Pandas, and the two words share three letters.  :ninja:

[quote="ashpotter"]WTH. I can't edit my post. Anyway, the powers are not invincible. No one can be invincible. Everyone is able to be killed by another on the other side.

About the superpower thing: Ninjas have elemental powers (making a wall of water, walking on water, fireballs). Pirates would have Physical powers (quick regeneration, being able to stretch their limbs)

I think Ninjas would win with a 100% chance if they catch the pirates off guard. By the time the pirates have realized what's happening, half of them are dead. I think Ninjas are better at offence then defence. They are best at evasion.[/quote]

Yes, I quite agree. But you said (or someone stated) earlier, Ninjas had the power to heal themselves. Obviously, like you said, the pirates can do that too. Being able to stretch their limbs? It reminds me of Clayface  :wink: Nobody is invincible, but the Ninjas have a better chance..The pirates don't have any long ranged weapons, not even their guns can go that far.

I think Ryu Hayabusa shows us that ninjas would win easily.

Pirates are kinda pointless. They're drunk, and they mutany every other day.  With such a disorganized caste system, how could they possibly stand up to this:

[url]http://halo.wikia.com/wiki/Image:MJOLNIR_Mk.VI-Hayabusa.jpg#file[/url]

Pirates are based purely on chaos.  They used their disorderly ways to commit barbaric acts and are successful in doing so only because their sly nature.  Ninjas, however, are based on a system structured through discipline.  In a battle, pirates may, at times, gain the upper-hand but ninjas' persistence would not let it be for long.  :wink:

Though just because a friend of mine think ninjas are cool, I would have to say pirates for the win.  :P

However, ninjas rely on chaos more than pirates. The more chaotic a scene is, the easier their task will be.

[quote="ProtestanT"][quote="D3m0n Shad0w"][quote="ashpotter"]Who would win in the ultimate battle between Ninjas and Pirates?

Who would win in most cases?

Which one is cooler

Which one is manlier

What if both teams had super powers. (Pirates of Carribean, Naruto Powers, One Piece... You get the picture) Then who would win?

Heres the pros

Ninjas: Fast. Vertisile. Have healing abilities. Have shuriken. Can hide very well.
Have short dagger/swords.
Pirates: Strong. Great in numbers. Can get powered up by alcohol. Have guns. Have long swords.[/quote]

I think this debate is from DragonFable, because recently there was a war: Ninjas vs. Pirates. More people joined the Ninja side, and the Ninjas won. Even though the pirates bribed every person who came to their side with like 10,000 gold. Goes to show that Ninjas are cooler (popular), and manlier, (not sure about that one, maybe pirates because they get drunk easily).
Ninjas would win. They have lightning quick skills, and they can dump all their water on the pirate's gun powder. Having no more guns, the pirates draw out their swords. The pirates charge at them, and the ninjas sit there, drinking tea and playing a game of soccer. When the pirates are 2 centimeters apart the ninjas just stab them with the shurikans.
Oh, and TV shows like Naruto show that Ninjas are really...weird. Like people in Pokemon or something..[/quote]
Where do ninjas get all their water? Pirates control the waters...and the beer :D[/quote]

Haha ninjas are lightning quick that they run to the ocean shores that pirates can't see them, and then they can dump the water on the pirates' gun powder. See?  :D

[quote]Pirates are kinda pointless. They're drunk, and they mutany every other day. With such a disorganized caste system, how could they possibly stand up to this:

http://halo.wikia.com/wiki/Image:MJOLNIR_Mk.VI-Hayabusa.jpg#file[/quote]

Very true. Pirates are so drunk they can't tie their shoes. They can't stand up to such 1337ness. Although ninjas rely on chaos, ninjas still have the power to be able to sense things in the dark. Pirates cannot. See where I'm going?

[quote="D3m0n Shad0w"][/quote][quote="ProtestanT"][quote="D3m0n Shad0w"][quote="ashpotter"]Who would win in the ultimate battle between Ninjas and Pirates?

Who would win in most cases?

Which one is cooler

Which one is manlier

What if both teams had super powers. (Pirates of Carribean, Naruto Powers, One Piece... You get the picture) Then who would win?

Heres the pros

Ninjas: Fast. Vertisile. Have healing abilities. Have shuriken. Can hide very well.
Have short dagger/swords.
Pirates: Strong. Great in numbers. Can get powered up by alcohol. Have guns. Have long swords.[/quote]

I think this debate is from DragonFable, because recently there was a war: Ninjas vs. Pirates. More people joined the Ninja side, and the Ninjas won. Even though the pirates bribed every person who came to their side with like 10,000 gold. Goes to show that Ninjas are cooler (popular), and manlier, (not sure about that one, maybe pirates because they get drunk easily).
Ninjas would win. They have lightning quick skills, and they can dump all their water on the pirate's gun powder. Having no more guns, the pirates draw out their swords. The pirates charge at them, and the ninjas sit there, drinking tea and playing a game of soccer. When the pirates are 2 centimeters apart the ninjas just stab them with the shurikans.
Oh, and TV shows like Naruto show that Ninjas are really...weird. Like people in Pokemon or something..[/quote]
Where do ninjas get all their water? Pirates control the waters...and the beer :D[/quote]

Haha ninjas are lightning quick that they run to the ocean shores that pirates can't see them, and then they can dump the water on the pirates' gun powder. See?  :D

[quote]Pirates are kinda pointless. They're drunk, and they mutany every other day. With such a disorganized caste system, how could they possibly stand up to this:

http://halo.wikia.com/wiki/Image:MJOLNIR_Mk.VI-Hayabusa.jpg#file[/quote][quote="D3m0n Shad0w"]

Very true. Pirates are so drunk they can't tie their shoes. They can't stand up to such 1337ness. Although ninjas rely on chaos, ninjas still have the power to be able to sense things in the dark. Pirates cannot. See where I'm going?[/quote]
You might not know this, but pirates can be lightning quick too.
Also, in the same way, pirates can also run to the ninjas and not be seen at night and murder ninjas up close (although it would be very weird if this happens)

edit-another thing:
Statistics show that there about 30 million pirates left in the world with 500 pirates to a ship (60000 ships). There are also about 30000 ninjas left in the world. I doubt that each ninja can capture 2 ships and 1000 pirates, especially if the pirates join up into one big force. Therefore pirates win by sheer numbers (1000:1 ratio), if not by skill.

Guys, guys we all know it's Superman!

But seriously I have no idea what this topic is about.

[quote="K81o7"][quote="ashpotter"]
I think this Fat means moer about massiveness then accually fat. A person with 200 pounds of muscle will beat a person with 200 pounds of fat[/quote]

Well I already said ninjas are probably more muscular.  I think, for example, that 100 pounds of muscle beats 75 pounds of muscle and 50 pounds of fat.  If we're talking about a fight, more fat only hurts a person.  If we're talking about manliness, I think fat is a bad measure...[/quote]

The likelyhoods are that the pirates have as much muscle as possible, because if they got any nonmuscly fat they're probably useless on the ship. Same with the ninjas, if they've got nonmuscly fat they can't be speedy.

True. But it seems that pirates are always bigger than ninjas. But not necccearily stronger.

Sorry about all those random posts, ragnarok23 hacked my account, again.
I actually took the time to change my password this time  :|

[quote="pianoforte"]No. Pirates = White. Ninja = Yellow.[/quote]

RACIST!!!! :P  :D

JK.

but yea, if youre a double agent, you can be a ninja pirate

NINJA NINJA NINJA

I LOVE NINJAS!!!!!!!!!!!!

 :ninja:  :ninja:  :ninja:

Ninjas rule the world!

For all you ninja fans, go to askaninja.com, this guy is hilarious!

pirates pirates pirates!!!!

they got cool birds..

Cool! Sorry for the revive, but which one is better? PIRATES or NINJAS???

Ugh I’ll tell you the truth. I revived for post #26, etc.


analytic geometry
Asymptote
trigonometry
vector
geometry
angle bisector
similar triangles

Today, I was working this problem:
Point N is on the hypotenuse BC of $ \triangle ABC$ such that <CAN = 45. Given AC = 8, AB = 6, find AN.
The solution is based on the fact that the projection of A to BC (H) is closer to B than N, while in my diagram, I drew it closer to C.
Is this truth a common fact, or am I just missing something?

Not sure if this actually changes anything, but any comments would be appreciated.

o_O Stewart bashing it would seem to be comparatively painless, but if you either a) algebraically bash it or b) draw it out carefully, you can clearly see that H is between B and N.

[hide="solution with stewarts & angle bisecting"][asy]unitsize(20); pair A,B,C,D,N,E; A=origin; B=(0,3); C=(4,0); D=(-1,-1); E=(1,1); N=(12/7,12/7); dot(A); dot(B); dot(C); dot(N); draw(A--B); draw(B--C); draw(A--C); draw(A--N); label("\small{$A$}",origin,D); label("\small{$B$}",B,D); label("\small{$C$}",C,D); label("\small{$N$}",N,E);[/asy]

Note that $ \overline{AN}$ bisects $ \angle BAC$.  Thus, we can apply the angle bisector theorem to $ \triangle ABC$ to find $ BN=\frac{30}{7}$ and $ CN=\frac{40}7$.

Then, we can apply Stewart's Theorem to find that $ (BC)(AN)^{2}+(BN)(CN)(BC)=(AB)(NC)^{2}+(AC)(NB)^{2}$

$ 10(AN)^{2}+\left(\frac{30}7\right) \left(\frac{40}7\right)10= 6\left(\frac{40}7\right)^{2}+8\left(\frac{30}7\right)^{2}$

$ AN=\frac{4\sqrt{30}}7$, I think.

w00t, first picture I've made![/hide]

You can easily prove that H lies between B and N by drawing AH and then using similar triangles to find some lengths, along with the lengths found using the angle bisector theorem.

[quote="b-flat"][hide="solution with stewarts & angle bisecting"][asy]unitsize(20); pair A,B,C,D,N,E; A=origin; B=(0,3); C=(4,0); D=(-1,-1); E=(1,1); N=(12/7,12/7); dot(A); dot(B); dot(C); dot(N); draw(A--B); draw(B--C); draw(A--C); draw(A--N); label("\small { $A$} ",origin,D); label("\small { $B$} ",B,D); label("\small { $C$} ",C,D); label("\small { $N$} ",N,E);[/asy]

Note that $ \overline{AN}$ bisects $ \angle BAC$.  Thus, we can apply the angle bisector theorem to $ \triangle ABC$ to find $ BN=\frac{30}{7}$ and $ CN=\frac{40}7$.

Then, we can apply Stewart's Theorem to find that $ (BC)(AN)^{2}+(BN)(CN)(BC)=(AB)(NC)^{2}+(AC)(NB)^{2}$

$ 10(AN)^{2}+\left(\frac{30}7\right) \left(\frac{40}7\right)10= 6\left(\frac{40}7\right)^{2}+8\left(\frac{30}7\right)^{2}$

$ AN=\frac{4\sqrt{30}}7$, I think.

w00t, first picture I've made![/hide]

You can easily prove that H lies between B and N by drawing AH and then using similar triangles to find some lengths, along with the lengths found using the angle bisector theorem.[/quote]

I got something different as in $ \frac{24\sqrt{2}}{7}$, bflat look at your Stewarts formula, you got the formula messed up on the right side.

[quote="b-flat"]Then, we can apply Stewart's Theorem to find that $ (BC)(AN)^{2}+(BN)(CN)(BC)=(AB)(NC)^{2}+(AC)(NB)^{2}$[/quote]
Thats not the right formula

[quote="b-flat"]Huh?  I double checked it with the one in the wiki, so either I'm missing something obvious, or the wiki is wrong...

