Problem: A B C D A B C D ω ω C D ‾ C D M M A M ‾ A M ω ω P P M M A P A P
Answer Choices:
A. 5 12 1 2 5 ​ ​ 5 10 1 0 5 ​ ​ 5 9 9 5 ​ ​ 5 8 8 5 ​ ​ 2 5 15 1 5 2 5 ​ ​ Solution:
Call the midpoint of A B ‾ A B N N N M ‾ N M N P ‾ N P ∠N P M = 9 0 ∘ ∠N P M = 9 0 ∘
Using the Pythagorean theorem, A M = 5 2 A M = 2 5 ​ ​ A P A P △ A P N ∼ △ A P N ∼ △ A N M △ A N M
A P A N = A N A M A N A P ​ = A M A N ​ A P 1 / 2 = 1 / 2 5 / 2 1 / 2 A P ​ = 5 ​ / 2 1 / 2 ​ A P = ( B ) 5 10 A P = ( B ) 1 0 5 ​ ​ ​
OR
Let N N A B ‾ A B ∠A N M = 9 0 ∘ ∠A N M = 9 0 ∘ ∠N P M = 9 0 ∘ ∠N P M = 9 0 ∘
We have the following diagram:
Since A N = 1 2 A N = 2 1 ​ N M = 1 N M = 1 A M = 5 2 A M = 2 5 ​ ​
Let A P = x A P = x P M = 5 2 − x P M = 2 5 ​ ​ − x △ A N P △ A N P N P 2 = ( 1 2 ) 2 − x 2 N P 2 = ( 2 1 ​ ) 2 − x 2 △ M N P △ M N P N P 2 = 1 2 − ( 5 2 − x ) 2 N P 2 = 1 2 − ( 2 5 ​ ​ − x ) 2 N P 2 N P 2
( 1 2 ) 2 − x 2 = 1 2 − ( 5 2 − x ) 2 ( 2 1 ​ ) 2 − x 2 = 1 2 − ( 2 5 ​ ​ − x ) 2 1 4 − x 2 = 1 − 5 4 + 5 x − x 2 4 1 ​ − x 2 = 1 − 4 5 ​ + 5 ​ x − x 2 1 2 = 5 x 2 1 ​ = 5 ​ x
Finally, dividing both sides by 5 5 ​ x = 1 2 5 = x = 2 5 ​ 1 ​ = ( B ) 5 10 ( B ) 1 0 5 ​ ​ ​
The problems on this page are the property of the MAA's American Mathematics Competitions
.jpg)
