ASUPRA UNOR PROBLEME PROPUSE LA CONCURSURI INTERJUDEŢENE
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1 DIDTI MTHEMTI, Vol. 33(15), pp SUPR UNOR PROLEME PROPUSE L ONURSURI INTERJUDEŢENE Dniel Văăreţu strt. This pper will present the solutions of two prolems whih were proposed to Grigore Moisil nd Mrin Ţrină inter-ounties Romnin mthemtil ompetitions. In Romni, these ompetitions re onsidered mongst the most diffiult ones, eing good opportunity for students to trin themselves nd to self-ssess their potentil in view of their prtiiption in the finl stge of the Ntionl Mthemtis Olympid. s n evidene for the lst sttement stys the ft tht the prolems proposed in these ompetitions re nnully pulished in the Mthemtis Review/ Gzet Mtemtiă. We lso hve to mention the ft tht the students results in these ompetitions re speil riteri for their dmission to the Fulty of Mthemtis nd Informtis of eş-olyi University from luj-npo. MS Key words. Mthemtil ompetitions, geometry of tringle 1 Fie un triunghi suţitunghi orere şi O entrul erului său irumsris. Se onsideră dreptele, respetiv perpendiulre în O pe dreptele O, O respetiv O, unde,,, şi,. Notăm u H, H respetiv H ortoentrele triunghiurilor, respetiv. Să se demonstreze ă segmentele H, H şi H pot fi lturile unui triunghi. Dniel Văăreţu onursul Interjudeţen de Mtemtiă şi Informtiă Grigore Moisil ediţi XXVI-, 11, Târgu Mureş Soluţi 1. m(ô) = m(ô) = 9 m( ) m( ) = m( ) nlog m( ) = m(ĉ). Dei este ntiprlelă l. nlog este ntiprlelă l şi este ntiprlelă l. Ne intereseză de fpt ii ă triunghiurile, şi sunt semene u triunghiul (u eleşi măsuri de unghiuri). vem: tg O + tg O + tg O OH = tg + tg + tg OH = tg O + tg O + tg O tg + tg + tg
2 114 D. Văăreţu OH = tg O + tg O + tg O. tg + tg + tg H O H H dunăm memru u memru şi oţinem: O + OH + OH = tg O + tg O + tg O tg + tg + tg tg ( O + O ) + tg ( O + O ) + tg ( O + O ) + tg + tg + tg O + OH + OH = OH + tg O + tg O + tg O tg + tg + tg unde m nott u H ortoentrul triunghiului şi u,, mijloele segmentelor [ ], [ ] şi [ ] re sunt de fpt şi mijloele lturilor,,.
3 3 supr unor proleme propuse l onursuri interjudeţene 115 Dr triunghiul este triunghiul medin, dei re unghiurile de măsuri m(â), m( ), m(ĉ). Dei tg O + tg O + tg O = tg + tg + tg pentru ă este vetorul OH, unde H este ortoentrul triunghiului re oinide u O. m oţinut dei: OH = OH + OH + OH din relţi lui Sylvester O + O + O = OH + OH + OH H + H + H = diă segmentele H, H şi H pot form un triunghi. Soluţi. H = R os, unde R este rz erului irumsris triunghiului re este semene u triunghiul (invers semene diă este ntiprlelă l ). Din semănre lor vem: R R = R h unde R = O este rz erului irumsris triunghiului şi înălţime orespunzătore lui în triunghiul, ir h este înălţime din triunghiului şi nlog: R = R H = R os h h H = R h os, H = R h os. H + H > H R os + R h h os os > R h + os > os. h h h Dr h = sin, h = sin, h = sin. Rezultă: os os sin + os sin > os sin os R sin sin + os R sin sin > os R sin sin sin os + sin os > sin os
4 116 D. Văăreţu 4 sin + sin > sin sin( + ) os( ) > sin os os( ) > os os( ) os > sin + sin > ( π ) ( sin sin π ) < os os > devărt. nlog se demonstreză H + H > H şi H + H > H. Fie O şi I entrul erului irumsris respetiv entrul erului însris triunghiului orere. Se onsideră dreptele, respetiv perpendiulre în O pe dreptele O, O respetiv O, unde,,, şi,. Fie H I şi O I ortoentrul respetiv entrul erului irumsris triunghiului I I I, unde I, I şi I sunt entrele erurilor însrise în triunghiurile, şi şi I entrul erului însris în triunghiul, unde, şi sunt mijloele lturilor, şi. Să se demonstreze ă: IH I = O I I. Dniel Văăreţu onursul Interjudeţen de Mtemtiă şi Informtiă Mrin Ţrină, ediţi XII-, 1, Turd Soluţie.
5 5 supr unor proleme propuse l onursuri interjudeţene 117 O Vetorul de poziţie l entrului erului însris în triunghiul este: r I = r + r + r = sin r + sin r + sin r. + + sin + sin + sin vem: m(â) = m(â) = m(â), m( ) = m( ) = m( ), m(ĉ) = m(ĉ) = m(ĉ), dei în zul triunghiurilor, şi vem: sin O I + sin O I + sin O I O I I = sin + sin + sin O I I = O I I = sin O I + sin O I + sin O I sin + sin + sin sin O I + sin O I + sin O I sin + sin + sin
6 118 D. Văăreţu 6 (m lut origine vetorilor puntul O I.) dunăm memru u memru şi vem din relţi lui Sylvester: = sin ( O I + O I I + = O I I + O I H I = O I I + O I I + O I I O I ) + sin ( O I + O I ) + sin ( O I + O I ) sin + sin + sin sin O I + sin O I + sin O I = O I I + O I I sin + sin + sin (pentru ă m(â ) = m(â), m( ) = m( ), m(ĉ ) = m(ĉ) şi,, sunt mijloele segmentelor, şi O I H I O I I = O I I IH I = O I I. ILIOGRFIE [1] onursul de Mtemtiă şi Informtiă Grigore Moisil, Ediţi XXVI-, Tg. Mureş, 1-3, prilie 11, prezentre de Eugeni Du şi Dorel I. Du, Gzet Mtemtiă seri, nr.5/11, pg [] onursul interjudeţen de Mtemtiă şi Informtiă Mrin Ţrină, Ediţi XII, Turd, 11-1 mi 1, prezentre de Dorel I. Du şi Gheorghe Loonţ, Gzet Mtemtiă seri, nr. 11/1, pg Fulty of Mthemtis nd omputer Siene eş-olyi University Str. Kogălnienu, no luj-npo, Romni e-mil: dvretu@yhoo.om Primit l redţie: Noiemrie 15
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