{"id":662,"date":"2012-07-26T18:20:50","date_gmt":"2012-07-26T18:20:50","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=662"},"modified":"2012-07-26T18:20:50","modified_gmt":"2012-07-26T18:20:50","slug":"sutazne-ulohy-kategorii-a-b-a-c-3","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=662","title":{"rendered":"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>62. ro\u010dn\u00edk matematickej olympi\u00e1dy 2012-2013<\/p>\n<p><span class=\"povinne\">aktualizovan\u00e9 5.5.2013 15:30<\/span><\/p>\n<p style=\"text-align: center;\"><script type=\"text\/javascript\"><!--\r\ngoogle_ad_client = \"pub-0508472359151885\";\r\n\/* 468x60, bola vytvoren\u00e1 12.11.2010 *\/\r\ngoogle_ad_slot = \"3127737040\";\r\ngoogle_ad_width = 468;\r\ngoogle_ad_height = 60;\r\n\/\/-->\r\n<\/script>\r\n<script type=\"text\/javascript\"\r\nsrc=\"http:\/\/pagead2.googlesyndication.com\/pagead\/show_ads.js\">\r\n<\/script><\/p>\n<p><!--more--><\/p>\n<table style=\"width: 98%; text-align: center;\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr class=\"a\" bgcolor=\"#c0c0c0\">\n<td>C<\/td>\n<td>B<\/td>\n<td>A<\/td>\n<\/tr>\n<tr align=\"center\">\n<td><span class=\"a\"><a href=\"#c1\">dom\u00e1ce kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#b1\">dom\u00e1ce kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#a1\">dom\u00e1ce kolo<\/a><\/span><\/td>\n<\/tr>\n<tr align=\"center\">\n<td><span class=\"a\"><a href=\"#c2\">\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#b2\">\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#a2\">\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<\/tr>\n<tr align=\"center\">\n<td><span class=\"a\"><a href=\"#c3\">krajsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#b3\">krajsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#a3\">krajsk\u00e9 kolo<\/a><\/span><\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\" bgcolor=\"#c0c0c0\"><\/td>\n<td align=\"center\"><span class=\"a\">celo\u0161t\u00e1tne kolo<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"background-color: #ffff99;\" colspan=\"3\" align=\"center\"><span class=\"rs\">Zadania \u00faloh dom\u00e1ceho kola vo form\u00e1te PDF:   <a href=\"\/62\/62abc.pdf\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" src=\"\/img\/down.png\" border=\"0\" alt=\"\" width=\"48\" height=\"48\" align=\"absmiddle\" \/><\/a> (ulo\u017ei\u0165 ako &#8230;)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"c1\"><\/a><\/p>\n<form> <input onclick=\"window.location='62\/62cir.pdf'\" name=\"buttc1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>C-I-1<\/strong><br \/>\n\u0160tvorcov\u00e1 tabu\u013eka je rozdelen\u00e1 na 16\u00d716 pol\u00ed\u010dok. Kobylka sa po nej pohybuje dvoma smermi: vpravo alebo dole, pri\u010dom strieda skoky o dve a o tri pol\u00ed\u010dka (t.j. \u017eiadne dva po sebe id\u00face skoky nie s\u00fa rovnako dlh\u00e9). Za\u010d\u00edna skokom d\u013a\u017eky dva z \u013eav\u00e9ho horn\u00e9ho pol\u00ed\u010dka. Ko\u013ek\u00fdmi r\u00f4znymi cestami sa m\u00f4\u017ee kobylka dosta\u0165 na prav\u00e9 doln\u00e9 pol\u00ed\u010dko? (Pod cestou m\u00e1me na mysli postupnos\u0165 pol\u00ed\u010dok, na ktor\u00e9 kobylka dosko\u010d\u00ed.)