{"id":43,"date":"2010-08-24T09:24:05","date_gmt":"2010-08-24T09:24:05","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=43"},"modified":"2010-08-24T09:24:05","modified_gmt":"2010-08-24T09:24:05","slug":"kategoria-abc","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=43","title":{"rendered":"Kateg\u00f3rie ABC"},"content":{"rendered":"

<\/a>59. ro\u010dn\u00edk matematickej olympi\u00e1dy 2009-2010<\/p>\n


\n

[ hana-code-insert ] 'etarget_text' is not found <\/div><\/p>\n\n\n\n\n\n\n
C<\/span><\/td>\nB<\/span><\/td>\nA<\/span><\/td>\n<\/tr>\n
dom\u00e1ce kolo<\/a><\/span><\/td>\ndom\u00e1ce kolo<\/a><\/span><\/td>\ndom\u00e1ce kolo<\/a><\/span><\/td>\n<\/tr>\n
\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<\/tr>\n
krajsk\u00e9 kolo<\/a><\/span><\/td>\nkrajsk\u00e9 kolo<\/a><\/span><\/td>\nkrajsk\u00e9 kolo<\/a><\/span><\/td>\n<\/tr>\n
 <\/td>\ncelo\u0161t\u00e1tne kolo<\/span><\/td>\n<\/tr>\n<\/table>\n

<\/a><\/p>\n

C-I-1<\/b>
\nErika a Kl\u00e1rka hrali hru \u201eslovn\u00fd logik\u201c s t\u00fdmito pravidlami: Hr\u00e1\u010d A si mysl\u00ed slovo zlo\u017een\u00e9 z piatich r\u00f4znych p\u00edsmen. Hr\u00e1\u010d B vyslov\u00ed \u013eubovo\u013en\u00e9 slovo zlo\u017een\u00e9 z piatich r\u00f4znych p\u00edsmen a hr\u00e1\u010d A mu prezrad\u00ed,
\nko\u013eko p\u00edsmen uh\u00e1dol na spr\u00e1vnej poz\u00edcii a ko\u013eko na nespr\u00e1vnej. P\u00edsmen\u00e1 pova\u017eujeme za r\u00f4zne, aj ke\u010f sa l\u00ed\u0161ia len d\u013a\u017e\u0148om alebo m\u00e4k\u010de\u0148om (napr\u00edklad p\u00edsmen\u00e1 A, \u00c1 s\u00fa r\u00f4zne). Keby si napr\u00edklad hr\u00e1\u010d A myslel
\nslovo LO\u010eKA a B by vyslovil slovo KOL\u00c1\u010c, odpovedal by mu hr\u00e1\u010d A, \u017ee uh\u00e1dol jedno p\u00edsmeno na spr\u00e1vnej poz\u00edcii a dve na nespr\u00e1vnej. Skr\u00e1tene by povedal \u201e1 + 2\u201c, lebo obe slov\u00e1 sa naozaj zhoduj\u00fa len
\nv p\u00edsmene O vr\u00e1tane poz\u00edcie (druh\u00e9 p\u00edsmeno z\u013eava) a v p\u00edsmen\u00e1ch K a L, ktor\u00fdch poz\u00edcie s\u00fa odli\u0161n\u00e9.
\nErika si myslela slovo zlo\u017een\u00e9 z piatich r\u00f4znych p\u00edsmen a Kl\u00e1rka vyslovila slov\u00e1 KAB\u00c1T, STRUK, SKOBA, CESTA a Z\u00c1PAL. Erika na tieto slov\u00e1 v danom porad\u00ed odpovedala 0+3, 0+2, 1+2, 2+0 a 1+2. Zistite, ktor\u00e9 slovo si Erika mohla myslie\u0165.<\/p>\n

(Peter Novotn\u00fd)<\/div>\n<\/p>\n

C-I-2<\/b>
\nVrcholom C<\/i> pravouholn\u00edka ABCD<\/i> ve\u010fte priamky p<\/i> a q<\/i>, ktor\u00e9 maj\u00fa s dan\u00fdm pravouholn\u00edkom spolo\u010dn\u00fd iba bod C<\/i>, pri\u010dom priamka p<\/i> m\u00e1 od bodu A<\/i> najv\u00e4\u010d\u0161iu mo\u017en\u00fa vzdialenos\u0165 a priamka q<\/i> vymedzuje s priamkami AB<\/i> a AD<\/i> trojuholn\u00edk s \u010do najmen\u0161\u00edm obsahom.<\/p>\n

