[quote="Centy"]Bands? Bands? Read my profile...I'm more into different composers.[/quote]

Me tooooooo!

Beethoven
Brahms
Schubert
Prokofieff
Ravel
...

:D

My top 10 favourite composers:

Rachmaninov
Mendelssohn
Tchaikovsky
Beethoven
Grieg
Chopin
Brahms
Mozart
Paganini
Dvorak

I'm a out and out Romantic Era fan. My favourite 10 pieces of music of all time are:

1. Mendelssohn - Violin Concerto in E minor
2. Rachmaninov - Piano Concerto No. 2 in C minor
3. Tchaikovsky - Violin Concerto in D major
4. Grieg - Piano Concerto in A minor
5. Tchaikovsky - Piano Concerto No. 1 in Bb minor
6. Grieg - Peer Gynt Incidental Music
7. Beethoven - Romance No. 2 in F major
8. Mendelssohn - A Midsummer Night's Dream Incidental Music
9. Beethoven - Symphony No. 6 in F major
10. Vaughan Williams - The Lark Ascending

Maybe it helps that I play the violin.  :)

[quote="nr1337"]Pink Floyd
Yes
Genesis
Spock's Beard
Dream Theater
Transatlantic
Led Zeppelin
AC/DC
Beatles
Queen
ELO
ELP
Eagles

Among countless others...[/quote]

Wow.  That is surprisingly close to my tastes (Spock's Beard, The Beatles, and Queen are my favorite bands; I am also a fan of ELO and the Eagles, and Yes, Genesis, Dream Theater, and Transatlantic are all bands on my "I really need to hear more of their music" list.)  Hooray for Prog Rock!

Also, add TMBG, Styx, and Trans-Siberian Orchestra to that list.

Bach was the favourite composer of DNA. Read Salmon of Doubt for his descriptipn of a symphony of Bach!

Why should we? He's nothing special, but that's just my opinion, of course.

So many places i've mentioned my favourites  :D

The Rasmus

Guns n roses

Mettalica

Eminem

Black Eyed Peas

Creed

Coheed and Cambria

[quote="Arne"]I can't really make a top 10 since there are just too many works I really really like, but here's a list of a few things I've been listening to (and enjoying!) a lot recently:

Beethoven - late string quartets
Brahms - piano trios and piano concerti
Prokofieff - piano sonatas nrs 6 - 7 - 8
Shostakovich - violin concerto nr 1
Tchaikovsky - violin concerto nr 1
Schubert - piano sonatas[/quote]

Tchaikovsky Violin Concerto No. 1! Have you seen a Violin Concerto No. 2?  :lol:

Wieniawski in my view is the violinst version of Chopin and he would have been just as popular as Chopin if it weren't for Paganini who sort of stole his limelight. Most of the pyrotechnics Wieniawski could do had already been done to the best a violinist could do in Paganini. He sort of made sure no virtuoso violinist could ever step into his shoes. I mean there are a lot of great violinst composers I could have mentioned along with Wieniawski - Vieuxtemps, Ole Bull, Joachim, Kriesler, Ysaye.

Still I like his Violin Concerto No. 2. He is indeed very much like Chopin. Their grander pieces involving orchestra defintiely show them not to be the greatest composer on his time (they rarely strayed away from their own virtuoso instrument) and yet their piano/violin pieces aren't overtly flashy (as say Paganini or Liszt). But they made his niche by composing lovely minituare pieces both drawing from their Polish heritage. Chopin I felt became more popular because he came before those great virtuoso pianists and also he composed with soft delicate melodies (much as Rachmaninov was to do.) Wieniawski never eally managed to compose crowd pleasing tunes and there's always an element of a vrituoso violinst showing off his skills in every work rather than something I can sit back and enjoy played by someone else in a concert hall.

[quote="Centy"]
Tchaikovsky Violin Concerto No. 1! Have you seen a Violin Concerto No. 2?  :lol:
[/quote]

No, don't know what I was thinking :blush:

I'm proud to say I can play the entire 2nd movement of Tchaikovsky's Violin Concerto. SHame about the other two.  :D

The only concerto I've been playing (I'm a pianist) is Saint Saens' second concerto. Wonderful.

Bach did write some music - that was what made DNA mad!

[quote="B-tone_Party"]Wow.  That is surprisingly close to my tastes (Spock's Beard, The Beatles, and Queen are my favorite bands; I am also a fan of ELO and the Eagles, and Yes, Genesis, Dream Theater, and Transatlantic are all bands on my "I really need to hear more of their music" list.)  Hooray for Prog Rock!
[/quote]

Yay, another progger!!  You can't go wront with Yes' [i]Fragile[/i].  It's one of their best albums, along with [i]Close to the Edge, Relayer,[/i] and [i]Going for the One[/i].  For Genesis, start with [i]Selling England by the Pound[/i] and [i]A Trick of the Tail[/i].  People are very different in what Dream Theater albums they like, though a lot of people say that either [i]Awake[/i] or [i]Metropolis Part 2: Scenes from a Memory[/i] are their favorite albums.  I've only heard [i]SMPTe[/i] from Transatlantic, but it was good.

Okay, I'll stop lecturing now... :)

I like all kinds of classical music. My favorite composer is Mozart. I began to appreciate his music after watching Amadeus.  My favorite piece is The Moldau, a symphonic poem by Smetana.

Hmmm...I'm not into music. I don't know what a band is. I don't listen to classical. However, there are some songs I like. I don't even know how to classify them, but these songs are just catchy. There's Palladio by Karl Jenkins, 100 years by Five for Fighting, and If We Hold On Together by...idk, but I heard it on Land Before Time...the movie with the dinosaurs and stuff.

[quote="filletwho"] There's Palladio by Karl Jenkins.[/quote]
Palladio was great :). Our orchestra just played it in the spring concert!

[quote="Centy"]My top 10 favourite composers:

Rachmaninov
Tchaikovsky
Chopin
[/quote]

Umm hmm, good stuff.  I love Rachmaninov's stuff, but his piano pieces are too hard for me to play...(small hands, *sigh*)

I also like music from today...currently my favorite band is OK go. The music video A Million Ways to be Cruel...ah hahaha best one ever.  :P

[quote="juicybooty911"]Jack Johnson!!! :thumbup:[/quote]

Oh yes, I spent a lot of time listening to him.  But...his songs all start to sound the same now.  Definitely overlistened.

[quote="epitomy01"]hey what bands do you guys like?
some stuff that i like:

****RADIOHEAD****
Muse
Coldplay
Travis
U2
Portishead
Sigur Ros
The Smashing Pumpkins
Bloc Party
Blur
Bright Eyes
Death Cab for Cutie
Enya
Fleetwood Mac
Incubus
Interpol
Joe Satriani
Keane
Led Zeppelin
Oasis
Pink Floyd
Powderfinger
R.E.M.
Sarah Mclachlan
Secret Garden
Silverchair
The Corrs[/quote]
There exists a music group R.E.M.? Cool. I like it
Anyways, I ike rap, and my favourite artist is EMINEM.

let's see...

mahler all the way!!!!!

i also like ticheli's stuff, he writes some neat things for wind ensemble

copland is great too, but him and ticheli have a very definitive style that makes al their music sound very similar.

beethoven's 6th is amazing.

dvorak's new world symphony, tchaikovsky's romeo and juliet, borodin's polovetsian dances, holst's first and second, vaughn williams english folk songs suite, the list goes on and on...

i also like kodaly's dances of galanta, our youth orchestra is playing it now and i'm the clarinet soloist  :lol:

jazz is awesome too! stan getz and charlie parker, coltrane, charles mingus.

woooooo music ftw!

The Alkaline Trio is pretty good.  And by pretty good, I mean the best band ever.

"FORT MINOR" IS THE SHIZZ Mike Shinoda roxers boxers

Btw, has anyone else seen the Chase Credit Card commercial with the man going through the different stages of life? The song on there is 100 Years...It's ridiculously good. I think I liked the commercial also because of the Mickey Mouse thing, but it's mostly the song.


BREAK DIED :mad:

WE SHUD TRY TO EMULATE MISSOURI FOR SPAM GUYZ

SPAM SPAM SPAM

LALALALA

1+2=3

YOUR POST WAS VERY WELL APPRECIATED...


Let $ A,B \in M_n{(R)}$ such that : $ A^{2007} \equal{} I_n$,$ B^{2008} \equal{} I_n$,$ AB \equal{} BA$
 Prove that : $ I_n \plus{} A \plus{} B \in GL_n(R)$
$ I_n$ is identity matrix.
 :spider:

Replace the hypothesis $ A^{2007}\equal{}I,B^{2008}\equal{}I$ with $ A^p\equal{}I,B^q\equal{}I$. The eigenvalues of $ A,B$ have a modulus equal to $ 1$. If $ det(I\plus{}A\plus{}B)\equal{}0$ then there exist $ \lambda\in{spectrum(A)},\mu\in{spectrum(B)}$ s.t. $ 1\plus{}\lambda\plus{}\mu\equal{}0$. Then $ \lambda\equal{}e^{2i\pi/3},\mu\equal{}e^{\minus{}2i\pi/3}$ up to an exchange. Thus $ e^{2ip\pi/3}\equal{}e^{\minus{}2iq\pi/3}\equal{}1$ and $ 3$ is a divisor of $ p$ and $ q$.
Here $ 3$ is a divisor of $ 2007$ but is not a divisor of $ 2008$.

[color=purple]
 Excuse me zeng zeng, can you explain what does $ SP_{c}(A)$ mean ?
  Thank you very much !
[/color]


geometry
geometric transformation
homothety
trigonometry
Iran Lemma
graphing lines
slope

In triangle $ABC$, $D$, $E$ and $F$ are the points of tangency of incircle with the center of $I$ to $BC$, $CA$ and $AB$ respectively. Let $M$ be the foot of the perpendicular from $D$ to $EF$.  $P$ is on $DM$ such that $DP = MP$. If $H$ is the orthocenter of $BIC$, prove that $PH$ bisects $ EF$.

[quote="khashi70"]In triangle $ ABC$ , $ D$ , $ E$ and $ F$ are the points of tangency of incircle with the center of $ I$ to $ BC$ , $ CA$ and $ AB$ respectively . Let $ M$ be the feet of perpendicular from $ D$ to $ EF$ and $ P$ is on $ DM$ such that : $ DP \equal{} MP$ . If $ H$ be the orthocenter of $ ABC$, prove that $ PH$ bisects $ EF$.[/quote]
Maybe you have made a typo, check it again.

[color=darkblue]Let $ CI$ cut $ BH$ at $ G$ and $ BI$ cut $ EF$ ate $ Q$. Then

$ \angle GIB = 180^\circ - \dfrac{\angle B + \angle C}{2} = 90^\circ - \dfrac{A}{2} = \angle EFA$.

Thus, the quadrilateral $ BFGI$ is cyclic and

$ \angle{BGI} = \angle BFI = 90^\circ$.

So, points $ B,G$ and $ H$ are collinear. Denote $ K \in HI \cap EF$. Let $ J$ be midpoint of $ EF$. By Menelaos theorem for $ \triangle DMK$ and points $ P,J,H$ it's enough to prove that

$ \dfrac{PM}{PN}\cdot \dfrac{HD}{HK}\cdot \dfrac{JK}{JM} = 1 \Longleftrightarrow \dfrac{HD}{HK} = \dfrac{JM}{JK} \ (*)$.

From $ AF = AE$ we get $ IJ \perp EF$, so $ JI \parallel DM$. Therefore

$ \dfrac{JM}{JK} = \dfrac{ID}{IK}$.

We rewrite the relation $ (*)$ as

$ \dfrac{HD}{HK} = \dfrac{ID}{IK}$

This is equivalent to prove that $ (D,I,K,H)$ is harmonical.
$ QGD$ is orthic triangle of $ HBC$. So $ GI$ is bisector of $ DGK$. From $ \angle IGH = 90^\circ$ we get that $ GH$ is exterior bisector of $ DGK$ and the result follows.[/color]

[img]http://pic.ipicture.ru/uploads/090516/R7pLFPWPV7.png[/img]

Sketch of my solution
1. The intersection points of the altitudes from B and C with the corresponding sides in triangle HBC are on the line EF (well known lemma)
2. H, I and D are collinear. (easy with the use of harmonical division in both triangles ABC and HBC)
3. HBC and DEF are similar. (angle chasing)
4. Let the orthocenter of DEF be H' and the midpoint of EF be N. Now from the well known property, H'D is equal to 2 times IN. Using 3, we find that AI/AD=IN/PD. Then the result follows from the fact that IN//PD.

[b]Here is another aproach for this problem:[/b]
Let $ \{K\}\equiv CI\cap FE$, $ \{G\}\equiv BI\cap EF$. It is very well- know that $ BK\bot CK$ and $ BG\bot CG$ (this result can be proved easily by [b]harmonic division[/b]). Hence $ \{H\}\equiv BK\cap CG$. Let $ J$ be the midpoint of $ EF$. Denote $ P$ the intersection by $ HJ$ and $ DM$. We will prove that $ P$ is the midpoint of $ DM$.
Indeed, let $ S$ be the projection of $ H$ onto $ EF$, in order to prove $ P$ is the midpoint of $ DM$, it is enough to prove $ H(MJYS) \equal{} \minus{} 1$, where $ \{Y\}\equiv HD\cap EF$, which is equivalent to proving $ \frac {\overline {JM}}{\overline {JY}} \equal{} \minus{} \frac {\overline {SM}}{\overline {SY}}$.

Note that $ J$ is the midpoint of $ EF\Longrightarrow IJ\bot EF\Longrightarrow IJ//DM$, which implies $ \frac {\overline {JY}}{\overline {JM}} \equal{} \frac {\overline {IY}}{\overline {ID}}$.

But $ (HIYD) \equal{} \minus{} 1\Longrightarrow \frac {\overline {IY}}{\overline {ID}} \equal{} \minus{} \frac {\overline {HY}}{\overline {HD}} \equal{} \minus{} \frac {\overline {HY}}{\overline {HY} \plus{} \overline {YD}} \equal{} \minus{} \frac {\overline {SY}}{\overline {SY} \plus{} \overline {YM}} \equal{} \minus{} \frac {\overline {SY}}{\overline {SM}}$
$ \Longrightarrow \frac {\overline {SY}}{\overline {SM}} \equal{} \minus{} \frac {\overline {JY}}{\overline {JM}}$.
Which implies that $ (MJYS) \equal{} \minus{} 1\Longrightarrow H(MDPS) \equal{} \minus{} 1$, but $ DM//HS\Longrightarrow P$ is the midpoint of $ DM$, which leads to the result of the problem.
Our proof is completed then.

Sorry for my bad English... :( Here is the proof without calculation.
Lemma:In a triangle ABC,let D be the tangent of incircle $ (I)$ to BC,$ (I_a)$ be the excircle of A wrt triangle ABC and M be midpoint of the altitude from A wrt triangle ABC,then $ D,I_a,M$ are collinear.
Proof:
Let $ (I_a)$ meet BC at N,$ NI_a$ intersect $ (I_a)$ at N and P
The homothety center A take $ (I)$ to $ (I_a)$ also take D to P,then A,D,P are collinear.
Then the homothety center D take the altitude from A wrt triangle ABC to NP,it also take M to $ I_a$
We conclude that $ D,I_a,M$ are collinear.

