[quote="khashi70"]Manam ye soale Dge mizaram , soal male china hast vali natarsid sakht nist :) :
Farz konid $ C,D$ do noghteye delkhah bar rooye nimdayereye $ C$ ke be ghotre $ AB$ hast bashand . momas be nimdayere dar noghteye $ B$ khate $ CD$ ro dar $ P$ ghat mikonad . farz konid $ O$ markaze nimdayere bashat . khate $ PO$ , $ CA$ va $ AD$ ra be tartib dar $ E$ va $ F$ ghat mikonad , neshan dahid $ OE \equal{} OF$ .[/quote]
khob bebin baraye inke neshoon bedim $ O$ vasate $ EF$e kafie neshoon bedim $ (EF,O\infty)\equal{}\minus{}1$ dge...
khob hala bebin avval az hame az $ P$ be nimdayere momas kon (be joz $ PB$) va esme noghte tamaso bezar $ S$,khob hala $ PB,PS$ be dayere momasando $ PDC$ ghate delkhah pas $ (BS,DC)\equal{}\minus{}1$ khob hala be vasile noghte $ A$ rooye dayere inaro tasvir kon roo khatte $ EF$ pas darim:
$ (OS',FE)\equal{}\minus{}1$
ke dar an $ S'$ mahale talaghie $ AS$ va $ EF$ hast...
khob pas kafie sabet konim $ S'\to\infty$ ya be ebarati $ AS\parallel EF$ ya be ebarati $ AS\parallel PO$...
inam vazehe cherake bebinid $ PO$ bar ghotbie $ P$ amoode,amma ghotbie $ P$ hamoon $ SB$e...
pas kafie sabet konim $ AS\perp SB$ ya be ebarati $ \angle ASB\equal{}90$ ke inam doroste cherake $ AB$ ghotre...

[quote="sepehr ahmadzadeh"]bacheha goftan marathone hendese ara hal mide inam marathone hendese..in soalo fek konam khelia dide bashid soale marhale sevvome hamin emsale vali be dalili ghashangi heife to marathone nabashe
mosallase$ ABC$ va dayeremohatie an mafrooz ast. noghate $ D,M,K$ betartib mahalle tamase dayere mohati,vasate zel va paye ertefaye zelee $ BC$ mibashad. noghteye $ X$ ra rooye zele $ BC$ tori ekhtiar mikonim ke $ \frac{XM}{XH}\equal{}\frac{DM}{DH}$. dayere beghotre $ XD$ ra $ w_{A}$ minamim be tarighe moshabeh davayere $ w_{B}$ va $ w_{C}$ ra niz tarif mikonim.sabet konid in $ 3$ dayere bar yekdigar momasand..[/quote]
[hide="idea"]
in $ 3$ dayere moteallegh be yek daste dayere hastand...[/hide]

mosallase $ ABC$ mafroooz ast.$ D$ ra noghtei rooye emtedade $ BC$ (nazdike $ C$) begirid.betorike $ AC\equal{}CD$ .farz konid $ P$ noghte taghatoe dovvome dayere mohitie $ ACD$ ba dayerei be ghotre $ BC$ bashad. $ E$ ra noghteye taghatoe $ BP$ ba $ AC$ va $ F$ ra noghteye taghatoe $ CP$ ba $ AB$ begirid. sabet konid $ E,F,D$ hamkhatand.

[quote="sepehr ahmadzadeh"]mosallase $ ABC$ mafroooz ast.$ D$ ra noghtei rooye emtedade $ BC$ (nazdike $ C$) begirid.betorike $ AC \equal{} CD$ .farz konid $ P$ noghte taghatoe dovvome dayere mohitie $ ACD$ ba dayerei be ghotre $ BC$ bashad. $ E$ ra noghteye taghatoe $ BP$ ba $ AC$ va $ F$ ra noghteye taghatoe $ CP$ ba $ AB$ begirid. sabet konid $ E,F,D$ hamkhatand.[/quote]
farz konid $ EF$ zele $ BC$ ra dar $ D'$ ghat konad,mikhaim sabet konim $ D,D'$ bar ham montabeghan...
khob hala mahalle talaghie $ AP$ ba zele $ BC$ ro $ K$ benamid,khob hala chon $ AK,BE,CF$ hamresand pas $ (BC,KD') \equal{}\minus{}1$...
pas baraye inke sabet konim $ D,D'$ bar ham montabeghan kafie sbaet konim $ (BC,KD) \equal{}\minus{}1$ dge...
khob bebinid avval az hame deghat konid ke $ BC$ ghotre pas $ \angle BPC \equal{} 90$ khob pas baraye esbate $ (BC,KD) \equal{}\minus{}1$ kafie sabet konim $ CE$ nimsaze $ \angle DEK$ hastesh (cherake dar in soorat $ BE$ mishe nimsaze kharejio midoonim paye azla ba paye nimsaze dakhelio khareji hamsaz misaze...)
khob pas kafie sabet konim $ \angle CED \equal{}\angle CEK$...
inam doroste cherake bebinid avval az hame too dayere mohitie $ \triangle ACD$ darim:

$ \boxed{\angle CED \equal{}\angle CAD (I)}$

hamchenin chon $ AECD$ mohatie pas:

$ \boxed{\angle CEK \equal{}\angle ADC (II)}$

hamchenin tebghe farze masale $ AC \equal{} CD$ pas:

$ \boxed{\angle DAC \equal{}\angle CDA (III)}$

khob hala az $ (I),(II),(III)$ natije mishe ke $ CE$ nimsaze yeni $ (BC,KD) \equal{}\minus{}1$ amma dashtim $ (BC,KD') \equal{}\minus{}1$ pas $ D,D'$ bar ham montabeghan yeni $ EF$ az $ D$ migzare...

[quote="goodgirl_2278"]in ham ye soal az china,ke vaghean ghashange!
dar mosallase ABC , az noghteye D paye nimsaze rase A , amoodhaye DE va DF ra be tartib bar azlae AB va AC rasm mikonim. dayereyi ke az E va A migozarad,BF ra dar H va L ghat mikonad. sabet konid BL,BE,LF azlae yek mosallase ghaemozavie hastand![/quote]
yeni $ H$ hich naghshi nadare inja?   :huh:

sharmande,havasam nabood! :oops:
C,H,E hamkhatand!

[quote="goodgirl_2278"]dar mosallase ABC , az noghteye D paye nimsaze rase A , amoodhaye DE va DF ra be tartib bar azlae AB va AC rasm mikonim. dayereyi ke az E va A migozarad,BF ra dar H va L ghat mikonad. sabet konid BL,BE,LF azlae yek mosallase ghaemozavie hastand![/quote]
man hanoozam soorate soalo dorost motevajeh nashodam,yeni dar vaghe $ H$ mahale talaghie $ BF,CE$ hastesh,va ma oomadim dayere mohitie $ AHE$ ro keshidim ta $ BF$ ro dar $ L$ ghat kone...?  :maybe:
omidvaram manzoore soal hamin bashe...
khob mikhaim sabet konim $ BL,LF,BE$ azlae ye mosalase ghaemozavian,yeni dar vaghe mikhaim sabet konim ke too rabete fisaghores sedgh mikonan,khob bebinid avval az hame ye ghazie maroof hast ke mige $ AH$ hamoon ertefae rase $ A$ hastesh...
khob bebinid age betoonim sabet konim $ DL\perp BF$ oon moghe masalamoon halle chera ke khahim dasht:

$ \angle BLD\equal{}90\Rightarrow BL^{2}\equal{}BD^{2}\minus{}DL^{2}$ (1)

$ \angle DLF\equal{}90\Rightarrow LF^{2}\equal{}DF^{2}\minus{}DL^{2}$ (2)

az tarafi darim:

$ \angle DEB\equal{}90\Rightarrow BE^{2}\equal{}BD^{2}\minus{}DE^{2}$ (3)

hamchenin $ D$ paye nimsaze pas $ DE\equal{}DF$ khob hala darim:

$ (1)\minus{}(2)\equal{}(3)$

pas:

$ BL^{2}\minus{}LF^{2}\equal{}BE^{2}\Rightarrow BE^{2}\plus{}LF^{2}\equal{}BL^{2}$ yeni hamoon chizi ke mikhastim,pas faghat kafie sabet konim $ DL\perp BF$...
--------------------------------------
khob farz konid $ K$ paye ertefae rase $ A$ bashe,pas chon $ \angle HKD\equal{}90$ pas baraye inke sabet konim $ DL\perp BF$ kafie sabet konim $ DKLH$ mohatie...
ya be ebarati $ BL.BH\equal{}BK.BD$ amma tebghe ghovvate noghte $ B$ nesbat be dayerey ke keshidim darim:
$ BL.BH\equal{}BE.BA$ pas kafie sabet konim:

$ BE.BA\equal{}BK.BD$

ke inam doroste cherake darim:

$ \angle BED\equal{}90\Rightarrow BE\equal{}\cos B.BD$

pas kafie sabet konim:

$ \cos B.BD.BA\equal{}BK.BD$

$ \Rightarrow\cos B.BA\equal{}BK$ ke inam ba tavajoh be mosalase ghaemozavie $ ABK$ doroste...

omdivaram soorate soalo dorost fahmide basham... :maybe:

[quote="sepehr ahmadzadeh"]khashi to yadeTE??? soorat soalo?? mage hamoon niss ke sare kelas be vakili dadim??????? :?: soorat soal ghalat niss????[/quote]
rahe halle vakili chetori bood?mese hamine ya fargh dare?

are babak hamin bood vali oon sabet kard ke $ AH$ ertefaae yani nemidoonest sare kelas behesh residim..!!!ghazie??? man nashnidaaaam :wink:  soale hendese khoob dari toam bezar

soale ziad sakhti niss vali foghoalde ghashangee..!!
mosalllase $ ABC$ mafrooz ast.noghteye $ O$, markaze dayere mohiti..amoood monnasefe $ OA$, khatte $ BC$ ra dar noghteye $ A'$ ghat mikonad. noghate $ B'$ va $ C'$ niz betore moshabeh tarif mishavad sabet konid $ A'$,$ B'$ va $ C'$ bar khatti gharar darand ke bar $ ON'$ amood ast..($ N'$ mozdavaj hamzavye markaze dayere nohnoghte) halat kollisham amoood bar $ OA$ va $ OB$ va $ OC$ mibashad betorike faseleye paye amooodha az $ O$ beyek fasele bashad vali dg bar $ ON'$ amoood nist :wink:

are,ba seva mishe behesh resid! manam be onvane ghazie nashnide boodam,mr.vakili ham hamintor! :roll:

ghazie na be in ma'na ke jai oomade bashe,khob man az ghabl midoonestam $ AH$ ertefae,azash estefade kardam,hamin... :huh:

inam 1 soale nesbatan maroof:
1 dayereye sabet darim,ba 2 noghteye sabete A,B ke hamkhat ba markaze dayereand.ghotre PQ az dayere motoghayere..makan hendesie noghteye 2vome barkhorde dayerehaye mohitie APO & BQO ra peyda konid!

ye soal az ibero american $ 2007$ ham hast ke kheli asooone avli age kesi nadide bebinee :lol:
mosalasse $ ABC$ ra darnazar begirid va $ I$ markaze dayere mohatie dakheli..dayerei be markaze $ I$ ra ke shoa'ash az $ r$ bishtar ast va az hichkodam az raasha nagozashte bast ra darnazar begirid.. farz konid in dayere $ AB$ ra dar $ X_{1}$ ke be $ B$ nazdik ast va $ BC$ ra dar $ X_{2}$va $ X_{3}$ ke be tartib nazdik be $ B$ va $ C$ mibashad va $ AC$ ra dar $ X_{4}$ ke be $ C$ nazdik ast ghaat konad.. hal farz konid $ X_{1}X_{2}$ va $ X_{3}X_{4}$ yekdigar ra dar $ K$ ghat konand sabet konid $ KA$ az vasate $ X_{2}X_{3}$ migozarad..

[quote="sepehr ahmadzadeh"]
mosalllase $ ABC$ mafrooz ast.noghteye $ O$, markaze dayere mohiti..amoood monnasefe $ OA$, khatte $ BC$ ra dar noghteye $ A'$ ghat mikonad. noghate $ B'$ va $ C'$ niz betore moshabeh tarif mishavad sabet konid $ A'$,$ B'$ va $ C'$ bar khatti gharar darand ke bar $ ON'$ amood ast..($ N'$ mozdavaj hamzavye markaze dayere nohnoghte)[/quote]
[hide="idea"]kafie sabet konim ghotbiashoon hamrese...[/hide]

[quote="sepehr ahmadzadeh"]mosalasse $ ABC$ ra darnazar begirid va $ I$ markaze dayere mohatie dakheli..dayerei be markaze $ I$ ra ke shoa'ash az $ r$ bishtar ast va az hichkodam az raasha nagozashte bast ra darnazar begirid.. farz konid in dayere $ AB$ ra dar $ X_{1}$ ke be $ B$ nazdik ast va $ BC$ ra dar $ X_{2}$va $ X_{3}$ ke be tartib nazdik be $ B$ va $ C$ mibashad va $ AC$ ra dar $ X_{4}$ ke be $ C$ nazdik ast ghaat konad.. hal farz konid $ X_{1}X_{2}$ va $ X_{3}X_{4}$ yekdigar ra dar $ K$ ghat konand sabet konid $ KA$ az vasate $ X_{2}X_{3}$ migozarad..[/quote]
soale kheili ghashangie:
khob mahalle talaghie $ AK,X_{2}X_{3}$ ro $ S$ benamid,.vase inke sabet konim $ S$ vasate $ X_{2}X_{3}$e kafie sabet konim $ IS\perp X_{2}X_{3}$ ya be ebarati $ IS\perp BC$ amma farz konid $ D$ mahale tamase dayere mohati ba zele $ BC$ bashe oon moghe midoonim $ ID\perp BC$ pas kafie sabet konim $ D\equiv S$ yeni darvaghe kafie begim $ AS$ hamoon $ AD$e,ya be ebarati $ X_{1}X_{2},X_{3}X_{4}$ rooye $ AD$ be ham miresand...
khob ino ham kafie too mosalase $ \triangle AX_{1}X_{4}$ ceva sinusi ro check konim,yeni kafie sabet konim:

$ \frac{\sin BAD}{\sin DAC}.\frac{\sin AX_{4}X_{3}}{\sin X_{3}X_{4}X_{1}}.\frac{\sin X_{4}X_{1}X_{2}}{\sin X_{2}X_{1}A}\equal{} 1$

khob hala deghat konid ke too dayere darim:

$ \frac{C'X_{3}}{\sin AX_{4}X_{3}}\equal{} 2r'$

pas dar vaghe ebarate bala moadele ebarate zire:

