{"id":116,"date":"2010-07-28T18:49:13","date_gmt":"2010-07-28T18:49:13","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=116"},"modified":"2010-07-28T18:49:13","modified_gmt":"2010-07-28T18:49:13","slug":"kategoria-abc-2","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=116","title":{"rendered":"Kateg\u00f3rie ABC"},"content":{"rendered":"

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

58. ro\u010dn\u00edk matematickej olympi\u00e1dy 2008-2009<\/p>\n

<\/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>
\nTom\u00e1\u0161, Jakub, Martin a Peter organizovali na n\u00e1mest\u00ed zbierku pre dobro\u010dinn\u00e9 \u00fa\u010dely. Za chv\u00ed\u013eu sa pri nich postupne zastavilo p\u00e4\u0165 okoloid\u00facich. Prv\u00fd dal Tom\u00e1\u0161ovi do jeho pokladni\u010dky 3 Sk, Jakubovi 2 Sk, Martinovi 1 Sk a Petrovi ni\u010d. Druh\u00fd dal jedn\u00e9mu z chlapcov 8 Sk a ostatn\u00fdm trom nedal ni\u010d. Tret\u00ed dal dvom chlapcom po 2 Sk a dvom ni\u010d. \u0160tvrt\u00fd dal dvom chlapcom po 4 Sk a dvom ni\u010d. Piaty dal dvom chlapcom po 8 Sk a dvom ni\u010d. Potom chlapci zistili, \u017ee ka\u017ed\u00fd z nich vyzbieral in\u00fa \u010diastku, pri\u010dom tieto tvoria \u0161tyri po sebe id\u00face prirodzen\u00e9 \u010d\u00edsla. Ktor\u00fd z chlapcov vyzbieral najmenej a ktor\u00fd najviac kor\u00fan?<\/p>\n

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

C-I-2<\/b>
\nPravouhl\u00e9mu trojuholn\u00edku ABC<\/i> s preponou AB<\/i> je op\u00edsan\u00e1 kru\u017enica. P\u00e4ty kolm\u00edc z bodov A<\/i>, B<\/i> na doty\u010dnicu k tejto kru\u017enici v bode C<\/i> ozna\u010dme D<\/i>, E<\/i>. Vyjadrite d\u013a\u017eku \u00fase\u010dky DE<\/i> pomocou d\u013a\u017eok odvesien trojuholn\u00edka ABC<\/i>.<\/p>\n

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

C-I-3<\/b>
\nN\u00e1jdite v\u0161etky \u0161tvorcifern\u00e9 \u010d\u00edsla n<\/i>, ktor\u00e9 maj\u00fa nasleduj\u00face tri vlastnosti: V z\u00e1pise \u010d\u00edsla n<\/i> s\u00fa dve r\u00f4zne cifry, ka\u017ed\u00e1 dvakr\u00e1t. \u010c\u00edslo n<\/i> je delite\u013en\u00e9 siedmimi. \u010c\u00edslo, ktor\u00e9 vznikne oto\u010den\u00edm poradia ci?er \u010d\u00edsla n<\/i>, je tie\u017e \u0161tvorcifern\u00e9 a delite\u013en\u00e9 siedmimi.<\/p>\n

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

C-I-4<\/b>
\nDan\u00fd je konvexn\u00fd p\u00e4\u0165uholn\u00edk ABCDE<\/i>. Na polpriamke BC<\/i> zostrojte tak\u00fd bod G<\/i>, aby obsah trojuholn\u00edka ABG<\/i> bol zhodn\u00fd s obsahom dan\u00e9ho p\u00e4\u0165uholn\u00edka.<\/p>\n

(Lucie R\u016f\u017ei\u010dkov\u00e1)<\/div>\n<\/p>\n

C-I-5<\/b>
\nZ mno\u017einy {1, 2, 3, …, 99} vyberte \u010do najv\u00e4\u010d\u0161\u00ed po\u010det \u010d\u00edsel tak, aby s\u00fa\u010det \u017eiadnych dvoch vybran\u00fdch \u010d\u00edsel nebol n\u00e1sobkom jeden\u00e1stich. (Vysvetlite, pre\u010do zvolen\u00fd v\u00fdber m\u00e1 po\u017eadovan\u00fa vlastnos\u0165 a pre\u010do \u017eiadny v\u00fdber v\u00e4\u010d\u0161ieho po\u010dtu \u010d\u00edsel nevyhovuje.)<\/p>\n