Return to the problem,apply the lemma to the orthic triangle of triangle BHC then P,H,M' are collinear (M' is the midpoint of EF).
Q.E.D

let $ K$ midpoint of $ EF$ and $ r$ radius of incircle of $ \triangle ABC$.
let $ PH$ cut $ IK$ at $ K'$ so $ \frac{IK'}{DP}\equal{}\frac{HI}{HD}$ ($ i$)(because of $ IK \parallel{} PD$)
it is easy to measure this lenght:
$ IK\equal{}r \sin\frac{A}{2}$ (1)
$ DM\equal{}BD\cos\frac{B}{2}\equal{}2r\cos\frac{c}{2}\cos\frac{B}{2}$ so $ DP\equal{}r\cos\frac{B}{2}\cos\frac{C}{2}$ (2)
with (1),(2): $ \frac{IK}{DP}\equal{}\frac{\sin\frac{A}{2}}{\cos\frac{B}{2}\cos\frac{C}{2}}$ (3).

and $ ID\equal{}BD \tan\frac{B}{2}$,$ DH\equal{} \cot\frac{C}{2}$ so
$ HI\equal{}BD(\cot\frac{C}{2}\minus{}\tan\frac{B}{2})$ so
$ \frac{HI}{DH}\equal{}1\minus{}\frac{\tan\frac{B}{2}}{\cot\frac{C}{2}}\equal{}1\minus{}\tan\frac{B}{2}\frac{C}{2}\equal{}\frac{\sin\frac{A}{2}}{\cos\frac{B}{2}\cos\frac{C}{2}}$ (4).

with (3),(4): $ \frac{IK}{DP}\equal{}\frac{HI}{HD}$ ($ ii$).

with ($ i$),($ ii$): $ PH$ passer from $ K$.
problem solved. :rotfl:  :rotfl:  :rotfl:  :rotfl:

[img]http://pic.ipicture.ru/uploads/090520/dgwLbFWYqV.png[/img]

let $T$ be the intersection point of $PH$ and $EF$, $S$ the intersection point of $IH$ and $EF$, and $R,T'$ be the feet of the perpendicular from $H$ and $I$ to the line $EF$ respectively. since $HR||DM$ and $P$ is the midpoint of $DM$, we have $H(RPDM)=-1$, so $(RTSM)=-1$. on the other side, $EF$ is the line joining feet of the altitudes of vertices $B$ and $C$ of triangle $HBC$ respectively, so $(HISD)=-1$ and therefore $(RT'SM)=-1$. we cocnlude that $T=T'$, hence $T$ is the midpoint of $EF$, and the proof is complete.

(Writing the Idea only, calculation is easy).Take $F=(1,0),E=(-1,0),I=(0,r),M=(a,0),D=(a,b)$  now so equation of circle $\odot(DEF)$ be $x^2+(y-r)^2=r^2+1$. Now note tangent at $(x_1,y_1)$ point be $xx_1+yy_1-r(y+y_1)=1$, using this we get $B,C$ in terms of $a,b,r$. Now easily we’ll get $H$ in terms of $a,b,r$. We’ve midpoint of $MD$ is $P=(a,\frac {b}{2})$ , now basically we need to show now slope of the line is $tan^{-1}(\frac {b}{2})$ and that’s quite easy to show. Note here we’re saying , solution is easy by CG because no stuffs involving any power is coming, and that makes our calculation easier. (I got a trig solution too, but there calculation is damn hard to do.)

I will provide a more straightforward and bashy solution.(but a synthetic solution like you have done is obviously more appreciated)

Applying the sine rule in $\triangle{BFD}$ we get
$\frac{FD}{sinB}=\frac{s-b}{cos\frac{B}{2}} \implies FD=2sin\frac{B}{2}(s-b)$

So $MD=FDcos\frac{C}{2}=2sin\frac{B}{2}cos\frac{C}{2}(s-b) \implies DP=sin\frac{B}{2}cos\frac{C}{2}(s-b)$

Let $K$ be the midpoint of $EF$.Then $IK=IEsin\frac{A}{2}=rsin\frac{A}{2}=(s-b)tan\frac{B}{2}sin\frac{A}{2}$

Now easy angle chasing gives $\angle{ICH}=\frac{A}{2}$ and $\angle{HIC}=90+\frac{C}{2}$

So applying sine rule in $\triangle{HIC}$ we get
$HI=\frac{HCsin\frac{A}{2}}{cos\frac{C}{2}}$

Also $HD=HCcos\frac{B}{2}$.

Now note that $\frac{IK}{PD}=\frac{(s-b)tan\frac{B}{2}sin\frac{A}{2}}{sin\frac{B}{2}cos\frac{C}{2}(s-b)}=\frac{sin\frac{A}{2}}{cos\frac{B}{2}cos\frac{C}{2}}$

Again $\frac{HI}{HD}=\frac{\frac{HCsin\frac{A}{2}}{cos\frac{C}{2}}}{HCcos\frac{B}{2}}=\frac{sin\frac{A}{2}}{cos\frac{B}{2}cos\frac{C}{2}}$

Thus $\frac{HI}{HD}=\frac{IK}{PD}$.Also note that $IK \parallel PD$,both being perpendicular to $EF$.

So $H,K,P$ are collinear as desired.

My solution:
Let $BI\cap EF=T$ note that $\angle BTF=\angle AFE-\angle ABI=\frac{C}{2},\angle ECI=\frac{C}{2}\longrightarrow IETC$ is cyclic so $\angle IEC=\angle ITC=90$ thus $C,T,H$ are collinear similarly $B,S,H$ are collinear now let the perpendicular lines from $D$ to $DE,DF$ intersect $EF$ at $K,L$ note that $BT||DL$ (because both of them are perpendicular to $HC$) and $\angle STB=\angle BTD=\frac{C}{2}$ thus $TL=TD$ so $TH$ is perpendicular bisector of $DL$ similarly $HS$ is perpendicular bisector of $DK$ so $H$ is circumcenter of $\triangle DKL$ now using [url=http://artofproblemsolving.com/community/c6h597515p3545809]this problem[/url] we get that $P,M,H$ are collinear.
DONE

Let $Q=BI \cap EF$, $N=AI \cap EF$, and $K=BI \cap DF$.
Since $\angle CIQ=\angle CBI + \angle ICB =  90^{\circ} - \angle IAE = \angle AEF = \angle CEQ$ and $90^{\circ}=\angle CDI = \angle IEC$ the pentagon $CDIEQ$ is cyclic with $CI$ as diameter, so $\angle IQC = 90^{\circ}$ and hence $Q \in CH$.
Denote by $h$ the homothety centered at $H$ sending $I$ to $D$, and let $R=h(Q)$. We claim that $h(N)=P$ from which the assertion follows since $N$ is also the midpoint of $EF$.
But obviously $h(IN)=DP$ since $IN \parallel DP$, so we just have to show that $h(QN)=RP$. Note first that $h(IQ)=DR$ so $IQ \parallel DR$ which means the pentagon $DKMQR$ is cyclic since $90^{\circ} = \angle QRD = \angle DKQ = \angle DMQ$. Note also that $KP$ is the perpendicular bisector of $DM$ since it is the midline of the right angled triangle $DFM$, so $ \angle DKR = 90^{\circ} - \angle KRD =  90^{\circ} - \angle KMD = \angle DKP$ which yields $R \in KP$, but $KP\parallel FM$ so we are done.

[b]BEAUTIFUL[/b] problem!

Let $E'$ and $F'$ be the points where the perpendiculars from $I$ meet $CH$ and $BH$, respectively. Now, we see that $\angle IDB = \angle BF'I = \angle BFI$ which are all right angles, so $IDBF'F$ is cyclic.

Note that $I$ is also the incenter of $\triangle DE'F'$. This is based upon a well-known property of an orthic triangle with respect to its orthocenter.

Because $CI$ bisects $\angle DCE$, then $$\angle EFD = \frac{\angle EID}{2} = \frac{180^{\circ} - \angle DCE}{2} = 90^{\circ} - \angle DCI = 90^{\circ} - \angle BCF' = \angle F'BC$$ meaning that since $\angle F'BC = \angle F'BD = 180^{\circ} - \angle DFF'$, then we see that $E$, $F$, and $F'$ are all collinear.

Finally, we see that $\angle HF'I = \angle IE'H - 90^{\circ}$ so $HE'IF'$ is a cyclic quadrilateral. Since $H$ is also on $DI$, then $H$ is the $D$-excenter of $\triangle DE'F'$. Since $IE=IF$, then letting $H_I$ be where $IA$ hits $EF$, then $H_IE=H_IF$ and $\angle IH_IE = 90^{\circ}$ meaning that $H_I$ is the place where the incircle of $\triangle DE'F'$ hits $E'F'$. Now, it is well-known that $P$, $H_I$, and $H$ are collinear, so we are done.

Let $U$ and $V$ be $EF \cap CI$ and $EF \cap BI$. By the Iran Lemma, we know that $BU \perp CI$ and $CV \perp BI$. Then $H$ is $BU \cap CV$.

Now, since $H$ is the orthocenter of $\triangle BIC$, we know that $H$ is the $D$-excenter of $\triangle DUV$. Then there is a well known lemma in EGMO that states the following:

$\textbf{Lemma.}$ In triangle $ABC$ with incenter $I$, $A$-excenter $I_A$, and $D$ and $X$ representing the assiociated tangency points on $BC$, lines $DI_a$ and $XI$ concur at the midpoint of the altitude from $A$.

This implies that line $HP$ passes through the tangency point of the incircle of $\triangle DUV$ on $UV$. Since $I$ is the incenter of $\triangle DUV$, the said point is therefore the foot of the altitude from $I$ to $UV$. Then this is clearly the midpoint of $EF$, so we're done. $\square$

Let $N$ be the midpoint of $EF$. Since $I$ bisects minor arc $EF$, $N$ is the foot from $I$ to $EF$. Letting $X=BI\cap EF$ and $Y=CI\cap EF$, we have by EGMO 1.45 [Right angles on incircle chord] that $X$ and $Y$ are the feet of the altitudes from $B,C$ in $\triangle HBC$. Note that since $\triangle DXY$ is the orthic triangle of $\triangle HBC$, $H$ is the $D$-excenter and $I$ is the incenter of $\triangle DXY$. We're done by EGMO 4.13 [Midpoints of altitudes].

Let \(Q=\overline{BI}\cap\overline{EF}\cap(BC)\) and \(R=\overline{CI}\cap\overline{EF}\cap(BC)\) by the Iran lemma. Since \(DQR\) is the orthic triangle of \(\triangle BIC\), \(I\) is its incenter and \(H\) is its \(D\)-excenter. If \(N\) denotes the midpoint of \(\overline{EF}\), \(N\) is tangency point of the incircle of \(\triangle DQR\) and \(\overline{QR}\), so the collinearity is by the well-known ``midpoints of altitudes'' lemma.

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A [b][i]new[/i][/b] solution with Alan Bu, Ankit Bisain, David Dong, Elliott Liu, Eric Shen, Evan Chen, Isaac Zhu, Jeffrey Chen, Kevin Wu, Luke Robitaille, Ram Goel, and Ryan Yang.

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pair C = (6.41, 2.63); label("$C$", C, dir(44.32));
// dependency level 1
// Do not change anything below, unless you are experienced in Asymptote.
path segA_B = A--B; draw(segA_B, c000000);
path segB_C = B--C; draw(segB_C, c0080FF);
path segC_A = C--A; draw(segC_A, c000000);
path abA_B_C = (B-100.0*(bisectorpoint(A, B, C)-B))--(B+100.0*(bisectorpoint(A, B, C)-B));
path abB_C_A = (C-100.0*(bisectorpoint(B, C, A)-C))--(C+100.0*(bisectorpoint(B, C, A)-C));
pair Mp = (B + C)/2;
// dependency level 2
pair I = intersectionpoint(abA_B_C, abB_C_A); label("$I$", I, dir(75.32));
path perC_abA_B_C = (C-100.0*(dir(abA_B_C).y, -dir(abA_B_C).x))--(C+100.0*(dir(abA_B_C).y, -dir(abA_B_C).x));
// dependency level 3
path perI_segA_B = (I-100.0*(dir(segA_B).y, -dir(segA_B).x))--(I+100.0*(dir(segA_B).y, -dir(segA_B).x));
path perI_segC_A = (I-100.0*(dir(segC_A).y, -dir(segC_A).x))--(I+100.0*(dir(segC_A).y, -dir(segC_A).x));
path perI_segB_C = (I-100.0*(dir(segB_C).y, -dir(segB_C).x))--(I+100.0*(dir(segB_C).y, -dir(segB_C).x));
// dependency level 4
pair E = intersectionpoint(segC_A, perI_segC_A); label("$E$", E, dir(63.54));
pair F = intersectionpoint(segA_B, perI_segA_B); label("$F$", F, dir(176.84));
pair D = intersectionpoint(segB_C, perI_segB_C); label("$D$", D, dir(268.65));
pair H = intersectionpoint(perI_segB_C, perC_abA_B_C); label("$H$", H, dir(89.57));
// dependency level 5
pair Q = -1.00000*F + 2.00000*B; label("$Q$", Q, dir(209.86));
pair R = -1.00000*E + 2.00000*C; label("$R$", R, dir(315.00));
path circI_F = Circle(I, abs(I-F));
pair circI_Fcenter = I; real circI_Frad = abs(I - F); draw(circI_F, c000000);
path lineF_C = (F-100.0*unit(C-F))--(C+100.0*unit(C-F));
path lineB_E = (B-100.0*unit(E-B))--(E+100.0*unit(E-B));
pair X = (F + E)/2; label("$X$", X, dir(142.60));
path lineE_F = (E-100.0*unit(F-E))--(F+100.0*unit(F-E));
path segE_F = E--F; draw(segE_F, c000000);
path segB_H = B--H; draw(segB_H, c0080FF);
path segH_C = H--C; draw(segH_C, c0080FF);
path segD_I = D--I; draw(segD_I, c20C020);
// dependency level 6
path lineQ_I = (Q-100.0*unit(I-Q))--(I+100.0*unit(I-Q));
path lineR_I = (R-100.0*unit(I-R))--(I+100.0*unit(I-R));
pair G = intersectionpoint(lineF_C, lineB_E); label("$G$", G, dir(125.41));
pair Xp = -1.00000*X + 2.00000*Mp; label("$X'$", Xp, dir(77.40));
path perD_lineE_F = (D-100.0*(dir(lineE_F).y, -dir(lineE_F).x))--(D+100.0*(dir(lineE_F).y, -dir(lineE_F).x));
path segQ_I = Q--I; draw(segQ_I, c0080FF);
path segI_R = I--R; draw(segI_R, c0080FF);
path segR_Q = R--Q; draw(segR_Q, c0080FF);
path segQ_D = Q--D; draw(segQ_D, c20C020);
path segD_R = D--R; draw(segD_R, c20C020);
path segB_Q = B--Q; draw(segB_Q, c000000);
path segC_R = C--R; draw(segC_R, c000000);
// dependency level 7
path perC_lineQ_I = (C-100.0*(dir(lineQ_I).y, -dir(lineQ_I).x))--(C+100.0*(dir(lineQ_I).y, -dir(lineQ_I).x));
path perB_lineR_I = (B-100.0*(dir(lineR_I).y, -dir(lineR_I).x))--(B+100.0*(dir(lineR_I).y, -dir(lineR_I).x));
pair M = intersectionpoint(lineE_F, perD_lineE_F); label("$M$", M, dir(336.32));
// dependency level 8
pair K = intersectionpoint(perC_lineQ_I, perB_lineR_I); label("$K$", K, dir(123.30));
pair P = (M + D)/2; label("$P$", P, dir(189.70));
path segD_M = D--M; draw(segD_M, c000000);
// dependency level 9
pair Kp = -1.00000*K + 2.00000*Mp; label("$K'$", Kp, dir(281.77));
path segH_P = H--P; draw(segH_P, cFF0000);
//dots
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(I); dot(K); dot(Kp); dot(M); dot(P); dot(Q); dot(R); dot(X); dot(Xp);
// clip the drawing view
clip((0.0, 0.0)--(0.0, 6.81)--(10.0, 6.81)--(10.0, 0.0)--cycle);
[/asy]

Define $Q$ and $R$ to be the reflections of $F$ and $E$ over $B$ and $C$, respectively. Define $G$ as the Gergonne point of $\triangle ABC$ (by tangents, $G$ is also the symmedian point of $\triangle DEF$). Also let $X$ be the midpoint of $EF$. By the Schwatt line of $\triangle DEF$, it suffices to show that $H\in XG$. By [url=https://artofproblemsolving.com/community/c6h2604132p22480483]a nice geometry problem 10[/url], we have that $XG\perp QR$.

