Problem: A B C D A B C D Ο Ο C D βΎ C D M M A M βΎ A M Ο Ο P P M M A P A P
Answer Choices:
A. 5 12 1 2 5 β β 5 10 1 0 5 β β 5 9 9 5 β β 5 8 8 5 β β 2 5 15 1 5 2 5 β β Solution:
Call the midpoint of A B βΎ A B N N N M βΎ N M N P βΎ N P β N P M = 9 0 β β N P M = 9 0 β
Using the Pythagorean theorem, A M = 5 2 A M = 2 5 β β A P A P β³ A P N βΌ β³ A P N βΌ β³ A N M β³ A N M
A P A N = A N A M A N A P β = A M A N β A P 1 / 2 = 1 / 2 5 / 2 1 / 2 A P β = 5 β / 2 1 / 2 β A P = ( B ) 5 10 A P = ( B ) 1 0 5 β β β
OR
Let N N A B βΎ A B β A N M = 9 0 β β A N M = 9 0 β β N P M = 9 0 β β N P M = 9 0 β
We have the following diagram:
Since A N = 1 2 A N = 2 1 β N M = 1 N M = 1 A M = 5 2 A M = 2 5 β β
Let A P = x A P = x P M = 5 2 β x P M = 2 5 β β β x β³ A N P β³ A N P N P 2 = ( 1 2 ) 2 β x 2 N P 2 = ( 2 1 β ) 2 β x 2 β³ M N P β³ M N P N P 2 = 1 2 β ( 5 2 β x ) 2 N P 2 = 1 2 β ( 2 5 β β β x ) 2 N P 2 N P 2
( 1 2 ) 2 β x 2 = 1 2 β ( 5 2 β x ) 2 ( 2 1 β ) 2 β x 2 = 1 2 β ( 2 5 β β β x ) 2 1 4 β x 2 = 1 β 5 4 + 5 x β x 2 4 1 β β x 2 = 1 β 4 5 β + 5 β x β x 2 1 2 = 5 x 2 1 β = 5 β x
Finally, dividing both sides by 5 5 β x = 1 2 5 = x = 2 5 β 1 β = ( B ) 5 10 ( B ) 1 0 5 β β β
The problems on this page are the property of the MAA's American Mathematics Competitions
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