{"id":191,"date":"2010-05-21T14:59:20","date_gmt":"2010-05-21T14:59:20","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=191"},"modified":"2010-05-21T14:59:20","modified_gmt":"2010-05-21T14:59:20","slug":"kategorie-abc-5","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=191","title":{"rendered":"Kateg\u00f3rie ABC"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>56. ro\u010dn\u00edk matematickej olympi\u00e1dy 2006-2007<\/p>\n<p style=\"text-align: center;\"><!--more--><br \/>\n<script type=\"text\/javascript\"><!--\r\ngoogle_ad_client = \"pub-0508472359151885\";\r\n\/* 180x150, bola vytvoren\u00e1 12.11.2010 *\/\r\ngoogle_ad_slot = \"2162838046\";\r\ngoogle_ad_width = 180;\r\ngoogle_ad_height = 150;\r\n\/\/-->\r\n<\/script>\r\n<script type=\"text\/javascript\"\r\nsrc=\"http:\/\/pagead2.googlesyndication.com\/pagead\/show_ads.js\">\r\n<\/script>\r\n<script type=\"text\/javascript\"><!--\r\ngoogle_ad_client = \"pub-0508472359151885\";\r\n\/* 180x150, bola vytvoren\u00e1 12.11.2010 *\/\r\ngoogle_ad_slot = \"2162838046\";\r\ngoogle_ad_width = 180;\r\ngoogle_ad_height = 150;\r\n\/\/-->\r\n<\/script>\r\n<script type=\"text\/javascript\"\r\nsrc=\"http:\/\/pagead2.googlesyndication.com\/pagead\/show_ads.js\">\r\n<\/script>\r\n<script type=\"text\/javascript\"><!--\r\ngoogle_ad_client = \"pub-0508472359151885\";\r\n\/* 180x150, bola vytvoren\u00e1 12.11.2010 *\/\r\ngoogle_ad_slot = \"2162838046\";\r\ngoogle_ad_width = 180;\r\ngoogle_ad_height = 150;\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<table style=\"width: 550px;\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\" align=\"center\">\n<tbody>\n<tr align=\"center\" bgcolor=\"#c0c0c0\">\n<td><span class=\"a\">C<\/span><\/td>\n<td><span class=\"a\">B<\/span><\/td>\n<td><span class=\"a\">A<\/span><\/td>\n<\/tr>\n<tr align=\"center\">\n<td><span class=\"a\"><a href=\"#c1\">dom\u00e1ce kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#b1\">dom\u00e1ce kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#a1\">dom\u00e1ce kolo<\/a><\/span><\/td>\n<\/tr>\n<tr align=\"center\">\n<td><span class=\"a\"><a href=\"#c2\">\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#b2\">\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#a2\">\u0161kolsk\u00e9 kolo<\/a><\/span><\/td>\n<\/tr>\n<tr align=\"center\">\n<td><span class=\"a\"><a href=\"#c3\">krajsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#b3\">krajsk\u00e9 kolo<\/a><\/span><\/td>\n<td><span class=\"a\"><a href=\"#a3\">krajsk\u00e9 kolo<\/a><\/span><\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\" bgcolor=\"#c0c0c0\"><\/td>\n<td align=\"center\"><span class=\"a\"><a href=\"#a4\">celo\u0161t\u00e1tne kolo<\/a><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"c1\"><\/a><\/p>\n<p><strong>C-I-1<\/strong><\/p>\n<p>Ur\u010dte v\u0161etky dvojice (a, b) prirodzen\u00fdch \u010d\u00edsel, pre ktor\u00e9 plat\u00ed\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"\/56\/56ca1.gif\" border=\"0\" alt=\"\" width=\"117\" height=\"18\" align=\"absmiddle\" \/>.