[color=#f00][b]Claim:[/b][/color] There exists a point $K\in XG$ such that $BK\perp RI$ and $CK\perp QI$.

[i]Proof.[/i] Reflect $X$ over the midpoint of $BC$ to get $X'$, the midpoint of $QR$. It suffices to show the existence of a point $K'$ on the perpendicular bisector of $QR$ such that $BK'\perp QI$ and $CK'\perp RI$. But such a point does exist: consider $K'$ as the radical center of circles of radius 0 centered at $Q$ and $R$, and the incircle of $\triangle ABC$. $\square$

[color=#f00][b]Claim:[/b][/color] $ID\perp BC$, $QD\perp CH$, and $RD\perp BH$.

[i]Proof.[/i] The first assertion is trivial. The second follows from $QD\parallel BI$ and $BI\perp CH$, and the third follows symmetrically from the second. $\square$

Now, since $\triangle BHC$ and $\triangle QIR$ are orthologic (since there is a point $D$ such that $ID\perp BC$, $QD\perp CH$, and $RD\perp BH$), then there is a point $K$ such that $BK\perp RI$, $CK\perp QI$, and $HK\perp QR$. Therefore, $H$ lies on the line through $X$, $G$, and $K$, so $H\in XG$, as desired. $\blacksquare$

Let $N$ be the midpoint of $EF$, $X$ and $Y$ be the feet from $C,B$ to $BI,CI$, respectively. Triangle $DXY$ is the orthic triangle of $HBC$ as $ID\perp BC$. and because $I$ is the orthocenter of $HBC$, it is the incenter of $DXY$. By Iran lemma, line $XY$ coincides with line $EF$, so $N$, the foot from $I$ to $EF$, is the $D$-intouch point. By midpoint of altitudes, it remains to prove that $H$ is the excenter, but this is obvious since $\angle IXH=\angle IYX=90^\circ$.

Suppose we are back in 2009 where the Iran Lemma and Midpoint of Altitude Lemma were not well-known...

Let $N$ be the midpoint of $\overline{EF}$ and let $CI$ meet $EF$ at $Y$.

[b]Claim:[/b] Pentagon $FBDIY$ is cyclic.

[i]Proof.[/i] We have $FBDI$ is cyclic as $\measuredangle BDI=\measuredangle BFI=90^\circ$. It suffice to show that $Y$ lies on this circle. To do this, we will use Barycentric Coordinates.

Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can quickly get $D=(0:s-c:s-b),E=(s-c:0:s-a),F=(s-b:s-a:0),I=(a:b:c)$.

The coordinates of $Y=\overline{EF}\cap\overline{CI}$ are given by the solution to the system

\begin{align*}
\overline{EF}:& \begin{vmatrix}s-c&0&s-a\\ s-b&s-a&0\\ x&y&z\\ \end{vmatrix}=0\\
\overline{CI}:& \begin{vmatrix}0&0&1\\ a&b&c\\ x&y&z\\ \end{vmatrix}=0\\
&x+y+z=1\\
\end{align*}
which we reduce to

\begin{align*}
z(s-c)+y(s-b)-x(s-a)=&0\\
ay=&bx\\
z=&1-x-y\\.
\end{align*}
Solving this is just a matter of careful algebra. Upon substituting the third equation into the first for $z$, expanding, and rearranging, we obtain

\begin{align*}
(1-x-y)(s-c)+y(s-b)-x(s-a)=&0\\
(s-c)-x(s-c)-y(s-c)+y(s-b)-x(s-a)=&0\\
-x(s-c+s-a)-y(s-c-(s-b))=&-(s-c)\\
-bx-y(b-c)=&-(s-c)\\
bx+y(b-c)=&s-c\\
\end{align*}for the first equation.

Hence, our system is now

\begin{align*}
bx+y(b-c)=&(s-c)\\
ay=&bx\\
z=&1-x-y\\.
\end{align*}
Substituting the second equation into the first, we have$$ay+y(b-c)=s-c \iff y(a+b-c)=\frac{a+b-c}2$$so $y=\frac {1b}{2b}$. Now it is trivial to get $x=\frac{a}{2b}$ and $z=\frac{-a+b}{2b}$ so$$y=\left(\frac{a}{2b},\frac{b}{2b},\frac{-a+b}{2b}   \right).$$
Now, we seek the equation of the circle $(DBFI)$. In Barycentric Coordinates, the general form for a circle is $-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$ for $u,v,w\in \mathbb{R}$. Since three points determine a circle, we will use $B=(0,1,0),D=(0:s-c:s-b),F=(s-b:s-a:0)$ for convenience.

Substituting these three points into the circle formula, we have

\begin{align*}
v=&0\\
-a^2(s-c)(s-b)+(v(s-c)+w(s-b))(s-c+s-b)=&0\\
-c^2(s-a)(s-b)+(u(s-b)+v(s-a))(s-a+s-b)=&0\\
\end{align*}which simplifies to$$w=\frac{a^2(s-c)(s-b)}{a(s-b)}=a(s-c)$$and$$u=\frac{c^2(s-a)(s-b)}{c(s-b)}=c(s-a)$$upon substituting $v=0$.

Hence, $(BDF)$ has equation$$-a^2yz-b^2xz-c^2xy+(xc(s-a)+za(s-c))(x+y+z)=0$$or$$-a^2yz-b^2xz-c^2xy+\frac{(xc(-a+b+c)+za(a+b-c))}{2}(x+y+z)=0.$$
Now all that remains is to check that $Y=\left(\frac{a}{2b},\frac{b}{2b},\frac{-a+b}{2b}   \right)$ lies on this circle. Indeed,

\begin{align*}
\frac{(xc(-a+b+c)+za(a+b-c))}{2}(x+y+z)=&a^2yz+b^2xz+c^2xy\\
\frac{a^2b(-a+b)+b^2a(-a+b)+c^2ab}{4b^2}=&\frac{ac(-a+b+c)+(a-b)(a+b-c)}{2\cdot 2b}\\
\frac{ab[a(b-a)+b(b-a)+c^2]}{b}=&a[c(-a+b+c)+(b-a)(a+b-c)]\\
(b+a)(b-a)+c^2=&-ac+cb+c^2+ab+b^2-bc-a^2\\
b^2-a^2+c^2=&c^2+b^2-a^2\\
\end{align*}
which is clearly true. $\square$

In particular, we learn that $\measuredangle BYI=\measuredangle BDI=90^\circ$. If we define $X$ as the intersection of lines $EF$ and $BI$, we learn that $\measuredangle CXI=90^\circ$ through similar reasoning. Therefore, lines $BY$ and $CX$ meet at $H$ and we conclude that $\triangle HBC$ has orthic triangle $\triangle DYX$ with orthocenter $I$.

By the duality of Orthocenters and Excenters, we see that $\triangle DXY$ has excentral triangle $\triangle HBC$ and incenter $I$. Furthermore, since $N$ is the midpoint of chord $EF$ in $(DEF)$ with center $I$, the incircle of $\triangle DXY$ touches $XY$ at $N$.

All that remains is to show that $P,N,H$ are colinear. To do so, we will prove the following lemma.

[b]Lemma:[/b] In $\triangle ABC$, let $H$ be the foot of the $A$-alititude and let $A_0$ be the midpoint of $\overline{AH}$. Let $D$ be the $A$-touchpoint and let $I_a$ be the $A$-excenter. Then $A_0,D,I_a$ are colinear.

[i]Proof.[/i] Set $\triangle ABC$ as the reference triangle with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We also have $D=(0:s-c:s-b)$ and $I_a=(-a:b:c)$. Foot of $A$-altitude has coordinates $H=\left(0:a^2+b^2-c^2:a^2-b^2+c^2\right)$. We can normalize by multiplying each component by $\frac{1}{2a^2}$. Applying the midpoint formula on $\overline{AH}$ gives $A_0=\left(2a^2:a^2+b^2-c^2:a^2-b^2+c^2\right)$.

We claim that $A_0,D,I_a$ are colinear. To prove this, we just need to verify$$\begin{vmatrix}-a&b&c\\ 0&s-c&s-b\\ 2a^2&a^2+b^2-c^2&a^2-b^2+c^2\end{vmatrix}=0$$which is true upon expansion.$\square$

Applying this lemma to $\triangle DXY$ where $P\mapsto A_0, N\mapsto D, H\mapsto I_a$, we are done! $\blacksquare$

Let $Z$ be the midpoint of $EF$ and $P'=DM\cap ZH$. From the ceva-menelaus configuration we have \[ -1=(H, I; EF\cap DH, D)\stackrel{Z}{=}(P', \infty; M, D) \] Means $P'$ is the midpoint of $DM$. Hence $P'\equiv P$.

Let $X = \overline{BI} \cap \overline{EF}$ and $Y = \overline{CI} \cap \overline{EF}$, and denote the midpoint of $\overline{EF}$ as $N$. By Iran Lemma, $X$ and $Y$ are the feet of the $C$ and $B$ altitudes in $\triangle BIC$, respectively. So, $\triangle DXY$ is the orthic triangle of acute $\triangle BHC$, hence $I$ is its incenter and $H$ is its $D$-excenter. Then by the Midpoint of altitudes lemma, we find that $P$, $N$, and $H$ are collinear since $N$ is the foot from $I$ onto $\overline{XY}$, as desired.

Let $BI\cap CH = K$, $CI\cap BH = L$, and let $N$ be the midpoint of $EF$. Note that $I$ is the antipode of $H$ in $(HKL)$, and $K$ and $L$ are on $EF$ by Iran Lemma. Then since $\triangle DEF$ and $\triangle HLK$ are homothetic, and if $M'$ is the reflection of $M$ over $N$, $DM'$ is the same cevian in $\triangle DEF$ as $HN$ in $\triangle HKL$ (isotomic to altitude), we have $PN\parallel DM'\parallel HN$ as desired.

Let $T_B$ and $T_C$ denote the feet from $B$ to $CH$ and the feet from $C$ to $BH$ respectively. Note that $BICH$ is an orthocentric system, so $B,I,T_B$ are collinear as well as $C,I,T_C.$ By that egmo lemma, $BI$ and $EF$ intersect at some point $K$ for which $KB\perp CK.$ Thus, $K=T_B$, since the intersections of $BI$ and $(BC)$ are $B$ and $T_B$.\\

Now, we have an orthic triangle configuration with $D,T_B,T_C$ in $\triangle HBC$. Hence, $I$ is the incenter of $\triangle DT_BT_C.$ Since $T_B$ and $T_C$ lie on $EF$, the foot from $I$ to $T_BT_C$ is just the midpoint of $EF$.\\

Hence, in $\triangle DT_BT_C$, the midpoint of the D-altitude, $P$, the intouchpoint to $T_BT_C$, which is the midpoint of $EF$, and the $D$-excenter, which is $H$, are collinear, hence $PH$ bisects $EF$ as desired.

My sol is crazy long bruh. Just noticed that this actually connects up with 2011 ISL G4.

 One first considers the following claim.

[color=#f00][b]Claim (Partial Iran Lemma) :   The feet of the perpendiculars from $B$ to $CI$ and $C$ to $BI$ respectively lie on $\overline{EF}$.[/b][/color]

    Let $C_1=\overline{EF} \cap \overline{CI}$. Then, note that
    \begin{align*}
        2 \measuredangle EC_1I &= 2 \measuredangle ECI +2 \measuredangle C_1EC\\
        &= \measuredangle ACB + 2 \measuredangle FEA\\
        &= \measuredangle ACB + \measuredangle FAE \\
        &= \measuredangle ACB + \measuredangle BAC\\
        &= \measuredangle ABC\\
        &= 2\measuredangle FBI
    \end{align*}
    Thus, $\measuredangle EC_1I= \measuredangle FBI$. This means that $FBIC_1$ is cyclic and thus $\measuredangle IC_1B = 90^\circ$ which means that indeed $C_1$ is the foot of the perpendicular from $B$ to $CI$. Similarly, the foot of the perpendicular from $C$ to $BI$ also lies on $EF$ and we have our claim.

Now, we have the following very important claim.
[color=#f00][b]Claim (Extended 11SLG4) :
    Consider a triangle $ABC$ with $D,E,F$ the feet of the perpendiculars from $A,B,C$ to the opposite sides. Let $H$ be its orthocenter and $Y_A$ be the $A-$Why Point of $\triangle ABC$. Let $P$ be the foot of the altitude from $D$ to $EF$ Then, the following points are collinear,
    1. $A$
    2.  Foot of the perpendicular from $H$ to $EF$ ($Z$)
    3. Midpoint of the altitude from $D$ to $EF$ ($M$)
    4. $Y_A$
[/b][/color]

    First, consider the inversion centred at $A$ with radius $\sqrt{AD \cdot AH}$. This clearly maps $D$ to $H$, $B$ to $F$ and $C$ to $E$ (and vice versa). Thus, $\overline{EF}$ maps to $(ABC)$ and $\overline{BC}$ maps to $(AEF)$ and thus the $A-$Ex Point $X_A$ maps to the $A-$Queue Point $Q_A$ under this inversion. Further, since $D$ and $H$ map to each other, this in fact means that the circle $(X_AH)$ remains fixed under this inversion ($Q_A$ clearly lies on this circle as $\measuredangle HQ_AX_A=90^\circ$). Further, $Z$ will map to the intersection of $(X_AH)$ and $(ABC)$. The claim is that this intersection is $Y_A$.\\

    Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. Let $G$ be the centroid of $\triangle ABC$ and $W$ the intersection of $\overline{AM}$ and $(ABC)$. It is well known (from 11SLG4) that $Y_A-D-G-A'$ (with $A'G=2DG$). Simply note that
    \[GM \cdot GW = \frac{AG \cdot GW}{2} = \frac{A'G\cdot GY_A}{GD \cdot GY_A} = GD \cdot GY_A\]
    Thus, $DMWY_A$ is cyclic. Now, let $X_A' = \overline{Y_AW} \cap \overline{BC}$. Then,
    \[X_A'D\cdot X_A'M = X_A'Y_A\cdot X_A'W = X_A'B \cdot X_A'C\]
    Thus, $X_A' \equiv X_A$ and we have that $X_A-Y_A-W$. Let $H_A$ be the $A-$Humpty Point. It is well known that $X_A-H-H_A$ and $HH_AMD$ is cyclic. Now, note that from this it follows that,
    \[\measuredangle X_AY_AD = \measuredangle WY_AD = \measuredangle AMD=\measuredangle H_AMD = \measuredangle H_AHA = \measuredangle X_AHD\]
    and thus, $X_AHDY_A$ is cyclic and $Y_A$ lies on $(X_AH)$ as desired.\\

    Thus, $Z$ maps to $Y_A$ in the previously described inversion and we have that $A-Z-Y_A$.\\

    Now, we deal with the other part of the collinearity. Let $N$ be the midpoint of $X_AD$. We already know that $X_AY_ADZ$ is cyclic. Further, since $B,C,H_A$ and $Y_A$ are cyclic and $(X_AD;BC)=-1$ and we also have,
    \[\measuredangle H_AY_AD = \measuredangle H_AA' =90^\circ\]
    $Y_A$ must be the $H_A-$Humpty Point of $\triangle X_ADH_A$. Further, this implies that $H_AY_A$ is the $H_A-$median of this triangle which implies $H_A-Y_A-N$. Thus, $\measuredangle NY_AD = 90^\circ$. Further, $MN \parallel X_AP$ by Midpoint Theorem, which implies that $NY_ADM$ must indeed be cyclic.\\

    Now,
    \[\measuredangle DY_AM = \measuredangle DNM = \measuredangle DX_AZ = \measuredangle DY_AZ\]
    This clearly implies that $Y_A-M-Z$ and we put these two together and conclude that indeed $A-Z-M-Y_A$ are collinear as claimed.

