{"id":149,"date":"2010-06-08T11:30:31","date_gmt":"2010-06-08T11:30:31","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=149"},"modified":"2010-06-08T11:30:31","modified_gmt":"2010-06-08T11:30:31","slug":"kategorie-abc-4","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=149","title":{"rendered":"Kateg\u00f3rie ABC"},"content":{"rendered":"

57. ro\u010dn\u00edk matematickej olympi\u00e1dy 2007-2008<\/p>\n

<\/a><\/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<\/a><\/span><\/td>\n<\/tr>\n<\/table>\n

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

C-I-1<\/b>
\nUr\u010dte najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo n<\/i>, pre ktor\u00e9 aj \u010d\u00edsla , , s\u00fa prirodzen\u00e9.<\/p>\n

C-I-2<\/b>
\n\u0160tvoruholn\u00edku ABCD<\/i> je vp\u00edsan\u00e1 kru\u017enica so stredom S<\/i>. Ur\u010dte rozdiel |Ð<\/font> ASD<\/em>| – |Ð<\/font> CSD<\/em>|, ak |Ð<\/font> ASB<\/em>| – |Ð<\/font> BSC<\/em>| = 40\u00b0.<\/p>\n

C-I-3<\/b>
\nM\u00e1me ur\u010dit\u00fd po\u010det krabi\u010diek a ur\u010dit\u00fd po\u010det gu\u013e\u00f4\u010dok. Ak d\u00e1me do ka\u017edej krabi\u010dky pr\u00e1ve jednu gu\u013e\u00f4\u010dku, ostane n\u00e1m n<\/i> gu\u013e\u00f4\u010dok. Ak ale d\u00e1me n<\/i> krabi\u010diek na bok, m\u00f4\u017eeme v\u0161etky gu\u013e\u00f4\u010dky rozmiestni\u0165 do zost\u00e1vaj\u00facich krabi\u010diek tak, \u017ee v ka\u017edej ich bude presne n<\/i>. Ko\u013eko m\u00e1me krabi\u010diek a ko\u013eko gu\u013e\u00f4\u010dok?<\/p>\n

C-I-4<\/b>
\nTangram je sklada\u010dka, ktor\u00e1 sa d\u00e1 vyrobi\u0165 z papiera rozrezan\u00edm vystrihnut\u00e9ho \u0161tvorca na sedem dielov pod\u013ea \u010diar vyzna\u010den\u00fdch na obr\u00e1zku. Predpokladajme, \u017ee d\u013a\u017eka strany \u0161tvorca je  cm.<\/p>\n

<\/p>\n

Rozhodnite, \u010di sa d\u00e1 z dielov tangramu zlo\u017ei\u0165 <\/p>\n

   a)<\/b>  obd\u013a\u017enik 2 cm × 4 cm,
\n   b)<\/b>  obd\u013a\u017enik  cm ×  .<\/p>\n

C-I-5<\/b>
\nV skupine n<\/i> \u013eud\u00ed (n<\/i> ≥ 4) sa niektor\u00ed poznaj\u00fa. Vz\u0165ah \u201epozna\u0165 sa\u201c<\/i> je vz\u00e1jomn\u00fd; ak osoba A<\/b> pozn\u00e1 osobu B<\/b>, tak aj B<\/b> pozn\u00e1 A<\/b>; dvojicu A<\/b>, B<\/b> potom naz\u00fdvame dvojica zn\u00e1mych.<\/p>\n

   a)<\/b>  Dok\u00e1\u017ete, \u017ee ak s\u00fa medzi ka\u017ed\u00fdmi \u0161tyrmi osobami aspo\u0148 \u0161tyri dvojice zn\u00e1mych, potom ka\u017ed\u00e9 dve osoby, ktor\u00e9 sa nepoznaj\u00fa, maj\u00fa spolo\u010dn\u00e9ho zn\u00e1meho.
\n   b)<\/b>  Zistite, pre ktor\u00e9 n<\/i> ≥ 4 existuje skupina os\u00f4b, v ktorej s\u00fa medzi ka\u017ed\u00fdmi \u0161tyrmi osobami aspo\u0148 tri dvojica zn\u00e1mych a s\u00fa\u010dasne sa niektor\u00e9 osoby nepoznaj\u00fa ani nemaj\u00fa spolo\u010dn\u00e9ho zn\u00e1meho.
\n   c)<\/b>  Rozhodnite, \u010di v skupine \u0161iestich os\u00f4b m\u00f4\u017eu by\u0165 v ka\u017edej \u0161tvorici pr\u00e1ve tri dvojice zn\u00e1mych a pr\u00e1ve tri dvojice nezn\u00e1mych.<\/p>\n

