{"id":480,"date":"2011-08-17T20:03:49","date_gmt":"2011-08-17T20:03:49","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=480"},"modified":"2011-08-17T20:03:49","modified_gmt":"2011-08-17T20:03:49","slug":"sutazne-ulohy-kategorii-a-b-a-c-2","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=480","title":{"rendered":"S\u00fa\u0165a\u017en\u00e9 \u00falohy kateg\u00f3ri\u00ed A, B a C"},"content":{"rendered":"<p class=\"hh2\"><a name=\"top\"><\/a>61. ro\u010dn\u00edk matematickej olympi\u00e1dy 2011-2012<\/p>\n<p><span class=\"povinne\">aktualizovan\u00e9 17.4.2012 21:10<\/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\"><\/td>\n<td align=\"center\"><span class=\"a\">celo\u0161t\u00e1tne kolo<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"background-color: #ffff99;\" colspan=\"3\" align=\"center\"><span class=\"rs\">Rie\u0161enia v\u0161etk\u00fdch k\u00f4l na stiahnutie vo form\u00e1te PDF:\u00a0\u00a0\u00a0<a href=\"\/61\/61abcr.pdf\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" src=\"\/img\/down.png\" border=\"0\" alt=\"\" width=\"48\" height=\"48\" align=\"absmiddle\" \/><\/a> (ulo\u017ei\u0165 ako &#8230;)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"c1\"><\/a><\/p>\n<form> <input onclick=\"window.location='61\/61cir.pdf'\" name=\"buttc1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>C-I-1<\/strong><br \/>\nN\u00e1jdite v\u0161etky troj\u010dleny<em> p(x) = ax<sup>2<\/sup> + bx + c<\/em> , ktor\u00e9 d\u00e1vaj\u00fa po delen\u00ed dvoj\u010dlenom <em>x + 1<\/em> zvy\u0161ok 2 a po delen\u00ed dvoj\u0161\u010denom <em>x + 2<\/em> zvy\u0161ok 1, pri\u010dom <em>p(1)<\/em> = 61.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-I-2<\/strong><br \/>\nD\u013a\u017eky str\u00e1n trojuholn\u00edka s\u00fa v metroch vyjadren\u00e9 cel\u00fdmi \u010d\u00edslami. Ur\u010dte ich, ak m\u00e1 trojuholn\u00edk obvod 72 m a ak je najdlh\u0161ia strana trojuholn\u00edka rozdelen\u00e1 bodom dotyku vp\u00edsanej kru\u017enice v pomere 3 : 4.<\/p>\n<div class=\"pri\">(Pavel Leischner)<\/div>\n<p><strong>C-I-3<\/strong><br \/>\nN\u00e1jdite v\u0161etky trojice prirodzen\u00fdch \u010d\u00edsel <em>a, b, c<\/em>, pre ktor\u00e9 plat\u00ed mno\u017einov\u00e1 rovnos\u0165<\/p>\n<p class=\"r\">{(<em>a, b<\/em>), (<em>a, c<\/em>), (<em>b, <\/em><em>c<\/em>), [<em>a, b<\/em>], [<em>a, c<\/em>], [<em>b, c<\/em>]} = {2, 3, 5, 60, 90,180},<\/p>\n<p>pri\u010dom (<em>x, y<\/em>) a [<em>x, y<\/em>] ozna\u010duje postupne najv\u00e4\u010d\u0161\u00ed spolo\u010dn\u00fd delite\u013e a najmen\u0161\u00ed spolo\u010dn\u00fd n\u00e1sobok \u010d\u00edsel <em>x<\/em> a <em>y<\/em>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>C-I-4<\/strong><br \/>\nRe\u00e1lne \u010d\u00edsla <em>a, b, c, d<\/em> vyhovuj\u00fa rovnici <em>ab<\/em> + <em>bc<\/em> + <em>cd<\/em> + <em>da<\/em> = 16.<br \/>\n<strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee medzi \u010d\u00edslami <em>a, b, c, d<\/em> sa n\u00e1jdu dve so s\u00fa\u010dtom najviac 4.<br \/>\n<strong>b)<\/strong> Ak\u00fa najmen\u0161iu hodnotu m\u00f4\u017ee ma\u0165 s\u00fa\u010det <em>a<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup>?