{"id":140,"date":"2010-06-05T19:57:08","date_gmt":"2010-06-05T19:57:08","guid":{"rendered":"http:\/\/matematika.okamzite.eu\/?p=140"},"modified":"2010-06-05T19:57:08","modified_gmt":"2010-06-05T19:57:08","slug":"kategorie-z4-z9-3","status":"publish","type":"post","link":"https:\/\/matematika.besaba.com\/?p=140","title":{"rendered":"Kateg\u00f3rie Z4-Z9"},"content":{"rendered":"

57. ro\u010dn\u00edk matematickej olympi\u00e1dy 2007-2008<\/p>\n

<\/a><\/p>\n\n\n\n\n\n
Z4<\/span><\/td>\nZ5<\/span><\/td>\nZ6<\/span><\/td>\nZ7<\/span><\/td>\nZ8<\/span><\/td>\nZ9<\/span><\/td>\n<\/tr>\n
1. kolo<\/a><\/span><\/td>\n1. kolo<\/a><\/span><\/td>\n1. kolo<\/a><\/span><\/td>\n1. kolo<\/a><\/span><\/td>\n1. kolo<\/a><\/span><\/td>\n1. kolo<\/a><\/span><\/td>\n<\/tr>\n
2. kolo<\/a><\/span><\/td>\n2. kolo<\/a><\/span><\/td>\n2. kolo<\/a><\/span><\/td>\n2. kolo<\/a><\/span><\/td>\n2. kolo<\/a><\/span><\/td>\n2. kolo<\/a><\/span><\/td>\n<\/tr>\n
 <\/td>\n3. kolo<\/a><\/span><\/td>\n<\/tr>\n<\/table>\n

<\/a><\/p>\n

Z4-I-1<\/b>
\nZ p\u00e4\u0165cifern\u00fdch \u010d\u00edsel 53 827<\/b> a 19 763<\/b> vy\u0161krtni spolu dve \u010d\u00edslice tak, aby s\u00fa\u010det vzniknut\u00fdch \u010d\u00edsel bol \u010do najv\u00e4\u010d\u0161\u00ed.<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z4-I-2<\/b>
\nNa pomaran\u010dov\u00fa limon\u00e1du potrebujeme \u0161\u0165avu z \u00f4smich pomaran\u010dov, dvoch citr\u00f3nov, 2 \u010dajov\u00e9 ly\u017ei\u010dky cukru a 6 decilitrov vody. Do d\u017eb\u00e1nu sme si naliali 9 decilitrov vody. Ko\u013eko mus\u00edme od\u0161\u0165avi\u0165 pomaran\u010dov, citr\u00f3nov, ko\u013eko prida\u0165 ly\u017ei\u010diek cukru, aby sme dostali rovnako kvalitn\u00fa limon\u00e1du?<\/p>\n

(S. Bodl\u00e1kov\u00e1)<\/div>\n<\/p>\n

Z4-I-3<\/b>
\nNa drevenom plote je 70 latiek. Pa\u013eko s Ani\u010dkou ich mali v\u0161etky ponatiera\u0165 farbou. Za\u010dali aj skon\u010dili obaja naraz. K\u00fdm Ani\u010dka natrela dve latky, pre\u0161li 4 min\u00faty a za 8 min\u00fat stihol Pa\u013eko ponatiera\u0165 3 latky. Ako dlho im trvalo natretie v\u0161etk\u00fdch latiek?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z4-I-4<\/b>
\nNa obr\u00e1zku s \u010das\u0165ou \u0161tvor\u010dekovej siete s\u00fa vyzna\u010den\u00e9 tri body. Ka\u017ed\u00e9 dva z t\u00fdch troch vyzna\u010den\u00fdch bodov tvoria v\u017edy dva zo \u0161tyroch vrcholov nejak\u00e9ho \u0161tvorca. \u0160tvorec KAMI<\/i> je najmen\u0161\u00ed z nich, ale z obr\u00e1zku sa n\u00e1m vymazali men\u00e1 vyzna\u010den\u00fdch bodov. Dokresli do \u0161tvor\u010dekovej siete cel\u00fd \u0161tvorec KAMI<\/i>.<\/p>\n

<\/div>\n
(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z4-I-5<\/b>
\nMartin a Jana si porovn\u00e1vali svoje Mikul\u00e1\u0161ske bal\u00ed\u010dky. Mali tam nasypan\u00e9 aj svoje ob\u013e\u00faben\u00e9 \u010dokol\u00e1dky. Martin ich v\u0161ak mal in\u00fd po\u010det ako Jana, tak venoval \u0161tvrtinu svojich \u010dokol\u00e1dok Jane. Jana si v\u0161etky svoje prepo\u010d\u00edtala a polovicu z nich venovala nasp\u00e4\u0165 Martinovi. Potom Martin zase venoval \u0161tvrtinu svojich Jane. Po n\u00e1slednom prepo\u010d\u00edtan\u00ed zistili, \u017ee maj\u00fa obaja po 9 \u010dokol\u00e1dok. Ko\u013eko \u010dokol\u00e1dok mal Martin p\u00f4vodne v bal\u00ed\u010dku? Ko\u013eko ich tam mala Jana? (Po\u010das po\u010d\u00edtania a pres\u00favania ani jednu \u010dokol\u00e1dku nezjedli.)<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z4-I-6<\/b>
\nDopl\u0148 na pr\u00e1zdne tehli\u010dky pyram\u00eddy zn\u00e1zornenej na obr\u00e1zku ch\u00fdbaj\u00face \u010d\u00edsla tak, aby platilo: na ka\u017edej tehli\u010dke (okrem t\u00fdch z najspodnej\u0161ieho riadku) je nap\u00edsan\u00e9 \u010d\u00edslo, ktor\u00e9 sa rovn\u00e1 polovici s\u00fa\u010dtu \u010d\u00edsel nap\u00edsan\u00fdch na dvoch s \u0148ou susediacich tehli\u010dk\u00e1ch z ni\u017e\u0161ieho riadku.<\/p>\n

<\/div>\n
(S. Bedn\u00e1\u0159ov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z4-II-1<\/b>
\nDopl\u0148 na pr\u00e1zdne tehli\u010dky pyram\u00eddy zn\u00e1zornenej na obr\u00e1zku ch\u00fdbaj\u00face \u010d\u00edsla tak, aby platilo: na ka\u017edej tehli\u010dke (okrem t\u00fdch z najspodnej\u0161ieho riadku) je nap\u00edsan\u00e9 \u010d\u00edslo, ktor\u00e9 sa rovn\u00e1 dvojn\u00e1sobku s\u00fa\u010dtu \u010d\u00edsel nap\u00edsan\u00fdch na dvoch s \u0148ou susediacich tehli\u010dk\u00e1ch z ni\u017e\u0161ieho riadku.<\/p>\n