$ \frac{\sin BAD}{\sin DAC}.\frac{C'X_{3}}{X_{2}X_{3}}.\frac{X_{2}X_{4}}{B'X_{2}}\equal{} 1$ (*)

khob hala biaid yekhorde dayeramoono baresi konim,noghte dovome talaghie in dayerero ba zele $ AB,AC$ be tartib $ B',C'$ benamido shoae in dayeraro barabare $ r'$ farz konid...
bebinid age az $ I$ be zele $ AB$ amood konimo paye amoode $ E$ benamim oon moghe too mosalase ghaemozavie $ IEX_{1}$ darim:

$ \sin IX_{1}E \equal{}\frac{IE}{IX_{1}}\equal{}\frac{r}{r'}$

khob be hamin tartib natije mishe:

$ \sin IX_{4}A \equal{}\frac{r}{r'}$

$ \sin IX_{2}C \equal{}\frac{r}{r'}$

pas az se rabete bala natije mishe ke:

$ \angle IX_{1}E \equal{}\angle IX_{4}A \equal{}\angle IX_{2}C \equal{}\alpha$

az tarafi mosalase $ IX_{2}X_{3}$ motasaviosagheine pas $ \angle X_{2}IX_{3}\equal{} 180\minus{}2\angle IX_{2}X_{3}\equal{} 180\minus{}2\alpha$

be tarighe moshabeh natije mishe ke:

$ \angle X_{2}IX_{3}\equal{}\angle X_{4}IC' \equal{}\angle B'IX_{1}\equal{} 180\minus{}2\alpha$

pas darim:

$ \triangle X_{2}IX_{3}\equal{}\triangle X_{4}IC' \equal{}\triangle B'IX_{1}$

cherake harkodoom do zel barabare $ r'$ va zavie beine $ 180\minus{}2\alpha$ daran...

khob pas az tasavie se mosalas natije mishe:

$ X_{2}X_{3}\equal{} X_{4}C' \equal{} B'X_{1}\equal{} x$

hala tebghe ghovvate noghte $ C$ nesbat be in dayere darim:

$ CX_{4}.CC' \equal{} CX_{3}.CX_{2}$

$ \Rightarrow CX_{4}(CX_{4}\plus{}x) \equal{} CX_{3}(CX_{3}\plus{}x)$

$ \Rightarrow CX_{4}^{2}\plus{}CX_{4}x \equal{} CX_{3}^{2}\plus{}CX_{3}x$

$ \Rightarrow (CX_{4}\minus{}CX_{3})(CX_{4}\plus{}CX_{3}\plus{}x) \equal{} 0$

amma paranteze docom ye ebarate mosbate pas paranteze avval bayad sefr beshe yeni:

$ CX_{4}\equal{} CX_{3}$

be tarighe moshabeh natije mishe $ BX_{2}\equal{} BX_{1}$.

khob hala deghat konid ke $ C'X_{4}\equal{} X_{3}X_{2}$ pas kamane $ C'X_{3}$ ba kamane $ X_{4}X_{2}$ barabare,pas darim:

$ C'X_{3}\equal{} X_{2}X_{4}$ va be tarighe moshabeh $ B'X_{2}\equal{} X_{1}X_{3}$ pas ebarate $ (*)$ ro mishe intori baznevisish kard:

$ \frac{\sin BAD}{\sin DAC}.\frac{X_{2}X_{4}^{2}}{X_{1}X_{3}^{2}}\equal{} 1$

ya be ebarati:

$ \frac{X_{2}X_{4}^{2}}{X_{1}X_{3}^{2}}\equal{}\frac{\sin DAC}{\sin BAD}$

amma deghat konid ke tebghe ghazie $ \sin$ha dar do mosalase $ ABD$ va $ ADC$ natije mishe ke:

$ \frac{\sin DAC}{\sin BAD}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

pas dar vaghe kafie sabet konim:

$ \frac{X_{2}X_{4}^{2}}{X_{1}X_{3}^{2}}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

amma deghat konid ke bazam too dayeramoon darim:

$ \frac{X_{2}X_{4}}{X_{1}X_{3}}\equal{}\frac{\sin X_{2}X_{3}X_{4}}{\sin X_{1}X_{2}X_{3}}\equal{}\frac{\sin X_{4}X_{3}C}{\sin X_{1}X_{2}B}$

amma deghat konid ke hamoontori ke ghablan goftim $ CX_{3}\equal{} CX_{4}$ pas darim:

$ \angle CX_{3}X_{4}\equal{}\angle CX_{4}X_{3}\equal{}\frac{180\minus{}C}{2}\equal{} 90\minus{}\frac{C}{2}$

pas $ \sin CX_{3}X_{4}\equal{}\cos{\frac{C}{2}}$

be tarighe moshabeh natije mishe:

$ \sin BX_{2}X_{1}\equal{}\cos{\frac{B}{2}}$

pas yeni dar vaghe ebarate hokmemoon moadele ineke sabet konim:

$ \frac{\cos^{2}{\frac{C}{2}}}{\cos^{2}{\frac{B}{2}}}\equal{}\frac{c(p\minus{}c)}{(b(p\minus{}b)}$

amma midoonim $ \cos 2x \equal{}\cos^{2}x\minus{}1$ pas $ \cos^{2}x \equal{}\cos 2x\plus{}1$ pas ebarate hokmo mishe intori neveshtesh:

$ \frac{\cos C\plus{}1}{\cos B\plus{}1}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

amma tebghe ghazie $ \cos$ha dar mosalase $ ABC$ darim:

$ \cos C \equal{}\frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{2ab}$

$ \cos B \equal{}\frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{2ac}$

pas ebarate hokmemoon mishe:

$ \frac{\frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{2ab}\plus{}1}{\frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{2ac}\plus{}1}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

$ \frac{\frac{(a\plus{}b)^{2}\minus{}c^{2}}{2ab}}{\frac{(a\plus{}c)^{2}\minus{}b^{2}}{2ac}}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

$ \frac{c(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)}{b(a\plus{}b\plus{}c)(a\plus{}c\minus{}b)}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

$ \frac{c(p\minus{}c)}{b(p\minus{}b)}\equal{}\frac{c(p\minus{}c)}{b(p\minus{}b)}$

yeni hamoon chizi ke mikhastim...

pas se khatii ke mikhastim hanmresan,pas $ S\equiv D$ pas $ IS\perp BC$ pas $ S$ vasate $ X_{2}X_{3}$ hastesh,pas $ AK$ parekhate $ X_{2}X_{3}$ ro nesf mikone...

[quote="sepehr ahmadzadeh"]mosalasse $ ABC$ ra darnazar begirid va $ I$ markaze dayere mohatie dakheli..dayerei be markaze $ I$ ra ke shoa'ash az $ r$ bishtar ast va az hichkodam az raasha nagozashte bast ra darnazar begirid.. farz konid in dayere $ AB$ ra dar $ X_{1}$ ke be $ B$ nazdik ast va $ BC$ ra dar $ X_{2}$va $ X_{3}$ ke be tartib nazdik be $ B$ va $ C$ mibashad va $ AC$ ra dar $ X_{4}$ ke be $ C$ nazdik ast ghaat konad.. hal farz konid $ X_{1}X_{2}$ va $ X_{3}X_{4}$ yekdigar ra dar $ K$ ghat konand sabet konid $ KA$ az vasate $ X_{2}X_{3}$ migozarad..[/quote]

farz konid $ D,E,F$ mahale momas dayere mohati dakheili ba $ BC,AC,AB$ be tartib bashad.
va hamchenin farz konid EF va BC dar D' yek digar ghat konand va $ X_{1}X_{4}$ va khati ke az $ K$ be movazate $ BC$ rasm mishavad yek digar ra $ D''$ ghat konand.ba tavajoh be movazi bodane $ DF$ ba $ KD''$ va $ EF$ ba $ X_{1}X_{4}$ darim ke $ A,D',D''$ ham khatand.
pas $ (KD'',KA,KX_{1},KX_{4})\equal{}\minus{}1$ pas $ D$ vasate $ X_{2}X_{3}$ gharar darad.

[quote="Amir.S"][quote="sepehr ahmadzadeh"]mosalasse $ ABC$ ra darnazar begirid va $ I$ markaze dayere mohatie dakheli..dayerei be markaze $ I$ ra ke shoa'ash az $ r$ bishtar ast va az hichkodam az raasha nagozashte bast ra darnazar begirid.. farz konid in dayere $ AB$ ra dar $ X_{1}$ ke be $ B$ nazdik ast va $ BC$ ra dar $ X_{2}$va $ X_{3}$ ke be tartib nazdik be $ B$ va $ C$ mibashad va $ AC$ ra dar $ X_{4}$ ke be $ C$ nazdik ast ghaat konad.. hal farz konid $ X_{1}X_{2}$ va $ X_{3}X_{4}$ yekdigar ra dar $ K$ ghat konand sabet konid $ KA$ az vasate $ X_{2}X_{3}$ migozarad..[/quote]

farz konid $ D,E,F$ mahale momas dayere mohati dakheili ba $ BC,AC,AB$ be tartib bashad.
va hamchenin farz konid EF va BC dar D' yek digar ghat konand va $ X_{1}X_{4}$ va khati ke az $ K$ be movazate $ BC$ rasm mishavad yek digar ra $ D''$ ghat konand.ba tavajoh be movazi bodane $ DF$ ba $ KD''$ va $ EF$ ba $ X_{1}X_{4}$ darim ke $ A,D',D''$ ham khatand.
pas $ (KD'',KA,KX_{1},KX_{4}) \equal{}\minus{}1$ pas $ D$ vasate $ X_{2}X_{3}$ gharar darad.[/quote]
khob alan nabayad sabet koni ke $ AK$ az $ D$ migzare?  :maybe:
ya inke ye nokte badihie ke man nemifahmamesh...

mosallase $ ABC$ mafrooz ast.noghate $ M$ va $ N$ nesbat be noghteye $ A'$,vasate zel'e $ BC$, motegharenand va darim:$ \angle BAN\equal{}\angle CAM$. khotoote $ CM$ va $ BM$,khatte $ AN$ ra be tartib dar noghate $ E$ va $ F$ ghaat mikonand.sabet konid momas bar dayereye mohitie $ ACF$ dar noghteye $ C$ va momas bar dayereye mohitie $ ABE$ dar noghte $ B$ va $ AM$ hamresand... be nazar soale sakhti miadd...!!albate man roo rahe hall ba seva fekr nakardam yani roosh seva nazadam shaiad ba seva golabi she

[quote="BaBaK Ghalebi"]...

khob alan nabayad sabet koni ke $ AK$ az $ D$ migzare?  :maybe:
ya inke ye nokte badihie ke man nemifahmamesh...[/quote]

noktash badihiye , negah kon ke in 2 dayare ham markazand va $ ID\perp X_{2}X_{3}$ pas D vasate $ X_{2}X_{3}$.
hal fazr kon ke $ AK$ va $ BC$ yekdigar ro dar $ Y$ ghat konand Choon $ X_{2}X_{3}\parallel KD''$ pas $ Y$ vasate $ X_{2}X_{3}$ ast darnatije $ Y\equiv D$ :wink:

[quote="sepehr ahmadzadeh"]mosallase $ ABC$ mafrooz ast.noghate $ M$ va $ N$ nesbat be noghteye $ A'$,vasate zel'e $ BC$, motegharenand va darim:$ \angle BAN \equal{}\angle CAM$. khotoote $ CM$ va $ BM$,khatte $ AN$ ra be tartib dar noghate $ E$ va $ F$ ghaat mikonand.sabet konid momas bar dayereye mohitie $ ACF$ dar noghteye $ C$ va momas bar dayereye mohitie $ ABE$ dar noghte $ B$ va $ AM$ hamresand... be nazar soale sakhti miadd...!!albate man roo rahe hall ba seva fekr nakardam yani roosh seva nazadam shaiad ba seva golabi she[/quote]

[b]rahe halle aval[/b]i ke be zahn mirese ceva sinusiye ke javab mide va asoon ham hast.
[b]rahe halle dovvom[/b]:Zaviye bazi
farz konid momas dar $ C$ bar dayereye mohitiye $ ACF$ , $ AM$ ra dar $ X$ ghat konad.darim :
$ \angle AXC \equal{} 180\minus{}(\angle CAX\plus{}\angle ACB\plus{}\angle BCM\plus{}\angle FCX \equal{}\angle A\plus{}\angle B\minus{}\angle BAN\minus{}\angle NBC\minus{}\angle FAC \equal{}\angle ABN\ (\star)$ .

darnatije $ \triangle ABN\sim\triangle ACX$  :ninja:  hal farz konid $ BN$ va $ CX$ yekdigar ra dar $ K$ ghat konand banabar $ \star$ darim ke  4zeli $ ABXK$ mohatist.
pas:$ \angle BKX \equal{}\angle BAX \equal{}\angle NAC$ pas 4zeli $ AKCN$ niz mohatist pas $ \angle BXA \equal{}\angle AKB \equal{}\angle ACN$ choon ke $ BM\parallel NC$ dar natije $ \angle(BM,AC) \equal{}\angle ANC \equal{}\angle AXB$ dar natije $ BM$ va $ AC$ yek digar ro rooye dayere mohitiye $ AKXB$ ghat mikonand dar natije$ \angle MBX \equal{}\angle CAX \equal{}\angle BAN$ dar natije XB bar dayere mohitiye $ ABE$ momas ast.


algebra
polynomial
Vieta
pen
function
induction
Divisibility Theory

[quote="Peter"]Let $ a$ and $ b$ be positive integers such that $ ab \plus{} 1$ divides $ a^{2} \plus{} b^{2}$. Show that
\[ \frac {a^{2} \plus{} b^{2}}{ab \plus{} 1}
\]
is the square of an integer.[/quote]
And futhermore
Try *Let $ p,q\in \mathbb Z^ \plus{} ,p\le q$
Find $ a,b$ satisfying $ \frac {(a \minus{} b)^2}{pab \minus{} q} \equal{} k (k\in \mathbb N)$

*$ \frac {(a \minus{} b)^2 \plus{} m}{pab \minus{} q} \equal{} k(m\mathbb N^*)$ so K  belongs to a finite set
They are nice,do you think that
P/s,I posted here but no reply http://www.mathlinks.ro/viewtopic.php?t=150373

Well let us assume that a^2 + b^2 / 1 + ab = q (where q is some non square integer)
converting it into a linear form we get

a^2 + b^2  -qab -q =0

Now let us consider the equation x^2 + y^2 - qxy - q =0.......(1)

Let (m,n) be a soln. for which m^2 + n^2 is the least . (m>n without any loss of generality)

Clearly m is the soln of the quadratic x^2 + n^2 -qxn - q =0.........(2)

let us consider the other root of (2) say p.

Now we have,

                        p + m = qn
                        [b]pm = n^2 - q > n^2[/b]

Now we have m > n
Thus clearly p <n

Therefore p^2 + n^2 < 2n^2 < m^2 + n^2

Which is clearly a contradiction of our original assumption.
Therefore the given expression with the given conditions must be a square.

P.S. - The method is known as infinite descent in case. And sorry for the crude typing!

