geometry
area
circumcircle
trapezoid
ratio
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

At first we note an important result:

[b]Lemma 1.[/b] If A and B are two points in the plane, and X and Y are two other points, and Z is the point on the line XY which divides the segment XY in the ratio XZ : ZY = p : q (with directed segments), then

$\left[ AZB\right] =\frac{q\cdot \left[ AXB\right] +p\cdot \left[ AYB\right] }{q+p}$,

where $\left[P_1P_2P_3\right]$ denotes the directed area of a triangle $P_1P_2P_3$. ("Directed" means that the area is taken positive if the triangle is orientated counterclockwise, and negative if it is orientated clockwise.)

[i]Proof.[/i] Let x, y, z be the directed distances from the points X, Y, Z to the line AB. ("Directed" means that e. g. the distance x is positive if the triangle AXB is orientated counterclockwise, and negative if it is orientated clockwise.) Then, since the area of a triangle equals $\frac12 \cdot$ sidelength $\cdot$ corresponding altitude, we have

$\left[ AXB\right] =\frac{1}{2}\cdot AB\cdot x$;
$\left[ AYB\right] =\frac{1}{2}\cdot AB\cdot y$;
$\left[ AZB\right] =\frac{1}{2}\cdot AB\cdot z$.

Thus, the areas [AXB], [AYB], [AZB] are proportional to the distances x, y, z, respectively. Hence, instead of proving $\left[ AZB\right] =\frac{q\cdot \left[ AXB\right] +p\cdot \left[ AYB\right] }{q+p}$, it is enough to show $z =\frac{q\cdot x +p\cdot y}{q+p}$. But this is clear if you consider the line AB as the x-axis of an Cartesian coordinate system (then x, y, z are the x-coordinates of the points X, Y, Z, respectively). Lemma 1 is proven.

Another important and well-known lemma:

[b]Lemma 2.[/b] If ABC is a triangle with sidelengths a = BC, b = CA, c = AB and circumradius R, then the area of triangle ABC equals $\frac{abc}{4R}$.

Now, in our cyclic quadrilateral ABCD, let a = AB, b = BC, c = CD, d = DA, e = AC, f = BD. Let R be the circumradius of the quadrilateral ABCD. And finally, call p : q = AE : EB = CF : FD.

We will apply Lemma 1 but work with [i]non-directed areas[/i]. In other words, in the following, the notation $\left[P_1P_2P_3\right]$ will mean the [i]non-directed[/i] area of a triangle $P_1P_2P_3$. Of course, one can apply Lemma 1 for non-directed areas only if the points X, Y, Z all lie on one and the same side of the line AB; but this will be always guarranteed in the following.

Applied to triangle ABD, whose sidelengths are AB = a, BD = f and DA = d, and whose circumradius is R, Lemma 2 yields

$\left[ ABD\right] =\frac{afd}{4R}$.

Since the point E lies on the line AB and we have AE : EB = p : q, Lemma 1 gives

$\left[ AED\right] =\frac{q\cdot \left[ AAD\right] +p\cdot \left[ ABD\right] }{q+p}=\frac{q\cdot 0+p\cdot \left[ ABD\right] }{q+p}=\frac{p\cdot \left[ ABD \right] }{q+p}$
$=\frac{p}{q+p}\cdot \left[ ABD\right] =\frac{p}{q+p}\cdot \frac{afd}{4R}$.

Similarly,

$\left[ AFD\right] =\frac{q}{q+p}\cdot \frac{ced}{4R}$.

Since the point P lies on the line EF and we have PE : PF = AB : CD = a : c, Lemma 1 yields

$\left[ APD\right] =\frac{c\cdot \left[ AED\right] +a\cdot \left[ AFD\right] }{c+a}=\frac{c\cdot \frac{p}{q+p}\cdot \frac{afd}{4R}+a\cdot \frac{q}{q+p} \cdot \frac{ced}{4R}}{c+a}=\frac{\left( pf+qe\right) acd}{4R\left( p+q\right) \left( c+a\right) }$.

Similarly,

$\left[ BPC\right] =\frac{\left( pf+qe\right) acb}{4R\left( p+q\right) \left( c+a\right) }$.

Thus,

$\frac{\left[ APD\right] }{\left[ BPC\right] }=\frac{d}{b}$,

what is independent from the choice of E and F indeed. Proof complete.

  Darij

If $BC$ and $DA$ are parallel, then it's trivial, because then $ABCD$ would be an isosceles trapezoid, so in fact, we would have $AE=CF$ and then it's easy to see that $d(P,BC)=d(P,DA)$.

So let's now assume that $BC$ and $DA$ intersect at a point $G$.
We have $AB: CD=AE: CF=EP: FP$. But since $ABCD$ is cyclic, the triangles $GAB$ and $GCD$ are similar with ratio $AB: CD$. But since $AE: CF$ is equal to the similtude ratio, the point $E$ in $GAB$ corresponds to the point $F$ in $GCD$. Thus, the triangles $GFC$ and $GEA$ are similar with ratio $AB: CD$ so we have $GE: GF=AB: CD$ and $\angle DEF = \angle CGE$ Thus, $GE: GF=EP: FP$ and hence, $GP$ is the angle bisector of $\angle AGB$ from which we conclude $d(P,AD)=d(P,BC)$, yielding that the areas in question have a same attitude, thus their ratio is equal to the ratio of the corresponding sides, i.e. $AD: BC$

$\Box$

[hide="Solution"]
Let $AD \cap BC = X$.

Since quadrilateral $ABCD$ is cyclic, then $\angle XBA=\angle XDC$, and we have $\dfrac{XB}{XD}=\dfrac{AB}{CD}=\dfrac{BE}{DF}$, so $\triangle XBE \sim \triangle XDF$.  Then, since $\dfrac{PE}{PF}=\dfrac{AB}{CD}=\dfrac{XB}{XD}=\dfrac{XE}{XF}$, by the converse of the Angle-Bisector Theorem, $\angle FXP=\angle EXP$.  Using this with the fact that $\angle BXE=\angle DXF$ (from the proven triangle similarity), we obtain that $\angle DXP=\angle BXP$.

Let the length of the altitude from $P$ to $AD$ and $BC$ be $d_1$ and $d_2$ respectively.  Then $d_1=XP\sin \angle DXP=XP\sin \angle BXP=d_2$.

Thus, $\dfrac{[APD]}{[BPC]}=\dfrac{(AD)d_1}{(BC)d_2}=\dfrac{AD}{BC}$, which is independent of the choice of $E$ and $F$.

QED [/hide]

[quote=1998 G2]Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.[/quote]

Tutorial for linearity :)

Animate $E$ linearly on side $AB$. Then $E \mapsto F$ is linear and since $P$ divides $EF$ in a fixed ratio, $P$ moves linearly. Thus, it suffices to show that the locus of $P$ passes through $Q=AD \cap BC$, to conclude that $\frac{[APD]}{[BPC]}$ is fixed.

Let $X$ and $Y$ be points on $AC$ and $BD$ such that $\frac{AX}{XC}=\frac{BY}{YD}=\frac{AB}{CD}$. When $E=A$ we have $P=X$ and when $E=B$ we have $P=Y$. Thus, $\overline{XY}$ is the locus of $P$. We want to show that $Q, X, Y$ are collinear. Notice that $\frac{QA}{QC}=\frac{QB}{QD}=\frac{AB}{CD}$ hence $X, Y$ lie on the internal bisector of angle $BQD$, proving the result. $\blacksquare$

[hide=Diagram][img width=50]https://media.discordapp.net/attachments/986049049834713139/996612113546104952/ISL1998G2.png?width=1052&height=1026[/img][/hide]

Let $AE=x,BE=ax,CF=y,DF=ay.$ It follows that $AB:CD=x:y$ so $PE:PF=x:y.$ Define $d(X,YZ)$ to the distance between point $X$ to line $YZ$ for any $X,Y,Z.$

We have $d(E,AD)=x\sin(\angle A),d(F,AD)=ay\sin (\angle C).$ Since $P$ is $\frac{x}{x+y}$ of the way from $E$ to $F$ we have \[d(P,AD)=\frac{yd(E,AD)+xd(F,AD)}{x+y}=\frac{xy}{x+y}\left(\sin(\angle A)+a\sin\angle C\right)\] Similarly, we have \[d(P,BC)=\frac{xy}{x+y}\left(\sin(\angle C)+a\sin\angle D\right)\] and since $ABCD$ is cyclic, we see that these two are actually equal. Thus, we have \[\frac{[APD]}{[BPC]}=\frac{AD\cdot d(P,AD)}{BC\cdot d(P,BC)}=\frac{AD}{BC}\] which is not dependant on the choice of $a$ as desired.

G2?

Let $\frac{AE}{AB}=\frac{CF}{CD}=k$. Note that
\[[APD]=[AED]\cdot \frac{CD}{AB+CD}+[AFD]\cdot \frac{AB}{AB+CD}\]
which simplifies to
\[[APD]=k\cdot [ABD]\cdot \frac{CD}{AB+CD}+(1-k)\cdot [ACD]\cdot \frac{AB}{AB+CD}.\]
Repeating similar calculations, we find that
\[\frac{[APD]}{[BPC]}=\frac{k\cdot [ABD]\cdot CD+(1-k)\cdot [ACD]\cdot AB}{(1-k)\cdot [ABC]\cdot CD+k\cdot [BDC]\cdot AB}.\]
It is then enough to show that
\[\frac{[ABD]\cdot CD}{[BDC]\cdot AB}=\frac{[ACD]\cdot AB}{[ABC]\cdot CD}.\]
But now we have $\frac{[ABD]\cdot CD}{[BDC]\cdot AB}=\frac{AB\cdot CD\cdot AD}{AB\cdot CD\cdot BC}=\frac{AD}{BC}$. The RHS is also equal to $\frac{AD}{BC}$, so we are done. $\blacksquare$

MMP >

Note that as $E$ moves along $AB$, $F$ moves linearly along $BC$. Furthemore, $P$ is a weighted mean of $E$ and $F$, so it also moves linearly as $E$ moves.\\

Let $P_1$ be the point on $AC$ such that $$\frac{AP_1}{P_1C}=\frac{AB}{CD}.$$ Similarly, let $P_2$ be the point on $BD$ such that $$\frac{BP_2}{P_2D}=\frac{AB}{CD}.$$ Finally, let $P_3$ be the intersection of $AD$ and $BC$ (possibly a point at infinity).\\

If $E=A,$ then $F=C$ and $P=P_1$. Similarly, if $E=B$, then $F=D$ and $P=P_2$. Therefore, since $P$ moves linearly, $P$ always lies on line $P_1P_2$. We want to show that the ratio of the heights from $P$ to $AD$ and $BC$ is constant. Thus, it suffices to show that $P_1,P_2,P_3$ are collinear.\\

Note that from similar triangles $\triangle P_3AB$ and $\triangle P_3CD,$ $$\frac{AB}{CD}=\frac{P_3A}{P_3C}.$$ Hence, $$\frac{AP_1}{P_1C}=\frac{AB}{CD}=\frac{P_3A}{P_3C},$$ so $P_1$ lies on the bisector of $\angle AP_3C$. Similarly, $P_2$ lies on the bisector of $\angle BP_3D$, so $P_1,P_2,P_3$ are collinear and we are done.


Is there constant k for:
$(x+y)(y+z)(z+x)<kxyz$ with $x,y,z>0$ :)

Obviously not.  Fix $x, y$ and let $z$ approach $\infty$.

oh $RHS \rightarrow \infty, LHS \rightarrow \infty$what is contradiction?

LHS goes like $z^{2}$, RHS goes like $z$, so LHS grows faster and must eventually pass...
Take $x=y=1$. It is $2(1+z)^{2}< kz$, so pick even $z=k$ and you find $2(1+k)^{2}< k^{2}$ clearly false. So there's no $k$.


inequalities
geometry
parallelogram
circumcircle
geometry proposed

Let ABCD be a parallelogram with <ABC acute. Let M,N be the midpoints of BC and CD. AM intersects BD at P and AN intersects BD at Q. Show that 9R(APQ)>5R(ABC)
Where R(XYZ) is the circumradius of the triangle XYZ.

Show also that the constants are best. :)
(It would be nice to have a solution for this problem, don't forget it )


Let $a_1,a_2,\ldots,a_n$ be positive real numbers that have product equal to 1, let
\[ A=\sqrt[m]{\frac{a_1}{a_1+k}}+\sqrt[m]{\frac{a_2}{a_2+k}}+\cdots+\sqrt[m]{\frac{a_n}{a_n+k}},\;\; m\in\mathbb N. \]
For each $k>0$, find $\min A$, $\inf A$.

The proof is not easy, can you finish it?


Woohoo! Anyone as excited as I am?

Go FF7 all the way!

Am I the only one who thinks the Final Fantasy Series went to like 10 or something? I thought it did...meh Final Fantasy is nothing compared to Dragon Quest (to those who recognize this series, good for you! You are a true gamer :))

I have the first Dragon Warrior (on NES). Revolutionary game.

Everquest series is the most truly remarkable game. (Its made by those Sony japanies guys you can't get anything better).

I must admit: Dragon Warrior I is amazing.

