The incircle of triangle ABC touches BC, CA, AB at A', B', C'. Prove that
$ AB'^2 \plus{} BC'^2 \plus{} CA'^2 \geq AB'.B'C' \plus{} BC'.C'A' \plus{} CA'.A'B' \geq B'C'^2 \plus{} C'A'^2 \plus{} A'B'^2$

Here is another recent topic with the same problem, which was locked:
[url]http://artofproblemsolving.com/Forum/viewtopic.php?t=194080[/url]

[b]@ lalalala:[/b] If you post contest problems on here, you [i]will[/i] be caught. It's AoPS.

And if you label your post "original," everyone will wonder if it actually is. Note that the linked thread above was "New for all!"

and can someone post a solution?   :maybe:


geometry
circumcircle
geometric transformation
homothety
reflection
analytic geometry
trigonometry

Homothety with center $ A$ and coefficient $ 2$ sends the midline $ M_bM_c$ into the sideline $ BC,$ $ A_0$ into the reflection $ A_3$ of $ A$ across $ BC$ and $ A_1$ into the vertex $ A''$ of the tangential triangle $ \triangle A''B''C''$ of $ \triangle ABC.$ Let $ A_2'$ be the image of $ A_2$ and define similarly the points $ B_2',C_2'.$  We'll show that $ A_2',B_2',C_2'$ are collinear, which yields the collinearity of $ A_2,B_2,C_2.$  Let us use barycentric coordinates with respect to $ \triangle ABC$ at this point.

$ A_3( \minus{} a^2: a^2 \plus{} b^2 \minus{} c^2: a^2 \plus{} c^2 \minus{} b^2) \ , \ A''( \minus{} a^2: b^2: c^2).$ The line connecting $ A'',A_3$ intersects $ BC$ at $ A_2'(0: a^2 \minus{} c^2: a^2 \minus{} b^2).$ By cyclic exchange we get that $ B_2'(b^2 \minus{} c^2: 0: b^2 \minus{} a^2) \ , \ C_2'(c^2 \minus{} b^2: c^2 \minus{} a^2: 0)$ $ \Longrightarrow$ $ A_2',B_2',C_2'$ lie on the trilinear polar of $ X_{523} \ (b^2 \minus{} c^2: c^2 \minus{} a^2: a^2 \minus{} b^2),$ i.e. the isogonal conjugate of the Euler's reflection point of $ \triangle ABC.$ Our proof is completed then.

My proof.(nice my friend)
Let  $ M_{Ca},M_{Ba}$ are the intersections of $ M_CA_1, M_BA_1$ with BC.
Then, [b][$ A_2, B_2, C_2$ are collinear] [/b]
$ \frac {{\overline {A_2 M_C } }}{{\overline {A_2 M_B } }}.\frac {{\overline {C_2 M_B } }}{{\overline {C_2 M_A } }}.\frac {{\overline {B_2 M_A } }}{{\overline {B_2 M_C } }} \equal{} 1$

  $ \Leftrightarrow \frac {{\overline {A_1 M_{Ca} } }}{{\overline {A_1 M_{Ba} } }}.\frac {{\overline {C_1 M_{Bc} } }}{{\overline {C_1 M_{Ac} } }}.\frac {{\overline {B_1 M_{Ab} } }}{{\overline {B_1 M_{Cb} } }} \equal{} 1$

  $ \Leftrightarrow \frac {{ \minus{} \frac {h}{2}(\cos B \minus{} \cos A)}}{{\frac {h}{2}(\cos C \minus{} \cos A)}}.\frac {{ \minus{} \frac {h}{2}(\cos A \minus{} \cos C)}}{{\frac {h}{2}(\cos B \minus{} \cos C)}}.\frac {{ \minus{} \frac {h}{2}(\cos C \minus{} \cos B)}}{{\frac {h}{2}(\cos A \minus{} \cos B)}} \equal{} 1$

Denote the points like luisgeometria.Let $ A_4$ is the reflection of $ A''$ across $ BC$.Then $ A_4,A_2',A$ are collinear.
$ \vec{AA_4} \equal{} 2\vec{AM_a} \minus{} \vec{AA''} \equal{} \vec{AB} \plus{} \vec{AC} \minus{} \vec{AA''}$
Let the circle center $ A''$ pass through $ B,C$ cut $ AB,AC$ at the second point $ M,N$.Easy to prove that $ A'',M,N$ are collinear and two triangle $ ABC$ and $ ANM$ are similar.So:
$ \vec{AM} \equal{} \frac {AM}{AB}.\vec{AB} \equal{} \frac {MN.AC}{BC.AB}.\vec{AB} \equal{} \frac {2MC.AC}{2M_aC.AB} \equal{} \frac {AC}{cosA.AB}.\vec{AB}$.
Similarly $ \vec{AN} \equal{} \frac {AB}{cosA.AC}.\vec{AC}$
Then $ \vec{AA''} \equal{} \frac {AC}{2cosA.AB}.\vec{AB} \plus{} \frac {AB}{2cosA.AC}.\vec{AC}$
$ \rightarrow \vec{AA_4} \equal{} (1 \minus{} \frac {AC}{2cosA.AB}).\vec{AB} \plus{} (1 \minus{} \frac {AB}{2cosA.AC}).\vec{AC}$
So $ AA_4$ cut $ BC$ at $ A_2'$ satisfies:
$ \frac {\overline{A_2'B}}{\overline{A_2'C}} \equal{} \minus{} \frac {2cosA.AC.AB \minus{} AB^2}{2.cosA.AB.AC \minus{} AC^2} \equal{} \minus{} \frac {AC^2 \minus{} BC^2}{AB^2 \minus{} BC^2}$
Set two same equalites then $ A_2',B_2',C_2'$ are collinear by Menelaus's theorem.

PS:I don't know much about barycentric coordinates but I think my solution are somehow like luisgeometria's one ...

[quote="mathVNpro"]Let $ ABC$ be a triangle and $ M_b,$ $ M_c$ respectively are the midpoints of $ AC$ and $ AB$. Denote $ (O_a)$ by the circumcircle of $ \triangle AM_bM_c$. Let $ A_0$ and $ A_1$ respectively are the projection of $ A$ onto $ BC$ and the intersection of the tangents at $ M_b,$ $ M_c$ of $ (O_a)$. Let $ A_2\equiv A_0A_1 \cap M_bM_c$. Define the same for $ B_2,$ $ C_2$. Prove that $ A_2,$ $ B_2$ and $ C_2$ are collinear.

Best regard,
[i]mathVNpro[/i][/quote]

Need not to take any more points ... Just a little Melelauyt's Theorem...

[color=blue][b]Solution :
[/b]
[img]http://img693.imageshack.us/img693/7813/74650804.png[/img]

$ \frac {\overline{A_{2}M_{b}}}{\overline{A_{2}M_{c}}} \equal{} \frac {\sin\widehat{A_{1}A_{0}M_{b}}}{\sin\widehat{A_{1}A_{0}M_{c}}} \ . \ \frac {\overline{A_{0}M_{b}}}{\overline{A_{0}M_{c}}} \equal{} \frac {\overline{A_1M_b} \ . \ \frac {\sin\widehat{A_1M_bA_0}}{\overline{A_1A_0}} }{\overline{A_1M_c} \ . \ \frac {\sin\widehat{A_1M_cA_0}}{\overline{A_1A_0}} } \ . \ \frac {\overline{AM_b}}{\overline{AM_c}} \equal{} \frac {\sin(\widehat{C} \minus{} \widehat{A})}{\sin(\widehat{B} \minus{} \widehat{A})} \ . \ \frac {b}{c}$

Use the Melelauyt's theorem for $ \triangle M_aM_bM_c$ , we end the proof .[/color]

[quote="mathVNpro"][quote="TRAN THAI HUNG"]
Let  $ M_{Ca},M_{Ba}$ are the intersections of $ M_CA_0, M_BA_0$ with BC.[/quote]
Seems like your $ M_{Ca}\equiv M_{Ba}\equiv A_0$ !!?[/quote]
Sorry, it is typo. I have corrected it. :)

which is $ A0,A1$？, I can't understand what [b]mathVNpro[/b] say, can somebody make a figure？

[quote="hophinhan"]
$ \frac {\overline{A_{2}M_{b}}}{\overline{A_{2}M_{c}}} \equal{} \frac {\sin\widehat{A_{1}A_{0}M_{b}}}{\sin\widehat{A_{1}A_{0}M_{c}}} \ . \ \frac {\overline{A_{0}M_{b}}}{\overline{A_{0}M_{c}}} \equal{} \frac {\overline{A_1M_b} \ . \ \frac {\sin\widehat{A_1M_bA_0}}{\overline{A_1A_0}} }{\overline{A_1M_c} \ . \ \frac {\sin\widehat{A_1M_cA_0}}{\overline{A_1A_0}} } \ . \ \frac {\overline{AM_b}}{\overline{AM_c}} \equal{} \frac {\sin(\widehat{C} \minus{} \widehat{A})}{\sin(\widehat{B} \minus{} \widehat{A})} \ . \ \frac {b}{c}$
[/quote]
[quote="livetolove212"]I found some interesting results:
[b]1.[/b] The line $ (A_2,B_2,C_2)$ passes through centroid $ G$ of triangle $ ABC$.
[b]2.[/b] Let $ X_1, X_2$ be the intersections of $ A_1M_b, A_1M_c$ and $ BC$, similarly we define $ Y_1, Y_2, Z_1,Z_2$. Then $ X_1Y_1Z_1$ and $ X_2Y_2Z_2$ are pedal triangles of two point $ P,Q$, respectively and $ HPOQ$ is a parallelogram. ($ H,O$ are the orthocenter and circumcenter of triangle ABC).[/quote]

[color=blue]Very nice results Linh  :) , this is my proof for (1)...check plz

[b]Solution :[/b]

[img]http://img101.imageshack.us/img101/4712/10779065.png[/img]

It's easy to see that : the line $ \overline{A_2B_2C_2}$ passes through centroid $ G$ of $ \triangle ABC$  ( also is the centroid of $ \triangle M_aM_bM_c$ ) $ \Longleftrightarrow \frac {\overline{M_cM_b}}{\overline{A_2M_c}} \plus{} \frac {\overline{M_cM_a}}{\overline{B_2M_c}} \equal{} 3$ $ (*)$

Because :$ \frac {\overline{A_{2}M_{b}}}{\overline{A_{2}M_{c}}} \equal{} \frac {\sin(C \minus{} A)}{\sin(B \minus{} A)}.\frac {\sin B}{\sin C}$

$ \Longrightarrow \frac {\overline{M_cM_b}}{\overline{A_2M_c}} \equal{} 1 \plus{} \frac {\sin(C \minus{} A).\sin B}{\sin(B \minus{} A).\sin C}$ .Similarly,  $ \frac {\overline{M_cM_a}}{\overline{B_2M_c}} \equal{} 1 \plus{} \frac {\sin(C \minus{} B).\sin A}{\sin(A \minus{} B).\sin C}$

$ (*)$  $ \Longleftrightarrow 2 \plus{} \frac {\sin(C \minus{} A).\sin B}{\sin(B \minus{} A).\sin C} \plus{} \frac {\sin(C \minus{} B).\sin A}{\sin(A \minus{} B).\sin C} \equal{} 3$

Which is true because :
 $ \sin(C \minus{} A). \sin B \minus{} \sin(C \minus{} B).\sin A$

$ \equal{}(\sin C . \cos A \minus{} \cos C . \sin A). \sin B \minus{} (\sin C . \cos B \minus{} \cos C . \sin B). \sin A \equal{}$

$ \equal{} \sin C.(\sin B. \cos A \minus{} \sin A. \cos B) \equal{} \sin C .\sin(B \minus{} A)$

[b]Done !!![/b]

2h17 AM[/color]

[quote="livetolove212"][b]1.[/b] The line $ (A_2,B_2,C_2)$ passes through centroid $ G$ of triangle $ ABC.$[/quote]
Nice remark dear Linh, but there's more to say about this line. The subject line $ \tau \equiv A_2B_2C_2$ passes through the centroid $ G$ and symmedian point $ K$ of $ \triangle ABC,$ i.e the Retrocenter of $ \triangle M_aM_bM_c.$ Keeping the same trend of my previous post, we have

$ A_2(b^2 \plus{} c^2 \minus{} 2a^2: c^2 \minus{} a^2: b^2 \minus{} a^2) \ , B_2 (c^2 \minus{} b^2: c^2 \plus{} a^2 \minus{} 2b^2: a^2 \minus{} b^2)$

$C_2  (b^2 \minus{} c^2: a^2 \minus{} c^2: a^2 \plus{} b^2 \minus{} 2c^2)$

$ \Longrightarrow \tau \equiv (b^2 \minus{} c^2)x \plus{} (c^2 \minus{} a^2)y \plus{} (a^2 \minus{} b^2)z \equal{} 0$

This line clearly contains $ G \ (1: 1: 1)$ and $ K \ (a^2: b^2: c^2)$

Therefore, $ \tau$ is also the trilinear polar of the Steiner point $ X_{99}$ of $ \triangle ABC.$

Let $ ABC$ be a triangle and $ M_b,$ $ M_c$ respectively are the midpoints of $ AC$ and $ AB$. Denote by $ (O_a)$  the circumcircle of $ \triangle AM_bM_c$. Let $ A_0$ and $ A_1$ respectively are the projection of $ A$ onto $ BC$ and the intersection of the tangents at $ M_b,$ $ M_c$ of $ (O_a)$. Let $ S$ be a arbitrary point lies on the [b]Euler line [/b]of triangle $ ABC$.Denote points $ A' \equal{} AS\cap BC,B' \equal{} BS\cap CA,C' \equal{} CS\cap AB$. Let $ A_2 \equal{} A_0A_1\cap B'C'$.Define the same for $ B_2,C_2$.
Prove that $ A_2,B_2,C_2$ lie on a straight line $ d$.
 :)

[quote=mathVNpro]Let $ ABC$ be a triangle and $ M_b,$ $ M_c$ respectively are the midpoints of $ AC$ and $ AB$. Denote $ (O_a)$ by the circumcircle of $ \triangle AM_bM_c$. Let $ A_0$ and $ A_1$ respectively are the projection of $ A$ onto $ BC$ and the intersection of the tangents at $ M_b,$ $ M_c$ of $ (O_a)$. Let $ A_2\equiv A_0A_1 \cap M_bM_c$. Define the same for $ B_2,$ $ C_2$. Prove that $ A_2,$ $ B_2$ and $ C_2$ are collinear.