<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>C-I-5<\/strong><br \/>\nDan\u00fd je rovnoramenn\u00fd trojuholn\u00edk so z\u00e1klad\u0148ou d\u013a\u017eky <em>a<\/em> a ramenami d\u013a\u017eky <em>b<\/em>. Pomocou nich vyjadrite polomer <em>R<\/em> kru\u017enice op\u00edsanej a polomer <em>r<\/em> kru\u017enice vp\u00edsanej tomuto trojuholn\u00edku. Potom uk\u00e1\u017ete, \u017ee plat\u00ed <em>R<\/em>\u22652<em>r<\/em>, a zistite, kedy nastane rovnos\u0165.<\/p>\n<div class=\"pri\">(Leo Bo\u010dek)<\/div>\n<p><strong>C-I-6<\/strong><br \/>\nNa hracej ploche <em>n<\/em> \u00d7 <em>n<\/em> tvorenej bielymi \u0161tvorcov\u00fdmi pol\u00ed\u010dkami sa Monika a Tamara striedaj\u00fa v \u0165ahoch jednou \ffig\u00farkou pri nasleduj\u00facej hre. Najsk\u00f4r Monika polo\u017e\u00ed \ffig\u00farku na \u013eubovo\u013en\u00e9 pol\u00ed\u010dko a toto pol\u00ed\u010dko zafarb\u00ed namodro. \u010ealej v\u017edy hr\u00e1\u010dka, ktor\u00e1 je na \u0165ahu, urob\u00ed s fi\fg\u00farkou <em>skok<\/em> na pol\u00ed\u010dko, ktor\u00e9 je doposia\u013e biele, a toto pol\u00ed\u010dko zafarb\u00ed namodro. Pritom pod <em>skokom<\/em> rozumieme be\u017en\u00fd \u0165ah \u0161achov\u00fdm jazdcom, t. j. presun fig\u00farky o dve pol\u00ed\u010dka zvislo alebo vodorovne a s\u00fa\u010dasne o jedno pol\u00ed\u010dko v druhom smere. Hr\u00e1\u010dka, ktor\u00e1 je na rade a u\u017e nem\u00f4\u017ee urobi\u0165 \u0165ah, prehr\u00e1va. Postupne pre <em>n<\/em> = 4, 5, 6 rozhodnite, ktor\u00e1 z hr\u00e1\u010dok m\u00f4\u017ee hra\u0165 tak, \u017ee vyhr\u00e1 nez\u00e1visle na \u0165ahoch druhej hr\u00e1\u010dky.<\/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 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<form> <input onclick=\"window.location='61\/61ciir.pdf'\" name=\"buttc2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>C-S-1<\/strong><br \/>\nN\u00e1jdite v\u0161etky dvojice prirodzen\u00fdch \u010d\u00edsel <em>a, b<\/em>, pre ktor\u00e9 plat\u00ed rovnos\u0165 mno\u017e\u00edn<\/p>\n<div class=\"pc\"><strong>{<em>a<\/em> \u00b7 [<em>a, b<\/em>], <em>b<\/em> \u00b7 (<em>a, b<\/em>)} = {45, 180}<\/strong>,<\/div>\n<p>pri\u010dom (<em>x, y<\/em>) ozna\u010duje najv\u00e4\u010d\u0161\u00ed spolo\u010dn\u00fd delite\u013ea [<em>x, y<\/em>] najmen\u0161\u00ed spolo\u010dn\u00fd n\u00e1sobok \u010d\u00edsel <em>x<\/em> a <em>y<\/em>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>C-S-2<\/strong><br \/>\nOzna\u010dme <em>S<\/em> stred z\u00e1kladne <em>AB<\/em> dan\u00e9ho rovnoramenn\u00e9ho trojuholn\u00edka <em>ABC<\/em>. Predpokladajme, \u017ee kru\u017enice vp\u00edsan\u00e9 trojuholn\u00edkom <em>ACS<\/em>, <em>BCS<\/em> sa dot\u00fdkaj\u00fa priamky <em>AB<\/em> v bodoch, ktor\u00e9 delia z\u00e1klad\u0148u <em>AB<\/em> na tri zhodn\u00e9 diely. Vypo\u010d\u00edtajte pomer |<em>AB<\/em>| : |<em>CS<\/em>|.