[quote="MysticTerminator"][quote="aalindsey100"]I know both but never played either of them just watched. My gamer friends love both so I hear about them all the time and I edit their game reviews grammer mistakes :).[/quote][/quote]
Grammer. heh.


LaTeX TeXnicCenter editors configuration

can you please help me?

I don't see those words on the bar on the top. I do see a blank bar near the top but when i click on the down arrow it gives me nothing....

but when i right click it gives a list that says navigator bar, output bar, status bar and a lot more.

am I at the right bar?

so i actually tired skipping that step..but nothing works after that.

thanks

It looks as if you haven't set up TexnicCenter completely. Go to Build, Define Output Profiles, Wizard button (bottom left corner) and follow the instructions. There's a [url=http://www.youtube.com/watch?v=-h8Oi1oCTQg]video[/url] to show you what to do.


function
invariant
complex analysis
advanced fields
advanced fields unsolved

I have two questions about meromorphic differential:

1. Meromorphic differentials on $ \mathbb{C}$ can be identified naturally with meromorphic functions on $ \mathbb{C}$.

2. the residue of a meromorphic differential at a simple pole is always nonzero.

[quote="foggy"]I have two questions about meromorphic differential:

1. Meromorphic differentials on $ \mathbb{C}$ can be identified naturally with meromorphic functions on $ \mathbb{C}$[/quote]Take the mapping $ f\mapsto (z\mapsto f(z)dz)$.
[quote]2. the residue of a meromorphic differential at a simple pole is always nonzero.[/quote]To answer this, we should make sure what definition of a residue of meromorphic differentials (on Riemannian surfaces) you would like to use.

If the residue is zero and it grows slowly enough to be a simple pole, it's a removable singularity.
Of course- I don't know the general theory you're working with; that's just coming from what I know about analytic functions on subsets of $ \mathbb{C}$

[quote="jmerry"]If the residue is zero and it grows slowly enough to be a simple pole, it's a removable singularity.[/quote]But how does the term $ dz$ come into play here? I think this depends pretty much on the definition of the (chart invariant) residue of a meromorphic differential. Note that this becomes sort of trivial on $ \mathbb C$ since here meromorphic differentials are identified with entire functions as pointed out in my first reply.

Here the residue $ res\{\omega,p\}$ of a meromorphic differential $ \omega\equal{}fdg$ at a point $ p$ is defined to be the ordinary residue

$ res\{(f \circ {\phi_\alpha}^{\minus{}1})(g \circ {\phi_\alpha}^{\minus{}1})',\phi_\alpha(p)\}$,

where $ \phi: U_\alpha \to V_\alpha$ is a holomorphic chart on $ C$ such that $ p \in U_\alpha$,

and $ C$ is a nonsingular projective curve in $ \mathbb{P}_2$.

Ok. Then the notion of residue and simple pole are invariant under local chart changes or - what turns out to be the same - under local biholomorphisms of Riemannian curves, we can assume WLOG that the meromorphic differential $ \omega$ is defined on the open unit disk $ \mathbb E$ and that $ p\equal{}0\in\mathbb E$. Representing $ \omega$ as $ f(z)dz$ with $ f$ meromorphic on $ \mathbb E$, $ f$ inherits from $ \omega$ to have a simple pole at $ 0$ and that the 'ordinary' residue of $ f$ at $ 0$ coincides with $ res\{\omega,0\}$. From here it should be easy I believe.


I have a question concerning the college application recommendations. I know the college only requires 2 teacher evaluations and 1 school evaluation. However, I am not sure if I can submit another one other than those three, to help the admission office get to me better. For example, I can certainly ask my high school principal to write one for me, but I am afraid the admission office will discard that recommendation. What do you think of it?
(Here, I mean those colleges that do not mention the optional recommendation, unlike Princeton and UChi)

Thanks.

Colleges won't discard the additional recommendation. In fact, they would love to have that extra recommendation if it actually helps to get to know you better. Is your principal actually going to say anything new that the counselor and teachers haven't already said?


modular arithmetic
algebra unsolved
algebra

Let $P$ be a non-constant polynomial with integer coefficients. Prove that $ \overline{0.P(1)P(2)\cdots P(n) \cdots}$ is not a rational number.

[quote="tdl"]Let $ P\in Z[x]$. Prove that: $ \overline{0,P(1)P(2)...P(n)...}\notin Q$[/quote]

Wrong.

$ P(x)\equal{}c$ implies $ \overline{0,P(1)P(2)...P(n)...}\in Q$

for all $ n\equal{}deg P(x)\in Z[x]$ we get
\[ \sum_{k\equal{}1}^{\infty}\frac{P(n)}{10^n} \in Q.\]

To pco: If $ P\neq$ const then is my problem true? :blush:
To Rust: I don't understand your idea, what is $ n,k$ in your post?

If I'm understanding this problem correctly, won't $ P(x) \equal{} x$ give $ 10/81$? That fraction is equal to $ 0.123456790123456790\ldots$.

[quote="tdl"]To pco: If $ P\neq$ const then is my problem true? :blush: [/quote]

Ok, with this restriction, the problem is true :

Let $ S$ the number $ 0.P(1)P(2)\cdots$
Let $ P(x)\equiv ax^p$ when $ x\rightarrow \plus{}\infty$
The length (in digits) of $ P(n)$ is then $ l(P(n))\equiv \log(a)\plus{}p\log(n)$ and the difference of lengths between $ P(n\plus{}1)$ and $ P(n)$ is :
$ l(P(n\plus{}1))\minus{}l(P(n))\equiv p\log(1\plus{}\frac{1}{n})\equiv \frac{p}{n}$

So, when $ n$ grows sufficiently, the length of $ P(n)$ increases slowly (by less than 1) and then all lengths are found.
If the required number $ \in\mathbb{Q}$, its decimal representation is periodic, with period $ a$.
So, it exists $ n_0$ such that :
$ 1)$ $ P(n_0)$ is in the periodic part of $ S$
$ 2)$ $ l(P(n_0\plus{}1))\equal{}l(P(n_0))\equal{}0\pmod{a}$
$ 3)$ $ P(n)$ is positive and strictly increasing $ \forall n\geq n_0$

Then, $ P(n_0)$ is a $ ka$-digits number in the periodic part.
And $ P(n_0\plus{}1)$ is an adjacent $ ka$-digits number in the periodic part.

So $ P(n_0)\equal{}P(n_0\plus{}1)$. But this is impossible (see point 3 above).

So the decimal representation of $ S$ is never periodic.

So $ S\notin \mathbb{Q}$


16 only...
for Q7
I was thinking if the car has five seats......
so I didn't write down my answer
then.............I know I got it><

Keep in the AMC Discussion Threads please.


wow.noasnwer after about 2 weeks.
y? :?:


AMC
AMC 12
AMC 12 B
AIME
conics
parabola
geometry

Fermat 2007 is going to be held tommorow. I was preparing for Amc so I didn't look at the Fermat papers yet. I think I'll be fine though. Good luck to everyone writing the contest this year! :)

Note: I think we may have to wait a day or so before discussing any problems.

for fermat, really? does anyone know y they made ppl write the fermat on tuesday this year isnt it usually wednesday..

We might have to. Fermat was originally scheduled for the 21 (I think), no idea why they changed it.

strange..anyway good luck to all PCF writers.

Notice that AMC12B is same day as this Wednesday. If they made AIME overlap with CMO and AMC overlap with Fermat they would not be very nice to the students.

Good luck to everyone writing Fermat/Pascal/Cayley! :)

[b]Please do not discuss PCF until Wednesday afternoon.[/b]

ya..thats what i was thinking this morning like it mustve been some overlap with like an american contest but i wasnt sure..

I am writing the Cayley tommorow, for the first time. The past three years I wrote the pascel. (Last year I was 221 out of 38000!!)

That is a very nice rank Jackdillon, you will do very well of Cayley, I'm sure. :)
Good luck to everyone writing the Pascal and Cayley (for Fermat writers, see my first post :P ).

Ahhh, Im slightly nervous. I go to school in 15 min., and write it right away...

omg

why is Fermat so difficult this year?

I wished such questions in 2006 or 2002 to come up.

I guessed almost everything in part C, and some questions in part B were hard as well.

i solved #23... it was not that hard, i just needed more time. i'm screwed anyway. good luck

I actually didn't find section C that hard, I got 21, 22, and 23 almost certainly right, but 24 and 25 were tough.

I'm not going to say any more about it until tomorrow, but I'd really like a solution for 24, I had narrowed it down to two possibilities and I took a guess.

lol dint know how to delete a post and tried to delete this post so just ignore this

Please do not discuss the contest till Wednesday.
If you are really anxious to share your scores talk to each other via PM.

nevermind, figured out my problem with the last one.

i dont think probability works for the last problem,try to use number theory. :wink:

I think for the last problem, you have to use possiblity first, and then it becomes very simple number theory problem.

the last problem is answered on the second page, I just didnt know how to turn GC2 + BC2/(B+G)C2 = 1/2 into an equation, but once I figured it out it all makes sense.

the question says the probability of the same 2 colored balls being at the end are 1/2..

P = number of ways outcome can happen / total amount of outcomes

so, we want our outcome to be 2 black balls or 2 gold balls

P = number of ways 2 black balls can be chosen + number of ways 2 gold balls can be chosen / total number of ways 2 balls can be chosen

and it is this probability which is equal to 1/2 which enables us to solve the problem.

i know that,she told me...and i got No15 wrong... :mad:

[quote="vincent_five"]cynthia(Yan) Li from Dr. norman bethune got perfect on fermat[/quote]
whoa she's owning this yr...
i hope to come back on olympiads. :yup:

[quote="rem"]i have to admit she has a pretty good shot at doing well as there are not that many ownage people this year, and nobody this year has imo experience.[/quote]

What about Lin Fei?

[quote="bryant124"][quote="jackdillon"]For number 23 on the cayley, the answer was c. I guessed a...[/quote]
why is everybody saying 23 for cayley is hard? isnt it simply pure calculation...i didnt even spend any time thinking about how to solve it...its just adding the three areas up and then multiplying by 2...[/quote]

I am saying its hard because I'm in grade 7, and havnt seen that type of problem before.

[quote="J-I-M-B-O"][quote="rem"]Well..... maybe... but you have to remember that comc and fermat are both a joke compared to the olympiads. They are also contests of completely different format: on the waterloo contests fast thinking is required but on olympiads it's the true skill that comes out.

i have to admit she has a pretty good shot at doing well as there are not that many ownage people this year, and only one person this year has imo experience.[/quote]

who's the other person with imo experience if not lin fei?[/quote]

there's only one!


inequalities function induction geometry quadratics

Prove that there exists a positive integer x,y,z such that:

2x  2 +3y 2 -5z 2 =1997[/tex][/hide][/url][/list][/list][/code][/quote][/u][/i][/b]

[quote="Myth"]Though Pre-Olympiad is not my business... :blush:
Just take $x=t-1$, $y=t+1$, $z=t$.[/quote]

And you couldn't find time to help me out with the proof of Jensen's?  :maybe:

[quote="Myth"]As I wrote before, Jensen's inequality is a direct consequence of definition of convex function, i.e. $af(x)+(1-a)f(y)\geq f(ax+(1-a)y)$, $a\in[0,1]$, and it can be easily proved by induction (just group first $n-1$ terms). Ok?[/quote]

Thanks!

I realli should be hanging around at getting started area but Myth.

Can you xplain why you subbed x y and z as those values or what gave u the idea of doing that?

[quote="nodrog2002"]I realli should be hanging around at getting started area but Myth.[/quote]
My english is very bad sometimes. What did you want to say?

[quote]Can you xplain why you subbed x y and z as those values or what gave u the idea of doing that?[/quote]
Since equation consists of quadratic terms and sum of coefficients is zero (2+3-5), very first idea, which came to my mind, was to decrease equation's degre by substituting linear functions instead of $x$, $y$ and $z$ with common leading coefficient. After that it was routine to get exact expressions like $t$, $t-1$ and $t+1$.
Hope my explanations are readable. :blush:  :P


[quote="ragnarok23"][quote="Ignite168"][hide]7 is prime so the lowest unit denominators that would add to a fractioin ending in 7 are 1 and 7. Since you can only have positive fractions and $\frac{6}{7}$ is below 1, you can only have denominators ending in 7 or more. But you can easily see $\frac{1}{7}+\frac{2}{7}+\frac{3}{7}=\frac{6}{7}$ meaning the answer is $\boxed{21}$.[/hide][/quote]
Aren't unit fractions fractions that have 1 as the denominator?[/quote]
How can $\frac{2}{7}$ be expressed as $\frac{1}{\text{something}}?$

[quote="bfrock2000"]the sum of three different positive unit fractions is 6/7. what is the least number that could be the sum of the denominators of these fractions?[/quote]
[hide]$\frac{1}{7}+\frac{2}{7}+\frac{3}{7}=\frac{6}{7}$

The answer is 21.[/hide]

[quote="ragnarok23"][quote="Ignite168"][hide]7 is prime so the lowest unit denominators that would add to a fractioin ending in 7 are 1 and 7. Since you can only have positive fractions and $\frac{6}{7}$ is below 1, you can only have denominators ending in 7 or more. But you can easily see $\frac{1}{7}+\frac{2}{7}+\frac{3}{7}=\frac{6}{7}$ meaning the answer is $\boxed{21}$.[/hide][/quote]
Aren't unit fractions fractions that have 1 as the EDIT: numerator?[/quote]

Then his problem would be trivial..

