{"id":829,"date":"2013-09-03T19:00:39","date_gmt":"2013-09-03T19:00:39","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=829"},"modified":"2015-01-03T20:10:12","modified_gmt":"2015-01-03T20:10:12","slug":"sutazne-ulohy-kategorie-a-b-a-c","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=829","title":{"rendered":"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3rie A, B a C"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>63. ro\u010dn\u00edk matematickej olympi\u00e1dy 2013-2014<br \/>S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3rie A, B a C<\/p>\n<p><span class=\"povinne\">aktualizovan\u00e9 3.1.2015 19:30<\/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\">\u00a0<\/td>\n<td align=\"center\"><span class=\"a\"><a href=\"#a4\">celo\u0161t\u00e1tne kolo<\/a><\/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=\"\/63\/63_ABC.pdf\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" alt=\"\" src=\"\/img\/down.png\" width=\"48\" height=\"48\" align=\"absmiddle\" border=\"0\" \/><\/a> (ulo\u017ei\u0165 ako &#8230;)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=1009\"><input name=\"buttc1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><a name=\"c1\"><\/a><\/p>\n<p><strong>C-I-1<\/strong><br \/> Ur\u010dte, ak\u00fa najmen\u0161iu hodnotu m\u00f4\u017ee nadob\u00fada\u0165 v\u00fdraz <em>V = (a &#8211; b)<sup>2<\/sup> + (b &#8211; c)<sup>2<\/sup> + (c &#8211; a)<sup>2<\/sup><\/em>, ak re\u00e1lne \u010d\u00edsla <em>a, b, c<\/em> sp\u013a\u0148aj\u00fa dvojicu podmienok<\/p>\n<p style=\"text-align: center;\"><em>a + 3b + c = 6,<br \/>-a + b &#8211; c = 2.<\/em><\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>C-I-2<\/strong><br \/> V rovine s\u00fa dan\u00e9 body <em>A<\/em>, <em>P<\/em> , <em>T<\/em> nele\u017eiace na jednej priamke. Zostrojte trojuholn\u00edk <em>ABC<\/em> tak, aby <em>P<\/em> bola p\u00e4ta jeho v\u00fd\u0161ky z vrcholu <em>A<\/em> a <em>T<\/em> bod dotyku strany <em>AB<\/em> s kru\u017enicou jemu vp\u00edsanou. Uve\u010fte diskusiu o po\u010dte rie\u0161en\u00ed vzh\u013eadom na polohu dan\u00fdch bodov.<\/p>\n<div class=\"pri\">(Pavel Leischner)<\/div>\n<p><strong>C-I-3<\/strong><br \/> \u010c\u00edslo <em>n<\/em> je s\u00fa\u010dinom troch r\u00f4znych prvo\u010d\u00edsel. Ak zv\u00e4\u010d\u0161\u00edme dve men\u0161ie z nich o 1 a najv\u00e4\u010d\u0161ie ponech\u00e1me nezmenen\u00e9, zv\u00e4\u010d\u0161\u00ed sa ich s\u00fa\u010din o 915. Ur\u010dte \u010d\u00edslo <em>n<\/em>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>C-I-4<\/strong><br \/> Vo \u0161tvorci <em>ABCD<\/em> ozna\u010dme <em>K<\/em> stred strany <em>AB<\/em> a <em>L<\/em> stred strany <em>AD<\/em>. \u00dase\u010dky <em>KD<\/em> a <em>LC<\/em> sa pret\u00ednaj\u00fa v bode <em>M<\/em> a rozde\u013euj\u00fa \u0161tvorec na dva trojuholn\u00edky a dva \u0161tvoruholn\u00edky. Vypo\u010d\u00edtajte ich obsahy, ak \u00fase\u010dka LM m\u00e1 d\u013a\u017eku 1 cm.<\/p>\n<div class=\"pri\">(Leo Bo\u010dek)<\/div>\n<p><strong>C-I-5<\/strong><br \/> Dok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 nep\u00e1rne prirodzen\u00e9 \u010d\u00edslo <em>n<\/em> je s\u00fa\u010det <strong><em>n<\/em><sup>4<\/sup> + 2<em>n<\/em> + 2013<\/strong> delite\u013en\u00fd \u010d\u00edslom 96.