Problem: △ A B C △ A B C A B = 80 A B = 8 0 B C = 45 B C = 4 5 A C = 75 A C = 7 5 ∠ B ∠ B A B ‾ A B P P B P ? B P ?
Answer Choices:
A. 18 1 8 19 1 9 20 2 0 21 2 1 22 2 2
Solution:
Let D D C C A B ‾ A B P P B D B D ∠ B ∠ B C D ‾ C D A D A D B D B D
From right triangles A D C A D C B D C B D C
A C 2 = A D 2 + C D 2 , A C 2 = A D 2 + C D 2 ,
B C 2 = B D 2 + C D 2 . B C 2 = B D 2 + C D 2 .
From above, we get:
A C 2 − B C 2 = A D 2 − B D 2 = ( A D − B D ) ( A D + B D ) . A C 2 − B C 2 = A D 2 − B D 2 = ( A D − B D ) ( A D + B D ) .
A D + B D = A B = 80 , A C = 75 , B C = 45 , ⟹ ( 7 5 2 − 4 5 2 ) = 80 ( A D − B D ) ⟹ ( 75 − 45 ) ( 75 + 45 ) = 80 ( A D − B D ) ⟹ 30 ⋅ 120 = 80 ( A D − B D ) ⟹ A D − B D = 45. A D + B D = A B = 8 0 , A C = 7 5 , B C = 4 5 , ⟹ ( 7 5 2 − 4 5 2 ) = 8 0 ( A D − B D ) ⟹ ( 7 5 − 4 5 ) ( 7 5 + 4 5 ) = 8 0 ( A D − B D ) ⟹ 3 0 ⋅ 1 2 0 = 8 0 ( A D − B D ) ⟹ A D − B D = 4 5 .
Solving
{ A D + B D = 80 , A D − B D = 45 , { A D + B D = 8 0 , A D − B D = 4 5 ,
we get
A D = 125 2 , B D = 35 2 . A D = 2 1 2 5 , B D = 2 3 5 .
We find C D C D △ B D C △ B D C
C D = B C 2 − B D 2 = 4 5 2 − ( 35 2 ) 2 = 6875 4 = 25 2 11 . C D = B C 2 − B D 2 = 4 5 2 − ( 2 3 5 ) 2 = 4 6 8 7 5 = 2 2 5 1 1 .
We use the Angle Bisector Theorem in △ B C D △ B C D D P D P B P B P ∠ B ∠ B
D P P C = B D B C = 35 2 45 = 7 18 ⇒ D P = 7 25 C D . P C D P = B C B D = 4 5 2 3 5 = 1 8 7 ⇒ D P = 2 5 7 C D .
Hence,
D P = 7 25 ⋅ 25 2 11 = 7 11 2 . D P = 2 5 7 ⋅ 2 2 5 1 1 = 2 7 1 1 .
Now, we can again use the Pythagorean Theorem in △ B D P △ B D P B P B P
B P 2 = B D 2 + D P 2 = ( 35 2 ) 2 + ( 7 11 2 ) 2 = 1225 4 + 539 4 = 1764 4 = 441. B P 2 = B D 2 + D P 2 = ( 2 3 5 ) 2 + ( 2 7 1 1 ) 2 = 4 1 2 2 5 + 4 5 3 9 = 4 1 7 6 4 = 4 4 1 .
so B P = (D) 21 B P = (D) 21
The problems on this page are the property of the MAA's American Mathematics Competitions