[url]http://www.artofproblemsolving.com/Wiki/index.php/Stewart%27s_Theorem[/url]

However, when I did it with coordinates, I get $ \frac{24\sqrt{2}}7$ as you are.[/quote]

[hide]The correct formula should be
\[ (BC)(AN)^{2}+(BN)(CN)(BC)=(AB)^{2}(NC)+(AC)^{2}(NB) \]
not
\[ (BC)(AN)^{2}+(BN)(CN)(BC)=(AB)(NC)^{2}+(AC)(NB)^{2}\]
This produces the correct answer of $ \frac{24\sqrt{2}}{7}$.[/hide]

EDIT:
[hide="to b-flat"]Good! You started to learn Asymptote. I recommend you learn the package cse5 because it saves you a lot of code on the 2000 character limit.
But, looking at your code, I'm not sure what you are trying to do.

For example,
[asy]import cse5; pair A=MP("A", D((0,0))),B=MP("B", D((0,3))),C=MP("C", D((4,0))),N=MP("N",D((12/7,12/7))); D(A--B--C--cycle); D(A--N); fill(arc(A,0.5,0,45)--A--(0.5,0)--cycle,red); MP("45^ { \circ} ",0.5*dir(22.5),E); MP("8",(A+C)/2); MP("6",(A+B)/2,W);[/asy]
might by what you wanted. But why do you need the points D, E, and what are you trying to label? I might be able to help you out.[/hide]

Another, nicer solution can be found using this hint:
[hide="Hint"]Use the Angle Bisector Theorem on $ \triangle ABC$ to find $ NC$. Then, use law of sine on $ \triangle ANC$:
$ AN/\sin C = NC/\sin 45^{\circ}$.
You know $ \sin C$ from right triangle $ \triangle ABC$ and you know $ NC$ from the Angle Bisector Theorem. So, now, you can solve for $ AN$.[/hide]

[quote="Altheman"]Err...$ AN$ is the angle bisector of angle $ A$. The problem should be trivial now.[/quote]

That's.......pretty clear?

See, the thing is, I know Stewart's, LoS will work, but my problem initially was that I drew H on the wrong side which would screw it up anyway I believe, and anyway, this was in Intro to Geo "before" we have learned Stewarts, LoS, LoC, etc

$ \angle BAH=\angle C$, by similar triangles. $ \angle C<\angle B$, because their opposite sides satisfy$ c<b$. Since $ \angle C+\angle B=90$, and $ \angle C<\angle B$, $ \angle C=\angle BAH<\angle 45=\angle BAN$. Therefore, since $ \angle BAH<\angle BAN$, $ BH<BN$.

Then I guess the rest of the solution was to find BN by Angle Bisector Theorem, and use similar triangles $ \triangle ABC ~\triangle HBC$ to find BH and AH, NH=BN-BH, then $ AN=\sqrt{AH^{2}+NH^{2}}$, right?

[quote="vishalarul"]
[hide="to b-flat"]Good! You started to learn Asymptote. I recommend you learn the package cse5 because it saves you a lot of code on the 2000 character limit.
But, looking at your code, I'm not sure what you are trying to do.

For example,
[asy]import cse5; pair A=MP("A", D((0,0))),B=MP("B", D((0,3))),C=MP("C", D((4,0))),N=MP("N",D((12/7,12/7))); D(A--B--C--cycle); D(A--N); fill(arc(A,0.5,0,45)--A--(0.5,0)--cycle,red); MP("45^ { \circ} ",0.5*dir(22.5),E); MP("8",(A+C)/2); MP("6",(A+B)/2,W);[/asy]
might by what you wanted. But why do you need the points D, E, and what are you trying to label? I might be able to help you out.[/hide][/quote]

I wasn't sure exactly how labeling points worked, and one way I could get the labels to go in the correct directions was to use the points D and E as vectors to "add" on to the current points, which seemed to move the label around.  And, I didn't know how to use angles either, so I just left those out.

Thanks, though for your help on the code!

Is there a way to "construct" the angle bisector?  The only way I knew was to actually calculate N...

Drop perpendiculars to $ AB$ and $ AC$ to hit at $ P,Q$, then $ APNQ$ is a square. Then, $ BP\equal{}\frac{3}{4}PN\equal{}\frac{3}{4}PA$ by similar triangles, and it's easy from there.


[b]K [/b]is a natural number. prove that: [b]k! 3[/b] divisible by[b] (k!) k 2+k+1 . [/b]

Umm.. I don't know if you typed it correctly, but at the moment, since $k^2 + k + 1 \ge 3$, $(k!)^{k^2+k+1}$ is divisible by $(k!)^3$... (not the other way around)

I think he wants to say that:

$(k^3)!$ divisible by $(k!)^{k^2+k+1}$

It's a Poland Olympiad and I think It's not too difficult.

Isn't this a simple application of a result by Legendre which states that the highest power of a prime p dividing n! equals $[n/p] + [n/p^2] + [n/p^3] + ...$?

this solved by Myth long time ago, to get full sol, try unsolved section....

No,I think It's easy If use the usual Lenma:
$(mn)!$ divisible by $m!(n!) ^m$


If you'd like to suggest further topics for the FAQ or correct errors please do so in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123574]this thread[/url].

[b]1. I don't have LaTeX on my computer. Can I type a formula on this forum?[/b]
Yes! All you need do is enclose your mathematics between dollar signs and the rest is done by AoPS - MathLinks server. So to get $ \sqrt {x^{2} + y^{2}}$ all you need do is type [code]$\sqrt{x^2+y^2}$[/code]You can see the code others have used by hovering the mouse over a formula or by clicking on the formula which will also allow you copy that code.
There is no need to put ordinary text between dollar signs.

[b]2. How do I find out what to type on the forum?[/b]
[[LaTeX:Commands]] shows you how to get powers, fractions, integrals etc and [[LaTeX:Symbols]] shows you how to get plus or minus, greater than or equal, Greek letters etc.
You can try it out in the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url] or the [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]Test Forum[/url]
The [url=http://www.codecogs.com/components/equationeditor/equationeditor.php]LaTeX Equation Editor[/url] will help you write your LaTeX code for a short formula.

[b]3. The margins in my LaTeX document are wrong. How can I change them?[/b]
The geometry package makes setting margins etc very easy. For example, [code]\usepackage[margin=1in]{geometry}[/code]puts a margin of 1in all the way round.
You can make the various margins different and even mix measurements like this [code]\usepackage[top=2.9cm,left=2in,bottom=1in,right=1in]{geometry}[/code]See the geometry package [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]user's manual[/url] for more details.

[b]4. How can I change the font size and colo(u)r in my document?[/b]
The font size of text in can be changed by using the following commands which got from the smallest to the largest: \tiny \scriptsize \footnotesize \normalsize \large \Large \LARGE \huge \Huge
So you can write $ \text{{\tiny tiny}{\large \ large}}$ but it looks awful so use very sparingly if at all.

To change the colo(u)r you need to use the color package with [code]\usepackage{color}[/code] and then you can type things like [code]\color{red}This is in red and so is $x+y+z$\\
\normalcolor This is not[/code]which works with math(s) and text. However, it's not available on the forum.

[b]5. Tell me about eqnarray[/b]
The only thing to say is that it is now obsolete! If you want to know why then [url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url] will explain.
The replacement is align and is very similar to use, though it doesn't need so many &'s, but does require [code]\usepackage{amsmath}[/code] in your document (see 8. below). It lines up a series of equations nicely.[code]\begin{align}
   x &= 1 \\
   y &= 2
\end{align}[/code] results in
\begin{align}x & = 1 \\
y & = 2 \end{align}
If you don't want the equations numbered then use align* instead of align.

[b]6. I use MiKTeX and I want to add a new package. How should I do this?[/b]
Suppose that you want to add the fancyhdr package (the same procedure works for hundreds of other packages). In the latest version of MiKTeX (version 2.5 or later) you may just have to add the line [code]\usepackage{fancyhdr}[/code] at the start of your document and MiKTeX downloads the package and installs it automatically.

Sometimes this doesn't work (you may not have given MiKTeX permission to do this) or you have an older version. In that case go to
Version 2.4: Start -> All Programs -> MiKTeX -> MiKTeX Package Manager
Version 2.5 (or 2.6): Start -> All Programs -> MiKTeX 2.5 (or 2.6) -> Browse Packages

Wait a while until MiKTeX loads all the package names. Find fancyhdr in the list, right-click, choose Install and in a few moments MiKTeX will tell you it's done and you can close the program and compile your file.

In Version 2.5 (or 2.6) you can set MiKTeX to automatically install new packages as needed. Start - All Programs -> MiKTeX 2.5 (or 2.6) -> Settings -> Install missing packages on-the-fly. I suggest you set this to Ask me first. Then when you compile a document which uses the package, MiKTeX will install it for you.

[b]7. How can I make a table?[/b]
[url=http://www.g32.org/latable/]LaTable[/url] is a small free program that allows you to design tables complete with merged cells, borders etc. It generates the LaTeX code as you type which you then copy to TeXnicCenter. It makes tables very easy to do.

[b]8. What packages are needed for all documents?[/b]
amsmath and similar were created to solve most people's mathematical needs so should be included in all documents. A template for your documents could look something like this: [code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}

\end{document}[/code]TexnicCenter, for example, provides a number of templates you can customise which you use via File, New from Template. The templates are found in something like C:\Program Files\TeXnicCenter\Templates\Documents but you can add more places using Tools->Options->Directories.

[b]9.  My opening quotation marks are backwards, how do I correct this?[/b]
Only use ' and " for closing quotation marks not for opening ones - use one backtick ` or two backticks`` instead. You should find it on the left of the keyboard along with two other symbols ¬ and ¦ but keyboards in different countries may have it in a different place. TeXnicCenter can do this for you automatically Tools -> Options -> General -> Replace quotation marks.

[b]10. How can I make an arc over AB?[/b]
You can use \stackrel{\frown}{AB} which looks like $ \stackrel{\frown}{AB}$ or use the arcs package. However that package has stopped working on recent LaTeX distributions so use the revised arcsaops.sty package below instead. Then \overarc{AB} gives $\text{\overarc{AB}}$.

A summary of some key components for getting started:

Start TeXnicCenter.

1.  Create a project by choosing the File/New Project menu item, and choose one of the various templates.  This will create a project with a main file containing standard code for the selected type of LaTex document.  You can add code to this document, but at some point it becomes unwieldy to shift through.

2.  Several output formats are available.  I suggest choosing the DVI output:  Select the Build menu item then Choose Output Type/Latex=>DVI.  This will enable the YAP previewer which is the program displaying the document after it's been built.

3.  Once a project is created, a main latex file is created by the system and is displayed in the TeXnicCenter editor.  Build it and review the build log at the bottom of the editor and verify no critical errors occurred; this would prevent YAP from starting up the first time.  Otherwise YAP displays the output of the main file.  From the YAP View Menu item, select Options/Reverse DVI Search, then select TeXnicCenter, then choose Apply.  Now double clicking on any word in the output document directs you to the  source code at the point corresponding to the location you clicked.  This allows for quick and efficient code development.

4.  If necessary, segregate components of the document into chapters or other components.  Do this by creating separate files for each component using the File/New/From Template option on the main menu.  Choose an appropriate file type.  This creates a stub file which is then added to the project.  The file name should appear in the Navigator window.  Add the relevant code to the chapter file.  The stub includes default newcommand lines for defining short cuts.  These should be edited as necessary or removed entirely to prevent "duplicate definition" error messages when multiple stubs are being loaded into the build.  Rename the stubs as necessary.