<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>C-I-2<\/strong><br \/>\nPre kladn\u00e9 re\u00e1lne \u010d\u00edsla <em>a, b, c, d<\/em> plat\u00ed<\/p>\n<p style=\"text-align: center;\"><em>a + b = c + d<\/em>,\u00a0\u00a0\u00a0\u00a0\u00a0<em>ad = bc<\/em>,\u00a0\u00a0\u00a0\u00a0\u00a0<em>ac + bd = 1<\/em>.<\/p>\n<p>Ak\u00fa najv\u00e4\u010d\u0161iu hodnotu m\u00f4\u017ee ma\u0165 s\u00fa\u010det <em>a + b + c + d<\/em>?<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>C-I-3<\/strong><br \/>\nDan\u00fd je obd\u013a\u017enik <em>ABCD<\/em> s obvodom <em>o<\/em>. V jeho rovine n\u00e1jdite mno\u017einu v\u0161etk\u00fdch bodov, ktor\u00fdch s\u00fa\u010det vzdialenost\u00ed od priamok <em>AB<\/em>, <em>BC<\/em>, <em>CD<\/em>, <em>DA<\/em> je rovn\u00fd [pmath size=10]{2\/3}[\/pmath]<em>o<\/em>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>C-I-4<\/strong><br \/>\nRozhodnite, \u010di z \u013eubovo\u013en\u00fdch siedmich vrcholov dan\u00e9ho pravideln\u00e9ho 19-uholn\u00edka mo\u017eno v\u017edy vybra\u0165 \u0161tyri, ktor\u00e9 s\u00fa vrcholmi lichobe\u017en\u00edka.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-I-5<\/strong><br \/>\nUr\u010dte v\u0161etky cel\u00e9 \u010d\u00edsla <em>n<\/em>, pre ktor\u00e9 2<em>n<\/em><sup>3<\/sup> &#8211; 3<em>n<\/em><sup>2<\/sup> + <em>n<\/em> + 3 je prvo\u010d\u00edslo.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>C-I-6<\/strong><br \/>\nVn\u00fatri pravideln\u00e9ho \u0161es\u0165uholn\u00edka <em>ABCDEF<\/em> s obsahom 30 cm<sup>2<\/sup> je zvolen\u00fd bod <em>M<\/em>. Obsahy trojuholn\u00edkov <em>ABM<\/em> a <em>BCM<\/em> s\u00fa postupne 3 cm<sup>2<\/sup> a 2 cm<sup>2<\/sup>. Ur\u010dte obsahy trojuholn\u00edkov <em>CDM<\/em>, <em>DEM<\/em>, <em>EFM<\/em> a <em>FAM<\/em>.<\/p>\n<div class=\"pri\">(Pavel Leischner)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"c2\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form> <input onclick=\"window.location='62\/62ciir.pdf'\" name=\"buttc2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>C-S-1<\/strong><br \/>\nDan\u00e9mu rovnostrann\u00e9mu trojuholn\u00edku vp\u00ed\u0161me a op\u00ed\u0161me kru\u017enicu. Ozna\u010dme <em>S<\/em> obsah vzniknut\u00e9ho medzikru\u017eia a <em>T<\/em> obsah kruhu, ktor\u00e9ho priemer je zhodn\u00fd s d\u013a\u017ekou strany dan\u00e9ho trojuholn\u00edka. Ktor\u00fd z obsahov <em>S<\/em>, <em>T<\/em> je v\u00e4\u010d\u0161\u00ed? Svoju odpove\u010f zd\u00f4vodnite.<\/p>\n<div class=\"pri\">(Leo Bo\u010dek)<\/div>\n<p><strong>C-S-2<\/strong><br \/>\nUr\u010dte v\u0161etky dvojice <em>a<\/em>, <em>b<\/em> cel\u00fdch kladn\u00fdch \u010d\u00edsel, pre ktor\u00e9 plat\u00ed<\/p>\n<div class=\"pc\"><strong><em>a<\/em> . [<em>a, b<\/em>] = 4. (<em>a, b<\/em>)<\/strong>,<\/div>\n<p>pri\u010dom symbol <strong>[<em>a<\/em>, <em>b<\/em>]<\/strong> ozna\u010duje najmen\u0161\u00ed spolo\u010dn\u00fd n\u00e1sobok a <strong>(<em>a<\/em>, <em>b<\/em>)<\/strong> najv\u00e4\u010d\u0161\u00ed spolo\u010dn\u00fd delite\u013e cel\u00fdch kladn\u00fdch \u010d\u00edsel <em>a<\/em>, <em>b<\/em>.