{"id":446,"date":"2011-03-14T09:26:49","date_gmt":"2011-03-14T09:26:49","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=446"},"modified":"2011-03-14T09:26:49","modified_gmt":"2011-03-14T09:26:49","slug":"sutazne-ulohy-kategorii-a-b-a-c","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=446","title":{"rendered":"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C"},"content":{"rendered":"

<\/a>54. ro\u010dn\u00edk matematickej olympi\u00e1dy 2004-2005<\/h3>\n\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
\u00a0<\/td>\ncelo\u0161t\u00e1tne kolo<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

C-I-1<\/strong>
\nNech a, b, c, d<\/em> s\u00fa tak\u00e9 re\u00e1lne \u010d\u00edsla, \u017ee a + d = b + c<\/em>. Dok\u00e1\u017ete nerovnos\u0165<\/p>\n

(a – b)(c – d) + (a – c)(b – d) + (d – a)(b – c) \u2265 0.<\/p>\n

C-I-2<\/strong>
\nZistite, pre ktor\u00e9 prirodzen\u00e9 \u010d\u00edsla n \u2265 2<\/em> je mo\u017en\u00e9 z mno\u017einy {1, 2, …, n-1} vybra\u0165 navz\u00e1jom r\u00f4zne p\u00e1rne \u010d\u00edsla tak, aby ich s\u00fa\u010det bol delite\u013en\u00fd \u010d\u00edslom n<\/em><\/strong>.<\/p>\n

C-I-3<\/strong>
\nV \u013eubovo\u013enom konvexnom \u0161tvoruholn\u00edku ABCD<\/em> ozna\u010dme E<\/em> stred strany\u00a0BC<\/em> a F<\/em> stred strany\u00a0AD<\/em>. Dok\u00e1\u017ete, \u017ee trojuholn\u00edky AED<\/em> a BFC<\/em> maj\u00fa rovnak\u00fd obsah pr\u00e1ve vtedy, ke\u010f s\u00fa strany AB<\/em> a CD<\/em> rovnobe\u017en\u00e9.<\/p>\n

C-I-4<\/strong>
\nTri \u0161tvormiestne \u010d\u00edsla k, l, m<\/em> maj\u00fa rovnak\u00fd tvar ABAB<\/em>, t.j. \u010d\u00edslica na mieste jednotiek je rovnak\u00e1 ako \u010d\u00edslica na mieste stoviek a \u010d\u00edslica na mieste desiatok je rovnak\u00e1 ako \u010d\u00edslica na mieste tis\u00edcok. \u010c\u00edslo l<\/em> m\u00e1 \u010d\u00edslicu na mieste jednotiek o\u00a02 v\u00e4\u010d\u0161iu a \u010d\u00edslicu na mieste desiatok o\u00a01 men\u0161iu ako \u010d\u00edslo k<\/em>. \u010c\u00edslo m<\/em> je s\u00fa\u010dtom \u010d\u00edsel k<\/em> a l<\/em> a je delite\u013en\u00e9 deviatimi. Ur\u010dte v\u0161etky tak\u00e9 \u010d\u00edsla k<\/em>.<\/p>\n

C-I-5<\/strong>
\nUr\u010dte po\u010det v\u0161etk\u00fdch troj\u00edc dvojmiestnych prirodzen\u00fdch \u010d\u00edsel a, b, c<\/em>, ktor\u00fdch s\u00fa\u010din abc<\/em> m\u00e1 z\u00e1pis, v ktorom s\u00fa v\u0161etky \u010d\u00edslice rovnak\u00e9. Trojice l\u00ed\u0161iace sa len porad\u00edm \u010d\u00edsel pova\u017eujeme za rovnak\u00e9, t.j. zapo\u010d\u00edtavame ich len raz.<\/p>\n

