{"id":6,"date":"2010-09-23T17:53:38","date_gmt":"2010-09-23T17:53:38","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=6"},"modified":"2010-09-23T17:53:38","modified_gmt":"2010-09-23T17:53:38","slug":"kategorie-abc","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=6","title":{"rendered":"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>60. ro\u010dn\u00edk matematickej olympi\u00e1dy 2010-2011<\/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<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\"><\/td>\n<td align=\"center\"><span class=\"a\">celo\u0161t\u00e1tne kolo<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><!--more--><\/p>\n<p><a name=\"c1\"><\/a><strong>C-I-1<\/strong><br \/>\n Lucia nap\u00edsala na tabu\u013eu dve nenulov\u00e9 \u010d\u00edsla. Potom medzi ne postupne vkladala znamienka plus, m\u00ednus, kr\u00e1t, delen\u00e9 a v\u0161etky pr\u00edklady spr\u00e1vne vypo\u010d\u00edtala. Medzi v\u00fdsledkami boli len dve r\u00f4zne hodnoty. Ak\u00e9 dve \u010d\u00edsla mohla Lucia na tabu\u013eu nap\u00edsa\u0165?<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>C-I-2<\/strong><br \/>\n Dok\u00e1\u017ete, \u017ee v\u00fdrazy <em>23x\u00a0+\u00a0y<\/em>, <em>19x\u00a0+\u00a03y<\/em> s\u00fa delite\u013en\u00e9 \u010d\u00edslom 50 pre tie ist\u00e9 dvojice prirodzen\u00fdch \u010d\u00edsel <em>x, y<\/em>.<\/p>\n<div class=\"pri\">(Jaroslav Zhouf)<\/div>\n<p><strong>C-I-3<\/strong><br \/>\n M\u00e1me \u0161tvorec <em>ABCD<\/em> so stranou d\u013a\u017eky 1\u00a0cm. Body <em>K<\/em> a <em>L<\/em> s\u00fa stredy str\u00e1n <em>DA<\/em> a <em>DC<\/em>. Bod <em>P<\/em> le\u017e\u00ed na strane <em>AB<\/em> tak, \u017ee |<em>BP<\/em>|\u00a0=\u00a02.|<em>AP<\/em>|. Bod <em>Q<\/em> le\u017e\u00ed na strane <em>BC<\/em> tak, \u017ee |<em>CQ<\/em>|\u00a0=\u00a02.|<em>BQ<\/em>|. \u00dase\u010dky <em>KQ<\/em> a <em>PL<\/em> sa pret\u00ednaj\u00fa v bode <em>X<\/em>. Obsahy \u0161tvoruholn\u00edkov <em>APXK<\/em>, <em>BQXP<\/em>, <em>QCLX<\/em> a <em>LDKX<\/em> ozna\u010d\u00edme postupne <em>S<sub>A<\/sub><\/em>, <em>S<sub>B<\/sub><\/em>, <em>S<sub>C<\/sub><\/em> a <em>S<sub>D<\/sub><\/em> (obr\u00e1zok).<\/p>\n<p class=\"pc\"><img loading=\"lazy\" decoding=\"async\" src=\"\/60\/60c1.gif\" alt=\"\" width=\"161\" height=\"159\" \/><\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee <em>S<sub>B<\/sub><\/em> =\u00a0<em>S<sub>D<\/sub><\/em>.<\/p>\n<p><strong>b)<\/strong> Vypo\u010d\u00edtajte rozdiel <em>S<sub>C<\/sub><\/em> =\u00a0<em>S<sub>A<\/sub><\/em>.<\/p>\n<p><strong>c)<\/strong> Vysvetlite, pre\u010do neplat\u00ed <em>S<sub>A<\/sub><\/em> +\u00a0<em>S<sub>C<\/sub><\/em> =\u00a0<em>S<sub>B<\/sub><\/em> +\u00a0<em>S<sub>D<\/sub><\/em>.