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-S-3<\/strong><br \/>\nRe\u00e1lne \u010d\u00edsla <em>p, q, r, s<\/em> sp\u013a\u0148aj\u00fa rovnosti<\/p>\n<div class=\"pc\"><em><strong>p<sup>2<\/sup> + q<sup>2<\/sup> + r<sup>2<\/sup> + s<sup>2<\/sup> = 4<\/strong><\/em><img loading=\"lazy\" decoding=\"async\" src=\"\/img\/space.png\" alt=\"\" width=\"30\" height=\"1\" \/>a<img loading=\"lazy\" decoding=\"async\" src=\"\/img\/space.png\" alt=\"\" width=\"30\" height=\"1\" \/><em><strong>pq + rs = 1<\/strong>.<\/em><\/div>\n<p>Dok\u00e1\u017ete, \u017ee niektor\u00e9 dve z t\u00fdchto \u0161tyroch \u010d\u00edsel sa l\u00ed\u0161ia najviac o 1 a niektor\u00e9 dve sa l\u00ed\u0161ia najmenej o 1.<\/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<form> <input onclick=\"window.location='61\/61ciiir.pdf'\" name=\"buttc3\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>C-II-1<\/strong><br \/>\nPre v\u0161etky re\u00e1lne \u010d\u00edsla <em>x, y, z<\/em> tak\u00e9, \u017ee <em>x<\/em> &lt; <em>y<\/em> &lt; <em>z<\/em>, dok\u00e1\u017ete nerovnos\u0165<\/p>\n<div class=\"r\">x<sup>2<\/sup> &#8211; y<sup>2<\/sup> + z<sup>2<\/sup> &gt; (x &#8211; y + z)<sup>2<\/sup>.<\/div>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>C-II-2<\/strong><br \/>\nJanko m\u00e1 tri karti\u010dky, na ka\u017edej je in\u00e1 nenulov\u00e1 cifra. S\u00fa\u010det v\u0161etk\u00fdch trojcifern\u00fdch \u010d\u00edsel, ktor\u00e9 mo\u017eno z t\u00fdchto karti\u010diek zostavi\u0165, je \u010d\u00edslo o 6 v\u00e4\u010d\u0161ie ako trojn\u00e1sobok jedn\u00e9ho z nich. Ak\u00e9 cifry s\u00fa na karti\u010dk\u00e1ch?<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>C-II-3<\/strong><br \/>\nNech <em>E<\/em> je stred strany <em>CD<\/em> rovnobe\u017en\u00edka <em>ABCD<\/em>, v ktorom plat\u00ed 2|<em>AB<\/em>| = 3|<em>BC<\/em>|. Dok\u00e1\u017ete, \u017ee ak sa d\u00e1 do \u0161tvoruholn\u00edka <em>ABCE<\/em> vp\u00edsa\u0165 kru\u017enica, dot\u00fdka sa t\u00e1to kru\u017enica strany <em>BC<\/em> v jej strede.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>C-II-4<\/strong><br \/>\nNa tabuli je nap\u00edsan\u00fdch prv\u00fdch <em>n<\/em> cel\u00fdch kladn\u00fdch \u010d\u00edsel. Mar\u00edna a Tamara sa striedaj\u00fa v \u0165ahoch pri nasleduj\u00facej hre. Najsk\u00f4r Mar\u00edna zotrie jedno z \u010d\u00edsel na tabuli. \u010ealej v\u017edy hr\u00e1\u010dka, ktor\u00e1 je na \u0165ahu, zotrie jedno z \u010d\u00edsel, ktor\u00e9 sa od predch\u00e1dzaj\u00faceho zotret\u00e9ho \u010d\u00edsla ani nel\u00ed\u0161i o 1, ani s n\u00edm nie je s\u00fadelite\u013en\u00e9. Hr\u00e1\u010dka, ktor\u00e1 je na \u0165ahu a nem\u00f4\u017ee u\u017e \u017eiadne \u010d\u00edslo zotrie\u0165, prehr\u00e1. Pre <em>n<\/em> =\u00a06 a pre <em>n<\/em> =\u00a012 rozhodnite, ktor\u00e1 z hr\u00e1\u010dok m\u00f4\u017ee hra\u0165 tak, \u017ee vyhr\u00e1 nez\u00e1visle na \u0165ahoch druhej hr\u00e1\u010dky.<\/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 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<form> <input onclick=\"window.location='61\/61bir.pdf'\" name=\"buttb1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>B-I-1<\/strong><br \/>\nMedzi v\u0161etk\u00fdmi desa\u0165cifern\u00fdmi \u010d\u00edslami delite\u013en\u00fdmi jeden\u00e1stimi, v ktor\u00fdch sa \u017eiadna cifra neopakuje, n\u00e1jdite najmen\u0161ie a najv\u00e4\u010d\u0161ie.