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-I-6<\/strong><br \/> \u0160achov\u00e9ho turnaja sa z\u00fa\u010dastnilo 8 hr\u00e1\u010dov a ka\u017ed\u00fd s ka\u017ed\u00fdm odohral jednu partiu. Za prvenstvo z\u00edskal hr\u00e1\u010d 1 bod, za rem\u00edzu pol bodu, za prehru \u017eiadny bod. Na konci turnaja mali v\u0161etci \u00fa\u010dastn\u00edci r\u00f4zne po\u010dty bodov. Hr\u00e1\u010d, ktor\u00fd skon\u010dil na 2. mieste, z\u00edskal rovnak\u00fd po\u010det bodov ako posledn\u00ed \u0161tyria dokopy. Ur\u010dte v\u00fdsledok partie medzi 4. a 6. hr\u00e1\u010dom v celkovom porad\u00ed.<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/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\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=1013\"><input name=\"buttc2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>C-S-1<\/strong><br \/>\nUr\u010dte, ak\u00e9 hodnoty m\u00f4\u017ee nadob\u00fada\u0165 v\u00fdraz <em>V = ab + bc + cd + da<\/em>, ak re\u00e1lne \u010d\u00edsla <em>a, b, c, d<\/em> sp\u013a\u0148aj\u00fa dvojicu podmienok<\/p>\n<div class=\"r\">2a \u2212 5b + 2c \u2212 5d = 4,<br \/>3a + 4b + 3c + 4d = 6.<\/div>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>C-S-2<\/strong><br \/>\n\u010c\u00edsla 1, 2, &#8230;, 10 rozde\u013ete na dve skupiny tak, aby najmen\u0161\u00ed spolo\u010dn\u00fd n\u00e1sobok s\u00fa\u010dinu v\u0161etk\u00fdch \u010d\u00edsel prvej skupiny a s\u00fa\u010dinu v\u0161etk\u00fdch \u010d\u00edsel druhej skupiny bol \u010do najmen\u0161\u00ed.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>C-S-3<\/strong><br \/>\nDan\u00fd je trojuholn\u00edk <em>ABC<\/em> s prav\u00fdm uhlom pri vrchole <em>C<\/em>. Stredom <em>I<\/em> kru\u017enice trojuholn\u00edku vp\u00edsanej vedieme rovnobe\u017eky so stranami <em>CA<\/em> a <em>CB<\/em>, ktor\u00e9 pretn\u00fa preponu postupne v bodoch <em>X<\/em> a <em>Y<\/em>. Dok\u00e1\u017ete, \u017ee plat\u00ed <strong>|<em>AX<\/em>|<sup>2<\/sup> + |<em>BY<\/em>|<sup>2<\/sup> = |<em>XY<\/em>|<sup>2<\/sup><\/strong>.<\/p>\n<div class=\"pri\">(Michal Rol\u00ednek)<\/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\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=1079\"><input name=\"buttc3\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>C-II-1<\/strong><br \/>\nN\u00e1jdite v\u0161etky trojice (nie nutne r\u00f4znych) cifier <em>a, b, c<\/em> tak\u00e9, \u017ee p\u00e4\u0165cifern\u00e9 \u010d\u00edsla <span style=\"text-decoration: overline;\">6abc3<\/span> a <span style=\"text-decoration: overline;\">3abc6<\/span> s\u00fa v pomere 63 : 36.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-II-2<\/strong><br \/>\n\u0160achov\u00e9ho turnaja sa z\u00fa\u010dastnilo 5 hr\u00e1\u010dov a ka\u017ed\u00fd s ka\u017ed\u00fdm odohral jednu partiu. Za prvenstvo z\u00edskal hr\u00e1\u010d 1 bod, za rem\u00edzu pol bodu, za prehru \u017eiadny bod. Poradie hr\u00e1\u010dov na turnaji sa ur\u010duje pod\u013ea po\u010dtu z\u00edskan\u00fdch bodov. Jedin\u00fdm \u010fal\u0161\u00edm krit\u00e9riom rozhoduj\u00facim o kone\u010dnom umiestnen\u00ed hr\u00e1\u010dov v pr\u00edpade rovnosti bodov je po\u010det v\u00fdhier (kto m\u00e1 viac v\u00fdhier, je na tom v umiestnen\u00ed lep\u0161ie). Na turnaji z\u00edskali v\u0161etci hr\u00e1\u010di rovnak\u00fd po\u010det bodov. Vojto porazil Petra a o prv\u00e9 miesto sa delil s Tom\u00e1\u0161om. Ako dopadla partia medzi Petrom a Martinom? <\/p>\n<div class=\"pri\">(Martin Pan\u00e1k)<\/div>\n<p><strong>C-II-3<\/strong><br \/>\nPre kladn\u00e9 re\u00e1lne \u010d\u00edsla <em>a, b, c<\/em> plat\u00ed <strong><em>c<\/em><sup>2<\/sup> + <em>ab<\/em> = <em>a<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup><\/strong>. Dok\u00e1\u017ete, \u017ee potom plat\u00ed aj <strong><em>c<\/em><sup>2<\/sup> + <em>ab<\/em> &#8804; <em>ac<\/em> + <em>bc<\/em><\/strong>.