C-I-6<\/strong>
\nV trojuholn\u00edku ABC<\/em> so stranou BC<\/em> d\u013a\u017eky 2\u00a0cm je bod K<\/em> stredom strany AB<\/em>. Body L<\/em> a M<\/em> rozde\u013euj\u00fa stranu AC<\/em> na tri zhodn\u00e9 \u00fase\u010dky. Trojuholn\u00edk KLM<\/em> je rovnoramenn\u00fd a pravouhl\u00fd. Ur\u010dte d\u013a\u017eky str\u00e1n AB<\/em>, AC<\/em> v\u0161etk\u00fdch tak\u00fdch trojuholn\u00edkov ABC<\/em>.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C-S-1<\/strong>
\nN\u00e1jdite v\u0161etky trojice cel\u00fdch \u010d\u00edsel x, y, z<\/strong>, pre ktor\u00e9 plat\u00ed<\/p>\n

x + yz = 2005,
\ny + xz = 2006.<\/p>\n

C-S-2<\/strong>
\nPre ktor\u00e9 prirodzen\u00e9 \u010d\u00edsla n<\/em> mo\u017eno z mno\u017einy {n, n+1, n+2, …, n2<\/sup><\/em>} vybra\u0165 \u0161tyri navz\u00e1jom r\u00f4zne \u010d\u00edsla a, b, c, d<\/em> tak, aby platilo ab = cd<\/em> ?<\/p>\n

C-S-3<\/strong>
\nJe dan\u00e1 \u00fase\u010dka AB<\/em>. Zostrojte bod C<\/em> tak, aby sa obsah trojuholn\u00edka ABC<\/em> rovnal 1\/8 obsahu S \u0161tvorca so stranou AB<\/em> a s\u00fa\u010det obsahov \u0161tvorcov so stranami AC<\/em> a BC<\/em> sa rovnal S.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C-II-1<\/strong>
\nUr\u010dte \u010d\u00edslice x, y, z<\/em> tak, aby platila rovnos\u0165<\/p>\n

\"rovnica\",<\/p>\n

kde \u00a0\"\"\u00a0 ozna\u010duje \u010d\u00edslo zlo\u017een\u00e9 zo z<\/em> jednotiek, y<\/em> desat\u00edn a x<\/em> stot\u00edn.<\/p>\n

C-II-2<\/strong>
\nKu ka\u017ed\u00e9mu prirodzen\u00e9mu \u010d\u00edslu n\u00a0>\u00a02 n\u00e1jdite aspo\u0148 jednu dvojicu r\u00f4znych prirodzen\u00fdch \u010d\u00edsel p, q<\/em> tak, aby \u010d\u00edslo 1\/n<\/em> bolo aritmetick\u00fdm priemerom \u010d\u00edsel 1\/p<\/em> a 1\/q<\/em>.<\/p>\n

C-II-3<\/strong>
\n\u013dubovo\u013en\u00fdm vn\u00fatorn\u00fdm bodom P<\/em> uhloprie\u010dky AC<\/em> dan\u00e9ho obd\u013a\u017enika ABCD<\/em> s\u00fa veden\u00e9 rovnobe\u017eky s jeho stranami tak, \u017ee pret\u00ednaj\u00fa \u00fase\u010dky AB, BC, CD<\/em> a DA<\/em> postupne v bodoch K, L, M<\/em> a N<\/em>. Dok\u00e1\u017ete, \u017ee<\/p>\n

\u00a0\u00a0\u00a0\u00a0a)<\/strong>\u00a0\u00a0priamky LM<\/em> a KN<\/em> s\u00fa rovnobe\u017eky,
\n\u00a0\u00a0\u00a0\u00a0b)<\/strong>\u00a0\u00a0vzdialenos\u0165 rovnobe\u017eiek LM<\/em> a KN<\/em> je kon\u0161tantn\u00e1 (nez\u00e1vis\u00ed na vo\u013ebe bodu P<\/em>),
\n\u00a0\u00a0\u00a0\u00a0c)<\/strong>\u00a0\u00a0pre obvod o<\/em> \u0161tvoruholn\u00edka KLMN<\/em> plat\u00ed nerovnos\u0165 o<\/em>\u00a0\u2265\u00a02|AC<\/em>|.<\/p>\n

C-II-4<\/strong>
\nPop\u00ed\u0161te kon\u0161trukciu lichobe\u017en\u00edka ABCD<\/em> so z\u00e1klad\u0148ami AB<\/em> a CD<\/em>, ktor\u00e9mu sa d\u00e1 op\u00edsa\u0165 kru\u017enica s polomerom r<\/em>\u00a0=\u00a05\u00a0cm, ke\u010f je dan\u00e1 vzdialenos\u0165 d<\/em>\u00a0=\u00a02\u00a0cm jej stredu od priese\u010dn\u00edka uhloprie\u010dok a |<\u00a0BAC<\/em>|\u00a0=\u00a070\u00b0.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

