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Language: Lithuanian Day: 1 S S A B S C AC = BC S S AB C S P PA = PB = PC n 3 n n 3 n (a, b, c) ab - c, bc - a, ca - b 2 n n ABC AB > AC Γ ABC H F A BC M BC Q Γ HQA = 90 K Γ HKQ = 90 AB C K Q Γ KQH FKM 4 30 7

Language: Lithuanian · Day: 1 Penktadienis, 2015 m. liepos 10 d. 1 uºdavinys. Baigtin¦ plok²tumos ta²ku aib¦ S vadinsime subalansuota , jei bet kuriems dviems aibes S ta²kams

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Page 1: Language: Lithuanian · Day: 1 Penktadienis, 2015 m. liepos 10 d. 1 uºdavinys. Baigtin¦ plok²tumos ta²ku aib¦ S vadinsime subalansuota , jei bet kuriems dviems aibes S ta²kams

Language: Lithuanian

Day: 1

P♥t♥s ♠ ♣♦s

♥②s t♥ ♣♦t♠♦s t S ♥s♠ s♥s♦t t r♠s ♠ss S t♠s A r B s tsrs t♦s s S ts C AC = BC ②s♠ S

♥tr ♥tr♦ ♦s ♠t♠ trs srt♥s s S ts A B r C s ♦ s S t P

♥♦ ②② PA = PB = PC

r♦②t s ♥ ♥tr♦ n > 3 ③st♦ s♥s♦t sr②t ②n t

st ss ♥tr♦ss n > 3 s rs ③st♦ s♥s♦t r ♥tr♥t ♥tr♦ sr②t ② n t

♥②s st ss ♥tr s↔ trts (a, b, c) s rs ♥s s↔

ab− c, bc− a, ca− b

②r t♦ ♣s♥s

t♦ ♣s♥s ②r ♥♠ ♥tr s↔ 2n r n ②r sss ♥♥♠s s↔s

♥②s ♦ts s♠ss tr♠♣s ABC r♠ AB > AC Γ ②r tr♠♣♦ ABC

♣rt♥s ♣srt♠s H t♦ tr♠♣♦ ♦rt♦♥trs ♦ F t♥s ♥st♦s A rt♥ BC♣r♥s M ②r rt♥s BC r♦ ts Q ②r t♦s ♣srt♠♦ Γ ts ∠HQA = 90 ♦ K t♦s ♣srt♠♦ Γ ts ∠HKQ = 90 t♦ t A B C K r Q ②rs srt♥ r ♣rs♦ ♣srt♠ Γ t♥t t♦ tr

r♦②t tr♠♣ KQH r FKM ♣rt♥ ♣srt♠ ↔ ♥s t→

♥ t♥♥ r srts s 4 ♥♦s r 30 ♠♥↔

♥s ♥②s s rt♥♠s 7 ts

Page 2: Language: Lithuanian · Day: 1 Penktadienis, 2015 m. liepos 10 d. 1 uºdavinys. Baigtin¦ plok²tumos ta²ku aib¦ S vadinsime subalansuota , jei bet kuriems dviems aibes S ta²kams

Language: Lithuanian

Day: 2

➆t♥s ♠ ♣♦s

♥②s r♠♣♦ ABC ♣rt♥s ♣srt♠s ②r Ω ♦ ts O ②r t♦ ♣srt♠♦ ♥trs♣srt♠s Γ r♦ ♥trs A rt tr♣→ BC t♦s D r E t♣ t B D E r C ②rs srt♥ r ♣rs♦ ts BC t♥t t♦ tr F r G ②r ♣srt♠ Γ r Ω s♥rt♦st r A F B C r G ♣rs♦ ♣srt♠ Ω t♥t t♦ tr K ②r ♥trss tr♣♦sAB r ♣rt♦ ♣ tr♠♣ BDF ♣srt♠♦ s♥rt♦s ts L ②r ♥trss tr♣♦s CA

r ♣rt♦ ♣ tr♠♣ CGE ♣srt♠♦ s♥rt♦s ts

r♠ tss FK r GL ②r srt♥♦s r rts t X r♦②t ts X ♣rs♦ts AO

♥②s R ②♠ s r s↔ st ss ♥s f : R → R t♥♥♥↔s ②②

f(

x+ f(x+ y))

+ f(xy) = x+ f(x+ y) + yf(x)

s ss rss s↔s x r y

♥②s s↔ s a1, a2, . . . tr t♦s s②s

1 6 aj 6 2015 s ss j > 1

k + ak 6= ℓ+ aℓ s ss 1 6 k < ℓ

r♦②t ③st♦ t♦ s t♠ s↔ b r N

n∑

j=m+1

(aj − b)

6 10072

s ss sss m r n t♥♥♥↔s ♥②② n > m > N

♥ t♥♥ r srts s 4 ♥♦s r 30 ♠♥↔♥s ♥②s s rt♥♠s 7 ts