<\/p>\n<div class=\"pri\">(Michal Rol\u00ednek)<\/div>\n<p><strong>C-II-4<\/strong><br \/>\nDan\u00fd je konvexn\u00fd \u0161tvoruholn\u00edk <em>ABCD<\/em> s bodom <em>E<\/em> vn\u00fatri strany <em>AB<\/em> tak, \u017ee plat\u00ed |<font size=\"+1\">&ang;<\/font><em>ADE<\/em>| = |<font size=\"+1\">&ang;<\/font><em>DEC<\/em>| = |<font size=\"+1\">&ang;<\/font><em>ECB<\/em>|. Obsahy trojuholn\u00edkov <em>AED<\/em> a <em>CEB<\/em> s\u00fa postupne 18 cm<sup>2<\/sup> a 8 cm<sup>2<\/sup>. Ur\u010dte obsah trojuholn\u00edka <em>ECD<\/em>. <\/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 align=\"left\" width=\"85%\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b1\"><\/a><a href=\"#top\"><img decoding=\"async\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=1008\"><input name=\"buttb1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>B-I-1<\/strong><br \/>\nKa\u017ed\u00e9mu vrcholu pravideln\u00e9ho 63-uholn\u00edka prirad\u00edme jedno z \u010d\u00edsel 1 alebo -1. Ku ka\u017edej jeho strane prip\u00ed\u0161eme s\u00fa\u010din \u010d\u00edsel v jej vrcholoch a v\u0161etky \u010d\u00edsla pri jednotliv\u00fdch stran\u00e1ch s\u010d\u00edtame. N\u00e1jdite najmen\u0161iu mo\u017en\u00fa nez\u00e1porn\u00fa hodnotu tak\u00e9ho s\u00fa\u010dtu.<\/p>\n<div class=\"pri\">(Pavel Cal\u00e1bek)<\/div>\n<p><strong>B-I-2<\/strong><br \/>Ur\u010dte v\u0161etky dvojice (<em>x, y<\/em>) re\u00e1lnych \u010d\u00edsel, pre ktor\u00e9 plat\u00ed nerovnos\u0165<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" alt=\"\" src=\"\/63\/63ba1.png\" \/>.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-I-3<\/strong><br \/>Nech <em>D<\/em> je \u013eubovo\u013en\u00fd vn\u00fatorn\u00fd bod strany <em>AB<\/em> trojuholn\u00edka <em>ABC<\/em>. Na polpriamkach <em>BC<\/em> a <em>AC<\/em> zvo\u013eme postupne body <em>E<\/em> a <em>F<\/em> tak, aby platilo |<em>BD<\/em>| = |<em>BE<\/em>| a |<em>AD<\/em>| = |<em>AF<\/em>|. Dok\u00e1\u017ete, \u017ee body <em>C, E, F<\/em> a stred <em>I<\/em> kru\u017enice vp\u00edsanej trojuholn\u00edku <em>ABC<\/em> le\u017eia na jednej kru\u017enici.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-I-4<\/strong><br \/>Dana nap\u00edsala na papier trojcifern\u00e9 \u010d\u00edslo, ktor\u00e9 po delen\u00ed siedmimi d\u00e1va zvy\u0161ok 2. Prehoden\u00edm prv\u00fdch dvoch cifier vzniklo trojcifern\u00e9 \u010d\u00edslo, ktor\u00e9 po delen\u00ed siedmimi d\u00e1va zvy\u0161ok 3. \u010c\u00edslo, ktor\u00e9 vznikne prehoden\u00edm posledn\u00fdch dvoch cifier p\u00f4vodn\u00e9ho \u010d\u00edsla, d\u00e1va po delen\u00ed siedmimi zvy\u0161ok 5. Ak\u00fd zvy\u0161ok po delen\u00ed siedmimi bude ma\u0165 \u010d\u00edslo, ktor\u00e9 vznikne prehoden\u00edm prvej a poslednej cifry Daninho \u010d\u00edsla?