Track / Stage 6 / 76 of 400 #1076 of 2000

Problem 1076

National olympiad, first round

Algebra Difficulty 6.1 Prove it

10.81*. On the sides $B C, C A$ , and $A B$ of an acute-angled triangle $A B C$ , points $A_{1}, B_{1}$ , and $C_{1}$ are taken. Prove that

$2\left(B_{1} C_{1} \cos \alpha+C_{1} A_{1} \cos \beta+A_{1} B_{1} \cos \gamma\right) \geqslant a \cos \alpha+b \cos \beta+c \cos \gamma$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

10.81. Let $B_{2} C_{2}$ be the projection of the segment $B_{1} C_{1}$ onto the side $B C$ . Then $B_{1} C_{1} \geqslant B_{2} C_{2}=B C-B C_{1} \cos \beta-C B_{1} \cos \gamma$ . Similarly, $A_{1} C_{1} \geqslant A C-A C_{1} \cos \alpha-C A_{1} \cos \gamma$ and $A_{1} B_{1} \geqslant A B-A B_{1} \cos \alpha-B A_{1} \cos \beta$ . Multiply these inequalities by $\cos \alpha, \cos \beta$ and $\cos \gamma$ respectively and add them. We get $B_{1} C_{1} \cos \alpha+C_{1} A_{1} \cos \beta+A_{1} B_{1} \cos \gamma \geqslant a \cos \alpha+b \cos \beta+$ $+c \cos \gamma-(a \cos \beta \cos \gamma+b \cos \alpha \cos \gamma+c \cos \alpha \cos \beta)$ . Since $c=a \cos \beta+b \cos \alpha$ , then $c \cos \gamma=a \cos \beta \cos \gamma+b \cos \alpha \cos \gamma$ . Writing three similar inequalities and adding them, we get $a \cos \beta \cos \gamma+b \cos \alpha \cos \gamma+c \cos \alpha \cos \beta=(a \cos \alpha+b \cos \beta+$ $+c \cos \gamma) / 2$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.