{"id":1568,"date":"2022-11-13T13:47:12","date_gmt":"2022-11-13T13:47:12","guid":{"rendered":"https:\/\/matematika.besaba.com\/?p=1568"},"modified":"2024-08-08T20:09:24","modified_gmt":"2024-08-08T19:09:24","slug":"sutazne-ulohy-kategorie-a-b-a-c-6","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=1568","title":{"rendered":"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3rie A, B a C"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>52. ro\u010dn\u00edk matematickej olympi\u00e1dy 2002-2003<\/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<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=174\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc1z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=269\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc1r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><a name=\"c1\"><\/a><\/p>\n<p><b>C-I-1<\/b><br \/>\nZ piatich jedni\u010diek, z piatich dvojok, z piatich trojok, z piatich \u0161tvoriek a z piatich p\u00e4tiek zostavte p\u00e4\u0165 navz\u00e1jom r\u00f4znych p\u00e4\u0165miestnych \u010d\u00edsel tak, aby ich s\u00fa\u010det bol \u010do najv\u00e4\u010d\u0161\u00ed.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><b>C-I-2<\/b><br \/>\nJe dan\u00fd trojuholn\u00edk <em>ABC<\/em> s ostr\u00fdmi vn\u00fatorn\u00fdmi uhlami pri vrcholoch <em>A<\/em> a <em>B<\/em>. Ozna\u010dme <em>Q<\/em> priese\u010dn\u00edk \u0165a\u017enice <em>AD<\/em> s v\u00fd\u0161kou <em>CP<\/em> a <em>E<\/em> p\u00e4tu kolmice z bodu <em>D<\/em> na stranu <em>AB<\/em>. \u010ealej nech <em>R<\/em> je tak\u00fd bod na polpriamke opa\u010dnej k <em>PC<\/em>, \u017ee |<em>PR<\/em>| = |<em>CQ<\/em>|. Dok\u00e1\u017ete, \u017ee priamky <em>AD<\/em> a <em>RE<\/em> s\u00fa r\u00f4znobe\u017en\u00e9 a \u017ee ich priese\u010dn\u00edk le\u017e\u00ed na kolmici k priamke <em>AB<\/em> prech\u00e1dzaj\u00facej bodom B.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><b>C-I-3<\/b><br \/>\nPredpokladajme, \u017ee ka\u017ed\u00e1 z dvoch b\u00e1nk A a B bude ma\u0165 po\u010das nasleduj\u00facich dvoch rokov st\u00e1lu ro\u010dn\u00fa \u00farokov\u00fa mieru. Keby sme ulo\u017eili 5\/6 na\u0161ich \u00faspor v banke A a zvy\u0161ok v banke B, vzr\u00e1stli by na\u0161e \u00faspory po jednom roku na 67 000 Sk a po dvoch rokoch na 74 900 Sk. Keby sme v\u0161ak ulo\u017eili 5\/6 na\u0161ich \u00faspor v banke B a zvy\u0161ok v banke A, vzr\u00e1stli by na\u0161e \u00faspory po jednom roku na 71 000 Sk. Na ak\u00fa \u010diastku by sa v takom pr\u00edpade zv\u00fd\u0161ili na\u0161e \u00faspory po dvoch rokoch?<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><b>C-I-4<\/b><br \/>\nZostrojte lichobe\u017en\u00edk <em>ABCD<\/em> s v\u00fd\u0161kou 3 cm a zhodn\u00fdmi stranami <em>BC<\/em>, <em>CD<\/em> a <em>DA<\/em>, pre ktor\u00fd plat\u00ed: Na z\u00e1kladni <em>AB<\/em> existuje bod <em>E<\/em> tak\u00fd, \u017ee \u00fase\u010dka <em>DE<\/em> m\u00e1 d\u013a\u017eku 5 cm a del\u00ed lichobe\u017en\u00edk na dve \u010dasti s rovnak\u00fdmi obsahmi.