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

C-I-6<\/b>
\nDok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 r\u00f4zne kladn\u00e9 \u010d\u00edsla a<\/i>, b<\/i> plat\u00ed<\/p>\n

.<\/div>\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>
\nDok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 nez\u00e1porn\u00e9 \u010d\u00edsla a, b, c<\/i> plat\u00ed<\/p>\n

(a + bc)(b + ac) ≥ ab(c + 1)2<\/sup><\/p>\n

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

C-S-2<\/b>
\nV pravouhlom trojuholn\u00edku ABC<\/i> ozna\u010d\u00edme P<\/i> p\u00e4tu v\u00fd\u0161ky z vrcholu C<\/i> na preponu AB<\/i>. Priese\u010dn\u00edk \u00fase\u010dky AB<\/i> s priamkou, ktor\u00e1 prech\u00e1dza vrcholom C<\/i> a stredom kru\u017enice vp\u00edsanej trojuholn\u00edku PBC<\/i>, ozna\u010d\u00edme D<\/i>. Dok\u00e1\u017ete, \u017ee \u00fase\u010dky AD<\/i> a AC<\/i> s\u00fa zhodn\u00e9.<\/p>\n

C-S-3<\/b>
\nKe\u010f ist\u00e9 dve prirodzen\u00e9 \u010d\u00edsla v rovnakom porad\u00ed s\u010d\u00edtame, od\u010d\u00edtame, vydel\u00edme a vyn\u00e1sob\u00edme a v\u0161etky \u0161tyri v\u00fdsledky s\u010d\u00edtame, dostaneme 2 009. Ur\u010dte tieto dve \u010d\u00edsla.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

C-II-1<\/b>
\nUva\u017eujme v\u00fdraz<\/p>\n

\"\"<\/p>\n

    a)<\/b>   Dok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 re\u00e1lne \u010d\u00edslo x<\/i> plat\u00ed V(x)<\/i> ≥ 3.
\n    b)<\/b>   N\u00e1jdite najv\u00e4\u010d\u0161iu hodnotu V(x)<\/i>.<\/p>\n

C-II-2<\/b>
\nV pravouhlom trojuholn\u00edku ABC<\/i> ozna\u010d\u00edme P<\/i> p\u00e4tu v\u00fd\u0161ky z vrcholu C<\/i> na preponu AB<\/i> a D, E<\/i> stredy kru\u017en\u00edc vp\u00edsan\u00fdch postupne trojuholn\u00edkom APC, CPB<\/i>. Dok\u00e1\u017ete, \u017ee stred kru\u017enice vp\u00edsanej trojuholn\u00edku ABC<\/i> je priese\u010dn\u00edkom v\u00fd\u0161ok trojuholn\u00edka CDE<\/i>.<\/p>\n

C-II-3<\/b>
\nZ mno\u017einy {1, 2, 3, …, 99} je vybran\u00fdch nieko\u013eko r\u00f4znych \u010d\u00edsel tak, \u017ee s\u00fa\u010det \u017eiadnych troch z nich nie je n\u00e1sobkom deviatich.<\/p>\n

    a)<\/b>   Dok\u00e1\u017ete, \u017ee medzi vybran\u00fdmi \u010d\u00edslami s\u00fa najviac \u0161tyri delite\u013en\u00e9 tromi.
\n    b)<\/b>   Uk\u00e1\u017ete, \u017ee vybran\u00fdch \u010d\u00edsel m\u00f4\u017ee by\u0165 26.<\/p>\n

C-II-4<\/b>
\nPravouhl\u00e9mu trojuholn\u00edku ABC<\/i> s preponou AB<\/i> a obsahom S<\/i> je op\u00edsan\u00e1 kru\u017enica. Doty\u010dnica k tejto kru\u017enici v bode C<\/i> pret\u00edna doty\u010dnice veden\u00e9 bodmi A<\/i> a B<\/i> v bodoch D<\/i> a E<\/i>. Vyjadrite d\u013a\u017eku \u00fase\u010dky DE<\/i> pomocou d\u013a\u017eky c<\/i> prepony a obsahu S<\/i>.<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