Best regard,
[i]mathVNpro[/i][/quote]

[quote=livetolove212]
[b]1.[/b] The line $ (A_2,B_2,C_2)$ passes through centroid $ G$ of triangle $ ABC$.
[/quote]

[b]Lemma :[/b] Given a $ \triangle ABC $ and a point $ P $. Let $ Q $ be the isotomcomplement of $ P $ WRT $ \triangle ABC $ and $ \tau_Q $ be the trilinear polar of $ Q $ WRT $ \triangle ABC $. Then the Newton line of the complete qual $ \{ \triangle ABC, \tau _Q \} $ is the trilinear polar $ \tau_P $ of $ P $ WRT $ \triangle ABC $.

[b]Proof :[/b] Let $ \triangle P_AP_BP_C $ be the cevian triangle of $ P $ WRT $ \triangle ABC $. Let $ \tau_P $ cuts $ BC, $ $ CA, $ $ AB $ at $ P_a, $ $ P_b, $ $ P_c $, resp and $ \tau_Q $ cuts $ BC, $ $ CA, $ $ AB $ at $ Q_a, $ $ Q_b, $ $ Q_c $, resp.  Let $ M_a, $ $ M_b, $ $ M_c $ be the midpoint of $ AQ_a, $ $ BQ_b, $ $ CQ_c $, resp. Since the anticevian triangle of $ Q $ WRT $ \triangle ABC $ (formed by $ AQ_a, $ $ BQ_b, $ $ CQ_c $) and $ \triangle P_AP_BP_C $ are homothetic (well-known), so we get $$  -1 = (A,B;P_B,P_c) = (A,C;P_C,P_b) = (A,Q_a;\infty, M_a) \Longrightarrow M_a \in P_bP_c \equiv \tau_P .$$ Similarly, we can prove $ M_b $ and $ M_c $ lie on $ \tau_P $, so $ \tau_P $ is the Newton line of the complete quadrilateral $ \{ \triangle ABC, \tau_Q \} $.
____________________________________________________________
[b]Back to the main problem :[/b]

Let $ A_3 $ be the point such that $ \triangle A_3BC $ $ \stackrel{+}{\sim} $ $ \triangle A_1M_bM_c $ and let $ \triangle X_aY_aZ_a $ be the cevian triangle of $ A_3 $ WRT $ \triangle ABC $. Since $ A_1 $ is the complement of $ A_3 $ WRT $ \triangle ABC $, so notice $ A_0, $ $ M_a $ are symmetry WRT the perpendicular bisector of $ M_bM_c $ we get $ M_bA_2 $ $ : $ $ M_cA_2 $ $ = $ $ CX_a $ $ : $ $ BX_a $ $ \Longrightarrow $ $ A, $ $ A_2, $ $ A_3, $ $ X_a $ are collinear. On the other hand, from $ \measuredangle CBA_3 $ $ = $ $ \measuredangle M_cM_bA_1 $ $ = $ $ \measuredangle BAC $ we get $ \odot (ABY_a) $ is tangent to $ BC $ at $ B $. Similarly, we can prove $ \odot (ACZ_a) $ is tangent to $ BC $ at $ C $, so from the problem [url=http://artofproblemsolving.com/community/c6h624551]AF perpendicular to Euler line[/url] get $ AX $ is perpendicular to the Euler line $ \mathcal{E} $ of $ \triangle ABC $ where $ X $ $ \equiv $ $ Y_aZ_a $ $ \cap $ $ BC $ $ \Longrightarrow $ $ X_a $ lies on the trilinear polar $ \epsilon $ of the infinity point with direction $ \perp $ $ \mathcal{E} $ WRT $ \triangle ABC $, hence notice $ A_2 $ is the midpoint of $ AX_a $ we conclude that $ A_2 $ lies on the Newton line $ \mathcal{N} $ of the complete quadrilateral $ \{ \triangle ABC, \epsilon \} $.

Analogously, we can prove $ B_2 $ and $ C_2 $ lie on $ \mathcal{N} $, so combine the lemma we conclude that $ A_2, $ $ B_2, $ $ C_2 $ lie on the trilinear polar of the Steiner point of $ \triangle ABC $ WRT $ \triangle ABC $ which passes through the Centroid, Symmedian point of $ \triangle ABC $.


geometry
circle
Locus
Locus problems
IMO Longlist
IMO Shortlist

In the plane, consider a circle with center $S$ and radius $1.$ Let $ABC$ be an arbitrary triangle having this circle as its incircle, and assume that $SA\leq SB\leq SC.$ Find the locus of

[b]a.)[/b] all vertices $A$ of such triangles;

[b]b.)[/b] all vertices $B$ of such triangles;

[b]c.)[/b] all vertices $C$ of such triangles.

The triangle $ABC$ is uniquely determined by the acute triangle $XYZ$ inscribed in $(S)$, where $X,Y,Z$ are the points where $BC,CA,AB$ touch $(S)$. Now let $M,N,P$ be the midpoints of $YZ,ZX,XY$ respectively. We have $SM\cdot SA=SN\cdot SB=SP\cdot SC=1$, so the ordering of $SA,SB,SC$ implies $SM\ge SN\ge SP$. If we find the loci of $M,N,P$, then the loci of $A,B,C$ will be the images of those (respectively) through the inversion wrt $(S)$.

The condition $SM\ge SN\ge SP$ says that, in the acute triangle $XYZ$, we have $\angle X\le \angle Y\le \angle Z$.

We can find acute triangles of the form $\angle X<\angle Y=\angle Z<\frac \pi 2$ with $\angle X$ taking any value in the interval $(0,\frac \pi 3]$, so $M$ is a the midpoint of a chord subtending an angle of no more that $\frac \pi 3$, which means that the locus of $M$ is the region $\frac 12\le SM<1$, so the locus of $A$ is the region $1<SA\le 2$.

Regarding $\angle Z$, this is the largest angle, so it must be in the interval $[\frac \pi 3,\frac \pi 2)$. We can easily find triangles with their largest angle taking any of these values, so the locus of $P$ is the region $0<SP\le \frac 12$, meaning that the locus of $C$ is $2\le SC$.

$\angle Y$ is the middle angle, and we have to find bounds for it. It can take values as close to $\frac \pi 2$ as we want (we take isosceles triangles s.t. $\angle Y=$ one of the other angles), but it must be $>\frac \pi 4$, because otherwise the largest angle would be $\ge \frac \pi 2$. The  same reasoning shows that it can take values as close to $\frac \pi 4$ as we want, so the locus of $N$ is $0<SN<\frac{\sqrt 2}2$, so the locus of $B$ is $\sqrt 2<SB$.


Μήπως θεωρείτε  χρήσιμο  για σας αλλά και για τους φοιτητές που θα ακολουθήσουν , να φτιάξω ένα στικυ για να συγκεντρώνετε όλα αυτά τα ωραία θέματα ;

 Αν σας αρέσει , να ανοίξω ένα με τον τίτλο :

  ΑΕΙ - ΑΣΚΗΣΕΙΣ ΓΙΑ ΦΟΙΤΗΤΕΣ

 ή κάπως έτσι ή όπως μου προτείνετε.

[color=red][b] Καλή επιτυχία στις εξετάσεις σας αυτή την περίοδο, αν και κάποιοι τελείωσαν ήδη ! [/b][/color]

Πολύ καλή ιδέα ένα τέτοιο sticky, νομίζω θα είναι πολύ χρήσιμο!

Cheers,

Παναγιώτης

Συμφωνώ απόλυτα αν και δεν είμαι φοιτητής πολλά χρόνια τώρα
Ρ.Μπόρης


What pins did you guys get from Nationals?

Im missing Alaska and Mississippi but I am not sure if I am missing more.
If you have the puerto rico pin please tell me(even though almost everyone has it).

I have Puerto Rico (wooden frog= pwnage)!!!

[quote="dingzhou"]i got all of them!!!!!!!![/quote]

How? I thought Virginia wasn't trading them. We (alabama) were going to nottrade so no one would get all 57, but we decided to be nice and trade, although we didnt much. Some girl told me that Alabama's were really rare, and people were like harassing us....

virginia traded pins, i have a few to prove it.
i have all of them, except delaware (did not have pins)
dingzhou had all of them (i can confirm this) including delaware; he bartered off about four of his pins for a delaware pin from last year.

I am not sure if I am missing more.

[url=http://www.math123xyz.com]Interactive Math Resources[/url]

I got about 35 out of my 37 pins from dinner Thursday. Everyone has about two hours to socialize. :D I just decided to trade pins.

I think i got everything besides

virginia
michigan
mississippi
delaware
arkansas

i didn't get any lol.

not even virginia =P.  we gave all our pins to one of our team members to trade

[quote="distracted523"][quote="tonypr"]What pins did you guys get from Nationals?

Im missing Alaska and Mississippi but[b] I am not sure if I am missing more.[/b]
If you have the puerto rico pin please tell me(even though almost everyone has it).[/quote]

On another note, how do you guys get so many pins? I've never been to nats, so I don't know how it works.[/quote]

you just go to a random person and say, where you from? and they answer i'm from blah...and you either go, hey i'll trade you blah or blah or blah ( whatever you don't need) for your blah pin. and they either go, sorry, i have your blah and blah and blah...pins already or hmm..i don't have blah pin, i'll trade you this one for it...and then you go to the next guy....it gets annoying when you ask the same guy for like the5th time.

Yep, dingzhou sums it up pretty well.

Me and one of my teammates were confronted by the same person 8 times the first day.  :)

I think we got all of them except for like Arkansas and Delaware. On Thursday, we set up a "trading post" on one of the tables right next to the entrance room.

Ah, I started off with 50 Hawaii pins, and ended with about 100 pins total, most of them duplicates, mainly because I also traded about 40 unused and shiny New Mexico quarters from the mint in Denver.

hey what did you guys do with your pins? did you all pin it on the mc hats?

Aw shucks, Andersonw - it's only a pin! If that poor guy or gal wanted your state so badly that he/she came up to you 8 times, couldn't one of your team just Give them your state pin (even if they had nothing you wanted?) -- I mean, towards the end, a few people asked me for some pins I had, and I really didn't need what they were offering, so I just gave them to them (or I traded for a pin I didn't actually need). You would not BELIEVE how happy a little thing like a 10 cent pin (or whatever) made them - and after all, why would anyone need more than one of any state? Technically, if everyone was given 57 pins to trade, I don't get how my whole team occasionally got "no, sorry" from a whole 'nother team -- I mean, if they didn't get the pins of our state from us, where did they get them? (assuming we only traded one to one with other states and therefore other states didn't have extras of ours to trade with) -- does this make sense?

but wish I saw your trading table - it would have made trading easier.. :blush:

I didn't get like any of the rare ones b/c I was to stingy to give away more than 1 pin for 1 pin.  Therefore I know I didn't get Arkansas, South Carolina, Korea, and maybe Delaware but I had 1 from last year.