By the first claim, $EF$ is the line joining the feet of the altitudes from $B$ and $C$ to the opposite sides. One now notices that by applying the above-proved claim on $\triangle BHC$, we must have $H, M$ and the foot of the perpendicular from $I$ to $EF$ collinear. But, clearly $AI$ is the perpendicular bisector of $EF$ meaning that, the perpendicular from $I$ to $EF$ passes through the midpoint of $EF$.\\

Thus, we have that $\overline{PH}$ bisects $\overline{EF}$ as was required.

Let $N$ be the midpoint of $[EF]$,$B'=(BI)\cap (EF)$and $C'= (CI)\cap(EF)$. By the  Iran lemma (provable by angle chase), the triangle $\triangle DB'C'$ with incenter $I$ is the orthic triangle of $\triangle HBC$. Now since $P$ is the midpoint of the $D-$altitude in $\triangle DB'C'$ and $H$ the $D-$excenter point, the collinearity is well known(4.3 midpoints of Altitudes EGMO).

$$\mathbb{Q.E.D.}$$

Let $K = BI \cap EF$ and $L = CI \cap EF$. Iran Lemma tells us $\triangle DKL$ is the orthic triangle of $\triangle HBC$. Hence Midpoint of the Altitude Lemma this triangle with altitude midpoint $P$, touch point $M$ (midpoint of $EF$), and excenter $K$ finishes. $\blacksquare$

alright here's the stock solution I guess

Let $X$ and $Y$ be the feet from $C$ to $\overline{BI}$ and from $B$ to $\overline{CI}$, which both lie on $\overline{EF}$. It follows that $DXY$ is the orthic triangle of $HBC$, i.e. $H$ is the $D$-excenter of triangle $DXY$ and $I$ is the incenter. Midpoint of altitudes lemma implies $K = \overline{PH} \cap \overline{EF}$ satisfies $\angle IKE = 90^\circ$, or $K$ is the midpoint of $\overline{EF}$.


Question 1.

$p(x) = 3x^3 - 2x^2+x-1$ has roots $\alpha, \beta, \gamma$

Find the roots.

Question 2.

$q(x) = 2x^4-3x^3+x^2+5x-7$ has roots $\alpha, \beta, \gamma, \delta$ Find the roots.

Question 3.

The roots of $2x^3-19x^2+57x-54=0$ for a geometric progression. Solve the equation.

I will post my solution for Q3.  :D

[hide]
let the roots in GP be $ap, ap^2, ap^3$

We know $a^3p^6 = 27 \implies ap^2=3$

Hence we know our middle root is $3$, the other two have to be $\frac{3}{p}$ and $3p$

We know that $\frac{3}{p}+3+3p = \frac{19}{2}$

Solve this, we get a quadratic $\implies (2p-3)(3p-2)=0$

WLOG take $p=\frac{3}{2}$ and we find out our roots, $2, 3, 4.5$[/hide]

[quote="sen"]Question 1.

$p(x) = 3x^3 - 2x^2+x-1$ has roots $\alpha, \beta, \gamma$

Find the roots.

Question 2.

$q(x) = 2x^4-3x^3+x^2+5x-7$ has roots $\alpha, \beta, \gamma, \delta$ Find the roots.
[/quote]

Yeah, I tried working on these two and I finally gave up and went to Mathematica... and um... the zeros aren't exactly what you would call pretty...

Are you sure you typed in the polynomials correctly?

An alternative for doing question 1 might be to use the procedure for solving for a general cubic equation, but that gets pretty ugly too.  Soooooooo yeah.  :)

yeh they are right.

its an exercise to determine ugly roots i guess, like surds and stuff. :D


[hide="10 to the 27th power written out"]
I'll write it out just for fun :D
$100,000,000,000,000,000,000,000$[/hide]

[quote="riddler"]easy question....
$\frac{10^{27}+2}{3}$
find the sum of the digits of the fraction above.[/quote]

[hide]$\frac{10^{27}+2}{3}$

$\frac{1000...0002}{3}$

$33333......33334$

How many 3's are there? The numerator loses one zero when the 2 comes in. and loses another digit when it's divided by three. So there are $27-1=26$ 3's in the final number. The sum of all the 3's are, of course, $3*26$ or $78$

Therefore the sum of the digits of the number must be 82[/hide]

[quote="riddler"]easy question....
$\frac{10^{27}+2}{3}$
find the sum of the digits of the fraction above.[/quote]

You can look at a pattern, you'll see that $10^x+2$ when divided by $3$ leaves an $x-1$ amount of $3$s plus a $4$, therefore, $10^{27}$ has $26$ threes and a $4$ so the sum is $3(26)+4=\boxed{82}$.

[quote="riddler"]yes you are all right, was this the correct forum to put this in?[/quote]

Yeah, or perhaps Mathcounts..

I think this problem was Mathcounts  ;)  I did something like this. Begin long division, that's all.

okay I get your point anyway I posted it for a quick solution. :P  :P  :P  :P


Can somebody please explain what a telescopic sum is ?
 :blush:

A telescopic sum is a sum that can be rewritten so the most of it's terms cancel out. For example we can rewrite a sum as two sums of consecutive funtions like $\sum_{k=1}^{n}(f(k+1)-f(k))$ which is after canceling it's $f(n+1)-f(1)$.

Can you explain more because I got lost at that point...

[quote="DanK"]How do we get from
$\sum\limits_{k = 2}^{2006} {\frac{2}{{k^2 - 1}}} = \sum\limits_{k = 1}^{2005} {\frac{1}{k}} - \sum\limits_{k = 3}^{2007} {\frac{1}{k}}$
to
$\sum\limits_{k = 2}^{2006} {\frac{2}{{k^2 - 1}}} = \frac{1}{1} + \frac{1}{2} - \frac{1}{{2006}} - \frac{1}{{2007}}$?
Is this some formula I don't know?[/quote]

No formula, just a little logic. The only four distinct values for $k$ that aren't cancelled out are 1, 2, 2006, 2007...

Can someone help me out?

How do we get from here:

"Thus\[ \sum\limits_{k = 2}^{2006} {\frac{2}{{k^2 - 1}}} = \sum\limits_{k = 2}^{2006} {\frac{1}{{k - 1}}} - \sum\limits_{k = 2}^{2006} {\frac{1}{{k + 1}}}  \]"

To here:

"Thus\[ \sum\limits_{k = 2}^{2006} {\frac{2}{{k^2 - 1}}} = \sum\limits_{k = 1}^{2005} {\frac{1}{k}} - \sum\limits_{k = 3}^{2007} {\frac{1}{k}}  \]"

does the negative and positive $1$ in the denomitor allow $k$ and the limit be added and subtracted by $1$?

I'll do my best to explain Vihang's solution.
[hide="Explanation"]
First of all, note that $\frac{2}{k^2-1}=\frac{1}{k-1}-\frac{1}{k+1}$.

Hence $\sum_{k=2}^{2006} \frac{2}{k^2-1} =\sum_{k=2}^{2006} \left(\frac{1}{k-1}- \frac{1}{k+1} \right)$

We can see why the sum telescopes if we start writing out a couple of terms.
$\sum_{k=2}^{2006} \left(\frac{1}{k-1} -\frac{1}{k+1} \right)$

$=\left(\frac11-\frac13\right) +\left(\frac12-\frac14\right) +\left(\frac13-\frac15\right) +\left(\frac14-\frac16\right) +... +\left(\frac{1}{2004} -\frac{1}{2006}\right) +\left(\frac{1}{2005} -\frac{1}{2007}\right)$

Note that $\frac13$, $\frac14$, $\frac15$... all cancel out and we are left with $\frac11+ \frac12 -\frac{1}{2006} -\frac{1}{2007}$.
[/hide]

Oh I see now...  so basically only the last two terms and first two terms are not cancelled out... That makes a lot of sense.  Thanks frt.

[quote="Slizzel"]A telescopic sum is a sum that can be rewritten so the most of it's terms cancel out. For example we can rewrite a sum as two sums of consecutive funtions like $\sum_{k=1}^{n}(f(k+1)-f(k))$ which is after canceling it's $f(n+1)-f(n)$.[/quote]Slizzel, I'm sure you meant $f(n+1)-f(1)$

[quote="E^(pi*i)=-1"]Find $\displaystyle\sum_{k=2}^{2006}\displaystyle\frac{2}{k^2-1}$[/quote]

[hide]Notice that $k^{2}-1=(k+1)(k-1)$.

We shall try to find a way of expressing $\frac{2}{k^{2}-1}$ in the form of $\frac{a}{k-1}+\frac{b}{k+1}$.

This sum of two fractions is equal to
$\frac{(a+b)k+(a-b)}{k^{2}-1}$.

So now we have to equate this into a system of equations
$a+b=0$ and $a-b=2$.

Solving these the solution is $(1,-1)$.

So this means that $\frac{2}{k^{2}-1}=\frac{1}{k-1}-\frac{1}{k+1}$.

So basically the sum that we are trying to find is
$\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2007}$.

Notice that this is a telescoping series.  We can show that the fractions $\frac{1}{3}$, $\frac{1}{4}$,...$\frac{1}{2005}$ will disappear.

So the answer is $\frac{1}{1}+\frac{1}{2}-\frac{1}{2006}-\frac{1}{2007}$, which equals $\frac{3017525}{2013021}$

So the answer is $\frac{3017525}{2013021}$.[/hide]


inequalities
algebra
system of equations
inequalities unsolved

Solve the following system of equations in real numbers x,y,z:

              3(x^2 + y^2 + z^2) = 1

              x^2.y^2 + y^2.z^2 + z^2.x^2 = xyz(x+y+z)^3

(I found a hint: use ineq.)

By Cauchy, we have 1-(x+y+z)2 >=1-3(x2 +y2 +z2)=0.

But using Rearrangment Inequality, we see that x2y2 +y2z2 +z2x2 -xyz(x+y+z)3 >=xyz(x+y+z)(1-(x+y+z)2)>=0. Note that this is not true if xyz<0 or (x+y+z)<0, in which case, however, x2y2 +y2z2 +z2x2  >=0>xyz(x+y+z)3. Hence the only solutions are x=y=z=1/3.

Apologize for the double posting, but I just realized that the solutions should be x=y=z=1/3 and x=y=z=-1/3. (They just have to be equal, not necessarily positive.)


geometry rectangle trigonometry rotation AMC AIME inequalities

In the adjoining figure, two circles of radii 6 and 8 are drawn with their centers 12 units apart.  At $P$, one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length.  Find the square of the length of $QP$.

[asy]unitsize(2.5mm);
defaultpen(linewidth(.8pt)+fontsize(12pt));
dotfactor=3;

pair O1=(0,0), O2=(12,0);
path C1=Circle(O1,8), C2=Circle(O2,6);
pair P=intersectionpoints(C1,C2)[0];
path C3=Circle(P,sqrt(130));
pair Q=intersectionpoints(C3,C1)[0];
pair R=intersectionpoints(C3,C2)[1];

draw(C1);
draw(C2);
//draw(O2--O1);
//dot(O1);
//dot(O2);
draw(Q--R);

label("$Q$",Q,N);
label("$P$",P,dir(80));
label("$R$",R,E);
//label("12",waypoint(O1--O2,0.4),S);[/asy]

I think this involves similar triangles.  By drawing perpendicular bisectors from the centers to the chords.

i think you have a typo..shouldnt it be $6^{2}$ and $8^{2}$

in the other thread, i proved that R is collinear with two centers. so you can work on it without pythagorean  :D

Another related problem:

given the two circles, find a ruler and compass construction for $QRP$...without doing a length brute force

I did the three pythagorean sums and it works pretty quickly as well.

[quote="Altheman"][quote="Elemennop"]Haha, i totally guessed R was collinear with the centers...

what's the proof for it?[/quote]

be careful, this statement is not generally true, it is just because 6*2=12[/quote]

yea, I figured that was true

the only guess at showing it is collinear is assume it isn't, then draw the extension through the centers, line L,  until it hits line QPR and use similar triangles to show that the extension of line QPR from the perpendicular intersection in the smaller circle out to the intersection with line L must have length QP/2, so then the only possible way it could intersect is when R is collinear with line L.

But I totally have no clue if that's even right

Found another approach. Similar, but different.  :)

[hide="Sketch of Solution"]Let the circles of radius $8$ and $6$ be centered at $A$ and $B,$ respectively. Let the midpoints of $QP$ and $PR$ be $N$ and $O.$ Dropping a perpendicular from $B$ to $AN$ (let the point be um... $K?$) gives a rectangle, etc.

Ok, now note that triangle $ABK$ is right. Let the midpoint of $AB$ (segment of length 12) be $M.$ Hence, $KM=6=BM=BP.$

By now obvious similar triangles, $3BO=3KN=AN,$ so it's a quick system of two linear equations to solve for the desired length.
[/hide]

I'm really sorry for reviving an old thread, but how do you solve this problem using a spiral similarity (if the circles have centers S and T, and they intersect at P and U, triangles SUT and QUR...)? :o

[quote="rnwang2"]I'm really sorry[/quote]
You should be.
[quote]for not really knowing how to use the very simple pythagorean theorem method, but how do you solve this problem using a spiral similarity (if the circles have centers S and T, and they intersect at P and U, triangles SUT and QUR...)? :o[/quote]
Ummm ok let M be the midpoint of ST so uhhh now noticing that M lies on the smaller circle (or not, it's pretty easy anyway?) you just need to find 12UP/UM.