B-I-1<\/b>
\nNa tabuli je nap\u00edsan\u00e9 \u0161tvorcifern\u00e9 \u010d\u00edslo delite\u013en\u00e9 \u00f4smimi, ktor\u00e9ho posledn\u00e1 cifra je 8. Keby sme posledn\u00fa cifru nahradili cifrou 7, z\u00edskali by sme \u010d\u00edslo delite\u013en\u00e9 deviatimi. Keby sme v\u0161ak posledn\u00fa cifru nahradili cifrou 9, z\u00edskali by sme \u010d\u00edslo delite\u013en\u00e9 siedmimi. Ur\u010dte \u010d\u00edslo, ktor\u00e9 je nap\u00edsan\u00e9 na tabuli.<\/p>\n

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

B-I-2<\/b>
\nUr\u010dte v\u0161etky trojice (x<\/i>, y<\/i>, z<\/i>) re\u00e1lnych \u010d\u00edsel, pre ktor\u00e9 plat\u00ed<\/p>\n

x2<\/sup> + xy = y2<\/sup> + z2<\/sup>,
z2<\/sup> + zy = y2<\/sup> + x2<\/sup>.<\/p>\n

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

B-I-3<\/b>
\nNa strane BC<\/i>, resp. CD<\/i> rovnobe\u017en\u00edka ABCD<\/i> ur\u010dte body E<\/i>, resp. F<\/i> tak, aby \u00fase\u010dky EF<\/i>, BD<\/i> boli rovnobe\u017en\u00e9 a trojuholn\u00edky ABE<\/i>, AEF<\/i> a AFD<\/i> mali rovnak\u00e9 obsahy.<\/p>\n

(Jaroslav Zhouf)<\/div>\n<\/p>\n

B-I-4<\/b>
\nNa pl\u00e1ne 7\u00d77 hr\u00e1me hru lode. Nach\u00e1dza sa na nej jedna lo\u010f 2\u00d73. M\u00f4\u017eeme sa sp\u00fdta\u0165 na \u013eubovo\u013en\u00e9 pol\u00ed\u010dko pl\u00e1nu, a ak lo\u010f zasiahneme, hra kon\u010d\u00ed. Ak nie, p\u00fdtame sa znova. Ur\u010dte najmen\u0161\u00ed po\u010det ot\u00e1zok, ktor\u00e9 potrebujeme, aby sme s istotou lo\u010f zasiahli.<\/p>\n

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

B-I-5<\/b>
\nTrojuholn\u00edku ABC<\/i> je op\u00edsan\u00e1 kru\u017enica k<\/i>. Os strany AB<\/i> pretne kru\u017enicu k<\/i> v bode K<\/i>, ktor\u00fd le\u017e\u00ed v polrovine opa\u010dnej k polrovine ABC<\/i>. Osi str\u00e1n AC<\/i> a BC<\/i> pretn\u00fa priamku CK<\/i> postupne v bodoch P<\/i> a Q<\/i>. Dok\u00e1\u017ete, \u017ee trojuholn\u00edky AKP<\/i> a KBQ<\/i> s\u00fa zhodn\u00e9.<\/p>\n

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

B-I-6<\/b>
\nN\u00e1jdite v\u0161etky dvojice cel\u00fdch \u010d\u00edsel (m<\/i>, n<\/i>), pre ktor\u00e9 je hodnota v\u00fdrazu<\/p>\n

<\/p>\n

cel\u00e9 kladn\u00e9 \u010d\u00edslo.<\/p>\n

(Vojtech B\u00e1lint)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

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

ax + y = 2,
x – y = 2a,
x + y = 1<\/p>\n

s nezn\u00e1mymi x, y<\/i> a re\u00e1lnym parametrom a<\/i>.<\/p>\n

B-S-2<\/b>
\nPre vn\u00fatorn\u00fd bod P<\/i> strany AB<\/i> ostrouhl\u00e9ho trojuholn\u00edka ABC<\/i> ozna\u010dme K<\/i> a L<\/i> p\u00e4ty kolm\u00edc z bodu P<\/i> na priamky AC<\/i> a BC<\/i>. Zostrojte tak\u00fd bod P<\/i>, pre ktor\u00fd priamka CP<\/i> rozpo\u013euje \u00fase\u010dku KL<\/i>.<\/p>\n