<\/div>\n
(S. Bedn\u00e1\u0159ov\u00e1, M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z4-II-2<\/b>
\nZ dan\u00fdch p\u00e4\u0165cifern\u00fdch \u010d\u00edsel 53827<\/i><\/b> a 19763<\/i><\/b> vy\u0161krtni spolu dve \u010d\u00edslice tak, aby rozdiel vzniknut\u00fdch \u010d\u00edsel bol \u010do najmen\u0161\u00ed.<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z4-II-3<\/b>
\nMartin narysoval obd\u013a\u017enik, ktor\u00fd sa d\u00e1 bezo zvy\u0161ku rozstriha\u0165 na 3 \u0161tvorce. Ak\u00e9 rozmery mohol ma\u0165 obd\u013a\u017enik, ak jeden zo \u0161tvorcov mal stranu d\u013a\u017eky 7 cm? Vypo\u010d\u00edtaj jeho obvod.<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z5-I-1<\/b>
\nN\u00e1\u0161 kuchynsk\u00fd st\u00f4l m\u00e1 obd\u013a\u017enikov\u00fa vrchn\u00fa dosku s rozmermi 90 cm × 140 cm. Chceme na\u0148 u\u0161i\u0165 obrus tak, aby na ka\u017edej strane stola presahoval rovnako.<\/p>\n

   a)<\/b>  Ko\u013eko l\u00e1tky \u0161\u00edrky 140 cm treba k\u00fapi\u0165, aby sme ju u\u017e nemuseli \u010falej striha\u0165?
\n   b)<\/b>  Ko\u013eko cm bude tento obrus na ka\u017edej strane presahova\u0165?<\/p>\n

(S. Bedn\u00e1\u0159ov\u00e1)<\/div>\n<\/p>\n

Z5-I-2<\/b>
\nDopl\u0148 na pr\u00e1zdne tehli\u010dky pyram\u00eddy zn\u00e1zornenej na obr\u00e1zku ch\u00fdbaj\u00face \u010d\u00edsla tak, aby platilo: na ka\u017edej tehli\u010dke (okrem t\u00fdch z najspodnej\u0161ieho riadku) je nap\u00edsan\u00e9 \u010d\u00edslo, ktor\u00e9 sa rovn\u00e1 polovici s\u00fa\u010dtu \u010d\u00edsel nap\u00edsan\u00fdch na dvoch s \u0148ou susediacich tehli\u010dk\u00e1ch z ni\u017e\u0161ieho riadku.<\/p>\n

<\/div>\n
(S. Bedn\u00e1\u0159ov\u00e1)<\/div>\n<\/p>\n

Z5-I-3<\/b>
\nV \u0161k\u00f4lke maj\u00fa stavebnicu pozost\u00e1vaj\u00facu z rovnako ve\u013ek\u00fdch molitanov\u00fdch kv\u00e1drov. Ke\u010f ich deti v\u0161etky polo\u017eia na seba, poskladaj\u00fa ich v\u017edy tak, aby na sebe le\u017eali kv\u00e1dre rovnak\u00fdmi stenami a v \u017eiadnom \u201eposchod\u00ed\u201c neboli kv\u00e1dre dva. Takto sa im postupne podarilo postavi\u0165 tri r\u00f4zne vysok\u00e9 ve\u017ee. Prv\u00e1 mala 120 cm, druh\u00e1 150 cm a tretia 130 cm. Ko\u013eko kv\u00e1drov mohla ma\u0165 stavebnica, z ktorej stavali?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z5-I-4<\/b>
\nTroj\u010dat\u00e1 pr\u00e1ve osl\u00e1vili svoje tretie narodeniny. O p\u00e4\u0165 rokov bude s\u00fa\u010det ich vekov rovn\u00fd dne\u0161n\u00e9mu veku ich mamy. Ko\u013eko rokov bude ma\u0165 ich mama o 5 rokov?<\/p>\n

(M. Krej\u010dov\u00e1)<\/div>\n<\/p>\n

Z5-I-5<\/b>
\n\u010c\u00edslo sa naz\u00fdva PREF\u00cdKAN\u00c9<\/i>, ak po\u010dn\u00fac jeho tre\u0165ou \u010d\u00edslicou (po\u010d\u00edtan\u00e9 z\u013eava) plat\u00ed, \u017ee ka\u017ed\u00e1 jeho \u010d\u00edslica je s\u00fa\u010dtom v\u0161etk\u00fdch \u010d\u00edslic le\u017eiacich na\u013eavo od nej.<\/p>\n

   a)<\/b>  Nap\u00ed\u0161te dve najv\u00e4\u010d\u0161ie PREF\u00cdKAN\u00c9<\/i> \u010d\u00edsla.
\n   b)<\/b>  Ko\u013eko je v\u0161etk\u00fdch \u0161tvorcifern\u00fdch PREF\u00cdKAN\u00ddCH<\/i> \u010d\u00edsel?<\/p>\n

(S. Bedn\u00e1\u0159ov\u00e1)<\/div>\n<\/p>\n

Z5-I-6<\/b>
\nDopl\u0148 do pr\u00e1zdnych pol\u00ed\u010dok prirodzen\u00e9 \u010d\u00edsla od 1 do 16 (ka\u017ed\u00e9 \u010d\u00edslo m\u00f4\u017ee\u0161 pou\u017ei\u0165 len raz) tak, aby platili matematick\u00e9 vz\u0165ahy.<\/p>\n

<\/p>\n

(M. Smitkov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z5-II-1<\/b>
\nPolovica det\u00ed 5.A chod\u00ed na tane\u010dn\u00fd kr\u00fa\u017eok. Diev\u010dat\u00e1 chodia v\u0161etky a z osemn\u00e1stich chlapcov chod\u00ed iba jedna tretina.<\/p>\n

  a)<\/i><\/b>  Ko\u013eko det\u00ed chod\u00ed do 5.A?
\n  b)<\/i><\/b>  Ko\u013eko diev\u010dat chod\u00ed do 5.A?<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n