5.  At the point in the main file where this component is to be located, enter:

\input{file_name_here}

Each component file can now be edited separately, and the document then re-built from the editor simply by pressing Ctrl-F5.  If there are no serious errors with the code, YAP displays the changes.

6.  Use the Navigator tree on the left of the TeXnicCenter to navigate to various components of the document.  Alternatively, each file name is located on a tab at the bottom of the edit menu.  Just click one and open it in the editor.

7.  When including graphics, use EPS files for DVI output and PDF, PNG, JPEG, or GIF graphics files for the PDF output.  A comprehensive guide for using graphic files in LaTex can be found at:

[url]ftp://cam.ctan.org/tex-archive/info/epslatex/english/epslatex.pdf[/url]

Graphic files slow the build and display of the output.  To improve performance of developing a LaTex document, enable "draft" on some or all included graphics  files.  This prevents full loading of graphics files; only an empty box with the file name and a caption is displayed. For example use the following syntax for excluding a single file in a \begin{figure} block:

\includegraphics[scale=1,draft]{testeps.eps}

Alternatively, all graphics files can be excluded by enabling "draft" in the package command at the top of the main file:

\usepackage[draft]{graphicx}

See [[LaTeX:Downloads]]. But you don't need to do ths if you only want to post on the forum. See LaTeX FAQ 1 above.

either im really dumb, or, well, im really dumb.

how do you write 2^2x in latex?

This is the only spot where I could find a place to ask this.

How do I get a triangle to appear around text?

Ask in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote forum[/url] with more detail of exactly what you are looking for.

How do you type:
1) 9P3 (or similar)
2) <= (the < underlined; nonstrict inequality)
3) Space between words

Don't put words inside your dollar signs. Space solved.

\leq $ \leq$

\binom{9}{3} $ \binom{9}{3}$

Or in the case that you do use text (maybe in a longer explanation) use the \text{} command thing.

[code]$\text{Hi my name is Dojo}$[/code]
$ \text{Hi my name is Dojo}$

You can also add color and bold to your text
[code]$\textbf{Hi my name is Dojo}$[/code]
$ \textbf{Hi my name is Dojo}$
[code]$\color{red}\text{Hi my name is Dojo}$[/code]
$ \color{red}\text{Hi my name is Dojo}$

While I'm quite sure you can use LaTeX text for explanations or in formulas, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143044]this[/url] post.

Thanks guys :) . But I meant 9[u]P[/u]3, not 9C3.

Generally, you would denote permutations as
[code]$9P3$[/code]
$ 9P3$
or
[code]P(9,3)[/code]
$ P(9,3)$

Either way, it's just simple text.

Or $ ^9P_3$ or $ _9P_3$
topofmath needs to make clear exactly what they want, illustrating, if necessary, with a screenshot or a link to an image.

Thanks I sort of get it.....

$\sqrt{x^2+y^2}$

It worked!!!  :)

How do you do a fraction over a fraction?

$\sqrt{\frac{a^2 + b^2 c^2}{3}}$ $\geq$ $\frac{a+b+c}{3}$ $\geq$ $\sqrt[3]{abc}$

$\sqrt{\frac{a^2 \times b^2 \times c^2 \div d^2}{7}}$ $\geq$ $\frac{\sqrt{abc}}{\frac{3}{2}}$ $\leq$ $\frac{\sqrt{frac{2}}{9}}$ $\frac{s^2}{pty^r \times 12^w}$

I know nothing about LaTex!!!
Could someone please help me!!

Um, I was just wondering, how do I write in $\LaTeX$ in classrooms? I know you have to include a ";", but that is as far as I know. I have tried typing " ;\frac{1}{2} " and " ;a^2 " and gotten unsuccessful results. Please help ASAP! Thanks in advance.

You need dollar signs. If mathjax is enabled, you don't need the semicolons

[quote="hiamp"]how do you get latex on your computer[/quote]
You can also try LaTeX equation editors like : http://www.codecogs.com/latex/eqneditor.php
It is free and you don't need to download it.

Does anybody know any [b]good[/b] $\LaTeX$ books and PDFs out there?

There are a lot in the stickies...
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=123&t=117360

[quote="malachiv"]I know the basics of typing in color, but how do you mix colors?[/quote]
[url=http://en.wikibooks.org/wiki/LaTeX/Colors]Here[/url] is a guide on colors.

[quote="snoopygenius37"]You could type in solid and glittery colors but no rainbow or swirly.[/quote]
Can you show me the LaTeX coding in code format?
[code]
$\text{Yolo}$[/code]

Yes,  here it is:
     [u]color you want[/u]    [b]text you want[/b]
            ^
$\color[u]{}[/u]{\text[b]{}[/b]}$
EXAMPLE:  $\color{blue}{\text{this is gold.}}$

Underline is color you want and bold id text you want.  $\color{blue}{\text{banana is not blue.}}$


function limit algebra domain integration vector probability

It is a local contest of the Ural State University, Ekaterinburg, Russia (the most significant University in the Ural/Siberian region).

Problems are given for 1 night to solve. Numbers in the round brackets are average points for particular problem.

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=33773
(3) Solve $(x+y)^n=x^n+y^n$ in $\mathbb{R}$ for all $n\in\mathbb{N}$.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=33763
(5) Let $\{a,b\}$ is an alphabet and $w$ is a word over this alphabet. Denote by $A(w)$ the number of occurances of $a$ in $w$ and by $B(w)$ the number of occurances of $b$ in $w$. Let $L$ be a language consisting of those words $w$, for which $(A(w),B(w))=1$. Is it true that $L$ is a [i]regular[/i] language? (I don't know what [i]regular[/i] language is)

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=33778
(3) Gods painted a regular hexagon with side $a$ on a plane. Also they gave you a ruler without any marks. You are to construct a segment of length $a/n$, $n\in\mathbb{N}$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=33774
(10) Exclude either $x$ or $y$ from the following system
\[
\left\{ \begin{array}{l}
 2x^3  + 3xy^2  - 4x^2 y^2  + y + 1 = 0 \\
 3x + 5x^3 y^2  - 2xy^5  - xy + 2 = 0 \\
 \end{array} \right.
\]

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=33775
(3) Prove that if continuous function $f:\mathbb{R}\to\mathbb{R}$ satisfies the equality $f(f(f(x)))\equiv x$, then $f(x)=x$.

Give an example of function $g:\mathbb{R}\to\mathbb{R}$ satisfying $g(g(g(x)))= x$ and $g(x)\neq x$.

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=33764
(2) Given 37 integer points in $\mathbb{R}^3$, no three of them are collinear. Prove that there are three points, those centroid is an integer point.

7. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sequence+varphi&t=31810#
(7) Consider initial sequence 1,1. Insert between any two neigbour terms their sum:
1,1 -> 1,2,1.
And again
1,2,1 -> 1,3,2,3,1 -> 1,4,3,5,2,5,3,4,1
And so on...
Prove that every natural number $n>1$ will be written exactly $\varphi(n)$ times, where $\varphi(n)$ is an Euler's totient function.

8. http://www.mathlinks.ro/Forum/viewtopic.php?t=33772
(30) Let $p(x)$ be the density of the set of natural numbers, which have a divisor in the set $\{x,x+1,...,2x\}$.

Prove that $\liminf_{x\to\infty}p(x)=0$.

9. http://www.mathlinks.ro/Forum/viewtopic.php?t=33765
(7) Find all $x\in\mathbb{R}$ s.t. $f_n(x)=\sin(7^n\pi x)$ converges and find its limit.

10. http://www.mathlinks.ro/Forum/viewtopic.php?t=33766
(5) Let $f(x)\in C^{\infty}$ on $(-a,a)$, and $f^{(n)}(x)$ converges uniformly on $(-a,a)$. It is known that $\lim_{n\to\infty}f^{(n)}(0)=1$. Find $\lim_{n\to\infty}f^{(n)}(x)$.

11. http://www.mathlinks.ro/Forum/viewtopic.php?t=33767
(3) Let $f(x):M\to \mathbb{R}$ be a continuous function, $M\subset[0,1]$. Prove that we can define $f$ on $[0,1]\setminus M$ s.t. $f$ will stay continuous on $M$.

12. http://www.mathlinks.ro/Forum/viewtopic.php?t=33768
(15) Let $a$ be a smooth regular curve with continuous curvature, and $a$ bounds a convex domain $D\subset \mathbb{R}^2$. Define $p(x,y)=\text{dist}(a,(x,y))$ for $(x,y)\not\in D$, and $p(x,y)=-\text{dist}(a,(x,y))$ for $(x,y)\in D$. Prove that for all small $R$ the following equality holds
\[
\int_{|p(x,y)|<R}p(x,y)dxdy=\frac{4}{3}\pi R^3.
\]

13. http://www.mathlinks.ro/Forum/viewtopic.php?t=33769
(4) Prove that the equation $y'=y^2+x$, $y(0)=0$ doesn't have a solution on $(0,3)$.

14. http://www.mathlinks.ro/Forum/viewtopic.php?t=33771
(6) It is known that all columns of $A$ are pairwise orthogonal. Prove that $|\det A|$ equals to product of lengths of vector-columns.

15. http://www.mathlinks.ro/Forum/viewtopic.php?t=33776
(6) Prove that every integer number can be represented as a sum of 5 cubes.

16. http://www.mathlinks.ro/Forum/viewtopic.php?t=33770
(13) Find the probability $p_n$ of the event that an arbitrary matrix consisting of 1 and 0 has the non-zero determinant. Find $\lim\limits_{n\to\infty}p_n$ if you can!

I will create separate topics for each problem soon.

BTW, I didn't find an appropriate section for university's contests.

[quote="Myth"]

BTW, I didn't find an appropriate section for university's contests.[/quote]

Try [url=http://www.mathlinks.ro/Forum/index.php?f=78]this.[/url]

Mikhail, I have created a new forum in that category, to include all the other competitions not discussed there.


Given the points K = (0,0), L = (7,−1), M = (9,3), P = (6,7), Q = (10,5), and R = (1, 2), triangles KLM and RP Q are congruent.  Find coordinates for the point in triangle PQR that corresponds to (6,2) in triangle KLM.

Thanks.

LOCK THIS TOPIC, its already in Intermediate Topics

[url]http://www.mathlinks.ro/viewtopic.php?t=298713[/url]


The police discover a murder victim at 6:00am. The body temperature of the victim is measured then and found to be $ 25^oc$ . A doctor arrives on the scne of the crime 40 minutes later and measures the body temperature again. It is found to be $ 22^oc$. The temperature of the room remains constant at $ 15^oc$. The doctor, knowing that normal body temperature is $ 37^oc$ able to estimate the time of deatrh of the victim.

If the cooling process is modelled by the equation $ \theta \equal{} \theta_0e^{ \minus{} kt}$, where  is the excess temeprature over the room temperature, estimate the time of death

[b]PLEASE TYPE A HINT OF WHERE I CAN START. HIDE YOUR SOLUTION[/b]

regards,
bos1234

[quote="bos1234"]The police discover a murder victim at 6:00am. The body temperature of the victim is measured then and found to be $ 25^oc$ . A doctor arrives on the scne of the crime 40 minutes later and measures the body temperature again. It is found to be $ 22^oc$. The temperature of the room remains constant at $ 15^oc$. The doctor, knowing that normal body temperature is $ 37^oc$ able to estimate the time of deatrh of the victim.

If the cooling process is modelled by the equation $ \theta \equal{} \theta_0e^{ \minus{} kt}$, where  is the excess temeprature over the room temperature, estimate the time of death[/quote]

[hide="Hint."]
-----
I will omit units of measure for clarity: temperature will be in degrees and time in minutes.