C-I-6<\/b>
\nKl\u00e1rka mala na papieri nap\u00edsan\u00e9 trojcifern\u00e9 \u010d\u00edslo. Ke\u010f ho spr\u00e1vne vyn\u00e1sobila deviatimi, dostala \u0161tvorcifern\u00e9 \u010d\u00edslo, ktor\u00e9 za\u010d\u00ednalo tou istou \u010d\u00edslicou ako p\u00f4vodn\u00e9 \u010d\u00edslo, prostredn\u00e9 dve \u010d\u00edslice boli rovnak\u00e9 a posledn\u00e1 \u010d\u00edslica bola s\u00fa\u010dtom \u010d\u00edslic p\u00f4vodn\u00e9ho \u010d\u00edsla. Ktor\u00e9 \u0161trorcifern\u00e9 \u010d\u00edslo mohla Kl\u00e1rka dosta\u0165?<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

C-S-1<\/b>
\nN\u00e1jdite v\u0161etky dvojice prirodzen\u00fdch \u010d\u00edsel a, b<\/i> v\u00e4\u010d\u0161\u00edch ako 1<\/i> tak, aby ich s\u00fa\u010det aj s\u00fa\u010din boli mocniny prvo\u010d\u00edsel.<\/p>\n

C-S-2<\/b>
\nV danom rovnobe\u017en\u00edku ABCD<\/i> je bod E<\/i> stred strany BC<\/i> a bod F<\/i> le\u017e\u00ed vn\u00fatri strany AB<\/i>. Obsah trojuholn\u00edka AFD<\/i> je 15 cm2<\/sup> a obsah trojuholn\u00edka FBE<\/i> je 14 cm2<\/sup>. Ur\u010dte obsah \u0161tvoruholn\u00edka FECD<\/i>.<\/p>\n

C-S-3<\/b>
\nV skupine \u0161iestich \u013eud\u00ed existuje pr\u00e1ve 11 dvoj\u00edc zn\u00e1mych. Vz\u0165ah „pozna\u0165 sa“ je vz\u00e1jomn\u00fd, t.j. ak osoba A pozn\u00e1 osobu B,tak aj B pozn\u00e1 A. Ke\u010f sa ktoko\u013evek zo skupiny dozvie nejak\u00fa spr\u00e1vu, povie ju v\u0161etk\u00fdm svojim zn\u00e1mym. Dok\u00e1\u017ete, \u017ee sa t\u00fdmto sp\u00f4sobom nakoniec spr\u00e1vu dozvedia v\u0161etci.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

C-II-1<\/b>
\nTrojuholn\u00edk ABC<\/i> sp\u013a\u0148a pri zvy\u010dajnom ozna\u010den\u00ed d\u013a\u017eok str\u00e1n podmienku a<\/i> ≤ b<\/i> ≤ c<\/i>. Vp\u00edsan\u00e1 kru\u017enica sa dot\u00fdka str\u00e1n AB<\/i>, BC<\/i> a AC<\/i> postupne v bodoch K<\/i>, L<\/i> a M<\/i>. Dok\u00e1\u017ete, \u017ee z \u00fase\u010diek AK<\/i>, BL<\/i> a CM<\/i> mo\u017eno zostroji\u0165 trojuholn\u00edk pr\u00e1ve vtedy, ke\u010f plat\u00ed b + c < 3a<\/i><\/b>.<\/p>\n