B-I-1<\/strong>
\nUr\u010dte v\u0161etky dvojice (a, b<\/em>) re\u00e1lnych \u010d\u00edsel, pre ktor\u00e9 m\u00e1 ka\u017ed\u00e1 z rovn\u00edc<\/p>\n

x2<\/sup> + ax + b = 0,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + (2a + 1)x + 2b + 1 = 0<\/p>\n

dva r\u00f4zne re\u00e1lne korene, pri\u010dom korene druhej rovnice s\u00fa prevr\u00e1ten\u00fdmi hodnotami kore\u0148ov prvej rovnice.<\/p>\n

B-I-2<\/strong>
\nDan\u00fd je rovnobe\u017en\u00edk ABCD<\/em>. Priamka veden\u00e1 bodom D<\/em> pret\u00edna \u00fase\u010dku AC<\/em> v bode G<\/em>, \u00fase\u010dku BC<\/em> v bode F<\/em> a polpriamku AB<\/em> v bode E<\/em> tak, \u017ee trojuholn\u00edky BEF<\/em> a CGF<\/em> maj\u00fa rovnak\u00fd obsah. Ur\u010dte pomer |AG|\u00a0:\u00a0|GC|.<\/p>\n

B-I-3<\/strong>
\nNa stole le\u017e\u00ed k<\/em> hrom\u00e1dok o 1, 2, 3, …, k<\/em> kame\u0148och, kde k\u00a0\u2265\u00a03<\/em>. V ka\u017edom kroku vyberieme tri \u013eubovo\u013en\u00e9 hrom\u00e1dky na stole, zl\u00fa\u010dime ich do jednej a prid\u00e1me k nej jeden kame\u0148, ktor\u00fd dovtedy na stole nebol. Dok\u00e1\u017ete, \u017ee ak po nieko\u013ek\u00fdch krokoch vznikne jedin\u00e1 hrom\u00e1dka, potom v\u00fdsledn\u00fd po\u010det kame\u0148ov nie je delite\u013en\u00fd tromi.<\/p>\n

B-I-4<\/strong>
\nOzna\u010dme V<\/em> priese\u010dn\u00edk v\u00fd\u0161ok a S<\/em> stred kru\u017enice op\u00edsanej trojuholn\u00edku ABC<\/em>, ktor\u00fd nie je rovnostrann\u00fd. Dok\u00e1\u017ete, \u017ee ak uhol pri vrchole C<\/em> m\u00e1 60\u00b0, potom os uhla ACB je osou \u00fase\u010dky VS<\/em>.<\/p>\n

B-I-5<\/strong>
\nV obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te rovnicu<\/p>\n

\"rovnica\",<\/p>\n

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

B-I-6<\/strong>
\nDo kru\u017enice k<\/em> s polomerom r<\/em> s\u00fa vp\u00edsan\u00e9 dve kru\u017enice k1<\/sub>, k2<\/sub><\/em> s polomerom \"\", ktor\u00e9 sa vz\u00e1jomne dot\u00fdkaj\u00fa. Kru\u017enica l<\/em> sa zvonka dot\u00fdka kru\u017en\u00edc k1<\/sub>, k2<\/sub><\/em> a s kru\u017enicou k<\/em> m\u00e1 vn\u00fatorn\u00fd dotyk. Kru\u017enica m<\/em> m\u00e1 vonkaj\u0161\u00ed dotyk s kru\u017enicami k2<\/sub>, l<\/em> a vn\u00fatorn\u00fd dotyk s kru\u017enicou k<\/em>. Vypo\u010d\u00edtajte polomery kru\u017en\u00edc l<\/em> a m<\/em>.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