<\/p>\n<p><strong>C-I-2<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky trojuholn\u00edky, ktor\u00e9 sa daj\u00fa rozreza\u0165 na lichobe\u017en\u00edky s d\u013a\u017ekami str\u00e1n 1\u00a0cm, 1\u00a0cm, 1\u00a0cm a 2\u00a0cm.<\/p>\n<p><strong>C-I-3<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky prirodzen\u00e9 \u010d\u00edsla, ktor\u00fdch z\u00e1pis neobsahuje nulu a m\u00e1 nasledovn\u00fa vlastnos\u0165:<br \/>\n\u25cf\u00a0\u00a0\u00a0ak v \u0148om vynech\u00e1me \u013eubovo\u013en\u00fa \u010d\u00edslicu, dostaneme \u010d\u00edslo, ktor\u00e9 je delite\u013eom p\u00f4vodn\u00e9ho \u010d\u00edsla.<\/p>\n<p><strong>C-I-4<\/strong><\/p>\n<p>Je dan\u00fd lichobe\u017en\u00edk <em>ABCD<\/em> so z\u00e1klad\u0148ami <em>AB<\/em> a <em>CD<\/em>. Ozna\u010dme <em>E<\/em> stred strany <em>AB<\/em>, <em>F<\/em> stred \u00fase\u010dky <em>DE<\/em> a <em>G<\/em> priese\u010dn\u00edk \u00fase\u010diek <em>BD<\/em> a <em>CE<\/em>. Vyjadrite obsah lichobe\u017en\u00edka <em>ABCD<\/em> pomocou jeho v\u00fd\u0161ky <em><strong>v<\/strong><\/em> a d\u013a\u017eky <em><strong>d<\/strong><\/em> \u00fase\u010dky <em>FG<\/em> za predpokladu, \u017ee body <em>A<\/em>,\u00a0<em>F<\/em>,\u00a0<em>C<\/em> le\u017eia na jednej priamke.<\/p>\n<p><strong>C-I-5<\/strong><\/p>\n<p>Zistite, pre ktor\u00e9 prirodzen\u00e9 \u010d\u00edslo <strong><em>n<\/em><\/strong> je podiel<\/p>\n<p class=\"pc\"><img loading=\"lazy\" decoding=\"async\" src=\"\/56\/56ca2.gif\" border=\"0\" alt=\"\" width=\"93\" height=\"36\" \/><\/p>\n<p><strong><em>a) <\/em><\/strong> \u010do najv\u00e4\u010d\u0161ie prirodzen\u00e9 \u010d\u00edslo,<\/p>\n<p><strong><em>b)<\/em><\/strong> \u010do najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo.<\/p>\n<p><strong>C-I-6<\/strong><\/p>\n<p>Je dan\u00fd ostrouhl\u00fd trojuholn\u00edk <em>ABC<\/em>, v ktorom <em>D<\/em> je p\u00e4ta v\u00fd\u0161ky z vrcholu <em>C<\/em> a <em>V<\/em> priese\u010dn\u00edk v\u00fd\u0161ok. Dok\u00e1\u017ete, \u017ee |<em>AD<\/em>|<strong>.<\/strong>|<em>BD<\/em>|\u00a0=\u00a0|<em>AB<\/em>|<strong>.<\/strong>|<em>VD<\/em>| pr\u00e1ve vtedy, ke\u010f |<em>CD<\/em>|\u00a0=\u00a0|<em>AB<\/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=\"c2\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>C-S-1<\/strong><\/p>\n<p>Ur\u010dte po\u010det v\u0161etk\u00fdch \u0161tvorcifern\u00fdch prirodzen\u00fdch \u010d\u00edsel, ktor\u00e9 s\u00fa delite\u013en\u00e9 \u0161iestimi a v ich z\u00e1pise sa vyskytuj\u00fa pr\u00e1ve dve jednotky.