<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>B-I-5<\/strong><br \/>V rovine s\u00fa dan\u00e9 body <em>A, T, U<\/em> tak, \u017ee uhol <em>ATU<\/em> je tup\u00fd. Zostrojte trojuholn\u00edk <em>ABC<\/em>, v ktorom <em>T, U<\/em> s\u00fa postupne body dotyku strany <em>BC<\/em> s kru\u017enicou trojuholn\u00edku vp\u00edsanou a prip\u00edsanou. (Prip\u00edsanou kru\u017enicou tu rozumieme kru\u017enicu, ktor\u00e1 sa okrem strany <em>BC<\/em> dot\u00fdka aj polpriamok opa\u010dn\u00fdch k polpriamkam <em>BA<\/em> a <em>CA<\/em>.)<\/p>\n<div class=\"pri\">(\u0160\u00e1rka Gergelitsov\u00e1)<\/div>\n<p><strong>B-I-6<\/strong><br \/>N\u00e1jdite najmen\u0161ie re\u00e1lne \u010d\u00edslo <em>r<\/em> tak\u00e9, \u017ee ty\u010d s d\u013a\u017ekou 1 mo\u017eno rozl\u00e1ma\u0165 na \u0161tyri \u010dasti d\u013a\u017eky nanajv\u00fd\u0161 <em>r<\/em> tak, \u017ee sa zo \u017eiadnych troch t\u00fdchto \u010dast\u00ed ned\u00e1 zlo\u017ei\u0165 trojuholn\u00edk.<\/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 align=\"left\" width=\"85%\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"b2\"><\/a><a href=\"#top\"><img decoding=\"async\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=1011\"><input name=\"buttb2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>B-S-1<\/strong><br \/>\nV obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te rovnicu<\/p>\n<div class=\"r\">2<sup>|x+1|<\/sup> \u2212 2<sup>x<\/sup> = 1 + |2<sup>x<\/sup> \u2212 1|.<\/div>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>B-S-2<\/strong><br \/>\nMno\u017eina <strong><em>M<\/em><\/strong> obsahuje 2014 r\u00f4znych re\u00e1lnych \u010d\u00edsel. S\u00fa\u010det ka\u017ed\u00fdch dvoch r\u00f4znych \u010d\u00edsel z mno\u017einy <strong><em>M<\/em><\/strong> je cel\u00e9 \u010d\u00edslo.<br \/>\n&nbsp;&nbsp;&nbsp;&nbsp;a) Rozhodnite, \u010di existuje tak\u00e1 mno\u017eina <strong><em>M<\/em><\/strong>, ktor\u00e1 neobsahuje \u017eiadne cel\u00e9 \u010d\u00edslo.<br \/>\n&nbsp;&nbsp;&nbsp;&nbsp;b) Rozhodnite, \u010di existuje tak\u00e1 mno\u017eina <strong><em>M<\/em><\/strong>, ktor\u00e1 obsahuje iracion\u00e1lne \u010d\u00edslo.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>B-S-3<\/strong><br \/>\nNa priamke <em>a<\/em>, na ktorej le\u017e\u00ed strana <em>BC<\/em> trojuholn\u00edka <em>ABC<\/em>, s\u00fa dan\u00e9 body dotyku v\u0161etk\u00fdch troch jemu prip\u00edsan\u00fdch kru\u017en\u00edc (body <em>B<\/em> a <em>C<\/em> nie s\u00fa zn\u00e1me). N\u00e1jdite na tejto priamke bod dotyku kru\u017enice vp\u00edsanej. <\/p>\n<div class=\"pri\">(\u0160\u00e1rka Gergelitsov\u00e1)<\/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\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><a href=\"http:\/\/skmo.sk\/dokument.php?id=1078\"><input name=\"buttb3\" type=\"button\" value=\"Rie\u0161enie\" \/><\/a><\/form>\n<p><strong>B-II-1<\/strong><br \/>\nV obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te s\u00fastavu rovn\u00edc<\/p>\n<div class=\"r\">x2 + 6(y + z) = 85,<br \/>y2 + 6(z + x) = 85,<br \/>z2 + 6(x + y) = 85.<\/div>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>B-II-2<\/strong><br \/>\nJanko nap\u00edsal na tabu\u013eu nieko\u013eko r\u00f4znych prvo\u010d\u00edsel (aspo\u0148 tri). Ke\u010f s\u010d\u00edtal \u013eubovo\u013en\u00e9 dve z nich a tento s\u00fa\u010det zmen\u0161il o 7, bolo v\u00fdsledn\u00e9 \u010d\u00edslo medzi nap\u00edsan\u00fdmi. Ktor\u00e9 \u010d\u00edsla mohli na tabuli by\u0165? <\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>B-II-3<\/strong><br \/>\nNad stranami <em>BC<\/em> a <em>AB<\/em> ostrouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em> s\u00fa zvonka zostrojen\u00e9 polkru\u017enice <em>k<\/em> a <em>l<\/em>. Ozna\u010dme postupne <em>D<\/em> a <em>E<\/em> priese\u010dn\u00edky v\u00fd\u0161ok z vrcholov <em>A<\/em> a <em>C<\/em> s polkru\u017enicami <em>k<\/em> a <em>l<\/em> (v\u00fd\u0161kami rozumieme priamky). Dok\u00e1\u017ete, \u017ee plat\u00ed |<em>BE<\/em>| = |<em>BD<\/em>|.