<\/p>\n<div class=\"pri\">(Jaroslav Zhouf)<\/div>\n<p><strong>B-I-2<\/strong><br \/>\nDan\u00fd je pravouhl\u00fd trojuholn\u00edk <em>ABC<\/em> s prav\u00fdm uhlom pri vrchole <em>C<\/em>, ktor\u00e9ho obsah ozna\u010dme <em>P<\/em>. Nech <em>F<\/em> je p\u00e4ta v\u00fd\u0161ky z vrcholu <em>C<\/em> na preponu <em>AB<\/em>. Na kolmiciach na priamku <em>AB<\/em>, ktor\u00e9 prech\u00e1dzaj\u00fa vrcholmi <em>A<\/em> a <em>B<\/em>, v polrovine opa\u010dnej k polrovine <em>ABC<\/em> uva\u017eujme postupne body <em>D<\/em> a <em>E<\/em>, pre ktor\u00e9 plat\u00ed |<em>AF<\/em>| = |<em>AD<\/em>| a |<em>BF<\/em>| = |<em>BE<\/em>|. Obsah trojuholn\u00edka <em>DEF<\/em> ozna\u010dme <em>Q<\/em>. Dok\u00e1\u017ete, \u017ee plat\u00ed <em>P<\/em>\u2265<em>Q<\/em>, a zistite, kedy nast\u00e1va rovnos\u0165.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-I-3<\/strong><br \/>\nN\u00e1jdite v\u0161etky dvojice re\u00e1lnych \u010d\u00edsel <em>x, y<\/em>, ktor\u00e9 vyhovuj\u00fa s\u00fastave rovn\u00edc<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"\/61\/61ba1.png\" alt=\"\" width=\"91\" height=\"49\" \/><\/p>\n<p>(Z\u00e1pis <img loading=\"lazy\" decoding=\"async\" src=\"\/54\/54bb2.gif\" alt=\"\" width=\"17\" height=\"15\" align=\"absMiddle\" \/> ozna\u010duje <em>doln\u00fa cel\u00fa \u010das\u0165 \u010d\u00edsla a<\/em>, t. j. najv\u00e4\u010d\u0161ie cel\u00e9 \u010d\u00edslo, ktor\u00e9 neprevy\u0161uje <em>a<\/em>.)<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>B-I-4<\/strong><br \/>\nDan\u00e9 s\u00fa dve r\u00f4znobe\u017eky <em>a, c<\/em> prech\u00e1dzaj\u00face bodom <em>P<\/em> a bod <em>B<\/em>, ktor\u00fd na nich nele\u017e\u00ed. Zostrojte pravouholn\u00edk <em>ABCD<\/em> s vrcholmi <em>A<\/em>, <em>C<\/em> a <em>D<\/em> postupne na priamkach <em>a<\/em>, <em>c<\/em> a <em>PB<\/em>.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>B-I-5<\/strong><br \/>\nV istom meste maj\u00fa vybudovan\u00fa sie\u0165 na \u0161\u00edrenie klebiet, v ktorej si ka\u017ed\u00fd klebetn\u00edk vymie\u0148a inform\u00e1cie s tromi klebetnicami a ka\u017ed\u00e1 klebetnica si vymie\u0148a inform\u00e1cie s tromi klebetn\u00edkmi. Inak sa klebety ne\u0161\u00edria.<br \/>\n<strong>a)<\/strong> Dok\u00e1\u017ete, \u017ee klebetn\u00edkov a klebetn\u00edc je rovnako ve\u013ea.<br \/>\n<strong>b)<\/strong> Predpokladajme, \u017ee sie\u0165 na \u0161\u00edrenie klebiet je s\u00favisl\u00e1 (klebety od \u013eubovo\u013en\u00e9ho klebetn\u00edka a \u013eubovo\u013enej klebetnice sa m\u00f4\u017eu dosta\u0165 ku v\u0161etk\u00fdm ostatn\u00fdm). Dok\u00e1\u017ete, \u017ee aj ke\u010f sa jeden klebetn\u00edk z mesta ods\u0165ahuje, zostane sie\u0165 s\u00favisl\u00e1.<\/p>\n<div class=\"pri\">(J\u00e1n Maz\u00e1k)<\/div>\n<p><strong>B-I-6<\/strong><br \/>\nAnna a Boris hraj\u00fa kartov\u00fa hru. Ka\u017ed\u00fd z nich m\u00e1 p\u00e4\u0165 kariet s hodnotami 1 a\u017e 5 (z ka\u017edej jednu). V ka\u017edom z piatich k\u00f4l obaja vylo\u017eia jednu kartu a kto m\u00e1 vy\u0161\u0161ie \u010d\u00edslo, z\u00edska bod. V pr\u00edpade kariet s rovnak\u00fdmi \u010d\u00edslami nez\u00edska bod nikto. Pou\u017eit\u00e9 karty sa do hry nevracaj\u00fa. Ten, kto z\u00edska na konci viac bodov, vyhral. Ko\u013eko percent zo v\u0161etk\u00fdch mo\u017en\u00fdch priebehov takej hry skon\u010d\u00ed v\u00fdhrou Anny?