C-II-2<\/b>
\nKl\u00e1rka urobila chybu pri p\u00edsomnom n\u00e1soben\u00ed dvoch dvojcifern\u00fdch \u010d\u00edsel, a tak jej vy\u0161lo \u010d\u00edslo o 400 men\u0161ie, ako bol spr\u00e1vny v\u00fdsledok. Pre kontrolu vydelila \u010d\u00edslo, ktor\u00e9 dostala, men\u0161\u00edm z n\u00e1soben\u00fdch \u010d\u00edsel. Tentoraz po\u010d\u00edtala spr\u00e1vne a vy\u0161iel jej ne\u00fapln\u00fd podiel 67 a zvy\u0161ok 56. Ktor\u00e9 \u010d\u00edsla Kl\u00e1rka n\u00e1sobila?<\/p>\n

C-II-3<\/b>
\nDok\u00e1\u017ete, \u017ee pokia\u013e v skupine \u0161iestich os\u00f4b existuje aspo\u0148 desa\u0165 dvoj\u00edc zn\u00e1mych, tak v nej mo\u017eno n\u00e1js\u0165 tri osoby, ktor\u00e9 sa poznaj\u00fa navz\u00e1jom. Vz\u0165ah „pozna\u0165 sa“ je vz\u00e1jomn\u00fd, t.j. ak osoba A pozn\u00e1 osobu B, tak aj B pozn\u00e1 A. Uk\u00e1\u017ete, \u017ee tak\u00e1 trojica existova\u0165 nemus\u00ed, ak v skupine \u0161iestich os\u00f4b je menej ako desa\u0165 dvoj\u00edc zn\u00e1mych.<\/p>\n

C-II-4<\/b>
\nN\u00e1jdite v\u0161etky trojice cel\u00fdch \u010d\u00edsel x, y, z<\/i>, pre ktor\u00e9 plat\u00ed<\/p>\n

.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-I-1<\/b>
\nN\u00e1jdite v\u0161etky prirodzen\u00e9 \u010d\u00edsla k<\/i>, pre ktor\u00e9 je z\u00e1pis \u010d\u00edsla 6k<\/sup> . 72007-k<\/sup><\/i> v desiatkovej s\u00fastave zakon\u010den\u00fd na<\/p>\n

   a)<\/b>  02;
\n   b)<\/b>  04.<\/p>\n

B-I-2<\/b>
\nV p\u00e1se medzi rovnobe\u017ekami p, q<\/i> s\u00fa dan\u00e9 dva r\u00f4zne body M<\/i> a N<\/i>. Zostrojte koso\u0161tvorec alebo \u0161tvorec, ktor\u00e9ho dve proti\u013eahl\u00e9 strany le\u017eia na priamkach p<\/i> a q<\/i> a body M<\/i> a N<\/i> le\u017eia na zvy\u0161n\u00fdch dvoch stran\u00e1ch (ka\u017ed\u00fd na jednej).<\/p>\n

B-I-3<\/b>
\nNech x<\/i> a y<\/i> s\u00fa re\u00e1lne \u010d\u00edsla, pre ktor\u00e9 plat\u00ed x3<\/sup> + y3<\/sup> ≤ 2<\/i>. Dok\u00e1\u017ete, \u017ee x + y ≤ 2<\/i>.<\/p>\n

B-I-4<\/b>
\nN\u00e1jdite v\u0161etky pravouhl\u00e9 trojuholn\u00edky s d\u013a\u017ekami str\u00e1n a, b, c<\/i> a d\u013a\u017ekami \u0165a\u017en\u00edc ta<\/sub>, tb<\/sub>, tc<\/sub><\/i>, pre ktor\u00e9 plat\u00ed a + ta<\/sub> = b + tb<\/sub><\/i>. Uva\u017eujte oba pr\u00edpady, ke\u010f AB<\/i> je<\/p>\n

   a)<\/b>  prepona,
   b)<\/b>  odvesna.<\/p>\n

B-I-5<\/b>
\nUr\u010dte v\u0161etky dvojice a, b<\/i> re\u00e1lnych \u010d\u00edsel, pre ktor\u00e9 m\u00e1 ka\u017ed\u00e1 z kvadratick\u00fdch rovn\u00edc<\/p>\n

ax2<\/sup> + 2bx +1 = 0,     bx2<\/sup> + 2ax + 1 = 0<\/p>\n

dva r\u00f4zne re\u00e1lne korene, pri\u010dom pr\u00e1ve jeden je obidvom rovniciam spolo\u010dn\u00fd.<\/p>\n