KYMATHMAN -- Korea?? Was that from someone on the State Dept. of Defense? Last I checked Korea wasn't one of our states or territories lol

yeah, the korea pins were given out by a girl from the state department.

i think i got a pin from hawaii for only one pin : )

Speaking of the Hawaii pins...

specifically to 12markkram34 -- frankly, I don't think what you guys did was very nice - or in the spirit of MATHCOUNTS - yes, I could understand if your coach lost the pins or something and everyone on your team only got 10 pins, but I think you said you had about 50!  Do you realize by "forcing people" to trade 2:1 or worse, especially members who only got 57 or less pins to trade, probably made them go home with less than 57 pins while you guys "proudly" got 100 or more .. I understand about negotiating, but personally I don't think I would feel very good about insisting someone give me 2 pins every time I traded!  I think doing so likely frustrated many of your fellow MATHCOUNTERS. What do others think?

Ok, sorry about that. One of our teammates who was at nats last year wanted to try to beat his record last year. So, he started trading with people right after lunch on the first day at the rate of two to one. So, he had ended up setting up a very bad sort of "market price" for the pins. Sorry about that. I don't see why he didnt just trade quarters like I did. About 35 of the extra 50 pins I ended with was from the quarters.

So once again, sorry about that and I hope that you accept our sincere apologies  :) .

the state department pins this year were: the standard pin (the eagle thingy), a US flag, or a pin with two flags connected together, a taiwan pin, a korea pin, a phillipines pin, and one person on the team was giving pens from singapore

I get offering a trade of "something else" like a quarter or a cool general pin or a pen, etc, IF the "other person" didn't want your state's pin and you needed what they had - but since it sounds like most wanted a pin from Hawaii, why not just trade a pin for a pin?

Ok, let me clear up some confusion. Although we did occasionally offer to trade one pin for two or a quarter for one, we didn't annoy people or hawk them. If they didn't want to trade one for two or for a quarter, we just said "Thanks anyways" and left. We never forced, persuaded, or begged people to trade with us.

Once more, sorry about that, but know what we weren't harassing for forcing anyone to trade with us.

you're right - "forcing" is a poor choice of word and if I said or implied that, I'm sorry - it's not what I meant to say. I just meant that I think you are missing the basic point -- which is this: Most other states and competitors were happy "following generally accepted trading practices" if you will, and merely wanted to give you one of their state's pins for one of yours. When you guys refused to do that and asked if they would give a pin for a quarter, for instance, you were getting one of Their pins, which you presumably wanted, but were denying that person the opportunity to get one of yours -- or at least, to obtain one in a fair and equal "one to one" trade -- by you saying to someone (indirectly or directly) - "if you want one of my state's pins, you have to give me multiple pins" -- "take it or leave it" -- in a sense you are "forcing that person's hand" and if they Really want to collect all the states (yours included) then you sort of leave them no choice but to "lose" - ie: go home with an incomplete pin collection (no Hawaii) or to offer you more pins, in which case, unless they were given More than 57 pins (and most weren't!!) - they Still lose in that they got Hawaii, but now don't have enough pins to trade with some other state(s) -- uh, do you get what I am saying? I don't know what grades your team is in, but all I'm suggesting is if you're lucky enough to come back next yr, you may want to rethink how you trade - that's all. Put it another way - you REALLY want State X but all you have is one Hawaii pin left and no extras of anything else, and you ask X to trade and they say "sorry, but we'll only trade you one X for 2 or 3 Hawaii's (which you don't have!!) - how would you feel??

Uhhhh...this isn't getting us anywhere useful. I'd like to say sorry once again, and I hope that we can simply not make bad mistakes like this in the future.

Truce?

ok - no prob -- maybe other "new nats" participants who read this post will learn something for next yr ... Summary: For fairest trades, just stick to 1 for 1 .. and all (except for the most greedy -  :o ) will be happy! Yes?

Ah, glad that we could finally come to an agreement.

I did.  Hmm, it appears that I got way too many Puerto Rico's cuz like when I asked if any people had it, like a line of Puerto Ricans showed up and started offering me their pins.  After I traded with the first person in line and said, thanks, good luck at the competition, the rest of the people were still standing their looking questioningly at me for a GA pin.  Ugh, awkard...

[quote="PI-Dimension"][quote="dnkywin"][quote="tinytim"]I had like 35 pins but then someone from my team or NC stole like 20 of them when I wasn't in my room. I know who did it but I will not say because that will be too antagonizing.

To the person who stole it(I know you have an aops account!!!):
[hide="If you look at this then you are admitting that you stole the pins"]I know you did it although you might think I don't and now I know that you know that I know you did it so live with the guilt FOREVER!!!(Or you can buy an airplane ticket to St. Louis and give them to me. Either way works.)[/hide][/quote]
*points to PI-Dimension* :whistling:[/quote]

*points to mewto55555*  :o

POINTING IS NOT POLITE

pwnd[/quote]

Points to...Nick

I didn't get that many pins: I was at nationals last year and the pins were pretty much the same. Or at least, the ones I traded for were. There were quite a few unique ones though, I think. I don't recall the Las Vegas one from last year.

yeah, i have 3 of some pins cuz they're the same every yr

Yeah, the Wyoming pin has been pretty much the same every year, save for a few color/size changes.

Mssiouri had the lamest pins ever. The pin didn't even say missouri on it! Unless you spend your time memorizing the shapes of every state, you couldn't even tell that it was a missouri pin!

[quote="mewto55555"][/quote][quote="PI-Dimension"][quote="dnkywin"][quote="tinytim"]I had like 35 pins but then someone from my team or NC stole like 20 of them when I wasn't in my room. I know who did it but I will not say because that will be too antagonizing.

To the person who stole it(I know you have an aops account!!!):
[hide="If you look at this then you are admitting that you stole the pins"]I know you did it although you might think I don't and now I know that you know that I know you did it so live with the guilt FOREVER!!!(Or you can buy an airplane ticket to St. Louis and give them to me. Either way works.)[/hide][/quote]
*points to PI-Dimension* :whistling:[/quote]

*points to mewto55555*  :o

POINTING IS NOT POLITE

pwnd[/quote][quote="mewto55555"]

Points to...Nick[/quote]

lol

@tinytim: The new one (the bronze won with S&L on it or something) did...

Hey, there are many pins that were different from last year.

Hey, nebraska got red text this year, no longer white, and no longer looks like pizza hut!
Some pins that were mass-gven away last year were pretty rare this year too.


You're on the right path.  Now re-write this as $ e^{3}\equal{}\frac{x}{x\minus{}1}$.  So multiply both sides by the denominator so we have $ e^{3}x\minus{}e^{3}\equal{}x$.  Rearrange it so $ (e^{3}\minus{}1)x\equal{}e^{3}$ and finally, $ \boxed{x\equal{}\frac{e^{3}}{e^{3}\minus{}1}}$.

oh i see! Thanks i figured out what i was not realizing.

Thanks!!

[quote="gauss1181"][quote="orangedude654"]Solve for x

ln(x)-ln(x-1)=3

So, I know how to do one of these problem, but usually at the end you can factor out an x. But i can't seem to do it on this one.

ln(x/x-1)=3 (by property of logs)

e raised to both sides

(x/x-1)=e^3

Then i get stuck???[/quote]
You then should do x=(e^3)x-e^3, so e^3=(e^3-1)x and x=e^3/(e^3-1)

p.s. do you need to download LaTeX to use it, or do you just need to enter tags/codes in your post?[/quote]

Just enter tags. $ \frac{\text{Like}}{\text{this.}}$ ;)

just enclose the latex in dollar signs

[code]$\frac{v^2}{r}$[/code]
yields

$ \frac {v^2}{r}$

[quote="gauss1181"]
You then should do x=(e^3)x-e^3, so e^3=(e^3-1)x and x=e^3/(e^3-1)

p.s. do you need to download LaTeX to use it, or do you just need to enter tags/codes in your post?[/quote]From your formula, that's not written in $ \text{\LaTeX}$ typesetting operators (formats).
Try our AOPS LaTeX help here---> [url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]


function
trigonometry
floor function
calculus
calculus computations

The second derivatice of a function is given by f''(x) = 0.5 + cosx - e^(-x). How many points of inflection does the function f(x) have on the interval 0<= x <= 20?

$ 0\equal{}.5\plus{}\cos{x}\minus{}e^{\minus{}x}\Rightarrow .5\plus{}\cos{x}\equal{}\frac{1}{e^{x}}$  Now find the solutions and perform the test.

I have no idea.  We can use a Taylor expansion to approximate solutions but I'm not sure how much simpler that will make the problem.

Well, this is quite easy. Since we've established then all we need to do is find the number of roots of:

$ \frac{1}{2}\plus{}\cos x\equal{}e^{\minus{}x}\, ,\qquad 0<x<20$

then we can do the following. We know that $ 0<e^{\minus{}x}\leq1$ for all positive $ x$ so it intersects with $ \frac{1}{2}\plus{}\cos x$ the same number of times $ \cos x$ intersects with $ 0$ on the same interval. Now, also $ \left\lfloor \frac{20}{\pi}\right\rfloor \equal{} 6$ so that's it! :)

[quote="milin"]We know that $ 0 < e^{ \minus{} x}\leq1$ for all positive $ x$ so it intersects with $ \frac {1}{2} \plus{} \cos x$ the same number of times $ \cos x$ intersects with $ 0$ on the same interval. [/quote]
We know that $ 0 < \frac 12(1\plus{}0.3\sin 20000000x)\leq1$ for all positive $ x$ so it intersects with $ \frac {1}{2} \plus{} \cos x$ the same number of times $ \cos x$ intersects with $ 0$ on the same interval. (i.e., a few million times :P) ;)


Hello,

If $ a\equal{}55^{7777777}$

which of this proposal is wrong , with proof:

1) $ 7|a$
2) $ 3\nmid a$
3) $ 21 |a$

[quote="Johan Gunardi"]1 is wrong, 2 is true, 3 is wrong.[/quote]

With proof Johan Gunardi  :lol:

I'll left to other users. Anyway, I think this topic should be moved to High School Basic.

[hide="This is really basic"]
The exponent is not important. $ 3\nmid55,7\nmid55$, so 1) and 3) are wrong.[/hide]


What is the units digit of $ (133^{13})^{3}$?

We only care about the units digits, so this is really asking for the units digit of $ (3^{13})^3 \equal{} 3^{39}$.

Note that we have:

$ 3^1 \equal{} 3$
$ 3^2 \equal{} 9$
$ 3^3 \equal{} 7$ (units digit)
$ 3^4 \equal{} 1$
$ 3^5 \equal{} 3$
$ 3^6 \equal{} 9$
etc.

Since $ 39 \equiv 3 \pmod 4$, we have $ \boxed{7}$.


Hi,

I really need to get this problem solved:

sin3.14X + cos3.14X = 0,     0 =< x =< 2,   what are the values of x?

ps: sorry for the english, however math is international, thank you.

Come on Gabriel(fr), we already have: $\tan{3.14x}=-1$...

I'll try $\pi = 3.14$.    ;)

$\sin(\pi x) + \cos(\pi x) = 0$, $0 \leq x \leq 2$, what are the values of $x$?

$\sin(\pi x) + \cos(\pi x) = 0$
$\sin(\pi x) + \sin(\frac{\pi}2 - \pi x) = 0$
$2\sin\left(\frac{\pi x + \frac{\pi}2 - \pi x}{2}\right)\cos\left(\frac{\pi x - \frac{\pi}2 + \pi x}{2}\right) = 0$
$\sin(\frac{\pi}{4})\cos\left(\pi x - \frac{\pi}{4}\right) = 0$
$\cos\left(\pi x - \frac{\pi}{4}\right) = 0$
$\pi x - \frac{\pi}{4} = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z})$
$x - \frac14 = \frac12 + k$
$x = \frac34 + k$
It's possible for $k = 0,1$, then $x = \frac34$ or $x = \frac74$.


algebra function domain calculus integration parameterization

what are the possible values a, b, c and d?
any short cut for this type of problem:

$ a+b+c+d = 10$
$ 2b+5c+6d = 10$

[quote="redsmartie"]Where did you find this problem. Maybe understanding that might solve it.
Normally, with 4 variables and only 2 equations, it's either impossible or very hard to solve.[/quote]
That when you want to solve over a field. He is asking about over a domain, i.e. the integers. So there is a finite number of solutions.

[quote="i_like_pie"]I assume you want integers?[/quote]

yes. all possible integral values of a, b, c and d.

[quote="redsmartie"]Where did you find this problem. Maybe understanding that might solve it.
Normally, with 4 variables and only 2 equations, it's either impossible or very hard to solve.[/quote]

Actually it's not the original problem. It came from finding integer solutions of an equation.

Here's the original question:
Find the coefficient of $ x^{16}$ of $ (1+x^{2}+x^{5}+x^{6})^{10}$.

It turned out that in general the the number of solutions in general, is $ \frac{10!}{a!b!c!d!}{1^{a}x^{2b}x^{5c}x^{6d}}$ where

Of course, from combinatorics, we know that
$ a+b+c+d = 10$ and
$ 2b+5c+6d = 16$.