(Leo Bo\u010dek)<\/div>\n<\/p>\n

C-I-3<\/b>
\nUr\u010dte v\u0161etky re\u00e1lne \u010d\u00edsla x<\/i>, ktor\u00e9 vyhovuj\u00fa rovnici 4x<\/i> – 2 = 5. (Symbol ozna\u010duje najv\u00e4\u010d\u0161ie cel\u00e9 \u010d\u00edslo, ktor\u00e9 nie je v\u00e4\u010d\u0161ie ako \u010d\u00edslo x<\/i>, tzv. doln\u00fa cel\u00fa \u010das\u0165 re\u00e1lneho \u010d\u00edsla x<\/i>.)<\/p>\n

(Jaroslav \u0160vr\u010dek)<\/div>\n<\/p>\n

C-I-4<\/b>
\nKru\u017enica k(S; r)<\/i> sa dot\u00fdka priamky AB<\/i> v bode A<\/i>. Kru\u017enica l(T; s)<\/i> sa dot\u00fdka priamky AB<\/i> v bode B<\/i> a pret\u00edna kru\u017enicu k<\/i> v krajn\u00fdch bodoch C, D<\/i> jej priemeru. Vyjadrite d\u013a\u017eku a<\/i> \u00fase\u010dky AB<\/i> pomocou polomerov r, s<\/i>. Dok\u00e1\u017ete \u010falej, \u017ee priese\u010dn\u00edk M<\/i> priamok CD, AB<\/i> je stredom \u00fase\u010dky AB<\/i>.<\/p>\n

(Leo Bo\u010dek)<\/div>\n<\/p>\n

C-I-5<\/b>
\nDok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 kladn\u00e9 re\u00e1lne \u010d\u00edsla a, b<\/i> plat\u00ed<\/p>\n

<\/p>\n

a pre ka\u017ed\u00fa z oboch nerovnost\u00ed zistite, kedy prech\u00e1dza v rovnos\u0165.<\/p>\n

(J\u00e1n Maz\u00e1k)<\/div>\n<\/p>\n

C-I-6<\/b>
\nN\u00e1jdite v\u0161etky prirodzen\u00e9 \u010d\u00edsla, ktor\u00e9 nie s\u00fa delite\u013en\u00e9 desiatimi a ktor\u00e9 maj\u00fa vo svojom dekadickom z\u00e1pise niekde ved\u013ea seba dve nuly, po ktor\u00fdch vy\u0161krtnut\u00ed sa p\u00f4vodn\u00e9 \u010d\u00edslo 89-kr\u00e1t zmen\u0161\u00ed.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

C-S-1<\/b>
\nAk zv\u00e4\u010d\u0161\u00edme \u010ditate\u013e aj menovate\u013e ist\u00e9ho zlomku o 1, dostaneme zlomok o hodnotu 1\/20<\/i> v\u00e4\u010d\u0161\u00ed. Ak urob\u00edme s v\u00e4\u010d\u0161\u00edm zlomkom rovnak\u00fa oper\u00e1ciu, dostaneme zlomok o hodnotu 1\/12<\/i> v\u00e4\u010d\u0161\u00ed, ako bola hodnota zlomku na za\u010diatku. Ur\u010dte v\u0161etky tri zlomky.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n

C-S-2<\/b>
\nKru\u017enice k(S; 6 cm)<\/i> a l(O; 4 cm)<\/i> maj\u00fa vn\u00fatorn\u00fd dotyk v bode B<\/i>. Ur\u010dte d\u013a\u017eky str\u00e1n trojuholn\u00edka ABC<\/i>, pri\u010dom bod A<\/i> je priese\u010dn\u00edk priamky OB<\/i> s kru\u017enicou k<\/i> a bod C<\/i> je priese\u010dn\u00edk kru\u017enice k<\/i> s doty\u010dnicou z bodu A<\/i> ku kru\u017enici l<\/i>.<\/p>\n

