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Difference between revisions of "1953 AHSME Problems/Problem 49"
 
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==Solution==
 
==Solution==
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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>
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<math>k</math> will be between <math>1</math> and <math>5</math> for <math>AC+BC</math> to be the smallest. If we mirror point <math>A</math> across the y-axis to <math>A'</math>, with coordinates <math>(-5,5),</math> the distance <math>A'C+BC</math> will be same as <math>AC+BC</math>. The minimum of <math>A'C+BC</math> will occur when <math>C</math> is on the straight line connecting <math>A'</math> and <math>B</math> (i.e., <math>C</math> lies on the line <math>A'B</math>). Therefore, <math>k</math> is the y-intercept of the line that passes through <math>A'</math> and <math>B</math>.
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The slope of the line is <math>\frac{1-5}{2-(-5)}=-\frac 47</math>. Using [[point-slope form]], the equation of the line is <math>y-5=-\frac 47(x+5)</math>. Letting <math>x=0</math> gives <math>y-5=-\frac{20}{7},</math> so <math>y=\frac{15}{7}</math>. Therefore, <math>k = 2\frac17 \Rightarrow \boxed{\textbf{(E) } 2\frac17}.</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 19:20, 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 the smallest. If we mirror point $A$ across the y-axis to $A'$, with coordinates $(-5,5),$ 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 $A'B$). Therefore, $k$ is the y-intercept of the line that passes through $A'$ and $B$.

The slope of the line is $\frac{1-5}{2-(-5)}=-\frac 47$. Using point-slope form, the equation of the line is $y-5=-\frac 47(x+5)$. Letting $x=0$ gives $y-5=-\frac{20}{7},$ so $y=\frac{15}{7}$. Therefore, $k = 2\frac17 \Rightarrow \boxed{\textbf{(E) } 2\frac17}.$

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 48 		Followed by
Problem 50
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All AHSME Problems and Solutions


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