[quote="A K Bhattacharyya"][b]pm = n^2 - q > n^2[/b][/quote]
Shouldn't it be [b]pm = n^2 - q [color=red]<[/color] n^2[/b]
But that's ok with the part that follows...
Rather it would be wrong otherwise.

[quote="A K Bhattacharyya"]Thus clearly p <n

Therefore p^2 + n^2 < 2n^2 < m^2 + n^2[/quote]
Why $ p < n$ implies $ p^2 < n^2$
$ p$ could be negative, so you might have to prove it otherwise like Naphthalin did...

EDIT: ... doesn't that show that there's actually no such solutions... square or not... ??

No it can't.

We are dealing with both positive roots remember ?

Yeah. it has to be < . Typing error there Sorry.

[quote="A K Bhattacharyya"]We are dealing with both positive roots remember ?[/quote]
We are dealing with positive $ m,n$, but $ p$ could be negative, which is the other root of that quadratic, you don't yet have both of it's roots positive, or am I missing something ??

My solution is the following.

Let   k = (a^2 + b^2) / (ab + 1) , and let  k  isn’t the square of an integer. The equality is equivalent to                                 a^2 + b^2 - kab = k .
The equality is symmetric by  a  and  b, so we may accept, that  a>=b.
The equality is a quadratic equation by  a . Let  a  and  a1  are its roots. From Viet’s theorem  a1 + a = kb , therefore  a1  is integer.
If  a1=0  , then   k = b^2 (contradiction).
If   a1 < 0 , then  a > kb , and therefore   a^2 + b^2 - kab = b^2 + a (a - kb) >= b^2 + a > b^2 + kb > k , which is impossible.
Thus   a1 > 0. From Viet’s theorem   a1 = (b^2 - k) / a <= (a^2 - k) / a < a.
If we take  b1 = b, then  (a1, b1)  is a new solution of  equation, for which  a1 + b1 < a + b. If we do the same for   (a1, b1), we will get a new solution  (a1, b2) st  a2 + b2 < a1 + b1. Continuing the operation, we will get   (a, b), (a1, b1) , (a2, b2),.... infinitely many solutions, for which   a + b > a1 + b1 > a2 + b2 > .... , which is impossible, because   ai, bi (i = 1, 2, 3, ....)  are  positive integers. Therefore  k  is a perfect square (Q. E. D.). We can notice that  a=2, b=8 (or  a=b^3) numbers satisfy to equality.

i think this problem is overanalyzed: by definition, since ab+1 divides a^2+b^2, then a^2+b^2-abc-c=0, for c the number that results when ab+1 divides a^2+b^2. by considering the factorization of this into (a-x)(a-y), it is easily seen that
xy= b^2-c
x+y= b*c
Now, since this deals with positive integers, b^2-c is less than or equal to b^2
                                                                      b*c is less than or equal to b
This gives the inequalities 0 less than or equal to c less than or equal to 1. since the quotient of the two is an integer by definition, that means c is either 0 or 1, both of which are squares. I dont see the need to prove it is a square when it suffices to show it only equals 0 or 1

[quote="Naphthalin"]the problem is not too hard because the method of vieta-jumping is much more known today than 20 years before.

first we take \[\frac{a^{2}+b^{2}}{ab+1}=k.\]
so we can write instead \[a^{2}+b^{2}-kab-k=0.\]
we assume $a>b$ ($a=b$ implies $\frac{2a^{2}}{a^{2}+1}\in\mathbb{Z}$ so $a=b=1$ and $k=1$.)
we take the equation as a quadratic function in $a$. if we assume that there is a solution $a,b$ (otherwise the problem would be trivial), we have the solutions $a_{1}, a_{2}$ with $a_{1}=a$. wmls the equations $a_{1}+a_{2}=kb$ and $a_{1}a_{2}=b^{2}-k$ hold (vieta).
we want to show $a_{2}\geq 0$. so we assume $a_{2}<0$ so $b^{2}-k<0$ so $k>b^{2}$. since $a_{1}>kb>b^{3}$ we get $ab+1>b^{4}+1$ which is a contradiction to \[ab+1 | b^{2}(a^{2}+b^{2})=b^{4}+1+(ab-1)(ab+1).\]
so i can assume $a_{2}\geq 0$. because of $a_{1}a_{2}=b^{2}-k<b^{2}$ and $a_{1}>b$ $a_{2}<b$ holds. so we have created a new solution $a,b$ with $a<b$. We can re-do this until one of the numbers is $0$. Since the $k$ doesn't change we get $k=\frac{a^{2}}{1}=a^{2}$ which is a perfect square.

Naphthalin[/quote]
sorry what is the vieta jumping method

[quote="mozilla"][hide]a brilliant solution by someone else (not by me)

without loss of generality we may assume that $ a < b$

let $ n \equal{} \frac {a^2 \plus{} b^2}{ab \plus{} 1}$

define the infinite integer sequence $ \cdots,x_{ \minus{} 1},x_0,x_1,x_2,\cdots$ as follows:
\[ x_0 \equal{} a,\quad x_1 \equal{} b,\quad x_{i \plus{} 1} \equal{} nx_i \minus{} x_{i \minus{} 1}\mbox{ for }i \equal{} 1,2,\cdots,\mbox{n}
\]
it's easy to prove by mathematical induction that
\[ x_i^2 \plus{} x_{i \minus{} 1}^2 \equal{} n(x_ix_{i \minus{} 1} \plus{} 1)\mbox{ for }k \equal{} \cdots, \minus{} 1,0,1,2,\cdots
\]
and by mathematical induction that
\[ \cdots < x_{ \minus{} 1} < x_0 < x_1 < x_2 < \cdots,\mbox{(i.e. }x_i\mbox{ is a strictly increasing sequence)}
\]
the key to proving this problem is that one of the term $ x_i$ must be $ 0$

to prove this, we take $ x_m$ to be the smallest positive term in the sequence such that $ x_m > 0$, by defination that this term is smallest, we have $ x_{m \minus{} 1}\leq 0$

suppose that $ x_{m \minus{} 1} < 0$, then in the following equation $ (*)$, L.H.S. is greater than 0 while R.H.S. is less than or equal to 0, which is a contradiction
\[ x_m^2 \plus{} x_{m \minus{} 1}^2 \equal{} n(x_mx_{m \minus{} 1} \plus{} 1)\qquad(*)
\]
therefore $ x_{m \minus{} 1} \equal{} 0$, substitute into equation $ (*)$ we have $ n \equal{} x_m^2 \equal{} \frac {a^2 \plus{} b^2}{ab \plus{} 1}$ Q.E.D.[/hide][/quote]
wow. forget vieta jumping, this one is the best. do you know who gave this solution?

[quote=Naphthalin]
we want to show $a_{2}\geq 0$. so we assume $a_{2}<0$ so $b^{2}-k<0$ so $k>b^{2}$. since $a_{1}>kb>b^{3}$ we get $ab+1>b^{4}+1$ which is a contradiction to \[ab+1 | b^{2}(a^{2}+b^{2})=b^{4}+1+(ab-1)(ab+1).\]
so i can assume $a_{2}\geq 0$. because of $a_{1}a_{2}=b^{2}-k<b^{2}$ and $a_{1}>b$ $a_{2}<b$ holds. so we have created a new solution $a,b$ with $a<b$. We can re-do this until one of the numbers is $0$. Since the $k$ doesn't change we get $k=\frac{a^{2}}{1}=a^{2}$ which is a perfect square.
Naphthalin[/quote]

How do you get $a_{1}>kb>b^{3}$?

[quote=Peter]Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^{2}+b^{2}$. Show that \[\frac{a^{2}+b^{2}}{ab+1}\] is the square of an integer.[/quote]

Use Vieta jumping

This is also the legendary problem 6, from the 1988 IMO.

[center]https://artofproblemsolving.com/community/c6h57282_one_of_my_favourites_problems_yeah[/center]

its also solvable by finding quadratic equation roots. :play_ball:

I tried to solve this problem by using elementary algebra.
However, by assumption X = a^2 if b=0, the solve for x is a^2 = -2

https://documentcloud.adobe.com/link/track?uri=urn:aaid:scds:US:ed532489-a454-48eb-b48e-128f876d3a24

https://twitter.com/globeconomia/status/1453053910353072133?s=20

[quote=Peter]Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^{2}+b^{2}$. Show that \[\frac{a^{2}+b^{2}}{ab+1}\] is the square of an integer.[/quote]

[url=https://artofproblemsolving.com/community/c6h57282p24599415]Look here[/url]


function
superior algebra
superior algebra unsolved

these questions are really important for me can you help me with one of these??

1))

hi.. Q is rationals,
a) let K be finite over Q_p , let L_1 / K and L_2 / K be
 finite totally ramified extensions of K with ramification
 degrees e_1 and e_2 , suppose that (e_1,e_2)=1 .
Find a prime element of L_1.L_2, and
show L_1.L_2 / K has totally ramified of degree e_1.e_2

b) what is the valuation rings of
Q_p( a , p_th root of p )
and Q_p( a , (2p-1)_th root of p ) where
a = p_th root of 1

2))

 Z is integers
let K be finite extension of Q_p .
show that the group of roots of unity of K
has order (q-1).p^s for some s>=0 where q is the order
 of the residue field

3))

Z is integers
let f: Z_2 to Z_2 be the function defined by
f(x)=1 if x is element of 2Z_p
f(x)=-1 if x is not element of 2Z_p
show that f is continous and find the mahler
 expansion of it

[quote="berti"][quote="slash"]2))

 Z is integers
let K be finite extension of Q_p .
show that the group of roots of unity of K
has order (q-1).p^s for some s>=0 where q is the order
 of the residue field
[/quote]I think this is wrong: take $K = \mathbb Q_p$, $p > 2$. Then $\mu_{q-1} \subset K$ (Hensel's lemma) and $\mu_2 \subset K$, but $2\cdot(q-1) \nmid (q-1)p^s$.[/quote]
Note that $\mu_{2} \subset \mu_{q-1}$.


algebra
polynomial
quadratics
group theory
abstract algebra
Galois Theory
number theory

In p-element fields $F_p$, where p is a prime number
1. If 3 is not a factor of n, then $f(x)=x^n-8$ is irreducible
2. $x^4-x^2+1$ is reducible
3. $x^4+1$ is reducible

I really need some help in this kind of problems, thanks!

I think the statements 2 and 3 are wrong in some fields...
E.g. $p=17$, then it has even a root: $2$.

sorry, I made a big mistake and you are right.  I have edited the original post, and what I actually want to prove is they are reducible in p-element fields.

[hide="hint"]When $u^2=-1$ can be solved, then we can factor it as $(x^2+u)(x^2-u)$.
When $v^2=2$ can be solved, then we can factor it as $(x^2+vx+1)(x^2-vx+1)$.
When $w^2=-2$ can be solved, then we can factor it as $(x^2+wx-1)(x^2-wx-1)$.
I will leave it to you that at least one of this happens everytime.[/hide]
The second one can be done in a very similar way :)
(and both come from the fact that the rational Galois group is $(\mathbb{Z}/2\mathbb{Z})^2$, but Galois groups of finite fields are cyclic)

Actually I have already reached as far as your hint said, but I just don't know how to prove the rest since I did not learn much about elementary number theory.  But I found a way using only field extension to solve this problem.  Since $8|p^2-1$ when p is an odd prime, $x^4+1|x^8-1|x^{p^2}-x$, all roots of $x^4+1$ lies in a 2-degree extension of $\mathbb{F}_p$, which implies that $x^4+1$ can be reduced to two quadratic polynomials.  This method is due to "Abstract Algebra", Dummit & Foote

Nice :)
For mine, no number theory is necessary, either (but it's helpfull in Numbertea):
The only thing that is necessary to show is that the product $ab$ (here $-2$) of two non-squares $a,b$ (here $-1,2$) is a square.
Here Euler's for example criterion helps: $c \neq 0$ is a square iff $c^{\frac{p-1}{2}}=1$ (one direction follows from Lagrange's theorem, and both directions follow from the fact that the multiplicative group is cyclic or by some combinatorics).
Or in another way (essentially the same): the nonzero squares build a multiplicative subgroup of order $2$ in the multiplicatie group, and the only coset are the nonsquares.


$x, y, z > 0$

prove that:

$(y*z^{2})^\frac{1}{3}+(x^{2}*z)^\frac{1}{3}+(y^{2}*x)^\frac{1}{3}\leq 3*(xyz)^\frac{1}{3}$

(sorry but I don't know how to type 3rd degree root in latex)

If X=1,Y=1,Z=27,then
L.S.=9+3+1=13
R.S.=3*3=9
I think your inequation is wrong...

If you reversed the inequality sign by accident then it can be proven with a simple application of AM-GM


floor function
function
geometry
rectangle
ceiling function
analytic geometry
algebra

Let $ a$ and $ b$ be integers greater than one which have no common divisors. Prove that
\[ \sum_{i \equal{} 1}^{b \minus{} 1} \lfloor \frac {ai}{b}\rfloor \equal{} \sum_{j \equal{} 1}^{a \minus{} 1} \lfloor \frac {bj}{a} \rfloor
\]
and find the value of this common sum.

I tried 3 and 5 and they don't seem to equal

[quote="Tootlol"][quote="ocha"]Which part dont you understand?
$ \lfloor x \rfloor$ is the floor function it is basically rounding down (i.e $ \lfloor 2.345 \rfloor \equal{} 2$)

In this question with $ (a,b) \equal{} (3,5)$ we have

$ \left \lfloor \frac {3}{5} \right \rfloor \plus{} \left \lfloor \frac {2(3)}{5} \right \rfloor \plus{} \left \lfloor \frac {3(3)}{5} \right \rfloor \plus{} \left \lfloor \frac {4(3)}{5} \right \rfloor \equal{} \left \lfloor \frac {5}{3} \right \rfloor \plus{} \left \lfloor \frac {2(5)}{3}\right \rfloor$
$ 0 \plus{} 1 \plus{} 1 \plus{} 2 \equal{} 1 \plus{} 3$
$ 4 \equal{} 4$

So it does hold for $ (3,5)$[/quote]

5/3 is 1.67 so it should be 2 right?[/quote]

No, $ \lfloor 1.67 \rfloor \equal{} 1$. It is not rounding to the nearest integer, it is rounding down, as ocha said.

As for the question itself, look at a rectangle with vertices $ (0, 0), (a, 0), (a, b), (0, b)$.