(Pavel Leischner)<\/div>\n<\/p>\n

C-S-3<\/b>
\nN\u00e1jdite v\u0161etky dvojice nez\u00e1porn\u00fdch cel\u00fdch \u010d\u00edsel a, b<\/i>, pre ktor\u00e9 plat\u00ed<\/p>\n

a2<\/sup> + b + 2 = a + b2<\/sup>.<\/p>\n

(J\u00e1n Maz\u00e1k)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

C-II-1<\/b>
\nDok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 cel\u00e9 \u010d\u00edsla n<\/i> a k<\/i> v\u00e4\u010d\u0161ie ako 1 je \u010d\u00edslo nk + 2<\/sup> – nk<\/sup><\/b><\/i> delite\u013en\u00e9
\ndvan\u00e1stimi.<\/p>\n

(Vojtech B\u00e1lint)<\/div>\n<\/p>\n

C-II-2<\/b>
\nDok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 \u010d\u00edsla a, b<\/i> z intervalu <1, ∞) plat\u00ed nerovnos\u0165<\/p>\n

(a2<\/sup> + 1)(b2<\/sup> + 1) – (a – 1)2<\/sup>(b – 1)2<\/sup> ≥ 4<\/p>\n

a zistite, kedy nastane rovnos\u0165.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n

C-II-3<\/b>
\nDan\u00e1 je kru\u017enica k<\/i> so stredom S<\/i>. Kru\u017enica l<\/i> m\u00e1 v\u00e4\u010d\u0161\u00ed polomer ako kru\u017enica k<\/i>, prech\u00e1dza jej stredom a pret\u00edna ju v bodoch M<\/i> a N<\/i>. Priamka, ktor\u00e1 prech\u00e1dza bodom N<\/i> a je rovnobe\u017en\u00e1 s priamkou MS<\/i>, vyt\u00edna na kru\u017eniciach tetivy NP<\/i> a NQ<\/i>. Dok\u00e1\u017ete, \u017ee trojuholn\u00edk MPQ<\/i> je rovnoramenn\u00fd.<\/p>\n

(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<\/p>\n

C-II-4<\/b>
\nUr\u010dte v\u0161etky dvojice re\u00e1lnych \u010d\u00edsel x, y<\/i>, ktor\u00e9 vyhovuj\u00fa s\u00fastave rovn\u00edc<\/p>\n

<\/p>\n

ak    a)<\/b>  p<\/i> = 2,
       b)<\/b>  p<\/i> = 3.<\/p>\n

Symbol ozna\u010duje najv\u00e4\u010d\u0161ie cel\u00e9 \u010d\u00edslo, ktor\u00e9 nie je v\u00e4\u010d\u0161ie ako dan\u00e9 re\u00e1lne \u010d\u00edslo x (tzv. doln\u00e1 cel\u00e1 \u010das\u0165 re\u00e1lneho \u010d\u00edsla x).<\/i><\/p>\n

(Jaroslav \u0160vr\u010dek)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-I-1<\/b>
\nNa stole s\u00fa tri k\u00f4pky z\u00e1paliek: v jednej 2 009, v druhej 2 010, v tretej 2 011. Hr\u00e1\u010d, ktor\u00fd je na \u0165ahu, zvol\u00ed dve k\u00f4pky a z ka\u017edej z nich odoberie po jednej z\u00e1palke. V hre sa pravidelne striedaj\u00fa dvaja hr\u00e1\u010di. Hra kon\u010d\u00ed, akon\u00e1hle niektor\u00e1 k\u00f4pka zmizne. Vyhr\u00e1va ten hr\u00e1\u010d, ktor\u00fd urobil posledn\u00fd \u0165ah. Op\u00ed\u0161te strat\u00e9giu jedn\u00e9ho z hr\u00e1\u010dov, ktor\u00e1 mu zaru\u010d\u00ed v\u00fdhru.<\/p>\n

(J\u00e1n Maz\u00e1k)<\/div>\n<\/p>\n

B-I-2<\/b>
\nNa tabuli je nap\u00edsan\u00e9 \u0161tvorcifern\u00e9 \u010d\u00edslo, ktor\u00e9 m\u00e1 presne \u0161es\u0165 kladn\u00fdch delite\u013eov, z ktor\u00fdch pr\u00e1ve dva s\u00fa jednocifern\u00e9 a pr\u00e1ve dva dvojcifern\u00e9. V\u00e4\u010d\u0161\u00ed z dvojcifern\u00fdch delite\u013eov je druhou mocninou prirodzen\u00e9ho \u010d\u00edsla. Ur\u010dte v\u0161etky \u010d\u00edsla, ktor\u00e9 m\u00f4\u017eu by\u0165 na tabuli nap\u00edsan\u00e9.<\/p>\n