<\/p>\n<div class=\"pri\">(E. Kov\u00e1\u010d)<\/div>\n<p><b>C-I-5<\/b><br \/>\nK prirodzen\u00e9mu \u010d\u00edslu <em>m<\/em> zap\u00edsan\u00e9mu rovnak\u00fdmi \u010d\u00edslicami sme pripo\u010d\u00edtali \u0161tvormiestne prirodzen\u00e9 \u010d\u00edslo <em>n<\/em>. Z\u00edskali sme \u0161tvormiestne \u010d\u00edslo s opa\u010dn\u00fdm porad\u00edm \u010d\u00edslic ako m\u00e1 \u010d\u00edslo <em>n<\/em>. Ur\u010dte v\u0161etky tak\u00e9 dvojice \u010d\u00edsel <em>m<\/em> a <em>n<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><b>C-I-6<\/b><br \/>\nV rovine je dan\u00e1 priamka <em>p<\/em> a kru\u017enica <em>k<\/em>. Zostrojte tak\u00fd trojuholn\u00edk <em>ABC<\/em>, aby <em>k<\/em> bola kru\u017enicou jemu vp\u00edsanou, aby jej stred le\u017eal v jednej \u0161tvrtine jeho \u0165a\u017enice na stranu <em>AB<\/em> a aby vrchol <em>C<\/em> le\u017eal na priamke <em>p<\/em>. Urobte diskusiu o po\u010dte rie\u0161en\u00ed v z\u00e1vislosti na vz\u00e1jomnej polohe priamky <em>p<\/em> a kru\u017enice <em>k<\/em>.<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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><a href=\"http:\/\/skmo.sk\/dokument.php?id=177\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc2z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=270\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc2r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><b>C-S-1<\/b><br \/>\nAk od \u013eubovo\u013en\u00e9ho aspo\u0148 dvojmiestneho prirodzen\u00e9ho \u010d\u00edsla odtrhneme \u010d\u00edslicu na mieste jednotiek, dostaneme \u010d\u00edslo o jednu \u010d\u00edslicu &#8222;krat\u0161ie&#8220;. N\u00e1jdite v\u0161etky p\u00f4vodn\u00e9 \u010d\u00edsla, ktor\u00e9 sa rovnaj\u00fa absol\u00fatnej hodnote rozdielu druhej mocniny &#8222;krat\u0161ieho&#8220; \u010d\u00edsla a druhej mocniny odtrhnutej \u010d\u00edslice.<\/p>\n<div class=\"pri\">(J. Zhouf)<\/div>\n<p><b>C-S-2<\/b><br \/>\nNa strane <em>CD<\/em> \u0161tvorca <em>ABCD<\/em> je zvolen\u00fd bod <em>E<\/em> tak, \u017ee uhol <em>DAE<\/em> m\u00e1 ve\u013ekos\u0165 30\u25e6. Bod <em>P<\/em> je p\u00e4tou kolmice vedenej bodom <em>B<\/em> na priamku <em>AE<\/em>, bod <em>Q<\/em> p\u00e4tou kolmice vedenej bodom <em>C<\/em> na priamku <em>BP<\/em>. Rozhodnite, \u010di je obsah lichobe\u017en\u00edka <em>PQCE<\/em> men\u0161\u00ed ako tretina obsahu \u0161tvorca <em>ABCD<\/em>. <\/p>\n<div class=\"pri\">(L. Bo\u010dek)<\/div>\n<p><b>C-S-3<\/b><br \/>\nZ piatich jednotiek, piatich dvojok, piatich trojok, piatich \u0161tvoriek a piatich p\u00e4tiek zostav\u00edme p\u00e4\u0165 p\u00e4\u0165miestnych \u010d\u00edsel, ktor\u00e9 sa \u010d\u00edtaj\u00fa odpredu rovnako ako odzadu (napr. 32&nbsp;223), a potom tieto \u010d\u00edsla s\u010d\u00edtame. Ak\u00fa najmen\u0161iu a ak\u00fa najv\u00e4\u010d\u0161iu hodnotu m\u00f4\u017ee ma\u0165 v\u00fdsledn\u00fd s\u00fa\u010det?