Z5-II-2<\/b>
\nMajo s\u010d\u00edtal \u0161tyri po sebe id\u00face dvojcifern\u00e9 \u010d\u00edsla a s\u00fa\u010det zaokr\u00fahlil na desiatky. Juro vzal tie ist\u00e9 \u010d\u00edsla, najprv ich zaokr\u00fahlil na desiatky a potom ich s\u010d\u00edtal. Jeho v\u00fdsledok bol o 10 v\u00e4\u010d\u0161\u00ed ako Majov. Ktor\u00e9 \u010d\u00edsla s\u010d\u00edtavali chlapci, ak ich v\u00fdsledky neboli v\u00e4\u010d\u0161ie ako 100? N\u00e1jdi v\u0161etky mo\u017en\u00e9 rie\u0161enia.<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z5-II-3<\/b>
\nMami\u010dka \u0161ije utierky z l\u00e1tky \u0161\u00edrky 120 cm. Hotov\u00e1 utierka m\u00e1 rozmery 60 cm × 38 cm. Pri strihan\u00ed l\u00e1tky treba po\u010d\u00edta\u0165 2 cm naviac na ka\u017edom okraji na za\u010distenie. Najmenej ko\u013eko centimetrov tejto l\u00e1tky treba k\u00fapi\u0165, aby z nej mami\u010dke vy\u0161lo 10 utierok?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z6-I-1<\/b>
\nJurko k\u00fapil dve \u010dokol\u00e1dy v obchode oproti \u0161kole. Mi\u0161ko si k\u00fapil tak\u00e9 ist\u00e9 dve \u010dokol\u00e1dy v obchode za \u0161kolou a Ivan si k\u00fapil tie\u017e tak\u00fa \u010dokol\u00e1du, ale v \u0161kolskom bufete. Spolu potom zistili, \u017ee priemerne ich spolu vy\u0161la jedna \u010dokol\u00e1da na 19,70 Sk. Tak\u00fdmto sp\u00f4sobom boli v\u0161etky tri n\u00e1kupy spolu o 6 Sk drah\u0161ie, ako keby chlapci nakupovali v\u0161etk\u00fdch 5 \u010dokol\u00e1d v obchode oproti \u0161kole a o 6,50 Sk lacnej\u0161ie, ako keby nak\u00fapili iba v obchode za \u0161kolou. Ko\u013eko st\u00e1li \u010dokol\u00e1dy v jednotliv\u00fdch obchodoch?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z6-I-2<\/b>
\nMi\u0161ko mal farebn\u00e9 n\u00e1lepky v tvare rovnoramenn\u00fdch pravouhl\u00fdch trojuholn\u00edkov dvoch ve\u013ekost\u00ed. Prv\u00fd druh mal ramen\u00e1 d\u013a\u017eky 5 cm, t\u00fdch bolo 9. Druh\u00fd druh mal najdlh\u0161iu stranu d\u013a\u017eky 10 cm a t\u00fdchto n\u00e1lepiek bolo 17. Najmenej ko\u013eko n\u00e1lepiek prv\u00e9ho druhu si m\u00e1 Mi\u0161ko e\u0161te dok\u00fapi\u0165, aby svojimi n\u00e1lepkami mohol \u00faplne oblepi\u0165 (zakry\u0165) steny kocky s hranou d\u013a\u017eky 10 cm?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z6-I-3<\/b>
\nV rovine maj\u00fa le\u017ea\u0165 body A, B, C, D<\/i> tak, aby platilo: |AB<\/i>| = 7 cm, |BC<\/i>| = 8 cm, |CD<\/i>| = 5 cm a |DA<\/i>| = 9 cm.<\/p>\n

   a)<\/b>  Ur\u010di najv\u00e4\u010d\u0161iu mo\u017en\u00fa vzdialenos\u0165 bodov A<\/i> a C<\/i>.
\n   b)<\/b>  Ur\u010di najmen\u0161iu mo\u017en\u00fa vzdialenos\u0165 bodov A<\/i> a C<\/i>.<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z6-I-4<\/b>
\nPri chudokrvnosti sa odpor\u00fa\u010da pi\u0165 mrkvovo-cviklov\u00fa \u0161\u0165avu, pri\u010dom cviklov\u00e1 \u0161\u0165ava m\u00e1 predstavova\u0165 len 1\/5 z objemu n\u00e1poja. Z dvoch kg mrkvy z\u00edskame v od\u0161\u0165avova\u010di 7,5 dl \u0161\u0165avy, z jedn\u00e9ho kg cvikly 6 dl \u0161\u0165avy.<\/p>\n

   a)<\/b>  Ak\u00e9 mno\u017estvo mrkvy potrebujeme na 25 dag cvikly, aby sme z\u00edskali spr\u00e1vne namie\u0161an\u00fa mrkvovo-cviklov\u00fa \u0161\u0165avu?
\n   b)<\/b>  Ak\u00e9 mno\u017estvo mrkvovo-cviklovej \u0161\u0165avy takto z\u00edskame?<\/p>\n

(S. Bedn\u00e1\u0159ov\u00e1)<\/div>\n<\/p>\n

Z6-I-5<\/b>
\nAk povie mimozem\u0161\u0165an v rozhovore o Vianociach „haf quin lina“<\/i>, znamen\u00e1 to „ve\u013ek\u00e9 zlat\u00e9 hviezdy“<\/i>; ak „kari lina mejk“<\/i>, znamen\u00e1 to „blikaj\u00face zlat\u00e9 kolieska“<\/i>; ak „esca haf kari“<\/i>, znamen\u00e1 to „ve\u013ek\u00e9 \u010derven\u00e9 kolieska“<\/i>. Ako sa povie „blikaj\u00face hviezdy“<\/i> ? (Zap\u00ed\u0161 svoju \u00favahu.)<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n

Z6-I-6<\/b>
\nZ trojcifern\u00fdch \u010d\u00edsel 532 a 179 vy\u0161krtni spolu dve \u010d\u00edslice tak, aby s\u00fa\u010din vzniknut\u00fdch \u010d\u00edsel bol \u010do najv\u00e4\u010d\u0161\u00ed.<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z6-II-1<\/b>
\nNa z\u00e1hrade p\u00e1na Kozla kvitlo nieko\u013eko \u010dere\u0161n\u00ed. Na ka\u017edej \u010dere\u0161ni sedeli tri \u0161korce a e\u0161te jeden sedel na plote. Pes p\u00e1na Kozla ich vypla\u0161il a \u0161korce uleteli. Za chv\u00ed\u013eu sa v\u0161etci vr\u00e1tili a usadili sa op\u00e4\u0165 na \u010dere\u0161niach. \u010cere\u0161\u0148a, pod ktorou spal pes, zostala pr\u00e1zdna a na ka\u017edej z ostatn\u00fdch sa usadili \u0161tyri \u0161korce. Ko\u013eko \u010dere\u0161n\u00ed m\u00e1 p\u00e1n Kozel a ko\u013eko bolo na z\u00e1hrade \u0161korcov?<\/p>\n