<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/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<form> <input onclick=\"window.location='61\/61biir.pdf'\" name=\"buttb2\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>B-S-1<\/strong><br \/>\nV obore cel\u00fdch \u010d\u00edsel vyrie\u0161te rovnicu<\/p>\n<div class=\"r\">x<sup>2<\/sup> + y<sup>2<\/sup> + x + y = 4.<\/div>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-S-2<\/strong><br \/>\nDan\u00fd je pravouhl\u00fd trojuholn\u00edk <em>ABC<\/em> s prav\u00fdm uhlom pri vrchole <em>C<\/em>. Nech <em>F<\/em> je p\u00e4ta v\u00fd\u0161ky z vrcholu <em>C<\/em> na preponu <em>AB<\/em>. Na kolmiciach na priamku <em>AB<\/em>, ktor\u00e9 prech\u00e1dzaj\u00fa vrcholmi <em>A<\/em> a <em>B<\/em>, s\u00fa v polrovine opa\u010dnej k polrovine <em>ABC<\/em> zvolen\u00e9 postupne body <em>D<\/em> a <em>E<\/em>, pre ktor\u00e9 plat\u00ed |<em>AF<\/em>| = |<em>AD<\/em>| a |<em>BF<\/em>| = |<em>BE<\/em>|. Ozna\u010dme \u010falej <em>R<\/em> stred \u00fase\u010dky <em>DE<\/em>. Dok\u00e1\u017ete, \u017ee plat\u00ed nerovnos\u0165 |RF| = |CF| a zistite, kedy nastane rovnos\u0165.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-S-3<\/strong><br \/>\nV istom meste maj\u00fa vybudovan\u00fa s\u00favisl\u00fa sie\u0165 na \u0161\u00edrenie klebiet (klebety od \u013eubovo\u013en\u00e9ho klebetn\u00edka a \u013eubovo\u013enej klebetnice sa m\u00f4\u017eu dosta\u0165 ku v\u0161etk\u00fdm ostatn\u00fdm). V nej si ka\u017ed\u00fd klebetn\u00edk vymie\u0148a inform\u00e1cie s dvoma klebetnicami a ka\u017ed\u00e1 klebetnica si vymie\u0148a inform\u00e1cie s troma klebetn\u00edkmi. Predpokladajme, \u017ee v uvedenej sieti sa n\u00e1jde tak\u00fd mu\u017e aj tak\u00e1 \u017eena, \u017ee po pr\u00edpadnom ods\u0165ahovan\u00ed ktorejko\u013evek z t\u00fdchto dvoch os\u00f4b prestane by\u0165 sie\u0165 s\u00favislou. N\u00e1jdite najmen\u0161\u00ed mo\u017en\u00fd po\u010det \u010dlenov tejto siete.<\/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 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<form> <input onclick=\"window.location='61\/61biiir.pdf'\" name=\"buttb3\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>B-II-1<\/strong><br \/>\nDan\u00fdch je 2 012 kladn\u00fdch \u010d\u00edsel men\u0161\u00edch ako 1, ktor\u00fdch s\u00fa\u010det je 7. Dok\u00e1\u017ete, \u017ee tieto \u010d\u00edsla sa daj\u00fa rozdeli\u0165 na \u0161tyri skupiny tak, aby s\u00fa\u010det \u010d\u00edsel v ka\u017edej skupine bol aspo\u0148 1.<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>B-II-2<\/strong><br \/>\nUr\u010dte, ko\u013ek\u00fdmi sp\u00f4sobmi mo\u017eno vrcholom pravideln\u00e9ho 9-uholn\u00edka <em>ABCDEFGHI<\/em> priradi\u0165 \u010d\u00edsla z mno\u017einy {17, 27, 37, 47, 57, 67, 77, 87, 97} tak, aby ka6d\u00e9 z nich bolo priraden\u00e9 in\u00e9mu vrcholu a aby s\u00fa\u010det \u010d\u00edsel priraden\u00fdch ka\u017ed\u00fdm trom susedn\u00fdm vrcholom bol delite\u013en\u00fd tromi.