<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=180\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc3z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=271\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc3r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><b>C-II-1<\/b><br \/>\nN\u00e1jdite najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>n<\/em>, pre ktor\u00e9 je s\u00fa\u010din <\/p>\n<div class=\"r\">2&nbsp;003&nbsp;\u00b7&nbsp;2&nbsp;004&nbsp;\u00b7&nbsp;2&nbsp;005&nbsp;\u00b7&nbsp;&#8230;&nbsp;\u00b7&nbsp;(2&nbsp;003&nbsp;+&nbsp;<em>n<\/em>)<\/div>\n<p> delite\u013en\u00fd v\u0161etk\u00fdmi dvojmiestnymi prvo\u010d\u00edslami.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><b>C-II-2<\/b><br \/>\nV rovine je dan\u00e1 \u00fase\u010dka <em>AP<\/em>. Zostrojte pravideln\u00fd \u0161es\u0165uholn\u00edk <em>ABCDEF<\/em> tak, aby bod <em>P<\/em> bol stredom jeho strany <em>DE<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><b>C-II-3<\/b><br \/>\nKeby Karol po\u017ei\u010dal jedn\u00e9mu zn\u00e1memu <em>p<\/em> tis\u00edc Sk s \u00farokom <em>p<\/em> % a druh\u00e9mu zn\u00e1memu <em>q<\/em> tis\u00edc Sk s \u00farokom <em>q<\/em> %, kde <em>p<\/em> a <em>q<\/em> s\u00fa cel\u00e9 \u010d\u00edsla, priniesli by mu obe p\u00f4\u017ei\u010dky tak\u00fd ist\u00fd zisk, ako keby jednej osobe po\u017ei\u010dal celkov\u00fa \u010diastku s \u00farokom (<em>p<\/em> + 2,4) %. Keby po\u017ei\u010dal jedn\u00e9mu zn\u00e1memu <em>p<\/em> tis\u00edc Sk s \u00farokom <em>2p<\/em> % a druh\u00e9mu zn\u00e1memu <em>q<\/em> tis\u00edc Sk s \u00farokom <em>2q<\/em> %, priniesli by mu tieto p\u00f4\u017ei\u010dky rovnak\u00fd zisk, ako keby jednej osobe po\u017ei\u010dal celkov\u00fa \u010diastku s \u00farokom (<em>p<\/em> + 5,8) %. Ur\u010dte \u010d\u00edsla <em>p<\/em> a <em>q<\/em>. <\/p>\n<div class=\"pri\">(J. \u0160im\u0161a, J. Zhouf)<\/div>\n<p><b>C-II-4<\/b><br \/>\nUr\u010dte d\u013a\u017eku ramien rovnoramenn\u00e9ho lichobe\u017en\u00edka so z\u00e1klad\u0148ami d\u013a\u017eok 10 a 12 tak, aby d\u013a\u017eky v\u0161etk\u00fdch jeho str\u00e1n aj uhloprie\u010dok boli vyjadren\u00e9 cel\u00fdmi \u010d\u00edslami.<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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><a href=\"http:\/\/skmo.sk\/dokument.php?id=173\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc1z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=266\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttc1r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>B-I-1<\/strong><br \/>\n<em>Palindr\u00f3mom<\/em> rozumieme prirodzen\u00e9 \u010d\u00edslo, ktor\u00e9 sa \u010d\u00edta rovnako odpredu aj odzadu, napr. 16 261. N\u00e1jdite najv\u00e4\u010d\u0161\u00ed \u0161tvormiestny palindr\u00f3m, ktor\u00e9ho druh\u00e1 mocnina je tie\u017e palindr\u00f3mom.<\/p>\n<div class=\"pri\">(E. Kov\u00e1\u010d)<\/div>\n<p><strong>B-I-2<\/strong><br \/>\nN\u00e1jdite v\u0161etky trojice re\u00e1lnych \u010d\u00edsel (<em>x, y, z<\/em>) vyhovuj\u00facich s\u00fastave rovn\u00edc<\/p>\n<div class=\"r\">x<sup>3<\/sup> + y<sup>3<\/sup> = 9z<sup>3<\/sup>,<br \/>x<sup>2<\/sup>y + y<sup>2<\/sup>x = 6z<sup>3<\/sup>.