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>C-S-3<\/strong><br \/>\nKa\u017ed\u00fd vrchol pravideln\u00e9ho dev\u00e4tn\u00e1s\u0165uholn\u00edka je ofarben\u00fd jednou zo \u0161iestich farieb. Dok\u00e1\u017ete, \u017ee niektor\u00fd tupouhl\u00fd trojuholn\u00edk m\u00e1 v\u0161etky vrcholy ofarben\u00e9 rovnakou farbou.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"c3\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>C-II-1<\/strong><br \/>\nV tane\u010dnej sa zi\u0161la skupina chlapcov a diev\u010dat. Ka\u017ed\u00fd z pr\u00edtomn\u00fdch 15 chlapcov pozn\u00e1 pr\u00e1ve 4 diev\u010dat\u00e1 a ka\u017ed\u00e9 diev\u010da pozn\u00e1 pr\u00e1ve 10 chlapcov. (Zn\u00e1mosti s\u00fa vz\u00e1jomn\u00e9.) Dok\u00e1\u017ete, \u017ee \u013eubovo\u013en\u00ed dvaja chlapci maj\u00fa aspo\u0148 dve spolo\u010dn\u00e9 zn\u00e1me.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>C-II-2<\/strong><br \/>\nVn\u00fatri rovnobe\u017en\u00edka <em>ABCD<\/em> je dan\u00fd bod <em>K<\/em> a v p\u00e1se medzi rovnobe\u017ekami <em>BC<\/em> a <em>AD<\/em> v polrovine opa\u010dnej k <em>CDA<\/em> je dan\u00fd bod <em>L<\/em>. Obsahy trojuholn\u00edkov <em>ABK<\/em>, <em>BCK<\/em>, <em>DAK<\/em> a <em>DCL<\/em> s\u00fa S<sub>ABK<\/sub> = 18cm<sup>2<\/sup>, S<sub>BCK<\/sub> = 8cm<sup>2<\/sup>, S<sub>DAK<\/sub> = 16cm<sup>2<\/sup>, S<sub>DCL<\/sub> = 36cm<sup>2<\/sup>. Vypo\u010d\u00edtajte obsahy trojuholn\u00edkov <em>CDK<\/em> a <em>ABL<\/em>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>C-II-3<\/strong><br \/>\nN\u00e1jdite v\u0161etky dvojice cel\u00fdch kladn\u00fdch \u010d\u00edsel <strong><em>a<\/em><\/strong> a <strong><em>b<\/em><\/strong>, pre ktor\u00e9 je \u010d\u00edslo <strong><em>a<sup>2<\/sup>+b<\/em><\/strong> o 62 v\u00e4\u010d\u0161ie ako \u010d\u00edslo <strong><em>b<sup>2<\/sup>+a<\/em><\/strong>.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-II-4<\/strong><br \/>\nUr\u010dte najmen\u0161ie cel\u00e9 kladn\u00e9 \u010d\u00edslo <em>v<\/em>, pre ktor\u00e9 plat\u00ed: Medzi \u013eubovo\u013en\u00fdmi <em>v<\/em> vrcholmi pravideln\u00e9ho dvadsa\u0165uholn\u00edka mo\u017eno n\u00e1js\u0165 tri, ktor\u00e9 s\u00fa vrcholmi pravouhl\u00e9ho rovnoramenn\u00e9ho trojuholn\u00edka.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b1\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form> <input onclick=\"window.location='62\/62bir.pdf'\" name=\"buttb1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>B-I-1<\/strong><br \/>\nUr\u010dte v\u0161etky trojice (<em>a, b, c<\/em>) prirodzen\u00fdch \u010d\u00edsel, pre ktor\u00e9 plat\u00ed<\/p>\n<p style=\"text-align: center;\">2<sup><em>a<\/em><\/sup> + 4<sup><em>b<\/em><\/sup> = 8<sup><em>c<\/em><\/sup>.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-I-2<\/strong><br \/>\nV obore re\u00e1lnych \u010d\u00edsel rie\u0161te rovnicu<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" class=\"aligncenter\" src=\"\/62\/62b1a.png\" alt=\"\" \/><\/p>\n<p>ak viete, \u017ee m\u00e1 aspo\u0148 jeden celo\u010d\u00edseln\u00fd kore\u0148. Pr\u00edpadn\u00e9 iracion\u00e1lne korene zap\u00ed\u0161te v jednoduchom tvare bez odmocn\u00edn z iracion\u00e1lnych \u010d\u00edsel.