{"id":1416,"date":"2020-03-06T18:20:32","date_gmt":"2020-03-06T18:20:32","guid":{"rendered":"http:\/\/matematika.besaba.com\/?p=1416"},"modified":"2023-11-02T18:52:40","modified_gmt":"2023-11-02T18:52:40","slug":"kategoria-abc-3","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=1416","title":{"rendered":"Kateg\u00f3ria ABC"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>51. ro\u010dn\u00edk matematickej olympi\u00e1dy 2001-2002<\/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><br \/>\n<!--more--><\/p>\n<p><a name=\"c1\"><\/a><\/p>\n<p><b>C-I-1<\/b><br \/>\nDok\u00e1\u017ete, \u017ee existuje jedin\u00e1 \u010d\u00edslica <em>c<\/em>, pre ktor\u00fa mo\u017eno n\u00e1js\u0165 jedin\u00e9 prirodzen\u00e9 \u010d\u00edslo <em>n<\/em> kon\u010diace \u010d\u00edslicou <em>c<\/em> a maj\u00face vlastnos\u0165, \u017ee \u010d\u00edslo <em>2n + 1<\/em> je druhou mocninou prvo\u010d\u00edsla.<\/p>\n<div class=\"pri\">(M. Kobl\u00ed\u017ekov\u00e1)<\/div>\n<p><b>C-I-2<\/b><br \/>\nV \u0161tvoruholn\u00edku <em>ABCD<\/em> sa uhloprie\u010dky pret\u00ednaj\u00fa v bode <em>P<\/em>, uhloprie\u010dka <em>AC<\/em> je rozdelen\u00e1 bodmi <em>P<\/em>, <em>N<\/em> a <em>M<\/em> na \u0161tyri zhodn\u00e9 \u00faseky (|<em>AP<\/em>| = |<em>PN<\/em>| = |<em>NM<\/em>| = |<em>MC<\/em>|) a uhloprie\u010dka <em>BD<\/em> je rozdelen\u00e1 bodmi L, K a P na \u0161tyri zhodn\u00e9 \u00faseky (|<em>BL<\/em>| = |<em>LK<\/em>| = |<em>KP<\/em>| = |<em>PD<\/em>|). Ur\u010dte pomer obsahov \u0161tvoruholn\u00edkov <em>KLMN<\/em> a <em>ABCD<\/em>.<\/p>\n<div class=\"pri\">(J. Zhouf)<\/div>\n<p><b>C-I-3<\/b><br \/>\nUr\u010dte v\u0161etky dvojice (<em>x, y<\/em>) cel\u00fdch \u010d\u00edsel, ktor\u00e9 s\u00fa rie\u0161en\u00edm nerovnice<\/p>\n<div class=\"pc\"><img decoding=\"async\" src=\"51\/51ca1.png\" alt=\"\" border=\"0\" \/><\/div>\n<div class=\"pri\">(J. Zhouf)<\/div>\n<p><b>C-I-4<\/b><br \/>\nJo\u017eko sa vracal z v\u00fdletu. Najprv cestoval vlakom a potom pokra\u010doval zo zast\u00e1vky na bicykli. Cel\u00e1 cesta mu trvala presne 1 hodinu 30 min\u00fat a pre\u0161iel pri nej vzdialenos\u0165 60 km. Vlak i\u0161iel priemernou r\u00fdchlos\u0165ou 50 km\/h. Ur\u010dte, ako dlho i\u0161iel Jo\u017eko na bicykli, ke\u010f jeho r\u00fdchlos\u0165 v km\/h je vyjadren\u00e1 prirodzen\u00fdm \u010d\u00edslom rovnako ako vzdialenos\u0165 meran\u00e1 v km, ktor\u00fa pre\u0161iel na bicykli.<\/p>\n<div class=\"pri\">(E. Kov\u00e1\u010d)<\/div>\n<p><b>C-I-5<\/b><br \/>\nZostrojte rovnoramenn\u00fd trojuholn\u00edk <em>ABC<\/em> so z\u00e1klad\u0148ou <em>BC<\/em> danej d\u013a\u017eky <em>a<\/em>, ak je dan\u00fd stred <em>P<\/em> strany <em>AB<\/em> a bod <em>Q<\/em> (<em>Q \u2260 P<\/em>), ktor\u00fd je p\u00e4tou v\u00fd\u0161ky z vrcholu <em>B<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><b>C-I-6<\/b><br \/>\nIst\u00fd panovn\u00edk pozval na oslavu svojich naroden\u00edn 28 rytierov. Ka\u017ed\u00fd z rytierov mal medzi ostatn\u00fdmi pr\u00e1ve troch nepriate\u013eov.