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All-Russian Olympiad Regional Round
2009 All-Russian Olympiad Regional Round
11.4
11.4
Part of
2009 All-Russian Olympiad Regional Round
Problems
(1)
point symmetric to circumcneter of (HOA' ) lies on midline
Source: All-Russian MO 2009 Regional 11. 4
8/27/2024
In an acute non-isosceles triangle
A
B
C
ABC
A
BC
, the altitude
A
A
′
AA'
A
A
′
is drawn and point
H
H
H
is the intersection point of the altitudes and and
O
O
O
is the center of the circumscribed circle. Prove that the point symmetric to the circumcenter of triangle
H
O
A
′
HOA'
H
O
A
′
wrt straight line
H
O
HO
H
O
, lies on a midline of triangle
A
B
C
ABC
A
BC
.
geometry
midline