B-I-6<\/b>
\nObd\u013a\u017enik 2 005 × 2 007 je rozdelen\u00fd na \u010dierne a biele jednotkov\u00e9 \u0161tvor\u010deky. Dok\u00e1\u017ete, \u017ee pre jednu z farieb (\u010diernu alebo bielu) existuje viac ako 95 800 pravouholn\u00edkov (zlo\u017een\u00fdch z jednotkov\u00fdch \u0161tvor\u010dekov), ktor\u00e9 sa navz\u00e1jom neprekr\u00fdvaj\u00fa a ktor\u00fdch v\u0161etky rohov\u00e9 \u0161tvor\u010deky maj\u00fa zvolen\u00fa farbu, pri\u010dom ka\u017ed\u00e1 z ich str\u00e1n je tvoren\u00e1 aspo\u0148 dvoma \u0161tvor\u010dekmi.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-S-1<\/b>
\nKe\u010f \u013eubovo\u013en\u00e9 prvo\u010d\u00edslo vydel\u00edme tridsiatimi, bude zvy\u0161kom \u010d\u00edslo 1 alebo prvo\u010d\u00edslo. Dok\u00e1\u017ete.<\/p>\n

B-S-2<\/b>
\nUr\u010dte v\u0161etky dvojice (a, b<\/i>) re\u00e1lnych \u010d\u00edsel, pre ktor\u00e9 maj\u00fa rovnice<\/p>\n

x2<\/sup> + (3a + b)x + 4a = 0;  x2<\/sup> + (3b + a)x + 4b = 0<\/p>\n

spolo\u010dn\u00fd re\u00e1lny kore\u0148.<\/p>\n

B-S-3<\/b>
\nV rovine s\u00fa dan\u00e9 dve rovnobe\u017eky p<\/i> a q<\/i>, bod A<\/i> na priamke p<\/i> a bod M<\/i> le\u017eiaci vn\u00fatri p\u00e1su medzi priamkami p<\/i> a q<\/i>. Zostrojte koso\u0161tvorec alebo \u0161tvorec ABCD<\/i> tak, aby strana AB<\/i> le\u017eala na priamke p<\/i>, strana CD<\/i> na priamke q<\/i> a aby uhloprie\u010dka BD<\/i> prech\u00e1dzala bodom M<\/i>.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-II-1<\/b>
\nUva\u017eujme dve kvadratick\u00e9 rovnice<\/p>\n

x2<\/sup> – ax – b = 0, x2<\/sup> – bx – a = 0<\/p>\n

s re\u00e1lnymi parametrami a, b<\/i>. Zistite, ak\u00fa najmen\u0161iu a ak\u00fa najv\u00e4\u010d\u0161iu hodnotu m\u00f4\u017ee nadobudn\u00fa\u0165 s\u00fa\u010det a + b<\/i>, ak existuje pr\u00e1ve jedno re\u00e1lne \u010d\u00edslo x<\/i>, ktor\u00e9 s\u00fa\u010dasne vyhovuje obom rovniciam. Ur\u010dte \u010falej v\u0161etky dvojice (a, b<\/i>) re\u00e1lnych parametrov, pre ktor\u00e9 tento s\u00fa\u010det tieto hodnoty nadob\u00fada.<\/p>\n

B-II-2<\/b>
\nV trojuholn\u00edku ABC<\/i> m\u00e1 uhol ve\u013ekos\u0165 20\u00b0. Vypo\u010d\u00edtajte ve\u013ekosti uhlov β a γ, ak plat\u00ed rovnos\u0165 a + 2va<\/sub> = b + 2va<\/sub><\/i><\/b>.<\/p>\n

B-II-3<\/b>
\nV rovine je dan\u00fd rovnobe\u017en\u00edk ABCD<\/i>, ktor\u00e9ho uhloprie\u010dka BD<\/i> je kolm\u00e1 na stranu AD<\/i>. Ozna\u010dme M<\/i> (M<\/i>≠A<\/i>) priese\u010dn\u00edk priamky AC<\/i> s kru\u017enicou s priemerom AD<\/i>. Dok\u00e1\u017ete, \u017ee os \u00fase\u010dky BM<\/i> prech\u00e1dza stredom strany CD<\/i>.<\/p>\n