<br \/>\n&nbsp;&nbsp;&nbsp;a)&nbsp;&nbsp;Uk\u00e1\u017ete, \u017ee panovn\u00edk m\u00f4\u017ee rytierov rozsadi\u0165 k dvom stolom tak, aby ka\u017ed\u00fd rytier sedel pri rovnakom stole najviac s jedn\u00fdm nepriate\u013eom.<br \/>\n&nbsp;&nbsp;&nbsp;b)&nbsp;&nbsp;Uk\u00e1\u017ete, \u017ee v pr\u00edpade \u013eubovo\u013en\u00e9ho tak\u00e9hoto rozsadenia sed\u00ed pri ka\u017edom stole najviac 16 rytierov.<\/p>\n<p><em>(Nepriate\u013estvo je vz\u00e1jomn\u00fd vz\u0165ah: Ak A je nepriate\u013eom B, tak aj B je nepriate\u013eom A.)<\/em><\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\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 \/>\nDo \u0161portov\u00e9ho kr\u00fa\u017eku chod\u00ed 21 chlapcov. Na posledn\u00fdch dvoch sch\u00f4dzkach nikto nech\u00fdbal, chlapci sa zaka\u017ed\u00fdm rozdelili do troch dru\u017estiev po sedem hr\u00e1\u010dov. Dok\u00e1\u017ete, \u017ee niektor\u00ed traja chlapci boli na oboch sch\u00f4dzkach spolu v jednom dru\u017estve.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><b>C-S-2<\/b><br \/>\nV rovine je dan\u00fd pravouhl\u00fd trojuholn\u00edk <em>ABC<\/em> tak\u00fd, \u017ee kru\u017enica k(<em>A<\/em>; |<em>AC<\/em>|) pret\u00edna preponu <em>AB<\/em> v jej strede <em>S<\/em>. Dok\u00e1\u017ete, \u017ee kru\u017enica op\u00edsan\u00e1 trojuholn\u00edku <em>BCS<\/em> je zhodn\u00e1 s kru\u017enicou <em>k<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<\/p>\n<p><b>C-S-3<\/b><br \/>\nUr\u010dte v\u0161etky dvojice prvo\u010d\u00edsiel (<em>p<\/em>, <em>q<\/em>) tak\u00e9, \u017ee <em>p<\/em> > <em>q<\/em> a \u010d\u00edslo <em>p<\/em><sup>2<\/sup> &#8211; <em>q<\/em><sup>2<\/sup> m\u00e1 najviac \u0161tyroch delite\u013eov.<\/p>\n<div class=\"pri\">(P. Cal\u00e1bek)<\/div>\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>51. ro\u010dn\u00edk matematickej olympi\u00e1dy 2001-2002 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 &nbsp; celo\u0161t\u00e1tne kolo<\/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":[27],"tags":[],"class_list":["post-1416","post","type-post","status-publish","format-standard","hentry","category-51-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Kateg\u00f3ria 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=1416\" \/>\n<meta property=\"og:locale\" content=\"sk_SK\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Kateg\u00f3ria ABC - Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"og:description\" content=\"51. ro\u010dn\u00edk matematickej olympi\u00e1dy 2001-2002 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 &nbsp; 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