Advanced book on Mathematics Olympiad

Geometry and Trigonometry 647

A

B P C

M

N

Figure 88

from that figure, we have

A(BMP )

A(ABC)

=

(

BP

BC

) 2

and

A(CN P )

A(ABC)

=

(

PC

BC

) 2

.

Adding up, we obtain

A(BMP )+A(CN P )

A(ABC)

=

BP^2 +PC^2

(BP+PC)^2

≥

1

2

.

The last inequality follows from the AM–GM inequality:BP^2 +PC^2 ≥ 2 BP·PC.

Note that in the degenerate case the inequality is even stronger, with^13 replaced by^12.

Let us now consider the general case, with the notation from Figure 89. By what we

just proved, we know that the following three inequalities hold:

S 1 +S 2 ≥

1

2

A(A 1 B 2 C),

S 1 +S 3 ≥

1

2

A(A 2 BC 1 ),

S 2 +S 3 ≥

1

2

A(AB 1 C 2 ).

Adding them up, we obtain

A

B C

S

S

AA

B

B

C

C

P

12

2

3

1

S 2

2

1

1

Figure 89