Problem: A B C A B C A D A D A E A E ∠B A C ∠B A C B D , D E B D , D E E C E C 2 , 3 2 , 3 6 6 △ A B C △ A B C
Answer Choices:
A. 2 10 2 1 0 ​
B. 11 1 1
C. 6 6 6 6 ​
D. 6 6
E. not uniquely determined by the given information
Solution:
In the adjoining figure let \Varangle B A C = 3 α , x = A B = 3 α , x = A B y = A D y = A D A B = 2 3 A B = 3 2 ​ A D A C = 1 2 A C A D ​ = 2 1 ​ A E A E = 3 x 2 = 2 3 x ​ A C = 2 y A C = 2 y △ A D B , △ A E D △ A D B , △ A E D △ A C E △ A C E
x 2 + y 2 − 4 2 x y = 9 4 x 2 + y 2 − 9 3 x y = 9 4 x 2 + 4 y 2 − 36 6 x y 2 x y x 2 + y 2 − 4 ​ = 3 x y 4 9 ​ x 2 + y 2 − 9 ​ = 6 x y 4 9 ​ x 2 + 4 y 2 − 3 6 ​ ​
The equality of the first and second expressions implies
3 x 2 − 2 y 2 = 12 3 x 2 − 2 y 2 = 1 2
The equality of the first and third expressions implies
3 x 2 − 4 y 2 = − 96 3 x 2 − 4 y 2 = − 9 6
Solving these two equations for x x y y
x = 2 10 y = 54 = 3 6 . ​ x = 2 1 0 ​ y = 5 4 ​ = 3 6 ​ . ​
Thus the sides are A B = 2 10 ≈ 6.3 , A C = 6 6 ≈ 14.7 A B = 2 1 0 ​ ≈ 6 . 3 , A C = 6 6 ​ ≈ 1 4 . 7 B C = 11 B C = 1 1
OR OR
In the adjoining figure let A B = a A B = a A D = b , A E = c A D = b , A E = c A C = d A C = d
a c = 2 3 , b d = 3 6 c a ​ = 3 2 ​ , d b ​ = 6 3 ​
Thus a = 2 c 3 a = 3 2 c ​ d = 2 b d = 2 b
b 2 + 6 = a c c 2 + 18 = b d b 2 + 6 c 2 + 1 8 ​ = a c = b d ​
Using the relations above in these equations yields
b 2 + 6 = 2 c 2 3 c 2 + 18 = 2 b 2 b 2 + 6 c 2 + 1 8 ​ = 3 2 c 2 ​ = 2 b 2 ​
Solving these equations for c 2 c 2 b 2 b 2
c 2 = 90 b 2 = 54. c 2 b 2 ​ = 9 0 = 5 4 . ​
Thus,
a = 2 ( 3 10 ) 3 = 2 10 , d = 2 ( 3 6 ) = 6 6 . a d ​ = 3 2 ( 3 1 0 ​ ) ​ = 2 1 0 ​ , = 2 ( 3 6 ​ ) = 6 6 ​ . ​
Angle Bisector: Let A D A D ∠C A B ∠C A B △ A B C △ A B C ( A D ) 2 + ( B D ) ( D C ) = ( A B ) ( A C ) ( A D ) 2 + ( B D ) ( D C ) = ( A B ) ( A C )
.jpg)