[quote="i_like_pie"][quote="ragnarok23"][quote="Ignite168"][hide]7 is prime so the lowest unit denominators that would add to a fractioin ending in 7 are 1 and 7. Since you can only have positive fractions and $\frac{6}{7}$ is below 1, you can only have denominators ending in 7 or more. But you can easily see $\frac{1}{7}+\frac{2}{7}+\frac{3}{7}=\frac{6}{7}$ meaning the answer is $\boxed{21}$.[/hide][/quote]
Aren't unit fractions fractions that have 1 as the denominator?[/quote]
How can $\frac{2}{7}$ be expressed as $\frac{1}{\text{something}}?$[/quote]

$\frac{1}{\frac{7}{2}}$ and he said denominator, not numerator.

is there a formula for adding unit fractions?

[quote="i_like_pie"][quote="robinhe"]For unit fractions, the numerator must be $1$ and the denominator must be a positive integer ($>1$?)

[hide]Take the biggest possible first fraction.  That is $1/2$.  What remains is $5/14$.  Again, take the biggest possible fraction that is possible, which is $1/3$.  $1/42$ remains so the answer is $\frac{1}{2}+\frac{1}{3}+\frac{1}{42}$[/hide][/quote]
^^ What he said. ^^

[hide="But the answer is..."]$47$ because we are finding the sum of the denominators.[/hide][/quote]

Whoops sorry :oops: I didn't really read the question.

[quote="bfrock2000"][quote="ProtestanT"][quote="Phelpedo"]A slightly harder problem: give a general method to find the smallest sum of denominators of two unit fractions that add up to $\frac{a}{b}$.[/quote]

thats actually a lot harder...not "slightly harder"[/quote]

i think its like (x+y)/xy=a/b when (1/x)+(1/y)=a/b
but im sure it reduces more, tell me if im wrong though.[/quote]
It doesn't reduce more. Unless...
[hide="Is this more?"](x+y)b=axy
axy-xb-yb=0
$axy-\frac{xb}{a}-\frac{yb}{a}+\frac{b^{2}}{a^{2}}=\frac{b^{2}}{a^{2}}$
Simon's
$(x-\frac{b}{a})(y-\frac{b}{a})$[/hide]
Also $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}$
Notice the pattern ;). It works all the way (I think)...

Could you please be more specific in what you are talking about inom?

Thanks


A frustum is the solid obtained by slicing a right-circular cone perpendiclar to its axis and removing the small cone above the slice. This leaves a shape with two circular faces and a curved surface. The original cone has base radius $ 6$ cm and height $ 8$ cm, and the curved surface area of the frustum is equal to the area of the two circles. What is the height of the frustum?

I suppose you had a go at the SMC too then didnt you?


Find  all positive integer values of n for which (n^4)+4 is a prime number.

Thanks a lot

[hide="to finish off..."]
So we either need $x^{2}+2x+2=1$, which yields $x=-1$, which isn't positive, or $x^{2}-2x+2=1$, which yields $x=1$. $x=1$ works, giving 5 as our value. Hence, 1 is the only such value[/hide]

well...Altheman pretty much got the whole problem, the step that I did was pretty trivial.

Another question?
Find positive integer n ,$n^{4}+4^{n}$ is a prime number.


inequalities
trigonometry
triangle inequality
geometry solved
geometry

in triangle $ABC$ we have $a^3=b^3+c^3$ prove that $60<\angle A <90$

Since $a > b > 0,\ a > c > 0$ is the largest side of the triangle $\triangle ABC$, the angle $\angle A$ is the largest angle of this triangle. The largest angle of a triangle cannot be less than $60^o$, because then the sum of the other 2 angles $\angle B + \angle C$ would be larger than $180^o - 60^o = 120^o$ and at least one of the angles $\angle B, \angle C$ would be larger than $60^o$ ,i.e., larger than the angle $\angle A$. Alternately,

$a^3 = b^3 + c^3 = (b + c)(b^2 - bc + c^2)$

Since $0 < a < b + c$ by triangle inequality, $\frac 1 a > \frac{1}{b + c}$ and

$a^2 > b^2 + c^2 - bc$

$b^2 + c^2 - 2bc \cos{\widehat A} > b^2  + c^2 - 2bc \cos{60^o}$

$-\cos{\widehat A} > -\cos{60^o},\ \ \cos{\widehat A} < \cos{60^o},\ \ \angle A > 60^o$

Assume that the angle $\angle A \ge 90^o$, i.e., $b^2 + c^2 \le a^2,\ \ c^2 \le a^2 - b^2$. Then

$ a^3 - b^3 = c^3 \le c(a^2 - b^2)$

$(a - b)(a^2 + ab + b^2) \le c(a + b)(a - b)$

Since $a$ is the largest side, $a > b > 0,\ a > c > 0$, we have $a - b > 0,\ c < a$. Therefore,

$a^2 + ab + b^2 \le c(a + b) < a^2 + bc$

$b(a + b) < bc$

$a + b < c$

which contradicts the triangle inequality $a + b > c$. Consequently, it must be $\angle A < 90^o$.


binomial coefficients
Diophantine Equations

Show that the equation ${n \choose k}=m^{l}$ has no integral solution with $l \ge 2$ and $4 \le k \le n-4$.

This problem has been given in a lecture of China IMO team training.It is really tough indeed.
We should prove http://www.mathlinks.ro/Forum/viewtopic.php?p=1414377#1414377 as a lemma.

As the source states, this has been proved by Erdös in 1951. See e.g. [url]http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pja/1195510942[/url]

we know that every natural number $n$ can be write the $n=P 1 P 2 . . . P k X^2 $ for wich$Pi$ is

prime number .we get $T ( n ) = [P 1 , . . . , P k] $ . and  $m a x {T ( n ) } = M n $ .

we know $c( n, k ) = n ( n - 1 ) . . . ( n - k + 1 ) / k . ( k + 1 ) . . . 2 . 1 $  so we get $max[M n,Mn-1,...,M
n-k+1]=Px$  so we get  $m a x (k) $ such that $k$is divisible by $Px$ is $S$. so iff ther are no $l$  wich  $Px $ divises $l$  We have a contradiction

Rsyzh.and iff ther exist$ l $ such that  $P x $ divises $l$  then $P x $ divises $S-l$ we know $S-l<=k$ and we obtain $( n, k )=c( n, k
).S-l/S-l$ and At the end  we obtain $V p x (c ( n, k ) ) = 1 $

Nice result. Here is my solution:
Assume WLOG $k \leq \frac{n}{2}$ as $\binom{n}{k}=\binom{n}{n-k}$. If $k \geq \sqrt{n}$, then by Sylvester Theorem $\binom{n}{k}$ is divisible by a prime factor larger than $k$ and hence larger than $\sqrt{n}$. Since it is larger than $\sqrt{n}$ we have $v_p(\binom{n}{k})=1$ hence the binomial coefficient cannot be a perfect power. Now suppose the other case $4 \leq k \leq \sqrt{n}$. I claim that there exist integers $a_1,a_2,...a_k, b_1,b_2,..b_k$ such that $\prod a_i=k!, a_ib_i=n-i+1$ and for each prime $2 \leq p \leq k$, there is at most one term among the $b_i$ divisible by $p$ (hence the $b_i$ are co-prime). Consider a prime $p$ between $2$ and $k$. Let $f_p$ be a number between $n$ and $n-k+1$ which has the maximal possible value of $v_p$. For all $a_i$ such that $n-i+1 \neq f_p$, set $v_p(a_i)=v_p(n-i+1)$. For the $a_i$ corresponding to $f_p$, adjust $v_p(a_i)$ such that it agrees with $\prod a_i=k!$. Now we can check that this doesn't make any $v_p(a_i)$ negative by using induction. Repeat this process for each $2 \leq p \leq k$ and it is straightforward to check that the $a_i, b_i$ thus produced are indeed integers and they satisfy all the conditions in our claim. Hence, the $b_i$ are co-prime. Now if $\binom{n}{k}=m^l$ for some $l \geq 2$ then all the $b_i$ are perfect $l^{th}$ powers. We note that either the $a_i$ are $1,2,3,...k$ in some order or two of the $a_i$ are equal. If two of the $a_i$ are equal we get $a(b^l-c^l)=d$ where $ab^l, ac^l$ are between $n$ and $n-\sqrt{n}$ and $a,d$ are between $1$ and $\sqrt{n}$. We can size-bound and show that $d \geq alc^{l-1}$ and note an easy size bounding contradiction as the size of $ac^l$ is on the order of $n$ while the the size of $a,d$ are lower than $\sqrt{n}$. Consider the other case that the $a_i$ are $1,2,3,...k$ in some order. Take $i,j,k$ such that $a_i=\lfloor \frac{k}{4} \rfloor, a_j=2a_i$ and $a_k=2a_j$. So the corresponding $a_ib_i$ will be $\lfloor \frac{k}{4} \rfloor a^l, \lfloor \frac{k}{4} \rfloor*2b^l, \lfloor \frac{k}{4} \rfloor*4c^l$ for some $a,b,c$ and $l \geq 2$, and these numbers are between $n$ and $n-k$ and hence between $n$ and $n-\sqrt{n}$. Now we see that $d=4*\lfloor \frac{k}{4} \rfloor^2*((ac)^l-(b^2)^l)$ is less than $\frac{nk}{2}$. So an easy size argument shows that $l \geq 3$ is not possible as $m^l-n^l$ cannot be lower than $m^l-(m-1)^l$ whose size is on the order of $lm^{l-1}$. So the only possible case remaining is that this number is a square. So consider $l=2$. Consider $a_r,a_s$ such that $a_r=1, a_s=4$ (as $k \geq 4$). Then, $a_rb_r$ and $a_sb_s$ are two distinct perfect squares (as the $b_i$ are perfect $l^{th}$ powers) between $n$ and $n-\sqrt{n}$. However, this is a clear contradiction because for two distinct positive integers $a,b$, the numbers $a^2,b^2$ are at least $2a-1$ apart. So, for $4 \leq k \leq n-4$, we have shown that $\binom{n}{k}$ cannot be a perfect power. Proved

Edit: After reading Erdos's proof of this, turns out that this solution is quite similar to Erdos's proof...


LaTeX
inequalities
inequalities proposed

let a,b,c are positive.prove that :
\[ \sqrt {a^4 \plus{} b^4 \plus{} c^4} \plus{} \sqrt {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\geq\sqrt {a^3b \plus{} b^3c \plus{} c^3a} \plus{} \sqrt {ab^3 \plus{} bc^3\plus{}ca^3}
\]

[quote="stvs_f"]let a,b,c are positive.prove that :
\[ \sqrt {a^4 \plus{} b^4 \plus{} c^4} \plus{} \sqrt {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\geq\sqrt {a^3b \plus{} b^3c \plus{} c^3a} \plus{} \sqrt {ab^3 \plus{} bc^3 \plus{} ca^3}
\]
[/quote]

It's true since
\[ \sum {a^4 } \plus{} \sum {a^2 b^2 } \ge \sum {ab\left( {a^2 \plus{} b^2 } \right)}\]
and
\[ \left( {\sum {a^4 } } \right)\left( {\sum {a^2 b^2 } } \right) \ge \left( {\sum {a^3 b} } \right)\left( {\sum {ab^3 } } \right)\].
;)

Hey guys,I am going to write a proof that i just got but without using latex since i am a new member and i am not familiar with it(but hopefully I will soon learn).So:
By squaring both sides of the inequality we get:

a^4+b^4+c^4+(ab)^2+(bc)^2+(ca)^2+2sqrt(a^4+b^4+c^4)((ab)^2+(bc)^2+(ca)^2) >=
ba^3+cb^3+ac^3+ab^3+bc^3+ca^3+2sqrt(ba^3+cb^3+ac^3)(ab^3+bc^3+ca^3).

Now by using the inequality x^2+y^2>=2xy we receive the following inequalities:

              a^4+(ab)^2>=2ba^3

              b^4+(bc)^2>=2cb^3

              c^4+(ca)^2>=2ac^3

              b^4+(ab)^2>=2ab^3

              c^4+(bc)^2>=2bc^3

              a^4+(ca)^2>=2ca^3

Now,by adding all these inequalities we get that:

            a^4+b^4+c^4+(ab)^2+(bc)^2+(ca)^2>=ba^3+cb^3+ac^3+ab^3+bc^3+ca^3

Therefore it now suffices to prove that the expressions inside the radicals in our desired inequality satisfy the inequality:

            (a^4+b^4+c^4)((ab)^2+(bc)^2+(ca)^2)>=(ba^3+cb^3+ac^3)(ab^3+bc^3+ca^3).

By applying the Cauchy-Scharz inequality for the numbers (a^2,b^2,c^2), (ab,bc,ca) we get that:

           (a^4+b^4+c^4)((ab)^2+(bc)^2+(ca)^2)>=(ba^3+cb^3+ac^3)^2

Therefore it now suffices to prove that:

         ba^3+cb^3+ac^3>=ab^3+bc^3+ca^3 for all positive real numbers a,b,c.