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>B-II-3<\/strong><br \/>\nPravouhl\u00e9mu trojuholn\u00edku <em>ABC<\/em> je vp\u00edsan\u00e1 kru\u017enica, ktor\u00e1 sa dot\u00fdka prepony <em>AB<\/em> v bode K. \u00dase\u010dku <em>AK<\/em> oto\u010d\u00edme o 90\u00b0 do polohy <em>AP<\/em> a \u00fase\u010dku <em>BK<\/em> oto\u010d\u00edme o 90\u00b0 do polohy <em>BQ<\/em> tak, aby body <em>P<\/em>, <em>Q<\/em> le\u017eali v polrovine opa\u010dnej k polrovine <em>ABC<\/em>.<\/p>\n<p>&nbsp;&nbsp;&nbsp;<strong>a)<\/strong>&nbsp;&nbsp;Dok\u00e1\u017ete, \u017ee obsahy trojuholn\u00edkov <em>ABC<\/em> a <em>PQK<\/em> s\u00fa rovnak\u00e9.<br \/>\n&nbsp;&nbsp;&nbsp;<strong>b)<\/strong>&nbsp;&nbsp;Dok\u00e1\u017ete, \u017ee obvod trojuholn\u00edka <em>ABC<\/em> nie je v\u00e4\u010d\u0161\u00ed ako obvod trojuholn\u00edka <em>PQK<\/em>. Kedy nastane rovnos\u0165 obvodov?<\/p>\n<div class=\"pri\">(Jaroslav Zhouf)<\/div>\n<p><strong>B-II-4<\/strong><br \/>\nN\u00e1jdite v\u0161etky re\u00e1lne \u010d\u00edsla <em>x, y<\/em>, ktor\u00e9 vyhovuj\u00fa s\u00fastave rovn\u00edc<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"\/61\/61bc1.png\" alt=\"\" width=\"108\" height=\"78\" \/><\/p>\n<p>(Z\u00e1pis <img loading=\"lazy\" decoding=\"async\" src=\"\/54\/54bb2.gif\" alt=\"\" width=\"17\" height=\"15\" align=\"absMiddle\" \/> ozna\u010duje <em>doln\u00fa cel\u00fa \u010das\u0165 \u010d\u00edsla a<\/em>, t. j. najv\u00e4\u010d\u0161ie cel\u00e9 \u010d\u00edslo, ktor\u00e9 neprevy\u0161uje <em>a<\/em>.)<\/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 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<form> <input onclick=\"window.location='61\/61air.pdf'\" name=\"butta1\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-I-1<\/strong><\/p>\n<p>Ozna\u010dme <em>n<\/em> s\u00fa\u010det v\u0161etk\u00fdch desa\u0165cifern\u00fdch \u010d\u00edsel, ktor\u00e9 maj\u00fa vo svojom dekadickom z\u00e1pise ka\u017ed\u00fa z cifier 0, 1, &#8230;, 9. Zistite zvy\u0161ok po delen\u00ed \u010d\u00edsla <em>n<\/em> sedemdesiatimi siedmimi.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-I-2<\/strong><\/p>\n<p>Na stretnut\u00ed bolo nieko\u013eko \u013eud\u00ed. Ka\u017ed\u00ed dvaja, ktor\u00ed sa nepoznali, mali medzi ostatn\u00fdmi pr\u00edtomn\u00fdmi pr\u00e1ve dvoch spolo\u010dn\u00fdch zn\u00e1mych. \u00da\u010dastn\u00edci A a B sa poznali, ale nemali ani jedn\u00e9ho spolo\u010dn\u00e9ho zn\u00e1meho. Dok\u00e1\u017ete, \u017ee A aj B mali medzi pr\u00edtomn\u00fdmi rovnak\u00fd po\u010det zn\u00e1mych. Uk\u00e1\u017ete tie\u017e, \u017ee na stretnut\u00ed mohlo by\u0165 pr\u00e1ve \u0161es\u0165 os\u00f4b.<\/p>\n<div class=\"pri\">(Vojtech B\u00e1lint)<\/div>\n<p><strong>A-I-3<\/strong><\/p>\n<p>Ozna\u010dme <em>S<\/em> stred kru\u017enice vp\u00edsanej, <em>T<\/em> \u0165a\u017eisko a <em>V<\/em> priese\u010dn\u00edk v\u00fd\u0161ok dan\u00e9ho rovnoramenn\u00e9ho trojuholn\u00edka, ktor\u00fd nie je rovnostrann\u00fd.