MathDB

3

Part of 2014 Moldova Team Selection Test

Problems(3)

Moldova TST triangle geometry

Source: Moldova TST 2014, First Day, Problem 3

3/4/2014
Let △ABC\triangle ABC be an acute triangle and ADAD the bisector of the angle ∠BAC\angle BAC with D∈(BC)D\in(BC). Let EE and FF denote feet of perpendiculars from DD to ABAB and ACAC respectively. If BF∩CE=KBF\cap CE=K and ⊙AKE∩BF=L\odot AKE\cap BF=L prove that DL⊥BFDL\perp BF.
geometrytrigonometrypower of a pointradical axisgeometry proposed
Cyclic quadrilateral bisectors

Source: Moldova TST 2014, Second Day, Problem 3

3/30/2014
Let ABCDABCD be a cyclic quadrilateral. The bisectors of angles BADBAD and BCDBCD intersect in point KK such that K∈BDK \in BD. Let MM be the midpoint of BDBD. A line passing through point CC and parallel to ADAD intersects AMAM in point PP. Prove that triangle △DPC\triangle DPC is isosceles.
geometrycircumcircletrapezoidgeometry proposed
Geometry

Source: Moldova TST 2014, Third Day, Problem 3

3/31/2014
Let △ABC\triangle ABC be a triangle with ∠A\angle A-acute. Let PP be a point inside △ABC\triangle ABC such that ∠BAP=∠ACP\angle BAP = \angle ACP and ∠CAP=∠ABP\angle CAP =\angle ABP. Let M,NM, N be the centers of the incircle of △ABP\triangle ABP and △ACP\triangle ACP, and RR the radius of the circumscribed circle of △AMN\triangle AMN. Prove that 1R=1AB+1AC+1AP.\displaystyle \frac{1}{R}=\frac{1}{AB}+\frac{1}{AC}+\frac{1}{AP}.
geometrycircumcircletrigonometrygeometry proposed