Im just wondering if there are short cuts to obtain the integral solutions of the system stated above except by fixing the value of $ a$ and solving $ c$ and $ d$.

For the details of the discussion click the link below:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159636

Im not sure if this question is appropriate here in this section, it could be on PreOlympliad level.

in integers, there are really only six different cases to check....

I will post the explanation at the other link.

[hide="hint"]Pretend you know the values of $ a$ and $ b$ and solve for $ c$ and $ d$[/hide]

[hide="my answer"]Since we have 2 equations and 4 unknowns, it'll turn out we'll have an answer in terms of 4-2=2 parameters.
Calling these parameters $ m$ and $ n$ (IE we can fill in any value for $ m$ and $ n$ we want)
I got the solutions to be:

$ a=m$

$ b=n$

$ c=50-6m-4n$

$ d=-40+5m+3n$

If you want answers where $ a,b,c,d$ are all positive, from the second equation $ c$ is even. also $ 5c \leq 10$ so $ c=0,2$

This quickly yields the only solutions as

$ a=8, b=0, c=2, d=0$

$ a=7, b=2, c=0, d=1$

$ a=5, b=5, c=0, d=0$
[/hide]

EDIT: In your second equation should it be 16? I solved for 10.


Hi all.
In Sweden there's an annual math competition for 7-9th graders. The qualification round is being held this Tuesday. Before each qualification round, a so called "poster problem" is posted on their website, and this problem I'd like to ask about.

The problem text is as follows (translated):
[i]The labyrinth below [linked below] consists of a number of quadratic and triangular chambers, with paths between them. Some of the chambers are blocked entirely. How should one walk through the labyrinth if one is to pass each chamber exactly once, if the walk starts and ends in the same chamber?[/i]

The labyrinth is found here: http://img103.imageshack.us/img103/1320/hmtposter1ji0.gif

So, question is: what are the possible ways of approaching this problem? I realise you must start by looking at the chambers you can only access in one way (such as corners, the triangular chambers), but you can only do that to an extent. When you've finished solving that, can you do anything but brute force it?

Whew!! That looks almost impossible! Doing brute force would take ages, but I don't think there is any other way. Maybe break it down into sections. I'll try to work on it.
 I'm sorry if this completely unhelpful. I don't think there is a clever way, and if there is I don't know what it is.

 You live in Sweden? Cool!! :D


[hide] By definition, the limit is $e$ which is approximately 2.718281828.[/hide]

the taylor series for $e^{x}$ is:
$\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$.
 :huuh:

[quote="t0rajir0u"][quote="bigboy82892"]Prove something about $\sum_{n=0}^{x}\frac{1}{n!}$ for any natural number $x\ge1$, if possible. [/quote]

Quite an open-ended question you've got there.

How about this:  It is integer for finitely many $x$.   :D[/quote]

That's not too hard, considering the fact that it
[hide="is a beast"] converges. [/hide]

Oh, come on, at least make it interesting:

Show that

$\lim_{n \to \infty}\left( 1+\frac{1}{n}\right)^{n}= \lim_{n \to \infty}\sum_{k=0}^{n}\frac{1}{k!}$

[quote="bigboy82892"]Prove something about $\sum_{n=0}^{x}\frac{1}{n!}$ for any natural number $x\ge1$, if possible. (Like an equation or a fact or something.)
Also, find the sum if $\lim_{x\rightarrow\infty}$.[/quote]
I said PROVE SOMETHING, NOT JUST TELLING ME!!! :huuh:

$\sum_{n=0}^{x}\frac{1}{n!}$

[hide="Proof that this is positive"]Just kidding

But a proof of what t0rajir0u suggested can be found [url=http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Equivalence_of_characterizations_1_and_2]here[/url]

[/hide]

[quote="i_like_pie"][quote="bigboy82892"]Prove something about $\sum_{n=0}^{x}\frac{1}{n!}$ for any natural number $x\ge1$, if possible. (Like an equation or a fact or something.)
Also, find the sum if $\lim_{x\rightarrow\infty}$.[/quote]
[hide]$e=\lim_{x\rightarrow\infty}\left(\sum_{n=0}^{x}\frac{1}{n!}\right)$[/hide][/quote]
I meant something like proveing what Ilikepie says here. PROVE IT. [b][i][u]PROVE IT!!!!![/u][/i][/b]Ok? Thanks.

Well, that is a definition of $e$, so that is the proof.  That's like saying "prove $x^{4}=x\times x\times x\times x$."  That's the definition of it, so it's automatically true.  That's why t0rajir0u suggested proving the equivalence of two different ways to define $e$...

[quote="bigboy82892"]
 PROVE IT. [b][i][u]PROVE IT!!!!![/u][/i][/b]Ok? Thanks.[/quote]

$f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(c)\cdot (x-c)^{n}}{n!}$
is the taylor series, centered at $c$ for any given function. if $f(x)=e^{x}$ and c=0, you get $f(x)=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

[quote="bigboy82892"]Prove something about $\sum_{n=0}^{x}\frac{1}{n!}$ for any natural number $x\ge1$, if possible. (Like an equation or a fact or something.)
Also, find the sum if $\lim_{x\rightarrow\infty}$.[/quote]

Can we say that it becomes $e$?

[hide="solution"]Its positive: since x is a natural number, and 0! is 1, then it is positive.[/hide]

Ok. Prove that $(1+\frac{1}{n\rightarrow\infty})^{n\rightarrow\infty}=\sum_{n=0}^{\infty}\frac{1}{n!}$ without using $e$, definition $e$, or postulates related to $e$ (exception: this equation).
EDIT: Oh yeah, and you can say $(1+\frac{1}{n\rightarrow\infty})^{n\rightarrow\infty}=e$, you have to prove the second part is also equal.

The Zuton Force has already provided a link to the usual proof.

[quote="t0rajir0u"][quote="t0rajir0u"]Let's pose a question then:

When written as a single fraction in lowest terms, when is the denominator $x!$?[/quote]

Someone want to answer my question, then?    :D[/quote]

Always?

Really?

But

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}= \frac{8}{3}$, and the denominator is not $6$ here.

Edit:  Hmm, it's an interesting question.  It's true for the first few even values, and trivially false for the odds (since the numerator when $x!$ is the denominator would be even).

Doesn't work for odd n, since the numerator and denominator will share a factor of 2. Works for even n though, I'm thinking of why and am guessing it might have to do with continued-fractions.

No, the convergents of $e$ do not have denominators resembling $x!$ at all, sorry.    :(

Are you sure it works for all even $n$?  I doubt it does.

Yes.  But just because it works for $n = 0, 2, 4, 6, 8$ (which are the cases that I have checked so far) doesn't mean that it works for all even $n$.

Doesn't work for 12, 14 and others  :maybe:


Consider the cascading arrangement of tanks shown, with $V_{1}= 200 \; gal$, $V_{2}= 200 \; gal$, and $V_{3}= 500 \; gal$ as the volumes of brine in the three tanks. Each tank initially contains $20 \; lb$ of salt. The inflow rates and outflow rates for tanks $1$ and $2$ are all $5 \frac{gal}{sec}$, but the outflow rate for tank $3$ is $10 \frac{gal}{sec}$. The flow into tanks $1$ and $2$ is pure water.

http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=7262

(a) Let $x(t)$, $y(t)$, and $z(t)$ be the amounts of salt in tanks $1$, $2$, and $3$, respectively. Set up the differential equation and the initial conditions that describe how the amounts of salt in tanks 1 and 2 are changing at any time $t$.

$\frac{dx}{dt}= 0-\left(5 \frac{gal}{sec}\right)\left(\frac{x}{200}\frac{lbs}{gal}\right)$

$\frac{dy}{dt}= 0-\left(5 \frac{gal}{sec}\right)\left(\frac{y}{200}\frac{lbs}{gal}\right)$

$x(0) = 20$
$y(0) = 20$

(b) Solve for $x(t)$ and $y(t)$.

$\frac{dx}{dt}=-\frac{5x}{200}$

$200\; dx =-5x\; dt$
$\frac{dx}{-5x}= \frac{dt}{200}$

$\frac{1}{5}\ln|-5x| = \frac{1}{200}t+C$

$\ln|-5x| = \frac{1}{40}t+C$

$x =-\frac{1}{5}ce^{\frac{1}{40}t}$, $x(0) = 20$

$y =-\frac{1}{5}ce^{\frac{1}{40}t}$, $y(0) = 20$

$x = 20e^{\frac{1}{40}t}$
$y = 20e^{\frac{1}{40}t}$

(c) Using the results from part[b](b)[/b], solve for $z(t)$.

$\frac{dz}{dt}= \frac{-5x-5y}{200}-\frac{z}{500}$

$\frac{dz}{dt}+\frac{z}{500}=-e^{\frac{1}{40}t}$

So then I would just solve this equation for $z$? Does this look correct?

Thanks

anybody think that the method/ differential equation to solve is correct?

Thanks


An arbitrary point $ P$ is chosen in the plane of $ \triangle ABC$. Prove that the circumcircles of the medial triangles of $ \triangle ABC$, $ \triangle PAB$, $ \triangle PBC$, and $ \triangle PCA$ are concurrent.

Let ⊙O1 ⊙02 meet at X
∠EXJ=∠EKH+∠HGJ=∠BPA=∠EFJ
=>  E,F,X,J concyclic
∠FXE+∠EXJ+∠GXJ=∠EAF+∠BPA+∠GBJ=180+∠ABC=180+∠FKG
∠FXG=360-180-∠FKG=180-∠FKG
=>  X,F,K,G  concyclic
done！

Here is an extension that I thought of...given the point X on the nine-point circle of ABC, construct the locus of all possible P.


Circles $\omega_{1}$ and $\omega_{2}$ intersect at $P$ and $Q$.  $AC$ and $BD$ are chords of $\omega_{1}$ and $\omega_{2}$, respectively, such that $P$ is on segment $AB$ and on ray $CD$.  Lines $AC$ and $BD$ intersect at $X$.  Let the line through $P$ parallel to $AC$ intersect $\omega_{2}$ again at $Y$, and let the line through $P$ parallel to $BD$ intersect $\omega_{1}$ again at $Z$.  Prove $Q, X, Y, Z$ are collinear.

Kwsta, although you posted this problem almost 2 years ago, i'll answer it because i saw it yesterday and i think it shouldn't be still "open".

I think it is an easy "angle chasing" problem. We'll work backword. Assume that $ XY\cap\omega_1 \equal{} T$. We'll prove that $ TP\parallel\ BD$, in other words, $ Z\equiv\ T$.
Draw $ AT, QP, YB$.
$ \widehat{ATQ} \equal{} \widehat{QPB} \equal{} \pi \minus{} \widehat{TYB}$ implying that $ AT\parallel\ BY$. So $ \widehat{TAC} \equal{} \widehat{PYB}$ because they are acute and have both sides parallel.
But, $ \widehat{ATC} \equal{} \widehat{TPC} \equal{} \widehat{TPD}$ and $ \widehat{PYB} \equal{} \widehat{PDB}\Longrightarrow\widehat{TPD} \equal{} \widehat{PDB}\Longrightarrow TP\parallel\ BD$.  So $ X,Y,Z$ are collinear.
Now, $ \widehat{TQP} \equal{} \pi \minus{} \widehat{TAP} \equal{} \pi \minus{} \widehat{ABY} \equal{} \pi \minus{} \widehat{PQY}$ so $ Q$ belongs to the same line, QED

Giwrgos

(In my drawing, $ X\equiv\ K$)


trigonometry
function
real analysis
real analysis unsolved

Suppose that $ b_k$ is a sequence decreasing to $ 0$. Show that $ \sum b_k\sin{kt}$ converges uniformly on $ [0,2\pi]$ if and only if $ kb_k\to 0$.
I think I have solved the 'only if' part, but struggle with 'if'.

if $ f(t)\equal{}\sum b_k \sin kt$ converges unif on $ [0, 2\pi]$

then $ f'(t)\equal{} \sum k b_k \cos kt$ converges

$ f'(0) \equal{} \sum k b_k < \infty \to  [ kb_k \to 0]$

legal?

No, not legal.

Consider $ b_k\equal{}k^{\minus{}3/2}.$ Then $ f(t)\equal{}\sum k^{\minus{}3/2}\sin kt$ is a continuous function but $ f'(0)\equal{}\sum k^{\minus{}1/2}\equal{}\infty.$

You can't assume that $ f$ is differentiable.

Now, it is true that in this example the Fourier series for $ f(t)$ does converge uniformly, so this isn't an counterexample to the theorem.

Hint in the form of a question: how could one possibly exploit the condition that $ b_k$ is a decreasing sequence? That suggests to me that one should employ summation by parts, differencing the $ b_k$ and hence summing the sines.


