MathDB

1

Part of 2011 Turkey Team Selection Test

Problems(2)

Incenters and an equation

Source: Turkish TST 2011 Problem 4

7/23/2011
Let DD be a point different from the vertices on the side BCBC of a triangle ABC.ABC. Let I, I1I, \: I_1 and I2I_2 be the incenters of the triangles ABC, ABDABC, \: ABD and ADC,ADC, respectively. Let EE be the second intersection point of the circumcircles of the triangles AI1IAI_1I and ADI2,ADI_2, and FF be the second intersection point of the circumcircles of the triangles AII2AII_2 and AI1D.AI_1D. Prove that if AI1=AI2,AI_1=AI_2, then EIFI⋅EDFD=EI12FI12. \frac{EI}{FI} \cdot \frac{ED}{FD}=\frac{{EI_1}^2}{{FI_1}^2}.
geometryincentercircumcircleratiogeometry proposed
Geometric inequality

Source: Turkish TST 2011 Problem 7

7/23/2011
Let KK be a point in the interior of an acute triangle ABCABC and ARBPCQARBPCQ be a convex hexagon whose vertices lie on the circumcircle Γ\Gamma of the triangle ABC.ABC. Let A1A_1 be the second point where the circle passing through KK and tangent to Γ\Gamma at AA intersects the line AP.AP. The points B1B_1 and C1C_1 are defined similarly. Prove that min⁡{PA1AA1, QB1BB1, RC1CC1}≤1. \min\left\{\frac{PA_1}{AA_1}, \: \frac{QB_1}{BB_1}, \: \frac{RC_1}{CC_1}\right\} \leq 1.
inequalitiesgeometrycircumcircleratiogeometric transformationhomothetyinequalities proposed