<\/div>\n<div class=\"pri\">(J. Zhouf)<\/div>\n<p><strong>B-I-3<\/strong><br \/>\nJe dan\u00fd trojuholn\u00edk so stranami d\u013a\u017eok <em>a, b, c<\/em> a obsahom <em>S<\/em>. Dok\u00e1\u017ete, \u017ee rovnos\u0165<\/p>\n<div class=\"r\">2c<sup>2<\/sup> = |a<sup>2<\/sup> &#8211; b<sup>2<\/sup>|<\/div>\n<p>plat\u00ed pr\u00e1ve vtedy, ke\u010f existuje trojuholn\u00edk so stranami d\u013a\u017eok <em>a, b, 2c<\/em> a obsahom 2<em>S<\/em>.<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<p><strong>B-I-4<\/strong><br \/>\n<em>Krokom<\/em> budeme rozumie\u0165 nahradenie usporiadanej trojice cel\u00fdch \u010d\u00edsel (<em>p, q, r<\/em>) trojicou (<em>r + 5q, 3r \u2212 5p, 2q \u2212 3p<\/em>). Rozhodnite, \u010di existuje cel\u00e9 \u010d\u00edslo <em>k<\/em> tak\u00e9, \u017ee z trojice (1, 3, 7) vznikne po kone\u010dnom po\u010dte krokov trojica (<em>k<\/em>, <em>k<\/em> + 1, <em>k<\/em> + 2).<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<p><strong>B-I-5<\/strong><br \/>\nV rovine je dan\u00fd pravouhl\u00fd lichobe\u017en\u00edk <em>ABCD<\/em> s dlh\u0161ou z\u00e1klad\u0148ou <em>AB<\/em> a prav\u00fdm uhlom pri vrchole <em>A<\/em>. Kru\u017enica <em>k<\/em><sub>1<\/sub> zostrojen\u00e1 nad stranou <em>AD<\/em> ako priemerom a kru\u017enica <em>k<\/em><sub>2<\/sub>, ktor\u00e1 prech\u00e1dza vrcholmi <em>B<\/em>, <em>C<\/em> a dot\u00fdka sa priamky <em>AB<\/em>, maj\u00fa vonkaj\u0161\u00ed dotyk v bode <em>P<\/em>. Dok\u00e1\u017ete, \u017ee uhly <em>CPD<\/em> a <em>ABC<\/em> s\u00fa zhodn\u00e9!<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><strong>B-I-6<\/strong><br \/>\nV kartezi\u00e1nskej s\u00fastave s\u00faradn\u00edc O<em>uv<\/em> zn\u00e1zornite mno\u017einu v\u0161etk\u00fdch bodov [<em>u, v<\/em>], kde <em>u<\/em> > 0, pre ktor\u00e9 m\u00e1 rovnica<\/p>\n<div class=\"r\">|x<sup>2<\/sup> &#8211; ux| + vx &#8211; 1 = 0<\/div>\n<p>s nezn\u00e1mou <em>x<\/em> pr\u00e1ve tri r\u00f4zne re\u00e1lne rie\u0161enia.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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><a href=\"http:\/\/skmo.sk\/dokument.php?id=176\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttb2z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=267\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttb2r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>B-S-1<\/strong><br \/>\nN\u00e1jdite najv\u00e4\u010d\u0161ie p\u00e4\u0165miestne prirodzen\u00e9 \u010d\u00edslo, ktor\u00e9 je delite\u013en\u00e9 \u010d\u00edslom 101 a ktor\u00e9 sa \u010d\u00edta odpredu rovnako ako odzadu. <\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><strong>B-S-2<\/strong><br \/>\nJe dan\u00fd konvexn\u00fd \u0161tvoruholn\u00edk <em>ABCD<\/em>. Ozna\u010dme <em>P<\/em> priese\u010dn\u00edk jeho uhloprie\u010dok a <em>Q<\/em> priese\u010dn\u00edk spojn\u00edc stredov jeho proti\u013eahl\u00fdch str\u00e1n. Ak bod <em>Q<\/em> le\u017e\u00ed na uhloprie\u010dke <em>BD<\/em>, je bod <em>P<\/em> stredom uhloprie\u010dky <em>AC<\/em>. Dok\u00e1\u017ete. <\/p>\n<div