Problem: △ A B C △ A B C A B = 30 , B C = 32 A B = 3 0 , B C = 3 2 A C = 34 A C = 3 4 X X B C ‾ B C I 1 I 1 I 2 I 2 △ A B X △ A B X △ A C X △ A C X △ A I 1 I 2 △ A I 1 I 2 X X B C ‾ B C
Solution:
First note that
∠ I 1 A I 2 = ∠ I 1 A X + ∠ X A I 2 = ∠ B A X 2 + ∠ C A X 2 = ∠ A 2 . ∠ I 1 A I 2 = ∠ I 1 A X + ∠ X A I 2 = 2 ∠ B A X + 2 ∠ C A X = 2 ∠ A .
This is a constant not depending on X X a = B C , b = A C , c = A B a = B C , b = A C , c = A B α = ∠ A X B α = ∠ A X B
∠ A I 1 B = 18 0 ∘ − ( ∠ I 1 A B + ∠ I 1 B A ) = 18 0 ∘ − 1 2 ( 18 0 ∘ − α ) = 9 0 ∘ + α 2 . ∠ A I 1 B = 1 8 0 ∘ − ( ∠ I 1 A B + ∠ I 1 B A ) = 1 8 0 ∘ − 2 1 ( 1 8 0 ∘ − α ) = 9 0 ∘ + 2 α .
Applying the Law of Sines to △ A B I 1 △ A B I 1
A I 1 A B = sin ( ∠ A B I 1 ) sin ( ∠ A I 1 B ) , implying that A I 1 = c sin B 2 cos α 2 A B A I 1 = sin ( ∠ A I 1 B ) sin ( ∠ A B I 1 ) , implying that A I 1 = cos 2 α c sin 2 B
Analogously, ∠ A I 2 C = 9 0 ∘ + ∠ A X C 2 = 18 0 ∘ − α 2 ∠ A I 2 C = 9 0 ∘ + 2 ∠ A X C = 1 8 0 ∘ − 2 α
A I 2 = b sin C 2 sin α 2 A I 2 = sin 2 α b sin 2 C
Because the area of △ I 1 A I 2 △ I 1 A I 2 1 2 ( A I 1 ) ( A I 2 ) sin ( ∠ I 1 A I 2 ) 2 1 ( A I 1 ) ( A I 2 ) sin ( ∠ I 1 A I 2 )
[ △ A I 1 I 2 ] = b c sin A 2 sin B 2 sin C 2 2 cos α 2 sin α 2 = b c sin A 2 sin B 2 sin C 2 sin α ≥ b c sin A 2 sin B 2 sin C 2 [ △ A I 1 I 2 ] = 2 cos 2 α sin 2 α b c sin 2 A sin 2 B sin 2 C = sin α b c sin 2 A sin 2 B sin 2 C ≥ b c sin 2 A sin 2 B sin 2 C
with equality when α = 9 0 ∘ α = 9 0 ∘ X X A A B C ‾ B C b c sin A 2 sin B 2 sin C 2 b c sin 2 A sin 2 B sin 2 C
sin A 2 = 1 − cos A 2 = 1 − b 2 + c 2 − a 2 2 b c 2 = ( a − b + c ) ( a + b − c ) 4 b c . sin 2 A = 2 1 − cos A = 2 1 − 2 b c b 2 + c 2 − a 2 = 4 b c ( a − b + c ) ( a + b − c ) .
Applying similar logic to sin B 2 sin 2 B sin C 2 sin 2 C
b c sin A 2 sin B 2 sin C 2 = b c ⋅ ( a − b + c ) ( b − c + a ) ( c − a + b ) 8 a b c = ( 32 − 34 + 30 ) ( 34 − 30 + 32 ) ( 30 − 32 + 34 ) 8 ⋅ 32 = 126 . b c sin 2 A sin 2 B sin 2 C = b c ⋅ 8 a b c ( a − b + c ) ( b − c + a ) ( c − a + b ) = 8 ⋅ 3 2 ( 3 2 − 3 4 + 3 0 ) ( 3 4 − 3 0 + 3 2 ) ( 3 0 − 3 2 + 3 4 ) = 1 2 6 .
The problems on this page are the property of the MAA's American Mathematics Competitions