{"id":281,"date":"2010-04-07T20:13:56","date_gmt":"2010-04-07T20:13:56","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=281"},"modified":"2010-04-07T20:13:56","modified_gmt":"2010-04-07T20:13:56","slug":"kategorie-abc-2","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=281","title":{"rendered":"Kateg\u00f3rie ABC"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>55. ro\u010dn\u00edk matematickej olympi\u00e1dy 2005-2006<\/p>\n<p><!--more--><\/p>\n<table align=center border=\"1\" cellpadding=\"0\" cellspacing=\"0\" width=\"98%\">\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\">&nbsp;<\/td>\n<td align=\"center\"><span class=\"a\">celo\u0161t\u00e1tne kolo<\/span><\/td>\n<\/tr>\n<\/table>\n<p><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><\/center><\/p>\n<p><a name=\"c1\"><\/a><\/p>\n<p><b>C-I-1<\/b><br \/>\n<strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 prirodzen\u00e9 \u010d\u00edslo <em>m<\/em> je rozdiel <em>m<sup>6<\/sup> \u2013\u00a0m<sup>2<\/sup><\/em> delite\u013en\u00fd \u010d\u00edslom 60.<br \/>\n<strong>b)<\/strong> Ur\u010dte v\u0161etky prirodzen\u00e9 \u010d\u00edsla <em>m<\/em>, pre ktor\u00e9 je rozdiel <em>m<sup>6<\/sup> \u2013\u00a0m<sup>2<\/sup><\/em> delite\u013en\u00fd \u010d\u00edslom 120.<\/p>\n<p><b>C-I-2<\/b><br \/>\nKa\u017ed\u00e9 dve z kru\u017en\u00edc <em>k, l, m<\/em> maj\u00fa vonkaj\u0161\u00ed dotyk a v\u0161etky tri maj\u00fa spolo\u010dn\u00fa doty\u010dnicu. Polomery kru\u017en\u00edc <em>k, l<\/em> s\u00fa 3\u00a0cm a 12\u00a0cm. Vypo\u010d\u00edtajte polomer kru\u017enice <em>m<\/em>. N\u00e1jdite v\u0161etky rie\u0161enia.<\/p>\n<p><b>C-I-3<\/b><br \/>\nUr\u010dte po\u010det v\u0161etk\u00fdch troj\u00edc navz\u00e1jom r\u00f4znych trojmiestnych prirodzen\u00fdch \u010d\u00edsel, ktor\u00fdch s\u00fa\u010det je delite\u013en\u00fd ka\u017ed\u00fdm z troch s\u010ditovan\u00fdch \u010d\u00edsel.<\/p>\n<p><b>C-I-4<\/b><br \/>\nJe dan\u00e9 prirodzen\u00e9 \u010d\u00edslo <em>n<\/em> (<em>n\u00a0\u2265\u00a02<\/em>) a re\u00e1lne \u010d\u00edsla <em>x<sub>1<\/sub>, x<sub>2<\/sub>, &#8230;, x<sub>n<\/sub><\/em>, pre ktor\u00e9 plat\u00ed<\/p>\n<p class=\"r\">x<sub>1<\/sub> x<sub>2<\/sub> =\u00a0x<sub>2<\/sub> x<sub>3<\/sub> =\u00a0&#8230;\u00a0=\u00a0x<sub>n-1<\/sub> x<sub>n<\/sub> =\u00a0x<sub>n<\/sub> x<sub>1<\/sub> =\u00a01.<\/p>\n<p>Dok\u00e1\u017ete, \u017ee \u00a0\u00a0<span class=\"r\">x<sub>1<\/sub><sup>2<\/sup> +\u00a0x<sub>2<\/sub><sup>2<\/sup> +\u00a0&#8230;\u00a0+\u00a0x<sub>n<\/sub><sup>2<\/sup> \u2265\u00a0n<\/span>.<\/p>\n<p><b>C-I-5<\/b><br \/>\nV ostrouhlom trojuholn\u00edku <em>ABC<\/em> ozna\u010dme <em>D<\/em> p\u00e4tu v\u00fd\u0161ky z vrcholu <em>C<\/em> a <em>P<\/em>, <em>Q<\/em> p\u00e4ty kolm\u00edc veden\u00fdch bodom <em>D<\/em> na strany <em>AC<\/em> a <em>BC<\/em>. Obsahy trojuholn\u00edkov <em>ADP, DCP, DBQ, CDQ<\/em> ozna\u010dme postupne <em>S<sub>1<\/sub>, S<sub>2<\/sub>, S<sub>3<\/sub>, S<sub>4<\/sub><\/em>. Vypo\u010d\u00edtajte <em>S<sub>1<\/sub> :\u00a0S<sub>3<\/sub><\/em>, ak <em>S<sub>1<\/sub> :\u00a0S<sub>2<\/sub> =\u00a02\u00a0:\u00a03<\/em>, <em>S<sub>3<\/sub> :\u00a0S<sub>4<\/sub> =\u00a03\u00a0:\u00a08<\/em>.