[hide="solution"]
Torajirou - thanks for the hint, i think i've got it
Consider our intigers $ a,b$ such that $ \gcd(a,b) \equal{} 1$

Now we prove that our sum $ \sum_{i \equal{} 1}^{b \minus{} 1} \left \lfloor \frac {ai}{b}\right \rfloor$ is "symmetrical" (if that's the correct way to describe it)

Consider
$ \left \lfloor \frac {kb}{a} \right \rfloor \plus{} \left \lfloor \frac {(a \minus{} k)b}{a} \right \rfloor$

$ \Rightarrow b \minus{} \left \lceil \frac {kb}{a} \right \rceil \plus{} \left \lfloor \frac {kb}{a} \right \rfloor \equal{} b \minus{} 1$

$ \Rightarrow \sum_{i \equal{} 1}^{a \minus{} 1} \left \lfloor \frac {bi}{a}\right \rfloor \equal{} \frac {(b \minus{} 1)(a \minus{} 1)}{2}$

Similarly $ \sum_{i \equal{} 1}^{b \minus{} 1} \left \lfloor \frac {ai}{b}\right \rfloor \equal{} \frac {(a \minus{} 1)(b \minus{} 1)}{2}$

$ \frac {(b \minus{} 1)(a \minus{} 1)}{2} \equal{} \frac {(a \minus{} 1)(b \minus{} 1)}{2}$

$ \Rightarrow \sum_{i \equal{} 1}^{a \minus{} 1} \left \lfloor \frac {bj}{a}\right \rfloor \equal{} \sum_{j \equal{} 1}^{b \minus{} 1} \left \lfloor \frac {ai}{b}\right \rfloor \qquad$ QED :lol:
[/hide]

Just wondering, how did t0rajir0u's hint apply to this problem?

Count the lattice points (integer coordinates) within that rectangle, in each of the triangles determined by its main diagonal. This is a result by Gauss.


I want to creat a intense magnetic field, a several Tesla for a short period of time, less than some miliseconds.

Analisying the $RLC$ circuit It could be made making $C$ small and $L$ small.
the current will be several kiloAmpere. so we can induce a high eletric field.

but, how can i find good values or another  better circuit?

because theorical i can make a infinity eletric field.

but, for pratical purpose, there is a easy way to create intense eletromagnetics fields with simples components?

A better question would be: in what region of space do you want this field? I mean, it's not hard to make one tesla using a transformer, but you'll have this field located in a ~1mm wide crack... Somehow I don't think this is what you want ;)

I want a high magnetic field only to induce a eletric field.

If i make a circuit RLC, the square of the frequencia is proportional to $1/L$ and $1/R$

So, if i make the frequency too high, the eletric field will be intense.

but to your question Djole.
I want to create it in a very small lenght.

and a transformer is not good enought. because i want just a pulse, very fast, to induce a eletric field.

i made some circuits RLC and i can create a potential of more than 3MV and a current of some Amperes in a very small time.

theorical is ok. but i made this circuit in Real-Life, and i couldnt detected this intense field. so i ask here if someone know a way to improve this experiment.

i'm more interested in the circular current induced by it.

c ya


Let $ p,q,r>0$ such that  $ 2p\equal{}q\plus{}r$ and $ q\neq r$.
Prove that
$ \frac{r^{p^{q\plus{}r}}}{q^qr^r}<1$.

[quote="pqrs"]Let $ p,q,r > 0$ such that  $ 2p \equal{} q \plus{} r$ and $ q\neq r$.
Prove that
$ \frac {r^{p^{q \plus{} r}}}{q^qr^r} < 1$.[/quote]
Hmm. What about $ r\equal{}5$, $ q \equal{} 3$ and $ p\equal{}4$?
Then we have $ \frac {r^{p^{q \plus{} r}}}{q^qr^r} \equal{} \frac{5^{4^8}}{5^5\cdot3^3} > 1$!


geometry perimeter conics ellipse Euler integration trigonometry

[url=http://books.google.com/books?id=w-iZ2LmIDUgC&pg=PA24&lpg=PA24&dq=perimeter+ellipse+euler+differential+equation&source=web&ots=8qkaj9VpGH&sig=d9EILJ9qB4fTverS-CIihtKa7Ks]Link shortened by moderator[/url]

If you go to this link, you can see #4. I only need part (a). You don't have to prove the hint (you can just use it.)

Can anyone solve this one?

Thanks.

Since that's a very annoying link, here's a transcription of the question:

Let $ y(x)\equal{}\int_0^{2\pi}\sqrt{\sin^2 s\plus{}x^2\cos^2 s}\,ds$ for $ |x|<1$.

(a) Verify that $ y$ satisfies the differential equation $ x(1\minus{}x^2)y''\minus{}(1\plus{}x^2)y'\plus{}xy\equal{}0$.

90 percent of the time, I don't even [i]look[/i] at problems that are behind a link.

Hah... do you go to McGill by any chance? I have this problem on my homework. I was able to do it but it took a TON of number-crunching, I wasn't able to find a smart solution for it, although I'm sure there is one. You'll need to know Leibniz's Rule for Differentiating Integrals. Assuming that $ f$ is a nicely-behaved function, \[ \frac{d}{dx}\int f(x, y) dy \equal{} \int \frac{\partial}{\partial x}f(x, y) dy.\] So you want to show that $ f(x) \equal{} \int^{2\pi}_0\sqrt{\sin^2s \plus{} x^2\cos^2s}ds$ is a solution, then differentiate the integral with respect to $ x$ and apply Leibniz's Rule.

Brutal force(?) with Leibniz's rule for integral shows that

$ x(1\minus{}x^2)y''\minus{}(1\plus{}x^2)y'\plus{}xy \equal{} \int_{0}^{2\pi} \frac{x \sin^4 s \minus{} x^3 \cos^4 s}{(\sin^2 s \plus{} x^2 \cos^2 s)^{3/2}} \, ds.$

Let $ I$ be the integral on the RHS of this equation.

Using the substitution $ t \equal{} \tan s$ with periodicity of trigonometric functions, $ I \equal{} 4x \int_{0}^{\infty} \frac{t^4 \minus{} x^2}{(t^2 \plus{} x^2)^{3/2} (t^2 \plus{} 1)^{3/2}} \, dt.$

Also, applying new substitution $ u \equal{} x/t$ gives $ I \equal{} \minus{}4x \int_{0}^{\infty} \frac{u^4 \minus{} x^2}{(u^2 \plus{} x^2)^{3/2} (u^2 \plus{} 1)^{3/2}} \, du.$

Therefore, $ I \equal{} \minus{}I$ and hence $ I \equal{} 0$.

p.s. Is [url=http://kin.naver.com/detail/detail.php?d1id=11&dir_id=110203&eid=bYBJxoMvzpNHz4QWsRoGu+cxTS0TAoac&l_url=L2xpc3QvbGlzdF9ub2Fuc3dlci5waHA/ZDFpZD0xMSZkaXJfaWQ9MTEwMjAzJnNvcnQ9d3JpdGVfdGltZSZwYWdlPTk=]this[/url] your post?


I was trying to solve a problem, and needed to show that given two sets $A,B$, such that

$A=\{a_1,\ldots,a_k\}$
$B=\{b_1,\ldots,b_{100-k}\}$

where $a_i,b_j \in \{1,2,\ldots,100\}$, and $\sum{a_i}=\sum{b_j}$; we can choose two elements (say $k_1,k_2$)of $A$ and two elements (say $m_1,m_2$) of $B$ such that $k_1+k_2=m_1+m_2$.

Is that possible?

thanks

I suppose that you mean $\{a_1, a_2, ..., a_k, b_1, ..., b_{100-k}\}=\{1, 2, ..., 100\}$, as otherwise there are obvious counterexamples.

Well, take the set $\{1, 2, ..., 100\}$ and color each number red if it belongs to $A$, and blue if it belongs to $B$. So we get a sequence of letters $R, B$, for example $RRBBBR...RR$ if $1, 2, 6$ are in $A$ and $3, 4, 5$ are in $B$... etc. Note that if the sequence contains a $RB$ block and a $BR$ block, then we have four numbers $a, a+1, b, b+1$ with $a, b+1$ in $A$ and $a+1, b$ in $B$, which both sum $a+b+1$... so this works unless the blocks share some elements (i.e. they're both part of a $RBR$ or a $BRB$ block).

Suppose WLOG the sequence starts with an $R$. It can't be all $R$, so we must have something like $RRR...RB$. If the next letter is also $B$ (if it starts $RRR...RBB$), then it can't have any other $R$, or we'd have two disjoint $RB, BR$ blocks. Then it must be of the form $RRR...RRBB...B$, so $2(1+2+...+a)=1+2+...+100$ for some $a$, which we can check to have no solutions.

The case left is $RRR...RBR$. Similarly, if it's $RRR...RBRR$ we can have no more $B$s, absurd. So it must be $RRR...RBRB$. Then we can have no more $R$s, so it must be $RRR...RBRBB...B$. Here $2(1+2+...+(a-1)+(a+1))=1+2+...+100$ which we can check to have no solutions. Done.


[hide]Let $ z\equal{}a\plus{}b\imath$, where $ a,b\in\mathbb{Z}$.  We wish to show that both $ a\equal{}0$ and $ b\equal{}0$.  Then, we know that:
$ \sqrt{(a\plus{}1)^2\plus{}b^2}\leq1\Rightarrow (a\plus{}1)^2\plus{}b^2\leq1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
$ \sqrt{(a^2\minus{}b^2\plus{}1)^2\plus{}(2ab)^2}\leq1\Rightarrow (a^2\minus{}b^2\plus{}1)^2\plus{}(2ab)^2\leq1\ \ \ \ \ \ (2)$

From $ (2)$, we know that at least one of $ a$ and $ b$ are 0, because otherwise, $ (2ab)^2>1$ and $ (a^2\minus{}b^2\plus{}1)^2\geq0$.
Assume, for the sake of contradiction, that $ a\neq0\Rightarrow a^2\geq1$ and $ b\equal{}0$.  However, this contradicts $ (2)$, because we would then have $ (a^2\minus{}b^2\plus{}1)^2\geq(2)^2\not\leq1$.
Assume, for the sake of contradiction, that $ a\equal{}0$ and $ b\neq0\Rightarrow b^2\geq1$.  However, this contradicts $ (1)$ because we have $ (a\plus{}1)^2\equal{}1$ and $ b^2\geq1$, so $ (a\plus{}1)^2\plus{}b^2\geq2$.

Thus, both $ a$ and $ b$ must be 0 and we are done (note that the third equation was not necessary).[/hide]

[quote="b-flat"][hide]Let $ z \equal{} a \plus{} b\imath$, where $ a,b\in\mathbb{Z}$.  We wish to show that both $ a \equal{} 0$ and $ b \equal{} 0$.  Then, we know that:
$ \sqrt {(a \plus{} 1)^2 \plus{} b^2}\leq1\Rightarrow (a \plus{} 1)^2 \plus{} b^2\leq1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
$ \sqrt {(a^2 \minus{} b^2 \plus{} 1)^2 \plus{} (2ab)^2}\leq1\Rightarrow (a^2 \minus{} b^2 \plus{} 1)^2 \plus{} (2ab)^2\leq1\ \ \ \ \ \ (2)$

From $ (2)$, we know that at least one of $ a$ and $ b$ are 0, because otherwise, $ (2ab)^2 > 1$ and $ (a^2 \minus{} b^2 \plus{} 1)^2\geq0$.
Assume, for the sake of contradiction, that $ a\neq0\Rightarrow a^2\geq1$ and $ b \equal{} 0$.  However, this contradicts $ (2)$, because we would then have $ (a^2 \minus{} b^2 \plus{} 1)^2\geq(2)^2\not\leq1$.
Assume, for the sake of contradiction, that $ a \equal{} 0$ and $ b\neq0\Rightarrow b^2\geq1$.  However, this contradicts $ (1)$ because we have $ (a \plus{} 1)^2 \equal{} 1$ and $ b^2\geq1$, so $ (a \plus{} 1)^2 \plus{} b^2\geq2$.

Thus, both $ a$ and $ b$ must be 0 and we are done (note that the third equation was not necessary).[/hide][/quote]

If I'm not mistaken, he talks about all complex numbers, i.e. those with noninteger $ a,b$ too.

[quote="bertram"]If I'm not mistaken, he talks about all complex numbers, i.e. those with noninteger $ a,b$ too.[/quote]
Yes, [b]Bertram[/b] ! $ z\equal{}a\plus{}i\cdot b$ , where $ a\in R$ and $ b\in R$ !!

Write $ z \equal{} r e^{i\theta}$ with $ r \geq 0$ and $ 0 \leq \theta < 2\pi$. Suppose $ r > 0$. Note that if the real part of $ z$, $ z^2$, or $ z^3$ is positive, then the corresponding inequality fails; also if they are imaginary and nonzero the inequality fails. This means that the real part must be negative. For $ \text{Re}(z) < 0$, we must have $ \pi/2 < \theta < 3\pi/2$. For $ \text{Re}(z^2) < 0$, we must have $ \pi/4 < \theta < 3\pi/4$ or $ 5\pi/4 < \theta < 7\pi/4$. Finally for $ \text{Re}(z^3) < 0$, we must have $ \pi/6 < \theta < \pi/2$ or $ 5\pi/6 < \theta < 7\pi/6$ or $ 3\pi/2 < \theta < 11\pi/6$. The first and third conditions together imply $ 5\pi/6 < \theta < 7\pi/6$, and then the second one cannot be satisfied. Contradiction. $ r \equal{} 0$, so $ z \equal{} 0$.

$ \{\begin{array}{c} \{v,u\}\subset\mathcal C\ ,\ |u|=1\\\
= = = = = ===== =\\\
(*)\ \|\begin{array}{c}
|v + u|\le 1 \\
 \\
|v^2 + u^2|\le 1 \\
 \\
|v^3 + u^3|\le 1\end{array}\end{array}\ \implies\ v = 0$

[color=darkblue][b][u]Proof.[/u][/b]

Denote $ z=\frac vu$ . Then the relations $ (*)$ becomes : $ \{\begin{array}{c}
|z + 1|\le 1 \\
 \\
|z^2 + 1|\le 1 \\
 \\
|z^3 + 1|\le 1\end{array}$ .

Denote $ r=|z|$ , i.e. $ |z|^2=z\cdot\overline z=r^2$ . Therefore,

$ \{\begin{array}{ccc}
|z + 1|\le 1 & \implies & z+\overline z+r^2\le 0\\\\
|z^2 + 1|\le 1 & \implies & z^2+\overline z^2+r^4\le 0\\\\
|z^3 + 1|\le 1 & \implies & z^3+\overline z^3+r^6\le 0\end{array}$ .

Suppose $ r>0$ . Then $ \{\begin{array}{c}
z+\overline z< 0\\\\
z^2+\overline z^2<0\\\\
z^3+\overline z^3<0\end{array}$ .

Observe that $ z^3+\overline z^3=(z+\overline z)(z^2+\overline z^2-r^2)>0$ ,

what is falsely. In conclusion, $ z=0$ , i.e. $ v=0$ .[/color]

Can we solve this by graphing on the complex plane?


trigonometry
function
geometry
circumcircle
geometry solved

let ABCD be a convex quadrilaterall.angle ABC+angle BCD<180.

