Problema 1
Scrieti numarul \( 25^{2008} \) ca suma a cinci numere naturale consecutive. (Traian Tamaian)
Problema 2
Aratati ca pentru orice numar natural \( n>5 \), multimea \( A=\{n, 2n+1, 3n+2, 4n+3, 6n+1\} \) contine cel putin un mumar compus. (Vasile Berinde)
Problema 3
Se considera un triunghi echililateral ABC cu lungimea laturii de 10 cm. Pe (AB), (BC), (CA) se iau punctele D, E, M astfel incat: AD=BE=CM=3 cm.
Daca \( AE\cap BM=\{Q\}, BM\cap CD=\{R\}, CD\cap AE=\{P\}, \) sa se arate ca triunghiul PQR este echilateral.
Problema 4
In triunghiul ABC, BC este cea mai mare latura si are lungimea a, iar AB si AC au in mod respectiv lungimile b si c.
Bisectoarea unghiului B intersecteaza pe AC in D, iar bisectoarea unghiului C intersecteaza pe AB in E.
Notam cu M piciorul perpendicularei duse din varful A pe BD, iar cu N piciorul perpendicularei duse din A pe CE.
Demonstrati ca: \( MN=\frac{b+c-a}{2} \). (Maria Mihet)
Concursul "Grigore Moisil", Satu-Mare, 3-5 aprilie
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mihai miculita
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Problema 1 \( 25^2008=(x-2)+(x-1)+x+(x+1)+(x+2)=5x \Rightarrow x=\frac{(5^2)^{2008}}{5}=\frac{5^{4016}}{5}=5^{4015} \Rightarrow \) \( 25^{2008}=(5^{4015}-2)+(5^{4015}-1)+5^{4015}+(5^{4015}+1)+(5^{4015}+2) \).
Porblema 2 Avem 5 cazuri:
1)\( n=5k \Rightarrow \)n numar compus (n>5).
2)\( n=5k+1 \Rightarrow 3n+2=3(5k+1)+2=15k+3+2=15k+5=5(3k+1) \), numar compus.
3)\( n=5k+2 \Rightarrow 2n+1=2(5k+2)+1=10k+4+1=10k+5=5(2k+1) \), numar compus.
4) \( n=5k+3 \Rightarrow 4n+3=4(5k+3)+3=20k+12+3=20k+15=5(4k+3) \), numar compus.
5) \( n=5k+4 \Rightarrow 6n+1=6(5k+4)+1=30k+24+1=30k+25=5(6k+5) \), numar compus.
Porblema 2 Avem 5 cazuri:
1)\( n=5k \Rightarrow \)n numar compus (n>5).
2)\( n=5k+1 \Rightarrow 3n+2=3(5k+1)+2=15k+3+2=15k+5=5(3k+1) \), numar compus.
3)\( n=5k+2 \Rightarrow 2n+1=2(5k+2)+1=10k+4+1=10k+5=5(2k+1) \), numar compus.
4) \( n=5k+3 \Rightarrow 4n+3=4(5k+3)+3=20k+12+3=20k+15=5(4k+3) \), numar compus.
5) \( n=5k+4 \Rightarrow 6n+1=6(5k+4)+1=30k+24+1=30k+25=5(6k+5) \), numar compus.
Problema 3
\( \triangle ABE\equiv \triangle CAD\Longrightarrow \angle BAE\equiv \angle ACD,\angle AEB\equiv CDA \)
\( \triangle ADC\equiv \triangle CMB\Longrightarrow \angle ADC\equiv \angle CMB, \angle ACD\equiv \angle MBC \)
\( \triangle ADP\equiv \triangle CMR (ULU)\Longrightarrow \angle APD\equiv \angle MRC(1) \)
\( \triangle BEQ\equiv \angle CMR (ULU)\Longrightarrow \angle BQE\equiv \angle MRC(2) \)
\( \angle ADP\equiv \angle QPR(3) \)
\( \angle CRM\equiv \angle PRQ(4) \)
\( \angle BQE\equiv \angle PQR(5) \)
- din \( (1),(2),(3),(4),(5)\Longrightarrow \angle QPR\equiv \angle PRQ\equiv \angle PQR\Longrightarrow \triangle PRQ \)echilateral
\( \triangle ABE\equiv \triangle CAD\Longrightarrow \angle BAE\equiv \angle ACD,\angle AEB\equiv CDA \)
\( \triangle ADC\equiv \triangle CMB\Longrightarrow \angle ADC\equiv \angle CMB, \angle ACD\equiv \angle MBC \)
\( \triangle ADP\equiv \triangle CMR (ULU)\Longrightarrow \angle APD\equiv \angle MRC(1) \)
\( \triangle BEQ\equiv \angle CMR (ULU)\Longrightarrow \angle BQE\equiv \angle MRC(2) \)
\( \angle ADP\equiv \angle QPR(3) \)
\( \angle CRM\equiv \angle PRQ(4) \)
\( \angle BQE\equiv \angle PQR(5) \)
- din \( (1),(2),(3),(4),(5)\Longrightarrow \angle QPR\equiv \angle PRQ\equiv \angle PQR\Longrightarrow \triangle PRQ \)echilateral