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>C-I-4<\/strong><br \/>\n V skupine <em>n<\/em> \u017eiakov sa spolu niektor\u00ed kamar\u00e1tia. Vieme, \u017ee ka\u017ed\u00fd m\u00e1 medzi ostatn\u00fdmi aspo\u0148 \u0161tyroch kamar\u00e1tov. U\u010dite\u013eka chce \u017eiakov rozdeli\u0165 do dvoch najviac \u0161tvor\u010dlenn\u00fdch skup\u00edn tak, aby ka\u017ed\u00fd mal v svojej skupine aspo\u0148 jedn\u00e9ho kamar\u00e1ta.<\/p>\n<p><strong>a)<\/strong> Uk\u00e1\u017ete, \u017ee v pr\u00edpade <em>n\u00a0=\u00a07<\/em> je mo\u017en\u00e9 \u017eiakov po\u017eadovan\u00fdm sp\u00f4sobom rozdeli\u0165.<\/p>\n<p><strong>b)<\/strong> Zistite, \u010di je mo\u017en\u00e9 \u017eiakov takto rozdeli\u0165 i v pr\u00edpade <em>n\u00a0=\u00a08<\/em>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>C-I-5<\/strong><br \/>\n Dok\u00e1\u017ete, \u017ee najmen\u0161\u00ed spolo\u010dn\u00fd n\u00e1sobok <strong>[<em>a, b<\/em>]<\/strong> a najv\u00e4\u010d\u0161\u00ed spolo\u010dn\u00fd delite\u013e <strong>(<em>a, b<\/em>)<\/strong> \u013eubovo\u013en\u00fdch dvoch kladn\u00fdch cel\u00fdch \u010d\u00edsel <em>a, b<\/em> vyhovuj\u00fa nerovnosti <strong><em>a<\/em>.(<em>a, b<\/em>)\u00a0+\u00a0<em>b<\/em>.[<em>a, b<\/em>]\u00a0\u2265\u00a02<em>ab<\/em><\/strong>. Zistite, kedy v tejto nerovnosti nastane rovnos\u0165.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-I-6<\/strong><br \/>\n Je dan\u00fd lichobe\u017en\u00edk <em>ABCD<\/em>. Stred z\u00e1kladne <em>AB<\/em> ozna\u010dme <em>P<\/em>. Uva\u017eujme rovnobe\u017eku so z\u00e1klad\u0148ou <em>AB<\/em>, ktor\u00e1 pret\u00edna \u00fase\u010dky <em>AD, PD, PC, BC<\/em> postupne v bodoch <em>K, L, M, N<\/em>.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee |<em>KL<\/em>|\u00a0=\u00a0|<em>MN<\/em>|.<\/p>\n<p><strong>b)<\/strong> Ur\u010dte polohu priamky <em>KL<\/em> tak, aby platilo |<em>KL<\/em>|\u00a0=\u00a0|<em>LM<\/em>|.<\/p>\n<div class=\"pri\">(Jaroslav Zhouf)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\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<p><strong>C-S-1<\/strong><br \/>\n Po okruhu behaj\u00fa dvaja atl\u00e9ti, ka\u017ed\u00fd inou kon\u0161tantnou r\u00fdchlos\u0165ou. Ke\u010f be\u017eia opa\u010dn\u00fdmi smermi, stret\u00e1vaj\u00fa sa ka\u017ed\u00fdch 10 min\u00fat, ke\u010f be\u017eia rovnak\u00fdm smerom, stret\u00e1vaj\u00fa sa ka\u017ed\u00fdch 40 min\u00fat. Za ak\u00fd \u010das zabehne okruh r\u00fdchlej\u0161\u00ed atl\u00e9t ?<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>C-S-2<\/strong><br \/>\n Dan\u00fd je \u0161tvorec so stranou d\u013a\u017eky 6 cm. N\u00e1jdite mno\u017einu stredov v\u0161etk\u00fdch prie\u010dok \u0161tvorca, ktor\u00e9ho delia na dva \u0161tvoruholn\u00edky, z ktor\u00fdch jeden m\u00e1 obsah 12 cm<sup>2<\/sup>.