E is the intersection of AB,CD.

prove that:

           angle ABC=angle ADC   ,  if and only if   AC=(CD.CE - AB.AE):AC

In the following we will denote $\angle EBC$ as $\alpha$, $\angle BEC$ as $\beta$, $\angle BCA$ as $\gamma$ and $\angle CDA$ as $\delta$.
Applying the theorem of sines to the triangles $\triangle ABC, \triangle ACE, \triangle ACD, \triangle ACE$ to obtain $AB, AE, CD, CE$ as $AC$ times a sine and divided for another sine, and then substituting we arrive at
\[ CD\cdot CE-AB\cdot AE=AC^2\left(\frac{\sin(\alpha+\beta+\gamma-\delta)\sin(\alpha+\gamma)}{\sin\delta\sin\beta}-\frac{\sin\gamma\sin(\alpha+\beta+\gamma)}{\sin\alpha\sin\beta}\right)  \]
Applying the identity $\sin\alpha\sin\beta+\sin\gamma\sin(\alpha+\beta+\gamma)=\sin(\beta+\gamma)\sin(\alpha+\gamma)$ we arrive at the conclusion that $CD\cdot CE-AB\cdot AE=AC^2$ if and only if
\[ \frac{\sin(\alpha+\beta+\gamma-\delta)}{\sin\delta}=\frac{\sin(\beta+\gamma)}{\sin\alpha}  \]
but the only solution $\delta$ to this equation having $\delta\in(0,\alpha+\beta+\gamma)$ is $\delta=\alpha$, because the first member of the equation above is a monotone function of $\delta$.

nice solution again,you sove the problems good.i tnink you are good in geometry. :D

A more synthetic solution maybe:

[color=blue][b]Problem.[/b] Let ABCD be a quadrilateral, and let E be the point of intersection of the lines AB and CD.

Prove that < ABC = < CDA holds if and only if $AC^2=AB\cdot AE+CD\cdot CE$. Hereby, we use directed angles modulo 180° and directed segments.[/color]

Note that, by using directed segments, we made the equation look more symmetric (the - sign became a + sign) and the condition < ABC + < BCD < 180° unnecessary; also, by using directed angles modulo 180°, we made the convexity of the quadrilateral ABCD insignificant.

Now, the [i]solution of the problem[/i]:

First, we show that if < ABC = < CDA, then $AC^2=AB\cdot AE+CD\cdot CE$. In fact, let F be the point of intersection of the lines BC and DA. Then, < EBF = < ABC = < CDA = < EDF; thus, the points E, F, B and D lie on one circle. Denote this circle by k, its center by O and its radius by R. Then, the quadrilateral DFBE is inscribed in the circle k, and its opposite sides intersect at the points $DF\cap BE=A$ and $FB\cap ED=C$. Hence, according to Theorem 1 of the first post in http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6 , the point A lies on the polar of the point C with respect to the circle k. In other words, if C' is the image of the point C under the inversion with respect to the circle k, then the point A lies on the perpendicular to the line OC at the point C'. Thus, < AC'O = 90° and < AC'C = 90°. Hence, the triangles AC'O and AC'C are right-angled, so that Pythagoras yields $AC^{\prime}\ ^2=OA^2-OC^{\prime}\ ^2$ and $AC^2=AC^{\prime}\ ^2+C^{\prime}C^2$.

Now, since the point C' is the image of the point C under the inversion with respect to the circle k, this point C' lies on the ray OC and satisfies $OC\cdot OC^{\prime}=R^2$ (since O is the center and R is the radius of the circle k).

Thus,

$AC^2=AC^{\prime}\ ^2+C^{\prime}C^2=\left(OA^2-OC^{\prime}\ ^2\right)+C^{\prime}C^2=OA^2+\left(C^{\prime}C^2-OC^{\prime}\ ^2\right)$
$=OA^2+\left(C^{\prime}C-OC^{\prime}\right)\cdot\left(C^{\prime}C+OC^{\prime}\right)=OA^2+\left(\left(OC-OC^{\prime}\right)-OC^{\prime}\right)\cdot OC$
$=OA^2+\left(OC-2\cdot OC^{\prime}\right)\cdot OC=OA^2+OC^2-2\cdot OC\cdot OC^{\prime}=OA^2+OC^2-2\cdot R^2$
$=\left(OA^2-R^2\right)+\left(OC^2-R^2\right)$.

But on the other hand, since O is the center and R is the radius of the circle k, the expression $OA^2-R^2$ is the power of the point A with respect to the circle k, and thus equals $AB\cdot AE$ (remember, we are using directed segments). Similarly, $OC^2-R^2=CD\cdot CE$. Thus, we obtain $AC^2=AB\cdot AE+CD\cdot CE$.

So we proved one direction of the problem: if < ABC = < CDA, then $AC^2=AB\cdot AE+CD\cdot CE$. Remains to show the other direction: if $AC^2=AB\cdot AE+CD\cdot CE$, then < ABC = < CDA.

But in fact, this trivially follows from the direction that we already proved: In fact, let D' be the point on the line CE satisfying < ABC = < CD'A. Then, applying the direction of the problem that we already proved, we get $AC^2=AB\cdot AE+CD^{\prime}\cdot CE$, what, compared with $AC^2=AB\cdot AE+CD\cdot CE$, yields CD' = CD. Since the points D and D' both lie on the line CE and the segments are directed, this equation yields that the points D and D' coincide, so that < ABC = < CD'A becomes < ABC = < CDA. Thus, the other direction is proven, too, and the problem is solved.

Ashegh, do you like my solution? I have done everything to make it both inversive and power-ful ;) .

  darij

Here is my solution:

  Point D moves on a circumcircle of ABC reflected over AC ( angle D = angle B)

  When point D coincides with A the circle k from Darij's solution is tangent to AC

  at A.  As D moves away from A,  CE*CD is power of C wrt k and AB*AE is

  power of A wrt k. Even though center and radius of k change, point B is fixed

  and still on  circumference of k.  So powers of fixed points A and C change

  the same value to keep the difference of powers to 0.

  Thank you.

   M.T.

thanks alot guys...

very nice solutions...


function
Support
geometry
algebra
polynomial
topology
projective geometry

For the second question, I must say that curves are determined up to birationality (not isomorphism) by their function field. If I am not wrong, every birational equivalence class has a unique nonsingular representant, so the statement seems to be true for smooth curves.

Reference: first chapter of Hartshorne (maybe sixth section)

[quote="alggeo"]Why is a curve minus a point affine?

Why is a curve completely determined by its function field?[/quote]

What is your definition of a curve?

$\mathbb P^{1}\sqcup\mathbb P^{1}$ (or even $\mathbb P^{1}\sqcup\mathrm{Spec}\,k$). owk

and irreducible

and minus a closed point

[quote="alggeo"]Why is a curve minus a point affine?
[/quote]

In short, let $C$ be a smooth curve and let $P$ be a point. Then $P$, as a divisor on $C$, is ample (it has positive degree and use Cor 3.3 from chapter IV in Hartshorne pag 308), so $nP$ is very ample for some $n>0$ meaning that $P$ is the support of a hyperplane section on $C$ for a suitable Veronese embedding.
So there exists $N>0$ such that $C$ is isomorphic to a curve in $\mathbb P^{N}$ and $P$ is the intersection of $C$ and some hyperplane. $C\setminus\{P\}$ then lies in the complement of a hyperplane and we know that this open set is isomorphic to $\mathbb A^{N}$.

I'll try to think of a more elementary proof. The idea to try to make $P$ the intersection of $C$ and a hypersurface in some projective space is natural, but I didn't get too far in an elementary fashion that way.

Unless somebody finds a counterexample, I think that this problem works if we remove the smoothness condition (we keep it irreducible). A projective resolution of singularities should do the trick.

[quote="owk"]Once you know that closed points define ample divisors, you can also use Serre's affinity criterion. owk[/quote]

Which criterion is that?
For the Serre's affinity criterion that I know it should be checked that $H^{1}(C\setminus\{P\},\mathcal I)=0$ for all coherent sheaves of ideals $\mathcal I$ of $\mathcal O(C\setminus\{P\})$. I think this would be hard. But I guess you talk about some other Serre criterion; a criterion that involves ample divisors.

Let $j\colon C\setminus\{P\}\to C$ be the inclusion. Extend $\mathcal I$ to a coherent sheaf $\tilde{\mathcal I}\subset j_{*}\mathcal I$ on $C$ (EGA I, 9.4.7). By ampleness, $H^{1}(C,\tilde{\mathcal I}(nP))=0$ for $n\gg0$. We have an exact sequence $0\to\tilde{\mathcal I}(nP)\to j_{*}\mathcal I\to Q\to0$ with a skyscraper quotient $Q$, and the corresponding long exact sequence tells us that $H^{1}(C,j_{*}\mathcal I)=0$ as well. Since $j$ is affine, $H^{1}(C\setminus\{P\},\mathcal I)=H^{1}(C,j_{*}\mathcal I)=0$. owk

Thanks owk.
I am so happy people are starting to discus algebraic geometry here because it forces me to learn harder things I am otherwise to lazy to learn by myself.

Here is a slightly different method, but still not elementary:

[b]Lemma:[/b] There exists on $C$ a rational non-constant function that is regular anywhere except at $P$.

Assume the lemma proved.
Denote that function by $f$. Since $C$ is projective, $f=\frac{u}{v}$ for some homogeneous polynomials $u,v$ of the same degree. The application $C\ni x\to [u(x): v(x)]$ defines a finite map $\varphi: C\to\mathbb P^{1}$. We have $\varphi^{-1}([1: 0])=P$, so $C\setminus\{P\}$ is $\varphi^{-1}(\mathbb A^{1})$. But finite morphisms are afine, so $C\setminus\{P\}$ is also affine.

I know no other way of proving the lemma besides Riemann-Roch:
$l(nP)>0$ for some $n$, so there exists a rational function $f$ such that $(f)+nP\geq 0$ meaning that $f$ cannot have poles different from $P$.


quadratics
number theory unsolved
number theory

Find all positive integers n that are quadratic residues modulo all primes greater than n


abstract algebra
algebra
polynomial
advanced fields
advanced fields unsolved

suppose 2^(k) + 1 is a prime. Prove that k has no prime divisors other than 2.
(Hint: if k=ab, b odd, consider 2^(k)+1 module 2^(a)+1)

We use a well-known property: $ m^b\plus{}1$ is divisible by $ m\plus{}1$. Take $ m\equal{}2^a$. $ 2^{ab}\plus{}1$ is divisible by $ 2^a\plus{}1$, contradiction.

This problem is very well-known. I'm sure it has been posted before.

that's my guess...but how to proof the property??? since it has to do with the assumption that b is odd.

[quote="Fermat's Little Turtle"]$ m \equal{} 3,b \equal{} 2$ gives $ 3 \plus{} 1 | 3^2 \plus{} 1$ so $ 4|10$ ??[/quote]$ b$ must be odd.

consider the polynomial $ f(x)\equal{}x^b\plus{}1$, $ b$ is odd. Since $ f(\minus{}1)\equal{}0$, then $ f(x)$ is divisible by $ x\plus{}1$, i.e., $ x^b\plus{}1$ is divisible by $ x\plus{}1$.


The roots $ x_i$($ i\equal{}\overline{1,4}$) for the ecuation $ x^4\plus{}ax^2\plus{}bx\plus{}c\equal{}0$,$ a,b,c\in C$ have the property that $ 2x_i^2\plus{}a$ is diffrent from $ 0$.Calculate the sum $ S\equal{}\sum_{i\equal{}1}^4 \frac{1}{2x_i^2\plus{}a}$

The equation is equivalent to $ (2x^2+a)^2=-4bx-4c+a^2$.Denote that $ y^{-1}=2x^2+a$ and the last relation becomes $ \frac{1}{y^2}=-4bx-4c+a^2$.If $ b=0$ we obtain $ S=0$.For $ b$ diffrent from $ 0$ we have $ x=-\frac{1+(4c-a^2)y^2}{4by^2}$.
Finaly $ [8ab^2+(4c-a^2)^2]y^4-8b^2y^3+2(4c-a^2)y^2+1=0$ from where $ S=\frac{8b^2}{8ab^2+(4c-a^2)^2}$.


let $ a,b,c,A,B,C$ in $ R$ such that
$ A\equal{}bc\minus{}a^2$  and  $ B\equal{}ac\minus{}b^2$ and $ C\equal{}ab\minus{}c^2$ and $ A^\minus{}{1} \plus{}B^{\minus{}1}\plus{}C^{\minus{}1}\equal{}0$
prove that
$ aA^{\minus{}2}\plus{}bB^{\minus{}2}\plus{}cC^{\minus{}2}\equal{}0$

$ \sum \frac{a}{(bc\minus{}a^2)^2}\equal{}(\sum \frac{a}{(bc\minus{}a^2)})(\sum \frac{1}{(bc\minus{}a^2)})$
$ \minus{}\sum (\frac{a}{(bc\minus{}a^2)})((\frac{1}{(ca\minus{}b^2)})\plus{}(\frac{1}{(ab\minus{}c^2)}))$
$ \equal{}\minus{}(\prod \frac{1}{bc\minus{}a^2})(\sum a((ca\minus{}b^2)\plus{}(ab\minus{}c^2))$
$ \equal{}\minus{}(\prod \frac{1}{bc\minus{}a^2})(\sum (ca(a\minus{}c)\plus{}ab(a\minus{}b)))$
$ \equal{}\minus{}(\prod \frac{1}{bc\minus{}a^2})(0)\equal{}0$


linear algebra
matrix
blogs
linear algebra unsolved

Prove or disprove that every nilpotent matrix with entries either $ 0$ or $ 1$ is the adjacency matrix of a forest (a disjoint union of trees).

[hide="Remark"] What I'm really interested in is whether this is equivalent to the nilpotent Jordan normal form theorem.  Can we prove directly that every nilpotent matrix can be written in a basis where it has non-negative entries?[/hide]

This is really strange, but each time you post a question, I have no idea what you actually mean.  :(  And I do pretty well understand what you write in your blog, it's written absolutely clear and well. Are you confused when you write questions?

How can a non-symmetric matrix be an adjacency matrix? If you mean a directed graph, I'd like to know what "forest"/"tree" means. It is clear that any nilpotent matrix with entries either $ 0$ or $ 1$ which has only $ 0$'s on the main diagonal (or you should allow loops) is the adjacency matrix of a directed graph which has no cycle (because the $ \left(i,j\right)$-th entry of the $ n$-th power of this matrix gives the number of paths from vertex $ i$ to vertex $ j$, and if there is a cycle, it yields paths of arbitrarily large length, thus preventing the adjacency matrix from becoming nilpotent). But the only [i]symmetric[/i] nilpotent matrix with nonnegative entries is the zero matrix...

  darij

[quote="darij grinberg"]Are you confused when you write questions?[/quote]
Probably  :oops:

What I was thinking of is the following:  any tree can be made into a directed graph by picking a root and orienting all of the edges away from the root.  But after reading your comment it is clear that this construction does not cover the set of [url=http://en.wikipedia.org/wiki/Directed_acyclic_graph]directed acyclic graphs[/url].  My apologies.

Next question (which is what I really wanted to ask):  the nilpotent Jordan normal form theorem states that every nilpotent matrix is similar to a direct sum of adjacency matrices of directed [url=http://en.wikipedia.org/wiki/Path_graph]path graphs[/url].  Can this be proven by graph-theoretic means?