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>B-I-3<\/strong><br \/>\nNech <em>V<\/em> je priese\u010dn\u00edk v\u00fd\u0161ok ostrouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em>. Priamka <em>CV<\/em> je spolo\u010dnou doty\u010dnicou kru\u017en\u00edc <em>k<\/em> a <em>l<\/em>, ktor\u00e9 sa zvonka dot\u00fdkaj\u00fa v bode <em>V<\/em> a pritom ka\u017ed\u00e1 z nich prech\u00e1dza jedn\u00fdm z vrcholov <em>A<\/em> a <em>B<\/em>. Ich priese\u010dn\u00edky s vn\u00fatrami str\u00e1n <em>AC<\/em> a <em>BC<\/em> ozna\u010dme <em>P<\/em> a <em>Q<\/em>. Dok\u00e1\u017ete, \u017ee polpriamka <em>VC<\/em> je osou uhla <em>PVQ<\/em> a \u017ee body <em>A, B, P, Q<\/em> le\u017eia na jednej kru\u017enici.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-I-4<\/strong><br \/>\nN\u00e1jdite najmen\u0161iu hodnotu zlomku<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" class=\"aligncenter\" src=\"\/62\/62b1b.png\" alt=\"\" \/><\/p>\n<p>pri\u010dom <em>n<\/em> je \u013eubovo\u013en\u00e9 prirodzen\u00e9 \u010d\u00edslo v\u00e4\u010d\u0161ie ako 2.<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>B-I-5<\/strong><br \/>\nV rovine je dan\u00e1 \u00fase\u010dka <em>AB<\/em>. Pre \u013eubovo\u013en\u00fd bod <em>X<\/em> tejto roviny, ktor\u00fd je r\u00f4zny od <em>A<\/em> aj <em>B<\/em>, ozna\u010dme <em>X<\/em><sub>A<\/sub>, resp. <em>X<\/em><sub>B<\/sub> obraz bodu <em>A<\/em>, resp. <em>B<\/em> v osovej s\u00famernosti pod\u013ea priamky <em>XB<\/em>, resp. <em>XA<\/em>. N\u00e1jdite v\u0161etky tak\u00e9 body <em>X<\/em>, ktor\u00e9 spolu s bodmi <em>X<\/em><sub>A<\/sub>, <em>X<\/em><sub>B<\/sub> tvoria vrcholy rovnostrann\u00e9ho trojuholn\u00edka.<\/p>\n<div class=\"pri\">(Pavel Cal\u00e1bek)<\/div>\n<p><strong>B-I-6<\/strong><br \/>\nJe dan\u00e9 prirodzen\u00e9 \u010d\u00edslo k &lt; 12. Vo vrcholoch pravideln\u00e9ho 12-uholn\u00edka s\u00fa nap\u00edsan\u00e9 \u010d\u00edsla 1, 2, . .. , 12 (ako na cifern\u00edku hod\u00edn). V jednom kroku m\u00f4\u017eeme bu\u010f vymeni\u0165 niektor\u00e9 dve proti\u013eahl\u00e9 \u010d\u00edsla, alebo zvoli\u0165 \u013eubovo\u013en\u00fdch <em>k<\/em> susedn\u00fdch vrcholov a v nich nap\u00edsan\u00e9 \u010d\u00edsla zv\u00e4\u010d\u0161i\u0165 o 1. Ozna\u010dme <em>T<\/em>(<em>k<\/em>) nasledovn\u00e9 tvrdenie: &#8222;Po kone\u010dnom po\u010dte krokov mo\u017eno dosta\u0165 v\u0161etk\u00fdch 12 \u010d\u00edsel rovnak\u00fdch.&#8220; Dok\u00e1\u017ete, \u017ee <em>T<\/em>(2) neplat\u00ed, <em>T<\/em>(5) plat\u00ed, a rozhodnite o platnosti <em>T<\/em>(3).<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b2\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form> <input onclick=\"window.location='62\/62biir.pdf'\" name=\"buttb2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>B-S-1<\/strong><br \/>\nDok\u00e1\u017ete, \u017ee \u017eiadna z rovn\u00edc<\/p>\n<div class=\"pc\"><strong>3<sup>2<em>x<\/em><\/sup> + 6<sup><em>y<\/em><\/sup> = 2013,\u00a0\u00a0\u00a0\u00a0\u00a0|3<sup>2<em>x<\/em><\/sup> &#8211; 6<sup><em>y<\/em><\/sup>| = 2013<\/strong><\/div>\n<p>nem\u00e1 rie\u0161enie v obore cel\u00fdch kladn\u00fdch \u010d\u00edsel.