MathDB

1

Part of 2005 Moldova Team Selection Test

Problems(2)

alpha+beta+gamma=1

Source: Moldova TST 2005

4/9/2005
In triangle ABCABC, M∈(BC)M\in(BC), BMBC=α\frac{BM}{BC}=\alpha, N∈(CA)N\in(CA), CNCA=β\frac{CN}{CA}=\beta, P∈(AB)P\in(AB), APAB=γ\frac{AP}{AB}=\gamma. Let AM∩BN={D}AM\cap BN=\{D\}, BN∩CP={E}BN\cap CP=\{E\}, CP∩AM={F}CP\cap AM=\{F\}. Prove that SDEF=SBMD+SCNE+SAPFS_{DEF}=S_{BMD}+S_{CNE}+S_{APF} iff α+β+γ=1\alpha+\beta+\gamma=1.
analytic geometrygeometry solvedgeometry
2 triangles

Source: Moldova TST 2005

4/10/2005
Let ABCABC and A1B1C1A_{1}B_{1}C_{1} be two triangles. Prove that aa1+bb1+cc1≤3R2r1\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}\leq\frac{3R}{2r_{1}}, where a=BCa = BC, b=CAb = CA, c=ABc = AB are the sidelengths of triangle ABCABC, where a1=B1C1a_{1}=B_{1}C_{1}, b1=C1A1b_{1}=C_{1}A_{1}, c1=A1B1c_{1}=A_{1}B_{1} are the sidelengths of triangle A1B1C1A_{1}B_{1}C_{1}, where RR is the circumradius of triangle ABCABC and r1r_{1} is the inradius of triangle A1B1C1A_{1}B_{1}C_{1}.
geometrycircumcircleinradiusinequalitiestrigonometrytrig identitiesLaw of Sines