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

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SOLUTION:  
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==Problem==
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The coordinates of <math>A,B</math> and <math>C</math> are <math>(5,5),(2,1)</math> and <math>(0,k)</math> respectively.
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The value of <math>k</math> that makes <math>\overline{AC}+\overline{BC}</math> as small as possible is:  
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<math>\textbf{(A)}\ 3 \qquad
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\textbf{(B)}\ 4\frac{1}{2} \qquad
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\textbf{(C)}\ 3\frac{6}{7} \qquad
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\textbf{(D)}\ 4\frac{5}{6}\qquad
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\textbf{(E)}\ 2\frac{1}{7}    </math>
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==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 <math>\boxed{(E) 2 1/7}.</math>
 
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>
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==See Also==
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{{AHSME 50p box|year=1953|num-b=48|num-a=50}}
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{{MAA Notice}}

Revision as of 19:08, 17 February 2020

Problem

The coordinates of $A,B$ and $C$ are $(5,5),(2,1)$ and $(0,k)$ respectively. The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4\frac{1}{2} \qquad \textbf{(C)}\ 3\frac{6}{7} \qquad \textbf{(D)}\ 4\frac{5}{6}\qquad \textbf{(E)}\ 2\frac{1}{7}$

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 $\boxed{(E) 2 1/7}.$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 48
Followed by
Problem 50
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


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