By bringing all the terms to the left hand-side we get:

         ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)>=0.

If you guys remember the problem 3 in IMO 2006 Slovenia..its actually the same expression and can be factored as:

        (b-c)(a-b)(a-c)(a+b+c)>=0.

Now,we can assume without loss of generality that a>=b>=c..and because we are talking about positive reals the above inequality is obvious now.

We can use Chebychev and inequality:\[ a^4  \plus{} b^4  \ge a^3 b \plus{} ab^3
\]


geometry
circumcircle
incenter
search
geometry proposed

could someone find a solution without many metric relations to the following?
Let $I$ and $O$ be the incenter and the circumcenter of the triangle $ABC$ . The excircle $w_{a}$ is tangent to $AB,\ AC,\ BC$ in $K,\ M,\ N$ respectively .If the midpoint $P$ of $KM$ is on the circumcircle of $ABC$ prove that $O$ is the midpoint of $IN$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2042120209&t=6158]Here[/url] there are many solutions to this nice problem!


function
limit
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Suppose $ f: (0,1)\to\mathbb{R}$ is positive and $ \lim_{x\to 0}f(x)\equal{}\infty$. Is it true that there exists a convex function $ g$ such that $ g\leq f$ and $ \lim_{x\to 0}g(x)\equal{}\infty$?

This seems to be really intuitively true (but when has that line ever worked?), since you could just let $ g\equiv 0$ on $ (\epsilon, 1)$ where the unpredictable behavior is and then let it slope up near the $ y$-axis in such a way as to "stay to the left" of $ f$. That $ \epsilon$ has to exist by definition of the limit. Other than that description, I can't seem to prove or disprove or construct anything more concrete.

Now what if we switch things up and say $ f: (0,\infty)\to\mathbb{R}$ (positive) and $ \lim_{x\to\infty}f(x)\equal{}\infty$. Does there exist a convex function $ g\leq f$ such that $ \lim_{x\to\infty}g(x)\equal{}\infty$? This one seems less obvious, but you should be able to do it in almost the same way by staying along 0 until you are far enough out to start going up.

It is impossible if $ f(x)\equal{}\sqrt x$.


Για τους $ n \times n$ πίνακες $ A,B,C$ έχουμε ότι $ (A\minus{}C)(B\minus{}C)\equal{}A\minus{}B$

ΝΔΟ $ AB\plus{}BC\plus{}CA\equal{}BA\plus{}AC\plus{}CB$

 :)

Θέτουμε Κ=A-C , L=B-C
Έτσι εχουμε KL=K-L ή (I-L)(K+I)=I άρα και (Κ+I)(I-L)=I άρα και LK=K-L τελικά KL=LK οπότε
(A-C)(B-C)=(B-C)(A-C) δηλαδή AB+BC+CA=BA+AC+CB


Let 1> a > 0 be an arbitrary real number, and
x_1 = a, x_2 = a^a, x_3 = a^a^a, ...
The general term is  x_{n+1}= a^{x_n}

Analyze the long-term behavior of the sequence x_n as n tends to infinity. Show that for certain values of a, the terms x_n tend to some limiting value. How does that limit depend on a? Also find the region of chaos.


Let $x_1$, $x_2$, $\cdots$, $x_n$ be positive real numbers, and let
\[ S = x_1 + x_2 + \cdots + x_n. \]
Prove that
\[ (1 + x_1)(1 + x_2) \cdots (1 + x_n) \leq 1 + S + \frac{S^2}{2!} + \frac{S^3}{3!} + \cdots + \frac{S^n}{n!} \]

This should be a simple application of Maclaurin.

Just write LS as:
1 + (x_1 +x_2+...+x_n) + (x_1*x_2 + x1*x_3 + ... +x_(n-1)*x_n) + ... + (x_1*x_2*...x_n).
Then compare each term a_i on LS with the term b_i on RS, the terms are counted from the left.
It's very easy to verify that a_i =< b_i for every i. This gets the required result.

[hide="Another Solution"]By AM-GM,

$(1+x_{1})(1+x_{2})\cdots (1+x_{n}) \le \left(\frac{x_{1}+\cdots+x_{n}+n}{n}\right) = \left(\frac{S+n}{n}\right)^{n}= \left(1+\frac{S}{n}\right)^{n}= \sum_{i = 0}^{n}\binom{n}{i}\frac{S^{n-i}}{n^{n-i}}$

It suffices to prove that $\binom{n}{k}\frac{S^{n-k}}{n^{n-k}}\le \frac{S^{n-k}}{(n-k)!}$ for $k \in (0, 1, 2, \cdots, n).$ Since, $n! \le n^{n-k}\cdot k!$ for any $n, k,$ we have:

$n! \le n^{n-k}\cdot k! \Leftrightarrow \frac{n!}{k! (n-k)!}\cdot (n-k)! \le n^{n-k}$
$\Leftrightarrow \frac{\binom{n}{k}}{n^{n-k}}\le \frac{1}{(n-k)!}$
$\Leftrightarrow \binom{n}{k}\frac{S^{n-k}}{n^{n-k}}\le \frac{S^{n-k}}{(n-k)!},$

as desired. $\blacksquare$[/hide]

@Karth how we can conclude that obviously $n!\leq n^{n-k}k!$?

Using $k+1,...,n-1,n $ do not exceed$n$.

When $n=1$, we have the equality case.
When $n\geq 2$, the LHS expanded are the $i^{\textrm{th}}$ elementary symmetric polynomials, $e_i$ added together where $0\leq i \leq n$ and for $i=0,1$, it is equal to $1, S$ on the RHS. For all $2\leq i \leq n-1$, we can show that the $i^{\textrm{th}}$ elementary symmetric polynomial multiplied by $i!$ is included in $S^i$ which is obvious since $S^i$ contains all the ways that terms of $e_i$ can be formed because for each term in $e_i$, it can be formed in $i!$ ways in $S^i$. Then, for $i=n,$ it is suffices to show that $n!x_1x_2\ldots x_n\leq S^n$ which follows from the fact that $n!\leq n^n$ for $n\geq 1. \blacksquare$

Solution by induction.

Fix positive reals $x_1,x_2,\ldots, x_n$.
We proceed by induction and base case is trivial, therefore it is left as an exercise to the reader.
Assume this is true for some $k\geq 1$:

\[ \prod_{i=1}^{k}(1 + x_1) \leq \sum_{i=0}^{k}  \frac{S_k^i}{i!} \]
We want to show that this is true also for $k+1$:
\[ \prod_{i=1}^{k+1}(1 + x_1) \leq \sum_{i=0}^{k+1}  \frac{S_{k+1}^i}{i!} \]
Here, $S_k=\sum_{i=1}^{k} x_i$ and $S_{k+1}=\sum_{i=1}^{k+1} x_i$.
Using inductive hypothesis, we want to show that $(1+x_{k+1})\left(\sum_{i=0}^{k}  \frac{S_k^i}{i!}\right)\leq \sum_{i=0}^{k+1}  \frac{S_{k+1}^i}{i!}$.

We claim that $\frac{S_k^{i+1}}{(i+1)!}+x_{k+1}\frac{S_k^i}{i!}\leq \frac{S_{k+1}^{i+1}}{(i+1)!}$ for every $k-1>i>1$, from which the inequality right above follows.
$$\frac{S_k^{i+1}}{(i+1)!}+x_{k+1}\frac{S_k^i}{i!}\leq \frac{S_{k+1}^{i+1}}{(i+1)!}\Longleftrightarrow S_k^{i+1}+(i+1)x_{k+1}S_k^i\leq S_{k+1}^{i+1}\Longleftrightarrow (i+1)x_{k+1}S_k^i\leq x_{k+1}\left(\sum_{j=0}^{i} S_{k}^{i-j}S_{k+1}^{j}\right),$$
which is true, since $S_{k}\leq S_{k+1}$. Therefore, induction is complete, meaning inducting up to any $n$, we get desired inequality.

(a)Key claim is that \[\frac{S^k}{k!} \geq x_1x_2\cdots x_k+x_1x_2\cdots x_{k-1}x_{k+1}+\cdots +x_{n+1-k}\cdots x_n\]
where the RHS is the $\binom{n}{k}$ products of $k$ distinct terms.

(b)Notice that upon expansion, $S^k$ will have $k!$ copies of each of these terms, since not only that $S^k$ has any product of $k$ distinct terms, their $k!$ internal permutations will also be present.

(c) Using the lemma, we add the terms together and we are done.

Expand the LHS. For each $0\le i\le n$ we get

$$\sum_{\text{sym}} x_1\dots x_i \le \frac{\binom{n}{i}}{n^i}S^i = \frac{n(n-1)\dots (n-i+1)}{n^i}\cdot\frac{S^i}{i!}\le \frac{S^i}{i!}$$

and summing up yields the desired.


function
geometry
geometric transformation

The function y =f(x ) is transformed to y = f(2x + 4). Identify the
horizontal expansion or compression factor, then the translation to the
graph of the function

$f\left(2x+4\right)$ would be stretched vertically (or "compressed", as you said, horizontally) and would move to the left $4$ units, relative to $f\left(x\right)$

Because sharkman overestimates the difficulty of his textbook problems


Princeton college MIT Harvard Stanford Duke Yale

Princeton is consistently ranked the top National University by US news. Its got a lot of name being associated with Einstein and recent developments in String theory often have the epicentre there! Besides, Princeton is often regarded as the BEST place for math with an illustrious history and great students nowadays - of course, you can dispute me!

I want to know how Princeton compares with MIT, Caltech, Harvard, Stanford etc. for engineering ( Computer Science or electrical) and sciences in general ? Is it really the top one ( along with Harvard) ? What about research in engineering and Sciences there? Is it top notch - is it of the level as places like say MIT and Caltech ??? How much 'energy' and 'excitement' for knowledge and research and innovation that MIT boasts of, does Princeton have? Princeton has pride in its teaching and I won't ask you about that - there's enough evidence that Princeton focuses a lot on undergraduate studies and teaching - going by that it is small ( a bit bigger than Caltech only!) and has greater no. of undergrads and also that everyone in the faculty teaches! So, forgetting the rankings, Is Princeton really good for Science and Engineering??? Of course, it's the hardest to get into along with MIT! Is it academically rigorous??? Is it a fun place with amazingly unique activities like those at MIT? Does Princeton have an 'innovative' and 'energetic' environment????

Maybe more questions will follow if this discussion catches on!

P.S - Math at Princeton has been and is greatly influenced by Romanians!

Is your desire a strong engineering school, or a strong pure science and pure mathematics school? Harvard has very strong math, and very strong science in at least some disciplines, but is rather weaker (compared to other schools in the United States) in engineering. MIT is of course very strong in engineering (and quite good in pure math and pure science in general). I'll think about where Princeton fits in that spectrum of comparison while you reply with what kind of emphasis seems most appealing to you.

Rushil - It  would be nice, if the set was  a "linearly ordered set" but as you know the answer to this "comparison" (Princeton $\le \ \or \ \ge$ MIT ) is, to put it mildly,  not that simple.

Best advice, I can give, is to factor what is "important" to you, and most likely it will be something other than pure academc grading. Depending on , say, who is teaching one particular class,  course  in one school can be better/worse than the other.  The culture in Princeton is different than MIT  etc,  but what's more, in long run, it may or may not even make a difference which you can measure.

FWIW,  (Real  FWIW - as it happened decades  ago - and it was for graduate school -Phyics) in my own case, I chose not to go to Princeton, the advice came from my own brother (who was a  Faculty in Princeton at that time)... but the reason had VERY little to do with which school was "better" (after all my brother certainly liked it there, otherwise he wouldn't the there :) )  or "ranked higher".  Last year, one of my  nephew (who likes, or at least  I think he likes :),  to take my advice)   was also deciding (for grad school - Math) between Princeton and MIT. He chose MIT, but again,  the reason was  personal.   He did found out quite a bit (by talking to other friends, visiting etc) so may be I can give some "perspective"  if you have some particular question (private message is ok too).

Practically speaking , even if you flip a coin, to decide between ,say , MIT, Caltech, Princeton, Stanford, Duke, Harvard,  Yale, any of the IIT's, (And many other schools ) you,  IMO,  would not go wrong!

[quote="Rushil"]
I want to know how Princeton compares with MIT, Caltech, Harvard, Stanford etc. for engineering ( Computer Science or electrical) and sciences in general ? Is it really the top one ( along with Harvard) ? What about research in engineering and Sciences there? Is it top notch - is it of the level as places like say MIT and Caltech ??? How much 'energy' and 'excitement' for knowledge and research and innovation that MIT boasts of, does Princeton have? Princeton has pride in its teaching and I won't ask you about that - there's enough evidence that Princeton focuses a lot on undergraduate studies and teaching - going by that it is small ( a bit bigger than Caltech only!) and has greater no. of undergrads and also that everyone in the faculty teaches! So, forgetting the rankings, Is Princeton really good for Science and Engineering??? Of course, it's the hardest to get into along with MIT! Is it academically rigorous??? Is it a fun place with amazingly unique activities like those at MIT? Does Princeton have an 'innovative' and 'energetic' environment????