I think I got quite nice solution:

Let be center of larger circle be $ O_{1}$ and center of smaller $ O_{2}$

Then $ \angle O_{1}QP \equal{} \angle O_{1}PQ$ Isosceles triangle
And $ \angle O_{2}PR \equal{} \angle O_{1}RP$

$ \angle QPR \equal{} \pi \Rightarrow \angle QPO_{1} \plus{} \angle O_{1}QO_{2} \plus{} \angle O_{2}PR \equal{} \pi$
$ \angle O_{1}PO_{2} \equal{} \pi \minus{} \angle O_{1}PQ \minus{} \angle O_{2}PR$

$ \angle QO_{1}R \equal{} \pi \minus{} \angle O_{1}QP \minus{} \angle O_{1}RP$ (Sum of the angles in triangle)

Therefore $ \angle QO_{1}R \equal{} \angle O_{1}PO_{2} \equal{} \alpha$

$ cos \alpha \equal{} \frac {144 \minus{} 64 \minus{} 36}{2*6*8} \equal{} \frac { \minus{} 44}{96}$ (Law of Cosines)

Let be $ |QR| \equal{} 2x$

Then by Law of Cosines
$ 4x^2 \equal{} |O_{1}R|^2 \plus{} |O_{1}Q|^2 \minus{} 2*|O_{1}R|*|O_{1}Q|*cos \alpha$
Solving this equation we get:
$ x^2 \equal{} 130$

Therefore square of the length $ |QR|$ is $ \boxed {130}$

[quote="TZF"][hide]
Solution involves the sketch attached.

From here we have
$ \sqrt {6^{2} \minus{} a^{2}} \plus{} 2\sqrt {6^{2} \minus{} a^{2}} \equal{} \sqrt {8^{2} \minus{} a^{2}}\to 3\sqrt {6^{2} \minus{} a^{2}} \equal{} \sqrt {8^{2} \minus{} a^{2}}$
$ 8a^{2} \equal{} 260$

The value we want is $ (2a)^{2} \equal{} 4a^{2} \equal{} 260/2 \equal{} \boxed{130}$

[/hide][/quote]

How did you draw the above graph?  Is it based on the graph the problem given?

[quote="TZF"]Haha wow I don't even remember this problem.

Let $ C$ and $ D$ be the centers of the circles with radii $ 8$ and $ 6$, respectively. Let $ S$ be the midpoint of $ QP$ and $ T$ be the midpoint of $ PR$. Note that $ S$ is the projection of $ C$ onto $ QP$ and $ T$ is the projection of $ D$ onto $ PR$ -- this is the motivation behind my approach with right triangles.

Then:

The lower-left point is $ T$
The lower-right point is $ S$
The upper-left point is $ D$
The upper-right point is $ C$

I computed $ (ST)^2$ instead of $ (PQ)^2$, because they are the same, since $ QP \equal{} PR$.

Rotate the given picture about 162 degrees clockwise to get the perspective of my picture.[/quote]

Nice, but what's projection and how do you know that CS and DT both perpendicular to QR?

$ \triangle CPQ$ is an isoceles triangle with apex $ C$, because $ CQ\equal{}CP \equal{} 8$. We know $ CS$ is the median to side $ PQ$ in this triangle. But in an isoceles triangle, the median that involves the apex is also the altitude that involves the apex. Thus, $ CS$ is also an altitude of $ \triangle CPQ$, and is therefore perpendicular to $ QP$.

Similarly for $ \triangle DPR$.

A projection of a point onto a linear structure (a line, a plane, etc) is the point on the destination structure closest to the point. It has the property that the segment connecting a point to its projection in a structure is perpendicular to all lines contained within that structure.

For example, the projection of (1,2,3) onto the x-y plane is (1,2,0). The segment connecting (1,2,3) and (1,2,0) is perpendicular to all lines in the x-y plane.

[quote="TZF"]$ \triangle CPQ$ is an isoceles triangle with apex $ C$, because $ CQ \equal{} CP \equal{} 8$. We know $ CS$ is the median to side $ PQ$ in this triangle. But in an isoceles triangle, the median that involves the apex is also the altitude that involves the apex. Thus, $ CS$ is also an altitude of $ \triangle CPQ$, and is therefore perpendicular to $ QP$.

Similarly for $ \triangle DPR$.

A projection of a point onto a linear structure (a line, a plane, etc) is the point on the destination structure closest to the point. It has the property that the segment connecting a point to its projection in a structure is perpendicular to all lines contained within that structure.

For example, the projection of (1,2,3) onto the x-y plane is (1,2,0). The segment connecting (1,2,3) and (1,2,0) is perpendicular to all lines in the x-y plane.[/quote]

I see, so projection is basically a vertical height from a point to a structure or plane.

[quote="TZF"]Not necessarily vertical.... for example, the projection of the origin onto the plane x+y+z=1 is the point (1/3, 1/3, 1/3).

The line connecting these is not vertical. However, it is is the unique line connecting the origin with a point on x+y+z=1 such that the line and the plane are perpendicular.[/quote]

BTW, if you draw the graph of x+y+z=1, do you just mark (1,0,0),(0,1,0),(0,0,1) and make a plane out of it?
but how do you know its projection is (1/3,1/3,1/3)?

If you have more question or want more examples, please start a new thread, since this thread is supposed to be dedicated to an AIME problem or something like that :)

-------------------

Sure, that's a totally legitimate way of drawing the plane. Any other three points on the plane work, too.

In general, you can use a particular case of the [url=http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality#Notable_special_cases]Cauchy-Schwarz Inequality[/url] -- see the one with the summations.

The projection of a point onto a structure is the point on the structure that is closest to the original point.

In this case, begin with a statement of Cauchy-Schwarz, where $ (x_1, x_2, x_3) \equal{} (x,y,z)$ and $ (y_1,y_2,y_3) \equal{} (1,1,1)$. You get this: $ (1 \cdot x \plus{} 1 \cdot y \plus{} 1 \cdot z)^2 \leq (x^2 \plus{} y^2 \plus{} z^2) (1^2 \plus{} 1/^2 \plus{} 1^2)$.

But we know that $ x \plus{} y \plus{} z \equal{} 1$ and that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} D^2$, where $ D$ is the distance from the origin to a point on the plane. Thus, $ \frac {1}{3} \leq D^2$. The equality case, which gives the minimum distance, happens when $ \frac {x_1}{y_1} \equal{} \frac {x_2}{y_2} \equal{} \frac {x_3}{y_3}$, or $ x \equal{} y \equal{} z$. Since $ x,y,z$ add to $ 1$, they must all be $ 1/3$

Let A be the center of the big circle, B be the center of the small circle. Let AB intersect circle B at X. Let the perp from A to PQ be C and the perp from B to PR be D. It is clear that P it the midpoint of CD. Thus, the perpendicular to CD at P must bisect AB, thus $ PX \perp CD$. By Stewart's, $ PX\equal{}\sqrt{14}$. Let the bisector of PBX intersect PX at F. Then $ FX\equal{}\frac{\sqrt{14}}{2}$ since PBX is isoceles, so by Pythag, $ BF^2\equal{}36\minus{}\frac{14}{4}$ so $ QP^2\equal{}(2PD)^2\equal{}4BF^2\equal{}130$.

[quote="4everwise"]Found another approach. Similar, but different.  :)

[hide="Sketch of Solution"]Let the circles of radius $ 8$ and $ 6$ be centered at $ A$ and $ B,$ respectively. Let the midpoints of $ QP$ and $ PR$ be $ N$ and $ O.$ Dropping a perpendicular from $ B$ to $ AN$ (let the point be um... $ K?$) gives a rectangle, etc.

Ok, now note that triangle $ ABK$ is right. Let the midpoint of $ AB$ (segment of length 12) be $ M.$ Hence, $ KM \equal{} 6 \equal{} BM \equal{} BP.$

By now obvious similar triangles, $ 3BO \equal{} 3KN \equal{} AN,$ so it's a quick system of two linear equations to solve for the desired length.
[/hide][/quote]
Before applying similar triangles, how do we prove that $ R$ is colinear with $ AB$ in the first place?
(edit: I need to prove the perpendicular bisector of $ QR$ intersect the midpoint of $ AB$ with A,B the centers of the two circles... any ideas?)

(Also, does anyone have a solution for the construction problem proposed by Altheman?)

[quote="Altheman"]let O_1 and O_2 be the centers of the circles, take the perpendicular bisectors of QP and PR respectively, they pass through O_1 and O_2 respectively, then the perpendicular bisector passes through the midpoint of O_1O_2, this just happens to lie on the smaller circle, now that means PR, and the perpendicular bisector subtend half the circle, it follows that they are collinear[/quote]

Sorry I don't understand Altheman's solution, could someone clarify?

[hide="Solution"]
 [geogebra]5e62d4b862a74fb851b266a022cfdba5302b8170[/geogebra]

Look at picture for the naming of the points.

Let $ QR \equal{} 4x$, so $ QP \equal{} PR \equal{} 2x$. $ TR \equal{} 2$ because if the circles were tangent they would be 14 apart, but they are 12 apart, so the shared part must be 2.

From $ O_1$ and $ O_2$, draw perpendiculars to the chord. They are both perpendicular bisectors so $ QA \equal{} AP \equal{} PB \equal{} BR \equal{} x$.

Drawing $ PT$, we conclude $ R$ is collinear with the diameter.
This is because $ \frac {RB}{RP} \equal{} \frac {RO_2}{RT}$ so $ \triangle RBO_2 \sim \triangle RPT$ and $ \angle RBO_2 \equal{} \angle RPT \equal{} 90^\circ$ and $ RT$ is therefore a diameter.

Extend $ PT$ to meet the circle at $ S$. $ \angle QPS \equal{} \angle RPT \equal{} 90^\circ$, so $ QS$ is a diameter too.

Let $ BO_2 \equal{} a$, then $ PT \equal{} 2a$ because $ \triangle RBO_2 \sim \triangle RPT$ with a scale factor of 2 and $ a*2 \equal{} 2a$

Also $ AO_1 \equal{} 3a$ because $ \triangle RBO_2 \sim \triangle RAO_1$ with a scale factor of 3 and $ a*3 \equal{} 3a$

And finally $ PS \equal{} 6a$ because $ \triangle QAO_1 \sim \triangle QPS$ with a scale factor of 2 and $ 3a*2 \equal{} 6a$

Power of a point on $ T$ gives us $ PT*TS \equal{} UT*TV$, so $ 8a^2 \equal{} 2*14$ so $ 4a^2 \equal{} 14$.

Pythagorean on $ \triangle RPT$ gives us $ (2a)^2 \plus{} (2x)^2 \equal{} 144$ and $ (2x)^2 \equal{} 144 \minus{} 4a^2 \equal{} 144 \minus{} 14 \equal{} 130$.

$ (2x)^2 \equal{} QP^2 \equal{} \boxed {130}$

[/hide]

[quote="ShocK n AwE"]
This is because $ \frac {RB}{RP} \equal{} \frac {RO_2}{O_2T}$ [/quote]

Slight typo $ \frac {RB}{RP} \equal{} \frac {RO_2}{RT}$.  But really nice solution! :D

[quote="Altheman"]then the perpendicular bisector passes through the midpoint of O_1O_2[/quote]
[i]Proof:[/i] Refer to diagram. We want to prove that $d=e$. By similar triangles, $a/b=c/d$ and $a/2b=c/(d+e)$. From the first equation we have $ad=bc$. From the second one we have $ad+ae=2bc$, but $ad=bc$ so this becomes $ae=bc$. Therefore $ad=bc=ae$, so $ad=ae$ giving $d=e$ as desired.

(Note: Of course the proof works with any angles, not just right angles.)

[color=darkred][b][u]PP[/u].[/b] Let two secant circles $w_1=\mathbb C\left(O_1,8\right)$ and $w_2=\mathbb C\left(O_2,6\right)$ , where $O_1O_2=12$ and $P\in w_1\cap w_2$ . Consider $Q\in w_1$

and $R\in w_2$ so that $P\in (QR)$ and the line $O_1O_2$ doesn't separate $Q$ and $R$ . Prove that $PQ=PR\implies PQ^2=130$ .[/color]

[color=black][b][u]Proof[/u].[/b] Let the midpoints $M$ , $X$ , $Y$ of $[O_1O_2]$ , $[PQ]$ , $[PR]$ respectively and $PX=PY=x$ , $PM=m$ . Thus, $PQ=PR=2x$

and observe that $O_2P=O_2M=6$ , i.e. the triangle $PO_2M$ is $O_2$-isosceles. Denote the midpoint $N$ of $[PM]$ , i.e. $O_2N\perp PM$ .

Denote $U\in O_1X\cap O_2N$  and $XO_1=u$ , $YO_2=v$ . Thus, $\boxed{u-v=m}$ , i.e. $O_1UPM$ is a parallelogram , i.e. $PU\parallel O_1O_@$ and

$R\in O_1O_2\cap w_2$ . Thus, $\boxed{u^2-v^2=8^2-6^2=2\cdot 14}\ (1)$ . Apply the  theorem of median $[PM]$ in $\triangle O_1PO_2$ and obtain $PM=\sqrt {14}$

i.e. $\boxed{m=\sqrt {14}}\ (*)$ . Therefore, from the relation $(1)$ obtain that $u^2-v^2=2m^2\iff$ $u+v=2m\iff$ $2u=3m\ \wedge\ 2v=m\iff$

$PQ^2=XY^2=O_1O_2^2-UO_1^2\iff$ $PQ^2=12^2-(u-v)^2=$ $12^2-m^2=144-13\iff$ $PQ^2=130$ .[/color]

But how can we sure that R is on the diameter?

[quote=maplestory]power of point R kills this problem in a few seconds[/quote]

Yes, but we need to prove that R is on the diameter (although since this was on an AIME, I guess it's okay to skip doing that on the test)

[quote=Elemennop]Haha, i totally guessed R was collinear with the centers...

what's the proof for it?[/quote]

LOC bash?

[quote=champion999]So many necroposts on this one thread...[/quote]

Lol yes

[color=darkred][b][u]PP[/u].[/b] Let two secant circles $w_1=\mathbb C\left(O_1,8\right)$ and $w_2=\mathbb C\left(O_2,6\right)$ , where $O_1O_2=12$ and $P\in w_1\cap w_2$ . Consider $Q\in w_1$

and $R\in w_2$ so that $P\in (QR)$ and the line $O_1O_2$ doesn't separate $Q$ and $R$ . Prove that $PQ=PR\implies PQ^2=130$ .[/color]

[color=black][b][u]Proof[/u].[/b] Let the midpoints $M$ , $X$ , $Y$ of $[O_1O_2]$ , $[PQ]$ , $[PR]$ respectively and $PX=PY=x$ , $PM=m$ . Thus, $PQ=PR=2x$

and observe that $O_2P=O_2M=6$ , i.e. the triangle $PO_2M$ is $O_2$-isosceles. Denote the midpoint $N$ of $[PM]$ , i.e. $O_2N\perp PM$ .

Denote $U\in O_1X\cap O_2N$  and $XO_1=u$ , $YO_2=v$ . Thus, $\boxed{u-v=m}$ , i.e. $O_1UPM$ is a parallelogram , i.e. $PU\parallel O_1O_@$ and

$R\in O_1O_2\cap w_2$ . Thus, $\boxed{u^2-v^2=8^2-6^2=2\cdot 14}\ (1)$ . Apply the  theorem of median $[PM]$ in $\triangle O_1PO_2$ and obtain $PM=\sqrt {14}$

i.e. $\boxed{m=\sqrt {14}}\ (*)$ . Therefore, from the relation $(1)$ obtain that $u^2-v^2=2m^2\iff$ $u+v=2m\iff$ $2u=3m\ \wedge\ 2v=m\iff$

$PQ^2=XY^2=O_1O_2^2-UO_1^2\iff$ $PQ^2=12^2-(u-v)^2=$ $12^2-m^2=144-14\iff$ $PQ^2=130$ .[/color]

The symmetrical of the circle containing $R$ (you may suppose this is the smaller one) about $P$ will intersect second time the other circle at $Q$. The problem can be solved on this sketch as well.

