MathDB

2

Part of 2001 Poland - Second Round

Problems(2)

Prove that DP=DR

Source: Polish Second Round 2001

3/6/2012
Points A,B,CA,B,C with AB<BCAB<BC lie in this order on a line. Let ABDEABDE be a square. The circle with diameter ACAC intersects the line DEDE at points PP and QQ with PP between DD and EE. The lines AQAQ and BDBD intersect at RR. Prove that DP=DRDP=DR.
geometry proposedgeometry
iff condition on AI+CD=AC where AD is the angle bisector

Source: Polish Second Round 2001

3/6/2012
In a triangle ABCABC, II is the incentre and DD the intersection point of AIAI and BCBC. Show that AI+CD=ACAI+CD=AC if and only if ∠B=60∘+13∠C\angle B=60^{\circ}+\frac{_1}{^3}\angle C.
geometryincentercircumcircleperpendicular bisectorgeometry proposed