<\/p>\n<div class=\"pri\">(Veronika Huc\u00edkov\u00e1)<\/div>\n<p><strong>B-II-4<\/strong><br \/>\nV ka\u017edom pol\u00ed\u010dku tabu\u013eky 8 \u00d7 8 je nap\u00edsan\u00e9 jedno nez\u00e1porn\u00e9 cel\u00e9 \u010d\u00edslo tak, \u017ee ka\u017ed\u00e9 dve \u010d\u00edsla, ktor\u00e9 s\u00fa na pol\u00ed\u010dkach s\u00famerne zdru\u017een\u00fdch pod\u013ea jednej \u010di druhej uhloprie\u010dky, s\u00fa rovnak\u00e9. S\u00fa\u010det v\u0161etk\u00fdch 64 \u010d\u00edsel je 1 000, s\u00fa\u010det 16 \u010d\u00edsel na uhloprie\u010dkach je 200. Dok\u00e1\u017ete, \u017ee s\u00fa\u010det \u010d\u00edsel v ka\u017edom riadku aj st\u013apci tabu\u013eky je nanajv\u00fd\u0161 300. Plat\u00ed rovnak\u00fd<br \/>\nz\u00e1ver aj pre \u010d\u00edslo 299? <\/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 align=\"left\" width=\"85%\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a1\"><\/a><a href=\"#top\"><img decoding=\"async\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><input onclick=\"window.location='http:\/\/skmo.sk\/dokument.php?id=992'\" type=\"button\" name=\"butta1\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-I-1<\/strong><br \/> \u010c\u00edslo <em>n<\/em> je s\u00fa\u010dinom troch (nie nutne r\u00f4znych) prvo\u010d\u00edsel. Ke\u010f zv\u00e4\u010d\u0161\u00edme ka\u017ed\u00e9 z nich o 1, zv\u00e4\u010d\u0161\u00ed sa ich s\u00fa\u010din o 963. Ur\u010dte p\u00f4vodn\u00e9 \u010d\u00edslo <em>n<\/em>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-I-2<\/strong><br \/>Pre \u013eubovo\u013en\u00e9 kladn\u00e9 re\u00e1lne \u010d\u00edsla x, y, z dok\u00e1\u017ete nerovnos\u0165<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" alt=\"\" src=\"\/63\/63aa1.png\" align=\"absmiddle\" \/>, pri\u010dom <em>m<\/em>=min <img decoding=\"async\" alt=\"\" src=\"\/63\/63aa2.png\" align=\"absmiddle\" \/>.<\/p>\n<p>Zistite tie\u017e, kedy v dok\u00e1zanej nerovnosti nastane rovnos\u0165.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek, Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-I-3<\/strong><br \/>Ozna\u010dme <em>I<\/em> stred kru\u017enice vp\u00edsanej do dan\u00e9ho trojuholn\u00edka <em>ABC<\/em>. Predpokladajme, \u017ee kolmica na priamku <em>CI<\/em> veden\u00e1 bodom <em>I<\/em> pret\u00edna priamku <em>AB<\/em> v bode <em>M<\/em>. Dok\u00e1\u017ete, \u017ee kru\u017enica op\u00edsan\u00e1 trojuholn\u00edku <em>ABC<\/em> pret\u00edna \u00fase\u010dku <em>CM<\/em> v jej vn\u00fatornom bode <em>N<\/em> a \u017ee priamky <em>NI<\/em> a <em>MC<\/em> s\u00fa navz\u00e1jom kolm\u00e9.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>A-I-4<\/strong><br \/>Ozna\u010dme <em>l<\/em>(<em>n<\/em>) najv\u00e4\u010d\u0161ieho nep\u00e1rneho delite\u013ea \u010d\u00edsla <em>n<\/em>. Ur\u010dte hodnotu s\u00fa\u010dtu<\/p>\n<p class=\"pc\"><strong><em>l<\/em>(1) + <em>l<\/em>(2) + <em>l<\/em>(3)+ &#8230; + <em>l<\/em>(2<sup>2013<\/sup>)<\/strong>.