(Peter Novotn\u00fd)<\/div>\n<\/p>\n

B-I-3<\/b>
\nV rovine je dan\u00e1 \u00fase\u010dka AB<\/i>. Zostrojte rovnobe\u017en\u00edk ABCD<\/i>, pre ktor\u00e9ho stredy str\u00e1n AB, CD, DA<\/i> ozna\u010den\u00e9 postupne K, L, <\/i>M plat\u00ed: body A, B, L, D<\/i> le\u017eia na jednej kru\u017enici a aj body K, L, D, M<\/i> le\u017eia na jednej kru\u017enici.<\/p>\n

(Jaroslav \u0160vr\u010dek)<\/div>\n<\/p>\n

B-I-4<\/b>
\nN\u00e1jdite 2 009 po sebe id\u00facich \u0161tvorcifern\u00fdch \u010d\u00edsel, ktor\u00fdch s\u00fa\u010det je s\u00fa\u010dinom troch po sebe id\u00facich prirodzen\u00fdch \u010d\u00edsel.<\/p>\n

(Radek Horensk\u00fd)<\/div>\n<\/p>\n

B-I-5<\/b>
\nVn\u00fatri krat\u0161ieho obl\u00faka AB<\/i> kru\u017enice op\u00edsanej rovnostrann\u00e9mu trojuholn\u00edku ABC<\/i> zvol\u00edme bod D<\/i>. Tetiva CD<\/i> pret\u00edna stranu AB<\/i> v bode E<\/i>. Dok\u00e1\u017ete, \u017ee trojuholn\u00edk so stranami d\u013a\u017eok |AE<\/i>|, |BE<\/i>|, |CE<\/i>| je podobn\u00fd trojuholn\u00edku ABD<\/i>.<\/p>\n

(Pavel Leischner)<\/div>\n<\/p>\n

B-I-6<\/b>
\nRe\u00e1lne \u010d\u00edsla a, b<\/i> maj\u00fa t\u00fato vlastnos\u0165: rovnica x2<\/sup> – ax + b – 1 = 0<\/i> m\u00e1 v mno\u017eine re\u00e1lnych \u010d\u00edsel dva r\u00f4zne korene, ktor\u00fdch rozdiel je kladn\u00fdm kore\u0148om rovnice x2<\/sup> – ax + b + 1 = 0<\/i>.<\/p>\n

    a)<\/b>   Dok\u00e1\u017ete nerovnos\u0165 b > 3<\/i>,
\n    b)<\/b>   Pomocou b<\/i> vyjadrite korene oboch rovn\u00edc.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-S-1<\/b>
\nUr\u010dte v\u0161etky hodnoty re\u00e1lnych parametrov p, q<\/i>, pre ktor\u00e9 m\u00e1 ka\u017ed\u00e1 z rovn\u00edc<\/p>\n

x(x – p) = 3 + q,   x(x + p) = 3 – q <\/p>\n

v obore re\u00e1lnych \u010d\u00edsel dva r\u00f4zne korene, ktor\u00fdch aritmetick\u00fd priemer je jedn\u00fdm z kore\u0148ov zvy\u0161nej rovnice.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n

B-S-2<\/b>
\nDan\u00e9 s\u00fa d\u013a\u017eky odvesien a<\/i>=|BC<\/i>|, b<\/i>=|AC<\/i>| pravouhl\u00e9ho trojuholn\u00edka ABC, pri\u010dom a>b<\/i>. Ozna\u010dme D<\/i> stred prepony AB<\/i> a E<\/i> (E≠C<\/i>) priese\u010dn\u00edk strany BC<\/i> s kru\u017enicou op\u00edsanou trojuholn\u00edku ADC<\/i>. Vypo\u010d\u00edtajte obsah trojuholn\u00edka EAD<\/i>.<\/p>\n

(Pavel Novotn\u00fd)<\/div>\n<\/p>\n

B-S-3<\/b>
\nUr\u010dte v\u0161etky dvojice cel\u00fdch kladn\u00fdch \u010d\u00edsel m, n<\/i>, pre ktor\u00e9 plat\u00ed 37 + 27m<\/sup> = n3<\/sup><\/i><\/b>.<\/p>\n