Best regards,
sunken rock

Unique solution :o

[hide=Solution]Suppose the center of the left circle is $A$ and the center of the right circle is $B$. Drop perpendiculars from $A$ to $QP$ and $B$ to $PR$ at points $C$ and $D$ respectively. Let $\angle APC = \angle AQP = \beta$ and $\angle BPR = \angle BRP = \angle \alpha$. By Law of Cosines in $\triangle APB$, $$8^2+6^2+2\cdot 8\cdot 6\cos(180-\beta-\alpha)=12^2$$ $$\implies \cos(\beta + \alpha) = \frac{44}{2\cdot 8\cdot 6}$$ $$\implies \frac{11}{24}=\cos\beta \cos \alpha - \sin \beta \sin \alpha.$$ Now since $QP=PR$, we can find that $4\cos \beta =3\cos \alpha$. This makes our equation $$\frac{11}{24}=\frac{4}{3}\cos^2\beta - \sqrt{(1-\cos^2\beta)(1-\frac{16}{9}\cos^2\beta)}$$ Multiplying the equation by $24$, $$11 = 32\cos^2 \beta -8\sqrt{16\cos^4 \beta - 25\cos^2\beta + 9}$$ Now let $\cos^2\beta = y$ and we get $$64(16y^2-25y+9)=(32y-11)^2$$ Solving for $y$ gives $y=\frac{455}{896}$ which means that $$\sqrt{\frac{455}{896}}=\cos \beta = \frac{\left(\frac{QP}{2}\right)}{8}$$ $$\implies QP^2=\boxed{130}$$[/hide]

Since so many people have asked about this recently, here is the full proof of the collinearity of $A,$ $B$, $R$:

[hide = A, B, R collinear proof]
I'm just going to copy the diagram from the wiki:

[asy]
size(0,5cm);
pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);
draw(b--r--n--b--a--m--n);
draw(a--q--m);
draw(circumcircle(origin,q,p));
draw(circumcircle((14,0),p,r));
draw(rightanglemark(a,m,n,24));
draw(rightanglemark(b,n,r,24));
label("$A$",a,S);
label("$B$",b,S);
label("$M$",m,NE);
label("$N$",n,NE);
label("$P$",p,N);
label("$Q$",q,NW);
label("$R$",r,E);
label("$12$",(14,0),SW);
label("$6$",(23,0),S);
label("$T$", t , NW);
[/asy]

Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so $QM = MP = PN = NR$, since the problem told us $QP = PR$.

We will show that $R$ lies on $AB$.

Let $T$ be the intersection of circle centered at $B$ with $AB$. Then $BT = TA = 6$.

Let $P$' be the foot of the perpendicular from $T$ to $MN$. Then $TP'$ is a midline (or midsegment) in trapezoid $AMNB$, so $P'$ coincides with $P$ (they are both supposed to be the midpoint of $MN$). In other words, since $\angle TP'N = 90^\circ$, then $ \angle TPN = 90^\circ$.

Thus, $\angle TPR$ subtends a $90^\circ \times 2 = 180^\circ$ degree arc. So arc $TR$ in circle $B$ is $180^\circ$, so $TR$ is a diameter, as desired. Thus $A$, $B$, $R$ are collinear. [/hide]

Note this collinearity only follows from the fact that $6$ is half of $12$ in the problem statement. The collinearity is untrue in general.

[b]Solution:[/b] Notice that $R$ is collinear with $X$ and $Y$ (center of the first and second circle, respectively). Let the length of the altitude from $X$ to $QP$ be $3r$ and the length of the altitude from $Y$ to $PR$ be $r$ (by similar triangles). By the Pythagorean theorem, the foot of the altitude from $X$ to $P$ is $\sqrt{36-r^2}$. So, we have that $64 = 9r^2 + 36-r^2 \Rightarrow r=\frac{\sqrt{14}}{2}$. We are trying to find $QP = 4\sqrt{36-r^2} = \sqrt{144 - 4r^2} = \sqrt{144-14} = \sqrt{130}$. Squaring, we get our final answer of $\boxed{130}$.

[u]Motivation:[/u] The mere setup of this problem contributes to how you would attack this. I believe this is a really overused idea, but probably was original in 1983. Long story short, similar triangles motivates and finishes.

[i]Comments:[/i] When solving this, I assumed that $R$ was collinear to $A$ and $B$. However, this isn't too bad to prove (see post above).

Let $O$, be the center of the circle. We have that $AO$, is tangent to $BC.$ The distance from $O$, to $BC$, and $A$ is $1$ away from $BC$, let $D$ be the point through $A$ that is parallel to $BC.$ We have that $O$ is $3$ away from $AD.$ So the answer is $\frac{7}{25}$, giving us $\boxed{175}.$


modular arithmetic
number theory
greatest common divisor
number theory proposed

Find $ (k,m,n)$ $ \in \mathbb {N}^3$ such that $ 5^m\plus{}7^n\equal{}k^3$.

by checking modulo $ 8$ we can find that $ 2|m$ and $ 2 \nmid n$
then by checking modulo $ 7$
(note that $ k^3 \equiv \pm 1 \pmod {7}$)
we have $ 5^m \equiv 2^m \equiv \pm 1 \pmod {7}$
since $ m$ is even and $ 7 \equiv \minus{}1 \pmod {4}$ so the congruency  $ 2^m \equiv  \minus{}1 \pmod {7}$ can't hold for any $ m$  because of this we must have $ 2^m \equiv 1 \pmod {7}$ which implies $ 3 |m$
now we put $ m\equal{}3l$ and we rewrite the equation in this form :
$ 7^n\equal{}(k\minus{}5^l)(k^2\plus{}25^l\plus{}5^lk)$
since $ (5,k)\equal{}1$
so the gcd of the two expressions on the right side must be $ 1$ or $ 3$
obviously it can't be $ 3$ since $ 3 \nmid 7$
also both of the two expressions must be a power of $ 7$ and because of what has been mentioned above we must have
$ k\minus{}5^l\equal{}1 \ \ ; \ \  k^2\plus{}25^l\plus{}5^lk\equal{}7^n$
which is a clear contradiction just see that in the second equation we have

$ k^2 \equiv 7^n \pmod{5}$

also we had proven that $ n$ is odd (at the beginning of the proof) so we get

$ \left(\frac{7^n}{5} \right)\equal{}1 \implies \left(\frac{7}{5} \right)\equal{}1 \implies \left(\frac{2}{5} \right)\equal{}1$

but we know that $ \left(\frac{2}{5} \right)\equal{}\minus{}1$
so there doesn't exist any solution for this equation .
we r done!

[quote="shoki"]by checking modulo $ 8$ we can find that $ 2|m$ and $ 2 \nmid n$
then by checking modulo $ 7$
(note that $ k^3 \equiv \pm 1 \pmod {7}$)
we have $ 5^m \equiv 2^m \equiv \pm 1 \pmod {7}$
since $ m$ is even and $ 7 \equiv \minus{} 1 \pmod {4}$ so the congruency  $ 2^m \equiv \minus{} 1 \pmod {7}$ can't hold for any $ m$  because of this we must have $ 2^m \equiv 1 \pmod {7}$ which implies $ 3 |m$
now we put $ m \equal{} 3l$ and we rewrite the equation in this form :
$ 7^n \equal{} (k \minus{} 5^l)(k^2 \plus{} 25^l \plus{} 5^lk)$
since $ (5,k) \equal{} 1$
so the gcd of the two expressions on the right side must be $ 1$ or $ 3$
obviously it can't be $ 3$ since $ 3 \nmid 7$
also both of the two expressions must be a power of $ 7$ and because of what has been mentioned above we must have
$ k \minus{} 5^l \equal{} 1 \ \ ; \ \ k^2 \plus{} 25^l \plus{} 5^lk \equal{} 7^n$
which is a clear contradiction just see that in the second equation we have

$ k^2 \equiv 7^n \pmod{5}$

also we had proven that $ n$ is odd (at the beginning of the proof) so we get

$ \left(\frac {7^n}{5} \right) \equal{} 1 \implies \left(\frac {7}{5} \right) \equal{} 1 \implies \left(\frac {2}{5} \right) \equal{} 1$

but we know that $ \left(\frac {2}{5} \right) \equal{} \minus{} 1$
so there doesn't exist any solution for this equation .
we r done![/quote]
I have not read your solution carefully but I think your answer is not right. Let try $ (k,m,n)\equal{}(2,0,1)$, you will see!  :wink:

[quote="aadil"][quote]$ \mathbb{N}^{3}$[/quote]
what do you mean by this[/quote]

$ \mathbb N^3\equal{}\mathbb N\times\mathbb N\times\mathbb N$ is the set of all triples $ (a,b,c)$ where $ a\in\mathbb N, b\in\mathbb N, c\in\mathbb N$

[quote="mathVNpro"][quote="shoki"]by checking modulo $ 8$ we can find that $ 2|m$ and $ 2 \nmid n$
then by checking modulo $ 7$
(note that $ k^3 \equiv \pm 1 \pmod {7}$)
we have $ 5^m \equiv 2^m \equiv \pm 1 \pmod {7}$
since $ m$ is even and $ 7 \equiv \minus{} 1 \pmod {4}$ so the congruency  $ 2^m \equiv \minus{} 1 \pmod {7}$ can't hold for any $ m$  because of this we must have $ 2^m \equiv 1 \pmod {7}$ which implies $ 3 |m$
now we put $ m \equal{} 3l$ and we rewrite the equation in this form :
$ 7^n \equal{} (k \minus{} 5^l)(k^2 \plus{} 25^l \plus{} 5^lk)$
since $ (5,k) \equal{} 1$
so the gcd of the two expressions on the right side must be $ 1$ or $ 3$
obviously it can't be $ 3$ since $ 3 \nmid 7$
also both of the two expressions must be a power of $ 7$ and because of what has been mentioned above we must have
$ k \minus{} 5^l \equal{} 1 \ \ ; \ \ k^2 \plus{} 25^l \plus{} 5^lk \equal{} 7^n$
which is a clear contradiction just see that in the second equation we have

$ k^2 \equiv 7^n \pmod{5}$

also we had proven that $ n$ is odd (at the beginning of the proof) so we get

$ \left(\frac {7^n}{5} \right) \equal{} 1 \implies \left(\frac {7}{5} \right) \equal{} 1 \implies \left(\frac {2}{5} \right) \equal{} 1$

but we know that $ \left(\frac {2}{5} \right) \equal{} \minus{} 1$
so there doesn't exist any solution for this equation .
we r done![/quote]
I have not read your solution carefully but I think your answer is not right. Let try $ (k,m,n) \equal{} (2,0,1)$, you will see!  :wink:[/quote]
plz the next time u mention $ N$ tell us whether $ 0 \in N$ or not!
thx!(as i know $ 0 \notin N$!) :)

even if $ 0 \in N$ then the problem won't be hard since as i had proved in my first post we must have $ k\minus{}5^l\equal{}1$
(no matter if $ l\equal{}0$ or not)
and so if $ l\equal{}0$ then $ k$ must be two and as u mentioned it's the only solution ,obviously!


geometry
circumcircle
trigonometry
geometry solved

Let $A'C,B'A,C'B$ be the external bisector of $\angle ACB,\angle BAC,\angle ABC$ of $\triangle ABC$ and $AA',BB',CC'$are perdencular to $A'C,AB',BC'$as the graph.
Suppose that $d$ denote the diameter of $\triangle A'B'C'$'s circucircle and $r,p$ the radiu of $\triangle ABC$'s inscribed cirlcle
and semi-perimeter.Prove
$r^{2}+p^{2}=d^{2}$

You also posted it at http://www.aoshoo.com/bbs1/dispbbs.asp?boardID=43&ID=5541&page=1.

So you want to show that the diameter d of the circumcircle of triangle A'B'C' equals $\sqrt{r^{2}+s^{2}}$. Equivalently, the radius of this circumcircle equals $\frac12\sqrt{r^{2}+s^{2}}$. This follows from the results in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=5958 : Your points A', B', C' are the points $Z_{a}$, $X_{b}$, $Y_{c}$ from that topic, and thus, as it was shown there, they lie on the second Spieker circle of triangle ABC, which has the radius $\frac12\sqrt{r^{2}+s^{2}}$. Hence, the circumcircle of triangle A'B'C' has radius $\frac12\sqrt{r^{2}+s^{2}}$.

  darij

[color=darkblue]Prove easily that for the triangle $A'B'C'$ there are the following relations :
$\frac{A'B'}{\sqrt{c(p-c)}}=\frac{B'C'}{\sqrt{a(p-a)}}=\frac{C'A'}{\sqrt{b(p-b)}}=\sqrt{\frac{p^{2}+r^{2}}{4Rr}}$ and $A'=90^{\circ}-\frac{A}{2}$ a.s.o., where $2p=a+b+c\ .$
Therefore, if the length of the diameter of the circumcircle of the triangle $A'B'C'$ is $d$, then
$B'C'=d\sin A'$ $\Longleftrightarrow$ $\sqrt{a(p-a)}\cdot\sqrt{\frac{p^{2}+r^{2}}{4Rr}}=d\cdot \cos \frac{A}{2}$ $\Longleftrightarrow$ $d^{2}=p^{2}+r^{2}\ .$[/color]

thanks to all.i'll study it.
darij your post is really $long \ldots$
but i enjoy it. :roll:


algebra
polynomial
inequalities
trigonometry
algebra unsolved

$\frac{1}{xy}=\frac{x}{z}+1$
$\frac{1}{yz}=\frac{y}{x}+1$
$\frac{1}{zx}=\frac{z}{y}+1$

Solve for x, y and z.

I tried a lot since last night. But I can’t find any way. I tried with polynomial and inequality but I don’t think they helped me towards the solution! Should I use trigonometry? I don’t think so. Anyhow if u can solve it plz tell me.

Well, i solved it just one hour ago. It was not so difficult. But it is a virtual truth that once you solve a problem then that will seem easy to you but you will say tough when you could not solve this! ;)

I think the solution is:-

$X = Y = Z = \frac{1}{\sqrt2}$

is this right ?????

right to stan17.

Anyhow my soln is followed:

first show that 3positive is equivalent to 3negetive AND 2positive 1negetive is equivalent to 1positive 2negetive.

Show second case is impossible.

For the first one deal with only positve.

Order x,y,z and apply some inequality.