Take $ 6: 00$ to be time $ t \equal{} 0$. You know that the ambient temperature is $ 15$ and that the body temperature at $ t \equal{} 40$ is $ 22$. You can then find the difference in temperature $ \theta$ and, knowing that the temperature at $ t \equal{} 0$ was $ 25$, the initial difference in temperature $ \theta_0$. Using Newton's Law of Cooling which you posted above ($ \theta \equal{} \theta_0e^{ \minus{} kt}$), you can find $ k$.

Using $ k$, $ \theta_0$, as well as the fact that the temperature at the time of death was $ 37$ which gives you $ \theta$, you can find $ t$, the time of death.

You should get the following answer:
[hide="Answer."]
=====
$ k\approx 0.008916...$
$ t\approx \minus{} 88.42307... \implies \boxed{4: 32\ \text{AM}}$
=====
[/hide]
-----
[/hide]

ok my friend i shall try with your hints

thanks for helping


function
calculus
derivative
calculus computations

function $ f(x,y)$ is continuous together with its derivatives in $ x$ and $ y$ and satisfies conditions:
$ f(0,0)\equal{}0; \left |\frac{\partial f}{\partial x}  \right | \leq 2004 |x\minus{}y|;  \left |\frac{\partial f}{\partial y}  \right | \leq 2003 |x\minus{}y|$
Prove that, $ |f(2004,2003)| \leq 2004^3$


Consider a triangle ABC inscribed in a circle (O). A point P is chosen on arc AC opposite of B such that PX=CX, where X is the intersection point of BP and AC. M is an arbitary point chosen on BC extended with PM=CM. prove that A,P and M are collinear

It is obvious- but then again, I am in high school  :D
[hide]Since PX=CX, P and C are symmetric w.r.t the perpendicular bisector of AB, and since PM=CM M lies on the perpendicular bisector of PC so since B,C, and M are collinear, A,P, and M are collinear.[/hide]

is middle school and high school the same thing or do we have the high school rooms closed?


For both questions below, z = 3 - 4i and w = 8 + 3i.

Write each expression in the standard form a + bi.

[color=red]NOTE: For question 1, the w on the right side has a bar over it.[/color]

[1] w - w

[color=blue]NOTE: For question 2, BOTH z and w lie under ONE BAR TOGETHER.[/color]

[2] z - w

$\overline{a+bi}= a-bi$

The "bar" over the number means to find the complex conjugate of that number.

$\overline{z}$ is known as the [i]complex conjugate[/i] of $z$. It is the unique complex number with the same real part as $z$, and an imaginary part with equal absolute value but opposite sign.

[size=67]My #500 post![/size]


[quote="Altunshukurlu"]Find all real numbers a for which the function f(x)= {ax +sin x} is periodic. {x} = the fractional part of x .[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=307480


In a triangle $ABC$ with $AB\not = AC$, the incircle touches the sides $BC, CA, AB$ at $D, E, F$ , respectively. Line $AD$ meets the incircle again at $P$ . The line $EF$ and the line through $P$ perpendicular to $AD$ meet at $Q$. Line $AQ$ intersects $DE$ at $X$ and $DF$ at $Y$ . Prove that $A$ is the midpoint of $XY$.

See the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113152]Looks easy[/url].

We can clearly see that $ AQ\parallel{}BC$. Now, consider the shaded triangles.
consider the tangents to the incircle, call them a',b',c' coming from A,B,C resp.
Then, $ XA/AY\equal{}XA/BD*BD/DC*DC/AY\equal{}a'/b'*b'/c'*c'/a'\equal{}1$ so A is the midpoint as desired.
[img]19454[/img]

I have another solution to this problem:
Suppose the line through P perpendicular to AD meets the line through A perpendicular to ID at Q.
We need to prove that E, F, Q are collinear. Let H be the intersection of PQ and (I). Therefore we have D, I, H are collinear. Hence, DH is perpendicular to AQ. Invert around D, with power DP.DA=DH.DK (K is the intersection of DI and AQ), that maps A to P, K to H. Therefore AQ is mapped into (DHK) or in the other word, AQ is maped in to (I). So Q is maped into Q* lies on (I). E and F are maped into itself.
DFEQ* is concyclic. So E, F, Q are collinear.
This leads to the fact that XY are parrallel to BC.(1)
We have PFDE are harmonic. So (DP,DC,DF,DE)=-1 or (DA,DC,DY,DX)=-1(2)
Combine (1) and (2) we have A is the midpoint of XY.
Proof is completed

[quote="N.T.TUAN"]In a triangle $ ABC$ with $ AB\not \equal{} AC$, the incircle touches the sides $ BC, CA, AB$ at $ D, E, F$ , respectively. Line $ AD$ meets the incircle again at $ P$ . The line $ EF$ and the line through $ P$ perpendicular to $ AD$ meet at $ Q$. Line $ AQ$ intersects $ DE$ at $ X$ and $ DF$ at $ Y$ . Prove that $ A$ is the midpoint of $ XY$.[/quote]

I am sorry if having anyone have posted the same solution with me before.
Here is my solution for this problem :
Let $ M$ is midpoint of $ FE$.
Easy to prove that quarilateral $ APMQ$ is cyclic.
$ \Rightarrow \angle PAQ\equal{}\angle PMF\equal{}\angle PEF\plus{} \angle MPE$
Addition , we have $ PQ$ is the symedian . So $ \angle FED\equal{}\angle FPD\equal{}\angle MPE$
$ \Rightarrow \angle PAQ\equal{}\angle PEF\plus{} \angle FED\equal{}\angle PED\equal{}\angle BDF$
$ \Rightarrow AQ // BC \Rightarrow AX\equal{}AF\equal{}AE\equal{}AY$
Q.E.D

well know lemma.
$ A^{*}$ is inversed point with respect to circle $ C$ with center $ O$.
A line through $ A$ meets $ C$ at two points $ X, Y$.
Then, $ O,X,Y,A^{*}$ are on the same circle.

using this lemma,
$ AI\perp EF$and ,$ AI \cap EF=Y$
Then, it is easy to elicit conclusion .

I hope that my solution is not duplicate.

Here I first invoke a lemma

$\textbf{Lemma}$ $\triangle{ABC}$ is inscribed in circle $O$. Let $P$ be the intersection of tangent lines of circle $O$ at $B,C$, i.e, the pole of $BC$ with respect to circle $O$. It is well-known that $AD$ is symmedian of $\triangle ABC$. $AP$ meets with circle $O$ at $D(\ne A)$. Line which is perpendicular to $AP$ and pass through $D$ meets with circle $O$ at point $Y$ and with line $BC$ at $Q$. $AY$, which is a diameter of circle $O$, meets with $BC$ at $X$. Then $B,X,C,Q$ are harmonic.

$\textbf{proof}$ Let $M$ be the midpoint of $BC$. It is sufficient to prove $PC \times PB = PX \times PM $. In turn it is sufficient to show $PX \times PM = PY \times PD$, which mean that $M,X,D,Y$ are cyclic.
 Let's denote the intersection of $AM$ with circle $O$ as $Z$.
 Since $AD$ is the symmedian line, it follows that $\angle{DMB}= \angle{ZMC} = \angle{AMB} = \angle{ACD} = \angle{AYD} $. Therfore $M,X,Y,D$ are cyclic and we're done.

Let' move to the original problem.

$AD$ meets $EF$ at $Z$. $PQ$ meets $DX$ at $T$ and $DY$ at $S$ and with the incircle at $N$ again. $DN$ meets $FE$ at $W$.

It is well observed that if $\triangle{DEF}$ and $\triangle{DYX}$ are similar, then $DA$ becomes median of $\triangle{DXY}$ so the proof becomes done. Therefore it is suffice to prove $X,E,F,Y$ are cyclic.

It is well known that $A,P,Z,D$ are harmonic. Then it is followed that $X,T,E,D$ and $Y,S,F,D$ are also harmonic since they are $Q-pencil$.

If $N'$ denotes $XF \cap YE$, then it is well known that $QN'$ pass through $S,T$.

$F,W,E,Q$ are harmonic from the lemma.
Here $S,N',T,Q$ are also harmonic.
Therefore $N'W$ passes through $D$, which means $N'=N$, in turn, $XF,EY,DW,PQ$ are all concurrent at $N$.

Since $\angle{NFD}=\angle{NED}=90^\circ$, $N$ is orthocenter of $\triangle{DXY}$.

Therefore $E,X,Y,F$ are cyclic and we are done.

Let $R$ be the antipode of $D$ wrt the inscribed circle and let the inersection of $AD$ ,$EF$ be $S$.
Claim:$AQ$ is parallel to $BC$
Proof:
$AD$ is the symmedian $\implies$ $$(P,D;F,E)=-1=R(P,D;F,E)=(Q,S;F,E)=(F,E;S,Q)=A(F,E;S,Q)=(B,C;M,Q^{'})$$
where $M$ is the intersection of $AS$,$BC$ and $Q^{'}$ is the intersection of $AQ$,$BC$.
Now it is well known $M$ is the midpoint of $BC$ so $Q^{'}$ is at infinity so $AQ$ is parallel to $BC$.
[u]--------------------------------------------[/u]
Now  $YFEX$ is cyclic and so $FE$ and $XY$ are antiparallel but $AD$ is the symmedian in $\triangle YDX$ so we are done

I found a solution using only angle chasing and power of point.

Let line $DI$ intersect the incircle at $R$ and the line parallel to $BC$ passing through $A$ at $S$. $DR$ is the diameter so $\angle DPR = \angle ASR = 90^{\circ}$ which means $APRS$ is cyclic and and $S$ is on the circumcircle $AFIE$. That means that the parallel to $BC$ through $A$, $PR$ and $EF$ all pass through point $Q$ being radical axes of three circles $APRS$, $AFIES$ and the incircle.

 Because $AQ \parallel BC$ we have $\angle AXE = \angle CDE = \angle DPE$ so $APEX$ is cyclic. Similarly $AYFP$ is cyclic. Now the circumcircles of $DPY$ and $DPX$ are tangent to $AQ$ because $\angle AYP = \angle AFP  = \angle PDY$ and similarly $\angle AXP = \angle PDX$, so $$AY^{2} = AP \cdot AD = AX^{2}$$

Dear MLs,

This makes A the midpoint of XY.

M.T.

Dear Mathlinkers,
this nice problem is the result of

1. http://www.artofproblemsolving.com/community/c6t48f6h1310999_parallel

2. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=113152

Sincerely
Jean-Louis

It's enough to prove that $(AQ)\parallel (BC)$ and then project on $(AQ)$ from $D$ .
To prove this, just remark that $APMQ$ is cyclic (where M denote the midpoint of $EF$ ) , and that $(PD)$ is the symmedian. And the rest is only Angle chasing.

Let $PQ \cap \odot(DEF) = \{ P, S \} $. We apply Pascal for $DPSFFE$. We have that $DP \cap FF = \{ A \}$ ; $  PS \cap EF = \{ Q \}$ ; $DE \cap SF = \{ X' \}$. We have that $A,Q, X'$ are collinear but $DE \cap AQ = X$.

So $X = X'$. We have that $F,S,X$ are collinear. But $\angle SFD = \angle FPD = 90$ so $XF \perp DY$. In analog way $YE  \perp XD$.

So $YFEX$ is a cyclic quadrilateral.

We have that $DA$ is symedian in $DEF$ and because $EF$ and $XY$ are antiparalel we have that $DA$ is median in $DXY$ so $A$ is midpoint of $XY$.

We first prove a claim.