class=\"pri\">(E. Kov\u00e1\u010d)<\/div>\n<p><strong>B-S-3<\/strong><br \/>\nKo\u013eko r\u00f4znych v\u00fdsledkov m\u00f4\u017eeme dosta\u0165, ak s\u010d\u00edtame ka\u017ed\u00e9 dve z dan\u00fdch piatich r\u00f4znych prirodzen\u00fdch \u010d\u00edsel? Pre ka\u017ed\u00fd mo\u017en\u00fd po\u010det uve\u010fte pr\u00edklad takej p\u00e4tice \u010d\u00edsel. <\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=179\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttb3z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=268\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"buttb3r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>B-II-1<\/strong><br \/>\nUr\u010dte najv\u00e4\u010d\u0161\u00ed po\u010det po sebe id\u00facich p\u00e4\u0165miestnych prirodzen\u00fdch \u010d\u00edsel, medzi ktor\u00fdmi nie je \u017eiadny palindr\u00f3m (t. j. \u010d\u00edslo, ktor\u00e9 sa \u010d\u00edta odpredu rovnako ako odzadu). <\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><strong>B-II-2<\/strong><br \/>\nV rovine je dan\u00fd pravouhl\u00fd trojuholn\u00edk <em>ABC<\/em>. Nech <em>K<\/em> je \u013eubovo\u013en\u00fd bod prepony <em>AB<\/em>. Kru\u017enica zostrojen\u00e1 nad \u00fase\u010dkou <em>CK<\/em> ako nad priemerom pretne odvesny <em>BC<\/em> a <em>CA<\/em> vo vn\u00fatorn\u00fdch bodoch, ktor\u00e9 ozna\u010d\u00edme postupne <em>L<\/em> a <em>M<\/em>. Rozhodnite, pre ktor\u00fd bod <em>K<\/em> m\u00e1 \u0161tvoruholn\u00edk <em>ABLM<\/em> najmen\u0161\u00ed mo\u017en\u00fd obsah. <\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><strong>B-II-3<\/strong><br \/>\nUr\u010dte v\u0161etky re\u00e1lne \u010d\u00edsla <em>p<\/em>, pre ktor\u00e9 m\u00e1 rovnica <\/p>\n<div class=\"r\">(x \u2212 1)<sup>2<\/sup> = 3|x| \u2212 px<\/div>\n<p>pr\u00e1ve tri r\u00f4zne rie\u0161enia v obore re\u00e1lnych \u010d\u00edsel. <\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><strong>B-II-4<\/strong><br \/>\nV rovine je dan\u00fd pravouhl\u00fd lichobe\u017en\u00edk <em>ABCD<\/em> s dlh\u0161ou z\u00e1klad\u0148ou <em>AB<\/em> a prav\u00fdm uhlom pri vrchole <em>A<\/em>. Ozna\u010dme <em>k<sub>1<\/sub><\/em> kru\u017enicu zostrojen\u00fa nad stranou <em>AD<\/em> ako nad priemerom a <em>k<sub>2<\/sub><\/em> kru\u017enicu, ktor\u00e1 prech\u00e1dza bodmi <em>B<\/em>, <em>C<\/em> a dot\u00fdka sa priamky <em>AB<\/em>. Ak maj\u00fa kru\u017enice <em>k<sub>1<\/sub><\/em>, <em>k<sub>2<\/sub><\/em> vonkaj\u0161\u00ed dotyk v bode <em>P<\/em>, je priamka <em>BC<\/em> doty\u010dnicou kru\u017enice op\u00edsanej trojuholn\u00edku <em>CDP<\/em>. Dok\u00e1\u017ete. <\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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><a href=\"http:\/\/skmo.sk\/dokument.php?id=172\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"butta1z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=262\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"butta1r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>A-I-1<\/strong><br \/>\nPostupnos\u0165 