<\/p>\n<p><strong>C-S-2<\/strong><\/p>\n<p>Kru\u017enica <em>k<\/em> so stredom <em>S<\/em> je op\u00edsan\u00e1 pravideln\u00e9mu \u0161es\u0165uholn\u00edku <em>ABCDEF<\/em>. Doty\u010dnica v bode\u00a0<em>A<\/em> ku kru\u017enici\u00a0<em>k<\/em> pret\u00edna priamku\u00a0<em>SB<\/em> v bode\u00a0<em>K<\/em> a doty\u010dnica v bode\u00a0<em>B<\/em> pret\u00edna priamku\u00a0<em>SC<\/em> v bode\u00a0<em>L<\/em>. Dok\u00e1\u017ete, \u017ee \u0161tvoruholn\u00edku <em>KLCB<\/em> sa d\u00e1 op\u00edsa\u0165 kru\u017enica, ktor\u00e1 je zhodn\u00e1 s kru\u017enicou\u00a0<em>k<\/em>.<\/p>\n<p><strong>C-S-3<\/strong><\/p>\n<p>Ur\u010dte v\u0161etky dvojice <strong><em>(a,\u00a0b)<\/em><\/strong> prirodzen\u00fdch \u010d\u00edsel, ktor\u00fdch rozdiel <em>a\u00a0&#8211;\u00a0b<\/em> je piatou mocninou niektor\u00e9ho prvo\u010d\u00edsla a pre ktor\u00e9 plat\u00ed <img loading=\"lazy\" decoding=\"async\" src=\"\/56\/56cb1.gif\" alt=\"\" width=\"118\" height=\"14\" align=\"bottom\" \/>.<\/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=\"c3\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>C-II-1<\/strong><\/p>\n<p>V rovine s\u00fa dan\u00e9 dva r\u00f4zne body <em>L<\/em>, <em>M<\/em> a kru\u017enica\u00a0<em>k<\/em>. Zostrojte trojuholn\u00edk\u00a0<em>ABC<\/em> s \u010do najv\u00e4\u010d\u0161\u00edm obsahom tak, aby jeho vrchol\u00a0<em>C<\/em> le\u017eal na kru\u017enici\u00a0<em>k<\/em>, bod\u00a0<em>L<\/em> bol stredom strany\u00a0<em>AC<\/em> a bod\u00a0<em>M<\/em> stredom strany\u00a0<em>BC<\/em>.<\/p>\n<p><strong>C-II-2<\/strong><\/p>\n<p>Nech <em>p, q, r<\/em> s\u00fa prirodzen\u00e9 \u010d\u00edsla, pre ktor\u00e9 plat\u00ed <img loading=\"lazy\" decoding=\"async\" src=\"\/56\/56cc1.gif\" alt=\"\" width=\"148\" height=\"18\" align=\"middle\" \/>.<br \/>\na) Ur\u010dte, ak\u00e9 hodnoty m\u00f4\u017ee nadob\u00fada\u0165 s\u00fa\u010det <em>p + q + r<\/em>.<br \/>\nb) Ur\u010dte po\u010det v\u0161etk\u00fdch usporiadan\u00fdch troj\u00edc (<em>p, q, r<\/em>) prirodzen\u00fdch \u010d\u00edsel, ktor\u00e9 vyhovuj\u00fa danej rovnici.<\/p>\n<p><strong>C-II-3<\/strong><\/p>\n<p>Rovnoramenn\u00e9mu lichobe\u017en\u00edku <em>ABCD<\/em> so z\u00e1klad\u0148ami <em>AB, CD<\/em> sa d\u00e1 vp\u00edsa\u0165 kru\u017enica so stredom\u00a0<em>O<\/em>. Ur\u010dte obsah\u00a0<em>S<\/em> lichobe\u017en\u00edka, ak s\u00fa dan\u00e9 d\u013a\u017eky \u00fase\u010diek <em>OB<\/em> a <em>OC<\/em>.<\/p>\n<p><strong>C-II-4<\/strong><\/p>\n<p>Ur\u010dte najv\u00e4\u010d\u0161ie dvojcifern\u00e9 \u010d\u00edslo\u00a0<em>k<\/em> s nasleduj\u00facou vlastnos\u0165ou:<\/p>\n<p>existuje prirodzen\u00e9 \u010d\u00edslo\u00a0<em>N<\/em>, z ktor\u00e9ho po \u0161krtnut\u00ed prvej \u010d\u00edslice z\u013eava dostaneme \u010d\u00edslo <em>k<\/em>-kr\u00e1t men\u0161ie. (Po \u0161krtnut\u00ed \u010d\u00edslice m\u00f4\u017ee z\u00e1pis \u010d\u00edsla za\u010d\u00edna\u0165 jednou \u010di nieko\u013ek\u00fdmi nulami.) K ur\u010den\u00e9mu \u010d\u00edslu\u00a0<em>k<\/em> potom n\u00e1jdite najmen\u0161ie vyhovuj\u00face \u010d\u00edslo <em>N<\/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=\"b1\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>B-I-1<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky dvojice (<em>a, b<\/em>) cel\u00fdch \u010d\u00edsel, ktor\u00e9 vyhovuj\u00fa rovnici<br \/>\n<strong><em>a<sup>2<\/sup> +\u00a07ab\u00a0+\u00a06b<sup>2<\/sup> +\u00a05a\u00a0+\u00a04b\u00a0+\u00a03\u00a0=\u00a00<\/em><\/strong>.