[quote="t0rajir0u"]Next question (which is what I really wanted to ask):  the nilpotent Jordan normal form theorem states that every nilpotent matrix is similar to a direct sum of adjacency matrices of directed [url=http://en.wikipedia.org/wiki/Path_graph]path graphs[/url].  Can this be proven by graph-theoretic means?[/quote]

Sorry, I fail to understand this...
The nilpotent Jordan normal form theorem is the one stating that every nilpotent matrix over a field is similar to a strictly upper triangular one, right? How is that a direct sum? And what do you mean by "direct sum" anyway?

  darij

In more concrete terms, every nilpotent matrix is similar to a block diagonal matrix whose blocks are [url=http://mathworld.wolfram.com/JordanBlock.html]Jordan blocks[/url] with zero entries on the diagonal and $ 1$ entries on the superdiagonal.  [url=http://en.wikipedia.org/wiki/Direct_sum]Direct sum[/url] here just means that we construct a block matrix whose diagonal blocks are the two matrices in question and whose non-diagonal blocks are zeroes.

In graph-theoretic terms, every nilpotent matrix is similar to the adjacency matrix of a disjoint union of directed path graphs.

One of the reasons I've been wondering this is a discussion on [url=http://math.ucr.edu/home/baez/FUN.html#mathematics]John Baez's[/url] website about the extent to which linear algebra is really "just combinatorics," and the other is that the construction of a Jordan block has always bothered me because I didn't have a good intuition for the sense in which a Jordan block with zero entries on the diagonal is a "prototypical" nilpotent matrix until I started looking at it in graph-theoretic terms.  Path graphs, on the other hand, are a very good candidate for a "prototypical" acyclic directed graph, although since the problem here is an isospectral one that is clearly not covered by graph automorphism I wonder whether I am asking the right question.

If it makes you feel any better, I understood that you had to be talking about digraphs.

But I've got an example for you:

\[ \begin{pmatrix}0&1&1&0\\ 0&0&0&1\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}\]

That's not a forest, is it?


Eight people go to a restauarant and each orders some different dish for dinner.  The waiter is inexperienced.  He forgot who ordered what and does not recognize any of the dishes, so he can't ask the patrons.  He starts to serve the dishes randomly.  If you are served third and you see the two people who got served before you frown as in they didn't get what they wanted (but you can't see the dish they got). What is the probability that you will NOT receive the dish you ordered?

This seems to be easy, but then it gets harder. I got an answer, but it doesn't agree with the answer book. PS. SingTheSorrow gave me this question, and I can't figure it out and nor can he.

Hint, which is probably what you already have tried to do, but I'll say it anyway:

[hide]

Just split it up into two cases: 1) neither of the first two people got your meal and 2) one of the first two people got your meal. Remember to not include the options where either of the first two got their own. I make it around 90% (I won't give it exactly so as to not give it away).

[/hide]

So, this question has two difficulties, really.  The first is to apply conditional probability correctly, but the second is to count cases correctly.  I would suggest you get as close as you have to to actually writing out the 8*7*6 possibilities for the first 3 plates in order to calculate it.  (Obviously, you don't have to write out all those possibilities, but you should be very careful with your counting and not rely on any fancy formulae or anything too "obvious," because those are likely to make your life more difficult.)

What I think this is is conditional probability. P(A|B) = P(A :cap: B) / P(B)... so P(B) is there probability that the first two don't get their meals correctly, and P(A) is the probability that the thrid guy doesn't get his meal. So, after finding that... I don't get what I want.

P(B) = 55/56 for me. Is that right???

P(A and B) = These are out of the possible arrangements of A, B, and C, where they are the corresponding people to be served, and C is the third where C doesn't get the right dish and A and B get the wrong dish.
For case 1) (Where 1 and 2 don't get 3rd's meal.) (6 * 6 * 5) / (8 * 7 *5) = 9/14
For case 2) (Where 1 or 2 gets the 3rd's meal.) (1 * 6 * 6) + (6 * 1 * 6) = 2*6*6. (2*6*6)/(8*7*5) = 9/35

9/14 + 9/35 = 441/490 = 63/70 = 9/10. And 9/10 / 55/56 = 504 / 550 = 252 / 275... which apparently isn't right...

Is my reasoning off?

I got 33/56, is that the answer?

3 cases:

1) 1st guy got the 3rd guy's meal: 1/8
2) 2nd guy got the 3rd guy's meal: 6/8 * 1/7 = 3/28
3) none of the first two guys got the 3rd guy's meal: 6/8 * 5/7 * 4/6 = 5/14

add: 1/8 + 3/28 + 5/14 = 33/56

I would suggest separating the cases in which 1 of the first 2 guys got the 3rd guy's meal, because their probabilities are different.

i need to learn LaTeX...

I'm so sick of this problem. :P

But you guys, I don't think the denominator shouldnt start w/8*7*6 because you are given that Persons A and B haven't received their food...so the denominator would be 7*6*5......Oh I seriously I have no clue how to do this problem.

[quote="white_horse_king88"]
P(B) = 55/56 for me. Is that right???
[/quote]

No. That is not right. Even here you have to split things up into more than 1 case. You seem to have calculated 1 - the probability that *both* are right. You don't want this. You want the probability that both are wrong.

And the other thing that I think you are missing the whole way through: you will need to consider the case where 1 gets 2s meal etc differently to where 1 doesn't get 2s meal.

If anyone would like to check their answer, the first 6 digits after the decimal place are [hide]879844[/hide].  (This won't help you reconstruct a solution, I'm afraid.)

You should be able to break it down into 5 cases or so, depending on which dishes the first and second people got.

The answer supposedly is:  [hide]227/258[/hide]  Which is weird.  It's from the WPI Math Meet this year...

OK, its been a few days, its fill in the gaps time :)
For each of these cases, find a) the probability that happens, b) the probability the third person then doesn't get his own meal (assuming that case happened).

Case 1: First person gets meal 2, second person gets meal 1.
a)
b)
Case 2: First person gets meal 2, second person gets meal 3.
a)
b)
Case 3: First person gets meal 4,5,6,7 or 8, second person gets meal 3.
a)
b)
Case 4: First person gets meal 3, second person gets meal 1.
a)
b)
Case 5: First person gets meal 3, second person gets meal 4,5,6,7 or 8.
a)
b)
Now from those results:
Total probability that one of the above cases occurs:

Total probability that one of the above cases occurs and the 3rd person doesn't get his own meal:

Final answer:


[quote="lajanugen"]An exam paper consists of 5 multiple choice questions, each with 4 different choices. 2000 students take the exam, and each student chooses exactly one answer per question. Find the smallest value of n for which it is possible for the students' answer sheets to have the following property: among any n of the students' answer sheets, there exist 4 of them among which any two have at most 3 common answers[/quote]
Smallest value of $ n$? If we choose $ n\equal{}4$, and 4 answer sheets are $ aaaaa, bbbbb, ccccc, ddddd$ ($ a,b,c,d$ are 4 different choices for each question) then the problem has solve!?\\
(Please tell me if I have any mistakes!)

You can't assume that the four answer sheets will be like that.

(and a china #6 should definitely be a lot harder than that)


quadratics
geometry
trigonometry
probability
number theory
complex numbers
\/closed

Hi!
I am planning on buying one of the Art of Problem Solving books, but I am not sure if I should buy the 3 Introduction books or the 2 volume books. The price doesn't matter.
Can anybody help me with this please?
What are the differences?
Which way will help me more?
I looked at all the table of contents, and it seemed like that the topics covered in the 3 Introduction books were also the topics in the 2 Volume books, but it was just covered faster in the Volume books.
Thanks for your help!

Jukjukenator

Tell us a little bit more about your math background, and your intended use of the books, and we'll be able to advise you better.

Ok.
Well, since I moved 1.5 years ago from Germany to the US my Algebra isn't too strong, but I have a strong background in arithmetic (can be compared to the number theory in the US) which is going much deeper into Number theory than what you guys learn in the US.
I am right now working on quadratic equations, and complex numbers in Algebra 2.
I know the German style Geometry (don't worry about what that is), but I am still not too strong at the American geometry (geometry you guys learn ;)). I also have basic trigonometry knowledge (believe it or not).
I hope this is not too confusing since there are probably thousands of grammar mistakes, but if you are confused; just ask specifically! ;)

Jukjuk

PS: I just want to use the book to kind of review my algebra skills (mainly algebra 2), and maybe learn a few new things. I also want to gain a lot more Geometry knowledge, maybe even my trigonometry knowledge, and my probability knowledge.
The main purpose will be to raise my scores in math competitions, and get used to American math. ;)

I suggest starting with AoPS Volume 1.  If any subject area in that book gives you trouble, then try the intro book in that subject.

I would definitely start with the AoPS volume 1 and 2 books.  A lot of the material in the Intro books would probably be review to you.  The volume 1 book has a lot of good geometry stuff, too.

I'd stick with the Volume 1/2 books. I started 'real' math in 10th grade and improved greatly over this year.

[quote="jukjukenator"]Why do you guys think that I shouldn't use the Introduction books, but buy the Volumes??[/quote]

At first I would do the volumes to get better at problem solving in general, and then I would work more in depth on specific topics using the Intro books.

in march, i believe, AoPS is coming out with an intro to algebra book. And in May/June the Intermediate Algebra is coming out. Since you say algebra is you weakness, those might be useful. but definetely buy the two volume books.


Let $a;b;c>0$ satisfying $a+b+c=abc$.Prove that:
$\frac{a}{b^{3}}+\frac{b}{c^{3}}+\frac{c}{a^{3}}\geq 1$

Do not discuss it for the same reason I said.


Let $ t \geq 0 $ , $ x_n = \frac{1+t+t^2+..+t^n}{n+1} $ .
  Prove that : $ ^{n-1}\sqrt{x_{n-1}} \leq ^n\sqrt{x_n} $ for all $ n \geq 1 $

It doesn't hold for $n=1$;

$1\leq \frac {1+t}{2}$ is not generally true.  I think we at least need $t\geq 1$.

[quote="Myth"]No we no need it, since inequalities for $t$ and $1/t$ are the same ;)
Forget about $n=1$ :)[/quote]
I don't think it is the same.please think about it again. :)

For more understoodable explanation : We need to show that :
 $ x_1 \leq \sqrt{x_2} \leq ^3\sqrt{x_3} \leq ^4{x_4} \leq .... $

Ican't prove $ f'(t) > 0 $
  How can you do with $ 0 < t < 1 $
  In short , this problem is still unsolved ! :(


Hello! I've been asking several different users of this forum about this, and I decided to simply post a topic about that. Do you have any suggestions from what books should everybody learn? Of course links/pdf versions/ Latex versions are all more than acceptable :wink:

The best advice is to [b]practice[/b] as much as possible, do as many problems that you can yourself. It doesn't matter which problems you do as long as they are challenging but you can still solve them.

Yeah, that's right, but when posting that topic I was thinking of sth more specific for example one or two of books/pdf versions you read and which had a great influence on you or/and simply deeply engraved their content into your mind.

Ok. I read a few pdfs bu they were not of large importance to me. By the way, in the resources section on this forum http://www.artofproblemsolving.com/Forum/index.php?f=121 you can find some very useful pdfs that you can view online for free. However, the thing that helped me most in preparing for math contests and improving my problem-solving skills is just doing lots and lots and lots of problems of various olympiads (you can find many problems and solutions here: http://www.artofproblemsolving.com/Forum/resources.php). Make sure you read the solution to the problem and understand what specific idea was used to solve the problem. Then maybe that idea could be used to solve some problem you encounter on an olympiad! :wink:


Alright I need some bigtime help here:

A 20-foot ladder rests vertically against a wall.  A bug starts at the bottom of the ladder and climbs up at a rate of 1 foot per minute.  At the same time, the foot of the ladder is being pulled along the ground at a rate of 1 foot per minute until the top reaches the ground.  The position of the bug after 1 minutes is shown in the following figure.  Let [i]x[/i] be the distance of the bug from the wall at time [i]t[/i].

a)  Find the function x(t).  This function gives the distance of the bug from the wall as a function of time, t.

b)  Using your answer in part (a) and your graphing calculator, what is the farthest distance the bug will be from the wall in its march up the ladder?  Some calculator work may be involved.

c)  At the time when the bug is farthest from the wall, how high is the top of the ladder from the ground?

Any help is much appreciated guys!

-Rick Jr.

This is a double posting, it should be deleted. :roll:

Japanese Communities Moderator

kunny

[quote="rickurukus"]what the - kunny who are you to talk you -

[/quote]

Look this!
[url]http://www.mathlinks.ro/Forum/index.php?f=368[/url]

Hey, you.
You have posted the very same problem, haven't you!
But you have deleted it! :mad:

By the way, Is this problem your home work?
If so, no one would not help with you! :wink:

To moderators;

Recently I have seen such a double posting, so you should keep them under observation than usual.In this trouble, I have copies posted by rickurukus, in the difference of two minutes, however  I would not like to quarrel with child.

Japanese Communities Moderator

kunny


linear algebra
matrix
trigonometry
complex numbers

Let $ t \in (0,\pi)$ be a fixed real number. Find all matrices $ X \in M_2(\mathbb{R})$ such that: $ X^n\equal{}\left(\begin{array}{cc}
\cos t & \minus{}\sin t \\
\sin t & \cos t \end{array}
\right)$

Let $ X$ be the set of all matrices in $ M_n(\mathbb{R})$ that are of the form $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}.$ Then the mapping which sends that matrix to the complex number $ a\plus{}bi$ is a ring isomormphism (hence a field isomorphism). That means that the set $ X$ may be regarded as a copy of the complex numbers and $ X$ has all of the algebraic properties of the complex numbers.

So, let's rephrase the question: Find all complex numbers $ z$ such that $ z^n\equal{}e^{it}.$

The answer to this is well known: $ z\equal{}e^{\frac{it\plus{}2\pi ik}{n}}$ where $ k\in\{0,1,2,\dots,n\minus{}1\}$ We can write that as $ z\equal{}\zeta^ke^{it/n}$ where $ \zeta\equal{}e^{2\pi i/n}.$

Re-expressed as matrices, we have
\[ \begin{pmatrix}\cos\left(\frac{t\plus{}2\pi k}{n}\right)& \minus{}\sin\left(\frac{t\plus{}2\pi k}{n}\right)\\ \sin\left(\frac{t\plus{}2\pi k}{n}\right)& \cos\left(\frac{t\plus{}2\pi k}{n}\right)\end{pmatrix}\]

While I wasn't sure from your question whether you wanted to fix $ n$ or to give the answer for all $ n,$ the interpretation of what I've said should be clear in either case.

But is it clear that the solution set is contained in that particular subalgebra?  It seems like the proper approach is to reason from eigenvalues first.