cel\u00fdch \u010d\u00edsel <img decoding=\"async\" src=\"52\/52aa2.png\" align=\"middle\" alt=\"\" \/> s prv\u00fdm \u010dlenom <em>x<sub>1<\/sub><\/em>&nbsp;=&nbsp;1 sp\u013a\u0148a podmienku<\/p>\n<div class=\"r\">x<sub>n<\/sub>&nbsp;=&nbsp;&#177;&nbsp;x<sub>n\u22121<\/sub>&nbsp;&#177; \u00b7 \u00b7 \u00b7 &#177;&nbsp;x<sub>1<\/sub><\/div>\n<p>s vhodnou vo\u013ebou znamienok &#8222;+&#8220; a &#8222;\u2212&#8220; pre \u013eubovo\u013en\u00e9 <em>n<\/em>&nbsp;>&nbsp;1, napr\u00edklad <em>x<sub>2<\/sub><\/em> = \u2212<em>x<sub>1<\/sub><\/em>, <em>x<sub>3<\/sub><\/em> = \u2212<em>x<sub>2<\/sub><\/em> + <em>x<sub>1<\/sub><\/em>, <em>x<sub>4<\/sub><\/em> = <em>x<sub>3<\/sub><\/em> \u2212 <em>x<sub>2<\/sub><\/em> \u2212 <em>x<sub>1<\/sub><\/em>, &#8230; Pre dan\u00e9 <em>n<\/em> ur\u010dte v\u0161etky mo\u017en\u00e9 hodnoty <em>x<sub>n<\/sub><\/em>. <\/p>\n<div class=\"pri\">(J. F\u00f6ldes)<\/div>\n<p><strong>A-I-2<\/strong><br \/>\nNa priamke <em>p<\/em> s\u00fa dan\u00e9 r\u00f4zne body <em>A, B, C<\/em> v tomto porad\u00ed, kde |<em>AB<\/em>|b&nbsp;=b&nbsp;1 a |<em>BC<\/em>|b&nbsp;=b&nbsp;<em>h<\/em>. Uva\u017eujme kru\u017enice <em>k<sub>A<\/sub><\/em>, <em>k<sub>B<\/sub><\/em>, <em>k<sub>C<\/sub><\/em>, ktor\u00e9 sa dot\u00fdkaj\u00fa priamky <em>p<\/em> postupne v bodoch <em>A, B, C<\/em>. Kru\u017enice <em>k<sub>A<\/sub><\/em>, <em>k<sub>B<\/sub><\/em> maj\u00fa pritom vonkaj\u0161\u00ed dotyk v bode <em>P<\/em> a kru\u017enice <em>k<sub>B<\/sub><\/em>, <em>k<sub>C<\/sub><\/em> maj\u00fa vonkaj\u0161\u00ed dotyk v bode <em>Q<\/em>. Ur\u010dte v\u0161etky tak\u00e9 hodnoty polomeru kru\u017enice <em>k<sub>B<\/sub><\/em>, pre ktor\u00e9 je trojuholn\u00edk <em>BPQ<\/em> rovnoramenn\u00fd. <\/p>\n<div class=\"pri\">(J. Zhouf)<\/div>\n<p><strong>A-I-3<\/strong><br \/>\nUr\u010dte v\u0161etky mo\u017en\u00e9 hodnoty v\u00fdrazu <\/p>\n<div class=\"pc\"><img decoding=\"async\" src=\"52\/52aa1.png\" alt=\"\" \/><\/div>\n<p>kde <em>a, b, c<\/em> s\u00fa d\u013a\u017eky str\u00e1n trojuholn\u00edka.<\/p>\n<div class=\"pri\">(P. Ka\u0148ovsk\u00fd)<\/div>\n<p><strong>A-I-4<\/strong><br \/>\nUr\u010dte v\u0161etky prirodzen\u00e9 \u010d\u00edsla <em>n<\/em>>1 tak\u00e9, \u017ee v niektorej \u010d\u00edselnej s\u00fastave so z\u00e1kladom <em>z<\/em>&#8805;5 plat\u00ed nasledovn\u00e9 krit\u00e9rium delite\u013enosti:<br \/>Trojmiestne \u010d\u00edslo (<em>abc<\/em>)<sub>z<\/sub> je delite\u013en\u00e9 \u010d\u00edslom <em>n<\/em> pr\u00e1ve vtedy, ke\u010f je \u010d\u00edslom <em>n<\/em> delite\u013en\u00e9 \u010d\u00edslo <em>c&nbsp;+&nbsp;3b&nbsp;\u2212&nbsp;4a<\/em>.