B-S-1<\/strong>
\nNa stole le\u017e\u00ed 54 k\u00f4pok s 1, 2, 3, …, 54 kame\u0148mi. V ka\u017edom kroku vyberieme \u013eubovo\u013en\u00fa k\u00f4pku, povedzme s k<\/em> kame\u0148mi, a odoberieme ju cel\u00fa zo stola spolu s k<\/em> kame\u0148mi z ka\u017edej tej k\u00f4pky, v ktorej je aspo\u0148 k<\/em> kame\u0148ov. Napr\u00edklad po prvom kroku, v ktorom vyberieme k\u00f4pku s 52 kame\u0148mi, zostan\u00fa na stole k\u00f4pky s 1, 2, 3, …, 51, 1 a 2 kame\u0148mi. Predpokladajme, \u017ee po ur\u010ditom po\u010dte krokov zostane na stole jedin\u00e1 k\u00f4pka.
\nZd\u00f4vodnite, ko\u013eko kame\u0148ov v nej m\u00f4\u017ee by\u0165.<\/p>\n

B-S-2<\/strong>
\nNech ABC je pravouhl\u00fd trojuholn\u00edk so stranami a<\/em>\u00a0<\u00a0b<\/em>\u00a0<\u00a0c<\/em>. Ozna\u010dme Q<\/em> stred odvesny BC<\/em> a S<\/em> stred prepony AB<\/em>. Priese\u010dn\u00edk osi \u00fase\u010dky AB<\/em> s odvesnou CA<\/em> ozna\u010dme R<\/em>. Dok\u00e1\u017ete, \u017ee |RQ<\/em>| = |RS<\/em>| pr\u00e1ve vtedy, ke\u010f<\/p>\n

a2<\/sup> : b2<\/sup> : c2<\/sup> = 1 : 2 : 3<\/p>\n

B-S-3<\/strong>
\nV obore re\u00e1lnych \u010d\u00edsel rie\u0161te rovnicu<\/p>\n

\"rovnica\"<\/p>\n

kde \"\" ozna\u010duje najv\u00e4\u010d\u0161ie cel\u00e9 \u010d\u00edslo, ktor\u00e9 neprevy\u0161uje \u010d\u00edslo a<\/em>.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

B-II-1<\/strong>
\nKru\u017enica k<\/em>1<\/sub> s polomerom 1 m\u00e1 vonkaj\u0161\u00ed dotyk s kru\u017enicou k<\/em>2<\/sub> s polomerom 2. Ka\u017ed\u00e1 z kru\u017en\u00edc k<\/em>1<\/sub>, k<\/em>2<\/sub> m\u00e1 vn\u00fatorn\u00fd dotyk s kru\u017enicou k<\/em>3<\/sub> s polomerom 3. Vypo\u010d\u00edtajte polomer kru\u017enice k<\/em>, ktor\u00e1 m\u00e1 s kru\u017enicami k<\/em>1<\/sub>, k<\/em>2<\/sub> vonkaj\u0161\u00ed dotyk a s kru\u017enicou k<\/em>3<\/sub> vn\u00fatorn\u00fd dotyk.<\/p>\n

B-II-2<\/strong>
\nNa jednej internetovej str\u00e1nke prebieha hlasovanie o najlep\u0161ieho hokejistu sveta posledn\u00e9ho desa\u0165ro\u010dia. Po\u010det hlasov pre jednotliv\u00fdch hr\u00e1\u010dov sa uv\u00e1dza po zaokr\u00fahlen\u00ed v cel\u00fdch percent\u00e1ch. Po Jo\u017ekovom hlasovan\u00ed pre Miroslava \u0160atana sa jeho zisk 7% nezmenil. Najmenej ko\u013eko \u013eud\u00ed vr\u00e1tane Jo\u017eka hlasovalo? Predpoklad\u00e1me, \u017ee ka\u017ed\u00fd \u00fa\u010dastn\u00edk ankety hlasoval pr\u00e1ve raz, a to pre jedin\u00e9ho hr\u00e1\u010da.<\/p>\n

B-II-3<\/strong>
\nNech ABC je ostrouhl\u00fd trojuholn\u00edk. Ozna\u010dme K a L p\u00e4ty v\u00fd\u0161ok z vrcholov A a B, M stred strany AB a V priese\u010dn\u00edk v\u00fd\u0161ok trojuholn\u00edka ABC. Dok\u00e1\u017ete, \u017ee os uhla KML prech\u00e1dza stredom \u00fase\u010dky VC.<\/p>\n