<\/p>\n<p><b>C-I-6<\/b><br \/>\nRozhodnite, ktor\u00e9 z \u010d\u00edsel <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ca1.gif\" border=\"0\" alt=\"\" width=\"121\" height=\"21\" align=\"absmiddle\" \/>, <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ca2.gif\" border=\"0\" alt=\"\" width=\"120\" height=\"21\" align=\"absmiddle\" \/> je v\u00e4\u010d\u0161ie, ak <em>p<\/em> a <em>q<\/em> s\u00fa r\u00f4zne kladn\u00e9 \u010d\u00edsla.<\/p>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"c2\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/table>\n<p><b>C-S-1<\/b><br \/>\nNa hokejovom turnaji sa z\u00fa\u010dastnili \u0161tyri dru\u017estv\u00e1, pri\u010dom ka\u017ed\u00e9 zohralo s ka\u017ed\u00fdm pr\u00e1ve jeden z\u00e1pas. V \u017eiadnych dvoch z\u00e1pasoch nepadlo rovnako ve\u013ea g\u00f3lov, ale po\u010det g\u00f3lov strelen\u00fdch v ka\u017edom z\u00e1pase del\u00ed celkov\u00fd po\u010det g\u00f3lov strelen\u00fdch na turnaji. Ko\u013eko najmenej g\u00f3lov mohlo na turnaji padn\u00fa\u0165?<\/p>\n<div class=\"pri\">(M. Pan\u00e1k)<\/div>\n<\/p>\n<p><b>C-S-2<\/b><br \/>\nVrchol <em>C<\/em> \u0161tvorcov <em>ABCD<\/em> a <em>CJKL<\/em> je vn\u00fatorn\u00fdm bodom \u00fase\u010dky <em>AK<\/em> aj \u00fase\u010dky <em>DJ<\/em>. Body <em>E, F, G<\/em> a <em>H<\/em> s\u00fa postupne stredy \u00fase\u010diek <em>BC, BK, DK<\/em> a <em>DC<\/em>. Vyjadrite obsah \u0161tvoruholn\u00edka <em>EFGH<\/em> pomocou obsahov <em>S<\/em> a <em>T<\/em> \u0161tvorcov <em>ABCD<\/em> a <em>CJKL<\/em>.<\/p>\n<div class=\"pri\">(P. Leischner)<\/div>\n<\/p>\n<p><b>C-S-3<\/b><br \/>\nKru\u017enice <em>k, l, m<\/em> sa dot\u00fdkaj\u00fa spolo\u010dnej doty\u010dnice v troch r\u00f4znych bodoch a ich stredy le\u017eia na jednej priamke. Kru\u017enice <em>k<\/em> a <em>l<\/em>, a tie\u017e kru\u017enice <em>l<\/em> a <em>m<\/em>, maj\u00fa vonkaj\u0161\u00ed dotyk. Ur\u010dte polomer kru\u017enice <em>l<\/em>, ak polomery kru\u017en\u00edc <em>k<\/em> a <em>m<\/em> s\u00fa <em>3\u00a0cm<\/em> a <em>12\u00a0cm<\/em>.<\/p>\n<div class=\"pri\">(L. Bo\u010dek)<\/div>\n<\/p>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"c3\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/table>\n<p><b>C-II-1<\/b><br \/>\nZ\u00e1klad\u0148a <em>AB<\/em> lichobe\u017en\u00edka <em>ABCD<\/em> je trikr\u00e1t dlh\u0161ia ako z\u00e1klad\u0148a <em>CD<\/em>. Ozna\u010dme <em>M<\/em> stred strany <em>AB<\/em> a <em>P<\/em> priese\u010dn\u00edk \u00fase\u010dky <em>DM<\/em> s uhloprie\u010dkou <em>AC<\/em>. Vypo\u010d\u00edtajte pomer obsahov trojuholn\u00edka <em>CDP<\/em> a \u0161tvoruholn\u00edka <em>MBCP<\/em>.