<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<p><strong>A-I-5<\/strong><br \/>\nV rovine s\u00fa dan\u00e9 tri r\u00f4zne body <em>K, L, M<\/em>, ktor\u00e9 v tomto porad\u00ed le\u017eia na priamke. V tejto rovine n\u00e1jdite mno\u017einu v\u0161etk\u00fdch vrcholov <em>C<\/em> \u0161tvorcov <em>ABCD<\/em> tak\u00fdch, \u017ee bod <em>K<\/em> le\u017e\u00ed na strane <em>AB<\/em>, bod <em>L<\/em> na uhloprie\u010dke <em>BD<\/em> a bod <em>M<\/em> na strane <em>CD<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160im\u0161a)<\/div>\n<p><strong>A-I-6<\/strong><br \/>\nHr\u00e1\u010di A a B hraj\u00fa na doske zlo\u017eenej zo \u0161iestich pol\u00ed o\u010d\u00edslovan\u00fdch 1, 2, . . . , 6 nasleduj\u00facu hru. Na za\u010diatku je umiestnen\u00e1 na pole s \u010d\u00edslom 2 fig\u00farka a potom sa h\u00e1d\u017ee be\u017enou hracou kockou. Ak padne \u010d\u00edslo delite\u013en\u00e9 tromi, posunie sa fig\u00farka na pole s \u010d\u00edslom o jedna men\u0161\u00edm, inak na pole s \u010d\u00edslom o jedna v\u00e4\u010d\u0161\u00edm. Hra kon\u010d\u00ed v\u00ed\u0165azstvom hr\u00e1\u010da A (resp. B), ak sa dostane fig\u00farka na pole s \u010d\u00edslom 1 (resp. 6). S akou pravdepodobnos\u0165ou zv\u00ed\u0165az\u00ed hr\u00e1\u010d A?<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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><a href=\"http:\/\/skmo.sk\/dokument.php?id=175\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"butta2z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=263\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"butta2r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>A-S-1<\/strong><br \/>\nHovor\u00edme, \u017ee tri navz\u00e1jom r\u00f4zne prirodzen\u00e9 \u010d\u00edsla tvoria s\u00fa\u010dtov\u00fa trojicu, ak s\u00fa\u010det prv\u00fdch dvoch z nich sa rovn\u00e1 tretiemu \u010d\u00edslu. Zistite, ak\u00fd najv\u00e4\u010d\u0161\u00ed po\u010det s\u00fa\u010dtov\u00fdch troj\u00edc sa m\u00f4\u017ee nach\u00e1dza\u0165 v mno\u017eine dvadsiatich prirodzen\u00fdch \u010d\u00edsel. <\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<p><strong>A-S-2<\/strong><br \/>\nV rovine s\u00fa dan\u00e9 kru\u017enice <em>k<sub>1<\/sub>(S<sub>1<\/sub>, r<sub>1<\/sub>)<\/em> a <em>k<sub>2<\/sub>(S<sub>2<\/sub>, r<sub>2<\/sub>)<\/em> tak, \u017ee <em>S<sub>2<\/sub><\/em> \u2208 <em>k<sub>1<\/sub><\/em> a <em>r<sub>1<\/sub><\/em> > <em>r<sub>2<\/sub><\/em>. Spolo\u010dn\u00e9 doty\u010dnice oboch kru\u017en\u00edc sa dot\u00fdkaj\u00fa kru\u017enice <em>k<sub>1<\/sub><\/em> v bodoch <em>P<\/em> a <em>Q<\/em>. Dok\u00e1\u017ete, \u017ee priamka <em>PQ<\/em> sa dot\u00fdka kru\u017enice <em>k<sub>2<\/sub><\/em>.<\/p>\n<div class=\"pri\">(J. F\u00f6ldes)<\/div>\n<p><strong>A-S-3<\/strong><br \/>\nZistite, pre ktor\u00e9 re\u00e1lne \u010d\u00edsla <em>p<\/em> maj\u00fa rovnice <\/p>\n<div class=\"r\">x<sup>3<\/sup> + x<sup>2<\/sup> \u2212 36x \u2212 p = 0,<br \/>x<sup>3<\/sup> \u2212 2x<sup>2<\/sup> \u2212 px + 2p = 0<\/div>\n<p> spolo\u010dn\u00fd kore\u0148. <\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td align=\"left\" width=\"85%\">\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><a href=\"http:\/\/skmo.sk\/dokument.php?id=178\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"butta3z\" type=\"button\" value=\"Zadanie\" \/><\/a><a href=\"http:\/\/skmo.sk\/dokument.php?id=264\" target=\"_blank\" rel=\"noopener noreferrer\"><input name=\"butta3r\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>A-II-1<\/strong><br \/>\nN\u00e1jdite z\u00e1klady z v\u0161etk\u00fdch \u010d\u00edseln\u00fdch s\u00fastav, v ktor\u00fdch je \u0161tvormiestne \u010d\u00edslo (1001)<sup>z<\/sup> delite\u013en\u00e9 dvojmiestnym \u010d\u00edslom (41)<sup>z<\/sup>.