[b][color=#00f]Claim:[/color][/b] In $\triangle ABC$ with altitudes $\overline{BE}$ and $\overline{CF}$, $\overline{BC}$-midpoint $M$ and orthocenter $H$, if $Q$ is the circumcenter of $AEF$—hence also the midpoint of $\overline{AH}$—then $\overline{MQ} \perp \overline{EF}$.
[color=#00f]Proof:[/color] By a $\times 2$ homothety at $H$ this is equivalent to $\overline{AO} \perp \overline{EF}$ where $O$ is the circumcenter of $ABC$. This follows since $O$ and $H$ are isogonal conjugates, and $\overline{AH} \perp \overline{BC}$. $\blacksquare$

Let $T$ be the foot of the altitude from $A$ to $\overline{DI}$ and let $D'$ be the $D$-antipode. By radical axis on $(EFPD')$, $(AEFIT)$, $(APD'T)$, it follows that $\overline{AQ} \parallel \overline{BC}$ (note: this is also a Sharygin 2018 problem).

Now, $\measuredangle AXD=\measuredangle CDE=\measuredangle DFE$ and $\measuredangle AYD=\measuredangle DEF$, so $\triangle DEF \sim \triangle AYX$. Furthermore, since $\overline{AI} \perp \overline{EF}$, it follows from the claim (and a straightforward phantom points argument) that $E$ and $F$ are in fact the feet of the altitudes from $Y$ and $X$ to the opposite sides of $\triangle AYX$. Hence $(AEFT)$ is the nine-point circle of $\triangle AXY$, so $A$ is the midpoint of $\overline{XY}$. $\blacksquare$


find the greatest value and the least  value of x+y , where x and y are real numbers with x>=-2 and y>=-3 and :

 x-2sqrt(x+2)=2sqrt(y+3)-y.


inequalities
algebra
polynomial
function
inequalities proposed

Let $ a,b,c\geq 0$. Prove that
$ \sum_{cyc} a^6 \plus{} 2\sum_{cyc} a^5 b \plus{} \sum_{cyc} a^4 b^2 \plus{} 2\sum_{cyc} a^3 b^2 c \geq 2\sum_{cyc} a^3 b^3 \plus{} 2\sum_{cyc} a^2 b^4 \plus{} 2\sum_{cyc} a^2 b^3 c$.
The equality holds if and only if
$ (a: b: c) \equal{} (1: 1: 1)$, $ (0: k: 1)$, $ (1: 0: k)$, or $ (k: 1: 0)$ where $ k$ is the real root of $ x^3 \plus{} x^2 \equal{} 1$.

Suppose that $ c$ is the minimum of $ a,b, and c$
We will perform the following substitution
$ a\equal{}c\plus{}x,b\equal{}c\plus{}y$and see the inequality as a polynomial function on c, mainly f(c)
We arrive at a fourth degree inequality on c.
Then, we differientiate by c, and prove that it is an positive function.
the deritave is $ 64c^3(x^2\minus{}xy\plus{}y^2)\plus{}24c^2(4x^3\plus{}x^2y\minus{}5xy^2\plus{}4y^3)\plus{}48c(x^4\plus{}x^3y\plus{}x^2y^2\minus{}xy^3\plus{}y^4)\plus{}4(2x^5\plus{}3x^4y\minus{}4x^2y^3\minus{}xy^4\plus{}2y^5)$
Then we can claim that $ f(a,b,c)>f(a\minus{}c,b\minus{}c,0)\minus{}(1)$
and let us return to the original inequality and put $ c\equal{}0$(by (1))
and $ b\equal{}1$(by homogenousity) and $ a\equal{}x$
then the inequality becomes$ (x^3\plus{}x^2\minus{}1)^2>0,$ which is trivial.


trigonometry
geometry
trig identities
Law of Sines
geometry unsolved

Let $\ ABC$ be a triangle where $\ M$ is the midpoint of $\ (BC)$, $\ m(<MAC)=30$ and $\ m(<MCA)=105$.
Compute $\ m(<ABC)$.
                                           $\ Nicolae\ Talau,\ Craiova$

$m(<ABC)=30$  my solution uses that 105=60+45 in c angle and i consider the projection of c on am and denote this with $l$ thus  m being  the midpoin we can easily find AB and apply in AMB the law of sines and we find the solution above

[hide="Hint"]

Let $P$ be a point outside $ABC$ such that $\measuredangle PAC=30^\circ$. Look at triangle $PCM$.[/hide]


IMO Shortlist
circles
geometry
Intersection

Let $D_1$, $D_2$, ..., $D_n$ be closed discs in the plane. (A closed disc is the region limited by a circle, taken jointly with this circle.) Suppose that every point in the plane is contained in at most $2003$ discs $D_i$. Prove that there exists a disc $D_k$ which intersects at most $7\cdot 2003 - 1 = 14020$ other discs $D_i$.

Here is a problem closely related to this, namely Problem 4 of the German pre-TST 2004, written in December 2003:

In the plane, consider $n$ circular disks $K_1$, $K_2$, ..., $K_n$ with equal radius $r$. Assume that each point of the plane is contained in not more than $2003$ of these circular disks. Show that each circular disk $K_i$ intersects not more than $14020$ other circular disks.

Does my memory cheat me or did Orl delete a discussion about this problem ?!

  dg

There was definitely a thread on this problem, Darij, I've no idea where it's gone either... :?

By the way, what's with the trend of posting up the IMO 2003 logo in all the threads?

[quote="darij grinberg"][quote="Valiowk"]There was definitely a thread on this problem, Darij, I've no idea where it's gone either... :?[/quote]

Orl says that there was only spam... I remember that there was a discussion about generalizations of the problem...

[quote="Valiowk"]By the way, what's with the trend of posting up the IMO 2003 logo in all the threads?[/quote]

That's definitely Orl. He uses the logo to characterize ISL problems.

  Darij[/quote]

Actually I do not delete mathematical content of any kind. And I cannot remember any mathematical content in this thread. Just "kueh" posted kind of 1 line hint like "the problem is easy..."

The problem with lots of spam messages is that users usually don't tend to publish a solution anymore.  But I think right now we have just spam in this thread again and no solution will be posted. Old state re-established...

And I want to say that I just like the IMO logo.  :)

Well, here is a solution of the problem:

First, as an auxiliary result, we will solve the following problem, which is the Problem 4 of the German pre-TST 2004, written in December 2003 (the official solution is online at http://www.bundeswettbewerb-mathematik.de/imo/aufgaben/aufgaben.htm ):

[color=blue][b]Auxiliary problem.[/b] In the plane, consider n circular disks $ K_1 $, $ K_2 $, ..., $ K_n $ with equal radius r. Assume that each point of the plane is contained in not more than 2003 of these circular disks. Show that each circular disk $ K_i $ intersects not more than 14020 other circular disks.[/color]

Here's my original solution of this problem, i. e. the solution I gave during the TST (well... not exactly the solution I gave; I have corrected a little bagatelle mistake which even the proof-reader didn't notice :) ).

In the following, [i]disk[/i] will always mean [i]closed circular disk[/i] ("closed" means that the points on the periphery of the circle belong the disk, too). If XY is a segment, then the [i]disk with diameter XY[/i] will mean the disk corresponding to the circle with diameter XY.

To [i]cover[/i] a figure will mean covering it without gaps (but maybe with overlaps).

First we prove:

[b]Lemma 1.[/b] For every positive number r, a disk of radius 2r can be covered by 7 disks of radius r.

[i]Proof.[/i] (See Fig. 1 below.) Let O be the center of a disk K with radius 2r, and let ABCDEF be a regular hexagon whose vertices lie on the periphery k of the disk K. Then, AB = BC = CD = DE = EF = FA = 2r (since it is well-known that the sidelength of a regular hexagon equals its circumradius). The disk with center O and radius r and the disks with diameters AB, BC, CD, DE, EF and FA will be called the [i]little disks[/i]. The radii of these little disks are all r (in fact, this is clear for the disk with center O and radius r, and concerning the disks with diameters AB, BC, CD, DE, EF and FA, it follows from $ \frac{AB}{2}=\frac{BC}{2}=\frac{CD}{2}=\frac{DE}{2}=\frac{EF}{2}=\frac{FA}{2}=\frac{2r}{2}=r $). We will now show that these 7 little disks cover the disk K. This will clearly prove Lemma 1.

Let P be a point inside the disk K; then we must show that this point P lies in one of our 7 little disks. Because of the rotational symmetry of the regular hexagon ABCDEF and because of the symmetrical construction of our disks, we can WLOG assume that the point P lies inside the circular sector AOB (including the arc AB of the circle k and the segments AO and BO). The triangle AOB is equilateral with sidelength AB = AO = BO = 2r; if A' and B' are the midpoints of the segments AO and BO, then $ OA^{\prime }=OB^{\prime }=\frac{2r}{2}=r $, so that these points A' and B' lie on the circle with center O and radius r. If the point P lies inside (or on the boundary) of triangle A'OB', then it follows that P lies inside the disk with center O and radius r, and the proof is done.

Hence, it remains only to consider the case when the point P lies outside the triangle A'OB'.

Since the triangle AOB is equilateral, symmetry yields $ BA^{\prime }\perp AO $, so that < AA'B = 90?. Similarly, < AB'B = 90?. Thus, the points A' and B' lie on the circle with diameter AB. Therefore, if the point P lies inside (or on the boundary) of the quadrilateral AA'B'B, then the point P lies inside the disk with diameter AB, and the proof is done again.

So it remains to consider the case when the point P lies outside the triangle A'OB' and outside the quadrilateral AA'B'B. Then, the points P and O lie in different halfplanes with respect to the line AB. Then, since the point P lies inside the disk K, the angle < APB is greater or equal to the obtuse chordal angle of the chord AB in the circle k. But the obtuse chordal angle of the chord AB in the circle k equals 180? - the acute chordal angle of the chord AB; the acute chordal angle equals $ \frac12 $ times the central angle of the chord AB, and the central angle equals < AOB = 60? (since the triangle AOB is equilateral). Thus, the obtuse chordal angle of the chord AB is $ 180^{\circ }-\frac{1}{2}\cdot 60^{\circ }=180^{\circ }-30^{\circ }=150^{\circ } $. And thus, $ \measuredangle APB\geq 150^{\circ } $. Hence, also $ \measuredangle APB\geq 90^{\circ } $, and thus the point P lies inside the disk with diameter AB. So we are done again.

Thus, every point P inside the disk K lies inside one of the 7 little disks. In other words, the 7 little disks cover the disk K. This proves Lemma 1.

Now to the problem. The disk $ K_i $ intersects a disk $ K_j $ if and only if $ O_iO_j \leq r+r=2r $, where for every k, the point $ O_k $ is the center of the disk $ K_k $. Hence, the disk $ K_i $ intersects the disk $ K_j $ if and only if the point $ O_j $ lies inside the disk with center $ O_i $ and radius 2r. In this case, after Lemma 1, the point $ O_j $ must lie inside one of 7 certain disks with radius r which cover the disk with center $ O_i $ and radius 2r. But each of these 7 disks can contain at most 2003 different points $ O_k $, since else the center of this disk would have a distance $ \leq r $ to more than 2003 points $ O_k $ and thus, itself, would lie inside more than 2003 disks $ K_k $, what contradicts the problem hypothesis. Hence, the disk with center $ O_i $ and radius 2r contains at most $ 7 \cdot 2003 $ points $ O_k $, i. e. the disk $ K_i $ intersects at most $ 7 \cdot 2003 $ disks $ K_k $. Since one of these disks $ K_k $ is $ K_i $ itself (every disk intersects itself), it follows that the disk $ K_i $ intersects at most $ 7 \cdot 2003 - 1 = 14021 - 1 = 14020 $ [i]other[/i] disks $ K_k $. And this for every i. This completes the solution of the auxiliary problem.