<\/p>\n<p><b>C-II-2<\/b><br \/>\nAk re\u00e1lne \u010d\u00edsla <em>a, b, c, d<\/em> sp\u013a\u0148aj\u00fa rovnosti\u00a0\u00a0<span class=\"r\">a<sup>2<\/sup> + b<sup>2<\/sup> = b<sup>2<\/sup> + c<sup>2<\/sup> = c<sup>2<\/sup> + d<sup>2<\/sup> = 1<\/span>; plat\u00ed nerovnos\u0165 <span class=\"r\">ab + ac + ad + bc + bd + cd\u00a0\u2264\u00a03.<\/span><\/p>\n<p>Dok\u00e1\u017ete a zistite, kedy za dan\u00fdch podmienok nastane rovnos\u0165.<\/p>\n<p><b>C-II-3<\/b><br \/>\nKru\u017enice <em>k, l<\/em> s vonkaj\u0161\u00edm dotykom le\u017eia obe v obd\u013a\u017eniku <em>ABCD<\/em>, ktor\u00e9ho obsah je 72\u00a0cm<sup>2<\/sup>. Kru\u017enica <em>k<\/em> sa pritom dot\u00fdka str\u00e1n <em>CD, DA a AB<\/em>, zatia\u013e \u010do kru\u017enica <em>l<\/em> sa dot\u00fdka str\u00e1n <em>AB<\/em> a <em>BC<\/em>. Ur\u010dte polomery kru\u017en\u00edc <em>k<\/em> a <em>l<\/em>, ak viete, \u017ee polomer kru\u017enice <em>k<\/em> je v centimetroch vyjadren\u00fd cel\u00fdm \u010d\u00edslom.<\/p>\n<p><b>C-II-4<\/b><br \/>\nN\u00e1jdite v\u0161etky dvojice prvo\u010d\u00edsel <em>p, q<\/em>, pre ktor\u00e9 plat\u00ed\u00a0\u00a0<span class=\"r\">p + q<sup>2<\/sup> = q + 145p<sup>2<\/sup><\/span>.<\/p>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b1\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/table>\n<p><strong>B-I-1<\/strong><\/p>\n<p>Ur\u010dte v\u0161etky hodnoty celo\u010d\u00edseln\u00e9ho parametra <em>a<\/em>, pre ktor\u00e9 m\u00e1 rovnica <strong>(<em>x\u00a0+\u00a0a<\/em>).(<em>x\u00a0+\u00a02a<\/em>)\u00a0=\u00a0<em>3a<\/em><\/strong> aspo\u0148 jeden celo\u010d\u00edseln\u00fd kore\u0148.<\/p>\n<p><strong>B-I-2<\/strong><\/p>\n<p>V danom trojuholn\u00edku <em>ABC<\/em> ozna\u010dme <em>D<\/em> ten bod polpriamky <em>CA<\/em>, pre ktor\u00fd plat\u00ed |<em>CD<\/em>|\u00a0=\u00a0|<em>CB<\/em>|. \u010ealej ozna\u010dme postupne <em>E, F<\/em> stredy \u00fase\u010diek <em>AD<\/em> a <em>BC<\/em>. Dok\u00e1\u017ete, \u017ee |<span style=\"font-family: Symbol;\">\u00d0<\/span> <em>BAC<\/em>|\u00a0=\u00a02.|<span style=\"font-family: Symbol;\">\u00d0<\/span> <em>CEF<\/em>| pr\u00e1ve vtedy, ke\u010f |<em>AB<\/em>|\u00a0=\u00a0|<em>BC<\/em>|.<\/p>\n<p><strong>B-I-3<\/strong><\/p>\n<p>Rozhodnite, \u010di nerovnos\u0165<\/p>\n<div><img decoding=\"async\" style=\"border: 0pt none;\" src=\"\/55\/55ba1.gif\" border=\"0\" alt=\"\" \/><\/div>\n<p>plat\u00ed pre \u013eubovo\u013en\u00e9 kladn\u00e9 \u010d\u00edsla <em>a, b, c, d<\/em>, ktor\u00e9 vyhovuj\u00fa podmienke<\/p>\n<p><strong>a)<\/strong> <em>ab\u00a0=\u00a0cd\u00a0=\u00a01<\/em>;<\/p>\n<p><strong>b)<\/strong> <em>ac\u00a0=\u00a0bd\u00a0=\u00a01<\/em>.<\/p>\n<p><strong>B-I-4<\/strong><\/p>\n<p>Ka\u017ed\u00fa z hviezdi\u010diek v z\u00e1pisoch dvan\u00e1s\u0165miestnych \u010d\u00edsel <em>A\u00a0=\u00a0*88\u00a0888\u00a0888\u00a0888<\/em>, <em>B\u00a0=\u00a0*11\u00a0111\u00a0111\u00a0111<\/em> nahra\u010fte nejakou \u010d\u00edslicou tak, aby v\u00fdraz |<em>14A\u00a0\u201313B<\/em>| mal \u010do najmen\u0161iu hodnotu.