<\/p>\n<p><strong>a) <\/strong>Dok\u00e1\u017ete, \u017ee bod <em>S<\/em> je vn\u00fatorn\u00fdm bodom \u00fase\u010dky <em>TV<\/em>.<\/p>\n<p><strong>b)<\/strong> Ur\u010dte pomer d\u013a\u017eok str\u00e1n dan\u00e9ho trojuholn\u00edka, ak je bod <em>S<\/em> stredom \u00fase\u010dky <em>TV<\/em>.<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-I-4<\/strong><\/p>\n<p>Nech <em>p, q<\/em> s\u00fa dve r\u00f4zne prvo\u010d\u00edsla, <em>m, n<\/em> prirodzen\u00e9 \u010d\u00edsla a s\u00fa\u010det<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"\/61\/61aa2.png\" alt=\"\" width=\"115\" height=\"41\" \/><\/p>\n<p>je cel\u00e9 \u010d\u00edslo. Dok\u00e1\u017ete, \u017ee plat\u00ed nerovnos\u0165<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"\/61\/61aa1.png\" alt=\"\" width=\"87\" height=\"40\" \/><\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-I-5<\/strong><\/p>\n<p>Dan\u00e9 s\u00fa dve zhodn\u00e9 kru\u017enice <em>k<\/em><sub>1<\/sub>, <em>k<\/em><sub>2<\/sub> s polomerom rovn\u00fdm vzdialenosti ich stredov. Ich priese\u010dn\u00edky ozna\u010dme <em>A<\/em> a <em>B<\/em>. Na kru\u017enici <em>k<\/em><sub>2<\/sub> zvo\u013eme bod <em>C<\/em> tak, \u017ee \u00fase\u010dka <em>BC<\/em> pretne kru\u017enicu <em>k<\/em><sub>1<\/sub> v bode r\u00f4znom od <em>B<\/em>, ktor\u00fd ozna\u010d\u00edme <em>L<\/em>. Priamka <em>AC<\/em> pretne kru\u017enicu <em>k<\/em><sub>1<\/sub> v bode r\u00f4znom od <em>A<\/em>, ktor\u00fd ozna\u010d\u00edme <em>K<\/em>. Dok\u00e1\u017ete, \u017ee priamka, na ktorej le\u017e\u00ed \u0165a\u017enica z vrcholu <em>C<\/em> trojuholn\u00edka <em>KLC<\/em>, prech\u00e1dza pevn\u00fdm bodom nez\u00e1visl\u00fdm od polohy bodu <em>C<\/em>.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>A-I-6<\/strong><\/p>\n<p>N\u00e1jdite najv\u00e4\u010d\u0161ie re\u00e1lne \u010d\u00edslo <em>k<\/em> tak\u00e9, \u017ee nerovnos\u0165<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"\/61\/61aa3.png\" alt=\"\" width=\"188\" height=\"46\" \/><\/p>\n<p>plat\u00ed pre v\u0161etky dvojice kladn\u00fdch re\u00e1lnych \u010d\u00edsel <em>a, b<\/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 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<form> <input onclick=\"window.location='61\/61aiir.pdf'\" name=\"buttas\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-S-1<\/strong><br \/>\nV obore re\u00e1lnych \u010d\u00edsel vyrie\u0161te s\u00fastavu rovn\u00edc<\/p>\n<div class=\"r\">y + 3x = 4x<sup>3<\/sup>,<br \/>\nx + 3y = 4y<sup>3<\/sup>.<\/div>\n<div class=\"pri\">(Pavel Cal\u00e1bek)<\/div>\n<p><strong>A-S-2<\/strong><br \/>\nV rovine uva\u017eujme lichobe\u017en\u00edk ABCD so z\u00e1klad\u0148ami <em>AB<\/em> a <em>CD<\/em> a ozna\u010dme <em>M<\/em> stred jeho uhloprie\u010dky <em>AC<\/em>. Dok\u00e1\u017ete, \u017ee ak maj\u00fa trojuholn\u00edky <em>ABM<\/em> a <em>ACD<\/em> rovnak\u00e9 obsahy, tak s\u00fa priamky <em>DM<\/em> a <em>BC<\/em> rovnobe\u017en\u00e9.<\/p>\n<div class=\"pri\">(Jaroslav \u0160vr\u010dek)<\/div>\n<p><strong>A-S-3<\/strong><br \/>\nN\u00e1jdite v\u0161etky prirodzen\u00e9 \u010d\u00edsla <em>n<\/em>, pre ktor\u00e9 je s\u00fa\u010din <strong><em>(2<sup>n<\/sup> + 1)(3<sup>n<\/sup> + 2)<\/em><\/strong> delite\u013en\u00fd \u010d\u00edslom <strong><em>5<sup>n<\/sup><\/em><\/strong>.