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Let A_1A_2A_3A_4 be a tetrahedron, G its centroid, and A'_1, A'_2, A'_3, and A'_4 the points where the circumsphere of A_1A_2A_3A_4 intersects GA_1,GA_2,GA_3, and GA_4, respectively. Prove that

GA_1 \cdot GA_2 \cdot GA_3 \cdot GA_ \cdot4 \leq GA'_1 \cdot GA'_2 \cdot GA'_3 \cdot GA'_4

and

\frac{1}{GA'_1} + \frac{1}{GA'_2} + \frac{1}{GA'_3} + \frac{1}{GA'_4} \leq \frac{1}{GA_1} + \frac{1}{GA_2} + \frac{1}{GA_3} +...

Slični zadaci

Find all x,y and z in positive integer: z + y^{2} + x^{3} = xyz and x = \gcd(y,z).
Let the sides of two rectangles be \{a,b\} and \{c,d\}, respectively, with a < c \leq d < b and ab < cd. Prove that the first rectangle can be placed within the second one if and only if

\left(b^2 - a^2\right)^2 \leq \left(bc - ad \right)^2 + \left(bd - ac \right)^2.
Let ABCD be a convex quadrilateral. The perpendicular bisectors of its sides AB and CD meet at Y. Denote by X a point inside the quadrilateral ABCD such that \measuredangle ADX = \measuredangle BCX < 90^{\circ} and \measuredangle DAX = \measuredangle CBX < 90^{\circ}. Show that \measuredangle AYB = 2\cdot\measuredangle ADX.
Let A_1A_2A_3...A_n be a regular n-gon. Let B_1 and B_n be the midpoints of its sides A_1A_2 and A_{n-1}A_n. Also, for every i\in\left\{2;\;3;\;4;\;...;\;n-1\right\}, let S be the point of intersection of the lines A_1A_{i+1} and A_nA_i, and let B_i be the point of intersection of the angle bisector bisector of the angle \measuredangle A_iSA_{i+1} with the segment A_iA_{i+1}.

Prove that: \sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}.
Let ABCD be a convex quadrilateral and let P and Q be points in ABCD such that PQDA and QPBC are cyclic quadrilaterals. Suppose that there exists a point E on the line segment PQ such that \angle PAE = \angle QDE and \angle PBE = \angle QCE. Show that the quadrilateral ABCD is cyclic.

Proposed by John Cuya, Peru
There is given a convex quadrilateral ABCD. Prove that there exists a point P inside the quadrilateral such that
\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}
if and only if the diagonals AC and BD are perpendicular.

Proposed by Dukan Dukic, Serbia