B-II-4<\/b>
\nHokejov\u00fd turnaj sa hr\u00e1 syst\u00e9mom „ka\u017ed\u00fd s ka\u017ed\u00fdm“. V priebehu turnaja sa ka\u017ed\u00e1 dvojica dru\u017estiev stretne pr\u00e1ve raz. Turnaj sa odohr\u00e1va po jednotliv\u00fdch kol\u00e1ch. Pri p\u00e1rnom po\u010dte dru\u017estiev odohr\u00e1 ka\u017ed\u00e9 v jednom kole jedno stretnutie, pri nep\u00e1rnom po\u010dte m\u00e1 v ka\u017edom kole jedno dru\u017estvo vo\u013eno. Za rem\u00edzu dostane ka\u017ed\u00fd zo s\u00faperov po jednom bode. Ak sa stretnutie neskon\u010d\u00ed rem\u00edzou, dostane v\u00ed\u0165az dva body, porazen\u00fd nez\u00edska \u017eiadny bod. O porad\u00ed v tabu\u013eke rozhoduje predov\u0161etk\u00fdm po\u010det bodov, pri rovnosti bodov potom sk\u00f3re. Po odohrat\u00ed nieko\u013ek\u00fdch k\u00f4l nemala \u017eiadna dvojica dru\u017estiev ten ist\u00fd po\u010det bodov. Dok\u00e1\u017ete, \u017ee v tom pr\u00edpade u\u017e posledn\u00fd v tabu\u013eke stratil n\u00e1dej na celkov\u00e9 prvenstvo. \u00dalohu rie\u0161te pre turnaj<\/p>\n

   a)<\/b>  desiatich dru\u017estiev,
\n   b)<\/b>  jeden\u00e1stich dru\u017estiev.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-I-1<\/b>
\nN\u00e1jdite v\u0161etky trojice re\u00e1lnych \u010d\u00edsel a, b, c<\/i> s nasledovnou vlastnos\u0165ou:
\nKa\u017ed\u00e1 z rovn\u00edc<\/p>\n

x3<\/sup> + (a + 1)x2<\/sup> + (b + 3)x + (c + 2) = 0
\nx3<\/sup> + (a + 2)x2<\/sup> + (b + 1)x + (c + 3) = 0
\nx3<\/sup> + (a + 3)x2<\/sup> + (b + 2)x + (c + 1) = 0<\/p>\n

m\u00e1 v obore re\u00e1lnych \u010d\u00edsel tri r\u00f4zne korene, spolu je to ale len p\u00e4\u0165 r\u00f4znych \u010d\u00edsel.<\/p>\n

A-I-2<\/b>
\nV rovine ja dan\u00e1 \u00fase\u010dka AV<\/i> a ostr\u00fd uhol ve\u013ekosti a<\/font>. Ur\u010dte mno\u017einu stredov kru\u017en\u00edc op\u00edsan\u00fdch v\u0161etk\u00fdm t\u00fdm trojuholn\u00edkom ABC<\/i> s vn\u00fatorn\u00fdm uhlom a<\/font> pri vrchole A<\/i>, ktor\u00fdch v\u00fd\u0161ky sa pret\u00ednaj\u00fa v bode V<\/i>.<\/p>\n

A-I-3<\/b>
\nMno\u017einu M<\/b> tvor\u00ed 2n<\/i> navz\u00e1jom r\u00f4znych kladn\u00fdch \u010d\u00edsel, kde n<\/i> ≥ 2. Uva\u017eujme n<\/i> obd\u013a\u017enikov, ktor\u00fdch rozmery s\u00fa \u010d\u00edsla z M<\/b>, pri\u010dom ka\u017ed\u00fd prvok z M<\/b> je pou\u017eit\u00fd pr\u00e1ve jedenkr\u00e1t. Ur\u010dte, ak\u00e9 rozmery maj\u00fa tieto obd\u013a\u017eniky, ak je s\u00fa\u010det ich obsahov<\/p>\n

   a)<\/b>  najv\u00e4\u010d\u0161\u00ed mo\u017en\u00fd;
   b)<\/b>  najmen\u0161\u00ed mo\u017en\u00fd.<\/p>\n