Anyhow, if u r interested in full soln then i may type out that later.

actually my inernet line was facing a big trouble. sorry for delay! ;)


Let be $ x,y\ge 0$ . Prove that :

$ 8x^3y\plus{}y^3\plus{}2x\ge 2xy(2x\plus{}y\plus{}1)$

Nice inequality! It is found by adding
$ x^2 \plus{} \frac {y^2}{8x} \plus{} \frac {y^2}{8x} \plus{} \frac {1}{4y} \plus{} \frac {1}{4y} \plus{} \frac {1}{4y} \plus{} \frac {1}{4y} \ge \frac {7}{4}$
$ \frac {y^2}{8x} \plus{} \frac {y^2}{8x} \plus{} x^2 \plus{} \frac {1}{4} \ge y$
$ 2x^2 \plus{} \frac {1}{2} \ge 2x$

Equality when $ x\equal{}\frac{1}{2}$, $ y\equal{}1$


function
induction
algebra
domain
algebra unsolved

Find all $ f: \mathbb{R} \rightarrow  \mathbb{R}$ such that
1) $ f(1)\equal{}1$
2) for all non-zero reals $x$ and $y$, \[ f(x\plus{}y)\equal{}f(x)\plus{}f(y).\]
3) for all non-zero real $x$, \[ f(x)\cdot f\left( \frac 1x \right) \equal{}1.\]

$ f(1\plus{}1)\equal{}f(1)\plus{}f(1)$
$ f(2)\equal{}2$
$ f(2\plus{}1)\equal{}f(2)\plus{}f(1)$
$ f(3)\equal{}3$

By induction, we can see that $ f(n)\equal{}f(n\minus{}1)\plus{}f(1)\equal{}n$, $ \forall n \in \mathbb{N}$

$ f(2\minus{}1)\equal{}f(2)\plus{}f(\minus{}1)$
$ 1\equal{}2\plus{}f(\minus{}1)$
$ f(\minus{}1)\equal{}\minus{}1$

$ f(x\minus{}(x\minus{}1))\equal{}f(x)\plus{}f(1\minus{}x)$
$ 1\minus{}x\equal{}f(1\minus{}x)$
Thus:
$ f((1\minus{}x)\minus{}1)\equal{}f(1\minus{}x)\plus{}f(\minus{}1)$
$ f(\minus{}x)\equal{}1\minus{}x\minus{}1\equal{}\minus{}x$
And therefore:
$ f(x)\equal{}x$,  $ \forall x \neq 0 \in \mathbb{Z}$

Now, since $ f(x)f\left (\frac{1}{x}\right )\equal{}1$ and $ f(x)\equal{}x$
$ f\left(\frac{1}{x}\right)\equal{}\frac{1}{x}$,  $ \forall x \neq 0 \in\mathbb{Z}$

Clearly for any non-zero integer $ a$, $ \frac{a}{x}$ is equal to a sum of multiple $ \left(\frac{1}{x}\right )$'s, and from $ f\left (\frac{1}{x}\plus{}\frac{1}{x}\right )\equal{}\frac{2}{x}$ using induction we can see that $ f\left (\frac{a}{x}\right )\equal{}\frac{a}{x}$.

Thus, $ \forall x\in\mathbb{Q}$, $ f(x)\equal{}x$

Thus the only (almost-) continuous function (except at 0 which is not in the domain) that satisfies it is $ f(x)\equal{}x$, but multiple non-continuous solutions exist. (I suppose $ f(x)\equal{}\frac{x^2}{x}$ satisfies all of the conditions...)

If we assume that the restriction on 0 only applies to the third line, and not the second, then $ f(x)\equal{}x$ is the only continuous solution.

$ f(x)\equal{}x$ for $ x \in \mathbb{Q}$ follows from first and second conditions by Cauchy function.But for proving it for $ \mathbb{R}$ you need some kind of continuity, bound  or something like that

[quote="birzhan"]Find all $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$ f(1) \equal{} 1$
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$
$ f(x)*f(1/x) \equal{} 1$
for all $ x,y \in \mathbb{R} \setminus 0$[/quote]

We obviously have $ f(x)\equal{}x$ $ \forall x\in\mathbb Q$ (Cauchy's equation).

We also know that $ f(x)\neq 0$ $ \forall x\neq 0$

Let then $ x\notin\{0,1\}$.
$ \frac{1}{x(x\minus{}1)}\plus{}\frac{1}{x}\equal{}\frac{1}{x\minus{}1}$ $ \implies$ $ f(\frac{1}{x(x\minus{}1)})\plus{}f(\frac{1}{x})\equal{}f(\frac{1}{x\minus{}1})$

$ \implies$ $ \frac{1}{f(x^2)\minus{}f(x)}\plus{}\frac{1}{f(x)}\equal{}\frac{1}{f(x)\minus{}1}$

$ \implies$ $ \frac{1}{f(x^2)\minus{}f(x)}\equal{}\frac{1}{f(x)(f(x)\minus{}1)}$

$ \implies$ $ f(x^2)\equal{}f(x)^2$ $ \forall x\notin\{0,1\}$

$ \implies$ $ f(x^2)\equal{}f(x)^2$ $ \forall x$

$ \implies$ $ f(x)>0$ $ \forall x>0$

$ \implies$ $ f(x)$ is an increasing function ($ f(x\plus{}y)\equal{}f(x)\plus{}f(y)>f(x)$ $ \forall y>0$)

$ \implies$ $ f(x)\equal{}x$ $ \forall x\in\mathbb R$


inequalities
trigonometry
geometry
circumcircle
function
geometric transformation
reflection

let $ABC$ be an acute triangle,let $AD,BE ,CF$ be his altitudes ,prove the inequalty:
$DE+DF\leq BC$

Denote: $H=AD\cap BE\cap CF$
We have: $DE=HC\cdot\sin C=2\cdot R\cos C\sin C=R\cdot\sin 2C,\,DF=R\cdot\sin 2B$

Thus, we have to prove that: $\sin 2B+\sin 2C\le 2\sin A\iff\cos (B-C)\le 1,$ what is truly.

$\triangle DEF$ is orthic triangle of $\triangle ABC$.We know that circumcircle of $\triangle DEF$ is 9-points circle of $\triangle ABC$ and its radii $R_{DEF}=\frac{R_{ABC}}{2}=\frac{R}{2}$.In $\triangle DEF$, we apply sinus theorem: $\frac{DE}{sinDFE}=R\Longrightarrow \frac{DE}{sin(180^{o}-2C)}=R\Longrightarrow \frac{DE}{sin2C}=R$.
Similarly, $\frac{DF}{sin2B}=R$So, $DE+DF=R.(sin2B+sin2C)$.

Now, in $\triangle ABC$, we apply sinus theorem: $BC=2R.sinA$
Clearly, we need to prove $sin2B+sin2C\leq 2.sinA$ for proof of $DE+DF\leq BC$
Lets consider the function $f(x)=sinx$ for all $0<x<\pi$.Because of $f''(x)=-sinx<0$, $f$ is concave.So, according to Jensen's inequality: $\frac{f(2B)+f(2C)}{2}\leq f(\frac{2B+2C}{2}) \Longrightarrow sin2B+sin2C\leq 2.sinA$.

NOTE: ...and we easily prove that in any actue $\triangle ABC$, $Perimeter(DEF)\leq u$
Lokman GÖKÇE

Reflect the diagram over $BC$. $E$ maps to $E'$, etc

Since $\angle FDA=\angle ABE=\angle ACF=\angle ADE$. $F$, $D$, $E'$ are collinear. Note that $BFECE'F'$ is cyclic with $BC$ a diameter. $FE'=FD+DE$ is a chord in that circle. A diameter is the longest chord in a circle, it follows that $BC\ge FE'=FD+DE$.


Given in a $ \Delta ABC$ that $ cosAcosB\plus{}sinAsinBsinC\equal{}1$ show that $ a: b: c\equal{}1: 1: \sqrt{2}$

The Best solution....!!!  :lol:
\[ 1 \ge Cos(A\minus{}B) \equal{} cosAcosB \plus{} sinAsinB \ge cosAcosB \plus{} sinAsinBsinC \equal{} 1\] since the Sine function is always positive and less than $ 1$ for a triangle. (Actually, even the triangle property doesn't matter, it is true in every case)
Therefore, equality exists everywhere
$ cos(A\minus{}B)\equal{} 1 \Longrightarrow A\minus{}B \equal{} 0 \Longrightarrow A\equal{}B$ AND
$ sinC\equal{}1 \Longrightarrow C\equal{}90^{\circ}$

The result immediately follows.... :D
This proof shows the beauty of inequalities!

Note: A proof by cosine rule exists but it is quite cumbersome....So I would suggest not to use that and keep it for the worst case scenario....


A solid object is sitting on the xy plane so that its cross sectional area at height z, sliced parallel to the xy-plane, is
$[\pi]$  x  (6-z) squared over 4

Find its volume. Hint: It has the same cross sectional area as a certain cone.

Can we put a solution with calculus involved, or is this a non-calculus exam problem?

Calculus makes this too easy.
Otherwise, smash and bash with Riemann sums maybe?


[b]1. [/b]Sea $ABC$ un triángulo y sean $A',B',C'$ puntos en $BC,CA,AB$ respectivamente tales que $A'B'C'$ es semejante a $ABC$. Pruebe que el ortocentro de $A'B'C'$ es el circuncentro de $ABC$.

[b]2.[/b] Considere la secuencia $(a_{n})_{n\ge 0}$,
donde
$a_{0}=a_{1}=97$
 y
$a_{n+1}=a_{n-1}a_{n}+\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}$
 para todo $n\ge 1$.
 Pruebe que $2+\sqrt{2+2a_{n}}$  es un cuadrado perfecto pra todo $n\ge 0$.

[b]3.[/b] Los enteros positivos son coloreados de 2006 colores. Pruebe que ó bien existen dos colores digamos azul y rojo tales que hay progresiones aritmeticas arbitrariamente largas coloreadas de azul y tambien de  rojo, ó existe un color, digamos negro, tal que existen bloques arbitrariamente largos de enteros consecutivos coloreados de negro.

[b]4.[/b] Sea $p$ un número primo impar dado. Un entero positivo $a$ es llamado una semip-potencia si $a^{p}-1|b^{p}-a$ para algun entero $b$. Pruebe que Existen infinitas semip-potencias, e infinitos números que no son semip-potencias.

Bono: Sea $d(n)$ el número de semip-potencias menores o iguales que $n$ pruebe que si el limite $lim_{n\rightarrow \infty}\frac{d(n)}{n}$ existe, entonces es menor que $\frac{1}{p}$.

Disfruten los problemas, probablemente no han visto la mayoría, luego sabran por que cuando los discutamos.

Veran que mucha teoria se puede desarrollar con ellos....

Chao!!!

Espero solucioens y mucha discucion dentro de una semana...

posteamos las soluciones aqui o a tu correo?

Al correo  de Pascual ..
....
aki solo discutimos .. algunas aclaraciones respecto a la spreguntas ..
mas no las soluciones  ;)

Carlos ...

[quote="Pascual2005"]
[b]3.[/b] Los enteros positivos son coloreados de 2006 colores. Pruebe que ó bien existen dos colores digamos azul y rojo tales que hay progresiones aritmeticas arbitrariamente largas coloreadas de azul y tambien de  rojo, ó existe un color, digamos negro, tal que existen bloques arbitrariamente largos de enteros consecutivos coloreados de negro.
[/quote]

Para el problema 3, las progreciones aritméticas arbitrariamente largas son cada una de rojo y azul o existen unas arbitrariamente grandes azules y otras arbitrariamente grandes rojas?

En el 4, $p$ puede ser primo impar cualquiera :?:

en el 4 $p$ es un primo impar dado.

en el 3, tanto azules como rojas

[quote="CarlosBr"] Al correo de Pascual ..
....
aki solo discutimos .. algunas aclaraciones respecto a la spreguntas ..
mas no las soluciones

Carlos ...[/quote]
mmm y el correo de Pascual es....
¿O en todo caso mando PM?

ok, la discusion esta abierta asi que posteare las soluciones que recibi:

Posteare las soluciones de Gabriel por haber sido el que mas envio...

[b]1[/b][i]Pascual[/i]
O triângulo $A'B'C'$ é semelhante ao triângulo $ABC$ logo $\angle X=\angle X'$ para todo $X \in \{A,B,C\}$. Suponhamos que $\angle A=\angle A', \angle B=\angle B', \angle C=\angle C'$ e $O$ seja o ortocentro de $A'B'C'$ logo

$\angle C'OB'=\angle B'+\angle C'=180^{o}-\angle A$ analogamente $\angle C'OA'=180^{o}-\angle B, \angle A'OB'=180^{o}-\angle C$ portanto os quadriláteros $AC'OB'$, $A'OC'B$, $A'CB'O$ são todos inscritiveis e consequentemente

$\angle C'AO = \angle C'B'O=\angle C'A'O=\angle C'BO=90^{o}-\angle C'$

$\angle OAB' = \angle OC'B'=\angle OA'B'=\angle OCB'=90^{o}-\angle B'$

$\angle OBA'=\angle A'C'O=\angle OB'A'=\angle A'CO=90^{o}-\angle A'$

portanto

$AO=BO=CO$ o que evidencia que $O$ coincide com o circuncentro de $ABC$.

[i]
tambien resuelto por Jossy, y Miguel Acosta, parcialmente resuelto por Edson Laura Galvez[/i]

[b]2[/b][i]Titu Andreescu[/i]
Temos que $a_{n+1}-a_{n}a_{n-1}=\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}$ elevando ambos os membros ao quadrado obtemos

$a_{n+1}^{2}+a_{n}^{2}+a_{n-1}^{2}-2a_{n+1}a_{n}a_{n-1}=1$ percebemos que o primeiro membro da equação é constante para todo $n \ge 2$ portanto

$a_{n+1}^{2}+a_{n}^{2}+a_{n-1}^{2}-2a_{n+1}a_{n}a_{n-1}=a_{n}^{2}+a_{n-1}^{2}+a_{n-2}^{2}-2a_{n}a_{n-1}a_{n-2}$
$a_{n+1}^{2}-a_{n-2}^{2}=2a_{n}a_{n-1}(a_{n+1}-a_{n-2})$
$(a_{n+1}-a_{n-2})(a_{n+1}+a_{n-2})=2a_{n}a_{n-1}(a_{n+1}-a_{n-2})$
$a_{n+1}=2a_{n}a_{n-1}-a_{n-2}$

mas sabemos que $a_{0}=a_{1}=97$ e $a_{2}=97^{2}+(97^{2}-1)=18817$ portanto $a_{3}$ também será um numero inteiro e indutivamente podemos dizer que todo $a_{n}$ é um numero inteiro positivo. Agora vamos provar que $2a_{n+1}+2$ é um quadrado perfeito.Temos que

$a_{n+1}=(a_{n}+1)(a_{n-1}+1)+(a_{n}-1)(a_{n-1}-1)+2\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}-a_{n}a_{n-1}-\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}-2 ={(\sqrt{(a_{n}+1)(a_{n-1}+1)}+\sqrt{(a_{n}-1)(a_{n-1}-1)})}^{2}-a_{n+1}-2$
$2a_{n+1}+2={(\sqrt{(a_{n}+1)(a_{n-1}+1)}+\sqrt{(a_{n}-1)(a_{n-1}-1)})}^{2}$ e como

$\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}=\sqrt{(a_{n}+1)(a_{n-1}+1)(a_{n}-1)(a_{n-1}-1)}=a_{n+1}-a_{n}a_{n-1}$
é um número inteiro temos que $2a_{n+1}+2$ é um quadrado perfeito.

