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I Soros Olympiad 1994-95 (Rus + Ukr)
10.2
OA+OB+OC<BC+CA if AB<=BC<=CA (I Soros Olympiad 1994-95 Ukraine R2 10.2)
OA+OB+OC<BC+CA if AB<=BC<=CA (I Soros Olympiad 1994-95 Ukraine R2 10.2)
Source:
June 6, 2024
geometry
geometric inequality
Problem Statement
Given a triangle
A
B
C
ABC
A
BC
and a point
O
O
O
inside it, it is known that
A
B
≤
B
C
≤
C
A
AB\le BC\le CA
A
B
≤
BC
≤
C
A
. Prove that
O
A
+
O
B
+
O
C
<
B
C
+
C
A
.
OA+OB+OC<BC+CA.
O
A
+
OB
+
OC
<
BC
+
C
A
.
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