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-S-2<\/strong><br \/>\nDo pol\u00ed\u010dok \u0161tvor\u010dekovej mrie\u017eky 11\u00d711 sme postupne z\u013eava doprava a zhora nadol zap\u00edsali \u010d\u00edsla 1, 2, &#8230;, 121. \u0160tvorcovou doskou 3\u00d73 sme v\u0161etk\u00fdmi mo\u017en\u00fdmi sp\u00f4sobmi zakryli presne dev\u00e4\u0165 pol\u00ed\u010dok. V ko\u013ek\u00fdch pr\u00edpadoch bol s\u00fa\u010det deviatich zakryt\u00fdch \u010d\u00edsel druhou mocninou cel\u00e9ho \u010d\u00edsla?<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>B-S-3<\/strong><br \/>\nUva\u017eujme dve kru\u017enice so stredmi <em>S<\/em><sub>1<\/sub> a <em>S<\/em><sub>2<\/sub> tak\u00e9, \u017ee ich spolo\u010dn\u00e9 vn\u00fatorn\u00e9 doty\u010dnice pret\u00ednaj\u00fa ich spolo\u010dn\u00e9 vonkaj\u0161ie doty\u010dnice v \u0161tyroch bodoch. Dok\u00e1\u017ete, \u017ee tieto \u0161tyri priese\u010dn\u00edky le\u017eia na T\u00e1lesovej kru\u017enici nad priemerom <em>S<\/em><sub>1<\/sub><em>S<\/em><sub>2<\/sub>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b3\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>B-II-1<\/strong><br \/>\nPre \u013eubovo\u013en\u00e9 re\u00e1lne \u010d\u00edsla <em>k<\/em> \u2260 \u00b1\u00061, <em>p<\/em> \u2260 0 a <em>q<\/em> dok\u00e1\u017ete tvrdenie: Rovnica<\/p>\n<div class=\"r\">x<sup>2<\/sup> + px + q = 0<\/div>\n<p>m\u00e1 v obore re\u00e1lnych \u010d\u00edsel dva korene, z ktor\u00fdch jeden je <em>k<\/em>-n\u00e1sobkom druh\u00e9ho, pr\u00e1ve vtedy, ke\u010f plat\u00ed <em>kp<\/em><sup>2<\/sup> = (<em>k<\/em> + 1)<sup>2<\/sup><em>q<\/em>.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>B-II-2<\/strong><br \/>\nObec m\u00e1 100 obyvate\u013eov. Vieme, \u017ee ka\u017ed\u00fd z nich m\u00e1 v obci pr\u00e1ve troch zn\u00e1mych. (Zn\u00e1mosti s\u00fa vz\u00e1jomn\u00e9.)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0<strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee v obci existuje skupina 25 os\u00f4b, medzi ktor\u00fdmi sa \u017eiadne dve nepoznaj\u00fa.<br \/>\n\u00a0\u00a0\u00a0\u00a0<strong>b)<\/strong> N\u00e1jdite najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>n<\/em> s vlastnos\u00bbou, \u017ee v \u013eubovo\u013enej skupine <em>n<\/em> os\u00f4b ka\u017edej takej obce existuje dvojica zn\u00e1mych.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>B-II-3<\/strong><br \/>\nUr\u010dte v\u0161etky trojice (<em>a, b, c<\/em>) cel\u00fdch kladn\u00fdch \u010d\u00edsel, pre ktor\u00e9 plat\u00ed<\/p>\n<div class=\"r\">2<sup>a+2b+1<\/sup> + 4<sup>a<\/sup> + 16<sup>b<\/sup> = 4<sup>c<\/sup>.