<\/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=\"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> <input onclick=\"window.location='61\/61aiiir.pdf'\" name=\"butta3\" type=\"button\" value=\"Rie\u0161enie\" \/><\/form>\n<p><strong>A-II-1<\/strong><br \/>\nOzna\u010dme <em>S<sub>n<\/sub><\/em> s\u00fa\u010det v\u0161etk\u00fdch <em>n<\/em>-cifern\u00fdch \u010d\u00edsel, ktor\u00fdch dekadick\u00fd z\u00e1pis obsahuje iba cifry 1, 2, 3, ka\u017ed\u00fa aspo\u0148 raz. N\u00e1jdite v\u0161etky cel\u00e9 \u010d\u00edsla <em>n<\/em>\u22653, pre ktor\u00e9 je \u010d\u00edslo <em>S<sub>n<\/sub><\/em> delite\u013en\u00e9 siedmimi.<\/p>\n<div class=\"pri\">(Pavel Novotn\u00fd)<\/div>\n<p><strong>A-II-2<\/strong><br \/>\nDan\u00e9 je cel\u00e9 \u010d\u00edslo <em>a<\/em> v\u00e4\u010d\u0161ie ako 1.N\u00e1jdite aritmetick\u00fa postupnos\u0165 s prv\u00fdm \u010dlenom <em>a<\/em>, ktor\u00e1 obsahuje pr\u00e1ve dve z \u010d\u00edsel a<sup>2<\/sup>, a<sup>3<\/sup>, a<sup>4<\/sup>, a<sup>5<\/sup> a m\u00e1 \u010do najv\u00e4\u010d\u0161iu diferenciu. (<em>Nepredpoklad\u00e1me, \u017ee diferencia je nutne celo\u010d\u00edseln\u00e1.<\/em>)<\/p>\n<div class=\"pri\">(Jarom\u00edr \u0160im\u0161a)<\/div>\n<p><strong>A-II-3<\/strong><br \/>\nDo kru\u017enice je vp\u00edsan\u00fd \u0161es\u0165uholn\u00edk <em>ABCDEF<\/em>, v ktorom plat\u00ed <em>AB<\/em>\u22a5<em>BD<\/em>,|<em>BC<\/em>| = |<em>EF<\/em>|. Predpokladajme, \u017ee priamky <em>BC<\/em>, <em>EF<\/em> pret\u00ednaj\u00fa polpriamku <em>AD<\/em> postupne v bodoch <em>P<\/em>, <em>Q<\/em>. Ozna\u010dme <em>S<\/em> stred uhloprie\u010dky <em>AD<\/em> a <em>K<\/em>, <em>L<\/em> stredy kru\u017en\u00edc vp\u00edsan\u00fdch trojuholn\u00edkom <em>BPS<\/em>, <em>EQS<\/em>. Dok\u00e1\u017ete, \u017ee trojuholn\u00edk <em>KLD<\/em> je pravouhl\u00fd.<\/p>\n<div class=\"pri\">(Tom\u00e1\u0161 Jur\u00edk)<\/div>\n<p><strong>A-II-4<\/strong><br \/>\nPredpokladajme, \u017ee pre kladn\u00e9 re\u00e1lne \u010d\u00edsla <em>a, b, c, d<\/em> plat\u00ed<\/p>\n<p class=\"pc\"><em><strong>ab + cd = ac + bd = 4<\/strong><\/em><img loading=\"lazy\" decoding=\"async\" src=\"\/img\/space.png\" alt=\"\" width=\"30\" height=\"1\" \/>a<img loading=\"lazy\" decoding=\"async\" src=\"\/img\/space.png\" alt=\"\" width=\"30\" height=\"1\" \/><em><strong>ad + bc = 5<\/strong>.<\/em><\/p>\n<p>N\u00e1jdite najmen\u0161iu mo\u017en\u00fa hodnotu s\u00fa\u010dtu <em>a + b + c + d<\/em> a zistite, ktor\u00e9 vyhovuj\u00face \u0161tvorice <em>a, b, c, d<\/em> ju dosahuj\u00fa.<\/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 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>61. ro\u010dn\u00edk matematickej olympi\u00e1dy 2011-2012 aktualizovan\u00e9 17.4.2012 21:10<\/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":[10],"tags":[],"class_list":["post-480","post","type-post","status-publish","format-standard","hentry","category-61-rocnik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - 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