Find all f:R+->R+ such that
f(x)^2 + f(y)^2 = f(x+y) [ f(f(x)) + f(y) ]

May I know if there is a reason why the subject is "Modified post"?
(just wondering)

Otherwise, here is my solution:

Letting $x = 0$,
 $f(0)=f(y)$ unless $f(0)=0$

(1) If $f(0) = f(y)$, then
\[f(x)=k\]
where $k$ is a constant.

(2) If $f(0)=0$, letting $y=0$, we get
\[f(x) = f(f(x))\]
unless f(x) = 0, which is a constant.

Further, when $f(0)=0$,

\[ f(x+y) = \frac{f(x)^{2} + f(y)^{2}}{f(x)+f(y)}\]
\[ f(\delta x+0) = \frac{f(\delta x)^{2} + f(0)^{2}}{f(\delta x)+f(0)}
\[ f(\delta x) = f(\delta x)\]
where delta x is a very small rational number approaching 0.

\[ f(\delta x + \delta x) = \frac{f(\delta x)^{2} + f(\delta x)^{2}}{f(\delta x)+f(\delta x)}=f(\delta x) \]
Thus, $f(x) = k$, where $k$ is a constant. (not completely sure about this last statement but sounds correct to me)

Therefore, in all cases, $f(x) = k$.

First the problem was $f$:$R$-->$R$
And Now $f$:$R^+-{0}$-->$R^+-{0}$
$x=0$ is not allowed!

We have:
$f(x)^2+f(y)^2=f(x+y)(f(f(x)+f(y))=f(x+y)(f(f(y)+f(x))$
As$f>0$ then$f(f(x))-f(x)=f(f(y))-f(y)=cont=k$
It seems to me its already discussed!

yes, x=0 is not allowed.
I don't think that this problem is discussed before.
I will post my solution when I find the time.
-Ali


probability MIT college percent AMC USA(J)MO USAMO

[quote="samus"]I've also been wondering: Is U of M a good university in reputation for firms such as DE Shaw?[/quote]

People in states with a name that begins with "M" have to learn to distinguish among the "U of MI," "U of MD," "U of MN," "U of MO," "U of ME," "U of MT," and whatever other "U of M" there might be. That helps get more accurate answers for questions like yours.

[quote="samus"]How do these courses prepare you to be a quant?:
Linear and Nonlinear Programming
Probability Theory
Optimization
Programming Courses
-Any other courses that prepare you?
Thanks?

I've also been wondering: Is U of M a good university in reputation for firms such as DE Shaw?[/quote]

Those courses would be great.  Make sure you learn how to code.

As for schools outside the major ones like MIT, it really varies by department.  Michigan's math department, for example, has a good reputation.

[quote="samus"]How do these courses prepare you to be a quant?:
Linear and Nonlinear Programming
Probability Theory
Optimization
Programming Courses
-Any other courses that prepare you?
Thanks?

I've also been wondering: Is U of M a good university in reputation for firms such as DE Shaw?[/quote]

There are many linear and nonlinear problems in finance.

There are tons and tons of probability problems in finance.  For instance, options models are based in stochastic processes.

There is nothing in finance that couldn't use a little optimization.  That's the whole game.

Programming enables you to use all of the above with greater efficiency.

[quote="samus"]So in all, what courses do you think will prepare you most for this career (specifics, please)?
Thanks![/quote]

It's not a matter of what courses you take.  It's a matter of learning how to think on a high problem solving level and use the tools of technology to aide you.  I don't remember ever being asked what classes I took in any interview I ever had while working in finance.

[b]Edit: Come to think of it, I can't recall being asked about what classes I took in college in any interview ever.[/b]

I will note that with your problem solving credentials, you're somewhat of an extreme case.  If you don't have those creds, a bunch of econ classes aren't going to get you an interview.  Definitely get some coding skills, and have something to prove it (classwork or summer jobs).

Are there any quantitative trading jobs that are non-coding? Because I am only in the Silver divison of the USACO.
I am okay with coding, but don't like it.
What I mean by courses to take is not econ courses, but math courses. What ones do you use?

I would suggest not stressing so much about career paths this early.  Enjoy the educational exploration.  The skills you enjoy learning will lead you to careers you will enjoy.

In regards to math classes (and economics), I would recommend studying game theory.  Not only is it a lot of fun (in my opinion), but trading is a game.  Outside of probability and ordinary math skills, I think I used game theory as much as any other area of problem solving while working in finance, though experiences vary widely.

If you're starting to get up to USAMO qualification level in your math problem solving skills, you don't need to stress about whether or not you are capable.

I don't know much about being silver on the USACO, but I think you're on a fine path so long as you continue to improve your skills.

[quote="MCrawford"]Enjoy the educational exploration[/quote]

That is encouraging.
That's [i]the most[/i] important reason why I want to do math.
Thanks!

Both the administrators gave some really good advice. Programming seems to be an essential skill for many employers.  If you have trouble with programming there are a couple of math courses I would really recommend to help you understand some of the concepts.  Taking courses in graph theory, probability, discrete math, number theory, and linear and abstract algebra really help...

Specifically about careers in IT, employers seem to look for candidates who have skills and experience in a large range of languages. But in order to learn these languages, you need a good understanding of several fundamental ones.  For beginners, I HIGHLY recommend you learn C as a first language. It'll make life easier. It's not used as often as it used to be, but C++, Java, C# and many other commonly used languages are all derived from it. Also, if you're really career-minded, learn Java, SQL and C# (the hideous Microsoft language unfortunately used everywhere).

One concern about IT careers is that because of heavy competition and outsourcing, many of the positions available seem to require a lot of experience, and a larger range of knowledge every year. But most importantly, do what the two admins said, don't stress! [/quote]

[quote="samus"]Also, I was wondering:
How much percent of the people in DE Shaw made the USAMO?
Top 250 in Putnam?

Beacause I am pretty intimidated by the job descriptions.
I only go a 3 on the AIME this year, and aiming to qualify for the USAMO. But now I only am getting 125ish on the AMC 12s, and 8ish on AIMEs, only about a 200-210 range, 20-30 away from the cutoffs.
I'm preparing though! :blush:[/quote]

At this point, I think around half of their quants have IMO gold medals.  But I'll note that there are many, many more Shaw-like funds than there used to be.  In other words, you don't have to be an IMO math person to get an excellent hedge fund job.  Shaw can afford to be more selective because it is the largest and most well-known hedge fund in the math/science community.

[quote="samus"]How do these courses prepare you to be a quant?:
Linear and Nonlinear Programming
Probability Theory
Optimization
Programming Courses
-Any other courses that prepare you?
Thanks?

I've also been wondering: Is U of M a good university in reputation for firms such as DE Shaw?[/quote]

are you in U of M now? DE shaw basically wants the cream of the crop. As long as you come from a top 25 and can make that stellar GPA and kill it in other areas (top awards) then you will get an interview.

Just because you come from MIT doesn't mean you'll get hired. You gotta do lots of extra stuff on top.

[quote="qvc122"]Well, it is very intimidating to read the hiring description from DE Shaw website. One of their relative recent hiring was someone with one IMO gold and Putnam fellow twice who graduated from Harvard.

By the way, rrusczyk, what kind of customers does DE Shaw do business with, institutional or individual? :)[/quote]

Sorry I didn't answer this earlier - didn't see it.  This is a hard question to answer now.  For example, they own FAO Schwarz.  Also, it's been 8 years since I was there.  Most of the work then was proprietary trading with institutional investors.

I like perl first, then scheme, then C, then Objective-C/Java/Python.  Scheme will make you bettern, even if you never need it. Perl is widely used to tie applications together and convert formats. C is ,well, C. And Java is widely used. Remember, language doesn't matter as much as understanding how to code.


Hi!

Prove that if $ a,b,c\geq 0$ then

\[ \frac{a^2}{\sqrt{a^2\plus{}b^2}}\plus{}\frac{b^2}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{c^2}{\sqrt{c^2\plus{}a^2}}\geq\frac{a\plus{}b\plus{}c}{\sqrt{2}}.\]

[quote="yuyang"]Hi!

Prove that if $ a,b,c\geq 0$ then
\[ \frac {a^2}{\sqrt {a^2 \plus{} b^2}} \plus{} \frac {b^2}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {c^2}{\sqrt {c^2 \plus{} a^2}}\geq\frac {a \plus{} b \plus{} c}{\sqrt {2}}.\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265475


calculus
trigonometry
calculus computations

1 - What is the minimum distance between the points $ A$ and $ C$ if the path need to be inside the region $ ABCD$ ?

Consider the angle $ AuB > \tan ^{\minus{}1}\left(\frac{\sqrt{\delta ^2\plus{}2 r \delta }}{r}\right)$

I don't know if you really need calculus of variation here - the shortest path from $ C$ to $ A$ goes along a tangent line to inner circle and continues along the circle.

Idea of proof: let $ E$ be the point where the path described above meets the inner circle. Draw a line through $ u$ and $ E$ and let $ F$ be its intersection with outer circle. Now split the problem into two parts:
1) Find the shortest path from $ A$ to the segment $ EF$;
2) Find the shortest path from $ C$ to the segment $ EF$.

The reason I Said

[quote]Consider the angle $ AuB > \tan ^{\minus{}1}\left(\frac{\sqrt{\delta ^2\plus{}2 r \delta }}{r}\right)$[/quote]

is to the reader think about the tangent line.

You can divide the problem, but, you didn't proove nothing. For that you need calculos of variations.

Is quite intuitive that the distance between a 2 points is minimum is a line. But to prove that is another story.

For this problem, the path is a straight line plus a arc of a circuference.

Alright, I'll say a little more:
1) Any path from $ A$ to the segment $ EF$ (within $ ABCD$) is at least as long as its radial projection onto the arc $ AE$.
2) Any path from $ C$ to the segment $ EF$ is at least as long as its orthogonal projection onto $ CE$.
Therefore, any path from $ A$ to $ C$ (within $ ABCD$) is at least as long as the length of arc $ AE$ plus the length of segment $ EC$.


function
Putnam
algebra unsolved
algebra

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=27129

Pierre.

The set of zeros of the function $f(x)=0$ is not countable even if this function is continuous...

Pierre.

[quote="ali"]I have a method for solving it, but i need to know if the set of zero's of a continous
function is countable. I can't prove it just from the definition of continouty.
It might be even just wrong![/quote]

You can prove that every closed resl set is the set of zeros of some continuous function, actually, so yes, it is wrong... :)

[quote="grobber"]
You can prove that every closed resl set is the set of zeros of some continuous function[/quote]

Even more precise....
http://www.mathlinks.ro/Forum/viewtopic.php?t=19616&highlight=closed+continuous&start=15

Pierre.

[quote="Kapi"]Find all continuous function $f:R \to R$ satisfies:$f(f(x))=f(x)+x \forall x \in R$[/quote]

There were some problems like this on Putnam competitions (something with $2x$, but the same idea).  The principle is this:  composition with $f$, i.e. $g(x) \mapsto g(f(x))$ is a linear operator on the (2-dimensional space of) functions of the form $Ax + Bf(x)$.  Then look at the action of $f$ in a basis of eigenvectors and use continuity or injectivity.


Find all $a,b,c,d \in N^*$  (a <> b <> c <> d) such that:
$abcd - 1$ is divisible by $(a - 1)(b - 1)(c - 1)(d - 1)$

I think the source is not so unknown...it is just the same problem as :
http://www.kalva.demon.co.uk/imo/isoln/isoln921.html

but with four variables.

Pierre.

An interesting related question might be:
For any positive integer $k$ , if exists some positive integers $a_1, a_2, ... a_k$ each two of them different and:
$a_1 a_2 ... a_k-1$ is divisible by $(a_1-1)(a_2-1) ... (a_k-1)$ ,
which are the possible values for:
$\frac{a_1 a_2 ... a_k-1}{(a_1-1)(a_2-1) ... (a_k-1)}$   ?
We have to check all the integers between $2$ and $\frac{(2k)!!-1}{(2k-1)!!}$ .


inequalities
search
inequalities unsolved

Let $ a,b,c>0$ satisfy $ abc\equal{}1$. Solve that:
$ \frac{a}{1\plus{}b\plus{}c}\plus{}\frac{b}{1\plus{}c\plus{}a}\plus{}\frac{c}{1\plus{}a\plus{}b}\geq 1$

we can prove this one using AM-GM just simplify it,we are left to prove that:

$ \sum a\plus{}\sum ab\leq\sum a^{2}\plus{}\sum a^{3}$

which is true because we have:

$ 3\sum ab\leq 2\sum a^{3}\plus{}3$

and

$ 3\sum a\leq\sum a^{3}\plus{}6$

so adding these two inequalities we get that:

$ 3 LHS\leq 3\sum a^{3}\plus{}9$

so its sufficient to show that:

$ 3\sum a^{3}\plus{}9\leq 3\sum a^{3}\plus{}3\sum a^{2}$

$ \Rightarrow 3\leq\sum a^{2}$

which is true by AM-GM:

$ ab\plus{}bc\plus{}ca\geq 3\sqrt [3]{a^{2}b^{2}c^{2}}\equal{} 3$...