<\/div>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-II-4<\/strong><br \/>\nV rovine s\u00fa dan\u00e9 kru\u017enice <em>m<\/em>, <em>n<\/em>, ktor\u00e9 sa pret\u00ednaj\u00fa v bodoch <em>K<\/em>, <em>L<\/em>. Doty\u010dnica v bode <em>K<\/em> ku kru\u017enici <em>m<\/em> pret\u00edna kru\u017enicu <em>n<\/em> v bode <em>A<\/em> \u2260 <em>K<\/em>, doty\u010dnica v bode <em>L<\/em> ku kru\u017enici <em>n<\/em> pret\u00edna kru\u017enicu <em>m<\/em> v bode <em>C<\/em> \u2260 <em>K<\/em>. Bod <em>B<\/em> \u2260 <em>L<\/em> je priese\u010dn\u00edk priamky <em>AL<\/em> s kru\u017enicou <em>m<\/em> a bod <em>D<\/em> \u2260 <em>K<\/em> je priese\u010dn\u00edk priamky <em>CK<\/em> s kru\u017enicou <em>n<\/em>. Dok\u00e1\u017ete, \u017ee \u0161tvoruholn\u00edk <em>ABCD<\/em> je rovnobe\u017en\u00edk.<\/p>\n<div class=\"pri\">(Pavel Leischner)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a1\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form> <input onclick=\"window.location='62\/62air.pdf'\" name=\"butta1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-I-1<\/strong><br \/>\nN\u00e1jdite v\u0161etky dvojice prvo\u010d\u00edsel <em>p, q<\/em>, pre ktor\u00e9 existuje prirodzen\u00e9 \u010d\u00edslo <em>a<\/em> tak\u00e9, \u017ee<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" class=\"aligncenter\" src=\"\/62\/62a1a.png\" alt=\"\" \/><\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k, R\u00f3bert T\u00f3th)<\/div>\n<p><strong>A-I-2<\/strong><br \/>\nDve kru\u017enice <em>k<\/em><sub>1<\/sub>(<em>S<\/em><sub>1<\/sub>, <em>r<\/em><sub>1<\/sub>) a <em>k<\/em><sub>2<\/sub>(<em>S<\/em><sub>2<\/sub>, <em>r<\/em><sub>2<\/sub>) sa zvonka dot\u00fdkaj\u00fa a le\u017eia vo \u0161tvorci <em>ABC D<\/em> so stranou <em>a<\/em> tak, \u017ee\u00a0<em> <\/em><em>k<\/em><sub>1<\/sub> sa dot\u00fdka str\u00e1n <em>AD<\/em> a <em>CD<\/em> a <em>k<\/em><sub>2<\/sub> sa dot\u00fdka str\u00e1n <em>BC<\/em> a <em>CD<\/em>. Dok\u00e1\u017ete, \u017ee aspo\u0148 jeden z trojuholn\u00edkov <em>AS<\/em><sub>1<\/sub><em>S<\/em><sub>2<\/sub>, <em>BS<\/em><sub>1<\/sub><em>S<\/em><sub>2<\/sub> m\u00e1 obsah najviac [pmath size=10]{3\/16}[\/pmath]<em>a<\/em><sup>2<\/sup>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>A-I-3<\/strong><br \/>\nOzna\u010dme <em>p(n)<\/em> po\u010det v\u0161etk\u00fdch <em>n<\/em>-cifern\u00fdch \u010d\u00edsel zlo\u017een\u00fdch len z cifier 1, 2, 3, 4, 5, v ktor\u00fdch sa ka\u017ed\u00e9 dve susedn\u00e9 cifry l\u00ed\u0161ia aspo\u0148 o 2. Dok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 prirodzen\u00e9 \u010d\u00edslo <em>n<\/em> plat\u00ed<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" class=\"aligncenter\" src=\"\/62\/62a1b.png\" alt=\"\" \/><\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-I-4<\/strong><br \/>\nN\u00e1jdite v\u0161etky funkcie <strong><em>f<\/em>: R  {0} \u2192 R<\/strong> tak\u00e9, \u017ee pre v\u0161etky nenulov\u00e9 \u010d\u00edsla <em>x, y<\/em> plat\u00ed<\/p>\n<div class=\"pc\"><strong><em>x<\/em> \u2219 <em>f(xy)<\/em> + <em>f(\u2014y)<\/em> = <em>x<\/em> \u2219 <em>f(x)<\/em><\/strong>.