<\/p>\n<div class=\"pri\">(Michal Rol\u00ednek)<\/div>\n<p><strong>A-I-5<\/strong><br \/>Ko\u013ek\u00fdmi r\u00f4znymi sp\u00f4sobmi mo\u017eno vydl\u00e1\u017edi\u0165 plochu 3\u00d710 dla\u017edicami 2\u00d71, ak je dovolen\u00e9 kl\u00e1s\u0165 ich v oboch navz\u00e1jom kolm\u00fdch smeroch?<\/p>\n<div class=\"pri\">(Stanislava Soj\u00e1kov\u00e1)<\/div>\n<p><strong>A-I-6<\/strong><br \/>V rovine dan\u00e9ho trojuholn\u00edka <em>ABC<\/em> ur\u010dte v\u0161etky body, ktor\u00fdch obrazy v osov\u00fdch s\u00famernostiach pod\u013ea priamok <em>AB, BC, CA<\/em> tvoria vrcholy rovnostrann\u00e9ho trojuholn\u00edka.<\/p>\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 align=\"left\" width=\"85%\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a name=\"a2\"><\/a><a href=\"#top\"><img decoding=\"async\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><input onclick=\"window.location='http:\/\/skmo.sk\/dokument.php?id=991'\" type=\"button\" name=\"butta2\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-S-1<\/strong><br \/>Dok\u00e1\u017ete, \u017ee pre ka\u017ed\u00e9 cel\u00e9 \u010d\u00edslo <em>n<\/em>\u22653 je 2<em>n<\/em>-cifern\u00e9 \u010d\u00edslo s dekadick\u00fdm z\u00e1pisom druhou mocninou niektor\u00e9ho cel\u00e9ho \u010d\u00edsla.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" alt=\"\" src=\"\/63\/63ab2.png\" width=\"112\" height=\"33\" \/><\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>A-S-2<\/strong><br \/>Ozna\u010dme <em>M<\/em> stred strany <em>AB<\/em> \u013eubovo\u013en\u00e9ho trojuholn\u00edka <em>ABC<\/em>. Dok\u00e1\u017ete, \u017ee rovnos\u0165 |&ang;<em>ABC<\/em>| + |&ang;<em>ACM<\/em>| = 90\u00b0 plat\u00ed pr\u00e1ve vtedy, ke\u010f je trojuholn\u00edk <em>ABC<\/em> rovnoramenn\u00fd so z\u00e1klad\u0148ou <em>AB<\/em> alebo pravouhl\u00fd s preponou <em>AB<\/em>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-S-3<\/strong><br \/>D\u013a\u017eky str\u00e1n pravouholn\u00edka s\u00fa cel\u00e9 \u010d\u00edsla <em>x<\/em> a <em>y<\/em> v\u00e4\u010d\u0161ie ako 1. V pravouholn\u00edku vyzna\u010d\u00edme rozdelenie na <em>x\u00b7y<\/em> jednotkov\u00fdch \u0161tvorcov a potom z neho zvinut\u00edm a zlepen\u00edm dvoch proti\u013eahl\u00fdch str\u00e1n zhotov\u00edme pl\u00e1\u0161\u0165 rota\u010dn\u00e9ho valca. Ka\u017ed\u00e9 dva vrcholy jednotkov\u00fdch \u0161tvorcov na pl\u00e1\u0161ti spoj\u00edme \u00fase\u010dkou. Ko\u013eko z t\u00fdchto \u00fase\u010diek prech\u00e1dza vn\u00fatorn\u00fdmi bodmi tohto valca? V pr\u00edpade x > y rozhodnite, kedy bude tento po\u010det v\u00e4\u010d\u0161\u00ed &#8211; ke\u010f bude obvod podstavy valca rovn\u00fd <em>x<\/em>, alebo <em>y<\/em>?<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/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\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><input onclick=\"window.location='http:\/\/skmo.sk\/dokument.php?id=996'\" type=\"button\" name=\"butta2\" value=\"Zadanie\" \/><input onclick=\"window.location='http:\/\/skmo.sk\/dokument.php?id=997'\" type=\"button\" name=\"butta2\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-II-1<\/strong><br \/>\nN\u00e1jdite v\u0161etky cel\u00e9 kladn\u00e9 \u010d\u00edsla, ktor\u00e9 nie s\u00fa mocninou \u010d\u00edsla 2 a ktor\u00e9 sa rovnaj\u00fa s\u00fa\u010dtu trojn\u00e1sobku svojho najv\u00e4\u010d\u0161ieho nep\u00e1rneho delite\u013ea a p\u00e4\u0165n\u00e1sobku svojho najmen\u0161ieho nep\u00e1rneho delite\u013ea v\u00e4\u010d\u0161ieho ako 1.