(Martin Pan\u00e1k)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-II-1<\/b>
\nKru\u017enica l(T;s)<\/i> prech\u00e1dza stredom kru\u017enice k(S;2cm)<\/i>. Kru\u017enica m(U;t)<\/i> sa zvonka dot\u00fdka kru\u017en\u00edc k<\/i> a l<\/i>, pri\u010dom US<\/i> ^<\/font> ST<\/i>. Polomery s<\/i> a t<\/i> vyjadren\u00e9 v centimetroch s\u00fa cel\u00e9 \u010d\u00edsla. Ur\u010dte ich.<\/p>\n

(Pavel Leischner)<\/div>\n<\/p>\n

B-II-2<\/b>
\nV matematickej s\u00fa\u0165a\u017ei bolo zadan\u00fdch 7 \u00faloh a za ka\u017ed\u00fa z nich mohol s\u00fa\u0165a\u017eiaci z\u00edska\u0165 0, 1 alebo 2 body. S\u00fa\u0165a\u017ee sa z\u00fa\u010dastnilo 60 \u017eiakov. Za ka\u017ed\u00fa \u00falohu bolo udelen\u00fdch aspo\u0148 95 bodov. Dok\u00e1\u017ete, \u017ee medzi s\u00fa\u0165a\u017eiacimi n\u00e1jdeme dvoch tak\u00fdch, \u017ee ka\u017ed\u00fa z \u00faloh vyrie\u0161il aspo\u0148 jeden z nich za 2 body.<\/p>\n

(J\u00e1n Maz\u00e1k)<\/div>\n<\/p>\n

B-II-3<\/b>
\nV rovine je dan\u00fd rovnobe\u017en\u00edk ABCD<\/i>. Ozna\u010dme postupne K, L, M<\/i> stredy str\u00e1n AB, CD, AD<\/i>. Predpokladajme, \u017ee body A, B, L, D<\/i> le\u017eia na jednej kru\u017enici a s\u00fa\u010dasne aj body K, L, D, M<\/i> le\u017eia na jednej kru\u017enici. Dok\u00e1\u017ete, \u017ee |AC<\/i>| = 2\u00b7|AD<\/i>|.<\/p>\n

(Jaroslav \u0160vr\u010dek)<\/div>\n<\/p>\n

B-II-4<\/b>
\n\u010c\u00edslo n<\/i> je s\u00fa\u010dinom \u0161tyroch prvo\u010d\u00edsel. Ak ka\u017ed\u00e9 z t\u00fdchto prvo\u010d\u00edsel zv\u00e4\u010d\u0161\u00edme o 1 a vzniknut\u00e9 \u0161tyri \u010d\u00edsla vyn\u00e1sob\u00edme, dostaneme \u010d\u00edslo o 2886 v\u00e4\u010d\u0161ie ako p\u00f4vodn\u00e9 \u010d\u00edslo n<\/i>. Ur\u010dte v\u0161etky tak\u00e9 n<\/i>.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-I-1<\/b>
\nV obore re\u00e1lnych \u010d\u00edsel rie\u0161te s\u00fastavu rovn\u00edc<\/p>\n

<\/p>\n

(Radek Horensk\u00fd)<\/div>\n<\/p>\n

A-I-2<\/b>
\nDo koso\u0161tvorca ABCD<\/i> je vp\u00edsan\u00e1 kru\u017enica. Uva\u017eujme jej \u013eubovo\u013en\u00fa doty\u010dnicu pret\u00ednaj\u00facu obe strany BC, CD<\/i> a ozna\u010dme postupne R, S<\/i> jej priese\u010dn\u00edky s priamkami AB, AD<\/i>. Dok\u00e1\u017ete, \u017ee hodnota s\u00fa\u010dinu |BR<\/i>|.|DS<\/i>| od vo\u013eby doty\u010dnice nez\u00e1vis\u00ed.<\/p>\n

(Leo Bo\u010dek)<\/div>\n<\/p>\n

A-I-3<\/b>
\nNa tabuli s\u00fa nap\u00edsan\u00e9 \u010d\u00edsla 1, 2, …, 33. V jednom kroku zvol\u00edme na tabuli dve \u010d\u00edsla, z ktor\u00fdch jedno je delite\u013eom druh\u00e9ho, obe zotrieme a na tabu\u013eu nap\u00ed\u0161eme ich (celo\u010d\u00edseln\u00fd) podiel. Takto pokra\u010dujeme, k\u00fdm na tabuli nezostan\u00fa iba \u010d\u00edsla, z ktor\u00fdch \u017eiadne nie je delite\u013eom in\u00e9ho. (V jednom kroku m\u00f4\u017eeme zotrie\u0165 aj dve rovnak\u00e9 \u010d\u00edsla a nahradi\u0165 ich \u010d\u00edslom 1.) Ko\u013eko najmenej \u010d\u00edsel m\u00f4\u017ee na tabuli zosta\u0165?<\/p>\n