<\/p>\n<p>(Prie\u010dka \u0161tvorca je \u00fase\u010dka, ktorej krajn\u00e9 body le\u017eia na stran\u00e1ch \u0161tvorca.)<\/p>\n<div class=\"pri\">(Pavel Leischner)<\/div>\n<p><strong>C-S-3<\/strong><br \/>\n Nech <em>x, y<\/em> s\u00fa tak\u00e9 kladn\u00e9 cel\u00e9 \u010d\u00edsla, \u017ee obe \u010d\u00edsla <em>3x\u00a0+\u00a05y<\/em> a <em>5x\u00a0+\u00a02y<\/em> s\u00fa delite\u013en\u00e9 \u010d\u00edslom 60.<\/p>\n<p>Zd\u00f4vodnite, pre\u010do \u010d\u00edslo 60 del\u00ed aj  s\u00fa\u010det <em>2x\u00a0+\u00a03y<\/em>.<\/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 width=\"85%\" align=\"left\">\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<p><strong>C-II-1<\/strong><br \/>\nNa tabuli s\u00fa nap\u00edsan\u00e9 pr\u00e1ve tri (nie nutne r\u00f4zne) re\u00e1lne \u010d\u00edsla. Vieme, \u017ee s\u00fa\u010det \u013eubovo\u013en\u00fdch dvoch z nich je tam nap\u00edsan\u00fd tie\u017e. Ur\u010dte v\u0161etky trojice tak\u00fdch \u010d\u00edsel.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>C-II-2<br \/>\n<\/strong>N\u00e1jdite v\u0161etky kladn\u00e9 cel\u00e9 \u010d\u00edsla <em>n<\/em>, pre ktor\u00e9 je \u010d\u00edslo <em>n<\/em><sup>2<\/sup>+6<em>n<\/em> druhou mocninou cel\u00e9ho \u010d\u00edsla.<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>C-II-3<\/strong><br \/>\nV lichobe\u017en\u00edku <em>ABCD<\/em> m\u00e1 z\u00e1klad\u0148a <em>AB<\/em> d\u013a\u017eku 18cm a z\u00e1klad\u0148a <em>CD<\/em> d\u013a\u017eku 6cm. Pre bod <em>E<\/em> strany <em>AB<\/em> plat\u00ed 2|<em>AE<\/em>| = |<em>EB<\/em>|. Body <em>K, L, M<\/em>, ktor\u00e9 s\u00fa postupne \u0165a\u017eiskami trojuholn\u00edkov <em>ADE<\/em>, <em>CDE<\/em>, <em>BCE<\/em>, tvoria vrcholy rovnostrann\u00e9ho trojuholn\u00edka.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee priamky <em>KM<\/em> a <em>CM<\/em> zvieraj\u00fa prav\u00fd uhol.<\/p>\n<p><strong>b)<\/strong> Vypo\u010d\u00edtajte d\u013a\u017eky ramien lichobe\u017en\u00edka <em>ABCD<\/em>.<\/p>\n<div class=\"pri\">(Pavel Cal\u00e1bek)<\/div>\n<p><strong>C-II-4<\/strong><br \/>\nNech <em>x, y, z<\/em> s\u00fa kladn\u00e9 re\u00e1lne \u010d\u00edsla. Uk\u00e1\u017ete, \u017ee aspo\u0148 jedno z \u010d\u00edsel <em>x + y + z &#8211; xyz<\/em> a <em>xy + yz + zx &#8211; 3<\/em> je nez\u00e1porn\u00e9.<\/p>\n<div class=\"pri\">(Stanislava Soj\u00e1kov\u00e1)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\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<p><strong>B-I-1<\/strong><br \/>\n V obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te s\u00fastavu rovn\u00edc<\/p>\n<p class=\"pc\"><img loading=\"lazy\" decoding=\"async\" src=\"\/60\/60b1.gif\" alt=\"\" width=\"116\" height=\"79\" \/><\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>B-I-2<\/strong><br \/>\n Uva\u017eujme vn\u00fatorn\u00fd bod <em>P<\/em> dan\u00e9ho obd\u013a\u017enika <em>ABCD<\/em> a ozna\u010dme postupne <em>Q, R<\/em> obrazy bodu P v s\u00famernostiach pod\u013ea stredov <em>A, C<\/em>. Predpokladajme, \u017ee priamka <em>QR<\/em> pretne strany <em>AB<\/em> a <em>BC<\/em> vo vn\u00fatorn\u00fdch bodoch <em>M<\/em> a <em>N<\/em>. Zostrojte mno\u017einu v\u0161etk\u00fdch bodov <em>P<\/em>, pre ktor\u00e9 plat\u00ed |<em>MN<\/em>|\u00a0=\u00a0|<em>AB<\/em>|.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-I-3<\/strong><br \/>\n Nech <em>a, b, c<\/em> s\u00fa re\u00e1lne \u010d\u00edsla, ktor\u00fdch s\u00fa\u010det je 6. Dok\u00e1\u017ete, \u017ee aspo\u0148 jedno z \u010d\u00edsel <em>ab\u00a0+\u00a0bc<\/em>, <em>bc\u00a0+\u00a0ca<\/em>, <em>ca\u00a0+\u00a0ab<\/em> nie je v\u00e4\u010d\u0161ie ako 8.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>B-I-4<\/strong><br \/>\n N\u00e1jdite v\u0161etky cel\u00e9 \u010d\u00edsla <em>n<\/em>, pre ktor\u00e9 m\u00e1 zlomok\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"\/60\/60b2.gif\" alt=\"\" width=\"68\" height=\"41\" align=\"middle\" \/> celo\u010d\u00edseln\u00fa hodnotu.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>B-I-5<\/strong><br \/>\n Zaoberajme sa ot\u00e1zkou, ktor\u00e9 trojuholn\u00edky <em>ABC<\/em> s ostr\u00fdmi uhlami pri vrcholoch <em>A<\/em> a <em>B<\/em> maj\u00fa nasledovn\u00fa vlastnos\u0165:<\/p>\n<p>Ak vedieme stredom v\u00fd\u0161ky z vrcholu <em>C<\/em> tri priamky rovnobe\u017en\u00e9 so stranami trojuholn\u00edka <em>ABC<\/em>, pretn\u00fa ich tieto priamky v \u0161iestich bodoch le\u017eiacich na jednej kru\u017enici.<\/p>\n<p><strong>a)<\/strong> Uk\u00e1\u017ete, \u017ee vyhovuje ka\u017ed\u00fd trojuholn\u00edk s prav\u00fdm uhlom pri vrchole <em>C<\/em>.<\/p>\n<p><strong>b)<\/strong> Vysvetlite, pre\u010do \u017eiadny in\u00fd trojuholn\u00edk <em>ABC<\/em> nevyhovuje.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>B-I-6<\/strong><br \/>\n Ur\u010dte po\u010det desa\u0165cifern\u00fdch \u010d\u00edsel, v ktor\u00fdch je mo\u017en\u00e9 vy\u0161krtn\u00fa\u0165 dve susedn\u00e9 \u010d\u00edslice a dosta\u0165 tak \u010d\u00edslo 99-kr\u00e1t men\u0161ie.<\/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 width=\"85%\" align=\"left\">\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<p><strong>B-S-1<\/strong><br \/>\n V obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te rovnicu<\/p>\n<div>[pmath size=12]{sqrt{x+3}}+{sqrt{x}}=p[\/pmath]<\/div>\n<p>s nezn\u00e1mou <em>x<\/em> a re\u00e1lnym parametrom <em>p<\/em>.