B-S-3<\/b>
\n\u010c\u00edslo nazveme „magick\u00fdm“<\/i> pr\u00e1ve vtedy, ke\u010f sa d\u00e1 vyjadri\u0165 ako s\u00fa\u010det trojcifern\u00e9ho \u010d\u00edsla m<\/i> a trojcifern\u00e9ho \u010d\u00edsla m\u00b4<\/i> zap\u00edsan\u00e9ho rovnak\u00fdmi \u010d\u00edslicami v opa\u010dnom porad\u00ed. Niektor\u00e9 „magick\u00e9“<\/i> \u010d\u00edsla mo\u017eno takto vyjadri\u0165 viacer\u00fdmi sp\u00f4sobmi; napr\u00edklad 1554 = 579 + 975 = 777 + 777. Ur\u010dte v\u0161etky „magick\u00e9“<\/i> \u010d\u00edsla, ktor\u00e9 maj\u00fa tak\u00fdch vyjadren\u00ed m + m\u00b4<\/i> \u010do najviac. (Na poradie m<\/i> a m\u00b4<\/i> neberieme oh\u013ead.)<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

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

x + y = 1,
x – y = a,
-4ax + 4y = z2<\/sup> + 4<\/p>\n

s nezn\u00e1mymi x, y, z a re\u00e1lnym parametrom a.<\/p>\n

B-II-2<\/b>
\nNa pl\u00e1ne 5\u00d75 hr\u00e1me hru lode. Zo \u0161tyroch pol\u00ed\u010dok pl\u00e1nu je vytvoren\u00e1 jedna lo\u010f niektor\u00e9ho z tvarov na obr\u00e1zku. M\u00f4\u017eeme sa sp\u00fdta\u0165 na \u013eubovo\u013en\u00e9 pol\u00ed\u010dko pl\u00e1nu, a ak lo\u010f zasiahneme, hra kon\u010d\u00ed.<\/p>\n

\"\"<\/p>\n

    a)<\/b>   Navrhnite osem pol\u00ed\u010dok, na ktor\u00e9 sa sta\u010d\u00ed sp\u00fdta\u0165, aby sme mali istotu z\u00e1sahu lode.
\n    b)<\/b>   Zd\u00f4vodnite, pre\u010do \u017eiadnych sedem ot\u00e1zok tak\u00fa istotu ned\u00e1va.<\/p>\n

B-II-3<\/b>
\nJe dan\u00fd ostrouhl\u00fd trojuholn\u00edk ABC<\/i>, ktor\u00fd nie je rovnoramenn\u00fd. Ozna\u010dme K<\/i> priese\u010dn\u00edk osi uhla ACB<\/i> s osou strany AB<\/i>. Priamka CK<\/i> pret\u00edna v\u00fd\u0161ky z vrcholov A<\/i> a B<\/i> v bodoch, ktor\u00e9 ozna\u010d\u00edme postupne P<\/i> a Q<\/i>. Predpokladajme, \u017ee trojuholn\u00edky AKP<\/i> a BKQ<\/i> maj\u00fa rovnak\u00fd obsah. Ur\u010dte ve\u013ekos\u0165 uhla ACB<\/i>.<\/p>\n

B-II-4<\/b>
\nK \u013eubovo\u013en\u00e9mu prirodzen\u00e9mu \u010d\u00edslu ur\u010d\u00edme jeho zvy\u0161ky po delen\u00ed ka\u017ed\u00fdm z desiatich prirodzen\u00fdch \u010d\u00edsel 2, 3, 4, …, 11 a t\u00fdchto desa\u0165 zvy\u0161kov (niektor\u00e9 m\u00f4\u017eu by\u0165 nulov\u00e9) s\u010d\u00edtame. Ur\u010dte v\u0161etky tak\u00e9 e\u00edsla men\u0161ie ako 25 000, ktor\u00e9 maj\u00fa uveden\u00fd s\u00fa\u010det \u010do najmen\u0161\u00ed. (Nulu za prirodzen\u00e9 \u010d\u00edslo nepova\u017eujeme).<\/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

2sinx.cos(x + y) + siny = 1,
2siny.cos(y + x) + sinx = 1.<\/p>\n

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

A-I-2<\/b>
\nDan\u00fd je tetivov\u00fd \u0161tvoruholn\u00edk ABCD<\/i>. Dok\u00e1\u017ete, \u017ee spojnica priese\u010dn\u00edkov v\u00fd\u0161ok trojuholn\u00edka ABC<\/i> s priese\u010dn\u00edkom v\u00fd\u0161ok trojuholn\u00edka ABD<\/i> je rovnobe\u017en\u00e1 s priamkou CD<\/i>.<\/p>\n