Maybe more questions will follow if this discussion catches on!

P.S - Math at Princeton has been and is greatly influenced by Romanians![/quote]

As for engineering, unless things have changed a lot, I wouldn't put Harvard in the same list as the others you have (Princeton, MIT, Caltech).

Princeton is very good for science and engineering research.  It's culture isn't quite as innovative and mathy-friendly as I hear MIT's is, but it is probably more diverse in types of people that are there (outside the science/engineering school, of course).  Basically, I'd pick Princeton over MIT if you think there's a fair chance you'd want to do serious study in the arts in addition to your science studies.

No offence, but it seems your views have put me in more of a fix than clarifying stuff!! Well, that happens when you try to compare top universities... Let me takle the responses one by one : -

tokenadult - my situation regarding my possible major is a bit tricky.... My fields of SUPER interest and those which I enjoy - including minors - are[b] Aerospace Engineering [/b](particularly Aircraft Design) - I simply ADORE aircraft! , [b]Theoretical Computer Science [/b]- love the things we can achieve - although my interest in this has fostered just recently!  , [b]Atomic Physics + Nanotechnology [/b]- great prospects, love the things we can achieve with this - have personal experience of research at IISc , Bangalore -so like it! , [b]Theoretical Physics [/b]- I like to think of theories - have some ideas of my own , [b]Math[/b] - my favorite subject - though I don't think I'll major in this! , I like International /Public affairs  -like spending time on this - discussing, also like History - particularly the analytical part - love discussing policies etc - since my Dad is a top Public Servant (IAS -if you know, Gyan!) in India- so like discussing policies with him - maybe I'll take some courses in this but probably NOT major!!

Well - this is already a long post - will answer other question in next posts - will be easier to read then!

Gyan, I haven't quite understood your point. Are you referring to the Graduate school or the undergraduate? Also, you say that the culture in Princeton is different from MIT. Can you explain specifically, in what ways is it different! And that 'perspective', I'd love to receive your PM if you do not wish to discuss it here!

rrusczyk - Great to learn from a Princeton undergrad! You're right, Princeton is proud of the fact that even though it is a top research institute, it has a majority of undergrads! That's one thing that attracts me to Princeton  - this above fact seems to indicate that Princeton is much better for undergrads - with focus on them and even research major done by undergrads! Actually, my dad told that in a Princeton blog, the faculty were against even a Finance department for they felt that finance etc. were professional stuff not related to 'teaching' and 'education'. Princeton is definitely 'undergrad - centered' as compared to say MIT or Stanford!

regarding your second question, I have highlighted my interests in the above post! Please see what you'd suggest after my post. Your major was Math right! Princeton has a HUGE name for Math  - near;y every modern day Great Mathematician is at Princeton - take Andrew Wiles! I am definitely interested in arts and other subjects but I have one other view also - I don't want to go to an 'all-round' university (which by he way even MIT is! ) if it in any way sacrifice my primary interest! MIT keeps saying that we are a 'fun' place and the kinds of courses and activities they have cl;early indicates that! Caltech says we are totally research oriented and rigorous! Dunno about Harvard for their site doesn't say much! What about Berkeley? How would you, (basically my mind is between MIT and Princeton for MIT has always been my dream!) compare MIT with Princeton precisely! Is the 'energy' at MIT also at Princeton? Princeton doesn't carry a 'lot' of name in engineering research like say MIT and Caltech etc.... so does that mean the engineering department is a bit 'lower' than say MIT???  I'd prefer a specific answer!

[quote="rrusczyk"][quote="Gyan"]
FWIW,  (Real  FWIW - as it happened decades  ago - and it was for graduate school -Phyics) in my own case, I chose not to go to Princeton, the advice came from my own brother (who was a  Faculty in Princeton at that time)... but the reason had VERY little to do with which school was "better" (after all my brother certainly liked it there, otherwise he wouldn't the there :) )  or "ranked higher".[/quote]

Funny you should say this - as a Princeton undergrad, I would say from what I saw that Princeton was not a great choice for graduate students for reasons which have VERY little to do with academic.  (Mainly, it sure seemed like the school was structured socially and in other ways much more around undergrads than, say, Stanford, where I went briefly as a grad student.)[/quote]

I know at least two people currently going to Princeton who would say that it's still that way...

[quote="Rushil"]How good is the engineering research at Princeton? Is it really path breaking and remarkable and 'fun' like that at MIT and Caltech??? Or is it just good research but not really great!? Is it innovative - I mean really amazing!?

Also, what about undergraduate research - especially in Pure Science and Math and Engineering?? IS it great or it just OK? Is it 'fun; like some topic covered by MIT students e.g. a research on Las vegas Casinos ???? Or is it just normal stuff - my question is very specific to engineering!!!

A more specific question - How good and how much research is being done on Nanotechnology at Princeton? Are undergrads given a chance to be part of this??[/quote]

These research questions may be answered more accurately if you actually browse princeton's website and email questions to students and professors there.

well, before I do that, maybe princeton 'people' - undergrads, grads, alumni, faculty! - here may well answer these  - or maybe people who know much about admissions and particarly, Princeton and MIT and Caltech etc.!!

What I want to know, is whether it's true that most of the undergrads there are preppy, conservative, snobby jerks.

because I think it sounds like a great school for the stuff I'm interested in, but it sounds like the culture there would be something I wouldn't do to well with.

I hope not to get into both MIT and Princeton... just one..
it would be to hard to decide

Yes, I too would like to know about the 'true' culture at Princeton! Is it  more of a 'snobby' type as in say Harvard or a more 'intellectual -equal ' and 'exciting' type as in say MIT???

Another thing that is 'attractive' about Princeton is this :- Princeton asks specifically for an essay on why you want to pursue a degree in engineering at Princeton - this indicates ( or seems to indicate) that Princeton's enginnering is coveted and really good!! Nobody else asks specifically for such an essay on just engineering - only Princeton asks ONLY for engineering!! Is this true!!???

Princeton's engineering program is not as good as MITs.  As for being particularly coveted, I expect that the acceptance rates for engineering are slightly higher than people applying to other programs - that's the way it is at most Ivy League schools, possibly barring Cornell (which is an anomaly in terms of acceptance rates, anyways...).  Cornell is a better school than Princeton for engineering, for that matter.  If you want to study engineering, the only reason to pick Princeton would be the name (and Cornell is a lot, lot easier to get into, for that matter)

You should rethink what you're asking us.  First of all, the quality of research done by faculty doesn't really affect you as an undergrad.  Unless you are an absolute genius, it will take you roughly four years to get to the point where you could do anything remotely original (and that's with a lot of guidance), much less understand state of the art research done by some of the best mathematical minds in the world.  Second of all, you are asking us to make sweeping generalizations as to the relative merit of universities that are by and large indistinguishable, except in some cases like MIT Engineering > [insert Ivy League here] engineering.  Third, you are asking us to characterize an ENTIRE student body of in some cases 6000+ students with a single adjective.  When I visited Princeton, I saw your typically socially awkward kids (as I recall, one kid playing pool with himself - VERY BADLY), your lacoste wearing wasp-y kids (there's a fair number there, I'll admit - might have something to do with the eating clubs) and a bunch of down to earth, normal people (well, except that they were a lot brighter than normal).  But I can't sum it up in a single word.

Blahblahblah, on what basis and which factors (important ones) did you consider in comparing the engineering at MIT, Princeton and Cornell? Was it the teaching and faculty or the possibilities and 'goodness' of research in engineering and development of technologies?

Regarding the engineering acceptance, I think I should correct you! At any university, even though they do not admit you to a particular department (except at Berkeley!) , it is obvious that engineering acceptance is lower than other fields? After all, why would Princeton ask separately why you want to do engineering! It's obvious that acceptance is lower as engineering is more coveted anywhere? The question is , at Princeton, do they ask so much separately for an engineering essay coz their engineering is really good !( obviously lower than MIT - but again, as in my first para, in what ways is it lower!?)!

It's obvious that you cant characterise the whole student body anywhere! But my question is different! Just as MIT is known for being different and fun and 'uncommon' - both in way of studies and research and what is developed as well as activities,  would be an appropriate word to describe MIT, I'm asking what such 'adjective' can be applied to Princeton!? Just as MIT is known for the above, what is Princeton known for!?

Another thing: - Since the coming of Pres. Tilghman, Princeton has focused a lot on engineering  - tale the recent purchase of one of the FASTEST supercomputers by Princeton/// Princeton has focused a lot recently on its future focus on engineering,. Although it might not come to the level of MIT, engineering will surely be the next big thing at Princeton with their new vision - it's known that US univs are known for achieving their goals!  :lol:  (I'm an Indian citizen!)

This is becoming wonderful!Let's hope I can learn more about Princeton and share with you!@

Just a last question ( unrelated) - IS Stanford a really coveted place in US ( i.e. for US Students ) as compared to Ivies and MIT and Caltech and Berkeley?

Rushil,

I studied computer engineering as an undergraduate (I'm now a grad student in math).  From the people I talked to (professors, friends), they definitely agree that Princeton is in the 2nd tier of engineering schools. The very best schools are MIT,Berkeley, Stanford, Illinois, Georgia Tech, and Cal Tech. The next level of schools includes Southern Cal, Michigan, Carnegie Mellon, Purdue, Cornell, Texas, Rice, Princeton (and others). The point is that there are many schools that are great places to get an engineering education besides MIT or Princeton.

My roommate went to Rice, and is very happy there. I went to visit him this summer, and it is a really nice engineering school.  He received excellent financial aid and is getting top notch training in signal processing. Another friend choose Southern Cal (he also got into Stanford, Purdue), since they are a top school in computer graphics. I also know a top student at my school who went to Stanford but is unhappy there. He is going elsewhere to do his phd.

I suggest you do more research.  Some of the weaker pure science/mathematics schools (eg Illinois, Georgia tech) are excellent in engineering, while top math schools such as Harvard or Princeton don't have top notch engineering schools.

Uh, thanks for correcting me...

I meant what I said, so I'll repeat it.  At any Ivy League school (except maybe Cornell), you will INCREASE your chances of acceptance by saying you want to do engineering.   Engineering is LESS coveted at these schools, because saying that you got an engineering degree from Princeton as opposed to Georgia Tech (or say, State U) makes a lot less difference than saying you got a math or sociology degree from Princeton as opposed to Georgia Tech (or again, State U).  Just because you think engineering is the end all be all of degrees doesn't mean others do.  I certainly don't.

As for Stanford, I'm biased because it's my top choice for grad school (Princeton is a close second).  Really, it is solid in everything and its reputation is second to none on the West Coast.  Their financial aid is not very good for undergrads (which is why I didn't apply there for my undergrad).  So yes, acceptance there is very sought after, but if you want to work in, say, New York, you're better off going somewhere on the East Coast.

In any event, however, this thread is becoming absurd.  You are making ridiculous extrapolations from almost no information and asking us to support or debunk them - why don't you do some of your own research?

[quote]Just because you think engineering is the end all be all of degrees doesn't mean others do. I certainly don't. [/quote]

I don't either. I wish I had realized this 3-4 years ago.  :(  I have too much math to learn in too short of a time.  :?

At the same time, that's not a putdown of engineering, either.  I don't consider it more or less prestigious on its own than a variety of other degrees.

Agreed. Rushil, do some research on your own. reputation isnt everything, generalization will never be what it really is. Look at each college's individual engineering programs (or whatver you want to look into). Asking people will only affirm generalizations or stereotypes. everyone has a different perspective. you have to find it yourself. Havard isn't "snobby white kids" all the time. MIT isn't always math/science. Cornell isn't always about looking at gorges. Stanford isn't always about being laid-back. Look at majors, programs, despcriptions. "good" "great" decent" have infinite meanings when talking about colleges, but usually only one to a single person. Find facts.

also, use more periods instead of exclamations.

I'm thinking of continuing in pure maths.
Any advice?
Also what tests should I give and what should my score be on those tests?

P.S. I know about SAT,PSAT.

I'd say:
MIT>Princeton>Harvard>Caltech = Stanford

The previous post is silly and unhelpful -- all of the colleges named have excellent math departments with high-quality faculty and student bodies.  The differences between them in style, size, location and so on are far, far more significant than the differences between the qualities of their math departments (or their educational quality in general).

As a matter of policy, I allow silly posts from time to time as long as they are generally on topic and don't violate the terms of service here. Whether or not those posts are helpful I allow each reader to judge.

Fair enough -- I'm just fufilling part of my higher duty to snark and criticize ;-)


algorithm
number theory
relatively prime
Euclidean algorithm

How could I figure out whether 2^2001 +1 and 2001! -1 are relatively prime?? I'm assuming that they are, as I used that fact to solve a problem, but how can I actually find that they are relatively prime? How about for the general equation 2^x +1 and x! -1. From a few trials it seems that they are always relatively prime, but I want to prove it.

hm...  looking at 2^x +1 and x! - 1

when x = 29, those 2 are not relatively prime..

and 29 is a factor of 2001

does that imply anything?

Well, if you assumed the general case of that theorem in order to show it for 2001, I'm afraid your proof wasn't valid.  Try x = 1 or 29.  Remember that "X is true" and "Assuming Y, X is true" do not imply Y, so the fact that you used their relative primeness to prove something does not mean that they need be relatively prime -- there might be a proof that doesn't use that fact.