<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>B-S-2<\/strong><\/p>\n<p>Pozd\u013a\u017e kru\u017enice je rozmiestnen\u00fdch 16 re\u00e1lnych \u010d\u00edsel so s\u00fa\u010dtom 7.<\/p>\n<p><strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee existuje \u00fasek piatich susedn\u00fdch \u010d\u00edsel so s\u00fa\u010dtom aspo\u0148 2.<\/p>\n<p><strong>b)<\/strong> Ur\u010dte najmen\u0161ie <em>k<\/em> tak\u00e9, \u017ee v op\u00edsanej situ\u00e1cii mo\u017eno v\u017edy n\u00e1js\u0165 \u00fasek <em>k<\/em> susedn\u00fdch \u010d\u00edsel so s\u00fa\u010dtom aspo\u0148 3.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>B-S-3<\/strong><br \/>\n Zvonka dan\u00e9ho trojuholn\u00edka <em>ABC<\/em> s\u00fa zostrojen\u00e9 \u0161tvorce <em>ACDE<\/em>, <em>BCGF<\/em>. Dok\u00e1\u017ete, \u017ee |<em>AG<\/em>| = |<em>BD<\/em>|. \u010ealej uk\u00e1\u017ete, \u017ee stredy oboch \u0161tvorcov spolu so stredmi \u00fase\u010diek <em>AB<\/em> a <em>DG<\/em> s\u00fa vrcholmi \u0161tvorca.<\/p>\n<div class=\"pri\">(Pavel Leischner)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\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<p><strong>B-II-1<\/strong><br \/>\n S\u00fa\u010din kladn\u00fdch re\u00e1lnych \u010d\u00edsel <em>a, b, c<\/em> je 60 a ich s\u00fa\u010det je 15. Dok\u00e1\u017ete nerovnos\u0165<\/p>\n<p class=\"r\">(a + b)(a + c) \u2265 60<\/p>\n<p>a zistite, pre ktor\u00e9 tak\u00e9 \u010d\u00edsla <em>a, b, c<\/em> nastane rovnos\u0165.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>B-II-2<\/strong><br \/>\n N\u00e1jdite v\u0161etky dvojice kladn\u00fdch cel\u00fdch \u010d\u00edsel <em>a, b<\/em>, pre ktor\u00e9 \u010d\u00edslo <em>b<\/em> je delite\u013en\u00e9 \u010d\u00edslom <em>a<\/em> a s\u00fa\u010dasne \u010d\u00edslo <em>3a+4<\/em> je delite\u013en\u00e9 \u010d\u00edslom <em>b+1<\/em>.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>B-II-3<\/strong><br \/>\n Nech <em>M, N<\/em> s\u00fapostupne vn\u00fatorn\u00e9 body str\u00e1n <em>AB<\/em>, <em>BC<\/em> rovnostrann\u00e9ho trojuholn\u00edka <em>ABC<\/em>, pre ktor\u00e9 plat\u00ed |<em>AM<\/em>| : |<em>MB<\/em>| = |<em>BN<\/em>| : |<em>NC<\/em>| = 2 : 1. Ozna\u010dme <em>P<\/em> priese\u010dn\u00edk priamok <em>AN<\/em> a <em>CM<\/em>. Dok\u00e1\u017ete, \u017ee priamky <em>BP<\/em> a <em>AN<\/em> s\u00fa navz\u00e1jom kolm\u00e9.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-II-4<\/strong><br \/>\n Zap\u00ed\u0161eme v\u0161etky p\u00e4\u0165cifern\u00e9 \u010d\u00edsla, v ktor\u00fdch sa ka\u017ed\u00e1 z ci\ffier 4, 5, 6, 7, 8 vyskytuje pr\u00e1ve raz. Potom jedno (\u013eubovo\u013en\u00e9 z nich) \u0161krtneme a v\u0161etky zvy\u0161n\u00e9 s\u010d\u00edtame. Ak\u00e9 s\u00fa mo\u017en\u00e9 hodnoty cifern\u00e9ho s\u00fa\u010dtu tak\u00e9ho v\u00fdsledku?