Now, let's solve the actual problem:

[color=blue][b]ISL 2003 problem C2.[/b] Let $ D_1 $, $ D_2 $, ..., $ D_n $ be closed discs in the plane. (A closed disc is the region limited by a circle, taken jointly with this circle.) Suppose that every point in the plane is contained in at most 2003 discs $ D_i $. Prove that there exists a disc $ D_k $ which intersects at most $ 7\cdot 2003 - 1 = 14020 $ other discs $ D_i $.[/color]

[i]Solution.[/i] This ISL problem can be reduced to the auxiliary problem as follows:

Let $ O_1 $, $ O_2 $, ..., $ O_n $ be the centers and $ r_1 $, $ r_2 $, ..., $ r_n $ the radii of the discs $ D_1 $, $ D_2 $, ..., $ D_n $. Let $ r_u $ be the smallest radius among the radii $ r_1 $, $ r_2 $, ..., $ r_n $.

Now, for every i, define a new disc $ K_i $ as follows:

- Let $ K_u=D_u $.
- If $ i\neq u $, then the disc $ K_i $ is defined as follows: Let the segment $ O_uO_i $ meet the periphery of the disc $ D_i $ at a point $ S_i $. Since the radius $ r_u $ is minimal, we have $ S_iO_i = r_i \geq r_u $, and hence we can find a point $ P_i $ on the segment $ S_iO_i $ such that $ S_iP_i=r_u $. Then, call $ K_i $ the disc with center $ P_i $ and radius $ r_u $.

For every i, we have:

[b](1)[/b] The disc $ K_i $ will completely lie inside the disc $ D_i $.
[b](2)[/b] If the disc $ D_i $ intersects the disc $ D_u $, then the disc $ K_i $ intersects the disk $ D_u $, too.

[i]Proof of (1):[/i] This is trivial for i = u, and for $ i\neq u $, this is clear since the disk $ K_i $ touches the disc $ D_i $ at $ S_i $.

[i]Proof of (2):[/i] This is clear for i = u, and for $ i\neq u $, this follows from the simple observation that if the disc $ D_i $ intersects the disc $ D_u $, the point $ S_i $ lies inside the disc $ D_u $, and hence this point $ S_i $ is common to the discs $ D_u $ and $ K_i $.

All discs $ K_1 $, $ K_2 $, ..., $ K_n $ have the same radius $ r_u $. Also, from the problem hypothesis we know that every point in the plaine is contained in at most 2003 discs $ D_i $; hence, after (1), we can conclude that every point in the plane is contained in at most 2003 discs $ K_i $. Thus, we can apply the problem I posted before, and see that every of the discs $ K_i $ intersects at most 14020 other discs $ K_j $. In particular, this is true for the disc $ K_u=D_u $. Thus, the disc $ D_u $ intersects at most 14020 other discs $ K_j $. Now, using (2), this implies that the disc $ D_u $ intersects at most 14020 other discs $ D_j $. And the ISL 2003 problem is solved.

  Darij

A simpler solution exists to this problem (Darij's solution is incredibly hard to come up with in my opinion).

WLOG suppose $D_1$ is a disk with minimal radius, and WLOG suppose it has radius $1$. Let $O_i$ be the center of disk $i$. I claim that $k=1$ satisfies the desired conditions. Suppose this is not true, so $D_1$ meets at least $7\cdot 2003$ disks $D_i$.

First, suppose that $\ge 2003$ disks $D_j$ have their centers $O_j$ inside $D_1$ for $j>1$. Then since the radius of each disk is $\ge 1$, each disk $D_j$ contains $O_1$, a contradiction. Therefore $<2003$ disks $D_j$ have centers inside $D_1$. Now ignore all such disks and consider only the disks with centers outside of $D_1$ which meet $D_1$ at some point, so there are at least $6\cdot 2003+1$ of these disks $D_i$.

Draw six rays emanating from $O_1$, each ray $60^{\circ}$ apart from the previous ray, which meet $D_1$ at six points $A,B,C,D,E,F$, such that none of the $O_i$ lie on any of the six rays drawn. These rays divide the plane into six regions, and each $O_i$ for $i>1$ lies in one of the six regions, so some region, say the region contained in $\angle AO_1B$, has at least $2004$  centers $O_i$ within it. WLOG suppose these centers are $O_2,O_3,... O_{2005}$.

[b]Lemma:[/b] $D_2,D_3, .. D_{2005}$ each contain point $X$, where $XAOB$ is a parallelogram.

[b]Proof:[/b] We have two cases for each $2\le i\le 2005$.

[i]Case 1:[/i]$1\le O_1O_i< 2$. Then it's not hard to see $O_iX \le 1$, so it's clear that $X$ lies inside $D_i$.

[i]Case 2:[/i]$O_1O_i\ge 2$. Let $O_1O_i=r+1$, so that the radius of $D_i$ is at least $r$. Clearly $\angle O_iO_1X$ is at most thirty degrees, so by Law of Cosines, $O_iX^2 \le (r+1)^2 + 3 - 3(r+1) = r^2-r+1 \le r^2$, hence $X$ lies inside disk $D_i$.

In either case, $D_i$ contains point $X$, so $2004$ disks contain point $X$, a contradiction, so we are done.

I have another solution to this problem If you want it email me I'll send it for you I can't write it in here
ida_4444@yahoo.com

Difficult problem. Solved with Ankan Bhattacharya and Luke Robitaille.

Let $m=2003$. Without loss of generality, let the minimal disk radius be $1$, and pick a disk with this radius. Suppose it has center $O$. We show that at most $7m-1$ disks intersect $(O)$. Clearly, at most $m-1$ additional disks contain the point $O$, so it suffices to show that at most $6m$ disks intersect $(O)$ but don't contain $O$.

Let $\omega$ be a circle (not disk) centered at $O$ with radius $\sqrt{3}$.

[b]Lemma:[/b]  Any disk $D$ intersecting the disk $(O)$ but not containing the point $O$ intersects $\omega$ in an arc with angle at least $\pi/3$.

[i]Proof:[/i]  By minimality, the radius of $D$ is at least $1$. By taking a subdisk of $D$ with radius $1$, we may WLOG assume that $D$ has radius exactly $1$. Note that the center $A$ of $D$ is at least $1$ away from $O$, and at most $2$ away from $O$.

Work in Cartesian coordinates with $O=(0,0)$, $A=(a,0)$, and let one of the two intersection of $\omega$ with $D$ be $X=(x,y)$. Then, $x^2+y^2=3$ and $(x-a)^2+y^2=1$. This implies \[x=\frac{a^2+2}{2a},\] and since $1\le a\le 2$, we have that $x\le 3/2$, which easily implies the result. $\blacksquare$

We now finish by noting that if $k$ disks intersect $(O)$ but don't contain $O$, then the total arc angle covered on $\omega$ is at least $k\pi/3$, so the average amount of disks a random point on $\omega$ is in is at least $k/6$, and this has to be at most $m$, so we're done.

[b]Remark:[/b]  There are several ideas to solve this problem. [list]     [*]Noting that the minimal disk probably satisfies the problem condition. This comes from examples where one can force every disk to intersect $7m$ others, except for the smallest one. Furthermore, the number $6$ probably represents the number of unit disks we can surround around another unit disk without interior overlap, and minimal disk works nicely with this.     [*]At this point, we want a global area or arc-length argument, and most natural area arguments give too weak of a bound. Furthermore, blocking these approaches are disks that are very close to the minimal disk. We get rid of these by doing the optimization of focusing only on disks that don't contain $O$, and we've turned $7$ into $6$ now!     [*]The "equality case" of six disks surrounding $D$ motivates looking at arc length (since none of the area estimates will be tight here), and magically radius $\sqrt{3}$ works (for a while I was doing horrible miscomputations and thought radius $2$ was the right way to go). [/list]

A good problem to showcase the powerful-ness of specific selection. \\

[hide = A rant about this problem.]
Back in Aug 2016 (around 4 years before today, in my final year of participating in Olympiads), my teacher sent me a file titled $\text{SLoneandtwo}$ in an email with the subject $\text{Obvious Primary-School Combinatorics}$ ($\textit{soal kombi gampang SD}$, in Indonesian.) \\

Obviously, (as the file title suggests) SL C1 and C2 problems are a big challenge (especially for a student in early grade 9 who probably still doesn't know what double counting means) --- and out of those 9 problems, I was most scared of this problem. After all, I expected the presence of $\textit{discs}$ and a $7 \times 2003$ to be major roadblocks. \\

Conclusively, as I'm able to solve this after initially being scared of the problem statement and condition, it made me appreciate the journey of getting better day-by-day, until getting to where I'm at now.
[/hide]
--------
$\color{green} \rule{9.3cm}{2pt}$ \\
$\color{green} \clubsuit$ $\boxed{\textbf{Classifying and Bounding Intersecting Circles.}}$ $\color{green} \clubsuit$ \\
$\color{green} \rule{9.3cm}{2pt}$ \\
Consider a disc $D$ with radius $r$ and center $O$. Pick a direction $d$, and plot six points $P_1,P_2,\ldots,P_6$ so that
\[ |OP_i| = \sqrt{3}r \]
and $\overrightarrow{OP_i}$'s direction towards $O$ is $d + 60^{\circ} \cdot (i-1)$. \\

Then, we Claim that each disc of radius $r$ which intersects $D$ must contain one of the seven points
\[ O, P_1, P_2,\ldots,P_6 \]
$\color{green} \rule{25cm}{0.2pt}$ \\
$\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ \\
Consider the center $O'$ of the disc which intersects $D$ (name it $D'$). Since the whole plane is divided into six regions, we can assume Without Loss of Generality that it lies inside the sector $OP_1P_2$. Furthermore, since that disc intersects $D$, we know that
\[ |OO'| = c \leq r + r = 2r \]
First, if $c \leq r$, we know that $O'$ lies inside $D$; and that implies $O$ also lies inside $D'$. \\

Otherwise, let $r < c \leq 2r$. As we know that
\[ \angle O'OP_1 + \angle O'OP_2 = 60^{\circ} \]
we can again WLOG that $\theta = \angle O'OP_1 \leq 30^{\circ}$. This will imply
\[ \cos{(\theta)} \geq \cos{30} = \dfrac{\sqrt{3}}{2} \]
Now, we would like to prove that $|P_1O'| \leq r$, as that would imply $P_1$ to lie inside $D'$. By the cosine law on $\triangle O'OP_1$, we get
\begin{align*}
|O'P_1| &= \sqrt{c^2+3r^2-2 \cos{\theta} \cdot \sqrt{3}r \cdot c} \\
&\leq \sqrt{c^2+3r^2-3rc} \\
&= \sqrt{ \left( c - \dfrac{3}{2}r \right)^2 + \dfrac{3}{4}r^2} \\
&\leq \sqrt{\dfrac{1}{4}r^2 + \dfrac{3}{4}r^2} \\
&= r
\end{align*}
We are done for the first part. $\blacksquare$ \\
--------
$\color{blue} \rule{6.5cm}{2pt}$ \\
$\color{blue} \clubsuit$ $\color{blue} \boxed{\textbf{The Ultra-Convenient Shrink.}}$ $\color{blue} \clubsuit$ \\
$\color{blue} \rule{6.5cm}{2pt}$ \\
Pick the disc with the smallest radius, and let it be $D$. $\blacksquare$ $\blacksquare$ \\