Edit: Ooops, I was beaten to it!  Actually, though, I can give you an idea for a proof: show that every prime factor of $2^{2001}+1$ is less than 2001.  (I don't know if this is true, either, but it would be sufficient for the theorem.)

[quote]show that every prime factor of $2^{2001}+1$ is less than 2001....[/quote]

But ...But ... isn't something like 3033169,  which is obviously a prime,   is greater than 2001, and  is a factor of $2^{2001}+1$  ...:)

Seems like there is no general theorem that would cover the question.
Neither do I see any good trick (though I may be just too stupid).
If you don't mind using computers, the good old Euclidean algorithm
is your best bet. This all makes me curious about what the original
problem you needed it for was.

fedja


what is the muttere guys
please verify the qestion  :mad:


geometry
circumcircle
incenter
trapezoid
geometric transformation
angle bisector
IMO Shortlist

Let $ABC$ be a triangle. Consider a circle which is tangent to the sides $AB$ and $BC$ of the triangle $ABC$ and is also tangent to the circumcircle of triangle $ABC$ at a point $J$. Let $I$ be the incenter of triangle $ABC$. Prove that the line $JI$ is an angle bisector of the angle $AJC$; in other words, show that $\measuredangle AJI=\measuredangle CJI$.

I think I've got it, but I'll use something which has been proved before on the forum without proof (you might try searching for it, although no good key words come to my mind): Let this smaller circle touch $AB,BC$ in $M,N$ respectively. Then $I$ is the midpoint of $MN$.

Now let's define some more points. Let $I_a,I_c$ be the excenters of $ABC$ corresponding to $A,C$ respectively. Also, let $T$ be the point where the external bisector from $B$ cuts the circumcircle the second time (it cuts the circle at $A$ and $T$). Also, let $T_a,T_c$ be the points where the internal bisectors of $A,C$ respectively cut the circumcircle the second time. Finally (two more :)), let $M',N'$ be the points where $JM,JN$ cut $BT$ (the external bisector from $B$).

Since $T$ is the midpoint of the arc $ABC$, it suffices to show that $J,I,T$ are collinear. Since $M'N'NM$ is a trapezoid ($MN\|M'N'$) and $I$ is the midpoint of $MN$, we need to show that $T$ is the midpoint of $M'N'$. Call this sentence $(*)$.

We have $\frac{NI}{N'I_a}=\frac{IT_a}{T_aI_a}=\frac{IT_c}{T_cI_c}=\frac{MI}{M'I_c}$, and since $MI=NI$, we also get $N'I_a=M'I_c$, meaning that $(*)$ is equivalent to $T$ being the midpoint of $I_aI_c$; call this $(**)$ (actually, we need a bit more work, since we also have to show that both $I_a,I_c$ are outside the segment [$M'N'$], but this is not hard to do).

$(**)$ is not hard to show, since if $I_b$ is the excenter corresponding to $B$, then $ABC$ is the orthic triangle of $I_aI_bI_c$, the circumcircle of $ABC$ is the nine-point circle of $I_aI_bI_c$, so it must cut $I_aI_c$ in the foot of the perpendicular from $I_b$ and the midpoint of $I_aI_c$. Since the foot of the perpendicular is $B$, the midpoint must be $T$, so we have proved $(**)$, Q.E.D.

To prove that $I$ is the midpoint of $MN$ it's enough to combine the two theorems:
$Caysey$ for $A,B,k,C$ were k is the smaller circle;
$Transversal$ for $M,N,I$ and we get the conclusion.

Yes, that's the usual proof, but I kind of (:?) remember someone posting a nice inversive proof. Was that just a dream of mine? :)

[quote="hardsoul"]A circle which is tangent to the sides AB and BC of the triangle ABC with incenter I is also tangent to its circumcircle at point J. Prove that JI is an angle bissector of AJC.[/quote]

Another solution: In http://www.mathlinks.ro/Forum/viewtopic.php?t=14440 , post #5, Zortech showed that, if P is the midpoint of the arc CBA on the circumcircle of triangle ABC, then the points J, I and P are collinear (with Zortech's notations, this is the collinearity of the points M, I and P). But since the point P is the midpoint of the arc CBA, we have CP = AP. Since equal chords in a circle correspond to equal chordal angles, the equation CP = AP yields that the chordal angles of the chords CP and AP in the circumcircle of triangle ABC are equal; in other words, < CJP = < AJP. In other words, < CJI = < AJI. Hence, the line JI is an angle bisector of the angle AJC.

  darij

Very nice problem!

For names of points, look at the picture.

Draw the common tangent at $ J$ of both circles. It's impossible that both $ BA$ and $ BC$ are parallel to this common tangent, so WLOG, say that $ BC$ intersects this comon tangent in $ S$.
We can easiley see by the alternating segment theorem that $ \angle CJS = \angle CAJ \hspace{1cm}(1)$.

Now, because of $ (1)$, we have that
\[ \angle BAJ = \angle \pi-\angle PCJ = \angle CPJ+\angle PJC \]

\[ = 2 \angle PJC+\angle CJS = 2 \angle PJC+\angle CAJ \]
This yields the following statement:
\[ 2 \angle PJC = \angle BAJ \hspace{1cm}(2) \]
In $ \Delta PIC$, we can easiley deduce that $ \angle BPQ = \frac{\pi}{2}-\frac{\angle ABC}{2}= \angle PCI+\angle PIC$
Hence, using $ (2)$, we find that $ \angle PIC = \frac{\angle BAC}{2}= \angle PJC$ so $ PIJC$ is cyclic.

We can now continue in two ways:
[b]Way 1:[/b] Give the same proof for the fact that $ IQAJ$ is cyclic. (You then also have to treat the case where $ BA \parallel SJ$)
[b]Way 2: [/b]Use the famous IMO Shortlist problem: $ I$ lies on $ PQ$. Then we can proceed:
It now follows that $ \angle PIJ = \pi-\angle BCJ = \angle QAJ$ so $ IQAJ$ is also cyclic.

The conclusion follows after both ways for proving that $ IQAJ$ is cyclic: $ \angle IJA = \angle BQI = \angle BPI = \angle IJC$

$ \mathbb{QED}$

[img]http://img130.imageshack.us/img130/73/mathlinksyoshtangentcirjh5.jpg[/img]

[color=indigo]Denote: $O,\,O_{1}$ be the centers of circumcircle of $\triangle ABC$ and of circle which tangents to $BA,\,BC$ and $(O),$ and $T=JP\cap (O)$
We have: $\begin{cases}OT=OJ\\ O_{1}T=O_{1}P\\ J\in OO_{1}\end{cases}\implies \widehat{JPO_{1}}=\widehat{JTO}\implies O_{1}P\parallel OT\implies OT\perp BC$
$\implies T$ is the midpoint of arc $BC$

Thus $\widehat{PJC}=\widehat{TJC}=\widehat{TAC}=\frac{A}{2}\quad (1)$

On the other hand $\begin{cases}\widehat{BPQ}=90^\circ-\frac{B}{2}=\frac{A+C}{2}\\ \widehat{BPQ}=\widehat{PCI}+\widehat{PIC}=\frac{C}{2}+\widehat{PIC}\end{cases}\implies \widehat{PIC}=\frac{A}{2}\quad (2)$

From $(1)$ and $(2)$ we have: $\widehat{PJC}=\widehat{PIC}\implies PIJC$ is cyclic.
Similarly, we also have $QIJA$ is cyclic

And hence, we have: $\widehat{IJA}=\widehat{IQB}=\widehat{IPB}=\widehat{IJC}$[/color]
[color=red]
@ [b]To Jan:[/b] [size=150]Good luck at IMO 2007![/size] [/color] :lol:

[size=134][color=purple][b]Very nice problem, but it is well-known ![/b][/color][/size]
[quote="hardsoul"][color=darkred]Let $ ABC$ be a triangle. Consider a circle $ \mathcal C(I_{1})$ which is tangent to the sides $ AB$ and $ BC$ of $ \triangle\ ABC$ and which is internal tangent to the circumcircle $ w$ of $ \triangle\ ABC$ at a point $ J$. Let $ I$ be the incenter of $ \triangle\ ABC$. Prove that the line $ JI$ is the bisector of the angle $ \widehat{AJC}$, i.e. $ \widehat{AJI}\equiv\widehat{CJI}$.[/color][/quote]
[color=darkblue][b]Proof[/b] (similarly to [b]April[/b]'s). Denote the second intersections $ R$, $ S$ of the circle $ w$ with the lines $ JP$, $ JQ$ respectively.

From http://www.mathlinks.ro/Forum/viewtopic.php?t=155800 results $ \{\begin{array}{c}m(\widehat{CJN})=m(\widehat{BJN})=\frac{A}{2}\\\ m(\widehat{AJM})=m(\widehat{BJM})=\frac{C}{2}\end{array}$.

Apply the [b]Pascal's theorem[/b] to the cyclic hexagon $ ABCMJN$, i.e. $ \{\begin{array}{c}Q\in AB\cap MJ\\\ P\in BC\cap JN\\\ I\in CM\cap NA\end{array}\|$ $ \implies$ $ I\in PQ$. Thus,

$ \{\begin{array}{c}m(\widehat{CIP})=m(\widehat{BPQ})-m(\widehat{BCI})=(90^{\circ}-\frac{B}{2})-\frac{C}{2}=\frac{A}{2}=m(\widehat{CJP})\implies \widehat{CIP}\equiv\widehat{CJP}\\\ m(\widehat{AIQ})=m(\widehat{BQP})-m(\widehat{BAI})=(90^{\circ}-\frac{B}{2})-\frac{A}{2}=\frac{C}{2}=m(\widehat{AJQ})\implies \widehat{AIQ}\equiv\widehat{AJQ}\end{array}$ $ \implies$

the quadrilaterals $ CPIJ$ and $ AQIJ$ are cyclically $ \implies$ $ \widehat{CJI}\equiv\widehat{BPQ}\equiv\widehat{BQP}\equiv\widehat{AJI}$ $ \implies$ $ \widehat{CJI}\equiv\widehat{AJI}$.

[/color]

Using the above drawing (post #7):
We know that:
 -1) $P, I, Q$ are collinear, $BP=BQ$ or $m(\widehat {BPQ})= \frac{\angle A+\angle C}{2} \; ( \; 1 \;)$
 -2) $JP$ is the angle bisector of $\widehat {BJC}$, i.e. $m(\widehat {CJP})=m(\widehat{CAI})=m\left(\frac{\widehat A}{2}\right) \; ( \; 2 \;)$.
$m(\widehat{PCI})=m\left(\frac{\widehat C}{2}\right)$, so $m(\widehat{PIC})=m(\widehat{BPI})-m(\widehat{PCI})=m\left(\frac{\hat A}{2}\right) \; (\; 3 \; )$, from the relations $(2)$ and $(3)$ getting that $IPCJ$ is cyclic, hence ${m(\widehat{CJI})=m(\widehat{BPQ})}$, in a similar way $IJAQ$ is cyclic and $m(\widehat{AJI})=m(\widehat{BQP})$, done.

[quote="jayme"]Dear Mathlinkers,
for a proof and more, see
http://perso.orange.fr    vol. 4 A new mixtilinear incircle adventure p. 18.
Sincerely
Jean-Louis[/quote]

Dear jayme...... you need to be a little less secretive and take out time to what you want to say. if you just always put , visit my website people will not appreciate your work too much.. i don't understand much of the material there and cannot find the paper.... put the link please..... and the translations too....

[quote="grobber"]I think I've got it, but I'll use something which has been proved before on the forum without proof (you might try searching for it, although no good key words come to my mind): Let this smaller circle touch $AB,BC$ in $M,N$ respectively. Then $I$ is the midpoint of $MN$.

[/quote]
A proof of this theorem is given in nathan altshiller court's book .... see the book here....
http://www.kiemcun.downarchive.com/ebooks/126325-college-geometry-an-introduction-to-the-modern.html


It's now permitted to discuss the 2009 AIME I that was held on 3/17/09.

Did anyone think that this years AIME was rather hard? One kid that got 9 last year only got a few this year. But mainly I really hate the arithmetic errors I make.:wallbash_red:  How did everyone else do?

Yes, this test was more computationally heavy and more prone to arithmetic errors. For example, I thought $ 4^4\equal{}1024$ on #14 so I mistakenly reduced the cases for # of 4's to {0, 1}. I eventually got # of 4's = 0, but due to my mistake I never considered # of 4's = 5.

I got a 9, but I solved 12 questions. (The 3 wrong were all arithmetic or counting errors).

On number 12, which I solved with trig, I thought

$ \frac{210}{1225} \equal{} \frac{2}{5}$ =(

IMHO, computation is one of those things that should be natural by now. If you get lots of problems wrong because of computation (i.e. me a year ago), then you just need to learn how to be more careful.

[quote="BOGTRO"]The answers are posted in another topic.

I'm really mad. This test was extremely focused on computation. I would have gotten a 9, but I got a 5 due to really stupid arithmetic errors.[/quote]

i would have gotten a 12, but got 6 for same reasons

[quote="BOGTRO"]The answers are posted in another topic.