<\/div>\n<div class=\"pri\">(Pavel Cal\u00e1bek)<\/div>\n<p><strong>A-I-5<\/strong><br \/>\nOzna\u010dme <em>I<\/em> stred kru\u017enice vp\u00edsanej trojuholn\u00edku <em>ABC<\/em>. Kru\u017enica, ktor\u00e1 prech\u00e1dza vrcholom <em>B<\/em> a dot\u00fdka sa priamky <em>AI<\/em> v bode <em>I<\/em>, pret\u00edna strany <em>AB<\/em>, <em>BC<\/em> postupne v bodoch <em>P<\/em>, <em>Q<\/em>. Priese\u010dn\u00edk priamky <em>QI<\/em> so stranou <em>AC<\/em> ozna\u010dme <em>R<\/em>. Dok\u00e1\u017ete, \u017ee plat\u00ed<\/p>\n<div class=\"pc\"><strong>|<em>AR<\/em>| \u2219 |<em>BQ<\/em>| = |<em>PI<\/em>|<sup>2<\/sup><\/strong>.<\/div>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>A-I-6<\/strong><br \/>\nV obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te s\u00fastavu rovn\u00edc<\/p>\n<div class=\"pc\"><strong>sin<sup>2<\/sup> <em>x<\/em> + cos<sup>2<\/sup> <em>y<\/em> = tg<sup>2<\/sup> <em>z<\/em>,<br \/>\nsin<sup>2<\/sup> <em>y<\/em> + cos<sup>2<\/sup> <em>z<\/em> = tg<sup>2<\/sup> <em>x<\/em>,<br \/>\nsin<sup>2<\/sup> <em>z<\/em> + cos<sup>2<\/sup> <em>x<\/em> = tg<sup>2<\/sup> <em>y<\/em><\/strong>.<\/div>\n<div class=\"pri\">(Pavel Cal\u00e1bek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a2\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form> <input onclick=\"window.location='62\/62aiir.pdf'\" name=\"butta2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-S-1<\/strong><br \/>\nV obd\u013a\u017eniku <em>ABCD<\/em> so stranami |<em>AB<\/em>| = 9,\u00a0|<em>BC<\/em>| = 8 le\u017eia navz\u00e1jom sa dot\u00fdkaj\u00face\u00a0kru\u017enice <em>k<\/em><sub>1<\/sub>(<em>S<\/em><sub>1, <\/sub><em>r<\/em><sub>1)<\/sub> a <em>k<\/em><sub>2<\/sub>(<em>S<\/em><sub>2<\/sub>,\u00a0<em>r<\/em><sub>2<\/sub>) tak, \u017ee <em>k<\/em><sub>1<\/sub> sa dot\u00fdka str\u00e1n <em>AD<\/em> a <em>CD<\/em>, <em>k<\/em><sub>2<\/sub> sa dot\u00fdka str\u00e1n\u00a0<em>AB<\/em> a <em>BC<\/em>.<\/p>\n<p>a) Dok\u00e1\u017ete, \u017ee <em>r<\/em><sub>1<\/sub> + <em>r<\/em><sub>2<\/sub> = 5.<br \/>\nb) Ur\u010dte najmen\u0161iu a najv\u00e4\u010d\u0161iu mo\u017en\u00fa hodnotu obsahu trojuholn\u00edka <em>AS<\/em><sub>1<\/sub><em>S<\/em><sub>2<\/sub>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-S-2<\/strong><br \/>\nNa ka\u017edej z <em>n+1<\/em> stien <em>n<\/em>-bok\u00e9ho ihlana je nap\u00edsan\u00e9 \u010d\u00edslo 0. V ka\u017edom kroku zvol\u00edme niektor\u00fd vrchol a \u010d\u00edsla na v\u0161etk\u00fdch sten\u00e1ch obsahuj\u00facich tento vrchol zv\u00e4\u010d\u0161\u00edme o 1 alebo ich v\u0161etky zmen\u0161\u00edme o 1. Dok\u00e1\u017ete, \u017ee nem\u00f4\u017ee nasta\u0165 situ\u00e1cia, v ktorej by na v\u0161etk\u00fdch sten\u00e1ch ihlana bolo nap\u00edsan\u00e9 \u010d\u00edslo 1.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>A-S-3<\/strong><br \/>\nUr\u010dte v\u0161etky trojice re\u00e1lnych \u010d\u00edsel <em>a, b, c<\/em>, ktor\u00e9 sp\u013a\u0148aj\u00fa podmienky<\/p>\n<div class=\"pc\"><strong><em>a<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup> = 26, \u00a0\u00a0\u00a0\u00a0<em>a<\/em> + <em>b<\/em> = 5\u00a0 \u00a0\u00a0a\u00a0\u00a0 \u00a0<em>b<\/em> + <em>c<\/em> \u2265 7<\/strong>.