(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z6-II-2<\/b>
\nDan\u00fd je trojuholn\u00edk ABC<\/i>, v ktorom p\u00e4ta P<\/i> kolmice z bodu C<\/i> na priamku AB<\/i> le\u017e\u00ed vo vn\u00fatri \u00fase\u010dky AB<\/i>. Z bodu P<\/i> s\u00fa veden\u00e9 kolmice p, q<\/i> na priamky AC<\/i> a BC<\/i> (v uvedenom porad\u00ed). Ozna\u010dme S<\/i> priese\u010dn\u00edk priamky BC<\/i> a priamky q<\/i> a nech T<\/i> je priese\u010dn\u00edk priamky AC<\/i> a priamky p<\/i>. Vypo\u010d\u00edtaj ve\u013ekos\u0165 uhla ACB<\/i>, ak vie\u0161, \u017ee |Ð<\/font> APT<\/em>| + |Ð<\/font> BPS<\/em>| = 20\u00b0.<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z6-II-3<\/b>
\nNa obr\u00e1zku je zaokr\u00fah\u013eovacia s\u010d\u00edtacia pyram\u00edda. Do ka\u017edej tehli\u010dky (okrem t\u00fdch z najspodnej\u0161ieho riadku) patr\u00ed s\u00fa\u010det \u010d\u00edsel nap\u00edsan\u00fdch na dvoch s \u0148ou susediacich tehli\u010dk\u00e1ch z ni\u017e\u0161ieho riadku, ale pr\u00edslu\u0161ne zaokr\u00fahlen\u00fd: s\u00fa\u010dty v druhom riadku odspodu zaokr\u00fah\u013eujeme na desiatky, v tre\u0165om na stovky a v najvy\u0161\u0161om \u0161tvrtom na tis\u00edce. Dopl\u0148te do pr\u00e1zdnych tehli\u010diek pyram\u00eddy najv\u00e4\u010d\u0161ie mo\u017en\u00e9 cel\u00e9 \u010d\u00edsla.<\/p>\n

<\/div>\n
(M. Dillingerov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z7-I-1<\/b>
\n\u010c\u00edslo nazveme trochu ne\u0161\u0165astn\u00e9<\/i>, ak je n\u00e1sobkom \u010d\u00edsla 13. \u010c\u00edslo, ktor\u00e9 je n\u00e1sobkom \u010d\u00edsla 17, nazveme trochu usmievav\u00e9<\/i>. Vezmime v\u0161etky prirodzen\u00e9 \u010d\u00edsla od 1 do 1 000 000, ktor\u00e9 nekon\u010dia ani 0 ani 5. Ko\u013eko z nich je trochu ne\u0161\u0165astn\u00fdch<\/i> a z\u00e1rove\u0148 trochu usmievav\u00fdch<\/i>?<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n

Z7-I-2<\/b>
\nVl\u00e1da Tramt\u00e1rie sa rozhodla, \u017ee svoje \u00fazemie rozdel\u00ed na \u0161es\u0165 krajov. Vybrala preto \u0161es\u0165 najv\u00fdznamnej\u0161\u00edch miest (krajsk\u00e9 mest\u00e1) a ka\u017ed\u00e9mu chce priradi\u0165 kraj pod\u013ea nasleduj\u00faceho pravidla: ka\u017ed\u00e9 miesto v krajine patr\u00ed do kraja toho krajsk\u00e9ho mesta, ku ktor\u00e9mu to m\u00e1 vzdu\u0161nou \u010diarou najbli\u017e\u0161ie. Prekreslite si vo vhodnej mierke mapu Tramt\u00e1rie a narysujte do nej hranice krajov.
\n(Krajsk\u00e9 mest\u00e1 s\u00fa ozna\u010den\u00e9 p\u00edsmenami A<\/i> \u2013 F<\/i>, hrub\u00e1 \u010diara ozna\u010duje hranice Tramt\u00e1rie. \u0160tvorcov\u00e1 sie\u0165 m\u00e1 iba u\u013eah\u010dova\u0165 orient\u00e1ciu na mape a \u017eiadnym sp\u00f4sobom neovplyv\u0148uje hranice krajov!)<\/p>\n

<\/div>\n
(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z7-I-3<\/b>
\nO dvan\u00e1stej st\u00e1li na parkovisku \u010desk\u00e9, nemeck\u00e9 a franc\u00fazske aut\u00e1 v pomere: \u010desk\u00e9 k nemeck\u00fdm 9:4, nemeck\u00e9 k franc\u00fazskym 2:3. V priebehu hodiny odi\u0161lo jeden\u00e1s\u0165 a pri\u0161lo p\u00e4\u0165 \u010desk\u00fdch \u00e1ut, odi\u0161lo jedno a pri\u0161lo jeden\u00e1s\u0165 nemeck\u00fdch \u00e1ut a odi\u0161li tri a pri\u0161lo \u0161es\u0165 franc\u00fazskych \u00e1ut. Ak\u00fd je pomer \u010desk\u00fdch, nemeck\u00fdch a franc\u00fazskych \u00e1ut stojacich o 13:00 na parkovisku, ak o 12:00 tam bolo dvan\u00e1s\u0165 franc\u00fazskych \u00e1ut?<\/p>\n

(\u0160. Pt\u00e1\u010dkov\u00e1)<\/div>\n<\/p>\n

Z7-I-4<\/b>
\n\u00dase\u010dky AM, BM, CM<\/i> a DM<\/i> usporiadan\u00e9 ako na obr\u00e1zku s\u00fa rovnakej d\u013a\u017eky. Uhly, ktor\u00e9 zvieraj\u00fa, maj\u00fa ve\u013ekosti 20\u00b0, 20\u00b0, 50\u00b0, 50\u00b0, 70\u00b0 a a<\/font>. Zisti ve\u013ekos\u0165 uhla, ktor\u00fd zvieraj\u00fa priamky AB<\/i> a CD<\/i>.
\n(Obr\u00e1zok je nepresn\u00fd, nevyplat\u00ed sa ti mera\u0165.)<\/p>\n