<\/p>\n<p><strong>B-I-5<\/strong><\/p>\n<p>Kruh so stredom <em>S<\/em> a polomerom <em>r<\/em> je rozdelen\u00fd na \u0161tyri \u010dasti dvoma tetivami, z ktor\u00fdch jedna m\u00e1 d\u013a\u017eku <em>r<\/em> a druh\u00e1 m\u00e1 od stredu <em>S<\/em> vzdialenos\u0165 <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ba2.gif\" border=\"0\" alt=\"\" width=\"10\" height=\"17\" align=\"absmiddle\" \/>. Dok\u00e1\u017ete, \u017ee absol\u00fatna hodnota rozdielu obsahov t\u00fdch dvoch \u010dast\u00ed, ktor\u00e9 maj\u00fa spolo\u010dn\u00fd pr\u00e1ve jeden bod a pritom \u017eiadna neobsahuje stred <em>S<\/em>, je rovn\u00fd jednej \u0161estine obsahu kruhu.<\/p>\n<p><strong>B-I-6<\/strong><\/p>\n<p>Ur\u010dte najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>n<\/em> s nasledovnou vlastnos\u0165ou: Ak zvol\u00edme \u013eubovo\u013ene <em>n<\/em> r\u00f4znych prirodzen\u00fdch \u010d\u00edsel men\u0161\u00edch ako <em>2005<\/em>, s\u00fa medzi nimi dve tak\u00e9, \u017ee podiel s\u00fa\u010dtu a rozdielu ich druh\u00fdch mocn\u00edn je v\u00e4\u010d\u0161\u00ed ako <em>3<\/em>.<\/p>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b2\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>B-S-1<\/strong><\/p>\n<p>Dok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 kladn\u00e9 \u010d\u00edsla <em>a, b, c<\/em> plat\u00ed nerovnos\u0165<\/p>\n<div><img decoding=\"async\" style=\"border: 0pt none;\" src=\"\/55\/55bb1.gif\" border=\"0\" alt=\"\"  \/>.<\/div>\n<p>Zistite, kedy nast\u00e1va rovnos\u0165.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><strong>B-S-2<\/strong><br \/>\nNa prepone <em>AB<\/em> pravouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em> uva\u017eujme tak\u00e9 body <em>P<\/em> a <em>Q<\/em>, \u017ee |<em>AP<\/em>|\u00a0=\u00a0|<em>AC<\/em>| a |<em>BQ<\/em>|\u00a0=\u00a0|<em>BC<\/em>|. Ozna\u010dme <em>M<\/em> priese\u010dn\u00edk kolmice z vrcholu <em>A<\/em> na priamku <em>CP<\/em> a kolmice z vrcholu <em>B<\/em> na priamku <em>CQ<\/em>. Dok\u00e1\u017ete, \u017ee priamky <em>PM<\/em> a <em>QM<\/em> s\u00fa navz\u00e1jom kolm\u00e9.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><strong>B-S-3<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky dvojice cel\u00fdch \u010d\u00edsel <em>a, b<\/em>, pre ktor\u00e9 \u017eiadna z rovn\u00edc<\/p>\n<div class=\"r\">x<sup>2<\/sup> + ax + b = 0;<br \/>\ny<sup>2<\/sup> + by + a = 0<\/div>\n<p>nem\u00e1 dva r\u00f4zne re\u00e1lne korene.<\/p>\n<div class=\"pri\">(E. Kov\u00e1\u010d)<\/div>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b3\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>B-II-1<\/strong><\/p>\n<p>Ur\u010dte v\u0161etky dvojice prvo\u010d\u00edsel <em>p, q<\/em>, pre ktor\u00e9 plat\u00ed\u00a0\u00a0<span class=\"r\">p + q<sup>2<\/sup> = q + p<sup>3<\/sup><\/span>.