<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<p><strong>A-II-2<\/strong><br \/>\nVn\u00fatri strany <em>AB<\/em> dan\u00e9ho ostrouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em> n\u00e1jdite bod <em>S<\/em> tak, aby trojuholn\u00edk <em>SXY<\/em>, kde <em>X<\/em> a <em>Y<\/em> s\u00fa postupne stredy kru\u017en\u00edc op\u00edsan\u00fdch trojuholn\u00edkom <em>ASC<\/em> a <em>BSC<\/em>, mal najmen\u0161\u00ed mo\u017en\u00fd obsah.<\/p>\n<div class=\"pri\">(P. \u010cernek)<\/div>\n<p><strong>A-II-3<\/strong><br \/>\nV obore re\u00e1lnych \u010d\u00edsel rie\u0161te s\u00fastavu rovn\u00edc <\/p>\n<div class=\"r\">log<sub>x<\/sub> (y + z) = p,<br \/>log<sub>y<\/sub> (z + x) = p,<br \/>log<sub>z<\/sub> (x + y) = p<\/div>\n<p>s nezn\u00e1mymi <em>x, y, z<\/em> a nez\u00e1porn\u00fdm celo\u010d\u00edseln\u00fdm parametrom <em>p<\/em>.<\/p>\n<div class=\"pri\">(J. \u0160vr\u010dek)<\/div>\n<p><strong>A-II-4<\/strong><br \/>\nPostupnos\u0165 <img decoding=\"async\" src=\"52\/52aa2.png\" align=\"middle\" alt=\"\" \/> s prv\u00fdm \u010dlenom <em>x<sub>1<\/sub><\/em>&nbsp;=&nbsp;1 sp\u013a\u0148a pre ka\u017ed\u00e9 <em>n<\/em>&nbsp;>&nbsp;1 podmienku<\/p>\n<div class=\"pc\"><img decoding=\"async\" src=\"52\/52ba1.png\" alt=\"\" \/><\/div>\n<p>s vhodnou vo\u013ebou znamienok \u201d +\u201c a \u201d \u2212\u201c v exponentoch mocn\u00edn.<br \/>\na) Rozhodnite, \u010di niektor\u00fd \u010dlen takej postupnosti mus\u00ed by\u0165 v\u00e4\u010d\u0161\u00ed ako 1 000.<br \/>\nb) Zistite najmen\u0161iu mo\u017en\u00fa hodnotu \u010dlena <em>x<sub>1 000 000<\/sub><\/em>.<br \/>\nc) Dok\u00e1\u017ete, \u017ee nerovnos\u0165 <em>x<sub>n<\/sub><\/em> < 4 nem\u00f4\u017ee plati\u0165 pre dev\u00e4\u0165 \u010dlenov <em>x<sub>n<\/sub><\/em> takej postupnosti.<\/p>\n<div class=\"pri\">(J. F\u00f6ldes)<\/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 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>52. ro\u010dn\u00edk matematickej olympi\u00e1dy 2002-2003 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":"closed","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":[36],"tags":[],"class_list":["post-1568","post","type-post","status-publish","format-standard","hentry","category-52-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3rie A, B a C - Matematick\u00e1 olympi\u00e1da<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/matematika.besaba.com\/?p=1568\" \/>\n<meta property=\"og:locale\" content=\"sk_SK\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3rie A, B a C - Matematick\u00e1 olympi\u00e1da\" \/>\n<meta property=\"og:description\" content=\"52. ro\u010dn\u00edk matematickej olympi\u00e1dy 2002-2003 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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