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>A-II-2<\/strong><br \/>\nV rovine s\u00fa dan\u00e9 dve kru\u017enice <em>k<sub>1<\/sub><\/em>(<em>S<sub>1<\/sub>, r<sub>1<\/sub><\/em>) a <em>k<sub>2<\/sub><\/em>(<em>S<sub>2<\/sub>, r<sub>2<\/sub><\/em>), pri\u010dom |<em>S<sub>1<\/sub>S<sub>2<\/sub><\/em>| > <em>r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub><\/em>. N\u00e1jdite mno\u017einu v\u0161etk\u00fdch bodov <em>X<\/em>, ktor\u00e9 nele\u017eia na priamke <em>S<sub>1<\/sub>S<sub>2<\/sub><\/em> a maj\u00fa t\u00fa vlastnos\u0165, \u017ee \u00fase\u010dky <em>S<sub>1<\/sub>X<\/em>, <em>S<sub>2<\/sub>X<\/em> pret\u00ednaj\u00fa postupne kru\u017enice <em>k<sub>1<\/sub>, k<sub>2<\/sub><\/em> v bodoch, ktor\u00fdch vzdialenosti od priamky <em>S<sub>1<\/sub>S<sub>2<\/sub><\/em> sa rovnaj\u00fa.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-II-3<\/strong><br \/>\nN\u00e1jdite v\u0161etky trojice re\u00e1lnych \u010d\u00edsel <em>x, y<\/em> a <em>z<\/em>, pre ktor\u00e9 plat\u00ed<\/p>\n<p class=\"r\">x(y<sup>2<\/sup> + 2z<sup>2<\/sup>) = y(z<sup>2<\/sup> + 2x<sup>2<\/sup>) = z(x<sup>2<\/sup> + 2y<sup>2<\/sup>)<\/p>\n<div class=\"pri\">(Michal Rol\u00ednek)<\/div>\n<p><strong>A-II-4<\/strong><br \/>\nVolejbalov\u00e9ho turnaja sa z\u00fa\u010dastnilo \u0161es\u0165 dru\u017estiev, ka\u017ed\u00e9 hralo proti ka\u017ed\u00e9mu pr\u00e1ve raz. V jednotliv\u00fdch piatich kol\u00e1ch prebiehali v tom istom \u010dase v\u017edy tri z\u00e1pasy na troch kurtoch 1, 2 a 3. Ko\u013eko bolo mo\u017enost\u00ed pre rozpis tak\u00e9ho turnaja? Rozpisom rozumieme tabu\u013eku 3\u00d75, v ktorej pre i<span style=\"font-family: Symbol;\">\u00ce<\/span>{1, 2, 3} a j<span style=\"font-family: Symbol;\">\u00ce<\/span>{1, 2, 3, 4, 5} je na poz\u00edcii (i, j) uveden\u00e1 dvojica dru\u017estiev (bez ur\u010denia poradia), ktor\u00e9 hrali proti sebe v j-tom kole na kurte \u010d\u00edslo i. Namiesto dekadick\u00e9ho z\u00e1pisu v\u00fdsledn\u00e9ho \u010d\u00edsla sta\u010d\u00ed uvies\u0165 jeho rozklad na s\u00fa\u010din prvo\u010d\u00edsel.<\/p>\n<div class=\"pri\">(Martin Pan\u00e1k)<\/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=\"a4\"><\/a><a href=\"#top\"><img decoding=\"async\" alt=\"\" src=\"\/img\/top.gif\" align=\"right\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<form><input onclick=\"window.location='http:\/\/skmo.sk\/dokument.php?id=1071'\" type=\"button\" name=\"butta2\" value=\"Rie\u0161enie\" \/><input onclick=\"window.location='http:\/\/skmo.sk\/dokument.php?id=1072'\" type=\"button\" name=\"butta2\" value=\"Poradie\" \/><\/form>\n<p><strong>A-III-1<\/strong><br \/>\nNech <em>n<\/em> je cel\u00e9 kladn\u00e9 \u010d\u00edslo. Ozna\u010dme v\u0161etky jeho kladn\u00e9 delitele d<sub>1<\/sub>, d<sub>2<\/sub>, &#8230;, d<sub>k<\/sub> tak, aby platilo d<sub>1<\/sub> < d<sub>2<\/sub> < ... < d<sub>k<\/sub> (\u010di\u017ee d<sub>1<\/sub> = 1 a d<sub>k<\/sub> = <em>n<\/em>). Ur\u010dte v\u0161etky tak\u00e9 hodnoty <em>n<\/em>, pre ktor\u00e9 plat\u00ed d<sub>5<\/sub> \u2212 d<sub>3<\/sub> = 50 a 11d<sub>5<\/sub> + 8d<sub>7<\/sub> = 3<em>n<\/em>. <\/p>\n<div class=\"pri\">(Mat\u00fa\u0161 Harminc)<\/div>\n<p><strong>A-III-2<\/strong><br \/>\nV rovine, v ktorej je dan\u00e1 \u00fase\u010dka <em>AB<\/em>, uva\u017eujme trojuholn\u00edky <em>XYZ<\/em> tak\u00e9, \u017ee <em>X<\/em> je vn\u00fatorn\u00fdm bodom \u00fase\u010dky <em>AB<\/em>, trojuholn\u00edky <em>XBY<\/em> a <em>XZA<\/em> s\u00fa podobn\u00e9 (&#8710;<em>XBY<\/em> <font size=\"+1\">\u223c<\/font> &#8710;<em>XZA<\/em>) a body <em>A, B, Y, Z<\/em> le\u017eia v tomto porad\u00ed na kru\u017enici. N\u00e1jdite mno\u017einu stredov v\u0161etk\u00fdch \u00fase\u010diek <em>YZ<\/em>.