<\/p>\n<p><strong>B-I-2<\/strong><\/p>\n<p>Je dan\u00e1 kru\u017enica <em>k<\/em> s priemerom <em>AB<\/em>. K \u013eubovo\u013en\u00e9mu bodu <em>Y<\/em> kru\u017enice <em>k<\/em>, <em>Y<\/em> \u2260\u00a0<em>A<\/em>, zostrojme na polpriamke <em>AY<\/em> bod <em>X<\/em>, pre ktor\u00fd plat\u00ed |<em>AX<\/em>|\u00a0=\u00a0|<em>YB<\/em>|. Ur\u010dte mno\u017einu v\u0161etk\u00fdch tak\u00fdch bodov <em>X<\/em>.<\/p>\n<p><strong>B-I-3<\/strong><\/p>\n<p>N\u00e1jdite najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>k<\/em> tak\u00e9, \u017ee ka\u017ed\u00e1 <em>k<\/em>-prvkov\u00e1 mno\u017eina trojmiestnych po dvoch nes\u00fadelite\u013en\u00fdch \u010d\u00edsel obsahuje aspo\u0148 jedno prvo\u010d\u00edslo.<\/p>\n<p><strong>B-I-4<\/strong><\/p>\n<p>V \u013eubovo\u013enom trojuholn\u00edku <em>ABC<\/em> ozna\u010dme <em>T<\/em> \u0165a\u017eisko, <em>D<\/em> stred strany <em>AC<\/em> a <em>E<\/em> stred strany <em>BC<\/em>. N\u00e1jdite v\u0161etky pravouhl\u00e9 trojuholn\u00edky <em>ABC<\/em> s preponou <em>AB<\/em>, pre ktor\u00e9 je \u0161tvoruholn\u00edk <em>CDTE<\/em> doty\u010dnicov\u00fd.<\/p>\n<p><strong>B-I-5<\/strong><\/p>\n<p>N\u00e1jdite v\u0161etky dvojice (<em>p, q<\/em>) re\u00e1lnych \u010d\u00edsel tak\u00e9, \u017ee polyn\u00f3m <strong><em>x<sup>2<\/sup> +\u00a0px\u00a0+\u00a0q<\/em><\/strong> je delite\u013eom polyn\u00f3mu<br \/>\n<strong><em>x<sup>4<\/sup> +\u00a0px<sup>2<\/sup> +\u00a0q<\/em><\/strong>.<\/p>\n<p><strong>B-I-6<\/strong><\/p>\n<p>Je dan\u00e1 \u00fase\u010dka <em>AA<\/em><sub>0<\/sub> a priamka <em>p<\/em>. Zostrojte trojuholn\u00edk s vrcholom <em>A<\/em> a v\u00fd\u0161kou <em>AA<\/em><sub>0<\/sub>, ktor\u00e9ho \u0165a\u017eisko a stred op\u00edsanej kru\u017enice le\u017eia na priamke <em>p<\/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>Ur\u010dte v\u0161tky dvojice re\u00e1lnych \u010d\u00edsel <em>a, b<\/em>, pre ktor\u00e9 je polyn\u00f3m <strong><em>x<sup>4<\/sup> + ax<sup>2<\/sup> + b<\/em><\/strong> delite\u013en\u00fd polyn\u00f3mom <strong><em>x<sup>2<\/sup> + bx + a<\/em><\/strong>.<\/p>\n<p><strong>B-S-2<\/strong><\/p>\n<p>V trojuholn\u00edku <em>ABC<\/em> ozna\u010dme <em>D<\/em> stred strany <em>BC<\/em>, <em>E<\/em> stred strany <em>AC<\/em> a <em>T<\/em> \u0165a\u017eisko. Dok\u00e1\u017ete, \u017ee ak je strana <em>BC<\/em> dlh\u0161ia ako strana <em>AC<\/em>, m\u00e1 kru\u017enica vp\u00edsan\u00e1 trojuholn\u00edku <em>BDT<\/em> men\u0161\u00ed polomer ako kru\u017enica vp\u00edsan\u00e1 trojuholn\u00edku <em>ATE<\/em>.