MathDB

3

Part of 1997 Polish MO Finals

Problems(2)

medians, tetrahedron and inequality

Source: Polish MO 1997

7/31/2005
In a tetrahedron ABCDABCD, the medians of the faces ABDABD, ACDACD, BCDBCD from DD make equal angles with the corresponding edges ABAB, ACAC, BCBC. Prove that each of these faces has area less than or equal to the sum of the areas of the other two faces. [hide="Comment"]Equivalent version of the problem: ABCDABCD is a tetrahedron. DEDE, DFDF, DGDG are medians of triangles DBCDBC, DCADCA, DABDAB. The angles between DEDE and BCBC, between DFDF and CACA, and between DGDG and ABAB are equal. Show that: area DBCDBC ≤\leq area DCADCA + area DABDAB.
geometry3D geometrytetrahedroninequalitiestrigonometry
n points on a unit circle

Source:

11/11/2005
Given any nn points on a unit circle show that at most n23\frac{n^2}{3} of the segments joining two points have length >2> \sqrt{2}.
combinatorics unsolvedcombinatorics