MathDB

3

Part of 2013 Sharygin Geometry Olympiad

Problems(4)

Concurrency of perpendiculars

Source: Sharygin First Round 2013, Problem 3

4/8/2013
Let ABCABC be a right-angled triangle (∠B=90∘\angle B = 90^\circ). The excircle inscribed into the angle AA touches the extensions of the sides ABAB, ACAC at points A1,A2A_1, A_2 respectively; points C1,C2C_1, C_2 are defined similarly. Prove that the perpendiculars from A,B,CA, B, C to C1C2,A1C1,A1A2C_1C_2, A_1C_1, A_1A_2 respectively, concur.
trigonometrygeometrytrig identitiesLaw of Sinesgeometry proposed
(AOB)+ (COD) <= 2(AOD)+ 2(BOC), with areas in convex ABCD

Source: Sharygin 2013 Final 9.3

8/19/2018
Each sidelength of a convex quadrilateral ABCDABCD is not less than 11 and not greater than 22. The diagonals of this quadrilateral meet at point OO. Prove that SAOB+SCOD≤2(SAOD+SBOC)S_{AOB}+ S_{COD} \le 2(S_{AOD}+ S_{BOC}).
geometryInequalityareas
each vertex of a convex polygon is projected to all nonadjacent sidelines ...

Source: Sharygin 2013 Final 8.3

8/16/2018
Each vertex of a convex polygon is projected to all nonadjacent sidelines. Can it happen that each of these projections lies outside the corresponding side?
geometryconvex polygonprojections
Prove the concurrency

Source: 10.3 Final Round of Sharygin geometry Olympiad 2013

8/8/2013
Let XX be a point inside triangle ABCABC such that XA.BC=XB.AC=XC.ACXA.BC=XB.AC=XC.AC. Let I1,I2,I3I_1, I_2, I_3 be the incenters of XBC,XCA,XABXBC, XCA, XAB. Prove that AI1,BI2,CI3AI_1, BI_2, CI_3 are concurrent.
Of course, the most natural way to solve this is the Ceva sin theorem, but there is an another approach that may surprise you;), try not to use the Ceva theorem :))
geometryincenter3D geometrytetrahedronsphere