MathTime

Cerinta problemei e echivalenta cu a arata AF^2+EC^2=FC^2+AE^2 . Pentru usurinta voi nota \widehat{CAD}=x , \widehat{BAC}=y , \widehat{ADC}=d , \widehat{ABC}=b . Folosindu-ne de teorema cosinusului in triunghiurile EAC si FAC , ne ramane de aratat ca AE \cos{\dfrac{y}{2}}=AF cos{\dfrac{x}{2}} (*) Cu...

Pp ca \triangle{DEC} nu e isoscel. Fie M pe AC a.i. \triangle {DCM} sa fie isoscel \Rightarrow \widehat{DMA}=20 Aplicand teorema sinusurilor in ADM: \dfrac{AM}{DM}=\dfrac{AM}{MC}=\dfrac{sin30}{sin10} . Dar aplicand teorema sinusurilor in ABC: \dfrac{AB}{BC}=\dfrac{sin80}{sin20} \Rightarrow \dfrac{AM...

Cerinta este echivalenta cu a arata ca $S_{NAB}$$=S_{MAD}$. Dar $S_{NAB}=S_{DAB}=\dfrac{S_{ABCD}}{2}$, iar $S_{MAD}=S_{CAD}=\dfrac{S_{ABCD}}{2}$. q.e.d.

$DOCA-inscriptibil$$\Rightarrow$$\widehat{DAO}=\widehat{DCO}$.

$\triangle{BOA}-isoscel$$\Rightarrow$ $\widehat{DBO}=\widehat{DCO}$$\Rightarrow$$\widehat{DBC}=\widehat{DCB}$$\Rightarrow$$\triangle{DBC}-isoscel$ q.e.d.

Din ipoteza \Rightarrow coordonatele baricentrice ale punctului P_({\dfrac{1}{7},\dfrac{4}{7},\dfrac{2}{7}) Fie M mijlocul lui BC si raportam planul la un reper cartezian aflat in M. Fara a pierde generalitatea, consideram lungimea laturii BC de 2 si astfel coordonatele carteziene ale varfurilor tri...

In \triangle{A_1B_1C_1} din teorema sinusurilor obtinem: A_1B_1=2R\sin{\dfrac{A+B}{2}}=2R\cos{\dfrac{C}{2}} si analog A_1C_1=2R\sin{\dfrac{A+C}{2}}=2R\cos{\dfrac{B}{2}} si observam ca \widehat{B_1A_1C_1}=\dfrac{B+C}{2} . S_{A_1B_1C_1}=2R^2\cos{\dfrac{A}{2}} \cos{\dfrac{B}{2}} \cos{\dfrac{C}{2}} , ia...

O solutie trigonometrica: Aplicand teorema sinusurilor in \triangle{DMB} , obtinem: DM=\dfrac{\sin{30} MB}{\sin{70}} Aplicand din nou teorema sinusurilor in \triangle{AMB} , obtinem: AM=\dfrac{\sin{40} MB}{\sin{40}} \Rightarrow \dfrac{DM}{MA}=\dfrac {\sin{30}}{\sin{70}} Din teorema sinusurilor aplic...

Egalezi fractia cu un $a\in\Bbb{Z}$ si il scoti pe $x$in functie de $a$.

Cu teorema sinusurilor in triunghiul \triangle{A_{1}B_{1}C_{1}} , obtinem (relatii cunoscute): A_{1}B_{1}=2R\cos{\dfrac{C}{2}} , B_{1}C_{1}=2R\cos{\dfrac{A}{2}} , C_{1}A_{1}=2R\cos{\dfrac{B}{2}} . Aplicand teorema sinusurilor in \triangle{AA_{1}B si analoagele obtinem: AA_{1}=2R\sin\dfrac{2B+A}{2} ,...

Cu ocazia zilei de nastere, domnule profesor Mihai Miculita, va urez la multi ani si multa sanatate!