<\/div>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a3\"><\/a><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form> <input onclick=\"window.location='62\/62aiiir.pdf'\" name=\"butta3\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-II-1<\/strong><br \/>\nDan\u00fdch je 21 r\u00f4znych cel\u00fdch \u010d\u00edsel tak\u00fdch, \u017ee s\u00fa\u010det \u013eubovo\u013en\u00fdch jeden\u00e1stich z nich je v\u00e4\u010d\u0161\u00ed ako s\u00fa\u010det desiatich zvy\u0161n\u00fdch \u010d\u00edsel.<\/p>\n<p>a) Dok\u00e1\u017ete, \u017ee ka\u017ed\u00e9 z dan\u00fdch \u010d\u00edsel je v\u00e4\u010d\u0161ie ako 100.<br \/>\nb) Ur\u010dte v\u0161etky tak\u00e9 skupiny 21 r\u00f4znych cel\u00fdch \u010d\u00edsel, ktor\u00e9 obsahuj\u00fa \u010d\u00edslo 101.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-II-2<\/strong><br \/>\nNech <em>A, B<\/em> s\u00fa mno\u017einy cel\u00fdch kladn\u00fdch \u010d\u00edsel tak\u00e9, \u017ee s\u00fa\u010det \u013eubovo\u013en\u00fdch dvoch r\u00f4znych \u010d\u00edsel z <em>A<\/em> patr\u00ed do <em>B<\/em> a podiel \u013eubovo\u013en\u00fdch dvoch r\u00f4znych \u010d\u00edsel z <em>B<\/em> (v\u00e4\u010d\u0161ie delen\u00e9 men\u0161\u00edm) patr\u00ed do A. Ur\u010dte najv\u00e4\u010d\u0161\u00ed mo\u017en\u00fd po\u010det prvkov mno\u017einy <em>A<\/em> \u222a <em>B<\/em>.<\/p>\n<div class=\"pri\">(Martin Pan\u00e1k)<\/div>\n<p><strong>A-II-3<\/strong><br \/>\nV pravouhlom trojuholn\u00edku <em>ABC<\/em> s preponou <em>AB<\/em> a odvesnami d\u013a\u017eok |<em>AC<\/em>| = 4 a\u00a0|<em>BC<\/em>| = 3 le\u017eia navz\u00e1jom sa dot\u00fdkaj\u00face\u00a0kru\u017enice <em>k<\/em><sub>1<\/sub>(<em>S<\/em><sub>1, <\/sub><em>r<\/em><sub>1)<\/sub> a <em>k<\/em><sub>2<\/sub>(<em>S<\/em><sub>2<\/sub>,\u00a0<em>r<\/em><sub>2<\/sub>) tak, \u017ee <em>k<\/em><sub>1<\/sub> sa dot\u00fdka str\u00e1n <em>AB<\/em> a <em>AC<\/em> a <em>k<\/em><sub>2<\/sub> sa dot\u00fdka str\u00e1n\u00a0<em>AB<\/em> a <em>BC<\/em>. Ur\u010dte polomery <em>r<\/em><sub>1<\/sub> a <em>r<\/em><sub>2<\/sub>, ak plat\u00ed 4<em>r<\/em><sub>1<\/sub> = 9<em>r<\/em><sub>2<\/sub>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-II-4<\/strong><br \/>\nDok\u00e1\u017ete, \u017ee kladn\u00e9 \u010d\u00edsla <em>a, b, c<\/em> s\u00fa d\u013a\u017ekami str\u00e1n trojuholn\u00edka pr\u00e1ve vtedy, ke\u010f s\u00fastava rovn\u00edc<\/p>\n<p class=\"pc\"><strong><em>a<\/em>(<em>yz<\/em> + <em>x<\/em>) = <em>b<\/em>(<em>xz<\/em> + <em>y<\/em>) = <em>c<\/em>(<em>xy<\/em> + <em>z<\/em>),\u00a0\u00a0\u00a0\u00a0\u00a0<em>x<\/em> + <em>y<\/em> + <em>z<\/em> = 1<\/strong><\/p>\n<p>s nezn\u00e1mymi <em>x, y, z<\/em> m\u00e1 rie\u0161enie v obore kladn\u00fdch re\u00e1lnych \u010d\u00edsel.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<!--CusAds0-->\n<div style=\"font-size: 0px; height: 0px; line-height: 0px; margin: 0; padding: 0; clear: both;\"><\/div>","protected":false},"excerpt":{"rendered":"<p>62. ro\u010dn\u00edk matematickej olympi\u00e1dy 2012-2013 aktualizovan\u00e9 5.5.2013 15:30<\/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":[11],"tags":[],"class_list":["post-662","post","type-post","status-publish","format-standard","hentry","category-62-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - 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