B-II-4<\/strong>
\nN\u00e1jdite v\u0161etky trojice re\u00e1lnych \u010d\u00edsel x, y, z, pre ktor\u00e9 plat\u00ed<\/p>\n

\"rovnica\"\u00a0,<\/p>\n

kde \"\" ozna\u010duje najv\u00e4\u010d\u0161ie cel\u00e9 \u010d\u00edslo, ktor\u00e9 neprevy\u0161uje \u010d\u00edslo a<\/em>.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

A-I-1<\/strong>
\nNepr\u00e1zdnu podmno\u017einu prirodzen\u00fdch \u010d\u00edsel nazveme malou<\/em>, ke\u010f m\u00e1 menej prvkov, ako je jej najmen\u0161\u00ed prvok. Ur\u010dte po\u010det v\u0161etk\u00fdch t\u00fdch mal\u00fdch mno\u017e\u00edn M<\/em>, ktor\u00e9 s\u00fa podmno\u017einami mno\u017einy {1,\u00a02,\u00a03,\u00a0…,\u00a0100} a maj\u00fa nasledovn\u00fa vlastnos\u0165: ak do M<\/em> patria dve r\u00f4zne \u010d\u00edsla x<\/em> a y<\/em>, potom do M<\/em> patr\u00ed aj \u010d\u00edslo |x\u00a0–\u00a0y<\/em>|.<\/p>\n

A-I-2<\/strong>
\nNech M<\/em> je \u013eubovo\u013en\u00fd vn\u00fatorn\u00fd bod krat\u0161ieho obl\u00faka CD<\/em> kru\u017enice op\u00edsanej \u0161tvorcu ABCD<\/em>. Ozna\u010dme P<\/em>, R<\/em> priese\u010dn\u00edky priamky AM<\/em> postupne s\u00a0\u00fase\u010dkami BD<\/em>, CD<\/em> a podobne Q<\/em>, S<\/em> priese\u010dn\u00edky priamky BM<\/em> s\u00a0\u00fase\u010dkami AC<\/em>, DC<\/em>. Dok\u00e1\u017ete, \u017ee priamky PS<\/em> a QR<\/em> s\u00fa navz\u00e1jom kolm\u00e9.<\/p>\n

A-I-3<\/strong>
\nNech k<\/em> je \u013eubovo\u013en\u00e9 prirodzen\u00e9 \u010d\u00edslo. Uva\u017eujme dvojice (a, b)<\/em> cel\u00fdch \u010d\u00edsel, pre ktor\u00e9 maj\u00fa kvadratick\u00e9 rovnice<\/p>\n

x2<\/sup> – 2ax + b = 0,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y2<\/sup> + 2ay + b = 0<\/p>\n

re\u00e1lne korene (nie nutne r\u00f4zne), ktor\u00e9 mo\u017eno ozna\u010di\u0165 x1,2<\/sub><\/em> resp. y1,2<\/sub><\/em> v takom porad\u00ed, \u017ee plat\u00ed rovnos\u0165 x1<\/sub>y1<\/sub>\u00a0–\u00a0x2<\/sub>y2<\/sub>\u00a0=\u00a04k<\/em><\/strong>.<\/p>\n

\u00a0\u00a0\u00a0\u00a0a)<\/strong>\u00a0\u00a0Pre dan\u00e9 k<\/em> ur\u010dte najv\u00e4\u010d\u0161iu mo\u017en\u00fa hodnotu b<\/em> zo v\u0161etk\u00fdch tak\u00fdch dvoj\u00edc (a, b)<\/em>.
\n\u00a0\u00a0\u00a0\u00a0b)<\/strong>\u00a0\u00a0Pre k<\/em> = 2004 ur\u010dte po\u010det v\u0161etk\u00fdch tak\u00fdch dvoj\u00edc (a, b)<\/em>.
\n\u00a0\u00a0\u00a0\u00a0c)<\/strong>\u00a0\u00a0Pre dan\u00e9 k<\/em> vypo\u010d\u00edtajte s\u00fa\u010det \u010d\u00edsel b<\/em> zo v\u0161etk\u00fdch tak\u00fdch dvoj\u00edc (a, b)<\/em>, pri\u010dom ka\u017ed\u00e9 \u010d\u00edslo b<\/em> sa pripo\u010d\u00edta to\u013ekokr\u00e1t, v ko\u013ek\u00fdch dvojiciach (a, b)<\/em> vystupuje.<\/p>\n