<\/p>\n<div class=\"pri\">(Michal Rol\u00ednek, Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>A-III-3<\/strong><br \/>\nMajme \u0161achovnicu 8 \u00d7 8 a ku ka\u017edej &#8222;hrane&#8220;, ktor\u00e1 odde\u013euje dve jej pol\u00ed\u010dka, nap\u00ed\u0161me prirodzen\u00e9 \u010d\u00edslo, ktor\u00e9 ud\u00e1va po\u010det sp\u00f4sobov, ako mo\u017eno cel\u00fa \u0161achovnicu rozreza\u0165 na obd\u013a\u017eni\u010dky 2 \u00d7 1 tak, aby doty\u010dn\u00e1 hrana bola s\u00fa\u010das\u0165ou rezu. Ur\u010dte posledn\u00fa cifru s\u00fa\u010dtu v\u0161etk\u00fdch takto nap\u00edsan\u00fdch \u010d\u00edsel. <\/p>\n<div class=\"pri\">(Michal Rol\u00ednek)<\/div>\n<p><strong>A-III-4<\/strong><br \/>\nDo kina pri\u0161lo 234 div\u00e1kov. Ur\u010dte, pre ktor\u00e9 <em>n<\/em> &#8805; 4 sa mohlo sta\u0165, \u017ee div\u00e1kov bolo mo\u017en\u00e9 rozsadi\u0165 do <em>n<\/em> radov tak, aby ka\u017ed\u00fd div\u00e1k v <em>i<\/em>-tom rade sa poznal pr\u00e1ve s <em>j<\/em> div\u00e1kmi v <em>j<\/em>-tom rade pre \u013eubovo\u013en\u00e9 <em>i, j<\/em> \u2208 {1, 2, &#8230; , <em>n<\/em>}, <em>i<\/em> &#8800; <em>j<\/em>. (Vz\u0165ah zn\u00e1mosti je vz\u00e1jomn\u00fd.)<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>A-III-5<\/strong><br \/>\nDan\u00fd je ostrouhl\u00fd trojuholn\u00edk <em>ABC<\/em>. Ozna\u010dme <em>k<\/em> kru\u017enicu s priemerom <em>AB<\/em>. Kru\u017enica, ktor\u00e1 sa dot\u00fdka osi uhla <em>BAC<\/em> v bode <em>A<\/em> a prech\u00e1dza bodom <em>C<\/em>, pret\u00edna kru\u017enicu <em>k<\/em> v bode <em>P<\/em>, <em>P<\/em> &#8800; <em>A<\/em>. Kru\u017enica, ktor\u00e1 sa dot\u00fdka osi uhla <em>ABC<\/em> v bode <em>B<\/em> a prech\u00e1dza bodom <em>C<\/em>, pret\u00edna kru\u017enicu <em>k<\/em> v bode <em>Q<\/em>, <em>Q<\/em> &#8800; <em>B<\/em>. Dok\u00e1\u017ete, \u017ee priese\u010dn\u00edk priamok <em>AQ<\/em> a <em>BP<\/em> le\u017e\u00ed na osi uhla <em>ACB<\/em>. <\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>A-III-6<\/strong><br \/>\nPre \u013eubovo\u013en\u00e9 nez\u00e1porn\u00e9 re\u00e1lne \u010d\u00edsla <em>a<\/em> a <em>b<\/em> dok\u00e1\u017ete nerovnos\u0165<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" alt=\"\" src=\"\/63\/63a31.png\" \/><\/p>\n<p>a zistite, kedy nast\u00e1va rovnos\u0165. <\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk, Jarom\u00edr \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 href=\"#top\"><img decoding=\"async\" alt=\"\" src=\"\/img\/top.gif\" 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>63. ro\u010dn\u00edk matematickej olympi\u00e1dy 2013-2014S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3rie A, B a C aktualizovan\u00e9 3.1.2015 19:30<\/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":[12],"tags":[],"class_list":["post-829","post","type-post","status-publish","format-standard","hentry","category-63-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - 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