(Peter Novotn\u00fd)<\/div>\n<\/p>\n

A-I-4<\/b>
\nV \u013eubovo\u013enom ostrouhlom r\u00f4znostrannom trojuholn\u00edku ABC<\/i> ozna\u010dme O, V<\/i> a S<\/i> postupne stred kru\u017enice op\u00edsanej, priese\u010dn\u00edk v\u00fd\u0161ok a stred kru\u017enice vp\u00edsanej. Dok\u00e1\u017ete, \u017ee os \u00fase\u010dky OV<\/i> prech\u00e1dza bodom S<\/i> pr\u00e1ve vtedy, ke\u010f jeden vn\u00fatorn\u00fd uhol trojuholn\u00edka ABC<\/i> m\u00e1 ve\u013ekos\u0165 60\u00b0.<\/p>\n

(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<\/p>\n

A-I-5<\/b>
\nV n\u00e1dr\u017ei je r0<\/sub><\/i> r\u00fdb, spolo\u010dn\u00fd \u00falovok n<\/i> ryb\u00e1rov. Prich\u00e1dzaj\u00fa pre svoj podiel jednotlivo. Ka\u017ed\u00fd si mysl\u00ed, \u017ee sa dostavil ako prv\u00fd, a aby si vzal presne n<\/i>-tinu aktu\u00e1lneho po\u010dtu r\u00fdb v n\u00e1dr\u017ei, mus\u00ed predt\u00fdm jednu z r\u00fdb pusti\u0165 sp\u00e4\u0165 do mora. Ur\u010dte najmen\u0161ie mo\u017en\u00e9 \u010d\u00edslo r0<\/sub><\/i> v z\u00e1vislosti od dan\u00e9ho n<\/i> ≥ 2, ke\u010f aj posledn\u00fd ryb\u00e1r si aspo\u0148 jednu rybu odnesie.<\/p>\n

(Dag Hrub\u00fd)<\/div>\n<\/p>\n

A-I-6<\/b>
\nPre dan\u00e9 prvo\u010d\u00edslo p<\/i> ur\u010dte po\u010det (v\u0161etk\u00fdch) usporiadan\u00fdch troj\u00edc (a, b, c) \u010d\u00edsel z mno\u017einy {1, 2, 3, …, 2p<\/i>2<\/sup>}, ktor\u00e9 sp\u013a\u0148aj\u00fa vz\u0165ah<\/p>\n

<\/p>\n

pri\u010dom [x, y<\/i>] ozna\u010duje najmen\u0161\u00ed spolo\u010dn\u00fd n\u00e1sobok \u010d\u00edsel x<\/i> a y<\/i>.<\/p>\n

(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-S-1<\/b>
\nV obore re\u00e1lnych \u010d\u00edsel rie\u0161te s\u00fastavu rovn\u00edc<\/p>\n

<\/p>\n

A-S-2<\/b>
\nN\u00e1jdite v\u0161etky mo\u017en\u00e9 hodnoty podielu<\/p>\n

<\/p>\n

pri\u010dom r<\/i> je polomer kru\u017enice op\u00edsanej a ρ<\/i> polomer kru\u017enice vp\u00edsanej pravouhl\u00e9mu trojuholn\u00edku s odvesnami d\u013a\u017eok a<\/i> a b<\/i>.<\/p>\n

A-S-3<\/b>
\nNa tabuli s\u00fa nap\u00edsan\u00e9 \u010d\u00edsla 1, 2, …, 33. V jednom kroku zvol\u00edme nieko\u013eko \u010d\u00edsel nap\u00edsan\u00fdch na tabuli (aspo\u0148 dve), ktor\u00fdch s\u00fa\u010din je druhou mocninou prirodzen\u00e9ho \u010d\u00edsla, zvolen\u00e9 \u010d\u00edsla zotrieme a na tabu\u013eu nap\u00ed\u0161eme druh\u00fa odmocninu z ich s\u00fa\u010dinu. Takto pokra\u010dujeme, a\u017e na tabuli ostan\u00fa iba tak\u00e9 \u010d\u00edsla, \u017ee s\u00fa\u010din \u017eiadnych z nich nie je druhou mocninou. Ko\u013eko najmenej \u010d\u00edsel m\u00f4\u017ee na tabuli osta\u0165?<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-II-1<\/b>
\nDok\u00e1\u017ete, \u017ee rovnica x2<\/sup> + p|x| = qx – 1<\/b><\/i> s re\u00e1lnymi parametrami p, q<\/i> m\u00e1 v obore re\u00e1lnych \u010d\u00edsel \u0161tyri rie\u0161enia pr\u00e1ve vtedy, ke\u010f plat\u00ed p + |q| + 2 < 0<\/i>.<\/p>\n