A-I-4<\/strong>
\nDan\u00e9 aritmetick\u00e9 postupnosti \"\" a \"\" maj\u00fa rovnak\u00fd prv\u00fd \u010dlen a nasledovn\u00fa vlastnos\u0165: existuje index k<\/em> (k<\/em>\u00a0>\u00a01), pre ktor\u00fd platia rovnosti<\/p>\n

\"rovnice\".<\/p>\n

N\u00e1jdite v\u0161etky tak\u00e9 indexy k<\/em>.<\/p>\n

A-I-5<\/strong>
\nV lichobe\u017en\u00edku ABCD<\/em>, kde AB<\/em>\u2551CD<\/em>, plat\u00ed |AB<\/em>|\u00a0=\u00a02|CD<\/em>|. Ozna\u010dme E<\/em> stred ramena BC<\/em>. Dok\u00e1\u017ete, \u017ee rovnos\u0165 |AB<\/em>|\u00a0=\u00a0|BC<\/em>| plat\u00ed pr\u00e1ve vtedy, ke\u010f \u0161tvoruholn\u00edk AECD<\/em> je doty\u010dnicov\u00fd.<\/p>\n

A-I-6<\/strong>
\nN\u00e1jdite v\u0161etky funkcie f<\/em>: <0,\u00a0+\u221e)\u00a0\u2192\u00a0<0,\u00a0+\u221e), ktor\u00e9 spl\u0148uj\u00fa z\u00e1rove\u0148 tri nasledovn\u00e9 podmienky:<\/p>\n

\u00a0\u00a0\u00a0\u00a0a)<\/strong>\u00a0\u00a0Pre \u013eubovo\u013en\u00e9 nez\u00e1porn\u00e9 \u010d\u00edsla x, y<\/em> tak\u00e9, \u017ee x\u00a0+\u00a0y\u00a0>\u00a00<\/em>, plat\u00ed rovnos\u0165\u00a0\u00a0\"A-I-5\";
\n\u00a0\u00a0\u00a0\u00a0b)<\/strong>\u00a0\u00a0f<\/em>(1)\u00a0=\u00a00;
\n\u00a0\u00a0\u00a0\u00a0c)<\/strong>\u00a0\u00a0f<\/em>(x<\/em>)\u00a0>\u00a00 pre \u013eubovo\u013en\u00e9 x<\/em>\u00a0>\u00a01.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

A-S-1<\/strong>
\nUr\u010dte po\u010det v\u0161etk\u00fdch nekone\u010dn\u00fdch aritmetick\u00fdch postupnost\u00ed cel\u00fdch \u010d\u00edsel, ktor\u00e9 maj\u00fa medzi svojimi prv\u00fdmi desiatimi \u010dlenmi obe \u010d\u00edsla 1 a 2\u00a0005.<\/p>\n

A-S-2<\/strong>
\nV rovnobe\u017en\u00edku ABCD<\/em> plat\u00ed |AB<\/em>|\u00a0>\u00a0|BC<\/em>|. Ozna\u010dme K, L, M<\/em> a N<\/em> postupne body dotyku kru\u017en\u00edc vp\u00edsan\u00fdch trojuholn\u00edkom ACD, BCD, ABC<\/em> a ABD<\/em> s pr\u00edslu\u0161nou uhloprie\u010dkou AC<\/em>, resp. BD<\/em>. Dok\u00e1\u017ete, \u017ee KLMN<\/em> je obd\u013a\u017enik.<\/p>\n

A-S-3<\/strong>
\nZistite, pre ktor\u00e9 prirodzen\u00e9 \u010d\u00edsla k<\/em> m\u00e1 s\u00fastava nerovn\u00edc<\/p>\n

\"A-S-3\"<\/p>\n

s nezn\u00e1mou x<\/em> pr\u00e1ve (k + 1)2<\/sup><\/em> rie\u0161en\u00ed v obore cel\u00fdch \u010d\u00edsel.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