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><strong>B-II-2<\/strong><\/p>\n<p>Obd\u013a\u017enik <em>ABCD<\/em> so stranami d\u013a\u017eok |<em>AB<\/em>| = 2008 a |<em>BC<\/em>| = 2006 je rozdelen\u00fd na 2008\u00a0\u00d7\u00a02006 jednotkov\u00fdch \u0161tvorcov a tie s\u00fa striedavo ofarben\u00e9 \u010diernou, sivou a bielou farbou podobne ako obd\u013a\u017enik na obr\u00e1zku: \u0161tvorce pri vrcholoch <em>A<\/em> a <em>B<\/em> s\u00fa \u010dierne, \u0161tvorce pri vrcholoch <em>C<\/em> a <em>D<\/em> s\u00fa biele. Ur\u010dte obsah tej \u010dasti trojuholn\u00edka <em>ABC<\/em>, ktor\u00e1 je siv\u00e1.<\/p>\n<div><img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55bc1.gif\" border=\"0\" alt=\"\" width=\"270\" height=\"193\" align=\"middle\" \/><\/div>\n<div class=\"pri\">(P. Novotn\u00fd)<\/div>\n<p><strong>B-II-3<\/strong><\/p>\n<p>V lichobe\u017en\u00edku <em>ABCD<\/em>, ktor\u00e9ho z\u00e1klad\u0148a <em>AB<\/em> m\u00e1 dvakr\u00e1t v\u00e4\u010d\u0161iu d\u013a\u017eku ako z\u00e1klad\u0148a <em>CD<\/em>, ozna\u010dme E stred ramena <em>BC<\/em>. Dok\u00e1\u017ete, \u017ee kru\u017enica op\u00edsan\u00e1 trojuholn\u00edku <em>CDE<\/em> prech\u00e1dza stredom uhloprie\u010dky <em>AC<\/em> pr\u00e1ve vtedy, ke\u010f strany <em>AB<\/em> a <em>BC<\/em> s\u00fa navz\u00e1jom kolm\u00e9.<\/p>\n<div class=\"pri\">(P. Leischner)<\/div>\n<p><strong>B-II-4<\/strong><\/p>\n<p>Dok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 re\u00e1lne \u010d\u00edsla <em>a, b, c<\/em> z intervalu &lt;0; 1&gt; plat\u00ed<\/p>\n<p class=\"r\">1\u00a0\u2264\u00a0a + b + c + 2(ab + bc + ca) + 3(1 &#8211; a)(1 &#8211; b)(1 &#8211; c)\u00a0\u2264\u00a09<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a1\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>A-I-1<\/strong><\/p>\n<p>V obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te rovnicu <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55aa1.gif\" alt=\"\" width=\"220\" height=\"21\" align=\"absMiddle\" \/> .<\/p>\n<p><strong>A-I-2<\/strong><\/p>\n<p>Nech <em>ABCD<\/em> je tetivov\u00fd \u0161tvoruholn\u00edk s navz\u00e1jom kolm\u00fdmi uhloprie\u010dkami. Ozna\u010dme postupne <em>p, q<\/em> kolmice z bodov <em>D, C<\/em> na priamku <em>AB<\/em> a \u010falej <em>X<\/em> priese\u010dn\u00edk priamok <em>AC<\/em> a <em>p<\/em> a <em>Y<\/em> priese\u010dn\u00edk priamok <em>BD<\/em> a <em>q<\/em>. Dok\u00e1\u017ete, \u017ee <em>XYCD<\/em> je koso\u0161tvorec alebo \u0161tvorec.<\/p>\n<p><strong>A-I-3<\/strong><\/p>\n<p>Postupnos\u0165 <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55aa2.gif\" border=\"0\" alt=\"\" width=\"45\" height=\"20\" align=\"absmiddle\" \/> <em>nenulov\u00fdch<\/em> cel\u00fdch \u010d\u00edsel m\u00e1 t\u00fa vlastnos\u0165, \u017ee pre ka\u017ed\u00e9 <em>n\u00a0\u2265\u00a00<\/em> plat\u00ed <em>a<sub>n+1<\/sub> =\u00a0a<sub>n<\/sub> \u2013\u00a0b<sub>n<\/sub><\/em>, kde <em>b<sub>n<\/sub><\/em> je \u010d\u00edslo, ktor\u00e9 m\u00e1 rovnak\u00e9 znamienko ako \u010d\u00edslo <em>a<sub>n<\/sub><\/em>, ale opa\u010dn\u00e9 poradie \u010d\u00edslic (z\u00e1pis \u010d\u00edsla <em>b<sub>n<\/sub><\/em> m\u00f4\u017ee na rozdiel od z\u00e1pisu \u010d\u00edsla <em>a<sub>n<\/sub><\/em> za\u010d\u00edna\u0165 jednou alebo viacer\u00fdmi nulami). Napr\u00edklad pre <em>a<sub>0<\/sub> =\u00a01\u00a0210<\/em> je <em>a<sub>1<\/sub> =\u00a01\u00a0089<\/em>, <em>a<sub>2<\/sub> =\u00a0\u20138\u00a0712<\/em>, <em>a<sub>3<\/sub> =\u00a0\u2013\u00a06\u00a0534<\/em>.