MathDB

6

Part of 2012 Sharygin Geometry Olympiad

Problems(4)

Find angle

Source: Sharygin Geometry Olympiad 2012 - Problem 6

4/28/2012
Point C1C_{1} of hypothenuse ACAC of a right-angled triangle ABCABC is such that BC=CC1BC = CC_{1}. Point C2C_{2} on cathetus ABAB is such that AC2=AC1AC_{2} = AC_{1}; point A2A_{2} is defined similarly. Find angle AMCAMC, where MM is the midpoint of A2C2A_{2}C_{2}.
symmetrygeometrycircumcircleinradiusincentertrigonometrytrapezoid
prove that the orthocenter of one triangle lies on the circumcircle of another

Source: 2012 Sharygin Geometry Olympiad Final Round 8.6

8/3/2018
Let ω\omega be the circumcircle of triangle ABCABC. A point B1B_1 is chosen on the prolongation of side ABAB beyond point B so that AB1=ACAB_1 = AC. The angle bisector of ∠BAC\angle BAC meets ω\omega again at point WW. Prove that the orthocenter of triangle AWB1AWB_1 lies on ω\omega .
(A.Tumanyan)
geometrycircumcircleorthocenter
calculate the circumradius

Source: 2012 Sharygin Geometry Olympiad Final Round 9.6

8/3/2018
Let ABCABC be an isosceles triangle with BC=aBC = a and AB=AC=bAB = AC = b. Segment ACAC is the base of an isosceles triangle ADCADC with AD=DC=aAD = DC = a such that points DD and BB share the opposite sides of AC. Let CMCM and CNCN be the bisectors in triangles ABCABC and ADCADC respectively. Determine the circumradius of triangle CMNCMN.
(M.Rozhkova)
geometrycircumcircleisosceles
inequality with sums of segments in space, starting with a tetrahedron

Source: 2012 Sharygin Geometry Olympiad Final Round 10.6

8/3/2018
Consider a tetrahedron ABCDABCD. A point XX is chosen outside the tetrahedron so that segment XDXD intersects face ABCABC in its interior point. Let A′,B′A' , B' , and C′C' be the projections of DD onto the planes XBC,XCAXBC, XCA, and XABXAB respectively. Prove that A′B′+B′C′+C′A′≤DA+DB+DCA' B' + B' C' + C' A' \le DA + DB + DC.
(V.Yassinsky)
geometry3D geometrytetrahedroninequalities