<\/p>\n<p><strong>B-S-3<\/strong><\/p>\n<p>N\u00e1jdite najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>n<\/em>, pre ktor\u00e9 je podiel <img loading=\"lazy\" decoding=\"async\" src=\"\/56\/56bb1.gif\" alt=\"\" width=\"51\" height=\"31\" align=\"absmiddle\" \/> prirodzen\u00e9 \u010d\u00edslo.<\/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=\"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 re\u00e1lne \u010d\u00edsla <em>a, b, c<\/em> tak, aby polyn\u00f3m <strong><em>x<sup>4<\/sup> +\u00a0ax<sup>2<\/sup> +\u00a0bx\u00a0+\u00a0c<\/em><\/strong> bol delite\u013en\u00fd polyn\u00f3mom <strong><em>x<sup>2<\/sup> +\u00a0x\u00a0+\u00a01<\/em><\/strong> a pritom s\u00fa\u010det <strong><em>a<sup>2<\/sup> +\u00a0b<sup>2<\/sup> +\u00a0c<sup>2<\/sup><\/em><\/strong> bol \u010do najmen\u0161\u00ed.<\/p>\n<p><strong>B-II-2<\/strong><\/p>\n<p>Dan\u00fd je trojuholn\u00edk <em>ABC<\/em> so stranou <em>BC<\/em> d\u013a\u017eky 22\u00a0cm a stranou <em>AC<\/em> d\u013a\u017eky 19\u00a0cm, ktor\u00e9ho \u0165a\u017enice <em>t<sub>a<\/sub>, t<sub>b<\/sub><\/em> s\u00fa navz\u00e1jom kolm\u00e9. Vypo\u010d\u00edtajte d\u013a\u017eku strany <em>AB<\/em>.<\/p>\n<p><strong>B-II-3<\/strong><\/p>\n<p>Prirodzen\u00e9 \u010d\u00edslo nazveme <em>vlnit\u00fdm<\/em>, ak pre ka\u017ed\u00e9 tri po sebe id\u00face \u010d\u00edslice <em>a, b, c<\/em> jeho dekadick\u00e9ho z\u00e1pisu plat\u00ed (<em>a\u00a0&#8211;\u00a0b<\/em>)(<em>b\u00a0&#8211;\u00a0c<\/em>)\u00a0&lt;\u00a00. Dok\u00e1\u017ete, \u017ee z \u010d\u00edslic 0,\u00a01,\u00a0&#8230;,\u00a09 je mo\u017en\u00e9 zostavi\u0165 viac ako 25\u00a0000 desaci\u0165cifern\u00fdch <em>vlnit\u00fdch \u010d\u00edsel<\/em>, z ktor\u00fdch ka\u017ed\u00e9 obsahuje v\u0161etky \u010d\u00edslice od nuly po deviatku (\u010d\u00edslica 0 nesmie by\u0165 na prvom mieste).<\/p>\n<p><strong>B-II-4<\/strong><\/p>\n<p>Je dan\u00fd ostrouhl\u00fd trojuholn\u00edk <em>ABC<\/em>. Pre \u013eubovo\u013en\u00fd bod <em>L<\/em> jeho strany <em>AB<\/em> ozna\u010dme <em>K, M<\/em> p\u00e4ty kolm\u00edc z bodu <em>L<\/em> na strany <em>AC, BC<\/em>. Zistite, pre ktor\u00fa polohu bodu <em>L<\/em> je \u00fase\u010dka <em>KM<\/em> najkrat\u0161ia.<\/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=\"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<\/p>\n<p class=\"r\">4x<sup>4<\/sup> &#8211;\u00a012x<sup>3<\/sup> &#8211;\u00a07x<sup>2<\/sup> +\u00a022x\u00a0+\u00a014\u00a0=\u00a00,<\/p>\n<p>ak viete, \u017ee m\u00e1 \u0161tyri r\u00f4zne re\u00e1lne korene, pri\u010dom s\u00fa\u010det dvoch z nich je 1.<\/p>\n<p><strong>A-I-2<\/strong><\/p>\n<p>Kru\u017enica vp\u00edsan\u00e1 dan\u00e9mu trojuholn\u00edku <em>ABC<\/em> sa dot\u00fdka str\u00e1n<em> BC, CA, AB<\/em> postupne v bodoch <em>K, L, M<\/em>. Ozna\u010dme <em>P<\/em> priese\u010dn\u00edk osi vn\u00fatorn\u00e9ho uhla pri vrchole <em>C<\/em> s priamkou <em>MK<\/em>. Dok\u00e1\u017ete, \u017ee priamky <em>AP<\/em> a <em>LK<\/em> s\u00fa rovnobe\u017en\u00e9.