[quote="nktp"]My solution:
By Cauchy-Schwarz: $ \sum\frac{a}{1\plus{}b\plus{}c}.\sum a(1\plus{}b\plus{}c)\geq (\sum a)^{2}$
We need to prove: $ (\sum a)^{2}\geq\sum a(1\plus{}b\plus{}c)\Leftrightarrow\sum a^{2}\geq\sum a$
By Cauchy-Schwarz: $ 3\sum a^{2}\geq (\sum a)^{2}\geq 3\sum a$. It is OK![/quote]
nice,now try to prove the other one...

[quote="BaBaK Ghalebi"]also another similar one holds:
with $ abc\geq 1$ prove that:
$ \frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}c\plus{}a}\leq 1$[/quote]

it was also discussed here
http://www.mathlinks.ro/viewtopic.php?search_id=1785282224&t=156173
http://www.mathlinks.ro/Forum/viewtopic.php?p=344767#344767


Find all integer solutions of x^{4}+y^{2}=z^{4}


conics
ellipse
geometry
perimeter
parallelogram
rhombus
geometry unsolved

This problem concerns inscribing an equalateral polygon of n sides within
an ellipse of any eccentricity such that the vertices of the polygon
touch exactly the perimeter of the ellipse and the sides of the
polygon create equal length chords to the ellipse. In an example of
one that I worked out through trial and error ( and many hours ) if
you had an ellipse with a major axis 360 and a minor axis 240 and
wanted to inscribe a twelve equal sided figure within it I have found
that the lengths of the sides if this figure will be 78.248688. Now,
is there such a formula that can work for any ellipse and any
equalateral polygon? There is such a solution for an ellipse of zero
eccentricity ( a circle ) namely,
Sin (180/n)*2 *r
But what about when the eccentricity is greater?

I look forward to hearing from you all soon. Thankyou Taylew.

Wow I guess this problem is far too hard for the guys here to solve, sorry. Does any one know of an advanced math forum ( probably an American forum )that could possibly solve it?

Unless you impose aditional conditions, such as the equilateral polygon inscribed in an ellipse is symmetrical WRT the ellipse major  or minor axis (or both), the problem does not have a unique solution. Consider a line $ a$ through the ellipse center O, forming an arbitrary angle $ \phi$ with the ellipse major axis, and another line $ b \perp a$ at O. The lines a, b intersect the ellipse at A, C and  B, D, respectively. The segments AC, BD cut each other in half, ABCD is a parallelogram with perpendicular diagonals, rhombus, solving the problem for n = 4, but changing with the arbitrary angle $ \phi.$


Ten families have an average of 2 children per family. If exactly two of these families are childless, what is the average number of children in the families with children? Express your answer as a decimal to the nearest tenth.

There are 20 children total. Among the 8 families with children, there are an average of 20/8=2.5 children.


geometry
geometric transformation
rotation

In how many ways can four people sit around a circular table? Seating arrangements are considered different only if at least one person has at least one different neighbor.

Well, if you draw a diagram with 4 people A,B,C, and D Person A can have neibors B and C or C and D or B and D so three.

Can you explain that a little more clearly?

We have 4 people around a circular table, so there are $ 3!$ or $ 6$ ways to arrange them. However, rotations can be the same, so we divide by $ 2$ to get $ \boxed{3}$.

[quote="iin77"]I believe this solution is more in-depth.

There are four seats, so there's 4 choices for first seat, 3 for second, 2 for third, 1 for last.  $ 4*3*2*1 \equal{} 24$.  But since it's a circular table, there's the potential of rotating the table into 4 different positions.  You must divide by 4 to compensate.  $ 24/4 \equal{} 6$.  But you can also "flip" the table, meaning the people can be sitting next to the same people either counterclockwise or clockwise from each other.  So you must divide by 2 to compensate.  $ 6/2 \equal{} \boxed3$.

From this, you can see that a circular table with $ n$ seats will have a total of $ \frac {(n \minus{} 1)!}2$ seating arrangements.  This makes it quite easy to do this question in... 1 second.[/quote]

Actually, no, it's only $ \frac {(n \minus{} 1)!}2$ if the table is reversible (you can flip it over) which you obviously can't do. THus you use the circular forumula $ (n\minus{}1)!\equal{}6$

[quote="problem"]Seating arrangements are considered different only if at least one person has at least one different neighbor.[/quote]

you can flip because everyone still has the same neighbor, so the answer is actually 3.

[quote="mewto55555"][quote="iin77"]I believe this solution is more in-depth.

There are four seats, so there's 4 choices for first seat, 3 for second, 2 for third, 1 for last.  $ 4*3*2*1 \equal{} 24$.  But since it's a circular table, there's the potential of rotating the table into 4 different positions.  You must divide by 4 to compensate.  $ 24/4 \equal{} 6$.  But you can also "flip" the table, meaning the people can be sitting next to the same people either counterclockwise or clockwise from each other.  So you must divide by 2 to compensate.  $ 6/2 \equal{} \boxed3$.

From this, you can see that a circular table with $ n$ seats will have a total of $ \frac {(n \minus{} 1)!}2$ seating arrangements.  This makes it quite easy to do this question in... 1 second.[/quote]

Actually, no, it's only $ \frac {(n \minus{} 1)!}2$ if the table is reversible (you can flip it over) which you obviously can't do. THus you use the circular forumula $ (n \minus{} 1)! \equal{} 6$[/quote]

As it says in my post, they can be seated next to the same people in either a clockwise or a counterclockwise fashion.  I realize you can't literally flip the table, "flip" is simply the term I have chosen to use to describe this.  You see, the sentence "Seating arrangements are considered different only if at least one person has at least one different neighbor," causes the circular table to become analagous to a ring.

And I know for a fact the answer is 3, because I have gotten this correct many times.


geometry
circumcircle
geometric transformation
reflection

To April:

I have a solution to this problem.
[b] HINT[/b]
  From $ C$ we draw a parallel to $ SN$ which meets line $ AB$ at $ K$ ect. I will send the solution later , if nobody answers , because this time I;m very busy.

 But if you have any solution based on polars and harmonics ,it is wecome ! I prefare this kind of solution to this problem , if it is short.

 Babis

@stergiu&april:
denote point $ D'$ on the circumcircle s.t $ SD'$ is tangent to the circle...
now we know that $ S(DD',BC)=-1$ so $ A(DD',BC)=-1$ now with respect to line $ SO$ we get that $ (OX,MN)=-1$ where $ X$ is the intersection of $ AD',SO$ now its sufficient to show that $ X \rightarrow \infty$ because in this case we get that $ (O\infty,MN)=-1$ so $ O$ must be the midpoint of $ MN$ and we are done...
hence its sufficient to show that $ X\rightarrow\infty$ i.e. $ AD' \parallel SO$:

we know that $ SO$ is perpendicular to the polar of point $ S$ which is line $ DD'$,so we have $ \boxed{SO \bot DD' (I)}$...
and also $ AD$ is a diameter so $ \angle AD'D=90$ so $ \boxed{AD \bot DD' (II)}$...

form $ (I),(II)$ we get that $ SO \parallel AD'$ as we wanted...

[hide="A idea?"]
Reflect $ B$ over $ O$ to get $ E$.  Prove $ N$ lies on $ DE$.[/hide]

[color=red]Thank you for all interest. And now, I'll present a solution I have in mind.[/color]
[quote="April"][color=darkred]Let be given a triangle $ ABC$ and $ AD$ the diameter of it's circumcircle $ (O).$ The tangent line at $ D$ of $ (O)$ intersect $ BC$ at $ S.$ Let $ M=SO\cap AB,\,N=SO\cap AC.$
Prove that $ OM=ON$[/color][/quote]
[color=darkblue][b]Proof.[/b] Draw a parallel line to $ SN,$ which meets $ AD,\,AB$ at $ P,\,Q,$ respectively. And denote $ R$ be the middle point of $ BC.$
We will show that $ P$ is the midpoint of $ CQ$

We have $ \angle ORS=\angle ODS=90^\circ\implies O,\,R,\,D,\,S$ are concylic.
So $ \angle ODR=\angle OSR=\angle QCS\implies C,\,D,\,P,\,R$ are concylic.
$ \implies\angle PRC=180^\circ-\angle PDC=180^\circ-\angle ABC=\angle QBC$
$ \implies PR\parallel BQ,$ hence $ P$ is middle point of $ CQ\quad QED$
[/color]

April , very nice! As I told you in my first message , this is exactly the solution that I have in my archieve.
 I think that now we have nice solutions to this also nice problem.

  [u]Babis[/u]


trigonometry algorithm calculus derivative function

xsin(x) = 9

ive approximated one value to be around 23, but I cannot think of how to solve mathematically

hello, your equation can only be solved by a numerical way, discuss the funktion $ f(x)\equal{}x\cdot\sin(x)\minus{}9$ for all real $ x$.
Sonnhard.

hello, $ f^'(x)=\sin(x)+x\cos(x)$.
Sonnhard.

Hi Sonnhard  , I think we should discuss $ x\cdot f(x) \equal{} 9$ ,where $ f(x) \equal{} \sin x$.

[i] It may be the same as you wrote Sonnhard ,I'm not good in trigonometry [/i]

[b]ee"[/b][/i]

Unless you're hinting at Newton's method or some other root approximation algorithm, I'm not sure how useful $ f'(x)$ is.

And enndb0x: You might want to consider not posting unless you have new insight into a problem...

So? Does anyone have a solution, or just ideas and corrections?

I dont quite understand how the derivative can help here either, if anything it is more complex.

I tried a method of solving, but I just ended up proving the sin^2 + cos^2 = 1 identity :/

looking at the triangle of the function, we can derive these equations using sine law

$ \frac {1}{x} \equal{} \frac {\sin{x}}{9} \equal{} \frac {\cos{x}}{\sqrt {x^2 \minus{} 81}}$

I think im going somewhere in the right direction, but then again it doesnt really leave us with anything to solve other than another trig function of x multipled by some form of x

hello, from $ \sin(x)\plus{}x\cdot\cos(x)\equal{}0$ we get $ \tan(x)\equal{}\minus{}x$, with this equation can we derive the local extrema of $ f(x)$ with the help of the newton iteration.
Sonnhard.

hello, here you can find something for the newton method.
http://en.wikipedia.org/wiki/Newton's_method
Sonnhard.

So essentially the only way to solve such a problem is by brute forcing?

Not to call newtons method brute forcing, but essentially to me it is. The method is very intriguing though, thank you for linking me to it.

hello, yes, you can't solve the equation $ x\sin(x)\equal{}9$ for $ x$ with any known elementary function,the only way is a numerical method, like Newton's method.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, one solution i have found is the following
$ x \approx 13.3090086732243185265556384119$
Sonnhard.[/quote]

also

$ x \approx \minus{}13.3090086732243185265556384119$

:)

those are the only two solutions I think, if you think of the graph of sine

I'm no good at this, so please tell me just at which interval of a does this have solution?
xsinx=a

$ x \sin x$ is a continuous function and at the value $ 2n\pi \plus{} \frac{\pi}{2}$ for $ n \in \bf N$ it takes the value $ 2n \pi \plus{} \frac{\pi}{2}$, which gets arbitrarily large, while at the value $ 2n \pi \plus{} \frac{3\pi}{2}$ it takes the value $ \minus{}2n\pi \minus{} \frac{3\pi}{2}$, which gets arbitrarily small (i.e., large and negative).  Thus it takes all real values (and in fact takes every value infinitely many times).  Just thinking about what the graph looks like will probably help.


I think its mean to be co - she but I'm not sure. I say something more like cow - she.

I read it "cow-chi"
The sure fire way to figure out would be to look it up on a bibliographical dictionary with pronounciation.

Every American mathematician I know pronounces the name "co-she".

He was a great mathematician!

co-she as in the word (ok its not really a word) co followed by the word she. Long e sound, in other words :)

yes, co-she is the way people usually pronounce it.

same here. "cow-chi" and "couch-ie" is the same pronounciation, I think. At least they're both humorous :)

After u learn to pronounce Cauchy (its supposed to be co-she), then learn how to pronounce Schwarz and Boudenovski (ok, im 200% sure i spelled that wrong...)

shwarts and boudenovski? How's that? (From the movie commercial)

[quote="Tare"]So it sounds like "Kosher" except for the r?[/quote]

Koshe?  No, that's more of a koa-shu sound, whereas cow-she would me more like cow-she, eh?

i dont understand what you mean fierytycoon, but the spelling is right.  Its pretty straightforward to pronounce.