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

A-I-3<\/b>
\nN\u00e1jdite v\u0161etky dvojice prirodzen\u00fdch \u010d\u00edsel x<\/i>, y<\/i> tak\u00e9, \u017ee    je prvo\u010d\u00edslo.<\/p>\n

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

A-I-4<\/b>
\nUva\u017eujme nekone\u010dn\u00fa aritmetick\u00fa postupnos\u0165<\/p>\n

a, a + d, a + 2d, …,      (*)<\/p>\n

kde a<\/i>, d<\/i> s\u00fa prirodzen\u00e9 (t. j. kladn\u00e9 cel\u00e9) \u010d\u00edsla.<\/p>\n

    a)<\/b>   N\u00e1jdite pr\u00edklad postupnosti (*), ktor\u00e1 obsahuje nekone\u010dne ve\u013ea k<\/i>-tych mocn\u00edn prirodzen\u00fdch \u010d\u00edsel pre v\u0161etky k<\/i> = 2, 3, …
\n    b)<\/b>   N\u00e1jdite pr\u00edklad postupnosti (*), ktor\u00e1 neobsahuje \u017eiadnu k<\/i>-tu mocninu prirodzen\u00e9ho \u010d\u00edsla pre \u017eiadne k<\/i> = 2, 3, …
\n    c)<\/b>   N\u00e1jdite pr\u00edklad postupnosti (*), ktor\u00e1 neobsahuje \u017eiadnu druh\u00fa mocninu prirodzen\u00e9ho \u010d\u00edsla, ale obsahuje nekone\u010dne ve\u013ea tret\u00edch mocn\u00edn prirodzen\u00fdch \u010d\u00edsel.
\n    d)<\/b>   Dok\u00e1\u017ete, \u017ee pre v\u0161etky prirodzen\u00e9 \u010d\u00edsla a<\/i>, d<\/i>, k<\/i> (k<\/i> < 1) plat\u00ed: Postupnos\u0165 (*) bu\u010f neobsahuje \u017eiadnu k<\/i>-tu mocninu prirodzen\u00e9ho \u010d\u00edsla, alebo obsahuje nekone\u010dne ve\u013ea k<\/i>-tych mocn\u00edn prirodzen\u00fdch \u010d\u00edsel.<\/p>\n

(Jaroslav Zhouf)<\/div>\n<\/p>\n

A-I-5<\/b>
\nV ka\u017edom vrchole pravideln\u00e9ho 2008-uholn\u00edka le\u017e\u00ed jedna minca. Vyberieme dve mince a premiestnime ka\u017ed\u00fa z nich do susedn\u00e9ho vrcholu tak, \u017ee jedna sa posunie v smere a druh\u00e1 proti smeru chodu hodinov\u00fdch ru\u010di\u010diek. Rozhodnite, \u010di je mo\u017en\u00e9 t\u00fdmto sp\u00f4sobom v\u0161etky mince postupne presun\u00fa\u0165:<\/p>\n

    a)<\/b>   na 8 k\u00f4pok po 251 minciach,
    b)<\/b>   na 251 k\u00f4pok po 8 minciach.<\/p>\n

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

A-I-6<\/b>
\nDan\u00fd je trojuholn\u00edk ABC<\/i>. Vn\u00fatri str\u00e1n AC<\/i>, BC<\/i> s\u00fa dan\u00e9 body E<\/i>, D<\/i> tak, \u017ee |AE<\/i>| = |BD<\/i>|. Ozna\u010dme M<\/i> stred strany AB<\/i> a P<\/i> priese\u010dn\u00edk priamok AD<\/i> a BE<\/i>. Dok\u00e1\u017ete, \u017ee obraz bodu P<\/i> v stredovej s\u00famernosti so stredom M<\/i> le\u017e\u00ed na osi uhla ACB<\/i>.<\/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

A-S-1<\/b>
\nZistite, pre ktor\u00e9 dvojice kladn\u00fdch cel\u00fdch \u010d\u00edsel m<\/em> a n<\/em> plat\u00ed \"\".<\/p>\n