I'm really mad. This test was extremely focused on computation. I would have gotten a 9, but I got a 5 due to really stupid arithmetic errors.[/quote]

11 to 6 for me

actually, probably 10 to 6

i was hoping to make usamo this year, now i only will if lots of other people fail and someone gets 150/6...or 144/6

I was hoping I'd qualify this year...

I would have had I not made exactly 4 stupid mistakes, which would've made my index 36 points higher.

The worst one: I got the answer, put down the wrong number........

 :wallbash_red:  :wallbash_red:  :wallbash_red:

[quote="BOGTRO"]The answers are posted in another topic.[/quote]

Sorry, but where are the answers found?  I can't seem to find them.  Thanks.

hmm...
i do not mess up that much anymore at computation, so i got a 9 (if i don't i'm getting a rescore. my #9 was erased like 4 times.)

but i'm too slow, so I only got to #9.

Should've solved #11,#12,#13 if i had more time (but not #10, no me gusta circular combo.)


Are there any good sites for me to LEARN the material in this section? Or learning Math in general? I wan't to be able to do the questions here but they are just too foriegn to me... >.<

You've already found it! Search the AoPS Wiki for a quick overview on all the major concepts that you don't understand. Alternatively, you could purchase the AoPS books which really help.


function
algebra
floor function
calculus
calculus computations

A function f(x)is defined as f(x)= x^m sin(1/x) x ≠0

0 if x=0

The least value of m for which f`(x) (differential of f(x)) is continuous at x=0 is

Q2 on the interval I= [-2, 2] the function f(x) =(x+1) e^-[1/mod(x) +1/x] If x≠0 ,
0 if x=0

Is continuous for all values of x ε I (prove it). [] denotes greatest integer function

Q3 consider f(x) = [2(sin(x) –sin3x) +mod (sin x –sin3x)/ 2(sin(x) –sin3x)-mod (sin x –sin3x)]
Where [] denotes greatest integer function

x≠ pie/2
for x έ (0, pie). f(pie/2)=3

prove that f is continuous and differentiable at x=pie/2
_________________
GOD has given us one face but we created another

3.If u try to draw graph for $ g(x)=sin(x)-sin(3x)$ u will see that this function is $ +ve$ in interval $ (\frac{\pi}{4},\frac{3\pi}{4})$
so for this interval excluding $ x=\frac{\pi]}{2}$ we could write f(x) as
$ f(x)=[\frac{2(sinx-sin3x)+sinx-sin3x}{2(sinx-sin3x)-(sinx+sin3x)}]$
$ =3$
thus at $ x=\frac{\pi}{2}$ f(x) is continuous and differentiable as for this interval f(x) is a straight line ie. $ f(x)=3$
thank u

[quote="MUN"][quote="kabi"]1. $ m \equal{} 1$[/quote]
ans is 3[/quote]
sorry  :(  i overlooked $ f'(x)$
yes the answer will be 3.

[quote="kabi"]hello
Is second question $ f(x) \equal{} (x \plus{} 1)e^{ \minus{} [\frac {1}{\mod x} \plus{} \frac {1}{x}]}$?
[hide="solution"]
 $ f(x) \equal{} (x \plus{} 1)e^{[ \minus{} \frac {2}{x}]}$ if $ x > 0$
        =$ 0$ if $ x \equal{} 0$
        =$ (x \plus{} 1)$ if $ x < 0$
Thus it is non-continuous and not differential at $ x \equal{} 0$  as $ LHL \equal{} 1$
 [/hide]
Thank u[/quote] answer is cont pl give view abut 3 question

It could be continuous only if $ f(x)\equal{}xe^{\minus{}(\frac{1}{mode x}\plus{}\frac{1}{x})}$
in ur question $ f(0)\equal{}0$ as given but $ LHL\equal{}1$ so it is not continuous.

[quote="kabi"]It could be continuous only if $ f(x) \equal{} xe^{ \minus{} (\frac {1}{mode x} \plus{} \frac {1}{x})}$
in ur question $ f(0) \equal{} 0$ as given but $ LHL \equal{} 1$ so it is not continuous.[/quote]


Are these new archives so that we can view the old forums for classes that we previously were in?  If so, that sounds awesome!

Thanks.

[quote="RC-7th"]For some strange reason, I can view the Interm Counting Archive. While I did participate in one of the classes, I didn't participate in the one that the forum belongs to.[/quote]

The archive includes posts from more than one Interm Counting class.

[quote="MithsApprentice"]Would it be possible to archive all of the classes we've taken so far? (Including those that have been over for like a year now)

And also, could something like this be done with class files too?

To save confusion, you could add an edition and time taught after the class title.

So something like Olympiad Problem Solving 3.04 would mean that it's the fourth time the class has been taught since it's been changed and the class has been changed twice previously (aka this is the third edition).[/quote]

We don't plan to do this with the files, but we'll look into back classes.  It will be a while before we do that, though.

[quote="ml2"]So, the stuff in the archives now, will be there forever?[/quote]

To be clear...forever is a long time.  We don't want to be legally bound to...forever.

Consider it an [b]indefinite[/b] opportunity to learn about the problems you couldn't solve on your own from an expanded group of peers.


Find:

i) $\tan\frac{\pi}{16}+\tan\frac{5 \pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$

ii) $\tan^{2}\frac{\pi}{16}+\tan^{2}\frac{5 \pi}{16}+\tan^{2}\frac{9\pi}{16}+\tan^{2}\frac{13\pi}{16}$

I did it in a pretty rough way. Oh, yeah, no calculators.

hm.. this relation is interesting!!

we can try to look for integers solutions for ${\tan^{n}\left(\frac{\pi }{s}\right)+\tan^{n}\left(\frac{5 \pi }{s}\right)+\tan^{n}\left(\frac{9 \pi }{s}\right)+\tan^{n}\left(\frac{13 \pi }{s}\right)}=K$

where k,s,n are integers

We may be able to. We should do this after we solved the above problem. Actually I have found a quite astounding solution to it.

[hide=" A Sketch of It"]

Express $\tan 4 \theta$ in terms of $\tan \theta$ and find a polynomial with the tangent values as its roots.[/hide]

very nice solution Banished Traitor

the second part should be achievable through Newtonian sums
I don't exactly remember how to do them, but they would get sums of squares of roots


geometry
rectangle
quadratics
algebra
polynomial
function
3D geometry

Here are the problems of our 1st TST:

1st independent study

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=47918
Consider a circle $O_1$ with radius $R$ and a point $A$ outside the circle. It is known that $\angle BAC=60^\circ$, where $AB$ and $AC$ are tangent to $O_1$. We construct infinitely many circles $O_i$ $(i=1,2\dots\>)$ such that for $I >1$, $O_i$ is tangent to $O_{i-1}$ and $O_{i+1}$, that they share the same tangent lines $AB$ and $AC$ with respect to $A$, and that none of the $O_i$ are larger than $O_1$. Find the total area of these circles.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47920
Does there exist an positive integer $n$, so that for any positive integer $m<1002$, there exists an integer $k$ so that \[\displaystyle \frac{m}{1002} < \frac{k}{n} < \frac {m+1}{1003}\] holds? If $n$ does not exist, prove it; if $n$ exists, determine the minimum value of it.

2nd independent study
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=47921
More than three quarters of the circumference of a circle is colored black. Prove that there exists a rectangle such that all of its vertices are black.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47923
The absolute value of every number in the sequence $\{a_n\}$ is smaller than 2005, and \[a_{n+6}=a_{n+4}+a_{n+2}-a_n.\] holds for all positive integers n. Prove that $\{a_n\}$ is periodic.

3rd independent study
1. http://www.mathlinks.ro/Forum/viewtopic.phpt=47924
Prove that there exists infinitely many positive integers $n$ such that $n, n+1$, and $n+2$ can be written as the sum of two perfect squares.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47926
Let $ABCD$ be a convex quadrilateral. Is it possible to find a point $P$ such that the segments drawn between $P$ and the midpoints of the sides of $ABCD$ divide the quadrilateral into four sections of equal area? If $P$ exists, is it unique?

Final exam-1st day
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=47927
Let $f(x)=Ax^2+Bx+C$, $g(x)=ax^2+bx+c$ be two quadratic polynomial functions with real coefficients that satisfy the relation \[|f(x)| \ge |g(x)|\] for all real $x$. Prove that $|b^2-4ac| \le |B^2-4AC|.$

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47935
$P$ is a point in the interior of $\triangle ABC$, and $\angle ABP = \angle PCB = 10^\circ$.
(a) If $\angle PBC = 10^\circ$ and $\angle ACP = 20^\circ$, what is the value of $\angle BAP$?
(b)If $\angle PBC = 20^\circ$ and $\angle ACP = 10^\circ$, what is the value of $\angle BAP$?

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=47936
$n$ teams take part in a tournament, in which every two teams compete exactly once, and that no draws are possible. It is known that for any two teams, there exists another team which defeated both of the two teams. Find all $n$ for which this is possible.

Final exam-2nd day
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=30865
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. Prove that \[\displaystyle \sqrt[3]{\frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a} \le \frac{1}{abc}.\]

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=47937
Show that for any tetrahedron, the condition that opposite edges have the same length is equivalent to the condition that the segment drawn between the midpoints of any two opposite edges is perpendicular to the two edges.

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=47938
Find all positive integer triples $(x,y,z)$ such that
$x<y<z$, $\gcd (x,y)=6$, $\gcd (y,z)=10$, $\gcd (x,z)=8$, and lcm$(x,y,z)=2400$.

Unfortunately, I do not have the problems of our national olympiad. I will try to make them, as well as the problems from the other TSTs (two more) as soon as possible.

I failed to get in the top 6 in the last TST, and so I did not make it to Mexico......

Any solutions, ideas, or comments to the problems would be greatly appreciated. Please forgive my poor English.

The so-called "independent study" is actually an exam with two problems in a time limit of two hours. Sometimes we may get to discuss the problems with our instructors during classes in the selection camp. I also wonder why they call it "independent study"...

Can you tell us something about how Taiwan's Math Olympiad System work?

Of course. The math season usually begins with the MO in the 1st semester, divided into 3 rounds; the school round, the local round and the national round. The school round is held at a date decided by each school; both the problems and the results are provided and decided by each school itself. The school winners advance to the local round in October or November, and the top 5 winners in the local round advance to the national round. Medals are not awarded in the national round, but rather 1st diploma, 2nd diploma and 3rd diploma.

The students receiving diplomas in the national olympiad and a number of students selected from a few prestigious schools take part in the APMO selection camp in mid-February. Here participants for the APMO are selected.

After the APMO is held in mid-March, the top ten students in the APMO and the top fifteen in the APMO camp (discluding the top ten in the APMO) are eligible to take part in the 1st TST in late March or early April. Past IMO team members are not allowed to take part in the APMO camp or the 1st TST; they advance to the 2nd TST directly. However, they are eligible for the APMO.

Fifteen students are selected from the 1st TST; they and the past IMO team members take part in the 2nd TST in mid-April. 12 students are selected for the 3rd TST in late April; out of them the IMO team will be selected.

The IMO team will be subjected to a two-month training; I can't give you the details because I haven't been there!


[hide]a.) The big hose goest at a rate of 1/3 pph (pools per hour), and the little hose goes at the rate of 1/6 pph.  To find the pph for both hoses put together, just add them up to get 1/2 pph.  That mean it will take 2hours for the pool to fill up using both hoses.

b.)  When 30 minutes has past the pool has been 1/4 (1/2 of 1/2) of the way filled up.  In 2 hours, if no more water is added the pool will be empty with the leak.  That means the leak will go at -1/8 pph.  If you add that to the rate of both hoses, you get 3/8 pph.  Since the pool has already been filled up 1/4 of the way, only 3/4th of the pool need to filled up, taking it 2 more hours for the pool to be filled up.[/hide]

[quote="rrusczyk"]My big hose can fill my pool in 3 hours.  My little pool can fill it in 6 hours.  [/quote]

I'll take it we're talking about little hose?  :D  ;)
Hehe, a pool filling a pool. That'd be interesting to see.  :lol:  :cool:

[hide]a) 1/3+1/6=1/2, so 2 hrs.

b) leak is 1/8of the pool per hour, 1/2-1/8=3/8. To fill 3/4=6/8 at this rate, it will take 2hrs.[/hide]

My big hose can fill my pool in 3 hours. My little hose can fill it in 6 hours.

a) How long will it take to fill the pool using both hoses?

b) 30 minutes after turning on both hoses, my pool springs a leak. The leak allows water to leave the pool so fast that if I turned off the hoses right when the leak started, the pool will be empty 2 hours later. However, I don't turn off the hoses and I don't worry about the leak. How much longer (starting from the 30 minute point) will it take for me to fill the pool?

[hide]

1/6 + 1/3 = 1/x

1/6 + 2/6 = 1/x

3/6 = 1/x

1/2 = 1/x

x=2

30 minutes after turning on both hoses, the pool would be 1/4 full. That means the pool would drain 1/8 of the pool in an hour.

1/3 + 1/6 - 1/8 = 1/x

8/24 + 4/24 - 3/24 = 1/x

9/24 = 1/x

x = 24/9

x = 8/3

But that would be if the leak was sprung when the pool is empty. The pool is 1/4 full, so you will need to fill up just 3/4 of the pool.

