Problem: β³ A B C β³ 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 βΎ B C I 1 I 1 β I 2 I 2 β β³ A B X β³ A B X β³ A C X β³ A C X β³ A I 1 I 2 β³ A I 1 β I 2 β X X B C βΎ B C
Solution:
First note that
β I 1 A I 2 = β I 1 A X + β X A I 2 = β B A X 2 + β C A X 2 = β A 2 . β I 1 β A I 2 β = β I 1 β A X + β X A I 2 β = 2 β B A X β + 2 β C A X β = 2 β A β .
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 Ξ± = β A X B Ξ± = β A X B
β A I 1 B = 18 0 β β ( β I 1 A B + β I 1 B A ) = 18 0 β β 1 2 ( 18 0 β β Ξ± ) = 9 0 β + Ξ± 2 . β A I 1 β B = 1 8 0 β β ( β I 1 β A B + β I 1 β B A ) = 1 8 0 β β 2 1 β ( 1 8 0 β β Ξ± ) = 9 0 β + 2 Ξ± β .
Applying the Law of Sines to β³ A B I 1 β³ A B I 1 β
A I 1 A B = sin β‘ ( β A B I 1 ) sin β‘ ( β A I 1 B ) , implying that A I 1 = c sin β‘ B 2 cos β‘ Ξ± 2 A B A I 1 β β = sin ( β A I 1 β B ) sin ( β A B I 1 β ) β , implying that A I 1 β = cos 2 Ξ± β c sin 2 B β β
Analogously, β A I 2 C = 9 0 β + β A X C 2 = 18 0 β β Ξ± 2 β A I 2 β C = 9 0 β + 2 β A X C β = 1 8 0 β β 2 Ξ± β
A I 2 = b sin β‘ C 2 sin β‘ Ξ± 2 A I 2 β = sin 2 Ξ± β b sin 2 C β β
Because the area of β³ I 1 A I 2 β³ I 1 β A I 2 β 1 2 ( A I 1 ) ( A I 2 ) sin β‘ ( β I 1 A I 2 ) 2 1 β ( A I 1 β ) ( A I 2 β ) sin ( β I 1 β A I 2 β )
[ β³ A I 1 I 2 ] = b c sin β‘ A 2 sin β‘ B 2 sin β‘ C 2 2 cos β‘ Ξ± 2 sin β‘ Ξ± 2 = b c sin β‘ A 2 sin β‘ B 2 sin β‘ C 2 sin β‘ Ξ± β₯ b c sin β‘ A 2 sin β‘ B 2 sin β‘ C 2 [ β³ A I 1 β I 2 β ] = 2 cos 2 Ξ± β sin 2 Ξ± β b c sin 2 A β sin 2 B β sin 2 C β β β = sin Ξ± b c sin 2 A β sin 2 B β sin 2 C β β β₯ b c sin 2 A β sin 2 B β sin 2 C β β
with equality when Ξ± = 9 0 β Ξ± = 9 0 β X X A A B C βΎ B C b c sin β‘ A 2 sin β‘ B 2 sin β‘ C 2 b c sin 2 A β sin 2 B β sin 2 C β
sin β‘ A 2 = 1 β cos β‘ A 2 = 1 β b 2 + c 2 β a 2 2 b c 2 = ( a β b + c ) ( a + b β c ) 4 b c . sin 2 A β = 2 1 β cos A β β = 2 1 β 2 b c b 2 + c 2 β a 2 β β β = 4 b c ( a β b + c ) ( a + b β c ) β β .
Applying similar logic to sin β‘ B 2 sin 2 B β sin β‘ C 2 sin 2 C β
b c sin β‘ A 2 sin β‘ B 2 sin β‘ C 2 = b c β
( a β b + c ) ( b β c + a ) ( c β a + b ) 8 a b c = ( 32 β 34 + 30 ) ( 34 β 30 + 32 ) ( 30 β 32 + 34 ) 8 β
32 = 126 . b c sin 2 A β sin 2 B β sin 2 C β β = b c β
8 a b c ( a β b + c ) ( b β c + a ) ( c β a + b ) β = 8 β
3 2 ( 3 2 β 3 4 + 3 0 ) ( 3 4 β 3 0 + 3 2 ) ( 3 0 β 3 2 + 3 4 ) β = 1 2 6 β . β
The problems on this page are the property of the MAA's American Mathematics Competitions