Perform the procedure above to yield the seven points. We Claim that $\textbf{any}$ disc in the plane which intersects $D$ must intersect one of the seven points (not only the ones with radius equal to $D$, as depicted above.) \\
$\color{blue} \rule{6.5cm}{0.2pt}$ \\
If this is proven, since $O$ lies inside at most $2003-1$ other discs (aside from $D$) and $P_i$ lies inside at most $2003$ discs, the number of discs which intersect $D$ must not be more than $(2003-1)+6 \cdot 2003$. This is because any intersecting disc must be one of the discs specified in the previous sentence. \\

So, $D$ satisfies the problem statement. \\
$\color{blue} \rule{25cm}{0.2pt}$ \\
$\color{blue} \spadesuit$ $\color{blue} \boxed{\textbf{Proof of the Claim.}}$ $\color{blue} \spadesuit$ \\
Let the intersecting disc be $I_D$. (I'm just slamming random variable names in there) \\

Since $I_D$ intersects $D$, there must exist a point $T$ (it stands for $\textit{tangent}$) belonging to $I_D$ which is inside $D$ (i.e. just pick the intersection point, or the point $OO_D \cap I_D$ where $O_D$ is the center of $I_D$.) \\

Next, construct another disc $T_D$ with radius equal to $D$ which is internally tangent to $I_D$ at the point $T$. It is apparent that $T_D$ lies inside $I_D$, since the radius of $D$ is smaller than $I_D$'s. \\

Finally, note that by $\color{green} \clubsuit$ $\boxed{\textbf{Classifying and Bounding Intersecting Circles}}$ $\color{green} \clubsuit$, we know that $T_D$ contains one of the seven points. This implies that $I_D$ contains that point too, by the previous paragraph. \\

We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

--------
[hide = Motivation: Selectively picking the circle and seven points.]
$\color{green} \rule{25cm}{2pt}$ \\
$\color{red} \text{Sec 1: The Oddball.}$ Similar to $\#11$, I decided to purposefully ignore the second statement of the problem, which I initially deemed to be un-process-able. Instead, I focused on individual discs more than individual points' inclusion on disks. \\

Not long after that, I noted that the bigger the disc, the more vulnerable it is to smaller discs encroaching around it, while never being able to produce a point with noteworthy intersections. Since no other parameter is in play, I just opted for the obvious minimal-radius disc. \\

While there are some other choices to pick from (such as chaining discs until getting the most favorable one; or picking one at the convex hull), but these will guarantee a small disc after quite a while --- why not just pick the oddball and try pulling out something from there? \\

$\color{green} \rule{25cm}{2pt}$ \\
$\color{red} \text{Sec 2: Minimal Damage that needs Undoing.}$ After sniping on the smallest disc, I wanted to see if the remaining still can cause some unseen damage to $D$'s reputation. Here, I was forced to pull out a conclusion about the number $7$, when the normal way out is to consider six regions surrounding $D$. \\

This clearly reminds me to distinguish the $\textit{surrounding regions}$ into six; and here I (embarrassingly) brainstormed ways to factor in a $7$ into the equation (like forming trees out of the regions, to artifically spike up six into seven), but I couldn't find any (of course ...) \\

After putting in one point inside each region (preferably along the region's rays or exactly in the middle ray between the region's rays), I noted that $\textbf{almost all}$ discs intersects the outer points, but there's no indication as to how the inner ones will ever reach a faraway point across the rays. Here I added the emergency point $O$ as an addition to the six, and it's set. \\

$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{blue} \text{Sec 2.5: Precisely Specifying the Specs.}$ A final error I had to undo was the choice of $P_i$; I initially set it to be
\[ |OP_i| = 2r \]
The fatal flaw in this was that there is a disc with center
\[ O' = \ \text{the circumcenter of} \ P_1P_2O \]
with radius $r$ that does not contain any of the seven. This also generates an infinite family of discs (with centers around $O'$), which of course contains a singular point of intersection, but not wide enough to reach out to the designated $\textbf{pegs}$. \\

The only way I thought that could fix this if $P_1P_2O$ is small enough so that its circumradius is equal to $r$; and I could achieve that by setting $|OP_i| = \sqrt{3}r$. The beautiful thing about this is its second intersection lies exactly $2r$ away from $O$, and that would imply that any circle with radius $r$ will mostly lie on $O_r(P_1) \cup O_r(P_2)$. \\

[hide = Notes.]
To avoid fakesolving with presumptuous claims involving convex sets, I opted to categorize the distances $|OO'|$, $|O'P_i$ and $|OP_i|$ by using central angle and
\[ 0 \leq c = |OO'| \leq 2r \]
as reference. However, the green Claim is just equivalent to Darij's first Claim --- and my first initial thought was also that the seven circles' union encompasses all possible $O'$ you could place. \\
(We can rephrase the casework as attempting to include $O'$ into the already fixed seven circles. A natural way would be to consider the locus $O'$ --- which is a disk of radius $2r$ --- and make sure it lies wholly on the seven unions. I attempted this by algorithmically seeking the closest $P_i$ to $O$ by specifying the smallest central angle.) \\

This happens nicely if $I_D$ has a radius of exactly $r$ --- and we haven't considered what happens otherwise.
[/hide]

Finally, the (possibly most unmotivated) Claim about shrinking big circles came into play. If a big circle has a center in the region $OP_1P_2$, it's reasonable to expect it to hang over its mini-version, which hangs over one of the seven points. \\

So, as long as we can scale it so that the circle with radius $r$ lies strictly inside it, we're set. Since I knew that tangents are a good way (geometrically) to ensure complete inclusion, all damage is undone and the proof is complete.
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
[/hide]

ISL Marabot solve

Let's define $O_i =$ center of $D_i$ and $R_i= $ radius of $D_i$ for all $1\leq i \leq n$.
WLOG, let, $R_1 \leq R_2 \leq \dots \leq R_n$ and $R_1=1$
FTSOC, let's assume that $D_1$ intersects with at least $7\cdot 2003$ other disks.

First, there can be at most $2002$ disks other than $D_1$ which contains $O_1$.
So, there are at least $6\cdot 2003 +1$ disks that intersects with $D_1$ but doesn’t contain $O_1$.
Now, we consider $\omega$ the circle with radius $\sqrt 3$ and center $O_1$. And, consider the set $S$ of six evenly separated point on the circumference of $\omega$.

$\textbf{Claim:}$ If $D_i$ intersects with $D_1$ but doesn’t contain $O_1$ then, $D_i$ contains at least one point from $S$.
$\textbf{Proof:}$ Here we consider a disk $E_i$ that is on the same direction and have same distance from $O_1$ as $D_i$ but have a radius of $1$. Let's call its center $M_i$. Clearly, $E_i$ also intersects $D_1$, doesn't contain $O_1$ and $E_i \subseteq D_i$ [As, $1\leq R_i$]
Let, $P_i$ be the intersection of ray $\overrightarrow{O_1M_i}$ and boundary of $D_1$ and $Q_i$ is the reflection of $O_1$ over $P_i$. So, $M_i$ must lie on segment $P_iQ_i$. [As radius of $E_i = 1$]
Let, $X_i,Y_i$ be the intersections of circles with center $P_i, Q_i$ and radius $1$.
[url=https://ibb.co/N932d2y][img]https://i.ibb.co/Jvdjwjs/Cam-Scanner-08-10-2022-13-20-1.jpg[/img][/url]
It's easy to show that $O_1X_i=O_1Y_i=\sqrt 3$ so, $X_i,Y_i\in \omega$. And, $\angle X_iO_1Y_i = 60^\circ$ so the minor arc $X_iY_i$ is $\frac16$ of the circumference of $\omega$. So minor arc $X_iY_i$ must contains at least  one point from $S$.
But, $X_iP_iQ_i$ is equilateral triangle. So, $M_iX_i \leq 1\implies X_i\in E_i$ as the same $Y_i\in E_i$.
So, $E_i$ contains minor arc $X_iY_i$ of $\omega$. Hence, $E_i$ contains at least one point from $S$. So $D_i$ contains at least one point from $S$. [As, $E_i\subseteq D_i$]
And our claim is proved.

Now, by our claim and using PHP, at least one point of $S$ must be contained in $\left\lceil\frac{6\cdot 2003+1}{6}\right\rceil = 2004$ disks. And this gives the contradiction we wanted.   $\blacksquare$

We claim that the smallest disk satisfies the conditions described in the problem. Let the smallest disk be $ D$ and the center of $ D$ be $ O$. Let its radius be $1$. Now, let $A_1$, $A_2$, $\dots$, and $A_6$ be six distinct points a distance of $\sqrt3$ from $ O$, and $\angle A_iOA_{i+1}=60^\circ$.

Clearly, there can only be $2003-1$ other disks containing $O$. Consider any other disk, $D_1$ with center $O_i$. If $O$ is not contained in $D_1$, then note that $1\le O_1O\le 2$. WLOG, assume that $\angle A_1OO_1\le 30^\circ$ then by cosine law, $O_1A_1\le 1$. Therefore any such disk contains one of $A_1$, $A_2$, $\dots$, $A_6$.

Thus, there are at most $6\cdot 2003$ disks not containing $O$ intersecting $O$. We are done.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let $I$ be an ideal of a noetherian ring and assume that it contains a regular element (non-zero divisor). Prove that $I$ is generated by its regular elements.

In any ring in which the ideal $(0)$ is primary (so, in particular, in Noetherian rings), the set of zero divisors is precisely the union of all the prime ideals $\frak p_{i},\ i=\overline{1,n}$ belonging to $(0)$. This means that our ideal $I$ is contained in the union $J\cup\bigcup_{i=1}^{n}\frak p_{i}$, where $J$ is the ideal generated by the regular elements of $I$. On the other hand, we know that $I$ does indeed contain regular elements, so it's not contained in $\bigcup_{i=1}^{n}\frak p_{i}$. As is well-known, an ideal contained in a finite union of ideals at most one of which is not prime is contained in one of those ideals. Coupled with the previous observations this gives $I\subseteq J\Rightarrow I=J$ (by the very definition of $J,\ J\subseteq I$).


i'm really getting stuck with this one.Can anyone help me? :(

[hide="An idea that gives insubstantial progress..."]Define $ f(t) \equal{} |t \minus{} a_1| x_1 \plus{} |t \minus{} a_2| x_2 \plus{} |t \minus{} a_3| x_3 \plus{} |t \minus{} a_4|x_4 \minus{} 1$. Our equations tell us that $ f(a_1) \equal{} f(a_2) \equal{} f(a_3) \equal{} f(a_4) \equal{} 0$.  [/hide]

[hide]
Haven't double-checked this and this seems a bit too easy, but by symmetry one may wlog $ a_1 > a_2 > a_3 > a_4$ [if $ \sigma$ is a permutation such that $ a_{\sigma(1)} > \cdots > a_{\sigma(4)}$, swap variables $ i \to \sigma(i)$]; then the difference of the first two equations is simply $ (a_1 \minus{} a_2)(x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4) \equal{} 0 \Longrightarrow x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 0$. It turns out that the other equations are equivalent in a similar manner, so it suffices to take a triple $ (x_2,x_3,x_4)$ satisfying the first equation and then take $ x_1 \equal{} \minus{}x_2 \minus{} x_3 \minus{} x_4$.
[/hide]


I find this list instructive.  A competent understanding of these biases is a great help in clearing up one's thoughts.  :)

http://en.wikipedia.org/wiki/List_of_cognitive_biases

You didn't really ask any question or start a discussion, so I'll just give some random comments:

- Those lists are kinda redundant. There are some things on there that look really similar to each other, and there are some things on there on multiple lists.

