MathDB

2

Part of 1991 Iran MO (2nd round)

Problems(2)

Tetragonal ABCD [Iran Second Round 1991]

Source:

11/30/2010
Let ABCDABCD be a tetragonal.
(a) If the plane (P)(P) cuts ABCD,ABCD, find the necessary and sufficient condition such that the area formed from the intersection of the plane (P)(P) and the tetragonal be a parallelogram. Prove that the problem has three solutions in this case.
(b) Consider one of the solutions of (a). Find the situation of the plane (P)(P) for which the parallelogram has maximum area.
(c) Find a plane (P)(P) for which the parallelogram be a lozenge and then find the length side of his lozenge in terms of the length of the edges of ABCD.ABCD.
geometryparallelogramgeometry proposed
Inequalituy with the incenter [Iran Second Round 1991]

Source:

11/30/2010
Triangle ABCABC is inscribed in circle C.C. The bisectors of the angles A,BA,B and CC meet the circle CC again at the points A′,B′,C′A', B', C'. Let II be the incenter of ABC,ABC, prove that IA′IA+IB′IB+IC′IC≥3\frac{IA'}{IA} + \frac{IB'}{IB}+\frac{IC'}{IC} \geq 3,IA′+IB′+IC′≥IA+IB+IC, IA'+IB'+IC' \geq IA+IB+IC
geometryincentercircumcircleinequalitiesgeometry proposed