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee postupnos\u0165 (<em>a<sub>n<\/sub><\/em>) je periodick\u00e1.<\/p>\n<p><strong>b)<\/strong> Zistite, ak\u00e9 najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo m\u00f4\u017ee by\u0165 <em>a<sub>0<\/sub><\/em>.<\/p>\n<p><strong>A-I-4<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky kubick\u00e9 rovnice <em>P(x)\u00a0=\u00a00<\/em>, ktor\u00e9 maj\u00fa aspo\u0148 dva r\u00f4zne re\u00e1lne korene, z ktor\u00fdch jeden je \u010d\u00edslo 7, a ktor\u00e9 pre ka\u017ed\u00e9 re\u00e1lne \u010d\u00edslo <em>t<\/em> vyhovuj\u00fa podmienke: Ak <em>P(t)\u00a0=\u00a00<\/em> potom <em>P(t\u00a0+\u00a01)\u00a0=\u00a01<\/em>.<\/p>\n<p><strong>A-I-5<\/strong><\/p>\n<p>S\u00fa dan\u00e9 \u00fase\u010dky d\u013a\u017eok <em>a, b, c, d<\/em>. Dok\u00e1\u017ete, \u017ee konvexn\u00e9 \u0161tvoruholn\u00edky <em>ABCD<\/em> so stranami d\u013a\u017eok <em>a, b, c, d<\/em> (pri zvy\u010dajnom ozna\u010den\u00ed) existuj\u00fa a pritom uhloprie\u010dky ka\u017ed\u00e9ho z nich zvieraj\u00fa ten ist\u00fd uhol pr\u00e1ve vtedy, ke\u010f plat\u00ed rovnos\u0165 <em>a<sup>2<\/sup> +\u00a0c<sup>2<\/sup> =\u00a0b<sup>2<\/sup> +\u00a0d<sup>2<\/sup><\/em>.<\/p>\n<p><strong>A-I-6<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky usporiadan\u00e9 dvojice <strong>(x, y)<\/strong> prirodzen\u00fdch \u010d\u00edsel, pre ktor\u00e9 plat\u00ed \u00a0<em><strong>x<sup>2<\/sup> +\u00a0y<sup>2<\/sup> =\u00a02005.(x\u00a0\u2013y)<\/strong><\/em>.<\/p>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a2\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>A-S-1<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky dvojice cel\u00fdch \u010d\u00edsel <em>x<\/em> a <em>y<\/em>, pre ktor\u00e9 plat\u00ed <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ab1.gif\" border=\"0\" alt=\"\" width=\"192\" height=\"27\" align=\"absmiddle\" \/>.<\/p>\n<div class=\"pri\">(J. Morav\u010d\u00edk)<\/div>\n<p><strong>A-S-2<\/strong><\/p>\n<p>Dan\u00fd je rovnostrann\u00fd trojuholn\u00edk <em>ABC<\/em> s obsahom <em>S<\/em> a jeho vn\u00fatorn\u00fd bod <em>M<\/em>. Ozna\u010dme postupne <em>A<sub>1<\/sub>, B<sub>1<\/sub>, C<sub>1<\/sub><\/em> tie body str\u00e1n <em>BC<\/em>, <em>CA<\/em> a <em>AB<\/em>, pre ktor\u00e9 plat\u00ed <em>MA<sub>1<\/sub>||AB<\/em>, <em>MB<sub>1<\/sub>||BC<\/em> a <em>MC<sub>1<\/sub>||CA<\/em>. Priese\u010dn\u00edky os\u00ed \u00fase\u010diek <em>MA<sub>1<\/sub>, MB<sub>1<\/sub><\/em> a <em>MC<sub>1<\/sub><\/em> tvoria vrcholy trojuholn\u00edka s obsahom <em>T<\/em>. Dok\u00e1\u017ete, \u017ee plat\u00ed <em>S\u00a0=\u00a03T<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><strong>A-S-3<\/strong><\/p>\n<p>V obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te rovnicu \u00a0<img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ab2.gif\" border=\"0\" alt=\"\" width=\"178\" height=\"28\" align=\"middle\" \/>.