A-II-1<\/strong>
\nAk je s\u00fa\u010din kladn\u00fdch re\u00e1lnych \u010d\u00edsel a, b, c<\/em> rovn\u00fd 1, plat\u00ed nerovnos\u0165<\/p>\n

\"A-II-1\"<\/p>\n

Dok\u00e1\u017ete a zistite, kedy nast\u00e1va rovnos\u0165.<\/p>\n

A-II-2<\/strong>
\nV obore cel\u00fdch \u010d\u00edsel rie\u0161te s\u00fastavu rovn\u00edc<\/p>\n

x(y + z + 1) = y2<\/sup> + z2<\/sup> – 5,
\ny(z + x + 1) = z2<\/sup> + x2<\/sup> – 5,
\nz(x + y + 1) = x2<\/sup> + y2<\/sup> – 5.<\/p>\n

A-II-3<\/strong>
\nV rovine je dan\u00fd rovnoramenn\u00fd trojuholn\u00edk KLM<\/em> so z\u00e1klad\u0148ou KL<\/em>. Uva\u017eujme \u013eubovo\u013en\u00e9 dve kru\u017enice k<\/em> a l<\/em>, ktor\u00e9 maj\u00fa vonkaj\u0161\u00ed dotyk a ktor\u00e9 sa dot\u00fdkaj\u00fa priamok KM<\/em> a LM<\/em> postupne v bodoch K<\/em> a L<\/em>. Ur\u010dte mno\u017einu dotykov\u00fdch bodov T<\/em> v\u0161etk\u00fdch tak\u00fdch kru\u017en\u00edc k<\/em> a l<\/em>.<\/p>\n

A-II-4<\/strong>
\nN\u00e1jdite v\u0161etky dvojice prirodzen\u00fdch \u010d\u00edsel, ktor\u00fdch s\u00fa\u010det m\u00e1 posledn\u00fa \u010d\u00edslicu 3, rozdiel je prvo\u010d\u00edslo a s\u00fa\u010din je druhou mocninou prirodzen\u00e9ho \u010d\u00edsla.<\/p>\n\n\n\n
\n
\n<\/td>\n		<\/a>\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

A-III-1<\/strong>
\nUva\u017eujme \u013eubovo\u013en\u00e9 aritmetick\u00e9 postupnosti re\u00e1lnych \u010d\u00edsel(xi)1 i=1 a(yi)1 i=1,ktor\u00e9 maj\u00farovnak\u00fdprv\u00fd\u00e8lenasp\u00e5\u00f2aj\u00fapreniektor\u00e9 k> 1rovnosti xk 1yk 1 =42;xkyk =30a xk+1yk+1 =16: N\u00e1jditev\u00b9etkytak\u00e9postupnosti,prektor\u00e9jeindex k najv\u00e4\u00e8\u00b9\u00edmo\u00ben\u00fd.(J.\u00a9im\u00b9a) 2. Zistite,prektor\u00e9 m existujepr\u00e1ve215 podmno\u00be\u00edn X mno\u00beiny f1; 2; 3;:::; 47g svlast- nos\u00bbou:\u00c8\u00edslo m jenajmen\u00b9\u00edprvokmno\u00beiny X apreka\u00bed\u00e9 x 2 X plat\u00edbu\u00ef x + m 2 X, alebo x + m> 47.(R.Ku\u00e8era) 3. Vlichobe\u00ben\u00edku ABCD (AB k CD)ozna\u00e8me E stredramena BC.Aks\u00faoba \u00b9tvoruholn\u00edky ABED a AECD doty\u00e8nicov\u00e9,sp\u00e5\u00f2aj\u00fad\u00e5\u00bekystr\u00e1nlichobe\u00ben\u00edka ABCD ozna\u00e8en\u00e9zvy\u00e8ajn\u00fdmsp\u00f4sobomrovnosti a + c = b 3 + d a 1 a + 1 c = 3 b : Dok\u00e1\u00bete.(R.Horensk\u00fd) 4. Vrovinejedan\u00fdostrouhl\u00fdtrojuholn\u00edk AKL.Uva\u00beujme\u00b5ubovo\u00b5n\u00fdpravouholn\u00edk ABCD,ktor\u00fdjetrojuholn\u00edku AKL op\u00edsan\u00fdtak,\u00beebod K le\u00be\u00ednastrane BC abod L le\u00be\u00ednastrane CD.Ur\u00e8temno\u00beinupriese\u00e8n\u00edkov S uhloprie\u00e8ok AC, BD v\u00b9etk\u00fdchtak\u00fdch pravouholn\u00edkov ABCD.(J.\u00a9im\u00b9a) 5<\/p>\n