<\/div>\n
(M. Raabov\u00e1)<\/div>\n<\/p>\n

Z7-I-5<\/b>
\nV\u0161etky pol\u00ed\u010dka na \u0161achovnici 4\u00d74 vyfarbite \u0161tyrmi r\u00f4znymi farbami a vp\u00ed\u0161te do nich p\u00edsmen\u00e1 L<\/b>, E<\/b>, T<\/b>, O<\/b> tak, aby v ka\u017edom riadku aj v ka\u017edom st\u013apci boli zast\u00fapen\u00e9 v\u0161etky farby aj v\u0161etky p\u00edsmen\u00e1. Ka\u017ed\u00e9 pol\u00ed\u010dko bude cel\u00e9 jednofarebn\u00e9 a bude obsahova\u0165 pr\u00e1ve jedno p\u00edsmeno.
\nKa\u017ed\u00e9 p\u00edsmeno mus\u00ed by\u0165 nap\u00edsan\u00e9 na pol\u00ed\u010dku ka\u017edej farby a na ka\u017edej farbe musia by\u0165 postupne umiestnen\u00e9 v\u0161etky p\u00edsmen\u00e1. N\u00e1jdi jedno rie\u0161enie.<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n

Z7-I-6<\/b>
\nNa papieri je nap\u00edsan\u00fdch nieko\u013eko po sebe id\u00facich prirodzen\u00fdch \u010d\u00edsel. Je medzi nimi 12 tak\u00fdch, ktor\u00e9 s\u00fa n\u00e1sobkom piatich a 10 tak\u00fdch, ktor\u00e9 s\u00fa n\u00e1sobkom siedmich.<\/p>\n

   a)<\/b>  Ko\u013eko prirodzen\u00fdch \u010d\u00edsel je nap\u00edsan\u00fdch na papieri?
\n   b)<\/b>  N\u00e1jdite jednu postupnos\u0165 prirodzen\u00fdch \u010d\u00edsel, ktor\u00e1 odpoved\u00e1 vy\u0161\u0161ie op\u00edsan\u00fdm podmienkam.<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z7-II-1<\/b>
\nDan\u00fd je rovnobe\u017en\u00edk ABCD<\/i> (|AB<\/i>| ¹<\/font> |BC<\/i>|) s vn\u00fatorn\u00fdm uhlom ve\u013ekosti 72\u00b0 pri vrchole A<\/i>. Jedn\u00fdm vrcholom tohto rovnobe\u017en\u00edka vedieme dve priamky, ktor\u00e9 rozde\u013euj\u00fa rovnobe\u017en\u00edk na tri rovnoramenn\u00e9 trojuholn\u00edky. Ur\u010dte ve\u013ekosti vn\u00fatorn\u00fdch uhlov t\u00fdchto trojuholn\u00edkov.<\/p>\n

L. Hozov\u00e1<\/div>\n<\/p>\n

Z7-II-2<\/b>
\nKr\u00e1\u013e Leniv\u00e9ho kr\u00e1\u013eovstva vydal v nede\u013eu 1. apr\u00edla 2007 dekr\u00e9t, ktor\u00fdm vyradil zo v\u0161etk\u00fdch nasleduj\u00facich t\u00fd\u017ed\u0148ov piatky. Od tej doby v jeho kr\u00e1\u013eovstve nasleduje v\u017edy po \u0161tvrtku sobota a t\u00fd\u017ede\u0148 m\u00e1 iba \u0161es\u0165 dn\u00ed. Ktor\u00fd de\u0148 v t\u00fd\u017edni pripadne v Lenivom kr\u00e1\u013eovstve na 9. apr\u00edla 2008?
(Nezabudnite, \u017ee rok 2008 je priestupn\u00fd!)<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z7-II-3<\/b>
\nU Nov\u00e1kovcov napiekli svadobn\u00e9 kol\u00e1\u010de. \u0160tvrtinu odviezli pr\u00edbuzn\u00fdm na Moravu, \u0161estinu rozdali kolegom v pr\u00e1ci a dev\u00e4tinu dali susedom. Keby im zostali o tri kol\u00e1\u010de viacej, bola by to polovica p\u00f4vodn\u00e9ho po\u010dtu. Ko\u013eko kol\u00e1\u010dov napiekli?<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z8-I-1<\/b>
\nN\u00e1jdite v\u0161etky \u0161tvorcifern\u00e9 \u010d\u00edsla delite\u013en\u00e9 tromi, ktor\u00e9 po vyn\u00e1soben\u00ed \u010d\u00edslom 17 d\u00e1vaj\u00fa \u010d\u00edslo kon\u010diace troj\u010d\u00edsl\u00edm 519. <\/p>\n

(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z8-I-2<\/b>
\nN\u00e1jdite v\u0161etky trojice prirodzen\u00fdch \u010d\u00edsel men\u0161\u00edch ako 10, pre ktor\u00e9 plat\u00ed, \u017ee ich s\u00fa\u010din je sedemn\u00e1sobok ich s\u00fa\u010dtu.<\/p>\n

(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z8-I-3<\/b>
\nJano si k\u00fapil sedemm\u00ed\u013eov\u00e9 \u010di\u017emy. Jeho kamar\u00e1t Honza z \u010ciech si k\u00fapil lietaj\u00faci koberec. Potom sa obaja z\u00fa\u010dastnili na rozpr\u00e1vkov\u00fdch 12-hodinov\u00fdch pretekoch. Po\u010das pretekov boli hladn\u00ed, a tak sa obaja zastavili najes\u0165. Jedenie ka\u017ed\u00e9mu trvalo hodinu. Keby sa Honza nezastavil na „vep\u0159o-knedlo-zelo“, predbehol by Jana o 51 rozpr\u00e1vkov\u00fdch m\u00ed\u013e. Keby sa Jano nezastavil na bryndzov\u00e9 halu\u0161ky, predbehol by Honzu o  28 rozpr\u00e1vkov\u00fdch m\u00ed\u013e. Ako \u010faleko od seba by skon\u010dili, keby nejedol ani jeden z nich? Kto z nich by bol prv\u00fd?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z8-I-4<\/b>
\nV Tramt\u00e1rii maj\u00fa 5 lek\u00e1rskych fak\u00falt, z ktor\u00fdch ka\u017ed\u00e1 m\u00f4\u017ee do prv\u00e9ho ro\u010dn\u00edka prija\u0165 presne 200 \u0161tudentov. Prij\u00edmacie sk\u00fa\u0161ky na jednotliv\u00e9 fakulty sa konaj\u00fa v r\u00f4zne dni, preto si \u0161tudenti m\u00f4\u017eu poda\u0165 prihl\u00e1\u0161ku na viacero fak\u00falt. P\u00fdtali sme sa na fakult\u00e1ch, ko\u013eko im pri\u0161lo prihl\u00e1\u0161ok na \u0161kolsk\u00fd rok 2007\/2008. Z\u00edskali sme tieto odpovede:<\/p>\n