A-I-4<\/b>
\nUr\u010dte po\u010det kone\u010dn\u00fdch rast\u00facich postupnost\u00ed prirodzen\u00fdch \u010d\u00edsel a1<\/sub>, a2<\/sub>, …, ak<\/sub><\/i> v\u0161etk\u00fdch mo\u017en\u00fdch d\u013a\u017eok k<\/i>, pre ktor\u00e9 plat\u00ed a1<\/sub><\/i> = 1, ai<\/sub><\/i> ½<\/font> ai+1<\/sub><\/i> pre v\u0161etky i<\/i> \u00ce<\/font> {1, 2, …, k<\/i> – 1} a ak<\/sub><\/i> = 969 969.<\/p>\n

A-I-5<\/b>
\nJe dan\u00e1 kru\u017enica k<\/i>, bod O<\/i>, ktor\u00fd na nej nele\u017e\u00ed, a priamka p<\/i>, ktor\u00e1 ju nepret\u00edna. Uva\u017eujme \u013eubovo\u013en\u00fa kru\u017enicu l<\/i>, ktor\u00e1 m\u00e1 vonkaj\u0161\u00ed dotyk s kru\u017enicou k<\/i> a dot\u00fdka sa aj priamky p<\/i>. Pr\u00edslu\u0161n\u00e9 body dotyku ozna\u010dme A<\/i> a B<\/i>. Ak body O, A, B<\/i> nele\u017eia na jednej priamke, zostroj\u00edme kru\u017enicu m<\/i> op\u00edsan\u00fa trojuholn\u00edku OAB<\/i>. Dok\u00e1\u017ete, \u017ee v\u0161etky tak\u00e9 kru\u017enice m<\/i> maj\u00fa \u010fal\u0161\u00ed spolo\u010dn\u00fd bod r\u00f4zny od bodu O<\/i> alebo sa dot\u00fdkaj\u00fa tej istej priamky.<\/p>\n

A-I-6<\/b>
\nDok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 prirodzen\u00e9 \u010d\u00edslo n<\/i> existuje prirodzen\u00e9 \u010d\u00edslo a<\/i> tak\u00e9, \u017ee 1 < a<\/i> < 5n<\/sup> a 5n<\/sup> ½<\/font> (a<\/i>3<\/sup> – a<\/i> + 1).<\/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

x2<\/sup> – y = z2<\/sup>,
y2<\/sup> – z = x2<\/sup>,
z2<\/sup> – x = y2<\/sup>.<\/p>\n

A-S-2<\/b>
\nPodstavy hranola s\u00fa tvoren\u00e9 dvoma zhodn\u00fdmi konvexn\u00fdmi n<\/i>-uholn\u00edkmi. Po\u010det v<\/i> vrcholov tohto telesa, po\u010det s<\/i> jeho stenov\u00fdch uhloprie\u010dok a po\u010det t<\/i> jeho telesov\u00fdch uhloprie\u010dok tvoria v istom porad\u00ed prv\u00e9 tri \u010dleny aritmetickej postupnosti. Pre ktor\u00e9 n<\/i> to plat\u00ed?
\n(Pozn\u00e1mka: Steny hranola s\u00fa bo\u010dn\u00e9 steny aj podstavy. Telesov\u00e1 uhloprie\u010dka je \u00fase\u010dka sp\u00e1jaj\u00faca dva vrcholy hranola, ktor\u00e9 nele\u017eia v rovnakej stene.)<\/p>\n