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

A-S-2<\/b>
\nNech ABC je ostrouhl\u00fd trojuholn\u00edk, v ktorom vn\u00fatorn\u00fd uhol pri vrchole A<\/em> m\u00e1 ve\u013ekos\u0165 45\u00b0. Ozna\u010dme D<\/em> p\u00e4tu v\u00fd\u0161ky z vrcholu C<\/em>. Uva\u017eujme \u010falej \u013eubovo\u013en\u00fd vn\u00fatorn\u00fd bod P<\/em> v\u00fd\u0161ky CD<\/em>. Dok\u00e1\u017ete tvrdenie: Priamky AP<\/em> a BC<\/em> s\u00fa navz\u00e1jom kolm\u00e9 pr\u00e1ve vtedy, ke\u010f \u00fase\u010dky AP<\/em> a BC<\/em> s\u00fa zhodn\u00e9.<\/p>\n

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

A-S-3<\/b>
\nUr\u010dte v\u0161etky prirodzen\u00e9 \u010d\u00edsla, ktor\u00fdmi mo\u017eno kr\u00e1ti\u0165 niektor\u00fd zo zlomkov tvaru \"\", kde p<\/em> a q<\/em> s\u00fa nes\u00fadelite\u013en\u00e9 cel\u00e9 \u010d\u00edsla.<\/p>\n

(Vojtech B\u00e1lint)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

A-II-1<\/b>
\nIst\u00e9 \u0161tvorcifern\u00e9 prirodzen\u00e9 \u010d\u00edslo je delite\u013en\u00e9 siedmimi. Ak zap\u00ed\u0161eme jeho \u010d\u00edslice v opa\u010dnom porad\u00ed, dostaneme v\u00e4\u010d\u0161ie \u0161tvorcifern\u00e9 \u010d\u00edslo, ktor\u00e9 je tie\u013e delite\u013en\u00e9 siedmimi. Navy\u0161e po delen\u00ed \u010d\u00edslom 37 d\u00e1vaj\u00fa obe spomenut\u00e9 \u0161tvorcifern\u00e9 \u010d\u00edsla rovnak\u00fd zvy\u0161ok. Ur\u010dte p\u00f4vodn\u00e9 \u0161tvorcifern\u00e9 \u010d\u00edslo.<\/p>\n

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

A-II-2<\/b>
\nNa odvesn\u00e1ch d\u013a\u017eok a, b<\/em> pravouhl\u00e9ho trojuholn\u00edka le\u017eia postupne stredy dvoch kru\u017en\u00edc ka<\/sub>, kb<\/sub><\/em>. Obe kru\u017enice sa dot\u00fdkaj\u00fa prepony a prech\u00e1dzaj\u00fa vrcholom oproti prepone. Polomery uveden\u00fdch kru\u017en\u00edc ozna\u010dme r<\/font>a<\/sub>, r<\/font>b<\/sub>. Ur\u010dte najv\u00e4\u010d\u0161ie kladn\u00e9 re\u00e1lne \u010d\u00edslo p<\/em> tak\u00e9, \u017ee nerovnos\u0165<\/p>\n

\"\"<\/p>\n

plat\u00ed pre v\u0161etky pravouhl\u00e9 trojuholn\u00edky.<\/p>\n

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

A-II-3<\/b>
\nUr\u010dte ve\u013ekosti vn\u00fatorn\u00fdch uhlov a, b, g<\/em><\/font> trojuholn\u00edka, pre ktor\u00e9 plat\u00ed<\/p>\n

2sin b<\/em><\/font> . sin(a + b<\/em><\/font>) – cos a<\/em><\/font> = 1;
\n2sin g<\/em><\/font> . sin(b + g<\/em><\/font>) – cos b<\/em><\/font> = 0.<\/p>\n

(Jaroslav \u0160vr\u010dekt)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n\n
<\/div>","protected":false},"excerpt":{"rendered":"

58. ro\u010dn\u00edk matematickej olympi\u00e1dy 2008-2009<\/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":[7],"tags":[],"class_list":["post-116","post","type-post","status-publish","format-standard","hentry","category-58-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=116\" \/>\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=\"58. ro\u010dn\u00edk matematickej olympi\u00e1dy 2008-2009\" \/>\n<meta property=\"og:url\" content=\"https:\/\/matematika.besaba.com\/?p=116\" \/>\n<meta property=\"og:site_name\" content=\"Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"article:published_time\" content=\"2010-07-28T18:49:13+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=\"10 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=116\",\"url\":\"https:\/\/matematika.besaba.com\/?p=116\",\"name\":\"Kateg\u00f3rie ABC - 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