1. fakulta:<\/i> „Dostali sme p\u00e4\u0165kr\u00e1t viac prihl\u00e1\u0161ok, ako sme mali vo\u013en\u00fdch miest.“
\n2. fakulta:<\/i> „U n\u00e1s po\u010det uch\u00e1dza\u010dov prev\u00fd\u0161il kapacitu o 320%.“
\n3. fakulta:<\/i> „Na na\u0161u fakultu sa hl\u00e1silo o 520 uch\u00e1dza\u010dov viac, ako sme mali miest.“
\n4. fakulta:<\/i> „U n\u00e1s na ka\u017ed\u00e9 vo\u013en\u00e9 miesto pripadli v priemere 3 prihl\u00e1\u0161ky.“
\n5. fakulta:<\/i> „K n\u00e1m sa hl\u00e1silo o tri \u0161tvrtiny z\u00e1ujemcov viac, ako sme mali miest.“<\/p>\n

V akademickom roku 2007\/2008 nakoniec \u0161t\u00fadium za\u010dalo 1000 medikov. Zo \u0161tatistiky vypl\u00fdva, \u017ee z\u00e1ujemca o \u0161t\u00fadium medic\u00edny podal na lek\u00e1rske fakulty v priemere 2,5 prihl\u00e1\u0161ky. Ko\u013eko z\u00e1ujemcov sa nedostalo na \u017eiadnu z lek\u00e1rskych fak\u00falt Tramt\u00e1rie?<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z8-I-5<\/b>
\nP\u00e1n Poleno a p\u00e1n \u010criepok vyr\u00e1bali vchodov\u00e9 dvere tvaru obd\u013a\u017enika s obsahom 3 m2<\/sup>. R\u00e1m, uhloprie\u010dky a dve \u010fal\u0161ie prie\u010dky, ktor\u00e9 sp\u00e1jali vrcholy obd\u013a\u017enika so stredmi proti\u013eahl\u00fdch str\u00e1n boli z kovov\u00fdch ty\u010d\u00ed (pozri obr\u00e1zok). P\u00e1n Poleno vyplnil drevom \u0161tyri tmav\u00e9 \u010dasti dver\u00ed a p\u00e1n \u010criepok zost\u00e1vaj\u00face \u010dasti dver\u00ed zasklil. Ko\u013eko metrov \u0161tvorcov\u00fdch dreva potreboval p\u00e1n Poleno na v\u00fdpl\u0148 dver\u00ed?
(Hr\u00fabku kovov\u00fdch ty\u010d\u00ed zanedbajte.)<\/p>\n

<\/div>\n
(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z8-I-6<\/b>
\nUprostred n\u00e1mestia v Koc\u00farkove je \u0161tvorcov\u00fd tr\u00e1vnat\u00fd z\u00e1hon. Ke\u010f Koc\u00farkov\u010dania zistili, \u017ee zabudli urobi\u0165 chodn\u00edk, tak z ka\u017ed\u00e9ho kraja z\u00e1honu na\u0148 ubrali 2 metre. Pred polo\u017een\u00edm z\u00e1mkovej dla\u017eby (a \u0161trku pod \u0148u) bolo treba pod cel\u00fa plochu chodn\u00edka urobi\u0165 0,5 m hlbok\u00fd v\u00fdkop. Odkopan\u00edm tr\u00e1vy a hliny sa z\u00e1hon zmen\u0161il o 1200 m2<\/sup>.<\/p>\n

   a)<\/b>  Ak\u00fd obsah m\u00e1 teraz tr\u00e1vnat\u00fd z\u00e1hon?
\n   b)<\/b>  Ko\u013eko m3<\/sup> \u0161trku je pod dla\u017ebou, ak je povrch dla\u017eby zarovno s tr\u00e1vnat\u00fdm z\u00e1honom a v\u00fd\u0161ka dla\u017edice je 8 cm?<\/p>\n

(M. Smitkov\u00e1, M. Dillingerov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z8-II-1<\/b>
\nDru\u017estvo chce v sez\u00f3ne vyhra\u0165 v\u0161etk\u00fdch svojich z\u00e1pasov. V prvej tretine z nich v\u0161ak vyhralo iba 55% z\u00e1pasov.<\/p>\n

  a)<\/b>  Ko\u013eko percent zvy\u0161n\u00fdch z\u00e1pasov by muselo dru\u017estvo vyhra\u0165, aby dosiahlo zam\u00fd\u0161\u013ean\u00fd cie\u013e?
\n  b)<\/b>  Keby dru\u017estvo vyhralo v\u0161etky zvy\u0161n\u00e9 z\u00e1pasy, ko\u013eko percent svojich z\u00e1pasov by v celej sez\u00f3ne vyhralo?<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n

Z8-II-2<\/b>
\nAk\u00fa \u010das\u0165 obsahu nerovnoramenn\u00e9ho lichobe\u017en\u00edka KLMN<\/i> (KL<\/i> ║ MN<\/i>) tvor\u00ed obsah trojuholn\u00edka ABC<\/i>, kde A<\/i> je stred z\u00e1kladne KL<\/i>, B<\/i> je stred z\u00e1kladne MN<\/i> a C<\/i> je stred ramena KN<\/i>?<\/p>\n

(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z8-II-3<\/b>
\nAby prirodzen\u00e9 \u010d\u00edslo prin\u00e1\u0161alo Liborovi \u0161\u0165astie, mus\u00ed by\u0165 jeho druh\u00e1 mocnina delite\u013en\u00e1 \u010d\u00edslami sedem, osem, dev\u00e4\u0165 a desa\u0165. N\u00e1jdite v\u0161etky prirodzen\u00e9 \u010d\u00edsla men\u0161ie ako 1000, ktor\u00e9 Liborovi nosia \u0161\u0165astie.<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z9-I-1<\/b>
\nN\u00e1jdite v\u0161etky \u0161tvorcifern\u00e9 \u010d\u00edsla kon\u010diace \u010d\u00edslicou 9, ktor\u00e9 s\u00fa delite\u013en\u00e9 ka\u017edou svojou \u010d\u00edslicou.<\/p>\n