A-S-3<\/b>
\nV rovine je dan\u00fd uhol XSY<\/i> a kru\u017enica k<\/i> so stredom S<\/i>. Uva\u017eujme \u013eubovo\u013en\u00fd trojuholn\u00edk ABC<\/i> s vp\u00edsanou kru\u017enicou k<\/i>, ktor\u00e9ho vrcholy A<\/i> a B<\/i> le\u017eia postupne na polpriamkach SX<\/i> a SY<\/i>. Ur\u010dte mno\u017einu vrcholov C<\/i> v\u0161etk\u00fdch tak\u00fdch trojuholn\u00edkov ABC<\/i>.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-II-1<\/b>
\nN\u00e1jdite v\u0161etky \u0161tvorice p, q, r, s<\/i> navz\u00e1jom r\u00f4znych re\u00e1lnych \u010d\u00edsel, pre ktor\u00e9 s\u00fa p, q<\/i> kore\u0148mi rovnice<\/p>\n

x2<\/sup> + rx + s – 1 = 0<\/p>\n

a r, s<\/i> kore\u0148mi rovnice<\/p>\n

px2<\/sup> +p(q – 1)x + 12 = 0.<\/p>\n

A-II-2<\/b>
\nV tabu\u013eke n<\/i>×n<\/i>, pri\u010dom n<\/i>≥2, s\u00fa po riadkoch nap\u00edsan\u00e9 v\u0161etky \u010d\u00edsla 1, 2, …, n<\/i>2<\/sup> v tomto porad\u00ed (v prvom riadku s\u00fa za sebou nap\u00edsan\u00e9 \u010d\u00edsla 1, 2, …, n<\/i>, v druhom riadku n<\/i>+1, n<\/i>+2, …, 2n<\/i>, at\u010f.). V jednom kroku m\u00f4\u017eeme zvoli\u0165 \u013eubovo\u013en\u00e9 dve \u010d\u00edsla na susedn\u00fdch pol\u00ed\u010dkach (t.j. na tak\u00fdch, ktor\u00e9 maj\u00fa spolo\u010dn\u00fa stranu),a ak je ich aritmetick\u00fd priemer cel\u00e9 \u010d\u00edslo, obe nahrad\u00edme t\u00fdmto priemerom. Pre ktor\u00e9 n<\/i> mo\u017eno po kone\u010dnom po\u010dte krokov dosta\u0165 tabu\u013eku, v ktorej s\u00fa v\u0161etky \u010d\u00edsla rovnak\u00e9?<\/p>\n

A-II-3<\/b>
\nDan\u00fd je ostrouhl\u00fd trojuholn\u00edk ABC s p\u00e4tami v\u00fd\u0161ok D, E, F<\/i> le\u017eiacimi postupne na stran\u00e1ch AB<\/i>, BC<\/i>, CA<\/i>. Obraz bodu F<\/i> v stredovej s\u00famernosti pod\u013ea stredu strany AB<\/i> le\u017e\u00ed na priamke DE<\/i>. Ur\u010dte ve\u013ekos\u0165 uhla BAC<\/i>.<\/p>\n

A-II-4<\/b>
\nDok\u00e1\u017ete, \u017ee pre nez\u00e1porn\u00e9 re\u00e1lne \u010d\u00edsla x, y<\/i> sp\u013a\u0148aj\u00face vz\u0165ah x2<\/sup> + y6<\/sup> = 2<\/b><\/i> plat\u00ed x2<\/sup> + 2 ≥ 3xy<\/b><\/i>.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-III-1<\/b>
\nUr\u010dte koe?cienty p, q, r<\/i> polyn\u00f3mu f(x) = x3<\/sup> + px2<\/sup> + qx + r<\/i>, ak viete, \u017ee s\u00fa to
\nnenulov\u00e9 navz\u00e1jom r\u00f4zne cel\u00e9 \u010d\u00edsla a \u017ee f(p) = p3<\/sup><\/i>, f(q) = q3<\/sup><\/i>. <\/p>\n