<\/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 width=\"85%\" align=\"left\">\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<p><strong>A-I-1<\/strong><br \/>\n Korene rovnice <strong><em>ax<\/em><sup>4<\/sup> +\u00a0<em>bx<\/em><sup>2<\/sup> +\u00a0<em>a<\/em> =\u00a01<\/strong> v obore re\u00e1lnych \u010d\u00edsel s\u00fa \u0161tyri po sebe id\u00face \u010dleny rast\u00facej aritmetickej postupnosti. Jeden z t\u00fdchto \u010dlenov je s\u00fa\u010dasne rie\u0161en\u00edm rovnice <strong><em>bx<\/em><sup>2<\/sup> +\u00a0<em>ax<\/em> +\u00a0<em>a<\/em> =\u00a01<\/strong>. Ur\u010dte v\u0161etky mo\u017en\u00e9 hodnoty re\u00e1lnych parametrov <em>a, b<\/em>.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>A-I-2<\/strong><br \/>\n Nech <em>k, n<\/em> s\u00fa prirodzen\u00e9 \u010d\u00edsla. Z platnosti tvrdenia &#8222;\u010d\u00edslo (<em>n<\/em>-1)(<em>n<\/em>+1) je delite\u013en\u00e9 \u010d\u00edslom <em>k<\/em>&#8220; Adam us\u00fadil, \u017ee pr\u00e1ve jedno z \u010d\u00edsel <em>n<\/em>-1, <em>n<\/em>+1 je delite\u013en\u00e9 <em>k<\/em>. Ur\u010dte v\u0161etky prirodzen\u00e9 \u010d\u00edsla <em>k<\/em>, pre ktor\u00e9 je Adamova \u00favaha spr\u00e1vna pre ka\u017ed\u00e9 prirodzen\u00e9 <em>n<\/em>.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>A-I-3<\/strong><br \/>\n S\u00fa dan\u00e9 kru\u017enice <em>k, l<\/em>, ktor\u00e9 sa pret\u00ednaj\u00fa v bodoch <em>A, B<\/em>. Ozna\u010dme postupne <em>K, L<\/em> dotykov\u00e9 body ich spolo\u010dnej doty\u010dnice zvolenej tak, \u017ee bod <em>B<\/em> je vn\u00fatorn\u00fdm bodom trojuholn\u00edka <em>AKL<\/em>. Na kru\u017eniciach <em>k<\/em> a <em>l<\/em> zvo\u013eme postupne body <em>N<\/em> a <em>M<\/em> tak, aby bod <em>A<\/em> bol vn\u00fatorn\u00fdm bodom \u00fase\u010dky <em>MN<\/em>. Dok\u00e1\u017ete, \u017ee \u0161tvoruholn\u00edk <em>KLMN<\/em> je tetivov\u00fd pr\u00e1ve vtedy, ke\u010f priamka <em>MN<\/em> je doty\u010dnicou kru\u017enice op\u00edsanej trojuholn\u00edku <em>AKL<\/em>.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>A-I-4<\/strong><br \/>\n M\u00e1me <em>6n<\/em> \u017eet\u00f3nov a\u017e na farbu zhodn\u00fdch, po tri z ka\u017edej z <em>2n<\/em> farieb. Pre ka\u017ed\u00e9 prirodzen\u00e9 \u010d\u00edslo <em>n<\/em>>1 ur\u010dte po\u010det <em>p<sub>n<\/sub><\/em> v\u0161etk\u00fdch rozdelen\u00ed t\u00fdchto <em>6n<\/em> \u017eet\u00f3nov na dve k\u00f4pky po <em>3n<\/em> \u017eet\u00f3noch, pri ktor\u00fdch \u017eiadne tri \u017eet\u00f3ny tej istej farby nie s\u00fa v tej istej k\u00f4pke. Dok\u00e1\u017ete, \u017ee \u010d\u00edslo <em>p<sub>n<\/sub><\/em> je nep\u00e1rne pr\u00e1ve vtedy, ke\u010f <em>n\u00a0=\u00a02<sup>k<\/sup><\/em> pre vhodn\u00e9 prirodzen\u00e9 <em>k<\/em>.