MathDB

6

Part of 2019 All-Russian Olympiad

Problems(3)

ARMO 2019 9.6

Source: All-Russian Math Olympiad 2019

7/12/2019
There is point DD on edge ACAC isosceles triangle ABCABC with base BCBC. There is point KK on the smallest arc CDCD of circumcircle of triangle BCDBCD. Ray CKCK intersects line parallel to line BCBC through AA at point TT. Let MM be midpoint of segment DTDT. Prove that ∠AKT=∠CAM\angle AKT=\angle CAM.
ARMO
Problem 6

Source: All-Russian Olympiad 2019 grade 10 Problem 6

4/24/2019
Let LL be the foot of the internal bisector of ∠B\angle B in an acute-angled triangle ABC.ABC. The points DD and EE are the midpoints of the smaller arcs ABAB and BCBC respectively in the circumcircle ω\omega of △ABC.\triangle ABC. Points PP and QQ are marked on the extensions of the segments BDBD and BEBE beyond DD and EE respectively so that ∡APB=∡CQB=90∘.\measuredangle APB=\measuredangle CQB=90^{\circ}. Prove that the midpoint of BLBL lies on the line PQ.PQ.
geometrycircumcircle
2019 All Russian MO Grade 11 P6

Source:

5/1/2019
In the segment ACAC of an isosceles triangle △ABC\triangle ABC with base BCBC is chosen a point DD. On the smaller arc CDCD of the circumcircle of △BCD\triangle BCD is chosen a point KK. Line CKCK intersects the line through AA parallel to BCBC at TT. MM is the midpoint of segment DTDT. Prove that ∠AKT=∠CAM\angle AKT=\angle CAM.
(A.Kuznetsov)
geometrycircumcirclemoving points