<\/p>\n<p><strong>A-I-3<\/strong><\/p>\n<p>Ak s\u00fa <em>x, y, z<\/em> re\u00e1lne \u010d\u00edsla z intervalu &lt;-1, 1&gt; vyhovuj\u00face podmienke <em>xy\u00a0+\u00a0yz\u00a0+\u00a0zx\u00a0=\u00a01<\/em>, potom plat\u00ed<\/p>\n<p class=\"pc\"><img loading=\"lazy\" decoding=\"async\" src=\"\/56\/56aa1.gif\" border=\"0\" alt=\"\" width=\"328\" height=\"27\" \/><\/p>\n<p>Dok\u00e1\u017ete a zistite, kedy nastane rovnos\u0165.<\/p>\n<p><strong>A-I-4<\/strong><\/p>\n<p>Zistite, pre ktor\u00e9 prirodzen\u00e9 \u010d\u00edsla <em>n<\/em> je mo\u017en\u00e9 mno\u017einu M\u00a0=\u00a0{1,\u00a02,\u00a0&#8230;\u00a0,<em>n<\/em>} rozdeli\u0165<\/p>\n<p><strong><em>a)<\/em><\/strong> na dve,<\/p>\n<p><strong><em>b)<\/em><\/strong> na tri<\/p>\n<p>navz\u00e1jom disjunktn\u00e9 podmno\u017einy s rovnak\u00fdm po\u010dtom prvkov tak, aby ka\u017ed\u00e1 z nich obsahovala aj aritmetick\u00fd priemer v\u0161etk\u00fdch svojich prvkov.<\/p>\n<p><strong>A-I-5<\/strong><\/p>\n<p>V rovine je dan\u00e1 kru\u017enica <em>k<\/em> so stredom <em>S<\/em> a bod <em>A<\/em> \u2260\u00a0<em>S<\/em>. Ur\u010dte mno\u017einu stredov kru\u017en\u00edc op\u00edsan\u00fdch v\u0161etk\u00fdm trojuholn\u00edkom <em>ABC<\/em>, ktor\u00fdch strana <em>BC<\/em> je priemerom kru\u017enice <em>k<\/em>.<\/p>\n<p><strong>A-I-6<\/strong><\/p>\n<p>Ur\u010dte v\u0161etky funkcie <em>f<\/em> :\u00a0<em>Z<\/em> \u2192\u00a0<em>Z<\/em> tak\u00e9, \u017ee pre v\u0161etky cel\u00e9 \u010d\u00edsla <em>x, y<\/em> plat\u00ed\u00a0\u00a0<em><strong>f\u00a0(f(x)\u00a0+\u00a0y)\u00a0=\u00a0x\u00a0+\u00a0f(y\u00a0+\u00a02006)<\/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>Ur\u010dte v\u0161etky re\u00e1lne \u010d\u00edsla <em>s<\/em>, pre ktor\u00e9 m\u00e1 rovnica<\/p>\n<p class=\"r2s\">4x<sup>4<\/sup> &#8211;\u00a020x<sup>3<\/sup> +\u00a0sx<sup>2<\/sup> +\u00a022x-\u00a02\u00a0=\u00a00<\/p>\n<p>\u0161tyri r\u00f4zne re\u00e1lne korene, pri\u010dom s\u00fa\u010din dvoch z nich je rovn\u00fd \u010d\u00edslu <strong>-2<\/strong>.<\/p>\n<p><strong>A-S-2<\/strong><\/p>\n<p>Uva\u017eujme mno\u017einu <em>{1,\u00a02,\u00a04,\u00a05,\u00a08,\u00a010,\u00a016,\u00a020,\u00a032,\u00a040,\u00a080,\u00a0160}<\/em> a v\u0161etky jej trojprvkov\u00e9 podmno\u017einy. Rozhodnite, \u010di je viac t\u00fdch, ktor\u00e9 maj\u00fa s\u00fa\u010din svojich prvkov v\u00e4\u010d\u0161\u00ed ako <strong>2006<\/strong>, alebo t\u00fdch, ktor\u00e9 maj\u00fa s\u00fa\u010din svojich prvkov men\u0161\u00ed ako <strong>2006<\/strong>.<\/p>\n<p><strong>A-S-3<\/strong><\/p>\n<p>Dan\u00fd je lichobe\u017en\u00edk <em>ABCD<\/em> s prav\u00fdm uhlom pri vrchole <em>A<\/em> a z\u00e1klad\u0148ou <em>AB<\/em>, v ktorom plat\u00ed |<em>AB<\/em>|\u00a0&gt;\u00a0|<em>CD<\/em>|\u00a0\u2265\u00a0|<em>DA<\/em>|. Ozna\u010dme <em>S<\/em> priese\u010dn\u00edk os\u00ed jeho vn\u00fatorn\u00fdch uhlov pri vrcholoch <em>A,\u00a0B<\/em> a\u00a0<em>T<\/em> priese\u010dn\u00edk os\u00ed vn\u00fatorn\u00fdch uhlov pri vrcholoch <em>C,\u00a0D<\/em>. Podobne ozna\u010dme <em>U,\u00a0V<\/em> priese\u010dn\u00edky os\u00ed vn\u00fatorn\u00fdch uhlov pri vrcholoch <em>A, D<\/em>, resp. <em>B, C<\/em>.