Agora suponhamos que ${2a_{n}+2}=4b_{n}^{2}$ e $2a_{n-1}+2=4b_{n-1}^{2}$ para $b_{n},b_{n-1}\in \mathbb{Z}$ consequentemente

$\sqrt{2a_{n+1}+2}=(\sqrt{(a_{n}+1)(a_{n-1}+1)}+\sqrt{(a_{n}-1)(a_{n-1}-1)}) =2b_{n}b_{n-1}+2\sqrt{(b_{n}^{2}-1)(b_{n-1}^{2}-1)}$
$\sqrt{2a_{n+1}+2}+2=[{(\sqrt{(b_{n}+1)(b_{n-1}+1)}+\sqrt{(b_{n}-1)(b_{n-1}-1)})}]$

o mesmo raciocínio segue quanto ao fato de sabermos que $\sqrt{(b_{n}^{2}-1)(b_{n-1}^{2}-1)}$ é inteiro e como $(b_{n-1}+1)$ e $(b_{n-1}-1)$ ou $(b_{n}-1)$ e $(b_{n}+1)$ são de mesma paridade concluímos que

$\sqrt{2a_{n+1}+2}=(2k)^{2}$ para $k \in \mathbb{Z}$

[i]Resuelto parcialmente por Miguel Acosta[/i]

[b]4[/b][i]Pascual[/i]
Lema(1)-Todos os números da forma $k^{p}$, donde $k$ é um numero inteiro, são semip-potencias.

Prova(1)-Basta fazermos $a=k^{p}$ e $b=(a^{p}-1)+\sqrt[p]{a}$ dai

$a^{p}-1|b^{p}-a=\sum_{i=0}^{p-1}\left(\begin{array}{cc}p\\\i\end{array}\right){(a^{p}-1)}^{i}.k^{p-i}$

Lema(2)-Todos os numeros da forma $pk+1$, donde $gcd(p,k)=1$ não são semip-potencias.

Prova(2)- De fato temos que $a=pk+1$ e $a^{p}-1=\sum_{i=0}^{p-1}\left(\begin{array}{cc}p\\\i\end{array}\right){(pk)}^{i}$ e como todo
$\left(\begin{array}{cc}p\\\i\end{array}\right)$ é divisível por $p$ temos que

$a^{p}-1=p^{2}m$ para $m \in \mathbb{Z}$ mas sabemos que $b=c+1$ para todo $c \ge 0$ portanto

$b^{p}-a=b^{p}-pk-1=c^{p}+\sum_{i=1}^{p-1}\left(\begin{array}{cc}p\\\i\end{array}\right){(c)}^{i}-pk$ mas $a^{p}-1|b^{p}-a$ portanto necessariamente $c=pn n \in \mathbb{Z}$ consequentemente

$c^{p}+\sum_{i=1}^{p-1}\left(\begin{array}{cc}p\\\i\end{array}\right){(c)}^{i}-pk=p^{2}n-pk=p(pn-k)$ conquanto temos que $p.pm|p(pn-k) \implies pm|pn-k$ mas como $gcd(p,k)=1$ isto nos leva a uma contradição.

(no he revisado esta solución, la mia es un poco diferente, colgare unos comentarios acerca de estos problemas luego, por el momento los dejo discutirlos)

Cuelgo tambien los trabajos que recibi!!! asi ustedes los discuten!!!

La tabla de posiciones las cuelgo mañana, gracias  a quienes participaron!!!

Eu estava um pouco ocupado quando escrevi as soluções, portanto, infelizmente, houve alguns erros ortográficos:

para o número '2' por favor considere $\sqrt{2a_{n+1}+2}+2=(2k)^{2}$ para $k \in \mathbb{Z_{+}}$

para o número "4' a última parte é $p^{2}no-pk$ para $o \in \mathbb{Z_{+}}$

Tomando $a_{0}=a_{1}=x$, y hallando $a_{2},a_{3},a_{4}$ obtenía los polinomios de Chebyshev, ¿esto se cumple para todo $a_{n}$ :?: .

Yo tengo una solucion diferente para el 1. Es mas larga y menos elegante pero bueno.
Vamos a probar que poder hacer una dilatacion espiral de $A'B'C'$ con centro en $H$ (donde $H$ es el otrocentro de $A'B'C'$) que cae sobre el triangulo medial de $ABC$. Con eso se acaba el problema pues claramente el ortocentro del triangulo medial es el circuncentro $ABC$.
Sean $A_{1}, B_{1}, C_{1}$ los puntos medios de $BC, CA, AB$ respectivamente. Si $A'$ esta sobre $BA_{1}$ $\implies B', C'$ estan sobre $CB_{1}, AC_{1}$ respectivamente. Y si $A'$ esta del otro lado los otros dos tambien cambian. (Si no es asi $A'B'C'$ no seria semejante a $A_{1}B_{1}C_{1}$ pues al mover los puntos medios no si no quedan asi crecerian 2 y uno se volveria mas chiquito o al reves).
Ahora supongamos sin perdida de generalidad que $A'$ esta sobre $BA_{1}$ y $B', C'$ estan sobre $CB_{1}, AC_{1}$ (si estan del otro lado es lo mismo).
Los puntos despues de la dilatacion espiral de $A'B'C'$ los llamaremos $A''B''C''$ respectivamente.
Para la dilatacion espiral es suficiente que $A'A''H$~$B'B''H$~$C'C''H$
Entonces probaremos primero que $\angle HA'C = \angle HB'A =\angle HC'B$
Sea $\alpha = \angle B'A'C$.
$\implies \angle HA'C =\angle HA'B'+\alpha = 90-\angle B+\alpha$ por lo que H es el ortocentro de $A'B'C'$
Ahora como $\angle A'B'C = \angle A+\angle B-\alpha$ tenemos que:
$\angle HB'A = 180-\left(\angle HB'A'+\angle A'B'C\right)= 180-(90-\angle A+\angle A+\angle B-\alpha) = 90-\angle B+\alpha$
Con lo que queda probado que:
$\angle HA'C = \angle HB'A$ y analogamente obtenemos que $\angle HA'C = \angle HC'B$ por lo que los 3 angulos que queriamos son iguales. Consideremos ahora los pies de las perpendiculares a $BC, CA , AB$ por $H$. Esos 3 puntos nos forman la dilatacion espiral que queremos. Entonces los pies van a ser $A'' , B'' , C''$ respectivamente.
Ahora como $H$ es el centro de la dilatacion espiral, es el otrocentro de $A''B''C''$. Entonces $A''H$ es perpendicular a $B''C''$ y tambien a $BC$, por lo que tenemos que:
$BC$ es paralela a $B''C''$. Analogamente $AC$ paralelo $A''C''$ y $AB$paralelo a $A''B''$. Entonces $A''B''C''$ tiene lados correspondientes paralelos a $ABC$ por lo que es el triangulo medial.

mmm el uncio que falta es el 3...

Este problema es original mio, mi solucion es usar Van der Waerden:
Si los enteros se colorean con finitos colroes hay progresiones aritmeticas de un color arbitrariamente largas.

Ahora considere la secuencia de colroes en un bloque, y considere a esto como una coloracion del bloque (un bloque son $k$ enteros consecutivos). Aplicando de nuevo Vanderwaerden a los bloques, hallamos que hay una progresiond e bloques arbitrariamente larga tal que los bloques estan identicamente coloreados.

Esto implica que los bloques deben ser monocromaticos...

Chao!!!

aun el bono se encunetra sin resolver!!!

dare 7 puntos a quien envie una solucion del bono!
!!!

Solución Chen:

1.- sea $H$ el ortocentro de $A'B'C'$  si $x, y, ,z$ son sus angulos respectiv.
$A'HC' =180-y=180-C'BA'$
Entonces el cuadrilstero BC'HA' es inscriptible, entonces $HBA=HA'C'=90-z$
Analogamente $HAB=90-z \Rightarrow HA=HB$
Analogamente $HA=HC$
Luego $H$ es circuncentro de $ABC$

2.-Haciendo $b_{n}= \sqrt{\frac{a_{n}+1}{2}}$ entonces $a_{n}=2{b_{n}}^{2}-1$
Reemplazando en la expresion, obtenemos la misma expresion pero con $b_{n}$ en vez de $a_{n}$,
Haciendo $c_{n}= \sqrt{\frac{b_{n}+1}{2}}$ entonces $b_{n}=2{c_{n}}^{2}-1$
Reemplazando en la nueva expresion, obtenemos la misma expresion pero con $c_{n}$ en vez de $b_{n}$..
Pero ahora c_0=2, c_1=2
Es facil probar por induccion que $c_{n}\ge 2$, entonces con eso podemos demostrar que $c_{n+1}>c_{n}$, $\ge 1$
$(c_{n+1}-c_{n-1}c{n})^{2}=(1-{c_{n}}^{2})(1-{c_{n-1}}^{2})$ ;$n \ge1$
Entonces:
$(c_{n}-c_{n-2}c{n-1})^{2}=(1-{c_{n}-1}^{2})(1-{c_{n-2}}^{2})$ ;$n \ge2$
Restando tenemos: $(c_{n+1}-c_{n-2})(c_{n+1}+c_{n-2}-2c_{n}c_{n-1})=0$. el primer termino es positivo ($c_{n+1}>c_{n}\ge c{n-2}$). Asi que :
$c_{n+1}=-c_{n-2}+2c_{n}c_{n-1}$ ;$n \ge2$
Luego, como $c_{0}, c_{1}$ y $c_{2}$ son enteros positivos, entonces por induccion la sucesion entera lo sera. pero $a_{n}=2{b_{n}}^{2}-1$ entonces $\sqrt{2+2a_{n}}=2b_{n}$ Luego lo que te piden es de mostrar que $2+2b_{n}$ que es igual a $4{c_{n}}^{2}$ es cuadrado perfecto, y eso es cierto porque $c_{n}$ es entero.


Show that the equation $\dfrac{A^2}{x-a} + \dfrac{B^2}{x-b} + ..... \dfrac{H}{x-h} = k$ has no imaginary roots. All constants above are real.

I assume you mean "H^2"

Suppose it has an imaginary root x = p + qi.  Then $\dfrac{A^2}{p-a + qi} = \dfrac{A^2(p-a-qi)}{(p-a)^2 +q^2}$.

The imaginary part of the LHS is $-q(A^2 + \dots + H^2)$, while the imaginary part of the RHS is $0$.  So this is an identity.

So either A = ... = H = 0 or q = 0.


Three rods are connected at one end such that they can swing freely around this point. The rod lengths are $57$, $65$ and $73$ cms.

[u]$a$[/u]
If you swing the rods about the hinge such that their ends form an equilateral triangle, how long is the side of this triangle?

[u]$b$[/u]
And another question: If you want to make the largest triangle by suitably arranging the rods, what is the area of this triangle?

I don't think that I quite understand this problem, how would you make an equilateral triangle with $3$ rods with lengths $57,65,$ and $73$.  :huh:

Imagine a point inside an equilateral triangle. Then connect it to the vertices. You get three segments, whose lengths should be $57, 65, 73$.


Let $ a,b,c >0 ; a\plus{}b\plus{}c \ge 6$. Prove that:

$ \sqrt{a^2 \plus{} \frac{1}{b\plus{}c}}\plus{}\sqrt{b^2 \plus{} \frac{1}{c\plus{}a}}\plus{}\sqrt{c^2 \plus{} \frac{1}{a\plus{}b}} \ge \frac{3\sqrt{17}}{2}$

By the Cauchy Schwartz we have $ \sum {\sqrt {a^2 \plus{} \frac {1}{{b \plus{} c}}} \ge \sqrt {\left( {\sum a } \right)^2 \plus{} \left( {\sum {\frac {1}{{\sqrt {b \plus{} c} }}} } \right)^2 } }$
And AM-GM
$ \sqrt {\left( {\sum a } \right)^2 \plus{} \left( {\sum {\frac {1}{{\sqrt {b \plus{} c} }}} } \right)^2 } \ge \sqrt {\left( {\sum a } \right)^2 \plus{} \frac {9}{{\sqrt [3]{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}}}} \ge \sqrt {\left( {\sum a } \right)^2 \plus{} \frac {{27}}{{2\left( {\sum a } \right)}}} \ge \frac {{3\sqrt {17} }}{2}$
Equality holds for a=b=c=2

[quote="Honey_S"]By the Cauchy Schwartz we have $ \sum {\sqrt {a^2 \plus{} \frac {1}{{b \plus{} c}}} \ge \sqrt {\left( {\sum a } \right)^2 \plus{} \left( {\sum {\frac {1}{{\sqrt {b \plus{} c} }}} } \right)^2 } }$
[/quote]
I think, it's Minkowsky. :wink:


Message says it all.  If someone would mail me one or a photocopy as we missed it this year and I would like to keep my records complete.  You can email at shnidei AT cogeco DOT ca.  I could reimburse you postage via paypal.
Thanks in advance
Steve

I THINK you can order exams from the AMC website.  I'm not sure though.

Contact me at the AMC office and I can tell you how you can get the official contest document and solutions.

Steve Dunbar
MAA Director, American Mathematics Competitions


Is there any way I can search for all posts made by [User A] on [User B]'s blog?

If I want to find a specific entry in my blog, I generally go to Google, type in the name of my blog and title of the entry I'm looking for. I suppose you could do the same thing with the name of [User A] instead of the title.


calculus
integration
logarithms
real analysis
real analysis unsolved

$ \sum_{n \equal{} 0}^{\infty} \frac {1}{n!} \left\{\sum_{k \equal{} 0}^{n} (k \plus{} 1)\int\limits^1_0 2^{ \minus{} (k \plus{} 1)x}dx\right\} \equal{} \;\; ?$

[hide="What I get:"]
$ \equal{} \frac { \minus{} 1}{\log 2} \sum_{n \equal{} 0}^{\infty} \frac {1}{n!} \left(\left[\sum_{k \equal{} 0}^{n} \frac {1}{2^{k \plus{} 1}}\right] \minus{} (n \plus{} 1)\right) \\
 \\
 \\
\equal{} \frac {1}{\log 2} \left( \sum_{n \equal{} 0}^{\infty} \frac {2^{ \minus{} (n \plus{} 1)}}{n!} \minus{} \sum_{n \equal{} 0}^{\infty} \frac {1}{(n \minus{} 1)!}\right)$

Now what do I do?[/hide]

[hide="The answer is:"]
$ \frac{1}{\log2}\left(\frac{1}{2}\sqrt{e} \minus{} e\right)$

So I guess there's some series expansion involved.[/hide]

Remember:

$ e^x \equal{} \sum^\infty_{n\equal{}0}\frac{x^n}{n!}$

Thanks. Now I get the answer as $ \frac{1}{\log2}\left(\frac{1}{2}\sqrt{e} \minus{} e\right)$.


function
number theory unsolved
number theory

Find all functions $f : N$ → $N$ such that :
                       $f(f(f(n)))+6f(n) = 3f(f(n))+4n+2001$
for all $\large n \in N$.

The same comments as in http://www.mathlinks.ro/Forum/viewtopic.php?t=140438 hold here, too.


In a certain fictional country, families stop having children precisely when they produce both a boy and a girl (at least one of each).  Assuming boys and girls are equally likely, how many children does the average family have?

[hide]Consider a family with one child.  Without loss of generality, assume that child to be a boy.  The expected value of the number of children that family will have (not including the first child) is

$E=\frac{1}{2}+\frac{1}{4}(2)+\frac{1}{8}(3)\ldots$
$E=\frac{1}{2}+\frac{1}{2}(\frac{1}{2}(2)+\frac{1}{4}(3)\ldots$
$E=\frac{1}{2}+\frac{1}{2}(E+\frac{1}{2}+\frac{1}{4}\ldots$
$E=\frac{1}{2}+\frac{1}{2}(E+1)$
$2E=1+E+1$
$E=2$

Then the expected value of the number of children that a family will have including the first child is $\fbox{3}$[/hide]