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-I-5<\/strong><br \/>\n Na ka\u017edej stene kocky je nap\u00edsan\u00e9 pr\u00e1ve jedno cel\u00e9 \u010d\u00edslo. V jednom kroku zvol\u00edme \u013eubovo\u013en\u00e9 dve susedn\u00e9 steny kocky a \u010d\u00edsla na nich nap\u00edsan\u00e9 zv\u00e4\u010d\u0161\u00edme o\u00a01. Ur\u010dte nutn\u00fa a posta\u010duj\u00facu podmienku pre za\u010diato\u010dn\u00e9 o\u010d\u00edslovanie stien kocky, aby po kone\u010dnom po\u010dte vhodn\u00fdch krokov boli na v\u0161etk\u00fdch sten\u00e1ch rovnak\u00e9 \u010d\u00edsla.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>A-I-6<\/strong><br \/>\n Dok\u00e1\u017ete, \u017ee v ka\u017edom trojuholn\u00edku <em>ABC<\/em> s ostr\u00fdm uhlom pri vrchole <em>C<\/em> (pri zvy\u010dajnom ozna\u010den\u00ed d\u013a\u017eok str\u00e1n a ve\u013ekost\u00ed vn\u00fatorn\u00fdch uhlov) plat\u00ed nerovnos\u0165 <strong>(<em>a<\/em><sup>2<\/sup> +\u00a0<em>b<\/em><sup>2<\/sup>)cos(<em>\u03b1<\/em> &#8211;\u00a0<em>\u03b2<\/em>)\u00a0\u2264\u00a02<em>ab<\/em><\/strong>. Zistite, kedy nastane rovnos\u0165.<\/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 width=\"85%\" align=\"left\">\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<p><strong>A-S-1<\/strong><br \/>\n Ur\u010dte v\u0161etky re\u00e1lne \u010d\u00edsla <em>c<\/em>, ktor\u00e9 mo\u017eno s oboma kore\u0148mi kvadratickej rovnice<\/p>\n<p style=\"text-align: center;\">[pmath size=13]{x^2} + {5\/2}x + c = 0[\/pmath]<\/p>\n<p>usporiada\u0165 do troj\u010dlennej aritmetickej postupnosti.<\/p>\n<div class=\"pri\">(Pavel Cal\u00e1bek, Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>A-S-2<\/strong><br \/>\n Nech <em>P, Q, R<\/em> s\u00fa body prepony <em>AB<\/em> pravouhl\u00e9ho trojuholn\u00edka <em>ABC<\/em>, pre ktor\u00e9 plat\u00ed |<em>AP<\/em>| = |<em>PQ<\/em>| = |<em>QR<\/em>| = |<em>RB<\/em>| = [pmath size=6]{1\/4}[\/pmath]|<em>AB<\/em>|. Dok\u00e1\u017ete, \u017ee priese\u010dn\u00edk <em>M<\/em> kru\u017en\u00edc op\u00edsan\u00fdch trojuholn\u00edkom <em>APC<\/em> a <em>BRC<\/em>, ktor\u00fd je r\u00f4zny od bodu <em>C<\/em>, je toto\u017en\u00fd so stredom <em>S<\/em> \u00fase\u010dky <em>CQ<\/em>.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<p><strong>A-S-3<\/strong><br \/>\n Dok\u00e1\u017ete, \u017ee pre \u013eubovo\u013en\u00e9 dve r\u00f4zne prvo\u010d\u00edsla <em>p<\/em>, <em>q<\/em> v\u00e4\u010d\u0161ie ako 2 plat\u00ed nerovnos\u0165<\/p>\n<div style=\"text-align: center;\">[pmath size=13]delim{|}{{p\/q}-{q\/p}}{|}>{4\/sqrt{pq}}[\/pmath].<\/div>\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 width=\"85%\" align=\"left\">\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<p><strong>A-II-1<\/strong><br \/>\n Rozhodnite, \u010di medzi v\u0161etk\u00fdmi osemcifern\u00fdmi n\u00e1sobkami \u010d\u00edsla 4 je viac t\u00fdch, ktor\u00e9 vo svojom dekadickom z\u00e1pise obsahuj\u00fa cifru 1, alebo t\u00fdch, ktor\u00e9 cifru 1 neobsahuj\u00fa.