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee priamky <em>UV<\/em> a <em>AB<\/em> s\u00fa rovnobe\u017en\u00e9.<\/p>\n<p><strong>b)<\/strong> Dok\u00e1\u017ete, \u017ee priese\u010dn\u00edk\u00a0<em>E<\/em> polpriamky <em>DT<\/em> s priamkou <em>AB<\/em> a body <em>S, T, B<\/em> le\u017eia na jednej kru\u017enici.<\/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=\"a3\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>A-II-1<\/strong><\/p>\n<p>Zistite, ak\u00fd je najmen\u0161\u00ed mo\u017en\u00fd obsah trojuholn\u00edka <em>ABC<\/em>, ktor\u00e9ho v\u00fd\u0161ky sp\u013a\u0148aj\u00fa nerovnosti <em>v<sub>a<\/sub><\/em> \u2265\u00a03\u00a0cm, <em>v<sub>b<\/sub><\/em> \u2265\u00a04\u00a0cm, <em>v<sub>c<\/sub><\/em> \u2265\u00a05\u00a0cm.<\/p>\n<p><strong>A-II-2<\/strong><\/p>\n<p>Nech <em>a, b<\/em> s\u00fa re\u00e1lne \u010d\u00edsla. Ak m\u00e1 rovnica<\/p>\n<div class=\"r\">x<sup>4<\/sup> &#8211;\u00a04x<sup>3<\/sup> +\u00a04x<sup>2<\/sup> +\u00a0ax\u00a0+\u00a0b\u00a0=\u00a00<\/div>\n<p>dva r\u00f4zne re\u00e1lne korene tak\u00e9, \u017ee ich s\u00fa\u010det sa rovn\u00e1 ich s\u00fa\u010dinu, tak plat\u00ed <em>a\u00a0+\u00a0b\u00a0&gt;\u00a00<\/em> a pritom dan\u00e1 rovnica nem\u00e1 \u017eiadne in\u00e9 re\u00e1lne korene. Dok\u00e1\u017ete.<\/p>\n<p><strong>A-II-3<\/strong><\/p>\n<p>Nech <em>M<\/em> je \u013eubovo\u013en\u00fd vn\u00fatorn\u00fd bod prepony <em>AB<\/em> pravouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em>. Ozna\u010dme <em>S, S<sub>1<\/sub>, S<sub>2<\/sub><\/em> stredy kru\u017en\u00edc op\u00edsan\u00fdch postupne trojuholn\u00edkom <em>ABC, AMC, BMC<\/em>.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee body <em>M, C, S<sub>1<\/sub>, S<sub>2<\/sub><\/em> a <em>S<\/em> le\u017eia na jednej kru\u017enici.<\/p>\n<p><strong>b)<\/strong> Pre ktor\u00fa polohu bodu <em>M<\/em> m\u00e1 t\u00e1to kru\u017enica najmen\u0161\u00ed polomer?<\/p>\n<p><strong>A-II-4<\/strong><\/p>\n<p>Nech <em>p, q<\/em> s\u00fa dan\u00e9 prirodzen\u00e9 \u010d\u00edsla, pri\u010dom <em>p\u00a0&lt;\u00a0q<\/em>. Ur\u010dte najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>m<\/em> s vlastnos\u0165ou: S\u00fa\u010det v\u0161etk\u00fdch zlomkov v z\u00e1kladnom tvare, ktor\u00e9 maj\u00fa menovate\u013ea <em>m<\/em> a ktor\u00fdch hodnoty le\u017eia v otvorenom intervale (<em>p,\u00a0q<\/em>), je aspo\u0148 <em>56(q<sup>2<\/sup> &#8211;\u00a0p<sup>2<\/sup>)<\/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=\"a4\"><\/a><a href=\"#top\">\u25b2 hore \u25b2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>A-III-1<\/strong><\/p>\n<p>Na niektor\u00e9 pol\u00ed\u010dko \u0161tvorcovej \u0161achovnice <em>n<\/em>\u00d7<em>n<\/em> (<em>n<\/em> 2)postav\u00edme fig\u00farku a potom ju pos\u00favame striedavo <em>&#8222;\u0161ikmo&#8220;<\/em> a <em>&#8222;priamo&#8220;<\/em>. <em>&#8222;\u0160ikmo&#8220;<\/em> znamen\u00e1 na pol\u00ed\u010dko, ktor\u00e9 m\u00e1 s predch\u00e1dzaj\u00facim spolo\u010dn\u00fd pr\u00e1ve jeden bod. <em>&#8222;Priamo&#8220;<\/em> znamen\u00e1 na susedn\u00e9 pol\u00ed\u010dko, ktor\u00e9 m\u00e1 s predch\u00e1dzaj\u00facim spolo\u010dn\u00fa stranu. Ur\u010dte v\u0161etky <em>n<\/em>, pre ktor\u00e9 existuje v\u00fdchodiskov\u00e9 pol\u00ed\u010dko a tak\u00e1 postupnos\u0165 \u0165ahov za\u010d\u00ednaj\u00faca <em>&#8222;\u0161ikmo&#8220;<\/em>, \u017ee fig\u00farka prejde cel\u00fa \u0161achovnicu a na ka\u017edom pol\u00ed\u010dku sa ocitne pr\u00e1ve raz.<\/p>\n<p><strong>A-III-2<\/strong><\/p>\n<p>Nech <em>a, b<\/em> s\u00fa re\u00e1lne \u010d\u00edsla. Ak m\u00e1 rovnica<\/p>\n<div class=\"r\">x<sup>4<\/sup> &#8211;\u00a04x<sup>3<\/sup> +\u00a04x<sup>2<\/sup> +\u00a0ax\u00a0+\u00a0b\u00a0=\u00a00<\/div>\n<p>dva r\u00f4zne re\u00e1lne korene tak\u00e9, \u017ee ich s\u00fa\u010det sa rovn\u00e1 ich s\u00fa\u010dinu, tak plat\u00ed <em>a\u00a0+\u00a0b\u00a0&gt;\u00a00<\/em> a pritom dan\u00e1 rovnica nem\u00e1 \u017eiadne in\u00e9 re\u00e1lne korene. Dok\u00e1\u017ete.<\/p>\n<p><strong>A-III-3<\/strong><\/p>\n<p>Nech <em>M<\/em> je \u013eubovo\u013en\u00fd vn\u00fatorn\u00fd bod prepony <em>AB<\/em> pravouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em>. Ozna\u010dme <em>S, S<sub>1<\/sub>, S<sub>2<\/sub><\/em> stredy kru\u017en\u00edc op\u00edsan\u00fdch postupne trojuholn\u00edkom <em>ABC, AMC, BMC<\/em>.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee body <em>M, C, S<sub>1<\/sub>, S<sub>2<\/sub><\/em> a <em>S<\/em> le\u017eia na jednej kru\u017enici.<\/p>\n<p><strong>b)<\/strong> Pre ktor\u00fa polohu bodu <em>M<\/em> m\u00e1 t\u00e1to kru\u017enica najmen\u0161\u00ed polomer?<\/p>\n<p><strong>A-III-4<\/strong><\/p>\n<p>Nech <em>p, q<\/em> s\u00fa dan\u00e9 prirodzen\u00e9 \u010d\u00edsla, pri\u010dom <em>p\u00a0&lt;\u00a0q<\/em>. Ur\u010dte najmen\u0161ie prirodzen\u00e9 \u010d\u00edslo <em>m<\/em> s vlastnos\u0165ou: S\u00fa\u010det v\u0161etk\u00fdch zlomkov v z\u00e1kladnom tvare, ktor\u00e9 maj\u00fa menovate\u013ea <em>m<\/em> a ktor\u00fdch hodnoty le\u017eia v otvorenom intervale (<em>p,\u00a0q<\/em>), je aspo\u0148 <em>56(q<sup>2<\/sup> &#8211;\u00a0p<sup>2<\/sup>)<\/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 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>56. ro\u010dn\u00edk matematickej olympi\u00e1dy 2006-2007<\/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":[5],"tags":[],"class_list":["post-191","post","type-post","status-publish","format-standard","hentry","category-56-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - 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