In some cases, it certainly isn't. These are all solutions when $ \sin t\neq 0$, but we get many more in the cases $ X^n\equal{}\pm I$. Those solutions are a 2-dimensional variety.


algebra
polynomial
function
complex analysis
real analysis
real analysis unsolved

here is an easy one, but its cute.

let a sequence of polynomials $(p_{n})$ converge uniformly to $f$ on $\mathbb{R}$. Is $f$ necesarrily differentiable?

At first, $f$ is plainly continuous.
But, even better, one can prove that $f$ is a polynomial function.
Once we'll have proved it, we'll clearly have proved that the answer to that question is "yes."

Now, to the proof.
Cauchy's criterion on uniform convergence yields that there exists an $N \in \mathbb{N}$ such that for any $n \geq N$, for any $x \in \mathbb{R}, \quad |p_{n}(x)-p_{N}(x)| \leq 1.$
Thus for any $n \geq N,$ $p_{N}-p_{n}$ is constant, so that there exists an $a_{n}$ such that $p_{n}= p_{N}+a_{n}.$
And now it's not hard to get that $f = p_{N}+a,$ where $a$ is the limit of $(a_{n}),$ is a polynomial function, qed.

This has some of the flavor of complex analysis- if we asked about a sequence of entire functions converging uniformly in $\mathbb{C}$, we would find that the limit was entire, and all but finitely many functions would be the limit plus a constant.

If you want nontrivial convergence, you should ask about uniform convergence on compact sets: for each compact interval $K=[a,b]$, $f_{n}$ converges uniformly to $f$ on $K$. That leads to a very different answer.

[quote="jmerry"] and all but finitely many functions would be the limit plus a constant.
[/quote]
I've never seen this result. Is it called something? Is it easy to prove? Is it true for locally uniform convergence?

The proof? Liouville's theorem. I've never seen it stated elsewhere, but it's trivial.
It's also a good reason why we don't care about global uniform convergence for analytic functions.

yes, mathmanman and the others are right. even i gave the same proof.

@jmerry - i will think about your question

And if you dropped the insistence on global uniform convergence and asked only for uniform convergence on some bounded interval, there would be sequences of polynomials that converge uniformly to a non-differentiable continuous functions. (Examine one of the constructive proofs of the the Weierstrass approximation theorem.)


if $ xy \plus{} yz \plus{} zx \equal{} 1$
such that $ x,y,z$ are non negative reals

then find the min. of $ x \plus{} y \plus{} z$

oooh!
such a silly me! :blush:

anyway...thanks for it


search
function
geometry
geometry theorems

hi  i  would  very  much like  to  know about symedians and all  their properties  they  seem  very  useful  sometimes. maybe a  nice little  do  it yourself course like  the one on isogonal  conjugates  by  darij .  thanks in advance  :)

dear marco ,i couldnt find the mining of symedian...

the symedian  is the isogonal  conjugate  of the median  and it  has  a lot  of cool  properties  tha  are  used.  can someone  cite  them?  or maybe post  some  problems  that  use  symedians  ?'

  many  thanks  :)

[quote="mathmanman"]What if you use the search function of this forum? :)
(With keywords : "symedian" and/or "symmedian")[/quote]

dear mathmanman u really helped me... ;)

thanks alot... :D


Θεωρούμε ακολουθία κανονικών πενταγώνων Π1 , Π2 , Π3 ,...... και ακολουθία κύκλων Κ1 , Κ2 , Κ3 ,......που ορίζονται εξής :
Το Π1 είναι κανονικόν πεντάγωνον και δια τυχαίο φυσικό ν ≥ 1 , ο κύκλος Κν είναι εγγεγραμμένος στο Πν , ενώ το Πν+1 έχει ως κορυφές τα σημεία επαφής του Κν και Πν .
1) Να βρεθεί σημείο το οποίο να περιέχεται είς όλα τα Πν .
2) Να αποδειχθεί ότι δεν υπάρχουν περισσότερα του ενός σημεία περιεχόμενα είς ολα τα Πν .

[hide="solution"]1) Έστω $A$ το κέντρο του πρώτου κύκλου .

Οι κύκλοι $K_{n}$ είναι ομόκεντροι άρα το $A$ ανήκει σε κάθε κύκλο

Κάθε πεντάγωνο $P_{n}$ περιέχει τον κύκλο $K_{n+1}$ άρα το σημείο
$\\ A$ ανήκει σε κάθε $P_{n}$ αφού ανήκει σε κάθε κύκλο .

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

2) Έστω ότι υπάρχει σημείο $B$ διαφορετικό του $A$ που ανήκει
σε κάθε πεντάγωνο .Τότε θα ανήκει και σε κάθε κύκλο .

Έστω $2a_{n}$ η πλευρά του $n$ - οστού πενταγώνου και $r_{n}$
η ακτίνα του περιγεγραμένου του κύκλου , $n =1,2,...$ .

Ο εγγεγραμένος κύκλος εφάπτεται στα μέσα των πλευρών του
αντίστοιχου πενταγώνου άρα προκύπτουν έυκολα τα εξής :

$\\ \bullet \frac{a_{n+1}}{a_{n}}= \frac{r_{n+1}}{r_{n}}, \,\, (1)$

$\\ \bullet{r}^{2}_{n+1}={r}^{2}_{n}-{a}^{2}_{n}, \,\, (2)$

Από την $(1), \,\, \prod_{k=1}^{n}\frac{a_{k+1}}{a_{k}}= \prod_{k=1}^{n}\frac{r_{k+1}}{r_{k}}\Rightarrow a_{n+1}= \frac{a_{1}r_{n+1}}{r_{1}}$

και αντικαθιστώντας στην $(2)$ παίρνουμε $\frac{{r}^{2}_{n+1}}{{r}^{2}_{n}}= 1-\frac{{a}^{2}_{1}}{{r}^{2}_{1}}$

άρα $\prod_{k=1}^{n}\frac{{r}^{2}_{k+1}}{{r}^{2}_{k}}= \left(1-\frac{{a}^{2}_{1}}{{r}^{2}_{1}}\right)^{n}\Rightarrow \frac{{r}^{2}_{n+1}}{{r}^{2}_{1}}= \left(1-\frac{{a}^{2}_{1}}{{r}^{2}_{1}}\right)^{n}$.

Έιναι  $\lim_{n \to \infty}{r}^{2}_{n}= \lim_{n \to \infty}\left(1-\frac{{a}^{2}_{1}}{{r}^{2}_{1}}\right)^{n}=0 = \lim_{n \to \infty}{r}_{n}$

Aν λοιπόν $d = AB$ η απόσταση των σημείων τότε αφού
$\\{r}_{n}\to 0$ υπάρχει ένα $n_{0}$ τέτοιο ώστε $r_{n}<d$ για $n \geq n_{0}$
δηλαδή το $B$ δεν ανήκει σε όλους τους κύκλους - άτοπο.[/hide]

 :)

Μπορεί να την έκανα να φαίνεται πιό
πολύπλοκη στο 2 ή πιο απλή στο 1 - sorry .
Πιθανόν να υπάρχει λάθος στους δείκτες .   :D
Καλή η άσκηση Νίκο .
Αν έχεις πρόσβαση σε τέτοια θέματα βάλε κι άλλα .

a. re ramanujan me prolaves....... :chief:
8eorontas gnosto oti o egegramenos kai o perigegramenos kiklos enos kanoniku poligonu ine omokentri...prokipti oti oli i kikli ine omokentri...
epipleon epidi ta poligonap1,p3,p5...ine omio8eta opos kai ta p2,p4,p6...8a ine kai ayta omokentra me to idio kentro kai ara apodikniete kai to 1+2......

1966?????eleos......araxniasmeni.....
vsk logo...simetrias ta pentagona ine san na sigklinun sto kentro tu protu kiklu(simio A sti lisi tu ramanujan...)....
niko otan les politexnio...se pio ma8ima enois?euklidia geomettria?

 :D

[quote="RAKAR"]a. re ramanujan me prolaves....... :chief:
[/quote]

 :P

[quote="RAKAR"]
niko otan les politexnio...se pio ma8ima enois?euklidia geomettria?
[/quote]

Palaiotera (den kserw akrivws mexri pote-sigoura prin to 80 pou mpikan oi desmes), gia na mpeis se mia sxoli den ginotan to mpaxalo auto pou ginetai twra, pou dilwneis 100 sxoles kai an den peraseis stis prwtes 99 pigaineis stin 100h kai xairesai (kai xairetai kai to kratos pou dilwneis foititis kai oxi anergos) ALLA kathe sxoli ekane tis dikes tou eksetaseis, me apotelesma an esi itheles parapanw apo 2,3 sxoles eprepe na dwseis isarithmes eksetaseis. Gi'auto kai palia osoi mpainane kserane ti thelane (Ante na dwseis twra se 20 diaforetika tmimata eisagwgikes eksetaseis). H Gewmetria, h Algevra kai h Trigwnometria (3 diaforetika mathimata) itan ipoxrewtika mathimata apo ta 6 pou edinan gia auto to pedio (ta alla itan fisiki,ximeia,ekthesi). Endiaferon e?

Alexandros

η αληθεια ειναι οτι το σχολειο με το πανεπιστημιο εχουν πολυ μεγαλη διαφορα.ειχα δει ενα βιβλιο γεωμετριας που διαβαζαν οι υποψηφιοι των φυσικομαθηματικων και πολυτεχνιακων σχολων πριν 30 χρονια.ηταν υψηλου επιπεδου γνωσεις

Ενα τέτοιο βιβλίο έχω μπροστά μου kosta και εγώ .
Βρίσκω αρκετά καλές ασκήσεις .


Functional Analysis
advanced fields
advanced fields unsolved

X- Banach space, Y-space, $(A_{n})_{n}\subset L(X,Y)$. Provet that: If $(A_{n}x)_{n}$ is a basic sequence, $\forall x \in X$ then $sup_{n}\parallel A_{n}\parallel <+\infty$

It helps to include a definition: $(e_{n})$ is a basic sequence iff for any $x$ in the closed span of $(e_{n})$ there exists a unique sequence $c_{n}$ such that $x=\sum_{n}c_{n}e_{n}$.

Consider $X=Y=\ell_{2}$, and let $S\colon X\to Y$ be the shift operator $(Sx)_{n}=x_{n+1}$. Define $A_{n}=2^{n}S^{n}$. Looks like a counterexample. :maybe:


(1) Find integers a,b,c such that a^1/2 + b^1/2 = c^1/2 and a^2 + b =c.
Prove the solution is unique.

(2) Consider a^1/n + b^1/n = c^1/n. Prove a^n + b <> c for all a,b,c,n

(1) (4,9,25) is a solution, but is it unique?

(1) consider the triplet $ (a,b,c)\equal{}(n^2,(\frac{n^4 \minus{} n^2} {2n})^2, (\frac{n^4 \plus{} n^2}{2n})^2)$. $ n\plus{}\frac{n^4 \minus{} n^2}{2n} \equal{} \frac{n^4 \plus{} n^2}{2n}$, and $ n^4 \plus{} (\frac{n^4 \minus{} n^2} {2n})^2 \equal{} (\frac{n^4 \plus{} n^2}{2n})^2$, so there are infinitely many solutions  :wink:


[hide]
We have 5 pairs of terms whose angle arguments add up to 90 (pretend there is a sin^2(0) expression.  From the pythagorean identity, we have:
$\sin^2 \theta + \cos^2 \theta=1$
Now:
$\sin(90- \theta = \cos \theta  $
so
$\sin^2 \theta + \cos^2 \theta=\sin^2 \theta + \sin^2(90- \theta)= 1$
Thus we have 5(1)=5
[/hide]

[quote="AntonioMainenti"][hide]$5-\frac{\sin100^\circ\cos90^\circ}{2\sin10^\circ}$

This is a direct application of problem 226b in The USSR Olympiad Problem Book, which is to prove that
\[\sum_{k=1}^n\sin^2k\alpha=\frac{n+1}{2}-\frac{\sin(n+1)\alpha\cos{n\alpha}}{2\sin\alpha}\][/hide][/quote]

[hide]
And, of course, since $\cos 90^\circ=0$ you have $5$ like the other answers. That theorem is not needed on this problem.  All you need is $\sin{(90-\alpha)}=\cos \alpha$ and $\sin^2 \theta + \cos^2 \theta=1$ (like the others said) to solve this one.[/hide]


In the diagnostic test, there is this question:

[quote]I have 7 books I want to arrange on a shelf. Two of them are math books, and one is a physics book. How many ways are there for me to arrange the books if I want to put the math books next to each other, and put both of them to the left of the physics book?[/quote]

Does "to the left" mean immediately left (with no spaces) or just somewhere left of the physics book?

Normally, this English phrase would mean "somewhere left" in your terminology. If you find it ambiguous and want to clarify it, you are welcome to propose your version of the question to the Alcumus administrators either through the bug report system or in this thread and see what they will say :).


Segments BA, ED, and BD are tangents of Circle O. BD= 8 + 6rt3. Angle B measures 60 degrees. How far would segment CE be from the center of circle O?

Is there supposed to be a diagram or something?


function
complex analysis
complex analysis unsolved

I think the following illustrates very well how strong an hypothesis analyticity is.

Considering a uniformly bounded family of analytic functions on an open subset $U \subseteq \mathbb{C}$ which is uniformly bounded and pointwise convergent; show that the convergence is uniform on the compact subsets of $U$.

This is not really about analyticity: the same is true for any uniformly bounded family of functions that are [i]equicontinuous[/i] on compact subsets. For analytic functions uniform boundedness implies equicontinuity on compact subsets, by the Cauchy formula. This implication holds for harmonic functions as well.

yep, a uniformly bounded sequence of analytic functions is equicontinuous ( Cauchy Formula for example). And if $K$ is a compact and $\epsilon > 0$, one can find $\mu > 0$ such that $|f_{k}(x)-f_{k}(y)| < \epsilon$ for $|x-y| < \mu$, $k > 0$. Then cover $K$ with a finite number of balls $B(a_{i}, \mu)$ and use simple convergence on the $a_{i}$ to conclude.


Figure out this sequence
What are the next 3 numbers

1
11
21
1211

[hide="hint"]Think outside the box[/hide]

This is a cool brain teaser.
[hide]
Each number describes its predacessor.
1::one
11::one one
21::two ones
1211::1 two, one one
111221::1 one, 1 two, 2 ones
The next 3 terms are:
312211
13112121
11132112111211

[/hide]

Good job.

My friend showed that to me...Unfortunately, I couldn't solve it...

[quote="eryaman"]Isn't this called the "see and say" sequence?[/quote]

Yeah, that's usually what it is called.


Find the minimum possible value of x^2 + y^2 given that x and y are real numbers satisfying:

xy(x^2 - y^2) = x^2 + y^2 and x=/= 0.

Try the substitution: $ x\equal{}r\cos\theta$, $ y\equal{}r\sin\theta$.

ummm.... please elaborate?

another hint?

I don't think Lagrange Multipliers are supposed to be used in Pre-Olympiad, especially when there's a much more elementary solution.