(P. Tlust\u00fd)<\/div>\n<\/p>\n

Z9-I-2<\/b>
\nPeter sa p\u00fdtal babi\u010dky, ko\u013eko rokov m\u00e1 dedko. Babi\u010dka mu takto odpovedala: „To vie\u0161, u\u017e d\u00e1vno nem\u00e1me p\u00e4\u0165desiat, ale zato e\u0161te nem\u00e1me osemdesiat rokov. Ak vyn\u00e1sob\u00ed\u0161 s\u00fa\u010det m\u00f4jho a dedkovho veku ich rozdielom a k v\u00fdsledku pripo\u010d\u00edta\u0161 oba na\u0161e veky, dostane\u0161 492.“
\n„Aha,“ povedal po chv\u00edli Peter, „tak to m\u00e1 dedko …“
\nKo\u013eko rokov m\u00e1 Petrov dedko, ak viete, \u017ee je star\u0161\u00ed ako Petrova babi\u010dka?<\/p>\n

(M. Raabov\u00e1)<\/div>\n<\/p>\n

Z9-I-3<\/b>
\nStredom rota\u010dn\u00e9ho valca s v\u00fd\u0161kou v<\/i> a podstavou, ktorej polomer je r<\/i>, bol vyv\u0155tan\u00fd valcov\u00fd otvor. Objem takto vytvoren\u00e9ho „dut\u00e9ho valca“ je polovi\u010dn\u00fd ako objem p\u00f4vodn\u00e9ho valca. Vyjadrite hr\u00fabku steny dut\u00e9ho valca pomocou polomeru r<\/i>.<\/p>\n

<\/div>\n
(M. Krej\u010dov\u00e1)<\/div>\n<\/p>\n

Z9-I-4<\/b>
\nMinul\u00fa divadeln\u00fa sez\u00f3nu sa pred\u00e1vali vstupenky za jednotn\u00fa cenu 160 Sk. Pre tohtoro\u010dn\u00fa sez\u00f3nu sa sedadl\u00e1 v h\u013eadisku rozdelili do dvoch kateg\u00f3ri\u00ed. Miesta I. kateg\u00f3rie stoja 180 Sk a miesta II. kateg\u00f3rie 155 Sk. Ak sa v\u0161etky sedadl\u00e1 v h\u013eadisku vypredaj\u00fa, bude celkov\u00e1 tr\u017eba rovnak\u00e1 ako minul\u00fa sez\u00f3nu pri vypredanom h\u013eadisku. Riadite\u013e divadla st\u00e1le nie je spokojn\u00fd a pre bud\u00facu sez\u00f3nu pl\u00e1nuje zmenu: z najhor\u0161\u00edch miest s\u00fa\u010dasnej II. kateg\u00f3rie urob\u00ed III. kateg\u00f3riu. Aby sa v\u0161ak tr\u017eba za vypredan\u00e9 h\u013eadisko nezmenila, tak rozhodol, \u017ee vstupenky bud\u00fa st\u00e1\u0165 180 Sk (I. kateg\u00f3ria), 160 Sk (II. kateg\u00f3ria) a 130 Sk (III. kateg\u00f3ria). V akom pomere bud\u00fa v bud\u00facej sez\u00f3ne po\u010dty sedadiel jednotliv\u00fdch kateg\u00f3ri\u00ed?<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z9-I-5<\/b>
\nJurko k\u00fapil dve \u010dokol\u00e1dy v obchode oproti \u0161kole. Mi\u0161ko si k\u00fapil tak\u00e9 ist\u00e9 dve \u010dokol\u00e1dy v obchode za \u0161kolou a Ivan si k\u00fapil tie\u017e tak\u00fa \u010dokol\u00e1du, ale v \u0161kolskom bufete. Tak\u00fdmto sp\u00f4sobom boli v\u0161etky tri n\u00e1kupy spolu o 6 Sk drah\u0161ie, ako keby chlapci nakupovali v\u0161etk\u00fdch 5 \u010dokol\u00e1d v obchode oproti \u0161kole a o 6,50 Sk lacnej\u0161ie, ako keby nak\u00fapili iba v obchode za \u0161kolou. V \u0161kolskom bufete pred\u00e1vaj\u00fa jednu \u010dokol\u00e1du za 19,50 Sk. Ko\u013eko st\u00e1li v\u0161etky \u010dokol\u00e1dy spolu? Ko\u013eko stoj\u00ed \u010dokol\u00e1da v obchode za \u0161kolou?<\/p>\n

(M. Dillingerov\u00e1)<\/div>\n<\/p>\n

Z9-I-6<\/b>
\nV rovine je dan\u00fd \u0161tvoruholn\u00edk ABCD<\/i>. Zostrojte bod K<\/i>, ktor\u00fd je vrcholom rovnobe\u017en\u00edka BCDK<\/i>, a bod L<\/i>, ktor\u00fd je vrcholom rovnobe\u017en\u00edka CDAL<\/i>. Uk\u00e1\u017ete, \u017ee priamka KL<\/i> prech\u00e1dza stredom strany AB<\/i> dan\u00e9ho \u0161tvoruholn\u00edka ABCD<\/i>.<\/p>\n

(J. \u0160vr\u010dek)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z9-II-1<\/b>
\nBarbora si nap\u00edsala dve r\u00f4zne cel\u00e9 \u010d\u00edsla. Potom ich s\u010d\u00edtala, od\u010d\u00edtala, vyn\u00e1sobila a vydelila. Dostala \u0161tyri v\u00fdsledky, ktor\u00fdch s\u00fa\u010det bol -100<\/b>. Ke\u010f vynechala v\u00fdsledok s\u010d\u00edtania a spo\u010d\u00edtala ostatn\u00e9 tri v\u00fdsledky, dostala tie\u017e s\u00fa\u010det -100<\/b>. Ktor\u00e9 cel\u00e9 \u010d\u00edsla mohla Barbora p\u00f4vodne nap\u00edsa\u0165?<\/p>\n