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a3\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>A-II-1<\/strong><br \/>\nN\u00e1jdite v\u0161etky dvojice tak\u00fdch cel\u00fdch \u010d\u00edsel <em>a, b<\/em>, \u017ee s\u00fa\u010det <em>a + b<\/em> je kore\u0148om rovnice \u00a0<em><strong>x<sup>2<\/sup> +\u00a0ax\u00a0+\u00a0b\u00a0=\u00a00<\/strong><\/em>.<\/p>\n<p><strong>A-II-2<\/strong><\/p>\n<p>Postupnos\u0165 re\u00e1lnych \u010d\u00edsel <img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ac1.gif\" border=\"0\" alt=\"\" width=\"45\" height=\"19\" align=\"absmiddle\" \/> sp\u013a\u0148a pre ka\u017ed\u00e9 n\u00a0\u2265\u00a01 rovnos\u0165<\/p>\n<p class=\"pc\"><img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ac2.gif\" border=\"0\" alt=\"\" width=\"201\" height=\"39\" \/><\/p>\n<p>a naviac plat\u00ed <em>a<sub>11<\/sub><\/em> =\u00a04, <em>a<sub>22<\/sub><\/em> =\u00a02, <em>a<sub>33<\/sub><\/em> =\u00a01. Dok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 prirodzen\u00e9 \u010d\u00edslo <em>k<\/em> je s\u00fa\u010det<\/p>\n<p class=\"pc\"><img loading=\"lazy\" decoding=\"async\" src=\"\/55\/55ac3.gif\" border=\"0\" alt=\"\" width=\"144\" height=\"21\" \/><\/p>\n<p>druhou mocninou prirodzen\u00e9ho \u010d\u00edsla.<\/p>\n<p><strong>A-II-3<\/strong><\/p>\n<p>Dan\u00fd je trojuholn\u00edk <em>ABC<\/em> a vn\u00fatri neho bod <em>P<\/em>. Ozna\u010dme <em>X<\/em> priese\u010dn\u00edk priamky <em>AP<\/em> so stranou <em>BC<\/em> a <em>Y<\/em> priese\u010dn\u00edk priamky <em>BP<\/em> so stranou <em>AC<\/em>. Dok\u00e1\u017ete, \u017ee \u0161tvoruholn\u00edk <em>ABXY<\/em> je tetivov\u00fd pr\u00e1ve vtedy, ke\u010f druh\u00fd priese\u010dn\u00edk (r\u00f4zny od bodu <em>C<\/em>) kru\u017en\u00edc op\u00edsan\u00fdch trojuholn\u00edkom <em>ACX<\/em> a <em>BCY<\/em> le\u017e\u00ed na priamke <em>CP<\/em>.<\/p>\n<p><strong>A-II-4<\/strong><\/p>\n<p>V obore re\u00e1lnych \u010d\u00edsel rie\u0161te s\u00fastavu rovn\u00edc<\/p>\n<p class=\"r\">sin<sup>2<\/sup> x + cos<sup>2<\/sup> y = y<sup>2<\/sup>;<br \/>\nsin<sup>2<\/sup> y + cos<sup>2<\/sup> x = x<sup>2<\/sup><\/p>\n<table style=\"width: 100%;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"88%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a href=\"#top\">\u25b2 hore \u25b2<\/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>55. ro\u010dn\u00edk matematickej olympi\u00e1dy 2005-2006<\/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":[4],"tags":[],"class_list":["post-281","post","type-post","status-publish","format-standard","hentry","category-55-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - 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