(Jarom\u00edr \u0160im\u0161a)<\/div>\n<\/p>\n

A-II-2<\/b>
\nDan\u00fd je rovnobe\u017en\u00edk ABCD<\/i> s tup\u00fdm uhlom ABC<\/i>. Na jeho uhloprie\u010dke AC<\/i> v polrovine BDC<\/i> zvo\u013eme bod P<\/i> tak, aby platilo |BPD<\/i>| = |ABC<\/i>|. Dok\u00e1\u017ete, \u017ee priamka CD<\/i> je doty\u010dnicou ku kru\u017enici op\u00edsanej trojuholn\u00edku BCP<\/i> pr\u00e1ve vtedy, ke\u010f \u00fase\u010dky AB<\/i> a BD<\/i> s\u00fa zhodn\u00e9.<\/p>\n

(Jaroslav \u0160vr\u010dek)<\/div>\n<\/p>\n

A-II-3<\/b>
\nUr\u010dte v\u0161etky cel\u00e9 kladn\u00e9 \u010d\u00edsla m, n<\/i> tak\u00e9, \u017ee n<\/i> del\u00ed 2m – 1<\/i> a m<\/i> del\u00ed 2n – 1<\/i>.<\/p>\n

(Tom\u00e1\u0161 Szaniszlo)<\/div>\n<\/p>\n

A-II-4<\/b>
\nV \u013eubovo\u013enom trojuholn\u00edku ABC<\/i> ozna\u010dme O<\/i> stred kru\u017enice vp\u00edsanej, P<\/i> stred kru\u017enice prip\u00edsanej ku strane BC<\/i> a D<\/i> priese\u010dn\u00edk osi uhla CAB<\/i> so stranou BC<\/i>. Dok\u00e1\u017ete, \u017ee plat\u00ed<\/p>\n

<\/p>\n

(Kru\u017enica prip\u00edsan\u00e1 ku strane BC<\/i> je tak\u00e1 kru\u017enica, ktor\u00e1 sa dot\u00fdka jednak strany BC<\/i>, jednak oboch polpriamok opa\u010dn\u00fdch k polpriamkam BA<\/i> a CA<\/i>.)<\/p>\n

(Pavel Leischner)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n\n
<\/div>","protected":false},"excerpt":{"rendered":"

59. ro\u010dn\u00edk matematickej olympi\u00e1dy 2009-2010<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[8],"tags":[],"class_list":["post-43","post","type-post","status-publish","format-standard","hentry","category-59-rocnik"],"yoast_head":"\nKateg\u00f3rie ABC - Matematick\u00e1 olympi\u00e1da<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/matematika.besaba.com\/?p=43\" \/>\n<meta property=\"og:locale\" content=\"sk_SK\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Kateg\u00f3rie ABC - Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"og:description\" content=\"59. ro\u010dn\u00edk matematickej olympi\u00e1dy 2009-2010\" \/>\n<meta property=\"og:url\" content=\"https:\/\/matematika.besaba.com\/?p=43\" \/>\n<meta property=\"og:site_name\" content=\"Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"article:published_time\" content=\"2010-08-24T09:24:05+00:00\" \/>\n<meta name=\"author\" content=\"admin\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Autor\" \/>\n\t<meta name=\"twitter:data1\" content=\"admin\" \/>\n\t<meta name=\"twitter:label2\" content=\"Predpokladan\u00fd \u010das \u010d\u00edtania\" \/>\n\t<meta name=\"twitter:data2\" content=\"11 min\u00fat\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/matematika.besaba.com\/?p=43\",\"url\":\"https:\/\/matematika.besaba.com\/?p=43\",\"name\":\"Kateg\u00f3rie ABC - 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