(\u0160. \u010cern\u00ed\u010dkov\u00e1)<\/div>\n<\/p>\n

Z9-II-2<\/b>
\nZ kociek o hrane 1 cm sme postavili kv\u00e1der. Keby sme z kv\u00e1dra odobrali jeden st\u013apec (t.j. zvisl\u00fd kom\u00ednik<\/i>), zvy\u0161ok stavby by sa skladal zo 602 kociek. Keby sme miesto toho odobrali jeden riadok hornej vrstvy, zostala by n\u00e1m stavba zo 605 kociek. Ak\u00e9 rozmery m\u00e1 kv\u00e1der?<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z9-II-3<\/b>
\nDan\u00fd je \u0161tvorec ABCD<\/i> so stranou d\u013a\u017eky a<\/i> a \u00fase\u010dka KL<\/i> d\u013a\u017eky 5a<\/i> tak, \u017ee A<\/i> ≡ K<\/i> a strana AB<\/i> le\u017e\u00ed na \u00fase\u010dke KL<\/i>. \u0160tvorec ABCD<\/i> oto\u010d\u00edme okolo prav\u00e9ho doln\u00e9ho bodu o prav\u00fd uhol a tak\u00e9to ot\u00e1\u010dania rob\u00edme dovtedy, k\u00fdm cel\u00e1 strana AB<\/i> po prv\u00fd raz op\u00e4\u0165 nesplynie s \u010das\u0165ou \u00fase\u010dky KL<\/i> a bod B<\/i> s bodom L<\/i>.<\/p>\n

  a)<\/b>  Narysujte po akej dr\u00e1he sa bude pohybova\u0165 stred S<\/i> strany AB<\/i>.
\n  b)<\/b>  Vypo\u010d\u00edtajte d\u013a\u017eku krivky, po ktorej sa bod S<\/i> pohyboval.<\/p>\n

(M. Volfov\u00e1)<\/div>\n<\/p>\n

Z9-II-4<\/b>
\nSeversk\u00fdch pretekov ps\u00edch z\u00e1prahov sa z\u00fa\u010dastnilo spolu 315 dvojz\u00e1prahov a trojz\u00e1prahov. Do cie\u013ea dorazilo v stanovenom limite 60% v\u0161etk\u00fdch dvojz\u00e1prahov a jedna tretina v\u0161etk\u00fdch trojz\u00e1prahov, \u010do predstavovalo polovicu v\u0161etk\u00fdch psov na \u0161tarte. Ko\u013eko dvojz\u00e1prahov a ko\u013eko trojz\u00e1prahov \u0161tartovalo?<\/p>\n

(L. \u010cern\u00ed\u010dek)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		<\/a>▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n

Z9-III-1<\/b>
\nV Dlhej Lehote volili starostu. Kandidovali dvaja ob\u010dania: Ing. Schopn\u00fd a jeho man\u017eelka Dr. Schopn\u00e1. V obci boli tri volebn\u00e9 miestnosti. V prvej i druhej miestnosti dostala viac hlasov Dr. Schopn\u00e1. V prvej bol pomer hlasov 7:5, v druhej 5:3. V tretej volebnej miestnosti bol pomer hlasov 3:7 v prospech Ing. Schopn\u00e9ho. Vo\u013eby nakoniec skon\u010dili nerozhodne, obaja kandid\u00e1ti toti\u017e z\u00edskali rovnak\u00fd po\u010det hlasov. V akom pomere boli po\u010dty odovzdan\u00fdch platn\u00fdch hlasovac\u00edch l\u00edstkov v jednotliv\u00fdch volebn\u00fdch miestnostiach, ak vieme, \u017ee v prvej a druhej miestnosti odovzdal platn\u00fd hlas rovnak\u00fd po\u010det ob\u010danov?<\/p>\n

(L. \u0160im\u016fnek)<\/div>\n<\/p>\n

Z9-III-2<\/b>
\nJe dan\u00fd rovnoramenn\u00fd lichobe\u017en\u00edk ABCD<\/i> (AB<\/i>║CD<\/i>), kde |AB<\/i>| > |CD<\/i>|. Bodom A<\/i> sa daj\u00fa vies\u0165 dve priamky tak, aby rozdelili lichobe\u017en\u00edk na tri rovnoramenn\u00e9 trojuholn\u00edky. Ur\u010dite ve\u013ekosti uhlov lichobe\u017en\u00edka ABCD<\/i>.<\/p>\n

(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z9-III-3<\/b>
\nN\u00e1jdite v\u0161etky prirodzen\u00e9 \u010d\u00edsla x, y<\/i>, pre ktor\u00e9 plat\u00ed:<\/p>\n

1 + x + y + xy = 2008.<\/p>\n

(L. Hozov\u00e1)<\/div>\n<\/p>\n

Z9-III-4<\/b>
\nDan\u00fd je \u0161tvorec ABCD<\/i> so stranou d\u013a\u017eky a<\/i> a obd\u013a\u017enik KLMN<\/i> so stranami d\u013a\u017eok |KL<\/i>| = 3a<\/i> a |LM<\/i>| = a<\/i>. Na za\u010diatku je \u0161tvorec umiestnen\u00fd tak, \u017ee A<\/i> ≡ N<\/i> a strana AB<\/i> le\u017e\u00ed na strane NM<\/i>. \u0160tvorec ABCD<\/i> oto\u010d\u00edme okolo bodu B<\/i> tak, \u017ee strana BC<\/i> bude le\u017ea\u0165 na strane MN<\/i>. \u0160tvorec sa ot\u00e1\u010dan\u00edm okolo svojich vrcholov (pozri obr\u00e1zok) pohybuje po obvode obd\u013a\u017enika KLMN<\/i> tak dlho, a\u017e sa op\u00e4\u0165 dostane do svojej p\u00f4vodnej polohy.<\/p>\n

  a)<\/b>   Narysujte dr\u00e1hu, po ktorej sa bude pohybova\u0165 bod T<\/i>, ktor\u00fd je stredom strany AB<\/i>.
\n  b)<\/b>   Ur\u010dte obsah plochy ohrani\u010denej krivkou, ktor\u00fa opisuje bod T<\/i>.<\/p>\n

<\/div>\n
(M. Dillingerov\u00e1)<\/div>\n<\/p>\n\n\n
\n
\n<\/td>\n		▲ hore ▲<\/a><\/td>\n<\/tr>\n<\/table>\n\n
<\/div>","protected":false},"excerpt":{"rendered":"

57. ro\u010dn\u00edk matematickej olympi\u00e1dy 2007-2008<\/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":[6],"tags":[],"class_list":["post-140","post","type-post","status-publish","format-standard","hentry","category-57-rocnik"],"yoast_head":"\nKateg\u00f3rie Z4-Z9 - Matematick\u00e1 olympi\u00e1da<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/matematika.besaba.com\/?p=140\" \/>\n<meta property=\"og:locale\" content=\"sk_SK\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Kateg\u00f3rie Z4-Z9 - 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