(3/4)(8/3)

2 hours

[/hide]


function
linear algebra
matrix
induction
ratio
logarithms
linear algebra unsolved

If $\Gamma$ denotes the restriction of Gamma function on $(0,\infty),$ let us consider the determinant $n\times n$ \[D_{n}(x): =\begin{array}{|llcl|}\Gamma(x)&\Gamma(x+1)&\ldots &\Gamma(x+n-1)\\ \Gamma(x+1)&\Gamma(x+2)&\ldots&\Gamma(x+n)\\ \vdots&\vdots&\vdots&\vdots\\ \Gamma(x+n-1)&\Gamma(x+n)&\ldots &\Gamma(x+2n-2)\\ \end{array}\; .\] [b]When $x>0 ,$ find bounds for $\frac{D_{n}\left(x+0.5\right)}{D_{n}(x+1)}, \; \; (n\ge 2) .$ [/b]

A quick conjecture. (Don't trust me to have this right):

$D_{n}(x)=\Gamma(x)^{n}(n-1)!\,x^{n-1}(x+1)^{n-2}(x+2)^{n-3}\cdots (x+n-2)$

Added after jmerry's post:

That $(n-1)!$ should be $\prod_{k=1}^{n-1}k!.$ Fix that and we agree.

By row elimination (subtract $x+n-2$ times the second to last row from the last, $x+n-3$ times the third to last from the second to last, ...) we have
$D_{n}(x)=\begin{vmatrix}\Gamma(x)&\Gamma(x+1)& \Gamma(x+2) &\cdots&\Gamma(x+n-1)\\ 0&\Gamma(x+1)&2\Gamma(x+2)&\cdots&(n-1)\Gamma(x+n-1)\\ 0&\Gamma(x+2)&2\Gamma(x+3)&\cdots&(n-1)\Gamma(x+n)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&\Gamma(x+n-1)&2\Gamma(x+n)&\cdots&(n-1)\Gamma(x+2n-3) \end{vmatrix}$
$=(n-1)!\Gamma(x)D_{n-1}(x+1)$.

By induction on $n$, this gives us $D_{n}(x)=\prod_{k=0}^{n-1}k!\cdot\prod_{k=0}^{n-1}\Gamma(x+k)$

Without working out the details, this does give us that the ratio is asymptotically comparable to $x^{-\frac{n}{2}}$.

[quote="jmerry"] ...$D_{n}(x)=\prod_{k=0}^{n-1}k!\cdot\prod_{k=0}^{n-1}\Gamma(x+k)$
Without working out the details, this does give us that the ratio is asymptotically comparable to $x^{-\frac{n}{2}}$.[/quote]
Indeed.I have used the log-convexity of $\Gamma: (0,\infty)\to (0,\infty).$ Because  $\log{\Gamma }$ is a convex function, writing this condition ,
e.g.at the points  $x,\; x+0.5,\; x+1 ,$ we obtain \[(*)\; \; \; \;\frac{1}{\sqrt{x+0.5}}\le \frac{\Gamma(x+0.5)}{\Gamma(x+1)}\le \frac{1}{\sqrt{x}}\; \; ,\; x>0,\] (which in fact it's well known in literature). From (*) I  give \[\frac{1}{\sqrt{(x+0.5)_{n}}}\le \frac{D_{n}(x+0.5)}{D_{n}(x+1)}\le \frac{1}{\sqrt{(x)_{n}}}\] where $(y)_{n}: =y(y+1)\cdots(y+n-1) =y^{n}+\ldots$.


geometry
rectangle
trigonometry
circumcircle
incenter
trig identities
Law of Cosines

Hello,

These two problems do not have any solutions (source: Paul Yiu's Notes on Euclidean Geometry) so any help would be greatly appreciated. Thanks!

1) $ ABPQ$ is a rectangle constructed on the hypotenuse of a right triangle $ ABC$, with $ C$ as the right angle. $ X$ and $ Y$ are the intersections of $ AB$ with $ CP$ and $ CQ$ respectively. If $ ABPQ$ is a square, prove that $ XY^2 \equal{} BX \cdot AY$. Also prove that if $ AB \equal{} \sqrt{2} \cdot AQ$, then $ AX^2 \plus{} BY^2 \equal{} AB^2$.

2) In a triangle ABC, $ A,B,C$ denote the angles at the respective vertices. If $ \angle C \equal{} 2\angle B$, prove that $ c^2 \equal{} (a \plus{} b)b$. (The only hint provided was that the relevant rule was the Law of Cosines, but it gets very messy for me.)

Thanks again!

For the second use law of sines and the relations $ sin2x$=$ 2sinx*cosx$ and $ sin3x$=$ 3sinx\minus{}4sin^3x$. :)

thanks for your reply! I have an idea for a geometric proof.  If V is the foot of the angle bisector, the circumcircle of BCV is tangent to AC. Thanks a lot! Do you know how to do the first problem?

Oh, never mind, I got it. Is it because $ \angle BVC \equal{} 180 \minus{} 2B$, so the arc subtended by $ BCA \equal{} 180 \minus{} \angle BVC \equal{} 2B$, which shows that the line $ AC$ is tangent to the circumcircle. I think that makes sense. Thanks so much! Can anyone do the first problem? :(

[hide="For Square"]Extend $ CB$ and $ CA$ to hit $ PQ$ at $ D$ and $ E$ respectively. Then we have that $ \frac {BX}{XY} = \frac {DP}{PQ}$ and $ \frac {XY}{AY} = \frac {PQ}{QE}$ (easy to see triangle similarities).

Also,

$ \triangle DPB\sim \triangle AQE \longrightarrow \frac {DP}{PB} = \frac {AQ}{QE} \iff DP*QE = PB*AQ \iff DP*QE = PQ^{2} \iff \frac {DP}{PQ} = \frac {PQ}{QE} \iff \frac {BX}{XY} = \frac {XY}{AY}$

hence we are done.
[/hide]

[hide="For rectangle"]Once again we extend $ CB$ and $ CA$ to hit $ PQ$ at $ D$ and $ E$ respectively.

Proceeding as before, we have $ \triangle PBD \sim \triangle QEA$ so that $ PB^2 = PD*QE \iff \frac {AB^2}{2} = PD*QE \iff XY^2 = 2AY*BX$.

Now consider
$ AX^2 + BY^2 = (AY + XY)^2 + (BX + XY)^2
= AY^2 + BX^2 + 2XY^2 + 2AY*XY + 2BX*XY
= AY^2 + BX^2 + XY^2 + 2AY*BX + 2BX*XY + 2AY*XY
= (AY + XY + BX)^2 = AB^2$

done[/hide]

Thank you resurrection! I have more questions, if anyone can answer.
a) The incenter of a right triangle is equidistant from the midpoint of the hypotenuse and the vertex of the right angle. Show that the triangle contains a $ 30^{\circ}$ angle.
b) The incircle of triangle $ ABC$ touches the sides $ AC$ and $ AB$ at $ Y$ and $ Z$, respectively. Suppose $ BY \equal{} CZ$. Show that the triangle is isosceles.

Thank you for any help!

$ \Delta CIP \cong \Delta MIN \Rightarrow CP\equal{}MN$

$ CP \equal{} p\minus{}c \;, MN\equal{}c/2 \minus{} (p\minus{}b)\equal{}(b\minus{}a)/2 \Rightarrow 2p \minus{} 2c \equal{} b\minus{}a \Rightarrow c\equal{}2a$

Oh, I did not think of using $ p \minus{} c$. Nice! Thanks!


[quote="Devendra Patel"]Could someone pls give a simple explanation for inmtensity and brightness of light with regard to frequency and wavelength of light[/quote]

The intensity of a source is defined as the time-average of its Poynting vector.

$I=\langle S \rangle =\frac{P}{A}$

The brightness of a source is also known as its [i]specific intensity[/i].

$I_{\nu}=\frac{\mathrm{d}I}{\mathrm{d}\nu}$


1. Given a set of 9 points of which no 3 points are coar, how many lines are needed so that each point is connected by a line to every other point?

2. Cncentric circles have radii of 4 and 7. idthe area of the portion between the circles.

[hide] I think the answer to this problem is $ \frac{8\times9}{2}=\boxed{36}$. [/hide]

I agree with lokesh.

[hide="1"]First, connect the first point to the others. This takes $ 8$ lines to do. Then, connect any other point to all others except for the first one, which takes $ 7$ lines. Each time, you need $ 1$ less line, and the total number of lines is $ 8\triangle$, or $ \boxed{36}$.[/hide]

[hide="*Note..."]I_like_pie's argument is equivalent to answering "How many ways can we choose 2 points from the 9 and then draw in their line?"

Thus, our answer is $ \binom{9}{2}= 36$[/hide]

What is that triangle in i_like_pie's post? It doesn't seem to mean change in this example...

That triangle refers to the triangular numbers, specifically the 8th triangular number.

Such a number is defined as follows:
$ T_{n}\equal{}1\plus{}2\plus{}3\plus{}...\plus{}n$
Which also equals $ \frac{n(n\plus{}1)}{2}$ and $ \binom{n\plus{}1}{2}$

This is why I_like_pie, after adding $ 1\plus{}2\plus{}...\plus{}8$ said that this is the 8th triangular number.  This also explains why my explanation of $ \binom{9}{2}$ is also equal to 36.


i think it`s wrong try $a=1$, $b=0,1$ , $c=0,1$

[quote="andyciup"]Let $a,b,c$ be positive real numbers.If $a+b+c\geq 3$,prove that
\[\displaystyle a+b+c\geq\sqrt{4(ab+bc+ca)-3abc}\]
[/quote]

$\Leftrightarrow 3abc\geq 2\sum{ab}-\sum{a^{2}}$

but from schur we have that $\frac{9abc}{a+b+c}\geq 2\sum{ab}-\sum{a^{2}}$

so it remains us to prove that : $3abc\geq \frac{9abc}{a+b+c}\Leftrightarrow 1\geq \frac{3}{a+b+c}\Leftrightarrow a+b+c\geq 3$, which is now true. :)


Given, in quadrilateral $ ABCD$, that $ AB\equal{}CD$ and $ \angle ABC \cong \angle ADC$, prove or disprove that $ ABCD$ is a parallelogram.  If it is not a parallelogram, give a counter-example.

[hide]
$ \angle ABC \cong \angle ADC$
$ AB \equal{} CD$
$ AC$ is same for both triangles

$ \Rightarrow \bigtriangleup ABC \cong \bigtriangleup ADC$

Hence we have a parallelogram
[/hide]

EDIT; oops :oops: side-side-angle only means ABC and ADC are similar

[hide]
 [geogebra]069e35ce2a780175982eb74dde5b319375398bf7[/geogebra]

Drag the points around to see more examples!
[/hide]

[hide="A quadrilateral is a parallelogram iff"]
1) AB=CD, AD=BC
2) same, but parallel
3) AB=CD, $ AB\parallel CD$
4) its diagonals bisect each other
5) $ \angle A\plus{}\angle B\equal{}180^\circ$
6) I can't remember...
And for SSA, you need to prove that the angles opposing the other sides are of same type (example, both acute)
[/hide]

[quote="Bugi"]A quadrilateral is a parallelogram iff
5) $ \angle A \plus{} \angle B \equal{} 180^\circ$
[/quote]Are you sure about this? It seems that it is sufficient but not necessary, e.g. trapezoid.

you would need to show that

$ \angle A \plus{} \angle B \equal{} 180^\circ$

and

$ \angle A \plus{} \angle C \equal{} 180^\circ$

this can show by the con. of same side interior thrm that the sides are parallel. You cant have just one or like 10000th user said, it would just be a trapazoid

---

I think ffao's beautiful applet (can u pm me on how you made that) shows that it can either be parallel or not, for now enough info is given...but knowing amie15, there probably is an answer

I mistaken. It can't be a proof(for angles).

[quote="AIME15"]You make it with geogebra.  When you're posintg a message, it's the last one of the tags.

Thanks ffao![/quote]

I mean how would u make the points be able to move?

All points that you independently choose with the "New Point" command are movable. If they depend on other points, then you can't move them directly.


haha i was just doing that problem today. i didn't get the solution they gave with that problem. does anyone have a better solution?

yeah I didn't get it either  :blush: . Somehow, it was one of the 5 I missed. pretty lame. I tried to think about it, but I couldn't understand the content.

what are the choices, it says closest hour (6750/5000)

[quote="now a ranger"]Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath he cuts is 28 inches wide, but he overlaps each cut by 4 inches to make sure that no grass is missed. He walks at the rate of 5000 feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?[/quote]

[hide]using this method, he will cut a bunch of 24 inch pieces, and one 28 inch one .  so we have $24x+28$.  So $24x+28=(90)(12)=1080$.

$24x=1052$
$x=43\frac{5}{6}$

So he needs to walk across the thing $43\frac{5}{6}+1\approx45$ times.  And hes walking a length of 150 feet.  So we have $\frac{(150)(45)}{5000}=1.35$[/hide]

is that right? i dont have an answer key.

I think this problem is poorly worded, but yeah the answer is 1.35 hours.