<\/p>\n<div class=\"pri\">(J\u00e1nMaz\u00e1k)<\/div>\n<p><strong>A-II-2<\/strong><br \/>\n Dan\u00fd je trojuholn\u00edk <em>ABC<\/em> s obsahom <em>S<\/em>. Vn\u00fatri trojuholn\u00edka, ktor\u00e9ho vrcholmi s\u00fa stredy str\u00e1n trojuholn\u00edka <em>ABC<\/em>, je \u013eubovo\u013ene zvolen\u00fd bod <em>O<\/em>. Ozna\u010dme <em>A\u00b4<\/em>, <em>B\u00b4<\/em>, <em>C\u00b4<\/em> postupne obrazy bodov <em>A<\/em>, <em>B<\/em>, <em>C<\/em> v stredovej s\u00famernosti pod\u013ea bodu <em>O<\/em>. Dok\u00e1\u017ete, \u017ee \u0161es\u0165uholn\u00edk <em>AC\u00b4BA\u00b4CB\u00b4<\/em> m\u00e1 obsah <em>2S<\/em>.<\/p>\n<div class=\"pri\">(PavelLeischner)<\/div>\n<p><strong>A-II-3<\/strong><br \/>\n Ur\u010dte v\u0161etky dvojice (<em>m, n<\/em>) kladn\u00fdch cel\u00fdch \u010d\u00edsel, pre ktor\u00e9 je \u010d\u00edslo <em>4(mn\u00a0+\u00a01)<\/em> delite\u013en\u00e9 \u010d\u00edslom (<em>m\u00a0+\u00a0n<\/em>)<sup>2<\/sup>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>A-II-4<\/strong><br \/>\n Nech <em>M<\/em> je mno\u017eina \u0161iestich navz\u00e1jom r\u00f4znych kladn\u00fdch cel\u00fdch \u010d\u00edsel, ktor\u00fdch s\u00fa\u010det je 60. V\u0161etky ich nap\u00ed\u0161eme na steny kocky, na ka\u017ed\u00fa pr\u00e1ve jedno z nich. V jednom kroku zvol\u00edme \u013eubovo\u013en\u00e9 tri steny kocky, ktor\u00e9 maj\u00fa spolo\u010dn\u00fd vrchol, a ka\u017ed\u00e9 z \u010d\u00edsel na t\u00fdchto troch sten\u00e1ch zv\u00e4\u010d\u0161\u00edme o 1. Ur\u010dte po\u010det v\u0161etk\u00fdch tak\u00fdch mno\u017e\u00edn <em>M<\/em>, ktor\u00fdch \u010d\u00edsla mo\u017eno nap\u00edsa\u0165 na steny kocky uveden\u00fdm sp\u00f4sobom tak, \u017ee pokone\u010dnom po\u010dte vhodn\u00fdch krokov bud\u00fa na v\u0161etk\u00fdch sten\u00e1ch rovnak\u00e9 \u010d\u00edsla.<\/p>\n<div class=\"pri\">(Peter Novotn\u00fd)<\/div>\n<table style=\"width: 100%; height: 25px;\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td width=\"85%\" align=\"left\">\n<hr \/>\n<\/td>\n<td class=\"h\" width=\"12%\"><a href=\"#top\"><img decoding=\"async\" src=\"\/img\/top.gif\" alt=\"\" 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>60. ro\u010dn\u00edk matematickej olympi\u00e1dy 2010-2011 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 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":[9],"tags":[],"class_list":["post-6","post","type-post","status-publish","format-standard","hentry","category-60-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - 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