- I think outside of these fields of study, "bias" is a bit of a charged word with negative connotation. But I don't think all of these biases are actually bad. For example, under Probability and Belief, they list the "primacy effect" which is the tendency to put more weight on events that are chronologically closer (I think they mean in the future). But this isn't unreasonable, because the farther along in the future you go, the more likely it is that unforeseen factors will influence the situation. That is, our abilities to influence and to predict the future are much greater for the near future than for the distant future. So if I'm making a decision about the future, I might be well-justified in putting extra weight on considerations involving the near future and less on the distant future.

This isn't always true, though. It would be stupid for me to spend all of my money on worldly pleasures, because in January or February or something, I will have to pay my tuition for the second semester. So, I'm an idiot if I value pleasure in the near future highly and claim that the trouble I may have later is less weighty because some other events may intervene and save me. I don't even play the lottery.

I suppose the word "bias" is somewhat negative.  Of course, most of these are really just simplifying heuristics evolved to speed up the decision-making process, and actually work fairly well most of the time.

The thing, though, is that in modern society, many thought "shortcuts," as it were, produce inaccurate results, and in such cases it's useful to understand what the natural tendencies of the brain are so that subjects can be treated more carefully :)

There are a couple of reasons why the list repeats itself:
- Psychologists have given different names to two related, but distinct, phenomena (ex.: confirmation and congruence biases), and
- The list itself is not well-managed, being on a Wiki and everything.

Francis Bacon's Four Idols: Idols of the Tribe, Cave, Marketplace, Theater.

Had to read that as enrichment reading in Government.


4.  The diagram below represents a simplified version of the buffering action of hemoglobin as a buffer and the uptake and release of oxygen (Reference: I.H. Segel, “Biochemical Calculations”, 2nd ed., Wiley, 1976, p. 88).  Hemoglobin is the oxygen carrier in blood that transports oxygen from our lungs to tissues.  Diffusion due to partial pressure differences is one mechanism by which oxygen transport occurs.

the picture is on

During the 3/23/07 lecture, we discussed the relevant equilibrium reactions involving the four different forms hemoglobin.
a.  Calculate the ratio of the acid and conjugate base forms of oxyhemoglobin and deoxyhemoglobin at pH 7.4..
b.  Of the four forms of hemoglobin, what form of hemoglobin is present in the highest amount in the blood in our lungs?  Give reasons.
c.  Once hemoglobin is oxygenated, which equilibrium reaction is affected?  In which direction does this reaction shift?  Give reasons.
d.  The reaction in part c increases the [H+] and forces another equilibrium reaction to shift which releases CO2 to the atmosphere.  Which equilibrium reaction is affected and in which direction does this reaction shift?  Give reasons.  What happens to the blood pH when this equilibrium reaction shifts?
e.  Of the four forms of hemoglobin, what form of hemoglobin is present in the highest amount in the blood in our tissues?  When O2 is released, which equilibrium reaction is affected?  In which direction does this reaction shift?


我去年開始創作一個???<[i]數學國度[/i]>([i]<Math Country>[/i])的漫畫系列，但是我畫畫的水平實在太差，?在準備改爲?説，我已經寫了幾個故事，是關於解答數學題的。
你?這個系列有什麽建議？

崇拜? :D
请问具体什么时候出版？价格如何？（强烈建议早点在大陆出版，好让我去买 :) ）

想??其程度如何?
?會是大學的數學?... :?

我?爲此書（以?我其他的作?）收?任何費用，所以沒有價格這回事 :) ，我的打算是把它放到我的網?上，?寫一部分就上傳一部分。
?幾個故事的程度是高二。
還有就是qxy?此書報有很大的期望，我?常感激，一定會寫出令人滿?的作?。但是我沒什麽時間，所以會寫的很慢，有消?我會在這裡跟貼。 ;)

也?了一個月了..
請?寫好了嗎..?

[b]
你的网站是多少呀？
跟我们说说?[/b]


number theory
relatively prime
number theory solved

can any body solve this?

for positive integers n let T(n) be defined as the sum of digits in the decimal representation of n.

1> Find a positive integer N s.t.  T(kN) is even for all integers k : 1<= k<=1992
2> show that there exists no positive integer N  s.t. T(kN) is even for all positive integers k

well, the fist part is obvious :D ...Take N=10000001 by example
the second part, I think more at the paper...

There must be another solution, since my solution is just horrible for b). Of course, suppose that n has the property that all its multiples have even sum of digits. We may assume that n is relatively prime with 10 (otherwise we multiply it with powers of 2 and 5 and of course the zeroes that appear do not affect the argument). Then there is a multiple of n, lets say A that is written only with 1. Then A must have an even number of 1. It is easy to check (but it took me one hour!) that 19A has odd sum of digits.

excellent solution for b, but I forgot to add another clause to part a

find N so that T(kN) is even for 1<=k<=1992 and T(1993N) is odd.

Do you think this can be done??[/hide]

Thanks!!  :)
Yes, I can! But, Give me a time, ok!
see ya!

Yeah! There is a short solution for a!
The main idea is to find a integer k such that [tex]1992k < 10^n < 1993k[/tex] for a integer n. See that 5.1992 is very close of 10000. So if you may a further investication must have that k=502 is the searched number.
ence N=502000502 is the solution!  :D

Sorry, I made a mistake....Becose K=502 isn't a solution! :(
But if you take k=5020003 you will have a satisfactory solution,
so N=50200030005020003 (ooww! This is a really big number!! ) :D


A contractor estimated that one of his two bricklayers would take 9 hours to build a certain wall and the other 10 hours. However, he knew from experience that when they worked together, 10 fewer bricks got laid per hour. Since he was in hurry, he put both men on the job and found it took exactly 5 hours to build the wall.
How  many bricks did it contain ?

The first bricklayer gets $ \dfrac{1}{9}$ of the job done in an hour, and the second gets $ \dfrac{1}{10}$ of the job done in an hour. Technically, the two get $ \dfrac{1}{9}\plus{}\dfrac{1}{10}\equal{}\dfrac{19}{90}$ done per hour, but they actually get 10 less bricks per hour. The pair should have gotten $ \dfrac{95}{90}$ of the job done, which is $ \dfrac{1}{18}$ extra on the job. This means that $ \dfrac{1}{18}$ of the job is $ 5\times 10\equal{}50$ bricks. There are a total of $ 900$ bricks to be layed.


here are the links to the pages
[url=http://www.mathlinks.ro/viewtopic.php?t=78567]INPho 2006 Results[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=78568]INCho 2006 Results[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=77001]INMO 2006 Result[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=78569]INBO 2006 Results[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=78566]INAO 2006 Results[/url]
Will be updated soon

here are the results
[url=http://www.hbcse.tifr.res.in/olympiads/nse07/nsePres]NSEP 2007[/url]
[url=http://www.hbcse.tifr.res.in/olympiads/nse07/nseCres]NSEC 2007[/url]
[url=http://www.hbcse.tifr.res.in/olympiads/nse07/nseBres]NSEB 2007[/url]
[url=http://www.hbcse.tifr.res.in/olympiads/nse07/nseAresjun]NSEA JUNIOR 2007[/url]
[url=http://www.hbcse.tifr.res.in/olympiads/nse07/nseAressen]NSEA SENIOR 2007[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=180122]Pre-Results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=182217]Post-Results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=188318]POST MARKS DISCUSSION[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=184038]INOI 2008 Pre-results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=187009]INOI 2008 Post-Results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=181060]INMO 2008 Pre-Paper Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=184278]INMO 2008 Post-Paper,Pre-Results Discussion[/url]
[url=http://www.mathlinks.ro/index.php?f=362]INMO 2008 Post  Results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=186189]INChO 2008 Pre-Results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=186188]INAO 2008 Pre-Results Discussion[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=186175]INPhO 2008 Pre-Results Discussion[/url]


I'm trying to generate more interest in the math team at our school, to help build a strong program that students will enjoy and be able to learn a lot, etc etc.

but to do this, we need to raise money and get funds. so..does anyone know of any good methods to do this, and how to ask businesses for money, and so on..?

eh, larger teams cost more money to send to math competitions  ;)

I would suggest you do a search of the forum -- there have been discussions along these lines in the past.  I suggest a couple of things: find out how sports teams and other extracurricular events are funded at your school.  Also, talk to the math department.  If you feel you have to go to businesses, you may have to offer them something (that is, most businesses won't just give you money).  If you were to publish something, you could sell advertising space, for instance.  Or, possibly, get t-shirts printed with the name of the sponsor on them somewhere.  I don't think either of those are really attractive options, honestly.

Pizza sales are great fundraisers.  My school was able to set up a deal with the local pizzaria.  We get pies for $7 each and free delivery in exchange for agreeing only to use the one pizzaria for all of our fundraisers. We then sell the pizza for$2 per slice.  We easily make $50-$100 in 15 minutes of selling with minimal effort.  As far as recruiting, I reccomend brightly colored posters and providing some sort of food - chips, cookies, something.  Also advertise that you are providing food.  I know that it seems superficial, but it was just what my math team needed to make it look cool enough for interested students to join without the fear of being rejected by the rest of the student body.


A mass is attached to the wall by a spring of constant $ k$. When the spring is at its natural length, the mass is given a certain initial velocity, resulting in oscillations of amplitude $ A$. If the spring is replaced by a spring of constant $ 2k$, and the mass is given the same initial velocity, what is the amplitude of the resulting oscillation?

$ W \equal{} \frac 12 mv^2 \equal{} \frac 12 kA^2 \Longrightarrow A_2 \equal{} \frac {A}{\sqrt {2}}$

Aha. Once again, I was too stupid to realize that displacement was the amplitude...


Let $ A,B \in M_n{\mathbb{(C)}}$ are Hermitian matrices such that  : $ A^2B^3\plus{}B^3A^2\equal{}2AB^3A$
Prove that : $ AB\equal{}BA$

Let $ U\equal{}[u_{ij}]\equal{}B^3$ (a hermitian matrix). We may assume that $ A\equal{}diag(\lambda_1,\cdots,\lambda_n)$ where $ \lambda_i$ is a real number. $ A^2U\plus{}UA^2\equal{}2AUA$ implies: $ \lambda_i^2u_{ij}\plus{}u_{ij}\lambda_j^2\equal{}2\lambda_i\lambda_ju_{ij}$ that is $ u_{ij}(\lambda_i\minus{}\lambda_j)\equal{}0$ that is $ AU\equal{}UA$.
$ A,B^3$ are simultaneously diagonalizable and we may assume that $ A,B^3$ are diagonal real matrices. Moreover $ x\in\mathbb{R}\rightarrow{x^3}$ is one to one. Therefore $ B,B^3$ have exactly the same eigenvectors.-The result is absolutely false if we consider $ B,B^2$ (think to an orthogonal projector)- Conclusion: $ AB\equal{}BA$.

A precision: line 4  $ A,B^3$ are simultaneously diagonalizable with a unitary matrix.
A correction: line 6-7  Think to an orthogonal symmetry.


A house and store were sold for $ \$12000$ each. The house was sold at a loss of $ 20\%$ of the cost, and the store at a gain of $ 20\%$ of the cost. The entire transaction resulted in:

$ \textbf{(A)}\ \text{no loss or gain} \qquad\textbf{(B)}\ \text{loss of } \$1000 \qquad\textbf{(C)}\ \text{gain of } \$1000 \qquad\textbf{(D)}\ \text{gain of }\$2000 \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
$ 12000\equal{}H\minus{}\frac{1}{5}H \implies H\equal{}15000$.  $ 12000\equal{}S\plus{}\frac{1}{5}S \implies S\equal{}10000 \implies H\plus{}S\equal{}25000$.

Thus, the sale resulted in a loss of $ \$1000$ or $ \boxed{\textbf{(B)}}$.
[/hide]