Difference between revisions of "1953 AHSME Problems/Problem 49"

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SOLUTION:  
 
SOLUTION:  
k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is (E).
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k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is <math>\boxed{(E) 2 1/7}.</math>

Revision as of 11:29, 25 November 2018

SOLUTION: k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is $\boxed{(E) 2 1/7}.$