[color=cyan]It's really more like cow-shee than co-shee.  The "au" isn't quite an o sound, more of an ow.  Buniakowski is much closer to the spelling of the name darkquantum was getting at, for the variously-named Cauchy-Schwarz-Buniakowski inequality, known in the United States usually just as Cauchy-Schwarz.  Go look it up :).[/color]

i think its schwartz and buniakowsky, as in Cauchy-Schwartz-Buniakowsky Inequality (as Cauchy's inequality is known in foreign countries (at least Romania for sure)).

Yes, but it's spelled Schwarz and not Schwartz.

[quote="shobber"][quote="Fierytycoon"]Ok, thanks, so co-she. Now how about Schwarz? Right now I am pronouncing it like "Schwartz" with the 't'.[/quote]
Indeed, how do i pronounc $\text{Schwarz}$?[/quote]

"shh-warts"?


linear algebra
matrix
Gauss
algorithm
quadratics
algebra
quadratic formula

5x-y+z = 29
3x+2y-3z = 20
x-3y+2z = 3

I dont completely understand how to solve problems like these. Even after looking at the steps.

hello, multiplying the first equation by $ \minus{}3$ and the second by 5 and adding these two equations we have
$ 5x\minus{}y\plus{}z\equal{}29$
  $ 13y\minus{}18z\equal{}13$
and multiplying the third equation by $ \minus{}5$ and adding this to the first equation
our system looks like
$ 5x\minus{}y\plus{}z\equal{}29$
  $ 13y\minus{}18z\equal{}13$
  $ 14y\minus{}9z\equal{}14$
Multiplying the last equation by $ \minus{}2$ and adding to the second equation we get $ \minus{}15y\equal{}\minus{}15$. From here we have $ y\equal{}1$.
Sonnhard.

To generally solve systems of equations, you would first look to see if there are any "eliminatable" variables. Note that the elimination method is very slow. Also, you might want to write variables in terms of other variables, and substitute them in. In the end, [b]it's just a lot of jiggling with the variables[/b], keep performing algebraic operations with the equations to isolate the variables.

[hide="The Method"]
Here, $ \boxed{C}$ will represent the equation $ C$, or a form of it.

$ \boxed{A.} \; 5x \minus{} y \plus{} z \equal{} 29$
$ \boxed{B.} \; 3x \plus{} 2y \minus{} 3z \equal{} 20$
$ \boxed{C.} \; x \minus{} 3y \plus{} 2z \equal{} 3$

Get rid of one variable first. I'm choosing x.
Multiply one equation by a constant so that the coefficient of x in that equation is the same as the one in another equation.

So, starting with equation C and multiplying so that the coefficient of x is the same as the coefficient of x in equation B, we rewrite equation C, multiplying both sides by 3.

$ \boxed{A.} \; 5x \minus{} y \plus{} z \equal{} 29$
$ \boxed{B.} \; 3x \plus{} 2y \minus{} 3z \equal{} 20$
$ \boxed{C.} \; 3x \minus{} 9y \plus{} 6z \equal{} 9$

Alright, now we can subtract B and C.
$ \boxed{B \minus{} C.} \; 3x \plus{} 2y \minus{} 3z \minus{} (3x \minus{} 9y \plus{} 6z) \equal{} 20 \minus{} 9$
$ \boxed{B \minus{} C.} \; 11y \minus{} 9z \equal{} 11$.

Now, we do this one more time... Now from C to A (So we can subtract later).

$ \boxed{A.} \; 5x \minus{} y \plus{} z \equal{} 29$
$ \boxed{C.} \; 5x \minus{} 15y \plus{} 10z \equal{} 15$

$ \boxed{A \minus{} C.} \; 14y \minus{} 9z \equal{} 14$.

There! Now we have these two equations.
$ \boxed{A \minus{} C.} \; 14y \minus{} 9z \equal{} 14$
$ \boxed{B \minus{} C.} \; 11y \minus{} 9z \equal{} 11$

Yey! We can subtract one from the other!

$ \boxed{A \minus{} B.} \; y \equal{} 1$
Also, from there we plug into anyone of the equations to get $ z \equal{} 0$.

Now, just plug in and solve for x!

$ \boxed{C.} \; x \minus{} 3y \plus{} 2z \equal{} 3$
$ x \minus{} 3(1) \plus{} 2(0) \equal{} 3$
$ x \equal{} 6$.

So, our solutions are $ (6, 1, 0)$.

This is just one example of how you may do it. I realize there are much quicker and more efficient ways.
[/hide]

EDIT: Indeed, matrices make life a LOT easier...

hello, the best method is the so called Gauss algorithm, see here:
http://en.wikipedia.org/wiki/Gaussian_elimination
Sonnhard.

Apart from substitution, elimination, graphing, and calculator work, you can also use some methods involving matrices like

Cramer's Rule

Solving with inverses

5x-y+z = 29
3x+2y-3z = 20
x-3y+2z = 3

Okay, in case you don't understand any of the other solutions, this one might be helpful.
If you take the first equation;

5x - y + z = 29
You can add y to get;
5x + z = 29 + y
You can also subtract 5x to get;
z = 29 + y - 5x.
So we can substitute this equation whenever we use z.

If we take the next equation;
3x+2y-3z = 20
We know what z equals, so we can substitute (29 + y - 5x) for z to get;
3x + 2y - 3( 29 + y - 5x) = 20
Then if we multiply the equation by 3 we can get;
3x + 2y - 87 - 3y + 15x = 20
Then we can combine like terms to get;
18x - y - 87 = 20
We can add 87 to get;
18x - y = 107
We can add y to get;
18x = 107 + y
We can subtract 107 to get;
18x - 107 = y
Now we know what y equals, but we need to plug in this equation for the equation equal to z.
z = 29 + y - 5x
We can plug in 18x - 107 for y to get;
z = 29 + 18x - 107 - 5x
We can combine like terms to get;
z =  13x - 78

So now we have the following equations left to mess with;

x = x
y = 18x - 107
z = 13x - 78
x-3y+2z = 3

So plug in x for x to get;
x - 3y + 2z = 3
And plug in 18x - 107 for y to get;
x - 3(18x - 107) + 2z = 3
And plug in 13x - 78 to get;
x - 3(18x - 107) + 2(13x - 78) = 3
Now we can multiply everything out to get;
x - 54x + 321 + 26x - 156 = 3.
Add 156 to get;
x - 54x + 321 + 26x = 159
Subtract 321 to get;
x - 54x + 26x = -162
Then combine like terms to get;
-27x = -162
Divide both sides by -27 to get;
x = 6

Then you can plug in 6 for the equations that equal y and z, and then you can check your work to see if it is right.

I hope that helped, because I had to type a lot.

substitution would take a long time for this problem
i would suggest using textangle's solution unless you find it confusing and refuse to use anything except substitution

[url=http://en.wikipedia.org/wiki/Gauss%E2%80%93Seidel_method]This[/url] may help you.

[quote="moogra"]1 way is to plug it into a ti-83+/84+ calculator (matrix).
Go to matrix, make it matrix [A] and make it 3x4
type 5 -1 1 29 on the first row
3 2 -3 20 on the second row
1 -3 2 3 on the third row
then 2nd mode to quit
then do rref([A] to solve the matrix (answers are on the right side)[/quote]

RREF stands for [url=http://en.wikipedia.org/wiki/Reduced_row_echelon_form#Reduced_row_echelon_form]reduced row echelon form[/url].  You'll learn it in Alg 2, I think.

You could also use the Elimination Method.

i think the elimination method is better!
 :D

Yeah, it is very effective and fast, the only difficulty is finding where to eliminate.

In general, different problems will have different preferred methods of solving. If a variable has been isolated in one equation, you might want to use substitution, but maybe the variables can cancel out, so you want to eliminate. It all depends on the problem.

Another way of saying this: Would you use the quadratic formula when you could factor? :wink:

For a great narrated and visual explanation of solving these type of equations (2 linear equations), check out

http://www.mathematics.com.au/watch.html?lesson=SNSSSLESLEJELLHR

and also

http://www.mathematics.com.au/watch.html?lesson=JKOWFCWYCDUESVPO

The first link show the elimination method, the second link shows the substitution method in action.


Wow I forgot how specific people are on aops. :wink:
It's the same theme right?

Yeah I had some extra time so I made this!

Now you know what I do in my spare time.


Find all primes $ p$ such that $ 2p^{2} \plus{} 1$ is also a prime.

[hide]First, $ 3$ is a solution.

For $ p \ne 3$

$ p \equal{} 1 (mod 3)$ or $ p\equal{}2 (mod 3)$

Both mean

$ 2p^2\plus{}1 \equal{} 0 (mod 3)$

Therefore $ 3$ is the only solution.[/hide]


induction
number theory unsolved
number theory

Let $a_1,a_2,...$ be a sequence and $a_1=2,a_2=5$ and\[a_{n+2}=(2-n^2)a_{n+1}+(2+n^2)a_n\]for every $n\geq 1$. Are there integers $p,q,r$ so that $a_pa_q=a_r$ ?

A straightforward induction leads to $a_n = 2 \mod [3]$ for all $n$.
Thus $a_pa_q =1 \neq 2 = a_r \mod [3]$ so that the desired equality cannot occur at all.

Pierre.


Sea $ABCD$ un cuadrilátero cíclico, $P$ la intersección de las diagonales $\overline{AC}$ y $\overline{BD}$, $O$ circuncentro del $\triangle APB$ y $H$ ortocentro del $\triangle CPD$. Muestre que $O$, $P$ y $H$ están alineados.

Un problema verdaderamente facil.
Supongamos que ABP=x, entonces por ser O circuncentro tenemos que OPA=90-x. Ahora como ABP=ABD=ACD=x,  vemos que OP es perpendicular a PC, lo cual significa que es altura y por lo tanto pasa por H (el ortocentro de  CPD.

Otra solucion, como ABCD es ciclico, imaginames que el triangulo APB lo montamos sobre el CPD (teniendo en comun el vertice P y ademas aciendo que AB y CD sean antiparalelas). Ya colocado asi, es facil ver que la linea que pasa por el Circuncentro de APB es la sialtura (o reflejado de la altura con respecto de la bisectriz), de manera que sabemos que dicha linea pasa por el Ortocentro de CPD, lo cual queriamos demostrar.

Sí, es fácil, está a nivel básico, por eso es que sale tan rápido.

Saludos.


geometry
incenter
conics
hyperbola
Asymptote
geometry unsolved

Internal bisector of $\angle (\ell_b,\ell_c)$ cuts $r$ at the incenter $I.$ Construct incircle $(I)$ centered at $I$ and tangent to $\ell_b,\ell_c.$

Animate the third side $\ell_a$ tangent to $(I)$ $\Longrightarrow$ $B \mapsto C$ is then a homography between $\ell_b$ and $\ell_c.$ Since the directions $CH \perp \ell_c$ and $BH \perp \ell_b$ remain fixed, the locus of $H$ is a hyberbola $\mathcal{H}$ with asymptotes perpendicular to $\ell_b,\ell_c,$ therefore $H \equiv \mathcal{H} \cap r,$ giving rise to at most two solution triangles $\triangle ABC.$ These intersections are then the double points of the homography that $\mathcal{H}$ induces on $r,$ that is, construct five points $H_1,H_2,H_3,H_4,H_5$ of $\mathcal{H}$ (5 positions of $\ell_a$), then $H$ are the double points of the homography $H_1(H_2,H_3,H_4) \mapsto H_5(H_2,H_3,H_4)$ induced on $r.$ Once $H$ is determined, the construction of $\triangle ABC$ is straightforward; $\ell_a$ is the incirle tangent perpendicular to $AH,$ leaving $A,H$ in the same side.

A different construction by Angel Montesdeoca:

https://groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/22273


Here's a question that no one got anywhere on and I would like to see the solution to:

Prove that all positive integers can be represented as the sum of three triangle numbers (0 included).

It looks more like number theory. If I remember correctly, this was equivalent to the fact that any natural number is the sum of four squares.

it can be derived from which numbers are expressible as the sum of three squares.

If i remember correctly, a positive integer N cannot be written as the sum of three squares iff N is of the form 4^k(8t+7).

So if you have a number n, consider 8n+3 = x'^2 + y'^2 + z'^2, then looking mod 4 we have that x',y',z' are odd.

so x' = 2x+1, y'=2y+1, z'=2z+1 and plugging in we have:

8n+3 = 4(x^2+x+y^2+y+z^2+z)+3, so that 2n = x^2+x+y^2+y+z^2+z

then n = x(x+1)/2 + y(y+1)/2 + z(z+1)/2, three triangular numbers

of course proving stuff about sums of three squares is not easy and uses cubic reciprocity i beilieve.

Sorry! I don't know what I was thinking :(.


MATHCOUNTS
geometry
calculus
AMC 8
AMC 10

[quote="Dnas"]Thanks!