A-III-5<\/strong>
\nDok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 re\u00e1lne \u010d\u00edsla p, q, r, s<\/em> za podmienok q<\/em> = 1 a s<\/em> = 1 plat\u00ed: Kvadratick\u00e9 rovnice x<\/em>2<\/sup> + px<\/em> + q<\/em> =0, x<\/em>2<\/sup> + rx<\/em> + s<\/em> =0 maj\u00fa v obore re\u00e1lnych \u010d\u00edsel spolo\u010dn\u00fd kore\u0148 a ich \u010fal\u0161ie korene s\u00fa navz\u00e1jom prevr\u00e1ten\u00e9 \u010d\u00edsla pr\u00e1ve vtedy, ke\u010f koeficienty p, q, r, s<\/em> sp\u013a\u0148aj\u00fa rovnosti pr<\/em> =(q<\/em> +1)(s<\/em> +1) a p<\/em>(q<\/em> +1)s<\/em> = r<\/em>(s<\/em> +1)q.<\/em> (Dvojn\u00e1sobn\u00fd kore\u0148 kvadratickej rovnice po\u010d\u00edtame dvakr\u00e1t.)(J. \u0160im\u0161a)<\/p>\n

A-III-6
\n<\/strong>Rozhodnite, \u010di pre ka\u017ed\u00e9 poradie \u010d\u00edsel 1, \u00a02, \u00a03, …, 15 mo\u017eno tieto \u010d\u00edsla zap\u00edsa\u0165 najviac \u0161tyrmi r\u00f4znymi farbami tak, aby v\u0161etky \u010d\u00edsla rovnakej farby tvorili v danom porad\u00ed monot\u00f3nnu (t.j. rast\u00facu alebo klesaj\u00facu) postupnos\u0165.
\n(Jedno\u010dlenn\u00e1 postupnos\u0165 je monot\u00f3nna.)<\/em>(J. \u0160im\u0161a)<\/p>\n\n\n\n
\n
\n<\/td>\n		\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n
<\/div>","protected":false},"excerpt":{"rendered":"

54. ro\u010dn\u00edk matematickej olympi\u00e1dy 2004-2005 C B A dom\u00e1ce kolo dom\u00e1ce kolo dom\u00e1ce kolo \u0161kolsk\u00e9 kolo \u0161kolsk\u00e9 kolo \u0161kolsk\u00e9 kolo krajsk\u00e9 kolo krajsk\u00e9 kolo krajsk\u00e9 kolo \u00a0 celo\u0161t\u00e1tne kolo C-I-1 Nech a, b, c, d s\u00fa tak\u00e9 re\u00e1lne \u010d\u00edsla, \u017ee a + d = b + c. Dok\u00e1\u017ete nerovnos\u0165 (a – b)(c – d) + … Pokra\u010dova\u0165 na S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C<\/span> →<\/span><\/a><\/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":[3],"tags":[],"class_list":["post-446","post","type-post","status-publish","format-standard","hentry","category-54-rocnik"],"yoast_head":"\nS\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C - 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=446\" \/>\n<meta property=\"og:locale\" content=\"sk_SK\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C - Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"og:description\" content=\"54. ro\u010dn\u00edk matematickej olympi\u00e1dy 2004-2005 C B A dom\u00e1ce kolo dom\u00e1ce kolo dom\u00e1ce kolo \u0161kolsk\u00e9 kolo \u0161kolsk\u00e9 kolo \u0161kolsk\u00e9 kolo krajsk\u00e9 kolo krajsk\u00e9 kolo krajsk\u00e9 kolo \u00a0 celo\u0161t\u00e1tne kolo C-I-1 Nech a, b, c, d s\u00fa tak\u00e9 re\u00e1lne \u010d\u00edsla, \u017ee a + d = b + c. 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