A-III-2<\/b>
\nV ostrouhlom trojuholn\u00edku ABC<\/i>, v ktorom |AC<\/i>| = |BC<\/i>|, ozna\u010dme D<\/i> a E<\/i> p\u00e4ty v\u00fd\u0161ok z vrcholov A<\/i> a B<\/i>. Nech V<\/i> je priese\u010dn\u00edk v\u00fd\u0161ok trojuholn\u00edka ABC<\/i>, bod F<\/i> je priese\u010dn\u00edk priamok AB<\/i> a DE<\/i> a bod S<\/i> je stred strany AB<\/i>. \u010ealej nech K<\/i> je priese\u010dn\u00edk kru\u017en\u00edc op\u00edsan\u00fdch trojuholn\u00edkom BDS<\/i> a AES<\/i> r\u00f4zny od bodu S<\/i>.<\/p>\n

   a)<\/b>  Dok\u00e1\u017ete, \u017ee body D, E, V , K<\/i> le\u017eia na jednej kru\u017enici.
\n   b)<\/b>  Dok\u00e1\u017ete, \u017ee body F, V , K<\/i> le\u017eia na jednej priamke.<\/p>\n

A-III-3<\/b>
\nV tabu\u013eke n<\/i> \u00d7 n<\/i>, pri\u010dom n<\/i> = 2, s\u00fa po riadkoch nap\u00edsan\u00e9 v\u0161etky \u010d\u00edsla 1, 2, …, n2<\/sup> v tomto porad\u00ed (v prvom riadku s\u00fa za sebou nap\u00edsan\u00e9 \u010d\u00edsla 1, 2, …, n<\/i>, v druhom riadku n<\/i>+1, n<\/i>+2, …, 2n<\/i>, at\u010f.). V jednom kroku m\u00f4\u017eeme zvoli\u0165 \u013eubovo\u013en\u00e9 dve \u010d\u00edsla na susedn\u00fdch pol\u00ed\u010dkach (t.j. na tak\u00fdch, ktor\u00e9 maj\u00fa spolo\u010dn\u00fa stranu) a bu\u010f obidve s\u00fa\u010dasne zv\u00e4\u010d\u0161i\u0165 o 1 alebo obidve s\u00fa\u010dasne zmen\u0161i\u0165 o 1. Pre ktor\u00e9 n<\/i> mo\u017eno po kone\u010dnom po\u010dte krokov dosta\u0165 tabu\u013eku, v ktorej s\u00fa v\u0161etky \u010d\u00edsla rovn\u00e9 365?<\/p>\n

A-III-4<\/b>
\nDok\u00e1\u017ete, \u017ee pre \u017eiadne prirodzen\u00e9 \u010d\u00edslo n<\/i> nie je \u010d\u00edslo 27n<\/sup> – n27<\/sup> prvo\u010d\u00edslom.<\/p>\n

A-III-5<\/b>
\nNech x, y, z<\/i> s\u00fa kladn\u00e9 re\u00e1lne \u010d\u00edsla, ktor\u00fdch s\u00fa\u010din je 1. Dok\u00e1\u017ete, \u017ee ak k, m<\/i> s\u00fa kladn\u00e9 cel\u00e9 \u010d\u00edsla, pri\u010dom k<\/i> > m<\/i>, tak<\/p>\n

xk<\/sup> + yk<\/sup> + zk<\/sup> ≥ xm<\/sup> + ym<\/sup> + zm<\/sup>.<\/p>\n

A-III-6<\/b>
\nOzna\u010dme zvy\u010dajn\u00fdm sp\u00f4sobom d\u013a\u017eky str\u00e1n a \u0165a\u017en\u00edc dan\u00e9ho trojuholn\u00edka. N\u00e1jdite v\u0161etky mo\u017en\u00e9 hodnoty v\u00fdrazu<\/p>\n

   a)<\/b>            b)<\/b>  .<\/p>\n\n\n
\n
\n<\/td>\n		▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n\n
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57. ro\u010dn\u00edk matematickej olympi\u00e1dy 2007-2008<\/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":[6],"tags":[],"class_list":["post-149","post","type-post","status-publish","format-standard","hentry","category-57-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=149\" \/>\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=\"57. ro\u010dn\u00edk matematickej olympi\u00e1dy 2007-2008\" \/>\n<meta property=\"og:url\" content=\"https:\/\/matematika.besaba.com\/?p=149\" \/>\n<meta property=\"og:site_name\" content=\"Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"article:published_time\" content=\"2010-06-08T11:30:31+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=\"13 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=149\",\"url\":\"https:\/\/matematika.besaba.com\/?p=149\",\"name\":\"Kateg\u00f3rie ABC - 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