[quote]seems that most of the 9th/8th/7th graders have skipped a grade or are young for their grade.[/quote]
Not my case. I'm too old if anything.[/quote]

Yeah, a lot of kids who are really bright when it comes to math have moved from other countries, and so they're a grade or so behind where they should be for their age, based on education differences or language issues.  Although I guess there are more kids who skip/start early.

[quote="Caroline"]Yeah, a lot of kids who are really bright when it comes to math have moved from other countries, and so they're a grade or so behind where they should be for their age, based on education differences or language issues.  Although I guess there are more kids who skip/start early.[/quote]

Hi, Caroline, I asked about this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8057]issue of grade-skipping[/url] in one of the first threads I ever opened on this site, and I was surprised how few participants here have declared a full grade skip. (Feel free to post on the old thread if you want to add data.) My son, homeschooled, would be young in grade (unless other people in his grade held back or skipped) because of what time of the year his birthday is. I have NO interest in him declaring any grade placement other than the one he would have by default in the public school system. He can study what he is ready to study when he is ready to study it (which is the flexibility advantage of homeschooling) and I am happy to have him declared at his default grade level in math contests. He has two more years of eligibility for Mathcounts and for AMC 8, four more years for AMC 10, and six more years for AMC 12. He's got plenty of math to learn, and I hope he has sufficient time to learn it.

[quote="tokenadult"]

Hi, Caroline, I asked about this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8057]issue of grade-skipping[/url] in one of the first threads I ever opened on this site, and I was surprised how few participants here have declared a full grade skip. (Feel free to post on the old thread if you want to add data.) My son, homeschooled, would be young in grade (unless other people in his grade held back or skipped) because of what time of the year his birthday is. I have NO interest in him declaring any grade placement other than the one he would have by default in the public school system. He can study what he is ready to study when he is ready to study it (which is the flexibility advantage of homeschooling) and I am happy to have him declared at his default grade level in math contests. He has two more years of eligibility for Mathcounts and for AMC 8, four more years for AMC 10, and six more years for AMC 12. He's got plenty of math to learn, and I hope he has sufficient time to learn it.[/quote]

I personally never skipped, nor do I have any official data (i.e. numbers).  I understand and agree with you as far as your son goes - if you're homeschooling him, there's no point in "skipping," as that will just take away valuable study time before college and at a valid level for high school (and middle school) competitions.  I go to AAST, and there are a bunch of kids there who have been pushed ahead - two of my friends (both on math team) are 13 as freshmen (and won't be turning 14 soon.. they "should be" in seventh grade), and a bunch are 14 ("should be" eighth grade).

I'm far from an advocate for grade-skipping; I think it's best to enjoy your time in elementary/middle/high school, and learn as much as you can, along with others of your own maturity level.  But I guess if some parents feel it's best for their children, that's their decision.

I have a feeling I went off-topic.. I probably shouldn't post on here after midnight while attempting to do physics homework ;)

Yes, that's true. There are also some kids (mainly homeschoolers) who are way ahead (in schoolwork) than their peers but mark the usual grade level of someone their age on competitions. (Not that I've ever done that.)

Also, I'm willing to bet that almost everyone on this site is taking math at a more advanced level than their non-mathematically-motivated peers.

I've always wondered: what's to stop a homeschooler from saying he or she is repeating a year of school in order to gain eligibility for competitions.  Say, if  a 15-year-old homeschooler says he's an 8th grader so that he can compete in MATHCOUNTS another year.

And what about people who take a post-graduate year (13th grade)?  Are they still allowed to compete in math competitions?

[quote="mathfanatic"]I've always wondered: what's to stop a homeschooler from saying he or she is repeating a year of school in order to gain eligibility for competitions.[/quote]

It's not just homeschoolers. Careful examination of the AMC site will reveal a few instances of famous test-takers who seem to be in the same grade for two consecutive years, for reasons I don't entirely understand. (One person seemed to have a grade skip undone, dropping back down to his "natural" grade, when he moved to a different school, but I really don't know the specific circumstances of any of these cases.) That's one of the reasons I never declare a grade skip in the first place: it gets too confusing to keep two sets of records for different purposes. Looking at the national summer programs and what they offer, and at what is available to my son locally, I don't see any reason to grade-skip him. He can get plenty of challenging math between now and when he goes off to university somewhere.

I could have skipped, but my parents were against it - as well as I. I don't think skipping does much except to speed up your life. There is so much to learn and experience while you are younger, so I want to make the most of it.

My parents wanted me to skip a grade, but I refused. Hm.

[quote="mathfanatic"]I've always wondered: what's to stop a homeschooler from saying he or she is repeating a year of school in order to gain eligibility for competitions.  Say, if  a 15-year-old homeschooler says he's an 8th grader so that he can compete in MATHCOUNTS another year.
[/quote]

You can't do it with Mathcounts; they have a 3-year  rule.  But apparently, you can do it with AMC.

[quote="tokenadult"] Careful examination of the AMC site will reveal a few instances of famous test-takers who seem to be in the same grade for two consecutive years, for reasons I don't entirely understand. [/quote]

I've looked at a couple of those cases.  The usual cause is typically inconsistent or incomplete information provided to us by the test taker.

If errors on our site are pointed out to us, we will clean them up as soon as we can.

Steve Dunbar
AMC Director

Apparently someone revived an old [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8057&postdays=0&postorder=asc&start=0]Grade Skipping in Math Contests?[/url] poll.  Interesting.

I have basicly skipped a grade in math but not by choice since I was just put in it.  other wise I am in the grade I'm supposed to be.  I would like to skip a grade though but my parents won't let me.  They say it's a bad idea becuase they want me to just have it as easy as possible they are against me even studying ahead because it might "confuse" me on my current topics.  Many of my peers tell me to skip a grade and leave them alone :lol:.

[quote="BHorseMath"] they are against me even studying ahead because it might "confuse" me on my current topics.  Many of my peers tell me to skip a grade and leave them alone :lol:.[/quote]

Interesting.  My parents get mad at me when I work ahead in my history book, saying that I should just go to bed instead of staying up late doing something pointless.  I agree that the history homework we have to do is pointless, but I'd rather get it over with in one chunk.

[quote="Caroline"][quote="yif man12"]what are the basic math subjects for each grade so you can classify what "skipping" a grade in math is? i know it's probably different for different schools, but is there a typical one?[/quote]

If you're in a standard high school honors math track, you'll usually have algebra 1 in eighth grade, algebra 2 in ninth grade, geometry in tenth, pre-calc/trig in eleventh, and calculus in twelfth (though often algebra 2 and geometry are switched).[/quote]

That would be an honors track, as she said.  The standard track is shifted back one year (algebra 1 in 9th grade, and so on).

acctually i think what caroline wrote applys to more schools most people in my school have parent pushing them to study math, so 1/4 of the kids were learning geometry in 8th grade, and around have is 1/2 is taking algebra1 and a lil less 1/4 is taking algebra 1A, and there is a class for pre-algebra for those who never tries ,so for my school its more of how fast u learn and how much u know.


We have two parallel lines K and L. Triangle ABC has base BC on line L and point A on K. On line K, another point D is made 2006 units from point A. If the height of triangle ABC is equal to the square of the base, and the base is the seventh triangular number, what is the area of triangle DBC?

[hide]$\left[DBC\right] = \left[ABC\right] = \frac{BC \cdot h}{2}= \frac{BC^{3}}{2}$. $T_{n}= \dbinom{n+1}{2}$. Hence $\left[DBC\right] = \frac{28^{3}}{2}= 10976$.[/hide]

[hide]Since DBC and ABC have the same altitude, the ratio of their areas is the ratio of their bases, which is 1. The base is $\frac{7(8)}{2}=28$, thus the altitude is $28^{2}$; the area is $\frac{28^{3}}{2}=10976$[/hide]


$\sqrt[3]{i}$ can be expressed as $\frac{\sqrt{a}+bi}{c}$. Compute $a+b+c$.

Nice work furious. However, the problem may not have a unique answer. Do you remember DeMoivre's Theorem?
We know there are, in fact, three values of $\sqrt[3]{i}$. Assuming $a,b,c$ are integers, then one of the other two forms can also be expressed as $\frac{\sqrt{a}+bi}{c}$.

Well, we would have $i=\text{cis}\left(\frac{\pi}{2}+2\pi k\right)\Rightarrow i^{1/3}=\text{cis}\left(\frac{\pi}{6}+\frac{2\pi k}{3}\right).$  This produces 3 distinct values, namely $\text{cis}\left(\frac{\pi}{6}\right), \text{cis}\left(\frac{5\pi}{6}\right),$ and $\text{cis}\left(\frac{3\pi}{2}\right).$  These equal $\frac{\sqrt{3}+i}2, \frac{-\sqrt{3}+i}2,$ and $-i$, respectively.

The problem should probably have clarified about $a$, $b$, and $c$ a little more...but most likely they wanted what furious did.

And from what furious did, it would appear that he does remember de Moivre's :D

well, i used de moivre's. i, uh, just forgot about the $+2k\pi$ part.  :blush:

Are we assuming a,b,c are integers? Otherwise we could obtain many answers...

Yeah, sorry guys, I kinda assumed a,b,c would be integers in lowest terms etc. but you are correct that there are 3 answers.


1. In triangle ABC, AC=BC. If AC is extended its own length through C to D and segment DB is drawn, find the sum of Angle A+ Angle D.

[hide]
$\angle A = \angle B = \theta$
$\angle BCD = 2\theta \implies \angle D = 90-\theta$
$\therefore \angle A+\angle D = 90^\circ$
[/hide]


Find all pairs of nonnegative integers $(m,n)$ such that
                                              $(m+n-5)^2=9mn$

[quote="pavel kozlov"]This problem comes to the "almost Pell-equation"s $5x^2-y^2=4$ and (2nd variant) $x^2-5y^2=4$. ;)[/quote]
How ?:?

Let $d$ be $gcd(m,n)$. Evident, $d|5$.
1) If $d=1$, since $mn$ is a perfect square, then $m=a^2, n=b^2$ :arrow: $a^2+b^2-5=3ab$ :arrow: $(2a-3b)^2-5b^2=20$  :arrow: $5|2a-3b$ :arrow: $5|a+b$ :arrow: $a=5c-b$ and we get equation
$5(2c-b)^2-b^2=4$.
2) If $d=5$ we analogously get $n=5a^2, n=5b^2$ :arrow: $a^2+b^2-1=3ab$ :arrow:
$(2a-3b)^2-5b^2=4$.

Maybe we can also do it like this:
The equation is:$m^2-(7n+10)m+(n-5)^2=0$
First we assume that $m\geq n$,if $n\geq5$,then we can know $(n,\frac{(n-5)^2}{m}=7n+10-m)$ is also a solution for the equation.So we can get a sequence from each solution $(m,n)(m\geq n\geq5)$:
$a_1=m,a_2=n,a_{k+2}=\frac{a_{k+1}^2}{a_k}$,the sequence will be decreasing until $a_{k+1}<5$.
So we only need to solve the equation when $n=0,1,2,3,4$ we can get $(m,n)=(5,0)(16,1)(1,1)$
From each solution$(m_0,n_0)$ we can get a sequence $b_1=n_0,b_2=m_0,b_{k+2}=7b_{k+1}+10-b_k$.
And all solution is $(b_k,b_{k+1})$.
So the sequence is $b_1=0,b_2=5$ or $b_1=1,b_2=1$ just solve the sequence!(In the second sequence $b_3=16$)


Does anyone here like tables???
i mean PICNIC TABLES!!!!!!!!!!!!!!!!!
cuz i love them..
especially the ones made of wood and has holes in them.. :)
i named all the ones in my neighborhood.

[quote="Ignite168"]Wow, you're sure bored during the summer aren't you?[/quote]
lol actually it's an "inside joke"
i named all the tables about half a year ago.
do you find yourself naming objects that aren't YOURS????

go tables!!

grannies thign??
oh the grannies thing i randomly posted as a reply to one's of Ignite168's topics?
nah that was something i just randomly thought of...
Yanny sounds like Granny.

i also named my car
marine609.

well it's the name of my new car.
my friend thought it'd be fun to name it.. but he first named it gas eater, since it's an SUV and eats alot
but i decided to be optimistic (like always) and name it marine instead.
the 609 is just a random number... cuz i forgot my lisense plate number.

a topic about pinic table... WOW!!!!! what do you find soooooooooooo intresting in boring pinic tables?

piepiepiepie[color=green]pie[/color]piepiepiepie

i love theeeeeeeeeeeeee piepiepiepie[color=green]pie[/color]piepiepiepie on it

no offense but i think this is the "stupidess" forum... lol  :!:  :!:
who would want to talk about TABLES on a AOPS MATH WEBSITE???????????????????? ANSWER THAT PLEASE..