As Altheman said, try the substitution, using the identities $ \sin{2\theta} \equal{} 2\sin\theta\cos\theta$, $ \cos{2\theta} \equal{} \cos^2\theta \minus{} \sin^2\theta$, and $ \sin^2\theta \plus{} \cos^2\theta \equal{} 1$.

[hide="Solution"]This solution uses Altheman's substitution.

Substituting $ x \equal{} r\cos\theta$, and $ y \equal{} r\sin\theta$ into $ xy(x^2 \minus{} y^2) \equal{} x^2 \plus{} y^2$, we obtain

$ r^2\cos\theta\sin\theta(r^2\cos^2\theta \plus{} r^2\sin^2\theta) \equal{} r^2\cos^2\theta \plus{} r^2\sin^2\theta$

$ \frac {r^4 \sin 2\theta}{2}(\cos^2\theta \minus{} \sin^2\theta) \equal{} r^2$

$ \frac {r^4\sin 2\theta\cos 2\theta}{2} \equal{} r^2$

$ \frac {r^4\sin 4\theta}{4} \equal{} r^2$

$ r^2\sin 4\theta \equal{} 4$

$ \sin 4\theta$ is always less than or equal to $ 1$, so we maximize $ x^2 \plus{} y^2 \equal{} r^2$ by setting $ \sin 4\theta \equal{} 1 \Rightarrow \theta \equal{} \pi/8 \Rightarrow r^2 \equal{} \boxed{4}$.
The maximum is attained when $ (x,y) \equal{} \left(\pm 2\cos(\frac {\pi}{8}),\mp 2\sin(\frac {\pi}{8})\right)$.
[/hide]

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1072829802&t=72695   :)

[quote="Dr Sonnhard Graubner"]hello, use the method of Lagrange Multipliers, i.e. define
$ F(x,y,\lambda) \equal{} x^2 \plus{} y^2 \plus{} \lambda(xy(x^2 \minus{} y^2) \minus{} (x^2 \plus{} y^2))$
and calculate
$ F_x$
$ F_y$
$ F_\lambda$
and solve the system
$ F_x \equal{} 0$ and $ F_y \equal{} 0$ and $ F_\lambda \equal{} 0$
with respect to $ x,y,\lambda$
Sonnhard.[/quote]

You have to first verify that a maximum exists...

[quote="malcolm"]Done 7 out of 8 hopefully. Ive done half of no. 8 but I can probably get away with it, and the rest should be easy. Can't get number 3 though =/ What about you?

Managing to solve many past BMO questions? I've managed about 3 out of 20..[/quote]

wow, i havn't spent a huge amount of time on the assignment, but i think i've got 3 of them...
and i dont really understand q4 and 7 that well.

Yeah, number 4 and 7 are weird but easy once you understand them


[quote="DaReaper"]How many sides does a convex polygon with interior angles totaling 4860 degrees have?

**is there a specific formula, or what?**[/quote]
[hide]
The formula should be $ 180(n\minus{}2)\equal{}T$ where $ T$ is the sum of all the measures of the angles and $ n$ is the number of sides.
(triangle totals to $ 180$, quadrilateral totals to $ 360$, pentagon totals to $ 540$, hexagon totals to $ 720$, etc...)
Solving for $ n$ when plugging $ 4860$ for $ T$, there are $ 25$ sides.
[/hide]

actually when pluggging in 4860, i got the answer as $ 29$?

\[ 180n\minus{}360\equal{}4860\\
180n\equal{}5220\\
n\equal{}29\]

yea...i think protestant just made a calculation error

oops :D
I kinda rounded it to nearest 5 sides for some reason.


prove the given formula and say for which $ x$ does it hold
\[ \left(\frac{\arctan x}{x}\right)^2\equal{}1\minus{}\sum_{n\equal{}1}^{\infty}(\minus{}1)^{n\minus{}1}\left(1\plus{}\frac{1}{3}\plus{}\dots \plus{}\frac{1}{2n\plus{}1}\right)\frac{x^{2n}}{n\plus{}1}\]

there is not nothing new Idea

you know that :
$ arctan x \equal{} \sum_{n \equal{} 0}\frac {( \minus{} 1)^{n}x^{2n \plus{} 1}}{2n \plus{} 1}$
then
$ \frac {arcatn x}{x} \equal{} \sum_{n \equal{} 0}\frac {( \minus{} 1)^{n}x^{2n}}{2n \plus{} 1}$
and then
$ (\frac {arcatn x}{x})^{2} \equal{} (\sum_{n \equal{} 0}\frac {( \minus{} 1)^{n}x^{2n}}{2n \plus{} 1})^{2}$ and after you find coeficien for $ x^{2n}$ and you have done
this is true for $ |x|\leq 1$


inequalities
number theory
inequalities proposed

$ a,\ b,\ c$ are real numbers such that the sum of any two numbers is greater than or equal to the rest.

Prove that $ 3(a^4\plus{}b^4\plus{}c^4)\plus{}3abc(a\plus{}b\plus{}c)\geq ab(a^2\plus{}b^2)\plus{}bc(b^2\plus{}c^2)\plus{}ca(c^2\plus{}a^2)$.

[quote="kunny"]$ a,\ b,\ c$ are real numbers such that the sum of any two numbers is greater than or equal to the rest.

Prove that $ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} 3abc(a \plus{} b \plus{} c)\geq ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2)$.[/quote]
Ehh... Either I'm mistaken or $ a^4\plus{}b^4\plus{}c^4 \plus{} abc(a\plus{}b\plus{}c) \geqslant ab(a^2\plus{}b^2)\plus{}bc(b^2\plus{}c^2)\plus{}ca(c^2\plus{}a^2)$ holds for all real numbers $ a,b,c$..

[quote="Mathias_DK"][quote="kunny"]$ a,\ b,\ c$ are real numbers such that the sum of any two numbers is greater than or equal to the rest.

Prove that $ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} 3abc(a \plus{} b \plus{} c)\geq ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2)$.[/quote]
Ehh... Either I'm mistaken or $ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c) \geqslant ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2)$ holds for all real numbers $ a,b,c$..[/quote]
Yes, it is Schur's, if I'm not mistaken.

Yes, it is :

$ \sum_{cyc} a^2(a\minus{}b)(a\minus{}c) \ge 0$

Equality holds iff $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$ and cyclic permutations..

Thanks. I myself can't use Schur's inequality even now.

Here is my soluton:

$ \sum_{cyc} (a\plus{}b\minus{}c)(a^3\plus{}b^3\plus{}3abc\minus{}c^3)\geq 0.$

By the way, how do you pronounce Schur?

Pronounce Schur ? I don't understand what you mean..

btw, i'm Indonesian, so I can even pronounce every words formed by 26 basic letters easily..  :blush:

[quote="kunny"]Thanks. I myself can't use Schur's inequality even now.

Here is my soluton:

$ \sum_{cyc} (a \plus{} b \minus{} c)(a^3 \plus{} b^3 \plus{} 3abc \minus{} c^3)\geq 0.$

By the way, how do you pronounce Schur?[/quote]
Yeah, I was really surprised when my teacher pronounced "Muirhead"; he said it something that sounded like "mawrai"(???); and also when he taught me the "langrage's theorem" in number theory - it seemed to me like "lanraj" ....
 :D
Well, I know how to write "schur" it must be pronounced something like "scar"(it's my intuition) :rotfl:


Let $ a, b, c$ rel positive numbers. Prove or disprove that

$ \sum\frac {a^2 \plus{} 4ab \plus{} b^2}{c^2 \plus{} ab}\ge \frac {15}{2} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{2abc}$ .

________
 :?:

[quote="shaaam"]I think inequality is wrong.

[u]Counter Example:[/u]
Try $ a \equal{} b \equal{} (2008!)^{2008}$ $ c \equal{} (\frac {1}{ab})^{200000008}$ If you observe this term (in right hand side):
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{2abc}$ will get so big, Since we chose $ c$ very small, the denominator will be small..
and we chose $ a,b$ big too (although, relatively small to $ c$, because i want to make denominator very small)

Therefore, right hand side will be bigger than left hand side.[/quote]

---------------------
Traditional solutions exist. I look for traditional and elegant solutions for my next book. Thanks for the contribution. I still wait...  :)

[quote="Ligouras"]Let $ a, b, c$ rel positive numbers. Prove or disprove that

$ \sum\frac {a^2 \minus{} 4ab \plus{} b^2}{c^2 \plus{} ab}\ge \frac {15}{2} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{2abc}$ .

________
 :?:[/quote]

[quote="Ligouras"]Traditional solutions exist. I look for traditional and elegant solutions for my next book. Thanks for the contribution. I still wait...  :)[/quote]

Are you sure? Try $ a\equal{}b\equal{}c$.  :wink:

[quote="Ligouras"][quote="Honey_S"][quote="Ligouras"]Let $ a, b, c$ rel positive numbers. Prove or disprove that

$ \sum\frac {a^2 \minus{} 4ab \plus{} b^2}{c^2 \plus{} ab}\ge \frac {15}{2} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{2abc}$ .

________
 :?:[/quote]

[quote="Ligouras"]Traditional solutions exist. I look for traditional and elegant solutions for my next book. Thanks for the contribution. I still wait...  :)[/quote]

Are you sure? Try $ a \equal{} b \equal{} c$.  :wink:[/quote]

-------------
You sure, I am interested of elegant solutions (aesthetically and visually...)[/quote]

I mean your inequality is wrong. Please try with $ a \equal{} b \equal{} c$, then
\[ LHS \equal{} \minus{} 3,RHS \equal{} 9
\]

\[ LHS < RHS
\]
Maybe you mean
\[ \sum\frac {a^2 \plus{} 4ab \plus{} b^2}{c^2 \plus{} ab}\ge \frac {15}{2} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{2abc}
\]
Anyway, be more careful.  :)

[quote="Honey_S"][/quote][quote="Ligouras"][quote="Honey_S"][quote="Ligouras"]Let $ a, b, c$ rel positive numbers. Prove or disprove that

$ \sum\frac {a^2 - 4ab + b^2}{c^2 + ab}\ge \frac {15}{2} + \frac {a^3 + b^3 + c^3}{2abc}$ .

________
 :?:[/quote]

[quote="Ligouras"]Traditional solutions exist. I look for traditional and elegant solutions for my next book. Thanks for the contribution. I still wait...  :)[/quote]

Are you sure? Try $ a = b = c$.  :wink:[/quote]

-------------
You sure, I am interested of elegant solutions (aesthetically and visually...)[/quote][quote="Honey_S"]

I mean your inequality is wrong. Please try with $ a = b = c$, then
\[ LHS = - 3,RHS = 9
\]

\[ LHS < RHS
\]
Maybe you mean
\[ \sum\frac {a^2 + 4ab + b^2}{c^2 + ab}\ge \frac {15}{2} + \frac {a^3 + b^3 + c^3}{2abc}
\]
Anyway, be more careful.  :)[/quote]
For last $ a\rightarrow 0,b=c=1$

[quote="Honey_S"][/quote][quote="Ligouras"][quote="Honey_S"][quote="Ligouras"]Let $ a, b, c$ rel positive numbers. Prove or disprove that

$ \sum\frac {a^2 - 4ab + b^2}{c^2 + ab}\ge \frac {15}{2} + \frac {a^3 + b^3 + c^3}{2abc}$ .

________
 :?:[/quote]

[quote="Ligouras"]Traditional solutions exist. I look for traditional and elegant solutions for my next book. Thanks for the contribution. I still wait...  :)[/quote]

Are you sure? Try $ a = b = c$.  :wink:[/quote]

-------------
You sure, I am interested of elegant solutions (aesthetically and visually...)[/quote][quote="Honey_S"]

I mean your inequality is wrong. Please try with $ a = b = c$, then
\[ LHS = - 3,RHS = 9
\]

\[ LHS < RHS
\]
Maybe you mean
\[ \sum\frac {a^2 + 4ab + b^2}{c^2 + ab}\ge \frac {15}{2} + \frac {a^3 + b^3 + c^3}{2abc}
\]
Anyway, be more careful.  :)[/quote]

------------
Forgive me, Excuse me and thanks. Have a good 2009


algebra function domain conics hyperbola

find all real numbers $ x$ and $ y$  such that :
$ 2x^2\minus{}xy\minus{}5x\plus{}y\plus{}4\equal{}0$ and $ x\geq y \geq1$
from moroccan TST 2002

If we make the substitutions: $ x \equal{} 2a \plus{} b \plus{} 1$ and $ y \equal{} b \minus{} 1$, then the domain restriction is equivalently: $ b\ge 2$ and $ a\ge \minus{} 1$. The relation becomes:
\[ 8a^2 \plus{} 6ab \plus{} b^2 \plus{} 1 \equal{} 0
\]
If $ a \equal{} \minus{} 1$, then $ b \equal{} 3$. Otherwise, $ a > \minus{} 1$. Since $ b$ is real, $ \triangle_b \equal{} 4(a^2 \minus{} 1)\ge 0$. So $ a\ge 1$. But then $ 0 \equal{} 8a^2 \plus{} 6ab \plus{} b^2 \plus{} 1\ge 8 \plus{} 12 \plus{} 4 \plus{} 1 \equal{} 25 > 0$, which is a contradiction. Hence the only solution is $ (x,y) \equal{} (2,2)$.

[quote="greatestmaths"]find all real numbers $ x$ and $ y$  such that :
$ 2x^2 \minus{} xy \minus{} 5x \plus{} y \plus{} 4 \equal{} 0$ and $ x\geq y \geq1$
[/quote]
More way:
$ 2x^2 \minus{} xy \minus{} 5x \plus{} y \plus{} 4 \equal{} 0\Leftrightarrow2x^2 \minus{} 5x \plus{} 4 \equal{} y(x \minus{} 1).$
Hence, $ x > 1.$
Thus, $ y \equal{} \frac {2x^2 \minus{} 5x \plus{} 4}{x \minus{} 1}.$
But $ \frac {2x^2 \minus{} 5x \plus{} 4}{x \minus{} 1}\leq x\Leftrightarrow(x \minus{} 2)^2\leq0.$
Id est, $ x \equal{} 2$ and $ y \equal{} 2.$

One comment: this function is a hyperbola in the xy-plane. The border of the domain restriction is tangent to the hyperbola.


Let x,y,z be three integers. Show that, if x^3 + y^3 + z^3 is dividible by 9, then, at least one of the numbers x,y and z is dividible by 3. :)

[quote="h_s_potter2002"][hide]Any number cubed mod 9 is equivelent to 1, 8, or 0 mod 9.  So if you have $x^3+y^3+z^3$, the only way to have it divisible by 9 is if $x^3 \equiv y^3 \equiv z^3 \equiv 0 \mod (9)$, or if out of $x^3$, $y^3$, or $z^3$, one is equivelent to 8 mod 9, one is equivelent to 1 mod 9, and if one is equivelent to 0 mod 9.  Either way, you have one number equivelent to 0 mod 9.[/hide][/quote]You forgot about the case where all three numbers are $\equiv 0 \pmod 9$.  Minor detail, but its still something.  :D

Bah, don't worry lol
You made a good job :)